I am working through Tom Kibbles' Classical Mechanics and am struggling to derive an equation in the excercises on calculus of variations.
The question is to find the geodesics of a cone with semi-vertical angle $\alpha$. Let the cone by specified by the distance r from the vertex and azimuth angle $\phi$ about the axis. You are told to prove:
$$ r\frac{d^2r}{d\phi^2} - 2\left ( \frac{dr}{d\phi } \right )^2 -r^2sin^2\alpha = 0 $$
My Attempt:
I started with $ ds = \sqrt{dr^2 + r^2sin^2(\alpha) d\phi^2} $
Then my function to maximize was $ F = \sqrt{\left ( \frac{dr}{d\phi } \right )^2 + r^2sin^2(\alpha)}$
However the problem I am having is when I calculate $\frac{\partial F}{\partial r} $ and $ \frac{\partial F}{\partial r'}$ I get the answer above without the factor of 2 infront of $r'^2$.
My equations for $\frac{\partial F}{\partial r} $ and $ \frac{\partial F}{\partial r'}$ where:
$$ \frac{\partial F}{\partial r} = \frac{rsin^2(\alpha)}{\sqrt{\left ( \frac{dr}{d\phi} \right )^2+r^2sin^2(\alpha)}}$$
$$ \frac{\partial F}{\partial r} = \frac{\frac{dr}{d\phi}}{\sqrt{\left ( \frac{dr}{d\phi} \right )^2+r^2sin^2(\alpha)}}$$
Any advice on where I may have made a mistake would be appreciated thank you.
The partial derivatives $\frac{\partial F}{\partial r},\frac{\partial F}{\partial r'}$ are well developed. The problem perhaps appears on $\frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)$ which gives
$$ \frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)=\frac{\sin ^2(\alpha ) r(\phi ) \left(r(\phi ) r''(\phi )-r'(\phi )^2\right)}{\left(\sin ^2(\alpha ) r(\phi )^2+r'(\phi )^2\right)^{3/2}} $$ so following
$$ \frac{\partial F}{\partial r}-\frac{d}{d\phi}\left(\frac{\partial F}{\partial r'}\right)=\frac{\sin ^2(\alpha ) r(\phi ) \left(\sin ^2(\alpha ) r(\phi )^2-r(\phi ) r''(\phi )+2 r'(\phi )^2\right)}{\left(\sin ^2(\alpha ) r(\phi )^2+r'(\phi )^2\right)^{3/2}}=0 $$