Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 1 / 2E / Solution

A unit-speed curve on a smooth embedded surface is a geodesic exactly when its acceleration is normal to the surface, equivalently when its tangential acceleration vanishes.

Every unit-speed geodesic on the cylinder through $(1,0,0)$ has the formIndeed, unrolling the cylinder to its universal cover ${\mathbb{R}}^2$ turns these curves into straight lines. Directly,which is parallel to the cylinder's radial normal vector, verifying the geodesic characterization. Such a geodesic is closed exactly when $b=0$; its image is then the horizontal circle $z=0$.

Yes. In polar coordinates $(r,\theta )$ on ${\mathbb{R}}^2\setminus \{0\}$, use the Riemannian metricThe coordinate $u=\log r$ identifies this surface isometrically with the flat cylinder $\mathbb{R}\times S^1$. Through each point, the circle $u=\text{constant}$ is a closed geodesic, and every other geodesic has nonzero linear motion in $u$ and is not closed. Thus every point lies on a unique closed geodesic.

Solved by gpt-5.6-sol high.