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So your dx's are positive. In this situation, your dx's are negative. The heights are always going to be the same. They're always going to be f of x, but here your change in x is a negative change in x when you go from b to a, and that's why you get a negative integral. In either case here, our path changes, but our ds's are going to be positive. And the way I've drawn the surface, it's above the xy plane, the f of xy is also going to be positive. So that also kind of gives the same intuition that this should be the exact same area.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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They're always going to be f of x, but here your change in x is a negative change in x when you go from b to a, and that's why you get a negative integral. In either case here, our path changes, but our ds's are going to be positive. And the way I've drawn the surface, it's above the xy plane, the f of xy is also going to be positive. So that also kind of gives the same intuition that this should be the exact same area. But let's prove it to ourselves. So let's start off with our first parameterization, just like we did in the last video. We have x is equal to x of t, y is equal to y of t, and we're dealing with this from t goes from a to b.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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So that also kind of gives the same intuition that this should be the exact same area. But let's prove it to ourselves. So let's start off with our first parameterization, just like we did in the last video. We have x is equal to x of t, y is equal to y of t, and we're dealing with this from t goes from a to b. And we know we're going to need the derivatives of these, so let me write that down right now. We can write dx dt is equal to x prime of t, and dy dt is equal to y prime of t. This is nothing groundbreaking I've done so far, but we know the integral over c of f of xy, f is a scalar field, not a vector field, ds is equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, times the square root of dx dt squared, which is the same thing as x prime of t squared, plus dy dt squared, which is the same thing as y prime of t squared, plus y prime of t squared, all that under the radical, times dt. This integral is exactly that, given this parameterization.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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We have x is equal to x of t, y is equal to y of t, and we're dealing with this from t goes from a to b. And we know we're going to need the derivatives of these, so let me write that down right now. We can write dx dt is equal to x prime of t, and dy dt is equal to y prime of t. This is nothing groundbreaking I've done so far, but we know the integral over c of f of xy, f is a scalar field, not a vector field, ds is equal to the integral from t is equal to a to t is equal to b of f of x of t, y of t, times the square root of dx dt squared, which is the same thing as x prime of t squared, plus dy dt squared, which is the same thing as y prime of t squared, plus y prime of t squared, all that under the radical, times dt. This integral is exactly that, given this parameterization. I have so much trouble saying that. Now let's do the minus c version. I'll do that in this orange color.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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This integral is exactly that, given this parameterization. I have so much trouble saying that. Now let's do the minus c version. I'll do that in this orange color. So the minus c version, actually let me do the minus c version down here. The minus c version, we have x is equal to, you remember this actually just from up here, this was from the last video, x is equal to a plus b minus t. x is equal to a plus b minus t. y is, sorry, x doesn't equal a plus b minus t. x is equal to x of a plus b minus t. Got ahead of myself. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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I'll do that in this orange color. So the minus c version, actually let me do the minus c version down here. The minus c version, we have x is equal to, you remember this actually just from up here, this was from the last video, x is equal to a plus b minus t. x is equal to a plus b minus t. y is, sorry, x doesn't equal a plus b minus t. x is equal to x of a plus b minus t. Got ahead of myself. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b. And this is just exactly what we did in that last video. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. Same curve, just going in a different direction as t increases from a to b. But let's get the derivatives.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. And then t goes from a to b. And this is just exactly what we did in that last video. x is equal to x of a plus b minus t. y is equal to y of a plus b minus t. Same curve, just going in a different direction as t increases from a to b. But let's get the derivatives. So I'll do it in the derivative color maybe. So dx dt for this path. It's going to be a little different.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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But let's get the derivatives. So I'll do it in the derivative color maybe. So dx dt for this path. It's going to be a little different. We have to do the chain rule now. Derivative of the inside with respect to t. Well, these are constants. Minus t, derivative of minus t with respect to t is minus 1.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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It's going to be a little different. We have to do the chain rule now. Derivative of the inside with respect to t. Well, these are constants. Minus t, derivative of minus t with respect to t is minus 1. So it's minus 1 times the derivative of the outside with respect to the inside. Well, that's just x prime of a plus b minus t. Or we could rewrite this as this is just minus x prime of a plus b minus t. dy dt, same logic. Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Minus t, derivative of minus t with respect to t is minus 1. So it's minus 1 times the derivative of the outside with respect to the inside. Well, that's just x prime of a plus b minus t. Or we could rewrite this as this is just minus x prime of a plus b minus t. dy dt, same logic. Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1. Times the derivative of the outside with respect to the inside. So y prime of a plus b minus t. Same thing as minus y prime a plus b minus t. So given all of that, what is this integral going to be equal to? The integral of minus c of the scalar field f of x, y, ds.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Derivative of the inside is minus 1 with respect to t. Derivative of minus t is just minus 1. Times the derivative of the outside with respect to the inside. So y prime of a plus b minus t. Same thing as minus y prime a plus b minus t. So given all of that, what is this integral going to be equal to? The integral of minus c of the scalar field f of x, y, ds. What is this going to be equal to? Well, it's going to be the integral from, you could almost pattern match it, t is equal to a to t is equal to b, of f of x. But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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The integral of minus c of the scalar field f of x, y, ds. What is this going to be equal to? Well, it's going to be the integral from, you could almost pattern match it, t is equal to a to t is equal to b, of f of x. But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking. Hopefully it's not too confusing. And once again, y is no longer y of t. y is y of a plus b minus t. And then times the square root, I'll just switch colors, times the square root of dx dt squared. What's dx dt squared?
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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But now x is no longer x of t. x now equals x of a plus b minus t. It's a little bit hairy, but I don't think anything here is groundbreaking. Hopefully it's not too confusing. And once again, y is no longer y of t. y is y of a plus b minus t. And then times the square root, I'll just switch colors, times the square root of dx dt squared. What's dx dt squared? dx dt squared is just this thing squared. Or this thing squared. If I have minus anything squared, that's the same thing as the anything squared, right?
