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So you'd wanna mark somewhere here, two, four. And that represents an input-output pair. And if you do that with negative one, one, you go negative one, one. And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve. And the implication for doing this is that we typically think of what is on the x-axis as being where the inputs live. You know, this would be, we think of it as the input one, and this is the input two, and so on. And then you think of the output as being the height, the height of the graph above each point.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve. And the implication for doing this is that we typically think of what is on the x-axis as being where the inputs live. You know, this would be, we think of it as the input one, and this is the input two, and so on. And then you think of the output as being the height, the height of the graph above each point. But this is kind of a consequence of the fact where we're just listing all of the pairs here. So now if we go to the world of multivariable functions, here, I'm not gonna show the graph right now. Let's just think we've got three-dimensional space at our disposal to do with what we will.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And then you think of the output as being the height, the height of the graph above each point. But this is kind of a consequence of the fact where we're just listing all of the pairs here. So now if we go to the world of multivariable functions, here, I'm not gonna show the graph right now. Let's just think we've got three-dimensional space at our disposal to do with what we will. We still wanna understand the relationship between inputs and outputs of this guy. But in this case, inputs are something that we think of as a pair of points. So you might have a pair of points like one, two.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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Let's just think we've got three-dimensional space at our disposal to do with what we will. We still wanna understand the relationship between inputs and outputs of this guy. But in this case, inputs are something that we think of as a pair of points. So you might have a pair of points like one, two. And the output there is gonna be one squared plus two squared. And what that equals is five. So how do we visualize that?
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So you might have a pair of points like one, two. And the output there is gonna be one squared plus two squared. And what that equals is five. So how do we visualize that? Well, if we wanna pair these things together, the natural way to do that is to think of a triplet of some kind. So in this case, you'd wanna plot the triplet one, two, five. And to do that in three dimensions, we'll take a look over here.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So how do we visualize that? Well, if we wanna pair these things together, the natural way to do that is to think of a triplet of some kind. So in this case, you'd wanna plot the triplet one, two, five. And to do that in three dimensions, we'll take a look over here. We think of going one in the x direction. This axis here is the x axis. So we wanna move a distance one there.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And to do that in three dimensions, we'll take a look over here. We think of going one in the x direction. This axis here is the x axis. So we wanna move a distance one there. And we wanna go two in the y direction. So we kind of think of going a distance two there. And then five up.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So we wanna move a distance one there. And we wanna go two in the y direction. So we kind of think of going a distance two there. And then five up. And then that's gonna give us some kind of point. So we think of this point in space, and that's a given input-output pair. But we could do this for a lot.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And then five up. And then that's gonna give us some kind of point. So we think of this point in space, and that's a given input-output pair. But we could do this for a lot. A couple different points that you might get if you start plotting various different ones will look something like this. And of course, there's infinitely many that you can do, and it'll take forever if you try to just draw each one in three dimensions. But what's very nice here is that, here, I'll get rid of those lines.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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But we could do this for a lot. A couple different points that you might get if you start plotting various different ones will look something like this. And of course, there's infinitely many that you can do, and it'll take forever if you try to just draw each one in three dimensions. But what's very nice here is that, here, I'll get rid of those lines. If you imagine doing this for all of the infinitely many pairs of inputs that you could possibly have, you end up drawing a surface. So in this case, the surface kind of looks like a three-dimensional parabola. That's no coincidence.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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But what's very nice here is that, here, I'll get rid of those lines. If you imagine doing this for all of the infinitely many pairs of inputs that you could possibly have, you end up drawing a surface. So in this case, the surface kind of looks like a three-dimensional parabola. That's no coincidence. It has to do with the fact that we're using x squared and y squared here. And now, the inputs, like one, two, we think of as being on the xy-plane. So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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That's no coincidence. It has to do with the fact that we're using x squared and y squared here. And now, the inputs, like one, two, we think of as being on the xy-plane. So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph. So it's very similar to two dimensions. We think of the input as being on one axis, and the height gives the output there. So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right?
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So you think of the inputs living here, and then what corresponds to the output is that height of a given point above the graph. So it's very similar to two dimensions. We think of the input as being on one axis, and the height gives the output there. So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right? So I'll draw it in red here. Let's say that we have our function, but I'm gonna change it so that it outputs one half of x squared plus y squared. What's gonna be the shape of the graph for that function?
