Differential Geometry Question

[latentcorpse]

Problem1.3. Describe the one-sheeted hyperboloid as a surface of revolution;
that is, find a positive function f : R \rightarrow R such that
x(u, v)= \left[ \begin {array}{c} f \left( u \right) {\it cos}\nu <br /> \\\noalign{\medskip}f \left( u \right) {\it sin}\nu <br /> \\\noalign{\medskip}\nu\end {array} \right] parameterises the hyperboloid.

So far all I have is the equaiton of the hyperboloid is x^2+y^2-z^2=1 and no clue how to proceed. Help please?

 

[Dick]

How about putting your u,v expressions for x, y and z into the xy form and trying to solve for f(u)?

 

[latentcorpse]

i'm not too sure what you mean, in particular by the "xy form" you talk about?

could f(u)=x^2+y^2-z^2-1. if so then what is \nu

or am i waffling?

 

[Dick]

Waffling. x=f(u)*cos(v), y=f(u)*sin(v) etc. Put them into x^2+y^2-z^2=1.

 

[latentcorpse]

ok ill give that a bash. it's pretty of confusing of them to call that matrix x, no?

 

[Dick]

ok ill give that a bash. it's pretty of confusing of them to call that matrix x, no?

Probably should have called it X with a vector on it.

 

[latentcorpse]

ok. working that through i get f(u)=\sqrt{1+u^2}. is that the end of the question? it seems awfully short and yet i appear to have found an f as required.

also, how did you know to procede this way?

 

[Dick]

Sure. I knew because X(u,v) and x^2+y^2-z^2=1 are supposed to describe the same surface. X(u,v) had better satisfy the second equation.