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So as kind of a first pass here, I'm gonna do a little more simplifying, but I'm gonna make a substitution that'll make this easier for us. So I see S over H here, and they're both to the same power, so I feel it might be a little bit easier if I just substitute U in for S divided by H. And what that'll let me do is rewrite this first equation here as 200 thirds, 200 thirds times U to the power 1 third, and that's equal to 20 times lambda. And then likewise, what that means for this guy is, well, this is H over S, not S over H, so that one's gonna be 100 thirds, not times U to the power 2 thirds, but times U to the negative 2 thirds because we swapped the H and S here. So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think. We kind of want to treat H and S in the same package. Now let me go ahead and put all the constants together, and I'm gonna take this guy and multiply it by three divided by 200, multiply both sides of that just to cancel out what's on the left side here. And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda.
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Lagrange multiplier example, part 2.mp3
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So that's U to the negative 2 thirds, and this is just to make it a little bit cleaner, I think. We kind of want to treat H and S in the same package. Now let me go ahead and put all the constants together, and I'm gonna take this guy and multiply it by three divided by 200, multiply both sides of that just to cancel out what's on the left side here. And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda. And then similarly over here, I'm gonna take this whole equation and multiply it by 3 over 100, and what that's gonna leave me with is that U to the negative 2 thirds, U to the negative 2 thirds is equal to, let's see, this 2000 when we divide by 100 becomes 20, and that 20 times 3 is 60, so that'll be 60 times lambda. Alright, so now I want a way to simplify each of these, and what I notice is they look pretty similar on each side, you know, it's something related to U equal to lambda, so if I can get this in a form where I'm really isolating U, that would be great. And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that.
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Lagrange multiplier example, part 2.mp3
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And what that's gonna give me, and I'll go ahead and write it over here, kind of all over the place, U to the 1 third is equal to, let's see, 3 over 200, so that 20 almost cancels out with the 200, it just leaves a 10, so that's gonna give me 3 tenths of lambda. And then similarly over here, I'm gonna take this whole equation and multiply it by 3 over 100, and what that's gonna leave me with is that U to the negative 2 thirds, U to the negative 2 thirds is equal to, let's see, this 2000 when we divide by 100 becomes 20, and that 20 times 3 is 60, so that'll be 60 times lambda. Alright, so now I want a way to simplify each of these, and what I notice is they look pretty similar on each side, you know, it's something related to U equal to lambda, so if I can get this in a form where I'm really isolating U, that would be great. And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that. So when I do that to the top part, like I said, that U to the 1 third times U to the 2 thirds ends up being U, and then on the right side, we have our 3 tenths lambda, but now U to the 2 thirds. And then on the bottom here, when I multiply it by U to the 2 thirds, that right side becomes 1, because it cancels out with U to the negative 2 thirds, and the right side is 60 times lambda times U to the 2 thirds. Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side.
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Lagrange multiplier example, part 2.mp3
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And the way I'm gonna do this is I'm gonna multiply each one of them by U to the 2 thirds, so I'm gonna multiply it into this guy, and I'm gonna multiply it into that guy, because on the top, it's gonna turn this into just U, which will be nice, and on the bottom, it'll cancel out that U entirely, so it feels like it'll make both of these nicer, even though it might make the right side a little uglier, those right sides will still kind of be the same, and we'll take advantage of that. So when I do that to the top part, like I said, that U to the 1 third times U to the 2 thirds ends up being U, and then on the right side, we have our 3 tenths lambda, but now U to the 2 thirds. And then on the bottom here, when I multiply it by U to the 2 thirds, that right side becomes 1, because it cancels out with U to the negative 2 thirds, and the right side is 60 times lambda times U to the 2 thirds. Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side. So in this case, I'm gonna multiply that top by 10, which will get it to 3, and then by another 20 to make that constant 60. So I'm gonna multiply this entire top equation by 200, and what that gives me is that 200 times U is equal to 60 times lambda times U to the 2 thirds. And now these two equations, these two equations have the same right side.
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Lagrange multiplier example, part 2.mp3
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Now these right sides look very similar, and the left sides are quite simple, so I'm gonna multiply this top one by whatever it takes to make it look exactly like that right side. So in this case, I'm gonna multiply that top by 10, which will get it to 3, and then by another 20 to make that constant 60. So I'm gonna multiply this entire top equation by 200, and what that gives me is that 200 times U is equal to 60 times lambda times U to the 2 thirds. And now these two equations, these two equations have the same right side. So this is the same as saying 200 times U is equal to, well, 1, because each one of those expressions equals the same complicated thing. And now 200 times U, well that's S divided by H. So this is the same thing as saying 200 times S over H equals 1, which we can write more succinctly as H is equal to 200 times S. Great, so I'm gonna go ahead and circle that. H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget.
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Lagrange multiplier example, part 2.mp3
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And now these two equations, these two equations have the same right side. So this is the same as saying 200 times U is equal to, well, 1, because each one of those expressions equals the same complicated thing. And now 200 times U, well that's S divided by H. So this is the same thing as saying 200 times S over H equals 1, which we can write more succinctly as H is equal to 200 times S. Great, so I'm gonna go ahead and circle that. H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget. I'll go ahead and kind of write that down again. That our 20 times H, I think, 20 times the hours of labor plus $2,000 per ton of steel is equal to our budget of $20,000. And now we can just substitute in.
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Lagrange multiplier example, part 2.mp3
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H is equal to 200 times S. And now what we apply that to is the constraint, is the 200 times H plus 2,000 times S equals our budget. I'll go ahead and kind of write that down again. That our 20 times H, I think, 20 times the hours of labor plus $2,000 per ton of steel is equal to our budget of $20,000. And now we can just substitute in. Instead of H, I'm gonna write 200 S. So that's 200, sorry, 20 times 200 S, 200 S, plus 2,000 times S is equal to 20,000. And now this right side, 20 times 200 is equal to 4,000. And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000.
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Lagrange multiplier example, part 2.mp3
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And now we can just substitute in. Instead of H, I'm gonna write 200 S. So that's 200, sorry, 20 times 200 S, 200 S, plus 2,000 times S is equal to 20,000. And now this right side, 20 times 200 is equal to 4,000. And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000. 6,000 times S is equal to 20,000. And when those cancel out, what that gives us is S is equal to 20 divided by six, which is the same as 10 divided by three. So that's how many tons of steel we should get.
