Let $\alpha:I \rightarrow \mathbb{R}^2$ an curve parametrized by its length and $c\in I$. The evolvent of $\alpha$ that passes through the point $c$ is defined as:
$$\beta(s)=\alpha(s)-(s-c)\alpha'(s), \ \ s\in I, \ \ s\gt c$$
Its geometrical interpretation is the following: given a curve, it consists in unwinding a string that has been previously wound. The evolvent are the extreme points of that string.
I don't understand the geometrical interpretation at all. Is there easier to understand explanation of what the curve $\beta$ represents?
In order to fix ideas assume that the curve $\gamma$ has curvature $\kappa(s)>0$. Its tangent $\alpha'(s)$ then turns counterclockwise as $s$ increases. Let a $c\in I$ be given. We now put a pencil tip at the point $C:=\alpha(c)\in\gamma$. The pencil is bound to one end of a string, and this string winds along $\gamma$ all the way for $s\geq c$.
When we unwind this string from $\gamma$, starting at $C$, and keeping the unwound part tight all the time, the pencil will draw a new curve $\varepsilon$ called an evolvent of $\gamma$. This curve has a cusp at $C$, but you don't see this cusp, since we consider only parameter values $s> c$. A parametrization $s\mapsto\beta(s)$ of $\varepsilon$ is obtained as follows: During the unwinding, at each moment the unwound part of the string connects the pencil at $\beta(s)$ with the corresponding point $\alpha(s)\in\gamma$. The length of this segment is $\ =s-c$; furthermore the segment is tangent to $\gamma$ at $\alpha(s)$. A moments reflection about the prevalent signs then shows that $\varepsilon$ possesses the parametrization $$\beta(s)=\alpha(s)-(s-c)\alpha'(s)\qquad(s> c)\ .$$ (Note that $\alpha'(s)$ is a unit vector.)