[{"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sometimes when you're doing a large multi-part proof like this, it's easy to lose your bearings. We're trying to prove the divergence theorem. We started off just rewriting the flux across the surface and rewriting the triple integral of the divergence. We said, if we can prove that each of these components are equal to each other, we will have our proof done. What I said is that we're going to use the fact that this is a type I region to prove this part, the fact that it's a type II region to prove this part, and the fact that it's a type III region to prove this part. In particular, I'm going to show the fact that if it is a type I region, we can prove this, and you can use the exact same argument to prove the other two, in which case the divergence theorem would be correct. We're focused on this part right here, and in particular we're focused on evaluating the surface integral."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We said, if we can prove that each of these components are equal to each other, we will have our proof done. What I said is that we're going to use the fact that this is a type I region to prove this part, the fact that it's a type II region to prove this part, and the fact that it's a type III region to prove this part. In particular, I'm going to show the fact that if it is a type I region, we can prove this, and you can use the exact same argument to prove the other two, in which case the divergence theorem would be correct. We're focused on this part right here, and in particular we're focused on evaluating the surface integral. To evaluate that surface integral, we broke up the entire surface into three surfaces, that upper bound on z, the lower bound on z, and kind of the side of, if you imagine, some type of a funky cylinder. These don't have to be flat tops and bottoms. For a type I region, you don't even have to have this side surface."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're focused on this part right here, and in particular we're focused on evaluating the surface integral. To evaluate that surface integral, we broke up the entire surface into three surfaces, that upper bound on z, the lower bound on z, and kind of the side of, if you imagine, some type of a funky cylinder. These don't have to be flat tops and bottoms. For a type I region, you don't even have to have this side surface. This is only for the case where the two surfaces don't touch each other within this domain right over here. We broke it up into three surfaces, but we said, look, the normal vector along the side right over here is never going to have a k component. When you take the dot product with the k unit vector, this is just going to cross out."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "For a type I region, you don't even have to have this side surface. This is only for the case where the two surfaces don't touch each other within this domain right over here. We broke it up into three surfaces, but we said, look, the normal vector along the side right over here is never going to have a k component. When you take the dot product with the k unit vector, this is just going to cross out. Our surface integral is simplified to the surface integral over S2 and the surface integral of S1. In the last video, we evaluated the surface integral of S2, or at least we turned it into a double integral over the domain. Now we're going to do the same exact thing with S1."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When you take the dot product with the k unit vector, this is just going to cross out. Our surface integral is simplified to the surface integral over S2 and the surface integral of S1. In the last video, we evaluated the surface integral of S2, or at least we turned it into a double integral over the domain. Now we're going to do the same exact thing with S1. Let's just remind ourselves what we are concerned with. We want to re-express S1, and S1 and this integral, the surface integral, I should say we want to re-express the surface integral over S1. This is R of x, y, and z."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we're going to do the same exact thing with S1. Let's just remind ourselves what we are concerned with. We want to re-express S1, and S1 and this integral, the surface integral, I should say we want to re-express the surface integral over S1. This is R of x, y, and z. Up here we just wrote R, but this is making it explicit that R is a function of x, y, and z times k dot n dS. The way we evaluate any surface integral is we want to make sure or we want to have a parameterization for our actual surface. Let's introduce a parameterization for a surface."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is R of x, y, and z. Up here we just wrote R, but this is making it explicit that R is a function of x, y, and z times k dot n dS. The way we evaluate any surface integral is we want to make sure or we want to have a parameterization for our actual surface. Let's introduce a parameterization for a surface. Let's say that S1, and I'll use the letter O, it's kind of a weird one to use, it looks like a zero, and we're already using it for a normal vector. Let's use E. The parameterization of our surface could be x times i plus y times j plus, and it's a type 2 region, our surface is a function of x and y, plus F1, which is a function of x, y, times k. That's the F1 right over here. That's our lower bound on our region."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's introduce a parameterization for a surface. Let's say that S1, and I'll use the letter O, it's kind of a weird one to use, it looks like a zero, and we're already using it for a normal vector. Let's use E. The parameterization of our surface could be x times i plus y times j plus, and it's a type 2 region, our surface is a function of x and y, plus F1, which is a function of x, y, times k. That's the F1 right over here. That's our lower bound on our region. This is 4 for all of the x, y pairs that are a member of our domain in question. We have our parameterization, and now we can think about how we can write this right over here, the n dS. n times dS."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's our lower bound on our region. This is 4 for all of the x, y pairs that are a member of our domain in question. We have our parameterization, and now we can think about how we can write this right over here, the n dS. n times dS. We've done this multiple, multiple, multiple times, which is actually the same thing as dS. We've done this multiple times. This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "n times dS. We've done this multiple, multiple, multiple times, which is actually the same thing as dS. We've done this multiple times. This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA. We want to make sure we get the order right. I'm going to claim that this is going to be the partial of our parameterization with respect to y crossed with the partial of our parameterization with respect to x, and then we have times dA. We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to the cross product of the parameterization in one direction with respect to one parameter, and then that crossed with respect to the other parameter, and then that times dA. We want to make sure we get the order right. I'm going to claim that this is going to be the partial of our parameterization with respect to y crossed with the partial of our parameterization with respect to x, and then we have times dA. We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region. If we're going in the y direction, the partial with respect to y is like that. The partial with respect to x is like that. If you use the right-hand rule, your thumb will point downward like that."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We need to make sure this has the right orientation because for this bottom surface, remember, we need to be pointed straight down, outward from the region. If we're going in the y direction, the partial with respect to y is like that. The partial with respect to x is like that. If you use the right-hand rule, your thumb will point downward like that. I could draw my right hand. My index finger could go like that. My middle finger would bend like this."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you use the right-hand rule, your thumb will point downward like that. I could draw my right hand. My index finger could go like that. My middle finger would bend like this. I don't really care what my other two fingers do. Then my thumb would go downwards. This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "My middle finger would bend like this. I don't really care what my other two fingers do. Then my thumb would go downwards. This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time. We'll see. We'll just get a negative value. Let's just work it through just to be a little bit more convincing."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the right ordering, which is a different ordering, or it's the opposite ordering as we did last time. We'll see. We'll just get a negative value. Let's just work it through just to be a little bit more convincing. This business right over here is going to be equal to we're going to divide our i, j, and k unit vectors. It's going to be all of that times dA. The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's just work it through just to be a little bit more convincing. This business right over here is going to be equal to we're going to divide our i, j, and k unit vectors. It's going to be all of that times dA. The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y. Partial with respect to x is going to be 1, 0. Partial of F1 with respect to x. Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The partial of E with respect to y is 0, 1, 0, 1, partial of F1 with respect to y. Partial with respect to x is going to be 1, 0. Partial of F1 with respect to x. Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j. On our k component, you're going to have 0 times 0 minus 1 times k. Minus k and then all of that times dA. The reason why I didn't even worry about what these things are going to be is I'm going to have to take the drop product with k. This whole thing, all of this business right over here, I can now express in the xy domain. I can now write is going to be equal to the double integral."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then when you evaluate this whole thing, let me draw a little dotted line here, this is going to be equal to some business times i minus some other business times j. On our k component, you're going to have 0 times 0 minus 1 times k. Minus k and then all of that times dA. The reason why I didn't even worry about what these things are going to be is I'm going to have to take the drop product with k. This whole thing, all of this business right over here, I can now express in the xy domain. I can now write is going to be equal to the double integral. Let me do it in that same purple color because that was the original color for that surface integral. It's going to be the double integral over the domain, our parameters domain in the xy plane of R of x. Let me write this a little bit cleaner."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I can now write is going to be equal to the double integral. Let me do it in that same purple color because that was the original color for that surface integral. It's going to be the double integral over the domain, our parameters domain in the xy plane of R of x. Let me write this a little bit cleaner. R of xy, instead of z, I'm going to write z over that surface is F1 of x and y so that we have everything in terms of our parameters times all of this business, k dotted this. What is k dotted this? The dot product of k and negative k is negative 1, so we're just going to be left with negative dA."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write this a little bit cleaner. R of xy, instead of z, I'm going to write z over that surface is F1 of x and y so that we have everything in terms of our parameters times all of this business, k dotted this. What is k dotted this? The dot product of k and negative k is negative 1, so we're just going to be left with negative dA. We'll put the dA out front here and we are left with a negative dA right over here. We have now expressed this surface integral as the sum of two double integrals, as the sum of this, the sum of that, and that right over there. Actually, let me just rewrite it so that we can make everything clear."}, {"video_title": "Divergence theorem proof (part 4)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The dot product of k and negative k is negative 1, so we're just going to be left with negative dA. We'll put the dA out front here and we are left with a negative dA right over here. We have now expressed this surface integral as the sum of two double integrals, as the sum of this, the sum of that, and that right over there. Actually, let me just rewrite it so that we can make everything clear. This surface integral over our entire surface of R times k dot n dS is equal to the double integral and I'll do this in a new color, is equal to the double integral in the domain D of this thing minus this thing dA. I'll write R of x, y, and F2 of x, y, and that's that, minus this, minus R of x, y, and F2 of x, sorry, F1, be careful here, F1 of x, y, that's this thing right over here, all of that times dA. Now, we just showed this is equal to this."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "We're looking at this formula and trying to understand why it corresponds to curvature, why it tells you how much a curve actually curves. And the first thing we did is we noticed that this numerator corresponds to a certain cross product, the cross product between the first derivative and the second derivative of the function parameterizing the curve. And the way we started to understand that is we say, well, the function parameterizing the curve, s of t, produces vectors whose tips trace out that curve itself. And now if you think about how one tip moves to the next, the direction that it needs to go for that tip to move to the next one, that's what the first derivative tells you. And when you treat that in an infinitesimal way, this is why you get tangent vectors along the curve. And there's an entire series of videos on that for the derivative of a position vector-valued function. And they explain why the first derivative of a parametric function gives you tangent vectors to that curve."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And now if you think about how one tip moves to the next, the direction that it needs to go for that tip to move to the next one, that's what the first derivative tells you. And when you treat that in an infinitesimal way, this is why you get tangent vectors along the curve. And there's an entire series of videos on that for the derivative of a position vector-valued function. And they explain why the first derivative of a parametric function gives you tangent vectors to that curve. But if we draw all of those tangent vectors just in their own space, their own little s prime of t space, you get all of these vectors, and we're rooting them at the same point to be able to relate them more easily. The way that you move from the tip of one of those to the next one is given by the second derivative. It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And they explain why the first derivative of a parametric function gives you tangent vectors to that curve. But if we draw all of those tangent vectors just in their own space, their own little s prime of t space, you get all of these vectors, and we're rooting them at the same point to be able to relate them more easily. The way that you move from the tip of one of those to the next one is given by the second derivative. It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function. And specifically, if you have a circumstance where the tangent vectors are just turning, the only thing they're doing is purely turning around, which is when your curve is actually curving, this corresponds to a case when the second derivative function is pretty much perpendicular as a vector. The vector it produces is perpendicular to the first derivative vector. So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "It kind of plays the same role for the first derivative as the first derivative plays for the original parametric function. And specifically, if you have a circumstance where the tangent vectors are just turning, the only thing they're doing is purely turning around, which is when your curve is actually curving, this corresponds to a case when the second derivative function is pretty much perpendicular as a vector. The vector it produces is perpendicular to the first derivative vector. So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are. But there's a bit of a catch. The original formula for curvature, the whole reason we're doing it with respect to arc length and not with respect to the parameter t, is that curvature doesn't really care about how you parametrize the function. If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "So this is loosely why the cross product is kind of a good measure of curving because it tells you how perpendicular these guys are. But there's a bit of a catch. The original formula for curvature, the whole reason we're doing it with respect to arc length and not with respect to the parameter t, is that curvature doesn't really care about how you parametrize the function. If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle. The curvature should just always be the same. But this is a problem if you think back to the cross product that we're now looking at, where you're taking the cross product between the first derivative and the second derivative because if you were traveling along this curve twice as quickly, what that would mean is your first derivative vector, so I'll kind of draw it again over here, would be twice as long to indicate that you're going twice as fast. And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "If you imagine zipping along it really quickly so your first derivative vectors are all super long, it shouldn't matter compared to crawling along it like a turtle. The curvature should just always be the same. But this is a problem if you think back to the cross product that we're now looking at, where you're taking the cross product between the first derivative and the second derivative because if you were traveling along this curve twice as quickly, what that would mean is your first derivative vector, so I'll kind of draw it again over here, would be twice as long to indicate that you're going twice as fast. And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long. And as a result, the parallelogram that they trace out, and I should actually, kind of going off screen here, the parallelogram that they trace out would be actually four times as big, right, because both of the vectors get scaled up. So the way that we really want to be thinking about this is not the tangent vector due to the derivative, but normalizing this. And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And similarly, your second derivative vector to kind of keep up with that changing rate would also be twice as long. And as a result, the parallelogram that they trace out, and I should actually, kind of going off screen here, the parallelogram that they trace out would be actually four times as big, right, because both of the vectors get scaled up. So the way that we really want to be thinking about this is not the tangent vector due to the derivative, but normalizing this. And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole. So if you imagine instead kind of cutting off the vector to make sure that it's got a unit length, a length of one, and what that means is you're taking the derivative vector and dividing it by its own magnitude, by the magnitude of that derivative vector, and then similarly we'll want to scale everything else down. So you're taking this and kind of scaling it down by what you need. The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And this should kind of make sense because we're thinking in terms of unit tangent vectors for the curvature as a whole. So if you imagine instead kind of cutting off the vector to make sure that it's got a unit length, a length of one, and what that means is you're taking the derivative vector and dividing it by its own magnitude, by the magnitude of that derivative vector, and then similarly we'll want to scale everything else down. So you're taking this and kind of scaling it down by what you need. The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are. And what this guy would be, by the way, then, this is the second derivative vector, not normalized with respect to itself, but we're still dividing, you know, the thing we're dividing by as we scale everything down is still just the size of that first derivative vector. So this cross product, if we take the cross product between S prime normalized, S double prime, no, no, no, sorry, S prime itself, and I should be saying vectors for all of these, these are all vectors. If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "The resulting parallelogram they trace out is a more pure measurement of how perpendicular they are without caring about how long they are. And what this guy would be, by the way, then, this is the second derivative vector, not normalized with respect to itself, but we're still dividing, you know, the thing we're dividing by as we scale everything down is still just the size of that first derivative vector. So this cross product, if we take the cross product between S prime normalized, S double prime, no, no, no, sorry, S prime itself, and I should be saying vectors for all of these, these are all vectors. If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first. And the reason we don't really care about the second derivative being normalized is if it was the case that, you know, the second derivative was really, really strong and wasn't necessarily a unit vector, that's fine. That's just telling us that the tangent vector turns much more quickly and the curvature should be higher. And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "If we take the cross product between that and S double prime scaled down by that same value that's still S prime, so it's not normalized, this is just scaled down by S prime, this here is a more pure measurement of how perpendicular the second derivative vector is to the first. And the reason we don't really care about the second derivative being normalized is if it was the case that, you know, the second derivative was really, really strong and wasn't necessarily a unit vector, that's fine. That's just telling us that the tangent vector turns much more quickly and the curvature should be higher. And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is. And now if you think back to the, I'm not sure if it was the last video or the one before that, but I talked about how when you take, when we're looking for this derivative of the tangent vector with respect to arc length, the way that you compute this is to first take its derivative with respect to the parameter, which is something we can actually do because everything is expressed in terms of that parameter, and then dividing it by the, basically the change in arc length with respect to that parameter, which is the size of that first derivative function. So if this whole thing is the derivative of the tangent vector with respect to T, what that means is when we take this and we divide that whole thing by the derivative, by S prime, that should give us curvature. And in fact, that's just worth writing on its own here."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And in fact, it turns out that this whole expression is the derivative of the unit tangent vector, T, that unit tangent vector that I've talked about a lot, with respect to the parameter T. So whatever the parameter of your original function is. And now if you think back to the, I'm not sure if it was the last video or the one before that, but I talked about how when you take, when we're looking for this derivative of the tangent vector with respect to arc length, the way that you compute this is to first take its derivative with respect to the parameter, which is something we can actually do because everything is expressed in terms of that parameter, and then dividing it by the, basically the change in arc length with respect to that parameter, which is the size of that first derivative function. So if this whole thing is the derivative of the tangent vector with respect to T, what that means is when we take this and we divide that whole thing by the derivative, by S prime, that should give us curvature. And in fact, that's just worth writing on its own here. That's curvature. Curvature is equal to, and what I'm gonna do is I'm gonna take, since we see this three different times, we see the magnitude of S prime here, magnitude of S prime here, magnitude of S prime here, since we see that three times, I'm gonna go over here and I'm gonna put that on the denominator, but cubing it. So S prime, that derivative vector, the magnitude of that derivative vector cubed."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And in fact, that's just worth writing on its own here. That's curvature. Curvature is equal to, and what I'm gonna do is I'm gonna take, since we see this three different times, we see the magnitude of S prime here, magnitude of S prime here, magnitude of S prime here, since we see that three times, I'm gonna go over here and I'm gonna put that on the denominator, but cubing it. So S prime, that derivative vector, the magnitude of that derivative vector cubed. And then on the top, we still have S prime, that vector S prime cross product with S double prime. That vector. And this is, I mean, you could think of this as yet another formula for curvature."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "So S prime, that derivative vector, the magnitude of that derivative vector cubed. And then on the top, we still have S prime, that vector S prime cross product with S double prime. That vector. And this is, I mean, you could think of this as yet another formula for curvature. I think I've given you like four at this point. Or you could think of it as just kind of the same thing. And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here?"}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And this is, I mean, you could think of this as yet another formula for curvature. I think I've given you like four at this point. Or you could think of it as just kind of the same thing. And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here? If we take X prime squared plus Y prime squared, and if we think of the square root of that, so kind of taking it to the 1 1\u20442 power, that would be the magnitude of the derivative. And I kind of showed that in some of the previous videos. And what we're doing is we're cubing that."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And if we look back up to our original one that I was trying to justify, this is just the spelled out version of it because what is this bottom component here? If we take X prime squared plus Y prime squared, and if we think of the square root of that, so kind of taking it to the 1 1\u20442 power, that would be the magnitude of the derivative. And I kind of showed that in some of the previous videos. And what we're doing is we're cubing that. So that whole formula that's the very explicit, you know, in terms of X and Ys, what's going on, is really just expressing this idea. You take the cross product between the first derivative and the second derivative, and then because you're normalizing it, you know, normalizing with respect to the first derivative, you want to scale down the second derivative by that same amount just so that the parallelogram we're thinking kind of shrinks and everything stays in proportion. And then once again, we're dividing about that S prime basically because curvature is supposed to be with respect to S and not with respect to T. So that's a way of kind of getting a correction factor for how wrong you're going to be if you just think in terms of the parameter T instead of steps in terms of the arc length."}, {"video_title": "Curvature formula, part 5.mp3", "Sentence": "And what we're doing is we're cubing that. So that whole formula that's the very explicit, you know, in terms of X and Ys, what's going on, is really just expressing this idea. You take the cross product between the first derivative and the second derivative, and then because you're normalizing it, you know, normalizing with respect to the first derivative, you want to scale down the second derivative by that same amount just so that the parallelogram we're thinking kind of shrinks and everything stays in proportion. And then once again, we're dividing about that S prime basically because curvature is supposed to be with respect to S and not with respect to T. So that's a way of kind of getting a correction factor for how wrong you're going to be if you just think in terms of the parameter T instead of steps in terms of the arc length. So hopefully this makes the original formula a little bit less random-seeming, you know, in this two-dimensional case. And it also gives another strong conceptual tool for understanding yet another way that you can think about how much a curve itself actually curves. And I think I've probably done enough videos here going through all the different formulas for what curvature should be, and then in the next one or two I'll go through some specific examples just to see what it looks like to compute that."}, {"video_title": "Directional derivative.mp3", "Sentence": "And that's a way to extend the idea of a partial derivative. And partial derivatives, if you'll remember, have to do with functions with some kind of multivariable input, and I'll just use two inputs because that's the easiest to think about. And it could be some single variable output. It could also deal with vector variable outputs. We haven't gotten to that yet. So we'll just think about a single variable ordinary real number output that's, you know, an expression of x and y. And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane."}, {"video_title": "Directional derivative.mp3", "Sentence": "It could also deal with vector variable outputs. We haven't gotten to that yet. So we'll just think about a single variable ordinary real number output that's, you know, an expression of x and y. And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane. So this would be the x axis, this is y. And, you know, vaguely in your mind, you're thinking that somehow this outputs to a line. This outputs to just the real numbers."}, {"video_title": "Directional derivative.mp3", "Sentence": "And the partial derivative, one of the ways that I said you could think about it is to take a look at the input space, your x and y plane. So this would be the x axis, this is y. And, you know, vaguely in your mind, you're thinking that somehow this outputs to a line. This outputs to just the real numbers. And maybe you're thinking about a transformation that takes it there, or maybe you're just thinking, okay, this is the input space, that's the output. And when you take the partial derivative at some kind of point, so I'll write it out like partial derivative of f with respect to x at a point like one, two, you think about that point, you know, one, y is equal to two. And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction."}, {"video_title": "Directional derivative.mp3", "Sentence": "This outputs to just the real numbers. And maybe you're thinking about a transformation that takes it there, or maybe you're just thinking, okay, this is the input space, that's the output. And when you take the partial derivative at some kind of point, so I'll write it out like partial derivative of f with respect to x at a point like one, two, you think about that point, you know, one, y is equal to two. And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction. And you see what the resulting nudge is in the output space. And the ratio between the size of that resulting nudge and the original one, the ratio between, you know, partial f and partial x, is the value that you want. And when you did it with respect to y, you know, you were thinking about traveling in a different direction."}, {"video_title": "Directional derivative.mp3", "Sentence": "And if you're taking it with respect to x, you think about just nudging it a little bit in that x direction. And you see what the resulting nudge is in the output space. And the ratio between the size of that resulting nudge and the original one, the ratio between, you know, partial f and partial x, is the value that you want. And when you did it with respect to y, you know, you were thinking about traveling in a different direction. Maybe you nudge it straight up, and you're wondering, okay, how does that influence the output? And the question here, with directional derivatives, what if you have some vector, v, and I'll give a little vector hat on top of it, that, you know, I don't know, let's say it's negative one, two, is the vector. So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction."}, {"video_title": "Directional derivative.mp3", "Sentence": "And when you did it with respect to y, you know, you were thinking about traveling in a different direction. Maybe you nudge it straight up, and you're wondering, okay, how does that influence the output? And the question here, with directional derivatives, what if you have some vector, v, and I'll give a little vector hat on top of it, that, you know, I don't know, let's say it's negative one, two, is the vector. So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction. So it's gonna be something that ends up there. This is your vector, v, at least if you're thinking of v as stemming from the original point. And you're wondering, what does a nudge in that direction do to the function itself?"}, {"video_title": "Directional derivative.mp3", "Sentence": "So you'd be thinking about that as a step of negative one in the x direction, and then two more in the y direction. So it's gonna be something that ends up there. This is your vector, v, at least if you're thinking of v as stemming from the original point. And you're wondering, what does a nudge in that direction do to the function itself? And remember, with these original, you know, nudges in the x direction, nudges in the y, you're not really thinking of it as, you know, this is kind of a large step. You're really thinking of it as something itty, itty, bitty, bitty, bitty. You know, it's not that, but it's really something very, very small."}, {"video_title": "Directional derivative.mp3", "Sentence": "And you're wondering, what does a nudge in that direction do to the function itself? And remember, with these original, you know, nudges in the x direction, nudges in the y, you're not really thinking of it as, you know, this is kind of a large step. You're really thinking of it as something itty, itty, bitty, bitty, bitty. You know, it's not that, but it's really something very, very small. And formally, you'd be thinking about the limit, as this gets really, really, really small, approaching zero, and this gets really, really small, approaching zero, what does the ratio of the two approach? And similarly, with the y, you're not thinking of it as something, this is pretty sizable, but it would be something really, really small. And the directional derivative is similar."}, {"video_title": "Directional derivative.mp3", "Sentence": "You know, it's not that, but it's really something very, very small. And formally, you'd be thinking about the limit, as this gets really, really, really small, approaching zero, and this gets really, really small, approaching zero, what does the ratio of the two approach? And similarly, with the y, you're not thinking of it as something, this is pretty sizable, but it would be something really, really small. And the directional derivative is similar. You're not thinking of the actual vector actually taking a step along that, but you'd be thinking of taking a step along, say, h, multiplied by that vector. And h might represent some really, really small numbers. I know maybe this here is like 0.001."}, {"video_title": "Directional derivative.mp3", "Sentence": "And the directional derivative is similar. You're not thinking of the actual vector actually taking a step along that, but you'd be thinking of taking a step along, say, h, multiplied by that vector. And h might represent some really, really small numbers. I know maybe this here is like 0.001. And when you're doing this formally, you'd just be thinking the limit as h goes to zero. So the directional derivative is saying, when you take a slight nudge in the direction of that vector, what is the resulting change to the output? And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here."}, {"video_title": "Directional derivative.mp3", "Sentence": "I know maybe this here is like 0.001. And when you're doing this formally, you'd just be thinking the limit as h goes to zero. So the directional derivative is saying, when you take a slight nudge in the direction of that vector, what is the resulting change to the output? And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here. So it's kind of like you took negative one nudge in the x direction, and then two nudges in the y direction. You know, so for whatever your nudge in the v direction there, you take a negative one step by x, and then two of them up by y. So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here."}, {"video_title": "Directional derivative.mp3", "Sentence": "And one way to think about this is you say, well, that slight nudge of the vector, if we actually expand things out, and we look at the definition itself, it'll be negative h, negative one times that component, and then two h here. So it's kind of like you took negative one nudge in the x direction, and then two nudges in the y direction. You know, so for whatever your nudge in the v direction there, you take a negative one step by x, and then two of them up by y. So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here. So this is the directional derivative in the direction of v. And there's a whole bunch of other notations, too. You know, I think there's like derivative of f with respect to that vector is one way people think about it. Some people just write like partial with a little subscript vector."}, {"video_title": "Directional derivative.mp3", "Sentence": "So when we actually write this out, the notation, by the way, is you take that same nabla from the gradient, but then you put the vector down here. So this is the directional derivative in the direction of v. And there's a whole bunch of other notations, too. You know, I think there's like derivative of f with respect to that vector is one way people think about it. Some people just write like partial with a little subscript vector. There's a whole bunch of different notations, but this is the one I like. You think that nabla with a little f down there, with a little v for your vector of f, and it's still a function of x and y. And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video."}, {"video_title": "Directional derivative.mp3", "Sentence": "Some people just write like partial with a little subscript vector. There's a whole bunch of different notations, but this is the one I like. You think that nabla with a little f down there, with a little v for your vector of f, and it's still a function of x and y. And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video. And for this particular example, a good guess that you might have is to say, well, we take a negative step in the x direction, so you think of it as whatever the change that's caused by such a step in the x direction, you do the negative of that, and then it's two steps in the y direction, so whatever the change caused by a tiny step in the y direction, let's just take two of those. Two times partial f, partial y. And this is actually how you calculate it."}, {"video_title": "Directional derivative.mp3", "Sentence": "And the reason I like this is it's indicative of how you end up calculating it, which I'll talk about at the end of the video. And for this particular example, a good guess that you might have is to say, well, we take a negative step in the x direction, so you think of it as whatever the change that's caused by such a step in the x direction, you do the negative of that, and then it's two steps in the y direction, so whatever the change caused by a tiny step in the y direction, let's just take two of those. Two times partial f, partial y. And this is actually how you calculate it. And if I was gonna be more general, you know, let's say we've got a vector w. I'm gonna keep it abstract and just call it a and b as its components, rather than the specific numbers. You would say that the directional derivative in the direction of w, whatever that is, of f, is equal to a times the partial derivative of f with respect to x plus b times the partial derivative of f with respect to y. And this is it."}, {"video_title": "Directional derivative.mp3", "Sentence": "And this is actually how you calculate it. And if I was gonna be more general, you know, let's say we've got a vector w. I'm gonna keep it abstract and just call it a and b as its components, rather than the specific numbers. You would say that the directional derivative in the direction of w, whatever that is, of f, is equal to a times the partial derivative of f with respect to x plus b times the partial derivative of f with respect to y. And this is it. This is the formula that you would use for the directional derivative. And again, the way that you're thinking about this is you're really saying, you know, you take a little nudge that's a in the x direction and b in the y direction, so this should kind of make sense. And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient."}, {"video_title": "Directional derivative.mp3", "Sentence": "And this is it. This is the formula that you would use for the directional derivative. And again, the way that you're thinking about this is you're really saying, you know, you take a little nudge that's a in the x direction and b in the y direction, so this should kind of make sense. And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient. And this is because it makes it much more compact, more general if you're dealing with other dimensions. So I'll just write it over here. If you look at this expression, it looks like a dot product."}, {"video_title": "Directional derivative.mp3", "Sentence": "And sometimes you see this written not with respect to the partial derivatives themselves and the actual components a and b, but with respect to the gradient. And this is because it makes it much more compact, more general if you're dealing with other dimensions. So I'll just write it over here. If you look at this expression, it looks like a dot product. If you take the dot product of the vectors, a, b, and the one that has the partial derivatives in it, so what's lined up with a is the partial derivative with respect to x, partial f, partial x, and what's lined up with b is the partial derivative with respect to y. And you look at this and you say, hey, a, b, I mean, that's just the original vector, right? That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component."}, {"video_title": "Directional derivative.mp3", "Sentence": "If you look at this expression, it looks like a dot product. If you take the dot product of the vectors, a, b, and the one that has the partial derivatives in it, so what's lined up with a is the partial derivative with respect to x, partial f, partial x, and what's lined up with b is the partial derivative with respect to y. And you look at this and you say, hey, a, b, I mean, that's just the original vector, right? That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component. That's just the gradient. That is the gradient of f. And here, you know, it's nabla without that little w at the bottom. And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it."}, {"video_title": "Directional derivative.mp3", "Sentence": "That's w. That's the vector w. And then you're dotting this with, well, partial derivative with respect to x in one component, the other partial derivative in the other component. That's just the gradient. That is the gradient of f. And here, you know, it's nabla without that little w at the bottom. And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it. So this is really what you'll see in a textbook or see as the compact way of writing it. And you can see how this is more flexible for dimensions. So if we were talking about something that has like a five-dimensional input and the vector, the direction you move, has five different components, this is flexible."}, {"video_title": "Directional derivative.mp3", "Sentence": "And this is why we use this notation, because it's so suggestive of the way that you ultimately calculate it. So this is really what you'll see in a textbook or see as the compact way of writing it. And you can see how this is more flexible for dimensions. So if we were talking about something that has like a five-dimensional input and the vector, the direction you move, has five different components, this is flexible. When you expand it, the gradient would have five components and the vector itself would have five components. So this is the directional derivative and how you calculate it. And the way you interpret, you're thinking of moving along that vector by a tiny nudge, by a tiny, you know, little value multiplied by that vector, and saying, how does that change the output and what's the ratio of the resulting change?"}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we said if we can just prove that each of these parts are equal to each other, we essentially have proven that that is equal to that. Because here in yellow is another way of writing the flux across the surface. And here in green is another way of writing the triple integral over our region of the divergence of f. What I'm going to do in this video and probably the next video is prove that these two are equivalent to each other. And I'm going to prove it using the fact that our original region is a simple solid region, or in particular, is a type 1 region. And then that's essentially going to be it because you can use the exact same argument and the fact that it is a type 2 region to prove this. And the exact same argument and the fact that it is a type 3 region to prove that. So I'm going to assume it's a type 1, which I can."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'm going to prove it using the fact that our original region is a simple solid region, or in particular, is a type 1 region. And then that's essentially going to be it because you can use the exact same argument and the fact that it is a type 2 region to prove this. And the exact same argument and the fact that it is a type 3 region to prove that. So I'm going to assume it's a type 1, which I can. It's a type 1, 2, and 3 region. So given the fact that it's type 1, I'm now going to prove this relationship right over here. And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'm going to assume it's a type 1, which I can. It's a type 1, 2, and 3 region. So given the fact that it's type 1, I'm now going to prove this relationship right over here. And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region. So let's get going. So type 1 region, just to remind ourselves, a type 1 region is a region that is equal to the set of all x, y's, and z's, such that the xy pairs are a member of a domain in the xy plane. And z is bounded by two functions."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then I'll leave it up to you to do the exact same argument with a type 2 and a type 3 region. So let's get going. So type 1 region, just to remind ourselves, a type 1 region is a region that is equal to the set of all x, y's, and z's, such that the xy pairs are a member of a domain in the xy plane. And z is bounded by two functions. z's lower bound is f1 of x and y. And that's going to be less than or equal to z. And z's upper bound, we can call it f2 of x, y."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And z is bounded by two functions. z's lower bound is f1 of x and y. And that's going to be less than or equal to z. And z's upper bound, we can call it f2 of x, y. And then let me close the set notation right over here. And let me just draw a general version of a type 1 region. So let me draw my x, y, and z axes."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And z's upper bound, we can call it f2 of x, y. And then let me close the set notation right over here. And let me just draw a general version of a type 1 region. So let me draw my x, y, and z axes. So this is my z-axis. This is my x-axis. And there is my y-axis."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw my x, y, and z axes. So this is my z-axis. This is my x-axis. And there is my y-axis. And so we might have a region D here. So our region, I'll draw it as a little circle right over here. This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And there is my y-axis. And so we might have a region D here. So our region, I'll draw it as a little circle right over here. This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say. So this might define an f1, which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want. So every x, y there, when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be, you might get a surface that looks something like this. It doesn't have to be flat."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is our region D. And for any x, y in our region D, you can evaluate the function f. You can figure out an f1, I should say. So this might define an f1, which we can kind of imagine as a surface or a little thing that's at the bottom of a cylinder if you want. So every x, y there, when you evaluate or when you figure out what the corresponding f1 of those points in this domain would be, you might get a surface that looks something like this. It doesn't have to be flat. But hopefully this gives the idea. It doesn't have to be completely flat. It can be curved or whatever else."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It doesn't have to be flat. But hopefully this gives the idea. It doesn't have to be completely flat. It can be curved or whatever else. But this just shows that every x, y, when you evaluate it right over here, it gets associated with a point, this lower bound surface right over here. And I'll draw a dotted line to show that's only for the x, y's in this domain. And then we have an upper bound surface that might be up here."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It can be curved or whatever else. But this just shows that every x, y, when you evaluate it right over here, it gets associated with a point, this lower bound surface right over here. And I'll draw a dotted line to show that's only for the x, y's in this domain. And then we have an upper bound surface that might be up here. Give me any x, y. When I evaluate f2, I get this surface up here. And once again, they don't have to look the same."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we have an upper bound surface that might be up here. Give me any x, y. When I evaluate f2, I get this surface up here. And once again, they don't have to look the same. This could be like a dome, or it could be slanted, or who knows what it might be. But this will give you the general idea. And then z fills up the region."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And once again, they don't have to look the same. This could be like a dome, or it could be slanted, or who knows what it might be. But this will give you the general idea. And then z fills up the region. Remember, the region isn't just the surface of the figure. It's the entire volume inside of it. So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then z fills up the region. Remember, the region isn't just the surface of the figure. It's the entire volume inside of it. So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region. And this way I drew it, it looks like a cylinder. But it doesn't have to be a cylinder like this. And these two surfaces might touch each other, in which case there would be no side of the cylinder."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when z varies between that surface and that surface, for any given x, y in our domain, we fill up the entire region. And this way I drew it, it looks like a cylinder. But it doesn't have to be a cylinder like this. And these two surfaces might touch each other, in which case there would be no side of the cylinder. They could be lumpier than this. They might be inclined in some way. But hopefully this is a good generalization of a type I region."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And these two surfaces might touch each other, in which case there would be no side of the cylinder. They could be lumpier than this. They might be inclined in some way. But hopefully this is a good generalization of a type I region. Now, a type I region, you can kind of think of it, it can be broken up into three parts. It can be broken up into surface, or the surfaces of a type I region, I should say, can be broken up into three parts. It can be broken up into, let's call that surface one."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But hopefully this is a good generalization of a type I region. Now, a type I region, you can kind of think of it, it can be broken up into three parts. It can be broken up into surface, or the surfaces of a type I region, I should say, can be broken up into three parts. It can be broken up into, let's call that surface one. Let's call this right over here surface two, the top of the cylinder, or whatever kind of lumpy top it might be. And let's call the side, if these two surfaces don't touch each other, let's call that surface three. There might not necessarily even be a surface three if these two touch each other in the case of a sphere."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It can be broken up into, let's call that surface one. Let's call this right over here surface two, the top of the cylinder, or whatever kind of lumpy top it might be. And let's call the side, if these two surfaces don't touch each other, let's call that surface three. There might not necessarily even be a surface three if these two touch each other in the case of a sphere. But let's just assume that there actually is a surface three. So if we're evaluating the surface integral, we'll think about this triple integral in a second. But let's think about how we can rewrite this surface integral right over here."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "There might not necessarily even be a surface three if these two touch each other in the case of a sphere. But let's just assume that there actually is a surface three. So if we're evaluating the surface integral, we'll think about this triple integral in a second. But let's think about how we can rewrite this surface integral right over here. Instead of this entire surface is s1 plus s2 plus s3. So we can essentially break this up into three separate surface integrals. So let's do that."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's think about how we can rewrite this surface integral right over here. Instead of this entire surface is s1 plus s2 plus s3. So we can essentially break this up into three separate surface integrals. So let's do that. So remember, we're just focusing on this part right over here. So the surface integral of r times k dot n, the dot product of k and n, ds can be rewritten as the surface integral over s2 of r times k dot n, ds, plus the surface integral over s1 of r times k dot n, ds, plus the surface integral over surface three of the same thing, r times k dot n, ds. Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's do that. So remember, we're just focusing on this part right over here. So the surface integral of r times k dot n, the dot product of k and n, ds can be rewritten as the surface integral over s2 of r times k dot n, ds, plus the surface integral over s1 of r times k dot n, ds, plus the surface integral over surface three of the same thing, r times k dot n, ds. Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other. And for a type one situation right over here, the normal vector at any given point on this side of the cylinder for this type one region, if there is this in-between region, there always isn't. In a sphere, there wouldn't be the surface, in which case this would be 0. But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, the way I've drawn it, and this is actually the case, s3 is if those surfaces don't touch each other. And for a type one situation right over here, the normal vector at any given point on this side of the cylinder for this type one region, if there is this in-between region, there always isn't. In a sphere, there wouldn't be the surface, in which case this would be 0. But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component. The normal vector will always be pointing flat out. It will only have an i and j component. So if you take this normal vector right over here, does not have a k component, and you are dotting it with a k vector, then the dot product of two things that are orthogonal, the k vector, goes like that, you're going to get 0."}, {"video_title": "Divergence theorem proof (part 2)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But if there is this surface in a type one region, the one that essentially connects the boundaries of the top and the bottom, then the normal vector will never have a k component. The normal vector will always be pointing flat out. It will only have an i and j component. So if you take this normal vector right over here, does not have a k component, and you are dotting it with a k vector, then the dot product of two things that are orthogonal, the k vector, goes like that, you're going to get 0. So this thing is going to be 0, because k dot n is going to be 0 in this situation for this surface. k dot n is going to be equal to 0. So this part right over here simplifies to this right over here."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see if we can apply some of our new tools to solve some line integrals. So let's say we have a line integral along a closed curve. I'm going to define the path in a second. Of x squared plus y squared times dx plus 2xy times dy. And then our curve C is going to be defined by the parameterization x is equal to cosine of t and y is equal to sine of t. And this is valid for t between 0 and 2pi. So this is essentially a circle, a unit circle in the xy plane. We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Of x squared plus y squared times dx plus 2xy times dy. And then our curve C is going to be defined by the parameterization x is equal to cosine of t and y is equal to sine of t. And this is valid for t between 0 and 2pi. So this is essentially a circle, a unit circle in the xy plane. We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process. So the first thing you might say is, hey, this looks like a line integral, but you have a dx and dy. I don't see a dot dr here. It's not clear to me that this is some type of even a vector line integral."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We know how to solve these, but let's see if we can use some of our discoveries in the last couple of videos to maybe simplify this process. So the first thing you might say is, hey, this looks like a line integral, but you have a dx and dy. I don't see a dot dr here. It's not clear to me that this is some type of even a vector line integral. I don't see any vectors. What I want to do first, and the reason why I wanted to show you this example, is just to show you that this is just another form of writing, really, a vector line integral. And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's not clear to me that this is some type of even a vector line integral. I don't see any vectors. What I want to do first, and the reason why I wanted to show you this example, is just to show you that this is just another form of writing, really, a vector line integral. And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there. I'm just going to write it's x of t times i plus y of t times j. We've seen several videos now that we can write dr dt as being equal to dx dt times i plus dy dt times j. We've seen this multiple times."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to show you that, you just have to realize if I have some r of t, this is our curve, and I won't even write these functions in there. I'm just going to write it's x of t times i plus y of t times j. We've seen several videos now that we can write dr dt as being equal to dx dt times i plus dy dt times j. We've seen this multiple times. And we've seen multiple times we want to get the differential dr. We could just multiply everything times dt. And normally, I just put a dt here and a dt there and get rid of this dt. But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've seen this multiple times. And we've seen multiple times we want to get the differential dr. We could just multiply everything times dt. And normally, I just put a dt here and a dt there and get rid of this dt. But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt. So dr, you can imagine, is equal to dx times the unit vector i plus dy times the unit vector j. So put that aside, and you might already see a pattern here. So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing?"}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But if you multiply everything times dt, if you view this as the differentials as actual numbers you can multiply, normally you can treat them like that, then you just get rid of all of the dt. So dr, you can imagine, is equal to dx times the unit vector i plus dy times the unit vector j. So put that aside, and you might already see a pattern here. So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing? What is this thing over here? What is f dot dr going to be? Dot products, you just multiply the corresponding components of our vectors and then add them up."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we define our vector field f of x, y as being equal to x squared plus y squared i plus 2xyj, what is this thing? What is this thing over here? What is f dot dr going to be? Dot products, you just multiply the corresponding components of our vectors and then add them up. So it's going to be, if you take this f and dot it with that dr, you're going to get the i component, x squared plus y squared times that dx plus the j component, 2xy times that dy. That's the dot product. Notice, this thing right here, that thing right here, is identical to that thing right there."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Dot products, you just multiply the corresponding components of our vectors and then add them up. So it's going to be, if you take this f and dot it with that dr, you're going to get the i component, x squared plus y squared times that dx plus the j component, 2xy times that dy. That's the dot product. Notice, this thing right here, that thing right here, is identical to that thing right there. So our line integral, just to put it in a form that we're familiar with, this is the same exact thing as the line integral over this curve C, this closed curve C, of this f, maybe I'll write it in that magenta color, or that actually is more of a purple or pink color, f dot this dr. That's what this line integral is, it's just a different way of writing. So now that you've seen it, in the future if you see it in this differential form, you'll immediately know, okay, there's one vector field, that this is its x component, this is its y component, dotting with dr. This is the x component of dr, the i component, and this is the y component, or the j component of the dr."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Notice, this thing right here, that thing right here, is identical to that thing right there. So our line integral, just to put it in a form that we're familiar with, this is the same exact thing as the line integral over this curve C, this closed curve C, of this f, maybe I'll write it in that magenta color, or that actually is more of a purple or pink color, f dot this dr. That's what this line integral is, it's just a different way of writing. So now that you've seen it, in the future if you see it in this differential form, you'll immediately know, okay, there's one vector field, that this is its x component, this is its y component, dotting with dr. This is the x component of dr, the i component, and this is the y component, or the j component of the dr. So you immediately know what the vector field is that we're taking a line integral of. This is the x, that's the y. Let's ask ourselves a question, is f conservative?"}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the x component of dr, the i component, and this is the y component, or the j component of the dr. So you immediately know what the vector field is that we're taking a line integral of. This is the x, that's the y. Let's ask ourselves a question, is f conservative? So is f equal to the gradient of some scalar field, we'll call it capital F. Is this the case? So let's assume it is, and see if we can solve for a scalar field whose gradient really is f. Then we know that f is conservative. And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's ask ourselves a question, is f conservative? So is f equal to the gradient of some scalar field, we'll call it capital F. Is this the case? So let's assume it is, and see if we can solve for a scalar field whose gradient really is f. Then we know that f is conservative. And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0. And we'd be done. So if we can show this, then the answer to this question, or this question, is going to be 0. We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if f is conservative, and this is the whole reason we want to do it, that means that any closed loop, any line integral over a closed curve of f, is going to be equal to 0. And we'd be done. So if we can show this, then the answer to this question, or this question, is going to be 0. We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives. So let's see if we can find an f whose gradient is equal to that right there. So in order for f's gradient to be that, that means that the partial derivative of our capital F with respect to x has got to be equal to that right there. It's got to be equal to x squared plus y squared."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We don't even have to mess with the cosine of t's and the sine of t's and all of that, and actually we don't even have to take anti-derivatives. So let's see if we can find an f whose gradient is equal to that right there. So in order for f's gradient to be that, that means that the partial derivative of our capital F with respect to x has got to be equal to that right there. It's got to be equal to x squared plus y squared. And that also tells us that the partial derivative of capital F with respect to y has got to be equal to 2xy. And just as a review, if I have the gradient of any function of any scalar field is equal to the partial of f with respect to x times i plus the partial of capital F with respect to y times j. So that's why I'm just pattern matching."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's got to be equal to x squared plus y squared. And that also tells us that the partial derivative of capital F with respect to y has got to be equal to 2xy. And just as a review, if I have the gradient of any function of any scalar field is equal to the partial of f with respect to x times i plus the partial of capital F with respect to y times j. So that's why I'm just pattern matching. I'm just saying, well, gee, if this is the gradient of that, then this must be that, which I wrote down right here. And this must be that, which I wrote down here. So let's see if I can find an f that satisfies both of these constraints."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's why I'm just pattern matching. I'm just saying, well, gee, if this is the gradient of that, then this must be that, which I wrote down right here. And this must be that, which I wrote down here. So let's see if I can find an f that satisfies both of these constraints. So we can just take the anti-derivative with respect to x on both sides. And if we take the anti-derivative with respect to x on both sides, remember, you just treat y like a constant or y squared like a constant. It's just a number."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see if I can find an f that satisfies both of these constraints. So we can just take the anti-derivative with respect to x on both sides. And if we take the anti-derivative with respect to x on both sides, remember, you just treat y like a constant or y squared like a constant. It's just a number. So then we could say that f is equal to the anti-derivative of x squared is x to the third over 3. And then the anti-derivative of y squared, remember, with respect to x. So you just treat it like a number."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's just a number. So then we could say that f is equal to the anti-derivative of x squared is x to the third over 3. And then the anti-derivative of y squared, remember, with respect to x. So you just treat it like a number. That could just be the number k. Or this could be the number 5. So this is just going to be that times x. So plus x times y squared."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you just treat it like a number. That could just be the number k. Or this could be the number 5. So this is just going to be that times x. So plus x times y squared. And then there could be some function of y here. So plus some, I don't know, I'll call it g of y. Because there could have been some function of y here."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So plus x times y squared. And then there could be some function of y here. So plus some, I don't know, I'll call it g of y. Because there could have been some function of y here. If it's a pure function of y, when you take the derivative or the partial with respect to x, this would have disappeared. So it would reappear when we take the anti-derivative. So let me, just to be clear, let me make it clear that f is going to be a function of x and y."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because there could have been some function of y here. If it's a pure function of y, when you take the derivative or the partial with respect to x, this would have disappeared. So it would reappear when we take the anti-derivative. So let me, just to be clear, let me make it clear that f is going to be a function of x and y. So this, we just took the, I guess you could say the anti-derivative with respect to x. Let's see if we take the anti-derivative with respect to y and then we can reconcile the two. So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me, just to be clear, let me make it clear that f is going to be a function of x and y. So this, we just took the, I guess you could say the anti-derivative with respect to x. Let's see if we take the anti-derivative with respect to y and then we can reconcile the two. So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here. So remember, you just treat x like it's just some number. It could be a k, it could be an m, it could be a 5. It's just some number."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So based on this, f of xy is going to have to look like, so let's take the anti-derivative with respect to y here. So remember, you just treat x like it's just some number. It could be a k, it could be an m, it could be a 5. It's just some number. So if x is just some, the anti-derivative of 2y is y squared. And if x is just a number there, the anti-derivative of this with respect to y is just going to be xy squared. Don't believe me?"}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's just some number. So if x is just some, the anti-derivative of 2y is y squared. And if x is just a number there, the anti-derivative of this with respect to y is just going to be xy squared. Don't believe me? Take the partial of this with respect to y. Treat x like a constant, you'll get 2 times xy. And then you, with no exponent there."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Don't believe me? Take the partial of this with respect to y. Treat x like a constant, you'll get 2 times xy. And then you, with no exponent there. And of course, since we took the anti-derivative with respect to x, there might be some function of x here. We were just basing it off of that information. Now, given that, this information says f of xy is going to have to look something like this."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you, with no exponent there. And of course, since we took the anti-derivative with respect to x, there might be some function of x here. We were just basing it off of that information. Now, given that, this information says f of xy is going to have to look something like this. This information tells us f of xy is going to have to look something like that. Let's see if there is an f of xy that looks like both of them, essentially. So let's see."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, given that, this information says f of xy is going to have to look something like this. This information tells us f of xy is going to have to look something like that. Let's see if there is an f of xy that looks like both of them, essentially. So let's see. We have, on this one, we have xy squared here. We have an xy squared there. So good."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see. We have, on this one, we have xy squared here. We have an xy squared there. So good. That looks good. And then over here, we have an f of x. We have something that's a pure function of x."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So good. That looks good. And then over here, we have an f of x. We have something that's a pure function of x. And here we have something that is a pure function of x. So these two things could be the same thing. And then, here we have a pure function of y that might be there."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have something that's a pure function of x. And here we have something that is a pure function of x. So these two things could be the same thing. And then, here we have a pure function of y that might be there. But it didn't really show up anywhere over here. So we could just say, hey, that's going to be 0. 0 is a pure function of y."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then, here we have a pure function of y that might be there. But it didn't really show up anywhere over here. So we could just say, hey, that's going to be 0. 0 is a pure function of y. You could have something called g of y is equal to 0. And then we get that f of xy is equal to x to the third over 3 plus xy squared. And the gradient of this is going to be equal to f. And we've already established that."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "0 is a pure function of y. You could have something called g of y is equal to 0. And then we get that f of xy is equal to x to the third over 3 plus xy squared. And the gradient of this is going to be equal to f. And we've already established that. But just to hit the point home, let's take the gradient of it. Just so you don't believe this little stuff that I did right there. Let's take the gradient."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the gradient of this is going to be equal to f. And we've already established that. But just to hit the point home, let's take the gradient of it. Just so you don't believe this little stuff that I did right there. Let's take the gradient. The gradient of f is equal to, and sometimes people put a little vector there because you're getting a vector out of it. You could put a little vector on top of that gradient sign. The gradient of f is going to be what?"}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's take the gradient. The gradient of f is equal to, and sometimes people put a little vector there because you're getting a vector out of it. You could put a little vector on top of that gradient sign. The gradient of f is going to be what? The partial of this with respect to x times i. So the partial of this with respect to x, the derivative here is 3 divided by 3 is 1. So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The gradient of f is going to be what? The partial of this with respect to x times i. So the partial of this with respect to x, the derivative here is 3 divided by 3 is 1. So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y. Well, the partial with respect to y of this is 0. The partial with respect to y of this is 2xy to the first. So it's 2xy times j."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So just x squared plus the derivative of this with respect to x is y squared times i plus the partial with respect to y. Well, the partial with respect to y of this is 0. The partial with respect to y of this is 2xy to the first. So it's 2xy times j. And this is exactly equal to f, our f that we wrote up there. So we've established that f can definitely be written, f is definitely the gradient of some potential scalar function there. So f is conservative."}, {"video_title": "Example of closed line integral of conservative field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's 2xy times j. And this is exactly equal to f, our f that we wrote up there. So we've established that f can definitely be written, f is definitely the gradient of some potential scalar function there. So f is conservative. And that tells us that this closed loop line integral of f is going to be equal to 0. And we are done. We've explored the actual parameterization of the path."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "So this over here is the input space, it's just a copy of the xy-plane. And the output space is also two-dimensional, so the output space in this case, also the two-dimensional plane. And what I'm gonna do, I'm just gonna first play an example of one of these transformations and then go through the details of the underlying function and how you can understand the transformation as a result. So here's what it looks like, here's what we're gonna be going towards. Very complicated, a lot of points moving, lots of different things happening here. And what's common with this sort of thing when you're thinking about moving from two dimensions to two dimensions, given that it's really the same space, the xy-plane, you often just think about the input and output space all at once and instead just watch a copy of that plane move on to itself. And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "So here's what it looks like, here's what we're gonna be going towards. Very complicated, a lot of points moving, lots of different things happening here. And what's common with this sort of thing when you're thinking about moving from two dimensions to two dimensions, given that it's really the same space, the xy-plane, you often just think about the input and output space all at once and instead just watch a copy of that plane move on to itself. And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you. When I think about transformations, it's usually a very vague thought in the back of my mind somewhere, but it helps to understand what's really going on with the function. I'll talk about that more at the end, but first let's just go into what this function is. So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "And by the way, when I say watch, I don't mean that you'll always have an animation like this just sort of sitting in front of you. When I think about transformations, it's usually a very vague thought in the back of my mind somewhere, but it helps to understand what's really going on with the function. I'll talk about that more at the end, but first let's just go into what this function is. So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output. So just to help start understanding this, let's take a relatively simple point like the origin. So here the origin, which is zero, zero, and let's think about what happens to that. f of zero, zero."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "So the one that I told the computer to animate here is f of xy, as the input, is equal to x squared plus y squared as the x component of the output, and x squared minus y squared as the y component of the output. So just to help start understanding this, let's take a relatively simple point like the origin. So here the origin, which is zero, zero, and let's think about what happens to that. f of zero, zero. Well, x and y are both zero, so that top is zero. And same with the bottom. That bottom also equals zero, which means it's taking the point zero, zero to itself."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "f of zero, zero. Well, x and y are both zero, so that top is zero. And same with the bottom. That bottom also equals zero, which means it's taking the point zero, zero to itself. And if you watch the transformation, what this means is that the point zero stays fixed. It's like you can hold your thumb down on it and nothing really happens to it. And in fact, we call this a fixed point of the function as a whole."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "That bottom also equals zero, which means it's taking the point zero, zero to itself. And if you watch the transformation, what this means is that the point zero stays fixed. It's like you can hold your thumb down on it and nothing really happens to it. And in fact, we call this a fixed point of the function as a whole. And that kind of terminology doesn't really make sense unless you're thinking of the function as a transformation. So let's look at another example here. Let's take a point like one, one."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "And in fact, we call this a fixed point of the function as a whole. And that kind of terminology doesn't really make sense unless you're thinking of the function as a transformation. So let's look at another example here. Let's take a point like one, one. f of one, one. So in the input space, let's just kind of start this thing over so we're only looking at the input. In the input space, one, one is sitting right here."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "Let's take a point like one, one. f of one, one. So in the input space, let's just kind of start this thing over so we're only looking at the input. In the input space, one, one is sitting right here. And we're wondering where that's gonna move. So when we plug it in, x squared plus y squared, it's gonna be one squared plus one squared. And then the bottom, x squared minus y squared, one squared minus y squared."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "In the input space, one, one is sitting right here. And we're wondering where that's gonna move. So when we plug it in, x squared plus y squared, it's gonna be one squared plus one squared. And then the bottom, x squared minus y squared, one squared minus y squared. Whoop, minus one squared. I'm plugging things in here. So that's two, zero."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then the bottom, x squared minus y squared, one squared minus y squared. Whoop, minus one squared. I'm plugging things in here. So that's two, zero. Two, zero. Which means we expect this point to move over to two, zero in some way. So if we watch the transformation, we expect to watch that point move over to here."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "So that's two, zero. Two, zero. Which means we expect this point to move over to two, zero in some way. So if we watch the transformation, we expect to watch that point move over to here. And again, it can be hard to follow because there's a lot of moving parts. But if you're careful as you watch it, the point will actually land right there. And you can, in principle, do this for any given point and understand how it moves from one to another."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "So if we watch the transformation, we expect to watch that point move over to here. And again, it can be hard to follow because there's a lot of moving parts. But if you're careful as you watch it, the point will actually land right there. And you can, in principle, do this for any given point and understand how it moves from one to another. But you might ask, hey Grant, what is the point of all of this? We have other ways of visualizing functions that are more precise and kind of less confusing, to be honest. Like vector fields are a great way for functions like this."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "And you can, in principle, do this for any given point and understand how it moves from one to another. But you might ask, hey Grant, what is the point of all of this? We have other ways of visualizing functions that are more precise and kind of less confusing, to be honest. Like vector fields are a great way for functions like this. Graphs were a great way for functions with one input and one output. Why think in terms of transformations? And the main reason is conceptual."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "Like vector fields are a great way for functions like this. Graphs were a great way for functions with one input and one output. Why think in terms of transformations? And the main reason is conceptual. It's not like you'll have an animation sitting in front of you. And it's not like you're gonna, by hand, evaluate a bunch of points and think of how they move. But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the main reason is conceptual. It's not like you'll have an animation sitting in front of you. And it's not like you're gonna, by hand, evaluate a bunch of points and think of how they move. But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding. Things like derivatives or the variations of the derivative that you're gonna learn with multivariable calculus. There's different ways of understanding it in terms of like stretching or squishing space and things like this that doesn't really have a good analog in terms of graphs or vector fields. So it adds a new color to your understanding."}, {"video_title": "Transformations, part 2   Multivariable calculus   Khan Academy.mp3", "Sentence": "But there's a lot of different concepts in math and with functions where when you understand it in terms of a transformation, it gives you a more nuanced understanding. Things like derivatives or the variations of the derivative that you're gonna learn with multivariable calculus. There's different ways of understanding it in terms of like stretching or squishing space and things like this that doesn't really have a good analog in terms of graphs or vector fields. So it adds a new color to your understanding. Also, transformations are a super important part of linear algebra. There will come a point when you start learning the connection between linear algebra and multivariable calculus. And if you have a strong conception of transformations both in the context of linear algebra and in the context of multivariable calculus, you'll be in a much better position to understand the connection between those two fields."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in order to evaluate a surface integral, we had to take the parameterization, take its partial with respect to s and with respect to t. We did that in the first video. Then we had to take its cross product. We did that in the second video. Now we're ready to take the magnitude of the cross product. We have to take the magnitude of the cross product. And then we can evaluate it inside of a double integral. And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we're ready to take the magnitude of the cross product. We have to take the magnitude of the cross product. And then we can evaluate it inside of a double integral. And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career. So this is kind of exciting. So this was the cross product right here. Now let's take the magnitude of this thing."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we will have solved, or we would have computed an actual surface integral, something you see very few times in your education career. So this is kind of exciting. So this was the cross product right here. Now let's take the magnitude of this thing. And you might remember, the magnitude of any vector is kind of a Pythagorean theorem. And in this case, it's going to be kind of the distance formula, the Pythagorean theorem in three dimensions. So the magnitude, this right here is equal to, just as a reminder, is equal to this right here."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's take the magnitude of this thing. And you might remember, the magnitude of any vector is kind of a Pythagorean theorem. And in this case, it's going to be kind of the distance formula, the Pythagorean theorem in three dimensions. So the magnitude, this right here is equal to, just as a reminder, is equal to this right here. It's equal to the partial of r with respect to s cross with this partial of r with respect to t. Let me copy it and paste it. That is equal to that right there, put an equal sign. These two quantities are equal."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the magnitude, this right here is equal to, just as a reminder, is equal to this right here. It's equal to the partial of r with respect to s cross with this partial of r with respect to t. Let me copy it and paste it. That is equal to that right there, put an equal sign. These two quantities are equal. Now we want to figure out the magnitude. So if we want to take the magnitude of this thing, so if we are interested in taking the magnitude of that thing, that's going to be equal to, well, this is just a scalar that's multiplying everything. So let's just write the scalar out there."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These two quantities are equal. Now we want to figure out the magnitude. So if we want to take the magnitude of this thing, so if we are interested in taking the magnitude of that thing, that's going to be equal to, well, this is just a scalar that's multiplying everything. So let's just write the scalar out there. So b plus a cosine of s times the magnitude of this thing right here. And the magnitude of this thing right here is going to be the sum of, you can imagine it's the square root of this thing dotted with itself. Or you could say it's the sum of the squares of each of these terms to the 1 half power."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's just write the scalar out there. So b plus a cosine of s times the magnitude of this thing right here. And the magnitude of this thing right here is going to be the sum of, you can imagine it's the square root of this thing dotted with itself. Or you could say it's the sum of the squares of each of these terms to the 1 half power. So let me write it like that. So it's equal to, let me write the sum of the squares. So if you square this, you get a squared cosine squared of s sine squared of t. That's that term."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or you could say it's the sum of the squares of each of these terms to the 1 half power. So let me write it like that. So it's equal to, let me write the sum of the squares. So if you square this, you get a squared cosine squared of s sine squared of t. That's that term. Plus, let me color code it. That's that term. Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you square this, you get a squared cosine squared of s sine squared of t. That's that term. Plus, let me color code it. That's that term. Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term. And then finally, I'll do another color, this term squared. So plus a squared sine squared of s. That's going to be all of this business to the 1 half power. This right here is the same thing as the magnitude of this right here."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus, I'll do the magenta, plus that term squared, plus a squared cosine squared of s cosine squared of t. That's that term. And then finally, I'll do another color, this term squared. So plus a squared sine squared of s. That's going to be all of this business to the 1 half power. This right here is the same thing as the magnitude of this right here. And this is just a scalar that's multiplying by both of these terms. So let's see if we can do anything interesting here, if this can be simplified in any way. We have a squared cosine squared of s. We have an a squared cosine squared of s here."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right here is the same thing as the magnitude of this right here. And this is just a scalar that's multiplying by both of these terms. So let's see if we can do anything interesting here, if this can be simplified in any way. We have a squared cosine squared of s. We have an a squared cosine squared of s here. So let's factor that out from both of these terms and see what happens. So I'm just going to rewrite the second part. So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have a squared cosine squared of s. We have an a squared cosine squared of s here. So let's factor that out from both of these terms and see what happens. So I'm just going to rewrite the second part. So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power. Now what is this? Well, we have sine squared of t plus cosine squared of t. That's nice. That's equal to 1, the most basic of trig identities."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be a squared cosine squared of s times sine squared of t plus cosine squared of t. Sine squared of t plus cosine squared of t. And then you're going to have this plus a squared sine squared of s. And of course, all of that is to the 1 half power. Now what is this? Well, we have sine squared of t plus cosine squared of t. That's nice. That's equal to 1, the most basic of trig identities. So this expression right here simplifies to a squared cosine squared of s plus this over here. a squared sine squared of s. And then all of that to the 1 half power. You might immediately recognize, you can factor out an a squared."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's equal to 1, the most basic of trig identities. So this expression right here simplifies to a squared cosine squared of s plus this over here. a squared sine squared of s. And then all of that to the 1 half power. You might immediately recognize, you can factor out an a squared. This is equal to a squared times cosine squared of s plus sine squared of s. And all of that to the 1 half power. I'm just focusing on this term right here. I'll write this in a second."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You might immediately recognize, you can factor out an a squared. This is equal to a squared times cosine squared of s plus sine squared of s. And all of that to the 1 half power. I'm just focusing on this term right here. I'll write this in a second. But once again, cosine squared plus sine squared of anything is going to be equal to 1. As long as it's the same anything, it's equal to 1. So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll write this in a second. But once again, cosine squared plus sine squared of anything is going to be equal to 1. As long as it's the same anything, it's equal to 1. So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a. So all of this, all of that crazy business right here, just simplifies to a. So this cross product here simplifies to this times a, which is a pretty neat and clean simplification. So let me rewrite this."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this term is a squared to the 1 half power, or the square root of a squared, which is just going to be equal to a. So all of this, all of that crazy business right here, just simplifies to a. So this cross product here simplifies to this times a, which is a pretty neat and clean simplification. So let me rewrite this. So this, that simplifies to a times that. And what's that? a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me rewrite this. So this, that simplifies to a times that. And what's that? a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far. It's nice when you do something so beastly. And eventually it gets to something reasonably simple. And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "a times b, so it's ab plus a squared cosine of s. Plus a squared cosine of s. So already we've gotten pretty far. It's nice when you do something so beastly. And eventually it gets to something reasonably simple. And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined. So s going from 0 to 2 pi, and t going from 0 to 2 pi over this region. So we want to integrate this over that region. So that region, we're going to vary s from 0 to 2 pi, so ds."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just to review what we had to do, what our mission was several videos ago, is we want to evaluate what this thing is over the region over which the surface is defined. So s going from 0 to 2 pi, and t going from 0 to 2 pi over this region. So we want to integrate this over that region. So that region, we're going to vary s from 0 to 2 pi, so ds. And then we're going to vary t from 0 to 2 pi, dt. And this is what we're evaluating. We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that region, we're going to vary s from 0 to 2 pi, so ds. And then we're going to vary t from 0 to 2 pi, dt. And this is what we're evaluating. We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization. So this is what we can put in there. Things are getting simple all of a sudden. Or simpler, ab plus a squared cosine of s. And what is this equal to?"}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're evaluating the magnitude of the cross product of these two partial derivatives of our original parameterization. So this is what we can put in there. Things are getting simple all of a sudden. Or simpler, ab plus a squared cosine of s. And what is this equal to? So this is going to be equal to, well, we just take the antiderivative of the inside with respect to s. So the antiderivative, so let me do the outside of our integral. So we're still going to have to deal with the 0 to 2 pi. And our dt right here."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or simpler, ab plus a squared cosine of s. And what is this equal to? So this is going to be equal to, well, we just take the antiderivative of the inside with respect to s. So the antiderivative, so let me do the outside of our integral. So we're still going to have to deal with the 0 to 2 pi. And our dt right here. But the antiderivative with respect to s right here is going to be, ab is just a constant, so it's going to be abs plus, what's the antiderivative of cosine of s? It's sine of s. So plus a squared sine of s. And we're going to evaluate it from 0 to 2 pi. And what is this going to be equal to?"}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And our dt right here. But the antiderivative with respect to s right here is going to be, ab is just a constant, so it's going to be abs plus, what's the antiderivative of cosine of s? It's sine of s. So plus a squared sine of s. And we're going to evaluate it from 0 to 2 pi. And what is this going to be equal to? Let's put our boundaries out again. Or the t integral that we're going to have to do in a second. 0 to 2 pi dt."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what is this going to be equal to? Let's put our boundaries out again. Or the t integral that we're going to have to do in a second. 0 to 2 pi dt. When you put 2 pi here, you're going to get ab times 2 pi, or 2 pi ab. So you're going to have 2 pi ab plus a squared sine of 2 pi. Sine of 2 pi is 0, so there's not going to be any term there."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "0 to 2 pi dt. When you put 2 pi here, you're going to get ab times 2 pi, or 2 pi ab. So you're going to have 2 pi ab plus a squared sine of 2 pi. Sine of 2 pi is 0, so there's not going to be any term there. And then minus 0 times ab, which is 0. And then you're going to have minus a squared sine of 0, which is also 0. So all of the other terms are all 0."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sine of 2 pi is 0, so there's not going to be any term there. And then minus 0 times ab, which is 0. And then you're going to have minus a squared sine of 0, which is also 0. So all of the other terms are all 0. So that's what we're left with. It's simplified nicely. So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all of the other terms are all 0. So that's what we're left with. It's simplified nicely. So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt. And we need to evaluate that from 0 to 2 pi, which is equal to, so we put 2 pi in there. You have a 2 pi for t. It'll be a 2 pi times 2 pi ab, or we should say 2 pi squared times ab minus 0 times this thing. Well, that's just going to be 0, so we're going to have to write it down."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now we just have to take the antiderivative of this with respect to t. And this is a constant in t. So this is going to be equal to, take the antiderivative with respect to t, 2 pi abt. And we need to evaluate that from 0 to 2 pi, which is equal to, so we put 2 pi in there. You have a 2 pi for t. It'll be a 2 pi times 2 pi ab, or we should say 2 pi squared times ab minus 0 times this thing. Well, that's just going to be 0, so we're going to have to write it down. So we're done. This is the surface area of the torus. So this is equal to, this is exciting."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's just going to be 0, so we're going to have to write it down. So we're done. This is the surface area of the torus. So this is equal to, this is exciting. It just kind of snuck up on us. This is equal to 4 pi squared ab, which is kind of a neat formula. Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is equal to, this is exciting. It just kind of snuck up on us. This is equal to 4 pi squared ab, which is kind of a neat formula. Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle. We're squaring it, which kind of makes sense, because we're taking the product of, you can kind of imagine the product of these two circles. I'm speaking in very abstract, general terms, but that kind of feels good. And then we're taking the product of those two radiuses."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because it's a very neat and clean, it has the 2 pi, which is kind of the diameter of a circle. We're squaring it, which kind of makes sense, because we're taking the product of, you can kind of imagine the product of these two circles. I'm speaking in very abstract, general terms, but that kind of feels good. And then we're taking the product of those two radiuses. Remember, let me just copy this thing down here. Let me, actually, let me copy this thing, because this is our exciting result. Let me copy this."}, {"video_title": "Example of calculating a surface integral part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're taking the product of those two radiuses. Remember, let me just copy this thing down here. Let me, actually, let me copy this thing, because this is our exciting result. Let me copy this. So copy. So all of this work that we did simplified to this, which is exciting. We now know that if you have a torus where the radius of the cross section is a, and the radius from the center of the torus to the middle of the cross sections is b, that the surface area of that torus is going to be 4 pi squared times a times b, which I think is a pretty, pretty neat outcome."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of writing the surface as a capital sigma, I've written it as a capital S. Instead of writing d lowercase sigma, I wrote d uppercase S. But this is still a surface integral of the function y. And the surface we care about is x plus y squared minus z is equal to zero. X between zero and one, y between zero and two. Now this one might be a little bit more straightforward than the last one we did, or at least I hope it's a little bit more straightforward. Because we can explicitly define z in terms of x and y. And actually we could even explicitly define x in terms of y and z. But I'll do it the other way."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now this one might be a little bit more straightforward than the last one we did, or at least I hope it's a little bit more straightforward. Because we can explicitly define z in terms of x and y. And actually we could even explicitly define x in terms of y and z. But I'll do it the other way. It's a little bit easier for me to visualize. So if you add z to both sides of this equation right over here, you get x plus y squared is equal to z. Or z is equal to x plus y squared."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I'll do it the other way. It's a little bit easier for me to visualize. So if you add z to both sides of this equation right over here, you get x plus y squared is equal to z. Or z is equal to x plus y squared. And this is actually pretty straightforward. This surface is pretty easy to visualize, or we can give our best attempt at visualizing it. So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or z is equal to x plus y squared. And this is actually pretty straightforward. This surface is pretty easy to visualize, or we can give our best attempt at visualizing it. So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one. So maybe this is x equals one, and y between zero and two. So let's say this is one, this is two in the y area. So this is, we essentially care about the surface over this, over this region of the xy plane."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if that is our z-axis, and that is our x-axis, and that is the y-axis, we care about the region x between zero and one. So maybe this is x equals one, and y between zero and two. So let's say this is one, this is two in the y area. So this is, we essentially care about the surface over this, over this region of the xy plane. And then we can think about what the surface actually looks like. This isn't the surface, this is just kind of the range of x and y's that we actually care about. And so let's think about the surface."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is, we essentially care about the surface over this, over this region of the xy plane. And then we can think about what the surface actually looks like. This isn't the surface, this is just kind of the range of x and y's that we actually care about. And so let's think about the surface. When x and y are zero, z is zero. So we're gonna be sitting, let me do this in a, I'm doing it in green. Z is going to be right over there."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's think about the surface. When x and y are zero, z is zero. So we're gonna be sitting, let me do this in a, I'm doing it in green. Z is going to be right over there. And now as y increases, or if we, when x is equal to zero, if we're just talking about the zy plane, z is going to be equal to y squared. So z is going to be equal to y squared. So this might be z is equal to four, this is z is equal to two, one, three."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Z is going to be right over there. And now as y increases, or if we, when x is equal to zero, if we're just talking about the zy plane, z is going to be equal to y squared. So z is going to be equal to y squared. So this might be z is equal to four, this is z is equal to two, one, three. So z is going to do something like this. It's going to be a parabola in the zy plane. It's gonna look something like that."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this might be z is equal to four, this is z is equal to two, one, three. So z is going to do something like this. It's going to be a parabola in the zy plane. It's gonna look something like that. Now when y is equal to zero, z is just equal to x. So as x goes to one, z will also go to one. So z will go like this."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's gonna look something like that. Now when y is equal to zero, z is just equal to x. So as x goes to one, z will also go to one. So z will go like this. The scales of the axes aren't, they're not drawn to scale. The z is a little bit more compressed than the x or y, the way I've drawn them. And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So z will go like this. The scales of the axes aren't, they're not drawn to scale. The z is a little bit more compressed than the x or y, the way I've drawn them. And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there. And then this point, when y is equal to two, and x is equal to one, you have z is equal to five. It's gonna look something like this. Something like this."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then from this point right over here, you add the y squared, and so you get something that looks, something that looks, so this is, this is this point there. And then this point, when y is equal to two, and x is equal to one, you have z is equal to five. It's gonna look something like this. Something like this. And then you're going to have the straight line like that. And this point is right over there. And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Something like this. And then you're going to have the straight line like that. And this point is right over there. And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y. And so one way you could think about it, y could be maybe the mass density of this surface. And so when you multiply y times each ds, you're essentially figuring out the mass of that little chunk, and then you're figuring out the mass of this entire surface. And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this surface is the surface that we are going to take, or the surface over which we're going to, we're going to, we're going to evaluate the surface integral of the function y. And so one way you could think about it, y could be maybe the mass density of this surface. And so when you multiply y times each ds, you're essentially figuring out the mass of that little chunk, and then you're figuring out the mass of this entire surface. And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense. So this part, this part of the surface is more dense, is more dense than as y becomes lower and lower. And then that would actually give us the mass. But with that out of the way, let's actually evaluate it."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so one way you can imagine as we go more and more in that direction, as y is increasing, this thing is getting more and more dense. So this part, this part of the surface is more dense, is more dense than as y becomes lower and lower. And then that would actually give us the mass. But with that out of the way, let's actually evaluate it. And so as you know, the first step is to figure out a parameterization. And it should be pretty straightforward because we can write z explicitly in terms of x and y. And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But with that out of the way, let's actually evaluate it. And so as you know, the first step is to figure out a parameterization. And it should be pretty straightforward because we can write z explicitly in terms of x and y. And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could. But let me, so let's just write, let me do that. So let me just write x is equal to, x is equal to, and in the spirit of using different notation, instead of using s and t, I'll use u and v. x is equal to u. Let's say y is equal to v. And then z is going to be equal to u plus v squared."}, {"video_title": "Surface integral ex2 part 1 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we can actually use x and y as the actual parameters, or if we wanna just substitute it with different parameters, we could. But let me, so let's just write, let me do that. So let me just write x is equal to, x is equal to, and in the spirit of using different notation, instead of using s and t, I'll use u and v. x is equal to u. Let's say y is equal to v. And then z is going to be equal to u plus v squared. And so our surface written as a vector position function or position vector function, our surface, we can write it as r, which is going to be a function of u and v. And it's going to be equal to ui plus v plus vj plus u plus v squared, u plus v squared k. And then u is going to be between zero and one because x is just equal to u or u is equal to x. So u is going to be equal to, is going to be between zero and one. And then v is going to be, v is going to be between zero and two."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we saw that if we had some curve in the xy plane, and we just parameterized it in a very general sense like this, we could generate another parameterization that essentially is the same curve but goes in the opposite direction. It starts here and it goes here as t goes from a to b, as opposed to the first parameterization. We started with t equal a over here, and it went up like that. And the question I want to answer in this video is how a line integral of a scalar field over this curve, so this is my scalar field, it's a function of x and y, how a line integral over a scalar field over this curve relates to a line integral of that same scalar field over the reverse curve, over the curve going in the other direction. So the question is, does it even matter whether we move in this direction or that direction when we're taking the line integral of a scalar field? And in the next video, we'll talk about whether it matters on a vector field. Let's see if we can get a little intuition to our answer before we even prove our answer."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the question I want to answer in this video is how a line integral of a scalar field over this curve, so this is my scalar field, it's a function of x and y, how a line integral over a scalar field over this curve relates to a line integral of that same scalar field over the reverse curve, over the curve going in the other direction. So the question is, does it even matter whether we move in this direction or that direction when we're taking the line integral of a scalar field? And in the next video, we'll talk about whether it matters on a vector field. Let's see if we can get a little intuition to our answer before we even prove our answer. So let me draw a little diagram here. Actually, let me do it a little bit lower because I think I'm going to need a little bit more real estate. So let me draw the y-axis."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see if we can get a little intuition to our answer before we even prove our answer. So let me draw a little diagram here. Actually, let me do it a little bit lower because I think I'm going to need a little bit more real estate. So let me draw the y-axis. That is the x-axis. Let me draw the vertical axis just like that. That is z."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw the y-axis. That is the x-axis. Let me draw the vertical axis just like that. That is z. Let me draw a scalar field here. So I'll just draw it as some surface. I'll draw part of it."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is z. Let me draw a scalar field here. So I'll just draw it as some surface. I'll draw part of it. That is my scalar field. That is f of x, y right there. For any point on the xy-plane, we can associate a height that defines this surface, this scalar field."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll draw part of it. That is my scalar field. That is f of x, y right there. For any point on the xy-plane, we can associate a height that defines this surface, this scalar field. And let me put a curve down there. So let's say that this is the curve C, just like that. And the way we define it first, we start over here and we move in that direction."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "For any point on the xy-plane, we can associate a height that defines this surface, this scalar field. And let me put a curve down there. So let's say that this is the curve C, just like that. And the way we define it first, we start over here and we move in that direction. That was our curve C. And we know from several videos ago that the way to visualize what this line integral means is we're essentially trying to figure out the area of a curtain that has this curve as its base, and then its ceiling is defined by this surface, by the scalar field. So we're literally just trying to find the area of this curvy piece of paper or wall or whatever you want to view it. That's what this thing is."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the way we define it first, we start over here and we move in that direction. That was our curve C. And we know from several videos ago that the way to visualize what this line integral means is we're essentially trying to figure out the area of a curtain that has this curve as its base, and then its ceiling is defined by this surface, by the scalar field. So we're literally just trying to find the area of this curvy piece of paper or wall or whatever you want to view it. That's what this thing is. Now, if we take the same integral but we take the reverse curve, instead of going in that direction, we're now going in the opposite direction. We're not taking a curve where we're going from the top to the bottom. But the idea is still the same."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what this thing is. Now, if we take the same integral but we take the reverse curve, instead of going in that direction, we're now going in the opposite direction. We're not taking a curve where we're going from the top to the bottom. But the idea is still the same. I don't know which one's C, which one's minus C. I could have defined this path going from that way as C, and then the minus C path would have started here and gone back up. So it seems in either case, no matter what I'm doing, I'm going to try to figure out the area of this curved piece of paper. So my intuition tells me that either of these are going to give me the area of this curved piece of paper."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But the idea is still the same. I don't know which one's C, which one's minus C. I could have defined this path going from that way as C, and then the minus C path would have started here and gone back up. So it seems in either case, no matter what I'm doing, I'm going to try to figure out the area of this curved piece of paper. So my intuition tells me that either of these are going to give me the area of this curved piece of paper. So maybe they should be equal to each other. I haven't proved anything very rigorously yet, but it seems that they should be equal to each other. In this case, let me just make it very clear."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So my intuition tells me that either of these are going to give me the area of this curved piece of paper. So maybe they should be equal to each other. I haven't proved anything very rigorously yet, but it seems that they should be equal to each other. In this case, let me just make it very clear. I'm taking a ds, a little change in distance, and I'm multiplying it by the height to find kind of a differential of the area. And then I'm going to add a bunch of these together to get the whole area. Here I'm doing the same thing."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In this case, let me just make it very clear. I'm taking a ds, a little change in distance, and I'm multiplying it by the height to find kind of a differential of the area. And then I'm going to add a bunch of these together to get the whole area. Here I'm doing the same thing. I'm taking a little ds. And remember, the ds is always going to be positive the way we've parameterized it. It's the hardest word in the English language for me to say."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Here I'm doing the same thing. I'm taking a little ds. And remember, the ds is always going to be positive the way we've parameterized it. It's the hardest word in the English language for me to say. So here too, we're taking a ds, and we're going to multiply it by the height. So once again, we should take the area. And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's the hardest word in the English language for me to say. So here too, we're taking a ds, and we're going to multiply it by the height. So once again, we should take the area. And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative. That equals the negative of the integral from b to a of f of x dx. 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."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I want to actually differentiate that relative to when you take a normal integral from a to b of, let's say, f of x dx, we know that when we switch the boundaries of the integration, that it makes the integral negative. That equals the negative of the integral from b to a of f of x dx. 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? The each increment, the right boundary is going to be higher than the left boundary. So your dx's are positive."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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? The each increment, the right boundary is going to be higher than the left boundary. 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."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 the surface, it's above the xy 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."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 the surface, it's above the xy 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 like we did in the last video. 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."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 like we did in the last video. 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 me write that down right now. We can write dx dt is equal to x prime of t, and 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 equal to b of f of x of t, y of t, times the square root of dx dt squared, which is the same thing as x prime of t squared, plus dy dt squared, which is the same thing as y prime of t squared, plus y prime of t squared, all that under the radical, times dt. This integral is exactly that, given this parameterization."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 me write that down right now. We can write dx dt is equal to x prime of t, and 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 equal to b of f of x of t, y of t, times the square root of dx dt squared, which is the same thing as x prime of t squared, plus dy dt squared, which is the same thing as y prime of t squared, plus y prime of t squared, all that under the radical, times dt. This integral is exactly that, given this parameterization. I have so much trouble saying that. Now let's do the minus c version. I'll do that in this orange color."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This integral is exactly that, given this parameterization. I have so much trouble saying that. Now let's do the minus c version. I'll do that in this orange color. So the minus c version, actually let me do the minus c version down here. The minus c version, we have x is equal to, you remember this actually just from up here, this was from the last video, x is equal to a plus b minus t. x is equal to a plus b minus t. y is, sorry, x doesn't equal a plus b minus t. x is equal to x of a plus b minus t. Got ahead of myself. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do that in this orange color. So the minus c version, actually let me do the minus c version down here. The minus c version, we have x is equal to, you remember this actually just from up here, this was from the last video, x is equal to a plus b minus t. x is equal to a plus b minus t. y is, sorry, x doesn't equal a plus b minus t. x is equal to x of a plus b minus t. Got ahead of myself. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b. And this is just exactly what we did in that last video. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. Same curve, just going in a different direction as t increases from a to b. But let's get the derivatives."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b. And this is just exactly what we did in that last video. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. Same curve, just going in a different direction as t increases from a to b. But let's get the derivatives. So I'll do it in the derivative color maybe. So dx dt for this path. It's going to be a little different."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's get the derivatives. So I'll do it in the derivative color maybe. So dx dt for this path. It's going to be a little different. We have to do the chain rule now. Derivative of the inside with respect to t. Well, these are constants. Minus t, derivative of minus t with respect to t is minus 1."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be a little different. We have to do the chain rule now. Derivative of the inside with respect to t. Well, these are constants. Minus t, derivative of minus t with respect to t is minus 1. So it's minus 1 times the derivative of the outside with respect to the inside. Well, that's just x prime of a plus b minus t. Or we could rewrite this as this is just minus x prime of a plus b minus t. dy dt, same logic. Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Minus t, derivative of minus t with respect to t is minus 1. So it's minus 1 times the derivative of the outside with respect to the inside. Well, that's just x prime of a plus b minus t. Or we could rewrite this as this is just minus x prime of a plus b minus t. dy dt, same logic. Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1. Times the derivative of the outside with respect to the inside. So y prime of a plus b minus t. Same thing as minus y prime a plus b minus t. So given all of that, what is this integral going to be equal to? The integral of minus c of the scalar field f of x, y, ds."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1. Times the derivative of the outside with respect to the inside. So y prime of a plus b minus t. Same thing as minus y prime a plus b minus t. So given all of that, what is this integral going to be equal to? The integral of minus c of the scalar field f of x, y, ds. What is this going to be equal to? Well, it's going to be the integral from, you could almost pattern match it, t is equal to a to t is equal to b, of f of x. But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The integral of minus c of the scalar field f of x, y, ds. What is this going to be equal to? Well, it's going to be the integral from, you could almost pattern match it, t is equal to a to t is equal to b, of f of x. But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking. Hopefully it's not too confusing. And once again, y is no longer y of t. y is y of a plus b minus t. And then times the square root, I'll just switch colors, times the square root of dx dt squared. What's dx dt squared?"}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking. Hopefully it's not too confusing. And once again, y is no longer y of t. y is y of a plus b minus t. And then times the square root, I'll just switch colors, times the square root of dx dt squared. What's dx dt squared? dx dt squared is just this thing squared. Or this thing squared. If I have minus anything squared, that's the same thing as the anything squared, right?"}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What's dx dt squared? dx dt squared is just this thing squared. Or this thing squared. If I have minus anything squared, that's the same thing as the anything squared, right? This is equal to minus x prime of a plus b minus t squared. Which is the same thing as just x prime of a plus b minus t squared, right? You lose that minus information when you square it."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I have minus anything squared, that's the same thing as the anything squared, right? This is equal to minus x prime of a plus b minus t squared. Which is the same thing as just x prime of a plus b minus t squared, right? You lose that minus information when you square it. So that's going to be equal to x prime of a plus b minus t squared, the whole result function, square it, plus dy dt squared. By the same logic, that's going to be, you lose the negative when you square it, y prime of a plus b minus t squared. Let me extend the radical."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You lose that minus information when you square it. So that's going to be equal to x prime of a plus b minus t squared, the whole result function, square it, plus dy dt squared. By the same logic, that's going to be, you lose the negative when you square it, y prime of a plus b minus t squared. Let me extend the radical. And then all of that dt. So that's the surface integral over the line integral. We're not doing surface integrals yet."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me extend the radical. And then all of that dt. So that's the surface integral over the line integral. We're not doing surface integrals yet. This is the line integral over the curve c. This is the line integral over the curve minus c. They don't look equal just yet. This looks a lot more convoluted than that one does. So let's see if we can simplify it a little bit."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're not doing surface integrals yet. This is the line integral over the curve c. This is the line integral over the curve minus c. They don't look equal just yet. This looks a lot more convoluted than that one does. So let's see if we can simplify it a little bit. And we can simplify it, perhaps, by making a substitution. Let me get a nice substitution color. Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see if we can simplify it a little bit. And we can simplify it, perhaps, by making a substitution. Let me get a nice substitution color. Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral. Well, actually, let's just figure out what's du. So du dt, the derivative of u with respect to t is just going to be equal to minus 1. Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral. Well, actually, let's just figure out what's du. So du dt, the derivative of u with respect to t is just going to be equal to minus 1. Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt. And let's figure out our boundaries of integration. When t is equal to a, what is u? u is equal to a plus b minus a, which is equal to b."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt. And let's figure out our boundaries of integration. When t is equal to a, what is u? u is equal to a plus b minus a, which is equal to b. And then when t is equal to b, u is equal to a plus b minus b, which is equal to a. So if we do the substitution on this crazy, hairy-looking integral, I should simplify a little bit. And it changes our, so this integral is going to be the same thing as the integral from u."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "u is equal to a plus b minus a, which is equal to b. And then when t is equal to b, u is equal to a plus b minus b, which is equal to a. So if we do the substitution on this crazy, hairy-looking integral, I should simplify a little bit. And it changes our, so this integral is going to be the same thing as the integral from u. When t is a, u is b. When t is b, u is a. And f of x of, this thing right here is just u."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it changes our, so this integral is going to be the same thing as the integral from u. When t is a, u is b. When t is b, u is a. And f of x of, this thing right here is just u. So let's simplify it a good bit. And y of, this thing right here, is just u. y of u times the square root of x prime of u squared plus y prime of u squared. Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And f of x of, this thing right here is just u. So let's simplify it a good bit. And y of, this thing right here, is just u. y of u times the square root of x prime of u squared plus y prime of u squared. Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du. So instead of dt, we have to put a minus du here. So this is times minus du. Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du. So instead of dt, we have to put a minus du here. So this is times minus du. Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that. So we're going from b to a of this thing. Like that. And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that. So we're going from b to a of this thing. Like that. And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them. And I said at the beginning of this video, if you swap the, for just a standard, regular, run-of-the-mill integral, if you swap, if you have something going from b to a of f of x dx, or du, and maybe I should write it this way, f of u du, this is equal to the minus of the integral from a to b of f of u du. And we did that by the logic that I graphed up here. That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them. And I said at the beginning of this video, if you swap the, for just a standard, regular, run-of-the-mill integral, if you swap, if you have something going from b to a of f of x dx, or du, and maybe I should write it this way, f of u du, this is equal to the minus of the integral from a to b of f of u du. And we did that by the logic that I graphed up here. That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve. So let's do that. Let's swap the boundaries of integration right here. And if we do that, that'll negate this negative or make it a positive."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve. So let's do that. Let's swap the boundaries of integration right here. And if we do that, that'll negate this negative or make it a positive. So this is going to be equal to the integral from a to b. I'm dropping the negative sign because I swapped these two things. So I'm going to take the negative of a negative, which is a positive, of f of x of u y of u times the square root of x prime of u squared plus y prime of u squared du. Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we do that, that'll negate this negative or make it a positive. So this is going to be equal to the integral from a to b. I'm dropping the negative sign because I swapped these two things. So I'm going to take the negative of a negative, which is a positive, of f of x of u y of u times the square root of x prime of u squared plus y prime of u squared du. Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds. Now, how does this compare to when we take the regular curve? How does this compare to that? Let me copy and paste it to see."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds. Now, how does this compare to when we take the regular curve? How does this compare to that? Let me copy and paste it to see. No, I'm using the wrong tool. Let me copy and paste it to see how they compare. And then let me paste it down here."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me copy and paste it to see. No, I'm using the wrong tool. Let me copy and paste it to see how they compare. And then let me paste it down here. Edit, paste. So how do these two things compare? Let's take a close look."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then let me paste it down here. Edit, paste. So how do these two things compare? Let's take a close look. Well, they actually look pretty similar. Over here, for the minus curve, we have a bunch of u's. Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's take a close look. Well, they actually look pretty similar. Over here, for the minus curve, we have a bunch of u's. Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places. These integrals are the exact same integrals. If you make a u substitution here, if you just make the substitution u is equal to t, this thing is going to be the integral from a to b of what's going to be this exact same thing, of f of x of u, y of u, times the square root of x prime of u squared plus y prime of u squared du. These two things are identical."}, {"video_title": "Scalar field line integral independent of path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places. These integrals are the exact same integrals. If you make a u substitution here, if you just make the substitution u is equal to t, this thing is going to be the integral from a to b of what's going to be this exact same thing, of f of x of u, y of u, times the square root of x prime of u squared plus y prime of u squared du. These two things are identical. So we did all the substitution and everything, but we got the exact same integrals. So hopefully that satisfies you that it doesn't matter what direction we go on the curve, as long as the shape of the curve is the same. It doesn't matter if we go forward or backward on the curve."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So here I'm gonna talk about the Laplacian. Laplacian. And the Laplacian is a certain operator in the same way that the divergence or the gradient or the curl or even just the derivative are operators, the things that take in some kind of function and give you another function. So in this case, let's say we have a multivariable function like f that just takes in a two-dimensional input, f of x, y. So you might imagine its graph as being something like this where the input space is this x, y plane here. So each of the points x, y is a point here and then the output is just given by the height of that graph. So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So in this case, let's say we have a multivariable function like f that just takes in a two-dimensional input, f of x, y. So you might imagine its graph as being something like this where the input space is this x, y plane here. So each of the points x, y is a point here and then the output is just given by the height of that graph. So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y. So it's gonna give you a new function that takes in a two-dimensional input and just outputs a number. And it's kind of like a second derivative because the way that it's defined is that you take the divergence of the gradient of your function f. So that's kind of how it's defined, the divergence of the gradient of f. And a more common notation that you might see here is to take that upside-down triangle, nabla, dot product with nabla of f. So remember, if f is a scalar-valued function, then the gradient of f gives you a vector field, a certain vector field, but the divergence of a vector field gives you another scalar-valued function. So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y. So it's gonna give you a new function that takes in a two-dimensional input and just outputs a number. And it's kind of like a second derivative because the way that it's defined is that you take the divergence of the gradient of your function f. So that's kind of how it's defined, the divergence of the gradient of f. And a more common notation that you might see here is to take that upside-down triangle, nabla, dot product with nabla of f. So remember, if f is a scalar-valued function, then the gradient of f gives you a vector field, a certain vector field, but the divergence of a vector field gives you another scalar-valued function. So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean. Because the gradient, if you'll remember, gives you the slope of steepest ascent. So it's a vector field in the input space of x and each one of the vectors points in the direction that you should walk, such that if this graph is kind of like a hill on top of you, it tells you the direction you should go to increase your direction the most rapidly. And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean. Because the gradient, if you'll remember, gives you the slope of steepest ascent. So it's a vector field in the input space of x and each one of the vectors points in the direction that you should walk, such that if this graph is kind of like a hill on top of you, it tells you the direction you should go to increase your direction the most rapidly. And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other. So with the specific graph that I have pictured here, when you have kind of the top of a hill, all of the points around it, the direction that you should walk is towards the top of that hill. Whereas when you have kind of like the bottom, a little gully here, all of the directions you should walk to increase the value of the function most rapidly are directly away from that valley, which you might call a local minimum. So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other. So with the specific graph that I have pictured here, when you have kind of the top of a hill, all of the points around it, the direction that you should walk is towards the top of that hill. Whereas when you have kind of like the bottom, a little gully here, all of the directions you should walk to increase the value of the function most rapidly are directly away from that valley, which you might call a local minimum. So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly. And now let's think about what the divergence is supposed to represent. So now the divergence, and again, if this feels unfamiliar, maybe go back and take a look at the divergence videos. But the divergence has you imagining that this vector field corresponds to some kind of fluid flow."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly. And now let's think about what the divergence is supposed to represent. So now the divergence, and again, if this feels unfamiliar, maybe go back and take a look at the divergence videos. But the divergence has you imagining that this vector field corresponds to some kind of fluid flow. So you imagine little like water molecules and at any given moment, they're moving along the vector that they're attached to. So for example, if you had a water molecule that started off kind of here, it would start by going along that vector and then kind of follow the ones near it. And it looks like it kind of ends up in this spot."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "But the divergence has you imagining that this vector field corresponds to some kind of fluid flow. So you imagine little like water molecules and at any given moment, they're moving along the vector that they're attached to. So for example, if you had a water molecule that started off kind of here, it would start by going along that vector and then kind of follow the ones near it. And it looks like it kind of ends up in this spot. And a lot of the water molecules seem to kind of converge over there. Whereas over here, the water molecules tend to go away when they're following those vectors away from this point. And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "And it looks like it kind of ends up in this spot. And a lot of the water molecules seem to kind of converge over there. Whereas over here, the water molecules tend to go away when they're following those vectors away from this point. And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away. So over here, divergence is positive. Whereas the opposite case, where all of the water molecules are kind of coming in towards a point, that's where divergence is negative. And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away. So over here, divergence is positive. Whereas the opposite case, where all of the water molecules are kind of coming in towards a point, that's where divergence is negative. And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out. And at least from this picture, it doesn't seem like the ones going out are doing so at a faster rate or slower than they are here. This would be roughly zero divergence. So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out. And at least from this picture, it doesn't seem like the ones going out are doing so at a faster rate or slower than they are here. This would be roughly zero divergence. So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here. Points of high divergence, points where it diverges a lot here, why are those vectors pointing away? And if we pull up the graph again, the reason they're pointing away is because the direction of steepest ascent is kind of uphill everywhere. You're in a valley."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here. Points of high divergence, points where it diverges a lot here, why are those vectors pointing away? And if we pull up the graph again, the reason they're pointing away is because the direction of steepest ascent is kind of uphill everywhere. You're in a valley. Whereas in the opposite circumstance, where divergence is highly negative because points are converging towards it, why are they pointing towards it? Well, this is a gradient field. So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "You're in a valley. Whereas in the opposite circumstance, where divergence is highly negative because points are converging towards it, why are they pointing towards it? Well, this is a gradient field. So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill. So in other words, the divergence of the gradient is very high at points that are kind of like minima, at points where everyone around them tends to be higher. But the divergence of the gradient is low at points that look more like maximum points, where when you evaluate the function at all of the points around that input point, they give something smaller. So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill. So in other words, the divergence of the gradient is very high at points that are kind of like minima, at points where everyone around them tends to be higher. But the divergence of the gradient is low at points that look more like maximum points, where when you evaluate the function at all of the points around that input point, they give something smaller. So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y. It will be very positive when f evaluated at that point tends to give a smaller value than f evaluated at neighbors of that point. But it'll be very negative if when you evaluate f at that point, it tends to be bigger than its neighbors. And this should feel kind of analogous to the second derivative in ordinary calculus, when you have some kind of graph of just a single variable function."}, {"video_title": "Laplacian intuition.mp3", "Sentence": "So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y. It will be very positive when f evaluated at that point tends to give a smaller value than f evaluated at neighbors of that point. But it'll be very negative if when you evaluate f at that point, it tends to be bigger than its neighbors. And this should feel kind of analogous to the second derivative in ordinary calculus, when you have some kind of graph of just a single variable function. The second derivative, you know, the second derivative of x will be low, it'll be negative at points where it kind of looks like a local maximum. But over here, the second derivative of x would be positive at points that kind of look like a local minimum. So in that way, the Laplacian is sort of an analog of the second derivative for scalar-valued multivariable functions."}, {"video_title": "Divergence example.mp3", "Sentence": "And the picture of this vector field is here. This is what that vector field looks like. And what I'd like to do is compute and interpret the divergence of v. So, the divergence of v as a function of x and y. And in the last couple videos, I explained that the formula for this, and hopefully it's more than just a formula, but something you have an intuition for, is the partial derivative of p with respect to x. And by p, I mean that first component. So if you're thinking about this as being p of x, y, and q of x, y. So it could use any letters, right?"}, {"video_title": "Divergence example.mp3", "Sentence": "And in the last couple videos, I explained that the formula for this, and hopefully it's more than just a formula, but something you have an intuition for, is the partial derivative of p with respect to x. And by p, I mean that first component. So if you're thinking about this as being p of x, y, and q of x, y. So it could use any letters, right? And p and q are common, but the upshot is it's the partial derivative of the first component with respect to the first variable, plus the partial derivative of that second component with respect to that second variable, y. And as we actually plug this in and start computing, the partial derivative of p with respect to x of this guy with respect to x, x looks like a variable, y looks like a constant, the derivative is y, that constant. And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y."}, {"video_title": "Divergence example.mp3", "Sentence": "So it could use any letters, right? And p and q are common, but the upshot is it's the partial derivative of the first component with respect to the first variable, plus the partial derivative of that second component with respect to that second variable, y. And as we actually plug this in and start computing, the partial derivative of p with respect to x of this guy with respect to x, x looks like a variable, y looks like a constant, the derivative is y, that constant. And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y. And then x just looks like a constant, so nothing happens there. So on the whole, the divergence evidently just depends on the y value. It's three times y."}, {"video_title": "Divergence example.mp3", "Sentence": "And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y. And then x just looks like a constant, so nothing happens there. So on the whole, the divergence evidently just depends on the y value. It's three times y. So what that should mean is if we look, for example, let's say we look at all points where y equals zero, we'd expect the divergence to be zero, that fluid neither goes towards nor away from each point. So y equals zero corresponds with this x-axis of points. So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here."}, {"video_title": "Divergence example.mp3", "Sentence": "It's three times y. So what that should mean is if we look, for example, let's say we look at all points where y equals zero, we'd expect the divergence to be zero, that fluid neither goes towards nor away from each point. So y equals zero corresponds with this x-axis of points. So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here. And wherever you look, I mean here it's kind of only flowing in by a little bit, and I guess it's flowing out just by that same amount, and that all cancels out. Whereas, let's say we take a look at y equals, I don't know, like three. So in this case, the divergence should equal nine."}, {"video_title": "Divergence example.mp3", "Sentence": "So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here. And wherever you look, I mean here it's kind of only flowing in by a little bit, and I guess it's flowing out just by that same amount, and that all cancels out. Whereas, let's say we take a look at y equals, I don't know, like three. So in this case, the divergence should equal nine. So we'd expect there to be positive divergence when y is positive. So if we go up and y is equal to one, two, three, and if we look at a point, I don't know, let's say around here, and we kind of consider the region around it, you can kind of see how the vectors leaving it seem to be bigger, so the fluid kind of flowing out of this region is pretty rapid, whereas the fluid flowing into it is less rapid. So on the whole, in a region around this point, the fluid, I guess, is going away."}, {"video_title": "Divergence example.mp3", "Sentence": "So in this case, the divergence should equal nine. So we'd expect there to be positive divergence when y is positive. So if we go up and y is equal to one, two, three, and if we look at a point, I don't know, let's say around here, and we kind of consider the region around it, you can kind of see how the vectors leaving it seem to be bigger, so the fluid kind of flowing out of this region is pretty rapid, whereas the fluid flowing into it is less rapid. So on the whole, in a region around this point, the fluid, I guess, is going away. And you know, you look anywhere where y is positive, and if you kind of look around here, the same is true. Where fluid does flow into it, it seems, but the vectors kind of going out of this region are larger, so you'd expect, on the whole, for things to diverge away from that point. And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points."}, {"video_title": "Divergence example.mp3", "Sentence": "So on the whole, in a region around this point, the fluid, I guess, is going away. And you know, you look anywhere where y is positive, and if you kind of look around here, the same is true. Where fluid does flow into it, it seems, but the vectors kind of going out of this region are larger, so you'd expect, on the whole, for things to diverge away from that point. And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points. So you go down to, you know, I guess I said y equals negative four, but really, I'm thinking of anything where y is negative. And let's say we take a look at like this point here. Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large."}, {"video_title": "Divergence example.mp3", "Sentence": "And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points. So you go down to, you know, I guess I said y equals negative four, but really, I'm thinking of anything where y is negative. And let's say we take a look at like this point here. Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large. It's flowing out of it just kind of in a lackadaisical way. So it kind of makes sense, just looking at the picture, that divergence tends to be negative when y is negative. And what's surprising, what I wouldn't have been able to tell just looking at the picture, is that the divergence only depends on the y value."}, {"video_title": "Divergence example.mp3", "Sentence": "Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large. It's flowing out of it just kind of in a lackadaisical way. So it kind of makes sense, just looking at the picture, that divergence tends to be negative when y is negative. And what's surprising, what I wouldn't have been able to tell just looking at the picture, is that the divergence only depends on the y value. That once you compute everything, it's only dependent on the y value here, and that as you go kind of left and right on the diagram there, as we look left and right, the value of that divergence doesn't change. That's kind of surprising. It makes a little bit of sense."}, {"video_title": "2d curl intuition.mp3", "Sentence": "Curl is one of those very cool vector calculus concepts and you'll be pretty happy that you've learned it once you have, if for no other reason because it's kind of artistically pleasing. And there's two different versions. There's a two-dimensional curl and a three-dimensional curl. And naturally enough, I'll start talking about the two-dimensional version and kind of build our way up to the 3D one. And in this particular video, I just wanna lay down the intuition for what's visually going on. And curl has to do with the fluid flow interpretation of vector fields. Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that."}, {"video_title": "2d curl intuition.mp3", "Sentence": "And naturally enough, I'll start talking about the two-dimensional version and kind of build our way up to the 3D one. And in this particular video, I just wanna lay down the intuition for what's visually going on. And curl has to do with the fluid flow interpretation of vector fields. Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that. But just as a reminder, you kind of imagine that each point in space is a particle, like an air molecule or a water molecule. And since what a vector field does is associate each point in space with some kind of vector, and remember, we don't always draw every single vector, we just draw a small subsample, but in principle, every single point in space has a vector attached to it. You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to."}, {"video_title": "2d curl intuition.mp3", "Sentence": "Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that. But just as a reminder, you kind of imagine that each point in space is a particle, like an air molecule or a water molecule. And since what a vector field does is associate each point in space with some kind of vector, and remember, we don't always draw every single vector, we just draw a small subsample, but in principle, every single point in space has a vector attached to it. You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to. So as it moves to a different location in space and that velocity vector changes, it might be turning or it might be accelerating, and that velocity might change, and you end up getting kind of a trajectory for your point. And since every single point is moving in this way, you can start thinking about a flow, kind of a global view of the vector field. And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play."}, {"video_title": "2d curl intuition.mp3", "Sentence": "You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to. So as it moves to a different location in space and that velocity vector changes, it might be turning or it might be accelerating, and that velocity might change, and you end up getting kind of a trajectory for your point. And since every single point is moving in this way, you can start thinking about a flow, kind of a global view of the vector field. And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play. And at any given moment, if you look at the movement of one of these blue dots, it's moving along the vector that it's attached to at that point, or if that vector's not pictured, you know the vector that would be attached to it at that point. And as we get kind of a feel for what's going on in this entire flow, I want you to notice a couple particular regions. First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there."}, {"video_title": "2d curl intuition.mp3", "Sentence": "And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play. And at any given moment, if you look at the movement of one of these blue dots, it's moving along the vector that it's attached to at that point, or if that vector's not pictured, you know the vector that would be attached to it at that point. And as we get kind of a feel for what's going on in this entire flow, I want you to notice a couple particular regions. First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there. And I'll go ahead and start playing the animation over here. And what's most notable about this region is that there's counterclockwise rotation, and this corresponds to an idea that the vector field has a curl here, and I'll go very specifically into what curl means, but just right now you should have the idea that in a region where there's counterclockwise rotation, we want to say the curl is positive. Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl."}, {"video_title": "2d curl intuition.mp3", "Sentence": "First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there. And I'll go ahead and start playing the animation over here. And what's most notable about this region is that there's counterclockwise rotation, and this corresponds to an idea that the vector field has a curl here, and I'll go very specifically into what curl means, but just right now you should have the idea that in a region where there's counterclockwise rotation, we want to say the curl is positive. Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl. Here, I'll start it over here. And in contrast, if you look at a place where there's no rotation, where like at the center here, you have some points coming in from the top right and from the bottom left, and then going out from the other corners, but there's no net rotation. If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating."}, {"video_title": "2d curl intuition.mp3", "Sentence": "Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl. Here, I'll start it over here. And in contrast, if you look at a place where there's no rotation, where like at the center here, you have some points coming in from the top right and from the bottom left, and then going out from the other corners, but there's no net rotation. If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating. These are regions where you think of them as having zero curl. So with that as a general idea, you know, clockwise rotation regions correspond to positive curl, counterclockwise rotation regions correspond to negative curl, and then no rotation corresponds to zero curl, in the next video, I'm gonna start going through what this means in terms of the underlying function defining the vector field, and how we can start looking at the partial differential information of that function to quantify this intuition of fluid rotation. And what's neat is that it's not just about fluid rotation, right?"}, {"video_title": "2d curl intuition.mp3", "Sentence": "If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating. These are regions where you think of them as having zero curl. So with that as a general idea, you know, clockwise rotation regions correspond to positive curl, counterclockwise rotation regions correspond to negative curl, and then no rotation corresponds to zero curl, in the next video, I'm gonna start going through what this means in terms of the underlying function defining the vector field, and how we can start looking at the partial differential information of that function to quantify this intuition of fluid rotation. And what's neat is that it's not just about fluid rotation, right? If you have vector fields in other contexts, and you just imagine that they represent a fluid, even though they don't, this idea of rotation and curling actually has certain importance, in ways that you totally wouldn't expect. The gradient turns out to relate to the curl, even though you wouldn't necessarily think the gradient has something to do with fluid rotation. In electromagnetism, this idea of fluid rotation has a certain importance, even though fluids aren't actually involved."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "And here, I want to go through a slightly more intricate example. So I'll go ahead and get rid of this vector field. And in this example, the x component of the output will be y times z. The y component of the output will be x times z. And the z component of the output will be x times y. So I'll just show this vector field, and then we can start to get a feel for how the function that I just wrote relates to the vectors that you're seeing. So you see some of the vectors are kind of pointing away from the origin."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "The y component of the output will be x times z. And the z component of the output will be x times y. So I'll just show this vector field, and then we can start to get a feel for how the function that I just wrote relates to the vectors that you're seeing. So you see some of the vectors are kind of pointing away from the origin. Some of them are pointing in towards the origin. So how can we understand this vector field in terms of the function itself? And a good step is to just zero in on one of the components."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you see some of the vectors are kind of pointing away from the origin. Some of them are pointing in towards the origin. So how can we understand this vector field in terms of the function itself? And a good step is to just zero in on one of the components. So in this case, I'll choose the z component of the output, which is made up of x times y, and kind of start to understand what that should be. The z component will represent how much the vectors point up or down. This is the z axis."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "And a good step is to just zero in on one of the components. So in this case, I'll choose the z component of the output, which is made up of x times y, and kind of start to understand what that should be. The z component will represent how much the vectors point up or down. This is the z axis. And the x, y plane here, so I'll point the z axis straight in our face. This is the x axis, this is the y axis. The values of x and y are gonna completely determine that z component."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "This is the z axis. And the x, y plane here, so I'll point the z axis straight in our face. This is the x axis, this is the y axis. The values of x and y are gonna completely determine that z component. So I'm gonna go off to the side here and just draw myself a little x, y plane for reference. So this is my x value, this is my y value. And I wanna understand the meaning of the term x times y."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "The values of x and y are gonna completely determine that z component. So I'm gonna go off to the side here and just draw myself a little x, y plane for reference. So this is my x value, this is my y value. And I wanna understand the meaning of the term x times y. So when both x and y are positive, their product is positive. And when both of them are negative, their product is also positive. If x is negative and y is positive, their product is negative."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "And I wanna understand the meaning of the term x times y. So when both x and y are positive, their product is positive. And when both of them are negative, their product is also positive. If x is negative and y is positive, their product is negative. But if x is positive and y is negative, their product is also negative. So what this should mean in terms of our vector field is that when we're in this first quadrant, vectors tend to point up in the z direction. Same over here in the third quadrant."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "If x is negative and y is positive, their product is negative. But if x is positive and y is negative, their product is also negative. So what this should mean in terms of our vector field is that when we're in this first quadrant, vectors tend to point up in the z direction. Same over here in the third quadrant. But over in the other two, they should tend to point down. So let's focus in on that first quadrant and try to look at what's going on. So you see like this vector here applies this vector."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "Same over here in the third quadrant. But over in the other two, they should tend to point down. So let's focus in on that first quadrant and try to look at what's going on. So you see like this vector here applies this vector. And all of them generally point upwards. They have a positive z component. So that seems in line with what we were predicting."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you see like this vector here applies this vector. And all of them generally point upwards. They have a positive z component. So that seems in line with what we were predicting. Whereas over here, which corresponds to the fourth quadrant of the x, y plane, the z component of each vector tends to be down. You know, and they're doing other things in terms of the x and y component. It's not just z component action."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "So that seems in line with what we were predicting. Whereas over here, which corresponds to the fourth quadrant of the x, y plane, the z component of each vector tends to be down. You know, and they're doing other things in terms of the x and y component. It's not just z component action. But right now, we're just focusing on up and down. And if you look over in the third quadrant, they tend to be pointing up. And that corresponds to the fact that x times y will be positive."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "It's not just z component action. But right now, we're just focusing on up and down. And if you look over in the third quadrant, they tend to be pointing up. And that corresponds to the fact that x times y will be positive. And you look and it all starts to align that way. And because I chose a rather symmetric function, you could imagine doing this where you analyze also the y component here. You analyze the x component in terms of y and z."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "And that corresponds to the fact that x times y will be positive. And you look and it all starts to align that way. And because I chose a rather symmetric function, you could imagine doing this where you analyze also the y component here. You analyze the x component in terms of y and z. And it's actually gonna look very similar for understanding when the x component of a vector tends to be positive, like up here, or when it turns out to be negative, like over here. And same with when the y component of a vector tends to be negative, or if the y component tends to be positive. And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "You analyze the x component in terms of y and z. And it's actually gonna look very similar for understanding when the x component of a vector tends to be positive, like up here, or when it turns out to be negative, like over here. And same with when the y component of a vector tends to be negative, or if the y component tends to be positive. And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it. And just like with two-dimensional vector fields, a kind of neat thing to do is imagine that this represents a fluid flow. So imagine like maybe air around you, flowing in towards the origin here, flowing out away from the origin there, you know, kind of be rotating around here. And later on in multivariable calculus, you'll learn about various ways that you can study just the function itself and just the variables, and get a feel for how that fluid itself would behave, even though it's a very complicated thing to think about."}, {"video_title": "3d vector field example   Multivariable calculus   Khan Academy.mp3", "Sentence": "And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it. And just like with two-dimensional vector fields, a kind of neat thing to do is imagine that this represents a fluid flow. So imagine like maybe air around you, flowing in towards the origin here, flowing out away from the origin there, you know, kind of be rotating around here. And later on in multivariable calculus, you'll learn about various ways that you can study just the function itself and just the variables, and get a feel for how that fluid itself would behave, even though it's a very complicated thing to think about. It's even complicated to draw or use graphing software with. But just with analytic tools, you can get very powerful results. And these kind of things come up in physics all the time, because you're thinking in three-dimensional space, and it doesn't just have to be fluid flow, it could be a force field, like an electric force field, or a gravitational force field, where each vector tells you how a particle tends to get pushed."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I've restated Stokes' Theorem. And what I want to do in this video is make sure that we get our orientation right. Because when we think about a normal vector, when we think about a normal vector to a surface, there are actually two normal vectors. They were based on the way I've drawn it right over here. There could be the one that might pop outward like this. Or there might be the one that pops inward just like that. Both of those would be normal to the surface right there."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "They were based on the way I've drawn it right over here. There could be the one that might pop outward like this. Or there might be the one that pops inward just like that. Both of those would be normal to the surface right there. And also, when we think about a path that goes around the boundary of a surface, there's two ways to think about that path. We could be going, based on how I've oriented it right now, we could go in a counterclockwise orientation or direction. Or we could go in a clockwise orientation or direction."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Both of those would be normal to the surface right there. And also, when we think about a path that goes around the boundary of a surface, there's two ways to think about that path. We could be going, based on how I've oriented it right now, we could go in a counterclockwise orientation or direction. Or we could go in a clockwise orientation or direction. So in order to make sure we're using Stokes' Theorem correctly, we need to make sure we understand which convention it is using. And the way we think about it is, whatever the normal direction we pick, and so let's say we pick this normal direction right over here, the one I am drawing in yellow. So if we pick this as our normal vector, so we're essentially saying maybe that's the top."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or we could go in a clockwise orientation or direction. So in order to make sure we're using Stokes' Theorem correctly, we need to make sure we understand which convention it is using. And the way we think about it is, whatever the normal direction we pick, and so let's say we pick this normal direction right over here, the one I am drawing in yellow. So if we pick this as our normal vector, so we're essentially saying maybe that's the top. One way of doing it is that's the top of our surface. Then the positive orientation that we need to traverse the path in is the one that if your head was pointed in the direction of the normal vector, and you were to walk along that path, the inside or the surface itself would be to your left. And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we pick this as our normal vector, so we're essentially saying maybe that's the top. One way of doing it is that's the top of our surface. Then the positive orientation that we need to traverse the path in is the one that if your head was pointed in the direction of the normal vector, and you were to walk along that path, the inside or the surface itself would be to your left. And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector. I'm wearing a big arrow hat right over there. And if I'm walking around the boundary, the actual surface needs to be to my left. So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector. I'm wearing a big arrow hat right over there. And if I'm walking around the boundary, the actual surface needs to be to my left. So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that. Then that's the convention that we use when we're thinking about Stokes' Theorem. If we oriented this thing differently, or if we said that no, no, no, no, no, this is not the normal vector, this is not essentially the top that we want to pick, if we wanted to pick it the other way, if we wanted to go in that direction, if we wanted that to be our normal vector, in order to be consistent, we would have to now do the opposite. I would now have to have my head going in that direction."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that. Then that's the convention that we use when we're thinking about Stokes' Theorem. If we oriented this thing differently, or if we said that no, no, no, no, no, this is not the normal vector, this is not essentially the top that we want to pick, if we wanted to pick it the other way, if we wanted to go in that direction, if we wanted that to be our normal vector, in order to be consistent, we would have to now do the opposite. I would now have to have my head going in that direction. And then I would have to walk, once again, and this might be a little bit harder to visualize, I would have to be walking in the direction that the surface is to my left. And now in this situation, instead of the surface looking like a hill to me, the surface would look like some type of a bowl, or some type of a valley, or something like that to me. And the way that I would have to do it now, and it's a little bit harder to visualize the upside down cell, but the upside down cell would have to walk in this direction, in order for the bowl, or the dip, to be to my left."}, {"video_title": "Orienting boundary with surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I would now have to have my head going in that direction. And then I would have to walk, once again, and this might be a little bit harder to visualize, I would have to be walking in the direction that the surface is to my left. And now in this situation, instead of the surface looking like a hill to me, the surface would look like some type of a bowl, or some type of a valley, or something like that to me. And the way that I would have to do it now, and it's a little bit harder to visualize the upside down cell, but the upside down cell would have to walk in this direction, in order for the bowl, or the dip, to be to my left. So it's just important to keep this in mind, in order for this to be consistent with this, right over here. Put your head in the direction of the normal vector, or you can kind of view that as the top of the direction that the top of the surface is going in. And then the contour, or the direction that you would have to traverse the boundary, in order for this to be true, is the direction with which the surface is to your left."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "If you're computing the determinant of the guy that we have pictured there in the upper right, you start by taking this upper left component and then multiplying it by the determinant of the submatrix, the submatrix whose rows are not the row of I and whose columns are not the row of I. So what that looks like over here is we're gonna take that univector I and then multiply it by a certain little determinant. And what this subdeterminant involves is multiplying this partial partial y by r, which means taking the partial derivative with respect to y of the multivariable function r and then subtracting off the partial derivative with respect to z of q. So we're subtracting off partial derivative with respect to z of the multivariable function q. And then that, so that's the first thing that we do. And then as a second part, we take this j and we're gonna subtract off. So you're kind of thinking plus, minus, plus for the elements in this top row."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "So we're subtracting off partial derivative with respect to z of the multivariable function q. And then that, so that's the first thing that we do. And then as a second part, we take this j and we're gonna subtract off. So you're kind of thinking plus, minus, plus for the elements in this top row. So we're gonna subtract off j multiplied by another subdeterminant. And then this one is gonna involve, you know, this column that it's not part of and this column that it's not part of. And you imagine those guys as a two by two matrix."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "So you're kind of thinking plus, minus, plus for the elements in this top row. So we're gonna subtract off j multiplied by another subdeterminant. And then this one is gonna involve, you know, this column that it's not part of and this column that it's not part of. And you imagine those guys as a two by two matrix. And its determinant involves taking the partial derivative with respect to x of r, so that's kind of the diagonal, partial partial x of r and then subtracting off the partial derivative with respect to z of p. So partial partial z of p. And then that's just two out of three of the things we need to do for our overall determinant because the last part we're gonna add, we're gonna add that top right component, k, multiplied by the, you know, the submatrix whose columns involve the column it's not part of and whose rows involve the rows that it's not part of. So k multiplied by the determinant of this guy is going to be, let's see, partial partial x of q. So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "And you imagine those guys as a two by two matrix. And its determinant involves taking the partial derivative with respect to x of r, so that's kind of the diagonal, partial partial x of r and then subtracting off the partial derivative with respect to z of p. So partial partial z of p. And then that's just two out of three of the things we need to do for our overall determinant because the last part we're gonna add, we're gonna add that top right component, k, multiplied by the, you know, the submatrix whose columns involve the column it's not part of and whose rows involve the rows that it's not part of. So k multiplied by the determinant of this guy is going to be, let's see, partial partial x of q. So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression. And, you know, here we're writing it with ijk notation. If you were writing it as a column vector, I guess I didn't erase some of these guys, but if you were writing this as a column vector, it would look like saying the curl of your vector valued function v as a function of x, y, and z is equal to, and then what I'd put in for this first component would be what's up there. So that would be your partial with respect to y of r minus partial of q with respect to z."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression. And, you know, here we're writing it with ijk notation. If you were writing it as a column vector, I guess I didn't erase some of these guys, but if you were writing this as a column vector, it would look like saying the curl of your vector valued function v as a function of x, y, and z is equal to, and then what I'd put in for this first component would be what's up there. So that would be your partial with respect to y of r minus partial of q with respect to z. So partial of q with respect to z. And I won't copy it down for all of the other ones, but in principle, you know, you'd kind of, whatever this j component is, and I guess we're subtracting it, so you'd subtract there. You'd copy that as the next component, and then over here."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "So that would be your partial with respect to y of r minus partial of q with respect to z. So partial of q with respect to z. And I won't copy it down for all of the other ones, but in principle, you know, you'd kind of, whatever this j component is, and I guess we're subtracting it, so you'd subtract there. You'd copy that as the next component, and then over here. But oftentimes, when you're computing curl, you kind of switch to using this ijk notation. My personal preference, I typically default to column vectors, and other people will write in terms of ij and k. It doesn't really matter, as long as you know how to go back and forth between the two. One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "You'd copy that as the next component, and then over here. But oftentimes, when you're computing curl, you kind of switch to using this ijk notation. My personal preference, I typically default to column vectors, and other people will write in terms of ij and k. It doesn't really matter, as long as you know how to go back and forth between the two. One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula. If you kind of look back to the videos on 2D curl and what its formula is, that is what we have here. And in fact, all the other components kind of look like mirrors of that, but you're using slightly different operators and slightly different functions. But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula. If you kind of look back to the videos on 2D curl and what its formula is, that is what we have here. And in fact, all the other components kind of look like mirrors of that, but you're using slightly different operators and slightly different functions. But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane. And what's happening with these other guys is kind of similar, right? Rotation that happens purely in the xz plane is gonna correspond with a rotation vector in the y direction, the direction perpendicular to the x, let's see, so the xz plane over here. And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane. And what's happening with these other guys is kind of similar, right? Rotation that happens purely in the xz plane is gonna correspond with a rotation vector in the y direction, the direction perpendicular to the x, let's see, so the xz plane over here. And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane. Now, when you compute it, you're not always thinking about, oh, you know, this corresponds to rotation in that plane, and this corresponds to rotation in that plane. You're just kind of computing it to get a formula out, but I think it's kind of nice to recognize that all the intuition that we put into the two-dimensional curl does show up here. And another thing I wanna emphasize is this is not a formula to be memorized."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane. Now, when you compute it, you're not always thinking about, oh, you know, this corresponds to rotation in that plane, and this corresponds to rotation in that plane. You're just kind of computing it to get a formula out, but I think it's kind of nice to recognize that all the intuition that we put into the two-dimensional curl does show up here. And another thing I wanna emphasize is this is not a formula to be memorized. I would not, if I were you, try to sit down and memorize this long expression. The only thing that you need to remember, the only thing, is that curl is represented as this del cross v, this nabla symbol cross product with the vector-valued function v. Because from there, whatever your components are, you can kind of go through the process that I just did, and the more you do it, the quicker it becomes. It's kind of long, but it doesn't take that long."}, {"video_title": "3d curl formula, part 2.mp3", "Sentence": "And another thing I wanna emphasize is this is not a formula to be memorized. I would not, if I were you, try to sit down and memorize this long expression. The only thing that you need to remember, the only thing, is that curl is represented as this del cross v, this nabla symbol cross product with the vector-valued function v. Because from there, whatever your components are, you can kind of go through the process that I just did, and the more you do it, the quicker it becomes. It's kind of long, but it doesn't take that long. And it's certainly much more fault-tolerant than trying to remember something that has as many moving parts as the formula that you see here. And in the next video, I'll go through an actual example of that. I'll have functions for p, q, and r, and walk through that process in a more concrete context."}, {"video_title": "Gradient.mp3", "Sentence": "And in this video, I'm only going to describe how you compute the gradient, and in the next couple ones, I'm going to give the geometric interpretation. And I hate doing this, I hate showing the computation before the geometric intuition, since usually it should go the other way around. But the gradient is one of those weird things where the way that you compute it actually seems kind of unrelated to the intuition. And you'll see that. We'll connect them in the next few videos, but to do that, we need to know what both of them actually are. So, on the computation side of things, let's say you have some sort of function, and I'm just going to make it a two-variable function. And let's say it's f of x, y equals x squared sine of y."}, {"video_title": "Gradient.mp3", "Sentence": "And you'll see that. We'll connect them in the next few videos, but to do that, we need to know what both of them actually are. So, on the computation side of things, let's say you have some sort of function, and I'm just going to make it a two-variable function. And let's say it's f of x, y equals x squared sine of y. The gradient is a way of packing together all the partial derivative information of a function. So let's just start by computing the partial derivatives of this guy. So partial of f with respect to x is equal to..."}, {"video_title": "Gradient.mp3", "Sentence": "And let's say it's f of x, y equals x squared sine of y. The gradient is a way of packing together all the partial derivative information of a function. So let's just start by computing the partial derivatives of this guy. So partial of f with respect to x is equal to... So we look at this and we consider x the variable and y the constant. Well, in that case, sine of y is also a constant, you know, as far as x is concerned. The derivative of x is 2x, so we see that this will be 2x times that constant sine of y."}, {"video_title": "Gradient.mp3", "Sentence": "So partial of f with respect to x is equal to... So we look at this and we consider x the variable and y the constant. Well, in that case, sine of y is also a constant, you know, as far as x is concerned. The derivative of x is 2x, so we see that this will be 2x times that constant sine of y. Whereas the partial derivative with respect to y... Now we look up here and we say x is considered a constant, so x squared is also considered a constant. So this is just a constant times sine of y, so that's going to equal that same constant times the cosine of y, which is the derivative of sine. So now what the gradient does is it just puts both of these together in a vector."}, {"video_title": "Gradient.mp3", "Sentence": "The derivative of x is 2x, so we see that this will be 2x times that constant sine of y. Whereas the partial derivative with respect to y... Now we look up here and we say x is considered a constant, so x squared is also considered a constant. So this is just a constant times sine of y, so that's going to equal that same constant times the cosine of y, which is the derivative of sine. So now what the gradient does is it just puts both of these together in a vector. And specifically, let me all change colors here, you denote it with a little upside-down triangle. The name of that symbol is nabla, but you often just pronounce it del, you'd say del f or gradient of f. And what this equals is a vector that has those two partial derivatives in it. So the first one is the partial derivative with respect to x, 2x times sine of y."}, {"video_title": "Gradient.mp3", "Sentence": "So now what the gradient does is it just puts both of these together in a vector. And specifically, let me all change colors here, you denote it with a little upside-down triangle. The name of that symbol is nabla, but you often just pronounce it del, you'd say del f or gradient of f. And what this equals is a vector that has those two partial derivatives in it. So the first one is the partial derivative with respect to x, 2x times sine of y. And the bottom one, partial derivative with respect to y, x squared cosine of y. And notice, maybe I should emphasize, this is actually a vector-valued function, right? So maybe I'll give it a little bit more room here and emphasize that it's got an x and a y."}, {"video_title": "Gradient.mp3", "Sentence": "So the first one is the partial derivative with respect to x, 2x times sine of y. And the bottom one, partial derivative with respect to y, x squared cosine of y. And notice, maybe I should emphasize, this is actually a vector-valued function, right? So maybe I'll give it a little bit more room here and emphasize that it's got an x and a y. This is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector. So you could also imagine doing this with three different variables, then you would have three partial derivatives and a three-dimensional output. And the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x and partial of f with respect to y."}, {"video_title": "Gradient.mp3", "Sentence": "So maybe I'll give it a little bit more room here and emphasize that it's got an x and a y. This is a function that takes in a point in two-dimensional space and outputs a two-dimensional vector. So you could also imagine doing this with three different variables, then you would have three partial derivatives and a three-dimensional output. And the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x and partial of f with respect to y. And in some sense, you know, we call these partial derivatives, I like to think of the gradient as the full derivative because it kind of captures all of the information that you need. So a very helpful mnemonic device with the gradient is to think about this triangle, this nabla symbol, as being a vector full of partial derivative operators. And by operator, I just mean, you know, here, let's like partial with respect to x, something where you could give it a function and it gives you another function."}, {"video_title": "Gradient.mp3", "Sentence": "And the way you might write this more generally is we could go down here and say the gradient of any function is equal to a vector with its partial derivatives, partial of f with respect to x and partial of f with respect to y. And in some sense, you know, we call these partial derivatives, I like to think of the gradient as the full derivative because it kind of captures all of the information that you need. So a very helpful mnemonic device with the gradient is to think about this triangle, this nabla symbol, as being a vector full of partial derivative operators. And by operator, I just mean, you know, here, let's like partial with respect to x, something where you could give it a function and it gives you another function. So you give this guy, you know, the function f, and it gives you this expression, this multivariable function as a result. So the nabla symbol is this vector full of different partial derivative operators, and in this case, it might just be two of them. And this is kind of a weird thing, right, because it's like, what, this is a vector, it's got like operators in it, that's not what I thought vectors do."}, {"video_title": "Gradient.mp3", "Sentence": "And by operator, I just mean, you know, here, let's like partial with respect to x, something where you could give it a function and it gives you another function. So you give this guy, you know, the function f, and it gives you this expression, this multivariable function as a result. So the nabla symbol is this vector full of different partial derivative operators, and in this case, it might just be two of them. And this is kind of a weird thing, right, because it's like, what, this is a vector, it's got like operators in it, that's not what I thought vectors do. But you can kind of see where it's going. It's really just a, you could think of it as a memory trick, but it's in some sense a little bit deeper than that. And really, when you take this triangle and you say, okay, let's take this triangle, and you can kind of imagine multiplying it by f, really it's like an operator taking in this function and it's going to give you another function."}, {"video_title": "Gradient.mp3", "Sentence": "And this is kind of a weird thing, right, because it's like, what, this is a vector, it's got like operators in it, that's not what I thought vectors do. But you can kind of see where it's going. It's really just a, you could think of it as a memory trick, but it's in some sense a little bit deeper than that. And really, when you take this triangle and you say, okay, let's take this triangle, and you can kind of imagine multiplying it by f, really it's like an operator taking in this function and it's going to give you another function. It's like you take this triangle and you put an f in front of it, and you can imagine, like, this part gets multiplied, quote-unquote, multiplied with f, this part gets, quote-unquote, multiplied with f, but really you're just saying you take the partial derivative with respect to x and then with y, and on and on. And the reason for doing this, this symbol comes up a lot in other contexts. There are two other operators that you're going to learn about called the divergence and the curl."}, {"video_title": "Gradient.mp3", "Sentence": "And really, when you take this triangle and you say, okay, let's take this triangle, and you can kind of imagine multiplying it by f, really it's like an operator taking in this function and it's going to give you another function. It's like you take this triangle and you put an f in front of it, and you can imagine, like, this part gets multiplied, quote-unquote, multiplied with f, this part gets, quote-unquote, multiplied with f, but really you're just saying you take the partial derivative with respect to x and then with y, and on and on. And the reason for doing this, this symbol comes up a lot in other contexts. There are two other operators that you're going to learn about called the divergence and the curl. We'll get to those later, all in due time. But it's useful to think about this vector-ish thing of partial derivatives. And I mean, one weird thing about it, you could say, okay, so this nabla symbol is a vector of partial derivative operators."}, {"video_title": "Gradient.mp3", "Sentence": "There are two other operators that you're going to learn about called the divergence and the curl. We'll get to those later, all in due time. But it's useful to think about this vector-ish thing of partial derivatives. And I mean, one weird thing about it, you could say, okay, so this nabla symbol is a vector of partial derivative operators. What's its dimension? And it's like how many dimensions you've got, because if you had a three-dimensional function, that would mean that you should treat this like it's got three different operators as part of it. And you know, I'd kind of finish this off down here."}, {"video_title": "Gradient.mp3", "Sentence": "And I mean, one weird thing about it, you could say, okay, so this nabla symbol is a vector of partial derivative operators. What's its dimension? And it's like how many dimensions you've got, because if you had a three-dimensional function, that would mean that you should treat this like it's got three different operators as part of it. And you know, I'd kind of finish this off down here. And if you had something that was 100-dimensional, it would have 100 different operators in it. And that's fine. It's really just, again, kind of a memory trick."}, {"video_title": "Gradient.mp3", "Sentence": "And you know, I'd kind of finish this off down here. And if you had something that was 100-dimensional, it would have 100 different operators in it. And that's fine. It's really just, again, kind of a memory trick. So with that, that's how you compute the gradient. Not too much to it. It's pretty much just partial derivatives, but you smack them into a vector."}, {"video_title": "Gradient.mp3", "Sentence": "It's really just, again, kind of a memory trick. So with that, that's how you compute the gradient. Not too much to it. It's pretty much just partial derivatives, but you smack them into a vector. Where it gets fun and where it gets interesting is with the geometric interpretation. I'll get to that in the next couple videos. It's also a super important tool for something called the directional derivative."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And here, I'm gonna go ahead and talk about how you actually compute it. So 3D curl is the kind of thing that you take with regards to a three-dimensional vector field. So something that takes in a three-dimensional point as its input, and then it's gonna output a three-dimensional vector. And it's common to write the component functions as P, Q, and R. So each one of these is a scalar-valued function that takes in a three-dimensional point and just outputs a number. So it'll be that same 3D point with the coordinates X, Y, and Z. X, Y, and Z. So when you have a three-dimensional vector field like this, the image you might have in mind would be something like this, where every point in three-dimensional space has a vector attached to it. And when you actually look at it, there's quite a lot going on."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And it's common to write the component functions as P, Q, and R. So each one of these is a scalar-valued function that takes in a three-dimensional point and just outputs a number. So it'll be that same 3D point with the coordinates X, Y, and Z. X, Y, and Z. So when you have a three-dimensional vector field like this, the image you might have in mind would be something like this, where every point in three-dimensional space has a vector attached to it. And when you actually look at it, there's quite a lot going on. But in principle, all that's really happening is that each point in space is associated with a vector, and the point in space is the input and the vector is the output, and you're just kind of gluing them together. And naturally, because between the three dimensions of the input and the three dimensions of the output, we have six dimensions going on, the picture that you're looking at becomes quite messy. So the question is, how do you compute this curl value that I've been talking about?"}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And when you actually look at it, there's quite a lot going on. But in principle, all that's really happening is that each point in space is associated with a vector, and the point in space is the input and the vector is the output, and you're just kind of gluing them together. And naturally, because between the three dimensions of the input and the three dimensions of the output, we have six dimensions going on, the picture that you're looking at becomes quite messy. So the question is, how do you compute this curl value that I've been talking about? Curl of your vector-valued function. And just as a quick reminder, what this is supposed to be is you're gonna have some kind of fluid flow induced by this vector field, where you're imagining air flowing along each vector. And what you want is a function that tells you at any given point, what is the rotation induced by that fluid flow around that point?"}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So the question is, how do you compute this curl value that I've been talking about? Curl of your vector-valued function. And just as a quick reminder, what this is supposed to be is you're gonna have some kind of fluid flow induced by this vector field, where you're imagining air flowing along each vector. And what you want is a function that tells you at any given point, what is the rotation induced by that fluid flow around that point? And because rotation is described with a three-dimensional vector, you're expecting this to be vector-valued. It'll be something that equals a vector output. And if that doesn't make sense, if that doesn't quite jive, maybe go check out the video on how to represent three-dimensional rotation with a vector."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And what you want is a function that tells you at any given point, what is the rotation induced by that fluid flow around that point? And because rotation is described with a three-dimensional vector, you're expecting this to be vector-valued. It'll be something that equals a vector output. And if that doesn't make sense, if that doesn't quite jive, maybe go check out the video on how to represent three-dimensional rotation with a vector. So what you have here, is gonna be something that takes as its input, x, y, and z. It takes a three-dimensional point, and what it outputs is a vector describing rotation. And there's actually another notation that's quite, quite helpful when it comes to computing this, where you take nabla, that upside-down triangle we used in divergence and gradient, and you imagine taking the cross product between that and your vector v. And as a reminder, this nabla, you imagine it as if it's a vector containing partial differential operators."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And if that doesn't make sense, if that doesn't quite jive, maybe go check out the video on how to represent three-dimensional rotation with a vector. So what you have here, is gonna be something that takes as its input, x, y, and z. It takes a three-dimensional point, and what it outputs is a vector describing rotation. And there's actually another notation that's quite, quite helpful when it comes to computing this, where you take nabla, that upside-down triangle we used in divergence and gradient, and you imagine taking the cross product between that and your vector v. And as a reminder, this nabla, you imagine it as if it's a vector containing partial differential operators. And that's the kind of thing where when you say it out loud, it sounds kind of fancy. A vector full of partial differential operators. But all it really means is, I'm just gonna write a bunch of symbols, and this partial partial x is something that wants to take in a function, a multivariable function, and tell you its partial derivative."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And there's actually another notation that's quite, quite helpful when it comes to computing this, where you take nabla, that upside-down triangle we used in divergence and gradient, and you imagine taking the cross product between that and your vector v. And as a reminder, this nabla, you imagine it as if it's a vector containing partial differential operators. And that's the kind of thing where when you say it out loud, it sounds kind of fancy. A vector full of partial differential operators. But all it really means is, I'm just gonna write a bunch of symbols, and this partial partial x is something that wants to take in a function, a multivariable function, and tell you its partial derivative. And strictly speaking, this doesn't really make sense. Like, hey, how can a vector contain these partial differential operators? But as a series of symbolic movements, it's actually quite helpful."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "But all it really means is, I'm just gonna write a bunch of symbols, and this partial partial x is something that wants to take in a function, a multivariable function, and tell you its partial derivative. And strictly speaking, this doesn't really make sense. Like, hey, how can a vector contain these partial differential operators? But as a series of symbolic movements, it's actually quite helpful. Because when you're multiplying these guys by a thing, it's not really multiplication. You're really gonna be giving it some kind of multivariable function, like p, q, or r, the component functions of our vector field and evaluating it. So just as a warm up for how to do this, let's see what this looks like in the case of two dimensions, where we already kind of know the formula for two dimensional curl."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "But as a series of symbolic movements, it's actually quite helpful. Because when you're multiplying these guys by a thing, it's not really multiplication. You're really gonna be giving it some kind of multivariable function, like p, q, or r, the component functions of our vector field and evaluating it. So just as a warm up for how to do this, let's see what this looks like in the case of two dimensions, where we already kind of know the formula for two dimensional curl. So what that would look like is you have a smaller, more two dimensional, just partial partial x, partial partial y, you know, del operator. And you're gonna take the cross product between that and a two dimensional vector that's just the component functions p and q. And in this case, p and q would be just functions of x and y."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So just as a warm up for how to do this, let's see what this looks like in the case of two dimensions, where we already kind of know the formula for two dimensional curl. So what that would look like is you have a smaller, more two dimensional, just partial partial x, partial partial y, you know, del operator. And you're gonna take the cross product between that and a two dimensional vector that's just the component functions p and q. And in this case, p and q would be just functions of x and y. So I'm kind of overloading notation, right? Over here I have a two dimensional vector field that I'm saying p and y are scalar valued functions with a two dimensional input. But over here, I'm also using p and q to represent ones with a three dimensional input."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And in this case, p and q would be just functions of x and y. So I'm kind of overloading notation, right? Over here I have a two dimensional vector field that I'm saying p and y are scalar valued functions with a two dimensional input. But over here, I'm also using p and q to represent ones with a three dimensional input. So you should think of these as separate, but it's common to use the same names. And this is just kind of gonna illustrate the broader, kind of more complicated point. So when you compute something like this, the cross product, you typically think of it as taking these diagonal components and multiplying them."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "But over here, I'm also using p and q to represent ones with a three dimensional input. So you should think of these as separate, but it's common to use the same names. And this is just kind of gonna illustrate the broader, kind of more complicated point. So when you compute something like this, the cross product, you typically think of it as taking these diagonal components and multiplying them. So that would be your partial partial x, quote unquote, multiplied with q, which really means you're taking the partial derivative of q with respect to x. And then you subtract off this diagonal component here, which is, let's see, so partial partial, oh sorry, this should be a y. This should be partial partial y."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So when you compute something like this, the cross product, you typically think of it as taking these diagonal components and multiplying them. So that would be your partial partial x, quote unquote, multiplied with q, which really means you're taking the partial derivative of q with respect to x. And then you subtract off this diagonal component here, which is, let's see, so partial partial, oh sorry, this should be a y. This should be partial partial y. So okay, sorry about that. You're taking partial partial y of p, and that's what you're subtracting off. So partial partial y of p. So just the partial derivative of that p function with respect to y."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "This should be partial partial y. So okay, sorry about that. You're taking partial partial y of p, and that's what you're subtracting off. So partial partial y of p. So just the partial derivative of that p function with respect to y. And hopefully this is something you recognize. This is the two dimensional curl. And it's something we got an intuition for."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So partial partial y of p. So just the partial derivative of that p function with respect to y. And hopefully this is something you recognize. This is the two dimensional curl. And it's something we got an intuition for. I want it to be more than just a formula. But hopefully this is kind of reassuring that when you take that del operator, that nabla symbol, and cross product with the vector valued function itself, it gives you a sense of curl. Now when we do this in the three dimensional case, we're gonna take a three dimensional cross product between this three dimensional vector-ish thing and this three dimensional function."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And it's something we got an intuition for. I want it to be more than just a formula. But hopefully this is kind of reassuring that when you take that del operator, that nabla symbol, and cross product with the vector valued function itself, it gives you a sense of curl. Now when we do this in the three dimensional case, we're gonna take a three dimensional cross product between this three dimensional vector-ish thing and this three dimensional function. And now would be a good time, by the way, if you're not terribly comfortable with the cross product of how to compute it or how to interpret it and things like that, now would probably be a good time to go find the videos that Sal does on this and build up that intuition for what a cross product actually is and how to compute it. Because at this point, I'm gonna assume that you know how to compute it because we're doing it in kind of an absurd context of partial differential operators and functions. So it's important to have that foundation."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "Now when we do this in the three dimensional case, we're gonna take a three dimensional cross product between this three dimensional vector-ish thing and this three dimensional function. And now would be a good time, by the way, if you're not terribly comfortable with the cross product of how to compute it or how to interpret it and things like that, now would probably be a good time to go find the videos that Sal does on this and build up that intuition for what a cross product actually is and how to compute it. Because at this point, I'm gonna assume that you know how to compute it because we're doing it in kind of an absurd context of partial differential operators and functions. So it's important to have that foundation. So the way you compute a thing like this is you construct a determinant. So I'm gonna go down here. Determinant of a certain three by three matrix."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So it's important to have that foundation. So the way you compute a thing like this is you construct a determinant. So I'm gonna go down here. Determinant of a certain three by three matrix. And the top row of that is all of the unit vectors in various directions of three dimensional space. So these i, j, and k guys, i represents the unit vector in the x direction. So that would be i is equal to, you know, x component is one, but then the other components are zero."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "Determinant of a certain three by three matrix. And the top row of that is all of the unit vectors in various directions of three dimensional space. So these i, j, and k guys, i represents the unit vector in the x direction. So that would be i is equal to, you know, x component is one, but then the other components are zero. And then similarly, j and k represent the unit vectors in the y and z direction. And again, if that doesn't quite make sense why I'm putting them up there or what we're about to do, maybe check out that cross product video. So we put those in the top rows as vectors."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So that would be i is equal to, you know, x component is one, but then the other components are zero. And then similarly, j and k represent the unit vectors in the y and z direction. And again, if that doesn't quite make sense why I'm putting them up there or what we're about to do, maybe check out that cross product video. So we put those in the top rows as vectors. And this is kind of the trick to computing the cross product because again, it's like, what does it mean to put a vector inside a matrix? But it's a notational trick. And then we're gonna take the first vector that we're doing the cross product with and put its components in the next row."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "So we put those in the top rows as vectors. And this is kind of the trick to computing the cross product because again, it's like, what does it mean to put a vector inside a matrix? But it's a notational trick. And then we're gonna take the first vector that we're doing the cross product with and put its components in the next row. So what that would look like is the next row has a partial, partial y, and a partial, oh, sorry, God, I keep messing up here. That's an x. You do whatever the first component is first and then the second component second, and the third component, the z, partial, partial z. I don't know why I'm making that little mistake."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "And then we're gonna take the first vector that we're doing the cross product with and put its components in the next row. So what that would look like is the next row has a partial, partial y, and a partial, oh, sorry, God, I keep messing up here. That's an x. You do whatever the first component is first and then the second component second, and the third component, the z, partial, partial z. I don't know why I'm making that little mistake. And then for the last row, you put in the second vector, which in this case is vector valued function, p, q, and r. P, which is a multivariable function, q, and r. And first it's worth stepping back and looking at this. This is kind of an absurd thing. Usually when we talk about matrices and taking the determinant, all of the components are numbers because you're multiplying numbers together."}, {"video_title": "3d curl formula, part 1.mp3", "Sentence": "You do whatever the first component is first and then the second component second, and the third component, the z, partial, partial z. I don't know why I'm making that little mistake. And then for the last row, you put in the second vector, which in this case is vector valued function, p, q, and r. P, which is a multivariable function, q, and r. And first it's worth stepping back and looking at this. This is kind of an absurd thing. Usually when we talk about matrices and taking the determinant, all of the components are numbers because you're multiplying numbers together. But here we have got like a notational trick layered on top of a notational trick so that one of the rows is vectors, one of the rows is like partial differential operators, and then the last one, each one of these is a multivariable function. So it seems like this absurd, convoluted, as far away from a matrix full of numbers thing as you can get, but it's actually very helpful for computation because if you go through the process of computing this determinant and saying what could that mean, the thing that plops out is gonna be the formula for three-dimensional curl. And at the risk of having a video that runs too long, I'll call things in in here, but continue going through that operation in the next video."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So in the last video, I introduced this thing called the second partial derivative test. And if you have some kind of multivariable function, or really just a two variable function is what this applies to, something that's f of x, y and it outputs a number, when you're looking for places where it has a local maximum or a local minimum, the first step, as I talked about a few videos ago, is to find where the gradient equals zero. And sometimes you'll hear these called critical points or stable points, but inputs where the gradient equals zero. And that's really just a way of compactly writing the fact that all the partial derivatives are equal to zero. Now when you find a point like this, in order to test whether it's a local maximum or a local minimum or a saddle point, without actually looking at the graph, because you don't always have the ability to do that at your disposal, the first step is to compute this long value. And this is the thing I wanna give intuition behind. Where you take all three second partial derivatives, the second partial derivative with respect to x, the second partial derivative with respect to y, and the mixed partial derivative, where first you do it with respect to x, then you do it with respect to y."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And that's really just a way of compactly writing the fact that all the partial derivatives are equal to zero. Now when you find a point like this, in order to test whether it's a local maximum or a local minimum or a saddle point, without actually looking at the graph, because you don't always have the ability to do that at your disposal, the first step is to compute this long value. And this is the thing I wanna give intuition behind. Where you take all three second partial derivatives, the second partial derivative with respect to x, the second partial derivative with respect to y, and the mixed partial derivative, where first you do it with respect to x, then you do it with respect to y. And you compute this value where you evaluate each one of those at your critical point, and you multiply the two pure second partial derivatives, and then subtract off the square of the mixed partial derivative. And again, I'll give intuition for that in a reason, but right now we just kinda take it, oh, all right, I guess you compute this number. And if that value h, if that value h is greater than zero, what it tells you, what it tells you is that you definitely have either a maximum or a minimum."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "Where you take all three second partial derivatives, the second partial derivative with respect to x, the second partial derivative with respect to y, and the mixed partial derivative, where first you do it with respect to x, then you do it with respect to y. And you compute this value where you evaluate each one of those at your critical point, and you multiply the two pure second partial derivatives, and then subtract off the square of the mixed partial derivative. And again, I'll give intuition for that in a reason, but right now we just kinda take it, oh, all right, I guess you compute this number. And if that value h, if that value h is greater than zero, what it tells you, what it tells you is that you definitely have either a maximum or a minimum. So you definitely have either a maximum or a minimum. And then to determine which one, you just have to look at the concavity in one direction. So you'll look at the second partial derivative with respect to x, for example, and if that was positive, that would tell you when you look in the x direction, there's a positive concavity."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And if that value h, if that value h is greater than zero, what it tells you, what it tells you is that you definitely have either a maximum or a minimum. So you definitely have either a maximum or a minimum. And then to determine which one, you just have to look at the concavity in one direction. So you'll look at the second partial derivative with respect to x, for example, and if that was positive, that would tell you when you look in the x direction, there's a positive concavity. If it was negative, it would mean a negative concavity. And so that means a positive value for that second partial derivative would mean a local minimum, and a negative value would mean a local maximum. So that's what it means if this value h turns out to be greater than zero."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So you'll look at the second partial derivative with respect to x, for example, and if that was positive, that would tell you when you look in the x direction, there's a positive concavity. If it was negative, it would mean a negative concavity. And so that means a positive value for that second partial derivative would mean a local minimum, and a negative value would mean a local maximum. So that's what it means if this value h turns out to be greater than zero. And if this value h turns out to be less than zero, strictly less than zero, then you definitely have a saddle point, saddle point, which is neither a maximum nor a minimum. It's kind of like there's disagreement in different directions over whether it should be a maximum or a minimum. And if h equals zero, the test isn't good enough."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So that's what it means if this value h turns out to be greater than zero. And if this value h turns out to be less than zero, strictly less than zero, then you definitely have a saddle point, saddle point, which is neither a maximum nor a minimum. It's kind of like there's disagreement in different directions over whether it should be a maximum or a minimum. And if h equals zero, the test isn't good enough. You would have to do something else to figure it out. So why does this work? Why does this seemingly random conglomeration of second partial derivatives give you a test that lets you determine what type of stable point you're looking at?"}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And if h equals zero, the test isn't good enough. You would have to do something else to figure it out. So why does this work? Why does this seemingly random conglomeration of second partial derivatives give you a test that lets you determine what type of stable point you're looking at? Well, let's just understand each term individually. So this second partial derivative with respect to x, since you're taking both partial derivatives with respect to x, you're basically treating the entire multivariable function as if x is the only variable and y was just some constant. So it's like you're only looking at movement in the x direction."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "Why does this seemingly random conglomeration of second partial derivatives give you a test that lets you determine what type of stable point you're looking at? Well, let's just understand each term individually. So this second partial derivative with respect to x, since you're taking both partial derivatives with respect to x, you're basically treating the entire multivariable function as if x is the only variable and y was just some constant. So it's like you're only looking at movement in the x direction. So in terms of a graph, let's say we've got like this graph here, you can imagine slicing this with a plane that represents movement purely in the x direction. So that'll be a constant y value slice. And you take a look at the curve where this slice intersects your graph."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So it's like you're only looking at movement in the x direction. So in terms of a graph, let's say we've got like this graph here, you can imagine slicing this with a plane that represents movement purely in the x direction. So that'll be a constant y value slice. And you take a look at the curve where this slice intersects your graph. And in the one that I have pictured here, it looks like it's a positive concavity. So this term right here kind of tells you x concavity. So it's kind of like the, what is the concavity as far as the variable x is concerned?"}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And you take a look at the curve where this slice intersects your graph. And in the one that I have pictured here, it looks like it's a positive concavity. So this term right here kind of tells you x concavity. So it's kind of like the, what is the concavity as far as the variable x is concerned? And then symmetrically, this over here, when you take the partial derivative with respect to y two times in a row, it's like you're ignoring the fact that x is even a variable. And you're looking purely at what movement in the y direction looks like, which on the graph that I have pictured here, also happens to give you kind of this positive concavity parabola look. But the point is that the curve on the graph that results from looking at movement purely in the y direction can be analyzed just looking at this partial derivative with respect to y twice in a row."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So it's kind of like the, what is the concavity as far as the variable x is concerned? And then symmetrically, this over here, when you take the partial derivative with respect to y two times in a row, it's like you're ignoring the fact that x is even a variable. And you're looking purely at what movement in the y direction looks like, which on the graph that I have pictured here, also happens to give you kind of this positive concavity parabola look. But the point is that the curve on the graph that results from looking at movement purely in the y direction can be analyzed just looking at this partial derivative with respect to y twice in a row. So that term kind of tells you y concavity, y concavity. Now, first of all, notice what happens if these disagree. If say, x thought there should be positive concavity and y thought there should be negative concavity."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "But the point is that the curve on the graph that results from looking at movement purely in the y direction can be analyzed just looking at this partial derivative with respect to y twice in a row. So that term kind of tells you y concavity, y concavity. Now, first of all, notice what happens if these disagree. If say, x thought there should be positive concavity and y thought there should be negative concavity. Here, I'll write that down, what that means. If x thinks there's positive concavity, we have here some kind of positive number that I'll just write as like a plus sign in parentheses. And then this here, y concavity would be some kind of negative number."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "If say, x thought there should be positive concavity and y thought there should be negative concavity. Here, I'll write that down, what that means. If x thinks there's positive concavity, we have here some kind of positive number that I'll just write as like a plus sign in parentheses. And then this here, y concavity would be some kind of negative number. So I'll just put like a negative sign in parentheses. So that would mean this very first term would be a positive times a negative. And that first term would be negative."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And then this here, y concavity would be some kind of negative number. So I'll just put like a negative sign in parentheses. So that would mean this very first term would be a positive times a negative. And that first term would be negative. And now the thing that we're subtracting off, I'll get to the intuition behind this mixed partial derivative term in a moment. But for now, you can notice that it's something squared. It's something that's always a positive term."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And that first term would be negative. And now the thing that we're subtracting off, I'll get to the intuition behind this mixed partial derivative term in a moment. But for now, you can notice that it's something squared. It's something that's always a positive term. So you're always subtracting off a positive term, which means if this initial one is negative, the entire term h is definitely gonna be negative. So it's gonna put you over into this saddle point territory, which makes sense because if the x direction and the y direction disagree on concavity, that should be a saddle point. The quintessential example here is when you have, when you have the function f of x, y is equal to x squared minus y squared, x squared minus y squared."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "It's something that's always a positive term. So you're always subtracting off a positive term, which means if this initial one is negative, the entire term h is definitely gonna be negative. So it's gonna put you over into this saddle point territory, which makes sense because if the x direction and the y direction disagree on concavity, that should be a saddle point. The quintessential example here is when you have, when you have the function f of x, y is equal to x squared minus y squared, x squared minus y squared. And the graph of that, by the way, the graph of that would look like this, where, let's see, so orienting myself here, moving in the x direction, you have kind of positive concavity, which corresponds to the positive coefficient in front of x squared. And in the y direction, it looks like negative concavity, corresponding to that negative coefficient in front of the y squared. So when there's disagreement among these, the test ensures that we're gonna have a saddle point."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "The quintessential example here is when you have, when you have the function f of x, y is equal to x squared minus y squared, x squared minus y squared. And the graph of that, by the way, the graph of that would look like this, where, let's see, so orienting myself here, moving in the x direction, you have kind of positive concavity, which corresponds to the positive coefficient in front of x squared. And in the y direction, it looks like negative concavity, corresponding to that negative coefficient in front of the y squared. So when there's disagreement among these, the test ensures that we're gonna have a saddle point. Now, what about if they agree, right? What if either it's the case that x thinks there should be positive concavity and y thinks there should be positive concavity, or they both agree that there should be negative concavity? In either one of these cases, when you multiply them together, they're positive."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So when there's disagreement among these, the test ensures that we're gonna have a saddle point. Now, what about if they agree, right? What if either it's the case that x thinks there should be positive concavity and y thinks there should be positive concavity, or they both agree that there should be negative concavity? In either one of these cases, when you multiply them together, they're positive. So it's kind of like saying, if you look purely in the x direction or purely in the y direction, they agree that there should be definitely positive concavity or definitely negative concavity. So that entire first term is going to be positive. So it's kind of like a clever way of capturing whether or not the x directions and y directions agree."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "In either one of these cases, when you multiply them together, they're positive. So it's kind of like saying, if you look purely in the x direction or purely in the y direction, they agree that there should be definitely positive concavity or definitely negative concavity. So that entire first term is going to be positive. So it's kind of like a clever way of capturing whether or not the x directions and y directions agree. However, the reason that it's not enough is because in either case, we're still subtracting off something that's always a positive term. So when you have this agreement between the x direction and the y direction, it then turns into a battle between this xy agreement and whatever's going on with this mixed partial derivative term. And the stronger that mixed partial derivative term, the bigger this negative number, so the more it's pulling the entire value h towards being negative."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So it's kind of like a clever way of capturing whether or not the x directions and y directions agree. However, the reason that it's not enough is because in either case, we're still subtracting off something that's always a positive term. So when you have this agreement between the x direction and the y direction, it then turns into a battle between this xy agreement and whatever's going on with this mixed partial derivative term. And the stronger that mixed partial derivative term, the bigger this negative number, so the more it's pulling the entire value h towards being negative. So let me see if I can give a little bit of reasoning behind why this mixed partial derivative term is trying to pull things towards being a saddle point. Let's take a look at the very simple function, f of xy, xy, is equal to x times y. So what that looks like graphically, f of xy equals x times y, is this."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And the stronger that mixed partial derivative term, the bigger this negative number, so the more it's pulling the entire value h towards being negative. So let me see if I can give a little bit of reasoning behind why this mixed partial derivative term is trying to pull things towards being a saddle point. Let's take a look at the very simple function, f of xy, xy, is equal to x times y. So what that looks like graphically, f of xy equals x times y, is this. It looks like a saddle point. So let's go ahead and look at its partial derivatives. So the first partial derivative is partial with respect to x and partial with respect to y."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So what that looks like graphically, f of xy equals x times y, is this. It looks like a saddle point. So let's go ahead and look at its partial derivatives. So the first partial derivative is partial with respect to x and partial with respect to y. Well, when you do it with respect to x, x looks like a variable, y looks like a constant, it's just that constant y. And when you do it with respect to y, it goes the other way around. Y looks like the variable, x looks like the constant, so the derivative is that constant x."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "So the first partial derivative is partial with respect to x and partial with respect to y. Well, when you do it with respect to x, x looks like a variable, y looks like a constant, it's just that constant y. And when you do it with respect to y, it goes the other way around. Y looks like the variable, x looks like the constant, so the derivative is that constant x. Now when you take the second partial derivatives, if you do it with respect to x twice in a row, you're differentiating this with respect to x, that looks like a constant, so you get zero. And similarly, if you do it with respect to y twice in a row, you're doing this, and the derivative of x with respect to y, x looks like a constant, goes to zero. But the important term, the one that we're getting an intuition about here, this mixed partial derivative, first with respect to x, then with respect to y, well, you can view it in two ways."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "Y looks like the variable, x looks like the constant, so the derivative is that constant x. Now when you take the second partial derivatives, if you do it with respect to x twice in a row, you're differentiating this with respect to x, that looks like a constant, so you get zero. And similarly, if you do it with respect to y twice in a row, you're doing this, and the derivative of x with respect to y, x looks like a constant, goes to zero. But the important term, the one that we're getting an intuition about here, this mixed partial derivative, first with respect to x, then with respect to y, well, you can view it in two ways. Either you take the derivative of this expression with respect to y, in which case it's one, or you think of taking the derivative of this expression with respect to x, in which case it's also one. So it's kind of like this function is a very pure way to take a look at what this mixed partial derivative term looks like. And the higher the coefficient here, if I had put a coefficient of three here, that would mean that the mixed partial derivative would ultimately end up being three."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "But the important term, the one that we're getting an intuition about here, this mixed partial derivative, first with respect to x, then with respect to y, well, you can view it in two ways. Either you take the derivative of this expression with respect to y, in which case it's one, or you think of taking the derivative of this expression with respect to x, in which case it's also one. So it's kind of like this function is a very pure way to take a look at what this mixed partial derivative term looks like. And the higher the coefficient here, if I had put a coefficient of three here, that would mean that the mixed partial derivative would ultimately end up being three. So notice, the reason that this looks like a saddle isn't because the x and y directions disagree. In fact, if you take a look at pure movement in the x direction, it just looks like a constant. The height of the graph along this plane, along this line here, is just a constant, which corresponds to the fact that the second partial derivative with respect to x is equal to zero."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And the higher the coefficient here, if I had put a coefficient of three here, that would mean that the mixed partial derivative would ultimately end up being three. So notice, the reason that this looks like a saddle isn't because the x and y directions disagree. In fact, if you take a look at pure movement in the x direction, it just looks like a constant. The height of the graph along this plane, along this line here, is just a constant, which corresponds to the fact that the second partial derivative with respect to x is equal to zero. And then likewise, if you cut it with a plane representing a constant x value, meaning movement purely in the y direction, the height of the graph doesn't really change along there. It's constantly zero, which corresponds to the fact that this other partial derivative is zero. The reason that the whole thing looks like a saddle is because when you cut it with a diagonal plane here, a diagonal plane, it looks like it has negative concavity."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "The height of the graph along this plane, along this line here, is just a constant, which corresponds to the fact that the second partial derivative with respect to x is equal to zero. And then likewise, if you cut it with a plane representing a constant x value, meaning movement purely in the y direction, the height of the graph doesn't really change along there. It's constantly zero, which corresponds to the fact that this other partial derivative is zero. The reason that the whole thing looks like a saddle is because when you cut it with a diagonal plane here, a diagonal plane, it looks like it has negative concavity. But if you were to chop it in another direction, it would look like it has positive concavity. So in fact, this xy term is kind of like a way of capturing whether there's disagreement in the diagonal directions. And one thing that might be surprising at first is that you only need one of these second partial derivatives in order to determine all of the information about the diagonal directions."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "The reason that the whole thing looks like a saddle is because when you cut it with a diagonal plane here, a diagonal plane, it looks like it has negative concavity. But if you were to chop it in another direction, it would look like it has positive concavity. So in fact, this xy term is kind of like a way of capturing whether there's disagreement in the diagonal directions. And one thing that might be surprising at first is that you only need one of these second partial derivatives in order to determine all of the information about the diagonal directions. Because you could imagine, you know, maybe there's disagreement between movement along one certain vector and movement along another, and you would have to account for infinitely many directions and look at all of them. And yet evidently, it's the case that you only really need to take a look at this mixed partial derivative term. You know, along with the original pure second partial derivatives with respect to x twice and with respect to y twice."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "And one thing that might be surprising at first is that you only need one of these second partial derivatives in order to determine all of the information about the diagonal directions. Because you could imagine, you know, maybe there's disagreement between movement along one certain vector and movement along another, and you would have to account for infinitely many directions and look at all of them. And yet evidently, it's the case that you only really need to take a look at this mixed partial derivative term. You know, along with the original pure second partial derivatives with respect to x twice and with respect to y twice. But still, looking at only three different terms to take into account possible disagreement in infinitely many directions actually feels like quite the surprise. And if you want the full rigorous justification for why this is the case, why this second partial derivative test works and kind of an airtight argument, I've put that in an article that you can find that kind of goes into the dirty details for those who are interested. But if you just want the intuition, I think it's fine to think about the fact that this mixed partial derivative is telling you how much your function looks like the graph of f of xy equals x times y, which is the graph that kind of captures all of the diagonal disagreement."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "You know, along with the original pure second partial derivatives with respect to x twice and with respect to y twice. But still, looking at only three different terms to take into account possible disagreement in infinitely many directions actually feels like quite the surprise. And if you want the full rigorous justification for why this is the case, why this second partial derivative test works and kind of an airtight argument, I've put that in an article that you can find that kind of goes into the dirty details for those who are interested. But if you just want the intuition, I think it's fine to think about the fact that this mixed partial derivative is telling you how much your function looks like the graph of f of xy equals x times y, which is the graph that kind of captures all of the diagonal disagreement. And then when you let that term, that mixed partial derivative term, kind of compete with the agreement between the x and y directions, you know, if they agree very strongly, you have to subtract off a very strong amount in order to pull it back to being negative. So this battle back and forth, if it's pulled to be very negative, that'll give you a saddle point. If it doesn't pull hard enough, then the agreement between the x and y directions wins out and it's either a local maximum or a local minimum."}, {"video_title": "Second partial derivative test intuition.mp3", "Sentence": "But if you just want the intuition, I think it's fine to think about the fact that this mixed partial derivative is telling you how much your function looks like the graph of f of xy equals x times y, which is the graph that kind of captures all of the diagonal disagreement. And then when you let that term, that mixed partial derivative term, kind of compete with the agreement between the x and y directions, you know, if they agree very strongly, you have to subtract off a very strong amount in order to pull it back to being negative. So this battle back and forth, if it's pulled to be very negative, that'll give you a saddle point. If it doesn't pull hard enough, then the agreement between the x and y directions wins out and it's either a local maximum or a local minimum. So hopefully that sheds a little bit of light on why this term makes sense and why it's a reasonable way to combine the three different second partial derivatives available to you. And again, if you want the full details, I've written that up in an article form. I'll see you next video."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "No obvious way to directly take the antiderivative of sine cubed theta, but if we had some mixtures of sines and cosines there, then we could start essentially doing u-substitution, which at this point you probably can do in your head. So what we could do is we could write this as a product. So we could write this as sine of theta, so I'll do this part right over here. This is sine of theta, sine of theta times sine squared theta times sine squared theta, and sine squared theta can be rewritten as one minus cosine squared theta. So this is the same thing as sine of theta times sine of theta times one minus cosine squared theta, and if we multiply this out, this gives us sine theta, sine theta minus sine theta cosine squared theta, and this is much easier for us to integrate, although it looks like a more complicated expression because it's easy to take the antiderivative of sine theta, and now it's easy to take the antiderivative of this because we have the derivative of cosine theta sitting right over here, so this is gonna be cosine cubed theta over three. So essentially we're doing u-substitution right over here, but I'll save that for a second. Let's rewrite all of these in a way that's easy to take the antiderivative of it."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is sine of theta, sine of theta times sine squared theta times sine squared theta, and sine squared theta can be rewritten as one minus cosine squared theta. So this is the same thing as sine of theta times sine of theta times one minus cosine squared theta, and if we multiply this out, this gives us sine theta, sine theta minus sine theta cosine squared theta, and this is much easier for us to integrate, although it looks like a more complicated expression because it's easy to take the antiderivative of sine theta, and now it's easy to take the antiderivative of this because we have the derivative of cosine theta sitting right over here, so this is gonna be cosine cubed theta over three. So essentially we're doing u-substitution right over here, but I'll save that for a second. Let's rewrite all of these in a way that's easy to take the antiderivative of it. Cosine squared theta, we know this is a common trig identity, that's the same thing as 1 1.5, this is the same thing as 1 1.5 of one plus cosine of two theta, and once again, this is much, much easier to take the antiderivative of, so I'll write plus, plus I'll write plus 1 1.5 plus 1 1.5 cosine of two theta, and now we, all of this is actually quite easy to take the antiderivative of, and so I'll just rewrite it again. So minus four cosine theta plus four cosine theta sine theta minus cosine theta sine squared theta d theta, just was able to sneak it in, and so that's our integral between zero and two pi. So let's just take the antiderivative in every one of these steps."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's rewrite all of these in a way that's easy to take the antiderivative of it. Cosine squared theta, we know this is a common trig identity, that's the same thing as 1 1.5, this is the same thing as 1 1.5 of one plus cosine of two theta, and once again, this is much, much easier to take the antiderivative of, so I'll write plus, plus I'll write plus 1 1.5 plus 1 1.5 cosine of two theta, and now we, all of this is actually quite easy to take the antiderivative of, and so I'll just rewrite it again. So minus four cosine theta plus four cosine theta sine theta minus cosine theta sine squared theta d theta, just was able to sneak it in, and so that's our integral between zero and two pi. So let's just take the antiderivative in every one of these steps. It's starting to get a little bit messy, I'll try to write a little bit neater. The antiderivative of sine of theta is cosine of, is negative cosine of theta, negative cosine of theta. If you take the derivative of cosine theta, you get negative sine theta, then the negatives cancel out, you get that right over there."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's just take the antiderivative in every one of these steps. It's starting to get a little bit messy, I'll try to write a little bit neater. The antiderivative of sine of theta is cosine of, is negative cosine of theta, negative cosine of theta. If you take the derivative of cosine theta, you get negative sine theta, then the negatives cancel out, you get that right over there. Then over here, we have the derivative of cosine theta, which is negative sine theta, so we can essentially kind of treat this, we can kind of make the substitution that u is cosine theta, that's essentially what we're doing in our head. So the antiderivative of this is going to be equal to plus cosine cubed theta over three, and then the antiderivative of 1 1\u20442 with respect to theta is just going to be plus 1 1\u20442 theta. The antiderivative of cosine two theta, well we want the derivative of this thing sitting someplace, the derivative of this thing over here is two, so if we put a two here, we can't just multiply by two arbitrarily, we'd have to multiply and divide by two, so we could put a two here, and then we could also, but we'd also have to divide by two, so then that would become a four, and we haven't changed this, notice this is now 2\u20444 cosine of two theta, the exact same thing as 1 1\u20442 cosine two theta, but this is useful because now, the way I've written it here, we have the derivative of two theta right over here, and so we can just say, well we'll just take the antiderivative of this whole thing, which is going to be sine of two theta, but we still have the one, so the antiderivative of this part right over here is sine of two theta, and then we have the 1\u20444 out there, so plus 1\u20444 sine of two theta, and then the antiderivative of cosine theta is just sine theta, so minus four sine theta, minus four sine theta, antiderivative of this right over here, we can kind of pick whichever way we want to do it, but we could say, well the derivative of sine theta is cosine theta, so this is going to be the same thing as four, sine squared theta, sine squared theta over two, over two, or instead of saying over two, instead of writing that four, I'll just divide the four by the two, and I will get a two, I will get a two, so let me erase that, and put a two right over here, and you could work it out yourself, you were to take the derivative of this thing right over here, it'd be the derivative of sine theta, if you just use chain rule, which would be cosine theta, and then times four sine of theta, so that's exactly what we have right over here, and then we have this last part, the derivative of sine theta is cosine theta, and so once again, just like we've done before, the antiderivative of this whole thing is going to be negative sine cubed of theta over three, and we need to evaluate this entire expression between zero, zero, and two pi, so let's see how it evaluates."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you take the derivative of cosine theta, you get negative sine theta, then the negatives cancel out, you get that right over there. Then over here, we have the derivative of cosine theta, which is negative sine theta, so we can essentially kind of treat this, we can kind of make the substitution that u is cosine theta, that's essentially what we're doing in our head. So the antiderivative of this is going to be equal to plus cosine cubed theta over three, and then the antiderivative of 1 1\u20442 with respect to theta is just going to be plus 1 1\u20442 theta. The antiderivative of cosine two theta, well we want the derivative of this thing sitting someplace, the derivative of this thing over here is two, so if we put a two here, we can't just multiply by two arbitrarily, we'd have to multiply and divide by two, so we could put a two here, and then we could also, but we'd also have to divide by two, so then that would become a four, and we haven't changed this, notice this is now 2\u20444 cosine of two theta, the exact same thing as 1 1\u20442 cosine two theta, but this is useful because now, the way I've written it here, we have the derivative of two theta right over here, and so we can just say, well we'll just take the antiderivative of this whole thing, which is going to be sine of two theta, but we still have the one, so the antiderivative of this part right over here is sine of two theta, and then we have the 1\u20444 out there, so plus 1\u20444 sine of two theta, and then the antiderivative of cosine theta is just sine theta, so minus four sine theta, minus four sine theta, antiderivative of this right over here, we can kind of pick whichever way we want to do it, but we could say, well the derivative of sine theta is cosine theta, so this is going to be the same thing as four, sine squared theta, sine squared theta over two, over two, or instead of saying over two, instead of writing that four, I'll just divide the four by the two, and I will get a two, I will get a two, so let me erase that, and put a two right over here, and you could work it out yourself, you were to take the derivative of this thing right over here, it'd be the derivative of sine theta, if you just use chain rule, which would be cosine theta, and then times four sine of theta, so that's exactly what we have right over here, and then we have this last part, the derivative of sine theta is cosine theta, and so once again, just like we've done before, the antiderivative of this whole thing is going to be negative sine cubed of theta over three, and we need to evaluate this entire expression between zero, zero, and two pi, so let's see how it evaluates. So first let's evaluate everything at two pi, so this evaluated at two pi is negative one, this evaluated at two pi is one third, this evaluated at two pi is just going to be pi, this evaluated at two pi is zero, cosine of four pi is going to be zero, this evaluated at two pi is going to be zero, this evaluated at two pi is going to be zero, this evaluated at two pi is going to be zero, So that's a nice simplification. So that's everything evaluated at 2 pi. And from that, we're going to have to subtract everything evaluated at 0."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The antiderivative of cosine two theta, well we want the derivative of this thing sitting someplace, the derivative of this thing over here is two, so if we put a two here, we can't just multiply by two arbitrarily, we'd have to multiply and divide by two, so we could put a two here, and then we could also, but we'd also have to divide by two, so then that would become a four, and we haven't changed this, notice this is now 2\u20444 cosine of two theta, the exact same thing as 1 1\u20442 cosine two theta, but this is useful because now, the way I've written it here, we have the derivative of two theta right over here, and so we can just say, well we'll just take the antiderivative of this whole thing, which is going to be sine of two theta, but we still have the one, so the antiderivative of this part right over here is sine of two theta, and then we have the 1\u20444 out there, so plus 1\u20444 sine of two theta, and then the antiderivative of cosine theta is just sine theta, so minus four sine theta, minus four sine theta, antiderivative of this right over here, we can kind of pick whichever way we want to do it, but we could say, well the derivative of sine theta is cosine theta, so this is going to be the same thing as four, sine squared theta, sine squared theta over two, over two, or instead of saying over two, instead of writing that four, I'll just divide the four by the two, and I will get a two, I will get a two, so let me erase that, and put a two right over here, and you could work it out yourself, you were to take the derivative of this thing right over here, it'd be the derivative of sine theta, if you just use chain rule, which would be cosine theta, and then times four sine of theta, so that's exactly what we have right over here, and then we have this last part, the derivative of sine theta is cosine theta, and so once again, just like we've done before, the antiderivative of this whole thing is going to be negative sine cubed of theta over three, and we need to evaluate this entire expression between zero, zero, and two pi, so let's see how it evaluates. So first let's evaluate everything at two pi, so this evaluated at two pi is negative one, this evaluated at two pi is one third, this evaluated at two pi is just going to be pi, this evaluated at two pi is zero, cosine of four pi is going to be zero, this evaluated at two pi is going to be zero, this evaluated at two pi is going to be zero, this evaluated at two pi is going to be zero, So that's a nice simplification. So that's everything evaluated at 2 pi. And from that, we're going to have to subtract everything evaluated at 0. So cosine of 0, well, that's going to be, once again, 1. And we have a negative sign, so it's negative 1. Then you're going to have plus 1 3rd."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And from that, we're going to have to subtract everything evaluated at 0. So cosine of 0, well, that's going to be, once again, 1. And we have a negative sign, so it's negative 1. Then you're going to have plus 1 3rd. And then you're going to have 0. And then all of these other things are going to be 0. And so if you simplify it, you get this is going to be equal to negative 1 plus 1 3rd plus pi."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then you're going to have plus 1 3rd. And then you're going to have 0. And then all of these other things are going to be 0. And so if you simplify it, you get this is going to be equal to negative 1 plus 1 3rd plus pi. And then we have plus 1 minus 1 3rd. Well, that cancels with that. That cancels with that."}, {"video_title": "Evaluating line integral directly - part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so if you simplify it, you get this is going to be equal to negative 1 plus 1 3rd plus pi. And then we have plus 1 minus 1 3rd. Well, that cancels with that. That cancels with that. And we deserve a drum roll now. It all simplified, just like when we used Stokes' theorem in the four videos. Actually, I think it was a little bit simpler to just directly evaluate the line integral over here."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And graphically, this has the interpretation that if you have a graph of f, setting its derivative equal to zero means that you're looking for places where it's got a flat tangent line. So in the graph that I drew, it would be these two flat tangent lines. And then once you find these points, so for example, here you have one solution that I'll call x1, and then here you have another solution, x2, you can ask yourself the question, are these maxima or are they minima, right? Because both of these can have flat tangent lines. So when you do find this, and you want to understand is it a maximum or a minimum, if you're just looking at the graph, we can tell. You can tell that this point here is a local maximum, and this point here is a local minimum. But if you weren't looking at the graph, there's a nice test that will tell you the answer."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "Because both of these can have flat tangent lines. So when you do find this, and you want to understand is it a maximum or a minimum, if you're just looking at the graph, we can tell. You can tell that this point here is a local maximum, and this point here is a local minimum. But if you weren't looking at the graph, there's a nice test that will tell you the answer. You basically look for the second derivative, and in this case, because the concavity is down, that second derivative is gonna be less than zero. And then over here, because the concavity is up, that second derivative is greater than zero. And by getting this information of the concavity, you can make a conclusion that when the concavity is down, you're at a local maximum."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "But if you weren't looking at the graph, there's a nice test that will tell you the answer. You basically look for the second derivative, and in this case, because the concavity is down, that second derivative is gonna be less than zero. And then over here, because the concavity is up, that second derivative is greater than zero. And by getting this information of the concavity, you can make a conclusion that when the concavity is down, you're at a local maximum. When the concavity is up, you're at a local minimum. And in the case where the second derivative is zero, it's undetermined. You would have to do more tests to figure it out."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And by getting this information of the concavity, you can make a conclusion that when the concavity is down, you're at a local maximum. When the concavity is up, you're at a local minimum. And in the case where the second derivative is zero, it's undetermined. You would have to do more tests to figure it out. It's unknown. So in the multivariable world, the situation is very similar. As I've talked about in previous videos, what you do is you'd have some kind of function, and let's say it's a two-variable function, and instead of looking for where the derivative equals zero, you're gonna be looking for where the gradient of your function is equal to the zero vector, which we might make bold to emphasize that that's a vector."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "You would have to do more tests to figure it out. It's unknown. So in the multivariable world, the situation is very similar. As I've talked about in previous videos, what you do is you'd have some kind of function, and let's say it's a two-variable function, and instead of looking for where the derivative equals zero, you're gonna be looking for where the gradient of your function is equal to the zero vector, which we might make bold to emphasize that that's a vector. And that corresponds with finding flat tangent planes. And if that seems unfamiliar, go back and take a look at the video where I introduced the idea of multivariable maxima and minima. But the subject of this video is gonna be on what is analogous to this second derivative test, where in the single-variable world, you just find the second derivative and check if it's greater than or less than zero."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "As I've talked about in previous videos, what you do is you'd have some kind of function, and let's say it's a two-variable function, and instead of looking for where the derivative equals zero, you're gonna be looking for where the gradient of your function is equal to the zero vector, which we might make bold to emphasize that that's a vector. And that corresponds with finding flat tangent planes. And if that seems unfamiliar, go back and take a look at the video where I introduced the idea of multivariable maxima and minima. But the subject of this video is gonna be on what is analogous to this second derivative test, where in the single-variable world, you just find the second derivative and check if it's greater than or less than zero. How can we, in the multivariable world, do something similar to figure out if you have a local minimum, a local maximum, or that new possibility of a saddle point that I talked about in the last video? So there is another test, and it's called the second partial derivative test. And I'll get to the specifics of that at the very end of this video."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "But the subject of this video is gonna be on what is analogous to this second derivative test, where in the single-variable world, you just find the second derivative and check if it's greater than or less than zero. How can we, in the multivariable world, do something similar to figure out if you have a local minimum, a local maximum, or that new possibility of a saddle point that I talked about in the last video? So there is another test, and it's called the second partial derivative test. And I'll get to the specifics of that at the very end of this video. But to set the landscape, I wanna actually talk through a specific example where we're finding when the gradient equals zero, just to see what that looks like and just to have some concrete formulas to deal with. So the function that you're looking at right now is f of x, y is equal to x to the fourth minus four x squared plus y squared, okay? So that's the function that we're dealing with."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And I'll get to the specifics of that at the very end of this video. But to set the landscape, I wanna actually talk through a specific example where we're finding when the gradient equals zero, just to see what that looks like and just to have some concrete formulas to deal with. So the function that you're looking at right now is f of x, y is equal to x to the fourth minus four x squared plus y squared, okay? So that's the function that we're dealing with. And in order to find where its tangent plane is flat, we're looking for where the gradient equals zero. And remember, this is just really a way of unpacking the requirements that both partial derivatives, the partial derivative of f with respect to x at some point, and we'll kind of write it in as we're looking for the x and y where this is zero, and also where the partial derivative of f with respect to y at that same point, x, y, is equal to zero. So the idea is that this is gonna give us some kind of system of equations that we can solve for x and y."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So that's the function that we're dealing with. And in order to find where its tangent plane is flat, we're looking for where the gradient equals zero. And remember, this is just really a way of unpacking the requirements that both partial derivatives, the partial derivative of f with respect to x at some point, and we'll kind of write it in as we're looking for the x and y where this is zero, and also where the partial derivative of f with respect to y at that same point, x, y, is equal to zero. So the idea is that this is gonna give us some kind of system of equations that we can solve for x and y. So let's go ahead and actually do that. In this case, the partial derivative with respect to x, we look up here, and the only places where x shows up, we have x to the fourth minus four x squared. So that x to the fourth turns into four times x cubed minus four x squared, that becomes minus eight x, and then y, y just looks like a constant, so we're adding a constant, and nothing changes here."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So the idea is that this is gonna give us some kind of system of equations that we can solve for x and y. So let's go ahead and actually do that. In this case, the partial derivative with respect to x, we look up here, and the only places where x shows up, we have x to the fourth minus four x squared. So that x to the fourth turns into four times x cubed minus four x squared, that becomes minus eight x, and then y, y just looks like a constant, so we're adding a constant, and nothing changes here. So the first requirement is that this portion is equal to zero. Now the second part, where we're looking for the partial derivative with respect to y, the only place where y shows up is this y squared term, so the partial derivative with respect to y is just two y, and we're setting that equal to zero. And I chose a simple example where these partial derivative equations, you know, this one nicely only includes x, and this one nicely only includes y, but that's not always the case."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So that x to the fourth turns into four times x cubed minus four x squared, that becomes minus eight x, and then y, y just looks like a constant, so we're adding a constant, and nothing changes here. So the first requirement is that this portion is equal to zero. Now the second part, where we're looking for the partial derivative with respect to y, the only place where y shows up is this y squared term, so the partial derivative with respect to y is just two y, and we're setting that equal to zero. And I chose a simple example where these partial derivative equations, you know, this one nicely only includes x, and this one nicely only includes y, but that's not always the case. You can imagine if you intermingle the variables a little bit more, these will actually kind of intermingle x's and y's, and it'll be a harder thing to solve. But I just want something where we can actually start to find the solutions. So if we actually solve this system, this equation here, the two y equals zero, just gives us the fact that y has to equal zero."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And I chose a simple example where these partial derivative equations, you know, this one nicely only includes x, and this one nicely only includes y, but that's not always the case. You can imagine if you intermingle the variables a little bit more, these will actually kind of intermingle x's and y's, and it'll be a harder thing to solve. But I just want something where we can actually start to find the solutions. So if we actually solve this system, this equation here, the two y equals zero, just gives us the fact that y has to equal zero. So that's nice enough, right? And then the second equation, that four x cubed minus eight x equals zero, let's go ahead and rewrite that, where I'm gonna factor out one of the x's and factor out a four. So this is four x multiplied by x squared minus two has to equal zero."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So if we actually solve this system, this equation here, the two y equals zero, just gives us the fact that y has to equal zero. So that's nice enough, right? And then the second equation, that four x cubed minus eight x equals zero, let's go ahead and rewrite that, where I'm gonna factor out one of the x's and factor out a four. So this is four x multiplied by x squared minus two has to equal zero. So there's two different ways that this can equal zero, right? Either x itself is equal to zero, so that would be one solution, x is equal to zero, or x squared minus two is zero, which would mean x is plus or minus the square root of two. So we have x is plus or minus the square root of two."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So this is four x multiplied by x squared minus two has to equal zero. So there's two different ways that this can equal zero, right? Either x itself is equal to zero, so that would be one solution, x is equal to zero, or x squared minus two is zero, which would mean x is plus or minus the square root of two. So we have x is plus or minus the square root of two. So the solution to the system of equations, we know that no matter what, y has to equal zero, and then one of three different things can happen. X equals zero, x equals positive square root of two, or x equals negative square root of two. So this gives us three separate solutions."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So we have x is plus or minus the square root of two. So the solution to the system of equations, we know that no matter what, y has to equal zero, and then one of three different things can happen. X equals zero, x equals positive square root of two, or x equals negative square root of two. So this gives us three separate solutions. And I'll go ahead and write them down. Our three solutions as ordered pairs are gonna be either zero, zero, for when x is zero and y is zero. You have square root of two, zero, and then you have negative square root of two, zero."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So this gives us three separate solutions. And I'll go ahead and write them down. Our three solutions as ordered pairs are gonna be either zero, zero, for when x is zero and y is zero. You have square root of two, zero, and then you have negative square root of two, zero. These are the three different points, the three different values for x and y that satisfy the two requirements that both partial derivatives are zero. What that should mean on the graph then is when we look at those three different inputs, all of those have flat tangent planes. So the first one, zero, zero, if we kind of look above, I guess we're kind of inside the graph here, zero, zero, is right at the origin."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "You have square root of two, zero, and then you have negative square root of two, zero. These are the three different points, the three different values for x and y that satisfy the two requirements that both partial derivatives are zero. What that should mean on the graph then is when we look at those three different inputs, all of those have flat tangent planes. So the first one, zero, zero, if we kind of look above, I guess we're kind of inside the graph here, zero, zero, is right at the origin. And we can see, just looking at the graph, that that's actually a saddle point. You know, this is neither a local maximum nor a local minimum. It doesn't look like a peak or like a valley."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So the first one, zero, zero, if we kind of look above, I guess we're kind of inside the graph here, zero, zero, is right at the origin. And we can see, just looking at the graph, that that's actually a saddle point. You know, this is neither a local maximum nor a local minimum. It doesn't look like a peak or like a valley. And then the other two, where we kind of move along the x-axis, and I guess it turns out that this point here is directly below x equals positive square root of two, and this other minimum is directly below x equals negative square root of two. I wouldn't have been able to guess that just looking at the graph, but we just figured it out. And we can see visually that both of those are local minima."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "It doesn't look like a peak or like a valley. And then the other two, where we kind of move along the x-axis, and I guess it turns out that this point here is directly below x equals positive square root of two, and this other minimum is directly below x equals negative square root of two. I wouldn't have been able to guess that just looking at the graph, but we just figured it out. And we can see visually that both of those are local minima. But the question is, how could we have figured that out once we find these solutions if you didn't have the graph to look at immediately? How could you have figured out that zero, zero corresponds to a saddle point and that both of these other solutions correspond to local minima? Well, following the idea of the single variable second derivative test, what you might do is take the second partial derivative of our function and see how that might influence concavity."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And we can see visually that both of those are local minima. But the question is, how could we have figured that out once we find these solutions if you didn't have the graph to look at immediately? How could you have figured out that zero, zero corresponds to a saddle point and that both of these other solutions correspond to local minima? Well, following the idea of the single variable second derivative test, what you might do is take the second partial derivative of our function and see how that might influence concavity. For example, if we take the second partial derivative with respect to x, and I'll try to squeeze it up here, second partial derivative of the function with respect to x, we're doing that twice, we're taking the second derivative of this expression with respect to x, so we bring down that three, and that's gonna become 12, because three times four times x squared, 12 times x squared minus eight, minus eight. So what this means, whoop, kinda moved that around, what this means in terms of the graph is that if we move purely in the x direction, which means we kind of cut it with a plane representing a constant y value, and we look at the slice of the graph itself, this expression will tell us the concavity at every given point. So these bottom two points here correspond to plus and minus x equals the square root of two."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "Well, following the idea of the single variable second derivative test, what you might do is take the second partial derivative of our function and see how that might influence concavity. For example, if we take the second partial derivative with respect to x, and I'll try to squeeze it up here, second partial derivative of the function with respect to x, we're doing that twice, we're taking the second derivative of this expression with respect to x, so we bring down that three, and that's gonna become 12, because three times four times x squared, 12 times x squared minus eight, minus eight. So what this means, whoop, kinda moved that around, what this means in terms of the graph is that if we move purely in the x direction, which means we kind of cut it with a plane representing a constant y value, and we look at the slice of the graph itself, this expression will tell us the concavity at every given point. So these bottom two points here correspond to plus and minus x equals the square root of two. So if we go over here and think about the case where x equals the square root of two, and we plug that into the expression, what are we gonna get? Well, we're gonna get 12 multiplied by, if x equals square root of two, then x squared is equal to two, so that's 12 times two minus eight, so that's 24 minus eight, and we're gonna get 16, which is a positive number, which is why you have positive concavity at each of these points. So as far as the x direction is concerned, it feels like, ah, yes, both of these have positive concavity, so they should look like local minima."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So these bottom two points here correspond to plus and minus x equals the square root of two. So if we go over here and think about the case where x equals the square root of two, and we plug that into the expression, what are we gonna get? Well, we're gonna get 12 multiplied by, if x equals square root of two, then x squared is equal to two, so that's 12 times two minus eight, so that's 24 minus eight, and we're gonna get 16, which is a positive number, which is why you have positive concavity at each of these points. So as far as the x direction is concerned, it feels like, ah, yes, both of these have positive concavity, so they should look like local minima. And then if you plug in zero, if instead we went over here and we said x equals zero, then when you plug that in, you'd have 12 times zero minus eight, and instead of 16, you would be getting negative eight. So because you have a negative amount, that gives you this negative concavity on the graph, which is why, as far as x is concerned, the origin looks like a local maximum. So let's actually write that down."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So as far as the x direction is concerned, it feels like, ah, yes, both of these have positive concavity, so they should look like local minima. And then if you plug in zero, if instead we went over here and we said x equals zero, then when you plug that in, you'd have 12 times zero minus eight, and instead of 16, you would be getting negative eight. So because you have a negative amount, that gives you this negative concavity on the graph, which is why, as far as x is concerned, the origin looks like a local maximum. So let's actually write that down. If we kinda go down here, and we're analyzing each one of these, and we think about what does it look like from the perspective of each variable, as far as x is concerned, that origin should look like a max, and then each of these two points should look like minima. This is kind of what the variable x thinks. And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "So let's actually write that down. If we kinda go down here, and we're analyzing each one of these, and we think about what does it look like from the perspective of each variable, as far as x is concerned, that origin should look like a max, and then each of these two points should look like minima. This is kind of what the variable x thinks. And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two. And because it's positive, it's telling you that, as far as y is concerned, there's positive concavity everywhere. And on the graph, what that would mean, what that would mean, if you just look at things where you're kind of slicing with a constant x value to see pure movement in the y direction, there's always gonna be positive concavity. And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And then the variable y, if we do something similar, and we take the second partial derivative with respect to y, yeah, I'll go ahead and write that over here because this'll be pretty quick, second partial derivative with respect to y, we're taking the derivative of this expression with respect to y, and that's just a constant, that's just two. And because it's positive, it's telling you that, as far as y is concerned, there's positive concavity everywhere. And on the graph, what that would mean, what that would mean, if you just look at things where you're kind of slicing with a constant x value to see pure movement in the y direction, there's always gonna be positive concavity. And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity. So as far as y is concerned, everything looks like a local minimum. So we kind of go down here, and you'd say, everything looks like a local minimum. Minimum, minimum, and minimum."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "And here, I've only drawn the plane where x is constantly equal to zero, but if you imagine kind of sliding that plane around left and right, you're always getting positive concavity. So as far as y is concerned, everything looks like a local minimum. So we kind of go down here, and you'd say, everything looks like a local minimum. Minimum, minimum, and minimum. So it might be tempting here to think that you're done, to think you found all the information you need to, because you say, well, in the x and y direction, they disagree about whether that origin should be a maximum or a minimum, which is why it looks like a saddle point. And then they agree, they agree on the other two points that both of them should look like a minimum, which is why, you know, which is why you could say, you think you might say, both of these guys look like a minimum. However, that's actually not enough."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "Minimum, minimum, and minimum. So it might be tempting here to think that you're done, to think you found all the information you need to, because you say, well, in the x and y direction, they disagree about whether that origin should be a maximum or a minimum, which is why it looks like a saddle point. And then they agree, they agree on the other two points that both of them should look like a minimum, which is why, you know, which is why you could say, you think you might say, both of these guys look like a minimum. However, that's actually not enough. There are cases, there are examples that I could draw where doing this kind of analysis would lead you to the wrong conclusion. You would conclude that certain points are, you know, a local minimum, when in fact, they're a saddle point. And the basic reason is that you need to take into account the information given by that other second partial derivative, because in the multivariable world, you can take the partial derivative with respect to one variable, and then with respect to another."}, {"video_title": "Warm up to the second partial derivative test.mp3", "Sentence": "However, that's actually not enough. There are cases, there are examples that I could draw where doing this kind of analysis would lead you to the wrong conclusion. You would conclude that certain points are, you know, a local minimum, when in fact, they're a saddle point. And the basic reason is that you need to take into account the information given by that other second partial derivative, because in the multivariable world, you can take the partial derivative with respect to one variable, and then with respect to another. And you have to take into account this mixed partial derivative term in order to make full conclusions. And I'm a little bit afraid that this video might be running long, so I'll cut it short here, and then I will give you the second partial derivative test in its full glory, accounting for this mixed partial derivative term in the next video. And I'll also, you know, give intuition for where this comes in, why it comes in, why this simple analysis that we did in this case is close, and it does give intuition, but it's not quite full, and it won't give you the right conclusion always."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we really need to express it in terms of a double integral with the parameters, in the domain of the parameters. And the first thing I'm going to do is rewrite this part right over here using our parameters. And we already know that n, our normal vector, times our surface differential can also be written as kind of a vector version of our surface differential that points in the same direction as our normal vector. This is going to be the same thing, and we need to make sure that we get the order on the cross product right. As the partial derivative, and I'm going to confirm this in a second, the partial derivative of the parameterization with respect to one of the parameters crossed with the partial derivative of the parameterization with respect to the other parameter. And then that whole thing, I'm not going to take the absolute value because I need a vector right over here, times the differentials of the parameters, d theta, dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be the same thing, and we need to make sure that we get the order on the cross product right. As the partial derivative, and I'm going to confirm this in a second, the partial derivative of the parameterization with respect to one of the parameters crossed with the partial derivative of the parameterization with respect to the other parameter. And then that whole thing, I'm not going to take the absolute value because I need a vector right over here, times the differentials of the parameters, d theta, dr. And we can swap these two things around depending on what will make our eventual double integral easier. But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we can't swap these two things around because this would actually change the direction of the vector. So we need to make sure that this is popping us out in the right direction. So let's think about the direction that the partial with respect to r will take us. So as r increases, we're going to be moving radially outward from the center of our surface. As r increases, let me do this in a different color, as r increases, we'll be moving radially outward. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So as r increases, we're going to be moving radially outward from the center of our surface. As r increases, let me do this in a different color, as r increases, we'll be moving radially outward. So this quantity will be a vector that looks something like that. And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things, and we could use the right-hand rule, you could imagine, take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear, this is the orange vector right over here. Take your index finger in the direction of that yellow vector."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then as theta increases, we'll be going roughly in that direction. And so if we take the cross product of those two things, and we could use the right-hand rule, you could imagine, take your right hand, point your index finger in the direction of that yellow vector. Let me make it clear, this is the orange vector right over here. Take your index finger in the direction of that yellow vector. So this is my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Take your index finger in the direction of that yellow vector. So this is my index finger, my shakily drawn yellow index finger. Put your middle finger in the direction of the orange vector. So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do. And then your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it, which is exactly the direction we need it to point in."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So my middle finger, you bend it and put it in the direction of the orange vector, and we don't care what the other two fingers do. And then your thumb will be in the direction of the cross product. So your thumb will point outward like that. That's my best attempt at drawing it, which is exactly the direction we need it to point in. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. This is actually the right order. If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's my best attempt at drawing it, which is exactly the direction we need it to point in. We need it to point upward here in order to be oriented properly with the direction that we are actually traversing the path. This is actually the right order. If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. Let's evaluate this cross product. So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If when we did this, we got the thumb pointing into it or below the plane, then we would actually have to swap these orders. So with that out of the way, let's actually evaluate this cross product. Let's evaluate this cross product. So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta. I like to set up the matrix to take the cross product. So we'll put our i, j, and k components just like that. And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the cross product of the partial of our parameterization with respect to r crossed with the partial of our parameterization with respect to theta. I like to set up the matrix to take the cross product. So we'll put our i, j, and k components just like that. And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to r is just going to be a negative sine theta. Negative sine theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then first I will write the partial with respect to r. So the i component, if you take the derivative of this with respect to r, it's just going to be cosine theta. The derivative of this with respect to r is just sine theta. And the derivative of this with respect to r is just going to be a negative sine theta. Negative sine theta. And then if we take, and then we're going to cross it with this. So the derivative of this with respect to theta is going to be negative r sine of theta. So it's negative r sine of theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Negative sine theta. And then if we take, and then we're going to cross it with this. So the derivative of this with respect to theta is going to be negative r sine of theta. So it's negative r sine of theta. Derivative of our j component with respect to theta will be r cosine theta. r cosine theta. And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's negative r sine of theta. Derivative of our j component with respect to theta will be r cosine theta. r cosine theta. And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta. Negative r cosine theta. Is that right? Negative derivative of sine theta is cosine theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the derivative of k, of our k component, or our z component with respect to theta is going to be negative r cosine theta. Negative r cosine theta. Is that right? Negative derivative of sine theta is cosine theta. So it's negative r cosine theta. And now we just evaluate this determinant over here. So this is going to be equal to our i component."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Negative derivative of sine theta is cosine theta. So it's negative r cosine theta. And now we just evaluate this determinant over here. So this is going to be equal to our i component. It's going to be, ignore that row and that column. And we get sine times negative r cosine theta. So we're going to get, I'll just do this in a new color."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to our i component. It's going to be, ignore that row and that column. And we get sine times negative r cosine theta. So we're going to get, I'll just do this in a new color. So we're going to get negative r, that wasn't a new color, I'll do it in purple. We will get negative r cosine theta sine theta. Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to get, I'll just do this in a new color. So we're going to get negative r, that wasn't a new color, I'll do it in purple. We will get negative r cosine theta sine theta. Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus. So it's going to be plus r cosine theta sine theta. Plus r cosine theta sine theta. And it's always nice when things cancel out like this."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Cosine theta sine theta minus, well this is going to be a negative number, so when you subtract a negative number it's going to be plus. So it's going to be plus r cosine theta sine theta. Plus r cosine theta sine theta. And it's always nice when things cancel out like this. This plus this is just zero. Negative of it plus the positive of it. So that all cancels out to zero."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's always nice when things cancel out like this. This plus this is just zero. Negative of it plus the positive of it. So that all cancels out to zero. We don't have an i component. Now let's go to the j component. And remember we need to do our little checkerboard pattern."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that all cancels out to zero. We don't have an i component. Now let's go to the j component. And remember we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be, ignore this column, that row. Cosine theta times negative r cosine theta is negative r cosine squared theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And remember we need to do our little checkerboard pattern. So it's going to be minus j. And it's going to be, ignore this column, that row. Cosine theta times negative r cosine theta is negative r cosine squared theta. I just multiplied those two. And then from that I'm going to subtract this times that. And so this times that, the negative cancels out, we get r sine squared theta."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Cosine theta times negative r cosine theta is negative r cosine squared theta. I just multiplied those two. And then from that I'm going to subtract this times that. And so this times that, the negative cancels out, we get r sine squared theta. So this is minus, let me make sure, I'm going to subtract the product of these two. I'm going to get this positive and it's going to be r sine squared theta. This is always the hard part."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this times that, the negative cancels out, we get r sine squared theta. So this is minus, let me make sure, I'm going to subtract the product of these two. I'm going to get this positive and it's going to be r sine squared theta. This is always the hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually I'll just distribute this negative sign just for fun."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is always the hard part. You can make a lot of careless mistakes here. And this looks like we might be able to simplify this in a second, but I'll wait. Actually I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k component."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually I'll just distribute this negative sign just for fun. So if we distribute the negative sign, these all become positive. Helps simplify things a little bit. And now let's worry about the k component. K component I will be doing in purple. I'll do it in blue. K component, ignore this row, ignore this column."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now let's worry about the k component. K component I will be doing in purple. I'll do it in blue. K component, ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared theta. And then from that I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "K component, ignore this row, ignore this column. So plus k times cosine theta times r cosine theta is r cosine squared theta. And then from that I'm going to subtract negative r sine theta times sine theta. So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here we can factor out, let me just rewrite it."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's going to be negative r sine squared theta, but I'm subtracting it. So it's going to be plus r sine squared theta. And this looks like we're going to simplify it as well. And so this piece right over here we can factor out, let me just rewrite it. This can be written as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1, so this is just r times j."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this piece right over here we can factor out, let me just rewrite it. This can be written as r times cosine squared theta plus sine squared theta. Basic trig identity. That just evaluates to 1, so this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That just evaluates to 1, so this is just r times j. And this over here simplifies for the exact same reason. This is r times cosine squared theta plus sine squared theta. This also is just 1. So this just simplifies to r times k. And so this whole cross product, all of this business right over here, simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral, our original surface integral, we can write as the double integral. And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This also is just 1. So this just simplifies to r times k. And so this whole cross product, all of this business right over here, simplified to, quite luckily, is equal to r times our j unit vector plus r times our k unit vector. And so now we can write our surface integral, our original surface integral, we can write as the double integral. And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now it's going to be over our parameter domain, the r theta domain. So it's the double integral of, we still have the curl of f, and we're going to have to evaluate the curl of f. So I'll just write curl of our vector field f dotted with this business, rj plus rk, and then we have our two parameters."}, {"video_title": "Stokes example part 3 Surface to double integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we might want to change the order in which we integrate, but we'll give ourselves that option a little bit later. Double integral. Now it's going to be over our parameter domain, the r theta domain. So it's the double integral of, we still have the curl of f, and we're going to have to evaluate the curl of f. So I'll just write curl of our vector field f dotted with this business, rj plus rk, and then we have our two parameters. And we might want to switch the order. So maybe we could write d theta dr, and then if we do it in this order, theta goes from 0 to 2 pi and r is going from 0 to 1. But if we swap these two, then obviously we're going to have to swap these two as well."}, {"video_title": "Curvature of a cycloid.mp3", "Sentence": "So let's do another curvature example. This time, I'll just take a two-dimensional curve. So it'll have two different components, x of t and y of t. And the specific components here will be t minus the sine of t, t minus sine of t, and then one minus cosine of t, one minus cosine of t. And this is actually the curve, if you watched the very first video that I did about curvature, introducing it, this is that curve. This is the curve that I said, imagine that it's a road and you're driving along it, and if your steering wheel gets stuck, you're thinking of the circle that you trace out as a result, and at various different points, you're going to be turning at various different amounts, so the circle that your car ends up tracing out would be of varying sizes. So if the curvature is high, if you're steering a lot, radius of curvature is low, and things like that. Here, let's actually compute it. And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power."}, {"video_title": "Curvature of a cycloid.mp3", "Sentence": "This is the curve that I said, imagine that it's a road and you're driving along it, and if your steering wheel gets stuck, you're thinking of the circle that you trace out as a result, and at various different points, you're going to be turning at various different amounts, so the circle that your car ends up tracing out would be of varying sizes. So if the curvature is high, if you're steering a lot, radius of curvature is low, and things like that. Here, let's actually compute it. And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power. So this was a formula, and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit tangent vector with respect to arc length, and if you need to, you can just look up a formula like this, but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So first thing to do is just find x prime, y double prime, y prime, and x prime."}, {"video_title": "Curvature of a cycloid.mp3", "Sentence": "And in the last example, I walked through thinking in terms of the derivative of the unit tangent vector with respect to arc length, but in this case, instead of doing that, I just want to show what it looks like when we take the explicit formula that looks like x prime times y double prime minus y prime times x double prime, and then all of that divided by, all of that divided by x prime squared plus y prime squared, and I'm writing x prime and y prime and such, and all of these you should think of as taking in the variable t, I'm just being a little too lazy to write it, and you take that to the 3 halves power. So this was a formula, and I'm not a huge fan of memorizing formulas and then hoping to apply them later. I really do think the one thing you should take away from curvature is the idea that it's the derivative of the unit tangent vector with respect to arc length, and if you need to, you can just look up a formula like this, but it's worth pointing out that it makes some things easier to compute because finding the tangent vector and everything can be kind of like reinventing the wheel when you already have the results here. So first thing to do is just find x prime, y double prime, y prime, and x prime. So let's go ahead and write those out. So the first derivative of x of t, if we go up here, that's t minus sine of t, so its derivative is 1 minus cosine of t, and the derivative of the y component of 1 minus cosine t, y prime of t, is going to be derivative of cosine is negative sine, so negative derivative of that is sine, and that 1 goes to a constant. And then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of t, so now we're taking the derivative of this, which actually we just did because by coincidence, the first derivative of x is also the y component, so that also equals sine of t. And then y double prime is just the derivative of sine here, so that's just going to be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was 1 minus cosine of t, multiplied by y double prime is cosine of t, we subtract off from that y prime, which is sine of t, multiplied by x double prime, x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared, so x prime was 1 minus cosine of t squared, plus y prime squared, so y prime was just sine, so that's just going to be sine squared of t, and that whole thing to the power 3 halves."}, {"video_title": "Curvature of a cycloid.mp3", "Sentence": "So first thing to do is just find x prime, y double prime, y prime, and x prime. So let's go ahead and write those out. So the first derivative of x of t, if we go up here, that's t minus sine of t, so its derivative is 1 minus cosine of t, and the derivative of the y component of 1 minus cosine t, y prime of t, is going to be derivative of cosine is negative sine, so negative derivative of that is sine, and that 1 goes to a constant. And then when we take the second derivatives of those guys, so maybe change the color for the second derivative here, x double prime of t, so now we're taking the derivative of this, which actually we just did because by coincidence, the first derivative of x is also the y component, so that also equals sine of t. And then y double prime is just the derivative of sine here, so that's just going to be cosine, cosine of t. So now, when we just plug those four values in for kappa, for our curvature, what we get is x prime was 1 minus cosine of t, multiplied by y double prime is cosine of t, we subtract off from that y prime, which is sine of t, multiplied by x double prime, x double prime is also sine of t, so I could just say sine of t squared, and the whole thing is divided by x prime squared, so x prime was 1 minus cosine of t squared, plus y prime squared, so y prime was just sine, so that's just going to be sine squared of t, and that whole thing to the power 3 halves. And that's your answer, right? You apply the formula, you get the answer. So for example, when I was drawing this curve and kind of telling the computer to draw out the appropriate circle, I didn't go through the entire find the unit tangent vector, differentiate it with respect to arc length process, even though that's, you know, decently easy to do in the case of things like circles or helixes, but instead I just went to that formula, I looked it up because I had forgotten, and I found the radius of curvature that way."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's try to think of a parameterization. And let me just copy and paste this entire drawing just so that I can use it down below as we parameterize it. So let me copy it, and then go all the way down here and let me paste it. OK, that is our shape again, our surface. And then let me go to the layer that I wanted to get on. And then let me start evaluating it. So what we want to care about is the integral over surface 3 of z ds."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "OK, that is our shape again, our surface. And then let me go to the layer that I wanted to get on. And then let me start evaluating it. So what we want to care about is the integral over surface 3 of z ds. And surface 3 here, we see that the x and y values essentially take on all of the possible x and y values inside of the unit circle, including the boundary. And then the z values are going to be a function of the x values. We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what we want to care about is the integral over surface 3 of z ds. And surface 3 here, we see that the x and y values essentially take on all of the possible x and y values inside of the unit circle, including the boundary. And then the z values are going to be a function of the x values. We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x. It's a subset that's kind of above the unit circle in the xy plane, or kind of the subset that intersects with our cylinder and kind of chops it. So let's think about x and y's first. So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We know that this plane, that this top surface right over here, S3, it is a subset of the plane z. z is equal to 1 minus x. It's a subset that's kind of above the unit circle in the xy plane, or kind of the subset that intersects with our cylinder and kind of chops it. So let's think about x and y's first. So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it. So I'm going to redraw kind of a top view. So that is my y-axis, and this is my x-axis. And the x's and y's can take on any value."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So first, so x, so let's think about it in terms of polar coordinates, because that's probably the easiest way to think about it. So I'm going to redraw kind of a top view. So that is my y-axis, and this is my x-axis. And the x's and y's can take on any value. They essentially have to fill the unit circle. So if you were to kind of project this top surface down onto the xy plane, you would get this orange surface, that bottom surface, which looked like this. It was essentially the unit circle, just like that."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the x's and y's can take on any value. They essentially have to fill the unit circle. So if you were to kind of project this top surface down onto the xy plane, you would get this orange surface, that bottom surface, which looked like this. It was essentially the unit circle, just like that. Let me draw it a little bit neater than that. I can do a better job. So let me draw the unit circle as neatly as I can."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It was essentially the unit circle, just like that. Let me draw it a little bit neater than that. I can do a better job. So let me draw the unit circle as neatly as I can. So there's my unit circle. And so we can have one parameter that essentially says how far around the unit circle we're going. So essentially, that would be our angle."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw the unit circle as neatly as I can. So there's my unit circle. And so we can have one parameter that essentially says how far around the unit circle we're going. So essentially, that would be our angle. And let's use theta, because that's just for fun. We haven't used theta as a parameter yet. That's theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So essentially, that would be our angle. And let's use theta, because that's just for fun. We haven't used theta as a parameter yet. That's theta. But if we had x's and y's as just a function of theta, and we had a fixed radius, that would essentially just give us the points on the outside of the unit circle. But we need to be able to have all of the xy's that are outside and inside the unit circle. So we actually have to have two parameters."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's theta. But if we had x's and y's as just a function of theta, and we had a fixed radius, that would essentially just give us the points on the outside of the unit circle. But we need to be able to have all of the xy's that are outside and inside the unit circle. So we actually have to have two parameters. We need to not only vary this angle, but we also need to vary the radius. So we would want to trace out the outside of that unit circle, and maybe we'd want to shorten it a little bit, and then trace it out again, and then shorten it some more, and then trace it out again. And so you want to actually have a variable radius as well."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we actually have to have two parameters. We need to not only vary this angle, but we also need to vary the radius. So we would want to trace out the outside of that unit circle, and maybe we'd want to shorten it a little bit, and then trace it out again, and then shorten it some more, and then trace it out again. And so you want to actually have a variable radius as well. And so you could have how far out you're going. You could call that r. So for example, if r is fixed, and you change your ranges of theta, then you would essentially get all of those points right over there. And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so you want to actually have a variable radius as well. And so you could have how far out you're going. You could call that r. So for example, if r is fixed, and you change your ranges of theta, then you would essentially get all of those points right over there. And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle. And so let's do that. So r is going to go between 0 and 1. r is going to be between 0 and 1. And our theta is going to go all the way around."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you would do that for all of the r's, and from r0 all the way to r1, and you would essentially fill up the entire unit circle. And so let's do that. So r is going to go between 0 and 1. r is going to be between 0 and 1. And our theta is going to go all the way around. So our theta is going to go between 0 and 2 pi. This is, let me write this down. I wrote 0 instead of theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And our theta is going to go all the way around. So our theta is going to go between 0 and 2 pi. This is, let me write this down. I wrote 0 instead of theta. Our theta is going to be greater than or equal to 0, less than or equal to 2 pi. And now we're ready to parameterize it. x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I wrote 0 instead of theta. Our theta is going to be greater than or equal to 0, less than or equal to 2 pi. And now we're ready to parameterize it. x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta. So x is going to be r cosine theta. y is going to be r sine theta. That's going to be the y value, r sine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x of r and theta is going to be equal to, so whatever r is, it's going to be r cosine theta. So x is going to be r cosine theta. y is going to be r sine theta. That's going to be the y value, r sine theta. And now z is essentially just a function of x. z is going to be equal to 1 minus x, but x is just r cosine theta. So there you have it. We have our parameterization of this surface right over here."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's going to be the y value, r sine theta. And now z is essentially just a function of x. z is going to be equal to 1 minus x, but x is just r cosine theta. So there you have it. We have our parameterization of this surface right over here. The x's and y's can take all the values of the unit circle, but then the z is up here based on a function of, well, based really on a function of x. It's 1 minus x. And so that will give us all of the possible points right over here on the surface."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have our parameterization of this surface right over here. The x's and y's can take all the values of the unit circle, but then the z is up here based on a function of, well, based really on a function of x. It's 1 minus x. And so that will give us all of the possible points right over here on the surface. You pick an x and y, and then the z is going to pop us right here someplace on that surface. And we can write it as a position vector function. Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so that will give us all of the possible points right over here on the surface. You pick an x and y, and then the z is going to pop us right here someplace on that surface. And we can write it as a position vector function. Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here. Let's call it p for position vector function. And so p, our surface p, we can write it as, actually, I just call it surface 3. So surface 3, I'll do it in that same purple color too, so we know we're talking about this."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of calling that position vector function r, since we've already used r for radius, I will call it, I don't know, let's call it, I'm just going to pick a random letter here. Let's call it p for position vector function. And so p, our surface p, we can write it as, actually, I just call it surface 3. So surface 3, I'll do it in that same purple color too, so we know we're talking about this. Surface 3 as a position vector function, as a function of theta and r. Maybe I'll write r and theta, because that's how I think of things. r and theta is going to be equal to r cosine theta i plus r sine of theta j plus 1 minus r cosine theta. Need to get some real estate here."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So surface 3, I'll do it in that same purple color too, so we know we're talking about this. Surface 3 as a position vector function, as a function of theta and r. Maybe I'll write r and theta, because that's how I think of things. r and theta is going to be equal to r cosine theta i plus r sine of theta j plus 1 minus r cosine theta. Need to get some real estate here. 1 minus r cosine of theta k. And now we are ready to start doing all of the business of evaluating the actual surface integral. So the first thing we do is take the cross product of this, the partial of this with respect to r, and the partial of this with respect to theta. And so let's just get down to business."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Need to get some real estate here. 1 minus r cosine of theta k. And now we are ready to start doing all of the business of evaluating the actual surface integral. So the first thing we do is take the cross product of this, the partial of this with respect to r, and the partial of this with respect to theta. And so let's just get down to business. Let's take the cross product. And so we have our i unit vector, we have our j unit vector, and we have our k unit vector. And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's just get down to business. Let's take the cross product. And so we have our i unit vector, we have our j unit vector, and we have our k unit vector. And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space. And so the partial of this with respect to r. So let's take the partial of this with respect to r. I'll do it in blue. The partial of this with respect to r is just the cosine theta i so this is just cosine theta. I said I was going to do it in blue, and that's not blue."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this might get a little bit involved, but we'll try our best to just work through it, give myself a little bit more space. And so the partial of this with respect to r. So let's take the partial of this with respect to r. I'll do it in blue. The partial of this with respect to r is just the cosine theta i so this is just cosine theta. I said I was going to do it in blue, and that's not blue. So this is going to be cosine theta. The partial of this with respect to r is sine theta and the partial of this This term right over here with respect to r is negative cosine theta. Now let's take the partial with respect to theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I said I was going to do it in blue, and that's not blue. So this is going to be cosine theta. The partial of this with respect to r is sine theta and the partial of this This term right over here with respect to r is negative cosine theta. Now let's take the partial with respect to theta. The partial of this with respect to theta is negative r sine of theta. The partial of this with respect to theta is r cosine theta. And the partial here, this is 0, and then this would be negative r sine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's take the partial with respect to theta. The partial of this with respect to theta is negative r sine of theta. The partial of this with respect to theta is r cosine theta. And the partial here, this is 0, and then this would be negative r sine theta. Oh, no, let me be careful. This is going to be, you have a negative r. So the derivative of cosine theta with respect to theta is negative sine theta. So the negatives are going to cancel out, and so you're going to have r sine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the partial here, this is 0, and then this would be negative r sine theta. Oh, no, let me be careful. This is going to be, you have a negative r. So the derivative of cosine theta with respect to theta is negative sine theta. So the negatives are going to cancel out, and so you're going to have r sine theta. And now we can actually evaluate this determinant to figure out the cross product of the partial of this with respect to r and the partial of this with respect to theta. I'm not writing it down just to kind of save some real estate here. And so we have, actually, maybe I will write it down just to be clear what we're doing."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the negatives are going to cancel out, and so you're going to have r sine theta. And now we can actually evaluate this determinant to figure out the cross product of the partial of this with respect to r and the partial of this with respect to theta. I'm not writing it down just to kind of save some real estate here. And so we have, actually, maybe I will write it down just to be clear what we're doing. The partial of S3 with respect to r crossed with the partial of S3 with respect to theta is equal to, now our i component is going to be sine theta times r sine theta. So that's going to be r sine squared theta minus r cosine theta times negative cosine theta. So that's plus r cosine squared theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we have, actually, maybe I will write it down just to be clear what we're doing. The partial of S3 with respect to r crossed with the partial of S3 with respect to theta is equal to, now our i component is going to be sine theta times r sine theta. So that's going to be r sine squared theta minus r cosine theta times negative cosine theta. So that's plus r cosine squared theta. All of that times i. And a simplification might be popping out here at you. And then you have minus the j component."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's plus r cosine squared theta. All of that times i. And a simplification might be popping out here at you. And then you have minus the j component. The j component is going to be cosine theta times r sine theta. So it's r cosine theta sine theta. And then we're going to subtract from that."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you have minus the j component. The j component is going to be cosine theta times r sine theta. So it's r cosine theta sine theta. And then we're going to subtract from that. See, the negative sines cancel out. So you're going to subtract r sine theta cosine theta, or r cosine theta sine theta. Well, this is interesting because these are the negatives of each other."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to subtract from that. See, the negative sines cancel out. So you're going to subtract r sine theta cosine theta, or r cosine theta sine theta. Well, this is interesting because these are the negatives of each other. r cosine theta sine theta minus r cosine theta sine theta. This just evaluates to 0. So we have no j component."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, this is interesting because these are the negatives of each other. r cosine theta sine theta minus r cosine theta sine theta. This just evaluates to 0. So we have no j component. And then finally, for our k component, we have cosine theta times r cosine theta. So we have r cosine squared theta minus r sine theta times sine theta. Or minus negative r sine theta times sine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have no j component. And then finally, for our k component, we have cosine theta times r cosine theta. So we have r cosine squared theta minus r sine theta times sine theta. Or minus negative r sine theta times sine theta. So this would give you a negative, but we're going to have to subtract it, so it gives you a positive. So plus r sine squared theta k. And so this simplifies quite nicely because this is going to be equal to this term up here. You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or minus negative r sine theta times sine theta. So this would give you a negative, but we're going to have to subtract it, so it gives you a positive. So plus r sine squared theta k. And so this simplifies quite nicely because this is going to be equal to this term up here. You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1. So that's just r times i. So this is equal to r times i. And we do the same thing over here."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can factor out an r. This is r times sine squared theta plus cosine squared theta, which is just that part simplifies to 1. So that's just r times i. So this is equal to r times i. And we do the same thing over here. This also simplifies. This is actually the same thing. This also simplifies to r. So this whole thing simplifies."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we do the same thing over here. This also simplifies. This is actually the same thing. This also simplifies to r. So this whole thing simplifies. Let me write it this way. This is also r sine squared theta plus cosine squared theta also simplifies to r. So you have r times k. And so if you want the magnitude of this business, so let me make it clear. So the magnitude, we'll go back to the magenta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This also simplifies to r. So this whole thing simplifies. Let me write it this way. This is also r sine squared theta plus cosine squared theta also simplifies to r. So you have r times k. And so if you want the magnitude of this business, so let me make it clear. So the magnitude, we'll go back to the magenta. The magnitude of, I don't feel like rewriting it all. I'll just copy and paste it. Edit, copy, and paste."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the magnitude, we'll go back to the magenta. The magnitude of, I don't feel like rewriting it all. I'll just copy and paste it. Edit, copy, and paste. The magnitude of all of this business is going to be equal to the square root of this squared, which is r squared, plus this squared, which is r squared, which simplifies to this is 2r squared. So you take the square root of both of those. You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Edit, copy, and paste. The magnitude of all of this business is going to be equal to the square root of this squared, which is r squared, plus this squared, which is r squared, which simplifies to this is 2r squared. So you take the square root of both of those. You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values. r only takes on positive values. So it's the square root of 2 times r, which is very nice because now we can evaluate ds. ds is going to be this business times dr d theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You get the square root of 2 times r. So this is equal to the square root of 2 times, it would be the absolute value of r, but we know that r only takes on positive values. r only takes on positive values. So it's the square root of 2 times r, which is very nice because now we can evaluate ds. ds is going to be this business times dr d theta. So let's do it. So our surface integral, the thing that we were dealing with from the beginning, that thing right over there. So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds is going to be this business times dr d theta. So let's do it. So our surface integral, the thing that we were dealing with from the beginning, that thing right over there. So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration. So one on the outside, and then I'll do one on the inside. I'll do the inside one in pink. z is equal to 1 minus r cosine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our surface integral, the surface S3 of z ds is now equal to, so I'm going to use different colors for the different variables of integration. So one on the outside, and then I'll do one on the inside. I'll do the inside one in pink. z is equal to 1 minus r cosine theta. So z is equal to 1 minus r cosine theta. It involves both, so I'll use a different color. Minus r cosine theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "z is equal to 1 minus r cosine theta. So z is equal to 1 minus r cosine theta. It involves both, so I'll use a different color. Minus r cosine theta. And then I just have to integrate relative to both of the variables. 1 minus r cosine theta. Oh, no, I have to do the times ds."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Minus r cosine theta. And then I just have to integrate relative to both of the variables. 1 minus r cosine theta. Oh, no, I have to do the times ds. ds is this thing. It's this thing times d theta d dr. So let's see."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Oh, no, I have to do the times ds. ds is this thing. It's this thing times d theta d dr. So let's see. Let's write this down. Times square root of 2r. So let's write that down."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see. Let's write this down. Times square root of 2r. So let's write that down. So this times the square root of 2r. Times the square root of 2. And we can write the square root of 2 out front since it's a constant."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's write that down. So this times the square root of 2r. Times the square root of 2. And we can write the square root of 2 out front since it's a constant. So let me just do that. Simplify this thing. Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we can write the square root of 2 out front since it's a constant. So let me just do that. Simplify this thing. Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way. So let's do that. Let's do dr d theta. So dr d theta."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Square root of 2 times r. And now d theta dr. Or we could write dr d theta either way. So let's do that. Let's do dr d theta. So dr d theta. We could do it either way. It's going to be about the same level of complexity. And so first we're going to integrate with respect to r. And let me do the colors the same way, actually."}, {"video_title": "Surface integral ex3 part 3 Top surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So dr d theta. We could do it either way. It's going to be about the same level of complexity. And so first we're going to integrate with respect to r. And let me do the colors the same way, actually. So dr d theta. If I've got colors, I might as well use them. And actually, I just realized that I'm way out of time."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "Hello everyone. So I have here the graph of a two variable function and I'd like to talk about how you can interpret the partial derivative of that function. So specifically, the function that you're looking at is f of x, y is equal to x squared times y plus sine of y. And the question is, if I take the partial derivative of this function, so maybe I'm looking at the partial derivative of f with respect to x, and let's say I want to do this at negative one, one. So I'll be looking at the partial derivative at a specific point. How do you interpret that on this whole graph? So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "And the question is, if I take the partial derivative of this function, so maybe I'm looking at the partial derivative of f with respect to x, and let's say I want to do this at negative one, one. So I'll be looking at the partial derivative at a specific point. How do you interpret that on this whole graph? So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there. So negative one, move up one, and it's the point that's sitting on the graph. And the first thing you might do is you say, well, when we're taking the partial derivative with respect to x, we're gonna pretend that y is a constant. So let's actually just go ahead and evaluate that."