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Lagrange multiplier example, part 2.mp3
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And I'm just gonna go ahead and kind of write, so this here is 4,000 S. So the entire right side of the equation simplifies to 6,000. 6,000 times S is equal to 20,000. And when those cancel out, what that gives us is S is equal to 20 divided by six, which is the same as 10 divided by three. So that's how many tons of steel we should get. S is 10 over three. Then when we apply that to the fact that H is 200 times S, that's gonna mean that H is equal to 200 times that value, 10 over three, which is equal to 2,000, 2,000 thirds. 2,000 thirds, that's how many hours of labor we want.
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Lagrange multiplier example, part 2.mp3
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So a vector field is a function, I'll just do a two dimensional example here, is gonna be something that has a two dimensional input, and then the output has the same number of dimensions, that's the important part. And each of these components in the output is gonna depend somehow on the input variables, so the example I have in mind will be x times y as that first component, and then y squared minus x squared as that second component. And you can compute the partial derivative of a guy like this, right? You'll take the partial derivative with respect to one of the input variables, I'll choose x, it's always a nice one to start with, partial derivative with respect to x. And if we were to actually compute it in this case, it's another, it's a function of x and y, what you do is you take the partial derivative component wise, so you go to each component, and the first one you say, okay, x looks like a variable, y looks like a constant, the derivative will just be that constant, and then the partial derivative of the second component, that y squared looks like a constant, derivative of negative x squared with respect to x, negative two x. So analytically, if you know how to take a partial derivative, you already know how to take a partial derivative of vector-valued functions, and hence vector fields, but the fun part, the important part here, how do you actually interpret this? And this has everything to do with visualizing it in some way, so the vector field, the reason we call it a vector field, is you kind of take the whole xy-plane, and you're gonna fill it with vectors, and concretely, what I mean by that, you'll take a given input, what's an input you wanna look at?
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Partial derivatives of vector fields.mp3
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You'll take the partial derivative with respect to one of the input variables, I'll choose x, it's always a nice one to start with, partial derivative with respect to x. And if we were to actually compute it in this case, it's another, it's a function of x and y, what you do is you take the partial derivative component wise, so you go to each component, and the first one you say, okay, x looks like a variable, y looks like a constant, the derivative will just be that constant, and then the partial derivative of the second component, that y squared looks like a constant, derivative of negative x squared with respect to x, negative two x. So analytically, if you know how to take a partial derivative, you already know how to take a partial derivative of vector-valued functions, and hence vector fields, but the fun part, the important part here, how do you actually interpret this? And this has everything to do with visualizing it in some way, so the vector field, the reason we call it a vector field, is you kind of take the whole xy-plane, and you're gonna fill it with vectors, and concretely, what I mean by that, you'll take a given input, what's an input you wanna look at? Like, I'll say maybe one, two, or yeah, let's do that, let's do one, two, which would mean you kinda go x equals one, and then y equals two, this input point, and we wanna associate that with the output vector in some way, and so let's just compute what it should equal, so when we plug in x equals one and y equals two, x times y becomes two, y squared minus x squared becomes two squared minus one squared, so four minus one is three, so we have this vector two, three that we want to associate with that input point, and vector fields, you just attach the two points, you just, I'm gonna take the vector two, three and attach it to this guy, so it should have an x component of two, and then a y component of three, so it's gonna end up looking something like, let's see, so, y component of three, something like this, so that'll be the vector, and we attach it to that point, and in principle, you do this to all of the different points, and if you did, what you'd get would be something like this, and remember, when we represent these, especially with computers, it tends to lie where each represented vector is much, much shorter than it should be in reality, but you just wanna squish them all onto the same page so they don't overrun each other, and here, color is supposed to give a general vague sense of relative length, so ones that are blue should be thought of as much shorter than the ones that are yellow, but that doesn't really give a specific thought for how long they should be, but for partial derivatives, we actually care a lot about the specifics, and if you think back to how we interpret partial derivatives in a lot of other contexts, what we want to do is imagine this partial x here as a slight nudge in the x direction, right, so this was our original input, and you might imagine just nudging it a little bit, and the size of that nudge as a number would be your partial x, so then the question is, what's the resulting change to the output, and because the output is a vector, the change in the output is also gonna be a vector, so what we want is to say there's gonna be some other vector attached to this point, right, it's gonna look very similar, maybe it looks like, maybe it looks something like this, so something similar, but maybe a little bit different, and you wanna take that difference in vector form, and I'll describe what I mean by that in just a moment, and then divide it by the size of that original nudge, so to be much more specific about what I mean here, if you're comparing two different vectors, and they're rooted in two different spots, I think a good way to start is to just move them to a new space where they're rooted in the same spot, so in this case, I'm just gonna kinda draw a separate space over here, and be thinking of this as a place for these vectors to live, and I'm gonna put them both on this plane, but I'm gonna root them each in the origin, so this first one that has components two, three, let's give it a name, let's call this guy V1, so that'll be V1, and then the nudged output, the second one, I'll call V2, V2, and let's say V2 is also in this space, and I might exaggerate the difference just so that we can see it here, let's say it was different in some way, in reality, if it's a small nudge, it'll be different in only a very small way, but let's say these were our two vectors, the difference between these guys is gonna be a vector that connects the tips, and I'm gonna call that guy partial V, and the way that you can be thinking about this is to say that V1, V1, that original guy, plus that tiny nudge, the difference between them, is equal to V2, the nudged output, and in terms of tip to tail with vectors, you're seeing that, kinda the green vector plus that blue vector is the same as that pink vector that connects the tail of the original one to the tip of the new one, so when we're thinking of a partial derivative, you're basically saying, hey, what happens if we take this, the size of the nudge of the output, and then we divide it by that nudge of the input, so let's say you were thinking of that original nudge as being, I don't know, like of a size 1 1 2, like 0.5, as the change in the X direction, then that would mean, when you go over here and you say, what's that DV, that changing vector V, divided by DX, you'd be dividing it by 0.5, and in principle, you'd be thinking of, that would mean that you're kind of scaling this by two, as if to say this little DV is 1 1 2 of some other vector, and that other vector is what the partial derivative is, so this other vector here, the full blue guy, would be DV, you know, scaled down, or scaled up, however you wanna think about it, by that partial X, and that's what makes it such that, you know, in principle, if this partial X change was really small, it was like 1 1 100th, and the output nudge also was really small, it was like 1 1 100th, or you know, something on that order, it wouldn't be specifically that, then the DV DX, that change, would still be a normal sized vector, and the direction that it points is still kind of an indication of the direction that this green vector should change as you're scooting over. So just to be concrete, and you know, actually compute this guy, let's say we were to take this partial derivative, partial V with respect to X, and evaluate it at that point 1 2, that we're just dealing with, 1 2. What that would mean, Y is equal to 2, so that first component is 2, and then X is equal to 1, so that next one should be negative 2, and we can, I guess we can see just how wrong my drawing was to start here, I was just kind of guessing what the pink vector would be, but I guess it changes in the direction of 2 negative 2, so that should be something, here I'll kind of erase the, what turns out to be the wrong direction here, but, get rid of this guy, and I guess the change should be in the direction kind of 2 as the X component, and then negative 2, that's a negative 2 as the Y component, so the derivative vector should look something like this, which means all corresponding little DV nudges will be slight changes, will be slight changes on that, so these will be your DVs, something in that direction, and what that means in our vector field then, as you move in the X direction, and consider the various vectors attached to each point, as you're kind of passing through the point 1, 2, the way that the vectors are changing should be somehow down and to the right, the tip should move down and to the right, so if this starts highly up and to the right, then it should be getting kind of shorter, but then longer to the right, so the V2, if I were to have drawn it more accurately here, you know, what that nudged output should look like, it would really be something that's kind of like, I don't know, like this, where it's getting shorter in the Y direction, but then longer in the X direction, as per that blue nudging arrow, and in the next video, I'll kind of go through more examples of how you might think of this, how you think of it in terms of what each component means, which becomes very important for later topics, like divergence and curl, and I'll see you next video.
