For each surface given by the vector function,
r of u comma v,
identify the best description of the shape.
Looking at the equations of the parametrized surfaces,
we'll first write the surface using Cartesian coordinates,
and then hopefully recognize the form of the equation
to identify the shape.
But then we'll also verify the work graphically.
Looking at the first vector function,
we have x equals v cosine u,
y equals v,
and z equals v sine u.
And now we need to eliminate the parameters, u and v.
Notice x squared plus z squared is equal to
the square of v cosine u
plus the square of v sine u.
And therefore, x squared plus z squared is equal to
v squared cosine squared u
plus v squared sine squared u.
If we factor out v squared from the right side,
we have x squared plus z squared equals
v squared times the quantity
cosine squared u
plus sine squared u.
And we should recognize that cosine squared u
plus sine squared u is equal to one,
which gives us x squared plus z squared equals v squared.
And because y is equal to v,
we can substitute y squared for v squared,
which gives us the equation in Cartesian coordinates,
x squared plus z squared equals y squared,
or if we want,
x squared plus z squared minus y squared equals zero.
And we should recognize this equation as a cone.
If you need a review on the forms
of the equations in Cartesian coordinates
that result in various quadric surfaces,
I would recommend going to this site, here,
and review the notes.
Looking at the second vector function,
it looks similar,
but now we have x equals v cosine u.
Y equals u,
not y equals v, like we have here.
And then z is equal to v sine u.
So for these parametric equations,
the approach we use on the left is not going to work
because when we get down to
x squared plus z squared equals v squared,
notice here y is equal to u not v,
and therefore we can't perform a substitution for v squared.
Let's use the fact that we know that
sine theta divided by cosine theta
is equal to tangent theta,
and therefore z divided by x is equal to v sine u,
divided by v cosine u.
V divided by v is equal to one,
and sine u divided by cosine u is equal to tangent u.
And because y is equal to u,
we can write this as z divided by x equals tangent y.
And now we have the equation using Cartesian coordinates.
If we multiply both sides by x,
we have z equals x times tangent y.
We're probably not going to recognize
the name of this surface,
so let's go ahead and take a look at the graph.
Here's the graph of the surface.
It's a very famous surface.
It's called a screw or a helicoid.
[birds chirping]
I do want to mention,
if you tried to graph this using the Cartesian equation,
the graph would not look correct.
And that's because of the tangent function here.
Remember, because x equals v cosine u,
and z equals v sine u,
which we also have over here,
we know that x squared plus z squared must equal v squared.
So the Cartesian equation
will only give us the correct graph
if we have the restriction
that x squared plus z squared equals v squared.
V would given over some interval
based on the vector function r of u comma v.
So looking at our work so far,
we got the first vector function,
so we select c here.
We know the second vector function is a screw or a helicoid,
which is b.
And now let's look at the last two vector functions.
For the third vector function,
we have x equals v squared,
y equals u,
and z equals v.
Looking at the parametric equations,
because x is equal to v squared,
and z is equal to v,
we can substitute z for v,
giving us the equation in Cartesian coordinates,
x equals z squared.
Notice how we do not have to use the equation,
y equals u,
which means y could be any real number given by u
in the vector function,
r of u comma v as shown here.
So this is the equation in Cartesian coordinates,
which we should recognize as a cylinder.
If we look at the graph,
here it is,
we have a parabolic cylinder.
Which is a.
And for the last vector function,
we have x equals u,
y equals v,
and z equals u plus v.
So here we know that z is equal to u plus v,
but x is equal to u and y is equal to v,
which gives us the equation in Cartesian coordinates,
z equals x plus y.
Or if we get this equal to zero,
we can write this as zero equals x plus y minus z,
which we should recognize as a plane,
which is d.
Let's go ahead and graph this.
Here's the graph of the plane.
I hope you found this helpful.