= Solution
For a timelike tangent $v=dx/du$, put $L=\sqrt{-g(v,v)}>0$ and $\mathscr L=L^2/2=-g(v,v)/2$, and use the <Euler-Lagrange equations>. For the quadratic <geodesic Lagrangian>, they reduce to
$$
\frac d{du}(g_{\mu\nu}v^\nu)-\frac12\partial_\mu g_{\alpha\beta}v^\alpha v^\beta=0,
\qquad \nabla_vv=0.
$$
Here the first equation has been multiplied by an irrelevant minus sign. For the length Lagrangian, its derivatives are $L^{-1}$ times those of $\mathscr L$, and differentiating that factor yields instead
$$
\boxed{\nabla_vv=\frac{\dot L}{L}\,v.}
$$
This is the <unparametrized geodesic equation>, the key to why <timelike length and energy have the same geodesic images>. Define <proper time> locally by $d\tau/du=L>0$ and put $w=dx/d\tau=v/L$. The product rule then gives $\nabla_ww=L^{-2}[\nabla_vv-(\dot L/L)v]=0$ and $g(w,w)=-1$. Conversely an affinely parametrized timelike <geodesic> has constant $L$, by <metric compatibility>, and therefore satisfies both equations.
Thus \b[the two variational problems have the same oriented timelike geodesic images, after reparametrization]. They do not have exactly the same parametrized solutions with an arbitrary fixed $u$: in <Minkowski spacetime>, $t(u)=u^2$, $\mathbf x=0$, $u>0$, satisfies the length equation but not the quadratic equation. The timelike assumption excludes $L=0$, where this argument and the length derivative would fail.
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