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This one represents the constant y value one, but you could imagine it, you could imagine sliding the plane back and forth, and that would represent various different y values. So for the general partial derivative, you can imagine whichever one you want, but this one is y equals one. And I'll go ahead and slice the actual graph at that point, and draw a red line. And this red line is basically all the points on the graph where y is equal to one. So I'll just kind of emphasize that, where y is equal to one, and this is y is equal to one. So when we're looking at that, we can actually interpret the partial derivative as a slope, because we're looking at the point here, we're asking how the function changes as we move in the x direction, and from single variable calculus, you might be familiar with thinking of that as the slope of a line. And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at.
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Partial derivatives and graphs.mp3
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And this red line is basically all the points on the graph where y is equal to one. So I'll just kind of emphasize that, where y is equal to one, and this is y is equal to one. So when we're looking at that, we can actually interpret the partial derivative as a slope, because we're looking at the point here, we're asking how the function changes as we move in the x direction, and from single variable calculus, you might be familiar with thinking of that as the slope of a line. And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at. But let's do this with the partial derivative with respect to y, let's erase what we've got going on here, and I'll go ahead and move the graph back to what it was, get rid of these guys. So now we're no longer slicing with respect to y, but instead, you know, let's say we slice it with a constant x value. So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one.
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Partial derivatives and graphs.mp3
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And to be a little more concrete about this, I could say, you know, you're starting here, you consider some nudge over there, just some tiny step, I'm drawing it as a sizable one, but you imagine that as a really small step, as your dx, and then the distance to your function here, the change in the value of your function as your, I said dx, but I should say partial x or del x, partial f. And as that tiny nudge gets smaller and smaller, this change here is gonna correspond with what the tangent line does, and that's why you have this rise over run appealing for the slope, and you look at that value, and the line itself looks like it has a slope of about negative two, so it should actually make sense that we get negative two over here, given what we're looking at. But let's do this with the partial derivative with respect to y, let's erase what we've got going on here, and I'll go ahead and move the graph back to what it was, get rid of these guys. So now we're no longer slicing with respect to y, but instead, you know, let's say we slice it with a constant x value. So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one. And now when we take the partial derivative, we're gonna interpret it as a slice, as the slope of this resulting curve. So that slope ends up looking like this, that's our blue line, and let's go ahead and evaluate the partial derivative of f with respect to y. So I'll go over here, use a different color.
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Partial derivatives and graphs.mp3
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So this here is the x-axis, this plane represents the constant value x equals negative one, and we could slice the graph there, we could kind of slice it, I'll draw the red line again that represents the curve, and this time, that curve represents the value x, or equals negative one, it's all the points on the graph where x equals negative one. And now when we take the partial derivative, we're gonna interpret it as a slice, as the slope of this resulting curve. So that slope ends up looking like this, that's our blue line, and let's go ahead and evaluate the partial derivative of f with respect to y. So I'll go over here, use a different color. So the partial derivative of f with respect to y, partial y, so we go up here, and it says, okay, I see x squared times y, it's considering x squared to be a constant now. So it looks at that and says, x, you're a constant, y, you're the variable, constant times a variable, the derivative is just equal to that constant, so that x squared. And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y.
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Partial derivatives and graphs.mp3
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So I'll go over here, use a different color. So the partial derivative of f with respect to y, partial y, so we go up here, and it says, okay, I see x squared times y, it's considering x squared to be a constant now. So it looks at that and says, x, you're a constant, y, you're the variable, constant times a variable, the derivative is just equal to that constant, so that x squared. And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y. And if we actually want to evaluate this at our point negative one, one, what we'd get is negative one squared plus cosine of one. And I'm not sure what the cosine of one is, but it's something a little bit positive, and the ultimate result that we see here is gonna be one plus something, I don't know what it is, but it's something positive, and that should make sense, because we look at the slope here, and it's a little bit more than one, not sure exactly, but it's a little bit more than one. So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph.
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Partial derivatives and graphs.mp3
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And over here, sine of y, the derivative of that with respect to y is cosine y, cosine y. And if we actually want to evaluate this at our point negative one, one, what we'd get is negative one squared plus cosine of one. And I'm not sure what the cosine of one is, but it's something a little bit positive, and the ultimate result that we see here is gonna be one plus something, I don't know what it is, but it's something positive, and that should make sense, because we look at the slope here, and it's a little bit more than one, not sure exactly, but it's a little bit more than one. So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph. And in other contexts, that might not be the case. Maybe it's something that has a multi-dimensional output, we'll talk about that later, when you have a vector valued function, what its partial derivative looks like. But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again.
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Partial derivatives and graphs.mp3
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So you'll often hear about people talking about the partial derivative as being the slope of the slice of a graph, which is great, if you're looking at a function that has a two variable input and a one variable output so that we can think about its graph. And in other contexts, that might not be the case. Maybe it's something that has a multi-dimensional output, we'll talk about that later, when you have a vector valued function, what its partial derivative looks like. But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again. Now, you're looking at your point here, and you say we're gonna take a tiny step in the y direction, and I'll call that partial y. And you say that makes some kind of change, it causes a change in the function, which you'll call partial f. And as you imagine this getting really, really small, and the resulting change also getting really small, the rise over run of that is gonna give you the slope of the tangent line. So this is just one way of interpreting that ratio, the change in the output that corresponds to a little nudge in the input.
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Partial derivatives and graphs.mp3
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But maybe it's also something that has, you know, a hundred inputs, and you certainly can't visualize the graph, but the general idea of saying, well, if you take a tiny step in a direction, here, I'll actually walk through it in this graph context again. Now, you're looking at your point here, and you say we're gonna take a tiny step in the y direction, and I'll call that partial y. And you say that makes some kind of change, it causes a change in the function, which you'll call partial f. And as you imagine this getting really, really small, and the resulting change also getting really small, the rise over run of that is gonna give you the slope of the tangent line. So this is just one way of interpreting that ratio, the change in the output that corresponds to a little nudge in the input. But later on, we'll talk about different ways that you can do that, so I think graphs are very useful. When I move that, the text doesn't move. I think graphs are very useful for thinking about these things, but they're not the only way.