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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What's dx dt squared? dx dt squared is just this thing squared. Or this thing squared. If I have minus anything squared, that's the same thing as the anything squared, right? This is equal to minus x prime of a plus b minus t squared. Which is the same thing as just x prime of a plus b minus t squared, right? You lose that minus information when you square it.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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If I have minus anything squared, that's the same thing as the anything squared, right? This is equal to minus x prime of a plus b minus t squared. Which is the same thing as just x prime of a plus b minus t squared, right? You lose that minus information when you square it. So that's going to be equal to x prime of a plus b minus t squared, the whole result function, square it, plus dy dt squared. By the same logic, that's going to be, you lose the negative when you square it, y prime of a plus b minus t squared. Let me extend the radical.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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You lose that minus information when you square it. So that's going to be equal to x prime of a plus b minus t squared, the whole result function, square it, plus dy dt squared. By the same logic, that's going to be, you lose the negative when you square it, y prime of a plus b minus t squared. Let me extend the radical. And then all of that dt. So that's the surface integral over the line integral. We're not doing surface integrals yet.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Let me extend the radical. And then all of that dt. So that's the surface integral over the line integral. We're not doing surface integrals yet. This is the line integral over the curve c. This is the line integral over the curve minus c. They don't look equal just yet. This looks a lot more convoluted than that one does. So let's see if we can simplify it a little bit.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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We're not doing surface integrals yet. This is the line integral over the curve c. This is the line integral over the curve minus c. They don't look equal just yet. This looks a lot more convoluted than that one does. So let's see if we can simplify it a little bit. And we can simplify it, perhaps, by making a substitution. Let me get a nice substitution color. Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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So let's see if we can simplify it a little bit. And we can simplify it, perhaps, by making a substitution. Let me get a nice substitution color. Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral. Well, actually, let's just figure out what's du. So du dt, the derivative of u with respect to t is just going to be equal to minus 1. Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Let's let u equal to a plus b minus t. So first we're going to have to figure out the boundaries of our integral. Well, actually, let's just figure out what's du. So du dt, the derivative of u with respect to t is just going to be equal to minus 1. Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt. And let's figure out our boundaries of integration. When t is equal to a, what is u? u is equal to a plus b minus a, which is equal to b.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Or we could say that du, if we multiply both sides by the differential dt, is equal to minus dt. And let's figure out our boundaries of integration. When t is equal to a, what is u? u is equal to a plus b minus a, which is equal to b. And then when t is equal to b, u is equal to a plus b minus b, which is equal to a. So if we do the substitution on this crazy, hairy-looking integral, I should simplify a little bit. And it changes our, so this integral is going to be the same thing as the integral from u.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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u is equal to a plus b minus a, which is equal to b. And then when t is equal to b, u is equal to a plus b minus b, which is equal to a. So if we do the substitution on this crazy, hairy-looking integral, I should simplify a little bit. And it changes our, so this integral is going to be the same thing as the integral from u. When t is a, u is b. When t is b, u is a. And f of x of, this thing right here is just u.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And it changes our, so this integral is going to be the same thing as the integral from u. When t is a, u is b. When t is b, u is a. And f of x of, this thing right here is just u. So let's simplify it a good bit. And y of, this thing right here, is just u. y of u times the square root of x prime of u squared plus y prime of u squared. Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And f of x of, this thing right here is just u. So let's simplify it a good bit. And y of, this thing right here, is just u. y of u times the square root of x prime of u squared plus y prime of u squared. Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du. So instead of dt, we have to put a minus du here. So this is times minus du. Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Instead of dt, we have to write a, or we could write, if we multiply both sides of this by minus, we have dt is equal to minus du. So instead of dt, we have to put a minus du here. So this is times minus du. Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that. So we're going from b to a of this thing. Like that. And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Or just so we don't think this is a subtraction, let's just put that negative sign out here in the front, just like that. So we're going from b to a of this thing. Like that. And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them. And I said at the beginning of this video, if you swap the, for just a standard, regular, run-of-the-mill integral, if you swap, if you have something going from b to a of f of x dx, or du, and maybe I should write it this way, f of u du, this is equal to the minus of the integral from a to b of f of u du. And we did that by the logic that I graphed up here. That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And just to make the boundaries of integration make a little bit more sense, because we know that a is less than b, let's swap them. And I said at the beginning of this video, if you swap the, for just a standard, regular, run-of-the-mill integral, if you swap, if you have something going from b to a of f of x dx, or du, and maybe I should write it this way, f of u du, this is equal to the minus of the integral from a to b of f of u du. And we did that by the logic that I graphed up here. That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve. So let's do that. Let's swap the boundaries of integration right here. And if we do that, that'll negate this negative or make it a positive.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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That here, when you switch the order, your du's will become the negatives of each other when you actually visualize it, when you're actually finding the area of the curve. So let's do that. Let's swap the boundaries of integration right here. And if we do that, that'll negate this negative or make it a positive. So this is going to be equal to the integral from a to b. I'm dropping the negative sign because I swapped these two things. So I'm going to take the negative of a negative, which is a positive, of f of x of u y of u times the square root of x prime of u squared plus y prime of u squared du. Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And if we do that, that'll negate this negative or make it a positive. So this is going to be equal to the integral from a to b. I'm dropping the negative sign because I swapped these two things. So I'm going to take the negative of a negative, which is a positive, of f of x of u y of u times the square root of x prime of u squared plus y prime of u squared du. Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds. Now, how does this compare to when we take the regular curve? How does this compare to that? Let me copy and paste it to see.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Now remember, everything we just did with the substitution, this was all equal to, just to remember what we're doing, this was the integral of the minus curve of our scalar field, f of x y ds. Now, how does this compare to when we take the regular curve? How does this compare to that? Let me copy and paste it to see. No, I'm using the wrong tool. Let me copy and paste it to see how they compare. And then let me paste it down here.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Let me copy and paste it to see. No, I'm using the wrong tool. Let me copy and paste it to see how they compare. And then let me paste it down here. Edit, paste. So how do these two things compare? Let's take a close look.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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And then let me paste it down here. Edit, paste. So how do these two things compare? Let's take a close look. Well, they actually look pretty similar. Over here, for the minus curve, we have a bunch of u's. Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Let's take a close look. Well, they actually look pretty similar. Over here, for the minus curve, we have a bunch of u's. Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places. These integrals are the exact same integrals. If you make a u substitution here, if you just make the substitution u is equal to t, this thing is going to be the integral from a to b of what's going to be this exact same thing, of f of x of u, y of u, times the square root of x prime of u squared plus y prime of u squared du. These two things are identical.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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Over here, for the positive curve, we have a bunch of t's, but they're in the exact same places. These integrals are the exact same integrals. If you make a u substitution here, if you just make the substitution u is equal to t, this thing is going to be the integral from a to b of what's going to be this exact same thing, of f of x of u, y of u, times the square root of x prime of u squared plus y prime of u squared du. These two things are identical. So we did all the substitution and everything, but we got the exact same integrals. So hopefully that satisfies you that it doesn't matter what direction we go on the curve, as long as the shape of the curve is the same. It doesn't matter if we go forward or backward on the curve.
