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And now the second requirement of classifying those points, that's what the second derivative test is all about. Once you find something where the gradient equals zero, you want to be able to determine, is it a local maximum, is it a local minimum, or is it a saddle point? So let's go ahead and work through this example. The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line. We take both partial derivatives. So the partial derivative with respect to x is, well, this first term, when we take the derivative of three x squared times y with respect to x, that two hops down, so we have six times x times y. Y cubed, well, y looks like a constant, so y cubed looks like a constant, minus three x squared, so that two comes down, so we're subtracting off six times x, six times x. And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned.
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Second partial derivative test example, part 1.mp3
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The first thing we're gonna need to do if we're solving for when the gradient equals zero, and remember, when we say equals zero, we really mean the zero vector, but it's just a convenient way of putting it all on one line. We take both partial derivatives. So the partial derivative with respect to x is, well, this first term, when we take the derivative of three x squared times y with respect to x, that two hops down, so we have six times x times y. Y cubed, well, y looks like a constant, so y cubed looks like a constant, minus three x squared, so that two comes down, so we're subtracting off six times x, six times x. And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned. And we do partial of f with respect to y. Then this first term looks like some sort of constant, three x squared, x looks like a constant, so some kind of constant times y, so the whole thing looks like three x squared. The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared.
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Second partial derivative test example, part 1.mp3
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And then again, this three y squared term, y looks like a constant, so everything here looks like a constant with zero derivative as far as the x direction is concerned. And we do partial of f with respect to y. Then this first term looks like some sort of constant, three x squared, x looks like a constant, so some kind of constant times y, so the whole thing looks like three x squared. The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared. And then this next term only has an x, so it looks like a constant as far as y is concerned. And then this last term, we take down the two, because we're differentiating y squared, and you'll get negative six y, negative six times y. So when we are finding the critical points, the first step is to set both of these guys equal to zero.
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Second partial derivative test example, part 1.mp3
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The second term, minus y cubed, excuse me, looks like minus three y squared, when we take the derivative, minus three y squared. And then this next term only has an x, so it looks like a constant as far as y is concerned. And then this last term, we take down the two, because we're differentiating y squared, and you'll get negative six y, negative six times y. So when we are finding the critical points, the first step is to set both of these guys equal to zero. So this first one, when we do set it equal to zero, we can simplify a bit by factoring out six x. So this really looks like six x multiplied by y minus one. And then that's what we're setting equal to zero.
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Second partial derivative test example, part 1.mp3
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So when we are finding the critical points, the first step is to set both of these guys equal to zero. So this first one, when we do set it equal to zero, we can simplify a bit by factoring out six x. So this really looks like six x multiplied by y minus one. And then that's what we're setting equal to zero. And what this equation tells us is that either it's the six x term that equals zero, in which case that would mean x is equal to zero, or it's the case that y minus one equals zero, in which case that would mean that y equals one. So at least one of these things has to be true. That's kind of the first requirement that we've found.
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Second partial derivative test example, part 1.mp3
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And then that's what we're setting equal to zero. And what this equation tells us is that either it's the six x term that equals zero, in which case that would mean x is equal to zero, or it's the case that y minus one equals zero, in which case that would mean that y equals one. So at least one of these things has to be true. That's kind of the first requirement that we've found. Let me scroll down a little bit here. And for the second equation, when we set it equal to zero, it's not immediately straightforward how you would solve for this in a nice way in terms of x and y, but because we've already solved one, we can kind of plug them in and say, for example, if it was the case that x is equal to zero, and we kind of want to see what that turns our equation into, then we would have, well, three x squared is nothing. That would be zero.
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Second partial derivative test example, part 1.mp3
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That's kind of the first requirement that we've found. Let me scroll down a little bit here. And for the second equation, when we set it equal to zero, it's not immediately straightforward how you would solve for this in a nice way in terms of x and y, but because we've already solved one, we can kind of plug them in and say, for example, if it was the case that x is equal to zero, and we kind of want to see what that turns our equation into, then we would have, well, three x squared is nothing. That would be zero. And we'd just be left with negative three y squared minus six y is equal to zero. And that we can factor out a bit. So I'm gonna factor out a negative three y.
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Second partial derivative test example, part 1.mp3
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That would be zero. And we'd just be left with negative three y squared minus six y is equal to zero. And that we can factor out a bit. So I'm gonna factor out a negative three y. So I'll factor out negative three y, which means that first term just has a y remaining, and then that second term has a two, a positive two, since I factored out negative three. So positive two, and that equals zero. So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two.
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Second partial derivative test example, part 1.mp3
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So I'm gonna factor out a negative three y. So I'll factor out negative three y, which means that first term just has a y remaining, and then that second term has a two, a positive two, since I factored out negative three. So positive two, and that equals zero. So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two. So that's the first situation, where we plug in x equals zero. Now, alternatively, there's the possibility that y equals one, so we could say y equals one. And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared.