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Partial derivatives and graphs.mp3
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Let's now parameterize our surface, and then we can figure out what ds would actually look like. And so I will define my position vector function for our surface as r, and I'm going to make it a function of two parameters, because we're going to have to define a surface right over here. And I can actually use x and y as my parameters, because the surface can be defined as a function of x and y. So I'll say that my parameterization is going to be a function of x and y, and in my i direction it's going to be x times i, in my j direction it's going to be y times j, and then in the k direction, well that's just going to be z, and z is a function of x and y. It's going to be z as a function of x and y. And whenever you do a parameterization of a surface, you have to think about, well, what are the constraints on the domain for your parameters? And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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So I'll say that my parameterization is going to be a function of x and y, and in my i direction it's going to be x times i, in my j direction it's going to be y times j, and then in the k direction, well that's just going to be z, and z is a function of x and y. It's going to be z as a function of x and y. And whenever you do a parameterization of a surface, you have to think about, well, what are the constraints on the domain for your parameters? And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here. We could call this the domain of the parameters. So it has to be a member of R, and actually we assume that up here. And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And so the constraints of the domain for my parameters, I'm going to say that every pair of x, y, every coordinate, x, y coordinate, it has to be a member, this is a symbol for member, it has to be a member of this little region right over here. We could call this the domain of the parameters. So it has to be a member of R, and actually we assume that up here. And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface. An x, y that is not a member of R, then we're not going to consider the z of that x, y as part of the surface. Only the z of x, y's where the x, y's are part of this region. So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And actually I should have written this as the coordinates of x, y, the x, y pairs that are a member of R, that's going to help define our surface. An x, y that is not a member of R, then we're not going to consider the z of that x, y as part of the surface. Only the z of x, y's where the x, y's are part of this region. So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be. And we need to think about this a little bit carefully. So first I'll just write something, and then we can confirm whether this really will be the case. ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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So now that we have a parameterization for our surface, we're ready to start thinking about what ds might be. And we need to think about this a little bit carefully. So first I'll just write something, and then we can confirm whether this really will be the case. ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain. So you could view it as the partial of R, of this parameterization with respect to x, crossed with the partial of R with respect to y. And then that whole thing, and we actually want this to be a vector, not just the absolute value or the magnitude of this vector. We actually want this to be a vector.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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ds, and we've seen this before about why this is the case, is going to be the cross product of the partial derivative of this with respect to each of the parameters times the little chunk of area in that domain. So you could view it as the partial of R, of this parameterization with respect to x, crossed with the partial of R with respect to y. And then that whole thing, and we actually want this to be a vector, not just the absolute value or the magnitude of this vector. We actually want this to be a vector. That thing times, we could put in some order dx, dy, or we could write dy, dx. And if we want to be general, just to say that it's a little differential of our region right over here, we can just write, instead of writing dx, dy, or dy, dx, we will write da. And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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We actually want this to be a vector. That thing times, we could put in some order dx, dy, or we could write dy, dx. And if we want to be general, just to say that it's a little differential of our region right over here, we can just write, instead of writing dx, dy, or dy, dx, we will write da. And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in. Because remember, if we're traversing a boundary like this, we want to make sure that the surface is oriented the right way. And the way we think about it, if you were to twist a cap like this, the cap would move up. Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And the reason why I said we need to be careful is we need to make sure based on how we've parameterized this position vector function, based on how we've parameterized it, whether this cross product really points in the right direction, the direction that we need to be oriented in. Because remember, if we're traversing a boundary like this, we want to make sure that the surface is oriented the right way. And the way we think about it, if you were to twist a cap like this, the cap would move up. Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up. And so we need to make sure that this vector, which really defines the orientation of the surface, is going to be pointed, is definitely going to be pointed up, or above the surface as opposed to going below the surface. And so let's think about that a little bit. The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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Or if you were to walk around this boundary in this direction with the surface to your left, your head would point up. And so we need to make sure that this vector, which really defines the orientation of the surface, is going to be pointed, is definitely going to be pointed up, or above the surface as opposed to going below the surface. And so let's think about that a little bit. The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface. And the partial with respect to y, as y gets bigger, it's going to go in that direction, in that direction along the surface. If I take r cross y, and we could use the right-hand rule here, we put our index finger in the direction of the first thing we're taking the cross product of, our middle finger in the direction of the second thing, so just like that, we bend our middle finger, we don't care what the other fingers are doing, so I'll just draw them right there, then the thumb will go in the direction of the cross product. So in this case, the thumb is going to point up.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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The partial derivative with respect to x, well, as x gets bigger, it's going to go in that direction, it's going to go in that direction along the surface. And the partial with respect to y, as y gets bigger, it's going to go in that direction, in that direction along the surface. If I take r cross y, and we could use the right-hand rule here, we put our index finger in the direction of the first thing we're taking the cross product of, our middle finger in the direction of the second thing, so just like that, we bend our middle finger, we don't care what the other fingers are doing, so I'll just draw them right there, then the thumb will go in the direction of the cross product. So in this case, the thumb is going to point up. And that's exactly what we wanted to happen. So this actually is the right ordering. The partial with respect to y cross the partial with respect to x actually would not have been right.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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So in this case, the thumb is going to point up. And that's exactly what we wanted to happen. So this actually is the right ordering. The partial with respect to y cross the partial with respect to x actually would not have been right. That would have given us the other orientation. We would have done that if this boundary actually went the other way around. But this is the right orientation given the way that we are going to traverse the boundary.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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The partial with respect to y cross the partial with respect to x actually would not have been right. That would have given us the other orientation. We would have done that if this boundary actually went the other way around. But this is the right orientation given the way that we are going to traverse the boundary. Now with that out of the way, let's actually calculate this cross product. So let me just rewrite it. So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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But this is the right orientation given the way that we are going to traverse the boundary. Now with that out of the way, let's actually calculate this cross product. So let me just rewrite it. So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors. I think that's a helpful way of thinking about it. So i, j, and k, actually, fine, I'll use this magenta color, i, j, and k, let me close, let me write this line right over here. And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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So this is going to be equal to, I'll just focus on the cross product right now, the cross r, the partial of r with respect to x crossed with the partial of r with respect to y is going to be equal to, and we've done this many times already, I'll just do it in kind of more general terms now, is going to be equal to the determinant of this matrix, i, j, and k, and let me do those in different colors. I think that's a helpful way of thinking about it. So i, j, and k, actually, fine, I'll use this magenta color, i, j, and k, let me close, let me write this line right over here. And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x. And so the partial of the i component with respect to x is just going to be 1. The partial of the j component with respect to x is going to be 0. And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And now we want to write the x components of the partial with respect, or we want to write the different components of the partial with respect to x. And so the partial of the i component with respect to x is just going to be 1. The partial of the j component with respect to x is going to be 0. And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x. And so r sub x, or the partial with respect to x, is the vector 1i plus 0j plus z sub xk. And we'll do the same thing for this piece right over here. The partial with respect to y, its i component, well, this is going to be 0.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And the partial of this with respect to x, well, we can just write that as a partial derivative of the function z with respect to x. And so r sub x, or the partial with respect to x, is the vector 1i plus 0j plus z sub xk. And we'll do the same thing for this piece right over here. The partial with respect to y, its i component, well, this is going to be 0. Its j component is going to be 1. The partial of this with respect to y is 1. And the partial of this with respect to y is the partial of z with respect to y.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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The partial with respect to y, its i component, well, this is going to be 0. Its j component is going to be 1. The partial of this with respect to y is 1. And the partial of this with respect to y is the partial of z with respect to y. And actually, I forgot to write k right over here. I forgot to write k right over here in our parameterization. So now with all of this set up, we are ready to figure out what the cross product is.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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And the partial of this with respect to y is the partial of z with respect to y. And actually, I forgot to write k right over here. I forgot to write k right over here in our parameterization. So now with all of this set up, we are ready to figure out what the cross product is. The cross product is going to be equal to, so our i component is going to be 0 times the partial of z with respect to y minus 1 times the partial of z with respect to x. So we get negative partial of z with respect to x. And then checkerboard pattern, we'll have minus j times, so ignore that column, that row, 1 times the partial with respect to y, so that's z sub y, the partial of z with respect to y, minus 0 times this.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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So now with all of this set up, we are ready to figure out what the cross product is. The cross product is going to be equal to, so our i component is going to be 0 times the partial of z with respect to y minus 1 times the partial of z with respect to x. So we get negative partial of z with respect to x. And then checkerboard pattern, we'll have minus j times, so ignore that column, that row, 1 times the partial with respect to y, so that's z sub y, the partial of z with respect to y, minus 0 times this. So we're just left with that right over there. And then finally, we're going to have plus k. And here we have 1 times 1 minus 0 times 0. So it's going to be k times 1.
