curve evolution length formula I run to the following problem which says if you have a smooth curve that is evolving over time (say finite length at the beginning) then 
$$\frac{d}{dt}(curve \; length \; at \; time \; t)=-\int_{curve} k\cdot v \; ds,$$
where $k$ is curvature of the curve and $v$ is velocity of point on curve.  $ds$ represents integration by parametrization by arclength. 
I have tried proving this but I can not get out without re parametrazing curves. Is there some neater way to do this?
 A: Here's a nice crib-sheet for differential geometry of curves in space. 

 Let $\gamma: [a,b]\times [0,1] \to \mathbb{R}^3$ be a family of curves.  The rate of change of arc-length is:
$$ \frac{d}{dt}(\mathrm{arc length})=\frac{d}{dt}\int_{a}^{b} ||\gamma'(s,t)|| ds 
= \int_{a}^{b}  \frac{d}{dt} ||\gamma'(s,t)|| ds $$ 
I will ignore the last term since there's no dependence on endpoints in your question. The size of the tangent vector is inner product of $\gamma' = \frac{d\gamma}{ds}$ with itself (square-root).
$$ ||\gamma'(s,t)|| = ( \gamma'\cdot \gamma')^{1/2}$$
Differentiate both sides with respect to time and use the chain rule from calculus:
$$ \frac{d}{dt} ||\gamma'(s,t)|| = \frac{\frac{d\gamma'}{dt}\cdot \gamma'}{\sqrt{ \gamma'\cdot \gamma'}} = \frac{dv}{ds} \cdot T $$

where $T$ is the unit tangent vector.  $$ T =  \frac{\gamma'}{ ( \gamma'\cdot \gamma')^{1/2}} $$  Velocity is relative to the deformation parameter $s$ not the curve paramter $t$.  Since partial derivatives commute, we can relate $\gamma', v$:
$$ \frac{d \gamma'}{dt} = \frac{\partial^2 \gamma}{\partial s \partial t} = \frac{\partial}{\partial s} \frac{\partial \gamma}{\partial t} = \frac{dv}{ds}$$
Then... integration by parts:
$$\frac{d}{dt}(\mathrm{arc length})= \int  \frac{dv}{ds} \cdot T ds = -\int v \cdot \frac{dT}{ds}ds= -
\int  v \cdot (\kappa N) ds $$
Curvature is the derivative of the unit tangent vector $\fbox{$\frac{dT}{ds} = \kappa N$}$.  It was missing in your problem that curvature points in the direction of the  normal.


For small pieces of arc, the curvature is constant and the curve can be approximate by the arc of a circle.  The arc-length is just $ds = r d\theta$.

The maximum potential for growth is when the arc moves outward in the normal direction.  If each curve moves tangentially, the arc-length doesn't change.   In general, the arc grows according to the normal component of the velocity, $\Delta (ds) = (v \Delta t \cdot N)d\theta $.

Globally we can imagine the curve swept out by $\gamma(\cdot,t)$.  Our change in arc-length is $\frac{d^2A}{dt^2}$.  This area should grow the most if it expands out in the normal direction, so it should be proportional to $v\cdot N$.  Sharper turns should grow faster and we just showed it should be proportional to curvature $k$.  So we get $\frac{d}{dt}(\mathrm{arclength}) =-\int v\cdot (kN) ds$.

A: This is not as comprehensive an answer as John Mangual's.
Consider it, rather, supplemental information to provide intuition,
drawn from the textbook, Discrete and Computational Geometry:



