## Geodesic on a cone, calculus of variations

I have to find the geodesics over a cone. I've used cylindrical coordinates. So, I've defined:

$$x=r \cos\theta$$
$$y=r \sin \theta$$
$$z=Ar$$

Then I've defined the arc lenght:
$$ds^2=dr^2+r^2d\theta^2+A^2dr^2$$

So, the arclenght:
$$ds=\int_{r_1}^{r_2}\sqrt { 1+A^2+r^2 \left ( \frac{d\theta}{dr}\right )^2 }dr$$

And using Euler-Lagrange equation:
$$\frac{\partial f}{\partial \theta}=0$$
$$\frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}$$
$$\frac{d}{dr}\frac{\partial f}{\partial \dot \theta}=0 \rightarrow \frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}=K$$

The differential equation which I've arrived is non linear. I don't know if what I've done is fine. I know the problem can also be done using spherical coordinates instead of cylindrical, but I've choossen to do it this way.

 Quote by Telemachus I have to find the geodesics over a cone. I've used cylindrical coordinates. So, I've defined: $$x=r \cos\theta$$ $$y=r \sin \theta$$ $$z=Ar$$ Then I've defined the arc lenght: $$ds^2=dr^2+r^2d\theta^2+A^2dr^2$$ So, the arclenght: $$ds=\int_{r_1}^{r_2}\sqrt { 1+A^2+r^2 \left ( \frac{d\theta}{dr}\right )^2 }dr$$ And using Euler-Lagrange equation: $$\frac{\partial f}{\partial \theta}=0$$ $$\frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}$$ $$\frac{d}{dr}\frac{\partial f}{\partial \dot \theta}=0 \rightarrow \frac{\partial f}{\partial \dot \theta}=-\frac{r^2\frac{d\theta}{dr}}{\sqrt{1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2}}=K$$ The differential equation which I've arrived is non linear. I don't know if what I've done is fine. I know the problem can also be done using spherical coordinates instead of cylindrical, but I've choossen to do it this way.
Is it separable? I didn't simplify your expression, but it looks like you might have

$d\theta=g(r)dr$
 Yes, you're right: $$r^2\frac{d\theta}{dr}=K \sqrt{ 1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2 }$$ $$r^4 \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2+r^2\left ( \frac{d\theta}{dr}\right )^2 \right ]$$ $$r^4 \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2+r^2\left ( \frac{d \theta}{dr}\right )^2 \right ]$$ $$(r^4-r^2\gamma) \left ( \frac{d\theta}{dr} \right )^2=\gamma \left [ 1+A^2 \right ]$$ $$d\theta=\sqrt{ \frac { \eta }{r^2(r^2-\gamma) }}dr$$ eta and gamma are constants. Anyway, I wanted to know if what I did was ok. I don't care that much about the integral :P Thank you algebrat.