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Stokes' theorem proof part 2 Multivariable Calculus Khan Academy.mp3
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In the last couple videos, I talked about the local linearization of a function. And in terms of graphs, there's a nice interpretation here where if you imagine the graph of a function and you want to approximate it near a specific point, so you picture that point somewhere on the graph, and it doesn't have to be there, you know, it could choose to be kind of anywhere else along the graph, but if you have some sort of point and you want to approximate the function near there, you can have another function whose graph is just a flat plane, and specifically a plane which is tangent to your original graph at that point. And that's kind of visually how you think about the local linearization. And what I'm gonna start doing here in this next video and in the ones following, is talking about quadratic approximations. So quadratic approximations. And these basically take this to the next level. And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas.
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What do quadratic approximations look like.mp3
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And what I'm gonna start doing here in this next video and in the ones following, is talking about quadratic approximations. So quadratic approximations. And these basically take this to the next level. And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas. But graphically, instead of having a plane that's flat, you have a few more parameters to deal with, and you can give yourself some kind of surface that hugs the graph a little bit more closely. It's still gonna be simpler in terms of formulas, it can still be notably simpler than the original function, but this actually hugs it closely. And as we move around the point that it's approximating near, the way that it hugs it can look pretty different.
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What do quadratic approximations look like.mp3
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And first I'll show what they look like graphically, and then I'll show you what that actually means in formulas. But graphically, instead of having a plane that's flat, you have a few more parameters to deal with, and you can give yourself some kind of surface that hugs the graph a little bit more closely. It's still gonna be simpler in terms of formulas, it can still be notably simpler than the original function, but this actually hugs it closely. And as we move around the point that it's approximating near, the way that it hugs it can look pretty different. And if you want to think graphically what a quadratic approximation is, you can basically say if you slice this surface, this kind of ghostly white surface, in any direction, it'll look like a parabola of some kind. And notice that given that we're dealing in multiple dimensions, that can make things look pretty complicated. Like this right here, if you slice it in this direction, oh, moving things about.
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What do quadratic approximations look like.mp3
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And as we move around the point that it's approximating near, the way that it hugs it can look pretty different. And if you want to think graphically what a quadratic approximation is, you can basically say if you slice this surface, this kind of ghostly white surface, in any direction, it'll look like a parabola of some kind. And notice that given that we're dealing in multiple dimensions, that can make things look pretty complicated. Like this right here, if you slice it in this direction, oh, moving things about. If you look at it from this angle, it kind of looks like a concave up parabola. But if you were looking at it from another direction, it kind of looks concave down. And all in all, you get a surface that actually has quite a bit of information carried within it.
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What do quadratic approximations look like.mp3
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Like this right here, if you slice it in this direction, oh, moving things about. If you look at it from this angle, it kind of looks like a concave up parabola. But if you were looking at it from another direction, it kind of looks concave down. And all in all, you get a surface that actually has quite a bit of information carried within it. And you can see that by hugging the graph very closely, this approximation is gonna be, well, it's gonna be even closer. Because near the point where you're approximating, you can go out, you can take a couple steps away, and the approximation is still gonna be very close to what the graph is. And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself.
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What do quadratic approximations look like.mp3
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And all in all, you get a surface that actually has quite a bit of information carried within it. And you can see that by hugging the graph very closely, this approximation is gonna be, well, it's gonna be even closer. Because near the point where you're approximating, you can go out, you can take a couple steps away, and the approximation is still gonna be very close to what the graph is. And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself. So this is gonna be something that, although it takes more information to describe than a local linearization, it gives us a much closer approximation. So a linear function, which, you know, one that just draws a plane like this, in terms of actual functions, what this means, so I'll kinda write linear. This is gonna be some kind of function of x and y.
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What do quadratic approximations look like.mp3
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And it's only when you step really far away from the original point, that the approximation starts to deviate away from the graph itself. So this is gonna be something that, although it takes more information to describe than a local linearization, it gives us a much closer approximation. So a linear function, which, you know, one that just draws a plane like this, in terms of actual functions, what this means, so I'll kinda write linear. This is gonna be some kind of function of x and y. And what it looks like is some kind of constant, which I'll say a, plus another constant times the variable x, plus another constant times the variable y. This is sort of the basic form of linear functions. And technically, this isn't linear.
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What do quadratic approximations look like.mp3
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This is gonna be some kind of function of x and y. And what it looks like is some kind of constant, which I'll say a, plus another constant times the variable x, plus another constant times the variable y. This is sort of the basic form of linear functions. And technically, this isn't linear. If one was gonna be really pedantic, they would say that that's actually affine, because strictly speaking, linear functions shouldn't have this constant term. It should be purely x's and y's. But in the context of approximations, people usually would call this the linear term.
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What do quadratic approximations look like.mp3
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And technically, this isn't linear. If one was gonna be really pedantic, they would say that that's actually affine, because strictly speaking, linear functions shouldn't have this constant term. It should be purely x's and y's. But in the context of approximations, people usually would call this the linear term. So a quadratic term, what this is gonna look like, quadratic, we are allowed to have all the same terms as that linear one. So you can have, you know, a constant, you can have these two linear terms, bx and cy. And then you're allowed to have anything that has two variables multiplied into it.
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What do quadratic approximations look like.mp3
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But in the context of approximations, people usually would call this the linear term. So a quadratic term, what this is gonna look like, quadratic, we are allowed to have all the same terms as that linear one. So you can have, you know, a constant, you can have these two linear terms, bx and cy. And then you're allowed to have anything that has two variables multiplied into it. So maybe I'll have d times x squared. And then you can also have something times xy. This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two.
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What do quadratic approximations look like.mp3
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And then you're allowed to have anything that has two variables multiplied into it. So maybe I'll have d times x squared. And then you can also have something times xy. This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two. But really, it's just saying, anytime you have two variables multiplied in. And then we can add some other constant, say f times y squared, where, you know, now we're multiplying two y's into it. So all of these guys, these are what you would call your quadratic terms.
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What do quadratic approximations look like.mp3
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This is considered a quadratic term, which is a little bit weird at first, because usually we think of quadratics as associated with that exponent two. But really, it's just saying, anytime you have two variables multiplied in. And then we can add some other constant, say f times y squared, where, you know, now we're multiplying two y's into it. So all of these guys, these are what you would call your quadratic terms. Things that, you know, either x squared, y squared, or x times y. Anything that has two variables in it. So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space.
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What do quadratic approximations look like.mp3
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So all of these guys, these are what you would call your quadratic terms. Things that, you know, either x squared, y squared, or x times y. Anything that has two variables in it. So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space. And if you choose the most optimal one, you'll get one that's tangent to your curve at the specific point. And kind of, it'll depend on where that point is, you'll get different planes, but they're all tangent. So what we're going to do, in the next couple videos, is talk about how you tweak all of these six different constants, so that you can get functions that really closely hug the curve.