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Second partial derivative test example, part 1.mp3
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So what this whole situation would imply is that either negative three y equals zero, which would mean y equals zero, or it would be the case that y plus two equals zero, which would mean that y is equal to negative two. So that's the first situation, where we plug in x equals zero. Now, alternatively, there's the possibility that y equals one, so we could say y equals one. And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared. And then the rest of it becomes, let's see, minus three times one squared. So minus three, we're plugging in one for y. And then we subtract off six, plugging in that one for y again.
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Second partial derivative test example, part 1.mp3
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And what that gives us in the entire equation, we still have that three x squared, because we're kind of solving for x now, three x squared. And then the rest of it becomes, let's see, minus three times one squared. So minus three, we're plugging in one for y. And then we subtract off six, plugging in that one for y again. And that whole thing is equal to three x squared, then minus three minus six. So I'm subtracting off nine. So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three.
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Second partial derivative test example, part 1.mp3
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And then we subtract off six, plugging in that one for y again. And that whole thing is equal to three x squared, then minus three minus six. So I'm subtracting off nine. So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three. And what that implies then, since this whole thing has to equal zero, what that implies is that x squared minus three is equal to zero. So we have x is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found.
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Second partial derivative test example, part 1.mp3
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So from here, I can factor out a little bit, and this will be three multiplied by x squared minus three. And what that implies then, since this whole thing has to equal zero, what that implies is that x squared minus three is equal to zero. So we have x is equal to plus or minus the square root of three. And maybe I should kind of specify, these are distinct things that we found. One of them was in the circumstance where x equals zero, and then the other was what we found in the circumstance where y equals one. So this gives us a grand total of three different critical points. Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two.
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Second partial derivative test example, part 1.mp3
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And maybe I should kind of specify, these are distinct things that we found. One of them was in the circumstance where x equals zero, and then the other was what we found in the circumstance where y equals one. So this gives us a grand total of three different critical points. Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another two possibilities here where if y is equal to one, when y is equal to one, we'll have x is positive or negative square root of three.
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Second partial derivative test example, part 1.mp3
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Because in the first situation where x equals zero, the critical points that we have, well, both of them are gonna have an x coordinate of zero in them, x coordinate of zero, and the two corresponding y coordinates are zero or negative two. So you have zero or negative two. There's kind of two possibilities. And then there's another two possibilities here where if y is equal to one, when y is equal to one, we'll have x is positive or negative square root of three. So we have positive square root of three and y equals one, and then we have negative square root of three and y equals one. So these are the critical points. Critical points.
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Second partial derivative test example, part 1.mp3
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In the last video, I started to talk about the formula for curvature. And just to remind everyone of where we are, you imagine that you have some kind of curve in, let's say, two-dimensional space, just for the sake of being simple. And let's say this curve is parameterized by a function s of t. So every number t corresponds to some point on the curve. For the curvature, you start thinking about unit tangent vectors. At every given point, what does the unit tangent vector look like? And the curvature itself, which is denoted by this sort of Greek letter kappa, is gonna be the rate of change of those unit vectors, kind of how quickly they're turning in direction, not with respect to the parameter t, but with respect to arc length, ds. And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds.
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Curvature formula, part 2.mp3
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For the curvature, you start thinking about unit tangent vectors. At every given point, what does the unit tangent vector look like? And the curvature itself, which is denoted by this sort of Greek letter kappa, is gonna be the rate of change of those unit vectors, kind of how quickly they're turning in direction, not with respect to the parameter t, but with respect to arc length, ds. And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds. And you're wondering, as you take a tiny step like that, does the unit tangent vector turn a lot or does it turn a little bit? And the little schematic that I said you might have in mind is just a completely separate space where for each one of these unit tangent vectors, you go ahead and put them in that space, saying, okay, so this one would look something like this. This one is pointed down and to the right, so it would look something like this.
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Curvature formula, part 2.mp3
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And so what I mean by arc length here is just a tiny step, you could think the size of a tiny step along the curve would be ds. And you're wondering, as you take a tiny step like that, does the unit tangent vector turn a lot or does it turn a little bit? And the little schematic that I said you might have in mind is just a completely separate space where for each one of these unit tangent vectors, you go ahead and put them in that space, saying, okay, so this one would look something like this. This one is pointed down and to the right, so it would look something like this. This one is pointed very much down. And you're wondering, basically, as you take tiny little steps of size ds, what does this change to the unit tangent vector? And that change, it's gonna be some kind of vector.
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Curvature formula, part 2.mp3
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This one is pointed down and to the right, so it would look something like this. This one is pointed very much down. And you're wondering, basically, as you take tiny little steps of size ds, what does this change to the unit tangent vector? And that change, it's gonna be some kind of vector. And because the curvature is really just a value, a number that we want, all we care about is the size of that vector. The size of the change to the tangent vector as you take a tiny step in ds. Now this is pretty abstract, right?
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Curvature formula, part 2.mp3
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And that change, it's gonna be some kind of vector. And because the curvature is really just a value, a number that we want, all we care about is the size of that vector. The size of the change to the tangent vector as you take a tiny step in ds. Now this is pretty abstract, right? I've got these two completely separate things that are not the original function that you have to think about. You have to think about this unit tangent vector function. And then you also have to think about this notion of arc length.