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Partial derivatives of vector fields.mp3
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I think I did it over several videos because it was a bit hairy. And I'll write our position vector valued function first. So we have r as a function of our two parameters s and t. And then I'll review a little bit of what all the terms, what the s, the t, and the a's and the b's represent. But it's equal to b plus a cosine of s. And once again, we saw this several videos ago. So you might want to watch the videos on parameterizing surfaces with two parameters to figure out how we got here. Times the sine of t. I'm going to put the s terms and the t terms in different colors. Times our i unit vector.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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But it's equal to b plus a cosine of s. And once again, we saw this several videos ago. So you might want to watch the videos on parameterizing surfaces with two parameters to figure out how we got here. Times the sine of t. I'm going to put the s terms and the t terms in different colors. Times our i unit vector. I'll put the vectors or the unit vectors in this orange color. Plus, doing the same yellow. Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Times our i unit vector. I'll put the vectors or the unit vectors in this orange color. Plus, doing the same yellow. Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction. Plus a sine of s. Times the k unit vector, the unit vector in the z direction. And in order to generate the torus or the donut shape, this is true for our parameters. So we don't wrap multiple times around the torus.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Plus b plus a cosine of s times cosine of t. Times the j unit vector, the unit vector in the y direction. Plus a sine of s. Times the k unit vector, the unit vector in the z direction. And in order to generate the torus or the donut shape, this is true for our parameters. So we don't wrap multiple times around the torus. For s being between 0 and 2 pi. And for t being between 0 and 2 pi. And just as a bit of a review of where all of this came from.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So we don't wrap multiple times around the torus. For s being between 0 and 2 pi. And for t being between 0 and 2 pi. And just as a bit of a review of where all of this came from. And I'm going to have to do what my plan is for this video over several videos. But let's review where all of this came from. Let me draw a donut.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And just as a bit of a review of where all of this came from. And I'm going to have to do what my plan is for this video over several videos. But let's review where all of this came from. Let me draw a donut. My best effort at a donut right here. That looks like a donut or a torus. And you can imagine a torus or this donut shape is kind of the product of two circles.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Let me draw a donut. My best effort at a donut right here. That looks like a donut or a torus. And you can imagine a torus or this donut shape is kind of the product of two circles. You have this circle that's kind of the cross section of the donut at any point. You can take it there. You can take it over there.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And you can imagine a torus or this donut shape is kind of the product of two circles. You have this circle that's kind of the cross section of the donut at any point. You can take it there. You can take it over there. And then you have the circle that kind of wraps around all of these other circles. Or these other circles wrap around it. And so when we derived this formula up here, this parameterization.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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You can take it over there. And then you have the circle that kind of wraps around all of these other circles. Or these other circles wrap around it. And so when we derived this formula up here, this parameterization. A was the radius of these cross sectional circles. That's A. That's what these A terms were.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And so when we derived this formula up here, this parameterization. A was the radius of these cross sectional circles. That's A. That's what these A terms were. And B was the distance from the center of our torus. B was the distance from the center of our torus out to the center of these cross sections. So this was B.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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That's what these A terms were. And B was the distance from the center of our torus. B was the distance from the center of our torus out to the center of these cross sections. So this was B. So you can imagine that B is kind of the radius of the big circle up to the midpoint of the cross section. And A is the radius of the cross sectional circles. And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So this was B. So you can imagine that B is kind of the radius of the big circle up to the midpoint of the cross section. And A is the radius of the cross sectional circles. And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle. So it's an angle from 0 to 2 pi to say where we are in that circle. And T tells us how much we've rotated around the larger circle. So T was telling us how much we rotated around the larger circle.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And when we parameterized it, the parameter S was essentially telling us how far are we wrapping around this circle. So it's an angle from 0 to 2 pi to say where we are in that circle. And T tells us how much we've rotated around the larger circle. So T was telling us how much we rotated around the larger circle. So if you think about it, you can specify any point on this donut or on this surface or on this torus by telling you an S or a T. And so that's why we picked that as the parameterization. Now, the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral. And the surface integral we're going to compute will tell us the surface area of this torus.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So T was telling us how much we rotated around the larger circle. So if you think about it, you can specify any point on this donut or on this surface or on this torus by telling you an S or a T. And so that's why we picked that as the parameterization. Now, the whole reason why I'm even revisiting this stuff that we saw several videos ago is we're going to actually use it to compute an actual surface integral. And the surface integral we're going to compute will tell us the surface area of this torus. So this surface right here is sigma, right like that. It's being represented by this position vector-valued function that is parameterized by these two parameters right there. And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And the surface integral we're going to compute will tell us the surface area of this torus. So this surface right here is sigma, right like that. It's being represented by this position vector-valued function that is parameterized by these two parameters right there. And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface. Here this capital sigma does not represent sum. It represents a surface of a bunch of the little d sigmas, a bunch of the little chunks of the surface. And just as a review, you can imagine each d sigma is a little patch of the surface right there.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And if we wanted to figure out the surface area, if we just kind of set it as a surface integral, we saw in I think the last video, or at least the last vector calculus video I did, that this is a surface integral over the surface. Here this capital sigma does not represent sum. It represents a surface of a bunch of the little d sigmas, a bunch of the little chunks of the surface. And just as a review, you can imagine each d sigma is a little patch of the surface right there. That is a d sigma. And we're taking, it's a double integral here because we want to add up all the d sigmas in two directions. You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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And just as a review, you can imagine each d sigma is a little patch of the surface right there. That is a d sigma. And we're taking, it's a double integral here because we want to add up all the d sigmas in two directions. You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus. So that's why it's a double integral. And this is just going to give you the surface area, which is the whole point of this video and probably the next one or two videos. But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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You can imagine one kind of rotating this way around the torus, and then the other direction is going in the other direction around the torus. So that's why it's a double integral. And this is just going to give you the surface area, which is the whole point of this video and probably the next one or two videos. But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there. But here we're just multiplying it by 1. And we saw in the last video that this is a way of expressing an idea, but you really can't do much computation with this. But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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But if you wanted to also multiply these sigmas times some other value, there's some scalar field that this is in that you cared about, you could put that other value right there. But here we're just multiplying it by 1. And we saw in the last video that this is a way of expressing an idea, but you really can't do much computation with this. But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos. This is the same thing as the double integral over the region over which our parameters are defined. So it's this region over here where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we don't have to.