 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/25i/b/i/solution

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

Because the profile is parametrized by arc length,

Y^{'2} + Z^{'2} = 1.

ϕ_{u} = (Y^{'} cos v, Y^{'} sin v, Z^{'}), ϕ_{v} = (- Y sin v, Y cos v, 0),

I = d u^{2} + Y (u)^{2} d v^{2} .

With the orientation determined by

ϕ_{u} \times ϕ_{v}

N = (- Z^{'} cos v, - Z^{'} sin v, Y^{'}) .

e = N \cdot ϕ_{uu} = Y^{'} Z^{''} - Z^{'} Y^{''}, f = 0, g = N \cdot ϕ_{vv} = Y Z^{'},

and hence

II = (Y^{'} Z^{''} - Z^{'} Y^{''}) d u^{2} + Y Z^{'} d v^{2} .

Differentiating

Y^{'2} + Z^{'2} = 1

yields

Y^{'} Y^{''} + Z^{'} Z^{''} = 0

, from which

(Y^{'} Z^{''} - Z^{'} Y^{''}) Z^{'} = - Y^{''} .

Therefore

K = \frac{e g}{EG} = - \frac{Y ^{''}}{Y},

while

H = \frac{1}{2} (Y^{'} Z^{''} - Z^{'} Y^{''} + \frac{Z ^{'}}{Y})

for the chosen orientation.

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