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "So first, let's consider where the point negative one, one is if we're looking above, this is our x-axis, this is our y-axis, the point negative one, one is sitting right there. So negative one, move up one, and it's the point that's sitting on the graph. And the first thing you might do is you say, well, when we're taking the partial derivative with respect to x, we're gonna pretend that y is a constant. So let's actually just go ahead and evaluate that. So when you're doing this, it looks, it says x squared looks like a variable, y looks like a constant, sine of y also looks like a constant. So this is gonna be, we differentiate x squared, and that's two times x times y, which is like a constant, and then the derivative of a constant there is zero. And we're evaluating this whole thing at x is equal to negative one, and y is equal to one."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "So let's actually just go ahead and evaluate that. So when you're doing this, it looks, it says x squared looks like a variable, y looks like a constant, sine of y also looks like a constant. So this is gonna be, we differentiate x squared, and that's two times x times y, which is like a constant, and then the derivative of a constant there is zero. And we're evaluating this whole thing at x is equal to negative one, and y is equal to one. So when we actually plug that in, it'll be two times negative one, multiplied by one, which is two, negative two, excuse me. But what does that mean, right? We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph?"}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "And we're evaluating this whole thing at x is equal to negative one, and y is equal to one. So when we actually plug that in, it'll be two times negative one, multiplied by one, which is two, negative two, excuse me. But what does that mean, right? We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph? Well, first of all, treating y as a constant is basically like slicing the whole graph with a plane that represents a constant y value. So this is the y-axis, and the plane that cuts it perpendicularly, that represents a constant y value. This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "We evaluate this, and maybe you're thinking this is kind of a slight nudge in the x direction, this is the resulting nudge of f. What does that mean for the graph? Well, first of all, treating y as a constant is basically like slicing the whole graph with a plane that represents a constant y value. So this is the y-axis, and the plane that cuts it perpendicularly, that represents a constant y value. This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values. So for the general partial derivative, you can imagine whichever one you want, but this one is y equals one. And I'll go ahead and slice the actual graph at that point, and draw a red line. And this red line is basically all the points on the graph where y is equal to one."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values. So for the general partial derivative, you can imagine whichever one you want, but this one is y equals one. And I'll go ahead and slice the actual graph at that point, and draw a red line. And this red line is basically all the points on the graph where y is equal to one. So I'll just kind of emphasize that, where y is equal to one, and this is y is equal to one. So when we're looking at that, we can actually interpret the partial derivative as a slope, because we're looking at the point here, we're asking how the function changes as we move in the x direction, and from single variable calculus, you might be familiar with thinking of that as the slope of a line. And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "And this red line is basically all the points on the graph where y is equal to one. So I'll just kind of emphasize that, where y is equal to one, and this is y is equal to one. So when we're looking at that, we can actually interpret the partial derivative as a slope, because we're looking at the point here, we're asking how the function changes as we move in the x direction, and from single variable calculus, you might be familiar with thinking of that as the slope of a line. And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at. But let's do this with the partial derivative with respect to y, let's erase what we've got going on here, and I'll go ahead and move the graph back to what it was, get rid of these guys. So now we're no longer slicing with respect to y, but instead, you know, let's say we slice it with a constant x value. So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at. But let's do this with the partial derivative with respect to y, let's erase what we've got going on here, and I'll go ahead and move the graph back to what it was, get rid of these guys. So now we're no longer slicing with respect to y, but instead, you know, let's say we slice it with a constant x value. So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one. And now when we take the partial derivative, we're gonna interpret it as a slice, as the slope of this resulting curve. So that slope ends up looking like this, that's our blue line, and let's go ahead and evaluate the partial derivative of f with respect to y. So I'll go over here, use a different color."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one. And now when we take the partial derivative, we're gonna interpret it as a slice, as the slope of this resulting curve. So that slope ends up looking like this, that's our blue line, and let's go ahead and evaluate the partial derivative of f with respect to y. So I'll go over here, use a different color. So the partial derivative of f with respect to y, partial y, so we go up here, and it says, okay, I see x squared times y, it's considering x squared to be a constant now. So it looks at that and says, x, you're a constant, y, you're the variable, constant times a variable, the derivative is just equal to that constant, so that x squared. And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "So I'll go over here, use a different color. So the partial derivative of f with respect to y, partial y, so we go up here, and it says, okay, I see x squared times y, it's considering x squared to be a constant now. So it looks at that and says, x, you're a constant, y, you're the variable, constant times a variable, the derivative is just equal to that constant, so that x squared. And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y. And if we actually want to evaluate this at our point negative one, one, what we'd get is negative one squared plus cosine of one. And I'm not sure what the cosine of one is, but it's something a little bit positive, and the ultimate result that we see here is gonna be one plus something, I don't know what it is, but it's something positive, and that should make sense, because we look at the slope here, and it's a little bit more than one, not sure exactly, but it's a little bit more than one. So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y. And if we actually want to evaluate this at our point negative one, one, what we'd get is negative one squared plus cosine of one. And I'm not sure what the cosine of one is, but it's something a little bit positive, and the ultimate result that we see here is gonna be one plus something, I don't know what it is, but it's something positive, and that should make sense, because we look at the slope here, and it's a little bit more than one, not sure exactly, but it's a little bit more than one. So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph. And in other contexts, that might not be the case. Maybe it's something that has a multi-dimensional output, we'll talk about that later, when you have a vector valued function, what its partial derivative looks like. But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph. And in other contexts, that might not be the case. Maybe it's something that has a multi-dimensional output, we'll talk about that later, when you have a vector valued function, what its partial derivative looks like. But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again. Now, you're looking at your point here, and you say we're gonna take a tiny step in the y direction, and I'll call that partial y. And you say that makes some kind of change, it causes a change in the function, which you'll call partial f. And as you imagine this getting really, really small, and the resulting change also getting really small, the rise over run of that is gonna give you the slope of the tangent line. So this is just one way of interpreting that ratio, the change in the output that corresponds to a little nudge in the input."}, {"video_title": "Partial derivatives and graphs.mp3", "Sentence": "But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again. Now, you're looking at your point here, and you say we're gonna take a tiny step in the y direction, and I'll call that partial y. And you say that makes some kind of change, it causes a change in the function, which you'll call partial f. And as you imagine this getting really, really small, and the resulting change also getting really small, the rise over run of that is gonna give you the slope of the tangent line. So this is just one way of interpreting that ratio, the change in the output that corresponds to a little nudge in the input. But later on, we'll talk about different ways that you can do that, so I think graphs are very useful. When I move that, the text doesn't move. I think graphs are very useful for thinking about these things, but they're not the only way."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's now parameterize our surface, and then we can figure out what ds would actually look like. And so I will define my position vector function for our surface as r, and I'm going to make it a function of two parameters, because we're going to have to define a surface right over here. And I can actually use x and y as my parameters, because the surface can be defined as a function of x and y. So I'll say that my parameterization is going to be a function of x and y, and in my i direction it's going to be x times i, in my j direction it's going to be y times j, and then in the k direction, well that's just going to be z, and z is a function of x and y. It's going to be z as a function of x and y. And whenever you do a parameterization of a surface, you have to think about, well, what are the constraints on the domain for your parameters? And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'll say that my parameterization is going to be a function of x and y, and in my i direction it's going to be x times i, in my j direction it's going to be y times j, and then in the k direction, well that's just going to be z, and z is a function of x and y. It's going to be z as a function of x and y. And whenever you do a parameterization of a surface, you have to think about, well, what are the constraints on the domain for your parameters? And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here. We could call this the domain of the parameters. So it has to be a member of R, and actually we assume that up here. And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here. We could call this the domain of the parameters. So it has to be a member of R, and actually we assume that up here. And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface. An x, y that is not a member of R, then we're not going to consider the z of that x, y as part of the surface. Only the z of x, y's where the x, y's are part of this region. So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface. An x, y that is not a member of R, then we're not going to consider the z of that x, y as part of the surface. Only the z of x, y's where the x, y's are part of this region. So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be. And we need to think about this a little bit carefully. So first I'll just write something, and then we can confirm whether this really will be the case. ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be. And we need to think about this a little bit carefully. So first I'll just write something, and then we can confirm whether this really will be the case. ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain. So you could view it as the partial of R, of this parameterization with respect to x, crossed with the partial of R with respect to y. And then that whole thing, and we actually want this to be a vector, not just the absolute value or the magnitude of this vector. We actually want this to be a vector."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain. So you could view it as the partial of R, of this parameterization with respect to x, crossed with the partial of R with respect to y. And then that whole thing, and we actually want this to be a vector, not just the absolute value or the magnitude of this vector. We actually want this to be a vector. That thing times, we could put in some order dx, dy, or we could write dy, dx. And if we want to be general, just to say that it's a little differential of our region right over here, we can just write, instead of writing dx, dy, or dy, dx, we will write da. And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We actually want this to be a vector. That thing times, we could put in some order dx, dy, or we could write dy, dx. And if we want to be general, just to say that it's a little differential of our region right over here, we can just write, instead of writing dx, dy, or dy, dx, we will write da. And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in. Because remember, if we're traversing a boundary like this, we want to make sure that the surface is oriented the right way. And the way we think about it, if you were to twist a cap like this, the cap would move up. Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in. Because remember, if we're traversing a boundary like this, we want to make sure that the surface is oriented the right way. And the way we think about it, if you were to twist a cap like this, the cap would move up. Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up. And so we need to make sure that this vector, which really defines the orientation of the surface, is going to be pointed, is definitely going to be pointed up, or above the surface as opposed to going below the surface. And so let's think about that a little bit. The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up. And so we need to make sure that this vector, which really defines the orientation of the surface, is going to be pointed, is definitely going to be pointed up, or above the surface as opposed to going below the surface. And so let's think about that a little bit. The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface. And the partial with respect to y, as y gets bigger, it's going to go in that direction, in that direction along the surface. If I take r cross y, and we could use the right-hand rule here, we put our index finger in the direction of the first thing we're taking the cross product of, our middle finger in the direction of the second thing, so just like that, we bend our middle finger, we don't care what the other fingers are doing, so I'll just draw them right there, then the thumb will go in the direction of the cross product. So in this case, the thumb is going to point up."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface. And the partial with respect to y, as y gets bigger, it's going to go in that direction, in that direction along the surface. If I take r cross y, and we could use the right-hand rule here, we put our index finger in the direction of the first thing we're taking the cross product of, our middle finger in the direction of the second thing, so just like that, we bend our middle finger, we don't care what the other fingers are doing, so I'll just draw them right there, then the thumb will go in the direction of the cross product. So in this case, the thumb is going to point up. And that's exactly what we wanted to happen. So this actually is the right ordering. The partial with respect to y cross the partial with respect to x actually would not have been right."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in this case, the thumb is going to point up. And that's exactly what we wanted to happen. So this actually is the right ordering. The partial with respect to y cross the partial with respect to x actually would not have been right. That would have given us the other orientation. We would have done that if this boundary actually went the other way around. But this is the right orientation given the way that we are going to traverse the boundary."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The partial with respect to y cross the partial with respect to x actually would not have been right. That would have given us the other orientation. We would have done that if this boundary actually went the other way around. But this is the right orientation given the way that we are going to traverse the boundary. Now with that out of the way, let's actually calculate this cross product. So let me just rewrite it. So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is the right orientation given the way that we are going to traverse the boundary. Now with that out of the way, let's actually calculate this cross product. So let me just rewrite it. So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors. I think that's a helpful way of thinking about it. So i, j, and k, actually, fine, I'll use this magenta color, i, j, and k, let me close, let me write this line right over here. And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors. I think that's a helpful way of thinking about it. So i, j, and k, actually, fine, I'll use this magenta color, i, j, and k, let me close, let me write this line right over here. And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x. And so the partial of the i component with respect to x is just going to be 1. The partial of the j component with respect to x is going to be 0. And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x. And so the partial of the i component with respect to x is just going to be 1. The partial of the j component with respect to x is going to be 0. And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x. And so r sub x, or the partial with respect to x, is the vector 1i plus 0j plus z sub xk. And we'll do the same thing for this piece right over here. The partial with respect to y, its i component, well, this is going to be 0."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x. And so r sub x, or the partial with respect to x, is the vector 1i plus 0j plus z sub xk. And we'll do the same thing for this piece right over here. The partial with respect to y, its i component, well, this is going to be 0. Its j component is going to be 1. The partial of this with respect to y is 1. And the partial of this with respect to y is the partial of z with respect to y."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The partial with respect to y, its i component, well, this is going to be 0. Its j component is going to be 1. The partial of this with respect to y is 1. And the partial of this with respect to y is the partial of z with respect to y. And actually, I forgot to write k right over here. I forgot to write k right over here in our parameterization. So now with all of this set up, we are ready to figure out what the cross product is."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the partial of this with respect to y is the partial of z with respect to y. And actually, I forgot to write k right over here. I forgot to write k right over here in our parameterization. So now with all of this set up, we are ready to figure out what the cross product is. The cross product is going to be equal to, so our i component is going to be 0 times the partial of z with respect to y minus 1 times the partial of z with respect to x. So we get negative partial of z with respect to x. And then checkerboard pattern, we'll have minus j times, so ignore that column, that row, 1 times the partial with respect to y, so that's z sub y, the partial of z with respect to y, minus 0 times this."}, {"video_title": "Stokes' theorem proof part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now with all of this set up, we are ready to figure out what the cross product is. The cross product is going to be equal to, so our i component is going to be 0 times the partial of z with respect to y minus 1 times the partial of z with respect to x. So we get negative partial of z with respect to x. And then checkerboard pattern, we'll have minus j times, so ignore that column, that row, 1 times the partial with respect to y, so that's z sub y, the partial of z with respect to y, minus 0 times this. So we're just left with that right over there. And then finally, we're going to have plus k. And here we have 1 times 1 minus 0 times 0. So it's going to be k times 1."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "In the last couple videos, I talked about the local linearization of a function. And in terms of graphs, there's a nice interpretation here where if you imagine the graph of a function and you want to approximate it near a specific point, so you picture that point somewhere on the graph, and it doesn't have to be there, you know, it could choose to be kind of anywhere else along the graph, but if you have some sort of point and you want to approximate the function near there, you can have another function whose graph is just a flat plane, and specifically a plane which is tangent to your original graph at that point. And that's kind of visually how you think about the local linearization. And what I'm gonna start doing here in this next video and in the ones following, is talking about quadratic approximations. So quadratic approximations. And these basically take this to the next level. And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And what I'm gonna start doing here in this next video and in the ones following, is talking about quadratic approximations. So quadratic approximations. And these basically take this to the next level. And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas. But graphically, instead of having a plane that's flat, you have a few more parameters to deal with, and you can give yourself some kind of surface that hugs the graph a little bit more closely. It's still gonna be simpler in terms of formulas, it can still be notably simpler than the original function, but this actually hugs it closely. And as we move around the point that it's approximating near, the way that it hugs it can look pretty different."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas. But graphically, instead of having a plane that's flat, you have a few more parameters to deal with, and you can give yourself some kind of surface that hugs the graph a little bit more closely. It's still gonna be simpler in terms of formulas, it can still be notably simpler than the original function, but this actually hugs it closely. And as we move around the point that it's approximating near, the way that it hugs it can look pretty different. And if you want to think graphically what a quadratic approximation is, you can basically say if you slice this surface, this kind of ghostly white surface, in any direction, it'll look like a parabola of some kind. And notice that given that we're dealing in multiple dimensions, that can make things look pretty complicated. Like this right here, if you slice it in this direction, oh, moving things about."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And as we move around the point that it's approximating near, the way that it hugs it can look pretty different. And if you want to think graphically what a quadratic approximation is, you can basically say if you slice this surface, this kind of ghostly white surface, in any direction, it'll look like a parabola of some kind. And notice that given that we're dealing in multiple dimensions, that can make things look pretty complicated. Like this right here, if you slice it in this direction, oh, moving things about. If you look at it from this angle, it kind of looks like a concave up parabola. But if you were looking at it from another direction, it kind of looks concave down. And all in all, you get a surface that actually has quite a bit of information carried within it."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "Like this right here, if you slice it in this direction, oh, moving things about. If you look at it from this angle, it kind of looks like a concave up parabola. But if you were looking at it from another direction, it kind of looks concave down. And all in all, you get a surface that actually has quite a bit of information carried within it. And you can see that by hugging the graph very closely, this approximation is gonna be, well, it's gonna be even closer. Because near the point where you're approximating, you can go out, you can take a couple steps away, and the approximation is still gonna be very close to what the graph is. And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And all in all, you get a surface that actually has quite a bit of information carried within it. And you can see that by hugging the graph very closely, this approximation is gonna be, well, it's gonna be even closer. Because near the point where you're approximating, you can go out, you can take a couple steps away, and the approximation is still gonna be very close to what the graph is. And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself. So this is gonna be something that, although it takes more information to describe than a local linearization, it gives us a much closer approximation. So a linear function, which, you know, one that just draws a plane like this, in terms of actual functions, what this means, so I'll kinda write linear. This is gonna be some kind of function of x and y."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself. So this is gonna be something that, although it takes more information to describe than a local linearization, it gives us a much closer approximation. So a linear function, which, you know, one that just draws a plane like this, in terms of actual functions, what this means, so I'll kinda write linear. This is gonna be some kind of function of x and y. And what it looks like is some kind of constant, which I'll say a, plus another constant times the variable x, plus another constant times the variable y. This is sort of the basic form of linear functions. And technically, this isn't linear."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "This is gonna be some kind of function of x and y. And what it looks like is some kind of constant, which I'll say a, plus another constant times the variable x, plus another constant times the variable y. This is sort of the basic form of linear functions. And technically, this isn't linear. If one was gonna be really pedantic, they would say that that's actually affine, because strictly speaking, linear functions shouldn't have this constant term. It should be purely x's and y's. But in the context of approximations, people usually would call this the linear term."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And technically, this isn't linear. If one was gonna be really pedantic, they would say that that's actually affine, because strictly speaking, linear functions shouldn't have this constant term. It should be purely x's and y's. But in the context of approximations, people usually would call this the linear term. So a quadratic term, what this is gonna look like, quadratic, we are allowed to have all the same terms as that linear one. So you can have, you know, a constant, you can have these two linear terms, bx and cy. And then you're allowed to have anything that has two variables multiplied into it."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "But in the context of approximations, people usually would call this the linear term. So a quadratic term, what this is gonna look like, quadratic, we are allowed to have all the same terms as that linear one. So you can have, you know, a constant, you can have these two linear terms, bx and cy. And then you're allowed to have anything that has two variables multiplied into it. So maybe I'll have d times x squared. And then you can also have something times xy. This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "And then you're allowed to have anything that has two variables multiplied into it. So maybe I'll have d times x squared. And then you can also have something times xy. This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two. But really, it's just saying, anytime you have two variables multiplied in. And then we can add some other constant, say f times y squared, where, you know, now we're multiplying two y's into it. So all of these guys, these are what you would call your quadratic terms."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two. But really, it's just saying, anytime you have two variables multiplied in. And then we can add some other constant, say f times y squared, where, you know, now we're multiplying two y's into it. So all of these guys, these are what you would call your quadratic terms. Things that, you know, either x squared, y squared, or x times y. Anything that has two variables in it. So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "So all of these guys, these are what you would call your quadratic terms. Things that, you know, either x squared, y squared, or x times y. Anything that has two variables in it. So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space. And if you choose the most optimal one, you'll get one that's tangent to your curve at the specific point. And kind of, it'll depend on where that point is, you'll get different planes, but they're all tangent. So what we're going to do, in the next couple videos, is talk about how you tweak all of these six different constants, so that you can get functions that really closely hug the curve."}, {"video_title": "What do quadratic approximations look like.mp3", "Sentence": "So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space. And if you choose the most optimal one, you'll get one that's tangent to your curve at the specific point. And kind of, it'll depend on where that point is, you'll get different planes, but they're all tangent. So what we're going to do, in the next couple videos, is talk about how you tweak all of these six different constants, so that you can get functions that really closely hug the curve. Right, and as you, they're all going to depend on the original point, because as you move that point around, what it takes to hug the curve is going to be different. It's going to have to do with partial differential information about your original function, the function whose graph this is. And it's going to look pretty similar to the local linearization case, just, you know, added complexity."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "So we've just computed a vector-valued partial derivative of a vector-valued function, but the question is, what does this mean? What does this jumble of symbols actually mean in a more intuitive geometric setting? And that has everything to do with how you visualize the function, and with this specific function, given that the input is two-dimensional, but the output is three-dimensional, and the output is more dimensions than the input, it's nice to visualize it as a parametric surface. And the way that I do that, maybe you could call this visualizing it as a transformation also, because what I want to do is basically think of the TS plane, think of the TS plane where all these input values live, and kind of think of how that's going to map into three-dimensional space. But when I do that, I'm actually going to cheat a little bit. Rather than having a separate plane off there as the TS plane, I'm going to kind of overwrite onto the XY plane itself, and plop the TS plane down like this. And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And the way that I do that, maybe you could call this visualizing it as a transformation also, because what I want to do is basically think of the TS plane, think of the TS plane where all these input values live, and kind of think of how that's going to map into three-dimensional space. But when I do that, I'm actually going to cheat a little bit. Rather than having a separate plane off there as the TS plane, I'm going to kind of overwrite onto the XY plane itself, and plop the TS plane down like this. And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three. So each tick mark on the graph here corresponds with a half, so this is one, that's two, and then up here is three. Same with S, S also ranges from zero to three. And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three. So each tick mark on the graph here corresponds with a half, so this is one, that's two, and then up here is three. Same with S, S also ranges from zero to three. And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier. It's, you could call it laziness. But the benefit here is, what we can do is watch each point, and each one of these points you're thinking of as corresponding to some kind of TS pair, an input point, which is just a pair of numbers, and we're going to watch each one of those points move to the corresponding output. The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier. It's, you could call it laziness. But the benefit here is, what we can do is watch each point, and each one of these points you're thinking of as corresponding to some kind of TS pair, an input point, which is just a pair of numbers, and we're going to watch each one of those points move to the corresponding output. The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it. And what that looks like when we animate this, actually, is each one of those points in our square of TS plane moves to the corresponding output, and you end up with a certain surface. And just to make it a little more concrete what's actually going on here, let's focus in on just one point, and we'll focus in on this point not just for the function visualization, but for the partial derivative as well. And the function, or the point, rather, that I care about is going to be at one, one."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it. And what that looks like when we animate this, actually, is each one of those points in our square of TS plane moves to the corresponding output, and you end up with a certain surface. And just to make it a little more concrete what's actually going on here, let's focus in on just one point, and we'll focus in on this point not just for the function visualization, but for the partial derivative as well. And the function, or the point, rather, that I care about is going to be at one, one. So this point right here represents the pair of TS, where each one of them is equal to one. One, one. And you can start by predicting where you think this is going to get output."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And the function, or the point, rather, that I care about is going to be at one, one. So this point right here represents the pair of TS, where each one of them is equal to one. One, one. And you can start by predicting where you think this is going to get output. And to do that, you just plug it into the function. This is kind of what the visualization means, is we're plugging this in, and for T and S, we're going to plug in just one and one. So that top part is going to look like one squared minus one squared, which becomes zero."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And you can start by predicting where you think this is going to get output. And to do that, you just plug it into the function. This is kind of what the visualization means, is we're plugging this in, and for T and S, we're going to plug in just one and one. So that top part is going to look like one squared minus one squared, which becomes zero. That middle part is going to be one times one, which is one. And then over here, we're going to have one times one squared, which is one, minus one times one squared, which is again one. And you can probably see, because of the symmetry there, those also cancel out, you get zero."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "So that top part is going to look like one squared minus one squared, which becomes zero. That middle part is going to be one times one, which is one. And then over here, we're going to have one times one squared, which is one, minus one times one squared, which is again one. And you can probably see, because of the symmetry there, those also cancel out, you get zero. Which means the output corresponding with this input should be the vector zero, one, zero. A vector that's of unit length pointing in the y direction. So if we look here, you know, this is the x-axis, this here is the y-axis."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And you can probably see, because of the symmetry there, those also cancel out, you get zero. Which means the output corresponding with this input should be the vector zero, one, zero. A vector that's of unit length pointing in the y direction. So if we look here, you know, this is the x-axis, this here is the y-axis. So you would think it should be a vector that looks kind of like this. Unit vector in the y direction. And the point of the surface is what corresponds to the tip of that vector."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "So if we look here, you know, this is the x-axis, this here is the y-axis. So you would think it should be a vector that looks kind of like this. Unit vector in the y direction. And the point of the surface is what corresponds to the tip of that vector. Right, this is how we visualize parametric things. You just think of the tip of the vector as kind of moving through space and drawing out the thing. In this case, the thing it's drawing is a surface."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And the point of the surface is what corresponds to the tip of that vector. Right, this is how we visualize parametric things. You just think of the tip of the vector as kind of moving through space and drawing out the thing. In this case, the thing it's drawing is a surface. So if we watch that animation again and we let things play forward, that dot corresponding with the input one, one does indeed land at the tip of that vector. So at least for that value, you can see that I'm not lying to you with the animation. And in principle, you could do that for every single point, right?"}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "In this case, the thing it's drawing is a surface. So if we watch that animation again and we let things play forward, that dot corresponding with the input one, one does indeed land at the tip of that vector. So at least for that value, you can see that I'm not lying to you with the animation. And in principle, you could do that for every single point, right? If any given input point, you kind of plug it through the function and you draw the vector in three-dimensional space, as you watch this animation, it'll land at the tip of that vector. So, now if we want to start thinking about what the partial derivative means, remember, this little dt, this partial t is telling you to nudge it in the t direction. So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here?"}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And in principle, you could do that for every single point, right? If any given input point, you kind of plug it through the function and you draw the vector in three-dimensional space, as you watch this animation, it'll land at the tip of that vector. So, now if we want to start thinking about what the partial derivative means, remember, this little dt, this partial t is telling you to nudge it in the t direction. So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here? Well, the t direction, I'm saying, is in this direction here where, you know, this represents one, two, three of t values. And this line here represents the constant value for s, so this will be s constantly equaling one, which you can know because it's passing through the point one, one. And then otherwise, you're just letting t range freely."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here? Well, the t direction, I'm saying, is in this direction here where, you know, this represents one, two, three of t values. And this line here represents the constant value for s, so this will be s constantly equaling one, which you can know because it's passing through the point one, one. And then otherwise, you're just letting t range freely. And if we watch how this gets transformed under the transformation, under the mapping to the parametric surface, you can get a feel for what varying the input t does in the output space. So this whole pink line is basically telling you what happens if you let s constantly equal one, but you let the variable t, the input t, vary freely, and you get a certain curve in three-dimensional space. And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And then otherwise, you're just letting t range freely. And if we watch how this gets transformed under the transformation, under the mapping to the parametric surface, you can get a feel for what varying the input t does in the output space. So this whole pink line is basically telling you what happens if you let s constantly equal one, but you let the variable t, the input t, vary freely, and you get a certain curve in three-dimensional space. And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one. So if instead of thinking about movement of t as a whole, you start thinking about nudges, this whole partial t is something where we're just imagining a tiny, tiny little movement in the t direction, not really that much, just a tiny little move, and it's like you're recording its value as partial t. So maybe if you're being concrete, you'd say partial t would be something like 0.01. And really, it's gonna be a limiting variable that gets smaller and smaller, but I find it's kind of nice to think about an actual value like 1.100. And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one. So if instead of thinking about movement of t as a whole, you start thinking about nudges, this whole partial t is something where we're just imagining a tiny, tiny little movement in the t direction, not really that much, just a tiny little move, and it's like you're recording its value as partial t. So maybe if you're being concrete, you'd say partial t would be something like 0.01. And really, it's gonna be a limiting variable that gets smaller and smaller, but I find it's kind of nice to think about an actual value like 1.100. And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve. And it'll be tangent to that curve, right? The vector that tells you how you move just a tiny, tiny little bit, it'll be tangent in some way. And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve. And it'll be tangent to that curve, right? The vector that tells you how you move just a tiny, tiny little bit, it'll be tangent in some way. And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger. So the actual derivative isn't gonna be just some tiny little nudge that's hardly, hardly visible, but it's gonna be that nudge vector scaled appropriately. In this case, it would be divided by 1.100, or multiplied by 100, and it would be something that remains tangent to the curve, but maybe it's pointing big. And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right?"}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger. So the actual derivative isn't gonna be just some tiny little nudge that's hardly, hardly visible, but it's gonna be that nudge vector scaled appropriately. In this case, it would be divided by 1.100, or multiplied by 100, and it would be something that remains tangent to the curve, but maybe it's pointing big. And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right? The ratio of the nudge sizes is bigger. So if you were to have a very long partial derivative vector that's still tangent, but really goes out there, that would tell you that as you vary t, you're zipping along super quickly. And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right? The ratio of the nudge sizes is bigger. So if you were to have a very long partial derivative vector that's still tangent, but really goes out there, that would tell you that as you vary t, you're zipping along super quickly. And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right. You're moving positively in the z direction, you're moving, not z, sorry, the y, positively in the y direction up there. And the z direction is actually negative, isn't it? This curve kind of goes down as far as z is concerned."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right. You're moving positively in the z direction, you're moving, not z, sorry, the y, positively in the y direction up there. And the z direction is actually negative, isn't it? This curve kind of goes down as far as z is concerned. So before even computing it, if I were to tell you that I'm gonna plug in the value one, one to this partial derivative that we computed in the last video, you would say, oh, well, just looking at the picture, you can kind of tell that the x value is gonna be something positive, something greater than zero. The y value is also gonna be something positive. And again, that's because, you know, the movement is to the right, so positive x."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "This curve kind of goes down as far as z is concerned. So before even computing it, if I were to tell you that I'm gonna plug in the value one, one to this partial derivative that we computed in the last video, you would say, oh, well, just looking at the picture, you can kind of tell that the x value is gonna be something positive, something greater than zero. The y value is also gonna be something positive. And again, that's because, you know, the movement is to the right, so positive x. It's moving up, so positive y. But the z value should actually be a little bit negative, right? Because as you look at this curve, it's going down, in a sense."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And again, that's because, you know, the movement is to the right, so positive x. It's moving up, so positive y. But the z value should actually be a little bit negative, right? Because as you look at this curve, it's going down, in a sense. And with that being our prediction, if you start plugging in one, one to t and s, what you'll see is that, you know, two times one is two. S equals one, so that's just one. And then over here, this looks like one squared minus two times one times one."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "Because as you look at this curve, it's going down, in a sense. And with that being our prediction, if you start plugging in one, one to t and s, what you'll see is that, you know, two times one is two. S equals one, so that's just one. And then over here, this looks like one squared minus two times one times one. So this will be one minus two. That's negative one. So it is, in fact, that kind of positive, positive, negative pattern that you're seeing."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And then over here, this looks like one squared minus two times one times one. So this will be one minus two. That's negative one. So it is, in fact, that kind of positive, positive, negative pattern that you're seeing. And maybe even from this curve, you can get a feel for why the movement in the x direction is twice as much as the movement in the y. It's moving more to the right than it is up in the y direction. And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "So it is, in fact, that kind of positive, positive, negative pattern that you're seeing. And maybe even from this curve, you can get a feel for why the movement in the x direction is twice as much as the movement in the y. It's moving more to the right than it is up in the y direction. And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface. And the corresponding movement, the direction that nudges in the t direction take you, will give you some vector in three-dimensional space. And that's the interpretation. That is the meaning of the partial derivative of the vector-valued function here."}, {"video_title": "Partial derivative of a parametric surface, part 1.mp3", "Sentence": "And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface. And the corresponding movement, the direction that nudges in the t direction take you, will give you some vector in three-dimensional space. And that's the interpretation. That is the meaning of the partial derivative of the vector-valued function here. And again, it's not the tiny, the actual nudge vector itself, right? When you nudge the input and you get just a little smidgen in the output space here, but it's that divided by the size of the nudge. So that's why you'll get kind of normal-sized vectors rather than tiny vectors."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So let's say you have a function that's got a single input, t, and then it outputs a vector. And the vector's gonna depend on t, so the x component will be t times the cosine of t, and then the y component will be t times the sine of t. This is what's called a parametric function. And I should maybe say one parameter parametric function. One parameter. And parameter is just kind of a fancy word for input. Parameter. So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "One parameter. And parameter is just kind of a fancy word for input. Parameter. So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional. So you might think, when you visualize something like this, oh, it's got a single input, and it's got a two-dimensional output, let's graph it. Let's put those three numbers together and plot them. But what turns out to be even better is to look just in the output space."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional. So you might think, when you visualize something like this, oh, it's got a single input, and it's got a two-dimensional output, let's graph it. Let's put those three numbers together and plot them. But what turns out to be even better is to look just in the output space. So in this case, the output space is two-dimensional, so I'll go ahead and draw a coordinate plane here. And let's just evaluate this function at a couple different points and see what it looks like. So maybe the easiest place to evaluate it would be zero."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "But what turns out to be even better is to look just in the output space. So in this case, the output space is two-dimensional, so I'll go ahead and draw a coordinate plane here. And let's just evaluate this function at a couple different points and see what it looks like. So maybe the easiest place to evaluate it would be zero. So f of zero f of zero is equal to, and then in both cases, it'll be zero times something, so zero times cosine of zero is just zero, and then zero times sine of zero is also just zero. So that input corresponds to the output. You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So maybe the easiest place to evaluate it would be zero. So f of zero f of zero is equal to, and then in both cases, it'll be zero times something, so zero times cosine of zero is just zero, and then zero times sine of zero is also just zero. So that input corresponds to the output. You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it. So let's take a different point just to see what else could happen. And I'm gonna choose pi halves. Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it. So let's take a different point just to see what else could happen. And I'm gonna choose pi halves. Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of. So t is pi halves, cosine of pi halves. And you start thinking, okay, what's the cosine of pi halves, what's the sine of pi halves? And maybe you go off and draw a little unit circle while you're writing things out."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of. So t is pi halves, cosine of pi halves. And you start thinking, okay, what's the cosine of pi halves, what's the sine of pi halves? And maybe you go off and draw a little unit circle while you're writing things out. Oops. This is the problem with talking while writing. Sine of pi halves."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "And maybe you go off and draw a little unit circle while you're writing things out. Oops. This is the problem with talking while writing. Sine of pi halves. And you know, it's, if you go off and scribble that little unit circle, and you say pi halves is gonna bring us a quarter of the way around over here, and cosine of pi halves is measuring the x component of that so that just cancels out to zero. And then sine of pi halves is the y component of that, so that ends up equaling one. Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sine of pi halves. And you know, it's, if you go off and scribble that little unit circle, and you say pi halves is gonna bring us a quarter of the way around over here, and cosine of pi halves is measuring the x component of that so that just cancels out to zero. And then sine of pi halves is the y component of that, so that ends up equaling one. Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component. And what that would look like, you know, the y component of pi halves, it's about 1.7 up there and there's no x component, so you might get a vector like this. And if you imagine doing this at all the different input points, you might get a bunch of different vectors off doing different things. And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component. And what that would look like, you know, the y component of pi halves, it's about 1.7 up there and there's no x component, so you might get a vector like this. And if you imagine doing this at all the different input points, you might get a bunch of different vectors off doing different things. And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch. So we just wanna trace the points that correspond to the output, the tips of each vector. And what I'll do here, I'll show a little animation. Let me just clear the board a bit."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch. So we just wanna trace the points that correspond to the output, the tips of each vector. And what I'll do here, I'll show a little animation. Let me just clear the board a bit. An animation where I'll let t range between zero and 10. Let's write that down. So the value t is gonna start at zero and then it's gonna go to 10."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "Let me just clear the board a bit. An animation where I'll let t range between zero and 10. Let's write that down. So the value t is gonna start at zero and then it's gonna go to 10. And we'll just see what vectors does that output and what curve does the tip of that vector trace out. So here it goes, all of the values just kind of ranging zero to 10. And you end up getting this spiral shape."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So the value t is gonna start at zero and then it's gonna go to 10. And we'll just see what vectors does that output and what curve does the tip of that vector trace out. So here it goes, all of the values just kind of ranging zero to 10. And you end up getting this spiral shape. And you can maybe think about why this cosine of t, sine of t scaled by the value t itself would give you this spiral. But what it means is that when you, we saw that zero goes here, evidently. It's the case that 10 outputs here."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "And you end up getting this spiral shape. And you can maybe think about why this cosine of t, sine of t scaled by the value t itself would give you this spiral. But what it means is that when you, we saw that zero goes here, evidently. It's the case that 10 outputs here. And a disadvantage of drawing things like this, you're not quite sure of what the interim values are. You could kind of guess maybe one goes somewhere here, two goes somewhere here, and you're kind of hoping that they're evenly spaced as you move along. But you don't get that information."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "It's the case that 10 outputs here. And a disadvantage of drawing things like this, you're not quite sure of what the interim values are. You could kind of guess maybe one goes somewhere here, two goes somewhere here, and you're kind of hoping that they're evenly spaced as you move along. But you don't get that information. You lose the input information. You get the shape of the curve. And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "But you don't get that information. You lose the input information. You get the shape of the curve. And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate. But just to show where it might matter, I'll animate the same thing again, another function that draws the same curve, but it starts going really quickly. And then it slows down as you go on. So that function is not quite the t cosine t, t sine of t that I originally had written."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate. But just to show where it might matter, I'll animate the same thing again, another function that draws the same curve, but it starts going really quickly. And then it slows down as you go on. So that function is not quite the t cosine t, t sine of t that I originally had written. And in fact, it would mean that, let's just erase these guys. When it starts slowly, you can interpret that as saying, okay, maybe, well, actually it started quickly, didn't it? So one would be really far off here."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So that function is not quite the t cosine t, t sine of t that I originally had written. And in fact, it would mean that, let's just erase these guys. When it starts slowly, you can interpret that as saying, okay, maybe, well, actually it started quickly, didn't it? So one would be really far off here. And then two, it could have zipped along here. And then three, still going really fast. And then maybe by the time you get to the end, it's just going very slowly, just kind of seven is here, eight is here, and it's hardly making any progress before it gets to 10."}, {"video_title": "Parametric curves   Multivariable calculus   Khan Academy.mp3", "Sentence": "So one would be really far off here. And then two, it could have zipped along here. And then three, still going really fast. And then maybe by the time you get to the end, it's just going very slowly, just kind of seven is here, eight is here, and it's hardly making any progress before it gets to 10. So you can have two different functions draw the same curve. And the fancy word here is parameterize. So functions will parameterize a curve if when you draw just in the output space, you get that curve."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And hopefully we'll be able to use that to understand or get a better intuition behind what exactly it means to take a derivative of a position vector valued function. So let's say my first parametrization, I have x of t is equal to t, and let's say that y of t is equal to t squared, and this is true for t is greater than or equal to 0 and less than or equal to 2. And if I want to write this as a position vector valued function, let me write this x1, let me call that y1, and let me write my position vector valued function, I could say r1, because I'm numbering them because I'm going to do a different version of this exact same curve with a slightly different parametrization. So r1 of t we could say is x1 of t times i, the unit vector i, so we'll just say t times i, plus, right, this is just x of t right here, or x1 of t, I'm numbering them because I'll later have an x2 of t, plus t squared times j. And if I wanted to graph this, I'm going to be very careful graphing it, because I really want to understand what the derivative means here. Let me get my draw, try to draw it roughly to scale. So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So r1 of t we could say is x1 of t times i, the unit vector i, so we'll just say t times i, plus, right, this is just x of t right here, or x1 of t, I'm numbering them because I'll later have an x2 of t, plus t squared times j. And if I wanted to graph this, I'm going to be very careful graphing it, because I really want to understand what the derivative means here. Let me get my draw, try to draw it roughly to scale. So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2. And so at t equals 0, both my x and y coordinates are at 0, or this is just going to be the 0 vector, so this is where we are at t equals 0. At t equals 1, this is going to be 1 times i, so we're going to be just like that, plus 1 times j, right, 1 squared is j, so we're going to be right there. And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2. And so at t equals 0, both my x and y coordinates are at 0, or this is just going to be the 0 vector, so this is where we are at t equals 0. At t equals 1, this is going to be 1 times i, so we're going to be just like that, plus 1 times j, right, 1 squared is j, so we're going to be right there. And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this. So this is what, just to make it clear what we're doing, that's r1 of 2, right, this is r1 of 0, this is r1 of 1, but the bottom line is the path looks like this, it's a parabola. So the path will look like that, just like that. Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this. So this is what, just to make it clear what we're doing, that's r1 of 2, right, this is r1 of 0, this is r1 of 1, but the bottom line is the path looks like this, it's a parabola. So the path will look like that, just like that. Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing. So it's going to be a parabola, well, let me get rid of that other point too, just because I didn't draw it exactly where it needs to be, it needs to be right there, and my parabola, or my part of my parabola is going to look something like that. Alright, good enough. So this is the first parametrization."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing. So it's going to be a parabola, well, let me get rid of that other point too, just because I didn't draw it exactly where it needs to be, it needs to be right there, and my parabola, or my part of my parabola is going to look something like that. Alright, good enough. So this is the first parametrization. Now I'm going to do this exact same curve, but I'm going to do it slightly differently. So let's say, I'll do it in different colors, so x2 of t, let's say it equals 2t, and y2 of t, let's say it's equal to 2t squared, or we could alternatively write that, that's the same thing as 4t squared, just raising both of these guys to the second power, that's equal to 4t squared, I could write it either way, and then let's say my second parametrization, and let's say instead of going from t equals 0 to 2, we're going to go from, t goes from 0 to 1, t goes from 0 to 1, but we're going to see, we're going to cover the exact same path, and our second vector, or position vector valued function, r2 of t is going to be equal to 2t times i, plus, I could say 2t squared, or 4t squared times j. And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here?"}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the first parametrization. Now I'm going to do this exact same curve, but I'm going to do it slightly differently. So let's say, I'll do it in different colors, so x2 of t, let's say it equals 2t, and y2 of t, let's say it's equal to 2t squared, or we could alternatively write that, that's the same thing as 4t squared, just raising both of these guys to the second power, that's equal to 4t squared, I could write it either way, and then let's say my second parametrization, and let's say instead of going from t equals 0 to 2, we're going to go from, t goes from 0 to 1, t goes from 0 to 1, but we're going to see, we're going to cover the exact same path, and our second vector, or position vector valued function, r2 of t is going to be equal to 2t times i, plus, I could say 2t squared, or 4t squared times j. And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here? 1 half times 2 is 1, and then we're going to get the point, 1 half squared is 1 fourth times 4 is 1, so when t is equal to 1 half, we're going to be at the point 1,1, and when t is equal to 1, we're going to be at the point 2,4. So notice the curve is exactly, the path we go is exactly the same, but before we even do the derivatives, these two paths are identical, I want to think about something, let's pretend that our parameter t really is time, that t is equal to time, and that tends to be the most common, that's why they call it t, it doesn't have to be time, but let's say it is time, so what's happening here? In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here? 1 half times 2 is 1, and then we're going to get the point, 1 half squared is 1 fourth times 4 is 1, so when t is equal to 1 half, we're going to be at the point 1,1, and when t is equal to 1, we're going to be at the point 2,4. So notice the curve is exactly, the path we go is exactly the same, but before we even do the derivatives, these two paths are identical, I want to think about something, let's pretend that our parameter t really is time, that t is equal to time, and that tends to be the most common, that's why they call it t, it doesn't have to be time, but let's say it is time, so what's happening here? In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so. In this situation, we have a dot moving along the same curve, but it's able to cover the same curve in only 1 second, in half a second it gets here, it took this guy 1 second to get here, in 1 second, this guy is all the way over here, this guy takes 2 seconds to go over here, so in this second parametrization, even though the path is the same, the curves are the same, the dot is faster, and I want you to keep that in mind when we think about the derivatives of both of these position vector valued functions. Just remember, the dot is moving faster, for every second it's getting further along the curve than here, that's why it only took them 1 second. Now let's look at the derivatives of both of these guys."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so. In this situation, we have a dot moving along the same curve, but it's able to cover the same curve in only 1 second, in half a second it gets here, it took this guy 1 second to get here, in 1 second, this guy is all the way over here, this guy takes 2 seconds to go over here, so in this second parametrization, even though the path is the same, the curves are the same, the dot is faster, and I want you to keep that in mind when we think about the derivatives of both of these position vector valued functions. Just remember, the dot is moving faster, for every second it's getting further along the curve than here, that's why it only took them 1 second. Now let's look at the derivatives of both of these guys. The derivative here, if I were to write r1' of t, let me do it in a different color, I already used the orange, let me do it in blue, r1' of t, so the derivative now is going to be, remember it's just the derivative of each of these times the unit vectors, so the derivative of t with respect to t, that's just 1, so it's 1 times i, so I'll just write 1i, plus, I didn't have to write the 1 there, plus the derivative of t squared with respect to t is 2t, plus 2tj, and let me take the derivative over here, r2' of t, the derivative of 2t with respect to t is 2, so 2i, plus the derivative of 4t squared is 8t, 2 times 4, it is 8t, just like that. Now, the question is, what do their respective derivative vectors look like at different points? So let's look at, I don't know, let's see how fast they're moving when time is equal to 1."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's look at the derivatives of both of these guys. The derivative here, if I were to write r1' of t, let me do it in a different color, I already used the orange, let me do it in blue, r1' of t, so the derivative now is going to be, remember it's just the derivative of each of these times the unit vectors, so the derivative of t with respect to t, that's just 1, so it's 1 times i, so I'll just write 1i, plus, I didn't have to write the 1 there, plus the derivative of t squared with respect to t is 2t, plus 2tj, and let me take the derivative over here, r2' of t, the derivative of 2t with respect to t is 2, so 2i, plus the derivative of 4t squared is 8t, 2 times 4, it is 8t, just like that. Now, the question is, what do their respective derivative vectors look like at different points? So let's look at, I don't know, let's see how fast they're moving when time is equal to 1. So let's take a specific point, this is just the general formula, but let's figure out what the derivative is at a specific point. So let's take r1 when time is equal to 1, and I want to take the specific point on the curve, not the specific point in time. So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's look at, I don't know, let's see how fast they're moving when time is equal to 1. So let's take a specific point, this is just the general formula, but let's figure out what the derivative is at a specific point. So let's take r1 when time is equal to 1, and I want to take the specific point on the curve, not the specific point in time. So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second. So r1 of 1 is equal to, I'm taking the derivative there, is equal to 1i, it's not dependent on t at all, so it's 1i plus 2 times 1j, so plus 2j. So at this point, the derivative of our position vector function is going to be 1i plus 2j, so we can draw it like this. So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second. So r1 of 1 is equal to, I'm taking the derivative there, is equal to 1i, it's not dependent on t at all, so it's 1i plus 2 times 1j, so plus 2j. So at this point, the derivative of our position vector function is going to be 1i plus 2j, so we can draw it like this. So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that. So our derivative right there, I'll do it in the same color that I wrote it in, in this green color, is going to look like this. It's going to look like that. And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that. So our derivative right there, I'll do it in the same color that I wrote it in, in this green color, is going to look like this. It's going to look like that. And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is. This right here, just to be clear, is r1 prime, it's a vector, so it's telling us the instantaneous change in our position vector with respect to t, or time, when time is equal to 1 second. That's this thing right here. Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is. This right here, just to be clear, is r1 prime, it's a vector, so it's telling us the instantaneous change in our position vector with respect to t, or time, when time is equal to 1 second. That's this thing right here. Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy. We already said it only takes him, he's here at time is equal to 1 half second. So let's take, let me do it in, I'll do it in the same color. So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy. We already said it only takes him, he's here at time is equal to 1 half second. So let's take, let me do it in, I'll do it in the same color. So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j. So what does this look like? The instantaneous derivative here, and this is the derivative, we have to be very clear. So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j. So what does this look like? The instantaneous derivative here, and this is the derivative, we have to be very clear. So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing. You get something that looks like that. I didn't draw it as neatly as I would like to, but let's notice something. Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing. You get something that looks like that. I didn't draw it as neatly as I would like to, but let's notice something. Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed. The length is equal to the speed. If you imagine time, t being time, and these parametrizations are representing a dot moving along these curves, so in this case, the particle only takes a second to go there, so at this point in its path, it's moving much faster than this particle is. If you think about it, this vector right here, if you imagine this is a position vector, this is velocity."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Both of these vectors are going in the exact same direction, they're both tangential to the path, to our curve, but this vector is going, its length, its magnitude, is much larger than this vector's magnitude, and that makes sense because I hinted at it when we first talked about these vector-valued position functions and their derivatives, is that the length, you can kind of view it as the speed. The length is equal to the speed. If you imagine time, t being time, and these parametrizations are representing a dot moving along these curves, so in this case, the particle only takes a second to go there, so at this point in its path, it's moving much faster than this particle is. If you think about it, this vector right here, if you imagine this is a position vector, this is velocity. This right here is velocity. Velocity is speed plus direction. Speed is just how fast are you going, velocity is how fast you're going in what direction."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you think about it, this vector right here, if you imagine this is a position vector, this is velocity. This right here is velocity. Velocity is speed plus direction. Speed is just how fast are you going, velocity is how fast you're going in what direction. I'm going this fast, and you could calculate it using Pythagorean theorem, but I just want to give you the intuition right here. I'm going that fast in this direction. Here I'm going this fast."}, {"video_title": "Vector valued function derivative example   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Speed is just how fast are you going, velocity is how fast you're going in what direction. I'm going this fast, and you could calculate it using Pythagorean theorem, but I just want to give you the intuition right here. I'm going that fast in this direction. Here I'm going this fast. I'm going even faster. That's my magnitude, but I'm still going in the same direction. Hopefully you have a gut feeling now what the derivative of these position vectors really are."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But what I want to do in this video is show you how to set the boundaries when the figure is a little bit more complicated. And if we have time, we'll try to do it where we change the order of integration. So let's say I have the surface, let me just make something up, 2x plus 3z plus y is equal to 6. Let's draw that surface. Looks something like this. This will be my x-axis. It's going to be my z-axis."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's draw that surface. Looks something like this. This will be my x-axis. It's going to be my z-axis. That's going to be my y-axis. Draw them out. x, y, and z."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be my z-axis. That's going to be my y-axis. Draw them out. x, y, and z. And I care about the surface in the kind of positive octant, right, because when you're dealing with three dimensions, we have instead of four quadrants, we have eight octants, but we want the octant where all x, y, and z is positive, which is the one I drew here. So let's see. Let me draw some."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x, y, and z. And I care about the surface in the kind of positive octant, right, because when you're dealing with three dimensions, we have instead of four quadrants, we have eight octants, but we want the octant where all x, y, and z is positive, which is the one I drew here. So let's see. Let me draw some. What is the x-intercept? When y and z are 0, so we're right here, that's the x-intercept. 2x is equal to 6, so x is equal to 3."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw some. What is the x-intercept? When y and z are 0, so we're right here, that's the x-intercept. 2x is equal to 6, so x is equal to 3. So 1, 2, 3. So that's the x-intercept. The y-intercept, when x and z are 0, we're on the y-axis, so y will be equal to 6."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "2x is equal to 6, so x is equal to 3. So 1, 2, 3. So that's the x-intercept. The y-intercept, when x and z are 0, we're on the y-axis, so y will be equal to 6. So we have 1, 2, 3, 4, 5, 6. So y-intercept. And then finally, the z-intercept."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The y-intercept, when x and z are 0, we're on the y-axis, so y will be equal to 6. So we have 1, 2, 3, 4, 5, 6. So y-intercept. And then finally, the z-intercept. When x and y are 0, we're on the z-axis. 3z will be equal to 6. So z is equal to 1, 2."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then finally, the z-intercept. When x and y are 0, we're on the z-axis. 3z will be equal to 6. So z is equal to 1, 2. So the figure that I care about will look something like this. It'll be this inclined surface. It will look something like that."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So z is equal to 1, 2. So the figure that I care about will look something like this. It'll be this inclined surface. It will look something like that. And this positive octant. So this is the surface defined by this function. Let's say that I care about the volume."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It will look something like that. And this positive octant. So this is the surface defined by this function. Let's say that I care about the volume. And I'm going to make it a little bit more complicated. We could say, oh, well, what's the volume between the surface and the xy-plane? But I'm going to make it a little bit more complicated."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that I care about the volume. And I'm going to make it a little bit more complicated. We could say, oh, well, what's the volume between the surface and the xy-plane? But I'm going to make it a little bit more complicated. Let's say the volume above this surface and the surface z is equal to 2. So essentially, the surface z, so the volume we care about is going to look something like this. Let me see if I can pull off drawing this."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I'm going to make it a little bit more complicated. Let's say the volume above this surface and the surface z is equal to 2. So essentially, the surface z, so the volume we care about is going to look something like this. Let me see if I can pull off drawing this. So if we go up 2 here. So if we go up 2 here. And then we have, let me draw the top in a different color."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me see if I can pull off drawing this. So if we go up 2 here. So if we go up 2 here. And then we have, let me draw the top in a different color. Let me draw the top in green. So this is along the zy-plane. And then the other edge is going to look something like this."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we have, let me draw the top in a different color. Let me draw the top in green. So this is along the zy-plane. And then the other edge is going to look something like this. Let me make sure I can draw it. This is the hardest part. We'll go up 2 here."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the other edge is going to look something like this. Let me make sure I can draw it. This is the hardest part. We'll go up 2 here. And then this is along the zx-plane. And we'd have another line connecting these two. So this green triangle, this is part of the plane z is equal to 2."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll go up 2 here. And then this is along the zx-plane. And we'd have another line connecting these two. So this green triangle, this is part of the plane z is equal to 2. But what we can do is, what the volume we care about is the volume between this top green plane and this tilted plane defined by 2x plus 3z plus y is equal to 6. So this area in between. Let me see if I can make it a little bit clearer."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this green triangle, this is part of the plane z is equal to 2. But what we can do is, what the volume we care about is the volume between this top green plane and this tilted plane defined by 2x plus 3z plus y is equal to 6. So this area in between. Let me see if I can make it a little bit clearer. Because the visualization, as I say, is often the hardest part. So we'd have kind of a front wall here. And then the back wall would be this wall back here."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me see if I can make it a little bit clearer. Because the visualization, as I say, is often the hardest part. So we'd have kind of a front wall here. And then the back wall would be this wall back here. And then there'd be another wall here. And then the base of it, the base, I'll do in magenta, will be this plane. So the base is that plane."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the back wall would be this wall back here. And then there'd be another wall here. And then the base of it, the base, I'll do in magenta, will be this plane. So the base is that plane. That's the bottom part. Anyway, I don't know if I should have made it that messy, because we're going to have to draw dv's and d-volumes on it. But anyway, let's try our best."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the base is that plane. That's the bottom part. Anyway, I don't know if I should have made it that messy, because we're going to have to draw dv's and d-volumes on it. But anyway, let's try our best. So if we're going to figure out the volume, and actually since we're doing a triple integral and we want to show that we have to use a triple integral, instead of just doing a volume, let's do the mass of something of variable density. So let's say the density in this volume that we care about, the density function, it's a function of x, y, and z. It can be anything."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But anyway, let's try our best. So if we're going to figure out the volume, and actually since we're doing a triple integral and we want to show that we have to use a triple integral, instead of just doing a volume, let's do the mass of something of variable density. So let's say the density in this volume that we care about, the density function, it's a function of x, y, and z. It can be anything. That's not the point of what I'm trying to teach here. But I'll just define something. Let's say it's x squared, y, z."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It can be anything. That's not the point of what I'm trying to teach here. But I'll just define something. Let's say it's x squared, y, z. Our focus here is really just to set up the integrals. So the first thing I like to do is I visualize. What we're going to do is we're going to set up a little cube in the volume under consideration."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say it's x squared, y, z. Our focus here is really just to set up the integrals. So the first thing I like to do is I visualize. What we're going to do is we're going to set up a little cube in the volume under consideration. So if I had a, let me do it in a bold color so that you can see it, so if I have a cube, maybe I'll do it in brown. It's not as bold, but it's different enough from the other colors. So if I had a little cube here in the volume under consideration, that's a little cube."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What we're going to do is we're going to set up a little cube in the volume under consideration. So if I had a, let me do it in a bold color so that you can see it, so if I have a cube, maybe I'll do it in brown. It's not as bold, but it's different enough from the other colors. So if I had a little cube here in the volume under consideration, that's a little cube. You could consider that dv, the volume of that cube, is kind of a volume differential. And that is equal to dx. No, sorry, this is dy."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I had a little cube here in the volume under consideration, that's a little cube. You could consider that dv, the volume of that cube, is kind of a volume differential. 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 length, dy times dx, dx times dz. That's the volume of that little cube."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "No, sorry, this is dy. Let me do this in yellow or green even better. So dy, which is this length, 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 dm. The mass differential is going to be equal to that times that."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 dm. The mass differential is going to be equal to that times that. So x squared yz times this. dy dx 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The mass differential is going to be equal to that times that. So x squared yz times this. dy dx 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 way the last couple triple integrals we did, we integrated with respect to z first. So let's do that. So we're going to integrate with respect to z first."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's do it traditionally. The way the last couple triple integrals we did, we integrated with respect to z first. 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 them up, we're going to sum up all of the cubes in the z axis, so going up and down first. 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 them up, we're going to sum up all of the cubes in the z axis, so going up and down first. 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. So what is the bottom of the column, the bottom bound? Well, it's the surface. It's the surface defined right here."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when you sum up, up and down, these cubes are going to turn to columns. So what is the bottom of the column, the bottom bound? Well, it'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?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 to 2 minus 2 thirds x minus y over 3. This is the same thing as that, but when we're talking about z, explicitly defining a z, this is how we get it. We just algebraically manipulate it."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 to 2 minus 2 thirds x minus y over 3. This is the same thing as that, but when we're talking about z, explicitly defining a z, this is how we get it. We just algebraically manipulate it. So the bottom boundary, and you visualize it, right? The bottom of these columns, because we're going to go up and down, we're going to add up all the columns in an up and down direction, right? You can imagine summing them."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just algebraically manipulate it. So the bottom boundary, and you visualize it, right? The bottom of these columns, because we're going to go up and down, we're going to add up all the columns in an 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 thirds x minus y over 3. And then what's the upper bound?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can imagine summing them. The bottom boundary is going to be this surface. z is equal to 2 minus 2 thirds x 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 that green plane was? Well, it's z is equal to 2."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 that green plane was? Well, it's z is equal to 2. And that's this plane, this surface right here. z is equal to 2. And of course, what is the volume of that column?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, it's z is equal to 2. And 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 y z times the volume differential, but we're integrating with respect to z first. Let me write dz there. And let's say we want to integrate with respect to x for next."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And of course, what is the volume of that column? Well, it's going to be the density function, x squared y z times the volume differential, but we're integrating with respect to z first. Let me write dz there. And let's say we want to integrate with respect to x for next. In the last couple videos, I integrated with respect to y next. Well, let's do x, just to show you it really doesn't matter. So we're going to integrate with respect to x."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say we want to integrate with respect to x for next. In the last couple videos, I integrated with respect to y next. Well, 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, where the top boundary is that plane. Let's see if I can draw it decently."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, where 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. And now we want to integrate with respect to x."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see if I can draw it decently. The top boundary is that plane. The bottom boundary is this surface. And 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 x."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And 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 x. 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've already integrated with respect to z."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, this surface is defined all the way to x. 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've 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?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've 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? So I will do that, because that actually does help simplify things. So if we twist it, so if we take this y and flip it out like that, x like that, we'll get it in the traditional way that we learned when we first learned algebra. The xy axis."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what does that look like? So I will do that, because that actually does help simplify things. So if we twist it, so if we take this y and flip it out like that, x like that, we'll get it in 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The xy axis. So this is x, this is y. And this point is what? Or this point. What is that? That's x is equal to 3. So it's 1, 2, 3."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or this point. What is that? 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And so the xy axis, kind of the domain, you could view it that way, looks something like that. And 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 look like this. Where the base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So 1, 2, 3, 4, 5, 6. And so the xy axis, kind of the domain, you could view it that way, looks something like that. And 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 look like this. Where the base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down. 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?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Where the base of the column, where the column's going to pop out of your screen in the z direction, but the base of each column is going to be dx like that, and then dy up and down. 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."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And what's our top bound? I realize I'm already pushing it."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And what's our top bound? I realize I'm already pushing it. Well, our top bound is this relation, but it has to be in terms of x. So what is this relation? So you could view it as kind of saying, well, if z is equal to 0, what is this line?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I realize I'm already pushing it. Well, our top bound is this relation, but it has to be in terms of x. So what is this relation? 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 if z is equal to 0, we have 2x plus y is equal to 6. We want the relationship in terms of x."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 if 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, or x is equal to 3 minus y over 2. x is equal to 3 minus y over 2. And then finally, we're going to integrate with respect to y, and this is the easy part. So we've figured out, we've integrated up and down to get a column."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We want the relationship in terms of x. So we get 2x is equal to 6 minus y, or x is equal to 3 minus y over 2. x is equal to 3 minus y over 2. And then finally, we're going to integrate with respect to y, and this is the easy part. So we've figured out, we've integrated up and down to get a column. These are the bases of the column, and so we've integrated in the x direction. Now we just have to go up and down with respect to y, or in the xy plane with respect to y. So what is the y bottom boundary?"}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we've figured out, we've integrated up and down to get a column. These are the bases of the column, and so we've integrated in the x direction. Now we just have to go up and down with respect to y, or in the xy plane with respect to y. So what is the y bottom boundary? Well, it's 0. y is equal to 0. And the top boundary is y is equal to 6. And there you have it."}, {"video_title": "Triple integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what is the y bottom boundary? Well, it's 0. y is equal to 0. And the top boundary is y is equal to 6. And there you have it. We have set up the integral, and now it's just a matter of chugging through it mechanically. But I've run out of time, and I don't want this video to get rejected. So I'll leave you there."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've taken partial derivatives of non-vector valued functions before, where we only vary one of the variables. We only take it with respect to one variable. You hold the other one constant. We're going to do the exact same thing here. And we've taken regular derivatives of vector valued functions in the past, and those just ended up being the regular derivative of each of the terms. And we're going to see it's going to be the same thing here with the partial derivative. So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to do the exact same thing here. And we've taken regular derivatives of vector valued functions in the past, and those just ended up being the regular derivative of each of the terms. And we're going to see it's going to be the same thing here with the partial derivative. So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result. So I'm going to define it as being equal to the limit as delta s approaches 0 of r of s plus delta s. We're only finding the limit with respect to a change in s comma t. We're holding t, you can imagine, constant for a given t, minus r of s and t, all of that over delta s. Now, if you do a little bit of algebra here, you literally, r of s plus delta s comma t, that's the same thing as x of s plus delta st, i, plus y of s plus delta st, j, plus z, all of that minus this thing. If you do a little bit of algebra with that, and if you don't believe me, try it out, this is going to be equal to the limit of delta s approaching 0, and I'm going to write it small because it can take up a lot of space, of x of s plus delta s comma t, minus x of s and t. I think you know where I'm going, this is all a little bit monotonous to write it all out, but never hurts. Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's define the partial derivative of r with respect to s. And everything I do with respect to s, you could just swap it with t, and you're going to get the same exact result. So I'm going to define it as being equal to the limit as delta s approaches 0 of r of s plus delta s. We're only finding the limit with respect to a change in s comma t. We're holding t, you can imagine, constant for a given t, minus r of s and t, all of that over delta s. Now, if you do a little bit of algebra here, you literally, r of s plus delta s comma t, that's the same thing as x of s plus delta st, i, plus y of s plus delta st, j, plus z, all of that minus this thing. If you do a little bit of algebra with that, and if you don't believe me, try it out, this is going to be equal to the limit of delta s approaching 0, and I'm going to write it small because it can take up a lot of space, of x of s plus delta s comma t, minus x of s and t. I think you know where I'm going, this is all a little bit monotonous to write it all out, but never hurts. Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y. This limit as delta approaches 0 applies to every term I'm writing out here. y of s plus delta s comma t, minus y of s comma t, all of that over delta s times j. And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times s, or divided by delta s, times i, and then I'll do it in different colors, so it's less monotonous, plus y. This limit as delta approaches 0 applies to every term I'm writing out here. y of s plus delta s comma t, minus y of s comma t, all of that over delta s times j. And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition. If you literally just put s plus delta s in place for s, you evaluate all this, do a little algebra, you're going to get this exact same thing. And this hopefully pops out at you as, gee, we're just taking the partial derivative of each of these functions with respect to s. And these functions right here, this x of s and t, this is a non-vector valued function. This y, this is also a non-vector valued function."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then finally, plus z of s plus delta s comma t, minus z of s and t, all of that over delta s times the z unit vector, k. And this all comes out of this definition. If you literally just put s plus delta s in place for s, you evaluate all this, do a little algebra, you're going to get this exact same thing. And this hopefully pops out at you as, gee, we're just taking the partial derivative of each of these functions with respect to s. And these functions right here, this x of s and t, this is a non-vector valued function. This y, this is also a non-vector valued function. z is also a non-vector valued function. When you put them all together, it becomes a vector valued function, because we're multiplying the first one times a vector, the second one times another vector, the third one times another vector. But independently, these functions are non-vector valued."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This y, this is also a non-vector valued function. z is also a non-vector valued function. When you put them all together, it becomes a vector valued function, because we're multiplying the first one times a vector, the second one times another vector, the third one times another vector. But independently, these functions are non-vector valued. So this is just the definition of the regular partial derivatives, where you're taking the limit as delta s approaches 0 in each of these cases. So this is the exact same thing. This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But independently, these functions are non-vector valued. So this is just the definition of the regular partial derivatives, where you're taking the limit as delta s approaches 0 in each of these cases. So this is the exact same thing. This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy. But it's going to come out, the whole reason I'm even doing this video is it's going to give us some good tools or in our toolkit for the videos that I'm about to do on surface integrals. So I'm going to do one thing here that's a little pseudo-mathy. And that's really because differentials are these things that are very hard to define rigorously."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to, this is the exact same thing as the partial derivative of x with respect to s times i, plus the partial derivative of y with respect to s times j, plus the partial derivative of z with respect to s times k. I'm going to do one more thing here, and this is pseudo-mathy. But it's going to come out, the whole reason I'm even doing this video is it's going to give us some good tools or in our toolkit for the videos that I'm about to do on surface integrals. So I'm going to do one thing here that's a little pseudo-mathy. And that's really because differentials are these things that are very hard to define rigorously. But I think it'll give you the intuition of what's going on. So this thing right here, I'm going to say this is also equal to, and you're not going to see this in any math textbook. And hardcore mathematicians are going to kind of cringe when they see me do this."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that's really because differentials are these things that are very hard to define rigorously. But I think it'll give you the intuition of what's going on. So this thing right here, I'm going to say this is also equal to, and you're not going to see this in any math textbook. And hardcore mathematicians are going to kind of cringe when they see me do this. But I like to do it because I think it'll give you the intuition on what's going on when we take our surface integrals. So I'm going to say that this whole thing right here, that that is equal to r of s plus the differential of s, a super small change in s, t minus r of s and t, all of that over that same smooth, super small change in s. So hopefully you understand at least why I view things this way. When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And hardcore mathematicians are going to kind of cringe when they see me do this. But I like to do it because I think it'll give you the intuition on what's going on when we take our surface integrals. So I'm going to say that this whole thing right here, that that is equal to r of s plus the differential of s, a super small change in s, t minus r of s and t, all of that over that same smooth, super small change in s. So hopefully you understand at least why I view things this way. When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small. And in my head, that's how I imagine differentials. When someone writes the derivative of y with respect to x, and let's say that they say that is 2. And we've done a little bit of math with differentials before."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When I take the limit as delta s approaches 0, these delta s's are going to get super duper duper small. And in my head, that's how I imagine differentials. When someone writes the derivative of y with respect to x, and let's say that they say that is 2. And we've done a little bit of math with differentials before. You can imagine multiplying both sides by dx, and you could get dy is equal to 2 dx. We've done this throughout calculus. The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we've done a little bit of math with differentials before. You can imagine multiplying both sides by dx, and you could get dy is equal to 2 dx. We've done this throughout calculus. The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x. So if you have a super small change in x, your change in y is going to be still super small, but it's going to be 2 times that. I guess that's the best way to view it. But in general, I view differentials as super small changes in a variable."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The way I imagine it is super small change in y. Infinitely small change in y is equal to 2 times, you can imagine an equally small change in x. So if you have a super small change in x, your change in y is going to be still super small, but it's going to be 2 times that. I guess that's the best way to view it. But in general, I view differentials as super small changes in a variable. So with that out of the way, and me explaining to you that many mathematicians would cringe at what I just wrote, hopefully this gives you a little, you know, this isn't like some crazy thing I did. I'm just saying, oh, delta s is delta s approaches 0. I kind of imagine that as ds."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But in general, I view differentials as super small changes in a variable. So with that out of the way, and me explaining to you that many mathematicians would cringe at what I just wrote, hopefully this gives you a little, you know, this isn't like some crazy thing I did. I'm just saying, oh, delta s is delta s approaches 0. I kind of imagine that as ds. And the whole reason I did that is if you take this side and that side and multiply both sides times this differential ds, then what happens? The left hand side, you get the partial of r with respect to s is equal to this, oh, let me, times ds. I'll do ds in maybe pink."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I kind of imagine that as ds. And the whole reason I did that is if you take this side and that side and multiply both sides times this differential ds, then what happens? The left hand side, you get the partial of r with respect to s is equal to this, oh, let me, times ds. I'll do ds in maybe pink. Times ds, this is just a regular differential, super small change in s. This was kind of a partial with respect to s. That's going to be equal to, well, if you multiply this side of the equation times ds, this guy's going to disappear. So it's going to be r of s plus our super small change in s. t minus r of s and t. And let me put a little square around this. This is going to be valuable for us in the next video."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do ds in maybe pink. Times ds, this is just a regular differential, super small change in s. This was kind of a partial with respect to s. That's going to be equal to, well, if you multiply this side of the equation times ds, this guy's going to disappear. So it's going to be r of s plus our super small change in s. t minus r of s and t. And let me put a little square around this. This is going to be valuable for us in the next video. We're going to actually think about what this means and how to visualize this on a surface. As you can imagine, this is a vector right here, right? You have two vector valued functions."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be valuable for us in the next video. We're going to actually think about what this means and how to visualize this on a surface. As you can imagine, this is a vector right here, right? You have two vector valued functions. You're taking the difference. And we're going to visualize it in the next video. It's going to really help us with surface integrals."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You have two vector valued functions. You're taking the difference. And we're going to visualize it in the next video. It's going to really help us with surface integrals. By the same exact logic, we can do everything we did here with s, we can do it with t as well. So we can define the partial. I'll draw a little."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to really help us with surface integrals. By the same exact logic, we can do everything we did here with s, we can do it with t as well. So we can define the partial. I'll draw a little. I can define the partial of r with respect. Let me do it in a different color, completely different color. It's orange."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll draw a little. I can define the partial of r with respect. Let me do it in a different color, completely different color. It's orange. The partial of r with respect to t. The definition is just right here. The limit as delta t approaches 0 of r of s t plus delta t minus r of s and t. In this situation, we're holding the s. You can imagine it constant. We're finding its change in t. All of that over delta t. And the same thing falls out."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's orange. The partial of r with respect to t. The definition is just right here. The limit as delta t approaches 0 of r of s t plus delta t minus r of s and t. In this situation, we're holding the s. You can imagine it constant. We're finding its change in t. All of that over delta t. And the same thing falls out. This is equal to the partial of x with respect to t, i, plus y with respect to t, j, plus z with respect to t, k. Same exact thing. You just kind of swap the s's and the t's. And by that same logic, you'd have the same result but in terms of t. If you do this pseudo mathy thing that I did up here, then you would get the partial of r with respect to t times a super small change in t, dt, or t differential, you could imagine, is equal to r of s t plus dt minus r of s and t. So let's box these two guys away."}, {"video_title": "Partial derivatives of vector-valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're finding its change in t. All of that over delta t. And the same thing falls out. This is equal to the partial of x with respect to t, i, plus y with respect to t, j, plus z with respect to t, k. Same exact thing. You just kind of swap the s's and the t's. And by that same logic, you'd have the same result but in terms of t. If you do this pseudo mathy thing that I did up here, then you would get the partial of r with respect to t times a super small change in t, dt, or t differential, you could imagine, is equal to r of s t plus dt minus r of s and t. So let's box these two guys away. And in the next video, we're going to actually visualize what these mean. And sometimes when you kind of do a bunch of silly math like this, you're always like, hey, whatever it's all about. Remember, all I did is I said, what does it mean to take the derivative of this with respect to s or t, played around with it a little bit."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In all of the double integrals we've done so far, the boundaries on x and y were fixed. Now we'll see what happens when the boundaries on x and y are variable. So let's say I have the same surface. And I'm not going to draw it the way it looks. I'll just kind of draw it figuratively. But the problem we're actually going to do is z. And this is the exact same one we've been doing all along."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'm not going to draw it the way it looks. I'll just kind of draw it figuratively. But the problem we're actually going to do is z. And this is the exact same one we've been doing all along. But the point of here isn't to show you how to integrate. The point of here is to show you how to visualize and think about these problems. And frankly, in double integral problems, the hardest part is figuring out the boundaries."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is the exact same one we've been doing all along. But the point of here isn't to show you how to integrate. The point of here is to show you how to visualize and think about these problems. And frankly, in double integral problems, the hardest part is figuring out the boundaries. Once you do that, the integration is pretty straightforward. It's really not any harder than single variable integration. So let's say that's our surface."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And frankly, in double integral problems, the hardest part is figuring out the boundaries. Once you do that, the integration is pretty straightforward. It's really not any harder than single variable integration. So let's say that's our surface. z is equal to xy squared. And let me draw the axes again. So that's my x-axis."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that's our surface. z is equal to xy squared. And let me draw the axes again. So that's my x-axis. That's my z-axis. That's my y-axis, x, y, and z. And you saw what this graph looked like several videos ago."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's my x-axis. That's my z-axis. That's my y-axis, x, y, and z. And you saw what this graph looked like several videos ago. I took out the whole grapher and we rotated and things. I'm not going to draw the graph the way it looks. I'm just going to draw it fairly abstractly as just an abstract surface."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you saw what this graph looked like several videos ago. I took out the whole grapher and we rotated and things. I'm not going to draw the graph the way it looks. I'm just going to draw it fairly abstractly as just an abstract surface. Because the point here is really to figure out the boundaries of integration. Before I actually even draw the surface, I'm going to draw the boundaries. So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just going to draw it fairly abstractly as just an abstract surface. Because the point here is really to figure out the boundaries of integration. Before I actually even draw the surface, I'm going to draw the boundaries. So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain. Now let's do something else. Let's figure out, let's say that x goes from 0 to 1. So x goes from 0 to 1."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the first time we did this problem, we said, OK, x goes from 0 to 2, y goes from 0 to 1, and then we figured out the volume above that bounded domain. Now let's do something else. Let's figure out, let's say that x goes from 0 to 1. So x goes from 0 to 1. And let's say that the volume that we want to figure out under the surface, it's not from a fixed y to an upper bound y. Or the kind of, well, I'll show you. It's actually a curve."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So x goes from 0 to 1. And let's say that the volume that we want to figure out under the surface, it's not from a fixed y to an upper bound y. Or the kind of, well, I'll show you. It's actually a curve. So this is all on the xy-plane, everything I'm drawing here. And this curve, we could view it two ways. We could say y is a function of x, y is equal to x squared."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's actually a curve. So this is all on the xy-plane, everything I'm drawing here. And this curve, we could view it two ways. We could say y is a function of x, y is equal to x squared. Or we could write x is equal to square root of y. We don't have to write plus or minus or anything like that, because we're in the first quadrant. So this is the area above which we want to figure out the volume."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could say y is a function of x, y is equal to x squared. Or we could write x is equal to square root of y. We don't have to write plus or minus or anything like that, because we're in the first quadrant. So this is the area above which we want to figure out the volume. Let me, yeah, it doesn't hurt to color it in, just so we can really hone in on what we care about. So that's the area above which we want to figure out the volume. Or you could kind of say that's our bounded domain."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the area above which we want to figure out the volume. Let me, yeah, it doesn't hurt to color it in, just so we can really hone in on what we care about. So that's the area above which we want to figure out the volume. Or you could kind of say that's our bounded domain. And so x goes from 0 to 1, and then this point is going to be what? That point's going to be 1 comma 1, right? 1 is equal to 1 squared, 1 is equal to square root of 1."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or you could kind of say that's our bounded domain. And so x goes from 0 to 1, and then this point is going to be what? That point's going to be 1 comma 1, right? 1 is equal to 1 squared, 1 is equal to square root of 1. So this point is y is equal to 1. That's y is equal to 1. And then I'm not going to draw this surface exactly."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 is equal to 1 squared, 1 is equal to square root of 1. So this point is y is equal to 1. That's y is equal to 1. And then I'm not going to draw this surface exactly. I'm just trying to give you a sense of what the volume of the figure we're trying to calculate is. So the top of, if this is just some arbitrary surface, let me do it in a different color. So this is the top."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then I'm not going to draw this surface exactly. I'm just trying to give you a sense of what the volume of the figure we're trying to calculate is. So the top of, if this is just some arbitrary surface, let me do it in a different color. So this is the top. This line is going vertical in the z direction. Actually, I could draw it like this, like it's a curve. And then this curve back here is going to be like a wall."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the top. This line is going vertical in the z direction. Actually, I could draw it like this, like it's a curve. And then this curve back here is going to be like a wall. That's going to be like a wall. And maybe I'll paint this side of the wall, just so you can see what it kind of looks like. Trying my best."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then this curve back here is going to be like a wall. That's going to be like a wall. And maybe I'll paint this side of the wall, just so you can see what it kind of looks like. Trying my best. I think you get an idea. Let me make it a little darker. This is actually more of an exercise in art than in math, in many ways."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Trying my best. I think you get an idea. Let me make it a little darker. This is actually more of an exercise in art than in math, in many ways. You get the idea. And then the boundary here is like this. And this top isn't flat."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is actually more of an exercise in art than in math, in many ways. You get the idea. And then the boundary here is like this. And this top isn't flat. It could be a curved surface. I do it a little like that, but it's a curved surface. And we know, in the example we're about to do, that this surface right here, z is equal to x squared."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this top isn't flat. It could be a curved surface. I do it a little like that, but it's a curved surface. And we know, in the example we're about to do, that this surface right here, z is equal to x squared. So we want to figure out the volume under this. So how do we do it? Well, let's think about it."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we know, in the example we're about to do, that this surface right here, z is equal to x squared. So we want to figure out the volume under this. So how do we do it? Well, let's think about it. We could actually use the intuition that I just gave you. We're essentially just going to take a dA, which is a little small square down here. And that little area, that's the same thing as a dx."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, let's think about it. We could actually use the intuition that I just gave you. We're essentially just going to take a dA, which is a little small square down here. And that little area, that's the same thing as a dx. Let me use a darker color. As a dx times a dy. And then we just have to multiply it times f of xy, which is this, for each area."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that little area, that's the same thing as a dx. Let me use a darker color. As a dx times a dy. And then we just have to multiply it times f of xy, which is this, for each area. And then sum them all up. And then we could take a sum in the x direction first or the y direction first. Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we just have to multiply it times f of xy, which is this, for each area. And then sum them all up. And then we could take a sum in the x direction first or the y direction first. Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane. So let me rotate up like that. I'm just going to draw our xy plane, because that's what matters. Because the hard part here is just figuring out our bounds of integration."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now before doing that, just to make sure that you have the intuition, because the boundaries are the hard part, let me just draw our xy plane. So let me rotate up like that. I'm just going to draw our xy plane, because that's what matters. Because the hard part here is just figuring out our bounds of integration. So the curve is just y is equal to x squared. So it'll look something like that. y is equal to x squared."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because the hard part here is just figuring out our bounds of integration. So the curve is just y is equal to x squared. So it'll look something like that. y is equal to x squared. This is the point y is equal to 1. This is y-axis. This is the x-axis."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "y is equal to x squared. This is the point y is equal to 1. This is y-axis. This is the x-axis. This is the point x is equal to 1. x is equal to 1. There you go. That's not an x, that's a 1."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the x-axis. This is the point x is equal to 1. x is equal to 1. There you go. That's not an x, that's a 1. This is the x-axis. Anyway, so we want to figure out how do we sum up this dx times dy, or this da, along this domain. So let's draw it."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's not an x, that's a 1. This is the x-axis. Anyway, so we want to figure out how do we sum up this dx times dy, or this da, along this domain. So let's draw it. Let's visually draw it. And it doesn't hurt to do this when you actually have to do the problem, because this frankly is the hard part. A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's draw it. Let's visually draw it. And it doesn't hurt to do this when you actually have to do the problem, because this frankly is the hard part. A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1. So this area here is the same thing as this area here. So its base is dx, and its height is dy. And then you can imagine that we're looking at this thing from above."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "A lot of calculus teachers will just have you set up the integral and then say, OK, well the rest is easy, or the rest is calc 1. So this area here is the same thing as this area here. So its base is dx, and its height is dy. And then you can imagine that we're looking at this thing from above. So the surface is up here someplace, 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 the volume above this column, first of all, is this area times dx dy."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you can imagine that we're looking at this thing from above. So the surface is up here someplace, 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 the volume above this column, first of all, is this area times dx dy. 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. So it's going to be xy squared times dx dy."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we want to sum up the volume above this column, first of all, is this area times dx dy. 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. So it's going to be 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 to sum in the x direction first. So we want to sum that dx, sum one here, sum here, et cetera, et cetera."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be 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 to sum in the x direction first. So we want to sum that dx, sum one here, sum here, et cetera, et cetera. 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?"}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we want to sum that dx, sum one here, sum here, et cetera, et cetera. 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 as a function of y, right?"}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 as a function of y, right? 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. x is equal to the square root of y."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what is this curve if we were to write x as a function of y, right? 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. x is equal to the square root of y. That's interesting. I think this is 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's easy."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Our lower bound is x is equal to the square root of y. x is equal to the square root of y. That's interesting. I think this is 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's easy. The upper bound is x is equal to 1. The upper bound is x is equal to 1. But the lower bound is x is equal to the square root of y, right?"}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But it makes sense, because for this row that we're adding up right here, the upper bound's easy. The upper bound is x is equal to 1. The upper bound is x is equal to 1. But the lower bound is x is equal to the square root of y, right? Because as you go back, like, oh, I bump into the curve. And what's the curve? Well, the curve is x is equal to the square root of y, because we don't know which y we picked."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But the lower bound is x is equal to the square root of y, right? Because as you go back, like, oh, I bump into the curve. And what's 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, right?"}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, right? And remember, there's a whole volume above what I'm drawing right here, right? 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."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we want to add up the dy's, right? And remember, there's a whole volume above what I'm drawing right here, right? 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."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 up these rectangles? Well, the lower bound is y is equal to 0, right? So we're going to go from y is equal to 0 to what?"}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "My dx's and dy's look too similar. So now, what is the lower bound on the y-axis, if I'm summing up these rectangles? Well, the lower bound is y is equal to 0, right? So we're going to go from y is equal to 0 to what? What is the upper bound? To y is equal to 1. And there you have it."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to go from y is equal to 0 to what? What is the upper bound? To y is equal to 1. And there you have it. Let me rewrite that integral. So the double integral is going to be from x is equal to the square root of y to x is equal to 1, xy squared, dx, dy. And then the y bounds, y goes from 0 to y to 1."}, {"video_title": "Double integrals 5   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And there you have it. Let me rewrite that integral. So the double integral is going to be from x is equal to the square root of y to x is equal to 1, xy squared, dx, dy. And then the y bounds, y goes from 0 to y to 1. I've just realized I've run out of time. In the next video, we'll evaluate this, and then we'll do it in the other order. See you soon."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just trying to draw a bit of an arbitrary path. And let's say we go in a counterclockwise direction, like that, along our path. And we could call this path, so we're in a counterclockwise direction, we could call that path C. And let's say we also have a vector field. And our vector field is going to be a little unusual, I'll call it P. P of xy, it only has an i component, or all of its vectors are only multiples of the i unit vector. So it's P, capital P of xy, times the unit vector i. There is no j component. So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And our vector field is going to be a little unusual, I'll call it P. P of xy, it only has an i component, or all of its vectors are only multiples of the i unit vector. So it's P, capital P of xy, times the unit vector i. There is no j component. So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector. Or they could be negative multiples, so they could also go in that direction. But they don't go diagonal, or they don't go up, they all go left to right, or right to left. That's what this vector field would look like."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you had to visualize this vector field, all of the vectors only go, they're all multiples of the i unit vector. Or they could be negative multiples, so they could also go in that direction. But they don't go diagonal, or they don't go up, they all go left to right, or right to left. That's what this vector field would look like. Now what I'm interested in doing is figuring out the line integral over a closed loop, the closed loop C, or the closed path C, of P dot dr. Which is just our standard kind of way of solving for a line integral. And we've seen what dr is in the past."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what this vector field would look like. Now what I'm interested in doing is figuring out the line integral over a closed loop, the closed loop C, or the closed path C, of P dot dr. Which is just our standard kind of way of solving for a line integral. And we've seen what dr is in the past. dr is equal to dx times i, plus dy times the j unit vector. And you might say, isn't it dx dt times dt? Let me write that."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we've seen what dr is in the past. dr is equal to dx times i, plus dy times the j unit vector. And you might say, isn't it dx dt times dt? Let me write that. Can't dr be written as dx dt times dt i, plus dy dt times dt j? And it could, but if you imagine these differentials could cancel out, and you'll just look for the dx and the dy. And we've seen that multiple times."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write that. Can't dr be written as dx dt times dt i, plus dy dt times dt j? And it could, but if you imagine these differentials could cancel out, and you'll just look for the dx and the dy. And we've seen that multiple times. And I'm going to leave it in this form, because hopefully, if we're careful, we won't have to deal with the third parameter t. So let's just look at it in this form right here, with just the dx's and the dy's. So this integral can be rewritten as the line integral, the curve C, actually let me just do it over down here. The line integral over the path, or the curve C, of P dot dr."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we've seen that multiple times. And I'm going to leave it in this form, because hopefully, if we're careful, we won't have to deal with the third parameter t. So let's just look at it in this form right here, with just the dx's and the dy's. So this integral can be rewritten as the line integral, the curve C, actually let me just do it over down here. The line integral over the path, or the curve C, of P dot dr. So we take the product of each of the coefficients, let's say the coefficient of the i component. So we get P, I'll do it in green, actually let me do it in that purple color. So we get P of xy times dx, plus, well there's no 0 times j times dy."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The line integral over the path, or the curve C, of P dot dr. So we take the product of each of the coefficients, let's say the coefficient of the i component. So we get P, I'll do it in green, actually let me do it in that purple color. So we get P of xy times dx, plus, well there's no 0 times j times dy. 0 times dy is just going to be 0. So this, our line integral simplified to this right here. This is equal to this original integral up here."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we get P of xy times dx, plus, well there's no 0 times j times dy. 0 times dy is just going to be 0. So this, our line integral simplified to this right here. This is equal to this original integral up here. So we're literally just taking the line integral around this path. Now I said that if we play our cards right, we're not going to have to deal with the third variable t, that we might be able to just solve this integral only in terms of x. And so let's see if we can do that."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to this original integral up here. So we're literally just taking the line integral around this path. Now I said that if we play our cards right, we're not going to have to deal with the third variable t, that we might be able to just solve this integral only in terms of x. And so let's see if we can do that. So let's look at our minimum and maximum x points. That looks like our minimum x point. Let's call that a."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's see if we can do that. So let's look at our minimum and maximum x points. That looks like our minimum x point. Let's call that a. And let's call that our maximum x point. Let's call that b. What we could do is we can break up this curve into two functions of x. Y is functions of x."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's call that a. And let's call that our maximum x point. Let's call that b. What we could do is we can break up this curve into two functions of x. Y is functions of x. So this bottom one right here we could call as y1 of x. This is just a standard curve. When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What we could do is we can break up this curve into two functions of x. Y is functions of x. So this bottom one right here we could call as y1 of x. This is just a standard curve. When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When we just start dealing with standard calculus, this is just, you can imagine this is f of x, and it's a function of x. And this is y2 of x. Just like that. So you can imagine two paths. One path defined by y1 of x. Let me do that in a different color, a little magenta. One path defined by y1 of x as we go from x is equal to a to x is equal to b."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can imagine two paths. One path defined by y1 of x. Let me do that in a different color, a little magenta. One path defined by y1 of x as we go from x is equal to a to x is equal to b. And then another path defined by y2 of x as we go from x is equal to b to x is equal to a. That is our curve. So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "One path defined by y1 of x as we go from x is equal to a to x is equal to b. And then another path defined by y2 of x as we go from x is equal to b to x is equal to a. That is our curve. So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x. And I could just say p of x and y, but we know along this path y is a function of x. It's y1 of x. So we say x and y1 of x."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what we could do is we could rewrite this integral, which is the same thing as that integral, as this is equal to the integral, we'll first do this first path, of x going from a to b of p of x. And I could just say p of x and y, but we know along this path y is a function of x. It's y1 of x. So we say x and y1 of x. Wherever we see a y, we substitute it with y1 of x dx. So that covers that first path. I'll do it in the same color."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we say x and y1 of x. Wherever we see a y, we substitute it with y1 of x dx. So that covers that first path. I'll do it in the same color. So this is, we could imagine this is c1. This is kind of the first half of our curve. Well, it's not exactly necessarily the half, but that takes us right from that point to that point."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it in the same color. So this is, we could imagine this is c1. This is kind of the first half of our curve. Well, it's not exactly necessarily the half, but that takes us right from that point to that point. And then we want to complete the circle. Maybe I'll do that. And that is going to be, I'll do that in yellow."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, it's not exactly necessarily the half, but that takes us right from that point to that point. And then we want to complete the circle. Maybe I'll do that. And that is going to be, I'll do that in yellow. That's going to be equal to, sorry, we're going to have to add these two, plus the integral from x is equal to b to x is equal to a of, and I'll do it in that same color, of p of x. And now y is going to be y2 of x. Wherever you see a y, you can substitute with y2 of x along this curve."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that is going to be, I'll do that in yellow. That's going to be equal to, sorry, we're going to have to add these two, plus the integral from x is equal to b to x is equal to a of, and I'll do it in that same color, of p of x. And now y is going to be y2 of x. Wherever you see a y, you can substitute with y2 of x along this curve. y2 of x dx. This is already getting interesting, and you might already see where I'm going with this. So this is the curve c2."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Wherever you see a y, you can substitute with y2 of x along this curve. y2 of x dx. This is already getting interesting, and you might already see where I'm going with this. So this is the curve c2. I think you'd appreciate if you take the union of c1 and c2. We've got our whole curve. So let's see if we can simplify this integral a little bit."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the curve c2. I think you'd appreciate if you take the union of c1 and c2. We've got our whole curve. So let's see if we can simplify this integral a little bit. Well, one thing we want to do, we might want to make their endpoints the same. So if you swap a and b here, it just turns the integral negative. So you make this into a b, that into an a, and then make that plus sign into a minus sign."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see if we can simplify this integral a little bit. Well, one thing we want to do, we might want to make their endpoints the same. So if you swap a and b here, it just turns the integral negative. So you make this into a b, that into an a, and then make that plus sign into a minus sign. And now we can rewrite this whole thing as being equal to the integral from a to b of this thing, of p of x and y1 of x, minus this thing, minus p of x and y2 of x. And then all of that times dx. Maybe I'll write it in a third color."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you make this into a b, that into an a, and then make that plus sign into a minus sign. And now we can rewrite this whole thing as being equal to the integral from a to b of this thing, of p of x and y1 of x, minus this thing, minus p of x and y2 of x. And then all of that times dx. Maybe I'll write it in a third color. All of that times dx. Now, I'm going to do something a little bit arbitrary, but I think you'll appreciate why I did this by the end of this video. And it's just a very simple operation."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe I'll write it in a third color. All of that times dx. Now, I'm going to do something a little bit arbitrary, but I think you'll appreciate why I did this by the end of this video. And it's just a very simple operation. What I'm going to do is I'm going to swap these two. So I'm essentially going to multiply this whole thing by negative 1, or essentially multiply and divide by negative 1. So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's just a very simple operation. What I'm going to do is I'm going to swap these two. So I'm essentially going to multiply this whole thing by negative 1, or essentially multiply and divide by negative 1. So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral. I'm multiplying by negative 1 twice. So if I swap these two things, if I multiply the inside times negative 1, so this is going to be equal to, we do the outside of the integral, a to b. If I multiply the inside, I'll do a dx out here."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I can multiply this by negative 1, and then multiply the outside by negative 1, and I will not have changed the integral. I'm multiplying by negative 1 twice. So if I swap these two things, if I multiply the inside times negative 1, so this is going to be equal to, we do the outside of the integral, a to b. If I multiply the inside, I'll do a dx out here. If I multiply the inside of the integral by negative 1, these two guys switch. So this becomes p of x of y2 of x, and then you're going to have minus p of x and y1 of x. My handwriting is getting a little messy."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I multiply the inside, I'll do a dx out here. If I multiply the inside of the integral by negative 1, these two guys switch. So this becomes p of x of y2 of x, and then you're going to have minus p of x and y1 of x. My handwriting is getting a little messy. But I can't just multiply just the inside by minus 1. I don't want to change the integral, so I multiply the inside by minus 1. Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "My handwriting is getting a little messy. But I can't just multiply just the inside by minus 1. I don't want to change the integral, so I multiply the inside by minus 1. Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent. You could say this is the negative of that. Either way, I think you appreciate that I haven't changed the integral at all numerically. I multiply the inside and the outside by minus 1."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me multiply the outside by minus 1, and since I multiply by minus 1 twice, these two things are equivalent. You could say this is the negative of that. Either way, I think you appreciate that I haven't changed the integral at all numerically. I multiply the inside and the outside by minus 1. Now the next step I'm going to do might look a little bit foreign to you, but I think you'll appreciate it. It might be obvious to you if you've recently done some double integrals. This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I multiply the inside and the outside by minus 1. Now the next step I'm going to do might look a little bit foreign to you, but I think you'll appreciate it. It might be obvious to you if you've recently done some double integrals. This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated. Let me make it very clear. This is y is equal to y2 of x minus this function evaluated at y is equal to y1 of x, and then of course all of that times dx. This statement and what we saw right here and this statement right here are completely identical."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This thing can be rewritten as minus the integral from a to b of the function p of x, y, evaluated at y2 of x minus it evaluated. Let me make it very clear. This is y is equal to y2 of x minus this function evaluated at y is equal to y1 of x, and then of course all of that times dx. This statement and what we saw right here and this statement right here are completely identical. Then if we assume that a partial derivative of capital P with respect to y exists, hopefully you'll realize, and I'll focus on this a little bit because I don't want to confuse you in this step. Let me write the outside of the integral. This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This statement and what we saw right here and this statement right here are completely identical. Then if we assume that a partial derivative of capital P with respect to y exists, hopefully you'll realize, and I'll focus on this a little bit because I don't want to confuse you in this step. Let me write the outside of the integral. This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do. Let me do the outside, dx. This right here, if we assume that capital P has a partial derivative, this right here is the exact same thing. This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be equal to, this is kind of a neat outcome, and we're starting to build up to a very neat outcome, which we'll probably have to take the next video to do. Let me do the outside, dx. This right here, if we assume that capital P has a partial derivative, this right here is the exact same thing. This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent. To realize they're equivalent, you probably just have to start here and then go to that. We're used to seeing this. We're used to seeing a double integral like this."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right here is the exact same thing as the partial derivative of P with respect to y, dy, the antiderivative of that, from y1 of x to y2 of x. I want to make you feel comfortable that these two things are equivalent. To realize they're equivalent, you probably just have to start here and then go to that. We're used to seeing this. We're used to seeing a double integral like this. Then the very first step, we say, okay, to solve this double integral, we start on the inside integral right there. We say, okay, let's take the antiderivative of this with respect to y. If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're used to seeing a double integral like this. Then the very first step, we say, okay, to solve this double integral, we start on the inside integral right there. We say, okay, let's take the antiderivative of this with respect to y. If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x. You're going to evaluate that from y is equal to y2 of x and you're going to subtract from that y is equal to y1 of x. Normally, we start with something like this and we go to something like this. This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you take the antiderivative of the partial of P with respect to y, you're going to end up with P. Since this is a definite integral, the boundaries are going to be in terms of x. You're going to evaluate that from y is equal to y2 of x and you're going to subtract from that y is equal to y1 of x. Normally, we start with something like this and we go to something like this. This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral. Hopefully, you realize that this is true, that this is just, we're kind of going in the reverse direction that we normally do. If you do realize that, then we've just established a pretty neat outcome because what is this right here? What is this right here?"}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is kind of unusual that we started, we kind of solved, we started with the solution of the definite integral and then we slowly built back to the definite integral. Hopefully, you realize that this is true, that this is just, we're kind of going in the reverse direction that we normally do. If you do realize that, then we've just established a pretty neat outcome because what is this right here? What is this right here? Let me go back, let me see if I can fit everything. I have some function, I have some function and I'm assuming that the partial of P with respect to y exists, but I have some function defined over the xy plane. You can imagine if we're dealing in three dimensions now."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What is this right here? Let me go back, let me see if I can fit everything. I have some function, I have some function and I'm assuming that the partial of P with respect to y exists, but I have some function defined over the xy plane. You can imagine if we're dealing in three dimensions now. Let me draw a little bit neater. That's y, that's x, that's z, so that's y, that's x. This, you can imagine, is some surface."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can imagine if we're dealing in three dimensions now. Let me draw a little bit neater. That's y, that's x, that's z, so that's y, that's x. This, you can imagine, is some surface. It just happens to be the partial of P with respect to x. It's some surface on the xy plane like that. What are we doing?"}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This, you can imagine, is some surface. It just happens to be the partial of P with respect to x. It's some surface on the xy plane like that. What are we doing? We're taking the double integral under that surface, around this region. The region's boundaries in terms of y are defined by y2 and y1 of x, by y2 and y1 of x. You literally have that curve."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What are we doing? We're taking the double integral under that surface, around this region. The region's boundaries in terms of y are defined by y2 and y1 of x, by y2 and y1 of x. You literally have that curve. That's y2 on top, y1 on the bottom. We're essentially taking the volume above this. If you imagine with the base, the whole floor of this is going to be the area inside of this curve."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You literally have that curve. That's y2 on top, y1 on the bottom. We're essentially taking the volume above this. If you imagine with the base, the whole floor of this is going to be the area inside of this curve. Then the height is going to be the function, partial of P with respect to y. The height is going to be this function, partial of P with respect to y. It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you imagine with the base, the whole floor of this is going to be the area inside of this curve. Then the height is going to be the function, partial of P with respect to y. The height is going to be this function, partial of P with respect to y. It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way. The really neat outcome here is, if you call this region R, we've just simplified this line integral. This was a special one. It only had an x component, the vector field."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be a little hard for me to draw, but this is essentially some type of a volume if you want to visualize it that way. The really neat outcome here is, if you call this region R, we've just simplified this line integral. This was a special one. It only had an x component, the vector field. We've just simplified this line integral to being equivalent to, and maybe I should write this line integral because that's what's the really neat outcome. We've just established that this thing right here, which is the same as our original one, so let me write that. The closed line integral around the curve C of P of x, y, dx."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It only had an x component, the vector field. We've just simplified this line integral to being equivalent to, and maybe I should write this line integral because that's what's the really neat outcome. We've just established that this thing right here, which is the same as our original one, so let me write that. The closed line integral around the curve C of P of x, y, dx. We've just established that that's the same thing as the double integral over the region R. This is the region R. Over the region R of the partial of P with respect to y. The partial of P with respect to y. We could write dy dx, or we could write da, or whatever you want to write, but this is the double integral over that region."}, {"video_title": "Green's theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The closed line integral around the curve C of P of x, y, dx. We've just established that that's the same thing as the double integral over the region R. This is the region R. Over the region R of the partial of P with respect to y. The partial of P with respect to y. We could write dy dx, or we could write da, or whatever you want to write, but this is the double integral over that region. The neat thing here is, using a vector field that only had an x component, we were able to connect its line integral to the double integral over a region. Oh, and I forgot something very important. We had a negative sign out here, so this was a minus sign out here."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "Hey everyone, so in the next couple videos I'm gonna be talking about a different sort of optimization problem, something called a constrained optimization problem. And an example of this is something where you might see, you might be asked to maximize some kind of multivariable function. And let's just say it was the function f of x, y is equal to x squared times y. But that's not all you're asked to do. You're subject to a certain constraint where you're only allowed values of x and y on a certain set. And I'm just gonna say the set of all values of x and y such that x squared plus y squared equals one. And this is something you might recognize as the unit circle, this particular constraint that I've put on here."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "But that's not all you're asked to do. You're subject to a certain constraint where you're only allowed values of x and y on a certain set. And I'm just gonna say the set of all values of x and y such that x squared plus y squared equals one. And this is something you might recognize as the unit circle, this particular constraint that I've put on here. This is the unit circle. So one way that you might think about a problem like this, you know, you're maximizing a certain two variable function, is to first think of the graph of that function. That's what I have pictured here, is the graph of f of x, y equals x squared times y."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "And this is something you might recognize as the unit circle, this particular constraint that I've put on here. This is the unit circle. So one way that you might think about a problem like this, you know, you're maximizing a certain two variable function, is to first think of the graph of that function. That's what I have pictured here, is the graph of f of x, y equals x squared times y. And now this constraint, x squared plus y squared, is basically just a subset of the x, y plane. So if we look at it head on here, and we look at the x, y plane, this circle represents all of the points x, y such that this holds. And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "That's what I have pictured here, is the graph of f of x, y equals x squared times y. And now this constraint, x squared plus y squared, is basically just a subset of the x, y plane. So if we look at it head on here, and we look at the x, y plane, this circle represents all of the points x, y such that this holds. And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph. So this is showing you basically the values where this constraint holds, and also what they look like when graphed. So a way you can think about a problem like this is that you're looking on this circle, this kind of projected circle onto the graph, and looking for the highest points. And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "And what I've actually drawn here isn't the circle on the x, y plane, but I've projected it up onto the graph. So this is showing you basically the values where this constraint holds, and also what they look like when graphed. So a way you can think about a problem like this is that you're looking on this circle, this kind of projected circle onto the graph, and looking for the highest points. And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one. And then the low points would be, you know, around that point and around over here. Now, this is good, and I think this is a nice way to sort of wrap your head around what this problem is asking. But there's actually a better way to visualize it in terms of finding the actual solution."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "And you might notice kind of here, there's sort of a peak on that wiggly circle, and over here there's another one. And then the low points would be, you know, around that point and around over here. Now, this is good, and I think this is a nice way to sort of wrap your head around what this problem is asking. But there's actually a better way to visualize it in terms of finding the actual solution. And that's to look only in the x, y plane, rather than trying to graph things, and just limit our perspective to the input space. So what I have here are the contour lines for f of x, y equals x squared plus y squared. And if you're unfamiliar with contour lines, or contour map, I have a video on that."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "But there's actually a better way to visualize it in terms of finding the actual solution. And that's to look only in the x, y plane, rather than trying to graph things, and just limit our perspective to the input space. So what I have here are the contour lines for f of x, y equals x squared plus y squared. And if you're unfamiliar with contour lines, or contour map, I have a video on that. You can go back and take a look. It's gonna be pretty crucial for the next couple videos to have a feel for that. But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right?"}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "And if you're unfamiliar with contour lines, or contour map, I have a video on that. You can go back and take a look. It's gonna be pretty crucial for the next couple videos to have a feel for that. But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right? So if you looked at all the values of x and y where this is true, you'd find yourself on one of these lines. And each line represents a different possible value for what this constant here actually is. So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right?"}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "But basically, each one of these lines represents a certain constant value for f. So for example, one of them might represent all of the values of x and y, where f of x, y is equal to, you know, two, right? So if you looked at all the values of x and y where this is true, you'd find yourself on one of these lines. And each line represents a different possible value for what this constant here actually is. So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right? So this here is something that I'm gonna vary, where I'm gonna be able to change what the constant we're setting f equal to is, and look at how the contour line changes as a result. So for example, if I put it around here-ish, what you're looking at is the contour line, the contour line for f of x, y equals 0.1. So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "So what I'm gonna do here is I'm actually gonna just zoom in on one particular contour line, right? So this here is something that I'm gonna vary, where I'm gonna be able to change what the constant we're setting f equal to is, and look at how the contour line changes as a result. So for example, if I put it around here-ish, what you're looking at is the contour line, the contour line for f of x, y equals 0.1. So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1. But on the other hand, I could also shift this guy up, and maybe I'll shift it up, I'm gonna set it to where that constant is actually equal to one. So that would be kind of an alternative. I'll just kinda separate over here."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "So all of the values on these two blue lines here tell you what values of x and y satisfy 0.1. But on the other hand, I could also shift this guy up, and maybe I'll shift it up, I'm gonna set it to where that constant is actually equal to one. So that would be kind of an alternative. I'll just kinda separate over here. That would be the line where f of x, y is set equal to one itself. And the main thing I wanna highlight here is that at some values, like 0.1, this contour line intersects with the circle. It intersects with our constraint."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "I'll just kinda separate over here. That would be the line where f of x, y is set equal to one itself. And the main thing I wanna highlight here is that at some values, like 0.1, this contour line intersects with the circle. It intersects with our constraint. And let's just think about what that means. If there's a point, x and y, on that intersection there, that basically gives us a pair of numbers, x and y, such that this is true, that fact, that f of x, y equals 0.1, and also that x squared plus y squared equals one. So it means this is something that actually exists and is possible."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "It intersects with our constraint. And let's just think about what that means. If there's a point, x and y, on that intersection there, that basically gives us a pair of numbers, x and y, such that this is true, that fact, that f of x, y equals 0.1, and also that x squared plus y squared equals one. So it means this is something that actually exists and is possible. In fact, we can see that there's four different pairs of numbers where that's true, where they intersect here, where they intersect over here, and then the other two kind of symmetrically on that side. But on the other hand, if we look at this other world, where we shift up to the line f of x, y equals one, this never intersects with the constraint. So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "So it means this is something that actually exists and is possible. In fact, we can see that there's four different pairs of numbers where that's true, where they intersect here, where they intersect over here, and then the other two kind of symmetrically on that side. But on the other hand, if we look at this other world, where we shift up to the line f of x, y equals one, this never intersects with the constraint. So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint. They're off of that circle, x squared plus y squared equals one. So what that tells us, as we try to maximize this function, subject to this constraint, is that we can never get as high as one. 0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "So what that means is x, y, the pairs of numbers that satisfy this guy, are off the constraint. They're off of that circle, x squared plus y squared equals one. So what that tells us, as we try to maximize this function, subject to this constraint, is that we can never get as high as one. 0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle. In fact, you could play around with it and increase it a little bit more. And if I go to 0.3 instead, and I go over here and I say 0.3, that's also possible. And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "0.1 would be achievable, and in fact, if we kind of go back to that and we look at 0.1, if I upped that value and changed it to the line where instead what you're looking at is 0.2, that's also possible because it intersects with the circle. In fact, you could play around with it and increase it a little bit more. And if I go to 0.3 instead, and I go over here and I say 0.3, that's also possible. And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle. And the key here, the key observation, is that that maximum value happens when these guys are tangent. And in the next video, I'll start going into the details of how we can use that observation, this notion of tangency, to solve the problem, to find the actual value of x and y that maximizes this subject to the constraint. But in the interim, I kind of want you to mull on that and think a little bit about how you might use that."}, {"video_title": "Constrained optimization introduction.mp3", "Sentence": "And what we're basically trying to do is find the maximum value that we can put here, the maximum value so that if we look at the line that represents f of x, y equals that value, it still intersects with the circle. And the key here, the key observation, is that that maximum value happens when these guys are tangent. And in the next video, I'll start going into the details of how we can use that observation, this notion of tangency, to solve the problem, to find the actual value of x and y that maximizes this subject to the constraint. But in the interim, I kind of want you to mull on that and think a little bit about how you might use that. What does tangency mean here? How can you take advantage of certain other notions that we've learned about in multivariable calculus, like hint, hint, the gradient, to actually solve something like this? So with that, I will see you next video."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "In the last couple videos, I talked about this multivariable chain rule, and I gave some justification, and it might have been considered a little bit hand-wavy by some. I was doing a lot of things that looked kind of like taking a derivative with respect to t, and then multiplying that by an infinitesimal quantity dt, and thinking of canceling those out. And some people might say, ah, but this isn't really a fraction, that's a derivative, that's a differential operator, and you're treating it incorrectly. And while that's true, the intuitions underlying a lot of this actually matches with the formal argument pretty well. So what I wanna do here is just talk about what the formal argument behind the multivariable chain rule is. And just to remind ourselves of the setup, of where we are, you're thinking of v as a vector-valued function. So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And while that's true, the intuitions underlying a lot of this actually matches with the formal argument pretty well. So what I wanna do here is just talk about what the formal argument behind the multivariable chain rule is. And just to remind ourselves of the setup, of where we are, you're thinking of v as a vector-valued function. So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space. In the simplest case, you might just think of that as a two-dimensional space, maybe it's three-dimensional space, or it could be 100-dimensional, you don't have to literally be visualizing it. And then f, our function f, somehow takes that 100-dimensional space, or two-dimensional, or three-dimensional, whatever it is, and then maps it onto the number line. So the overall effect of the composition function is to just take a real number to a real number."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So this is something that takes as an input t that lives on a number line, and then v maps this to some kind of high-dimensional space. In the simplest case, you might just think of that as a two-dimensional space, maybe it's three-dimensional space, or it could be 100-dimensional, you don't have to literally be visualizing it. And then f, our function f, somehow takes that 100-dimensional space, or two-dimensional, or three-dimensional, whatever it is, and then maps it onto the number line. So the overall effect of the composition function is to just take a real number to a real number. So it's a single-variable function, so that's why we're taking this ordinary derivative, rather than a partial derivative, or gradient, or anything like that. But because it goes through a multi-dimensional space, and you have this intermediary multivariable nature to it, that's why you have a gradient and a vector-valued derivative. So with the formal argument, the first thing you might do is just write out the formal definition of a derivative."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So the overall effect of the composition function is to just take a real number to a real number. So it's a single-variable function, so that's why we're taking this ordinary derivative, rather than a partial derivative, or gradient, or anything like that. But because it goes through a multi-dimensional space, and you have this intermediary multivariable nature to it, that's why you have a gradient and a vector-valued derivative. So with the formal argument, the first thing you might do is just write out the formal definition of a derivative. And in this case, it's a limit. Definitions of derivatives are always gonna be some kind of limit as a variable goes to zero. And here, you're loosely thinking about h as being dt."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So with the formal argument, the first thing you might do is just write out the formal definition of a derivative. And in this case, it's a limit. Definitions of derivatives are always gonna be some kind of limit as a variable goes to zero. And here, you're loosely thinking about h as being dt. And you could write delta t, but it's common to use h just because that can be used for whatever your differential quantity is. So that's on the denominator, because you're thinking of it as dt. And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And here, you're loosely thinking about h as being dt. And you could write delta t, but it's common to use h just because that can be used for whatever your differential quantity is. So that's on the denominator, because you're thinking of it as dt. And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function. And now, what do you do as you're trying to reason about what this should equal? And a good place to start, actually, is to look back to the intuition that I was giving for the multivariable chain rule in the first place. You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And the top is whatever the change to this whole function is when you nudge that input by t. And what I mean by that is you'll take f of v, not of t, but of t plus h, that kind of nudged output value, and you're wondering how different that is from f of v of t, your original value, v of t. So this is what happens when you just apply the formal definition of the derivative, the ordinary derivative, to your composition function. And now, what do you do as you're trying to reason about what this should equal? And a good place to start, actually, is to look back to the intuition that I was giving for the multivariable chain rule in the first place. You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector. And the way that you're thinking about that is you take the vector value derivative and multiply it by dt. It's the proportionality constant between the size of your nudge and the resulting vector. And loosely, you might imagine those dts crossing out as if they were fractions."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "You imagine nudging your input by some dt, some tiny change, and I was saying, oh, so that causes a change in the intermediary space of some kind of, you know, you could call it dv, a change in the vector. And the way that you're thinking about that is you take the vector value derivative and multiply it by dt. It's the proportionality constant between the size of your nudge and the resulting vector. And loosely, you might imagine those dts crossing out as if they were fractions. Doesn't really matter. And then you say, what does this change, does this change by a dv cause for f? And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality?"}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And loosely, you might imagine those dts crossing out as if they were fractions. Doesn't really matter. And then you say, what does this change, does this change by a dv cause for f? And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality? You say, well, in this intermediary space, we had to deal with the vector value derivative of v. So it might be a good thing to just write down that definition, right? Write down the fact that the definition for the vector value derivative of v, again, it looks almost identical. All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And by definition, the resulting nudge to the output space of f is the directional derivative in the direction of whatever your vector nudge is of the function f. So this is the loose intuition, and where does that carry over to formality? You say, well, in this intermediary space, we had to deal with the vector value derivative of v. So it might be a good thing to just write down that definition, right? Write down the fact that the definition for the vector value derivative of v, again, it looks almost identical. All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero. h we're still thinking of as being dt. So it kind of sits on the bottom. But here, you're just wondering how your vector changes."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "All these derivative definitions really do look kind of the same, because what you're doing is you're taking the limit as h goes to zero. h we're still thinking of as being dt. So it kind of sits on the bottom. But here, you're just wondering how your vector changes. And the difference, even though, you know, we're kind of writing this the same way, and it looks almost identical notationally, is on the numerator here, this v of t plus h, and this v of t, these are vectors. So this is kind of a vector minus a vector. So when you take the limit, you're getting a limiting vector, something in your high-dimensional space."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "But here, you're just wondering how your vector changes. And the difference, even though, you know, we're kind of writing this the same way, and it looks almost identical notationally, is on the numerator here, this v of t plus h, and this v of t, these are vectors. So this is kind of a vector minus a vector. So when you take the limit, you're getting a limiting vector, something in your high-dimensional space. It's not just a number. And now, another way to write this, one that's more helpful, more conducive to manipulation, is to say, not that it equals the limit of this value, and I'm gonna go ahead and just kind of copy this value here, kind of down here, and say the value of our derivative actually equals this, subject to some kind of error, which I'll just write as e of h, like an error function of h. And what you should be thinking is that that error function goes to zero as h goes to zero. This is just writing things so that we're able to manipulate it a little bit more easily."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So when you take the limit, you're getting a limiting vector, something in your high-dimensional space. It's not just a number. And now, another way to write this, one that's more helpful, more conducive to manipulation, is to say, not that it equals the limit of this value, and I'm gonna go ahead and just kind of copy this value here, kind of down here, and say the value of our derivative actually equals this, subject to some kind of error, which I'll just write as e of h, like an error function of h. And what you should be thinking is that that error function goes to zero as h goes to zero. This is just writing things so that we're able to manipulate it a little bit more easily. So I'll give ourselves some room here. And what you can do with this is multiply all sides by h. So this is our vector value derivative, just rewriting it, multiplied by h. And you're thinking of this h as a dt, so maybe in the back of your mind, you're kind of thinking of canceling this dt with the h. And what it equals is this top, this numerator here, which was v of t plus h, minus v of t. And in the back of your mind, you might be thinking this whole thing represents, you know, dv, a change in v, so the idea of canceling out that dt with the h really does kind of come through here. But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "This is just writing things so that we're able to manipulate it a little bit more easily. So I'll give ourselves some room here. And what you can do with this is multiply all sides by h. So this is our vector value derivative, just rewriting it, multiplied by h. And you're thinking of this h as a dt, so maybe in the back of your mind, you're kind of thinking of canceling this dt with the h. And what it equals is this top, this numerator here, which was v of t plus h, minus v of t. And in the back of your mind, you might be thinking this whole thing represents, you know, dv, a change in v, so the idea of canceling out that dt with the h really does kind of come through here. But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function. And there's actually another way that I'm gonna write this. There's a very useful convention in analysis where I'll take something like this, and instead I'll write it as little o of h. And this isn't literally a function, it's just a stand-in to say whatever this is, whatever function that represents, it satisfies the property that when we take that function and divide it by h, that will go to zero as h goes to zero, right? Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "But the difference between the more hand-waving argument before of canceling those out, and what we're doing here, is now we're accounting for that error function, that, in this case, it's now multiplied by h, because everything was multiplied by h error function. And there's actually another way that I'm gonna write this. There's a very useful convention in analysis where I'll take something like this, and instead I'll write it as little o of h. And this isn't literally a function, it's just a stand-in to say whatever this is, whatever function that represents, it satisfies the property that when we take that function and divide it by h, that will go to zero as h goes to zero, right? Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero. So now what I do is I use this entire expression to write this v of t plus h. And the reason I wanna do that, if we kind of scroll back up, is because we see v of t plus h showing up in the original definition we care about. So this is just a way of starting to get a grapple on that a little bit more firmly. So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "Which is true here, because you imagine taking this and dividing it by h, and that would be, this h cancels out, and you just have your error function is gonna go to zero. So now what I do is I use this entire expression to write this v of t plus h. And the reason I wanna do that, if we kind of scroll back up, is because we see v of t plus h showing up in the original definition we care about. So this is just a way of starting to get a grapple on that a little bit more firmly. So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term. You know, we're evaluating it at whatever that t is, but we're multiplying it by the value of that nudge, that linear term. And then the rest of the stuff is just some little o of h. And maybe you'd say, shouldn't you be subtracting off that little o of h? And it's not an actual function, it just represents anything that shrinks."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So what I'd write, I'd say that that v of t plus h, v of t plus h, that nudged output value, is equal to the original value that I have, v of t, plus, and it's gonna be plus this derivative term, and you can kind of think that it's almost like a Taylor polynomial, where this is our first order term. You know, we're evaluating it at whatever that t is, but we're multiplying it by the value of that nudge, that linear term. And then the rest of the stuff is just some little o of h. And maybe you'd say, shouldn't you be subtracting off that little o of h? And it's not an actual function, it just represents anything that shrinks. And maybe I should say it's the absolute value, like the magnitude, because in this case, this is a vector-valued quantity, you know that error is a vector. So it's the size of that vector divided by the size of h goes to zero. So this is the main tool that we're gonna end up using."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And it's not an actual function, it just represents anything that shrinks. And maybe I should say it's the absolute value, like the magnitude, because in this case, this is a vector-valued quantity, you know that error is a vector. So it's the size of that vector divided by the size of h goes to zero. So this is the main tool that we're gonna end up using. This is the way to represent v of t plus h. And now if we go back up to the original definition of the vector-valued derivative, and I'll go ahead and copy that. Go ahead and copy that guy. A little bit of debris."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "So this is the main tool that we're gonna end up using. This is the way to represent v of t plus h. And now if we go back up to the original definition of the vector-valued derivative, and I'll go ahead and copy that. Go ahead and copy that guy. A little bit of debris. So copied that original definition for the ordinary derivative of the composition function. And now when I write things in according to all the manipulations that we just did, this is really, it's still a limit, as h goes to zero. But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "A little bit of debris. So copied that original definition for the ordinary derivative of the composition function. And now when I write things in according to all the manipulations that we just did, this is really, it's still a limit, as h goes to zero. But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there. It's the value of v of t plus the derivative at our point times the size of h. So again, it's kind of like a Taylor polynomial, this is your linear term. And then it's plus something that we don't care about, something that's gonna get really small as h goes small, and really small in comparison to h, more importantly. And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "But what we put on the inside here is it's f of, now instead of writing v of t plus h, I'm gonna use everything that I did up there. It's the value of v of t plus the derivative at our point times the size of h. So again, it's kind of like a Taylor polynomial, this is your linear term. And then it's plus something that we don't care about, something that's gonna get really small as h goes small, and really small in comparison to h, more importantly. And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge. And all of that is divided by h. Now the point here is when you look at this limit, because we're taking it as h goes to zero, we'll basically be able to ignore this o of h component, because as h goes to zero, this gets very, very small in comparison to h. So everything that's on the inside here is basically just the v of t plus this vector value. And this is h times some kind of vector. But if you think back, I made a video on the formal definition of the directional derivative."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And from that, you subtract off f of v of t. Kind of running off the edge, I always keep running off the edge. And all of that is divided by h. Now the point here is when you look at this limit, because we're taking it as h goes to zero, we'll basically be able to ignore this o of h component, because as h goes to zero, this gets very, very small in comparison to h. So everything that's on the inside here is basically just the v of t plus this vector value. And this is h times some kind of vector. But if you think back, I made a video on the formal definition of the directional derivative. And if you remember it, or if you kind of go back and take a look now, this is exactly the formal definition of the directional derivative. We're taking h to go to zero. The thing we're multiplying it by is a certain vector quantity."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "But if you think back, I made a video on the formal definition of the directional derivative. And if you remember it, or if you kind of go back and take a look now, this is exactly the formal definition of the directional derivative. We're taking h to go to zero. The thing we're multiplying it by is a certain vector quantity. That vector is the nudge to your original value. And then we're dividing everything by h. So by definition, this entire thing is the directional derivative in the direction of the derivative of the function at t. I'm writing v prime t instead of getting the whole dv dt down there. All of that of f evaluated at where?"}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "The thing we're multiplying it by is a certain vector quantity. That vector is the nudge to your original value. And then we're dividing everything by h. So by definition, this entire thing is the directional derivative in the direction of the derivative of the function at t. I'm writing v prime t instead of getting the whole dv dt down there. All of that of f evaluated at where? Well, the place that we're starting is just v of t. So that's v of t. And that's it, that's the answer. Because when you evaluate the directional derivative, the way that you do that, you take the gradient of f, evaluate it at whatever point you're starting at, in this case, it's the output of v of t, and you take the dot product between that and the vector value derivative, well, I mean, the dot product between that and whatever your vector is, which in this case is the vector value derivative of v. And that's the multivariable chain rule. And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "All of that of f evaluated at where? Well, the place that we're starting is just v of t. So that's v of t. And that's it, that's the answer. Because when you evaluate the directional derivative, the way that you do that, you take the gradient of f, evaluate it at whatever point you're starting at, in this case, it's the output of v of t, and you take the dot product between that and the vector value derivative, well, I mean, the dot product between that and whatever your vector is, which in this case is the vector value derivative of v. And that's the multivariable chain rule. And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged. Because the reason we thought to use the vector value derivative was because of that intuition. And the reason for all the manipulation that I did is just because I wanted to be able to express what a nudge to the input of v looks like, and what that looks like is the original value plus a certain vector here. This was the resulting nudge in the intermediary space."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "And if you look back through the line of reasoning, it all really did match the thoughts of kind of nudging, nudging, and seeing how that nudged. Because the reason we thought to use the vector value derivative was because of that intuition. And the reason for all the manipulation that I did is just because I wanted to be able to express what a nudge to the input of v looks like, and what that looks like is the original value plus a certain vector here. This was the resulting nudge in the intermediary space. I wanted to express that in a formal way. And sure, we have this kind of O of H term that expresses something that shrinks really fast. But once you express it like that, you just end up plopping out the original, the definition of the directional derivative."}, {"video_title": "More formal treatment of multivariable chain rule.mp3", "Sentence": "This was the resulting nudge in the intermediary space. I wanted to express that in a formal way. And sure, we have this kind of O of H term that expresses something that shrinks really fast. But once you express it like that, you just end up plopping out the original, the definition of the directional derivative. So I hope that gives kind of a satisfying reason for those of you who are a little bit more rigor-inclined for why the multivariable chain rule works. I should also maybe mention there's a more general multivariable chain rule for vector-valued functions. I'll get to that at another point when I talk about the connections between multivariable calculus and linear algebra."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I've rewritten Stokes' theorem right over here. What I want to focus on in this video is the question of orientation. Because there are two different orientations for our boundary curve, we could go in that direction like that, or we could go in the opposite direction. We could be going like that. And there are also two different orientations for this normal vector. The normal vector might pop out like that, or it could actually go in to the surface like that. So we have to make sure that our orientations are consistent."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could be going like that. And there are also two different orientations for this normal vector. The normal vector might pop out like that, or it could actually go in to the surface like that. So we have to make sure that our orientations are consistent. And what I want to do is give you two different ways of thinking about it, and you might think of others, but these are the ones that work for me. In order for Stokes' theorem to hold, we have to make sure that we're not actually picking the negative of one or the other orientations. And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have to make sure that our orientations are consistent. And what I want to do is give you two different ways of thinking about it, and you might think of others, but these are the ones that work for me. In order for Stokes' theorem to hold, we have to make sure that we're not actually picking the negative of one or the other orientations. And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left. So over here, he would have to go in this direction, in order to keep the surface to his left. So he would have to go, he would have to go just like that. If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so the easiest way for me to remember it is if our normal vector, if the normal vector is, let's say it goes in that direction, and if you have some hypothetical person traversing the boundary of our surface, and their direction is pointed, and their head is pointed in the same direction as the normal vector, so this is the normal vector, so their head is pointed in the exact same direction as the normal vector, so you could say maybe their body, or really their head, so that's them, then the direction that you would have to actually traverse the boundary is the direction that would allow this person to keep the surface to their left. So over here, he would have to go in this direction, in order to keep the surface to his left. So he would have to go, he would have to go just like that. If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface. So if we had a surface, so this surface looks very similar. If this surface, this is a very similar looking surface that I'm drawing right over here, just to give an idea of some of the contours, but if we said that the normal vector for this surface, we orient it in the opposite way. So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we oriented the surface differently, so let me redraw the surface right over here, draw a similar surface. So if we had a surface, so this surface looks very similar. If this surface, this is a very similar looking surface that I'm drawing right over here, just to give an idea of some of the contours, but if we said that the normal vector for this surface, we orient it in the opposite way. So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction. Because once again, if I draw my little character right over here, his head is pointed in the direction of the normal vector. He is now upside down, so let me draw him. So this is him, this is him running right over here."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we said that the normal vector here was actually pointing downward like that, then we would have to, in order for Stokes' Theorem to hold, we would have to traverse the boundary in a different direction. Because once again, if I draw my little character right over here, his head is pointed in the direction of the normal vector. He is now upside down, so let me draw him. So this is him, this is him running right over here. I could draw a better job. This is him running right over here. Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is him, this is him running right over here. I could draw a better job. This is him running right over here. Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind. It would actually go down. Here it looks like a hill to him. But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, in order to keep, and from his point of view, this would kind of look like some type of a pool or a ditch of some kind. It would actually go down. Here it looks like a hill to him. But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction. So depending on the orientation of your normal vector, or which is really the orientation of your actual surface, will dictate how you need to traverse the path. Now, another way to think about it, and this idea was introduced by one of the viewers on YouTube but it's a valid way of thinking about it, is to imagine that the surface is a bottle cap. And so let me draw some type of a bottle over here."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But since he's upside down, in order for him to keep the boundary to his left, he would have to now go in, he would have to now go in the other direction. So depending on the orientation of your normal vector, or which is really the orientation of your actual surface, will dictate how you need to traverse the path. Now, another way to think about it, and this idea was introduced by one of the viewers on YouTube but it's a valid way of thinking about it, is to imagine that the surface is a bottle cap. And so let me draw some type of a bottle over here. So I'll draw, let me draw a bottle. So this is, you can imagine some type of a glass soda bottle. So, but what we really care about is the cap of the bottle."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let me draw some type of a bottle over here. So I'll draw, let me draw a bottle. So this is, you can imagine some type of a glass soda bottle. So, but what we really care about is the cap of the bottle. So, make it feel like it's glass. So there, that's our bottle. And let me draw, let me draw its cap."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So, but what we really care about is the cap of the bottle. So, make it feel like it's glass. So there, that's our bottle. And let me draw, let me draw its cap. Let me draw the cap of the bottle because that's what we care about. We can kind of imagine that being the surface. So this is the cap of our bottle."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let me draw, let me draw its cap. Let me draw the cap of the bottle because that's what we care about. We can kind of imagine that being the surface. So this is the cap of our bottle. And you just need to think about, well, which way would I have to twist the cap in order to make the cap move up, in order to take the cap off? And you could think of the normal vector as the direction that the cap would move, is the direction that the cap would move. And the twisting is the direction that you would have to traverse the path."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the cap of our bottle. And you just need to think about, well, which way would I have to twist the cap in order to make the cap move up, in order to take the cap off? And you could think of the normal vector as the direction that the cap would move, is the direction that the cap would move. And the twisting is the direction that you would have to traverse the path. So you would have to twist the bottle that way. Or you can think about the other way. If you twisted the bottle the other way, then the cap would move down."}, {"video_title": "Orientation and stokes   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the twisting is the direction that you would have to traverse the path. So you would have to twist the bottle that way. Or you can think about the other way. If you twisted the bottle the other way, then the cap would move down. So the normal vector is the direction that the cap would move. And the direction that you would traverse the boundary is how you would actually twist it. So either of these are ways of thinking about it, but they're important to keep in mind, especially once the shapes start getting a little bit more convoluted and oriented in strange ways."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we'll start with an example of a torus, or more commonly known as a donut shape. And we know what a donut looks like. Let me draw it in a suitable, well I don't have any suitable donut colors, so I'll just use green. So a donut looks something like this. So it has a hole in the center, and maybe the other side of the donut looks something like that. And we could shade it in like that, and then maybe shade it in like that. That is what a donut looks like."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So a donut looks something like this. So it has a hole in the center, and maybe the other side of the donut looks something like that. And we could shade it in like that, and then maybe shade it in like that. That is what a donut looks like. So how do we construct that using two parameters? So what we want to do is you can just visualize that a donut, if we were to draw some axes here, so that's our donut, let me draw some axes. So let's say I have the z-axis that goes straight up and down."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is what a donut looks like. So how do we construct that using two parameters? So what we want to do is you can just visualize that a donut, if we were to draw some axes here, so that's our donut, let me draw some axes. So let's say I have the z-axis that goes straight up and down. So the way I've drawn it here, the donut's a little at a tilt so the z-axis, I'll tilt it a little bit. So our z-axis goes straight through the center of the donut. So that right there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say I have the z-axis that goes straight up and down. So the way I've drawn it here, the donut's a little at a tilt so the z-axis, I'll tilt it a little bit. So our z-axis goes straight through the center of the donut. So that right there. This is going to be an exercise in drawing more than anything else. So that is my z-axis, and then you can imagine the z-axis goes from there, and then this coming out of here will be my x-axis, that right there is my x-axis, and then maybe my y-axis comes out like that. And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that right there. This is going to be an exercise in drawing more than anything else. So that is my z-axis, and then you can imagine the z-axis goes from there, and then this coming out of here will be my x-axis, that right there is my x-axis, and then maybe my y-axis comes out like that. And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this. If I were to just slice it in the xz-axis, it would look something like that. I'm slicing, this is the xz-axis. That would be the slice."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the whole reason why I drew it this way is that if you imagine the cross-section of this donut, I'll draw it a little bit neater, but the cross-section of this donut in the xz-axis is going to look something like this. If I were to just slice it in the xz-axis, it would look something like that. I'm slicing, this is the xz-axis. That would be the slice. It would trace out, and we're thinking about not a full donut, just the surface of a donut. So it would trace out a circle like this. If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That would be the slice. It would trace out, and we're thinking about not a full donut, just the surface of a donut. So it would trace out a circle like this. If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there. And if you go out here, you're going to get circles. It's just a bunch of circles. So if you think about it, it's a bunch of circles rotated around the z-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you were to cut the donut in the positive zy-axis, it's going to trace out a circle that looks something like that right there. And if you go out here, you're going to get circles. It's just a bunch of circles. So if you think about it, it's a bunch of circles rotated around the z-axis. If you think of it that way, it'll give us some good intuition for the best way to parameterize this thing. So let's do it that way. So let's start off with just the zy-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you think about it, it's a bunch of circles rotated around the z-axis. If you think of it that way, it'll give us some good intuition for the best way to parameterize this thing. So let's do it that way. So let's start off with just the zy-axis. I'll draw it a little bit neater than I've done here. So that is the z-axis, and that is y, just like that. And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's start off with just the zy-axis. I'll draw it a little bit neater than I've done here. So that is the z-axis, and that is y, just like that. And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis. I didn't draw it that neatly here, but I think you can visualize. So it sits right there on the y-axis. And let's say that it is a distance b away from the center of the torus, or from the z-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say that the center of these circles, let's say it lies on, it can lie, when you cross the zy-axis, the center sits on the y-axis. I didn't draw it that neatly here, but I think you can visualize. So it sits right there on the y-axis. And let's say that it is a distance b away from the center of the torus, or from the z-axis. It's a distance of b. It's always going to be a distance of b. If you imagine the top of the donut, let me draw the top of the donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say that it is a distance b away from the center of the torus, or from the z-axis. It's a distance of b. It's always going to be a distance of b. If you imagine the top of the donut, let me draw the top of the donut. If you're looking down on a donut, so let me draw a donut right here. So if you're looking down on a donut, it just looks something like that. The z-axis is just going to be popping straight out."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you imagine the top of the donut, let me draw the top of the donut. If you're looking down on a donut, so let me draw a donut right here. So if you're looking down on a donut, 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 donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 donut. It'll look just like this. And if you imagine the cross-section, this circle right here will look just like that. This circle right here, that top part of the circle, if you're looking down, would look just like that."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm sitting on the z-axis looking down at the donut. It'll look just like this. And if you imagine the cross-section, this circle right here will look just like that. This circle right here, that top part of the circle, if you're looking down, would look just like that. And this distance, this distance b, is the distance from the z-axis to the center of each of these circles. 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. And it's just going to keep going to the center of the circles."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This circle right here, that top part of the circle, if you're looking down, would look just like that. And this distance, this distance b, is the distance from the z-axis to the center of each of these circles. 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. And 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."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And 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. All of, from the center of our torus to the center of our circle that defies this 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 of length a."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's going to be b. All of, from the center of our torus to the center of our circle that defies this 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 of length a. So these circles have radius of length a. So this distance right here is a. This distance right here is a."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And from b, we can imagine we have 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 to look at these circles, these circles have radius a."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This distance right here is a. This distance right there is a. That distance right there is a. If I were to 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 xz-plane. So you can imagine the x-axis coming out."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I were to 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 xz-plane. So you can imagine the x-axis coming out. Let me do that in the same color. You can imagine the x-axis coming out here. So this is the xz-plane."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can imagine the x-axis coming out. Let me do that in the same color. You can imagine the x-axis coming out here. So this is the xz-plane. So one parameter is going to be the angle between our radius and the xz-plane. We're going to call that angle, or that parameter, s. And so as s goes between 0 and 2 pi, when 0 is just going to be, you're 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."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the xz-plane. So one parameter is going to be the angle between our radius and the xz-plane. We're going to call that angle, or that parameter, s. And so as s goes between 0 and 2 pi, when 0 is just going to be, you're 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 just that circle right there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 just 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."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What I just drew is just 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. This one's getting a little bit messy. Let me draw another top view. As you can see, this is all about visualization."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll call this one t. And I'll take the top view again. This one's getting a little bit messy. Let me draw another top view. As you can see, this is all about visualization. So let's say this is my x-axis, that is my y-axis. And we said we started here on the zy-plane. We are b away from the z-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "As you can see, this is all about visualization. So let's say this is my x-axis, that is my y-axis. And we said we started here on the zy-plane. We are b away from the z-axis. So that distance is b. In this diagram, the z-axis is just popping out at us. It's popping out of the page."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We are b away from the z-axis. So that distance is b. In this diagram, the z-axis is just popping out at us. It's popping out of the page. We're looking straight down. It's just like the same view is right there. And what I just drew, when s is equal to 0 radians, we're going to be out here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's popping out of the page. We're looking straight down. It's just like the same view is right there. And what I just drew, when s is equal to 0 radians, we're going to be out here. We're going to be exactly 1 radius further along the y-axis. Then we're going to rotate. As we rotate around, we're going to rotate and then come all the way over here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what I just drew, when s is equal to 0 radians, we're going to be out here. We're going to be exactly 1 radius further along the y-axis. Then we're going to rotate. As we rotate around, we're going to rotate and then come all the way over here. That's when we're right over there. And then come back down. So if you looked on the top of the circle, it's going to look like that."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "As we rotate around, we're going to rotate and then come all the way over here. That's when we're right over there. And then come back down. So if you looked on the top of the circle, it's going to look like that. Now, to make the donut, we're going to have to rotate this whole setup around the z-axis. Remember, the z-axis is popping out. It's looking straight up at us."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you looked on the top of the circle, it's going to look like that. Now, to make the donut, we're going to have to rotate this whole setup around the z-axis. Remember, the z-axis is popping out. It's looking straight up at us. It's coming out of your video screen. Now, to rotate it, we're going to rotate this circle around the z-axis. And to do that, we'll define a parameter that tells us how much we have rotated it."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's looking straight up at us. It's coming out of your video screen. Now, to rotate it, we're going to rotate this circle around the z-axis. And to do that, we'll define a parameter that tells us how much we have rotated it. So this is when we've rotated 0 radians. At some point, we're going to be over here. And we would have rotated it."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to do that, we'll define a parameter that tells us how much we have rotated it. So this is when we've rotated 0 radians. At some point, we're going to be over here. And we would have rotated it. This is b as well. And our circle is going to be looking like this. This is maybe this point on our donut right there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we would have rotated it. This is b as well. And our circle is going to be looking like this. This is maybe this point on our donut right there. At that point, we would have rotated it, let's say, t radians. So this parameter of how much have we rotated it around the z-axis, how much have we gone around that way, we're going to call that t. And t is also going to vary between 0 and 2 pi. And I want to make this very clear."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is maybe this point on our donut right there. At that point, we would have rotated it, let's say, t radians. So this parameter of how much have we rotated it around the z-axis, how much have we gone around that way, we're going to call that t. And t is also going to vary between 0 and 2 pi. And I want to make this very clear. Let's actually draw the domain that we're mapping from to our surface so that we understand this, I guess, fully. So let me draw some, and then we'll talk about how we can actually parameterize that into a position vector-valued function. So right here, let's call that the t-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I want to make this very clear. Let's actually draw the domain that we're mapping from to our surface so that we understand this, I guess, fully. So let me draw some, and then we'll talk about how we can actually parameterize that into a position vector-valued function. So right here, let's call that the t-axis. Let's remember how much we rotated around the z-axis right there. And let's call this down here our s-axis. And I think this will help things out a good bit."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So right here, let's call that the t-axis. Let's remember how much we rotated around the z-axis right there. And let's call this down here our s-axis. And I think this will help things out a good bit. So when s is equal to 0, and we vary just t. So they're both going to vary between 0 and 2 pi. So this right here is 0. This right here is 2 pi."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I think this will help things out a good bit. So when s is equal to 0, and we vary just t. So they're both going to vary between 0 and 2 pi. So this right here is 0. This right here is 2 pi. Let me do some things in between. This is pi. This would be pi over 2, obviously."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right here is 2 pi. Let me do some things in between. This is pi. This would be pi over 2, obviously. This would be 3 pi over 4. You do the same thing on the t-axis. It's going to go up to 2 pi."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This would be pi over 2, obviously. This would be 3 pi over 4. You do the same thing on the t-axis. It's going to go up to 2 pi. Let's do that. So we're going to go up to 2 pi. I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to go up to 2 pi. Let's do that. So we're going to go up to 2 pi. I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward. So that's 2 pi. This is pi. This is pi over 2."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I really want you to visualize this, because then the parameterization, I think, will be fairly straightforward. So that's 2 pi. This is pi. This is pi over 2. And then this is 3 pi over 4. So let's think about what it looks like if you just hold s constant at 0, and we just vary t between 0 and 2 pi. And let me do that in magenta right here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is pi over 2. And then this is 3 pi over 4. So let's think about what it looks like if you just hold s constant at 0, and we just vary t between 0 and 2 pi. And let me do that in magenta right here. So we're holding s constant, and we're just varying the parameter 2 pi. So this, if you think about it, should just form a curve in three dimensions, not a surface, because we're only varying one parameter right here. So let's think about what this is."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let me do that in magenta right here. So we're holding s constant, and we're just varying the parameter 2 pi. So this, if you think about it, should just form a curve in three dimensions, not a surface, because we're only varying one parameter right here. So let's think about what this is. Remember, s is, let me draw my axes. So that is my x-axis. That is my y-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's think about what this is. Remember, s is, let me draw my axes. So that is my x-axis. That is my y-axis. And then this is my, I'm getting messier and messier, that is my z-axis. Actually, let me draw it a little bit bigger than that. I think it'll help all of our visualizations."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is my y-axis. And then this is my, I'm getting messier and messier, that is my z-axis. Actually, let me draw it a little bit bigger than that. I think it'll help all of our visualizations. All right. So this is my x-axis. That is my y-axis."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I think it'll help all of our visualizations. All right. So this is my x-axis. That is my y-axis. And then my z-axis goes up like that. Now remember, when s is equal to 0, that means we haven't rotated around this circle at all. That means we're out here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is my y-axis. And then my z-axis goes up like that. Now remember, when s is equal to 0, that means we haven't rotated around this circle at all. That means we're out here. We're going to be b away and then a away again. We haven't rotated around this at all. We're setting s is equal to 0."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That means we're out here. We're going to be b away and then a away again. We haven't rotated around this at all. We're setting s is equal to 0. So essentially, we're going to be b away. So this is going to be a distance of b away. And then we're going to be another a away."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're setting s is equal to 0. So essentially, we're going to be b away. So this is going to be a distance of b away. And then we're going to be another a away. The radius, the b is the center of the circle. And then we're going to be another a away. We're going to be right over there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to be another a away. The radius, the b is the center of the circle. And then we're going to be another a away. We're going to be right over there. So this is a plus b away. And then we're going to vary t. Remember, t was how much we've gone around the z-axis. These were top views over here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to be right over there. So this is a plus b away. And then we're going to vary t. Remember, t was how much we've gone around the z-axis. These were top views over here. So this line right here in our st domain, we can say, when we map it or when you parameterize it, it'll correspond to the curve that's essentially the outer edge of our donut. If this is the top view of the donut, it will be the outer edge of our donut, just like that. So let me draw the outer edge."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These were top views over here. So this line right here in our st domain, we can say, when we map it or when you parameterize it, it'll correspond to the curve that's essentially the outer edge of our donut. If this is the top view of the donut, it will be the outer edge of our donut, just like that. So let me draw the outer edge. And to do that a little bit better, let me draw the axes in both the positive and the negative domain. It might make my graph a little bit easier to visualize things. Positive and negative domain."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw the outer edge. And to do that a little bit better, let me draw the axes in both the positive and the negative domain. It might make my graph a little bit easier to visualize things. Positive and negative domain. This is negative z right there. So this line in our t-s plane, I guess we could say, this magenta line, where we hold s at 0 radians and we increase t. This is t is 0. This is t is equal to 2 pi."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Positive and negative domain. This is negative z right there. So this line in our t-s plane, I guess we could say, this magenta line, where we hold s at 0 radians and we increase t. This is t is 0. This is t is equal to 2 pi. That's t is equal to pi. This is t is equal to 3 pi over 2, all the way back to t is equal to 2 pi. This line corresponds to that line as we increase t and hold s constant at 0."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is t is equal to 2 pi. That's t is equal to pi. This is t is equal to 3 pi over 2, all the way back to t is equal to 2 pi. This line corresponds to that line as we increase t and hold s constant at 0. Now let's do another point. Let's say when s is at pi. Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This line corresponds to that line as we increase t and hold s constant at 0. Now let's do another point. Let's say when s is at pi. Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles. So we're right over there. And now let's hold it constant at pi and then rotate it around to form our donut. So we're going to form the inside of our donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Remember, when s is at pi, we've gone exactly, pi is 180 degrees, when s is at pi, we've gone exactly 180 degrees around the circle, around each of these circles. So we're right over there. And now let's hold it constant at pi and then rotate it around to form our donut. So we're going to form the inside of our donut. So when s is at pi, and we're going to take t from 0. So when s is pi and t is 0, this was the center of our circle. We're going to be a below that."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to form the inside of our donut. So when s is at pi, and we're going to take t from 0. So when s is pi and t is 0, this was the center of our circle. We're going to be a below that. We're going to be right over there. And then as we vary, as we increase t, so as we move up along holding s at pi, and we increase t, we're going to trace out the inside of our donut. That will look something like that."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to be a below that. We're going to be right over there. And then as we vary, as we increase t, so as we move up along holding s at pi, and we increase t, we're going to trace out the inside of our donut. That will look something like that. That was my best shot at drawing it. And then we could do that multiple times. When s is pi over 2, I want to do multiple different colors."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That will look something like that. That was my best shot at drawing it. And then we could do that multiple times. When s is pi over 2, I want to do multiple different colors. When s is pi over 2, up here we've rotated exactly 90 degrees, right? Pi over 2 is 90 degrees at this point. And then if we vary t, we're essentially tracing out the top of the donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When s is pi over 2, I want to do multiple different colors. When s is pi over 2, up here we've rotated exactly 90 degrees, right? Pi over 2 is 90 degrees at this point. And then if we vary t, we're essentially tracing out the top of the donut. So let me make sure I draw it. So the cross section, the top of the donut, we're going to start off right over here. So when s is pi over 2 and you vary it right, and then you vary t, I'm having trouble drawing straight lines, and then you vary t, it's going to look like this."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if we vary t, we're essentially tracing out the top of the donut. So let me make sure I draw it. So the cross section, the top of the donut, we're going to start off right over here. So when s is pi over 2 and you vary it right, and then you vary t, I'm having trouble drawing straight lines, and then you vary t, it's going to look like this. That's the top of that circle right there. The top of this circle is going to be right there. Top of this circle is going to be right over there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when s is pi over 2 and you vary it right, and then you vary t, I'm having trouble drawing straight lines, and then you vary t, it's going to look like this. That's the top of that circle right there. The top of this circle is going to be right there. Top of this circle is going to be right over there. Top of this circle is going to be right over there. So then I just connect the dots. It's going to look something like that."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Top of this circle is going to be right over there. Top of this circle is going to be right over there. So then I just connect the dots. It's going to look something like that. That is the top of our donut. If I was doing this top view, it would be the top of the donut just like that. And if I want to do the bottom of the donut, just to make the picture clear, if I make the bottom of the donut, the bottom of the donut would be, let's see, if I kick s as 3 pi over 4 and I vary t, that's the bottoms of our donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to look something like that. That is the top of our donut. If I was doing this top view, it would be the top of the donut just like that. And if I want to do the bottom of the donut, just to make the picture clear, if I make the bottom of the donut, the bottom of the donut would be, let's see, if I kick s as 3 pi over 4 and I vary t, that's the bottoms of our donut. So let me draw the circle. So it's right there. The circle is right there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if I want to do the bottom of the donut, just to make the picture clear, if I make the bottom of the donut, the bottom of the donut would be, let's see, if I kick s as 3 pi over 4 and I vary t, that's the bottoms of our donut. So let me draw the circle. So it's right there. The circle is right there. You wouldn't even be able to see the whole thing if this wasn't transparent. So you'd be tracing out the bottom of the donut just like that. I know that this graph is becoming a little confusing, but hopefully you get the idea."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The circle is right there. You wouldn't even be able to see the whole thing if this wasn't transparent. So you'd be tracing out the bottom of the donut just like that. I know that this graph is becoming a little confusing, but hopefully you get the idea. When s is 2 pi again, you're going to be back to the outside of the donut again. That's also going to be in purple. So that's what happens when we hold the s constant at certain values and vary the t. Now let's do the opposite."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I know that this graph is becoming a little confusing, but hopefully you get the idea. When s is 2 pi again, you're going to be back to the outside of the donut again. That's also going to be in purple. So that's what happens when we hold the s constant at certain values and vary the t. Now let's do the opposite. What happens if we hold t at 0 and we vary the s? So if we hold t at 0 and we vary the s. So t is 0. That means we haven't rotated it all yet."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's what happens when we hold the s constant at certain values and vary the t. Now let's do the opposite. What happens if we hold t at 0 and we vary the s? So if we hold t at 0 and we vary the s. So t is 0. That means we haven't rotated it all yet. So we're in the zy plane. So t is 0. And s will start at 0 and it'll go to 2 pi."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That means we haven't rotated it all yet. So we're in the zy plane. So t is 0. And s will start at 0 and it'll go to 2 pi. This is this point. Sorry, pi over 2. That's that point over there."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And s will start at 0 and it'll go to 2 pi. This is this point. Sorry, pi over 2. That's that point over there. Then it'll go to pi. This point is the same thing as that point. Then it'll go to 3 pi over 4."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's that point over there. Then it'll go to pi. This point is the same thing as that point. Then it'll go to 3 pi over 4. And then it'll come back all the way to 2 pi. So this line corresponds to this circle right there. We could keep doing these."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then it'll go to 3 pi over 4. And then it'll come back all the way to 2 pi. So this line corresponds to this circle right there. We could keep doing these. If we pick when s is pi over 2. Sorry, when t is pi over 2. Let me do a different color."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could keep doing these. If we pick when s is pi over 2. Sorry, when t is pi over 2. Let me do a different color. That's not different enough. When t is pi over 2, just like that, we would have rotated around the z-axis 90 degrees. So now we're over here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me do a different color. That's not different enough. When t is pi over 2, just like that, we would have rotated around the z-axis 90 degrees. So now we're over here. And now when we vary s, s will start off over here and it'll go all the way around like that. So this line corresponds to that circle. We could keep doing it like this."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now we're over here. And now when we vary s, s will start off over here and it'll go all the way around like that. So this line corresponds to that circle. We could keep doing it like this. When t is equal to pi, that means we've gone all the way around the circle like that. And now when we vary s from 0 to pi over 2, we're going to start all the way over here. And then we're going to vary all the way."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could keep doing it like this. When t is equal to pi, that means we've gone all the way around the circle like that. And now when we vary s from 0 to pi over 2, we're going to start all the way over here. And then we're going to vary all the way. We're going to go down and hit all those contours that we talked about before. And I'll do one more just to kind of make the scaffold clear, this dark purple. Hopefully you can see it."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to vary all the way. We're going to go down and hit all those contours that we talked about before. And I'll do one more just to kind of make the scaffold clear, this dark purple. Hopefully you can see it. When t is 3 pi over 4, we've rotated all the way. So we're on the xz-plane. And then when you vary s, s will start off over here."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully you can see it. When t is 3 pi over 4, we've rotated all the way. So we're on the xz-plane. And then when you vary s, s will start off over here. And as you increase s, you're going to go around the circle just like that. And of course, when you get all the way back full circle, t over pi over 2, that's the same thing. You're back over here again."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when you vary s, s will start off over here. And as you increase s, you're going to go around the circle just like that. And of course, when you get all the way back full circle, t over pi over 2, that's the same thing. You're back over here again. So this is going to be, we can even shade it the same color. And hopefully you're getting a sense now of the parameterization. I haven't done any math yet."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're back over here again. So this is going to be, we can even shade it the same color. And hopefully you're getting a sense now of the parameterization. I haven't done any math yet. I haven't actually showed you how to mathematically represent it as a vector-valued function. But hopefully you're getting a sense of what it means to parameterize by two parameters. And just to get an idea of what these areas on our xst-plane correspond to onto this surface in R3, this little square right here, let's see what it's bounded by."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I haven't done any math yet. I haven't actually showed you how to mathematically represent it as a vector-valued function. But hopefully you're getting a sense of what it means to parameterize by two parameters. And just to get an idea of what these areas on our xst-plane correspond to onto this surface in R3, this little square right here, let's see what it's bounded by. It's this little square. I want to make sure I pick a square that I can draw neatly. So this square right here, that it is between, when you look at t, it's between this, t is equal to 0 and pi over 2."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just to get an idea of what these areas on our xst-plane correspond to onto this surface in R3, this little square right here, let's see what it's bounded by. It's this little square. I want to make sure I pick a square that I can draw neatly. So this square right here, that it is between, when you look at t, it's between this, t is equal to 0 and pi over 2. So between 0 is t between 0 and pi over 2. And s is between 0 and pi over 2. So this right here is this part of our torus."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this square right here, that it is between, when you look at t, it's between this, t is equal to 0 and pi over 2. So between 0 is t between 0 and pi over 2. And s is between 0 and pi over 2. So this right here is this part of our torus. If you're looking at it from an outer edge, or sorry, from the top, it would look like that right there. You can imagine we've transformed this square. I haven't even shown you how to do it mathematically yet, but we've transformed this square to this part of the donut."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this right here is this part of our torus. If you're looking at it from an outer edge, or sorry, from the top, it would look like that right there. You can imagine we've transformed this square. I haven't even shown you how to do it mathematically yet, but we've transformed this square to this part of the donut. Now, I think we've done about as much as I can do on the visualization side. I'll stop this video here. In the next video, we're going to actually talk about how do we actually parameterize using these two parameters."}, {"video_title": "Introduction to parametrizing a surface with two parameters   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I haven't even shown you how to do it mathematically yet, but we've transformed this square to this part of the donut. Now, I think we've done about as much as I can do on the visualization side. I'll stop this video here. In the next video, we're going to actually talk about how do we actually parameterize using these two parameters. Remember, s takes us around each of these circles, and then t takes us around the z-axis. And if you take all of the combinations of s and t, you're going to have every point along the surface of this torus or this donut. How do you actually go from an s and a t that goes from 0 to 2 pi in both cases, and turn it into a three-dimensional position vector valued function that would define this surface?"}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I needed a break. But where we left off, we were in the home stretch. We were evaluating the third surface integral. And we were setting it up as a double integral with respect to r and theta. And we just had to set up the bounds. And we know that r takes on values between 0 and 1. And theta takes on values between 0 and 2 pi."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we were setting it up as a double integral with respect to r and theta. And we just had to set up the bounds. And we know that r takes on values between 0 and 1. And theta takes on values between 0 and 2 pi. So r, we're going to integrate with respect to r first. r takes on values between 0 and 1. And theta takes on values between 0 and 2 pi."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And theta takes on values between 0 and 2 pi. So r, we're going to integrate with respect to r first. r takes on values between 0 and 1. And theta takes on values between 0 and 2 pi. And so now we are ready to integrate. So let's do the first part. Let's do this inside part right over here."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And theta takes on values between 0 and 2 pi. And so now we are ready to integrate. So let's do the first part. Let's do this inside part right over here. And I'm just going to rewrite the outside part. So this is going to be equal to we have our square root of 2 times the integral from 0 to 2 pi of, and we have d theta right over here. So that's the outside."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's do this inside part right over here. And I'm just going to rewrite the outside part. So this is going to be equal to we have our square root of 2 times the integral from 0 to 2 pi of, and we have d theta right over here. So that's the outside. This inside part right over here, we can rewrite it as if we distribute the r, it's r minus r squared cosine of theta. Now we're going to integrate with respect to r. So when we integrate with respect to r, cosine of theta is just a constant. So if you integrate this with respect to r, you get, and I'll do it in that pink color, if you integrate with respect to r, the antiderivative of r is r squared over 2."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the outside. This inside part right over here, we can rewrite it as if we distribute the r, it's r minus r squared cosine of theta. Now we're going to integrate with respect to r. So when we integrate with respect to r, cosine of theta is just a constant. So if you integrate this with respect to r, you get, and I'll do it in that pink color, if you integrate with respect to r, the antiderivative of r is r squared over 2. So it's r squared over 2. And the antiderivative of r squared is r cubed over 3 minus r cubed over 3 cosine theta. This is just a constant."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you integrate this with respect to r, you get, and I'll do it in that pink color, if you integrate with respect to r, the antiderivative of r is r squared over 2. So it's r squared over 2. And the antiderivative of r squared is r cubed over 3 minus r cubed over 3 cosine theta. This is just a constant. r cubed over 3 cosine theta. And we're going to evaluate that from 0 to 1. So when you would evaluate it at 1, you get 1 half minus 1 third cosine theta."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just a constant. r cubed over 3 cosine theta. And we're going to evaluate that from 0 to 1. So when you would evaluate it at 1, you get 1 half minus 1 third cosine theta. So you get 1 half, I'll just do it right over here, 1 half minus 1 third cosine theta. And then minus both of these evaluated at 0, well, that's just going to be 0. 0 squared minus 0 squared times whatever."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when you would evaluate it at 1, you get 1 half minus 1 third cosine theta. So you get 1 half, I'll just do it right over here, 1 half minus 1 third cosine theta. And then minus both of these evaluated at 0, well, that's just going to be 0. 0 squared minus 0 squared times whatever. It's all going to be 0. So this business right over here just evaluates to 1 half minus 1 third cosine of theta. And so we get the integral."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "0 squared minus 0 squared times whatever. It's all going to be 0. So this business right over here just evaluates to 1 half minus 1 third cosine of theta. And so we get the integral. This is all going to be equal now to the square root of 2 times the integral from theta 0 to 2 pi of 1 half minus 1 third cosine of theta d theta. And this is equal to square root of 2 times the antiderivative of 1 half is 1 half 1 half theta. And the antiderivative of cosine theta is sine theta."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we get the integral. This is all going to be equal now to the square root of 2 times the integral from theta 0 to 2 pi of 1 half minus 1 third cosine of theta d theta. And this is equal to square root of 2 times the antiderivative of 1 half is 1 half 1 half theta. And the antiderivative of cosine theta is sine theta. So minus 1 third sine theta. And we're evaluating it from 0 to 2 pi. When you evaluate these at 2 pi, you have, let me just write it all out."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the antiderivative of cosine theta is sine theta. So minus 1 third sine theta. And we're evaluating it from 0 to 2 pi. When you evaluate these at 2 pi, you have, let me just write it all out. It's a home stretch. I don't want to make a careless mistake. We have square root of 2 times, let's evaluate it at 2 pi."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When you evaluate these at 2 pi, you have, let me just write it all out. It's a home stretch. I don't want to make a careless mistake. We have square root of 2 times, let's evaluate it at 2 pi. 1 half times 2 pi is pi minus 1 third times sine of 2 pi. Well, that's just going to be 0. And when you evaluate it at 0, 1 half times 0 is 0."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have square root of 2 times, let's evaluate it at 2 pi. 1 half times 2 pi is pi minus 1 third times sine of 2 pi. Well, that's just going to be 0. And when you evaluate it at 0, 1 half times 0 is 0. Sine of 0 is 0. So that all comes out to 0. So all of this business simplifies to pi."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And when you evaluate it at 0, 1 half times 0 is 0. Sine of 0 is 0. So that all comes out to 0. So all of this business simplifies to pi. And we are done. We have evaluated surface 3. It is square root of 2, or the third surface integral, or the surface integral over surface 3."}, {"video_title": "Surface integral ex3 part 4 Home stretch   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all of this business simplifies to pi. And we are done. We have evaluated surface 3. It is square root of 2, or the third surface integral, or the surface integral over surface 3. It is square root of 2 pi. And we are done. So this part right over here is square root of 2, or root 2, times pi."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "In the last video, I started introducing the intuition for the Laplacian operator in the context of the function with this graph and with the gradient field pictured below it, and here I'd like to go through the computation involved in that. So the function that I had there was defined, it's a two variable function, and it's defined as f of x, y is equal to three plus the cosine of x divided by two multiplied by the sine of y divided by two, y divided by two. And then the Laplacian, which we define with this right side up triangle, is an operator of f, and it's defined to be the divergence, so kind of this nabla dot, times the gradient, which is just nabla, of f. So two different things going on, it's kind of like a second derivative. And the first thing we need to do is take the gradient of f. And the way we do that, we kind of imagine expanding this upside down triangle as a vector full of partial differential operators, partial partial x and partial partial y. And with the gradient, you just kind of imagine multiplying that by the function. So if you imagine multiplying that by the function, what it looks like is just a vector full of partial derivatives. You're taking the partial of f with respect to x and the partial of f with respect to y."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "And the first thing we need to do is take the gradient of f. And the way we do that, we kind of imagine expanding this upside down triangle as a vector full of partial differential operators, partial partial x and partial partial y. And with the gradient, you just kind of imagine multiplying that by the function. So if you imagine multiplying that by the function, what it looks like is just a vector full of partial derivatives. You're taking the partial of f with respect to x and the partial of f with respect to y. Those are the two different components of this vector valued function that is the gradient. And in our specific example, when we take the partial derivative of f with respect to x, what we get, so we look over here, three just looks like a constant, so nothing happens. Cosine of x halves the derivative of that with respect to x, we kind of take out that 1 1.5."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "You're taking the partial of f with respect to x and the partial of f with respect to y. Those are the two different components of this vector valued function that is the gradient. And in our specific example, when we take the partial derivative of f with respect to x, what we get, so we look over here, three just looks like a constant, so nothing happens. Cosine of x halves the derivative of that with respect to x, we kind of take out that 1 1.5. So 1 1.5, and the derivative of cosine is negative sine. So that's negative sine of x over two. And sine of y over two, well, y just looks like a constant, so sine of y over two is just some other constant."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "Cosine of x halves the derivative of that with respect to x, we kind of take out that 1 1.5. So 1 1.5, and the derivative of cosine is negative sine. So that's negative sine of x over two. And sine of y over two, well, y just looks like a constant, so sine of y over two is just some other constant. So in our derivative, we just keep that constant in there, that sine of y over two. And then for the second component, the partial derivative of f with respect to y three still looks like a constant, because it is a constant. Now cosine of x over two looks like a constant, because as far as y is concerned, x is a constant, so cosine of x is a constant."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "And sine of y over two, well, y just looks like a constant, so sine of y over two is just some other constant. So in our derivative, we just keep that constant in there, that sine of y over two. And then for the second component, the partial derivative of f with respect to y three still looks like a constant, because it is a constant. Now cosine of x over two looks like a constant, because as far as y is concerned, x is a constant, so cosine of x is a constant. But then the sine of y has a derivative of cosine, and we also take out that 1 1.5. So you take out that 1 1.5 when you take the derivative of the inside, and then the derivative of the outside is cosine of whatever was in there. So in this case, y over two."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "Now cosine of x over two looks like a constant, because as far as y is concerned, x is a constant, so cosine of x is a constant. But then the sine of y has a derivative of cosine, and we also take out that 1 1.5. So you take out that 1 1.5 when you take the derivative of the inside, and then the derivative of the outside is cosine of whatever was in there. So in this case, y over two. And we're multiplying it by that original constant, cosine of x over two. So still we have our cosine of x over two, since it was a constant times a certain variable thing. X over two."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So in this case, y over two. And we're multiplying it by that original constant, cosine of x over two. So still we have our cosine of x over two, since it was a constant times a certain variable thing. X over two. So that's the gradient. And then the next step here is to take the divergence of that. So with the divergence, we're gonna imagine taking that del operator and dot-producting with this guy."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "X over two. So that's the gradient. And then the next step here is to take the divergence of that. So with the divergence, we're gonna imagine taking that del operator and dot-producting with this guy. So if I scroll down to give some room here, we're gonna take that vector that's kind of, the same vector, the partial partial x, and I say vector, but vector-ish thing, partial y. And now we're gonna take the dot-product with this entire guy. So I'll go ahead and just copy it over."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So with the divergence, we're gonna imagine taking that del operator and dot-producting with this guy. So if I scroll down to give some room here, we're gonna take that vector that's kind of, the same vector, the partial partial x, and I say vector, but vector-ish thing, partial y. And now we're gonna take the dot-product with this entire guy. So I'll go ahead and just copy it over. Just kind of copy it over here. And let's see. So we'll need a little bit more room to evaluate this."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So I'll go ahead and just copy it over. Just kind of copy it over here. And let's see. So we'll need a little bit more room to evaluate this. So here, when you imagine taking the dot-product, you kind of multiply these top components together. So we're taking the partial derivative with respect to x of this whole guy, and when you do that, what you get, you still have that 1 1\u20442, and then the derivative of negative sine of x over two. So that 1 1\u20442 gets pulled out when you're kind of taking the derivative of the inside, and the derivative of negative sine is negative cosine."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So we'll need a little bit more room to evaluate this. So here, when you imagine taking the dot-product, you kind of multiply these top components together. So we're taking the partial derivative with respect to x of this whole guy, and when you do that, what you get, you still have that 1 1\u20442, and then the derivative of negative sine of x over two. So that 1 1\u20442 gets pulled out when you're kind of taking the derivative of the inside, and the derivative of negative sine is negative cosine. So negative cosine of that stuff on the inside, that x over two. And of course, we still multiply it by this. This looks like a constant, the sine of y over two, and we multiply by that."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So that 1 1\u20442 gets pulled out when you're kind of taking the derivative of the inside, and the derivative of negative sine is negative cosine. So negative cosine of that stuff on the inside, that x over two. And of course, we still multiply it by this. This looks like a constant, the sine of y over two, and we multiply by that. Sine of y over two. And then we add that, because it's kind of like a dot-product, you add that to what it looks like when you multiply these next two components. So we're gonna add, and you have that 1 1\u20442, and then cosine of y over two."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "This looks like a constant, the sine of y over two, and we multiply by that. Sine of y over two. And then we add that, because it's kind of like a dot-product, you add that to what it looks like when you multiply these next two components. So we're gonna add, and you have that 1 1\u20442, and then cosine of y over two. When we differentiate that, you also pull out the 1 1\u20442. So again, you have that pulled out 1 1\u20442, and the derivative of cosine is negative sine. So now we're taking negative sine of, and then that stuff on the inside, y over two."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So we're gonna add, and you have that 1 1\u20442, and then cosine of y over two. When we differentiate that, you also pull out the 1 1\u20442. So again, you have that pulled out 1 1\u20442, and the derivative of cosine is negative sine. So now we're taking negative sine of, and then that stuff on the inside, y over two. And we continue multiplying by the constant. As far as y is concerned, cosine of x over two is a constant, so we multiply it by that. Cosine of x over two."}, {"video_title": "Laplacian computation example.mp3", "Sentence": "So now we're taking negative sine of, and then that stuff on the inside, y over two. And we continue multiplying by the constant. As far as y is concerned, cosine of x over two is a constant, so we multiply it by that. Cosine of x over two. And then that, so that is the divergence of that gradient field. So the divergence of the gradient of our original function gives us the Laplacian. And in fact, we could simplify this further, because both of these terms kind of look identical."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "So in the last video, I introduced the vector form of the multivariable chain rule. And just to remind ourselves, I'm saying you have some kind of function f, and in this case I said it comes from a 100 dimensional space. So you might imagine, well, I can't imagine a 100 dimensional space, but in principle, you're just thinking of some area that's 100 dimensions. It could be two if you wanted to think more concretely, two dimensions. And it's a scalar valued function, so it just outputs to a number line, some kind of number line that I'll think of as f as its output. And what we're gonna do is we compose it with a vector valued function. So some function that takes in a single number t, and then it outputs into that super high dimensional space."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "It could be two if you wanted to think more concretely, two dimensions. And it's a scalar valued function, so it just outputs to a number line, some kind of number line that I'll think of as f as its output. And what we're gonna do is we compose it with a vector valued function. So some function that takes in a single number t, and then it outputs into that super high dimensional space. So you're thinking, you go from the single variable t to some very high dimensional space that we think of as full of vectors, and then you take from that over to a single variable, over to a number. And you know, the way you'd write that out is you'd say f composed with the output of v. So f composed with v of t. And what we're interested in doing is taking its derivative. So the derivative of that composition is, and I told you, and we kind of walked through where this comes from, the gradient of f evaluated at v of t, evaluated at your original output, dot product with the derivative of v, the vectorized derivative."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "So some function that takes in a single number t, and then it outputs into that super high dimensional space. So you're thinking, you go from the single variable t to some very high dimensional space that we think of as full of vectors, and then you take from that over to a single variable, over to a number. And you know, the way you'd write that out is you'd say f composed with the output of v. So f composed with v of t. And what we're interested in doing is taking its derivative. So the derivative of that composition is, and I told you, and we kind of walked through where this comes from, the gradient of f evaluated at v of t, evaluated at your original output, dot product with the derivative of v, the vectorized derivative. And what that means, you know, for v, you're just taking the derivative of every component. So when you take this and you take the derivative with respect to t, all that means is that each component, you're taking the derivative of it. The x1 dt, the x2 dt, on and on until d, and then the 100th component, dt."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "So the derivative of that composition is, and I told you, and we kind of walked through where this comes from, the gradient of f evaluated at v of t, evaluated at your original output, dot product with the derivative of v, the vectorized derivative. And what that means, you know, for v, you're just taking the derivative of every component. So when you take this and you take the derivative with respect to t, all that means is that each component, you're taking the derivative of it. The x1 dt, the x2 dt, on and on until d, and then the 100th component, dt. So this was the vectorized form of the multivariable chain rule. And what I want to do here is show how this looks a lot like a directional derivative. And if you haven't watched the video on the directional derivative, maybe go back, take a look, kind of remind yourself."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "The x1 dt, the x2 dt, on and on until d, and then the 100th component, dt. So this was the vectorized form of the multivariable chain rule. And what I want to do here is show how this looks a lot like a directional derivative. And if you haven't watched the video on the directional derivative, maybe go back, take a look, kind of remind yourself. But in principle, you say if you're in the input space of f and you nudge yourself along some kind of vector v, and maybe just because I'm using v there, I'll instead say some kind of vector w. So not a function, just a vector. And you're wondering, hey, how much does that result in a change to the output of f? That's answered by the directional derivative."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "And if you haven't watched the video on the directional derivative, maybe go back, take a look, kind of remind yourself. But in principle, you say if you're in the input space of f and you nudge yourself along some kind of vector v, and maybe just because I'm using v there, I'll instead say some kind of vector w. So not a function, just a vector. And you're wondering, hey, how much does that result in a change to the output of f? That's answered by the directional derivative. And you write directional derivative in the direction of w of f, the directional derivative of f, and I should say at some point, some input point, p for that input point, and it's a vector in this case, like a 100-dimensional vector. And the way you evaluate it is you take the gradient of f, this is why we use the Nabla notation in the first place, it's an indicative of how we compute it, the gradient of f, evaluated at that same input point, same input point, same input vector p. So here, just to be clear, you'd be thinking of whatever vector to your input point, that's p. But then the nudge, the nudge away from that input point is w. And you take the dot product between that and the vector itself, the vector that represents your nudge direction. But that looks a lot like the multivariable chain rule up here, except instead of w, you're taking the derivative, the vector value derivative of v. So this whole thing, you could say, is the directional derivative in the direction of the derivative of t, and it's kind of confusing, directional derivative in the direction of a derivative of f. And at what point are you taking this?"}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "That's answered by the directional derivative. And you write directional derivative in the direction of w of f, the directional derivative of f, and I should say at some point, some input point, p for that input point, and it's a vector in this case, like a 100-dimensional vector. And the way you evaluate it is you take the gradient of f, this is why we use the Nabla notation in the first place, it's an indicative of how we compute it, the gradient of f, evaluated at that same input point, same input point, same input vector p. So here, just to be clear, you'd be thinking of whatever vector to your input point, that's p. But then the nudge, the nudge away from that input point is w. And you take the dot product between that and the vector itself, the vector that represents your nudge direction. But that looks a lot like the multivariable chain rule up here, except instead of w, you're taking the derivative, the vector value derivative of v. So this whole thing, you could say, is the directional derivative in the direction of the derivative of t, and it's kind of confusing, directional derivative in the direction of a derivative of f. And at what point are you taking this? At what point are you taking this directional derivative? Well, it's wherever the output of v is. So this is very compact, it's saying quite a bit here, but a way that you could be thinking about this is v of t, I'm gonna kind of erase here, v of t, as you're zooming all about and as you shift t, it kind of moves you through this space in some way."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "But that looks a lot like the multivariable chain rule up here, except instead of w, you're taking the derivative, the vector value derivative of v. So this whole thing, you could say, is the directional derivative in the direction of the derivative of t, and it's kind of confusing, directional derivative in the direction of a derivative of f. And at what point are you taking this? At what point are you taking this directional derivative? Well, it's wherever the output of v is. So this is very compact, it's saying quite a bit here, but a way that you could be thinking about this is v of t, I'm gonna kind of erase here, v of t, as you're zooming all about and as you shift t, it kind of moves you through this space in some way. And each one of these output points here represents the vector, v of t at some point. The derivative of that, what does this derivative represent? That's the tangent vector to that motion."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "So this is very compact, it's saying quite a bit here, but a way that you could be thinking about this is v of t, I'm gonna kind of erase here, v of t, as you're zooming all about and as you shift t, it kind of moves you through this space in some way. And each one of these output points here represents the vector, v of t at some point. The derivative of that, what does this derivative represent? That's the tangent vector to that motion. So you're zipping about through that space, the tangent vector to your motion, that's how we interpret v prime of t, the derivative of v with respect to t. And why should that make sense? Why should the directional derivative in the direction of v prime of t, this change to the intermediary function v, have anything to do with the multivariable chain rule? Well, remember what we're asking when we say d dt of this composition, is we're saying we take a tiny nudge to t, so that tiny change here in the value t, and we're wondering what change that results in after the composition."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "That's the tangent vector to that motion. So you're zipping about through that space, the tangent vector to your motion, that's how we interpret v prime of t, the derivative of v with respect to t. And why should that make sense? Why should the directional derivative in the direction of v prime of t, this change to the intermediary function v, have anything to do with the multivariable chain rule? Well, remember what we're asking when we say d dt of this composition, is we're saying we take a tiny nudge to t, so that tiny change here in the value t, and we're wondering what change that results in after the composition. Well, at a given point, that tiny nudge in t causes a change in the direction of v prime of t. That's kind of the whole meaning of this vector value derivative. You change t by a little bit, and that's gonna tell you how you move in the output space. But then you say, okay, so I've moved a little bit in this intermediary, 100-dimensional space."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "Well, remember what we're asking when we say d dt of this composition, is we're saying we take a tiny nudge to t, so that tiny change here in the value t, and we're wondering what change that results in after the composition. Well, at a given point, that tiny nudge in t causes a change in the direction of v prime of t. That's kind of the whole meaning of this vector value derivative. You change t by a little bit, and that's gonna tell you how you move in the output space. But then you say, okay, so I've moved a little bit in this intermediary, 100-dimensional space. How does that influence the output of f based on the behavior of just the multivariable function f? Well, that's what the directional derivative is asking. It says you take a nudge in the direction of some vector."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "But then you say, okay, so I've moved a little bit in this intermediary, 100-dimensional space. How does that influence the output of f based on the behavior of just the multivariable function f? Well, that's what the directional derivative is asking. It says you take a nudge in the direction of some vector. In this case, I wrote v prime of t over here. More generally, you could say any vector w, you take a nudge in that direction. And more importantly, the size of v prime of t matters here."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "It says you take a nudge in the direction of some vector. In this case, I wrote v prime of t over here. More generally, you could say any vector w, you take a nudge in that direction. And more importantly, the size of v prime of t matters here. If you're moving really quickly, you would expect that change to be larger. So the fact that v prime of t would be larger is helpful. And the directional derivative is telling you the size of the change in f as a ratio of the proportion of that directional vector that you went along."}, {"video_title": "Multivariable chain rule and directional derivatives.mp3", "Sentence": "And more importantly, the size of v prime of t matters here. If you're moving really quickly, you would expect that change to be larger. So the fact that v prime of t would be larger is helpful. And the directional derivative is telling you the size of the change in f as a ratio of the proportion of that directional vector that you went along. Another notation for the directional derivative is to say partial f and then partial whatever that vector is basically saying you take a size of that nudge along that vector as a proportion of the vector itself, and then you consider the change to the output and you're taking the ratio. So I think this is a very beautiful way of understanding the multivariable chain rule because it gives this image of, you know, you're thinking of v of t and you're thinking of zipping along in some way, and the direction and value of your velocity as you zip along is what determines the change in the output of the function f. So hopefully that helps give a better understanding both of the directional derivative and of the multivariable chain rule. It's one of those nice little interpretations."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And before we jump into it, I just want to give a quick review of how you think about the determinant itself, just in an ordinary linear algebra context. So if I'm taking the determinant of some kind of matrix, let's say, three, zero, one, two, one, two, something like this. To compute the determinant, you take these diagonal terms here. So you take three multiplied by that two, and then you subtract off the other diagonal. Subtract off one multiplied by zero. And in this case, that evaluates to six. But there is, of course, much more than just a computation going on here."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "So you take three multiplied by that two, and then you subtract off the other diagonal. Subtract off one multiplied by zero. And in this case, that evaluates to six. But there is, of course, much more than just a computation going on here. There's a really nice geometric intuition. Namely, if we think of this matrix, three, zero, one, two, as a linear transformation, as something that's going to take this first basis vector over to the coordinates three, zero, and that second basis vector over to the coordinates one, two, you know, thinking about the columns, you can think of the determinant as measuring how much this transformation stretches or squishes space. And in particular, you'll notice how I have this yellow region highlighted."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "But there is, of course, much more than just a computation going on here. There's a really nice geometric intuition. Namely, if we think of this matrix, three, zero, one, two, as a linear transformation, as something that's going to take this first basis vector over to the coordinates three, zero, and that second basis vector over to the coordinates one, two, you know, thinking about the columns, you can think of the determinant as measuring how much this transformation stretches or squishes space. And in particular, you'll notice how I have this yellow region highlighted. And this region starts off as the unit square, a square with side lengths one, so its area is one. And there's nothing special about this particular region. It's just nice as a canonical shape with an area of one so that we can compare it to what happens after the transformation."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And in particular, you'll notice how I have this yellow region highlighted. And this region starts off as the unit square, a square with side lengths one, so its area is one. And there's nothing special about this particular region. It's just nice as a canonical shape with an area of one so that we can compare it to what happens after the transformation. Ask, how much does that area get stretched out? And the answer is, it gets stretched out by a factor of the determinant. That's kind of what the determinant means, is that all areas, if you were to drop any kind of shape, not just that one square, are gonna get stretched out by a factor of six."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "It's just nice as a canonical shape with an area of one so that we can compare it to what happens after the transformation. Ask, how much does that area get stretched out? And the answer is, it gets stretched out by a factor of the determinant. That's kind of what the determinant means, is that all areas, if you were to drop any kind of shape, not just that one square, are gonna get stretched out by a factor of six. And we can actually verify, looking at this parallelogram that the square turned into, it has a base of three, and then the height is two. And three times two is six. And that has everything to do with the fact that this three showed up here and this two showed up there."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "That's kind of what the determinant means, is that all areas, if you were to drop any kind of shape, not just that one square, are gonna get stretched out by a factor of six. And we can actually verify, looking at this parallelogram that the square turned into, it has a base of three, and then the height is two. And three times two is six. And that has everything to do with the fact that this three showed up here and this two showed up there. So now, let's think about what this might mean in the context of what I've been describing in the last couple videos. And if you'll remember, we had a multivariable function, something that you can write out as F1 with two inputs, and then the second component, F2, also with two inputs. And the function that I was looking at, that we were kind of analyzing to learn about the Jacobian, had the first component, X plus sine of Y. X plus sine Y."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And that has everything to do with the fact that this three showed up here and this two showed up there. So now, let's think about what this might mean in the context of what I've been describing in the last couple videos. And if you'll remember, we had a multivariable function, something that you can write out as F1 with two inputs, and then the second component, F2, also with two inputs. And the function that I was looking at, that we were kind of analyzing to learn about the Jacobian, had the first component, X plus sine of Y. X plus sine Y. And the second component was Y plus the sine of X. And the idea was that this function is not at all linear. It's gonna make everything very curvy and complicated."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And the function that I was looking at, that we were kind of analyzing to learn about the Jacobian, had the first component, X plus sine of Y. X plus sine Y. And the second component was Y plus the sine of X. And the idea was that this function is not at all linear. It's gonna make everything very curvy and complicated. However, if we zoom in around a particular region, which is what this outer yellow box represents, zooming in, it will look like a linear transformation. In fact, I can kind of play this forward, and we see that even though everything is crazy, inside that zoomed-in version, things loosely look like a linear function. And you'll notice I have this inner yellow box highlighted."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "It's gonna make everything very curvy and complicated. However, if we zoom in around a particular region, which is what this outer yellow box represents, zooming in, it will look like a linear transformation. In fact, I can kind of play this forward, and we see that even though everything is crazy, inside that zoomed-in version, things loosely look like a linear function. And you'll notice I have this inner yellow box highlighted. And this yellow box inside corresponds to the unit square that I was showing in the last animation. And again, it's just a placeholder as something to watch to see how much the area of any kind of blob in that region gets stretched. So in this particular case, when you play out the animation, areas don't really change that much."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And you'll notice I have this inner yellow box highlighted. And this yellow box inside corresponds to the unit square that I was showing in the last animation. And again, it's just a placeholder as something to watch to see how much the area of any kind of blob in that region gets stretched. So in this particular case, when you play out the animation, areas don't really change that much. They get stretched out a little bit, but it's not that dramatic. So if we know the matrix that describes the transformation that this looks like zoomed in, the determinant of that matrix will tell us the factor by which areas tend to get stretched out. And in particular, you can think of this little yellow box and the factor by which it gets stretched."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "So in this particular case, when you play out the animation, areas don't really change that much. They get stretched out a little bit, but it's not that dramatic. So if we know the matrix that describes the transformation that this looks like zoomed in, the determinant of that matrix will tell us the factor by which areas tend to get stretched out. And in particular, you can think of this little yellow box and the factor by which it gets stretched. And as a reminder, the matrix describing that zoomed-in transformation is the Jacobian. It is this thing that kind of holds all of the partial differential information. You take the partial derivative of f with respect to x, sorry, partial of f1 of that first component, and then the partial derivative of the second component with respect to x."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And in particular, you can think of this little yellow box and the factor by which it gets stretched. And as a reminder, the matrix describing that zoomed-in transformation is the Jacobian. It is this thing that kind of holds all of the partial differential information. You take the partial derivative of f with respect to x, sorry, partial of f1 of that first component, and then the partial derivative of the second component with respect to x. And then on the other column, we have the partial derivative of that first component with respect to y, and the partial derivative of that second component with respect to y. And if you, let's see, we'll close this off, close off this matrix. And if you evaluate each one of these partial derivatives at a particular point, at whatever point we happen to zoom in on, in this case it was negative two, one, once you plug that into all of these, you get some matrix that's just full of numbers."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "You take the partial derivative of f with respect to x, sorry, partial of f1 of that first component, and then the partial derivative of the second component with respect to x. And then on the other column, we have the partial derivative of that first component with respect to y, and the partial derivative of that second component with respect to y. And if you, let's see, we'll close this off, close off this matrix. And if you evaluate each one of these partial derivatives at a particular point, at whatever point we happen to zoom in on, in this case it was negative two, one, once you plug that into all of these, you get some matrix that's just full of numbers. And what turns out to be a very useful thing later on in multivariable calc concepts is to take the determinant of that matrix, to kind of analyze how much space is getting stretched or squished in that region. So in the last video, we worked this out for the specific example here, where that top left function turned out just to be the constant function one, right? Because we were taking the partial derivative of this guy with respect to x, and that was one."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And if you evaluate each one of these partial derivatives at a particular point, at whatever point we happen to zoom in on, in this case it was negative two, one, once you plug that into all of these, you get some matrix that's just full of numbers. And what turns out to be a very useful thing later on in multivariable calc concepts is to take the determinant of that matrix, to kind of analyze how much space is getting stretched or squished in that region. So in the last video, we worked this out for the specific example here, where that top left function turned out just to be the constant function one, right? Because we were taking the partial derivative of this guy with respect to x, and that was one. And likewise, in the bottom right, that was also a constant function of one. And then the others were cosine functions. This one was cosine x, because we were taking the partial derivative of this second component here with respect to x."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "Because we were taking the partial derivative of this guy with respect to x, and that was one. And likewise, in the bottom right, that was also a constant function of one. And then the others were cosine functions. This one was cosine x, because we were taking the partial derivative of this second component here with respect to x. And then the top right of our matrix was cosine of y. And these are in general functions of x and y, because you're gonna plug in whatever the input point you're zooming in on. And when we're thinking about the determinant here, let's just go ahead and take the determinant in this form, in the form as a function."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "This one was cosine x, because we were taking the partial derivative of this second component here with respect to x. And then the top right of our matrix was cosine of y. And these are in general functions of x and y, because you're gonna plug in whatever the input point you're zooming in on. And when we're thinking about the determinant here, let's just go ahead and take the determinant in this form, in the form as a function. So I'm gonna ask about the determinant of this matrix, or maybe you think of it as a matrix-valued function. And in this case, we do the same thing. I mean, procedurally, you know how to take a determinant."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And when we're thinking about the determinant here, let's just go ahead and take the determinant in this form, in the form as a function. So I'm gonna ask about the determinant of this matrix, or maybe you think of it as a matrix-valued function. And in this case, we do the same thing. I mean, procedurally, you know how to take a determinant. We take these diagonals, so that's just gonna be one times one, and then we subtract off the product of the other diagonal. Subtract off cosine of x, multiplied by cosine of y. And as an example, let's plug in this point here that we're zooming in on, negative two, one."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "I mean, procedurally, you know how to take a determinant. We take these diagonals, so that's just gonna be one times one, and then we subtract off the product of the other diagonal. Subtract off cosine of x, multiplied by cosine of y. And as an example, let's plug in this point here that we're zooming in on, negative two, one. So I'm gonna plug in x is equal to negative two, and y is equal to one. And when you plug in cosine of negative two, that's gonna come out to be approximately negative 0.42. And when you plug in cosine of y, cosine of one in this case, that's gonna come out to be about 0.54."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And as an example, let's plug in this point here that we're zooming in on, negative two, one. So I'm gonna plug in x is equal to negative two, and y is equal to one. And when you plug in cosine of negative two, that's gonna come out to be approximately negative 0.42. And when you plug in cosine of y, cosine of one in this case, that's gonna come out to be about 0.54. And when we multiply those, we can take one minus the product of those. It's gonna be about negative 0.227. And that's all stuff that you can plug into your calculator if you want."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And when you plug in cosine of y, cosine of one in this case, that's gonna come out to be about 0.54. And when we multiply those, we can take one minus the product of those. It's gonna be about negative 0.227. And that's all stuff that you can plug into your calculator if you want. And what that means is that the total determinant evaluated at that point, the Jacobian determinant, at the point negative two, one, is about 1.0, sorry, 1.227. So that's telling you that areas tend to get stretched out by this factor around that point. And that kind of lines up with what we see."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And that's all stuff that you can plug into your calculator if you want. And what that means is that the total determinant evaluated at that point, the Jacobian determinant, at the point negative two, one, is about 1.0, sorry, 1.227. So that's telling you that areas tend to get stretched out by this factor around that point. And that kind of lines up with what we see. We see that areas get stretched out maybe a little bit, but not that much, right? It's only by a factor of about 1.2. And now let's contrast this if instead we zoom in at the point where x is equal to zero and y is equal to one."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And that kind of lines up with what we see. We see that areas get stretched out maybe a little bit, but not that much, right? It's only by a factor of about 1.2. And now let's contrast this if instead we zoom in at the point where x is equal to zero and y is equal to one. So I'm gonna go over here, and all I'm gonna change, all I'm gonna change is that x is equal to zero and y will still equal one. And what that means is that cosine of x, instead of being negative 0.42, instead of what's cosine of zero, that's actually precisely equal to one. We don't have to approximate on this one."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "And now let's contrast this if instead we zoom in at the point where x is equal to zero and y is equal to one. So I'm gonna go over here, and all I'm gonna change, all I'm gonna change is that x is equal to zero and y will still equal one. And what that means is that cosine of x, instead of being negative 0.42, instead of what's cosine of zero, that's actually precisely equal to one. We don't have to approximate on this one. Which means when we multiply them, one times 0.54, well that, that's gonna now be about 0.54. So this one, once we actually perform the subtraction, instead when you take one minus 0.54, that's gonna give us 0.46. So even before watching, because this determinant of the Jacobian around the point zero one is less than one, this is telling us we should expect areas to get squished down."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "We don't have to approximate on this one. Which means when we multiply them, one times 0.54, well that, that's gonna now be about 0.54. So this one, once we actually perform the subtraction, instead when you take one minus 0.54, that's gonna give us 0.46. So even before watching, because this determinant of the Jacobian around the point zero one is less than one, this is telling us we should expect areas to get squished down. Precisely, they should be squished by a factor of 0.46. And let's see if this looks right, right? We're looking at the zoomed in version around that point, and areas should tend to contract around that."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "So even before watching, because this determinant of the Jacobian around the point zero one is less than one, this is telling us we should expect areas to get squished down. Precisely, they should be squished by a factor of 0.46. And let's see if this looks right, right? We're looking at the zoomed in version around that point, and areas should tend to contract around that. And indeed they do. You see it got squished down, it looks like by a fair bit. And from our calculation, we can conclude that they got scaled down precisely by a factor of 0.46."}, {"video_title": "The Jacobian Determinant.mp3", "Sentence": "We're looking at the zoomed in version around that point, and areas should tend to contract around that. And indeed they do. You see it got squished down, it looks like by a fair bit. And from our calculation, we can conclude that they got scaled down precisely by a factor of 0.46. That's what the determinant means. So like I said, this is actually a very nice notion throughout multivariable calculus, is that you look at a tiny little local neighborhood around a point, and if you just wanna get a general feel for does this function as a transformation tend to stretch out that region or to squish it together, you know, how much do areas change in that little neighborhood, that's exactly what this Jacobian determinant is, you know, built to solve. So with that, I'll see you guys next video."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Although this should hopefully be second nature to you at this point. If it's not, you might want to review the definite integration videos. But if I have some function, this is the xy-plane, that's the x-axis, that's the y-axis. And I have some function, let's call that, you know, this is y is equal to some function of x. You give me an x and I'll give you a y. If I wanted to figure out the area under this curve between, let's say, x is equal to a and x is equal to b. So this is the area I want to figure out."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I have some function, let's call that, you know, this is y is equal to some function of x. You give me an x and I'll give you a y. If I wanted to figure out the area under this curve between, let's say, x is equal to a and x is equal to b. So this is the area I want to figure out. This area right here. What I do is I split it up into a bunch of columns or a bunch of rectangles. Let me draw one of those rectangles where you could view."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the area I want to figure out. This area right here. What I do is I split it up into a bunch of columns or a bunch of rectangles. Let me draw one of those rectangles where you could view. And there's different ways to do this, but this is just a review. Where you could review, that's maybe one of the rectangles. The area of the rectangle is just base times height, right?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw one of those rectangles where you could view. And there's different ways to do this, but this is just a review. Where you could review, that's maybe one of the rectangles. The area of the rectangle is just base times height, right? We're going to make these rectangles really skinny and just sum up an infinite number of them. So we want to make them infinitely small, but let's just call the base of this rectangle dx. And then the height of this rectangle is going to be f of x at that point."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The area of the rectangle is just base times height, right? We're going to make these rectangles really skinny and just sum up an infinite number of them. So we want to make them infinitely small, but let's just call the base of this rectangle dx. And then the height of this rectangle is going to be f of x at that point. It's going to be f of, if this is x naught or whatever, you could just call it f of x. That's the height of that rectangle. And if we wanted to take the sum of all of these rectangles, there's just going to be a bunch of them."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the height of this rectangle is going to be f of x at that point. It's going to be f of, if this is x naught or whatever, you could just call it f of x. That's the height of that rectangle. And if we wanted to take the sum of all of these rectangles, there's just going to be a bunch of them. One there, one there. Then we'll get the area. And if we have infinite number of these rectangles and they're infinitely skinny, we have exactly the area under that curve."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we wanted to take the sum of all of these rectangles, there's just going to be a bunch of them. One there, one there. Then we'll get the area. And if we have infinite number of these rectangles and they're infinitely skinny, we have exactly the area under that curve. That's the intuition behind the definite integral. And the way we write that, it's the definite integral. We're going to take the sums of these rectangles from x is equal to a to x is equal to b."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we have infinite number of these rectangles and they're infinitely skinny, we have exactly the area under that curve. That's the intuition behind the definite integral. And the way we write that, it's the definite integral. We're going to take the sums of these rectangles from x is equal to a to x is equal to b. And the sum, or the areas that we're summing up, are going to be the height is f of x and the width is d of x. It's going to be f of x times d of x. This is equal to the area under the curve, f of x, y is equal to f of x, from x is equal to a to x is equal to b."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to take the sums of these rectangles from x is equal to a to x is equal to b. And the sum, or the areas that we're summing up, are going to be the height is f of x and the width is d of x. It's going to be f of x times d of x. This is equal to the area under the curve, f of x, y is equal to f of x, from x is equal to a to x is equal to b. And that's just a little bit of review. But hopefully you'll now see the parallel of how we extend this to taking the volume under a surface. First of all, what is a surface?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to the area under the curve, f of x, y is equal to f of x, from x is equal to a to x is equal to b. And that's just a little bit of review. But hopefully you'll now see the parallel of how we extend this to taking the volume under a surface. First of all, what is a surface? If we're thinking in three dimensions, a surface is going to be a function of x and y. So we can write a surface as, instead of saying that y is a function of x, we can write a surface as z is equal to a function of x and y. So you can kind of view it as the domain."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "First of all, what is a surface? If we're thinking in three dimensions, a surface is going to be a function of x and y. So we can write a surface as, instead of saying that y is a function of x, we can write a surface as z is equal to a function of x and y. So you can kind of view it as the domain. The domain is all of the set of valid things that you can input into a function. So now, before our domain was just, at least for most of what we dealt with, was just the x-axis, or kind of the real number line in the x direction. Now, our domain is the xy-plane."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can kind of view it as the domain. The domain is all of the set of valid things that you can input into a function. So now, before our domain was just, at least for most of what we dealt with, was just the x-axis, or kind of the real number line in the x direction. Now, our domain is the xy-plane. We can give it any x and any y, and we'll just deal with reals right now, but I don't want to get too technical. And then it'll pop out another number, and if we wanted to graph it, it'll be our height. And so that could be the height of a surface."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, our domain is the xy-plane. We can give it any x and any y, and we'll just deal with reals right now, but I don't want to get too technical. And then it'll pop out another number, and if we wanted to graph it, it'll be our height. And so that could be the height of a surface. So let me just show you what a surface looks like, in case you don't remember, and we'll actually figure out the volume under this surface. So this is a surface. I'll tell you its function in a second, but it's pretty neat to look at."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so that could be the height of a surface. So let me just show you what a surface looks like, in case you don't remember, and we'll actually figure out the volume under this surface. So this is a surface. I'll tell you its function in a second, but it's pretty neat to look at. But as you can see, it's a surface. It's like a piece of paper that's bent. Let me rotate it to its traditional form."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll tell you its function in a second, but it's pretty neat to look at. But as you can see, it's a surface. It's like a piece of paper that's bent. Let me rotate it to its traditional form. So this is the x direction, this is the y direction, and the height is a function of where we are in the xy-plane. So how do we figure out the volume under a surface like this? How do we figure out the volume?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me rotate it to its traditional form. So this is the x direction, this is the y direction, and the height is a function of where we are in the xy-plane. So how do we figure out the volume under a surface like this? How do we figure out the volume? It seems like a bit of a stretch, given what we've learned from this. But what if, and I'm just going to draw an abstract surface here, let me draw some axes. Let's say that's my x-axis, that's my y-axis, that's my z-axis."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "How do we figure out the volume? It seems like a bit of a stretch, given what we've learned from this. But what if, and I'm just going to draw an abstract surface here, let me draw some axes. Let's say that's my x-axis, that's my y-axis, that's my z-axis. I practice these videos ahead of time, so I'm often wondering what I'm about to draw. So that's x, that's y, and that's z. And let's say I have some surface."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that's my x-axis, that's my y-axis, that's my z-axis. I practice these videos ahead of time, so I'm often wondering what I'm about to draw. So that's x, that's y, and that's z. And let's say I have some surface. I'll just draw something. I don't know what it is. Some surface."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say I have some surface. I'll just draw something. I don't know what it is. Some surface. This is our surface. z is a function of x and y. So if you give me a coordinate in the xy-plane, say here, I'll put it into the function and it'll give us a z value."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Some surface. This is our surface. z is a function of x and y. So if you give me a coordinate in the xy-plane, say here, I'll put it into the function and it'll give us a z value. And I'll plot it there and it'll be a point on the surface. So what we want to figure out is the volume under the surface, and we have to specify bounds. From here we said x is equal to a to x is equal to b."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you give me a coordinate in the xy-plane, say here, I'll put it into the function and it'll give us a z value. And I'll plot it there and it'll be a point on the surface. So what we want to figure out is the volume under the surface, and we have to specify bounds. From here we said x is equal to a to x is equal to b. So let's make a square bound first, because this keeps it a lot simpler. So let's say that the region of the x and y region of this part of the surface under which we want to calculate the volume, let's say the shadow, if the sun was right above the surface, the shadow would be right there. Let me try my best to draw this neatly."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "From here we said x is equal to a to x is equal to b. So let's make a square bound first, because this keeps it a lot simpler. So let's say that the region of the x and y region of this part of the surface under which we want to calculate the volume, let's say the shadow, if the sun was right above the surface, the shadow would be right there. Let me try my best to draw this neatly. So this is what we're going to try to figure out the volume of. So if we wanted to draw it in the xy-plane, you can kind of view it as the projection of the surface in the xy-plane or the shadow of the surface in the xy-plane. What are the bounds?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me try my best to draw this neatly. So this is what we're going to try to figure out the volume of. So if we wanted to draw it in the xy-plane, you can kind of view it as the projection of the surface in the xy-plane or the shadow of the surface in the xy-plane. What are the bounds? You can almost view it as what are the bounds of the domain? Let's say that this right here, that's 0, 0 in the xy-plane. Let's say that this is y is equal to a."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What are the bounds? You can almost view it as what are the bounds of the domain? Let's say that this right here, that's 0, 0 in the xy-plane. Let's say that this is y is equal to a. That's this line right here, y is equal to a. And let's say that this line right here is x is equal to b. I hope you get that. This is the xy-plane."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that this is y is equal to a. That's this line right here, y is equal to a. And let's say that this line right here is x is equal to b. I hope you get that. This is the xy-plane. If we have a constant x, it would be a line like that, a constant y, a line like that, and then we have the area in between it. So how do we figure out the volume under this? Well, if I just wanted to figure out the area of, let's just say, this sliver."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the xy-plane. If we have a constant x, it would be a line like that, a constant y, a line like that, and then we have the area in between it. So how do we figure out the volume under this? Well, if I just wanted to figure out the area of, let's just say, this sliver. Let's say we had a constant y. So let's say I had some sliver. I don't want to confuse you."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, if I just wanted to figure out the area of, let's just say, this sliver. Let's say we had a constant y. So let's say I had some sliver. I don't want to confuse you. Let's say that I have some constant y. I just want to give you the intuition. Let's say, I don't know what that is, it's an arbitrary y. But for some constant y, what if I could just figure out the area under the curve there?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I don't want to confuse you. Let's say that I have some constant y. I just want to give you the intuition. Let's say, I don't know what that is, it's an arbitrary y. But for some constant y, what if I could just figure out the area under the curve there? How would I figure out just the area under that curve? It'll be a function of which y I pick, right? Because if I pick a y here, it'll be a different area."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But for some constant y, what if I could just figure out the area under the curve there? How would I figure out just the area under that curve? It'll be a function of which y I pick, right? Because if I pick a y here, it'll be a different area. If I pick a y there, it'll be a different area. But I could view this now as a very similar problem to this one up here. I could have my dx's."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because if I pick a y here, it'll be a different area. If I pick a y there, it'll be a different area. But I could view this now as a very similar problem to this one up here. I could have my dx's. Let me pick a vibrant color so you can see it. Let's say that's a dx, right? That's a change in x."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I could have my dx's. Let me pick a vibrant color so you can see it. Let's say that's a dx, right? That's a change in x. And then the height, this height, is going to be a function of the x I have and the y I've picked. Although I'm assuming to some degree that that's a constant y. So what would be the area of this sheet of paper?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a change in x. And then the height, this height, is going to be a function of the x I have and the y I've picked. Although I'm assuming to some degree that that's a constant y. So what would be the area of this sheet of paper? It's kind of a constant y. It's a sheet of paper within this volume. You can kind of view it."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what would be the area of this sheet of paper? It's kind of a constant y. It's a sheet of paper within this volume. You can kind of view it. Well, we said the height of each of these rectangles is f of xy. That's the height. It depends which x and y we pick down here."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can kind of view it. Well, we said the height of each of these rectangles is f of xy. That's the height. It depends which x and y we pick down here. And then its width is going to be dx. And then if we integrated it from x is equal to 0, which was back here, all the way to x is equal to b, what would it look like? It would look like that."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It depends which x and y we pick down here. And then its width is going to be dx. And then if we integrated it from x is equal to 0, which was back here, all the way to x is equal to b, what would it look like? It would look like that. x is going from 0 to b. Fair enough. And this would actually give us a function of y."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would look like that. x is going from 0 to b. Fair enough. And this would actually give us a function of y. This would give us an expression so that if I would know the area of this kind of sliver of the volume for any given value of y. If you give me a y, I can tell you the area of the sliver that corresponds to that y. Now what could I do?"}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this would actually give us a function of y. This would give us an expression so that if I would know the area of this kind of sliver of the volume for any given value of y. If you give me a y, I can tell you the area of the sliver that corresponds to that y. Now what could I do? If I know the area of any given sliver, what if I multiplied the area of that sliver times dy? This is a dy. Let me do it in a vibrant color."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now what could I do? If I know the area of any given sliver, what if I multiplied the area of that sliver times dy? This is a dy. Let me do it in a vibrant color. So dy, a very small change in y. If I multiply this area times a small dy, then all of a sudden I have a sliver of volume. Hopefully that makes some sense."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me do it in a vibrant color. So dy, a very small change in y. If I multiply this area times a small dy, then all of a sudden I have a sliver of volume. Hopefully that makes some sense. I'm making that little cut that I took the area of, I'm making it three dimensional. So what would be the volume of that sliver? The volume of that sliver will be this function of y times dy or this whole thing times dy."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully that makes some sense. I'm making that little cut that I took the area of, I'm making it three dimensional. So what would be the volume of that sliver? The volume of that sliver will be this function of y times dy or this whole thing times dy. So it would be the integral from 0 to b of f of x, y dx. That gives us the area of this blue sheet. Now if I multiply this whole thing times dy, I get this volume."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The volume of that sliver will be this function of y times dy or this whole thing times dy. So it would be the integral from 0 to b of f of x, y dx. That gives us the area of this blue sheet. Now if I multiply this whole thing times dy, I get this volume. It gets some depth. This little area that I'm shading right here gives depth of that sheet. Now if I added all of those sheets that now have depth, if I took the infinite sum, so if I took the integral of this from my lower y bound, from 0 to my upper y bound, then, at least based on our intuition here, maybe I will have figured out the volume under this surface."}, {"video_title": "Double integral 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now if I multiply this whole thing times dy, I get this volume. It gets some depth. This little area that I'm shading right here gives depth of that sheet. Now if I added all of those sheets that now have depth, if I took the infinite sum, so if I took the integral of this from my lower y bound, from 0 to my upper y bound, then, at least based on our intuition here, maybe I will have figured out the volume under this surface. But anyway, I didn't want to confuse you, but that's the intuition of what we're going to do. I think you're going to find out that actually calculating the volumes are pretty straightforward, especially when you have fixed x and y bounds. That's what we're going to do in the next video."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "You produce some little trinket that people enjoy buying. And the main costs that you have are labor, you know, the workers that you have creating these, and steel. And let's just say that your labor costs are $20 per hour, $20 each hour, and then your steel costs are $2,000. $2,000, keep the numbers kind of related to each other, $2,000 for every ton of steel. And then you've had your analysts work a little bit on trying to model the revenues you can make with your widgets as a function of hours of labor and tons of steel. Now let's say the revenue model that they've come up with, the revenue as a function of hours of labor, and then S for steel, let's say, is equal to about 100 times the hours of labor to the power 2 3rds multiplied by the tons of steel to the power 1 3rd. If you put in a given amount of labor and a given amount of steel, this is about how much money you're gonna expect to earn."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "$2,000, keep the numbers kind of related to each other, $2,000 for every ton of steel. And then you've had your analysts work a little bit on trying to model the revenues you can make with your widgets as a function of hours of labor and tons of steel. Now let's say the revenue model that they've come up with, the revenue as a function of hours of labor, and then S for steel, let's say, is equal to about 100 times the hours of labor to the power 2 3rds multiplied by the tons of steel to the power 1 3rd. If you put in a given amount of labor and a given amount of steel, this is about how much money you're gonna expect to earn. And of course you wanna earn as much as you can, but let's say you actually have a budget for how much you're able to spend on all these things, and your budget, the budget is $20,000. You're willing to spend $20,000 and you wanna make as much money as you can according to this model based on that. Now this is exactly the kind of problem that the Lagrange multiplier technique is made for."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "If you put in a given amount of labor and a given amount of steel, this is about how much money you're gonna expect to earn. And of course you wanna earn as much as you can, but let's say you actually have a budget for how much you're able to spend on all these things, and your budget, the budget is $20,000. You're willing to spend $20,000 and you wanna make as much money as you can according to this model based on that. Now this is exactly the kind of problem that the Lagrange multiplier technique is made for. We're trying to maximize some kind of function and we have a constraint. Now right now the constraint isn't written as a formula, but we can pretty easily write it as a formula because what makes up our budget? Well, it's gonna be the number of hours of labor multiplied by 20, so that's gonna be $20 per hour multiplied by the number of hours you put in plus $2,000 per tons of steel times the tons of steel that you put in."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "Now this is exactly the kind of problem that the Lagrange multiplier technique is made for. We're trying to maximize some kind of function and we have a constraint. Now right now the constraint isn't written as a formula, but we can pretty easily write it as a formula because what makes up our budget? Well, it's gonna be the number of hours of labor multiplied by 20, so that's gonna be $20 per hour multiplied by the number of hours you put in plus $2,000 per tons of steel times the tons of steel that you put in. So the constraint is basically that you have to have these values equal $20,000. I mean, you could say less than, right? You could say you're not willing to go any more than that, but intuitively and in reality, it's gonna be the case that in order to maximize your revenues, you should squeeze every dollar that you have available and actually hit this constraint."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "Well, it's gonna be the number of hours of labor multiplied by 20, so that's gonna be $20 per hour multiplied by the number of hours you put in plus $2,000 per tons of steel times the tons of steel that you put in. So the constraint is basically that you have to have these values equal $20,000. I mean, you could say less than, right? You could say you're not willing to go any more than that, but intuitively and in reality, it's gonna be the case that in order to maximize your revenues, you should squeeze every dollar that you have available and actually hit this constraint. So this right here is the constraint of our problem. And let's go ahead and give this guy a name, the function that we're dealing with a name. And I'm gonna call it G of HS, which is gonna be that guy."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "You could say you're not willing to go any more than that, but intuitively and in reality, it's gonna be the case that in order to maximize your revenues, you should squeeze every dollar that you have available and actually hit this constraint. So this right here is the constraint of our problem. And let's go ahead and give this guy a name, the function that we're dealing with a name. And I'm gonna call it G of HS, which is gonna be that guy. And now if you'll remember in the last few videos, the way we visualize something like this is to think about the set of all possible inputs. So in this case, you might be thinking about the HS plane, you know, the number of hours of labor on one axis, the number of tons of steel on another. And this constraint, well, in this case, it's a linear function."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "And I'm gonna call it G of HS, which is gonna be that guy. And now if you'll remember in the last few videos, the way we visualize something like this is to think about the set of all possible inputs. So in this case, you might be thinking about the HS plane, you know, the number of hours of labor on one axis, the number of tons of steel on another. And this constraint, well, in this case, it's a linear function. So this constraint is gonna give us some kind of line that tells us which pairs of S and H are gonna achieve that constraint. And then the revenue function that we're dealing with will have certain contours. You know, maybe revenues of $10,000 have a certain contour that looks like this, and revenues of $100,000 have a certain contour that looks like this."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "And this constraint, well, in this case, it's a linear function. So this constraint is gonna give us some kind of line that tells us which pairs of S and H are gonna achieve that constraint. And then the revenue function that we're dealing with will have certain contours. You know, maybe revenues of $10,000 have a certain contour that looks like this, and revenues of $100,000 have a certain contour that looks like this. But what we want is to find which value is barely touching the constraint curve, just tangent to it at a given point, because that's gonna be the contour line where if you up the value by just a little bit, it would no longer intersect with that curve. There would no longer be values of H and S that satisfy this constraint. And the way to think about finding that tangency is to consider the vector perpendicular to the tangent line to the curve at that point, which fortunately is represented by, let's see, let me make some room for myself here, represented by the gradient of our R function, the function whose contours this is, the revenue."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "You know, maybe revenues of $10,000 have a certain contour that looks like this, and revenues of $100,000 have a certain contour that looks like this. But what we want is to find which value is barely touching the constraint curve, just tangent to it at a given point, because that's gonna be the contour line where if you up the value by just a little bit, it would no longer intersect with that curve. There would no longer be values of H and S that satisfy this constraint. And the way to think about finding that tangency is to consider the vector perpendicular to the tangent line to the curve at that point, which fortunately is represented by, let's see, let me make some room for myself here, represented by the gradient of our R function, the function whose contours this is, the revenue. And what it means for this to be tangent to the constraint line is that there's gonna be another vector, the gradient of G, of our constraint function, that points in the same direction, that's proportional to that. And typically the way you write this is to say that the gradient of this function is proportional to the gradient of G, and this proportionality constant is called our Lagrange multiplier. It's called the Lagrange multiplier."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "And the way to think about finding that tangency is to consider the vector perpendicular to the tangent line to the curve at that point, which fortunately is represented by, let's see, let me make some room for myself here, represented by the gradient of our R function, the function whose contours this is, the revenue. And what it means for this to be tangent to the constraint line is that there's gonna be another vector, the gradient of G, of our constraint function, that points in the same direction, that's proportional to that. And typically the way you write this is to say that the gradient of this function is proportional to the gradient of G, and this proportionality constant is called our Lagrange multiplier. It's called the Lagrange multiplier. So let's go ahead and start working it out. Let's first compute the gradient of R. So the gradient of R is gonna be the partial derivative of R with respect to its first variable, which is H. So partial derivative with respect to H. And the second component is its partial derivative with respect to that second variable, S, with respect to S. And in this case, that first partial derivative, if we treat H as a variable and S as a constant, then that 2 3rds gets brought down. So that'll be 100 times 2 3rds times H, H to the power of, well, we've gotta subtract one from 2 3rds when we bring it down, so that'll be negative 1 3rd, multiplied by S to the 1 3rd."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "It's called the Lagrange multiplier. So let's go ahead and start working it out. Let's first compute the gradient of R. So the gradient of R is gonna be the partial derivative of R with respect to its first variable, which is H. So partial derivative with respect to H. And the second component is its partial derivative with respect to that second variable, S, with respect to S. And in this case, that first partial derivative, if we treat H as a variable and S as a constant, then that 2 3rds gets brought down. So that'll be 100 times 2 3rds times H, H to the power of, well, we've gotta subtract one from 2 3rds when we bring it down, so that'll be negative 1 3rd, multiplied by S to the 1 3rd. And then the second component here, the partial derivative with respect to S, is gonna be 100 times, well, now by treating S as the variable, we take down that 1 3rd, so that's 1 3rd, H to the 2 3rds just looks like a constant as far as S is concerned, and then we take S to the 1 3rd minus one, which is negative 2 3rds, negative 2 3rds. Great, so that's the gradient of R, and now we need the gradient of G, and that one's a lot easier, actually, because G is just a linear function. So when we take the gradient of G, which is its partial derivative with respect to H, partial H, and its partial derivative with respect to S, partial S, well, the partial with respect to H is just 20."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "So that'll be 100 times 2 3rds times H, H to the power of, well, we've gotta subtract one from 2 3rds when we bring it down, so that'll be negative 1 3rd, multiplied by S to the 1 3rd. And then the second component here, the partial derivative with respect to S, is gonna be 100 times, well, now by treating S as the variable, we take down that 1 3rd, so that's 1 3rd, H to the 2 3rds just looks like a constant as far as S is concerned, and then we take S to the 1 3rd minus one, which is negative 2 3rds, negative 2 3rds. Great, so that's the gradient of R, and now we need the gradient of G, and that one's a lot easier, actually, because G is just a linear function. So when we take the gradient of G, which is its partial derivative with respect to H, partial H, and its partial derivative with respect to S, partial S, well, the partial with respect to H is just 20. The function looks like 20 times H plus something that's a constant. So that ends up being 20, and then the partial with respect to S, likewise, it's just 2,000, because it's just some constant multiplied by S plus a bunch of other stuff that looks like constants. So that's great."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "So when we take the gradient of G, which is its partial derivative with respect to H, partial H, and its partial derivative with respect to S, partial S, well, the partial with respect to H is just 20. The function looks like 20 times H plus something that's a constant. So that ends up being 20, and then the partial with respect to S, likewise, it's just 2,000, because it's just some constant multiplied by S plus a bunch of other stuff that looks like constants. So that's great. And this means when we set the gradient of R equal to the gradient of G, the pair of equations that we get, and let me just write it all out again, is we have this top one, which I'll call 2 3rds times, and let's go ahead and do a little simplifying while I'm rewriting things here. So H to the 1 3rd is really one over, H to the negative 1 3rd, sorry, is one over H to the 1 3rd, and that's S to the 1 3rd. So all of this, that first component, is being set equal to the first component of the gradient of G, which is 20 times lambda, times this Lagrange multiplier."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "So that's great. And this means when we set the gradient of R equal to the gradient of G, the pair of equations that we get, and let me just write it all out again, is we have this top one, which I'll call 2 3rds times, and let's go ahead and do a little simplifying while I'm rewriting things here. So H to the 1 3rd is really one over, H to the negative 1 3rd, sorry, is one over H to the 1 3rd, and that's S to the 1 3rd. So all of this, that first component, is being set equal to the first component of the gradient of G, which is 20 times lambda, times this Lagrange multiplier. Because we're not setting the gradients equal to each other, we're just setting them proportional to each other. So that's the first equation. And then the second one, I'll go ahead and do some simplifying while I rewrite that one also."}, {"video_title": "Lagrange multiplier example, part 1.mp3", "Sentence": "So all of this, that first component, is being set equal to the first component of the gradient of G, which is 20 times lambda, times this Lagrange multiplier. Because we're not setting the gradients equal to each other, we're just setting them proportional to each other. So that's the first equation. And then the second one, I'll go ahead and do some simplifying while I rewrite that one also. That's gonna be 100 3rds, and then H to the 2 3rds, so times H to the 2 3rds, divided by S to the 2 3rds, because S to the negative 2 3rds is the same as one over S to the 2 3rds. All of that is equal to 2,000. 2,000 times lambda."}, {"video_title": "Local linearization.mp3", "Sentence": "In the last couple videos, I showed how you can take a function, just a function with two inputs, and find the tangent plane to its graph. And the way that you think about this, you first find a point, some kind of input point, which is, I'll just write abstractly as x-naught and y-naught, and then you see where that point ends up on the graph, and you want to find a new function, a new function which we were calling L, and maybe you say L sub f, which also is a function of x and y, and you want the graph of that function to be a plane tangent to the graph. Now this often goes by another name. This will go under the name local linearization. Local linearization. It's kind of a long word. Zation."}, {"video_title": "Local linearization.mp3", "Sentence": "This will go under the name local linearization. Local linearization. It's kind of a long word. Zation. And what this basically means, the word local means you're looking at a specific input point. So in this case, it's at a specific input point, x-naught, y-naught. And the idea of a linearization, a linearization means you're approximating the function with something simpler, with something that's actually linear."}, {"video_title": "Local linearization.mp3", "Sentence": "Zation. And what this basically means, the word local means you're looking at a specific input point. So in this case, it's at a specific input point, x-naught, y-naught. And the idea of a linearization, a linearization means you're approximating the function with something simpler, with something that's actually linear. And I'll tell you what I mean by linear in just a moment. But the whole idea here is that we don't really care about, you know, tangent planes in an abstract 3D space to some kind of graph. The whole reason for doing this is that this is a really good way to approximate a function, which is potentially a very complicated function, with something that's much easier, something that has constant partial derivatives."}, {"video_title": "Local linearization.mp3", "Sentence": "And the idea of a linearization, a linearization means you're approximating the function with something simpler, with something that's actually linear. And I'll tell you what I mean by linear in just a moment. But the whole idea here is that we don't really care about, you know, tangent planes in an abstract 3D space to some kind of graph. The whole reason for doing this is that this is a really good way to approximate a function, which is potentially a very complicated function, with something that's much easier, something that has constant partial derivatives. Now my goal of this video is going to be to show how we write this local linearization here in vector form, because it'll be both more compact and hopefully easier to remember, and also it's more general. It'll apply to things that have more than just two input variables, like this one does. So just to remind us of where we were and what we got to in the last couple videos, I'll write it a little bit more abstractly this time, rather than a specific example."}, {"video_title": "Local linearization.mp3", "Sentence": "The whole reason for doing this is that this is a really good way to approximate a function, which is potentially a very complicated function, with something that's much easier, something that has constant partial derivatives. Now my goal of this video is going to be to show how we write this local linearization here in vector form, because it'll be both more compact and hopefully easier to remember, and also it's more general. It'll apply to things that have more than just two input variables, like this one does. So just to remind us of where we were and what we got to in the last couple videos, I'll write it a little bit more abstractly this time, rather than a specific example. The way you do this local linearization is first you find the partial derivative of f with respect to x, which I'll write with the subscript notation. And you evaluate that at x sub o, or x naught, y naught. You evaluate it at the point about which you're approximating."}, {"video_title": "Local linearization.mp3", "Sentence": "So just to remind us of where we were and what we got to in the last couple videos, I'll write it a little bit more abstractly this time, rather than a specific example. The way you do this local linearization is first you find the partial derivative of f with respect to x, which I'll write with the subscript notation. And you evaluate that at x sub o, or x naught, y naught. You evaluate it at the point about which you're approximating. And then you multiply that by x minus that constant. So the only variable right here, everything is a constant, but the only variable part is that x. And then we add to that, basically doing the same thing with y."}, {"video_title": "Local linearization.mp3", "Sentence": "You evaluate it at the point about which you're approximating. And then you multiply that by x minus that constant. So the only variable right here, everything is a constant, but the only variable part is that x. And then we add to that, basically doing the same thing with y. You take the partial derivative with respect to y, you evaluate it at the input point, the point about which you are linearizing, and then you multiply it by y minus y sub o. And then to this entire thing, because you want to make sure that when you evaluate this function at the input point itself, you see when you plug in x naught and y naught, this term goes to zero, because x naught minus x naught is zero, this term goes to zero, and this is the whole reason we kind of paired up these terms and organized the constants in this way. This way you can just think about adding whatever the function itself evaluates to at that point."}, {"video_title": "Local linearization.mp3", "Sentence": "And then we add to that, basically doing the same thing with y. You take the partial derivative with respect to y, you evaluate it at the input point, the point about which you are linearizing, and then you multiply it by y minus y sub o. And then to this entire thing, because you want to make sure that when you evaluate this function at the input point itself, you see when you plug in x naught and y naught, this term goes to zero, because x naught minus x naught is zero, this term goes to zero, and this is the whole reason we kind of paired up these terms and organized the constants in this way. This way you can just think about adding whatever the function itself evaluates to at that point. And this will ensure that your linearization actually equals the function itself at the local point, because hopefully if you're approximating it near a point, then at that point it's actually equal. So what do I mean by this word linear? The word linear has a very precise formulation, especially in the context of linear algebra, and admittedly, this is not actually a linear function in the technical sense."}, {"video_title": "Local linearization.mp3", "Sentence": "This way you can just think about adding whatever the function itself evaluates to at that point. And this will ensure that your linearization actually equals the function itself at the local point, because hopefully if you're approximating it near a point, then at that point it's actually equal. So what do I mean by this word linear? The word linear has a very precise formulation, especially in the context of linear algebra, and admittedly, this is not actually a linear function in the technical sense. But loosely what it means, and the reason people call it linear, is that this x term here, this variable term, doesn't have anything fancy going on with it. It's just being multiplied by a constant. And similarly, this y term, it's just being multiplied by a constant."}, {"video_title": "Local linearization.mp3", "Sentence": "The word linear has a very precise formulation, especially in the context of linear algebra, and admittedly, this is not actually a linear function in the technical sense. But loosely what it means, and the reason people call it linear, is that this x term here, this variable term, doesn't have anything fancy going on with it. It's just being multiplied by a constant. And similarly, this y term, it's just being multiplied by a constant. It's not squared, there's no square root, it's not in an exponent or anything like that. And although there is a more technical meaning of the word linear, this is all it really means in this context. This is all you need to think about."}, {"video_title": "Local linearization.mp3", "Sentence": "And similarly, this y term, it's just being multiplied by a constant. It's not squared, there's no square root, it's not in an exponent or anything like that. And although there is a more technical meaning of the word linear, this is all it really means in this context. This is all you need to think about. Each variable is just multiplied by a constant. Now you might see this in a more complicated form, or what's at first a more complicated form, using vectors. So first of all, let's think about how we would start describing everything going on here with vectors."}, {"video_title": "Local linearization.mp3", "Sentence": "This is all you need to think about. Each variable is just multiplied by a constant. Now you might see this in a more complicated form, or what's at first a more complicated form, using vectors. So first of all, let's think about how we would start describing everything going on here with vectors. So the input, rather than talking about the input as being a pair of points, what I want to say is that there's some vector, some vector that has these as its components, and we just want to capture that all, and I want to give that a name. And kind of unfortunately, the name that we give this, it's very common to just call it x, and maybe a bold-faced x, and that would be easier to do typing than it is writing. So I'll just kind of try to emphasize bold-faced x equals this vector, and that's confusing, because x is already one of the input variables that's just a number."}, {"video_title": "Local linearization.mp3", "Sentence": "So first of all, let's think about how we would start describing everything going on here with vectors. So the input, rather than talking about the input as being a pair of points, what I want to say is that there's some vector, some vector that has these as its components, and we just want to capture that all, and I want to give that a name. And kind of unfortunately, the name that we give this, it's very common to just call it x, and maybe a bold-faced x, and that would be easier to do typing than it is writing. So I'll just kind of try to emphasize bold-faced x equals this vector, and that's confusing, because x is already one of the input variables that's just a number. But I'll try to emphasize it, just making it bold. You'll see this in writing a lot. x is this input vector."}, {"video_title": "Local linearization.mp3", "Sentence": "So I'll just kind of try to emphasize bold-faced x equals this vector, and that's confusing, because x is already one of the input variables that's just a number. But I'll try to emphasize it, just making it bold. You'll see this in writing a lot. x is this input vector. And then similarly, the specified input about which we are approximating, you would call, let's say I'll make it a nice bold-faced x, not. We'll do that not to just kind of indicate that it's a constant of some kind. And what that is, it's a vector containing the two numbers x-not, y-not."}, {"video_title": "Local linearization.mp3", "Sentence": "x is this input vector. And then similarly, the specified input about which we are approximating, you would call, let's say I'll make it a nice bold-faced x, not. We'll do that not to just kind of indicate that it's a constant of some kind. And what that is, it's a vector containing the two numbers x-not, y-not. So this is just our starting to write things in a more vectorized way, and the convenience here is that then if you're dealing with a function with three input variables, or four, or 100, you could still just write it as this bold-faced x with the understanding that the vector has a lot more components. So now, let's take a look at these first two terms in our linearization. We can start thinking of this as a dot product, actually."}, {"video_title": "Local linearization.mp3", "Sentence": "And what that is, it's a vector containing the two numbers x-not, y-not. So this is just our starting to write things in a more vectorized way, and the convenience here is that then if you're dealing with a function with three input variables, or four, or 100, you could still just write it as this bold-faced x with the understanding that the vector has a lot more components. So now, let's take a look at these first two terms in our linearization. We can start thinking of this as a dot product, actually. So let me first just kind of move this guy out of the way and give ourselves some room. So he's gonna just go up there, just the same guy. And now I want to think about writing this other term here as a dot product."}, {"video_title": "Local linearization.mp3", "Sentence": "We can start thinking of this as a dot product, actually. So let me first just kind of move this guy out of the way and give ourselves some room. So he's gonna just go up there, just the same guy. And now I want to think about writing this other term here as a dot product. And what that looks like is we have the two partial derivatives, f sub x and f sub y, indicating the partial derivatives with respect to x and y, and each one of them is evaluated, let's see, I'll do it evaluating at our bold-faced x-not. And then this one is also evaluated at that bold-faced x-not. So really, you're thinking about this as being a vector that contains two different variables, you're just packing it into a single symbol."}, {"video_title": "Local linearization.mp3", "Sentence": "And now I want to think about writing this other term here as a dot product. And what that looks like is we have the two partial derivatives, f sub x and f sub y, indicating the partial derivatives with respect to x and y, and each one of them is evaluated, let's see, I'll do it evaluating at our bold-faced x-not. And then this one is also evaluated at that bold-faced x-not. So really, you're thinking about this as being a vector that contains two different variables, you're just packing it into a single symbol. And the dot product here is against, you know, the first component is x minus x-not, so I'd write that as x minus x-not, the number, and then similarly, y minus, let's see, I'll do it in the same color, y-not, the number. But we can write each one of these in a more compact form, where this, the vector that has the partial derivatives, that's the gradient. And if that feels unfamiliar, maybe go back and check out the videos on the gradient, but this whole vector is basically just saying take the gradient and evaluate it at that vector input, you know, x-not."