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Curvature formula, part 2.mp3
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Now this is pretty abstract, right? I've got these two completely separate things that are not the original function that you have to think about. You have to think about this unit tangent vector function. And then you also have to think about this notion of arc length. And the reason, by the way, that I'm using an s here as well as here for the parameterization of the curve is because they're actually quite related. And I'll get to that a little bit below. And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair.
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Curvature formula, part 2.mp3
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And then you also have to think about this notion of arc length. And the reason, by the way, that I'm using an s here as well as here for the parameterization of the curve is because they're actually quite related. And I'll get to that a little bit below. And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair. So we've got cosine of t is the x component and then sine of t is the y component, sine of t. And just to make it so that it's not completely boring, let's multiply both of these components by a constant r. And what this means, you might recognize this, cosine sine pair. What this means is that in the xy plane, you're actually drawing a circle with radius r. So this would be some kind of circle with a radius r. And while I go through this example, I also wanna make a note of what things would look like a little bit more abstractly. If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component.
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Curvature formula, part 2.mp3
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And to make it clear what this means, I'm gonna go ahead and go through an example here where let's say our parameterization with respect to t is a cosine sine pair. So we've got cosine of t is the x component and then sine of t is the y component, sine of t. And just to make it so that it's not completely boring, let's multiply both of these components by a constant r. And what this means, you might recognize this, cosine sine pair. What this means is that in the xy plane, you're actually drawing a circle with radius r. So this would be some kind of circle with a radius r. And while I go through this example, I also wanna make a note of what things would look like a little bit more abstractly. If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component. And the reason I wanna do this is because the concrete version is gonna be helpful and simple and something we can deal with, but almost so simple as to not be indicative of just how complicated the normal circumstance is. But the more general circumstance is so complicated, I think it'll actually confuse things a little bit too much. So it'll be good to kind of go through both of them in parallel.
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Curvature formula, part 2.mp3
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If we just had s of t equals, not specific functions that I lay down, but just any general function for the x component and for the y component. And the reason I wanna do this is because the concrete version is gonna be helpful and simple and something we can deal with, but almost so simple as to not be indicative of just how complicated the normal circumstance is. But the more general circumstance is so complicated, I think it'll actually confuse things a little bit too much. So it'll be good to kind of go through both of them in parallel. And the first step is to figure out what is this unit tangent vector? What is that function that at every given point gives you a unit tangent vector to the curve? And the first thing for that is to realize that we already have a notion of what should give the tangent vector.
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Curvature formula, part 2.mp3
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So it'll be good to kind of go through both of them in parallel. And the first step is to figure out what is this unit tangent vector? What is that function that at every given point gives you a unit tangent vector to the curve? And the first thing for that is to realize that we already have a notion of what should give the tangent vector. The derivative of our vector valued function as a function of t, the direction in which it points is in the tangent direction. So if I go over here and if I compute this derivative and I say s prime of t, which involves just taking the derivative of both components, so the derivative of cosine is negative sine of t multiplied by r, and the derivative of sine is cosine of t multiplied by r. And more abstractly, this is just gonna be any time you have two different component functions, you just take the derivative of each one. And hopefully you've seen this.
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Curvature formula, part 2.mp3
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And the first thing for that is to realize that we already have a notion of what should give the tangent vector. The derivative of our vector valued function as a function of t, the direction in which it points is in the tangent direction. So if I go over here and if I compute this derivative and I say s prime of t, which involves just taking the derivative of both components, so the derivative of cosine is negative sine of t multiplied by r, and the derivative of sine is cosine of t multiplied by r. And more abstractly, this is just gonna be any time you have two different component functions, you just take the derivative of each one. And hopefully you've seen this. If not, maybe take a look at the videos on taking the derivative of a position vector valued function. And this we can interpret as that tangent vector, but it might not be a unit vector, right? We want a unit tangent vector, and this only promises us the direction.
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Curvature formula, part 2.mp3
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And hopefully you've seen this. If not, maybe take a look at the videos on taking the derivative of a position vector valued function. And this we can interpret as that tangent vector, but it might not be a unit vector, right? We want a unit tangent vector, and this only promises us the direction. So what we do to normalize it, what we do to normalize it and get a unit tangent vector function, maybe a different color, and get a unit vector tangent function, which I'll call capital T of lowercase t. That's kind of confusing, right? Capital T is for tangent vector, lowercase t is the parameter. So I'll try to keep that straight.
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Curvature formula, part 2.mp3
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We want a unit tangent vector, and this only promises us the direction. So what we do to normalize it, what we do to normalize it and get a unit tangent vector function, maybe a different color, and get a unit vector tangent function, which I'll call capital T of lowercase t. That's kind of confusing, right? Capital T is for tangent vector, lowercase t is the parameter. So I'll try to keep that straight. It's sort of standard notation, but there is the potential to confuse with this. What that's gonna be is your vector value derivative, but normalized. So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector.
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Curvature formula, part 2.mp3
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So I'll try to keep that straight. It's sort of standard notation, but there is the potential to confuse with this. What that's gonna be is your vector value derivative, but normalized. So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector. What we get, I'm giving myself even more room here, is the square root of sine squared, negative sine squared is just gonna be sine squared. So sine squared of t multiplied by r squared. And then over here, cosine squared times r squared.