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So just to give an example of what the consequence of this is, I want you to think about what might happen if we changed our multivariable function a little bit, and we multiplied everything by half, right? So I'll draw it in red here. Let's say that we have our function, but I'm gonna change it so that it outputs one half of x squared plus y squared. What's gonna be the shape of the graph for that function? And what it means is the height of every point above this xy-plane is gonna have to get cut in half. So it's actually just a modification of what we already have, but everything kind of sloops on down to be about half of what it was. So in this case, instead of that height being five, it would be 2.5.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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What's gonna be the shape of the graph for that function? And what it means is the height of every point above this xy-plane is gonna have to get cut in half. So it's actually just a modification of what we already have, but everything kind of sloops on down to be about half of what it was. So in this case, instead of that height being five, it would be 2.5. You could imagine, let's say we did this, you know, it was even more extreme. Instead of saying one half, you cut it down by like 1 12th. Maybe I'll use the same color.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So in this case, instead of that height being five, it would be 2.5. You could imagine, let's say we did this, you know, it was even more extreme. Instead of saying one half, you cut it down by like 1 12th. Maybe I'll use the same color. By 1 12th. That would mean that everything, you know, sloops very flat, very flat and close to the xy-plane. So a graph being very close to the xy-plane like this corresponds with very small outputs.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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Maybe I'll use the same color. By 1 12th. That would mean that everything, you know, sloops very flat, very flat and close to the xy-plane. So a graph being very close to the xy-plane like this corresponds with very small outputs. And one thing that I'd like to caution you against, it's very tempting to try to think of every multivariable function as a graph, because we're so used to graphs in two dimensions, and we're so used to trying to find analogies between two dimensions and three dimensions directly. But the only reason that this works is because if you take the number of dimensions in the input, two dimensions, and then the number of dimensions in the output, one dimension, it was reasonable to fit all of that into three, which we could do. But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So a graph being very close to the xy-plane like this corresponds with very small outputs. And one thing that I'd like to caution you against, it's very tempting to try to think of every multivariable function as a graph, because we're so used to graphs in two dimensions, and we're so used to trying to find analogies between two dimensions and three dimensions directly. But the only reason that this works is because if you take the number of dimensions in the input, two dimensions, and then the number of dimensions in the output, one dimension, it was reasonable to fit all of that into three, which we could do. But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output. That would require a five-dimensional graph, but we're not very good at visualizing things like that. So there's lots of other methods, and I think it's very important to kind of open your mind to what those might be. In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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But imagine you have a multivariable function with, you know, a three-dimensional input and a two-dimensional output. That would require a five-dimensional graph, but we're not very good at visualizing things like that. So there's lots of other methods, and I think it's very important to kind of open your mind to what those might be. In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space. That's called a contour map. Couple other ones, like parametric functions, you just look in the output space. Things like a vector space, you kind of look in the input space, but get all the outputs.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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In particular, another one that I'm gonna go through soon lets us think about 3D graphs, but kind of in a two-dimensional setting, and we're just gonna look at the input space. That's called a contour map. Couple other ones, like parametric functions, you just look in the output space. Things like a vector space, you kind of look in the input space, but get all the outputs. There's lots of different ways. I'll go over those in the next few videos. And that's three-dimensional graphs.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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This is saying that there's no net flux across this surface right over here. Or if you sum up all of the divergences in this volume, you are getting zero. So why is that? Well, the simple way to think about it is when we took the divergence of F, this vector field F is hard to visualize, but the divergence of F is fairly easy to visualize. The divergence is equal to two times X. So over here, you're going to get, as you go further and further in this direction, as X becomes larger, your divergence becomes more and more positive. So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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Well, the simple way to think about it is when we took the divergence of F, this vector field F is hard to visualize, but the divergence of F is fairly easy to visualize. The divergence is equal to two times X. So over here, you're going to get, as you go further and further in this direction, as X becomes larger, your divergence becomes more and more positive. So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there. And that's also true, obviously, as you go higher, because you're just changing the Z, you're not changing the X. So all over here, you have positive divergence. Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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So you have very, you have kind of a divergence of two right over here, you have a divergence of one along that line, and you have a divergence of zero right there. And that's also true, obviously, as you go higher, because you're just changing the Z, you're not changing the X. So all over here, you have positive divergence. Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence. I guess you could say in the positive X side of our octant. But then as you go in that side, on the other side, you have negative divergence. And this diagram is symmetric with respect to the ZY plane.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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Over there, you have positive divergence, and not just along that plane, but if you go in the X direction as well, this kind of, this whole region of space, you have positive divergence. I guess you could say in the positive X side of our octant. But then as you go in that side, on the other side, you have negative divergence. And this diagram is symmetric with respect to the ZY plane. And so those divergences cancel them out. You would have had a positive flux across the surface, or a positive value right over here. Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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And this diagram is symmetric with respect to the ZY plane. And so those divergences cancel them out. You would have had a positive flux across the surface, or a positive value right over here. Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one. So let's just think about that region. So that region that would have been cut off right over here, that would have been cut off right over there. And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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Instead of calculating it for this region, we had calculated it for a region that was just between X is zero and one. So let's just think about that region. So that region that would have been cut off right over here, that would have been cut off right over there. And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane. Now if we care about this, if we care about this volume, so we're essentially eliminating the rest of it. So let me try to eliminate it as best as I can, change the colors. So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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And so the back, I guess you could say the back wall of that, the back wall of this would have been the ZY plane. Now if we care about this, if we care about this volume, so we're essentially eliminating the rest of it. So let me try to eliminate it as best as I can, change the colors. So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first. So if we eliminate the back part of it, and we're just dealing with, we're just dealing with it when X is positive, then our entire solution that we did in the last video would have been the exact same, except now X is going to vary between, instead of negative one and one, it'll vary between zero and one. And so our bounds of integration, X is going to go between zero and one, zero and one. And then in that situation, our final answer, this part, this would be between zero and one, that would all be zero, and we would be left with 3 1⁄2 minus 1 1⁄2.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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So if I eliminate that part of it over there, and I'll maybe even what we see, all of that, I should have deleted that first. So if we eliminate the back part of it, and we're just dealing with, we're just dealing with it when X is positive, then our entire solution that we did in the last video would have been the exact same, except now X is going to vary between, instead of negative one and one, it'll vary between zero and one. And so our bounds of integration, X is going to go between zero and one, zero and one. And then in that situation, our final answer, this part, this would be between zero and one, that would all be zero, and we would be left with 3 1⁄2 minus 1 1⁄2. 3 1⁄2 minus 1 1⁄2 is one, minus 1 6, which is just going to be 5 6. And so when you just think about this part of it, this side of it, this one that I've just drawn, you had a positive flux of 5 6. On the other side, you had a negative flux of 5 6, and then they canceled it out.
