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Instead, they found this term as being significant in various other formulas and circumstances, and I think electromagnetism might be where it originally came about. But then in trying to understand this formula, they realized that you can give a fluid flow interpretation that gives a very deep understanding of what's going on beyond just the symbols themselves. So let me go ahead and walk through this example in terms of the formula representing the vector field. It's a relatively straightforward formula, actually. So p and q, that x component is going to be negative y, and the y component, q, is equal to x. So when we apply our 2D curl formula and apply the partial of q with respect to x, so partial of the second component with respect to x is just one, and then we subtract off the partial of p with respect to y, which up here is negative one, because p is just equal to negative y. So the 2D curl is equal to two, and in particular, it's a constant two that doesn't depend on x and y, which is pretty unusual.
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2d curl nuance.mp3
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It's a relatively straightforward formula, actually. So p and q, that x component is going to be negative y, and the y component, q, is equal to x. So when we apply our 2D curl formula and apply the partial of q with respect to x, so partial of the second component with respect to x is just one, and then we subtract off the partial of p with respect to y, which up here is negative one, because p is just equal to negative y. So the 2D curl is equal to two, and in particular, it's a constant two that doesn't depend on x and y, which is pretty unusual. Most times that you apply 2D curl to a vector field, you're going to get some kind of function of x and y. But the fact that this is constant tells us that when we look over at this fluid flow, the sense in which curl, the formula for curl, wants to say that rotation happens around the center is just as strong as it's supposed to happen over here on the right, or anywhere on the plane for that matter. So if we're playing this, and if you imagine the paddle wheel in the center, evidently it would be rotating just as quickly as the paddle on the right, even though it might, I don't know, to me that feels a little unintuitive, because the one on the right I'm thinking, okay, you know, there's, maybe there's a little bit more torque on the right side than there is on the left, and that's kind of a counterbalancing act, but the idea that that's actually the same as the very clear-cut, I see the counterclockwise rotation with my eyes example in the center, does seem a little unusual.
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2d curl nuance.mp3
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So I have talked a lot about different ways that you can visualize multivariable functions, functions that'll have some kind of multidimensional input or output. These include three-dimensional graphs, which are very common, contour maps, vector fields, parametric functions, but here I want to talk about one of my all-time favorite ways to think about functions, which is as a transformation. So anytime you have some sort of function, if you're thinking very abstractly, I like to think that there's some sort of input space, and I'll draw it as a blob, even though that could be the real number line, so it should be a line, or it could be three-dimensional space, and then there's some kind of output space, and again, I just very vaguely think about it as this blob, but that could be, again, the real number line, the xy-plane, three-dimensional space, and the function is just some way of taking inputs to outputs. And every time that we're trying to visualize something, like with a graph or a contour map, you're just trying to associate input- output pairs. If f inputs 3 gets mapped to the vector 1, 2, it's a question of how do you associate the number 3 with that vector 1, 2? And the thought behind transformations is that we're just going to watch the actual points of the input space move to the output space. And I'll start with a simple example that's just a one-dimensional function.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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And every time that we're trying to visualize something, like with a graph or a contour map, you're just trying to associate input- output pairs. If f inputs 3 gets mapped to the vector 1, 2, it's a question of how do you associate the number 3 with that vector 1, 2? And the thought behind transformations is that we're just going to watch the actual points of the input space move to the output space. And I'll start with a simple example that's just a one-dimensional function. It'll have a single variable input, and it'll have a single variable output. So let's consider the function f of x is equal to x squared minus 3. And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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And I'll start with a simple example that's just a one-dimensional function. It'll have a single variable input, and it'll have a single variable output. So let's consider the function f of x is equal to x squared minus 3. And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs. I just want to say, how do the inputs move to those outputs? So as an example, if you go to 0, when you plug in 0, you're going to get negative 3, you know, 0 squared minus 3 is equal to negative 3. So somehow we want to watch 0 move to negative 3.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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And of course, the way we're used to visualizing something like this, it'll be as a graph, and you might kind of be thinking of something roughly parabolic that's squished down by 3, but here, I don't want to think in terms of graphs. I just want to say, how do the inputs move to those outputs? So as an example, if you go to 0, when you plug in 0, you're going to get negative 3, you know, 0 squared minus 3 is equal to negative 3. So somehow we want to watch 0 move to negative 3. And then similarly, let's say you plug in 1, and you get 1 squared minus 3 is negative 2, so somehow we want to watch 1 move to negative 2. And just to list another example here, let's say you're plugging in 3 itself, so 3 squared minus 3 is 9 minus 3 is 6. So somehow in this transformation, we want to watch 3 move to the number 6.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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So somehow we want to watch 0 move to negative 3. And then similarly, let's say you plug in 1, and you get 1 squared minus 3 is negative 2, so somehow we want to watch 1 move to negative 2. And just to list another example here, let's say you're plugging in 3 itself, so 3 squared minus 3 is 9 minus 3 is 6. So somehow in this transformation, we want to watch 3 move to the number 6. And with a little animation, we can watch this happen. We can actually watch what it looks like for all these numbers to move to their corresponding outputs. So here we go.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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So somehow in this transformation, we want to watch 3 move to the number 6. And with a little animation, we can watch this happen. We can actually watch what it looks like for all these numbers to move to their corresponding outputs. So here we go. Each number will move over and land on its output. And I'll clear up the board here. So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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So here we go. Each number will move over and land on its output. And I'll clear up the board here. So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves. And I'll play it again. Here, let's just watch where each number from the input space moves over to the output. And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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So I kept track of what the original input numbers are by just kind of writing them on top here, and that was a way of just watching how it moves. And I'll play it again. Here, let's just watch where each number from the input space moves over to the output. And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions. So now, let me consider a function that has a one-dimensional input and a two-dimensional output. And specifically, it'll be f of x is equal to cosine of x cosine of x and then the y component will be x sine of y. Sorry, x sine of x.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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And with single variable functions, this is a little bit nice because it gives this sense of inputs moving to outputs, but where it gets fun is in the context of multivariable functions. So now, let me consider a function that has a one-dimensional input and a two-dimensional output. And specifically, it'll be f of x is equal to cosine of x cosine of x and then the y component will be x sine of y. Sorry, x sine of x. So just to think about a couple examples, if you plug in something like 0 and think about where 0 ought to go, you'd have f of 0 is equal to cosine of 0 is 1 and then 0 times anything is 0. So somehow, we're going to watch 0 move over to the point 1, 0. So this is where we expect 0 to land.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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Sorry, x sine of x. So just to think about a couple examples, if you plug in something like 0 and think about where 0 ought to go, you'd have f of 0 is equal to cosine of 0 is 1 and then 0 times anything is 0. So somehow, we're going to watch 0 move over to the point 1, 0. So this is where we expect 0 to land. And let's think about pi. So, f of pi, then cosine of pi is negative 1. This is going to be pi multiplied by sine of pi is 0, so that'll again be 0.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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So this is where we expect 0 to land. And let's think about pi. So, f of pi, then cosine of pi is negative 1. This is going to be pi multiplied by sine of pi is 0, so that'll again be 0. So, you know, this little guy is where 0 lands and we expect that this is going to be where the value pi lands. And if we watch this take place, and we actually watch each element of the input space move over to the output space, we get something like this. And again, this is just kind of a nice way to think about what's actually going on.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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This is going to be pi multiplied by sine of pi is 0, so that'll again be 0. So, you know, this little guy is where 0 lands and we expect that this is going to be where the value pi lands. And if we watch this take place, and we actually watch each element of the input space move over to the output space, we get something like this. And again, this is just kind of a nice way to think about what's actually going on. You might ask questions about whether the space ends up getting stretched or squished. And notice that this is also what a parametric plot of this function would look like. If you interpret it as a parametric function, this is what you get in the end.