}, {"video_title": "Local linearization.mp3", "Sentence": "So really, you're thinking about this as being a vector that contains two different variables, you're just packing it into a single symbol. And the dot product here is against, you know, the first component is x minus x-not, so I'd write that as x minus x-not, the number, and then similarly, y minus, let's see, I'll do it in the same color, y-not, the number. But we can write each one of these in a more compact form, where this, the vector that has the partial derivatives, that's the gradient. And if that feels unfamiliar, maybe go back and check out the videos on the gradient, but this whole vector is basically just saying take the gradient and evaluate it at that vector input, you know, x-not. And then the second component here, that's telling you you've got x and y minus x-not, y-not. So what you're basically doing is taking the, you know, bold-faced input, the variable vector, x, and then you're subtracting off, you know, x-not, where x-not is some kind of constant. So this right here, this is just vector terms where you're thinking of this as being a vector with two components, and this one is a vector with two components."}, {"video_title": "Local linearization.mp3", "Sentence": "And if that feels unfamiliar, maybe go back and check out the videos on the gradient, but this whole vector is basically just saying take the gradient and evaluate it at that vector input, you know, x-not. And then the second component here, that's telling you you've got x and y minus x-not, y-not. So what you're basically doing is taking the, you know, bold-faced input, the variable vector, x, and then you're subtracting off, you know, x-not, where x-not is some kind of constant. So this right here, this is just vector terms where you're thinking of this as being a vector with two components, and this one is a vector with two components. But if your function happened to be something more complicated with, you know, 100 input variables, this would be the same thing you write down. You would just understand that when you expand this, there's gonna be 100 different components in the vector. And this is what a linear term looks like in vector terminology, because this dot product is telling you that all of the components of that bold-faced x vector, the, that expands into, you know, not bold-faced x, y, z, whatever else it expands to, all of those are just being multiplied by some kind of constant."}, {"video_title": "Local linearization.mp3", "Sentence": "So this right here, this is just vector terms where you're thinking of this as being a vector with two components, and this one is a vector with two components. But if your function happened to be something more complicated with, you know, 100 input variables, this would be the same thing you write down. You would just understand that when you expand this, there's gonna be 100 different components in the vector. And this is what a linear term looks like in vector terminology, because this dot product is telling you that all of the components of that bold-faced x vector, the, that expands into, you know, not bold-faced x, y, z, whatever else it expands to, all of those are just being multiplied by some kind of constant. So we take that whole thing, that's how you simplify the first couple terms here. And then, of course, we just add on the value of the function itself. So you would take that as the linear term."}, {"video_title": "Local linearization.mp3", "Sentence": "And this is what a linear term looks like in vector terminology, because this dot product is telling you that all of the components of that bold-faced x vector, the, that expands into, you know, not bold-faced x, y, z, whatever else it expands to, all of those are just being multiplied by some kind of constant. So we take that whole thing, that's how you simplify the first couple terms here. And then, of course, we just add on the value of the function itself. So you would take that as the linear term. And now, I kind of like to add it on to the front, actually, where you think about taking the function itself and evaluating it at that constant input, x-not. Because that way, you can kind of think this is your constant term, this is your constant term, and then the rest of the stuff here is your linear term. The rest of your stuff is your linear."}, {"video_title": "Local linearization.mp3", "Sentence": "So you would take that as the linear term. And now, I kind of like to add it on to the front, actually, where you think about taking the function itself and evaluating it at that constant input, x-not. Because that way, you can kind of think this is your constant term, this is your constant term, and then the rest of the stuff here is your linear term. The rest of your stuff is your linear. Because later on, if we start adding other terms, like a quadratic term or more complicated things, you can kind of keep adding them on the end. So this right here is the expression that you will often see for the local linearization. And the only place where the actual variable shows up, the variable vector, is right here, is this guy."}, {"video_title": "Local linearization.mp3", "Sentence": "The rest of your stuff is your linear. Because later on, if we start adding other terms, like a quadratic term or more complicated things, you can kind of keep adding them on the end. So this right here is the expression that you will often see for the local linearization. And the only place where the actual variable shows up, the variable vector, is right here, is this guy. Because, you know, when you evaluate the function f at a specified input, that's just a constant. When you evaluate the gradient at that input, it's just a constant. And we're subtracting off that specified input, that's just a constant."}, {"video_title": "Local linearization.mp3", "Sentence": "And the only place where the actual variable shows up, the variable vector, is right here, is this guy. Because, you know, when you evaluate the function f at a specified input, that's just a constant. When you evaluate the gradient at that input, it's just a constant. And we're subtracting off that specified input, that's just a constant. So this is the only place where your variable shows up. So once all is said and done, and once you do your computations, this is a very simple function. And the important part is maybe this is much simpler than the function f itself, which allows you to, you know, maybe compute something more quickly if you're writing a program that needs to, you know, deal with some kind of complicated function, but runtime is an issue."}, {"video_title": "Local linearization.mp3", "Sentence": "And we're subtracting off that specified input, that's just a constant. So this is the only place where your variable shows up. So once all is said and done, and once you do your computations, this is a very simple function. And the important part is maybe this is much simpler than the function f itself, which allows you to, you know, maybe compute something more quickly if you're writing a program that needs to, you know, deal with some kind of complicated function, but runtime is an issue. Or maybe it's a function that you never knew in the first place, but you were able to approximate its value at a point and approximate its gradient. So this is what lets you approximate the function as a whole near that point. So again, this might look very abstract, but if you just kind of unravel everything and think back to where it came from and look at the specific example of a, you know, tangent plane, hopefully it all makes a little bit of sense and you see that this is really just the simplest possible function that evaluates to the same value as f when you input this point, and whose partial derivatives all evaluate to the same values as those of f at that specified point."}, {"video_title": "Local linearization.mp3", "Sentence": "And the important part is maybe this is much simpler than the function f itself, which allows you to, you know, maybe compute something more quickly if you're writing a program that needs to, you know, deal with some kind of complicated function, but runtime is an issue. Or maybe it's a function that you never knew in the first place, but you were able to approximate its value at a point and approximate its gradient. So this is what lets you approximate the function as a whole near that point. So again, this might look very abstract, but if you just kind of unravel everything and think back to where it came from and look at the specific example of a, you know, tangent plane, hopefully it all makes a little bit of sense and you see that this is really just the simplest possible function that evaluates to the same value as f when you input this point, and whose partial derivatives all evaluate to the same values as those of f at that specified point. And if you want to see more examples of this and what it looks like and maybe how you can use it to approximate certain functions, I have an article on that that you can go check out. And it would be particularly good to kind of go in with a piece of paper and sort of work through the examples yourself as you work through it. And with that said, I will see you next video."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the plane that kind of slants down, slants down like that. It's the intersection of that plane and the cylinder, actually I shouldn't even call it a cylinder because if you just have x squared plus y squared is equal to one, it would essentially be like a pole, an infinite pole that keeps going up forever and keeps going down forever. So it would never have a top or a bottom. But we've sliced that pole with y plus z is equal to two to get and where they intersect, we get our path C. We also have this vector field defined in this way and we're asked to evaluate the line integral of F dot dr over this path with this orientation. And we could directly solve this line integral. We've done it before and actually we'll do it again to show that we're getting the same answer. But this is about Stokes' Theorem."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we've sliced that pole with y plus z is equal to two to get and where they intersect, we get our path C. We also have this vector field defined in this way and we're asked to evaluate the line integral of F dot dr over this path with this orientation. And we could directly solve this line integral. We've done it before and actually we'll do it again to show that we're getting the same answer. But this is about Stokes' Theorem. So let's see if we can apply Stokes' Theorem to this circumstance. So Stokes' Theorem tells us that this is going to be the same thing as the surface integral over a piecewise smooth surface bounded by this boundary right over here. And actually we'll pick the simplest of all the, of all of the surfaces bounded by this."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is about Stokes' Theorem. So let's see if we can apply Stokes' Theorem to this circumstance. So Stokes' Theorem tells us that this is going to be the same thing as the surface integral over a piecewise smooth surface bounded by this boundary right over here. And actually we'll pick the simplest of all the, of all of the surfaces bounded by this. I'll pick the portion of the plane that is bounded by C. So S is going to be the portion, portion of the plane, of the plane Y plus Z is equal to two bounded, bounded by C. So it's gonna be the surface integral over this. We could have picked any other surface too that's piecewise smooth bounded by this, but this is gonna be the easiest for us to work with. We have information on it."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually we'll pick the simplest of all the, of all of the surfaces bounded by this. I'll pick the portion of the plane that is bounded by C. So S is going to be the portion, portion of the plane, of the plane Y plus Z is equal to two bounded, bounded by C. So it's gonna be the surface integral over this. We could have picked any other surface too that's piecewise smooth bounded by this, but this is gonna be the easiest for us to work with. We have information on it. It's a nice flat, it's at an incline, but it's a nice flat surface that we're gonna be able to operate on analytically. And so we're gonna take the surface integral of the curl, the curl of our vector field dotted with, and we could say dot ds if we think of the vector differential, or we could say dotted with the normal vector, dotted with the normal vector at any point, ds times the little section of our surface, a little surface differential right over there. And we've talked already about orientation, so we wanna make sure that we get our orientation right."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have information on it. It's a nice flat, it's at an incline, but it's a nice flat surface that we're gonna be able to operate on analytically. And so we're gonna take the surface integral of the curl, the curl of our vector field dotted with, and we could say dot ds if we think of the vector differential, or we could say dotted with the normal vector, dotted with the normal vector at any point, ds times the little section of our surface, a little surface differential right over there. And we've talked already about orientation, so we wanna make sure that we get our orientation right. We are traversing it in this direction, and if we think about the little man analogy, if when he's walking in this direction, if he's walking with his head upwards, then the surface will be to his left, and so we want the normal vector that would also go upwards. So the normal vector would need to go, the unit normal vector would need to look something like that. The other analogy, if you're twisting a bottle cap like this then the bottle cap will move in that direction, in the direction of this normal vector."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we've talked already about orientation, so we wanna make sure that we get our orientation right. We are traversing it in this direction, and if we think about the little man analogy, if when he's walking in this direction, if he's walking with his head upwards, then the surface will be to his left, and so we want the normal vector that would also go upwards. So the normal vector would need to go, the unit normal vector would need to look something like that. The other analogy, if you're twisting a bottle cap like this then the bottle cap will move in that direction, in the direction of this normal vector. So we've set it all up. Now all we have to do is evaluate, now we have to do is evaluate this integral. To evaluate it, we have to do a couple of things."}, {"video_title": "Stokes example part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The other analogy, if you're twisting a bottle cap like this then the bottle cap will move in that direction, in the direction of this normal vector. So we've set it all up. Now all we have to do is evaluate, now we have to do is evaluate this integral. To evaluate it, we have to do a couple of things. We one, have to come up with a parameterization for our surface, which shouldn't be too difficult. We've done it before when we evaluated surface integrals. And we also have to calculate what the curl of f is going to be, and then we just need to evaluate the double integral after all of that."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "All right. So in the last video, I set up the scaffolding for the quadratic approximation, which I'm calling Q, of a function, an arbitrary two-variable function, which I'm calling F. And the form that we have right now looks like quite a lot, actually. We have six different terms. Now, the first three were just basically stolen from the local linearization formula and written in their full abstractness. It almost makes it seem a little bit more complicated than it is. And then these next three terms are basically the quadratic parts. We have what is basically x squared."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "Now, the first three were just basically stolen from the local linearization formula and written in their full abstractness. It almost makes it seem a little bit more complicated than it is. And then these next three terms are basically the quadratic parts. We have what is basically x squared. We take it as x minus x-naught squared so that we don't mess with anything previously once we plug in x equals x-naught. But basically, we think of this as x-squared. And then this here is basically x times y, but of course, we're matching each one of them with the corresponding x-naught, y-naught."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "We have what is basically x squared. We take it as x minus x-naught squared so that we don't mess with anything previously once we plug in x equals x-naught. But basically, we think of this as x-squared. And then this here is basically x times y, but of course, we're matching each one of them with the corresponding x-naught, y-naught. And then this term is the y-squared. And the question at hand is how do we fill in these constants, the coefficients in front of each one of these quadratic terms to make it so that this guy q hugs the graph of f as closely as possible? And I showed that in the very first video, kind of what that hugging means."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And then this here is basically x times y, but of course, we're matching each one of them with the corresponding x-naught, y-naught. And then this term is the y-squared. And the question at hand is how do we fill in these constants, the coefficients in front of each one of these quadratic terms to make it so that this guy q hugs the graph of f as closely as possible? And I showed that in the very first video, kind of what that hugging means. Now, in formulas, the goal here, I should probably state what it is that we want is for the second partial derivatives of q. So for example, if we take the partial derivative with respect to x twice in a row, we want it to be the case that if you take that guy and you evaluate it at the point of interest, the point about which we are approximating, it should be the same as when you take the second partial derivative of f, or the corresponding second partial derivative, I should say, since there's multiple different second partial derivatives, and you evaluate it at that same point. And of course, we want this to be true not just with the second partial derivative with respect to x twice in a row, but if we did it with the other ones."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And I showed that in the very first video, kind of what that hugging means. Now, in formulas, the goal here, I should probably state what it is that we want is for the second partial derivatives of q. So for example, if we take the partial derivative with respect to x twice in a row, we want it to be the case that if you take that guy and you evaluate it at the point of interest, the point about which we are approximating, it should be the same as when you take the second partial derivative of f, or the corresponding second partial derivative, I should say, since there's multiple different second partial derivatives, and you evaluate it at that same point. And of course, we want this to be true not just with the second partial derivative with respect to x twice in a row, but if we did it with the other ones. Like for example, let's say we took the partial derivative first with respect to x, and then with respect to y. This is called the mixed partial derivative. We want it to be the case that when we evaluate that at the point of interest, it's the same as taking the mixed partial derivative of f, with respect to x, and then with respect to y, and we evaluate it at that same point."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And of course, we want this to be true not just with the second partial derivative with respect to x twice in a row, but if we did it with the other ones. Like for example, let's say we took the partial derivative first with respect to x, and then with respect to y. This is called the mixed partial derivative. We want it to be the case that when we evaluate that at the point of interest, it's the same as taking the mixed partial derivative of f, with respect to x, and then with respect to y, and we evaluate it at that same point. At that same point. And remember, for almost all functions that you deal with, when you take this second partial derivative where we mix two of the variables, it doesn't matter the order in which you take them. You could take it first with respect to x, then y, or you could do it first with respect to y, and then with respect to x."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "We want it to be the case that when we evaluate that at the point of interest, it's the same as taking the mixed partial derivative of f, with respect to x, and then with respect to y, and we evaluate it at that same point. At that same point. And remember, for almost all functions that you deal with, when you take this second partial derivative where we mix two of the variables, it doesn't matter the order in which you take them. You could take it first with respect to x, then y, or you could do it first with respect to y, and then with respect to x. Usually, these guys are equal. There are some functions for which this isn't true, but we're gonna basically assume that we're dealing with functions where this is. So, that's the only mixed partial derivative that we have to take into account."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "You could take it first with respect to x, then y, or you could do it first with respect to y, and then with respect to x. Usually, these guys are equal. There are some functions for which this isn't true, but we're gonna basically assume that we're dealing with functions where this is. So, that's the only mixed partial derivative that we have to take into account. And I'll just kinda get rid of that guy there. And then of course, the final one, final one, just to have it on record here, is that we want the partial derivative when we take it with respect to y two times in a row, and we evaluate that at the same point. There's kind of a lot of, this is, there's a lot of writing that goes on with these things, and that's just kind of par for the course when it comes to multivariable calculus, but you take the partial derivative with respect to y at both of them, and you want it to be the same value at this point."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, that's the only mixed partial derivative that we have to take into account. And I'll just kinda get rid of that guy there. And then of course, the final one, final one, just to have it on record here, is that we want the partial derivative when we take it with respect to y two times in a row, and we evaluate that at the same point. There's kind of a lot of, this is, there's a lot of writing that goes on with these things, and that's just kind of par for the course when it comes to multivariable calculus, but you take the partial derivative with respect to y at both of them, and you want it to be the same value at this point. So, even though there's a lot going on here, all I'm basically saying is all the second partial differential information should be the same for q as it is for f. So, let's actually go up and take a look at our function and start thinking about what its partial derivatives are, what its first and second partial derivatives are. And to do that, let me first just kind of clear up some of the board here, just to make it so we can actually start computing what this second partial derivative is. So, let's go ahead and do it."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "There's kind of a lot of, this is, there's a lot of writing that goes on with these things, and that's just kind of par for the course when it comes to multivariable calculus, but you take the partial derivative with respect to y at both of them, and you want it to be the same value at this point. So, even though there's a lot going on here, all I'm basically saying is all the second partial differential information should be the same for q as it is for f. So, let's actually go up and take a look at our function and start thinking about what its partial derivatives are, what its first and second partial derivatives are. And to do that, let me first just kind of clear up some of the board here, just to make it so we can actually start computing what this second partial derivative is. So, let's go ahead and do it. This first, this partial derivative with respect to x twice, what we'll do is I'll take one of those out and think partial derivative with respect to x, and then on the inside, I'm gonna put what the partial derivative of this entire expression with respect to x is. Well, we just take it one term at a time. This first term here is a constant, so that goes to zero."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, let's go ahead and do it. This first, this partial derivative with respect to x twice, what we'll do is I'll take one of those out and think partial derivative with respect to x, and then on the inside, I'm gonna put what the partial derivative of this entire expression with respect to x is. Well, we just take it one term at a time. This first term here is a constant, so that goes to zero. The second term here actually has the variable x in it. And when we take its partial derivative, since this is a linear term, it's just gonna be that constant sitting in front of it. So, it'll be that constant, which is the value of the partial derivative of f with respect to x, evaluated at the point of interest."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "This first term here is a constant, so that goes to zero. The second term here actually has the variable x in it. And when we take its partial derivative, since this is a linear term, it's just gonna be that constant sitting in front of it. So, it'll be that constant, which is the value of the partial derivative of f with respect to x, evaluated at the point of interest. And that's just a constant. All right, so that's there. This next term has no x's in it, so that's just gonna go to zero."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, it'll be that constant, which is the value of the partial derivative of f with respect to x, evaluated at the point of interest. And that's just a constant. All right, so that's there. This next term has no x's in it, so that's just gonna go to zero. This term is interesting because it's got an x in it. So, when we take its derivative with respect to x, that two comes down. So, this will be two times a, whatever the constant a is, multiplied by x minus x naught."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "This next term has no x's in it, so that's just gonna go to zero. This term is interesting because it's got an x in it. So, when we take its derivative with respect to x, that two comes down. So, this will be two times a, whatever the constant a is, multiplied by x minus x naught. That's what the derivative of this component is with respect to x. Then this over here, this also has an x, but it's just showing up basically as a linear term. And when we treat y as a constant, since we're taking the partial derivative with respect to x, what that ends up being is b, multiplied by that, what looks like a constant, as far as x is concerned, y minus y naught."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, this will be two times a, whatever the constant a is, multiplied by x minus x naught. That's what the derivative of this component is with respect to x. Then this over here, this also has an x, but it's just showing up basically as a linear term. And when we treat y as a constant, since we're taking the partial derivative with respect to x, what that ends up being is b, multiplied by that, what looks like a constant, as far as x is concerned, y minus y naught. And then the last term doesn't have any x's in it. So, that is the first partial derivative with respect to x. And now we do it again."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And when we treat y as a constant, since we're taking the partial derivative with respect to x, what that ends up being is b, multiplied by that, what looks like a constant, as far as x is concerned, y minus y naught. And then the last term doesn't have any x's in it. So, that is the first partial derivative with respect to x. And now we do it again. Now we take the partial derivative with respect to x, and I'll, hmm, maybe I should actually clear up even more of this guy. And now when we take the partial derivative of this expression with respect to x, f sub x of x naught, y naught, that's just a constant, so that goes to zero. Two times a times x, that's gonna, we take the derivative with respect to x, and we're just gonna get two times a."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And now we do it again. Now we take the partial derivative with respect to x, and I'll, hmm, maybe I should actually clear up even more of this guy. And now when we take the partial derivative of this expression with respect to x, f sub x of x naught, y naught, that's just a constant, so that goes to zero. Two times a times x, that's gonna, we take the derivative with respect to x, and we're just gonna get two times a. And this last term doesn't have any x's in it, so that also goes to zero. So, conveniently, when we take the second partial derivative of q with respect to x, we just get a constant. It's this constant, two a."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "Two times a times x, that's gonna, we take the derivative with respect to x, and we're just gonna get two times a. And this last term doesn't have any x's in it, so that also goes to zero. So, conveniently, when we take the second partial derivative of q with respect to x, we just get a constant. It's this constant, two a. And since we want it to be the case, we want that this entire thing is equal to, well, what do we want? We want it to be the second partial derivative of f, you know, both times with respect to x. So, here I'm gonna use the subscript notation."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "It's this constant, two a. And since we want it to be the case, we want that this entire thing is equal to, well, what do we want? We want it to be the second partial derivative of f, you know, both times with respect to x. So, here I'm gonna use the subscript notation. Over here I'm using the kind of Leibniz notation. But here, just second partial derivative with respect to x. We want it to match whatever that looks like when we evaluate it at the point of interest."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, here I'm gonna use the subscript notation. Over here I'm using the kind of Leibniz notation. But here, just second partial derivative with respect to x. We want it to match whatever that looks like when we evaluate it at the point of interest. So, what we could do to make that happen, to make sure that two a is equal to this guy, is we set a equal to 1 1 2 of that second partial derivative evaluated at the point of interest. Okay, so this is something we kind of tuck away. We remember this, this is, we have solved for one of the constants."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "We want it to match whatever that looks like when we evaluate it at the point of interest. So, what we could do to make that happen, to make sure that two a is equal to this guy, is we set a equal to 1 1 2 of that second partial derivative evaluated at the point of interest. Okay, so this is something we kind of tuck away. We remember this, this is, we have solved for one of the constants. So, now let's start thinking about another one of them. Like, let's actually don't have to scroll off, because let's say we just want to take the mixed partial derivative here, where if instead of taking it with respect to x twice, we wanted to, let's see, I'll kind of erase this. We wanted to first do it with respect to x, and then do it with respect to y."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "We remember this, this is, we have solved for one of the constants. So, now let's start thinking about another one of them. Like, let's actually don't have to scroll off, because let's say we just want to take the mixed partial derivative here, where if instead of taking it with respect to x twice, we wanted to, let's see, I'll kind of erase this. We wanted to first do it with respect to x, and then do it with respect to y. Then we can just kind of edit what we have over here, and we say we already took it with respect to x. So, now as our second go, as our second go, we're gonna be taking it with respect to y. So, in that case, instead of getting two a, let's kind of figure out what it is that we get."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "We wanted to first do it with respect to x, and then do it with respect to y. Then we can just kind of edit what we have over here, and we say we already took it with respect to x. So, now as our second go, as our second go, we're gonna be taking it with respect to y. So, in that case, instead of getting two a, let's kind of figure out what it is that we get. When we take the derivative of this whole guy with respect to y, well, this looks like a constant. This here also looks like a constant, since we're doing it with respect to y, and no y's show up. And the partial derivative of this just ends up being b."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, in that case, instead of getting two a, let's kind of figure out what it is that we get. When we take the derivative of this whole guy with respect to y, well, this looks like a constant. This here also looks like a constant, since we're doing it with respect to y, and no y's show up. And the partial derivative of this just ends up being b. So, again, we just get a constant. This time it's b, not, you know, two, well, previously it was two a, but now it's just b. And this time we want it to equal the mixed partial derivative."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And the partial derivative of this just ends up being b. So, again, we just get a constant. This time it's b, not, you know, two, well, previously it was two a, but now it's just b. And this time we want it to equal the mixed partial derivative. So, instead of saying f sub xx, I'm gonna say fxy, which basically says you take the partial derivative first with respect to x, and then with respect to y. We want this guy to equal the value of that mixed partial derivative evaluated at that point. So, that gives us another fact."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And this time we want it to equal the mixed partial derivative. So, instead of saying f sub xx, I'm gonna say fxy, which basically says you take the partial derivative first with respect to x, and then with respect to y. We want this guy to equal the value of that mixed partial derivative evaluated at that point. So, that gives us another fact. That means we can just basically set b equal to that. And this is another fact, another constant that we can record. And now for c, now for c, when we're trying to figure out what that should be, the reasoning is almost identical."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, that gives us another fact. That means we can just basically set b equal to that. And this is another fact, another constant that we can record. And now for c, now for c, when we're trying to figure out what that should be, the reasoning is almost identical. It's pretty much symmetric. We did everything that we did for the case x, and instead we do it for taking the partial derivative with respect to y twice in a row. And I encourage you to do that for yourself."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And now for c, now for c, when we're trying to figure out what that should be, the reasoning is almost identical. It's pretty much symmetric. We did everything that we did for the case x, and instead we do it for taking the partial derivative with respect to y twice in a row. And I encourage you to do that for yourself. It'll definitely solidify everything that we're doing here, because it can seem kind of like a lot in a lot of computations, but you're gonna get basically the same conclusion you did for the constant a. It's gonna be the case that you have the constant c is equal to 1 1\u20442 of the second partial derivative of f with respect to y. So, you're differentiating with respect to y twice, evaluated at the point of interest."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And I encourage you to do that for yourself. It'll definitely solidify everything that we're doing here, because it can seem kind of like a lot in a lot of computations, but you're gonna get basically the same conclusion you did for the constant a. It's gonna be the case that you have the constant c is equal to 1 1\u20442 of the second partial derivative of f with respect to y. So, you're differentiating with respect to y twice, evaluated at the point of interest. So, this is gonna be kind of the third fact. And the way that you get to that conclusion, again, it's gonna be almost identical to the way that we found this one for x. Now, when you plug in these values for a, b, and c, and these are constants, even though there's, you know, we've written them as formulas, they are constants."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "So, you're differentiating with respect to y twice, evaluated at the point of interest. So, this is gonna be kind of the third fact. And the way that you get to that conclusion, again, it's gonna be almost identical to the way that we found this one for x. Now, when you plug in these values for a, b, and c, and these are constants, even though there's, you know, we've written them as formulas, they are constants. When you plug those in to this full formula, you're going to get the quadratic approximation. It'll have six separate terms, one that corresponds to the constant, two that correspond to the linear fact, and then three which correspond to the various quadratic terms. And if you wanna dig into more details and kind of go through an example or two on this, I do have an article on quadratic approximations, and hopefully you can kind of step through and do some of the computations yourself as you go."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "Now, when you plug in these values for a, b, and c, and these are constants, even though there's, you know, we've written them as formulas, they are constants. When you plug those in to this full formula, you're going to get the quadratic approximation. It'll have six separate terms, one that corresponds to the constant, two that correspond to the linear fact, and then three which correspond to the various quadratic terms. And if you wanna dig into more details and kind of go through an example or two on this, I do have an article on quadratic approximations, and hopefully you can kind of step through and do some of the computations yourself as you go. But in all of this, even though there's a lot of formulas going on, it can be pretty notationally heavy, I want you to think back to that original graphical intuition. Here, let me actually pull up the graphical intuition here. So, if you're approximating a function near a specific point, the quadratic approximation looks like this curve where, if you were to chop it in any direction, it would be a parabola, but it's hugging the graph pretty closely, so it gives us a pretty close approximation."}, {"video_title": "Quadratic approximation formula, part 2.mp3", "Sentence": "And if you wanna dig into more details and kind of go through an example or two on this, I do have an article on quadratic approximations, and hopefully you can kind of step through and do some of the computations yourself as you go. But in all of this, even though there's a lot of formulas going on, it can be pretty notationally heavy, I want you to think back to that original graphical intuition. Here, let me actually pull up the graphical intuition here. So, if you're approximating a function near a specific point, the quadratic approximation looks like this curve where, if you were to chop it in any direction, it would be a parabola, but it's hugging the graph pretty closely, so it gives us a pretty close approximation. So, even though there's a lot of formulas that go on to get us that, the ultimate visual, and I think the ultimate intuition, is actually a pretty sensible one. You're just hoping to find something that hugs the function nice and closely. And with that, I will see you next video."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "I think this is really the hard thing to do here. But the way we can parameterize the surface of a torus, which is the surface of this donut, is to say, hey, let's take a point and let's rotate it around a circle. It could be any circle. I picked a circle in the zy-plane. And how far it's gone around that circle, we'll parameterize that by s. And s can go between 0 all the way to 2 pi. And then we're going to rotate this circle around itself. Or I guess, even a better way to say it, we're going to rotate the circle around the z-axis."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "I picked a circle in the zy-plane. And how far it's gone around that circle, we'll parameterize that by s. And s can go between 0 all the way to 2 pi. And then we're going to rotate this circle around itself. Or I guess, even a better way to say it, we're going to rotate the circle around the z-axis. And the center of the circles are always going to keep a distance b away. And so these were top views right there. And then we defined our second parameter, t, which tells us how far the entire circle has rotated around the z-axis."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Or I guess, even a better way to say it, we're going to rotate the circle around the z-axis. And the center of the circles are always going to keep a distance b away. And so these were top views right there. And then we defined our second parameter, t, which tells us how far the entire circle has rotated around the z-axis. And those are our two parameter definitions. And then here we tried to visualize what happens. This is kind of the domain that our parameterization is going to be defined on."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And then we defined our second parameter, t, which tells us how far the entire circle has rotated around the z-axis. And those are our two parameter definitions. And then here we tried to visualize what happens. This is kind of the domain that our parameterization is going to be defined on. s goes between 0 and 2 pi. So when t is 0, we haven't rotated out of the zy-plane. s is at 0, goes all the way to 2 pi over there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "This is kind of the domain that our parameterization is going to be defined on. s goes between 0 and 2 pi. So when t is 0, we haven't rotated out of the zy-plane. s is at 0, goes all the way to 2 pi over there. Then when t goes to 2 pi, we've kind of moved our circle. We've moved it along. We've rotated around the z-axis a bit."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "s is at 0, goes all the way to 2 pi over there. Then when t goes to 2 pi, we've kind of moved our circle. We've moved it along. We've rotated around the z-axis a bit. And then this line in our st domain corresponds to that circle in three dimensions, or in our xyz space. Now, given that, hopefully we visualize it pretty well. Let's think about actually how to define a position vector valued function that is essentially this parameterization."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "We've rotated around the z-axis a bit. And then this line in our st domain corresponds to that circle in three dimensions, or in our xyz space. Now, given that, hopefully we visualize it pretty well. Let's think about actually how to define a position vector valued function that is essentially this parameterization. So let's first do the z, because that's pretty straightforward. So let's look at this view right here. What's our z going to be as a function?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Let's think about actually how to define a position vector valued function that is essentially this parameterization. So let's first do the z, because that's pretty straightforward. So let's look at this view right here. What's our z going to be as a function? So our x, our y's, and our z's should all be a function of s and t. That's what it's all about. Any position in space should be a function of picking a particular t and a particular s. And we saw that over here. This point right here, let me actually do that with a couple of points, this point right there that corresponds to that point right there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "What's our z going to be as a function? So our x, our y's, and our z's should all be a function of s and t. That's what it's all about. Any position in space should be a function of picking a particular t and a particular s. And we saw that over here. This point right here, let me actually do that with a couple of points, this point right there that corresponds to that point right there. Let me pick another one. This point right here corresponds to this point right over there. I could do a few more."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "This point right here, let me actually do that with a couple of points, this point right there that corresponds to that point right there. Let me pick another one. This point right here corresponds to this point right over there. I could do a few more. Let me pick this point right here. So s is still 0. That's going to be this outer edge way out over there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "I could do a few more. Let me pick this point right here. So s is still 0. That's going to be this outer edge way out over there. I'll pick one more just to define this square. This point right over here where we haven't rotated t at all, but we've gone a quarter way around the circle is that point right there. So for any s and t, we're mapping it to a point in xyz space."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "That's going to be this outer edge way out over there. I'll pick one more just to define this square. This point right over here where we haven't rotated t at all, but we've gone a quarter way around the circle is that point right there. So for any s and t, we're mapping it to a point in xyz space. So our z's, our x's, and our y's should all be a function of s and t. So the first one to think about is just the z. And I think this will be pretty straightforward. So z as a function of s and t is going to equal what?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So for any s and t, we're mapping it to a point in xyz space. So our z's, our x's, and our y's should all be a function of s and t. So the first one to think about is just the z. And I think this will be pretty straightforward. So z as a function of s and t is going to equal what? Well, if you take any circle, remember s is how much the angle between our radius and the xy plane. So I could even draw it over here. Let me do it in another color."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So z as a function of s and t is going to equal what? Well, if you take any circle, remember s is how much the angle between our radius and the xy plane. So I could even draw it over here. Let me do it in another color. So I'm running out of colors. So let's say that this is a radius right there. That angle, we said that is s. So if I were to draw that circle out just like that, we can do a little bit of trigonometry."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Let me do it in another color. So I'm running out of colors. So let's say that this is a radius right there. That angle, we said that is s. So if I were to draw that circle out just like that, we can do a little bit of trigonometry. This is the angle is s. We know the radius is a, the radius of our circle. We define that. So z is just going to be the distance above the xy plane."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "That angle, we said that is s. So if I were to draw that circle out just like that, we can do a little bit of trigonometry. This is the angle is s. We know the radius is a, the radius of our circle. We define that. So z is just going to be the distance above the xy plane. It's going to be this distance right there. And that's straightforward trigonometry. That's going to be a. I mean, we could do Sohcahtoa and all of that."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So z is just going to be the distance above the xy plane. It's going to be this distance right there. And that's straightforward trigonometry. That's going to be a. I mean, we could do Sohcahtoa and all of that. You might want to review the videos. But the sine, you could view it this way. So if this is z right there, you could say that the sine of s, Sohcahtoa is the opposite over the hypotenuse, is equal to z over a."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "That's going to be a. I mean, we could do Sohcahtoa and all of that. You might want to review the videos. But the sine, you could view it this way. So if this is z right there, you could say that the sine of s, Sohcahtoa is the opposite over the hypotenuse, is equal to z over a. Multiply both sides by a. You have a sine of s is equal to z. That tells us how much above the xy plane we are."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So if this is z right there, you could say that the sine of s, Sohcahtoa is the opposite over the hypotenuse, is equal to z over a. Multiply both sides by a. You have a sine of s is equal to z. That tells us how much above the xy plane we are. And that's just some simple trigonometry. So z of s and t, it's only going to be a function of s. It's going to be a times the sine of s. Not too bad. Now let's see if we can figure out what x and y are going to be."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "That tells us how much above the xy plane we are. And that's just some simple trigonometry. So z of s and t, it's only going to be a function of s. It's going to be a times the sine of s. Not too bad. Now let's see if we can figure out what x and y are going to be. Remember, z doesn't matter. It doesn't matter how much we've rotated around the z-axis. What matters is how much we've rotated around the circle."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Now let's see if we can figure out what x and y are going to be. Remember, z doesn't matter. It doesn't matter how much we've rotated around the z-axis. What matters is how much we've rotated around the circle. If s is 0, we're just going to be in the xy plane. z is going to be 0. If s is pi over 2 or up here, then we're going to be traveling around the top of the donut."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "What matters is how much we've rotated around the circle. If s is 0, we're just going to be in the xy plane. z is going to be 0. If s is pi over 2 or up here, then we're going to be traveling around the top of the donut. And we're going to be exactly a above the xy plane, or z is going to be equal to a. Hopefully that makes reasonable sense to you. Now let's think about what happens as we rotate around. Remember, these two are top views."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "If s is pi over 2 or up here, then we're going to be traveling around the top of the donut. And we're going to be exactly a above the xy plane, or z is going to be equal to a. Hopefully that makes reasonable sense to you. Now let's think about what happens as we rotate around. Remember, these two are top views. We are looking down. We're looking down on this donut. So the center of each of these circles is b away from the origin, or from the z-axis."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Remember, these two are top views. We are looking down. We're looking down on this donut. So the center of each of these circles is b away from the origin, or from the z-axis. It's always b away. So our x and y coordinates, so if we go to the center of the circle here, we're going to be b away. And then this is going to be b away right over there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So the center of each of these circles is b away from the origin, or from the z-axis. It's always b away. So our x and y coordinates, so if we go to the center of the circle here, we're going to be b away. And then this is going to be b away right over there. So let's think about where we are in the xy plane, or how far the part of our, I guess you could imagine, if you were to project our point into the xy plane, how far is that going to be from our origin? Well, it's always going to be, remember, let's go back to this drawing here. This might be the most instructive."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And then this is going to be b away right over there. So let's think about where we are in the xy plane, or how far the part of our, I guess you could imagine, if you were to project our point into the xy plane, how far is that going to be from our origin? Well, it's always going to be, remember, let's go back to this drawing here. This might be the most instructive. This is just one particular circle on the zy plane, but could be any of them. If this is the z-axis over here, that is the z-axis. This distance right here is always going to be b."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "This might be the most instructive. This is just one particular circle on the zy plane, but could be any of them. If this is the z-axis over here, that is the z-axis. This distance right here is always going to be b. We know that for sure. And so what is this distance going to be? We're at b to the center, and then we're going to have some angle s. And so depending on that angle s, this distance onto, I guess, the xy plane, if we're sitting on the xy plane, how far are we from the z-axis, or the projection onto the xy plane, or the y position?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "This distance right here is always going to be b. We know that for sure. And so what is this distance going to be? We're at b to the center, and then we're going to have some angle s. And so depending on that angle s, this distance onto, I guess, the xy plane, if we're sitting on the xy plane, how far are we from the z-axis, or the projection onto the xy plane, or the y position? I'm saying it as many ways as possible. I think you're visualizing it. If z is a sine of theta, this distance right here, this little shorter distance right here, that's going to be a cosine of theta."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "We're at b to the center, and then we're going to have some angle s. And so depending on that angle s, this distance onto, I guess, the xy plane, if we're sitting on the xy plane, how far are we from the z-axis, or the projection onto the xy plane, or the y position? I'm saying it as many ways as possible. I think you're visualizing it. If z is a sine of theta, this distance right here, this little shorter distance right here, that's going to be a cosine of theta. s is that angle right there. This distance right here is going to be a cosine of s. So if we talk about just straight distance from the origin along the xy plane, our distance is always going to be b plus a cosine of s. When s is out here, then it's actually going to become a negative number, and that makes sense because our distance is going to be less than b. We're going to be at that point right there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "If z is a sine of theta, this distance right here, this little shorter distance right here, that's going to be a cosine of theta. s is that angle right there. This distance right here is going to be a cosine of s. So if we talk about just straight distance from the origin along the xy plane, our distance is always going to be b plus a cosine of s. When s is out here, then it's actually going to become a negative number, and that makes sense because our distance is going to be less than b. We're going to be at that point right there. So if you look at the top views over here, no matter where we are, that is b. Let's say we've rotated a little bit. That distance right here, if you look along the xy plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point, depending on our s's and t's."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "We're going to be at that point right there. So if you look at the top views over here, no matter where we are, that is b. Let's say we've rotated a little bit. That distance right here, if you look along the xy plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point, depending on our s's and t's. Now, as we rotate around, if we're at a point here, let's say we're at a point there, and that point, we already said, is b plus a cosine of s away from the origin on the xy plane, what are the x and y coordinates of that? Well, remember, this is a top-down. We're sitting on the z-axis looking straight down on the xy plane right now."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "That distance right here, if you look along the xy plane, that is always going to be b plus a cosine of s. That's what that distance is to any given point, depending on our s's and t's. Now, as we rotate around, if we're at a point here, let's say we're at a point there, and that point, we already said, is b plus a cosine of s away from the origin on the xy plane, what are the x and y coordinates of that? Well, remember, this is a top-down. We're sitting on the z-axis looking straight down on the xy plane right now. We're looking down on the donut. So what are your x and y's going to be? Well, you draw another right triangle right here."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "We're sitting on the z-axis looking straight down on the xy plane right now. We're looking down on the donut. So what are your x and y's going to be? Well, you draw another right triangle right here. This angle right here is t. This distance right here is going to be this times the sine of our angle. So this right here, which is essentially our x, this is going to be our x coordinate. x as a function of s and t is going to be equal to the sine of t, t is our angle right there, times this radius, times b plus a cosine of s. Because remember, how far we are depends on how much around the circle we are."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Well, you draw another right triangle right here. This angle right here is t. This distance right here is going to be this times the sine of our angle. So this right here, which is essentially our x, this is going to be our x coordinate. x as a function of s and t is going to be equal to the sine of t, t is our angle right there, times this radius, times b plus a cosine of s. Because remember, how far we are depends on how much around the circle we are. When we're over here, we're much further away. Here we're exactly b away if you're looking only on the xy plane. And then over here, we're b minus a away if we're on the xy plane."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "x as a function of s and t is going to be equal to the sine of t, t is our angle right there, times this radius, times b plus a cosine of s. Because remember, how far we are depends on how much around the circle we are. When we're over here, we're much further away. Here we're exactly b away if you're looking only on the xy plane. And then over here, we're b minus a away if we're on the xy plane. So that's x as a function of s and t. And actually, the way I defined it right here, our positive x-axis would actually go in this direction. So this is x positive, this is x in the negative direction. I could have flipped the sines, but hopefully this actually makes sense, that that would be the positive x, this is the negative x."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And then over here, we're b minus a away if we're on the xy plane. So that's x as a function of s and t. And actually, the way I defined it right here, our positive x-axis would actually go in this direction. So this is x positive, this is x in the negative direction. I could have flipped the sines, but hopefully this actually makes sense, that that would be the positive x, this is the negative x. It depends on whether you're using a right-handed or left-handed coordinate system. But hopefully that makes sense. We're just saying, OK, what is this distance right here?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "I could have flipped the sines, but hopefully this actually makes sense, that that would be the positive x, this is the negative x. It depends on whether you're using a right-handed or left-handed coordinate system. But hopefully that makes sense. We're just saying, OK, what is this distance right here? That is b plus a cosine of s. We got that from this right here when we're taking a view just kind of a cut of the torus. That's how far we are in kind of the xy direction at any point, or kind of radially out without thinking about the height. And then if you want the x coordinate, you multiply it times the cosine, or sorry, you multiply it times the sine of t, the way I've set it up here."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "We're just saying, OK, what is this distance right here? That is b plus a cosine of s. We got that from this right here when we're taking a view just kind of a cut of the torus. That's how far we are in kind of the xy direction at any point, or kind of radially out without thinking about the height. And then if you want the x coordinate, you multiply it times the cosine, or sorry, you multiply it times the sine of t, the way I've set it up here. And the y coordinate is going to be this right here, the way we've set up this triangle. So y as a function of s and t is going to be equal to the cosine of t times this radius, b plus a cosine of s. And so our parameterization, and 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 SOH CAH TOA, cosine of t, CAH is equal to adjacent, which is y, right?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And then if you want the x coordinate, you multiply it times the cosine, or sorry, you multiply it times the sine of t, the way I've set it up here. And the y coordinate is going to be this right here, the way we've set up this triangle. So y as a function of s and t is going to be equal to the cosine of t times this radius, b plus a cosine of s. And so our parameterization, and 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 SOH CAH TOA, cosine of t, CAH 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. So let me copy and paste all of our takeaways. And we're done with our parameterization."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "I mean, if you say that this is our y coordinate right here, you just do SOH CAH TOA, cosine of t, CAH 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. So 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. Find a nice color, maybe pink. So let's say our position vector function, 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."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "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. Find a nice color, maybe pink. So let's say our position vector function, 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 in this case, remember, the way I defined it, the positive x direction is going to be here."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So let's say our position vector function, 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 in 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."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And in 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. Remember, the j unit vector will just go just like that. 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 a unit vector in the z direction."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "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. Remember, the j unit vector will just go just like that. 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 a 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."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "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 a 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. Actually, let's do it. So in that case, so when r of pi over 2, what do we get?"}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "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. Actually, let's do it. So in that case, so when r of pi over 2, what do we get? It's going to be b plus a times cosine of pi over 2. Cosine of pi over 2 is 0, right? Cosine of 90 degrees."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So in that case, so when r of pi over 2, what do we get? It's going to be b plus a times cosine of pi over 2. Cosine of pi over 2 is 0, right? Cosine of 90 degrees. So it's going to be b, this whole thing is going to be 0, times sine of pi over 2. Sine of pi over 2 is just 1. So it's going to be b times i plus, once again, cosine of pi over 2 is 0."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Cosine of 90 degrees. So it's going to be b, this whole thing is going to be 0, times sine of pi over 2. Sine of pi over 2 is just 1. So it's going to be b times i plus, once again, cosine of pi over 2 is 0. So this term right here is going to be b. And then cosine of pi over 2 is 0. So it's going to be 0j."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So it's going to be b times i plus, once again, cosine of pi over 2 is 0. So this term right here is going to be b. And then cosine of pi over 2 is 0. So it's going to be 0j. So it's going to be plus 0j. And then finally, pi over 2, well, there's no t here. Sine of pi over 2 is 1."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "So it's going to be 0j. So it's going to be plus 0j. And then finally, pi over 2, well, there's no t here. Sine of pi over 2 is 1. So plus a times k. So there's actually no j direction. So this is going to be equal to b times i plus a times k. So the point that it specifies, according to this parameterization, or the vector it specifies, is b times i plus a times k. So b times i will get us right out there. And then a times k will get us right over there."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "Sine of pi over 2 is 1. So plus a times k. So there's actually no j direction. So this is going to be equal to b times i plus a times k. So the point that it specifies, according to this parameterization, or the vector it specifies, is b times i plus a times k. So b times i will get us right out there. And then a times k will get us right over there. So the position vector being specified is right over there, just as we predicted. That dot, that point right there, corresponds to that point, just like that. And of course, I picked kind of points that was easy to calculate, but this whole, when you take every s and t in this domain right here, you're going to transform it to this surface."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And then a times k will get us right over there. So the position vector being specified is right over there, just as we predicted. That dot, that point right there, corresponds to that point, just like that. And of course, I picked kind of points that was easy to calculate, but this whole, when you take every s and t in this domain right here, you're going to transform it to this surface. And this is the transformation right here. And of course, we have to specify that s is between, we could write it multiple ways. We could, well, we'll just write s is between 2 pi and 0."}, {"video_title": "Determining a position vector-valued function for a parametrization of two parameters   Khan Academy.mp3", "Sentence": "And of course, I picked kind of points that was easy to calculate, but this whole, when you take every s and t in this domain right here, you're going to transform it to this surface. And this is the transformation right here. And of course, we have to specify that s is between, we could write it multiple ways. We could, well, we'll just write s is between 2 pi and 0. And we could also say t is between 2 pi and 0. And you could, you know, we're kind of overlapping one extra time at 2 pi, so maybe we can get rid of one of these equal signs if you like, although that won't change the area any if you're taking the surface area. But hopefully this gives you at least a gut sense, or more than a gut sense, of how to parameterize these things and what we're even doing."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "How do you describe rotation in three dimensions? So for example, I have here a globe, and it's rotating in some way, and there's a certain direction that it's rotating, and a speed with which it's rotating, and the question is, how could you give me some numerical information that perfectly describes that rotation? So you give me some numbers, and I can tell you the speed and the direction and everything associated with this rotation. Well, before talking about that, let's remind ourselves of how we talk about two-dimensional rotation. So I have here a little pie creature, and I set him to start rotating about, and the way that we can describe this, we pretty much need to just give a rate to it, and you might give that rate as a number of rotations, number of rotations per second, some unit of time, so rotations per second. And in this case, I think I programmed him so that he's gonna do one rotation for every five seconds, so his rotational rate would be 0.2. But that's a little bit ambiguous, if you just say, hey, this little pie creature is rotating at 0.2 rotations per second, someone could say, well, is it clockwise or counterclockwise?"}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "Well, before talking about that, let's remind ourselves of how we talk about two-dimensional rotation. So I have here a little pie creature, and I set him to start rotating about, and the way that we can describe this, we pretty much need to just give a rate to it, and you might give that rate as a number of rotations, number of rotations per second, some unit of time, so rotations per second. And in this case, I think I programmed him so that he's gonna do one rotation for every five seconds, so his rotational rate would be 0.2. But that's a little bit ambiguous, if you just say, hey, this little pie creature is rotating at 0.2 rotations per second, someone could say, well, is it clockwise or counterclockwise? So there's some ambiguity, and the convention that people have adopted is to say, well, if I give you a positive number, if the number is positive, then that's gonna tell you that the nature of the rotation is counterclockwise, but if I give you a negative number, if instead you see something that's a negative number of rotations per second, that would be rotation the other way, going clockwise. And that's the convention, that's just what people have decided on. And with this, it's very nice, because given a single number, just one number, and it could be positive or negative, you can perfectly describe two-dimensional rotation."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "But that's a little bit ambiguous, if you just say, hey, this little pie creature is rotating at 0.2 rotations per second, someone could say, well, is it clockwise or counterclockwise? So there's some ambiguity, and the convention that people have adopted is to say, well, if I give you a positive number, if the number is positive, then that's gonna tell you that the nature of the rotation is counterclockwise, but if I give you a negative number, if instead you see something that's a negative number of rotations per second, that would be rotation the other way, going clockwise. And that's the convention, that's just what people have decided on. And with this, it's very nice, because given a single number, just one number, and it could be positive or negative, you can perfectly describe two-dimensional rotation. And there's a minor nuance here, usually in physics and math, we don't actually use rotations per unit second, but instead you describe things in terms of the number of radians per unit second. Radians. And just as a quick reminder of what that means, if you imagine some kind of circle, and it could be any circle, the size doesn't really matter, and if you draw the radius to that, and then ask the question, how far along the circumference would I have to go, such that the arc length, that sort of sub-portion of the circumference, is exactly as long as the radius."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "And with this, it's very nice, because given a single number, just one number, and it could be positive or negative, you can perfectly describe two-dimensional rotation. And there's a minor nuance here, usually in physics and math, we don't actually use rotations per unit second, but instead you describe things in terms of the number of radians per unit second. Radians. And just as a quick reminder of what that means, if you imagine some kind of circle, and it could be any circle, the size doesn't really matter, and if you draw the radius to that, and then ask the question, how far along the circumference would I have to go, such that the arc length, that sort of sub-portion of the circumference, is exactly as long as the radius. So if this was R, you'd want to know how far you have to go before that arc length is also R. And then that, that angle, that amount of turning that you can do, determines one radian. And because there's exactly two pi radians for every rotation to convert between rotations per unit second and radians per unit second, you would just multiply this guy by two pi. So this, you know, it would be whatever the number you have there, times two pi."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "And just as a quick reminder of what that means, if you imagine some kind of circle, and it could be any circle, the size doesn't really matter, and if you draw the radius to that, and then ask the question, how far along the circumference would I have to go, such that the arc length, that sort of sub-portion of the circumference, is exactly as long as the radius. So if this was R, you'd want to know how far you have to go before that arc length is also R. And then that, that angle, that amount of turning that you can do, determines one radian. And because there's exactly two pi radians for every rotation to convert between rotations per unit second and radians per unit second, you would just multiply this guy by two pi. So this, you know, it would be whatever the number you have there, times two pi. And the specific numbers aren't too important. The main upshot here is that with a single number, positive or negative, you can perfectly describe two-dimensional rotation. But if we look over here at the 3D case, there's actually more information than just one number that we're gonna need to know."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "So this, you know, it would be whatever the number you have there, times two pi. And the specific numbers aren't too important. The main upshot here is that with a single number, positive or negative, you can perfectly describe two-dimensional rotation. But if we look over here at the 3D case, there's actually more information than just one number that we're gonna need to know. First of all, you want to know the axis around which it's rotating. So the line that you can draw, such that all rotation happens around that line. And then you want to describe the actual rate at which it's going."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "But if we look over here at the 3D case, there's actually more information than just one number that we're gonna need to know. First of all, you want to know the axis around which it's rotating. So the line that you can draw, such that all rotation happens around that line. And then you want to describe the actual rate at which it's going. You know, is it slow rotation or is it fast? So you need to know a direction along with a magnitude. And you might say to yourself, hey, direction, magnitude, sounds like we could use a vector."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "And then you want to describe the actual rate at which it's going. You know, is it slow rotation or is it fast? So you need to know a direction along with a magnitude. And you might say to yourself, hey, direction, magnitude, sounds like we could use a vector. And in fact, that's what we do. You use some kind of vector whose length is gonna correspond to the rate at which it's rotating, typically in radians per second. It's called the angular velocity."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "And you might say to yourself, hey, direction, magnitude, sounds like we could use a vector. And in fact, that's what we do. You use some kind of vector whose length is gonna correspond to the rate at which it's rotating, typically in radians per second. It's called the angular velocity. And then the direction describes the axis of rotation itself. But similar to how in two dimensions, there was an ambiguity between clockwise and counterclockwise, if this was the only convention we had, it would be ambiguous whether you should use this vector or if you should use one pointing in the opposite direction. And the way I've chosen to draw these guys, by the way, it doesn't matter where they are."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "It's called the angular velocity. And then the direction describes the axis of rotation itself. But similar to how in two dimensions, there was an ambiguity between clockwise and counterclockwise, if this was the only convention we had, it would be ambiguous whether you should use this vector or if you should use one pointing in the opposite direction. And the way I've chosen to draw these guys, by the way, it doesn't matter where they are. Remember, a vector, it just has a magnitude and a direction and you can put it anywhere in space. I figured it was natural enough to just kind of put them around the poles just so that you could see them on the axis of rotation itself. So the question is, what vector do you use?"}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "And the way I've chosen to draw these guys, by the way, it doesn't matter where they are. Remember, a vector, it just has a magnitude and a direction and you can put it anywhere in space. I figured it was natural enough to just kind of put them around the poles just so that you could see them on the axis of rotation itself. So the question is, what vector do you use? Do you use the one pointing in this direction or do you use this green one pointing in the opposite direction? And for this, we have a convention known as the right-hand rule. So I'll go ahead and bring in a picture here to illustrate the right-hand rule."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "So the question is, what vector do you use? Do you use the one pointing in this direction or do you use this green one pointing in the opposite direction? And for this, we have a convention known as the right-hand rule. So I'll go ahead and bring in a picture here to illustrate the right-hand rule. What you imagine doing is taking the fingers of your right hand and curling them around in the direction of rotation. And what I mean by that is the tips of your fingers will be pointing kind of the direction that the surface of the sphere would move. Then when you stick out your thumb, that's the direction, that is the choice of vector which should describe that rotation."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "So I'll go ahead and bring in a picture here to illustrate the right-hand rule. What you imagine doing is taking the fingers of your right hand and curling them around in the direction of rotation. And what I mean by that is the tips of your fingers will be pointing kind of the direction that the surface of the sphere would move. Then when you stick out your thumb, that's the direction, that is the choice of vector which should describe that rotation. So in the specific example we have here, when you stick out your right thumb, that corresponds to the white vector, not the green one. But if you did things the other way around, if, oh, move this a little bit. So let's see, get him to stay in place."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "Then when you stick out your thumb, that's the direction, that is the choice of vector which should describe that rotation. So in the specific example we have here, when you stick out your right thumb, that corresponds to the white vector, not the green one. But if you did things the other way around, if, oh, move this a little bit. So let's see, get him to stay in place. If you move things the other way around such that the rotation were going kind of in the opposite direction, then when you imagine curling the fingers of your right hand around that direction, your thumb is gonna point according to the green vector. But with the original rotation that I started illustrating, it's the white vector. The white vector is the one to go with."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "So let's see, get him to stay in place. If you move things the other way around such that the rotation were going kind of in the opposite direction, then when you imagine curling the fingers of your right hand around that direction, your thumb is gonna point according to the green vector. But with the original rotation that I started illustrating, it's the white vector. The white vector is the one to go with. And this is actually pretty cool, right? Because you're packing a lot of information into that vector. It tells you what the axis is, it tells you the speed of rotation via its magnitude, and then the choice of which direction along that axis tells you whether the globe is going one way or if it's going the other."}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "The white vector is the one to go with. And this is actually pretty cool, right? Because you're packing a lot of information into that vector. It tells you what the axis is, it tells you the speed of rotation via its magnitude, and then the choice of which direction along that axis tells you whether the globe is going one way or if it's going the other. So with just three numbers, the three-dimensional coordinates of this vector, you can perfectly describe any one given three-dimensional rotation. And the reason I'm talking about this, by the way, in a series of videos about curl is because what I'm about to talk about is three-dimensional curl, which relates to fluid flow in three dimensions and how that induces a rotation at every single point in space. And what's going to happen is you're gonna associate a vector with every single point in space to answer the question, what rotation at that point is induced by the certain fluid flow?"}, {"video_title": "Describing rotation in 3d with a vector.mp3", "Sentence": "It tells you what the axis is, it tells you the speed of rotation via its magnitude, and then the choice of which direction along that axis tells you whether the globe is going one way or if it's going the other. So with just three numbers, the three-dimensional coordinates of this vector, you can perfectly describe any one given three-dimensional rotation. And the reason I'm talking about this, by the way, in a series of videos about curl is because what I'm about to talk about is three-dimensional curl, which relates to fluid flow in three dimensions and how that induces a rotation at every single point in space. And what's going to happen is you're gonna associate a vector with every single point in space to answer the question, what rotation at that point is induced by the certain fluid flow? And I'm getting a little bit ahead of myself here. So for right now, you just need to focus on a single point of rotation and a single vector corresponding to that. But it's important to kind of get your head around how exactly we represent this rotation with a vector before moving on to the notably more cognitively intensive subject of three-dimensional curl."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And when I say that, I mean that let's say I were to take this line integral along the path c of f dot dr. And let's say my path looks like this. That's my x and y-axis, y and x. And let's say my path looks something like this. I start there and I go over there to point c, just like, or to my end point. The curve here is c. And so I would evaluate this line integral, this vector field, along this path. This would be a path independent vector field, or we call that a conservative vector field. If this thing is equal to the same integral over a different path that has the same end points."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I start there and I go over there to point c, just like, or to my end point. The curve here is c. And so I would evaluate this line integral, this vector field, along this path. This would be a path independent vector field, or we call that a conservative vector field. If this thing is equal to the same integral over a different path that has the same end points. So let's call this c1. So this is c1 and this is c2. This vector field is conservative if I start at the same point but I take a different path."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If this thing is equal to the same integral over a different path that has the same end points. So let's call this c1. So this is c1 and this is c2. This vector field is conservative if I start at the same point but I take a different path. Let's say I go something like that. If I take a different path, I start at the same point, but I still get the same value. These integrals, what this is telling me is that all it cares about to evaluate these integrals is my starting point and my ending point."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This vector field is conservative if I start at the same point but I take a different path. Let's say I go something like that. If I take a different path, I start at the same point, but I still get the same value. These integrals, what this is telling me is that all it cares about to evaluate these integrals is my starting point and my ending point. It doesn't care what I do in between. It doesn't care how I get from my starting point to my end point. These two integrals have the same start point and same end point."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These integrals, what this is telling me is that all it cares about to evaluate these integrals is my starting point and my ending point. It doesn't care what I do in between. It doesn't care how I get from my starting point to my end point. These two integrals have the same start point and same end point. So irregardless of their actual path, they're going to be the same. That's what it means for f to be a conservative field or what it means for this integral to be path independent. So before I prove or I show you the conditions, let's build up our tool kit a little bit."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These two integrals have the same start point and same end point. So irregardless of their actual path, they're going to be the same. That's what it means for f to be a conservative field or what it means for this integral to be path independent. So before I prove or I show you the conditions, let's build up our tool kit a little bit. And so you may or may not have already seen the multivariable chain rule. And I'm not going to prove it in this video, but I think it'll be pretty intuitive for you. So maybe it doesn't need to have a proof, or I'll prove it eventually, but I really just want to give you the intuition."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So before I prove or I show you the conditions, let's build up our tool kit a little bit. And so you may or may not have already seen the multivariable chain rule. And I'm not going to prove it in this video, but I think it'll be pretty intuitive for you. So maybe it doesn't need to have a proof, or I'll prove it eventually, but I really just want to give you the intuition. And all that says is that if I have some function, let's say I have f of x and y, but x and y are then functions of, let's say, a third variable t. So f of x of t and y of t. That the derivative of f with respect to t, this is going to this multivariable, I have two variables here in x and y. This is going to be equal to the partial of f with respect to x, how fast does f change as x changes, times the derivative of x with respect to t. This is a single variable function right here. So you can take a regular derivative."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So maybe it doesn't need to have a proof, or I'll prove it eventually, but I really just want to give you the intuition. And all that says is that if I have some function, let's say I have f of x and y, but x and y are then functions of, let's say, a third variable t. So f of x of t and y of t. That the derivative of f with respect to t, this is going to this multivariable, I have two variables here in x and y. This is going to be equal to the partial of f with respect to x, how fast does f change as x changes, times the derivative of x with respect to t. This is a single variable function right here. So you can take a regular derivative. So times how fast x changes with respect to t. This is a standard derivative, this is partial derivative, because at that level we're dealing with two variables. And we're not done. Plus how fast f changes with respect to y, how fast the partial of f with respect to y, times the derivative of y with respect to t. So dy dt."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can take a regular derivative. So times how fast x changes with respect to t. This is a standard derivative, this is partial derivative, because at that level we're dealing with two variables. And we're not done. Plus how fast f changes with respect to y, how fast the partial of f with respect to y, times the derivative of y with respect to t. So dy dt. I'm not going to prove it, but I think it makes pretty good intuition. This is saying, as I move a little bit dt, how much of a df do I get? Or how fast does f change with respect to t?"}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus how fast f changes with respect to y, how fast the partial of f with respect to y, times the derivative of y with respect to t. So dy dt. I'm not going to prove it, but I think it makes pretty good intuition. This is saying, as I move a little bit dt, how much of a df do I get? Or how fast does f change with respect to t? It says, well, there's two ways that f can change. It can change with respect to x, and it can change with respect to y. So why don't I add those two things together as they are both changing with respect to t. That's all it's saying."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or how fast does f change with respect to t? It says, well, there's two ways that f can change. It can change with respect to x, and it can change with respect to y. So why don't I add those two things together as they are both changing with respect to t. That's all it's saying. And if you kind of imagined that you could cancel out this partial x with this dx, and this partial y with this dy, you could kind of imagine the partial of f with respect to t on the x side of things, and then plus the partial of f with respect to t on the y, in the y dimension. And then that'll give you the total change of f with respect to t. Kind of a hand wavy argument there, but at least to me, this is a pretty intuitive formula. So that's our toolkit right there, the multivariable chain rule."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So why don't I add those two things together as they are both changing with respect to t. That's all it's saying. And if you kind of imagined that you could cancel out this partial x with this dx, and this partial y with this dy, you could kind of imagine the partial of f with respect to t on the x side of things, and then plus the partial of f with respect to t on the y, in the y dimension. And then that'll give you the total change of f with respect to t. Kind of a hand wavy argument there, but at least to me, this is a pretty intuitive formula. So that's our toolkit right there, the multivariable chain rule. We're going to put that aside for a second. Now, let's say I have some vector field f. And it's different than this f, so I'll do it in a different color, magenta. I have some vector field f that is a function of x and y."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's our toolkit right there, the multivariable chain rule. We're going to put that aside for a second. Now, let's say I have some vector field f. And it's different than this f, so I'll do it in a different color, magenta. I have some vector field f that is a function of x and y. And let's say that it happens to be the gradient of some scalar field. Let's say it equals the gradient of some scalar field. I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I have some vector field f that is a function of x and y. And let's say that it happens to be the gradient of some scalar field. Let's say it equals the gradient of some scalar field. I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y. So I could write f is also, let me write it, I don't want to write it on a new line. I could also write up here, capital F is also a function of x and y. And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y. So I could write f is also, let me write it, I don't want to write it on a new line. I could also write up here, capital F is also a function of x and y. And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector. This is the definition of the gradient right here. This is the definition of a gradient. And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector. This is the definition of the gradient right here. This is the definition of a gradient. And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y. The gradient of f of x, y is going to be a vector field that tells you the direction of steepest descent at any point. So it'll be defined on the x, y plane. So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y. The gradient of f of x, y is going to be a vector field that tells you the direction of steepest descent at any point. So it'll be defined on the x, y plane. So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis. So the gradient of it, if you take any point on the x, y plane, it'll tell you the direction you need to travel to go into the steepest descent. And for this gradient field, it's going to be something like this. And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis. So the gradient of it, if you take any point on the x, y plane, it'll tell you the direction you need to travel to go into the steepest descent. And for this gradient field, it's going to be something like this. And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here. Anyway, I don't want to get too involved in that. And the whole point of this isn't to really get the intuition behind gradients. There are other videos on this."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here. Anyway, I don't want to get too involved in that. And the whole point of this isn't to really get the intuition behind gradients. There are other videos on this. The point of this is to get a test to see whether something is path independent, whether a vector field is path independent, whether it's conservative. And it turns out that if this exists, and I'm going to prove it now, if this exists, if f is the gradient of some scalar field, if f is equal to the gradient of some scalar field, then f is conservative. Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "There are other videos on this. The point of this is to get a test to see whether something is path independent, whether a vector field is path independent, whether it's conservative. And it turns out that if this exists, and I'm going to prove it now, if this exists, if f is the gradient of some scalar field, if f is equal to the gradient of some scalar field, then f is conservative. Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point. Now let me see if I can prove that to you. So let's start with the assumption that f can be written this way, as the gradient of that lowercase f can be written as the gradient of some uppercase F. So in that case, our integral, let's define our path first. So our position vector function, we always need one of those to do a line integral or a vector line integral."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point. Now let me see if I can prove that to you. So let's start with the assumption that f can be written this way, as the gradient of that lowercase f can be written as the gradient of some uppercase F. So in that case, our integral, let's define our path first. So our position vector function, we always need one of those to do a line integral or a vector line integral. r of t is going to be equal to x of t times i plus y of t times j, for t going between a and b. We've seen this multiple times. This is just a very general definition of pretty much any path in two dimensions."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our position vector function, we always need one of those to do a line integral or a vector line integral. r of t is going to be equal to x of t times i plus y of t times j, for t going between a and b. We've seen this multiple times. This is just a very general definition of pretty much any path in two dimensions. And then we're going to say f of x, y is going to be equal to this. It's going to be the partial derivative of uppercase F with respect to x. So we're assuming that this exists, that this is true."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just a very general definition of pretty much any path in two dimensions. And then we're going to say f of x, y is going to be equal to this. It's going to be the partial derivative of uppercase F with respect to x. So we're assuming that this exists, that this is true. With respect to times i plus the partial of uppercase F with respect to y times j. Now given this, what is lowercase f dot dr going to equal over this path right here? This path is defined by this position function right there."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're assuming that this exists, that this is true. With respect to times i plus the partial of uppercase F with respect to y times j. Now given this, what is lowercase f dot dr going to equal over this path right here? This path is defined by this position function right there. Well, it's going to be equal to, well, we need to figure out what dr is, and we've done that in multiple videos. I'll do that on the right over here. dr, we've seen it multiple times."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This path is defined by this position function right there. Well, it's going to be equal to, well, we need to figure out what dr is, and we've done that in multiple videos. I'll do that on the right over here. dr, we've seen it multiple times. I'll just solve it out again. dr dt by definition was equal to dx dt times i plus dy dt times j. That's what dr dt is."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dr, we've seen it multiple times. I'll just solve it out again. dr dt by definition was equal to dx dt times i plus dy dt times j. That's what dr dt is. So if we want to figure out what dr is, the differential dr, if we want to play with differentials in this way, multiply both sides times dt. And actually I'm going to treat dt as, well, I'll multiply it, I'll distribute it. It's dx dt times dt i plus dy dt times dt j."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what dr dt is. So if we want to figure out what dr is, the differential dr, if we want to play with differentials in this way, multiply both sides times dt. And actually I'm going to treat dt as, well, I'll multiply it, I'll distribute it. It's dx dt times dt i plus dy dt times dt j. So if we're taking the dot product of f with dr, what are we going to get? We are going to get, and this is, so we're going to take, so this is going to be the integral over the curve from, I'll write the c right there. We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's dx dt times dt i plus dy dt times dt j. So if we're taking the dot product of f with dr, what are we going to get? We are going to get, and this is, so we're going to take, so this is going to be the integral over the curve from, I'll write the c right there. We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt. I'm going to write this dt in a different color. Times dt plus the partial of uppercase F with respect to y times this, we're multiplying the j components, right? We take the dot product, multiply the i components, and then add that to what you get from the product of the j components."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt. I'm going to write this dt in a different color. Times dt plus the partial of uppercase F with respect to y times this, we're multiplying the j components, right? We take the dot product, multiply the i components, and then add that to what you get from the product of the j components. So this j component is partial of uppercase F with respect to y, and then we have times, let me switch to a yellow, dy dt times that dt right over there. And then we can factor out the dt. Or actually, just so I don't have to even write it again."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We take the dot product, multiply the i components, and then add that to what you get from the product of the j components. So this j component is partial of uppercase F with respect to y, and then we have times, let me switch to a yellow, dy dt times that dt right over there. And then we can factor out the dt. Or actually, just so I don't have to even write it again. Right now I wrote it without, well let me write it again. So this is equal to the integral. And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or actually, just so I don't have to even write it again. Right now I wrote it without, well let me write it again. So this is equal to the integral. And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b. And so this is going to be equal to, I'll write it in blue, the partial of uppercase F with respect to x times dx dt plus, I'm distributing this dt out, plus the partial of uppercase F with respect to y dy dt, all of that times dt. This is equivalent to that. Now, you might realize why I talked about the multivariable chain rule."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b. And so this is going to be equal to, I'll write it in blue, the partial of uppercase F with respect to x times dx dt plus, I'm distributing this dt out, plus the partial of uppercase F with respect to y dy dt, all of that times dt. This is equivalent to that. Now, you might realize why I talked about the multivariable chain rule. What is this right here? What is that right there? You can do some pattern matching."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, you might realize why I talked about the multivariable chain rule. What is this right here? What is that right there? You can do some pattern matching. That is the same thing as the derivative of uppercase F with respect to t. Look at this. Let me copy and paste this, just so you appreciate it. Let me copy and paste that."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can do some pattern matching. That is the same thing as the derivative of uppercase F with respect to t. Look at this. Let me copy and paste this, just so you appreciate it. Let me copy and paste that. Copy it and then let me paste it. So this is our definition, or this is our, I won't say definition, one can actually prove it. You don't have to start from there."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me copy and paste that. Copy it and then let me paste it. So this is our definition, or this is our, I won't say definition, one can actually prove it. You don't have to start from there. But this is our multivariable chain rule right here. The derivative of any function with respect to t is the partial of that function with respect to x times dx dt plus the partial of that function with respect to y dy dt. I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You don't have to start from there. But this is our multivariable chain rule right here. The derivative of any function with respect to t is the partial of that function with respect to x times dx dt plus the partial of that function with respect to y dy dt. I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y. This and this are identical if you just replace this lowercase f with an uppercase F. So this in blue right here, so this whole expression is equal to the integral from a to b, t is equal to a to t is equal to b, of, in blue here, the regular derivative of F with respect to t dt. And how do you evaluate, let me write the dt in green. How do you evaluate something like this?"}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y. This and this are identical if you just replace this lowercase f with an uppercase F. So this in blue right here, so this whole expression is equal to the integral from a to b, t is equal to a to t is equal to b, of, in blue here, the regular derivative of F with respect to t dt. And how do you evaluate, let me write the dt in green. How do you evaluate something like this? And I just want to make a point. This is just this from the multivariable chain rule. And how do you evaluate a definite integral like this?"}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "How do you evaluate something like this? And I just want to make a point. This is just this from the multivariable chain rule. And how do you evaluate a definite integral like this? Well you take the antiderivative of the inside with respect to dt. So what is this going to be equal to? You take the antiderivative of the inside."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And how do you evaluate a definite integral like this? Well you take the antiderivative of the inside with respect to dt. So what is this going to be equal to? You take the antiderivative of the inside. The antiderivative of the inside, that's just dF. I'm sorry, that's just F. So this is equal to F of t. And let me be clear here. We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You take the antiderivative of the inside. The antiderivative of the inside, that's just dF. I'm sorry, that's just F. So this is equal to F of t. And let me be clear here. We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways. And this could be just written as F of t. This is the same thing as F of t. These are all equivalent, depending on whether you want to include the x's and the y's only, or the t's only, or them both. Because both of the x's and y's are functions of t. So this is the derivative of F with respect to t. If this was just in terms of t, this is the derivative of that with respect to t. We take its antiderivative. We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways. And this could be just written as F of t. This is the same thing as F of t. These are all equivalent, depending on whether you want to include the x's and the y's only, or the t's only, or them both. Because both of the x's and y's are functions of t. So this is the derivative of F with respect to t. If this was just in terms of t, this is the derivative of that with respect to t. We take its antiderivative. We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b. And so this is equal to, and this is the home stretch, this is equal to F of b minus F of a. And if you want to think about it in these terms, this is the same thing. This is equal to F of x of b y of b minus F of x of a y of a."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b. And so this is equal to, and this is the home stretch, this is equal to F of b minus F of a. And if you want to think about it in these terms, this is the same thing. This is equal to F of x of b y of b minus F of x of a y of a. These are equivalent. You give me any point on the xy plane, an x and a y, and it tells me where I am. This is my capital F. It gives me a height just like that."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to F of x of b y of b minus F of x of a y of a. These are equivalent. You give me any point on the xy plane, an x and a y, and it tells me where I am. This is my capital F. It gives me a height just like that. But what this tells me, this associates a value with every point on the xy plane. This whole exercise, remember, this is the same thing as that. This is our whole thing that we were trying to prove."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is my capital F. It gives me a height just like that. But what this tells me, this associates a value with every point on the xy plane. 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 or 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 the ending point, t of b, or t is equal to b, which is x of b, y of b."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 or 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 the ending point, t of b, or t is equal to b, which is x of b, y of b. 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."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the point x of a, y of a, and the ending point, t of b, or t is equal to b, which is x of b, y of b. 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 of each other, but it's the same idea. If the vector field F is the gradient of some scalar field uppercase F, then we can say that F is conservative, or that the line integral of F dot dr is path independent."}, {"video_title": "Path independence for line integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 of each other, but it's the same idea. If the vector field F is the gradient of some scalar field uppercase 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 and ending points are the same. Hopefully you found that useful. And we'll do some examples with that."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so I can rewrite that. So this is the triple integral over our region, which we're assuming is a type one region, of the partial of r with respect to z, which you could write like this, doesn't matter, partial of r with respect to z. And I'll dv. And we can rewrite this as, we can assume we're going to integrate with respect to z first, so I'm going to integrate with respect to z. We'll do that in another color. I'm going to integrate with respect to z first. The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we can rewrite this as, we can assume we're going to integrate with respect to z first, so I'm going to integrate with respect to z. We'll do that in another color. I'm going to integrate with respect to z first. The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2. So we're going to integrate from f1 of x, y to f2 of x, y. So I'm going to take f2 of x, y, and I'm going to integrate the partial of r with respect to z. So let me do that in that same yellow color."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2. So we're going to integrate from f1 of x, y to f2 of x, y. So I'm going to take f2 of x, y, and I'm going to integrate the partial of r with respect to z. So let me do that in that same yellow color. Partial of r with respect to z, and then I have dz. And then I'll have to integrate with respect to y and x, or with respect to x and y. So it's dx dy, or dy dx."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me do that in that same yellow color. Partial of r with respect to z, and then I have dz. And then I'll have to integrate with respect to y and x, or with respect to x and y. So it's dx dy, or dy dx. I can just write that as da. So what you could think of it, we can evaluate the yellow part, and then we're going to have the double integral over the x, y domain. So this is just going to be over the x, y domain."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's dx dy, or dy dx. I can just write that as da. So what you could think of it, we can evaluate the yellow part, and then we're going to have the double integral over the x, y domain. So this is just going to be over the x, y domain. So let me put some brackets here, just to make it clear what we're going to do. So all we're doing is we're integrating with respect to z first, and we have the bounds there. Well, this is pretty straightforward."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is just going to be over the x, y domain. So let me put some brackets here, just to make it clear what we're going to do. So all we're doing is we're integrating with respect to z first, and we have the bounds there. Well, this is pretty straightforward. This is all going to be equal to, I'll write the outside first, the double integral over the domain. And I have the da right over here. Actually, let me give myself some real estate da."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, this is pretty straightforward. This is all going to be equal to, I'll write the outside first, the double integral over the domain. And I have the da right over here. Actually, let me give myself some real estate da. Well, what's the antiderivative of this? This is just r. And this is just r, or r of x, y, z, evaluated when z is f1, or when z is f2. And from that, we evaluate when z is f1."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually, let me give myself some real estate da. Well, what's the antiderivative of this? This is just r. And this is just r, or r of x, y, z, evaluated when z is f1, or when z is f2. And from that, we evaluate when z is f1. So this is just going to be r of x, y, and z, and we evaluate when z is equal to that. And from that, we subtract when z is equal to that. So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And from that, we evaluate when z is f1. So this is just going to be r of x, y, and z, and we evaluate when z is equal to that. And from that, we subtract when z is equal to that. So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y. And from that, we need to subtract r when z is this, minus r of x, y, f1 of x, y. And then make sure that we got our parentheses. Now, this is exactly what we saw in the last video."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y. And from that, we need to subtract r when z is this, minus r of x, y, f1 of x, y. And then make sure that we got our parentheses. Now, this is exactly what we saw in the last video. It is exactly that, which shows that this is exactly this. So when we assumed it was a type I region, we got that this is exactly equal to this. You do the exact same argument with the type II region to show that this is equal to this."}, {"video_title": "Divergence theorem proof (part 5)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, this is exactly what we saw in the last video. It is exactly that, which shows that this is exactly this. So when we assumed it was a type I region, we got that this is exactly equal to this. You do the exact same argument with the type II region to show that this is equal to this. Type III regions show this is equal to that. And you have your divergence theorem proved. And we can consider ourselves done."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "Now harmonic functions are a very special kind of multivariable function, and they're defined in terms of the Laplacian, which I've been talking about in the last few videos. So the Laplacian, which we denote with this upper right side up triangle, is an operator that you might take on a multivariable function. So it might have two inputs. It could have 100 inputs, just some kind of multivariable function with a scalar output. And I talked about in the last few videos, but as a reminder, it's defined as the divergence of the gradient of f. And it's kind of like the second derivative. It's sort of the way to extend the idea of the second derivative into multiple dimensions. Now what a harmonic function is, is one where the Laplacian is equal to 0."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "It could have 100 inputs, just some kind of multivariable function with a scalar output. And I talked about in the last few videos, but as a reminder, it's defined as the divergence of the gradient of f. And it's kind of like the second derivative. It's sort of the way to extend the idea of the second derivative into multiple dimensions. Now what a harmonic function is, is one where the Laplacian is equal to 0. And it's equal to 0 at every possible input point. And sometimes the way that people write this to distinguish it, they'll make it kind of a triple equal sign, maybe saying like equivalent to 0. And this is really just a way of emphasizing that it's equal to 0 at all possible input points."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "Now what a harmonic function is, is one where the Laplacian is equal to 0. And it's equal to 0 at every possible input point. And sometimes the way that people write this to distinguish it, they'll make it kind of a triple equal sign, maybe saying like equivalent to 0. And this is really just a way of emphasizing that it's equal to 0 at all possible input points. It's not an equation that you're solving for the specific x and y where it equals 0. It's a statement about the function. And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And this is really just a way of emphasizing that it's equal to 0 at all possible input points. It's not an equation that you're solving for the specific x and y where it equals 0. It's a statement about the function. And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function. If you just have some single variable function of x, and you're looking at its second derivative, which is kind of the analog of the Laplacian, what does it mean if that's equal to 0? Well, we can integrate it. We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0?"}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function. If you just have some single variable function of x, and you're looking at its second derivative, which is kind of the analog of the Laplacian, what does it mean if that's equal to 0? Well, we can integrate it. We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0? The only functions are the constant ones. So c is just going to mean some constant here. And if you integrate that again, say what function has as its derivative a constant?"}, {"video_title": "Harmonic Functions.mp3", "Sentence": "We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0? The only functions are the constant ones. So c is just going to mean some constant here. And if you integrate that again, say what function has as its derivative a constant? Well, it's going to be that constant times x plus some other constant, some other constant k. So basically linear functions. So if you're thinking of a graph, it's just something that's got a line passing through it like that. And this should kind of make sense if you think of the geometric interpretation for the second derivative."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And if you integrate that again, say what function has as its derivative a constant? Well, it's going to be that constant times x plus some other constant, some other constant k. So basically linear functions. So if you're thinking of a graph, it's just something that's got a line passing through it like that. And this should kind of make sense if you think of the geometric interpretation for the second derivative. Because if you're just looking at a random arbitrary function that's kind of curving as it does, the second derivative is negative when this concavity is down. So this right here would be a point where the second derivative, it's not 0, it's negative. And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And this should kind of make sense if you think of the geometric interpretation for the second derivative. Because if you're just looking at a random arbitrary function that's kind of curving as it does, the second derivative is negative when this concavity is down. So this right here would be a point where the second derivative, it's not 0, it's negative. And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive. So if we're saying that the second derivative has to always be 0, then it can't curve down, and it can't curve up, and it can't do that anywhere. So basically there's no curving allowed. So whatever direction it starts at, it's not allowed to curve."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive. So if we're saying that the second derivative has to always be 0, then it can't curve down, and it can't curve up, and it can't do that anywhere. So basically there's no curving allowed. So whatever direction it starts at, it's not allowed to curve. So it just sticks straight like that. But once we extend this to the idea of a multivariable function, things can get a lot more interesting than just a straight line. So an example, I've got the graph here of a multivariable function that happens to be harmonic."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So whatever direction it starts at, it's not allowed to curve. So it just sticks straight like that. But once we extend this to the idea of a multivariable function, things can get a lot more interesting than just a straight line. So an example, I've got the graph here of a multivariable function that happens to be harmonic. So the graph that you're looking at, this is of a two variable function. And the function specifically is f of x, y, x, y, is equal to e to the x multiplied by sine of y. And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So an example, I've got the graph here of a multivariable function that happens to be harmonic. So the graph that you're looking at, this is of a two variable function. And the function specifically is f of x, y, x, y, is equal to e to the x multiplied by sine of y. And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern. Because as we're moving in the positive x direction, this here is the positive x direction, you have this exponential shape. And this corresponds with the fact that over here, we've got an e to the x. So as you move x, it kind of looks like e to the x."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern. Because as we're moving in the positive x direction, this here is the positive x direction, you have this exponential shape. And this corresponds with the fact that over here, we've got an e to the x. So as you move x, it kind of looks like e to the x. And it's being multiplied by something that is a function of y. So if you're holding y constant, this just looks like a constant. But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So as you move x, it kind of looks like e to the x. And it's being multiplied by something that is a function of y. So if you're holding y constant, this just looks like a constant. But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down. It's sort of like a negative e to the x look. But if you imagine moving in the y direction, so instead of the pure x direction like that, if we imagine ourselves moving with the input going along, let's see what it would be. It would be this way, positive y direction."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down. It's sort of like a negative e to the x look. But if you imagine moving in the y direction, so instead of the pure x direction like that, if we imagine ourselves moving with the input going along, let's see what it would be. It would be this way, positive y direction. You have this sort of sinusoidal shape. And that should make sense, because you've got the sine of y. And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "It would be this way, positive y direction. You have this sort of sinusoidal shape. And that should make sense, because you've got the sine of y. And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here. It's going way up and way down. But if e to the x was really small, it hardly even looks like it's wiggling over here. It pretty much looks flat."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here. It's going way up and way down. But if e to the x was really small, it hardly even looks like it's wiggling over here. It pretty much looks flat. So that's the graph that we're looking at. And I'm telling you right now, I claim that this is harmonic. This is a function whose Laplacian is equal to 0."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "It pretty much looks flat. So that's the graph that we're looking at. And I'm telling you right now, I claim that this is harmonic. This is a function whose Laplacian is equal to 0. And what that would mean is that as we go over here and we say we evaluate the Laplacian of f, which, just to remind you, there's a different formula, rather than thinking divergence of gradient, that turns out to be completely the same as saying you take the second derivative of that function with respect to x, its first input, and you add that. Let's see, second derivative with respect to x, you add that to the second derivative of your function with respect to the next variable. And you keep doing this for all of the different variables that there are."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "This is a function whose Laplacian is equal to 0. And what that would mean is that as we go over here and we say we evaluate the Laplacian of f, which, just to remind you, there's a different formula, rather than thinking divergence of gradient, that turns out to be completely the same as saying you take the second derivative of that function with respect to x, its first input, and you add that. Let's see, second derivative with respect to x, you add that to the second derivative of your function with respect to the next variable. And you keep doing this for all of the different variables that there are. But this is just a two-variable function, so you do this twice. The claim is that this is always equal to 0. So I might say kind of equivalent."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And you keep doing this for all of the different variables that there are. But this is just a two-variable function, so you do this twice. The claim is that this is always equal to 0. So I might say kind of equivalent. At every possible input, it's equal to 0. And I think I'll leave that as something for you to compute. It might be kind of good practice to kind of get a feel for computing Laplacian."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So I might say kind of equivalent. At every possible input, it's equal to 0. And I think I'll leave that as something for you to compute. It might be kind of good practice to kind of get a feel for computing Laplacian. But what I want to do is interpret what does this actually mean, right? Because you can plug it through and you can see, ah, yes, at all possible inputs, it will be 0. But what does that mean?"}, {"video_title": "Harmonic Functions.mp3", "Sentence": "It might be kind of good practice to kind of get a feel for computing Laplacian. But what I want to do is interpret what does this actually mean, right? Because you can plug it through and you can see, ah, yes, at all possible inputs, it will be 0. But what does that mean? Because in the single-variable context, once we started thinking about the geometric interpretation of a second derivative as this concavity, it sort of made sense that forcing it to be 0 will give us a straight line. But clearly, that's not the case. This is much more complicated than a straight line."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "But what does that mean? Because in the single-variable context, once we started thinking about the geometric interpretation of a second derivative as this concavity, it sort of made sense that forcing it to be 0 will give us a straight line. But clearly, that's not the case. This is much more complicated than a straight line. And for that, I want to give a kind of a different way that you can think about the single-variable second derivative. On the one hand, you can think of, let's say, negative second derivative as being this concavity where it's kind of frowning down. But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "This is much more complicated than a straight line. And for that, I want to give a kind of a different way that you can think about the single-variable second derivative. On the one hand, you can think of, let's say, negative second derivative as being this concavity where it's kind of frowning down. But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here. And if you go a little bit to the left, the neighbor is less than it. And if you go a little bit to the right, that other neighbor is also less than it. So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here. And if you go a little bit to the left, the neighbor is less than it. And if you go a little bit to the right, that other neighbor is also less than it. So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it. And if you do a similar thing over at a positive concavity point, where it's kind of smiling up, you say, well, its neighbor to the right has a greater value, and its neighbor to the left has a greater value. So at some point where the second derivative, instead of being less than 0, happens to be greater than 0, that means that the neighbors tend to be greater than the point itself. And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it. And if you do a similar thing over at a positive concavity point, where it's kind of smiling up, you say, well, its neighbor to the right has a greater value, and its neighbor to the left has a greater value. So at some point where the second derivative, instead of being less than 0, happens to be greater than 0, that means that the neighbors tend to be greater than the point itself. And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum. But let's say you're looking at a graph. Let's say you're looking at a function at a point where it's concave up, but it's not this idealized local minimum kind of circumstance. So instead, you might be looking at a point like this."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum. But let's say you're looking at a graph. Let's say you're looking at a function at a point where it's concave up, but it's not this idealized local minimum kind of circumstance. So instead, you might be looking at a point like this. And if you look at its neighbor to the left, that'll have some value that's actually less than your original guy. So the neighbor looks like it's less than it on the left. But if you move that same distance to the right, its neighbor is greater."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So instead, you might be looking at a point like this. And if you look at its neighbor to the left, that'll have some value that's actually less than your original guy. So the neighbor looks like it's less than it on the left. But if you move that same distance to the right, its neighbor is greater. But you would say, on average, if you took the average value of the neighbors, the neighbor on the right kind of outbalances the neighbor on the left. And you would say, on average, its neighbors are greater than the point itself. So let's say that input point there was like x sub o."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "But if you move that same distance to the right, its neighbor is greater. But you would say, on average, if you took the average value of the neighbors, the neighbor on the right kind of outbalances the neighbor on the left. And you would say, on average, its neighbors are greater than the point itself. So let's say that input point there was like x sub o. That would mean that the second derivative of your function at that point is greater than 0. So with this positive concavity, you can also think of it as a measure of, on average, are the neighbors greater than your original point or less than it? And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So let's say that input point there was like x sub o. That would mean that the second derivative of your function at that point is greater than 0. So with this positive concavity, you can also think of it as a measure of, on average, are the neighbors greater than your original point or less than it? And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world. So if we look at a function like this, and let's say we're looking at it kind of from a bird's eye view. So we've got our xy-plane. This over here is the x-axis, and this up here is the y-axis."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world. So if we look at a function like this, and let's say we're looking at it kind of from a bird's eye view. So we've got our xy-plane. This over here is the x-axis, and this up here is the y-axis. And let's say that we're looking at some specific input point. With the Laplacian, you want to start thinking about a circle of points around it, all of its neighbors. And in fact, think of a perfect circle, so all of the points that are a specified distance away."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "This over here is the x-axis, and this up here is the y-axis. And let's say that we're looking at some specific input point. With the Laplacian, you want to start thinking about a circle of points around it, all of its neighbors. And in fact, think of a perfect circle, so all of the points that are a specified distance away. The question the Laplacian is asking is, hey, are those neighbor points, on average, greater than or less than your original point? And this is actually how I introduced the Laplacian in the original video where I was giving kind of the intuition for the Laplacian. You're asking, do the points around a given input happen to be greater than it or less than it?"}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And in fact, think of a perfect circle, so all of the points that are a specified distance away. The question the Laplacian is asking is, hey, are those neighbor points, on average, greater than or less than your original point? And this is actually how I introduced the Laplacian in the original video where I was giving kind of the intuition for the Laplacian. You're asking, do the points around a given input happen to be greater than it or less than it? And if you're looking at a point where the Laplacian of your function happens to be greater than 0 at some point, that would mean all of the neighbors tend to be, on average, greater than your point. Whereas if you're looking at a point where the Laplacian of your function is less than 0, then all of those neighbors, on average, would be less than your point. So in particular, if the Laplacian was less than 0, your point might look like a local maximum."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "You're asking, do the points around a given input happen to be greater than it or less than it? And if you're looking at a point where the Laplacian of your function happens to be greater than 0 at some point, that would mean all of the neighbors tend to be, on average, greater than your point. Whereas if you're looking at a point where the Laplacian of your function is less than 0, then all of those neighbors, on average, would be less than your point. So in particular, if the Laplacian was less than 0, your point might look like a local maximum. Or if the Laplacian was greater than 0, it might look like a local minimum, because all of its neighbors would be greater than where it is. But for harmonic functions, what makes them so special is that you're saying the value of the function itself, or the value of the Laplacian of the function at every possible point, is equal to 0. So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So in particular, if the Laplacian was less than 0, your point might look like a local maximum. Or if the Laplacian was greater than 0, it might look like a local minimum, because all of its neighbors would be greater than where it is. But for harmonic functions, what makes them so special is that you're saying the value of the function itself, or the value of the Laplacian of the function at every possible point, is equal to 0. So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy. So the height of the graph above those neighbors will, on average, be the same. So if we look around the graph, what that should mean is, let's say you're looking at an input point, the output of this guy. And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy. So the height of the graph above those neighbors will, on average, be the same. So if we look around the graph, what that should mean is, let's say you're looking at an input point, the output of this guy. And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that. And no matter where you look, that should average out. And again, I encourage you to take a look at this function and actually evaluate the Laplacian to see that it's 0. But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that. And no matter where you look, that should average out. And again, I encourage you to take a look at this function and actually evaluate the Laplacian to see that it's 0. But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center. That's not something you can easily tell just looking at that formula. But with what's not that hard a computation, you can make this conclusion, which is pretty far reaching. And this comes up all the time in physics."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center. That's not something you can easily tell just looking at that formula. But with what's not that hard a computation, you can make this conclusion, which is pretty far reaching. And this comes up all the time in physics. For example, heat is one where maybe you want to describe how the heat at a certain point in a room is related to the average value of the heat of all of the points kind of around it. And in fact, it comes up in all sorts of circumstances where you have some point in physical space. And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And this comes up all the time in physics. For example, heat is one where maybe you want to describe how the heat at a certain point in a room is related to the average value of the heat of all of the points kind of around it. And in fact, it comes up in all sorts of circumstances where you have some point in physical space. And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it. So whenever you're sort of relating neighbors to your original point, the Laplacian comes in. And harmonic functions have this tendency to correspond to some notion of stability. And I won't go deeper into that now."}, {"video_title": "Harmonic Functions.mp3", "Sentence": "And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it. So whenever you're sort of relating neighbors to your original point, the Laplacian comes in. And harmonic functions have this tendency to correspond to some notion of stability. And I won't go deeper into that now. This is really, that really starts to get into the topic of partial differential equations. But at least in the context of just multivariable calculus, I wanted to shed some light on interpreting this operator and kind of interpreting the physical and geometric properties that that implies about a function. And with that, I will see you next video."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "And what I'm going to focus on in this first video, because it will take us several videos to do it, is just the parameterization of this surface right over here. And as you'll see, this is often the hardest part, because it takes a little bit of visualization. And then after that, it's kind of mechanical, but it can be kind of hairy at the same time. So it's worth going through. So first, let's think about how we can parameterize, and I have trouble even saying the word, how we can parameterize this unit sphere as a function of two parameters. So let's think about it a little bit. So first, let's just think about the unit sphere."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So it's worth going through. So first, let's think about how we can parameterize, and I have trouble even saying the word, how we can parameterize this unit sphere as a function of two parameters. So let's think about it a little bit. So first, let's just think about the unit sphere. I'm going to take a side view of the unit sphere. So let's take the unit sphere. So this right over here is our z-axis."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So first, let's just think about the unit sphere. I'm going to take a side view of the unit sphere. So let's take the unit sphere. So this right over here is our z-axis. That's our z-axis. And then over here, I'm going to draw, this is going to be not just the x or the y-axis. This is going to be the entire xy-plane viewed from the side."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So this right over here is our z-axis. That's our z-axis. And then over here, I'm going to draw, this is going to be not just the x or the y-axis. This is going to be the entire xy-plane viewed from the side. That is the xy-plane. Now, our sphere, our unit sphere, might look something like this. The unit sphere itself is not too hard to visualize."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "This is going to be the entire xy-plane viewed from the side. That is the xy-plane. Now, our sphere, our unit sphere, might look something like this. The unit sphere itself is not too hard to visualize. It might look something like that. The radius, let me make it very clear, the radius at any point is 1. So this length right over here is 1."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "The unit sphere itself is not too hard to visualize. It might look something like that. The radius, let me make it very clear, the radius at any point is 1. So this length right over here is 1. That length right over there is 1. And this is a sphere, not just a circle. So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So this length right over here is 1. That length right over there is 1. And this is a sphere, not just a circle. So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it. So that's shading it in. It kind of makes it look a little bit more spherical. Now, let's attempt to parameterize this."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it. So that's shading it in. It kind of makes it look a little bit more spherical. Now, let's attempt to parameterize this. And as a first step, let's just think, if we didn't have to think above and below the xy-plane, if we just thought about where this unit sphere intersected the xy-plane, how we could parameterize that. So let's just think about it. So it intersects the xy-plane."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Now, let's attempt to parameterize this. And as a first step, let's just think, if we didn't have to think above and below the xy-plane, if we just thought about where this unit sphere intersected the xy-plane, how we could parameterize that. So let's just think about it. So it intersects the xy-plane. It intersects it there and there and actually everywhere. So it intersects it right over there. So let's just draw the xy-plane and think about that intersection."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So it intersects the xy-plane. It intersects it there and there and actually everywhere. So it intersects it right over there. So let's just draw the xy-plane and think about that intersection. And then we could think about what happens as we go above and below the xy-plane. So on the xy-plane, this little region where we just shaded in, so let me draw. So now you could view this as almost a top view."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So let's just draw the xy-plane and think about that intersection. And then we could think about what happens as we go above and below the xy-plane. So on the xy-plane, this little region where we just shaded in, so let me draw. So now you could view this as almost a top view. The z-axis is now going to be pointing straight out at you, straight out of the screen. So that's x. Let me draw it."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So now you could view this as almost a top view. The z-axis is now going to be pointing straight out at you, straight out of the screen. So that's x. Let me draw it. So that's x. And then this right over there is y. So this thing that we were viewing sideways, now we're viewing it from the top."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Let me draw it. So that's x. And then this right over there is y. So this thing that we were viewing sideways, now we're viewing it from the top. And so now our unit sphere is going to look something like this, viewed from above. And this, what I just drew, this dotted circle right over here, this is going to be where our unit sphere intersects what I labeled that y. That should be x. I don't want to confuse you already."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So this thing that we were viewing sideways, now we're viewing it from the top. And so now our unit sphere is going to look something like this, viewed from above. And this, what I just drew, this dotted circle right over here, this is going to be where our unit sphere intersects what I labeled that y. That should be x. I don't want to confuse you already. Let me clear that. So this is our x-axis. This is our x-axis."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "That should be x. I don't want to confuse you already. Let me clear that. So this is our x-axis. This is our x-axis. So this little dotted blue circle, this is where our unit sphere intersects the xy-plane. And so using this, we can start to think about how to parameterize at least our x and y values, our x and y coordinates, as a function of a first parameter. So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "This is our x-axis. So this little dotted blue circle, this is where our unit sphere intersects the xy-plane. And so using this, we can start to think about how to parameterize at least our x and y values, our x and y coordinates, as a function of a first parameter. So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us. So essentially if we're rotating around that z-axis, viewed from above, we can imagine an angle, and I'll call that angle, I will call that angle s, which is essentially saying how much we're rotating from the x-axis towards the y-axis. You could think about it in the xy-plane or in a plane that is parallel to the xy-plane, or you could say going around the z-axis, the z-axis popping straight up at us. And the radius here is always 1."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So the first parameter, we can think of something that is, so this is the z-axis popping straight out at us. So essentially if we're rotating around that z-axis, viewed from above, we can imagine an angle, and I'll call that angle, I will call that angle s, which is essentially saying how much we're rotating from the x-axis towards the y-axis. You could think about it in the xy-plane or in a plane that is parallel to the xy-plane, or you could say going around the z-axis, the z-axis popping straight up at us. 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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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, I'm going to introduce another parameter. And this new parameter I'm going to introduce is t. t is how much we have 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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Now let's think about if we go above and below it. And to figure out how far above or below it, I'm going to introduce another parameter. And this new parameter I'm going to introduce is t. t is how much we have 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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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, and then that works right over there when we're in the xy-plane."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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, and then that works right over there when we're in the xy-plane. So the radius over here is going to be 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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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, and then that works right over there when we're in the xy-plane. So the radius over here is going to be 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."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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. 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 that we're at, what would now be the parameterization using s of x and y?"}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "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. 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 that we're at, what would now be the parameterization using s of x and y? Well, it's the exact same thing, except now our radius isn't a fixed one. It is now a function of t. So we're now a little bit higher. So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Our radius right over here is cosine of t. And so given that, I guess, altitude that we're at, what would now be the parameterization using s of x and y? Well, it's the exact same thing, except now our radius isn't a fixed one. It is now a function of t. So we're now a little bit higher. So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around. And in this case, s has gone all the way around here. So it's going to be cosine of t times cosine of s. And then our y-coordinate is going to be our radius, which is cosine of t, times sine of s. Same exact logic here, except now we have a different radius. Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So now our x-coordinate is going to be our radius, which is cosine of t. That's just our radius, times cosine of s. How much we've angled around. And in this case, s has gone all the way around here. So it's going to be cosine of t times cosine of s. And then our y-coordinate is going to be our radius, which is cosine of t, times sine of s. Same exact logic here, except now we have a different radius. Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit. And this looks very confusing, but you just have to say, at any given level we are, we're parallel to the x-axis. We're kind of tracing out another circle where another plane intersects our unit sphere. We're now then rotating around with s. And so our radius will change."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Our radius is no longer 1, times sine of s. Running out of space, let me scroll to the right a little bit. And this looks very confusing, but you just have to say, at any given level we are, we're parallel to the x-axis. We're kind of tracing out another circle where another plane intersects our unit sphere. We're now then rotating around with s. And so our radius will change. It's a function of how much above or below the xy-plane we've rotated. So this is just our radius instead of 1. And then s is how much we've rotated around the z-axis."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "We're now then rotating around with s. And so our radius will change. It's a function of how much above or below the xy-plane we've rotated. So this is just our radius instead of 1. And then s is how much we've rotated around the z-axis. Same there for the y-coordinate. And then the z-coordinate is pretty straightforward. It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "And then s is how much we've rotated around the z-axis. Same there for the y-coordinate. And then the z-coordinate is pretty straightforward. It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude. It is what our altitude actually is. And we can go straight to this diagram right over here. Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "It's going to be completely a function of t. It's not dependent on how much we've rotated around here at any given altitude. It is what our altitude actually is. And we can go straight to this diagram right over here. Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down. So z is going to be equal to sine of t. So now every point on this sphere can be described as a function of t and s. And we have to think about over what range will they be defined. Well, s is going to go at any given level, you could think. For any given t, s is going to go all the way around."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Our z-coordinate is just going to be the sine of t. So our z is equal to sine of t. So let me write that down. So z is going to be equal to sine of t. So now every point on this sphere can be described as a function of t and s. And we have to think about over what range will they be defined. Well, s is going to go at any given level, you could think. For any given t, s is going to go all the way around. We see that right over here. At any given level viewed from above, s is going to go all the way around. So thinking about it in radians, s is going to be between 0 and 2 pi."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "For any given t, s is going to go all the way around. We see that right over here. At any given level viewed from above, s is going to go all the way around. So thinking about it in radians, s is going to be between 0 and 2 pi. And t is essentially our altitude in the z direction. So t can go all the way down here, which would be negative pi over 2. So t can be between negative pi over 2."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So thinking about it in radians, s is going to be between 0 and 2 pi. And t is essentially our altitude in the z direction. So t can go all the way down here, which would be negative pi over 2. So t can be between negative pi over 2. And it can go all the way up to pi over 2. It doesn't need to go all the way back down again. And so it goes all the way back."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "So t can be between negative pi over 2. And it can go all the way up to pi over 2. It doesn't need to go all the way back down again. And so it goes all the way back. It always goes only up to pi over 2. And then we have our parameterization. Let me write this down in a form that you might recognize even more."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "And so it goes all the way back. It always goes only up to pi over 2. And then we have our parameterization. Let me write this down in a form that you might recognize even more. If we wanted to write our surface as a position vector function, we could write it like this. We could write it r is a function of s and t. And it is equal to our x component, our i component, is going to be cosine of t, cosine of s, i, and then plus our y component is cosine of t, sine of s, plus our z component, which is the sine, which is just, oh, I forgot our j vector. j plus the z component, which is just sine of t, sine of t, k. And we're done."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "Let me write this down in a form that you might recognize even more. If we wanted to write our surface as a position vector function, we could write it like this. We could write it r is a function of s and t. And it is equal to our x component, our i component, is going to be cosine of t, cosine of s, i, and then plus our y component is cosine of t, sine of s, plus our z component, which is the sine, which is just, oh, I forgot our j vector. j plus the z component, which is just sine of t, sine of t, k. And we're done. And these are the ranges that those parameters will take on. So that's just the first step. We've parameterized this surface."}, {"video_title": "Surface integral example part 1 Parameterizing the unit sphere   Khan Academy.mp3", "Sentence": "j plus the z component, which is just sine of t, sine of t, k. And we're done. And these are the ranges that those parameters will take on. So that's just the first step. We've parameterized this surface. Now we're going to have to actually set up the surface integral. It's going to involve a little bit of taking a cross product, which can get hairy. And then we can actually evaluate the integral itself."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So F dot dr. Well, we remember what F is. F is all the way up here. It's i, j, and k components. We're just p, q, and r, so that's kind of easy to remember. But let's think about what dr is equal to. Let's think about what dr is. And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're just p, q, and r, so that's kind of easy to remember. But let's think about what dr is equal to. Let's think about what dr is. And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here. So dr is the same thing as dr dt times dt. So we really just have to figure out what the derivative of r with respect to t, and this is where we're going to have to break out a little bit of the chain rule. So let me write this down."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to have to break out a little bit of our three-dimensional, our multivariable chain rule right over here. So dr is the same thing as dr dt times dt. So we really just have to figure out what the derivative of r with respect to t, and this is where we're going to have to break out a little bit of the chain rule. So let me write this down. So dr dt is going to be equal to the derivative of x with respect to t times i, plus the derivative of y with respect to t. I'm just taking the derivative with respect to t, j. And now we're going to have to break the, because z is a function of x, which is a function of t, and z is also a function of y, which is a function of t, we're going to have to break out our multivariable chain rule. So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me write this down. So dr dt is going to be equal to the derivative of x with respect to t times i, plus the derivative of y with respect to t. I'm just taking the derivative with respect to t, j. And now we're going to have to break the, because z is a function of x, which is a function of t, and z is also a function of y, which is a function of t, we're going to have to break out our multivariable chain rule. So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here. The derivative of z with respect to t, you really, the way I conceptualize this, what's all the different ways that z can change from a change in t? Well, it could change because x is changing due to a change in t, so we could have, so z could change due to x, the partial of z with respect to x, when x changes with respect to t. But then that's not the only way that z can change. We have to add to that how z can change with respect to y."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we want to take the derivative of z with respect to t, I'll do it separately here, and then I'll write it down down here. The derivative of z with respect to t, you really, the way I conceptualize this, what's all the different ways that z can change from a change in t? Well, it could change because x is changing due to a change in t, so we could have, so z could change due to x, the partial of z with respect to x, when x changes with respect to t. But then that's not the only way that z can change. We have to add to that how z can change with respect to y. So partial of z with respect to y times how fast or how y is changing with respect to t. And this is just our multivariable chain rule. And so this is dz dt, so I'll just write it right over here. And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have to add to that how z can change with respect to y. So partial of z with respect to y times how fast or how y is changing with respect to t. And this is just our multivariable chain rule. And so this is dz dt, so I'll just write it right over here. And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer. So it's going to be the partial of z with respect to x, dx dt, plus the, actually let me write it this way, plus the partial of z with respect to y, dy dt, and then we're going to multiply everything times our k. We are going to multiply everything times our k. So with this out of the way, so if we wanted what dr is, dr is just going to be this whole thing times dt. So let's do that. So now we can rewrite our line integral right over here."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'll use slightly different notation that's consistent with what we were doing before, and it'll help make things a little bit clearer. So it's going to be the partial of z with respect to x, dx dt, plus the, actually let me write it this way, plus the partial of z with respect to y, dy dt, and then we're going to multiply everything times our k. We are going to multiply everything times our k. So with this out of the way, so if we wanted what dr is, dr is just going to be this whole thing times dt. So let's do that. So now we can rewrite our line integral right over here. And we're going to now go into the t domain. And so t is going to go between a and b, and f dot dr, remember f's components were just the functions p, q, and r. And each of those were functions of x, y, and z. And z is a function of x and y, so we'll have to think about all of that in a little bit."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now we can rewrite our line integral right over here. And we're going to now go into the t domain. And so t is going to go between a and b, and f dot dr, remember f's components were just the functions p, q, and r. And each of those were functions of x, y, and z. And z is a function of x and y, so we'll have to think about all of that in a little bit. Use a little more of our multivariable chain rule. But when we take the dot products, we're just going to take the corresponding components and multiply them. So it is going to be, actually let me just copy and paste."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And z is a function of x and y, so we'll have to think about all of that in a little bit. Use a little more of our multivariable chain rule. But when we take the dot products, we're just going to take the corresponding components and multiply them. So it is going to be, actually let me just copy and paste. So f is, let me rewrite it down here. Our vector field f, and I'm going to write it a little bit shorter, our vector field f is p times i plus q times j plus r times k. So when we take our dot product of f dot dr, we're essentially taking the dot product of this and this, and we have to throw a dt at the end. So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it is going to be, actually let me just copy and paste. So f is, let me rewrite it down here. Our vector field f, and I'm going to write it a little bit shorter, our vector field f is p times i plus q times j plus r times k. So when we take our dot product of f dot dr, we're essentially taking the dot product of this and this, and we have to throw a dt at the end. So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt. And then we have to multiply all of this times dt. We can't forget that part right over there. So we're going to multiply all of this right here times dt."}, {"video_title": "Stokes' theorem proof part 5   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to get p times dx dt plus q times dy dt plus r times all of this business over here, which is the partial of z with respect to x dx dt plus the partial of z with respect to y dy dt. And then we have to multiply all of this times dt. We can't forget that part right over there. So we're going to multiply all of this right here times dt. Now I'm going to leave you there in this video just because I'm afraid of making careless mistakes. What we're going to do now is we're going to rearrange this whole thing. Recognize that this is the same thing as this right over here."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say I have a path in the xy-plane that's essentially the unit circle. This is essentially the unit circle, so it's my y-axis. This is my x-axis. And our path is going to be the unit circle. It's going to be the unit circle, and we're going to traverse it just like that. We're going to traverse it clockwise. I think you get the idea."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And our path is going to be the unit circle. It's going to be the unit circle, and we're going to traverse it just like that. We're going to traverse it clockwise. I think you get the idea. And so you can see its equation is the unit circle. So the equation of this is x squared plus y squared is equal to 1, has a radius of 1, unit circle. And what we're concerned with is the line integral."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I think you get the idea. And so you can see its equation is the unit circle. So the equation of this is x squared plus y squared is equal to 1, has a radius of 1, unit circle. And what we're concerned with is the line integral. The line integral over this curve C. It's a closed curve C. It's actually going in that direction. Of 2y dx minus 3x dy. So we are probably tempted to use Green's Theorem and why not?"}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what we're concerned with is the line integral. The line integral over this curve C. It's a closed curve C. It's actually going in that direction. Of 2y dx minus 3x dy. So we are probably tempted to use Green's Theorem and why not? So let's try. This is our path. So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we are probably tempted to use Green's Theorem and why not? So let's try. This is our path. So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area. And of course the region is that, what I just showed you. Now, you may or may not remember, I made, well, there's a slight subtle thing in this which would give you the wrong answer. In the last video we said that Green's Theorem applies when we're going counterclockwise."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So Green's Theorem tells us that the integral of some curve F dot dr, F dot dr over some path where F is equal to, where F is, let me write it a little bit neater, where F of xy is equal to P of xyi plus Q of xyj, that this integral is equal to the double integral over the region, this would be the region under question in this example, over the region of the partial of Q with respect to x, minus the partial of P with respect to y, all of that da, the differential of area. And of course the region is that, what I just showed you. Now, you may or may not remember, I made, well, there's a slight subtle thing in this which would give you the wrong answer. In the last video we said that Green's Theorem applies when we're going counterclockwise. Notice even on this little thing on the integral, I made it go counterclockwise. In our example, the curve goes clockwise. The region is to our right."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video we said that Green's Theorem applies when we're going counterclockwise. Notice even on this little thing on the integral, I made it go counterclockwise. In our example, the curve goes clockwise. The region is to our right. Green's Theorem, this applies when the region is to our left. So in this situation, when the region is to our right and we're going clockwise, so this is counterclockwise. So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The region is to our right. Green's Theorem, this applies when the region is to our left. So in this situation, when the region is to our right and we're going clockwise, so this is counterclockwise. So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this. So in our example, we're going to have the integral of C and we're going to go in the clockwise direction. So maybe I'll draw it like that, of F dot dr. This is going to be equal to the double integral over the region."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in our example, where we're going clockwise, so this is to our right, Green's Theorem is going to be the negative of this. So in our example, we're going to have the integral of C and we're going to go in the clockwise direction. So maybe I'll draw it like that, of F dot dr. This is going to be equal to the double integral over the region. And we could just swap these two, the partial of P with respect to y, minus the partial of Q with respect to x, da. So let's do that. So this is going to be equal to, in this example, the integral over the region."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be equal to the double integral over the region. And we could just swap these two, the partial of P with respect to y, minus the partial of Q with respect to x, da. So let's do that. So this is going to be equal to, in this example, the integral over the region. Let's just keep it abstract for now. We could start setting the boundaries, but let's just keep the region abstract. And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to, in this example, the integral over the region. Let's just keep it abstract for now. We could start setting the boundaries, but let's just keep the region abstract. And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this. The dr contributes those components. The F contributes these two components. So this is P of xy, that is P of xy, and then this is Q of xy."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what is the partial of P with respect to, let's just remember, this right here is our, I think we could recognize right now that this, if we take F dot dr, we're going to get this. The dr contributes those components. The F contributes these two components. So this is P of xy, that is P of xy, and then this is Q of xy. And we've seen it, I don't want to go into the whole dot dr and take the dot product over and over again. I think you can see that this is the dot product of two vectors. This is the x component of F, y component of F. This is the x component of dr, y component of dr."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is P of xy, that is P of xy, and then this is Q of xy. And we've seen it, I don't want to go into the whole dot dr and take the dot product over and over again. I think you can see that this is the dot product of two vectors. This is the x component of F, y component of F. This is the x component of dr, y component of dr. So let's take the partial of P with respect to y. Take the derivative of this with respect to y, you get 2. The derivative of 2y is just 2."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the x component of F, y component of F. This is the x component of dr, y component of dr. So let's take the partial of P with respect to y. Take the derivative of this with respect to y, you get 2. The derivative of 2y is just 2. So you get 2, and then minus the derivative of Q with respect to x. The derivative of this with respect to x is minus 3. So we're going to get minus 3, and then all of that dA."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The derivative of 2y is just 2. So you get 2, and then minus the derivative of Q with respect to x. The derivative of this with respect to x is minus 3. So we're going to get minus 3, and then all of that dA. And this is equal to the integral over the region. What's this? This is 2 minus minus 3."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to get minus 3, and then all of that dA. And this is equal to the integral over the region. What's this? This is 2 minus minus 3. That's the same thing as 2 plus 3. So it's the integral over the region of 5 dA. 5 is just a constant, so we can take it out of the integral."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is 2 minus minus 3. That's the same thing as 2 plus 3. So it's the integral over the region of 5 dA. 5 is just a constant, so we can take it out of the integral. So this is going to turn out to be quite a simple problem. So this is going to be equal to 5 times the double integral over the region r dA. Now what is this thing?"}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "5 is just a constant, so we can take it out of the integral. So this is going to turn out to be quite a simple problem. So this is going to be equal to 5 times the double integral over the region r dA. Now what is this thing? What is this thing right here? It looks very abstract, but we can solve this. This is just the area of the region."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now what is this thing? What is this thing right here? It looks very abstract, but we can solve this. This is just the area of the region. That's what that double integral represents. You just sum up all the little dAs. That's a dA."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just the area of the region. That's what that double integral represents. You just sum up all the little dAs. That's a dA. You sum up the infinite sums of those little dAs over the region. Well, what's the area of this unit circle? Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a dA. You sum up the infinite sums of those little dAs over the region. Well, what's the area of this unit circle? Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? It's a unit circle."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Here we just break out a little bit of 9th grade, actually even earlier than that, pre-algebra or middle school geometry. Area is equal to pi r squared. What's our radius? It's a unit circle. Our radius is 1. Length is 1. So the area here is pi."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's a unit circle. Our radius is 1. Length is 1. So the area here is pi. So this thing over here, that whole thing is just equal to pi. So the answer to our line integral is just 5 pi, which is pretty straightforward. We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the area here is pi. So this thing over here, that whole thing is just equal to pi. So the answer to our line integral is just 5 pi, which is pretty straightforward. We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1. But that would have been super hairy and a huge pain. We just have to realize, no, this is just the area. The other interesting thing is I challenge you to solve the same integral without using Green's Theorem."}, {"video_title": "Green's theorem example 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could have taken the trouble of setting up a double integral where we take the antiderivative with respect to y first and write y is equal to the negative square root of 1 minus x squared, y is equal to the positive square root, x goes from minus 1 to 1. But that would have been super hairy and a huge pain. We just have to realize, no, this is just the area. The other interesting thing is I challenge you to solve the same integral without using Green's Theorem. The way we actually take generating a parameterization for this curve, going in that direction, taking the derivatives of x of t and y of t, multiplying by the appropriate thing, and then taking the antiderivative, way hairier than what we just did using Green's Theorem to get 5 pi. And remember, the reason why it wasn't minus 5 pi here is because we're going in a clockwise direction. If we were going in a counterclockwise direction, we could have applied the straight-up Green's Theorem and we would have gotten minus 5 pi."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 could make it do the negative x and y axis, you 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, it's going to be just a straight line that looks something like that. And then for any given x, you're going to have a parabola in y. Y is going to look something like that. I'm just going to do it in the positive quadrant, it's going to look something like that."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, it's going to be just a straight line that looks something like that. And then for any given x, you're going to have a parabola in y. Y is going to look something like that. I'm just going to do it in the positive quadrant, it's going to look something like that. 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 that looks something like that. Maybe I'll do another attempt at drawing it."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just going to do it in the positive quadrant, it's going to look something like that. 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 that looks something like that. Maybe I'll do 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. I'm going to start at the point 2,0, x is equal to 2, y is 0."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe I'll do 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. I'm going to start at the point 2,0, x is 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 is going to have a radius 2. I'm going to go counterclockwise in that circle. This is on the xy plane, just to be able to visualize it properly."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm going to start at the point 2,0, x is 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 is going to have a radius 2. I'm going to go counterclockwise in that circle. This is on the xy plane, just to be able to visualize it properly. So this right here is a point 0,2. Then I'm going to come back along the y-axis. This is my path."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is on the xy plane, just to be able to visualize it properly. So this right here is a point 0,2. Then I'm going to come back along the y-axis. This is my path. I'm going to come back along the y-axis, just like that. And then I'm going to take a left here, and then I'm going to take another left here, and then come back along the x-axis. So what I drew in these two shades of green, that is my contour."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is my path. I'm going to come back along the y-axis, just like that. And then I'm going to take a left here, and then I'm going to take another left here, and then come back along the x-axis. So what I drew 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 x, y is equal to x plus y squared. And I want to find the surface area of its walls. So you'll have this wall right here, whose base is the x-axis."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what I drew 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 x, y is equal to x plus y squared. And I want to find the surface area of its walls. So you'll have this wall right here, whose base is the x-axis. Then you're going to have this wall, which is along the curve. It's going to look something like a funky wall on that curved side right there. I'll try my best effort to try to... Actually, it's going to be curving way up like that."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you'll have this wall right here, whose base is the x-axis. Then you're going to have this wall, which is along the curve. It's going to look something like a funky wall on that curved side right there. I'll try my best effort to try to... Actually, it's going to be curving way up like that. It's going to be curved up like that. And then along the y-axis, along the y-axis when x is equal to 0, it's going to have like a half of a parabolic wall right there. I'll do that back wall along the y-axis."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll try my best effort to try to... Actually, it's going to be curving way up like that. It's going to be curved up like that. And then along the y-axis, along the y-axis when x is equal to 0, it's going to have like a half of a parabolic wall right there. I'll do that back wall along the y-axis. I'll do that in orange. Actually, I was already using orange. I'll use magenta."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do that back wall along the y-axis. I'll do that in orange. Actually, I was already using 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, curvy curtain or wall."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, curvy curtain or wall. I'll do that maybe in blue. That goes along this curve right here, this part of a circle of radius 2. Hopefully you get that visualization."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you have this weird, curvy, curvy curtain or wall. I'll do that maybe in blue. That goes along this curve right here, this part of a circle of radius 2. Hopefully you get that visualization. It's a little harder. I'm not using any graphic program this time. But I want to figure out the surface area, the combined surface area of these three walls."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully you get that visualization. It's a little harder. I'm not using any graphic program this time. But I want to figure out the surface area, the combined surface area of these three walls. In very simple notation, we could say, well, the surface area of those walls, of this wall plus that wall plus that wall, is going to be equal to the line integral along this curve or along this contour, however you want to call it, of f of x, y, so that's x plus y squared, ds, where ds is just a little length along our contour. And since this is a closed loop, we'll call this a closed line integral. And you'll sometimes see this notation right here."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I want to figure out the surface area, the combined surface area of these three walls. In very simple notation, we could say, well, the surface area of those walls, of this wall plus that wall plus that wall, is going to be equal to the line integral along this curve or along this contour, however you want to call it, of f of x, y, so that's x plus y squared, ds, where ds is just a little length along our contour. And since this is a closed loop, we'll call this a closed line integral. And you'll sometimes see this notation right here. That notation, often you'll see that in physics books. You'll be dealing with a lot more. You put a circle on that integral sign."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you'll sometimes see this notation right here. That notation, often you'll see that in physics books. You'll be dealing with a lot more. You put a circle on that integral sign. And all that means is that the contour we're dealing with is a closed contour. We get back to where we started from. But how do we solve this thing?"}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You put a circle on that integral sign. And all that means is that the contour we're dealing with is a closed contour. We get back to where we started from. But how do we solve this thing? A good place to start is to just define the contour itself. And just to simplify it, we're going to divide it into three pieces and essentially just do three separate line integrals because this isn't a very continuous contour. So the first part, let's do this first part of the curve as we're going along a circle of radius 2."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But how do we solve this thing? A good place to start is to just define the contour itself. And just to simplify it, we're going to divide it into three pieces and essentially just do three separate line integrals because this isn't a very continuous contour. So the first part, let's do this first part of the curve as we're going along a circle of radius 2. And that's pretty easy to construct. If we have x is equal to... Let me do that in each part of the contour in a different color. So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the first part, let's do this first part of the curve as we're going along a circle of radius 2. And that's pretty easy to construct. If we have x is equal to... Let me do that in each part of the contour in a different color. So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video. If we say that t is greater than or equal to 0 and is less than or equal to pi over 2, t is essentially going to be the angle that we're going along this circle right here. And this will actually describe this path. And how I constructed this is a little confusing."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I do an orange, this part of the contour, if we say that x is equal to 2 cosine of t and y is equal to 2 sine of t, if we say that t... and this is really just building off just what we saw in the last video. If we say that t is greater than or equal to 0 and is less than or equal to pi over 2, t is essentially going to be the angle that we're going along this circle right here. And this will actually describe this path. And how I constructed this is a little confusing. You might want to review the video on parametric equations. So this is the first part of our path. So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And how I constructed this is a little confusing. You might want to review the video on parametric equations. So this is the first part of our path. So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt. So let's get that out of the way right now. So if we say dx, dt is going to be equal to minus 2 sine of t. dy, dt is going to be equal to 2 cosine of t. Just took the derivatives of these. We've seen that many times before."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we just wanted to find the surface area of that wall right there, we know we're going to have to find dx, dt, and dy, dt. So let's get that out of the way right now. So if we say dx, dt is going to be equal to minus 2 sine of t. dy, dt is going to be equal to 2 cosine of t. Just took the derivatives of these. We've seen that many times before. So if we want this orange wall's surface area, we can take the integral. And if any of this is confusing, there are two videos before this where we kind of derive this formula. But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've seen that many times before. So if we want this orange wall's surface area, we can take the integral. And if any of this is confusing, there are two videos before this where we kind of derive this formula. But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared. And then times the ds. So x plus y squared will give the height of each little block. And then we want to get the width of each little block, which is ds."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we could take the integral from t is equal to 0 to pi over 2 of our function of x plus y squared. And then times the ds. So x plus y squared will give the height of each little block. And then we want to get the width of each little block, which is ds. But we know that we can rewrite ds as the square root of dx of the derivative of x with respect to t squared. So that is minus 2 sine of t squared plus the derivative of y with respect to t squared. So plus 2 cosine of t squared."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we want to get the width of each little block, which is ds. But we know that we can rewrite ds as the square root of dx of the derivative of x with respect to t squared. So that is minus 2 sine of t squared plus the derivative of y with respect to t squared. So plus 2 cosine of t squared. dt. This will give us the orange section. And then we can worry about the other two walls."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So plus 2 cosine of t squared. dt. This will give us the orange section. And then we can worry about the other two walls. And so how can we simplify this? Well, this is going to be equal to the integral from 0 to pi over 2 of x plus y squared. And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we can worry about the other two walls. And so how can we simplify this? Well, this is going to be equal to the integral from 0 to pi over 2 of x plus y squared. And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down. So it's 2 cosine of t plus y, which is 2 sine of t. And we're going to square everything. And then all of that times this crazy radical. Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually, let me write everything in terms of t. So x is equal to 2 cosine of t. So let me write that down. So it's 2 cosine of t plus y, which is 2 sine of t. And we're going to square everything. And then all of that times this crazy radical. Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad. This is going to be equal to 4 sine squared of t plus 4 cosine squared of t. We can factor a 4 out. Now we don't want to forget the dt. This over here, let me just simplify this expression so I don't have to keep rewriting it."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Right now it looks like a hard antiderivative or integral to solve, but I think we'll find out it's not too bad. This is going to be equal to 4 sine squared of t plus 4 cosine squared of t. We can factor a 4 out. Now we don't want to forget the dt. This over here, let me just simplify this expression so I don't have to keep rewriting it. That is the same thing as the square root of 4 times sine squared of t plus cosine squared of t. We know what that is. That's just 1. So this whole thing just simplifies to the square root of 4, which is just 2."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This over here, let me just simplify this expression so I don't have to keep rewriting it. That is the same thing as the square root of 4 times sine squared of t plus cosine squared of t. We know what that is. That's just 1. So this whole thing just simplifies to the square root of 4, which is just 2. So this whole thing simplifies to 2, which is nice for us solving our antiderivative. It simplifies things a lot. So this whole thing simplifies down to, I'll do it over here."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing just simplifies to the square root of 4, which is just 2. So this whole thing simplifies to 2, which is nice for us solving our antiderivative. It simplifies things a lot. So this whole thing simplifies down to, I'll do it over here. I don't want to waste too much space. I have two more walls to figure out. The integral from t is equal to 0 to pi over 2."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing simplifies down to, I'll do it over here. I don't want to waste too much space. I have two more walls to figure out. The integral from t is equal to 0 to pi over 2. And I want to make it very clear. I just chose the simplest parametrization I could for x and y, but I could have picked other parametrizations and I would have had to change t accordingly. So as long as you're consistent with how you do it, it should all work out."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The integral from t is equal to 0 to pi over 2. And I want to make it very clear. I just chose the simplest parametrization I could for x and y, but I could have picked other parametrizations and I would have had to change t accordingly. So as long as you're consistent with how you do it, it should all work out. There isn't just one parametrization for this curve. It's kind of depending on how fast you want to go along the curve. Watch the parametric functions videos if you want a little bit more depth on that."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So as long as you're consistent with how you do it, it should all work out. There isn't just one parametrization for this curve. It's kind of depending on how fast you want to go along the curve. Watch the parametric functions videos if you want a little bit more depth on that. But anyway, this thing simplifies. We have a 2 here, 2 times cosine of t. That's 4 cosine of t. And then here we have 2 sine squared sine of t squared. So that's 4 sine squared of t. And then we have to multiply it times this 2 again."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Watch the parametric functions videos if you want a little bit more depth on that. But anyway, this thing simplifies. We have a 2 here, 2 times cosine of t. That's 4 cosine of t. And then here we have 2 sine squared sine of t squared. So that's 4 sine squared of t. And then we have to multiply it times this 2 again. So that gives us an 8. 8 times sine squared of t dt. And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's 4 sine squared of t. And then we have to multiply it times this 2 again. So that gives us an 8. 8 times sine squared of t dt. And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u. So we can reuse this identity. I could write a t here. Sine squared of t is equal to 1 half times 1 minus cosine of 2t."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then sine squared of t, that looks like a tough thing to find the antiderivative for, but we can remember that sine squared of really anything, we could say sine squared of u, is equal to 1 half times 1 minus cosine of 2u. So we can reuse this identity. I could write a t here. Sine squared of t is equal to 1 half times 1 minus cosine of 2t. Let me rewrite it that way, because that will make the integral a lot easier to solve. So we get the integral from 0 to pi over 2. And actually I could break up, well, I won't break it up."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sine squared of t is equal to 1 half times 1 minus cosine of 2t. Let me rewrite it that way, because that will make the integral a lot easier to solve. So we get the integral from 0 to pi over 2. And actually I could break up, well, I won't break it up. 4 cosine of t plus 8 times this thing. 8 times this thing. This is the same thing as sine squared of t. So 8 times this, 8 times 1 half is 4."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually I could break up, well, I won't break it up. 4 cosine of t plus 8 times this thing. 8 times this thing. This is the same thing as sine squared of t. So 8 times this, 8 times 1 half is 4. 4 times 1 minus cosine of 2t. Just use a little trig identity there. And all of that dt."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the same thing as sine squared of t. So 8 times this, 8 times 1 half is 4. 4 times 1 minus cosine of 2t. Just use a little trig identity there. And all of that dt. Now this should be reasonably straightforward to get the antiderivative of. Let's just take it. The antiderivative of this is."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And all of that dt. Now this should be reasonably straightforward to get the antiderivative of. Let's just take it. The antiderivative of this is. Antiderivative of cosine of t. That's just 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."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The antiderivative of this is. Antiderivative of cosine of t. That's just 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 2t."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 2t. Let's see, it's going to be sine of 2t. Let's see, sine of 2t. The derivative of sine of 2t is 2 cosine of 2t."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the antiderivative of minus 4 cosine of 2t. Let's see, it's going to be sine of 2t. Let's see, 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 there. And now it should work out. What's the derivative of minus 2 sine of t?"}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 there. And now it should work out. 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."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 figured out our antiderivative. Now we evaluate it from 0 to pi over 2."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the derivative of sine of 2t with respect to 2t is cosine of 2t. So there we go. We figured out our antiderivative. Now we evaluate it from 0 to pi over 2. And what do we get? We get 4 sine of pi. Let me write this down."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we evaluate it from 0 to pi over 2. And what do we get? We get 4 sine of pi. Let me write this down. 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."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write this down. 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 pi. And then all of that minus, all of this evaluated at 0. That's actually pretty straightforward because sine of 0 is 0."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's just 2 pi minus 2 sine of 2 times pi over 2. Sine of pi. And then all of that minus, all of 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 all the 0's work out nicely."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 all 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, same thing. That is 1."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all 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, same thing. That is 1. And then sine of pi is 0. That's 180 degrees. So this whole thing cancels out."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is 1. 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."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, since they go along the axes, are much, much, much easier. But we're going to have to find different parametrizations for this."}, {"video_title": "Line integral example 2 (part 1)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, since 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 my videos have been running a little long."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And what I want to do is build up to the formal definition of the directional derivative of that same function in the direction of some vector v. And v with a little thing on top, this will be some vector in the input space. And I have another video on the formal definition of the partial derivative if you want to check that out. And just to really quickly go through here, I've drawn this diagram before, but it's worth drawing again. You think of your input space, which is the xy-plane. And you think of it somehow mapping over to the real number line, which is where your output f lives. And when you're taking the partial derivative at a point a, b, you're looking over here and you say, maybe that's your point, some point a, b. And you imagine nudging it slightly in the x direction and saying, hey, how does that influence the function?"}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "You think of your input space, which is the xy-plane. And you think of it somehow mapping over to the real number line, which is where your output f lives. And when you're taking the partial derivative at a point a, b, you're looking over here and you say, maybe that's your point, some point a, b. And you imagine nudging it slightly in the x direction and saying, hey, how does that influence the function? So maybe this is where a, b lands, and maybe the result is a nudge that's a little bit negative. That would be a negative partial derivative. And you think of the size of that nudge as partial x and the size of the resulting nudge in the output space as partial f. So the way that you read this formal definition is you think of this variable h. You could say delta x, but h seems to be the common variable people use."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And you imagine nudging it slightly in the x direction and saying, hey, how does that influence the function? So maybe this is where a, b lands, and maybe the result is a nudge that's a little bit negative. That would be a negative partial derivative. And you think of the size of that nudge as partial x and the size of the resulting nudge in the output space as partial f. So the way that you read this formal definition is you think of this variable h. You could say delta x, but h seems to be the common variable people use. You think of it as that change in your input space, that slight nudge. And you look at how that influences the function when you only change the x component here. You're only changing the x component with that nudge."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And you think of the size of that nudge as partial x and the size of the resulting nudge in the output space as partial f. So the way that you read this formal definition is you think of this variable h. You could say delta x, but h seems to be the common variable people use. You think of it as that change in your input space, that slight nudge. And you look at how that influences the function when you only change the x component here. You're only changing the x component with that nudge. And you say, what's the change in f? What's that partial f? So I'm going to write this in a slightly different way using vector notation."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "You're only changing the x component with that nudge. And you say, what's the change in f? What's that partial f? So I'm going to write this in a slightly different way using vector notation. Instead, I'm going to say partial f, partial x. And instead of saying the input is a, b, I'm going to say it's just a and then make it clear that that's a vector. And this will be a two-dimensional vector."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "So I'm going to write this in a slightly different way using vector notation. Instead, I'm going to say partial f, partial x. And instead of saying the input is a, b, I'm going to say it's just a and then make it clear that that's a vector. And this will be a two-dimensional vector. So I'll put that little arrow on top to indicate that it's a vector. And if we rewrite this definition, we'd be thinking the limit as h goes to 0 of something divided by h. But that thing, now that we're writing in terms of vector notation, is going to be f of. So it's going to be our original starting point, a, but plus what?"}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And this will be a two-dimensional vector. So I'll put that little arrow on top to indicate that it's a vector. And if we rewrite this definition, we'd be thinking the limit as h goes to 0 of something divided by h. But that thing, now that we're writing in terms of vector notation, is going to be f of. So it's going to be our original starting point, a, but plus what? I mean, up here, it was clear we could just add it to the first component. But if I'm not writing in terms of components and I have to think in terms of vector addition, really what I'm adding is that h times the vector, the unit vector, in the x direction. And it's common to use this little i with a hat to represent the unit vector in the x direction."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "So it's going to be our original starting point, a, but plus what? I mean, up here, it was clear we could just add it to the first component. But if I'm not writing in terms of components and I have to think in terms of vector addition, really what I'm adding is that h times the vector, the unit vector, in the x direction. And it's common to use this little i with a hat to represent the unit vector in the x direction. So when I'm adding these, it's really the same. This h is only going to go to that first component. And the second component is multiplied by 0."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And it's common to use this little i with a hat to represent the unit vector in the x direction. So when I'm adding these, it's really the same. This h is only going to go to that first component. And the second component is multiplied by 0. And what we subtract off is the value of the function at that original input, that original two-dimensional input that I'm just thinking of as a vector here. And when I write it like this, it's actually much clearer how we might extend this idea to moving in different directions. Because now, all of the information about what direction you're moving is captured with this vector here, what you multiply your nudge by as you're adding the input."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And the second component is multiplied by 0. And what we subtract off is the value of the function at that original input, that original two-dimensional input that I'm just thinking of as a vector here. And when I write it like this, it's actually much clearer how we might extend this idea to moving in different directions. Because now, all of the information about what direction you're moving is captured with this vector here, what you multiply your nudge by as you're adding the input. So let's just rewrite that over here in the context of directional derivative. What you would say is that the directional derivative in the direction of some vector, any vector, of f, evaluate it at a point. And we'll think about that input point as being a vector itself."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "Because now, all of the information about what direction you're moving is captured with this vector here, what you multiply your nudge by as you're adding the input. So let's just rewrite that over here in the context of directional derivative. What you would say is that the directional derivative in the direction of some vector, any vector, of f, evaluate it at a point. And we'll think about that input point as being a vector itself. A, I'll get rid of this guy. It's also going to be a limit. And as always with these things, we think of some, not I mean always, but with derivatives, you think of some variable as going to 0."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And we'll think about that input point as being a vector itself. A, I'll get rid of this guy. It's also going to be a limit. And as always with these things, we think of some, not I mean always, but with derivatives, you think of some variable as going to 0. And then that's going to be on the denominator. And the change in the function that we're looking for is going to be f evaluated at that initial input vector plus h, that scaling value, that little nudge of a value, multiplied by the vector whose direction we care about. And then you subtract off the value of f at that original input."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And as always with these things, we think of some, not I mean always, but with derivatives, you think of some variable as going to 0. And then that's going to be on the denominator. And the change in the function that we're looking for is going to be f evaluated at that initial input vector plus h, that scaling value, that little nudge of a value, multiplied by the vector whose direction we care about. And then you subtract off the value of f at that original input. So this right here is the formal definition for the directional derivative. And you see how it's much easier to write in vector notation because you're thinking of your input as a vector and your output as just some nudge by something. So let's take a look at what that would feel like over here."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And then you subtract off the value of f at that original input. So this right here is the formal definition for the directional derivative. And you see how it's much easier to write in vector notation because you're thinking of your input as a vector and your output as just some nudge by something. So let's take a look at what that would feel like over here. Instead of thinking of dx and a nudge purely in the x direction, I'll erase these guys, you would think of this point as being a, as being a vector valued a. So just to make clear how it's a vector, you'd be thinking of it starting at the origin and the tip represents that point. And then h times v, you know, maybe v is some vector often, you know, direction that's neither purely x nor purely y."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "So let's take a look at what that would feel like over here. Instead of thinking of dx and a nudge purely in the x direction, I'll erase these guys, you would think of this point as being a, as being a vector valued a. So just to make clear how it's a vector, you'd be thinking of it starting at the origin and the tip represents that point. And then h times v, you know, maybe v is some vector often, you know, direction that's neither purely x nor purely y. But when you scale it down, you know, you scale it down, it'll just be a tiny little nudge, you know, that's going to be h, that tiny little value, scaling your vector v. So that tiny little nudge. And what you wonder is, hey, what's the resulting nudge to the output? And the ratio between the size of that resulting nudge to the output and the original guy there is your directional derivative."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And then h times v, you know, maybe v is some vector often, you know, direction that's neither purely x nor purely y. But when you scale it down, you know, you scale it down, it'll just be a tiny little nudge, you know, that's going to be h, that tiny little value, scaling your vector v. So that tiny little nudge. And what you wonder is, hey, what's the resulting nudge to the output? And the ratio between the size of that resulting nudge to the output and the original guy there is your directional derivative. And more importantly, as you take the limit for that original nudge getting really, really small, that's going to be your directional derivative. And you can probably anticipate there's a way to interpret this as the slope of a graph. That's what I'm going to talk about next video."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "And the ratio between the size of that resulting nudge to the output and the original guy there is your directional derivative. And more importantly, as you take the limit for that original nudge getting really, really small, that's going to be your directional derivative. And you can probably anticipate there's a way to interpret this as the slope of a graph. That's what I'm going to talk about next video. But you actually have to be a little bit careful because we call this the directional derivative. But notice, if you scale the value v by 2, you know, if you go over here and you start plugging in 2 times v and seeing how that influences things, it'll be twice the change. Because here, even if you're scaling by the same value h, it's going to double the initial nudge that you have."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "That's what I'm going to talk about next video. But you actually have to be a little bit careful because we call this the directional derivative. But notice, if you scale the value v by 2, you know, if you go over here and you start plugging in 2 times v and seeing how that influences things, it'll be twice the change. Because here, even if you're scaling by the same value h, it's going to double the initial nudge that you have. And that's going to double the resulting nudge out here, even though the denominator h doesn't stay changed. So when you're taking the ratio, what you're considering is the size of your initial nudge actually might be influenced. So some authors, they'll actually change this definition and they'll throw a little, you know, absolute value of the original vector just to make sure that when you scale it by something else, it doesn't influence things and you only care about the direction."}, {"video_title": "Directional derivative, formal definition.mp3", "Sentence": "Because here, even if you're scaling by the same value h, it's going to double the initial nudge that you have. And that's going to double the resulting nudge out here, even though the denominator h doesn't stay changed. So when you're taking the ratio, what you're considering is the size of your initial nudge actually might be influenced. So some authors, they'll actually change this definition and they'll throw a little, you know, absolute value of the original vector just to make sure that when you scale it by something else, it doesn't influence things and you only care about the direction. But I actually don't like that. I think there's some usefulness in the definition as it is right here and that there's kind of a good interpretation to be had for when, if you double the size of your vector, why that should double the size of your derivative. But I'll get to that in following videos."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "So just as a reminder of where we are, we've got this very nonlinear transformation, and we showed that if you zoom in on a specific point while that transformation is happening, it looks a lot like something linear. And we reasoned that you can figure out what linear transformation that looks like by taking the partial derivatives of your given function, the one that I defined up here, and then turning that into a matrix. And what I want to do here is basically just finish up what I was talking about by computing all of those partial derivatives. So first of all, let me just rewrite the function back on the screen so we have it in a convenient place to look at. The first component is x plus sine of y, sine of y, and then y plus sine of x was the second component. So what I want to do here is just compute all of those partial derivatives to show what kind of thing this looks like. So let's go ahead and get rid of this word."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "So first of all, let me just rewrite the function back on the screen so we have it in a convenient place to look at. The first component is x plus sine of y, sine of y, and then y plus sine of x was the second component. So what I want to do here is just compute all of those partial derivatives to show what kind of thing this looks like. So let's go ahead and get rid of this word. Then I'll go ahead and kind of redraw the matrix here. So for that upper left component, we're taking the partial derivative with respect to x of the first component. So we look up at this first component, and the partial derivative with respect to x is just one, since there's one times x plus something that has nothing to do with x."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "So let's go ahead and get rid of this word. Then I'll go ahead and kind of redraw the matrix here. So for that upper left component, we're taking the partial derivative with respect to x of the first component. So we look up at this first component, and the partial derivative with respect to x is just one, since there's one times x plus something that has nothing to do with x. And then below that, we take the partial derivative of the second component with respect to x down here. And that guy, the y, well, that looks like a constant, so nothing happens, and the derivative of sine of x becomes cosine of x. And then up here, we're taking the partial derivative with respect to y of the first component, that upper one here."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "So we look up at this first component, and the partial derivative with respect to x is just one, since there's one times x plus something that has nothing to do with x. And then below that, we take the partial derivative of the second component with respect to x down here. And that guy, the y, well, that looks like a constant, so nothing happens, and the derivative of sine of x becomes cosine of x. And then up here, we're taking the partial derivative with respect to y of the first component, that upper one here. And for that, partial derivative of x with respect to y is zero, and partial derivative of sine of y with respect to y is cosine of y. And then finally, the partial derivative of the second component with respect to y looks like one, because it's just one times y plus some constant. And this is the general Jacobian, as a function of x and y, but if we want to understand what happens around the specific point that started off at, well, I think I recorded it here, at negative two, one, we plug that into each one of these values."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "And then up here, we're taking the partial derivative with respect to y of the first component, that upper one here. And for that, partial derivative of x with respect to y is zero, and partial derivative of sine of y with respect to y is cosine of y. And then finally, the partial derivative of the second component with respect to y looks like one, because it's just one times y plus some constant. And this is the general Jacobian, as a function of x and y, but if we want to understand what happens around the specific point that started off at, well, I think I recorded it here, at negative two, one, we plug that into each one of these values. So when we plug in negative two, one, so go ahead and just kind of, again, rewrite it to remember, we're plugging in negative two, one as our specific point. And that matrix as a function, kind of a matrix-valued function, becomes one. And then next we have cosine, but we're plugging in negative two for x, cosine of negative two."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "And this is the general Jacobian, as a function of x and y, but if we want to understand what happens around the specific point that started off at, well, I think I recorded it here, at negative two, one, we plug that into each one of these values. So when we plug in negative two, one, so go ahead and just kind of, again, rewrite it to remember, we're plugging in negative two, one as our specific point. And that matrix as a function, kind of a matrix-valued function, becomes one. And then next we have cosine, but we're plugging in negative two for x, cosine of negative two. And if you're curious, that is approximately equal to, I calculated this earlier, negative 0.42, if you just want to think in terms of a number there. Then for the upper right, we have cosine again, but now we're plugging in the value for y, which is one, and cosine of one is approximately equal to 0.54. And then bottom right, that's just another constant, one."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "And then next we have cosine, but we're plugging in negative two for x, cosine of negative two. And if you're curious, that is approximately equal to, I calculated this earlier, negative 0.42, if you just want to think in terms of a number there. Then for the upper right, we have cosine again, but now we're plugging in the value for y, which is one, and cosine of one is approximately equal to 0.54. And then bottom right, that's just another constant, one. So that is the matrix, just as a matrix full of numbers. And just as kind of a gut check, we can take a look at the linear transformation this was supposed to look like. And notice how the first basis vector, the thing it got turned into, which is this vector here, does look like it has coordinates one and negative 0.42, right, it's got this rightward component that's about as long as the vector itself started, and then this downward component, which I think that's pretty believable that that's negative 0.42."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "And then bottom right, that's just another constant, one. So that is the matrix, just as a matrix full of numbers. And just as kind of a gut check, we can take a look at the linear transformation this was supposed to look like. And notice how the first basis vector, the thing it got turned into, which is this vector here, does look like it has coordinates one and negative 0.42, right, it's got this rightward component that's about as long as the vector itself started, and then this downward component, which I think that's pretty believable that that's negative 0.42. And then likewise, this second column is telling us what happened to that second basis vector, which is the one that looks like this. And again, its y component is about as long as how it started, right, a length of one. And then the rightward component is around half of that."}, {"video_title": "Computing a Jacobian matrix.mp3", "Sentence": "And notice how the first basis vector, the thing it got turned into, which is this vector here, does look like it has coordinates one and negative 0.42, right, it's got this rightward component that's about as long as the vector itself started, and then this downward component, which I think that's pretty believable that that's negative 0.42. And then likewise, this second column is telling us what happened to that second basis vector, which is the one that looks like this. And again, its y component is about as long as how it started, right, a length of one. And then the rightward component is around half of that. And we actually see that in the diagram. But this is something you compute, so again, it's pretty straightforward. You just take all of the possible partial derivatives, and you organize them into a grid like this."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What I want to do in this video is to show that we didn't have to use Stokes' Theorem, that we could have just evaluated this line integral. And the thing to keep in mind is, in this case, it's kind of a toss-up which one is actually simpler to do, but Stokes' Theorem is valuable because sometimes, if you're faced with a line integral, it's simpler to use Stokes' Theorem and evaluate the surface integral. Or sometimes, if you're faced with a surface integral, it's simpler to use Stokes' Theorem and evaluate the line integral. So let's try to figure out this line integral. And hopefully, we're gonna get the same answer if we do everything correctly. So the first thing we wanna do is find a parameterization for our path right over here, this intersection of the plane y plus z is equal to two. And essentially, you can imagine this hollow pipe that intersects the xy-axis at the unit circle that goes up and down forever."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's try to figure out this line integral. And hopefully, we're gonna get the same answer if we do everything correctly. So the first thing we wanna do is find a parameterization for our path right over here, this intersection of the plane y plus z is equal to two. And essentially, you can imagine this hollow pipe that intersects the xy-axis at the unit circle that goes up and down forever. And we get this path right over here with this orientation. And since we're only parameterizing a path, it's only gonna deal with one parameter. And so let's think about it a little bit."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And essentially, you can imagine this hollow pipe that intersects the xy-axis at the unit circle that goes up and down forever. And we get this path right over here with this orientation. And since we're only parameterizing a path, it's only gonna deal with one parameter. And so let's think about it a little bit. We've done this many times before, but it doesn't hurt to go through the exercise again. So that is our y-axis. This is our x-axis."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's think about it a little bit. We've done this many times before, but it doesn't hurt to go through the exercise again. So that is our y-axis. This is our x-axis. That is our x-axis. And the x and y values are gonna take on every value on the unit circle. And then the z value is gonna tell us how far above the unit circle we need to be to actually be on this path."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is our x-axis. That is our x-axis. And the x and y values are gonna take on every value on the unit circle. And then the z value is gonna tell us how far above the unit circle we need to be to actually be on this path. So x and y are gonna take on all the values on the unit circle. And we've done that many times before. The easiest way to think about that is let's introduce a parameter called theta."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the z value is gonna tell us how far above the unit circle we need to be to actually be on this path. So x and y are gonna take on all the values on the unit circle. And we've done that many times before. The easiest way to think about that is let's introduce a parameter called theta. Let's introduce a parameter called theta that essentially measures the angle with the positive x-axis and theta, we're just gonna sweep it all around, all the way around the unit circle. So theta is going to go between zero and two pi. So zero is less than theta, which is less than or equal to two pi."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The easiest way to think about that is let's introduce a parameter called theta. Let's introduce a parameter called theta that essentially measures the angle with the positive x-axis and theta, we're just gonna sweep it all around, all the way around the unit circle. So theta is going to go between zero and two pi. So zero is less than theta, which is less than or equal to two pi. And in that situation, x is just going to be, this is just the unit circle definition of trig functions, it's just going to be equal to cosine of theta. Y is going to be sine of theta. And then z, how high we have to go, we can use this constraint to help us figure it out."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So zero is less than theta, which is less than or equal to two pi. And in that situation, x is just going to be, this is just the unit circle definition of trig functions, it's just going to be equal to cosine of theta. Y is going to be sine of theta. And then z, how high we have to go, we can use this constraint to help us figure it out. Y plus z is equal to two, or we could say that z is equal to two minus y. And if y is sine theta, then z is going to be equal to two minus sine theta. And so we're done, that's our parameterization."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then z, how high we have to go, we can use this constraint to help us figure it out. Y plus z is equal to two, or we could say that z is equal to two minus y. And if y is sine theta, then z is going to be equal to two minus sine theta. And so we're done, that's our parameterization. If we wanted to write it as a position vector function, we could write r, which is going to be a function of theta, is equal to cosine of theta i plus sine of theta j plus two minus sine of theta k. And now we're ready to at least attempt to evaluate this line integral. We need to figure out what f dot dr is, and to do that we need to figure out what dr is. And we just have to remind ourselves, dr is the same thing as dr d theta times d theta, which is equal to, so if you take the derivative of this with respect to theta, derivative of cosine theta is negative sine theta, negative sine theta i, derivative of sine theta is cosine theta plus cosine theta j, and derivative of two minus sine theta is going to be negative cosine theta, negative cosine theta k, and then all of that times, we still have this d theta to worry about."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we're done, that's our parameterization. If we wanted to write it as a position vector function, we could write r, which is going to be a function of theta, is equal to cosine of theta i plus sine of theta j plus two minus sine of theta k. And now we're ready to at least attempt to evaluate this line integral. We need to figure out what f dot dr is, and to do that we need to figure out what dr is. And we just have to remind ourselves, dr is the same thing as dr d theta times d theta, which is equal to, so if you take the derivative of this with respect to theta, derivative of cosine theta is negative sine theta, negative sine theta i, derivative of sine theta is cosine theta plus cosine theta j, and derivative of two minus sine theta is going to be negative cosine theta, negative cosine theta k, and then all of that times, we still have this d theta to worry about. This is dr d theta, and then we have, let me write this down, so then we still have to write our d theta, just like that. And now we're ready to take the dot product of f with dr. So let's think about this a little bit."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we just have to remind ourselves, dr is the same thing as dr d theta times d theta, which is equal to, so if you take the derivative of this with respect to theta, derivative of cosine theta is negative sine theta, negative sine theta i, derivative of sine theta is cosine theta plus cosine theta j, and derivative of two minus sine theta is going to be negative cosine theta, negative cosine theta k, and then all of that times, we still have this d theta to worry about. This is dr d theta, and then we have, let me write this down, so then we still have to write our d theta, just like that. And now we're ready to take the dot product of f with dr. So let's think about this a little bit. F dot dr, f, f dot, I'll write dr in this color, f dot dr is going to be equal to, so we look first at our i components. We have negative y squared times negative sine of theta. That's going to be, well, the negatives are gonna cancel out, so you're gonna get y squared times sine theta, y squared sine theta, that's from the i component, plus, now we're gonna have x times cosine theta, plus x times cosine theta, x times cosine theta, and then we're gonna have plus z squared times negative cosine theta."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's think about this a little bit. F dot dr, f, f dot, I'll write dr in this color, f dot dr is going to be equal to, so we look first at our i components. We have negative y squared times negative sine of theta. That's going to be, well, the negatives are gonna cancel out, so you're gonna get y squared times sine theta, y squared sine theta, that's from the i component, plus, now we're gonna have x times cosine theta, plus x times cosine theta, x times cosine theta, and then we're gonna have plus z squared times negative cosine theta. So that's going to be negative z squared, negative z squared times cosine theta, times cosine theta, and then all of that times d theta, all of that business times d theta, and if we're actually going to evaluate the integral, and now we have it in kind of our, we have it, if we're gonna evaluate it over that path that we care about, now we have it in our theta domain, so we could just say this is a simple integral from theta going from zero to two pi. Actually, we're not fully in our theta domain just yet. We still have it expressed in terms of y, x's, and z's, so we have to express those in terms of theta, so let's do that."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's going to be, well, the negatives are gonna cancel out, so you're gonna get y squared times sine theta, y squared sine theta, that's from the i component, plus, now we're gonna have x times cosine theta, plus x times cosine theta, x times cosine theta, and then we're gonna have plus z squared times negative cosine theta. So that's going to be negative z squared, negative z squared times cosine theta, times cosine theta, and then all of that times d theta, all of that business times d theta, and if we're actually going to evaluate the integral, and now we have it in kind of our, we have it, if we're gonna evaluate it over that path that we care about, now we have it in our theta domain, so we could just say this is a simple integral from theta going from zero to two pi. Actually, we're not fully in our theta domain just yet. We still have it expressed in terms of y, x's, and z's, so we have to express those in terms of theta, so let's do that. So that's going to be equal to the integral from zero to two pi, and actually, let me give myself some more space because I have a feeling this might take up a lot of horizontal real estate. This is going to be the integral from zero to two pi, y squared, well, y is just sine theta, so it's going to be sine squared theta times another sine theta, so this is going to be sine cubed theta. Let me write it in a new color or in blue."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We still have it expressed in terms of y, x's, and z's, so we have to express those in terms of theta, so let's do that. So that's going to be equal to the integral from zero to two pi, and actually, let me give myself some more space because I have a feeling this might take up a lot of horizontal real estate. This is going to be the integral from zero to two pi, y squared, well, y is just sine theta, so it's going to be sine squared theta times another sine theta, so this is going to be sine cubed theta. Let me write it in a new color or in blue. So this is sine squared theta times another sine theta, so this is going to be sine cubed theta, sine cubed theta. Actually, let me color code it. So that's sine cubed theta, put some parentheses here, and then x is cosine theta times a cosine theta, so this is going to be plus cosine squared theta, and then z squared, this is actually going to get a little bit involved, so let's think about it, what z squared is."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write it in a new color or in blue. So this is sine squared theta times another sine theta, so this is going to be sine cubed theta, sine cubed theta. Actually, let me color code it. So that's sine cubed theta, put some parentheses here, and then x is cosine theta times a cosine theta, so this is going to be plus cosine squared theta, and then z squared, this is actually going to get a little bit involved, so let's think about it, what z squared is. I'll do it up here. Z squared is just going to be four minus, what is this, four minus four sine theta plus sine squared theta, plus sine squared theta, and so we're going to take, it's going to be negative z squared times cosine theta, so negative z squared is going to be equal to negative four plus four sine theta minus sine squared theta, so that's negative z squared, and we're going to multiply that times cosine theta. I'll do it in orange."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's sine cubed theta, put some parentheses here, and then x is cosine theta times a cosine theta, so this is going to be plus cosine squared theta, and then z squared, this is actually going to get a little bit involved, so let's think about it, what z squared is. I'll do it up here. Z squared is just going to be four minus, what is this, four minus four sine theta plus sine squared theta, plus sine squared theta, and so we're going to take, it's going to be negative z squared times cosine theta, so negative z squared is going to be equal to negative four plus four sine theta minus sine squared theta, so that's negative z squared, and we're going to multiply that times cosine theta. I'll do it in orange. So this whole, all of this business right over here is going to be equal to cosine theta times all of this, cosine theta times all of this right over here, so it's going to be negative four cosine theta, plus four cosine theta sine theta, And then minus cosine theta sine squared theta. Sine squared theta. And looks like we're done, at least for this step."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it in orange. So this whole, all of this business right over here is going to be equal to cosine theta times all of this, cosine theta times all of this right over here, so it's going to be negative four cosine theta, plus four cosine theta sine theta, And then minus cosine theta sine squared theta. Sine squared theta. And looks like we're done, at least for this step. D d theta. And so now we just have to evaluate this integral. So we saw actually setting up our final integral, getting to a simple one-dimensional definite integral is actually much simpler in this case, but the actual integral we have to evaluate is a little bit more complicated."}, {"video_title": "Evaluating line integral directly - part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And looks like we're done, at least for this step. D d theta. And so now we just have to evaluate this integral. So we saw actually setting up our final integral, getting to a simple one-dimensional definite integral is actually much simpler in this case, but the actual integral we have to evaluate is a little bit more complicated. We might have to break out a few of our trig identity tools in order to solve it properly, but we can solve it. But I'll leave you there. In the next video, we'll just work on actually evaluating this integral."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw some arbitrary function right here. That's my function f of x. And let's say we want to find the area between x is equal to a, so that's x equal to a and x is equal to b. We saw this many, many, many videos ago, that the way you can think about it is you take super small widths of x, or super small changes in x. We could call them delta x's, but because they're so small, we're going to call them a dx. We're going to call them a dx, super infinitesimally small changes in x. And then you multiply them times the value of f of x at that point."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We saw this many, many, many videos ago, that the way you can think about it is you take super small widths of x, or super small changes in x. We could call them delta x's, but because they're so small, we're going to call them a dx. We're going to call them a dx, super infinitesimally small changes in x. And then you multiply them times the value of f of x at that point. So you multiply it times the height at that point, which is the value of f of x. So you get f of x times each of these infinitesimally small bases. That will give you the area of this infinitesimally narrow rectangle right there."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you multiply them times the value of f of x at that point. So you multiply it times the height at that point, which is the value of f of x. So you get f of x times each of these infinitesimally small bases. That will give you the area of this infinitesimally narrow rectangle right there. And since each of these guys are infinitely small, you're going to have an infinite number of these rectangles in order to fill this space. You're going to have an infinite number of these. And so the tool we used was the definite integral."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That will give you the area of this infinitesimally narrow rectangle right there. And since each of these guys are infinitely small, you're going to have an infinite number of these rectangles in order to fill this space. You're going to have an infinite number of these. And so the tool we used was the definite integral. The definite integral is an infinite sum of these infinitely small areas, or these infinitely small rectangles. And the notations that we used would say we'd go from a to b. And we've done many videos on how do you evaluate these things, but I just want to remind you conceptually what this is saying."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so the tool we used was the definite integral. The definite integral is an infinite sum of these infinitely small areas, or these infinitely small rectangles. And the notations that we used would say we'd go from a to b. And we've done many videos on how do you evaluate these things, but I just want to remind you conceptually what this is saying. This is conceptually saying let's take a small change in x, multiply it times the height at that point. So small change in x, multiply it times the height at that point. And you're going to have an infinite number of these because these x's are super small."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we've done many videos on how do you evaluate these things, but I just want to remind you conceptually what this is saying. This is conceptually saying let's take a small change in x, multiply it times the height at that point. So small change in x, multiply it times the height at that point. And you're going to have an infinite number of these because these x's are super small. They're infinitely small, so you're going to have an infinite number of those. So take an infinite sum of all of those from x is equal to a to x is equal to b. And that's just our standard definite integral."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you're going to have an infinite number of these because these x's are super small. They're infinitely small, so you're going to have an infinite number of those. So take an infinite sum of all of those from x is equal to a to x is equal to b. And that's just our standard definite integral. Now what I want to do in this video is extend this, broaden this a little bit to solve a, I guess maybe you could say a harder or a broader class of problems. Let's say that we are, let's go to three dimensions now. And I'll just draw the xy plane first."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that's just our standard definite integral. Now what I want to do in this video is extend this, broaden this a little bit to solve a, I guess maybe you could say a harder or a broader class of problems. Let's say that we are, let's go to three dimensions now. And I'll just draw the xy plane first. Maybe I'll keep this just to kind of make the analogy clear. So let me draw it. I'm going to kind of flatten this so we have some perspective."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'll just draw the xy plane first. Maybe I'll keep this just to kind of make the analogy clear. So let me draw it. I'm going to kind of flatten this so we have some perspective. So let's say that this right here is the y-axis, kind of going behind the screen. And you can imagine if I just pushed on this and knocked it down. So that's the y-axis, and that is my x-axis right there."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm going to kind of flatten this so we have some perspective. So let's say that this right here is the y-axis, kind of going behind the screen. And you can imagine if I just pushed on this and knocked it down. So that's the y-axis, and that is my x-axis right there. That is my x-axis. And let's say I have some path in the xy plane. So, and in order to really define a path in the xy plane, I'll have to parametrize, or parametrize, I have trouble saying that, both the x and y variables."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the y-axis, and that is my x-axis right there. That is my x-axis. And let's say I have some path in the xy plane. So, and in order to really define a path in the xy plane, I'll have to parametrize, or parametrize, I have trouble saying that, both the x and y variables. So let's say that x is equal to, let me switch colors. I'm using that orange too much. Let's say that x is equal to some function of some parameter t. And let's say y is equal to some other function of that same parameter t. And let's say we're going to start, we're going to have t go from, t is going to be greater than or equal to a, and then less than or equal to b."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So, and in order to really define a path in the xy plane, I'll have to parametrize, or parametrize, I have trouble saying that, both the x and y variables. So let's say that x is equal to, let me switch colors. I'm using that orange too much. Let's say that x is equal to some function of some parameter t. And let's say y is equal to some other function of that same parameter t. And let's say we're going to start, we're going to have t go from, t is going to be greater than or equal to a, and then less than or equal to b. Now this will define a path in the xy plane, and if this seems confusing, you might want to review the videos on parametric equations. But essentially, you know, when t is equal to a, you're going to have x is equal to, so t is equal to a, you're going to have x is equal to g of a, and you're going to have y is equal to h of a. So you're going to have this point right here, so maybe it might be, I don't know, I'll just draw an ordinary, a random point here, maybe that's, when t is equal to a, you're going to have, you're going to plot the coordinate point g of a, that's going to be our x coordinate, this is g of a right here, and then our y coordinate is going to be h of a. Alright, you just put t is equal to a in each of these equations, and then you get a value for x and y."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that x is equal to some function of some parameter t. And let's say y is equal to some other function of that same parameter t. And let's say we're going to start, we're going to have t go from, t is going to be greater than or equal to a, and then less than or equal to b. Now this will define a path in the xy plane, and if this seems confusing, you might want to review the videos on parametric equations. But essentially, you know, when t is equal to a, you're going to have x is equal to, so t is equal to a, you're going to have x is equal to g of a, and you're going to have y is equal to h of a. So you're going to have this point right here, so maybe it might be, I don't know, I'll just draw an ordinary, a random point here, maybe that's, when t is equal to a, you're going to have, you're going to plot the coordinate point g of a, that's going to be our x coordinate, this is g of a right here, and then our y coordinate is going to be h of a. Alright, you just put t is equal to a in each of these equations, and then you get a value for x and y. So this coordinate right here would be h of a. And then you would keep incrementing t larger and larger until you get to b, but you're going to get a series of points that are going to look something like, let's say it looks something like that. That right there is a curve, or it's a path in the xy plane."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you're going to have this point right here, so maybe it might be, I don't know, I'll just draw an ordinary, a random point here, maybe that's, when t is equal to a, you're going to have, you're going to plot the coordinate point g of a, that's going to be our x coordinate, this is g of a right here, and then our y coordinate is going to be h of a. Alright, you just put t is equal to a in each of these equations, and then you get a value for x and y. So this coordinate right here would be h of a. And then you would keep incrementing t larger and larger until you get to b, but you're going to get a series of points that are going to look something like, let's say it looks something like that. That right there is a curve, or it's a path in the xy plane. And you know, you're saying, how does that relate to that right now? What are we doing? Well, let me just write a c here for saying that's our curve or that's our path."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That right there is a curve, or it's a path in the xy plane. And you know, you're saying, how does that relate to that right now? What are we doing? Well, let me just write a c here for saying that's our curve or that's our path. Now, let's say I have another function that associates every point in the xy plane with some value. So let's say I have some function f of xy. What it does is it associates every point on the xy plane with some value."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, let me just write a c here for saying that's our curve or that's our path. Now, let's say I have another function that associates every point in the xy plane with some value. So let's say I have some function f of xy. What it does is it associates every point on the xy plane with some value. So let me plot f of xy. So let me make a vertical axis here. We could do it in a different color."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What it does is it associates every point on the xy plane with some value. So let me plot f of xy. So let me make a vertical axis here. We could do it in a different color. Call it the f of xy axis. Maybe we could even call it the z axis if you want to. But it's some vertical axis right there."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could do it in a different color. Call it the f of xy axis. Maybe we could even call it the z axis if you want to. But it's some vertical axis right there. And for every point, so if you give me an x and a y, and you put it into my f of xy function, it's going to give you some point. So I could just draw some type of a surface that f of xy represents. And this will all become a lot more concrete when I do some concrete examples."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But it's some vertical axis right there. And for every point, so if you give me an x and a y, and you put it into my f of xy function, it's going to give you some point. So I could just draw some type of a surface that f of xy represents. And this will all become a lot more concrete when I do some concrete examples. So let's say that f of xy looks something like this. I'm going to try my best to draw it. I'll do it in a different color."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this will all become a lot more concrete when I do some concrete examples. So let's say that f of xy looks something like this. I'm going to try my best to draw it. I'll do it in a different color. Let's say f of xy, it's some surface. I'll draw part of it. It's some surface that looks something like that."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it in a different color. Let's say f of xy, it's some surface. I'll draw part of it. It's some surface that looks something like that. That is f of xy. And remember, all this is, is you give me an x and you give me a y. You pop it into f of xy."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's some surface that looks something like that. That is f of xy. And remember, all this is, is you give me an x and you give me a y. You pop it into f of xy. It's going to give me some third value that we're going to plot in this vertical axis right here. Examples, f of xy, it could be, I'm not saying this is the particular case, it could be x plus y. It could be f of xy."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You pop it into f of xy. It's going to give me some third value that we're going to plot in this vertical axis right here. Examples, f of xy, it could be, I'm not saying this is the particular case, it could be x plus y. It could be f of xy. These are just examples. It could be x times y. If x is 1, y is 2. f of xy will be 1 times 2."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It could be f of xy. These are just examples. It could be x times y. If x is 1, y is 2. f of xy will be 1 times 2. But let's say when you plot for every point on the xy plane, when you plot f of xy, you get this surface up here. And we want to do something interesting. We want to figure out not the area under this curve."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If x is 1, y is 2. f of xy will be 1 times 2. But let's say when you plot for every point on the xy plane, when you plot f of xy, you get this surface up here. And we want to do something interesting. We want to figure out not the area under this curve. This was very simple when we did it the first time. I want to find the area, if you imagine a curtain or a fence that goes along this curve, this path, you can imagine this being a very straight linear path going just along the x axis from a to b. Now we have this kind of crazy curvy path that's going along the xy plane."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We want to figure out not the area under this curve. This was very simple when we did it the first time. I want to find the area, if you imagine a curtain or a fence that goes along this curve, this path, you can imagine this being a very straight linear path going just along the x axis from a to b. Now we have this kind of crazy curvy path that's going along the xy plane. And you can imagine if you drew a wall or curtain or a fence that went straight up from this to my f of xy. Let me do my best effort to draw that. Let me draw it."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we have this kind of crazy curvy path that's going along the xy plane. And you can imagine if you drew a wall or curtain or a fence that went straight up from this to my f of xy. Let me do my best effort to draw that. Let me draw it. It's going to go up to there. Maybe this point corresponds to there. And when you draw that curtain up, it's going to intersect it something like that."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw it. It's going to go up to there. Maybe this point corresponds to there. And when you draw that curtain up, it's going to intersect it something like that. Let's say it looks something like that. This point right here corresponds to that point right there. So if you imagine you have a curtain, f of xy is the roof, and what I've drawn here, this curve, this kind of shows you the bottom of a wall."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And when you draw that curtain up, it's going to intersect it something like that. Let's say it looks something like that. This point right here corresponds to that point right there. So if you imagine you have a curtain, f of xy is the roof, and what I've drawn here, this curve, this kind of shows you the bottom of a wall. This is some kind of crazy wall. And let me say this point, it corresponds to... Actually, let me draw it a little bit different. I want to draw it..."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you imagine you have a curtain, f of xy is the roof, and what I've drawn here, this curve, this kind of shows you the bottom of a wall. This is some kind of crazy wall. And let me say this point, it corresponds to... Actually, let me draw it a little bit different. I want to draw it... This point will correspond to some point up here. So when you trace where it intersects, it'll look something like that. I don't know."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to draw it... This point will correspond to some point up here. So when you trace where it intersects, it'll look something like that. I don't know. Something like that. I'm trying my best to help you visualize this. Maybe I'll shade this in to make it a little solid."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I don't know. Something like that. I'm trying my best to help you visualize this. Maybe I'll shade this in to make it a little solid. Let's say f of xy is a little transparent. You can see, but you have this curvy-looking wall here. And the whole point of this video is how can we figure out the area of this curvy-looking wall?"}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe I'll shade this in to make it a little solid. Let's say f of xy is a little transparent. You can see, but you have this curvy-looking wall here. And the whole point of this video is how can we figure out the area of this curvy-looking wall? This curvy-looking wall. That's essentially the wall or the fence that happens if you go from this curve and jump up and hit the ceiling at this f of xy. So let's think a little bit about how we can do it."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the whole point of this video is how can we figure out the area of this curvy-looking wall? This curvy-looking wall. That's essentially the wall or the fence that happens if you go from this curve and jump up and hit the ceiling at this f of xy. So let's think a little bit about how we can do it. If we just use the analogy of what we did previously, we could say, well, look, let's make a little change in distance of our curve. Let's call that ds. That's a little change in distance of my curve right there."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's think a little bit about how we can do it. If we just use the analogy of what we did previously, we could say, well, look, let's make a little change in distance of our curve. Let's call that ds. That's a little change in distance of my curve right there. And if I multiply that change in distance of the curve times f of xy at that point, I'm going to get the area of that little rectangle right there. So if I take ds, my change in my, you can imagine the arc length of this curve at that point. So let me write ds is equal to super small change in arc length of our path or of our curve."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a little change in distance of my curve right there. And if I multiply that change in distance of the curve times f of xy at that point, I'm going to get the area of that little rectangle right there. So if I take ds, my change in my, you can imagine the arc length of this curve at that point. So let me write ds is equal to super small change in arc length of our path or of our curve. That's our ds. So you can imagine the area of that little rectangle right there along my curvy wall is going to be ds times the height at that point. Well, that's f of xy."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me write ds is equal to super small change in arc length of our path or of our curve. That's our ds. So you can imagine the area of that little rectangle right there along my curvy wall is going to be ds times the height at that point. Well, that's f of xy. And then if I take the sum, because these are infinitely narrow, these ds's have infinitely small width. If I were to take the infinite sum of all of those guys from t is equal to a to t is equal to b, from t is equal to a, I keep taking the sum of those rectangles to t is equal to b right there, that will give me my area. I'm just using the exact same logic as I did up there."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's f of xy. And then if I take the sum, because these are infinitely narrow, these ds's have infinitely small width. If I were to take the infinite sum of all of those guys from t is equal to a to t is equal to b, from t is equal to a, I keep taking the sum of those rectangles to t is equal to b right there, that will give me my area. I'm just using the exact same logic as I did up there. I'm not being very mathematically rigorous, but I want to give you the intuition of what we're doing. We're really kind of just bending the base of this thing to get a curvy wall instead of a straight, direct wall like we had up here. But you're saying, Sal, this is all abstract, and how can I even calculate something like this?"}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just using the exact same logic as I did up there. I'm not being very mathematically rigorous, but I want to give you the intuition of what we're doing. We're really kind of just bending the base of this thing to get a curvy wall instead of a straight, direct wall like we had up here. But you're saying, Sal, this is all abstract, and how can I even calculate something like this? This makes no sense to me. I have an s here. I have an x and a y. I have a t. What can I do with this?"}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But you're saying, Sal, this is all abstract, and how can I even calculate something like this? This makes no sense to me. I have an s here. I have an x and a y. I have a t. What can I do with this? Let's see if we can make some headway. And I promise you when we do it with a tangible problem, the end product of this video is going to be a little bit hairy to look at. But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I have an x and a y. I have a t. What can I do with this? Let's see if we can make some headway. And I promise you when we do it with a tangible problem, the end product of this video is going to be a little bit hairy to look at. But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with. But let's see if we can get all of this in terms of t. First of all, let's focus just on this ds. Let me re-pick up the xy-axis. If I were to re-flip the xy, let me switch colors."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But when we do it with an actual problem, it'll actually, I think, be very concrete, and you'll see it's not too hard to deal with. But let's see if we can get all of this in terms of t. First of all, let's focus just on this ds. Let me re-pick up the xy-axis. If I were to re-flip the xy, let me switch colors. This is getting a little monotonous. If I were to re-flip the xy-axis like that, I'm actually only doing that same green, so you know we're dealing with the same xy-axis. That's my y-axis."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I were to re-flip the xy, let me switch colors. This is getting a little monotonous. If I were to re-flip the xy-axis like that, I'm actually only doing that same green, so you know we're dealing with the same xy-axis. That's my y-axis. That is my x-axis. And so this path right here, if I were to just draw it straight up like this, it would look something like this. It would look something like this."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's my y-axis. That is my x-axis. And so this path right here, if I were to just draw it straight up like this, it would look something like this. It would look something like this. That's my path, my arc. This is when t is equal to a. So this is t is equal to a."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would look something like this. That's my path, my arc. This is when t is equal to a. So this is t is equal to a. This is t is equal to b. Same thing, I just kind of picked it back up so we can visualize it. And we say that we have some change in arc length."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is t is equal to a. This is t is equal to b. Same thing, I just kind of picked it back up so we can visualize it. And we say that we have some change in arc length. Let me switch colors. Let's say that this one right here. Let's say that's some small change in arc length, and we're calling that ds."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we say that we have some change in arc length. Let me switch colors. Let's say that this one right here. Let's say that's some small change in arc length, and we're calling that ds. Now, is there some way to relate ds to infinitely small changes in x or y? Well, if we think about it, and this is all a little bit hand-wavy. I'm not being mathematically rigorous, but I think it will give you the correct intuition."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that's some small change in arc length, and we're calling that ds. Now, is there some way to relate ds to infinitely small changes in x or y? Well, if we think about it, and this is all a little bit hand-wavy. I'm not being mathematically rigorous, but I think it will give you the correct intuition. If you imagine, you can figure out the length of ds if you know the lengths of these super small changes in x and super small changes in y. So if this distance right here is dx, infinitesimally small change in x. This distance right here is dy, infinitesimally small change in y."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm not being mathematically rigorous, but I think it will give you the correct intuition. If you imagine, you can figure out the length of ds if you know the lengths of these super small changes in x and super small changes in y. So if this distance right here is dx, infinitesimally small change in x. This distance right here is dy, infinitesimally small change in y. Then we can figure out ds from the Pythagorean theorem. You can say that ds is going to be, it's the hypotenuse of this triangle. It's equal to the square root of dx squared plus dy squared."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This distance right here is dy, infinitesimally small change in y. Then we can figure out ds from the Pythagorean theorem. You can say that ds is going to be, it's the hypotenuse of this triangle. It's equal to the square root of dx squared plus dy squared. That seems to make things a little bit, we can get rid of the ds all of a sudden. Let's rewrite this little expression here using this sense of what ds, ds is really the square root of dx squared plus dy squared. I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's equal to the square root of dx squared plus dy squared. That seems to make things a little bit, we can get rid of the ds all of a sudden. Let's rewrite this little expression here using this sense of what ds, ds is really the square root of dx squared plus dy squared. I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense. We can say that this integral, the area of this curvy curtain, is going to be the integral from t is equal to a to t is equal to b of f of xy. Instead of writing ds, we can write this, times the square root of dx squared plus dy squared. Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm not being very rigorous, and actually it's very hard to be rigorous with differentials, but intuitively I think it makes a lot of sense. We can say that this integral, the area of this curvy curtain, is going to be the integral from t is equal to a to t is equal to b of f of xy. Instead of writing ds, we can write this, times the square root of dx squared plus dy squared. Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this. We have it in terms of t here, but we only have it in terms of x's and y's here. We need to get everything in terms of t. We know x and y are both functions of t, so we can actually rewrite it like this. We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we at least got rid of this big capital S, but we still haven't solved the problem of how do you solve something, an integral, a definite integral that looks like this. We have it in terms of t here, but we only have it in terms of x's and y's here. We need to get everything in terms of t. We know x and y are both functions of t, so we can actually rewrite it like this. We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is. We have that, and then we have this part right here, I'll do it in orange, square root of dx squared plus dy squared. We still don't have things in terms of t. We need a dt someplace here in order to be able to evaluate this integral, and we'll see that in the next video when I do a concrete problem. I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can rewrite it as from t is equal to a to t is equal to b. F of xy, we can write it, f is a function of x, which is a function of t, and f is also a function of y, which is also a function of t. You give me a t, I'll be able to give you an x or a y, and once you give me an x or a y, I can figure out what f is. We have that, and then we have this part right here, I'll do it in orange, square root of dx squared plus dy squared. We still don't have things in terms of t. We need a dt someplace here in order to be able to evaluate this integral, and we'll see that in the next video when I do a concrete problem. I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from. One thing we can do is if we allow ourselves to algebraically manipulate differentials, what we could do is let us multiply and divide by dt. One way to think about it, you could rewrite, let me just do this orange part right here, let's do a little side right here. If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I really want to give you a sense for the end product, the formula we're going to get at the end product of this video, where it comes from. One thing we can do is if we allow ourselves to algebraically manipulate differentials, what we could do is let us multiply and divide by dt. One way to think about it, you could rewrite, let me just do this orange part right here, let's do a little side right here. If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared. Let's say we just multiply it times dt over dt. 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 were to bring this part inside of the square root sign, 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."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you take this orange part and write it in pink, you have dx squared, and then you have plus dy squared. Let's say we just multiply it times dt over dt. 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 were to bring this part inside of the square root sign, 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 leaving this over here. 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 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 then all of that times dt."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we were to bring this part inside of the square root sign, 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 leaving this over here. 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 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 then all of that times dt. I didn't do anything, you could just take the square root of this and you get 1 over dt. 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. dx squared over dt squared is just dx over dt squared, same thing with the y's."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to, and I'll do it very slowly just to 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 then all of that times dt. I didn't do anything, you could just take the square root of this and you get 1 over dt. 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. dx squared over dt squared is just dx over dt squared, same thing with the y's. Now all of a sudden this starts to look pretty interesting. Let's substitute this expression with this one, we said that these are equivalent. I'll switch colors just for the sake of it."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dx squared over dt squared is just dx over dt squared, same thing with the y's. Now all of a sudden this starts to look pretty interesting. Let's substitute this expression with this one, we said that these are equivalent. I'll switch colors just for the sake of it. We have the integral from t is equal to a, let me get our drawing back, from t is equal to a to t is equal to b, of f of x of t, 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 over dt? dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll switch colors just for the sake of it. We have the integral from t is equal to a, let me get our drawing back, from t is equal to a to t is equal to b, of f of x of t, 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 over dt? dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form. The square root, and we take the derivative of x with respect to t squared, plus the derivative of y with respect to t squared, and then all of that times dt. This might look like some strange and convoluted formula, but this is actually something that we know how to deal with. We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dx over dt is the same thing, let me write this down, dx over dt is the same thing, actually I should write it here, is the same thing as g prime of t. x is a function of t, the function I wrote is g prime of t, and then this dy over dt is the same thing as h prime of t, we could say that this function of t. I just want to make that clear, we know these two functions, so we can just take their derivatives with respect to t, but I'm just going to leave it in that form. The square root, and we take the derivative of x with respect to t squared, plus the derivative of y with respect to t squared, and then all of that times dt. This might look like some strange and convoluted formula, but this is actually something that we know how to deal with. We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis. We've taken this strange line integral, that's in terms of the arc length of the line, and x's and y's, and we've put everything in terms of t, and I'm going to show you that in the next video. Everything is going to be expressed in terms of t, so this just turns into a simple, definite integral. Hopefully that didn't confuse you too much, I think you're going to see in the next video that this right here is actually a very straightforward thing to implement."}, {"video_title": "Introduction to the line integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've now simplified this strange, this arc length problem, or this line integral, that's essentially what we're doing, we're taking an integral over a curve, or over a line, as opposed to just an interval on the x axis. We've taken this strange line integral, that's in terms of the arc length of the line, and x's and y's, and we've put everything in terms of t, and I'm going to show you that in the next video. Everything is going to be expressed in terms of t, so this just turns into a simple, definite integral. Hopefully that didn't confuse you too much, I think you're going to see in the next video that this right here is actually a very straightforward thing to implement. Just to remind you where it all came from, I think I got the parentheses right, this right here was just a change in our arc length, that whole thing right there was just a change in arc length, and this is just the height of our function at that point, and we're just summing it, doing an infinite sum of infinitely small lengths. This was a change in our arc length times the height, this is going to have an infinitely narrow width, and you're going to take an infinite number of these rectangles to get the area of this entire fence, or this entire curtain. That's what this definite integral will give us, and we'll actually apply it in the next video."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So type two regions, type two region is a region, I'll call it R2, that's the set of all X, Ys, and Zs in three dimensions, such that, and now instead of thinking of our domain in terms of XY coordinates, we're gonna think of them in terms of YZ coordinates, such that our YZ pairs are a member of some domain, I'll call it D2, since we're talking about type two regions, and X is bounded below by some function of YZ, so I'll call it G1, G1 of YZ is less than or equal to X, which is less than or equal to some other function of YZ, G2 of YZ. And so you'll immediately see a very similar way of thinking about it, but instead of having Z vary between two functions of X and Y, as we had in a type one region, we now have X varying between two functions of Y and Z. Now let's think about some of the shapes we explored. We saw that these two right up here, the sphere and the cylinder, were type one regions, but this dumbbell, the way that I oriented it here, was not a type one region. Let's think about which of these are type two regions and what might not be a type two region. So first let's think about the sphere. So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We saw that these two right up here, the sphere and the cylinder, were type one regions, but this dumbbell, the way that I oriented it here, was not a type one region. Let's think about which of these are type two regions and what might not be a type two region. So first let's think about the sphere. So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane. So YZ plane is this business right over here, so this will be our domain. Don't want to make it more spherical than that. So our domain is this right over here in the YZ plane, that is our D2."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I have my axes right over here, let me scroll down a little bit, so I got my axes, and so over here our domain, we can still construct our sphere, but our domain is now going to be in the YZ plane. So YZ plane is this business right over here, so this will be our domain. Don't want to make it more spherical than that. So our domain is this right over here in the YZ plane, that is our D2. And now the lower bound, in order to construct a region, the solid region of the sphere, or the globe or whatever you want to call it, the lower bound on X would be kind of the back half of the sphere, the one that's away from us right over here. So the lower bound, so let me see how well I can wireframe it at first, the lower bound, I can do a better job than that, so it might go do something like that, and then do something like that, but this is if the domain right over here was transparent, but all we might catch, we'll just catch a glimpse of it in the back right over here. So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our domain is this right over here in the YZ plane, that is our D2. And now the lower bound, in order to construct a region, the solid region of the sphere, or the globe or whatever you want to call it, the lower bound on X would be kind of the back half of the sphere, the one that's away from us right over here. So the lower bound, so let me see how well I can wireframe it at first, the lower bound, I can do a better job than that, so it might go do something like that, and then do something like that, but this is if the domain right over here was transparent, but all we might catch, we'll just catch a glimpse of it in the back right over here. So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us. So if I were to do some contours, it might look something like this, and then look something like this, and then we would color in this entire region right over here. And X can take on all the values above that magenta surface and below this green surface, and essentially it would fill up the globe for every YZ point in our domain. So a sphere is both a type I and a type II region."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's the side of the sphere that's facing away from us, and then the upper bound on X would be the side of the sphere that's facing us. So if I were to do some contours, it might look something like this, and then look something like this, and then we would color in this entire region right over here. And X can take on all the values above that magenta surface and below this green surface, and essentially it would fill up the globe for every YZ point in our domain. So a sphere is both a type I and a type II region. Actually, we're going to see it's going to be a type III region as well. What about this cylinder right over here? Can we construct it or think about it in a way that it would actually be a type II region?"}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So a sphere is both a type I and a type II region. Actually, we're going to see it's going to be a type III region as well. What about this cylinder right over here? Can we construct it or think about it in a way that it would actually be a type II region? So let's try to do that. So let me paste it. So what if we had a domain, what if our domain was something like this?"}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Can we construct it or think about it in a way that it would actually be a type II region? So let's try to do that. So let me paste it. So what if we had a domain, what if our domain was something like this? It was a rectangle in the YZ plane. So it was a rectangle in the YZ plane. So this is our domain, a rectangle in the YZ plane."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what if we had a domain, what if our domain was something like this? It was a rectangle in the YZ plane. So it was a rectangle in the YZ plane. So this is our domain, a rectangle in the YZ plane. So that would be my D2. And what if the lower bound was kind of the back side of the cylinder? So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is our domain, a rectangle in the YZ plane. So that would be my D2. And what if the lower bound was kind of the back side of the cylinder? So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder. And so if we just saw the outside of it, it would look something like that. It's facing away from us, so we barely see it. If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the back side of the cylinder, try to draw it as good as I can, the back side of the cylinder. And so if we just saw the outside of it, it would look something like that. It's facing away from us, so we barely see it. If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that. So that over there would be our G1. And then our G2 would be the front side of the cylinder. The G2 could be the front side of the cylinder."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we could see through the cylinder or see through the little flat cut of the cylinder, it would look something like that. So that over there would be our G1. And then our G2 would be the front side of the cylinder. The G2 could be the front side of the cylinder. So let me color it in as best as I can. So G2 would be the front side of the cylinder. And X can vary above G1 and below G2, and it would fill up this entire cylinder."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The G2 could be the front side of the cylinder. So let me color it in as best as I can. So G2 would be the front side of the cylinder. And X can vary above G1 and below G2, and it would fill up this entire cylinder. So we see that this same cylinder that we also saw was a type I region can also be a type II region. Now what about this hourglass thing that we saw could not be a type I region? Can this be a type II region?"}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And X can vary above G1 and below G2, and it would fill up this entire cylinder. So we see that this same cylinder that we also saw was a type I region can also be a type II region. Now what about this hourglass thing that we saw could not be a type I region? Can this be a type II region? Well, let's think about it. Let's think about it. I'll do it the same way."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Can this be a type II region? Well, let's think about it. Let's think about it. I'll do it the same way. We can construct a domain. So maybe our domain, it's in the Y, well, it should be in the YZ plane if we're talking about type II regions, or if we want to think of it as a type II region. So our domain could be this kind of flat hourglass shape that's in the YZ plane."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it the same way. We can construct a domain. So maybe our domain, it's in the Y, well, it should be in the YZ plane if we're talking about type II regions, or if we want to think of it as a type II region. So our domain could be this kind of flat hourglass shape that's in the YZ plane. So our domain could be a region that looks something like this in the YZ plane. So this is kind of flattened out. So this is our domain right over there."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our domain could be this kind of flat hourglass shape that's in the YZ plane. So our domain could be a region that looks something like this in the YZ plane. So this is kind of flattened out. So this is our domain right over there. And then the lower bound on X, G1, could be a surface. It's a function of Y and Z that is kind of the backside of our hourglass. You can see, I'll try to show the contours from the underside right over there."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is our domain right over there. And then the lower bound on X, G1, could be a surface. It's a function of Y and Z that is kind of the backside of our hourglass. You can see, I'll try to show the contours from the underside right over there. So that could be our G1. And then our G2 could be the front side of the hourglass. So my best attempt to draw the front side of the hourglass."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can see, I'll try to show the contours from the underside right over there. So that could be our G1. And then our G2 could be the front side of the hourglass. So my best attempt to draw the front side of the hourglass. And I could color it in. And so the way I somewhat confusingly drew it just now, you see that this hourglass oriented the way it is would actually be a type II region. Now, if we were to rotate it like this, so let me draw it like this."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So my best attempt to draw the front side of the hourglass. And I could color it in. And so the way I somewhat confusingly drew it just now, you see that this hourglass oriented the way it is would actually be a type II region. Now, if we were to rotate it like this, so let me draw it like this. So if we were to make it like this, so that the top of my hourglass is facing us, I'll try my best to draw it. So let's say the top intersects the X axis right over there. This is the bottom of my hourglass right over there."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, if we were to rotate it like this, so let me draw it like this. So if we were to make it like this, so that the top of my hourglass is facing us, I'll try my best to draw it. So let's say the top intersects the X axis right over there. This is the bottom of my hourglass right over there. And then it bends in and then comes back out like that. So it bends in and comes back out just like that. For the same reasons that this was not a type I region, this now would not be a type II region."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the bottom of my hourglass right over there. And then it bends in and then comes back out like that. So it bends in and comes back out just like that. For the same reasons that this was not a type I region, this now would not be a type II region. For any XY, you're going to see that there's multiple points. There's multiple, sorry, for any ZY, you can see there could be multiple X points that are associated with the different points of this hourglass. You can't just have a simple lower and upper bound functions right over here."}, {"video_title": "Type II regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "For the same reasons that this was not a type I region, this now would not be a type II region. For any XY, you're going to see that there's multiple points. There's multiple, sorry, for any ZY, you can see there could be multiple X points that are associated with the different points of this hourglass. You can't just have a simple lower and upper bound functions right over here. So this right over here is not a type II region. You could show a rationale, or this is going to be a type I region. You could create a region over here in the XY plane and have an upper and lower bound functions for Z, so you could be type I."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what I want to do is think about the value of the line integral, let me write this down, the value of the line integral of F dot dr, where F is the vector field that I've drawn in magenta in each of these diagrams, and obviously it's different in each of these diagrams. The only part of the vector field that I've drawn is the part that's along the surface. I could have drawn the part of the vector field that's off the surface, but we're only going to be concerned with what's going on on the surface. So the vector field could be defined in this entire three-dimensional space, and here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about, remember we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the vector field could be defined in this entire three-dimensional space, and here as well. And these are obviously different vector fields, and we can see that visually based on how we drew it. And the contour that we care about, remember we're going to take a line integral, so the path matters. The path that we care about is the counterclockwise boundary of our surface. So it's going to be this right over here, the counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation, it's going to be counterclockwise. We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The path that we care about is the counterclockwise boundary of our surface. So it's going to be this right over here, the counterclockwise boundary of our surface is what we're going to be taking F dot dr along. So this right over here, and let me draw the orientation, it's going to be counterclockwise. We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example, and obviously the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to do it in every one of these situations, every one of these surfaces and every one of these Fs. And what I really want to think about is how the value of F dot dr over that contour, how it might change from example to example, and obviously the only difference between each of these is what the vector field F is doing. So first let's think about this example right over here. This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So it's going to be we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This part of the contour, this bottom part right over here, our vector field is going in the exact same direction as our line, as our contour. So it's going to be we're going to get positive values of F dot dr down here, and we're going to keep summing them up. We're taking an integral. Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then as we go up the curve, or as we go kind of uphill right over here, we see that our vector field is going essentially orthogonal. It's going perpendicular to our contour. So our F dot dr, we're not going to get any value from that. F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "F dot dr and all of these, this part of the contour, it's all going to be 0. So we're not going to get anything right over there. Actually, let me just write nothing. So we get nothing right over there. Or maybe I'll write 0. We'll get 0 right over there. Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we get nothing right over there. Or maybe I'll write 0. We'll get 0 right over there. Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. So we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then up here, when we're at this part of the contour, our vector field is going in the exact opposite direction as our path. Up here our path is going, I guess, from the right to the left, while our vector field is going from the left to the right. So we're actually going to get negative values. We are going to get negative values up here, and they're going to sum up to a reasonable negative value. If the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We are going to get negative values up here, and they're going to sum up to a reasonable negative value. If the vector field is constant, and I kind of drew it like it is, and if this length is equal to this length, then these two values are going to cancel out. When you add this positive sum to this negative sum, they're going to be 0. Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F, in this example right over here, it might all cancel out. You'll get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then once again, when you go downhill again, the vector field is perpendicular to our actual path, so we're going to get 0. So based on the way I've described it, your line integral of F dot dr for this version of F, in this example right over here, it might all cancel out. You'll get something, if we make the assumptions that I made, this might be equal to 0. So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in this example, F dot dr could be equal to 0. Now let's think about what's going on in this situation. In this one, just like the last one, as we go along the bottom, the vector field is going in the exact same direction as our contour, so we're going to get positive values. Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part. But then up here, our vector field has switched directions, and once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. Then as we go down here, it's not going to add anything because our vector field is perpendicular to our path, so we're going to get 0. But notice, now these two ends don't cancel out with each other."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Once we go up the hill, the vector field is going perpendicular to our path, so it's not going to add anything really to it, so we're just going to get 0 along this part. But then up here, our vector field has switched directions, and once again, it's going in the exact same direction as our path, so we're going to get more positive values right over there. Then as we go down here, it's not going to add anything because our vector field is perpendicular to our path, so we're going to get 0. But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. What was the difference between this version of F, this vector field, and this vector field right over here?"}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But notice, now these two ends don't cancel out with each other. We're going to get a positive value. We are going to get a positive value. What was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is it had some curl. There's a little bit of spinning going on."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What was the difference between this version of F, this vector field, and this vector field right over here? Well, this vector field switched directions so that the top part didn't cancel out with the bottom part. Or another way to think about it is it had some curl. There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "There's a little bit of spinning going on. If this was describing the velocity of a fluid, and if you were to put a stick right over there on the surface, the stick would spin. It has some spin or some curl, however you want to describe it. This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation, and we also have what looks like a positive curl. Now let's think about this one."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right over here has no curl. If you put a stick right over here, it would just flow with the fluid, but the stick itself would not spin. So we got a positive value for the line integral in this situation, and we also have what looks like a positive curl. Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction so we'll get more positive values. Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's think about this one. In this situation, as we go along this part of our contour, our vector field F is going in the exact same direction so we're going to get positive values. Now as we go uphill, our vector field F, it's also kind of turned in that direction so we'll get more positive values. Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour So we'll get even more positive values. In this situation, the value of our line integral of F dot dr is even more positive. More, more positive."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, as we go in that direction, our vector field F is still going in the direction of our contour, we're going to get more positive values. And as we go down, once again the vector field F is going in the direction of our contour So we'll get even more positive values. In this situation, the value of our line integral of F dot dr is even more positive. More, more positive. And we see that the actual vector field along the surface, and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "More, more positive. And we see that the actual vector field along the surface, and remember, the vector field might be doing all sorts of crazy things off the surface. Actually, let me draw that in that same magenta color. It might be doing all sorts of crazy things off the surface. But what we really care about is what's happening on the surface. And because this vector field is, I guess you could say, curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It might be doing all sorts of crazy things off the surface. But what we really care about is what's happening on the surface. And because this vector field is, I guess you could say, curling, or it's spinning along the surface, it allows it to go with the boundary along all the points, and we get a very positive value for this line integral. So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have a higher curl, so more curl, more curl, it looks like, is leading to a more positive line integral. Now let's think about what's happening in this situation right over here. This situation down here, our vector field is going in the same direction as our path, so we're going to get positive values. Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us, so it's negative right over here. We're going in the exact opposite direction of our path. And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Just like the first situation, as we go up the hill like this, or up the surface like that, our vector field is perpendicular to our surface, so it's really not going to add anything to our line integral. And then as we go along this top part, this first part of the top part right over here, the vector field is going against us, so it's negative right over here. We're going in the exact opposite direction of our path. And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing, because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then right as we get to the end, the vector field flips direction, and we get a little bit of positive right over here, so it's a little bit of it going in the same direction. And then we go back downhill. When we go back downhill, it adds nothing, because our vector field is going perpendicular to our path. So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that if we take the line integral, but more positive than that. And one other way to think about it is we have a little bit of curl going on right over here, our vector field switched directions right around there, or I guess you could say that it's spinning right around there, so if you put a stick, if that was in the water, it would start spinning. But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the big difference between this case and the case up here is the case up here, or actually the big difference between this, well actually I could compare between these two or these two, but the difference between this one and this one is that at least this part of the vector field has switched direction, so we get a little bit of positive value. And one way to think about it, this one is going to be less positive than that if we take the line integral, but more positive than that. And one other way to think about it is we have a little bit of curl going on right over here, our vector field switched directions right around there, or I guess you could say that it's spinning right around there, so if you put a stick, if that was in the water, it would start spinning. But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface. Over here you had curl going on over a larger portion of our surface, and so up here you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But everywhere else there isn't a lot of curl, so you have some curl, but it's over a little small region of the surface. Over here you had curl going on over a larger portion of our surface, and so up here you had a more positive curl, more positive line integral. Here you have a curl over less of the surface, and you're going to have a less positive line integral. Now let's think about this one over here. This vector field along the surface, there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's think about this one over here. This vector field along the surface, there is some curl. There's some curl going on right over there. If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl. But then it switches direction again, so you have curl there as well. It's actually curl in the opposite direction. To some degree, if you were to sum all of this up, maybe it would cancel out."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you put a stick in the water, if you view that as the velocity of water, the stick would spin, so you have some curl. But then it switches direction again, so you have curl there as well. It's actually curl in the opposite direction. To some degree, if you were to sum all of this up, maybe it would cancel out. It makes sense that it would cancel out, because when you take the line integral around the whole thing, just like this first situation, it looks like it'll add up to zero. Because even though you have some curl, the curls cancel out each other, and so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "To some degree, if you were to sum all of this up, maybe it would cancel out. It makes sense that it would cancel out, because when you take the line integral around the whole thing, just like this first situation, it looks like it'll add up to zero. Because even though you have some curl, the curls cancel out each other, and so when you go to this top part of the surface, the vector field is going in the exact same direction as this bottom part of the surface. If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. It would be negative up there, and then as we go down, it would be zero. This thing right over here also looks just like the first one, because the curls essentially cancel out. We switch direction twice."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you were to take your line integral, the same one that we care about, just like the first situation, it would be positive down here, zero as we go up the curve, and then as we go down here, the vector field has switched directions twice, so it's still going against the path of our contour, just like this first situation. It would be negative up there, and then as we go down, it would be zero. This thing right over here also looks just like the first one, because the curls essentially cancel out. We switch direction twice. Over here, our line integral might also be zero. The whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We switch direction twice. Over here, our line integral might also be zero. The whole reason why I went through this exercise is to give you an intuition of why it might make sense that if we have more curl happening over more of this surface, why that might make the value of this line integral be larger. Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface. Let's think about this. It could be a surface integral. We're going to go over the surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully, it starts to give you an intuition that maybe, just maybe, the value of this line integral, the value of F dot dr over that contour, over this contour that's going in the counterclockwise direction, we'll talk more about orientation in future videos, maybe this is equal to the sum of the curls over the surface. Let's think about this. It could be a surface integral. We're going to go over the surface. What we care about is the curl of F. We don't just care about the curl of F generally, because F might be spinning in a direction in a way that's off the surface. We care about how much it's curling on the surface. What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to go over the surface. What we care about is the curl of F. We don't just care about the curl of F generally, because F might be spinning in a direction in a way that's off the surface. We care about how much it's curling on the surface. What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself. Then multiply that times the surface itself. Another way of writing this, and this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral might be. We saw that when we compared these three examples."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What we'd want to do is we'd want to take the curl of F and dot it with the normal vector at any point of the surface, and then multiply that times the surface itself. Then multiply that times the surface itself. Another way of writing this, and this is just to say that the more surface where we have more curling going on, the more that the line integral, the value of the line integral might be. We saw that when we compared these three examples. Another way of writing all of this is the surface integral. Let me write the surface in that same brown color. The surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface."}, {"video_title": "Stokes' theorem intuition   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We saw that when we compared these three examples. Another way of writing all of this is the surface integral. Let me write the surface in that same brown color. The surface integral of the curl of F, which would just be another vector that tells us how much we are spinning generally, but we want to care how much we're spinning along the surface. We're dotting it with the normal vector. Another way to write this whole thing is to say dot ds. If you take essentially the sum across the entire surface of how much we're curling, how much we're spinning along that surface, then maybe, just maybe, this will be equal to the value of the line integral as we go around the boundary of the surface."}, {"video_title": "2d curl formula.mp3", "Sentence": "So after introducing the idea of fluid rotation in a vector field like this, let's start tightening up our grasp on this intuition to get something that we can actually apply formulas to. So a vector field like the one that I had there that's two dimensional is given by a function that has a two dimensional input and a two dimensional output. And it's common to write the components of that output as the functions p and q. So each one of those p and q takes in two different variables as its input, p and q. And what I wanna do here is talk about this idea of curl. And you might write it down as just curl, curl of v, the vector field, which takes in the same inputs that the vector field does. And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function."}, {"video_title": "2d curl formula.mp3", "Sentence": "So each one of those p and q takes in two different variables as its input, p and q. And what I wanna do here is talk about this idea of curl. And you might write it down as just curl, curl of v, the vector field, which takes in the same inputs that the vector field does. And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function. And you give it a function and it gives you a new function, the derivative. Here, you think of this 2D curl as like an operator. You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued."}, {"video_title": "2d curl formula.mp3", "Sentence": "And because this is the two dimensional example, I might write just to distinguish it from three dimensional curl, which is something we'll get later on, 2D curl of v. So you're kind of thinking of this as a differential thing in the same way that you have a derivative, d dx, it's gonna take in some kind of function. And you give it a function and it gives you a new function, the derivative. Here, you think of this 2D curl as like an operator. You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued. And the reason it's scalar valued is because at every given point, you want it to give you a number. So if I look back at the vector field that I have here, we want that at a point like this where there's a lot of counterclockwise rotation happening around it, for the curl function to return a positive number. But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number."}, {"video_title": "2d curl formula.mp3", "Sentence": "You give it a function, a vector field function, and it gives you another function, which in this case will be scalar valued. And the reason it's scalar valued is because at every given point, you want it to give you a number. So if I look back at the vector field that I have here, we want that at a point like this where there's a lot of counterclockwise rotation happening around it, for the curl function to return a positive number. But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number. So let's start thinking about what that should mean. And a good way to understand this two-dimensional curl function and start to get a feel for it is to imagine the quintessential 2D curl scenario. Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space."}, {"video_title": "2d curl formula.mp3", "Sentence": "But at a point like this where there's some counter, where there's clockwise rotation happening around it, we want the curl to return a negative number. So let's start thinking about what that should mean. And a good way to understand this two-dimensional curl function and start to get a feel for it is to imagine the quintessential 2D curl scenario. Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space. And let's say there's no vector attached to it, as in the values p and q at x and y are zero. And then let's say that to the right of it, you have a vector pointing straight up. Above it, in the vector field, you have a vector pointing straight to the left."}, {"video_title": "2d curl formula.mp3", "Sentence": "Well, let's say you have a point, and this here is gonna be our point x, y, sitting off somewhere in space. And let's say there's no vector attached to it, as in the values p and q at x and y are zero. And then let's say that to the right of it, you have a vector pointing straight up. Above it, in the vector field, you have a vector pointing straight to the left. To its left, you have one pointing straight down. And below it, you have one pointing straight to the right. So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero."}, {"video_title": "2d curl formula.mp3", "Sentence": "Above it, in the vector field, you have a vector pointing straight to the left. To its left, you have one pointing straight down. And below it, you have one pointing straight to the right. So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero. So this function q that corresponds to the y component, the up and down component of each vector, when you evaluate it at this point to the right of our x, y point, q is gonna be greater than zero. Whereas if you evaluate it to the left over here, q would be less than zero, less than zero, in our kind of perfect curl will be positive example. And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive."}, {"video_title": "2d curl formula.mp3", "Sentence": "So in terms of the functions, what that means is this vector to its right, whatever point it's evaluated at, that's gonna be q is greater than zero. So this function q that corresponds to the y component, the up and down component of each vector, when you evaluate it at this point to the right of our x, y point, q is gonna be greater than zero. Whereas if you evaluate it to the left over here, q would be less than zero, less than zero, in our kind of perfect curl will be positive example. And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive. And then above it, above it here, when you evaluate p at that point, it would have to be negative. Whereas p, if you did it on the left and right points, would be equal to zero because there's kind of no x component. And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero."}, {"video_title": "2d curl formula.mp3", "Sentence": "And then these bottom guys, if you start thinking about what this means for, you'd have a rightward vector below and a leftward vector above, the one below it, whatever point you're evaluating that at, p, which gives us the kind of left-right component of these vectors, since it's the first component of the output, would have to be positive. And then above it, above it here, when you evaluate p at that point, it would have to be negative. Whereas p, if you did it on the left and right points, would be equal to zero because there's kind of no x component. And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero. So this is just the very specific, almost contrived scenario that I'm looking at. And I wanna say, hey, if this should have positive curl, maybe if we look at the information, the partial derivative information to be specific, about p and q in a scenario like this, it'll give us a way to quantify the idea of curl. And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative."}, {"video_title": "2d curl formula.mp3", "Sentence": "And similarly, q, if you did it on the top and bottom points since there's no up and down component of those vectors, would also be zero. So this is just the very specific, almost contrived scenario that I'm looking at. And I wanna say, hey, if this should have positive curl, maybe if we look at the information, the partial derivative information to be specific, about p and q in a scenario like this, it'll give us a way to quantify the idea of curl. And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative. So we would expect that the partial derivative of p with respect to y, so as we change that y component moving up in the plane and look at the x component of the vectors, that should be negative. That should be negative in circumstances where we want positive curl. So all of this we're looking at cases, you know, the quintessential case where curl is positive."}, {"video_title": "2d curl formula.mp3", "Sentence": "And first let's look at p. So p starts positive, and then as y increases, as the y value of our input increases, it goes from being positive to zero to negative. So we would expect that the partial derivative of p with respect to y, so as we change that y component moving up in the plane and look at the x component of the vectors, that should be negative. That should be negative in circumstances where we want positive curl. So all of this we're looking at cases, you know, the quintessential case where curl is positive. So evidently this is a fact that corresponds to positive curl. Whereas q, let's take a look at q. It starts negative when you're at the left, and then becomes zero, then it becomes positive."}, {"video_title": "2d curl formula.mp3", "Sentence": "So all of this we're looking at cases, you know, the quintessential case where curl is positive. So evidently this is a fact that corresponds to positive curl. Whereas q, let's take a look at q. It starts negative when you're at the left, and then becomes zero, then it becomes positive. So here as x increases, q increases. So we're expecting that a partial derivative of q with respect to x should be positive. Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl."}, {"video_title": "2d curl formula.mp3", "Sentence": "It starts negative when you're at the left, and then becomes zero, then it becomes positive. So here as x increases, q increases. So we're expecting that a partial derivative of q with respect to x should be positive. Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl. And in fact, it turns out, these guys tell us all you need to know. We can say as a formula that the 2D curl, 2D curl of our vector field v as a function of x and y is equal to the partial derivative of q with respect to x, partial derivative of q with respect to x, and then I'm gonna subtract off the partial of p with respect to y, because I want when this is negative for that to correspond with more positive 2D curl. So I'm gonna subtract off partial of p with respect to y."}, {"video_title": "2d curl formula.mp3", "Sentence": "Or at the very least, that situations where the partial derivative of q with respect to x is positive corresponds to positive two-dimensional curl. And in fact, it turns out, these guys tell us all you need to know. We can say as a formula that the 2D curl, 2D curl of our vector field v as a function of x and y is equal to the partial derivative of q with respect to x, partial derivative of q with respect to x, and then I'm gonna subtract off the partial of p with respect to y, because I want when this is negative for that to correspond with more positive 2D curl. So I'm gonna subtract off partial of p with respect to y. And this right here is the formula for two-dimensional curl. Which basically, you can think of it as a measure at any given point, you're asking how much does the surrounding information to that point look like this setup, like this perfect counterclockwise rotation setup. And the more it looks like this setup, the more this value will be positive."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "One of the most fundamental ideas in all of physics is the idea of work. And when you first learn work, you just say, oh, that's just force times distance. But then later on, when you learn a little bit about vectors, you realize that the force isn't always going in the same direction as your displacement. So you learn that work is really the magnitude of the force in the direction, or the component of the force in the direction, in direction of displacement. Displacement is just distance with some direction. In direction of displacement times the magnitude of the displacement. Or you could just kind of say times the distance displaced."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you learn that work is really the magnitude of the force in the direction, or the component of the force in the direction, in direction of displacement. Displacement is just distance with some direction. In direction of displacement times the magnitude of the displacement. Or you could just kind of say times the distance displaced. I'm not being too particular about it. And the classic example, maybe you have an ice cube or some type of block. I just say ice so that it's not a lot of friction."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or you could just kind of say times the distance displaced. I'm not being too particular about it. And the classic example, maybe you have an ice cube or some type of block. I just say ice so that it's not a lot of friction. Maybe it's standing on a bigger sheet of lake or ice or something. And maybe you're pulling on that ice cube at an angle. Let's say you're pulling at an angle like that."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I just say ice so that it's not a lot of friction. Maybe it's standing on a bigger sheet of lake or ice or something. And maybe you're pulling on that ice cube at an angle. Let's say you're pulling at an angle like that. That is my force right there. Let's say my force is equal to, well, that's my force vector, let's say the magnitude of my force vector. So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say you're pulling at an angle like that. That is my force right there. Let's say my force is equal to, well, that's my force vector, let's say the magnitude of my force vector. So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons. And let's say the direction of my force vector, any vector has to have a magnitude and a direction. And the direction, let's say it has a 30 degree angle, let's say it's 60 degree angle above horizontal. So that's the direction I'm pulling in."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'll say the magnitude, so I'm going to put two brackets around there, the magnitude of my force vector we can say is let's say it's 10 newtons. And let's say the direction of my force vector, any vector has to have a magnitude and a direction. And the direction, let's say it has a 30 degree angle, let's say it's 60 degree angle above horizontal. So that's the direction I'm pulling in. And then let's say I displace it. This is all a bit of review, hopefully. If you're displacing it, let's say you displace it 5 newtons."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the direction I'm pulling in. And then let's say I displace it. This is all a bit of review, hopefully. If you're displacing it, let's say you displace it 5 newtons. So let's say the displacement, that's the displacement vector right there, and the magnitude of it is equal to 5 meters. So you've learned from the definition of work, you can't just say, oh, I'm pulling with 10 newtons of force and I'm moving at 5 meters. You can't just multiply the 10 newtons times the 5 meters."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you're displacing it, let's say you displace it 5 newtons. So let's say the displacement, that's the displacement vector right there, and the magnitude of it is equal to 5 meters. So you've learned from the definition of work, you can't just say, oh, I'm pulling with 10 newtons of force and I'm moving at 5 meters. You can't just multiply the 10 newtons times the 5 meters. You have to find the magnitude of the component going in the same direction as my displacement. So what I essentially need to do is, if you imagine the length of this vector being 10, that's the total force, but you need to figure out the length of the vector, the component of the force going in the same direction as my displacement. And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can't just multiply the 10 newtons times the 5 meters. You have to find the magnitude of the component going in the same direction as my displacement. So what I essentially need to do is, if you imagine the length of this vector being 10, that's the total force, but you need to figure out the length of the vector, the component of the force going in the same direction as my displacement. And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees. Or that's equal to, cosine of 60 degrees is 1 half, so that's just equal to 5. So this magnitude, the magnitude of the force going in the same direction of the displacement in this case, is 5 newtons. And then you can figure out the work."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And a little simple trigonometry, you know that this is 10 times the cosine of 60 degrees. Or that's equal to, cosine of 60 degrees is 1 half, so that's just equal to 5. So this magnitude, the magnitude of the force going in the same direction of the displacement in this case, is 5 newtons. And then you can figure out the work. You could say that the work is equal to 5 newtons times, I'll just write a dot for times, I don't want you to think it's cross product, times 5 meters, which is 25 newton meters, or you could even say 25 joules of work have been done. And this is all really a review of somewhat basic physics, but just think about what happened here. What was the work?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you can figure out the work. You could say that the work is equal to 5 newtons times, I'll just write a dot for times, I don't want you to think it's cross product, times 5 meters, which is 25 newton meters, or you could even say 25 joules of work have been done. And this is all really a review of somewhat basic physics, but just think about what happened here. What was the work? If I write it in the abstract. The work is equal to the 5 newtons, that was the magnitude of my force vector, so it's the magnitude of my force vector, times the cosine of this angle. Times the cosine of the, so let's call that theta, let's say it a little generally."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What was the work? If I write it in the abstract. The work is equal to the 5 newtons, that was the magnitude of my force vector, so it's the magnitude of my force vector, times the cosine of this angle. Times the cosine of the, so let's call that theta, let's say it a little generally. So times the cosine of the angle. This is the amount of my force in the direction of the displacement, the cosine of the angle between them, times the magnitude of the displacement. So times the magnitude of the displacement."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times the cosine of the, so let's call that theta, let's say it a little generally. So times the cosine of the angle. This is the amount of my force in the direction of the displacement, the cosine of the angle between them, times the magnitude of the displacement. So times the magnitude of the displacement. Or if I wanted to rewrite that, I could just write that as the magnitude of the displacement times the magnitude of the force times the cosine of theta. And I've done multiple videos of this in the linear algebra playlist and the physics playlist where I talk about the dot product and the cross product and all of that. But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So times the magnitude of the displacement. Or if I wanted to rewrite that, I could just write that as the magnitude of the displacement times the magnitude of the force times the cosine of theta. And I've done multiple videos of this in the linear algebra playlist and the physics playlist where I talk about the dot product and the cross product and all of that. But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors. And if the dot product is a completely foreign concept to you, you might want to watch, I think I've made multiple, four or five videos on the dot product and its intuition and how it compares. But just to give you a little bit of that intuition right here, the dot product, when I take f dot d or d dot f, what it's giving me is I'm multiplying the magnitude. Well, I could just read this out."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is the dot product of d and f, of the vectors d and f. So in general, if you're trying to find the work for a constant displacement and you have a constant force, you just take the dot product of those two vectors. And if the dot product is a completely foreign concept to you, you might want to watch, I think I've made multiple, four or five videos on the dot product and its intuition and how it compares. But just to give you a little bit of that intuition right here, the dot product, when I take f dot d or d dot f, what it's giving me is I'm multiplying the magnitude. Well, I could just read this out. But the idea of the dot product is take how much of this vector is going in the same direction as this vector. And in this case, it's this much. And then multiply the two magnitudes."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, I could just read this out. But the idea of the dot product is take how much of this vector is going in the same direction as this vector. And in this case, it's this much. And then multiply the two magnitudes. And that's what we did right here. So the work is going to be the force vector dot, taking the dot product of the force vector with the displacement vector, and this, of course, is a scalar value. And we'll work out some examples in the future where you'll see that that's true."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then multiply the two magnitudes. And that's what we did right here. So the work is going to be the force vector dot, taking the dot product of the force vector with the displacement vector, and this, of course, is a scalar value. And we'll work out some examples in the future where you'll see that that's true. So this is all a review of fairly elementary physics. Now let's take a more complex example, but it's really the same idea. Let's define a vector field."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we'll work out some examples in the future where you'll see that that's true. So this is all a review of fairly elementary physics. Now let's take a more complex example, but it's really the same idea. Let's define a vector field. So let's say that I have a vector field f, and we're going to think about what this means in a second. It's a function of x and y, and it's equal to some scalar function of x and y times the i unit vector, or the horizontal unit vector, plus some other function, scalar function of x and y, times the vertical unit vector. So what would something like this be?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's define a vector field. So let's say that I have a vector field f, and we're going to think about what this means in a second. It's a function of x and y, and it's equal to some scalar function of x and y times the i unit vector, or the horizontal unit vector, plus some other function, scalar function of x and y, times the vertical unit vector. So what would something like this be? This is a vector field. This is a vector field in two-dimensional space, or on the xy-plane. This is a vector field on xy-plane."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what would something like this be? This is a vector field. This is a vector field in two-dimensional space, or on the xy-plane. This is a vector field on xy-plane. Or you can even say on or on r2. Either way, I don't want to get too much into the mathiness of it, but what does this do? Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is a vector field on xy-plane. Or you can even say on or on r2. Either way, I don't want to get too much into the mathiness of it, but what does this do? Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line. All right, there we go. That's my y-axis, and that's my x-axis. I'm just drawing the first quadrant, but you could go negative in either direction if you like."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, if I were to draw my xy-plane, so that is my, I'm having trouble drawing a straight line. All right, there we go. That's my y-axis, and that's my x-axis. I'm just drawing the first quadrant, but you could go negative in either direction if you like. What does this thing do? Well, it's essentially saying, look, you give me any x, any y. You give any xy in the xy-plane, and these are going to end up with some numbers, right?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just drawing the first quadrant, but you could go negative in either direction if you like. What does this thing do? Well, it's essentially saying, look, you give me any x, any y. You give any xy in the xy-plane, and these are going to end up with some numbers, right? These are going to, with xy here, you're going to get some value. And when you put xy here, you're going to get some value, so you're going to get some combination of the i and j unit vector. So you're going to get some vector."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You give any xy in the xy-plane, and these are going to end up with some numbers, right? These are going to, with xy here, you're going to get some value. And when you put xy here, you're going to get some value, so you're going to get some combination of the i and j unit vector. So you're going to get some vector. So what this does is defines a vector that's associated with every point on the xy-plane. So you could say, hey, if I take this point on the xy-plane and I were to pop it into this, I'll get something times i plus something times j, and when you add those two, maybe I get a vector that looks like something like that. And you could do it on every point."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you're going to get some vector. So what this does is defines a vector that's associated with every point on the xy-plane. So you could say, hey, if I take this point on the xy-plane and I were to pop it into this, I'll get something times i plus something times j, and when you add those two, maybe I get a vector that looks like something like that. And you could do it on every point. I'm just taking random samples. Maybe when I go here, the vector looks something like that, maybe when I go here, the vector looks like this, maybe when I go here, the vector looks like that, maybe when I go up here, the vector goes like that. I'm just randomly picking points."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you could do it on every point. I'm just taking random samples. Maybe when I go here, the vector looks something like that, maybe when I go here, the vector looks like this, maybe when I go here, the vector looks like that, maybe when I go up here, the vector goes like that. I'm just randomly picking points. It defines a vector on all of the xy-coordinates where these functions, these scalar functions, are properly defined, and that's why it's called a vector field. It defines what a potential, maybe force would be, or some other type of, well, we could say force, at any point. If you happen to have something there, maybe that's what the function is."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just randomly picking points. It defines a vector on all of the xy-coordinates where these functions, these scalar functions, are properly defined, and that's why it's called a vector field. It defines what a potential, maybe force would be, or some other type of, well, we could say force, at any point. If you happen to have something there, maybe that's what the function is. And I could keep doing this forever and filling in all the gaps, but I think you get the idea. It associates a vector with every point on the xy-plane. Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you happen to have something there, maybe that's what the function is. And I could keep doing this forever and filling in all the gaps, but I think you get the idea. It associates a vector with every point on the xy-plane. Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field. It could be a gravitation field, it could be an electric field, it could be a magnetic field, and this could be essentially telling you how much force there would be on some particle in that field. That's exactly what this would describe. Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, this is called a vector field, so it probably makes a lot of sense that this could be used to describe any type of field. It could be a gravitation field, it could be an electric field, it could be a magnetic field, and this could be essentially telling you how much force there would be on some particle in that field. That's exactly what this would describe. Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at. So let's say it moves in a path that moves something like this. And let's say that this path, or this curve, is defined by a position vector function. So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, let's say that in this field I have some particle traveling on the xy-plane, and let's say it starts here, and by virtue of all of these crazy forces that are acting on it, and maybe it's on some tracks or something, so it won't always move exactly in the direction that the field is trying to move it at. So let's say it moves in a path that moves something like this. And let's say that this path, or this curve, is defined by a position vector function. So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j. That's r of t right there. Well, in order for this to be a finite path, this is true before t is greater than or equal to a and less than or equal to b. This is the path that the particle just happens to take due to all of these wacky forces."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that that's defined by r of t, which is just x of t times i plus y of t times our unit vector j. That's r of t right there. Well, in order for this to be a finite path, this is true before t is greater than or equal to a and less than or equal to b. This is the path that the particle just happens to take due to all of these wacky forces. So when the particle is right here, maybe the vector field acting on it, maybe it's putting a force like that. But since the thing is on some type of tracks, it moves in this direction. And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the path that the particle just happens to take due to all of these wacky forces. So when the particle is right here, maybe the vector field acting on it, maybe it's putting a force like that. But since the thing is on some type of tracks, it moves in this direction. And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks. Now, everything I've done in this video is to build up to a fundamental question. What was the work done on the particle by the field? To answer that question, we could zoom in a little bit."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when it's here, maybe the vector field is like that, but it moves in that direction because it's on some type of tracks. Now, everything I've done in this video is to build up to a fundamental question. What was the work done on the particle by the field? To answer that question, we could zoom in a little bit. Let's say I'm going to zoom in on only a little small snippet of our path. And let's just try to figure out what the work is done in a very small part of our path because it's constantly changing. The field is changing directions."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "To answer that question, we could zoom in a little bit. Let's say I'm going to zoom in on only a little small snippet of our path. And let's just try to figure out what the work is done in a very small part of our path because it's constantly changing. The field is changing directions. My object is changing directions. So let's say when I'm here, and let's say I move a small amount of my path. So let's say I move, this is an infinitesimally small dr."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The field is changing directions. My object is changing directions. So let's say when I'm here, and let's say I move a small amount of my path. So let's say I move, this is an infinitesimally small dr. I have a differential. It's a differential vector, infinitely small displacement. And let's say over the course of that, the vector field is acting in this local area."