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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But a way that you can express this so that you can actually take the integral, you say this is the same thing, and we saw this in the last several videos. This is the same thing as the double integral over the region over which our parameters are defined. So it's this region over here where s and t go from 0 to 2 pi of whatever function this is. We just have a 1 here, so we don't have to. We could just write a 1 if we like. It doesn't change much. Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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We just have a 1 here, so we don't have to. We could just write a 1 if we like. It doesn't change much. Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt. So we saw this in the last video. What we're going to do now is actually compute this. That's the whole point of this video.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Times, and this is what we learned, times the magnitude of the partial derivative of r with respect to s. The magnitude of that crossed with the partial derivative of r with respect to t. ds, you can take it in either order, but ds dt. So we saw this in the last video. What we're going to do now is actually compute this. That's the whole point of this video. We're going to take the cross product of these two vectors. So let's figure out these vectors. Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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That's the whole point of this video. We're going to take the cross product of these two vectors. So let's figure out these vectors. Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral. You're going to see it's a pretty hairy problem, and this is probably the reason that very few people ever see an actual surface integral get computed. But let's do it anyway. So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Then in the next video we're going to take the cross product, and then the video after that we'll actually evaluate this double integral. You're going to see it's a pretty hairy problem, and this is probably the reason that very few people ever see an actual surface integral get computed. But let's do it anyway. So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video. This term is what? We just want to hold t constant and take the partial with respect to just s. So this up here, if we distribute it, the sine of t times b, that's just going to be a constant in terms of s, so we can ignore that. Then you have sine of t times this over here.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So the partial derivative of r with respect to s, so this term right here, we'll do the cross product in the next video. This term is what? We just want to hold t constant and take the partial with respect to just s. So this up here, if we distribute it, the sine of t times b, that's just going to be a constant in terms of s, so we can ignore that. Then you have sine of t times this over here. So sine of t and a is a constant, and you take the derivative of cosine of s, that's negative sine of s. So it's going to be, so the derivative of this with respect to s, so the partial with respect to s, is going to be minus a. I'll write in green the sine of t so you know that's where it came from. Sine of t and then sine of s. The derivative of this is negative sine of s, that's where that negative came from, and then I'll write the sine of s right there. Sine of s times the unit vector i.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Then you have sine of t times this over here. So sine of t and a is a constant, and you take the derivative of cosine of s, that's negative sine of s. So it's going to be, so the derivative of this with respect to s, so the partial with respect to s, is going to be minus a. I'll write in green the sine of t so you know that's where it came from. Sine of t and then sine of s. The derivative of this is negative sine of s, that's where that negative came from, and then I'll write the sine of s right there. Sine of s times the unit vector i. That's the partial of just this x term with respect to s. And then we'll do the same thing with the y term or the j term. So plus, same logic, b times cosine of t with respect to s, we take the partial just to become zero. So you're left with a, well it's going to be a minus a again, right?
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Sine of s times the unit vector i. That's the partial of just this x term with respect to s. And then we'll do the same thing with the y term or the j term. So plus, same logic, b times cosine of t with respect to s, we take the partial just to become zero. So you're left with a, well it's going to be a minus a again, right? Because when you take the derivative of the cosine of s, it's going to be negative sine of s. So you have a, let me do it. So you're going to have a minus a, this cosine of t, minus a cosine of t, that's the constant terms, sine of s. Just taking partial derivatives. Sine of s j.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So you're left with a, well it's going to be a minus a again, right? Because when you take the derivative of the cosine of s, it's going to be negative sine of s. So you have a, let me do it. So you're going to have a minus a, this cosine of t, minus a cosine of t, that's the constant terms, sine of s. Just taking partial derivatives. Sine of s j. And then finally we take the derivative of this with respect to s, and that's pretty straight forward. It's just going to be a cosine of s. So plus a cosine of s k. Now hopefully you didn't find this confusing. The negative signs, because the derivative of cosines are negative signs.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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Sine of s j. And then finally we take the derivative of this with respect to s, and that's pretty straight forward. It's just going to be a cosine of s. So plus a cosine of s k. Now hopefully you didn't find this confusing. The negative signs, because the derivative of cosines are negative signs. So negative sine of s, so that's why it's negative sine of s times the constants. Negative sine of s times the constants. The constant cosine of t, sine of t. So hopefully this makes some sense.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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The negative signs, because the derivative of cosines are negative signs. So negative sine of s, so that's why it's negative sine of s times the constants. Negative sine of s times the constants. The constant cosine of t, sine of t. So hopefully this makes some sense. This is a review of taking a partial derivative. Now let's do the same thing with respect to t. Let's do the same thing with respect to t, and I'll do that in a different color. So now we want to take the partial of r with respect to t. So the partial of r with respect to t is equal to, so now this whole term over here is a constant, and so it's going to be that whole term times the derivative of this with respect to t, which is just cosine of t. So it's going to be b plus a cosine of s times cosine of t. Cosine of t i.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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The constant cosine of t, sine of t. So hopefully this makes some sense. This is a review of taking a partial derivative. Now let's do the same thing with respect to t. Let's do the same thing with respect to t, and I'll do that in a different color. So now we want to take the partial of r with respect to t. So the partial of r with respect to t is equal to, so now this whole term over here is a constant, and so it's going to be that whole term times the derivative of this with respect to t, which is just cosine of t. So it's going to be b plus a cosine of s times cosine of t. Cosine of t i. And then plus, and it's actually going to be a minus, because when you take the derivative of this with respect to t, it's going to be minus sine of t. So let me put them, it's going to be negative, and then let me leave some space for this term right here, negative sine of t, and you're going to have this constant out there, that's a constant in t. B plus a cosine of s, that's just that term right there, the derivative of cosine t is negative sine of t times j, and then the partial of this with respect to t, this is just a constant in t, so the partial is going to be zero. So I'll write plus zero k. Plus zero, let me do all my vectors in that same color, plus zero times the unit vector k. So that gives us our partial derivatives. Now we have to take the cross product, then find the magnitude of the cross product, and then evaluate this double integral, and I'll do that in the next couple of videos.