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Curvature formula, part 2.mp3
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So we have to divide by whatever its magnitude is as a function of t. And in this case, in our specific example, that magnitude, if we take the magnitude of negative sine of t, r, multiplied by r, and then cosine of t, cosine of t multiplied by r. So we're taking the magnitude of this whole vector. What we get, I'm giving myself even more room here, is the square root of sine squared, negative sine squared is just gonna be sine squared. So sine squared of t multiplied by r squared. And then over here, cosine squared times r squared. Cosine squared of t times r squared. We can bring that r squared outside of the radical and just sort of factor it out, turning it into an r. And on the inside, we have sine squared plus cosine squared. I'm being too lazy to write down the t's right now, because no matter what the t is, that whole value just equals one.
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Curvature formula, part 2.mp3
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In the last video, we finished off with these two results. We started off just thinking about what it means to take the partial derivative of a vector-valued function. And I got to these kind of, you might call them, bizarre results. What was the whole point in getting here, Sal? And the whole point is so that I can give you the tools you need to understand what a surface integral is. So let's just think about, let's draw the ST plane and then see how it gets transformed into this surface R. So let's do that. Let's say that is the t-axis.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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What was the whole point in getting here, Sal? And the whole point is so that I can give you the tools you need to understand what a surface integral is. So let's just think about, let's draw the ST plane and then see how it gets transformed into this surface R. So let's do that. Let's say that is the t-axis. And let's say that that is the s-axis. And let's say that our vector-valued function, or our position vector-valued function, is defined from s's between a and b. So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Let's say that is the t-axis. And let's say that that is the s-axis. And let's say that our vector-valued function, or our position vector-valued function, is defined from s's between a and b. So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface. And if you map each of these points, you will eventually get the surface R. So let me draw R in three dimensions. The surface in 3D. So that is our x-axis.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So between a and b, I'm just picking arbitrary boundaries, and between t being equal to c and d. So if we were to, so the area under question, if you take any t and any s in this rectangle right here, it will be mapped to part of that surface. And if you map each of these points, you will eventually get the surface R. So let me draw R in three dimensions. The surface in 3D. So that is our x-axis. That is our y-axis. And then that is the z-axis. And just as a bit of a reminder, it might look something like this.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So that is our x-axis. That is our y-axis. And then that is the z-axis. And just as a bit of a reminder, it might look something like this. If we were to this point right here, where s is equal to a and t is equal to c, remember, we're going to draw the surface indicated by the position vector function s, R of s and t. So this point right here, when s is a and t is c, maybe it maps to that point right there. When you take a and c and you put it into this thing over here, you're just going to get the vector that points at that. So you could say it'll give you a position vector that'll point right at that position right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And just as a bit of a reminder, it might look something like this. If we were to this point right here, where s is equal to a and t is equal to c, remember, we're going to draw the surface indicated by the position vector function s, R of s and t. So this point right here, when s is a and t is c, maybe it maps to that point right there. When you take a and c and you put it into this thing over here, you're just going to get the vector that points at that. So you could say it'll give you a position vector that'll point right at that position right there. And then let's say that this line right here, if we were to hold s constant at a and just vary t from c to d, maybe that looks something like this. I'm just drawing some arbitrary contour there. Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So you could say it'll give you a position vector that'll point right at that position right there. And then let's say that this line right here, if we were to hold s constant at a and just vary t from c to d, maybe that looks something like this. I'm just drawing some arbitrary contour there. Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that. I don't know. I'm just trying to show you an example. So this point right here would correspond to that point right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Maybe if we hold t constant at c and vary s from a to b, maybe that'll look something like that. I don't know. I'm just trying to show you an example. So this point right here would correspond to that point right there. When you put it into the vector-valued function R, you would get a vector that points to that point, just like that. And this point right here in purple, when you evaluate R of s and t, it'll give you a vector that points right there to that point over there. We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So this point right here would correspond to that point right there. When you put it into the vector-valued function R, you would get a vector that points to that point, just like that. And this point right here in purple, when you evaluate R of s and t, it'll give you a vector that points right there to that point over there. We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible. So maybe I'll do it in this bluish color. If we hold t at d and vary s from a to b, we're going to start here. This is when t is d and s is a.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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We could do a couple of other points just to get an idea of what the surface looks like, although I'm trying to keep things as general as possible. So maybe I'll do it in this bluish color. If we hold t at d and vary s from a to b, we're going to start here. This is when t is d and s is a. And when you vary it, maybe you get something like that. I don't know. So this point right here would correspond to a vector that points to that point right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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This is when t is d and s is a. And when you vary it, maybe you get something like that. I don't know. So this point right here would correspond to a vector that points to that point right there. And then finally, this line, or if we hold s at b and vary t between c and d, we're going to go from that point to that point. So it's going to look something like this. We're going to go from this point to that point.