 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2019/ii/paper-4/25h/a/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

Write

q (v) = f^{'} (v)^{2} + g^{'} (v)^{2} .

The regularity of the profile makes

q > 0

. The coordinate tangent vectors are

ϕ_{u} = (- f sin u, f cos u, 0), ϕ_{v} = (f^{'} cos u, f^{'} sin u, g^{'}),

E = f^{2}, F = 0, G = q^{2} .

N = \frac{1}{q} (- g^{'} cos u, - g^{'} sin u, f^{'}) .

Taking scalar products with the second derivatives of

ϕ

e = \frac{f g ^{'}}{q}, f_{II} = 0, g_{II} = \frac{f ^{'} g ^{''} - g ^{'} f ^{''}}{q} .

Thus the coordinate directions are principal directions, with principal curvatures

k_{1} = \frac{g ^{'}}{f q}, k_{2} = \frac{f ^{'} g ^{''} - g ^{'} f ^{''}}{q ^{3}} .

H = \frac{1}{2} (\frac{g ^{'}}{f q} + \frac{f ^{'} g ^{''} - g ^{'} f ^{''}}{q ^{3}}), K = \frac{g ^{'} ( f ^{'} g ^{''} - g ^{'} f ^{''} )}{f q ^{4}} .

Reversing the orientation of a surface reverses the sign of

H

but leaves

K

unchanged.

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