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Why we got zero flux in divergence theorem example 1 Multivariable Calculus Khan Academy.mp3
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In the last video, we took a look at this function, f of x, y equals x to the fourth minus four x squared plus y squared, which has the graph that you're looking at on the left. And we looked for all of the points where the gradient is equal to zero, which basically means both partial derivatives are equal to zero. And we solved that and we found that there were three different points, the origin zero, zero, and then square root of two, zero, and negative square root of two, zero, which corresponds to this origin here, which is a saddle point, and then these two local minima. And it seemed like we had a reasonable explanation for why this is a saddle point and why both of those are local minima. Because we took the second partial derivative with respect to x, I was kind of all over the board here, second partial derivative with respect to x, and found that when you evaluate it at x equals zero, you get a negative number, kind of indicating a negative concavity, so it should look like a maximum. And then when you do the same with the second partial derivative of y, man, I always do this, I always leave out the squared on that lower term there. Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y.
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Second partial derivative test.mp3
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And it seemed like we had a reasonable explanation for why this is a saddle point and why both of those are local minima. Because we took the second partial derivative with respect to x, I was kind of all over the board here, second partial derivative with respect to x, and found that when you evaluate it at x equals zero, you get a negative number, kind of indicating a negative concavity, so it should look like a maximum. And then when you do the same with the second partial derivative of y, man, I always do this, I always leave out the squared on that lower term there. Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y. So that's why, you know, this origin point looks like a saddle point, because the x direction and y direction disagree. And when you do this with the other points, they kind of both agree that it should look like a minimum. But I said that's not enough.
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Second partial derivative test.mp3
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Okay, so second partial derivative with respect to y, you get two as a constant, a positive, and that kind of indicates that it looks like a minimum to y. So that's why, you know, this origin point looks like a saddle point, because the x direction and y direction disagree. And when you do this with the other points, they kind of both agree that it should look like a minimum. But I said that's not enough. I said that you need to take into account the mixed partial derivative term. And to see why that's true, let me go ahead and pull up another example for you here. So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually.
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Second partial derivative test.mp3
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But I said that's not enough. I said that you need to take into account the mixed partial derivative term. And to see why that's true, let me go ahead and pull up another example for you here. So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually. But when we get the equation for this function, the equation is f of x, y is equal to x squared plus y squared minus four times x, y. Now let's go ahead and analyze the partial differential information of this function. We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y.
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Second partial derivative test.mp3
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So, the graph of the function that you're looking at right now, it clearly has a saddle point at the origin that we can see visually. But when we get the equation for this function, the equation is f of x, y is equal to x squared plus y squared minus four times x, y. Now let's go ahead and analyze the partial differential information of this function. We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y. And when we do the partial derivative with respect to y, very similarly, we're gonna get two y when we differentiate y squared, two y, and now we subtract minus four x, because x looks like the constant and y looks like the variable, minus four x. Now when we plug in, x and y are each equal to zero, you know, we plug in the origin point to both of these functions, we see that they're equal to zero, because x is zero, y is zero, this guy goes to zero. Similarly over here, that goes to zero.
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Second partial derivative test.mp3
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We'll just take its partial derivatives, so the partial with respect to x is equal to, so we've got, when we differentiate this term, we get two x, two x, y looks like a constant, we do nothing, and then this last term looks like negative four times y, because y looks like a constant, so negative four y. And when we do the partial derivative with respect to y, very similarly, we're gonna get two y when we differentiate y squared, two y, and now we subtract minus four x, because x looks like the constant and y looks like the variable, minus four x. Now when we plug in, x and y are each equal to zero, you know, we plug in the origin point to both of these functions, we see that they're equal to zero, because x is zero, y is zero, this guy goes to zero. Similarly over here, that goes to zero. So we will indeed get a flat tangent plane at the origin. But now let's take a look at the second partial derivatives. If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us.
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Second partial derivative test.mp3
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Similarly over here, that goes to zero. So we will indeed get a flat tangent plane at the origin. But now let's take a look at the second partial derivatives. If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us. And then similarly, when we take the second partial derivative with respect to y, always forget that squared on the bottom, we also get a constant positive two, because this x does nothing for us when we take the derivative with respect to y. So your constant positive two. So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum.
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Second partial derivative test.mp3
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If we do the second partial derivatives purely in terms of x and y, so if we take the pure second partial derivative of f with respect to x squared, what we get, we look at this expression, we differentiate it with respect to x, and we get a constant positive two, because that y does nothing for us. And then similarly, when we take the second partial derivative with respect to y, always forget that squared on the bottom, we also get a constant positive two, because this x does nothing for us when we take the derivative with respect to y. So your constant positive two. So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum. But when we look at the graph, this isn't true. It's not a local minimum, it's a saddle point. So what this tells us is that these two second partial derivatives aren't enough, we need more information.