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So this point right here would correspond to a vector that points to that point right there. And then finally, this line, or if we hold s at b and vary t between c and d, we're going to go from that point to that point. So it's going to look something like this. We're going to go from this point to that point. We're holding s at b, varying t from c to d. Maybe it looks something like that. So our surface, we went from this nice area, this rectangular area in the t, s plane, and it gets transformed into this wacky looking surface. And we could even draw some other things right here.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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We're going to go from this point to that point. We're holding s at b, varying t from c to d. Maybe it looks something like that. So our surface, we went from this nice area, this rectangular area in the t, s plane, and it gets transformed into this wacky looking surface. And we could even draw some other things right here. Let's say we have a, let's say we get some arbitrary value. Let me pick a new color, I'll do it in white or a new non-color. And let's say if we hold s at that constant value and we vary t, maybe that will look something like this.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And we could even draw some other things right here. Let's say we have a, let's say we get some arbitrary value. Let me pick a new color, I'll do it in white or a new non-color. And let's say if we hold s at that constant value and we vary t, maybe that will look something like this. Maybe that will look something like, oh, I don't know. Maybe it'll look something like that. So you get an idea of what this surface might look like.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And let's say if we hold s at that constant value and we vary t, maybe that will look something like this. Maybe that will look something like, oh, I don't know. Maybe it'll look something like that. So you get an idea of what this surface might look like. Now, given this, I want to think about what these quantities are. And then when we visualize what these quantities are, we'll be able to kind of use these results of the last video to do something that I think will be useful. So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So you get an idea of what this surface might look like. Now, given this, I want to think about what these quantities are. And then when we visualize what these quantities are, we'll be able to kind of use these results of the last video to do something that I think will be useful. So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here. Let's say this is the point, that is the point s, t. If you were to put those values in here, maybe it maps to, and I want to make sure I'm consistent with everything I've drawn, maybe it maps to this point right here. Maybe it maps to that point right there. So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So let's say that we pick arbitrary s and t. So this is the point, let me just pick it right here. Let's say this is the point, that is the point s, t. If you were to put those values in here, maybe it maps to, and I want to make sure I'm consistent with everything I've drawn, maybe it maps to this point right here. Maybe it maps to that point right there. So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general. I could call this a, well, I already used a and b. I could call this x and y. This would be r of x and y would map to that point right there. Now, so that's that right there, or that right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So this right here, this point right here, that is r of s and t. For a particular s and t, I mean, I could put little subscripts here, but I want to be general. I could call this a, well, I already used a and b. I could call this x and y. This would be r of x and y would map to that point right there. Now, so that's that right there, or that right there. Now let's see what happens if we move just in the s direction. We could view that as s. Now let us move forward by a differential, by a super small amount of s. So this right here, let's call that s plus a super small differential in s. That's right there. So that's that point.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Now, so that's that right there, or that right there. Now let's see what happens if we move just in the s direction. We could view that as s. Now let us move forward by a differential, by a super small amount of s. So this right here, let's call that s plus a super small differential in s. That's right there. So that's that point. Let me do that in a better color, in this yellow. So that point right there is the point s plus my differential of s. I could write delta s, but I want a super small change in s, comma t. And what is that going to get mapped to? Well, if we apply these two points in r, we're going to get something that maybe is right over there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So that's that point. Let me do that in a better color, in this yellow. So that point right there is the point s plus my differential of s. I could write delta s, but I want a super small change in s, comma t. And what is that going to get mapped to? Well, if we apply these two points in r, we're going to get something that maybe is right over there. And I want to be very clear, this right here, that is r of s plus ds, comma t. That's what that is. That's the point when we just shift s by a super small differential. This distance here you can view as ds.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Well, if we apply these two points in r, we're going to get something that maybe is right over there. And I want to be very clear, this right here, that is r of s plus ds, comma t. That's what that is. That's the point when we just shift s by a super small differential. This distance here you can view as ds. It's a super small change in s. And then when we map it, or transform it, or put that point into our vector-valued function. Let me copy and paste the original vector-valued function, just so we have a good image of what we're talking about this whole time. Let me put it right down there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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This distance here you can view as ds. It's a super small change in s. And then when we map it, or transform it, or put that point into our vector-valued function. Let me copy and paste the original vector-valued function, just so we have a good image of what we're talking about this whole time. Let me put it right down there. Just to be clear what's going on, when we took this little blue point right here, this s and t, and we put the s and t values here, we get a vector that points to that point on the surface right there. When you add a ds to your s values, you get a vector that points at that yellow point right there. So going back to the results we got in the last presentation, or the last video, what is this?
