Let $S\subseteq \mathbb{R}^3$ be a regular surface with Riemannian metric $g$.
Moreover let $p\in S$, $v\in T_pS$ and $c$ a geodesic with $c(0)=p$ and $c'(0)=v$.
I want to show that for any constant $\alpha\in\mathbb{R}$ the curve $c_\alpha(t):=c(\alpha t)$ is also a geodesic with respect to the boundary conditions $c_\alpha(0)=p$ and $c_\alpha'(0)=\alpha v$.
My attempt is to verify that the energy functionals are the same.
But why is
$$E[c_\alpha]=\frac{1}{2}\int_Ig_{c_\alpha(t)}(c'_\alpha(t),c'_\alpha(t))dt=\frac{1}{2}\int_I\alpha^2g_{c(\alpha t)}(c'(\alpha t),c'(\alpha t))dt$$
the same as
$$E[c]=\frac{1}{2}\int_I g_{c(t)}(c'(t),c'(t))dt\quad\text{?}$$
I tried to use integration by substitution, but this would change the integration interval $I$.
I would appreciate any help. Thanks for your attention!