In the last video, we started to talk about how to
parameterize a torus, or a doughnut shape.
And the two parameters we were using, and I spent a lot of
time trying to visualize it, because this is all
I think this is really the hard thing to do here.
But the way we can parameterize a torus, which is the surface
of this doughnut, is to say say hey, let's take a point let's
rotate it around a circle.
It could be any circle.
I picked a circle in the z-y plane.
And how far it's gone around that circle, we'll parameterize
that by s, and s can go between 0 all the way to 2 pi, and then
we're going to rotate this circle around itself.
Or I guess a better way to say it, we're going to rotate the
circle around the z-axis, and it's all at the center of the
circle, so we're always going to keep a distance b away.
And so these were top views right there.
And then we defined our second parameter t, which tells us how
far the entire circle has rotated around the
And those were our two parameter definitions.
And then here we tried to visualize what happens.
This is kind of the domain that our parameterization
is going to be defined on.
s goes between 0 and 2 pi, so when t is 0, we haven't
rotated out of the z-y plane.
s is at 0, goes all the way to 2 pi over there.
Then when t goes to 2 pi, we've kind of moved our circle.
We've moved it along, we've rotated around
the z-axis a bit.
And then this line in our s-t domain corresponds to that
circle in 3 dimensions, or in our x-y-z space.
Now given that, hopefully we visualize it pretty well.
Let's think about actually how to define a position
vector-valued function that is essentially this
So let's first to do the z, because that's
So let's look at this view right here.
What's our z going to be as a function?
So our x's, our y's, and our z's should all be
a function of s and t.
That's what it's all about.
Any position in space should be a function of picking a
particular t and a particular s.
And we saw that over here.
This point right here, let me actually do that with
a couple of points.
This point right there, that corresponds to that
point, right there.
Then we pick another one.
This point right here, corresponds to this
point, right over there.
I can do a few more.
Let me pick.
This point right here, so s is still 0.
That's going to be this outer edge, way out over there.
I'll pick one more, just to define this square.
This point right over here, where we haven't rotated t at
all, but we've gone a quarter way around the circle, is
that point right there.
So for any s and t we're mapping it to a point
in x-y-z space.
So our z's, our x's, and our y's should all be
a function of s and t.
So the first one to think about is just the z.
I think this will be pretty straightforward.
So z as a function of s and t is going to equal what?
Well, if you take any circle, remember s is how the angle
between our radius and the x-y plane.
So I can even draw it over here.
Let me do it in another color.
I'm running out of colors.
So let's say that this is a radius, right there.
That angle, we said, that is s.
So if I were to draw that circle out, just like
that, we can do a little bit of trigonometry.
The angle is s.
We know the radius is a, the radius of our
circle, we defined that.
So z is just going to be the distance above the x-y plane.
It's going to be this distance, right there.
And that's straightforward trigonometry.
That's going to be a, I mean, we can do SOCATOA and all
of that, you might want to review the videos.
But the sine, you can view it this way.
So if this is z right there, you could say that the sine of
s, SOCATOA is the opposite over the hypotenuse, is
equal to a z over a.
Multiply both sides by a, you have a sine s is equal to z.
That tells us how much above the x-y plane we are.
Just some simple trigonometry.
So z of s and t, it's only going to be a function of s.
It's going to be a times the sine of s.
Not too bad.
Now see if we can figure out what x and y are going to be.
Remember, z doesn't matter.
Doesn't matter how much we've rotated around the z-axis.
What matters is, how much we've rotated around the circle.
If s is 0, we're just going to be in the x-y plane,
z is going to be zero.
If s is pi over 2, up here, then we're going to be
traveling around the top of the doughnut.
And we're going to be exactly a above the x-y plane, or z
is going to be equal to a.
Hopefully that makes reasonable sense to you.
Now let's think about what happens as we rotate around.
Remember, these two are top views.
We are looking down on this doughnut.
So the center of each of these circles is b away from the
origin, or from the z-axis, what we're rotating around.
It's always b away.
So our x-coordinate, or our x- and y-coordinate, so if we go
to the center of the circle here, we're going to be b away,
and this is going to be b away, right over there.
So let's think about where we are in the x-y plane, or how
far the part of our, what we're, I guess you could
imagine, if you were to project our point into the x-y
plane, how far is that going to be from our origin?
Well, it's always going to be, remember, let's go
back to this drawing here.
This might be the most instructive.
This is just one particular circle on the z-y plane, but
it could be any of them.
If this is the z-axis, over here, this distance right here
is always going to be b.
We know that for sure.
And so what is this distance going to be?
We're at b to the center, and then we're going to have some
angle s, and so depending on that angle s, this distance
onto, I guess, the x-y plane, you know, if we're sitting on
the x-y plane, how far are we from the z-axis, or the
projection onto the x-y plane.
Or you can, you know, the x or the y position.
I'm saying it as many ways as possible.
I think you're visualizing it.
If z is a sine of theta, this distance right here, this
little shorter distance right here, that's going to
be a cosine if s.
s is that angle right there.
This distance right here is going to be a cosine of s.
So if we talk about just straight distance from the
origin, along the x-y plane, our distance is always going
to be b plus a cosine of s.
When s is out here, then it's actually going to become a
negative number, and that makes sense, because our distance
is going to be less than b.
We're going to be at that point right there.
So if you look at this top views over here, no matter
where we are, that is b.
And let's say we've rotated a little bit.
