# Calculus of Variations (Geodesics on a Cone)

1. Aug 27, 2015

### B3NR4Y

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

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.
$s = \int_{a}^{b} \sqrt{d\rho^{2} + \rho d\theta{2} + dz^{2} } = \int_{a}^{b} \sqrt{1 + \rho \theta ^{'2} + z^{'2}} d\rho$, 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?

Last edited: Aug 27, 2015
2. Aug 27, 2015

### TSny

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

Last edited: Aug 28, 2015
3. Aug 28, 2015

### 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.