= Solution
Since $\gamma'=r_uu'+r_vv'$, we have $\gamma'\cdot r_u=Eu'+Fv'$. The normal-acceleration characterization of a <geodesic> gives
$$
\frac d{ds}(Eu'+Fv')=\gamma'\cdot(r_{uu}u'+r_{uv}v').
$$
Expanding 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 $r_v\cdot r_{uv}=G_u/2$. Therefore
$$
\boxed{\frac d{ds}(Eu'+Fv')=\tfrac12E_u(u')^2+F_uu'v'+\tfrac12G_u(v')^2.}
$$
Applying the same calculation to $\gamma'\cdot r_v=Fu'+Gv'$ gives
$$
\boxed{\frac d{ds}(Fu'+Gv')=\tfrac12E_v(u')^2+F_vu'v'+\tfrac12G_v(v')^2.}
$$
These are also the <Euler-Lagrange equations> of the quadratic metric energy.
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