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In the last video, we started to talk about how to

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parameterize a torus, or a doughnut shape.

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And the two parameters we were using, and I spent a lot of

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time trying to visualize it, because this is all

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about visualization.

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I think this is really the hard thing to do here.

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But the way we can parameterize a torus, which is the surface

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of this doughnut, is to say say hey, let's take a point let's

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rotate it around a circle.

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It could be any circle.

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I picked a circle in the z-y plane.

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And how far it's gone around that circle, we'll parameterize

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that by s, and s can go between 0 all the way to 2 pi, and then

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we're going to rotate this circle around itself.

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Or I guess a better way to say it, we're going to rotate the

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circle around the z-axis, and it's all at the center of the

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circle, so we're always going to keep a distance b away.

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And so these were top views right there.

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And then we defined our second parameter t, which tells us how

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far the entire circle has rotated around the

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z-axis access.

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And those were our two parameter definitions.

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And then here we tried to visualize what happens.

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This is kind of the domain that our parameterization

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is going to be defined on.

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s goes between 0 and 2 pi, so when t is 0, we haven't

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rotated out of the z-y plane.

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s is at 0, goes all the way to 2 pi over there.

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Then when t goes to 2 pi, we've kind of moved our circle.

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We've moved it along, we've rotated around

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the z-axis a bit.

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And then this line in our s-t domain corresponds to that

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circle in 3 dimensions, or in our x-y-z space.

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Now given that, hopefully we visualize it pretty well.

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Let's think about actually how to define a position

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vector-valued function that is essentially this

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parameterization.

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So let's first to do the z, because that's

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pretty straightforward.

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So let's look at this view right here.

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What's our z going to be as a function?

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So our x's, our y's, and our z's should all be

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a function of s and t.

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That's what it's all about.

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Any position in space should be a function of picking a

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particular t and a particular s.

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And we saw that over here.

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This point right here, let me actually do that with

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a couple of points.

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This point right there, that corresponds to that

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point, right there.

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Then we pick another one.

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This point right here, corresponds to this

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point, right over there.

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I can do a few more.

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Let me pick.

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This point right here, so s is still 0.

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That's going to be this outer edge, way out over there.

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I'll pick one more, just to define this square.

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This point right over here, where we haven't rotated t at

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all, but we've gone a quarter way around the circle, is

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that point right there.

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So for any s and t we're mapping it to a point

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in x-y-z space.

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So our z's, our x's, and our y's should all be

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a function of s and t.

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So the first one to think about is just the z.

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I think this will be pretty straightforward.

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So z as a function of s and t is going to equal what?

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Well, if you take any circle, remember s is how the angle

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between our radius and the x-y plane.

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So I can even draw it over here.

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Let me do it in another color.

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I'm running out of colors.

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So let's say that this is a radius, right there.

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That angle, we said, that is s.

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So if I were to draw that circle out, just like

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that, we can do a little bit of trigonometry.

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The angle is s.

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We know the radius is a, the radius of our

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circle, we defined that.

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So z is just going to be the distance above the x-y plane.

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It's going to be this distance, right there.

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And that's straightforward trigonometry.

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That's going to be a, I mean, we can do SOCATOA and all

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of that, you might want to review the videos.

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But the sine, you can view it this way.

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So if this is z right there, you could say that the sine of

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s, SOCATOA is the opposite over the hypotenuse, is

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equal to a z over a.

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Multiply both sides by a, you have a sine s is equal to z.

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That tells us how much above the x-y plane we are.

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Just some simple trigonometry.

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So z of s and t, it's only going to be a function of s.

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It's going to be a times the sine of s.

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Not too bad.

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Now see if we can figure out what x and y are going to be.

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Remember, z doesn't matter.

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Doesn't matter how much we've rotated around the z-axis.

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What matters is, how much we've rotated around the circle.

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If s is 0, we're just going to be in the x-y plane,

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z is going to be zero.

