 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2024/ii/paper-1/26j/c/solution

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

Put

W = 1 + h_{x}^{2} + h_{y}^{2}

. An upward-pointing Gauss map is

N = \frac{( - h _{x} , - h _{y} , 1 )}{W} .

v, w

d ϕ_{t} = d ϕ + t d N

gives

\frac{d}{d t}_{t = 0} I^{S_{t}} (v, w) = ⟨ d N (v), d ϕ (w)⟩ + ⟨ d ϕ (v), d N (w)⟩ .

The Weingarten map is self-adjoint and

S = - d N

, so the right side is

- 2 I I^{S} (v, w)

. Hence

I I^{S} = - \frac{1}{2} \frac{d}{d t}_{t = 0} I^{S_{t}} .

Alternatively differentiating

⟨ N, ϕ_{i} ⟩ = 0

gives

I I_{ij} = - ⟨ N_{i}, ϕ_{j} ⟩ = ⟨ N, ϕ_{ij} ⟩ .

Therefore

I I^{S} = \frac{1}{W} (h_{xx} h_{x y} h_{x y} h_{yy}) .

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