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Or how fast does f change with respect to t? It says, well, there's two ways that f can change. It can change with respect to x, and it can change with respect to y. So why don't I add those two things together as they are both changing with respect to t. That's all it's saying. And if you kind of imagined that you could cancel out this partial x with this dx, and this partial y with this dy, you could kind of imagine the partial of f with respect to t on the x side of things, and then plus the partial of f with respect to t on the y, in the y dimension. And then that'll give you the total change of f with respect to t. Kind of a hand wavy argument there, but at least to me, this is a pretty intuitive formula. So that's our toolkit right there, the multivariable chain rule.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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So why don't I add those two things together as they are both changing with respect to t. That's all it's saying. And if you kind of imagined that you could cancel out this partial x with this dx, and this partial y with this dy, you could kind of imagine the partial of f with respect to t on the x side of things, and then plus the partial of f with respect to t on the y, in the y dimension. And then that'll give you the total change of f with respect to t. Kind of a hand wavy argument there, but at least to me, this is a pretty intuitive formula. So that's our toolkit right there, the multivariable chain rule. We're going to put that aside for a second. Now, let's say I have some vector field f. And it's different than this f, so I'll do it in a different color, magenta. I have some vector field f that is a function of x and y.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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So that's our toolkit right there, the multivariable chain rule. We're going to put that aside for a second. Now, let's say I have some vector field f. And it's different than this f, so I'll do it in a different color, magenta. I have some vector field f that is a function of x and y. And let's say that it happens to be the gradient of some scalar field. Let's say it equals the gradient of some scalar field. I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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I have some vector field f that is a function of x and y. And let's say that it happens to be the gradient of some scalar field. Let's say it equals the gradient of some scalar field. I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y. So I could write f is also, let me write it, I don't want to write it on a new line. I could also write up here, capital F is also a function of x and y. And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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I'll call that capital F. And this is gradient, which means that this is, capital F is also a function of x and y. So I could write f is also, let me write it, I don't want to write it on a new line. I could also write up here, capital F is also a function of x and y. And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector. This is the definition of the gradient right here. This is the definition of a gradient. And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And the gradient, all that means is that the vector field f of x, y, lowercase f of x, y, is equal to the partial derivative of uppercase F with respect to x times the i unit vector plus the partial of uppercase F with respect to y times the j unit vector. This is the definition of the gradient right here. This is the definition of a gradient. And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y. The gradient of f of x, y is going to be a vector field that tells you the direction of steepest descent at any point. So it'll be defined on the x, y plane. So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And if you imagine that uppercase F is some type of surface, so if uppercase F is some type of surface like that, just trying to do my, so this is uppercase F of x, y. The gradient of f of x, y is going to be a vector field that tells you the direction of steepest descent at any point. So it'll be defined on the x, y plane. So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis. So the gradient of it, if you take any point on the x, y plane, it'll tell you the direction you need to travel to go into the steepest descent. And for this gradient field, it's going to be something like this. And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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So on the x, y plane, it'll tell you, so let me draw, that's the vertical axis, maybe that's the x-axis, that's the y-axis. So the gradient of it, if you take any point on the x, y plane, it'll tell you the direction you need to travel to go into the steepest descent. And for this gradient field, it's going to be something like this. And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here. Anyway, I don't want to get too involved in that. And the whole point of this isn't to really get the intuition behind gradients. There are other videos on this.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And maybe over here it starts going in that direction, because you would descend towards this little minimum point right here. Anyway, I don't want to get too involved in that. And the whole point of this isn't to really get the intuition behind gradients. There are other videos on this. The point of this is to get a test to see whether something is path independent, whether a vector field is path independent, whether it's conservative. And it turns out that if this exists, and I'm going to prove it now, if this exists, if f is the gradient of some scalar field, if f is equal to the gradient of some scalar field, then f is conservative. Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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There are other videos on this. The point of this is to get a test to see whether something is path independent, whether a vector field is path independent, whether it's conservative. And it turns out that if this exists, and I'm going to prove it now, if this exists, if f is the gradient of some scalar field, if f is equal to the gradient of some scalar field, then f is conservative. Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point. Now let me see if I can prove that to you. So let's start with the assumption that f can be written this way, as the gradient of that lowercase f can be written as the gradient of some uppercase F. So in that case, our integral, let's define our path first. So our position vector function, we always need one of those to do a line integral or a vector line integral.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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Or you could say it doesn't matter what path we follow when we take a line integral over f. It just matters about our starting point and our ending point. Now let me see if I can prove that to you. So let's start with the assumption that f can be written this way, as the gradient of that lowercase f can be written as the gradient of some uppercase F. So in that case, our integral, let's define our path first. So our position vector function, we always need one of those to do a line integral or a vector line integral. r of t is going to be equal to x of t times i plus y of t times j, for t going between a and b. We've seen this multiple times. This is just a very general definition of pretty much any path in two dimensions.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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So our position vector function, we always need one of those to do a line integral or a vector line integral. r of t is going to be equal to x of t times i plus y of t times j, for t going between a and b. We've seen this multiple times. This is just a very general definition of pretty much any path in two dimensions. And then we're going to say f of x, y is going to be equal to this. It's going to be the partial derivative of uppercase F with respect to x. So we're assuming that this exists, that this is true.