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If s is pi over 2, up here, then we're going to be

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traveling around the top of the doughnut.

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And we're going to be exactly a above the x-y plane, or z

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is going to be equal to a.

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Hopefully that makes reasonable sense to you.

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Now let's think about what happens as we rotate around.

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Remember, these two are top views.

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We are looking down on this doughnut.

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So the center of each of these circles is b away from the

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origin, or from the z-axis, what we're rotating around.

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It's always b away.

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So our x-coordinate, or our x- and y-coordinate, so if we go

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to the center of the circle here, we're going to be b away,

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and this is going to be b away, right over there.

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So let's think about where we are in the x-y plane, or how

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far the part of our, what we're, I guess you could

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imagine, if you were to project our point into the x-y

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plane, how far is that going to be from our origin?

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Well, it's always going to be, remember, let's go

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back to this drawing here.

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This might be the most instructive.

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This is just one particular circle on the z-y plane, but

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it could be any of them.

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If this is the z-axis, over here, this distance right here

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is always going to be b.

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We know that for sure.

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And so what is this distance going to be?

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We're at b to the center, and then we're going to have some

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angle s, and so depending on that angle s, this distance

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onto, I guess, the x-y plane, you know, if we're sitting on

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the x-y plane, how far are we from the z-axis, or the

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projection onto the x-y plane.

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Or you can, you know, the x or the y position.

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I'm saying it as many ways as possible.

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I think you're visualizing it.

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If z is a sine of theta, this distance right here, this

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little shorter distance right here, that's going to

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be a cosine if s.

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s is that angle right there.

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This distance right here is going to be a cosine of s.

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So if we talk about just straight distance from the

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origin, along the x-y plane, our distance is always going

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to be b plus a cosine of s.

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When s is out here, then it's actually going to become a

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negative number, and that makes sense, because our distance

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is going to be less than b.

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We're going to be at that point right there.

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So if you look at this top views over here, no matter

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where we are, that is b.

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And let's say we've rotated a little bit.

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That distance right here, if you're looking along the x-y

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plane, that is always going to be b plus a cosine of s.

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That's what that distance is to any given point.

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We're depending on our s's and t's.

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Now, as we rotate around, if we're at a point here, let's

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say we're at a point there, and that point, we already said, is

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b plus a cosine of s, away from the origin, on the xy plane.

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What are the x and y coordinates of that?

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Well, remember.

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This is a top-down.

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We're sitting on the z-axis looking straight down

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the x-y plane right now.

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We're looking down on the doughnut.

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So what are our x's and y's going to be?

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Well, you draw another right triangle right here.

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You have another right triangle.

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This angle right here is t.

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This distance right here is going to be this times

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the sine of our angle.

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So this right here, which is essentially our x, this is

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going to be our x-coordinate, x as a function of s and t, os

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going to be equal to the sine of t, t is our angle right

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there, times this radius.

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Times, we could write it either way, times

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b plus a cosine of s.

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Because remember, how far we are depends on how much around

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the circle we are, right?

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When we're over here, we're much further away.

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Here we're exactly b away, if you're looking only

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on the x-y plane.

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And then over here, we're b minus a away, if

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we're on the x-y plane.

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So that's x as a function of s and t.

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And actually, the way I defined it right here, our positive

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x-axis would actually go in this direction.

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So this is x positive, this is x in the negative direction.

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I could've flipped the signs, but hopefully, you know, this

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actually make sense that that would be the positive x,

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this is the negative x.

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Depends on whether using a right-handed or left-handed

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coordinate system, but hopefully that makes sense.

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We're just saying, OK, what is this distance right here that

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is b plus a cosine of s?

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We got that from this right here, when we're taking a view,

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just a cut of the torus.

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That's how far we are, in kind of the x-y direction at any

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point, or kind of radially out, without thinking

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about the height.

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And then if you want the x-coordinate, you multiply it

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times the sine of t, the way I've had it up here, and the

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y-coordinate is going to be this, right here, the way

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we've set up this triangle.

