Description of surface, vector calculus

ohlala191785
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Homework Statement


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∏.

Homework 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)>.

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!
 
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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.
 
ohlala191785 said:

Homework Statement


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∏.


Homework 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)>.

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

The Attempt at a Solution


In the x-y plane, this is a circle.

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?

x = v cos(u) so z = 45x. This is a slanted plane?
Yes. And ##x## and ##z## must be on that plane, no matter what ##y## is.
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!

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?
 
So the description of the surface just an ellipse? If you project the circle onto the slanted plane, it should look like an ellipse.
 
ohlala191785 said:
So the description of the surface just an ellipse? If you project the circle onto the slanted plane, it should look like an ellipse.
I suppose you could describe it as an elliptical disc.
 
Oh OK.
Thank you for your help.
 
There are two things I don't understand about this problem. First, when finding the nth root of a number, there should in theory be n solutions. However, the formula produces n+1 roots. Here is how. The first root is simply ##\left(r\right)^{\left(\frac{1}{n}\right)}##. Then you multiply this first root by n additional expressions given by the formula, as you go through k=0,1,...n-1. So you end up with n+1 roots, which cannot be correct. Let me illustrate what I mean. For this...

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