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Scalar field line integral independent of path direction Multivariable Calculus Khan Academy.mp3
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So here I'm gonna talk about the Laplacian. Laplacian. And the Laplacian is a certain operator in the same way that the divergence or the gradient or the curl or even just the derivative are operators, the things that take in some kind of function and give you another function. So in this case, let's say we have a multivariable function like f that just takes in a two-dimensional input, f of x, y. So you might imagine its graph as being something like this where the input space is this x, y plane here. So each of the points x, y is a point here and then the output is just given by the height of that graph. So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y.
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Laplacian intuition.mp3
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So in this case, let's say we have a multivariable function like f that just takes in a two-dimensional input, f of x, y. So you might imagine its graph as being something like this where the input space is this x, y plane here. So each of the points x, y is a point here and then the output is just given by the height of that graph. So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y. So it's gonna give you a new function that takes in a two-dimensional input and just outputs a number. And it's kind of like a second derivative because the way that it's defined is that you take the divergence of the gradient of your function f. So that's kind of how it's defined, the divergence of the gradient of f. And a more common notation that you might see here is to take that upside-down triangle, nabla, dot product with nabla of f. So remember, if f is a scalar-valued function, then the gradient of f gives you a vector field, a certain vector field, but the divergence of a vector field gives you another scalar-valued function. So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean.
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Laplacian intuition.mp3
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So the Laplacian of f is denoted with a right-side up triangle and it's gonna give you a new scalar-valued function of x and y. So it's gonna give you a new function that takes in a two-dimensional input and just outputs a number. And it's kind of like a second derivative because the way that it's defined is that you take the divergence of the gradient of your function f. So that's kind of how it's defined, the divergence of the gradient of f. And a more common notation that you might see here is to take that upside-down triangle, nabla, dot product with nabla of f. So remember, if f is a scalar-valued function, then the gradient of f gives you a vector field, a certain vector field, but the divergence of a vector field gives you another scalar-valued function. So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean. Because the gradient, if you'll remember, gives you the slope of steepest ascent. So it's a vector field in the input space of x and each one of the vectors points in the direction that you should walk, such that if this graph is kind of like a hill on top of you, it tells you the direction you should go to increase your direction the most rapidly. And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other.
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Laplacian intuition.mp3
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So this is the sense in which it's a second derivative, but let's see if we can kind of understand intuitively what this should mean. Because the gradient, if you'll remember, gives you the slope of steepest ascent. So it's a vector field in the input space of x and each one of the vectors points in the direction that you should walk, such that if this graph is kind of like a hill on top of you, it tells you the direction you should go to increase your direction the most rapidly. And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other. So with the specific graph that I have pictured here, when you have kind of the top of a hill, all of the points around it, the direction that you should walk is towards the top of that hill. Whereas when you have kind of like the bottom, a little gully here, all of the directions you should walk to increase the value of the function most rapidly are directly away from that valley, which you might call a local minimum. So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly.
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Laplacian intuition.mp3
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And if that seems unfamiliar or doesn't make sense, maybe go take a look at that video on gradients and graphs and how they relate to each other. So with the specific graph that I have pictured here, when you have kind of the top of a hill, all of the points around it, the direction that you should walk is towards the top of that hill. Whereas when you have kind of like the bottom, a little gully here, all of the directions you should walk to increase the value of the function most rapidly are directly away from that valley, which you might call a local minimum. So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly. And now let's think about what the divergence is supposed to represent. So now the divergence, and again, if this feels unfamiliar, maybe go back and take a look at the divergence videos. But the divergence has you imagining that this vector field corresponds to some kind of fluid flow.
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Laplacian intuition.mp3
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So let's temporarily get rid of the graph just so we can look at the gradient field here pretty clearly. And now let's think about what the divergence is supposed to represent. So now the divergence, and again, if this feels unfamiliar, maybe go back and take a look at the divergence videos. But the divergence has you imagining that this vector field corresponds to some kind of fluid flow. So you imagine little like water molecules and at any given moment, they're moving along the vector that they're attached to. So for example, if you had a water molecule that started off kind of here, it would start by going along that vector and then kind of follow the ones near it. And it looks like it kind of ends up in this spot.
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Laplacian intuition.mp3
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But the divergence has you imagining that this vector field corresponds to some kind of fluid flow. So you imagine little like water molecules and at any given moment, they're moving along the vector that they're attached to. So for example, if you had a water molecule that started off kind of here, it would start by going along that vector and then kind of follow the ones near it. And it looks like it kind of ends up in this spot. And a lot of the water molecules seem to kind of converge over there. Whereas over here, the water molecules tend to go away when they're following those vectors away from this point. And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away.
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Laplacian intuition.mp3
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And it looks like it kind of ends up in this spot. And a lot of the water molecules seem to kind of converge over there. Whereas over here, the water molecules tend to go away when they're following those vectors away from this point. And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away. So over here, divergence is positive. Whereas the opposite case, where all of the water molecules are kind of coming in towards a point, that's where divergence is negative. And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out.
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Laplacian intuition.mp3
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And when they go away like that, when you have a whole bunch of vectors kind of pointed away, that's an indication that divergence is positive because they're diverging away. So over here, divergence is positive. Whereas the opposite case, where all of the water molecules are kind of coming in towards a point, that's where divergence is negative. And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out. And at least from this picture, it doesn't seem like the ones going out are doing so at a faster rate or slower than they are here. This would be roughly zero divergence. So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here.
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Laplacian intuition.mp3
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And in another area, let's say it was kind of like the center point where you have some water molecules that looks like they're coming in, but other ones are going out. And at least from this picture, it doesn't seem like the ones going out are doing so at a faster rate or slower than they are here. This would be roughly zero divergence. So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here. Points of high divergence, points where it diverges a lot here, why are those vectors pointing away? And if we pull up the graph again, the reason they're pointing away is because the direction of steepest ascent is kind of uphill everywhere. You're in a valley.
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Laplacian intuition.mp3
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So now let's think about what it might mean when you take the divergence of the gradient field of F. So let me kind of clear up the markings I made on top of it here. Points of high divergence, points where it diverges a lot here, why are those vectors pointing away? And if we pull up the graph again, the reason they're pointing away is because the direction of steepest ascent is kind of uphill everywhere. You're in a valley. Whereas in the opposite circumstance, where divergence is highly negative because points are converging towards it, why are they pointing towards it? Well, this is a gradient field. So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill.