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So y as a function of s and t is going to be equal to the

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cosine of t times this radius.

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b plus a cosine of s.

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And so our parameterization, and you know, just play with

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this triangle, and hopefully it'll make sense.

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I mean, if you say that this is our y-coordinate right here,

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you just do SOCATOA, cosine of t, CA is equal to adjacent,

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which is y, right, this is the angle right here,

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over the hypotenuse.

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Over b plus a cosine of s.

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Multiply both sides of the equation times this, and you

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get y of s of t is equal to cosine of t times this

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thing, right there.

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Let me copy and paste all of our takeaways.

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And we're done with our parameterization.

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We could leave it just like this, but if we want to

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represent it as a position vector-valued function, we

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can define it like this.

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Find a nice color, maybe pink.

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So let's say our position vector-valued function is r.

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It's going to be a function of two parameters, s and t, and

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it's going to be equal to its x-value.

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Let me do that in the same color.

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So it's going to be, I'll do this part first.

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b plus a cosine of s times sine of t, and that's going to go in

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the x-direction, so we'll say that's times i.

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And this case, remember, the way I defined it,

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the positive x-direction is going to be here.

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So the i-unit vector will look like that.

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i will go in that direction, the way I've defined things.

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And then plus our y-value is going to be b plus a cosine of

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s times cosine of t in the y-unit vector direction.

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Remember, the j-unit vector will just go just like that.

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That's our j-unit vector.

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And then, finally, we'll throw in the z, which was actually

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the most straightforward.

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plus a sine of s times the k-unit vector, which is the

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unit vector in the z-direction.

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So times the k-unit vector.

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And so you give me, now, any s and t within this domain right

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here, and you put it into this position vector-valued

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function, it'll give you the exact position vector that

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specifies the appropriate point on the torus.

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So if you pick, let's just make sure we understand

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what we're doing.

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If you pick that point right there, where s and t are both

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equal to pi over 2, and you might even want to go

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through the exercise.

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Take pi over 2 in all of these.

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Actually, let's do it.

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So in that case, so when r of pi over 2, what do we get?

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It's going to be b plus a times cosine of pi over 2.

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Cosine of pi over 2 is 0, right?

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Cosine of 90 degrees.

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So it's going to be b, right, this whole thing is going to be

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0, times sine of pi over 2.

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Sine of pi over 2 is just 1.

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So it's going to be b times i plus, once again, cosine of pi

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over 2 is 0, so this term right here is going to be b, and then

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cosine of pi over 2 is 0, so it's going to be 0 j.

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So it's going to be plus 0 j.

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And then finally, pi over 2, well, there's no t here,

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sine of pi over 2 is 1.

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So plus a times k.

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So there's actually no j-direction.

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So this is going to be equal to b times i plus a times k.

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So the point that it specifies, according to this

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parameterization, or the vector [UNINTELLIGIBLE], is b times

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i plus a times k.

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So b times i will get us right out there, and then a times k

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ill get us right over there.

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So the position of the vector being specified

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is right over there.

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Just as we predicted.

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That dot, that point right there, corresponds to that

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point, just like that.

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Of course, I picked points it was easy to calculate, but this

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whole, when you take every s and t in this domain right

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here, you're going to transform it to this surface.

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And this is the transformation, right here.

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And of course, we have to specify that s is between, we

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could write it multiple ways.

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s is between 2 pi and 0, and we could also say t

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is between 2 pi and 0.

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And you could, you know, we're kind of overlapping one extra

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time at 2 pi, so maybe we can get rid of one of these equal

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signs if you like, although that won't necessarily change

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the area any, if you're taking the surface area.

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But hopefully this gives you at least a gut sense, or more than

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a gut sense, of how to parameterize these things, and

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what we're even doing, because it's going to be really

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important when we start talking about surface integrals.

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And the hardest thing about doing all of this is

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just the visualization.

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