 Solution

For fixed endpoint values, independent variations of $u$ and $v$ give the Euler-Lagrange equationsThe sphere parametrization has metric $d{s}^2=a^2(d{\theta }^2+{\sin }^2\theta d{\varphi }^2)$, obtained by differentiating its position vector and taking its Euclidean squared length. The displayed functional is therefore one half the integral of squared speed, the geodesic energy. Its fixed-endpoint extremals are affinely parametrized geodesics; energy stationarity chooses a constant-speed parametrization.

Its explicit equations, after dividing by $a^2$, areThe second gives ${\sin }^2\theta \dot{\varphi }=h$. Since the Lagrangian has no explicit time dependence, its energy $\dot{q}\cdot L_{\dot{q}}-L=L$ is conserved. ThusThe same second identity follows by differentiating it and substituting the two equations. These polar-coordinate formulas hold on a nonsingular coordinate patch; the geodesic curve itself can pass through a pole in another chart.