 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/ii/paper-3/25i/c/i/solution

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

Use angular coordinate

v

. The two surfaces have parametrizations

R (u, v) = (e^{u} cos v, e^{u} sin v, u),

and

S (s, v) = (cosh s cos v, cosh s sin v, s) .

K_{R} (u) = - \frac{1}{( 1 + e ^{2 u} ) ^{2}}, K_{S} (s) = - \frac{1}{c o s h ^{4} s} .

Since

s > 0

, the change of coordinate

u = lo g (sinh s)

(0, \infty)

R

and satisfies

1 + e^{2 u} = cosh^{2} s

. Therefore

ϕ (S (s, v)) = R (lo g (sinh s), v)

K_{R} \circ ϕ = K_{S}

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