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What do quadratic approximations look like.mp3
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So you can see, this gives us a lot more control, because previously, as we tweaked the constants, a, b, and c, you're able to get yourself, you know, that gives you control over all sorts of planes in space. And if you choose the most optimal one, you'll get one that's tangent to your curve at the specific point. And kind of, it'll depend on where that point is, you'll get different planes, but they're all tangent. So what we're going to do, in the next couple videos, is talk about how you tweak all of these six different constants, so that you can get functions that really closely hug the curve. Right, and as you, they're all going to depend on the original point, because as you move that point around, what it takes to hug the curve is going to be different. It's going to have to do with partial differential information about your original function, the function whose graph this is. And it's going to look pretty similar to the local linearization case, just, you know, added complexity.
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What do quadratic approximations look like.mp3
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So we've just computed a vector-valued partial derivative of a vector-valued function, but the question is, what does this mean? What does this jumble of symbols actually mean in a more intuitive geometric setting? And that has everything to do with how you visualize the function, and with this specific function, given that the input is two-dimensional, but the output is three-dimensional, and the output is more dimensions than the input, it's nice to visualize it as a parametric surface. And the way that I do that, maybe you could call this visualizing it as a transformation also, because what I want to do is basically think of the TS plane, think of the TS plane where all these input values live, and kind of think of how that's going to map into three-dimensional space. But when I do that, I'm actually going to cheat a little bit. Rather than having a separate plane off there as the TS plane, I'm going to kind of overwrite onto the XY plane itself, and plop the TS plane down like this. And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three.
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Partial derivative of a parametric surface, part 1.mp3
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And the way that I do that, maybe you could call this visualizing it as a transformation also, because what I want to do is basically think of the TS plane, think of the TS plane where all these input values live, and kind of think of how that's going to map into three-dimensional space. But when I do that, I'm actually going to cheat a little bit. Rather than having a separate plane off there as the TS plane, I'm going to kind of overwrite onto the XY plane itself, and plop the TS plane down like this. And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three. So each tick mark on the graph here corresponds with a half, so this is one, that's two, and then up here is three. Same with S, S also ranges from zero to three. And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier.
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Partial derivative of a parametric surface, part 1.mp3
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And this isn't the full TS plane, this is actually supposed to represent just values of T that range from zero up to three. So each tick mark on the graph here corresponds with a half, so this is one, that's two, and then up here is three. Same with S, S also ranges from zero to three. And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier. It's, you could call it laziness. But the benefit here is, what we can do is watch each point, and each one of these points you're thinking of as corresponding to some kind of TS pair, an input point, which is just a pair of numbers, and we're going to watch each one of those points move to the corresponding output. The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it.
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Partial derivative of a parametric surface, part 1.mp3
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And the reason that I'm plopping it inside three-dimensional space to start with, kind of overwriting the XY plane with the TS plane, is just to make the animating a little bit easier. It's, you could call it laziness. But the benefit here is, what we can do is watch each point, and each one of these points you're thinking of as corresponding to some kind of TS pair, an input point, which is just a pair of numbers, and we're going to watch each one of those points move to the corresponding output. The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it. And what that looks like when we animate this, actually, is each one of those points in our square of TS plane moves to the corresponding output, and you end up with a certain surface. And just to make it a little more concrete what's actually going on here, let's focus in on just one point, and we'll focus in on this point not just for the function visualization, but for the partial derivative as well. And the function, or the point, rather, that I care about is going to be at one, one.
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Partial derivative of a parametric surface, part 1.mp3
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The output is a three-dimensional value, a three-dimensional vector or point, however you want to think about it. And what that looks like when we animate this, actually, is each one of those points in our square of TS plane moves to the corresponding output, and you end up with a certain surface. And just to make it a little more concrete what's actually going on here, let's focus in on just one point, and we'll focus in on this point not just for the function visualization, but for the partial derivative as well. And the function, or the point, rather, that I care about is going to be at one, one. So this point right here represents the pair of TS, where each one of them is equal to one. One, one. And you can start by predicting where you think this is going to get output.
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Partial derivative of a parametric surface, part 1.mp3
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And the function, or the point, rather, that I care about is going to be at one, one. So this point right here represents the pair of TS, where each one of them is equal to one. One, one. And you can start by predicting where you think this is going to get output. And to do that, you just plug it into the function. This is kind of what the visualization means, is we're plugging this in, and for T and S, we're going to plug in just one and one. So that top part is going to look like one squared minus one squared, which becomes zero.
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Partial derivative of a parametric surface, part 1.mp3
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And you can start by predicting where you think this is going to get output. And to do that, you just plug it into the function. This is kind of what the visualization means, is we're plugging this in, and for T and S, we're going to plug in just one and one. So that top part is going to look like one squared minus one squared, which becomes zero. That middle part is going to be one times one, which is one. And then over here, we're going to have one times one squared, which is one, minus one times one squared, which is again one. And you can probably see, because of the symmetry there, those also cancel out, you get zero.
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Partial derivative of a parametric surface, part 1.mp3
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So that top part is going to look like one squared minus one squared, which becomes zero. That middle part is going to be one times one, which is one. And then over here, we're going to have one times one squared, which is one, minus one times one squared, which is again one. And you can probably see, because of the symmetry there, those also cancel out, you get zero. Which means the output corresponding with this input should be the vector zero, one, zero. A vector that's of unit length pointing in the y direction. So if we look here, you know, this is the x-axis, this here is the y-axis.
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Partial derivative of a parametric surface, part 1.mp3
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And you can probably see, because of the symmetry there, those also cancel out, you get zero. Which means the output corresponding with this input should be the vector zero, one, zero. A vector that's of unit length pointing in the y direction. So if we look here, you know, this is the x-axis, this here is the y-axis. So you would think it should be a vector that looks kind of like this. Unit vector in the y direction. And the point of the surface is what corresponds to the tip of that vector.
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Partial derivative of a parametric surface, part 1.mp3
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So if we look here, you know, this is the x-axis, this here is the y-axis. So you would think it should be a vector that looks kind of like this. Unit vector in the y direction. And the point of the surface is what corresponds to the tip of that vector. Right, this is how we visualize parametric things. You just think of the tip of the vector as kind of moving through space and drawing out the thing. In this case, the thing it's drawing is a surface.
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Partial derivative of a parametric surface, part 1.mp3
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And the point of the surface is what corresponds to the tip of that vector. Right, this is how we visualize parametric things. You just think of the tip of the vector as kind of moving through space and drawing out the thing. In this case, the thing it's drawing is a surface. So if we watch that animation again and we let things play forward, that dot corresponding with the input one, one does indeed land at the tip of that vector. So at least for that value, you can see that I'm not lying to you with the animation. And in principle, you could do that for every single point, right?
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Partial derivative of a parametric surface, part 1.mp3
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In this case, the thing it's drawing is a surface. So if we watch that animation again and we let things play forward, that dot corresponding with the input one, one does indeed land at the tip of that vector. So at least for that value, you can see that I'm not lying to you with the animation. And in principle, you could do that for every single point, right? If any given input point, you kind of plug it through the function and you draw the vector in three-dimensional space, as you watch this animation, it'll land at the tip of that vector. So, now if we want to start thinking about what the partial derivative means, remember, this little dt, this partial t is telling you to nudge it in the t direction. So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here?
