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

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.