I encountered the following formula to calculate the time variation of the arc length element $ds$ of a moving curve:
$ \partial_t ds = ds \,v_n k \,,$
where $v_n$ is the normal velocity of the curve and $k$ is the curvature.
Intuitively it makes sense, since if the curve is moving with a normal velocity and a curvature different from 0, then it will expand (or shrink), as in the following (terrible) drawing:
However, I would like to have a formal proof of that, could please anybody point me in the right direction?
HINT: If you take a normal variation of the original (let's assume arclength-parametrized) curve $\alpha$, you get $$\beta_t(s) = \alpha(s) + tv(s)N(s),$$ where $T,N$ is the usual Frenet frame of the curve (so, in this case, $N$ points inward and your $v$ is negative). Differentiating with respect to $s$ gives \begin{align*} \beta_t'(s) &= \alpha'(s) + tv'(s)N(s) + tv(s)(-\kappa(s)T(s)) \\&= (1-tv(s)\kappa(s))T(s) + tv'(s)N(s). \end{align*} Now the element of arclength of $\beta_t$ is given by $\|\beta_t'(s)\|ds$, and so we want $$\frac{\partial}{\partial t}\Big|_{t=0} \|\beta_t'(s)\|.$$ Since $\|\beta_t'(s)\|^2 = (1-tv(s)\kappa(s))^2 + (tv'(s))^2$, $$\frac{\partial}{\partial t}\Big|_{t=0} \|\beta_t'(s)\|^2 = -2v(s)\kappa(s).$$ Therefore, by the chain rule, since $\|\beta_0'(s)\| = 1$, we have $$\frac{\partial}{\partial t}\Big|_{t=0} \|\beta_t'(s)\| = -v(s)\kappa(s).$$