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say I move, this is an infinitesimally small dr. I have a differential. It's a differential vector, infinitely small displacement. And let's say over the course of that, the vector field is acting in this local area. Let's say it looks something like that. It's providing a force that looks something like that. So that's the vector field in that area, or the force directed on that particle right when it's at that point."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say over the course of that, the vector field is acting in this local area. Let's say it looks something like that. It's providing a force that looks something like that. So that's the vector field in that area, or the force directed on that particle right when it's at that point. There's an infinitesimally small amount of time and space, so you could say, OK, over that little small point, we have this constant force. What was the work done over this small period? You could say, what's the small interval of work?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the vector field in that area, or the force directed on that particle right when it's at that point. There's an infinitesimally small amount of time and space, so you could say, OK, over that little small point, we have this constant force. What was the work done over this small period? You could say, what's the small interval of work? You could say, d work, or a differential of work. Well, by the same exact logic that we did with the simple problem, it's the magnitude of the force in the direction of our displacement times the magnitude of our displacement. And we know what that is just from this example up here."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could say, what's the small interval of work? You could say, d work, or a differential of work. Well, by the same exact logic that we did with the simple problem, it's the magnitude of the force in the direction of our displacement times the magnitude of our displacement. And we know what that is just from this example up here. That's the dot product. It's the dot product of the force and our super small displacement. So that's equal to the dot product of our force and our super small displacement."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we know what that is just from this example up here. That's the dot product. It's the dot product of the force and our super small displacement. So that's equal to the dot product of our force and our super small displacement. Now, just by doing this, we're just figuring out the work over maybe a really small, super small dr. But what we want to do is we want to sum them all up. We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's equal to the dot product of our force and our super small displacement. Now, just by doing this, we're just figuring out the work over maybe a really small, super small dr. But what we want to do is we want to sum them all up. We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in. We will do a line integral over, I mean, you could think of it two ways. You could write just dw there, but we could say, we'll do a line integral that says along this curve C, we could call that C or along R, whatever you want to say it, of dw. That'll give us the total work."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We want to sum up all of the dr's to figure out the total work done, and that's where the integral comes in. We will do a line integral over, I mean, you could think of it two ways. You could write just dw there, but we could say, we'll do a line integral that says along this curve C, we could call that C or along R, whatever you want to say it, of dw. That'll give us the total work. So let's say work is equal to that. Or we could also write it over the integral, over the same curve, of F dot dr. And this might seem like a really, gee, this is really abstract, Sal. How do we actually calculate something like this?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That'll give us the total work. So let's say work is equal to that. Or we could also write it over the integral, over the same curve, of F dot dr. And this might seem like a really, gee, this is really abstract, Sal. 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?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 think about, 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, times the i unit vector, plus y prime of t times the j unit vector. And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or what is F dot dr? Well, actually, to answer that, let's think about, 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, times the i unit vector, plus y prime of t times the j unit vector. And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous. We get dr is equal to x prime of t dt times the unit vector i, plus y prime of t times the differential dt times the unit vector j. So this is our dr right here. That is our dr. And remember what our vector field was."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we just wanted dr, we could multiply both sides if we're being a little bit, maybe more hand-wavy with the differentials, not too rigorous. We get dr is equal to x prime of t dt times the unit vector i, plus y prime of t times the differential dt times the unit vector j. So this is our dr right here. That is our dr. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And then we'll see that the dot product is actually not so crazy."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is our dr. And remember what our vector field was. It was this thing up here. Let me copy and paste it. And then we'll see that the dot product is actually not so crazy. So let me copy, and then let me paste it down here. So what's this integral going to look like? This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is super fundamental to pretty much any serious physics that you might eventually find yourself doing."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we'll see that the dot product is actually not so crazy. So let me copy, and then let me paste it down here. So what's this integral going to look like? This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is 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, well, let's just say from t is equal to a to t is equal to b. a is where we started off on the path. t is equal to a to t is equal to b. You could imagine it as being time, as a particle moving as time increases."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This integral right here that seems, gives us the total work done by the field on the particle as it moves along that path, which is 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, well, let's just say from t is equal to a to t is equal to b. a is where we started off on the path. t is equal to a to t is equal to b. You could imagine it as being time, as a particle moving as time increases. And then what is f dot dr? 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 vector and add them up."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could imagine it as being time, as a particle moving as time increases. And then what is f dot dr? 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 vector and add them up. So this is going to be the integral from t equals a to t equals b of p of x, really, instead of writing xy, it's x of t, right? x is a function of t. y is a function of t. So that's that. Times this thing right here, times this component, right?"}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, if you remember from just what the dot product is, you can essentially just take the product of the corresponding components of your vector and add them up. So this is going to be the integral from t equals a to t equals b of p of x, really, instead of writing xy, it's x of t, right? x is a function of t. y is 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 dt. And then that plus, we're going to do the same thing with the q function."}, {"video_title": "Line integrals and vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times this thing right here, times this component, right? We're multiplying the i components. So times x prime of t dt. 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 the y component, or the j component, y prime of t dt. And we're done. This might still seem a little bit abstract, but we're going to see in the next video that this is actually, everything is now in terms of t, so this is just a straight up integration with respect to dt."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The last video was very abstract in general and I used f of x and g of t and h of t. What I want to do in this video is do an actual example. So let's say I have f of x, y is equal to x, y. Let's say that we have a path in the x, y plane or a curve in the x, y plane. I'm going to define my curve, say my curve is going to be defined by x being equal to cosine of t and y being equal to sine of t. We have to define what are our boundaries on our t. We're going to go from t is equal to 0 or t is going to be greater than or equal to 0 and then less than or equal to, we're going to deal in radians, pi over 2. If this was degrees, that would be 90 degrees. So that's our curve and immediately you might already know what this type of a curve looks like and I'm going to draw that really fast right here and then we'll try to visualize this. I've actually graphed it ahead of time so that we can visualize this."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm going to define my curve, say my curve is going to be defined by x being equal to cosine of t and y being equal to sine of t. We have to define what are our boundaries on our t. We're going to go from t is equal to 0 or t is going to be greater than or equal to 0 and then less than or equal to, we're going to deal in radians, pi over 2. If this was degrees, that would be 90 degrees. So that's our curve and immediately you might already know what this type of a curve looks like and I'm going to draw that really fast right here and then we'll try to visualize this. I've actually graphed it ahead of time so that we can visualize this. So this curve right here, if I were to just draw it in the standard x, y plane, let me do that in a different color so we can make the curve green. Let's say that is y and this is right here x. So when t is equal to 0, x is going to be equal to cosine of 0."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I've actually graphed it ahead of time so that we can visualize this. So this curve right here, if I were to just draw it in the standard x, y plane, let me do that in a different color so we can make the curve green. Let's say that is y and this is right here x. So when t is equal to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1. Y is going to be equal to sine of 0, which is 0. So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when t is equal to 0, x is going to be equal to cosine of 0. Cosine of 0 is 1. Y is going to be equal to sine of 0, which is 0. So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0. So we're going to be right there. That's at t is equal to 0. When t is equal to pi of 2, what's going to happen?"}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So t is equal to 0, we're going to be at x is equal to 1, that's cosine of 0 and y is sine of 0 or y is going to be 0. So we're going to be right there. That's at t is equal to 0. When t is equal to pi of 2, what's going to happen? Cosine of pi over 2, that's the angle, cosine of pi over 2 is 0. Sine of pi over 2 is 1. So we're going to be at the point 0, 1."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When t is equal to pi of 2, what's going to happen? Cosine of pi over 2, that's the angle, cosine of pi over 2 is 0. Sine of pi over 2 is 1. So we're going to be at the point 0, 1. So this is when we're at t is equal to pi over 2. You might recognize what we're going to draw is actually the first quadrant of the unit circle. When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to be at the point 0, 1. So this is when we're at t is equal to pi over 2. You might recognize what we're going to draw is actually the first quadrant of the unit circle. When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2. You can try it out for yourself, but we're just going to have a curve that looks like this. It's going to be the top right of a circle, of the unit circle. It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When t is equal to pi over 4 or 45 degrees, we're going to be at square root of 2, square root of 2. You can try it out for yourself, but we're just going to have a curve that looks like this. It's going to be the top right of a circle, of the unit circle. It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2. That's what this curve looks like. But in the last, our goal isn't here just to graph a parametric equation. What we want to do is raise a fence out of this kind of base and rise it to this surface."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to have radius 1 and we're going to go in that direction from t is equal to 0 to t is equal to pi over 2. That's what this curve looks like. But in the last, our goal isn't here just to graph a parametric equation. What we want to do is raise a fence out of this kind of base and rise it to this surface. So let's see if we can do that or at least visualize it first. And then we'll use the tools we used in the last video. So right here I've graphed this function and I've rotated it a little bit so you can see the other case."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What we want to do is raise a fence out of this kind of base and rise it to this surface. So let's see if we can do that or at least visualize it first. And then we'll use the tools we used in the last video. So right here I've graphed this function and I've rotated it a little bit so you can see the other case. This right here, let me get some dark colors out, that right there is the x-axis. That's the x-axis. That in the back is the y-axis and the vertical axis is the z-axis."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So right here I've graphed this function and I've rotated it a little bit so you can see the other case. This right here, let me get some dark colors out, that right there is the x-axis. That's the x-axis. That in the back is the y-axis and the vertical axis is the z-axis. And this is actually 2, this is 1 right here. y equals 1 is right there. So this is it graphed that way."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That in the back is the y-axis and the vertical axis is the z-axis. And this is actually 2, this is 1 right here. y equals 1 is right there. So this is it graphed that way. So if I were to graph this contour on the xy-plane, it would be under this graph and it would go something like this. Let me see if I can draw it. It would look something like this."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is it graphed that way. So if I were to graph this contour on the xy-plane, it would be under this graph and it would go something like this. Let me see if I can draw it. It would look something like this. This would be on the xy-plane. This is the same exact graph, f of x is equal to xy. Let me make that clear."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would look something like this. This would be on the xy-plane. This is the same exact graph, f of x is equal to xy. Let me make that clear. This is f of x, f of xy is equal to xy. That's both of these. I just rotated it."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me make that clear. This is f of x, f of xy is equal to xy. That's both of these. I just rotated it. In this situation, this is now, that right there is now the x-axis. I rotated to the left, you can kind of imagine. That right there is the x-axis."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I just rotated it. In this situation, this is now, that right there is now the x-axis. I rotated to the left, you can kind of imagine. That right there is the x-axis. That right there is the y-axis. It was rotated closer to me. That's the z-axis."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That right there is the x-axis. That right there is the y-axis. It was rotated closer to me. That's the z-axis. And then this curve, if I were to draw it in this rotation, is going to look like this. When t is equal to 0, we're at x is equal to 1, y is equal to 0. That's going to form a unit circle, something like, or a quarter of a unit circle like that."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the z-axis. And then this curve, if I were to draw it in this rotation, is going to look like this. When t is equal to 0, we're at x is equal to 1, y is equal to 0. That's going to form a unit circle, something like, or a quarter of a unit circle like that. And when t is equal to pi over 2, we're going to get there. What we want to do is find the area of the curtain that's defined. Let's raise a curtain from this curve up to f of xy."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's going to form a unit circle, something like, or a quarter of a unit circle like that. And when t is equal to pi over 2, we're going to get there. What we want to do is find the area of the curtain that's defined. Let's raise a curtain from this curve up to f of xy. If we keep raising walls from this up to x of y, we're going to have a wall that looks something like that. Let me shade it in, color it in so it looks a little bit more substantive. A wall that looks something like that."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's raise a curtain from this curve up to f of xy. If we keep raising walls from this up to x of y, we're going to have a wall that looks something like that. Let me shade it in, color it in so it looks a little bit more substantive. A wall that looks something like that. If I were to try to do it here, this would be under the ceiling, but the wall would look something like that. We want to find the area of that. We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "A wall that looks something like that. If I were to try to do it here, this would be under the ceiling, but the wall would look something like that. We want to find the area of that. We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations. In the last video, we came up with a, well, you could argue whether it's simple, but the idea is, well, let's just take small arc lengths, change in arc lengths, and multiply them by the height at that point. Those small change in arc lengths, we call them ds, and then the height is just f of xy at that point. We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We want to find the area of this right here, where the base is defined by this curve, and then the ceiling is defined by this surface here, xy, which I graphed and rotated in two situations. In the last video, we came up with a, well, you could argue whether it's simple, but the idea is, well, let's just take small arc lengths, change in arc lengths, and multiply them by the height at that point. Those small change in arc lengths, we call them ds, and then the height is just f of xy at that point. We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there. What we said is, well, to figure out the area of that, we're just going to take the integral from t is equal to 0 to t is equal to pi over 2. It doesn't make a lot of sense when I write it like this, of f of xy times, or let me even better, instead of writing f of xy, let me just write the actual function. Let's get a little bit more concrete."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll take an infinite sum of these from t is equal to 0 to t equal to pi over 2, and then that should give us the area of this wall right there. What we said is, well, to figure out the area of that, we're just going to take the integral from t is equal to 0 to t is equal to pi over 2. It doesn't make a lot of sense when I write it like this, of f of xy times, or let me even better, instead of writing f of xy, let me just write the actual function. Let's get a little bit more concrete. f of xy is xy times, so the particular xy times the little change in our arc length at that point. I'm going to be very hand wavy here. This is all a little bit of review of the last video."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's get a little bit more concrete. f of xy is xy times, so the particular xy times the little change in our arc length at that point. I'm going to be very hand wavy here. This is all a little bit of review of the last video. We figured out in the last video, this change in arc length right here, this change in arc length, ds, we figured out that we could rewrite that as the square root of dx versus, or the derivative of x with respect to t squared plus the derivative of y with respect to t squared, and then all of that times dt. I'm just rebuilding the formula that we got in the last video. This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is all a little bit of review of the last video. We figured out in the last video, this change in arc length right here, this change in arc length, ds, we figured out that we could rewrite that as the square root of dx versus, or the derivative of x with respect to t squared plus the derivative of y with respect to t squared, and then all of that times dt. I'm just rebuilding the formula that we got in the last video. This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy. You know what? Right from the get-go, we want everything eventually to be in terms of t. Instead of writing x times y, let's substitute the parametric form. Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This expression can be rewritten as the integral from t is equal to 0 to t is equal to pi over 2 times xy. You know what? Right from the get-go, we want everything eventually to be in terms of t. Instead of writing x times y, let's substitute the parametric form. Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve. That's how we define x in terms of the parameter t. Then times y, which we're saying is sine of t. That's our y. All I did is rewrote xy in terms of t times ds. ds is this."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of x, let's write cosine of t. Let me write cosine of t. That is x. x is equal to cosine of t on this curve. That's how we define x in terms of the parameter t. Then times y, which we're saying is sine of t. That's our y. All I did is rewrote xy in terms of t times ds. ds is this. It's the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared. All of that times dt. Now we just have to find these two derivatives."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds is this. It's the square root of the derivative of x with respect to t squared plus the derivative of y with respect to t squared. All of that times dt. Now we just have to find these two derivatives. It might seem really hard, but it's very easy for us to find the derivative of x with respect to t and the derivative of y with respect to t. I can do it right down here. We'll lose our graphs for a little bit. We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t?"}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we just have to find these two derivatives. It might seem really hard, but it's very easy for us to find the derivative of x with respect to t and the derivative of y with respect to t. I can do it right down here. We'll lose our graphs for a little bit. We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t? That's minus sine of t. The derivative of y with respect to t, the derivative of sine of anything is the cosine of that anything. It's cosine of t. We could substitute these back into this equation. Remember, we're just trying to find the area of this curtain."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We know that the derivative of x with respect to t is just going to be what's the derivative of cosine of t? That's minus sine of t. The derivative of y with respect to t, the derivative of sine of anything is the cosine of that anything. It's cosine of t. We could substitute these back into this equation. Remember, we're just trying to find the area of this curtain. It has our curve here as kind of its base and has this function, this surface, as its ceiling. We go back down here. Let me rewrite this whole thing."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Remember, we're just trying to find the area of this curtain. It has our curve here as kind of its base and has this function, this surface, as its ceiling. We go back down here. Let me rewrite this whole thing. This becomes the integral from t is equal to 0 to t is equal to pi over 2. I don't like this color. Cosine times sine, that's just the xy, times ds, which is this expression right here."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me rewrite this whole thing. This becomes the integral from t is equal to 0 to t is equal to pi over 2. I don't like this color. Cosine times sine, that's just the xy, times ds, which is this expression right here. Now we can write this as, I'll go switch back to that color I don't like. The square root of, the derivative of x with respect to t is minus sine of t. We're going to square it. Plus derivative of y with respect to t. That's cosine of t. We're going to square it."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Cosine times sine, that's just the xy, times ds, which is this expression right here. Now we can write this as, I'll go switch back to that color I don't like. The square root of, the derivative of x with respect to t is minus sine of t. We're going to square it. Plus derivative of y with respect to t. That's cosine of t. We're going to square it. Let me make my radic a little bit bigger. Then all of that times dt. This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus derivative of y with respect to t. That's cosine of t. We're going to square it. Let me make my radic a little bit bigger. Then all of that times dt. This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing. Let me rewrite. Minus, let me do this in the side right here. Minus sine of t squared plus cosine of t squared."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This still might seem like a really hard integral until you realize that this right here, when you take a negative number and you square it, this is the same thing. Let me rewrite. Minus, let me do this in the side right here. Minus sine of t squared plus cosine of t squared. This is equivalent to, this is the same thing as sine of t squared plus cosine of t squared. You lose the sine information when you square something. It just becomes a positive."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Minus sine of t squared plus cosine of t squared. This is equivalent to, this is the same thing as sine of t squared plus cosine of t squared. You lose the sine information when you square something. It just becomes a positive. These two things are equivalent. This is the most basic trig identity. This comes straight out of the unit circle definition."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It just becomes a positive. These two things are equivalent. 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. All this stuff under the radical sign is just equal to 1. We're getting the square root of 1, which is just 1."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This comes straight out of the unit circle definition. Sine squared plus cosine squared, this is just equal to 1. All this stuff under the radical sign is just equal to 1. We're getting the square root of 1, which is just 1. All of this stuff right here will just become 1. This whole crazy integral simplifies a good bit. It just equals the square root of t equals 0 to t is equal to pi over 2."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're getting the square root of 1, which is just 1. All of this stuff right here will just become 1. This whole crazy integral simplifies a good bit. It just equals the square root of t equals 0 to t is equal to pi over 2. I'm going to switch these around just because it will make it a little easier in the next step. Sine of t times cosine of t dt. All I did, this whole thing equals 1."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It just equals the square root of t equals 0 to t is equal to pi over 2. I'm going to switch these around just because it will make it a little easier in the next step. 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 will make the next step a little bit easier to explain. This integral, you say sine times cosine, what's the antiderivative of that?"}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All I did, this whole thing equals 1. Got rid of it and I just switched the order of that. It will make the next step a little bit easier to explain. This integral, you say sine times cosine, what's the antiderivative of that? The first thing you should recognize is I have a function or an expression here. I have its derivative. The derivative of sine is cosine of t. You might be able to do u substitution in your head."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This integral, you say sine times cosine, what's the antiderivative of that? The first thing you should recognize is I have a function or an expression here. I have its derivative. The derivative of sine is cosine of t. You might be able to do 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. If you have something as derivative, you define that something is u. 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. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The derivative of sine is cosine of t. You might be able to do 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. If you have something as derivative, you define that something is u. 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. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt. Notice right here I have a u and then cosine of t dt, this thing right here, that thing is equal to d of u. Then we just have to redefine the boundaries. When t is equal to 0, this thing is going to turn into the integral."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. Or if you multiply both sides by the differential dt, if we're not going to be too rigorous, you get du is equal to cosine of t dt. Notice right here I have a u and then cosine of t dt, this thing right here, that thing is equal to d of u. Then we just have to redefine the boundaries. When t is equal to 0, this thing is going to turn into the integral. Instead of t is equal to 0, when t is equal to 0, what is u equal to? Sine of 0 is 0. This goes from u is equal to 0."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When t is equal to 0, this thing is going to turn into the integral. Instead of t is equal to 0, when t is equal to 0, what is u equal to? Sine of 0 is 0. This goes from u is equal to 0. When t is pi over 2, sine of pi over 2 is 1. When t is pi over 2, u is equal to 1. From u is equal to 0 to u is equal to 1."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This goes from u is equal to 0. When t is pi over 2, sine of pi over 2 is 1. When t is pi over 2, u is equal to 1. From u is equal to 0 to u is equal to 1. Just redid the boundaries in terms of u. Then we have instead of sine of t, I'm going to write u. Instead of cosine of t dt, I'm just going to write du."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "From u is equal to 0 to u is equal to 1. Just redid the boundaries in terms of u. Then we have instead of sine of t, I'm going to write u. Instead of cosine of t dt, I'm just going to write du. This is a super easy integral in terms of u. This is just equal to the antiderivative of u is u 1 half times u squared. We just raised the exponent and then divided by that raised exponent."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of cosine of t dt, I'm just going to write du. This is a super easy integral in terms of u. This is just equal to the antiderivative of u is u 1 half times u squared. We just raised the exponent and then divided by that raised exponent. 1 half u squared. We're going to evaluate it from 0 to 1. This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just raised the exponent and then divided by that raised exponent. 1 half u squared. We're going to evaluate it from 0 to 1. This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared. Which is equal to 1 half times 1 minus 0, which is equal to 1 half. We did all that work and we got a nice simple answer. The area of this curtain, we just performed a line integral."}, {"video_title": "Line integral example 1   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be equal to 1 half times 1 squared minus 1 half times 0 squared. Which is equal to 1 half times 1 minus 0, which is equal to 1 half. We did all that work and we got a nice simple answer. The area of this curtain, we just performed a line integral. The area of this curtain along this curve right here is 1 half. If this was in centimeters, it would be 1 half centimeter squared. I think that was a pretty neat application of the line integral."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "So first of all, this idea of a critical point basically means anywhere where the gradient equals zero. So you're looking for places where the gradient of your function at some kind of input, some specified input x and y that you're solving for is equal to zero. And as I've talked about in the last couple videos, the reason you might want to do this is because you're hoping to maximize the function or to maybe minimize the function. And now the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something where the gradient equals zero, you want to be able to determine, is it a local maximum, is it a local minimum, or is it a saddle point? So let's go ahead and work through this example. The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And now the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something where the gradient equals zero, you want to be able to determine, is it a local maximum, is it a local minimum, or is it a saddle point? So let's go ahead and work through this example. The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line. We take both partial derivatives. So the partial derivative with respect to x is, well, this first term, when we take the derivative of three x squared times y with respect to x, that two hops down, so we have six times x times y. Y cubed, well, y looks like a constant, so y cubed looks like a constant, minus three x squared, so that two comes down, so we're subtracting off six times x, six times x. And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line. We take both partial derivatives. So the partial derivative with respect to x is, well, this first term, when we take the derivative of three x squared times y with respect to x, that two hops down, so we have six times x times y. Y cubed, well, y looks like a constant, so y cubed looks like a constant, minus three x squared, so that two comes down, so we're subtracting off six times x, six times x. And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned. And we do partial of f with respect to y. Then this first term looks like some sort of constant, three x squared, x looks like a constant, so some kind of constant times y, so the whole thing looks like three x squared. The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned. And we do partial of f with respect to y. Then this first term looks like some sort of constant, three x squared, x looks like a constant, so some kind of constant times y, so the whole thing looks like three x squared. The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared. And then this next term only has an x, so it looks like a constant as far as y is concerned. And then this last term, we take down the two, because we're differentiating y squared, and you'll get negative six y, negative six times y. So when we are finding the critical points, the first step is to set both of these guys equal to zero."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared. And then this next term only has an x, so it looks like a constant as far as y is concerned. And then this last term, we take down the two, because we're differentiating y squared, and you'll get negative six y, negative six times y. So when we are finding the critical points, the first step is to set both of these guys equal to zero. So this first one, when we do set it equal to zero, we can simplify a bit by factoring out six x. So this really looks like six x multiplied by y minus one. And then that's what we're setting equal to zero."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "So when we are finding the critical points, the first step is to set both of these guys equal to zero. So this first one, when we do set it equal to zero, we can simplify a bit by factoring out six x. So this really looks like six x multiplied by y minus one. And then that's what we're setting equal to zero. And what this equation tells us is that either it's the six x term that equals zero, in which case that would mean x is equal to zero, or it's the case that y minus one equals zero, in which case that would mean that y equals one. So at least one of these things has to be true. That's kind of the first requirement that we've found."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And then that's what we're setting equal to zero. And what this equation tells us is that either it's the six x term that equals zero, in which case that would mean x is equal to zero, or it's the case that y minus one equals zero, in which case that would mean that y equals one. So at least one of these things has to be true. That's kind of the first requirement that we've found. Let me scroll down a little bit here. And for the second equation, when we set it equal to zero, it's not immediately straightforward how you would solve for this in a nice way in terms of x and y, but because we've already solved one, we can kind of plug them in and say, for example, if it was the case that x is equal to zero, and we kind of want to see what that turns our equation into, then we would have, well, three x squared is nothing. That would be zero."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "That's kind of the first requirement that we've found. Let me scroll down a little bit here. And for the second equation, when we set it equal to zero, it's not immediately straightforward how you would solve for this in a nice way in terms of x and y, but because we've already solved one, we can kind of plug them in and say, for example, if it was the case that x is equal to zero, and we kind of want to see what that turns our equation into, then we would have, well, three x squared is nothing. That would be zero. And we'd just be left with negative three y squared minus six y is equal to zero. And that we can factor out a bit. So I'm gonna factor out a negative three y."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "That would be zero. And we'd just be left with negative three y squared minus six y is equal to zero. And that we can factor out a bit. So I'm gonna factor out a negative three y. So I'll factor out negative three y, which means that first term just has a y remaining, and then that second term has a two, a positive two, since I factored out negative three. So positive two, and that equals zero. So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "So I'm gonna factor out a negative three y. So I'll factor out negative three y, which means that first term just has a y remaining, and then that second term has a two, a positive two, since I factored out negative three. So positive two, and that equals zero. So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two. So that's the first situation, where we plug in x equals zero. Now, alternatively, there's the possibility that y equals one, so we could say y equals one. And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two. So that's the first situation, where we plug in x equals zero. Now, alternatively, there's the possibility that y equals one, so we could say y equals one. And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared. And then the rest of it becomes, let's see, minus three times one squared. So minus three, we're plugging in one for y. And then we subtract off six, plugging in that one for y again."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared. And then the rest of it becomes, let's see, minus three times one squared. So minus three, we're plugging in one for y. And then we subtract off six, plugging in that one for y again. And that whole thing is equal to three x squared, then minus three minus six. So I'm subtracting off nine. So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And then we subtract off six, plugging in that one for y again. And that whole thing is equal to three x squared, then minus three minus six. So I'm subtracting off nine. So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three. And what that implies then, since this whole thing has to equal zero, what that implies is that x squared minus three is equal to zero. So we have x is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three. And what that implies then, since this whole thing has to equal zero, what that implies is that x squared minus three is equal to zero. So we have x is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found. One of them was in the circumstance where x equals zero, and then the other was what we found in the circumstance where y equals one. So this gives us a grand total of three different critical points. Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "And maybe I should kind of specify, these are distinct things that we found. One of them was in the circumstance where x equals zero, and then the other was what we found in the circumstance where y equals one. So this gives us a grand total of three different critical points. Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another two possibilities here where if y is equal to one, when y is equal to one, we'll have x is positive or negative square root of three."}, {"video_title": "Second partial derivative test example, part 1.mp3", "Sentence": "Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another two possibilities here where if y is equal to one, when y is equal to one, we'll have x is positive or negative square root of three. So we have positive square root of three and y equals one, and then we have negative square root of three and y equals one. So these are the critical points. Critical points."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "In the last video, I started to talk about the formula for curvature. And just to remind everyone of where we are, you imagine that you have some kind of curve in, let's say, two-dimensional space, just for the sake of being simple. And let's say this curve is parameterized by a function s of t. So every number t corresponds to some point on the curve. For the curvature, you start thinking about unit tangent vectors. At every given point, what does the unit tangent vector look like? And the curvature itself, which is denoted by this sort of Greek letter kappa, is gonna be the rate of change of those unit vectors, kind of how quickly they're turning in direction, not with respect to the parameter t, but with respect to arc length, ds. And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "For the curvature, you start thinking about unit tangent vectors. At every given point, what does the unit tangent vector look like? And the curvature itself, which is denoted by this sort of Greek letter kappa, is gonna be the rate of change of those unit vectors, kind of how quickly they're turning in direction, not with respect to the parameter t, but with respect to arc length, ds. And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds. And you're wondering, as you take a tiny step like that, does the unit tangent vector turn a lot or does it turn a little bit? And the little schematic that I said you might have in mind is just a completely separate space where for each one of these unit tangent vectors, you go ahead and put them in that space, saying, okay, so this one would look something like this. This one is pointed down and to the right, so it would look something like this."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds. And you're wondering, as you take a tiny step like that, does the unit tangent vector turn a lot or does it turn a little bit? And the little schematic that I said you might have in mind is just a completely separate space where for each one of these unit tangent vectors, you go ahead and put them in that space, saying, okay, so this one would look something like this. This one is pointed down and to the right, so it would look something like this. This one is pointed very much down. And you're wondering, basically, as you take tiny little steps of size ds, what does this change to the unit tangent vector? And that change, it's gonna be some kind of vector."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "This one is pointed down and to the right, so it would look something like this. This one is pointed very much down. And you're wondering, basically, as you take tiny little steps of size ds, what does this change to the unit tangent vector? And that change, it's gonna be some kind of vector. And because the curvature is really just a value, a number that we want, all we care about is the size of that vector. The size of the change to the tangent vector as you take a tiny step in ds. Now this is pretty abstract, right?"}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And that change, it's gonna be some kind of vector. And because the curvature is really just a value, a number that we want, all we care about is the size of that vector. The size of the change to the tangent vector as you take a tiny step in ds. Now this is pretty abstract, right? I've got these two completely separate things that are not the original function that you have to think about. You have to think about this unit tangent vector function. And then you also have to think about this notion of arc length."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "Now this is pretty abstract, right? I've got these two completely separate things that are not the original function that you have to think about. You have to think about this unit tangent vector function. And then you also have to think about this notion of arc length. And the reason, by the way, that I'm using an s here as well as here for the parameterization of the curve is because they're actually quite related. And I'll get to that a little bit below. And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And then you also have to think about this notion of arc length. And the reason, by the way, that I'm using an s here as well as here for the parameterization of the curve is because they're actually quite related. And I'll get to that a little bit below. And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair. So we've got cosine of t is the x component and then sine of t is the y component, sine of t. And just to make it so that it's not completely boring, let's multiply both of these components by a constant r. And what this means, you might recognize this, cosine sine pair. What this means is that in the xy plane, you're actually drawing a circle with radius r. So this would be some kind of circle with a radius r. And while I go through this example, I also wanna make a note of what things would look like a little bit more abstractly. If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair. So we've got cosine of t is the x component and then sine of t is the y component, sine of t. And just to make it so that it's not completely boring, let's multiply both of these components by a constant r. And what this means, you might recognize this, cosine sine pair. What this means is that in the xy plane, you're actually drawing a circle with radius r. So this would be some kind of circle with a radius r. And while I go through this example, I also wanna make a note of what things would look like a little bit more abstractly. If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component. And the reason I wanna do this is because the concrete version is gonna be helpful and simple and something we can deal with, but almost so simple as to not be indicative of just how complicated the normal circumstance is. But the more general circumstance is so complicated, I think it'll actually confuse things a little bit too much. So it'll be good to kind of go through both of them in parallel."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component. And the reason I wanna do this is because the concrete version is gonna be helpful and simple and something we can deal with, but almost so simple as to not be indicative of just how complicated the normal circumstance is. But the more general circumstance is so complicated, I think it'll actually confuse things a little bit too much. So it'll be good to kind of go through both of them in parallel. And the first step is to figure out what is this unit tangent vector? What is that function that at every given point gives you a unit tangent vector to the curve? And the first thing for that is to realize that we already have a notion of what should give the tangent vector."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "So it'll be good to kind of go through both of them in parallel. And the first step is to figure out what is this unit tangent vector? What is that function that at every given point gives you a unit tangent vector to the curve? And the first thing for that is to realize that we already have a notion of what should give the tangent vector. The derivative of our vector valued function as a function of t, the direction in which it points is in the tangent direction. So if I go over here and if I compute this derivative and I say s prime of t, which involves just taking the derivative of both components, so the derivative of cosine is negative sine of t multiplied by r, and the derivative of sine is cosine of t multiplied by r. And more abstractly, this is just gonna be any time you have two different component functions, you just take the derivative of each one. And hopefully you've seen this."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And the first thing for that is to realize that we already have a notion of what should give the tangent vector. The derivative of our vector valued function as a function of t, the direction in which it points is in the tangent direction. So if I go over here and if I compute this derivative and I say s prime of t, which involves just taking the derivative of both components, so the derivative of cosine is negative sine of t multiplied by r, and the derivative of sine is cosine of t multiplied by r. And more abstractly, this is just gonna be any time you have two different component functions, you just take the derivative of each one. And hopefully you've seen this. If not, maybe take a look at the videos on taking the derivative of a position vector valued function. And this we can interpret as that tangent vector, but it might not be a unit vector, right? We want a unit tangent vector, and this only promises us the direction."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "And hopefully you've seen this. If not, maybe take a look at the videos on taking the derivative of a position vector valued function. And this we can interpret as that tangent vector, but it might not be a unit vector, right? We want a unit tangent vector, and this only promises us the direction. So what we do to normalize it, what we do to normalize it and get a unit tangent vector function, maybe a different color, and get a unit vector tangent function, which I'll call capital T of lowercase t. That's kind of confusing, right? Capital T is for tangent vector, lowercase t is the parameter. So I'll try to keep that straight."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "We want a unit tangent vector, and this only promises us the direction. So what we do to normalize it, what we do to normalize it and get a unit tangent vector function, maybe a different color, and get a unit vector tangent function, which I'll call capital T of lowercase t. That's kind of confusing, right? Capital T is for tangent vector, lowercase t is the parameter. So I'll try to keep that straight. It's sort of standard notation, but there is the potential to confuse with this. What that's gonna be is your vector value derivative, but normalized. So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "So I'll try to keep that straight. It's sort of standard notation, but there is the potential to confuse with this. What that's gonna be is your vector value derivative, but normalized. So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector. What we get, I'm giving myself even more room here, is the square root of sine squared, negative sine squared is just gonna be sine squared. So sine squared of t multiplied by r squared. And then over here, cosine squared times r squared."}, {"video_title": "Curvature formula, part 2.mp3", "Sentence": "So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector. What we get, I'm giving myself even more room here, is the square root of sine squared, negative sine squared is just gonna be sine squared. So sine squared of t multiplied by r squared. And then over here, cosine squared times r squared. Cosine squared of t times r squared. We can bring that r squared outside of the radical and just sort of factor it out, turning it into an r. And on the inside, we have sine squared plus cosine squared. I'm being too lazy to write down the t's right now, because no matter what the t is, that whole value just equals one."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we finished off with these two results. We started off just thinking about what it means to take the partial derivative of a vector-valued function. And I got to these kind of, you might call them, bizarre results. What was the whole point in getting here, Sal? And the whole point is so that I can give you the tools you need to understand what a surface integral is. So let's just think about, let's draw the ST plane and then see how it gets transformed into this surface R. So let's do that. Let's say that is the t-axis."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What was the whole point in getting here, Sal? And the whole point is so that I can give you the tools you need to understand what a surface integral is. So let's just think about, let's draw the ST plane and then see how it gets transformed into this surface R. So let's do that. Let's say that is the t-axis. And let's say that that is the s-axis. And let's say that our vector-valued function, or our position vector-valued function, is defined from s's between a and b. So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say that is the t-axis. And let's say that that is the s-axis. And let's say that our vector-valued function, or our position vector-valued function, is defined from s's between a and b. So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface. And if you map each of these points, you will eventually get the surface R. So let me draw R in three dimensions. The surface in 3D. So that is our x-axis."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface. And if you map each of these points, you will eventually get the surface R. So let me draw R in three dimensions. The surface in 3D. So that is our x-axis. That is our y-axis. And then that is the z-axis. And just as a bit of a reminder, it might look something like this."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that is our x-axis. That is our y-axis. And then that is the z-axis. And just as a bit of a reminder, it might look something like this. If we were to this point right here, where s is equal to a and t is equal to c, remember, we're going to draw the surface indicated by the position vector function s, R of s and t. So this point right here, when s is a and t is c, maybe it maps to that point right there. When you take a and c and you put it into this thing over here, you're just going to get the vector that points at that. So you could say it'll give you a position vector that'll point right at that position right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just as a bit of a reminder, it might look something like this. If we were to this point right here, where s is equal to a and t is equal to c, remember, we're going to draw the surface indicated by the position vector function s, R of s and t. So this point right here, when s is a and t is c, maybe it maps to that point right there. When you take a and c and you put it into this thing over here, you're just going to get the vector that points at that. So you could say it'll give you a position vector that'll point right at that position right there. And then let's say that this line right here, if we were to hold s constant at a and just vary t from c to d, maybe that looks something like this. I'm just drawing some arbitrary contour there. Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you could say it'll give you a position vector that'll point right at that position right there. And then let's say that this line right here, if we were to hold s constant at a and just vary t from c to d, maybe that looks something like this. I'm just drawing some arbitrary contour there. Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that. I don't know. I'm just trying to show you an example. So this point right here would correspond to that point right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that. I don't know. I'm just trying to show you an example. So this point right here would correspond to that point right there. When you put it into the vector-valued function R, you would get a vector that points to that point, just like that. And this point right here in purple, when you evaluate R of s and t, it'll give you a vector that points right there to that point over there. We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this point right here would correspond to that point right there. When you put it into the vector-valued function R, you would get a vector that points to that point, just like that. And this point right here in purple, when you evaluate R of s and t, it'll give you a vector that points right there to that point over there. We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible. So maybe I'll do it in this bluish color. If we hold t at d and vary s from a to b, we're going to start here. This is when t is d and s is a."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible. So maybe I'll do it in this bluish color. If we hold t at d and vary s from a to b, we're going to start here. This is when t is d and s is a. And when you vary it, maybe you get something like that. I don't know. So this point right here would correspond to a vector that points to that point right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is when t is d and s is a. And when you vary it, maybe you get something like that. I don't know. So this point right here would correspond to a vector that points to that point right there. And then finally, this line, or if we hold s at b and vary t between c and d, we're going to go from that point to that point. So it's going to look something like this. We're going to go from this point to that point."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this point right here would correspond to a vector that points to that point right there. And then finally, this line, or if we hold s at b and vary t between c and d, we're going to go from that point to that point. So it's going to look something like this. We're going to go from this point to that point. We're holding s at b, varying t from c to d. Maybe it looks something like that. So our surface, we went from this nice area, this rectangular area in the t, s plane, and it gets transformed into this wacky looking surface. And we could even draw some other things right here."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to go from this point to that point. We're holding s at b, varying t from c to d. Maybe it looks something like that. So our surface, we went from this nice area, this rectangular area in the t, s plane, and it gets transformed into this wacky looking surface. And we could even draw some other things right here. Let's say we have a, let's say we get some arbitrary value. Let me pick a new color, I'll do it in white or a new non-color. And let's say if we hold s at that constant value and we vary t, maybe that will look something like this."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we could even draw some other things right here. Let's say we have a, let's say we get some arbitrary value. Let me pick a new color, I'll do it in white or a new non-color. And let's say if we hold s at that constant value and we vary t, maybe that will look something like this. Maybe that will look something like, oh, I don't know. Maybe it'll look something like that. So you get an idea of what this surface might look like."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say if we hold s at that constant value and we vary t, maybe that will look something like this. Maybe that will look something like, oh, I don't know. Maybe it'll look something like that. So you get an idea of what this surface might look like. Now, given this, I want to think about what these quantities are. And then when we visualize what these quantities are, we'll be able to kind of use these results of the last video to do something that I think will be useful. So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you get an idea of what this surface might look like. Now, given this, I want to think about what these quantities are. And then when we visualize what these quantities are, we'll be able to kind of use these results of the last video to do something that I think will be useful. So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here. Let's say this is the point, that is the point s, t. If you were to put those values in here, maybe it maps to, and I want to make sure I'm consistent with everything I've drawn, maybe it maps to this point right here. Maybe it maps to that point right there. So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here. Let's say this is the point, that is the point s, t. If you were to put those values in here, maybe it maps to, and I want to make sure I'm consistent with everything I've drawn, maybe it maps to this point right here. Maybe it maps to that point right there. So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general. I could call this a, well, I already used a and b. I could call this x and y. This would be r of x and y would map to that point right there. Now, so that's that right there, or that right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general. I could call this a, well, I already used a and b. I could call this x and y. This would be r of x and y would map to that point right there. Now, so that's that right there, or that right there. Now let's see what happens if we move just in the s direction. We could view that as s. Now let us move forward by a differential, by a super small amount of s. So this right here, let's call that s plus a super small differential in s. That's right there. So that's that point."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, so that's that right there, or that right there. Now let's see what happens if we move just in the s direction. We could view that as s. Now let us move forward by a differential, by a super small amount of s. So this right here, let's call that s plus a super small differential in s. That's right there. So that's that point. Let me do that in a better color, in this yellow. So that point right there is the point s plus my differential of s. I could write delta s, but I want a super small change in s, comma t. And what is that going to get mapped to? Well, if we apply these two points in r, we're going to get something that maybe is right over there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's that point. Let me do that in a better color, in this yellow. So that point right there is the point s plus my differential of s. I could write delta s, but I want a super small change in s, comma t. And what is that going to get mapped to? Well, if we apply these two points in r, we're going to get something that maybe is right over there. And I want to be very clear, this right here, that is r of s plus ds, comma t. That's what that is. That's the point when we just shift s by a super small differential. This distance here you can view as ds."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, if we apply these two points in r, we're going to get something that maybe is right over there. And I want to be very clear, this right here, that is r of s plus ds, comma t. That's what that is. That's the point when we just shift s by a super small differential. This distance here you can view as ds. It's a super small change in s. And then when we map it, or transform it, or put that point into our vector-valued function. Let me copy and paste the original vector-valued function, just so we have a good image of what we're talking about this whole time. Let me put it right down there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This distance here you can view as ds. It's a super small change in s. And then when we map it, or transform it, or put that point into our vector-valued function. Let me copy and paste the original vector-valued function, just so we have a good image of what we're talking about this whole time. Let me put it right down there. Just to be clear what's going on, when we took this little blue point right here, this s and t, and we put the s and t values here, we get a vector that points to that point on the surface right there. When you add a ds to your s values, you get a vector that points at that yellow point right there. So going back to the results we got in the last presentation, or the last video, what is this?"}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me put it right down there. Just to be clear what's going on, when we took this little blue point right here, this s and t, and we put the s and t values here, we get a vector that points to that point on the surface right there. When you add a ds to your s values, you get a vector that points at that yellow point right there. So going back to the results we got in the last presentation, or the last video, what is this? r of s plus delta s, or r of s plus ds, the differential of s, t, well that is that right there. That is the vector that points to that position. This right here is the vector that points to this blue position."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So going back to the results we got in the last presentation, or the last video, what is this? r of s plus delta s, or r of s plus ds, the differential of s, t, well that is that right there. That is the vector that points to that position. This right here is the vector that points to this blue position. So what is the difference of those two vectors? And this is a bit of, this is just basic vector math you might remember. The difference of these two vectors, head to tails."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right here is the vector that points to this blue position. So what is the difference of those two vectors? And this is a bit of, this is just basic vector math you might remember. The difference of these two vectors, head to tails. The difference of these two vectors is going to be this vector. If you subtract this vector from that vector, you're going to get that vector right there. You're going to get that vector right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The difference of these two vectors, head to tails. The difference of these two vectors is going to be this vector. If you subtract this vector from that vector, you're going to get that vector right there. You're going to get that vector right there. A vector that looks just like that. So that's what this is equal to, that vector. And it makes sense."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're going to get that vector right there. A vector that looks just like that. So that's what this is equal to, that vector. And it makes sense. This blue vector plus the orange vector, this blue vector right here plus the orange vector is equal to this vector. Makes complete sense, heads to tails. So that's what that represents."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it makes sense. This blue vector plus the orange vector, this blue vector right here plus the orange vector is equal to this vector. Makes complete sense, heads to tails. So that's what that represents. Now let's do the same thing in the t direction. I'm running out of colors. I'll have to go back to the pink or maybe the magenta."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's what that represents. Now let's do the same thing in the t direction. I'm running out of colors. I'll have to go back to the pink or maybe the magenta. So if we have that s and t, now if we go up a little bit in that direction, let's say that that is t. So this is the point s, t plus a super small change in t. That's that point right there. This distance right there is dt. You can view it that way."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll have to go back to the pink or maybe the magenta. So if we have that s and t, now if we go up a little bit in that direction, let's say that that is t. So this is the point s, t plus a super small change in t. That's that point right there. This distance right there is dt. You can view it that way. If you put s and t plus dt into our vector value function, what are you going to get? You're going to get a vector that maybe points to this point right here. Maybe I'll draw it right here."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can view it that way. If you put s and t plus dt into our vector value function, what are you going to get? You're going to get a vector that maybe points to this point right here. Maybe I'll draw it right here. Maybe it points to this point right here. A vector that points right there. So that will be mapped to a vector that points to that position right over there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe I'll draw it right here. Maybe it points to this point right here. A vector that points right there. So that will be mapped to a vector that points to that position right over there. Now by the same argument that we did on the s side, this point or the vector that points to that, that is r of s, t plus dt. That is the exact same thing as that right there. And of course, this we already saw."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that will be mapped to a vector that points to that position right over there. Now by the same argument that we did on the s side, this point or the vector that points to that, that is r of s, t plus dt. That is the exact same thing as that right there. And of course, this we already saw. This is the same thing as that over there. So what is that vector minus this blue vector? The magenta vector minus the blue vector."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And of course, this we already saw. This is the same thing as that over there. So what is that vector minus this blue vector? The magenta vector minus the blue vector. Well, once again, this is fairly, hopefully a bit of a review of adding vectors. It's going to be a vector that looks like this. I'll do it in white."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The magenta vector minus the blue vector. Well, once again, this is fairly, hopefully a bit of a review of adding vectors. It's going to be a vector that looks like this. I'll do it in white. It's going to be a vector that looks like that. This thing is going to be a vector that looks just like that. And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it in white. It's going to be a vector that looks like that. This thing is going to be a vector that looks just like that. And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector. So it makes sense if the purple vector minus the blue vector is going to be equal to this white vector. So something interesting is going on here. I have these two."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector. So it makes sense if the purple vector minus the blue vector is going to be equal to this white vector. So something interesting is going on here. I have these two. This is a vector that is kind of going along this parameterized surface as we changed our s by a super small amount. And then this is a vector that is going along our surface if we change our t by a super small amount. Now, you may or may not remember this."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I have these two. This is a vector that is kind of going along this parameterized surface as we changed our s by a super small amount. And then this is a vector that is going along our surface if we change our t by a super small amount. Now, you may or may not remember this. And I've done several videos where I show this to you. But the magnitude, if I take two vectors and I take their cross product. So if I take the cross product of a and b and I take the magnitude of the resulting vector."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, you may or may not remember this. And I've done several videos where I show this to you. But the magnitude, if I take two vectors and I take their cross product. So if I take the cross product of a and b and I take the magnitude of the resulting vector. Remember, when you take the cross product, you get a third vector that is perpendicular to both of these. But if you were just to take the magnitude of that vector, that is equal to the area of parallelogram defined by a and b. What do I mean by that?"}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I take the cross product of a and b and I take the magnitude of the resulting vector. Remember, when you take the cross product, you get a third vector that is perpendicular to both of these. But if you were just to take the magnitude of that vector, that is equal to the area of parallelogram defined by a and b. What do I mean by that? Well, if that is vector a and that is vector b, if you were to just take the cross product of those two, you're going to get a third vector that's perpendicular to both of them. They kind of pop out of the page. That would be a cross b."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What do I mean by that? Well, if that is vector a and that is vector b, if you were to just take the cross product of those two, you're going to get a third vector that's perpendicular to both of them. They kind of pop out of the page. That would be a cross b. But the magnitude of this. So if you just take a cross product, you're going to get a vector. But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That would be a cross b. But the magnitude of this. So if you just take a cross product, you're going to get a vector. But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b. And I've proved that in the linear algebra videos. Maybe I'll prove it again in this. I mean, it's because it's the, well, I won't go into that in too detail."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b. And I've proved that in the linear algebra videos. Maybe I'll prove it again in this. I mean, it's because it's the, well, I won't go into that in too detail. I've done it before. I don't want to make this video too long. So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I mean, it's because it's the, well, I won't go into that in too detail. I've done it before. I don't want to make this video too long. So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there. And then another parallel version of b is right over there. So this is the parallelogram defined by a and b. So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there. And then another parallel version of b is right over there. So this is the parallelogram defined by a and b. So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors. So if I take the parallel to that one, it'll look something like this. And then a parallel to the orange one, it'll look something like that. So if I take the cross product of that and that, I am going to get the area of that parallelogram."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors. So if I take the parallel to that one, it'll look something like this. And then a parallel to the orange one, it'll look something like that. So if I take the cross product of that and that, I am going to get the area of that parallelogram. Now you might say, hey, this is a surface. You're taking a straight up parallelogram. But remember, these are super small changes."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I take the cross product of that and that, I am going to get the area of that parallelogram. Now you might say, hey, this is a surface. You're taking a straight up parallelogram. But remember, these are super small changes. So you can imagine a surface can be broken up into super small changes in parallelograms, or into infinitely many parallelograms. And the more parallelograms you have, the better approximation of the surface you're going to have. And this is no different than when we first took integrals."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But remember, these are super small changes. So you can imagine a surface can be broken up into super small changes in parallelograms, or into infinitely many parallelograms. And the more parallelograms you have, the better approximation of the surface you're going to have. And this is no different than when we first took integrals. We approximated the area under a curve with a bunch of rectangles. The more rectangles we had, the better. So let's call this little change in our surface d sigma."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is no different than when we first took integrals. We approximated the area under a curve with a bunch of rectangles. The more rectangles we had, the better. So let's call this little change in our surface d sigma. For a little change, for a little amount of our surface. And we could even say that the surface area of the surface will be the infinite sum of all of these infinitely small d sigmas. And there's actually a little notation for that."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's call this little change in our surface d sigma. For a little change, for a little amount of our surface. And we could even say that the surface area of the surface will be the infinite sum of all of these infinitely small d sigmas. And there's actually a little notation for that. So surface area is equal to, we could integrate over the surface. And the notation usually is a capital sigma for surface, as opposed to a region. So you're integrating over the surface."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And there's actually a little notation for that. So surface area is equal to, we could integrate over the surface. And the notation usually is a capital sigma for surface, as opposed to a region. So you're integrating over the surface. And you do a double integral, because you're going in two directions. The surface is kind of a folded two-dimensional structure. And you're going to take the infinite sum of all of the d sigmas."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you're integrating over the surface. And you do a double integral, because you're going in two directions. The surface is kind of a folded two-dimensional structure. And you're going to take the infinite sum of all of the d sigmas. So this would be the surface area of this. So that's what a d sigma is. Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you're going to take the infinite sum of all of the d sigmas. So this would be the surface area of this. So that's what a d sigma is. Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors. So let me write it here. This is all, it's not rigorous mathematics. The whole point here is to give you the intuition of what a surface integral is all about."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors. So let me write it here. This is all, it's not rigorous mathematics. The whole point here is to give you the intuition of what a surface integral is all about. So we can write that d sigma is equal to the cross product of the orange vector and the white vector. The orange vector is this. But we could also write it like this."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The whole point here is to give you the intuition of what a surface integral is all about. So we can write that d sigma is equal to the cross product of the orange vector and the white vector. The orange vector is this. But we could also write it like this. This was the result from the last video. I'll write it in orange. So the partial of r with respect to s ds."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we could also write it like this. This was the result from the last video. I'll write it in orange. So the partial of r with respect to s ds. And d sigma is going to be the magnitude of the cross product, not just the cross product. The cross product by itself will just give you a vector. And that's going to be useful when we start doing vector-valued surface integrals."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the partial of r with respect to s ds. And d sigma is going to be the magnitude of the cross product, not just the cross product. The cross product by itself will just give you a vector. And that's going to be useful when we start doing vector-valued surface integrals. But just think about it this way. So this orange vector is the same thing as that. And we're going to take the cross product of that with this white vector."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that's going to be useful when we start doing vector-valued surface integrals. But just think about it this way. So this orange vector is the same thing as that. And we're going to take the cross product of that with this white vector. This white vector is the same thing as that, which we saw, which is the same thing as this. The partial of r with respect to t dt. And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to take the cross product of that with this white vector. This white vector is the same thing as that, which we saw, which is the same thing as this. The partial of r with respect to t dt. And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here. Now, you may or may not remember that if you take these. So let's just be clear. This and this, these are vectors."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here. Now, you may or may not remember that if you take these. So let's just be clear. This and this, these are vectors. When you take the partial derivative of a vector-valued function, you're still getting a vector. This ds, this is a number. That's a number and that's a number."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This and this, these are vectors. When you take the partial derivative of a vector-valued function, you're still getting a vector. This ds, this is a number. That's a number and that's a number. And you might remember when we, in the linear algebra, or whenever you first saw taking cross products, taking the cross product of some scalar multiple, you can take the scalars out. So if we take this number and that number, we essentially factor them out of the cross product. This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a number and that's a number. And you might remember when we, in the linear algebra, or whenever you first saw taking cross products, taking the cross product of some scalar multiple, you can take the scalars out. So if we take this number and that number, we essentially factor them out of the cross product. This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt. So I wrote this here. Hey, maybe our surface area, if we were to take the sum of all of these little d sigmas, but there's no obvious way to evaluate that. But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt. So I wrote this here. Hey, maybe our surface area, if we were to take the sum of all of these little d sigmas, but there's no obvious way to evaluate that. But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's. So if you take all of the ds's, all of the dt's. So this is a ds times a dt, right? A ds times a dt, ds times a dt is right there."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's. So if you take all of the ds's, all of the dt's. So this is a ds times a dt, right? A ds times a dt, ds times a dt is right there. If we multiply this times the partial derivatives of the cross product of the partial derivatives, this times this is going to give us this area. So if we summed up all of this times this, or this times this, if we summed them up over this entire region, we will get all of the parallelograms in this region. We will get the surface area."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "A ds times a dt, ds times a dt is right there. If we multiply this times the partial derivatives of the cross product of the partial derivatives, this times this is going to give us this area. So if we summed up all of this times this, or this times this, if we summed them up over this entire region, we will get all of the parallelograms in this region. We will get the surface area. So we can write, I know this is all a little bit convoluted, and you need to kind of ponder it a little bit. Surface integrals, at least in my head, are one of the hardest things to really visualize, but it'll all hopefully make sense. So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We will get the surface area. So we can write, I know this is all a little bit convoluted, and you need to kind of ponder it a little bit. Surface integrals, at least in my head, are one of the hardest things to really visualize, but it'll all hopefully make sense. So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here. And of course, we're also going to have to take this cross product in here. We know how to do that. That's a double integral."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here. And of course, we're also going to have to take this cross product in here. We know how to do that. That's a double integral. So we're going to take the double integral over this, we could call it this region or this area right here. That area is the same thing as that whole area right over there of this thing. I'll just write it in yellow."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a double integral. So we're going to take the double integral over this, we could call it this region or this area right here. That area is the same thing as that whole area right over there of this thing. I'll just write it in yellow. Of the cross product of the partial of r with respect to s and the partial of r with respect to t, ds and dt. And so you literally just take, and it seems very convoluted how you're going to actually evaluate it, but we were able to express this thing called a surface, well, this is a very simple surface integral, in something that we can actually calculate. And in the next few videos, I'm going to show you examples of actually calculating it."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll just write it in yellow. Of the cross product of the partial of r with respect to s and the partial of r with respect to t, ds and dt. And so you literally just take, and it seems very convoluted how you're going to actually evaluate it, but we were able to express this thing called a surface, well, this is a very simple surface integral, in something that we can actually calculate. And in the next few videos, I'm going to show you examples of actually calculating it. Now this right here will only give you the surface area. But what if at every point here, so over here, what we've done in both of these expressions is we're just figuring out the surface area of each of these parallelograms and then adding them all up. That's what we're doing."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in the next few videos, I'm going to show you examples of actually calculating it. Now this right here will only give you the surface area. But what if at every point here, so over here, what we've done in both of these expressions is we're just figuring out the surface area of each of these parallelograms and then adding them all up. That's what we're doing. But what if associated with each of those little parallelograms we had some value, where that value is defined by some third function, f of x, y, z. So every parallelogram, it's super small, it's around a point, you could kind of say it's maybe the center of it, it doesn't have to be the center. It may be the center of it is some point in three-dimensional space."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what we're doing. But what if associated with each of those little parallelograms we had some value, where that value is defined by some third function, f of x, y, z. So every parallelogram, it's super small, it's around a point, you could kind of say it's maybe the center of it, it doesn't have to be the center. It may be the center of it is some point in three-dimensional space. And if you use some other function, f of x, y, and z, you'll get the value of that point. And what we want to do is figure out what happens if for every one of those parallelograms we were to multiply it times the value of the function at that point. So we could write it this way."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It may be the center of it is some point in three-dimensional space. And if you use some other function, f of x, y, and z, you'll get the value of that point. And what we want to do is figure out what happens if for every one of those parallelograms we were to multiply it times the value of the function at that point. So we could write it this way. So this is where you can imagine the function is just 1. We're just multiplying each of the parallelograms by 1. But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we could write it this way. So this is where you can imagine the function is just 1. We're just multiplying each of the parallelograms by 1. But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma. That's going to be the exact same thing. Where this is each of the little parallelograms, we're just going to multiply it by f of x, y, and z there. So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we could imagine we're multiplying each of the little parallelograms by f of x, y, and z d sigma. That's going to be the exact same thing. Where this is each of the little parallelograms, we're just going to multiply it by f of x, y, and z there. So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt. And of course, we're integrating with respect to s and t. Hopefully, we can express this function in terms of s and t. And we should be able to, because we have a parametrization there. Wherever we see an x there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super convoluted and hard. And the visualizations for this, of why you'd want to do this, it has applications in physics."}, {"video_title": "Introduction to the surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to integrate it over the area, over that region of f of x, y, and z, and then times the magnitude of the partial of r with respect to s, crossed with the partial of r with respect to t, ds dt. And of course, we're integrating with respect to s and t. Hopefully, we can express this function in terms of s and t. And we should be able to, because we have a parametrization there. Wherever we see an x there, it's really x is a function of s and t. y is a function of s and t. z is a function of s and t. And this might look super convoluted and hard. And the visualizations for this, of why you'd want to do this, it has applications in physics. It's a little hard to visualize. It's easier just to visualize the straight up surface area. But we're going to see in the next few videos that it's a little hairy to calculate these problems, but they're not too hard to do."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I have some region, so this is my region right over here. We'll call it r. And let's call the boundary of my region, let's call that c. And if I have some vector field in this region, so let me draw a vector field like this. If I draw a vector field just like that, our two-dimensional divergence theorem, which we really derived from Green's theorem, told us that the flux across our boundary of this region, so let me write that out. The flux across the boundary. So the flux is essentially going to be the vector field. It's going to be our vector field f dotted with the normal outward-facing vector. So the normal vector at any point is this outward-facing vector."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The flux across the boundary. So the flux is essentially going to be the vector field. It's going to be our vector field f dotted with the normal outward-facing vector. So the normal vector at any point is this outward-facing vector. So our vector field dotted with the normal-facing vector at our boundary times our little chunk of the boundary. If we were to sum them all up over the entire boundary, that's the same thing as summing up over the entire region. So let's sum up over the entire region."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the normal vector at any point is this outward-facing vector. So our vector field dotted with the normal-facing vector at our boundary times our little chunk of the boundary. If we were to sum them all up over the entire boundary, that's the same thing as summing up over the entire region. So let's sum up over the entire region. So summing up over this entire region, each little chunk of area, dA, we could call that dx dy if we wanted to, if we're dealing in the xy domain right over here, but each little chunk of area times the divergence of f, which is really saying how much is that vector field pulling apart. So it's times the divergence of f. And hopefully it made intuitive sense. The way that I drew this vector field right over here, you see everything's kind of coming out."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's sum up over the entire region. So summing up over this entire region, each little chunk of area, dA, we could call that dx dy if we wanted to, if we're dealing in the xy domain right over here, but each little chunk of area times the divergence of f, which is really saying how much is that vector field pulling apart. So it's times the divergence of f. And hopefully it made intuitive sense. The way that I drew this vector field right over here, you see everything's kind of coming out. We could almost call this a source right here, where the vector field seems like it's popping out of there. This has positive divergence right over here. And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The way that I drew this vector field right over here, you see everything's kind of coming out. We could almost call this a source right here, where the vector field seems like it's popping out of there. This has positive divergence right over here. And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector. So it makes sense. You have positive divergence, and this is going to be a positive value. The vector field is going, for the most part, in the direction of the normal vector."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so because of this, you actually see that the vector field at the boundary is actually going in the direction of the normal vector, pretty close to the direction of the normal vector. So it makes sense. You have positive divergence, and this is going to be a positive value. The vector field is going, for the most part, in the direction of the normal vector. So the larger this is, the larger that is. So hopefully some intuitive sense. If you had another vector field, so let me draw another region, that looked like this."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The vector field is going, for the most part, in the direction of the normal vector. So the larger this is, the larger that is. So hopefully some intuitive sense. If you had another vector field, so let me draw another region, that looked like this. So I could draw a couple of situations. So one where there's very limited divergence, maybe it's just a constant vector. The vector field doesn't really change as you go in any given direction."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you had another vector field, so let me draw another region, that looked like this. So I could draw a couple of situations. So one where there's very limited divergence, maybe it's just a constant vector. The vector field doesn't really change as you go in any given direction. Over here, you'll get positive fluxes. I don't know what the plural of flux is. You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The vector field doesn't really change as you go in any given direction. Over here, you'll get positive fluxes. I don't know what the plural of flux is. You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector. But here you'll get a negative flux. So stuff is coming in here. If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You'll get positive fluxes because the vector field seems to be going in roughly the same direction as our normal vector. But here you'll get a negative flux. So stuff is coming in here. If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out. So your net flux will be close to 0. Stuff is coming in and stuff is coming out. Here you're just saying, hey, stuff is constantly coming out of this surface."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you imagine your vector field is essentially some type of mass density times volume, and we've thought about that before, this is showing how much stuff is coming in, and then stuff is coming out. So your net flux will be close to 0. Stuff is coming in and stuff is coming out. Here you're just saying, hey, stuff is constantly coming out of this surface. So hopefully this gives you a sense that here you have very low divergence and you would have a low flux, total aggregate flux going through your boundary. Here you have a high divergence and you would have a high aggregate flux. I could draw another situation."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Here you're just saying, hey, stuff is constantly coming out of this surface. So hopefully this gives you a sense that here you have very low divergence and you would have a low flux, total aggregate flux going through your boundary. Here you have a high divergence and you would have a high aggregate flux. I could draw another situation. So this is my region R. And let's say that we have negative divergence, or we could even call it convergence. Convergence isn't an actual technical term, but you can imagine if the vector field is converging within R, well, the divergence is going to be negative in this situation. It's actually converging, which is the opposite of diverging."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I could draw another situation. So this is my region R. And let's say that we have negative divergence, or we could even call it convergence. Convergence isn't an actual technical term, but you can imagine if the vector field is converging within R, well, the divergence is going to be negative in this situation. It's actually converging, which is the opposite of diverging. So the divergence is negative in this situation. And also the flux across the boundary is going to be negative, because as we see here, the way I drew it, across most of this boundary, the vector field is going in the opposite direction. It's going in the opposite direction as our normal vector at any point."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's actually converging, which is the opposite of diverging. So the divergence is negative in this situation. And also the flux across the boundary is going to be negative, because as we see here, the way I drew it, across most of this boundary, the vector field is going in the opposite direction. It's going in the opposite direction as our normal vector at any point. So hopefully this gives you a sense of why there's this connection between the divergence over the region and the flux across the boundary. Well, now we're just going to extend this to three dimensions. And it's the exact same reasoning."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going in the opposite direction as our normal vector at any point. So hopefully this gives you a sense of why there's this connection between the divergence over the region and the flux across the boundary. Well, now we're just going to extend this to three dimensions. And it's the exact same reasoning. If we have a, and I'll define it a little bit more precisely in future videos, a simple solid region. So let me just draw it. And I'm going to try to draw it in three dimensions."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's the exact same reasoning. If we have a, and I'll define it a little bit more precisely in future videos, a simple solid region. So let me just draw it. And I'm going to try to draw it in three dimensions. So let's say it looks something like that. And one way to think about it is this is going to be a region that doesn't bend back on itself. If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'm going to try to draw it in three dimensions. So let's say it looks something like that. And one way to think about it is this is going to be a region that doesn't bend back on itself. If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves. And there are some that you might not be able to imagine that would also not make the case. But even if you had ones that bent back on themselves, you could separate them out into other ones that don't. So here's just a simple solid region over here."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you have a region that bends back on itself, and we'll think about it in multiple ways, but out of all of the volumes of three dimensions that you can imagine, these are the ones that don't bend back on themselves. And there are some that you might not be able to imagine that would also not make the case. But even if you had ones that bent back on themselves, you could separate them out into other ones that don't. So here's just a simple solid region over here. I'll make it look three dimensional. So maybe if it was transparent, you would see it like that. And then you see the front of it like that."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So here's just a simple solid region over here. I'll make it look three dimensional. So maybe if it was transparent, you would see it like that. And then you see the front of it like that. So it's this kind of elliptical, circular, blob-looking thing. So that could be the back of it. And then if you go to the front, it could look something like that."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you see the front of it like that. So it's this kind of elliptical, circular, blob-looking thing. So that could be the back of it. And then if you go to the front, it could look something like that. So this is our simple solid region. I'll call it R still, but we're dealing with a three dimensional. We are now dealing with a three dimensional region."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if you go to the front, it could look something like that. So this is our simple solid region. I'll call it R still, but we're dealing with a three dimensional. We are now dealing with a three dimensional region. And now the boundary of this is no longer a line. We're now in three dimensions. The boundary is a surface."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We are now dealing with a three dimensional region. And now the boundary of this is no longer a line. We're now in three dimensions. The boundary is a surface. So I'll call that S. S is the boundary of R. And now let's throw on a vector field here. Now this is a vector field in three dimensions. And now let's imagine that we actually have positive divergence of our vector field within this region right over here."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The boundary is a surface. So I'll call that S. S is the boundary of R. And now let's throw on a vector field here. Now this is a vector field in three dimensions. And now let's imagine that we actually have positive divergence of our vector field within this region right over here. So we have positive divergence. So you could imagine that it's kind of the vector field within the region. It's a source of the vector field."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now let's imagine that we actually have positive divergence of our vector field within this region right over here. So we have positive divergence. So you could 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 of the vector field. 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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's a source of the vector field. Or the vector field is diverging out of the vector field. 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 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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 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, and then multiply that times a little chunk of surface."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, and then multiply that times a little chunk of surface. So multiply that times a little chunk of surface. And then sum it up along the whole surface. So sum it up."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the flux across the surface, you would take your vector field, dot it with the normal vector at the surface, and then multiply that times a little chunk of 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. Is going to be equal to, if we were to sum up the divergence of f, if we were to sum up across the whole volume."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So sum it up. So it's going to be a surface integral. So this is flux across the surface. Is going to be equal to, if we were to sum up the divergence of f, 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, what is the divergence of f at each point? And then multiply it times the volume of that little chunk."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Is going to be equal to, if we were to sum up the divergence of f, 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, what is the divergence of f at each point? And then multiply it times the volume of that little chunk. And multiply it times the volume of that chunk, in the sense of how much is it totally diverging in that volume. And then you sum it up. That should be equal to the flux."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then multiply it times the volume of that little chunk. And multiply it times the volume of that chunk, in the sense of how much is it totally diverging in that volume. And then you sum it up. That should be equal to the flux. It's completely analogous to what 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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That should be equal to the flux. It's completely analogous to what 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 the surface. Here we were summing the divergence in the region."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 the 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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "3D divergence theorem intuition   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And now in the next few videos, we can do some worked examples just so you feel comfortable computing or manipulating these integrals. And then we'll do a couple of proof videos where we actually prove the divergence theorem."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "It's a scalar valued function. The question is, how do we take the derivative of an expression like this? And there's a certain method called a partial derivative which is very similar to ordinary derivatives and I kinda wanna show how they're secretly the same thing. So to do that, let me just remind ourselves of how we interpret the notation for ordinary derivatives. So if you have something like f of x is equal to x squared, and let's say you wanna take its derivative, and I'll use the Leibniz notation here, df dx, and let's evaluate it at two, let's say. I really like this notation because it's suggestive of what's going on. If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So to do that, let me just remind ourselves of how we interpret the notation for ordinary derivatives. So if you have something like f of x is equal to x squared, and let's say you wanna take its derivative, and I'll use the Leibniz notation here, df dx, and let's evaluate it at two, let's say. I really like this notation because it's suggestive of what's going on. If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that. Then we go to the input, x equals one, two. This little dx here, I like to interpret as just a little nudge in the x direction, and it's kinda the size of that nudge. And then df, df, is the resulting change in the output after you make that initial little nudge."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "If we sketch out a graph, so you know, this axis represents our output, this over here represents our input, and x squared has a certain parabolic shape to it, something like that. Then we go to the input, x equals one, two. This little dx here, I like to interpret as just a little nudge in the x direction, and it's kinda the size of that nudge. And then df, df, is the resulting change in the output after you make that initial little nudge. So it's this resulting change. And when you're thinking in terms of graphs, this is slope, you kind of have this rise over run for your ratio between the tiny change of the output that's caused by a tiny change in the input, and of course this is dependent on where you start. Over here we have x equals two."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And then df, df, is the resulting change in the output after you make that initial little nudge. So it's this resulting change. And when you're thinking in terms of graphs, this is slope, you kind of have this rise over run for your ratio between the tiny change of the output that's caused by a tiny change in the input, and of course this is dependent on where you start. Over here we have x equals two. But you could also think about this without graphs if you really wanted to. You might just think about, you know, your input space is just a number line, and your output space also is just a number line, the output of f over here. And really you're just thinking of somehow mapping numbers from here onto the second line."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Over here we have x equals two. But you could also think about this without graphs if you really wanted to. You might just think about, you know, your input space is just a number line, and your output space also is just a number line, the output of f over here. And really you're just thinking of somehow mapping numbers from here onto the second line. And in that case, your initial nudge, your initial little dx, would be some nudge on that number line. And you're wondering how that influences the function itself. So maybe that causes a nudge that's, you know, four times as big."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And really you're just thinking of somehow mapping numbers from here onto the second line. And in that case, your initial nudge, your initial little dx, would be some nudge on that number line. And you're wondering how that influences the function itself. So maybe that causes a nudge that's, you know, four times as big. And that would mean your derivative is four at that point. So the reason that I'm talking about this is because over in the multivariable world, we can pretty much do the same thing. You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output?"}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So maybe that causes a nudge that's, you know, four times as big. And that would mean your derivative is four at that point. So the reason that I'm talking about this is because over in the multivariable world, we can pretty much do the same thing. You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output? But this time the way that you might visualize it, you'd be thinking of your input space, here I'll draw it down here, as the xy plane. So this time this is not gonna be graphing the function, this is every point on the plane is an input. And let's say you were evaluating this at a point like one, two."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "You know, you could write df dx, and interpret that as saying, hey, how does a tiny change in the input in the x direction influence the output? But this time the way that you might visualize it, you'd be thinking of your input space, here I'll draw it down here, as the xy plane. So this time this is not gonna be graphing the function, this is every point on the plane is an input. And let's say you were evaluating this at a point like one, two. Okay? So in that case, so you'd go over to the input that's one, and then two. And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output?"}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And let's say you were evaluating this at a point like one, two. Okay? So in that case, so you'd go over to the input that's one, and then two. And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output? And in this case, the output, I mean, it's still just a number. So maybe we go off to the side here, and we draw just like a number line as our output. And somehow we're thinking about the function as mapping points on the plane to the number line."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And then you'd say, okay, so this tiny nudge in the input, this tiny change dx, how does that influence the output? And in this case, the output, I mean, it's still just a number. So maybe we go off to the side here, and we draw just like a number line as our output. And somehow we're thinking about the function as mapping points on the plane to the number line. So you'd say, okay, that's your dx, how much does it change the output? And maybe this time it changes it negatively, it depends on your function. And that would be your df."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And somehow we're thinking about the function as mapping points on the plane to the number line. So you'd say, okay, that's your dx, how much does it change the output? And maybe this time it changes it negatively, it depends on your function. And that would be your df. And you can also do this with the y variable, right? There's no reason that you can't say df dy, and evaluate at that same point, one, two, and interpret totally the same way. Except this time your dy would be a change in the y direction."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And that would be your df. And you can also do this with the y variable, right? There's no reason that you can't say df dy, and evaluate at that same point, one, two, and interpret totally the same way. Except this time your dy would be a change in the y direction. So maybe I should really emphasize here that that dx is a change in the x direction here, and that dy is a change in the y direction. And maybe when you change your f according to y, it does something different, right? Maybe the output increases, and it increases by a lot, it's more sensitive to y."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Except this time your dy would be a change in the y direction. So maybe I should really emphasize here that that dx is a change in the x direction here, and that dy is a change in the y direction. And maybe when you change your f according to y, it does something different, right? Maybe the output increases, and it increases by a lot, it's more sensitive to y. 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 ds, and dx, df."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Maybe the output increases, and it increases by a lot, it's more sensitive to y. 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 ds, and dx, df. And a lot of people came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multivariable 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."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "But first, there's kind of an annoying thing associated with partial derivatives, where we don't write them with ds, and dx, df. And a lot of people came up with this new notation, mostly just to emphasize to the reader of your equation that it's a multivariable 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, because it only cares about the x direction, neither does this, this only cares about the y direction, 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."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "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, because it only cares about the x direction, neither does this, this only cares about the y direction, 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, ah, little remnants. So if you're actually evaluating something like this, here, I'll write it again up here."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "I'm gonna go ahead and clear the board over here. I think the one-dimensional analogy is something we probably have already. So, ah, little remnants. So if you're actually evaluating something like this, here, I'll write it 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."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So if you're actually evaluating something like this, here, I'll write it 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, because when you're only moving in the x direction, this is kind of how the multivariable function sees the world, and I'll just keep a little note that we're evaluating this whole thing at x equals one."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "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, because when you're only moving in the x direction, this is kind of how the multivariable 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."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "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, because when you're only moving in the x direction, this is kind of how the multivariable 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, the derivative of x squared times two is gonna be four x, because x squared goes to two x, and then the derivative of a constant, sine 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 just for practice, let's also do that with the derivative with respect to y."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "You're asking how it changes as you shift around x, and you know how to do this. This is just taking the derivative, the derivative of x squared times two is gonna be four x, because x squared goes to two x, and then the derivative of a constant, sine 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 just for practice, let's also do that with the 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."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And just for practice, let's also do that with the 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 sine of y. Sine of y. And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "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 sine of y. Sine of y. And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two. When you take the derivative, this is just one times y, so the derivative is one. This over here, the derivative is cosine, cosine of y. Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And you're saying, oh, I'm keeping track of this at y equals two, so that's kind of your evaluating at y equals two. When you take the derivative, this is just one times y, so the derivative is one. This over here, the derivative is cosine, cosine of y. Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two. I'm not sure what the value of cosine of two is off the top of my head, but that would be your answer. And this is a partial derivative at a point, but a lot of times, you're not asked to just compute it at a point. What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Again, we're evaluating this whole thing at y equals two, so your overall answer, it would be one plus cosine of two. I'm not sure what the value of cosine of two is off the top of my head, but that would be your answer. And this is a partial derivative at a point, but a lot of times, you're not asked to just compute it at a point. What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer. So let me just kind of go over how you would do that. It's actually very similar, but this time, instead of plugging in the constant ahead of time, we just have to pretend that it's a constant. So let me make a little bit of space for ourselves here."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "What you want is a general formula that tells you, hey, plug in any point x, y, and it should spit out the answer. So let me just kind of go over how you would do that. It's actually very similar, but this time, instead of plugging in the constant ahead of time, we just have to pretend that it's a constant. So let me make a little bit of space for ourselves here. Really. We don't need any of this anymore. I'm gonna leave the partial partial f, partial partial y."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So let me make a little bit of space for ourselves here. Really. We don't need any of this anymore. I'm gonna leave the partial partial f, partial partial y. And we want this as a more general function of x and y. Well, we kind of do the same thing. We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "I'm gonna leave the partial partial f, partial partial y. And we want this as a more general function of x and y. Well, we kind of do the same thing. We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative. But now we'd write x squared, and then kind of emphasize that it's a constant value of y, plus the sine, and again, I'll say y. And here, I'm writing the variable y, but we have to pretend like it's a constant. Now, you're pretending that you plug in two or something like that, and you still just take the derivative."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "We're gonna say that this is a derivative with respect to x, and I'm using partials just to kind of emphasize that it's a partial derivative. But now we'd write x squared, and then kind of emphasize that it's a constant value of y, plus the sine, and again, I'll say y. And here, I'm writing the variable y, but we have to pretend like it's a constant. Now, you're pretending that you plug in two or something like that, and you still just take the derivative. So in this case, the derivative of x squared times a constant is just two x times that constant, two x times that constant. And over here, the derivative of a constant is always zero. So that's just always gonna be zero."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Now, you're pretending that you plug in two or something like that, and you still just take the derivative. So in this case, the derivative of x squared times a constant is just two x times that constant, two x times that constant. And over here, the derivative of a constant is always zero. So that's just always gonna be zero. So this is your partial derivative as a more general formula. If you plugged in one, two to this, you'd get what we had before. And similarly, if you're doing this with partial f, partial y, we write down all of the same things."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So that's just always gonna be zero. So this is your partial derivative as a more general formula. If you plugged in one, two to this, you'd get what we had before. And similarly, if you're doing this with partial f, partial y, we write down all of the same things. Now, you're taking it with respect to y, and I'm just gonna copy this formula here, actually. But this time, we're considering all of the x's to be constants. So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And similarly, if you're doing this with partial f, partial y, we write down all of the same things. Now, you're taking it with respect to y, and I'm just gonna copy this formula here, actually. But this time, we're considering all of the x's to be constants. So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant. So this is gonna be x squared. And over here, you're taking the derivative of sine of y. There's no x's in there."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "So in this case, when you take the derivative with respect to y of some kind of constant, you know, constant squared is a constant, times y, it's just gonna equal that constant. So this is gonna be x squared. And over here, you're taking the derivative of sine of y. There's no x's in there. So that remains the sine of y. If you're being clear. And now, this is a more general formula."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "There's no x's in there. So that remains the sine of y. If you're being clear. And now, this is a more general formula. If you plugged in one, two, you would get one. Oh, sorry, that's cosine of y. Cosine of y. Because we're taking a derivative."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "And now, this is a more general formula. If you plugged in one, two, you would get one. Oh, sorry, that's cosine of y. Cosine of y. Because we're taking a derivative. So if you plugged in one, two, you know, you would get one plus the cosine of one, which is what we had before. So this, this is really what you'll see for how to compute a partial derivative. You pretend that one of the variables is constant, and you take an ordinary derivative."}, {"video_title": "Partial derivatives, introduction.mp3", "Sentence": "Because we're taking a derivative. So if you plugged in one, two, you know, you would get one plus the cosine of one, which is what we had before. So this, this is really what you'll see for how to compute a partial derivative. You pretend that one of the variables is constant, and you take an ordinary derivative. And in the back of your mind, you're thinking this is because you're just moving in one direction for the input, and you're seeing how that influences things, and then, you know, you might move in one direction for another input, and see how that influences things. In the next video, I'll show you what this means in terms of graphs and slopes. But it's important to understand that graphs and slopes are not the only way to understand derivatives, because as soon as you start thinking about vector-valued functions, or functions with inputs of higher dimensions than just two, you can no longer think in terms of graphs."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "Or if you prefer my kindergartener drawing skills, you can just look at the helix over here while we work it through. And the part where we left off, we have this tangent vector function, this unit tangent vector function for our curve, so at every given value t, whatever point that corresponds to on the curve, this function is going to give us the vector that is of unit length and tangent to the curve. And the ultimate goal for curvature is to find the derivative of that unit tangent vector with respect to arc length, and specifically we want its magnitude. And what that typically requires, and I talked about this in the videos on the formula, is you take its derivative with respect to the parameter t, because that's the thing you can actually do, and that might not correspond to unit length. If you nudge the parameter t, it might not nudge you a corresponding length on the curve, but you correct that by dividing out by the derivative of the parameterization function with respect to t. And that's actually arc length, the magnitude of the derivative of the parameterization with respect to t. Boy, is that a hard word. So with our given function, let's go ahead and start doing that. And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And what that typically requires, and I talked about this in the videos on the formula, is you take its derivative with respect to the parameter t, because that's the thing you can actually do, and that might not correspond to unit length. If you nudge the parameter t, it might not nudge you a corresponding length on the curve, but you correct that by dividing out by the derivative of the parameterization function with respect to t. And that's actually arc length, the magnitude of the derivative of the parameterization with respect to t. Boy, is that a hard word. So with our given function, let's go ahead and start doing that. And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply. So our unit tangent vector function, that first component where we're dividing by the square root of 26 divided by 5, instead I'm just going to write that as negative 5 times the sine of t, so I'm just kind of moving that 5 up into the numerator, divided by the root of 26. And I'll do the same thing for cosine, where we move that 5 up into the numerator. Cosine of t divided by the root of 26."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And I think first, given that we have this kind of fractions of fractions form, I'm just going to start by writing it a little bit more simply. So our unit tangent vector function, that first component where we're dividing by the square root of 26 divided by 5, instead I'm just going to write that as negative 5 times the sine of t, so I'm just kind of moving that 5 up into the numerator, divided by the root of 26. And I'll do the same thing for cosine, where we move that 5 up into the numerator. Cosine of t divided by the root of 26. And then that last part, 1 5th divided by the square root of 26 5ths, gives us just 1 divided by the square root of 26. All right, so the first step in using our curvature formula is going to be to take the derivative of this guy, right? We need the derivative of the tangent vector function."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "Cosine of t divided by the root of 26. And then that last part, 1 5th divided by the square root of 26 5ths, gives us just 1 divided by the square root of 26. All right, so the first step in using our curvature formula is going to be to take the derivative of this guy, right? We need the derivative of the tangent vector function. So we go ahead and start doing that. We see that d big T, d little t, so tangent vector function parameter, is equal to, and we just take the derivative of each component. So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "We need the derivative of the tangent vector function. So we go ahead and start doing that. We see that d big T, d little t, so tangent vector function parameter, is equal to, and we just take the derivative of each component. So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26. Similarly, 5 cosine goes to negative 5 sine, since the derivative of cosine is negative sine. Negative 5 sine divided by 26, or square root of 26. And that last component is just a constant, so derivative is nothing, it's zero."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "So negative 5 sine goes to negative 5 cosine, and we divide out by that constant, square root of 26. Similarly, 5 cosine goes to negative 5 sine, since the derivative of cosine is negative sine. Negative 5 sine divided by 26, or square root of 26. And that last component is just a constant, so derivative is nothing, it's zero. Next step, we're going to take the magnitude of that vector that we just found. So we're trying to find the magnitude of the derivative of the tangent vector function. So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And that last component is just a constant, so derivative is nothing, it's zero. Next step, we're going to take the magnitude of that vector that we just found. So we're trying to find the magnitude of the derivative of the tangent vector function. So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys. So the square of that first component is going to be 25 multiplied by cosine squared, cosine squared of T, all divided by 26, right, because the square root of the square, the square of the square root of 26 is 26. And then we add to that 25 times sine squared, sine squared of T, also divided by 26. And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "So we say, all right, magnitude of what we just found, d big T, d little t, involves, so this magnitude will be the square root, make the little tick there, square root of the sum of the squares of these guys. So the square of that first component is going to be 25 multiplied by cosine squared, cosine squared of T, all divided by 26, right, because the square root of the square, the square of the square root of 26 is 26. And then we add to that 25 times sine squared, sine squared of T, also divided by 26. And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair. The reason we love things involving circles, this always happens, nice cancellation. This just becomes one, so what we're left with on the whole is root 25 over 26, pretty nice, root 25 over 26. So for our curvature equation, we go up and we can start plugging that in."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And from that, we can factor out, factor out 25 over 26 inside that radical, because both terms involve multiplying by 25 and dividing by 26, and what we're left with is a nice and friendly cosine sine pair. The reason we love things involving circles, this always happens, nice cancellation. This just becomes one, so what we're left with on the whole is root 25 over 26, pretty nice, root 25 over 26. So for our curvature equation, we go up and we can start plugging that in. So we just found that numerator and we found that it was the square root of 25 divided by 26, the entire thing, 25 over 26. And we already found the magnitude of the derivative itself, that's one of the things we needed to do to find the tangent vector. That's where this 26 over 5 came from."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "So for our curvature equation, we go up and we can start plugging that in. So we just found that numerator and we found that it was the square root of 25 divided by 26, the entire thing, 25 over 26. And we already found the magnitude of the derivative itself, that's one of the things we needed to do to find the tangent vector. That's where this 26 over 5 came from. I deleted it from where we did last video to make room, but if you look at the last video, you can see where we got that square root of 26 divided by 5, and I'll actually write that as square root of 26 divided by 25. That's how we originally found it, I'm just putting the 5 back under the radical. And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right?"}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "That's where this 26 over 5 came from. I deleted it from where we did last video to make room, but if you look at the last video, you can see where we got that square root of 26 divided by 5, and I'll actually write that as square root of 26 divided by 25. That's how we originally found it, I'm just putting the 5 back under the radical. And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right? One is 25 over 26, the other is 26 over 25. So if we put everything under the radical there, I'm going to say equals the square root, and we're going to have 25 over 26 divided by 26 over 25. And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And it's tempting if you aren't looking too closely to think these guys cancel out, but it's actually they're the opposite of each other, right? One is 25 over 26, the other is 26 over 25. So if we put everything under the radical there, I'm going to say equals the square root, and we're going to have 25 over 26 divided by 26 over 25. And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared. And the root of that whole thing just gives us 25 over 26. And that is our curvature. That right there is the answer, that's the curvature."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "And when we flip that bottom and multiply, what we get is 25 squared divided by 26 squared. And the root of that whole thing just gives us 25 over 26. And that is our curvature. That right there is the answer, that's the curvature. So it's a little bit greater than 1. No, no, sorry, so it's a little bit less than 1, which means that you're curving a little bit less than you would if it was a circle with radius 1, which kind of makes sense if we look at the image here, because if the helix were completely flattened out, if you imagine squishing this down onto the xy-plane, you'd just be going around a circle with radius 1. But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "That right there is the answer, that's the curvature. So it's a little bit greater than 1. No, no, sorry, so it's a little bit less than 1, which means that you're curving a little bit less than you would if it was a circle with radius 1, which kind of makes sense if we look at the image here, because if the helix were completely flattened out, if you imagine squishing this down onto the xy-plane, you'd just be going around a circle with radius 1. But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight. So the curvature should go down a little bit because it's becoming a little bit more straight. The radius of curvature will go up. So that's the curvature of a helix."}, {"video_title": "Curvature of a helix, part 2.mp3", "Sentence": "But by kind of pulling that spring and pulling it so that there's a z component, you're making it a little bit more straight. So the curvature should go down a little bit because it's becoming a little bit more straight. The radius of curvature will go up. So that's the curvature of a helix. And that's a pretty good example of how you can find the curvature by walking through directly the idea of finding dT, dS, and, you know, getting that unit tangent vector, getting that little unit arc length. And in the next example, I think I'll go through one where you just use the formula, where it's something a little bit more complicated than thinking about this, and you turn to the formula itself. See you then."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say surface 2 can be represented by the vector position function. I'll just call it t, t for 2. So t, t, which is a vector, it's going to be a function of x and y. Those are going to be our parameters. And it's going to be equal to, and we can do this once again, because we're dealing with our surface is a function of x and y. So it's going to be equal to x times i plus y times j plus f2 of x, y times k for all the x, y's that are a member of our domain. Now with that out of the way, we can re-express what k dot n ds is."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Those are going to be our parameters. And it's going to be equal to, and we can do this once again, because we're dealing with our surface is a function of x and y. So it's going to be equal to x times i plus y times j plus f2 of x, y times k for all the x, y's that are a member of our domain. Now with that out of the way, we can re-express what k dot n ds is. So let me write this over here. k dot n ds is equal to, and we could put parentheses here that we're going to take this dot product. That's at least how I like to think about it."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now with that out of the way, we can re-express what k dot n ds is. So let me write this over here. k dot n ds is equal to, and we could put parentheses here that we're going to take this dot product. That's at least how I like to think about it. This is the exact same thing as k dotted with the cross product of the partial of t with respect to x crossed with the partial of t with respect to y times a little chunk of our area, times da, a little chunk of our area in the x, y domain. We've done this multiple times as we evaluated surface integrals, and we got the intuition for why this works. So we're essentially just evaluating the surface integral."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's at least how I like to think about it. This is the exact same thing as k dotted with the cross product of the partial of t with respect to x crossed with the partial of t with respect to y times a little chunk of our area, times da, a little chunk of our area in the x, y domain. We've done this multiple times as we evaluated surface integrals, and we got the intuition for why this works. So we're essentially just evaluating the surface integral. And the one thing we want to make sure is make sure this has the right orientation. Because remember, in order for the divergence theorem to be true, the way we've defined it is all the normal vectors have to be outward facing. So for this top surface, the normal vector has to be pointing straight up."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're essentially just evaluating the surface integral. And the one thing we want to make sure is make sure this has the right orientation. Because remember, in order for the divergence theorem to be true, the way we've defined it is all the normal vectors have to be outward facing. So for this top surface, the normal vector has to be pointing straight up. Or not necessarily straight up, at least upwards. If this is a curve, it wouldn't be necessarily straight up, but it needs to be kind of outward facing like that. On the sides, it would be outward facing like that."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So for this top surface, the normal vector has to be pointing straight up. Or not necessarily straight up, at least upwards. If this is a curve, it wouldn't be necessarily straight up, but it needs to be kind of outward facing like that. On the sides, it would be outward facing like that. And down here, it would be outward facing, going in the general downwards direction. So let's just make sure that this is upwards facing. If we're changing with respect to x, we're going in this direction."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "On the sides, it would be outward facing like that. And down here, it would be outward facing, going in the general downwards direction. So let's just make sure that this is upwards facing. If we're changing with respect to x, we're going in this direction. Changing with respect to y, we're going in that direction. Take the right hand rule with the cross product. Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we're changing with respect to x, we're going in this direction. Changing with respect to y, we're going in that direction. Take the right hand rule with the cross product. Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up. So it goes in the right direction. So this would be an upward pointing vector. So we got the right orientation for our surface."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Index finger there, middle finger there, your left thumb, or your right thumb, I should say, will go straight up. So it goes in the right direction. So this would be an upward pointing vector. So we got the right orientation for our surface. Now let's think about what this is. And it's important to realize we could calculate all of the components of this, but then we're just going to take the dot product of that with k. So it's really we care about the k component only. But I'll work it out."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we got the right orientation for our surface. Now let's think about what this is. And it's important to realize we could calculate all of the components of this, but then we're just going to take the dot product of that with k. So it's really we care about the k component only. But I'll work it out. So this is equal to k times a matrix, i, j, k, of the partial of t with respect to x. Well, the partial of t with respect to x, I'll do this in blue, is going to be 1, 0, 1, 0, and the partial of f2 with respect to x. And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I'll work it out. So this is equal to k times a matrix, i, j, k, of the partial of t with respect to x. Well, the partial of t with respect to x, I'll do this in blue, is going to be 1, 0, 1, 0, and the partial of f2 with respect to x. And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y. And then, of course, we have to multiply times dA. And this is all going to be equal to k, unit vector k, dotted with, and I don't even have to even work it all out, it's going to be something times the i unit vector minus something, checkerboard pattern, minus something times something else, necessarily, times our j unit vector, plus, and now we can think about the k unit vector. So the k unit vector is going to be 1 times 1 minus 0."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the partial of t with respect to y is going to be 0, 1, 0, 1, and the partial of f2 with respect to y. And then, of course, we have to multiply times dA. And this is all going to be equal to k, unit vector k, dotted with, and I don't even have to even work it all out, it's going to be something times the i unit vector minus something, checkerboard pattern, minus something times something else, necessarily, times our j unit vector, plus, and now we can think about the k unit vector. So the k unit vector is going to be 1 times 1 minus 0. So it's just going to be plus the k unit vector. We know it's 1 times the k unit vector. And so when you take this dot, and of course, we have our dA out right here."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the k unit vector is going to be 1 times 1 minus 0. So it's just going to be plus the k unit vector. We know it's 1 times the k unit vector. And so when you take this dot, and of course, we have our dA out right here. But when you take this dot product, you only are left with the k components. And it's essentially just 1 times 1. You end up with a scalar quantity of 1."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so when you take this dot, and of course, we have our dA out right here. But when you take this dot product, you only are left with the k components. And it's essentially just 1 times 1. You end up with a scalar quantity of 1. All of this business just simplified to dA. So now we can rewrite our surface integral. And we're going to rewrite it in the xy domain now, in our parameters domain."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You end up with a scalar quantity of 1. All of this business just simplified to dA. So now we can rewrite our surface integral. And we're going to rewrite it in the xy domain now, in our parameters domain. So our surface integral right up here. So this will be good for this video. And then we'll do the same thing with this surface here, just making sure that we get the orientation right."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to rewrite it in the xy domain now, in our parameters domain. So our surface integral right up here. So this will be good for this video. And then we'll do the same thing with this surface here, just making sure that we get the orientation right. So this surface integral, S2, and I'll even rewrite a little bit. S2, which is a function, r is a function of x, y, and z, times k dot n ds. I just rewrote all of this right up here."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we'll do the same thing with this surface here, just making sure that we get the orientation right. So this surface integral, S2, and I'll even rewrite a little bit. S2, which is a function, r is a function of x, y, and z, times k dot n ds. I just rewrote all of this right up here. Is equivalent to the double integral over our parameters domain, which is just d, of r of x, y, z times all of this business. All of this business just simplified to dA. And since I want to write it in terms of my parameters, I'll write it as r of x, y."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I just rewrote all of this right up here. Is equivalent to the double integral over our parameters domain, which is just d, of r of x, y, z times all of this business. All of this business just simplified to dA. And since I want to write it in terms of my parameters, I'll write it as r of x, y. And while we're on that surface, z is equal to F2. So it's x, y, and F2 of x, y. And then all of this business we just saw simplified to dA."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And since I want to write it in terms of my parameters, I'll write it as r of x, y. And while we're on that surface, z is equal to F2. So it's x, y, and F2 of x, y. And then all of this business we just saw simplified to dA. So you might be saying, hey, Sal, it didn't look like you simplified it a lot. But at least now put it in terms of a double integral, instead of a surface integral. So at least in my mind, that is a simplification."}, {"video_title": "Divergence theorem proof (part 3)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then all of this business we just saw simplified to dA. So you might be saying, hey, Sal, it didn't look like you simplified it a lot. But at least now put it in terms of a double integral, instead of a surface integral. So at least in my mind, that is a simplification. In the next video, we're going to do the exact same thing with this, just making sure that our vectors are oriented properly. And we could just introduce a negative sign to make sure that they are. And then we're going to think about the triple integrals and try to simplify those."}, {"video_title": "Fluid flow and vector fields   Multivariable calculus   Khan Academy.mp3", "Sentence": "So imagine that we're sitting in the coordinate plane, and that I draw for you a whole bunch of little droplets, droplets of water, and then these are going to start flowing in some way. How would you describe this flow mathematically? So, at every given point, the particles are moving in some different way over here, they're kind of moving down and to the left. Here, they're moving kind of quickly up, over here, they're moving more slowly down. So what you might want to do is assign a vector to every single point in space, and a common attribute of the way that fluids flow, this isn't necessarily obvious, but if you look at a given point in space, let's say like right here, every time that a particle passes through it, it's with roughly the same velocity. So you might think over time, that velocity would change, and sometimes it does, a lot of times there's some fluid flow where it depends on time, but for many cases, you can just say, at this point in space, whatever particle is going through it, it'll have this velocity vector. So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field."}, {"video_title": "Fluid flow and vector fields   Multivariable calculus   Khan Academy.mp3", "Sentence": "Here, they're moving kind of quickly up, over here, they're moving more slowly down. So what you might want to do is assign a vector to every single point in space, and a common attribute of the way that fluids flow, this isn't necessarily obvious, but if you look at a given point in space, let's say like right here, every time that a particle passes through it, it's with roughly the same velocity. So you might think over time, that velocity would change, and sometimes it does, a lot of times there's some fluid flow where it depends on time, but for many cases, you can just say, at this point in space, whatever particle is going through it, it'll have this velocity vector. So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field. So this here is a little bit of a cleaner drawing than what I have, and as I mentioned in the last video, it's common for these vectors not to be drawn to scale, but to all have the same length, just to get a sense of direction, and here you can see each particle is flowing roughly along that vector, so whatever one it's closest to, it's moving in that direction, and this is not just a really good way of understanding fluid flow, but it goes the other way around it's a really good way of understanding vector fields themselves. So sometimes, you might just be given some new vector field, and to get a feel for what it's all about, how to interpret it, what special properties it might have, it's actually helpful, even if it's not meant to represent a fluid, to imagine that it does, and think of all the particles, and think of how they would move along it. For example, this particular one, as you play the animation, as you let the particles move along the vectors, there's no change in the density, at no point do a bunch of particles go inward, or a bunch of particles go outward, it stays kind of constant, and that turns out to have a certain mathematical significance down the road, you'll see this later on, as we study a certain concept called divergence, and over here, you see this vector field, and you might want to understand what it's all about, and it's kind of helpful to think of a fluid that pushes outward from everywhere, and is kind of decreasing in density around the center, and that also has a certain mathematical significance."}, {"video_title": "Fluid flow and vector fields   Multivariable calculus   Khan Academy.mp3", "Sentence": "So over here, they might be pretty high upwards, whereas here, it's kind of a smaller vector downwards, even though, and here I'll play the animation a little bit more here, and if you imagine doing this at all of the different points in space, and assigning a vector to describe the motion of each fluid particle at each different point, what you end up getting is a vector field. So this here is a little bit of a cleaner drawing than what I have, and as I mentioned in the last video, it's common for these vectors not to be drawn to scale, but to all have the same length, just to get a sense of direction, and here you can see each particle is flowing roughly along that vector, so whatever one it's closest to, it's moving in that direction, and this is not just a really good way of understanding fluid flow, but it goes the other way around it's a really good way of understanding vector fields themselves. So sometimes, you might just be given some new vector field, and to get a feel for what it's all about, how to interpret it, what special properties it might have, it's actually helpful, even if it's not meant to represent a fluid, to imagine that it does, and think of all the particles, and think of how they would move along it. For example, this particular one, as you play the animation, as you let the particles move along the vectors, there's no change in the density, at no point do a bunch of particles go inward, or a bunch of particles go outward, it stays kind of constant, and that turns out to have a certain mathematical significance down the road, you'll see this later on, as we study a certain concept called divergence, and over here, you see this vector field, and you might want to understand what it's all about, and it's kind of helpful to think of a fluid that pushes outward from everywhere, and is kind of decreasing in density around the center, and that also has a certain mathematical significance. And it might also lead you to ask certain other questions, like if you look at the fluid flow that we started with in this video, you might ask a couple questions about it, like it seems to be rotating around some points, in this case counterclockwise, but it's rotating clockwise around others still, does that have any kind of mathematical significance? Does the fact that there seem to be the same number of particles roughly in this area, but they're slowly spilling out there, what does that imply for the function that represents this whole vector field? And you'll see a lot of this later on, especially when I talk about divergence and curl, but here I just wanted to give a little warm up to that, as we're just visualizing multivariable functions."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's work on the curl of f. So the curl of f is going to be equal to, and I just remember it as the determinant. So we have our i, j, k components. It's really, you can imagine, it's the del operator crossed with the actual vector. So the del operator, and I'll write this in a different color just to ease the monotony. So it's the partial with respect to x, partial with respect to y, partial with respect to z, and then our vector field, I copy and pasted it right over here. It is just equal to negative y squared is our i component, x is our j component, and z squared is our k component. And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the del operator, and I'll write this in a different color just to ease the monotony. So it's the partial with respect to x, partial with respect to y, partial with respect to z, and then our vector field, I copy and pasted it right over here. It is just equal to negative y squared is our i component, x is our j component, and z squared is our k component. And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y. Well, the z squared is just a constant with respect to y, so the partial of z squared with respect to y is just going to be zero. So it's going to be zero minus the partial of x with respect to z. Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, this is going to be equal to i, is going to be equal to i times the partial of z squared with respect to y. Well, the z squared is just a constant with respect to y, so the partial of z squared with respect to y is just going to be zero. So it's going to be zero minus the partial of x with respect to z. Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero. So that's nice simplification. And then we're gonna have minus j, we need our little checkerboard pattern, so we put a negative in front of the j, minus j, and so we'll have the partial of z squared with respect to x, well, that's zero again, and then minus the partial of negative y squared with respect to z. Well, that's zero again."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, once again, this is just a constant when you think in terms of z, so that's just going to be zero. So that's nice simplification. And then we're gonna have minus j, we need our little checkerboard pattern, so we put a negative in front of the j, minus j, and so we'll have the partial of z squared with respect to x, well, that's zero again, and then minus the partial of negative y squared with respect to z. Well, that's zero again. And then finally, we have our k component, k, so plus k, and k, we're gonna have the partial of x with respect to x, well, that actually gives us a value, that's just going to be one, minus the partial of negative y squared with respect to y. So the partial of negative y squared with respect to y is negative two y, and we're subtracting that, so it's going to be plus two y. So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's zero again. And then finally, we have our k component, k, so plus k, and k, we're gonna have the partial of x with respect to x, well, that actually gives us a value, that's just going to be one, minus the partial of negative y squared with respect to y. So the partial of negative y squared with respect to y is negative two y, and we're subtracting that, so it's going to be plus two y. So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y. And so if we go back to this right up here, if we go back up to that, we are going to get, we rewrite the integral, so from zero to one, then that's our r, our r parameter's gonna go from zero to one, theta's gonna go from zero to two pi, and now curl of f has simplified to, and I won't skip any steps here, although it's tempting, it's one plus two y, and actually, instead of writing two y, let me write it in terms of the parameters. We saw it up here, y was r sine theta, if I remember correctly, right? Y was r sine theta, so let me write y that way."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So curl of f simplifies to just, oh, this is just zero up here, it's just one plus two y times k, or k times one plus two y. And so if we go back to this right up here, if we go back up to that, we are going to get, we rewrite the integral, so from zero to one, then that's our r, our r parameter's gonna go from zero to one, theta's gonna go from zero to two pi, and now curl of f has simplified to, and I won't skip any steps here, although it's tempting, it's one plus two y, and actually, instead of writing two y, let me write it in terms of the parameters. We saw it up here, y was r sine theta, if I remember correctly, right? Y was r sine theta, so let me write y that way. Two times r sine theta k, and we're gonna dot this, we're gonna take the dot product of that with this right over here, with r times j plus r times k, d theta dr. And so when we take the dot product, this thing only has a k component. The j component is zero, so when you take the dot product with this j component, you're gonna get zero, and neither of them actually even have an i component, and so the inside is just going to simplify to, this piece right over here is going to simplify to, we just have to think about the k components, because everything else is zero, so it's gonna be r times this, and then we're done. So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Y was r sine theta, so let me write y that way. Two times r sine theta k, and we're gonna dot this, we're gonna take the dot product of that with this right over here, with r times j plus r times k, d theta dr. And so when we take the dot product, this thing only has a k component. The j component is zero, so when you take the dot product with this j component, you're gonna get zero, and neither of them actually even have an i component, and so the inside is just going to simplify to, this piece right over here is going to simplify to, we just have to think about the k components, because everything else is zero, so it's gonna be r times this, and then we're done. So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral. We just have to evaluate this thing. And so first we take the antiderivative with respect to theta. So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's gonna be r plus two r squared sine theta, d theta dr. D theta dr. And once again, theta goes from zero to two pi, and r goes from zero to one, and now this is just a straight up double integral. We just have to evaluate this thing. And so first we take the antiderivative with respect to theta. So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first. So the antiderivative of r with respect to theta is just r theta. You can just view r as a constant. And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the antiderivative with respect to theta is going to give us, so this is going to be giving, so we're gonna focus on theta first. So the antiderivative of r with respect to theta is just r theta. You can just view r as a constant. And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta. So this is gonna be negative two r squared cosine of theta, and we're gonna evaluate it from zero to two pi, and then we have the outside integral, which I will recolor in yellow. Recolor in yellow. So we'll still have to integrate with respect to r, and r is gonna go from zero to one."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the antiderivative of this, antiderivative of sine of theta is negative cosine of theta. So this is gonna be negative two r squared cosine of theta, and we're gonna evaluate it from zero to two pi, and then we have the outside integral, which I will recolor in yellow. Recolor in yellow. So we'll still have to integrate with respect to r, and r is gonna go from zero to one. But inside right over here, if we evaluate all of this business right over here at two pi, we get two pi r, two pi r, that's that right over there, minus, cosine of two pi is just one, so it's minus two r squared, and then from that we are going to subtract from that, we're going to subtract, this evaluated at zero. Well r times zero is just zero, and then cosine of zero is one, so it's just minus two r squared, or negative two r squared. Negative two r squared."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we'll still have to integrate with respect to r, and r is gonna go from zero to one. But inside right over here, if we evaluate all of this business right over here at two pi, we get two pi r, two pi r, that's that right over there, minus, cosine of two pi is just one, so it's minus two r squared, and then from that we are going to subtract from that, we're going to subtract, this evaluated at zero. Well r times zero is just zero, and then cosine of zero is one, so it's just minus two r squared, or negative two r squared. Negative two r squared. And this negative and this negative, you get a positive, and then you have a negative two r squared, and then a plus two r squared is just going to cancel out. That and that cancel out, and so this whole thing has simplified quite nicely to a simple definite integral, zero to one of two pi, two pi r dr. And the antiderivative of this is just going to be pi r squared. So we're just going to evaluate pi r squared from zero to one."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Negative two r squared. And this negative and this negative, you get a positive, and then you have a negative two r squared, and then a plus two r squared is just going to cancel out. That and that cancel out, and so this whole thing has simplified quite nicely to a simple definite integral, zero to one of two pi, two pi r dr. And the antiderivative of this is just going to be pi r squared. So we're just going to evaluate pi r squared from zero to one. When you evaluate it at one, you get pi. When you evaluate it at zero, you just get zero. So you get pi minus zero, which is equal to, and now we deserve a drum roll, because we've been doing a lot of work over many videos, this is equal to pi."}, {"video_title": "Stokes example part 4 Curl and final answer   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're just going to evaluate pi r squared from zero to one. When you evaluate it at one, you get pi. When you evaluate it at zero, you just get zero. So you get pi minus zero, which is equal to, and now we deserve a drum roll, because we've been doing a lot of work over many videos, this is equal to pi. So just to remind ourselves what we've done over the last few videos, we had this line integral that we were trying to figure out, and instead of directly evaluating the line integral, which we could do, and I encourage you to do so, and if I have time, I might do it in the next video. Instead of directly evaluating that line integral, we use Stokes' Theorem to say, oh, we could actually instead say that that's the same thing as a surface integral over a piecewise smooth boundary, or over a piecewise smooth surface that this path is the boundary of. And so we evaluated this surface integral, and eventually with a good bit of, little bit of calculation, we got to evaluating it to be equal to pi."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've seen this before. dS is going to be equal to the magnitude of the cross product of the partial of R with respect to U crossed with the partial of R with respect to V, dU, dV. So first, let's take the cross product, and we'll do that with a 3 by 3 matrix. I'll do that right over here. We set up, and I'll just kind of fill in what R sub U and R sub V is in the actual determinant right over here. So first, we have our components, I, J, and K. And now first, let's think about what R sub U is, the partial of R with respect to U. Well, its I component is going to be 1, the partial of U with respect to U is just 1."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do that right over here. We set up, and I'll just kind of fill in what R sub U and R sub V is in the actual determinant right over here. So first, we have our components, I, J, and K. And now first, let's think about what R sub U is, the partial of R with respect to U. Well, its I component is going to be 1, the partial of U with respect to U is just 1. So its I component is 1. Its J component is going to be 0. Partial of V with respect to U is 0."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, its I component is going to be 1, the partial of U with respect to U is just 1. So its I component is 1. Its J component is going to be 0. Partial of V with respect to U is 0. V does not change with respect to U. So this is going to be 0. This should be parentheses around here."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Partial of V with respect to U is 0. V does not change with respect to U. So this is going to be 0. This should be parentheses around here. Partial of this with respect to U is just going to be 1 again. This is the partial of V squared with respect to U is just 0. So this is just 1 again."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This should be parentheses around here. Partial of this with respect to U is just going to be 1 again. This is the partial of V squared with respect to U is just 0. So this is just 1 again. And then r sub v, the partial of r with respect to v, the i component is going to be 0. j component is going to be 1. And the partial of u plus v squared with respect to v is going to be 2v. So it's a pretty straightforward determinant."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is just 1 again. And then r sub v, the partial of r with respect to v, the i component is going to be 0. j component is going to be 1. And the partial of u plus v squared with respect to v is going to be 2v. So it's a pretty straightforward determinant. So let's try to evaluate. So the i component, it's going to be i. So we cross out this column, this row."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's a pretty straightforward determinant. So let's try to evaluate. So the i component, it's going to be i. So we cross out this column, this row. It's going to be 0 times 2v minus 1 times 1. So essentially, it's just going to be negative 1 times i. So we're going to have negative i."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we cross out this column, this row. It's going to be 0 times 2v minus 1 times 1. So essentially, it's just going to be negative 1 times i. So we're going to have negative i. So this is equal to negative i. And then the j component, and we're going to have to put a negative out front, because remember, we do that checkerboard pattern. So cross out that row, that column."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're going to have negative i. So this is equal to negative i. And then the j component, and we're going to have to put a negative out front, because remember, we do that checkerboard pattern. So cross out that row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v minus 0 times 1."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So cross out that row, that column. 1 times 2v is 2v. Let me make sure I got that. 1 times 2v is 2v minus 0 times 1. So this is going to be 2v. But since it was the j component, which is going to be negative, it's going to be negative 2vj. Let me make sure I did that right."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 times 2v is 2v minus 0 times 1. So this is going to be 2v. But since it was the j component, which is going to be negative, it's going to be negative 2vj. Let me make sure I did that right. That column, that row, 1 times 2v is 2v minus 0 is 2v. The checkerboard pattern, you'd have a negative j. So you have negative 2vj."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me make sure I did that right. That column, that row, 1 times 2v is 2v minus 0 is 2v. The checkerboard pattern, you'd have a negative j. So you have negative 2vj. And then we have, find the k component. Cross out that row, that column, 1 times 1 minus 0. So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have negative 2vj. And then we have, find the k component. Cross out that row, that column, 1 times 1 minus 0. So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root. I'm just taking the magnitude of this part right over here, the actual cross product. It's going to be of negative 1 squared, which is just 1, plus negative 2v squared, which is 4v squared, plus 1 squared, which is just 1. So this whole thing is going to evaluate 2."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be plus k. So if we want the magnitude of this, this whole thing right over here is just going to be the square root. I'm just taking the magnitude of this part right over here, the actual cross product. It's going to be of negative 1 squared, which is just 1, plus negative 2v squared, which is 4v squared, plus 1 squared, which is just 1. So this whole thing is going to evaluate 2. 2 is going to evaluate 2. We have 2 plus 2v squared du dv. And if we want to, actually, I almost made a mistake."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing is going to evaluate 2. 2 is going to evaluate 2. We have 2 plus 2v squared du dv. And if we want to, actually, I almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want, maybe it'll help us a little bit if we factor out a 2 right over here."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we want to, actually, I almost made a mistake. That would have been a disaster. 2 plus 4v squared du dv. And if we want, maybe it'll help us a little bit if we factor out a 2 right over here. This is the same thing as 2 times 1 plus 2v squared du dv. If you factor out the 2, you get this is equal to the square root of 2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to evaluate the surface integral."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we want, maybe it'll help us a little bit if we factor out a 2 right over here. This is the same thing as 2 times 1 plus 2v squared du dv. If you factor out the 2, you get this is equal to the square root of 2 times the square root of 1 plus 2v squared du dv. And I now think we are ready to evaluate the surface integral. So let's do it. All right. So let me just write this thing down here."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I now think we are ready to evaluate the surface integral. So let's do it. All right. So let me just write this thing down here. So I'm going to write everything that has to do with v, I'm going to write in purple. So I'm just going to write the ds part right over here. It is the square root of 2 times the square root of 1 plus 2v squared."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me just write this thing down here. So I'm going to write everything that has to do with v, I'm going to write in purple. So I'm just going to write the ds part right over here. It is the square root of 2 times the square root of 1 plus 2v squared. And then we're going to have du, which I'll write in green, and then dv. So this is just the ds part. Now we have the y right over here."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It is the square root of 2 times the square root of 1 plus 2v squared. And then we're going to have du, which I'll write in green, and then dv. So this is just the ds part. Now we have the y right over here. y is just equal to v. So I'll write that in purple. So y is just equal to v. Let me make it very clear. And all of this is ds."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we have the y right over here. y is just equal to v. So I'll write that in purple. So y is just equal to v. Let me make it very clear. And all of this is ds. ds. All of this is ds. And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And all of this is ds. ds. All of this is ds. And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x. It goes between 0 and 1. And then v is the same thing as y. And y goes between, or v goes between 0 and 2."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now I can write the bounds in terms of u and v. And so the u part, u is the same thing as x. It goes between 0 and 1. And then v is the same thing as y. And y goes between, or v goes between 0 and 2. And I now think we're ready to evaluate. And so the u and v variables are not so mixed up. So we can actually separate out these two integrals, make this a product of two single integrals."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And y goes between, or v goes between 0 and 2. And I now think we're ready to evaluate. And so the u and v variables are not so mixed up. So we can actually separate out these two integrals, make this a product of two single integrals. The first thing, if you look at it with respect to du, all this stuff in purple is just a constant with respect to u. So we can take it out of the du integral. We can take all of this purple stuff out of this du integral."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we can actually separate out these two integrals, make this a product of two single integrals. The first thing, if you look at it with respect to du, all this stuff in purple is just a constant with respect to u. So we can take it out of the du integral. We can take all of this purple stuff out of this du integral. And so this double integral simplifies to the integral from 0 to 2 of, I'll write it as the square root of 2 v times the square root of 1 plus 2 v squared. And then so I factor it out all of this stuff. And then you have times the integral from 0 to 1 du."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can take all of this purple stuff out of this du integral. And so this double integral simplifies to the integral from 0 to 2 of, I'll write it as the square root of 2 v times the square root of 1 plus 2 v squared. And then so I factor it out all of this stuff. And then you have times the integral from 0 to 1 du. And then times d. And then you have dv. Now, if this was really complicated, I could say, OK, this is just going to be a function of u. It's a constant with respect to v. And you could factor this whole thing out and separate the integrals."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you have times the integral from 0 to 1 du. And then times d. And then you have dv. Now, if this was really complicated, I could say, OK, this is just going to be a function of u. It's a constant with respect to v. And you could factor this whole thing out and separate the integrals. But this is even easier. This integral is just going to evaluate to 1. So this whole thing just evaluates to 1."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's a constant with respect to v. And you could factor this whole thing out and separate the integrals. But this is even easier. This integral is just going to evaluate to 1. So this whole thing just evaluates to 1. And so we've simplified this into one single integral. So this simplifies to, and I could even take the square root of 2 out front, the square root of 2 times the integral from v is equal to 0 to v is equal to 2 of v times the square root of 1 plus 2 v squared dv. And so we really are in the home stretch of evaluating the surface integral."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing just evaluates to 1. And so we've simplified this into one single integral. So this simplifies to, and I could even take the square root of 2 out front, the square root of 2 times the integral from v is equal to 0 to v is equal to 2 of v times the square root of 1 plus 2 v squared dv. And so we really are in the home stretch of evaluating the surface integral. And here, this is basic. This is actually a little bit of u substitution that we can do in our head. You have a function, or this kind of this embedded function right here, 1 plus 2v squared."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we really are in the home stretch of evaluating the surface integral. And here, this is basic. This is actually a little bit of u substitution that we can do in our head. You have a function, or this kind of this embedded function right here, 1 plus 2v squared. What's the derivative of 1 plus 2v squared? Well, it would be 4v. We have something almost 4v here."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You have a function, or this kind of this embedded function right here, 1 plus 2v squared. What's the derivative of 1 plus 2v squared? Well, it would be 4v. We have something almost 4v here. We can make this 4v by multiplying it by 4 here and then dividing it by 4 out here. This doesn't change the value of the integral. And so now, this part right over here, it's pretty straightforward to take the antiderivative."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have something almost 4v here. We can make this 4v by multiplying it by 4 here and then dividing it by 4 out here. This doesn't change the value of the integral. And so now, this part right over here, it's pretty straightforward to take the antiderivative. The antiderivative of this is going to be, we have this embedded function's derivative right over here. So we can kind of just treat it like an x or a v. Take the antiderivative with respect to this thing right over here. And we get, this is essentially 1 plus 2v squared to the 1 half power."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so now, this part right over here, it's pretty straightforward to take the antiderivative. The antiderivative of this is going to be, we have this embedded function's derivative right over here. So we can kind of just treat it like an x or a v. Take the antiderivative with respect to this thing right over here. And we get, this is essentially 1 plus 2v squared to the 1 half power. We increment it by 1, so it's 1 plus 2v squared to the 3 halves power. And then you divide by 3 halves, or you multiply by 2 thirds. So times 2 thirds."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we get, this is essentially 1 plus 2v squared to the 1 half power. We increment it by 1, so it's 1 plus 2v squared to the 3 halves power. And then you divide by 3 halves, or you multiply by 2 thirds. So times 2 thirds. So this is the antiderivative of that. And then, of course, you still have all of this stuff out front, square root of 2 over 4. Square root of 2 over 4."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So times 2 thirds. So this is the antiderivative of that. And then, of course, you still have all of this stuff out front, square root of 2 over 4. Square root of 2 over 4. And we are going to evaluate this from 0 to 2. And actually, just to simplify it, let me factor out the 2 thirds so we don't have to worry about that at 2 and 0. So I'm going to factor out the 2 thirds."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Square root of 2 over 4. And we are going to evaluate this from 0 to 2. And actually, just to simplify it, let me factor out the 2 thirds so we don't have to worry about that at 2 and 0. So I'm going to factor out the 2 thirds. So times 2 over 3. And actually, this will cancel out. That becomes a 1."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'm going to factor out the 2 thirds. So times 2 over 3. And actually, this will cancel out. That becomes a 1. That becomes a 2. And so we are left, this stuff over here is 1 sixth times, and now if you evaluate this at 2, you have 2v squared. That's going to be 2 times 4 is 8, plus 1 is 9."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That becomes a 1. That becomes a 2. And so we are left, this stuff over here is 1 sixth times, and now if you evaluate this at 2, you have 2v squared. That's going to be 2 times 4 is 8, plus 1 is 9. 9 to the 3 halves power. So 9 to the 1 half is 3. 3 to the 3rd is 27."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's going to be 2 times 4 is 8, plus 1 is 9. 9 to the 3 halves power. So 9 to the 1 half is 3. 3 to the 3rd is 27. So it's going to be equal to 27. And then minus this thing evaluated at 0. Well, this is evaluated at 0, so it's going to be 1."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "3 to the 3rd is 27. So it's going to be equal to 27. And then minus this thing evaluated at 0. Well, this is evaluated at 0, so it's going to be 1. 1 to the 3 halves is just 1. So minus 1. And so this gives us, oh, I almost made a mistake."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, this is evaluated at 0, so it's going to be 1. 1 to the 3 halves is just 1. So minus 1. And so this gives us, oh, I almost made a mistake. There should be a square root of 2 up here. Square root of 2 over 6 times 27 minus 1. So drum roll, this gives us the value of our surface integral."}, {"video_title": "Surface integral ex2 part 2 Evaluating integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this gives us, oh, I almost made a mistake. There should be a square root of 2 up here. Square root of 2 over 6 times 27 minus 1. So drum roll, this gives us the value of our surface integral. Let's see, 27 minus 1 is 26. So we get 26 times the square root of 2 over 6. And we can simplify a little bit more."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "Hello everyone. So now that we have an intuition for what divergence is trying to represent, let's start actually drilling in on a formula. And the first thing I want to do is just limit our perspective to functions that only have an x component, or rather where the y component of the output is just zero. So this is some kind of vector field, and if there's only an x component, what this means it's gonna look like is all of the vectors only go left or right, and there's kind of no up or down involved in any of them. So in this case, let's start thinking about what positive divergence of your vector fields might look like near some point x, y. So if you have your point, you know, this is the point x, y, somewhere sitting off in space. Two cases where the divergence of this might look positive are one, where nothing happens at the point, right?"}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "So this is some kind of vector field, and if there's only an x component, what this means it's gonna look like is all of the vectors only go left or right, and there's kind of no up or down involved in any of them. So in this case, let's start thinking about what positive divergence of your vector fields might look like near some point x, y. So if you have your point, you know, this is the point x, y, somewhere sitting off in space. Two cases where the divergence of this might look positive are one, where nothing happens at the point, right? So in this case, p would be equal to zero at our point, but then to the left of it, things are moving to the left, meaning p, the x component of our vector-valued function, is negative, right? That's why the x component of this vector is negative, but then to the right, vectors would be moving off to the right, so over here, p would be positive. So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "Two cases where the divergence of this might look positive are one, where nothing happens at the point, right? So in this case, p would be equal to zero at our point, but then to the left of it, things are moving to the left, meaning p, the x component of our vector-valued function, is negative, right? That's why the x component of this vector is negative, but then to the right, vectors would be moving off to the right, so over here, p would be positive. So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive. So as you're changing in the x direction, p should be increasing. So a positive divergence here seems to correspond to a positive partial derivative of p with respect to x. And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "So this would be an example of kind of a positive divergence circumstance where only the x component is responsible, and what you'll notice here, this would be p starts negative, goes zero, then becomes positive. So as you're changing in the x direction, p should be increasing. So a positive divergence here seems to correspond to a positive partial derivative of p with respect to x. And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x. And once you do, hopefully it makes sense why this specific positive divergence example corresponds with a positive partial derivative of p. But remember, this isn't the only way that a positive divergence might look. You could have another circumstance where, let's say, your point, x, y, actually has a vector attached to it. So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "And if that seems a little unfamiliar, if you're not sure how to think about, you know, partial derivatives of a component of a vector field, I have a video on that, and you can kind of take a look and refresh yourself how you might think about this partial derivative of p with respect to x. And once you do, hopefully it makes sense why this specific positive divergence example corresponds with a positive partial derivative of p. But remember, this isn't the only way that a positive divergence might look. You could have another circumstance where, let's say, your point, x, y, actually has a vector attached to it. So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there. But another way that positive divergence might look is that you have things coming in towards that point and things going away, but the things going away are bigger than the ones coming in. But again, this kind of exhibits the idea of p increasing in value. You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "So this here, again, represents our point, x, y, and in this specific example, this would be kind of p is positive, p of x, y is positive at your point there. But another way that positive divergence might look is that you have things coming in towards that point and things going away, but the things going away are bigger than the ones coming in. But again, this kind of exhibits the idea of p increasing in value. You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger. So once again, we have this idea of positive partial derivative of p with respect to x, because changes in x, as you increase x, it causes an increase in p, seems to correspond to positive divergence. And you can even look at it if you go the other way, where you have a little bit of negative component to p here, so p is a little bit negative, but to the left of your point, it's really negative, and then to the right, it's not nearly as negative. And in this case, it's kind of like, as you're moving to the right, as x is increasing, you start off very negative, and then only kind of negative, and then barely negative."}, {"video_title": "Divergence formula, part 1.mp3", "Sentence": "You know, p starts off small, it's a positive but small component, and then it gets bigger, and then it gets even bigger. So once again, we have this idea of positive partial derivative of p with respect to x, because changes in x, as you increase x, it causes an increase in p, seems to correspond to positive divergence. And you can even look at it if you go the other way, where you have a little bit of negative component to p here, so p is a little bit negative, but to the left of your point, it's really negative, and then to the right, it's not nearly as negative. And in this case, it's kind of like, as you're moving to the right, as x is increasing, you start off very negative, and then only kind of negative, and then barely negative. And once again, that corresponds to an increase in the value of p as x increases. So what you'd expect is that a partial derivative of p, that x component of the output with respect to x, is gonna be somewhere involved in the formula for the divergence of our vector field at a point x, y. And in the next video, I'm gonna go a similar line of reasoning to see what should go on with that y component."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to do that, let's imagine. So let me draw my axes. So that's my z-axis. That is my x-axis. And then that is my y-axis. And now let's imagine a region in the xy-plane. So let me draw it like this."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is my x-axis. And then that is my y-axis. And now let's imagine a region in the xy-plane. So let me draw it like this. So let's say this is my region in the xy-plane. I will call that region R. And I also have a boundary of that region. And let's say we care about the direction that we traverse the boundary."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw it like this. So let's say this is my region in the xy-plane. I will call that region R. And I also have a boundary of that region. And let's say we care about the direction that we traverse the boundary. And let's say we're going to traverse it in a counterclockwise direction. So we have this path that goes around this region. We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say we care about the direction that we traverse the boundary. And let's say we're going to traverse it in a counterclockwise direction. So we have this path that goes around this region. We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction. And let's say that we also have a vector field F that essentially its i component is just going to be a function of x and y. And its j component is only going to be a function of x and y. And let's say it has no k components."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can call that C. So we'll call that C. And we're going to traverse it in the counterclockwise direction. And let's say that we also have a vector field F that essentially its i component is just going to be a function of x and y. And its j component is only going to be a function of x and y. And let's say it has no k components. So the vector field on this region, it might look something like this. I'm just drawing random things. And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say it has no k components. So the vector field on this region, it might look something like this. I'm just drawing random things. And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher. So that vector, it wouldn't change as you change your z component. And all of the vectors would essentially be parallel to, or if z is 0, actually sitting on the xy-plane. Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if you go off that region, if you go in the z direction, it's just going to look the same as you go higher and higher. So that vector, it wouldn't change as you change your z component. And all of the vectors would essentially be parallel to, or if z is 0, actually sitting on the xy-plane. Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour. Let me draw that a little bit neater. The line integral over the contour C of F dot dr, F dot lowercase dr, where dr is obviously going along the contour. So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now given this, let's think about what Stokes' theorem would tell us about the value of the line integral over the contour. Let me draw that a little bit neater. The line integral over the contour C of F dot dr, F dot lowercase dr, where dr is obviously going along the contour. So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here. It should be equal to the double integral over the surface. Well, this region is really just a surface that's sitting in the xy-plane. So it should really just be the double integral."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we take Stokes' theorem, then this quantity right over here should be equal to this quantity right over here. It should be equal to the double integral over the surface. Well, this region is really just a surface that's sitting in the xy-plane. So it should really just be the double integral. Let me write that in that same. It'll be the double integral over our region, which is really just the same thing as our surface of the curl of F dot n. So let's just think about what the curl of F dot n is. And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it should really just be the double integral. Let me write that in that same. It'll be the double integral over our region, which is really just the same thing as our surface of the curl of F dot n. So let's just think about what the curl of F dot n is. And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there. So instead of ds, I'll just write da. But let's think about what curl of F dot n would actually be. So let's work on curl of F first."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then d of s would just be a little chunk of our region, a little chunk of our flattened surface right over there. So instead of ds, I'll just write da. But let's think about what curl of F dot n would actually be. So let's work on curl of F first. So the curl of F, and the way I always remember it is we're going to take the determinant of this ijk, partial with respect to x, partial with respect to y, partial with respect to z. This is just the definition of taking the curl. We're figuring out how much this vector field would cause something to spin."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's work on curl of F first. So the curl of F, and the way I always remember it is we're going to take the determinant of this ijk, partial with respect to x, partial with respect to y, partial with respect to z. This is just the definition of taking the curl. We're figuring out how much this vector field would cause something to spin. And then we want the i component, which is our function p, which is just a function of x and y, j component, which is just the function q. And there was no z component over here, so 0. And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're figuring out how much this vector field would cause something to spin. And then we want the i component, which is our function p, which is just a function of x and y, j component, which is just the function q. And there was no z component over here, so 0. And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0. That's just going to be 0 minus the partial of q with respect to z. Well, what's the partial of q with respect to z? Well, q isn't a function of z at all, so that's also going to be 0."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this is going to be equal to, well, if we look at the i component, it's going to be the partial of y of 0. That's just going to be 0 minus the partial of q with respect to z. Well, what's the partial of q with respect to z? Well, q isn't a function of z at all, so that's also going to be 0. Let me write this out, just so it's not too confusing. So our i component is going to be partial of 0 with respect to y. Well, that's just going to be 0 minus the partial of q with respect to z."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, q isn't a function of z at all, so that's also going to be 0. Let me write this out, just so it's not too confusing. So our i component is going to be partial of 0 with respect to y. Well, that's just going to be 0 minus the partial of q with respect to z. Well, the partial of q with respect to z is just going to be 0. So we have a 0i component, and then we want to subtract the j component. And then the j component, partial of 0 with respect to x is 0."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's just going to be 0 minus the partial of q with respect to z. Well, the partial of q with respect to z is just going to be 0. So we have a 0i component, and then we want to subtract the j component. And then the j component, partial of 0 with respect to x is 0. And then from that, you're going to subtract the partial of p with respect to z. Well, once again, p is not a function of z at all, so that's going to be 0 again. And then you have plus k times the partial of q with respect to x."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the j component, partial of 0 with respect to x is 0. And then from that, you're going to subtract the partial of p with respect to z. Well, once again, p is not a function of z at all, so that's going to be 0 again. And then you have plus k times the partial of q with respect to x. Remember, this is just the partial derivative operator. So partial of q with respect to x. And from that, we're going to subtract the partial of p with respect to y."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you have plus k times the partial of q with respect to x. Remember, this is just the partial derivative operator. So partial of q with respect to x. And from that, we're going to subtract the partial of p with respect to y. So the curl of f just simplifies to this right over here. Now, what is n? What is the unit normal vector?"}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And from that, we're going to subtract the partial of p with respect to y. So the curl of f just simplifies to this right over here. Now, what is n? What is the unit normal vector? Well, we're in the xy plane, so the unit normal vector is just going to be straight up in the z direction. It's going to have a magnitude of 1. So in this case, our unit normal vector is just going to be the k vector."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What is the unit normal vector? Well, we're in the xy plane, so the unit normal vector is just going to be straight up in the z direction. It's going to have a magnitude of 1. So in this case, our unit normal vector is just going to be the k vector. So we're essentially just going to take curl of f is this, and our unit normal vector is just going to be equal to the k unit vector. It's going to go straight up. So what happens if we take the curl of f dot k?"}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in this case, our unit normal vector is just going to be the k vector. So we're essentially just going to take curl of f is this, and our unit normal vector is just going to be equal to the k unit vector. It's going to go straight up. So what happens if we take the curl of f dot k? If we just dot this with k? We're just dotting this with this. Well, we're just going to end up with this part right over here."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what happens if we take the curl of f dot k? If we just dot this with k? We're just dotting this with this. Well, we're just going to end up with this part right over here. So curl of f dot the unit normal vector is just going to be equal to this business. It's just going to be equal to the partial of q with respect to x minus the partial of p with respect to y. And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, we're just going to end up with this part right over here. So curl of f dot the unit normal vector is just going to be equal to this business. It's just going to be equal to the partial of q with respect to x minus the partial of p with respect to y. And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem. This thing right over here just boiled down to Green's theorem. So we see that Green's theorem is really just a special case. Let me write theorem a little bit neater."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is neat, because using Stokes' theorem in this special case where we're dealing with a flattened out surface in the xy plane, in this situation, this just boiled down to Green's theorem. This thing right over here just boiled down to Green's theorem. So we see that Green's theorem is really just a special case. Let me write theorem a little bit neater. We see that Green's theorem is really just a special case of Stokes' theorem where our surface is flattened out and it's in the xy plane. So that should make us feel pretty good. Although we still have not proven Stokes' theorem."}, {"video_title": "Green's and Stokes' theorem relationship   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write theorem a little bit neater. We see that Green's theorem is really just a special case of Stokes' theorem where our surface is flattened out and it's in the xy plane. So that should make us feel pretty good. Although we still have not proven Stokes' theorem. But the one thing that I do like about this is seeing that Green's theorem and Stokes' theorem is consistent is now it starts to make sense of this right over here. When we first learned Green's theorem, we're like, what's going on over here? But now this is telling us this is just taking the curl in this region along this surface."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 x, y, and we're always multiplying f of x, y times 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 multiple problems. So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 x, y, and we're always multiplying f of x, y times 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 multiple problems. So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1. This part we can call, let me make a pointer, c2. And then this point right here, c3. So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can imagine this whole contour, this whole path we call c, but we could call this part that we figured out in the last video c1. This part we can call, let me make a pointer, c2. And then this point right here, c3. So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals. This will be equal to the line integral along the path c1 of f of x, y, ds, plus the line integral along c2 of f of x, y, ds, plus the line integral, and you might have guessed it, along c3 of f of x, y, ds. And in the last video, we got as far as figuring out this first part, this first curvy wall right here. Its surface area, we figured out, was 4 plus 2 pi."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we could redefine, or we can break up this line integral, this closed line integral, into three non-closed line integrals. This will be equal to the line integral along the path c1 of f of x, y, ds, plus the line integral along c2 of f of x, y, ds, plus the line integral, and you might have guessed it, along c3 of f of x, y, ds. And in the last video, we got as far as figuring out this first part, this first curvy wall right here. Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other two parts. So let's do c2, let's do this line integral next. And in order to do it, we need to do another parametrization of x and y."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Its surface area, we figured out, was 4 plus 2 pi. Now we've got to figure out the other two parts. So let's do c2, let's do this line integral next. And in order to do it, we need to do another parametrization 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."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in order to do it, we need to do another parametrization 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 parametrization. x is equal to 0. So if we're along the y-axis, x is definitely equal to 0."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So as long as we're there, x is definitely going to be equal to 0. So that's my parametrization. x is equal to 0. So 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."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So 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. And eventually, when t is equal to 2, we're at that point right there."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And eventually, when t is equal to 2, we're at that point right there. So that's our parametrization. And so let's evaluate this line. And actually, we could do our derivatives too, if we like."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And eventually, when t is equal to 2, we're at that point right there. So that's our parametrization. And so let's evaluate this line. And actually, we could do our derivatives too, if we like. What's the derivative? Let me write it right here. What's dx dt?"}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually, we could do our derivatives too, if we like. What's the derivative? Let me write it right here. What's dx dt? Pretty straightforward. Derivative of 0 is 0. And dy dt is equal to the derivative of this."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What's dx dt? Pretty straightforward. Derivative of 0 is 0. And dy dt is equal to the derivative of this. It's just minus 1. 2 minus t, derivative of minus t is just minus 1. And so let's just break it up."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And dy dt is equal to the derivative of this. It's just minus 1. 2 minus t, derivative of minus t is just minus 1. And so let's just break it up. So we have this thing right here. So we have the integral along C2. But instead of writing C2, I'll leave C2 there."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's just break it up. So we have this thing right here. So we have the integral along C2. But instead of writing C2, I'll leave C2 there. But we'll say we're going from t is equal to 0 to 2 of f of xy. So f of xy is this thing right here, is x plus y squared. And then times ds."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But instead of writing C2, I'll leave C2 there. But we'll say we're going from t is equal to 0 to 2 of f of xy. So f of xy is this thing right here, is x plus y squared. And then times ds. ds times ds. And we know from the last several videos, ds can be rewritten as the square root of dx dt squared. So 0 squared plus dy dt squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then times ds. ds times ds. And we know from the last several videos, ds can be rewritten as the square root of dx dt squared. So 0 squared plus dy dt squared. So minus 1 squared is 1. All of that times dt. And obviously, this is pretty nice."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So 0 squared plus dy dt squared. So minus 1 squared is 1. All of that times dt. And obviously, this is pretty nice. It gets very nice and clean. This is 0 plus 1 square root. This just becomes 1."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And obviously, this is pretty nice. It gets very nice and clean. This is 0 plus 1 square root. This just becomes 1. And then what is x? x, if we write it in terms of our parametrization, is always going to be equal to 0. And then y squared is going to be 2 minus t squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This just becomes 1. And then what is x? x, if we write it in terms of our parametrization, is always going to be equal to 0. And then y squared is going to be 2 minus t squared. So this is going to be 2 minus t squared. So this whole crazy thing simplified to, we're going to go from t is equal to 0 to t is equal to 2. The x disappears in our parametrization."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then y squared is going to be 2 minus t squared. So this is going to be 2 minus t squared. So this whole crazy thing simplified to, we're going to go from t is equal to 0 to t is equal to 2. The x disappears in our parametrization. x just stays 0 regardless of what t is. And then you have y squared. But y is the same thing as 2 minus t. So 2 minus t squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The x disappears in our parametrization. x just stays 0 regardless of what t is. And then you have y squared. But y is the same thing as 2 minus t. So 2 minus t squared. And then you have your dt sitting out there. This is pretty straightforward. I always find it easier when you're finding an antiderivative of this."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But y is the same thing as 2 minus t. So 2 minus t squared. And then you have your dt sitting out there. This is pretty straightforward. I always find it easier when you're finding an antiderivative of this. Although you can do this in your head. I like to just actually multiply out this binomial. So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I always find it easier when you're finding an antiderivative of this. Although you can do this in your head. I like to just actually multiply out this binomial. So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared. Just like that. dt. And this is pretty straightforward."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to the antiderivative from t is equal to 0 to t is equal to 2 of 4 minus 4t plus t squared. Just like that. dt. And this is pretty straightforward. This is going to be the antiderivative of this is 4t minus 2t squared. When you take the derivative, 2 times minus 2 is minus 4t. And then you have plus 1 third t to the third."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is pretty straightforward. This is going to be the antiderivative of this is 4t minus 2t squared. When you take the derivative, 2 times minus 2 is minus 4t. And then you have plus 1 third t to the third. These are just simple antiderivatives. And we need to evaluate it from 0 to 2. And so let's evaluate it at 2."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you have plus 1 third t to the third. These are just simple antiderivatives. And we need to evaluate it from 0 to 2. And so let's evaluate it at 2. 4 times 2 is 8. Let me pick a new color. 4 times 2 is 8 minus 2 times 2 squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's evaluate it at 2. 4 times 2 is 8. Let me pick a new color. 4 times 2 is 8 minus 2 times 2 squared. So 2 times 4. So minus 8 plus 1 third times 2 to the third power. So 1 third times 8."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "4 times 2 is 8 minus 2 times 2 squared. So 2 times 4. So minus 8 plus 1 third times 2 to the third power. So 1 third times 8. So these cancel out. We have 8 minus 8. And we just have 8 thirds."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So 1 third times 8. So these cancel out. We have 8 minus 8. And we just have 8 thirds. So this just becomes 8 thirds. And then we have to put a 0 in. Minus 0 evaluated here."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we just have 8 thirds. So this just becomes 8 thirds. And then we have to put a 0 in. Minus 0 evaluated here. But this is just going to be 0. We have 4 times 0, 2 times 0. All of these are going to be 0."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Minus 0 evaluated here. But this is just going to be 0. We have 4 times 0, 2 times 0. All of these are going to be 0. So minus 0. So just like that, we found our surface area of our second wall. So this turned out being this right here is 8 thirds."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All of these are going to be 0. So minus 0. So just like that, we found our surface area of our second wall. So this turned out being this right here is 8 thirds. And now we have our last wall. And then we could just add them up. So we have our last wall."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this turned out being this right here is 8 thirds. And now we have our last wall. And then we could just add them up. So we have our last wall. We're looking. So let's just do another parametrization. I want to have the graph there."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have our last wall. We're looking. So let's just do another parametrization. I want to have the graph there. Well, maybe I can paste it again. Edit. So there's the graph again."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to have the graph there. Well, maybe I can paste it again. Edit. So there's the graph again. And now we're going to do our last wall. So our last wall is this one right here, which is we could write it. This was C3."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So there's the graph again. And now we're going to do our last wall. So our last wall is this one right here, which is we could write it. This was C3. Let me switch colors here. So this is C. We're going to go along contour C3 of f of x, y, ds, which is the same thing as let's do a parametrization. Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This was C3. Let me switch colors here. So this is C. We're going to go along contour C3 of f of x, y, ds, which is the same thing as let's do a parametrization. Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2. And then this whole time that we're along the x-axis, y is going to be equal to 0. That's a pretty straightforward parametrization. So this is going to be equal to."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Along this curve, if we just say x is equal to t, very straightforward, for t is greater than or equal to 0, less than or equal to 2. And then this whole time that we're along the x-axis, y is going to be equal to 0. That's a pretty straightforward parametrization. So this is going to be equal to. We're going to go from t is equal to 0 to t is equal to 2 of f of x, y, which is I'll write in terms of x right now, x and y, x plus y squared, times ds. Now, what is dx? Well, let me write ds right here, times ds."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to. We're going to go from t is equal to 0 to t is equal to 2 of f of x, y, which is I'll write in terms of x right now, x and y, x plus y squared, times ds. Now, what is dx? Well, let me write ds right here, times ds. That's what we're dealing with. Now, we know what ds is. ds is equal to the square root of dx dt squared plus dy dt squared times dt."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, let me write ds right here, times ds. That's what we're dealing with. Now, we know what ds is. ds is equal to the square root of dx dt squared plus dy dt squared times dt. We proved that in the first video. Or we didn't rigorously prove it, but we got the sense of why this is true. And what's the derivative of x with respect to t?"}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds is equal to the square root of dx dt squared plus dy dt squared times dt. We proved that in the first video. Or we didn't rigorously prove it, but we got the sense of why this is true. And what's the derivative of x with respect to t? Well, it's just 1. So this is just going to be a 1. 1 squared, same thing."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what's the derivative of x with respect to t? Well, it's just 1. So this is just going to be a 1. 1 squared, same thing. And derivative of y with respect to t is 0. So this is a 0. 1 plus 0 is 1."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 squared, same thing. And derivative of y with respect to t is 0. So this is a 0. 1 plus 0 is 1. Square root of 1 is 1. So this thing just becomes dt. ds is going to be equal to dt in this case."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 plus 0 is 1. Square root of 1 is 1. So this thing just becomes dt. ds is going to be equal to dt in this case. So this just becomes a dt. And then our x is going to be equal to a t. That's our definition of our parametrization. And y is 0, so we can ignore it."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds is going to be equal to dt in this case. So this just becomes a dt. And then our x is going to be equal to a t. That's our definition of our parametrization. And y is 0, so we can ignore it. So this was a super simple integral. So this simplified down to, we're going to go from 0 to 2 of t dt, which is equal to the anti derivative of t is just 1 half t squared. And we're going to go 0 to 2, which is equal to 1 half times 2 squared."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And y is 0, so we can ignore it. So this was a super simple integral. So this simplified down to, we're going to go from 0 to 2 of t dt, which is equal to the anti derivative of t is just 1 half t squared. And we're going to go 0 to 2, which is equal to 1 half times 2 squared. 2 squared is 4 times 1 half is 2. And then minus 1 half times 0 squared. So this third wall's area right there is just 2, pretty straightforward."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to go 0 to 2, which is equal to 1 half times 2 squared. 2 squared is 4 times 1 half is 2. And then minus 1 half times 0 squared. So this third wall's area right there is just 2, pretty straightforward. So that right there, the area there is just 2. And so to answer our question, what was this line integral evaluated over this closed path of f of x, y? Well, we just add up these numbers."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this third wall's area right there is just 2, pretty straightforward. So that right there, the area there is just 2. And so to answer our question, what was this line integral evaluated over this closed path of f of x, y? Well, we just add up these numbers. We have 4 plus 2 pi plus 8 thirds plus 2. Well, what is this? 8 thirds is the same thing as 2 and 2 thirds."}, {"video_title": "Line integral example 2 (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, we just add up these numbers. We have 4 plus 2 pi plus 8 thirds plus 2. Well, what is this? 8 thirds is the same thing as 2 and 2 thirds. So we have 4 plus 2 and 2 thirds is 6 and 2 thirds plus another 2 is 8 and 2 thirds. So this whole thing becomes 8 and 2 thirds, if you write it as a mixed number, plus 2 pi. And we're done."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Draw my y-axis, that is my x-axis. Let's say the path looks like this. It looks something like this. It's the same one we had in the last video. It might not look exactly like it. Let me see what I did in the last video. It looked like that in the last video, but close enough."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's the same one we had in the last video. It might not look exactly like it. Let me see what I did in the last video. It looked like that in the last video, but close enough. Let's say we're dealing with the exact same curve as the last video. We could call that curve C. In the last video, we dealt with a vector field that only had vectors in the i direction. Let's deal with another vector field that only has vectors in the j direction, or the vertical direction."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It looked like that in the last video, but close enough. Let's say we're dealing with the exact same curve as the last video. We could call that curve C. In the last video, we dealt with a vector field that only had vectors in the i direction. Let's deal with another vector field that only has vectors in the j direction, or the vertical direction. Let's say that Q, the vector field Q of x, y, is equal to capital Q of x, y times j. We are going to concern ourselves with the closed line integral around the path C of Q dot dr. We've seen it already. dr can be rewritten as dx times i plus dy times j."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's deal with another vector field that only has vectors in the j direction, or the vertical direction. Let's say that Q, the vector field Q of x, y, is equal to capital Q of x, y times j. We are going to concern ourselves with the closed line integral around the path C of Q dot dr. We've seen it already. dr can be rewritten as dx times i plus dy times j. If we were to take the dot product of these two, this line integral is going to be the exact same thing. This is going to be the same thing as the closed line integral over C of Q dot dr. Q only has a j component. If you take its zero i, so zero times dx, that's just zero."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dr can be rewritten as dx times i plus dy times j. If we were to take the dot product of these two, this line integral is going to be the exact same thing. This is going to be the same thing as the closed line integral over C of Q dot dr. Q only has a j component. If you take its zero i, so zero times dx, that's just zero. Then you're going to have Q x, y times dy. It had no i component. This is just going to be Q of x and y. Q of x and y times dy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you take its zero i, so zero times dx, that's just zero. Then you're going to have Q x, y times dy. It had no i component. This is just going to be Q of x and y. Q of x and y times dy. That's the dot product. There was no i component. That's why we lose the dx."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just going to be Q of x and y. Q of x and y times dy. That's the dot product. There was no i component. That's why we lose the dx. Let's see if there's any way that we can solve this line integral without having to resort to a third parameter, t. Just like we did in the last video. It will be almost identical. We're just dealing with y's now instead of x's."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's why we lose the dx. Let's see if there's any way that we can solve this line integral without having to resort to a third parameter, t. Just like we did in the last video. It will be almost identical. We're just dealing with y's now instead of x's. What we can do is say, what's our minimum y and our maximum y? Our minimum y, let's say it's right here. Let's call that a."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're just dealing with y's now instead of x's. What we can do is say, what's our minimum y and our maximum y? Our minimum y, let's say it's right here. Let's call that a. Let's say our maximum y that we attain is right over there. Let's call that b. Just like in the last, I forgot to tell you the direction of the curve."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's call that a. Let's say our maximum y that we attain is right over there. Let's call that b. Just like in the last, I forgot to tell you the direction of the curve. This is the same path as last time. We're going in a counterclockwise direction. It's the exact same curve, exact same path."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Just like in the last, I forgot to tell you the direction of the curve. This is the same path as last time. We're going in a counterclockwise direction. It's the exact same curve, exact same path. We're going in that direction. In the last video, we broke it up into two functions of x's. Two y's, a function of x."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's the exact same curve, exact same path. We're going in that direction. In the last video, we broke it up into two functions of x's. Two y's, a function of x. Now we want to deal with y's. Let's break it up into two functions of y. If we break this path into two paths, those are kind of our extreme points."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Two y's, a function of x. Now we want to deal with y's. Let's break it up into two functions of y. If we break this path into two paths, those are kind of our extreme points. Let's call this path right here. Let's call that path right there. Let's call that y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we break this path into two paths, those are kind of our extreme points. Let's call this path right here. Let's call that path right there. Let's call that y. Let's call that x of. Or along this path, x is equal to, I could just write this path 2 or I'll call it c2. We could say it's x is equal to x2 of y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's call that y. Let's call that x of. Or along this path, x is equal to, I could just write this path 2 or I'll call it c2. We could say it's x is equal to x2 of y. That's that path. Then the first path, it doesn't have to be the first path, depending where you start. You can start anywhere."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could say it's x is equal to x2 of y. That's that path. Then the first path, it doesn't have to be the first path, depending where you start. You can start anywhere. Let's say this one in magenta. We'll call that path 1. We could say that that's defined as x is equal to x1 of y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can start anywhere. Let's say this one in magenta. We'll call that path 1. We could say that that's defined as x is equal to x1 of y. It's a little confusing when you have x as a function of y, but it's really completely analogous to what we did in the last video. We're literally just swapping x's and y's. We're now expressing x as a function of y instead of y as a function of x."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could say that that's defined as x is equal to x1 of y. It's a little confusing when you have x as a function of y, but it's really completely analogous to what we did in the last video. We're literally just swapping x's and y's. We're now expressing x as a function of y instead of y as a function of x. We have these two curves. You can imagine just flipping it, and we're doing the exact same thing that we did in the last video. It's just now in terms of y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're now expressing x as a function of y instead of y as a function of x. We have these two curves. You can imagine just flipping it, and we're doing the exact same thing that we did in the last video. It's just now in terms of y. If you look at it this way, this line integral can be rewritten. This could be rewritten as being equal to the integral. Let's just do c2 first."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's just now in terms of y. If you look at it this way, this line integral can be rewritten. This could be rewritten as being equal to the integral. Let's just do c2 first. This is the integral from b to a. We start at b, and we go to a. We're coming back from a high y to a low y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's just do c2 first. This is the integral from b to a. We start at b, and we go to a. We're coming back from a high y to a low y. The integral from b to a of q of, instead of having an x there, we know along this curve right here, x is equal to, we want everything in terms of y. Here, x is equal to x2 of y. q of x2 of y, maybe I'm using too many colors here, but I think you get the idea. dy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're coming back from a high y to a low y. The integral from b to a of q of, instead of having an x there, we know along this curve right here, x is equal to, we want everything in terms of y. Here, x is equal to x2 of y. q of x2 of y, maybe I'm using too many colors here, but I think you get the idea. dy. This is the part of the line integral just over this left-hand curve. Then we're going to add to that the line integral, or really just the regular integral now, from y is equal to a to y is equal to b of q of, and instead of x being equal to x2, now x is equal to x1 of y. It's equal to this curve, this other function."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dy. This is the part of the line integral just over this left-hand curve. Then we're going to add to that the line integral, or really just the regular integral now, from y is equal to a to y is equal to b of q of, and instead of x being equal to x2, now x is equal to x1 of y. It's equal to this curve, this other function. x1 of y, y, dy. We can do exactly what we did in the previous video. We don't like the larger number on the bottom, so let's swap these two around."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's equal to this curve, this other function. x1 of y, y, dy. We can do exactly what we did in the previous video. We don't like the larger number on the bottom, so let's swap these two around. If you swap these two, if you make this into an a and this into a b, that makes it the negative of the integral. When you swap the two, change the direction. This is exactly what we did in the last video, so hopefully it's nothing too fancy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We don't like the larger number on the bottom, so let's swap these two around. If you swap these two, if you make this into an a and this into a b, that makes it the negative of the integral. When you swap the two, change the direction. This is exactly what we did in the last video, so hopefully it's nothing too fancy. Now that we have the same boundaries of integration, these two definite integrals, we can just write them as one definite integral. This is going to be equal to the integral from a to b. I'll write this one first, since it's positive. q of x1 of y, y minus this one."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is exactly what we did in the last video, so hopefully it's nothing too fancy. Now that we have the same boundaries of integration, these two definite integrals, we can just write them as one definite integral. This is going to be equal to the integral from a to b. I'll write this one first, since it's positive. q of x1 of y, y minus this one. We have the minus sign here. Minus q of x2 of y and y, dy. Let me do that in that neutral color."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "q of x1 of y, y minus this one. We have the minus sign here. Minus q of x2 of y and y, dy. Let me do that in that neutral color. dy, that's multiplied by all of these things. I distributed out the dy. I think you get the idea."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me do that in that neutral color. dy, that's multiplied by all of these things. I distributed out the dy. I think you get the idea. This is identical to what we did in the last video. This could be rewritten as this is equal to the integral from a to b of, and inside of the integral we're evaluating the function q of xy from the boundaries, the upper boundary, where the upper boundary is going to be from x is equal to x1 of y and the lower boundary is x is equal to x2 of y. All the x's, we substitute it with that and then we get some expression."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I think you get the idea. This is identical to what we did in the last video. This could be rewritten as this is equal to the integral from a to b of, and inside of the integral we're evaluating the function q of xy from the boundaries, the upper boundary, where the upper boundary is going to be from x is equal to x1 of y and the lower boundary is x is equal to x2 of y. All the x's, we substitute it with that and then we get some expression. From that, we subtract this with x, substitute it as x2 of y. That's exactly what we did, just like I said in the last video. We're going in the reverse direction that we normally go in definite integrals."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All the x's, we substitute it with that and then we get some expression. From that, we subtract this with x, substitute it as x2 of y. That's exactly what we did, just like I said in the last video. We're going in the reverse direction that we normally go in definite integrals. We normally get to this and then the next step is we get this. Now we're going in the reverse direction, but it's all the same difference. All of that times dy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going in the reverse direction that we normally go in definite integrals. We normally get to this and then the next step is we get this. Now we're going in the reverse direction, but it's all the same difference. All of that times dy. Just like we saw in the last video, this expression right here. Actually, let me draw that dy a little further out so it doesn't get all congested. Let me do that dy out here."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All of that times dy. Just like we saw in the last video, this expression right here. Actually, let me draw that dy a little further out so it doesn't get all congested. Let me do that dy out here. This expression, this entire expression, is the same exact thing. That entire expression is the exact same thing as the integral from x is equal to I can just write it here. Let me write it in the same colors."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me do that dy out here. This expression, this entire expression, is the same exact thing. That entire expression is the exact same thing as the integral from x is equal to I can just write it here. Let me write it in the same colors. x2 of y to x1 of y of the partial. I'll do it in the orange color. Of the partial of q with respect to x dx."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write it in the same colors. x2 of y to x1 of y of the partial. I'll do it in the orange color. Of the partial of q with respect to x dx. This, I want to make it very clear, this is, at least in my mind, the first part. When I first saw it, it was a little confusing. If you just saw an integral like this, if this is the inside of a double integral, and it is."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Of the partial of q with respect to x dx. This, I want to make it very clear, this is, at least in my mind, the first part. When I first saw it, it was a little confusing. If you just saw an integral like this, if this is the inside of a double integral, and it is. The outside is what we saw there. It's the integral from a to b dy. If you just saw this in a double integral, what you would do is you would take the anti-derivative of this."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you just saw an integral like this, if this is the inside of a double integral, and it is. The outside is what we saw there. It's the integral from a to b dy. If you just saw this in a double integral, what you would do is you would take the anti-derivative of this. The anti-derivative of this with respect to x. The anti-derivative of the partial of q with respect to x. With respect to x is going to be just q of xy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you just saw this in a double integral, what you would do is you would take the anti-derivative of this. The anti-derivative of this with respect to x. The anti-derivative of the partial of q with respect to x. With respect to x is going to be just q of xy. Since it's a definite integral, you would evaluate it at x1 of y. Then subtract from that this function evaluated at x2 of y, which is exactly what we did. Hopefully, you appreciate that."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "With respect to x is going to be just q of xy. Since it's a definite integral, you would evaluate it at x1 of y. Then subtract from that this function evaluated at x2 of y, which is exactly what we did. Hopefully, you appreciate that. Then we got our result, which is very similar to the last result. What does this double integral represent? What does it represent?"}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully, you appreciate that. Then we got our result, which is very similar to the last result. What does this double integral represent? What does it represent? It represents anything. If you have any double integral that goes from, if you imagine this is some function. Let me draw it in three dimensions."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What does it represent? It represents anything. If you have any double integral that goes from, if you imagine this is some function. Let me draw it in three dimensions. This is really almost a review of what we did in the last video. That's the y-axis. That's our x-axis."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw it in three dimensions. This is really almost a review of what we did in the last video. That's the y-axis. That's our x-axis. That's our z-axis. This is some function of x and y. It's some surface you could imagine on the xy-plane."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's our x-axis. That's our z-axis. This is some function of x and y. It's some surface you could imagine on the xy-plane. It's some surface. We could call that the partial of q with respect to x. What this double integral is, this is essentially defining a region."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's some surface you could imagine on the xy-plane. It's some surface. We could call that the partial of q with respect to x. What this double integral is, this is essentially defining a region. You could view this dx times dy as a small differential of area. The region under question, the boundary points are from y going from, at the bottom, it goes from x2 of y, which we saw was a curve that looks something like this. That's the lower y."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What this double integral is, this is essentially defining a region. You could view this dx times dy as a small differential of area. The region under question, the boundary points are from y going from, at the bottom, it goes from x2 of y, which we saw was a curve that looks something like this. That's the lower y. Over here, if we draw it in two dimensions, this was the lower y curve. The upper y curve is x1 of y. The upper y curve looks something like that."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the lower y. Over here, if we draw it in two dimensions, this was the lower y curve. The upper y curve is x1 of y. The upper y curve looks something like that. The upper y curve goes something like that. x varies from the lower y curve to the upper y curve. That's what we're doing right here."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The upper y curve looks something like that. The upper y curve goes something like that. x varies from the lower y curve to the upper y curve. That's what we're doing right here. Then y varies from a to b. This is essentially saying, let's take the double integral over this region, over this region right here, of this function. It's essentially the volume, this is the ceiling, and this boundary is essentially the wall."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what we're doing right here. Then y varies from a to b. This is essentially saying, let's take the double integral over this region, over this region right here, of this function. It's essentially the volume, this is the ceiling, and this boundary is essentially the wall. It's the volume of that room. I don't know what it would look like when it comes up here, but you can imagine something like that. It would be the volume of that."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's essentially the volume, this is the ceiling, and this boundary is essentially the wall. It's the volume of that room. I don't know what it would look like when it comes up here, but you can imagine something like that. It would be the volume of that. That's what we're taking. This is the identical result we got in the last video. This is a pretty neat thing."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would be the volume of that. That's what we're taking. This is the identical result we got in the last video. This is a pretty neat thing. All of a sudden, this vector, and q of xy, I didn't draw it out like I did the last time. This q of xy only has things in the j direction. If I were to draw it, it's a vector field."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is a pretty neat thing. All of a sudden, this vector, and q of xy, I didn't draw it out like I did the last time. This q of xy only has things in the j direction. If I were to draw it, it's a vector field. The vectors only go up and down. They have no horizontal component to them. What we saw, when you start with a vector field like this, you take the line integral around this closed loop."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I were to draw it, it's a vector field. The vectors only go up and down. They have no horizontal component to them. What we saw, when you start with a vector field like this, you take the line integral around this closed loop. I'll rewrite it right here. You take this line integral around this closed loop of q dot dr, which is equal to the integral around the closed loop of q of xy dy. We just figured out that that's equivalent to the double integral over the region."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What we saw, when you start with a vector field like this, you take the line integral around this closed loop. I'll rewrite it right here. You take this line integral around this closed loop of q dot dr, which is equal to the integral around the closed loop of q of xy dy. We just figured out that that's equivalent to the double integral over the region. This is the region. That's exactly what we're doing over here. If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just figured out that that's equivalent to the double integral over the region. This is the region. That's exactly what we're doing over here. If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b. You might want to review the double integral videos if that confuses you. We're taking the double integral over the region of the partial of q with respect to x. You could write dx dy, or you could even write a little da, the differential of area."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I just gave you the region and you had to define it, you might say, x is going from this function to that function, and y is going from a to b. You might want to review the double integral videos if that confuses you. We're taking the double integral over the region of the partial of q with respect to x. You could write dx dy, or you could even write a little da, the differential of area. That's what we can imagine as a da, which is the same thing as a dx dy. If we combine that with the last video, this is kind of the neat bringing it all together part, the result of the last video was this. If I had a function that's defined completely in terms of x, we had this right here."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could write dx dy, or you could even write a little da, the differential of area. That's what we can imagine as a da, which is the same thing as a dx dy. If we combine that with the last video, this is kind of the neat bringing it all together part, the result of the last video was this. If I had a function that's defined completely in terms of x, we had this right here. We had that result. Actually, let me copy and paste both of these to a nice, clean part of my whiteboard, and then we can do the exciting conclusion. Let me copy and paste that."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I had a function that's defined completely in terms of x, we had this right here. We had that result. Actually, let me copy and paste both of these to a nice, clean part of my whiteboard, and then we can do the exciting conclusion. Let me copy and paste that. That's what we got in the last video. In this video, we got this result. I'll actually just copy and paste that part right there."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me copy and paste that. That's what we got in the last video. In this video, we got this result. I'll actually just copy and paste that part right there. You might already predict where this is going. Then let me paste it over here. This is the result from this video."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll actually just copy and paste that part right there. You might already predict where this is going. Then let me paste it over here. This is the result from this video. Now, let's think about an arbitrary vector field that is defined as, let's say I have an arbitrary vector field. I'll do it in pink. Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the result from this video. Now, let's think about an arbitrary vector field that is defined as, let's say I have an arbitrary vector field. I'll do it in pink. Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj. You can almost imagine F being the addition of our vector fields P and Q that we did in the last two videos. Q was this video, and we did P in the video before that. This is really any arbitrary vector field."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say F is a vector field defined over the xy-plane, and F is equal to P of xyi plus Q of xyj. You can almost imagine F being the addition of our vector fields P and Q that we did in the last two videos. Q was this video, and we did P in the video before that. This is really any arbitrary vector field. Let's say we wanted to take the line integral of this vector field along some path. It could be the same one we've done, which has been a very arbitrary one. It's really any arbitrary path."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is really any arbitrary vector field. Let's say we wanted to take the line integral of this vector field along some path. It could be the same one we've done, which has been a very arbitrary one. It's really any arbitrary path. Let me draw some arbitrary path over here. Let's say that is my arbitrary path, my arbitrary curve. Let's say it goes in that counterclockwise direction, just like that."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's really any arbitrary path. Let me draw some arbitrary path over here. Let's say that is my arbitrary path, my arbitrary curve. Let's say it goes in that counterclockwise direction, just like that. I'm interested in what the closed line integral around that path of F dot dr is. We've seen it multiple times. dr is equal to dx times i plus dy times j."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say it goes in that counterclockwise direction, just like that. I'm interested in what the closed line integral around that path of F dot dr is. We've seen it multiple times. dr is equal to dx times i plus dy times j. This line integral can be rewritten as this is equal to the line integral around the path C. F dot dr, that's going to be this term times dx. That's P of xy times dx plus this term, Q of xy times dy. This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dr is equal to dx times i plus dy times j. This line integral can be rewritten as this is equal to the line integral around the path C. F dot dr, that's going to be this term times dx. That's P of xy times dx plus this term, Q of xy times dy. This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy. What are these things? This is what we figured out in the first video. This is what we figured out just now in this video."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This whole thing, this is the same thing as the line integral of P of xy dx plus the line integral of Q of xy dy. What are these things? This is what we figured out in the first video. This is what we figured out just now in this video. This thing right here is the exact same thing as that over there. This is going to be equal to the double integral over this region of the minus partial of P with respect to y. Instead of a dy dx, we can say over the differential of area."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is what we figured out just now in this video. This thing right here is the exact same thing as that over there. This is going to be equal to the double integral over this region of the minus partial of P with respect to y. Instead of a dy dx, we can say over the differential of area. Plus this result, this thing right here is exactly what we just proved. It's exactly what we just showed in this video. That's plus, I'll leave it up there, maybe I'll do it in the yellow."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of a dy dx, we can say over the differential of area. Plus this result, this thing right here is exactly what we just proved. It's exactly what we just showed in this video. That's plus, I'll leave it up there, maybe I'll do it in the yellow. Plus the double integral over the same region of the partial of Q with respect to x, da. That's just dy dx or dx dy, you can switch the order, it's the differential of area. Now we can add these two integrals."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's plus, I'll leave it up there, maybe I'll do it in the yellow. Plus the double integral over the same region of the partial of Q with respect to x, da. That's just dy dx or dx dy, you can switch the order, it's the differential of area. Now we can add these two integrals. What do we get? This is equal to, and this is kind of our big grand conclusion, maybe magenta is called for here. The double integral of over the region of, I'll write this one first because it's positive, that one's negative."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we can add these two integrals. What do we get? This is equal to, and this is kind of our big grand conclusion, maybe magenta is called for here. The double integral of over the region of, I'll write this one first because it's positive, that one's negative. Over the region of the partial of Q with respect to x minus the partial of P with respect to y, d, the differential of area. This is our big takeaway. The line integral of the closed loop of F dot dr is equal to the double integral of this expression."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The double integral of over the region of, I'll write this one first because it's positive, that one's negative. Over the region of the partial of Q with respect to x minus the partial of P with respect to y, d, the differential of area. This is our big takeaway. The line integral of the closed loop of F dot dr is equal to the double integral of this expression. Just remember, we're taking the function that was associated with the x component, or the i component, we're taking the partial with respect to y. The function that was associated with the y component, we're taking the partial with respect to x. That first one we're taking the negative of, that's a good way to remember it."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The line integral of the closed loop of F dot dr is equal to the double integral of this expression. Just remember, we're taking the function that was associated with the x component, or the i component, we're taking the partial with respect to y. The function that was associated with the y component, we're taking the partial with respect to x. That first one we're taking the negative of, that's a good way to remember it. This result right here, this is, maybe I should write it in green, this is Green's Theorem. It's a neat way to relate a line integral of a vector field that has these partial derivatives, assuming it has these partial derivatives, to the region, to a double integral of the region. Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That first one we're taking the negative of, that's a good way to remember it. This result right here, this is, maybe I should write it in green, this is Green's Theorem. It's a neat way to relate a line integral of a vector field that has these partial derivatives, assuming it has these partial derivatives, to the region, to a double integral of the region. Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero. That's still true. That tells us that if F is conservative, this thing right here must be equal to zero. That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Just as a little bit of a side note, we've seen in several videos before, if F is conservative, we've learned that if F is conservative, which means it's the gradient of some function, that it's path independent, that the closed integral around any path is equal to zero. That's still true. That tells us that if F is conservative, this thing right here must be equal to zero. That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region. You could think of situations where they cancel each other out, but really over any region, that's the only way that this is going to be true, that these two things are going to be equal to zero. Then you could say partial of Q with respect to X minus the partial of P with respect to Y has to be equal to zero, or these two things have to equal each other. Or, this is kind of a corollary to Green's Theorem, kind of a low-hanging fruit you could have figured out."}, {"video_title": "Green's theorem proof (part 2)   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the only way that you're always going to enforce that this whole integral is going to be equal to zero over any region. You could think of situations where they cancel each other out, but really over any region, that's the only way that this is going to be true, that these two things are going to be equal to zero. Then you could say partial of Q with respect to X minus the partial of P with respect to Y has to be equal to zero, or these two things have to equal each other. Or, this is kind of a corollary to Green's Theorem, kind of a low-hanging fruit you could have figured out. The partial of Q with respect to X is equal to the partial of P with respect to Y. When you study exact equations in differential equation, you'll see this a lot more. I won't go into too much, but conservative fields, the differential form of what you see in the line integral, if it's conservative, it would be an exact equation."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 realize that I can scroll things, which is quite useful, because now I have a lot more board space. So how do I evaluate this integral?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see how to evaluate it now. And look at this. I actually realize that I can scroll things, which is quite useful, because now I have a lot more board space. So how do I 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."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So how do I 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 xy squared with respect to x?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 xy 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."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the antiderivative of xy 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."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And then let me draw the outside of the integral. This is y is equal to 0 to y is equal to 1 dy. Now, if x is equal to 1, this expression becomes y squared over 2 minus."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But you'll see that it's actually not that bad once you evaluate them. And then let me draw the outside of the integral. This is y is equal to 0 to y is equal to 1 dy. Now, if x is equal to 1, this expression becomes 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. So it's y to the third over 3."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, if x is equal to 1, this expression becomes 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. So it's y to the third over 3. Fair enough. And now I take the integral with respect to y. So now I sum up all of these rectangles in the y direction."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's y to the third over 3. Fair enough. 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?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 y to the third over 3? Well, the antiderivative of y squared, it's y cubed."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 y to the third over 3? Well, the antiderivative of y squared, it's y cubed. 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 someplace? No, I think this is correct. y to the fourth over 12."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, the antiderivative of y squared, it's y cubed. 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 someplace? No, I think this is correct. y to the fourth over 12. Oh wait, how did I get a 3 here? That's where I messed up. This is a 2, right?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "y to the fourth over 12. Oh wait, 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, right?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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, right? 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, 8."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I don't know how I ended up, right? 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, 8. Got to make sure I don't make those careless mistakes. That's the tough part. I just want to make sure that you got that too."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when I take the integral of this, it's 4 times 2, 8. Got to make sure I don't make those careless mistakes. That's the tough part. I just want to make sure that you got that too. I hate it when I do that. But I don't want to re-record the whole video. So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I just want to make sure that you got that too. I hate it when I do that. But I don't want to re-record the whole video. So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. And now I evaluate this at 1 and 0. And that gives us what? 1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when I evaluated this, this is right, and then I take the antiderivative of y to the third over 2, I get y to the fourth over 8. And now I evaluate this at 1 and 0. And that gives us what? 1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0. So it's going to be 0 minus 0. So you don't have to worry about that. So what's 1 sixth minus 1 eighth?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 sixth minus 1 eighth minus, well both of these when you evaluate them at 0 are going to be 0. So it's going to be 0 minus 0. So you don't have to worry about that. So what's 1 sixth minus 1 eighth? Let's see. 24, that's 4 minus 3 over 24, which is equal to 1 24th, is the volume of our figure. Is the volume of this figure."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's 1 sixth minus 1 eighth? Let's see. 24, that's 4 minus 3 over 24, which is equal to 1 24th, is the volume of our figure. Is the volume of this figure. So this time, the way we just did it, we took the integral with respect to x first, and then we did it with respect to y. Let's do it the other way around. So let me erase some things, and hopefully I won't make these careless mistakes again."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Is the volume of this figure. So this time, the way we just did it, we took the integral with respect to x first, and then we did it with respect to y. Let's do it the other way around. So let me erase some things, and hopefully I won't make these careless mistakes again. I'll keep this figure, but I'll even erase this one. Just because, let me erase all of this stuff. We have room to work with."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me erase some things, and hopefully I won't make these careless mistakes again. I'll keep this figure, but I'll even erase this one. Just because, let me erase all of this stuff. We have room to work with. So I kept that figure, but let me redraw just the xy-plane, just so we get the visualization right. It's more important to visualize the xy-plane in these problems than it is to visualize the whole thing in 3D. That's the y-axis, that's the x-axis, y, x."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have room to work with. So I kept that figure, but let me redraw just the xy-plane, just so we get the visualization right. It's more important to visualize the xy-plane in these problems than it is to visualize the whole thing in 3D. That's the y-axis, that's the x-axis, y, x. Our upper bound, you can say, is the graph y is equal to x squared, or you can view it as the bound x is equal to square root of y. That is, x is equal to 1, that's y is equal to 1. And we care about the volume above the shaded region."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the y-axis, that's the x-axis, y, x. Our upper bound, you can say, is the graph y is equal to x squared, or you can view it as the bound x is equal to square root of y. That is, x is equal to 1, that's y is equal to 1. And we care about the volume above the shaded region. That shaded region is this yellow region right here. And let's draw our dA, I'll draw a little square actually, and I'll do it in magenta. So that's our little dA."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we care about the volume above the shaded region. That shaded region is this yellow region right here. And let's draw our dA, I'll draw a little square actually, and I'll do it in magenta. So that's our little dA. And the height is dy, that's a y, and its width is dx. So the volume above this little square, that's the same thing as this little square, just like we said before, the volume above it is equal to the value of the function, the height is the value of the function, which is xy squared. And then we multiply it times the area of the base."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's our little dA. And the height is dy, that's a y, and its width is dx. So the volume above this little square, that's the same thing as this little square, just like we said before, the volume above it is equal to the value of the function, the height is the value of the function, which is xy squared. And then we multiply it times the area of the base. Well, the area of the base, you could say it's dA, but we know it's really dy times dx. I didn't have to write that times there, you could ignore it. And I wrote the y first just because we're going to integrate with respect to y first."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we multiply it times the area of the base. Well, the area of the base, you could say it's dA, but we know it's really dy times dx. I didn't have to write that times there, you could ignore it. And I wrote the y first just because we're going to integrate with respect to y first. We're going to sum in the y direction. So what does summing in the y direction mean? It means we're going to add that square to that square to that square."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I wrote the y first just because we're going to integrate with respect to y first. We're going to sum in the y direction. So what does summing in the y direction mean? It means we're going to add that square to that square to that square. Excuse me. So we're adding all the dy's together, right? So my question to you is, what is the upper bound on the y?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It means we're going to add that square to that square to that square. Excuse me. So we're adding all the dy's together, right? So my question to you is, what is the upper bound on the y? Well, once again, we bump into the curve, right? So the curve is the upper bound when we go upwards. And what is the upper bound on the curve?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So my question to you is, what is the upper bound on the y? Well, once again, we bump into the curve, right? So the curve is the upper bound when we go upwards. And what is the upper bound on the curve? Well, we're holding x fixed. So for any given x, what is this point? Well, that's going to be x squared, right?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what is the upper bound on the curve? Well, we're holding x fixed. So for any given x, what is this point? Well, that's going to be x squared, right? Because this is the graph of y equals x squared. So our upper bound is y is equal to x squared. And what's our lower bound?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's going to be x squared, right? Because this is the graph of y equals x squared. So our upper bound is y is equal to x squared. And what's our lower bound? We can keep adding these squares down here. We're adding all the little changes in y. So what's our lower bound?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what's our lower bound? We can keep adding these squares down here. We're adding all the little changes in y. So what's our lower bound? Well, our lower bound is just 0. That was pretty straightforward. So this expression, as it is written right now, is the volume above this rectangle."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's our lower bound? Well, our lower bound is just 0. That was pretty straightforward. So this expression, as it is written right now, is the volume above this rectangle. Let me draw it. It's the volume above this rectangle. These are the same rectangles."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this expression, as it is written right now, is the volume above this rectangle. Let me draw it. It's the volume above this rectangle. These are the same rectangles. Now what we want to do is add up all the dx's together, and we'll get the volume above the entire surface. So that rectangle, now we'll add it to another dx1, dx1, like that. So what's the upper and lower bounds on the x's?"}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These are the same rectangles. Now what we want to do is add up all the dx's together, and we'll get the volume above the entire surface. So that rectangle, now we'll add it to another dx1, dx1, like that. So what's the upper and lower bounds on the x's? We're going from x is equal to 0, right? We don't bump into the graph when we go all the way down. So we go for x is equal to 0."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the upper and lower bounds on the x's? We're going from x is equal to 0, right? We don't bump into the graph when we go all the way down. So we go for x is equal to 0. And then our upper bound is x is equal to 1. x is equal to 0, x is equal to 1. And in general, one way to think about it is, when you're doing this outer, kind of the last sum or the last integral, you really shouldn't have variable boundaries at this point, right? Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we go for x is equal to 0. And then our upper bound is x is equal to 1. x is equal to 0, x is equal to 1. And in general, one way to think about it is, when you're doing this outer, kind of the last sum or the last integral, you really shouldn't have variable boundaries at this point, right? Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. But our final answer is going to have a number. So if you had some variables here, you probably did something wrong. It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because our final answer has to be a number, assuming that we're not dealing with something very, very abstract. But our final answer is going to have a number. So if you had some variables here, you probably did something wrong. It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first. When I go upwards, I bump into the curve. What is that upward bound if x is constant? Oh, it's x squared."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's always useful, I think, to draw that little da and then figure out, OK, I'm summing to dy first. When I go upwards, I bump into the curve. What is that upward bound if x is constant? Oh, it's x squared. y is equal to x squared. If I go down, I bump into the x-axis where y is equal to 0. And so forth and so on."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Oh, it's x squared. y is equal to x squared. If I go down, I bump into the x-axis where y is equal to 0. And so forth and so on. So now let's just evaluate this and confirm that we get the same answer. So we're integrating with respect to y first. So that's xy to the third over 3 at x squared and 0."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so forth and so on. So now let's just evaluate this and confirm that we get the same answer. So we're integrating with respect to y first. So that's xy to the third over 3 at x squared and 0. And then we have our outer integral. x goes from 0 to 1 dx. If we substitute x squared in for y, x squared to the third power is x to the sixth."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's xy to the third over 3 at x squared and 0. And then we have our outer integral. x goes from 0 to 1 dx. If we substitute x squared in for y, x squared to the third power is x to the sixth. x to the sixth times x. Let me write that. So we have x times x squared to the third power over 3."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we substitute x squared in for y, x squared to the third power is x to the sixth. x to the sixth times x. Let me write that. So we have x times x squared to the third power over 3. That equals x to the seventh. x to the squared to the third, you multiply the exponents and then you add these. x to the seventh over 3 minus this evaluated when y is 0."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have x times x squared to the third power over 3. That equals x to the seventh. x to the squared to the third, you multiply the exponents and then you add these. x to the seventh over 3 minus this evaluated when y is 0. But that's just going to be 0, right? And then we evaluate that from 0 to 1 dx. We're almost there."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x to the seventh over 3 minus this evaluated when y is 0. But that's just going to be 0, right? And then we evaluate that from 0 to 1 dx. We're almost there. Increment the exponent. You get x to the eighth over 8. And we already have a 3 down there, so it's over 24."}, {"video_title": "Double integrals 6   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're almost there. Increment the exponent. You get x to the eighth over 8. And we already have a 3 down there, so it's over 24. And you evaluate that from 0 to 1. And I think we get the same answer. When you evaluate it at 1, you get 1 24th minus 0."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "What I want to do in the next few videos is try to see what happens to a line integral, either a line integral over a scalar field or a vector field. But what happens to that line integral when we change the direction of our path? So let's say when I say change direction, let's say that I have some curve C that looks something like this. We draw the x and y axis. That's my y axis. That is my x axis. And let's say my parametrization starts there and then as t increases, ends up over there, just like that."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "We draw the x and y axis. That's my y axis. That is my x axis. And let's say my parametrization starts there and then as t increases, ends up over there, just like that. So it's moving in that direction. And when I say I reverse the path, we could define another curve, let's call it minus C, that looks something like this. That is my y axis."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "And let's say my parametrization starts there and then as t increases, ends up over there, just like that. So it's moving in that direction. And when I say I reverse the path, we could define another curve, let's call it minus C, that looks something like this. That is my y axis. That is my x axis. And it looks exactly the same, but it starts up here. And then as t increases, it goes down to the starting point of the other curve."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "That is my y axis. That is my x axis. And it looks exactly the same, but it starts up here. And then as t increases, it goes down to the starting point of the other curve. So it's the exact same shape of a curve, but it goes in the opposite direction. So what I'm going to do in this video is just understand how we can construct a parametrization like this and hopefully understand it pretty well. And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "And then as t increases, it goes down to the starting point of the other curve. So it's the exact same shape of a curve, but it goes in the opposite direction. So what I'm going to do in this video is just understand how we can construct a parametrization like this and hopefully understand it pretty well. And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field. So let's just say this parametrization right here, this is defined in the basic way that we've always defined them. Let's say that this is x is equal to x of t, y is equal to y of t. And let's say this is from t is equal, or t, let me write it this way, t starts at a, so t is greater than or equal to a, and it goes up to b. So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "And in the next two videos after this, we'll try to see what this actually does to the line integral, one for a scalar field and then one for a vector field. So let's just say this parametrization right here, this is defined in the basic way that we've always defined them. Let's say that this is x is equal to x of t, y is equal to y of t. And let's say this is from t is equal, or t, let me write it this way, t starts at a, so t is greater than or equal to a, and it goes up to b. So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. And then when t is equal to b up here, this is really just a review of what we've seen before, really just a review of parametrization. When t is equal to b up here, this is the point x of b, y of b. Nothing new there."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "So in this example, this was when t is equal to a, and the point right here is the coordinate x of a, y of a. And then when t is equal to b up here, this is really just a review of what we've seen before, really just a review of parametrization. When t is equal to b up here, this is the point x of b, y of b. Nothing new there. Now given these functions, how can we construct another parametrization here that has the same shape but starts here? So I want this to be t is equal to a. Let me switch colors."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Nothing new there. Now given these functions, how can we construct another parametrization here that has the same shape but starts here? So I want this to be t is equal to a. Let me switch colors. Let me switch to maybe magenta. I want this to be t is equal to a, and as t increases, I want this to be t equals b. So I want to move in the opposite direction."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Let me switch colors. Let me switch to maybe magenta. I want this to be t is equal to a, and as t increases, I want this to be t equals b. So I want to move in the opposite direction. So when t is equal to a, I want my coordinate to still be x of b, y of b. When t is equal to a, I want a b in each of these functions. And when t is equal to b, I want the coordinate to be x of a, y of a."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "So I want to move in the opposite direction. So when t is equal to a, I want my coordinate to still be x of b, y of b. When t is equal to a, I want a b in each of these functions. And when t is equal to b, I want the coordinate to be x of a, y of a. Notice they're opposites now. Here t is equal to a, x of a, y of a. Here t is equal to b, our end point."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "And when t is equal to b, I want the coordinate to be x of a, y of a. Notice they're opposites now. Here t is equal to a, x of a, y of a. Here t is equal to b, our end point. Now I'm at this coordinate x of a, y of a. So how do I construct that? Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Here t is equal to b, our end point. Now I'm at this coordinate x of a, y of a. So how do I construct that? Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. So what if we define our x in this case for our minus c curve. What if we say x is equal to x of, I'm going to say x of, I'm talking about the same exact function. Actually, maybe I should write it in that same exact color."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Well, if you think about it, when t is equal to a, we want both of these functions to evaluate it at b. So what if we define our x in this case for our minus c curve. What if we say x is equal to x of, I'm going to say x of, I'm talking about the same exact function. Actually, maybe I should write it in that same exact color. x of, but instead of putting t in there, instead of putting a straight up t in there, what if I put an a plus b minus t in there? What happens? Well, let me do it for the y as well."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Actually, maybe I should write it in that same exact color. x of, but instead of putting t in there, instead of putting a straight up t in there, what if I put an a plus b minus t in there? What happens? Well, let me do it for the y as well. y is equal to y of a plus b minus t. I'm using slightly different shades of yellow. It might be a little disconcerting. Anyway, what happens when we define this?"}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Well, let me do it for the y as well. y is equal to y of a plus b minus t. I'm using slightly different shades of yellow. It might be a little disconcerting. Anyway, what happens when we define this? When t is equal to a, and let's say that this parametrization is also for t starts at a and then goes up to b. So let's just experiment and confirm that this parametrization really is the same thing as this thing, but it goes in an opposite direction, or at least confirm in our minds intuitively. So when t is equal to a, x will be equal to x of a plus b minus a."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Anyway, what happens when we define this? When t is equal to a, and let's say that this parametrization is also for t starts at a and then goes up to b. So let's just experiment and confirm that this parametrization really is the same thing as this thing, but it goes in an opposite direction, or at least confirm in our minds intuitively. So when t is equal to a, x will be equal to x of a plus b minus a. This is when t is equal to a. So minus t or minus a, which is equal to what? Well, a minus a cancel out."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "So when t is equal to a, x will be equal to x of a plus b minus a. This is when t is equal to a. So minus t or minus a, which is equal to what? Well, a minus a cancel out. That's equal to x of b. Similarly, when t is equal to a, y will be equal to y of a plus b minus a. The a's cancel out, so it's equal to y of b."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Well, a minus a cancel out. That's equal to x of b. Similarly, when t is equal to a, y will be equal to y of a plus b minus a. The a's cancel out, so it's equal to y of b. So that worked. When t is equal to a, my parametrization evaluates to the coordinate x of b, y of b. When t is equal to a, x of b, y of b."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "The a's cancel out, so it's equal to y of b. So that worked. When t is equal to a, my parametrization evaluates to the coordinate x of b, y of b. When t is equal to a, x of b, y of b. Then we can do the exact same thing when t is equal to b. I'll do it over here because I don't want to lose this. Let me just draw a line here. I'm still dealing with this parametrization over here."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "When t is equal to a, x of b, y of b. Then we can do the exact same thing when t is equal to b. I'll do it over here because I don't want to lose this. Let me just draw a line here. I'm still dealing with this parametrization over here. Actually, let me scroll over to the right just so that I don't get confused. When t is equal to b, what is x equal? x is equal to x of a plus b minus b. a plus b minus b when t is equal to b."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "I'm still dealing with this parametrization over here. Actually, let me scroll over to the right just so that I don't get confused. When t is equal to b, what is x equal? x is equal to x of a plus b minus b. a plus b minus b when t is equal to b. So that's equal to x of a. Then when t is equal to b, y is equal to y of a plus b minus b. Of course, that's going to be equal to y of a."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "x is equal to x of a plus b minus b. a plus b minus b when t is equal to b. So that's equal to x of a. Then when t is equal to b, y is equal to y of a plus b minus b. Of course, that's going to be equal to y of a. The endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b. We saw that down here. As t increases, this value is going to decrease."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "Of course, that's going to be equal to y of a. The endpoints work, and if you think about it intuitively, as t increases, so when t is at a, this thing is going to be x of b, y of b. We saw that down here. As t increases, this value is going to decrease. We start at x of b, y of b, and as t increases, this value is going to decrease to a. It starts from b and it goes to a. This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "As t increases, this value is going to decrease. We start at x of b, y of b, and as t increases, this value is going to decrease to a. It starts from b and it goes to a. This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that. It just goes in a completely opposite direction. Now, with that out of the way, if you accept what I've told you, that these are really the same parametrizations, just opposite directions. I shouldn't say same parametrizations."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "This one obviously starts at a and it goes to b. Hopefully that should give you the intuition why this is the exact same curve as that. It just goes in a completely opposite direction. Now, with that out of the way, if you accept what I've told you, that these are really the same parametrizations, just opposite directions. I shouldn't say same parametrizations. Same curve going in an opposite direction or the same path going in the opposite direction. In the next video, I'm going to see what happens when we evaluate this line integral, f of x, ds, versus this line integral. This is a line integral of a scalar field using this curve or this path."}, {"video_title": "Parametrization of a reverse path   Khan Academy.mp3", "Sentence": "I shouldn't say same parametrizations. Same curve going in an opposite direction or the same path going in the opposite direction. In the next video, I'm going to see what happens when we evaluate this line integral, f of x, ds, versus this line integral. This is a line integral of a scalar field using this curve or this path. What happens if we take a line integral over the same scalar field but we do it over this reverse path? That's what we're going to do in the next video. The video after that, we'll do it for vector fields."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "There's S, the tons of steel that you're using, H, the hours of labor, and then lambda, this Lagrange multiplier we introduced that's basically a proportionality constant between the gradient vectors of the revenue function and the constraint function. And always the third equation that we're dealing with here to solve this is the constraint itself. That gives us another equation that'll help solve for H and S, and ultimately lambda, if that's something you want as well. So as kind of a first pass here, I'm gonna do a little more simplifying, but I'm gonna make a substitution that'll make this easier for us. So I see S over H here, and they're both to the same power, so I feel it might be a little bit easier if I just substitute U in for S divided by H. And what that'll let me do is rewrite this first equation here as 200 thirds, 200 thirds times U to the power 1 third, and that's equal to 20 times lambda. And then likewise, what that means for this guy is, well, this is H over S, not S over H, so that one's gonna be 100 thirds, not times U to the power 2 thirds, but times U to the negative 2 thirds because we swapped the H and S here. So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "So as kind of a first pass here, I'm gonna do a little more simplifying, but I'm gonna make a substitution that'll make this easier for us. So I see S over H here, and they're both to the same power, so I feel it might be a little bit easier if I just substitute U in for S divided by H. And what that'll let me do is rewrite this first equation here as 200 thirds, 200 thirds times U to the power 1 third, and that's equal to 20 times lambda. And then likewise, what that means for this guy is, well, this is H over S, not S over H, so that one's gonna be 100 thirds, not times U to the power 2 thirds, but times U to the negative 2 thirds because we swapped the H and S here. So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think. We kind of want to treat H and S in the same package. Now let me go ahead and put all the constants together, and I'm gonna take this guy and multiply it by three divided by 200, multiply both sides of that just to cancel out what's on the left side here. And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think. We kind of want to treat H and S in the same package. Now let me go ahead and put all the constants together, and I'm gonna take this guy and multiply it by three divided by 200, multiply both sides of that just to cancel out what's on the left side here. And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda. And then similarly over here, I'm gonna take this whole equation and multiply it by 3 over 100, and what that's gonna leave me with is that U to the negative 2 thirds, U to the negative 2 thirds is equal to, let's see, this 2000 when we divide by 100 becomes 20, and that 20 times 3 is 60, so that'll be 60 times lambda. Alright, so now I want a way to simplify each of these, and what I notice is they look pretty similar on each side, you know, it's something related to U equal to lambda, so if I can get this in a form where I'm really isolating U, that would be great. And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda. And then similarly over here, I'm gonna take this whole equation and multiply it by 3 over 100, and what that's gonna leave me with is that U to the negative 2 thirds, U to the negative 2 thirds is equal to, let's see, this 2000 when we divide by 100 becomes 20, and that 20 times 3 is 60, so that'll be 60 times lambda. Alright, so now I want a way to simplify each of these, and what I notice is they look pretty similar on each side, you know, it's something related to U equal to lambda, so if I can get this in a form where I'm really isolating U, that would be great. And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that. So when I do that to the top part, like I said, that U to the 1 third times U to the 2 thirds ends up being U, and then on the right side, we have our 3 tenths lambda, but now U to the 2 thirds. And then on the bottom here, when I multiply it by U to the 2 thirds, that right side becomes 1, because it cancels out with U to the negative 2 thirds, and the right side is 60 times lambda times U to the 2 thirds. Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that. So when I do that to the top part, like I said, that U to the 1 third times U to the 2 thirds ends up being U, and then on the right side, we have our 3 tenths lambda, but now U to the 2 thirds. And then on the bottom here, when I multiply it by U to the 2 thirds, that right side becomes 1, because it cancels out with U to the negative 2 thirds, and the right side is 60 times lambda times U to the 2 thirds. Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side. So in this case, I'm gonna multiply that top by 10, which will get it to 3, and then by another 20 to make that constant 60. So I'm gonna multiply this entire top equation by 200, and what that gives me is that 200 times U is equal to 60 times lambda times U to the 2 thirds. And now these two equations, these two equations have the same right side."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side. So in this case, I'm gonna multiply that top by 10, which will get it to 3, and then by another 20 to make that constant 60. So I'm gonna multiply this entire top equation by 200, and what that gives me is that 200 times U is equal to 60 times lambda times U to the 2 thirds. And now these two equations, these two equations have the same right side. So this is the same as saying 200 times U is equal to, well, 1, because each one of those expressions equals the same complicated thing. And now 200 times U, well that's S divided by H. So this is the same thing as saying 200 times S over H equals 1, which we can write more succinctly as H is equal to 200 times S. Great, so I'm gonna go ahead and circle that. H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "And now these two equations, these two equations have the same right side. So this is the same as saying 200 times U is equal to, well, 1, because each one of those expressions equals the same complicated thing. And now 200 times U, well that's S divided by H. So this is the same thing as saying 200 times S over H equals 1, which we can write more succinctly as H is equal to 200 times S. Great, so I'm gonna go ahead and circle that. H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget. I'll go ahead and kind of write that down again. That our 20 times H, I think, 20 times the hours of labor plus $2,000 per ton of steel is equal to our budget of $20,000. And now we can just substitute in."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget. I'll go ahead and kind of write that down again. That our 20 times H, I think, 20 times the hours of labor plus $2,000 per ton of steel is equal to our budget of $20,000. And now we can just substitute in. Instead of H, I'm gonna write 200 S. So that's 200, sorry, 20 times 200 S, 200 S, plus 2,000 times S is equal to 20,000. And now this right side, 20 times 200 is equal to 4,000. And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "And now we can just substitute in. Instead of H, I'm gonna write 200 S. So that's 200, sorry, 20 times 200 S, 200 S, plus 2,000 times S is equal to 20,000. And now this right side, 20 times 200 is equal to 4,000. And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000. 6,000 times S is equal to 20,000. And when those cancel out, what that gives us is S is equal to 20 divided by six, which is the same as 10 divided by three. So that's how many tons of steel we should get."}, {"video_title": "Lagrange multiplier example, part 2.mp3", "Sentence": "And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000. 6,000 times S is equal to 20,000. And when those cancel out, what that gives us is S is equal to 20 divided by six, which is the same as 10 divided by three. So that's how many tons of steel we should get. S is 10 over three. Then when we apply that to the fact that H is 200 times S, that's gonna mean that H is equal to 200 times that value, 10 over three, which is equal to 2,000, 2,000 thirds. 2,000 thirds, that's how many hours of labor we want."}, {"video_title": "Partial derivatives of vector fields.mp3", "Sentence": "So a vector field is a function, I'll just do a two dimensional example here, is gonna be something that has a two dimensional input, and then the output has the same number of dimensions, that's the important part. And each of these components in the output is gonna depend somehow on the input variables, so the example I have in mind will be x times y as that first component, and then y squared minus x squared as that second component. And you can compute the partial derivative of a guy like this, right? You'll take the partial derivative with respect to one of the input variables, I'll choose x, it's always a nice one to start with, partial derivative with respect to x. And if we were to actually compute it in this case, it's another, it's a function of x and y, what you do is you take the partial derivative component wise, so you go to each component, and the first one you say, okay, x looks like a variable, y looks like a constant, the derivative will just be that constant, and then the partial derivative of the second component, that y squared looks like a constant, derivative of negative x squared with respect to x, negative two x. So analytically, if you know how to take a partial derivative, you already know how to take a partial derivative of vector-valued functions, and hence vector fields, but the fun part, the important part here, how do you actually interpret this? And this has everything to do with visualizing it in some way, so the vector field, the reason we call it a vector field, is you kind of take the whole xy-plane, and you're gonna fill it with vectors, and concretely, what I mean by that, you'll take a given input, what's an input you wanna look at?"}, {"video_title": "Partial derivatives of vector fields.mp3", "Sentence": "You'll take the partial derivative with respect to one of the input variables, I'll choose x, it's always a nice one to start with, partial derivative with respect to x. And if we were to actually compute it in this case, it's another, it's a function of x and y, what you do is you take the partial derivative component wise, so you go to each component, and the first one you say, okay, x looks like a variable, y looks like a constant, the derivative will just be that constant, and then the partial derivative of the second component, that y squared looks like a constant, derivative of negative x squared with respect to x, negative two x. So analytically, if you know how to take a partial derivative, you already know how to take a partial derivative of vector-valued functions, and hence vector fields, but the fun part, the important part here, how do you actually interpret this? And this has everything to do with visualizing it in some way, so the vector field, the reason we call it a vector field, is you kind of take the whole xy-plane, and you're gonna fill it with vectors, and concretely, what I mean by that, you'll take a given input, what's an input you wanna look at? Like, I'll say maybe one, two, or yeah, let's do that, let's do one, two, which would mean you kinda go x equals one, and then y equals two, this input point, and we wanna associate that with the output vector in some way, and so let's just compute what it should equal, so when we plug in x equals one and y equals two, x times y becomes two, y squared minus x squared becomes two squared minus one squared, so four minus one is three, so we have this vector two, three that we want to associate with that input point, and vector fields, you just attach the two points, you just, I'm gonna take the vector two, three and attach it to this guy, so it should have an x component of two, and then a y component of three, so it's gonna end up looking something like, let's see, so, y component of three, something like this, so that'll be the vector, and we attach it to that point, and in principle, you do this to all of the different points, and if you did, what you'd get would be something like this, and remember, when we represent these, especially with computers, it tends to lie where each represented vector is much, much shorter than it should be in reality, but you just wanna squish them all onto the same page so they don't overrun each other, and here, color is supposed to give a general vague sense of relative length, so ones that are blue should be thought of as much shorter than the ones that are yellow, but that doesn't really give a specific thought for how long they should be, but for partial derivatives, we actually care a lot about the specifics, and if you think back to how we interpret partial derivatives in a lot of other contexts, what we want to do is imagine this partial x here as a slight nudge in the x direction, right, so this was our original input, and you might imagine just nudging it a little bit, and the size of that nudge as a number would be your partial x, so then the question is, what's the resulting change to the output, and because the output is a vector, the change in the output is also gonna be a vector, so what we want is to say there's gonna be some other vector attached to this point, right, it's gonna look very similar, maybe it looks like, maybe it looks something like this, so something similar, but maybe a little bit different, and you wanna take that difference in vector form, and I'll describe what I mean by that in just a moment, and then divide it by the size of that original nudge, so to be much more specific about what I mean here, if you're comparing two different vectors, and they're rooted in two different spots, I think a good way to start is to just move them to a new space where they're rooted in the same spot, so in this case, I'm just gonna kinda draw a separate space over here, and be thinking of this as a place for these vectors to live, and I'm gonna put them both on this plane, but I'm gonna root them each in the origin, so this first one that has components two, three, let's give it a name, let's call this guy V1, so that'll be V1, and then the nudged output, the second one, I'll call V2, V2, and let's say V2 is also in this space, and I might exaggerate the difference just so that we can see it here, let's say it was different in some way, in reality, if it's a small nudge, it'll be different in only a very small way, but let's say these were our two vectors, the difference between these guys is gonna be a vector that connects the tips, and I'm gonna call that guy partial V, and the way that you can be thinking about this is to say that V1, V1, that original guy, plus that tiny nudge, the difference between them, is equal to V2, the nudged output, and in terms of tip to tail with vectors, you're seeing that, kinda the green vector plus that blue vector is the same as that pink vector that connects the tail of the original one to the tip of the new one, so when we're thinking of a partial derivative, you're basically saying, hey, what happens if we take this, the size of the nudge of the output, and then we divide it by that nudge of the input, so let's say you were thinking of that original nudge as being, I don't know, like of a size 1 1 2, like 0.5, as the change in the X direction, then that would mean, when you go over here and you say, what's that DV, that changing vector V, divided by DX, you'd be dividing it by 0.5, and in principle, you'd be thinking of, that would mean that you're kind of scaling this by two, as if to say this little DV is 1 1 2 of some other vector, and that other vector is what the partial derivative is, so this other vector here, the full blue guy, would be DV, you know, scaled down, or scaled up, however you wanna think about it, by that partial X, and that's what makes it such that, you know, in principle, if this partial X change was really small, it was like 1 1 100th, and the output nudge also was really small, it was like 1 1 100th, or you know, something on that order, it wouldn't be specifically that, then the DV DX, that change, would still be a normal sized vector, and the direction that it points is still kind of an indication of the direction that this green vector should change as you're scooting over. So just to be concrete, and you know, actually compute this guy, let's say we were to take this partial derivative, partial V with respect to X, and evaluate it at that point 1 2, that we're just dealing with, 1 2. What that would mean, Y is equal to 2, so that first component is 2, and then X is equal to 1, so that next one should be negative 2, and we can, I guess we can see just how wrong my drawing was to start here, I was just kind of guessing what the pink vector would be, but I guess it changes in the direction of 2 negative 2, so that should be something, here I'll kind of erase the, what turns out to be the wrong direction here, but, get rid of this guy, and I guess the change should be in the direction kind of 2 as the X component, and then negative 2, that's a negative 2 as the Y component, so the derivative vector should look something like this, which means all corresponding little DV nudges will be slight changes, will be slight changes on that, so these will be your DVs, something in that direction, and what that means in our vector field then, as you move in the X direction, and consider the various vectors attached to each point, as you're kind of passing through the point 1, 2, the way that the vectors are changing should be somehow down and to the right, the tip should move down and to the right, so if this starts highly up and to the right, then it should be getting kind of shorter, but then longer to the right, so the V2, if I were to have drawn it more accurately here, you know, what that nudged output should look like, it would really be something that's kind of like, I don't know, like this, where it's getting shorter in the Y direction, but then longer in the X direction, as per that blue nudging arrow, and in the next video, I'll kind of go through more examples of how you might think of this, how you think of it in terms of what each component means, which becomes very important for later topics, like divergence and curl, and I'll see you next video."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I think I did it over several videos because it was a bit hairy. And I'll write our position vector valued function first. So we have r as a function of our two parameters s and t. And then I'll review a little bit of what all the terms, what the s, the t, and the a's and the b's represent. But it's equal to b plus a cosine of s. And once again, we saw this several videos ago. So you might want to watch the videos on parameterizing surfaces with two parameters to figure out how we got here. Times the sine of t. I'm going to put the s terms and the t terms in different colors. Times our i unit vector."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But it's equal to b plus a cosine of s. And once again, we saw this several videos ago. So you might want to watch the videos on parameterizing surfaces with two parameters to figure out how we got here. Times the sine of t. I'm going to put the s terms and the t terms in different colors. Times our i unit vector. I'll put the vectors or the unit vectors in this orange color. Plus, doing the same yellow. Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times our i unit vector. I'll put the vectors or the unit vectors in this orange color. Plus, doing the same yellow. Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction. Plus a sine of s. Times the k unit vector, the unit vector in the z direction. And in order to generate the torus or the donut shape, this is true for our parameters. So we don't wrap multiple times around the torus."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction. Plus a sine of s. Times the k unit vector, the unit vector in the z direction. And in order to generate the torus or the donut shape, this is true for our parameters. So we don't wrap multiple times around the torus. For s being between 0 and 2 pi. And for t being between 0 and 2 pi. And just as a bit of a review of where all of this came from."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we don't wrap multiple times around the torus. For s being between 0 and 2 pi. And for t being between 0 and 2 pi. And just as a bit of a review of where all of this came from. And I'm going to have to do what my plan is for this video over several videos. But let's review where all of this came from. Let me draw a donut."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just as a bit of a review of where all of this came from. And I'm going to have to do what my plan is for this video over several videos. But let's review where all of this came from. Let me draw a donut. My best effort at a donut right here. That looks like a donut or a torus. And you can imagine a torus or this donut shape is kind of the product of two circles."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw a donut. My best effort at a donut right here. That looks like a donut or a torus. And you can imagine a torus or this donut shape is kind of the product of two circles. You have this circle that's kind of the cross section of the donut at any point. You can take it there. You can take it over there."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you can imagine a torus or this donut shape is kind of the product of two circles. You have this circle that's kind of the cross section of the donut at any point. You can take it there. You can take it over there. And then you have the circle that kind of wraps around all of these other circles. Or these other circles wrap around it. And so when we derived this formula up here, this parameterization."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can take it over there. And then you have the circle that kind of wraps around all of these other circles. Or these other circles wrap around it. And so when we derived this formula up here, this parameterization. A was the radius of these cross sectional circles. That's A. That's what these A terms were."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so when we derived this formula up here, this parameterization. A was the radius of these cross sectional circles. That's A. That's what these A terms were. And B was the distance from the center of our torus. B was the distance from the center of our torus out to the center of these cross sections. So this was B."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's what these A terms were. And B was the distance from the center of our torus. B was the distance from the center of our torus out to the center of these cross sections. So this was B. So you can imagine that B is kind of the radius of the big circle up to the midpoint of the cross section. And A is the radius of the cross sectional circles. And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this was B. So you can imagine that B is kind of the radius of the big circle up to the midpoint of the cross section. And A is the radius of the cross sectional circles. And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle. So it's an angle from 0 to 2 pi to say where we are in that circle. And T tells us how much we've rotated around the larger circle. So T was telling us how much we rotated around the larger circle."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle. So it's an angle from 0 to 2 pi to say where we are in that circle. And T tells us how much we've rotated around the larger circle. So T was telling us how much we rotated around the larger circle. So if you think about it, you can specify any point on this donut or on this surface or on this torus by telling you an S or a T. And so that's why we picked that as the parameterization. Now, the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral. And the surface integral we're going to compute will tell us the surface area of this torus."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So T was telling us how much we rotated around the larger circle. So if you think about it, you can specify any point on this donut or on this surface or on this torus by telling you an S or a T. And so that's why we picked that as the parameterization. Now, the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral. And the surface integral we're going to compute will tell us the surface area of this torus. So this surface right here is sigma, right like that. It's being represented by this position vector-valued function that is parameterized by these two parameters right there. And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the surface integral we're going to compute will tell us the surface area of this torus. So this surface right here is sigma, right like that. It's being represented by this position vector-valued function that is parameterized by these two parameters right there. And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface. Here this capital sigma does not represent sum. It represents a surface of a bunch of the little d sigmas, a bunch of the little chunks of the surface. And just as a review, you can imagine each d sigma is a little patch of the surface right there."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface. Here this capital sigma does not represent sum. It represents a surface of a bunch of the little d sigmas, a bunch of the little chunks of the surface. And just as a review, you can imagine each d sigma is a little patch of the surface right there. That is a d sigma. And we're taking, it's a double integral here because we want to add up all the d sigmas in two directions. You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just as a review, you can imagine each d sigma is a little patch of the surface right there. That is a d sigma. And we're taking, it's a double integral here because we want to add up all the d sigmas in two directions. You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus. So that's why it's a double integral. And this is just going to give you the surface area, which is the whole point of this video and probably the next one or two videos. But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus. So that's why it's a double integral. And this is just going to give you the surface area, which is the whole point of this video and probably the next one or two videos. But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there. But here we're just multiplying it by 1. And we saw in the last video that this is a way of expressing an idea, but you really can't do much computation with this. But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there. But here we're just multiplying it by 1. And we saw in the last video that this is a way of expressing an idea, but you really can't do much computation with this. But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos. This is the same thing as the double integral over the region over which our parameters are defined. So it's this region over here where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we don't have to."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos. This is the same thing as the double integral over the region over which our parameters are defined. So it's this region over here where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we don't have to. We could just write a 1 if we like. It doesn't change much. Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just have a 1 here, so we don't have to. We could just write a 1 if we like. It doesn't change much. Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt. So we saw this in the last video. What we're going to do now is actually compute this. That's the whole point of this video."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt. So we saw this in the last video. What we're going to do now is actually compute this. That's the whole point of this video. We're going to take the cross product of these two vectors. So let's figure out these vectors. Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the whole point of this video. We're going to take the cross product of these two vectors. So let's figure out these vectors. Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral. You're going to see it's a pretty hairy problem, and this is probably the reason that very few people ever see an actual surface integral get computed. But let's do it anyway. So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral. You're going to see it's a pretty hairy problem, and this is probably the reason that very few people ever see an actual surface integral get computed. But let's do it anyway. So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video. This term is what? We just want to hold t constant and take the partial with respect to just s. So this up here, if we distribute it, the sine of t times b, that's just going to be a constant in terms of s, so we can ignore that. Then you have sine of t times this over here."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video. This term is what? We just want to hold t constant and take the partial with respect to just s. So this up here, if we distribute it, the sine of t times b, that's just going to be a constant in terms of s, so we can ignore that. Then you have sine of t times this over here. So sine of t and a is a constant, and you take the derivative of cosine of s, that's negative sine of s. So it's going to be, so the derivative of this with respect to s, so the partial with respect to s, is going to be minus a. I'll write in green the sine of t so you know that's where it came from. Sine of t and then sine of s. The derivative of this is negative sine of s, that's where that negative came from, and then I'll write the sine of s right there. Sine of s times the unit vector i."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then you have sine of t times this over here. So sine of t and a is a constant, and you take the derivative of cosine of s, that's negative sine of s. So it's going to be, so the derivative of this with respect to s, so the partial with respect to s, is going to be minus a. I'll write in green the sine of t so you know that's where it came from. Sine of t and then sine of s. The derivative of this is negative sine of s, that's where that negative came from, and then I'll write the sine of s right there. Sine of s times the unit vector i. That's the partial of just this x term with respect to s. And then we'll do the same thing with the y term or the j term. So plus, same logic, b times cosine of t with respect to s, we take the partial just to become zero. So you're left with a, well it's going to be a minus a again, right?"}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sine of s times the unit vector i. That's the partial of just this x term with respect to s. And then we'll do the same thing with the y term or the j term. So plus, same logic, b times cosine of t with respect to s, we take the partial just to become zero. So you're left with a, well it's going to be a minus a again, right? Because when you take the derivative of the cosine of s, it's going to be negative sine of s. So you have a, let me do it. So you're going to have a minus a, this cosine of t, minus a cosine of t, that's the constant terms, sine of s. Just taking partial derivatives. Sine of s j."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you're left with a, well it's going to be a minus a again, right? Because when you take the derivative of the cosine of s, it's going to be negative sine of s. So you have a, let me do it. So you're going to have a minus a, this cosine of t, minus a cosine of t, that's the constant terms, sine of s. Just taking partial derivatives. Sine of s j. And then finally we take the derivative of this with respect to s, and that's pretty straight forward. It's just going to be a cosine of s. So plus a cosine of s k. Now hopefully you didn't find this confusing. The negative signs, because the derivative of cosines are negative signs."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sine of s j. And then finally we take the derivative of this with respect to s, and that's pretty straight forward. It's just going to be a cosine of s. So plus a cosine of s k. Now hopefully you didn't find this confusing. The negative signs, because the derivative of cosines are negative signs. So negative sine of s, so that's why it's negative sine of s times the constants. Negative sine of s times the constants. The constant cosine of t, sine of t. So hopefully this makes some sense."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The negative signs, because the derivative of cosines are negative signs. So negative sine of s, so that's why it's negative sine of s times the constants. Negative sine of s times the constants. The constant cosine of t, sine of t. So hopefully this makes some sense. This is a review of taking a partial derivative. Now let's do the same thing with respect to t. Let's do the same thing with respect to t, and I'll do that in a different color. So now we want to take the partial of r with respect to t. So the partial of r with respect to t is equal to, so now this whole term over here is a constant, and so it's going to be that whole term times the derivative of this with respect to t, which is just cosine of t. So it's going to be b plus a cosine of s times cosine of t. Cosine of t i."}, {"video_title": "Example of calculating a surface integral part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The constant cosine of t, sine of t. So hopefully this makes some sense. This is a review of taking a partial derivative. Now let's do the same thing with respect to t. Let's do the same thing with respect to t, and I'll do that in a different color. So now we want to take the partial of r with respect to t. So the partial of r with respect to t is equal to, so now this whole term over here is a constant, and so it's going to be that whole term times the derivative of this with respect to t, which is just cosine of t. So it's going to be b plus a cosine of s times cosine of t. Cosine of t i. And then plus, and it's actually going to be a minus, because when you take the derivative of this with respect to t, it's going to be minus sine of t. So let me put them, it's going to be negative, and then let me leave some space for this term right here, negative sine of t, and you're going to have this constant out there, that's a constant in t. B plus a cosine of s, that's just that term right there, the derivative of cosine t is negative sine of t times j, and then the partial of this with respect to t, this is just a constant in t, so the partial is going to be zero. So I'll write plus zero k. Plus zero, let me do all my vectors in that same color, plus zero times the unit vector k. So that gives us our partial derivatives. Now we have to take the cross product, then find the magnitude of the cross product, and then evaluate this double integral, and I'll do that in the next couple of videos."}, {"video_title": "Saddle points.mp3", "Sentence": "So you're looking just for where it's tangent, where you can find a tangent plane that's flat. But this will give you some other points, like the little local minima here, the bumps, where the value of the function at that point is higher than all of the neighbor points, you know, if you walk in any direction, you're going downhill, so that's another thing you're gonna, you know, incidentally pick up by looking for places where this tangent plane is flat. But there's also a really interesting new possibility that comes up in the context of multivariable functions. And this is the idea of a saddle point. So let me pull up another graph here, this guy. And the function that you're looking at, here, I'll write it down, the function that you're looking at is f of x, y is equal to x squared minus y squared, okay? So now let's think about what the tangent plane at the origin of this entire graph would be."}, {"video_title": "Saddle points.mp3", "Sentence": "And this is the idea of a saddle point. So let me pull up another graph here, this guy. And the function that you're looking at, here, I'll write it down, the function that you're looking at is f of x, y is equal to x squared minus y squared, okay? So now let's think about what the tangent plane at the origin of this entire graph would be. Now, the tangent plane to this graph at the origin is actually flat, here's what it looks like. And to convince yourself of this, let's go ahead and actually compute the partial derivatives of this function and evaluate each one at the origin. So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x."}, {"video_title": "Saddle points.mp3", "Sentence": "So now let's think about what the tangent plane at the origin of this entire graph would be. Now, the tangent plane to this graph at the origin is actually flat, here's what it looks like. And to convince yourself of this, let's go ahead and actually compute the partial derivatives of this function and evaluate each one at the origin. So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x. And the other partial derivative, the partial with respect to y, with respect to y, we take the derivative of this negative y squared and we ignore the x because it looks like a constant as far as y is concerned, and we get negative two y. Now, if we plug in the point, the origin, into any one of these, you know, we plug in the point x, y is equal to zero, zero, then what do each of these go to? Well, the top one, x equals zero, so this guy goes to zero."}, {"video_title": "Saddle points.mp3", "Sentence": "So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x. And the other partial derivative, the partial with respect to y, with respect to y, we take the derivative of this negative y squared and we ignore the x because it looks like a constant as far as y is concerned, and we get negative two y. Now, if we plug in the point, the origin, into any one of these, you know, we plug in the point x, y is equal to zero, zero, then what do each of these go to? Well, the top one, x equals zero, so this guy goes to zero. And similarly over here, y is zero, so this goes to zero. So both partial derivatives are zero. And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement."}, {"video_title": "Saddle points.mp3", "Sentence": "Well, the top one, x equals zero, so this guy goes to zero. And similarly over here, y is zero, so this goes to zero. So both partial derivatives are zero. And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement. And one way of seeing this is to chop the graph. So if we imagine chopping it with a plane that represents a constant x value, and we kind of chop off the graph there, what you'll see, here I'll get rid of the tangent plane, what you'll see is that the curve where this intersects the graph, so let me trace that out, the curve where it intersects the graph basically has a local maximum at that origin point. The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective."}, {"video_title": "Saddle points.mp3", "Sentence": "And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement. And one way of seeing this is to chop the graph. So if we imagine chopping it with a plane that represents a constant x value, and we kind of chop off the graph there, what you'll see, here I'll get rid of the tangent plane, what you'll see is that the curve where this intersects the graph, so let me trace that out, the curve where it intersects the graph basically has a local maximum at that origin point. The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective. But now let's imagine chopping it in a different direction. So if instead we have the full graph and instead of chopping it with a constant x value, we chop it with a constant y value, then in that case, we look at the curve, and if we trace out the curve where this constant y value intersects the graph, let's see what it would look like, it's also kind of this parabolic shape, again, the tangent line is flat because it looks like a local minimum of that curve. So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat."}, {"video_title": "Saddle points.mp3", "Sentence": "The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective. But now let's imagine chopping it in a different direction. So if instead we have the full graph and instead of chopping it with a constant x value, we chop it with a constant y value, then in that case, we look at the curve, and if we trace out the curve where this constant y value intersects the graph, let's see what it would look like, it's also kind of this parabolic shape, again, the tangent line is flat because it looks like a local minimum of that curve. So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat. But notice, this is neither a local maximum nor a local minimum, because from one direction, from one direction, it looked like it was a local maximum. Yeah, I'll get rid of that guy. It looks like it's a local maximum when you look on the curve there."}, {"video_title": "Saddle points.mp3", "Sentence": "So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat. But notice, this is neither a local maximum nor a local minimum, because from one direction, from one direction, it looked like it was a local maximum. Yeah, I'll get rid of that guy. It looks like it's a local maximum when you look on the curve there. But from another direction, if you chop it in another way, it looks like a local minimum. And if we look at the equations, this kind of makes sense, because if you're just thinking about movements in the x direction, the entire function looks like x squared plus some kind of constant. So the graph of that would look like an x squared parabola shape that has a local minimum."}, {"video_title": "Saddle points.mp3", "Sentence": "It looks like it's a local maximum when you look on the curve there. But from another direction, if you chop it in another way, it looks like a local minimum. And if we look at the equations, this kind of makes sense, because if you're just thinking about movements in the x direction, the entire function looks like x squared plus some kind of constant. So the graph of that would look like an x squared parabola shape that has a local minimum. But if you're thinking of pure movements in the y direction, and you're just focused on that negative y squared component, the graph that you get for negative y squared is gonna look like an upside down parabola. Here, I'll draw that again. It's gonna look like an upside down parabola, and that's got a local maximum."}, {"video_title": "Saddle points.mp3", "Sentence": "So the graph of that would look like an x squared parabola shape that has a local minimum. But if you're thinking of pure movements in the y direction, and you're just focused on that negative y squared component, the graph that you get for negative y squared is gonna look like an upside down parabola. Here, I'll draw that again. It's gonna look like an upside down parabola, and that's got a local maximum. So it's kind of like the x and y directions disagree over whether this point, whether this point where you have a flat tangent plane should be a local maximum or a local minimum. And this is new to multivariable calculus. This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum."}, {"video_title": "Saddle points.mp3", "Sentence": "It's gonna look like an upside down parabola, and that's got a local maximum. So it's kind of like the x and y directions disagree over whether this point, whether this point where you have a flat tangent plane should be a local maximum or a local minimum. And this is new to multivariable calculus. This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum. It can't disagree, because there's only one input variable. There's only one, you know, x as the input variable for your graph. But once we have two, it's possible that they disagree."}, {"video_title": "Saddle points.mp3", "Sentence": "This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum. It can't disagree, because there's only one input variable. There's only one, you know, x as the input variable for your graph. But once we have two, it's possible that they disagree. And this kind of point has a special name. And the name is kind of after this graph that you're looking at. It's called a saddle point."}, {"video_title": "Saddle points.mp3", "Sentence": "But once we have two, it's possible that they disagree. And this kind of point has a special name. And the name is kind of after this graph that you're looking at. It's called a saddle point. Saddle point. And this is one of those rare times where I actually kind of like the terminology that mathematicians have given something, because this looks like a saddle, the sort of thing that you would put on a horse's back before riding it. So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point."}, {"video_title": "Saddle points.mp3", "Sentence": "It's called a saddle point. Saddle point. And this is one of those rare times where I actually kind of like the terminology that mathematicians have given something, because this looks like a saddle, the sort of thing that you would put on a horse's back before riding it. So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point. And if you're just looking at the graph, it's fine. You can recognize it visually. But oftentimes, if you're just given the formula of a function, it's some long thing."}, {"video_title": "Saddle points.mp3", "Sentence": "So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point. And if you're just looking at the graph, it's fine. You can recognize it visually. But oftentimes, if you're just given the formula of a function, it's some long thing. Without looking at the graph, how would you be able to tell, just by doing certain computations to the formula, whether or not it's a saddle point? And that comes down to something called the second partial derivative test, which I will talk about in the next few videos. See you then."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And so the first thing I'm going to do is figure out what d sigma is in terms of s and t, in terms of our parameters. So we can turn this whole thing into a double integral with respect to, or a double integral in the st plane. And remember, d sigma right over here, it's just a little chunk of the surface. It's a little area of the surface right over there. And we saw in previous videos, the ones where we learned what a surface integral even is, we saw that d sigma right over there, it is equivalent to the magnitude of the cross product of the partial of our parameterization with respect to one parameter, crossed with the parameterization with respect to the other parameter, times the differentials of each of the parameters. So this is what we're going to use right here. And it's a pretty simple looking statement."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "It's a little area of the surface right over there. And we saw in previous videos, the ones where we learned what a surface integral even is, we saw that d sigma right over there, it is equivalent to the magnitude of the cross product of the partial of our parameterization with respect to one parameter, crossed with the parameterization with respect to the other parameter, times the differentials of each of the parameters. So this is what we're going to use right here. And it's a pretty simple looking statement. But as we'll see, taking cross products tend to get a little bit hairy, especially cross products of three dimensional vectors. But we'll do it step by step. But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And it's a pretty simple looking statement. But as we'll see, taking cross products tend to get a little bit hairy, especially cross products of three dimensional vectors. But we'll do it step by step. But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant. So cosine of t isn't going to change. The partial of, or the derivative of cosine of s with respect to s is negative sine of s. So this is going to be equal to, I'll put the negative out front, negative cosine of t sine of s. I'm going to keep everything that has a t involved purple. Sine of s. And let me make, I don't know, I'll make the vectors orange."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant. So cosine of t isn't going to change. The partial of, or the derivative of cosine of s with respect to s is negative sine of s. So this is going to be equal to, I'll put the negative out front, negative cosine of t sine of s. I'm going to keep everything that has a t involved purple. Sine of s. And let me make, I don't know, I'll make the vectors orange. I, and then we'll plus, and once again, we're going to take the derivative with respect to s. Cosine of t is just a constant. Derivative of sine of s with respect to s is cosine of s. So it's going to be plus cosine of t cosine of s. Cosine of t cosine of s. J. And then plus the derivative of this with respect to s, well, this is just a constant."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Sine of s. And let me make, I don't know, I'll make the vectors orange. I, and then we'll plus, and once again, we're going to take the derivative with respect to s. Cosine of t is just a constant. Derivative of sine of s with respect to s is cosine of s. So it's going to be plus cosine of t cosine of s. Cosine of t cosine of s. J. And then plus the derivative of this with respect to s, well, this is just a constant. The derivative of 5 with respect to s would just be 0. This is the same thing. This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And then plus the derivative of this with respect to s, well, this is just a constant. The derivative of 5 with respect to s would just be 0. This is the same thing. This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0. So we could write even 0k. We write 0, I'll just write 0k right over there. And that's nice to see because that'll make our cross product a little bit more straightforward."