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We saw several videos ago that we can parameterize a torus or

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a doughnut shape as a position vector-valued function

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of two parameters.

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And this is the outcome that we had.

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I think I did it over several videos because

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it was a bit hairy.

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And I'll write our position vector-valued function first.

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So we have r as a function of our two parameters s and t.

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And then I'll review a little bit of what all the terms--

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what the s, the t, and the a's and the b's represent.

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But it's equal to b plus a cosine of s.

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And once again, we saw this several videos ago.

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So you might want to watch the videos on parameterizing

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surfaces with two parameters to figure out how we got here.

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Times the sine of t.

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I'm going to put the s terms and the t terms

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in different colors.

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Times our i unit vector.

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I'll put the vectors, or the unit vectors

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in this orange color.

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Plus-- I'll do it in the same yellow.

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Plus b plus a cosine of s times cosine of t times the j unit

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vector-- the unit factor in the y direction.

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Plus a sine of s times the k unit vector or the unit

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vector in the z direction.

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And in order to generate the torus or the doughnut shape,

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this is true for our parameters-- so we don't wrap

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multiple times around the torus-- for s being between 0

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and 2 pi and for t being between 0 and 2 pi.

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And just as a bit of review, we're all of this came from--

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and I'm going to have to do what my plan is for this

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video over several videos.

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But let's review where all of this came from.

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Let me draw a doughnut.

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My best effort at a doughnut right here.

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That looks like a doughnut or a torus.

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And you can imagine a torus, or this doughnut shape is kind of

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the product of two circles.

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You have the circle that's kind of the cross section of

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the doughnut at any point.

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You could take it there.

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You could take it over there.

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And then you have the circle that kind of wraps around all

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of these other circles or these other circles wrap around it.

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And so, when we derived this formula up here or this

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parameterization, a was the radius of these cross

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sectional circles.

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That's a.

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That's what these a terms were.

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And b was the distance from the center of our torus out to the

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center of these cross sections.

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So this was b.

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So you can imagine that b is kind of the radius of the big

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circle up to the midpoint of the, I guess, cross section.

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And a is the radius of the cross sectional circles.

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And when we parameterized it, the parameter s was essentially

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telling us how far-- s was telling us how far or where are

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we wrapping around this circle.

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So it's an angle from 0 to 2 pi to say where

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we are on that circle.

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And t tells us how much we've rotated around

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the larger circle.

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So if you think about it, you can specify any point on this

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doughnut or on this surface or on this torus by telling

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you an s or a t.

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And so that's why we picked that as the parameterization.

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Now, the whole reason why I'm even revisiting this stuff that

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we saw several videos ago is we're going to actually use it

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to compute an actual surface integral.

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And the surface integral we're going to compute will tell us

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the surface area of this torus.

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So this surface right here is sigma, like that and it's being

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represented by this position vector-valued function.

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That is parameterized by these two parameters right there.

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And if we wanted to figure out the surface area, if we just

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kind of set it as the surface integral we saw in, I think,

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the last video at least the last vector calculus video I

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did that this is a surface integral over the surface.

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Here this capital Sigma does not represent a sum, it

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represents a surface of a bunch of the little d sigmas--

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a bunch of the little chunks of the surface.

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And just as a review, you can imagine each d sigma

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is a little patch of the surface right there.

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That is a d sigma.

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It's a double integral here because we want to add up

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all of the d sigmas in 2 directions.

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You can imagine one kind of rotating this way around the

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torus and then the other direction is going in the other

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direction around the torus.

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So that's why it's a double integral.

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And this is just going to give you the surface area, which is

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the whole point of this video and probably the next

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one or two videos.

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But if you wanted to also multiply these sigmas times

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some other value-- there's some scalar field that this is in

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that you cared about-- you could put that other

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value right there.

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But here we're just multiplying it by 1.

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And we saw in the last video that it's a way of expressing

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an idea, but you really can't do much computation with this.