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Transformations, part 1 Multivariable calculus Khan Academy.mp3
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In the last video, we were given a multivariable function and asked to find and classify all of its critical points. So critical points just means finding where the gradient is equal to zero. And we found four different points for that. I have them down here. They were zero, zero, zero, negative two, square root of three and one, and negative square root of three and one. So then, the next step is to classify those. And that requires the second partial derivative test.
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Second partial derivative test example, part 2.mp3
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I have them down here. They were zero, zero, zero, negative two, square root of three and one, and negative square root of three and one. So then, the next step is to classify those. And that requires the second partial derivative test. So what I'm gonna go ahead and do is copy down the partial derivatives, since we already computed those, copy. And then just kind of paste them down here where we can start to use them for the second partial derivatives. So let me clean things up a little bit.
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Second partial derivative test example, part 2.mp3
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And that requires the second partial derivative test. So what I'm gonna go ahead and do is copy down the partial derivatives, since we already computed those, copy. And then just kind of paste them down here where we can start to use them for the second partial derivatives. So let me clean things up a little bit. And we don't need this simplification of it. So we've got our partial derivatives. Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function.
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Second partial derivative test example, part 2.mp3
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So let me clean things up a little bit. And we don't need this simplification of it. So we've got our partial derivatives. Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function. That's just kind of the first thing to do. So let's go ahead and do it. The second partial derivative of the function with respect to x twice in a row.
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Second partial derivative test example, part 2.mp3
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Now, since we know we want to apply the second partial derivative test, we've got to first just compute all of the different second partial derivatives of our function. That's just kind of the first thing to do. So let's go ahead and do it. The second partial derivative of the function with respect to x twice in a row. We'll take the partial derivative with respect to x and then do it with respect to x again. So this first term looks like six times a variable times a constant. So it'll just be six times that constant.
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Second partial derivative test example, part 2.mp3
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The second partial derivative of the function with respect to x twice in a row. We'll take the partial derivative with respect to x and then do it with respect to x again. So this first term looks like six times a variable times a constant. So it'll just be six times that constant. And then the second term, the derivative of negative six x is just negative six. Moving right along, when we do the second partial derivative with respect to y twice in a row, we take the partial derivative with respect to y and then do it again. So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it.
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Second partial derivative test example, part 2.mp3
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So it'll just be six times that constant. And then the second term, the derivative of negative six x is just negative six. Moving right along, when we do the second partial derivative with respect to y twice in a row, we take the partial derivative with respect to y and then do it again. So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it. The derivative of negative three y squared is negative six times y. And then the derivative of negative six y is just negative six. And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y.
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Second partial derivative test example, part 2.mp3
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So this x squared term looks like nothing, looks like a constant as far as y is concerned, so we ignore it. The derivative of negative three y squared is negative six times y. And then the derivative of negative six y is just negative six. And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y. The order doesn't really matter in this case since it's a perfectly ordinary polynomial function. So we could do it either way, but I'm just gonna choose to take a look at this guy and differentiate it with respect to y. So the derivative of the first term with respect to y is six x, six x.
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Second partial derivative test example, part 2.mp3
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And then we can't forget that last crucially important mixed partial derivative term, which is the partial derivative of f, where first we do it with respect to x and then with respect to y. The order doesn't really matter in this case since it's a perfectly ordinary polynomial function. So we could do it either way, but I'm just gonna choose to take a look at this guy and differentiate it with respect to y. So the derivative of the first term with respect to y is six x, six x. And then that second term looks like a constant with respect to y. So that's all we have. So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression.
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Second partial derivative test example, part 2.mp3
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So the derivative of the first term with respect to y is six x, six x. And then that second term looks like a constant with respect to y. So that's all we have. So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression. And to remind you of what that is, that expression is, we take the second partial derivative with respect to x twice, and I'll just write it with a kind of shorter notation using subscripts. And we multiply that by the second partial derivative with respect to x, and then we subtract off, subtract off the mixed partial derivative term squared. So let's go ahead and do that for each of our points.
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Second partial derivative test example, part 2.mp3
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So now what we're gonna do is plug in each of the critical points to the special second partial derivative test expression. And to remind you of what that is, that expression is, we take the second partial derivative with respect to x twice, and I'll just write it with a kind of shorter notation using subscripts. And we multiply that by the second partial derivative with respect to x, and then we subtract off, subtract off the mixed partial derivative term squared. So let's go ahead and do that for each of our points. So when we do this at the point zero, zero, zero, zero, what we end up getting, plugging that into the partial derivative with respect to x twice, six times zero is zero, so there's just negative six. So that gives us negative six multiplied by, when we plug it into this partial derivative with respect to y squared, again, that y goes to zero, so we're left with just negative six. And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero.
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Second partial derivative test example, part 2.mp3
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So let's go ahead and do that for each of our points. So when we do this at the point zero, zero, zero, zero, what we end up getting, plugging that into the partial derivative with respect to x twice, six times zero is zero, so there's just negative six. So that gives us negative six multiplied by, when we plug it into this partial derivative with respect to y squared, again, that y goes to zero, so we're left with just negative six. And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero. So we're subtracting off zero squared. And that entire thing equals negative six times negative six is 36. 36.
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Second partial derivative test example, part 2.mp3
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And then we subtract off the mixed partial derivative term, which in this case is zero, because when we plug in x equals zero, we get zero. So we're subtracting off zero squared. And that entire thing equals negative six times negative six is 36. 36. And we'll get to analyzing what it means that that's positive in just a moment, but let's just kind of get all of them on the board so we can kind of start doing this with all of them. If we do this with zero and negative two, zero and negative two, then once we plug in y equals negative two to this expression, this time I'll write it out, six times negative two minus six, so that's negative 12 minus six, we'll get negative 18. Negative 18.