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Second partial derivative test.mp3
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So this would suggest that, you know, there's positive concavity in the x direction, there's positive concavity in the y direction, so it would suggest that, you know, it looks like an upward smiley face from all directions, and it should be a local minimum. But when we look at the graph, this isn't true. It's not a local minimum, it's a saddle point. So what this tells us is that these two second partial derivatives aren't enough, we need more information. And what it kind of comes down to is that this last term here, minus four x, what, oh, actually, I think I made a mistake. I think I meant to make this plus four x, y. So let's see.
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Second partial derivative test.mp3
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So what this tells us is that these two second partial derivatives aren't enough, we need more information. And what it kind of comes down to is that this last term here, minus four x, what, oh, actually, I think I made a mistake. I think I meant to make this plus four x, y. So let's see. Plus four x, y, which would influence these. It actually won't make a difference, it still gives a saddle point. But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum.
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Second partial derivative test.mp3
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So let's see. Plus four x, y, which would influence these. It actually won't make a difference, it still gives a saddle point. But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum. And just to give a loose intuition for what's going on here, if instead of writing four here, I wrote, I'm gonna write the variable p, okay, and p is just gonna be some number, and I'm gonna move that variable around. I'm gonna basically let it range from zero up to four. So right now, as you're looking at it, it's sitting at four.
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Second partial derivative test.mp3
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But anyway, we've got this plus four x, y term that evidently makes a difference, that evidently kind of influences whether this is a local minimum or a maximum. And just to give a loose intuition for what's going on here, if instead of writing four here, I wrote, I'm gonna write the variable p, okay, and p is just gonna be some number, and I'm gonna move that variable around. I'm gonna basically let it range from zero up to four. So right now, as you're looking at it, it's sitting at four. So I'm gonna pull it back and kind of let it range back to zero just to see how this influences the graph. And we see that once we pull it back to zero, we get something where it kind of reflects what you would expect, where from the x direction, it's a positive smiley face, from the y direction, it's also a positive smiley face, and everything's good. It looks like a local minimum, and it is.
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Second partial derivative test.mp3
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So right now, as you're looking at it, it's sitting at four. So I'm gonna pull it back and kind of let it range back to zero just to see how this influences the graph. And we see that once we pull it back to zero, we get something where it kind of reflects what you would expect, where from the x direction, it's a positive smiley face, from the y direction, it's also a positive smiley face, and everything's good. It looks like a local minimum, and it is. And even as you let p range more and more, and here, p is around, I'm guessing right now it's around 1.5, you get something that's still a local minimum. It's a positive concavity in all directions. But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point.
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Second partial derivative test.mp3
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It looks like a local minimum, and it is. And even as you let p range more and more, and here, p is around, I'm guessing right now it's around 1.5, you get something that's still a local minimum. It's a positive concavity in all directions. But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point. And again, this is entirely the coefficient in front of the xy term. It has nothing to do with the x squared or the y squared. So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point.
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Second partial derivative test.mp3
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But there's a critical point here where as you're moving p, at some point, it kind of passes over and turns it into a saddle point. And again, this is entirely the coefficient in front of the xy term. It has nothing to do with the x squared or the y squared. So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point. And in a moment, it'll become clear that that critical point happens when p is equal to two. So right here, it's gonna be when p equals two, it kind of passes from making things a local minimum to a saddle point. And let me show you the test which will tell us why this is true.
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Second partial derivative test.mp3
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So at some point, it kind of passes over, and from that point on, everything is going to be a saddle point. And in a moment, it'll become clear that that critical point happens when p is equal to two. So right here, it's gonna be when p equals two, it kind of passes from making things a local minimum to a saddle point. And let me show you the test which will tell us why this is true. So the full reasoning behind this test is something that I'll get to in later videos, but right now, I just want to kind of have it on the table and teach you what it is and how to use it. So this is called the second partial derivative test. Second partial derivative test.
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Second partial derivative test.mp3
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And let me show you the test which will tell us why this is true. So the full reasoning behind this test is something that I'll get to in later videos, but right now, I just want to kind of have it on the table and teach you what it is and how to use it. So this is called the second partial derivative test. Second partial derivative test. I'll just write derivative test since I'm a slow writer. And basically, what it says is if you found a point where the gradient of your function at this point, and I'll write it kind of x-naught, y-naught is our point. If you found where it equals zero, then calculate the following value.
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Second partial derivative test.mp3
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Second partial derivative test. I'll just write derivative test since I'm a slow writer. And basically, what it says is if you found a point where the gradient of your function at this point, and I'll write it kind of x-naught, y-naught is our point. If you found where it equals zero, then calculate the following value. You'll take the second partial derivative with respect to x twice. So here, I'm just using that subscript notation which is completely the same as saying second partial derivative with respect to x twice, just different choice in notation. And you evaluate it at this point, x-naught, y-naught.
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Second partial derivative test.mp3
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If you found where it equals zero, then calculate the following value. You'll take the second partial derivative with respect to x twice. So here, I'm just using that subscript notation which is completely the same as saying second partial derivative with respect to x twice, just different choice in notation. And you evaluate it at this point, x-naught, y-naught. Then you multiply that by the second partial derivative with respect to y, evaluate it at that same point, y-naught. And then you subtract off the mixed partial derivative, the one where first you do it with respect to x, then with respect to y, or in the other order, it doesn't really matter, as long as you take it with respect to both variables. You subtract off that guy squared.