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Example of calculating a surface integral part 1 Multivariable Calculus Khan Academy.mp3
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So you're looking just for where it's tangent, where you can find a tangent plane that's flat. But this will give you some other points, like the little local minima here, the bumps, where the value of the function at that point is higher than all of the neighbor points, you know, if you walk in any direction, you're going downhill, so that's another thing you're gonna, you know, incidentally pick up by looking for places where this tangent plane is flat. But there's also a really interesting new possibility that comes up in the context of multivariable functions. And this is the idea of a saddle point. So let me pull up another graph here, this guy. And the function that you're looking at, here, I'll write it down, the function that you're looking at is f of x, y is equal to x squared minus y squared, okay? So now let's think about what the tangent plane at the origin of this entire graph would be.
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Saddle points.mp3
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And this is the idea of a saddle point. So let me pull up another graph here, this guy. And the function that you're looking at, here, I'll write it down, the function that you're looking at is f of x, y is equal to x squared minus y squared, okay? So now let's think about what the tangent plane at the origin of this entire graph would be. Now, the tangent plane to this graph at the origin is actually flat, here's what it looks like. And to convince yourself of this, let's go ahead and actually compute the partial derivatives of this function and evaluate each one at the origin. So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x.
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Saddle points.mp3
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So now let's think about what the tangent plane at the origin of this entire graph would be. Now, the tangent plane to this graph at the origin is actually flat, here's what it looks like. And to convince yourself of this, let's go ahead and actually compute the partial derivatives of this function and evaluate each one at the origin. So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x. And the other partial derivative, the partial with respect to y, with respect to y, we take the derivative of this negative y squared and we ignore the x because it looks like a constant as far as y is concerned, and we get negative two y. Now, if we plug in the point, the origin, into any one of these, you know, we plug in the point x, y is equal to zero, zero, then what do each of these go to? Well, the top one, x equals zero, so this guy goes to zero.
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Saddle points.mp3
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So the partial derivative with respect to x, we look here and x squared is the only spot where an x shows up, so it's two x. And the other partial derivative, the partial with respect to y, with respect to y, we take the derivative of this negative y squared and we ignore the x because it looks like a constant as far as y is concerned, and we get negative two y. Now, if we plug in the point, the origin, into any one of these, you know, we plug in the point x, y is equal to zero, zero, then what do each of these go to? Well, the top one, x equals zero, so this guy goes to zero. And similarly over here, y is zero, so this goes to zero. So both partial derivatives are zero. And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement.
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Saddle points.mp3
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Well, the top one, x equals zero, so this guy goes to zero. And similarly over here, y is zero, so this goes to zero. So both partial derivatives are zero. And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement. And one way of seeing this is to chop the graph. So if we imagine chopping it with a plane that represents a constant x value, and we kind of chop off the graph there, what you'll see, here I'll get rid of the tangent plane, what you'll see is that the curve where this intersects the graph, so let me trace that out, the curve where it intersects the graph basically has a local maximum at that origin point. The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective.
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Saddle points.mp3
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And what that means is if you are standing at the origin and you move in any direction, there's no slope to your movement. And one way of seeing this is to chop the graph. So if we imagine chopping it with a plane that represents a constant x value, and we kind of chop off the graph there, what you'll see, here I'll get rid of the tangent plane, what you'll see is that the curve where this intersects the graph, so let me trace that out, the curve where it intersects the graph basically has a local maximum at that origin point. The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective. But now let's imagine chopping it in a different direction. So if instead we have the full graph and instead of chopping it with a constant x value, we chop it with a constant y value, then in that case, we look at the curve, and if we trace out the curve where this constant y value intersects the graph, let's see what it would look like, it's also kind of this parabolic shape, again, the tangent line is flat because it looks like a local minimum of that curve. So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat.
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Saddle points.mp3
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The tangent line of the curve at that point in the y direction is flat, and it's because it looks like a local maximum from that perspective. But now let's imagine chopping it in a different direction. So if instead we have the full graph and instead of chopping it with a constant x value, we chop it with a constant y value, then in that case, we look at the curve, and if we trace out the curve where this constant y value intersects the graph, let's see what it would look like, it's also kind of this parabolic shape, again, the tangent line is flat because it looks like a local minimum of that curve. So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat. But notice, this is neither a local maximum nor a local minimum, because from one direction, from one direction, it looked like it was a local maximum. Yeah, I'll get rid of that guy. It looks like it's a local maximum when you look on the curve there.
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Saddle points.mp3
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So because it's flat in one direction and it's flat in the other direction, the tangent plane of the graph as a whole is indeed gonna be flat. But notice, this is neither a local maximum nor a local minimum, because from one direction, from one direction, it looked like it was a local maximum. Yeah, I'll get rid of that guy. It looks like it's a local maximum when you look on the curve there. But from another direction, if you chop it in another way, it looks like a local minimum. And if we look at the equations, this kind of makes sense, because if you're just thinking about movements in the x direction, the entire function looks like x squared plus some kind of constant. So the graph of that would look like an x squared parabola shape that has a local minimum.
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Saddle points.mp3
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It looks like it's a local maximum when you look on the curve there. But from another direction, if you chop it in another way, it looks like a local minimum. And if we look at the equations, this kind of makes sense, because if you're just thinking about movements in the x direction, the entire function looks like x squared plus some kind of constant. So the graph of that would look like an x squared parabola shape that has a local minimum. But if you're thinking of pure movements in the y direction, and you're just focused on that negative y squared component, the graph that you get for negative y squared is gonna look like an upside down parabola. Here, I'll draw that again. It's gonna look like an upside down parabola, and that's got a local maximum.
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Saddle points.mp3
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So the graph of that would look like an x squared parabola shape that has a local minimum. But if you're thinking of pure movements in the y direction, and you're just focused on that negative y squared component, the graph that you get for negative y squared is gonna look like an upside down parabola. Here, I'll draw that again. It's gonna look like an upside down parabola, and that's got a local maximum. So it's kind of like the x and y directions disagree over whether this point, whether this point where you have a flat tangent plane should be a local maximum or a local minimum. And this is new to multivariable calculus. This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum.
