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A relatively easy differential geometry question concerning principle curvatures

  1. Sep 26, 2011 #1
    1. The problem statement, all variables and given/known data

    Show the principal curvatures on x sin z - y cos z = 0 are +-1/(1 + x^2 + y^2)
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Sep 26, 2011 #2
    no attempt?
     
  4. Sep 27, 2011 #3
    Got a reply from the OP:

    Decided to post that here along with my help for future reference, in case other people google his question and arrive at this page.

    We can rewrite (as noted by the OP) the equation to x tan(z) = y.
    Using this, we can parametrize the surface as follows:
    [tex]x(u,v) = (u, u \tan v, v)[/tex]
    Note: x is now not the first coordinate, but rather the position vector

    The only economical way I know of calculating the principal values, is using the formulae [itex]k_{1,2} = H \pm \sqrt{H^2 - K}[/itex].

    So we need to calculate H and K. This can be easily done using the parametrization of the surface and the following:
    [tex]H \propto En - 2Fm + Gl \qquad \qquad K = \frac{ln-m^2}{EG-F^2}[/tex]
    where [itex]E:= x_u \cdot x_u \qquad F:= x_u \cdot x_v \qquad G = x_v \cdot x_v[/itex]
    and [itex]l := x_{uu} \cdot \xi \qquad m := x_{uv} \cdot \xi \qquad n := x_{vv} \cdot \xi[/itex]
    where xi is the normal [itex]\xi = x_u \times x_v[/itex].

    I've only given H up to something proportional to it, because you'll only have to prove that H = 0, which you can expect by comparing the formulae for k_1 and k_2 given above with what you know will be the results.
     
    Last edited: Sep 27, 2011
  5. Sep 27, 2011 #4
    Mr Vodka:

    What you are showing me is what I finally figured out last night. You have basically set this out as a Monge Patch (using the nomenclature of the text) which then allows calculation of the fundamental forms prior to solving for k1 and k2. This requires some plowing through, which I am in the middle - end of doing, albeit with much greater confidence as a result of now reading your note. I consider my finally figuring this out a result of thinking about my reply to you. I am very grateful. While I am not Erdos and thus can not offer a financial reward, a caramel apple appears in order. Are you near Albuquerque NM?

    Again, grateful thanks for allowing my life to continue (though I will still need that final push).

    Regards,

    Jeff
     
  6. Sep 27, 2011 #5
    No caramel apple for me, I'm a European citizen ;) But you're welcome!

    If you get stuck with the calculations, let me know.
     
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