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This is just a very general definition of pretty much any path in two dimensions. And then we're going to say f of x, y is going to be equal to this. It's going to be the partial derivative of uppercase F with respect to x. So we're assuming that this exists, that this is true. With respect to times i plus the partial of uppercase F with respect to y times j. Now given this, what is lowercase f dot dr going to equal over this path right here? This path is defined by this position function right there.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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So we're assuming that this exists, that this is true. With respect to times i plus the partial of uppercase F with respect to y times j. Now given this, what is lowercase f dot dr going to equal over this path right here? This path is defined by this position function right there. Well, it's going to be equal to, well, we need to figure out what dr is, and we've done that in multiple videos. I'll do that on the right over here. dr, we've seen it multiple times.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This path is defined by this position function right there. Well, it's going to be equal to, well, we need to figure out what dr is, and we've done that in multiple videos. I'll do that on the right over here. dr, we've seen it multiple times. I'll just solve it out again. dr dt by definition was equal to dx dt times i plus dy dt times j. That's what dr dt is.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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dr, we've seen it multiple times. I'll just solve it out again. dr dt by definition was equal to dx dt times i plus dy dt times j. That's what dr dt is. So if we want to figure out what dr is, the differential dr, if we want to play with differentials in this way, multiply both sides times dt. And actually I'm going to treat dt as, well, I'll multiply it, I'll distribute it. It's dx dt times dt i plus dy dt times dt j.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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That's what dr dt is. So if we want to figure out what dr is, the differential dr, if we want to play with differentials in this way, multiply both sides times dt. And actually I'm going to treat dt as, well, I'll multiply it, I'll distribute it. It's dx dt times dt i plus dy dt times dt j. So if we're taking the dot product of f with dr, what are we going to get? We are going to get, and this is, so we're going to take, so this is going to be the integral over the curve from, I'll write the c right there. We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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It's dx dt times dt i plus dy dt times dt j. So if we're taking the dot product of f with dr, what are we going to get? We are going to get, and this is, so we're going to take, so this is going to be the integral over the curve from, I'll write the c right there. We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt. I'm going to write this dt in a different color. Times dt plus the partial of uppercase F with respect to y times this, we're multiplying the j components, right? We take the dot product, multiply the i components, and then add that to what you get from the product of the j components.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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We could write it in terms of the endpoints of t. Once we feel good that we have everything in terms of t. But it's going to be equal to this dot that, which is equal to the partial, I'll try to stay color consistent, the partial of uppercase F with respect to x times that, times dx dt. I'm going to write this dt in a different color. Times dt plus the partial of uppercase F with respect to y times this, we're multiplying the j components, right? We take the dot product, multiply the i components, and then add that to what you get from the product of the j components. So this j component is partial of uppercase F with respect to y, and then we have times, let me switch to a yellow, dy dt times that dt right over there. And then we can factor out the dt. Or actually, just so I don't have to even write it again.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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We take the dot product, multiply the i components, and then add that to what you get from the product of the j components. So this j component is partial of uppercase F with respect to y, and then we have times, let me switch to a yellow, dy dt times that dt right over there. And then we can factor out the dt. Or actually, just so I don't have to even write it again. Right now I wrote it without, well let me write it again. So this is equal to the integral. And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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Or actually, just so I don't have to even write it again. Right now I wrote it without, well let me write it again. So this is equal to the integral. And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b. And so this is going to be equal to, I'll write it in blue, the partial of uppercase F with respect to x times dx dt plus, I'm distributing this dt out, plus the partial of uppercase F with respect to y dy dt, all of that times dt. This is equivalent to that. Now, you might realize why I talked about the multivariable chain rule.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And let's say we have it in terms of t. We've written everything in terms of t. So t goes from a to b. And so this is going to be equal to, I'll write it in blue, the partial of uppercase F with respect to x times dx dt plus, I'm distributing this dt out, plus the partial of uppercase F with respect to y dy dt, all of that times dt. This is equivalent to that. Now, you might realize why I talked about the multivariable chain rule. What is this right here? What is that right there? You can do some pattern matching.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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Now, you might realize why I talked about the multivariable chain rule. What is this right here? What is that right there? You can do some pattern matching. That is the same thing as the derivative of uppercase F with respect to t. Look at this. Let me copy and paste this, just so you appreciate it. Let me copy and paste that.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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You can do some pattern matching. That is the same thing as the derivative of uppercase F with respect to t. Look at this. Let me copy and paste this, just so you appreciate it. Let me copy and paste that. Copy it and then let me paste it. So this is our definition, or this is our, I won't say definition, one can actually prove it. You don't have to start from there.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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Let me copy and paste that. Copy it and then let me paste it. So this is our definition, or this is our, I won't say definition, one can actually prove it. You don't have to start from there. But this is our multivariable chain rule right here. The derivative of any function with respect to t is the partial of that function with respect to x times dx dt plus the partial of that function with respect to y dy dt. I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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You don't have to start from there. But this is our multivariable chain rule right here. The derivative of any function with respect to t is the partial of that function with respect to x times dx dt plus the partial of that function with respect to y dy dt. I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y. This and this are identical if you just replace this lowercase f with an uppercase F. So this in blue right here, so this whole expression is equal to the integral from a to b, t is equal to a to t is equal to b, of, in blue here, the regular derivative of F with respect to t dt. And how do you evaluate, let me write the dt in green. How do you evaluate something like this?
