Past exam of the mathematics course of the year 1 of the natural sciences course of the University of Cambridge / 2001 / Paper 2 / 3 / b / Solution

On the circular cylinder, $\rho =R$ is constant, so the given line element reduces to $d{s}^2=R^{2}d{\varphi }^2+d{z}^2$. Unroll the circular cylinder with $u=R\varphi$. The line element is then $d{s}^2=d{u}^2+d{z}^2$, so the usual Euclidean geometry formula measures arc length in the $(u,z)$ plane.

The endpoint has infinitely many lifts $(R({\varphi }_1+2\pi k),z_1)$, with $k\in \mathbb{Z}$. For each lift, the straight line segment is shortest; its length is $\sqrt{R^2({\varphi }_1+2\pi k{)}^2+z_1^2}$. Comparing the lifts gives the unrestricted shortest path on a circular cylinder:If $\delta$ is a minimizing angular increment, a shortest geodesic is parametrized by $(\rho ,\varphi ,z)=(R,\delta t,z_{1}t)$, $0\le t\le 1$. This is a helix when both increments are nonzero, and a circular arc or vertical generator in the limiting cases. A nonminimal lift also gives a geodesic, but it is only shortest within its fixed winding class.

For $0\le {\varphi }_1\le 2\pi$, put $m=\min ({\varphi }_1,2\pi -{\varphi }_1)$. Then $\boxed{L_{\min }=\sqrt{R^2{m}^2+z_1^2}}$. At ${\varphi }_1=\pi$, both winding directions give distinct shortest paths of equal length.