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Saddle points.mp3
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It's gonna look like an upside down parabola, and that's got a local maximum. So it's kind of like the x and y directions disagree over whether this point, whether this point where you have a flat tangent plane should be a local maximum or a local minimum. And this is new to multivariable calculus. This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum. It can't disagree, because there's only one input variable. There's only one, you know, x as the input variable for your graph. But once we have two, it's possible that they disagree.
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Saddle points.mp3
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This is something that doesn't come up in single variable calculus, because when you're looking at the graph of a function, you know, you're looking at some kind of graph, if the tangent line is zero, you know, if the tangent line is completely flat at some point, either it's a local maximum or it's a local minimum. It can't disagree, because there's only one input variable. There's only one, you know, x as the input variable for your graph. But once we have two, it's possible that they disagree. And this kind of point has a special name. And the name is kind of after this graph that you're looking at. It's called a saddle point.
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Saddle points.mp3
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But once we have two, it's possible that they disagree. And this kind of point has a special name. And the name is kind of after this graph that you're looking at. It's called a saddle point. Saddle point. And this is one of those rare times where I actually kind of like the terminology that mathematicians have given something, because this looks like a saddle, the sort of thing that you would put on a horse's back before riding it. So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point.
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Saddle points.mp3
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It's called a saddle point. Saddle point. And this is one of those rare times where I actually kind of like the terminology that mathematicians have given something, because this looks like a saddle, the sort of thing that you would put on a horse's back before riding it. So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point. And if you're just looking at the graph, it's fine. You can recognize it visually. But oftentimes, if you're just given the formula of a function, it's some long thing.
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Saddle points.mp3
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So one thing that this means for us is we're gonna try to figure out ways to find the absolute maximum or the minimum of a function as we're trying to optimize a function that might represent, you know, represent like profits of your company or a cost function in a machine learning setting or something like that, is we're gonna have to be able to recognize if a point is a saddle point. And if you're just looking at the graph, it's fine. You can recognize it visually. But oftentimes, if you're just given the formula of a function, it's some long thing. Without looking at the graph, how would you be able to tell, just by doing certain computations to the formula, whether or not it's a saddle point? And that comes down to something called the second partial derivative test, which I will talk about in the next few videos. See you then.
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Saddle points.mp3
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And so the first thing I'm going to do is figure out what d sigma is in terms of s and t, in terms of our parameters. So we can turn this whole thing into a double integral with respect to, or a double integral in the st plane. And remember, d sigma right over here, it's just a little chunk of the surface. It's a little area of the surface right over there. And we saw in previous videos, the ones where we learned what a surface integral even is, we saw that d sigma right over there, it is equivalent to the magnitude of the cross product of the partial of our parameterization with respect to one parameter, crossed with the parameterization with respect to the other parameter, times the differentials of each of the parameters. So this is what we're going to use right here. And it's a pretty simple looking statement.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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It's a little area of the surface right over there. And we saw in previous videos, the ones where we learned what a surface integral even is, we saw that d sigma right over there, it is equivalent to the magnitude of the cross product of the partial of our parameterization with respect to one parameter, crossed with the parameterization with respect to the other parameter, times the differentials of each of the parameters. So this is what we're going to use right here. And it's a pretty simple looking statement. But as we'll see, taking cross products tend to get a little bit hairy, especially cross products of three dimensional vectors. But we'll do it step by step. But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And it's a pretty simple looking statement. But as we'll see, taking cross products tend to get a little bit hairy, especially cross products of three dimensional vectors. But we'll do it step by step. But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant. So cosine of t isn't going to change. The partial of, or the derivative of cosine of s with respect to s is negative sine of s. So this is going to be equal to, I'll put the negative out front, negative cosine of t sine of s. I'm going to keep everything that has a t involved purple. Sine of s. And let me make, I don't know, I'll make the vectors orange.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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But before we even take the cross product, we first have to take the partial of this with respect to s, and then the partial of this with respect to t. So first let's take the partial with respect to s. The partial of r with respect to s. So right over here, all the stuff with the t in it, you can just view that as a constant. So cosine of t isn't going to change. The partial of, or the derivative of cosine of s with respect to s is negative sine of s. So this is going to be equal to, I'll put the negative out front, negative cosine of t sine of s. I'm going to keep everything that has a t involved purple. Sine of s. And let me make, I don't know, I'll make the vectors orange. I, and then we'll plus, and once again, we're going to take the derivative with respect to s. Cosine of t is just a constant. Derivative of sine of s with respect to s is cosine of s. So it's going to be plus cosine of t cosine of s. Cosine of t cosine of s. J. And then plus the derivative of this with respect to s, well, this is just a constant.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Sine of s. And let me make, I don't know, I'll make the vectors orange. I, and then we'll plus, and once again, we're going to take the derivative with respect to s. Cosine of t is just a constant. Derivative of sine of s with respect to s is cosine of s. So it's going to be plus cosine of t cosine of s. Cosine of t cosine of s. J. And then plus the derivative of this with respect to s, well, this is just a constant. The derivative of 5 with respect to s would just be 0. This is the same thing. This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And then plus the derivative of this with respect to s, well, this is just a constant. The derivative of 5 with respect to s would just be 0. This is the same thing. This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0. So we could write even 0k. We write 0, I'll just write 0k right over there. And that's nice to see because that'll make our cross product a little bit more straightforward.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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This is a constant with, this does not change with respect to s. So it does, so our derivative, our partial here with respect to s is just 0. So we could write even 0k. We write 0, I'll just write 0k right over there. And that's nice to see because that'll make our cross product a little bit more straightforward. Now let's take the partial with respect to t. Now let's take the partial with respect to t. And we get, so the derivative of this with respect to t, now cosine of s is a constant. Derivative of cosine t with respect to t is negative sine of t. So this is going to be negative sine of t cosine of s. I'll do it in that, I'll use this blue i. And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And that's nice to see because that'll make our cross product a little bit more straightforward. Now let's take the partial with respect to t. Now let's take the partial with respect to t. And we get, so the derivative of this with respect to t, now cosine of s is a constant. Derivative of cosine t with respect to t is negative sine of t. So this is going to be negative sine of t cosine of s. I'll do it in that, I'll use this blue i. And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this. This is a painful problem. J plus derivative of sine of t with respect to t. We're taking the partial with respect to t. It's just cosine of t. So plus cosine of t. And now times the k unit vector. Now we're ready to take the cross product of these two characters right over here.