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Second partial derivative test.mp3
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And you evaluate it at this point, x-naught, y-naught. Then you multiply that by the second partial derivative with respect to y, evaluate it at that same point, y-naught. And then you subtract off the mixed partial derivative, the one where first you do it with respect to x, then with respect to y, or in the other order, it doesn't really matter, as long as you take it with respect to both variables. You subtract off that guy squared. So what you do is you compute this entire value and it's gonna give you some kind of number, and let's give it a name. Let's name it h. And if it's the case that h is greater than zero, then you have either a max or a min. You're not sure which one yet.
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Second partial derivative test.mp3
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You subtract off that guy squared. So what you do is you compute this entire value and it's gonna give you some kind of number, and let's give it a name. Let's name it h. And if it's the case that h is greater than zero, then you have either a max or a min. You're not sure which one yet. And it's a max or a min. And you can tell whether it's a maximum or a minimum basically by looking at one of these partial derivative with respect to x twice or with respect to y twice and kind of getting a feel for the concavity there. If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum.
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Second partial derivative test.mp3
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You're not sure which one yet. And it's a max or a min. And you can tell whether it's a maximum or a minimum basically by looking at one of these partial derivative with respect to x twice or with respect to y twice and kind of getting a feel for the concavity there. If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum. And the fact that this entire value h is greater than zero is what you need to tell you that you can just do that. You can look at the concavity with respect to one of those guys and that'll tell you the information you need about the entire graph. But if h is less than zero, if h is less than zero, then you definitely have a saddle point.
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Second partial derivative test.mp3
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If this was positive, it would indicate kind of a smiley face concavity and it would be a local minimum. And the fact that this entire value h is greater than zero is what you need to tell you that you can just do that. You can look at the concavity with respect to one of those guys and that'll tell you the information you need about the entire graph. But if h is less than zero, if h is less than zero, then you definitely have a saddle point. Saddle point. And if h is purely equal to zero, if you get that unlucky case, then you don't know. Then the second partial derivative test isn't enough to determine.
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Second partial derivative test.mp3
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But if h is less than zero, if h is less than zero, then you definitely have a saddle point. Saddle point. And if h is purely equal to zero, if you get that unlucky case, then you don't know. Then the second partial derivative test isn't enough to determine. But almost all cases, you'll find it either is purely greater than zero or purely less than zero. So as an example, let's see what that looks like in the case of the specific function we started with, where p is some constant that I was letting kind of range from zero to four when I was animating it here. But you should just think of p as being some number.
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Second partial derivative test.mp3
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Then the second partial derivative test isn't enough to determine. But almost all cases, you'll find it either is purely greater than zero or purely less than zero. So as an example, let's see what that looks like in the case of the specific function we started with, where p is some constant that I was letting kind of range from zero to four when I was animating it here. But you should just think of p as being some number. Well, in that case, this value h that we plug in, and let's say we're plugging in at the origin, right? We're analyzing at the origin. Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row.
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Second partial derivative test.mp3
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But you should just think of p as being some number. Well, in that case, this value h that we plug in, and let's say we're plugging in at the origin, right? We're analyzing at the origin. Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row. And both of those, when we computed those, were just constants two. They were equal to two everywhere. And in particular, they're equal to two at the origin.
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Second partial derivative test.mp3
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Well, we've already calculated the partial derivative with respect to x twice in a row and y twice in a row. And both of those, when we computed those, were just constants two. They were equal to two everywhere. And in particular, they're equal to two at the origin. So we can go ahead and just plug in those. And we see that this is two times two. And then now we need to subtract off the mixed partial derivative squared.
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Second partial derivative test.mp3
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And in particular, they're equal to two at the origin. So we can go ahead and just plug in those. And we see that this is two times two. And then now we need to subtract off the mixed partial derivative squared. So if we go ahead and compute that, where we take the derivative with respect to x and then y, or y and then x, let's say we started with the partial derivative with respect to x. When we take this derivative with respect to y, we're gonna get this constant term that's sitting in front of the y. But really, it's whatever this constant p happened to equal.
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Second partial derivative test.mp3
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And then now we need to subtract off the mixed partial derivative squared. So if we go ahead and compute that, where we take the derivative with respect to x and then y, or y and then x, let's say we started with the partial derivative with respect to x. When we take this derivative with respect to y, we're gonna get this constant term that's sitting in front of the y. But really, it's whatever this constant p happened to equal. And you might be able to see that just looking at this function, that when you take the mixed partial derivative, it's gonna be the coefficient in front of the xy term. Because it's kind of like, first you do the derivative with respect to x and the x goes away, and then with respect to y, and that y goes away. And you're just left with a constant.
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Second partial derivative test.mp3
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But really, it's whatever this constant p happened to equal. And you might be able to see that just looking at this function, that when you take the mixed partial derivative, it's gonna be the coefficient in front of the xy term. Because it's kind of like, first you do the derivative with respect to x and the x goes away, and then with respect to y, and that y goes away. And you're just left with a constant. So what you end up getting here in the second partial derivative test, when we take that value, which is p, which might equal four or zero or whatever we happen to have it as, and we square that. We square that. That's gonna be the value that we analyze.