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Let me put it right down there. Just to be clear what's going on, when we took this little blue point right here, this s and t, and we put the s and t values here, we get a vector that points to that point on the surface right there. When you add a ds to your s values, you get a vector that points at that yellow point right there. So going back to the results we got in the last presentation, or the last video, what is this? r of s plus delta s, or r of s plus ds, the differential of s, t, well that is that right there. That is the vector that points to that position. This right here is the vector that points to this blue position.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So going back to the results we got in the last presentation, or the last video, what is this? r of s plus delta s, or r of s plus ds, the differential of s, t, well that is that right there. That is the vector that points to that position. This right here is the vector that points to this blue position. So what is the difference of those two vectors? And this is a bit of, this is just basic vector math you might remember. The difference of these two vectors, head to tails.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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This right here is the vector that points to this blue position. So what is the difference of those two vectors? And this is a bit of, this is just basic vector math you might remember. The difference of these two vectors, head to tails. The difference of these two vectors is going to be this vector. If you subtract this vector from that vector, you're going to get that vector right there. You're going to get that vector right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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The difference of these two vectors, head to tails. The difference of these two vectors is going to be this vector. If you subtract this vector from that vector, you're going to get that vector right there. You're going to get that vector right there. A vector that looks just like that. So that's what this is equal to, that vector. And it makes sense.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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You're going to get that vector right there. A vector that looks just like that. So that's what this is equal to, that vector. And it makes sense. This blue vector plus the orange vector, this blue vector right here plus the orange vector is equal to this vector. Makes complete sense, heads to tails. So that's what that represents.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And it makes sense. This blue vector plus the orange vector, this blue vector right here plus the orange vector is equal to this vector. Makes complete sense, heads to tails. So that's what that represents. Now let's do the same thing in the t direction. I'm running out of colors. I'll have to go back to the pink or maybe the magenta.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So that's what that represents. Now let's do the same thing in the t direction. I'm running out of colors. I'll have to go back to the pink or maybe the magenta. So if we have that s and t, now if we go up a little bit in that direction, let's say that that is t. So this is the point s, t plus a super small change in t. That's that point right there. This distance right there is dt. You can view it that way.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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I'll have to go back to the pink or maybe the magenta. So if we have that s and t, now if we go up a little bit in that direction, let's say that that is t. So this is the point s, t plus a super small change in t. That's that point right there. This distance right there is dt. You can view it that way. If you put s and t plus dt into our vector value function, what are you going to get? You're going to get a vector that maybe points to this point right here. Maybe I'll draw it right here.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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You can view it that way. If you put s and t plus dt into our vector value function, what are you going to get? You're going to get a vector that maybe points to this point right here. Maybe I'll draw it right here. Maybe it points to this point right here. A vector that points right there. So that will be mapped to a vector that points to that position right over there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Maybe I'll draw it right here. Maybe it points to this point right here. A vector that points right there. So that will be mapped to a vector that points to that position right over there. Now by the same argument that we did on the s side, this point or the vector that points to that, that is r of s, t plus dt. That is the exact same thing as that right there. And of course, this we already saw.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So that will be mapped to a vector that points to that position right over there. Now by the same argument that we did on the s side, this point or the vector that points to that, that is r of s, t plus dt. That is the exact same thing as that right there. And of course, this we already saw. This is the same thing as that over there. So what is that vector minus this blue vector? The magenta vector minus the blue vector.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And of course, this we already saw. This is the same thing as that over there. So what is that vector minus this blue vector? The magenta vector minus the blue vector. Well, once again, this is fairly, hopefully a bit of a review of adding vectors. It's going to be a vector that looks like this. I'll do it in white.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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The magenta vector minus the blue vector. Well, once again, this is fairly, hopefully a bit of a review of adding vectors. It's going to be a vector that looks like this. I'll do it in white. It's going to be a vector that looks like that. This thing is going to be a vector that looks just like that. And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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I'll do it in white. It's going to be a vector that looks like that. This thing is going to be a vector that looks just like that. And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector. So it makes sense if the purple vector minus the blue vector is going to be equal to this white vector. So something interesting is going on here. I have these two.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And you can imagine, if you take the blue vector plus the white vector, the blue vector plus this white vector is going to equal this purple vector. So it makes sense if the purple vector minus the blue vector is going to be equal to this white vector. So something interesting is going on here. I have these two. This is a vector that is kind of going along this parameterized surface as we changed our s by a super small amount. And then this is a vector that is going along our surface if we change our t by a super small amount. Now, you may or may not remember this.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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I have these two. This is a vector that is kind of going along this parameterized surface as we changed our s by a super small amount. And then this is a vector that is going along our surface if we change our t by a super small amount. Now, you may or may not remember this. And I've done several videos where I show this to you. But the magnitude, if I take two vectors and I take their cross product. So if I take the cross product of a and b and I take the magnitude of the resulting vector.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Now, you may or may not remember this. And I've done several videos where I show this to you. But the magnitude, if I take two vectors and I take their cross product. So if I take the cross product of a and b and I take the magnitude of the resulting vector. Remember, when you take the cross product, you get a third vector that is perpendicular to both of these. But if you were just to take the magnitude of that vector, that is equal to the area of parallelogram defined by a and b. What do I mean by that?