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0. So we could write even 0k. We write 0, I'll just write 0k right over there. And that's nice to see because that'll make our cross product a little bit more straightforward. Now let's take the partial with respect to t. Now let's take the partial with respect to t. And we get, so the derivative of this with respect to t, now cosine of s is a constant. Derivative of cosine t with respect to t is negative sine of t. So this is going to be negative sine of t cosine of s. I'll do it in that, I'll use this blue i. And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And that's nice to see because that'll make our cross product a little bit more straightforward. Now let's take the partial with respect to t. Now let's take the partial with respect to t. And we get, so the derivative of this with respect to t, now cosine of s is a constant. Derivative of cosine t with respect to t is negative sine of t. So this is going to be negative sine of t cosine of s. I'll do it in that, I'll use this blue i. And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this. This is a painful problem. J plus derivative of sine of t with respect to t. We're taking the partial with respect to t. It's just cosine of t. So plus cosine of t. And now times the k unit vector. Now we're ready to take the cross product of these two characters right over here."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this. This is a painful problem. J plus derivative of sine of t with respect to t. We're taking the partial with respect to t. It's just cosine of t. So plus cosine of t. And now times the k unit vector. Now we're ready to take the cross product of these two characters right over here. And to take the cross products, let me write this down. So we're going to take the cross product of that with that. Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Now we're ready to take the cross product of these two characters right over here. And to take the cross products, let me write this down. So we're going to take the cross product of that with that. Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff. So maybe I'll take up about that much space so that I have space to work in. And I'll write my unit vectors up here. i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff. So maybe I'll take up about that much space so that I have space to work in. And I'll write my unit vectors up here. i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors. Take the determinant of this 3 by 3 matrix. The first row are just our unit vectors. The second row is the first vector that I'm taking the cross product of."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors. Take the determinant of this 3 by 3 matrix. The first row are just our unit vectors. The second row is the first vector that I'm taking the cross product of. So this is going to be negative. I'm just going to rewrite this right over here. So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "The second row is the first vector that I'm taking the cross product of. So this is going to be negative. I'm just going to rewrite this right over here. So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations. And then you have the next vector. That's the third row. Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations. And then you have the next vector. That's the third row. Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going. It's good practice. And even if you have to watch this whole thing to see how it's done, try to then do it again on your own. Because this is one of those things you really got to do yourself to really have it sit in."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going. It's good practice. And even if you have to watch this whole thing to see how it's done, try to then do it again on your own. Because this is one of those things you really got to do yourself to really have it sit in. Negative sine of t, sine of s. And then finally, cosine of t. So let's take the determinant now. So first we'll think about our I component. You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Because this is one of those things you really got to do yourself to really have it sit in. Negative sine of t, sine of s. And then finally, cosine of t. So let's take the determinant now. So first we'll think about our I component. You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here. So it's going to be I times something. I'll put the something in parentheses there. Normally you see the something in front of the I, but you can swap them there."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here. So it's going to be I times something. I'll put the something in parentheses there. Normally you see the something in front of the I, but you can swap them there. So it's going to be I times something. Ignore this column, that row. This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Normally you see the something in front of the I, but you can swap them there. So it's going to be I times something. Ignore this column, that row. This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater. Cosine squared of t, cosine of s. And then from that we would need to subtract 0 times this. But that's just going to be 0, so we're just left with that. Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater. Cosine squared of t, cosine of s. And then from that we would need to subtract 0 times this. But that's just going to be 0, so we're just left with that. Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices. Positive, negative, positive. So you write a negative J, negative coefficient, I guess, in front of the J, times something. And so ignore J's column, J's row."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices. Positive, negative, positive. So you write a negative J, negative coefficient, I guess, in front of the J, times something. And so ignore J's column, J's row. And so you have negative cosine of t, sine of s, times cosine of t. Well, that's going to be negative cosine squared of t times sine of s. Let me make sure I'm doing that right. Ignore that and that. It's going to be that times that."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And so ignore J's column, J's row. And so you have negative cosine of t, sine of s, times cosine of t. Well, that's going to be negative cosine squared of t times sine of s. Let me make sure I'm doing that right. Ignore that and that. It's going to be that times that. So negative cosine sine of s minus 0 times that. And so that's just going to be 0, so we can ignore it. And you have a negative times a negative here, so they can both become a positive."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "It's going to be that times that. So negative cosine sine of s minus 0 times that. And so that's just going to be 0, so we can ignore it. And you have a negative times a negative here, so they can both become a positive. And then finally you have the K component. And once again, you go back to positive there. Positive, negative, positive on the coefficients."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And you have a negative times a negative here, so they can both become a positive. And then finally you have the K component. And once again, you go back to positive there. Positive, negative, positive on the coefficients. That's just evaluating a 3 by 3 matrix. And then you have plus K times. And now this might get a little bit more involved, because we won't have the 0 to help us out."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Positive, negative, positive on the coefficients. That's just evaluating a 3 by 3 matrix. And then you have plus K times. And now this might get a little bit more involved, because we won't have the 0 to help us out. Ignore this row, ignore this column. Take the determinant of this sub 2 by 2. You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And now this might get a little bit more involved, because we won't have the 0 to help us out. Ignore this row, ignore this column. Take the determinant of this sub 2 by 2. You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out. So it's going to be cosine of t times sine squared of s. And then from that, we're going to subtract the product of these two things. But that product is going to be negative. So you subtract a negative."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out. So it's going to be cosine of t times sine squared of s. And then from that, we're going to subtract the product of these two things. But that product is going to be negative. So you subtract a negative. That's the same thing as adding a positive. So plus, and you're going to have cosine t sine t again. Plus cosine t. Let me scroll to the right a little bit."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So you subtract a negative. That's the same thing as adding a positive. So plus, and you're going to have cosine t sine t again. Plus cosine t. Let me scroll to the right a little bit. Plus cosine t sine t again. And that's times cosine squared of s. Now this is already looking pretty hairy, but it already looks like a simplification there. And that's where the colors are helpful."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Plus cosine t. Let me scroll to the right a little bit. Plus cosine t sine t again. And that's times cosine squared of s. Now this is already looking pretty hairy, but it already looks like a simplification there. And that's where the colors are helpful. I actually now have trouble doing math in anything other than kind of multiple pastel colors, because this actually makes it much easier to see some patterns. And so what we can do over here is we can factor out the cosine t sine t. And so this is equal to cosine t sine t times sine squared s plus cosine squared s. And this we know, the definition of the unit circle, this is just going to be equal to 1. So that was a significant simplification."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And that's where the colors are helpful. I actually now have trouble doing math in anything other than kind of multiple pastel colors, because this actually makes it much easier to see some patterns. And so what we can do over here is we can factor out the cosine t sine t. And so this is equal to cosine t sine t times sine squared s plus cosine squared s. And this we know, the definition of the unit circle, this is just going to be equal to 1. So that was a significant simplification. And so now we get our cross product. We get our cross product being equal to, let me just rewrite it all. Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So that was a significant simplification. And so now we get our cross product. We get our cross product being equal to, let me just rewrite it all. Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector. So that was pretty good, but we're still not done. We need to figure out the magnitude of this thing. Remember, d sigma simplified to the magnitude of this thing times ds dt."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector. So that was pretty good, but we're still not done. We need to figure out the magnitude of this thing. Remember, d sigma simplified to the magnitude of this thing times ds dt. So let's figure out what the magnitude of that is. And this is really the home stretch. So I'm really crossing my fingers that I don't make any careless mistakes now."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "Remember, d sigma simplified to the magnitude of this thing times ds dt. So let's figure out what the magnitude of that is. And this is really the home stretch. So I'm really crossing my fingers that I don't make any careless mistakes now. So the magnitude of all of this business is going to be equal to the square root. And I'm just going to have to square each of these terms and then add them up. The square root of the sum of the squares of each of those terms."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So I'm really crossing my fingers that I don't make any careless mistakes now. So the magnitude of all of this business is going to be equal to the square root. And I'm just going to have to square each of these terms and then add them up. The square root of the sum of the squares of each of those terms. So the square of this is going to be cosine squared squared is cosine to the fourth cosine of the fourth t cosine squared s plus cosine to the fourth t sine squared s. And I already see a pattern jumping out. Sine squared s plus cosine squared t sine squared t. Now, the first pattern I see is just this first part right over here. We can factor out a cosine to the fourth t. Then we get something like this happening again."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "The square root of the sum of the squares of each of those terms. So the square of this is going to be cosine squared squared is cosine to the fourth cosine of the fourth t cosine squared s plus cosine to the fourth t sine squared s. And I already see a pattern jumping out. Sine squared s plus cosine squared t sine squared t. Now, the first pattern I see is just this first part right over here. We can factor out a cosine to the fourth t. Then we get something like this happening again. So let's do that. So these first two terms are equal to cosine to the fourth t times cosine squared s plus sine squared s, which once again we know is just 1. So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "We can factor out a cosine to the fourth t. Then we get something like this happening again. So let's do that. So these first two terms are equal to cosine to the fourth t times cosine squared s plus sine squared s, which once again we know is just 1. So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them. So let's factor those out. So this is going to be equal to everything I'm doing is under the radical sign. So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them. So let's factor those out. So this is going to be equal to everything I'm doing is under the radical sign. So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1. All of this is under the radical sign. Maybe I'll keep drawing the radical signs here to make it clear that all of this is still on the radical sign. And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1. All of this is under the radical sign. Maybe I'll keep drawing the radical signs here to make it clear that all of this is still on the radical sign. And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward. So all of this is just going to be equal to cosine of t. So going back to what we wanted before, we want to rewrite what d sigma is. It's just cosine t ds dt. So let me write that down."}, {"video_title": "Surface integral example part 2 Calculating the surface differential   Khan Academy.mp3", "Sentence": "And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward. So all of this is just going to be equal to cosine of t. So going back to what we wanted before, we want to rewrite what d sigma is. It's just cosine t ds dt. So let me write that down. So d sigma, and then we can use this for the next part. d sigma is equal to cosine of t ds dt. And I'll see you in the next part."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "Three dimensional graphs are a way that we represent a certain kind of multivariable function, the kind that has two inputs, or rather a two dimensional input, and then a one dimensional output of some kind. So the one that I have pictured here is f of x, y equals x squared plus y squared. And before talking exactly about this graph, I think it'll be helpful if by analogy we take a look at two dimensional graphs and kind of remind ourselves how those work, what it is that we do, because it's pretty much the same thing in three dimensions, but it takes a little bit more visualization. So with two dimensional graphs, you have some kind of function. You know, let's say you have f of x is equal to x squared. And any time that you're visualizing a function, you're trying to understand the relationship between the inputs and the outputs. And here, those are both just numbers."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So with two dimensional graphs, you have some kind of function. You know, let's say you have f of x is equal to x squared. And any time that you're visualizing a function, you're trying to understand the relationship between the inputs and the outputs. And here, those are both just numbers. So you know, you input a number like two, and it's gonna output four. You know, you input negative one, it's gonna output one, and you're trying to understand all of the possible input-output pairs. And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And here, those are both just numbers. So you know, you input a number like two, and it's gonna output four. You know, you input negative one, it's gonna output one, and you're trying to understand all of the possible input-output pairs. And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible. And the way we go about this with graphs is you think of just plotting these actual pairs, right? So you're gonna plot the point. Let's say we're gonna plot the point two, four."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible. And the way we go about this with graphs is you think of just plotting these actual pairs, right? So you're gonna plot the point. Let's say we're gonna plot the point two, four. So we might kind of mark our graph to here. One, two, three, four. So you'd wanna mark somewhere here, two, four."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "Let's say we're gonna plot the point two, four. So we might kind of mark our graph to here. One, two, three, four. So you'd wanna mark somewhere here, two, four. And that represents an input-output pair. And if you do that with negative one, one, you go negative one, one. And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you'd wanna mark somewhere here, two, four. And that represents an input-output pair. And if you do that with negative one, one, you go negative one, one. And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve. And the implication for doing this is that we typically think of what is on the x-axis as being where the inputs live. You know, this would be, we think of it as the input one, and this is the input two, and so on. And then you think of the output as being the height, the height of the graph above each point."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve. And the implication for doing this is that we typically think of what is on the x-axis as being where the inputs live. You know, this would be, we think of it as the input one, and this is the input two, and so on. And then you think of the output as being the height, the height of the graph above each point. But this is kind of a consequence of the fact where we're just listing all of the pairs here. So now if we go to the world of multivariable functions, here, I'm not gonna show the graph right now. Let's just think we've got three-dimensional space at our disposal to do with what we will."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then you think of the output as being the height, the height of the graph above each point. But this is kind of a consequence of the fact where we're just listing all of the pairs here. So now if we go to the world of multivariable functions, here, I'm not gonna show the graph right now. Let's just think we've got three-dimensional space at our disposal to do with what we will. We still wanna understand the relationship between inputs and outputs of this guy. But in this case, inputs are something that we think of as a pair of points. So you might have a pair of points like one, two."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "Let's just think we've got three-dimensional space at our disposal to do with what we will. We still wanna understand the relationship between inputs and outputs of this guy. But in this case, inputs are something that we think of as a pair of points. So you might have a pair of points like one, two. And the output there is gonna be one squared plus two squared. And what that equals is five. So how do we visualize that?"}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you might have a pair of points like one, two. And the output there is gonna be one squared plus two squared. And what that equals is five. So how do we visualize that? Well, if we wanna pair these things together, the natural way to do that is to think of a triplet of some kind. So in this case, you'd wanna plot the triplet one, two, five. And to do that in three dimensions, we'll take a look over here."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So how do we visualize that? Well, if we wanna pair these things together, the natural way to do that is to think of a triplet of some kind. So in this case, you'd wanna plot the triplet one, two, five. And to do that in three dimensions, we'll take a look over here. We think of going one in the x direction. This axis here is the x axis. So we wanna move a distance one there."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And to do that in three dimensions, we'll take a look over here. We think of going one in the x direction. This axis here is the x axis. So we wanna move a distance one there. And we wanna go two in the y direction. So we kind of think of going a distance two there. And then five up."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So we wanna move a distance one there. And we wanna go two in the y direction. So we kind of think of going a distance two there. And then five up. And then that's gonna give us some kind of point. So we think of this point in space, and that's a given input-output pair. But we could do this for a lot."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then five up. And then that's gonna give us some kind of point. So we think of this point in space, and that's a given input-output pair. But we could do this for a lot. A couple different points that you might get if you start plotting various different ones will look something like this. And of course, there's infinitely many that you can do, and it'll take forever if you try to just draw each one in three dimensions. But what's very nice here is that, here, I'll get rid of those lines."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "But we could do this for a lot. A couple different points that you might get if you start plotting various different ones will look something like this. And of course, there's infinitely many that you can do, and it'll take forever if you try to just draw each one in three dimensions. But what's very nice here is that, here, I'll get rid of those lines. If you imagine doing this for all of the infinitely many pairs of inputs that you could possibly have, you end up drawing a surface. So in this case, the surface kind of looks like a three-dimensional parabola. That's no coincidence."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "But what's very nice here is that, here, I'll get rid of those lines. If you imagine doing this for all of the infinitely many pairs of inputs that you could possibly have, you end up drawing a surface. So in this case, the surface kind of looks like a three-dimensional parabola. That's no coincidence. It has to do with the fact that we're using x squared and y squared here. And now, the inputs, like one, two, we think of as being on the xy-plane. So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "That's no coincidence. It has to do with the fact that we're using x squared and y squared here. And now, the inputs, like one, two, we think of as being on the xy-plane. So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph. So it's very similar to two dimensions. We think of the input as being on one axis, and the height gives the output there. So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right?"}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph. So it's very similar to two dimensions. We think of the input as being on one axis, and the height gives the output there. So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right? So I'll draw it in red here. Let's say that we have our function, but I'm gonna change it so that it outputs one half of x squared plus y squared. What's gonna be the shape of the graph for that function?"}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right? So I'll draw it in red here. Let's say that we have our function, but I'm gonna change it so that it outputs one half of x squared plus y squared. What's gonna be the shape of the graph for that function? And what it means is the height of every point above this xy-plane is gonna have to get cut in half. So it's actually just a modification of what we already have, but everything kind of sloops on down to be about half of what it was. So in this case, instead of that height being five, it would be 2.5."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "What's gonna be the shape of the graph for that function? And what it means is the height of every point above this xy-plane is gonna have to get cut in half. So it's actually just a modification of what we already have, but everything kind of sloops on down to be about half of what it was. So in this case, instead of that height being five, it would be 2.5. You could imagine, let's say we did this, you know, it was even more extreme. Instead of saying one half, you cut it down by like 1 12th. Maybe I'll use the same color."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So in this case, instead of that height being five, it would be 2.5. You could imagine, let's say we did this, you know, it was even more extreme. Instead of saying one half, you cut it down by like 1 12th. Maybe I'll use the same color. By 1 12th. That would mean that everything, you know, sloops very flat, very flat and close to the xy-plane. So a graph being very close to the xy-plane like this corresponds with very small outputs."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "Maybe I'll use the same color. By 1 12th. That would mean that everything, you know, sloops very flat, very flat and close to the xy-plane. So a graph being very close to the xy-plane like this corresponds with very small outputs. And one thing that I'd like to caution you against, it's very tempting to try to think of every multivariable function as a graph, because we're so used to graphs in two dimensions, and we're so used to trying to find analogies between two dimensions and three dimensions directly. But the only reason that this works is because if you take the number of dimensions in the input, two dimensions, and then the number of dimensions in the output, one dimension, it was reasonable to fit all of that into three, which we could do. But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "So a graph being very close to the xy-plane like this corresponds with very small outputs. And one thing that I'd like to caution you against, it's very tempting to try to think of every multivariable function as a graph, because we're so used to graphs in two dimensions, and we're so used to trying to find analogies between two dimensions and three dimensions directly. But the only reason that this works is because if you take the number of dimensions in the input, two dimensions, and then the number of dimensions in the output, one dimension, it was reasonable to fit all of that into three, which we could do. But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output. That would require a five-dimensional graph, but we're not very good at visualizing things like that. So there's lots of other methods, and I think it's very important to kind of open your mind to what those might be. In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output. That would require a five-dimensional graph, but we're not very good at visualizing things like that. So there's lots of other methods, and I think it's very important to kind of open your mind to what those might be. In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space. That's called a contour map. Couple other ones, like parametric functions, you just look in the output space. Things like a vector space, you kind of look in the input space, but get all the outputs."}, {"video_title": "Introduction to 3d graphs   Multivariable calculus   Khan Academy.mp3", "Sentence": "In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space. That's called a contour map. Couple other ones, like parametric functions, you just look in the output space. Things like a vector space, you kind of look in the input space, but get all the outputs. There's lots of different ways. I'll go over those in the next few videos. And that's three-dimensional graphs."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is saying that there's no net flux across this surface right over here. Or if you sum up all of the divergences in this volume, you are getting zero. So why is that? Well, the simple way to think about it is when we took the divergence of F, this vector field F is hard to visualize, but the divergence of F is fairly easy to visualize. The divergence is equal to two times X. So over here, you're going to get, as you go further and further in this direction, as X becomes larger, your divergence becomes more and more positive. So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, the simple way to think about it is when we took the divergence of F, this vector field F is hard to visualize, but the divergence of F is fairly easy to visualize. The divergence is equal to two times X. So over here, you're going to get, as you go further and further in this direction, as X becomes larger, your divergence becomes more and more positive. So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there. And that's also true, obviously, as you go higher, because you're just changing the Z, you're not changing the X. So all over here, you have positive divergence. Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there. And that's also true, obviously, as you go higher, because you're just changing the Z, you're not changing the X. So all over here, you have positive divergence. Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence. I guess you could say in the positive X side of our octant. But then as you go in that side, on the other side, you have negative divergence. And this diagram is symmetric with respect to the ZY plane."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence. I guess you could say in the positive X side of our octant. But then as you go in that side, on the other side, you have negative divergence. And this diagram is symmetric with respect to the ZY plane. And so those divergences cancel them out. You would have had a positive flux across the surface, or a positive value right over here. Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this diagram is symmetric with respect to the ZY plane. And so those divergences cancel them out. You would have had a positive flux across the surface, or a positive value right over here. Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one. So let's just think about that region. So that region that would have been cut off right over here, that would have been cut off right over there. And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one. So let's just think about that region. So that region that would have been cut off right over here, that would have been cut off right over there. And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane. Now if we care about this, if we care about this volume, so we're essentially eliminating the rest of it. So let me try to eliminate it as best as I can, change the colors. So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane. Now if we care about this, if we care about this volume, so we're essentially eliminating the rest of it. So let me try to eliminate it as best as I can, change the colors. So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first. So if we eliminate the back part of it, and we're just dealing with, we're just dealing with it when X is positive, then our entire solution that we did in the last video would have been the exact same, except now X is going to vary between, instead of negative one and one, it'll vary between zero and one. And so our bounds of integration, X is going to go between zero and one, zero and one. And then in that situation, our final answer, this part, this would be between zero and one, that would all be zero, and we would be left with 3 1\u20442 minus 1 1\u20442."}, {"video_title": "Why we got zero flux in divergence theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first. So if we eliminate the back part of it, and we're just dealing with, we're just dealing with it when X is positive, then our entire solution that we did in the last video would have been the exact same, except now X is going to vary between, instead of negative one and one, it'll vary between zero and one. And so our bounds of integration, X is going to go between zero and one, zero and one. And then in that situation, our final answer, this part, this would be between zero and one, that would all be zero, and we would be left with 3 1\u20442 minus 1 1\u20442. 3 1\u20442 minus 1 1\u20442 is one, minus 1 6, which is just going to be 5 6. And so when you just think about this part of it, this side of it, this one that I've just drawn, you had a positive flux of 5 6. On the other side, you had a negative flux of 5 6, and then they canceled it out."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "In the last video, we took a look at this function, f of x, y equals x to the fourth minus four x squared plus y squared, which has the graph that you're looking at on the left. And we looked for all of the points where the gradient is equal to zero, which basically means both partial derivatives are equal to zero. And we solved that and we found that there were three different points, the origin zero, zero, and then square root of two, zero, and negative square root of two, zero, which corresponds to this origin here, which is a saddle point, and then these two local minima. And it seemed like we had a reasonable explanation for why this is a saddle point and why both of those are local minima. Because we took the second partial derivative with respect to x, I was kind of all over the board here, second partial derivative with respect to x, and found that when you evaluate it at x equals zero, you get a negative number, kind of indicating a negative concavity, so it should look like a maximum. And then when you do the same with the second partial derivative of y, man, I always do this, I always leave out the squared on that lower term there. Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And it seemed like we had a reasonable explanation for why this is a saddle point and why both of those are local minima. Because we took the second partial derivative with respect to x, I was kind of all over the board here, second partial derivative with respect to x, and found that when you evaluate it at x equals zero, you get a negative number, kind of indicating a negative concavity, so it should look like a maximum. And then when you do the same with the second partial derivative of y, man, I always do this, I always leave out the squared on that lower term there. Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y. So that's why, you know, this origin point looks like a saddle point, because the x direction and y direction disagree. And when you do this with the other points, they kind of both agree that it should look like a minimum. But I said that's not enough."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y. So that's why, you know, this origin point looks like a saddle point, because the x direction and y direction disagree. And when you do this with the other points, they kind of both agree that it should look like a minimum. But I said that's not enough. I said that you need to take into account the mixed partial derivative term. And to see why that's true, let me go ahead and pull up another example for you here. So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But I said that's not enough. I said that you need to take into account the mixed partial derivative term. And to see why that's true, let me go ahead and pull up another example for you here. So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually. But when we get the equation for this function, the equation is f of x, y is equal to x squared plus y squared minus four times x, y. Now let's go ahead and analyze the partial differential information of this function. We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually. But when we get the equation for this function, the equation is f of x, y is equal to x squared plus y squared minus four times x, y. Now let's go ahead and analyze the partial differential information of this function. We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y. And when we do the partial derivative with respect to y, very similarly, we're gonna get two y when we differentiate y squared, two y, and now we subtract minus four x, because x looks like the constant and y looks like the variable, minus four x. Now when we plug in, x and y are each equal to zero, you know, we plug in the origin point to both of these functions, we see that they're equal to zero, because x is zero, y is zero, this guy goes to zero. Similarly over here, that goes to zero."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y. And when we do the partial derivative with respect to y, very similarly, we're gonna get two y when we differentiate y squared, two y, and now we subtract minus four x, because x looks like the constant and y looks like the variable, minus four x. Now when we plug in, x and y are each equal to zero, you know, we plug in the origin point to both of these functions, we see that they're equal to zero, because x is zero, y is zero, this guy goes to zero. Similarly over here, that goes to zero. So we will indeed get a flat tangent plane at the origin. But now let's take a look at the second partial derivatives. If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "Similarly over here, that goes to zero. So we will indeed get a flat tangent plane at the origin. But now let's take a look at the second partial derivatives. If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us. And then similarly, when we take the second partial derivative with respect to y, always forget that squared on the bottom, we also get a constant positive two, because this x does nothing for us when we take the derivative with respect to y. So your constant positive two. So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us. And then similarly, when we take the second partial derivative with respect to y, always forget that squared on the bottom, we also get a constant positive two, because this x does nothing for us when we take the derivative with respect to y. So your constant positive two. So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum. But when we look at the graph, this isn't true. It's not a local minimum, it's a saddle point. So what this tells us is that these two second partial derivatives aren't enough, we need more information."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum. But when we look at the graph, this isn't true. It's not a local minimum, it's a saddle point. So what this tells us is that these two second partial derivatives aren't enough, we need more information. And what it kind of comes down to is that this last term here, minus four x, what, oh, actually, I think I made a mistake. I think I meant to make this plus four x, y. So let's see."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So what this tells us is that these two second partial derivatives aren't enough, we need more information. And what it kind of comes down to is that this last term here, minus four x, what, oh, actually, I think I made a mistake. I think I meant to make this plus four x, y. So let's see. Plus four x, y, which would influence these. It actually won't make a difference, it still gives a saddle point. But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So let's see. Plus four x, y, which would influence these. It actually won't make a difference, it still gives a saddle point. But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum. And just to give a loose intuition for what's going on here, if instead of writing four here, I wrote, I'm gonna write the variable p, okay, and p is just gonna be some number, and I'm gonna move that variable around. I'm gonna basically let it range from zero up to four. So right now, as you're looking at it, it's sitting at four."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum. And just to give a loose intuition for what's going on here, if instead of writing four here, I wrote, I'm gonna write the variable p, okay, and p is just gonna be some number, and I'm gonna move that variable around. I'm gonna basically let it range from zero up to four. So right now, as you're looking at it, it's sitting at four. So I'm gonna pull it back and kind of let it range back to zero just to see how this influences the graph. And we see that once we pull it back to zero, we get something where it kind of reflects what you would expect, where from the x direction, it's a positive smiley face, from the y direction, it's also a positive smiley face, and everything's good. It looks like a local minimum, and it is."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So right now, as you're looking at it, it's sitting at four. So I'm gonna pull it back and kind of let it range back to zero just to see how this influences the graph. And we see that once we pull it back to zero, we get something where it kind of reflects what you would expect, where from the x direction, it's a positive smiley face, from the y direction, it's also a positive smiley face, and everything's good. It looks like a local minimum, and it is. And even as you let p range more and more, and here, p is around, I'm guessing right now it's around 1.5, you get something that's still a local minimum. It's a positive concavity in all directions. But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "It looks like a local minimum, and it is. And even as you let p range more and more, and here, p is around, I'm guessing right now it's around 1.5, you get something that's still a local minimum. It's a positive concavity in all directions. But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point. And again, this is entirely the coefficient in front of the xy term. It has nothing to do with the x squared or the y squared. So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point. And again, this is entirely the coefficient in front of the xy term. It has nothing to do with the x squared or the y squared. So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point. And in a moment, it'll become clear that that critical point happens when p is equal to two. So right here, it's gonna be when p equals two, it kind of passes from making things a local minimum to a saddle point. And let me show you the test which will tell us why this is true."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point. And in a moment, it'll become clear that that critical point happens when p is equal to two. So right here, it's gonna be when p equals two, it kind of passes from making things a local minimum to a saddle point. And let me show you the test which will tell us why this is true. So the full reasoning behind this test is something that I'll get to in later videos, but right now, I just want to kind of have it on the table and teach you what it is and how to use it. So this is called the second partial derivative test. Second partial derivative test."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And let me show you the test which will tell us why this is true. So the full reasoning behind this test is something that I'll get to in later videos, but right now, I just want to kind of have it on the table and teach you what it is and how to use it. So this is called the second partial derivative test. Second partial derivative test. I'll just write derivative test since I'm a slow writer. And basically, what it says is if you found a point where the gradient of your function at this point, and I'll write it kind of x-naught, y-naught is our point. If you found where it equals zero, then calculate the following value."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "Second partial derivative test. I'll just write derivative test since I'm a slow writer. And basically, what it says is if you found a point where the gradient of your function at this point, and I'll write it kind of x-naught, y-naught is our point. If you found where it equals zero, then calculate the following value. You'll take the second partial derivative with respect to x twice. So here, I'm just using that subscript notation which is completely the same as saying second partial derivative with respect to x twice, just different choice in notation. And you evaluate it at this point, x-naught, y-naught."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "If you found where it equals zero, then calculate the following value. You'll take the second partial derivative with respect to x twice. So here, I'm just using that subscript notation which is completely the same as saying second partial derivative with respect to x twice, just different choice in notation. And you evaluate it at this point, x-naught, y-naught. Then you multiply that by the second partial derivative with respect to y, evaluate it at that same point, y-naught. And then you subtract off the mixed partial derivative, the one where first you do it with respect to x, then with respect to y, or in the other order, it doesn't really matter, as long as you take it with respect to both variables. You subtract off that guy squared."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And you evaluate it at this point, x-naught, y-naught. Then you multiply that by the second partial derivative with respect to y, evaluate it at that same point, y-naught. And then you subtract off the mixed partial derivative, the one where first you do it with respect to x, then with respect to y, or in the other order, it doesn't really matter, as long as you take it with respect to both variables. You subtract off that guy squared. So what you do is you compute this entire value and it's gonna give you some kind of number, and let's give it a name. Let's name it h. And if it's the case that h is greater than zero, then you have either a max or a min. You're not sure which one yet."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "You subtract off that guy squared. So what you do is you compute this entire value and it's gonna give you some kind of number, and let's give it a name. Let's name it h. And if it's the case that h is greater than zero, then you have either a max or a min. You're not sure which one yet. And it's a max or a min. And you can tell whether it's a maximum or a minimum basically by looking at one of these partial derivative with respect to x twice or with respect to y twice and kind of getting a feel for the concavity there. If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "You're not sure which one yet. And it's a max or a min. And you can tell whether it's a maximum or a minimum basically by looking at one of these partial derivative with respect to x twice or with respect to y twice and kind of getting a feel for the concavity there. If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum. And the fact that this entire value h is greater than zero is what you need to tell you that you can just do that. You can look at the concavity with respect to one of those guys and that'll tell you the information you need about the entire graph. But if h is less than zero, if h is less than zero, then you definitely have a saddle point."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum. And the fact that this entire value h is greater than zero is what you need to tell you that you can just do that. You can look at the concavity with respect to one of those guys and that'll tell you the information you need about the entire graph. But if h is less than zero, if h is less than zero, then you definitely have a saddle point. Saddle point. And if h is purely equal to zero, if you get that unlucky case, then you don't know. Then the second partial derivative test isn't enough to determine."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But if h is less than zero, if h is less than zero, then you definitely have a saddle point. Saddle point. And if h is purely equal to zero, if you get that unlucky case, then you don't know. Then the second partial derivative test isn't enough to determine. But almost all cases, you'll find it either is purely greater than zero or purely less than zero. So as an example, let's see what that looks like in the case of the specific function we started with, where p is some constant that I was letting kind of range from zero to four when I was animating it here. But you should just think of p as being some number."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "Then the second partial derivative test isn't enough to determine. But almost all cases, you'll find it either is purely greater than zero or purely less than zero. So as an example, let's see what that looks like in the case of the specific function we started with, where p is some constant that I was letting kind of range from zero to four when I was animating it here. But you should just think of p as being some number. Well, in that case, this value h that we plug in, and let's say we're plugging in at the origin, right? We're analyzing at the origin. Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But you should just think of p as being some number. Well, in that case, this value h that we plug in, and let's say we're plugging in at the origin, right? We're analyzing at the origin. Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row. And both of those, when we computed those, were just constants two. They were equal to two everywhere. And in particular, they're equal to two at the origin."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row. And both of those, when we computed those, were just constants two. They were equal to two everywhere. And in particular, they're equal to two at the origin. So we can go ahead and just plug in those. And we see that this is two times two. And then now we need to subtract off the mixed partial derivative squared."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And in particular, they're equal to two at the origin. So we can go ahead and just plug in those. And we see that this is two times two. And then now we need to subtract off the mixed partial derivative squared. So if we go ahead and compute that, where we take the derivative with respect to x and then y, or y and then x, let's say we started with the partial derivative with respect to x. When we take this derivative with respect to y, we're gonna get this constant term that's sitting in front of the y. But really, it's whatever this constant p happened to equal."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And then now we need to subtract off the mixed partial derivative squared. So if we go ahead and compute that, where we take the derivative with respect to x and then y, or y and then x, let's say we started with the partial derivative with respect to x. When we take this derivative with respect to y, we're gonna get this constant term that's sitting in front of the y. But really, it's whatever this constant p happened to equal. And you might be able to see that just looking at this function, that when you take the mixed partial derivative, it's gonna be the coefficient in front of the xy term. Because it's kind of like, first you do the derivative with respect to x and the x goes away, and then with respect to y, and that y goes away. And you're just left with a constant."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "But really, it's whatever this constant p happened to equal. And you might be able to see that just looking at this function, that when you take the mixed partial derivative, it's gonna be the coefficient in front of the xy term. Because it's kind of like, first you do the derivative with respect to x and the x goes away, and then with respect to y, and that y goes away. And you're just left with a constant. So what you end up getting here in the second partial derivative test, when we take that value, which is p, which might equal four or zero or whatever we happen to have it as, and we square that. We square that. That's gonna be the value that we analyze."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And you're just left with a constant. So what you end up getting here in the second partial derivative test, when we take that value, which is p, which might equal four or zero or whatever we happen to have it as, and we square that. We square that. That's gonna be the value that we analyze. So in the case where p was equal to zero, if we go over here and we scale so that p is completely equal to zero, then our entire value h, h would equal four. And because h is positive, it's definitely a maximum or a minimum. And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "That's gonna be the value that we analyze. So in the case where p was equal to zero, if we go over here and we scale so that p is completely equal to zero, then our entire value h, h would equal four. And because h is positive, it's definitely a maximum or a minimum. And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum. Because positive concavity gives local minimum. But in the other case, where let's say we let p range such that p is all the way equal to four, in formulas, what that means for us, when we let p equal four, is we're taking two times two minus four squared. So we're taking two times two minus 16."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum. Because positive concavity gives local minimum. But in the other case, where let's say we let p range such that p is all the way equal to four, in formulas, what that means for us, when we let p equal four, is we're taking two times two minus four squared. So we're taking two times two minus 16. What that would imply, sorry about getting kind of scrunched on the board here, is that h is equal to, let's see, four minus 16, negative 12. So I'm just gonna erase this to kind of clear up some room. So when p equals four, this is negative 12."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So we're taking two times two minus 16. What that would imply, sorry about getting kind of scrunched on the board here, is that h is equal to, let's see, four minus 16, negative 12. So I'm just gonna erase this to kind of clear up some room. So when p equals four, this is negative 12. And in fact, this kind of explains the crossover point for when it goes from being local minimum to a saddle point. It's gonna be at that point where this entire expression is equal to zero. And you can see that happens when p equals two."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "So when p equals four, this is negative 12. And in fact, this kind of explains the crossover point for when it goes from being local minimum to a saddle point. It's gonna be at that point where this entire expression is equal to zero. And you can see that happens when p equals two. So over here, the crossover point, when it kind of goes from being a local minimum to a saddle point, is at p equals two. And when p perfectly equals two, let's see, so about here, the second partial derivative test isn't gonna be enough to tell us anything. It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "And you can see that happens when p equals two. So over here, the crossover point, when it kind of goes from being a local minimum to a saddle point, is at p equals two. And when p perfectly equals two, let's see, so about here, the second partial derivative test isn't gonna be enough to tell us anything. It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point. And in this particular case, that corresponds to the fact that the graph is perfectly flat in one direction, and a minimum in another direction. In other cases, it might mean something different, and I'll probably make a video just about that special case when this whole value is equal to zero. But for now, all that I wanna emphasize is what this test is, where you take all three second partial derivatives, and you kind of multiply together the two pure second partial derivatives, where you do x and then x, and the one where you do y and then y."}, {"video_title": "Second partial derivative test.mp3", "Sentence": "It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point. And in this particular case, that corresponds to the fact that the graph is perfectly flat in one direction, and a minimum in another direction. In other cases, it might mean something different, and I'll probably make a video just about that special case when this whole value is equal to zero. But for now, all that I wanna emphasize is what this test is, where you take all three second partial derivatives, and you kind of multiply together the two pure second partial derivatives, where you do x and then x, and the one where you do y and then y. You multiply those together, and then you subtract off that mixed partial derivative squared. And in the next video, I'll try to give a little bit more intuition for, you know, where this whole formula comes from, why it's not completely random, and why taking this and analyzing whether it's greater than zero or less than zero is a reasonable thing to do for analyzing whether a point that you're looking at is a local minimum or a local maximum or a saddle point. See you then."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say we take the line integral over some curve C. I'll define the curve in a second. Of x squared plus y squared dx plus 2xy dy. Well, this might look very familiar. This was very similar to what we saw last time, except last time we had a closed line integral. So this is not a closed line integral. And our curve C, the parameterization, is x is equal to cosine of t, y is equal to sine of t. So far, it looks like sint. Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This was very similar to what we saw last time, except last time we had a closed line integral. So this is not a closed line integral. And our curve C, the parameterization, is x is equal to cosine of t, y is equal to sine of t. So far, it looks like sint. Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video. But instead of t going from 0 to 2 pi, we're going to have t go from 0 to pi. t is greater than or equal to 0 is less than or equal to pi. So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video. But instead of t going from 0 to 2 pi, we're going to have t go from 0 to pi. t is greater than or equal to 0 is less than or equal to pi. So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis. That is my x-axis. So now our path isn't all the way around the unit circle. Our path, our curve C now, just starts at t is equal to 0."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis. That is my x-axis. So now our path isn't all the way around the unit circle. Our path, our curve C now, just starts at t is equal to 0. You can imagine t is almost the angle. t is equal to 0, and we're going to go all the way to pi. So that's what our path is right now in this example."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Our path, our curve C now, just starts at t is equal to 0. You can imagine t is almost the angle. t is equal to 0, and we're going to go all the way to pi. So that's what our path is right now in this example. So it's not a curved path. It's not a closed path. So we can't just show that f in this example, and we're going to re-look at what f looks like."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's what our path is right now in this example. So it's not a curved path. It's not a closed path. So we can't just show that f in this example, and we're going to re-look at what f looks like. Hey, if that's a conservative vector field, if it's a closed loop, it equals 0. This isn't a closed loop, so we can't apply that. But let's see if we can apply some of our other tools."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we can't just show that f in this example, and we're going to re-look at what f looks like. Hey, if that's a conservative vector field, if it's a closed loop, it equals 0. This isn't a closed loop, so we can't apply that. But let's see if we can apply some of our other tools. So like we saw in the last video, this might look a little bit foreign to you. But if you say that f is equal to that times i plus that times j, then it might look a little bit more familiar. If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's see if we can apply some of our other tools. So like we saw in the last video, this might look a little bit foreign to you. But if you say that f is equal to that times i plus that times j, then it might look a little bit more familiar. If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j. And dr, I don't even have to look at this right now, dr you can always write it as dx times i plus dy times j. You'll immediately see if you take the dot product of these two things, if you take f dot dr, they're both vector valued, vector valued differential, vector valued field, or vector valued function. If you take f dot dr, you'll get this right here."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j. And dr, I don't even have to look at this right now, dr you can always write it as dx times i plus dy times j. You'll immediately see if you take the dot product of these two things, if you take f dot dr, they're both vector valued, vector valued differential, vector valued field, or vector valued function. If you take f dot dr, you'll get this right here. You'll get what we have inside of the integral. You'll get that right there. That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you take f dot dr, you'll get this right here. You'll get what we have inside of the integral. You'll get that right there. That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that. So our integral, we can rewrite it as this, along this curve of f dot dr, where this is our f. Now, we still might want to ask ourselves, is this a conservative field? Or does it have a potential? Is f equal to the gradient of some function capital F?"}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that. So our integral, we can rewrite it as this, along this curve of f dot dr, where this is our f. Now, we still might want to ask ourselves, is this a conservative field? Or does it have a potential? Is f equal to the gradient of some function capital F? You could write the gradient like that, because it creates a vector. This is a vector too. Is this true?"}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Is f equal to the gradient of some function capital F? You could write the gradient like that, because it creates a vector. This is a vector too. Is this true? And we saw in the last video it is. So we'll redo it a little bit fast this time. Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Is this true? And we saw in the last video it is. So we'll redo it a little bit fast this time. Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0. But if this is true, then we know that the integral is path independent. And we'll know that this is going to be equal to capital F. If we say that t is going from, well, in this case, t is going from 0 to pi, we could say that this is going to be equal to capital F of pi minus capital F of 0. Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0. But if this is true, then we know that the integral is path independent. And we'll know that this is going to be equal to capital F. If we say that t is going from, well, in this case, t is going from 0 to pi, we could say that this is going to be equal to capital F of pi minus capital F of 0. Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here. We could also write that this is equal to f of x of pi, y of pi, minus f of x of 0, y of 0. That's what I mean when I say f of pi. If we were to write f purely as a function of t. But we know that this capital F is going to be a function."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here. We could also write that this is equal to f of x of pi, y of pi, minus f of x of 0, y of 0. That's what I mean when I say f of pi. If we were to write f purely as a function of t. But we know that this capital F is going to be a function. It's a scalar function defined on x, y. So we could say f of x of pi, y of pi. These are the t's now."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we were to write f purely as a function of t. But we know that this capital F is going to be a function. It's a scalar function defined on x, y. So we could say f of x of pi, y of pi. These are the t's now. These are all equivalent things. So if it is path dependent, we can find our f. We can just evaluate this thing by just taking our f, evaluating it at these two points. At this point and at that point right there."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These are the t's now. These are all equivalent things. So if it is path dependent, we can find our f. We can just evaluate this thing by just taking our f, evaluating it at these two points. At this point and at that point right there. Because it would be path independent. If this is a conservative, if this has a potential function, if this is the gradient of another scalar field, then this is a conservative vector field. And its line integral is path independent."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "At this point and at that point right there. Because it would be path independent. If this is a conservative, if this has a potential function, if this is the gradient of another scalar field, then this is a conservative vector field. And its line integral is path independent. It's only dependent on that point and that point. So let's see if we can find our f. So I'm going to do exactly what we did in the last video. If you watched that last video, it might be a little bit monotonous."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And its line integral is path independent. It's only dependent on that point and that point. So let's see if we can find our f. So I'm going to do exactly what we did in the last video. If you watched that last video, it might be a little bit monotonous. But I'll do it a little bit faster here. So we know that the partial of f with respect to x is going to have to be equal to this right here. So that's x squared plus y squared."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you watched that last video, it might be a little bit monotonous. But I'll do it a little bit faster here. So we know that the partial of f with respect to x is going to have to be equal to this right here. So that's x squared plus y squared. Which tells us if we take the antiderivative with respect to x, that f of x, y is going to have to be equal to x to the third over 3 plus xy squared. y squared is just a constant in terms of x. Plus f of y."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's x squared plus y squared. Which tells us if we take the antiderivative with respect to x, that f of x, y is going to have to be equal to x to the third over 3 plus xy squared. y squared is just a constant in terms of x. Plus f of y. There might be some function of y that when you take the partial with respect to x, it just disappears. And then we know that the partial of f with respect to y has got to be equal to that thing or that thing. We're saying that this is the gradient of f. So this has to be the partial with respect to y."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus f of y. There might be some function of y that when you take the partial with respect to x, it just disappears. And then we know that the partial of f with respect to y has got to be equal to that thing or that thing. We're saying that this is the gradient of f. So this has to be the partial with respect to y. 2xy. And you might want to watch the other video. I go through this a little bit slower in that one."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're saying that this is the gradient of f. So this has to be the partial with respect to y. 2xy. And you might want to watch the other video. I go through this a little bit slower in that one. So the antiderivative of this with respect to y. So we get f of x, y would be equal to xy squared plus some function of x. Now, we did this in the last video."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I go through this a little bit slower in that one. So the antiderivative of this with respect to y. So we get f of x, y would be equal to xy squared plus some function of x. Now, we did this in the last video. These two things have to be the same thing in order for the gradient of capital F to be lowercase f. And we have xy squared, xy squared. We have a function of x. We have a function purely of x."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, we did this in the last video. These two things have to be the same thing in order for the gradient of capital F to be lowercase f. And we have xy squared, xy squared. We have a function of x. We have a function purely of x. And then we don't have a function purely of y here. So this thing right here must be 0. So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have a function purely of x. And then we don't have a function purely of y here. So this thing right here must be 0. So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared. So we know that lowercase f is definitely conservative. It is path independent. It has its potential."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared. So we know that lowercase f is definitely conservative. It is path independent. It has its potential. It is the gradient of this thing right here. And so to solve our integral, this was a 0. To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It has its potential. It is the gradient of this thing right here. And so to solve our integral, this was a 0. To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0. Evaluate at both points and then subtract the two. So let's do that. So let me just figure out."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0. Evaluate at both points and then subtract the two. So let's do that. So let me just figure out. So x was cosine of t, y is sine of t. Let me rewrite it down here. So x is equal to cosine of t, y is equal to sine of t. So x of 0 is equal to cosine of 0, which is equal to 1. x of pi is equal to cosine of pi, which is equal to minus 1. y of 0 is sine of 0, which is 0. y of pi, which is equal to sine of pi, which is equal to 0. So f of x of pi, y of pi, this is the same thing."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me just figure out. So x was cosine of t, y is sine of t. Let me rewrite it down here. So x is equal to cosine of t, y is equal to sine of t. So x of 0 is equal to cosine of 0, which is equal to 1. x of pi is equal to cosine of pi, which is equal to minus 1. y of 0 is sine of 0, which is 0. y of pi, which is equal to sine of pi, which is equal to 0. So f of x of pi, y of pi, this is the same thing. So let me rewrite this. Our integral is simplified to our integral along that path of f dot dr is going to be equal to capital F of x of pi. x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So f of x of pi, y of pi, this is the same thing. So let me rewrite this. Our integral is simplified to our integral along that path of f dot dr is going to be equal to capital F of x of pi. x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0. And so what is this equal to? Well, just remember, this right here, it's the same thing as that right there. That is x of pi."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0. And so what is this equal to? Well, just remember, this right here, it's the same thing as that right there. That is x of pi. That is y of pi. That's that term right there. You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is x of pi. That is y of pi. That's that term right there. You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear. So this is just straightforward to evaluate. What is F of minus 1, 0? x is minus 1, y is 0."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear. So this is just straightforward to evaluate. What is F of minus 1, 0? x is minus 1, y is 0. So it's going to be minus 1 to the third power, that's our x, over 3. So it's minus 1 over 3. It's going to be minus 1 third plus minus 1 times 0 squared."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x is minus 1, y is 0. So it's going to be minus 1 to the third power, that's our x, over 3. So it's minus 1 over 3. It's going to be minus 1 third plus minus 1 times 0 squared. So that's just going to be a 0. In both cases, the y is 0, so this term's going to disappear. So we can ignore that."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be minus 1 third plus minus 1 times 0 squared. So that's just going to be a 0. In both cases, the y is 0, so this term's going to disappear. So we can ignore that. And then we have minus F of 1, 0. So you put a 1 here, 1 to the third over 3, that is 1 third plus 1 times 0 squared, that's just 0. So this is going to be equal to minus 1 third minus 1 third is equal to minus 2 thirds."}, {"video_title": "Second example of line integral of conservative vector field   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we can ignore that. And then we have minus F of 1, 0. So you put a 1 here, 1 to the third over 3, that is 1 third plus 1 times 0 squared, that's just 0. So this is going to be equal to minus 1 third minus 1 third is equal to minus 2 thirds. And we're done. And once again, because this is a conservative vector field and it's path independent, we really didn't have to mess with the cosines of t's and sines of t's when we actually took our antiderivative. We just had to find the potential function and evaluate it at the two endpoints to get the answer of our line integral, minus 2 thirds."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll write ds in magenta. So first let's conceptualize what this is even saying, and then let's manipulate it a little bit to see if we can come up with an interesting conclusion. We actually will manipulate it, we'll use Green's theorem, and we're going to actually come up with a two-dimensional version of the divergence theorem, which all sounds very complicated, but hopefully we can get a little bit of an intuition for it as to why it is actually a little bit of common sense. So first let's just think about this. Let me draw a coordinate plane here. So let me do it in white. So this right over here, that's our y-axis, that over there is our x-axis."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So first let's just think about this. Let me draw a coordinate plane here. So let me do it in white. So this right over here, that's our y-axis, that over there is our x-axis. Let me draw ourselves my curve. So my curve might look something like, I'll do it in the blue color. So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this right over here, that's our y-axis, that over there is our x-axis. Let me draw ourselves my curve. So my curve might look something like, I'll do it in the blue color. So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that. And now we have our vector field. And just a reminder, we've seen this multiple times. My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So my curve might look something like this, my contour, going in the positive counterclockwise direction, just like that. And now we have our vector field. And just a reminder, we've seen this multiple times. My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point. So some function of x and y times i, plus some other scalar function of x and y times j. And so if you give me any point, there will be an associated vector with it. Any point, there's an associated vector, depending on how we define this function."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "My vector field will associate a vector with any point on the xy-plane, and it can be defined as some function of x and y, actually I'll call that p, some function of x and y times the i unit vector, so it says what the i component of the vector field is for any x and y point, and then what the j component, or what we multiply the j component by, or the vertical component by, for any x and y point. So some function of x and y times i, plus some other scalar function of x and y times j. And so if you give me any point, there will be an associated vector with it. Any point, there's an associated vector, depending on how we define this function. But this expression right over here, we're taking a line integral. We care specifically about the points along this curve, along this contour right over here. And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Any point, there's an associated vector, depending on how we define this function. But this expression right over here, we're taking a line integral. We care specifically about the points along this curve, along this contour right over here. And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces. So if we just take f dot n, so let's just think about a point on this curve. So a point on this curve, maybe this point right over here. So associated with that point, there is a vector."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's think about what this is actually, what this piece right over here is actually telling us before we sum up all of these infinitesimally small pieces. So if we just take f dot n, so let's just think about a point on this curve. So a point on this curve, maybe this point right over here. So associated with that point, there is a vector. And that's what the vector field does. So f might look something like that. Right over there."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So associated with that point, there is a vector. And that's what the vector field does. So f might look something like that. Right over there. So that might be f at that point. And then we're going to dot it with the unit normal vector at that point. So the unit normal vector might look something like that."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Right over there. So that might be f at that point. And then we're going to dot it with the unit normal vector at that point. So the unit normal vector might look something like that. That would be n hat at that point. This is the vector field at that point. When you take the dot product, you get a scalar quantity."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the unit normal vector might look something like that. That would be n hat at that point. This is the vector field at that point. When you take the dot product, you get a scalar quantity. You essentially just get a number, and you might remember it, and there are several videos where we go into detail about this. But that tells you how much those two vectors go together. It's essentially, if they're completely orthogonal to each other, you're going to get zero."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When you take the dot product, you get a scalar quantity. You essentially just get a number, and you might remember it, and there are several videos where we go into detail about this. But that tells you how much those two vectors go together. It's essentially, if they're completely orthogonal to each other, you're going to get zero. And if they go completely in the same direction, it's essentially you're just going to multiply their magnitudes times each other. And since you have a unit normal vector here, what this is essentially going to give you is the magnitude of the vector field f that goes in the normal direction. So think of it this way."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's essentially, if they're completely orthogonal to each other, you're going to get zero. And if they go completely in the same direction, it's essentially you're just going to multiply their magnitudes times each other. And since you have a unit normal vector here, what this is essentially going to give you is the magnitude of the vector field f that goes in the normal direction. So think of it this way. So let's think about the component of this that goes in the normal direction. It might look something like that. This would be the component that goes in the tangential direction."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So think of it this way. So let's think about the component of this that goes in the normal direction. It might look something like that. This would be the component that goes in the tangential direction. So this expression right over here is going to give us the magnitude of this vector. Let me write this down. This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This would be the component that goes in the tangential direction. So this expression right over here is going to give us the magnitude of this vector. Let me write this down. This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector. And then we're going to multiply that times a very infinitely small length of our contour, of our curve, right around that point. So we're going to multiply that thing times this right over here. And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant?"}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right over here is the magnitude of the component of f in the normal direction, or in the same direction as that unit normal vector. And then we're going to multiply that times a very infinitely small length of our contour, of our curve, right around that point. So we're going to multiply that thing times this right over here. And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant? Or what could be the intuition for what this expression is even measuring? And to think about that, I always visualize this is in two dimensions. You will later do things like this in three dimensions."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so you might say, well, OK, I kind of get what that is saying, but how could this ever be physically relevant? Or what could be the intuition for what this expression is even measuring? And to think about that, I always visualize this is in two dimensions. You will later do things like this in three dimensions. Imagine this is a two-dimensional universe, and we're studying gases. And so you have all these gas particles in a two-dimensional universe, so they only can have kind of an x and y coordinate. And this vector field is essentially telling you the velocity at any point there."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You will later do things like this in three dimensions. Imagine this is a two-dimensional universe, and we're studying gases. And so you have all these gas particles in a two-dimensional universe, so they only can have kind of an x and y coordinate. And this vector field is essentially telling you the velocity at any point there. So this is the velocity of the particles at this point. This is the velocity of the particles at that point. That is the velocity of the particles at that point."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this vector field is essentially telling you the velocity at any point there. So this is the velocity of the particles at this point. This is the velocity of the particles at that point. That is the velocity of the particles at that point. And so when you take f right on this curve, that's the velocity of the particles at that point. They're going in that direction. When you dot it with n, it tells you essentially the speed going straight out, right at that point."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is the velocity of the particles at that point. And so when you take f right on this curve, that's the velocity of the particles at that point. They're going in that direction. When you dot it with n, it tells you essentially the speed going straight out, right at that point. And when you multiply that times ds, you're essentially saying at any given moment, how fast or at any given, at that point right over there on the curve, how fast are the particles exiting the curve? And so if you were to sum up all of them, which is essentially what this integral is doing, with this line integral, it's essentially saying how fast are the particles exiting this contour? Or even entering the contour if you get a negative number."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When you dot it with n, it tells you essentially the speed going straight out, right at that point. And when you multiply that times ds, you're essentially saying at any given moment, how fast or at any given, at that point right over there on the curve, how fast are the particles exiting the curve? And so if you were to sum up all of them, which is essentially what this integral is doing, with this line integral, it's essentially saying how fast are the particles exiting this contour? Or even entering the contour if you get a negative number. But since we're taking the unit vector that goes, the unit normal vector that is outward pointing, it's saying how fast are they exiting this thing? If you get a negative number, that means there might be some net entrance. So this whole expression is, if you take that analogy, it doesn't have to have that physical representation."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or even entering the contour if you get a negative number. But since we're taking the unit vector that goes, the unit normal vector that is outward pointing, it's saying how fast are they exiting this thing? If you get a negative number, that means there might be some net entrance. So this whole expression is, if you take that analogy, it doesn't have to have that physical representation. How fast are particles, our two-dimensional gas particle, exiting the contour? And in the future, you can do it in three dimensions where you have a surface and you can say how fast are things exiting that surface? And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole expression is, if you take that analogy, it doesn't have to have that physical representation. How fast are particles, our two-dimensional gas particle, exiting the contour? And in the future, you can do it in three dimensions where you have a surface and you can say how fast are things exiting that surface? And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector. So let's rewrite it using what we know about how to construct a normal vector. So if we rewrite it, our integral becomes this. We have our vector field F dotted with the normal vector."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's start, now that we have hopefully a decent conceptual understanding of what this could represent, let's play around with it a little bit, because we know how to define a normal vector. So let's rewrite it using what we know about how to construct a normal vector. So if we rewrite it, our integral becomes this. We have our vector field F dotted with the normal vector. The normal vector we can write this way. A normal vector we saw was dy times i minus dx times j. And then we had to divide it by its magnitude in order to make it a unit normal vector."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have our vector field F dotted with the normal vector. The normal vector we can write this way. A normal vector we saw was dy times i minus dx times j. And then we had to divide it by its magnitude in order to make it a unit normal vector. The magnitude was this right over here, dx squared plus dy squared, which is the same thing as ds, which is the exact same thing as that little mini arc length, that infinitely small length of our contour. So we're going to divide it by ds. And then we're going to multiply it times ds."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we had to divide it by its magnitude in order to make it a unit normal vector. The magnitude was this right over here, dx squared plus dy squared, which is the same thing as ds, which is the exact same thing as that little mini arc length, that infinitely small length of our contour. So we're going to divide it by ds. And then we're going to multiply it times ds. And ds is just a scalar quantity, and so we can actually even multiply this thing times ds before taking the dot product or vice versa. But these two things are going to cancel out. And so we're essentially left with F dot this thing right over here."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to multiply it times ds. And ds is just a scalar quantity, and so we can actually even multiply this thing times ds before taking the dot product or vice versa. But these two things are going to cancel out. And so we're essentially left with F dot this thing right over here. But we have F defined right over here, so let's take the dot product. So I'll just write the line integral symbol again. We're going in the counterclockwise direction."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we're essentially left with F dot this thing right over here. But we have F defined right over here, so let's take the dot product. So I'll just write the line integral symbol again. We're going in the counterclockwise direction. And when we evaluate, and now let's pick a color that I have not used. Well, I've used many colors. So I'll do yellow again."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going in the counterclockwise direction. And when we evaluate, and now let's pick a color that I have not used. Well, I've used many colors. So I'll do yellow again. So now let's evaluate F dot this business. So dot product, fairly straightforward. You take the product of the x components, or essentially the magnitude of the x components."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'll do yellow again. So now let's evaluate F dot this business. So dot product, fairly straightforward. You take the product of the x components, or essentially the magnitude of the x components. So it's going to be P of xy times dy. And plus the product of the magnitudes of the y components, or the j components. So it's going to be plus Q of xy times minus dx, or times negative dx."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You take the product of the x components, or essentially the magnitude of the x components. So it's going to be P of xy times dy. And plus the product of the magnitudes of the y components, or the j components. So it's going to be plus Q of xy times minus dx, or times negative dx. Well, that's going to give us negative Q of xy times dx. So this is kind of an interesting statement. We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be plus Q of xy times minus dx, or times negative dx. Well, that's going to give us negative Q of xy times dx. So this is kind of an interesting statement. We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem. Let me rewrite it here just so we remember. So the definition, when we learned about Green's Theorem, it told us if we're taking a line integral over this contour, and there's multiple ways to write it, but one way that we often see it, and we've explored it already in our videos, is if we were to say if you have m times dx plus n times dy, this is equal to, and I'm just restating Green's Theorem right over here, this is equal to the double integral over the region that this contour surrounds of, and whatever function you have here times dy, you take the partial of that with respect to x. So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We've seen something like not too different than this before, and when we saw even just the definition of Green's Theorem. Let me rewrite it here just so we remember. So the definition, when we learned about Green's Theorem, it told us if we're taking a line integral over this contour, and there's multiple ways to write it, but one way that we often see it, and we've explored it already in our videos, is if we were to say if you have m times dx plus n times dy, this is equal to, and I'm just restating Green's Theorem right over here, this is equal to the double integral over the region that this contour surrounds of, and whatever function you have here times dy, you take the partial of that with respect to x. So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area. So I'll just write, I could write da here. I'll write dx dy, or it could be dy dx. Actually, let me just write da here, since we're speaking in generalizations."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you take the partial of this with respect to x, the partial of n with respect to x, and from that you subtract whatever was on the dx side, so the partial of m with respect to y, and we could say times dx dy, or you could say da, for the infinitesimally small little chunk of area. So I'll just write, I could write da here. I'll write dx dy, or it could be dy dx. Actually, let me just write da here, since we're speaking in generalizations. Where da is an infinitely small little chunk of area. So this right over here, this is just a restatement of Green's Theorem. So we already know this."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually, let me just write da here, since we're speaking in generalizations. Where da is an infinitely small little chunk of area. So this right over here, this is just a restatement of Green's Theorem. So we already know this. This is a restatement of Green's Theorem, and how can we apply it here? Well, it's the same thing. We have a little bit of sign differences, but we can apply Green's Theorem right over here."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we already know this. This is a restatement of Green's Theorem, and how can we apply it here? Well, it's the same thing. We have a little bit of sign differences, but we can apply Green's Theorem right over here. This is going to be equal to the double integral over the region that this contour surrounds, and then what we want to do is we want to look at whatever is the function that's being multiplied times the dy, and in this case, this is the function that's being multiplied times the dy, and we want to take the partial of that with respect to x. So we're going to take the partial of p with respect to x. And then from that, we are going to subtract the other function, whatever is being multiplied times dx."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have a little bit of sign differences, but we can apply Green's Theorem right over here. This is going to be equal to the double integral over the region that this contour surrounds, and then what we want to do is we want to look at whatever is the function that's being multiplied times the dy, and in this case, this is the function that's being multiplied times the dy, and we want to take the partial of that with respect to x. So we're going to take the partial of p with respect to x. And then from that, we are going to subtract the other function, whatever is being multiplied times dx. We're going to take the partial of that with respect to y. So here, we're going to take the partial of this whole thing with respect to y, but we have a negative out here. So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then from that, we are going to subtract the other function, whatever is being multiplied times dx. We're going to take the partial of that with respect to y. So here, we're going to take the partial of this whole thing with respect to y, but we have a negative out here. So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive. So then this is going to be equal to the double integral over the region, and maybe you might already see where this is going, maybe getting a little bit excited, the partial of p with respect to x plus the partial of q with respect to y, da. Now, what is this telling me, Sal? Well, look at this thing right over here."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be minus partial of q with respect to y, and then we have da, and obviously, these two negatives, subtracting a negative just gives you a positive. So then this is going to be equal to the double integral over the region, and maybe you might already see where this is going, maybe getting a little bit excited, the partial of p with respect to x plus the partial of q with respect to y, da. Now, what is this telling me, Sal? Well, look at this thing right over here. p was originally the function that is telling us the magnitude in the x direction. q was telling us the magnitude in the y direction. We're taking the partial of this with respect to x."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, look at this thing right over here. p was originally the function that is telling us the magnitude in the x direction. q was telling us the magnitude in the y direction. We're taking the partial of this with respect to x. We're taking the partial of this with respect to y, and we're summing them. This is essentially, or this is exactly, the divergence of f. And if that doesn't make any sense, go watch the video on divergence. This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're taking the partial of this with respect to x. We're taking the partial of this with respect to y, and we're summing them. This is essentially, or this is exactly, the divergence of f. And if that doesn't make any sense, go watch the video on divergence. This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing. This thing that we saw, this original expression that we started studying, which is essentially saying, what's the speed at which the particles are exiting this surface? We now get it in terms of this little expression, and we'll interpret it in an intuitive way in a little bit. So this is equal to the divergence of f times dA over the whole region, so the double integral."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This right over here is the divergence of f. This is the divergence, by definition, really, this is the divergence of our vector field f. And so we have a very interesting thing. This thing that we saw, this original expression that we started studying, which is essentially saying, what's the speed at which the particles are exiting this surface? We now get it in terms of this little expression, and we'll interpret it in an intuitive way in a little bit. So this is equal to the divergence of f times dA over the whole region, so the double integral. So we're summing up the divergences of f times the infinitely small little chunk of area, and we're summing them up over the whole region. Now, why does that make intuitive sense? And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is equal to the divergence of f times dA over the whole region, so the double integral. So we're summing up the divergences of f times the infinitely small little chunk of area, and we're summing them up over the whole region. Now, why does that make intuitive sense? And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is. Divergence is a measure of whether things are expanding or diverging or kind of contracting. If you have a point over here where around that the particles are kind of moving away from each other, you would have a positive divergence here. If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And for you to realize why it makes intuitive sense, you just have to remind yourself what the divergence is. Divergence is a measure of whether things are expanding or diverging or kind of contracting. If you have a point over here where around that the particles are kind of moving away from each other, you would have a positive divergence here. If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here. So this should make a lot of sense. You take any little infinitely small area in this, and then you multiply that times the divergence there. So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If you have a point, sometimes called a sink, where all of the particles are kind of condensing or converging, you would have a negative divergence here. So this should make a lot of sense. You take any little infinitely small area in this, and then you multiply that times the divergence there. So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region. So that makes a lot of sense. The more diverging that's going on through your region, obviously the more stuff is going to be exiting your boundary over here. So it actually makes complete sense, or hopefully it makes a little bit of complete sense, why if you were to see how fast are things exiting the surface, it's really the two-dimensional flux, how fast are things exiting the surface, and you take the sum over all of them, that's going to be the same thing as summing all of the divergences over this area that the contour is surrounding."}, {"video_title": "2D divergence theorem   Line integrals and Green's theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the more divergence here, you're going to get a larger number, and then you sum them up across the entire region. So that makes a lot of sense. The more diverging that's going on through your region, obviously the more stuff is going to be exiting your boundary over here. So it actually makes complete sense, or hopefully it makes a little bit of complete sense, why if you were to see how fast are things exiting the surface, it's really the two-dimensional flux, how fast are things exiting the surface, and you take the sum over all of them, that's going to be the same thing as summing all of the divergences over this area that the contour is surrounding. So hopefully that makes a little bit of sense to you, and it's actually another way of kind of thinking about Green's theorem. We also have just explored what we've just said, this expression, that the divergence summed over this region over here is the same thing as the F dot N over the contour. That essentially is the two-dimensional divergence theorem."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the first type of region, and it's appropriately named, we will call a type one region. Type one region. At first I'll give a formal definition. Hopefully the formal definition makes some intuitive sense. But then I'll draw a couple of type one regions, and then I'll show you what would not be a type one region, because sometimes that's the more important question. So type one region, maybe a type one region R, is the set, and these little curly brackets mean set, is the set of all x, y's, and z's. It's the set of all points in three dimensions, such that the x and y's are part of some domain."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully the formal definition makes some intuitive sense. But then I'll draw a couple of type one regions, and then I'll show you what would not be a type one region, because sometimes that's the more important question. So type one region, maybe a type one region R, is the set, and these little curly brackets mean set, is the set of all x, y's, and z's. It's the set of all points in three dimensions, such that the x and y's are part of some domain. So the x and y's are part of some domain, are a member. That's what this little symbol represents. Are a member of some domain, and z can essentially vary between two functions of x and y."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's the set of all points in three dimensions, such that the x and y's are part of some domain. So the x and y's are part of some domain, are a member. That's what this little symbol represents. Are a member of some domain, and z can essentially vary between two functions of x and y. So let me write it over here. So f1 of x, y is kind of the lower bound on z. So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Are a member of some domain, and z can essentially vary between two functions of x and y. So let me write it over here. So f1 of x, y is kind of the lower bound on z. So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y. And let me close the curly brackets to show that this was all a set. This is a set of x, y's and z's, and right here we are defining that set. So what would be a reasonable type one region?"}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be less than or equal to z, which is less than or equal to another function of x and y, which is going to be less than or equal to f2 of x and y. And let me close the curly brackets to show that this was all a set. This is a set of x, y's and z's, and right here we are defining that set. So what would be a reasonable type one region? Well, a very simple type one region is a sphere. So let me draw a sphere right over here. So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what would be a reasonable type one region? Well, a very simple type one region is a sphere. So let me draw a sphere right over here. So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here. So I'll do it in blue. So let me draw my best attempt at drawing that domain. So this is the domain D right over here for a sphere."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in a sphere, where it intersects the x, y plane, that's essentially this domain D right over here. So I'll do it in blue. So let me draw my best attempt at drawing that domain. So this is the domain D right over here for a sphere. This is a sphere centered at zero, but you can make the same argument for a sphere anywhere else. So that is my domain. And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the domain D right over here for a sphere. This is a sphere centered at zero, but you can make the same argument for a sphere anywhere else. So that is my domain. And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere. So you really can't see it well right over here, but these contours right over here would be on the bottom half. And I can even color in this part right over here. The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then f1 of x, y, which is the lower bound of z, will be the bottom half of the sphere. So you really can't see it well right over here, but these contours right over here would be on the bottom half. And I can even color in this part right over here. The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere. So it will look something like that. So this thing that I'm drawing right over here is definitely a type one region. As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The bottom surface of our sphere would be f1 of x, y, and f2 of x, y, as you can imagine, will be the top half of the sphere, the top hemisphere. So it will look something like that. So this thing that I'm drawing right over here is definitely a type one region. As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region. Another example of a type one region, and actually this might even be more obvious. Let me draw some axes again, and let me draw some type of a cylinder. And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "As we'll see, this could be a type one, type two, or type three region, but it's definitely a type one region. Another example of a type one region, and actually this might even be more obvious. Let me draw some axes again, and let me draw some type of a cylinder. And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane. So this is the bottom of the cylinder, it's right over here. And once again, it doesn't have to be centered around the z axis, but I'll do it that way just for this video. Actually I can draw it a little bit better than that."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And just to make it clear that our domain, where the x, y plane does not have to be inside of our region, let's imagine a cylinder that is below, that is, well actually I'll draw it above, that is above the x, y plane. So this is the bottom of the cylinder, it's right over here. And once again, it doesn't have to be centered around the z axis, but I'll do it that way just for this video. Actually I can draw it a little bit better than that. So this is the bottom surface of our cylinder, and then the top surface of our cylinder might be right over here. And these things actually don't even have to be flat, they could actually be curvy in some way. And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually I can draw it a little bit better than that. So this is the bottom surface of our cylinder, and then the top surface of our cylinder might be right over here. And these things actually don't even have to be flat, they could actually be curvy in some way. And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane. And then for each of those x, y pairs, f1 of x, y defines the bottom boundary of our region, so f1 of x, y is going to be this right over here. So you give me any of these x, y's in this domain D, and then you evaluate the function at those points, and it will correspond to this surface right over here. And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in this situation, so in this cylinder, let me draw it a little bit neater, in this cylinder right over here, our domain, all of the values that the x and y's can take on, so our domain is going to be this region right over here in the x, y plane. And then for each of those x, y pairs, f1 of x, y defines the bottom boundary of our region, so f1 of x, y is going to be this right over here. So you give me any of these x, y's in this domain D, and then you evaluate the function at those points, and it will correspond to this surface right over here. And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here. And we're saying that z will take on all the values in between, and so it is really this whole solid, it's really this entire solid area. Likewise, over here, z could take on any value between this magenta surface and this green surface, so it would essentially fill up our entire volume, so it would become a solid region. Now, you might be wondering, what would not be a type I region?"}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then f2 of x, y, once again, give me any one of those x, y points in our domain, and you evaluate f2 at those points, and it will give you this surface up here. And we're saying that z will take on all the values in between, and so it is really this whole solid, it's really this entire solid area. Likewise, over here, z could take on any value between this magenta surface and this green surface, so it would essentially fill up our entire volume, so it would become a solid region. Now, you might be wondering, what would not be a type I region? So let's think about that. So it would essentially be something that we could not define in this way, and I'll try my best to draw it, but you can imagine a shape that does something funky like this. So there's like one big, I guess you could imagine a sideways dumbbell."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, you might be wondering, what would not be a type I region? So let's think about that. So it would essentially be something that we could not define in this way, and I'll try my best to draw it, but you can imagine a shape that does something funky like this. So there's like one big, I guess you could imagine a sideways dumbbell. So a sideways dumbbell, I'll maybe curve it out a little bit. So maybe it's, so this is kind of the top of the dumbbell, and then it, or an hourglass, I guess you could say, or a dumbbell, it would look something like that. So I'm trying my best to draw it."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So there's like one big, I guess you could imagine a sideways dumbbell. So a sideways dumbbell, I'll maybe curve it out a little bit. So maybe it's, so this is kind of the top of the dumbbell, and then it, or an hourglass, I guess you could say, or a dumbbell, it would look something like that. So I'm trying my best to draw it. It would look something like that, and the reason why this is not definable in this way, it becomes obvious if you kind of look at a cross-section of it. There's no way to define only two functions, that's a lower bound and an upper bound, in terms of z. So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'm trying my best to draw it. It would look something like that, and the reason why this is not definable in this way, it becomes obvious if you kind of look at a cross-section of it. There's no way to define only two functions, that's a lower bound and an upper bound, in terms of z. So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on. Let me see how well I can draw this. So you say my xy values, let me try to draw this whole thing a little bit better, a better attempt. So you might say, okay, for something like a dumbbell, let me clear out that part as well."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So even if you say, hey, maybe my domain, my domain will be all of the xy values that can be taken on. Let me see how well I can draw this. So you say my xy values, let me try to draw this whole thing a little bit better, a better attempt. So you might say, okay, for something like a dumbbell, let me clear out that part as well. For something like a dumbbell, let me erase that. So for something like a dumbbell, maybe my domain is right over here. So these are all the xy values that you can take on."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you might say, okay, for something like a dumbbell, let me clear out that part as well. For something like a dumbbell, let me erase that. So for something like a dumbbell, maybe my domain is right over here. So these are all the xy values that you can take on. But in order to have a dumbbell shape, for any one xy, z is going to take on, there's not just an upper and a lower bound, and z doesn't take on all values in between. Let me just draw it more clearly. So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So these are all the xy values that you can take on. But in order to have a dumbbell shape, for any one xy, z is going to take on, there's not just an upper and a lower bound, and z doesn't take on all values in between. Let me just draw it more clearly. So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that. And then it goes below the xy plane, and it does kind of a similar thing. It goes below the xy plane and looks something like that. So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our dumbbell, maybe it's centered on the z axis, this is the middle of our dumbbell, and then it comes out like that, and then up here, the z axis, so it looks like that. And then it goes below the xy plane, and it does kind of a similar thing. It goes below the xy plane and looks something like that. So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface. Down here is the bottom surface. But notice, z can't take on every value in between. You kind of have to break this up if you wanted to be able to do something like that."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So notice, for any given xy, what would be, if you attempted to make it a type I region, you would say, well, maybe this is the top surface, and maybe you would say down here is the bottom surface. Down here is the bottom surface. But notice, z can't take on every value in between. You kind of have to break this up if you wanted to be able to do something like that. You would have to break this up into two separate regions, where this would be the bottom region, and then this right over here would be another top region. So this dumbbell shape itself is not a type I region, but you could actually break it up into two separate type I regions. So hopefully that helps out."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You kind of have to break this up if you wanted to be able to do something like that. You would have to break this up into two separate regions, where this would be the bottom region, and then this right over here would be another top region. So this dumbbell shape itself is not a type I region, but you could actually break it up into two separate type I regions. So hopefully that helps out. And actually, another way to think about it, and this might be an easier way, if we were to look at it from this direction, and if we were to just think about the zy, if we were to just think about what's happening on the zy plane, so that's z, and this is y right over here, our dumbbell shape would look something like this. My best attempt to draw our dumbbell shape. So if you get a given x or y, maybe x is even zero, and you're sitting right here on the y-axis, notice z is not, even up here, cannot be a function of just y."}, {"video_title": "Type I regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So hopefully that helps out. And actually, another way to think about it, and this might be an easier way, if we were to look at it from this direction, and if we were to just think about the zy, if we were to just think about what's happening on the zy plane, so that's z, and this is y right over here, our dumbbell shape would look something like this. My best attempt to draw our dumbbell shape. So if you get a given x or y, maybe x is even zero, and you're sitting right here on the y-axis, notice z is not, even up here, cannot be a function of just y. On this top part, there's two possible z values that we need to take on for that given y. Two possible z values for that given y. So you can't define it simply in terms of just one lower bound function and one upper bound function."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So if you take this guy, how do you compute the curl of that vector valued function? So what you do, as I mentioned in the last video, is you imagine taking this del operator and taking the cross product between that and your vector valued function. And what that means when you expand it is that del operator, you just kind of fill it with partial differential operators, you could say, but really it's just the symbol partial partial x, partial partial y, partial partial z. And these are things that are just waiting to take in some kind of function. So we're gonna take the cross product between that and the function that we have defined here. So let me just actually copy it over. Copy it over here."}, {"video_title": "3d curl computation example.mp3", "Sentence": "And these are things that are just waiting to take in some kind of function. So we're gonna take the cross product between that and the function that we have defined here. So let me just actually copy it over. Copy it over here. And a little residue. And to compute this cross product, we take a certain determinant. So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky."}, {"video_title": "3d curl computation example.mp3", "Sentence": "Copy it over here. And a little residue. And to compute this cross product, we take a certain determinant. So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky. So the top row, just like we would have with any other cross product that we're computing, is gonna have i, j, and k, these unit vectors in three dimensional space. And the second row here is gonna have all of these partial differential operators since that's the first vector in our cross product. So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So I'm gonna write over here determinant, and it's gonna be of a three by three matrix, but really it's kind of like a quote unquote matrix because each component has something funky. So the top row, just like we would have with any other cross product that we're computing, is gonna have i, j, and k, these unit vectors in three dimensional space. And the second row here is gonna have all of these partial differential operators since that's the first vector in our cross product. So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of. And then that third row is gonna be the functions that we have, so the first component here is x, y. The second component is cosine of z, cosine of z. And then that final component is z squared plus y, z squared plus y."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So that's partial partial x, partial partial y, and again, all of these are just kind of waiting to be given a function that they can take the derivative of. And then that third row is gonna be the functions that we have, so the first component here is x, y. The second component is cosine of z, cosine of z. And then that final component is z squared plus y, z squared plus y. So I'll give some room here, maybe make it more visible. So this is the determinant we need to compute, and this is gonna be broken up into three different parts. The first one, we take this top part, i, and multiply it by the determinant of this sub matrix."}, {"video_title": "3d curl computation example.mp3", "Sentence": "And then that final component is z squared plus y, z squared plus y. So I'll give some room here, maybe make it more visible. So this is the determinant we need to compute, and this is gonna be broken up into three different parts. The first one, we take this top part, i, and multiply it by the determinant of this sub matrix. So when we do that, this sub-determinant, we're taking partial derivative with respect to y of z squared plus y. Now as far as y is concerned, z looks like a constant. So z squared is a constant, and the partial derivative of this entire guy is just one."}, {"video_title": "3d curl computation example.mp3", "Sentence": "The first one, we take this top part, i, and multiply it by the determinant of this sub matrix. So when we do that, this sub-determinant, we're taking partial derivative with respect to y of z squared plus y. Now as far as y is concerned, z looks like a constant. So z squared is a constant, and the partial derivative of this entire guy is just one. So that'll look like one. And then we're subtracting off the partial derivative with respect to z of cosine of z. And that just looks the same as a derivative of cosine z, which is negative sine."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So z squared is a constant, and the partial derivative of this entire guy is just one. So that'll look like one. And then we're subtracting off the partial derivative with respect to z of cosine of z. And that just looks the same as a derivative of cosine z, which is negative sine. So that's negative sine of z. So that's the first part. And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants."}, {"video_title": "3d curl computation example.mp3", "Sentence": "And that just looks the same as a derivative of cosine z, which is negative sine. So that's negative sine of z. So that's the first part. And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants. So we're gonna subtract off j, multiply it by its own little sub-determinant, and this time the sub-determinant is gonna involve the two columns that it's not part of. So you're imagining this first column and this second column as being part of a matrix. So the first thing you do is you take this partial derivative with respect to x of z squared plus y."}, {"video_title": "3d curl computation example.mp3", "Sentence": "And then as the next part, we're gonna take j, but we're subtracting, because you're always kind of thinking plus, minus, plus when you're doing these determinants. So we're gonna subtract off j, multiply it by its own little sub-determinant, and this time the sub-determinant is gonna involve the two columns that it's not part of. So you're imagining this first column and this second column as being part of a matrix. So the first thing you do is you take this partial derivative with respect to x of z squared plus y. Well, no x's show up there, right? That's z squared and y each look like constants as far as x is concerned. So that's zero."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So the first thing you do is you take this partial derivative with respect to x of z squared plus y. Well, no x's show up there, right? That's z squared and y each look like constants as far as x is concerned. So that's zero. Then we take the partial with respect to z of x times y. And again, there's no z that shows up there, so that's also zero. So we're kind of subtracting off zero."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So that's zero. Then we take the partial with respect to z of x times y. And again, there's no z that shows up there, so that's also zero. So we're kind of subtracting off zero. And then finally, we're adding this last component. We're gonna add that last component, k, multiplied by the determinant of this sub-matrix of the columns that it's not part of. So this involves partial derivative with respect to x of cosine z."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So we're kind of subtracting off zero. And then finally, we're adding this last component. We're gonna add that last component, k, multiplied by the determinant of this sub-matrix of the columns that it's not part of. So this involves partial derivative with respect to x of cosine z. Well, no x's show up there, so that's just zero. So that's just a zero. And then we're subtracting off the partial with respect to y of x times y."}, {"video_title": "3d curl computation example.mp3", "Sentence": "So this involves partial derivative with respect to x of cosine z. Well, no x's show up there, so that's just zero. So that's just a zero. And then we're subtracting off the partial with respect to y of x times y. Well, x looks like a constant, y looks like the variable, so that partial derivative is just x. So we're subtracting off x, which means if we simplify this, so the curl of our vector field, curl of our vector field as a whole, as this function of x, y, and z, is equal to, and that first component, the i component, we've got one minus negative sine of z, so minus minus sine of z, that's one plus sine of z. And then the j component, we're subtracting off, but at zero."}, {"video_title": "3d curl computation example.mp3", "Sentence": "And then we're subtracting off the partial with respect to y of x times y. Well, x looks like a constant, y looks like the variable, so that partial derivative is just x. So we're subtracting off x, which means if we simplify this, so the curl of our vector field, curl of our vector field as a whole, as this function of x, y, and z, is equal to, and that first component, the i component, we've got one minus negative sine of z, so minus minus sine of z, that's one plus sine of z. And then the j component, we're subtracting off, but at zero. Usually if you were subtracting off, you'd have to make sure to remember to flip those, but both of those are zero, so the entire j component here, or the y component of the output is zero. And then finally we're adding, the k component is zero minus x, so that entire thing is just negative x. And that's the curl of the function."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "Hey everyone. So here and in the next few videos, I'm gonna be talking about tangent planes, tangent planes of graphs. And I'll specify that this is tangent planes of graphs and not of some other thing, because in different contexts of multivariable calculus, you might be taking a tangent plane of, say, a parametric surface or something like that, but here I'm just focused on graphs. So in the single variable world, a common problem that people like to ask in calculus is you have some sort of curve, and you wanna find, at a given point, what the tangent line to that curve is, what the tangent line is. And you'll find the equation for that tangent line, and this gives you various information, kind of how to, let's say you wanted to approximate the function around that point, and it turns out to be a nice, simple approximation. And in the multivariable world, it's actually pretty similar. In terms of geometric intuition, it's almost identical."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "So in the single variable world, a common problem that people like to ask in calculus is you have some sort of curve, and you wanna find, at a given point, what the tangent line to that curve is, what the tangent line is. And you'll find the equation for that tangent line, and this gives you various information, kind of how to, let's say you wanted to approximate the function around that point, and it turns out to be a nice, simple approximation. And in the multivariable world, it's actually pretty similar. In terms of geometric intuition, it's almost identical. You'll have some kind of graph of a function, like the one that I have here, and then instead of having a tangent line, because a line is a very one-dimensional thing, and here it's a very two-dimensional surface, instead you'll have some kind of tangent plane. So this is something where it's just gonna barely be kissing the graph, in the same way that the tangent line just barely kisses the function graph in the one-dimensional circumstance. And it could be at various different points, rather than just being at that point."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "In terms of geometric intuition, it's almost identical. You'll have some kind of graph of a function, like the one that I have here, and then instead of having a tangent line, because a line is a very one-dimensional thing, and here it's a very two-dimensional surface, instead you'll have some kind of tangent plane. So this is something where it's just gonna barely be kissing the graph, in the same way that the tangent line just barely kisses the function graph in the one-dimensional circumstance. And it could be at various different points, rather than just being at that point. You could kind of move it around, and say that, okay, it'll just barely be kissing the graph of this function, but at different points. And usually the way that a problem like this would be framed, if you're trying to find such a tangent plane, is first, you think about the specified input that you want. So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here?"}, {"video_title": "What is a tangent plane.mp3", "Sentence": "And it could be at various different points, rather than just being at that point. You could kind of move it around, and say that, okay, it'll just barely be kissing the graph of this function, but at different points. And usually the way that a problem like this would be framed, if you're trying to find such a tangent plane, is first, you think about the specified input that you want. So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here? Maybe you'd name it like X sub O, and then you're gonna find the graph of the function that corresponds to kind of just kissing the graph at that input point. Over here in the multivariable world, kind of move things about, you'll choose some kind of input point, like this little red dot, and that could be at various different spots. It doesn't have to be where I put it."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "So in the same way that over in the single variable world, what you might do is say, okay, what is the input value here? Maybe you'd name it like X sub O, and then you're gonna find the graph of the function that corresponds to kind of just kissing the graph at that input point. Over here in the multivariable world, kind of move things about, you'll choose some kind of input point, like this little red dot, and that could be at various different spots. It doesn't have to be where I put it. You could imagine putting it somewhere else. But once you decide on what input point you want, you see where that is on the graph. So you kind of go and say, oh, that input point corresponds to such and such a height."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "It doesn't have to be where I put it. You could imagine putting it somewhere else. But once you decide on what input point you want, you see where that is on the graph. So you kind of go and say, oh, that input point corresponds to such and such a height. So in this case, it actually looks like the graph is about zero at that point, so the output of the function would be zero. And what you want is a plane that's tangent right at that point. So you'll draw some kind of plane that's tangent right at that point."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "So you kind of go and say, oh, that input point corresponds to such and such a height. So in this case, it actually looks like the graph is about zero at that point, so the output of the function would be zero. And what you want is a plane that's tangent right at that point. So you'll draw some kind of plane that's tangent right at that point. And if we think about what this input point corresponds to, it's not X sub O, a single variable input like we have in the single variable world, but instead that red dot that you're seeing is gonna correspond to some kind of input pair, X sub O and Y sub O. So the ultimate goal over here in our multivariable circumstance is gonna be to find some kind of new function, so I'll write it down here, some kind of new function that I'll call L for linear that's gonna take in X and Y, and we want the graph of that function to be this plane. And you might specify that this is dependent on the original function that you have, and maybe you also specify that it's dependent on this input point in some way, but the basic idea is we're gonna be looking for a function whose graph is this plane tangent at a given point."}, {"video_title": "What is a tangent plane.mp3", "Sentence": "So you'll draw some kind of plane that's tangent right at that point. And if we think about what this input point corresponds to, it's not X sub O, a single variable input like we have in the single variable world, but instead that red dot that you're seeing is gonna correspond to some kind of input pair, X sub O and Y sub O. So the ultimate goal over here in our multivariable circumstance is gonna be to find some kind of new function, so I'll write it down here, some kind of new function that I'll call L for linear that's gonna take in X and Y, and we want the graph of that function to be this plane. And you might specify that this is dependent on the original function that you have, and maybe you also specify that it's dependent on this input point in some way, but the basic idea is we're gonna be looking for a function whose graph is this plane tangent at a given point. And in the next couple videos, I'm gonna talk through how you actually compute that. It might seem a little intimidating at first because how do you control a plane in three dimensions like this? But it's actually very similar to the single variable circumstance, and you just kind of take it one step at a time."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And that means that it's representing some kind of function that has a two-dimensional input and a one-dimensional output. So that might look something like f of x, y equals, and then just some expression that has a bunch of x's and y's in it. And graphs are great, but they're kind of clunky to draw. I mean, certainly you can't just scribble it down. It typically requires some kind of graphing software. And when you take a static image of it, it's not always clear what's going on. So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "I mean, certainly you can't just scribble it down. It typically requires some kind of graphing software. And when you take a static image of it, it's not always clear what's going on. So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper. And this is a very common way that you'll see if you're reading a textbook or if someone is drawing on a blackboard. It's known as a contour plot. And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "So here, I'm going to describe a way that you can represent these functions in these graphs two-dimensionally, just by scribbling down on a two-dimensional piece of paper. And this is a very common way that you'll see if you're reading a textbook or if someone is drawing on a blackboard. It's known as a contour plot. And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times. So I'm going to slice it with various planes that are all parallel to the xy-plane. And let's think for a moment about what these guys represent. So the bottom one here represents the value z is equal to negative 2."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the idea of a contour plot is that we're going to take this graph and slice it a bunch of times. So I'm going to slice it with various planes that are all parallel to the xy-plane. And let's think for a moment about what these guys represent. So the bottom one here represents the value z is equal to negative 2. So this is the z-axis over here. And when we fix that to be negative 2 and let x and y run freely, we get this whole plane. And if you let z increase, keep it constant, but let it increase by 1 to negative 1, we get a new plane still parallel to the xy-plane."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "So the bottom one here represents the value z is equal to negative 2. So this is the z-axis over here. And when we fix that to be negative 2 and let x and y run freely, we get this whole plane. And if you let z increase, keep it constant, but let it increase by 1 to negative 1, we get a new plane still parallel to the xy-plane. And it's got a, but its distance from the xy-plane is negative 1. And the rest of these guys, they're all still values, constant values of z. Now in terms of our graph, what that means is that these represent constant values of the graph itself."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And if you let z increase, keep it constant, but let it increase by 1 to negative 1, we get a new plane still parallel to the xy-plane. And it's got a, but its distance from the xy-plane is negative 1. And the rest of these guys, they're all still values, constant values of z. Now in terms of our graph, what that means is that these represent constant values of the graph itself. These represent constant values for the function itself. So because we always represent the output of the function as the height off of the xy-plane, these represent constant values for the output. So what that's going to look like, so what we do is we say, where do these slices cut into the graph?"}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "Now in terms of our graph, what that means is that these represent constant values of the graph itself. These represent constant values for the function itself. So because we always represent the output of the function as the height off of the xy-plane, these represent constant values for the output. So what that's going to look like, so what we do is we say, where do these slices cut into the graph? So I'm going to draw on all of the points where those slices cut into the graph. And these are called contour lines. We're still in three dimensions, so we're not done yet."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what that's going to look like, so what we do is we say, where do these slices cut into the graph? So I'm going to draw on all of the points where those slices cut into the graph. And these are called contour lines. We're still in three dimensions, so we're not done yet. So what I'm going to do is take all these contour lines, and I'm going to squish them down onto the xy-plane. So what that means, each of them has some kind of z component at the moment, and we're just going to chop it down and squish them all nice and flat onto the xy-plane. And now we have something two-dimensional, and it still represents some of the outputs of our function."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "We're still in three dimensions, so we're not done yet. So what I'm going to do is take all these contour lines, and I'm going to squish them down onto the xy-plane. So what that means, each of them has some kind of z component at the moment, and we're just going to chop it down and squish them all nice and flat onto the xy-plane. And now we have something two-dimensional, and it still represents some of the outputs of our function. Not all of them, it's not perfect, but it does give a very good idea. So I'm going to switch over to a two-dimensional graph here. And this is that same function that we were just looking at."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And now we have something two-dimensional, and it still represents some of the outputs of our function. Not all of them, it's not perfect, but it does give a very good idea. So I'm going to switch over to a two-dimensional graph here. And this is that same function that we were just looking at. Let's actually move it a little bit more central here. So this is the same function that we were just looking at, but each of these lines represents a constant output of the function. So it's important to realize we're still representing a function that has a two-dimensional input and a one-dimensional output."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And this is that same function that we were just looking at. Let's actually move it a little bit more central here. So this is the same function that we were just looking at, but each of these lines represents a constant output of the function. So it's important to realize we're still representing a function that has a two-dimensional input and a one-dimensional output. It's just that we're looking in the input space of that function as a whole. So this is still f of xy, and then some expression of those guys. But this line might represent the constant value of f when all of the values where it outputs 3."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "So it's important to realize we're still representing a function that has a two-dimensional input and a one-dimensional output. It's just that we're looking in the input space of that function as a whole. So this is still f of xy, and then some expression of those guys. But this line might represent the constant value of f when all of the values where it outputs 3. Over here, this also, both of these circles together give you all the values where f outputs 3. This one over here will tell you where it outputs 2. And you can't know this just looking at the contour plot."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "But this line might represent the constant value of f when all of the values where it outputs 3. Over here, this also, both of these circles together give you all the values where f outputs 3. This one over here will tell you where it outputs 2. And you can't know this just looking at the contour plot. So typically, if someone's drawing it, if it matters that you know the specific values, they'll mark it somehow. They'll let you know what value each line corresponds to. But as soon as you know that this line corresponds to 0, it tells you that every possible input point that sits somewhere on this line will evaluate to 0 when you pump it through the function."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And you can't know this just looking at the contour plot. So typically, if someone's drawing it, if it matters that you know the specific values, they'll mark it somehow. They'll let you know what value each line corresponds to. But as soon as you know that this line corresponds to 0, it tells you that every possible input point that sits somewhere on this line will evaluate to 0 when you pump it through the function. And this actually gives a very good feel for the shape of things. If you like thinking in terms of graphs, you can kind of imagine how these circles and everything would pop out of the page. You can also look."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "But as soon as you know that this line corresponds to 0, it tells you that every possible input point that sits somewhere on this line will evaluate to 0 when you pump it through the function. And this actually gives a very good feel for the shape of things. If you like thinking in terms of graphs, you can kind of imagine how these circles and everything would pop out of the page. You can also look. Notice how the lines are really close together over here, very, very close together, but they're a little more spaced over here. How do you interpret that? Well, over here, this means it takes a very, very small step to increase the value of the function by 1."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "You can also look. Notice how the lines are really close together over here, very, very close together, but they're a little more spaced over here. How do you interpret that? Well, over here, this means it takes a very, very small step to increase the value of the function by 1. Very small step, and it increases by 1. But over here, it takes a much larger step to increase the function by the same value. So over here, this kind of means steepness."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "Well, over here, this means it takes a very, very small step to increase the value of the function by 1. Very small step, and it increases by 1. But over here, it takes a much larger step to increase the function by the same value. So over here, this kind of means steepness. If you see a very short distance between contour lines, it's going to be very steep. But over here, it's much more shallow. And you can do things like this to kind of get a better feel for the function as a whole."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "So over here, this kind of means steepness. If you see a very short distance between contour lines, it's going to be very steep. But over here, it's much more shallow. And you can do things like this to kind of get a better feel for the function as a whole. The idea of a whole bunch of concentric circles usually corresponds to a maximum or a minimum. And you end up seeing these a lot. Another common thing people will do with contour plots as they represent them is color them."}, {"video_title": "Contour plots   Multivariable calculus   Khan Academy.mp3", "Sentence": "And you can do things like this to kind of get a better feel for the function as a whole. The idea of a whole bunch of concentric circles usually corresponds to a maximum or a minimum. And you end up seeing these a lot. Another common thing people will do with contour plots as they represent them is color them. So what that might look like is here, where warmer colors like orange correspond to high values, and cooler colors like blue correspond to low values. And the contour lines end up going along the division between red and green here, between light green and green. And that's another way where colors tell you the output, and then the contour lines themselves can be thought of as the borders between different colors."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And here, I want to talk about second partial derivatives. So I'm going to write some kind of multivariable function, let's say it's, I don't know, sine of x times y squared. Sine of x multiplied by y squared. And if you take the partial derivative, you have two options, given that there's two variables, you can go one way and say, what's the partial derivative, partial derivative of f with respect to x, and what you do for that, x looks like a variable, as far as this direction is concerned, y looks like a constant, so you differentiate this by saying the derivative of sine of x is cosine x, you know, you're differentiating with respect to x, and then that looks like it's multiplied by a constant, so you just continue multiplying that by a constant. But you could also go another direction, you could also say, you know, what's the partial derivative with respect to y, and in that case, you're considering y to be the variable, so here it looks at y, and says y squared looks like a variable, x looks like a constant, sine of x then just looks like sine of a constant, which is a constant, so that'll be that constant, sine of x, multiplied by the derivative of y squared, which is gonna be two times y. Two times y. And these are what you might call first partial derivatives, and there's some alternate notation here, df dy, you could also say f and then a little subscript y, and over here, similarly, you'd say f with a little subscript x."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And if you take the partial derivative, you have two options, given that there's two variables, you can go one way and say, what's the partial derivative, partial derivative of f with respect to x, and what you do for that, x looks like a variable, as far as this direction is concerned, y looks like a constant, so you differentiate this by saying the derivative of sine of x is cosine x, you know, you're differentiating with respect to x, and then that looks like it's multiplied by a constant, so you just continue multiplying that by a constant. But you could also go another direction, you could also say, you know, what's the partial derivative with respect to y, and in that case, you're considering y to be the variable, so here it looks at y, and says y squared looks like a variable, x looks like a constant, sine of x then just looks like sine of a constant, which is a constant, so that'll be that constant, sine of x, multiplied by the derivative of y squared, which is gonna be two times y. Two times y. And these are what you might call first partial derivatives, and there's some alternate notation here, df dy, you could also say f and then a little subscript y, and over here, similarly, you'd say f with a little subscript x. Now each of these two functions, these two partial derivatives that you get, are also multivariable functions, they take in two variables and they output a scalar, so we can do something very similar, where here, you might then apply the partial derivative with respect to x to that partial derivative of your original function with respect to x, right? It's just like a second derivative in ordinary calculus, but this time, we're doing it partial, so when you do it with respect to x, cosine x looks like cosine of a variable, the derivative of which is negative sine times that variable, and y squared here just looks like a constant, so it just stays constant at y squared, and similarly, you could go down a different branch of options here, and say what if you did your partial derivative with respect to y of that whole function, which itself is a partial derivative with respect to x? And if you did that, then y squared now looks like the variable, so you're gonna take the derivative of that, which is two y, and then what's in front of it just looks like a constant as far as the variable y is concerned, so that stays as cosine of x."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And these are what you might call first partial derivatives, and there's some alternate notation here, df dy, you could also say f and then a little subscript y, and over here, similarly, you'd say f with a little subscript x. Now each of these two functions, these two partial derivatives that you get, are also multivariable functions, they take in two variables and they output a scalar, so we can do something very similar, where here, you might then apply the partial derivative with respect to x to that partial derivative of your original function with respect to x, right? It's just like a second derivative in ordinary calculus, but this time, we're doing it partial, so when you do it with respect to x, cosine x looks like cosine of a variable, the derivative of which is negative sine times that variable, and y squared here just looks like a constant, so it just stays constant at y squared, and similarly, you could go down a different branch of options here, and say what if you did your partial derivative with respect to y of that whole function, which itself is a partial derivative with respect to x? And if you did that, then y squared now looks like the variable, so you're gonna take the derivative of that, which is two y, and then what's in front of it just looks like a constant as far as the variable y is concerned, so that stays as cosine of x. And the notation here, first of all, just as in single variable calculus, it's common to kind of do a abusive notation with this kind of thing and write partial squared of f divided by partial x squared. And this always, I don't know, when I first learned about these things, they always threw me off because here, this Leibniz notation, you have the great intuition of nudging the x and nudging the f, but you kind of lose that when you do this, but it makes sense if you think of this partial partial x as being an operator and you're just applying it twice. And over here, the way that that would look, it's a little bit funny, because you still have that partial squared f on top, but then on the bottom, you write partial y, partial x."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And if you did that, then y squared now looks like the variable, so you're gonna take the derivative of that, which is two y, and then what's in front of it just looks like a constant as far as the variable y is concerned, so that stays as cosine of x. And the notation here, first of all, just as in single variable calculus, it's common to kind of do a abusive notation with this kind of thing and write partial squared of f divided by partial x squared. And this always, I don't know, when I first learned about these things, they always threw me off because here, this Leibniz notation, you have the great intuition of nudging the x and nudging the f, but you kind of lose that when you do this, but it makes sense if you think of this partial partial x as being an operator and you're just applying it twice. And over here, the way that that would look, it's a little bit funny, because you still have that partial squared f on top, but then on the bottom, you write partial y, partial x. And I'm putting them in these order just because it's as if I wrote it that way. This reflects the fact that first I did the x derivative, then I did the y derivative. And you could do this on this side also."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And over here, the way that that would look, it's a little bit funny, because you still have that partial squared f on top, but then on the bottom, you write partial y, partial x. And I'm putting them in these order just because it's as if I wrote it that way. This reflects the fact that first I did the x derivative, then I did the y derivative. And you could do this on this side also. And this might feel tedious, but it's actually kind of worth doing for a result that we end up seeing here that I find a little bit surprising, actually. So here, if we go down the path of doing, in this case, like a partial derivative with respect to x, and you're thinking of this as being applied to your original partial derivative with respect to y, it looks here, it says sine of x looks like a variable, two y looks like a constant, so what we end up getting is derivative of sine of x, cosine x, multiplied by that two y. And a pretty cool thing worth pointing out here that maybe you take it for granted, maybe you think it's as surprising as I did when I first saw it, both of these turn out to be equal, right, even though it was a very different way that we got there, right?"}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And you could do this on this side also. And this might feel tedious, but it's actually kind of worth doing for a result that we end up seeing here that I find a little bit surprising, actually. So here, if we go down the path of doing, in this case, like a partial derivative with respect to x, and you're thinking of this as being applied to your original partial derivative with respect to y, it looks here, it says sine of x looks like a variable, two y looks like a constant, so what we end up getting is derivative of sine of x, cosine x, multiplied by that two y. And a pretty cool thing worth pointing out here that maybe you take it for granted, maybe you think it's as surprising as I did when I first saw it, both of these turn out to be equal, right, even though it was a very different way that we got there, right? You first take the partial derivative with respect to x, and you get cosine xy squared, which looks very different from sine x two y, and then when you take the derivative with respect to y, you know, you get a certain value, and when you go down the other path, you also get that same value. And maybe the way that you'd write this is that you'd say, let me just copy this guy over here, and what you might say is that the partial derivative of f, when you do it the other way around, when instead of doing x and then y, you do y and then x, partial x, that these guys are equal to each other, and that's a pretty cool result. And maybe in this case, given that the original function just looks like the product of two things, you can kind of reason through why it's the case, but what's surprising is that this turns out to be true for, I mean, not all functions, there's actually a certain criterion."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And a pretty cool thing worth pointing out here that maybe you take it for granted, maybe you think it's as surprising as I did when I first saw it, both of these turn out to be equal, right, even though it was a very different way that we got there, right? You first take the partial derivative with respect to x, and you get cosine xy squared, which looks very different from sine x two y, and then when you take the derivative with respect to y, you know, you get a certain value, and when you go down the other path, you also get that same value. And maybe the way that you'd write this is that you'd say, let me just copy this guy over here, and what you might say is that the partial derivative of f, when you do it the other way around, when instead of doing x and then y, you do y and then x, partial x, that these guys are equal to each other, and that's a pretty cool result. And maybe in this case, given that the original function just looks like the product of two things, you can kind of reason through why it's the case, but what's surprising is that this turns out to be true for, I mean, not all functions, there's actually a certain criterion. There's a special theorem, it's called Schwarz's theorem, where if the second partial derivatives of your function are continuous at the relevant point, that's the circumstance for this being true, but for all intents and purposes, the kind of functions you can expect to run into, this is the case. This order of partial derivatives doesn't matter truth turns out to hold, which is actually pretty cool. And I'd encourage you to play around with some other functions, just come up with any multivariable function, maybe a little bit more complicated than just multiplying two separate things there, and see that it's true, and maybe try to convince yourself why it's true in certain cases."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And maybe in this case, given that the original function just looks like the product of two things, you can kind of reason through why it's the case, but what's surprising is that this turns out to be true for, I mean, not all functions, there's actually a certain criterion. There's a special theorem, it's called Schwarz's theorem, where if the second partial derivatives of your function are continuous at the relevant point, that's the circumstance for this being true, but for all intents and purposes, the kind of functions you can expect to run into, this is the case. This order of partial derivatives doesn't matter truth turns out to hold, which is actually pretty cool. And I'd encourage you to play around with some other functions, just come up with any multivariable function, maybe a little bit more complicated than just multiplying two separate things there, and see that it's true, and maybe try to convince yourself why it's true in certain cases. I think that would actually be a really good exercise. And just before I go, one thing I should probably mention, a bit of notation that people will commonly use. With this second partial derivative, sometimes instead of saying partial squared F, partial X squared, they'll just write it as partial and then XX."}, {"video_title": "Symmetry of second partial derivatives.mp3", "Sentence": "And I'd encourage you to play around with some other functions, just come up with any multivariable function, maybe a little bit more complicated than just multiplying two separate things there, and see that it's true, and maybe try to convince yourself why it's true in certain cases. I think that would actually be a really good exercise. And just before I go, one thing I should probably mention, a bit of notation that people will commonly use. With this second partial derivative, sometimes instead of saying partial squared F, partial X squared, they'll just write it as partial and then XX. And over here, this would be partial, let's see, first you did it with X, then Y, so over here you do it first X and then Y, kind of the order of these reverses, because you're reading left to right, but when you do it with this, you're kind of reading right to left for how you multiply it in. Which would mean that this guy, let's see, this guy over here, you know, he would be partial, first you did the Y, and then you did the X. So those two, those two guys are just different notations for the same thing."}, {"video_title": "Divergence intuition, part 1.mp3", "Sentence": "Alright everyone, we've gotten to one of my all-time favorite multivariable calculus topics, divergence, and in the next few videos I'm going to describe what it is mathematically and how you compute it and all of that, but here I just want to give a very visual understanding of what it is that it's trying to represent. So I've got a human pictured in front of us, a vector field, and I've said before that a pretty neat way to understand vector fields is to think of them as representing a fluid flow, and what I mean by that is you can think of every single point in space as a particle, maybe like an air particle or a water particle, something to that effect, and since what a vector field does is it associates each point in space with some kind of vector, and remember I mean whenever we represent vector fields we only show a small subset of all of those vectors, but in principle you should be thinking of every one of those infinitely many points in space being associated with one of these vectors, and the fact that they're kind of smoothly changing as you traverse across space means that showing this very small, you know, finite subsample of those infinitely many vectors still gives a pretty good feel for what's going on. So if we have these fluid particles and you, you know, have a vector assigned to each one, kind of a natural thought you might have is to say what would happen if you let things progress over time where at any given instant the velocity of one of these particles is given by that vector connected to it, and as it moves, you know, it'll be touching a different vector so its velocity might turn, it might go in a different direction, and for each one it'll kind of traverse some path as determined by the vectors that it's touching as it goes, and when you think of all of them doing this at once it'll feel like a certain, a certain fluid flow, and for this, for this you don't actually have to imagine, I went ahead and put together an animation for you. So we'll put some water molecules, or dots, to represent a small sample of the water molecules throughout space here, and then I'm just gonna let it play where each one moves along the vector that it's closest to, and I'll just let it play forward here where each one is flowing along the vector that's touching the point where it is in that moment. So for example if we were to, you know, go back and maybe focus our attention on just one vector like this guy, one particle excuse me, he's attached to this vector so he'll be moving in that direction, but just for an instant because after he moves a little he'll be attached to a different vector. So if you kind of let it play and follow that particular dot after a little bit you'll find him, you know, elsewhere. I think, I think this is the one right, and now he's gonna be moving along this vector or whatever one is really attached to him."}, {"video_title": "Divergence intuition, part 1.mp3", "Sentence": "So we'll put some water molecules, or dots, to represent a small sample of the water molecules throughout space here, and then I'm just gonna let it play where each one moves along the vector that it's closest to, and I'll just let it play forward here where each one is flowing along the vector that's touching the point where it is in that moment. So for example if we were to, you know, go back and maybe focus our attention on just one vector like this guy, one particle excuse me, he's attached to this vector so he'll be moving in that direction, but just for an instant because after he moves a little he'll be attached to a different vector. So if you kind of let it play and follow that particular dot after a little bit you'll find him, you know, elsewhere. I think, I think this is the one right, and now he's gonna be moving along this vector or whatever one is really attached to him. And thinking about all of the particles all at once doing this gives a good sort of global view of the vector field, and if you're studying math you might start to ask some natural questions about the nature of that fluid flow. Like for example you might wonder if you were to just look in a certain region and count the number of water molecules that are inside that region, does that count of yours change as you play this, this animation, as you let this flow over time? And in this particular example you can look and it doesn't look like the count changes, certainly not by much, it's not increasing over time or decreasing over time, and in a little bit if I gave you the function that determines this vector field you will be able to tell me why it's the case that the number of molecules in that region doesn't tend to change."}, {"video_title": "Divergence intuition, part 1.mp3", "Sentence": "I think, I think this is the one right, and now he's gonna be moving along this vector or whatever one is really attached to him. And thinking about all of the particles all at once doing this gives a good sort of global view of the vector field, and if you're studying math you might start to ask some natural questions about the nature of that fluid flow. Like for example you might wonder if you were to just look in a certain region and count the number of water molecules that are inside that region, does that count of yours change as you play this, this animation, as you let this flow over time? And in this particular example you can look and it doesn't look like the count changes, certainly not by much, it's not increasing over time or decreasing over time, and in a little bit if I gave you the function that determines this vector field you will be able to tell me why it's the case that the number of molecules in that region doesn't tend to change. But if you were to look at another example, like a guy that looks like this, and if I were to say I want you to focus on what happens around the origin, in that little region around the origin, and you can probably predict how once I start playing it, once I put some water molecules in there and let them flow along the vectors that they flow along, the density inside that region around the origin decreases. So we put a whole bunch of vectors there, and I'll just play it for a quick instant, just kind of let it jump for an instant, and one thing that characterizes this field around the origin is that decrease in density, and what you might say if you want it to be suggestive of the operation that I'm leading to here, is that the water molecules tend to diverge away from the origin, so the kind of divergence of the vector field near that origin is positive, and you'll see what I mean mathematically by that in the next couple videos, but if we were to flip all of these vectors, right, and we were to flip them around, now if I were to ask about the density in that same region around the origin, you can probably see how it's going to increase, and when I play that fluid flow over just a short spurt of time, the density in that region increases, so these don't diverge away, they converge towards the origin, and that fact actually has some mathematical significance for the function representing this vector field around that point, and even if the vector field doesn't represent fluid flow, if it represents like a magnetic field or an electric field or things like that, there's a certain meaning to this idea of diverging away from a point or converging to a point, and another way that people sometimes think about this, if you look at that same kind of outward flowing vector field, is rather than thinking of a decrease in density, imagining that the fluid would have to constantly be repopulated around that point, so you're really thinking of the origin as a source of fluid, and if I had animated this better, a whole bunch of other points should be sources of fluid so that the density doesn't decrease everywhere, but the idea is that points of positive divergence, where things are diverging away, would have to have a source of that fluid in order to kind of keep things sustaining, and conversely, if you were to look at that kind of inward flow, or what you might call negative divergence example, and you were to play it, but it were to go continuously, you'd have to think of that center point as a sink, where all the fluid just sort of flows away, and that's actually a technical term. People will say the vector field has a sink at such and such point, or the electromagnetic field has a sink at such and such point, and that often has a certain significance, and if we go back to that original example here, where there is no change in fluid density, what you might notice, this feels a lot more like actual water than the other ones, because there is no change in density there, and if you can find a way to mathematically describe that lack of a change in density, that's a pretty good way to model water flow, and again, even if it's not water flow, but it's something like the electromagnetic field, there's often a significance to this no changing in density idea, so with that, I think I've jabbered on enough about the visuals of it, and in the next video, I'll tell you what divergence is mathematically, how you compute it, go through a couple examples, things like that."}, {"video_title": "Curvature intuition.mp3", "Sentence": "So what I'd like to do here is talk about curvature. Curvature. And I've drawn on the xy plane here a certain curve, so this is our x-axis, this is our y-axis, and this is a curve running through space. And I'd like you to imagine that this is a road of some kind and you're driving on it, and you're at a certain point. So let's say this point right here. And if you imagine what it feels like to drive along this road and where you need to have your steering wheel, you're turning it a little bit to the right, not a lot because it's kind of a gentle curve at this point, you're not curving a lot, but the steering wheel isn't straight, you are still turning on the road. And now, imagine that your steering wheel's stuck, that it's not gonna move, and however you're turning it, you're stuck in that situation."}, {"video_title": "Curvature intuition.mp3", "Sentence": "And I'd like you to imagine that this is a road of some kind and you're driving on it, and you're at a certain point. So let's say this point right here. And if you imagine what it feels like to drive along this road and where you need to have your steering wheel, you're turning it a little bit to the right, not a lot because it's kind of a gentle curve at this point, you're not curving a lot, but the steering wheel isn't straight, you are still turning on the road. And now, imagine that your steering wheel's stuck, that it's not gonna move, and however you're turning it, you're stuck in that situation. What's gonna happen, hopefully you're on an open field or something, because your car's gonna trace out some kind of circle. Your steering wheel can't do anything different, you're just turning at a certain rate, and that's gonna have you tracing out some giant circle. And this depends on where you are, right?"}, {"video_title": "Curvature intuition.mp3", "Sentence": "And now, imagine that your steering wheel's stuck, that it's not gonna move, and however you're turning it, you're stuck in that situation. What's gonna happen, hopefully you're on an open field or something, because your car's gonna trace out some kind of circle. Your steering wheel can't do anything different, you're just turning at a certain rate, and that's gonna have you tracing out some giant circle. And this depends on where you are, right? If you had been at a different point on the curve, where the curve was rotating a lot, let's say you were back a little bit towards the start, at the start here, you have to turn the steering wheel to the right, but you're turning it much more sharply to stay on this part of the curve than you were to be on this relatively straight part. And the circle that you draw as a result is much smaller. And this turns out to be a pretty nice way to think about a measure for just how much the curve actually curves."}, {"video_title": "Curvature intuition.mp3", "Sentence": "And this depends on where you are, right? If you had been at a different point on the curve, where the curve was rotating a lot, let's say you were back a little bit towards the start, at the start here, you have to turn the steering wheel to the right, but you're turning it much more sharply to stay on this part of the curve than you were to be on this relatively straight part. And the circle that you draw as a result is much smaller. And this turns out to be a pretty nice way to think about a measure for just how much the curve actually curves. And one way you could do this is you can think, okay, what is the radius of that circle, the circle that you would trace out if your steering wheel locked at any given point? And if you kind of follow the point along different parts of the curve and see, oh, what's the different circle that my car would trace out if it was stuck at that point? You get circles of varying different radii, right?"}, {"video_title": "Curvature intuition.mp3", "Sentence": "And this turns out to be a pretty nice way to think about a measure for just how much the curve actually curves. And one way you could do this is you can think, okay, what is the radius of that circle, the circle that you would trace out if your steering wheel locked at any given point? And if you kind of follow the point along different parts of the curve and see, oh, what's the different circle that my car would trace out if it was stuck at that point? You get circles of varying different radii, right? And this radius actually has a very special name. I'll call this R. This is called the radius of curvature. And you can kind of see how this is a good way to describe how much you're turning, radius of curvature."}, {"video_title": "Curvature intuition.mp3", "Sentence": "You get circles of varying different radii, right? And this radius actually has a very special name. I'll call this R. This is called the radius of curvature. And you can kind of see how this is a good way to describe how much you're turning, radius of curvature. You may have heard with a car descriptions of the turning radius. You know, if you have a car with a very good turning radius, it's very small, because what that means is if you turned it all the way, you could trace out just a very small circle. But a car with a bad turning radius, you know, you don't turn very much at all, so you'd have to trace out a much larger circle."}, {"video_title": "Curvature intuition.mp3", "Sentence": "And you can kind of see how this is a good way to describe how much you're turning, radius of curvature. You may have heard with a car descriptions of the turning radius. You know, if you have a car with a very good turning radius, it's very small, because what that means is if you turned it all the way, you could trace out just a very small circle. But a car with a bad turning radius, you know, you don't turn very much at all, so you'd have to trace out a much larger circle. And curvature itself isn't this R. It's not the radius of curvature. But what it is is it's the reciprocal of that, one over R. And there's a special symbol for it. It's kind of a K, and I'm not sure in handwriting how I'm gonna distinguish it from an actual K. Maybe give it a little curly there."}, {"video_title": "Curvature intuition.mp3", "Sentence": "But a car with a bad turning radius, you know, you don't turn very much at all, so you'd have to trace out a much larger circle. And curvature itself isn't this R. It's not the radius of curvature. But what it is is it's the reciprocal of that, one over R. And there's a special symbol for it. It's kind of a K, and I'm not sure in handwriting how I'm gonna distinguish it from an actual K. Maybe give it a little curly there. It's the Greek letter kappa, and this is curvature. And I want you to think for a second why we would take one over R. R is a perfectly fine description of how much the road curves. But why is it that you would think one divided by R instead of R itself?"}, {"video_title": "Curvature intuition.mp3", "Sentence": "It's kind of a K, and I'm not sure in handwriting how I'm gonna distinguish it from an actual K. Maybe give it a little curly there. It's the Greek letter kappa, and this is curvature. And I want you to think for a second why we would take one over R. R is a perfectly fine description of how much the road curves. But why is it that you would think one divided by R instead of R itself? And the reason, basically, is you want curvature to be a measure of how much it curves in the sense that more sharp turns should give you a higher number. So if you're at a point where you're turning the steering wheel a lot, you want that to result in a much higher number. But radius of curvature will be really small when you're turning it a lot."}, {"video_title": "Curvature intuition.mp3", "Sentence": "But why is it that you would think one divided by R instead of R itself? And the reason, basically, is you want curvature to be a measure of how much it curves in the sense that more sharp turns should give you a higher number. So if you're at a point where you're turning the steering wheel a lot, you want that to result in a much higher number. But radius of curvature will be really small when you're turning it a lot. But if you're at a point that's basically like a straight road, you know, there's some slight curve to it, but it's basically a straight road, you want the curvature to be a very small number. But in this case, the radius of curvature is very large. So it's pretty helpful to just have one divided by R as the measure of how much the road is turning."}, {"video_title": "Curvature intuition.mp3", "Sentence": "But radius of curvature will be really small when you're turning it a lot. But if you're at a point that's basically like a straight road, you know, there's some slight curve to it, but it's basically a straight road, you want the curvature to be a very small number. But in this case, the radius of curvature is very large. So it's pretty helpful to just have one divided by R as the measure of how much the road is turning. And in the next video, I'm gonna go ahead and start describing a little bit more mathematically how we capture this value. Because as a loose description, if you're just kind of drawing pictures, it's perfectly fine to say, oh, yeah, yeah, you imagine a circle that's kind of closely hugging the curve. It's what your steering wheel would do if you were locked."}, {"video_title": "Curvature intuition.mp3", "Sentence": "So it's pretty helpful to just have one divided by R as the measure of how much the road is turning. And in the next video, I'm gonna go ahead and start describing a little bit more mathematically how we capture this value. Because as a loose description, if you're just kind of drawing pictures, it's perfectly fine to say, oh, yeah, yeah, you imagine a circle that's kind of closely hugging the curve. It's what your steering wheel would do if you were locked. But in math, we will describe this curve parametrically. It'll be the output of a certain vector-valued function. And I wanna know how you can capture this idea, this one over R curvature idea in a certain formula."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I want to introduce you to a different way. Some people might have taught this first, but the way I taught it in the first double integral video is kind of the way that I always think about when I do the problems. But sometimes it's more useful to think about the way I'm about to show you. And maybe you won't see the difference. Or maybe you say, oh, Sal, those are just the exact same things. Someone actually emailed me and told me that I should make it so I could scroll things. And I said, oh, that's not too hard to do."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And maybe you won't see the difference. Or maybe you say, oh, Sal, those are just the exact same things. Someone actually emailed me and told me that I should make it so I could scroll things. And I said, oh, that's not too hard to do. So I just did that. I scrolled my drawing. But anyway, let's say we have a surface in three dimensions."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I said, oh, that's not too hard to do. So I just did that. I scrolled my drawing. But anyway, let's say we have a surface in three dimensions. It's a function of x and y. You give me a coordinate down here, and I'll tell you how high the surface is at that point. And we want to figure out the volume under that surface."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But anyway, let's say we have a surface in three dimensions. It's a function of x and y. You give me a coordinate down here, and I'll tell you how high the surface is at that point. And we want to figure out the volume under that surface. So we can very easily figure out the volume of a very small column underneath this surface. So this whole area, or this whole volume, is what we're trying to figure out, right? Between the dotted lines."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we want to figure out the volume under that surface. So we can very easily figure out the volume of a very small column underneath this surface. So this whole area, or this whole volume, is what we're trying to figure out, right? Between the dotted lines. I think you can see it. You have some experience visualizing this right now. So let's say that I have a little area here."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Between the dotted lines. I think you can see it. You have some experience visualizing this right now. So let's say that I have a little area here. We could call that dA. Let me see if I can draw this. Let's say we have a little area down here."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that I have a little area here. We could call that dA. Let me see if I can draw this. Let's say we have a little area down here. A little square, the xy-plane. And it's, depending on how you view it, this side of it is dx. Its length is dx."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say we have a little area down here. A little square, the xy-plane. And it's, depending on how you view it, this side of it is dx. Its length is dx. And its height, you could say, on that side is dy. Because it's a little small change in y there, and it's a little small change in x here. And its area, the area of this little square, is going to be dx times dy."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Its length is dx. And its height, you could say, on that side is dy. Because it's a little small change in y there, and it's a little small change in x here. And its area, the area of this little square, is going to be dx times dy. And then if we wanted to figure out the volume of the solid between this little area and the surface, we could just multiply this area times the function, right? Because the height at this point is going to be the value of the function, roughly at this point. This is going to be an approximation, and then we're going to take an infinite sum."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And its area, the area of this little square, is going to be dx times dy. And then if we wanted to figure out the volume of the solid between this little area and the surface, we could just multiply this area times the function, right? Because the height at this point is going to be the value of the function, roughly at this point. This is going to be an approximation, and then we're going to take an infinite sum. I think you know where this is going. But let me do that. Let me at least draw the little column that I want to show you."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be an approximation, and then we're going to take an infinite sum. I think you know where this is going. But let me do that. Let me at least draw the little column that I want to show you. So that's one end of it. That's another end of it. That's the front end of it."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me at least draw the little column that I want to show you. So that's one end of it. That's another end of it. That's the front end of it. That's the other end of it. So we have a little figure that looks something like that. A little column."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's the front end of it. That's the other end of it. So we have a little figure that looks something like that. A little column. Right? It intersects the top of the surface. And the volume of this column, not too difficult."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "A little column. Right? It intersects the top of the surface. And the volume of this column, not too difficult. It's going to be this little area down here, which is, we could call that da. Sometimes written like that. da."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the volume of this column, not too difficult. It's going to be this little area down here, which is, we could call that da. Sometimes written like that. da. It's a little bit small area. And we're going to multiply that area times the height of this column, and that's the function at that point. So it's f of x and y."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "da. It's a little bit small area. And we're going to multiply that area times the height of this column, and that's the function at that point. So it's f of x and y. And of course, we could have also written it as, this da is just dx times dy, or dy times dx. I'm going to write it in every different way. So we could also have written this as f of xy times dx times dy, and of course, since multiplication is associative, I could have also written it as f of xy times dy dx."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's f of x and y. And of course, we could have also written it as, this da is just dx times dy, or dy times dx. I'm going to write it in every different way. So we could also have written this as f of xy times dx times dy, and of course, since multiplication is associative, I could have also written it as f of xy times dy dx. These are all equivalent, and these all represent the volume of this column that's between this little area here and the surface. So now, if we wanted to figure out the volume of the entire surface, there's a couple of things we could do. We could add up all the volumes in the x direction, because between the lower x bound and the upper x bound, and then we'd have kind of a thin sheet, although it'll already have some depth, because we're not adding up just dx's, there's also a dy back there."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we could also have written this as f of xy times dx times dy, and of course, since multiplication is associative, I could have also written it as f of xy times dy dx. These are all equivalent, and these all represent the volume of this column that's between this little area here and the surface. So now, if we wanted to figure out the volume of the entire surface, there's a couple of things we could do. We could add up all the volumes in the x direction, because between the lower x bound and the upper x bound, and then we'd have kind of a thin sheet, although it'll already have some depth, because we're not adding up just dx's, there's also a dy back there. So we would have a volume of a figure that would extend from the lower x all the way to the upper x, go back dy, and come back here, if we wanted to sum up all the dx's. And if we wanted to do that, which expression would we use? Well, we would be summing with respect to x first, so we could use this expression, right?"}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could add up all the volumes in the x direction, because between the lower x bound and the upper x bound, and then we'd have kind of a thin sheet, although it'll already have some depth, because we're not adding up just dx's, there's also a dy back there. So we would have a volume of a figure that would extend from the lower x all the way to the upper x, go back dy, and come back here, if we wanted to sum up all the dx's. And if we wanted to do that, which expression would we use? Well, we would be summing with respect to x first, so we could use this expression, right? And actually, we could write it here, but it just becomes confusing if we're summing with respect to x, but we have the dy written here first. It's really not incorrect, but it just becomes a little ambiguous, are we summing with respect to x or y? But here, we could say, OK, if we want to sum up all the dx's first, let's do that."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, we would be summing with respect to x first, so we could use this expression, right? And actually, we could write it here, but it just becomes confusing if we're summing with respect to x, but we have the dy written here first. It's really not incorrect, but it just becomes a little ambiguous, are we summing with respect to x or y? But here, we could say, OK, if we want to sum up all the dx's first, let's do that. We're taking the sum with respect to x. And I'm going to write down the actual, normally you just write numbers here, but I'm going to say, well, the lower bound here is x is equal to a, and the upper bound here is x is equal to b. And that'll give us the volume of, you could imagine, a sheet with depth, right?"}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But here, we could say, OK, if we want to sum up all the dx's first, let's do that. We're taking the sum with respect to x. And I'm going to write down the actual, normally you just write numbers here, but I'm going to say, well, the lower bound here is x is equal to a, and the upper bound here is x is equal to b. And that'll give us the volume of, you could imagine, a sheet with depth, right? The sheet is going to be parallel to the x-axis, right? And then once we have that sheet, my video, I think that's the newspaper people trying to sell me something. Anyway, so once we have this sheet, I'll try to draw it here, too, I don't want to get this picture too muddied up."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that'll give us the volume of, you could imagine, a sheet with depth, right? The sheet is going to be parallel to the x-axis, right? And then once we have that sheet, my video, I think that's the newspaper people trying to sell me something. Anyway, so once we have this sheet, I'll try to draw it here, too, I don't want to get this picture too muddied up. But once we have that sheet, then we could integrate those, we could add up the dy's, right? Because this width right here is still dy, we could add up all the different dy's, and we would have the volume of the whole figure. So once we take this sum, then we could take this sum, where y is going from its bottom, which we said was c, from y is equal to c, to y's upper bound, to y is equal to d. Fair enough."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Anyway, so once we have this sheet, I'll try to draw it here, too, I don't want to get this picture too muddied up. But once we have that sheet, then we could integrate those, we could add up the dy's, right? Because this width right here is still dy, we could add up all the different dy's, and we would have the volume of the whole figure. So once we take this sum, then we could take this sum, where y is going from its bottom, which we said was c, from y is equal to c, to y's upper bound, to y is equal to d. Fair enough. And then once we evaluate this whole thing, we have the volume of this solid, or the volume under the surface. Now we could have gone the other way. I know this gets a little bit messy, but I think you get what I'm saying."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So once we take this sum, then we could take this sum, where y is going from its bottom, which we said was c, from y is equal to c, to y's upper bound, to y is equal to d. Fair enough. And then once we evaluate this whole thing, we have the volume of this solid, or the volume under the surface. Now we could have gone the other way. I know this gets a little bit messy, but I think you get what I'm saying. Let's start with that little dA we had originally. Let's start with that little dA. Instead of going this way, instead of summing up the dx's and getting this sheet, let's sum up the dy's first."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I know this gets a little bit messy, but I think you get what I'm saying. Let's start with that little dA we had originally. Let's start with that little dA. Instead of going this way, instead of summing up the dx's and getting this sheet, let's sum up the dy's first. So we could take, we're summing in the y direction first, so 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 summing the dy's first, we would take the sum, we would take the integral with respect to y, and 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."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Instead of going this way, instead of summing up the dx's and getting this sheet, let's sum up the dy's first. So we could take, we're summing in the y direction first, so 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 summing the dy's first, we would take the sum, we would take the integral with respect to y, and 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."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 small depths, and the lower bound is x is equal to a, and 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."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 small depths, and the lower bound is x is equal to a, and 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. Because the xy-plane here is our domain. So we're going to do a double integral, a two-dimensional integral."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. Because the xy-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. The reason why I want to show you this is, you see this in physics books all the time."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. The reason why I want to show you this is, you see this in physics books all the time. And 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, it actually is more confusing. 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."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The reason why I want to show you this is, you see this in physics books all the time. And 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, it actually is more confusing. 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 as just adding up all of these squares. Where we do it here, we're very ordered about it, right?"}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 as 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 domain, it leaves it a little bit ambiguous in terms of how we're going to sum up all of the da's."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 domain, it 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 could do it in polar coordinates. You could do it a ton of different ways."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 domain, it 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 could do it in polar coordinates. You could do it a ton of different ways. But I just wanted to show you that this is another way of having an intuition of the volume under a surface. And these are the exact same thing as this type of notation that you might see in a physics book. Sometimes they won't write a domain, sometimes they'd write over a surface."}, {"video_title": "Double integrals 4   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could do it a ton of different ways. But I just wanted to show you that this is another way of having an intuition of the volume under a surface. And these are the exact same thing as this type of notation that you might see in a physics book. 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see if we can use our knowledge of Green's theorem to solve some actual line integrals. And actually, before I show an example, I want to make one clarification on Green's theorem. All of the examples that I did is I had a region like this. And the inside of the region was to the left of what we traversed. So all my examples, I went counterclockwise. And so our region was to the left of, if you imagine walking along the path in that direction, it was always to our left. And that's the situation which Green's theorem would apply."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the inside of the region was to the left of what we traversed. So all my examples, I went counterclockwise. And so our region was to the left of, if you imagine walking along the path in that direction, it was always to our left. And that's the situation which Green's theorem would apply. So if you were to take a line integral along this path, the closed line integral, maybe we could even specify it like that. You'll see that in some textbooks. Along the curve C of F dot dr, this is what equals the double integral over this region r of the partial of q with respect to x minus the partial of p with respect to y d area."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that's the situation which Green's theorem would apply. So if you were to take a line integral along this path, the closed line integral, maybe we could even specify it like that. You'll see that in some textbooks. Along the curve C of F dot dr, this is what equals the double integral over this region r of the partial of q with respect to x minus the partial of p with respect to y d area. And just as a reminder, this q and p are coming from the components of F. F, in this situation, F would be written as F of xy is equal to p of xy times the i component plus q of xy times the j component. So this is a situation where the inside of the region is to the left of the direction that we're taking the path. If it was in the reverse, then we would put a minus sign right here."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Along the curve C of F dot dr, this is what equals the double integral over this region r of the partial of q with respect to x minus the partial of p with respect to y d area. And just as a reminder, this q and p are coming from the components of F. F, in this situation, F would be written as F of xy is equal to p of xy times the i component plus q of xy times the j component. So this is a situation where the inside of the region is to the left of the direction that we're taking the path. If it was in the reverse, then we would put a minus sign right here. If this arrow went the other way, we'd put a minus sign. And we can do that because we know that when we're taking line integrals through vector fields, if we reverse the direction, it becomes the negative of that. We showed that, I think, four or five videos ago."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If it was in the reverse, then we would put a minus sign right here. If this arrow went the other way, we'd put a minus sign. And we can do that because we know that when we're taking line integrals through vector fields, if we reverse the direction, it becomes the negative of that. We showed that, I think, four or five videos ago. With that said, it was convenient to write Green's theorem up here. Let's actually solve a problem. Let's say I have the line integral."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We showed that, I think, four or five videos ago. With that said, it was convenient to write Green's theorem up here. Let's actually solve a problem. Let's say I have the line integral. Let's say we're over a curve. I'll define the curve in a second. But let's say that the integral we're trying to solve is x squared minus y squared dx plus 2xy dy."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say I have the line integral. Let's say we're over a curve. I'll define the curve in a second. But let's say that the integral we're trying to solve is x squared minus y squared dx plus 2xy dy. And then our curve, they're giving us the boundary. The boundary is the region. I'll do it in a different color."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's say that the integral we're trying to solve is x squared minus y squared dx plus 2xy dy. And then our curve, they're giving us the boundary. The boundary is the region. I'll do it in a different color. So the curve is boundary of the region given by all of the points x, y such that x is greater than or equal to 0, less than or equal to 1. And then y is greater than or equal to 2x squared and less than or equal to 2x. So let's draw this region that we're dealing with right now."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll do it in a different color. So the curve is boundary of the region given by all of the points x, y such that x is greater than or equal to 0, less than or equal to 1. And then y is greater than or equal to 2x squared and less than or equal to 2x. So let's draw this region that we're dealing with right now. So let me draw my y-axis and then my x-axis right there. And let's see, x goes from 0 to 1. So if we make, that's obviously 0."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's draw this region that we're dealing with right now. So let me draw my y-axis and then my x-axis right there. 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 it's 1, 2. This is a line y is equal to 2x."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But if you're below 1, this is actually going to be smaller than that. So the upper boundary is 2x. So it's 1, 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And then this bottom curve, maybe I'll do that in yellow. And then the bottom curve right here is y is going to be greater than 2x squared."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The line y is equal to 2x looks something like that. That right there is y is equal to 2x. And then this bottom curve, 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. And of course, the region that they're talking about is this. But we're saying that the curve is the boundary of this region."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the bottom curve right here is y is going to be greater than 2x squared. It might look something like this. And of course, the region that they're talking about is this. But we're saying that the curve is the boundary of this region. And we're going to go in a counterclockwise direction. I have to specify that. Counterclockwise direction."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we're saying that the curve is the boundary of this region. And we're going to go in a counterclockwise direction. I have to specify that. Counterclockwise direction. So our curve, we could start at any point, really. But we're going to go like that and 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Counterclockwise direction. So our curve, we could start at any point, really. But we're going to go like that and 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. So our region, let's just do our region. This integral right here is going to go. I'll just do it the way y goes from y."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's define our region. So our region, let's just do our region. This integral right here is going to go. I'll just do it the way y goes from y. Let me do it. y varies from y is equal to 2x squared to y is equal to 2x. And so maybe we'll integrate with respect to y first."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll just do it the way y goes from y. Let me do it. y varies from y is equal to 2x squared to y is equal to 2x. 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. And now we just have to figure out what goes over here."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. And now we just have to figure out what goes over here. Green's theorem. This right here is our f. It would look like this in this situation. f of x, y is going to be equal to x squared minus y squared i plus 2xyj."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we just have to figure out what goes over here. Green's theorem. This right here is our f. It would look like this in this situation. f of x, y is going to be equal to x squared minus y squared i plus 2xyj. 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."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "f of x, y is going to be equal to x squared minus y squared i plus 2xyj. 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 x, y. So inside of here, we're just going to apply Green's theorem straight up. So the partial of q with respect to x."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this expression right here is our p of x, y. And this expression right here is our q of x, y. So inside of here, we're just going to apply Green's theorem straight up. So the partial of q with respect to x. 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 a 2y. And then from that, we're going to subtract the partial of p with respect to y."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the partial of q with respect to x. 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 a 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. So you have a minus 2y just like that."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. So you have a minus 2y just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus 2y. I'm just subtracting a negative."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have a minus 2y just like that. And so this simplifies to 2y minus minus 2y. That's 2y plus 2y. I'm just subtracting a negative. Or this inside, and just to save space, this inside, that's just 4y. So let me just, I don't want to have to rewrite the boundaries. So that right there is the same thing as 4y."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just subtracting a negative. Or this inside, and just to save space, this inside, that's just 4y. So let me just, I don't want to have to rewrite the boundaries. So that right there is the same thing as 4y. The partial of q with respect to y, 2y, minus the partial of p with respect to y, which is minus 2y. You subtract a negative, you get a positive. You have 4y."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that right there is the same thing as 4y. The partial of q with respect to y, 2y, minus the partial of p with respect to y, which is minus 2y. You subtract a negative, you get a positive. You have 4y. So let's take the antiderivative of the inside with respect to y. And we're going to get 2y squared. Let me do it a little bit lower."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You have 4y. So let's take the antiderivative of the inside with respect to y. And we're going to get 2y squared. Let me do it a little bit lower. We're going to get 2y squared. If you take the derivative of this with respect, the partial with respect to y, you're going to get 4y. And we're going to evaluate that from y is equal to 2x squared to y is equal to 2x."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me do it a little bit lower. We're going to get 2y squared. If you take the derivative of this with respect, the partial with respect to y, you're going to get 4y. And we're going to evaluate that from y is equal to 2x squared to y is equal to 2x. And of course, we still have the outside integral still there. x goes from 0 to 1 dx. This thing is going to be equal to the integral from 0 to 1."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to evaluate that from y is equal to 2x squared to y is equal to 2x. And of course, we still have the outside integral still there. x goes from 0 to 1 dx. This thing is going to be equal to the integral from 0 to 1. And then we evaluate it first at 2x. So you put 2x in here. 2x squared is 4x squared, right?"}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This thing is going to be equal to the integral from 0 to 1. And then we evaluate it first at 2x. So you put 2x in here. 2x squared is 4x squared, right? 2 squared x squared. So 4x squared times 2 is going to be 8x squared minus, put this guy in there. 2x squared squared is 4x to the fourth times 2 is 8x to the 4th."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "2x squared is 4x squared, right? 2 squared x squared. So 4x squared times 2 is going to be 8x squared minus, put this guy in there. 2x squared squared is 4x to the fourth times 2 is 8x to the 4th. Did I do that right? 2x squared, I'm going to put it there for y, substitute y with it. That squared is 4x to the fourth times 2 is 8x to the fourth."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "2x squared squared is 4x to the fourth times 2 is 8x to the 4th. Did I do that right? 2x squared, I'm going to put it there for y, substitute y with it. That squared is 4x to the fourth times 2 is 8x to the fourth. Very good. All right. Now dx, now this is just a straightforward one-dimensional integral."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That squared is 4x to the fourth times 2 is 8x to the fourth. Very good. All right. Now dx, now this is just a straightforward one-dimensional integral. This is going to be equal to the antiderivative of 8x squared is 8 thirds x to the third. And then the antiderivative of x to the fourth is minus 8 fifths x to the fifth. And we're going to have to evaluate that from 0 to 1."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now dx, now this is just a straightforward one-dimensional integral. This is going to be equal to the antiderivative of 8x squared is 8 thirds x to the third. And then the antiderivative of x to the fourth is minus 8 fifths x to the fifth. And we're going to have to evaluate that from 0 to 1. I'll give it a little line there sometimes. And when you put 1 in there, you get, I'll do it in a different color. We get 8 fifths times 1 to the third, which is 8 fifths, minus 8 fifths."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to have to evaluate that from 0 to 1. I'll give it a little line there sometimes. And when you put 1 in there, you get, I'll do it in a different color. We get 8 fifths times 1 to the third, which is 8 fifths, minus 8 fifths. And then we're going to have minus, when you put 0 in here, you're just going to get a bunch of 0's. Oh, sorry. I made a mistake."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We get 8 fifths times 1 to the third, which is 8 fifths, minus 8 fifths. And then we're going to have minus, when you put 0 in here, you're just going to get a bunch of 0's. Oh, sorry. I made a mistake. It's not a blunder. It's 8 thirds. 8 thirds times 1 to the third minus 8 fifths times 1 to the fifth."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I made a mistake. It's not a blunder. It's 8 thirds. 8 thirds times 1 to the third minus 8 fifths times 1 to the fifth. So that's minus 8 fifths. And then when you subtract the 0, so then minus, you evaluate 0 here, you're just going to get a bunch of 0's. You don't have to do anything else."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "8 thirds times 1 to the third minus 8 fifths times 1 to the fifth. So that's minus 8 fifths. And then when you subtract the 0, so then minus, you evaluate 0 here, you're just going to get a bunch of 0's. You don't have to do anything else. So now we just have to subtract these two fractions. So let's get a common denominator of 15. 8 thirds is the same thing if we multiply the numerator and denominator by 5."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You don't have to do anything else. So now we just have to subtract these two fractions. So let's get a common denominator of 15. 8 thirds is the same thing if we multiply the numerator and denominator by 5. That is 40 fifteenths. And if we multiply this numerator and denominator by 3, that's going to be 24 over 15. So minus 24 over 15."}, {"video_title": "Green's theorem example 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "8 thirds is the same thing if we multiply the numerator and denominator by 5. That is 40 fifteenths. And if we multiply this numerator and denominator by 3, that's going to be 24 over 15. So minus 24 over 15. And we get it being equal to 16 over 15. So using Green's theorem, we were able to find the answer to this integral up here. It's equal to 16 fifteenths."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "So far, when I've talked about the gradient of a function, and you know, let's think about this as a multivariable function with just two inputs, those are the easiest to think about. So maybe it's something like x squared plus y squared, very friendly function. When I've talked about the gradient, I've left open a mystery. We have the way of computing it, and the way that you think about computing it is you just take this vector and you just throw the partial derivatives in there, partial with respect to x and the partial with respect to y, with respect to y, and if it was a higher dimensional input, then the output would have as many variables as you need. If it was f of x, y, z, you'd have partial x, partial y, partial z. And this is the way to compute it. But then I gave you a graphical intuition."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "We have the way of computing it, and the way that you think about computing it is you just take this vector and you just throw the partial derivatives in there, partial with respect to x and the partial with respect to y, with respect to y, and if it was a higher dimensional input, then the output would have as many variables as you need. If it was f of x, y, z, you'd have partial x, partial y, partial z. And this is the way to compute it. But then I gave you a graphical intuition. I said that it points in the direction of steepest ascent. And maybe the way you think about that is you have your input space, which in this case is the xy plane, and you think of it as somehow mapping over to the number line, to your output space. And if you have a given point somewhere, the question is, of all the possible directions that you can move away from this point, all those different directions you can go, which one of them, you know, like this point will land somewhere on the function, and as you move in the various directions, maybe one of them nudges your output a little bit, one of them nudges it a lot, one of it slides it negative, you know, one of them slides it negative a lot."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "But then I gave you a graphical intuition. I said that it points in the direction of steepest ascent. And maybe the way you think about that is you have your input space, which in this case is the xy plane, and you think of it as somehow mapping over to the number line, to your output space. And if you have a given point somewhere, the question is, of all the possible directions that you can move away from this point, all those different directions you can go, which one of them, you know, like this point will land somewhere on the function, and as you move in the various directions, maybe one of them nudges your output a little bit, one of them nudges it a lot, one of it slides it negative, you know, one of them slides it negative a lot. Which one of these directions results in the greatest increase to your function? And this was the loose intuition. If you want to think in terms of graphs, we could look over at the graph of f of x squared, and this is the gradient field."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And if you have a given point somewhere, the question is, of all the possible directions that you can move away from this point, all those different directions you can go, which one of them, you know, like this point will land somewhere on the function, and as you move in the various directions, maybe one of them nudges your output a little bit, one of them nudges it a lot, one of it slides it negative, you know, one of them slides it negative a lot. Which one of these directions results in the greatest increase to your function? And this was the loose intuition. If you want to think in terms of graphs, we could look over at the graph of f of x squared, and this is the gradient field. All of these vectors in the xy plane are the gradients, and you kind of look from below, you can maybe see why each one of these points in the direction you should move to walk uphill on that graph as fast as you can. You know, if you're a mountain climber and you want to get to the top as quickly as possible, these tell you the direction that you should move to go as quickly. This is why you call it direction of steepest ascent."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "If you want to think in terms of graphs, we could look over at the graph of f of x squared, and this is the gradient field. All of these vectors in the xy plane are the gradients, and you kind of look from below, you can maybe see why each one of these points in the direction you should move to walk uphill on that graph as fast as you can. You know, if you're a mountain climber and you want to get to the top as quickly as possible, these tell you the direction that you should move to go as quickly. This is why you call it direction of steepest ascent. So back over here, I don't see the connection immediately, or at least when I was first learning about it, it wasn't clear why this combination of partial derivatives has anything to do with choosing the best direction. And now that we've learned about the directional derivative, I can give you a little bit of an intuition. So let's say instead of thinking about, you know, all the possible directions, and all of the possible changes to the output that they have, I'll fill in my line there."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "This is why you call it direction of steepest ascent. So back over here, I don't see the connection immediately, or at least when I was first learning about it, it wasn't clear why this combination of partial derivatives has anything to do with choosing the best direction. And now that we've learned about the directional derivative, I can give you a little bit of an intuition. So let's say instead of thinking about, you know, all the possible directions, and all of the possible changes to the output that they have, I'll fill in my line there. You know, let's say you just have, you've got your point where you're evaluating things, and then you just have a single vector. And let's actually make it a univector. Let's make it the case that this guy has a length of one."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "So let's say instead of thinking about, you know, all the possible directions, and all of the possible changes to the output that they have, I'll fill in my line there. You know, let's say you just have, you've got your point where you're evaluating things, and then you just have a single vector. And let's actually make it a univector. Let's make it the case that this guy has a length of one. So I'll go over here and I'll just think of that guy as being v, and say that v has a length of one. So this is our vector. We know now, having learned about the directional derivative that you can tell the rate at which the function changes as you move in this direction by taking the directional derivative of your function, and let's say this point, I don't know, what's a good name for this point?"}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "Let's make it the case that this guy has a length of one. So I'll go over here and I'll just think of that guy as being v, and say that v has a length of one. So this is our vector. We know now, having learned about the directional derivative that you can tell the rate at which the function changes as you move in this direction by taking the directional derivative of your function, and let's say this point, I don't know, what's a good name for this point? Just like a, b. a, b is this point. When you evaluate this at a, b, and the way that you do that is just dotting the gradient of f. I should say dotting it, evaluate it at that point, because gradient is a vector-valued function and we just want a specific vector here. So evaluating that at your point, a, b, together with whatever the vector is, whatever that value is."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "We know now, having learned about the directional derivative that you can tell the rate at which the function changes as you move in this direction by taking the directional derivative of your function, and let's say this point, I don't know, what's a good name for this point? Just like a, b. a, b is this point. When you evaluate this at a, b, and the way that you do that is just dotting the gradient of f. I should say dotting it, evaluate it at that point, because gradient is a vector-valued function and we just want a specific vector here. So evaluating that at your point, a, b, together with whatever the vector is, whatever that value is. And in this case, we're thinking of v as a univector. So this, this is how you tell the rate of change, and when I originally introduced the directional derivative, I gave kind of an indication why. You know, if you imagine dotting this together with, I don't know, let's say it was a vector that's like one, two."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "So evaluating that at your point, a, b, together with whatever the vector is, whatever that value is. And in this case, we're thinking of v as a univector. So this, this is how you tell the rate of change, and when I originally introduced the directional derivative, I gave kind of an indication why. You know, if you imagine dotting this together with, I don't know, let's say it was a vector that's like one, two. Really, you're thinking this vector represents one step in the x direction, two steps in the y direction, so the amount that it changes things should be one times the change caused by a pure step in the x direction, plus two times a change caused by a pure step in the y direction. So that was kind of the loose intuition. You can see the directional derivative video if you want a little bit more discussion on that."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "You know, if you imagine dotting this together with, I don't know, let's say it was a vector that's like one, two. Really, you're thinking this vector represents one step in the x direction, two steps in the y direction, so the amount that it changes things should be one times the change caused by a pure step in the x direction, plus two times a change caused by a pure step in the y direction. So that was kind of the loose intuition. You can see the directional derivative video if you want a little bit more discussion on that. And this is the formula that you have, but this starts to give us the key for how we could choose the direction of steepest descent, because now what we're really asking, when we say which one of these changes things the most, you know, maybe when you move in that direction, it changes f, you know, a little bit negatively, and we want to know, you know, maybe does another vector w, is the change caused by that gonna be positive? Is it gonna be as big as possible? What we're doing is we're saying, find the maximum for all unit vectors, so for all vectors v that satisfy the property that their length is one, find the maximum of the dot product between f, evaluated at that point, right, evaluated at whatever point we care about, and v. Find that maximum."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "You can see the directional derivative video if you want a little bit more discussion on that. And this is the formula that you have, but this starts to give us the key for how we could choose the direction of steepest descent, because now what we're really asking, when we say which one of these changes things the most, you know, maybe when you move in that direction, it changes f, you know, a little bit negatively, and we want to know, you know, maybe does another vector w, is the change caused by that gonna be positive? Is it gonna be as big as possible? What we're doing is we're saying, find the maximum for all unit vectors, so for all vectors v that satisfy the property that their length is one, find the maximum of the dot product between f, evaluated at that point, right, evaluated at whatever point we care about, and v. Find that maximum. Well, let's just think about what the dot product represents. So let's say we go over here and let's say, you know, let's say we evaluate the gradient vector, and it turns out that the gradient points in this direction, and maybe it's, you know, it doesn't have to be a unit vector, it might be something very long, like that. So if you imagine some vector v, you know, some unit vector v, let's say it was sticking off in this direction, the way that you interpret this dot product, the dot product between the gradient f and this new vector v, is you would project that vector directly, kind of a perpendicular projection onto your gradient vector, and you'd say, what's that length?"}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "What we're doing is we're saying, find the maximum for all unit vectors, so for all vectors v that satisfy the property that their length is one, find the maximum of the dot product between f, evaluated at that point, right, evaluated at whatever point we care about, and v. Find that maximum. Well, let's just think about what the dot product represents. So let's say we go over here and let's say, you know, let's say we evaluate the gradient vector, and it turns out that the gradient points in this direction, and maybe it's, you know, it doesn't have to be a unit vector, it might be something very long, like that. So if you imagine some vector v, you know, some unit vector v, let's say it was sticking off in this direction, the way that you interpret this dot product, the dot product between the gradient f and this new vector v, is you would project that vector directly, kind of a perpendicular projection onto your gradient vector, and you'd say, what's that length? You know, what's that length right there? And just as an example, it would be something a little bit less than one, right, because this is a unit vector, so as an example, let's say that was like 0.7. And then you'd multiply that by the length of the gradient itself, of the vector against which you're dotting, and maybe that guy, maybe the length of the entire gradient vector, just again as an example, maybe that's two."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "So if you imagine some vector v, you know, some unit vector v, let's say it was sticking off in this direction, the way that you interpret this dot product, the dot product between the gradient f and this new vector v, is you would project that vector directly, kind of a perpendicular projection onto your gradient vector, and you'd say, what's that length? You know, what's that length right there? And just as an example, it would be something a little bit less than one, right, because this is a unit vector, so as an example, let's say that was like 0.7. And then you'd multiply that by the length of the gradient itself, of the vector against which you're dotting, and maybe that guy, maybe the length of the entire gradient vector, just again as an example, maybe that's two. It doesn't have to be, it could be anything. But the way that you interpret this whole dot product then is to take the product of those two. You would take 0.7, the length of your projection, times the length of the original vector."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And then you'd multiply that by the length of the gradient itself, of the vector against which you're dotting, and maybe that guy, maybe the length of the entire gradient vector, just again as an example, maybe that's two. It doesn't have to be, it could be anything. But the way that you interpret this whole dot product then is to take the product of those two. You would take 0.7, the length of your projection, times the length of the original vector. And the question is, when is this maximized? What unit vector maximizes this? And if you start to imagine maybe swinging that unit vector around, so if instead of that guy, you were to use one that pointed a little bit more closely in the direction, then its projection would be a little bit longer."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "You would take 0.7, the length of your projection, times the length of the original vector. And the question is, when is this maximized? What unit vector maximizes this? And if you start to imagine maybe swinging that unit vector around, so if instead of that guy, you were to use one that pointed a little bit more closely in the direction, then its projection would be a little bit longer. Maybe that projection would be like 0.75 or something. If you take the unit vector that points directly in the same direction as that full vector, then the length of its projection is just the length of the vector itself. It would be one."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And if you start to imagine maybe swinging that unit vector around, so if instead of that guy, you were to use one that pointed a little bit more closely in the direction, then its projection would be a little bit longer. Maybe that projection would be like 0.75 or something. If you take the unit vector that points directly in the same direction as that full vector, then the length of its projection is just the length of the vector itself. It would be one. Because projecting it doesn't change what it is at all. So it shouldn't be too hard to convince yourself. And if you have shaky intuitions on the dot product, I'd suggest finding the videos we have on Khan Academy for those."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "It would be one. Because projecting it doesn't change what it is at all. So it shouldn't be too hard to convince yourself. And if you have shaky intuitions on the dot product, I'd suggest finding the videos we have on Khan Academy for those. Sal does a great job giving that deep intuition. But it should kind of make sense why the unit vector that points in the same direction as your gradient is gonna be what maximizes it. So the answer here, the answer to what vector maximizes this is gonna be, well, it's the gradient itself."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And if you have shaky intuitions on the dot product, I'd suggest finding the videos we have on Khan Academy for those. Sal does a great job giving that deep intuition. But it should kind of make sense why the unit vector that points in the same direction as your gradient is gonna be what maximizes it. So the answer here, the answer to what vector maximizes this is gonna be, well, it's the gradient itself. It is that gradient vector evaluated at the point we care about, except you'd normalize it because we're only considering unit vectors. So to do that, you just divide it by whatever its magnitude is. If its magnitude was already one, it stays one."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "So the answer here, the answer to what vector maximizes this is gonna be, well, it's the gradient itself. It is that gradient vector evaluated at the point we care about, except you'd normalize it because we're only considering unit vectors. So to do that, you just divide it by whatever its magnitude is. If its magnitude was already one, it stays one. If its magnitude was two, you're dividing it down by a half. So this is your answer. This is the direction of steepest descent."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "If its magnitude was already one, it stays one. If its magnitude was two, you're dividing it down by a half. So this is your answer. This is the direction of steepest descent. So I think one thing to notice here is the most fundamental fact is that the gradient is this tool for computing directional derivatives. You can think of that vector as something that you really want to dot against. And that's actually a pretty powerful thought."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "This is the direction of steepest descent. So I think one thing to notice here is the most fundamental fact is that the gradient is this tool for computing directional derivatives. You can think of that vector as something that you really want to dot against. And that's actually a pretty powerful thought. It's that the gradient, it's not just a vector. It's a vector that loves to be dotted together with other things. That's the fundamental."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And that's actually a pretty powerful thought. It's that the gradient, it's not just a vector. It's a vector that loves to be dotted together with other things. That's the fundamental. And as a consequence of this, as a consequence of that, the direction of steepest descent is that vector itself because anything, if you're saying what maximizes the dot product with that thing, it's, well, the vector that points in the same direction as that thing. And this can also give us an interpretation for the length of the gradient. We know the direction is the direction of steepest ascent, but what does the length mean?"}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "That's the fundamental. And as a consequence of this, as a consequence of that, the direction of steepest descent is that vector itself because anything, if you're saying what maximizes the dot product with that thing, it's, well, the vector that points in the same direction as that thing. And this can also give us an interpretation for the length of the gradient. We know the direction is the direction of steepest ascent, but what does the length mean? So let's give this guy a name. Let's give this normalized version of it a name. I'm just gonna call it W. So W will be the unit vector that points in the direction of the gradient."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "We know the direction is the direction of steepest ascent, but what does the length mean? So let's give this guy a name. Let's give this normalized version of it a name. I'm just gonna call it W. So W will be the unit vector that points in the direction of the gradient. If you take the directional derivative in the direction of W of F, what that means is the gradient of F dotted with that W. And if you kind of spell out what W means here, that means you're taking the gradient of the vector dotted with itself, but because it's W and not the gradient, we're normalizing. We're dividing that not by magnitude of F. That doesn't really make sense, but by the value of the gradient. And all of these, I'm just writing gradient of F, but maybe you should be thinking about gradient of F evaluated at AB, but I'm just being kind of lazy and just writing gradient of F. And the top, when you take the dot product with itself, what that means is the square of its magnitude."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "I'm just gonna call it W. So W will be the unit vector that points in the direction of the gradient. If you take the directional derivative in the direction of W of F, what that means is the gradient of F dotted with that W. And if you kind of spell out what W means here, that means you're taking the gradient of the vector dotted with itself, but because it's W and not the gradient, we're normalizing. We're dividing that not by magnitude of F. That doesn't really make sense, but by the value of the gradient. And all of these, I'm just writing gradient of F, but maybe you should be thinking about gradient of F evaluated at AB, but I'm just being kind of lazy and just writing gradient of F. And the top, when you take the dot product with itself, what that means is the square of its magnitude. But the whole thing is divided by the magnitude. So you can kind of cancel that out. You can say this doesn't need to be there."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "And all of these, I'm just writing gradient of F, but maybe you should be thinking about gradient of F evaluated at AB, but I'm just being kind of lazy and just writing gradient of F. And the top, when you take the dot product with itself, what that means is the square of its magnitude. But the whole thing is divided by the magnitude. So you can kind of cancel that out. You can say this doesn't need to be there. That exponent doesn't need to be there. And basically, the directional derivative, the directional derivative in the direction of the gradient itself has a value equal to the magnitude of the gradient. So this tells you when you're moving in that direction, in the direction of the gradient, the rate at which the function changes is given by the magnitude of the gradient."}, {"video_title": "Why the gradient is the direction of steepest ascent.mp3", "Sentence": "You can say this doesn't need to be there. That exponent doesn't need to be there. And basically, the directional derivative, the directional derivative in the direction of the gradient itself has a value equal to the magnitude of the gradient. So this tells you when you're moving in that direction, in the direction of the gradient, the rate at which the function changes is given by the magnitude of the gradient. So it's this really magical vector. It does a lot of things. It's the tool that lets you dot against other vectors to tell you the directional derivative."}, {"video_title": "Vector form of multivariable quadratic approximation.mp3", "Sentence": "Okay, so we are finally ready to express the quadratic approximation of a multivariable function in vector form. So I have the whole thing written out here where f is the function that we are trying to approximate, x-naught, y-naught is the constant point about which we are approximating, and then this entire expression is the quadratic approximation, which I've talked about in past videos, and if it seems very complicated or absurd or you're unfamiliar with it, and just dissecting it real quick, this over here is the constant term. This is just gonna evaluate to a constant, everything over here is the linear term, because it just involves taking a variable multiplied by a constant, and then the remainder, every one of these components will have two variables multiplied into it, so like x-squared comes up, and x times y, and y-squared comes up, so that's the quadratic term. Quadratic. Now to vectorize things, first of all, let's write down the input, the input variable x, y as a vector, and typically we'll do that with a bold-faced x to indicate that it's a vector, and its components are just gonna be the single variables x and y, the non-bold-faced, so this is the vector representing the variable input, and then correspondingly, a bold-faced x with a little subscript o, x-naught, is gonna be the constant input, the single point in space near which we are approximating, so when we write things like that, this constant term, simply enough, is gonna look like evaluating your function at that bold-faced x-naught, so that's probably the easiest one to handle. Now the linear term, this looks like a dot product, and if we kind of expand it out as the dot product, it looks like we're taking the partial derivative of f with respect to x, and then the partial derivative with respect to y, and we're evaluating both of those at that bold-faced x-naught input, x-naught as its input, now each one of those partial derivatives is multiplied by variable minus constant number, so this looks like taking the dot product here, I'm gonna erase the word linear, we're taking it with x minus x-naught, and y minus y-naught, this is just expressing the same linear term, but as a dot product, but the convenience here is that this is totally the same thing as saying the gradient of f, gradient of f, that's the vector that contains all the partial derivatives, evaluated at the special input, x-naught, and then we're taking the dot product between that and the variable vector, bold-faced x minus x-naught, since when you do this component-wise, bold-faced x minus x-naught, if we kind of think here, it'll be x the variable minus x-naught the constant, y the variable minus y-naught the constant, which is what we have up there, so this expression kind of vectorizes the whole linear term, and now the beef here, the hard part, how are we gonna vectorize this quadratic term? Now that's what I was leading to in the last couple videos, where I talked about how you express a quadratic form like this with a matrix, and the way that you do it, I'll just kind of scroll down to give us some room, the way that you do it is we'll have a matrix whose components are all of these constants, it'll be this 1 1 2 times the second partial derivative evaluated there, and I'm just gonna, for convenience's sake, I'm gonna just take 1 1 2 times the second partial derivative with respect to x, and leave it as understood that we're evaluating it at this point, and then on the other diagonal, you have 1 1 2 times the other kind of partial derivative with respect to y two times in a row, and then we're gonna multiply it by this constant here, but this term kind of gets broken apart into two different components, if you'll remember in the quadratic form video, it was always things where it was a, and then 2b and c as your constants for the quadratic form, so if we're interpreting this as two times something, then it gets broken down, and on one corner it shows up as f x y, and on the other one, kind of 1 1 2 f x y, so like both of these together are gonna constitute the entire mixed partial derivative, and then the way that we express the quadratic form is we're gonna multiply this by, well, by what?"}, {"video_title": "Vector form of multivariable quadratic approximation.mp3", "Sentence": "Quadratic. Now to vectorize things, first of all, let's write down the input, the input variable x, y as a vector, and typically we'll do that with a bold-faced x to indicate that it's a vector, and its components are just gonna be the single variables x and y, the non-bold-faced, so this is the vector representing the variable input, and then correspondingly, a bold-faced x with a little subscript o, x-naught, is gonna be the constant input, the single point in space near which we are approximating, so when we write things like that, this constant term, simply enough, is gonna look like evaluating your function at that bold-faced x-naught, so that's probably the easiest one to handle. Now the linear term, this looks like a dot product, and if we kind of expand it out as the dot product, it looks like we're taking the partial derivative of f with respect to x, and then the partial derivative with respect to y, and we're evaluating both of those at that bold-faced x-naught input, x-naught as its input, now each one of those partial derivatives is multiplied by variable minus constant number, so this looks like taking the dot product here, I'm gonna erase the word linear, we're taking it with x minus x-naught, and y minus y-naught, this is just expressing the same linear term, but as a dot product, but the convenience here is that this is totally the same thing as saying the gradient of f, gradient of f, that's the vector that contains all the partial derivatives, evaluated at the special input, x-naught, and then we're taking the dot product between that and the variable vector, bold-faced x minus x-naught, since when you do this component-wise, bold-faced x minus x-naught, if we kind of think here, it'll be x the variable minus x-naught the constant, y the variable minus y-naught the constant, which is what we have up there, so this expression kind of vectorizes the whole linear term, and now the beef here, the hard part, how are we gonna vectorize this quadratic term? Now that's what I was leading to in the last couple videos, where I talked about how you express a quadratic form like this with a matrix, and the way that you do it, I'll just kind of scroll down to give us some room, the way that you do it is we'll have a matrix whose components are all of these constants, it'll be this 1 1 2 times the second partial derivative evaluated there, and I'm just gonna, for convenience's sake, I'm gonna just take 1 1 2 times the second partial derivative with respect to x, and leave it as understood that we're evaluating it at this point, and then on the other diagonal, you have 1 1 2 times the other kind of partial derivative with respect to y two times in a row, and then we're gonna multiply it by this constant here, but this term kind of gets broken apart into two different components, if you'll remember in the quadratic form video, it was always things where it was a, and then 2b and c as your constants for the quadratic form, so if we're interpreting this as two times something, then it gets broken down, and on one corner it shows up as f x y, and on the other one, kind of 1 1 2 f x y, so like both of these together are gonna constitute the entire mixed partial derivative, and then the way that we express the quadratic form is we're gonna multiply this by, well, by what? Well, the first component is whatever the thing is that's squared here, so it's gonna be that x minus x naught, and then the second component is whatever the other thing squared is, which in this case is y minus y naught, and of course we take that same vector, but we put it in on the other side too, so let me make a little bit of room, so this is gonna be wide, so we're gonna take that same vector, and then kind of put it on its side, so it'll be x minus x naught as the first component, and then y minus y naught as the second component, but it's written horizontally, and this, if you multiply out the entire matrix, is gonna give us the same expression that you have up here, and if that seems unfamiliar, if that seems, you know, how do you go from there to there, check out the video on quadratic forms, or you can check out the article where I'm talking about the quadratic approximation as a whole, I kind of go through the computation there. Now this matrix right here is almost the Hessian matrix, this is why I made a video about the Hessian matrix, it's not quite because everything has a 1 1 2 multiplied into it, so I'm just gonna kind of take that out, and we'll remember we have to multiply a 1 1 2 in at some point, but otherwise it is the Hessian matrix, which we denote with a kind of bold-faced H, bold-faced H, and emphasize that it's the Hessian of F, the Hessian is something you take of a function, and like I said, remember each of these terms we should be thinking of as evaluated on the special input point, evaluating it at that special, you know, bold-faced x naught input point, I was just kind of too lazy to write it in, each time the x naught, y naught, x naught, y naught, x naught, y naught, all of that, but what we have then is we're multiplying it on the right by this whole vector is the variable vector, bold-faced x minus bold-faced x naught, that's what that entire vector is, and then we kind of have the same thing on the right, you know, bold-faced vector x minus x naught, except that we transpose it, we kind of put it on its side, and the way you denote that, you have a little T there for transpose, so this term captures all of the quadratic information that we need for the approximation, so just to put it all together, if we go back up when we put the constant term that we have, the linear term, and this quadratic form that we just found all together, what we get is that the quadratic approximation of f, which is a function, we'll think of it as a vector input, bold-faced x, it equals the function itself evaluated at, you know, whatever point we're approximating near, plus the gradient of f, which is kind of, it's vector analog of a derivative, evaluated at that point, so this is a constant vector, dot product with the variable vector x minus the constant vector x naught, that whole thing, plus 1 1 2, the, we'll just copy down this whole quadratic term up there, the variable minus the constant, multiplied by the Hessian, which is kind of like an extension of the second derivative to multivariable functions, and we're evaluating that, no, let's see, we're evaluating it at the constant, at the constant x naught, and then on the right side, we're multiplying it by the variable, x minus x naught, and this, this is the quadratic approximation in vector form and the important part is, now, it doesn't just have to be of a two-variable input, you could imagine plugging in a three-variable input or four-variable input, and all of these terms make sense, you know, you take the gradient of a four-variable function, you'll get a vector with four components, you take the Hessian of a four-variable function, you would get a four-by-four matrix, and all of these terms make sense, and I think it's also prettier to write it this way because it looks a lot more like a Taylor expansion in the single-variable world, you have, you know, a constant term, plus the value of a derivative times x minus a constant, plus 1 1 2, what's kind of like the second derivative term, what's kind of like taking an x squared, but this is how it looks in the vector world, so in that way, it's actually maybe a little bit more familiar than writing it out in the full, you know, component-by-component term, where it's easy to kind of get lost in the weeds there, so full vectorized form of the quadratic approximation of a scalar-valued multivariable function. Boy, is that a lot to say."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And more specifically, it's the Jacobian matrix, or sometimes the associated determinant. And here, I just want to talk about some of the background knowledge that I'm assuming. Because to understand the Jacobian, you do have to have a little bit of a background in linear algebra. And in particular, I want to make sure that everyone here understands how to think about matrices as transformations of space. And when I say transformations, let me just get kind of a matrix on here. I'll call it 2, 1, and negative 3, 1. You'll see why I'm coloring it like this in just a moment."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And in particular, I want to make sure that everyone here understands how to think about matrices as transformations of space. And when I say transformations, let me just get kind of a matrix on here. I'll call it 2, 1, and negative 3, 1. You'll see why I'm coloring it like this in just a moment. And when I say how to think about this as a transformation of space, I mean you can multiply a matrix by some kind of two-dimensional vector, some kind of two-dimensional x, y. And this is gonna give us a new two-dimensional vector. This is gonna bring us to, let's see, in this case, it'll be, I'll write kind of 2, 1, negative 3, 1, where what it gives us is 2x plus negative 3 times y, and then 1x plus 1 times y."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "You'll see why I'm coloring it like this in just a moment. And when I say how to think about this as a transformation of space, I mean you can multiply a matrix by some kind of two-dimensional vector, some kind of two-dimensional x, y. And this is gonna give us a new two-dimensional vector. This is gonna bring us to, let's see, in this case, it'll be, I'll write kind of 2, 1, negative 3, 1, where what it gives us is 2x plus negative 3 times y, and then 1x plus 1 times y. Right, this is a new two-dimensional vector somewhere else in space. And even if you know how to compute it, there's still room for a deeper geometric understanding of what it actually means to take a vector x, y to the vector 2x plus negative 3y and 1x plus 1y. And there's also still a deeper understanding in what we mean when we call this a linear transformation, a linear transformation."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "This is gonna bring us to, let's see, in this case, it'll be, I'll write kind of 2, 1, negative 3, 1, where what it gives us is 2x plus negative 3 times y, and then 1x plus 1 times y. Right, this is a new two-dimensional vector somewhere else in space. And even if you know how to compute it, there's still room for a deeper geometric understanding of what it actually means to take a vector x, y to the vector 2x plus negative 3y and 1x plus 1y. And there's also still a deeper understanding in what we mean when we call this a linear transformation, a linear transformation. So what I'm gonna do is just show you what this particular transformation looks like on the left here, where every single point on this blue grid, I'm gonna tell the computer, hey, if that point was x, y, I want you to take it to 2x plus negative 3y, 1x plus 1y. And here's what it looks like. So let me just kind of play it out here."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And there's also still a deeper understanding in what we mean when we call this a linear transformation, a linear transformation. So what I'm gonna do is just show you what this particular transformation looks like on the left here, where every single point on this blue grid, I'm gonna tell the computer, hey, if that point was x, y, I want you to take it to 2x plus negative 3y, 1x plus 1y. And here's what it looks like. So let me just kind of play it out here. All of the points in space move, and you end up in some final state here. And there are a couple important things to note. First of all, all of the grid lines remain parallel and evenly spaced."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "So let me just kind of play it out here. All of the points in space move, and you end up in some final state here. And there are a couple important things to note. First of all, all of the grid lines remain parallel and evenly spaced. And they're still lines, they didn't get curved in some way. And that's very, very special. That is the geometric way that you can think about this term, this idea of a linear transformation."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "First of all, all of the grid lines remain parallel and evenly spaced. And they're still lines, they didn't get curved in some way. And that's very, very special. That is the geometric way that you can think about this term, this idea of a linear transformation. I kind of like to think about it that lines stay lines. And in particular, the grid lines here, the ones that started off as, you know, kind of vertical and horizontal, they still remain parallel and they still remain evenly spaced. And the other thing to notice here is I have these two vectors highlighted, the green vector and the red vector."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "That is the geometric way that you can think about this term, this idea of a linear transformation. I kind of like to think about it that lines stay lines. And in particular, the grid lines here, the ones that started off as, you know, kind of vertical and horizontal, they still remain parallel and they still remain evenly spaced. And the other thing to notice here is I have these two vectors highlighted, the green vector and the red vector. And these are the ones that started off, if we kind of back things up, these are the ones that started off as the basis vectors, right? Let me kind of make a little bit more room here. The green vector is 1, 0, 1 in the x direction, 0 in the y direction."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And the other thing to notice here is I have these two vectors highlighted, the green vector and the red vector. And these are the ones that started off, if we kind of back things up, these are the ones that started off as the basis vectors, right? Let me kind of make a little bit more room here. The green vector is 1, 0, 1 in the x direction, 0 in the y direction. And then that red vertical vector here is 0, 1. Right, 0, 1. And if we notice where they land under this transformation, when the matrix is multiplied by every single vector in space, the place where the green vector lands, the one that started off as 1, 0, has coordinates 2, 1."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "The green vector is 1, 0, 1 in the x direction, 0 in the y direction. And then that red vertical vector here is 0, 1. Right, 0, 1. And if we notice where they land under this transformation, when the matrix is multiplied by every single vector in space, the place where the green vector lands, the one that started off as 1, 0, has coordinates 2, 1. And that corresponds very directly with the fact that the first column of our matrix is 2, 1. And then similarly, over here, the second vector, the one that started off at 0, 1, ends up at the coordinates negative 3, 1. And that's what corresponds with the fact that the next column is negative 3, 1."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And if we notice where they land under this transformation, when the matrix is multiplied by every single vector in space, the place where the green vector lands, the one that started off as 1, 0, has coordinates 2, 1. And that corresponds very directly with the fact that the first column of our matrix is 2, 1. And then similarly, over here, the second vector, the one that started off at 0, 1, ends up at the coordinates negative 3, 1. And that's what corresponds with the fact that the next column is negative 3, 1. And it's actually relatively simple to see why that's going to be true. Here, I'll go ahead and multiply this matrix that we had that was... See, now it's kind of easy to remember what the matrix is, right? I can just kind of read it off here as 2, 1, negative 3, 1."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And that's what corresponds with the fact that the next column is negative 3, 1. And it's actually relatively simple to see why that's going to be true. Here, I'll go ahead and multiply this matrix that we had that was... See, now it's kind of easy to remember what the matrix is, right? I can just kind of read it off here as 2, 1, negative 3, 1. But just to see why it's actually taking the basis vectors to the columns like this, if we do the multiplication by 1, 0, notice how it's going to take us to... So it's 2 times 1, that'll be 2, and then negative 3 times 0, so that'll just be 0. And over here, it's 1 times 1, so that's 1, and then 1 times 0."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "I can just kind of read it off here as 2, 1, negative 3, 1. But just to see why it's actually taking the basis vectors to the columns like this, if we do the multiplication by 1, 0, notice how it's going to take us to... So it's 2 times 1, that'll be 2, and then negative 3 times 0, so that'll just be 0. And over here, it's 1 times 1, so that's 1, and then 1 times 0. So again, we're adding 0. So the only terms that actually mattered because of the 0 down here was everything in that first column. But if we take that same matrix, 2, 1, negative 3, 1, and we multiply it by 0, 1 over here, by the second basis vector, what you're going to get is 2 times 0, so 0 plus that element in that second column, and then 1 times 0, so another 0, plus 1 times 1, plus that 1."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And over here, it's 1 times 1, so that's 1, and then 1 times 0. So again, we're adding 0. So the only terms that actually mattered because of the 0 down here was everything in that first column. But if we take that same matrix, 2, 1, negative 3, 1, and we multiply it by 0, 1 over here, by the second basis vector, what you're going to get is 2 times 0, so 0 plus that element in that second column, and then 1 times 0, so another 0, plus 1 times 1, plus that 1. It's kind of like that 0 knocks out all of the terms in other columns. And then like I said, geometrically, the meaning of a linear transformation is that grid lines remain parallel and evenly spaced. And when you start to think about it a little bit, if you can know where this green vector lands and where this red vector lands, that's going to lock into place where the entire grid has to go."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "But if we take that same matrix, 2, 1, negative 3, 1, and we multiply it by 0, 1 over here, by the second basis vector, what you're going to get is 2 times 0, so 0 plus that element in that second column, and then 1 times 0, so another 0, plus 1 times 1, plus that 1. It's kind of like that 0 knocks out all of the terms in other columns. And then like I said, geometrically, the meaning of a linear transformation is that grid lines remain parallel and evenly spaced. And when you start to think about it a little bit, if you can know where this green vector lands and where this red vector lands, that's going to lock into place where the entire grid has to go. And let me show you what I mean and how this corresponds with maybe a different definition that you've heard for what linear transformation means. If we have some kind of function L, and it's going to take in a vector and spit out a vector, it's said to be linear if it satisfies the property that when you take a constant times a vector, what it produces is that same constant times whatever would have happened if you applied that transformation to the vector not scaled, right? So here you're applying the transformation to a scaled vector, and evidently that's the same as scaling the transformation of the vector."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And when you start to think about it a little bit, if you can know where this green vector lands and where this red vector lands, that's going to lock into place where the entire grid has to go. And let me show you what I mean and how this corresponds with maybe a different definition that you've heard for what linear transformation means. If we have some kind of function L, and it's going to take in a vector and spit out a vector, it's said to be linear if it satisfies the property that when you take a constant times a vector, what it produces is that same constant times whatever would have happened if you applied that transformation to the vector not scaled, right? So here you're applying the transformation to a scaled vector, and evidently that's the same as scaling the transformation of the vector. And similarly, second property of linearity is that if you add two vectors, it doesn't really matter if you add them before or after the transformation. If you take the sum of the vectors then apply the transformation, that's the same as first applying the transformation to each one separately and then adding up the results. And one of the most important consequences of this formal definition of linearity is that it means if you take your function and apply it to some vector, x, y, well it can split up that vector as x times the first basis vector, x times one, zero, plus y, let's see, y, times that second basis vector, zero, one."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "So here you're applying the transformation to a scaled vector, and evidently that's the same as scaling the transformation of the vector. And similarly, second property of linearity is that if you add two vectors, it doesn't really matter if you add them before or after the transformation. If you take the sum of the vectors then apply the transformation, that's the same as first applying the transformation to each one separately and then adding up the results. And one of the most important consequences of this formal definition of linearity is that it means if you take your function and apply it to some vector, x, y, well it can split up that vector as x times the first basis vector, x times one, zero, plus y, let's see, y, times that second basis vector, zero, one. And because of these two properties of linearity, if I can split it up like this, it doesn't matter if I do the scaling and adding before the transformation, or if I do that scaling and adding after the transformation, and say that it's x times whatever the transformed version of one, zero is, and I'll show you geometrically what this means in just a moment, but I kind of want to get all the algebra on the screen, plus y times the transformed version of zero, one. Zero, one. So to be concrete, let's actually put in a value for x and y here and try to think about that specific vector geometrically."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And one of the most important consequences of this formal definition of linearity is that it means if you take your function and apply it to some vector, x, y, well it can split up that vector as x times the first basis vector, x times one, zero, plus y, let's see, y, times that second basis vector, zero, one. And because of these two properties of linearity, if I can split it up like this, it doesn't matter if I do the scaling and adding before the transformation, or if I do that scaling and adding after the transformation, and say that it's x times whatever the transformed version of one, zero is, and I'll show you geometrically what this means in just a moment, but I kind of want to get all the algebra on the screen, plus y times the transformed version of zero, one. Zero, one. So to be concrete, let's actually put in a value for x and y here and try to think about that specific vector geometrically. So maybe I'll put in something like vector two, one. So if we look over on the grid, we're going to be focusing on the point that's over here, two, one, and this particular point. And I'm going to play the transformation, and I want you to follow this point to see where it lands, and it's going to end up over here."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "So to be concrete, let's actually put in a value for x and y here and try to think about that specific vector geometrically. So maybe I'll put in something like vector two, one. So if we look over on the grid, we're going to be focusing on the point that's over here, two, one, and this particular point. And I'm going to play the transformation, and I want you to follow this point to see where it lands, and it's going to end up over here. Okay, so in terms of the old grid, right, the original one that we started with, it's now at the point one, three. This is where we've ended up. But importantly, I want you to notice how it's still two times that green vector, plus one times that red vector."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "And I'm going to play the transformation, and I want you to follow this point to see where it lands, and it's going to end up over here. Okay, so in terms of the old grid, right, the original one that we started with, it's now at the point one, three. This is where we've ended up. But importantly, I want you to notice how it's still two times that green vector, plus one times that red vector. So it's satisfying that property that it's still x times whatever the transformed version of that first basis vector is, plus y times the transformed version of that second basis vector. So that's all just a little overview. And the upshot, the main thing I want you to remember from all of this, is when you have some kind of matrix, you can think of it as a transformation of space that keeps grid lines parallel and evenly spaced, and that's a very special kind of transformation."}, {"video_title": "Jacobian prerequisite knowledge.mp3", "Sentence": "But importantly, I want you to notice how it's still two times that green vector, plus one times that red vector. So it's satisfying that property that it's still x times whatever the transformed version of that first basis vector is, plus y times the transformed version of that second basis vector. So that's all just a little overview. And the upshot, the main thing I want you to remember from all of this, is when you have some kind of matrix, you can think of it as a transformation of space that keeps grid lines parallel and evenly spaced, and that's a very special kind of transformation. That is a very restrictive property to have on a function from 2D points to other 2D points. And the convenient way to encode that is that the landing spot for that first basis vector, the one that started off one unit to the right, is represented with the first column of the matrix, and the landing spot for the second basis vector, the one that was pointing one unit up, is encoded with that second column. If this feels totally unfamiliar or you want to learn more about this, it's something that I've made other videos on in the past."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So continuing on with where we were in the last video, we're looking for this unit tangent vector function given the parameterization. So the specific example that I have is a function that parameterizes a circle with radius capital R, but I also kind of want to show in parallel what this looks like more abstractly. So here I'll just kind of write down in the abstract half what we did here, what we did for the unit tangent vector. So we actually have the same thing over here where the unit tangent vector should be the derivative function, which we know gives a tangent, right, it's just it might not be unit, but then we normalize it. We take the magnitude of that tangent vector function. And in our specific case with the circle, once we did this and we kind of took the X component squared, Y component squared, and simplified it all out, we got the function R. But in the general case, we might not be so lucky because the magnitude of this derivative is gonna be the square root of X prime of T squared, right? It's this, the X component of the derivative plus Y prime of T squared, just taking the magnitude of a vector here."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So we actually have the same thing over here where the unit tangent vector should be the derivative function, which we know gives a tangent, right, it's just it might not be unit, but then we normalize it. We take the magnitude of that tangent vector function. And in our specific case with the circle, once we did this and we kind of took the X component squared, Y component squared, and simplified it all out, we got the function R. But in the general case, we might not be so lucky because the magnitude of this derivative is gonna be the square root of X prime of T squared, right? It's this, the X component of the derivative plus Y prime of T squared, just taking the magnitude of a vector here. So when we take the entire function and divide it by that, what we get doesn't simplify as it did in the case of a circle. Instead, we have that X prime of T, which is the X component of our S prime of T, and we have to divide it by that entire magnitude, which was this whole expression, right? You have to divide it by that whole square root expression."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "It's this, the X component of the derivative plus Y prime of T squared, just taking the magnitude of a vector here. So when we take the entire function and divide it by that, what we get doesn't simplify as it did in the case of a circle. Instead, we have that X prime of T, which is the X component of our S prime of T, and we have to divide it by that entire magnitude, which was this whole expression, right? You have to divide it by that whole square root expression. I'm just gonna kind of write dot, dot, dot with the understanding that this square root expression is what goes up there. And similarly, over here, we'd have Y prime of T divided by that entire expression again, right? So simplification doesn't always happen."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "You have to divide it by that whole square root expression. I'm just gonna kind of write dot, dot, dot with the understanding that this square root expression is what goes up there. And similarly, over here, we'd have Y prime of T divided by that entire expression again, right? So simplification doesn't always happen. That was just kind of a lucky happenstance of our circle example. And then now what we want, once we have the unit tangent vector as a function of that same parameter, what we're hoping to find is the derivative of that unit tangent vector with respect to arc length, the arc length S, and to find its magnitude. That's gonna be what curvature is."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So simplification doesn't always happen. That was just kind of a lucky happenstance of our circle example. And then now what we want, once we have the unit tangent vector as a function of that same parameter, what we're hoping to find is the derivative of that unit tangent vector with respect to arc length, the arc length S, and to find its magnitude. That's gonna be what curvature is. But the way to do this is to take the derivative with respect to the parameter T, so D big T, D little t, and then divide it out by the derivative of our function S with respect to T, which we already found. And the reason I'm doing this, loosely, if you're just thinking of the notation, you might say, oh, you can kind of cancel out the DTs from each one. But another way to think about this is to say, when we have our tangent vector function as a function of T, the parameter T, we're not sure of what its change is with respect to S, right?"}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "That's gonna be what curvature is. But the way to do this is to take the derivative with respect to the parameter T, so D big T, D little t, and then divide it out by the derivative of our function S with respect to T, which we already found. And the reason I'm doing this, loosely, if you're just thinking of the notation, you might say, oh, you can kind of cancel out the DTs from each one. But another way to think about this is to say, when we have our tangent vector function as a function of T, the parameter T, we're not sure of what its change is with respect to S, right? That's something we don't know directly. But we do directly know its change with respect to a tiny change in that parameter. So then if we just kind of correct that by saying, hey, how much does the length of the curve change?"}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "But another way to think about this is to say, when we have our tangent vector function as a function of T, the parameter T, we're not sure of what its change is with respect to S, right? That's something we don't know directly. But we do directly know its change with respect to a tiny change in that parameter. So then if we just kind of correct that by saying, hey, how much does the length of the curve change? How far do you move along the curve as you change that parameter? And maybe if I go back up to the picture here, this Ds Dt is saying, for a tiny nudge in time, right, what is the ratio of the size of the movement there with respect to that tiny time? So the reason that this comes out to be a very large vector, right, it's not a tiny thing, is because you're taking the ratio."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So then if we just kind of correct that by saying, hey, how much does the length of the curve change? How far do you move along the curve as you change that parameter? And maybe if I go back up to the picture here, this Ds Dt is saying, for a tiny nudge in time, right, what is the ratio of the size of the movement there with respect to that tiny time? So the reason that this comes out to be a very large vector, right, it's not a tiny thing, is because you're taking the ratio. Maybe this tiny change was a just itty bitty smidgen vector, but you're dividing it by like one one millionth, or whatever the size of Dt that you're thinking. And in this specific case for our circle, we saw that the magnitude of this guy, you know, if we took the magnitude of that guy, it's gonna be equal to R, which is a little bit poetic, right, that the magnitude of the derivative is the same as the distance from the center. And what this means in our specific case, if we wanna apply this to our circle example, we take Dt, D big T, the tangent vector function, and I'll go ahead and write it here."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So the reason that this comes out to be a very large vector, right, it's not a tiny thing, is because you're taking the ratio. Maybe this tiny change was a just itty bitty smidgen vector, but you're dividing it by like one one millionth, or whatever the size of Dt that you're thinking. And in this specific case for our circle, we saw that the magnitude of this guy, you know, if we took the magnitude of that guy, it's gonna be equal to R, which is a little bit poetic, right, that the magnitude of the derivative is the same as the distance from the center. And what this means in our specific case, if we wanna apply this to our circle example, we take Dt, D big T, the tangent vector function, and I'll go ahead and write it here. We have the derivative of our tangent vector with respect to the parameter, and we go up and we look here, we say, okay, the unit tangent vector has the formula negative sine of T and cosine of T. So the derivative of negative sine of T is negative cosine. So over here, this guy should look like negative cosine of T. And the other component, the Y component, the derivative of cosine T, as we're differentiating our unit tangent vector function, is negative sine, negative sine of T. And what this implies is that the magnitude of that derivative of the tangent vector with respect to T, well, what's the magnitude of this vector? You've got a cosine, you've got a sine."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And what this means in our specific case, if we wanna apply this to our circle example, we take Dt, D big T, the tangent vector function, and I'll go ahead and write it here. We have the derivative of our tangent vector with respect to the parameter, and we go up and we look here, we say, okay, the unit tangent vector has the formula negative sine of T and cosine of T. So the derivative of negative sine of T is negative cosine. So over here, this guy should look like negative cosine of T. And the other component, the Y component, the derivative of cosine T, as we're differentiating our unit tangent vector function, is negative sine, negative sine of T. And what this implies is that the magnitude of that derivative of the tangent vector with respect to T, well, what's the magnitude of this vector? You've got a cosine, you've got a sine. There's nothing else in there. You're gonna end up with cosine squared plus sine squared. So this magnitude just equals one."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "You've got a cosine, you've got a sine. There's nothing else in there. You're gonna end up with cosine squared plus sine squared. So this magnitude just equals one. And when we do what we're supposed to over here and divide it by the magnitude of the derivative, right, we take this and we divide it by the magnitude of the derivative, the S, Dt. Well, we've already computed the magnitude of the derivative. That was R. That's how we got this R, is we took the derivative here and took its magnitude and found it."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So this magnitude just equals one. And when we do what we're supposed to over here and divide it by the magnitude of the derivative, right, we take this and we divide it by the magnitude of the derivative, the S, Dt. Well, we've already computed the magnitude of the derivative. That was R. That's how we got this R, is we took the derivative here and took its magnitude and found it. So we find that in the specific case of the circle, the curvature function that we want is just constantly equal to one over R, which is good and hopeful, right, because I said in the original video on curvature that it's defined as one divided by the radius of the circle that hugs your curve most closely. And if your curve is actually a circle, it's literally a circle, then the circle that hugs it most closely is itself, right? So I should hope that its curvature ends up being one divided by R. And in the more general case, if we take a look at what this ought to be, you can maybe imagine just how horrifying it's gonna be to compute this, right?"}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "That was R. That's how we got this R, is we took the derivative here and took its magnitude and found it. So we find that in the specific case of the circle, the curvature function that we want is just constantly equal to one over R, which is good and hopeful, right, because I said in the original video on curvature that it's defined as one divided by the radius of the circle that hugs your curve most closely. And if your curve is actually a circle, it's literally a circle, then the circle that hugs it most closely is itself, right? So I should hope that its curvature ends up being one divided by R. And in the more general case, if we take a look at what this ought to be, you can maybe imagine just how horrifying it's gonna be to compute this, right? We've got our tangent vector function, which itself, you know, is almost too long for me to write down. I just put these dot, dot, dots where you're filling in x prime of t squared plus y prime of t squared. And you're gonna have to take this, take its derivative with respect to t, right?"}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So I should hope that its curvature ends up being one divided by R. And in the more general case, if we take a look at what this ought to be, you can maybe imagine just how horrifying it's gonna be to compute this, right? We've got our tangent vector function, which itself, you know, is almost too long for me to write down. I just put these dot, dot, dots where you're filling in x prime of t squared plus y prime of t squared. And you're gonna have to take this, take its derivative with respect to t, right? It's not gonna get any simpler when you take its derivative. Take the magnitude of that and divide all of that by the magnitude of the derivative of your original function. And I think what I'll do, I'm not gonna go through all of that here."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And you're gonna have to take this, take its derivative with respect to t, right? It's not gonna get any simpler when you take its derivative. Take the magnitude of that and divide all of that by the magnitude of the derivative of your original function. And I think what I'll do, I'm not gonna go through all of that here. It's a little bit much, and I'm not sure how helpful it is to walk through all those steps. But for the sake of having it, for anyone who's curious, I think I'll put that into an article, and you can kind of go through the steps at your own pace and see what the formula comes out to be. And I'll just tell you right now, maybe kind of a spoiler alert, what that formula comes out to be is x prime, the derivative of that first component, multiplied by y double prime, the second derivative of that second component, minus y prime, first derivative of that second component, multiplied by x double prime."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And I think what I'll do, I'm not gonna go through all of that here. It's a little bit much, and I'm not sure how helpful it is to walk through all those steps. But for the sake of having it, for anyone who's curious, I think I'll put that into an article, and you can kind of go through the steps at your own pace and see what the formula comes out to be. And I'll just tell you right now, maybe kind of a spoiler alert, what that formula comes out to be is x prime, the derivative of that first component, multiplied by y double prime, the second derivative of that second component, minus y prime, first derivative of that second component, multiplied by x double prime. And all of that is divided by the, divided by the, kind of magnitude component, the x prime squared plus y prime squared. That whole thing to the 3 halves. And you can maybe see why you're gonna get terms like this, right, because when you're taking, when you're taking the derivative of, when you're taking the derivative of the unit tangent vector function, you have the square root term in it, the square root that has x primes and y primes."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And I'll just tell you right now, maybe kind of a spoiler alert, what that formula comes out to be is x prime, the derivative of that first component, multiplied by y double prime, the second derivative of that second component, minus y prime, first derivative of that second component, multiplied by x double prime. And all of that is divided by the, divided by the, kind of magnitude component, the x prime squared plus y prime squared. That whole thing to the 3 halves. And you can maybe see why you're gonna get terms like this, right, because when you're taking, when you're taking the derivative of, when you're taking the derivative of the unit tangent vector function, you have the square root term in it, the square root that has x primes and y primes. So that's where you're gonna get your x double prime, y double prime, as if the chain rule takes you down there. And you can maybe see why this whole x prime squared, y prime squared term is gonna maintain itself. And it turns out it comes in here at a 3 halves power."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And you can maybe see why you're gonna get terms like this, right, because when you're taking, when you're taking the derivative of, when you're taking the derivative of the unit tangent vector function, you have the square root term in it, the square root that has x primes and y primes. So that's where you're gonna get your x double prime, y double prime, as if the chain rule takes you down there. And you can maybe see why this whole x prime squared, y prime squared term is gonna maintain itself. And it turns out it comes in here at a 3 halves power. And what I'm gonna do in the next video, I'm gonna go ahead and describe kind of an intuition for why this formula isn't random. Why if you break down what this is saying, it really does give a feeling for the curvature, the amount that a curve curves, that we want to try to measure. So it's almost like this is a third way of thinking about it, right?"}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "And it turns out it comes in here at a 3 halves power. And what I'm gonna do in the next video, I'm gonna go ahead and describe kind of an intuition for why this formula isn't random. Why if you break down what this is saying, it really does give a feeling for the curvature, the amount that a curve curves, that we want to try to measure. So it's almost like this is a third way of thinking about it, right? The first one, I said you have whatever circle most closely hugs your curve, and you take one divided by its radius. And then the second way, you're thinking of this DTDS, the change in the unit tangent vector with respect to arc length and taking its magnitude. And of course, all of these are the same, but they're just kind of three different ways to think about it or things that you might plug in when you come across a function."}, {"video_title": "Curvature formula, part 3.mp3", "Sentence": "So it's almost like this is a third way of thinking about it, right? The first one, I said you have whatever circle most closely hugs your curve, and you take one divided by its radius. And then the second way, you're thinking of this DTDS, the change in the unit tangent vector with respect to arc length and taking its magnitude. And of course, all of these are the same, but they're just kind of three different ways to think about it or things that you might plug in when you come across a function. And I'll go through an example. I'll go through something where we're really computing the curvature of something that's not just a circle. But with that, I'll see you next video."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And what that means is we're starting to allow ourselves to use terms like x squared, x times y, and y squared. And quadratic basically just means anytime you have two variables multiplied together. So here you have two x's multiplied together, here it's an x multiplied with a y, and here, y squared, that kind of thing. So let's take a look at this local linearization. It seems like a lot, but once you actually kind of go through term by term, you realize it's a relatively simple function, and if you were to plug in numbers for the constant terms, it would come out as something relatively simple. Because this right here, where you're evaluating the function at the specific input point, that's just gonna be some kind of constant. That's just gonna output some kind of number."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "So let's take a look at this local linearization. It seems like a lot, but once you actually kind of go through term by term, you realize it's a relatively simple function, and if you were to plug in numbers for the constant terms, it would come out as something relatively simple. Because this right here, where you're evaluating the function at the specific input point, that's just gonna be some kind of constant. That's just gonna output some kind of number. And similarly, when you do that to the partial derivative, this little f sub x means partial derivative at that point, you're just getting another number. And over here, this is also just another number, but we've written it in the abstract form so that you can see what you would need to plug in for any function and for any possible input point. And the reason for having it like this, the reason that it comes out to this form is because of a few important properties that this linearization has."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "That's just gonna output some kind of number. And similarly, when you do that to the partial derivative, this little f sub x means partial derivative at that point, you're just getting another number. And over here, this is also just another number, but we've written it in the abstract form so that you can see what you would need to plug in for any function and for any possible input point. And the reason for having it like this, the reason that it comes out to this form is because of a few important properties that this linearization has. Let me move this stuff out of the way. We'll get back to it in a moment. But I just wanna emphasize a few properties that this has, because it's gonna be properties that we want our quadratic approximation to have as well."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And the reason for having it like this, the reason that it comes out to this form is because of a few important properties that this linearization has. Let me move this stuff out of the way. We'll get back to it in a moment. But I just wanna emphasize a few properties that this has, because it's gonna be properties that we want our quadratic approximation to have as well. First of all, when you actually evaluate this function at the desired point, at x naught, y naught, what do you get? Well, this constant term isn't influenced by the variable, so you'll just get that f evaluated at those points, x naught, y naught. And now the rest of the terms, when we plug in x here, this is the only place where you actually see the variable."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "But I just wanna emphasize a few properties that this has, because it's gonna be properties that we want our quadratic approximation to have as well. First of all, when you actually evaluate this function at the desired point, at x naught, y naught, what do you get? Well, this constant term isn't influenced by the variable, so you'll just get that f evaluated at those points, x naught, y naught. And now the rest of the terms, when we plug in x here, this is the only place where you actually see the variable. Maybe that's worth pointing out, right? We've got two variables here, and there's a lot going on, but the only places where you actually see those variables show up, where you have to plug in anything, is over here and over here. When you plug in x naught for our initial input, this entire term goes to zero, right?"}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And now the rest of the terms, when we plug in x here, this is the only place where you actually see the variable. Maybe that's worth pointing out, right? We've got two variables here, and there's a lot going on, but the only places where you actually see those variables show up, where you have to plug in anything, is over here and over here. When you plug in x naught for our initial input, this entire term goes to zero, right? And then similarly, when you plug in y naught over here, this entire term is gonna go to zero, which multiplies out to zero for everything. So what you end up with, you don't have to add anything else. This is just a fact."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "When you plug in x naught for our initial input, this entire term goes to zero, right? And then similarly, when you plug in y naught over here, this entire term is gonna go to zero, which multiplies out to zero for everything. So what you end up with, you don't have to add anything else. This is just a fact. And this is an important fact, because it tells you your approximation for the function at the point about which you are approximating actually equals the value of the function at that point, so that's very good. But we have a couple other important facts also, because this isn't just a constant approximation, this is doing a little bit more for us. If you were to take the partial derivative of this linearization with respect to x, what do you get?"}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "This is just a fact. And this is an important fact, because it tells you your approximation for the function at the point about which you are approximating actually equals the value of the function at that point, so that's very good. But we have a couple other important facts also, because this isn't just a constant approximation, this is doing a little bit more for us. If you were to take the partial derivative of this linearization with respect to x, what do you get? What do you get when you actually take this partial derivative? Well, if you look up at the original function, this constant term is nothing, so that just corresponds to a zero. Over here, this entire thing looks like a constant multiplied by x minus something, and if you differentiate this with respect to x, what you're gonna get is that constant term, which is the partial derivative of f evaluated at our specific point."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "If you were to take the partial derivative of this linearization with respect to x, what do you get? What do you get when you actually take this partial derivative? Well, if you look up at the original function, this constant term is nothing, so that just corresponds to a zero. Over here, this entire thing looks like a constant multiplied by x minus something, and if you differentiate this with respect to x, what you're gonna get is that constant term, which is the partial derivative of f evaluated at our specific point. And then the other term has no x's in it, it's just a y, which as far as x's concerned is a constant. So this whole thing would be zero, which means the partial derivative with respect to x is equal to the value of the partial derivative of our original function with respect to x at that point. Now notice, this is not saying that our linearization has the same partial derivative as f everywhere, it's just saying that its partial derivative happens to be a constant, and the constant that it is is the value of the partial derivative of f at that specific input point."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "Over here, this entire thing looks like a constant multiplied by x minus something, and if you differentiate this with respect to x, what you're gonna get is that constant term, which is the partial derivative of f evaluated at our specific point. And then the other term has no x's in it, it's just a y, which as far as x's concerned is a constant. So this whole thing would be zero, which means the partial derivative with respect to x is equal to the value of the partial derivative of our original function with respect to x at that point. Now notice, this is not saying that our linearization has the same partial derivative as f everywhere, it's just saying that its partial derivative happens to be a constant, and the constant that it is is the value of the partial derivative of f at that specific input point. And you can do pretty much the same thing, and you'll see that the partial derivative of the linearization with respect to y is a constant, and the constant that it happens to be is the value of the partial derivative of f evaluated at that desired point. So these are three facts. You know the value of the linearization at the point, and the value of its two different partial derivatives, and these kind of define the linearization itself."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "Now notice, this is not saying that our linearization has the same partial derivative as f everywhere, it's just saying that its partial derivative happens to be a constant, and the constant that it is is the value of the partial derivative of f at that specific input point. And you can do pretty much the same thing, and you'll see that the partial derivative of the linearization with respect to y is a constant, and the constant that it happens to be is the value of the partial derivative of f evaluated at that desired point. So these are three facts. You know the value of the linearization at the point, and the value of its two different partial derivatives, and these kind of define the linearization itself. Now what we're gonna do for the quadratic approximation is take this entire formula, and I'm just literally gonna copy it here, and then we're gonna add to it so that the second partial differential information of our approximation matches that of the original function. Okay, that's kind of a mouthful, but it'll become clear as I actually work it out. And let me just kind of clean it up at least a little bit here."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "You know the value of the linearization at the point, and the value of its two different partial derivatives, and these kind of define the linearization itself. Now what we're gonna do for the quadratic approximation is take this entire formula, and I'm just literally gonna copy it here, and then we're gonna add to it so that the second partial differential information of our approximation matches that of the original function. Okay, that's kind of a mouthful, but it'll become clear as I actually work it out. And let me just kind of clean it up at least a little bit here. So what we're gonna do is we're gonna extend this, and I'm gonna change its name because I don't want it to be a linear function anymore. What I want is for this to be a quadratic function, so instead I'm gonna call it q of x, y. And now we're gonna add some terms, and what I could do, what I could do is add a constant times x squared, since that's something we're allowed, plus some kind of constant times x, y, and then another constant times y squared."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And let me just kind of clean it up at least a little bit here. So what we're gonna do is we're gonna extend this, and I'm gonna change its name because I don't want it to be a linear function anymore. What I want is for this to be a quadratic function, so instead I'm gonna call it q of x, y. And now we're gonna add some terms, and what I could do, what I could do is add a constant times x squared, since that's something we're allowed, plus some kind of constant times x, y, and then another constant times y squared. But the problem there is that if I just add these as they are then it might mess things up when I plug in x-naught and y-naught, right? Well, it was very important that when you plug in those values that you get the original value of the function and that the partial derivatives of the approximation also match that of the function. And that could mess things up because once you start plugging in x-naught and y-naught over here, that might actually mess with the value."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And now we're gonna add some terms, and what I could do, what I could do is add a constant times x squared, since that's something we're allowed, plus some kind of constant times x, y, and then another constant times y squared. But the problem there is that if I just add these as they are then it might mess things up when I plug in x-naught and y-naught, right? Well, it was very important that when you plug in those values that you get the original value of the function and that the partial derivatives of the approximation also match that of the function. And that could mess things up because once you start plugging in x-naught and y-naught over here, that might actually mess with the value. So we're basically gonna do the same thing we did with the linearization where we put in, every time we have an x, we kind of attach it. We say x minus x-naught, just to make sure that we don't mess things up when we plug in x-naught. So instead, instead of what I had written there, what we're gonna add as our quadratic approximation is some kind of constant, and we'll fill in that constant in a moment, times x minus x-naught squared."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "And that could mess things up because once you start plugging in x-naught and y-naught over here, that might actually mess with the value. So we're basically gonna do the same thing we did with the linearization where we put in, every time we have an x, we kind of attach it. We say x minus x-naught, just to make sure that we don't mess things up when we plug in x-naught. So instead, instead of what I had written there, what we're gonna add as our quadratic approximation is some kind of constant, and we'll fill in that constant in a moment, times x minus x-naught squared. And then we're gonna add another constant multiplied by x minus x-naught times y minus y-naught. And then that times yet another constant, which I'll call c, multiplied by y minus y-naught squared. All right, this is quite a lot going on."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "So instead, instead of what I had written there, what we're gonna add as our quadratic approximation is some kind of constant, and we'll fill in that constant in a moment, times x minus x-naught squared. And then we're gonna add another constant multiplied by x minus x-naught times y minus y-naught. And then that times yet another constant, which I'll call c, multiplied by y minus y-naught squared. All right, this is quite a lot going on. This is a heck of a function. And these are three different constants that we're gonna try to fill in to figure out what they should be to most closely approximate the original function f. Now the important part of making this x minus x-naught and y minus y-naught is that when we plug in here, when we plug in x-naught for our variable x, and when we plug in y-naught for our variable y, all of this stuff is just gonna go to zero, and it's gonna cancel out. And moreover, when we take the partial derivatives, all of it's gonna go to zero as well."}, {"video_title": "Quadratic approximation formula, part 1.mp3", "Sentence": "All right, this is quite a lot going on. This is a heck of a function. And these are three different constants that we're gonna try to fill in to figure out what they should be to most closely approximate the original function f. Now the important part of making this x minus x-naught and y minus y-naught is that when we plug in here, when we plug in x-naught for our variable x, and when we plug in y-naught for our variable y, all of this stuff is just gonna go to zero, and it's gonna cancel out. And moreover, when we take the partial derivatives, all of it's gonna go to zero as well. And we'll see that in a moment. Maybe I'll just actually show that right now. Or rather, I think I'll call the video done here, and then start talking about how we fill in these constants in the next video."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say y is greater than or equal to 0, and is less than or equal to 4. And then let's say that z is greater than or equal to 0, and is less than or equal to 2. Now I know, using basic geometry, you could figure out, just multiply the width times the height times the depth, and you'd have the volume. But I want to do this example just so that you get used to what a triple integral looks like, how it relates to a double integral, and then later in the next video we could do something slightly more complicated. So let's just draw this volume. So this is my x-axis. This is my z-axis."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I want to do this example just so that you get used to what a triple integral looks like, how it relates to a double integral, and then later in the next video we could do something slightly more complicated. So let's just draw this volume. So this is my x-axis. This is my z-axis. This is the y, x, y, z. OK, so x is between 0 and 3, so that's x is equal to 0. This is x is equal to, say, 1, 2, 3. y is between 0 and 4, 1, 2, 3, 4. So the xy-plane, it looks something like this."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is my z-axis. This is the y, x, y, z. OK, so x is between 0 and 3, so that's x is equal to 0. This is x is equal to, say, 1, 2, 3. y is between 0 and 4, 1, 2, 3, 4. So the xy-plane, it looks 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 xy-plane. And then 1, 2."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the xy-plane, it looks 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 xy-plane. And then 1, 2. So this would be the top part, and I'll maybe do that in a slightly different color. So this is along the xz-axis. You'd have a boundary here, and then it would come in like this."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then 1, 2. So this would be the top part, and I'll maybe do that in a slightly different color. So this is along the xz-axis. You'd have a boundary here, and then it would come in like this. You'd have a boundary here, come in like that, boundary there, and then like that. So we want to figure out the volume of this cube. And you could do it."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You'd have a boundary here, and then it would come in like this. You'd have a boundary here, come in like that, boundary there, and then like that. 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 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."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you could do it. You could say, well, the depth is 3, 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. So what does a triple integral mean? 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."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could say it's 24 cubic units, whatever units we're doing. But let's do it as a triple integral. So what does a triple integral mean? 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 someplace 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."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 someplace 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 someplace in this larger cube, this larger rectangle, cubic rectangle, whatever you want to call it. So what's the volume of that cube?"}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's say we want to figure out the volume of this little small cube here. That's my cube. It's someplace 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. Its height is dz, right? The way I drew it, z is up and down."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the volume of that cube? Let's say that its width is dy, so that length right there is dy. Its height is dz, right? The way I drew it, z is up and down. And then its depth is dx. This is dx. So this is dx, this is dz, this is dy."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The way I drew it, z is up and down. And then its depth is dx. This is dx. So this is dx, this is dz, this is dy. So you could say that a small volume within this larger volume, you could call that dv, which is kind of the volume differential. And that would be equal to, you could say, it's just the width times the length times the height. dy, dx times dy times dz."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is dx, this is dz, this is dy. So you could say that a small volume within this larger volume, you could call that dv, which is kind of the volume differential. And that would be equal to, you could say, it's just the width times the length times the height. dy, 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 of that. But anyway, what can you do with it here?"}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dy, 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 of that. But anyway, what can you do with it here? Well, we can take the integral. All integrals help us do is it helps 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."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But anyway, what can you do with it here? Well, we can take the integral. All integrals help us do is it helps 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 taking the sum in the z direction, we'd have an integral."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 taking the sum in the z direction, we'd have an integral. And then what's the lowest z value? Well, it's z is equal to 0. And what's the upper bound?"}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, since we're going up and down, we're taking the sum in the z direction, we'd have an integral. And then what's the lowest z value? Well, it's z is equal to 0. And what's the upper bound? Like if you were to just keep adding these cubes and keep going up, you'd run into the upper bound. And what's the upper bound? It's z is equal to 2."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what's the upper bound? Like if you were to just keep adding these cubes and keep going up, you'd run into the upper bound. And what's the upper bound? It's z is equal to 2. And of course, you would take the sum of these dv's. I'll write dz first, just so it reminds us that we're going to take the integral with respect to z first. And let's say we'll do y next."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's z is equal to 2. And of course, you would take the sum of these dv's. I'll write dz first, just so it reminds us that we're going to take the integral with respect to z first. And let's say we'll do y next. And then we'll do x. So this integral, this value, as I have written it, will figure out the volume of a column given any x and y. It'll be a function of x and y, but since we're dealing with all constant here, it's actually going to be a constant value."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say we'll do y next. And then we'll do x. So this integral, this value, as I have written it, will figure out the volume of a column given any x and y. It'll be a function of x and y, but since we're dealing with all constant here, it's actually going to be a constant value. It'll be the constant value of the volume of one of these columns. And so essentially, it'll be 2 times dy dx. Because the height of one of these columns is 2."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It'll be a function of x and y, but since we're dealing with all constant here, it's actually going to be a constant value. It'll be the constant value of the volume of one of these columns. And so essentially, it'll be 2 times dy dx. Because the height of one of these columns is 2. And then its width and its depth is dy and dx. So then if we wanted to figure out the entire volume, so what we did just now is we figured out the height of a column. So then we could take those columns and sum them in the y direction, right?"}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because the height of one of these columns is 2. And then its width and its depth is dy and dx. So then if we wanted to figure out the entire volume, so what we did just now is we figured out the height of a column. So then we could take those columns and sum them in the y direction, right? So for summing in the y direction, we could just take another integral of this, sum in the y direction. And y goes from 0 to what? y goes from 0 to 4."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So then we could take those columns and sum them in the y direction, right? So for summing in the y direction, we could just take another integral of this, sum in the y direction. And y goes from 0 to what? y goes from 0 to 4. I wrote this integral a little bit too far to the left. It looks strange. But I think you get the idea."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "y goes from 0 to 4. I wrote this integral a little bit too far to the left. It looks strange. But I think you get the idea. y is equal to 4. y is equal to 0 to y is equal to 4. And then that'll give us the volume of kind of a sheet that is parallel to the zy plane. And then all we have left to do is add up a bunch of those sheets in the x direction."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But I think you get the idea. y is equal to 4. y is equal to 0 to y is equal to 4. And then that'll give us the volume of kind of a sheet that is parallel to the zy plane. And then all we have left to do is add up a bunch of those sheets in the x direction. We'll have the volume of our entire figure. So to add up those sheets, we would have to sum in the x direction, and we'd go from x is equal to 0 to x is equal to 3. And to evaluate this, it's actually fairly straightforward."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then all we have left to do is add up a bunch of those sheets in the x direction. We'll have the volume of our entire figure. So to add up those sheets, we would have to sum in the x direction, and we'd go from x is equal to 0 to x is equal to 3. And to evaluate this, it's actually fairly straightforward. So first we're taking the integral with respect to z. Well, we don't have anything written under here, but we can just assume that there's a 1, right? Because dz times dy times dx is the same thing as 1 times dz times dy dx."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to evaluate this, it's actually fairly straightforward. So first we're taking the integral with respect to z. Well, we don't have anything written under here, but we can just assume that there's a 1, right? Because dz times dy times dx is the same thing as 1 times dz times dy dx. So what's the value of this integral? Well, the antiderivative of 1 with respect to z is just z, right, because the derivative of z is 1. And you evaluate that from 2 to 0."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because dz times dy times dx is the same thing as 1 times dz times dy dx. So what's the value of this integral? Well, the antiderivative of 1 with respect to z is just z, right, because the derivative of z is 1. And you evaluate that from 2 to 0. So then you're left with 2 minus 0. So you're just left with 2. You're going to take the integral of that from y is equal to 0 to y is equal to 4 dy."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you evaluate that from 2 to 0. So then you're left with 2 minus 0. So you're just left with 2. You're going to take the integral of that from y is equal to 0 to y is equal to 4 dy. And then you have the x from x is equal to 0 to x is equal to 3 dx. And notice, when we just took the integral with respect to z, we ended up with a double integral. And this double integral is the exact integral we would have done in the previous videos on the double integral where you would have just said, well, z is a function of x and y."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're going to take the integral of that from y is equal to 0 to y is equal to 4 dy. And then you have the x from x is equal to 0 to x is equal to 3 dx. And notice, when we just took the integral with respect to z, we ended up with a double integral. And this double integral is the exact integral we would have done in the previous videos on the double integral where you would have just said, well, z is a function of x and y. So you could have written z is a function of x and y is always equal to 2. It's a constant function. It's independent of x and y."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this double integral is the exact integral we would have done in the previous videos on the double integral where you would have just said, well, z is a function of x and y. So you could have written z is a function of x and y is always equal to 2. It's a constant function. It's independent of x and y. But if you had defined z in this way and you wanted to figure out the volume under this surface where the surface is z is equal to 2, we would have ended up with this. So you see that what we're doing with the triple integral, it's really, really nothing different. And you might be wondering, well, why are we doing it at all?"}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's independent of x and y. But if you had defined z in this way and you wanted to figure out the volume under this surface where the surface is z is equal to 2, we would have ended up with this. So you see that what we're doing with the triple integral, it's really, really nothing different. And you might be wondering, well, why are we doing it at all? And I'll show you that in a second. But anyway, to evaluate this, you can take the antiderivative of this with respect to y. You get 2y."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And you might be wondering, well, why are we doing it at all? And I'll show you that in a second. But anyway, to evaluate this, you can take the antiderivative of this with respect to y. You get 2y. Let me scroll down a little bit. You get 2y, evaluating that at 4 and 0. And then so you get 2 times 4, so it's 8 minus 0."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You get 2y. Let me scroll down a little bit. You get 2y, evaluating that at 4 and 0. And then so you get 2 times 4, so it's 8 minus 0. And then you integrate that with respect to x from 0 to 3. So that's 8x from 0 to 3. So that'll be equal to 24 units cubed."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then so you get 2 times 4, so it's 8 minus 0. And then you integrate that with respect to x from 0 to 3. So that's 8x from 0 to 3. So that'll be equal to 24 units cubed. So I know the obvious question is, what is this good for? Well, when you have a kind of a constant value within the volume, you're right. You could have just done a double integral."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that'll be equal to 24 units cubed. So I know the obvious question is, what is this good for? Well, when you have a kind of a constant value within the volume, you're right. You could have just done a double integral. But what if I were told you, our goal is not to figure out the volume of this figure. Our goal is to figure out the mass of this figure. And even more, this volume, this area of space or whatever, its mass is not uniform."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could have just done a double integral. But what if I were told you, our goal is not to figure out the volume of this figure. Our goal is to figure out the mass of this figure. And even more, this volume, this area of space or whatever, its mass is not uniform. If its mass was uniform, you could just multiply its uniform density times its volume, and you would get its mass. But let's say the density changes. It could be a volume of some gas, or it could be even some material with different compounds in it."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And even more, this volume, this area of space or whatever, its mass is not uniform. If its mass was uniform, you could just multiply its uniform density times its volume, and you would get its mass. But let's say the density changes. It could be a volume of some gas, or it could be even some material with different compounds in it. So let's say that its density is a variable function of x, y, and z. So let's say that the density, this row, this thing that looks like a p is what you normally use in physics for density. So its density is a function of x, y, and z."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It could be a volume of some gas, or it could be even some material with different compounds in it. So let's say that its density is a variable function of x, y, and z. So let's say that the density, this row, this thing that looks like a p is what you normally use in physics for density. So its density is a function of x, y, and z. Let's just to make it simple, let's make it x times y times z. So if we wanted to figure out the mass of any small volume, it would be that volume times the density. Because the units of density are like kilograms per meter cubed."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So its density is a function of x, y, and z. Let's just to make it simple, let's make it x times y times z. So if we wanted to figure out the mass of any small volume, it would be that volume times the density. Because the units of density are like kilograms per meter cubed. So if you multiply it times meter cubed, you get kilograms. So we could say that the mass, I'll make up notation, d mass. This isn't a function."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Because the units of density are like kilograms per meter cubed. So if you multiply it times meter cubed, you get kilograms. So we could say that the mass, I'll make up notation, d mass. This isn't a function. Well, I don't want to write it in parentheses because it makes it look like a function. So a differential of mass, or a very small mass, is going to equal the density at that point, which would be x, y, z, times the volume of that small mass. And that volume of that small mass we could write as dv."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This isn't a function. Well, I don't want to write it in parentheses because it makes it look like a function. So a differential of mass, or a very small mass, is going to equal the density at that point, which would be x, y, z, times the volume of that small mass. And that volume of that small mass we could write as dv. And we know that dv is the same thing as the width times the height times the depth. dv doesn't always have to be dx times dy times dz. If we're doing other coordinates, if we're doing polar coordinates, it could be something slightly different."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that volume of that small mass we could write as dv. And we know that dv is the same thing as the width times the height times the depth. dv doesn't always have to be dx times dy times dz. If we're doing other coordinates, if we're doing polar coordinates, it could be something slightly different. And we'll do that eventually. But if we want to figure out the mass, since we're using rectangular coordinates, it would be the density function at that point times our differential of volume. So times dx, dy, dz."}, {"video_title": "Triple integrals 1   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we're doing other coordinates, if we're doing polar coordinates, it could be something slightly different. And we'll do that eventually. But if we want to figure out the mass, since we're using rectangular coordinates, it would be the density function at that point times our differential of volume. So times dx, dy, dz. And of course, we can change the order here. So when you want to figure out the volume, when you want to figure out the mass, which I will do in the next video, we essentially will have to integrate this function as opposed to just 1 over z, y, and x. And I'm going to do that in the next video."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So in the last video, I talked about curvature and the radius of curvature, and I described it purely geometrically, where I'm saying you imagine driving along a certain road, your steering wheel locks, and you're wondering what the radius of the circle that you draw with your car, you know, through whatever surrounding fields there are on the road as a result. And the special symbol that we have for this, for this idea of curvature, is a little kappa, and that's gonna be one divided by the radius. And the reason we do that is basically you want a large kappa, a large curvature, to correspond with a sharp turn. So sharp turn, small radius, large curvature. But the question is, how do we describe this in a more mathematical way? So I'm gonna go ahead and clear up, get rid of the circle itself and all of that radius, and just be looking at the curve itself in its own right. The way you typically describe a curve like this is parametrically."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So sharp turn, small radius, large curvature. But the question is, how do we describe this in a more mathematical way? So I'm gonna go ahead and clear up, get rid of the circle itself and all of that radius, and just be looking at the curve itself in its own right. The way you typically describe a curve like this is parametrically. So you'll have some kind of vector-valued function, s, that takes in a single parameter, t, and then it's gonna output the x and y coordinates, each as functions of t. And so this will be the x coordinate, the y coordinate. And in this specific case, I'll just tell you the curve that I drew happens to be parameterized by one minus the sine of t as the x-component function. Actually, no, it's t minus sine of t, and the bottom part is one minus cosine of t. That's the curve that I drew."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "The way you typically describe a curve like this is parametrically. So you'll have some kind of vector-valued function, s, that takes in a single parameter, t, and then it's gonna output the x and y coordinates, each as functions of t. And so this will be the x coordinate, the y coordinate. And in this specific case, I'll just tell you the curve that I drew happens to be parameterized by one minus the sine of t as the x-component function. Actually, no, it's t minus sine of t, and the bottom part is one minus cosine of t. That's the curve that I drew. And the way that you're thinking about this is for each value t, you get a certain vector that puts you at a point on the curve. And as t changes, the vector you get changes, but the tip of that vector kind of traces out the curve as a whole, right? You know, maybe."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "Actually, no, it's t minus sine of t, and the bottom part is one minus cosine of t. That's the curve that I drew. And the way that you're thinking about this is for each value t, you get a certain vector that puts you at a point on the curve. And as t changes, the vector you get changes, but the tip of that vector kind of traces out the curve as a whole, right? You know, maybe. And you can imagine just the vector drawing the curve as t varies. And the thought behind making curvature mathematical, here I'll kind of clear up some room for myself, is that you take the tangent vectors here, so you might imagine like a unit tangent vector at every given point, and you're wondering how quickly those guys change direction, right? So with the little schematic that I have drawn here, I might call this guy, I'm just gonna call this guy t1, for like the first tangent vector, t2, t3, and I haven't specified where they start or anything, I just want to give a feel for you've got various different tangent vectors, and I'm just gonna say that all of them, each one of those t sub somethings, has a unit length."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "You know, maybe. And you can imagine just the vector drawing the curve as t varies. And the thought behind making curvature mathematical, here I'll kind of clear up some room for myself, is that you take the tangent vectors here, so you might imagine like a unit tangent vector at every given point, and you're wondering how quickly those guys change direction, right? So with the little schematic that I have drawn here, I might call this guy, I'm just gonna call this guy t1, for like the first tangent vector, t2, t3, and I haven't specified where they start or anything, I just want to give a feel for you've got various different tangent vectors, and I'm just gonna say that all of them, each one of those t sub somethings, has a unit length. And the idea of curvature is to look at how quickly that unit tangent vector changes directions. So you know, you might imagine just a completely different space, so rather than rooting each vector on the curve, let's see what it would look like if you just kind of write each vector in its own right off in some other spot. So this guy here would be t1, and then t2 points a little bit down, and then t3 points kind of much more down."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So with the little schematic that I have drawn here, I might call this guy, I'm just gonna call this guy t1, for like the first tangent vector, t2, t3, and I haven't specified where they start or anything, I just want to give a feel for you've got various different tangent vectors, and I'm just gonna say that all of them, each one of those t sub somethings, has a unit length. And the idea of curvature is to look at how quickly that unit tangent vector changes directions. So you know, you might imagine just a completely different space, so rather than rooting each vector on the curve, let's see what it would look like if you just kind of write each vector in its own right off in some other spot. So this guy here would be t1, and then t2 points a little bit down, and then t3 points kind of much more down. So all of these, this would be t1, that guy is t2, and these are the same vectors, I'm just kind of drawing them all rooted at the same spot, so it's a little easier to see how they turn. And you want to say, okay, how much do you change as you go from t1 to t2, is that a large angle change? And as you go from t2 to t3, is that a large change as well?"}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So this guy here would be t1, and then t2 points a little bit down, and then t3 points kind of much more down. So all of these, this would be t1, that guy is t2, and these are the same vectors, I'm just kind of drawing them all rooted at the same spot, so it's a little easier to see how they turn. And you want to say, okay, how much do you change as you go from t1 to t2, is that a large angle change? And as you go from t2 to t3, is that a large change as well? And you can kind of see how if you have a curve, and let's say it's, if you have a curve that curves quite a bit, you know, it's doing something like that, then the unit vector, the unit tangent vector at this point, changes quite rapidly over a short distance to be something almost 90 degrees different. Whereas if you take the unit vector here and see how much does it change as you go from this point over to this point, it doesn't really change that much. So the thought behind curvature is we're going to take the rate of change of that unit tangent vector, so the rate of change of t, and I'm going to let capital T be a function that tells you whatever the unit tangent vector at each point is, and I'm not going to take the rate of change in terms of, you know, the parameter, little t, which is what we use to parametrize the curve, because it shouldn't matter how you parametrize the curve."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "And as you go from t2 to t3, is that a large change as well? And you can kind of see how if you have a curve, and let's say it's, if you have a curve that curves quite a bit, you know, it's doing something like that, then the unit vector, the unit tangent vector at this point, changes quite rapidly over a short distance to be something almost 90 degrees different. Whereas if you take the unit vector here and see how much does it change as you go from this point over to this point, it doesn't really change that much. So the thought behind curvature is we're going to take the rate of change of that unit tangent vector, so the rate of change of t, and I'm going to let capital T be a function that tells you whatever the unit tangent vector at each point is, and I'm not going to take the rate of change in terms of, you know, the parameter, little t, which is what we use to parametrize the curve, because it shouldn't matter how you parametrize the curve. Maybe you're driving along it really quickly or really slowly. Instead, what you want to take is the rate of change with respect to arc length, arc length. And I'm using the variable s here to denote arc length."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So the thought behind curvature is we're going to take the rate of change of that unit tangent vector, so the rate of change of t, and I'm going to let capital T be a function that tells you whatever the unit tangent vector at each point is, and I'm not going to take the rate of change in terms of, you know, the parameter, little t, which is what we use to parametrize the curve, because it shouldn't matter how you parametrize the curve. Maybe you're driving along it really quickly or really slowly. Instead, what you want to take is the rate of change with respect to arc length, arc length. And I'm using the variable s here to denote arc length. And what I mean by that is if you take just a tiny little step here, the distance of that step, the actual distance in the xy plane, you'd consider to be the arc length. And if you imagine it being really, really small, you're considering that a ds, a tiny change in the arc length. So this is the quantity that we care about."