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Second partial derivative test example, part 2.mp3
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36. And we'll get to analyzing what it means that that's positive in just a moment, but let's just kind of get all of them on the board so we can kind of start doing this with all of them. If we do this with zero and negative two, zero and negative two, then once we plug in y equals negative two to this expression, this time I'll write it out, six times negative two minus six, so that's negative 12 minus six, we'll get negative 18. Negative 18. Then when we plug it into the partial derivative of f with respect to y squared, again, I'll kind of write it out, we have negative six times y is equal to negative two minus six. So now we have negative six times negative two, so that's positive 12 minus six. So that will be a positive six that we plug in here.
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Second partial derivative test example, part 2.mp3
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Negative 18. Then when we plug it into the partial derivative of f with respect to y squared, again, I'll kind of write it out, we have negative six times y is equal to negative two minus six. So now we have negative six times negative two, so that's positive 12 minus six. So that will be a positive six that we plug in here. And then for the mixed partial derivative, again, x is equal to zero, so the mixed partial derivative is just gonna look like zero when we do this. So we're subtracting off zero squared and we get negative 18 times six. And geez, what's 18 times six?
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Second partial derivative test example, part 2.mp3
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So that will be a positive six that we plug in here. And then for the mixed partial derivative, again, x is equal to zero, so the mixed partial derivative is just gonna look like zero when we do this. So we're subtracting off zero squared and we get negative 18 times six. And geez, what's 18 times six? So that's gonna be 36 times three, so that's the same as 90 plus 18, so I think that's 108. Negative 108. And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative.
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Second partial derivative test example, part 2.mp3
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And geez, what's 18 times six? So that's gonna be 36 times three, so that's the same as 90 plus 18, so I think that's 108. Negative 108. And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative. So now, kind of moving right along, these examples can take quite a while. If we plug in square root of three, one, square root of three, one, what we get. Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero.
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Second partial derivative test example, part 2.mp3
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And the specific magnitude won't matter, it's gonna be the sign that's important, and this is definitely negative. So now, kind of moving right along, these examples can take quite a while. If we plug in square root of three, one, square root of three, one, what we get. Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero. And then for the partial derivative with respect to y squared, instead of plugging in negative two, now we're plugging in y equals one, so we have negative six times one minus six, so the whole thing is negative 12, so negative 12. And now for the mixed partial derivative term, which is six x, x is equal to the square root of three, so now we're subtracting off the square root of three squared, so what that equals is, this first part is just entirely zero, and we're subtracting off three, so that's negative three. And then we have square root of three, no, no, we don't, that's what we just did.
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Second partial derivative test example, part 2.mp3
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Now instead of plugging in y equals negative two, we're plugging in y equals one, so that'll be six times one minus six, so the whole thing is just zero. And then for the partial derivative with respect to y squared, instead of plugging in negative two, now we're plugging in y equals one, so we have negative six times one minus six, so the whole thing is negative 12, so negative 12. And now for the mixed partial derivative term, which is six x, x is equal to the square root of three, so now we're subtracting off the square root of three squared, so what that equals is, this first part is just entirely zero, and we're subtracting off three, so that's negative three. And then we have square root of three, no, no, we don't, that's what we just did. Now we have negative square root of three, one, and this will be very similar because this first term just had a y and we plugged in the y, so it's also gonna be zero for totally the same reasons, and same deal over here, the value of y didn't change, so that's also gonna be negative 12. Doesn't really matter because we're multiplying it by a zero, right? And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three.
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Second partial derivative test example, part 2.mp3
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And then we have square root of three, no, no, we don't, that's what we just did. Now we have negative square root of three, one, and this will be very similar because this first term just had a y and we plugged in the y, so it's also gonna be zero for totally the same reasons, and same deal over here, the value of y didn't change, so that's also gonna be negative 12. Doesn't really matter because we're multiplying it by a zero, right? And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three. So what does this second partial derivative test tell us? Once we express this term, if it's greater than zero, we have a max or a min. That's what the test tells us.
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Second partial derivative test example, part 2.mp3
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And then over here, now we're plugging in negative square root of three, and that's gonna have the same square, so again, we're just subtracting off three. So what does this second partial derivative test tell us? Once we express this term, if it's greater than zero, we have a max or a min. That's what the test tells us. And then if it's less than zero, if it's less than zero, we have a saddle point. So in this case, the only term that's greater than zero is this first one, is this first one, and to analyze whether it's a maximum or a minimum, notice that the partial derivative with respect to x twice in a row or with respect to y twice in a row was negative, which indicates a sort of negative concavity, meaning this corresponds to a maximum. So this guy corresponds to a local maximum.
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Second partial derivative test example, part 2.mp3
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That's what the test tells us. And then if it's less than zero, if it's less than zero, we have a saddle point. So in this case, the only term that's greater than zero is this first one, is this first one, and to analyze whether it's a maximum or a minimum, notice that the partial derivative with respect to x twice in a row or with respect to y twice in a row was negative, which indicates a sort of negative concavity, meaning this corresponds to a maximum. So this guy corresponds to a local maximum. Now all of the other three gave us negative numbers, so all of these other three give us saddle points. Saddle points. So the answer to the question, the original find and classify such and such points, is that we found four different critical points.
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Second partial derivative test example, part 2.mp3
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So this guy corresponds to a local maximum. Now all of the other three gave us negative numbers, so all of these other three give us saddle points. Saddle points. So the answer to the question, the original find and classify such and such points, is that we found four different critical points. Let's see. Four different critical points, zero, zero, zero, negative two, square root of three, one, and negative square root of three, one, and all of them are saddle points except for zero, zero, which is a local maximum. And all of that is something that we can tell without even looking at the graph of the function.