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So if I take the cross product of a and b and I take the magnitude of the resulting vector. Remember, when you take the cross product, you get a third vector that is perpendicular to both of these. But if you were just to take the magnitude of that vector, that is equal to the area of parallelogram defined by a and b. What do I mean by that? Well, if that is vector a and that is vector b, if you were to just take the cross product of those two, you're going to get a third vector that's perpendicular to both of them. They kind of pop out of the page. That would be a cross b.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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What do I mean by that? Well, if that is vector a and that is vector b, if you were to just take the cross product of those two, you're going to get a third vector that's perpendicular to both of them. They kind of pop out of the page. That would be a cross b. But the magnitude of this. So if you just take a cross product, you're going to get a vector. But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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That would be a cross b. But the magnitude of this. So if you just take a cross product, you're going to get a vector. But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b. And I've proved that in the linear algebra videos. Maybe I'll prove it again in this. I mean, it's because it's the, well, I won't go into that in too detail.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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But then if you take the magnitude of that vector, you're just saying how big is that vector, how long is that vector, that's going to be the area of the parallelogram defined by a and b. And I've proved that in the linear algebra videos. Maybe I'll prove it again in this. I mean, it's because it's the, well, I won't go into that in too detail. I've done it before. I don't want to make this video too long. So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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I mean, it's because it's the, well, I won't go into that in too detail. I've done it before. I don't want to make this video too long. So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there. And then another parallel version of b is right over there. So this is the parallelogram defined by a and b. So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So the parallelogram defined by a and b, you just imagine a, and then you take another kind of parallel version of a is right over there. And then another parallel version of b is right over there. So this is the parallelogram defined by a and b. So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors. So if I take the parallel to that one, it'll look something like this. And then a parallel to the orange one, it'll look something like that. So if I take the cross product of that and that, I am going to get the area of that parallelogram.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So going back to our surface example, if we were to take the cross product of this orange vector and this white vector, I'm going to get the surface area, I'm going to get the area of the parallelogram defined by these two vectors. So if I take the parallel to that one, it'll look something like this. And then a parallel to the orange one, it'll look something like that. So if I take the cross product of that and that, I am going to get the area of that parallelogram. Now you might say, hey, this is a surface. You're taking a straight up parallelogram. But remember, these are super small changes.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So if I take the cross product of that and that, I am going to get the area of that parallelogram. Now you might say, hey, this is a surface. You're taking a straight up parallelogram. But remember, these are super small changes. So you can imagine a surface can be broken up into super small changes in parallelograms, or into infinitely many parallelograms. And the more parallelograms you have, the better approximation of the surface you're going to have. And this is no different than when we first took integrals.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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But remember, these are super small changes. So you can imagine a surface can be broken up into super small changes in parallelograms, or into infinitely many parallelograms. And the more parallelograms you have, the better approximation of the surface you're going to have. And this is no different than when we first took integrals. We approximated the area under a curve with a bunch of rectangles. The more rectangles we had, the better. So let's call this little change in our surface d sigma.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And this is no different than when we first took integrals. We approximated the area under a curve with a bunch of rectangles. The more rectangles we had, the better. So let's call this little change in our surface d sigma. For a little change, for a little amount of our surface. And we could even say that the surface area of the surface will be the infinite sum of all of these infinitely small d sigmas. And there's actually a little notation for that.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So let's call this little change in our surface d sigma. For a little change, for a little amount of our surface. And we could even say that the surface area of the surface will be the infinite sum of all of these infinitely small d sigmas. And there's actually a little notation for that. So surface area is equal to, we could integrate over the surface. And the notation usually is a capital sigma for surface, as opposed to a region. So you're integrating over the surface.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And there's actually a little notation for that. So surface area is equal to, we could integrate over the surface. And the notation usually is a capital sigma for surface, as opposed to a region. So you're integrating over the surface. And you do a double integral, because you're going in two directions. The surface is kind of a folded two-dimensional structure. And you're going to take the infinite sum of all of the d sigmas.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So you're integrating over the surface. And you do a double integral, because you're going in two directions. The surface is kind of a folded two-dimensional structure. And you're going to take the infinite sum of all of the d sigmas. So this would be the surface area of this. So that's what a d sigma is. Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And you're going to take the infinite sum of all of the d sigmas. So this would be the surface area of this. So that's what a d sigma is. Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors. So let me write it here. This is all, it's not rigorous mathematics. The whole point here is to give you the intuition of what a surface integral is all about.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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Now, we just figured out, we just said, well, that d sigma can be represented, that value, that area of that little part of the surface, of that parallelogram, can be represented as a cross product of those two vectors. So let me write it here. This is all, it's not rigorous mathematics. The whole point here is to give you the intuition of what a surface integral is all about. So we can write that d sigma is equal to the cross product of the orange vector and the white vector. The orange vector is this. But we could also write it like this.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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The whole point here is to give you the intuition of what a surface integral is all about. So we can write that d sigma is equal to the cross product of the orange vector and the white vector. The orange vector is this. But we could also write it like this. This was the result from the last video. I'll write it in orange. So the partial of r with respect to s ds.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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But we could also write it like this. This was the result from the last video. I'll write it in orange. So the partial of r with respect to s ds. And d sigma is going to be the magnitude of the cross product, not just the cross product. The cross product by itself will just give you a vector. And that's going to be useful when we start doing vector-valued surface integrals.