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Second partial derivative test.mp3
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And you're just left with a constant. So what you end up getting here in the second partial derivative test, when we take that value, which is p, which might equal four or zero or whatever we happen to have it as, and we square that. We square that. That's gonna be the value that we analyze. So in the case where p was equal to zero, if we go over here and we scale so that p is completely equal to zero, then our entire value h, h would equal four. And because h is positive, it's definitely a maximum or a minimum. And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum.
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Second partial derivative test.mp3
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That's gonna be the value that we analyze. So in the case where p was equal to zero, if we go over here and we scale so that p is completely equal to zero, then our entire value h, h would equal four. And because h is positive, it's definitely a maximum or a minimum. And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum. Because positive concavity gives local minimum. But in the other case, where let's say we let p range such that p is all the way equal to four, in formulas, what that means for us, when we let p equal four, is we're taking two times two minus four squared. So we're taking two times two minus 16.
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Second partial derivative test.mp3
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And then by analyzing one of those second partial derivatives with respect to x or y, and seeing that it's positive concavity, we would see, oh, it's definitely a local minimum. Because positive concavity gives local minimum. But in the other case, where let's say we let p range such that p is all the way equal to four, in formulas, what that means for us, when we let p equal four, is we're taking two times two minus four squared. So we're taking two times two minus 16. What that would imply, sorry about getting kind of scrunched on the board here, is that h is equal to, let's see, four minus 16, negative 12. So I'm just gonna erase this to kind of clear up some room. So when p equals four, this is negative 12.
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Second partial derivative test.mp3
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So we're taking two times two minus 16. What that would imply, sorry about getting kind of scrunched on the board here, is that h is equal to, let's see, four minus 16, negative 12. So I'm just gonna erase this to kind of clear up some room. So when p equals four, this is negative 12. And in fact, this kind of explains the crossover point for when it goes from being local minimum to a saddle point. It's gonna be at that point where this entire expression is equal to zero. And you can see that happens when p equals two.
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Second partial derivative test.mp3
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So when p equals four, this is negative 12. And in fact, this kind of explains the crossover point for when it goes from being local minimum to a saddle point. It's gonna be at that point where this entire expression is equal to zero. And you can see that happens when p equals two. So over here, the crossover point, when it kind of goes from being a local minimum to a saddle point, is at p equals two. And when p perfectly equals two, let's see, so about here, the second partial derivative test isn't gonna be enough to tell us anything. It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point.
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Second partial derivative test.mp3
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And you can see that happens when p equals two. So over here, the crossover point, when it kind of goes from being a local minimum to a saddle point, is at p equals two. And when p perfectly equals two, let's see, so about here, the second partial derivative test isn't gonna be enough to tell us anything. It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point. And in this particular case, that corresponds to the fact that the graph is perfectly flat in one direction, and a minimum in another direction. In other cases, it might mean something different, and I'll probably make a video just about that special case when this whole value is equal to zero. But for now, all that I wanna emphasize is what this test is, where you take all three second partial derivatives, and you kind of multiply together the two pure second partial derivatives, where you do x and then x, and the one where you do y and then y.
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Second partial derivative test.mp3
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It can't tell us it's definitely a max, and it can't tell us that it's definitely a saddle point. And in this particular case, that corresponds to the fact that the graph is perfectly flat in one direction, and a minimum in another direction. In other cases, it might mean something different, and I'll probably make a video just about that special case when this whole value is equal to zero. But for now, all that I wanna emphasize is what this test is, where you take all three second partial derivatives, and you kind of multiply together the two pure second partial derivatives, where you do x and then x, and the one where you do y and then y. You multiply those together, and then you subtract off that mixed partial derivative squared. And in the next video, I'll try to give a little bit more intuition for, you know, where this whole formula comes from, why it's not completely random, and why taking this and analyzing whether it's greater than zero or less than zero is a reasonable thing to do for analyzing whether a point that you're looking at is a local minimum or a local maximum or a saddle point. See you then.