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Second partial derivative test example, part 2.mp3
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And one way to think about it is we want our x and y values to take on all of the values inside of the unit circle, what I'm shading in right over here. And then our z values can be a function of the y values. We can express this equation right here, z is equal to two minus y. And then we can figure out how high above to go to get our z value. And by doing that, we'll be able to essentially get to every point that sits on our surface. And so first let's think about how we can get every x and y value inside of the unit circle. And so let me, let's just focus on the xy plane.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And then we can figure out how high above to go to get our z value. And by doing that, we'll be able to essentially get to every point that sits on our surface. And so first let's think about how we can get every x and y value inside of the unit circle. And so let me, let's just focus on the xy plane. We're kind of rotated around a little bit so it looks a little bit more traditional. So this is our, this is my x-axis, and then my y-axis would look something like that. Let me draw a little bit different.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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And so let me, let's just focus on the xy plane. We're kind of rotated around a little bit so it looks a little bit more traditional. So this is our, this is my x-axis, and then my y-axis would look something like that. Let me draw a little bit different. This is my y-axis, y-axis. And then if I were to draw the unit circle, so kind of the base of this thing, or at least where it intersects the xy plane, actually this thing would keep going down if I wanted to draw the x squared plus y squared equals one. But if I draw where it intersects the xy plane, it is, we get the unit circle.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Let me draw a little bit different. This is my y-axis, y-axis. And then if I were to draw the unit circle, so kind of the base of this thing, or at least where it intersects the xy plane, actually this thing would keep going down if I wanted to draw the x squared plus y squared equals one. But if I draw where it intersects the xy plane, it is, we get the unit circle. So let me just draw it. So we get, that's my best attempt at drawing a unit circle. We get the unit circle.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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But if I draw where it intersects the xy plane, it is, we get the unit circle. So let me just draw it. So we get, that's my best attempt at drawing a unit circle. We get the unit circle. We need to think of using parameters so that we can get every x and y coordinate that's inside of the unit circle. And to think about that, I'll introduce one parameter that's essentially the angle with the x-axis, and I'll call that parameter theta. So theta is the angle with the x-axis.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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We get the unit circle. We need to think of using parameters so that we can get every x and y coordinate that's inside of the unit circle. And to think about that, I'll introduce one parameter that's essentially the angle with the x-axis, and I'll call that parameter theta. So theta is the angle with the x-axis. And so theta will essentially sweep things all the way around. So theta can go between zero and two pi. So theta will take on values between zero and two pi.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So theta is the angle with the x-axis. And so theta will essentially sweep things all the way around. So theta can go between zero and two pi. So theta will take on values between zero and two pi. And if we just fix the radius at some point, say radius one, that would only give us all of the points on the unit circle. But we want all the points inside of it too. So we need to vary the radius as well.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So theta will take on values between zero and two pi. And if we just fix the radius at some point, say radius one, that would only give us all of the points on the unit circle. But we want all the points inside of it too. So we need to vary the radius as well. So let's introduce another parameter. Let's call it r. That is the radius. So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So we need to vary the radius as well. So let's introduce another parameter. Let's call it r. That is the radius. So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius. And if you change radius a little bit more, you'll sweep out another circle. And if you vary radius between zero and one, you'll get all of the circles that will fill out this entire area. So the radius is going to go between zero and one.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So for any given r, if we keep changing theta, we would essentially sweep out a circle of that radius. And if you change radius a little bit more, you'll sweep out another circle. And if you vary radius between zero and one, you'll get all of the circles that will fill out this entire area. So the radius is going to go between zero and one. Another way of thinking about it is, for any given theta, if you keep varying the radiuses, you'll sweep out all the points on this line. And then as you change theta, it'll sweep out the entire circle. So either way you think about it.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So the radius is going to go between zero and one. Another way of thinking about it is, for any given theta, if you keep varying the radiuses, you'll sweep out all the points on this line. And then as you change theta, it'll sweep out the entire circle. So either way you think about it. So with that, let's actually define x and y in those terms. So we could say that x is equal to, x is going to be equal to, so the x value, whatever r is, the x value is going to be r cosine theta. It's going to be that component.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So either way you think about it. So with that, let's actually define x and y in those terms. So we could say that x is equal to, x is going to be equal to, so the x value, whatever r is, the x value is going to be r cosine theta. It's going to be that component. It's going to be r cosine theta. And then the y component, this is just basic trigonometry, is going to be, the y is just going to be r sine theta. Y is going to be r sine theta.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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It's going to be that component. It's going to be r cosine theta. And then the y component, this is just basic trigonometry, is going to be, the y is just going to be r sine theta. Y is going to be r sine theta. And then the z component, we already said z can be expressed as a function of y. Right over here, we can rewrite this as z is equal to two minus y. That'll tell us how high to go so that we end up on that plane.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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Y is going to be r sine theta. And then the z component, we already said z can be expressed as a function of y. Right over here, we can rewrite this as z is equal to two minus y. That'll tell us how high to go so that we end up on that plane. So if z is equal to two minus y, and if y is r sine theta, we can rewrite z, we can rewrite z as being equal to two minus r sine theta. So there we're done. That's our parameterization.
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Stokes example part 2 Parameterizing the surface Multivariable Calculus Khan Academy.mp3
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So I've said that if you have a vector field, a two-dimensional vector field with component functions p and q, that the divergence of this guy, the divergence of v, which is a scalar-valued function of x and y, is by definition the partial derivative of p with respect to x, plus the partial derivative of q with respect to y. And there's actually another notation for divergence that's kind of helpful for remembering the formula. And what it is, is you take this nabla symbol, that upside-down triangle that we also use for the gradient, and imagine taking the dot product between that and your vector-valued function. And as we did with the gradient, the loose mnemonic you have for this upside-down triangle, as you think of it as a vector full of partial differential operators. And that sounds fancy, but all it means is you kind of take this partial partial x, a thing that wants to take in a function and take its partial derivative, and that's its first component, and the second component is this partial partial y, a thing that wants to take in a function and take its partial derivative with respect to y. And loosely, this isn't really a vector. These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically.
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Divergence notation.mp3
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And as we did with the gradient, the loose mnemonic you have for this upside-down triangle, as you think of it as a vector full of partial differential operators. And that sounds fancy, but all it means is you kind of take this partial partial x, a thing that wants to take in a function and take its partial derivative, and that's its first component, and the second component is this partial partial y, a thing that wants to take in a function and take its partial derivative with respect to y. And loosely, this isn't really a vector. These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically. And you imagine taking the dot product with that and v, who has components, these scalar-valued functions, p of x, y, and q of x, y. And when you imagine doing this dot product, and you're kind of lining up terms, and the first one multiplied by the second, right, quote-unquote multiplied, because in this case, when I say this first component multiplied by p, I really mean you're taking that partial derivative operator, partial partial x, and evaluating it at p. That's kind of what multiplication looks like in this case. So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative.
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Divergence notation.mp3
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These aren't numbers or functions or things like that, but it's something you can write down, and it'll be kind of helpful symbolically. And you imagine taking the dot product with that and v, who has components, these scalar-valued functions, p of x, y, and q of x, y. And when you imagine doing this dot product, and you're kind of lining up terms, and the first one multiplied by the second, right, quote-unquote multiplied, because in this case, when I say this first component multiplied by p, I really mean you're taking that partial derivative operator, partial partial x, and evaluating it at p. That's kind of what multiplication looks like in this case. So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative. So we see we get the same thing over here. It's the same formula that we have, and it's just kind of a nice little, you could think of it as a mnemonic device for remembering what the divergence is. But another nice thing, this can apply to higher-dimensional functions as well, right?