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "And I'm using the variable s here to denote arc length. And what I mean by that is if you take just a tiny little step here, the distance of that step, the actual distance in the xy plane, you'd consider to be the arc length. And if you imagine it being really, really small, you're considering that a ds, a tiny change in the arc length. So this is the quantity that we care about. How much does that unit tangent vector change with respect to a tiny change in the arc length? You know, as we travel along, let's say it was a distance, of like 0.5, right? You want to know, did this unit vector change a lot or a little bit?"}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So this is the quantity that we care about. How much does that unit tangent vector change with respect to a tiny change in the arc length? You know, as we travel along, let's say it was a distance, of like 0.5, right? You want to know, did this unit vector change a lot or a little bit? But I should add something here. It's this tiny change in the vector, that would tell you what the vector connecting their two tips is. So this would be a vector-valued quantity, and curvature itself should just be a number."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "You want to know, did this unit vector change a lot or a little bit? But I should add something here. It's this tiny change in the vector, that would tell you what the vector connecting their two tips is. So this would be a vector-valued quantity, and curvature itself should just be a number. So what we really care about is the size of this guy. So what we really care about, the size of this, which is a vector-valued quantity, and that'll be an indication of just how much the curve curves. But if on a sharper-turned curve, you go over that same distance, and then suddenly the change in the tangent vectors goes by quite a bit, that would be telling you it's a high curvature."}, {"video_title": "Curvature formula, part 1.mp3", "Sentence": "So this would be a vector-valued quantity, and curvature itself should just be a number. So what we really care about is the size of this guy. So what we really care about, the size of this, which is a vector-valued quantity, and that'll be an indication of just how much the curve curves. But if on a sharper-turned curve, you go over that same distance, and then suddenly the change in the tangent vectors goes by quite a bit, that would be telling you it's a high curvature. And in the next video, I'm gonna talk through what that looks like, how you think about this tangent vector function, this unit tangent vector function, and what it looks like to take the derivative of that with respect to arc length. It can get a little convoluted, in terms of the symbols involved. And the constant picture you should have in the back of your mind is that circle, that circle that's hugging the curve very closely at a certain point."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "So I thought I'd make a little video here to spell out exactly how it is that we describe points and vectors in three dimensions. And before we do that, I think it'll be valuable if we start off by describing points and vectors in two dimensions. And I'm assuming if you're learning about multivariable calculus that a lot of you have already learned about this, and you might be saying, what's the point? I already know how to represent points and vectors in two dimensions. But there's a huge value in analogy here, because as soon as you start to compare two dimensions and three dimensions, you start to see patterns for how it could extend to other dimensions that you can't necessarily visualize, or when it might be useful to think about one dimension versus another. So in two dimensions, if you have some kind of point just off sitting there, we typically represent it, you've got an x-axis and a y-axis that are perpendicular to each other. And we represent this number with a pair, sorry, we represent this point with a pair of numbers."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "I already know how to represent points and vectors in two dimensions. But there's a huge value in analogy here, because as soon as you start to compare two dimensions and three dimensions, you start to see patterns for how it could extend to other dimensions that you can't necessarily visualize, or when it might be useful to think about one dimension versus another. So in two dimensions, if you have some kind of point just off sitting there, we typically represent it, you've got an x-axis and a y-axis that are perpendicular to each other. And we represent this number with a pair, sorry, we represent this point with a pair of numbers. So in this case, I don't know, it might be something like one, three. And what that would represent is it's saying that you have to move a distance of one along the x-axis, and then a distance of three up along the y-axis. So you know this, let's say that's the distance of one, that's the distance of three, it might not be exactly that the way I drew it, but let's say that those are the coordinates."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "And we represent this number with a pair, sorry, we represent this point with a pair of numbers. So in this case, I don't know, it might be something like one, three. And what that would represent is it's saying that you have to move a distance of one along the x-axis, and then a distance of three up along the y-axis. So you know this, let's say that's the distance of one, that's the distance of three, it might not be exactly that the way I drew it, but let's say that those are the coordinates. What this means is every point in two-dimensional space can be given a pair of numbers like this, and you think of them as instructions, where it's kind of telling you how far to walk in one way, how far to walk in another. But you can also think of the reverse, right? Every time that you have a pair of things, you know that you should be thinking two-dimensionally."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "So you know this, let's say that's the distance of one, that's the distance of three, it might not be exactly that the way I drew it, but let's say that those are the coordinates. What this means is every point in two-dimensional space can be given a pair of numbers like this, and you think of them as instructions, where it's kind of telling you how far to walk in one way, how far to walk in another. But you can also think of the reverse, right? Every time that you have a pair of things, you know that you should be thinking two-dimensionally. And that's actually a surprisingly powerful idea that I don't think I appreciated for a long time, how there's this back and forth between pairs of numbers and points in space, and it lets you visualize things you didn't think you could visualize, or it lets you understand things that are inherently visual just by kind of going back and forth. And in three dimensions, there's a similar dichotomy, but between triplets of points and points in three-dimensional space. So let me just plop down a point in three-dimensional space here."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "Every time that you have a pair of things, you know that you should be thinking two-dimensionally. And that's actually a surprisingly powerful idea that I don't think I appreciated for a long time, how there's this back and forth between pairs of numbers and points in space, and it lets you visualize things you didn't think you could visualize, or it lets you understand things that are inherently visual just by kind of going back and forth. And in three dimensions, there's a similar dichotomy, but between triplets of points and points in three-dimensional space. So let me just plop down a point in three-dimensional space here. And it's hard to get a feel for exactly where it is until you move things around. This is one thing that makes three dimensions hard, is you can't really draw it without moving it around or showing a difference in perspective in various ways. But we describe points like this, again, with a set of coordinates, but this time it's a triplet."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "So let me just plop down a point in three-dimensional space here. And it's hard to get a feel for exactly where it is until you move things around. This is one thing that makes three dimensions hard, is you can't really draw it without moving it around or showing a difference in perspective in various ways. But we describe points like this, again, with a set of coordinates, but this time it's a triplet. And this particular point, I happen to know, is one, two, five. And what those numbers are telling you is how far to move parallel to each axis. So just like with two dimensions, we have an x-axis and a y-axis, but now there's a third axis that's perpendicular to both of them and moves us into that third dimension, the z-axis."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "But we describe points like this, again, with a set of coordinates, but this time it's a triplet. And this particular point, I happen to know, is one, two, five. And what those numbers are telling you is how far to move parallel to each axis. So just like with two dimensions, we have an x-axis and a y-axis, but now there's a third axis that's perpendicular to both of them and moves us into that third dimension, the z-axis. And the first number in our coordinate is gonna tell us how far, whoop, can't move those guys, how far we need to move in the x direction as our first step. The second number, two in this case, tells us how far we have to move parallel to the y-axis for our second step. And then the third number tells us how far up we have to go to get to that point."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "So just like with two dimensions, we have an x-axis and a y-axis, but now there's a third axis that's perpendicular to both of them and moves us into that third dimension, the z-axis. And the first number in our coordinate is gonna tell us how far, whoop, can't move those guys, how far we need to move in the x direction as our first step. The second number, two in this case, tells us how far we have to move parallel to the y-axis for our second step. And then the third number tells us how far up we have to go to get to that point. And you can do this for any point in three-dimensional space, right? Any point that you have, you can give the instructions for how to move along the x and then how to move parallel to the y and how to move parallel to the z to get to that point, which means there's this back and forth between triplets of numbers and points in 3D. So whenever you come across a triplet of things, and you'll see this in the next video when we start talking about three-dimensional graphs, you'll know, just by virtue of the fact that it's a triplet, ah, yes, I should be thinking in three dimensions somehow."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then the third number tells us how far up we have to go to get to that point. And you can do this for any point in three-dimensional space, right? Any point that you have, you can give the instructions for how to move along the x and then how to move parallel to the y and how to move parallel to the z to get to that point, which means there's this back and forth between triplets of numbers and points in 3D. So whenever you come across a triplet of things, and you'll see this in the next video when we start talking about three-dimensional graphs, you'll know, just by virtue of the fact that it's a triplet, ah, yes, I should be thinking in three dimensions somehow. Just in the same way that whenever you have pairs, you should be thinking, ah, this is a very two-dimensional thing. So there's another context, though, where pairs of numbers come up. And that would be vectors."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "So whenever you come across a triplet of things, and you'll see this in the next video when we start talking about three-dimensional graphs, you'll know, just by virtue of the fact that it's a triplet, ah, yes, I should be thinking in three dimensions somehow. Just in the same way that whenever you have pairs, you should be thinking, ah, this is a very two-dimensional thing. So there's another context, though, where pairs of numbers come up. And that would be vectors. So a vector you might represent, you know, you typically represent it with an arrow. Whoa. Ah, help, help."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "And that would be vectors. So a vector you might represent, you know, you typically represent it with an arrow. Whoa. Ah, help, help. So vectors, so vectors we typically represent with some kind of arrow. Let's make this arrow a nice color. An arrow."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "Ah, help, help. So vectors, so vectors we typically represent with some kind of arrow. Let's make this arrow a nice color. An arrow. And if it's a vector from the origin to a simple point, the coordinates of that vector are just the same as those of the point. And the convention is to write those coordinates in a column. You know, it's not set in stone, but typically if you see numbers in a column, you should be thinking about it as a vector, some kind of arrow."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "An arrow. And if it's a vector from the origin to a simple point, the coordinates of that vector are just the same as those of the point. And the convention is to write those coordinates in a column. You know, it's not set in stone, but typically if you see numbers in a column, you should be thinking about it as a vector, some kind of arrow. And if it's a pair with parentheses around it, you just think about it as a point. And even though both of these are ways of representing the same pair of numbers, the main difference is that a vector, you could have started it at any point in space. It didn't have to start in the origin."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "You know, it's not set in stone, but typically if you see numbers in a column, you should be thinking about it as a vector, some kind of arrow. And if it's a pair with parentheses around it, you just think about it as a point. And even though both of these are ways of representing the same pair of numbers, the main difference is that a vector, you could have started it at any point in space. It didn't have to start in the origin. So if we have that same guy, but you know, if he starts here and he still has a rightward component of one and an upward component of three, we think of that as the same vector. And typically these are representing motion of some kind, whereas points are just representing like actual points in space. And the other big thing that you can do is you can add vectors together."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "It didn't have to start in the origin. So if we have that same guy, but you know, if he starts here and he still has a rightward component of one and an upward component of three, we think of that as the same vector. And typically these are representing motion of some kind, whereas points are just representing like actual points in space. And the other big thing that you can do is you can add vectors together. So you know, if you had another, let's say you have another vector that has a large x component, but a small negative y component like this guy. And what that means is you can kind of add by imagining that that second vector started at the tip of the first one. And then however you get from the origin to the new tip there, that's gonna be the resulting vector."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the other big thing that you can do is you can add vectors together. So you know, if you had another, let's say you have another vector that has a large x component, but a small negative y component like this guy. And what that means is you can kind of add by imagining that that second vector started at the tip of the first one. And then however you get from the origin to the new tip there, that's gonna be the resulting vector. So I'd say this is the sum of those two vectors. And you can't really do that with points as much. In order to think about adding points, you end up thinking about them as vectors."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then however you get from the origin to the new tip there, that's gonna be the resulting vector. So I'd say this is the sum of those two vectors. And you can't really do that with points as much. In order to think about adding points, you end up thinking about them as vectors. And the same goes with three dimensions. For a given point, if you draw an arrow from the origin up to that point, this arrow would be represented with that same triplet of numbers, but you typically do it in a column. I call this a column vector."}, {"video_title": "Representing points in 3d   Multivariable calculus   Khan Academy.mp3", "Sentence": "In order to think about adding points, you end up thinking about them as vectors. And the same goes with three dimensions. For a given point, if you draw an arrow from the origin up to that point, this arrow would be represented with that same triplet of numbers, but you typically do it in a column. I call this a column vector. It's not three, that's five. And the difference between the point and the arrow is you can think of the arrow or the vector as starting anywhere in space. It doesn't really matter as long as it's got those same components for how far does it move parallel to the x, how far does it move parallel to the y-axis, and how far does it move parallel to the z-axis."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "And the one I have in mind has a special name. It's a helix. And the first two components kind of make it look like a circle. It's gonna be cosine of t for the x component, sine of t for the y component, but this is three-dimensional, and it makes it a little different from a circle. I'm gonna have the last component be t divided by five. And what this looks like, we can visualize it pretty well. I'm gonna go on over to the graph of it here."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "It's gonna be cosine of t for the x component, sine of t for the y component, but this is three-dimensional, and it makes it a little different from a circle. I'm gonna have the last component be t divided by five. And what this looks like, we can visualize it pretty well. I'm gonna go on over to the graph of it here. So this shape is called a helix, and you can sort of see how, looking from the xy plane's perspective, it looks as if it's gonna draw a circle. And really, these lines should all line up when you're facing it, but it's due to the perspective where things farther away look smaller. But it would just be drawing a circle."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "I'm gonna go on over to the graph of it here. So this shape is called a helix, and you can sort of see how, looking from the xy plane's perspective, it looks as if it's gonna draw a circle. And really, these lines should all line up when you're facing it, but it's due to the perspective where things farther away look smaller. But it would just be drawing a circle. But then the z component, because z increases while your parameter t increases, you're kind of rising as if it's a spiral staircase. And now, before we compute curvature, to know what we're really going for, what this represents, you kind of imagine yourself, you know, maybe this isn't a road, but it's like a space freeway, right? And you're driving your spaceship along it, and you imagine that you get stuck at some point, or maybe not that you get stuck, but all of your instruments lock, and your steering wheel locks, or your joystick, or however you're steering, it all just locks up."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "But it would just be drawing a circle. But then the z component, because z increases while your parameter t increases, you're kind of rising as if it's a spiral staircase. And now, before we compute curvature, to know what we're really going for, what this represents, you kind of imagine yourself, you know, maybe this isn't a road, but it's like a space freeway, right? And you're driving your spaceship along it, and you imagine that you get stuck at some point, or maybe not that you get stuck, but all of your instruments lock, and your steering wheel locks, or your joystick, or however you're steering, it all just locks up. And you're gonna trace out a certain circle in space, right? And that circle might look something like this. So if you were turning however you were on the helix, but then you can't do anything different, you might trace out a giant circle."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "And you're driving your spaceship along it, and you imagine that you get stuck at some point, or maybe not that you get stuck, but all of your instruments lock, and your steering wheel locks, or your joystick, or however you're steering, it all just locks up. And you're gonna trace out a certain circle in space, right? And that circle might look something like this. So if you were turning however you were on the helix, but then you can't do anything different, you might trace out a giant circle. And what we care about is the radius of that circle, and if you take one divided by the radius of that circle you trace out, that's gonna be the curvature. That's gonna be the little kappa curvature. And of course, the way that we compute it, we don't directly talk about that circle at all, but it's actually a good thing to keep in the back of your mind."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "So if you were turning however you were on the helix, but then you can't do anything different, you might trace out a giant circle. And what we care about is the radius of that circle, and if you take one divided by the radius of that circle you trace out, that's gonna be the curvature. That's gonna be the little kappa curvature. And of course, the way that we compute it, we don't directly talk about that circle at all, but it's actually a good thing to keep in the back of your mind. The way that we compute it is to first find a unit tangent vector function with the same parameter. And what that means, you know, if you imagine your helix kind of spiraling through three-dimensional space, man, I am not as good an artist as the computer is when it comes to drawing a helix. But the unit tangent vector function would be something that gives you a tangent vector at every given point, you know, kind of the direction that you on your spaceship are traveling."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "And of course, the way that we compute it, we don't directly talk about that circle at all, but it's actually a good thing to keep in the back of your mind. The way that we compute it is to first find a unit tangent vector function with the same parameter. And what that means, you know, if you imagine your helix kind of spiraling through three-dimensional space, man, I am not as good an artist as the computer is when it comes to drawing a helix. But the unit tangent vector function would be something that gives you a tangent vector at every given point, you know, kind of the direction that you on your spaceship are traveling. And to do that, you take the derivative of your parameterization, that derivative, which is gonna give you a tangent vector, but it might not be a unit tangent vector, so you divide it by its own magnitude, and that'll give you a unit tangent vector. And then ultimately, the goal that we're shooting for is gonna be to find the derivative of this tangent vector function with respect to the arc length. So as a first step, we'll start by finding a derivative of our parameterization function."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "But the unit tangent vector function would be something that gives you a tangent vector at every given point, you know, kind of the direction that you on your spaceship are traveling. And to do that, you take the derivative of your parameterization, that derivative, which is gonna give you a tangent vector, but it might not be a unit tangent vector, so you divide it by its own magnitude, and that'll give you a unit tangent vector. And then ultimately, the goal that we're shooting for is gonna be to find the derivative of this tangent vector function with respect to the arc length. So as a first step, we'll start by finding a derivative of our parameterization function. So when we take that derivative, luckily there's not a lot of new things going on. Let's see, derivative, so S prime. From single-variable calculus, we just take the derivative of each component."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "So as a first step, we'll start by finding a derivative of our parameterization function. So when we take that derivative, luckily there's not a lot of new things going on. Let's see, derivative, so S prime. From single-variable calculus, we just take the derivative of each component. So cosine goes to negative sine, negative sine of t. Sine, its derivative is cosine, cosine of t. And then the derivative of t divided by five is just a constant, that's just one over five. Boy, it is hard to say the word derivative over and over. Say it five times."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "From single-variable calculus, we just take the derivative of each component. So cosine goes to negative sine, negative sine of t. Sine, its derivative is cosine, cosine of t. And then the derivative of t divided by five is just a constant, that's just one over five. Boy, it is hard to say the word derivative over and over. Say it five times. Okay, so that's S prime of t. And now what we need to do, we need to find the magnitude of S prime of t. So what that involves, as we're taking the magnitude, S prime of t as a vector, we take the square root of the sum of the squares of each of its components. So sine, negative sine squared, just looks like sine squared, sine squared of t, cosine squared, cosine squared of t, and then 1 5th squared, and that's just 1 25th. And you might notice I use a lot of these sine-cosine pairs in examples, partly because they draw circles, and lots of things are fun that involve drawing circles, but also because it has a tendency to let things simplify, especially if you're taking a magnitude, because sine squared plus cosine squared just equals one."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "Say it five times. Okay, so that's S prime of t. And now what we need to do, we need to find the magnitude of S prime of t. So what that involves, as we're taking the magnitude, S prime of t as a vector, we take the square root of the sum of the squares of each of its components. So sine, negative sine squared, just looks like sine squared, sine squared of t, cosine squared, cosine squared of t, and then 1 5th squared, and that's just 1 25th. And you might notice I use a lot of these sine-cosine pairs in examples, partly because they draw circles, and lots of things are fun that involve drawing circles, but also because it has a tendency to let things simplify, especially if you're taking a magnitude, because sine squared plus cosine squared just equals one. So this entire formula, this entire formula, boils down to the square root of one plus one divided by 25. And for this, you might kind of be thinking off to the side that that's 25 over 25 plus one over 25. So making even more room here, what that equals is the square root of 26 divided by 25."}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "And you might notice I use a lot of these sine-cosine pairs in examples, partly because they draw circles, and lots of things are fun that involve drawing circles, but also because it has a tendency to let things simplify, especially if you're taking a magnitude, because sine squared plus cosine squared just equals one. So this entire formula, this entire formula, boils down to the square root of one plus one divided by 25. And for this, you might kind of be thinking off to the side that that's 25 over 25 plus one over 25. So making even more room here, what that equals is the square root of 26 divided by 25. And just because 25 is already a square, and it kind of might make things look nice, I'm gonna write this as the square root of 26 divided by five, that square root of 25. So this whole thing is the magnitude of our derivative, right? And we think to ourselves, it's quite lucky that this came out to be a constant, because as we saw with the more general formula, it's often pretty nasty, and it can get pretty bad, but in this case, it's just a constant, which is nice, because as we go up, and we start to think about what our unit tangent vector function for the helix should be, we're just gonna take the derivative function and divide each term by that magnitude, right?"}, {"video_title": "Curvature of a helix, part 1.mp3", "Sentence": "So making even more room here, what that equals is the square root of 26 divided by 25. And just because 25 is already a square, and it kind of might make things look nice, I'm gonna write this as the square root of 26 divided by five, that square root of 25. So this whole thing is the magnitude of our derivative, right? And we think to ourselves, it's quite lucky that this came out to be a constant, because as we saw with the more general formula, it's often pretty nasty, and it can get pretty bad, but in this case, it's just a constant, which is nice, because as we go up, and we start to think about what our unit tangent vector function for the helix should be, we're just gonna take the derivative function and divide each term by that magnitude, right? So it's gonna look almost identical. It's gonna be negative sine of t, except now we're dividing by that magnitude, and that magnitude, of course, is root 26 over five. So we go up here, and we say we're dividing this by root 26 over five, that whole quantity, and then similarly, y component is cosine of t divided by the quantity of the root, 26 divided by five, and the last part is 1 5th, 1 5th, I'll put that in parentheses, divided by that same amount, root 26 over five."}, {"video_title": "2d curl nuance.mp3", "Sentence": "In the last couple videos, I've been talking about curl, where if we have a two-dimensional vector field V defined with component functions P and Q, I've said that the 2D curl of that function V gives you a new function that also takes in x and y as inputs, and its formula is the partial derivative of Q with respect to x minus the partial derivative of P with respect to y. And my hope is that this is more than just a formula, and that you can understand how this represents fluid rotation in two dimensions. But what I want to do here is show how the original intuition I gave for this formula might be a little oversimplified. Because, for example, if we look at this, the partial Q partial x component, I said that you can imagine that Q at some point starting off a little bit negative, so the y component of the output is a little negative, then as you move positively in the x direction, it goes to being 0, and then it goes to being a little bit positive. And with this particular picture, it's hopefully a little bit clear why this can correspond to counterclockwise rotation in the fluid. But this is only a very specific circumstance for what partial Q partial x being positive could look like. You know, it might also look like Q starting off a little bit positive, and then as you move in the x direction, it becomes even more positive, and then even more positive."}, {"video_title": "2d curl nuance.mp3", "Sentence": "Because, for example, if we look at this, the partial Q partial x component, I said that you can imagine that Q at some point starting off a little bit negative, so the y component of the output is a little negative, then as you move positively in the x direction, it goes to being 0, and then it goes to being a little bit positive. And with this particular picture, it's hopefully a little bit clear why this can correspond to counterclockwise rotation in the fluid. But this is only a very specific circumstance for what partial Q partial x being positive could look like. You know, it might also look like Q starting off a little bit positive, and then as you move in the x direction, it becomes even more positive, and then even more positive. And according to the formula, this should contribute as much to positive curl as this very clear-cut kind of whirlpool example. And to illustrate what this might look like, if we take a look at this vector field here, if we look in the center, this is kind of the clear-cut whirlpool counterclockwise rotation example. And if we play the fluid flow, the fluid does indeed rotate counterclockwise in the region."}, {"video_title": "2d curl nuance.mp3", "Sentence": "You know, it might also look like Q starting off a little bit positive, and then as you move in the x direction, it becomes even more positive, and then even more positive. And according to the formula, this should contribute as much to positive curl as this very clear-cut kind of whirlpool example. And to illustrate what this might look like, if we take a look at this vector field here, if we look in the center, this is kind of the clear-cut whirlpool counterclockwise rotation example. And if we play the fluid flow, the fluid does indeed rotate counterclockwise in the region. But contrast that with what goes on over here on the right. This doesn't look like rotation in that sense at all. Instead, the fluid particles are just kind of rushing up through it."}, {"video_title": "2d curl nuance.mp3", "Sentence": "And if we play the fluid flow, the fluid does indeed rotate counterclockwise in the region. But contrast that with what goes on over here on the right. This doesn't look like rotation in that sense at all. Instead, the fluid particles are just kind of rushing up through it. But in fact, the curl in this region is going to be just as strong as it is over here. And I'll show that with the formula, and kind of computing it through in just a moment. But the image that you might have in your mind is to imagine a paddle wheel of sorts, where let's say it's got arms kind of like that, and then you hold down with your thumb that middle portion."}, {"video_title": "2d curl nuance.mp3", "Sentence": "Instead, the fluid particles are just kind of rushing up through it. But in fact, the curl in this region is going to be just as strong as it is over here. And I'll show that with the formula, and kind of computing it through in just a moment. But the image that you might have in your mind is to imagine a paddle wheel of sorts, where let's say it's got arms kind of like that, and then you hold down with your thumb that middle portion. So even though the paddle wheel left to its own devices would just kind of fly up, I want to say, let's say you're holding that down with your thumb, but it's free to rotate. Then the vectors on its left are pointing up, but less strongly than the vectors on its right, which are even greater. So if you imagine that setup, and you kind of have your paddle wheel there, then when you play the fluid rotation, holding your thumb down, but letting the paddle wheel rotate freely, it's also going to rotate just as it would over here in the easier to see whirlpool example."}, {"video_title": "2d curl nuance.mp3", "Sentence": "But the image that you might have in your mind is to imagine a paddle wheel of sorts, where let's say it's got arms kind of like that, and then you hold down with your thumb that middle portion. So even though the paddle wheel left to its own devices would just kind of fly up, I want to say, let's say you're holding that down with your thumb, but it's free to rotate. Then the vectors on its left are pointing up, but less strongly than the vectors on its right, which are even greater. So if you imagine that setup, and you kind of have your paddle wheel there, then when you play the fluid rotation, holding your thumb down, but letting the paddle wheel rotate freely, it's also going to rotate just as it would over here in the easier to see whirlpool example. And in terms of the formula, this is because a situation like this one here, where q goes from being negative to zero to positive, should be treated just the same as a situation like this, as far as 2D curl is concerned, because this term in the 2D curl formula is going to come out the same for either one of these. And it's worth pointing out, by the way, curl isn't something that mathematicians and physicists came across trying to understand fluid flow. Instead, they found this term as being significant in various other formulas and circumstances, and I think electromagnetism might be where it originally came about."}, {"video_title": "2d curl nuance.mp3", "Sentence": "So if you imagine that setup, and you kind of have your paddle wheel there, then when you play the fluid rotation, holding your thumb down, but letting the paddle wheel rotate freely, it's also going to rotate just as it would over here in the easier to see whirlpool example. And in terms of the formula, this is because a situation like this one here, where q goes from being negative to zero to positive, should be treated just the same as a situation like this, as far as 2D curl is concerned, because this term in the 2D curl formula is going to come out the same for either one of these. And it's worth pointing out, by the way, curl isn't something that mathematicians and physicists came across trying to understand fluid flow. Instead, they found this term as being significant in various other formulas and circumstances, and I think electromagnetism might be where it originally came about. But then in trying to understand this formula, they realized that you can give a fluid flow interpretation that gives a very deep understanding of what's going on beyond just the symbols themselves. So let me go ahead and walk through this example in terms of the formula representing the vector field. It's a relatively straightforward formula, actually."}, {"video_title": "2d curl nuance.mp3", "Sentence": "Instead, they found this term as being significant in various other formulas and circumstances, and I think electromagnetism might be where it originally came about. But then in trying to understand this formula, they realized that you can give a fluid flow interpretation that gives a very deep understanding of what's going on beyond just the symbols themselves. So let me go ahead and walk through this example in terms of the formula representing the vector field. It's a relatively straightforward formula, actually. So p and q, that x component is going to be negative y, and the y component, q, is equal to x. So when we apply our 2D curl formula and apply the partial of q with respect to x, so partial of the second component with respect to x is just one, and then we subtract off the partial of p with respect to y, which up here is negative one, because p is just equal to negative y. So the 2D curl is equal to two, and in particular, it's a constant two that doesn't depend on x and y, which is pretty unusual."}, {"video_title": "2d curl nuance.mp3", "Sentence": "It's a relatively straightforward formula, actually. So p and q, that x component is going to be negative y, and the y component, q, is equal to x. So when we apply our 2D curl formula and apply the partial of q with respect to x, so partial of the second component with respect to x is just one, and then we subtract off the partial of p with respect to y, which up here is negative one, because p is just equal to negative y. So the 2D curl is equal to two, and in particular, it's a constant two that doesn't depend on x and y, which is pretty unusual. Most times that you apply 2D curl to a vector field, you're going to get some kind of function of x and y. But the fact that this is constant tells us that when we look over at this fluid flow, the sense in which curl, the formula for curl, wants to say that rotation happens around the center is just as strong as it's supposed to happen over here on the right, or anywhere on the plane for that matter. So if we're playing this, and if you imagine the paddle wheel in the center, evidently it would be rotating just as quickly as the paddle on the right, even though it might, I don't know, to me that feels a little unintuitive, because the one on the right I'm thinking, okay, you know, there's, maybe there's a little bit more torque on the right side than there is on the left, and that's kind of a counterbalancing act, but the idea that that's actually the same as the very clear-cut, I see the counterclockwise rotation with my eyes example in the center, does seem a little unusual."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So I have talked a lot about different ways that you can visualize multivariable functions, functions that'll have some kind of multidimensional input or output. These include three-dimensional graphs, which are very common, contour maps, vector fields, parametric functions, but here I want to talk about one of my all-time favorite ways to think about functions, which is as a transformation. So anytime you have some sort of function, if you're thinking very abstractly, I like to think that there's some sort of input space, and I'll draw it as a blob, even though that could be the real number line, so it should be a line, or it could be three-dimensional space, and then there's some kind of output space, and again, I just very vaguely think about it as this blob, but that could be, again, the real number line, the xy-plane, three-dimensional space, and the function is just some way of taking inputs to outputs. And every time that we're trying to visualize something, like with a graph or a contour map, you're just trying to associate input- output pairs. If f inputs 3 gets mapped to the vector 1, 2, it's a question of how do you associate the number 3 with that vector 1, 2? And the thought behind transformations is that we're just going to watch the actual points of the input space move to the output space. And I'll start with a simple example that's just a one-dimensional function."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "And every time that we're trying to visualize something, like with a graph or a contour map, you're just trying to associate input- output pairs. If f inputs 3 gets mapped to the vector 1, 2, it's a question of how do you associate the number 3 with that vector 1, 2? And the thought behind transformations is that we're just going to watch the actual points of the input space move to the output space. And I'll start with a simple example that's just a one-dimensional function. It'll have a single variable input, and it'll have a single variable output. So let's consider the function f of x is equal to x squared minus 3. And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "And I'll start with a simple example that's just a one-dimensional function. It'll have a single variable input, and it'll have a single variable output. So let's consider the function f of x is equal to x squared minus 3. And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs. I just want to say, how do the inputs move to those outputs? So as an example, if you go to 0, when you plug in 0, you're going to get negative 3, you know, 0 squared minus 3 is equal to negative 3. So somehow we want to watch 0 move to negative 3."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs. I just want to say, how do the inputs move to those outputs? So as an example, if you go to 0, when you plug in 0, you're going to get negative 3, you know, 0 squared minus 3 is equal to negative 3. So somehow we want to watch 0 move to negative 3. And then similarly, let's say you plug in 1, and you get 1 squared minus 3 is negative 2, so somehow we want to watch 1 move to negative 2. And just to list another example here, let's say you're plugging in 3 itself, so 3 squared minus 3 is 9 minus 3 is 6. So somehow in this transformation, we want to watch 3 move to the number 6."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So somehow we want to watch 0 move to negative 3. And then similarly, let's say you plug in 1, and you get 1 squared minus 3 is negative 2, so somehow we want to watch 1 move to negative 2. And just to list another example here, let's say you're plugging in 3 itself, so 3 squared minus 3 is 9 minus 3 is 6. So somehow in this transformation, we want to watch 3 move to the number 6. And with a little animation, we can watch this happen. We can actually watch what it looks like for all these numbers to move to their corresponding outputs. So here we go."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So somehow in this transformation, we want to watch 3 move to the number 6. And with a little animation, we can watch this happen. We can actually watch what it looks like for all these numbers to move to their corresponding outputs. So here we go. Each number will move over and land on its output. And I'll clear up the board here. So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So here we go. Each number will move over and land on its output. And I'll clear up the board here. So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves. And I'll play it again. Here, let's just watch where each number from the input space moves over to the output. And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves. And I'll play it again. Here, let's just watch where each number from the input space moves over to the output. And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions. So now, let me consider a function that has a one-dimensional input and a two-dimensional output. And specifically, it'll be f of x is equal to cosine of x cosine of x and then the y component will be x sine of y. Sorry, x sine of x."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions. So now, let me consider a function that has a one-dimensional input and a two-dimensional output. And specifically, it'll be f of x is equal to cosine of x cosine of x and then the y component will be x sine of y. Sorry, x sine of x. So just to think about a couple examples, if you plug in something like 0 and think about where 0 ought to go, you'd have f of 0 is equal to cosine of 0 is 1 and then 0 times anything is 0. So somehow, we're going to watch 0 move over to the point 1, 0. So this is where we expect 0 to land."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sorry, x sine of x. So just to think about a couple examples, if you plug in something like 0 and think about where 0 ought to go, you'd have f of 0 is equal to cosine of 0 is 1 and then 0 times anything is 0. So somehow, we're going to watch 0 move over to the point 1, 0. So this is where we expect 0 to land. And let's think about pi. So, f of pi, then cosine of pi is negative 1. This is going to be pi multiplied by sine of pi is 0, so that'll again be 0."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "So this is where we expect 0 to land. And let's think about pi. So, f of pi, then cosine of pi is negative 1. This is going to be pi multiplied by sine of pi is 0, so that'll again be 0. So, you know, this little guy is where 0 lands and we expect that this is going to be where the value pi lands. And if we watch this take place, and we actually watch each element of the input space move over to the output space, we get something like this. And again, this is just kind of a nice way to think about what's actually going on."}, {"video_title": "Transformations, part 1   Multivariable calculus   Khan Academy.mp3", "Sentence": "This is going to be pi multiplied by sine of pi is 0, so that'll again be 0. So, you know, this little guy is where 0 lands and we expect that this is going to be where the value pi lands. And if we watch this take place, and we actually watch each element of the input space move over to the output space, we get something like this. And again, this is just kind of a nice way to think about what's actually going on. You might ask questions about whether the space ends up getting stretched or squished. And notice that this is also what a parametric plot of this function would look like. If you interpret it as a parametric function, this is what you get in the end."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "In the last video, we were given a multivariable function and asked to find and classify all of its critical points. So critical points just means finding where the gradient is equal to zero. And we found four different points for that. I have them down here. They were zero, zero, zero, negative two, square root of three and one, and negative square root of three and one. So then, the next step is to classify those. And that requires the second partial derivative test."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "I have them down here. They were zero, zero, zero, negative two, square root of three and one, and negative square root of three and one. So then, the next step is to classify those. And that requires the second partial derivative test. So what I'm gonna go ahead and do is copy down the partial derivatives, since we already computed those, copy. And then just kind of paste them down here where we can start to use them for the second partial derivatives. So let me clean things up a little bit."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And that requires the second partial derivative test. So what I'm gonna go ahead and do is copy down the partial derivatives, since we already computed those, copy. And then just kind of paste them down here where we can start to use them for the second partial derivatives. So let me clean things up a little bit. And we don't need this simplification of it. So we've got our partial derivatives. Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So let me clean things up a little bit. And we don't need this simplification of it. So we've got our partial derivatives. Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function. That's just kind of the first thing to do. So let's go ahead and do it. The second partial derivative of the function with respect to x twice in a row."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function. That's just kind of the first thing to do. So let's go ahead and do it. The second partial derivative of the function with respect to x twice in a row. We'll take the partial derivative with respect to x and then do it with respect to x again. So this first term looks like six times a variable times a constant. So it'll just be six times that constant."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "The second partial derivative of the function with respect to x twice in a row. We'll take the partial derivative with respect to x and then do it with respect to x again. So this first term looks like six times a variable times a constant. So it'll just be six times that constant. And then the second term, the derivative of negative six x is just negative six. Moving right along, when we do the second partial derivative with respect to y twice in a row, we take the partial derivative with respect to y and then do it again. So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So it'll just be six times that constant. And then the second term, the derivative of negative six x is just negative six. Moving right along, when we do the second partial derivative with respect to y twice in a row, we take the partial derivative with respect to y and then do it again. So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it. The derivative of negative three y squared is negative six times y. And then the derivative of negative six y is just negative six. And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it. The derivative of negative three y squared is negative six times y. And then the derivative of negative six y is just negative six. And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y. The order doesn't really matter in this case since it's a perfectly ordinary polynomial function. So we could do it either way, but I'm just gonna choose to take a look at this guy and differentiate it with respect to y. So the derivative of the first term with respect to y is six x, six x."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y. The order doesn't really matter in this case since it's a perfectly ordinary polynomial function. So we could do it either way, but I'm just gonna choose to take a look at this guy and differentiate it with respect to y. So the derivative of the first term with respect to y is six x, six x. And then that second term looks like a constant with respect to y. So that's all we have. So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So the derivative of the first term with respect to y is six x, six x. And then that second term looks like a constant with respect to y. So that's all we have. So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression. And to remind you of what that is, that expression is, we take the second partial derivative with respect to x twice, and I'll just write it with a kind of shorter notation using subscripts. And we multiply that by the second partial derivative with respect to x, and then we subtract off, subtract off the mixed partial derivative term squared. So let's go ahead and do that for each of our points."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression. And to remind you of what that is, that expression is, we take the second partial derivative with respect to x twice, and I'll just write it with a kind of shorter notation using subscripts. And we multiply that by the second partial derivative with respect to x, and then we subtract off, subtract off the mixed partial derivative term squared. So let's go ahead and do that for each of our points. So when we do this at the point zero, zero, zero, zero, what we end up getting, plugging that into the partial derivative with respect to x twice, six times zero is zero, so there's just negative six. So that gives us negative six multiplied by, when we plug it into this partial derivative with respect to y squared, again, that y goes to zero, so we're left with just negative six. And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So let's go ahead and do that for each of our points. So when we do this at the point zero, zero, zero, zero, what we end up getting, plugging that into the partial derivative with respect to x twice, six times zero is zero, so there's just negative six. So that gives us negative six multiplied by, when we plug it into this partial derivative with respect to y squared, again, that y goes to zero, so we're left with just negative six. And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero. So we're subtracting off zero squared. And that entire thing equals negative six times negative six is 36. 36."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero. So we're subtracting off zero squared. And that entire thing equals negative six times negative six is 36. 36. And we'll get to analyzing what it means that that's positive in just a moment, but let's just kind of get all of them on the board so we can kind of start doing this with all of them. If we do this with zero and negative two, zero and negative two, then once we plug in y equals negative two to this expression, this time I'll write it out, six times negative two minus six, so that's negative 12 minus six, we'll get negative 18. Negative 18."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "36. And we'll get to analyzing what it means that that's positive in just a moment, but let's just kind of get all of them on the board so we can kind of start doing this with all of them. If we do this with zero and negative two, zero and negative two, then once we plug in y equals negative two to this expression, this time I'll write it out, six times negative two minus six, so that's negative 12 minus six, we'll get negative 18. Negative 18. Then when we plug it into the partial derivative of f with respect to y squared, again, I'll kind of write it out, we have negative six times y is equal to negative two minus six. So now we have negative six times negative two, so that's positive 12 minus six. So that will be a positive six that we plug in here."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "Negative 18. Then when we plug it into the partial derivative of f with respect to y squared, again, I'll kind of write it out, we have negative six times y is equal to negative two minus six. So now we have negative six times negative two, so that's positive 12 minus six. So that will be a positive six that we plug in here. And then for the mixed partial derivative, again, x is equal to zero, so the mixed partial derivative is just gonna look like zero when we do this. So we're subtracting off zero squared and we get negative 18 times six. And geez, what's 18 times six?"}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So that will be a positive six that we plug in here. And then for the mixed partial derivative, again, x is equal to zero, so the mixed partial derivative is just gonna look like zero when we do this. So we're subtracting off zero squared and we get negative 18 times six. And geez, what's 18 times six? So that's gonna be 36 times three, so that's the same as 90 plus 18, so I think that's 108. Negative 108. And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And geez, what's 18 times six? So that's gonna be 36 times three, so that's the same as 90 plus 18, so I think that's 108. Negative 108. And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative. So now, kind of moving right along, these examples can take quite a while. If we plug in square root of three, one, square root of three, one, what we get. Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative. So now, kind of moving right along, these examples can take quite a while. If we plug in square root of three, one, square root of three, one, what we get. Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero. And then for the partial derivative with respect to y squared, instead of plugging in negative two, now we're plugging in y equals one, so we have negative six times one minus six, so the whole thing is negative 12, so negative 12. And now for the mixed partial derivative term, which is six x, x is equal to the square root of three, so now we're subtracting off the square root of three squared, so what that equals is, this first part is just entirely zero, and we're subtracting off three, so that's negative three. And then we have square root of three, no, no, we don't, that's what we just did."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero. And then for the partial derivative with respect to y squared, instead of plugging in negative two, now we're plugging in y equals one, so we have negative six times one minus six, so the whole thing is negative 12, so negative 12. And now for the mixed partial derivative term, which is six x, x is equal to the square root of three, so now we're subtracting off the square root of three squared, so what that equals is, this first part is just entirely zero, and we're subtracting off three, so that's negative three. And then we have square root of three, no, no, we don't, that's what we just did. Now we have negative square root of three, one, and this will be very similar because this first term just had a y and we plugged in the y, so it's also gonna be zero for totally the same reasons, and same deal over here, the value of y didn't change, so that's also gonna be negative 12. Doesn't really matter because we're multiplying it by a zero, right? And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And then we have square root of three, no, no, we don't, that's what we just did. Now we have negative square root of three, one, and this will be very similar because this first term just had a y and we plugged in the y, so it's also gonna be zero for totally the same reasons, and same deal over here, the value of y didn't change, so that's also gonna be negative 12. Doesn't really matter because we're multiplying it by a zero, right? And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three. So what does this second partial derivative test tell us? Once we express this term, if it's greater than zero, we have a max or a min. That's what the test tells us."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three. So what does this second partial derivative test tell us? Once we express this term, if it's greater than zero, we have a max or a min. That's what the test tells us. And then if it's less than zero, if it's less than zero, we have a saddle point. So in this case, the only term that's greater than zero is this first one, is this first one, and to analyze whether it's a maximum or a minimum, notice that the partial derivative with respect to x twice in a row or with respect to y twice in a row was negative, which indicates a sort of negative concavity, meaning this corresponds to a maximum. So this guy corresponds to a local maximum."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "That's what the test tells us. And then if it's less than zero, if it's less than zero, we have a saddle point. So in this case, the only term that's greater than zero is this first one, is this first one, and to analyze whether it's a maximum or a minimum, notice that the partial derivative with respect to x twice in a row or with respect to y twice in a row was negative, which indicates a sort of negative concavity, meaning this corresponds to a maximum. So this guy corresponds to a local maximum. Now all of the other three gave us negative numbers, so all of these other three give us saddle points. Saddle points. So the answer to the question, the original find and classify such and such points, is that we found four different critical points."}, {"video_title": "Second partial derivative test example, part 2.mp3", "Sentence": "So this guy corresponds to a local maximum. Now all of the other three gave us negative numbers, so all of these other three give us saddle points. Saddle points. So the answer to the question, the original find and classify such and such points, is that we found four different critical points. Let's see. Four different critical points, zero, zero, zero, negative two, square root of three, one, and negative square root of three, one, and all of them are saddle points except for zero, zero, which is a local maximum. And all of that is something that we can tell without even looking at the graph of the function."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And one way to think about it is we want our x and y values to take on all of the values inside of the unit circle, what I'm shading in right over here. And then our z values can be a function of the y values. We can express this equation right here, z is equal to two minus y. And then we can figure out how high above to go to get our z value. And by doing that, we'll be able to essentially get to every point that sits on our surface. And so first let's think about how we can get every x and y value inside of the unit circle. And so let me, let's just focus on the xy plane."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we can figure out how high above to go to get our z value. And by doing that, we'll be able to essentially get to every point that sits on our surface. And so first let's think about how we can get every x and y value inside of the unit circle. And so let me, let's just focus on the xy plane. We're kind of rotated around a little bit so it looks a little bit more traditional. So this is our, this is my x-axis, and then my y-axis would look something like that. Let me draw a little bit different."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let me, let's just focus on the xy plane. We're kind of rotated around a little bit so it looks a little bit more traditional. So this is our, this is my x-axis, and then my y-axis would look something like that. Let me draw a little bit different. This is my y-axis, y-axis. And then if I were to draw the unit circle, so kind of the base of this thing, or at least where it intersects the xy plane, actually this thing would keep going down if I wanted to draw the x squared plus y squared equals one. But if I draw where it intersects the xy plane, it is, we get the unit circle."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw a little bit different. This is my y-axis, y-axis. And then if I were to draw the unit circle, so kind of the base of this thing, or at least where it intersects the xy plane, actually this thing would keep going down if I wanted to draw the x squared plus y squared equals one. But if I draw where it intersects the xy plane, it is, we get the unit circle. So let me just draw it. So we get, that's my best attempt at drawing a unit circle. We get the unit circle."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But if I draw where it intersects the xy plane, it is, we get the unit circle. So let me just draw it. So we get, that's my best attempt at drawing a unit circle. We get the unit circle. We need to think of using parameters so that we can get every x and y coordinate that's inside of the unit circle. And to think about that, I'll introduce one parameter that's essentially the angle with the x-axis, and I'll call that parameter theta. So theta is the angle with the x-axis."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We get the unit circle. We need to think of using parameters so that we can get every x and y coordinate that's inside of the unit circle. And to think about that, I'll introduce one parameter that's essentially the angle with the x-axis, and I'll call that parameter theta. So theta is the angle with the x-axis. And so theta will essentially sweep things all the way around. So theta can go between zero and two pi. So theta will take on values between zero and two pi."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So theta is the angle with the x-axis. And so theta will essentially sweep things all the way around. So theta can go between zero and two pi. So theta will take on values between zero and two pi. And if we just fix the radius at some point, say radius one, that would only give us all of the points on the unit circle. But we want all the points inside of it too. So we need to vary the radius as well."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So theta will take on values between zero and two pi. And if we just fix the radius at some point, say radius one, that would only give us all of the points on the unit circle. But we want all the points inside of it too. So we need to vary the radius as well. So let's introduce another parameter. Let's call it r. That is the radius. So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we need to vary the radius as well. So let's introduce another parameter. Let's call it r. That is the radius. So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius. And if you change radius a little bit more, you'll sweep out another circle. And if you vary radius between zero and one, you'll get all of the circles that will fill out this entire area. So the radius is going to go between zero and one."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius. And if you change radius a little bit more, you'll sweep out another circle. And if you vary radius between zero and one, you'll get all of the circles that will fill out this entire area. So the radius is going to go between zero and one. Another way of thinking about it is, for any given theta, if you keep varying the radiuses, you'll sweep out all the points on this line. And then as you change theta, it'll sweep out the entire circle. So either way you think about it."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the radius is going to go between zero and one. Another way of thinking about it is, for any given theta, if you keep varying the radiuses, you'll sweep out all the points on this line. And then as you change theta, it'll sweep out the entire circle. So either way you think about it. So with that, let's actually define x and y in those terms. So we could say that x is equal to, x is going to be equal to, so the x value, whatever r is, the x value is going to be r cosine theta. It's going to be that component."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So either way you think about it. So with that, let's actually define x and y in those terms. So we could say that x is equal to, x is going to be equal to, so the x value, whatever r is, the x value is going to be r cosine theta. It's going to be that component. It's going to be r cosine theta. And then the y component, this is just basic trigonometry, is going to be, the y is just going to be r sine theta. Y is going to be r sine theta."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be that component. It's going to be r cosine theta. And then the y component, this is just basic trigonometry, is going to be, the y is just going to be r sine theta. Y is going to be r sine theta. And then the z component, we already said z can be expressed as a function of y. Right over here, we can rewrite this as z is equal to two minus y. That'll tell us how high to go so that we end up on that plane."}, {"video_title": "Stokes example part 2 Parameterizing the surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Y is going to be r sine theta. And then the z component, we already said z can be expressed as a function of y. Right over here, we can rewrite this as z is equal to two minus y. That'll tell us how high to go so that we end up on that plane. So if z is equal to two minus y, and if y is r sine theta, we can rewrite z, we can rewrite z as being equal to two minus r sine theta. So there we're done. That's our parameterization."}, {"video_title": "Divergence notation.mp3", "Sentence": "So I've said that if you have a vector field, a two-dimensional vector field with component functions p and q, that the divergence of this guy, the divergence of v, which is a scalar-valued function of x and y, is by definition the partial derivative of p with respect to x, plus the partial derivative of q with respect to y. And there's actually another notation for divergence that's kind of helpful for remembering the formula. And what it is, is you take this nabla symbol, that upside-down triangle that we also use for the gradient, and imagine taking the dot product between that and your vector-valued function. And as we did with the gradient, the loose mnemonic you have for this upside-down triangle, as you think of it as a vector full of partial differential operators. And that sounds fancy, but all it means is you kind of take this partial partial x, a thing that wants to take in a function and take its partial derivative, and that's its first component, and the second component is this partial partial y, a thing that wants to take in a function and take its partial derivative with respect to y. And loosely, this isn't really a vector. These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically."}, {"video_title": "Divergence notation.mp3", "Sentence": "And as we did with the gradient, the loose mnemonic you have for this upside-down triangle, as you think of it as a vector full of partial differential operators. And that sounds fancy, but all it means is you kind of take this partial partial x, a thing that wants to take in a function and take its partial derivative, and that's its first component, and the second component is this partial partial y, a thing that wants to take in a function and take its partial derivative with respect to y. And loosely, this isn't really a vector. These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically. And you imagine taking the dot product with that and v, who has components, these scalar-valued functions, p of x, y, and q of x, y. And when you imagine doing this dot product, and you're kind of lining up terms, and the first one multiplied by the second, right, quote-unquote multiplied, because in this case, when I say this first component multiplied by p, I really mean you're taking that partial derivative operator, partial partial x, and evaluating it at p. That's kind of what multiplication looks like in this case. So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative."}, {"video_title": "Divergence notation.mp3", "Sentence": "These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically. And you imagine taking the dot product with that and v, who has components, these scalar-valued functions, p of x, y, and q of x, y. And when you imagine doing this dot product, and you're kind of lining up terms, and the first one multiplied by the second, right, quote-unquote multiplied, because in this case, when I say this first component multiplied by p, I really mean you're taking that partial derivative operator, partial partial x, and evaluating it at p. That's kind of what multiplication looks like in this case. So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative. So we see we get the same thing over here. It's the same formula that we have, and it's just kind of a nice little, you could think of it as a mnemonic device for remembering what the divergence is. But another nice thing, this can apply to higher-dimensional functions as well, right?"}, {"video_title": "Divergence notation.mp3", "Sentence": "So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative. So we see we get the same thing over here. It's the same formula that we have, and it's just kind of a nice little, you could think of it as a mnemonic device for remembering what the divergence is. But another nice thing, this can apply to higher-dimensional functions as well, right? If we have something that's, let's see, something that's a vector-valued function, and it's going to be a three-dimensional vector field, so it's got x, y, and z as its inputs, and its output then also has to have three dimensions, so it might be like p, q, and r, and all of these are functions of x and y. So that's p of x and y, q, oh, no, no, x, y, and z, right? So p of x, y, kind of got in the habit of two dimensions there."}, {"video_title": "Divergence notation.mp3", "Sentence": "But another nice thing, this can apply to higher-dimensional functions as well, right? If we have something that's, let's see, something that's a vector-valued function, and it's going to be a three-dimensional vector field, so it's got x, y, and z as its inputs, and its output then also has to have three dimensions, so it might be like p, q, and r, and all of these are functions of x and y. So that's p of x and y, q, oh, no, no, x, y, and z, right? So p of x, y, kind of got in the habit of two dimensions there. p of x, y, and z, q of x, y, and z, and then r of x, y, and z, and I haven't talked about three-dimensional divergence, but if you take this, and then you imagine doing your nabla dotted with the vector-valued function, it can still make sense. And in this case, that nabla you're thinking of as having three different components, right? It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z."}, {"video_title": "Divergence notation.mp3", "Sentence": "So p of x, y, kind of got in the habit of two dimensions there. p of x, y, and z, q of x, y, and z, and then r of x, y, and z, and I haven't talked about three-dimensional divergence, but if you take this, and then you imagine doing your nabla dotted with the vector-valued function, it can still make sense. And in this case, that nabla you're thinking of as having three different components, right? It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z. And the ordering of these, of the variables here, x, y, and z, is just whatever I have here. So even if they didn't have the names x, y, z, you kind of put them in the same order that they show up in your function. And when you imagine taking the dot product between this and your p as a function, q as a function, and r as a function, vector-valued output, what you'd get, and I'll write it over here, is you take that partial partial x and kind of multiply it with p, which means you're really evaluating at p, so partial x here, then you add partial partial y, and you're evaluating at q, because you're kind of imagining multiplying these second components, and then you'll add what happens when you multiply by these third components, where that's partial partial z by that last component."}, {"video_title": "Divergence notation.mp3", "Sentence": "It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z. And the ordering of these, of the variables here, x, y, and z, is just whatever I have here. So even if they didn't have the names x, y, z, you kind of put them in the same order that they show up in your function. And when you imagine taking the dot product between this and your p as a function, q as a function, and r as a function, vector-valued output, what you'd get, and I'll write it over here, is you take that partial partial x and kind of multiply it with p, which means you're really evaluating at p, so partial x here, then you add partial partial y, and you're evaluating at q, because you're kind of imagining multiplying these second components, and then you'll add what happens when you multiply by these third components, where that's partial partial z by that last component. And, you know, since I haven't talked about three-dimensional vector fields or three-dimensional divergence, this last term, maybe it's not a given that you'd have as strong an intuition for why this shows up in divergence as the other two, but it's actually quite similar. You're thinking about changes to the z component of a vector as the value z of the input, as you're kind of moving up and down in that direction changes. But this pattern will go for even higher dimensions that we can't visualize, four, five, 100, whatever you want, and that's what makes this notation here quite nice, is that it encapsulates that and gives a really compact way of describing this formula that has a simple pattern to it, but would otherwise kind of get out of hand to write."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "So for example, I'm gonna give you a function, some kind of function that takes in a 2D vector, x,y, and it's also gonna spit out a 2D vector. And the specific one I have in mind, this is just kind of arbitrary, is x plus sine of y, and then because I'm a sucker for symmetry, I'm gonna make it y plus sine of x. Though of course, this could be any arbitrary function, you don't need that kind of symmetry. So in the last video, I gave a little refresher on how to think about linear transformations and ideas from linear algebra, and how you encode a linear transformation using a matrix and kind of visualize it, I used this grid. And here, I wanna show what this function looks like as a transformation of space. As in, I'm gonna tell the computer, take every single point on this blue grid here, and if that point is x,y, I want you to move it over to the point x plus sine of y, y plus sine of x. And here's what that looks like."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "So in the last video, I gave a little refresher on how to think about linear transformations and ideas from linear algebra, and how you encode a linear transformation using a matrix and kind of visualize it, I used this grid. And here, I wanna show what this function looks like as a transformation of space. As in, I'm gonna tell the computer, take every single point on this blue grid here, and if that point is x,y, I want you to move it over to the point x plus sine of y, y plus sine of x. And here's what that looks like. Alright, so things get really wavy, really curly, this is not at all a linear transformation, right? All of the lines don't remain lines, they're no longer nice grid lines that are parallel and evenly spaced. In some sense, there is much, much more information that goes into nonlinear functions than into linear functions."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "And here's what that looks like. Alright, so things get really wavy, really curly, this is not at all a linear transformation, right? All of the lines don't remain lines, they're no longer nice grid lines that are parallel and evenly spaced. In some sense, there is much, much more information that goes into nonlinear functions than into linear functions. And because this is rather complicated, I think it might be easier to see what's going on if we just focus on a single individual point. So let me look at a point like, let's say, pi halves and zero, okay? So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "In some sense, there is much, much more information that goes into nonlinear functions than into linear functions. And because this is rather complicated, I think it might be easier to see what's going on if we just focus on a single individual point. So let me look at a point like, let's say, pi halves and zero, okay? So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same. And then for the bottom, y, well y is also zero, plus sine of x, sine of pi halves is one, I'll go ahead and write sine of pi halves, sine of pi halves, but you can think of that as just being one. So what that means on the transformation over here is if we look at the point that's at pi halves zero, and pi halves is a little above 1.5, so that's gonna be around here, we expect it to move to the point pi halves one. So it should just move vertically by one unit."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same. And then for the bottom, y, well y is also zero, plus sine of x, sine of pi halves is one, I'll go ahead and write sine of pi halves, sine of pi halves, but you can think of that as just being one. So what that means on the transformation over here is if we look at the point that's at pi halves zero, and pi halves is a little above 1.5, so that's gonna be around here, we expect it to move to the point pi halves one. So it should just move vertically by one unit. And if you just focus on that one point during the transformation, notice that's exactly what happens, it just moves vertically one point. And of course things are quite complicated because every point is doing that, right? The computer's taking in every point and moving it to where it should go."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "So it should just move vertically by one unit. And if you just focus on that one point during the transformation, notice that's exactly what happens, it just moves vertically one point. And of course things are quite complicated because every point is doing that, right? The computer's taking in every point and moving it to where it should go. So after having given the refresher on thinking about linear transformations and encoding them with matrices last time, something like this might feel completely intractable. You certainly have to store much more information than just four numbers to record where everything goes. But this function has a nice property, a property that we deal with all the time in multivariable calculus."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "The computer's taking in every point and moving it to where it should go. So after having given the refresher on thinking about linear transformations and encoding them with matrices last time, something like this might feel completely intractable. You certainly have to store much more information than just four numbers to record where everything goes. But this function has a nice property, a property that we deal with all the time in multivariable calculus. It's what we'd call locally linear. Locally linear. And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "But this function has a nice property, a property that we deal with all the time in multivariable calculus. It's what we'd call locally linear. Locally linear. And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that. And first of all, I'm gonna add some more grid lines. So they're really very close grid lines, right? We can see from the zoomed out picture."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that. And first of all, I'm gonna add some more grid lines. So they're really very close grid lines, right? We can see from the zoomed out picture. But this just makes it so that when we're zoomed in, we can see a little bit more of what's going on. And now when I play the animation, I'm gonna have this yellow box that's doing the zooming to follow the point at its center, right? So this box will be moving and we're always just gonna look at what it zoomed in on."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "We can see from the zoomed out picture. But this just makes it so that when we're zoomed in, we can see a little bit more of what's going on. And now when I play the animation, I'm gonna have this yellow box that's doing the zooming to follow the point at its center, right? So this box will be moving and we're always just gonna look at what it zoomed in on. Okay, so it's gonna be following what's going on around that point during the transformation. And we can see inside this zoomed version, it's still not linear, right? The lines get a little bit curved, but this looks a lot more like a linear function."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "So this box will be moving and we're always just gonna look at what it zoomed in on. Okay, so it's gonna be following what's going on around that point during the transformation. And we can see inside this zoomed version, it's still not linear, right? The lines get a little bit curved, but this looks a lot more like a linear function. It looks a lot more like the grid lines that started off horizontal and vertical are remaining parallel and evenly spaced. And in fact, let's say I zoom in even further to an even smaller yellow box here. And again, I'm gonna add in some more grid lines right around it."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "The lines get a little bit curved, but this looks a lot more like a linear function. It looks a lot more like the grid lines that started off horizontal and vertical are remaining parallel and evenly spaced. And in fact, let's say I zoom in even further to an even smaller yellow box here. And again, I'm gonna add in some more grid lines right around it. So they're very, very densely packed. And this is purely an artifact of visualizing things, right? I could choose to put points or lines or anything wherever I want."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "And again, I'm gonna add in some more grid lines right around it. So they're very, very densely packed. And this is purely an artifact of visualizing things, right? I could choose to put points or lines or anything wherever I want. And I just think showing the grid lines and only the grid lines and where they move gives sort of a feel for what the function is doing. So this time when I play it, and that zooming in box kind of tracks the point that we're looking at, as it goes, the neighborhood around it, all of the points around it really, really do look like a linear function. And the more you zoom in, the more it looks precisely like a certain linear function."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "I could choose to put points or lines or anything wherever I want. And I just think showing the grid lines and only the grid lines and where they move gives sort of a feel for what the function is doing. So this time when I play it, and that zooming in box kind of tracks the point that we're looking at, as it goes, the neighborhood around it, all of the points around it really, really do look like a linear function. And the more you zoom in, the more it looks precisely like a certain linear function. Oh, I guess I should have written an R over here, locally linear. So this raises the question, if we're looking around some specific point, which I'll call X naught and Y naught, this should correspond in some way to the linear transformation that it looks like around it. There should be some kind of matrix, some two by two matrix, that represents the linear transformation that this function, this much more complicated function, looks like around that point."}, {"video_title": "Local linearity for a multivariable function.mp3", "Sentence": "And the more you zoom in, the more it looks precisely like a certain linear function. Oh, I guess I should have written an R over here, locally linear. So this raises the question, if we're looking around some specific point, which I'll call X naught and Y naught, this should correspond in some way to the linear transformation that it looks like around it. There should be some kind of matrix, some two by two matrix, that represents the linear transformation that this function, this much more complicated function, looks like around that point. So this idea of zooming in is what we mean by local. And in the next video, I'm gonna show you what this matrix looks like in terms of partial derivatives for our original function. See you then."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's do that. So this integral right over here is the same thing as the integral from a to b, and then what I'm gonna do is I'm gonna group the things that are being multiplied by dx dt. So if I group them and then factor out a dx dt, so if I take this part right here and this part right over here, I'm essentially going to distribute the r, let me make it clear. I'm gonna distribute the r, and so I'm gonna group that and that right over there, I will be left with, and I'll factor out the dx dt, I will be left with p plus r times the partial of z with respect to x times dx dt. dx dt. And now I'll do the same thing for the dy dts. So that's that part right over there, and then the r is gonna get distributed, and this thing right over here."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm gonna distribute the r, and so I'm gonna group that and that right over there, I will be left with, and I'll factor out the dx dt, I will be left with p plus r times the partial of z with respect to x times dx dt. dx dt. And now I'll do the same thing for the dy dts. So that's that part right over there, and then the r is gonna get distributed, and this thing right over here. So it's gonna be plus q, q plus r times the partial, the r is gonna get distributed, r times the partial of z with respect to y, and I'm gonna factor out a dy dt. dy dt, and then we can't forget all of that, all of that is going to be multiplied, all of that is going to be multiplied by dt. Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's that part right over there, and then the r is gonna get distributed, and this thing right over here. So it's gonna be plus q, q plus r times the partial, the r is gonna get distributed, r times the partial of z with respect to y, and I'm gonna factor out a dy dt. dy dt, and then we can't forget all of that, all of that is going to be multiplied, all of that is going to be multiplied by dt. Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field. In fact, let me kind of copy and paste it. Let me copy and paste it. So, actually I don't know if I'm on the right layer of my work right over here."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field. In fact, let me kind of copy and paste it. Let me copy and paste it. So, actually I don't know if I'm on the right layer of my work right over here. So let me copy and paste. No, that didn't work. Let me try it one more time."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So, actually I don't know if I'm on the right layer of my work right over here. So let me copy and paste. No, that didn't work. Let me try it one more time. So if I try to copy, I'll go a layer down. I'm using an art program for this. Copy, and then I think this will work, and paste."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me try it one more time. So if I try to copy, I'll go a layer down. I'm using an art program for this. Copy, and then I think this will work, and paste. There we go. So this is a result that we had before. This is kind of a template to look at."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Copy, and then I think this will work, and paste. There we go. So this is a result that we had before. This is kind of a template to look at. But what is going on over here if we just look at this template? We see that we're in the t domain where we're integrating over t right over here, but then we have these things. We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is kind of a template to look at. But what is going on over here if we just look at this template? We see that we're in the t domain where we're integrating over t right over here, but then we have these things. We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt. Well, that's exactly what we're doing right over here. We can distribute the dt, and we have something that looks exactly like this right over here, where m is analogous to this piece right over here. Let me make it clear."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt. Well, that's exactly what we're doing right over here. We can distribute the dt, and we have something that looks exactly like this right over here, where m is analogous to this piece right over here. Let me make it clear. m, this piece right over here, looks a lot like m, m in our example right over here. It's being multiplied by dx dt, and then this dt, which you can distribute, and this piece right over here looks a lot like n. And so we can say that, well, we have something that looks like this. We can rewrite it like this and kind of go back into the, kind of deparameterize it."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me make it clear. m, this piece right over here, looks a lot like m, m in our example right over here. It's being multiplied by dx dt, and then this dt, which you can distribute, and this piece right over here looks a lot like n. And so we can say that, well, we have something that looks like this. We can rewrite it like this and kind of go back into the, kind of deparameterize it. So this thing is going to be equal to, it's going to be equal to now the line integral of c1. We are in the xy plane. We started with the curve c, but now we're going to go to c1."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can rewrite it like this and kind of go back into the, kind of deparameterize it. So this thing is going to be equal to, it's going to be equal to now the line integral of c1. We are in the xy plane. We started with the curve c, but now we're going to go to c1. This is c1. It's completely analogous. These are only functions of x's and y's."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We started with the curve c, but now we're going to go to c1. This is c1. It's completely analogous. These are only functions of x's and y's. Everything here is. So now this is going to be the line integral, the line integral over c1, and I can even draw it as like that if I like, of m dx, and that makes sense because if you multiply dt times dx dt, the dt's cancel out and you're just left with dx. So m times dx, so let me write it that way."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These are only functions of x's and y's. Everything here is. So now this is going to be the line integral, the line integral over c1, and I can even draw it as like that if I like, of m dx, and that makes sense because if you multiply dt times dx dt, the dt's cancel out and you're just left with dx. So m times dx, so let me write it that way. So it's going to be p plus r times the partial of z with respect to x, dx plus n, let me scroll to the right a little bit, plus n, which is q plus r times the partial of z with respect to y, dy. And then this is really interesting because this path that we are now concerned with, it's completely analogous. I hope you don't think I'm doing some voodoo here."}, {"video_title": "Stokes' theorem proof part 6   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So m times dx, so let me write it that way. So it's going to be p plus r times the partial of z with respect to x, dx plus n, let me scroll to the right a little bit, plus n, which is q plus r times the partial of z with respect to y, dy. And then this is really interesting because this path that we are now concerned with, it's completely analogous. I hope you don't think I'm doing some voodoo here. This statement is completely analogous to this statement where m could be this and n could be this. And so we can revert it back to now the path c1 that sits in the xy plane, not our original boundary c, but now we're just dealing with things, a boundary in the xy plane. So it reverts to this, but what's powerful about getting it to this point is we can now apply Green's theorem to this."}, {"video_title": "Differential of a vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last couple of videos, we saw that we can describe a curve by a position vector valued function. In very general terms, it would be the x position as a function of time times the unit vector in the horizontal direction plus the y position as a function of time times the unit vector in the vertical direction. And this will essentially describe this, if you can imagine a particle, and let's say the parameter t represents time, it will describe where the particle is at any given time. If we wanted a particular curve, we can say this only applies for some curve, we're dealing with r of t, and it's only applicable between t being greater than a and less than b. That would describe some curve in two dimensions. Let me just draw it here. This is all a review of the last two videos."}, {"video_title": "Differential of a vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we wanted a particular curve, we can say this only applies for some curve, we're dealing with r of t, and it's only applicable between t being greater than a and less than b. That would describe some curve in two dimensions. Let me just draw it here. This is all a review of the last two videos. This curve might look something like that, where this is where t is equal to a, that's where t is equal to b. So r of a will be this vector right here that ends at that point, and then as t, or if you can imagine the parameter being time, it doesn't have to be time, but that's a convenient one to visualize. Each corresponding as t gets larger and larger, we're specifying different points on the path."}, {"video_title": "Differential of a vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is all a review of the last two videos. This curve might look something like that, where this is where t is equal to a, that's where t is equal to b. So r of a will be this vector right here that ends at that point, and then as t, or if you can imagine the parameter being time, it doesn't have to be time, but that's a convenient one to visualize. Each corresponding as t gets larger and larger, we're specifying different points on the path. We saw that two videos ago, and in the last video, we thought about what does it mean to take the derivative of a vector-valued function, and we came up with this idea, and it wasn't an idea, we actually showed it to be true, we came up with a definition really, that the derivative, I could call it r prime of t, and it's going to be a vector. The derivative of a vector-valued function is once again going to be a derivative, but it is equal to the way we defined it, x prime of t times i plus y prime of t times j, or another way to write that, and I'll just write all the different ways just so you get familiar with it, dr dt is equal to dx dt. This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j."}, {"video_title": "Differential of a vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Each corresponding as t gets larger and larger, we're specifying different points on the path. We saw that two videos ago, and in the last video, we thought about what does it mean to take the derivative of a vector-valued function, and we came up with this idea, and it wasn't an idea, we actually showed it to be true, we came up with a definition really, that the derivative, I could call it r prime of t, and it's going to be a vector. The derivative of a vector-valued function is once again going to be a derivative, but it is equal to the way we defined it, x prime of t times i plus y prime of t times j, or another way to write that, and I'll just write all the different ways just so you get familiar with it, dr dt is equal to dx dt. This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j. If we wanted to think about the differential, one thing that we can think about, whenever I do the math of the differential, it's a little bit hand-wavy, I'm not being very rigorous, but if you imagine multiplying both sides of this equation by a very small dt, or this exact dt, you would get dr is equal to, I'll just leave it like this, dx dt times dt, I could make these cancel out, but I'll just write it like this first, times the unit vector i plus dy dt times the unit vector j, or we could rewrite this, and I'm just rewriting it in all of the different ways that one can rewrite it. You could also write this as dr is equal to x prime of t dt times the unit vector i, sorry, this was x prime of t dt, this is x prime of t right there, times the unit vector i plus y prime of t, that's just that right there, times dt times the unit vector j, and just to complete the trifecta, the other way that we could write this is that dr is equal to, if we just allowed these to cancel out, then we get is equal to dx times i plus dy times dy times j. And that actually makes a lot of intuitive sense, that if I look at any dr, so let's say I look at the change between this vector and this vector, let's say the super small change right there, that is our dr, and it's made up of, it's our dx, our change in x, is that right there, you can imagine it's that right there, times, but we're vectorizing it by multiplying it by the unit vector in the horizontal direction, plus dy times the unit vector in the vertical direction."}, {"video_title": "Differential of a vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j. If we wanted to think about the differential, one thing that we can think about, whenever I do the math of the differential, it's a little bit hand-wavy, I'm not being very rigorous, but if you imagine multiplying both sides of this equation by a very small dt, or this exact dt, you would get dr is equal to, I'll just leave it like this, dx dt times dt, I could make these cancel out, but I'll just write it like this first, times the unit vector i plus dy dt times the unit vector j, or we could rewrite this, and I'm just rewriting it in all of the different ways that one can rewrite it. You could also write this as dr is equal to x prime of t dt times the unit vector i, sorry, this was x prime of t dt, this is x prime of t right there, times the unit vector i plus y prime of t, that's just that right there, times dt times the unit vector j, and just to complete the trifecta, the other way that we could write this is that dr is equal to, if we just allowed these to cancel out, then we get is equal to dx times i plus dy times dy times j. And that actually makes a lot of intuitive sense, that if I look at any dr, so let's say I look at the change between this vector and this vector, let's say the super small change right there, that is our dr, and it's made up of, it's our dx, our change in x, is that right there, you can imagine it's that right there, times, but we're vectorizing it by multiplying it by the unit vector in the horizontal direction, plus dy times the unit vector in the vertical direction. So when you multiply this distance times the unit vector, you're essentially getting this vector, you're essentially getting this vector right here, and when you multiply this guy, and actually our change in y here is negative, you're going to get this vector right here. So when you add those together, you get your change in your actual position vector. So that was all a little bit of background, and this might be somewhat useful in a future video from now."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that we've got the curve R defined. So this is our curve R. It's x of t times i plus y of t times j. It's a curve in two dimensions on the xy plane. And let's graph it. Just graph it in kind of a generalized form. So that's our y-axis. This is our x-axis."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's graph it. Just graph it in kind of a generalized form. So that's our y-axis. This is our x-axis. Our curve R might look something like this. It might look something. Let me draw a little bit more of a, maybe it looks something like this."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is our x-axis. Our curve R might look something like this. It might look something. Let me draw a little bit more of a, maybe it looks something like this. Maybe that's just part of it. And as t increases, we're going in that direction right over there. What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw a little bit more of a, maybe it looks something like this. Maybe that's just part of it. And as t increases, we're going in that direction right over there. What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector. Obviously, if you can figure out a normal vector, you can just divide it by its magnitude and you will get the unit normal vector. So I want to figure out at any given point a vector that's popping straight out in that direction and has a magnitude 1. So that would be our unit normal vector."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector. Obviously, if you can figure out a normal vector, you can just divide it by its magnitude and you will get the unit normal vector. So I want to figure out at any given point a vector that's popping straight out in that direction and has a magnitude 1. So that would be our unit normal vector. And to do that, first we'll think about what a tangent vector is. And from a tangent vector, we can figure out the normal vector. And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that would be our unit normal vector. And to do that, first we'll think about what a tangent vector is. And from a tangent vector, we can figure out the normal vector. And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line. We're going to see a very similar thing when we do it right over here with this vector algebra. So the first thing I want to think about is how do we construct a tangent line? Well, you can imagine at some t, this is what our position vector is going to look like."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line. We're going to see a very similar thing when we do it right over here with this vector algebra. So the first thing I want to think about is how do we construct a tangent line? Well, you can imagine at some t, this is what our position vector is going to look like. So call that r1 right over there. And then if we allow t to go up a little bit, if t is time, we wait a little while, a few seconds, however we're measuring things. And then r2 might look something like this."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, you can imagine at some t, this is what our position vector is going to look like. So call that r1 right over there. And then if we allow t to go up a little bit, if t is time, we wait a little while, a few seconds, however we're measuring things. And then r2 might look something like this. This is when t has gotten a little bit larger. We're further down the path. And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then r2 might look something like this. This is when t has gotten a little bit larger. We're further down the path. And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors. The difference between these two vectors is you could view that as delta r. This vector plus that vector is equal to that vector. Or r2 minus r1 is going to give you this delta r right over here. And as that increment between r1 and r2 gets smaller and smaller and smaller, as we have a smaller and smaller t increment, the slope of that delta r is going to more and more approximate the slope of the tangent line, all the way to the point that if you have an infinitely small change in t, so you have a dt, so you go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors. The difference between these two vectors is you could view that as delta r. This vector plus that vector is equal to that vector. Or r2 minus r1 is going to give you this delta r right over here. And as that increment between r1 and r2 gets smaller and smaller and smaller, as we have a smaller and smaller t increment, the slope of that delta r is going to more and more approximate the slope of the tangent line, all the way to the point that if you have an infinitely small change in t, so you have a dt, so you go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector. So if you have a very small change in t, then your very small dr, I'll call it, because now we're talking about a differential, your very small differential right over here, that is a tangent vector. So dr is a tangent vector at any given point. And once again, all of this is a little bit of review."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And as that increment between r1 and r2 gets smaller and smaller and smaller, as we have a smaller and smaller t increment, the slope of that delta r is going to more and more approximate the slope of the tangent line, all the way to the point that if you have an infinitely small change in t, so you have a dt, so you go from r, then you change t a very small amount, that delta r, and we can kind of conceptualize that as dr, then does approximate the a tangent vector. So if you have a very small change in t, then your very small dr, I'll call it, because now we're talking about a differential, your very small differential right over here, that is a tangent vector. So dr is a tangent vector at any given point. And once again, all of this is a little bit of review. But dr we can write as dr is equal to dx times i plus the infinitesimally small change in x times the i unit vector plus the infinitesimally small change in y times the j unit vector. And you see that if I were to draw a curve, let me just draw another one. Actually, I don't even have to draw the axes."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And once again, all of this is a little bit of review. But dr we can write as dr is equal to dx times i plus the infinitesimally small change in x times the i unit vector plus the infinitesimally small change in y times the j unit vector. And you see that if I were to draw a curve, let me just draw another one. Actually, I don't even have to draw the axes. If our dr looks like that, if that is our dr, then we can break that down into its vertical and horizontal components. This right over here is dy, and that right over there is dx. And so we see that dx times i, actually this is dx times i, and this is dy times j. dy is the magnitude, j gives us the direction."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually, I don't even have to draw the axes. If our dr looks like that, if that is our dr, then we can break that down into its vertical and horizontal components. This right over here is dy, and that right over there is dx. And so we see that dx times i, actually this is dx times i, and this is dy times j. dy is the magnitude, j gives us the direction. dx is the magnitude, i tells us that we're moving in the horizontal direction. Over here, this actually would be a negative. This must be a negative value right over here, and this must be a positive value based on the way that I drew it."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we see that dx times i, actually this is dx times i, and this is dy times j. dy is the magnitude, j gives us the direction. dx is the magnitude, i tells us that we're moving in the horizontal direction. Over here, this actually would be a negative. This must be a negative value right over here, and this must be a positive value based on the way that I drew it. So that gives us a tangent vector. And now we want to, from that tangent vector, figure out a normal vector, a vector that is essentially perpendicular to this vector right over here. And there's actually going to be two vectors like that."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This must be a negative value right over here, and this must be a positive value based on the way that I drew it. So that gives us a tangent vector. And now we want to, from that tangent vector, figure out a normal vector, a vector that is essentially perpendicular to this vector right over here. And there's actually going to be two vectors like that. There's going to be the vector that kind of is perpendicular in the right direction, because we care about direction, or the vector that's perpendicular in the left direction. We can pick either one, but for this video, I'm going to focus on the one that goes in the right direction. We're going to see that that's going to be useful in the next video when we start doing a little bit of vector calculus."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And there's actually going to be two vectors like that. There's going to be the vector that kind of is perpendicular in the right direction, because we care about direction, or the vector that's perpendicular in the left direction. We can pick either one, but for this video, I'm going to focus on the one that goes in the right direction. We're going to see that that's going to be useful in the next video when we start doing a little bit of vector calculus. And so let's think about what that might be. And what I'll do to make it a little bit clearer, let me draw a dr again. I'll draw a dr like this."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to see that that's going to be useful in the next video when we start doing a little bit of vector calculus. And so let's think about what that might be. And what I'll do to make it a little bit clearer, let me draw a dr again. I'll draw a dr like this. This is our dr. And then this right over here, this right over there, we already said this is dy times i. And then this, sorry, that's dy times j. We're going in the vertical direction."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll draw a dr like this. This is our dr. And then this right over here, this right over there, we already said this is dy times i. And then this, sorry, that's dy times j. We're going in the vertical direction. dy times j. And then in a different color, this right, I already used that color. Let's see, I haven't used, well, I haven't used that blue yet."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going in the vertical direction. dy times j. And then in a different color, this right, I already used that color. Let's see, I haven't used, well, I haven't used that blue yet. So this right over here is dx times i. So we know from our algebra courses, you take the negative reciprocal. So there's going to be something about swapping these two things around and then taking the negative one."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see, I haven't used, well, I haven't used that blue yet. So this right over here is dx times i. So we know from our algebra courses, you take the negative reciprocal. So there's going to be something about swapping these two things around and then taking the negative one. But we have to figure out, well, we want the one that goes to the right. So which one should we use? So let's think about it a little bit."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So there's going to be something about swapping these two things around and then taking the negative one. But we have to figure out, well, we want the one that goes to the right. So which one should we use? So let's think about it a little bit. If we take dy times i, so we take this length, but in the i direction, we're going to get this. We're going to get that. So this is dy times i."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's think about it a little bit. If we take dy times i, so we take this length, but in the i direction, we're going to get this. We're going to get that. So this is dy times i. And then if we were to just take dx times j, that would take us down. Because dx, it must be negative here, since it's pointed to the left. So we have to swap the sine of dx to go upwards."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is dy times i. And then if we were to just take dx times j, that would take us down. Because dx, it must be negative here, since it's pointed to the left. So we have to swap the sine of dx to go upwards. So we swap the sine of dx to go upwards. Because obviously here, it was a negative sine. It went leftwards."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have to swap the sine of dx to go upwards. So we swap the sine of dx to go upwards. Because obviously here, it was a negative sine. It went leftwards. But we wanted to go upwards. So this is going to be negative dx times j. We're now moving in the vertical direction."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It went leftwards. But we wanted to go upwards. So this is going to be negative dx times j. We're now moving in the vertical direction. And that, at least visually, this isn't kind of a rigorous proof that I'm giving you. But this is hopefully a good visual representation that that does get you pretty close, just visually inspecting it, to what looks like the perpendicular line. It's consistent with what you learned in algebra class as well, that we're taking the negative reciprocal."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're now moving in the vertical direction. And that, at least visually, this isn't kind of a rigorous proof that I'm giving you. But this is hopefully a good visual representation that that does get you pretty close, just visually inspecting it, to what looks like the perpendicular line. It's consistent with what you learned in algebra class as well, that we're taking the negative reciprocal. We're swapping the x's and the y's, or the change in x and the change in y. And we're taking the negative of one of them. And so we have our normal line just like that, our normal vector."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's consistent with what you learned in algebra class as well, that we're taking the negative reciprocal. We're swapping the x's and the y's, or the change in x and the change in y. And we're taking the negative of one of them. And so we have our normal line just like that, our normal vector. So a normal vector is going to be dyi minus dxj. But then if we want to normalize it, we want to divide by that magnitude. So a normal vector is going to be dy times i minus dx times j."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we have our normal line just like that, our normal vector. So a normal vector is going to be dyi minus dxj. But then if we want to normalize it, we want to divide by that magnitude. So a normal vector is going to be dy times i minus dx times j. Now, if we want this to be a unit normal vector, we have to divide it by the magnitude of a. But what is the magnitude of a? The magnitude of a is going to be equal to the square root of, I'll start with the dx squared."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So a normal vector is going to be dy times i minus dx times j. Now, if we want this to be a unit normal vector, we have to divide it by the magnitude of a. But what is the magnitude of a? The magnitude of a is going to be equal to the square root of, I'll start with the dx squared. So it's the negative dx squared, which is just going to be dx squared. Same thing as positive dx squared. It's going to be dx squared plus dy squared."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The magnitude of a is going to be equal to the square root of, I'll start with the dx squared. So it's the negative dx squared, which is just going to be dx squared. Same thing as positive dx squared. It's going to be dx squared plus dy squared. I could have put the negative right in here. But then when you square it, that negative would disappear. But this thing right over here, and we saw this when we first started exploring arc length, this thing right over here is the exact same thing as ds."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be dx squared plus dy squared. I could have put the negative right in here. But then when you square it, that negative would disappear. But this thing right over here, and we saw this when we first started exploring arc length, this thing right over here is the exact same thing as ds. And I know there's no ds that we've shown right over here. But we've seen it multiple times. That when you're thinking about, if you think about the length of dr as ds, that's exactly what this thing right over here is."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this thing right over here, and we saw this when we first started exploring arc length, this thing right over here is the exact same thing as ds. And I know there's no ds that we've shown right over here. But we've seen it multiple times. That when you're thinking about, if you think about the length of dr as ds, that's exactly what this thing right over here is. So this can also be written as ds. The infinitesimally change in the arc length, but it's a scalar quantity. You're not concerned."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That when you're thinking about, if you think about the length of dr as ds, that's exactly what this thing right over here is. So this can also be written as ds. The infinitesimally change in the arc length, but it's a scalar quantity. You're not concerned. You're just concerned with the absolute distance. You're not concerned so much with the direction. Another way to view it is it's the magnitude right over here of dr."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're not concerned. You're just concerned with the absolute distance. You're not concerned so much with the direction. Another way to view it is it's the magnitude right over here of dr. So now we have everything we need to construct our unit normal vector, a unit normal vector at any point. And I'll now write n. And I'll put a hat on top of it to say that this is a unit normal vector. We'll have magnitude 1."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Another way to view it is it's the magnitude right over here of dr. So now we have everything we need to construct our unit normal vector, a unit normal vector at any point. And I'll now write n. And I'll put a hat on top of it to say that this is a unit normal vector. We'll have magnitude 1. It is going to be equal to a divided by this, or we could even write it this way. So we could write it as, there are multiple ways we can write it. We can write it as, I'll write it in this color, as dy times i minus dx times j."}, {"video_title": "Constructing a unit normal vector to a curve   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll have magnitude 1. It is going to be equal to a divided by this, or we could even write it this way. So we could write it as, there are multiple ways we can write it. We can write it as, I'll write it in this color, as dy times i minus dx times j. And then all of that times, or let me, not times, divided by ds, divided by the magnitude of this. So divided by ds. And obviously I could distribute it on each of these terms."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "It's got a two-dimensional input, two different coordinates to its input, and then a three-dimensional output. Specifically, it's a three-dimensional vector, and each one of these is some expression, it's a bunch of cosines and sines, that depends on the two input coordinates. And in the last video, we talked about how to visualize functions that have a single input, a single parameter, like t, and then a two-dimensional vector output, so some kind of expression of t and another expression of t. And this is sort of the three-dimensional analog of that. So what we're gonna do, we're just gonna visualize things in the output space, and we're gonna try to think of all the possible points that could be outputs. So, for example, let's just start off simple. Let's get a feel for this function by evaluating it at a simple pair of points. So let's say we evaluate this function, f, at t equals zero."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what we're gonna do, we're just gonna visualize things in the output space, and we're gonna try to think of all the possible points that could be outputs. So, for example, let's just start off simple. Let's get a feel for this function by evaluating it at a simple pair of points. So let's say we evaluate this function, f, at t equals zero. I think it would probably be pretty simple. And then s is equal to pi. So let's think about what this would be."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So let's say we evaluate this function, f, at t equals zero. I think it would probably be pretty simple. And then s is equal to pi. So let's think about what this would be. We go up and we say, okay, t of zero, cosine of zero is one, so this whole thing is gonna be one, same with this one. And sine of zero is zero, so this over here is gonna be zero, and this is also gonna be zero. Now cosine of pi is negative one, so this here is gonna be negative one, this one here is also gonna be negative one, and then sine of pi, just like sine of zero, is zero."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So let's think about what this would be. We go up and we say, okay, t of zero, cosine of zero is one, so this whole thing is gonna be one, same with this one. And sine of zero is zero, so this over here is gonna be zero, and this is also gonna be zero. Now cosine of pi is negative one, so this here is gonna be negative one, this one here is also gonna be negative one, and then sine of pi, just like sine of zero, is zero. So this whole thing actually ends up simplifying quite a bit so that the top is three times one plus negative one, one times negative one is negative one, and we get two, then we have three times zero plus zero, so the y component is just zero, and then the z component is also zero. So what that would mean is that this output is gonna be the point that's two along the x-axis, and there's nothing else to it, it's just two along the x-axis. So we'll go ahead and, whoop, remove the graph about, add that point there."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "Now cosine of pi is negative one, so this here is gonna be negative one, this one here is also gonna be negative one, and then sine of pi, just like sine of zero, is zero. So this whole thing actually ends up simplifying quite a bit so that the top is three times one plus negative one, one times negative one is negative one, and we get two, then we have three times zero plus zero, so the y component is just zero, and then the z component is also zero. So what that would mean is that this output is gonna be the point that's two along the x-axis, and there's nothing else to it, it's just two along the x-axis. So we'll go ahead and, whoop, remove the graph about, add that point there. So that's what would correspond to this one particular input, zero and pi. And you know, you could do this with a whole bunch, and you might add a couple other points based on other inputs that you find, but this will take forever to start to get a feel of the function as a whole. And another thing you can do is say, okay, maybe rather than thinking of evaluating at a particular point, imagine one of the inputs was constant."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So we'll go ahead and, whoop, remove the graph about, add that point there. So that's what would correspond to this one particular input, zero and pi. And you know, you could do this with a whole bunch, and you might add a couple other points based on other inputs that you find, but this will take forever to start to get a feel of the function as a whole. And another thing you can do is say, okay, maybe rather than thinking of evaluating at a particular point, imagine one of the inputs was constant. So let's imagine that s stayed constant at pi, okay? But then we let t range freely, so that means we're gonna have some kind of different output here, and we're gonna let t just be some kind of variable while the output is pi. So what that means is we keep all of these, this negative one, negative one, and zero for what sine of pi is, but the output now, it's gonna be three cosine of t, cosine of t, plus negative one times the cosine of t, so it's gonna be minus cosine of t. The next part, it's gonna still be three sine of t. This is no longer zero."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "And another thing you can do is say, okay, maybe rather than thinking of evaluating at a particular point, imagine one of the inputs was constant. So let's imagine that s stayed constant at pi, okay? But then we let t range freely, so that means we're gonna have some kind of different output here, and we're gonna let t just be some kind of variable while the output is pi. So what that means is we keep all of these, this negative one, negative one, and zero for what sine of pi is, but the output now, it's gonna be three cosine of t, cosine of t, plus negative one times the cosine of t, so it's gonna be minus cosine of t. The next part, it's gonna still be three sine of t. This is no longer zero. I should probably erase those guys, actually, so. We're no longer evaluating when t was zero. Okay, so three times sine of t, that's just still the function that we're dealing with."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what that means is we keep all of these, this negative one, negative one, and zero for what sine of pi is, but the output now, it's gonna be three cosine of t, cosine of t, plus negative one times the cosine of t, so it's gonna be minus cosine of t. The next part, it's gonna still be three sine of t. This is no longer zero. I should probably erase those guys, actually, so. We're no longer evaluating when t was zero. Okay, so three times sine of t, that's just still the function that we're dealing with. Three sine of t, and then minus one times sine of t, so minus sine of t. Keep drawing it in green just to be consistent. And then the bottom stays at zero. And this whole thing actually simplifies."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "Okay, so three times sine of t, that's just still the function that we're dealing with. Three sine of t, and then minus one times sine of t, so minus sine of t. Keep drawing it in green just to be consistent. And then the bottom stays at zero. And this whole thing actually simplifies. Three cosine t minus cosine t, that's just two cosine t. And then same deal for the other one. It's gonna be two sine of t. So this whole thing actually simplifies down to this. So this is, again, when we're letting s stay constant and t ranges freely."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "And this whole thing actually simplifies. Three cosine t minus cosine t, that's just two cosine t. And then same deal for the other one. It's gonna be two sine of t. So this whole thing actually simplifies down to this. So this is, again, when we're letting s stay constant and t ranges freely. And when you do that, what you're gonna end up getting is a circle that you draw. And you can maybe see why it's a circle, because you have this cosine sine pattern. It's a circle with radius two, and it should make sense that it runs through that first point that we evaluated."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So this is, again, when we're letting s stay constant and t ranges freely. And when you do that, what you're gonna end up getting is a circle that you draw. And you can maybe see why it's a circle, because you have this cosine sine pattern. It's a circle with radius two, and it should make sense that it runs through that first point that we evaluated. So that's what happens if you let just one of the variables run. But now let's do the same thing, but think instead of what happens is s varies and t stays constant. I encourage you to work it out for yourself."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "It's a circle with radius two, and it should make sense that it runs through that first point that we evaluated. So that's what happens if you let just one of the variables run. But now let's do the same thing, but think instead of what happens is s varies and t stays constant. I encourage you to work it out for yourself. I'll go ahead and just kind of draw it, because I kind of want to give the intuition here. So in that case, you're gonna get a circle that looks like this. So again, I encourage you to try to think through for the same reasons."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "I encourage you to work it out for yourself. I'll go ahead and just kind of draw it, because I kind of want to give the intuition here. So in that case, you're gonna get a circle that looks like this. So again, I encourage you to try to think through for the same reasons. Imagine that you let s run freely, keep t constant at zero. Why is it that you would get a circle that looks like this? And in fact, if you let both t and s run freely, a very nice way to visualize that is to imagine that this circle, which represents s running freely, sweeps throughout space as you start to let t run freely."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "So again, I encourage you to try to think through for the same reasons. Imagine that you let s run freely, keep t constant at zero. Why is it that you would get a circle that looks like this? And in fact, if you let both t and s run freely, a very nice way to visualize that is to imagine that this circle, which represents s running freely, sweeps throughout space as you start to let t run freely. And what you're gonna end up getting when you do that is a shape that goes like this. And this is a donut. We have a fancy word for this in mathematics."}, {"video_title": "Parametric surfaces   Multivariable calculus   Khan Academy.mp3", "Sentence": "And in fact, if you let both t and s run freely, a very nice way to visualize that is to imagine that this circle, which represents s running freely, sweeps throughout space as you start to let t run freely. And what you're gonna end up getting when you do that is a shape that goes like this. And this is a donut. We have a fancy word for this in mathematics. We call it a torus. But it turns out the function here is a fancy way of drawing the torus. And in another video, I'm gonna go through in more detail, if you were just given the torus, how you could find this function, how you can kind of get the intuitive feel for that."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's continue with our proof of Stokes' Theorem. And this time we're gonna focus on the other side of Stokes' Theorem. We're gonna try to figure out what is the line integral over the boundary C, where this is C right over here, the boundary of our surface, of F dot dr. What we're gonna see is that we're gonna get the exact same result as we have right up here. But before we do that, I'm gonna take a little bit of a detour to kind of build up to this. So let's just take this and put it to the side for now. Actually, let me actually just delete it right now. And what I'm actually gonna do is I'm gonna focus on this region down here, this region in the xy-plane."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But before we do that, I'm gonna take a little bit of a detour to kind of build up to this. So let's just take this and put it to the side for now. Actually, let me actually just delete it right now. And what I'm actually gonna do is I'm gonna focus on this region down here, this region in the xy-plane. And in particular, so this is path C, which is the boundary of our surface. I'm gonna focus on the path that is the boundary of this region, this path that sits in the xy-plane. And I will call that path C1."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what I'm actually gonna do is I'm gonna focus on this region down here, this region in the xy-plane. And in particular, so this is path C, which is the boundary of our surface. I'm gonna focus on the path that is the boundary of this region, this path that sits in the xy-plane. And I will call that path C1. And so one, we can think about a parameterization of just that path in the xy-plane. We could say that C1 could be parameterized as x is equal to x of t, and y is also a function of t. And t is obviously our parameter, and it can go between a and b. So maybe when t is equal to a, it sits right here, and then as t gets larger and larger and larger, it goes all the way around, and eventually, when t is equal to b, it gets to that exact same point."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I will call that path C1. And so one, we can think about a parameterization of just that path in the xy-plane. We could say that C1 could be parameterized as x is equal to x of t, and y is also a function of t. And t is obviously our parameter, and it can go between a and b. So maybe when t is equal to a, it sits right here, and then as t gets larger and larger and larger, it goes all the way around, and eventually, when t is equal to b, it gets to that exact same point. So that's our parameterization right there. And now, just to make the rest of this proof a little bit more understandable, I'm going to give you a little bit of a review of something. Imagine that we have some vector field g, and g, at minimum, is defined on the xy-plane."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So maybe when t is equal to a, it sits right here, and then as t gets larger and larger and larger, it goes all the way around, and eventually, when t is equal to b, it gets to that exact same point. So that's our parameterization right there. And now, just to make the rest of this proof a little bit more understandable, I'm going to give you a little bit of a review of something. Imagine that we have some vector field g, and g, at minimum, is defined on the xy-plane. So, and it could be defined other places, but let's say that g is equal to m of xy i plus n of xy, n of xy j. Now, what would the line integral, this is all a review, we've seen this a long time ago, what would be the line integral over the path c1, not c, but this path that sits in the xy-plane? What would be the line integral over the path c1, and I'll even, I like to write that sometimes, of, and I'm using g, so I don't get confused with f, our original vector field, of g, of g, our vector field here, along that path, g dot dr. G dot dr. Well, dr, dr is just going to be equal to dx, dxi, dxi plus dy, dyj."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Imagine that we have some vector field g, and g, at minimum, is defined on the xy-plane. So, and it could be defined other places, but let's say that g is equal to m of xy i plus n of xy, n of xy j. Now, what would the line integral, this is all a review, we've seen this a long time ago, what would be the line integral over the path c1, not c, but this path that sits in the xy-plane? What would be the line integral over the path c1, and I'll even, I like to write that sometimes, of, and I'm using g, so I don't get confused with f, our original vector field, of g, of g, our vector field here, along that path, g dot dr. G dot dr. Well, dr, dr is just going to be equal to dx, dxi, dxi plus dy, dyj. So, if you take the dot product of these two things right over here, you're going to get, you're going to get the line integral over our path c1, remember, c1 is this path down here. Let me do it in that same color, so you don't think I'm changing colors on you. The line integral over our path c1, but when you take this dot product, you have the, you multiply the x components and then add that to the product of the y components."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What would be the line integral over the path c1, and I'll even, I like to write that sometimes, of, and I'm using g, so I don't get confused with f, our original vector field, of g, of g, our vector field here, along that path, g dot dr. G dot dr. Well, dr, dr is just going to be equal to dx, dxi, dxi plus dy, dyj. So, if you take the dot product of these two things right over here, you're going to get, you're going to get the line integral over our path c1, remember, c1 is this path down here. Let me do it in that same color, so you don't think I'm changing colors on you. The line integral over our path c1, but when you take this dot product, you have the, you multiply the x components and then add that to the product of the y components. So you have m times dx, you have m times dx, m times dx, plus, plus n times dy, n times dy. I just took the dot product of g and dr, n times dy. And when you evaluate these things, the one way to think about it is that dx is the same thing, dx is the same thing as, let me write it up here in a different color, dx is the same thing as dx, the derivative of x with respect to t, dt, and same logic for y. dy is equal to the derivative of y with respect to t, dt."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The line integral over our path c1, but when you take this dot product, you have the, you multiply the x components and then add that to the product of the y components. So you have m times dx, you have m times dx, m times dx, plus, plus n times dy, n times dy. I just took the dot product of g and dr, n times dy. And when you evaluate these things, the one way to think about it is that dx is the same thing, dx is the same thing as, let me write it up here in a different color, dx is the same thing as dx, the derivative of x with respect to t, dt, and same logic for y. dy is equal to the derivative of y with respect to t, dt. One way to think about it, these dt's cancel out and you are just left with dx. And this is an important thing to think about because then this allows us to take this line integral into the domain of our parameter. So then this will be equal to, this will be equal to, this will be equal to the integral in the domain of our parameter."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And when you evaluate these things, the one way to think about it is that dx is the same thing, dx is the same thing as, let me write it up here in a different color, dx is the same thing as dx, the derivative of x with respect to t, dt, and same logic for y. dy is equal to the derivative of y with respect to t, dt. One way to think about it, these dt's cancel out and you are just left with dx. And this is an important thing to think about because then this allows us to take this line integral into the domain of our parameter. So then this will be equal to, this will be equal to, this will be equal to the integral in the domain of our parameter. So now we are in the t domain, and t is going to vary between a and b. We are in the t domain between a and b. This is going to be equal to m, m times, instead of writing dx, I'm gonna write dx, dt, dt."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So then this will be equal to, this will be equal to, this will be equal to the integral in the domain of our parameter. So now we are in the t domain, and t is going to vary between a and b. We are in the t domain between a and b. This is going to be equal to m, m times, instead of writing dx, I'm gonna write dx, dt, dt. So it's going to be dx, let me write it this way, dx, the derivative of x with respect to t, dt, that's the first expression, plus, plus n, and then the exact same thing, times dy, dt, n times dy, dt, n times dy, dt, dt, dt. These are all equivalent statements. Now, with all of that out of the way, and this is really, all of this is really just a reminder so that the rest of this proof becomes a little bit intuitive."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to be equal to m, m times, instead of writing dx, I'm gonna write dx, dt, dt. So it's going to be dx, let me write it this way, dx, the derivative of x with respect to t, dt, that's the first expression, plus, plus n, and then the exact same thing, times dy, dt, n times dy, dt, n times dy, dt, dt, dt. These are all equivalent statements. Now, with all of that out of the way, and this is really, all of this is really just a reminder so that the rest of this proof becomes a little bit intuitive. With that out of the way, let's come up with a parameterization for this path up here, for c. For c. Remember, we just did c1 down in the xy plane, now we're gonna do c that sits up here that kind of rises above the xy plane. Well, for c, x can still be, the parameterizations for x and y can still be the exact same thing because the x and y values are going to be the exact same thing. The x value, the x and y value there is the exact same thing as the x and y value there."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, with all of that out of the way, and this is really, all of this is really just a reminder so that the rest of this proof becomes a little bit intuitive. With that out of the way, let's come up with a parameterization for this path up here, for c. For c. Remember, we just did c1 down in the xy plane, now we're gonna do c that sits up here that kind of rises above the xy plane. Well, for c, x can still be, the parameterizations for x and y can still be the exact same thing because the x and y values are going to be the exact same thing. The x value, the x and y value there is the exact same thing as the x and y value there. The only difference is we now have a z component and we know, we defined it way up here. Our z component is going to be a function, is going to be a function of x and y. It tells us how high to go."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The x value, the x and y value there is the exact same thing as the x and y value there. The only difference is we now have a z component and we know, we defined it way up here. Our z component is going to be a function, is going to be a function of x and y. It tells us how high to go. So we can parameterize, we can parameterize c as, we can parameterize c as, maybe I'll write it as a vector. So let me write it, let me parameterize, so I'll write c as a vector. C, actually no, I'll write it, let me write it this way."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It tells us how high to go. So we can parameterize, we can parameterize c as, we can parameterize c as, maybe I'll write it as a vector. So let me write it, let me parameterize, so I'll write c as a vector. C, actually no, I'll write it, let me write it this way. Let me write c, c can be written as, let me do that purple color. C we can say is, x is x of t, and actually let me write this as a vector. So I will write it, and I'll use a vector r, not to be confused with this r right over here."}, {"video_title": "Stokes' theorem proof part 4   Multivariable Calculus   Khan Academy.mp3", "Sentence": "C, actually no, I'll write it, let me write it this way. Let me write c, c can be written as, let me do that purple color. C we can say is, x is x of t, and actually let me write this as a vector. So I will write it, and I'll use a vector r, not to be confused with this r right over here. So these are two different r's, but I'll just use r's, because that tends to be the convention. So in order to parameterize c, it's going to be the position vector r, which is going to be a function of t, and x is still just going to be x of t. X of t, i, plus y of t, j, and now we're going to have a z component, and z is going to be a function of x and y, which are in turn functions of t. So z is a function of x, which is a function of t, and a function of y, which is a function of t, k, that tells us how high above to essentially get each of those points, and then once again we know that t is between a and b. T is greater than or equal to a, and less than or equal to b. So we have that parameterization right there, and now we can start to think about, we can start to think about the line integral of f dot dr along this path."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Hey guys, there's one more thing I need to talk about before I can describe the vectorized form for the quadratic approximation of multivariable functions, which is a mouthful to say. So let's say you have some kind of expression that looks like a times x squared, and I'm thinking of x as a variable, times b times xy, y as another variable, plus c times y squared. And I'm thinking of a, b, and c as being constants and x and y as being variables. Now this kind of expression has a fancy name. It's called a quadratic form, quadratic form. And that always threw me off. I always kind of was like, what does form mean?"}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Now this kind of expression has a fancy name. It's called a quadratic form, quadratic form. And that always threw me off. I always kind of was like, what does form mean? You know, I know what a quadratic expression is, and quadratic typically means something is squared or you have two variables, but why do they call it a form? And basically it just means that the only things in here are quadratic. You know, it's not the case that you have like an x term sitting on its own, or like a constant out here, like two, and you're adding all of those together."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "I always kind of was like, what does form mean? You know, I know what a quadratic expression is, and quadratic typically means something is squared or you have two variables, but why do they call it a form? And basically it just means that the only things in here are quadratic. You know, it's not the case that you have like an x term sitting on its own, or like a constant out here, like two, and you're adding all of those together. Instead it's just you have purely quadratic terms. But of course mathematicians don't want to call it just a purely quadratic expression. Instead they have to give a fancy name to things so that it seems more intimidating than it needs to be."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "You know, it's not the case that you have like an x term sitting on its own, or like a constant out here, like two, and you're adding all of those together. Instead it's just you have purely quadratic terms. But of course mathematicians don't want to call it just a purely quadratic expression. Instead they have to give a fancy name to things so that it seems more intimidating than it needs to be. But anyway, so we have a quadratic form, and the question is, how can we express this in a vectorized sense? And for analogy, let's think about linear terms, where let's say you have a times x, plus b times y, and I'll throw another variable in there, another constant times another variable z. If you see something like this, where every variable is just being multiplied by a constant, and then you add terms like that to each other, we can express this nicely with vectors, where you pile all of the constants into their own vector, a vector containing a, b, and c, and you imagine the dot product between that and a vector that contains all of the variable components, x, y, and z."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Instead they have to give a fancy name to things so that it seems more intimidating than it needs to be. But anyway, so we have a quadratic form, and the question is, how can we express this in a vectorized sense? And for analogy, let's think about linear terms, where let's say you have a times x, plus b times y, and I'll throw another variable in there, another constant times another variable z. If you see something like this, where every variable is just being multiplied by a constant, and then you add terms like that to each other, we can express this nicely with vectors, where you pile all of the constants into their own vector, a vector containing a, b, and c, and you imagine the dot product between that and a vector that contains all of the variable components, x, y, and z. And the convenience here is then you can have just a symbol, like a v let's say, which represents this whole constant vector. And then you can write down, take the dot product between that, and then have another symbol, maybe a bold-faced x, which represents a vector that contains all of the variables. And this way, your notation just kinda looks like a constant times a variable, just like in the single variable world, when you have a constant number times a variable number, it's kinda like taking a constant vector times a variable vector."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "If you see something like this, where every variable is just being multiplied by a constant, and then you add terms like that to each other, we can express this nicely with vectors, where you pile all of the constants into their own vector, a vector containing a, b, and c, and you imagine the dot product between that and a vector that contains all of the variable components, x, y, and z. And the convenience here is then you can have just a symbol, like a v let's say, which represents this whole constant vector. And then you can write down, take the dot product between that, and then have another symbol, maybe a bold-faced x, which represents a vector that contains all of the variables. And this way, your notation just kinda looks like a constant times a variable, just like in the single variable world, when you have a constant number times a variable number, it's kinda like taking a constant vector times a variable vector. And the importance of writing things down like this is that v could be a vector that contains not just three numbers, but like 100 numbers, and then x would have 100 corresponding variables, and the notation doesn't become any more complicated. It's generalizable to higher dimensions. So the question is, can we do something similar like that with our quadratic form?"}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And this way, your notation just kinda looks like a constant times a variable, just like in the single variable world, when you have a constant number times a variable number, it's kinda like taking a constant vector times a variable vector. And the importance of writing things down like this is that v could be a vector that contains not just three numbers, but like 100 numbers, and then x would have 100 corresponding variables, and the notation doesn't become any more complicated. It's generalizable to higher dimensions. So the question is, can we do something similar like that with our quadratic form? Because you can imagine, let's say we started introducing the variable z, then you would have to have some other term, you know, some other constant times the xz quadratic term, and then some other constant times the z squared quadratic term, and another one for the yz quadratic term. And it would get out of hand, and as soon as you start introducing things like 100 variables, it would get seriously out of hand, because there's a lot of different quadratic terms. So we want a nice way to express this."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So the question is, can we do something similar like that with our quadratic form? Because you can imagine, let's say we started introducing the variable z, then you would have to have some other term, you know, some other constant times the xz quadratic term, and then some other constant times the z squared quadratic term, and another one for the yz quadratic term. And it would get out of hand, and as soon as you start introducing things like 100 variables, it would get seriously out of hand, because there's a lot of different quadratic terms. So we want a nice way to express this. And I'm just gonna kind of show you how we do it, and then we'll work it through to see why it makes sense. So usually, instead of thinking of b times xy, we actually think of this as two times some constant times xy, and this of course doesn't make a difference. You would just change what b represents, but you'll see why it's more convenient to write it this way in just a moment."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So we want a nice way to express this. And I'm just gonna kind of show you how we do it, and then we'll work it through to see why it makes sense. So usually, instead of thinking of b times xy, we actually think of this as two times some constant times xy, and this of course doesn't make a difference. You would just change what b represents, but you'll see why it's more convenient to write it this way in just a moment. So the vectorized way to describe a quadratic form like this is to take a matrix, a two by two matrix, since this is two dimensions, where a and c are on the diagonal, and then b is on the other diagonal. And we always think of these as being symmetric matrices. So if you imagine kind of reflecting the whole matrix about this line, you'll get the same numbers."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "You would just change what b represents, but you'll see why it's more convenient to write it this way in just a moment. So the vectorized way to describe a quadratic form like this is to take a matrix, a two by two matrix, since this is two dimensions, where a and c are on the diagonal, and then b is on the other diagonal. And we always think of these as being symmetric matrices. So if you imagine kind of reflecting the whole matrix about this line, you'll get the same numbers. So it's important that we have that kind of symmetry. And now what you do is you multiply the vector, the variable vector that's got xy on the right side of this matrix, and then you multiply it again, but you kind of, you turn it on its side. So instead of being a vertical vector, you transpose it to being a horizontal vector on the other side."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So if you imagine kind of reflecting the whole matrix about this line, you'll get the same numbers. So it's important that we have that kind of symmetry. And now what you do is you multiply the vector, the variable vector that's got xy on the right side of this matrix, and then you multiply it again, but you kind of, you turn it on its side. So instead of being a vertical vector, you transpose it to being a horizontal vector on the other side. And this is a little bit analogous to having two variables multiplied in. You have two vectors multiplied in, but on either side. And this is a good point, by the way, if you are uncomfortable with matrix multiplication, to maybe pause the video, go find the videos about matrix multiplication and kind of refresh or learn about that."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So instead of being a vertical vector, you transpose it to being a horizontal vector on the other side. And this is a little bit analogous to having two variables multiplied in. You have two vectors multiplied in, but on either side. And this is a good point, by the way, if you are uncomfortable with matrix multiplication, to maybe pause the video, go find the videos about matrix multiplication and kind of refresh or learn about that. Because moving forward, I'm just gonna assume that it's something you're familiar with. So going about computing this, first let's tackle this right multiplication here. We have a matrix multiplied by a vector."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And this is a good point, by the way, if you are uncomfortable with matrix multiplication, to maybe pause the video, go find the videos about matrix multiplication and kind of refresh or learn about that. Because moving forward, I'm just gonna assume that it's something you're familiar with. So going about computing this, first let's tackle this right multiplication here. We have a matrix multiplied by a vector. Well, the first component that we get, we're gonna multiply the top row by each corresponding term in the vector. So it'll be a times x, a times x, plus b times y, plus b times that second term y. And then similarly for the bottom term, we'll take the bottom row and multiply the corresponding term."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "We have a matrix multiplied by a vector. Well, the first component that we get, we're gonna multiply the top row by each corresponding term in the vector. So it'll be a times x, a times x, plus b times y, plus b times that second term y. And then similarly for the bottom term, we'll take the bottom row and multiply the corresponding term. So b times x, b times x, plus c times y, c times y. So that's what it looks like when we do that right multiplication. And of course, we've gotta keep our transposed vector over there on the left side."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And then similarly for the bottom term, we'll take the bottom row and multiply the corresponding term. So b times x, b times x, plus c times y, c times y. So that's what it looks like when we do that right multiplication. And of course, we've gotta keep our transposed vector over there on the left side. So now we have, this is just a two by one vector now. This is a one by two. You could think of it as a horizontal vector or a one by two matrix."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And of course, we've gotta keep our transposed vector over there on the left side. So now we have, this is just a two by one vector now. This is a one by two. You could think of it as a horizontal vector or a one by two matrix. But now when we multiply these guys, you just kind of line up the corresponding terms. You'll have x multiplied by that entire top expression. So x multiplied by ax plus by."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "You could think of it as a horizontal vector or a one by two matrix. But now when we multiply these guys, you just kind of line up the corresponding terms. You'll have x multiplied by that entire top expression. So x multiplied by ax plus by. Ax plus by. And then we add that to the second term y multiplied by the second term of this guy, which is bx plus cy. So y multiplied by bx plus cy."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So x multiplied by ax plus by. Ax plus by. And then we add that to the second term y multiplied by the second term of this guy, which is bx plus cy. So y multiplied by bx plus cy. And all of these are numbers, so we can simplify it. Once we start distributing, the first term is x times a times x. So that's ax squared."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So y multiplied by bx plus cy. And all of these are numbers, so we can simplify it. Once we start distributing, the first term is x times a times x. So that's ax squared. And then the next term is x times b times y. So that's b times xy. Over here we have y times b times x."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So that's ax squared. And then the next term is x times b times y. So that's b times xy. Over here we have y times b times x. So that's the same thing as b times xy. So that's kind of why we have, why it's convenient to write a two there because that naturally comes out of our expansion. And then the last term is y times c times y."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Over here we have y times b times x. So that's the same thing as b times xy. So that's kind of why we have, why it's convenient to write a two there because that naturally comes out of our expansion. And then the last term is y times c times y. So that's cy squared. So we get back the original quadratic form that we were shooting for. You know, ax squared plus 2bxy plus cy squared."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And then the last term is y times c times y. So that's cy squared. So we get back the original quadratic form that we were shooting for. You know, ax squared plus 2bxy plus cy squared. That's how this entire term expands. As you kind of work it through, you end up with the same quadratic expression. Now, the convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "You know, ax squared plus 2bxy plus cy squared. That's how this entire term expands. As you kind of work it through, you end up with the same quadratic expression. Now, the convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly. And instead of writing the whole matrix M, you could just let a letter like M represent that whole matrix. And then take the vector that represents the variable, maybe like a bold-faced x, and you would multiply it on the right. And then you transpose it and multiply it on the left."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Now, the convenience of this quadratic form being written with a matrix like this is that we can write this more abstractly. And instead of writing the whole matrix M, you could just let a letter like M represent that whole matrix. And then take the vector that represents the variable, maybe like a bold-faced x, and you would multiply it on the right. And then you transpose it and multiply it on the left. So typically you denote that by putting a little t as a superscript. So x transposed multiplied by the matrix from the left. And this expression, this is what a quadratic form looks like in vectorized form."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And then you transpose it and multiply it on the left. So typically you denote that by putting a little t as a superscript. So x transposed multiplied by the matrix from the left. And this expression, this is what a quadratic form looks like in vectorized form. And the convenience is the same as it was in the linear case. Just like v could represent something that had 100 different numbers in it, and x would have 100 different constants, you could do something similar here, where you can write that same expression even if the matrix M is super huge. Let's just see what this would look like in a three-dimensional circumstance."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "And this expression, this is what a quadratic form looks like in vectorized form. And the convenience is the same as it was in the linear case. Just like v could represent something that had 100 different numbers in it, and x would have 100 different constants, you could do something similar here, where you can write that same expression even if the matrix M is super huge. Let's just see what this would look like in a three-dimensional circumstance. So, actually I'll need more room. So I'll go down even further. So we have x transpose multiplied by the matrix multiplied by x, bold-faced x."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Let's just see what this would look like in a three-dimensional circumstance. So, actually I'll need more room. So I'll go down even further. So we have x transpose multiplied by the matrix multiplied by x, bold-faced x. And let's say instead this represented, you know, you have x, then y, then z, our transposed vector. And then our matrix, let's say it was A, B, C, D, E, F. And because it needs to be symmetric, whatever term is in this spot here needs to be the same as over here, kind of when you reflected about that diagonal. Similarly, C, that's gonna be the same term here, and E would be over here."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "So we have x transpose multiplied by the matrix multiplied by x, bold-faced x. And let's say instead this represented, you know, you have x, then y, then z, our transposed vector. And then our matrix, let's say it was A, B, C, D, E, F. And because it needs to be symmetric, whatever term is in this spot here needs to be the same as over here, kind of when you reflected about that diagonal. Similarly, C, that's gonna be the same term here, and E would be over here. So there's only really six free terms that you have, but it fills up this entire matrix. And then on the right side, we would multiply that by x, y, z. Now, I won't work it out in this video, but you can imagine actually multiplying this matrix by this vector, and then multiplying the corresponding vector that you get by this transposed vector."}, {"video_title": "Expressing a quadratic form with a matrix.mp3", "Sentence": "Similarly, C, that's gonna be the same term here, and E would be over here. So there's only really six free terms that you have, but it fills up this entire matrix. And then on the right side, we would multiply that by x, y, z. Now, I won't work it out in this video, but you can imagine actually multiplying this matrix by this vector, and then multiplying the corresponding vector that you get by this transposed vector. And you'll get some kind of quadratic form with three variables. And the point is, you'll get a very complicated one, but it's very simple to express things like this. So with that tool in hand, in the next video, I will talk about how we can use this notation to express the quadratic approximations for multivariable functions."}, {"video_title": "Partial derivatives of vector fields, component by component.mp3", "Sentence": "This is one of those things that's pretty good practice for some important concepts coming up in multivariable calc, and it's also just good to sit down and take a complicated thing and kind of break it down piece by piece. So a vector field like the one that I just showed is represented by a vector-valued function, and since it's two-dimensional, it'll have some kind of two-dimensional input, and the output will be a vector each of whose components is some kind of function of x and y, right? So I'll just write p of x, y for that x component, and q of x, y for that y component, and each of these are just scalar-valued functions. It's actually quite common to use p and q for these values. It's one of those things where sometimes you'll even see a theorem about vector calculus in terms of just p and q, kind of leaving it understood to the reader that, yeah, p and q always refer to the x and y components of the output of a vector field, and in this specific case, the function that I chose, it's actually the one that I used in the last video, p is equal to x times y, and q is equal to y squared minus x squared, and in the last video, I was talking about interpreting the partial derivative of v, the vector-valued function, with respect to one of the variables, which has its merits, and I think it's a good way to understand vector-valued functions in general, but here, that's not what I'm gonna do. It's actually, another useful skill is to just think in terms of each specific component, so if we just think of p and q, we have four possible partial derivatives at our disposal here, two of them with respect to p, so you can think about the partial derivative of p with respect to x, or the partial derivative of p with respect to y, and then similarly, q, you could think about partial derivative of q with respect to x, x, this should be a partial, or the partial derivative of q with respect to y, so four different values that you could be looking at and considering and understanding how they influence the change of the vector field as a whole, and in this specific example, let's actually compute these, so derivative of p with respect to x, p is this first component, we're taking the partial of this with respect to x, y looks like a constant, constant times x, derivative is just that constant. If we took the derivative with respect to y, the roles are reversed, and its partial derivative is x, because x looks like that constant, but q, its partial derivative with respect to x, y looks like a constant, negative x squared goes to negative two x, but then when you're taking it with respect to y, y squared now looks like a function whose derivative is two y, and negative x squared looks like the constant, so these are the four possible partial derivatives, but let's actually see if we can understand how they influence the function as a whole, what it means in terms of the picture that we're looking at up here, and in particular, let's focus on a point, a specific point, and let's do this one here, so it's something that's sitting on the x-axis, so this is where y equals zero, and x is something positive, so this is probably when x is around two-ish, let's say, so the value we want to look at is x, y, when x is two, and y is zero, so if we start plugging that in here, what that would mean, this guy goes to zero, this guy goes to two, this guy, negative two times x, is gonna be negative two, and then negative two times y is gonna be zero, and let's start by just looking at the partial derivative of p with respect to x, so what that means is that we're looking for how the x component of these vectors change as you move in the x direction, so for example, around this point, we're kind of thinking of moving in the x direction vaguely, so we want to look at the two neighboring vectors and consider what's going on with the x direction, but these vectors, this one points purely down, this one also points purely down, and so does this one, so no change is happening when it comes to the x component of these vectors, which makes sense, because the value at that point is zero, the partial derivative of p with respect to x is zero, so we wouldn't expect a change, but on the other hand, if we're looking at partial derivative of p with respect to y, this should be positive, so this should suggest that the change in the x component as you move in the y direction is positive, so we go up here, and now we're not looking at change in the x direction, not looking at change in the x direction, but instead we're wondering what happens as we move generally upwards, so we're gonna kind of compare it to these two guys, and in that case, the x component of this one is a little bit to the left, the one below it, it's a little bit to the left, then we get to our main guy here, and it's zero, the x component is zero, because it's pointing purely down, and up here it's pointing a little bit to the right, so as y increases, the x component of these vectors also increases, and again, that makes sense, because this partial derivative is positive, this too suggests that as you're changing y, the value of p, the x component of our function, I should probably keep that on screen, the x component of our vector-valued function is increasing, because that's positive."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we figured out how to construct a unit normal vector to a surface. And so now we can use that back in our original surface integral to try to simplify a little bit, or at least give us a clue of how we can calculate these things. And also think about different ways to represent this type of a surface integral. So if we just substitute what we came up as our normal vector, our unit normal vector right here, we will get, we will get, so once again, it's the surface integral of F dot, and F dot all of this business right over here, and I'm going to write it all in white just so it doesn't take me too much time. So the partial of r with respect to u crossed with the partial of r with respect to v over the magnitude of the same thing. Partial of r with respect to u crossed with the partial of r with respect to v. And now we've played with ds a lot. We know that another way to write ds, and I gave the intuition, hopefully, for that several videos ago when we first explored what a surface integral was all about, we know that ds can be represented as the magnitude of the partial of r with respect to u crossed with the partial of r with respect to v, du dv."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we just substitute what we came up as our normal vector, our unit normal vector right here, we will get, we will get, so once again, it's the surface integral of F dot, and F dot all of this business right over here, and I'm going to write it all in white just so it doesn't take me too much time. So the partial of r with respect to u crossed with the partial of r with respect to v over the magnitude of the same thing. Partial of r with respect to u crossed with the partial of r with respect to v. And now we've played with ds a lot. We know that another way to write ds, and I gave the intuition, hopefully, for that several videos ago when we first explored what a surface integral was all about, we know that ds can be represented as the magnitude of the partial of r with respect to u crossed with the partial of r with respect to v, du dv. And obviously the du dv can be written as dv du. You could write it as da, a little chunk of area in the uv plane or in the uv domain. And actually since now this integral is in terms of uv, we're no longer taking a surface integral, we're now taking a double integral over the uv domain."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We know that another way to write ds, and I gave the intuition, hopefully, for that several videos ago when we first explored what a surface integral was all about, we know that ds can be represented as the magnitude of the partial of r with respect to u crossed with the partial of r with respect to v, du dv. And obviously the du dv can be written as dv du. You could write it as da, a little chunk of area in the uv plane or in the uv domain. And actually since now this integral is in terms of uv, we're no longer taking a surface integral, we're now taking a double integral over the uv domain. So you could say kind of a region in uv. So I'll say r to say that it's a region in the uv plane that we're now thinking about. But there's probably a huge, or there should be, or I'm guessing there's a huge simplification that's popping out at you right now."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually since now this integral is in terms of uv, we're no longer taking a surface integral, we're now taking a double integral over the uv domain. So you could say kind of a region in uv. So I'll say r to say that it's a region in the uv plane that we're now thinking about. But there's probably a huge, or there should be, or I'm guessing there's a huge simplification that's popping out at you right now. We're dividing by the magnitude of the cross product of these two vectors, and then we're multiplying by the magnitude of the cross product of these two vectors. Those are just scalar quantities. You divide by something and multiply by something."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But there's probably a huge, or there should be, or I'm guessing there's a huge simplification that's popping out at you right now. We're dividing by the magnitude of the cross product of these two vectors, and then we're multiplying by the magnitude of the cross product of these two vectors. Those are just scalar quantities. You divide by something and multiply by something. Well, that's just the same thing as multiplying or dividing by 1. So these two characters cancel out, and our integral simplifies to the double integral over that region, the corresponding region in the uv plane, of f, of our vector field f dotted with this cross product. This is going to give us a vector right over here."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You divide by something and multiply by something. Well, that's just the same thing as multiplying or dividing by 1. So these two characters cancel out, and our integral simplifies to the double integral over that region, the corresponding region in the uv plane, of f, of our vector field f dotted with this cross product. This is going to give us a vector right over here. That's going to give us a vector. It gives us actually a normal vector, and then we divide by its magnitude. It gives you a unit normal vector."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is going to give us a vector right over here. That's going to give us a vector. It gives us actually a normal vector, and then we divide by its magnitude. It gives you a unit normal vector. So this, you're going to take the dot product of f with r, the partial of r with respect to u, crossed with the partial of r with respect to v, du dv. Let me scroll over to the right a little bit. du dv."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It gives you a unit normal vector. So this, you're going to take the dot product of f with r, the partial of r with respect to u, crossed with the partial of r with respect to v, du dv. Let me scroll over to the right a little bit. du dv. And we'll see in a few videos from now that this is essentially how we go about actually calculating these things. If you have a parametrization, you can then get everything in terms of a double integral, in terms of uv, this way. Now, the last thing I want to do is explore another way that you'll see a surface integral like this written."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "du dv. And we'll see in a few videos from now that this is essentially how we go about actually calculating these things. If you have a parametrization, you can then get everything in terms of a double integral, in terms of uv, this way. Now, the last thing I want to do is explore another way that you'll see a surface integral like this written. It all comes from really writing this part in a different way, but hopefully it gives you a little bit more intuition of what this thing is even saying. So I'm just going to rewrite this chunk right over here. I'm just going to rewrite that chunk."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, the last thing I want to do is explore another way that you'll see a surface integral like this written. It all comes from really writing this part in a different way, but hopefully it gives you a little bit more intuition of what this thing is even saying. So I'm just going to rewrite this chunk right over here. I'm just going to rewrite that chunk. And I'm going to use slightly different notation because it will hopefully help make a little bit more sense. So the partial of r with respect to u I can write as the partial of r with respect to u. And we're taking the cross product."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just going to rewrite that chunk. And I'm going to use slightly different notation because it will hopefully help make a little bit more sense. So the partial of r with respect to u I can write as the partial of r with respect to u. And we're taking the cross product. Let me make my u's a little bit more u-like so we don't confuse them with v's. And we're taking the cross product of that with the partial of r with respect to v. So very small changes in our vector, in our parametrization right here, our position vector, given a small change in v. Very small changes in the vector given a small change in u. And then we're multiplying that times du dv."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're taking the cross product. Let me make my u's a little bit more u-like so we don't confuse them with v's. And we're taking the cross product of that with the partial of r with respect to v. So very small changes in our vector, in our parametrization right here, our position vector, given a small change in v. Very small changes in the vector given a small change in u. And then we're multiplying that times du dv. Now du and dv are just scalar quantities. They're infinitesimally small, but for the sake of this argument, you can just view that they're not vectors. They're just scalar quantities."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're multiplying that times du dv. Now du and dv are just scalar quantities. They're infinitesimally small, but for the sake of this argument, you can just view that they're not vectors. They're just scalar quantities. And so you can essentially include them. If you have the cross product, if you have a cross b times some scalar value, x, you could rewrite this as x times a cross b. Or you could write this as a cross x times b because x is just a scalar value."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "They're just scalar quantities. And so you can essentially include them. If you have the cross product, if you have a cross b times some scalar value, x, you could rewrite this as x times a cross b. Or you could write this as a cross x times b because x is just a scalar value. It's just a number. So we could do the same thing over here. We can rewrite all of this business as, and I'm going to group the du where we have the partial with respect to u in the denominator, and I'll do the same thing with the v's."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Or you could write this as a cross x times b because x is just a scalar value. It's just a number. So we could do the same thing over here. We can rewrite all of this business as, and I'm going to group the du where we have the partial with respect to u in the denominator, and I'll do the same thing with the v's. And so you'll get the partial of r with respect to u times du times that scalar. So that'll give us a vector. And we're going to cross that with the partial of r with respect to v dv."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can rewrite all of this business as, and I'm going to group the du where we have the partial with respect to u in the denominator, and I'll do the same thing with the v's. And so you'll get the partial of r with respect to u times du times that scalar. So that'll give us a vector. And we're going to cross that with the partial of r with respect to v dv. Now, these might look notationally like two different things, but that just comes from the necessity of when we take partial derivative to say, oh, no, this vector function is defined, it's a function of multiple variables, and this is taking a derivative with respect to only one of them. So this is how much does our vector change when you have a very small change in u. But this is also an infinitesimally small change in u over here."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to cross that with the partial of r with respect to v dv. Now, these might look notationally like two different things, but that just comes from the necessity of when we take partial derivative to say, oh, no, this vector function is defined, it's a function of multiple variables, and this is taking a derivative with respect to only one of them. So this is how much does our vector change when you have a very small change in u. But this is also an infinitesimally small change in u over here. We're just using slightly different notation. So for the sake of this, this is a little bit loosey-goosey mathematics, but it'll hopefully give you the intuition for why this thing can be written in a different way. These are essentially the same quantity."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is also an infinitesimally small change in u over here. We're just using slightly different notation. So for the sake of this, this is a little bit loosey-goosey mathematics, but it'll hopefully give you the intuition for why this thing can be written in a different way. These are essentially the same quantity. So if you divide by something and multiply by something, you can cancel them out. If you divide by something and multiply by something, you can cancel them out. And all you're left with then is the differential of r, and since we lost the information that it's in the u direction, I'll write it here, the differential of r in the u direction."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "These are essentially the same quantity. So if you divide by something and multiply by something, you can cancel them out. If you divide by something and multiply by something, you can cancel them out. And all you're left with then is the differential of r, and since we lost the information that it's in the u direction, I'll write it here, the differential of r in the u direction. I don't want to get the notation confused. This is just the differential. This is just how much r changed."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And all you're left with then is the differential of r, and since we lost the information that it's in the u direction, I'll write it here, the differential of r in the u direction. I don't want to get the notation confused. This is just the differential. This is just how much r changed. This is not the partial derivative of r with respect to u. This right over here is how much does r change given per unit change, per small change in u. This just says a differential in the direction of, I guess, as u changes, this is how much that infinitely small change that just r changes."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just how much r changed. This is not the partial derivative of r with respect to u. This right over here is how much does r change given per unit change, per small change in u. This just says a differential in the direction of, I guess, as u changes, this is how much that infinitely small change that just r changes. This isn't change in r with respect to change in u. We're going to cross that with the partial of r in the v direction. Now this right over here, let's just conceptualize this."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This just says a differential in the direction of, I guess, as u changes, this is how much that infinitely small change that just r changes. This isn't change in r with respect to change in u. We're going to cross that with the partial of r in the v direction. Now this right over here, let's just conceptualize this. This goes back to our original visions of what a surface integral was all about. If we're on a surface, and I'll draw a surface, right? Let me draw another surface."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now this right over here, let's just conceptualize this. This goes back to our original visions of what a surface integral was all about. If we're on a surface, and I'll draw a surface, right? Let me draw another surface. I won't use the one that I had already drawn on. If we draw a surface, and for a very small change in u, and we're not going to think about the rate. We're thinking about the change in r. You're going in that direction."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw another surface. I won't use the one that I had already drawn on. If we draw a surface, and for a very small change in u, and we're not going to think about the rate. We're thinking about the change in r. You're going in that direction. If that thing looks like this, this is actually a distance moved on the surface. Because it's not, remember, this isn't the derivative. This is the differential."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're thinking about the change in r. You're going in that direction. If that thing looks like this, this is actually a distance moved on the surface. Because it's not, remember, this isn't the derivative. This is the differential. It's just a small change along the surface. That's that over there. This is a small change when you change v. It's also a change along the surface."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the differential. It's just a small change along the surface. That's that over there. This is a small change when you change v. It's also a change along the surface. If you take the cross product of these two things, you get a vector that is orthogonal. You get a vector that is normal to the surface. Its magnitude, and we saw this when we first learned about cross products, its magnitude is equal to the area that is defined by these two vectors."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is a small change when you change v. It's also a change along the surface. If you take the cross product of these two things, you get a vector that is orthogonal. You get a vector that is normal to the surface. Its magnitude, and we saw this when we first learned about cross products, its magnitude is equal to the area that is defined by these two vectors. Its magnitude is equal to area. In a lot of ways, you can really think of it as a unit normal vector times ds. The way that we would, I guess, notationally do this is we can call this, because this is kind of a ds, but it's a vector version of the ds."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Its magnitude, and we saw this when we first learned about cross products, its magnitude is equal to the area that is defined by these two vectors. Its magnitude is equal to area. In a lot of ways, you can really think of it as a unit normal vector times ds. The way that we would, I guess, notationally do this is we can call this, because this is kind of a ds, but it's a vector version of the ds. Over here, this is just an area right over here. This is just a scalar value. Now, we have a vector that points normally from the surface, but its magnitude is the same thing as that ds that we were just talking about."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The way that we would, I guess, notationally do this is we can call this, because this is kind of a ds, but it's a vector version of the ds. Over here, this is just an area right over here. This is just a scalar value. Now, we have a vector that points normally from the surface, but its magnitude is the same thing as that ds that we were just talking about. We can call this thing right over here, we can call this ds. The key difference here is this is a vector now. We'll call it ds with a little vector over it to know that this is this thing."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, we have a vector that points normally from the surface, but its magnitude is the same thing as that ds that we were just talking about. We can call this thing right over here, we can call this ds. The key difference here is this is a vector now. We'll call it ds with a little vector over it to know that this is this thing. This isn't the scalar ds that is just concerned with the area. When you view things this way, we just saw that this entire thing simplifies to ds. Then our whole surface integral can be rewritten."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll call it ds with a little vector over it to know that this is this thing. This isn't the scalar ds that is just concerned with the area. When you view things this way, we just saw that this entire thing simplifies to ds. Then our whole surface integral can be rewritten. Instead of writing it like this, we can write it as the integral, or the surface integral, those integral signs were too fancy, the surface integral of f dot, instead of saying a normal vector times a scalar quantity, a little chunk of area on the surface, we can now just call that the vector differential ds. I want to make it clear, these are two different things. This is a vector."}, {"video_title": "Vector representation of a surface integral   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then our whole surface integral can be rewritten. Instead of writing it like this, we can write it as the integral, or the surface integral, those integral signs were too fancy, the surface integral of f dot, instead of saying a normal vector times a scalar quantity, a little chunk of area on the surface, we can now just call that the vector differential ds. I want to make it clear, these are two different things. This is a vector. This is essentially what we're calling it. This right over here is a scalar times a normal vector. These are three different ways of really representing the same thing."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "So in the last couple videos, I talked about the multivariable chain rule, which I have written up here, and if you haven't seen those, go take a look. And here I want to write it out in vector notation, and this helps us generalize it a little bit when the intermediary space is a little bit higher dimensional. So instead of writing x of t and y of t as separate functions and just trying to emphasize, oh, they have the same input space, and whatever x takes in, that's the same number y takes in, it's better and a little bit cleaner if we say there's a vector-valued function that takes in a single number t, and then it outputs some kind of vector. In this case, you could say the components of v are x of t and y of t, and that's fine. But I want to talk about what this looks like if we start writing everything in vector notation. And just since we see dx dt here and dy dt here, you might start thinking, oh, well, we should take the derivative of that vector-valued function. The derivative of v with respect to t, and when we compute this, it's nothing more than just taking the derivatives of each component."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "In this case, you could say the components of v are x of t and y of t, and that's fine. But I want to talk about what this looks like if we start writing everything in vector notation. And just since we see dx dt here and dy dt here, you might start thinking, oh, well, we should take the derivative of that vector-valued function. The derivative of v with respect to t, and when we compute this, it's nothing more than just taking the derivatives of each component. So in this case, the derivative of x, so you'd write dx dt, and the derivative of y, dy dt. This is the vector-valued derivative. And now you might start to notice something here."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "The derivative of v with respect to t, and when we compute this, it's nothing more than just taking the derivatives of each component. So in this case, the derivative of x, so you'd write dx dt, and the derivative of y, dy dt. This is the vector-valued derivative. And now you might start to notice something here. Okay, so we've got one of those components multiplied by a certain value, and another component multiplied by a certain value. You might recognize this as a dot product. This would be the dot product between the vector that contains the derivatives, the partial derivatives, partial of f with respect to y, partial of f with respect to x. Oh, whoops, don't know why I wrote it that way."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "And now you might start to notice something here. Okay, so we've got one of those components multiplied by a certain value, and another component multiplied by a certain value. You might recognize this as a dot product. This would be the dot product between the vector that contains the derivatives, the partial derivatives, partial of f with respect to y, partial of f with respect to x. Oh, whoops, don't know why I wrote it that way. So up here, that's with respect to x, and then here to y. So this whole thing, we're taking the dot product with the vector that contains ordinary derivative, dx dt, and ordinary derivative, dy dt. And of course, both of these are special vectors."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "This would be the dot product between the vector that contains the derivatives, the partial derivatives, partial of f with respect to y, partial of f with respect to x. Oh, whoops, don't know why I wrote it that way. So up here, that's with respect to x, and then here to y. So this whole thing, we're taking the dot product with the vector that contains ordinary derivative, dx dt, and ordinary derivative, dy dt. And of course, both of these are special vectors. They're not just random. The left one, that's the gradient of f, gradient of f. And the right vector here, that's what we just wrote. That's the derivative of v with respect to t. Just for being quick, I'm gonna write that as v prime of t. That's saying completely the same thing as dv dt."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "And of course, both of these are special vectors. They're not just random. The left one, that's the gradient of f, gradient of f. And the right vector here, that's what we just wrote. That's the derivative of v with respect to t. Just for being quick, I'm gonna write that as v prime of t. That's saying completely the same thing as dv dt. And this right here is another way to write the multivariable chain rule. And maybe if you were being a little bit more exact, you would emphasize that when you take the gradient of f, the thing that you input into it is the output of that vector-valued function. You know, you're throwing in x of t and y of t. So you might emphasize that you take in that as an input, and then you multiply it by the derivative, the vector-valued derivative of v of t. And when I say multiply, I mean dot product."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "That's the derivative of v with respect to t. Just for being quick, I'm gonna write that as v prime of t. That's saying completely the same thing as dv dt. And this right here is another way to write the multivariable chain rule. And maybe if you were being a little bit more exact, you would emphasize that when you take the gradient of f, the thing that you input into it is the output of that vector-valued function. You know, you're throwing in x of t and y of t. So you might emphasize that you take in that as an input, and then you multiply it by the derivative, the vector-valued derivative of v of t. And when I say multiply, I mean dot product. These are vectors, and you're taking the dot product. And this should seem very familiar to the single-variable chain rule. And just to remind us, I'll throw it up here."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "You know, you're throwing in x of t and y of t. So you might emphasize that you take in that as an input, and then you multiply it by the derivative, the vector-valued derivative of v of t. And when I say multiply, I mean dot product. These are vectors, and you're taking the dot product. And this should seem very familiar to the single-variable chain rule. And just to remind us, I'll throw it up here. If you take the derivative of, you know, composition of two single-variable functions, f and g, you take the derivative of the outside, f prime, and throw in g, throw in what was the interior function, and you multiply it by the derivative of that interior function, g prime of t. And this is super helpful in single-variable calculus for computing a lot of derivatives. And over here, it has a very similar form, right? The gradient, which really serves the function of the true extension of the derivative for multivariable functions, for scalar-valued multivariable functions at least."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "And just to remind us, I'll throw it up here. If you take the derivative of, you know, composition of two single-variable functions, f and g, you take the derivative of the outside, f prime, and throw in g, throw in what was the interior function, and you multiply it by the derivative of that interior function, g prime of t. And this is super helpful in single-variable calculus for computing a lot of derivatives. And over here, it has a very similar form, right? The gradient, which really serves the function of the true extension of the derivative for multivariable functions, for scalar-valued multivariable functions at least. You take that derivative and throw in the inner function, which just happens to be a vector-valued function, but you throw it in there, and then you multiply it by the derivative of that. But multiplying vectors in this context means taking the dot product of the two. And this could mean, if you have a function with a whole bunch of different variables, so let's say you have, you know, some f of x, or not f of x, but f of, like, x1 and x2, and it takes in a whole bunch of different variables, and it goes out to x100."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "The gradient, which really serves the function of the true extension of the derivative for multivariable functions, for scalar-valued multivariable functions at least. You take that derivative and throw in the inner function, which just happens to be a vector-valued function, but you throw it in there, and then you multiply it by the derivative of that. But multiplying vectors in this context means taking the dot product of the two. And this could mean, if you have a function with a whole bunch of different variables, so let's say you have, you know, some f of x, or not f of x, but f of, like, x1 and x2, and it takes in a whole bunch of different variables, and it goes out to x100. And then what you throw into it is a vector-valued function that's something that's vector-valued, takes in a single variable, and, you know, in order to be able to compose them, it's gonna have a whole bunch of intermediary functions. And I'll, you know, you could write it as x1, x2, x3, all the way up to x100. And these are all functions at this point."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "And this could mean, if you have a function with a whole bunch of different variables, so let's say you have, you know, some f of x, or not f of x, but f of, like, x1 and x2, and it takes in a whole bunch of different variables, and it goes out to x100. And then what you throw into it is a vector-valued function that's something that's vector-valued, takes in a single variable, and, you know, in order to be able to compose them, it's gonna have a whole bunch of intermediary functions. And I'll, you know, you could write it as x1, x2, x3, all the way up to x100. And these are all functions at this point. These are component functions of your vector-valued v. This expression still makes sense, right? You can still take the gradient of f. It's gonna have 100 components. You can plug in any vector, any set of 100 different numbers, and in particular, the output of a vector-valued function with 100 different components is gonna work."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "And these are all functions at this point. These are component functions of your vector-valued v. This expression still makes sense, right? You can still take the gradient of f. It's gonna have 100 components. You can plug in any vector, any set of 100 different numbers, and in particular, the output of a vector-valued function with 100 different components is gonna work. And then you take the dot product with the derivative of this. And that's the more general version of the multivariable chain rule. And another cool way about writing it like this is you can interpret it in terms of the directional derivative."}, {"video_title": "Vector form of the multivariable chain rule.mp3", "Sentence": "You can plug in any vector, any set of 100 different numbers, and in particular, the output of a vector-valued function with 100 different components is gonna work. And then you take the dot product with the derivative of this. And that's the more general version of the multivariable chain rule. And another cool way about writing it like this is you can interpret it in terms of the directional derivative. And I think I'll do that in the next video. I won't do that here. So a certain way to interpret this with the directional derivative."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So I'm gonna go ahead and go through a specific example here. And just to remind you of kind of what all these terms are, how there's actually kind of a pattern to what's going on, this here represents, you could think of it as the constant term, where this is just gonna evaluate to some kind of number. These two terms are what you might call the linear term. Linear. Because if you actually look, the only places where the variable x and y comes up is here, where it's just being multiplied by a constant, and here, where it's just being multiplied by a constant. So it's just variables times constants in there. And then all of this stuff at the end, which is kind of the whole essence of a quadratic approximation, where you start to have things, like you get an x squared, and you get like x gets to be multiplied by y."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Linear. Because if you actually look, the only places where the variable x and y comes up is here, where it's just being multiplied by a constant, and here, where it's just being multiplied by a constant. So it's just variables times constants in there. And then all of this stuff at the end, which is kind of the whole essence of a quadratic approximation, where you start to have things, like you get an x squared, and you get like x gets to be multiplied by y. All of this stuff is the quadratic term. And though it seems like a lot now, you'll see in the context of an actual example, it's not necessarily as bad as it seems. So let's say we're looking at the function f of x, y, and let's say it's gonna be e to the x divided by two multiplied by sine of y."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And then all of this stuff at the end, which is kind of the whole essence of a quadratic approximation, where you start to have things, like you get an x squared, and you get like x gets to be multiplied by y. All of this stuff is the quadratic term. And though it seems like a lot now, you'll see in the context of an actual example, it's not necessarily as bad as it seems. So let's say we're looking at the function f of x, y, and let's say it's gonna be e to the x divided by two multiplied by sine of y. This is our multivariable function. And let's say we want to approximate this near some kind of point. And I'm gonna choose a point that's, you know, something that we can actually evaluate these at."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So let's say we're looking at the function f of x, y, and let's say it's gonna be e to the x divided by two multiplied by sine of y. This is our multivariable function. And let's say we want to approximate this near some kind of point. And I'm gonna choose a point that's, you know, something that we can actually evaluate these at. So like x, it would be convenient if that was zero. And then y, I'll go with pi halves, because that's something where I'll know how to evaluate sine, and where I'll know how to evaluate its derivatives, things like that. So we're trying to approximate this function near this point."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And I'm gonna choose a point that's, you know, something that we can actually evaluate these at. So like x, it would be convenient if that was zero. And then y, I'll go with pi halves, because that's something where I'll know how to evaluate sine, and where I'll know how to evaluate its derivatives, things like that. So we're trying to approximate this function near this point. Now, first things first, we're just gonna need to get all of the different partial derivatives and second partial derivatives. We know we're gonna need them, so let's just kind of start working it through and figuring out what all of them are. So, let's start with the partial derivative with respect to x."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So we're trying to approximate this function near this point. Now, first things first, we're just gonna need to get all of the different partial derivatives and second partial derivatives. We know we're gonna need them, so let's just kind of start working it through and figuring out what all of them are. So, let's start with the partial derivative with respect to x. So this is also a function of x, y. And we look up at the original function, the only place where x shows up is in this e to the x over two. The derivative of that is 1 1 2, we bring down that 1 1 2 times e to the x over two."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So, let's start with the partial derivative with respect to x. So this is also a function of x, y. And we look up at the original function, the only place where x shows up is in this e to the x over two. The derivative of that is 1 1 2, we bring down that 1 1 2 times e to the x over two. And this is being multiplied by something that looks like a constant, as far as x is concerned, sine of y. Now when we do the partial derivative with respect to y, what we get, this first part just looks like a constant, so we kind of keep that constant there, as far as y is concerned. And the derivative of sine is cosine."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "The derivative of that is 1 1 2, we bring down that 1 1 2 times e to the x over two. And this is being multiplied by something that looks like a constant, as far as x is concerned, sine of y. Now when we do the partial derivative with respect to y, what we get, this first part just looks like a constant, so we kind of keep that constant there, as far as y is concerned. And the derivative of sine is cosine. Cosine of y. And then now, let's start taking second partial derivatives. So I'll start by doing the one where we take the partial derivative with respect to x twice."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And the derivative of sine is cosine. Cosine of y. And then now, let's start taking second partial derivatives. So I'll start by doing the one where we take the partial derivative with respect to x twice. Now here, I'll actually do this, I'll do this in a different color. How about, let's do like yellow. Just to make clear which ones are the second partial derivatives."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So I'll start by doing the one where we take the partial derivative with respect to x twice. Now here, I'll actually do this, I'll do this in a different color. How about, let's do like yellow. Just to make clear which ones are the second partial derivatives. So, partial with respect to x twice, also a function of x, y, like all of these guys. And so let's look up at the original partial derivative with respect to x, and we're now gonna take its derivative, again with respect to x. This is the only place where x shows up, that 1 1 2 kind of comes down again, so now it's gonna be 1 1 4 times e to the x over two."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Just to make clear which ones are the second partial derivatives. So, partial with respect to x twice, also a function of x, y, like all of these guys. And so let's look up at the original partial derivative with respect to x, and we're now gonna take its derivative, again with respect to x. This is the only place where x shows up, that 1 1 2 kind of comes down again, so now it's gonna be 1 1 4 times e to the x over two. And we just keep that sine of y, cause it looks like we're just multiplying by a constant. Sine of y. Next, we'll do the mixed partial derivative, where you do first with respect to x, then with respect to y, or you could do it the other way, because with almost all functions, it kind of doesn't matter which order you take the two."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "This is the only place where x shows up, that 1 1 2 kind of comes down again, so now it's gonna be 1 1 4 times e to the x over two. And we just keep that sine of y, cause it looks like we're just multiplying by a constant. Sine of y. Next, we'll do the mixed partial derivative, where you do first with respect to x, then with respect to y, or you could do it the other way, because with almost all functions, it kind of doesn't matter which order you take the two. So I'll go ahead and just look at the one that was with respect to x, and now let's think of its derivative with respect to y. This whole 1 1 2 e to the x halves, looks like a constant, derivative of sine of y is cosine y. So we take that constant, the 1 1 2 e to the x halves, and then we multiply it by the derivative of sine of y, which is cosine of y."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Next, we'll do the mixed partial derivative, where you do first with respect to x, then with respect to y, or you could do it the other way, because with almost all functions, it kind of doesn't matter which order you take the two. So I'll go ahead and just look at the one that was with respect to x, and now let's think of its derivative with respect to y. This whole 1 1 2 e to the x halves, looks like a constant, derivative of sine of y is cosine y. So we take that constant, the 1 1 2 e to the x halves, and then we multiply it by the derivative of sine of y, which is cosine of y. And then finally, we take the second derivative, second partial derivative with respect to y twice in a row. So f with respect to y twice in a row. And for this one, let's take a look at the partial derivative with respect to y."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So we take that constant, the 1 1 2 e to the x halves, and then we multiply it by the derivative of sine of y, which is cosine of y. And then finally, we take the second derivative, second partial derivative with respect to y twice in a row. So f with respect to y twice in a row. And for this one, let's take a look at the partial derivative with respect to y. This part is the only part where y shows up, derivative of cosine is negative sine, and then e to the x halves just still looks like a constant. So we bring that negative out front, that constant e to the x halves, and it was negative sine, so that negative went out front. Sine of y."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And for this one, let's take a look at the partial derivative with respect to y. This part is the only part where y shows up, derivative of cosine is negative sine, and then e to the x halves just still looks like a constant. So we bring that negative out front, that constant e to the x halves, and it was negative sine, so that negative went out front. Sine of y. So that's all of the partial differential information that we're gonna need. And now we know we're gonna need to evaluate all of these guys, all of these partial derivatives at the specific point, because if we go up and look at the original function that we have, we're gonna need to evaluate f at this point, both of the partial derivatives at this point, the second partial derivatives. Oh, I'm realizing actually that I made a little bit of a mistake here."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Sine of y. So that's all of the partial differential information that we're gonna need. And now we know we're gonna need to evaluate all of these guys, all of these partial derivatives at the specific point, because if we go up and look at the original function that we have, we're gonna need to evaluate f at this point, both of the partial derivatives at this point, the second partial derivatives. Oh, I'm realizing actually that I made a little bit of a mistake here. This should be 1 1 2 out in front of each of these guys. That should be plus 1 1 2 of this second partial derivative and 1 1 2 of this second partial derivative. The mixed partial derivative, it's still one, but these guys should have a 1 1 2."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Oh, I'm realizing actually that I made a little bit of a mistake here. This should be 1 1 2 out in front of each of these guys. That should be plus 1 1 2 of this second partial derivative and 1 1 2 of this second partial derivative. The mixed partial derivative, it's still one, but these guys should have a 1 1 2. That was a mistake on my part. In either case, though, we're gonna need to evaluate all of these guys. So if we go back down, let's just start plugging in the point zero and pi halves to each one of these."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "The mixed partial derivative, it's still one, but these guys should have a 1 1 2. That was a mistake on my part. In either case, though, we're gonna need to evaluate all of these guys. So if we go back down, let's just start plugging in the point zero and pi halves to each one of these. So the function itself, when we plug in zero, e to the zero is one, and sine of pi halves, sine of pi halves is also one, so this entire thing just comes to one. If we do this for the next one, again, e to the zero is gonna be one, sine of y is also gonna be one, but now we have that 1 1 2 sitting there, so that'll end up as 1 1 2. If we look at the partial derivative with respect to y, cosine of pi halves is zero, so this entire thing is gonna be zero."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So if we go back down, let's just start plugging in the point zero and pi halves to each one of these. So the function itself, when we plug in zero, e to the zero is one, and sine of pi halves, sine of pi halves is also one, so this entire thing just comes to one. If we do this for the next one, again, e to the zero is gonna be one, sine of y is also gonna be one, but now we have that 1 1 2 sitting there, so that'll end up as 1 1 2. If we look at the partial derivative with respect to y, cosine of pi halves is zero, so this entire thing is gonna be zero. Moving right along, if we do, let's take a look at the second partial derivative with respect to x. Again, e to the zero will be one, and sine of pi halves will be one, so this ends up just being that 1 1 4. The mixed partial derivative here, if we have 1 1 2 by the pattern starting to continue, we've got the one."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "If we look at the partial derivative with respect to y, cosine of pi halves is zero, so this entire thing is gonna be zero. Moving right along, if we do, let's take a look at the second partial derivative with respect to x. Again, e to the zero will be one, and sine of pi halves will be one, so this ends up just being that 1 1 4. The mixed partial derivative here, if we have 1 1 2 by the pattern starting to continue, we've got the one. This one's actually zero, so cosine of pi halves is zero, so the whole thing will be zero. And then the last one, it'll be negative one times that one again, for sine of pi halves is one. So all of that just comes out to be negative one."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "The mixed partial derivative here, if we have 1 1 2 by the pattern starting to continue, we've got the one. This one's actually zero, so cosine of pi halves is zero, so the whole thing will be zero. And then the last one, it'll be negative one times that one again, for sine of pi halves is one. So all of that just comes out to be negative one. So this is, I mean, I kind of chose a convenient example, right, where all the derivatives look very similar to the thing itself, which is actually pretty common, so we get to leverage a lot of the work that we did earlier. So now we have these six different constants, and I kind of can't keep them all on the screen at the same time, but we've got the six different constants, so now we just plug each one of these in to the quadratic approximation. So if we make our quadratic approximation of our function, the first term is that constant term, so we take a look up, and we say, where does f of x, y go at this point, and it'll just be one."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So all of that just comes out to be negative one. So this is, I mean, I kind of chose a convenient example, right, where all the derivatives look very similar to the thing itself, which is actually pretty common, so we get to leverage a lot of the work that we did earlier. So now we have these six different constants, and I kind of can't keep them all on the screen at the same time, but we've got the six different constants, so now we just plug each one of these in to the quadratic approximation. So if we make our quadratic approximation of our function, the first term is that constant term, so we take a look up, and we say, where does f of x, y go at this point, and it'll just be one. I'm gonna have to do a lot of scrolling back and forth here, there's a lot of text to deal with. And the next thing is gonna be something times x minus zero, the kind of x-coordinate of our specified point, and that something is the first derivative with respect to x, so that's gonna be 1 1\u20442. So come back down here, we've got 1 1\u20442."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So if we make our quadratic approximation of our function, the first term is that constant term, so we take a look up, and we say, where does f of x, y go at this point, and it'll just be one. I'm gonna have to do a lot of scrolling back and forth here, there's a lot of text to deal with. And the next thing is gonna be something times x minus zero, the kind of x-coordinate of our specified point, and that something is the first derivative with respect to x, so that's gonna be 1 1\u20442. So come back down here, we've got 1 1\u20442. And then similarly, we're gonna have something multiplied by y minus the y-coordinate of the point about which we are approximating, and for that, we take a look at the partial derivative with respect to y, which was just zero, so that's pretty convenient. That's just gonna end up being zero. And then for the second partial derivative terms, maybe I'll actually be able to keep it on the same screen here, we're gonna have something multiplied by x minus its coordinate squared, and that something is whatever the partial derivative with respect to x twice is, which is 1\u20444."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So come back down here, we've got 1 1\u20442. And then similarly, we're gonna have something multiplied by y minus the y-coordinate of the point about which we are approximating, and for that, we take a look at the partial derivative with respect to y, which was just zero, so that's pretty convenient. That's just gonna end up being zero. And then for the second partial derivative terms, maybe I'll actually be able to keep it on the same screen here, we're gonna have something multiplied by x minus its coordinate squared, and that something is whatever the partial derivative with respect to x twice is, which is 1\u20444. So we go ahead and plug in 1\u20444, and then for the mixed partial derivative, we'll put it down here, it'll be something multiplied by x minus its constant, and then y minus that pi halves, and that something is the mixed partial derivative, which in this case is zero. Oh, and I'm realizing I made the same mistake again, it's not 1\u20444, it's 1\u20442. For the same reason that I made a mistake up here earlier where it's actually 1\u20442 multiplied by this second partial derivative, and 1\u20442 by the second partial derivative there."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And then for the second partial derivative terms, maybe I'll actually be able to keep it on the same screen here, we're gonna have something multiplied by x minus its coordinate squared, and that something is whatever the partial derivative with respect to x twice is, which is 1\u20444. So we go ahead and plug in 1\u20444, and then for the mixed partial derivative, we'll put it down here, it'll be something multiplied by x minus its constant, and then y minus that pi halves, and that something is the mixed partial derivative, which in this case is zero. Oh, and I'm realizing I made the same mistake again, it's not 1\u20444, it's 1\u20442. For the same reason that I made a mistake up here earlier where it's actually 1\u20442 multiplied by this second partial derivative, and 1\u20442 by the second partial derivative there. I guess I keep forgetting that. Good lesson, I suppose, that that's an easy thing to forget if you find yourself computing one of these, where I'll put it in here, multiply that guy by 1\u20442. It's similar to a Taylor expansion in single variable calculus, where you kind of have to remember what's what that squared term would be, has a 1\u20442 associated with it."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "For the same reason that I made a mistake up here earlier where it's actually 1\u20442 multiplied by this second partial derivative, and 1\u20442 by the second partial derivative there. I guess I keep forgetting that. Good lesson, I suppose, that that's an easy thing to forget if you find yourself computing one of these, where I'll put it in here, multiply that guy by 1\u20442. It's similar to a Taylor expansion in single variable calculus, where you kind of have to remember what's what that squared term would be, has a 1\u20442 associated with it. So for that same reason, now we're gonna have, and this time I won't forget it, will be 1\u20442 multiplied by something multiplied by the y minus pi halves minus that y coordinate of the point we're approximating near. And this time that something is negative one. So we can kind of plug in here negative one."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "It's similar to a Taylor expansion in single variable calculus, where you kind of have to remember what's what that squared term would be, has a 1\u20442 associated with it. So for that same reason, now we're gonna have, and this time I won't forget it, will be 1\u20442 multiplied by something multiplied by the y minus pi halves minus that y coordinate of the point we're approximating near. And this time that something is negative one. So we can kind of plug in here negative one. And now this is something we can simplify quite a bit because that one stays there, 1\u20442 of x minus zero, that's just x halves. This whole part cancels out to zero, so there's nothing there. Over here we have 1\u20442 times 1\u20444, 1\u20448 times x squared."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So we can kind of plug in here negative one. And now this is something we can simplify quite a bit because that one stays there, 1\u20442 of x minus zero, that's just x halves. This whole part cancels out to zero, so there's nothing there. Over here we have 1\u20442 times 1\u20444, 1\u20448 times x squared. So that's x squared divided by eight. This mixed partial derivative term is zero, so that's pretty nice. And then this last term here is just negative 1\u20442."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "Over here we have 1\u20442 times 1\u20444, 1\u20448 times x squared. So that's x squared divided by eight. This mixed partial derivative term is zero, so that's pretty nice. And then this last term here is just negative 1\u20442. So let's see, I'll write it down as negative 1\u20442 by y minus pi halves squared. By y minus pi halves squared. So that is the quadratic approximation."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And then this last term here is just negative 1\u20442. So let's see, I'll write it down as negative 1\u20442 by y minus pi halves squared. By y minus pi halves squared. So that is the quadratic approximation. And you can see this actually feels like a quadratic function. We've got up to x squared and up to y squared. And there's a sense in which this is a simpler function."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So that is the quadratic approximation. And you can see this actually feels like a quadratic function. We've got up to x squared and up to y squared. And there's a sense in which this is a simpler function. I mean, it looks like it's got more terms than the original one, which was e to the x halves sine of y. But if it's a computer that needs to compute these things, for example, it's much easier to deal with polynomials. That's a faster thing to do."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And there's a sense in which this is a simpler function. I mean, it looks like it's got more terms than the original one, which was e to the x halves sine of y. But if it's a computer that needs to compute these things, for example, it's much easier to deal with polynomials. That's a faster thing to do. Also, for theoretical purposes, it can be nice to deal with just a quadratic polynomial to make conclusions about things. We'll see that in the context of something called the second partial derivative test. But just to get a feel for what this means, let's pull up the graph of the relevant functions."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "That's a faster thing to do. Also, for theoretical purposes, it can be nice to deal with just a quadratic polynomial to make conclusions about things. We'll see that in the context of something called the second partial derivative test. But just to get a feel for what this means, let's pull up the graph of the relevant functions. So this here is the graph of the original function, e to the x halves times sine of y. And the point that we're approximating near was where x equals zero. So let's see how we get oriented."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "But just to get a feel for what this means, let's pull up the graph of the relevant functions. So this here is the graph of the original function, e to the x halves times sine of y. And the point that we're approximating near was where x equals zero. So let's see how we get oriented. X is equal to zero, and then y is equal to pi halves. So this is the point we're approximating near. And the quadratic approximation, when you plug everything in, has a graph that looks like this white surface here."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "So let's see how we get oriented. X is equal to zero, and then y is equal to pi halves. So this is the point we're approximating near. And the quadratic approximation, when you plug everything in, has a graph that looks like this white surface here. So if I get rid of that original graph, this is how we're approximating the function near that point and that does a pretty good job, right? Because even as you step pretty far away from that point, it's pretty closely hugging the original surface. If you go very far away, you know, it certainly doesn't get the oscillating nature of that sine component."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And the quadratic approximation, when you plug everything in, has a graph that looks like this white surface here. So if I get rid of that original graph, this is how we're approximating the function near that point and that does a pretty good job, right? Because even as you step pretty far away from that point, it's pretty closely hugging the original surface. If you go very far away, you know, it certainly doesn't get the oscillating nature of that sine component. And the exponential component grows faster than the quadratic one. But nearby, this actually gives a very good feel for the shape of the graph. And again, later on, we'll see how this is a pretty useful theoretical tool for drawing conclusions about qualitative features of the shape of the graph."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "If you go very far away, you know, it certainly doesn't get the oscillating nature of that sine component. And the exponential component grows faster than the quadratic one. But nearby, this actually gives a very good feel for the shape of the graph. And again, later on, we'll see how this is a pretty useful theoretical tool for drawing conclusions about qualitative features of the shape of the graph. The fact that this looks kind of like a saddle is gonna end up being kind of important in certain contexts. But before we get to any of that, in the next couple of videos, I'm gonna talk about a simpler or rather a more generalizable form of writing down this quadratic approximation using vector notation. Because right now we're just limited to, you know, two variables."}, {"video_title": "Quadratic approximation example.mp3", "Sentence": "And again, later on, we'll see how this is a pretty useful theoretical tool for drawing conclusions about qualitative features of the shape of the graph. The fact that this looks kind of like a saddle is gonna end up being kind of important in certain contexts. But before we get to any of that, in the next couple of videos, I'm gonna talk about a simpler or rather a more generalizable form of writing down this quadratic approximation using vector notation. Because right now we're just limited to, you know, two variables. And you can imagine how monstrous this might look if you were dealing even just with a three variable function, right? Where, think of all the different possible second partial derivatives of a three variable function or a four variable function. It would quickly get out of hand."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So where we left off, we had this two dimensional vector field V, and I have it pictured here as kind of a yellow vector field, and I just stuck it in three dimensions in kind of an awkward way where I put it on the xy plane and said, pretend this is in three dimensions. And then when you describe the rotation around each point, what we were familiar with is 2D curl, that's where you get this vector field. It's not quite a 3D vector field because you're only assigning points on the xy plane to three dimensional vectors, rather than every point in space to a vector, but we're getting there. So here, let's actually extend this to a fully three dimensional vector field. And first of all, let me just kind of clear up the board from the computations that we did in the last part. And as I do that, kind of start thinking about how you might want to extend the vector field that I have here that's pretty much two dimensional into three dimensions. And one idea you might have, so kind of get rid of the circles and the plane, is to take this vector field and then just kind of copy it into different slices."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So here, let's actually extend this to a fully three dimensional vector field. And first of all, let me just kind of clear up the board from the computations that we did in the last part. And as I do that, kind of start thinking about how you might want to extend the vector field that I have here that's pretty much two dimensional into three dimensions. And one idea you might have, so kind of get rid of the circles and the plane, is to take this vector field and then just kind of copy it into different slices. So, you might get something kind of like this. And I've drawn each slice a little bit sparser than the original one, so technically, that original one, if you look on the xy plane, I've pictured many more vectors, but it's really the same vector field. And all I've done here is said that every slice in space, just copy that same vector field."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And one idea you might have, so kind of get rid of the circles and the plane, is to take this vector field and then just kind of copy it into different slices. So, you might get something kind of like this. And I've drawn each slice a little bit sparser than the original one, so technically, that original one, if you look on the xy plane, I've pictured many more vectors, but it's really the same vector field. And all I've done here is said that every slice in space, just copy that same vector field. So if you look from above, you can maybe see how really it's just the same vector field kind of copied a bunch. And if you look at each slice, in the same way that on the xy plane, you've got this vector field sitting on a slice, every other part of space will have that. And even though there's only, what, like six or seven slices displayed here, in principle, you're thinking that every one of those infinitely many slices of space has a copy of this vector field."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And all I've done here is said that every slice in space, just copy that same vector field. So if you look from above, you can maybe see how really it's just the same vector field kind of copied a bunch. And if you look at each slice, in the same way that on the xy plane, you've got this vector field sitting on a slice, every other part of space will have that. And even though there's only, what, like six or seven slices displayed here, in principle, you're thinking that every one of those infinitely many slices of space has a copy of this vector field. And in a formula, what does that mean? Well, what it means is we're taking not just x and y as input points, but we're gonna start taking z in as well. So if I go, I'm gonna say that z is an input point as well."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And even though there's only, what, like six or seven slices displayed here, in principle, you're thinking that every one of those infinitely many slices of space has a copy of this vector field. And in a formula, what does that mean? Well, what it means is we're taking not just x and y as input points, but we're gonna start taking z in as well. So if I go, I'm gonna say that z is an input point as well. And I wanna be considering these as vectors in three dimensions, so rather than just saying that it's got x and y components, I'm gonna pretend like it has a z component, it has a z component that just happens to be zero for this case. And the fact that you have a z in the input, but the output doesn't depend on the z, corresponds to the fact that all the slices are the same. As you change the z direction, the vectors won't change at all, they're just carbon copies of each other."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So if I go, I'm gonna say that z is an input point as well. And I wanna be considering these as vectors in three dimensions, so rather than just saying that it's got x and y components, I'm gonna pretend like it has a z component, it has a z component that just happens to be zero for this case. And the fact that you have a z in the input, but the output doesn't depend on the z, corresponds to the fact that all the slices are the same. As you change the z direction, the vectors won't change at all, they're just carbon copies of each other. And the fact that this output has a z component, but it just happens to be zero, is what corresponds to the fact that it's very flat looking. You know, none of them point up or down in the z direction, they're all purely x and y. So as three-dimensional vector fields go, this one is only barely a three-dimensional vector field, it's kind of phoning it in as far as three-dimensional vector fields are concerned."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "As you change the z direction, the vectors won't change at all, they're just carbon copies of each other. And the fact that this output has a z component, but it just happens to be zero, is what corresponds to the fact that it's very flat looking. You know, none of them point up or down in the z direction, they're all purely x and y. So as three-dimensional vector fields go, this one is only barely a three-dimensional vector field, it's kind of phoning it in as far as three-dimensional vector fields are concerned. But it'll be quite good for our example here. Because now, if we start thinking of this as representing a three-dimensional fluid flow, so now rather than just kind of the fluid flow like the one I have pictured over here, where you've got water molecules moving in two dimensions and it's very easy to understand clockwise rotation, counterclockwise rotation, things like that, whereas over here, it's a very kind of chaotic three-dimensional fluid flow. But because it's so flat, if you view it from above, it's still loosely the same, just kind of counterclockwise over here on the right, and clockwise up there above."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So as three-dimensional vector fields go, this one is only barely a three-dimensional vector field, it's kind of phoning it in as far as three-dimensional vector fields are concerned. But it'll be quite good for our example here. Because now, if we start thinking of this as representing a three-dimensional fluid flow, so now rather than just kind of the fluid flow like the one I have pictured over here, where you've got water molecules moving in two dimensions and it's very easy to understand clockwise rotation, counterclockwise rotation, things like that, whereas over here, it's a very kind of chaotic three-dimensional fluid flow. But because it's so flat, if you view it from above, it's still loosely the same, just kind of counterclockwise over here on the right, and clockwise up there above. So if I were to draw like a column, you could think of this column as being, having a tornado of fluid flow, right? Where everything is kind of rotating together in that same direction. So if you were to assign a vector to each point in space to describe the kind of rotation happening around each one of those points in space, you would expect that those inside this column, inside this sort of counterclockwise rotating tornado, and I say counterclockwise, but if we viewed it from below, it would look clockwise."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "But because it's so flat, if you view it from above, it's still loosely the same, just kind of counterclockwise over here on the right, and clockwise up there above. So if I were to draw like a column, you could think of this column as being, having a tornado of fluid flow, right? Where everything is kind of rotating together in that same direction. So if you were to assign a vector to each point in space to describe the kind of rotation happening around each one of those points in space, you would expect that those inside this column, inside this sort of counterclockwise rotating tornado, and I say counterclockwise, but if we viewed it from below, it would look clockwise. So that's the tricky part about three dimensions and why we need to describe it with vectors. But you would expect these using your right-hand rule, where you curl the fingers of your right hand around the direction of rotation here, you would expect vectors that point up in the z direction, the positive z direction. And if I do that, if I show what all of the rotation vectors look like, you'll get this."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So if you were to assign a vector to each point in space to describe the kind of rotation happening around each one of those points in space, you would expect that those inside this column, inside this sort of counterclockwise rotating tornado, and I say counterclockwise, but if we viewed it from below, it would look clockwise. So that's the tricky part about three dimensions and why we need to describe it with vectors. But you would expect these using your right-hand rule, where you curl the fingers of your right hand around the direction of rotation here, you would expect vectors that point up in the z direction, the positive z direction. And if I do that, if I show what all of the rotation vectors look like, you'll get this. And maybe this is kind of a mess because there's a lot of things on the screen at this point. So for the moment, I'll kind of remove that original vector field and remove the xy plane, and just kind of focus on this new vector field that I have pictured here. Inside that column, where we had that tornado of rotation I was describing, all of the vectors point in the positive z direction."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And if I do that, if I show what all of the rotation vectors look like, you'll get this. And maybe this is kind of a mess because there's a lot of things on the screen at this point. So for the moment, I'll kind of remove that original vector field and remove the xy plane, and just kind of focus on this new vector field that I have pictured here. Inside that column, where we had that tornado of rotation I was describing, all of the vectors point in the positive z direction. But if we were to view it elsewhere, like over in this region, those are pointing in the negative z direction. And if you stick your thumb in the direction of all of these vectors in the negative z direction, that tells you the direction of, that tells you how the fluid, maybe you're thinking of it as air kind of rushing about the room, how that fluid rotates in three dimensions. So what curl is gonna do, here I'll kind of clear things up, I have the formula from last time that hopefully hasn't looked too in the way while I've been doing this, but it described curl for a two-dimensional vector field."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "Inside that column, where we had that tornado of rotation I was describing, all of the vectors point in the positive z direction. But if we were to view it elsewhere, like over in this region, those are pointing in the negative z direction. And if you stick your thumb in the direction of all of these vectors in the negative z direction, that tells you the direction of, that tells you how the fluid, maybe you're thinking of it as air kind of rushing about the room, how that fluid rotates in three dimensions. So what curl is gonna do, here I'll kind of clear things up, I have the formula from last time that hopefully hasn't looked too in the way while I've been doing this, but it described curl for a two-dimensional vector field. If we imagine this not just taking x and y as its inputs, because it's a three-dimensional vector field, but if we imagine it taking x, y, and z, so it's a proper three-dimensional vector field, the output is gonna tell you at every point in space what the rotation that corresponds to that point is. And in the next video, I'm gonna give you the formula and tell you how you actually compute this curl given an arbitrary function, but for right now, we're just getting the visual intuition, we're just trying to understand what it is that curl's going to represent. And in this vector field, this one that was just kind of copies of a 2D one put above, it's almost contrived because all of the rotation happens in these perfect, perfect tornado-like patterns that doesn't really change as you move up and down in the x, y direction."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "So what curl is gonna do, here I'll kind of clear things up, I have the formula from last time that hopefully hasn't looked too in the way while I've been doing this, but it described curl for a two-dimensional vector field. If we imagine this not just taking x and y as its inputs, because it's a three-dimensional vector field, but if we imagine it taking x, y, and z, so it's a proper three-dimensional vector field, the output is gonna tell you at every point in space what the rotation that corresponds to that point is. And in the next video, I'm gonna give you the formula and tell you how you actually compute this curl given an arbitrary function, but for right now, we're just getting the visual intuition, we're just trying to understand what it is that curl's going to represent. And in this vector field, this one that was just kind of copies of a 2D one put above, it's almost contrived because all of the rotation happens in these perfect, perfect tornado-like patterns that doesn't really change as you move up and down in the x, y direction. But more generally, you might have a more complicated-looking vector field. So I'll go ahead and kind of finally erase this since it's been a little bit in the way for a while, and erase this guy too. And if you think about just arbitrary three-dimensional vector fields, like let's say this one that I have here, so I don't know about you, but for me, it's really hard to think about the fluid flow associated with this."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And in this vector field, this one that was just kind of copies of a 2D one put above, it's almost contrived because all of the rotation happens in these perfect, perfect tornado-like patterns that doesn't really change as you move up and down in the x, y direction. But more generally, you might have a more complicated-looking vector field. So I'll go ahead and kind of finally erase this since it's been a little bit in the way for a while, and erase this guy too. And if you think about just arbitrary three-dimensional vector fields, like let's say this one that I have here, so I don't know about you, but for me, it's really hard to think about the fluid flow associated with this. I have a vague notion in my mind that okay, like fluid is flowing out from this corner and kind of flowing in here, but it's very hard to think about it all at once. And certainly if you start talking about rotation, it's hard to look at a given point and say, oh yeah, there's gonna be a general fluid rotation in some certain way, and I can give you the vector for that. But as a more loose and vague idea, I can say, okay, given that there's some kind of crazy air current fluid flow happening around here, I can maybe understand that at a specific point, you're gonna have some kind of rotation."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And if you think about just arbitrary three-dimensional vector fields, like let's say this one that I have here, so I don't know about you, but for me, it's really hard to think about the fluid flow associated with this. I have a vague notion in my mind that okay, like fluid is flowing out from this corner and kind of flowing in here, but it's very hard to think about it all at once. And certainly if you start talking about rotation, it's hard to look at a given point and say, oh yeah, there's gonna be a general fluid rotation in some certain way, and I can give you the vector for that. But as a more loose and vague idea, I can say, okay, given that there's some kind of crazy air current fluid flow happening around here, I can maybe understand that at a specific point, you're gonna have some kind of rotation. And here, I'll picture it as if there's like a ball or a globe sitting there in space, and maybe you imagine your new vector field and saying what kind of rotation is it gonna induce in that ball that's just floating there in space. So maybe you're imagining this as like a tennis ball that you're sort of holding in place in space using magnets or magic or something like that, and you're letting the wind sort of freely rotate it, and you're wondering what direction it tends to rotate. And then when it does, and once you have this rotation, you can describe that 3D rotation with some kind of vector."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "But as a more loose and vague idea, I can say, okay, given that there's some kind of crazy air current fluid flow happening around here, I can maybe understand that at a specific point, you're gonna have some kind of rotation. And here, I'll picture it as if there's like a ball or a globe sitting there in space, and maybe you imagine your new vector field and saying what kind of rotation is it gonna induce in that ball that's just floating there in space. So maybe you're imagining this as like a tennis ball that you're sort of holding in place in space using magnets or magic or something like that, and you're letting the wind sort of freely rotate it, and you're wondering what direction it tends to rotate. And then when it does, and once you have this rotation, you can describe that 3D rotation with some kind of vector. And in this case, it would be a vector that points out in that direction because we're kind of curling our fingers, curling our right hand fingers over in that direction. And if you don't understand how we describe 3D rotation with a vector, I have a video on that. Maybe go back and check out that video."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "And then when it does, and once you have this rotation, you can describe that 3D rotation with some kind of vector. And in this case, it would be a vector that points out in that direction because we're kind of curling our fingers, curling our right hand fingers over in that direction. And if you don't understand how we describe 3D rotation with a vector, I have a video on that. Maybe go back and check out that video. But the idea here is that when you have some sort of crazy fluid flow that's induced by some sort of vector field, and you do this at every point and say, hey, what's the rotation at every single point? That's gonna give you the curl. That is what the curl of a three-dimensional vector field is trying to represent."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "Maybe go back and check out that video. But the idea here is that when you have some sort of crazy fluid flow that's induced by some sort of vector field, and you do this at every point and say, hey, what's the rotation at every single point? That's gonna give you the curl. That is what the curl of a three-dimensional vector field is trying to represent. And if this feels confusing, if this feels like something that's hard to wrap your mind around, don't worry, we've all been there. 3D curl is one of the most complicated things in multivariable calculus that we have to describe. But I think the key to understanding it is to just kind of patiently think through and take the time to think about what 2D curl is and start thinking about how you extend that to three dimensions and slowly say, okay, okay, I kind of get it, and tornadoes of rotation, that sort of makes sense."}, {"video_title": "3d curl intuition, part 2.mp3", "Sentence": "That is what the curl of a three-dimensional vector field is trying to represent. And if this feels confusing, if this feels like something that's hard to wrap your mind around, don't worry, we've all been there. 3D curl is one of the most complicated things in multivariable calculus that we have to describe. But I think the key to understanding it is to just kind of patiently think through and take the time to think about what 2D curl is and start thinking about how you extend that to three dimensions and slowly say, okay, okay, I kind of get it, and tornadoes of rotation, that sort of makes sense. And if you understand how to represent three-dimensional rotation around a single point with a vector, then understanding three-dimensional curl comes down to thinking about doing that at every single point in space according to whatever rotation the wind flow around that point would induce. But like I said, it is complicated, and it's okay if it doesn't sink the first time. It certainly took me a while to really wrap my head around this 3D curl idea."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And we ended up seeing how specifying that orientation comes down to certain partial derivative information. And first let me just kind of repeat what the conclusion was, but I'll put it in more abstract terms since I did it in a very specific example last time. So basically, if you want some kind of function which gives you a plane that passes through a certain point, well first let's say what that point is, right? Let's say the point was x-naught, y-naught, and z-naught. So these are just constant values, and this is my way of abstractly describing a single point in space, using x-naught to represent a constant x value, y-naught to represent constant y value, that kind of thing. So what it is, is it's gonna be some sort of other constant, a, multiplied by x minus x-naught. So this white x here is the variable, and then x-naught is just a constant."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Let's say the point was x-naught, y-naught, and z-naught. So these are just constant values, and this is my way of abstractly describing a single point in space, using x-naught to represent a constant x value, y-naught to represent constant y value, that kind of thing. So what it is, is it's gonna be some sort of other constant, a, multiplied by x minus x-naught. So this white x here is the variable, and then x-naught is just a constant. Let me go ahead and make that parentheses there. Okay, so then we add to that b multiplied by, and then b is just some other constant, just like a is some other constant, multiplied by y minus y-naught. And then all of that you add z-naught."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So this white x here is the variable, and then x-naught is just a constant. Let me go ahead and make that parentheses there. Okay, so then we add to that b multiplied by, and then b is just some other constant, just like a is some other constant, multiplied by y minus y-naught. And then all of that you add z-naught. And now if you just presented this as it is, it's kind of a lot, right? There's five different constants going on. But really what this is saying is you want something where the partial derivative with respect to x is just some constant, and you wanna be able to specify what that constant is."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And then all of that you add z-naught. And now if you just presented this as it is, it's kind of a lot, right? There's five different constants going on. But really what this is saying is you want something where the partial derivative with respect to x is just some constant, and you wanna be able to specify what that constant is. And similarly, the partial derivative with respect to y is another constant. And you just wanna ensure that this passes through this point, x-naught, y-naught, z-naught. And if you imagine plugging in x, the variable, equals x-naught, the constant, this part goes to zero."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "But really what this is saying is you want something where the partial derivative with respect to x is just some constant, and you wanna be able to specify what that constant is. And similarly, the partial derivative with respect to y is another constant. And you just wanna ensure that this passes through this point, x-naught, y-naught, z-naught. And if you imagine plugging in x, the variable, equals x-naught, the constant, this part goes to zero. Similarly, plugging in y-naught, the constant, makes this part go to zero. So this is a way of specifying that when you evaluate the function at x-naught, y-naught, it equals z-naught. And that's what makes sure that the graph passes through that point."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And if you imagine plugging in x, the variable, equals x-naught, the constant, this part goes to zero. Similarly, plugging in y-naught, the constant, makes this part go to zero. So this is a way of specifying that when you evaluate the function at x-naught, y-naught, it equals z-naught. And that's what makes sure that the graph passes through that point. So with that said, let's start thinking about how you confine the tangent plane to a graph. And first of all, let's think about what that point is, how you specify such a point. Instead of specifying any three numbers in space, because you have to make sure the point is somewhere on the graph, you instead only specify two."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And that's what makes sure that the graph passes through that point. So with that said, let's start thinking about how you confine the tangent plane to a graph. And first of all, let's think about what that point is, how you specify such a point. Instead of specifying any three numbers in space, because you have to make sure the point is somewhere on the graph, you instead only specify two. You're basically gonna say what's the x-coordinate, and in this case, let's say the x-coordinate was like one, that'd be like one, and then the y-coordinate, which looks about like negative two. To make it easier, I'm just gonna say, let's say it is negative two. Then the z-coordinate is specified, because this is a graph."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Instead of specifying any three numbers in space, because you have to make sure the point is somewhere on the graph, you instead only specify two. You're basically gonna say what's the x-coordinate, and in this case, let's say the x-coordinate was like one, that'd be like one, and then the y-coordinate, which looks about like negative two. To make it easier, I'm just gonna say, let's say it is negative two. Then the z-coordinate is specified, because this is a graph. The z-coordinate is forced to be whatever the output of the function is at one, negative two. So this is gonna be whatever the output of our function is at one, negative two. And f here, f is gonna be whatever function gives us this graph."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Then the z-coordinate is specified, because this is a graph. The z-coordinate is forced to be whatever the output of the function is at one, negative two. So this is gonna be whatever the output of our function is at one, negative two. And f here, f is gonna be whatever function gives us this graph. So maybe I should write down the actual function that I'm using for this graph. In this case, f, which is a function of x and y, is equal to three minus 1 3rd of x squared minus y squared. Okay, so this is the function that we're using, and you evaluate it at that point, and this will give you your point in three-dimensional space that our linear function, that our tangent plane has to pass through."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And f here, f is gonna be whatever function gives us this graph. So maybe I should write down the actual function that I'm using for this graph. In this case, f, which is a function of x and y, is equal to three minus 1 3rd of x squared minus y squared. Okay, so this is the function that we're using, and you evaluate it at that point, and this will give you your point in three-dimensional space that our linear function, that our tangent plane has to pass through. So we can start writing out our function, right? We can say, okay, so our linear function has a function of x and y. It's gotta make sure it goes through that one and that negative two."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Okay, so this is the function that we're using, and you evaluate it at that point, and this will give you your point in three-dimensional space that our linear function, that our tangent plane has to pass through. So we can start writing out our function, right? We can say, okay, so our linear function has a function of x and y. It's gotta make sure it goes through that one and that negative two. So this is gonna be some constant a that we'll fill in in a moment, multiplied by x minus that one, plus, and then b, also a constant that we'll specify in a moment, times y minus that negative two. So it's minus a negative two. And then the whole, the thing that we add to it is f of one, negative two."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "It's gotta make sure it goes through that one and that negative two. So this is gonna be some constant a that we'll fill in in a moment, multiplied by x minus that one, plus, and then b, also a constant that we'll specify in a moment, times y minus that negative two. So it's minus a negative two. And then the whole, the thing that we add to it is f of one, negative two. And let's just go ahead and evaluate that. Let's say we plug in one and negative two. So if we go up here and we plug in, so that would be three minus x equals one, whoop, three minus 1 3rd of, if x equals one, 1 3rd of one squared."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And then the whole, the thing that we add to it is f of one, negative two. And let's just go ahead and evaluate that. Let's say we plug in one and negative two. So if we go up here and we plug in, so that would be three minus x equals one, whoop, three minus 1 3rd of, if x equals one, 1 3rd of one squared. So that's 1 3rd, one squared, minus, and then y is negative two. So that will be minus negative two squared. So that's three minus 1 3rd, minus four."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So if we go up here and we plug in, so that would be three minus x equals one, whoop, three minus 1 3rd of, if x equals one, 1 3rd of one squared. So that's 1 3rd, one squared, minus, and then y is negative two. So that will be minus negative two squared. So that's three minus 1 3rd, minus four. So the whole thing is equal to, let's see, three minus four is negative one, minus another 1 3rd is negative 4 3rds. Okay, so that's what we add to this entire thing. We add negative 4 3rds."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So that's three minus 1 3rd, minus four. So the whole thing is equal to, let's see, three minus four is negative one, minus another 1 3rd is negative 4 3rds. Okay, so that's what we add to this entire thing. We add negative 4 3rds. And maybe I should just kind of make clear the separation here. So this is our function, but we don't know what a and b are. Those are things that we need to plug in."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "We add negative 4 3rds. And maybe I should just kind of make clear the separation here. So this is our function, but we don't know what a and b are. Those are things that we need to plug in. Now the whole idea of a tangent plane is that the partial derivative with respect to x should match that of the original function. So if we go over to the graph here and start thinking about partial derivative information, if we want the partial derivative with respect to x, and you imagine moving purely in the x direction here, this intersects the graph along some kind of curve. And what the partial derivative with respect to x at this point tells you is the slope of the tangent line."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Those are things that we need to plug in. Now the whole idea of a tangent plane is that the partial derivative with respect to x should match that of the original function. So if we go over to the graph here and start thinking about partial derivative information, if we want the partial derivative with respect to x, and you imagine moving purely in the x direction here, this intersects the graph along some kind of curve. And what the partial derivative with respect to x at this point tells you is the slope of the tangent line. I'm kind of bad at drawing there. Is the slope of the tangent line in that direction of that point. So that's what the partial derivative with respect to x is telling you."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And what the partial derivative with respect to x at this point tells you is the slope of the tangent line. I'm kind of bad at drawing there. Is the slope of the tangent line in that direction of that point. So that's what the partial derivative with respect to x is telling you. And what you want when you look at the tangent plane is that the tangent plane also has that same slope. You know, you kind of, if I line things up here, you'd want it also to have that same slope. So you can specify over here and say a, we want a to be equal to the partial derivative of the function with respect to x evaluated at this one negative two, evaluated at that special point, one negative two."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So that's what the partial derivative with respect to x is telling you. And what you want when you look at the tangent plane is that the tangent plane also has that same slope. You know, you kind of, if I line things up here, you'd want it also to have that same slope. So you can specify over here and say a, we want a to be equal to the partial derivative of the function with respect to x evaluated at this one negative two, evaluated at that special point, one negative two. And then similarly, b, for pretty much the same reasons, and I'll draw it out here, so let's kind of, ah, ah, let's erase this line. So instead of intersecting it with that slice, let's see what movement in the y direction looks like. So in this case, it looks like a very steep slope, right?"}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So you can specify over here and say a, we want a to be equal to the partial derivative of the function with respect to x evaluated at this one negative two, evaluated at that special point, one negative two. And then similarly, b, for pretty much the same reasons, and I'll draw it out here, so let's kind of, ah, ah, let's erase this line. So instead of intersecting it with that slice, let's see what movement in the y direction looks like. So in this case, it looks like a very steep slope, right? Because in this case, the tangent line in that direction is a pretty steep slope. And now when we bring in, when we bring in the tangent plane, it should intersect with that constant x value plane along that same slope. Made it kind of messy there, but you can see that the line formed by intersecting these two planes should be that desired tangent."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So in this case, it looks like a very steep slope, right? Because in this case, the tangent line in that direction is a pretty steep slope. And now when we bring in, when we bring in the tangent plane, it should intersect with that constant x value plane along that same slope. Made it kind of messy there, but you can see that the line formed by intersecting these two planes should be that desired tangent. And what that corresponds to in formulas is that this b, which represents the partial derivative of L, L is the tangent plane function, that should be the same as if we take the partial derivative of f with respect to y at that point, at this point, one, negative two. And this is stuff that we can compute and that we can figure out. So let's just kind of start plugging that in."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Made it kind of messy there, but you can see that the line formed by intersecting these two planes should be that desired tangent. And what that corresponds to in formulas is that this b, which represents the partial derivative of L, L is the tangent plane function, that should be the same as if we take the partial derivative of f with respect to y at that point, at this point, one, negative two. And this is stuff that we can compute and that we can figure out. So let's just kind of start plugging that in. First, let me just copy this function because we're gonna need it. Copy, and now let's go on down here. I'm just gonna, let's paste it down here, kind of in the bottom, because that's what we'll need."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So let's just kind of start plugging that in. First, let me just copy this function because we're gonna need it. Copy, and now let's go on down here. I'm just gonna, let's paste it down here, kind of in the bottom, because that's what we'll need. So let's compute the partial derivative of f with respect to x. So we look down here, the only place where x shows up is in this negative 1 3rd of x squared context. So the partial derivative of f with respect to x is gonna be just the derivative of this little guy, which is negative, we bring down the two, negative 2 3rds of x."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "I'm just gonna, let's paste it down here, kind of in the bottom, because that's what we'll need. So let's compute the partial derivative of f with respect to x. So we look down here, the only place where x shows up is in this negative 1 3rd of x squared context. So the partial derivative of f with respect to x is gonna be just the derivative of this little guy, which is negative, we bring down the two, negative 2 3rds of x. So when we go ahead and plug in, you know, x equals one, to see what it looks like when we evaluate at this point, that's just gonna be equal to negative 2 3rds. So that tells us that a is gonna have to be negative 2 3rds. Now for similar reasons, let's go ahead and compute the partial derivative with respect to y."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "So the partial derivative of f with respect to x is gonna be just the derivative of this little guy, which is negative, we bring down the two, negative 2 3rds of x. So when we go ahead and plug in, you know, x equals one, to see what it looks like when we evaluate at this point, that's just gonna be equal to negative 2 3rds. So that tells us that a is gonna have to be negative 2 3rds. Now for similar reasons, let's go ahead and compute the partial derivative with respect to y. We look down here, well the only place that y shows up in the entire expression is this negative y squared. So the partial derivative of f with respect to y is equal to just negative 2y, negative 2y. And now when we plug in y equals negative 2, what we get is negative 2 multiplied coincidentally by negative 2, that didn't have to be the case that those were the same, and that whole thing equals 4."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "Now for similar reasons, let's go ahead and compute the partial derivative with respect to y. We look down here, well the only place that y shows up in the entire expression is this negative y squared. So the partial derivative of f with respect to y is equal to just negative 2y, negative 2y. And now when we plug in y equals negative 2, what we get is negative 2 multiplied coincidentally by negative 2, that didn't have to be the case that those were the same, and that whole thing equals 4. So the partial derivative of f with respect to y evaluated at this point, 1 negative 2, is equal to 4. So if we were to plug this information back up into our formula, we would replace a with negative 2 3rds. We would say negative 2 3rds."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And now when we plug in y equals negative 2, what we get is negative 2 multiplied coincidentally by negative 2, that didn't have to be the case that those were the same, and that whole thing equals 4. So the partial derivative of f with respect to y evaluated at this point, 1 negative 2, is equal to 4. So if we were to plug this information back up into our formula, we would replace a with negative 2 3rds. We would say negative 2 3rds. And we would replace b with 4. We would replace b with 4. And that would give us the full formula, the full formula for the tangent plane."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "We would say negative 2 3rds. And we would replace b with 4. We would replace b with 4. And that would give us the full formula, the full formula for the tangent plane. And this could be kind of a lot to look at at first, because we have to specify the input point, you know, 1 negative 2, and then we had to figure out where the function evaluates at that point. And then we had to figure out both of the partial derivatives with respect to x and with respect to y. But all in all, there's not actually a lot to remember for how you go about computing this."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "And that would give us the full formula, the full formula for the tangent plane. And this could be kind of a lot to look at at first, because we have to specify the input point, you know, 1 negative 2, and then we had to figure out where the function evaluates at that point. And then we had to figure out both of the partial derivatives with respect to x and with respect to y. But all in all, there's not actually a lot to remember for how you go about computing this. Looking at the graph actually makes things seem a lot more reasonable, because each of those terms has an actual meaning. If we look at the 1 and negative 2, that's just telling us the input, the kind of x and y coordinates of the input. And of course we have to evaluate that, because that tells us the z coordinate that'll put us on the graph corresponding to that point."}, {"video_title": "Computing a tangent plane.mp3", "Sentence": "But all in all, there's not actually a lot to remember for how you go about computing this. Looking at the graph actually makes things seem a lot more reasonable, because each of those terms has an actual meaning. If we look at the 1 and negative 2, that's just telling us the input, the kind of x and y coordinates of the input. And of course we have to evaluate that, because that tells us the z coordinate that'll put us on the graph corresponding to that point. And then to get us a tangent plane, you just need to specify the two bits of partial differential information, and that'll tell you kind of how this graph needs to be oriented. And once you start thinking of things in that way, you know, geometrically, even though there's a lot going on here, there's five different numbers you have to put in, each one of them feels like, yeah, yeah, of course you need that number, otherwise you couldn't specify a tangent plane. There's kind of a lot of information required to put it on the appropriate spot."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "It's a very nonlinear function, and we were picturing it as a transformation that takes every point x, y in space to the point x plus sine y, y plus sine of x. And moreover, we zoomed in on a specific point. And let me actually write down what point we zoomed in on. It was negative 2, 1. That's something we're going to want to record here. Negative 2, 1. And I added a couple extra grid lines around it just so we can see in detail what the transformation does to points that are in a neighborhood of that point."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "It was negative 2, 1. That's something we're going to want to record here. Negative 2, 1. And I added a couple extra grid lines around it just so we can see in detail what the transformation does to points that are in a neighborhood of that point. And over here, this square shows the zoomed-in version of that neighborhood. And what we saw was that even though the function as a whole, as a transformation, looks rather complicated, around that one point, it looks like a linear function. It's locally linear."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And I added a couple extra grid lines around it just so we can see in detail what the transformation does to points that are in a neighborhood of that point. And over here, this square shows the zoomed-in version of that neighborhood. And what we saw was that even though the function as a whole, as a transformation, looks rather complicated, around that one point, it looks like a linear function. It's locally linear. So what I'll show you here is what matrix is going to tell you the linear function that this looks like. And this is going to be some kind of 2x2 matrix. I'll make a lot of room for ourselves here."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "It's locally linear. So what I'll show you here is what matrix is going to tell you the linear function that this looks like. And this is going to be some kind of 2x2 matrix. I'll make a lot of room for ourselves here. It'll be a 2x2 matrix. And the way to think about it is to first go back to our original setup before the transformation and think of just a tiny step to the right, what I'm going to think of as a little partial x, a tiny step in the x direction. And what that turns into after the transformation is going to be some tiny step in the output space."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "I'll make a lot of room for ourselves here. It'll be a 2x2 matrix. And the way to think about it is to first go back to our original setup before the transformation and think of just a tiny step to the right, what I'm going to think of as a little partial x, a tiny step in the x direction. And what that turns into after the transformation is going to be some tiny step in the output space. And here, let me actually kind of draw on what that tiny step turned into. It's no longer purely in the x direction. It has some rightward component, but now also some downward component."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And what that turns into after the transformation is going to be some tiny step in the output space. And here, let me actually kind of draw on what that tiny step turned into. It's no longer purely in the x direction. It has some rightward component, but now also some downward component. And to be able to represent this in a nice way, what I'm going to do is instead of writing the entire function as something with a vector-valued output, I'm going to go ahead and represent this as two separate scalar-valued functions. I'm going to write the scalar-valued functions f1 of x, y. So I'm just giving a name to x plus sine y."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "It has some rightward component, but now also some downward component. And to be able to represent this in a nice way, what I'm going to do is instead of writing the entire function as something with a vector-valued output, I'm going to go ahead and represent this as two separate scalar-valued functions. I'm going to write the scalar-valued functions f1 of x, y. So I'm just giving a name to x plus sine y. And f2 of x, y. Again, all I'm doing is giving a name to the functions we already have written down. When I look at this vector, the consequence of taking a tiny dx step in the input space that corresponds to some 2D movement in the output space and the x component of that movement, right, if I was going to draw this out and say, hey, what's the x component of that movement?"}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "So I'm just giving a name to x plus sine y. And f2 of x, y. Again, all I'm doing is giving a name to the functions we already have written down. When I look at this vector, the consequence of taking a tiny dx step in the input space that corresponds to some 2D movement in the output space and the x component of that movement, right, if I was going to draw this out and say, hey, what's the x component of that movement? That's something we think of as a little partial change in f1. That's the x component of our output. And if we divide this, if we take, you know, partial f1 divided by the size of that initial tiny change, it basically scales it up to be a normal-sized vector, not a tiny nudge, but something that's more constant, that doesn't shrink as we zoom in further and further."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "When I look at this vector, the consequence of taking a tiny dx step in the input space that corresponds to some 2D movement in the output space and the x component of that movement, right, if I was going to draw this out and say, hey, what's the x component of that movement? That's something we think of as a little partial change in f1. That's the x component of our output. And if we divide this, if we take, you know, partial f1 divided by the size of that initial tiny change, it basically scales it up to be a normal-sized vector, not a tiny nudge, but something that's more constant, that doesn't shrink as we zoom in further and further. And then similarly, the change in the y direction, right, the vertical component of that step that was still caused by the dx, right, it's still caused by that initial step to the right, that is going to be the tiny partial change in f2. The y component of the output, because here we're all just looking in the output space, that was caused by a partial change in the x direction. And again, I kind of like to think about this, we're dividing by a tiny amount."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And if we divide this, if we take, you know, partial f1 divided by the size of that initial tiny change, it basically scales it up to be a normal-sized vector, not a tiny nudge, but something that's more constant, that doesn't shrink as we zoom in further and further. And then similarly, the change in the y direction, right, the vertical component of that step that was still caused by the dx, right, it's still caused by that initial step to the right, that is going to be the tiny partial change in f2. The y component of the output, because here we're all just looking in the output space, that was caused by a partial change in the x direction. And again, I kind of like to think about this, we're dividing by a tiny amount. This partial f2 is really a tiny, tiny nudge, but by dividing by the size of the initial tiny nudge that caused it, we're getting something that's basically a number, something that doesn't shrink when we consider more and more zoomed-in versions. So that, that's all what happens when we take a tiny step in the x direction. But another thing you could do, another thing you can consider, is a tiny step in the y direction, right?"}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And again, I kind of like to think about this, we're dividing by a tiny amount. This partial f2 is really a tiny, tiny nudge, but by dividing by the size of the initial tiny nudge that caused it, we're getting something that's basically a number, something that doesn't shrink when we consider more and more zoomed-in versions. So that, that's all what happens when we take a tiny step in the x direction. But another thing you could do, another thing you can consider, is a tiny step in the y direction, right? Because we want to know, hey, if you take a single step some tiny unit upward, what does that turn into after the transformation? And what that looks like, what that looks like is this vector that still has some upward component, but it also has a rightward component. And now I'm going to write its components as the second column of the matrix, because as we know, when you're representing a linear transformation with a matrix, the first column tells you where the first basis vector goes, and the second column shows where the second basis vector goes."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "But another thing you could do, another thing you can consider, is a tiny step in the y direction, right? Because we want to know, hey, if you take a single step some tiny unit upward, what does that turn into after the transformation? And what that looks like, what that looks like is this vector that still has some upward component, but it also has a rightward component. And now I'm going to write its components as the second column of the matrix, because as we know, when you're representing a linear transformation with a matrix, the first column tells you where the first basis vector goes, and the second column shows where the second basis vector goes. If that feels unfamiliar, either check out the refresher video, or maybe go and look at some of the linear algebra content. But to figure out the coordinates of this guy, we do basically the same thing. We say, first of all, the change in the x direction here, the x component of this nudge vector, that's going to be given as a partial change to f1, right, to the x component of the output."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And now I'm going to write its components as the second column of the matrix, because as we know, when you're representing a linear transformation with a matrix, the first column tells you where the first basis vector goes, and the second column shows where the second basis vector goes. If that feels unfamiliar, either check out the refresher video, or maybe go and look at some of the linear algebra content. But to figure out the coordinates of this guy, we do basically the same thing. We say, first of all, the change in the x direction here, the x component of this nudge vector, that's going to be given as a partial change to f1, right, to the x component of the output. Here we're looking in the output space, so we're dealing with f1 and f2. And we're asking what that change was that was caused by a tiny change in the y direction. So the change in f1 caused by some tiny step in the y direction divided by the size of that tiny step."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "We say, first of all, the change in the x direction here, the x component of this nudge vector, that's going to be given as a partial change to f1, right, to the x component of the output. Here we're looking in the output space, so we're dealing with f1 and f2. And we're asking what that change was that was caused by a tiny change in the y direction. So the change in f1 caused by some tiny step in the y direction divided by the size of that tiny step. And then the y component of our output here, the y component of the step in the output space that was caused by the initial tiny step upward in the input space, well, that is the change of f2, the second component of our output, as caused by dy, as caused by that little partial y. And of course, all of this is very specific to the point that we started at, right? We started at the point (-2,1)."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "So the change in f1 caused by some tiny step in the y direction divided by the size of that tiny step. And then the y component of our output here, the y component of the step in the output space that was caused by the initial tiny step upward in the input space, well, that is the change of f2, the second component of our output, as caused by dy, as caused by that little partial y. And of course, all of this is very specific to the point that we started at, right? We started at the point (-2,1). So each of these partial derivatives is something that really we're saying, don't take the function, evaluate it at the point (-2,1). And when you evaluate each one of these at the point (-2,1), you'll get some number. And that will give you a very concrete 2x2 matrix that's going to represent the linear transformation that this guy looks like once you've zoomed in."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "We started at the point (-2,1). So each of these partial derivatives is something that really we're saying, don't take the function, evaluate it at the point (-2,1). And when you evaluate each one of these at the point (-2,1), you'll get some number. And that will give you a very concrete 2x2 matrix that's going to represent the linear transformation that this guy looks like once you've zoomed in. So this matrix here that's full of all of the different partial derivatives has a very special name. It's called, as you may have guessed, the Jacobian. Or more fully, you'd call it the Jacobian matrix."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "And that will give you a very concrete 2x2 matrix that's going to represent the linear transformation that this guy looks like once you've zoomed in. So this matrix here that's full of all of the different partial derivatives has a very special name. It's called, as you may have guessed, the Jacobian. Or more fully, you'd call it the Jacobian matrix. And one way to think about it is that it carries all of the partial differential information, right? It's taking into account both of these components of the output and both possible inputs, and giving you kind of a grid of what all the partial derivatives are. But as I hope you see, it's much more than just a way of recording what all the partial derivatives are."}, {"video_title": "The Jacobian matrix.mp3", "Sentence": "Or more fully, you'd call it the Jacobian matrix. And one way to think about it is that it carries all of the partial differential information, right? It's taking into account both of these components of the output and both possible inputs, and giving you kind of a grid of what all the partial derivatives are. But as I hope you see, it's much more than just a way of recording what all the partial derivatives are. There's a reason for organizing it like this in particular. And it really does come down to this idea of local linearity. If you understand that the Jacobian matrix is fundamentally supposed to represent what a transformation looks like when you zoom in near a specific point, almost everything else about it will start to fall in place."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "Hello everyone. So I'm talking about how to find the tangent plane to a graph, and I think the first step of that is to just figure out how we control planes in three dimensions in the first place. So what I have pictured here is a red dot representing a point in three dimensions, and the coordinates of that point, easily enough, are one, two, three. So the x coordinate is one, the y coordinate's two, and the z coordinate is three. And then I have a plane that passes through it, and the goal of the video is going to be to find a function, a function that I'll call L, that takes in a two-dimensional input, whoop, x and y, and this function L should have this plane as its graph. Now the first thing to notice is that there's lots of different planes that could be passing through this point, right? At the moment it's one that's got a certain kind of angle, you could think of it going up in one direction, but you could give this a lot of different directions and get a lot of different planes that all pass through that one point."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So the x coordinate is one, the y coordinate's two, and the z coordinate is three. And then I have a plane that passes through it, and the goal of the video is going to be to find a function, a function that I'll call L, that takes in a two-dimensional input, whoop, x and y, and this function L should have this plane as its graph. Now the first thing to notice is that there's lots of different planes that could be passing through this point, right? At the moment it's one that's got a certain kind of angle, you could think of it going up in one direction, but you could give this a lot of different directions and get a lot of different planes that all pass through that one point. So we're going to need to find some way of distinguishing the specific one that we're looking at, which is this one right here, from other possible planes that can pass through it. And as we work through, you'll see how this is done in terms of partial derivatives. But as we are getting our head around what the formula for this graph can be, let's just start observing properties that it has."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "At the moment it's one that's got a certain kind of angle, you could think of it going up in one direction, but you could give this a lot of different directions and get a lot of different planes that all pass through that one point. So we're going to need to find some way of distinguishing the specific one that we're looking at, which is this one right here, from other possible planes that can pass through it. And as we work through, you'll see how this is done in terms of partial derivatives. But as we are getting our head around what the formula for this graph can be, let's just start observing properties that it has. The first property is that the graph actually passes through this point, one, two, three. And what that means, in terms of functions over here, is that if you evaluate it at the point one, two, the input pair where x is one and y is two, then it should equal three. It should equal three, because that's telling you that when you go to x equals one and y equals two, and then you say, what's the height of the graph above that point, it should be the z-coordinate of the desired point."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "But as we are getting our head around what the formula for this graph can be, let's just start observing properties that it has. The first property is that the graph actually passes through this point, one, two, three. And what that means, in terms of functions over here, is that if you evaluate it at the point one, two, the input pair where x is one and y is two, then it should equal three. It should equal three, because that's telling you that when you go to x equals one and y equals two, and then you say, what's the height of the graph above that point, it should be the z-coordinate of the desired point. So this right here is kind of fact number one that we can take into consideration. And beyond that, let's start thinking about what makes planes, what makes these kind of flat graphs different from the sort of curvy graphs that you might be used to in other multivariable functions. The main idea is that if you intersect it with another plane, so here I'm gonna intersect it with y equals two, this kind of constant plane."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "It should equal three, because that's telling you that when you go to x equals one and y equals two, and then you say, what's the height of the graph above that point, it should be the z-coordinate of the desired point. So this right here is kind of fact number one that we can take into consideration. And beyond that, let's start thinking about what makes planes, what makes these kind of flat graphs different from the sort of curvy graphs that you might be used to in other multivariable functions. The main idea is that if you intersect it with another plane, so here I'm gonna intersect it with y equals two, this kind of constant plane. So I'll go ahead and write that down. That plane that you're looking at is y equals two. And you can think of this as representing, you know, what does movement in the x direction look like?"}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "The main idea is that if you intersect it with another plane, so here I'm gonna intersect it with y equals two, this kind of constant plane. So I'll go ahead and write that down. That plane that you're looking at is y equals two. And you can think of this as representing, you know, what does movement in the x direction look like? As we move along the x direction, this kind of has a slope. The two planes intersect along a line. And that's one of the crucial features of a plane, is that if you intersect it with another plane, you just get a straight line, meaning the slope is constant as you move in the x direction."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And you can think of this as representing, you know, what does movement in the x direction look like? As we move along the x direction, this kind of has a slope. The two planes intersect along a line. And that's one of the crucial features of a plane, is that if you intersect it with another plane, you just get a straight line, meaning the slope is constant as you move in the x direction. But it's also constant, that same slope, if you move in the y direction. If I had chosen a different plane, you know, if instead I had chosen y equals one, which looks like this, then you get a line with the same slope. And no matter what constant value of y you choose, it's always intersecting that plane with a line that has the same slope."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And that's one of the crucial features of a plane, is that if you intersect it with another plane, you just get a straight line, meaning the slope is constant as you move in the x direction. But it's also constant, that same slope, if you move in the y direction. If I had chosen a different plane, you know, if instead I had chosen y equals one, which looks like this, then you get a line with the same slope. And no matter what constant value of y you choose, it's always intersecting that plane with a line that has the same slope. And now if you look back to the videos on partial derivatives, and in particular on how you interpret the partial derivative of a function with respect to its graph, what this is telling you is that when we take the partial derivative of L with respect to x, because constant y means you're moving in the x direction, this should just be some kind of constant, some kind of constant a. I'll kind of emphasize that as a constant value here. And the same goes in the other direction, right? Let's say instead of intersecting it with constant values of y, you say, well, what if you intersected it with a constant value of x, like x equals one?"}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And no matter what constant value of y you choose, it's always intersecting that plane with a line that has the same slope. And now if you look back to the videos on partial derivatives, and in particular on how you interpret the partial derivative of a function with respect to its graph, what this is telling you is that when we take the partial derivative of L with respect to x, because constant y means you're moving in the x direction, this should just be some kind of constant, some kind of constant a. I'll kind of emphasize that as a constant value here. And the same goes in the other direction, right? Let's say instead of intersecting it with constant values of y, you say, well, what if you intersected it with a constant value of x, like x equals one? Well, in that case, what you should get, because you're intersecting it with a plane, is another straight line. So these two planes are intersecting along a straight line, which means as you move in the y direction, this slope won't change. But also as you move in the x direction, if you imagine slicing it with a bunch of different planes, all representing different constant values of x, you would be getting a line with that same slope."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "Let's say instead of intersecting it with constant values of y, you say, well, what if you intersected it with a constant value of x, like x equals one? Well, in that case, what you should get, because you're intersecting it with a plane, is another straight line. So these two planes are intersecting along a straight line, which means as you move in the y direction, this slope won't change. But also as you move in the x direction, if you imagine slicing it with a bunch of different planes, all representing different constant values of x, you would be getting a line with that same slope. And that's telling you another powerful fact, that the partial derivative of L with respect to y, you know, if you're moving in the y direction, that's equal to some other constant, that I'm gonna call b. And now keep in mind, these are very powerful statements, because the partial derivative of L with respect to x is a function. This is a function of x and y."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "But also as you move in the x direction, if you imagine slicing it with a bunch of different planes, all representing different constant values of x, you would be getting a line with that same slope. And that's telling you another powerful fact, that the partial derivative of L with respect to y, you know, if you're moving in the y direction, that's equal to some other constant, that I'm gonna call b. And now keep in mind, these are very powerful statements, because the partial derivative of L with respect to x is a function. This is a function of x and y. And that might actually be worth emphasizing here, that this partial derivative of x with respect to y is something that you evaluate at, you know, a point in two-dimensional space. And we're saying that that's equal to some kind of constant value. Now that's a pretty powerful thing, right?"}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "This is a function of x and y. And that might actually be worth emphasizing here, that this partial derivative of x with respect to y is something that you evaluate at, you know, a point in two-dimensional space. And we're saying that that's equal to some kind of constant value. Now that's a pretty powerful thing, right? Because it's telling you, it's giving you control over the function at all possible input points, you know, for movement in a specified direction. And the same goes over here. This is telling you that a function is constantly equal to, you know, some value b."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "Now that's a pretty powerful thing, right? Because it's telling you, it's giving you control over the function at all possible input points, you know, for movement in a specified direction. And the same goes over here. This is telling you that a function is constantly equal to, you know, some value b. And we're not sure what this value b is. But just geometrically, we can kind of estimate what these things should be. So if I take back the plane representing a constant y value, and we say, what's this slope?"}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "This is telling you that a function is constantly equal to, you know, some value b. And we're not sure what this value b is. But just geometrically, we can kind of estimate what these things should be. So if I take back the plane representing a constant y value, and we say, what's this slope? You know, you're moving in the x direction, we've got a constant y value. What is the slope at which this plane intersects our graph? I would estimate this as about a slope of two."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So if I take back the plane representing a constant y value, and we say, what's this slope? You know, you're moving in the x direction, we've got a constant y value. What is the slope at which this plane intersects our graph? I would estimate this as about a slope of two. You know, you kind of go over one, and it goes up two. So what that would tell you is that, at least in the specific graph that we're looking at, this is at least approximately equal to two. And then similarly, if we look at a constant x value, and we say that represents movement in the y direction, what is the slope there?"}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "I would estimate this as about a slope of two. You know, you kind of go over one, and it goes up two. So what that would tell you is that, at least in the specific graph that we're looking at, this is at least approximately equal to two. And then similarly, if we look at a constant x value, and we say that represents movement in the y direction, what is the slope there? This looks to me like about one as a slope. You kind of move over one unit, you go up one unit. So I'd say down here that the constant value of the partial derivative with respect to y is about equal to one."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And then similarly, if we look at a constant x value, and we say that represents movement in the y direction, what is the slope there? This looks to me like about one as a slope. You kind of move over one unit, you go up one unit. So I'd say down here that the constant value of the partial derivative with respect to y is about equal to one. So we have three different facts here. The value of the function at the point one, two. The value of its partial derivative with respect to x everywhere."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So I'd say down here that the constant value of the partial derivative with respect to y is about equal to one. So we have three different facts here. The value of the function at the point one, two. The value of its partial derivative with respect to x everywhere. And the partial derivative with respect to y everywhere. And this information is actually going to be enough to tell us what the function as a whole should equal. Now specifically, this idea that the partial derivative with respect to x is constant tells us that the function, function L of x, y, is going to equal two times x plus something that doesn't have any x's in it."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "The value of its partial derivative with respect to x everywhere. And the partial derivative with respect to y everywhere. And this information is actually going to be enough to tell us what the function as a whole should equal. Now specifically, this idea that the partial derivative with respect to x is constant tells us that the function, function L of x, y, is going to equal two times x plus something that doesn't have any x's in it. Something that as far as x is concerned is a constant. Because the only thing who's derivative with respect to x is the constant two, is two x, plus something that's constant as far as x is concerned. And then similarly over here, if the partial with respect to y is the constant one, then that tells you that the whole function looks like, you know, this looks like a constant as far as y is concerned."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "Now specifically, this idea that the partial derivative with respect to x is constant tells us that the function, function L of x, y, is going to equal two times x plus something that doesn't have any x's in it. Something that as far as x is concerned is a constant. Because the only thing who's derivative with respect to x is the constant two, is two x, plus something that's constant as far as x is concerned. And then similarly over here, if the partial with respect to y is the constant one, then that tells you that the whole function looks like, you know, this looks like a constant as far as y is concerned. So once we bring in y, it's going to be one times y plus something that's constant as far as y is concerned. You know, this part is already constant as far as y is concerned, so anything that I add beyond here has to be constant as far as both x and y is concerned. So that part has to actually literally be a constant."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And then similarly over here, if the partial with respect to y is the constant one, then that tells you that the whole function looks like, you know, this looks like a constant as far as y is concerned. So once we bring in y, it's going to be one times y plus something that's constant as far as y is concerned. You know, this part is already constant as far as y is concerned, so anything that I add beyond here has to be constant as far as both x and y is concerned. So that part has to actually literally be a constant. So I'm just going to put in, I'm just going to put in c for that to represent constant. So this is, you can see, this is a very restrictive property on our function because the only place x can show up is as this linear term, and the only place y can show up is as this linear term. And when I use the word linear, you can pretty much interpret it as saying the term x shows up without an exponent or without anything fancy happening to it."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So that part has to actually literally be a constant. So I'm just going to put in, I'm just going to put in c for that to represent constant. So this is, you can see, this is a very restrictive property on our function because the only place x can show up is as this linear term, and the only place y can show up is as this linear term. And when I use the word linear, you can pretty much interpret it as saying the term x shows up without an exponent or without anything fancy happening to it. It's just x times a constant. That's pretty much what I mean by linear. It's got more precise formulations in other contexts, but as far as we're concerned here, you can just think of it as meaning variable times a constant."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "And when I use the word linear, you can pretty much interpret it as saying the term x shows up without an exponent or without anything fancy happening to it. It's just x times a constant. That's pretty much what I mean by linear. It's got more precise formulations in other contexts, but as far as we're concerned here, you can just think of it as meaning variable times a constant. So the question is, what should this c be? And you can imagine that we can, based on this property, based on the value at the point 1,2, we can uniquely determine c. And you can plug in x equals 1, y equals 2, know that this has to equal 3, and solve for c, which we could do, but I'm going to actually do something a little bit more convenient. I'm going to kind of shift around where the constants show up, and I'm going to say that the whole function should equal two times, and then I'm going to put a constant in with x. I'm going to say x minus 1."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "It's got more precise formulations in other contexts, but as far as we're concerned here, you can just think of it as meaning variable times a constant. So the question is, what should this c be? And you can imagine that we can, based on this property, based on the value at the point 1,2, we can uniquely determine c. And you can plug in x equals 1, y equals 2, know that this has to equal 3, and solve for c, which we could do, but I'm going to actually do something a little bit more convenient. I'm going to kind of shift around where the constants show up, and I'm going to say that the whole function should equal two times, and then I'm going to put a constant in with x. I'm going to say x minus 1. And then I'm going to do the same thing with y. I'm going to say plus 1, here's the partial derivative with respect to y, y minus, and then I'm going to say 2. And the reason I'm doing this, notice this doesn't change the partial derivative information, it's just, if we take the partial derivative with respect to x, this will still be 2, and when we take it with respect to y, this will still be 1. But the reason I'm putting these in here is because we're going to evaluate it at the point 1,2, so I want to make it look as easy as possible to evaluate at the point 1,2."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "I'm going to kind of shift around where the constants show up, and I'm going to say that the whole function should equal two times, and then I'm going to put a constant in with x. I'm going to say x minus 1. And then I'm going to do the same thing with y. I'm going to say plus 1, here's the partial derivative with respect to y, y minus, and then I'm going to say 2. And the reason I'm doing this, notice this doesn't change the partial derivative information, it's just, if we take the partial derivative with respect to x, this will still be 2, and when we take it with respect to y, this will still be 1. But the reason I'm putting these in here is because we're going to evaluate it at the point 1,2, so I want to make it look as easy as possible to evaluate at the point 1,2. And then from here, I'm just going to add another constant. So instead of saying c, because this is going to be slightly different from c, I'll call it k. But the idea is I'm just moving around constants. If you imagined distributing the multiplication here and having, you know, 2 times that negative 1, and 1 times that negative 2, you're just changing what the value of the constant stuck on the end here is."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "But the reason I'm putting these in here is because we're going to evaluate it at the point 1,2, so I want to make it look as easy as possible to evaluate at the point 1,2. And then from here, I'm just going to add another constant. So instead of saying c, because this is going to be slightly different from c, I'll call it k. But the idea is I'm just moving around constants. If you imagined distributing the multiplication here and having, you know, 2 times that negative 1, and 1 times that negative 2, you're just changing what the value of the constant stuck on the end here is. Now the important part, the reason that I'm writing it this way, which is only slightly different, is because then when I evaluate this at L of 1,2, this whole first part cancels out because plugging in x equals 1 means this whole part goes to 0. Same with the second part, because when I plug in y equals 2, this part goes to 0. So k, this other constant that I'm tagging on the end, is going to completely specify what happens when I evaluate this at the point 1,2."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "If you imagined distributing the multiplication here and having, you know, 2 times that negative 1, and 1 times that negative 2, you're just changing what the value of the constant stuck on the end here is. Now the important part, the reason that I'm writing it this way, which is only slightly different, is because then when I evaluate this at L of 1,2, this whole first part cancels out because plugging in x equals 1 means this whole part goes to 0. Same with the second part, because when I plug in y equals 2, this part goes to 0. So k, this other constant that I'm tagging on the end, is going to completely specify what happens when I evaluate this at the point 1,2. And of course, I want it to be the case that when I evaluate it at 1,2, I get 3. I want it to be the case when I evaluate it at 1,2, I get 3. So that tells me that this constant k here should just equal 3."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So k, this other constant that I'm tagging on the end, is going to completely specify what happens when I evaluate this at the point 1,2. And of course, I want it to be the case that when I evaluate it at 1,2, I get 3. I want it to be the case when I evaluate it at 1,2, I get 3. So that tells me that this constant k here should just equal 3. So notice, the way that I've written the function here is actually quite powerful. We have a lot of control. This term 2 was telling us the slope with respect to x."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So that tells me that this constant k here should just equal 3. So notice, the way that I've written the function here is actually quite powerful. We have a lot of control. This term 2 was telling us the slope with respect to x. So when you move purely in the x direction, and that was kind of illustrated here, purely in the x direction, that's telling us the slope with respect to x. And then this term 1 here was telling us the slope with respect to y. So when we moved purely in the y direction, that's telling us the slope there."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "This term 2 was telling us the slope with respect to x. So when you move purely in the x direction, and that was kind of illustrated here, purely in the x direction, that's telling us the slope with respect to x. And then this term 1 here was telling us the slope with respect to y. So when we moved purely in the y direction, that's telling us the slope there. And we could just turn those knobs. If we change the 2 and we change the 1, that's what's going to allow us to basically change what the slopes of the plane are. I'm going to say slopes plural because it's with respect to the x and the y direction."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So when we moved purely in the y direction, that's telling us the slope there. And we could just turn those knobs. If we change the 2 and we change the 1, that's what's going to allow us to basically change what the slopes of the plane are. I'm going to say slopes plural because it's with respect to the x and the y direction. And that'll give us control over various different planes to pass through. You know, if I'm looking at the 1, oh, geez, I don't know, let's say the 1 right here, then the movement in the y direction is very shallow. So that would be turning this knob lower."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "I'm going to say slopes plural because it's with respect to the x and the y direction. And that'll give us control over various different planes to pass through. You know, if I'm looking at the 1, oh, geez, I don't know, let's say the 1 right here, then the movement in the y direction is very shallow. So that would be turning this knob lower. And instead of 1, it might be.01. And if I were looking at movement in the x direction, you know, this looks actually negative. So this would tell you that it's going to be some kind of negative number."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So that would be turning this knob lower. And instead of 1, it might be.01. And if I were looking at movement in the x direction, you know, this looks actually negative. So this would tell you that it's going to be some kind of negative number. So you can kind of dial these knobs and that changes the different planes that pass through that same point. And then plugging in this 1, 2, and 3 tells us what point we're specifying. We're basically saying when you input x equals 1 and you input y equals 2, the whole thing should equal 3."}, {"video_title": "Controlling a plane in space.mp3", "Sentence": "So this would tell you that it's going to be some kind of negative number. So you can kind of dial these knobs and that changes the different planes that pass through that same point. And then plugging in this 1, 2, and 3 tells us what point we're specifying. We're basically saying when you input x equals 1 and you input y equals 2, the whole thing should equal 3. So this form right here is powerful enough that I want you to remember it for the next video. I want you to remember the idea of writing things down in this way where you specify the point it's passing through with its x coordinate, y coordinate, and z coordinate placed where they are, and then you tweak the slopes using these coefficients out front. So with that, I will see you next video."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "But I think you're really gonna like where this is going in the end. So one of the examples I showed, and I think this is a pretty nice prototypical example for constrained optimization problems, is that you're running a company and you have some kind of revenue function that's dependent on various choices you make in running the company. And I think I said the number of hours of labor you employ and the number of tons of steel you use, you know, if you were manufacturing something metallic. And you know, this might be modeled as some multivariable function of H and S. Right now we don't really care about the specifics. And you're trying to maximize this, right? That's kind of the whole point of this unit that we've been doing, is that you're trying to maximize some function, but you have a constraint. This is the real world."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And you know, this might be modeled as some multivariable function of H and S. Right now we don't really care about the specifics. And you're trying to maximize this, right? That's kind of the whole point of this unit that we've been doing, is that you're trying to maximize some function, but you have a constraint. This is the real world. You can't just spend infinite money, you have some kind of budget. Some sort of amount of money you spend as a function of those same choices you make. The hours of labor you employ, the tons of steel you use."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "This is the real world. You can't just spend infinite money, you have some kind of budget. Some sort of amount of money you spend as a function of those same choices you make. The hours of labor you employ, the tons of steel you use. And this again, it's gonna equal some multivariable function that tells you, you know, how much money you spend for a given amount of hours and given number of tons of steel. And you set this equal to some constant. This tells you the amount of money you're willing to spend."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "The hours of labor you employ, the tons of steel you use. And this again, it's gonna equal some multivariable function that tells you, you know, how much money you spend for a given amount of hours and given number of tons of steel. And you set this equal to some constant. This tells you the amount of money you're willing to spend. And our goal has been to maximize some function subject to a constraint like this. And the mental model you have in mind is that you're looking in the input space inside the XY plane, or I guess really it's the HS plane in this case, right? Your inputs are H and S and points in this plane tell you possible choices you can make for hours of labor and tons of steel."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "This tells you the amount of money you're willing to spend. And our goal has been to maximize some function subject to a constraint like this. And the mental model you have in mind is that you're looking in the input space inside the XY plane, or I guess really it's the HS plane in this case, right? Your inputs are H and S and points in this plane tell you possible choices you can make for hours of labor and tons of steel. And you think of this budget as some kind of curve in that plane, right? All the sets of H and S that equal $10,000 is gonna give you some kind of curve. And the core value we care about is that when you maximize this revenue, you know, when you set it equal to a constant, I'm gonna call M star."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "Your inputs are H and S and points in this plane tell you possible choices you can make for hours of labor and tons of steel. And you think of this budget as some kind of curve in that plane, right? All the sets of H and S that equal $10,000 is gonna give you some kind of curve. And the core value we care about is that when you maximize this revenue, you know, when you set it equal to a constant, I'm gonna call M star. This is like the maximum possible revenue. That's gonna give you a contour that's just barely tangent to the constraint curve. And if that seems unfamiliar, definitely take a look at the videos preceding this one."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And the core value we care about is that when you maximize this revenue, you know, when you set it equal to a constant, I'm gonna call M star. This is like the maximum possible revenue. That's gonna give you a contour that's just barely tangent to the constraint curve. And if that seems unfamiliar, definitely take a look at the videos preceding this one. But just to kind of continue the review, this gave us the really nice property that you look at the gradient vector for the thing you're trying to maximize, R, and that's gonna be proportional to the gradient vector for the constraint function, for this B, so gradient of B. And this is because gradients are perpendicular to contour lines. Again, this should feel mostly like review at this point."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And if that seems unfamiliar, definitely take a look at the videos preceding this one. But just to kind of continue the review, this gave us the really nice property that you look at the gradient vector for the thing you're trying to maximize, R, and that's gonna be proportional to the gradient vector for the constraint function, for this B, so gradient of B. And this is because gradients are perpendicular to contour lines. Again, this should feel mostly like review at this point. So the core idea was that we take this gradient of R and then make it proportional with some kind of proportionality constant, lambda, to the gradient of B, to the gradient of the constraint function. And up till this point, this value lambda has been wholly uninteresting. It's just been a proportionality constant, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "Again, this should feel mostly like review at this point. So the core idea was that we take this gradient of R and then make it proportional with some kind of proportionality constant, lambda, to the gradient of B, to the gradient of the constraint function. And up till this point, this value lambda has been wholly uninteresting. It's just been a proportionality constant, right? Because we couldn't guarantee that the gradient of R is equal to the gradient of B. All we care about is that they're pointing in the same direction. So we just had this constant sitting here, and all we really said is make sure it's not zero."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "It's just been a proportionality constant, right? Because we couldn't guarantee that the gradient of R is equal to the gradient of B. All we care about is that they're pointing in the same direction. So we just had this constant sitting here, and all we really said is make sure it's not zero. But here, we're gonna get to where this little guy actually matters. So if you'll remember, in the last video, I introduced this function called the Lagrangian, the Lagrangian, and it takes in multiple inputs. They'll be the same inputs that you have for your budget function and your revenue function, or more generally, the constraint and the thing you're trying to maximize."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "So we just had this constant sitting here, and all we really said is make sure it's not zero. But here, we're gonna get to where this little guy actually matters. So if you'll remember, in the last video, I introduced this function called the Lagrangian, the Lagrangian, and it takes in multiple inputs. They'll be the same inputs that you have for your budget function and your revenue function, or more generally, the constraint and the thing you're trying to maximize. It takes in those same variables, but also, as another one of its inputs, it takes in lambda. So it is a higher-dimensional function than both of these two, because we've got this extra lambda. And the way it's defined looks a little strange at first."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "They'll be the same inputs that you have for your budget function and your revenue function, or more generally, the constraint and the thing you're trying to maximize. It takes in those same variables, but also, as another one of its inputs, it takes in lambda. So it is a higher-dimensional function than both of these two, because we've got this extra lambda. And the way it's defined looks a little strange at first. It just seems kind of like this random hodgepodge of functions that we're putting together. But last time, I kind of walked through why this makes sense. You take the thing you're trying to maximize, and you subtract off this lambda multiplied by the constraint function, which is B of those inputs, minus, and then whatever this constant is here, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And the way it's defined looks a little strange at first. It just seems kind of like this random hodgepodge of functions that we're putting together. But last time, I kind of walked through why this makes sense. You take the thing you're trying to maximize, and you subtract off this lambda multiplied by the constraint function, which is B of those inputs, minus, and then whatever this constant is here, right? I'm gonna give it a name. I'm gonna call this constant lowercase b, so maybe we're thinking of it as $10,000, but it's whatever your actual budget is. So we think of that, and I'm just gonna emphasize here that that's a constant, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "You take the thing you're trying to maximize, and you subtract off this lambda multiplied by the constraint function, which is B of those inputs, minus, and then whatever this constant is here, right? I'm gonna give it a name. I'm gonna call this constant lowercase b, so maybe we're thinking of it as $10,000, but it's whatever your actual budget is. So we think of that, and I'm just gonna emphasize here that that's a constant, right? That this b is being treated as a constant right now. You know, we're thinking of h and s and lambda all as these variables, and this gives us some multivariable function. And if you'll remember from the last video, the reason for defining this function is it gives us a really nice, compact way to solve the constrained optimization problem."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "So we think of that, and I'm just gonna emphasize here that that's a constant, right? That this b is being treated as a constant right now. You know, we're thinking of h and s and lambda all as these variables, and this gives us some multivariable function. And if you'll remember from the last video, the reason for defining this function is it gives us a really nice, compact way to solve the constrained optimization problem. You set the gradient of L equal to zero, or really the zero vector, right? It'll be a vector with three components here. And when you do that, you'll find some solution, right, you'll find some solution, which I'll call h star, s star, and lambda."