That distance right here, if you're looking along the x-y
plane, that is always going to be b plus a cosine of s.
That's what that distance is to any given point.
We're depending on our s's and t's.
Now, as we rotate around, if we're at a point here, let's
say we're at a point there, and that point, we already said, is
b plus a cosine of s, away from the origin, on the xy plane.
What are the x and y coordinates of that?
This is a top-down.
We're sitting on the z-axis looking straight down
the x-y plane right now.
We're looking down on the doughnut.
So what are our x's and y's going to be?
Well, you draw another right triangle right here.
You have another right triangle.
This angle right here is t.
This distance right here is going to be this times
the sine of our angle.
So this right here, which is essentially our x, this is
going to be our x-coordinate, x as a function of s and t, os
going to be equal to the sine of t, t is our angle right
there, times this radius.
Times, we could write it either way, times
b plus a cosine of s.
Because remember, how far we are depends on how much around
the circle we are, right?
When we're over here, we're much further away.
Here we're exactly b away, if you're looking only
on the x-y plane.
And then over here, we're b minus a away, if
we're on the x-y plane.
So that's x as a function of s and t.
And actually, the way I defined it right here, our positive
x-axis would actually go in this direction.
So this is x positive, this is x in the negative direction.
I could've flipped the signs, but hopefully, you know, this
actually make sense that that would be the positive x,
this is the negative x.
Depends on whether using a right-handed or left-handed
coordinate system, but hopefully that makes sense.
We're just saying, OK, what is this distance right here that
is b plus a cosine of s?
We got that from this right here, when we're taking a view,
just a cut of the torus.
That's how far we are, in kind of the x-y direction at any
point, or kind of radially out, without thinking
about the height.
And then if you want the x-coordinate, you multiply it
times the sine of t, the way I've had it up here, and the
y-coordinate is going to be this, right here, the way
we've set up this triangle.
So y as a function of s and t is going to be equal to the
cosine of t times this radius.
b plus a cosine of s.
And so our parameterization, and you know, just play with
this triangle, and hopefully it'll make sense.
I mean, if you say that this is our y-coordinate right here,
you just do SOCATOA, cosine of t, CA is equal to adjacent,
which is y, right, this is the angle right here,
over the hypotenuse.
Over b plus a cosine of s.
Multiply both sides of the equation times this, and you
get y of s of t is equal to cosine of t times this
thing, right there.
Let me copy and paste all of our takeaways.
And we're done with our parameterization.
We could leave it just like this, but if we want to
represent it as a position vector-valued function, we
can define it like this.
Find a nice color, maybe pink.
So let's say our position vector-valued function is r.
It's going to be a function of two parameters, s and t, and
it's going to be equal to its x-value.
Let me do that in the same color.
So it's going to be, I'll do this part first.
b plus a cosine of s times sine of t, and that's going to go in
the x-direction, so we'll say that's times i.
And this case, remember, the way I defined it,
the positive x-direction is going to be here.
So the i-unit vector will look like that.
i will go in that direction, the way I've defined things.
And then plus our y-value is going to be b plus a cosine of
s times cosine of t in the y-unit vector direction.
Remember, the j-unit vector will just go just like that.
That's our j-unit vector.
And then, finally, we'll throw in the z, which was actually
the most straightforward.
plus a sine of s times the k-unit vector, which is the
unit vector in the z-direction.
So times the k-unit vector.
And so you give me, now, any s and t within this domain right
here, and you put it into this position vector-valued
function, it'll give you the exact position vector that
specifies the appropriate point on the torus.
So if you pick, let's just make sure we understand
what we're doing.
If you pick that point right there, where s and t are both
equal to pi over 2, and you might even want to go
through the exercise.
Take pi over 2 in all of these.
Actually, let's do it.
So in that case, so when r of pi over 2, what do we get?
It's going to be b plus a times cosine of pi over 2.
Cosine of pi over 2 is 0, right?
Cosine of 90 degrees.
So it's going to be b, right, this whole thing is going to be
0, times sine of pi over 2.
Sine of pi over 2 is just 1.
So it's going to be b times i plus, once again, cosine of pi
over 2 is 0, so this term right here is going to be b, and then
cosine of pi over 2 is 0, so it's going to be 0 j.
So it's going to be plus 0 j.
And then finally, pi over 2, well, there's no t here,
sine of pi over 2 is 1.
So plus a times k.
So there's actually no j-direction.
So this is going to be equal to b times i plus a times k.
So the point that it specifies, according to this
parameterization, or the vector [UNINTELLIGIBLE], is b times
i plus a times k.
So b times i will get us right out there, and then a times k
ill get us right over there.
So the position of the vector being specified
is right over there.
Just as we predicted.
That dot, that point right there, corresponds to that
point, just like that.
Of course, I picked points it was easy to calculate, but this
whole, when you take every s and t in this domain right
here, you're going to transform it to this surface.
And this is the transformation, right here.
And of course, we have to specify that s is between, we
could write it multiple ways.
s is between 2 pi and 0, and we could also say t
is between 2 pi and 0.
And you could, you know, we're kind of overlapping one extra
time at 2 pi, so maybe we can get rid of one of these equal
signs if you like, although that won't necessarily change
the area any, if you're taking the surface area.
But hopefully this gives you at least a gut sense, or more than
a gut sense, of how to parameterize these things, and
what we're even doing, because it's going to be really
important when we start talking about surface integrals.
And the hardest thing about doing all of this is
just the visualization.