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And then plus, now derivative of this with respect to t, derivative of cosine of t is negative sine of t. So once again, so now we have minus sine of t sine of s. My hand is already hurting from this. This is a painful problem. J plus derivative of sine of t with respect to t. We're taking the partial with respect to t. It's just cosine of t. So plus cosine of t. And now times the k unit vector. Now we're ready to take the cross product of these two characters right over here. And to take the cross products, let me write this down. So we're going to take the cross product of that with that. Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Now we're ready to take the cross product of these two characters right over here. And to take the cross products, let me write this down. So we're going to take the cross product of that with that. Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff. So maybe I'll take up about that much space so that I have space to work in. And I'll write my unit vectors up here. i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Is going to be equal to, and I'm going to set up this huge matrix, or it's really just a 3 by 3 matrix, but it's going to be huge because I'm going to take up a lot of space to have to write down all this stuff. So maybe I'll take up about that much space so that I have space to work in. And I'll write my unit vectors up here. i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors. Take the determinant of this 3 by 3 matrix. The first row are just our unit vectors. The second row is the first vector that I'm taking the cross product of.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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i, j, k, or at least this is how I like to remember how to take cross products of three dimensional vectors. Take the determinant of this 3 by 3 matrix. The first row are just our unit vectors. The second row is the first vector that I'm taking the cross product of. So this is going to be negative. I'm just going to rewrite this right over here. So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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The second row is the first vector that I'm taking the cross product of. So this is going to be negative. I'm just going to rewrite this right over here. So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations. And then you have the next vector. That's the third row. Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So it's going to be negative cosine of t, sine of s. And then you have cosine of t, cosine of s. And then you have 0, which will hopefully simplify our calculations. And then you have the next vector. That's the third row. Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going. It's good practice. And even if you have to watch this whole thing to see how it's done, try to then do it again on your own. Because this is one of those things you really got to do yourself to really have it sit in.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Negative sine of t, cosine of s. And I encourage you to do this on your own if you already know where this is going. It's good practice. And even if you have to watch this whole thing to see how it's done, try to then do it again on your own. Because this is one of those things you really got to do yourself to really have it sit in. Negative sine of t, sine of s. And then finally, cosine of t. So let's take the determinant now. So first we'll think about our I component. You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Because this is one of those things you really got to do yourself to really have it sit in. Negative sine of t, sine of s. And then finally, cosine of t. So let's take the determinant now. So first we'll think about our I component. You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here. So it's going to be I times something. I'll put the something in parentheses there. Normally you see the something in front of the I, but you can swap them there.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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You would essentially ignore this column, the first column in the first row, and then take the determinant of this submatrix right over here. So it's going to be I times something. I'll put the something in parentheses there. Normally you see the something in front of the I, but you can swap them there. So it's going to be I times something. Ignore this column, that row. This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Normally you see the something in front of the I, but you can swap them there. So it's going to be I times something. Ignore this column, that row. This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater. Cosine squared of t, cosine of s. And then from that we would need to subtract 0 times this. But that's just going to be 0, so we're just left with that. Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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This determinant is cosine of t, cosine of s times cosine of t, which is going to be cosine squared of t. Let me write it a little neater. Cosine squared of t, cosine of s. And then from that we would need to subtract 0 times this. But that's just going to be 0, so we're just left with that. Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices. Positive, negative, positive. So you write a negative J, negative coefficient, I guess, in front of the J, times something. And so ignore J's column, J's row.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Now we're going to do the J component, but you probably remember the checkerboard thing where you have to evaluate 3 by 3 matrices. Positive, negative, positive. So you write a negative J, negative coefficient, I guess, in front of the J, times something. And so ignore J's column, J's row. And so you have negative cosine of t, sine of s, times cosine of t. Well, that's going to be negative cosine squared of t times sine of s. Let me make sure I'm doing that right. Ignore that and that. It's going to be that times that.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And so ignore J's column, J's row. And so you have negative cosine of t, sine of s, times cosine of t. Well, that's going to be negative cosine squared of t times sine of s. Let me make sure I'm doing that right. Ignore that and that. It's going to be that times that. So negative cosine sine of s minus 0 times that. And so that's just going to be 0, so we can ignore it. And you have a negative times a negative here, so they can both become a positive.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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It's going to be that times that. So negative cosine sine of s minus 0 times that. And so that's just going to be 0, so we can ignore it. And you have a negative times a negative here, so they can both become a positive. And then finally you have the K component. And once again, you go back to positive there. Positive, negative, positive on the coefficients.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And you have a negative times a negative here, so they can both become a positive. And then finally you have the K component. And once again, you go back to positive there. Positive, negative, positive on the coefficients. That's just evaluating a 3 by 3 matrix. And then you have plus K times. And now this might get a little bit more involved, because we won't have the 0 to help us out.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Positive, negative, positive on the coefficients. That's just evaluating a 3 by 3 matrix. And then you have plus K times. And now this might get a little bit more involved, because we won't have the 0 to help us out. Ignore this row, ignore this column. Take the determinant of this sub 2 by 2. You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And now this might get a little bit more involved, because we won't have the 0 to help us out. Ignore this row, ignore this column. Take the determinant of this sub 2 by 2. You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out. So it's going to be cosine of t times sine squared of s. And then from that, we're going to subtract the product of these two things. But that product is going to be negative. So you subtract a negative.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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You have negative cosine t sine s times negative sine of t sine s. Well, that's going to be the negatives cancel out. So it's going to be cosine of t times sine squared of s. And then from that, we're going to subtract the product of these two things. But that product is going to be negative. So you subtract a negative. That's the same thing as adding a positive. So plus, and you're going to have cosine t sine t again. Plus cosine t. Let me scroll to the right a little bit.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So you subtract a negative. That's the same thing as adding a positive. So plus, and you're going to have cosine t sine t again. Plus cosine t. Let me scroll to the right a little bit. Plus cosine t sine t again. And that's times cosine squared of s. Now this is already looking pretty hairy, but it already looks like a simplification there. And that's where the colors are helpful.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Plus cosine t. Let me scroll to the right a little bit. Plus cosine t sine t again. And that's times cosine squared of s. Now this is already looking pretty hairy, but it already looks like a simplification there. And that's where the colors are helpful. I actually now have trouble doing math in anything other than kind of multiple pastel colors, because this actually makes it much easier to see some patterns. And so what we can do over here is we can factor out the cosine t sine t. And so this is equal to cosine t sine t times sine squared s plus cosine squared s. And this we know, the definition of the unit circle, this is just going to be equal to 1. So that was a significant simplification.