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And if you'll remember from the last video, the reason for defining this function is it gives us a really nice, compact way to solve the constrained optimization problem. You set the gradient of L equal to zero, or really the zero vector, right? It'll be a vector with three components here. And when you do that, you'll find some solution, right, you'll find some solution, which I'll call h star, s star, and lambda. And lambda, I'll give it that green lambda color, lambda star. You'll find some value that when you input this into the function, the gradient will equal zero. And of course, you might find multiple of these, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And when you do that, you'll find some solution, right, you'll find some solution, which I'll call h star, s star, and lambda. And lambda, I'll give it that green lambda color, lambda star. You'll find some value that when you input this into the function, the gradient will equal zero. And of course, you might find multiple of these, right? You might find multiple solutions to this problem, but what you do is for each one of them, you're gonna take a look at h star and s star, then you're gonna plug those into the revenue function, or the thing that you're trying to maximize, and typically you only get a handful. You get a number that you could actually plug each one of them into the revenue function, and you'll just check which one of them makes this function the highest, and whatever the highest value this function can achieve, that is m star, that is the maximum possible revenue subject to this budget. But it's interesting that when you solve this, you get some specific value of lambda, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "And of course, you might find multiple of these, right? You might find multiple solutions to this problem, but what you do is for each one of them, you're gonna take a look at h star and s star, then you're gonna plug those into the revenue function, or the thing that you're trying to maximize, and typically you only get a handful. You get a number that you could actually plug each one of them into the revenue function, and you'll just check which one of them makes this function the highest, and whatever the highest value this function can achieve, that is m star, that is the maximum possible revenue subject to this budget. But it's interesting that when you solve this, you get some specific value of lambda, right? There is a specific lambda star that will be associated with this solution. And like I said, this turns out not to just be some dummy variable. It's gonna carry information about how much we can increase the revenue if we increase that budget."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "But it's interesting that when you solve this, you get some specific value of lambda, right? There is a specific lambda star that will be associated with this solution. And like I said, this turns out not to just be some dummy variable. It's gonna carry information about how much we can increase the revenue if we increase that budget. And here, let me show you what I mean. So we've got this m star, and I'll just write it again. M star here, and what that equals, I'm saying that's the maximum possible revenue."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "It's gonna carry information about how much we can increase the revenue if we increase that budget. And here, let me show you what I mean. So we've got this m star, and I'll just write it again. M star here, and what that equals, I'm saying that's the maximum possible revenue. So that's gonna be the revenue when you evaluate it at h star, h star, and s star. And h star and s star, they are whatever the solution to this gradient of the Lagrangian equals zero equation is. You set this multivariable function equal to the zero vector."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "M star here, and what that equals, I'm saying that's the maximum possible revenue. So that's gonna be the revenue when you evaluate it at h star, h star, and s star. And h star and s star, they are whatever the solution to this gradient of the Lagrangian equals zero equation is. You set this multivariable function equal to the zero vector. You solve when each of its partial derivatives equals zero, and you'll get some kind of solution. So when you plug that solution in the revenue, that gives you the maximum possible revenue. But what we could do is consider this as a function of the budget."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "You set this multivariable function equal to the zero vector. You solve when each of its partial derivatives equals zero, and you'll get some kind of solution. So when you plug that solution in the revenue, that gives you the maximum possible revenue. But what we could do is consider this as a function of the budget. Now, this is the kind of thing that looks a little bit wacky if you're just looking at the formulas. But if you actually think about what it means in this context of kind of a revenue and a budget, I think it's actually pretty sensible, where really, if we consider this b no longer to be constant, but something that you could change, right? You're wondering, well, what if I had a $20,000 budget?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "But what we could do is consider this as a function of the budget. Now, this is the kind of thing that looks a little bit wacky if you're just looking at the formulas. But if you actually think about what it means in this context of kind of a revenue and a budget, I think it's actually pretty sensible, where really, if we consider this b no longer to be constant, but something that you could change, right? You're wondering, well, what if I had a $20,000 budget? Or what if I had a $15,000 budget? You wanna ask the question, what happens as you change this b? Well, the maximizing value, h star and s star, each one of those guys is gonna depend on b, right?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "You're wondering, well, what if I had a $20,000 budget? Or what if I had a $15,000 budget? You wanna ask the question, what happens as you change this b? Well, the maximizing value, h star and s star, each one of those guys is gonna depend on b, right? As you change what this constant is, it's gonna change the values at which the gradient of the Lagrangian equals zero. So I'm gonna rewrite this function as the revenue evaluated at h star and s star, but now I'm gonna consider that h star and s star each as functions of b, right? Because they depend on it in some way."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "Well, the maximizing value, h star and s star, each one of those guys is gonna depend on b, right? As you change what this constant is, it's gonna change the values at which the gradient of the Lagrangian equals zero. So I'm gonna rewrite this function as the revenue evaluated at h star and s star, but now I'm gonna consider that h star and s star each as functions of b, right? Because they depend on it in some way. As you change b, it changes the solution to this problem. It's very implicit, and it's kind of hard to think about, right? It's hard to think, okay, as I change this b, how much does that change h star?"}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "Because they depend on it in some way. As you change b, it changes the solution to this problem. It's very implicit, and it's kind of hard to think about, right? It's hard to think, okay, as I change this b, how much does that change h star? Well, that depends on what the definition of r was and everything there. But in principle, in this context, I think it's quite intuitive. You have a maximum possible revenue, and that depends on what your budget is."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "It's hard to think, okay, as I change this b, how much does that change h star? Well, that depends on what the definition of r was and everything there. But in principle, in this context, I think it's quite intuitive. You have a maximum possible revenue, and that depends on what your budget is. So what turns out to be a beautiful, absolutely beautiful, magical fact is that this lambda star is equal to the derivative of m star, the derivative of this maximum possible revenue with respect to b, with respect to the budget. And let me just show you what that actually means, right? So if, for example, let's say you did all of your calculations, and it turned out that lambda star was equal to 2.3."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "You have a maximum possible revenue, and that depends on what your budget is. So what turns out to be a beautiful, absolutely beautiful, magical fact is that this lambda star is equal to the derivative of m star, the derivative of this maximum possible revenue with respect to b, with respect to the budget. And let me just show you what that actually means, right? So if, for example, let's say you did all of your calculations, and it turned out that lambda star was equal to 2.3. Previously, that just seemed like some dummy number that you ignore, and you just look at whatever the associated values here are. But if you plug this in the computer and you see lambda star equals 2.3, what that means is for a tiny change in budget, like let's say your budget increases from 10,000 to 10,001, it goes up to $10,001. So you increase your budget by just a little bit, a little dB."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "So if, for example, let's say you did all of your calculations, and it turned out that lambda star was equal to 2.3. Previously, that just seemed like some dummy number that you ignore, and you just look at whatever the associated values here are. But if you plug this in the computer and you see lambda star equals 2.3, what that means is for a tiny change in budget, like let's say your budget increases from 10,000 to 10,001, it goes up to $10,001. So you increase your budget by just a little bit, a little dB. Then the ratio of the change in the maximizing revenue to that dB is about 2.3. So what that would mean is increasing your budget by $1 is gonna increase m star over here. It would mean that m star increases by about $2.30 for every dollar that you increase your budget."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "So you increase your budget by just a little bit, a little dB. Then the ratio of the change in the maximizing revenue to that dB is about 2.3. So what that would mean is increasing your budget by $1 is gonna increase m star over here. It would mean that m star increases by about $2.30 for every dollar that you increase your budget. And that's information you'd wanna know, right? If you see that this lambda star is a number bigger than one, you'd say, hey, maybe we should increase our budget. We increase it from $10,000 to 10,001, and we're making more money."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "It would mean that m star increases by about $2.30 for every dollar that you increase your budget. And that's information you'd wanna know, right? If you see that this lambda star is a number bigger than one, you'd say, hey, maybe we should increase our budget. We increase it from $10,000 to 10,001, and we're making more money. So maybe as long as lambda star is greater than one, you should keep doing whatever it takes to increase that budget. So this fact is quite surprising, I think, and it seems like it totally comes out of nowhere. So what I'm gonna do in the next video is prove this to you, is prove why this is true, why this lambda star value happens to be the rate of change for the maximum value of the thing we're trying to maximize with respect to this constant, with respect to whatever constant you set your constraint function equal to."}, {"video_title": "Meaning of Lagrange multiplier.mp3", "Sentence": "We increase it from $10,000 to 10,001, and we're making more money. So maybe as long as lambda star is greater than one, you should keep doing whatever it takes to increase that budget. So this fact is quite surprising, I think, and it seems like it totally comes out of nowhere. So what I'm gonna do in the next video is prove this to you, is prove why this is true, why this lambda star value happens to be the rate of change for the maximum value of the thing we're trying to maximize with respect to this constant, with respect to whatever constant you set your constraint function equal to. For right now, though, I just want you to kind of try to sit back and digest what this means in the context of this specific economic example. And even if you never looked into the proof or never understood it there, I think this is an interesting and even useful tidbit of knowledge to have about Lagrange multipliers. So with that, I'll see you next video."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Some type of substance. It gives us the mass density at any point in three dimensions. And let's say we have another function. This is a scalar function. It just gives us a number for any point in 3D. And then let's say we have another function, v, which is a vector function. It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is a scalar function. It just gives us a number for any point in 3D. And then let's say we have another function, v, which is a vector function. It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about. Now let's imagine another function. This might all look a little bit familiar, because we went through a very similar exercise in two dimensions when we talked about line integrals. Now we're just extending it to three dimensions."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It gives us a vector for any point in three dimensions, and this right over here tells us the velocity of that same fluid or gas or whatever we're talking about. Now let's imagine another function. This might all look a little bit familiar, because we went through a very similar exercise in two dimensions when we talked about line integrals. Now we're just extending it to three dimensions. Let's say we have a function f, and it is equal to the product of rho and v. So for any point in x, y, z, this will give us a vector, and then we'll multiply it times this scalar right over here for that same point in three dimensions. So it's equal to rho times v. Let me use that same color that I used for v before. Rho times v. And there's a couple of ways you could conceptualize this."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we're just extending it to three dimensions. Let's say we have a function f, and it is equal to the product of rho and v. So for any point in x, y, z, this will give us a vector, and then we'll multiply it times this scalar right over here for that same point in three dimensions. So it's equal to rho times v. Let me use that same color that I used for v before. Rho times v. And there's a couple of ways you could conceptualize this. You could view this as it obviously maintains the direction of the velocity, but now its magnitude, one way to think about it, is kind of the momentum density. If that doesn't make too much sense, you don't have to worry too much about it. Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Rho times v. And there's a couple of ways you could conceptualize this. You could view this as it obviously maintains the direction of the velocity, but now its magnitude, one way to think about it, is kind of the momentum density. If that doesn't make too much sense, you don't have to worry too much about it. Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense. Now, what I want to do is think about what it means given this function f to evaluate the surface integral over some surface. So we're going to evaluate over some surface. We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Hopefully, as we use these two functions and we think a little bit more about them relative to a surface, it'll make a little bit more conceptual sense. Now, what I want to do is think about what it means given this function f to evaluate the surface integral over some surface. So we're going to evaluate over some surface. We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface. So let's think about what this is saying. So first let me draw my axes. So I have my z-axis."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to evaluate f dot n, where n is the unit normal vector at each point on that surface, dS, d capital S, d surface. So let's think about what this is saying. So first let me draw my axes. So I have my z-axis. This could be my x-axis. And let's say that this right over here is my y-axis. And let's say my surface, I'll use that same color, my surface looks something like that."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I have my z-axis. This could be my x-axis. And let's say that this right over here is my y-axis. And let's say my surface, I'll use that same color, my surface looks something like that. So that is my surface. That is the surface in question. That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say my surface, I'll use that same color, my surface looks something like that. So that is my surface. That is the surface in question. That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring. It's completely analogous to what we did in the two-dimensional case with line integrals. So we have a dS. A dS is a little chunk of area of that surface."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is S. Now let's think about the units, and hopefully that'll give us a conceptual understanding of what this thing right over here is measuring. It's completely analogous to what we did in the two-dimensional case with line integrals. So we have a dS. A dS is a little chunk of area of that surface. So that is dS. So this is going to be area. And if we want to pick particular units, this could be square meters."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "A dS is a little chunk of area of that surface. So that is dS. So this is going to be area. And if we want to pick particular units, this could be square meters. And I think when you do particular units, it starts to make a little bit more concrete sense. Now, the normal vector at that dS, the normal vector is going to point right out of it. It's literally normal to that plane."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we want to pick particular units, this could be square meters. And I think when you do particular units, it starts to make a little bit more concrete sense. Now, the normal vector at that dS, the normal vector is going to point right out of it. It's literally normal to that plane. It's literally normal to that plane. It has a magnitude 1. So that is our unit normal vector."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's literally normal to that plane. It's literally normal to that plane. It has a magnitude 1. So that is our unit normal vector. And F is defined throughout this three-dimensional space. You give me any x, y, z, I'll know its mass density, I'll know its velocity, and I'll get some F. I'll get some F at any point in three-dimensional space, including on the surface, including right over here. So right over here, F might look something like this."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that is our unit normal vector. And F is defined throughout this three-dimensional space. You give me any x, y, z, I'll know its mass density, I'll know its velocity, and I'll get some F. I'll get some F at any point in three-dimensional space, including on the surface, including right over here. So right over here, F might look something like this. So that is F right at that point. So what does all this mean? Well, when you take the dot product of two vectors, it's essentially saying how much do they go together."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So right over here, F might look something like this. So that is F right at that point. So what does all this mean? Well, when you take the dot product of two vectors, it's essentially saying how much do they go together. And since N is a unit vector, since it has a magnitude 1, this is essentially saying what is the magnitude of the component of F that's going in the direction of N? Or what is the magnitude of the component of F that is normal to the surface? How much of F is normal to the surface?"}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, when you take the dot product of two vectors, it's essentially saying how much do they go together. And since N is a unit vector, since it has a magnitude 1, this is essentially saying what is the magnitude of the component of F that's going in the direction of N? Or what is the magnitude of the component of F that is normal to the surface? How much of F is normal to the surface? So the component of F that is normal to the surface might look something like that. And this right over here will essentially just give the magnitude of that. And it's just going to keep the units of F. And right over here just specifies the direction."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "How much of F is normal to the surface? So the component of F that is normal to the surface might look something like that. And this right over here will essentially just give the magnitude of that. And it's just going to keep the units of F. And right over here just specifies the direction. It has no units associated with it. It's dimensionless. F's units are going to be units of mass density."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's just going to keep the units of F. And right over here just specifies the direction. It has no units associated with it. It's dimensionless. F's units are going to be units of mass density. So it could be, let's say, it could be kilogram per meter cubed. Well, that's actually just the rho part. So it's mass density times velocity times meters per second."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "F's units are going to be units of mass density. So it could be, let's say, it could be kilogram per meter cubed. Well, that's actually just the rho part. So it's mass density times velocity times meters per second. Let me write in those colors just so we have clear what's happening here. So the units of F are going to be the units of rho, which are going to be kilogram per cubic meter. That's mass density."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's mass density times velocity times meters per second. Let me write in those colors just so we have clear what's happening here. So the units of F are going to be the units of rho, which are going to be kilogram per cubic meter. That's mass density. Times the units of V, which is meters per second. And we're going to multiply that times meters squared. So what you have is you have a meter and then a meter squared in the numerator."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's mass density. Times the units of V, which is meters per second. And we're going to multiply that times meters squared. So what you have is you have a meter and then a meter squared in the numerator. That's meters cubed in the numerator. And meters cubed in the denominator. That cancels out."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what you have is you have a meter and then a meter squared in the numerator. That's meters cubed in the numerator. And meters cubed in the denominator. That cancels out. And so the units that we get for this are kilogram per second. And so the way to conceptualize it, given how we've defined F, what we say F represents, the way to conceptualize this, this is saying how much mass, how much mass given this mass density, this velocity, is going directly out of this little ds, this little infinitesimally chunk of surface in a given amount of time. And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That cancels out. And so the units that we get for this are kilogram per second. And so the way to conceptualize it, given how we've defined F, what we say F represents, the way to conceptualize this, this is saying how much mass, how much mass given this mass density, this velocity, is going directly out of this little ds, this little infinitesimally chunk of surface in a given amount of time. And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time. And this is really the same idea we do with the line integrals. This is essentially the flux through a two-dimensional surface. So this is the flux through a 2D surface."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if we were to add up all the ds's, and this is what essentially that surface integral is, we're essentially saying how much mass in kilograms per second, that's what we picked, how much mass is traveling across this surface at any given moment in time. And this is really the same idea we do with the line integrals. This is essentially the flux through a two-dimensional surface. So this is the flux through a 2D surface. And this isn't like some crazy abstract thing. I mean, you could imagine something like water vapor in your bathroom. Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the flux through a 2D surface. And this isn't like some crazy abstract thing. I mean, you could imagine something like water vapor in your bathroom. Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it. And we've all seen water vapor in our bathroom when you have a ray of sunlight, and you can see how the particles are traveling, and you see they have a certain density at different points. And so you can imagine you care about the surface of your, maybe you have a window in the bathroom. Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Water vapor in your bathroom, and I like to imagine that because that's actually visible, especially when sunlight is shining through it. And we've all seen water vapor in our bathroom when you have a ray of sunlight, and you can see how the particles are traveling, and you see they have a certain density at different points. And so you can imagine you care about the surface of your, maybe you have a window in the bathroom. Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through. If an F was essentially the mass density of the water vapor times the velocity of the water vapor, then this thing right over here will essentially tell you the mass of water vapor that is traveling through that window at any given moment of time. Another way to think about it is imagine a river, and I'm going to conceptualize this river as kind of just a section of the river. I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe you have a window, and so if the surface was the window, and the window, let's say the window is open, so it's kind of a, there's nothing physical there, it's just kind of a rectangular surface that things can pass freely through. If an F was essentially the mass density of the water vapor times the velocity of the water vapor, then this thing right over here will essentially tell you the mass of water vapor that is traveling through that window at any given moment of time. Another way to think about it is imagine a river, and I'm going to conceptualize this river as kind of just a section of the river. I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature. And so we would know the density, maybe it's constant, you know the density and you know the velocity at any point, that's what F gives us. So that tells us, as we said, we could view that as the momentum density at any given point in time. And maybe our surface is some type of a net."}, {"video_title": "Conceptual understanding of flux in three dimensions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm conceptualizing it, this is kind of a river, obviously this would be the surface that we normally see, but obviously it has some depth, it's three-dimensional in nature. And so we would know the density, maybe it's constant, you know the density and you know the velocity at any point, that's what F gives us. So that tells us, as we said, we could view that as the momentum density at any given point in time. And maybe our surface is some type of a net. And the net doesn't even have to be rectangular, it could be some weird-shaped net, but I'll do it rectangular just because it's easy to draw. It's some type of net that in no way impedes the flow of the fluid, then once again, when you evaluate this integral, it will tell you the mass of fluid that is flowing through that net at any given moment of time. So hopefully this makes a little bit of conceptual sense now."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We have our usual setup here for this constrained optimization situation. We have a function we want to maximize, which I'm thinking of as revenues for some company, a constraint, which I'm thinking of as some kind of budget for that company, and as you know if you've gotten to this video, one way to solve this constrained optimization problem is to define this function here, the Lagrangian, which involves taking this function that you're trying to maximize, in this case the revenue, and subtracting a new variable, lambda, what's called the Lagrange multiplier, times this quantity, which is the budget function, you know, however much you spend as a function of your input parameters, minus the budget itself, which you might think of as $10,000 in our example. So that's all the usual setup, and the crazy fact, which I just declared, is that when you set this gradient equal to zero, and you find some solution, and there will be three variables in this solution, h star, s star, and lambda star, that this lambda star is not meaningless, it's not just a proportionality constant between these gradient vectors, but it will actually tell you how much the maximum possible revenue changes as a function of your budget. And the way to start writing all of that in formulas would be to make explicit the fact that if you consider this value, you know, the $10,000 that is your budget, which I'm calling b, a variable, and not a constant, then you have to acknowledge that h star and s star are dependent on b, right? It's a very implicit relationship, something that's kind of hard to think about at first, because as you change b, it changes what the Lagrangian is, which is going to change where its gradient equals zero, which changes what h star, s star, and lambda star are, but in principle, they are some function of that budget, of b. And the maximum possible revenue is whatever you get when you just plug in that solution to your function r, and the claim I made that I just pulled out of the hat is that lambda star, the lambda value that comes packaged in with these two when you set the gradient of the Lagrangian equal to zero, equals the derivative of this maximum value, thought of as a function of b, maybe I should emphasize that, you know, we're thinking of this maximum value as a function of b, with respect to b. So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue?"}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the way to start writing all of that in formulas would be to make explicit the fact that if you consider this value, you know, the $10,000 that is your budget, which I'm calling b, a variable, and not a constant, then you have to acknowledge that h star and s star are dependent on b, right? It's a very implicit relationship, something that's kind of hard to think about at first, because as you change b, it changes what the Lagrangian is, which is going to change where its gradient equals zero, which changes what h star, s star, and lambda star are, but in principle, they are some function of that budget, of b. And the maximum possible revenue is whatever you get when you just plug in that solution to your function r, and the claim I made that I just pulled out of the hat is that lambda star, the lambda value that comes packaged in with these two when you set the gradient of the Lagrangian equal to zero, equals the derivative of this maximum value, thought of as a function of b, maybe I should emphasize that, you know, we're thinking of this maximum value as a function of b, with respect to b. So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue? So in a sense, this lambda star tells you for every dollar that you increase the budget, how much can your revenue increase if you're always maximizing it? So, why on earth is this true, right? This just seems like it comes out of nowhere."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's kind of a mouthful, it takes a lot just to even phrase what's going on, but in the context of an economic example, it has a very clear, precise meaning, which is if you increase your budget by a dollar, right, if you increase it from $10,000 to $10,001, you're wondering for that tiny change in budget, that tiny db, what is the ratio of the resulting change in revenue? So in a sense, this lambda star tells you for every dollar that you increase the budget, how much can your revenue increase if you're always maximizing it? So, why on earth is this true, right? This just seems like it comes out of nowhere. Well, there are a couple clever observations that go into proving this. The first, the first is to notice what happens if we evaluate this Lagrangian function itself at this critical point, when you input h star, s star, and lambda star. And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This just seems like it comes out of nowhere. Well, there are a couple clever observations that go into proving this. The first, the first is to notice what happens if we evaluate this Lagrangian function itself at this critical point, when you input h star, s star, and lambda star. And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star. So now, I'm just asking if you plug that not into the gradient of the Lagrangian, but to the Lagrangian itself, what do you get? Well, you're going to get, we just look at its definition up here, r evaluated at h star and s star, right? And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And remember, the way that these guys are defined is that you look at all of the values where the gradient of the Lagrangian equals the zero vector, and then if you get multiple options, you know, sometimes when you set the gradient equal to zero, you get multiple solutions, and whichever one maximizes r, that is h star, s star, lambda star. So now, I'm just asking if you plug that not into the gradient of the Lagrangian, but to the Lagrangian itself, what do you get? Well, you're going to get, we just look at its definition up here, r evaluated at h star and s star, right? And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to. Okay, Grant, you might say, why does this tell us anything? You're just plugging in stars instead of the usual variables. But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we subtract off lambda star times b of h star, s star, minus the constant that is your budget, you know, something you might think of as $10,000, whatever you set little b equal to. Okay, Grant, you might say, why does this tell us anything? You're just plugging in stars instead of the usual variables. But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint. Remember, one of the cool parts about this Lagrangian function as a whole is that when you take its partial derivative with respect to lambda, all that's left is this constraint function minus the constraint portion. When you set the gradient of the Lagrangian equal to the zero vector, one component of that is to set the partial derivative with respect to lambda equal to zero. And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right?"}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But the key is that if you plug in h star and s star, this value has to equal zero because h star and s star have to satisfy the constraint. Remember, one of the cool parts about this Lagrangian function as a whole is that when you take its partial derivative with respect to lambda, all that's left is this constraint function minus the constraint portion. When you set the gradient of the Lagrangian equal to the zero vector, one component of that is to set the partial derivative with respect to lambda equal to zero. And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right? Which would be your budget achieves $10,000. When you plug in the appropriate h star and s star to this value, you're hitting this constrained amount of money that you can spend. So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if you remember from the Lagrangian video, all that really boils down to is the fact that the constraint holds, right? Which would be your budget achieves $10,000. When you plug in the appropriate h star and s star to this value, you're hitting this constrained amount of money that you can spend. So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero. So we can just kind of cancel all that out. And what's left, what's left here is the maximum possible revenue, right? So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So by virtue of how h star and s star are defined, the fact that they are solutions to the constrained optimization problem means this whole portion goes to zero. So we can just kind of cancel all that out. And what's left, what's left here is the maximum possible revenue, right? So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star. It equals the maximum possible value for the function you're trying to maximize. So ultimately what we want is to understand how that maximum value changes when you consider it a function of the budget. So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So evidently, when you evaluate the Lagrangian at this critical point, at h star, s star, and lambda star, it equals m star. It equals the maximum possible value for the function you're trying to maximize. So ultimately what we want is to understand how that maximum value changes when you consider it a function of the budget. So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget. Now this is an interesting thing to observe, because if we just look up at the definition of the Lagrangian, if you just look at this formula, if I told you to take the derivative of this with respect to little b, right? You know, how much does this change with respect to little b? You would notice that this goes to zero, it doesn't have a little b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So evidently, what we can look for is to just ask how the Lagrangian changes as you consider it a function of the budget. Now this is an interesting thing to observe, because if we just look up at the definition of the Lagrangian, if you just look at this formula, if I told you to take the derivative of this with respect to little b, right? You know, how much does this change with respect to little b? You would notice that this goes to zero, it doesn't have a little b. This would also go to zero. And all you'd be left with would be negative lambda times negative b, and the derivative of that with respect to b would be lambda. So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You would notice that this goes to zero, it doesn't have a little b. This would also go to zero. And all you'd be left with would be negative lambda times negative b, and the derivative of that with respect to b would be lambda. So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda. And that's compelling, but ultimately, it's not entirely right. That overlooks the fact that L is not actually defined as a function of b. When we defined the Lagrangian, we were considering b to be a constant."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you might say, oh yeah, of course, of course, this, the derivative of that Lagrangian with respect to b, once we work it all out, the only term that was left there was the lambda. And that's compelling, but ultimately, it's not entirely right. That overlooks the fact that L is not actually defined as a function of b. When we defined the Lagrangian, we were considering b to be a constant. So if you really want to consider this to be a function that involves b, the way we should write it, and I'll go ahead and erase this guy, the way we should write this Lagrangian is to say, you're a function of h star, which itself is dependent on b, and s star, which is also a function of b, as soon as we start considering b a variable and not a constant, we have to acknowledge that this critical point, h star, s star, and lambda star, depends on the value of b. So likewise, that lambda star, lambda star, is also going to be a function of b. And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "When we defined the Lagrangian, we were considering b to be a constant. So if you really want to consider this to be a function that involves b, the way we should write it, and I'll go ahead and erase this guy, the way we should write this Lagrangian is to say, you're a function of h star, which itself is dependent on b, and s star, which is also a function of b, as soon as we start considering b a variable and not a constant, we have to acknowledge that this critical point, h star, s star, and lambda star, depends on the value of b. So likewise, that lambda star, lambda star, is also going to be a function of b. And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself. The value of b itself here. So now, when we want to know what is the value of the Lagrangian at the critical point, h star, s star, lambda star, as a function of b, so that can be kind of confusing, what you basically have is this function that only really depends on one value, right? It only depends on b, but it kind of goes through a four variable function."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we can consider it as a fourth variable, so we're adding on yet another variable to this function, the value of b itself. The value of b itself here. So now, when we want to know what is the value of the Lagrangian at the critical point, h star, s star, lambda star, as a function of b, so that can be kind of confusing, what you basically have is this function that only really depends on one value, right? It only depends on b, but it kind of goes through a four variable function. And so just to make it explicit, this would equal the value of r as a function of h star and s star, and each one of those is a function of little b, right? So this term is saying what's your revenue evaluated on the maximizing h and s for the given budget, and then you subtract off lambda star, oh here, I should probably, I'm not going to have room here, am I? So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right?"}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It only depends on b, but it kind of goes through a four variable function. And so just to make it explicit, this would equal the value of r as a function of h star and s star, and each one of those is a function of little b, right? So this term is saying what's your revenue evaluated on the maximizing h and s for the given budget, and then you subtract off lambda star, oh here, I should probably, I'm not going to have room here, am I? So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right? So you have this large, kind of complicated multivariable function, it's defined in terms of h stars and s stars, which are themselves very implicit, right? We just say, by definition, these are whatever values make the gradient of L equal zero, so very hard to think about what that means concretely, but all of this is really just dependent on the single value little b. And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what you subtract off minus lambda star at b of h star and s star, but each of these guys is also a function of little b, little b, minus little b, right? So you have this large, kind of complicated multivariable function, it's defined in terms of h stars and s stars, which are themselves very implicit, right? We just say, by definition, these are whatever values make the gradient of L equal zero, so very hard to think about what that means concretely, but all of this is really just dependent on the single value little b. And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule. And at this point, if you don't know the multivariable chain rule, I have a video on that, definitely pause, go take a look, make sure that it all makes sense, but right here, I'm just going to be assuming that you know what the multivariable chain rule is. So what it is, is we take the partial, we're going to look at the partial derivatives with respect to all four of these inputs. So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And from here, if we want to evaluate the derivative of L, we want to evaluate the derivative of this Lagrangian with respect to little b, which is really the only thing it depends on, it's just via all of these other variables, we use the multivariable chain rule. And at this point, if you don't know the multivariable chain rule, I have a video on that, definitely pause, go take a look, make sure that it all makes sense, but right here, I'm just going to be assuming that you know what the multivariable chain rule is. So what it is, is we take the partial, we're going to look at the partial derivatives with respect to all four of these inputs. So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b. And this might seem like a very hard thing to think about, like how do we know how h star changes as b changes? But don't worry about it, you'll see something magic happen in just a moment. And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we'll start with the partial derivative of L with respect to h star, with respect to h star, and we're going to multiply that by the derivative of h star with respect to b. And this might seem like a very hard thing to think about, like how do we know how h star changes as b changes? But don't worry about it, you'll see something magic happen in just a moment. And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b. You can see how you really need to know what the multivariable chain rule is, right? This would all seem kind of out of the blue. So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we add in partial derivative of L with respect to that second variable, s star, with respect to whatever the second variable is, multiplied by the derivative of s star with respect to b. You can see how you really need to know what the multivariable chain rule is, right? This would all seem kind of out of the blue. So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b. And then finally, finally, we take the partial derivative of this Lagrangian with respect to that little b, which we're now considering a variable in there, right? We're no longer considering that b a constant, multiplied by, well something kind of silly, the derivative of b with respect to itself. So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what we now add in is the partial derivative of L with respect to that lambda star, with respect to lambda star, multiplied by the derivative of lambda star with respect to little b. And then finally, finally, we take the partial derivative of this Lagrangian with respect to that little b, which we're now considering a variable in there, right? We're no longer considering that b a constant, multiplied by, well something kind of silly, the derivative of b with respect to itself. So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from. You have to know the derivative of lambda star with respect to b, you have to somehow intimately be familiar with how this lambda star changes as you change b, and like I said, that's such an implicit relationship, right? We just said that lambda star is by definition whatever the solution to this gradient equation is. So somehow you're supposed to know how that changes when you slightly alter b over here."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now, if you're thinking that this is going to be horrifying to compute, I can understand where you're coming from. You have to know the derivative of lambda star with respect to b, you have to somehow intimately be familiar with how this lambda star changes as you change b, and like I said, that's such an implicit relationship, right? We just said that lambda star is by definition whatever the solution to this gradient equation is. So somehow you're supposed to know how that changes when you slightly alter b over here. Well, you don't really have to worry about things, because by definition h star, s star, and lambda star are whatever values make the gradient of L equal to 0. But if you think about that, what does it mean for the gradient of L to equal the 0 vector? Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So somehow you're supposed to know how that changes when you slightly alter b over here. Well, you don't really have to worry about things, because by definition h star, s star, and lambda star are whatever values make the gradient of L equal to 0. But if you think about that, what does it mean for the gradient of L to equal the 0 vector? Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0. When you take its derivative with respect to the second variable, that equals 0 as well. And with respect to this third variable, that's going to equal 0. By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, what it means is that when you take its derivative with respect to that first variable, h star, it equals 0. When you take its derivative with respect to the second variable, that equals 0 as well. And with respect to this third variable, that's going to equal 0. By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0. So we don't even have to worry about most of this equation. The only part that matters here is the partial derivative of L with respect to b that we're now considering a variable multiplied by, well, what's db db? What is the rate of change of a variable with respect to itself?"}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "By definition, h star, s star, and lambda star are whatever values make it the case so that when you plug them in, the partial derivative of the Lagrangian with respect to any one of those variables equals 0. So we don't even have to worry about most of this equation. The only part that matters here is the partial derivative of L with respect to b that we're now considering a variable multiplied by, well, what's db db? What is the rate of change of a variable with respect to itself? It's 1. It is 1. So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What is the rate of change of a variable with respect to itself? It's 1. It is 1. So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b. And now, there's something very subtle here, right? Because this might seem obvious. I'm saying the derivative of L with respect to b equals the derivative of L with respect to b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all of this stuff, this entire multivariable chain rule boils down to a single innocent-looking factor which is the partial derivative of L partial derivative of L with respect to little b. And now, there's something very subtle here, right? Because this might seem obvious. I'm saying the derivative of L with respect to b equals the derivative of L with respect to b. But maybe I should give a different notation here, right? Because here when I'm taking the derivative, really I'm considering L as a single variable function, right? I'm considering not what happens as you can freely change all four of these variables."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm saying the derivative of L with respect to b equals the derivative of L with respect to b. But maybe I should give a different notation here, right? Because here when I'm taking the derivative, really I'm considering L as a single variable function, right? I'm considering not what happens as you can freely change all four of these variables. Three of them are locked into place by b. So maybe I should really give that a different name. I should call that L star."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm considering not what happens as you can freely change all four of these variables. Three of them are locked into place by b. So maybe I should really give that a different name. I should call that L star. L star is a single variable function. Whereas this L is a multivariable function. This is the function where you can freely change the values of h and s and lambda and b as you put them in."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I should call that L star. L star is a single variable function. Whereas this L is a multivariable function. This is the function where you can freely change the values of h and s and lambda and b as you put them in. So if we kind of scroll up to look at its definition, which I've written all over I guess here, let me actually rewrite its definition, right? I think that'll be useful. I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the function where you can freely change the values of h and s and lambda and b as you put them in. So if we kind of scroll up to look at its definition, which I've written all over I guess here, let me actually rewrite its definition, right? I think that'll be useful. I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b. And this is now when I'm considering little b to be a variable. So this is the Lagrangian when you consider all four of these to be freely changing as you want. Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm going to rewrite that L, if I consider it as a four variable function of h s lambda and b that what that equals is r evaluated at h and s minus lambda multiplied by this constraint function b evaluated at h and s minus little b. And this is now when I'm considering little b to be a variable. So this is the Lagrangian when you consider all four of these to be freely changing as you want. Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place. So effectively it's just a single variable function with respect to b. So it's actually quite miraculous that the single variable derivative of that L here, I should L star with respect to b ends up being the same as the partial derivative of L. This L where you're free to change all the variables that these should be the same. Usually in any usual circumstance all of these other terms would have come into play somehow."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Whereas the thing up here that I'm considering a single variable function has three of its inputs locked into place. So effectively it's just a single variable function with respect to b. So it's actually quite miraculous that the single variable derivative of that L here, I should L star with respect to b ends up being the same as the partial derivative of L. This L where you're free to change all the variables that these should be the same. Usually in any usual circumstance all of these other terms would have come into play somehow. But what's special here is that by the definition of this L star the specific way in which these h star, s star, and lambda stars are locked into place happens to be one in which all of these partial derivatives go to zero. So that's pretty subtle and I think it's quite clever and what it leaves us with is that we just have to evaluate this partial derivative which is quite simple because we look down here and you say what's the partial derivative of L with respect to b? Well this R has no b's in it so don't need to care about that."}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Usually in any usual circumstance all of these other terms would have come into play somehow. But what's special here is that by the definition of this L star the specific way in which these h star, s star, and lambda stars are locked into place happens to be one in which all of these partial derivatives go to zero. So that's pretty subtle and I think it's quite clever and what it leaves us with is that we just have to evaluate this partial derivative which is quite simple because we look down here and you say what's the partial derivative of L with respect to b? Well this R has no b's in it so don't need to care about that. This term over here it's partial derivative is negative one, right, just because there's a b here and that's multiplied by the constant lambda so that all just equals lambda. But if we're in the situation where lambda is locked into place as a function of little b then we'd write lambda star as a function of little b. Right?"}, {"video_title": "Proof for the meaning of Lagrange multipliers   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well this R has no b's in it so don't need to care about that. This term over here it's partial derivative is negative one, right, just because there's a b here and that's multiplied by the constant lambda so that all just equals lambda. But if we're in the situation where lambda is locked into place as a function of little b then we'd write lambda star as a function of little b. Right? So if that feels a little notationally confusing I'm right there with you, but the important part here the important thing to remember is that we just started considering b as a variable right, and we were looking at the h star, s star, and lambda star as they depended on that variable. We made the observation that the Lagrangian evaluated at that critical point equals the revenue evaluated at that critical point. The rest of the stuff just cancels out."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "Hey everyone. So in the last video I was talking about divergence and kind of laying down the intuition that we need for it. Where you're imagining a vector field as representing some kind of fluid flow where particles move according to the vector that they're attached to in that point in time. And as they move to a different point, the vector they're attached to is different, so their velocity changes in some way. And the key question that we want to think about is if you have a given point somewhere in space, does fluid tend to flow towards that point or does it tend to flow more away from it? Does it diverge away from that point? And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "And as they move to a different point, the vector they're attached to is different, so their velocity changes in some way. And the key question that we want to think about is if you have a given point somewhere in space, does fluid tend to flow towards that point or does it tend to flow more away from it? Does it diverge away from that point? And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves. Because ultimately that's what I'm gonna get to, a formula for divergence. But I want it to be something that's not just plopped down in front of you, but something that you actually feel deep in your bones. So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "And what I want to do here is start kind of closing our grasp on that intuition a little bit more tightly as if we are trying to discover the formula for divergence ourselves. Because ultimately that's what I'm gonna get to, a formula for divergence. But I want it to be something that's not just plopped down in front of you, but something that you actually feel deep in your bones. So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output. And it's common to, whoop, it's common to write p and q as the functions for the components of the output. So p and q are each just scalar-valued functions, and you think of them as the components of your vector-valued output. And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "So a vector field like the one I have pictured above is given as a function, a multivariable function, with a two-dimensional input, since it's a two-dimensional vector field, and then some kind of two-dimensional output. And it's common to, whoop, it's common to write p and q as the functions for the components of the output. So p and q are each just scalar-valued functions, and you think of them as the components of your vector-valued output. And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function. This div operator, you think of as taking in a vector field of some kind, and you get a new function. And the new function you get will be scalar-valued. It'll be something that just takes in points in space and outputs a number."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "And the divergence is kind of like a derivative, where you might denote it by just div, and in the same way that your derivative, you have this operator, and what it does is it takes in a function, and what you get is a whole new function. This div operator, you think of as taking in a vector field of some kind, and you get a new function. And the new function you get will be scalar-valued. It'll be something that just takes in points in space and outputs a number. Because what you're thinking, the thing that it's trying to do is take in some specific point with x, y coordinates and just give you a single number to tell you, hey, does fluid tend to diverge away from it? How much? Or does it tend to flow towards it, and how much?"}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "It'll be something that just takes in points in space and outputs a number. Because what you're thinking, the thing that it's trying to do is take in some specific point with x, y coordinates and just give you a single number to tell you, hey, does fluid tend to diverge away from it? How much? Or does it tend to flow towards it, and how much? So this is the kind of the form of the thing that we're going for. So here what we're gonna do is just start thinking about cases where this divergence is positive or negative or zero and what that should look like. So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "Or does it tend to flow towards it, and how much? So this is the kind of the form of the thing that we're going for. So here what we're gonna do is just start thinking about cases where this divergence is positive or negative or zero and what that should look like. So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive. What might that look like? So one case would be where your point, nothing is happening at that point, the vector attached to it is zero, and everyone around it is going away. This is kind of the extreme example of positive divergence."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "So for example, let's say we want cases where the divergence of our vector field at a specific point, at a specific point x, y, is positive. What might that look like? So one case would be where your point, nothing is happening at that point, the vector attached to it is zero, and everyone around it is going away. This is kind of the extreme example of positive divergence. And I animated this in the last video where we have all of the vectors pointing away from the origin, and if you look at a region around that origin, all the fluid particles kind of go out of that region. And that's the quintessential positive divergence example. But it doesn't have to look like that."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "This is kind of the extreme example of positive divergence. And I animated this in the last video where we have all of the vectors pointing away from the origin, and if you look at a region around that origin, all the fluid particles kind of go out of that region. And that's the quintessential positive divergence example. But it doesn't have to look like that. It actually, I mean, you could have something where there is a little bit of movement at your point, and then maybe there's movement towards it as well from one side, and vectors are kind of going towards it, but they're going away from it even more rapidly on the other side. So if you think of any kind of actual region around your point, you're saying, sure, fluid is going into that region a little bit, but it's much more counterbalanced by how quickly it's going out. So these are the kind of situations you might see for positive divergence."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "But it doesn't have to look like that. It actually, I mean, you could have something where there is a little bit of movement at your point, and then maybe there's movement towards it as well from one side, and vectors are kind of going towards it, but they're going away from it even more rapidly on the other side. So if you think of any kind of actual region around your point, you're saying, sure, fluid is going into that region a little bit, but it's much more counterbalanced by how quickly it's going out. So these are the kind of situations you might see for positive divergence. Now negative divergence, negative divergence, let's think about what examples of that might look like. Divergence of V at a given point, and really it's something that takes in all points of the plane, but we're just looking at specific points. So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "So these are the kind of situations you might see for positive divergence. Now negative divergence, negative divergence, let's think about what examples of that might look like. Divergence of V at a given point, and really it's something that takes in all points of the plane, but we're just looking at specific points. So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it. And this is the thing where I animated where we took this and we flipped all of the vectors and said, ah, there, if you start playing the fluid flow, then the density in any region around the origin increases a lot. All of the fluid particles tend to converge towards that center. But again, this isn't the only example that you might have."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "So if the divergence is negative, well, the quintessential example here is that nothing happens at your point, but all of the vectors around it are kind of flowing in towards it. And this is the thing where I animated where we took this and we flipped all of the vectors and said, ah, there, if you start playing the fluid flow, then the density in any region around the origin increases a lot. All of the fluid particles tend to converge towards that center. But again, this isn't the only example that you might have. You could have a little bit of activity at your point itself, and maybe it is the case that things do flow away from it a little bit as you're going away. And some of the fluid particles are going away, and it's just the case that the fluid particles flowing in towards it from another direction heavily counterbalance that. Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "But again, this isn't the only example that you might have. You could have a little bit of activity at your point itself, and maybe it is the case that things do flow away from it a little bit as you're going away. And some of the fluid particles are going away, and it's just the case that the fluid particles flowing in towards it from another direction heavily counterbalance that. Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end. So kind of loosely, intuitively, this is what a negative divergence case might look like. And finally, another case that we wanna start thinking about as we're tightening our grasp on this intuition is what happens, or what does it look like if the divergence of your function at a specific point is zero? Right, if it's just absolutely zero."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "Because then if you're looking at any kind of region around your point, you say fluid particles are coming in quite rapidly, a lot of particles per time, but they're not leaving too rapidly around the other end. So kind of loosely, intuitively, this is what a negative divergence case might look like. And finally, another case that we wanna start thinking about as we're tightening our grasp on this intuition is what happens, or what does it look like if the divergence of your function at a specific point is zero? Right, if it's just absolutely zero. And one thing this could look like is, you know, you have something going on, but nothing really changes, and all of the fluid just kind of flows in and it flows out, and on the whole it balances. You know, if you take any kind of region, the amount flowing in is balanced with the amount flowing out. But it could also look like you have fluid flowing in, kind of from one dimension, but it's canceled out by flowing away from the point in a manner that sort of perfectly balances it in another direction."}, {"video_title": "Divergence intuition, part 2.mp3", "Sentence": "Right, if it's just absolutely zero. And one thing this could look like is, you know, you have something going on, but nothing really changes, and all of the fluid just kind of flows in and it flows out, and on the whole it balances. You know, if you take any kind of region, the amount flowing in is balanced with the amount flowing out. But it could also look like you have fluid flowing in, kind of from one dimension, but it's canceled out by flowing away from the point in a manner that sort of perfectly balances it in another direction. So these are the loose pictures that I want you to have in the back of your mind as we start looking for the actual formula for divergence. And what I'll do in the next video or two is start looking at these functions p and q and thinking about the partial derivative properties that they have that will correspond with, you know, these positive divergence images that you should have in your head, or the negative divergence images that you should have in your head. So with that, I'll see you next video."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "So somehow our whole function, our function takes things from this two-dimensional space and plugs it onto this output. T, you're thinking of as just another number line up here, so t, and then you've got two separate functions here, you know, x of t and y of t, x of t and y of t, each of which take that same value for a specific input. You know, it's not that they're acting on different inputs, x of some other input t and y of some other input, it's the same one, and then they move that somewhere to this output space, which itself gets moved over. And in this way, you're thinking of it as just a single variable function that goes from t and ultimately outputs f, it's just that there's multidimensional stuff happening in between. And now if we start thinking about the derivative of it, what does that mean? What does that mean for the conception of the picture that we have going on here? Well, that bottom part, that dt, you're thinking of as a tiny change to t, right?"}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "And in this way, you're thinking of it as just a single variable function that goes from t and ultimately outputs f, it's just that there's multidimensional stuff happening in between. And now if we start thinking about the derivative of it, what does that mean? What does that mean for the conception of the picture that we have going on here? Well, that bottom part, that dt, you're thinking of as a tiny change to t, right? So you're thinking of it as kind of a nudge. I'll draw it as a sizable line here for like moving from some original input over, but you might in principle think of it as a very, very tiny nudge in t. And over here, you'd say, well, that's gonna move your intermediary output in the xy-plane to, you know, maybe it'll move it in some amount. Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "Well, that bottom part, that dt, you're thinking of as a tiny change to t, right? So you're thinking of it as kind of a nudge. I'll draw it as a sizable line here for like moving from some original input over, but you might in principle think of it as a very, very tiny nudge in t. And over here, you'd say, well, that's gonna move your intermediary output in the xy-plane to, you know, maybe it'll move it in some amount. Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it. And then whatever that nudge in the output space, and it's a nudge in some direction, that's gonna correspond to some change in f, some change based on, you know, based on the differential properties of the multivariable function itself. And if we think about this, this change, you might break it into components and say this shift here has some kind of dx, some kind of shift in the x direction and some kind of dy, some shift in the y direction. But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "Again, imagine this is a very small nudge, I'm gonna give it some size here just so I can write into it. And then whatever that nudge in the output space, and it's a nudge in some direction, that's gonna correspond to some change in f, some change based on, you know, based on the differential properties of the multivariable function itself. And if we think about this, this change, you might break it into components and say this shift here has some kind of dx, some kind of shift in the x direction and some kind of dy, some shift in the y direction. But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt. So if I go over here, I might say that dx is caused by that dt, and the whole meaning of the derivative, the whole meaning of the single variable derivative would be that when we take dx dt, this is the factor that tells us, you know, a tiny nudge in t, how much does that change the x component? And if you want, you could think of this as kind of canceling out the dts and you're just left with x. But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "But you can actually reason about what these should be because it's not just an arbitrary change in x or an arbitrary change in y, it's the one that was caused by dt. So if I go over here, I might say that dx is caused by that dt, and the whole meaning of the derivative, the whole meaning of the single variable derivative would be that when we take dx dt, this is the factor that tells us, you know, a tiny nudge in t, how much does that change the x component? And if you want, you could think of this as kind of canceling out the dts and you're just left with x. But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes. And similarly, similarly, that change in y here, that change in y is gonna be somehow proportional to the change in t, and that proportion is given by the derivative of y with respect to t, that's the whole point of the derivative, oh, no, no, no, with respect to t. And again, you can kind of think of it as if you're canceling out the t's, and this is why the fractional writing, this Leibniz notation is actually pretty helpful. You know, people will say, oh, mathematicians would like shake their heads at the idea of treating these like fractions, but not only is it a useful thing to do because it is a helpful mnemonic, it's reflective of what you're gonna do when you make a very formal argument, and I think I'll do that in one of the following videos. I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "But really, you're saying there's a tiny nudge in t and that results in a change in x, and this derivative is what tells you the ratio between those sizes. And similarly, similarly, that change in y here, that change in y is gonna be somehow proportional to the change in t, and that proportion is given by the derivative of y with respect to t, that's the whole point of the derivative, oh, no, no, no, with respect to t. And again, you can kind of think of it as if you're canceling out the t's, and this is why the fractional writing, this Leibniz notation is actually pretty helpful. You know, people will say, oh, mathematicians would like shake their heads at the idea of treating these like fractions, but not only is it a useful thing to do because it is a helpful mnemonic, it's reflective of what you're gonna do when you make a very formal argument, and I think I'll do that in one of the following videos. I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do. I don't think mathematicians are shaking their head every time that a student or a teacher does this, but anyway, so this is kind of what gives you what that dx is, what that dy is, and then over here, if you're saying, how much does that change the ultimate output of the f, you could say, well, your nudge of size dx over here, you're wondering how much that changes the output of f, that's the meaning of the partial derivative, right? If we say we have the partial derivative with respect to x, what that means is that if you take a tiny nudge of size x, this is giving you the ratio between that and the ultimate change to the output that you want. You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "I'll describe this in a much more formal way that's a little bit more airtight than the kind of hand-waving, nudging around, but the intuition you get from just writing this as a fraction is basically the scaffolding for that formal argument, so it's a fine thing to do. I don't think mathematicians are shaking their head every time that a student or a teacher does this, but anyway, so this is kind of what gives you what that dx is, what that dy is, and then over here, if you're saying, how much does that change the ultimate output of the f, you could say, well, your nudge of size dx over here, you're wondering how much that changes the output of f, that's the meaning of the partial derivative, right? If we say we have the partial derivative with respect to x, what that means is that if you take a tiny nudge of size x, this is giving you the ratio between that and the ultimate change to the output that you want. You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out. And similarly, that's just, you might call this the change in f caused by x, like due to x, due to, I should say, dx. But that's not the only thing changing the value of f, right, that's not the only change happening in the input space. You also have another change in f, and this one I might say is due to dy, due to that tiny shift in y, and what that's gonna be, we know it's gonna be proportional to that tiny shift in y, and the proportionality constant, this is the meaning of the partial derivative, that when you nudge y in some way, it results in some kind of nudge in f, and the ratio between those two is what the derivative gives."}, {"video_title": "Multivariable chain rule intuition.mp3", "Sentence": "You could kind of think of it like this partial x is canceling out with that dx if you wanted, or you could just say, this is a tiny nudge in x, this is gonna result in some change in f, I'm not sure what, but the meaning of the derivative is the ratio between those two, and that's what lets you figure it out. And similarly, that's just, you might call this the change in f caused by x, like due to x, due to, I should say, dx. But that's not the only thing changing the value of f, right, that's not the only change happening in the input space. You also have another change in f, and this one I might say is due to dy, due to that tiny shift in y, and what that's gonna be, we know it's gonna be proportional to that tiny shift in y, and the proportionality constant, this is the meaning of the partial derivative, that when you nudge y in some way, it results in some kind of nudge in f, and the ratio between those two is what the derivative gives. So ultimately, if you put this all together, what you'd say is there's two different things causing an ultimate change to f. So if you put these together, and you wanna know what the total change in f is, so I might go over here and say the total change in f, one of them is caused by partial f, partial x, and I can multiply it by dx here, but really, we know that dx, the change there, was in turn caused by dt, so that in turn is caused by the change in the x component that was due to dt, that was of course of size dt, and then for similar reasons, the other way that this changes in the y direction is a partial of f with respect to y, but what caused that initial shift in y? You'd say, well, that was a shift in y that was due to t, and that size of dy dt times dt, you could think of it. So slight nudge in t causes a change in y, that change in y causes a change in f, and when you add those two together, that's everything that's going on, that's everything that influences the ultimate change in f. So then if you take this whole expression and you divide everything out by dt, so I kind of erase it from this side and put it over here, dt, this is your multivariable chain rule, and of course, I've just written the same thing again, but hopefully this gives a little bit of an intuition for how you're composing different nudges and why you wanna think about it that way."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "So let's say you have yourself some kind of multivariable function, and this time let's say it's got some very high dimensional input. So x1, x2, on and on and on and on, up to x sub n for some large number n. In the last couple videos I told you about the Laplacian operator, which is a way of taking in your scalar valued function f, and it gives you a new scalar valued function. It's kind of like a second derivative thing, because it takes the divergence of the gradient of your function f. So the gradient of f gives you a vector field, and the divergence of that gives you a scalar field. And what I want to show you here is another formula that you might commonly see for this Laplacian. So first let's kind of abstractly write out what the gradient of f will look like. So we start by taking this del operator, which is gonna be a vector full of partial differential operators, partial with respect to x1, partial with respect to x2, and kind of on and on and on, up to partial with respect to whatever that last input variable is. You take that whole thing, and then you just kind of imagine multiplying it by your function."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "And what I want to show you here is another formula that you might commonly see for this Laplacian. So first let's kind of abstractly write out what the gradient of f will look like. So we start by taking this del operator, which is gonna be a vector full of partial differential operators, partial with respect to x1, partial with respect to x2, and kind of on and on and on, up to partial with respect to whatever that last input variable is. You take that whole thing, and then you just kind of imagine multiplying it by your function. So what you end up getting is all the different partial derivatives of f, right? It's partial of f with respect to the first variable, and then kind of on and on and on, up until you get the partial derivative of f with respect to that last variable, x sub n. And the divergence of that, and just to save myself some writing, I'm gonna say you take that nabla operator, and then you imagine taking the dot product between that whole operator and this gradient vector that you have here. What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "You take that whole thing, and then you just kind of imagine multiplying it by your function. So what you end up getting is all the different partial derivatives of f, right? It's partial of f with respect to the first variable, and then kind of on and on and on, up until you get the partial derivative of f with respect to that last variable, x sub n. And the divergence of that, and just to save myself some writing, I'm gonna say you take that nabla operator, and then you imagine taking the dot product between that whole operator and this gradient vector that you have here. What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable. So it looks like the second partial derivative of f with respect to that first variable. So second partial derivative of f with respect to x1, that first variable. And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "What you end up getting is, well you start by multiplying the first components, which involves taking the partial derivative with respect to x1, that first variable, of the partial derivative of f with respect to that same variable. So it looks like the second partial derivative of f with respect to that first variable. So second partial derivative of f with respect to x1, that first variable. And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared. And you do this to all of them, and you're adding them all up until you find yourself doing it to the last one. So you've got plus and then a whole bunch of things, and you'll be taking the second partial derivative of f with respect to that last variable, partial of x sub n. This is another format in which you might see the Laplacian, and oftentimes it's written kind of compactly. So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "And then you imagine kind of adding what the product of these next two items will be, and for very similar reasons, that's gonna look like the second partial derivative of f with respect to that second variable, partial x2 squared. And you do this to all of them, and you're adding them all up until you find yourself doing it to the last one. So you've got plus and then a whole bunch of things, and you'll be taking the second partial derivative of f with respect to that last variable, partial of x sub n. This is another format in which you might see the Laplacian, and oftentimes it's written kind of compactly. So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable. So, you know, if you were thinking in terms of three variables, often x1, x2, x3, we instead write x, y, z, but it's common to more generally just say x sub i. So this here is kind of the alternate formula that you might see for the Laplacian. Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "So people will say the Laplacian of your function f is equal to, and then using sigma notation, you'd say the sum from i is equal to one up to, you know, one, two, three, up to n. So the sum from that up to n of your second partial derivatives, partial squared of f with respect to that ith variable. So, you know, if you were thinking in terms of three variables, often x1, x2, x3, we instead write x, y, z, but it's common to more generally just say x sub i. So this here is kind of the alternate formula that you might see for the Laplacian. Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video. But this formula is probably a little more straightforward when it comes to actual computations, and, oh wait, sorry, I forgot a squared there, didn't I? So, partial x squared. So this is second derivative."}, {"video_title": "Explicit Laplacian formula.mp3", "Sentence": "Personally, I always like to think about it as taking the divergence of the gradient of f, because you're thinking about the gradient field and the divergence of that kind of corresponds to maxima and minima of your original function, which is what I talked about in the initial intuition of Laplacian video. But this formula is probably a little more straightforward when it comes to actual computations, and, oh wait, sorry, I forgot a squared there, didn't I? So, partial x squared. So this is second derivative. Yeah, so summing these second partial derivatives. And you can probably see this is kind of a more straightforward way to compute a given example that you might come across, and it also makes clear how the Laplacian is kind of an extension of the idea of a second derivative. See you next video."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can now express this as a double integral over the domain of the parameters that we care about. And we're going to do that in this video. And then in the next series of videos, we'll do the same thing for this expression. We're actually going to do that using Green's Theorem. What we're going to do is we're going to see we get the same expressions, which will show us that Stokes' Theorem is true, at least for this special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're actually going to do that using Green's Theorem. What we're going to do is we're going to see we get the same expressions, which will show us that Stokes' Theorem is true, at least for this special class of surfaces that we are studying right here. But they're pretty general. Now let's now try to do that. So our surface integral, I'm just going to rewrite it down here. It's the surface integral over our surface of the curl of f. Let me go a little bit lower. So we have our surface integral of the curl of f dot ds."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's now try to do that. So our surface integral, I'm just going to rewrite it down here. It's the surface integral over our surface of the curl of f. Let me go a little bit lower. So we have our surface integral of the curl of f dot ds. Well, we've already figured out what our curl of f is here two videos ago. And we've almost figured out what ds is. ds is a cross product of these two vectors times da."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we have our surface integral of the curl of f dot ds. Well, we've already figured out what our curl of f is here two videos ago. And we've almost figured out what ds is. ds is a cross product of these two vectors times da. The cross product of these two vectors is this right over here. So we could just write that ds is going to be equal to this thing times da. This is the cross product of the partial of r with respect to x and the partial of r with respect to y."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "ds is a cross product of these two vectors times da. The cross product of these two vectors is this right over here. So we could just write that ds is going to be equal to this thing times da. This is the cross product of the partial of r with respect to x and the partial of r with respect to y. And then we have to multiply that times da right over there. So this expression is just going to be the dot product of the curl of f, which is this stuff up here, dotted with this stuff down here. And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the cross product of the partial of r with respect to x and the partial of r with respect to y. And then we have to multiply that times da right over there. So this expression is just going to be the dot product of the curl of f, which is this stuff up here, dotted with this stuff down here. And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this. We can actually consider this to be a scalar value. So let's do that. So this is going to be equal to."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And essentially, we're going to take the dot product of this vector and that vector and then multiply it times this. We can actually consider this to be a scalar value. So let's do that. So this is going to be equal to. And when we do this, all of these, we're now going to start operating in the domain of our parameters. So it's going to turn from a surface integral into a double integral over that domain, over that region that we care about. This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to. And when we do this, all of these, we're now going to start operating in the domain of our parameters. So it's going to turn from a surface integral into a double integral over that domain, over that region that we care about. This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far. We've turned them into double integrals over the domain of the parameters. So this is going to turn into a double integral over the domain of our parameters, which is the region r. It's the region r in the xy-plane right up here. And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the domain of our parameters, this region r. And this is how we've manipulated any of the surface integrals that we've come across so far. We've turned them into double integrals over the domain of the parameters. So this is going to turn into a double integral over the domain of our parameters, which is the region r. It's the region r in the xy-plane right up here. And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here. Let me see if I can show both of them on the screen at the same time. There you go. So first, let's think about the x components."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we can take the dot product of the curl of f with the dotted ds, which is all of this business right over here. Let me see if I can show both of them on the screen at the same time. There you go. So first, let's think about the x components. So you have that right over there. And then you have this right over here. You multiply the two, the negative."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So first, let's think about the x components. So you have that right over there. And then you have this right over here. You multiply the two, the negative. We can swap the order right over there. You have the partial of z with respect to x times, and we're going to swap this order right over here, the partial of q with respect to z minus the partial of r with respect to y. Now let's think about the j component."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You multiply the two, the negative. We can swap the order right over there. You have the partial of z with respect to x times, and we're going to swap this order right over here, the partial of q with respect to z minus the partial of r with respect to y. Now let's think about the j component. You have negative z sub y times all of this up here, at least the stuff that's multiplied times the j component. This negative can cancel out with that negative. So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's think about the j component. You have negative z sub y times all of this up here, at least the stuff that's multiplied times the j component. This negative can cancel out with that negative. So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z. Let me make that clear. That is an r right over here. And then we have the k component."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have plus z sub y, the partial of z with respect to y, times the partial of r with respect to x minus the partial of p with respect to z. Let me make that clear. That is an r right over here. And then we have the k component. And the k component's actually the easiest because it's just one here. So it's just going to be 1 times, and I'll just do it in that same color, 1 times the partial of q with respect to x minus the partial of p with respect to y. And then finally, we just have this dA over here."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we have the k component. And the k component's actually the easiest because it's just one here. So it's just going to be 1 times, and I'll just do it in that same color, 1 times the partial of q with respect to x minus the partial of p with respect to y. And then finally, we just have this dA over here. And this dA is multiplied by everything. So we'll put some parentheses, and we'll write dA. So we're done."}, {"video_title": "Stokes' theorem proof part 3   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then finally, we just have this dA over here. And this dA is multiplied by everything. So we'll put some parentheses, and we'll write dA. So we're done. We've expressed our surface integral as a double integral over the domain of our parameters. And what we're going to do in the next few videos is do the same thing with this using Green's theorem. We're going to see that we get the exact same value."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "They can be types one, type two, and type three, and for a lot of situations that aren't simple solid regions, you can break them up into simple solid regions, but let's just prove it for this case, for this case right over here. So let's just assume that our vector field F, our vector field F can be written as P, which is a function of X, Y, and Z, times I, plus Q, which is a function of X, Y, and Z, times J, plus R, which is a function of X, Y, and Z, times K. So let's think about what each of these sides of the equation would come out to be. Well, first of all, what is going to be F dot N? So let's think about that a little bit. F dot N, F dot N, F dot N, is going to be equal to this, this component right over here, times N's I component, plus this component right over here, times N's J component, plus this component here, times N's K component. So we could write it as, it could be written as P times, P times, or I'll just write P open parentheses, the dot product of I and N, I dotted N, I dotted N, and let me make sure I write I as a vector, as a unit vector. I want to be clear, what's going to happen right over here?"}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's think about that a little bit. F dot N, F dot N, F dot N, is going to be equal to this, this component right over here, times N's I component, plus this component right over here, times N's J component, plus this component here, times N's K component. So we could write it as, it could be written as P times, P times, or I'll just write P open parentheses, the dot product of I and N, I dotted N, I dotted N, and let me make sure I write I as a vector, as a unit vector. I want to be clear, what's going to happen right over here? If you take the dot product of I and N, you're just going to get the I component, the scaling factor of the I component, of the N normal vector, and we're just going to multiply that times P. So that's essentially the product of the X components, or I guess you could say the magnitude of the X components. And then to that, we are going to add Q times J, J dotted with N. And once again, when you dot J with N, you get the magnitude of the J component of the normal vector right over there. And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to be clear, what's going to happen right over here? If you take the dot product of I and N, you're just going to get the I component, the scaling factor of the I component, of the N normal vector, and we're just going to multiply that times P. So that's essentially the product of the X components, or I guess you could say the magnitude of the X components. And then to that, we are going to add Q times J, J dotted with N. And once again, when you dot J with N, you get the magnitude of the J component of the normal vector right over there. And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true. This right over here is going to be equal to P times the magnitude of the I component of N's normal vector, which is exactly what we want in a dot product. This is the same thing for the J component. This is the same thing for the K component."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then times, or plus, I should say, plus R times K dotted with N. K dotted with N. This isn't how we normally see it, but I think it's reasonable to say that this is actually true. This right over here is going to be equal to P times the magnitude of the I component of N's normal vector, which is exactly what we want in a dot product. This is the same thing for the J component. This is the same thing for the K component. And you could try it out. Define N is equal to, I don't know, M times I plus N times J plus O times K or something like that, and you'll see that this actually does work out fine. So how can we simplify this expression right up here?"}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the same thing for the K component. And you could try it out. Define N is equal to, I don't know, M times I plus N times J plus O times K or something like that, and you'll see that this actually does work out fine. So how can we simplify this expression right up here? Well, we can rewrite this as, we can rewrite the left-hand side as, so the surface integral of F, and let me write it multiple ways, F dot DS, which is equal to the surface integral, the surface integral of F dot N times the scalar DS, is equal to the double integral of the surface of all of this business right over here, is equal to the double integral over the surface of, let me just copy and paste that, of all of that business. So let me copy and then let me paste. So all of that business right over there."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So how can we simplify this expression right up here? Well, we can rewrite this as, we can rewrite the left-hand side as, so the surface integral of F, and let me write it multiple ways, F dot DS, which is equal to the surface integral, the surface integral of F dot N times the scalar DS, is equal to the double integral of the surface of all of this business right over here, is equal to the double integral over the surface of, let me just copy and paste that, of all of that business. So let me copy and then let me paste. So all of that business right over there. I just noticed that I forgot to put the little unit vector symbol, the little caret, right over there. Put some parentheses, and then we are left with our DS. And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all of that business right over there. I just noticed that I forgot to put the little unit vector symbol, the little caret, right over there. Put some parentheses, and then we are left with our DS. And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS. So I just broke it up, we were taking the integral of this sum, and so I just took it as, I just rewrote it as the sum of the integrals. So that's the left-hand side right over here. Now let's think about the right-hand side."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then this, all of this can be rewritten as, this can be written as the surface integral of P times this business, and I'll just do it in the same color, of P times the dot product of I and N, DS, plus the surface integral of Q times the dot product of J and N, DS, plus the surface integral of R times the dot product of K and N, forgot a caret, K and N, DS. So I just broke it up, we were taking the integral of this sum, and so I just took it as, I just rewrote it as the sum of the integrals. So that's the left-hand side right over here. Now let's think about the right-hand side. What is the divergence of F? And actually I'm gonna take some space up here. What is the divergence of F?"}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's think about the right-hand side. What is the divergence of F? And actually I'm gonna take some space up here. What is the divergence of F? Well the divergence of F, based on this expression of F, is just going to be, let me just write it over here real small, the divergence of F is going to be the partial of P with respect to, let me do this in a new color, just because I'm using the yellow too much. The divergence of F, the divergence of F is going to be the partial of P with respect to X, plus the partial of Q with respect to Y, plus the partial of R, plus the partial of R with respect to Z. So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What is the divergence of F? Well the divergence of F, based on this expression of F, is just going to be, let me just write it over here real small, the divergence of F is going to be the partial of P with respect to, let me do this in a new color, just because I'm using the yellow too much. The divergence of F, the divergence of F is going to be the partial of P with respect to X, plus the partial of Q with respect to Y, plus the partial of R, plus the partial of R with respect to Z. So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z. Well this thing once again, Instead of writing it as the triple integral of this sum, we could write it as the sum of triple integrals. So this thing right over here can be rewritten as the triple integral over our three dimensional region. Actually, let me copy and paste that so I don't have to keep rewriting it."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we can, this triple integral right over here could be written as the triple integral of the partial of P with respect to X plus the partial of Q with respect to Y, plus the partial of R with respect to Z. Well this thing once again, Instead of writing it as the triple integral of this sum, we could write it as the sum of triple integrals. So this thing right over here can be rewritten as the triple integral over our three dimensional region. Actually, let me copy and paste that so I don't have to keep rewriting it. So let me copy it. So it's going to be equal to the triple integral of the partial of p with respect to x dv plus, let me paste that again, triple integral of the partial of q with respect to y dv plus, once again, triple integral of the partial of r with respect to z dv. So we've essentially restated our divergence theorem."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually, let me copy and paste that so I don't have to keep rewriting it. So let me copy it. So it's going to be equal to the triple integral of the partial of p with respect to x dv plus, let me paste that again, triple integral of the partial of q with respect to y dv plus, once again, triple integral of the partial of r with respect to z dv. So we've essentially restated our divergence theorem. This is our surface integral, and the divergence theorem says that this needs to be equal to this business right over here. We've just written it a different way. And so what I'm going to do in order to prove it is just show that each of these corresponding terms are equal to each other."}, {"video_title": "Divergence theorem proof (part 1)   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we've essentially restated our divergence theorem. This is our surface integral, and the divergence theorem says that this needs to be equal to this business right over here. We've just written it a different way. And so what I'm going to do in order to prove it is just show that each of these corresponding terms are equal to each other. That these are equal to each other. And in particular, we're going to focus the proof on this, and we're going to use the fact that our region is a type 1 region. It's a type 1, type 2, and type 3."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "So I've talked about the partial derivative and how you compute it, how you interpret it in terms of graphs, but what I'd like to do here is give its formal definition. So it's the kind of thing, just to remind you, that applies to a function that has a multivariable input. So x, y, and I'll just emphasize that it could actually be a number of other inputs. You could have 100 inputs or something like that. And as with a lot of things here, I think it's helpful to take a look at the one-dimensional analogy and think about how we define the derivative, just the ordinary derivative, when you have a function that's just one variable. You know, this would be just something simple, f of x, and you know, maybe you're thinking in the back of your mind that it's a function like f of x equals x squared. And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "You could have 100 inputs or something like that. And as with a lot of things here, I think it's helpful to take a look at the one-dimensional analogy and think about how we define the derivative, just the ordinary derivative, when you have a function that's just one variable. You know, this would be just something simple, f of x, and you know, maybe you're thinking in the back of your mind that it's a function like f of x equals x squared. And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization. So you might be thinking of the graph of this function. You know, maybe it's some kind of curve. And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And the way to think about the definition of this is to just actually spell out how we interpret this df and dx and then slowly start to tighten it up into a formalization. So you might be thinking of the graph of this function. You know, maybe it's some kind of curve. And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value. So this is in the x direction. You've got your x-coordinate, your output, f of x, is what the y-axis represents here. And then you're thinking of df as being the resulting nudge here, the resulting change to the function."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And when you think of evaluating it at some point, you know, let's say you're evaluating it at a point a, you're imagining dx here as representing a slight nudge, just a slight nudge in the input value. So this is in the x direction. You've got your x-coordinate, your output, f of x, is what the y-axis represents here. And then you're thinking of df as being the resulting nudge here, the resulting change to the function. So when we formalize this, we're gonna be thinking of a fraction that's gonna represent df over dx. And I'll leave myself some room. You can probably anticipate why if you know where this is going."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And then you're thinking of df as being the resulting nudge here, the resulting change to the function. So when we formalize this, we're gonna be thinking of a fraction that's gonna represent df over dx. And I'll leave myself some room. You can probably anticipate why if you know where this is going. And instead of saying dx, I'll say h. So instead of thinking, you know, dx is that tiny nudge, you'll think h. And I'm not sure why h is used necessarily, but just having some kind of variable that you think of as getting small. Maybe all the other letters in the alphabet were taken. And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge?"}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "You can probably anticipate why if you know where this is going. And instead of saying dx, I'll say h. So instead of thinking, you know, dx is that tiny nudge, you'll think h. And I'm not sure why h is used necessarily, but just having some kind of variable that you think of as getting small. Maybe all the other letters in the alphabet were taken. And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge? So when you take, you know, from that input point plus that nudge, plus that little h, what's the difference between that and the original function just, or the original value of the function at that point? So this top part is really what's representing df. You know, this is what's representing df over here."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And now, when you actually say, what do we mean by the resulting change in f, we should be writing as well, where does it go after you nudge? So when you take, you know, from that input point plus that nudge, plus that little h, what's the difference between that and the original function just, or the original value of the function at that point? So this top part is really what's representing df. You know, this is what's representing df over here. But you don't do this for any actual value of h. You don't do it for any specific nudge. The whole point, largely the whole point of calculus is that you're considering the limit as h goes to zero of this. And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "You know, this is what's representing df over here. But you don't do this for any actual value of h. You don't do it for any specific nudge. The whole point, largely the whole point of calculus is that you're considering the limit as h goes to zero of this. And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change. It's not that it's any specific one. You're taking the limit. And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And this is what makes concrete the idea of, you know, a tiny little nudge or a tiny little resulting change. It's not that it's any specific one. You're taking the limit. And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this. Now, over in the multivariable world, very similar story. We can pretty much do the same thing and we're gonna look at our original fraction and just start to formalize what we think of each of these variables as representing. That partial x still is common to use the letter h just to represent a tiny nudge in the x direction."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And, you know, you could get into the formal definition of a limit, but it gives you room for rigor as soon as you start writing something like this. Now, over in the multivariable world, very similar story. We can pretty much do the same thing and we're gonna look at our original fraction and just start to formalize what we think of each of these variables as representing. That partial x still is common to use the letter h just to represent a tiny nudge in the x direction. And now if we think about what is that nudge, and here, let me draw it out actually. The way that I kind of like to draw this out is you think of your entire input space as, you know, the xy-plane. If it was more variables, this would be a high-dimensional space."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "That partial x still is common to use the letter h just to represent a tiny nudge in the x direction. And now if we think about what is that nudge, and here, let me draw it out actually. The way that I kind of like to draw this out is you think of your entire input space as, you know, the xy-plane. If it was more variables, this would be a high-dimensional space. And you're thinking of some point, you know, maybe you're thinking of it as a, b. Or maybe I should specify that actually. We're, you know, we're doing this at a specific point, how you define it."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "If it was more variables, this would be a high-dimensional space. And you're thinking of some point, you know, maybe you're thinking of it as a, b. Or maybe I should specify that actually. We're, you know, we're doing this at a specific point, how you define it. We're doing this at a very specific point, a, b. And when you're thinking of your tiny little change in x, you'd be thinking, you know, a tiny little nudge in the x direction, tiny little shift there. And the entire function maps that input space, whatever it is, to the real number line."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "We're, you know, we're doing this at a specific point, how you define it. We're doing this at a very specific point, a, b. And when you're thinking of your tiny little change in x, you'd be thinking, you know, a tiny little nudge in the x direction, tiny little shift there. And the entire function maps that input space, whatever it is, to the real number line. This is your output space. And you're saying, hey, how does that tiny nudge influence the output? I've drawn this diagram a lot, just this loose sketch."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And the entire function maps that input space, whatever it is, to the real number line. This is your output space. And you're saying, hey, how does that tiny nudge influence the output? I've drawn this diagram a lot, just this loose sketch. I think it's actually a pretty good model because once we start thinking of higher-dimensional outputs or things like that, it's pretty flexible. And you're thinking of this as your partial f, partial x. Sorry, you're changing the x direction."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "I've drawn this diagram a lot, just this loose sketch. I think it's actually a pretty good model because once we start thinking of higher-dimensional outputs or things like that, it's pretty flexible. And you're thinking of this as your partial f, partial x. Sorry, you're changing the x direction. And this is that resulting change to the function. But we go back up here and we say, well, what does that mean, right? If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "Sorry, you're changing the x direction. And this is that resulting change to the function. But we go back up here and we say, well, what does that mean, right? If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x. And the point b, the point b just remains unchanged, right? So this is you evaluating it kind of at the new point. And you have to say, what's the difference between that and the old evaluation where it was just at a and b?"}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "If h represents that tiny change to your x value, well, then you have to evaluate the function at the point a, but plus that h. And you're adding it to the x value, that first component, just because this is the partial derivative with respect to x. And the point b, the point b just remains unchanged, right? So this is you evaluating it kind of at the new point. And you have to say, what's the difference between that and the old evaluation where it was just at a and b? And that's it. That's the formal definition of your partial derivative. Except, oh, I mean, the most important part, right?"}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And you have to say, what's the difference between that and the old evaluation where it was just at a and b? And that's it. That's the formal definition of your partial derivative. Except, oh, I mean, the most important part, right? The most important part, given that this is calculus, is that we're not doing this for any specific value of h, but we're actually, actually, let me just move this guy, give a little bit of room here. Yes. But we're actually taking the limit here, limit as h goes to zero."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "Except, oh, I mean, the most important part, right? The most important part, given that this is calculus, is that we're not doing this for any specific value of h, but we're actually, actually, let me just move this guy, give a little bit of room here. Yes. But we're actually taking the limit here, limit as h goes to zero. And what this means is you're not considering any specific size of dx, any specific size of this. And clearly, this is h, considering the notation up here. But any size for that partial x."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "But we're actually taking the limit here, limit as h goes to zero. And what this means is you're not considering any specific size of dx, any specific size of this. And clearly, this is h, considering the notation up here. But any size for that partial x. You're imagining that nudge shrinking more and more and more and the resulting change shrinks more and more and more, and you're wondering what the ratio between them approaches. So that would be the partial derivative with respect to x. And just for practice, let's actually write out what the partial derivative with respect to y would be."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "But any size for that partial x. You're imagining that nudge shrinking more and more and more and the resulting change shrinks more and more and more, and you're wondering what the ratio between them approaches. So that would be the partial derivative with respect to x. And just for practice, let's actually write out what the partial derivative with respect to y would be. So I'll get rid of some of this one-dimensional analogy stuff here. Don't need that anymore. And let's just think about what the partial derivative with respect to a different variable would be."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And just for practice, let's actually write out what the partial derivative with respect to y would be. So I'll get rid of some of this one-dimensional analogy stuff here. Don't need that anymore. And let's just think about what the partial derivative with respect to a different variable would be. So if we were doing it as partial derivative of f with respect to y, now we're nudging slightly in the other direction, right? We're nudging in the y direction. And you'd be thinking, okay, so we're still gonna divide something by that nudge."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And let's just think about what the partial derivative with respect to a different variable would be. So if we were doing it as partial derivative of f with respect to y, now we're nudging slightly in the other direction, right? We're nudging in the y direction. And you'd be thinking, okay, so we're still gonna divide something by that nudge. And again, I'm just using the same variable. Maybe it would be clearer to write something like the change in y, or to go up here and write something like, you know, the change in x. And people will do that, but it's less common."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And you'd be thinking, okay, so we're still gonna divide something by that nudge. And again, I'm just using the same variable. Maybe it would be clearer to write something like the change in y, or to go up here and write something like, you know, the change in x. And people will do that, but it's less common. I think people just kinda want the standard go-to limiting variable. But this time, when you're considering what is the resulting change. Oh, and again, I always forget to write in."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And people will do that, but it's less common. I think people just kinda want the standard go-to limiting variable. But this time, when you're considering what is the resulting change. Oh, and again, I always forget to write in. We're evaluating this at a specific point, at a specific point, a, b. And as a result, maybe I'll give myself a little bit more room here. So we're taking this whole thing, dividing by h, but what is the resulting change in f?"}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "Oh, and again, I always forget to write in. We're evaluating this at a specific point, at a specific point, a, b. And as a result, maybe I'll give myself a little bit more room here. So we're taking this whole thing, dividing by h, but what is the resulting change in f? This time, you say f, the new value is still gonna be at a, but the change happens for that second variable. It's gonna be that b, b plus h. So you're adding that nudge to the y value. And as before, you subtract off."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "So we're taking this whole thing, dividing by h, but what is the resulting change in f? This time, you say f, the new value is still gonna be at a, but the change happens for that second variable. It's gonna be that b, b plus h. So you're adding that nudge to the y value. And as before, you subtract off. You see the difference between that and how you evaluate it at the original point. And again, the whole reason I moved this over and gave myself some room is because we're taking the limit as this h goes to zero. And the way that you're thinking about this is very similar."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And as before, you subtract off. You see the difference between that and how you evaluate it at the original point. And again, the whole reason I moved this over and gave myself some room is because we're taking the limit as this h goes to zero. And the way that you're thinking about this is very similar. It's just that when you change the input by adding h to the y value, you're shifting it upwards. So again, this is the partial derivative, the formal definition of the partial derivative. Looks very similar to the formal definition of the derivative, but I just always think about this as spelling out what we mean by partial y and partial f and kind of spelling out why it is that the Leibniz came up with this notation in the first place."}, {"video_title": "Formal definition of partial derivatives.mp3", "Sentence": "And the way that you're thinking about this is very similar. It's just that when you change the input by adding h to the y value, you're shifting it upwards. So again, this is the partial derivative, the formal definition of the partial derivative. Looks very similar to the formal definition of the derivative, but I just always think about this as spelling out what we mean by partial y and partial f and kind of spelling out why it is that the Leibniz came up with this notation in the first place. Well, I don't know if Leibniz came up with the partials, but the df dx portion, and this is good to keep in the back of your mind, especially as we introduce new notions, new types of multivariable derivatives, like the directional derivative. I think it helps clarify what's really going on in certain contexts. Great, see you next video."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And as you can see, we can kind of decompose it into three separate surfaces. The first surface is its base, which is really the filled-in unit circle down here. The second surface, which we have in blue, you can kind of view it as the side of it. You can view it as it's kind of like the side of a cylinder, but the cylinder has been cut by a plane up here. The plane that cuts it is the plane z is equal to 1 minus x. And obviously, the plane itself goes well beyond the shape. But where that plane cuts the cylinder kind of defines this shape."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "You can view it as it's kind of like the side of a cylinder, but the cylinder has been cut by a plane up here. The plane that cuts it is the plane z is equal to 1 minus x. And obviously, the plane itself goes well beyond the shape. But where that plane cuts the cylinder kind of defines this shape. So the blue surface is above the boundary of the unit circle and below the plane. And then the third surface is the subset of the plane, of that purple plane of z equals 1 minus x, that overlaps, that kind of forms the top of this cylinder. And so we can rewrite this surface integral as the sum of three surface integrals."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "But where that plane cuts the cylinder kind of defines this shape. So the blue surface is above the boundary of the unit circle and below the plane. And then the third surface is the subset of the plane, of that purple plane of z equals 1 minus x, that overlaps, that kind of forms the top of this cylinder. And so we can rewrite this surface integral as the sum of three surface integrals. It's the surface integral of z over S1 plus the surface integral of z over S2 plus the surface integral of z over S3. And we can just tackle each of these independently. So let's start with surface 1."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And so we can rewrite this surface integral as the sum of three surface integrals. It's the surface integral of z over S1 plus the surface integral of z over S2 plus the surface integral of z over S3. And we can just tackle each of these independently. So let's start with surface 1. And you might immediately want to start parameterizing things and all the rest, but there's actually a very fast way to handle this surface integral, especially because we're taking the surface integral of z. What value does z take on throughout this surface, throughout this filled in unit circle right over here? Well, that surface is on the xy plane."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So let's start with surface 1. And you might immediately want to start parameterizing things and all the rest, but there's actually a very fast way to handle this surface integral, especially because we're taking the surface integral of z. What value does z take on throughout this surface, throughout this filled in unit circle right over here? Well, that surface is on the xy plane. When we're on the xy plane, z is equal to 0. So for this entire surface, z is going to be equal to 0. You're essentially integrating 0."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "Well, that surface is on the xy plane. When we're on the xy plane, z is equal to 0. So for this entire surface, z is going to be equal to 0. You're essentially integrating 0. 0 times dS is just going to be 0. So this whole thing is going to evaluate to 0. So that's about as simple as evaluating a surface integral can get."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "You're essentially integrating 0. 0 times dS is just going to be 0. So this whole thing is going to evaluate to 0. So that's about as simple as evaluating a surface integral can get. But it's always important to at least keep a lookout for things like that. And it'll keep you from going down this wild goose chase. That wouldn't be a wild goose chase."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So that's about as simple as evaluating a surface integral can get. But it's always important to at least keep a lookout for things like that. And it'll keep you from going down this wild goose chase. That wouldn't be a wild goose chase. You would eventually get the answer anyway. But you don't have to waste so much time in parameterizing things and all the rest. Now let's tackle the other two surfaces."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "That wouldn't be a wild goose chase. You would eventually get the answer anyway. But you don't have to waste so much time in parameterizing things and all the rest. Now let's tackle the other two surfaces. And we'll focus on surface 2. So the x and y values, the valid x and y values, are the x and y values along the unit circle right over here. So we can really parameterize it the way that we would parameterize a traditional unit circle."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "Now let's tackle the other two surfaces. And we'll focus on surface 2. So the x and y values, the valid x and y values, are the x and y values along the unit circle right over here. So we can really parameterize it the way that we would parameterize a traditional unit circle. So we could set x is equal to, let's set it, and our radius is 1. So x is equal to cosine of, I'll use the parameter u, is equal to x is equal to cosine of u. And then let's say y is equal to sine of u."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So we can really parameterize it the way that we would parameterize a traditional unit circle. So we could set x is equal to, let's set it, and our radius is 1. So x is equal to cosine of, I'll use the parameter u, is equal to x is equal to cosine of u. And then let's say y is equal to sine of u. And then u is the angle between the positive x-axis and wherever we are essentially on that unit circle. So that right over there, that angle right over there is u. And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And then let's say y is equal to sine of u. And then u is the angle between the positive x-axis and wherever we are essentially on that unit circle. So that right over there, that angle right over there is u. And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle. So those are all the possible x and y values that we can take on. And then the z value is what takes us up above that boundary and gets it someplace along the surface. But this is interesting."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And so this would give us for u, so as long as u is between 0 and 2 pi, that we're essentially going around this unit circle. So those are all the possible x and y values that we can take on. And then the z value is what takes us up above that boundary and gets it someplace along the surface. But this is interesting. Because the z value, it can take on obviously a bunch of different values. But it always has to be below this plane right over here. So for this z value, I'm going to introduce a new parameter."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "But this is interesting. Because the z value, it can take on obviously a bunch of different values. But it always has to be below this plane right over here. So for this z value, I'm going to introduce a new parameter. Let's call z v. That's the second parameter. And v is going to be, it's definitely going to be greater than 0. z and v, they're the same thing. They're always going to be positive."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So for this z value, I'm going to introduce a new parameter. Let's call z v. That's the second parameter. And v is going to be, it's definitely going to be greater than 0. z and v, they're the same thing. They're always going to be positive. So they're definitely going to be greater than or equal to 0. But it's not less than or equal to some constant thing. This has kind of a variable roof on it."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "They're always going to be positive. So they're definitely going to be greater than or equal to 0. But it's not less than or equal to some constant thing. This has kind of a variable roof on it. And so it's always going to be less than or equal to this plane right over here. And so we could say, is less than or equal to 1 minus x. I mean, we could have said, so we know that z is less than or equal to 1 minus x. But if we use the new parameters, v is z."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "This has kind of a variable roof on it. And so it's always going to be less than or equal to this plane right over here. And so we could say, is less than or equal to 1 minus x. I mean, we could have said, so we know that z is less than or equal to 1 minus x. But if we use the new parameters, v is z. And 1 minus x is the same thing as 1 minus cosine u. And so now we have our parameterization. We are ready to actually evaluate the surface integral."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "But if we use the new parameters, v is z. And 1 minus x is the same thing as 1 minus cosine u. And so now we have our parameterization. We are ready to actually evaluate the surface integral. And to do that, first let's do the cross product. We want to figure out what ds is. And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "We are ready to actually evaluate the surface integral. And to do that, first let's do the cross product. We want to figure out what ds is. And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts. So ds, and this is all, let me do it in blue still, because we're talking about surface 2. ds is going to be equal to the magnitude of the cross product of our parameterization with the partial of our parameterization with respect to u, crossed with our partial of our parameterization with respect to v, du, dv. So let's write the partial of our parameterization with respect to u. And you might say, wait, where's our parameterization?"}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And we have to take the magnitude of the cross product of the partial of our parameterization with respect to u, cross with the partial with respect to v. Actually, let me just write all this stuff down instead of trying to do shortcuts. So ds, and this is all, let me do it in blue still, because we're talking about surface 2. ds is going to be equal to the magnitude of the cross product of our parameterization with the partial of our parameterization with respect to u, crossed with our partial of our parameterization with respect to v, du, dv. So let's write the partial of our parameterization with respect to u. And you might say, wait, where's our parameterization? Well, it's right over here. I just didn't write it in the traditional ijk form. But I can."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And you might say, wait, where's our parameterization? Well, it's right over here. I just didn't write it in the traditional ijk form. But I can. I could write r is equal to, maybe I call it r2, because we're talking about surface 2. I shouldn't use that orange color, because I used that for surface 1. I'll use that in a still bluish color."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "But I can. I could write r is equal to, maybe I call it r2, because we're talking about surface 2. I shouldn't use that orange color, because I used that for surface 1. I'll use that in a still bluish color. So r for surface 2 is equal to cosine of ui plus sine of uj plus vk. And this is the range that the u's and the v's can take on. If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "I'll use that in a still bluish color. So r for surface 2 is equal to cosine of ui plus sine of uj plus vk. And this is the range that the u's and the v's can take on. If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui. Derivative of sine of u with respect to u is plus cosine uj. And the derivative of v with respect to u is just 0. So that's our partial with respect to u."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "If I want to take the partial of r with respect to u, I would get, let's see, the derivative of cosine u with respect to u is negative sine ui. Derivative of sine of u with respect to u is plus cosine uj. And the derivative of v with respect to u is just 0. So that's our partial with respect to u. Our partial, we'll do this in a different color, just for the sake of it. Our partial with respect to v is equal to, well, this is going to be 0, this is going to be 0, and we're just going to have 1k. So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So that's our partial with respect to u. Our partial, we'll do this in a different color, just for the sake of it. Our partial with respect to v is equal to, well, this is going to be 0, this is going to be 0, and we're just going to have 1k. So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product. So we'll evaluate the cross product, and then we'll take the magnitude of that. So just this part, so just the cross product, the inside of that, before we take the magnitude. So copy and paste."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So it's just going to be equal to k. And so now we're ready to at least evaluate this cross product. So we'll evaluate the cross product, and then we'll take the magnitude of that. So just this part, so just the cross product, the inside of that, before we take the magnitude. So copy and paste. So just that is going to be equal to the determinant, the 3 by 3 determinant. Let me write my components i, j, k. And then for r sub u, we have this. Our i component is negative sine of u."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So copy and paste. So just that is going to be equal to the determinant, the 3 by 3 determinant. Let me write my components i, j, k. And then for r sub u, we have this. Our i component is negative sine of u. Our j component is cosine of u. It has no k components. We'll put a 0 right there."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "Our i component is negative sine of u. Our j component is cosine of u. It has no k components. We'll put a 0 right there. And then r sub v, no i component, no j component, and it has 1 for its k coefficient. And so we can just evaluate this. And so first we think about the i component."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "We'll put a 0 right there. And then r sub v, no i component, no j component, and it has 1 for its k coefficient. And so we can just evaluate this. And so first we think about the i component. Well, we'll ignore that column, that row. It's going to be cosine of u minus 0. So it's going to be cosine of u i."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "And so first we think about the i component. Well, we'll ignore that column, that row. It's going to be cosine of u minus 0. So it's going to be cosine of u i. And then we'll have minus j times, ignore column, ignore row, negative sine of u times 1 minus 0. So it's negative j times negative sine of u. So it's just going to be plus sine of uj."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So it's going to be cosine of u i. And then we'll have minus j times, ignore column, ignore row, negative sine of u times 1 minus 0. So it's negative j times negative sine of u. So it's just going to be plus sine of uj. And then for the k, we have this times 0 minus this times 0. So we're just going to have 0. So this is the cross product."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So it's just going to be plus sine of uj. And then for the k, we have this times 0 minus this times 0. So we're just going to have 0. So this is the cross product. And if we were to take the magnitude of this, so now we're ready to take the magnitude of this. It's going to be equal to, the magnitude of this is going to be equal to the square root of cosine of u squared plus sine of u squared. There is no k component."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "So this is the cross product. And if we were to take the magnitude of this, so now we're ready to take the magnitude of this. It's going to be equal to, the magnitude of this is going to be equal to the square root of cosine of u squared plus sine of u squared. There is no k component. And from the unit circle, this is the most basic trig identity. This is just going to be equal to 1. And so we have this part right over here just simplifies to 1, which is nice."}, {"video_title": "Surface integral ex3 part 1 Parameterizing the outside surface   Khan Academy.mp3", "Sentence": "There is no k component. And from the unit circle, this is the most basic trig identity. This is just going to be equal to 1. And so we have this part right over here just simplifies to 1, which is nice. And ds simplifies to du dv for at least this surface. And so now we are ready to attempt to evaluate the integral. But I'm almost running out of time."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "So what I'd like to do here and in the following few videos is talk about how you take the partial derivative of vector-valued functions. So the kind of thing I have in mind will be a function with a multivariable input. So this specific example have a two-variable input, T and S. You could think of that as a two-dimensional space as the input or just two separate numbers. And its output will be three-dimensional. The first component, T squared minus S squared. The Y component will be S times T. And that Z component will be T times S squared minus S times T squared. Minus S times T squared."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "And its output will be three-dimensional. The first component, T squared minus S squared. The Y component will be S times T. And that Z component will be T times S squared minus S times T squared. Minus S times T squared. And the way that you compute a partial derivative of a guy like this is actually relatively straightforward. If you were to just guess what it might mean, you'd probably guess right. It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "Minus S times T squared. And the way that you compute a partial derivative of a guy like this is actually relatively straightforward. If you were to just guess what it might mean, you'd probably guess right. It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that. Because each component is just a normal scalar-valued function. So you go up to the top one and you say T squared looks like a variable as far as T is concerned, and its derivative is 2T. But S squared looks like a constant, so its derivative is zero."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "It will look like partial of V with respect to one of its input variables, and I'll choose T with respect to T. And you just do it component-wise, which means you look at each component and you do the partial derivative to that. Because each component is just a normal scalar-valued function. So you go up to the top one and you say T squared looks like a variable as far as T is concerned, and its derivative is 2T. But S squared looks like a constant, so its derivative is zero. S times T, when S looks like a constant and when T looks like a variable, has a derivative of S, then T times S squared, when T is the variable and S is the constant, just looks like that constant, which is S squared, minus S times T squared. So now derivative of T squared is 2T, and that constant S stays in. So that's 2 times S times T. And that's how you compute it."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "But S squared looks like a constant, so its derivative is zero. S times T, when S looks like a constant and when T looks like a variable, has a derivative of S, then T times S squared, when T is the variable and S is the constant, just looks like that constant, which is S squared, minus S times T squared. So now derivative of T squared is 2T, and that constant S stays in. So that's 2 times S times T. And that's how you compute it. Probably relatively straightforward. The way you do it with respect to S is very similar. But where this gets fun and where this gets cool is how you interpret the partial derivative, right?"}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "So that's 2 times S times T. And that's how you compute it. Probably relatively straightforward. The way you do it with respect to S is very similar. But where this gets fun and where this gets cool is how you interpret the partial derivative, right? How you interpret this value that we just found. And what that means depends a lot on how you actually visualize the function. So what I'll go ahead and do in the next video, and in the next few ones, is talk about visualizing this function."}, {"video_title": "Computing the partial derivative of a vector-valued function.mp3", "Sentence": "But where this gets fun and where this gets cool is how you interpret the partial derivative, right? How you interpret this value that we just found. And what that means depends a lot on how you actually visualize the function. So what I'll go ahead and do in the next video, and in the next few ones, is talk about visualizing this function. It'll be as a parametric surface in three-dimensional space. That's why I've got my grapher program out here. And I think you'll find there's actually a very satisfying understanding of what this value means."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we saw that if our vector field is the gradient of a scalar field, then we call it conservative. So that tells us that F is a conservative vector field. And it also tells us, and this was the big takeaway from the last video, that the line integral of F between two points, so let me draw two points here. So let me say, let me draw my coordinates just so that we know we're on the xy plane. My axes, x-axis, y-axis. Let's say I have that point and that point, and I have two different paths between those two points. So I have path one that goes something like that."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me say, let me draw my coordinates just so that we know we're on the xy plane. My axes, x-axis, y-axis. Let's say I have that point and that point, and I have two different paths between those two points. So I have path one that goes something like that. So I'll call that c1, and it goes in that direction. And then I have, maybe in a different shade of green, c2. It goes like that."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I have path one that goes something like that. So I'll call that c1, and it goes in that direction. And then I have, maybe in a different shade of green, c2. It goes like that. They both start here and go to there. c2, we learned in the last video that the line integral is path independent between any two points. So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It goes like that. They both start here and go to there. c2, we learned in the last video that the line integral is path independent between any two points. So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr. If we have a potential in a region, and we're maybe everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So in this case, the line integral along c1 of F dot dr is going to be equal to the line integral of c2, over the path c2 of F dot dr. If we have a potential in a region, and we're maybe everywhere, then the line integral between any two points is independent of the path. That's the neat thing about a conservative field. Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video. And it's actually a pretty important extension. It might already be obvious to you. I've already written this here."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now what I want to do in this video is do a little bit of an extension of the takeaway of the last video. And it's actually a pretty important extension. It might already be obvious to you. I've already written this here. I could rearrange this equation a little bit. So let me do it. So let me rearrange this."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I've already written this here. I could rearrange this equation a little bit. So let me do it. So let me rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus, I'll just subtract this from both sides, minus the line integral c2 of F dot dr is going to be equal to 0. All I did is I took this takeaway from the last video, and I subtracted this from both sides."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me rearrange this. I'll just rewrite this in orange. So the line integral on path c1 dot dr minus, I'll just subtract this from both sides, minus the line integral c2 of F dot dr is going to be equal to 0. All I did is I took this takeaway from the last video, and I subtracted this from both sides. Now, we learned several videos ago that if we're dealing with a line integral of a vector field, not a scalar field, with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of F dot dr is equal to the negative of the line integral of minus c2 of F dot dr, where we denoted minus c2 as the same path as c2, but just in the opposite direction. So for example, minus c2, I would write like this."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All I did is I took this takeaway from the last video, and I subtracted this from both sides. Now, we learned several videos ago that if we're dealing with a line integral of a vector field, not a scalar field, with a vector field, the direction of the path is important. We learned that the line integral over, say, c2 of F dot dr is equal to the negative of the line integral of minus c2 of F dot dr, where we denoted minus c2 as the same path as c2, but just in the opposite direction. So for example, minus c2, I would write like this. So let me do it in a different color. So let's say this is minus c2. It would be a path just like c2."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So for example, minus c2, I would write like this. So let me do it in a different color. So let's say this is minus c2. It would be a path just like c2. I'm going to call this minus c2, but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would be a path just like c2. I'm going to call this minus c2, but instead of going in that direction, I'm now going to go in that direction. So ignore the old c2 arrows. We're now starting from there and coming back here. So this is minus c2. Or we could put the minus on the other side, and we could say that the negative of the c2 line integral along the path of c2 of F dot dr is equal to the line integral over the reverse path of F dot dr. All I did is I switched the negative on the other side, multiplied both sides by negative 1. So let's replace in this equation, we have the minus of the c2 path."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're now starting from there and coming back here. So this is minus c2. Or we could put the minus on the other side, and we could say that the negative of the c2 line integral along the path of c2 of F dot dr is equal to the line integral over the reverse path of F dot dr. All I did is I switched the negative on the other side, multiplied both sides by negative 1. So let's replace in this equation, we have the minus of the c2 path. We have that right there. So we could just replace this with this right there. So let me do that."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's replace in this equation, we have the minus of the c2 path. We have that right there. So we could just replace this with this right there. So let me do that. So I'll write this first part first. So the integral along the curve c1 of F dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This, we've established, is the same thing as this."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me do that. So I'll write this first part first. So the integral along the curve c1 of F dot dr, instead of minus the line integral along c2, I'm going to say plus the integral along minus c2. This, we've established, is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of F dot dr is equal to 0. Now there's something interesting."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This, we've established, is the same thing as this. The negative of this curve, or the line integral along this path, is the same thing as the line integral, the positive of the line integral along the reverse path. So we'll say plus the line integral of minus c2 of F dot dr is equal to 0. Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now there's something interesting. Let's look at what the combination of the path of c1 and minus c2 is. c1 starts over here. Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me get a nice, vibrant color. c1 starts over here at this point. It moves from this point along this curve c1 and ends up at this point. And then we do the minus c2. Minus c2 starts at this point and then goes and comes back to the original point. It completes a loop. So this is a closed line integral."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we do the minus c2. Minus c2 starts at this point and then goes and comes back to the original point. It completes a loop. So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is a closed line integral. So if you combine this, we could rewrite this. Remember, this is just a loop. By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path, and we could call a closed path maybe c1 plus minus c2 if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "By reversing this, instead of having two guys starting here and going there, I now can start here, go all the way there, and then come all the way back on this reverse path of c2. So this is equivalent to a closed line integral. So that is the same thing as an integral along a closed path, and we could call a closed path maybe c1 plus minus c2 if we wanted to be particular about the closed path. But this could be, I drew c1 and c2 or minus c2 arbitrarily. This could be any closed path where our vector field F has a potential or where it is the gradient of a scalar field or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of F dot dr. That's just a rewriting of that. And so that's going to be equal to 0."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this could be, I drew c1 and c2 or minus c2 arbitrarily. This could be any closed path where our vector field F has a potential or where it is the gradient of a scalar field or where it is conservative. And so this can be written as a closed path of c1 plus the reverse of c2 of F dot dr. That's just a rewriting of that. And so that's going to be equal to 0. And this is our takeaway for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so that's going to be equal to 0. And this is our takeaway for this video. This is, you can view it as a corollary. It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region or maybe over the entire xy-plane, and this is called the potential of F. This is a potential function. Oftentimes it'll be the negative of it, but it's easy to mess with negatives. But if we have a vector field that is the gradient of a scalar field, we call that vector field conservative."}, {"video_title": "Closed curve line integrals of conservative vector fields   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's kind of a low-hanging conclusion that you can make after this conclusion. So now we know that if we have a vector field that's the gradient of a scalar field in some region or maybe over the entire xy-plane, and this is called the potential of F. This is a potential function. Oftentimes it'll be the negative of it, but it's easy to mess with negatives. But if we have a vector field that is the gradient of a scalar field, we call that vector field conservative. That tells us that at any point in the region where this is valid, the line integral from one point to another is independent of the path. That's what we got from the last video. And because of that, a closed-loop line integral or a closed-line integral, so if we take some other place, if we take any other closed-line integral or we take the line integral of the vector field on any closed loop, it will become 0 because it is path independent."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And we ended up working out through some nice geometrical reasoning that we need to solve this system of equations, so there's nothing left to do but to just solve the system of equations. We'll start with this first one at the top, see what we can simplify. We notice there's an x term in each one, so we'll go ahead and cancel those out, which is basically a way of saying we're assuming that x is not zero, and we can kind of return to that to see if x equals zero could be a solution. So maybe we'll kind of write that down. We're assuming x is not equal to zero in order to cancel out, and we kind of can revisit whether that could give us another possible solution later, but that will be two times y is equal to lambda times two. And from here, the twos can cancel out, no worries about two equaling zero. And we know that y equals lambda."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "So maybe we'll kind of write that down. We're assuming x is not equal to zero in order to cancel out, and we kind of can revisit whether that could give us another possible solution later, but that will be two times y is equal to lambda times two. And from here, the twos can cancel out, no worries about two equaling zero. And we know that y equals lambda. So that's a nice simplified form for this equation. And for this next equation, we can use what we just found, that y is equal to lambda, to replace the lambda that we see. And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And we know that y equals lambda. So that's a nice simplified form for this equation. And for this next equation, we can use what we just found, that y is equal to lambda, to replace the lambda that we see. And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y. So that's two times y squared. And I'll leave it in that form because I see that in the next equation, I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and replace that with y squared."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And instead, if I replace this with a y, what I'm gonna get is that x squared is equal to y times two times y. So that's two times y squared. And I'll leave it in that form because I see that in the next equation, I see an x squared, I see a y squared, so it might be nice to be able to plug this guy right into it. So in that next equation, x squared, I'm gonna go ahead and replace that with y squared. So that's two y squared plus y squared equals one. And then from there, simplifies to three times y squared equals one, which in turn means y squared is equal to 1 3rd. And so y is equal to plus or minus the square root of 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "So in that next equation, x squared, I'm gonna go ahead and replace that with y squared. So that's two y squared plus y squared equals one. And then from there, simplifies to three times y squared equals one, which in turn means y squared is equal to 1 3rd. And so y is equal to plus or minus the square root of 1 3rd. Great, so this gives us y. And I'll go ahead and put a box around that, that we have found what y must be. Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And so y is equal to plus or minus the square root of 1 3rd. Great, so this gives us y. And I'll go ahead and put a box around that, that we have found what y must be. Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd. So two times 1 3rd. So if x squared is equal to 2 3rds, what that implies is that x is equal to plus or minus the square root of 2 3rds. And then there we go, that's another one of the solutions."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "Now if y squared is equal to 1 3rd, then when we look up here and we say, hmm, two times y squared, that's gonna be the same thing as two times 1 3rd. So two times 1 3rd. So if x squared is equal to 2 3rds, what that implies is that x is equal to plus or minus the square root of 2 3rds. And then there we go, that's another one of the solutions. And I could write down what lambda is, right? And we could, I mean, in this case it's easy because y equals lambda. But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And then there we go, that's another one of the solutions. And I could write down what lambda is, right? And we could, I mean, in this case it's easy because y equals lambda. But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want. And we just have that pesky little possibility that x equals zero to address. And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "But all we really want in their final form are x and y, since that's gonna give us the answer to the original constraint problem. So this, this gives us what we want. And we just have that pesky little possibility that x equals zero to address. And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions. Well, in this equation, that would make sense, since two times zero would equal zero. In this equation, that would mean that we're setting zero equal to lambda times two times y. Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And for that, we can take a look and say, if x equals zero, you know, let's go through the possibility that maybe that's one of the, one of the constraint solutions. Well, in this equation, that would make sense, since two times zero would equal zero. In this equation, that would mean that we're setting zero equal to lambda times two times y. Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero. So, so evidently, you know, if it was the case that x equals zero, that would have to imply from the second equation that y equals zero. But if x and y both equal zero, this constraint can't be satisfied. So none of this is possible, so we never even had to worry about this to start with."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "Well, since lambda equals y, that would mean that for this side to equal zero, y would have to equal zero. So, so evidently, you know, if it was the case that x equals zero, that would have to imply from the second equation that y equals zero. But if x and y both equal zero, this constraint can't be satisfied. So none of this is possible, so we never even had to worry about this to start with. But it's something you do need to check, just every time you're dividing by a variable, you're basically assuming that it's not equal to zero. So this right here gives us four possible solutions, four possible values for x and y that satisfy this constraint and which potentially maximize this. And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "So none of this is possible, so we never even had to worry about this to start with. But it's something you do need to check, just every time you're dividing by a variable, you're basically assuming that it's not equal to zero. So this right here gives us four possible solutions, four possible values for x and y that satisfy this constraint and which potentially maximize this. And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines. So just to make it explicit, the four points that we're dealing with, here, I'll write them all here. So x could be the square root of 2 3rds, square root of 2 3rds, and y could be the positive square root of 1 3rd. And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And remember, when I say potentially maximize, the whole idea of this Lagrange multiplier is that we were looking for where there's a point of tangency between the contour lines. So just to make it explicit, the four points that we're dealing with, here, I'll write them all here. So x could be the square root of 2 3rds, square root of 2 3rds, and y could be the positive square root of 1 3rd. And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd. Or maybe x is the positive square root of 2 3rds, and y is the negative square root of 1 3rd. Kind of monotonous, but just getting all of the different possibilities on the table here. x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And then we can basically just toggle, you know, maybe x is the negative square root of 2 3rds, and y is still the positive square root of 1 3rd. Or maybe x is the positive square root of 2 3rds, and y is the negative square root of 1 3rd. Kind of monotonous, but just getting all of the different possibilities on the table here. x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd. So these are the four points where the contour lines are tangent. And to find which one of these maximizes our function, here, let's go ahead and write down our function again. It gets easy to forget."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "x is negative square root of 2 3rds, and then y is positive, no, negative, that's the last one, square root of 1 3rd. So these are the four points where the contour lines are tangent. And to find which one of these maximizes our function, here, let's go ahead and write down our function again. It gets easy to forget. So the whole thing we're doing is maximizing f of x, y equals x squared times y. So let me just put that down again. We're looking at f of x, y is equal to x squared times y."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "It gets easy to forget. So the whole thing we're doing is maximizing f of x, y equals x squared times y. So let me just put that down again. We're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them is actually greatest. And the first thing to observe is x squared is always gonna be positive. So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "We're looking at f of x, y is equal to x squared times y. So we could just plug these values in and see which one of them is actually greatest. And the first thing to observe is x squared is always gonna be positive. So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say that neither of these can be the maximum, because it'll be some positive number, some x squared times a negative. Whereas I know that these guys are gonna produce a positive number. And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "So if I plug in a negative value for y, or if I plug in either this guy here, or this guy here, where the value for y is negative, the entire function would be negative. So I'm just gonna say that neither of these can be the maximum, because it'll be some positive number, some x squared times a negative. Whereas I know that these guys are gonna produce a positive number. And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd. And in fact, that's gonna be the same as what we get plugging in this other value. So either one of these maximizes the function. It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "And specifically, if we plug in, if we plug in f of, let's say, this top one, square root of 2 3rds, square root of 1 3rd, well, x squared is gonna be 2 3rds, and then y is square root of 1 3rd. And in fact, that's gonna be the same as what we get plugging in this other value. So either one of these maximizes the function. It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd. And that's the final answer. But I do wanna emphasize that the takeaway here is not the specific algebra that you work out going towards the end, but it's the whole idea of this Lagrange multiplier technique, to find the gradient of one function, find the gradient of the constraining function, and then set them proportional to each other. That's the key takeaway."}, {"video_title": "Finishing the intro lagrange multiplier example.mp3", "Sentence": "It's got two different maximizing points, and each one of them has a maximum value of 2 3rds times the square root of 1 3rd. And that's the final answer. But I do wanna emphasize that the takeaway here is not the specific algebra that you work out going towards the end, but it's the whole idea of this Lagrange multiplier technique, to find the gradient of one function, find the gradient of the constraining function, and then set them proportional to each other. That's the key takeaway. And then the rest of it is just making sure that we check our work and go through the minute details, which is important. It has its place. And coming up, I'll go through a few more examples."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in order to take a surface integral, we had to find the partial of our parameterization with respect to s and the partial with respect to t. And now we're ready to take the cross product, and then we can take the magnitude of the cross product. And then we can actually take this double integral and figure out the surface area. So let's just do it step by step. Here we can take the cross product, which is not a non-hairy operation. This is why you don't see many surface integrals actually get done or many examples done. So let's take the cross product of these two fellows. So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Here we can take the cross product, which is not a non-hairy operation. This is why you don't see many surface integrals actually get done or many examples done. So let's take the cross product of these two fellows. So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you. You might remember this is going to be equal to the determinant. The determinant, I'm going to write the unit vectors up here of the first rows i, j, and k. And then the next two rows are going to be the components of these guys. So let me copy and paste them."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the partial of r with respect to s crossed with, in magenta, the partial of r with respect to t. This will be a little bit of a review of cross products for you. You might remember this is going to be equal to the determinant. The determinant, I'm going to write the unit vectors up here of the first rows i, j, and k. And then the next two rows are going to be the components of these guys. So let me copy and paste them. You have that right there. Let me copy and paste. Put that guy right there."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me copy and paste them. You have that right there. Let me copy and paste. Put that guy right there. Then you have this fellow right there. Copy and paste. Put them right there."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Put that guy right there. Then you have this fellow right there. Copy and paste. Put them right there. And then you got this guy right here. This will save us some time. Copy and paste."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Put them right there. And then you got this guy right here. This will save us some time. Copy and paste. Put them right there. Then the last row is going to be this guy's components. Copy and paste."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Copy and paste. Put them right there. Then the last row is going to be this guy's components. Copy and paste. Put them right here. Almost done. This guy."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Copy and paste. Put them right here. Almost done. This guy. Copy and paste. Put them right there. Make sure we know that these are separate terms."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This guy. Copy and paste. Put them right there. Make sure we know that these are separate terms. And finally, we don't have to copy and paste it, but just since we did it for all of the other terms, I'll do it for that zero as well. So the cross product of these is literally the determinant of this matrix right here. It's this determinant."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Make sure we know that these are separate terms. And finally, we don't have to copy and paste it, but just since we did it for all of the other terms, I'll do it for that zero as well. So the cross product of these is literally the determinant of this matrix right here. It's this determinant. It's that determinant. And so, just as a bit of a refresher of taking determinants, this is going to be i times the sub-determinant right here, if you cross out this column and that row. It's going to be equal to i."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's this determinant. It's that determinant. And so, just as a bit of a refresher of taking determinants, this is going to be i times the sub-determinant right here, if you cross out this column and that row. It's going to be equal to i. You're not used to seeing the unit vector written first, but we can switch the order later. Times i times the sub-matrix right here, if you cross out this column and that row. So it's going to be this term times zero, which is just zero, minus this term times that term."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be equal to i. You're not used to seeing the unit vector written first, but we can switch the order later. Times i times the sub-matrix right here, if you cross out this column and that row. So it's going to be this term times zero, which is just zero, minus this term times that term. So minus this term times this term. The negative signs are going to cancel out, so this will be positive. So it would be i times this term times this term without a negative sign right there."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be this term times zero, which is just zero, minus this term times that term. So minus this term times this term. The negative signs are going to cancel out, so this will be positive. So it would be i times this term times this term without a negative sign right there. So i times this term, which is a cosine of s. It's really that term times that term minus that term times that term, but the negatives cancel out. That times that is zero, so that's how we can do it. So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it would be i times this term times this term without a negative sign right there. So i times this term, which is a cosine of s. It's really that term times that term minus that term times that term, but the negatives cancel out. That times that is zero, so that's how we can do it. So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product. Now it's going to be minus j. You remember when you take the determinant, you have to checkerboard of switching signs. So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's a cosine of s times b plus a cosine of s. I'll switch to the same color, sine of t. So we've got our i term for the cross product. Now it's going to be minus j. You remember when you take the determinant, you have to checkerboard of switching signs. So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term. And once again, when you have, oh sorry, you cross out this column and that row, so it's going to be that guy times that guy minus this guy times this guy. So it's going to be minus this guy times this guy, so it's going to be, let me do it in yellow, so the negative times negative that guy, b plus a cosine of s cosine of t times this guy, a cosine of s. We'll clean it up in a little bit. Immediately, oh we'll clean this up."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now it's going to be minus j times, so you cross out that row and that column, and it's going to be this term times this term, which is just zero, minus this term times this term. And once again, when you have, oh sorry, you cross out this column and that row, so it's going to be that guy times that guy minus this guy times this guy. So it's going to be minus this guy times this guy, so it's going to be, let me do it in yellow, so the negative times negative that guy, b plus a cosine of s cosine of t times this guy, a cosine of s. We'll clean it up in a little bit. Immediately, oh we'll clean this up. You see this negative and that negative will cancel out. We're just multiplying everything. And then finally, the k term."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Immediately, oh we'll clean this up. You see this negative and that negative will cancel out. We're just multiplying everything. And then finally, the k term. So plus, I'll go to the next line, plus k times, cross out that row, that column, it's going to be that times that minus that times that. So that looks like a kind of a beastly thing, but I think if we take step by step it shouldn't be too bad. So that times that, the negatives are going to cancel out."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then finally, the k term. So plus, I'll go to the next line, plus k times, cross out that row, that column, it's going to be that times that minus that times that. So that looks like a kind of a beastly thing, but I think if we take step by step it shouldn't be too bad. So that times that, the negatives are going to cancel out. So we have this term right here is going to be a sine of t sine of s, and then this term right here is b plus a cosine of s sine of t, so that's that times that and the negatives cancelled out, that's why I didn't put any negatives here, minus this times this. So this times this is going to be a negative number, but if you take the negative of it, it's going to be a positive value. So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that times that, the negatives are going to cancel out. So we have this term right here is going to be a sine of t sine of s, and then this term right here is b plus a cosine of s sine of t, so that's that times that and the negatives cancelled out, that's why I didn't put any negatives here, minus this times this. So this times this is going to be a negative number, but if you take the negative of it, it's going to be a positive value. So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done. Let's see if we can clean this up a little bit, especially if we can clean up this last term a bit. So let's see what we can do to simplify it. So our first term, so let's just multiply it out, I guess is the easiest way to do it."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's going to be plus that a cosine of t sine of s times that, times b plus a cosine of s cosine of t. Now you see why you don't see many examples of surface integrals being done. Let's see if we can clean this up a little bit, especially if we can clean up this last term a bit. So let's see what we can do to simplify it. So our first term, so let's just multiply it out, I guess is the easiest way to do it. Actually the easiest first step would just be to factor out the b plus a cosine of s, because that's in every term, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s. So let's just factor that out. This whole crazy thing can be written as b plus a cosine of s. So we factored it out times, I'll put in maybe some brackets here so you know it multiplies times every component. So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our first term, so let's just multiply it out, I guess is the easiest way to do it. Actually the easiest first step would just be to factor out the b plus a cosine of s, because that's in every term, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s, b plus a cosine of s. So let's just factor that out. This whole crazy thing can be written as b plus a cosine of s. So we factored it out times, I'll put in maybe some brackets here so you know it multiplies times every component. So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green. It's going to be a cosine of s sine of t. You're not used to seeing the i before it, so I'm going to write the i here. And then plus, we're factoring this guy out, so you're just going to be left with cosine of t, a cosine of s. Or we could write it as a cosine of s cosine of t. That's that right there, just putting it in the same order as that, times the unit vector j. And then when we factored this guy out, so we're not going to see that or that anymore."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the i component, we need to factor this guy out, is going to be a cosine of s sine of t. Let me write it in green. It's going to be a cosine of s sine of t. You're not used to seeing the i before it, so I'm going to write the i here. And then plus, we're factoring this guy out, so you're just going to be left with cosine of t, a cosine of s. Or we could write it as a cosine of s cosine of t. That's that right there, just putting it in the same order as that, times the unit vector j. And then when we factored this guy out, so we're not going to see that or that anymore. When we factor that out, we can multiply this out, and what do we get? So in green I'll write again. So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here?"}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when we factored this guy out, so we're not going to see that or that anymore. When we factor that out, we can multiply this out, and what do we get? So in green I'll write again. So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here? We have a sine of s times cosine squared of t. A sine of s times cosine squared of t, and all of that times the k unit vector. All of that times the k unit vector. And so things are looking a little bit more simplified, but you might see something jump out at you."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you multiply sine of t times this thing over here, because that's all that we have left after we factor out this thing, we get a sine of s sine squared of t. We have sine of t times sine of t. Sine squared of t, so that's that over there, plus what do we have over here? We have a sine of s times cosine squared of t. A sine of s times cosine squared of t, and all of that times the k unit vector. All of that times the k unit vector. And so things are looking a little bit more simplified, but you might see something jump out at you. You have a sine squared and a cosine squared. So somehow if I can just make that just sine squared plus cosine squared of t, those will simplify to 1, and we can. And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so things are looking a little bit more simplified, but you might see something jump out at you. You have a sine squared and a cosine squared. So somehow if I can just make that just sine squared plus cosine squared of t, those will simplify to 1, and we can. And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms. And this is our most fundamental trig identity from the unit circle. This is equal to 1. So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this term right here, if we just focus on that term, and this is all kind of algebraic manipulation, if we just focus on that term, this term right here can be rewritten as a sine of s, if we factor that out, times sine squared of t plus cosine squared of t times our unit vector k. I just factored out an a sine of s from both of these terms. And this is our most fundamental trig identity from the unit circle. This is equal to 1. So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far. We were able to figure out the cross product of these two, I guess, partial derivatives of the vector value to our original parameterization there. We were able to figure out what this thing, this thing right here, before we take the magnitude of it, it translates to this thing right here. Let me rewrite it."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this last term simplifies to a sine of s times k. So, so far, we've gotten pretty far. We were able to figure out the cross product of these two, I guess, partial derivatives of the vector value to our original parameterization there. We were able to figure out what this thing, this thing right here, before we take the magnitude of it, it translates to this thing right here. Let me rewrite it. Well, I don't need to rewrite it. You know that. Well, I'll rewrite it."}, {"video_title": "Example of calculating a surface integral part 2   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me rewrite it. Well, I don't need to rewrite it. You know that. Well, I'll rewrite it. So that's equal to, I'll rewrite it neatly, and we'll use this in the next video, b plus a cosine of s times, open bracket, a cosine of s sine of t times i plus, switch back to the blue, plus a cosine of s cosine of t times j plus, switch back to the blue, this thing, plus, this simplified nicely, a sine of s times k, times the unit vector k. This, this right here is this expression right there. And I'll finish this video since I'm already over 10 minutes. And in the next video, we're going to take the magnitude of it, and then if we have time, actually take this double integral, and we'll all be done."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "In the video on parametric surfaces, I give you guys this function here. It's a very complicated looking function. It's got a two-dimensional input and a three-dimensional output. And I talked about how you can think about it as drawing a surface in three-dimensional space. And that one came out to be the surface of a donut, which we also call a torus. So what I want to do here is talk about how you might think of this as a transformation. And first, let me just get straight what the input space here is."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "And I talked about how you can think about it as drawing a surface in three-dimensional space. And that one came out to be the surface of a donut, which we also call a torus. So what I want to do here is talk about how you might think of this as a transformation. And first, let me just get straight what the input space here is. So the input space, you could think about it as the entire t-s plane. Right, we might draw this as the entire t-axis and the s-axis, and just everything here and see where it maps. But you can actually go to just a small subset of that."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "And first, let me just get straight what the input space here is. So the input space, you could think about it as the entire t-s plane. Right, we might draw this as the entire t-axis and the s-axis, and just everything here and see where it maps. But you can actually go to just a small subset of that. So if you limit yourself to t going between zero, so between zero and let's say 2 pi, and then similarly with s going from zero up to 2 pi, and you imagine what, you know, that would be sort of a square region, just limiting yourself to that, you're actually going to get all of the points that you need to draw the torus. And the basic reason for that is that as t ranges from zero to 2 pi, cosine of t goes over its full range before it starts becoming periodic. Sine of t does the same."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "But you can actually go to just a small subset of that. So if you limit yourself to t going between zero, so between zero and let's say 2 pi, and then similarly with s going from zero up to 2 pi, and you imagine what, you know, that would be sort of a square region, just limiting yourself to that, you're actually going to get all of the points that you need to draw the torus. And the basic reason for that is that as t ranges from zero to 2 pi, cosine of t goes over its full range before it starts becoming periodic. Sine of t does the same. And same deal with s. If you let s range from zero to 2 pi, that covers a full period of cosine, a full period of sine. So you'll get no new information by going elsewhere. So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sine of t does the same. And same deal with s. If you let s range from zero to 2 pi, that covers a full period of cosine, a full period of sine. So you'll get no new information by going elsewhere. So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space. This is sort of cheating, but it's a little bit easier to do this than to imagine, you know, moving from some separate area into the space. At the very least for the animation efforts, it's easier to just start it off in 3D. So what I'm thinking about here, this square is representing that t-s plane."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what we can do is think about this portion of the t-s plane kind of as living inside three-dimensional space. This is sort of cheating, but it's a little bit easier to do this than to imagine, you know, moving from some separate area into the space. At the very least for the animation efforts, it's easier to just start it off in 3D. So what I'm thinking about here, this square is representing that t-s plane. And for this function, which is taking all of the points in this square as its input and outputs a point in three-dimensional space, you can think about how those points move to their corresponding output points. Okay, so I'll show that again. We start off with our t-s plane here."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what I'm thinking about here, this square is representing that t-s plane. And for this function, which is taking all of the points in this square as its input and outputs a point in three-dimensional space, you can think about how those points move to their corresponding output points. Okay, so I'll show that again. We start off with our t-s plane here. And then whatever your input point is, if you were to follow it, and you were to follow it through this whole transformation, the place where it lands would be the corresponding output of this function. And one thing I should mention is all of the interpolating values as you go in between these don't really matter. A function is really a very static thing."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "We start off with our t-s plane here. And then whatever your input point is, if you were to follow it, and you were to follow it through this whole transformation, the place where it lands would be the corresponding output of this function. And one thing I should mention is all of the interpolating values as you go in between these don't really matter. A function is really a very static thing. There's just an input and there's an output. And if I'm thinking in terms of a transformation actually moving it, there's a little bit of magic sauce that has to go into making an animation do this. And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "A function is really a very static thing. There's just an input and there's an output. And if I'm thinking in terms of a transformation actually moving it, there's a little bit of magic sauce that has to go into making an animation do this. And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other. It doesn't really matter. But the general idea of starting with a square and somehow warping that, however you do choose to warp it, is actually a pretty powerful thought. And as we get into multivariable calculus and you start thinking a little bit more deeply about surfaces, I think it really helps if you, you know, you think about what a slight little movement over here on your input space would look like."}, {"video_title": "Transformations, part 3   Multivariable calculus   Khan Academy.mp3", "Sentence": "And in this case, I kind of put it into two different phases to sort of roll up one side and roll up the other. It doesn't really matter. But the general idea of starting with a square and somehow warping that, however you do choose to warp it, is actually a pretty powerful thought. And as we get into multivariable calculus and you start thinking a little bit more deeply about surfaces, I think it really helps if you, you know, you think about what a slight little movement over here on your input space would look like. What happens to that tiny little movement or that tiny little traversal? What it looks like if you did that same movement somewhere on the output space. And you'll get lots of chances to wrap your mind about this and engage with the idea."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So in this video, I'm gonna introduce vector fields. Now these are a concept that come up all the time in multivariable calculus, and that's probably because they come up all the time in physics. You know, it comes up with fluid flow, with electrodynamics. You see them all over the place. And what a vector field is, is it's pretty much a way of visualizing functions that have the same number of dimensions in their input as in their output. So here I'm gonna write a function that's got a two-dimensional input, x and y, and then its output is gonna be a two-dimensional vector. And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "You see them all over the place. And what a vector field is, is it's pretty much a way of visualizing functions that have the same number of dimensions in their input as in their output. So here I'm gonna write a function that's got a two-dimensional input, x and y, and then its output is gonna be a two-dimensional vector. And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y. And then the second component, the y component of the output, will be x cubed minus nine x. I made them symmetric here, looking kind of similar. They don't have to be. I'm just kind of a sucker for symmetry."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And each of the components will somehow depend on x and y. I'll make the first one y cubed minus nine y. And then the second component, the y component of the output, will be x cubed minus nine x. I made them symmetric here, looking kind of similar. They don't have to be. I'm just kind of a sucker for symmetry. So if you imagine trying to visualize a function like this with, I don't know, like a graph, it would be really hard because you have two dimensions in the input, two dimensions in the output, so you'd have to somehow visualize this thing in four dimensions. So instead what we do, we look only in the input space. So that means we look only in the xy plane."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "I'm just kind of a sucker for symmetry. So if you imagine trying to visualize a function like this with, I don't know, like a graph, it would be really hard because you have two dimensions in the input, two dimensions in the output, so you'd have to somehow visualize this thing in four dimensions. So instead what we do, we look only in the input space. So that means we look only in the xy plane. So I'll draw these coordinate axes, and just mark it up. This here is our x-axis. This here is our y-axis."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So that means we look only in the xy plane. So I'll draw these coordinate axes, and just mark it up. This here is our x-axis. This here is our y-axis. And for each individual input point, like let's say one, two. So let's say we go to one, two. I'm gonna consider the vector that it outputs and attach that vector to the point."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "This here is our y-axis. And for each individual input point, like let's say one, two. So let's say we go to one, two. I'm gonna consider the vector that it outputs and attach that vector to the point. So let's walk through an example of what I mean by that. So if we actually evaluate f at one, two, x is equal to one, y is equal to two, so we plug in two cubed, whoops, two cubed minus nine times two up here in the x component, and then one cubed minus nine times y, nine times one, excuse me, down in the y component, two cubed is eight, nine times two is 18, so eight minus 18 is negative 10, negative 10, and then one cubed is one, nine times one is nine, so one minus nine is negative eight. Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "I'm gonna consider the vector that it outputs and attach that vector to the point. So let's walk through an example of what I mean by that. So if we actually evaluate f at one, two, x is equal to one, y is equal to two, so we plug in two cubed, whoops, two cubed minus nine times two up here in the x component, and then one cubed minus nine times y, nine times one, excuse me, down in the y component, two cubed is eight, nine times two is 18, so eight minus 18 is negative 10, negative 10, and then one cubed is one, nine times one is nine, so one minus nine is negative eight. Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen. It's a very, very large vector, so it's gonna be something here, and it ends up having to go off the screen. But the nice thing about vectors, it doesn't matter where they start, so instead we can start it here, and we still want it to have that negative 10 x component, and then negative eight, so negative one, two, three, four, five, six, seven, eight, negative eight as its y component there. So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "Now first, imagine that this was, if we just drew this vector where we count, starting from the origin, negative one, two, three, four, five, six, seven, eight, nine, 10, so it's gonna have this as its x component, and then negative eight, one, two, three, four, five, six, seven, we're gonna actually go off the screen. It's a very, very large vector, so it's gonna be something here, and it ends up having to go off the screen. But the nice thing about vectors, it doesn't matter where they start, so instead we can start it here, and we still want it to have that negative 10 x component, and then negative eight, so negative one, two, three, four, five, six, seven, eight, negative eight as its y component there. So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess. There'd be markings all over the place, and you'd have, you know, this one might have some huge vector attached to it, and this one would have some huge vector attached to it, and it would get really, really messy. But instead what we do, so I'm gonna just clear up the board here, we scale them down, this is common, you'll scale them down so that you're kind of lying about what the vectors themselves are, but you get a much better feel for what each thing corresponds to. And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So this is a really big vector, and a plan with the vector field is to do this at not just one, two, but at a whole bunch of different points and see what vectors attach to them, and if we drew them all according to their size, this would be a real mess. There'd be markings all over the place, and you'd have, you know, this one might have some huge vector attached to it, and this one would have some huge vector attached to it, and it would get really, really messy. But instead what we do, so I'm gonna just clear up the board here, we scale them down, this is common, you'll scale them down so that you're kind of lying about what the vectors themselves are, but you get a much better feel for what each thing corresponds to. And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length. You know, I made this one just kind of the same unit, this one the same unit, and over here, they all just have the same length, even though in reality, the length of the vectors output by this function can be wildly different. This is kind of common practice when vector fields are drawn or when some kind of software is drawing them for you. So there are ways of getting around this."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And another thing about this drawing that's not entirely faithful to the original function that we have is that all of these vectors are the same length. You know, I made this one just kind of the same unit, this one the same unit, and over here, they all just have the same length, even though in reality, the length of the vectors output by this function can be wildly different. This is kind of common practice when vector fields are drawn or when some kind of software is drawing them for you. So there are ways of getting around this. One way is to just use colors with your vectors, so I'll switch over to a different vector field here, and here, color is used to kind of give a hint of length. So it still looks organized because all of them have the same lengths, but the difference is that red and warmer colors are supposed to indicate this is a very long vector somehow, and then blue would indicate that it's very short. Another thing you could do is scale them to be roughly proportional to what they should be."}, {"video_title": "Vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So there are ways of getting around this. One way is to just use colors with your vectors, so I'll switch over to a different vector field here, and here, color is used to kind of give a hint of length. So it still looks organized because all of them have the same lengths, but the difference is that red and warmer colors are supposed to indicate this is a very long vector somehow, and then blue would indicate that it's very short. Another thing you could do is scale them to be roughly proportional to what they should be. So notice all the blue vectors scaled way down to basically be zero. Red vectors kind of stayed the same size. And even though in reality, this might be representing a function where the true vector here should be really long or the true vector here should be kind of medium length, it's still common for people to just shrink them down so it's a reasonable thing to view."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I have this region, this simple solid right over here. It's x can go between negative 1 and 1. z, this kind of arch part right over here, is going to be a function of x. That's the upper bound on z. The lower bound on z is just 0. And then y can go anywhere between 0, and then it's bounded here by this plane where we can express y as a function of z. y is 2 minus z along this plane right over here. And we're given this crazy vector field. It has natural logs and tangents in it."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The lower bound on z is just 0. And then y can go anywhere between 0, and then it's bounded here by this plane where we can express y as a function of z. y is 2 minus z along this plane right over here. And we're given this crazy vector field. It has natural logs and tangents in it. And we're asked to evaluate the surface integral, or I should say the flux of our vector field across the boundary of this region, across the surface of this region right over here. And surface integrals are messy as is, especially when you have a crazy vector field like this. But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It has natural logs and tangents in it. And we're asked to evaluate the surface integral, or I should say the flux of our vector field across the boundary of this region, across the surface of this region right over here. And surface integrals are messy as is, especially when you have a crazy vector field like this. But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem. The divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it, or I'll just call it over the region, of the divergence of f, dv, where dv is some combination of dx, dy, dz, the divergence times each little cubic volume, infinitesimal cubic volume. So times dv. So let's see if this simplifies things a bit."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But you could imagine that there might be a way to simplify this, perhaps using the divergence theorem. The divergence theorem tells us that the flux across the boundary of this simple solid region is going to be the same thing as the triple integral over the volume of it, or I'll just call it over the region, of the divergence of f, dv, where dv is some combination of dx, dy, dz, the divergence times each little cubic volume, infinitesimal cubic volume. So times dv. So let's see if this simplifies things a bit. So let's calculate the divergence of f first. So the divergence of f is going to be the partial of the x component, or the partial of the, you could say, the i component, or the x component with respect to x. Well, the derivative of this with respect to x is just x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's see if this simplifies things a bit. So let's calculate the divergence of f first. So the divergence of f is going to be the partial of the x component, or the partial of the, you could say, the i component, or the x component with respect to x. Well, the derivative of this with respect to x is just x. Derivative of this with respect to x, luckily, is just 0. This is a constant in terms of x. Now let's go over here, the partial of this with respect to y."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, the derivative of this with respect to x is just x. Derivative of this with respect to x, luckily, is just 0. This is a constant in terms of x. Now let's go over here, the partial of this with respect to y. The partial of this with respect to y is just x. And then this is just a constant in terms of y, so it's just going to be 0 when you take the derivative with respect to y. And then finally, the partial of this with respect to z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now let's go over here, the partial of this with respect to y. The partial of this with respect to y is just x. And then this is just a constant in terms of y, so it's just going to be 0 when you take the derivative with respect to y. And then finally, the partial of this with respect to z. Well, this is just a constant in terms of z. It doesn't change when z changes, so the partial with respect to z is just going to be 0 here. And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then finally, the partial of this with respect to z. Well, this is just a constant in terms of z. It doesn't change when z changes, so the partial with respect to z is just going to be 0 here. And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x. So now we can restate the flux across the surface as a triple integral of 2x. So let me just write 2x here. And let's think about the ordering."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so taking the divergence really, really, really simplified things, the divergence of f simplified down to 2x. So now we can restate the flux across the surface as a triple integral of 2x. So let me just write 2x here. And let's think about the ordering. So y can go between 0 and this plane that is a function of z. So let's write that down. So y is bounded below by 0 and above by this plane 2 minus z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's think about the ordering. So y can go between 0 and this plane that is a function of z. So let's write that down. So y is bounded below by 0 and above by this plane 2 minus z. And z is bounded below by 0 and above by these parabolas of 1 minus x squared. And then x is bounded below by negative 1 and bounded above by 1. So negative 1 is less than or equal to x is less than or equal to 1."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So y is bounded below by 0 and above by this plane 2 minus z. And z is bounded below by 0 and above by these parabolas of 1 minus x squared. And then x is bounded below by negative 1 and bounded above by 1. So negative 1 is less than or equal to x is less than or equal to 1. And so this is probably a good order of integration. We can integrate with respect to y first, and then we'll get a function of z. Then we can integrate with respect to z, and we'll get a function of x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So negative 1 is less than or equal to x is less than or equal to 1. And so this is probably a good order of integration. We can integrate with respect to y first, and then we'll get a function of z. Then we can integrate with respect to z, and we'll get a function of x. And then we can integrate with respect to x. So let's do it in that order. So first we'll integrate with respect to y."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then we can integrate with respect to z, and we'll get a function of x. And then we can integrate with respect to x. So let's do it in that order. So first we'll integrate with respect to y. So we have dy. y is bounded below at 0 and above by the plane 2 minus z. So this right over here is the plane y is equal to 0."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So first we'll integrate with respect to y. So we have dy. y is bounded below at 0 and above by the plane 2 minus z. So this right over here is the plane y is equal to 0. And this up over here is the plane y is equal to 2 minus z. Then we can integrate with respect to z. And z, once again, is bounded below by 0 and bounded above by these parabolas, 1 minus x squared."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this right over here is the plane y is equal to 0. And this up over here is the plane y is equal to 2 minus z. Then we can integrate with respect to z. And z, once again, is bounded below by 0 and bounded above by these parabolas, 1 minus x squared. And then finally we can integrate with respect to x. And x is bounded below by negative 1 and bounded above by 1. So let's do some integration here."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And z, once again, is bounded below by 0 and bounded above by these parabolas, 1 minus x squared. And then finally we can integrate with respect to x. And x is bounded below by negative 1 and bounded above by 1. So let's do some integration here. So the first thing, when we're integrating with respect to y, 2x is just a constant. So this expression right over here is just going to be 2x times y. And then we're going to evaluate it from 0 to 2 minus z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's do some integration here. So the first thing, when we're integrating with respect to y, 2x is just a constant. So this expression right over here is just going to be 2x times y. And then we're going to evaluate it from 0 to 2 minus z. So it's going to be 2x times 2 minus z minus 2x times 0. Well, that second part is just going to be 0. So this is going to be equal to, this is going to simplify as 2x times 2 minus z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to evaluate it from 0 to 2 minus z. So it's going to be 2x times 2 minus z minus 2x times 0. Well, that second part is just going to be 0. So this is going to be equal to, this is going to simplify as 2x times 2 minus z. And actually I'll just leave it like that. And then we're going to integrate this with respect to z. We're going to integrate this with respect to z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to, this is going to simplify as 2x times 2 minus z. And actually I'll just leave it like that. And then we're going to integrate this with respect to z. We're going to integrate this with respect to z. And that's going to go from 0 to 1 minus x squared. And then we have our dz there. And then after that we're going to integrate with respect to x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to integrate this with respect to z. And that's going to go from 0 to 1 minus x squared. And then we have our dz there. And then after that we're going to integrate with respect to x. Negative 1 to 1 dx. So let's take the antiderivative here with respect to z. This you really can just view as a constant."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then after that we're going to integrate with respect to x. Negative 1 to 1 dx. So let's take the antiderivative here with respect to z. This you really can just view as a constant. We can actually even bring it out front. But I'll leave it there. So this piece right over here, I'll do it in z's color."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This you really can just view as a constant. We can actually even bring it out front. But I'll leave it there. So this piece right over here, I'll do it in z's color. This piece right over here. See, we can leave the 2x out front. Actually I'll leave the 2x out front of the whole thing."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this piece right over here, I'll do it in z's color. This piece right over here. See, we can leave the 2x out front. Actually I'll leave the 2x out front of the whole thing. It's going to be 2x times, so the antiderivative of this with respect to z is going to be 2z. Antiderivative of this is negative z squared over 2. And we are going to evaluate this from 0 to 1 minus x squared."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually I'll leave the 2x out front of the whole thing. It's going to be 2x times, so the antiderivative of this with respect to z is going to be 2z. Antiderivative of this is negative z squared over 2. And we are going to evaluate this from 0 to 1 minus x squared. When we evaluate them at 0, we're just going to get 0 right over here. And so we really just have to worry about when z is equal to 1 minus x squared. Did I do that right?"}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we are going to evaluate this from 0 to 1 minus x squared. When we evaluate them at 0, we're just going to get 0 right over here. And so we really just have to worry about when z is equal to 1 minus x squared. Did I do that right? Yep. 2z and then minus z squared over 2. You take the derivative, you get negative z."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Did I do that right? Yep. 2z and then minus z squared over 2. You take the derivative, you get negative z. Take the derivative here, you just get 2. So that's right. So this is going to be equal to 2x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You take the derivative, you get negative z. Take the derivative here, you just get 2. So that's right. So this is going to be equal to 2x. Let me do it in that same color. It's going to be equal to 2x. 2x times, let me get this right."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to 2x. Let me do it in that same color. It's going to be equal to 2x. 2x times, let me get this right. Let me go into that pink color. 2x times 2z. Well z is going to be 1 minus x squared."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "2x times, let me get this right. Let me go into that pink color. 2x times 2z. Well z is going to be 1 minus x squared. So it's going to be 2 minus 2x squared. That was just 2 times that. And then minus, I'll just write 1 half, times this quantity squared."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well z is going to be 1 minus x squared. So it's going to be 2 minus 2x squared. That was just 2 times that. And then minus, I'll just write 1 half, times this quantity squared. So this quantity squared is going to be 1 minus 2z squared. Minus 2x squared plus x to the fourth. That's just some basic algebra right over there."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then minus, I'll just write 1 half, times this quantity squared. So this quantity squared is going to be 1 minus 2z squared. Minus 2x squared plus x to the fourth. That's just some basic algebra right over there. And then from that you're going to subtract this thing evaluated at 0, which is just going to be 0. So you just won't even think about that. And now we need to simplify this a little bit."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's just some basic algebra right over there. And then from that you're going to subtract this thing evaluated at 0, which is just going to be 0. So you just won't even think about that. And now we need to simplify this a little bit. And we are going to get, if we simplify this, we get 2 minus 2x squared. Minus 1 half. And then plus, so this is negative 1 half times negative 2x squared."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we need to simplify this a little bit. And we are going to get, if we simplify this, we get 2 minus 2x squared. Minus 1 half. And then plus, so this is negative 1 half times negative 2x squared. So it's going to be positive x squared. Positive x squared minus 1 half x to the fourth. Let's see, can we simplify this part?"}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then plus, so this is negative 1 half times negative 2x squared. So it's going to be positive x squared. Positive x squared minus 1 half x to the fourth. Let's see, can we simplify this part? Let me just make sure we know what we're doing here. So we have this 2x right over there. I want to make sure I got the signs right."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see, can we simplify this part? Let me just make sure we know what we're doing here. So we have this 2x right over there. I want to make sure I got the signs right. Yep, looks like I did. And now let's look at this. Let's see, can I simplify a little bit?"}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to make sure I got the signs right. Yep, looks like I did. And now let's look at this. Let's see, can I simplify a little bit? I have 2 minus 1 half, which is 3 halves. So I have 3 halves. That's that term and that term taken into account."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's see, can I simplify a little bit? I have 2 minus 1 half, which is 3 halves. So I have 3 halves. That's that term and that term taken into account. And then I have negative 2x squared plus x squared. So that's just going to result in negative x squared. If I take that term and that term."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's that term and that term taken into account. And then I have negative 2x squared plus x squared. So that's just going to result in negative x squared. If I take that term and that term. And then I have negative 1 half x to the fourth. Negative 1 half x to the fourth. And I'm multiplying this whole thing by 2x."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I take that term and that term. And then I have negative 1 half x to the fourth. Negative 1 half x to the fourth. And I'm multiplying this whole thing by 2x. And so that's going to give us 2x times 3 halves. I want to make sure I'm doing this slowly so I don't make any careless mistakes. The 2's cancel out."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'm multiplying this whole thing by 2x. And so that's going to give us 2x times 3 halves. I want to make sure I'm doing this slowly so I don't make any careless mistakes. The 2's cancel out. You get 3x. And then 2x times negative x squared is negative 2x to the third. And then 2x times this right over here."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The 2's cancel out. You get 3x. And then 2x times negative x squared is negative 2x to the third. And then 2x times this right over here. The 2 cancels out with the negative 1 half. You have negative x to the fifth. So all of this simplifies to this right over here."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then 2x times this right over here. The 2 cancels out with the negative 1 half. You have negative x to the fifth. So all of this simplifies to this right over here. So our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth. And then we have dx. And now we just take the antiderivative with respect to x, which is going to be 3 halves x squared minus 1 half because it's going to be 2 fourths x to the fourth."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all of this simplifies to this right over here. So our whole thing simplifies to an integral with respect to x. x will go from negative 1 to 1 of this business of 3x minus 2x to the third minus x to the fifth. And then we have dx. And now we just take the antiderivative with respect to x, which is going to be 3 halves x squared minus 1 half because it's going to be 2 fourths x to the fourth. Is that right? Because if you multiply it, you're going to get 2x to the third and then minus x to the sixth over 6. And it's going to go from 1 to negative 1 or negative 1 to 1."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we just take the antiderivative with respect to x, which is going to be 3 halves x squared minus 1 half because it's going to be 2 fourths x to the fourth. Is that right? Because if you multiply it, you're going to get 2x to the third and then minus x to the sixth over 6. And it's going to go from 1 to negative 1 or negative 1 to 1. So when you evaluate it at 1, I'll just write it out real fast. First when you evaluate it at 1, you get 3 halves minus 1 half minus 1 sixth. And then from that, we are going to subtract 3 halves minus 1 half plus 1 sixth."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's going to go from 1 to negative 1 or negative 1 to 1. So when you evaluate it at 1, I'll just write it out real fast. First when you evaluate it at 1, you get 3 halves minus 1 half minus 1 sixth. And then from that, we are going to subtract 3 halves minus 1 half plus 1 sixth. Actually, no, they're actually all going to cancel out. Is that right? Are they all going to cancel out?"}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then from that, we are going to subtract 3 halves minus 1 half plus 1 sixth. Actually, no, they're actually all going to cancel out. Is that right? Are they all going to cancel out? Yep, I think that's right. They all cancel out. So it's actually going to be plus or I should say minus 1 sixth right over here."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Are they all going to cancel out? Yep, I think that's right. They all cancel out. So it's actually going to be plus or I should say minus 1 sixth right over here. And then all of these cancel out. That cancels with that. That cancels with that because we're subtracting the negative 1 half and that cancels with that."}, {"video_title": "Divergence theorem example 1   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's actually going to be plus or I should say minus 1 sixth right over here. And then all of these cancel out. That cancels with that. That cancels with that because we're subtracting the negative 1 half and that cancels with that. And so we are actually left with 0. So after doing all of that work, this whole thing evaluates to 0, which was actually kind of a neat simplification. So this whole thing right over here evaluated very conveniently, evaluated to be equal to 0."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "And I defined, in the last video, the gradient to be a certain operator, an operator just means you take in a function and you output another function, and we use this upside down triangle. So it gives you another function that's also of x and y, but this time it has a vector valued output. And the two components of its output are the partial derivatives, partial of f with respect to x, and the partial of f with respect to y. So for a function like this we actually evaluated it. Let's take a look. The first one is taking the derivative with respect to x, so it looks at x and says you look like a variable to me, I'm gonna take your derivative, your two x, two x, but the y component just looks like a constant as far as the partial x is concerned, and the derivative of a constant is zero. But when you take the partial derivative with respect to y, things reverse, it looks at the x component and says you look like a constant, your derivative is zero, but it looks at the y component and says, ah, you look like a variable, your derivative is two y."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "So for a function like this we actually evaluated it. Let's take a look. The first one is taking the derivative with respect to x, so it looks at x and says you look like a variable to me, I'm gonna take your derivative, your two x, two x, but the y component just looks like a constant as far as the partial x is concerned, and the derivative of a constant is zero. But when you take the partial derivative with respect to y, things reverse, it looks at the x component and says you look like a constant, your derivative is zero, but it looks at the y component and says, ah, you look like a variable, your derivative is two y. So this ultimate function that we get, the gradient, which takes in a two variable input, x, y, some point on this plane, but outputs a vector, can nicely be visualized with a vector field. And I have another video on vector fields if you're feeling unsure, but I want you to just take a moment, pause if you need to, and guess, or try to think about what vector field this will look like. I'm gonna show you in a moment, but what's it gonna look like, the one that takes in x, y and outputs two x, two y?"}, {"video_title": "Gradient and graphs.mp3", "Sentence": "But when you take the partial derivative with respect to y, things reverse, it looks at the x component and says you look like a constant, your derivative is zero, but it looks at the y component and says, ah, you look like a variable, your derivative is two y. So this ultimate function that we get, the gradient, which takes in a two variable input, x, y, some point on this plane, but outputs a vector, can nicely be visualized with a vector field. And I have another video on vector fields if you're feeling unsure, but I want you to just take a moment, pause if you need to, and guess, or try to think about what vector field this will look like. I'm gonna show you in a moment, but what's it gonna look like, the one that takes in x, y and outputs two x, two y? All right, have you done it, have you thought about what it's gonna look like? Here's what we get. It's a bunch of vectors pointing away from the origin."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "I'm gonna show you in a moment, but what's it gonna look like, the one that takes in x, y and outputs two x, two y? All right, have you done it, have you thought about what it's gonna look like? Here's what we get. It's a bunch of vectors pointing away from the origin. And the basic reason for that is that if you have any given input point, and say it's got coordinates x, y, then the vector that that input point represents would, you know, if it went from the origin here, that's what that vector looks like, but the output is two times that vector. So when we attach that output to the original point, you get something that's two times that original vector, but pointing in the same direction, which is away from the origin. I kinda drew it poorly here."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "It's a bunch of vectors pointing away from the origin. And the basic reason for that is that if you have any given input point, and say it's got coordinates x, y, then the vector that that input point represents would, you know, if it went from the origin here, that's what that vector looks like, but the output is two times that vector. So when we attach that output to the original point, you get something that's two times that original vector, but pointing in the same direction, which is away from the origin. I kinda drew it poorly here. And of course, when we draw vector fields, we don't usually draw them to scale, you scale them down, just so that things don't look as cluttered, that's why everything here, they all look the same length, but color indicates length. So you should think of these red guys as being really long, the blue ones as being really short. So what does this have to do with the graph of the function?"}, {"video_title": "Gradient and graphs.mp3", "Sentence": "I kinda drew it poorly here. And of course, when we draw vector fields, we don't usually draw them to scale, you scale them down, just so that things don't look as cluttered, that's why everything here, they all look the same length, but color indicates length. So you should think of these red guys as being really long, the blue ones as being really short. So what does this have to do with the graph of the function? There's actually a really cool interpretation. So imagine that you are just walking along this graph, you know, you're a hiker, and this is a mountain, and you picture yourself at any old point on this graph, let's say, what color should I use? Let's say you're sitting at a point like this, and you say, what direction should I walk to increase my altitude the fastest?"}, {"video_title": "Gradient and graphs.mp3", "Sentence": "So what does this have to do with the graph of the function? There's actually a really cool interpretation. So imagine that you are just walking along this graph, you know, you're a hiker, and this is a mountain, and you picture yourself at any old point on this graph, let's say, what color should I use? Let's say you're sitting at a point like this, and you say, what direction should I walk to increase my altitude the fastest? You wanna get uphill as quickly as possible. And from that point, you might walk, you know, what looks like straight up there, you certainly wouldn't go around in this way, you wouldn't go down, so you might go straight up there. And if you project your point down onto the input space, so this is the point above which you are, that vector, the one that's gonna get you going uphill the fastest, the direction you should walk, for this graph, it should kind of make sense, is directly away from the origin."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "Let's say you're sitting at a point like this, and you say, what direction should I walk to increase my altitude the fastest? You wanna get uphill as quickly as possible. And from that point, you might walk, you know, what looks like straight up there, you certainly wouldn't go around in this way, you wouldn't go down, so you might go straight up there. And if you project your point down onto the input space, so this is the point above which you are, that vector, the one that's gonna get you going uphill the fastest, the direction you should walk, for this graph, it should kind of make sense, is directly away from the origin. Because here, I'll erase this, because once I start moving things, that won't stick. If you would look at things from the very bottom, any point that you are on the mountain, on the graph here, and when you wanna increase the fastest, you should just go directly away from the origin, because that's when it's the steepest. And all of these vectors are also pointing directly away from the origin."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "And if you project your point down onto the input space, so this is the point above which you are, that vector, the one that's gonna get you going uphill the fastest, the direction you should walk, for this graph, it should kind of make sense, is directly away from the origin. Because here, I'll erase this, because once I start moving things, that won't stick. If you would look at things from the very bottom, any point that you are on the mountain, on the graph here, and when you wanna increase the fastest, you should just go directly away from the origin, because that's when it's the steepest. And all of these vectors are also pointing directly away from the origin. So people will say, the gradient points in the direction of steepest ascent. That might even be worth writing down. So, direction of steepest ascent."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "And all of these vectors are also pointing directly away from the origin. So people will say, the gradient points in the direction of steepest ascent. That might even be worth writing down. So, direction of steepest ascent. And let's just see what that looks like in the context of another example. So I'll pull up another graph here. Pull up another graph, and it's vector field."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "So, direction of steepest ascent. And let's just see what that looks like in the context of another example. So I'll pull up another graph here. Pull up another graph, and it's vector field. So this graph, it's all negative values, it's all below the xy-plane, and it's got these two different peaks. And I've also drawn the gradient field, which is the word for the vector field representing the gradient, on top. And notice, near the peak, all of the vectors are pointing kind of in the uphill direction, sort of telling you to go towards that peak in some way."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "Pull up another graph, and it's vector field. So this graph, it's all negative values, it's all below the xy-plane, and it's got these two different peaks. And I've also drawn the gradient field, which is the word for the vector field representing the gradient, on top. And notice, near the peak, all of the vectors are pointing kind of in the uphill direction, sort of telling you to go towards that peak in some way. And as you get a feel around, you can see here, this very top one, like the point that it's stemming from corresponds with something just a little bit shy of the peak there. And everybody's telling you to go uphill. Each vector is telling you which way to walk to increase the altitude on the graph the fastest."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "And notice, near the peak, all of the vectors are pointing kind of in the uphill direction, sort of telling you to go towards that peak in some way. And as you get a feel around, you can see here, this very top one, like the point that it's stemming from corresponds with something just a little bit shy of the peak there. And everybody's telling you to go uphill. Each vector is telling you which way to walk to increase the altitude on the graph the fastest. It's the direction of steepest ascent. And that's what the direction means, but what does the length mean? Well, if you take a look, take a look at these red vectors here."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "Each vector is telling you which way to walk to increase the altitude on the graph the fastest. It's the direction of steepest ascent. And that's what the direction means, but what does the length mean? Well, if you take a look, take a look at these red vectors here. So red means that they should be considered very, very long. And the graph itself, the point they correspond to on the graph is just way off screen for us, because this graph gets really steep and really negative very fast. So the points these correspond to have really, really steep slopes, whereas these blue ones over here, you know, it's kind of a relatively shallow slope."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "Well, if you take a look, take a look at these red vectors here. So red means that they should be considered very, very long. And the graph itself, the point they correspond to on the graph is just way off screen for us, because this graph gets really steep and really negative very fast. So the points these correspond to have really, really steep slopes, whereas these blue ones over here, you know, it's kind of a relatively shallow slope. By the time you get into the peak, things start leveling off. So the length of the gradient vector actually tells you the steepness of that direction of steepest ascent. But one thing I want to point out here, it doesn't really make sense immediately looking at it why just throwing the partial derivatives into a vector is gonna give you this direction of steepest ascent."}, {"video_title": "Gradient and graphs.mp3", "Sentence": "So the points these correspond to have really, really steep slopes, whereas these blue ones over here, you know, it's kind of a relatively shallow slope. By the time you get into the peak, things start leveling off. So the length of the gradient vector actually tells you the steepness of that direction of steepest ascent. But one thing I want to point out here, it doesn't really make sense immediately looking at it why just throwing the partial derivatives into a vector is gonna give you this direction of steepest ascent. Ultimately it will, we're gonna talk through that, and I hope to make that connection pretty clear. But unless you're some kind of intuitive genius, I don't think that connection is at all obvious at first. But you will see it in due time."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "So I've written here three different functions. The first one is a multivariable function. It has a two variable input, x, y, and a single variable output that's x squared times y, that's just a number. And then the other two functions are each just regular old single variable functions. And what I want to do is start thinking about the composition of them. So I'm going to take as the first component the value of the function x of t. So you pump t through that and then you make that the first component of f. And the second component will be the value of the function y of t. So the image that you might have in your head for something like this is you can think of t as just living on a number line of some kind. Then you have x and y, which is just a plane."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "And then the other two functions are each just regular old single variable functions. And what I want to do is start thinking about the composition of them. So I'm going to take as the first component the value of the function x of t. So you pump t through that and then you make that the first component of f. And the second component will be the value of the function y of t. So the image that you might have in your head for something like this is you can think of t as just living on a number line of some kind. Then you have x and y, which is just a plane. So that'll be your x coordinate, your y coordinate, two dimensional space. And then you have your output, which is just whatever the value of f is. And for this whole function, for this whole composition of functions, you're thinking of xt, yt as taking a single point in t and kind of moving it over to two dimensional space somewhere, and then from there, our multivariable function takes that back down."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "Then you have x and y, which is just a plane. So that'll be your x coordinate, your y coordinate, two dimensional space. And then you have your output, which is just whatever the value of f is. And for this whole function, for this whole composition of functions, you're thinking of xt, yt as taking a single point in t and kind of moving it over to two dimensional space somewhere, and then from there, our multivariable function takes that back down. So this is just a single variable function, nothing too fancy going on in terms of where you start and where you end up, it's just what's happening in the middle. And what I want to know is what's the derivative of this function? If I take this, and it's just an ordinary derivative, not a partial derivative, because this is a single variable function, one variable input, one variable output, how do you take its derivative?"}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "And for this whole function, for this whole composition of functions, you're thinking of xt, yt as taking a single point in t and kind of moving it over to two dimensional space somewhere, and then from there, our multivariable function takes that back down. So this is just a single variable function, nothing too fancy going on in terms of where you start and where you end up, it's just what's happening in the middle. And what I want to know is what's the derivative of this function? If I take this, and it's just an ordinary derivative, not a partial derivative, because this is a single variable function, one variable input, one variable output, how do you take its derivative? And there's a special rule for this, it's called the chain rule, the multivariable chain rule, but you don't actually need it. So let's actually walk through this, showing that you don't need it. It's not that you'll never need it, it's just for computations like this, you could go without it."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "If I take this, and it's just an ordinary derivative, not a partial derivative, because this is a single variable function, one variable input, one variable output, how do you take its derivative? And there's a special rule for this, it's called the chain rule, the multivariable chain rule, but you don't actually need it. So let's actually walk through this, showing that you don't need it. It's not that you'll never need it, it's just for computations like this, you could go without it. It's a very useful theoretical tool, a very useful model to have in mind for what function composition looks like and implies for derivatives in the multivariable world. So let's just start plugging things in here. If I have f of x of t of y of t, the first thing I might do is write, okay, f, and instead of x of t, just write in cosine of t, since that's the function that I have for x of t. And then y, we replace that with sine of t. Sine of t. And of course I'm hoping to take the derivative of this."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "It's not that you'll never need it, it's just for computations like this, you could go without it. It's a very useful theoretical tool, a very useful model to have in mind for what function composition looks like and implies for derivatives in the multivariable world. So let's just start plugging things in here. If I have f of x of t of y of t, the first thing I might do is write, okay, f, and instead of x of t, just write in cosine of t, since that's the function that I have for x of t. And then y, we replace that with sine of t. Sine of t. And of course I'm hoping to take the derivative of this. And then from there, we can go to the definition of f, f of xy equals x squared times y, which means we take that first component squared, so we'll take that first component, cosine of t, and then square it. Square that guy. And then we'll multiply it by the second component, sine of t. Sine of t. And again, we're just taking this derivative."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "If I have f of x of t of y of t, the first thing I might do is write, okay, f, and instead of x of t, just write in cosine of t, since that's the function that I have for x of t. And then y, we replace that with sine of t. Sine of t. And of course I'm hoping to take the derivative of this. And then from there, we can go to the definition of f, f of xy equals x squared times y, which means we take that first component squared, so we'll take that first component, cosine of t, and then square it. Square that guy. And then we'll multiply it by the second component, sine of t. Sine of t. And again, we're just taking this derivative. And you might be wondering, okay, why am I doing this? You're just showing me how to take a first derivative, an ordinary derivative, but the pattern that we'll see is gonna lead us to the multivariable chain rule, and it's actually kind of surprising when you see it in this context, because it pops out in a way that you might not expect things to pop out. So continuing our chugging along, when you take the derivative of this, you do the product rule, left d right plus right d left."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "And then we'll multiply it by the second component, sine of t. Sine of t. And again, we're just taking this derivative. And you might be wondering, okay, why am I doing this? You're just showing me how to take a first derivative, an ordinary derivative, but the pattern that we'll see is gonna lead us to the multivariable chain rule, and it's actually kind of surprising when you see it in this context, because it pops out in a way that you might not expect things to pop out. So continuing our chugging along, when you take the derivative of this, you do the product rule, left d right plus right d left. So in this case, the left is cosine squared of t. We just leave that as it is. Cosine squared of t. And multiply it by the derivative of the right, d right. So that's gonna be cosine of t. Cosine of t. And then we add to that right, which is, you know, keep that right side unchanged, multiplied by the derivative of the left."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "So continuing our chugging along, when you take the derivative of this, you do the product rule, left d right plus right d left. So in this case, the left is cosine squared of t. We just leave that as it is. Cosine squared of t. And multiply it by the derivative of the right, d right. So that's gonna be cosine of t. Cosine of t. And then we add to that right, which is, you know, keep that right side unchanged, multiplied by the derivative of the left. And for that, we use the chain rule, the single variable chain rule, where you think of taking the derivative of the outside, so you plop that two down, like you're taking the derivative of two x, but you're just writing in cosine, instead of x. Cosine t. And then you multiply that by the derivative of, by the derivative of the inside, that's a tongue twister, which is negative sine of t. Negative sine of t. And I'm afraid I'm gonna run off the edge here, certainly with the many, many parentheses that I need. I'll go ahead and rewrite this, though. I'm gonna rewrite it anyway, because there's a certain pattern that I hope to make clear."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "So that's gonna be cosine of t. Cosine of t. And then we add to that right, which is, you know, keep that right side unchanged, multiplied by the derivative of the left. And for that, we use the chain rule, the single variable chain rule, where you think of taking the derivative of the outside, so you plop that two down, like you're taking the derivative of two x, but you're just writing in cosine, instead of x. Cosine t. And then you multiply that by the derivative of, by the derivative of the inside, that's a tongue twister, which is negative sine of t. Negative sine of t. And I'm afraid I'm gonna run off the edge here, certainly with the many, many parentheses that I need. I'll go ahead and rewrite this, though. I'm gonna rewrite it anyway, because there's a certain pattern that I hope to make clear. Here. So let me just rewrite this side. Just copy that down here."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "I'm gonna rewrite it anyway, because there's a certain pattern that I hope to make clear. Here. So let me just rewrite this side. Just copy that down here. I just wanna rewrite this guy. You might be wondering why, but it'll become clear in just a moment why I wanna do this. So in this case, I'm gonna write this as two times cosine of t, times sine of t. And then all of that multiplied by negative sine of t. Negative sine of t. So this is the derivative."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "Just copy that down here. I just wanna rewrite this guy. You might be wondering why, but it'll become clear in just a moment why I wanna do this. So in this case, I'm gonna write this as two times cosine of t, times sine of t. And then all of that multiplied by negative sine of t. Negative sine of t. So this is the derivative. This is the derivative of the composition of functions that ultimately was a single variable function, but it kind of went through two different variables. And I just wanna make an observation in terms of the partial derivatives of f. So let me just make a copy of this guy. Give ourselves a little bit of room down here."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "So in this case, I'm gonna write this as two times cosine of t, times sine of t. And then all of that multiplied by negative sine of t. Negative sine of t. So this is the derivative. This is the derivative of the composition of functions that ultimately was a single variable function, but it kind of went through two different variables. And I just wanna make an observation in terms of the partial derivatives of f. So let me just make a copy of this guy. Give ourselves a little bit of room down here. Just paste that over here. So let's look at the partial derivatives of f for a second here. So if I took the partial derivative with respect to x, partial partial x, which means y is treated as a constant, so I take the derivative of x squared to get two x, and then multiply it by that constant, which is just y."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "Give ourselves a little bit of room down here. Just paste that over here. So let's look at the partial derivatives of f for a second here. So if I took the partial derivative with respect to x, partial partial x, which means y is treated as a constant, so I take the derivative of x squared to get two x, and then multiply it by that constant, which is just y. And if I also do it with respect to y, get all of them in there. So now y looks like a variable. X looks like a constant."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "So if I took the partial derivative with respect to x, partial partial x, which means y is treated as a constant, so I take the derivative of x squared to get two x, and then multiply it by that constant, which is just y. And if I also do it with respect to y, get all of them in there. So now y looks like a variable. X looks like a constant. So x squared also looks like a constant. Constant times a variable. The derivative is just that constant."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "X looks like a constant. So x squared also looks like a constant. Constant times a variable. The derivative is just that constant. These two, their pattern comes up in the ultimate result that we got. And this is the whole reason that I rewrote it. If you look at this two xy, you can see that over here, where cosine corresponds to x, sine corresponds to y based on our original functions."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "The derivative is just that constant. These two, their pattern comes up in the ultimate result that we got. And this is the whole reason that I rewrote it. If you look at this two xy, you can see that over here, where cosine corresponds to x, sine corresponds to y based on our original functions. And then x squared here corresponds with squaring the x that we put in there. Then if we take the derivative of our two intermediary functions, the ordinary derivative of x with respect to t, that's derivative of cosine, negative sine of t. And then similarly, derivative of y, just the ordinary derivative, no partials going on here, with respect to t, that's equal to cosine. Derivative of sine is cosine."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "If you look at this two xy, you can see that over here, where cosine corresponds to x, sine corresponds to y based on our original functions. And then x squared here corresponds with squaring the x that we put in there. Then if we take the derivative of our two intermediary functions, the ordinary derivative of x with respect to t, that's derivative of cosine, negative sine of t. And then similarly, derivative of y, just the ordinary derivative, no partials going on here, with respect to t, that's equal to cosine. Derivative of sine is cosine. And these guys show up, right? You see negative sine over here. And you see cosine show up over here."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "Derivative of sine is cosine. And these guys show up, right? You see negative sine over here. And you see cosine show up over here. And we could generalize this. We could write it down and say, at least for this specific example, it looks like the derivative of the composition is this part, which is the partial of f with respect to y. Right, that's kind of what it looks like here once we've plugged in the intermediary functions."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "And you see cosine show up over here. And we could generalize this. We could write it down and say, at least for this specific example, it looks like the derivative of the composition is this part, which is the partial of f with respect to y. Right, that's kind of what it looks like here once we've plugged in the intermediary functions. Multiply by, this guy was the ordinary derivative of y with respect to t. So that was the ordinary derivative of y with respect to t. And then very similarly, this guy was the partial of f with respect to x, partial x, and we're multiplying it by the ordinary derivative of x of t, so over here, x of t, with respect to t. And of course, when I write this partial f, partial y, what I really mean is you plug in for x and y the two coordinate functions, x of t, y of t. So if I say partial f, partial y over here, what I really mean is you take that x squared and then you plug in x of t squared to get cosine, cosine squared. And same deal over here. You're always plugging things in, so you ultimately have a function of t. But this right here has a name."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "Right, that's kind of what it looks like here once we've plugged in the intermediary functions. Multiply by, this guy was the ordinary derivative of y with respect to t. So that was the ordinary derivative of y with respect to t. And then very similarly, this guy was the partial of f with respect to x, partial x, and we're multiplying it by the ordinary derivative of x of t, so over here, x of t, with respect to t. And of course, when I write this partial f, partial y, what I really mean is you plug in for x and y the two coordinate functions, x of t, y of t. So if I say partial f, partial y over here, what I really mean is you take that x squared and then you plug in x of t squared to get cosine, cosine squared. And same deal over here. You're always plugging things in, so you ultimately have a function of t. But this right here has a name. This is the multivariable chain rule. And it's important enough, I'll just kind of, I'll just write it out all on its own here. If we take the ordinary derivative with respect to t of a composition of a multivariable function, in this case, just two variables, x of t, y of t, where we're plugging in two intermediary functions, x of t, y of t, each of which just single variable, the result is that we take the partial derivative with respect to x and we multiply it by the derivative of x with respect to t, and then we add to that the partial derivative with respect to y multiplied by the derivative of y with respect to t. So this entire expression here is what you might call the simple version of the multivariable chain rule."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "You're always plugging things in, so you ultimately have a function of t. But this right here has a name. This is the multivariable chain rule. And it's important enough, I'll just kind of, I'll just write it out all on its own here. If we take the ordinary derivative with respect to t of a composition of a multivariable function, in this case, just two variables, x of t, y of t, where we're plugging in two intermediary functions, x of t, y of t, each of which just single variable, the result is that we take the partial derivative with respect to x and we multiply it by the derivative of x with respect to t, and then we add to that the partial derivative with respect to y multiplied by the derivative of y with respect to t. So this entire expression here is what you might call the simple version of the multivariable chain rule. And you get, there's a more general version, and we'll kind of build up to it, but this is the simplest example you can think of where you start with one dimension and then you move over to two dimensions somehow, and then you move from those two dimensions down to one. So this is that. And in the next video, I'm gonna talk about the intuition for why this is true."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "If we take the ordinary derivative with respect to t of a composition of a multivariable function, in this case, just two variables, x of t, y of t, where we're plugging in two intermediary functions, x of t, y of t, each of which just single variable, the result is that we take the partial derivative with respect to x and we multiply it by the derivative of x with respect to t, and then we add to that the partial derivative with respect to y multiplied by the derivative of y with respect to t. So this entire expression here is what you might call the simple version of the multivariable chain rule. And you get, there's a more general version, and we'll kind of build up to it, but this is the simplest example you can think of where you start with one dimension and then you move over to two dimensions somehow, and then you move from those two dimensions down to one. So this is that. And in the next video, I'm gonna talk about the intuition for why this is true. You know, here I just went through an example and showed, oh, it just happens to be true, it fills this pattern. But there's a very nice line of reasoning for where this comes about. And I'll also talk about a more generalized form where you'll see it."}, {"video_title": "Multivariable chain rule.mp3", "Sentence": "And in the next video, I'm gonna talk about the intuition for why this is true. You know, here I just went through an example and showed, oh, it just happens to be true, it fills this pattern. But there's a very nice line of reasoning for where this comes about. And I'll also talk about a more generalized form where you'll see it. We start using vector notation, it makes things look very clean. And I might even get around to a more formal argument for why this is true. So see you next video."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in this video I want to get a little bit more exacting and actually use units so that we really understand what's going on here. So I've drawn our path C and we're traversing it in the positive counterclockwise direction. And then I've taken a few sample points for F. At any point in the XY plane it associates a two dimensional vector, maybe at that point the two dimensional vector looks like that. Maybe at that point the two dimensional vector looks like that. And then N is of course the unit normal vector at any point on our curve. The outward, I should say the outward pointing unit vector at any point on our curve. Now in the last video I talked about F as being some type of a velocity function."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Maybe at that point the two dimensional vector looks like that. And then N is of course the unit normal vector at any point on our curve. The outward, I should say the outward pointing unit vector at any point on our curve. Now in the last video I talked about F as being some type of a velocity function. That at any point it gives you the velocity of the particles there. And that wasn't exactly right. In order to really understand what's happening here, in order to really conceptualize this as kind of flux through the boundary, the rate of mass exiting this boundary here, we actually have to introduce a density aspect to F. So right over here I've rewritten F. And I've rewritten it as a product of a scalar function and a vector function."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now in the last video I talked about F as being some type of a velocity function. That at any point it gives you the velocity of the particles there. And that wasn't exactly right. In order to really understand what's happening here, in order to really conceptualize this as kind of flux through the boundary, the rate of mass exiting this boundary here, we actually have to introduce a density aspect to F. So right over here I've rewritten F. And I've rewritten it as a product of a scalar function and a vector function. So the scalar part right over here, rho of XY, rho is the Greek letter, is a Greek letter often used to represent density of some kind. In this case it's mass density. So at any given XY point, it kind of, this tells us what the mass density is."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In order to really understand what's happening here, in order to really conceptualize this as kind of flux through the boundary, the rate of mass exiting this boundary here, we actually have to introduce a density aspect to F. So right over here I've rewritten F. And I've rewritten it as a product of a scalar function and a vector function. So the scalar part right over here, rho of XY, rho is the Greek letter, is a Greek letter often used to represent density of some kind. In this case it's mass density. So at any given XY point, it kind of, this tells us what the mass density is. Mass density will be in some mass, we're in a two dimensional world. So it's mass per area. And if we want to pick particular units for our example, once again this isn't the only way that this can be conceived of."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So at any given XY point, it kind of, this tells us what the mass density is. Mass density will be in some mass, we're in a two dimensional world. So it's mass per area. And if we want to pick particular units for our example, once again this isn't the only way that this can be conceived of. There's other applications, but this is the easiest way for my brain to process it. We can imagine this is kilogram per square meter. And this right over here is a velocity vector."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we want to pick particular units for our example, once again this isn't the only way that this can be conceived of. There's other applications, but this is the easiest way for my brain to process it. We can imagine this is kilogram per square meter. And this right over here is a velocity vector. It tells us what is the velocity of the particles of that point. So this is kind of saying how many, how much particles do you have in a kind of a point, how dense are they? And then this is how fast are they going and in what direction."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this right over here is a velocity vector. It tells us what is the velocity of the particles of that point. So this is kind of saying how many, how much particles do you have in a kind of a point, how dense are they? And then this is how fast are they going and in what direction. And this whole thing is a vector, it's a velocity vector. But the components right over here, M of XY, M of XY is just a number and you multiply that times a vector. So M of XY right over here is going to be a scalar function."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then this is how fast are they going and in what direction. And this whole thing is a vector, it's a velocity vector. But the components right over here, M of XY, M of XY is just a number and you multiply that times a vector. So M of XY right over here is going to be a scalar function. When you multiply it times I becomes a vector. That's going to give you a speed. And then N of XY is also going to give you a speed."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So M of XY right over here is going to be a scalar function. When you multiply it times I becomes a vector. That's going to give you a speed. And then N of XY is also going to give you a speed. And then it tells you a speed in the J direction, so it becomes a vector. A speed in the I direction becomes a vector as well. But these speeds, the units of speed, so let me write this over here, the speeds, the unit of speed."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then N of XY is also going to give you a speed. And then it tells you a speed in the J direction, so it becomes a vector. A speed in the I direction becomes a vector as well. But these speeds, the units of speed, so let me write this over here, the speeds, the unit of speed. So now we're talking about in particular M of XY and N of XY. That would be in units of distance per time. And so maybe for this example, we'll say the units are meters per second."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But these speeds, the units of speed, so let me write this over here, the speeds, the unit of speed. So now we're talking about in particular M of XY and N of XY. That would be in units of distance per time. And so maybe for this example, we'll say the units are meters per second. So let's think about what the units will be for this function. If we distribute the rho, if we distribute the rho, because really in any given XY, it really is just a number. So if we do that, we're going to get F is going, and I'm not going to keep writing F of XY."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so maybe for this example, we'll say the units are meters per second. So let's think about what the units will be for this function. If we distribute the rho, if we distribute the rho, because really in any given XY, it really is just a number. So if we do that, we're going to get F is going, and I'm not going to keep writing F of XY. We'll just understand that F, rho, M, and N are functions of XY. F is going to be equal to rho times M, rho times M times the unit vector I plus rho times N, N times the unit vector J. Now what are the units here?"}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we do that, we're going to get F is going, and I'm not going to keep writing F of XY. We'll just understand that F, rho, M, and N are functions of XY. F is going to be equal to rho times M, rho times M times the unit vector I plus rho times N, N times the unit vector J. Now what are the units here? What's rho times M? What units are we going to get there? And we're going to get the same units when we do rho times N. Well, we're going to have, if we pick these particular units, we're going to have kilograms per meter squared times meters per second, times meters per second."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now what are the units here? What's rho times M? What units are we going to get there? And we're going to get the same units when we do rho times N. Well, we're going to have, if we pick these particular units, we're going to have kilograms per meter squared times meters per second, times meters per second. So a little bit of dimensional analysis here. This meter in the numerator will cancel out with one of the meters in the denominator, and we are left with something kind of strange. Kilograms per meter second, which is essentially what the, if you view this vector has a magnitude in some direction, the magnitude component is going to have these units right over here."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to get the same units when we do rho times N. Well, we're going to have, if we pick these particular units, we're going to have kilograms per meter squared times meters per second, times meters per second. So a little bit of dimensional analysis here. This meter in the numerator will cancel out with one of the meters in the denominator, and we are left with something kind of strange. Kilograms per meter second, which is essentially what the, if you view this vector has a magnitude in some direction, the magnitude component is going to have these units right over here. And then we're going to take this and we're dotting it with N. N just only gives us a direction. It is a unitless vector. It is only specifying a direction at any point in the curve."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Kilograms per meter second, which is essentially what the, if you view this vector has a magnitude in some direction, the magnitude component is going to have these units right over here. And then we're going to take this and we're dotting it with N. N just only gives us a direction. It is a unitless vector. It is only specifying a direction at any point in the curve. And so when I take a dot product with this, it's going to give us essentially what is the magnitude, what is the magnitude of F going in the direction, going in the direction of N. So this right over here, when you take the dot, it'll say it's essentially the magnitude, a part of the magnitude of F going in N's direction. It's just going to have the same exact units as F. So the units of this part, you're going to have kilograms per meter second. And then we're going to, so let me make this very clear."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It is only specifying a direction at any point in the curve. And so when I take a dot product with this, it's going to give us essentially what is the magnitude, what is the magnitude of F going in the direction, going in the direction of N. So this right over here, when you take the dot, it'll say it's essentially the magnitude, a part of the magnitude of F going in N's direction. It's just going to have the same exact units as F. So the units of this part, you're going to have kilograms per meter second. And then we're going to, so let me make this very clear. So let's say we're focusing on this point right over here. F looks like that. Its magnitude, the length of that vector is going to be in kilograms per meter second."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we're going to, so let me make this very clear. So let's say we're focusing on this point right over here. F looks like that. Its magnitude, the length of that vector is going to be in kilograms per meter second. Then we have a normal vector right over there. And when you take the dot product, you're essentially saying what's the magnitude that's going in the normal direction? So essentially what's the magnitude, what's the magnitude of that vector right over there?"}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Its magnitude, the length of that vector is going to be in kilograms per meter second. Then we have a normal vector right over there. And when you take the dot product, you're essentially saying what's the magnitude that's going in the normal direction? So essentially what's the magnitude, what's the magnitude of that vector right over there? It's going to be in kilograms per meter second. And then we're multiplying it times ds. We're multiplying it times this infinitesimally small chunk of or a little segment of the curve."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So essentially what's the magnitude, what's the magnitude of that vector right over there? It's going to be in kilograms per meter second. And then we're multiplying it times ds. We're multiplying it times this infinitesimally small chunk of or a little segment of the curve. We're going to multiply that times ds. Well, what are the units for ds? It's going to be a unit of length."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're multiplying it times this infinitesimally small chunk of or a little segment of the curve. We're going to multiply that times ds. Well, what are the units for ds? It's going to be a unit of length. We'll just go with meters. So this right over here is going to be meters. So you're going to have, there's this whole integral."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be a unit of length. We'll just go with meters. So this right over here is going to be meters. So you're going to have, there's this whole integral. You're going to have kilogram per meter second times meters. So if you have kilograms per meter second and you were to multiply that times meters, you were to multiply that times meters, what do you get? Well, this meter is going to cancel with that meters."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you're going to have, there's this whole integral. You're going to have kilogram per meter second times meters. So if you have kilograms per meter second and you were to multiply that times meters, you were to multiply that times meters, what do you get? Well, this meter is going to cancel with that meters. And then you get something that kind of starts to make sense. You have kilogram per second. And so this hopefully makes it clear what's going on here."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, this meter is going to cancel with that meters. And then you get something that kind of starts to make sense. You have kilogram per second. And so this hopefully makes it clear what's going on here. This is telling us how much mass is crossing that little ds, that little section of the curve per second. And if you were to add up, and that's what integrals are all about, adding up an infinite number of these infinitesimally small ds's, if you were to add all of that up, you're going to get the value of this entire integral is going to be in kilograms per second. And it's essentially going to say how much mass is exiting this curve at any given point or at any given time."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this hopefully makes it clear what's going on here. This is telling us how much mass is crossing that little ds, that little section of the curve per second. And if you were to add up, and that's what integrals are all about, adding up an infinite number of these infinitesimally small ds's, if you were to add all of that up, you're going to get the value of this entire integral is going to be in kilograms per second. And it's essentially going to say how much mass is exiting this curve at any given point or at any given time. So this is mass, so this whole integral, so the whole integral, so the whole integral, let me rewrite it, of F dot n ds tells us the mass exiting the curve per second. And this should also be consistent. In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's essentially going to say how much mass is exiting this curve at any given point or at any given time. So this is mass, so this whole integral, so the whole integral, so the whole integral, let me rewrite it, of F dot n ds tells us the mass exiting the curve per second. And this should also be consistent. In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem. In the last video, we saw that this is equivalent to the double integral over the area of the divergence of F, which is essentially just, well, I could write it two ways, the divergence of F, and this right over here, that's just the partial of, let me write it down here so I have some space, the partial of the i component with respect to x, let me write it over here, I don't want to do this too fast and loose. So this right over here is going to be the partial of rho m, let me write it like this, rho m with respect to x plus the partial of the y component, rho n, with respect to y, times each little chunk of area, times each little chunk of area. Well, what are the units of this going to be right over here?"}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we saw that this is equivalent to, and this is where we kind of view it as a two-dimensional divergence theorem. In the last video, we saw that this is equivalent to the double integral over the area of the divergence of F, which is essentially just, well, I could write it two ways, the divergence of F, and this right over here, that's just the partial of, let me write it down here so I have some space, the partial of the i component with respect to x, let me write it over here, I don't want to do this too fast and loose. So this right over here is going to be the partial of rho m, let me write it like this, rho m with respect to x plus the partial of the y component, rho n, with respect to y, times each little chunk of area, times each little chunk of area. Well, what are the units of this going to be right over here? We know what rho m is, rho m gives us kilogram per meter second. But if we were to take, essentially, the derivative with respect to meters again, the units here are going to give, the units for either of these, the units for either of these characters are going to be kilograms per meter second per second, because we're taking the derivative with respect, sorry, per meter, we're taking the derivative with respect to another unit of distance. So you're going to take per meter, so you're going to have another meter right here in the denominator, that's going to be the units here, and then you're multiplying it times an area."}, {"video_title": "Conceptual clarification for 2D divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, what are the units of this going to be right over here? We know what rho m is, rho m gives us kilogram per meter second. But if we were to take, essentially, the derivative with respect to meters again, the units here are going to give, the units for either of these, the units for either of these characters are going to be kilograms per meter second per second, because we're taking the derivative with respect, sorry, per meter, we're taking the derivative with respect to another unit of distance. So you're going to take per meter, so you're going to have another meter right here in the denominator, that's going to be the units here, and then you're multiplying it times an area. You're multiplying it times an area, so that would be meter squared, that's right, this right over here is square meters, they cancel it out, and once again, this whole part right over here that you're summing up gives us kilograms per second. So you're having a bunch of kilograms per second and you're just adding them up throughout the entire, throughout this entire area right over here. So hopefully this makes a little bit more sense about kind of how to conceptualize this vector function f. If it confuses you, try your best to ignore it, I guess, but for me at least, this helped me out conceptual, having a stronger conception of what vector f could kind of represent."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Then the way to think about maximizing this function is to try to increase that value of c as much as you can without it falling off the circle. And the key observation is that happens when they're tangent. So you might kind of draw this out in a little sketch and say there's some curve representing your constraint, which in this case would be kind of where our circle is, and then the curve representing the contour would just kiss that curve, just barely touch it in some way. Now, that's pretty, but in terms of solving the problem, we still have some work to do. And the main tool we're gonna use here is the gradient. So let me go ahead and draw a lot more contour lines than there already are for x squared times y, so this is many of the contour lines, and I'll draw the gradient field, the gradient field of f. So I've made a video about the relationship between the gradient and contour lines, and the upshot of it is that these gradient vectors, every time they pass through a contour line, they're perpendicular to it. And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Now, that's pretty, but in terms of solving the problem, we still have some work to do. And the main tool we're gonna use here is the gradient. So let me go ahead and draw a lot more contour lines than there already are for x squared times y, so this is many of the contour lines, and I'll draw the gradient field, the gradient field of f. So I've made a video about the relationship between the gradient and contour lines, and the upshot of it is that these gradient vectors, every time they pass through a contour line, they're perpendicular to it. And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change. But again, there's a whole video on that that's worth checking out if this feels unfamiliar. For our purposes, what it means is that when we're considering this point of tangency, the gradient of f at that point is gonna be some vector perpendicular to both of the curves at that point, so that little vector represents the gradient of f at this point on the plane. And we can do something very similar to understand the other curve."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "And the basic reason for that is if you walk along the contour line, the function isn't changing value, so if you want it to change most rapidly, you know, it kind of makes sense you should walk in the perpendicular direction, so that no component of the walk that you're taking is, you know, useless, is along the line where the function doesn't change. But again, there's a whole video on that that's worth checking out if this feels unfamiliar. For our purposes, what it means is that when we're considering this point of tangency, the gradient of f at that point is gonna be some vector perpendicular to both of the curves at that point, so that little vector represents the gradient of f at this point on the plane. And we can do something very similar to understand the other curve. Right now, I've just written it as a constraint x squared plus y squared equals one, but, you know, to give that function a name, let's say that we've defined g of xy to be x squared plus y squared, x squared plus y squared. In that case, this constraint is pretty much just one of the contour lines for the function g. And we can take a look at that. If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "And we can do something very similar to understand the other curve. Right now, I've just written it as a constraint x squared plus y squared equals one, but, you know, to give that function a name, let's say that we've defined g of xy to be x squared plus y squared, x squared plus y squared. In that case, this constraint is pretty much just one of the contour lines for the function g. And we can take a look at that. If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared. And if we took a look at the gradient of g, and we go over and ask about the gradient of g, it has that same property, that every gradient vector, if it passes through a contour line, is perpendicular to it. So over on our drawing here, the gradient vector of g would also be perpendicular to both these curves. And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "If we go over here and we look at all of the other contour lines for this function g, and it should make sense that they're circles because this function is x squared plus y squared. And if we took a look at the gradient of g, and we go over and ask about the gradient of g, it has that same property, that every gradient vector, if it passes through a contour line, is perpendicular to it. So over on our drawing here, the gradient vector of g would also be perpendicular to both these curves. And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer. There's no reason that it would be the same length, but the important fact is that it's proportional. And the way that we're gonna write this in formulas is to say that the gradient of f evaluated, let's see, evaluated at whatever the maximizing value of x and y are. So we should give that a name probably."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "And, you know, maybe in this case, it's not as long as the gradient of f, or maybe it's longer. There's no reason that it would be the same length, but the important fact is that it's proportional. And the way that we're gonna write this in formulas is to say that the gradient of f evaluated, let's see, evaluated at whatever the maximizing value of x and y are. So we should give that a name probably. Maybe, you know, x sub m, y sub m. The specific values of x and y that are gonna be at this point that maximizes the function subject to our constraint. So that's gonna be related to the gradient of g. It's not gonna be quite equal, so I'll leave some room here. Related to the gradient of g evaluated at that same point, xm, ym."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So we should give that a name probably. Maybe, you know, x sub m, y sub m. The specific values of x and y that are gonna be at this point that maximizes the function subject to our constraint. So that's gonna be related to the gradient of g. It's not gonna be quite equal, so I'll leave some room here. Related to the gradient of g evaluated at that same point, xm, ym. And like I said, they're not equal. They're proportional. So we need to have some kind of proportionality constant in there."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Related to the gradient of g evaluated at that same point, xm, ym. And like I said, they're not equal. They're proportional. So we need to have some kind of proportionality constant in there. And you almost always use the variable lambda. And this guy has a fancy name. It's called a Lagrange multiplier."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So we need to have some kind of proportionality constant in there. And you almost always use the variable lambda. And this guy has a fancy name. It's called a Lagrange multiplier. Lagrange. Lagrange was one of those famous French mathematicians. I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "It's called a Lagrange multiplier. Lagrange. Lagrange was one of those famous French mathematicians. I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace. There's a whole bunch of things. Let's see, multiplier. Distracting myself talking here."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "I always get him confused with some of the other French mathematicians at the time, like Legendre or Laplace. There's a whole bunch of things. Let's see, multiplier. Distracting myself talking here. So Lagrange multiplier. So there's a number of things in multivariable calculus named after Lagrange. And this is one of the big ones."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Distracting myself talking here. So Lagrange multiplier. So there's a number of things in multivariable calculus named after Lagrange. And this is one of the big ones. This is a technique that he kind of developed or at the very least popularized. And the core idea is to just set these gradients equal to each other, because that represents when the contour line for one function is tangent to the contour line of another. So this, this is something that we can actually work with."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "And this is one of the big ones. This is a technique that he kind of developed or at the very least popularized. And the core idea is to just set these gradients equal to each other, because that represents when the contour line for one function is tangent to the contour line of another. So this, this is something that we can actually work with. And let's start working it out, right? Let's see what this translates to in formulas. So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So this, this is something that we can actually work with. And let's start working it out, right? Let's see what this translates to in formulas. So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be. And that's the gradient of x squared plus y squared. And the way that we take our gradient is it's gonna be a vector whose components are all the partial derivatives. So the first component is the partial derivative with respect to x."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So I already have g written here, so let's go ahead and just evaluate what the gradient of g should be. And that's the gradient of x squared plus y squared. And the way that we take our gradient is it's gonna be a vector whose components are all the partial derivatives. So the first component is the partial derivative with respect to x. So we treat x as a variable. Y looks like a constant. The derivative is two x."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So the first component is the partial derivative with respect to x. So we treat x as a variable. Y looks like a constant. The derivative is two x. The second component, the partial derivative with respect to y. So now we're treating y as the variable. X is the constant."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "The derivative is two x. The second component, the partial derivative with respect to y. So now we're treating y as the variable. X is the constant. So the derivative looks like two y. Okay, so that's the gradient of g. And then the gradient of f, gradient of f, is gonna look like gradient of, let's see, what is x? What is f?"}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "X is the constant. So the derivative looks like two y. Okay, so that's the gradient of g. And then the gradient of f, gradient of f, is gonna look like gradient of, let's see, what is x? What is f? It's x squared times y. So x squared times y. We do the same thing."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "What is f? It's x squared times y. So x squared times y. We do the same thing. First component, partial derivative with respect to x. X looks like a variable, so its derivative is two times x. And then that y looks like a constant when we're up here. But then partial derivative with respect to y."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "We do the same thing. First component, partial derivative with respect to x. X looks like a variable, so its derivative is two times x. And then that y looks like a constant when we're up here. But then partial derivative with respect to y. That y looks like a variable. That x squared just looks like a constant sitting in front of it. So that's what we get."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "But then partial derivative with respect to y. That y looks like a variable. That x squared just looks like a constant sitting in front of it. So that's what we get. And now if we kind of work out this Lagrange multiplier expression using these two vectors, what we have written, what we're gonna have written is that this vector, two xy x squared, is proportional with a proportionality constant lambda to the gradient vector for g. So two x, two y. And if you want, you can think about this as two separate equations. I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So that's what we get. And now if we kind of work out this Lagrange multiplier expression using these two vectors, what we have written, what we're gonna have written is that this vector, two xy x squared, is proportional with a proportionality constant lambda to the gradient vector for g. So two x, two y. And if you want, you can think about this as two separate equations. I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations. Two times xy is equal to lambda. Ah, gotta change colors a lot here. Lambda times two x. Gonna be a stickler for color."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "I mean, right now it's one equation with vectors, but really what this is saying is you've got two separate equations. Two times xy is equal to lambda. Ah, gotta change colors a lot here. Lambda times two x. Gonna be a stickler for color. Keep red all of the things associated with g. And then the second equation is that x squared is equal to lambda times two y. And this might seem like a problem because we have three unknowns, x, y, and this new lambda that we introduced. Kind of shot ourselves in the foot by giving ourselves a new variable to deal with."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Lambda times two x. Gonna be a stickler for color. Keep red all of the things associated with g. And then the second equation is that x squared is equal to lambda times two y. And this might seem like a problem because we have three unknowns, x, y, and this new lambda that we introduced. Kind of shot ourselves in the foot by giving ourselves a new variable to deal with. But we only have two equations. So in order to solve this, we're gonna need three equations. And the third equation is something that we've known the whole time."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "Kind of shot ourselves in the foot by giving ourselves a new variable to deal with. But we only have two equations. So in order to solve this, we're gonna need three equations. And the third equation is something that we've known the whole time. It's been part of the original problem. It's the constraint itself, x squared plus y. X squared plus y squared, excuse me, equals one. So that, that third equation, x squared plus y squared is equal to one."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "And the third equation is something that we've known the whole time. It's been part of the original problem. It's the constraint itself, x squared plus y. X squared plus y squared, excuse me, equals one. So that, that third equation, x squared plus y squared is equal to one. So these are the three equations that characterize our constrained optimization problem. The bottom one just tells you that we have to be on this unit circle here. So let me just highlight it."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So that, that third equation, x squared plus y squared is equal to one. So these are the three equations that characterize our constrained optimization problem. The bottom one just tells you that we have to be on this unit circle here. So let me just highlight it. We have to be on this unit circle. And then these top two tell us what's necessary in order for our contour lines, the contour of f and the contour of g, to be perfectly tangent with each other. So in the next video, I'll go ahead and solve this."}, {"video_title": "Lagrange multipliers, using tangency to solve constrained optimization.mp3", "Sentence": "So let me just highlight it. We have to be on this unit circle. And then these top two tell us what's necessary in order for our contour lines, the contour of f and the contour of g, to be perfectly tangent with each other. So in the next video, I'll go ahead and solve this. At this point, it's pretty much just algebra to deal with, but it's worth going through. And then the next couple ones, I'll talk about a way that you can encapsulate all three of these equations into one expression, and also a little bit about the interpretation of this lambda that we introduced. Because it's not actually just a dummy variable."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now that we've explored Stokes' theorem a little bit, I want to talk about the situations in which we can use it. And we'll see it's a pretty general theorem, but we do have to think about what type of surfaces and what type of boundaries of those surfaces we are actually dealing with. And in the case of Stokes, we need surfaces that are piecewise smooth. Smooth, piecewise smooth surfaces. So this surface right over here, it is actually smooth, not just even piecewise smooth. Sounds like a very fancy term, but all the smooth part means, all the smooth part means is that you have continuous derivatives. And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Smooth, piecewise smooth surfaces. So this surface right over here, it is actually smooth, not just even piecewise smooth. Sounds like a very fancy term, but all the smooth part means, all the smooth part means is that you have continuous derivatives. And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick. So this is continuous derivatives. And another way to think about that conceptually is that if you pick a direction on the surface, if you say that we go in that direction, the slope in that direction changes gradually. It doesn't jump around."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And since we're talking about a surface, we're gonna think about continuous partial derivatives, regardless of which direction you pick. So this is continuous derivatives. And another way to think about that conceptually is that if you pick a direction on the surface, if you say that we go in that direction, the slope in that direction changes gradually. It doesn't jump around. If you pick this direction right over here, the slope is changing gradually. So we have a continuous derivative. Now you're like, well, what does the piecewise mean?"}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It doesn't jump around. If you pick this direction right over here, the slope is changing gradually. So we have a continuous derivative. Now you're like, well, what does the piecewise mean? Well, the piecewise actually allows us to use Stokes' theorem with more surfaces. Because if we have a surface that looks like, let's say a surface that looks like this, let's say it looks like a cup. So this is the opening of the top of the cup."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now you're like, well, what does the piecewise mean? Well, the piecewise actually allows us to use Stokes' theorem with more surfaces. Because if we have a surface that looks like, let's say a surface that looks like this, let's say it looks like a cup. So this is the opening of the top of the cup. Let's say it has no opening on top. So we can see the backside of the cup. And this is the side of the cup."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the opening of the top of the cup. Let's say it has no opening on top. So we can see the backside of the cup. And this is the side of the cup. And then this right over here is the bottom of the cup. And if it was transparent, we could actually see through it. So a surface like this is not entirely smooth because it has edges."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is the side of the cup. And then this right over here is the bottom of the cup. And if it was transparent, we could actually see through it. So a surface like this is not entirely smooth because it has edges. There are points right over here. So this edge right over here, if we pick this, let's say we pick this direction to go. And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So a surface like this is not entirely smooth because it has edges. There are points right over here. So this edge right over here, if we pick this, let's say we pick this direction to go. And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically. It jumps. So the slope is not continuous at that edge. The slope jumps and we start going straight up."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if we go in this direction along the bottom, then right when we get to the edge, then all of a sudden the slope changes dramatically. It jumps. So the slope is not continuous at that edge. The slope jumps and we start going straight up. And so this entire surface is not smooth, but the piecewise actually gives us an out. This tells us that it's okay as long as we can break the surface up into pieces that are smooth. And so this cup, we can break it up."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The slope jumps and we start going straight up. And so this entire surface is not smooth, but the piecewise actually gives us an out. This tells us that it's okay as long as we can break the surface up into pieces that are smooth. And so this cup, we can break it up. And we've been doing this when we've been tackling surface integrals. We can break it up into the bottom part, which is a smooth surface. It has a continuous derivative."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so this cup, we can break it up. And we've been doing this when we've been tackling surface integrals. We can break it up into the bottom part, which is a smooth surface. It has a continuous derivative. And the sides, the sides, the side, which kind of wraps around, is also a smooth surface. So most things that you will encounter in a traditional calculus course actually do, especially surfaces, do fit piecewise smooth. And the things that don't actually are fairly hard to visualize."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It has a continuous derivative. And the sides, the sides, the side, which kind of wraps around, is also a smooth surface. So most things that you will encounter in a traditional calculus course actually do, especially surfaces, do fit piecewise smooth. And the things that don't actually are fairly hard to visualize. And I imagine these ultra pointy, fractally looking things where it's hard to break it up into pieces that are actually smooth. That's for the surface part, but we also have to care about the boundary in order to apply Stokes' theorem. And that's that right over there."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the things that don't actually are fairly hard to visualize. And I imagine these ultra pointy, fractally looking things where it's hard to break it up into pieces that are actually smooth. That's for the surface part, but we also have to care about the boundary in order to apply Stokes' theorem. And that's that right over there. The boundary needs to be a simple, which means it doesn't cross itself, a simple, closed, piecewise smooth boundary. So once again, simple and closed, that just means, so this is not a simple boundary if it's really crossing itself, if it intersects itself, although you could break it up into two simple boundaries. But something like this is a simple boundary."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that's that right over there. The boundary needs to be a simple, which means it doesn't cross itself, a simple, closed, piecewise smooth boundary. So once again, simple and closed, that just means, so this is not a simple boundary if it's really crossing itself, if it intersects itself, although you could break it up into two simple boundaries. But something like this is a simple boundary. So that's a simple boundary right over there. It also has to be closed, which really means it just loops in on itself. You just won't have something like that."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But something like this is a simple boundary. So that's a simple boundary right over there. It also has to be closed, which really means it just loops in on itself. You just won't have something like that. It actually has to close. It actually has to loop in on itself in order to use Stokes' theorem. And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You just won't have something like that. It actually has to close. It actually has to loop in on itself in order to use Stokes' theorem. And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this. And piecewise smooth just means that you can break it up into sections where the derivative is continuous. The way I've drawn this one, this one, and this one, the slope is changing gradually. So over there, the slope is like that."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And once again, it has to be piecewise smooth, but now we're talking about a path or a line or a curve like this. And piecewise smooth just means that you can break it up into sections where the derivative is continuous. The way I've drawn this one, this one, and this one, the slope is changing gradually. So over there, the slope is like that. And it's changing gradually as we go around this path. Something that is not smooth, a path that is not smooth might look something like this, might look something like that. And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So over there, the slope is like that. And it's changing gradually as we go around this path. Something that is not smooth, a path that is not smooth might look something like this, might look something like that. And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there. But we just have to be simple, close. So this is simple and closed, and it's not smooth, but it is piecewise smooth. We can break it up into this section of the path, which is that line right over there is smooth."}, {"video_title": "Conditions for stokes theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the places that aren't smooth are at the edges, not smooth there, not smooth there, not smooth there, not smooth there. But we just have to be simple, close. So this is simple and closed, and it's not smooth, but it is piecewise smooth. We can break it up into this section of the path, which is that line right over there is smooth. That line right over there is smooth. That line is smooth, and that line is smooth. And we've done that when we've evaluated line integrals."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In this video, I will attempt to prove, or actually this in the next several videos, attempt to prove a special case version of Stokes' Theorem, or essentially Stokes' Theorem for a special case. And I'm doing this because the proof will be a little bit simpler, but at the same time, it's pretty convincing. And the special case we're going to assume is that the surface we're dealing with is a function of x and y. So if you give me any particular x and y, it only determines one point on that surface. So a surface like this would be the case. So it's kind of a mapping of this region in the xy plane into three dimensions. So for any xy, we can figure out the height."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you give me any particular x and y, it only determines one point on that surface. So a surface like this would be the case. So it's kind of a mapping of this region in the xy plane into three dimensions. So for any xy, we can figure out the height. So essentially z is going to be a function of x and y, and we can get a point on the surface. So this proof would not apply to a surface that's like a sphere or something like that, where any point on the xy plane could actually determine two points on our surface. But this is a pretty good start."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So for any xy, we can figure out the height. So essentially z is going to be a function of x and y, and we can get a point on the surface. So this proof would not apply to a surface that's like a sphere or something like that, where any point on the xy plane could actually determine two points on our surface. But this is a pretty good start. The other thing that we are going to assume, we are going to assume that z, which is essentially a function of x and y, that this function of x and y has continuous second-order derivatives. So continuous second derivatives. And the reason why I'm going to make that assumption is it's going to help us in our proof later on."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is a pretty good start. The other thing that we are going to assume, we are going to assume that z, which is essentially a function of x and y, that this function of x and y has continuous second-order derivatives. So continuous second derivatives. And the reason why I'm going to make that assumption is it's going to help us in our proof later on. It's going to allow us to say that the partial of our surface, or the partial of z with respect to x, and then taking the derivative of that with respect to y, is going to be the same as the partial of z with respect to y, and then taking the derivative of that with respect to x. And in order to be able to make this statement, we have to assume that z, or this z right over here, z is a function of x and y, has continuous second-order derivatives. And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the reason why I'm going to make that assumption is it's going to help us in our proof later on. It's going to allow us to say that the partial of our surface, or the partial of z with respect to x, and then taking the derivative of that with respect to y, is going to be the same as the partial of z with respect to y, and then taking the derivative of that with respect to x. And in order to be able to make this statement, we have to assume that z, or this z right over here, z is a function of x and y, has continuous second-order derivatives. And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives. Now with that out of the way, let's think about what Stokes' Theorem tells us, and then we'll think about, for this particular case, how we can write it out, and then hopefully we will see that the two things are equal. So let me write it out. So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And over here, we've just written our vector field F that we're going to deal with when we're trying to play with Stokes' Theorem, and we'll assume that it has continuous first-order derivatives. Now with that out of the way, let's think about what Stokes' Theorem tells us, and then we'll think about, for this particular case, how we can write it out, and then hopefully we will see that the two things are equal. So let me write it out. So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here. I'll do it in blue. It's this path right over here. This is the boundary of our surface."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So Stokes' Theorem tells us that F dot dr over some path, and the path that we care about is essentially this path right over here. I'll do it in blue. It's this path right over here. This is the boundary of our surface. So this is C right over here. Stokes' Theorem tells us that this should be the same thing. This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is the boundary of our surface. So this is C right over here. Stokes' Theorem tells us that this should be the same thing. This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself. And so in this video, I want to focus, or probably this and the next video, I want to focus on the second half. I want to focus this. I want to see how we can express this given the assumptions we've made."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This should be equivalent to the surface integral over our surface of curl of F dot ds, dot dotted with the surface itself. And so in this video, I want to focus, or probably this and the next video, I want to focus on the second half. I want to focus this. I want to see how we can express this given the assumptions we've made. And then after that, we're gonna see how we can express this given the same assumptions, and then hopefully we'll find that we get them to be equal to each other. So let's just start figuring out what the curl of F is equal to. So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to see how we can express this given the assumptions we've made. And then after that, we're gonna see how we can express this given the same assumptions, and then hopefully we'll find that we get them to be equal to each other. So let's just start figuring out what the curl of F is equal to. So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components. So I, I'll do them in different colors, I, J, and K components. I, J, and K components. And then I need to write my del operator, or my partial operators, I guess I should call them."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the curl of F is equal to, you could view it as the del operator crossed with our vector field F, which is equal to, we can write our components. So I, I'll do them in different colors, I, J, and K components. I, J, and K components. And then I need to write my del operator, or my partial operators, I guess I should call them. So the partial with respect to X, the partial with respect to Y, partial with respect to Z, and then I have to write the I, J, and K components of my vector field F. And I will do that in green, well, I'll do it in blue. And so we have P, which is a function of X, Y, Z, Q, which is a function of X, Y, and Z, and then R, which is a function of X, Y, and Z. And so this is going to evaluate as, it's going to be I, I times, so blank out that column, that row, it's gonna be the partial of R with respect to Y, partial of R with respect to Y, minus the partial of Q with respect to Z, minus the partial of Q with respect to Z."}, {"video_title": "Stokes' theorem proof part 1   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then I need to write my del operator, or my partial operators, I guess I should call them. So the partial with respect to X, the partial with respect to Y, partial with respect to Z, and then I have to write the I, J, and K components of my vector field F. And I will do that in green, well, I'll do it in blue. And so we have P, which is a function of X, Y, Z, Q, which is a function of X, Y, and Z, and then R, which is a function of X, Y, and Z. And so this is going to evaluate as, it's going to be I, I times, so blank out that column, that row, it's gonna be the partial of R with respect to Y, partial of R with respect to Y, minus the partial of Q with respect to Z, minus the partial of Q with respect to Z. And then checkerboard pattern, minus J, let me make that hat a little bit better, minus J, and then times the partial of R with respect to X, partial of R with respect to X, minus the partial of P with respect to Z, minus the partial of P with respect to Z. And then finally, plus K, plus K, and that's going to be times the partial of X, sorry, the partial of Q with respect to X, partial of Q with respect to X, minus the partial of P with respect to Y, minus the partial of P with respect to Y. So we figured out the curl of F, and I'll leave you there, I'm trying to make shorter videos now."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And here, I'd like to do the same, but for three dimensions. So a three-dimensional vector field is given by a function, a certain multivariable function, that has a three-dimensional input, given with coordinates x, y, and z, and then a three-dimensional vector output that has expressions that are somehow dependent on x, y, and z. I'll just put dots in here for now, but we'll fill this in with an example in just a moment. And the way that this works, just like with the two-dimensional vector field, you're gonna choose a sample of various points in three-dimensional space. And for each one of those points, you consider what the output of the function is, and that's gonna be some three-dimensional vector, and you draw that vector off of the point itself. So to start off, let's take a very simple example, one where the vector that it outputs is actually just a constant. So in this case, I'll make that constant the vector one, zero, zero. So what this vector is, it's just got a unit length in the x direction, so this is the x-axis."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And for each one of those points, you consider what the output of the function is, and that's gonna be some three-dimensional vector, and you draw that vector off of the point itself. So to start off, let's take a very simple example, one where the vector that it outputs is actually just a constant. So in this case, I'll make that constant the vector one, zero, zero. So what this vector is, it's just got a unit length in the x direction, so this is the x-axis. So all of the vectors are gonna end up looking something like this, where it's a vector that has length one in the x direction. And when we do this at every possible point, well, not every possible point, but a sample of a whole bunch of points, whoops, we get a vector field that looks like this. At any given point in space, we get one of these little blue vectors, and all of them are the same."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what this vector is, it's just got a unit length in the x direction, so this is the x-axis. So all of the vectors are gonna end up looking something like this, where it's a vector that has length one in the x direction. And when we do this at every possible point, well, not every possible point, but a sample of a whole bunch of points, whoops, we get a vector field that looks like this. At any given point in space, we get one of these little blue vectors, and all of them are the same. They're just copies of each other, each pointing with unit length in the x direction. So as vector fields go, this is relatively boring, but we can make it a little bit more exciting if we make the input start to depend somehow on the actual input. So what I'll do to start, I'll just make the input y, zero, zero."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "At any given point in space, we get one of these little blue vectors, and all of them are the same. They're just copies of each other, each pointing with unit length in the x direction. So as vector fields go, this is relatively boring, but we can make it a little bit more exciting if we make the input start to depend somehow on the actual input. So what I'll do to start, I'll just make the input y, zero, zero. So they're still just gonna point in the x direction, but now it's gonna depend on the y value. So let's think for a second before I change the image what that's gonna mean. The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So what I'll do to start, I'll just make the input y, zero, zero. So they're still just gonna point in the x direction, but now it's gonna depend on the y value. So let's think for a second before I change the image what that's gonna mean. The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y. So as y increases value to one, two, three, the length of these vectors are gonna increase. It's gonna be a stronger vector in the x direction, a very strong vector in the x direction. And if y is negative, these vectors are gonna point in the opposite direction."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "The y-axis is this one here, so now the z-axis is pointing straight in our face, that's the y. So as y increases value to one, two, three, the length of these vectors are gonna increase. It's gonna be a stronger vector in the x direction, a very strong vector in the x direction. And if y is negative, these vectors are gonna point in the opposite direction. So let's see what that looks like. There we go. So in this vector field, color and length are used to indicate how the magnitude of the vector."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And if y is negative, these vectors are gonna point in the opposite direction. So let's see what that looks like. There we go. So in this vector field, color and length are used to indicate how the magnitude of the vector. So red vectors are very long, blue vectors are pretty short, and at zero we don't even see any because those are vectors with zero length. And just like with two-dimensional vector fields, when you draw them, you lie a little bit. This one should have a length of one, right?"}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So in this vector field, color and length are used to indicate how the magnitude of the vector. So red vectors are very long, blue vectors are pretty short, and at zero we don't even see any because those are vectors with zero length. And just like with two-dimensional vector fields, when you draw them, you lie a little bit. This one should have a length of one, right? Because when y is equal to one, this should have a unit length, but it's made really, really small. And this one up here, where y is five or six, should be a really long vector, but we're lying a little bit because if we actually drew them to scale, it would really clutter up the image. So a couple things to notice about this one."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "This one should have a length of one, right? Because when y is equal to one, this should have a unit length, but it's made really, really small. And this one up here, where y is five or six, should be a really long vector, but we're lying a little bit because if we actually drew them to scale, it would really clutter up the image. So a couple things to notice about this one. Since the output doesn't depend on x or z, if you move in the x direction, which is back and forth here, the vectors don't change. And if you move in the z direction, which is up and down, the vectors also don't change. They only change as you move in the y direction."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So a couple things to notice about this one. Since the output doesn't depend on x or z, if you move in the x direction, which is back and forth here, the vectors don't change. And if you move in the z direction, which is up and down, the vectors also don't change. They only change as you move in the y direction. Okay, so this is, we're starting to get a feel for how the output can depend on the input. Now let's do something a little bit different. Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "They only change as you move in the y direction. Okay, so this is, we're starting to get a feel for how the output can depend on the input. Now let's do something a little bit different. Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function. At a given point x, y, z, you output the vector itself, x, y, z. So let's think about what this would actually mean. And let's say you've got a given point, some point floating off in space."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "Let's say that all three of the components of the input depend on x, y, and z, but I'm just gonna make it kind of an identity function. At a given point x, y, z, you output the vector itself, x, y, z. So let's think about what this would actually mean. And let's say you've got a given point, some point floating off in space. What is the output vector for that? Well, the point has a certain x component, a certain y component, and a z component. And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And let's say you've got a given point, some point floating off in space. What is the output vector for that? Well, the point has a certain x component, a certain y component, and a z component. And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself. Let me just draw that here. From the origin to the point itself. And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the vector that corresponds to x, y, z is gonna be the one from the origin to that point itself. Let me just draw that here. From the origin to the point itself. And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point. But the main thing to take away here is it's gonna point directly away from the origin. And the farther away the point is, the longer this vector will be. So with that, let's take a look at the vector field itself."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "And because of how we do vector fields, you move this so that instead of stemming from the origin, it actually stems from the point. But the main thing to take away here is it's gonna point directly away from the origin. And the farther away the point is, the longer this vector will be. So with that, let's take a look at the vector field itself. Here we go. So again, you kind of lie when you draw these. Like the vectors, these red guys that are out at the end, they should be really long."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "So with that, let's take a look at the vector field itself. Here we go. So again, you kind of lie when you draw these. Like the vectors, these red guys that are out at the end, they should be really long. Because this vector should be as long as that point is away from the origin. But to give a cleaner vector field, you scale things down. And notice the blue ones that are close to the center here are actually really, really short guys."}, {"video_title": "3d vector fields, introduction   Multivariable calculus   Khan Academy.mp3", "Sentence": "Like the vectors, these red guys that are out at the end, they should be really long. Because this vector should be as long as that point is away from the origin. But to give a cleaner vector field, you scale things down. And notice the blue ones that are close to the center here are actually really, really short guys. And all of these are pointing directly away from the origin. And this is one of those vector fields that is actually pretty, a good one to have a strong intuition of. Because it comes up now and then, thinking about what the identity function looks like as a vector field itself."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the way we did it is we integrated with respect to x first. We said, well, let's pick a y and let's just figure out the area under the curve. And so we integrated with respect to x first, and then we integrated with respect to y. But we could have done it the other way around. So let's do that and just make sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we could have done it the other way around. So let's do that and just make sure we got the right answer. So let me erase a lot of this. So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y. But I will show you that we can integrate the other way around. That's good. We can get the same answer in two different ways."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So remember, our answer was 2 thirds when we integrated with respect to x first and then with respect to y. But I will show you that we can integrate the other way around. That's good. We can get the same answer in two different ways. So let me redraw that graph because I want to give you the intuition again. So that's my x-axis, y-axis, z-axis. And then this is my xy-plane down here."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can get the same answer in two different ways. So let me redraw that graph because I want to give you the intuition again. So that's my x-axis, y-axis, z-axis. And then this is my xy-plane down here. y goes from 0 to 1, x goes from 0 to 2. And this, you could do this x equals 1, this is x equals 2, this is y equals 1. And then the graph, I will do my best to draw it."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then this is my xy-plane down here. y goes from 0 to 1, x goes from 0 to 2. And this, you could do this x equals 1, this is x equals 2, this is y equals 1. And then the graph, I will do my best to draw it. It looks something, let me do it in a more, get some contrast going here. So the graph looks something like this. Let me see if I can draw it."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the graph, I will do my best to draw it. It looks something, let me do it in a more, get some contrast going here. So the graph looks something like this. Let me see if I can draw it. This side, it looks something like that. And then it comes down like that straight. And then the volume we care about is actually this volume underneath the graph."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me see if I can draw it. This side, it looks something like that. And then it comes down like that straight. And then the volume we care about is actually this volume underneath the graph. This is the top of the surface on that side. And we care about this volume underneath the surface. And then when we draw the bottom of the surface, let me do it in a darker color."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the volume we care about is actually this volume underneath the graph. This is the top of the surface on that side. And we care about this volume underneath the surface. And then when we draw the bottom of the surface, let me do it in a darker color. Looks something like this. It looks something like this, this is the bottom underneath the surface. I could even shade it a little bit just to show you that it's underneath the surface."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when we draw the bottom of the surface, let me do it in a darker color. Looks something like this. It looks something like this, this is the bottom underneath the surface. I could even shade it a little bit just to show you that it's underneath the surface. Hopefully that's a decent rendering of it. Let's look back at what we had before. It's like a page that I just flipped up at this point."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I could even shade it a little bit just to show you that it's underneath the surface. Hopefully that's a decent rendering of it. Let's look back at what we had before. It's like a page that I just flipped up at this point. And we care about this volume, kind of the colored area under there. So let's figure out how to do it. Last time we integrated with respect to x first, let's integrate with respect to y first."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's like a page that I just flipped up at this point. And we care about this volume, kind of the colored area under there. So let's figure out how to do it. Last time we integrated with respect to x first, let's integrate with respect to y first. So let's hold x constant. So if we hold x constant, what we could do is for a given x, let's pick an x. So if we pick a given x, let's pick the x here."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Last time we integrated with respect to x first, let's integrate with respect to y first. So let's hold x constant. So if we hold x constant, what we could do is for a given x, let's pick an x. So if we pick a given x, let's pick the x here. Then what we can do, for a given x, you can view that function of x and y. If x is a constant, let's say if x is 1, then z is just equal to y squared. That's easy to figure out the area under."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we pick a given x, let's pick the x here. Then what we can do, for a given x, you can view that function of x and y. If x is a constant, let's say if x is 1, then z is just equal to y squared. That's easy to figure out the area under. As we can see, that x isn't a constant, but we can treat it as a constant. So for example, for any given x, we would have a curve like this. And what we could do is we could try to figure out the area of this curve first."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's easy to figure out the area under. As we can see, that x isn't a constant, but we can treat it as a constant. So for example, for any given x, we would have a curve like this. And what we could do is we could try to figure out the area of this curve first. So how do we do that? Well, we just said we could kind of view this function up here as z is equal to xy squared, because that's exactly what it is. But we're holding x constant."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what we could do is we could try to figure out the area of this curve first. So how do we do that? Well, we just said we could kind of view this function up here as z is equal to xy squared, because that's exactly what it is. But we're holding x constant. We're treating it like a constant. To figure out that area, we could take a dy, a change in y, multiply it by the height, which is xy squared. So we take xy squared, multiply it by dy."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we're holding x constant. We're treating it like a constant. To figure out that area, we could take a dy, a change in y, multiply it by the height, which is xy squared. So we take xy squared, multiply it by dy. And if we want this entire area, we integrate it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we take xy squared, multiply it by dy. And if we want this entire area, we integrate it from y is equal to 0 to y is equal to 1. Fair enough. Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going. Let me pick a nice color. This green. So let's say that's our dx."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now once we have that area, if we want the volume underneath this entire surface, what we could do is we could multiply this area times dx and get some depth going. Let me pick a nice color. This green. So let's say that's our dx. So if we multiply that times dx, we would get some depth. Let me do a darker color, get some contrast. Sometimes I feel like that guy who paints on PBS."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say that's our dx. So if we multiply that times dx, we would get some depth. Let me do a darker color, get some contrast. Sometimes I feel like that guy who paints on PBS. dx. So now we have the volume of this. You could kind of view it the area under the curve times a dx, so we have some depth here."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Sometimes I feel like that guy who paints on PBS. dx. So now we have the volume of this. You could kind of view it the area under the curve times a dx, so we have some depth here. So it's times dx. And if we want to figure out the entire volume under this surface, between the surface and the xy plane, given this constraint to our domain, we just integrate from x is equal to 0 to 2. All right."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You could kind of view it the area under the curve times a dx, so we have some depth here. So it's times dx. And if we want to figure out the entire volume under this surface, between the surface and the xy plane, given this constraint to our domain, we just integrate from x is equal to 0 to 2. All right. So let's think about it. This area, this area in green here that we started with, that should be a function of x. We held x constant, but depending on which x you pick, this area is going to change."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All right. So let's think about it. This area, this area in green here that we started with, that should be a function of x. We held x constant, but depending on which x you pick, this area is going to change. So when we evaluate this magenta inner integral with respect to y, we should get a function of x. And then when we evaluate the whole thing, we'll get our volume. So let's do it."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We held x constant, but depending on which x you pick, this area is going to change. So when we evaluate this magenta inner integral with respect to y, we should get a function of x. And then when we evaluate the whole thing, we'll get our volume. So let's do it. Let's evaluate this inner integral. Hold x constant. What's the antiderivative of y squared?"}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's do it. Let's evaluate this inner integral. Hold x constant. What's the antiderivative of y squared? It's y to the third over 3. So it's y to the third over 3. The x is a constant, right?"}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What's the antiderivative of y squared? It's y to the third over 3. So it's y to the third over 3. The x is a constant, right? And we're going to evaluate that at 1 and at 0. And the outer integral is still with respect to x dx. This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The x is a constant, right? And we're going to evaluate that at 1 and at 0. And the outer integral is still with respect to x dx. This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third. That's 1. So it's x over 3 minus, when y is 0, then that whole thing just becomes 0, right? So this purple expression is just x over 3."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to, let's see, when you evaluate y is equal to 1, you get 1 to the third. That's 1. So it's x over 3 minus, when y is 0, then that whole thing just becomes 0, right? So this purple expression is just x over 3. And then we still have the outside integral from 0 to 2 dx. So given what x we have, the area of this green surface, that was where we started, given any given x, that area, I wanted something with some contrast. This area is x over 3, depending on which x you pick."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this purple expression is just x over 3. And then we still have the outside integral from 0 to 2 dx. So given what x we have, the area of this green surface, that was where we started, given any given x, that area, I wanted something with some contrast. This area is x over 3, depending on which x you pick. If x is 1, this area right here is 1 third, right? But now we're going to integrate underneath the entire surface and get our volume. And like I said, when you integrate it, it's a function of x."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This area is x over 3, depending on which x you pick. If x is 1, this area right here is 1 third, right? But now we're going to integrate underneath the entire surface and get our volume. And like I said, when you integrate it, it's a function of x. So let's do that. And this is just plain old vanilla standard integral. So what's the antiderivative of x?"}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And like I said, when you integrate it, it's a function of x. So let's do that. And this is just plain old vanilla standard integral. So what's the antiderivative of x? It's x squared over 2. We have a 1 third there. So it equals x squared over 2 times 3."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the antiderivative of x? It's x squared over 2. We have a 1 third there. So it equals x squared over 2 times 3. So x squared over 6. We're going to evaluate it at 2 and at 0. 2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it equals x squared over 2 times 3. So x squared over 6. We're going to evaluate it at 2 and at 0. 2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6. What is 4 6? Well, that's just the same thing as 2 thirds. So the volume under the surface is 2 thirds."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "2 squared over 6 is 4 6 minus 0 over 6, which is equal to 0, equals 4 6. What is 4 6? Well, that's just the same thing as 2 thirds. So the volume under the surface is 2 thirds. And if you watched the previous video, you will appreciate the fact that when we integrated the other way around, when we did it with respect to x first and then y, we got the exact same answer. So the universe is in proper working order. And I've surprisingly actually finished this video with extra time."}, {"video_title": "Double integrals 3   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the volume under the surface is 2 thirds. And if you watched the previous video, you will appreciate the fact that when we integrated the other way around, when we did it with respect to x first and then y, we got the exact same answer. So the universe is in proper working order. And I've surprisingly actually finished this video with extra time. So for fun, we can just spin this graph and just appreciate the fact that we have figured out the volume between this surface, xy squared, and the xy plane. Pretty neat. Anyway, I'll see you in the next video."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is F dot dr, where our path is this boundary right over here, this path C. We had expressed it in terms of a line integral around this boundary, around path C1, which is the boundary of region R. And the reason why this is going to be valuable to us is now we can directly apply Green's theorem to this to turn this into a double integral over this region right over here, the region that it is actually bounding. So let's actually do that. We're just applying Green's theorem here. So Green's theorem. Actually, let me box this off right over here. This was just something that we wrote to remind ourselves in the last video. So let me just box it off right over here."