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Divergence notation.mp3
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So you take that, and as per the dot product, you then add what happens if you take this partial operator, this partial partial y, and quote-unquote multiply it with q, which in the case of an operator means you kind of give it the function q, and it's going to take its partial derivative. So we see we get the same thing over here. It's the same formula that we have, and it's just kind of a nice little, you could think of it as a mnemonic device for remembering what the divergence is. But another nice thing, this can apply to higher-dimensional functions as well, right? If we have something that's, let's see, something that's a vector-valued function, and it's going to be a three-dimensional vector field, so it's got x, y, and z as its inputs, and its output then also has to have three dimensions, so it might be like p, q, and r, and all of these are functions of x and y. So that's p of x and y, q, oh, no, no, x, y, and z, right? So p of x, y, kind of got in the habit of two dimensions there.
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Divergence notation.mp3
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But another nice thing, this can apply to higher-dimensional functions as well, right? If we have something that's, let's see, something that's a vector-valued function, and it's going to be a three-dimensional vector field, so it's got x, y, and z as its inputs, and its output then also has to have three dimensions, so it might be like p, q, and r, and all of these are functions of x and y. So that's p of x and y, q, oh, no, no, x, y, and z, right? So p of x, y, kind of got in the habit of two dimensions there. p of x, y, and z, q of x, y, and z, and then r of x, y, and z, and I haven't talked about three-dimensional divergence, but if you take this, and then you imagine doing your nabla dotted with the vector-valued function, it can still make sense. And in this case, that nabla you're thinking of as having three different components, right? It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z.
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Divergence notation.mp3
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So p of x, y, kind of got in the habit of two dimensions there. p of x, y, and z, q of x, y, and z, and then r of x, y, and z, and I haven't talked about three-dimensional divergence, but if you take this, and then you imagine doing your nabla dotted with the vector-valued function, it can still make sense. And in this case, that nabla you're thinking of as having three different components, right? It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z. And the ordering of these, of the variables here, x, y, and z, is just whatever I have here. So even if they didn't have the names x, y, z, you kind of put them in the same order that they show up in your function. And when you imagine taking the dot product between this and your p as a function, q as a function, and r as a function, vector-valued output, what you'd get, and I'll write it over here, is you take that partial partial x and kind of multiply it with p, which means you're really evaluating at p, so partial x here, then you add partial partial y, and you're evaluating at q, because you're kind of imagining multiplying these second components, and then you'll add what happens when you multiply by these third components, where that's partial partial z by that last component.
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Divergence notation.mp3
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It's going to be, on the one hand, this partial partial x, I should say partial x there, partial x, and the second component is partial partial y, and the last component is partial partial z. And the ordering of these, of the variables here, x, y, and z, is just whatever I have here. So even if they didn't have the names x, y, z, you kind of put them in the same order that they show up in your function. And when you imagine taking the dot product between this and your p as a function, q as a function, and r as a function, vector-valued output, what you'd get, and I'll write it over here, is you take that partial partial x and kind of multiply it with p, which means you're really evaluating at p, so partial x here, then you add partial partial y, and you're evaluating at q, because you're kind of imagining multiplying these second components, and then you'll add what happens when you multiply by these third components, where that's partial partial z by that last component. And, you know, since I haven't talked about three-dimensional vector fields or three-dimensional divergence, this last term, maybe it's not a given that you'd have as strong an intuition for why this shows up in divergence as the other two, but it's actually quite similar. You're thinking about changes to the z component of a vector as the value z of the input, as you're kind of moving up and down in that direction changes. But this pattern will go for even higher dimensions that we can't visualize, four, five, 100, whatever you want, and that's what makes this notation here quite nice, is that it encapsulates that and gives a really compact way of describing this formula that has a simple pattern to it, but would otherwise kind of get out of hand to write.
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Divergence notation.mp3
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So for example, I'm gonna give you a function, some kind of function that takes in a 2D vector, x,y, and it's also gonna spit out a 2D vector. And the specific one I have in mind, this is just kind of arbitrary, is x plus sine of y, and then because I'm a sucker for symmetry, I'm gonna make it y plus sine of x. Though of course, this could be any arbitrary function, you don't need that kind of symmetry. So in the last video, I gave a little refresher on how to think about linear transformations and ideas from linear algebra, and how you encode a linear transformation using a matrix and kind of visualize it, I used this grid. And here, I wanna show what this function looks like as a transformation of space. As in, I'm gonna tell the computer, take every single point on this blue grid here, and if that point is x,y, I want you to move it over to the point x plus sine of y, y plus sine of x. And here's what that looks like.
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Local linearity for a multivariable function.mp3
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So in the last video, I gave a little refresher on how to think about linear transformations and ideas from linear algebra, and how you encode a linear transformation using a matrix and kind of visualize it, I used this grid. And here, I wanna show what this function looks like as a transformation of space. As in, I'm gonna tell the computer, take every single point on this blue grid here, and if that point is x,y, I want you to move it over to the point x plus sine of y, y plus sine of x. And here's what that looks like. Alright, so things get really wavy, really curly, this is not at all a linear transformation, right? All of the lines don't remain lines, they're no longer nice grid lines that are parallel and evenly spaced. In some sense, there is much, much more information that goes into nonlinear functions than into linear functions.
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Local linearity for a multivariable function.mp3
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And here's what that looks like. Alright, so things get really wavy, really curly, this is not at all a linear transformation, right? All of the lines don't remain lines, they're no longer nice grid lines that are parallel and evenly spaced. In some sense, there is much, much more information that goes into nonlinear functions than into linear functions. And because this is rather complicated, I think it might be easier to see what's going on if we just focus on a single individual point. So let me look at a point like, let's say, pi halves and zero, okay? So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same.
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Local linearity for a multivariable function.mp3
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In some sense, there is much, much more information that goes into nonlinear functions than into linear functions. And because this is rather complicated, I think it might be easier to see what's going on if we just focus on a single individual point. So let me look at a point like, let's say, pi halves and zero, okay? So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same. And then for the bottom, y, well y is also zero, plus sine of x, sine of pi halves is one, I'll go ahead and write sine of pi halves, sine of pi halves, but you can think of that as just being one. So what that means on the transformation over here is if we look at the point that's at pi halves zero, and pi halves is a little above 1.5, so that's gonna be around here, we expect it to move to the point pi halves one. So it should just move vertically by one unit.