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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I have the partial of uppercase F with respect to x dx dt plus the partial of uppercase F with respect to y. This and this are identical if you just replace this lowercase f with an uppercase F. So this in blue right here, so this whole expression is equal to the integral from a to b, t is equal to a to t is equal to b, of, in blue here, the regular derivative of F with respect to t dt. And how do you evaluate, let me write the dt in green. How do you evaluate something like this? And I just want to make a point. This is just this from the multivariable chain rule. And how do you evaluate a definite integral like this?
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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How do you evaluate something like this? And I just want to make a point. This is just this from the multivariable chain rule. And how do you evaluate a definite integral like this? Well you take the antiderivative of the inside with respect to dt. So what is this going to be equal to? You take the antiderivative of the inside.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And how do you evaluate a definite integral like this? Well you take the antiderivative of the inside with respect to dt. So what is this going to be equal to? You take the antiderivative of the inside. The antiderivative of the inside, that's just dF. I'm sorry, that's just F. So this is equal to F of t. And let me be clear here. We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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You take the antiderivative of the inside. The antiderivative of the inside, that's just dF. I'm sorry, that's just F. So this is equal to F of t. And let me be clear here. We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways. And this could be just written as F of t. This is the same thing as F of t. These are all equivalent, depending on whether you want to include the x's and the y's only, or the t's only, or them both. Because both of the x's and y's are functions of t. So this is the derivative of F with respect to t. If this was just in terms of t, this is the derivative of that with respect to t. We take its antiderivative. We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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We wrote before that F is a function, so our uppercase F is a function of x and y, which could also be written, since each of these are functions of t, of, could be written as F of x of t of y of t. I'm just rewriting it in different ways. And this could be just written as F of t. This is the same thing as F of t. These are all equivalent, depending on whether you want to include the x's and the y's only, or the t's only, or them both. Because both of the x's and y's are functions of t. So this is the derivative of F with respect to t. If this was just in terms of t, this is the derivative of that with respect to t. We take its antiderivative. We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b. And so this is equal to, and this is the home stretch, this is equal to F of b minus F of a. And if you want to think about it in these terms, this is the same thing. This is equal to F of x of b y of b minus F of x of a y of a.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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We're left just with F. And we have to evaluate it from t is equal to a to t is equal to b. And so this is equal to, and this is the home stretch, this is equal to F of b minus F of a. And if you want to think about it in these terms, this is the same thing. This is equal to F of x of b y of b minus F of x of a y of a. These are equivalent. You give me any point on the xy plane, an x and a y, and it tells me where I am. This is my capital F. It gives me a height just like that.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This is equal to F of x of b y of b minus F of x of a y of a. These are equivalent. You give me any point on the xy plane, an x and a y, and it tells me where I am. This is my capital F. It gives me a height just like that. But what this tells me, this associates a value with every point on the xy plane. This whole exercise, remember, this is the same thing as that. This is our whole thing that we were trying to prove.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This is my capital F. It gives me a height just like that. But what this tells me, this associates a value with every point on the xy plane. This whole exercise, remember, this is the same thing as that. This is our whole thing that we were trying to prove. That is equal to F dot dr. F dot dr, our vector field, which is the gradient of the capital F. Remember, F was equal to the gradient of F. We assume that it's the gradient of some function capital F. If that is the case, then we just did a little bit of calculus or algebra or whatever you want to call it, and we found that we can evaluate this integral by evaluating capital F at t is equal to b and then subtracting from that capital F at t is equal to a. But what that tells you is that this integral, the value of this integral, is only dependent at our starting point, t is equal to a. This is the point x of a, y of a, and the ending point, t of b, or t is equal to b, which is x of b, y of b.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This is our whole thing that we were trying to prove. That is equal to F dot dr. F dot dr, our vector field, which is the gradient of the capital F. Remember, F was equal to the gradient of F. We assume that it's the gradient of some function capital F. If that is the case, then we just did a little bit of calculus or algebra or whatever you want to call it, and we found that we can evaluate this integral by evaluating capital F at t is equal to b and then subtracting from that capital F at t is equal to a. But what that tells you is that this integral, the value of this integral, is only dependent at our starting point, t is equal to a. This is the point x of a, y of a, and the ending point, t of b, or t is equal to b, which is x of b, y of b. That integral is only dependent on these two values. How do I know that? Because to solve it, because I'm saying that this thing exists, I just had to evaluate that thing at those two points.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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This is the point x of a, y of a, and the ending point, t of b, or t is equal to b, which is x of b, y of b. That integral is only dependent on these two values. How do I know that? Because to solve it, because I'm saying that this thing exists, I just had to evaluate that thing at those two points. I didn't care about the curve in between. So this shows that if F is equal to the gradient, this is often called a potential function of capital F, although they're usually the negative of each other, but it's the same idea. If the vector field F is the gradient of some scalar field uppercase F, then we can say that F is conservative, or that the line integral of F dot dr is path independent.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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Because to solve it, because I'm saying that this thing exists, I just had to evaluate that thing at those two points. I didn't care about the curve in between. So this shows that if F is equal to the gradient, this is often called a potential function of capital F, although they're usually the negative of each other, but it's the same idea. If the vector field F is the gradient of some scalar field uppercase F, then we can say that F is conservative, or that the line integral of F dot dr is path independent. It doesn't matter what path we go on, as long as our starting and ending points are the same. Hopefully you found that useful. And we'll do some examples with that.
