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The last video was very abstract in general, and

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I used you know, f of x and g of t, and h of t.

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What I want to do in this video is do an actual example.

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Let's say I have f of xy.

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Let's say that f of xy is equal to xy.

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And let's say that we have a path in the xy plane, or

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a curve in the xy plane.

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And I'm going to define my curve by x being equal to

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cosine of t, and y being equal to sine of t.

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And we're going to go from-- you know, we have to define

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what are our boundaries in our t --and we're going to go from

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t is equal to 0-- or t is going to be greater than or equal to

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0 --and then less than or equal to.

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We're going to deal in radians, pi over 2.

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If this was degrees, that would be 90 degrees.

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So that's our curve.

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And immediately you might already know what this type

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of a curve looks like.

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And I'm going to draw that really fast right here in and

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we'll try to visualize this.

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I've actually drafted ahead of time so that

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we can visualize this.

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So this curve right here, if I were to just draw it in the

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standard xy plane-- do that in a different color so we can

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make the curve green; let's say that is y, and this is right

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here x --so when t is equal to 0, x is going to be

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equal to cosine of 0.

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Cosine of 0 is 1, y is going to be equal to sine of 0, which

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is 0, so t is equal to 0.

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We're going to be at x equal to 1, that's cosine of 0, and y is

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sine of 0, or y is going to be 0, so we're going

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to be right there.

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That's what t is equal to; t is equal to 0.

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When t is equal to pi of 2, what's going to happen?

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Cosine of pie over 2-- that's the angle; cosine

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of pi over 2 --is 0.

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Sine of pi over 2 is 1.

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We're going to be at the point 0, 1.

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So this is when we're at t is equal to pi over 2.

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You might recognize what we're going to draw is actually the

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first quadrant of the unit circle; when t is equal to pi

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over 4, or 45 degrees, we're going to be at square root

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of 2, square root of 2.

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You can try it out for yourself, but we're just

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going to have a curve that looks like this.

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It's going to be the top right of a circle,

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of the units circle.

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It's going to have radius 1.

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And we're going to go in that direction, from t is equal to

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0, to t is equal to pi over 2.

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That's what this curve looks like.

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But our goal isn't here just to graph a parametric equation.

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What we want to do is raise a fence out of this kind of base

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and rise it to this surface.

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So let's see if we can do that or at least visualize it first,

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and then we'll use the tools we used in the last video.

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So right here I've graphed this function, and I've rotated it a

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little bit so you can see [UNINTELLIGIBLE]

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case.

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This right here-- let me get some dark colors out --that

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right there is the x axis, that in the back is the y axis, and

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the vertical axis is the z axis.

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And this is actually 2, this is 1 right here, y equal 1

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is right there, so this is graphed that way.

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So if I were to graphs this contour in the xy plane, it

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would be under this graph and it would go like something like

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this--- let me see if I can draw it --it would look

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something like this.

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This would be on the xy plane.

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This is the same exact graph, f of x is equal to xy.

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This is f of x; f of xy is equal to xy.

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That's both of these, I just rotated it.

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In this situation that right there is now the x axis.

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I rotated to the left, you can kind of imagine.

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That right there is the x axis, that right there is the y

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axis-- it was rotated closer to me --that's the z axis.

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And then this curve, if I were to draw in this rotation, is

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going to look like this: when t is equal to 0, we're at x is

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equal to 1, y is equal to 0, and it's going to form a unit

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circle, or half or quarter of a unit circle like that.

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And when t is equal to pi over 2, we're going to get there.

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And what we want to do is find the area of the

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curtain that's defined.

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So let's see, let's raise a curtain from this

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curve up to f of xy.

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So if we keep raising walls from this up to xy, we're

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going to have a wall looks something like that.

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Let me shade it in, color it in so it looks a little

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bit more substantive.

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So a wall that looks something like that.

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If I were to try to do it here this would be under the

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ceiling, but the wall look something like

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that right there.

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We want to find the area of that.

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We want to find the area of this right here where the base

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is defined by this curves, and then the ceiling is defined by

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this surface here, xy, which I graphed and I rotated

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in two situations.

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Now in the last video we came up with a, well, you could

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argue whether it's simple, but the is, well, let's just take

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small arc lengths-- change in arc lengths, and multiply them

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by the height at that point.

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And those small change in arc lengths, we called them ds,

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and then the height is just f of xy at that point.

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And we'll take an infinite sum of these, from t is equal to 0

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to t will equal pi over 2, and then that should give us

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the area of this wall.

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So we said is, well, to figure out the area of that we're just

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going to take the integral from t is equal to o to t is equal

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to pi over 2-- it doesn't make a lot of sense when I write it

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like this --of f of xy times-- or let me even better, instead

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of writing f of xy, let me just write the actual function.

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Let's get a little bit more concrete.

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So f of xy is xy times-- so the particular xy --times the

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little change in our arc length at that point.

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I'm going to be very hand-wavy here.

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This is all a little bit review of the last video.

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And we figured out in the last video this change in arc length

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right here, ds, we figured out that we could rewrite that as

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the square root of the dx versus-- or the derivative of x

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with respect to t squared --plus the derivative of y with

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respect to t squared, and then all of that times dt.

