Help deriving geodesic of $S^2$ by considering small deviations For $s(t)$ the geodesic confined to the surface of a (3D) sphere, how does one get $\|\dot{s}\|^2 s + \ddot{s} = 0$ by setting $\frac{d}{d\delta} \left( \int \|\frac{d}{dt} \frac{s(t)+\delta h(t)}{\|s(t)+\delta h(t) \|}\|^2 dt \right) = 0$ at $\delta = 0$ for any arbitrary (but smooth) $h(t)$?
Is brute-force differentiation of the expression (horribly ugly and I have not succeeded in arriving at the desired expression) the only way to do it?
Thanks!
Edit: Just wanted to clarify that I am asking for how to carry out the calculation using, in particular, the $\frac{d}{d\delta} (...) = 0$ method.
(Thanks to all who have pointed out the ambiguities.)
Oh no! I have just realized that I foolishly left out bits of the derivand! (Was too bogged down by the formatting and clearly wasn't checking properly.) Would the question make more sense now? (Sorry :S)
 A: First a general piece of advice in response to your assessment of the derivative as "horribly ugly": In this sort of calculation, it simplifies things quite considerably if you keep in mind while differentiating that you're about to set $\delta=0$; then you can immediately drop everything proportional to $\delta$ that you're not differentiating.
Keeping that in mind, taking the derivative with respect to $\delta$ and setting $\delta=0$ yields:
$$
\begin{eqnarray}
\left.\frac{\mathrm d}{\mathrm d\delta}\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)^2\right|_{\delta=0}
&=&
\left.2\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)\frac{\mathrm d}{\mathrm d\delta}\left(\frac{\mathrm d}{\mathrm dt}\frac{s+\delta h}{|s+\delta h|}\right)\right|_{\delta=0}
\\
&=&
2\left(\frac{\mathrm d}{\mathrm dt}\frac{s}{|s|}\right)\frac{\mathrm d}{\mathrm dt}\left(\frac{h}{|s|}-\frac{(h\cdot s)s}{|s|^3}\right)
\\
&=&
2\dot s\frac{\mathrm d}{\mathrm dt}\left(h-(h\cdot s)s\right)\;,
\end{eqnarray}$$
where in the end I used $|s|=1$. We can use integration by parts to move the derivative to the first factor to conclude that $\ddot s\cdot (h-(h\cdot s)s)$ must vanish. The second factor is the component of $h$ orthogonal to $s$, so this implies $\ddot s=\lambda s$. We can find $\lambda$ by differentiating $0=s\cdot\dot s$, which yields $0=s\cdot\ddot s+\dot s\cdot\dot s=\lambda s\cdot s+\dot s\cdot\dot s=\lambda+\dot s\cdot\dot s$, so $\lambda=-\dot s\cdot\dot s$, and thus $\ddot s=-(\dot s\cdot\dot s)s$. 
