Critical Points of the Length Functional I am tasked with finding the critical points of the following length functional $$ L(\lambda) = \int_{I} ||\lambda^{'}||dt$$ My guess would be to solve the Euler-Lagrange Equations. In doing so we would end up with a system of ODEs (perhaps just 1) and the solution to those equations are the critical points of L($\lambda)$. Is this correct or am I on the wrong path entirely?
 A: Obtaining an ODE by calculus of variations isn't too bad! The length of the curve is:
$$ L(\lambda) = \int_I \left( \sum_i \dot \lambda_i \dot \lambda_i \right)^{\frac 1 2} dt$$
where the $i$-index is a coordinate index.
The Euler-Lagrange equations say
$$ \frac{\delta L}{\delta \lambda_i} = \frac{d}{dt} \left( \frac{\delta L}{\delta \dot \lambda_i}\right)$$
for each $i$, i.e.
$$0 = \frac{d}{dt} \left( \frac{\dot \lambda_i}{\left( \sum_{j}\dot \lambda_j \dot \lambda_j \right)^{\frac 1 2}}\right)$$

This also works if you introduce a Riemannian metric $g_{ij}(\lambda)$:
$$ L(\lambda) = \int_I \left( \sum_{ij} g_{ij}(\lambda) \dot \lambda_i \dot \lambda_j \right)^{\frac 1 2} dt$$
The Euler-Lagrange equations now say (for each $i$):
$$ \frac{ \sum_{jk} \partial_i g_{jk}(\lambda) \dot \lambda_j \dot \lambda_k}{2 \left( \sum_{jk} g_{jk}(\lambda) \dot \lambda_j \dot \lambda_k \right)^{\frac 1 2}} = \frac{d}{dt} \left( \frac{ \sum_j g_{ij}(\lambda) \dot \lambda_j}{\left( \sum_{jk} g_{jk}(\lambda) \dot \lambda_j \dot \lambda_k \right)^{\frac 1 2}} \right)$$
Note that if we choose the parameter $t$ so that the curve $\lambda(t)$ is of unit velocity, i.e. if we choose $t$ so that $\sum_{ij} g_{ij}(\lambda) \dot \lambda_i \dot \lambda_j = 1$ everywhere, then this reduces to
$$ \ddot \lambda_i + \frac 1 2 \sum_{jkl}(g^{-1})^{il} \left( \partial_j g_{lk} + \partial_k g_{jl} - \partial_l g_{jk}\right)\dot \lambda_j \dot \lambda_k = 0.$$
And this is the geodesic equation! So it's not so bad after all.
