Given a unit speed curve $\gamma:[a,b] \to \mathbb{R}^3$ and a (smooth) vector field $V(t)$ that is perpendicular to $\dot{\gamma}$ for any $t \in [a,b]$ and has length $|V| = 1$.
Define $\gamma_s : [a,b] \to \mathbb{R}^3$ by $\gamma_s(t) = \gamma(t) + sV(t)$, where $s \in \mathbb{R}$.
I've calculated an expression for the derivative of the length function with respect to the arclength parameter $s$ evaluated at $s = 0$: $$\frac{\mathrm{d}}{\mathrm{d}s} L(\gamma_s) \big |_{s=0} = \int_a^b -\ddot{\gamma}\cdot V \mathrm{d}t.$$
I'd like to know what choice of $V$ makes the length decrease as quickly as possible. As $V$ is unitary and perpendicular to $\dot{\gamma}$, I can see that it's either $V = \ddot{\gamma}$ or V = $-\ddot{\gamma}$.
However, I don't know how to relate my finding for $s=0$ to the entire set of curves. I only have an expression for $\frac{\mathrm{d}}{\mathrm{d}s}L(\gamma_s)$ when $s=0$.
Any help would be greatly appreciated :).