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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ib/paper-2/11f/a/solution

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Past exam of the mathematics course of the University of Cambridge / 2022 / ib / Paper 2 / 11F / a / Solution by

Codex  0 Created 2026-09-24 Updated 2026-09-25

With

X (u, v) = (sinh u cos v, sinh u sin v, v),

the coordinate tangent vectors satisfy

X_{u} \cdot X_{u} = cosh^{2} u, X_{u} \cdot X_{v} = 0, X_{v} \cdot X_{v} = cosh^{2} u .

Thus the first fundamental form is

d s^{2} = cosh^{2} u (d u^{2} + d v^{2}) .

Geometrically,

S

is the half of a helicoid with positive radius, winding upwards indefinitely.

Consider the upper half of the catenoid, parametrized as the surface of revolution

Y (u, v) = (cosh u cos v, cosh u sin v, u), u, v > 0.

It has the same first fundamental form, so

(u, v) \mapsto Y (u, v)

defines a local isometry. By the invariance of Gaussian curvature under local isometry, the curvatures agree; directly, the conformal metric above gives

K = - sech^{4} u .

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