Past exam of the mathematics course of the University of Cambridge / 2019 / ii / Paper 3 / 25H / b

Because $f^{\prime }(a)$ and $f^{\prime }(b)$ are nonzero, $r=f(v)$ is a valid one-sided local coordinate at each endpoint. Defineusing the inverse near $a$, and define $h_b$ analogously near $b$. The necessary and sufficient conditions are:

Indeed, the second condition makes the surface near the corresponding pole the graphwhich is a smooth regular graph over the horizontal tangent plane. Together with part a(i), these two pole charts cover the closure by regular surface charts. The surface is compact because it is the continuous image of the compact cylinder $[0,2\pi ]\times [a,b]$ after each boundary circle is collapsed to its pole.

Conversely, rotational invariance forces the tangent plane at a pole to be horizontal. The surface is therefore locally the graph of a smooth radial function of $(x,y)$. A smooth rotation-invariant function near the origin is a smooth function of $x^2+y^2$, which gives the displayed condition. The poles must be distinct, since otherwise a punctured neighbourhood of their common image would have two components and could not be a surface chart. This is the smooth endpoint criterion for a surface of revolution. In particular it implies the familiar first-order conditions $g^{\prime }(a)=g^{\prime }(b)=0$, but those conditions alone do not ensure smoothness of every order.

Letbe the Clairaut first integral for a surface of revolution from part a(ii). If $c\ne 0$, then $|\cos \theta |\le 1$ givesfor every $s$. Since $f$ tends to zero at both poles, the geodesic avoids nonempty open neighbourhoods of the poles.

If $c=0$, then $u^{\prime }(s)=0$ whenever the geodesic is away from a pole. It consequently lies in one meridian of a surface of revolution; after crossing a pole, the angular coordinate changes by $\pi$ but remains in the same plane through the rotation axis. A small surface neighbourhood of any point outside that meridian is disjoint from the geodesic. In either case there is a nonempty open set $U$ withThis proves that the geodesic image on a compact two-pole surface of revolution is not dense.