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So the partial of r with respect to s ds. And d sigma is going to be the magnitude of the cross product, not just the cross product. The cross product by itself will just give you a vector. And that's going to be useful when we start doing vector-valued surface integrals. But just think about it this way. So this orange vector is the same thing as that. And we're going to take the cross product of that with this white vector.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And that's going to be useful when we start doing vector-valued surface integrals. But just think about it this way. So this orange vector is the same thing as that. And we're going to take the cross product of that with this white vector. This white vector is the same thing as that, which we saw, which is the same thing as this. The partial of r with respect to t dt. And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And we're going to take the cross product of that with this white vector. This white vector is the same thing as that, which we saw, which is the same thing as this. The partial of r with respect to t dt. And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here. Now, you may or may not remember that if you take these. So let's just be clear. This and this, these are vectors.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And we saw if we take the magnitude of that, that's going to be equal to our little small change in area, or the area of this little parallelogram over here. Now, you may or may not remember that if you take these. So let's just be clear. This and this, these are vectors. When you take the partial derivative of a vector-valued function, you're still getting a vector. This ds, this is a number. That's a number and that's a number.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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This and this, these are vectors. When you take the partial derivative of a vector-valued function, you're still getting a vector. This ds, this is a number. That's a number and that's a number. And you might remember when we, in the linear algebra, or whenever you first saw taking cross products, taking the cross product of some scalar multiple, you can take the scalars out. So if we take this number and that number, we essentially factor them out of the cross product. This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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That's a number and that's a number. And you might remember when we, in the linear algebra, or whenever you first saw taking cross products, taking the cross product of some scalar multiple, you can take the scalars out. So if we take this number and that number, we essentially factor them out of the cross product. This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt. So I wrote this here. Hey, maybe our surface area, if we were to take the sum of all of these little d sigmas, but there's no obvious way to evaluate that. But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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This is going to be equal to the magnitude of the cross product of the partial of r with respect to s crossed with the partial of r with respect to t. And then all of that times these two guys over here, times ds and dt. So I wrote this here. Hey, maybe our surface area, if we were to take the sum of all of these little d sigmas, but there's no obvious way to evaluate that. But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's. So if you take all of the ds's, all of the dt's. So this is a ds times a dt, right? A ds times a dt, ds times a dt is right there.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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But we know that all of the d sigmas, they're the same thing as if you take all of the ds's and all of the dt's. So if you take all of the ds's, all of the dt's. So this is a ds times a dt, right? A ds times a dt, ds times a dt is right there. If we multiply this times the partial derivatives of the cross product of the partial derivatives, this times this is going to give us this area. So if we summed up all of this times this, or this times this, if we summed them up over this entire region, we will get all of the parallelograms in this region. We will get the surface area.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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A ds times a dt, ds times a dt is right there. If we multiply this times the partial derivatives of the cross product of the partial derivatives, this times this is going to give us this area. So if we summed up all of this times this, or this times this, if we summed them up over this entire region, we will get all of the parallelograms in this region. We will get the surface area. So we can write, I know this is all a little bit convoluted, and you need to kind of ponder it a little bit. Surface integrals, at least in my head, are one of the hardest things to really visualize, but it'll all hopefully make sense. So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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We will get the surface area. So we can write, I know this is all a little bit convoluted, and you need to kind of ponder it a little bit. Surface integrals, at least in my head, are one of the hardest things to really visualize, but it'll all hopefully make sense. So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here. And of course, we're also going to have to take this cross product in here. We know how to do that. That's a double integral.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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So we can say that this thing right over here, the sum of all of the little parallelograms on our surface, or the surface area, is going to be equal to, instead of taking the sum over the surface, let's take the sum of all the ds times dt's over this region right here. And of course, we're also going to have to take this cross product in here. We know how to do that. That's a double integral. So we're going to take the double integral over this, we could call it this region or this area right here. That area is the same thing as that whole area right over there of this thing. I'll just write it in yellow.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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That's a double integral. So we're going to take the double integral over this, we could call it this region or this area right here. That area is the same thing as that whole area right over there of this thing. I'll just write it in yellow. Of the cross product of the partial of r with respect to s and the partial of r with respect to t, ds and dt. And so you literally just take, and it seems very convoluted how you're going to actually evaluate it, but we were able to express this thing called a surface, well, this is a very simple surface integral, in something that we can actually calculate. And in the next few videos, I'm going to show you examples of actually calculating it.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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I'll just write it in yellow. Of the cross product of the partial of r with respect to s and the partial of r with respect to t, ds and dt. And so you literally just take, and it seems very convoluted how you're going to actually evaluate it, but we were able to express this thing called a surface, well, this is a very simple surface integral, in something that we can actually calculate. And in the next few videos, I'm going to show you examples of actually calculating it. Now this right here will only give you the surface area. But what if at every point here, so over here, what we've done in both of these expressions is we're just figuring out the surface area of each of these parallelograms and then adding them all up. That's what we're doing.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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And in the next few videos, I'm going to show you examples of actually calculating it. Now this right here will only give you the surface area. But what if at every point here, so over here, what we've done in both of these expressions is we're just figuring out the surface area of each of these parallelograms and then adding them all up. That's what we're doing. But what if associated with each of those little parallelograms we had some value, where that value is defined by some third function, f of x, y, z. So every parallelogram, it's super small, it's around a point, you could kind of say it's maybe the center of it, it doesn't have to be the center. It may be the center of it is some point in three-dimensional space.
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Introduction to the surface integral Multivariable Calculus Khan Academy.mp3
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