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Laplacian intuition.mp3
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You're in a valley. Whereas in the opposite circumstance, where divergence is highly negative because points are converging towards it, why are they pointing towards it? Well, this is a gradient field. So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill. So in other words, the divergence of the gradient is very high at points that are kind of like minima, at points where everyone around them tends to be higher. But the divergence of the gradient is low at points that look more like maximum points, where when you evaluate the function at all of the points around that input point, they give something smaller. So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y.
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Laplacian intuition.mp3
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So they're pointing towards that spot because that's where anywhere around it, you should walk towards that spot to go uphill. So in other words, the divergence of the gradient is very high at points that are kind of like minima, at points where everyone around them tends to be higher. But the divergence of the gradient is low at points that look more like maximum points, where when you evaluate the function at all of the points around that input point, they give something smaller. So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y. It will be very positive when f evaluated at that point tends to give a smaller value than f evaluated at neighbors of that point. But it'll be very negative if when you evaluate f at that point, it tends to be bigger than its neighbors. And this should feel kind of analogous to the second derivative in ordinary calculus, when you have some kind of graph of just a single variable function.
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Laplacian intuition.mp3
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So this Laplacian operator is kind of like a measure of how much of a minimum point is this x, y. It will be very positive when f evaluated at that point tends to give a smaller value than f evaluated at neighbors of that point. But it'll be very negative if when you evaluate f at that point, it tends to be bigger than its neighbors. And this should feel kind of analogous to the second derivative in ordinary calculus, when you have some kind of graph of just a single variable function. The second derivative, you know, the second derivative of x will be low, it'll be negative at points where it kind of looks like a local maximum. But over here, the second derivative of x would be positive at points that kind of look like a local minimum. So in that way, the Laplacian is sort of an analog of the second derivative for scalar-valued multivariable functions.
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Laplacian intuition.mp3
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And the picture of this vector field is here. This is what that vector field looks like. And what I'd like to do is compute and interpret the divergence of v. So, the divergence of v as a function of x and y. And in the last couple videos, I explained that the formula for this, and hopefully it's more than just a formula, but something you have an intuition for, is the partial derivative of p with respect to x. And by p, I mean that first component. So if you're thinking about this as being p of x, y, and q of x, y. So it could use any letters, right?
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Divergence example.mp3
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And in the last couple videos, I explained that the formula for this, and hopefully it's more than just a formula, but something you have an intuition for, is the partial derivative of p with respect to x. And by p, I mean that first component. So if you're thinking about this as being p of x, y, and q of x, y. So it could use any letters, right? And p and q are common, but the upshot is it's the partial derivative of the first component with respect to the first variable, plus the partial derivative of that second component with respect to that second variable, y. And as we actually plug this in and start computing, the partial derivative of p with respect to x of this guy with respect to x, x looks like a variable, y looks like a constant, the derivative is y, that constant. And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y.
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Divergence example.mp3
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So it could use any letters, right? And p and q are common, but the upshot is it's the partial derivative of the first component with respect to the first variable, plus the partial derivative of that second component with respect to that second variable, y. And as we actually plug this in and start computing, the partial derivative of p with respect to x of this guy with respect to x, x looks like a variable, y looks like a constant, the derivative is y, that constant. And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y. And then x just looks like a constant, so nothing happens there. So on the whole, the divergence evidently just depends on the y value. It's three times y.
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Divergence example.mp3
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And then the partial derivative of q, that second component with respect to y, we look here, y squared looks like a variable and its derivative is two times y, two times y. And then x just looks like a constant, so nothing happens there. So on the whole, the divergence evidently just depends on the y value. It's three times y. So what that should mean is if we look, for example, let's say we look at all points where y equals zero, we'd expect the divergence to be zero, that fluid neither goes towards nor away from each point. So y equals zero corresponds with this x-axis of points. So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here.
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Divergence example.mp3
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It's three times y. So what that should mean is if we look, for example, let's say we look at all points where y equals zero, we'd expect the divergence to be zero, that fluid neither goes towards nor away from each point. So y equals zero corresponds with this x-axis of points. So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here. And wherever you look, I mean here it's kind of only flowing in by a little bit, and I guess it's flowing out just by that same amount, and that all cancels out. Whereas, let's say we take a look at y equals, I don't know, like three. So in this case, the divergence should equal nine.
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Divergence example.mp3
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So at a given point here, evidently it's the case that the fluid kind of flowing in from above is balanced out by how much fluid flows away from it here. And wherever you look, I mean here it's kind of only flowing in by a little bit, and I guess it's flowing out just by that same amount, and that all cancels out. Whereas, let's say we take a look at y equals, I don't know, like three. So in this case, the divergence should equal nine. So we'd expect there to be positive divergence when y is positive. So if we go up and y is equal to one, two, three, and if we look at a point, I don't know, let's say around here, and we kind of consider the region around it, you can kind of see how the vectors leaving it seem to be bigger, so the fluid kind of flowing out of this region is pretty rapid, whereas the fluid flowing into it is less rapid. So on the whole, in a region around this point, the fluid, I guess, is going away.
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Divergence example.mp3
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So in this case, the divergence should equal nine. So we'd expect there to be positive divergence when y is positive. So if we go up and y is equal to one, two, three, and if we look at a point, I don't know, let's say around here, and we kind of consider the region around it, you can kind of see how the vectors leaving it seem to be bigger, so the fluid kind of flowing out of this region is pretty rapid, whereas the fluid flowing into it is less rapid. So on the whole, in a region around this point, the fluid, I guess, is going away. And you know, you look anywhere where y is positive, and if you kind of look around here, the same is true. Where fluid does flow into it, it seems, but the vectors kind of going out of this region are larger, so you'd expect, on the whole, for things to diverge away from that point. And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points.
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Divergence example.mp3
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So on the whole, in a region around this point, the fluid, I guess, is going away. And you know, you look anywhere where y is positive, and if you kind of look around here, the same is true. Where fluid does flow into it, it seems, but the vectors kind of going out of this region are larger, so you'd expect, on the whole, for things to diverge away from that point. And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points. So you go down to, you know, I guess I said y equals negative four, but really, I'm thinking of anything where y is negative. And let's say we take a look at like this point here. Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large.
