Euler-Lagrange implies that geodesics has constant norm of velocity

Question

I am trying to prove that geodesics has norm of velocity constant.

To do this, I applied that Euler-Lagrange equation to $$S(\gamma,\gamma') := \int_{0}^1 \|\gamma'(t)\|^2dt.$$

The Euler equation for this functional is:

$$\frac{d}{dt} \frac{\partial}{\partial \gamma'} \|\gamma'(t)\|^2 = 0.$$

How can I conclude that $\|\gamma'(t)\| = cte$?

Thanks a lot...

Answer 1

$$ 2 \frac{d}{dt} { \|\gamma'\|} = 0, $$

Integrand is a constant minimizing length. Hope brief reply ok , as it is all that is required.