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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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?
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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.
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Line integral example 2 (part 1) Multivariable Calculus Khan Academy.mp3
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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?
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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?
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Directional derivative, formal definition.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Computing a Jacobian matrix.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Evaluating line integral directly - part 1 Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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...
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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?
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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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.
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Introduction to the line integral Multivariable Calculus Khan Academy.mp3
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