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Divergence example.mp3
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And in contrast, you know, if you look at something where y is negative, let's say it was y is equal to negative four, it doesn't have to be three there, so that would be a divergence of negative 12, so you'd expect things to definitely be converging towards your specific points. So you go down to, you know, I guess I said y equals negative four, but really, I'm thinking of anything where y is negative. And let's say we take a look at like this point here. Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large. It's flowing out of it just kind of in a lackadaisical way. So it kind of makes sense, just looking at the picture, that divergence tends to be negative when y is negative. And what's surprising, what I wouldn't have been able to tell just looking at the picture, is that the divergence only depends on the y value.
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Divergence example.mp3
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Fluid flowing into it seems to be, according to large vectors, so it's flowing into it pretty quickly here, but when it's flowing out of it, it's less large. It's flowing out of it just kind of in a lackadaisical way. So it kind of makes sense, just looking at the picture, that divergence tends to be negative when y is negative. And what's surprising, what I wouldn't have been able to tell just looking at the picture, is that the divergence only depends on the y value. That once you compute everything, it's only dependent on the y value here, and that as you go kind of left and right on the diagram there, as we look left and right, the value of that divergence doesn't change. That's kind of surprising. It makes a little bit of sense.
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Divergence example.mp3
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Curl is one of those very cool vector calculus concepts and you'll be pretty happy that you've learned it once you have, if for no other reason because it's kind of artistically pleasing. And there's two different versions. There's a two-dimensional curl and a three-dimensional curl. And naturally enough, I'll start talking about the two-dimensional version and kind of build our way up to the 3D one. And in this particular video, I just wanna lay down the intuition for what's visually going on. And curl has to do with the fluid flow interpretation of vector fields. Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that.
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2d curl intuition.mp3
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And naturally enough, I'll start talking about the two-dimensional version and kind of build our way up to the 3D one. And in this particular video, I just wanna lay down the intuition for what's visually going on. And curl has to do with the fluid flow interpretation of vector fields. Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that. But just as a reminder, you kind of imagine that each point in space is a particle, like an air molecule or a water molecule. And since what a vector field does is associate each point in space with some kind of vector, and remember, we don't always draw every single vector, we just draw a small subsample, but in principle, every single point in space has a vector attached to it. You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to.
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2d curl intuition.mp3
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Now this is something that I've talked about in other videos, especially the ones on divergence if you watched that. But just as a reminder, you kind of imagine that each point in space is a particle, like an air molecule or a water molecule. And since what a vector field does is associate each point in space with some kind of vector, and remember, we don't always draw every single vector, we just draw a small subsample, but in principle, every single point in space has a vector attached to it. You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to. So as it moves to a different location in space and that velocity vector changes, it might be turning or it might be accelerating, and that velocity might change, and you end up getting kind of a trajectory for your point. And since every single point is moving in this way, you can start thinking about a flow, kind of a global view of the vector field. And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play.
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2d curl intuition.mp3
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You can think of each particle, each one of these water molecules or air molecules as moving over time in such a way that the velocity vector of its movement at any given point in time is the vector that it's attached to. So as it moves to a different location in space and that velocity vector changes, it might be turning or it might be accelerating, and that velocity might change, and you end up getting kind of a trajectory for your point. And since every single point is moving in this way, you can start thinking about a flow, kind of a global view of the vector field. And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play. And at any given moment, if you look at the movement of one of these blue dots, it's moving along the vector that it's attached to at that point, or if that vector's not pictured, you know the vector that would be attached to it at that point. And as we get kind of a feel for what's going on in this entire flow, I want you to notice a couple particular regions. First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there.
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2d curl intuition.mp3
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And for this particular example, this particular vector field that I have pictured, I'm gonna go ahead and put a blue dot at various points in space, and each one of these you can think of as representing a water molecule or something, and I'm just gonna let it play. And at any given moment, if you look at the movement of one of these blue dots, it's moving along the vector that it's attached to at that point, or if that vector's not pictured, you know the vector that would be attached to it at that point. And as we get kind of a feel for what's going on in this entire flow, I want you to notice a couple particular regions. First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there. And I'll go ahead and start playing the animation over here. And what's most notable about this region is that there's counterclockwise rotation, and this corresponds to an idea that the vector field has a curl here, and I'll go very specifically into what curl means, but just right now you should have the idea that in a region where there's counterclockwise rotation, we want to say the curl is positive. Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl.
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2d curl intuition.mp3
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First, let's take a look at this region over here on the right, kind of around here, and just kind of concentrate on what's going on there. And I'll go ahead and start playing the animation over here. And what's most notable about this region is that there's counterclockwise rotation, and this corresponds to an idea that the vector field has a curl here, and I'll go very specifically into what curl means, but just right now you should have the idea that in a region where there's counterclockwise rotation, we want to say the curl is positive. Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl. Here, I'll start it over here. And in contrast, if you look at a place where there's no rotation, where like at the center here, you have some points coming in from the top right and from the bottom left, and then going out from the other corners, but there's no net rotation. If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating.
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2d curl intuition.mp3
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Whereas, if you look at a region that also has rotation, but clockwise, going the other way, we think of that as being negative curl. Here, I'll start it over here. And in contrast, if you look at a place where there's no rotation, where like at the center here, you have some points coming in from the top right and from the bottom left, and then going out from the other corners, but there's no net rotation. If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating. These are regions where you think of them as having zero curl. So with that as a general idea, you know, clockwise rotation regions correspond to positive curl, counterclockwise rotation regions correspond to negative curl, and then no rotation corresponds to zero curl, in the next video, I'm gonna start going through what this means in terms of the underlying function defining the vector field, and how we can start looking at the partial differential information of that function to quantify this intuition of fluid rotation. And what's neat is that it's not just about fluid rotation, right?
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2d curl intuition.mp3
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If you were to just put a, you know, if you were to put like a twig somewhere in this water, it wouldn't really be rotating. These are regions where you think of them as having zero curl. So with that as a general idea, you know, clockwise rotation regions correspond to positive curl, counterclockwise rotation regions correspond to negative curl, and then no rotation corresponds to zero curl, in the next video, I'm gonna start going through what this means in terms of the underlying function defining the vector field, and how we can start looking at the partial differential information of that function to quantify this intuition of fluid rotation. And what's neat is that it's not just about fluid rotation, right? If you have vector fields in other contexts, and you just imagine that they represent a fluid, even though they don't, this idea of rotation and curling actually has certain importance, in ways that you totally wouldn't expect. The gradient turns out to relate to the curl, even though you wouldn't necessarily think the gradient has something to do with fluid rotation. In electromagnetism, this idea of fluid rotation has a certain importance, even though fluids aren't actually involved.