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Path independence for line integrals Multivariable Calculus Khan Academy.mp3
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And so I can rewrite that. So this is the triple integral over our region, which we're assuming is a type one region, of the partial of r with respect to z, which you could write like this, doesn't matter, partial of r with respect to z. And I'll dv. And we can rewrite this as, we can assume we're going to integrate with respect to z first, so I'm going to integrate with respect to z. We'll do that in another color. I'm going to integrate with respect to z first. The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And we can rewrite this as, we can assume we're going to integrate with respect to z first, so I'm going to integrate with respect to z. We'll do that in another color. I'm going to integrate with respect to z first. The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2. So we're going to integrate from f1 of x, y to f2 of x, y. So I'm going to take f2 of x, y, and I'm going to integrate the partial of r with respect to z. So let me do that in that same yellow color.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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The lower bound on z in our type one region, the lower bound is f1, the upper bound is f2. So we're going to integrate from f1 of x, y to f2 of x, y. So I'm going to take f2 of x, y, and I'm going to integrate the partial of r with respect to z. So let me do that in that same yellow color. Partial of r with respect to z, and then I have dz. And then I'll have to integrate with respect to y and x, or with respect to x and y. So it's dx dy, or dy dx.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So let me do that in that same yellow color. Partial of r with respect to z, and then I have dz. And then I'll have to integrate with respect to y and x, or with respect to x and y. So it's dx dy, or dy dx. I can just write that as da. So what you could think of it, we can evaluate the yellow part, and then we're going to have the double integral over the x, y domain. So this is just going to be over the x, y domain.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So it's dx dy, or dy dx. I can just write that as da. So what you could think of it, we can evaluate the yellow part, and then we're going to have the double integral over the x, y domain. So this is just going to be over the x, y domain. So let me put some brackets here, just to make it clear what we're going to do. So all we're doing is we're integrating with respect to z first, and we have the bounds there. Well, this is pretty straightforward.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So this is just going to be over the x, y domain. So let me put some brackets here, just to make it clear what we're going to do. So all we're doing is we're integrating with respect to z first, and we have the bounds there. Well, this is pretty straightforward. This is all going to be equal to, I'll write the outside first, the double integral over the domain. And I have the da right over here. Actually, let me give myself some real estate da.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Well, this is pretty straightforward. This is all going to be equal to, I'll write the outside first, the double integral over the domain. And I have the da right over here. Actually, let me give myself some real estate da. Well, what's the antiderivative of this? This is just r. And this is just r, or r of x, y, z, evaluated when z is f1, or when z is f2. And from that, we evaluate when z is f1.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Actually, let me give myself some real estate da. Well, what's the antiderivative of this? This is just r. And this is just r, or r of x, y, z, evaluated when z is f1, or when z is f2. And from that, we evaluate when z is f1. So this is just going to be r of x, y, and z, and we evaluate when z is equal to that. And from that, we subtract when z is equal to that. So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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And from that, we evaluate when z is f1. So this is just going to be r of x, y, and z, and we evaluate when z is equal to that. And from that, we subtract when z is equal to that. So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y. And from that, we need to subtract r when z is this, minus r of x, y, f1 of x, y. And then make sure that we got our parentheses. Now, this is exactly what we saw in the last video.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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So that's going to be equal to, so r of x, y, z, evaluated when z is equal to that is r of x, y, f2 of x, y. And from that, we need to subtract r when z is this, minus r of x, y, f1 of x, y. And then make sure that we got our parentheses. Now, this is exactly what we saw in the last video. It is exactly that, which shows that this is exactly this. So when we assumed it was a type I region, we got that this is exactly equal to this. You do the exact same argument with the type II region to show that this is equal to this.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now, this is exactly what we saw in the last video. It is exactly that, which shows that this is exactly this. So when we assumed it was a type I region, we got that this is exactly equal to this. You do the exact same argument with the type II region to show that this is equal to this. Type III regions show this is equal to that. And you have your divergence theorem proved. And we can consider ourselves done.
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Divergence theorem proof (part 5) Divergence theorem Multivariable Calculus Khan Academy.mp3
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Now harmonic functions are a very special kind of multivariable function, and they're defined in terms of the Laplacian, which I've been talking about in the last few videos. So the Laplacian, which we denote with this upper right side up triangle, is an operator that you might take on a multivariable function. So it might have two inputs. It could have 100 inputs, just some kind of multivariable function with a scalar output. And I talked about in the last few videos, but as a reminder, it's defined as the divergence of the gradient of f. And it's kind of like the second derivative. It's sort of the way to extend the idea of the second derivative into multiple dimensions. Now what a harmonic function is, is one where the Laplacian is equal to 0.
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Harmonic Functions.mp3
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It could have 100 inputs, just some kind of multivariable function with a scalar output. And I talked about in the last few videos, but as a reminder, it's defined as the divergence of the gradient of f. And it's kind of like the second derivative. It's sort of the way to extend the idea of the second derivative into multiple dimensions. Now what a harmonic function is, is one where the Laplacian is equal to 0. And it's equal to 0 at every possible input point. And sometimes the way that people write this to distinguish it, they'll make it kind of a triple equal sign, maybe saying like equivalent to 0. And this is really just a way of emphasizing that it's equal to 0 at all possible input points.
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Harmonic Functions.mp3
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Now what a harmonic function is, is one where the Laplacian is equal to 0. And it's equal to 0 at every possible input point. And sometimes the way that people write this to distinguish it, they'll make it kind of a triple equal sign, maybe saying like equivalent to 0. And this is really just a way of emphasizing that it's equal to 0 at all possible input points. It's not an equation that you're solving for the specific x and y where it equals 0. It's a statement about the function. And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function.