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Second partial derivative test.mp3
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So let's say we take the line integral over some curve C. I'll define the curve in a second. Of x squared plus y squared dx plus 2xy dy. Well, this might look very familiar. This was very similar to what we saw last time, except last time we had a closed line integral. So this is not a closed line integral. And our curve C, the parameterization, is x is equal to cosine of t, y is equal to sine of t. So far, it looks like sint. Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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This was very similar to what we saw last time, except last time we had a closed line integral. So this is not a closed line integral. And our curve C, the parameterization, is x is equal to cosine of t, y is equal to sine of t. So far, it looks like sint. Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video. But instead of t going from 0 to 2 pi, we're going to have t go from 0 to pi. t is greater than or equal to 0 is less than or equal to pi. So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Let me write sine of t. So far it looks very similar to the closed line integral example we did in the last video. But instead of t going from 0 to 2 pi, we're going to have t go from 0 to pi. t is greater than or equal to 0 is less than or equal to pi. So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis. That is my x-axis. So now our path isn't all the way around the unit circle. Our path, our curve C now, just starts at t is equal to 0.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So now we're essentially, our path, if I were to draw it on the xy plane, so that is my y-axis. That is my x-axis. So now our path isn't all the way around the unit circle. Our path, our curve C now, just starts at t is equal to 0. You can imagine t is almost the angle. t is equal to 0, and we're going to go all the way to pi. So that's what our path is right now in this example.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Our path, our curve C now, just starts at t is equal to 0. You can imagine t is almost the angle. t is equal to 0, and we're going to go all the way to pi. So that's what our path is right now in this example. So it's not a curved path. It's not a closed path. So we can't just show that f in this example, and we're going to re-look at what f looks like.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So that's what our path is right now in this example. So it's not a curved path. It's not a closed path. So we can't just show that f in this example, and we're going to re-look at what f looks like. Hey, if that's a conservative vector field, if it's a closed loop, it equals 0. This isn't a closed loop, so we can't apply that. But let's see if we can apply some of our other tools.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So we can't just show that f in this example, and we're going to re-look at what f looks like. Hey, if that's a conservative vector field, if it's a closed loop, it equals 0. This isn't a closed loop, so we can't apply that. But let's see if we can apply some of our other tools. So like we saw in the last video, this might look a little bit foreign to you. But if you say that f is equal to that times i plus that times j, then it might look a little bit more familiar. If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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But let's see if we can apply some of our other tools. So like we saw in the last video, this might look a little bit foreign to you. But if you say that f is equal to that times i plus that times j, then it might look a little bit more familiar. If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j. And dr, I don't even have to look at this right now, dr you can always write it as dx times i plus dy times j. You'll immediately see if you take the dot product of these two things, if you take f dot dr, they're both vector valued, vector valued differential, vector valued field, or vector valued function. If you take f dot dr, you'll get this right here.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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If we say that f of x, y, the vector field f, is equal to x squared plus y squared times i plus 2xy times j. And dr, I don't even have to look at this right now, dr you can always write it as dx times i plus dy times j. You'll immediately see if you take the dot product of these two things, if you take f dot dr, they're both vector valued, vector valued differential, vector valued field, or vector valued function. If you take f dot dr, you'll get this right here. You'll get what we have inside of the integral. You'll get that right there. That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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If you take f dot dr, you'll get this right here. You'll get what we have inside of the integral. You'll get that right there. That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that. So our integral, we can rewrite it as this, along this curve of f dot dr, where this is our f. Now, we still might want to ask ourselves, is this a conservative field? Or does it have a potential? Is f equal to the gradient of some function capital F?
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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That times that, you take the product of the i terms, that times that is equal to that, and add it to the product of the j terms, 2xy times dy, right like that. So our integral, we can rewrite it as this, along this curve of f dot dr, where this is our f. Now, we still might want to ask ourselves, is this a conservative field? Or does it have a potential? Is f equal to the gradient of some function capital F? You could write the gradient like that, because it creates a vector. This is a vector too. Is this true?
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Is f equal to the gradient of some function capital F? You could write the gradient like that, because it creates a vector. This is a vector too. Is this true? And we saw in the last video it is. So we'll redo it a little bit fast this time. Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Is this true? And we saw in the last video it is. So we'll redo it a little bit fast this time. Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0. But if this is true, then we know that the integral is path independent. And we'll know that this is going to be equal to capital F. If we say that t is going from, well, in this case, t is going from 0 to pi, we could say that this is going to be equal to capital F of pi minus capital F of 0. Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Because if this is true, we can't say this is a closed loop and say, oh, it's just going to be equal to 0. But if this is true, then we know that the integral is path independent. And we'll know that this is going to be equal to capital F. If we say that t is going from, well, in this case, t is going from 0 to pi, we could say that this is going to be equal to capital F of pi minus capital F of 0. Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here. We could also write that this is equal to f of x of pi, y of pi, minus f of x of 0, y of 0. That's what I mean when I say f of pi. If we were to write f purely as a function of t. But we know that this capital F is going to be a function.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Or if we want to write in terms of x and y, because f is going to be a function of x and y, we could write, and this right here, these are t's right here. We could also write that this is equal to f of x of pi, y of pi, minus f of x of 0, y of 0. That's what I mean when I say f of pi. If we were to write f purely as a function of t. But we know that this capital F is going to be a function. It's a scalar function defined on x, y. So we could say f of x of pi, y of pi. These are the t's now.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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If we were to write f purely as a function of t. But we know that this capital F is going to be a function. It's a scalar function defined on x, y. So we could say f of x of pi, y of pi. These are the t's now. These are all equivalent things. So if it is path dependent, we can find our f. We can just evaluate this thing by just taking our f, evaluating it at these two points. At this point and at that point right there.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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These are the t's now. These are all equivalent things. So if it is path dependent, we can find our f. We can just evaluate this thing by just taking our f, evaluating it at these two points. At this point and at that point right there. Because it would be path independent. If this is a conservative, if this has a potential function, if this is the gradient of another scalar field, then this is a conservative vector field. And its line integral is path independent.