Past exam of the geometry course of the University of Oxford / 2009 / A / 3

For a regular parametrization $r(u,v)$, the first fundamental form is the induced metricIt measures the lengths and angles of tangent vectors. A geodesic is a curve whose unit tangent has zero covariant derivative along itself. For an arc-length parametrized surface curve in Euclidean three-space, this means its acceleration is normal to the surface, or ${\gamma }^{\prime \prime }\cdot r_u={\gamma }^{\prime \prime }\cdot r_v=0$.

Since ${\gamma }^{\prime }=r_u{u}^{\prime }+r_v{v}^{\prime }$, we have ${\gamma }^{\prime }\cdot r_u=E{u}^{\prime }+F{v}^{\prime }$. The normal-acceleration characterization of a geodesic givesExpanding and differentiating the metric coefficients yields $r_u\cdot r_{uu}=E_u/2$, $r_v\cdot r_{uu}+r_u\cdot r_{uv}=F_u$, and $r_v\cdot r_{uv}=G_u/2$. ThereforeApplying the same calculation to ${\gamma }^{\prime }\cdot r_v=F{u}^{\prime }+G{v}^{\prime }$ givesThese are also the Euler-Lagrange equations of the quadratic metric energy.

For the surface of revolution, the arc-length condition on its profile gives $E=1$, $F=0$, $G=f(u{)}^2$. Thus the geodesic equations reduce toConsequentlythe Clairaut first integral for a surface of revolution.

On a meridian, take $v=b$ and $u=\pm s+u_0$. Both equations hold, so every meridian is a geodesic. On a cross-section $u=c$, arc length requires $v^{\prime }=\pm 1/f(c)$. The first equation then holds exactly when $f^{\prime }(c)=0$, while the second holds automatically. Hence