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2d curl intuition.mp3
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And here, I want to go through a slightly more intricate example. So I'll go ahead and get rid of this vector field. And in this example, the x component of the output will be y times z. The y component of the output will be x times z. And the z component of the output will be x times y. So I'll just show this vector field, and then we can start to get a feel for how the function that I just wrote relates to the vectors that you're seeing. So you see some of the vectors are kind of pointing away from the origin.
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3d vector field example Multivariable calculus Khan Academy.mp3
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The y component of the output will be x times z. And the z component of the output will be x times y. So I'll just show this vector field, and then we can start to get a feel for how the function that I just wrote relates to the vectors that you're seeing. So you see some of the vectors are kind of pointing away from the origin. Some of them are pointing in towards the origin. So how can we understand this vector field in terms of the function itself? And a good step is to just zero in on one of the components.
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3d vector field example Multivariable calculus Khan Academy.mp3
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So you see some of the vectors are kind of pointing away from the origin. Some of them are pointing in towards the origin. So how can we understand this vector field in terms of the function itself? And a good step is to just zero in on one of the components. So in this case, I'll choose the z component of the output, which is made up of x times y, and kind of start to understand what that should be. The z component will represent how much the vectors point up or down. This is the z axis.
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3d vector field example Multivariable calculus Khan Academy.mp3
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And a good step is to just zero in on one of the components. So in this case, I'll choose the z component of the output, which is made up of x times y, and kind of start to understand what that should be. The z component will represent how much the vectors point up or down. This is the z axis. And the x, y plane here, so I'll point the z axis straight in our face. This is the x axis, this is the y axis. The values of x and y are gonna completely determine that z component.
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3d vector field example Multivariable calculus Khan Academy.mp3
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This is the z axis. And the x, y plane here, so I'll point the z axis straight in our face. This is the x axis, this is the y axis. The values of x and y are gonna completely determine that z component. So I'm gonna go off to the side here and just draw myself a little x, y plane for reference. So this is my x value, this is my y value. And I wanna understand the meaning of the term x times y.
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3d vector field example Multivariable calculus Khan Academy.mp3
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The values of x and y are gonna completely determine that z component. So I'm gonna go off to the side here and just draw myself a little x, y plane for reference. So this is my x value, this is my y value. And I wanna understand the meaning of the term x times y. So when both x and y are positive, their product is positive. And when both of them are negative, their product is also positive. If x is negative and y is positive, their product is negative.
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3d vector field example Multivariable calculus Khan Academy.mp3
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And I wanna understand the meaning of the term x times y. So when both x and y are positive, their product is positive. And when both of them are negative, their product is also positive. If x is negative and y is positive, their product is negative. But if x is positive and y is negative, their product is also negative. So what this should mean in terms of our vector field is that when we're in this first quadrant, vectors tend to point up in the z direction. Same over here in the third quadrant.
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3d vector field example Multivariable calculus Khan Academy.mp3
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If x is negative and y is positive, their product is negative. But if x is positive and y is negative, their product is also negative. So what this should mean in terms of our vector field is that when we're in this first quadrant, vectors tend to point up in the z direction. Same over here in the third quadrant. But over in the other two, they should tend to point down. So let's focus in on that first quadrant and try to look at what's going on. So you see like this vector here applies this vector.
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3d vector field example Multivariable calculus Khan Academy.mp3
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Same over here in the third quadrant. But over in the other two, they should tend to point down. So let's focus in on that first quadrant and try to look at what's going on. So you see like this vector here applies this vector. And all of them generally point upwards. They have a positive z component. So that seems in line with what we were predicting.
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3d vector field example Multivariable calculus Khan Academy.mp3
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So you see like this vector here applies this vector. And all of them generally point upwards. They have a positive z component. So that seems in line with what we were predicting. Whereas over here, which corresponds to the fourth quadrant of the x, y plane, the z component of each vector tends to be down. You know, and they're doing other things in terms of the x and y component. It's not just z component action.
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3d vector field example Multivariable calculus Khan Academy.mp3
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So that seems in line with what we were predicting. Whereas over here, which corresponds to the fourth quadrant of the x, y plane, the z component of each vector tends to be down. You know, and they're doing other things in terms of the x and y component. It's not just z component action. But right now, we're just focusing on up and down. And if you look over in the third quadrant, they tend to be pointing up. And that corresponds to the fact that x times y will be positive.
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3d vector field example Multivariable calculus Khan Academy.mp3
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It's not just z component action. But right now, we're just focusing on up and down. And if you look over in the third quadrant, they tend to be pointing up. And that corresponds to the fact that x times y will be positive. And you look and it all starts to align that way. And because I chose a rather symmetric function, you could imagine doing this where you analyze also the y component here. You analyze the x component in terms of y and z.
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3d vector field example Multivariable calculus Khan Academy.mp3
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And that corresponds to the fact that x times y will be positive. And you look and it all starts to align that way. And because I chose a rather symmetric function, you could imagine doing this where you analyze also the y component here. You analyze the x component in terms of y and z. And it's actually gonna look very similar for understanding when the x component of a vector tends to be positive, like up here, or when it turns out to be negative, like over here. And same with when the y component of a vector tends to be negative, or if the y component tends to be positive. And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it.
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3d vector field example Multivariable calculus Khan Academy.mp3
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You analyze the x component in terms of y and z. And it's actually gonna look very similar for understanding when the x component of a vector tends to be positive, like up here, or when it turns out to be negative, like over here. And same with when the y component of a vector tends to be negative, or if the y component tends to be positive. And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it. And just like with two-dimensional vector fields, a kind of neat thing to do is imagine that this represents a fluid flow. So imagine like maybe air around you, flowing in towards the origin here, flowing out away from the origin there, you know, kind of be rotating around here. And later on in multivariable calculus, you'll learn about various ways that you can study just the function itself and just the variables, and get a feel for how that fluid itself would behave, even though it's a very complicated thing to think about.
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3d vector field example Multivariable calculus Khan Academy.mp3
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And overall, it's a very complicated image to look at, but you can slowly, piece by piece, get a feel for it. And just like with two-dimensional vector fields, a kind of neat thing to do is imagine that this represents a fluid flow. So imagine like maybe air around you, flowing in towards the origin here, flowing out away from the origin there, you know, kind of be rotating around here. And later on in multivariable calculus, you'll learn about various ways that you can study just the function itself and just the variables, and get a feel for how that fluid itself would behave, even though it's a very complicated thing to think about. It's even complicated to draw or use graphing software with. But just with analytic tools, you can get very powerful results. And these kind of things come up in physics all the time, because you're thinking in three-dimensional space, and it doesn't just have to be fluid flow, it could be a force field, like an electric force field, or a gravitational force field, where each vector tells you how a particle tends to get pushed.