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Local linearity for a multivariable function.mp3
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So if that's what I'm plugging in, x is pi halves, so at the top here, x stays the same, it's pi halves, and then sine of y would be sine of zero, so that x component is gonna completely stay the same. And then for the bottom, y, well y is also zero, plus sine of x, sine of pi halves is one, I'll go ahead and write sine of pi halves, sine of pi halves, but you can think of that as just being one. So what that means on the transformation over here is if we look at the point that's at pi halves zero, and pi halves is a little above 1.5, so that's gonna be around here, we expect it to move to the point pi halves one. So it should just move vertically by one unit. And if you just focus on that one point during the transformation, notice that's exactly what happens, it just moves vertically one point. And of course things are quite complicated because every point is doing that, right? The computer's taking in every point and moving it to where it should go.
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Local linearity for a multivariable function.mp3
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So it should just move vertically by one unit. And if you just focus on that one point during the transformation, notice that's exactly what happens, it just moves vertically one point. And of course things are quite complicated because every point is doing that, right? The computer's taking in every point and moving it to where it should go. So after having given the refresher on thinking about linear transformations and encoding them with matrices last time, something like this might feel completely intractable. You certainly have to store much more information than just four numbers to record where everything goes. But this function has a nice property, a property that we deal with all the time in multivariable calculus.
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Local linearity for a multivariable function.mp3
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The computer's taking in every point and moving it to where it should go. So after having given the refresher on thinking about linear transformations and encoding them with matrices last time, something like this might feel completely intractable. You certainly have to store much more information than just four numbers to record where everything goes. But this function has a nice property, a property that we deal with all the time in multivariable calculus. It's what we'd call locally linear. Locally linear. And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that.
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Local linearity for a multivariable function.mp3
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But this function has a nice property, a property that we deal with all the time in multivariable calculus. It's what we'd call locally linear. Locally linear. And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that. And first of all, I'm gonna add some more grid lines. So they're really very close grid lines, right? We can see from the zoomed out picture.
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Local linearity for a multivariable function.mp3
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And what that means is if I was to take our initial setup and then zoom in on a given point, so I'm gonna zoom in around this point on the left here, and this box kind of in the upper right just shows the zoomed in version of that. And first of all, I'm gonna add some more grid lines. So they're really very close grid lines, right? We can see from the zoomed out picture. But this just makes it so that when we're zoomed in, we can see a little bit more of what's going on. And now when I play the animation, I'm gonna have this yellow box that's doing the zooming to follow the point at its center, right? So this box will be moving and we're always just gonna look at what it zoomed in on.
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Local linearity for a multivariable function.mp3
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We can see from the zoomed out picture. But this just makes it so that when we're zoomed in, we can see a little bit more of what's going on. And now when I play the animation, I'm gonna have this yellow box that's doing the zooming to follow the point at its center, right? So this box will be moving and we're always just gonna look at what it zoomed in on. Okay, so it's gonna be following what's going on around that point during the transformation. And we can see inside this zoomed version, it's still not linear, right? The lines get a little bit curved, but this looks a lot more like a linear function.
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Local linearity for a multivariable function.mp3
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So this box will be moving and we're always just gonna look at what it zoomed in on. Okay, so it's gonna be following what's going on around that point during the transformation. And we can see inside this zoomed version, it's still not linear, right? The lines get a little bit curved, but this looks a lot more like a linear function. It looks a lot more like the grid lines that started off horizontal and vertical are remaining parallel and evenly spaced. And in fact, let's say I zoom in even further to an even smaller yellow box here. And again, I'm gonna add in some more grid lines right around it.
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Local linearity for a multivariable function.mp3
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The lines get a little bit curved, but this looks a lot more like a linear function. It looks a lot more like the grid lines that started off horizontal and vertical are remaining parallel and evenly spaced. And in fact, let's say I zoom in even further to an even smaller yellow box here. And again, I'm gonna add in some more grid lines right around it. So they're very, very densely packed. And this is purely an artifact of visualizing things, right? I could choose to put points or lines or anything wherever I want.
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Local linearity for a multivariable function.mp3
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And again, I'm gonna add in some more grid lines right around it. So they're very, very densely packed. And this is purely an artifact of visualizing things, right? I could choose to put points or lines or anything wherever I want. And I just think showing the grid lines and only the grid lines and where they move gives sort of a feel for what the function is doing. So this time when I play it, and that zooming in box kind of tracks the point that we're looking at, as it goes, the neighborhood around it, all of the points around it really, really do look like a linear function. And the more you zoom in, the more it looks precisely like a certain linear function.
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Local linearity for a multivariable function.mp3
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I could choose to put points or lines or anything wherever I want. And I just think showing the grid lines and only the grid lines and where they move gives sort of a feel for what the function is doing. So this time when I play it, and that zooming in box kind of tracks the point that we're looking at, as it goes, the neighborhood around it, all of the points around it really, really do look like a linear function. And the more you zoom in, the more it looks precisely like a certain linear function. Oh, I guess I should have written an R over here, locally linear. So this raises the question, if we're looking around some specific point, which I'll call X naught and Y naught, this should correspond in some way to the linear transformation that it looks like around it. There should be some kind of matrix, some two by two matrix, that represents the linear transformation that this function, this much more complicated function, looks like around that point.
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Local linearity for a multivariable function.mp3
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And the more you zoom in, the more it looks precisely like a certain linear function. Oh, I guess I should have written an R over here, locally linear. So this raises the question, if we're looking around some specific point, which I'll call X naught and Y naught, this should correspond in some way to the linear transformation that it looks like around it. There should be some kind of matrix, some two by two matrix, that represents the linear transformation that this function, this much more complicated function, looks like around that point. So this idea of zooming in is what we mean by local. And in the next video, I'm gonna show you what this matrix looks like in terms of partial derivatives for our original function. See you then.