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Partial derivative of a parametric surface, part 1.mp3
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And in principle, you could do that for every single point, right? If any given input point, you kind of plug it through the function and you draw the vector in three-dimensional space, as you watch this animation, it'll land at the tip of that vector. So, now if we want to start thinking about what the partial derivative means, remember, this little dt, this partial t is telling you to nudge it in the t direction. So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here? Well, the t direction, I'm saying, is in this direction here where, you know, this represents one, two, three of t values. And this line here represents the constant value for s, so this will be s constantly equaling one, which you can know because it's passing through the point one, one. And then otherwise, you're just letting t range freely.
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Partial derivative of a parametric surface, part 1.mp3
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So what does movement, just not even nudges, but just movement in general look like in the t direction for our little snippet of the t-s plane here? Well, the t direction, I'm saying, is in this direction here where, you know, this represents one, two, three of t values. And this line here represents the constant value for s, so this will be s constantly equaling one, which you can know because it's passing through the point one, one. And then otherwise, you're just letting t range freely. And if we watch how this gets transformed under the transformation, under the mapping to the parametric surface, you can get a feel for what varying the input t does in the output space. So this whole pink line is basically telling you what happens if you let s constantly equal one, but you let the variable t, the input t, vary freely, and you get a certain curve in three-dimensional space. And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one.
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Partial derivative of a parametric surface, part 1.mp3
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And then otherwise, you're just letting t range freely. And if we watch how this gets transformed under the transformation, under the mapping to the parametric surface, you can get a feel for what varying the input t does in the output space. So this whole pink line is basically telling you what happens if you let s constantly equal one, but you let the variable t, the input t, vary freely, and you get a certain curve in three-dimensional space. And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one. So if instead of thinking about movement of t as a whole, you start thinking about nudges, this whole partial t is something where we're just imagining a tiny, tiny little movement in the t direction, not really that much, just a tiny little move, and it's like you're recording its value as partial t. So maybe if you're being concrete, you'd say partial t would be something like 0.01. And really, it's gonna be a limiting variable that gets smaller and smaller, but I find it's kind of nice to think about an actual value like 1.100. And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve.
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Partial derivative of a parametric surface, part 1.mp3
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And, you know, if you had a different constant for s, it would be another curve, and maybe you can kind of see on the grid lines what shape those other curves would have, and they're all, in a sense, you know, parallel-ish to this curve corresponding to s equals one. So if instead of thinking about movement of t as a whole, you start thinking about nudges, this whole partial t is something where we're just imagining a tiny, tiny little movement in the t direction, not really that much, just a tiny little move, and it's like you're recording its value as partial t. So maybe if you're being concrete, you'd say partial t would be something like 0.01. And really, it's gonna be a limiting variable that gets smaller and smaller, but I find it's kind of nice to think about an actual value like 1.100. And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve. And it'll be tangent to that curve, right? The vector that tells you how you move just a tiny, tiny little bit, it'll be tangent in some way. And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger.
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Partial derivative of a parametric surface, part 1.mp3
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And then if you let this whole thing undergo the transformation, and we kind of watch, watch the input point, watch the line representing t, that little nudge, that little nudge is gonna get maybe stretched or squished, and it's gonna result in some kind of vector pointing along that curve. And it'll be tangent to that curve, right? The vector that tells you how you move just a tiny, tiny little bit, it'll be tangent in some way. And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger. So the actual derivative isn't gonna be just some tiny little nudge that's hardly, hardly visible, but it's gonna be that nudge vector scaled appropriately. In this case, it would be divided by 1.100, or multiplied by 100, and it would be something that remains tangent to the curve, but maybe it's pointing big. And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right?
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Partial derivative of a parametric surface, part 1.mp3
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And that vector, that output nudge, is what you're thinking of as your tiny change to the output vector, that partial v. And when you divide it by the tiny value, right, if your tiny value was 0.01, and you divide it by that, it's gonna become something bigger. So the actual derivative isn't gonna be just some tiny little nudge that's hardly, hardly visible, but it's gonna be that nudge vector scaled appropriately. In this case, it would be divided by 1.100, or multiplied by 100, and it would be something that remains tangent to the curve, but maybe it's pointing big. And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right? The ratio of the nudge sizes is bigger. So if you were to have a very long partial derivative vector that's still tangent, but really goes out there, that would tell you that as you vary t, you're zipping along super quickly. And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right.
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Partial derivative of a parametric surface, part 1.mp3
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And the larger it is, the longer it is, that's telling you that as you let t vary, and you're kind of moving along this pink curve, tiny nudges in t correspond with larger movements, right? The ratio of the nudge sizes is bigger. So if you were to have a very long partial derivative vector that's still tangent, but really goes out there, that would tell you that as you vary t, you're zipping along super quickly. And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right. You're moving positively in the z direction, you're moving, not z, sorry, the y, positively in the y direction up there. And the z direction is actually negative, isn't it? This curve kind of goes down as far as z is concerned.
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Partial derivative of a parametric surface, part 1.mp3
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And if we just look at this, you know, this particular one, the simming off of one, you kind of get a feel for the curve around that point, and you say, okay, okay, in that curve, you're moving positively in the x direction, right, you're moving to the right. You're moving positively in the z direction, you're moving, not z, sorry, the y, positively in the y direction up there. And the z direction is actually negative, isn't it? This curve kind of goes down as far as z is concerned. So before even computing it, if I were to tell you that I'm gonna plug in the value one, one to this partial derivative that we computed in the last video, you would say, oh, well, just looking at the picture, you can kind of tell that the x value is gonna be something positive, something greater than zero. The y value is also gonna be something positive. And again, that's because, you know, the movement is to the right, so positive x.
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Partial derivative of a parametric surface, part 1.mp3
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This curve kind of goes down as far as z is concerned. So before even computing it, if I were to tell you that I'm gonna plug in the value one, one to this partial derivative that we computed in the last video, you would say, oh, well, just looking at the picture, you can kind of tell that the x value is gonna be something positive, something greater than zero. The y value is also gonna be something positive. And again, that's because, you know, the movement is to the right, so positive x. It's moving up, so positive y. But the z value should actually be a little bit negative, right? Because as you look at this curve, it's going down, in a sense.
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Partial derivative of a parametric surface, part 1.mp3
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And again, that's because, you know, the movement is to the right, so positive x. It's moving up, so positive y. But the z value should actually be a little bit negative, right? Because as you look at this curve, it's going down, in a sense. And with that being our prediction, if you start plugging in one, one to t and s, what you'll see is that, you know, two times one is two. S equals one, so that's just one. And then over here, this looks like one squared minus two times one times one.
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Partial derivative of a parametric surface, part 1.mp3
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Because as you look at this curve, it's going down, in a sense. And with that being our prediction, if you start plugging in one, one to t and s, what you'll see is that, you know, two times one is two. S equals one, so that's just one. And then over here, this looks like one squared minus two times one times one. So this will be one minus two. That's negative one. So it is, in fact, that kind of positive, positive, negative pattern that you're seeing.
