Let $c(t)$ be a regular $C^2([a,b])$ curve parameterized by arc length. It's curvature satisfies $\kappa(t)\ne 0$ and $\kappa'(t)\ne 0$.
I want to show that the length of the evolute $e(t) = c(t)+\frac{1}{\kappa(t)}N(t)$ is given by $$L(e) = \left\lvert\frac1{\kappa(b)}-\frac1{\kappa(a)}\right\rvert$$ I have already found the tangent vector of $e$, and am stuck on this particular integral: $$\int_a^b \frac{\lvert\kappa'(t)\rvert}{\kappa^2(t)}dt$$ The absolute value is giving me all the trouble.
As $c$ is only $C^2$, by the Frenet-Equations for planar curves, $\kappa'$ could be non-continuous as $c'''$ could be non-continuous. So I can't guarantee that $\kappa'$ won't change signs. Or can I?
Is $\kappa'$, in fact, continuous for some reason? Is there some other way to crack that absolute value?
Here is a trick. The Darboux's theorem state that if $f$ is a differentiable function on an interval, then $f'(I)$ is an interval $J\subset \mathbb{R}$. Thus, if $k$ is differentiable and $k'$ does not vanish, then $k'$ must have constant sign.
This does not, by the way, guarantee that $k'$ is locally integrable. I think you should look closely to the assumptions: the curvature $\kappa$ is defined for a $C^2$ parametrized curve but assuming $k$ is differentiable is closely related to assuming that the curve is three-time differentiable.
I would say that the good assumptions for your exercice are that your curve is $C^3$, or that it is $C^2$, three times differentiable, with $k'$ locally integrable.