# Description of surface, vector calculus

1. Feb 14, 2013

### ohlala191785

1. The problem statement, all variables and given/known data
Consider the surface parameterized by (v cos(u), v sin(u), 45 v cos(u)), where u and v both vary from 0 to 2∏.

2. Relevant equations
(v cos(u), v sin(u), 45 v cos(u))
I think this is supposed to be a vector function? As in r(u,v) = <v cos(u), v sin(u), 45 v cos(u)>.

3. The attempt at a solution
In the x-y plane, this is a circle. x = v cos(u) so z = 45x. This is a slanted plane? So I thought the surface would be an ellipse, since the coefficient of x is 45 and the circle would be very squashed. Does this seem to be part of a cylinder because the cross section of a cylinder, if the plane is slanted, would it be a portion of an elliptical paraboloid? Or could it be a portion of a cone or hyperboloid? One of the answer options is an ellipsoid, but I don't think that's right because when I graphed this on a computer, it showed a flat surface.

There are so many possibilities!

Last edited: Feb 14, 2013
2. Feb 14, 2013

### haruspex

Take v to be fixed, initially. Varying u should give you a curve, and yes, it's an ellipse. Describe the plane the ellipse lies in and where it is centred. Then imagine varying v over its given range.

3. Feb 14, 2013

### LCKurtz

Yes, that's exactly what it is and how you describe a surface.

Yes, that is correct for fixed $v$. Its equation is $x^2+ y^2 = v^2$. But if you let $v$ vary from $0$ to $2\pi$ what do you get?

Yes. And $x$ and $z$ must be on that plane, no matter what $y$ is.
There aren't that many possibilities. You know all the points on the surface must be on the plane $z=45x$ and your computer shows the surface is flat. Don't those agree? If you project the $xy$ "shadow" in the $xy$ plane up onto the slanted plane, what do you get?

4. Feb 17, 2013

### ohlala191785

So the description of the surface just an ellipse? If you project the circle onto the slanted plane, it should look like an ellipse.

5. Feb 17, 2013

### haruspex

I suppose you could describe it as an elliptical disc.

6. Feb 18, 2013

Oh OK.