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Partial derivative of a parametric surface, part 1.mp3
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And then over here, this looks like one squared minus two times one times one. So this will be one minus two. That's negative one. So it is, in fact, that kind of positive, positive, negative pattern that you're seeing. And maybe even from this curve, you can get a feel for why the movement in the x direction is twice as much as the movement in the y. It's moving more to the right than it is up in the y direction. And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface.
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Partial derivative of a parametric surface, part 1.mp3
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So it is, in fact, that kind of positive, positive, negative pattern that you're seeing. And maybe even from this curve, you can get a feel for why the movement in the x direction is twice as much as the movement in the y. It's moving more to the right than it is up in the y direction. And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface. And the corresponding movement, the direction that nudges in the t direction take you, will give you some vector in three-dimensional space. And that's the interpretation. That is the meaning of the partial derivative of the vector-valued function here.
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Partial derivative of a parametric surface, part 1.mp3
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And again, in principle, you can imagine doing this not just at the point, one, one, but at any given point, maybe any given point along this curve or any given point along the surface. And the corresponding movement, the direction that nudges in the t direction take you, will give you some vector in three-dimensional space. And that's the interpretation. That is the meaning of the partial derivative of the vector-valued function here. And again, it's not the tiny, the actual nudge vector itself, right? When you nudge the input and you get just a little smidgen in the output space here, but it's that divided by the size of the nudge. So that's why you'll get kind of normal-sized vectors rather than tiny vectors.
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Partial derivative of a parametric surface, part 1.mp3
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So let's say you have a function that's got a single input, t, and then it outputs a vector. And the vector's gonna depend on t, so the x component will be t times the cosine of t, and then the y component will be t times the sine of t. This is what's called a parametric function. And I should maybe say one parameter parametric function. One parameter. And parameter is just kind of a fancy word for input. Parameter. So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional.
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Parametric curves Multivariable calculus Khan Academy.mp3
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One parameter. And parameter is just kind of a fancy word for input. Parameter. So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional. So you might think, when you visualize something like this, oh, it's got a single input, and it's got a two-dimensional output, let's graph it. Let's put those three numbers together and plot them. But what turns out to be even better is to look just in the output space.
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Parametric curves Multivariable calculus Khan Academy.mp3
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So in this case, t is our single parameter, and what makes it a parametric function is that we think about it as drawing a curve, and its output is multidimensional. So you might think, when you visualize something like this, oh, it's got a single input, and it's got a two-dimensional output, let's graph it. Let's put those three numbers together and plot them. But what turns out to be even better is to look just in the output space. So in this case, the output space is two-dimensional, so I'll go ahead and draw a coordinate plane here. And let's just evaluate this function at a couple different points and see what it looks like. So maybe the easiest place to evaluate it would be zero.
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Parametric curves Multivariable calculus Khan Academy.mp3
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But what turns out to be even better is to look just in the output space. So in this case, the output space is two-dimensional, so I'll go ahead and draw a coordinate plane here. And let's just evaluate this function at a couple different points and see what it looks like. So maybe the easiest place to evaluate it would be zero. So f of zero f of zero is equal to, and then in both cases, it'll be zero times something, so zero times cosine of zero is just zero, and then zero times sine of zero is also just zero. So that input corresponds to the output. You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it.
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Parametric curves Multivariable calculus Khan Academy.mp3
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So maybe the easiest place to evaluate it would be zero. So f of zero f of zero is equal to, and then in both cases, it'll be zero times something, so zero times cosine of zero is just zero, and then zero times sine of zero is also just zero. So that input corresponds to the output. You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it. So let's take a different point just to see what else could happen. And I'm gonna choose pi halves. Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of.
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Parametric curves Multivariable calculus Khan Academy.mp3
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You could think of it as a vector that's infinitely small, or just the point at the origin, however you wanna go about it. So let's take a different point just to see what else could happen. And I'm gonna choose pi halves. Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of. So t is pi halves, cosine of pi halves. And you start thinking, okay, what's the cosine of pi halves, what's the sine of pi halves? And maybe you go off and draw a little unit circle while you're writing things out.
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Parametric curves Multivariable calculus Khan Academy.mp3
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Of course, the reason I'm choosing pi halves of all numbers is that it's something I know how to take the sine and the cosine of. So t is pi halves, cosine of pi halves. And you start thinking, okay, what's the cosine of pi halves, what's the sine of pi halves? And maybe you go off and draw a little unit circle while you're writing things out. Oops. This is the problem with talking while writing. Sine of pi halves.
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Parametric curves Multivariable calculus Khan Academy.mp3
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And maybe you go off and draw a little unit circle while you're writing things out. Oops. This is the problem with talking while writing. Sine of pi halves. And you know, it's, if you go off and scribble that little unit circle, and you say pi halves is gonna bring us a quarter of the way around over here, and cosine of pi halves is measuring the x component of that so that just cancels out to zero. And then sine of pi halves is the y component of that, so that ends up equaling one. Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component.
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Parametric curves Multivariable calculus Khan Academy.mp3
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Sine of pi halves. And you know, it's, if you go off and scribble that little unit circle, and you say pi halves is gonna bring us a quarter of the way around over here, and cosine of pi halves is measuring the x component of that so that just cancels out to zero. And then sine of pi halves is the y component of that, so that ends up equaling one. Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component. And what that would look like, you know, the y component of pi halves, it's about 1.7 up there and there's no x component, so you might get a vector like this. And if you imagine doing this at all the different input points, you might get a bunch of different vectors off doing different things. And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch.
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Parametric curves Multivariable calculus Khan Academy.mp3
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Which means that the vector as a whole is gonna be zero for the x component and then pi halves for the y component. And what that would look like, you know, the y component of pi halves, it's about 1.7 up there and there's no x component, so you might get a vector like this. And if you imagine doing this at all the different input points, you might get a bunch of different vectors off doing different things. And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch. So we just wanna trace the points that correspond to the output, the tips of each vector. And what I'll do here, I'll show a little animation. Let me just clear the board a bit.
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Parametric curves Multivariable calculus Khan Academy.mp3
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And if you were to draw it, you don't wanna just draw the arrows themselves because that'll clutter things up a whole bunch. So we just wanna trace the points that correspond to the output, the tips of each vector. And what I'll do here, I'll show a little animation. Let me just clear the board a bit. An animation where I'll let t range between zero and 10. Let's write that down. So the value t is gonna start at zero and then it's gonna go to 10.
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Parametric curves Multivariable calculus Khan Academy.mp3
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Let me just clear the board a bit. An animation where I'll let t range between zero and 10. Let's write that down. So the value t is gonna start at zero and then it's gonna go to 10. And we'll just see what vectors does that output and what curve does the tip of that vector trace out. So here it goes, all of the values just kind of ranging zero to 10. And you end up getting this spiral shape.
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Parametric curves Multivariable calculus Khan Academy.mp3
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So the value t is gonna start at zero and then it's gonna go to 10. And we'll just see what vectors does that output and what curve does the tip of that vector trace out. So here it goes, all of the values just kind of ranging zero to 10. And you end up getting this spiral shape. And you can maybe think about why this cosine of t, sine of t scaled by the value t itself would give you this spiral. But what it means is that when you, we saw that zero goes here, evidently. It's the case that 10 outputs here.
