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 $\frac{d}{dt}(\mathrm{arclength}) =-\int v\cdot (kN) ds$.