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And that's where the colors are helpful. I actually now have trouble doing math in anything other than kind of multiple pastel colors, because this actually makes it much easier to see some patterns. And so what we can do over here is we can factor out the cosine t sine t. And so this is equal to cosine t sine t times sine squared s plus cosine squared s. And this we know, the definition of the unit circle, this is just going to be equal to 1. So that was a significant simplification. And so now we get our cross product. We get our cross product being equal to, let me just rewrite it all. Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So that was a significant simplification. And so now we get our cross product. We get our cross product being equal to, let me just rewrite it all. Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector. So that was pretty good, but we're still not done. We need to figure out the magnitude of this thing. Remember, d sigma simplified to the magnitude of this thing times ds dt.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Our cross product, r sub s crossed with r sub t, is going to be equal to cosine squared t cosine s times our i unit vector plus cosine squared t sine of s times our j unit vector plus, all we have left, because this is just 1, cosine t sine t plus cosine t sine t times our k unit vector. So that was pretty good, but we're still not done. We need to figure out the magnitude of this thing. Remember, d sigma simplified to the magnitude of this thing times ds dt. So let's figure out what the magnitude of that is. And this is really the home stretch. So I'm really crossing my fingers that I don't make any careless mistakes now.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Remember, d sigma simplified to the magnitude of this thing times ds dt. So let's figure out what the magnitude of that is. And this is really the home stretch. So I'm really crossing my fingers that I don't make any careless mistakes now. So the magnitude of all of this business is going to be equal to the square root. And I'm just going to have to square each of these terms and then add them up. The square root of the sum of the squares of each of those terms.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So I'm really crossing my fingers that I don't make any careless mistakes now. So the magnitude of all of this business is going to be equal to the square root. And I'm just going to have to square each of these terms and then add them up. The square root of the sum of the squares of each of those terms. So the square of this is going to be cosine squared squared is cosine to the fourth cosine of the fourth t cosine squared s plus cosine to the fourth t sine squared s. And I already see a pattern jumping out. Sine squared s plus cosine squared t sine squared t. Now, the first pattern I see is just this first part right over here. We can factor out a cosine to the fourth t. Then we get something like this happening again.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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The square root of the sum of the squares of each of those terms. So the square of this is going to be cosine squared squared is cosine to the fourth cosine of the fourth t cosine squared s plus cosine to the fourth t sine squared s. And I already see a pattern jumping out. Sine squared s plus cosine squared t sine squared t. Now, the first pattern I see is just this first part right over here. We can factor out a cosine to the fourth t. Then we get something like this happening again. So let's do that. So these first two terms are equal to cosine to the fourth t times cosine squared s plus sine squared s, which once again we know is just 1. So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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We can factor out a cosine to the fourth t. Then we get something like this happening again. So let's do that. So these first two terms are equal to cosine to the fourth t times cosine squared s plus sine squared s, which once again we know is just 1. So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them. So let's factor those out. So this is going to be equal to everything I'm doing is under the radical sign. So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So this whole expression is simplified to cosine to the fourth t plus cosine squared t sine squared t. Now we can attempt to simplify this again because this term and this term both have a cosine squared t in them. So let's factor those out. So this is going to be equal to everything I'm doing is under the radical sign. So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1. All of this is under the radical sign. Maybe I'll keep drawing the radical signs here to make it clear that all of this is still on the radical sign. And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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So this is going to be equal to a cosine squared t times cosine squared t. And when you factor out a cosine squared t here, you just have a plus a sine squared t. And that's nice because that, once again, simplified to 1. All of this is under the radical sign. Maybe I'll keep drawing the radical signs here to make it clear that all of this is still on the radical sign. And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward. So all of this is just going to be equal to cosine of t. So going back to what we wanted before, we want to rewrite what d sigma is. It's just cosine t ds dt. So let me write that down.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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And then this is really, really useful for us because the square root of cosine squared of t is just going to be cosine of t. So all of that business actually finally simplified to something pretty straightforward. So all of this is just going to be equal to cosine of t. So going back to what we wanted before, we want to rewrite what d sigma is. It's just cosine t ds dt. So let me write that down. So d sigma, and then we can use this for the next part. d sigma is equal to cosine of t ds dt. And I'll see you in the next part.
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Surface integral example part 2 Calculating the surface differential Khan Academy.mp3
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Three dimensional graphs are a way that we represent a certain kind of multivariable function, the kind that has two inputs, or rather a two dimensional input, and then a one dimensional output of some kind. So the one that I have pictured here is f of x, y equals x squared plus y squared. And before talking exactly about this graph, I think it'll be helpful if by analogy we take a look at two dimensional graphs and kind of remind ourselves how those work, what it is that we do, because it's pretty much the same thing in three dimensions, but it takes a little bit more visualization. So with two dimensional graphs, you have some kind of function. You know, let's say you have f of x is equal to x squared. And any time that you're visualizing a function, you're trying to understand the relationship between the inputs and the outputs. And here, those are both just numbers.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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So with two dimensional graphs, you have some kind of function. You know, let's say you have f of x is equal to x squared. And any time that you're visualizing a function, you're trying to understand the relationship between the inputs and the outputs. And here, those are both just numbers. So you know, you input a number like two, and it's gonna output four. You know, you input negative one, it's gonna output one, and you're trying to understand all of the possible input-output pairs. And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And here, those are both just numbers. So you know, you input a number like two, and it's gonna output four. You know, you input negative one, it's gonna output one, and you're trying to understand all of the possible input-output pairs. And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible. And the way we go about this with graphs is you think of just plotting these actual pairs, right? So you're gonna plot the point. Let's say we're gonna plot the point two, four.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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And the fact that we can do this, that we can give a pretty good intuitive feel for every possible input-output pair, is pretty incredible. And the way we go about this with graphs is you think of just plotting these actual pairs, right? So you're gonna plot the point. Let's say we're gonna plot the point two, four. So we might kind of mark our graph to here. One, two, three, four. So you'd wanna mark somewhere here, two, four.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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Let's say we're gonna plot the point two, four. So we might kind of mark our graph to here. One, two, three, four. So you'd wanna mark somewhere here, two, four. And that represents an input-output pair. And if you do that with negative one, one, you go negative one, one. And when you do this for every possible input-output pair, what you end up getting, and I might not draw this super well, is some kind of smooth curve.
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Introduction to 3d graphs Multivariable calculus Khan Academy.mp3
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