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Parametric curves Multivariable calculus Khan Academy.mp3
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And you end up getting this spiral shape. And you can maybe think about why this cosine of t, sine of t scaled by the value t itself would give you this spiral. But what it means is that when you, we saw that zero goes here, evidently. It's the case that 10 outputs here. And a disadvantage of drawing things like this, you're not quite sure of what the interim values are. You could kind of guess maybe one goes somewhere here, two goes somewhere here, and you're kind of hoping that they're evenly spaced as you move along. But you don't get that information.
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Parametric curves Multivariable calculus Khan Academy.mp3
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It's the case that 10 outputs here. And a disadvantage of drawing things like this, you're not quite sure of what the interim values are. You could kind of guess maybe one goes somewhere here, two goes somewhere here, and you're kind of hoping that they're evenly spaced as you move along. But you don't get that information. You lose the input information. You get the shape of the curve. And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate.
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Parametric curves Multivariable calculus Khan Academy.mp3
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But you don't get that information. You lose the input information. You get the shape of the curve. And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate. But just to show where it might matter, I'll animate the same thing again, another function that draws the same curve, but it starts going really quickly. And then it slows down as you go on. So that function is not quite the t cosine t, t sine of t that I originally had written.
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Parametric curves Multivariable calculus Khan Academy.mp3
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And if you just want an analytical way of describing curves, you find some parametric function that does it and you don't really care about the rate. But just to show where it might matter, I'll animate the same thing again, another function that draws the same curve, but it starts going really quickly. And then it slows down as you go on. So that function is not quite the t cosine t, t sine of t that I originally had written. And in fact, it would mean that, let's just erase these guys. When it starts slowly, you can interpret that as saying, okay, maybe, well, actually it started quickly, didn't it? So one would be really far off here.
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Parametric curves Multivariable calculus Khan Academy.mp3
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So that function is not quite the t cosine t, t sine of t that I originally had written. And in fact, it would mean that, let's just erase these guys. When it starts slowly, you can interpret that as saying, okay, maybe, well, actually it started quickly, didn't it? So one would be really far off here. And then two, it could have zipped along here. And then three, still going really fast. And then maybe by the time you get to the end, it's just going very slowly, just kind of seven is here, eight is here, and it's hardly making any progress before it gets to 10.
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Parametric curves Multivariable calculus Khan Academy.mp3
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So one would be really far off here. And then two, it could have zipped along here. And then three, still going really fast. And then maybe by the time you get to the end, it's just going very slowly, just kind of seven is here, eight is here, and it's hardly making any progress before it gets to 10. So you can have two different functions draw the same curve. And the fancy word here is parameterize. So functions will parameterize a curve if when you draw just in the output space, you get that curve.
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Parametric curves Multivariable calculus Khan Academy.mp3
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And hopefully we'll be able to use that to understand or get a better intuition behind what exactly it means to take a derivative of a position vector valued function. So let's say my first parametrization, I have x of t is equal to t, and let's say that y of t is equal to t squared, and this is true for t is greater than or equal to 0 and less than or equal to 2. And if I want to write this as a position vector valued function, let me write this x1, let me call that y1, and let me write my position vector valued function, I could say r1, because I'm numbering them because I'm going to do a different version of this exact same curve with a slightly different parametrization. So r1 of t we could say is x1 of t times i, the unit vector i, so we'll just say t times i, plus, right, this is just x of t right here, or x1 of t, I'm numbering them because I'll later have an x2 of t, plus t squared times j. And if I wanted to graph this, I'm going to be very careful graphing it, because I really want to understand what the derivative means here. Let me get my draw, try to draw it roughly to scale. So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So r1 of t we could say is x1 of t times i, the unit vector i, so we'll just say t times i, plus, right, this is just x of t right here, or x1 of t, I'm numbering them because I'll later have an x2 of t, plus t squared times j. And if I wanted to graph this, I'm going to be very careful graphing it, because I really want to understand what the derivative means here. Let me get my draw, try to draw it roughly to scale. So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2. And so at t equals 0, both my x and y coordinates are at 0, or this is just going to be the 0 vector, so this is where we are at t equals 0. At t equals 1, this is going to be 1 times i, so we're going to be just like that, plus 1 times j, right, 1 squared is j, so we're going to be right there. And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So let's say that this is, so we get 1, 2, 3, 4, and then let me draw my x axis, that's good enough, and my x axis I want it to be roughly to scale 1 and 2. And so at t equals 0, both my x and y coordinates are at 0, or this is just going to be the 0 vector, so this is where we are at t equals 0. At t equals 1, this is going to be 1 times i, so we're going to be just like that, plus 1 times j, right, 1 squared is j, so we're going to be right there. And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this. So this is what, just to make it clear what we're doing, that's r1 of 2, right, this is r1 of 0, this is r1 of 1, but the bottom line is the path looks like this, it's a parabola. So the path will look like that, just like that. Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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And then at t is equal to 2, we're going to be at 2i, so 2i, you can imagine 2 times i would be this vector right there, right, 2 times i, plus 4, right, 2 squared is 4, 4 times j, so plus this, plus 4 times j, right, that's 4 times j, if you add these two vectors, heads to tails, you're going to get a vector that goes, that end point is right there, the vector is going to look something like this. So this is what, just to make it clear what we're doing, that's r1 of 2, right, this is r1 of 0, this is r1 of 1, but the bottom line is the path looks like this, it's a parabola. So the path will look like that, just like that. Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing. So it's going to be a parabola, well, let me get rid of that other point too, just because I didn't draw it exactly where it needs to be, it needs to be right there, and my parabola, or my part of my parabola is going to look something like that. Alright, good enough. So this is the first parametrization.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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Now, that's in my first parametrization of it, actually let me draw it a little bit more carefully, and I want to get rid of this arrow, just because I want to be a nice clean drawing. So it's going to be a parabola, well, let me get rid of that other point too, just because I didn't draw it exactly where it needs to be, it needs to be right there, and my parabola, or my part of my parabola is going to look something like that. Alright, good enough. So this is the first parametrization. Now I'm going to do this exact same curve, but I'm going to do it slightly differently. So let's say, I'll do it in different colors, so x2 of t, let's say it equals 2t, and y2 of t, let's say it's equal to 2t squared, or we could alternatively write that, that's the same thing as 4t squared, just raising both of these guys to the second power, that's equal to 4t squared, I could write it either way, and then let's say my second parametrization, and let's say instead of going from t equals 0 to 2, we're going to go from, t goes from 0 to 1, t goes from 0 to 1, but we're going to see, we're going to cover the exact same path, and our second vector, or position vector valued function, r2 of t is going to be equal to 2t times i, plus, I could say 2t squared, or 4t squared times j. And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here?
