We have previously shown that if $\gamma_0(t): [0, 1] \to \Sigma^2 \subset \mathbb{R}^3$ is the shortest path joining points $p := \gamma_0(0)$ and $q := \gamma_0(1)$, then one must have
$$ \nabla_{\gamma_0(t)} \frac{\gamma_0'(t)}{||\gamma_0'(t)||} = 0. $$
Now consider another similar Calculus of Variations problem. Define the kinetic energy of a path $\gamma(t): [0, 1] \to \Sigma^2 \subset \mathbb{R}^3$ to be
$$ E(\gamma) := \frac{1}{2} \int_{t=0}^{t=1} ||\gamma'(t)||^2 dt. $$
Show that if $\gamma_0: [0, 1] \to \Sigma^2 \subset \mathbb{R}^3$ is the path with the lowest kinetic energy among all paths joining $p := \gamma_0(0)$ and $q := \gamma_1(0)$, then we have
$$ \nabla_{\gamma_0(t)} \gamma_0'(t) = 0. $$