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So I'm just rebuilding the formula that we

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got in the last video.

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So this expression can be rewritten as the integral

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from t is equal to 0 to t is equal to pi over 2 times xy.

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But you know what?

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Right from the get go we want everything eventually

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be in terms of t.

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So instead of writing x times y, let's substitute

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the parametric form.

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So instead of x let's write cosine of t.

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That is x.

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x is equal to cosine of t on this curve.

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That's how we define x, in terms of the parameter t.

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And then times y, which we're saying is sine of t.

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That's our y; all I did is rewrote xy in

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terms of t times ds.

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ds is this; it's the square root of the derivative of x

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with respect to t squared plus the derivative of y with

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respect to t squared.

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All of that times dt.

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And now we just have to find these two derivatives.

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And it might seem really hard, but it's very easy for us to

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find the derivative of x with respect to t and the derivative

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of y with respect to t.

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I can do it right or down here.

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Let me lose our graphs for a little bit.

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We know that the derivative of x with respect to t is just

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going to be: what's the derivative of cosine of t?

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Well, it's minus sine of t.

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And the derivative of y was respect to t?

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Derivative of a sine of anything is the cosine

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of that anything.

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So it's cosine of t.

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And we can substitute these back into this equation.

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So remember, we're just trying to find the area of this

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curtain that has our curve here as kind of its base, and

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has this function, this surface as it's ceiling.

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So we go back down here, and let me rewrite

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this whole thing.

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So this becomes the integral from t is equal to o to t is

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equal to pi over 2-- I don't like this color --of cosine of

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t, sine of t, cosine times sine-- that's just the xy

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--times ds, which is this expression right here.

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And now we can write this as-- I'll go switch back to that

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color I don't like --the derivative of x with respect to

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t is minus sine of t, and we're going to square it, plus the

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derivative of y with respect to t, that's cosine of t, and

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we're going to square it-- let me make my radical a little bit

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bigger --and then all of that times dt.

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Now this still might seem like a really hard integral until

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you realize that this right here, and when you take a

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negative number and you squared it, this is the same thing.

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Let me rewrite, do this in the side right here.

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Minus sine of t squared plus the cosine of t squared, this

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is equivalent to sine of t squared plus cosine

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of t squared.

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You lose the sign information when you square something;

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it just becomes a positive.

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So these two things are equivalent.

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And this is the most basic trig identity.

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This comes straight out of the unit circle definition: sine

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squared plus cosine squared, this is just equal to 1.

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So all this stuff under the radical sign is

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just equal to 1.

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And we're taking the square root of 1 which is just 1.

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So all of this stuff right here will just become 1.

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And so this whole crazy integral simplifies a good bit

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and just equals the square root of t equals 0 to t is equal to

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pi over 2 of-- and I'm going to switch these around just

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because it will make it a little easier in the next step

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--of sine of t times cosine of t, dt.

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All I did, this whole thing equals 1, got rid of it,

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and I just switched the order of that.

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It'll make the next up a little bit easier to explain.

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Now this integral-- You say sine times cosine, what's

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the antiderivative of that?

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And the first thing you should recognize is, hey, I have a

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function or an expression here, and I have its derivative.

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The derivative of sine is cosine of t.

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So you might be able to a u substitution in your head;

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it's a good skill to be able to do in your head.

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But I'll do it very explicitly here.

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So if you have something that's derivative, you

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define that something as u.

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So you say u is equal to sine of t and then du, dt, the

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derivative of u with respect to t is equal to cosine of t.

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Or if you multiply both sides by the differential dt, if

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we're not going to be too rigorous, you get du is

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equal to cosine of t, dt.

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And notice right here I have a u.

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And then cosine of t, dt, this thing right here, that

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thing is equal to d of u.

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And then we just have to redefine the boundaries.

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When t is equal to 0-- I mean so this thing is going to turn

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into the integral --instead of t is equal to 0, when t is

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equal to 0 what is u equal to?

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Sine of 0 is 0, so this goes from u is equal to 0.

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When t is pie over 2 sine of pi over 2 is 1.

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So when t is pi over 2, u is equal to 1.

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So from is equal to 0 to u is equal to 1.

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Just redid the boundaries in terms of u.

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And then we have instead of sign of t, I'm

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going to write u.

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And instead of cosine of t, dt, I'm just going to write du.

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And then this is a super-easy integral in terms of u.

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This is just equal to: the antiderivative of u is u 1/2

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times u squared-- we just raised the exponent and then

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divided by that raised exponent --so 1/2 u squared, and we're

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going to evaluate it from 0 to 1.

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And so this is going to be equal to 1/2 times 1 squared

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minus 1/2 times 0 squared, which is equal to 1/2 times 1

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minus 0, which is equal to 1/2.

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So we did all that work and we got a nice simple answer.

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The area of this a curtain-- we just performed a line integral

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--the area of this curtain along this curve right here

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is-- let me do it in a darker color --on 1/2.

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You know, if this was in centimeters, it would be

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1/2 centimeters squared.

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So I think that was you know, a pretty neat application

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of the line integral.

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