Calculus of Variations (Geodesics on a Cone)

[B3NR4Y]

Homework Statement

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

Homework Equations

\frac{\partial F}{\partial y} - \frac{d}{dx} (\frac{\partial F}{\partial y'}) = 0 and ds^{2} = d\rho^{2} + \rho d\theta^{2} + dz^{2}

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.
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, I have an equation for z so z' is easy to find, it's just λ. \int_a^b \sqrt{1 + \rho \theta^{'2} + \lambda^{2} } d\rho.

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 \frac{\rho \theta ^{'}}{\sqrt{1+\rho \theta^{'2} + \lambda^{2}}} = C, some algebra gets me to \theta^{'} = \sqrt{\frac{C+C\lambda^{2}}{\rho^{2}(\rho^{2} - C)}} which integrating with respect to rho yields \theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho

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 \theta =\int \sqrt{\frac{C(1+\lambda ^2)}{\rho ^{2}(\rho ^{2}-C})} d\rho. Boy is this a hard integral to do, wolfram spits out some pretty crazy stuff. Where should I start integrating this?

 

[TSny]

Where should I start integrating this?

Try letting $\rho = \sqrt{C} \sec u$ and express the integral in terms of the variable $u$.

 

[haruspex]

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.