Past exam of the mathematics course of the University of Cambridge / 2014 / ib / Paper 2 / 15C / Solution

For fixed endpoints, the Euler-Lagrange equation isLet the sphere have radius $R$ and write the speed as $v_0\sin \theta$, with $v_0>0$. The infinitesimal travel time in the interior $0<\theta <\pi$ isThis is the flat-cylinder travel metric for sine-latitude speed. IntroduceThe travel-time metric becomes $(R/v_0)\sqrt{d{z}^2+d{\varphi }^2}$, an ordinary Euclidean metric on a cylinder because $\varphi$ is periodic. After choosing a lift of $\varphi$, shortest paths are straight segments in $(\varphi ,z)$. For a route that is a graph over $\varphi$, the Euler-Lagrange equation for $\sqrt{1+(dz/d\varphi {)}^2}$ readsThus $z^{\prime }=A$ is constant and $z=A\varphi +B$. Transforming back,This proves the requested form for nonmeridional routes, including the equator when $A=B=0$. It is also globally minimizing for its chosen azimuthal lift: Euclidean arc length is at least the straight-line separation, and the shortest lift minimizes that separation among angular differences differing by $2\pi$.

There is a coordinate exception to the printed universal form. When the shortest angular separation is zero, the fastest path is the meridian $\varphi =\text{constant}$; it is a vertical straight line on the cylinder and cannot be represented by finite $A,B$ as a graph over $\varphi$. The poles correspond to $z=\pm \infty$ and cannot be reached in finite time with this speed law.