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So this is the first parametrization. Now I'm going to do this exact same curve, but I'm going to do it slightly differently. So let's say, I'll do it in different colors, so x2 of t, let's say it equals 2t, and y2 of t, let's say it's equal to 2t squared, or we could alternatively write that, that's the same thing as 4t squared, just raising both of these guys to the second power, that's equal to 4t squared, I could write it either way, and then let's say my second parametrization, and let's say instead of going from t equals 0 to 2, we're going to go from, t goes from 0 to 1, t goes from 0 to 1, but we're going to see, we're going to cover the exact same path, and our second vector, or position vector valued function, r2 of t is going to be equal to 2t times i, plus, I could say 2t squared, or 4t squared times j. And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here? 1 half times 2 is 1, and then we're going to get the point, 1 half squared is 1 fourth times 4 is 1, so when t is equal to 1 half, we're going to be at the point 1,1, and when t is equal to 1, we're going to be at the point 2,4. So notice the curve is exactly, the path we go is exactly the same, but before we even do the derivatives, these two paths are identical, I want to think about something, let's pretend that our parameter t really is time, that t is equal to time, and that tends to be the most common, that's why they call it t, it doesn't have to be time, but let's say it is time, so what's happening here? In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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And if I were to graph this guy right here, it would look like, let me draw my axes again, it's going to look the same, but it's I think useful to draw it, because I'm going to draw the derivatives and all of that on it later, so let's see, 1, 2, 3, 4, 1, 2, and then let's see what happens, when t is equal to 0, or r of 0, all of these are going to be 0, we've just got the 0 vector, x and y are both equal to 0, and then t is equal to 1 half, so when t is equal to 1 half, what are we going to get here? 1 half times 2 is 1, and then we're going to get the point, 1 half squared is 1 fourth times 4 is 1, so when t is equal to 1 half, we're going to be at the point 1,1, and when t is equal to 1, we're going to be at the point 2,4. So notice the curve is exactly, the path we go is exactly the same, but before we even do the derivatives, these two paths are identical, I want to think about something, let's pretend that our parameter t really is time, that t is equal to time, and that tends to be the most common, that's why they call it t, it doesn't have to be time, but let's say it is time, so what's happening here? In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so. In this situation, we have a dot moving along the same curve, but it's able to cover the same curve in only 1 second, in half a second it gets here, it took this guy 1 second to get here, in 1 second, this guy is all the way over here, this guy takes 2 seconds to go over here, so in this second parametrization, even though the path is the same, the curves are the same, the dot is faster, and I want you to keep that in mind when we think about the derivatives of both of these position vector valued functions. Just remember, the dot is moving faster, for every second it's getting further along the curve than here, that's why it only took them 1 second. Now let's look at the derivatives of both of these guys.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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In this first parametrization, when we go from 0 to 2 seconds, we cover this path, you can imagine after 1 second, the dot moves here, then it moves there, you can imagine a dot moving along this curve, and it takes 2 seconds to do so. In this situation, we have a dot moving along the same curve, but it's able to cover the same curve in only 1 second, in half a second it gets here, it took this guy 1 second to get here, in 1 second, this guy is all the way over here, this guy takes 2 seconds to go over here, so in this second parametrization, even though the path is the same, the curves are the same, the dot is faster, and I want you to keep that in mind when we think about the derivatives of both of these position vector valued functions. Just remember, the dot is moving faster, for every second it's getting further along the curve than here, that's why it only took them 1 second. Now let's look at the derivatives of both of these guys. The derivative here, if I were to write r1' of t, let me do it in a different color, I already used the orange, let me do it in blue, r1' of t, so the derivative now is going to be, remember it's just the derivative of each of these times the unit vectors, so the derivative of t with respect to t, that's just 1, so it's 1 times i, so I'll just write 1i, plus, I didn't have to write the 1 there, plus the derivative of t squared with respect to t is 2t, plus 2tj, and let me take the derivative over here, r2' of t, the derivative of 2t with respect to t is 2, so 2i, plus the derivative of 4t squared is 8t, 2 times 4, it is 8t, just like that. Now, the question is, what do their respective derivative vectors look like at different points? So let's look at, I don't know, let's see how fast they're moving when time is equal to 1.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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Now let's look at the derivatives of both of these guys. The derivative here, if I were to write r1' of t, let me do it in a different color, I already used the orange, let me do it in blue, r1' of t, so the derivative now is going to be, remember it's just the derivative of each of these times the unit vectors, so the derivative of t with respect to t, that's just 1, so it's 1 times i, so I'll just write 1i, plus, I didn't have to write the 1 there, plus the derivative of t squared with respect to t is 2t, plus 2tj, and let me take the derivative over here, r2' of t, the derivative of 2t with respect to t is 2, so 2i, plus the derivative of 4t squared is 8t, 2 times 4, it is 8t, just like that. Now, the question is, what do their respective derivative vectors look like at different points? So let's look at, I don't know, let's see how fast they're moving when time is equal to 1. So let's take a specific point, this is just the general formula, but let's figure out what the derivative is at a specific point. So let's take r1 when time is equal to 1, and I want to take the specific point on the curve, not the specific point in time. So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So let's look at, I don't know, let's see how fast they're moving when time is equal to 1. So let's take a specific point, this is just the general formula, but let's figure out what the derivative is at a specific point. So let's take r1 when time is equal to 1, and I want to take the specific point on the curve, not the specific point in time. So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second. So r1 of 1 is equal to, I'm taking the derivative there, is equal to 1i, it's not dependent on t at all, so it's 1i plus 2 times 1j, so plus 2j. So at this point, the derivative of our position vector function is going to be 1i plus 2j, so we can draw it like this. So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So this point on the curve here is when time is equal to 1 second, this point over here, which is the exact corresponding point, is when time is equal to 1 half second. So r1 of 1 is equal to, I'm taking the derivative there, is equal to 1i, it's not dependent on t at all, so it's 1i plus 2 times 1j, so plus 2j. So at this point, the derivative of our position vector function is going to be 1i plus 2j, so we can draw it like this. So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that. So our derivative right there, I'll do it in the same color that I wrote it in, in this green color, is going to look like this. It's going to look like that. And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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So if we do 1i, it's like this, 1i, and then 2j, just 2j is like that. So our derivative right there, I'll do it in the same color that I wrote it in, in this green color, is going to look like this. It's going to look like that. And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is. This right here, just to be clear, is r1 prime, it's a vector, so it's telling us the instantaneous change in our position vector with respect to t, or time, when time is equal to 1 second. That's this thing right here. Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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And notice, it looks like, at least its direction is, let me do it a little bit straighter, its direction looks tangent to the curve, it's going in the direction that my particle is moving, remember my particle is going from here to there, so it's going in the direction, and I'm going to think about in a second what this length of this derivative vector is. This right here, just to be clear, is r1 prime, it's a vector, so it's telling us the instantaneous change in our position vector with respect to t, or time, when time is equal to 1 second. That's this thing right here. Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy. We already said it only takes him, he's here at time is equal to 1 half second. So let's take, let me do it in, I'll do it in the same color. So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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Now, let's take the exact same position here on our curve, but that's going to occur at a different time for this guy. We already said it only takes him, he's here at time is equal to 1 half second. So let's take, let me do it in, I'll do it in the same color. So here, we have r2, we're going to evaluate it at 1 half, because this is at time is equal to 1 half second, and this is going to be equal to 2i, this isn't dependent at all on time, so 2i plus 8 times the time, so time right here is 1 half, so 8 times 1 half is 4, so plus 4j. So what does this look like? The instantaneous derivative here, and this is the derivative, we have to be very clear. So 2i, let me draw some more, so 2i maybe gets us about that far, plus 4j will get us up to right around there, plus 4j is that vector, so when you add those two heads to tails, you get this thing.
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Vector valued function derivative example Multivariable Calculus Khan Academy.mp3
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