A sphere in $\mathbb{R}^3$ is parametrically given by $$ x = a \sin(\theta) \cos(\phi) \\ y = a \sin(θ) \sin(\phi)\\ z = a \cos(θ) $$ where $a > 0$ is the radius of the sphere, $θ$ is the polar angle $(0 < θ < π)$ and $\phi$ is the azimuthal angle $(0 < \phi < 2π)$ .
Find a differential equation for geodesics $\phi = \phi(\alpha) $ on the sphere. Show that for arbitrary constants $\phi_0$ and $\alpha$, the function $\phi(\theta) - \phi_0 = \sin^{-1} (\alpha \cos (\theta))$ satisfies the differential equation.
My solution So, I think I know how the solution goes, however I am stuck on finding derivative in the middle, and after that I do not know how to proceed. So first we need basically to find geodesics. We do that knowing that $ds^2=dx^2+dy^2+dz^2$. Then we get that $ds^2 = a^2d\theta^2 + a^2\sin^2(\theta)\, d\phi^2$. (You can trust me with that). Hence we need to find a minimum of an integral $$\int_A^B\sqrt{a^2d\theta^2 + a^2\sin^2(\theta)\, d\phi^2} = a \int_A^B\sqrt{1+\sin^2(\theta)\phi^{'2}(x)}\,d\theta$$ since $\phi=\phi(\theta) \rightarrow d\phi = \phi'(\theta)\,d\theta$. Hence to find a minimum to that integral we need to find a solution to Euler Lagrange equation $\dfrac{dF}{d\phi} - \dfrac{d}{dx}\left(\dfrac{d\phi}{d\phi'}\right)=0$. $\dfrac{dF}{d\phi} =0$ since our equation is independent of $\phi$.
So we get that $$\frac{dF}{d{\phi'}} =\text{const}$$ Now could anyone please help me to find $$\frac{dF}{d{\phi'}}$$ with an explanation please.