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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At this point and at that point right there. Because it would be path independent. If this is a conservative, if this has a potential function, if this is the gradient of another scalar field, then this is a conservative vector field. And its line integral is path independent. It's only dependent on that point and that point. So let's see if we can find our f. So I'm going to do exactly what we did in the last video. If you watched that last video, it might be a little bit monotonous.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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And its line integral is path independent. It's only dependent on that point and that point. So let's see if we can find our f. So I'm going to do exactly what we did in the last video. If you watched that last video, it might be a little bit monotonous. But I'll do it a little bit faster here. So we know that the partial of f with respect to x is going to have to be equal to this right here. So that's x squared plus y squared.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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If you watched that last video, it might be a little bit monotonous. But I'll do it a little bit faster here. So we know that the partial of f with respect to x is going to have to be equal to this right here. So that's x squared plus y squared. Which tells us if we take the antiderivative with respect to x, that f of x, y is going to have to be equal to x to the third over 3 plus xy squared. y squared is just a constant in terms of x. Plus f of y.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So that's x squared plus y squared. Which tells us if we take the antiderivative with respect to x, that f of x, y is going to have to be equal to x to the third over 3 plus xy squared. y squared is just a constant in terms of x. Plus f of y. There might be some function of y that when you take the partial with respect to x, it just disappears. And then we know that the partial of f with respect to y has got to be equal to that thing or that thing. We're saying that this is the gradient of f. So this has to be the partial with respect to y.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Plus f of y. There might be some function of y that when you take the partial with respect to x, it just disappears. And then we know that the partial of f with respect to y has got to be equal to that thing or that thing. We're saying that this is the gradient of f. So this has to be the partial with respect to y. 2xy. And you might want to watch the other video. I go through this a little bit slower in that one.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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We're saying that this is the gradient of f. So this has to be the partial with respect to y. 2xy. And you might want to watch the other video. I go through this a little bit slower in that one. So the antiderivative of this with respect to y. So we get f of x, y would be equal to xy squared plus some function of x. Now, we did this in the last video.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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I go through this a little bit slower in that one. So the antiderivative of this with respect to y. So we get f of x, y would be equal to xy squared plus some function of x. Now, we did this in the last video. These two things have to be the same thing in order for the gradient of capital F to be lowercase f. And we have xy squared, xy squared. We have a function of x. We have a function purely of x.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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Now, we did this in the last video. These two things have to be the same thing in order for the gradient of capital F to be lowercase f. And we have xy squared, xy squared. We have a function of x. We have a function purely of x. And then we don't have a function purely of y here. So this thing right here must be 0. So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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We have a function purely of x. And then we don't have a function purely of y here. So this thing right here must be 0. So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared. So we know that lowercase f is definitely conservative. It is path independent. It has its potential.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So we've solved our capital F of xy must be equal to x to the 3 over 3 plus xy squared. So we know that lowercase f is definitely conservative. It is path independent. It has its potential. It is the gradient of this thing right here. And so to solve our integral, this was a 0. To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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It has its potential. It is the gradient of this thing right here. And so to solve our integral, this was a 0. To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0. Evaluate at both points and then subtract the two. So let's do that. So let me just figure out.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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To solve our integral, we just have to figure out x of pi, y of pi, x of 0, y of 0. Evaluate at both points and then subtract the two. So let's do that. So let me just figure out. So x was cosine of t, y is sine of t. Let me rewrite it down here. So x is equal to cosine of t, y is equal to sine of t. So x of 0 is equal to cosine of 0, which is equal to 1. x of pi is equal to cosine of pi, which is equal to minus 1. y of 0 is sine of 0, which is 0. y of pi, which is equal to sine of pi, which is equal to 0. So f of x of pi, y of pi, this is the same thing.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So let me just figure out. So x was cosine of t, y is sine of t. Let me rewrite it down here. So x is equal to cosine of t, y is equal to sine of t. So x of 0 is equal to cosine of 0, which is equal to 1. x of pi is equal to cosine of pi, which is equal to minus 1. y of 0 is sine of 0, which is 0. y of pi, which is equal to sine of pi, which is equal to 0. So f of x of pi, y of pi, this is the same thing. So let me rewrite this. Our integral is simplified to our integral along that path of f dot dr is going to be equal to capital F of x of pi. x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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So f of x of pi, y of pi, this is the same thing. So let me rewrite this. Our integral is simplified to our integral along that path of f dot dr is going to be equal to capital F of x of pi. x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0. And so what is this equal to? Well, just remember, this right here, it's the same thing as that right there. That is x of pi.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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x of pi is minus 1. x of pi is minus 1. y of pi is equal to 0 minus capital F of x of 0 is 1, comma, y of 0 is 0. And so what is this equal to? Well, just remember, this right here, it's the same thing as that right there. That is x of pi. That is y of pi. That's that term right there. You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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That is x of pi. That is y of pi. That's that term right there. You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear. So this is just straightforward to evaluate. What is F of minus 1, 0? x is minus 1, y is 0.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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You can imagine this whole F of minus 1, 0, that's the same thing as F of pi, if you think in terms of just t. That could be a little confusing, so I want to make that clear. So this is just straightforward to evaluate. What is F of minus 1, 0? x is minus 1, y is 0. So it's going to be minus 1 to the third power, that's our x, over 3. So it's minus 1 over 3. It's going to be minus 1 third plus minus 1 times 0 squared.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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x is minus 1, y is 0. So it's going to be minus 1 to the third power, that's our x, over 3. So it's minus 1 over 3. It's going to be minus 1 third plus minus 1 times 0 squared. So that's just going to be a 0. In both cases, the y is 0, so this term's going to disappear. So we can ignore that.
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Second example of line integral of conservative vector field Multivariable Calculus Khan Academy.mp3
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