I'm trying to derive the extremal solutions to the Lagrangian for arclength on the unit sphere by setting up the Euler-Lagrange equations.
Starting from
$$L = \sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2} \ $$
I see not long after that
$$\frac{\partial L}{\partial \dot\theta}=\frac{\dot\theta}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\theta}=\frac{\dot\phi^2\sin(\theta)\cos(\theta)}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\dot\phi}=\frac{\sin^2(\theta)\dot\phi}{\sqrt{\dot \theta^2 + \sin^2 \theta \ \dot \phi^2}}\\ \frac{\partial L}{\partial\phi}=0$$
I'm pretty sure I've understood the planar case, and that had taught me to recognize that in the last two equations, I could then say that $$\frac{\partial L}{\partial\dot\phi}=C_1$$ a constant $C_1$.
Then one can rewrite
$$\frac{\partial L}{\partial\theta}=C_1\dot\phi\cot(\theta)\\ \frac{\partial L}{\partial\dot\theta}=C_1\frac{\dot\theta}{\sin^2(\theta)\dot\phi}$$
but then $$\frac{d}{dt}\frac{\partial L}{\partial\dot\theta}=\frac{\partial L}{\partial \theta}$$ still seems to be a second order differential equation in both variables, and I feel like I've run aground.
How do you go from here? Maybe I'm supposed to do more at the step where I found $C_1$ to make some reductions?
The final goal is to see concretely that the answer is going to yield a single solution when I'm going from $(0,0)$ and $(\pi/2, \pi/2)$, say, and that there are infinitely many extremals when going between antipodal points.