= Solution
For $R=1$ and $z_1=0$, the <shortest path on a circular cylinder> has <length>
$$
\boxed{L_0(\phi_1)=\begin{cases}\phi_1,&0\leq\phi_1\leq\pi,\\2\pi-\phi_1,&\pi\leq\phi_1\leq2\pi.\end{cases}}
$$
Its graph is triangular, with endpoint values zero and maximum $\pi$ at $\phi_1=\pi$. In particular, increasing the angle to a full turn brings the endpoint back to the starting point.
For $R=1$ and $z_1=1$, the corresponding <length> is
$$
\boxed{L_1(\phi_1)=\sqrt{1+\min(\phi_1,2\pi-\phi_1)^2}.}
$$
This graph has endpoint values one, a maximum $\sqrt{1+\pi^2}$ at $\pi$, and symmetry about $\pi$. Each half is smooth, with horizontal slope at the outer endpoint; at $\pi$ the two halves meet at a corner, because the minimizing winding direction changes.
\Image[/past-exam-of-the-mathematics-course-of-the-year-1-of-the-natural-sciences-course-of-the-university-of-cambridge/2001/paper-2-q3-cylinder-shortest-paths.png]
{title=Unrolling a cylinder and choosing the shorter angular route}
{description=Original diagram. Left: two lifts of an endpoint at angular position $3\pi/2$ and height one; the negative angular increment $-\pi/2$ gives the shorter segment. Right: both requested shortest-length graphs for unit radius, with height differences zero and one.}
Back to article page
