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Mathematics course of the University of Cambridge

Past exam of the mathematics course of the University of Cambridge

2022

Paper 3

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
$$
K_R(u)=-\frac1{(1+e^{2u})^2},
\qquad
K_S(s)=-\frac1{\cosh^4s}.
$$
Since $s>0$, the change of coordinate
$$
u=\log(\sinh s)
$$
is a diffeomorphism from $(0,\infty)$ to $\mathbb R$ and satisfies $1+e^{2u}=\cosh^2s$. Therefore
$$
\boxed{
\phi(S(s,v))=R(\log(\sinh s),v)
}
$$
is a diffeomorphism and obeys $K_R\circ\phi=K_S$. This is the <curvature-matching diffeomorphism between two surfaces of revolution>.

Solution

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

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

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

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