Past exam of the mathematics course of the University of Cambridge / 2022 / ii / Paper 3 / 25I / c / i

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- Solution i

Solution

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) .

The supplied surface-of-revolution formula gives

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)

is a diffeomorphism from

(0, \infty)

to

R

and satisfies

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

. Therefore

ϕ (S (s, v)) = R (lo g (sinh s), v)

is a diffeomorphism and obeys

K_{R} \circ ϕ = K_{S}

. This is the curvature-matching diffeomorphism between two surfaces of revolution.

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