Consider the arclength functional $$ S[y] = \int_a^b \sqrt{1 + (y')^2} \ dx $$ The functional derivative is given by $$ \frac{\delta S}{\delta y} = \frac{\partial L}{\partial y} - \frac{d}{dx} \left( \frac{\partial L}{\partial y'} \right) $$ with $L(y, y', x) = \sqrt{1 + (y')^2}$. We compute \begin{align} \frac{\delta S}{\delta y} &= 0 - \frac{d}{dx} \left( \frac{y'}{\sqrt{1 + (y')^2}}\right) \\&= -\frac{\sqrt{1 + (y')^2} \ y'' - \frac{(y')^2 y''}{\sqrt{1 + (y')^2}}}{1 + (y')^2} \\&= - \frac{(1 + (y')^2)y'' - (y')^2 y''}{(1 + (y')^2)^{\frac{3}{2}}} \\&= -\frac{y''}{(1 + (y')^2)^{\frac{3}{2}}} \\&= -k \end{align} where $k$ is the signed curvature of the graph of $y(x)$. This makes sense: curvature is the local obstruction to a path being as short as possible.
What happens if we generalize to $$ S[x^\mu] = \int_a^b \sqrt{\dot{x}^\mu g_{\mu\nu}\dot{x}^\nu} \ d\lambda $$ That is, what interpretation can we give to $$ K_\mu \equiv \frac{\delta S}{\delta x^\mu} = \frac{\partial L}{\partial x^\mu} - \frac{d}{dx} \left(\frac{\partial L}{\partial \dot{x}^\mu} \right) $$ with $L = \sqrt{\dot{x}^\mu g_{\mu\nu}\dot{x}^\nu}$?