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Harmonic Functions.mp3
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And this is really just a way of emphasizing that it's equal to 0 at all possible input points. It's not an equation that you're solving for the specific x and y where it equals 0. It's a statement about the function. And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function. If you just have some single variable function of x, and you're looking at its second derivative, which is kind of the analog of the Laplacian, what does it mean if that's equal to 0? Well, we can integrate it. We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0?
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Harmonic Functions.mp3
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And to get our head around this, because it's kind of a, as you're just starting to learn about the Laplacian, it's hard to just immediately see the intuition for what this means, let's think about what it means for a single variable function. If you just have some single variable function of x, and you're looking at its second derivative, which is kind of the analog of the Laplacian, what does it mean if that's equal to 0? Well, we can integrate it. We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0? The only functions are the constant ones. So c is just going to mean some constant here. And if you integrate that again, say what function has as its derivative a constant?
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Harmonic Functions.mp3
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We can take the antiderivative and say that means that the single derivative of f, well, let's see, what functions have a derivative that's 0? The only functions are the constant ones. So c is just going to mean some constant here. And if you integrate that again, say what function has as its derivative a constant? Well, it's going to be that constant times x plus some other constant, some other constant k. So basically linear functions. So if you're thinking of a graph, it's just something that's got a line passing through it like that. And this should kind of make sense if you think of the geometric interpretation for the second derivative.
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Harmonic Functions.mp3
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And if you integrate that again, say what function has as its derivative a constant? Well, it's going to be that constant times x plus some other constant, some other constant k. So basically linear functions. So if you're thinking of a graph, it's just something that's got a line passing through it like that. And this should kind of make sense if you think of the geometric interpretation for the second derivative. Because if you're just looking at a random arbitrary function that's kind of curving as it does, the second derivative is negative when this concavity is down. So this right here would be a point where the second derivative, it's not 0, it's negative. And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive.
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Harmonic Functions.mp3
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And this should kind of make sense if you think of the geometric interpretation for the second derivative. Because if you're just looking at a random arbitrary function that's kind of curving as it does, the second derivative is negative when this concavity is down. So this right here would be a point where the second derivative, it's not 0, it's negative. And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive. So if we're saying that the second derivative has to always be 0, then it can't curve down, and it can't curve up, and it can't do that anywhere. So basically there's no curving allowed. So whatever direction it starts at, it's not allowed to curve.
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Harmonic Functions.mp3
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And over here, when the concavity is up, and it's got a sort of bowl shape, that's where the second derivative is positive. So if we're saying that the second derivative has to always be 0, then it can't curve down, and it can't curve up, and it can't do that anywhere. So basically there's no curving allowed. So whatever direction it starts at, it's not allowed to curve. So it just sticks straight like that. But once we extend this to the idea of a multivariable function, things can get a lot more interesting than just a straight line. So an example, I've got the graph here of a multivariable function that happens to be harmonic.
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Harmonic Functions.mp3
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So whatever direction it starts at, it's not allowed to curve. So it just sticks straight like that. But once we extend this to the idea of a multivariable function, things can get a lot more interesting than just a straight line. So an example, I've got the graph here of a multivariable function that happens to be harmonic. So the graph that you're looking at, this is of a two variable function. And the function specifically is f of x, y, x, y, is equal to e to the x multiplied by sine of y. And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern.
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Harmonic Functions.mp3
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So an example, I've got the graph here of a multivariable function that happens to be harmonic. So the graph that you're looking at, this is of a two variable function. And the function specifically is f of x, y, x, y, is equal to e to the x multiplied by sine of y. And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern. Because as we're moving in the positive x direction, this here is the positive x direction, you have this exponential shape. And this corresponds with the fact that over here, we've got an e to the x. So as you move x, it kind of looks like e to the x.
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Harmonic Functions.mp3
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And as we're looking at the graph here, hopefully it makes a little bit of sense why this is sort of an e to the x sine of y pattern. Because as we're moving in the positive x direction, this here is the positive x direction, you have this exponential shape. And this corresponds with the fact that over here, we've got an e to the x. So as you move x, it kind of looks like e to the x. And it's being multiplied by something that is a function of y. So if you're holding y constant, this just looks like a constant. But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down.
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Harmonic Functions.mp3
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So as you move x, it kind of looks like e to the x. And it's being multiplied by something that is a function of y. So if you're holding y constant, this just looks like a constant. But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down. It's sort of like a negative e to the x look. But if you imagine moving in the y direction, so instead of the pure x direction like that, if we imagine ourselves moving with the input going along, let's see what it would be. It would be this way, positive y direction.
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Harmonic Functions.mp3
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But notice, if that was a negative constant, if sine of y at some point happens to be negative, then your whole exponential function actually kind of goes down. It's sort of like a negative e to the x look. But if you imagine moving in the y direction, so instead of the pure x direction like that, if we imagine ourselves moving with the input going along, let's see what it would be. It would be this way, positive y direction. You have this sort of sinusoidal shape. And that should make sense, because you've got the sine of y. And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here.
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Harmonic Functions.mp3
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It would be this way, positive y direction. You have this sort of sinusoidal shape. And that should make sense, because you've got the sine of y. And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here. It's going way up and way down. But if e to the x was really small, it hardly even looks like it's wiggling over here. It pretty much looks flat.
