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Calculus of Variations (Geodesics on a Cone)

  1. Aug 27, 2015 #1

    B3NR4Y

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    1. The problem statement, all variables and given/known data
    Find the geodesics on the cone whose equation in cylindrical-polar coordinates is z = λρ [Let the required curve have the form φ=φ(ρ)] check your result for the case λ→0

    2. Relevant equations
    [itex] \frac{\partial F}{\partial y} - \frac{d}{dx} (\frac{\partial F}{\partial y'}) = 0 [/itex] and [itex] ds^{2} = d\rho^{2} + \rho d\theta^{2} + dz^{2} [/itex]

    3. The attempt at a solution
    The integral of ds gives the length of a curve, a geodesic is a curve with the shortest length between two points on a curved surface. So the Euler Equation is necessary.
    [itex] s = \int_{a}^{b} \sqrt{d\rho^{2} + \rho d\theta{2} + dz^{2} } = \int_{a}^{b} \sqrt{1 + \rho \theta ^{'2} + z^{'2}} d\rho [/itex], I have an equation for z so z' is easy to find, it's just λ. [itex] \int_a^b \sqrt{1 + \rho \theta^{'2} + \lambda^{2} } d\rho [/itex].

    From here I just used the Euler-Equation, noting that the lack of a θ means that ∂F/∂θ' = constant by the Euler-Equation, so [itex]\frac{\rho \theta ^{'}}{\sqrt{1+\rho \theta^{'2} + \lambda^{2}}} = C [/itex], some algebra gets me to [itex] \theta^{'} = \sqrt{\frac{C+C\lambda^{2}}{\rho^{2}(\rho^{2} - C)}} [/itex] which integrating with respect to rho yields [itex]\theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho[/itex]

    I'm not sure if this is correct or not. I assume if they want me to check the result for the case λ=0, that case will look familiar to me. It doesn't.

    I made a boo boo, rho is squared in the line element. I'll work it out, but I feel like I still need help.

    So I fixed my mistake, and have decided [itex] \theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho [/itex]. Boy is this a hard integral to do, wolfram spits out some pretty crazy stuff. Where should I start integrating this?
     
    Last edited: Aug 27, 2015
  2. jcsd
  3. Aug 27, 2015 #2

    TSny

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    Try letting ##\rho = \sqrt{C} \sec u## and express the integral in terms of the variable ##u##.
     
    Last edited: Aug 28, 2015
  4. Aug 28, 2015 #3

    haruspex

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    You might find it a bit easier if you leave C as you originally introduced it, so that every C in your integral becomes C2.
    A hypergeometric substitution for rho might then suggest itself.

    Edit: TSny's response did not appear for me until I submitted my own, hours later. Strange. Yes, sec will work as well or better.
     
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