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Local linearity for a multivariable function.mp3
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And so let's do that. So this integral right over here is the same thing as the integral from a to b, and then what I'm gonna do is I'm gonna group the things that are being multiplied by dx dt. So if I group them and then factor out a dx dt, so if I take this part right here and this part right over here, I'm essentially going to distribute the r, let me make it clear. I'm gonna distribute the r, and so I'm gonna group that and that right over there, I will be left with, and I'll factor out the dx dt, I will be left with p plus r times the partial of z with respect to x times dx dt. dx dt. And now I'll do the same thing for the dy dts. So that's that part right over there, and then the r is gonna get distributed, and this thing right over here.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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I'm gonna distribute the r, and so I'm gonna group that and that right over there, I will be left with, and I'll factor out the dx dt, I will be left with p plus r times the partial of z with respect to x times dx dt. dx dt. And now I'll do the same thing for the dy dts. So that's that part right over there, and then the r is gonna get distributed, and this thing right over here. So it's gonna be plus q, q plus r times the partial, the r is gonna get distributed, r times the partial of z with respect to y, and I'm gonna factor out a dy dt. dy dt, and then we can't forget all of that, all of that is going to be multiplied, all of that is going to be multiplied by dt. Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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So that's that part right over there, and then the r is gonna get distributed, and this thing right over here. So it's gonna be plus q, q plus r times the partial, the r is gonna get distributed, r times the partial of z with respect to y, and I'm gonna factor out a dy dt. dy dt, and then we can't forget all of that, all of that is going to be multiplied, all of that is going to be multiplied by dt. Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field. In fact, let me kind of copy and paste it. Let me copy and paste it. So, actually I don't know if I'm on the right layer of my work right over here.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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Now this right over here is interesting because it's starting to look very similar, it's starting to look very similar to what we had up here where we just had our theoretical vector field. In fact, let me kind of copy and paste it. Let me copy and paste it. So, actually I don't know if I'm on the right layer of my work right over here. So let me copy and paste. No, that didn't work. Let me try it one more time.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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So, actually I don't know if I'm on the right layer of my work right over here. So let me copy and paste. No, that didn't work. Let me try it one more time. So if I try to copy, I'll go a layer down. I'm using an art program for this. Copy, and then I think this will work, and paste.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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Let me try it one more time. So if I try to copy, I'll go a layer down. I'm using an art program for this. Copy, and then I think this will work, and paste. There we go. So this is a result that we had before. This is kind of a template to look at.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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Copy, and then I think this will work, and paste. There we go. So this is a result that we had before. This is kind of a template to look at. But what is going on over here if we just look at this template? We see that we're in the t domain where we're integrating over t right over here, but then we have these things. We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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This is kind of a template to look at. But what is going on over here if we just look at this template? We see that we're in the t domain where we're integrating over t right over here, but then we have these things. We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt. Well, that's exactly what we're doing right over here. We can distribute the dt, and we have something that looks exactly like this right over here, where m is analogous to this piece right over here. Let me make it clear.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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We have some function that's a function of x and y times dx dt, and then some function, the function of x and y times dy dt, and we're integrating with respect to dt. Well, that's exactly what we're doing right over here. We can distribute the dt, and we have something that looks exactly like this right over here, where m is analogous to this piece right over here. Let me make it clear. m, this piece right over here, looks a lot like m, m in our example right over here. It's being multiplied by dx dt, and then this dt, which you can distribute, and this piece right over here looks a lot like n. And so we can say that, well, we have something that looks like this. We can rewrite it like this and kind of go back into the, kind of deparameterize it.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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Let me make it clear. m, this piece right over here, looks a lot like m, m in our example right over here. It's being multiplied by dx dt, and then this dt, which you can distribute, and this piece right over here looks a lot like n. And so we can say that, well, we have something that looks like this. We can rewrite it like this and kind of go back into the, kind of deparameterize it. So this thing is going to be equal to, it's going to be equal to now the line integral of c1. We are in the xy plane. We started with the curve c, but now we're going to go to c1.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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We can rewrite it like this and kind of go back into the, kind of deparameterize it. So this thing is going to be equal to, it's going to be equal to now the line integral of c1. We are in the xy plane. We started with the curve c, but now we're going to go to c1. This is c1. It's completely analogous. These are only functions of x's and y's.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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We started with the curve c, but now we're going to go to c1. This is c1. It's completely analogous. These are only functions of x's and y's. Everything here is. So now this is going to be the line integral, the line integral over c1, and I can even draw it as like that if I like, of m dx, and that makes sense because if you multiply dt times dx dt, the dt's cancel out and you're just left with dx. So m times dx, so let me write it that way.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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These are only functions of x's and y's. Everything here is. So now this is going to be the line integral, the line integral over c1, and I can even draw it as like that if I like, of m dx, and that makes sense because if you multiply dt times dx dt, the dt's cancel out and you're just left with dx. So m times dx, so let me write it that way. So it's going to be p plus r times the partial of z with respect to x, dx plus n, let me scroll to the right a little bit, plus n, which is q plus r times the partial of z with respect to y, dy. And then this is really interesting because this path that we are now concerned with, it's completely analogous. I hope you don't think I'm doing some voodoo here.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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So m times dx, so let me write it that way. So it's going to be p plus r times the partial of z with respect to x, dx plus n, let me scroll to the right a little bit, plus n, which is q plus r times the partial of z with respect to y, dy. And then this is really interesting because this path that we are now concerned with, it's completely analogous. I hope you don't think I'm doing some voodoo here. This statement is completely analogous to this statement where m could be this and n could be this. And so we can revert it back to now the path c1 that sits in the xy plane, not our original boundary c, but now we're just dealing with things, a boundary in the xy plane. So it reverts to this, but what's powerful about getting it to this point is we can now apply Green's theorem to this.
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Stokes' theorem proof part 6 Multivariable Calculus Khan Academy.mp3
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In the last couple of videos, we saw that we can describe a curve by a position vector valued function. In very general terms, it would be the x position as a function of time times the unit vector in the horizontal direction plus the y position as a function of time times the unit vector in the vertical direction. And this will essentially describe this, if you can imagine a particle, and let's say the parameter t represents time, it will describe where the particle is at any given time. If we wanted a particular curve, we can say this only applies for some curve, we're dealing with r of t, and it's only applicable between t being greater than a and less than b. That would describe some curve in two dimensions. Let me just draw it here. This is all a review of the last two videos.
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Differential of a vector valued function Multivariable Calculus Khan Academy.mp3
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If we wanted a particular curve, we can say this only applies for some curve, we're dealing with r of t, and it's only applicable between t being greater than a and less than b. That would describe some curve in two dimensions. Let me just draw it here. This is all a review of the last two videos. This curve might look something like that, where this is where t is equal to a, that's where t is equal to b. So r of a will be this vector right here that ends at that point, and then as t, or if you can imagine the parameter being time, it doesn't have to be time, but that's a convenient one to visualize. Each corresponding as t gets larger and larger, we're specifying different points on the path.
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Differential of a vector valued function Multivariable Calculus Khan Academy.mp3
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This is all a review of the last two videos. This curve might look something like that, where this is where t is equal to a, that's where t is equal to b. So r of a will be this vector right here that ends at that point, and then as t, or if you can imagine the parameter being time, it doesn't have to be time, but that's a convenient one to visualize. Each corresponding as t gets larger and larger, we're specifying different points on the path. We saw that two videos ago, and in the last video, we thought about what does it mean to take the derivative of a vector-valued function, and we came up with this idea, and it wasn't an idea, we actually showed it to be true, we came up with a definition really, that the derivative, I could call it r prime of t, and it's going to be a vector. The derivative of a vector-valued function is once again going to be a derivative, but it is equal to the way we defined it, x prime of t times i plus y prime of t times j, or another way to write that, and I'll just write all the different ways just so you get familiar with it, dr dt is equal to dx dt. This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j.