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Harmonic Functions.mp3
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And depending on what e to the x is, the amplitude of that sine wave is going to get really high at some points here. It's going way up and way down. But if e to the x was really small, it hardly even looks like it's wiggling over here. It pretty much looks flat. So that's the graph that we're looking at. And I'm telling you right now, I claim that this is harmonic. This is a function whose Laplacian is equal to 0.
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Harmonic Functions.mp3
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It pretty much looks flat. So that's the graph that we're looking at. And I'm telling you right now, I claim that this is harmonic. This is a function whose Laplacian is equal to 0. And what that would mean is that as we go over here and we say we evaluate the Laplacian of f, which, just to remind you, there's a different formula, rather than thinking divergence of gradient, that turns out to be completely the same as saying you take the second derivative of that function with respect to x, its first input, and you add that. Let's see, second derivative with respect to x, you add that to the second derivative of your function with respect to the next variable. And you keep doing this for all of the different variables that there are.
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Harmonic Functions.mp3
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This is a function whose Laplacian is equal to 0. And what that would mean is that as we go over here and we say we evaluate the Laplacian of f, which, just to remind you, there's a different formula, rather than thinking divergence of gradient, that turns out to be completely the same as saying you take the second derivative of that function with respect to x, its first input, and you add that. Let's see, second derivative with respect to x, you add that to the second derivative of your function with respect to the next variable. And you keep doing this for all of the different variables that there are. But this is just a two-variable function, so you do this twice. The claim is that this is always equal to 0. So I might say kind of equivalent.
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Harmonic Functions.mp3
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And you keep doing this for all of the different variables that there are. But this is just a two-variable function, so you do this twice. The claim is that this is always equal to 0. So I might say kind of equivalent. At every possible input, it's equal to 0. And I think I'll leave that as something for you to compute. It might be kind of good practice to kind of get a feel for computing Laplacian.
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Harmonic Functions.mp3
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So I might say kind of equivalent. At every possible input, it's equal to 0. And I think I'll leave that as something for you to compute. It might be kind of good practice to kind of get a feel for computing Laplacian. But what I want to do is interpret what does this actually mean, right? Because you can plug it through and you can see, ah, yes, at all possible inputs, it will be 0. But what does that mean?
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Harmonic Functions.mp3
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It might be kind of good practice to kind of get a feel for computing Laplacian. But what I want to do is interpret what does this actually mean, right? Because you can plug it through and you can see, ah, yes, at all possible inputs, it will be 0. But what does that mean? Because in the single-variable context, once we started thinking about the geometric interpretation of a second derivative as this concavity, it sort of made sense that forcing it to be 0 will give us a straight line. But clearly, that's not the case. This is much more complicated than a straight line.
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Harmonic Functions.mp3
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But what does that mean? Because in the single-variable context, once we started thinking about the geometric interpretation of a second derivative as this concavity, it sort of made sense that forcing it to be 0 will give us a straight line. But clearly, that's not the case. This is much more complicated than a straight line. And for that, I want to give a kind of a different way that you can think about the single-variable second derivative. On the one hand, you can think of, let's say, negative second derivative as being this concavity where it's kind of frowning down. But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here.
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Harmonic Functions.mp3
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This is much more complicated than a straight line. And for that, I want to give a kind of a different way that you can think about the single-variable second derivative. On the one hand, you can think of, let's say, negative second derivative as being this concavity where it's kind of frowning down. But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here. And if you go a little bit to the left, the neighbor is less than it. And if you go a little bit to the right, that other neighbor is also less than it. So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it.
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Harmonic Functions.mp3
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But another way you could maybe think about this is saying that all of the neighbors of your point, if you go a little bit to the left, you've got an input point here. And if you go a little bit to the left, the neighbor is less than it. And if you go a little bit to the right, that other neighbor is also less than it. So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it. And if you do a similar thing over at a positive concavity point, where it's kind of smiling up, you say, well, its neighbor to the right has a greater value, and its neighbor to the left has a greater value. So at some point where the second derivative, instead of being less than 0, happens to be greater than 0, that means that the neighbors tend to be greater than the point itself. And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum.
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Harmonic Functions.mp3
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So it's kind of a way of saying, hey, if you look at the neighbors of your input, so if you happen to be making the claim that f double prime at some particular input, like x sub o, is less than 0, it's saying that all of the neighbors of x sub o, all of the neighbors of that point, are less than it. And if you do a similar thing over at a positive concavity point, where it's kind of smiling up, you say, well, its neighbor to the right has a greater value, and its neighbor to the left has a greater value. So at some point where the second derivative, instead of being less than 0, happens to be greater than 0, that means that the neighbors tend to be greater than the point itself. And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum. But let's say you're looking at a graph. Let's say you're looking at a function at a point where it's concave up, but it's not this idealized local minimum kind of circumstance. So instead, you might be looking at a point like this.
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Harmonic Functions.mp3
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And even if you're looking at a circumstance that isn't this idealized, it happens to be a local minimum. But let's say you're looking at a graph. Let's say you're looking at a function at a point where it's concave up, but it's not this idealized local minimum kind of circumstance. So instead, you might be looking at a point like this. And if you look at its neighbor to the left, that'll have some value that's actually less than your original guy. So the neighbor looks like it's less than it on the left. But if you move that same distance to the right, its neighbor is greater.
