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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Divergence theorem example 1 Divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Gradient and graphs.mp3
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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?
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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?
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Gradient and graphs.mp3
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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?
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Gradient and graphs.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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?
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Multivariable chain rule.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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?
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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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.
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Conceptual clarification for 2D divergence theorem Multivariable Calculus Khan Academy.mp3
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