Welcome to a lesson on parameterized surfaces.
We've already looked at how to parameterize
a plane curve or a space curve using the parameter t,
as we see here for a plane curve,
and here for a space curve.
But to parameterize a surface, we will need two parameters
and three independent variables.
So our parameters will be u and v,
and the dependent variables will be x, y and z,
so x, y, and z depend on u and v,
and our two parameters u and v
define a region r in the uv-plane,
where the vector r of u v will sweep out the surface.
So let's take a look at some examples
of how to parameterize a surface.
Here we have the plane,
two x minus three y plus z is equal to six.
So if we let x equal u and y equal v,
we can solve this equation for z,
and then replace x and y with u and v,
let's go ahead and do that.
So we have z would be equal to
six minus two x plus three y,
which in terms of u and v would be
six minus two u plus three v.
So we can now write the surface as a vector valued function
in terms of u and v, where the x component would be u,
the y component would be v, and the z component
in terms of u and v would be six minus two u plus three v.
That one was pretty straight forward,
let's go and take a look at a couple more examples.
For our second example we have x-squared plus the quantity
y minus two squared is equal to four.
We can see from the graph that it's a cylinder,
so to parameterize this surface,
we'll going to take advantage of a trig identity.
Remember that cosine squared theta
plus sine squared theta is equal to one.
So what we're going to do is set this equation equal to one,
so we'll divide everything by four,
that'll give us x squared over four
plus the quantity y minus two squared
over four is equal to one.
So now if you let x equal two cosine u,
again we want to use u and v,
and then y minus two equal two sine u,
it would satisfy the given equation.
Let's go ahead and solve this equation for y,
so we have y equals two sine u plus two,
and notice there's no v in this equation
because this surface is not bounded vertically,
so we can just let z equal v,
and we've now parameterized this surface.
The vector valued function would be
equal to two cosine u,
two sine u plus two, and v.
Let's take a look at one more type.
Here we have a sphere, x-squared plus y-squared
plus z-squared is equal to nine.
So we use sprinkled coordinates
to parameterize this surface.
Since our radius is equal to three,
we know that rho is equal to three,
so we can go straight to the parameterization.
We'll just let u equal phi,
and v equal theta.
So we'll have three sine u cosine v,
three sine u sine v,
and the z component would be three cosine u.
Now take a look at a couple of problems
where we're given the parameterization,
and we want to determine the rectangular equation.
So here we know that x would be equal to u,
y would be equal to v,
and z would have to be equal to
the square root of u-squared plus v-squared.
So again, this first one here is pretty straight forward,
we can just perform a substitution,
because we know that u is equal to x,
so we have the square root of x-squared,
and v is equal to y, so v-squared would be y-squared,
and now we have the equation in rectangular form.
Let's take a look at another one
that's a little bit more involved.
So from this parameterization,
we know that x is equal to three u sine v,
y would be equal to three u cosine v,
and z would be equal to u-squared.
So here we're going to use another trig identity.
If we solve these first two for sine v and cosine v,
we can then use the identity that cosine squared v
plus sine squared v must equal one.
Solving this equation for sine v,
we'd have x over three u equal sine v,
and here we'd have cosine v
equals y over three u,
and we'll leave this equation as z equals u-squared.
So again, since we know that sine squared v
plus cosine squared v is equal to one,
must also be true that x over three u squared
plus y over three u squared
must equal one.
Well this is going to give us x-squared over nine u-squared
plus y-squared over nine u-squared is equal to one.
Now let's go ahead and multiply
everything by nine u-squared,
so that gives us x-squared plus y-squared
equals nine u-squared, but guess what,
u-squared is equal to z.
So we know that x-squared plus y-squared must equal z,
and we have our rectangular equation
with a given parameterization.
I hope you found these examples helpful.
Thank you for watching.