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3d vector field example Multivariable calculus Khan Academy.mp3
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I've restated Stokes' Theorem. And what I want to do in this video is make sure that we get our orientation right. Because when we think about a normal vector, when we think about a normal vector to a surface, there are actually two normal vectors. They were based on the way I've drawn it right over here. There could be the one that might pop outward like this. Or there might be the one that pops inward just like that. Both of those would be normal to the surface right there.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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They were based on the way I've drawn it right over here. There could be the one that might pop outward like this. Or there might be the one that pops inward just like that. Both of those would be normal to the surface right there. And also, when we think about a path that goes around the boundary of a surface, there's two ways to think about that path. We could be going, based on how I've oriented it right now, we could go in a counterclockwise orientation or direction. Or we could go in a clockwise orientation or direction.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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Both of those would be normal to the surface right there. And also, when we think about a path that goes around the boundary of a surface, there's two ways to think about that path. We could be going, based on how I've oriented it right now, we could go in a counterclockwise orientation or direction. Or we could go in a clockwise orientation or direction. So in order to make sure we're using Stokes' Theorem correctly, we need to make sure we understand which convention it is using. And the way we think about it is, whatever the normal direction we pick, and so let's say we pick this normal direction right over here, the one I am drawing in yellow. So if we pick this as our normal vector, so we're essentially saying maybe that's the top.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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Or we could go in a clockwise orientation or direction. So in order to make sure we're using Stokes' Theorem correctly, we need to make sure we understand which convention it is using. And the way we think about it is, whatever the normal direction we pick, and so let's say we pick this normal direction right over here, the one I am drawing in yellow. So if we pick this as our normal vector, so we're essentially saying maybe that's the top. One way of doing it is that's the top of our surface. Then the positive orientation that we need to traverse the path in is the one that if your head was pointed in the direction of the normal vector, and you were to walk along that path, the inside or the surface itself would be to your left. And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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So if we pick this as our normal vector, so we're essentially saying maybe that's the top. One way of doing it is that's the top of our surface. Then the positive orientation that we need to traverse the path in is the one that if your head was pointed in the direction of the normal vector, and you were to walk along that path, the inside or the surface itself would be to your left. And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector. I'm wearing a big arrow hat right over there. And if I'm walking around the boundary, the actual surface needs to be to my left. So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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And so if my head is pointed in the direction of the normal vector, so this is me right over here, my head is pointed in the direction of the normal vector. I'm wearing a big arrow hat right over there. And if I'm walking around the boundary, the actual surface needs to be to my left. So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that. Then that's the convention that we use when we're thinking about Stokes' Theorem. If we oriented this thing differently, or if we said that no, no, no, no, no, this is not the normal vector, this is not essentially the top that we want to pick, if we wanted to pick it the other way, if we wanted to go in that direction, if we wanted that to be our normal vector, in order to be consistent, we would have to now do the opposite. I would now have to have my head going in that direction.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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So I need to be, this is me walking right over here, I need to be walking in the counterclockwise direction, just like that. Then that's the convention that we use when we're thinking about Stokes' Theorem. If we oriented this thing differently, or if we said that no, no, no, no, no, this is not the normal vector, this is not essentially the top that we want to pick, if we wanted to pick it the other way, if we wanted to go in that direction, if we wanted that to be our normal vector, in order to be consistent, we would have to now do the opposite. I would now have to have my head going in that direction. And then I would have to walk, once again, and this might be a little bit harder to visualize, I would have to be walking in the direction that the surface is to my left. And now in this situation, instead of the surface looking like a hill to me, the surface would look like some type of a bowl, or some type of a valley, or something like that to me. And the way that I would have to do it now, and it's a little bit harder to visualize the upside down cell, but the upside down cell would have to walk in this direction, in order for the bowl, or the dip, to be to my left.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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I would now have to have my head going in that direction. And then I would have to walk, once again, and this might be a little bit harder to visualize, I would have to be walking in the direction that the surface is to my left. And now in this situation, instead of the surface looking like a hill to me, the surface would look like some type of a bowl, or some type of a valley, or something like that to me. And the way that I would have to do it now, and it's a little bit harder to visualize the upside down cell, but the upside down cell would have to walk in this direction, in order for the bowl, or the dip, to be to my left. So it's just important to keep this in mind, in order for this to be consistent with this, right over here. Put your head in the direction of the normal vector, or you can kind of view that as the top of the direction that the top of the surface is going in. And then the contour, or the direction that you would have to traverse the boundary, in order for this to be true, is the direction with which the surface is to your left.
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Orienting boundary with surface Multivariable Calculus Khan Academy.mp3
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If you're computing the determinant of the guy that we have pictured there in the upper right, you start by taking this upper left component and then multiplying it by the determinant of the submatrix, the submatrix whose rows are not the row of I and whose columns are not the row of I. So what that looks like over here is we're gonna take that univector I and then multiply it by a certain little determinant. And what this subdeterminant involves is multiplying this partial partial y by r, which means taking the partial derivative with respect to y of the multivariable function r and then subtracting off the partial derivative with respect to z of q. So we're subtracting off partial derivative with respect to z of the multivariable function q. And then that, so that's the first thing that we do. And then as a second part, we take this j and we're gonna subtract off. So you're kind of thinking plus, minus, plus for the elements in this top row.
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3d curl formula, part 2.mp3
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So we're subtracting off partial derivative with respect to z of the multivariable function q. And then that, so that's the first thing that we do. And then as a second part, we take this j and we're gonna subtract off. So you're kind of thinking plus, minus, plus for the elements in this top row. So we're gonna subtract off j multiplied by another subdeterminant. And then this one is gonna involve, you know, this column that it's not part of and this column that it's not part of. And you imagine those guys as a two by two matrix.
