length of a curve evolving under a vector field Suppose we have a family of closed curves $\gamma_t: [0,1] \rightarrow \mathbb{R}^n$ where $\gamma_0$ is given and $\gamma_t$ is defined to be the curve formed by allowing $\gamma_0$ to evolve under a vector field V(t). That is to say solve $\dot{x} = V(x,t)$, call the solution, for initial condition $x_0$ $\phi(x_0,t)$, then $\gamma_t(s) = \phi(\gamma_0(s),t)$ 
Let L(t) be the length of $\gamma_t$. I believe, from a more general result, that:
$$\frac{dL}{dt}_{t=0} = -\int_{s=0}^1{<\ddot{\gamma}_0,V>}$$
Is this true? I cannot prove it, although I can get something similar if $L$ is replaced by $E$, the curve's energy.
One thing I read indicated that $dE/dt_{t=0} = L(0) \times dL/dt_{t=0}$: I can't see why? Is it true?
The original result is here: https://arxiv.org/pdf/0705.3827.pdf
page 6 Lemma 2.7 and Corollary 2.9. 
No proof is offerend: it seems the author expects it to be obvious
 A: Instead of $s$, I will use $\tau$ to parametrize your curve $\gamma$ and reserve $s$ for its arc length. We will use $\partial_\tau, \partial_t$ as a short hand for $\frac{\partial}{\partial\tau}$, $\frac{\partial}{\partial \tau}$.
I will also restrict the derivation to the case $n = 2$. The formula for general $n$ is similar. In $2$-dimension, we have
$$\begin{align} L(t) &= \int_0^1 \sqrt{(\partial_\tau x)^2 + (\partial_\tau y)^2} d\tau\\
\implies \frac{dL(t)}{dt}
&= \int_0^1 \partial_t \sqrt{(\partial_\tau x)^2 + (\partial_\tau y)^2}d\tau
= \int_0^1 \frac{(\partial_\tau x)(\partial_t\partial_\tau y) + (\partial_\tau y)(\partial_t\partial_\tau y)}{\sqrt{(\partial_\tau x)^2 + (\partial_\tau y)^2}}d\tau\\
&= \int_0^1\left(\frac{dx}{ds}(\partial_\tau \partial_t x) + \frac{dy}{ds}(\partial_\tau \partial_t y)\right) d\tau
\stackrel{\text{I by P}}{=}
- \int_0^1 \left((\partial_t x)(\partial_\tau\frac{dx}{ds}) + (\partial_t y)(\partial_\tau\frac{dx}{ds})\right)d\tau\\
&= -\int_0^{L(t)} \left\langle V, \frac{d^2\gamma}{ds^2} \right\rangle ds
\end{align}
$$
As you can see, what you suspect does not work in general. If you want it
to work, you should parametrize your curve $\gamma$ at $t = 0$ by its arc-length.
