 Solution

ID: past-exam-of-the-geometry-course-of-the-university-of-oxford/2009/a/3/iii/solution

For the surface of revolution, the arc-length condition on its profile gives

E = 1

F = 0

G = f (u)^{2}

. Thus the geodesic equations reduce to

u^{''} = f f^{'} (v^{'})^{2}, \frac{d}{d s} (f^{2} v^{'}) = 0.

Consequently

f (u)^{2} v^{'} = constant,

On a meridian, take

v = b

and

u = \pm s + u_{0}

. Both equations hold, so every meridian is a geodesic. On a cross-section

u = c

, arc length requires

v^{'} = \pm 1/ f (c)

. The first equation then holds exactly when

f^{'} (c) = 0

, while the second holds automatically. Hence

u = c is a geodesic ⟺ f^{'} (c) = 0.

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