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So Green's theorem. Actually, let me box this off right over here. This was just something that we wrote to remind ourselves in the last video. So let me just box it off right over here. When you apply Green's theorem, you get that this is going to be the exact same thing as the double integral over the region that C1 bounds, that region R that's in our xy-plane, of the partial of this with respect to x. So the partial with respect to x of this business. So I'll do that in that same green color."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me just box it off right over here. When you apply Green's theorem, you get that this is going to be the exact same thing as the double integral over the region that C1 bounds, that region R that's in our xy-plane, of the partial of this with respect to x. So the partial with respect to x of this business. So I'll do that in that same green color. q plus r times the partial of z with respect to y minus the partial with respect to y of p plus r times the partial of z with respect to x. p plus r times the partial of z with respect to x. And then dA, a little differential of our region. And so let's actually calculate this."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'll do that in that same green color. q plus r times the partial of z with respect to y minus the partial with respect to y of p plus r times the partial of z with respect to x. p plus r times the partial of z with respect to x. And then dA, a little differential of our region. And so let's actually calculate this. You take the partials of each of these expressions. And we'll see that if we expand them and then simplify, we'll get something very similar, actually, hopefully identical to this right over here. And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so let's actually calculate this. You take the partials of each of these expressions. And we'll see that if we expand them and then simplify, we'll get something very similar, actually, hopefully identical to this right over here. And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. So let's do it. So let's just apply the partial derivative operator. So first, we want to take the partial derivative with respect to q."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that'll show that this line integral for the special case is the same thing as this surface integral, which will prove Stokes' theorem for this special case. So let's do it. So let's just apply the partial derivative operator. So first, we want to take the partial derivative with respect to q. And we need to remind ourselves, and we did this way up here, when we first thought about it, we saw that q, well, p, q, and r, they're each functions of x, y, and z. So we're assuming that's how they're represented. And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So first, we want to take the partial derivative with respect to q. And we need to remind ourselves, and we did this way up here, when we first thought about it, we saw that q, well, p, q, and r, they're each functions of x, y, and z. So we're assuming that's how they're represented. And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x. But we know that we've assumed that z is itself a function of x and y. So if we're taking the partial with respect to x over here, we have to think, first, well, how can q directly change with respect to x? And then how can it change due to something else changing due to x?"}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And if z was not a function of x, then if we were to take the partial of q with respect to x, we would just write it as a partial of q with respect to x. But we know that we've assumed that z is itself a function of x and y. So if we're taking the partial with respect to x over here, we have to think, first, well, how can q directly change with respect to x? And then how can it change due to something else changing due to x? And so that other thing that could change due to x is z. y is independent of x. But z is a function of x. So let's keep that in mind."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then how can it change due to something else changing due to x? And so that other thing that could change due to x is z. y is independent of x. But z is a function of x. So let's keep that in mind. So we're really going to do the multivariable chain rule. So when we try to take the derivative of this part, we're taking, essentially, of this whole function with respect to x, we have to think about how will q change directly with respect to x. And to that, we have to add how q could change due to changes in other variables due to changes in x."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's keep that in mind. So we're really going to do the multivariable chain rule. So when we try to take the derivative of this part, we're taking, essentially, of this whole function with respect to x, we have to think about how will q change directly with respect to x. And to that, we have to add how q could change due to changes in other variables due to changes in x. And the other only variable that q is a function of that could change due to a change in x is z. So q could also change to z because gz has changed due to x. So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to that, we have to add how q could change due to changes in other variables due to changes in x. And the other only variable that q is a function of that could change due to a change in x is z. So q could also change to z because gz has changed due to x. So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x. If we rewrote q so that z was substituted with x's and y's because it is a function of x and y's, then we would just have to write this first term right over here. But we're assuming this is expressed as a function of x, y, and z. And z itself is a function of x."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So operated here is the partial with respect to x plus the partial of q with respect to z times a partial of z with respect to x. If we rewrote q so that z was substituted with x's and y's because it is a function of x and y's, then we would just have to write this first term right over here. But we're assuming this is expressed as a function of x, y, and z. And z itself is a function of x. And so that's why we had to use the multivariable chain rule. Now let's move to the next part. And both of these might have some x's in them."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And z itself is a function of x. And so that's why we had to use the multivariable chain rule. Now let's move to the next part. And both of these might have some x's in them. So we have to use the product rule right over here. So first we can take the derivative of r with respect to x and then multiply that times z sub y. Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And both of these might have some x's in them. So we have to use the product rule right over here. So first we can take the derivative of r with respect to x and then multiply that times z sub y. Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus. So if we take the derivative of this with respect to x, same exact logic. r can change directly due to x. And it can change due to y."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Then we have to take the derivative of z sub y with respect to x and multiply that times r. So this is going to be plus. So if we take the derivative of this with respect to x, same exact logic. r can change directly due to x. And it can change due to y. And it can change due to z. And multiply that times how z could change due to x. Once again, you could use this as the multivariable chain rule in action here."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it can change due to y. And it can change due to z. And multiply that times how z could change due to x. Once again, you could use this as the multivariable chain rule in action here. But of course, we take the derivative of the first term times the second term. Do the second term in magenta. Times the second term."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Once again, you could use this as the multivariable chain rule in action here. But of course, we take the derivative of the first term times the second term. Do the second term in magenta. Times the second term. So the partial of z with respect to y. Plus the derivative of the second term, which is the partial of z with respect to y. And then taking the partial of that with respect to x, which we could just write as that, times the first term."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Times the second term. So the partial of z with respect to y. Plus the derivative of the second term, which is the partial of z with respect to y. And then taking the partial of that with respect to x, which we could just write as that, times the first term. Times our r. So that's the partial with respect to x of all of this business right over here. And then we need to subtract the partial of this with respect to y. And we're going to use the exact same logic."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then taking the partial of that with respect to x, which we could just write as that, times the first term. Times our r. So that's the partial with respect to x of all of this business right over here. And then we need to subtract the partial of this with respect to y. And we're going to use the exact same logic. So then we're going to subtract. And I'll put it in parentheses like that. So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to use the exact same logic. So then we're going to subtract. And I'll put it in parentheses like that. So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet. p could change directly due to y. So this is the partial of p with respect to y. But it could also change due to z changing because of y."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So p could change directly due to y. I will circle p. Let me do it with a color that I haven't used yet. p could change directly due to y. So this is the partial of p with respect to y. But it could also change due to z changing because of y. So plus the partial with respect to z times the partial of z with respect to y. Plus, and I'll do r maybe in that same color, plus the derivative of r. Well, we already figured that. But actually now it's with respect to x, not with respect to y. I have to be careful."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But it could also change due to z changing because of y. So plus the partial with respect to z times the partial of z with respect to y. Plus, and I'll do r maybe in that same color, plus the derivative of r. Well, we already figured that. But actually now it's with respect to x, not with respect to y. I have to be careful. So it's going to be the partial of r with respect to y plus the partial of r with respect to z times the partial of z with respect to y times z sub x. Plus now we take the derivative of the second term times the first term. The derivative of the partial of z with respect to x then with respect to y is going to be z."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But actually now it's with respect to x, not with respect to y. I have to be careful. So it's going to be the partial of r with respect to y plus the partial of r with respect to z times the partial of z with respect to y times z sub x. Plus now we take the derivative of the second term times the first term. The derivative of the partial of z with respect to x then with respect to y is going to be z. So the partial of z with respect to x and we're going to take the partial of that with respect and then we're going to multiply that times r. Then we're going to multiply that times r. Now let's see if we can expand this out and hopefully things simplify. Just a reminder, I'm just working on the inside of this double integral. I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The derivative of the partial of z with respect to x then with respect to y is going to be z. So the partial of z with respect to x and we're going to take the partial of that with respect and then we're going to multiply that times r. Then we're going to multiply that times r. Now let's see if we can expand this out and hopefully things simplify. Just a reminder, I'm just working on the inside of this double integral. I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit. Let me rewrite it a little bit. This is equal to the partial of q. I'll try to color code the same way. This is just algebra at this point."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll rewrite the outside, the double integral and the dA once I get this all cleaned up a little bit. Let me rewrite it a little bit. This is equal to the partial of q. I'll try to color code the same way. This is just algebra at this point. The partial of q with respect to x plus the partial of q with respect to z times the partial of z with respect to x plus the partial of r with respect to x times the partial of z with respect to y plus the partial of r with respect to z times the partial of z with respect to x times the partial of z with respect to y and then we have this term right over here, which I'll just do in purple, plus the partial of z with respect to y and then with respect to x times r. Now we're going to subtract all of this business right over here. So minus the partial of p with respect to y minus the partial of p with respect to z times the partial of z with respect to y and then we're going to subtract from that minus the partial of r with respect to y times the partial of z with respect to x minus the partial of r where we need to go. The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is just algebra at this point. The partial of q with respect to x plus the partial of q with respect to z times the partial of z with respect to x plus the partial of r with respect to x times the partial of z with respect to y plus the partial of r with respect to z times the partial of z with respect to x times the partial of z with respect to y and then we have this term right over here, which I'll just do in purple, plus the partial of z with respect to y and then with respect to x times r. Now we're going to subtract all of this business right over here. So minus the partial of p with respect to y minus the partial of p with respect to z times the partial of z with respect to y and then we're going to subtract from that minus the partial of r with respect to y times the partial of z with respect to x minus the partial of r where we need to go. The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x. And then finally, this term right over here minus this, because we have that negative out there, minus the partial of z with respect to x and then with respect to y, r. Now let's see if we can simplify things. So the first thing, this and this look to be the same. We just can commute the order in which we actually multiply."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The partial of r with respect to z times the partial of z with respect to y times the partial of z with respect to x. And then finally, this term right over here minus this, because we have that negative out there, minus the partial of z with respect to x and then with respect to y, r. Now let's see if we can simplify things. So the first thing, this and this look to be the same. We just can commute the order in which we actually multiply. But these are the exact same terms. So that is going to cancel out with that. And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just can commute the order in which we actually multiply. But these are the exact same terms. So that is going to cancel out with that. And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. And so this simplified things a good bit. And let's see if I can write it in a way, let me group terms in a way that might start to make sense. And actually, I'm going to try to see if I can make them similar to this."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And because we assumed that the second, because we assumed way up here that we have continuous second derivatives of the function z, z is a function of x and y, that that is equal to that, we can now say that these two right over here are going to be the negatives of each other or that they are going to cancel out. And so this simplified things a good bit. And let's see if I can write it in a way, let me group terms in a way that might start to make sense. And actually, I'm going to try to see if I can make them similar to this. So I have all the terms that have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms that have a z sub x in it."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually, I'm going to try to see if I can make them similar to this. So I have all the terms that have a z sub x and the z sub y and then the rest of them. So z sub x I'll do in blue. So you have the terms that have a z sub x in it. So you have this term right over here and this term right over here. We can factor out the z sub x and we get the partial of z with respect to x times the partial of q with respect to z minus the partial of r with respect to y. And then let's do the, and I want to get the colors the same way too."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have the terms that have a z sub x in it. So you have this term right over here and this term right over here. We can factor out the z sub x and we get the partial of z with respect to x times the partial of q with respect to z minus the partial of r with respect to y. And then let's do the, and I want to get the colors the same way too. I did yellow next. Plus I have all the terms with the partial of z with respect to y. So it's that term and that term right over here."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then let's do the, and I want to get the colors the same way too. I did yellow next. Plus I have all the terms with the partial of z with respect to y. So it's that term and that term right over here. So it becomes plus the partial of z with respect to y times the partial of r with respect to x minus the partial of p with respect to z. And then we have these last two terms. And I used the color green up there."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's that term and that term right over here. So it becomes plus the partial of z with respect to y times the partial of r with respect to x minus the partial of p with respect to z. And then we have these last two terms. And I used the color green up there. So I'll use the color green again. So for these two terms, I'll just write plus the partial of q with respect to x minus the partial of p with respect to y. So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I used the color green up there. So I'll use the color green again. So for these two terms, I'll just write plus the partial of q with respect to x minus the partial of p with respect to y. So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this. And we don't want to forget this was all because this was all a simplification of our double integral over the region dA. This is what we have been able to, using Green's theorem and the multivariable chain rule and whatever else, we've been able to say that that line integral around the boundary of our surface is the same thing as this. And now we can compare that to what our surface integral was."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our double integrals has kind of, I can't really say simplified, but we can rewrite it like this. And we don't want to forget this was all because this was all a simplification of our double integral over the region dA. This is what we have been able to, using Green's theorem and the multivariable chain rule and whatever else, we've been able to say that that line integral around the boundary of our surface is the same thing as this. And now we can compare that to what our surface integral was. Let's see if I have space. So copy, and then let me see if I have space up here to paste it. Well, it doesn't look like I actually have much space to paste it, although I'll try anyway."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we can compare that to what our surface integral was. Let's see if I have space. So copy, and then let me see if I have space up here to paste it. Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. So if I paste it, you see that they are identical. They are identical. Our line integral is identical to this."}, {"video_title": "Stokes' theorem proof part 7   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, it doesn't look like I actually have much space to paste it, although I'll try anyway. So if I paste it, you see that they are identical. They are identical. Our line integral is identical to this. We get the exact same thing. So our line integral, f dot dr around this pat c simplified to this, and our surface integral simplified to this. So using the assumptions we had, they both simplified to the same thing."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And to do that, I will assume that our surface can be parameterized by the position vector function, r. And r is a function of two parameters. It's a function of u, and it is a function of v. You give me a u and a v, and that will essentially specify a point on this two-dimensional surface right over here. It could be bent, so it kind of exists in three-dimensional space. But a u and a v will specify a given point on this surface. Now, let's think about what the directions of r, the partial of r with respect to u looks like, and what the direction of the partial of r with respect to v looks like. So let's say that we're at some point u, v. So for some u, v, if you find the position vector, it takes us to that point on the surface right over there. Now, let's say that we increment u just a little bit."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But a u and a v will specify a given point on this surface. Now, let's think about what the directions of r, the partial of r with respect to u looks like, and what the direction of the partial of r with respect to v looks like. So let's say that we're at some point u, v. So for some u, v, if you find the position vector, it takes us to that point on the surface right over there. Now, let's say that we increment u just a little bit. And as we increment u just a little bit, we're going to get to another point on our surface. And let's say that other point on the surface is right over there. So what would this r sub u vector look like?"}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, let's say that we increment u just a little bit. And as we increment u just a little bit, we're going to get to another point on our surface. And let's say that other point on the surface is right over there. So what would this r sub u vector look like? Well, its magnitude is essentially going to be dependent on how fast it's happening, how fast we're moving towards that point. But its direction is going to be in that direction. It's going to be towards that point."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what would this r sub u vector look like? Well, its magnitude is essentially going to be dependent on how fast it's happening, how fast we're moving towards that point. But its direction is going to be in that direction. It's going to be towards that point. It's going to be along the surface. We're going from one point of the surface to another. It's essentially going to be tangent to the surface at that point."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be towards that point. It's going to be along the surface. We're going from one point of the surface to another. It's essentially going to be tangent to the surface at that point. And I could draw it a little bit bigger. It would look something like that, r sub u. So I've just zoomed in right over here."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's essentially going to be tangent to the surface at that point. And I could draw it a little bit bigger. It would look something like that, r sub u. So I've just zoomed in right over here. Now, let's go back to this point. And now, let's make v a little bit bigger. Now, let's say if we make v a little bit bigger, we go to this point right over here."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I've just zoomed in right over here. Now, let's go back to this point. And now, let's make v a little bit bigger. Now, let's say if we make v a little bit bigger, we go to this point right over here. So then our position vector r would point to this point. And so what would r sub v look like? Well, its magnitude, once again, would be dependent on how fast we're going there."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, let's say if we make v a little bit bigger, we go to this point right over here. So then our position vector r would point to this point. And so what would r sub v look like? Well, its magnitude, once again, would be dependent on how fast we're going there. But the direction is what's interesting. The direction would also be tangential to the surface. We're going from one point on the surface to another as we change v. So r sub v might look something like that."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, its magnitude, once again, would be dependent on how fast we're going there. But the direction is what's interesting. The direction would also be tangential to the surface. We're going from one point on the surface to another as we change v. So r sub v might look something like that. And these two aren't necessarily perpendicular to each other. In fact, the way I drew them, they're not perpendicular. So r sub v is looking like this."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going from one point on the surface to another as we change v. So r sub v might look something like that. And these two aren't necessarily perpendicular to each other. In fact, the way I drew them, they're not perpendicular. So r sub v is looking like this. But they're both tangential to the plane. They're both essentially telling us right at that point, what is the tangent, what is the slope in the u direction or in the v direction? Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So r sub v is looking like this. But they're both tangential to the plane. They're both essentially telling us right at that point, what is the tangent, what is the slope in the u direction or in the v direction? Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane. And so you can imagine a plane that looks something like this. If you took linear combinations of these two things, you would get a plane that both of these would lie on. Now, we've done this before, but I'll revisit it."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, when you have two vectors that are tangential to the plane and they're not the same vector, these are actually already specifying, these are already kind of determining a plane. And so you can imagine a plane that looks something like this. If you took linear combinations of these two things, you would get a plane that both of these would lie on. Now, we've done this before, but I'll revisit it. What happens when I take the cross product of r sub u and r sub v? What happens when I take the cross product? Well, first, this is going to give us another vector."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, we've done this before, but I'll revisit it. What happens when I take the cross product of r sub u and r sub v? What happens when I take the cross product? Well, first, this is going to give us another vector. It's going to give us a vector that is perpendicular to both, to r sub u and r sub v. Or another way to think about it is this plane, when you take the cross product, this plane is essentially a tangential plane to the surface. And if something's going to be perpendicular to both of these characters, it's going to have to be normal to them, or it's definitely going to be perpendicular to both of them, but it's going to be normal to this plane, which is essentially going to be perpendicular to the surface itself. So this right over here is going to be a normal vector."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, first, this is going to give us another vector. It's going to give us a vector that is perpendicular to both, to r sub u and r sub v. Or another way to think about it is this plane, when you take the cross product, this plane is essentially a tangential plane to the surface. And if something's going to be perpendicular to both of these characters, it's going to have to be normal to them, or it's definitely going to be perpendicular to both of them, but it's going to be normal to this plane, which is essentially going to be perpendicular to the surface itself. So this right over here is going to be a normal vector. Well, let me just write it this way. A normal vector. I'm not saying the normal vector, because you could have different normal vectors of different magnitudes."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this right over here is going to be a normal vector. Well, let me just write it this way. A normal vector. I'm not saying the normal vector, because you could have different normal vectors of different magnitudes. This is a normal vector when you take the cross product. And we can even think about what direction it's pointing in. And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm not saying the normal vector, because you could have different normal vectors of different magnitudes. This is a normal vector when you take the cross product. And we can even think about what direction it's pointing in. And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector. So in this case, r sub u. So let me see if I can draw this. I'm literally looking at my hand and trying to draw it."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so when you have something cross something else, the easiest way I remember how to do it is you point your right thumb in the direction of the first vector. So in this case, r sub u. So let me see if I can draw this. I'm literally looking at my hand and trying to draw it. So you put your right thumb, so this is a right hand rule essentially, in the direction of the first vector. And then you put your index finger in the direction of the second vector right over here. So this is the second vector."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm literally looking at my hand and trying to draw it. So you put your right thumb, so this is a right hand rule essentially, in the direction of the first vector. And then you put your index finger in the direction of the second vector right over here. So this is the second vector. So that's the direction of my index finger. So my index finger is going to look something like that. And then you bend your middle finger inward, and that'll tell you the direction of the cross product."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the second vector. So that's the direction of my index finger. So my index finger is going to look something like that. And then you bend your middle finger inward, and that'll tell you the direction of the cross product. So if I bend my middle finger inward, it will look something like this. And of course, my other two fingers are just going to be folded in like that, and they're not really relevant. But my other two fingers and my hand looks like that."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then you bend your middle finger inward, and that'll tell you the direction of the cross product. So if I bend my middle finger inward, it will look something like this. And of course, my other two fingers are just going to be folded in like that, and they're not really relevant. But my other two fingers and my hand looks like that. And so that tells us the direction. The direction is going to be like that. It's going to be upward facing."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But my other two fingers and my hand looks like that. And so that tells us the direction. The direction is going to be like that. It's going to be upward facing. That's important because you're two normal vectors. Or there's two directions of normalcy, I guess you could say. One is going out like that, outwards."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be upward facing. That's important because you're two normal vectors. Or there's two directions of normalcy, I guess you could say. One is going out like that, outwards. Or I guess in the upward direction, one would be going downwards, or going, I guess you could say, into the surface. But the way I've set it up right now, this would be going outwards. It would be a normal vector to the surface."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "One is going out like that, outwards. Or I guess in the upward direction, one would be going downwards, or going, I guess you could say, into the surface. But the way I've set it up right now, this would be going outwards. It would be a normal vector to the surface. Now, in order to go from a normal vector to the unit normal vector, we just have to normalize this. We just have to divide this by its magnitude. So now we have our drum roll, the unit vector."}, {"video_title": "Constructing a unit normal vector to a surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It would be a normal vector to the surface. Now, in order to go from a normal vector to the unit normal vector, we just have to normalize this. We just have to divide this by its magnitude. So now we have our drum roll, the unit vector. And it's going to essentially be a function of u and v. You give me a u or v, and I'll give you that unit normal vector. It is going to be equal to the partial of r with respect to u crossed with the partial of r with respect to v. That just now, that gives us a normal vector, but it hasn't been normalized. So we want to divide by the magnitude of the exact same thing, r sub u crossed with r sub v. And we're done."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "So in the last video, I described how to interpret three-dimensional graphs. And I have another three-dimensional graph here. It's a very bumpy guy. And this happens to be the graph of the function f of x, y is equal to cosine of x multiplied by the sine of y. And you know, I could also say like that this graph represents, I could also say that this graph represents z is equal to that whole value because we think about the output of the function as the z-coordinate of each point. And what I want to do here is describe how you can interpret the relationship between this graph and these functions that you know by taking slices of it. So for example, let's say that we took a slice with this plane here."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "And this happens to be the graph of the function f of x, y is equal to cosine of x multiplied by the sine of y. And you know, I could also say like that this graph represents, I could also say that this graph represents z is equal to that whole value because we think about the output of the function as the z-coordinate of each point. And what I want to do here is describe how you can interpret the relationship between this graph and these functions that you know by taking slices of it. So for example, let's say that we took a slice with this plane here. And what this plane here is is it represents the value x equals zero. And you can kind of see that because this is the x-axis. So when you're at zero on the x-axis, you know, you pass through the origin."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "So for example, let's say that we took a slice with this plane here. And what this plane here is is it represents the value x equals zero. And you can kind of see that because this is the x-axis. So when you're at zero on the x-axis, you know, you pass through the origin. And then the values of y and z can go freely. So you end up with this, this plane. And let's say you want to just consider where this cuts through the graph, okay?"}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "So when you're at zero on the x-axis, you know, you pass through the origin. And then the values of y and z can go freely. So you end up with this, this plane. And let's say you want to just consider where this cuts through the graph, okay? So we'll limit our graph just down to the point where it cuts it. And I'm gonna draw a little red line over that spot. Now what you might notice here, that red line looks like a sinusoidal wave."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "And let's say you want to just consider where this cuts through the graph, okay? So we'll limit our graph just down to the point where it cuts it. And I'm gonna draw a little red line over that spot. Now what you might notice here, that red line looks like a sinusoidal wave. In fact, it looks exactly like the sine function itself. You know, it passes through the origin. It starts by going up."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "Now what you might notice here, that red line looks like a sinusoidal wave. In fact, it looks exactly like the sine function itself. You know, it passes through the origin. It starts by going up. And this makes sense if we start to plug things into the original form here. Because if you take, you know, if you take f and you plug in x equals zero, but then we still let y range freely, what it means is you're looking at cosine of zero multiplied by sine of y. Sine of y."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "It starts by going up. And this makes sense if we start to plug things into the original form here. Because if you take, you know, if you take f and you plug in x equals zero, but then we still let y range freely, what it means is you're looking at cosine of zero multiplied by sine of y. Sine of y. And what is cosine of zero? Cosine of zero evaluates to one. So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sine of y. And what is cosine of zero? Cosine of zero evaluates to one. So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with. And let's try this at a different point. Let's see what would happen if instead of plugging in x equals zero, let's imagine that we plugged in y equals zero. And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "So this whole function should look just like sine of y, in that when we let y run freely, the output, which is still represented by the z-coordinate, will give us this graph that's just a normal two-dimensional graph that we're probably familiar with. And let's try this at a different point. Let's see what would happen if instead of plugging in x equals zero, let's imagine that we plugged in y equals zero. And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero. So now I'm gonna write over on the other side, we have fx will still run freely, y is gonna be fixed at zero, and what this means is we have cosine of x, so maybe you expect to see something that looks kinda like a cosine graph, and then sine of zero. Except, what is sine of zero? Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "And this time, before I graph it and before I show everything that goes on, let's just try to figure out purely from the formula here what it's gonna look like when we plug in y equals zero. So now I'm gonna write over on the other side, we have fx will still run freely, y is gonna be fixed at zero, and what this means is we have cosine of x, so maybe you expect to see something that looks kinda like a cosine graph, and then sine of zero. Except, what is sine of zero? Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero. So what you'd expect is that this is gonna look like a constant function that's constantly equal to zero. And let's see if that's what we get. So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sine of zero cancels out and just becomes zero, which, multiplied by cosine of x, means everything cancels out and becomes zero. So what you'd expect is that this is gonna look like a constant function that's constantly equal to zero. And let's see if that's what we get. So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely. I'm gonna chop off my graph at that point, and indeed, it chops it just at this straight line, the straight line that goes right along the x-axis. But let's say that we did a different constant value of y. Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "So I'm gonna slice it with y equals zero here, and you notice we look at the y-axis, we see when it's zero, and x and z both run freely. I'm gonna chop off my graph at that point, and indeed, it chops it just at this straight line, the straight line that goes right along the x-axis. But let's say that we did a different constant value of y. Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value. So in this case, what I've chosen is y is equal to pi halves, and it looks kinda like we've got a wave here, and it looks like a cosine wave. And you can probably see where this is going. This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "Rather than y equals zero, so rather than y equals zero, and we'll erase all of this, let's say that I cut things at some other value. So in this case, what I've chosen is y is equal to pi halves, and it looks kinda like we've got a wave here, and it looks like a cosine wave. And you can probably see where this is going. This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out. We've got cosine of x, and then y is held at a constant, sine of pi halves. Sine of pi halves, this just always equals one, so we could replace this with one, which means the function as a whole should look like cosine x. So again, the multivariable function, we've frozen y, and we're letting x range freely, and it ends up looking like a cosine function."}, {"video_title": "Interpreting graphs with slices   Multivariable calculus   Khan Academy.mp3", "Sentence": "This is when x is running freely, and if we start to imagine plugging this in, I'll just actually write it out. We've got cosine of x, and then y is held at a constant, sine of pi halves. Sine of pi halves, this just always equals one, so we could replace this with one, which means the function as a whole should look like cosine x. So again, the multivariable function, we've frozen y, and we're letting x range freely, and it ends up looking like a cosine function. And I think a really good way to understand a given three-dimensional graph when you see it, so let's say you look back at the original graph, and we don't have anything going on. Get rid of that little line. So you've got this graph, and it looks wavy and bumpy and a little bit hard to understand at first, but if you just think in terms of holding one variable constant, it boils down always into a normal two-dimensional graph."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In the last video, we had this rectangle, and we used a triple integral to figure out its volume. And I know you were probably thinking, well, I could have just used my basic geometry to multiply the height times the width times the depth. And that's true, because this was a constant-valued function. And then even once we evaluated, once we integrated with respect to z, we ended up with a double integral, which is exactly what you would have done in the last several videos, when we just learned the volume under a surface. But then we added a twist at the end of the video. We said, fine, you could have figured out the volume within this rectangular domain, I guess, very straightforward using things you already knew. But what if our goal was not to figure out the volume?"}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then even once we evaluated, once we integrated with respect to z, we ended up with a double integral, which is exactly what you would have done in the last several videos, when we just learned the volume under a surface. But then we added a twist at the end of the video. We said, fine, you could have figured out the volume within this rectangular domain, I guess, very straightforward using things you already knew. But what if our goal was not to figure out the volume? Our goal was to figure out the mass of this volume. And even more, the mass, the material that we're taking the volume of, whether it's a volume of gas or a volume of some solid, that its density is not constant. So now the mass becomes kind of interesting to calculate."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But what if our goal was not to figure out the volume? Our goal was to figure out the mass of this volume. And even more, the mass, the material that we're taking the volume of, whether it's a volume of gas or a volume of some solid, that its density is not constant. So now the mass becomes kind of interesting to calculate. And so we defined a density function. And rho, this p-looking thing with a curvy bottom, that gives us the density at any given point. At the end of the last video, we said, well, what is mass?"}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now the mass becomes kind of interesting to calculate. And so we defined a density function. And rho, this p-looking thing with a curvy bottom, that gives us the density at any given point. At the end of the last video, we said, well, what is mass? Mass is just density times volume. Or you could view it another way. Density is the same thing as mass divided by volume."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "At the end of the last video, we said, well, what is mass? Mass is just density times volume. Or you could view it another way. Density is the same thing as mass divided by volume. So the mass around a very, very small point, and we call that d mass, the differential of the mass, is going to equal the density at that point, or the rough density at exactly that point, times the volume differential around that point. Times the volume of this little small cube. And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Density is the same thing as mass divided by volume. So the mass around a very, very small point, and we call that d mass, the differential of the mass, is going to equal the density at that point, or the rough density at exactly that point, times the volume differential around that point. Times the volume of this little small cube. And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance. So what if we wanted, the density was, that our density function is defined to be x, y, and z, and we wanted to figure out the mass of this volume. And let's say that our x, y, and z coordinates, their values, let's say they're in meters, and let's say this density is in kilograms per meter cubed. So our answer is going to be in kilograms, if that was the case."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we, as we saw it on the last video, if you're using rectangular coordinates, this volume differential could just be the x distance times the y distance times the z distance. So what if we wanted, the density was, that our density function is defined to be x, y, and z, and we wanted to figure out the mass of this volume. And let's say that our x, y, and z coordinates, their values, let's say they're in meters, and let's say this density is in kilograms per meter cubed. So our answer is going to be in kilograms, if that was the case. And those are kind of the traditional SI units. So let's figure out the mass of this variably dense volume. So all we do is we have the same integral up here."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So our answer is going to be in kilograms, if that was the case. And those are kind of the traditional SI units. So let's figure out the mass of this variably dense volume. So all we do is we have the same integral up here. So the differential of mass is going to be this value. So let's write that down. So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So all we do is we have the same integral up here. So the differential of mass is going to be this value. So let's write that down. So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first. But you could do that, you could actually switch the order, maybe we'll do that in the next video. We'll do dz first, then we'll do dy, then we'll do dx. And then we just define, so this is, once again, this is just the mass at any small differential of volume."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's, let me, it is x, I want to make sure I don't run a space, x, y, z times, and I'm going to integrate with respect to dz first. But you could do that, you could actually switch the order, maybe we'll do that in the next video. We'll do dz first, then we'll do dy, then we'll do dx. And then we just define, so this is, once again, this is just the mass at any small differential of volume. And if we integrate with z first, we said z goes from what, the boundaries on z were 0 to 2, the boundaries on y were 0 to 4, and the boundaries on x, x went from 0 to 3. And how do we evaluate this? Well, what is the anti-derivative of, we're integrating with respect to z first, so what's the anti-derivative of x, y, z with respect to z?"}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we just define, so this is, once again, this is just the mass at any small differential of volume. And if we integrate with z first, we said z goes from what, the boundaries on z were 0 to 2, the boundaries on y were 0 to 4, and the boundaries on x, x went from 0 to 3. And how do we evaluate this? Well, what is the anti-derivative of, we're integrating with respect to z first, so what's the anti-derivative of x, y, z with respect to z? Well, it's, let's see, this is just a constant, so it'll be x, y, z squared over 2. Right? Yeah, that's right."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, what is the anti-derivative of, we're integrating with respect to z first, so what's the anti-derivative of x, y, z with respect to z? Well, it's, let's see, this is just a constant, so it'll be x, y, z squared over 2. Right? Yeah, that's right. And then we'll evaluate that from 2 to 0. And so you get, I know I'm going to run out of space, so you're going to get 2 squared, which is 4, divided by 2, which is 2, so it's 2xy minus 0. So when you evaluate just this first integral, you get 2xy, and now you have the other two integrals left."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Yeah, that's right. And then we'll evaluate that from 2 to 0. And so you get, I know I'm going to run out of space, so you're going to get 2 squared, which is 4, divided by 2, which is 2, so it's 2xy minus 0. So when you evaluate just this first integral, you get 2xy, and now you have the other two integrals left. So I didn't write the other two integrals down, maybe I'll write it down. So then you're left with two integrals. You're left with dy and dx, and y goes from 0 to 4, and x goes from 0 to 3."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when you evaluate just this first integral, you get 2xy, and now you have the other two integrals left. So I didn't write the other two integrals down, maybe I'll write it down. So then you're left with two integrals. You're left with dy and dx, and y goes from 0 to 4, and x goes from 0 to 3. I'm definitely going to run out of space. And now you take the anti-derivative of this with respect to y. So what's the anti-derivative of this with respect to y?"}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're left with dy and dx, and y goes from 0 to 4, and x goes from 0 to 3. I'm definitely going to run out of space. And now you take the anti-derivative of this with respect to y. So what's the anti-derivative of this with respect to y? Let me erase some stuff, just so I don't get too messy. I was given the very good suggestion of making it scroll, but unfortunately I didn't make it scroll enough this time, so I can delete this stuff, I think. Oops, I deleted some of that, but you know what I deleted."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the anti-derivative of this with respect to y? Let me erase some stuff, just so I don't get too messy. I was given the very good suggestion of making it scroll, but unfortunately I didn't make it scroll enough this time, so I can delete this stuff, I think. Oops, I deleted some of that, but you know what I deleted. OK, so let's take the anti-derivative with respect to y, I'll start it up here. We're out of space. OK, so the anti-derivative of 2xy with respect to y is y squared over 2, 2's cancel out, so you get xy squared, and y goes from 0 to 4, and then we still have the outer integral to do."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Oops, I deleted some of that, but you know what I deleted. OK, so let's take the anti-derivative with respect to y, I'll start it up here. We're out of space. OK, so the anti-derivative of 2xy with respect to y is y squared over 2, 2's cancel out, so you get xy squared, and y goes from 0 to 4, and then we still have the outer integral to do. x goes from 0 to 3 dx. And when y is equal to 4, you get 16x, and then when y is 0, the whole thing is 0. So you have 16x integrated from 0 to 3 dx, and that is equal to 8x squared."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "OK, so the anti-derivative of 2xy with respect to y is y squared over 2, 2's cancel out, so you get xy squared, and y goes from 0 to 4, and then we still have the outer integral to do. x goes from 0 to 3 dx. And when y is equal to 4, you get 16x, and then when y is 0, the whole thing is 0. So you have 16x integrated from 0 to 3 dx, and that is equal to 8x squared. And we evaluate it from 0 to 3. When it's 3, 8 times 9 is 72, and 0 times 8 is 0. So the mass of our figure, the volume we figured out last time was 24 meters cubed, I erased it, but if you watched the last video, that's what we learned."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you have 16x integrated from 0 to 3 dx, and that is equal to 8x squared. And we evaluate it from 0 to 3. When it's 3, 8 times 9 is 72, and 0 times 8 is 0. So the mass of our figure, the volume we figured out last time was 24 meters cubed, I erased it, but if you watched the last video, that's what we learned. But its mass is 72 kilograms. And we did that by integrating this three-variable density function, this function of three variables, or in three dimensions, you can view it as a scalar field, right? At any given point, there is a value, but not really a direction, and that value is a density."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the mass of our figure, the volume we figured out last time was 24 meters cubed, I erased it, but if you watched the last video, that's what we learned. But its mass is 72 kilograms. And we did that by integrating this three-variable density function, this function of three variables, or in three dimensions, you can view it as a scalar field, right? At any given point, there is a value, but not really a direction, and that value is a density. But we integrated this scalar field in this volume. So that's kind of the new skill we learned with the triple integral. And in the next video, I'll show you how to set up a more complicated triple integrals."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "At any given point, there is a value, but not really a direction, and that value is a density. But we integrated this scalar field in this volume. So that's kind of the new skill we learned with the triple integral. And in the next video, I'll show you how to set up a more complicated triple integrals. But the real difficulty with triple integrals is, and I think you'll see that your calculus teacher will often do this. When you're doing triple integrals, unless you have a very easy figure like this, the evaluation, if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated, for example, a density function, the integral gets very hairy very fast, and it's often very difficult or very time consuming to evaluate it analytically, just using your traditional calculus skills. So you'll see that on a lot of calculus exams, when they start doing the triple integral, they just want you to set it up."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in the next video, I'll show you how to set up a more complicated triple integrals. But the real difficulty with triple integrals is, and I think you'll see that your calculus teacher will often do this. When you're doing triple integrals, unless you have a very easy figure like this, the evaluation, if you actually wanted to analytically evaluate a triple integral that has more complicated boundaries or more complicated, for example, a density function, the integral gets very hairy very fast, and it's often very difficult or very time consuming to evaluate it analytically, just using your traditional calculus skills. So you'll see that on a lot of calculus exams, when they start doing the triple integral, they just want you to set it up. And they take your word for it, that you've done so many integrals so far that you could take the antiderivative. And sometimes, if they really want to give you something more difficult, they'll just say, well, switch the order. This is the integral when we're doing with respect to z, then y, then x."}, {"video_title": "Triple integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you'll see that on a lot of calculus exams, when they start doing the triple integral, they just want you to set it up. And they take your word for it, that you've done so many integrals so far that you could take the antiderivative. And sometimes, if they really want to give you something more difficult, they'll just say, well, switch the order. This is the integral when we're doing with respect to z, then y, then x. We want you to rewrite this integral when you switch the order. And we will do that in the next video. See you soon."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's actually compute it. And I think it'll all become a lot more concrete. So let's say I have the surface z. And it's a function of x and y. And it equals xy squared. It's a surface in three-dimensional space. And I want to know the volume between this surface and the xy plane, and the domain in the xy plane that I care about is x is greater than or equal to 0 and less than or equal to 2, and y is greater than or equal to 0 and less than or equal to 1."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's a function of x and y. And it equals xy squared. It's a surface in three-dimensional space. And I want to know the volume between this surface and the xy plane, and the domain in the xy plane that I care about is x is greater than or equal to 0 and less than or equal to 2, and y is greater than or equal to 0 and less than or equal to 1. And let's see what that looks like, just so we have a good visualization of it. So I graphed it here. And we can rotate around."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I want to know the volume between this surface and the xy plane, and the domain in the xy plane that I care about is x is greater than or equal to 0 and less than or equal to 2, and y is greater than or equal to 0 and less than or equal to 1. And let's see what that looks like, just so we have a good visualization of it. So I graphed it here. And we can rotate around. This is z equals xy squared. And this is the bounding box, right? x goes from 0 to 2, y goes from 0 to 1."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we can rotate around. This is z equals xy squared. And this is the bounding box, right? x goes from 0 to 2, y goes from 0 to 1. So we literally want this. You can almost view it the volume, well, not almost, exactly view it as the volume under this surface, between this surface, the top surface, and the xy plane. And I'll rotate it around so you can get a little bit better sense of the actual volume."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x goes from 0 to 2, y goes from 0 to 1. So we literally want this. You can almost view it the volume, well, not almost, exactly view it as the volume under this surface, between this surface, the top surface, and the xy plane. And I'll rotate it around so you can get a little bit better sense of the actual volume. Let me rotate it. Now I should use the mouse for this. So it's this space underneath here."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I'll rotate it around so you can get a little bit better sense of the actual volume. Let me rotate it. Now I should use the mouse for this. So it's this space underneath here. It's like a makeshift shelter or something. And I can rotate it a little bit. So you can see whatever's under this, between the two surfaces, that's the volume."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's this space underneath here. It's like a makeshift shelter or something. And I can rotate it a little bit. So you can see whatever's under this, between the two surfaces, that's the volume. Whoops, I flipped it. There you go. So that's the volume that we care about."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you can see whatever's under this, between the two surfaces, that's the volume. Whoops, I flipped it. There you go. So that's the volume that we care about. Let's figure out how to do it. And we'll try to gather a little bit of the intuition as we go along. So I'm going to draw a not as impressive version of that graph."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that's the volume that we care about. Let's figure out how to do it. And we'll try to gather a little bit of the intuition as we go along. So I'm going to draw a not as impressive version of that graph. But I think it'll do the job for now. Let me draw my axes. That's my x-axis."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'm going to draw a not as impressive version of that graph. But I think it'll do the job for now. Let me draw my axes. That's my x-axis. That's my y-axis. That's my z-axis. x, y, z."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's my x-axis. That's my y-axis. That's my z-axis. x, y, z. And x is going from 0 to 2. Let's say that's 2. y is going from 0 to 1. And so we're taking the volume above this rectangle, the xy plane, and then the surface, I'm going to try my best to draw it."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x, y, z. And x is going from 0 to 2. Let's say that's 2. y is going from 0 to 1. And so we're taking the volume above this rectangle, the xy plane, and then the surface, I'm going to try my best to draw it. I'll draw it in a different color. I'm looking at the picture. At this end, it looks something like this."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we're taking the volume above this rectangle, the xy plane, and then the surface, I'm going to try my best to draw it. I'll draw it in a different color. I'm looking at the picture. At this end, it looks something like this. And then it has a straight line. Let's see if I can draw this surface going down like that. And then if I was really good, I could shade it."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "At this end, it looks something like this. And then it has a straight line. Let's see if I can draw this surface going down like that. And then if I was really good, I could shade it. It looks something like this. This surface looks something like that. And this right here is above this."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if I was really good, I could shade it. It looks something like this. This surface looks something like that. And this right here is above this. This is the bottom left corner. You can almost view it. So let me just say the yellow is the top of the surface."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this right here is above this. This is the bottom left corner. You can almost view it. So let me just say the yellow is the top of the surface. And then this is under the surface. So we care about this volume underneath here. Let me show you what the actual surface is."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me just say the yellow is the top of the surface. And then this is under the surface. So we care about this volume underneath here. Let me show you what the actual surface is. So this volume underneath here. I think you get the idea. So how do we do that?"}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me show you what the actual surface is. So this volume underneath here. I think you get the idea. So how do we do that? Well, in the last example, we said, well, let's pick an arbitrary y. And for that y, let's figure out the area under the curve. So if we fix some y, when you actually do the problem, you don't have to think about it in this much detail."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So how do we do that? Well, in the last example, we said, well, let's pick an arbitrary y. And for that y, let's figure out the area under the curve. So if we fix some y, when you actually do the problem, you don't have to think about it in this much detail. But I want to give you the intuition. So if we pick just an arbitrary y here. So on that y, we could think of it, if we have a fixed y, then the function of x and y, you can almost view it as a function of just x for that given y."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we fix some y, when you actually do the problem, you don't have to think about it in this much detail. But I want to give you the intuition. So if we pick just an arbitrary y here. So on that y, we could think of it, if we have a fixed y, then the function of x and y, you can almost view it as a function of just x for that given y. And so we're kind of figuring out the value of this, of the area under this curve. So this is kind of an up-down curve for a given y. So if we know a y, we can figure out then, for example, if y was 5, this function would become z equals 25x."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So on that y, we could think of it, if we have a fixed y, then the function of x and y, you can almost view it as a function of just x for that given y. And so we're kind of figuring out the value of this, of the area under this curve. So this is kind of an up-down curve for a given y. So if we know a y, we can figure out then, for example, if y was 5, this function would become z equals 25x. And then that's easy to figure out the value of the curve under. So we'll make the value under the curve as a function of y. We'll pretend like it's just a constant."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if we know a y, we can figure out then, for example, if y was 5, this function would become z equals 25x. And then that's easy to figure out the value of the curve under. So we'll make the value under the curve as a function of y. We'll pretend like it's just a constant. So let's do that. So if we have a dx, that's our change in x. And then our height of each of our rectangles is going to be a function, is going to be z."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We'll pretend like it's just a constant. So let's do that. So if we have a dx, that's our change in x. And then our height of each of our rectangles is going to be a function, is going to be z. The height is z, which is a function of x and y. So we can take the integral. So the area of each of these is going to be our function, xy squared, I'll do it here because I'll run out of space, xy squared times the width, which is dx."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then our height of each of our rectangles is going to be a function, is going to be z. The height is z, which is a function of x and y. So we can take the integral. So the area of each of these is going to be our function, xy squared, I'll do it here because I'll run out of space, xy squared times the width, which is dx. If we want the area of this slice for a given y, we just integrate along the x-axis. We're going to integrate from x is equal to 0 to x is equal to 2. From x is equal to 0 to 2."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the area of each of these is going to be our function, xy squared, I'll do it here because I'll run out of space, xy squared times the width, which is dx. If we want the area of this slice for a given y, we just integrate along the x-axis. We're going to integrate from x is equal to 0 to x is equal to 2. From x is equal to 0 to 2. Fair enough. Now, we just don't want to figure out the area under the curve at one slice for one particular y. We want to figure out the entire area of the curve."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "From x is equal to 0 to 2. Fair enough. Now, we just don't want to figure out the area under the curve at one slice for one particular y. We want to figure out the entire area of the curve. So what we do is, we say, OK, fine. The area under the curve, not the surface, under this curve for a particular y is this expression. Well, what if I gave it a little bit of depth?"}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We want to figure out the entire area of the curve. So what we do is, we say, OK, fine. The area under the curve, not the surface, under this curve for a particular y is this expression. Well, what if I gave it a little bit of depth? If I multiplied this area times dy, then it would give me a little bit of depth. We'd kind of have a three-dimensional slice of the volume that we care about. I know it's hard to imagine."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, what if I gave it a little bit of depth? If I multiplied this area times dy, then it would give me a little bit of depth. We'd kind of have a three-dimensional slice of the volume that we care about. I know it's hard to imagine. Let me bring that here. So if I had a slice here, this is what we just figured out, the area of a slice. And then I'm multiplying it by dy to give it a little bit of depth."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I know it's hard to imagine. Let me bring that here. So if I had a slice here, this is what we just figured out, the area of a slice. And then I'm multiplying it by dy to give it a little bit of depth. So you multiply it by dy to give it a little bit of depth. And then if we want the entire volume under the curve, we add all the dy's together, take the infinite sum of these infinitely small volumes, really, now. And so we will integrate from y is equal to 0 to y is equal to 1."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then I'm multiplying it by dy to give it a little bit of depth. So you multiply it by dy to give it a little bit of depth. And then if we want the entire volume under the curve, we add all the dy's together, take the infinite sum of these infinitely small volumes, really, now. And so we will integrate from y is equal to 0 to y is equal to 1. I know this graph is a little hard to understand, but you might want to re-watch the first video. I had a slightly easier to understand surface. So now how do we evaluate this?"}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so we will integrate from y is equal to 0 to y is equal to 1. I know this graph is a little hard to understand, but you might want to re-watch the first video. I had a slightly easier to understand surface. So now how do we evaluate this? Well, like we said, you evaluate from the inside and go outward. And it's almost like taking a partial derivative in reverse. So we're integrating here with respect to x."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now how do we evaluate this? Well, like we said, you evaluate from the inside and go outward. And it's almost like taking a partial derivative in reverse. So we're integrating here with respect to x. So we can treat y just like a constant, like it's like the number 5 or something like that. So it really doesn't change the integral. So what's the antiderivative of xy squared?"}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we're integrating here with respect to x. So we can treat y just like a constant, like it's like the number 5 or something like that. So it really doesn't change the integral. So what's the antiderivative of xy squared? Well, the antiderivative of xy squared, I want to make sure I'm color consistent. Well, the antiderivative of x is x to the 1 half, sorry, x squared over 2. And then y squared is just a constant, right?"}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what's the antiderivative of xy squared? Well, the antiderivative of xy squared, I want to make sure I'm color consistent. Well, the antiderivative of x is x to the 1 half, sorry, x squared over 2. And then y squared is just a constant, right? And then we don't have to worry about plus c, since this is a definite integral. And we're going to evaluate that at 2 and 0. And then we still have the outside integral with respect to y."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then y squared is just a constant, right? And then we don't have to worry about plus c, since this is a definite integral. And we're going to evaluate that at 2 and 0. And then we still have the outside integral with respect to y. So once we figure that out, we're going to integrate it from 0 to 1 with respect to dy. Now what does this evaluate? We put a 2 in here."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we still have the outside integral with respect to y. So once we figure that out, we're going to integrate it from 0 to 1 with respect to dy. Now what does this evaluate? We put a 2 in here. If you put a 2 in there, you get 2 squared over 2. Well, that's just 4 over 2. So it's 2y squared minus 0 squared over 2 times y squared."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We put a 2 in here. If you put a 2 in there, you get 2 squared over 2. Well, that's just 4 over 2. So it's 2y squared minus 0 squared over 2 times y squared. Well, that's just going to be 0, so it's minus 0. I won't write that down, because hopefully that's a little bit of second nature to you. We just evaluated this at the two endpoints, and I'm short for space."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's 2y squared minus 0 squared over 2 times y squared. Well, that's just going to be 0, so it's minus 0. I won't write that down, because hopefully that's a little bit of second nature to you. We just evaluated this at the two endpoints, and I'm short for space. So this evaluated at 2y squared, and now we evaluate the outside integral. 0, 1, dy. And this is an important thing to realize."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We just evaluated this at the two endpoints, and I'm short for space. So this evaluated at 2y squared, and now we evaluate the outside integral. 0, 1, dy. And this is an important thing to realize. When we evaluated this inside integral, remember what we were doing. We were trying to figure out for a given y what the area of this surface was. Well, not this surface, the area under the surface on this kind of for a given y."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is an important thing to realize. When we evaluated this inside integral, remember what we were doing. We were trying to figure out for a given y what the area of this surface was. Well, not this surface, the area under the surface on this kind of for a given y. For a given y, that surface kind of turns into a curve. And we've tried to figure out the area under that curve in the traditional sense. So this ended up being a function of y."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, not this surface, the area under the surface on this kind of for a given y. For a given y, that surface kind of turns into a curve. And we've tried to figure out the area under that curve in the traditional sense. So this ended up being a function of y. And that makes sense, because depending on which y we pick, we're going to get a different area here. Because obviously, depending on which y we pick, the area, kind of a wall, drops straight down. That area is going to change."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this ended up being a function of y. And that makes sense, because depending on which y we pick, we're going to get a different area here. Because obviously, depending on which y we pick, the area, kind of a wall, drops straight down. That area is going to change. So we got a function of y when we evaluated this. And now we just integrate with respect to y. And this is just plain old vanilla definite integration."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That area is going to change. So we got a function of y when we evaluated this. And now we just integrate with respect to y. And this is just plain old vanilla definite integration. What's the antiderivative of 2y squared? Well, that equals 2 times y to the third over 3, or 2 thirds y to the third. We evaluate that at 1 and 0, which is equal to, let's see, y of 1 to the third times 2 thirds, that's 2 thirds, minus 0 to the third times 2 thirds."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And this is just plain old vanilla definite integration. What's the antiderivative of 2y squared? Well, that equals 2 times y to the third over 3, or 2 thirds y to the third. We evaluate that at 1 and 0, which is equal to, let's see, y of 1 to the third times 2 thirds, that's 2 thirds, minus 0 to the third times 2 thirds. Well, that's just 0. So it equals 2 thirds. If our units were meters, it would be 2 thirds meters cubed, or cubic meters."}, {"video_title": "Double integrals 2   Double and triple integrals   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We evaluate that at 1 and 0, which is equal to, let's see, y of 1 to the third times 2 thirds, that's 2 thirds, minus 0 to the third times 2 thirds. Well, that's just 0. So it equals 2 thirds. If our units were meters, it would be 2 thirds meters cubed, or cubic meters. But there you go. That's how you evaluate a double integral. There really isn't a new skill here."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say I have some curve C, and it's described, it can be parametrized, 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 a, 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."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. You evaluate this function at b, and y of b."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is t is equal to b. This actual point right here will be x of b. That would be the x-coordinate. You evaluate this function at b, and y of b. 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."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You evaluate this function at b, and y of b. 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 a review here."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 a 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."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if I define a vector-valued function, and if you don't remember what those are, we'll have a little bit of a 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. 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."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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. 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 0 at the origin. And when you say it's a 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 even four, five, whatever, n-dimensional space."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "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 0 at the origin. And when you say it's a 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 even four, five, whatever, n-dimensional space. So when you say it's a position vector, you're literally saying, OK, this vector literally specifies that point in space. So let's see if we can describe this curve as a position vector-valued function. So we could say r of t, let me switch back to that pink color."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In this case, it's going to be in two-dimensional space, but it could be in three-dimensional space or really even four, five, whatever, n-dimensional space. So when you say it's a position vector, you're literally saying, OK, this vector literally specifies that point in space. So let's see if we can describe this curve as a position vector-valued function. So we could say r of t, let me switch back to that pink color. I'll just stay in green, is equal to x of t times the unit vector in the x direction. The unit vector gets a little caret on top, a little hat. That's like the arrow for it."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So we could say r of t, let me switch back to that pink color. I'll just stay in green, is equal to x of t times the unit vector in the x direction. The unit vector gets a little caret on top, a little hat. That's like the arrow for it. That just says it's a unit vector. Plus y of t times j. If I was dealing with a curve in three dimensions, I would have plus z of t times k, but we're dealing with two dimensions right here."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's like the arrow for it. That just says it's a unit vector. Plus y of t times j. If I was dealing with a curve in three dimensions, I would have plus z of t times k, but we're dealing with two dimensions right here. And so the way this works is you're just taking your, for any t, and still we're going to have t is greater than or equal to a and then less than or equal to b. And this is the exact same thing as that. Let me just redraw it."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I was dealing with a curve in three dimensions, I would have plus z of t times k, but we're dealing with two dimensions right here. And so the way this works is you're just taking your, for any t, and still we're going to have t is greater than or equal to a and then less than or equal to b. And this is the exact same thing as that. Let me just redraw it. So let me draw our coordinates right here, our axes. So that's the y-axis and this is the x-axis. So when you evaluate r of a, that's our starting point."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me just redraw it. So let me draw our coordinates right here, our axes. So that's the y-axis and this is the x-axis. So when you evaluate r of a, that's our starting point. So let me do that. So r of a, maybe I'll do it right over here. Our position vector-valued function, evaluated at t is equal to a, is going to be equal to x of a times our unit vector in the x direction plus y of a times our unit vector in the vertical direction, or in the y direction."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when you evaluate r of a, that's our starting point. So let me do that. So r of a, maybe I'll do it right over here. Our position vector-valued function, evaluated at t is equal to a, is going to be equal to x of a times our unit vector in the x direction plus y of a times our unit vector in the vertical direction, or in the y direction. What's that going to look like? Well, x of a is this thing right here. So it's x of a times the unit vector."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Our position vector-valued function, evaluated at t is equal to a, is going to be equal to x of a times our unit vector in the x direction plus y of a times our unit vector in the vertical direction, or in the y direction. What's that going to look like? Well, x of a is this thing right here. So it's x of a times the unit vector. So it's really, maybe the unit vector is this long. It has length 1. So now we're just going to have a length of x of a in that direction."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's x of a times the unit vector. So it's really, maybe the unit vector is this long. It has length 1. So now we're just going to have a length of x of a in that direction. And then same thing in y of a. It's going to be y of a length in that direction. But the bottom line, this vector right here, if you add these scaled values of these two unit vectors, you're going to get r of a looking something like this."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So now we're just going to have a length of x of a in that direction. And then same thing in y of a. It's going to be y of a length in that direction. But the bottom line, this vector right here, if you add these scaled values of these two unit vectors, you're going to get r of a looking something like this. It's going to be a vector that looks something like that. Just like that. It's a vector."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But the bottom line, this vector right here, if you add these scaled values of these two unit vectors, you're going to get r of a looking something like this. It's going to be a vector that looks something like that. Just like that. It's a vector. It's a position vector. That's why we're kneeling it at the origin, but drawing it in standard position. And that right there is r of a."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's a vector. It's a position vector. That's why we're kneeling it at the origin, but drawing it in standard position. And that right there is r of a. Now, what happens if a increases a little bit? What is r of a plus a little bit? And I don't know, we could call that r of a plus delta, or r of a plus h. We do it in different colors."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And that right there is r of a. Now, what happens if a increases a little bit? What is r of a plus a little bit? And I don't know, we could call that r of a plus delta, or r of a plus h. We do it in different colors. So let's say we increase a a little bit. r of a plus some small h. Well, that's just going to be x of a plus h times the unit vector i plus y times a plus h times the unit vector j. And what's that going to look like?"}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And I don't know, we could call that r of a plus delta, or r of a plus h. We do it in different colors. So let's say we increase a a little bit. r of a plus some small h. Well, that's just going to be x of a plus h times the unit vector i plus y times a plus h times the unit vector j. And what's that going to look like? Well, we're going to go a little bit further down the curve. That's like saying the coordinate x of a plus h and y plus a plus h. It might be that point right there. So it'll be a new unit vector."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what's that going to look like? Well, we're going to go a little bit further down the curve. That's like saying the coordinate x of a plus h and y plus a plus h. It might be that point right there. So it'll be a new unit vector. Sorry, it'll be a new vector, position vector, not a unit vector. These don't necessarily have length 1. That might be right here."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it'll be a new unit vector. Sorry, it'll be a new vector, position vector, not a unit vector. These don't necessarily have length 1. That might be right here. Let me do that same color as this. So it might be just like that. So that right here is r of a plus h. So you see, as you keep increasing your value of t until you get to b, these position vectors are going to keep specifying points along this curve."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That might be right here. Let me do that same color as this. So it might be just like that. So that right here is r of a plus h. So you see, as you keep increasing your value of t until you get to b, these position vectors are going to keep specifying points along this curve. So the curve, let me draw the curve in a different color. The curve looks something like this. It's meant to look exactly like the curve that I have up here."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So that right here is r of a plus h. So you see, as you keep increasing your value of t until you get to b, these position vectors are going to keep specifying points along this curve. So the curve, let me draw the curve in a different color. The curve looks something like this. It's meant to look exactly like the curve that I have up here. And for example, r of b is going to be a vector that looks like this. It's going to be a vector that looks like that. Let me draw it relatively straight."}, {"video_title": "Position vector valued functions   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's meant to look exactly like the curve that I have up here. And for example, r of b is going to be a vector that looks like this. It's going to be a vector that looks like that. Let me draw it relatively straight. That vector right there is r of b. So hopefully you realize that, look, these position vectors really are specifying the same points on this curve as this original, I guess, straight up parametrization that we did for this curve. And I just want to do that as a little bit of review, because we're now going to break in into the idea of actually taking a derivative of this vector-valued function."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "So I think I should probably start off by addressing the elephant in the living room here. I am sadly not Sal, but I'm still gonna teach you some math. My name is Grant. I'm pretty much a math enthusiast. I enjoy making animations of things when applicable, and boy is that applicable when it comes to multivariable calculus. So the first thing we gotta get straight is what is this word multivariable that separates calculus as we know it from the new topic that you're about to study? Well, I could say it's all about multivariable functions."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "I'm pretty much a math enthusiast. I enjoy making animations of things when applicable, and boy is that applicable when it comes to multivariable calculus. So the first thing we gotta get straight is what is this word multivariable that separates calculus as we know it from the new topic that you're about to study? Well, I could say it's all about multivariable functions. That doesn't really answer anything because what's a multivariable function? And basically, the kinds of functions that we're used to dealing with in the old world, in the ordinary calculus world, will have a single input, some kind of number as their input, and then an output's just a single number. And you would call this a single variable function basically because that guy there is the single variable."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Well, I could say it's all about multivariable functions. That doesn't really answer anything because what's a multivariable function? And basically, the kinds of functions that we're used to dealing with in the old world, in the ordinary calculus world, will have a single input, some kind of number as their input, and then an output's just a single number. And you would call this a single variable function basically because that guy there is the single variable. So then a multivariable function is something that handles multiple variables. So, you know, it's common to write it as like xy. It doesn't really matter what letters to use."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "And you would call this a single variable function basically because that guy there is the single variable. So then a multivariable function is something that handles multiple variables. So, you know, it's common to write it as like xy. It doesn't really matter what letters to use. And it could be, you know, xyz, x1, x2, x3, a whole bunch of things. But just to get started, we often think just two variables. And this will output something that depends on both of those."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "It doesn't really matter what letters to use. And it could be, you know, xyz, x1, x2, x3, a whole bunch of things. But just to get started, we often think just two variables. And this will output something that depends on both of those. Commonly, it'll output just a number. So you might imagine a number that depends on x and y in some way, like x squared plus y. But it could also output a vector, right?"}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "And this will output something that depends on both of those. Commonly, it'll output just a number. So you might imagine a number that depends on x and y in some way, like x squared plus y. But it could also output a vector, right? So you could also imagine something that's got multivariable input, f of x, y, and it outputs something that also has multiple variables, like, I mean, I'm just making stuff up here, 3x and, you know, 2y. And this isn't set in stone, but the convention is to usually think if there's multiple numbers that go into the output, think of it as a vector. If there's multiple numbers that go into the input, just kind of write them more sideways like this and think of them as a point in space."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "But it could also output a vector, right? So you could also imagine something that's got multivariable input, f of x, y, and it outputs something that also has multiple variables, like, I mean, I'm just making stuff up here, 3x and, you know, 2y. And this isn't set in stone, but the convention is to usually think if there's multiple numbers that go into the output, think of it as a vector. If there's multiple numbers that go into the input, just kind of write them more sideways like this and think of them as a point in space. Because, I mean, when you look at something like this, and you've got an x and you've got a y, you could think about those as two separate numbers. You know, here's your number line with the point x on it somewhere. Maybe that's 5, maybe that's 3, it doesn't really matter."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "If there's multiple numbers that go into the input, just kind of write them more sideways like this and think of them as a point in space. Because, I mean, when you look at something like this, and you've got an x and you've got a y, you could think about those as two separate numbers. You know, here's your number line with the point x on it somewhere. Maybe that's 5, maybe that's 3, it doesn't really matter. And then you've got another number line, and it's y. And you could think of them as separate entities. But it would probably be more accurate to call it multidimensional calculus, because really, instead of thinking of, you know, x and y as separate entities, whenever you see two things like that, you're going to be thinking about the xy-plane and thinking about just a single point."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Maybe that's 5, maybe that's 3, it doesn't really matter. And then you've got another number line, and it's y. And you could think of them as separate entities. But it would probably be more accurate to call it multidimensional calculus, because really, instead of thinking of, you know, x and y as separate entities, whenever you see two things like that, you're going to be thinking about the xy-plane and thinking about just a single point. And you'd think of this as a function that takes a point to a number, or a point to a vector. And a lot of people, when they start teaching multivariable calculus, they just jump into the calculus. And there's lots of fun things, partial derivatives, gradients, good stuff that you'll learn."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "But it would probably be more accurate to call it multidimensional calculus, because really, instead of thinking of, you know, x and y as separate entities, whenever you see two things like that, you're going to be thinking about the xy-plane and thinking about just a single point. And you'd think of this as a function that takes a point to a number, or a point to a vector. And a lot of people, when they start teaching multivariable calculus, they just jump into the calculus. And there's lots of fun things, partial derivatives, gradients, good stuff that you'll learn. But I think, first of all, I want to spend a couple videos just talking about the different ways we visualize the different types of multivariable functions. So, as a sneak peek, I'm just going to go through a couple of them really quickly right now, just so you kind of whet your appetite and see what I'm getting at. But the next few videos are going to go through them in much, much more detail."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "And there's lots of fun things, partial derivatives, gradients, good stuff that you'll learn. But I think, first of all, I want to spend a couple videos just talking about the different ways we visualize the different types of multivariable functions. So, as a sneak peek, I'm just going to go through a couple of them really quickly right now, just so you kind of whet your appetite and see what I'm getting at. But the next few videos are going to go through them in much, much more detail. So, first of all, graphs. When you have multivariable functions, graphs become three-dimensional. But these only really apply to functions that have some kind of two-dimensional input, which you might think about as living on this xy-plane."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "But the next few videos are going to go through them in much, much more detail. So, first of all, graphs. When you have multivariable functions, graphs become three-dimensional. But these only really apply to functions that have some kind of two-dimensional input, which you might think about as living on this xy-plane. And a single number is their output. And the height of the graph is going to correspond with that output. Like I said, you'll be able to learn much more about that in the dedicated video on it."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "But these only really apply to functions that have some kind of two-dimensional input, which you might think about as living on this xy-plane. And a single number is their output. And the height of the graph is going to correspond with that output. Like I said, you'll be able to learn much more about that in the dedicated video on it. But these functions also can be visualized just in two dimensions, flattening things out, where we visualize the entire input space and associate a color with each point. So this is the kind of thing where you'd have some function that's got a two-dimensional input, it would be f of x, y, and what we're looking at is the xy-plane, all of the input space, and this outputs just some number, maybe it's like x squared, but you know that, and maybe some complicated thing. And the color tells you roughly the size of that output, and the lines here, called contour lines, tell you which inputs all share a constant output value."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Like I said, you'll be able to learn much more about that in the dedicated video on it. But these functions also can be visualized just in two dimensions, flattening things out, where we visualize the entire input space and associate a color with each point. So this is the kind of thing where you'd have some function that's got a two-dimensional input, it would be f of x, y, and what we're looking at is the xy-plane, all of the input space, and this outputs just some number, maybe it's like x squared, but you know that, and maybe some complicated thing. And the color tells you roughly the size of that output, and the lines here, called contour lines, tell you which inputs all share a constant output value. And again, I'll go into much more detail there. These are really nice, much more convenient than three-dimensional graphs to just sketch out. Moving right along, I'm also going to talk about surfaces in three-dimensional space."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the color tells you roughly the size of that output, and the lines here, called contour lines, tell you which inputs all share a constant output value. And again, I'll go into much more detail there. These are really nice, much more convenient than three-dimensional graphs to just sketch out. Moving right along, I'm also going to talk about surfaces in three-dimensional space. They look like graphs, but they actually deal with a much different animal that you could think of it as mapping two dimensions, and I like to sort of spush it about, and we've got kind of a two-dimensional input that somehow moves into three dimensions, and you're just looking at what the output of that looks like, not really caring about how it gets there. These are called parametric surfaces. Another fun one is a vector field, where every input point is associated with some kind of vector, which is the output of the function there, so this would be a function with a two-dimensional input and then two-dimensional output, because each of these are two-dimensional vectors."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Moving right along, I'm also going to talk about surfaces in three-dimensional space. They look like graphs, but they actually deal with a much different animal that you could think of it as mapping two dimensions, and I like to sort of spush it about, and we've got kind of a two-dimensional input that somehow moves into three dimensions, and you're just looking at what the output of that looks like, not really caring about how it gets there. These are called parametric surfaces. Another fun one is a vector field, where every input point is associated with some kind of vector, which is the output of the function there, so this would be a function with a two-dimensional input and then two-dimensional output, because each of these are two-dimensional vectors. And the fun part with these guys is that you can just kind of imagine a fluid flowing, so here's a bunch of droplets like water, and they kind of flow along that, and that actually turns out to give insight about the underlying function. It's one of those beautiful aspects of multivariable calc, and we'll get lots of exposure to that. Again, I'm just sort of zipping through to whet your appetite."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Another fun one is a vector field, where every input point is associated with some kind of vector, which is the output of the function there, so this would be a function with a two-dimensional input and then two-dimensional output, because each of these are two-dimensional vectors. And the fun part with these guys is that you can just kind of imagine a fluid flowing, so here's a bunch of droplets like water, and they kind of flow along that, and that actually turns out to give insight about the underlying function. It's one of those beautiful aspects of multivariable calc, and we'll get lots of exposure to that. Again, I'm just sort of zipping through to whet your appetite. Don't worry if this doesn't make sense immediately. And one of my all-time favorite ways to think about multivariable functions is to just take the input space, in this case this is going to be a function that inputs points in two-dimensional space, and watch them move to their output. So this is going to be a function that also outputs in two dimensions, and I'm just going to watch every single point move over to where it's supposed to go."}, {"video_title": "Multivariable functions   Multivariable calculus   Khan Academy.mp3", "Sentence": "Again, I'm just sort of zipping through to whet your appetite. Don't worry if this doesn't make sense immediately. And one of my all-time favorite ways to think about multivariable functions is to just take the input space, in this case this is going to be a function that inputs points in two-dimensional space, and watch them move to their output. So this is going to be a function that also outputs in two dimensions, and I'm just going to watch every single point move over to where it's supposed to go. These can be kind of complicated to look at or to think about at first, but as you gain a little bit of thought and exposure to them, they're actually very nice, and it provides a beautiful connection with linear algebra. A lot of you out there, if you're studying multivariable calculus, you either are about to study linear algebra, or you just have, or maybe you're doing it concurrently, but understanding functions as transformations is going to be a great way to connect those two. So with that, I'll stop jabbering through these topics really quickly, and in the next few videos, I'll actually go through them in detail, and hopefully you can get a good feel for what linear algebra functions can actually feel like."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll call this region R sub three, R with a subscript three. It's going to be the set of all points in three dimensions, the set of all x, y's and z's, such that the x, z pairs are a member of a domain, I'll call this domain D three, D sub three, and, let's put a comma here, y is going to vary between two surfaces that are functions of x and z. So y is going to be greater than or equal to the surface, I'll call it H one of xz, is going to be less than or equal to y, less than or equal to y, which is going to be less than or equal to, y is going to be bounded from above by the surface H two of xz. And once again, let's close our set notation. So let's think about whether some of these regions that we already saw were type one and type two, whether they're type three, and then think about what would not be a type three region. So let's go to this sphere. What could be our domain?"}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And once again, let's close our set notation. So let's think about whether some of these regions that we already saw were type one and type two, whether they're type three, and then think about what would not be a type three region. So let's go to this sphere. What could be our domain? Well, the domain is a set of xz's, so it's going to be in the xz plane. So over here, our domain could be this region, right over here in the xz plane. So I'll color it in."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "What could be our domain? Well, the domain is a set of xz's, so it's going to be in the xz plane. So over here, our domain could be this region, right over here in the xz plane. So I'll color it in. It could be that region right over there in the xz plane. And then the lower bound on y will be the part that is behind the sphere in this direction right over here. This might be a little bit hard to visualize since I'm redrawing on top of it."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So I'll color it in. It could be that region right over there in the xz plane. And then the lower bound on y will be the part that is behind the sphere in this direction right over here. This might be a little bit hard to visualize since I'm redrawing on top of it. And then the upper bound on y is going to be this side right over here. It's going to be this side right over here. It's going to be the upper bound."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This might be a little bit hard to visualize since I'm redrawing on top of it. And then the upper bound on y is going to be this side right over here. It's going to be this side right over here. It's going to be the upper bound. So all of this is now going to be green. Let me redraw the sphere just to make it clear. So assuming that, let me just draw another coordinate axis."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's going to be the upper bound. So all of this is now going to be green. Let me redraw the sphere just to make it clear. So assuming that, let me just draw another coordinate axis. We have another coordinate axis. The back side of the sphere in the y direction, so I guess let's think of it this way. So this is the hemisphere mark."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So assuming that, let me just draw another coordinate axis. We have another coordinate axis. The back side of the sphere in the y direction, so I guess let's think of it this way. So this is the hemisphere mark. This is the halfway point for my sphere. And once again, that's kind of the boundary of our domain. And then the back side, let me do the front side first."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the hemisphere mark. This is the halfway point for my sphere. And once again, that's kind of the boundary of our domain. And then the back side, let me do the front side first. The front side, y's upper bound, that would be h2, would be all of this business right over here. So this would be h2 that I'm coloring in. h2 would be the side that is facing in that direction."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the back side, let me do the front side first. The front side, y's upper bound, that would be h2, would be all of this business right over here. So this would be h2 that I'm coloring in. h2 would be the side that is facing in that direction. And let me see how well I can color it. That didn't do a good thing. So h2 is all of this stuff out on this side of the sphere."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "h2 would be the side that is facing in that direction. And let me see how well I can color it. That didn't do a good thing. So h2 is all of this stuff out on this side of the sphere. And then h1 was the lower bound on y. So it's going to be that side right over there. I know I could draw it probably a little bit neater, but hopefully you get the point."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So h2 is all of this stuff out on this side of the sphere. And then h1 was the lower bound on y. So it's going to be that side right over there. I know I could draw it probably a little bit neater, but hopefully you get the point. It would be all of that side. And then y can vary between those two and essentially fill up the region. We make the exact same argument with the cylinder."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I know I could draw it probably a little bit neater, but hopefully you get the point. It would be all of that side. And then y can vary between those two and essentially fill up the region. We make the exact same argument with the cylinder. The cylinder can be defined the same way. So first of all, the sphere is a type 1, type 2, and type 3 region. It meets all of the constraints."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We make the exact same argument with the cylinder. The cylinder can be defined the same way. So first of all, the sphere is a type 1, type 2, and type 3 region. It meets all of the constraints. The cylinder, at least the way it was oriented there, actually any cylinder, actually will also be a type 3 region. Exact same argument. So let me draw my axes again."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It meets all of the constraints. The cylinder, at least the way it was oriented there, actually any cylinder, actually will also be a type 3 region. Exact same argument. So let me draw my axes again. And here, to make the cylinder that we've been making, our domain could be a rectangle in the xz plane. So our domain can be a rectangular region in the xz plane, just like that. And then the lower bound on y could be that side of the cylinder, the side facing in that direction right over there."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let me draw my axes again. And here, to make the cylinder that we've been making, our domain could be a rectangle in the xz plane. So our domain can be a rectangular region in the xz plane, just like that. And then the lower bound on y could be that side of the cylinder, the side facing in that direction right over there. That is the lower bound. And then the upper bound on y could be the side facing in that direction. The upper bound on y would be this side right over there."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the lower bound on y could be that side of the cylinder, the side facing in that direction right over there. That is the lower bound. And then the upper bound on y could be the side facing in that direction. The upper bound on y would be this side right over there. So once again, this is also a type 3 region. By the same logic, this one right over here, this hourglass, can be a type 3 region. The front side would be this front side right over here."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The upper bound on y would be this side right over there. So once again, this is also a type 3 region. By the same logic, this one right over here, this hourglass, can be a type 3 region. The front side would be this front side right over here. All of this, including that stuff that I... So it would be all of that. And then the back side, when you think of it in terms of y, the back side would be that part right over there."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The front side would be this front side right over here. All of this, including that stuff that I... So it would be all of that. And then the back side, when you think of it in terms of y, the back side would be that part right over there. And once again, this could also be a type 3 region. The domain would be kind of this cross... The boundary of the domain would be this cross section right over here."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then the back side, when you think of it in terms of y, the back side would be that part right over there. And once again, this could also be a type 3 region. The domain would be kind of this cross... The boundary of the domain would be this cross section right over here. So the boundary of our domain could be that cross section right over there. The lower bound on y would be the back half of this hourglass. And the upper bound on y would be the front half."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The boundary of the domain would be this cross section right over here. So the boundary of our domain could be that cross section right over there. The lower bound on y would be the back half of this hourglass. And the upper bound on y would be the front half. Let me do that magenta color because I've been using that, or actually that green color. So the upper bound on y would be this right half right over here. So what would not be a type 3 region?"}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And the upper bound on y would be the front half. Let me do that magenta color because I've been using that, or actually that green color. So the upper bound on y would be this right half right over here. So what would not be a type 3 region? Well, if we just rotated this around like that. So let me just draw something that is not a type 3 region just to show you that this definition does not include everything. So something that would not be a type 3 region for the same argument as we've seen before for type 2 and type 1 regions is an hourglass where it's along the y-axis, or at least it's oriented this way."}, {"video_title": "Type III regions in three dimensions   Divergence theorem   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what would not be a type 3 region? Well, if we just rotated this around like that. So let me just draw something that is not a type 3 region just to show you that this definition does not include everything. So something that would not be a type 3 region for the same argument as we've seen before for type 2 and type 1 regions is an hourglass where it's along the y-axis, or at least it's oriented this way. It actually does not have to be right centered on the y-axis. But an hourglass that looks like this, now all of a sudden the y can't be just expressed as being between two surfaces that are functions of x and z. You would have to break this up in order to do that."}, {"video_title": "2d curl example.mp3", "Sentence": "So let's compute the two-dimensional curl of a vector field. And the one I have in mind will have an x component of, let's see, not nine, but y cubed minus nine times y, and then the y component will be x cubed minus nine times x. And you can kind of see I'm just a sucker for symmetry when I choose examples. And I showed in the last video how the two-dimensional curl, the 2D curl of a vector field, of a vector field v, which is a function of x and y, is equal to the partial derivative of q, that second component, with respect to x, minus the partial derivative of p, that first component, with respect to y. And I went through the reasoning for why this is true, but just real quick, kind of in a nutshell here, this partial q, partial x, is because as you move from left to right, vectors tend to go from having a small or even negative y component to a positive y component. That corresponds to counterclockwise rotation. And similarly, this dp dy is because if vectors, as you move up and down, as you kind of increase the y value, go from being positive to zero to negative, or if they're decreasing, that also corresponds to counterclockwise rotation."}, {"video_title": "2d curl example.mp3", "Sentence": "And I showed in the last video how the two-dimensional curl, the 2D curl of a vector field, of a vector field v, which is a function of x and y, is equal to the partial derivative of q, that second component, with respect to x, minus the partial derivative of p, that first component, with respect to y. And I went through the reasoning for why this is true, but just real quick, kind of in a nutshell here, this partial q, partial x, is because as you move from left to right, vectors tend to go from having a small or even negative y component to a positive y component. That corresponds to counterclockwise rotation. And similarly, this dp dy is because if vectors, as you move up and down, as you kind of increase the y value, go from being positive to zero to negative, or if they're decreasing, that also corresponds to counterclockwise rotation. So taking the negative of that will tell you whether or not changes in the y direction around your point correspond with counterclockwise rotation. So in this particular case, when we start evaluating that, we start by looking at partial of q with respect to x. So we're looking at the second component and taking its partial derivative with respect to x."}, {"video_title": "2d curl example.mp3", "Sentence": "And similarly, this dp dy is because if vectors, as you move up and down, as you kind of increase the y value, go from being positive to zero to negative, or if they're decreasing, that also corresponds to counterclockwise rotation. So taking the negative of that will tell you whether or not changes in the y direction around your point correspond with counterclockwise rotation. So in this particular case, when we start evaluating that, we start by looking at partial of q with respect to x. So we're looking at the second component and taking its partial derivative with respect to x. And in this case, nothing but x's show up, so it's just like taking its derivative, and you'll get three x squared minus nine there. Three x squared minus nine. And that's the first part."}, {"video_title": "2d curl example.mp3", "Sentence": "So we're looking at the second component and taking its partial derivative with respect to x. And in this case, nothing but x's show up, so it's just like taking its derivative, and you'll get three x squared minus nine there. Three x squared minus nine. And that's the first part. And then we subtract off whatever the partial derivative of p with respect to y is. So we go up here, and it's entirely in terms of y, and kind of do the symmetry. We're just taking the same calculation."}, {"video_title": "2d curl example.mp3", "Sentence": "And that's the first part. And then we subtract off whatever the partial derivative of p with respect to y is. So we go up here, and it's entirely in terms of y, and kind of do the symmetry. We're just taking the same calculation. Three y squared, that derivative of y cubed, minus nine. So this right here is our two-dimensional curl. And let's go ahead and interpret what this means."}, {"video_title": "2d curl example.mp3", "Sentence": "We're just taking the same calculation. Three y squared, that derivative of y cubed, minus nine. So this right here is our two-dimensional curl. And let's go ahead and interpret what this means. And in fact, this vector field that I showed you is exactly the one that I used when I was kind of animating the intuition behind curl to start off with, where I had these specific parts where there's positive curl here and here, but negative curl up in these clockwise rotating areas. So we can actually see why that's the case here, and why I chose this specific function for something that'll have lots of good curl examples. Because if we look over in that region where there should be positive curl, that's where x is equal to three and y is equal to zero."}, {"video_title": "2d curl example.mp3", "Sentence": "And let's go ahead and interpret what this means. And in fact, this vector field that I showed you is exactly the one that I used when I was kind of animating the intuition behind curl to start off with, where I had these specific parts where there's positive curl here and here, but negative curl up in these clockwise rotating areas. So we can actually see why that's the case here, and why I chose this specific function for something that'll have lots of good curl examples. Because if we look over in that region where there should be positive curl, that's where x is equal to three and y is equal to zero. So I go over here and say if x is equal to three and y is equal to zero, this whole formula becomes, let's see, three times three squared. So three times three squared minus nine, minus nine, and then minus the quantity, now we're plugging in y here, so that's three times y squared is just zero, because y is equal to zero, minus nine, minus nine. And so this part is 27, that's three times nine is 27, minus nine gives us 18."}, {"video_title": "2d curl example.mp3", "Sentence": "Because if we look over in that region where there should be positive curl, that's where x is equal to three and y is equal to zero. So I go over here and say if x is equal to three and y is equal to zero, this whole formula becomes, let's see, three times three squared. So three times three squared minus nine, minus nine, and then minus the quantity, now we're plugging in y here, so that's three times y squared is just zero, because y is equal to zero, minus nine, minus nine. And so this part is 27, that's three times nine is 27, minus nine gives us 18. And then we're subtracting off a negative nine, so that's actually plus nine. So this whole thing is 27, it's actually quite positive. So this is a positive number, and that's why when we go over here and we're looking at the fluid flow, you have a counterclockwise rotation in that region."}, {"video_title": "2d curl example.mp3", "Sentence": "And so this part is 27, that's three times nine is 27, minus nine gives us 18. And then we're subtracting off a negative nine, so that's actually plus nine. So this whole thing is 27, it's actually quite positive. So this is a positive number, and that's why when we go over here and we're looking at the fluid flow, you have a counterclockwise rotation in that region. Whereas, let's say that we did all of this, but instead of x equals three and y equals zero, we looked at x is equal to zero and y is equal to three. So in that case, we would instead, so x equals zero, y equals three, let's take a look at where that is, x is zero, and then y, the tick marks here are each one half, so y equals three is right here, it's in that clockwise rotation area. So if I kind of play this, we've got the clockwise rotation, we're expecting a negative value."}, {"video_title": "2d curl example.mp3", "Sentence": "So this is a positive number, and that's why when we go over here and we're looking at the fluid flow, you have a counterclockwise rotation in that region. Whereas, let's say that we did all of this, but instead of x equals three and y equals zero, we looked at x is equal to zero and y is equal to three. So in that case, we would instead, so x equals zero, y equals three, let's take a look at where that is, x is zero, and then y, the tick marks here are each one half, so y equals three is right here, it's in that clockwise rotation area. So if I kind of play this, we've got the clockwise rotation, we're expecting a negative value. And let's see if that's what we get. We go over here, and I'm gonna evaluate this whole function again, but plugging in zero for x, so this is three times zero, times zero, minus nine, and then we're subtracting off three times y squared, so that's three times three squared, three squared, minus nine, and this whole part is zero minus nine, so that becomes negative nine, and over here, we're subtracting off nine, 27 minus nine, which is 18, so we're subtracting off 18. So the whole thing equals negative 27."}, {"video_title": "2d curl example.mp3", "Sentence": "So if I kind of play this, we've got the clockwise rotation, we're expecting a negative value. And let's see if that's what we get. We go over here, and I'm gonna evaluate this whole function again, but plugging in zero for x, so this is three times zero, times zero, minus nine, and then we're subtracting off three times y squared, so that's three times three squared, three squared, minus nine, and this whole part is zero minus nine, so that becomes negative nine, and over here, we're subtracting off nine, 27 minus nine, which is 18, so we're subtracting off 18. So the whole thing equals negative 27. So maybe I should say equals that, that equals negative 27. So because this is negative, that's what corresponds to the clockwise rotation that we have going on in that region, and if you went and you plugged in a bunch of different points, like you can perhaps see how if you plug in zero for x and zero for y, those nines cancel out, which is why over here, there's no general rotation around the origin when x and y are both equal to zero. And you can understand that every single point and the general rotation around every single point just by taking this formula that we found for 2D curl and plugging in the corresponding values of x and y."}, {"video_title": "The Lagrangian.mp3", "Sentence": "Now we've talked about Lagrange multipliers, this is a highly related concept. In fact, it's not really teaching anything new, this is just repackaging stuff that we already know. So to remind you of the setup, this is gonna be a constrained optimization problem setup. So we'll have some kind of multivariable function, f of x, y, and the one I have pictured here is, let's see, it's x squared times e to the y times y. So what I have shown here is a contour line for this function. So that is, we say what happens if we set this equal to some constant, and we ask about all values of x and y such that this holds, such that this function outputs that constant. And if I choose a different constant, then that contour line could look a little bit different."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So we'll have some kind of multivariable function, f of x, y, and the one I have pictured here is, let's see, it's x squared times e to the y times y. So what I have shown here is a contour line for this function. So that is, we say what happens if we set this equal to some constant, and we ask about all values of x and y such that this holds, such that this function outputs that constant. And if I choose a different constant, then that contour line could look a little bit different. It's kinda nice that it has similar shapes. So that's the function, and we're trying to maximize it. The goal is to maximize this guy."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And if I choose a different constant, then that contour line could look a little bit different. It's kinda nice that it has similar shapes. So that's the function, and we're trying to maximize it. The goal is to maximize this guy. And of course, it's not just that. The reason we call it a constrained optimization problem is because there's some kind of constraint, some kind of other function, g of x, y. In this case, x squared plus y squared."}, {"video_title": "The Lagrangian.mp3", "Sentence": "The goal is to maximize this guy. And of course, it's not just that. The reason we call it a constrained optimization problem is because there's some kind of constraint, some kind of other function, g of x, y. In this case, x squared plus y squared. And we want to say that this has to equal some specific amount. In this case, I'm gonna set it equal to four. So we say you can't look at any x, y to maximize this function."}, {"video_title": "The Lagrangian.mp3", "Sentence": "In this case, x squared plus y squared. And we want to say that this has to equal some specific amount. In this case, I'm gonna set it equal to four. So we say you can't look at any x, y to maximize this function. You are limited to the values of x and y that satisfy this property. And I talked about this in the last couple videos, and kind of the cool thing that we found was that you look through the various different contour lines of f, and the maximum will be achieved when that contour line is just perfectly parallel to this contour of g. And you know, a pretty classic example for what these sorts of things could mean or how it's used in practice is if this was, say, a revenue function for some kind of company. You're kind of modeling your revenues based on different choices you could make for running that company."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So we say you can't look at any x, y to maximize this function. You are limited to the values of x and y that satisfy this property. And I talked about this in the last couple videos, and kind of the cool thing that we found was that you look through the various different contour lines of f, and the maximum will be achieved when that contour line is just perfectly parallel to this contour of g. And you know, a pretty classic example for what these sorts of things could mean or how it's used in practice is if this was, say, a revenue function for some kind of company. You're kind of modeling your revenues based on different choices you could make for running that company. And the constraint that you'd have would be, let's say, a budget. So I'm just gonna go ahead and write budget, or b for budget here. So you're trying to maximize revenues, and then you have some sort of dollar limit for what you're willing to spend."}, {"video_title": "The Lagrangian.mp3", "Sentence": "You're kind of modeling your revenues based on different choices you could make for running that company. And the constraint that you'd have would be, let's say, a budget. So I'm just gonna go ahead and write budget, or b for budget here. So you're trying to maximize revenues, and then you have some sort of dollar limit for what you're willing to spend. And these, of course, are just kind of made-up functions. You'd never have a budget that looks like a circle in this kind of random configuration for your revenue. But in principle, you know what I mean, right?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "So you're trying to maximize revenues, and then you have some sort of dollar limit for what you're willing to spend. And these, of course, are just kind of made-up functions. You'd never have a budget that looks like a circle in this kind of random configuration for your revenue. But in principle, you know what I mean, right? So the way that we took advantage of this tangency property, and I think this is pretty clever. Let me just kind of redraw it over here. I'm looking at the point where the two functions are just tangent to each other, is that the gradient, the gradient vector for the thing we're maximizing, which in this case is r, is gonna be parallel, or proportional, to the gradient vector of the constraint, which in this case is b."}, {"video_title": "The Lagrangian.mp3", "Sentence": "But in principle, you know what I mean, right? So the way that we took advantage of this tangency property, and I think this is pretty clever. Let me just kind of redraw it over here. I'm looking at the point where the two functions are just tangent to each other, is that the gradient, the gradient vector for the thing we're maximizing, which in this case is r, is gonna be parallel, or proportional, to the gradient vector of the constraint, which in this case is b. It's gonna be proportional to the gradient of the constraint. And what this means is that if we were going to solve a set of equations, what you set up is you compute that gradient of r, which will involve two different partial derivatives, and you set it equal, not to the gradient of b, because it's not necessarily equal to the gradient of b, but it's proportional, with some kind of proportionality constant lambda. Now let me, that's kind of a squirrely lambda."}, {"video_title": "The Lagrangian.mp3", "Sentence": "I'm looking at the point where the two functions are just tangent to each other, is that the gradient, the gradient vector for the thing we're maximizing, which in this case is r, is gonna be parallel, or proportional, to the gradient vector of the constraint, which in this case is b. It's gonna be proportional to the gradient of the constraint. And what this means is that if we were going to solve a set of equations, what you set up is you compute that gradient of r, which will involve two different partial derivatives, and you set it equal, not to the gradient of b, because it's not necessarily equal to the gradient of b, but it's proportional, with some kind of proportionality constant lambda. Now let me, that's kind of a squirrely lambda. Lambda. That one doesn't look good either, does it? Why are lambdas so hard to draw?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "Now let me, that's kind of a squirrely lambda. Lambda. That one doesn't look good either, does it? Why are lambdas so hard to draw? All right, that looks fine. So the gradient of the revenue is proportional to the gradient of the budget. And we did a couple examples of solving this kind of thing."}, {"video_title": "The Lagrangian.mp3", "Sentence": "Why are lambdas so hard to draw? All right, that looks fine. So the gradient of the revenue is proportional to the gradient of the budget. And we did a couple examples of solving this kind of thing. This gives you two separate equations from the two partial derivatives, and then you use this right here, this budget constraint, as your third equation. And the Lagrangian, the point of this video, this Lagrangian function, is basically just a way to package up this equation along with this equation into a single entity. So it's not really adding new information, and if you're solving things by hand, it doesn't really do anything for you."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And we did a couple examples of solving this kind of thing. This gives you two separate equations from the two partial derivatives, and then you use this right here, this budget constraint, as your third equation. And the Lagrangian, the point of this video, this Lagrangian function, is basically just a way to package up this equation along with this equation into a single entity. So it's not really adding new information, and if you're solving things by hand, it doesn't really do anything for you. But what makes it nice is that it's something easier to hand a computer, and I'll show you what I mean. So I'm gonna define the Lagrangian itself, which we write with this kind of funky-looking script L. And it's a function with the same inputs that your revenue function, or the thing that you're maximizing has, along with lambda, along with that Lagrange multiplier. And the way that we define it, and I'm gonna need some extra room, so I'm gonna say it's equal to, and kind of define it down here, the revenue function, or whatever it is that you're maximizing, the function that you're maximizing, minus lambda, that Lagrange multiplier, so that's just another input to this new function that we're defining, multiplied by the constraint function, in this case, b, evaluated at x, y, minus whatever that constraint value is."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So it's not really adding new information, and if you're solving things by hand, it doesn't really do anything for you. But what makes it nice is that it's something easier to hand a computer, and I'll show you what I mean. So I'm gonna define the Lagrangian itself, which we write with this kind of funky-looking script L. And it's a function with the same inputs that your revenue function, or the thing that you're maximizing has, along with lambda, along with that Lagrange multiplier. And the way that we define it, and I'm gonna need some extra room, so I'm gonna say it's equal to, and kind of define it down here, the revenue function, or whatever it is that you're maximizing, the function that you're maximizing, minus lambda, that Lagrange multiplier, so that's just another input to this new function that we're defining, multiplied by the constraint function, in this case, b, evaluated at x, y, minus whatever that constraint value is. In this case, I put in four, so you'd write minus four. If we wanted to be more general, maybe we would write b for whatever your budget is. So over here, you're subtracting off little b."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And the way that we define it, and I'm gonna need some extra room, so I'm gonna say it's equal to, and kind of define it down here, the revenue function, or whatever it is that you're maximizing, the function that you're maximizing, minus lambda, that Lagrange multiplier, so that's just another input to this new function that we're defining, multiplied by the constraint function, in this case, b, evaluated at x, y, minus whatever that constraint value is. In this case, I put in four, so you'd write minus four. If we wanted to be more general, maybe we would write b for whatever your budget is. So over here, you're subtracting off little b. So this here is a new multivariable function, right? It's something where you could input x, y, and lambda, and just kind of plug it all in, and you'd get some kind of value. And remember, b, in this case, is a constant, so I'll go ahead and write that."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So over here, you're subtracting off little b. So this here is a new multivariable function, right? It's something where you could input x, y, and lambda, and just kind of plug it all in, and you'd get some kind of value. And remember, b, in this case, is a constant, so I'll go ahead and write that. This right here is not considered a variable, this is just some constant. Your variables are x, y, and lambda. And this would seem like a totally weird and random thing to do, if you just saw it out of context, or if it was unmotivated, but what's kind of neat, and we'll go ahead and work through this right now, is that when you take this, is that when you take the gradient of this function, called the Lagrangian, and you set it equal to zero, that's gonna encapsulate all three equations that you need."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And remember, b, in this case, is a constant, so I'll go ahead and write that. This right here is not considered a variable, this is just some constant. Your variables are x, y, and lambda. And this would seem like a totally weird and random thing to do, if you just saw it out of context, or if it was unmotivated, but what's kind of neat, and we'll go ahead and work through this right now, is that when you take this, is that when you take the gradient of this function, called the Lagrangian, and you set it equal to zero, that's gonna encapsulate all three equations that you need. And I'll show you what I mean by that. So let's just remember the gradient of L, that's a vector, it's got three different components, since L has three different inputs. You're gonna have the partial derivative of L with respect to x, you're gonna have the partial derivative of L with respect to y, and then finally, the partial derivative of L with respect to lambda, our Lagrange multiplier, which we're considering an input to this function."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And this would seem like a totally weird and random thing to do, if you just saw it out of context, or if it was unmotivated, but what's kind of neat, and we'll go ahead and work through this right now, is that when you take this, is that when you take the gradient of this function, called the Lagrangian, and you set it equal to zero, that's gonna encapsulate all three equations that you need. And I'll show you what I mean by that. So let's just remember the gradient of L, that's a vector, it's got three different components, since L has three different inputs. You're gonna have the partial derivative of L with respect to x, you're gonna have the partial derivative of L with respect to y, and then finally, the partial derivative of L with respect to lambda, our Lagrange multiplier, which we're considering an input to this function. And remember, whenever we write that a vector equals zero, really we mean the zero vector. Often you'll see it in bold, if it's in a textbook, but what we're really saying is, we set those three different functions, those three different partial derivatives, all equal to zero. So this is just a nice, closed-form, compact way of saying, set all of its partial derivatives equal to zero."}, {"video_title": "The Lagrangian.mp3", "Sentence": "You're gonna have the partial derivative of L with respect to x, you're gonna have the partial derivative of L with respect to y, and then finally, the partial derivative of L with respect to lambda, our Lagrange multiplier, which we're considering an input to this function. And remember, whenever we write that a vector equals zero, really we mean the zero vector. Often you'll see it in bold, if it's in a textbook, but what we're really saying is, we set those three different functions, those three different partial derivatives, all equal to zero. So this is just a nice, closed-form, compact way of saying, set all of its partial derivatives equal to zero. And let's go ahead and think about what those partial derivatives actually are. So this first one, the partial with respect to x, partial derivative of the Lagrangian with respect to x. It's kind of fun, you know, you have all these curly symbols, the curly D, the curly L, it makes it look like you're doing some truly advanced math."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So this is just a nice, closed-form, compact way of saying, set all of its partial derivatives equal to zero. And let's go ahead and think about what those partial derivatives actually are. So this first one, the partial with respect to x, partial derivative of the Lagrangian with respect to x. It's kind of fun, you know, you have all these curly symbols, the curly D, the curly L, it makes it look like you're doing some truly advanced math. But really, it's just kind of artificial fanciness, right? But anyway, so we take the partial derivative with respect to x, and what that equals is, well, it's whatever the partial derivative of R with respect to x is, minus, and then lambda, from x's perspective, lambda just looks like a constant. So it's gonna be lambda."}, {"video_title": "The Lagrangian.mp3", "Sentence": "It's kind of fun, you know, you have all these curly symbols, the curly D, the curly L, it makes it look like you're doing some truly advanced math. But really, it's just kind of artificial fanciness, right? But anyway, so we take the partial derivative with respect to x, and what that equals is, well, it's whatever the partial derivative of R with respect to x is, minus, and then lambda, from x's perspective, lambda just looks like a constant. So it's gonna be lambda. And then this inside the parentheses, the partial derivative of that with respect to x, well, it's gonna be whatever the partial derivative of B is with respect to x. But subtracting off that constant, that doesn't change the derivative. So this right here is the partial derivative of lambda with respect to x."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So it's gonna be lambda. And then this inside the parentheses, the partial derivative of that with respect to x, well, it's gonna be whatever the partial derivative of B is with respect to x. But subtracting off that constant, that doesn't change the derivative. So this right here is the partial derivative of lambda with respect to x. Now, if you set that equal to zero, and I know I've kind of run out of room on the right here, but if you set that equal to zero, that's the same as just saying that the partial derivative of R with respect to x equals lambda times the partial derivative of B with respect to x. And if you think about what's gonna happen when you unfold this property, that the gradient of R is proportional to the gradient of B, written up here, that's just the first portion of this, right? If we're setting the gradients equal, then the first component of that is to say that the partial derivative of R with respect to x is equal to lambda times the partial derivative of B with respect to x."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So this right here is the partial derivative of lambda with respect to x. Now, if you set that equal to zero, and I know I've kind of run out of room on the right here, but if you set that equal to zero, that's the same as just saying that the partial derivative of R with respect to x equals lambda times the partial derivative of B with respect to x. And if you think about what's gonna happen when you unfold this property, that the gradient of R is proportional to the gradient of B, written up here, that's just the first portion of this, right? If we're setting the gradients equal, then the first component of that is to say that the partial derivative of R with respect to x is equal to lambda times the partial derivative of B with respect to x. And then if you do this for y, if we take the partial derivative of this Lagrangian function with respect to y, it's very similar, right? It's gonna be, well, you just take the partial derivative of R with respect to y. In fact, it all looks just identical."}, {"video_title": "The Lagrangian.mp3", "Sentence": "If we're setting the gradients equal, then the first component of that is to say that the partial derivative of R with respect to x is equal to lambda times the partial derivative of B with respect to x. And then if you do this for y, if we take the partial derivative of this Lagrangian function with respect to y, it's very similar, right? It's gonna be, well, you just take the partial derivative of R with respect to y. In fact, it all looks just identical. Whatever R is, you take its partial derivative with respect to y, and then we subtract off. Lambda looks like a constant as far as y is concerned. And then that's multiplied by, well, what's the partial derivative of this term inside the parentheses with respect to y?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "In fact, it all looks just identical. Whatever R is, you take its partial derivative with respect to y, and then we subtract off. Lambda looks like a constant as far as y is concerned. And then that's multiplied by, well, what's the partial derivative of this term inside the parentheses with respect to y? Well, it's the partial of B with respect to y. And again, again, if you imagine setting that equal to zero, that's gonna be the same as setting this partial derivative term equal to lambda times this partial derivative term, right? You kind of just bring one to the other side."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And then that's multiplied by, well, what's the partial derivative of this term inside the parentheses with respect to y? Well, it's the partial of B with respect to y. And again, again, if you imagine setting that equal to zero, that's gonna be the same as setting this partial derivative term equal to lambda times this partial derivative term, right? You kind of just bring one to the other side. So this second component of our Lagrangian equals zero equation is just the second function that we've seen in a lot of these examples that we've been doing, where you set one of the gradient vectors proportional to the other one. And the only real difference here from stuff that we've seen already, and even then it's not that different, is that what happens when we take the partial derivative of this Lagrangian with respect to lambda, with respect, now I'll go ahead and give it that kind of green lambda color here. Well, when we take that partial derivative, if we kind of look up at the definition of the function, R, R never has a lambda in it, right?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "You kind of just bring one to the other side. So this second component of our Lagrangian equals zero equation is just the second function that we've seen in a lot of these examples that we've been doing, where you set one of the gradient vectors proportional to the other one. And the only real difference here from stuff that we've seen already, and even then it's not that different, is that what happens when we take the partial derivative of this Lagrangian with respect to lambda, with respect, now I'll go ahead and give it that kind of green lambda color here. Well, when we take that partial derivative, if we kind of look up at the definition of the function, R, R never has a lambda in it, right? It's purely a function of x and y. So that looks just like a constant when we're differentiating with respect to lambda. So that's just gonna be zero when we take its partial derivative."}, {"video_title": "The Lagrangian.mp3", "Sentence": "Well, when we take that partial derivative, if we kind of look up at the definition of the function, R, R never has a lambda in it, right? It's purely a function of x and y. So that looks just like a constant when we're differentiating with respect to lambda. So that's just gonna be zero when we take its partial derivative. And then this next component, b of x, y minus b, all of that just looks like a constant as far as lambda is concerned, right? There's x's, there's y, there's this constant b, but none of these things have lambdas in them. So when we take the partial derivative with respect to lambda, this just looks like some big constant times lambda itself."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So that's just gonna be zero when we take its partial derivative. And then this next component, b of x, y minus b, all of that just looks like a constant as far as lambda is concerned, right? There's x's, there's y, there's this constant b, but none of these things have lambdas in them. So when we take the partial derivative with respect to lambda, this just looks like some big constant times lambda itself. So what we're gonna get is, I guess, we're subtracting off, right? We're up here kind of writing a minus sign. We're subtracting off all the stuff that was in those parentheses, b of x, y minus b, that constant."}, {"video_title": "The Lagrangian.mp3", "Sentence": "So when we take the partial derivative with respect to lambda, this just looks like some big constant times lambda itself. So what we're gonna get is, I guess, we're subtracting off, right? We're up here kind of writing a minus sign. We're subtracting off all the stuff that was in those parentheses, b of x, y minus b, that constant. And this whole thing, if we set that whole thing equal to zero, well, that's pretty much the same as setting b of x, y minus b equal to zero. And that, that's really just the same as saying, hey, we're setting b of x, y equal to that little b, right? Setting this partial derivative of the Lagrangian with respect to the Lagrange multiplier equal to zero boils down to the constraint, right?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "We're subtracting off all the stuff that was in those parentheses, b of x, y minus b, that constant. And this whole thing, if we set that whole thing equal to zero, well, that's pretty much the same as setting b of x, y minus b equal to zero. And that, that's really just the same as saying, hey, we're setting b of x, y equal to that little b, right? Setting this partial derivative of the Lagrangian with respect to the Lagrange multiplier equal to zero boils down to the constraint, right? The third equation that we need to solve. So in that way, setting the gradient of this Lagrangian function equal to zero is just a very compact way of packaging three separate equations that we need to solve the constraint optimization problem. And I'll emphasize that in practice, if you actually see a function for r for the thing that you're maximizing and a function for the budget, it's much better, I think, to just directly think about these parallel gradients and kind of solve it from there."}, {"video_title": "The Lagrangian.mp3", "Sentence": "Setting this partial derivative of the Lagrangian with respect to the Lagrange multiplier equal to zero boils down to the constraint, right? The third equation that we need to solve. So in that way, setting the gradient of this Lagrangian function equal to zero is just a very compact way of packaging three separate equations that we need to solve the constraint optimization problem. And I'll emphasize that in practice, if you actually see a function for r for the thing that you're maximizing and a function for the budget, it's much better, I think, to just directly think about these parallel gradients and kind of solve it from there. Because if you construct the Lagrangian and then compute its gradient, all you're really doing is repackaging it up only to unpackage it again. But the point of this, kind of the reason that this is a very useful construct is that computers often have really fast ways of solving things like this, things like the gradient of some function equals zero. And the reason is because that's how you solve unconstrained maximization problems, right?"}, {"video_title": "The Lagrangian.mp3", "Sentence": "And I'll emphasize that in practice, if you actually see a function for r for the thing that you're maximizing and a function for the budget, it's much better, I think, to just directly think about these parallel gradients and kind of solve it from there. Because if you construct the Lagrangian and then compute its gradient, all you're really doing is repackaging it up only to unpackage it again. But the point of this, kind of the reason that this is a very useful construct is that computers often have really fast ways of solving things like this, things like the gradient of some function equals zero. And the reason is because that's how you solve unconstrained maximization problems, right? This is very similar to as if we just looked at this function L out of context and were asked, hey, what is its maximum value? What are the critical points that it has? And you said its gradient equal to zero."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And the reason is because that's how you solve unconstrained maximization problems, right? This is very similar to as if we just looked at this function L out of context and were asked, hey, what is its maximum value? What are the critical points that it has? And you said its gradient equal to zero. So kind of the whole point of this Lagrangian is that it turns our constrained optimization problem involving R and B and this new made-up variable lambda into an unconstrained optimization problem where we're just setting the gradient of some function equal to zero. So computers can often do that really quickly. So if you just hand the computer this function, it will be able to find you an answer."}, {"video_title": "The Lagrangian.mp3", "Sentence": "And you said its gradient equal to zero. So kind of the whole point of this Lagrangian is that it turns our constrained optimization problem involving R and B and this new made-up variable lambda into an unconstrained optimization problem where we're just setting the gradient of some function equal to zero. So computers can often do that really quickly. So if you just hand the computer this function, it will be able to find you an answer. Whereas it's harder to say, hey, computer, I want you to think about when the gradients are parallel and also consider this constraint function. It's just kind of a cleaner way to package it all up. So with that, I'll see you next video where I'm gonna talk about the significance of this lambda term, how it's not just a ghost variable, but it actually has a pretty nice interpretation for a given constraint problem."}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "This is on the XY plane. TS plane should just be some separate space over here, and we're imagining moving that separate space over into three dimensions, but that's harder to animate, so I'm just not gonna do it, and I'm gonna instead keep things inside the XY plane here, and we're thinking about the squares being T and S ranging each from zero to three, and what I said for partial derivative with respect to T is you imagine the line that represents movement in the T direction, and you see how that line gets mapped as all of the points move to their corresponding output, and the partial derivative vector gives you a certain tangent vector to the curve representing that line which corresponds to movement in the T direction, and the longer that is, the faster the movement, the more sensitive it is to nudges in the T direction, so in the S direction, let's say we were to take the partial derivative with respect to S, so I'll kind of clear this up here, also clear up this guy, and if you said instead, what if we were doing it with respect to S, right? Partial derivative of V, the vector value function with respect to S, and well, you do something very similar. You would say, okay, what is the line that corresponds to movement in the S direction, and the way I've drawn it, it's always gonna be perpendicular because we're in the TS plane, the T axis is perpendicular to the S plane, and in this case, this line represents T equals one, right? You're saying T constantly equals one, but you're letting S vary, and if you see how that line maps as you move everything from the input space over to the corresponding points in the output space, that line tells you what happens as you're varying the S value of the input, and I guess it kind of starts curving this way, and then it curves very much up and kind of goes off into the distance there, and again, the grid lines here really help because every time that you see the grid lines intersect, one of the lines represents movement in the T direction and the other represents movement in the S direction, and for partial derivatives, we think very similarly. You think of that partial S as representing, whoop, whoop, zoom on back here, that partial S you think of as representing a tiny movement in the S direction, just a little smidge and nudge, somehow nudging that guy along, and then the corresponding nudge you look for in the output space, you say, okay, if we nudge the input that much and we go over to the output, and maybe that tiny nudge correspond with one that's like three times bigger, I don't know, but it looked like it stretched things out, so that tiny nudge might turn into something that's still quite small, but maybe three times bigger, but it's a vector. What you do is you think of that vector as being your partial V, and you scale it by whatever the size of that partial S was, right?"}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "You would say, okay, what is the line that corresponds to movement in the S direction, and the way I've drawn it, it's always gonna be perpendicular because we're in the TS plane, the T axis is perpendicular to the S plane, and in this case, this line represents T equals one, right? You're saying T constantly equals one, but you're letting S vary, and if you see how that line maps as you move everything from the input space over to the corresponding points in the output space, that line tells you what happens as you're varying the S value of the input, and I guess it kind of starts curving this way, and then it curves very much up and kind of goes off into the distance there, and again, the grid lines here really help because every time that you see the grid lines intersect, one of the lines represents movement in the T direction and the other represents movement in the S direction, and for partial derivatives, we think very similarly. You think of that partial S as representing, whoop, whoop, zoom on back here, that partial S you think of as representing a tiny movement in the S direction, just a little smidge and nudge, somehow nudging that guy along, and then the corresponding nudge you look for in the output space, you say, okay, if we nudge the input that much and we go over to the output, and maybe that tiny nudge correspond with one that's like three times bigger, I don't know, but it looked like it stretched things out, so that tiny nudge might turn into something that's still quite small, but maybe three times bigger, but it's a vector. What you do is you think of that vector as being your partial V, and you scale it by whatever the size of that partial S was, right? So the result that you get is a tangent vector that's not puny, not a tiny nudge, but is actually a sizable tangent vector, and it's gonna kind of correspond to the rate at which tiny changes, not just tiny changes, but the rate at which changes in S cause movement in the output space. So let's actually compute it for this case, just kind of get some good practice computing things, and if we look up here, the T value, which used to be considered a variable when we were doing it with respect to T, but now that T value looks like a constant, so its derivative is zero, then negative S squared with respect to S has a derivative of negative two S. S T, S looks like a variable, T looks like a constant, the derivative is just that constant, T. Down here, T S squared, T looks like a constant, S looks like a variable, so two times T times S, and then over here, we're subtracting off, S is the variable, T squared looks like a constant, so that constant. And let's say we plug in the value, one, one, right?"}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "What you do is you think of that vector as being your partial V, and you scale it by whatever the size of that partial S was, right? So the result that you get is a tangent vector that's not puny, not a tiny nudge, but is actually a sizable tangent vector, and it's gonna kind of correspond to the rate at which tiny changes, not just tiny changes, but the rate at which changes in S cause movement in the output space. So let's actually compute it for this case, just kind of get some good practice computing things, and if we look up here, the T value, which used to be considered a variable when we were doing it with respect to T, but now that T value looks like a constant, so its derivative is zero, then negative S squared with respect to S has a derivative of negative two S. S T, S looks like a variable, T looks like a constant, the derivative is just that constant, T. Down here, T S squared, T looks like a constant, S looks like a variable, so two times T times S, and then over here, we're subtracting off, S is the variable, T squared looks like a constant, so that constant. And let's say we plug in the value, one, one, right? This red dot corresponds to one, one, so what we would get here, S is equal to one, so that's negative two, T is equal to one, so that's one, then two times one times one, oh, let's see, I'll write it, two times one times one minus one squared, minus one squared is gonna correspond to one, that's two minus one. So what we would expect for the tangent vector, the partial derivative vector, is the X component should be negative, and then the Y and Z components should each be positive. And if we go over and we take a look at what the movement along the curve actually is, that lines up, right?"}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "And let's say we plug in the value, one, one, right? This red dot corresponds to one, one, so what we would get here, S is equal to one, so that's negative two, T is equal to one, so that's one, then two times one times one, oh, let's see, I'll write it, two times one times one minus one squared, minus one squared is gonna correspond to one, that's two minus one. So what we would expect for the tangent vector, the partial derivative vector, is the X component should be negative, and then the Y and Z components should each be positive. And if we go over and we take a look at what the movement along the curve actually is, that lines up, right? Because you're moving, as you kind of zip along this curve, you're moving to the left, so the X component of the partial derivative should be negative, but you're moving upwards as far as Y is concerned, and you can also kind of see that the leftward movement is kind of twice as fast as the upward motion, the slope favors the X direction, and then as far as the Z component is concerned, you are in fact moving up. And maybe you could say, well, how do you know what you're moving? Are you moving that way, or is everything switched the other way around?"}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "And if we go over and we take a look at what the movement along the curve actually is, that lines up, right? Because you're moving, as you kind of zip along this curve, you're moving to the left, so the X component of the partial derivative should be negative, but you're moving upwards as far as Y is concerned, and you can also kind of see that the leftward movement is kind of twice as fast as the upward motion, the slope favors the X direction, and then as far as the Z component is concerned, you are in fact moving up. And maybe you could say, well, how do you know what you're moving? Are you moving that way, or is everything switched the other way around? And the benefit of animation here is we can say, ah, as S is ranging from zero up to three, this is the increasing direction, and you just keep your eye on what that direction is as we move things about, and that increasing direction does kind of correspond with moving along the curve this way. So you get a tangent vector in the other way. And one kind of nice thing about this then is the two different partial derivative vectors that we found, each one of them you could say is a tangent vector to the surface, right?"}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "Are you moving that way, or is everything switched the other way around? And the benefit of animation here is we can say, ah, as S is ranging from zero up to three, this is the increasing direction, and you just keep your eye on what that direction is as we move things about, and that increasing direction does kind of correspond with moving along the curve this way. So you get a tangent vector in the other way. And one kind of nice thing about this then is the two different partial derivative vectors that we found, each one of them you could say is a tangent vector to the surface, right? So the one that was a partial derivative with respect to T over here, is kind of goes in one direction, and the other one, it kind of gives you a different notion of what a tangent vector on the surface could be. And various different, you could have a notion of directional derivative too that kind of combines these in various ways, and that'll get you all the different ways that you can have a vector tangent to the surface. And later on, I'll talk about things like tangent planes, if you wanna express what a tangent plane is, and you kind of think of that as being defined in terms of two different vectors."}, {"video_title": "Partial derivative of a parametric surface, part 2.mp3", "Sentence": "And one kind of nice thing about this then is the two different partial derivative vectors that we found, each one of them you could say is a tangent vector to the surface, right? So the one that was a partial derivative with respect to T over here, is kind of goes in one direction, and the other one, it kind of gives you a different notion of what a tangent vector on the surface could be. And various different, you could have a notion of directional derivative too that kind of combines these in various ways, and that'll get you all the different ways that you can have a vector tangent to the surface. And later on, I'll talk about things like tangent planes, if you wanna express what a tangent plane is, and you kind of think of that as being defined in terms of two different vectors. But for now, that's really all you need to know about partial derivatives of parametric surfaces. And in the next couple videos, I'll talk about what partial derivatives of vector-valued functions can mean in other contexts, because it's not always a parametric surface, and maybe you're not always thinking about a curve that could be moved along, but you still wanna think, you know, how does this input nudge correspond to an output nudge, and what's the ratio between them? So with that, I'll see you next video."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we found a decent parameterization. And then given that parameterization, we were able to come up with ds for that surface, for surface 2, just simplified. All this business simplified to 1, so it just equaled du dv. And so now we are ready to evaluate the surface integral. This surface integral right over here, we're ready to evaluate the surface integral over surface 2 of z ds. And it's going to be equal to a double integral over u and v. And so let's write this. I'm going to do two different colors for the different variables of integration."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so now we are ready to evaluate the surface integral. This surface integral right over here, we're ready to evaluate the surface integral over surface 2 of z ds. And it's going to be equal to a double integral over u and v. And so let's write this. I'm going to do two different colors for the different variables of integration. So yellow for one, and maybe purple for the other. And we're taking the integral of z. And in our parameterization, z is equal to v. So this right over here is the same thing as v. So we can write v right over there."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm going to do two different colors for the different variables of integration. So yellow for one, and maybe purple for the other. And we're taking the integral of z. And in our parameterization, z is equal to v. So this right over here is the same thing as v. So we can write v right over there. And we already saw that ds is the same thing as du dv. Or we could even write that as dv du. We could just switch the order right over there."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And in our parameterization, z is equal to v. So this right over here is the same thing as v. So we can write v right over there. And we already saw that ds is the same thing as du dv. Or we could even write that as dv du. We could just switch the order right over there. And I'm going to choose to integrate with respect to dv first, to do dv on the inside integral, and then do du on the outside integral. And the reason why I'm choosing to integrate with respect to v first is based on the bounds of our parameters. v is bounded on the low end by 0."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We could just switch the order right over there. And I'm going to choose to integrate with respect to dv first, to do dv on the inside integral, and then do du on the outside integral. And the reason why I'm choosing to integrate with respect to v first is based on the bounds of our parameters. v is bounded on the low end by 0. It's bounded on the low end by 0. But on the high end, it's bounded by essentially a function of u. It's upper bound changes."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "v is bounded on the low end by 0. It's bounded on the low end by 0. But on the high end, it's bounded by essentially a function of u. It's upper bound changes. Because you see right over here, depending where we are, depending on what our x value is, essentially we have a different height that we need to get to. And since it's a function of u, we can integrate with respect to v, our boundaries are going to be 0 and 1 minus cosine of u. All of this business in magenta will give us a function of u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's upper bound changes. Because you see right over here, depending where we are, depending on what our x value is, essentially we have a different height that we need to get to. And since it's a function of u, we can integrate with respect to v, our boundaries are going to be 0 and 1 minus cosine of u. All of this business in magenta will give us a function of u. And then we'll be able to integrate with respect to u. And u just goes from 0 to 2 pi. And so that will give us a nice, straightforward number, assuming all of this works out OK. And so this is simplified to a straight up double integral."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "All of this business in magenta will give us a function of u. And then we'll be able to integrate with respect to u. And u just goes from 0 to 2 pi. And so that will give us a nice, straightforward number, assuming all of this works out OK. And so this is simplified to a straight up double integral. And now we're ready to compute. And so let me write the outside part. The outside part is from 0 to 2 pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so that will give us a nice, straightforward number, assuming all of this works out OK. And so this is simplified to a straight up double integral. And now we're ready to compute. And so let me write the outside part. The outside part is from 0 to 2 pi. It's du. And so the inside part, the antiderivative of v, is v squared over 2. And we're going to evaluate that from 0 to 1 minus cosine of u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The outside part is from 0 to 2 pi. It's du. And so the inside part, the antiderivative of v, is v squared over 2. And we're going to evaluate that from 0 to 1 minus cosine of u. And so this is going to be equal to, once again, the outside integral 0, 2 pi. I'll write du. I'll write du right over here."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're going to evaluate that from 0 to 1 minus cosine of u. And so this is going to be equal to, once again, the outside integral 0, 2 pi. I'll write du. I'll write du right over here. And so this is going to be equal to all of this. Let me just write 1 half. And actually, I can even write the 1 half out here."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'll write du right over here. And so this is going to be equal to all of this. Let me just write 1 half. And actually, I can even write the 1 half out here. I'll just write 1 half times 1 minus cosine u squared. Well, that's just going to be 1 squared minus 2 times the product of both of these. So minus 2 cosine of u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And actually, I can even write the 1 half out here. I'll just write 1 half times 1 minus cosine u squared. Well, that's just going to be 1 squared minus 2 times the product of both of these. So minus 2 cosine of u. Actually, let me give myself a little bit more real estate here. Let me give myself a little bit more real estate. 1 minus 2 cosine of u plus cosine of u squared minus this thing evaluated at 0, which is just going to be 0."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So minus 2 cosine of u. Actually, let me give myself a little bit more real estate here. Let me give myself a little bit more real estate. 1 minus 2 cosine of u plus cosine of u squared minus this thing evaluated at 0, which is just going to be 0. So we just get that right over there. And then we have du. And so now we can evaluate this."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 minus 2 cosine of u plus cosine of u squared minus this thing evaluated at 0, which is just going to be 0. So we just get that right over there. And then we have du. And so now we can evaluate this. We can integrate this with respect to u. So let's do that. And let me just take the 1 half on the outside just to simplify things."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And so now we can evaluate this. We can integrate this with respect to u. So let's do that. And let me just take the 1 half on the outside just to simplify things. So we have the 1 half out here. And so if you take the antiderivative of this with respect to u, you still have this 1 half out front. So this is equal to 1 half."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let me just take the 1 half on the outside just to simplify things. So we have the 1 half out here. And so if you take the antiderivative of this with respect to u, you still have this 1 half out front. So this is equal to 1 half. And we're going to take the antiderivative. So let's do it carefully. Actually, let me just simplify it so it's easier to take the antiderivative."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is equal to 1 half. And we're going to take the antiderivative. So let's do it carefully. Actually, let me just simplify it so it's easier to take the antiderivative. So it's going to be 1 half times the integral. I'll break this up into three different integrals. 1 half times the integral from 0 to 2 pi of 1 du, which is just du, minus 2 times the integral from 0 to 2 pi of cosine of u du."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Actually, let me just simplify it so it's easier to take the antiderivative. So it's going to be 1 half times the integral. I'll break this up into three different integrals. 1 half times the integral from 0 to 2 pi of 1 du, which is just du, minus 2 times the integral from 0 to 2 pi of cosine of u du. That's this term right over here. Plus the integral from 0 to 2 pi of cosine squared u. And cosine squared u, it's not so easy to take the antiderivative of that."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "1 half times the integral from 0 to 2 pi of 1 du, which is just du, minus 2 times the integral from 0 to 2 pi of cosine of u du. That's this term right over here. Plus the integral from 0 to 2 pi of cosine squared u. And cosine squared u, it's not so easy to take the antiderivative of that. So we'll use one of our trig identities. I always forget the formal name. I just think of it as the one that takes us from cosine squared to cosine of 2u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And cosine squared u, it's not so easy to take the antiderivative of that. So we'll use one of our trig identities. I always forget the formal name. I just think of it as the one that takes us from cosine squared to cosine of 2u. So this trig identity, this thing right over here, is the same thing. This comes straight out of our trigonometry class. This is 1 half plus 1 half cosine of 2 theta."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I just think of it as the one that takes us from cosine squared to cosine of 2u. So this trig identity, this thing right over here, is the same thing. This comes straight out of our trigonometry class. This is 1 half plus 1 half cosine of 2 theta. Plus 1 half cosine of 2u. So this last integral right over here, I can rewrite it as 1 half plus 1 half cosine of 2u. And then we have our final du."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is 1 half plus 1 half cosine of 2 theta. Plus 1 half cosine of 2u. So this last integral right over here, I can rewrite it as 1 half plus 1 half cosine of 2u. And then we have our final du. And now let me close the brackets. And all of that is times 1 half. So this thing right over here, cosine squared u, just a trig identity, takes us to that."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then we have our final du. And now let me close the brackets. And all of that is times 1 half. So this thing right over here, cosine squared u, just a trig identity, takes us to that. Now this is pretty easy to evaluate. The antiderivative of this is just going to be u evaluated at 2 pi and 0. So it essentially just, let me just write it out."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this thing right over here, cosine squared u, just a trig identity, takes us to that. Now this is pretty easy to evaluate. The antiderivative of this is just going to be u evaluated at 2 pi and 0. So it essentially just, let me just write it out. This part right over here is just u evaluated from 0 to 2 pi. It's 2 pi minus 0. It just gets evaluated and we get 2 pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it essentially just, let me just write it out. This part right over here is just u evaluated from 0 to 2 pi. It's 2 pi minus 0. It just gets evaluated and we get 2 pi. So out front, you have your 1 half and then times 2 pi. And then this right over here, the antiderivative, this is going to be equal to minus 2 times the antiderivative of cosine of u. Well, that's just sine of u. Evaluated from 0 to 2 pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It just gets evaluated and we get 2 pi. So out front, you have your 1 half and then times 2 pi. And then this right over here, the antiderivative, this is going to be equal to minus 2 times the antiderivative of cosine of u. Well, that's just sine of u. Evaluated from 0 to 2 pi. Well, sine of 2 pi is 0. Sine of 0 is 0. So this whole thing is going to evaluate to 0."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, that's just sine of u. Evaluated from 0 to 2 pi. Well, sine of 2 pi is 0. Sine of 0 is 0. So this whole thing is going to evaluate to 0. So we could say minus 0. And then we take the antiderivative of this right over here. The antiderivative of this is going to be 1 half u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing is going to evaluate to 0. So we could say minus 0. And then we take the antiderivative of this right over here. The antiderivative of this is going to be 1 half u. And the antiderivative of 1 half cosine of 2u. Well, if we had a 2 out front here, then that would be the derivative of sine of 2u. But we don't have a 2 out here."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The antiderivative of this is going to be 1 half u. And the antiderivative of 1 half cosine of 2u. Well, if we had a 2 out front here, then that would be the derivative of sine of 2u. But we don't have a 2 out here. But we can add a 2. Let me actually write it this way. We can add a 2 right over here."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But we don't have a 2 out here. But we can add a 2. Let me actually write it this way. We can add a 2 right over here. We can add a 2 and then divide by a 2. We can add a 2, and that's a little bit too confusing. Let me make it very clear."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can add a 2 right over here. We can add a 2 and then divide by a 2. We can add a 2, and that's a little bit too confusing. Let me make it very clear. 1 half cosine of 2u is equal to 1 half times 1 half. Let me write it this way. Is equal to 1 fourth times 2 cosine 2u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me make it very clear. 1 half cosine of 2u is equal to 1 half times 1 half. Let me write it this way. Is equal to 1 fourth times 2 cosine 2u. These are the exact same quantities. And the reason why I wrote it this way is this is clearly the derivative of sine of 2u. So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Is equal to 1 fourth times 2 cosine 2u. These are the exact same quantities. And the reason why I wrote it this way is this is clearly the derivative of sine of 2u. So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u. And we're going to evaluate that from 0 to 2 pi. And you confirm for yourself. You take the derivative of this, you do the chain rule, you get the 2 out front."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So when you take the antiderivative of this, it's the same thing as plus 1 fourth sine of 2u. And we're going to evaluate that from 0 to 2 pi. And you confirm for yourself. You take the derivative of this, you do the chain rule, you get the 2 out front. 2 times 1 fourth gives you 1 half. And the derivative of sine of 2u with respect to 2u is cosine of 2u. Now we need to evaluate this at 2 pi and at 0."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You take the derivative of this, you do the chain rule, you get the 2 out front. 2 times 1 fourth gives you 1 half. And the derivative of sine of 2u with respect to 2u is cosine of 2u. Now we need to evaluate this at 2 pi and at 0. When you evaluate it at 2 pi, you get to have 1 half times 2 pi, which is pi. So this is plus pi. Plus 1 fourth times sine of 2 times 2 pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we need to evaluate this at 2 pi and at 0. When you evaluate it at 2 pi, you get to have 1 half times 2 pi, which is pi. So this is plus pi. Plus 1 fourth times sine of 2 times 2 pi. Sine of 4 pi, that's just going to be 0. So this is going to evaluate to 0. And then minus 1 half times 0."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Plus 1 fourth times sine of 2 times 2 pi. Sine of 4 pi, that's just going to be 0. So this is going to evaluate to 0. And then minus 1 half times 0. And then 1 fourth sine of 2 times 0. This is all going to be 0 when you put 0 there. So this whole thing evaluated just 2 pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then minus 1 half times 0. And then 1 fourth sine of 2 times 0. This is all going to be 0 when you put 0 there. So this whole thing evaluated just 2 pi. And we're in the home stretch, at least for surface 2. And I'll switch back to surface 2's color now, now that we're near the end. So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this whole thing evaluated just 2 pi. And we're in the home stretch, at least for surface 2. And I'll switch back to surface 2's color now, now that we're near the end. So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi. So it's 1 half times 3 pi, which is equal to, we can have our drum roll, our mini drum roll, since we're not really done the entire problem right now. But it's equal to 3 pi over 2. So we're making pretty good headway."}, {"video_title": "Surface integral ex3 part 2 Evaluating the outside surface   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So the surface integral for surface 2 is going to be 1 half times 2 pi minus 0 plus pi. So it's 1 half times 3 pi, which is equal to, we can have our drum roll, our mini drum roll, since we're not really done the entire problem right now. But it's equal to 3 pi over 2. So we're making pretty good headway. This was 0. And now this part right over here is 3 pi over 2. And in the next video, we will try to tackle surface 3."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "And I brought up in the last video this pretty complicated formula, and to remind you of the circumstances, I was saying that curvature, denoted with this little kappa, is typically calculated as the derivative of the unit tangent vector function. So we'll think of some kind of function that gives unit tangent vectors at every single point, so you know, just a vector with a length of 1 lying tangent to the curve, and that's going to be some kind of function of the same parameter, so big T for unit tangent vector, little t for the parameter, hopefully that's not too confusing. Then we're taking the derivative of this, not with respect to T, the parameter, but with respect to arc length, S. And by arc length, I mean if you take a tiny step along the curve, and consider this to be of size dS, so S typically denotes length along a curve, this will be a tiny change in length, you're wondering how much that unit tangent vector changes. And specifically what I mean by that, if you kind of imagine tangent vectors off sitting in their own space somewhere, where each one of them has unit length, they're all just kind of of the same length, and let's say they're all stemming from the same point, so rather than drawing them stemming from the curve to show that they're tangent, I just want to show them in their own space, the derivative, the change in that tangent vector, would be some other vector that tells you how you move from one to the other, kind of tells you how much it's turning. So the curvature itself is given, not by that vector, because this would be a vector quantity, but by the absolute value of it. And I said in the last video, that it turns out that this quantity, when you work it out for a vector valued function, with components, with components x of t, x of t and y of t, that it happens to equal this complicated formula. And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "And specifically what I mean by that, if you kind of imagine tangent vectors off sitting in their own space somewhere, where each one of them has unit length, they're all just kind of of the same length, and let's say they're all stemming from the same point, so rather than drawing them stemming from the curve to show that they're tangent, I just want to show them in their own space, the derivative, the change in that tangent vector, would be some other vector that tells you how you move from one to the other, kind of tells you how much it's turning. So the curvature itself is given, not by that vector, because this would be a vector quantity, but by the absolute value of it. And I said in the last video, that it turns out that this quantity, when you work it out for a vector valued function, with components, with components x of t, x of t and y of t, that it happens to equal this complicated formula. And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point. So let's start by looking at the numerator here. This x prime of t, first derivative of the first component, times the second derivative of the second component, minus, and then kind of symmetrically, y prime times x double prime. You might be able to recognize this as a certain cross product."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "And what I'm gonna do here, I'm gonna break it down, and show why this formula isn't actually all that random, it's actually kind of sensible as a description for how much the curve really curves at a point. So let's start by looking at the numerator here. This x prime of t, first derivative of the first component, times the second derivative of the second component, minus, and then kind of symmetrically, y prime times x double prime. You might be able to recognize this as a certain cross product. So the cross product, if we take x prime, y prime, and these are still functions of t, as a vector crossed, with the vector containing the double primes. And if you're unfamiliar with cross products, or you're feeling a little shaky, and you kind of want to review, now would be a good time to pause the video, take a look at the cross product videos, and remind yourself both of how to compute it, and what the underlying intuition is. Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "You might be able to recognize this as a certain cross product. So the cross product, if we take x prime, y prime, and these are still functions of t, as a vector crossed, with the vector containing the double primes. And if you're unfamiliar with cross products, or you're feeling a little shaky, and you kind of want to review, now would be a good time to pause the video, take a look at the cross product videos, and remind yourself both of how to compute it, and what the underlying intuition is. Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal. You know, it kind of feels like a determinant. X double prime times y prime. But the way that you interpret this vector, well I'll get to the interpretation in just a moment."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "Because the way that we compute a cross product like this, is you take the components in the diagonal here, you know that rightward diagonal, and multiply them together, that's where you get your x prime, y double prime, and then you subtract off the components in the other diagonal. You know, it kind of feels like a determinant. X double prime times y prime. But the way that you interpret this vector, well I'll get to the interpretation in just a moment. First let me kind of write out what this is, in terms of our function s. This first vector is just the first derivative of s. So that's s prime of t, the first derivative of s. And we're crossing that with this one, which is the second derivative, vector value derivative, but the second derivative of s. So before we do anything else, let's just start to think about what do both of these vectors mean? How do you interpret the s prime and the s double prime? I'm gonna go ahead and give ourselves a little bit of room here."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "But the way that you interpret this vector, well I'll get to the interpretation in just a moment. First let me kind of write out what this is, in terms of our function s. This first vector is just the first derivative of s. So that's s prime of t, the first derivative of s. And we're crossing that with this one, which is the second derivative, vector value derivative, but the second derivative of s. So before we do anything else, let's just start to think about what do both of these vectors mean? How do you interpret the s prime and the s double prime? I'm gonna go ahead and give ourselves a little bit of room here. So I'll draw the curve again. We're on our xy plane, and you have some kind of curve. The function s itself is giving vectors whose tips trace out this curve."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "I'm gonna go ahead and give ourselves a little bit of room here. So I'll draw the curve again. We're on our xy plane, and you have some kind of curve. The function s itself is giving vectors whose tips trace out this curve. As t changes, the tips of these vectors trace out the curve. And now the first derivative, that first derivative vector, s prime of t, is telling you how that tip should move to go along the curve. As you go from one s vector to another s vector, what direction should that tip move?"}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "The function s itself is giving vectors whose tips trace out this curve. As t changes, the tips of these vectors trace out the curve. And now the first derivative, that first derivative vector, s prime of t, is telling you how that tip should move to go along the curve. As you go from one s vector to another s vector, what direction should that tip move? And what this means is that at every given point, when you kind of do this in a limiting fashion, and you only look at infinitesimal changes in the original vector, you always get some kind of tangent vector. So all of these are tangent vectors, not necessarily unit. You might have a very long tangent vector to indicate that you're traveling very quickly across that space."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "As you go from one s vector to another s vector, what direction should that tip move? And what this means is that at every given point, when you kind of do this in a limiting fashion, and you only look at infinitesimal changes in the original vector, you always get some kind of tangent vector. So all of these are tangent vectors, not necessarily unit. You might have a very long tangent vector to indicate that you're traveling very quickly across that space. And now, how would you think about the second derivative vector here, s double prime of t? Well, the way to do that, I like to think of all of the tangent vectors then just kind of living in their own space. If this here is representing s of t, I just want to look in isolation at what s prime of t looks like."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "You might have a very long tangent vector to indicate that you're traveling very quickly across that space. And now, how would you think about the second derivative vector here, s double prime of t? Well, the way to do that, I like to think of all of the tangent vectors then just kind of living in their own space. If this here is representing s of t, I just want to look in isolation at what s prime of t looks like. So each one of those vectors, you know maybe this first one is just this very long gargantuan indicating you're going very quickly. And then after that, you've got something else here. Maybe it's pointing a little bit down."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "If this here is representing s of t, I just want to look in isolation at what s prime of t looks like. So each one of those vectors, you know maybe this first one is just this very long gargantuan indicating you're going very quickly. And then after that, you've got something else here. Maybe it's pointing a little bit down. And you're kind of thinking about how all of these derivative vectors change, but I want them all rooted at the same point. I'll just see what happens when they're all rooted at the origin, because that gives us a way to think about what changing the vector should look like. So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "Maybe it's pointing a little bit down. And you're kind of thinking about how all of these derivative vectors change, but I want them all rooted at the same point. I'll just see what happens when they're all rooted at the origin, because that gives us a way to think about what changing the vector should look like. So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction. So this second derivative value is going to tell us how the tip of the first derivative should move. And then similarly over here, it tells us how the tip of that should move. And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "So in particular, as you're moving from this vector to this one, the tip should be kind of moving in this direction. So this second derivative value is going to tell us how the tip of the first derivative should move. And then similarly over here, it tells us how the tip of that should move. And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right. So if they were off, we'd kind of draw them on their own, rooted in their own location. You can maybe see how the second derivative vector is telling it to turn somehow in that direction. And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "And just as an example and kind of a hint as to why this has something to do with curvature, if you have a curve that turns very sharply, turns very sharply, you would have a tangent vector that starts by pointing up and to the right and then very quickly is pointing down, down and to the right. So if they were off, we'd kind of draw them on their own, rooted in their own location. You can maybe see how the second derivative vector is telling it to turn somehow in that direction. And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys. So the second derivative vector would be pulling perpendicular, perpendicular to that first derivative vector as a way of telling it to turn. So just to draw that on its own, if you have a first derivative vector and then the second derivative vector is perpendicular to it, it's telling it how it should turn in some way. But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "And if you capture not just those two, but infinitesimally what's going on as you move from one to the next, you're going to get this kind of turning motion for all of these guys. So the second derivative vector would be pulling perpendicular, perpendicular to that first derivative vector as a way of telling it to turn. So just to draw that on its own, if you have a first derivative vector and then the second derivative vector is perpendicular to it, it's telling it how it should turn in some way. But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way. So that would be an indication that not only is it turning, but it's getting smaller, meaning the trajectory based on S is probably slowing down. And if it was kind of turning it, but also pointing away, that would mean it's telling the first derivative vector to grow. So not only should you be turning on your curve, but you should also be speeding up."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "But if it was, let's say it was not purely perpendicular, but it was also pointing against, it would be telling that vector, that first derivative vector, to shrink in some way. So that would be an indication that not only is it turning, but it's getting smaller, meaning the trajectory based on S is probably slowing down. And if it was kind of turning it, but also pointing away, that would mean it's telling the first derivative vector to grow. So not only should you be turning on your curve, but you should also be speeding up. But the case we care about most is we're just trying to measure how perpendicular it is, right? And this is where the cross product comes in. Because if you think about how you interpret that cross product, how you interpret the cross product, it's basically the area, if you kind of take these two vectors, tip to tail as they are, and think about the parallelogram that they trace out, let's see, so this blue dotted line should be parallel to the blue vector, the area of this, of this traced out parallelogram for each of them, that's what tells you, that's how you interpret the cross product, the cross product between S prime and S double prime."}, {"video_title": "Curvature formula, part 4.mp3", "Sentence": "So not only should you be turning on your curve, but you should also be speeding up. But the case we care about most is we're just trying to measure how perpendicular it is, right? And this is where the cross product comes in. Because if you think about how you interpret that cross product, how you interpret the cross product, it's basically the area, if you kind of take these two vectors, tip to tail as they are, and think about the parallelogram that they trace out, let's see, so this blue dotted line should be parallel to the blue vector, the area of this, of this traced out parallelogram for each of them, that's what tells you, that's how you interpret the cross product, the cross product between S prime and S double prime. So this cross product, by giving you that area, is kind of a measure of just how perpendicular these vectors are, right? Because if one of them, if they point very much in the same direction, right, and they're only slightly perpendicular, that means that the parallelogram they trace out is gonna have a very small area, it's gonna be a smaller area in comparison. So with that, I don't want this video to run too long, so I'm gonna call it in here, but then I'll just continue on in the next one through the same line of reasoning to build to our original formula."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "And what this means is you're looking for the input points, the values of x and y and all of its other inputs, such that the output, f, is as great as it possibly can be. Now this actually comes up all the time in practice, because usually when you're dealing with a multivariable function, it's not just for fun and for dealing with abstract symbols, it's because it actually represents something. So maybe it represents like profits of a company. Maybe this is a function where you're considering all the choices you can make, like the wages you give your employees, or the prices of your goods, or the amount of debt that you raise for capital, all sorts of choices that you might make. And you want to know what values should you give to those choices, such that you maximize profits, you maximize a thing. And if you have a function that models these relationships, there are techniques, which I'm about to teach you, that you can use to maximize this. Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "Maybe this is a function where you're considering all the choices you can make, like the wages you give your employees, or the prices of your goods, or the amount of debt that you raise for capital, all sorts of choices that you might make. And you want to know what values should you give to those choices, such that you maximize profits, you maximize a thing. And if you have a function that models these relationships, there are techniques, which I'm about to teach you, that you can use to maximize this. Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task. So maybe you're trying to teach a computer how to understand audio, or how to read handwritten text. What you do is you find a function that basically tells it how wrong it is when it makes a guess. And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right?"}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "Another very common setting, more and more important these days, is that of machine learning and artificial intelligence, where often what you do is you assign something called a cost function to a task. So maybe you're trying to teach a computer how to understand audio, or how to read handwritten text. What you do is you find a function that basically tells it how wrong it is when it makes a guess. And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right? Instead of finding the maximum, you need to minimize a certain function. And if it minimizes this cost function, that means that it's doing a really good job at whatever task you've assigned it. So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "And if you do a good job designing that function, you just need to tell the computer to minimize, so that's kind of the flip side, right? Instead of finding the maximum, you need to minimize a certain function. And if it minimizes this cost function, that means that it's doing a really good job at whatever task you've assigned it. So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that. And a lot of time and research has gone into figuring out ways to basically apply these techniques, really quickly and efficiently. So first of all, on a conceptual level, let's just think about what it means to be finding the maximum of a multivariable function. So I have here the graph of a two-variable function."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "So a lot of the art and science of machine learning and artificial intelligence comes down to, well, one, finding this cost function and actually describing difficult tasks in terms of a function, but then applying the techniques that I'm about to teach you to have the computer minimize that. And a lot of time and research has gone into figuring out ways to basically apply these techniques, really quickly and efficiently. So first of all, on a conceptual level, let's just think about what it means to be finding the maximum of a multivariable function. So I have here the graph of a two-variable function. That's something that has a two-variable input that we're thinking of as the xy-plane, and then its output is the height of this graph. And if you're looking to maximize it, basically what you're finding is this peak, kind of the tallest mountain in the entire area. And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "So I have here the graph of a two-variable function. That's something that has a two-variable input that we're thinking of as the xy-plane, and then its output is the height of this graph. And if you're looking to maximize it, basically what you're finding is this peak, kind of the tallest mountain in the entire area. And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function. So how do you go about finding that? Well, this is perhaps the core observation in, well, calculus, not just multivariable calculus. This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "And you're looking for the input value, the point on the xy-plane, directly below that peak, because that tells you the values of the inputs that you should put in to maximize your function. So how do you go about finding that? Well, this is perhaps the core observation in, well, calculus, not just multivariable calculus. This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat. But let's say you did this at a different point, right? Because if you tried to find the tangent plane not at that point, but you kind of moved it about a bit to somewhere that's not quite a maximum, if the tangent plane has any kind of slope to it, what that's telling you is that if you take very small directions, kind of in the direction of that upward slope, you can increase the value of your function. So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "This is similar in the single-variable world, and there's similarities in other settings, but the core observation is that if you take a tangent plane at that peak, so let's just draw in a tangent plane at that peak, it's gonna be completely flat. But let's say you did this at a different point, right? Because if you tried to find the tangent plane not at that point, but you kind of moved it about a bit to somewhere that's not quite a maximum, if the tangent plane has any kind of slope to it, what that's telling you is that if you take very small directions, kind of in the direction of that upward slope, you can increase the value of your function. So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it. But if there's no slope to it, if it's flat, then that's a sign that no matter which direction you walk, you're not gonna be significantly increasing the value of your function. So what does this mean in terms of formulas? Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "So if there's any slope to the tangent plane, you know that you can walk in some direction to increase it. But if there's no slope to it, if it's flat, then that's a sign that no matter which direction you walk, you're not gonna be significantly increasing the value of your function. So what does this mean in terms of formulas? Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes. The slope of the plane in each direction, so you know this would be the slope in the x direction, and then if you look at it from another perspective, this would be the slope in the y direction, each one of those has to be zero. And that, in terms of partial derivatives, means the partial derivative of your function at whatever point you're dealing with, right, so I'll call it x-naught, y-naught, as the point where you're inputting this, has to be zero. And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "Well, if you kind of think back to how we compute tangent planes, and if you're not very comfortable with that, now would be a good time to take another look at those videos about tangent planes. The slope of the plane in each direction, so you know this would be the slope in the x direction, and then if you look at it from another perspective, this would be the slope in the y direction, each one of those has to be zero. And that, in terms of partial derivatives, means the partial derivative of your function at whatever point you're dealing with, right, so I'll call it x-naught, y-naught, as the point where you're inputting this, has to be zero. And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero. And both of these have to be true, because let's just take a look, I don't know, let's slide it over a little bit here. This tangent plane, if you look at the slope, if you imagine walking in the y direction, you're not increasing your value at all, the slope in the y direction would actually be zero, so that would mean the partial derivative with respect to y would be zero. But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "And then similarly, the partial derivative with respect to the other variable, with respect to y, at that same point, has to be zero. And both of these have to be true, because let's just take a look, I don't know, let's slide it over a little bit here. This tangent plane, if you look at the slope, if you imagine walking in the y direction, you're not increasing your value at all, the slope in the y direction would actually be zero, so that would mean the partial derivative with respect to y would be zero. But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps. So what this gives you here is gonna be a system of equations where you're solving for the value of x-naught and y-naught that satisfies both of these equations. And in future videos, I'll go through specific examples of this, for now I just wanna give a good conceptual understanding. But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "But with respect to x, when you're moving in the x direction here, the slope is clearly negative, because as you take positive steps in the x direction, the height of your tangent plane is decreasing, which corresponds to if you take tiny steps on your graph, then the height will decrease in a manner proportional to the size of those tiny steps. So what this gives you here is gonna be a system of equations where you're solving for the value of x-naught and y-naught that satisfies both of these equations. And in future videos, I'll go through specific examples of this, for now I just wanna give a good conceptual understanding. But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy. But for one thing, if you found the tangent plane at other little peaks, like this guy here, or this guy here, or all of the little bumps that go up, those tangent planes would also be flat. And those little bumps actually have a name, because this comes up a lot, they're called local minima, or local maxima, sorry, so those guys are called local maxima, maxima is just the plural of maximum, and local means that it's relative to a single point. So it's basically if you walk in any direction when you're on that little peak, you'll go downhill."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "But one very important thing to notice is that just because this condition is satisfied, meaning your tangent plane is flat, just because that's satisfied doesn't necessarily mean that you've found the maximum, that's just one requirement that it has to satisfy. But for one thing, if you found the tangent plane at other little peaks, like this guy here, or this guy here, or all of the little bumps that go up, those tangent planes would also be flat. And those little bumps actually have a name, because this comes up a lot, they're called local minima, or local maxima, sorry, so those guys are called local maxima, maxima is just the plural of maximum, and local means that it's relative to a single point. So it's basically if you walk in any direction when you're on that little peak, you'll go downhill. So relative to the neighbors of that little point, it is a maximum, but relative to the entire function, you know, these guys are the shorter mountains next to Mount Everest. But there's also another circumstance where you might find a flat tangent plane, and that's at the minima points, right, if you have the global minimum, the absolute smallest, or also just the local minima, these inverted peaks, you'll also find flat tangent planes. So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "So it's basically if you walk in any direction when you're on that little peak, you'll go downhill. So relative to the neighbors of that little point, it is a maximum, but relative to the entire function, you know, these guys are the shorter mountains next to Mount Everest. But there's also another circumstance where you might find a flat tangent plane, and that's at the minima points, right, if you have the global minimum, the absolute smallest, or also just the local minima, these inverted peaks, you'll also find flat tangent planes. So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum. And these requirements, by the way, often you'll see them written in a more succinct form, where instead of saying all the partial derivatives have to be zero, which is what you need to find, they'll write it in a different form where you say that the gradient of your function f, which of course is just the vector that contains all those partial derivatives, its first component is the partial derivative with respect to the first variable, its second component is the partial derivative with respect to the second variable, and if there was more variables, you would keep going. You'd say that this whole thing has to equal the zero vector, the vector that has nothing but zeros as its components. And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "So what that means, first of all, is that when you're minimizing a function, you also have to look for this requirement where all the partial derivatives are zero, but it mainly just means that your job isn't done once you've done this, you have to do more tests to check whether or not what you found is a local maximum, or a local minimum, or a global maximum. And these requirements, by the way, often you'll see them written in a more succinct form, where instead of saying all the partial derivatives have to be zero, which is what you need to find, they'll write it in a different form where you say that the gradient of your function f, which of course is just the vector that contains all those partial derivatives, its first component is the partial derivative with respect to the first variable, its second component is the partial derivative with respect to the second variable, and if there was more variables, you would keep going. You'd say that this whole thing has to equal the zero vector, the vector that has nothing but zeros as its components. And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement. You're just looking for where the gradient of your function is equal to the zero vector, and that way you can just write it on one line. But in practice, every time that you're expanding that out, what that means is you find all of the different partial derivatives. So this is really just a matter of notational convenience and using less space on a blackboard."}, {"video_title": "Multivariable maxima and minima.mp3", "Sentence": "And it's kind of a common abuse of notation, people will just call that zero vector zero, and maybe they'll emphasize it by making it bold, because the number zero is not a vector, and often making things bold emphasizes that you want to be referring to a vector, but this gives a very succinct way of describing the requirement. You're just looking for where the gradient of your function is equal to the zero vector, and that way you can just write it on one line. But in practice, every time that you're expanding that out, what that means is you find all of the different partial derivatives. So this is really just a matter of notational convenience and using less space on a blackboard. But whenever you see this, that the gradient equals zero, what you should be thinking of is the idea that the tangent plane, the tangent plane is completely flat. And as I just said, that's not enough, because you might also be finding local maxima or minima points, but in multivariable calculus, there's also another possibility, a place where the tangent plane is flat, but what you're looking at is neither a local maximum nor a local minimum. And this is the idea of a saddle point, which is new to multivariable calculus, and that's what I'll be talking about in the next video."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And what I want to do in this video is just get a little bit of a gut sense of what it means to take a derivative of a vector-valued function. In this case, it will be with respect to our parameter t. So let me draw some new stuff right here. So let's say I have the vector-valued function R of t, and this is no different than what I did in the last video, x of t times unit vector i plus y of t times the unit vector j. If we're dealing in three dimensions, we'd add a z of t times k, but let's keep things relatively simple. And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If we're dealing in three dimensions, we'd add a z of t times k, but let's keep things relatively simple. And let's say that this describes a curve, and let's say the curve we're dealing with between t is between a and b, and this curve will look something like... let me do my best effort to draw the curve. Let's draw some random curve here. So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say the curve looks something like that. This is when t is equal to a, so it's going to go in this direction. This is when t is equal to b right here. This is t is equal to a, so this right here would be x of a. This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is t is equal to a, so this right here would be x of a. This right here is y of a. And similarly, this up here, this is x of b, and this over here is y of b. Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video. It describes that point right there. I don't want to review that too much."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, we saw in the last video that the endpoints of these position vectors are what's describing this curve. So r of a we saw in the last video. It describes that point right there. I don't want to review that too much. But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I don't want to review that too much. But what I want to do is think about what is the difference between two points. So let's say that we take some random point here. Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say some random t here. Let's call that r of t. Actually, I'm going to do a different point just because I want to make it a little bit clearer. So let's say that that right there is r of some t, some particular t right there. That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is r of t. It's going to be a plus something. So that's some particular t. And let's say that we want to figure out, and let's say we increase t by a little bit, by h. So let's say that r of t plus h, well, if we view the parameter t as time, we can kind of view we've moved forward in time by some amount. So our little particle has moved a little bit. And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get?"}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And let's say that we're over here. So that is that right there in yellow is r of t plus h. Just a slightly larger value for h. Now, one question we might ask ourselves is how quickly is r changing with respect to t? So the first thing we might want to say is what's the difference between these two? If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I were to take, and I want to visualize it, if I were to take r, the position vector that we get by evaluating r at t plus h, and from that I were to subtract r of t, what do we get? Well, you might want to review some of your vector algebra, but we're essentially just going to get this vector. We're going to let me do it in a nice vibrant color. We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to get this vector right there that I'm doing in magenta. So that magenta vector right there is the vector r of t plus h minus r of t. And it should make sense because when you add vectors, you go heads to tails. You could alternatively write this as r of t plus this character right here, plus r of t plus h minus r of t. When you add two vectors, you're adding, let me make it very clear, I'm adding this vector to this vector right here. You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You put the tail of the second vector at the head of the first. So this is the first vector, and then I put the tail of the second there, and then the sum of those two, as we predicted, should be equal to this last one. It should be equal to r of t plus h. And we see that is the case, and even algebraically, you would see that obviously this guy and that guy are going to cancel out. So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So hopefully that satisfies you. And I want to be clear, this all of a sudden, this isn't a position vector. We're not saying that, hey, let's nail this guy's tail at the origin and use this guy to describe a unique position. Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that?"}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now all of a sudden, it's just kind of a pure vector. It's describing just a change between two other position vectors. So this guy is right out here, but this vector literally describes the change. But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But let's say we care, and how would this look algebraically if we were to expand it like that? So this is going to be equal to, what's r of t plus h? That's the same thing as x of t plus h times the unit vector i plus y of t plus h times the unit vector j. That's just that piece. That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's just that piece. That piece right there is that piece. Minus, so this piece, so minus, I'll do it in the second line. I could have done it out here, but I'm running out of space. Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I could have done it out here, but I'm running out of space. Minus x of t, right? R of t is just x of t times i plus, but I'll just distribute the minus sign. So it's minus y of t times j. Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's minus y of t times j. Actually, let me write it this way, plus this. So you realize that this is really just this guy right here. I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just evaluating at t. So you have x of t and y of t, and then later we can distribute. If you distribute this minus sign, you get a minus x of t and a minus y of t. And in vector addition, you might need a little review on this if you haven't seen it in a while, you can just add the corresponding components. You can add the x components and you can add the y components. So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is going to be equal to, let me rewrite it over here, because I think I'm going to need some space later on. So let me rewrite it over here. So I have R of t plus h minus R of t is equal to, now I'm just going to group the x and the y components. This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is equal to the x components added together, but this is a negative, so we're going to subtract this guy from that guy. So we have x of t plus h minus x of t, and then all of that times our unit vector in the x direction. And then we'll have plus y of t plus h minus y of t times the unit vector in the j direction. I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h?"}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I'm just rearranging things right now. So this will tell us what is our change between any two R's for a given change in distance. And our change in distance here is h between any two position vectors. Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now, what I set out at the beginning of this video, I said, well, I want to figure out the change, and we're going to think about the instantaneous change with respect to t. So I want to see, well, how much did this change over a period of h? Instead of writing h, we could have written delta t. It would have been the same thing. So I want to divide this by h. So I want to say, look, my vector has changed this much, but I want to say it's over a period of h. And this is analogous to when we do slope. We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We say rise over run over delta y or change in y over change in x. This is kind of the change in our function per change in x. Let's just divide everything, or I shouldn't say change in x, per change in t. So here our change in t is h. The difference between t plus h and t is just going to be h. And so we're going to divide everything by h. When you multiply a vector by some scalar or divide it by some scalar, you're just taking each of its components and multiplying or dividing by that scalar. And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we get that right there. So this, for any finite difference right here, h, this will tell us how much our vector changes per h. But if we want to find the instantaneous change, just like what we did when we first learned differential calculus, we said, okay, this is kind of analogous to a slope. This would be good. This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This would work out well for us if the path under question looked something like this, if it was a linear path. If our path looked something like this, we could just calculate this and we'll essentially have the average change in our position vectors. So you can imagine two position vectors. That's one of them. Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's one of them. Well, actually, they'd all be parallel. Well, they don't always. Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, the position vectors, they don't have to be parallel, but I won't. They don't have to be parallel. They could be like that. And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then this would just describe the change between these two per h, or how quickly are the position vectors changing per our change in our parameter. The h you could also consider as kind of a delta t. Sometimes people find the h simpler. Sometimes they find the delta t. But anyway, I'm concerned with the instantaneous. We're dealing with curves. We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do?"}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're dealing with curves. We're dealing with calculus. This would have been okay if we were just in an algebraic linear world. So what do we do? Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So what do we do? Well, maybe we can just take the limit as h approaches zero. So let me scroll this over. So let's just take the limit. Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's just take the limit. Let me do this in a nice, vibrant color. Let's take the limit as h approaches zero of both sides of this. So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t?"}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So here, too, I'm going to take the limit as h approaches zero. And here, too, I'm going to take the limit as h approaches zero. So I just want to say, well, what happens? How much do I change per a change in my parameter t? But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "How much do I change per a change in my parameter t? But what's kind of the instantaneous change? As the difference gets smaller and smaller and smaller. This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is exactly what we first learned when we learned about instantaneous slope or instantaneous velocity or slope of a tangent line. Well, this thing looks a little bit undefined to me right now. We haven't defined limits for vector-valued functions. We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We haven't defined derivatives for vector-valued functions. But lucky for us, all of this stuff here looks pretty familiar. This is actually the definition of our derivative. And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And these are scalar-valued functions right here. They're multiplied by vectors in order for us to get vector-valued functions. But this right here, by definition, this is the derivative. This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is x prime of t. Or this is dx dt. This right here is y prime of t. Or we could write that as dy dt. So all of a sudden, we can define, we can say, and I'm being a little hand-wavy here, but I want to give you the intuition more than anything. We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We can say that the derivative, we can call this expression right here as the derivative of my vector-valued function r with respect to t. Or we could call it dr dt. Notice I keep the vector signs there. This is its derivative, and all it's going to be equal to, all it's going to be equal to, r prime of t, is going to be equal to, well, this is just the derivative of x with respect to t, is equal to x prime of t times the x unit vector, the horizontal unit vector, plus y prime of t times the y unit vector, times j, the unit vector in the horizontal direction. That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's a pretty nice and simple outcome. But the harder thing, maybe, is to kind of visualize what it represents. So if we think about what happens, let me draw a big graph, just to get the visualization going in a healthy way. So let's say my curve looks something like this. That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's say my curve looks something like this. That's my curve. Let's say that this is, we want to figure out the instantaneous change at this point right here, so that is r of t. And then if we take r of t plus h, we saw this already, t plus h might be something like right there. So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is r of t plus h. Now, right now, the difference between these two, and this is just the numerator when you take the difference, or how fast we're changing from this vector to that vector, in terms of t, and it's hard to visualize here. And I'm going to do a whole video so we can think about the magnitude here. That might be some vector. Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Well, the difference between these two is just going to be that. But then when you divide it by h, it's going to be a larger vector. If we assume h is a small number, let's say h is less than 1, we're going to get a larger vector. But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is kind of the average change over this time. But as h gets smaller and smaller and smaller, this r prime of t, its direction is going to be tangential to the curve. And I think you can visualize that. As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "As these two guys get closer and closer and closer, the dr's get smaller, so the change, the dr, the difference between the two, the delta r's get smaller and smaller. You can imagine if h was even smaller, if it was right here. All of a sudden the difference between those two vectors is getting smaller. And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And it's getting more and more tangential to the curve. But then we're also dividing by a smaller h. So the actual derivative, as the limit as h approaches 0, it might be, maybe it's even a bigger number there. And actually, the magnitude of this vector, it's a little hard to visualize, it's going to be dependent on a parametrization for the curve. It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve. So the direction, this will be tangent to the curve. Or you could imagine that this vector is on the tangent line to the curve."}, {"video_title": "Derivative of a position vector valued function   Multivariable Calculus   Khan Academy.mp3", "Sentence": "It's not dependent on the shape of the curve. The direction of this vector is dependent on the shape of this curve. So the direction, this will be tangent to the curve. Or you could imagine that this vector is on the tangent line to the curve. The magnitude of it is a little bit harder to understand. I'll try to give you a little bit of intuition on that in the next video. But this is what I want you to understand right now, because we're going to be able to use this in the future when we do the line integral over vector-valued functions."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "It's defined over r2 over the xy-plane. So it's a function of x and y. It associates a vector with every point on the plane. And let's say my vector field is y times the unit vector i. Let's say minus x times the unit vector j. And so you can imagine if we were to draw our x and y axes. I'll do it over here."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And let's say my vector field is y times the unit vector i. Let's say minus x times the unit vector j. And so you can imagine if we were to draw our x and y axes. I'll do it over here. If we were to draw our x and y axes, this associates a vector, a force vector. Let's say this is actually a force vector with every point in our xy-plane. So this is x, this is y."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "I'll do it over here. If we were to draw our x and y axes, this associates a vector, a force vector. Let's say this is actually a force vector with every point in our xy-plane. So this is x, this is y. So if we're at the point, for example, 1, 0, what will the vector look like that's associated with that point? Well, at 1, 0, y is 0. So this will be 0i minus 1j."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So this is x, this is y. So if we're at the point, for example, 1, 0, what will the vector look like that's associated with that point? Well, at 1, 0, y is 0. So this will be 0i minus 1j. Minus 1j looks like this. So minus 1j will look like that. At x is equal to 2, I'm just picking points at random ones."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So this will be 0i minus 1j. Minus 1j looks like this. So minus 1j will look like that. At x is equal to 2, I'm just picking points at random ones. It'll be easy. y is still 0. And now the force vector here would be minus 2j."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "At x is equal to 2, I'm just picking points at random ones. It'll be easy. y is still 0. And now the force vector here would be minus 2j. So it would look something like this. Minus 2j, something like that. Likewise, if we were to go here where y is equal to 1 and x is equal to 0."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And now the force vector here would be minus 2j. So it would look something like this. Minus 2j, something like that. Likewise, if we were to go here where y is equal to 1 and x is equal to 0. When y is equal to 1, we have 1i minus 0j. So then our vector is going to look like that at this point. Our vector will look like that."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Likewise, if we were to go here where y is equal to 1 and x is equal to 0. When y is equal to 1, we have 1i minus 0j. So then our vector is going to look like that at this point. Our vector will look like that. If we were to go to 2, you could get the picture. You can keep plotting these points. You just want to get a sense of what it looks like."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Our vector will look like that. If we were to go to 2, you could get the picture. You can keep plotting these points. You just want to get a sense of what it looks like. If you go here, the vector is going to look like that. If you go maybe this point right here, the vector is going to look like that. I think you get the general idea."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "You just want to get a sense of what it looks like. If you go here, the vector is going to look like that. If you go maybe this point right here, the vector is going to look like that. I think you get the general idea. I could keep filling in the space for this entire field all over just to make it symmetric. If I was right over here, the vector is going to look like that. You get the idea."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "I think you get the general idea. I could keep filling in the space for this entire field all over just to make it symmetric. If I was right over here, the vector is going to look like that. You get the idea. I could just fill in all of the points if I had to. Now, in that field, I have some particle moving. And let's say its path is described by the curve C. And the parametrization of it is x of t is equal to cosine of t. And y of t is equal to sine of t. And the path will occur from t. Let's say 0 is less than or equal to t. It's less than or equal to 2 pi."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "You get the idea. I could just fill in all of the points if I had to. Now, in that field, I have some particle moving. And let's say its path is described by the curve C. And the parametrization of it is x of t is equal to cosine of t. And y of t is equal to sine of t. And the path will occur from t. Let's say 0 is less than or equal to t. It's less than or equal to 2 pi. You might already recognize what this would be. This parametrization is essentially a counterclockwise circle. So the path that this guy is going to go is going to start here."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And let's say its path is described by the curve C. And the parametrization of it is x of t is equal to cosine of t. And y of t is equal to sine of t. And the path will occur from t. Let's say 0 is less than or equal to t. It's less than or equal to 2 pi. You might already recognize what this would be. This parametrization is essentially a counterclockwise circle. So the path that this guy is going to go is going to start here. t, in this case, you can imagine it's the angle of the circle. But you could also imagine t as time. So time equals 0."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So the path that this guy is going to go is going to start here. t, in this case, you can imagine it's the angle of the circle. But you could also imagine t as time. So time equals 0. We're going to be over here. Then at time of pi over 2, we're going to have traveled a quarter of the circle to there. So we're moving in that direction."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So time equals 0. We're going to be over here. Then at time of pi over 2, we're going to have traveled a quarter of the circle to there. So we're moving in that direction. And then at time after pi seconds, we would have gotten right there. And then all the way after 2 pi seconds, we would have gotten all the way around the circle. So our path, our curve, is one counterclockwise rotation around the circle, so to speak."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So we're moving in that direction. And then at time after pi seconds, we would have gotten right there. And then all the way after 2 pi seconds, we would have gotten all the way around the circle. So our path, our curve, is one counterclockwise rotation around the circle, so to speak. So what is the work done by this field on this curve? So the work we learned in the previous video is equal to the line integral over this contour of our vector field dotted with the differential of our movement. Well, I haven't even defined r yet."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So our path, our curve, is one counterclockwise rotation around the circle, so to speak. So what is the work done by this field on this curve? So the work we learned in the previous video is equal to the line integral over this contour of our vector field dotted with the differential of our movement. Well, I haven't even defined r yet. I mean, I kind of have just the parametrization here. So we need to have a vector function. We need to have some r that defines this path."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Well, I haven't even defined r yet. I mean, I kind of have just the parametrization here. So we need to have a vector function. We need to have some r that defines this path. This is just a standard parametrization. But if I wanted to write it as a vector function of t, we would write that r of t is equal to x of t, which is cosine of t, times i plus y of t times j, which is just sine of t times j. And likewise, this is for 0 is less than or equal to t, which is less than or equal to 2 pi."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "We need to have some r that defines this path. This is just a standard parametrization. But if I wanted to write it as a vector function of t, we would write that r of t is equal to x of t, which is cosine of t, times i plus y of t times j, which is just sine of t times j. And likewise, this is for 0 is less than or equal to t, which is less than or equal to 2 pi. This and this are equivalent. The reason why I took the pain of doing this is so now I can take its vector function derivative and I can figure out its differential. And then I can actually take the dot product with this thing over here."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And likewise, this is for 0 is less than or equal to t, which is less than or equal to 2 pi. This and this are equivalent. The reason why I took the pain of doing this is so now I can take its vector function derivative and I can figure out its differential. And then I can actually take the dot product with this thing over here. So let's do all of that and actually calculate this line integral and figure out the work done by this field. And one thing might already pop in your mind. We're going in a counterclockwise direction."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And then I can actually take the dot product with this thing over here. So let's do all of that and actually calculate this line integral and figure out the work done by this field. And one thing might already pop in your mind. We're going in a counterclockwise direction. But at every point where we're passing through, it looks like the field is going exactly opposite the direction of our motion. For example, here we're moving upwards. The field is pulling us backwards."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "We're going in a counterclockwise direction. But at every point where we're passing through, it looks like the field is going exactly opposite the direction of our motion. For example, here we're moving upwards. The field is pulling us backwards. Here we're moving to the top left. The field is moving us to the bottom right. Here we're moving exactly to the left."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "The field is pulling us backwards. Here we're moving to the top left. The field is moving us to the bottom right. Here we're moving exactly to the left. The field is pulling us to the right. So it looks like the field is always doing the exact opposite of what we're trying to do. It's hindering our ability to move."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Here we're moving exactly to the left. The field is pulling us to the right. So it looks like the field is always doing the exact opposite of what we're trying to do. It's hindering our ability to move. So I'll give you a little intuition. This will probably deal with negative work. For example, if I lift something off the ground, I have to apply force to fight gravity."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "It's hindering our ability to move. So I'll give you a little intuition. This will probably deal with negative work. For example, if I lift something off the ground, I have to apply force to fight gravity. I'm doing positive work, but gravity's doing negative work on that. We're just going to do the math here just to make you comfortable with this idea. But it's interesting to think about what's exactly going on even here."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "For example, if I lift something off the ground, I have to apply force to fight gravity. I'm doing positive work, but gravity's doing negative work on that. We're just going to do the math here just to make you comfortable with this idea. But it's interesting to think about what's exactly going on even here. The field is, let me, the field, I'm doing that pink color. So let me stick to that. The field is pushing in that direction."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "But it's interesting to think about what's exactly going on even here. The field is, let me, the field, I'm doing that pink color. So let me stick to that. The field is pushing in that direction. So it's always going opposite the motion. But let's just do the math to make everything in the last video a little bit more concrete. So a good place to start is the derivative of our position vector function with respect to t. So let me switch."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "The field is pushing in that direction. So it's always going opposite the motion. But let's just do the math to make everything in the last video a little bit more concrete. So a good place to start is the derivative of our position vector function with respect to t. So let me switch. So we have dr dt, which we could also write as r prime of t. This is equal to the derivative of x of t with respect to t, which is minus sine of t times i, plus the derivative of y of t with respect to t. That's derivative of sine of t is just cosine of t times j. And if we want the differential, we just multiply everything times dt. So we get dr is equal to, we could write it this way."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So a good place to start is the derivative of our position vector function with respect to t. So let me switch. So we have dr dt, which we could also write as r prime of t. This is equal to the derivative of x of t with respect to t, which is minus sine of t times i, plus the derivative of y of t with respect to t. That's derivative of sine of t is just cosine of t times j. And if we want the differential, we just multiply everything times dt. So we get dr is equal to, we could write it this way. We could actually even just put the d, well, let me just do it. So it's minus sine of t dt. I'm just multiplying each of these terms by dt, distributive property, times the unit vector i, plus cosine of t dt times the unit vector j."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So we get dr is equal to, we could write it this way. We could actually even just put the d, well, let me just do it. So it's minus sine of t dt. I'm just multiplying each of these terms by dt, distributive property, times the unit vector i, plus cosine of t dt times the unit vector j. So we have this piece now. And then we want to take the dot product with this over here. But let me rewrite our vector field in terms of t, so to speak."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "I'm just multiplying each of these terms by dt, distributive property, times the unit vector i, plus cosine of t dt times the unit vector j. So we have this piece now. And then we want to take the dot product with this over here. But let me rewrite our vector field 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."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "But let me rewrite our vector field 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."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "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. y is a function of t. So it's sine of t. That's that. Sine of t times i, plus, or actually minus, x of t. x is a function of t. So minus cosine of t times j."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "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. y is a function of t. So it's sine of t. That's that. Sine of t times i, plus, or actually minus, x of t. x is a function of t. 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."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Sine of t times i, plus, or actually minus, x of t. x is a function of t. 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 2 pi of f dot dr. Now, when you take the dot product, you just multiply the corresponding components and add it up. So you take the product of the sine of t with the minus sine of t dt, I get minus sine squared t dt."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "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 2 pi of f dot dr. Now, when you take the dot product, you just multiply the corresponding components and add it up. So you take the product of the sine of t with the minus sine of t dt, I 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."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So you take the product of the sine of t with the minus sine of t dt, I 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. That was a wacky looking dt. And then you're going to have that plus. These two guys multiplied by each other."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Let me write that dt a little bit. That was a wacky looking 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 sign here. So plus, let me just change this to minus, minus cosine squared dt. And if we factor out a minus sign and a dt, what is this going to be equal to?"}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "These two guys multiplied by each other. So that's, well, there's a minus sign here. So plus, let me just change this to 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 2 pi of, we could say, sine squared plus, I want to put the t there, sine squared of t plus cosine squared of t. Actually, let me take the minus sign out to the front. So if we just factor the minus sign, so we put a minus there to make this a plus, make this a plus. So the minus sign out there."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "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 2 pi of, we could say, sine squared plus, I want to put the t there, sine squared of t plus cosine squared of t. Actually, let me take the minus sign out to the front. So if we just factor the minus sign, so we put a minus there to make this a plus, make this a plus. So the minus sign out there. And we factor dt out. I did a couple of steps in there, but I think you got it. And this is just algebra at this point."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So the minus sign out there. And we factor dt out. I did a couple of steps in there, but I think you got it. And 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. You could multiply this out and you'll get what we originally have, if that confuses you at all."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "And 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. You could multiply this out and you'll get what we 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."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "You could multiply this out and you'll get what we 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 2 pi of dt. And we've seen this before, we can say this is of 1, if you want to put something there. Then the antiderivative of 1 is just going to be equal to minus."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So this is just 1. So our whole integral has been reduced to the minus integral from 0 to 2 pi of dt. And we've seen this before, we can say this is of 1, if you want to put something there. Then the antiderivative of 1 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 2 pi to 0, or from 0 to 2 pi. So this is equal to minus that minus sign right there."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "Then the antiderivative of 1 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 2 pi to 0, or from 0 to 2 pi. So this is equal to minus that minus sign right there. 2 pi minus t at 0. So minus 0. So this is just equal to minus 2 pi."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So this is equal to minus that minus sign right there. 2 pi minus t at 0. So minus 0. So this is just equal to minus 2 pi. And there you have it. We figured out the work that this field did on the particle, or whatever, as whatever thing, was moving around in this counterclockwise fashion. And our intuition held up."}, {"video_title": "Using a line integral to find the work done by a vector field example   Khan Academy.mp3", "Sentence": "So this is just equal to minus 2 pi. And there you have it. We figured out the work that this field did on the particle, or whatever, as whatever thing, was moving around in this 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."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "And essentially what this is, it's just a way to package all the information of the second derivatives of a function. So let's say you have some kind of multivariable function, like, I don't know, like the example we had in the last video, e to the x halves multiplied by sine of y. So some kind of multivariable function. What the Hessian matrix is, and it's often denoted with an H, or kind of a bold-faced H, is it's a matrix, incidentally enough, that contains all the second partial derivatives of f. So the first component is gonna be the partial derivative of f with respect to x, kind of twice in a row. And everything in this first column, it's kind of like you first do it with respect to x. Because the next part is the second derivative where first you do it with respect to x, and then you do it with respect to y. So that's kind of the first column of the matrix."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "What the Hessian matrix is, and it's often denoted with an H, or kind of a bold-faced H, is it's a matrix, incidentally enough, that contains all the second partial derivatives of f. So the first component is gonna be the partial derivative of f with respect to x, kind of twice in a row. And everything in this first column, it's kind of like you first do it with respect to x. Because the next part is the second derivative where first you do it with respect to x, and then you do it with respect to y. So that's kind of the first column of the matrix. And then up here, it's the partial derivative where first you do it with respect to y, and then you do it with respect to x. And then over here it's where you do it with respect to y, both times in a row. So partial with respect to y, both times in a row."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "So that's kind of the first column of the matrix. And then up here, it's the partial derivative where first you do it with respect to y, and then you do it with respect to x. And then over here it's where you do it with respect to y, both times in a row. So partial with respect to y, both times in a row. So let's go ahead and actually compute this and think about what this would look like in the case of our specific function here. So in order to get all the second partial derivatives, we first should just kind of keep a record of the first partial derivatives. So the partial derivative of f with respect to x, the only place x shows up is in this e to the x halves."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "So partial with respect to y, both times in a row. So let's go ahead and actually compute this and think about what this would look like in the case of our specific function here. So in order to get all the second partial derivatives, we first should just kind of keep a record of the first partial derivatives. So the partial derivative of f with respect to x, the only place x shows up is in this e to the x halves. Kind of bring down that 1 1 2, e to the x halves, and sine of y just looks like a constant as far as x is concerned, sine of y. And then the partial derivative with respect to y, partial derivative of f with respect to y. Now e to the x halves looks like a constant and it's being multiplied by something that has a y in it, e to the x halves."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "So the partial derivative of f with respect to x, the only place x shows up is in this e to the x halves. Kind of bring down that 1 1 2, e to the x halves, and sine of y just looks like a constant as far as x is concerned, sine of y. And then the partial derivative with respect to y, partial derivative of f with respect to y. Now e to the x halves looks like a constant and it's being multiplied by something that has a y in it, e to the x halves. And the derivative of sine of y, since we're doing it with respect to y, is cosine of y, cosine of y. So these terms won't be included in the Hessian itself, but we're kind of just kind of keeping a record of them because now when we go in to fill in the matrix, this upper left component, we're taking the second partial derivative where we do it with respect to x, then x again. So up here is when we did it with respect to x."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "Now e to the x halves looks like a constant and it's being multiplied by something that has a y in it, e to the x halves. And the derivative of sine of y, since we're doing it with respect to y, is cosine of y, cosine of y. So these terms won't be included in the Hessian itself, but we're kind of just kind of keeping a record of them because now when we go in to fill in the matrix, this upper left component, we're taking the second partial derivative where we do it with respect to x, then x again. So up here is when we did it with respect to x. If we did it with respect to x again, we kind of bring down another half, so that becomes 1 4th by e to the x halves. And that sine of y just still looks like a constant. Sine of y."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "So up here is when we did it with respect to x. If we did it with respect to x again, we kind of bring down another half, so that becomes 1 4th by e to the x halves. And that sine of y just still looks like a constant. Sine of y. And then this mixed partial derivative where we do it with respect to x, then y. So we did it with respect to x here. When we differentiate this with respect to y, the 1 1 2 e to the x halves just looks like a constant, but then derivative of sine of y ends up as cosine of y."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "Sine of y. And then this mixed partial derivative where we do it with respect to x, then y. So we did it with respect to x here. When we differentiate this with respect to y, the 1 1 2 e to the x halves just looks like a constant, but then derivative of sine of y ends up as cosine of y. And then up here, it's gonna be the same thing, but let's kind of see how, when you do it in the other direction, when you do it first with respect to y, then x. So over here we did it first with respect to y. If we took this derivative with respect to x, you'd have the half would come down, so that would be 1 1 2 e to the x halves multiplied by cosine of y because that just looks like a constant since we're doing it with respect to x the second time."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "When we differentiate this with respect to y, the 1 1 2 e to the x halves just looks like a constant, but then derivative of sine of y ends up as cosine of y. And then up here, it's gonna be the same thing, but let's kind of see how, when you do it in the other direction, when you do it first with respect to y, then x. So over here we did it first with respect to y. If we took this derivative with respect to x, you'd have the half would come down, so that would be 1 1 2 e to the x halves multiplied by cosine of y because that just looks like a constant since we're doing it with respect to x the second time. So that would be cosine of y. And it shouldn't feel like a surprise that both of these terms turn out to be the same. With most functions, that's the case."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "If we took this derivative with respect to x, you'd have the half would come down, so that would be 1 1 2 e to the x halves multiplied by cosine of y because that just looks like a constant since we're doing it with respect to x the second time. So that would be cosine of y. And it shouldn't feel like a surprise that both of these terms turn out to be the same. With most functions, that's the case. Technically not all functions. You can come up with some crazy things where this won't be symmetric, where you'll have different terms than the diagonal. But for the most part, those you can kind of expect to be the same."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "With most functions, that's the case. Technically not all functions. You can come up with some crazy things where this won't be symmetric, where you'll have different terms than the diagonal. But for the most part, those you can kind of expect to be the same. And then this last term here where we do it with respect to y twice, we now think of taking the derivative of this whole term with respect to y, that e to the x halves looks like a constant, and derivative of cosine is negative sine, negative sine of y. So this whole thing, a matrix, each of whose components is a multivariable function, is the Hessian. This is the Hessian of f. And sometimes people will write it as Hessian of f, kind of specifying what function it's of."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "But for the most part, those you can kind of expect to be the same. And then this last term here where we do it with respect to y twice, we now think of taking the derivative of this whole term with respect to y, that e to the x halves looks like a constant, and derivative of cosine is negative sine, negative sine of y. So this whole thing, a matrix, each of whose components is a multivariable function, is the Hessian. This is the Hessian of f. And sometimes people will write it as Hessian of f, kind of specifying what function it's of. And you could think of this, I mean, you could think of it as a matrix-valued function, which feels kind of weird, but you plug in two different values, x and y, and you'll get a matrix. So it's this matrix-valued function. And the nice thing about writing it like this is that you can actually extend it so that rather than just for functions that have two variables, let's say you had a function, you know, kind of like clear this up, let's say you had a function that had three variables or four variables or kind of any number."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "This is the Hessian of f. And sometimes people will write it as Hessian of f, kind of specifying what function it's of. And you could think of this, I mean, you could think of it as a matrix-valued function, which feels kind of weird, but you plug in two different values, x and y, and you'll get a matrix. So it's this matrix-valued function. And the nice thing about writing it like this is that you can actually extend it so that rather than just for functions that have two variables, let's say you had a function, you know, kind of like clear this up, let's say you had a function that had three variables or four variables or kind of any number. So let's say it was a function of x, y, and z. Then you can follow this pattern and following down the first column here, the next term that you would get would be the second partial derivative of f, where first you do it with respect to x, and then you do it with respect to z. And then over here it would be the second partial derivative of f, where first you did it with respect to, first you did it with respect to y, and then you do it with respect to z. I'll clear up even more room here."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "And the nice thing about writing it like this is that you can actually extend it so that rather than just for functions that have two variables, let's say you had a function, you know, kind of like clear this up, let's say you had a function that had three variables or four variables or kind of any number. So let's say it was a function of x, y, and z. Then you can follow this pattern and following down the first column here, the next term that you would get would be the second partial derivative of f, where first you do it with respect to x, and then you do it with respect to z. And then over here it would be the second partial derivative of f, where first you did it with respect to, first you did it with respect to y, and then you do it with respect to z. I'll clear up even more room here. Because you'd have another column where you'd have the second partial derivative where this time in everything, first you do it with respect to z, and then with respect to x. And then over here you'd have the second partial derivative where first you do it with respect to z and then with respect to y. And then as the very last component, you'd have the second partial derivative where first you do it with respect to, well I guess you do it with respect to z twice."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then over here it would be the second partial derivative of f, where first you did it with respect to, first you did it with respect to y, and then you do it with respect to z. I'll clear up even more room here. Because you'd have another column where you'd have the second partial derivative where this time in everything, first you do it with respect to z, and then with respect to x. And then over here you'd have the second partial derivative where first you do it with respect to z and then with respect to y. And then as the very last component, you'd have the second partial derivative where first you do it with respect to, well I guess you do it with respect to z twice. So this whole thing, this three by three matrix would be the Hessian of a three variable function. And you can see how you could extend this pattern where if it was a four variable function, you would get a four by four matrix of all of the possible second partial derivatives. And if it was a 100 variable function, you would have a 100 by 100 matrix."}, {"video_title": "The Hessian matrix   Multivariable calculus   Khan Academy.mp3", "Sentence": "And then as the very last component, you'd have the second partial derivative where first you do it with respect to, well I guess you do it with respect to z twice. So this whole thing, this three by three matrix would be the Hessian of a three variable function. And you can see how you could extend this pattern where if it was a four variable function, you would get a four by four matrix of all of the possible second partial derivatives. And if it was a 100 variable function, you would have a 100 by 100 matrix. So the nice thing about having this is then we can talk about that by just referencing the symbol. And we'll see in the next video how this makes it very nice to express, for example, the quadratic approximation of any kind of multivariable function, not just a two variable function. And the symbols don't get way out of hand because you don't have to reference each one of these individual components."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "So here I want to talk about the gradient in the context of a contour map. So let's say we have a multivariable function, a two variable function, f of x, y, and this one is just going to equal x times y. So we can visualize this with a contour map just on the x, y plane. So what I'm going to do is I'm going to go over here, I'm going to draw my y axis and my x axis. Alright, so this right here represents x values, this represents y values, and this is entirely the input space. And I have a video on contour maps if you are unfamiliar with them or are feeling uncomfortable. And the contour map for x times y looks something like this."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "So what I'm going to do is I'm going to go over here, I'm going to draw my y axis and my x axis. Alright, so this right here represents x values, this represents y values, and this is entirely the input space. And I have a video on contour maps if you are unfamiliar with them or are feeling uncomfortable. And the contour map for x times y looks something like this. And each one of these lines represents a constant value. So you might be thinking that you have, you know, let's say you want to, the constant value for f of x times y is equal to two would be one of these lines, that would be what one of these lines represents. And a way you could think about that for this specific function is you're saying, hey, when is x times y equal to two?"}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And the contour map for x times y looks something like this. And each one of these lines represents a constant value. So you might be thinking that you have, you know, let's say you want to, the constant value for f of x times y is equal to two would be one of these lines, that would be what one of these lines represents. And a way you could think about that for this specific function is you're saying, hey, when is x times y equal to two? And that's kind of like the graph 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And a way you could think about that for this specific function is you're saying, hey, when is x times y equal to two? And that's kind of like the graph 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, it takes a look, x looks like a variable, y looks like a constant, the derivative of this whole thing is just equal to that constant, y."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, it takes a look, 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. Y looks like a variable, 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "So when we actually do this for our function, we take the partial derivative with respect to x, it takes a look, 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. Y looks like a variable, 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, but color represents length."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, but color represents length. So red here is super long, blue is gonna be kind of short. And one thing worth noticing, if you take a look at all of the given points around, if the vector is crossing a contour line, it's perpendicular to that contour line, wherever you go. You know, you go down here, this vector's perpendicular to the contour line."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And remember, we scaled down all the vectors, but color represents length. So red here is super long, blue is gonna be kind of short. And one thing worth noticing, if you take a look at all of the given points around, if the vector is crossing a contour line, it's perpendicular to that contour line, wherever you go. You know, you go down here, 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "You know, you go down here, 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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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. I'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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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. I'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."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, if you remember in the video about how to interpret the gradient in the context of a graph, I said it points in the direction of steepest ascent. So if you imagine all the possible vectors kind of pointing away from this point, the question is which direction should you move to increase the value of f the fastest? And there's two ways of thinking about that."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "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, if you remember in the video about how to interpret the gradient in the context of a graph, I said it points in the direction of steepest ascent. So if you imagine all the possible vectors kind of pointing away from this point, the question is which direction should you move to increase the value of f the fastest? And there's two ways of thinking about that. One is to look at all of these different directions and say which one increases x the most. But another way of doing it would be to get rid of them all and just take a look at another contour line that represents a slight increase. So let's say you're taking a look at a contour line, another contour line, something like this."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And there's two ways of thinking about that. One is to look at all of these different directions and say which one increases x the most. But another way of doing it would be to get rid of them all and just take a look at another contour line that represents a slight increase. So let's say you're taking a look at a contour line, another contour line, something like this. And maybe that represents something that's right next to it, like 2.1. That represents another value that's very close. And if I were a better artist and this was more representative, this would be, it would look like a line that's parallel to the original one."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "So let's say you're taking a look at a contour line, another contour line, something like this. And maybe that represents something that's right next to it, like 2.1. That represents another value that's very close. And if I were a better artist and this was more representative, this would be, it would look like a line that's parallel to the original one. Because if you change the output by just a little bit, the set of endpoints that look like it is pretty much the same but just shifted over a bit. So another way we can think about the gradient here is to say of all of the vectors that move from this output of two up to the value of 2.1, you're looking at all of the possible different vectors that do that, which one does it the fastest? And this time, instead of thinking of the fastest as constant length vectors, what increases it the most, we'll be thinking constant increase in the output, which one does it with the shortest distance?"}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And if I were a better artist and this was more representative, this would be, it would look like a line that's parallel to the original one. Because if you change the output by just a little bit, the set of endpoints that look like it is pretty much the same but just shifted over a bit. So another way we can think about the gradient here is to say of all of the vectors that move from this output of two up to the value of 2.1, you're looking at all of the possible different vectors that do that, which one does it the fastest? And this time, instead of thinking of the fastest as constant length vectors, what increases it the most, we'll be thinking constant increase in the output, which one does it with the shortest distance? And if you think of them as being roughly parallel lines, it shouldn't be hard to convince yourself that the shortest distance isn't gonna be any of those. It's gonna be the one that connects them pretty much perpendicular to the original line. Because if you think about these as lines, and the more you zoom in, the more they pretty much look like parallel lines, the path that connects one to the other is gonna be perpendicular to both of them."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "And this time, instead of thinking of the fastest as constant length vectors, what increases it the most, we'll be thinking constant increase in the output, which one does it with the shortest distance? And if you think of them as being roughly parallel lines, it shouldn't be hard to convince yourself that the shortest distance isn't gonna be any of those. It's gonna be the one that connects them pretty much perpendicular to the original line. Because if you think about these as lines, and the more you zoom in, the more they pretty much look like parallel lines, the path that connects one to the other is gonna be perpendicular to both of them. So because of this interpretation of the gradient as the direction of steepest descent, it's a natural consequence that every time it's on a contour line, wherever you're looking, it's actually perpendicular to that line. Because you can think about it as getting to the next contour line as fast as it can, increasing the function as fast as it can. And this is actually a very useful interpretation of the gradient in different contexts."}, {"video_title": "Gradient and contour maps.mp3", "Sentence": "Because if you think about these as lines, and the more you zoom in, the more they pretty much look like parallel lines, the path that connects one to the other is gonna be perpendicular to both of them. So because of this interpretation of the gradient as the direction of steepest descent, it's a natural consequence that every time it's on a contour line, wherever you're looking, it's actually perpendicular to that line. Because you can think about it as getting to the next contour line as fast as it can, increasing the function as fast as it can. And this is actually a very useful interpretation of the gradient in different contexts. So it's a good one to keep in the back of your mind. Gradient is always perpendicular to contour lines. Great, see you next video."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "And we landed at the idea that the divergence of v, you know, when you take the divergence of this vector-valued function, it should definitely have something to do with the partial derivative of p, that x component of the output with respect to x. And here I want to do the opposite and say, okay, what if we look at functions where that p, that first component, is zero, but then we have some kind of positive q component, some kind of positive, or maybe not positive, but some kind of non-zero, so positive or negative, y component. And what this would mean, instead of thinking about vectors just left and right, now we're looking at vectors that are purely up or down, kind of up or down. So kind of doing the same thing we did last time, if we start thinking about cases where the divergence of our function at a given point should be positive, and an example of that, you might be saying, nothing is happening at the point itself, so q itself would be zero, but then below it, things are kind of going away, so they're pointing down, and above it, things are going up. So in this case down here, q is a little bit less than zero, the y component of that vector is less than zero, and up here, q is greater than zero. So here we have the idea that as you're going from the bottom up, so the y value of your input is increasing as you're moving upwards in space, the value of q, this y component of the output, should also be increasing, because it goes from negative to zero to positive. So now you're starting to get this idea of partial q with respect to y, you know, as we change that y and move up in space, the value of q should be positive."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "So kind of doing the same thing we did last time, if we start thinking about cases where the divergence of our function at a given point should be positive, and an example of that, you might be saying, nothing is happening at the point itself, so q itself would be zero, but then below it, things are kind of going away, so they're pointing down, and above it, things are going up. So in this case down here, q is a little bit less than zero, the y component of that vector is less than zero, and up here, q is greater than zero. So here we have the idea that as you're going from the bottom up, so the y value of your input is increasing as you're moving upwards in space, the value of q, this y component of the output, should also be increasing, because it goes from negative to zero to positive. So now you're starting to get this idea of partial q with respect to y, you know, as we change that y and move up in space, the value of q should be positive. So positive divergence seems to correspond to a positive value here. And the thinking is actually going to be almost identical to what we did in the last video with the x component, because you can think of another circumstance where maybe you actually have a vector attached to your point and something's going on, and there even is some convergence towards it, where you have some fluid flow in towards the point, but it's just heavily outweighed by even higher divergence, even higher flow away from your point above it. And again, you have this idea of q starts off small, so here it's kind of q starts off small, maybe it's kind of like near zero, and then here q is something positive, and then here it's even more positive, and I'm sort of making up notation here, but I want the idea of kind of small, and then medium-sized, and then bigger."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "So now you're starting to get this idea of partial q with respect to y, you know, as we change that y and move up in space, the value of q should be positive. So positive divergence seems to correspond to a positive value here. And the thinking is actually going to be almost identical to what we did in the last video with the x component, because you can think of another circumstance where maybe you actually have a vector attached to your point and something's going on, and there even is some convergence towards it, where you have some fluid flow in towards the point, but it's just heavily outweighed by even higher divergence, even higher flow away from your point above it. And again, you have this idea of q starts off small, so here it's kind of q starts off small, maybe it's kind of like near zero, and then here q is something positive, and then here it's even more positive, and I'm sort of making up notation here, but I want the idea of kind of small, and then medium-sized, and then bigger. And once again, the idea of partial derivative of q with respect to y being greater than zero seems to correspond to positive divergence. And if you want, you can sketch out many more circumstances and think about, you know, what if the vector started off pointing down? What would positive and negative and zero divergence all look like?"}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "And again, you have this idea of q starts off small, so here it's kind of q starts off small, maybe it's kind of like near zero, and then here q is something positive, and then here it's even more positive, and I'm sort of making up notation here, but I want the idea of kind of small, and then medium-sized, and then bigger. And once again, the idea of partial derivative of q with respect to y being greater than zero seems to correspond to positive divergence. And if you want, you can sketch out many more circumstances and think about, you know, what if the vector started off pointing down? What would positive and negative and zero divergence all look like? But the upshot of it all, pretty much for the same reasons I went through in the last video, is this partial derivative with respect to y corresponds to the divergence. And when we combine this with our conclusions about the x component, that actually is all you need to know for the divergence. So just to write it all out, if we have a vector-valued function of x and y, and it's got both of its components, you've got p as the x component of the output, that first component of the output, and q, and we're looking at both of these at once, the way that we compute divergence, the definition of divergence of this vector-valued function is to say divergence of v as a function of x and y is actually equal to the partial derivative of p with respect to x plus the partial derivative of q with respect to y."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "What would positive and negative and zero divergence all look like? But the upshot of it all, pretty much for the same reasons I went through in the last video, is this partial derivative with respect to y corresponds to the divergence. And when we combine this with our conclusions about the x component, that actually is all you need to know for the divergence. So just to write it all out, if we have a vector-valued function of x and y, and it's got both of its components, you've got p as the x component of the output, that first component of the output, and q, and we're looking at both of these at once, the way that we compute divergence, the definition of divergence of this vector-valued function is to say divergence of v as a function of x and y is actually equal to the partial derivative of p with respect to x plus the partial derivative of q with respect to y. And that's it. That is the formula for divergence. And hopefully by now, this isn't just kind of a formula that I'm plopping down for you, but it's something that makes intuitive sense."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "So just to write it all out, if we have a vector-valued function of x and y, and it's got both of its components, you've got p as the x component of the output, that first component of the output, and q, and we're looking at both of these at once, the way that we compute divergence, the definition of divergence of this vector-valued function is to say divergence of v as a function of x and y is actually equal to the partial derivative of p with respect to x plus the partial derivative of q with respect to y. And that's it. That is the formula for divergence. And hopefully by now, this isn't just kind of a formula that I'm plopping down for you, but it's something that makes intuitive sense. When you see this term, this partial p with respect to x, you're thinking about, oh, yes, yes, because if you have flow that's kind of increasing as you move in the x direction, that's gonna correspond with movement away. And this partial derivative of q with respect to y term, hopefully you're thinking, ah, yes, as you're increasing the y component of your vector around your point, that corresponds with less flow in than there is out. So both of these correspond to that idea of divergence that we're going for."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "And hopefully by now, this isn't just kind of a formula that I'm plopping down for you, but it's something that makes intuitive sense. When you see this term, this partial p with respect to x, you're thinking about, oh, yes, yes, because if you have flow that's kind of increasing as you move in the x direction, that's gonna correspond with movement away. And this partial derivative of q with respect to y term, hopefully you're thinking, ah, yes, as you're increasing the y component of your vector around your point, that corresponds with less flow in than there is out. So both of these correspond to that idea of divergence that we're going for. And if you just add them up, this gives you everything you need to know. And one thing that's pretty neat and maybe kind of surprising is that the way we just came across this formula and started to think about it was in the simplified case where you have, you know, just pure movement in the x direction or pure movement in the y direction. But in reality, as we know, vector fields can look much more complicated."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "So both of these correspond to that idea of divergence that we're going for. And if you just add them up, this gives you everything you need to know. And one thing that's pretty neat and maybe kind of surprising is that the way we just came across this formula and started to think about it was in the simplified case where you have, you know, just pure movement in the x direction or pure movement in the y direction. But in reality, as we know, vector fields can look much more complicated. And maybe you have something where, you know, it's not just in the x direction. There's lots of things going on and you need to account for all of those. And evidently, just looking at the change in the x component with respect to x or the change in the y component of the output with respect to the y component of the input gives you all the information you need to know."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "But in reality, as we know, vector fields can look much more complicated. And maybe you have something where, you know, it's not just in the x direction. There's lots of things going on and you need to account for all of those. And evidently, just looking at the change in the x component with respect to x or the change in the y component of the output with respect to the y component of the input gives you all the information you need to know. And basically, what's going on here is that any fluid flow can just be broken down into the x and y components where you're just looking at each vector, you know, whatever vector you have, it could be broken down into its own x and y components. And if you want to think kind of concretely about the fluid flow idea, maybe you'd say that for your point, if you're looking at a point in space, you picture a very small box around it. And the reason you only need to think about x components and y components is that you're only really looking at, you know, what's going on on the left and right side, and then you can kind of calculate what the divergence according to fluid flowing in through those sides is."}, {"video_title": "Divergence formula, part 2.mp3", "Sentence": "And evidently, just looking at the change in the x component with respect to x or the change in the y component of the output with respect to the y component of the input gives you all the information you need to know. And basically, what's going on here is that any fluid flow can just be broken down into the x and y components where you're just looking at each vector, you know, whatever vector you have, it could be broken down into its own x and y components. And if you want to think kind of concretely about the fluid flow idea, maybe you'd say that for your point, if you're looking at a point in space, you picture a very small box around it. And the reason you only need to think about x components and y components is that you're only really looking at, you know, what's going on on the left and right side, and then you can kind of calculate what the divergence according to fluid flowing in through those sides is. And then you just look at kind of fluid flowing through the top or the bottom. And if you kind of shrink this box down, all you really care about is those two different directions. And anything else, anything that's kind of a diagonal into it is really just broken down into what's the y component there, what's the, you know, how's it contributing to movement up through that bottom part of the box, and then what's the x component, how's it contributing to movement through that side part of the box."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So what I want to talk about here is how to interpret the directional derivative in terms of graphs. So I have here the graph of a function, a multivariable function. It's f of x, y is equal to x squared times y. In the last couple videos, I talked about what the directional derivative is, how you can formally define it, how you compute it using the gradient. And generally the way that you, the setup that you might have is you'll have some kind of vector, and this is a vector in the input space. So in this case, it's gonna be in the xy plane. And in this case, I'll just take the vector, I'll take the vector one, one."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "In the last couple videos, I talked about what the directional derivative is, how you can formally define it, how you compute it using the gradient. And generally the way that you, the setup that you might have is you'll have some kind of vector, and this is a vector in the input space. So in this case, it's gonna be in the xy plane. And in this case, I'll just take the vector, I'll take the vector one, one. Okay. And the directional derivative, which we denote by kind of taking the gradient symbol, except you stick the name of that vector down in the lower part there, the directional derivative of your function, it'll still take the same inputs. This is kind of a measure of how the function changes when the input moves in that direction."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "And in this case, I'll just take the vector, I'll take the vector one, one. Okay. And the directional derivative, which we denote by kind of taking the gradient symbol, except you stick the name of that vector down in the lower part there, the directional derivative of your function, it'll still take the same inputs. This is kind of a measure of how the function changes when the input moves in that direction. So I'll show you what I mean. I mean, you can imagine slicing this graph by some kind of plane, but that plane doesn't necessarily have to be parallel to the x or y axes. You know, that's what we did for the partial derivative."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "This is kind of a measure of how the function changes when the input moves in that direction. So I'll show you what I mean. I mean, you can imagine slicing this graph by some kind of plane, but that plane doesn't necessarily have to be parallel to the x or y axes. You know, that's what we did for the partial derivative. We'd take a plane that represented a constant x value or a constant y value, but this is gonna be a plane that kind of tells you what movement in the direction of your vector looks like. And like I have a number of other times, I'm gonna go ahead and slice the graph along that plane. And just to make it clear, I'm gonna color in where the graph intersects that slice."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "You know, that's what we did for the partial derivative. We'd take a plane that represented a constant x value or a constant y value, but this is gonna be a plane that kind of tells you what movement in the direction of your vector looks like. And like I have a number of other times, I'm gonna go ahead and slice the graph along that plane. And just to make it clear, I'm gonna color in where the graph intersects that slice. And this vector here, this little v, you'd be thinking of as living on the xy plane, and it's determining the direction of this plane that we're slicing things with. So on the xy plane, you've got this vector, it's one, one. It kind of points in that diagonal direction."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "And just to make it clear, I'm gonna color in where the graph intersects that slice. And this vector here, this little v, you'd be thinking of as living on the xy plane, and it's determining the direction of this plane that we're slicing things with. So on the xy plane, you've got this vector, it's one, one. It kind of points in that diagonal direction. And you take the whole plane and you slice your graph. And if we want to interpret the directional derivative here, I'm gonna go ahead and fill in an actual value. So let's say we wanted to do it at like negative one, one."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "It kind of points in that diagonal direction. And you take the whole plane and you slice your graph. And if we want to interpret the directional derivative here, I'm gonna go ahead and fill in an actual value. So let's say we wanted to do it at like negative one, one. Negative one, negative one, because I guess I chose a plane that passes through the origin, so I've got to make sure that the point I'm evaluating it actually goes along this plane. But you could imagine one that points in the same direction, but you kind of slide it back and forth. So if we're doing this, we can interpret this as a slope, but you have to be very careful."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So let's say we wanted to do it at like negative one, one. Negative one, negative one, because I guess I chose a plane that passes through the origin, so I've got to make sure that the point I'm evaluating it actually goes along this plane. But you could imagine one that points in the same direction, but you kind of slide it back and forth. So if we're doing this, we can interpret this as a slope, but you have to be very careful. If you're gonna interpret this as a slope, it has to be the case that you're dealing with a unit vector that the magnitude of your vector is equal to one. I mean, it doesn't have to be, you can kind of account for it later, but it makes it more easy to think about if we're just thinking of a unit vector. So when I go over here, instead of saying that it's one, one, I'm gonna say it's whatever vector points in that same direction, but it has a unit length."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So if we're doing this, we can interpret this as a slope, but you have to be very careful. If you're gonna interpret this as a slope, it has to be the case that you're dealing with a unit vector that the magnitude of your vector is equal to one. I mean, it doesn't have to be, you can kind of account for it later, but it makes it more easy to think about if we're just thinking of a unit vector. So when I go over here, instead of saying that it's one, one, I'm gonna say it's whatever vector points in that same direction, but it has a unit length. And in this case, that happens to be square root of two over two for each of the components. And you can kind of think about why that would be true, Pythagorean theorem and all, but this is a vector with unit length, its magnitude is one, and it points in that direction. And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So when I go over here, instead of saying that it's one, one, I'm gonna say it's whatever vector points in that same direction, but it has a unit length. And in this case, that happens to be square root of two over two for each of the components. And you can kind of think about why that would be true, Pythagorean theorem and all, but this is a vector with unit length, its magnitude is one, and it points in that direction. And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is. And in this case, it'll be, whoop, moving things about when I add a point. It'll be this point here, and you kind of look from above, and you say, okay, that is kind of negative one, negative one. And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "And if we're evaluating this at a point like negative one, one, we can draw that on the graph, see where it actually is. And in this case, it'll be, whoop, moving things about when I add a point. It'll be this point here, and you kind of look from above, and you say, okay, that is kind of negative one, negative one. And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is. So the reason that the directional derivative is gonna give us this slope is because you think about this, another notation that might be kind of helpful for what this directional derivative is, some people will write partial f and then partial v. And you know, you can think about that as taking a slight nudge in the direction of v, right? So this would be a little nudge, a little partial nudge in the direction of v, and then you're saying, what change in the value of the function does that result in? You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "And if we want the slope at that point, you're kind of thinking of the tangent line here, tangent line to that curve, and we're wondering what its slope is. So the reason that the directional derivative is gonna give us this slope is because you think about this, another notation that might be kind of helpful for what this directional derivative is, some people will write partial f and then partial v. And you know, you can think about that as taking a slight nudge in the direction of v, right? So this would be a little nudge, a little partial nudge in the direction of v, and then you're saying, what change in the value of the function does that result in? You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line. So conceptually, that's kind of a nicer notation, but the reason we use this other notation, this nabla sub v one, is it's very indicative of how you compute things once you need to compute it. You take the gradient of f, just the vector-valued function gradient of f, and take the dot product with your vector, with the vector. So let's actually do that, just to see what this would look like."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "You know, the height of the graph tells you the value of the function, and as this initial change approaches zero, and the resulting change approaches zero as well, that ratio, the ratio of df to, or partial f to partial v, is gonna give you the slope of this tangent line. So conceptually, that's kind of a nicer notation, but the reason we use this other notation, this nabla sub v one, is it's very indicative of how you compute things once you need to compute it. You take the gradient of f, just the vector-valued function gradient of f, and take the dot product with your vector, with the vector. So let's actually do that, just to see what this would look like. And I'll go ahead and write it over here, I'll use a different color. So the gradient of f, first of all, gradient of f, is a vector full of partial derivatives. So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So let's actually do that, just to see what this would look like. And I'll go ahead and write it over here, I'll use a different color. So the gradient of f, first of all, gradient of f, is a vector full of partial derivatives. So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y. And when we actually evaluate this, we take a look, partial derivative of f with respect to x, x looks like the variable, y is just a constant, so its partial derivative is two times x times y. Two times x times y, but when we take the partial with respect to y, y now looks like a variable, x looks like a constant. Derivative of a constant times the variable is just that constant, x squared."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "So it'll be the partial derivative of f with respect to x, and the partial derivative of f with respect to y. And when we actually evaluate this, we take a look, partial derivative of f with respect to x, x looks like the variable, y is just a constant, so its partial derivative is two times x times y. Two times x times y, but when we take the partial with respect to y, y now looks like a variable, x looks like a constant. Derivative of a constant times the variable is just that constant, x squared. And if we were to evaluate this at the point negative one, negative one, now you could plug that in, two times negative one times negative one would be two, and then negative one squared would be one. So that would be our gradient at that point, which means if we want to evaluate gradient of f times v, we could go over here and say that that's two, one, because we evaluate the gradient at the point we care about, and then the dot product with v itself, in this case, root two over two, and root two over two. The answer that we'd get, we multiply the first two components together, two times root two over two is square root of two, and then here, we multiply the second components together, and that's gonna be one times root two over two, root two over two, and that would be our answer, that would be our slope."}, {"video_title": "Directional derivatives and slope.mp3", "Sentence": "Derivative of a constant times the variable is just that constant, x squared. And if we were to evaluate this at the point negative one, negative one, now you could plug that in, two times negative one times negative one would be two, and then negative one squared would be one. So that would be our gradient at that point, which means if we want to evaluate gradient of f times v, we could go over here and say that that's two, one, because we evaluate the gradient at the point we care about, and then the dot product with v itself, in this case, root two over two, and root two over two. The answer that we'd get, we multiply the first two components together, two times root two over two is square root of two, and then here, we multiply the second components together, and that's gonna be one times root two over two, root two over two, and that would be our answer, that would be our slope. But this only works if your vector is a unit vector, because, and I showed this in the last video where we talked about the formal definition of the directional derivative, if you scale v by two, and I could do it here, if instead of v, you're talking about two v, so I'll go ahead and make myself some room here. If you're taking the directional derivative along two v of f, the way that we're computing that, you're still taking the gradient of f, dot product with two times your vector, and dot products, you can pull out that two, this is just gonna double the value of the entire thing, dot v of this dotted with v, it's gonna be twice the value, the derivative will become twice the value, but you don't necessarily want that, because you'd say this plane that you slice it with, if instead of doing it in the direction of v, the unit vector, you did it in the direction of two times v, it's the same plane, it's the same slice you're taking, and you'd want that same slope, so that's gonna mess everything up. So this is super important if you're thinking about things in the context of slope, one thing that you could say is your formula for the slope of a graph in the direction of v, is you'd take your directional derivative, that dot product between f and v, and you just always make sure to divide it by the magnitude of v, divide it by that magnitude, and that'll always take care of what you want, that's basically a way of making sure that really you're taking the directional derivative in the direction of a certain unit vector, some people even go so far as to define the directional derivative to be this, to be something where you normalize out the length of that vector, I don't really like that, but I think that's because they're thinking of the slope context, they're thinking of rates of change as being the slope of a graph, and one thing I'd like to emphasize as always, graphical intuition is good, and visual intuition is always great, you should always be trying to find a way to think about things visually, but with multivariable functions, the graph isn't the only way, you can kind of more generally think about just a nudge in the v direction, and in the context where v doesn't have a length one, the nudge doesn't represent an actual size, but it's a certain scaling constant times that vector, you can look at the video on the formal definition for the directional derivative if you want more details on that, but I do think this is actually a good way to get a feel for what the directional derivative is all about."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me actually graph this. So let's say R of t, I'll draw it a little bit straighter than that. That's my y axis, that is my x axis. Let's say R of t, and this is for t is less than, let me write this. So this is for t is between a and b. So when t is equal to a, we're at this vector right here. So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let's say R of t, and this is for t is less than, let me write this. So this is for t is between a and b. So when t is equal to a, we're at this vector right here. So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there. And then as t increases, it traces out a curve, or the endpoints of our position vectors trace out a curve that looks something like that. So when t is equal to b, we get a position vector that points to that point right there. So this defines a path, and the path is going in this upward direction, just like that."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you actually substitute t is equal to a here, you would get a position vector that would point to that point over there. And then as t increases, it traces out a curve, or the endpoints of our position vectors trace out a curve that looks something like that. So when t is equal to b, we get a position vector that points to that point right there. So this defines a path, and the path is going in this upward direction, just like that. Now let's say that we have another position vector function. Let me call it R, I'll call it R, I don't know, R reverse, R sub R of t. Actually, let me just call it R of t, it'll make it more confusing if I try to differentiate. But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this defines a path, and the path is going in this upward direction, just like that. Now let's say that we have another position vector function. Let me call it R, I'll call it R, I don't know, R reverse, R sub R of t. Actually, let me just call it R of t, it'll make it more confusing if I try to differentiate. But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i. And instead of y of t, it's going to be y of a plus b minus t times i. And we've seen this in the last two videos. The path defined by this position vector function is going to look more like this."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But this is a different one, it's the green R. R of t, instead of being x of t times i, it's going to be x of a plus b minus t times i. And instead of y of t, it's going to be y of a plus b minus t times i. And we've seen this in the last two videos. The path defined by this position vector function is going to look more like this. Let me draw my axes. That's my y-axis, that is my x-axis, maybe I should label them y and x. This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The path defined by this position vector function is going to look more like this. Let me draw my axes. That's my y-axis, that is my x-axis, maybe I should label them y and x. This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b. So t is going to go from a to b. But here, when t is equal to a, you substitute it over here, you're going to get this vector. You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This path is going to look just like this, but instead of starting here and going there, when t is equal to 1, let me make it clear, this is also true for a is less than or equal to t, which is less than b. So t is going to go from a to b. But here, when t is equal to a, you substitute it over here, you're going to get this vector. You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction. So it's the same path in the opposite direction. And so when t is equal to b, you put that in here, you're actually going to get x of a and y of a there, the b's cancel out, and so you're going to point right like that. So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "You're going to start over here, you're going to start over there, and as you increment t, as you make it larger and larger and larger, you're going to trace out that same path, but in the opposite direction. So it's the same path in the opposite direction. And so when t is equal to b, you put that in here, you're actually going to get x of a and y of a there, the b's cancel out, and so you're going to point right like that. So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction. So what we're going to do in this video is to see what happens, how, I guess you could say, if I have some vector field, let's say I have some vector field, f of x, y equals p of x, y, i plus q of x, y, j, right? This is just a vector field over the x, y plane. How the line integral of this vector field over this path compares to the line integral of the same vector field over that path."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So these are the same, you can imagine the shape of these paths are the same, but we're going in the exact opposite direction. So what we're going to do in this video is to see what happens, how, I guess you could say, if I have some vector field, let's say I have some vector field, f of x, y equals p of x, y, i plus q of x, y, j, right? This is just a vector field over the x, y plane. How the line integral of this vector field over this path compares to the line integral of the same vector field over that path. How that compares to this. We'll call this the minus curve. So this is the positive curve, we're going to call this the minus curve."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "How the line integral of this vector field over this path compares to the line integral of the same vector field over that path. How that compares to this. We'll call this the minus curve. So this is the positive curve, we're going to call this the minus curve. So how does it going over the positive curve compare to going over the minus curve, f of f dot dr. So before I break into the math, let's just think about it a little bit. Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is the positive curve, we're going to call this the minus curve. So how does it going over the positive curve compare to going over the minus curve, f of f dot dr. So before I break into the math, let's just think about it a little bit. Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff. So you know, on every point in the x, y plane, it has a vector, it defines or maps a vector to every point in the x, y plane. But we really care about the points that are on the curve. So maybe on the curve, this is the vector field at the points on the curve."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me draw this vector field f. So maybe it looks, I'm just going to draw random stuff. So you know, on every point in the x, y plane, it has a vector, it defines or maps a vector to every point in the x, y plane. But we really care about the points that are on the curve. So maybe on the curve, this is the vector field at the points on the curve. And let me draw it over here too. So on the points on the curve where we care about, this is our vector field. That is our vector field."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So maybe on the curve, this is the vector field at the points on the curve. And let me draw it over here too. So on the points on the curve where we care about, this is our vector field. That is our vector field. And let's just get an intuition of what's going to be going. We're summing over the dot, we're taking each point along the line. We're going to take each point along the line."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is our vector field. And let's just get an intuition of what's going to be going. We're summing over the dot, we're taking each point along the line. We're going to take each point along the line. And we're summing, let me start over here. Each point along the line, let me do it in a different color. And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're going to take each point along the line. And we're summing, let me start over here. Each point along the line, let me do it in a different color. And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function. And dr you can kind of imagine as a small, an infinitesimally small vector going in the direction of our movement. And when we take this dot product here, it's essentially, it's going to be a scalar value, but the dot product, if you remember, it's the magnitude of f in the direction of dr times the magnitude of dr. So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And we're summing the dot product of the value of the vector field at that point, the dot product of that, with dr, or the differential of our position vector function. And dr you can kind of imagine as a small, an infinitesimally small vector going in the direction of our movement. And when we take this dot product here, it's essentially, it's going to be a scalar value, but the dot product, if you remember, it's the magnitude of f in the direction of dr times the magnitude of dr. So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful. So this little thing that I'm drawing right here, let's say that this is my path. This is f at that point. F at that point looks something like that."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So it's this, you can imagine it's the shadow of f onto dr. Let me zoom into that because I think it's useful. So this little thing that I'm drawing right here, let's say that this is my path. This is f at that point. F at that point looks something like that. And then dr at this point looks something like that. Let me do it in a different color. dr looks something like that."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "F at that point looks something like that. And then dr at this point looks something like that. Let me do it in a different color. dr looks something like that. So that is f. And so the dot product of these two says, okay, how much of f is going in the same direction as dr? And you can kind of imagine it as a shadow. So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dr looks something like that. So that is f. And so the dot product of these two says, okay, how much of f is going in the same direction as dr? And you can kind of imagine it as a shadow. So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product. In this case, we're going to get a positive number because this length is positive, this length is positive. That's going to be a positive number. Now what if our dr was going in the opposite direction as it is in this case?"}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So you take the f that's going in the same direction as dr, the magnitude of that times the magnitude of dr, that is the dot product. In this case, we're going to get a positive number because this length is positive, this length is positive. That's going to be a positive number. Now what if our dr was going in the opposite direction as it is in this case? So let me draw maybe that same part of the curve. We have our f. Our f will look something like that. That is our f. I'm drawing this exact same part of the curve."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now what if our dr was going in the opposite direction as it is in this case? So let me draw maybe that same part of the curve. We have our f. Our f will look something like that. That is our f. I'm drawing this exact same part of the curve. But now our dr isn't going in that direction. Our dr at this point is going to be going in the other direction. We're tracing the curve in the opposite direction."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That is our f. I'm drawing this exact same part of the curve. But now our dr isn't going in that direction. Our dr at this point is going to be going in the other direction. We're tracing the curve in the opposite direction. So our dr is now going to be going in that direction. So this is our dr. So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're tracing the curve in the opposite direction. So our dr is now going to be going in that direction. So this is our dr. So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr. So you can imagine that when you multiply the magnitude, you should get a negative number. Our direction is now opposite. The shadow of f onto the same direction as dr is going in the opposite direction as dr."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So if you do f dot dr, you're taking the shadow, or how much of f is going in the direction of dr, you take the shadow down here, it's going in the opposite direction of dr. So you can imagine that when you multiply the magnitude, you should get a negative number. Our direction is now opposite. The shadow of f onto the same direction as dr is going in the opposite direction as dr. So in this case, it's going in the same direction as dr. So the intuition is that maybe these two things are the negative of each other. And now we can do some math and try to see if that is definitely the case."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "The shadow of f onto the same direction as dr is going in the opposite direction as dr. So in this case, it's going in the same direction as dr. So the intuition is that maybe these two things are the negative of each other. And now we can do some math and try to see if that is definitely the case. So let us first figure out, let's write an expression for the differential dr. So in this case, dr dt is going to be equal to x prime of t times i plus y prime of t times j. In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to?"}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And now we can do some math and try to see if that is definitely the case. So let us first figure out, let's write an expression for the differential dr. So in this case, dr dt is going to be equal to x prime of t times i plus y prime of t times j. In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to? It's the derivative of x with respect to t. The derivative of this term with respect to t, that's the derivative of the inside, which is minus 1, or minus, times the derivative of the outside with respect to the inside. So the derivative of the outside is minus 1 times the derivative of the outside with respect to the inside. x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "In this other example, in the reverse case, our dr dt is going to be, what's it going to be equal to? It's the derivative of x with respect to t. The derivative of this term with respect to t, that's the derivative of the inside, which is minus 1, or minus, times the derivative of the outside with respect to the inside. So the derivative of the outside is minus 1 times the derivative of the outside with respect to the inside. x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj. So this is dr dt in this case. This is dr dt in that case. And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "x prime of a plus b minus t times i. of y of this term, the derivative of the inside, which is minus 1, times the derivative of the outside with respect to the inside, which is y prime of a plus b minus t. So this is going to be the derivative of the inside times y prime of a plus b minus tj. So this is dr dt in this case. This is dr dt in that case. And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt. I could multiply it down into each of these terms, but it keeps it simple just leaving it on the outside. Same logic over here. dr is equal to minus x. I changed my shade of green, but at least it's still green."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then if we wanted to write the differential dr in the forward curve example, it's going to be equal to x prime of t times i plus y prime of t times j times the scalar dt. I could multiply it down into each of these terms, but it keeps it simple just leaving it on the outside. Same logic over here. dr is equal to minus x. I changed my shade of green, but at least it's still green. a plus b minus ti minus y prime a plus b minus tj. And I'm multiplying both sides by dt. Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dr is equal to minus x. I changed my shade of green, but at least it's still green. a plus b minus ti minus y prime a plus b minus tj. And I'm multiplying both sides by dt. Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink. The pink one is going to be equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, dot this thing over here, which is, I'll just write it out here, I could simplify it later. x prime of ti plus y prime of tj. And then all of that times the scalar dt."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Now we're ready to express this as a function of t. So this curve right here, I'll do it in pink. The pink one is going to be equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, dot this thing over here, which is, I'll just write it out here, I could simplify it later. x prime of ti plus y prime of tj. And then all of that times the scalar dt. This will be a scalar value, and then we'll have another scalar value dt over there. Now what is this going to be equal to? If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then all of that times the scalar dt. This will be a scalar value, and then we'll have another scalar value dt over there. Now what is this going to be equal to? If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space. Dot, this is a vector, so dot this guy right here, dot dr. Dot minus x prime of a plus b minus ti minus y prime of, I'm using up too much space. Let me scroll, go back a little bit."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "If I were to take this reverse integral, the reverse integral is going to be the integral from, I'm going to need a little more space, from a to b, of f of not x of t, but x of a plus b minus t, y of a plus b minus t. I'm writing it small, so I have some space. Dot, this is a vector, so dot this guy right here, dot dr. Dot minus x prime of a plus b minus ti minus y prime of, I'm using up too much space. Let me scroll, go back a little bit. Actually, let me make it even simpler. Let me take this minus sign out of it. Let me put a plus, and I'll put the minus sign out front."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me scroll, go back a little bit. Actually, let me make it even simpler. Let me take this minus sign out of it. Let me put a plus, and I'll put the minus sign out front. So the minus sign is just a scalar value. So we could put that minus sign out, when you take a dot product, and if you multiply a scalar times a dot product, you could just take the scalar out. That's all I'm saying."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Let me put a plus, and I'll put the minus sign out front. So the minus sign is just a scalar value. So we could put that minus sign out, when you take a dot product, and if you multiply a scalar times a dot product, you could just take the scalar out. That's all I'm saying. So we could take that minus sign out to this part right here, and then you have x prime of a plus b minus t, i plus y prime of a plus b minus t, let me scroll over a little bit, tj, dt. So this is the forward. This is when we're following it along the forward curve."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "That's all I'm saying. So we could take that minus sign out to this part right here, and then you have x prime of a plus b minus t, i plus y prime of a plus b minus t, let me scroll over a little bit, tj, dt. So this is the forward. This is when we're following it along the forward curve. This is when we're following it along the reverse curve. Now, like we did with the scalar example, let's make a substitution. I want to make it very clear what I did."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "This is when we're following it along the forward curve. This is when we're following it along the reverse curve. Now, like we did with the scalar example, let's make a substitution. I want to make it very clear what I did. All I did here is I just took the dot product, but this negative sign, I just took it out. I just said this is the same thing as negative 1 times this thing, or negative 1 times this thing is the same thing as that. So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I want to make it very clear what I did. All I did here is I just took the dot product, but this negative sign, I just took it out. I just said this is the same thing as negative 1 times this thing, or negative 1 times this thing is the same thing as that. So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side. So let me make a substitution. u is equal to a plus b minus t. Then we get du is equal to minus dt. Just take the derivative of both sides, or you get dt is equal to minus du."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So let's make a substitution on this side, because I really just want to show you that this is the negative of that right there, because that's what our intuition was going for, so let me just focus on that side. So let me make a substitution. u is equal to a plus b minus t. Then we get du is equal to minus dt. Just take the derivative of both sides, or you get dt is equal to minus du. And then you get when t is equal to a, u is equal to a plus b minus a. So then u is equal to b. And then when t is equal to b, u is equal to a."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "Just take the derivative of both sides, or you get dt is equal to minus du. And then you get when t is equal to a, u is equal to a plus b minus a. So then u is equal to b. And then when t is equal to b, u is equal to a. When t is equal to b, a plus b minus b is a. u is equal to a. So this thing, using that substitution, simplifies to this is the whole point, that simplifies to minus the integral from when t is a, u is b. From b, when t is b, u is a."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "And then when t is equal to b, u is equal to a. When t is equal to b, a plus b minus b is a. u is equal to a. So this thing, using that substitution, simplifies to this is the whole point, that simplifies to minus the integral from when t is a, u is b. From b, when t is b, u is a. The integral from u is equal to b to u is equal to a of f of x of u, y of u, dot x prime of u times i, that's u right there, plus y prime of u times j. And then instead of a dt, I need to put a du. dt is equal to minus du."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "From b, when t is b, u is a. The integral from u is equal to b to u is equal to a of f of x of u, y of u, dot x prime of u times i, that's u right there, plus y prime of u times j. And then instead of a dt, I need to put a du. dt is equal to minus du. So I could write minus du here, or just to not make it confusing, I'll put the du here. And take the minus out front. I already have a minus out there, so they cancel out."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "dt is equal to minus du. So I could write minus du here, or just to not make it confusing, I'll put the du here. And take the minus out front. I already have a minus out there, so they cancel out. They will cancel out just like that. And so you might say, hey Sal, these two things look pretty similar to each other. They don't look like they're negative of each other."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "I already have a minus out there, so they cancel out. They will cancel out just like that. And so you might say, hey Sal, these two things look pretty similar to each other. They don't look like they're negative of each other. And I'd say, well, you're almost right, except this guy's limits of integration are reversed from this guy. So this thing right here, if we reverse the limits of integration, we have to then make it negative. So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "They don't look like they're negative of each other. And I'd say, well, you're almost right, except this guy's limits of integration are reversed from this guy. So this thing right here, if we reverse the limits of integration, we have to then make it negative. So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du. And now this is identical, this integral, this definite integral, is identical to that definite integral. We just have a different variable. We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "So this is equal to minus the integral from a to b of f dot, or the vector f of x of u, y of u, dot x prime of u i plus y prime of u j du. And now this is identical, this integral, this definite integral, is identical to that definite integral. We just have a different variable. We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters. If you go in the reverse direction, you're going to get the negative version of that. And that's because at any point we take the dot product, you're going in the opposite direction, so it'll be the negative of each other. But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "We're doing dt here, we're having du here, but we're going to get the same exact number for any a or b, and given this vector f and the position vector path r of t. So just to summarize everything up, when you're dealing with line integrals over vector fields, the direction matters. If you go in the reverse direction, you're going to get the negative version of that. And that's because at any point we take the dot product, you're going in the opposite direction, so it'll be the negative of each other. But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in. That the positive path has the same value as the negative path, and that's just because we're just trying to find the area of that curtain. Hopefully you found that mildly amusing."}, {"video_title": "Vector field line integrals dependent on path direction   Multivariable Calculus   Khan Academy.mp3", "Sentence": "But when you're dealing with a scalar field, we saw in the last video, that it doesn't matter which direction that you traverse the path in. That the positive path has the same value as the negative path, and that's just because we're just trying to find the area of that curtain. Hopefully you found that mildly amusing."}]