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Differential of a vector valued function Multivariable Calculus Khan Academy.mp3
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Each corresponding as t gets larger and larger, we're specifying different points on the path. We saw that two videos ago, and in the last video, we thought about what does it mean to take the derivative of a vector-valued function, and we came up with this idea, and it wasn't an idea, we actually showed it to be true, we came up with a definition really, that the derivative, I could call it r prime of t, and it's going to be a vector. The derivative of a vector-valued function is once again going to be a derivative, but it is equal to the way we defined it, x prime of t times i plus y prime of t times j, or another way to write that, and I'll just write all the different ways just so you get familiar with it, dr dt is equal to dx dt. This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j. If we wanted to think about the differential, one thing that we can think about, whenever I do the math of the differential, it's a little bit hand-wavy, I'm not being very rigorous, but if you imagine multiplying both sides of this equation by a very small dt, or this exact dt, you would get dr is equal to, I'll just leave it like this, dx dt times dt, I could make these cancel out, but I'll just write it like this first, times the unit vector i plus dy dt times the unit vector j, or we could rewrite this, and I'm just rewriting it in all of the different ways that one can rewrite it. You could also write this as dr is equal to x prime of t dt times the unit vector i, sorry, this was x prime of t dt, this is x prime of t right there, times the unit vector i plus y prime of t, that's just that right there, times dt times the unit vector j, and just to complete the trifecta, the other way that we could write this is that dr is equal to, if we just allowed these to cancel out, then we get is equal to dx times i plus dy times dy times j. And that actually makes a lot of intuitive sense, that if I look at any dr, so let's say I look at the change between this vector and this vector, let's say the super small change right there, that is our dr, and it's made up of, it's our dx, our change in x, is that right there, you can imagine it's that right there, times, but we're vectorizing it by multiplying it by the unit vector in the horizontal direction, plus dy times the unit vector in the vertical direction.
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Differential of a vector valued function Multivariable Calculus Khan Academy.mp3
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This is just a standard derivative, x of t is a scalar function, so this is a standard derivative, times i plus dy dt times j. If we wanted to think about the differential, one thing that we can think about, whenever I do the math of the differential, it's a little bit hand-wavy, I'm not being very rigorous, but if you imagine multiplying both sides of this equation by a very small dt, or this exact dt, you would get dr is equal to, I'll just leave it like this, dx dt times dt, I could make these cancel out, but I'll just write it like this first, times the unit vector i plus dy dt times the unit vector j, or we could rewrite this, and I'm just rewriting it in all of the different ways that one can rewrite it. You could also write this as dr is equal to x prime of t dt times the unit vector i, sorry, this was x prime of t dt, this is x prime of t right there, times the unit vector i plus y prime of t, that's just that right there, times dt times the unit vector j, and just to complete the trifecta, the other way that we could write this is that dr is equal to, if we just allowed these to cancel out, then we get is equal to dx times i plus dy times dy times j. And that actually makes a lot of intuitive sense, that if I look at any dr, so let's say I look at the change between this vector and this vector, let's say the super small change right there, that is our dr, and it's made up of, it's our dx, our change in x, is that right there, you can imagine it's that right there, times, but we're vectorizing it by multiplying it by the unit vector in the horizontal direction, plus dy times the unit vector in the vertical direction. So when you multiply this distance times the unit vector, you're essentially getting this vector, you're essentially getting this vector right here, and when you multiply this guy, and actually our change in y here is negative, you're going to get this vector right here. So when you add those together, you get your change in your actual position vector. So that was all a little bit of background, and this might be somewhat useful in a future video from now.
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Differential of a vector valued function Multivariable Calculus Khan Academy.mp3
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So let's say that we've got the curve R defined. So this is our curve R. It's x of t times i plus y of t times j. It's a curve in two dimensions on the xy plane. And let's graph it. Just graph it in kind of a generalized form. So that's our y-axis. This is our x-axis.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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And let's graph it. Just graph it in kind of a generalized form. So that's our y-axis. This is our x-axis. Our curve R might look something like this. It might look something. Let me draw a little bit more of a, maybe it looks something like this.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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This is our x-axis. Our curve R might look something like this. It might look something. Let me draw a little bit more of a, maybe it looks something like this. Maybe that's just part of it. And as t increases, we're going in that direction right over there. What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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Let me draw a little bit more of a, maybe it looks something like this. Maybe that's just part of it. And as t increases, we're going in that direction right over there. What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector. Obviously, if you can figure out a normal vector, you can just divide it by its magnitude and you will get the unit normal vector. So I want to figure out at any given point a vector that's popping straight out in that direction and has a magnitude 1. So that would be our unit normal vector.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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What I want to do in this video, and this is really more vector algebra than vector calculus, is think about at any given point here whether we can figure out a normal vector, in particular, a unit normal vector. Obviously, if you can figure out a normal vector, you can just divide it by its magnitude and you will get the unit normal vector. So I want to figure out at any given point a vector that's popping straight out in that direction and has a magnitude 1. So that would be our unit normal vector. And to do that, first we'll think about what a tangent vector is. And from a tangent vector, we can figure out the normal vector. And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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So that would be our unit normal vector. And to do that, first we'll think about what a tangent vector is. And from a tangent vector, we can figure out the normal vector. And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line. We're going to see a very similar thing when we do it right over here with this vector algebra. So the first thing I want to think about is how do we construct a tangent line? Well, you can imagine at some t, this is what our position vector is going to look like.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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And it really goes back to some of what you might have done in algebra 1 or algebra 2 of if you have a slope of the line, the negative reciprocal of that slope is going to be the slope of the perpendicular line. We're going to see a very similar thing when we do it right over here with this vector algebra. So the first thing I want to think about is how do we construct a tangent line? Well, you can imagine at some t, this is what our position vector is going to look like. So call that r1 right over there. And then if we allow t to go up a little bit, if t is time, we wait a little while, a few seconds, however we're measuring things. And then r2 might look something like this.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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Well, you can imagine at some t, this is what our position vector is going to look like. So call that r1 right over there. And then if we allow t to go up a little bit, if t is time, we wait a little while, a few seconds, however we're measuring things. And then r2 might look something like this. This is when t has gotten a little bit larger. We're further down the path. And so one way that you can approximate the slope of the tangent line or the slope between these two points for now is essentially the difference between these two vectors.
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Constructing a unit normal vector to a curve Multivariable Calculus Khan Academy.mp3
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