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Harmonic Functions.mp3
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So instead, you might be looking at a point like this. And if you look at its neighbor to the left, that'll have some value that's actually less than your original guy. So the neighbor looks like it's less than it on the left. But if you move that same distance to the right, its neighbor is greater. But you would say, on average, if you took the average value of the neighbors, the neighbor on the right kind of outbalances the neighbor on the left. And you would say, on average, its neighbors are greater than the point itself. So let's say that input point there was like x sub o.
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Harmonic Functions.mp3
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But if you move that same distance to the right, its neighbor is greater. But you would say, on average, if you took the average value of the neighbors, the neighbor on the right kind of outbalances the neighbor on the left. And you would say, on average, its neighbors are greater than the point itself. So let's say that input point there was like x sub o. That would mean that the second derivative of your function at that point is greater than 0. So with this positive concavity, you can also think of it as a measure of, on average, are the neighbors greater than your original point or less than it? And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world.
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Harmonic Functions.mp3
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So let's say that input point there was like x sub o. That would mean that the second derivative of your function at that point is greater than 0. So with this positive concavity, you can also think of it as a measure of, on average, are the neighbors greater than your original point or less than it? And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world. So if we look at a function like this, and let's say we're looking at it kind of from a bird's eye view. So we've got our xy-plane. This over here is the x-axis, and this up here is the y-axis.
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Harmonic Functions.mp3
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And the reason I'm saying this is because this idea of kind of comparing your neighbors to the original point is a much better way to contemplate the Laplacian in the multivariable world. So if we look at a function like this, and let's say we're looking at it kind of from a bird's eye view. So we've got our xy-plane. This over here is the x-axis, and this up here is the y-axis. And let's say that we're looking at some specific input point. With the Laplacian, you want to start thinking about a circle of points around it, all of its neighbors. And in fact, think of a perfect circle, so all of the points that are a specified distance away.
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Harmonic Functions.mp3
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This over here is the x-axis, and this up here is the y-axis. And let's say that we're looking at some specific input point. With the Laplacian, you want to start thinking about a circle of points around it, all of its neighbors. And in fact, think of a perfect circle, so all of the points that are a specified distance away. The question the Laplacian is asking is, hey, are those neighbor points, on average, greater than or less than your original point? And this is actually how I introduced the Laplacian in the original video where I was giving kind of the intuition for the Laplacian. You're asking, do the points around a given input happen to be greater than it or less than it?
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Harmonic Functions.mp3
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And in fact, think of a perfect circle, so all of the points that are a specified distance away. The question the Laplacian is asking is, hey, are those neighbor points, on average, greater than or less than your original point? And this is actually how I introduced the Laplacian in the original video where I was giving kind of the intuition for the Laplacian. You're asking, do the points around a given input happen to be greater than it or less than it? And if you're looking at a point where the Laplacian of your function happens to be greater than 0 at some point, that would mean all of the neighbors tend to be, on average, greater than your point. Whereas if you're looking at a point where the Laplacian of your function is less than 0, then all of those neighbors, on average, would be less than your point. So in particular, if the Laplacian was less than 0, your point might look like a local maximum.
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Harmonic Functions.mp3
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You're asking, do the points around a given input happen to be greater than it or less than it? And if you're looking at a point where the Laplacian of your function happens to be greater than 0 at some point, that would mean all of the neighbors tend to be, on average, greater than your point. Whereas if you're looking at a point where the Laplacian of your function is less than 0, then all of those neighbors, on average, would be less than your point. So in particular, if the Laplacian was less than 0, your point might look like a local maximum. Or if the Laplacian was greater than 0, it might look like a local minimum, because all of its neighbors would be greater than where it is. But for harmonic functions, what makes them so special is that you're saying the value of the function itself, or the value of the Laplacian of the function at every possible point, is equal to 0. So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy.
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Harmonic Functions.mp3
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So in particular, if the Laplacian was less than 0, your point might look like a local maximum. Or if the Laplacian was greater than 0, it might look like a local minimum, because all of its neighbors would be greater than where it is. But for harmonic functions, what makes them so special is that you're saying the value of the function itself, or the value of the Laplacian of the function at every possible point, is equal to 0. So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy. So the height of the graph above those neighbors will, on average, be the same. So if we look around the graph, what that should mean is, let's say you're looking at an input point, the output of this guy. And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that.
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Harmonic Functions.mp3
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So no matter what point you choose, those neighbors are going to be, on average, the same value as this guy. So the height of the graph above those neighbors will, on average, be the same. So if we look around the graph, what that should mean is, let's say you're looking at an input point, the output of this guy. And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that. And no matter where you look, that should average out. And again, I encourage you to take a look at this function and actually evaluate the Laplacian to see that it's 0. But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center.
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Harmonic Functions.mp3
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And if you looked at all of the circle of its neighbors and projected them up onto the graph, what it should mean is that the height of all the points on this circle, on average, are the same as that. And no matter where you look, that should average out. And again, I encourage you to take a look at this function and actually evaluate the Laplacian to see that it's 0. But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center. That's not something you can easily tell just looking at that formula. But with what's not that hard a computation, you can make this conclusion, which is pretty far reaching. And this comes up all the time in physics.
