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Differential Geometry Question

  1. Jan 27, 2009 #1
    Problem1.3. Describe the one-sheeted hyperboloid as a surface of revolution;
    that is, find a positive function [itex]f : R \rightarrow R [/itex] such that
    [itex]x(u, v)= \left[ \begin {array}{c} f \left( u \right) {\it cos}\nu
    \\\noalign{\medskip}f \left( u \right) {\it sin}\nu
    \\\noalign{\medskip}\nu\end {array} \right] [/itex] parameterises the hyperboloid.

    So far all I have is the equaiton of the hyperboloid is [itex]x^2+y^2-z^2=1[/itex] and no clue how to proceed. Help please?
     
  2. jcsd
  3. Jan 27, 2009 #2

    Dick

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    How about putting your u,v expressions for x, y and z into the xy form and trying to solve for f(u)?
     
  4. Jan 27, 2009 #3
    i'm not too sure what you mean, in particular by the "xy form" you talk about???

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

    or am i waffling?
     
  5. Jan 27, 2009 #4

    Dick

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    Waffling. x=f(u)*cos(v), y=f(u)*sin(v) etc. Put them into x^2+y^2-z^2=1.
     
  6. Jan 28, 2009 #5
    ok ill give that a bash. it's pretty of confusing of them to call that matrix x, no?
     
  7. Jan 28, 2009 #6

    Dick

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    Probably should have called it X with a vector on it.
     
  8. Jan 28, 2009 #7
    ok. working that through i get [itex]f(u)=\sqrt{1+u^2}[/itex]. 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?
     
  9. Jan 28, 2009 #8

    Dick

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    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.
     
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