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But a way that you can express this so that you can actually

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take the integral, you say this is the same thing-- and we saw

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this in the last several videos.

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This is the same thing as the double integral over the

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region over which our parameters are defined.

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So it's this region over here where s and t go from 0 to 2 pi

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of whatever function this is.

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We just have a 1 here, so we could just write a 1 if we

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like; it doesn't change much.

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Times-- and this is what we learned.

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Times the magnitude of the partial derivative of

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r with respect to s.

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The magnitude of that crossed with the partial derivative

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of r with respect to t ds.

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You could take it in either order, but ds dt.

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So we saw this in the last video.

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What we're going to do now is actually compute this.

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That's the whole point of this video.

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We're going to take the cross product of these two vectors.

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So let's figure out these vectors.

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Then in the next video we're going to take

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the cross product.

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And then the video after that we'll actually evaluate

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this double integral.

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And you're going to see it's a pretty hairy problem and this

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is probably the reason that very few people ever see an

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actual surface integral get computed.

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But let's do it anyway.

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So the partial derivative of r with respect to s-- so

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this term right here.

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We'll do the cross product in the next video.

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This term is what?

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We just want to hold t constant and took the partial

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with respect to just s.

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So this up here, if you distributed the sine of t times

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b-- that's just going to be a constant in terms of s,

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so we can ignore that.

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Then you have sine of t times this over here.

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So sine of t and a is a constant.

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And you take the derivative of cosine of s.

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That's negative sine of s.

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So the derivative of this with respect to s or the partial

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with respect to s is going to be minus a-- I'll write in

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green the sine of t, so you know that's where it came from.

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Sine of t and then sine of s.

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The derivative of this is negative sine of s.

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That's where that negative came from.

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And then I'll write the sine of s right there.

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Times the unit vector i.

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That's the partial of just this x term with respect to s.

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And then we'll do the same thing with the y

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term or the j term.

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So plus-- same logic-- b times cosine of t with respect to s.

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When you take the partial [INAUDIBLE]

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becomes 0 so you're left with a-- well, it's going to be a

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minus a again because when you take the derivative of the

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cosine of s it's going to be negative sine of s.

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Let me do it.

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You're going to have a minus a.

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This cosine of t.

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Minus a cosine of t.

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That's the constant terms.

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Sine of s.

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Just taking partial derivatives.

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Sine of s j.

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And then finally, we take the derivative of this

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with respect to s.

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And that's pretty straightforward.

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It's just going to be a cosine of s.

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So plus a cosine of s k.

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Now hopefully you didn't find this confusing.

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The negative sines because the derivative of cosines

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are negative sines.

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So negative sine of s.

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That's why it's negative sine of s times the constant.

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Negative sine of s times the constant-- the constant

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cosine of t sine of t.

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So hopefully this makes some sense just as a review of

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taking a partial derivative.

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Now let's do the same thing with respect to t.

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And I'll do that in a different color.

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So we're now going to take the partial of r with respect to t.

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So the partial of r with respect to t is equal to-- so

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now this whole term over here is a constant, and so it's

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going to be that whole term times the derivative of this

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with respect to t, which is just cosine of t.

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So it's going to be b plus a cosine of s

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times cosine of t i.

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And then, plus-- and it's actually going to be a minus

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because when you take the derivative of this with

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respect to t it's going to be minus sine of t.

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So it's going to be negative and then let me leave some

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space for this term right here.

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Negative sine of t.

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And you're going to have this constant out there.

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That's a constant in t.

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b plus a cosine of s.

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That's just that term right there.

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Derivative of cosine t is negative sine of t times j.

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And then the partial of this with respect to t-- this

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is just a constant in t.

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So the partial's going to be 0.

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So I'll write plus 0k.

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Let me do all my vectors in that same color.

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Plus 0 times the unit vector k.

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So that gives us our partial derivatives.

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Now we have to take their cross product, then find the

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magnitude of the cross product, and then evaluate this

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double integral.

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And I'll do that in the next couple of videos.

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