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Harmonic Functions.mp3
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But what's interesting is it's not at all clear, just looking at this e to the x times sine of y formula, that the average value of a circle of input points is always going to kind of equal the value of the point at the center. That's not something you can easily tell just looking at that formula. But with what's not that hard a computation, you can make this conclusion, which is pretty far reaching. And this comes up all the time in physics. For example, heat is one where maybe you want to describe how the heat at a certain point in a room is related to the average value of the heat of all of the points kind of around it. And in fact, it comes up in all sorts of circumstances where you have some point in physical space. And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it.
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Harmonic Functions.mp3
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And this comes up all the time in physics. For example, heat is one where maybe you want to describe how the heat at a certain point in a room is related to the average value of the heat of all of the points kind of around it. And in fact, it comes up in all sorts of circumstances where you have some point in physical space. And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it. So whenever you're sort of relating neighbors to your original point, the Laplacian comes in. And harmonic functions have this tendency to correspond to some notion of stability. And I won't go deeper into that now.
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Harmonic Functions.mp3
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And something about that point, maybe like the rate at which some property of it is changing, corresponds to the average value at points around it. So whenever you're sort of relating neighbors to your original point, the Laplacian comes in. And harmonic functions have this tendency to correspond to some notion of stability. And I won't go deeper into that now. This is really, that really starts to get into the topic of partial differential equations. But at least in the context of just multivariable calculus, I wanted to shed some light on interpreting this operator and kind of interpreting the physical and geometric properties that that implies about a function. And with that, I will see you next video.
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Harmonic Functions.mp3
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And what I'm going to focus on in this first video, because it will take us several videos to do it, is just the parameterization of this surface right over here. And as you'll see, this is often the hardest part, because it takes a little bit of visualization. And then after that, it's kind of mechanical, but it can be kind of hairy at the same time. So it's worth going through. So first, let's think about how we can parameterize, and I have trouble even saying the word, how we can parameterize this unit sphere as a function of two parameters. So let's think about it a little bit. So first, let's just think about the unit sphere.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So it's worth going through. So first, let's think about how we can parameterize, and I have trouble even saying the word, how we can parameterize this unit sphere as a function of two parameters. So let's think about it a little bit. So first, let's just think about the unit sphere. I'm going to take a side view of the unit sphere. So let's take the unit sphere. So this right over here is our z-axis.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So first, let's just think about the unit sphere. I'm going to take a side view of the unit sphere. So let's take the unit sphere. So this right over here is our z-axis. That's our z-axis. And then over here, I'm going to draw, this is going to be not just the x or the y-axis. This is going to be the entire xy-plane viewed from the side.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So this right over here is our z-axis. That's our z-axis. And then over here, I'm going to draw, this is going to be not just the x or the y-axis. This is going to be the entire xy-plane viewed from the side. That is the xy-plane. Now, our sphere, our unit sphere, might look something like this. The unit sphere itself is not too hard to visualize.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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This is going to be the entire xy-plane viewed from the side. That is the xy-plane. Now, our sphere, our unit sphere, might look something like this. The unit sphere itself is not too hard to visualize. It might look something like that. The radius, let me make it very clear, the radius at any point is 1. So this length right over here is 1.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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The unit sphere itself is not too hard to visualize. It might look something like that. The radius, let me make it very clear, the radius at any point is 1. So this length right over here is 1. That length right over there is 1. And this is a sphere, not just a circle. So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So this length right over here is 1. That length right over there is 1. And this is a sphere, not just a circle. So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it. So that's shading it in. It kind of makes it look a little bit more spherical. Now, let's attempt to parameterize this.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So I could even shade it in a little bit just to make it clear that this thing has some dimensionality to it. So that's shading it in. It kind of makes it look a little bit more spherical. Now, let's attempt to parameterize this. And as a first step, let's just think, if we didn't have to think above and below the xy-plane, if we just thought about where this unit sphere intersected the xy-plane, how we could parameterize that. So let's just think about it. So it intersects the xy-plane.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Now, let's attempt to parameterize this. And as a first step, let's just think, if we didn't have to think above and below the xy-plane, if we just thought about where this unit sphere intersected the xy-plane, how we could parameterize that. So let's just think about it. So it intersects the xy-plane. It intersects it there and there and actually everywhere. So it intersects it right over there. So let's just draw the xy-plane and think about that intersection.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So it intersects the xy-plane. It intersects it there and there and actually everywhere. So it intersects it right over there. So let's just draw the xy-plane and think about that intersection. And then we could think about what happens as we go above and below the xy-plane. So on the xy-plane, this little region where we just shaded in, so let me draw. So now you could view this as almost a top view.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So let's just draw the xy-plane and think about that intersection. And then we could think about what happens as we go above and below the xy-plane. So on the xy-plane, this little region where we just shaded in, so let me draw. So now you could view this as almost a top view. The z-axis is now going to be pointing straight out at you, straight out of the screen. So that's x. Let me draw it.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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So now you could view this as almost a top view. The z-axis is now going to be pointing straight out at you, straight out of the screen. So that's x. Let me draw it. So that's x. And then this right over there is y. So this thing that we were viewing sideways, now we're viewing it from the top.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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Let me draw it. So that's x. And then this right over there is y. So this thing that we were viewing sideways, now we're viewing it from the top. And so now our unit sphere is going to look something like this, viewed from above. And this, what I just drew, this dotted circle right over here, this is going to be where our unit sphere intersects what I labeled that y. That should be x. I don't want to confuse you already.
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Surface integral example part 1 Parameterizing the unit sphere Khan Academy.mp3
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