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3d curl formula, part 2.mp3
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So you're kind of thinking plus, minus, plus for the elements in this top row. So we're gonna subtract off j multiplied by another subdeterminant. And then this one is gonna involve, you know, this column that it's not part of and this column that it's not part of. And you imagine those guys as a two by two matrix. And its determinant involves taking the partial derivative with respect to x of r, so that's kind of the diagonal, partial partial x of r and then subtracting off the partial derivative with respect to z of p. So partial partial z of p. And then that's just two out of three of the things we need to do for our overall determinant because the last part we're gonna add, we're gonna add that top right component, k, multiplied by the, you know, the submatrix whose columns involve the column it's not part of and whose rows involve the rows that it's not part of. So k multiplied by the determinant of this guy is going to be, let's see, partial partial x of q. So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression.
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3d curl formula, part 2.mp3
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And you imagine those guys as a two by two matrix. And its determinant involves taking the partial derivative with respect to x of r, so that's kind of the diagonal, partial partial x of r and then subtracting off the partial derivative with respect to z of p. So partial partial z of p. And then that's just two out of three of the things we need to do for our overall determinant because the last part we're gonna add, we're gonna add that top right component, k, multiplied by the, you know, the submatrix whose columns involve the column it's not part of and whose rows involve the rows that it's not part of. So k multiplied by the determinant of this guy is going to be, let's see, partial partial x of q. So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression. And, you know, here we're writing it with ijk notation. If you were writing it as a column vector, I guess I didn't erase some of these guys, but if you were writing this as a column vector, it would look like saying the curl of your vector valued function v as a function of x, y, and z is equal to, and then what I'd put in for this first component would be what's up there. So that would be your partial with respect to y of r minus partial of q with respect to z.
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3d curl formula, part 2.mp3
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So that's partial partial x of q minus partial partial y of p. So partial derivative with respect to y of the multivariable function p. And that entire expression is the three dimensional curl of the function whose components are p, q, and r. So here we have our vector function, vector valued function v whose components are p, q, and r. And when you go through this whole process of imagining the cross product between the del operator, this nabla symbol, and the vector output p, q, and r, what you get is this whole expression. And, you know, here we're writing it with ijk notation. If you were writing it as a column vector, I guess I didn't erase some of these guys, but if you were writing this as a column vector, it would look like saying the curl of your vector valued function v as a function of x, y, and z is equal to, and then what I'd put in for this first component would be what's up there. So that would be your partial with respect to y of r minus partial of q with respect to z. So partial of q with respect to z. And I won't copy it down for all of the other ones, but in principle, you know, you'd kind of, whatever this j component is, and I guess we're subtracting it, so you'd subtract there. You'd copy that as the next component, and then over here.
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3d curl formula, part 2.mp3
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So that would be your partial with respect to y of r minus partial of q with respect to z. So partial of q with respect to z. And I won't copy it down for all of the other ones, but in principle, you know, you'd kind of, whatever this j component is, and I guess we're subtracting it, so you'd subtract there. You'd copy that as the next component, and then over here. But oftentimes, when you're computing curl, you kind of switch to using this ijk notation. My personal preference, I typically default to column vectors, and other people will write in terms of ij and k. It doesn't really matter, as long as you know how to go back and forth between the two. One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula.
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3d curl formula, part 2.mp3
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You'd copy that as the next component, and then over here. But oftentimes, when you're computing curl, you kind of switch to using this ijk notation. My personal preference, I typically default to column vectors, and other people will write in terms of ij and k. It doesn't really matter, as long as you know how to go back and forth between the two. One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula. If you kind of look back to the videos on 2D curl and what its formula is, that is what we have here. And in fact, all the other components kind of look like mirrors of that, but you're using slightly different operators and slightly different functions. But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane.
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3d curl formula, part 2.mp3
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One really quick thing that I want to highlight before doing an example of this is that the k component here, the z component of the output, is exactly the two-dimensional curl formula. If you kind of look back to the videos on 2D curl and what its formula is, that is what we have here. And in fact, all the other components kind of look like mirrors of that, but you're using slightly different operators and slightly different functions. But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane. And what's happening with these other guys is kind of similar, right? Rotation that happens purely in the xz plane is gonna correspond with a rotation vector in the y direction, the direction perpendicular to the x, let's see, so the xz plane over here. And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane.
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3d curl formula, part 2.mp3
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But if you think about rotation that happens purely in the xy plane, just two-dimensional rotation, and how in three dimensions, that's described with a vector in the k direction, and again, if that doesn't quite seem clear, maybe look back at the video on describing rotation with a three-dimensional vector and the right-hand rule, but vectors pointing in the pure z direction describe rotation in the xy plane. And what's happening with these other guys is kind of similar, right? Rotation that happens purely in the xz plane is gonna correspond with a rotation vector in the y direction, the direction perpendicular to the x, let's see, so the xz plane over here. And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane. Now, when you compute it, you're not always thinking about, oh, you know, this corresponds to rotation in that plane, and this corresponds to rotation in that plane. You're just kind of computing it to get a formula out, but I think it's kind of nice to recognize that all the intuition that we put into the two-dimensional curl does show up here. And another thing I wanna emphasize is this is not a formula to be memorized.
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3d curl formula, part 2.mp3
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And then similarly, this first component kind of tells you all the rotation happening in the yz plane, and the vectors in the i direction, the x direction of the output, kind of corresponds to rotation in that plane. Now, when you compute it, you're not always thinking about, oh, you know, this corresponds to rotation in that plane, and this corresponds to rotation in that plane. You're just kind of computing it to get a formula out, but I think it's kind of nice to recognize that all the intuition that we put into the two-dimensional curl does show up here. And another thing I wanna emphasize is this is not a formula to be memorized. I would not, if I were you, try to sit down and memorize this long expression. The only thing that you need to remember, the only thing, is that curl is represented as this del cross v, this nabla symbol cross product with the vector-valued function v. Because from there, whatever your components are, you can kind of go through the process that I just did, and the more you do it, the quicker it becomes. It's kind of long, but it doesn't take that long.
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3d curl formula, part 2.mp3
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And another thing I wanna emphasize is this is not a formula to be memorized. I would not, if I were you, try to sit down and memorize this long expression. The only thing that you need to remember, the only thing, is that curl is represented as this del cross v, this nabla symbol cross product with the vector-valued function v. Because from there, whatever your components are, you can kind of go through the process that I just did, and the more you do it, the quicker it becomes. It's kind of long, but it doesn't take that long. And it's certainly much more fault-tolerant than trying to remember something that has as many moving parts as the formula that you see here. And in the next video, I'll go through an actual example of that. I'll have functions for p, q, and r, and walk through that process in a more concrete context.
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3d curl formula, part 2.mp3
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