Length of Arc of Evolute equals Difference in Radii of Curvature

Theorem
Let $C$ be a curve defined by a real function which is twice differentiable.

Let the curvature of $C$ be non-constant.

The length of arc of the evolute $E$ of $C$ between any two points $Q_1$ and $Q_2$ of $C$ is equal to the difference between the radii of curvature at the corresponding points $P_1$ and $P_2$ of $C$.

Proof

 * CenterOfCurvature.png

Let $P = \tuple {x, y}$ be a general point on $C$.

Let $Q = \tuple {X, Y}$ be the center of curvature of $C$ at $P$.

From the above diagram:
 * $(1): \quad \begin{cases} x - X = \pm \rho \sin \psi \\

Y - y = \pm \rho \cos \psi \end{cases}$ where:
 * $\rho$ is the radius of curvature of $C$ at $P$
 * $\psi$ is the angle between the tangent to $C$ at $P$ and the $x$-axis.

Whether the sign is plus or minus depends on whether the curve is convex or concave.

For simplicity, let it be assumed that the curvature $k$ at each point under consideration on $C$ is positive.

The case for $k < 0$ can then be treated similarly.

Thus we have $k > 0$ and so $(1)$ can be written:
 * $(2): \quad \begin{cases} X = x - \rho \sin \psi \\

Y = y +\rho \cos \psi \end{cases}$

By definition of curvature:
 * $k = \dfrac {\d \psi} {\d s}$

and:
 * $\rho = \dfrac 1 k = \dfrac {\d s} {\d \psi}$

Hence:

and:

Differentiating $(2)$ $\psi$:

and:

Let $S$ be the length of arc of the evolute $E$ from a fixed point $Q_0$ on $E$ to the variable point $Q$ corresponding to $P$.

We have that:
 * $\d S^2 = \d X^2 + \d Y^2$

and so:

Choosing the direction of $S$ so that $S$ and $\rho$ increase together, this tells us:
 * $\dfrac {\d S} {\d \psi} = \dfrac {\d \rho} {\d \psi}$

Integrating $\psi$ gives:


 * $S = \rho + c$

where $c$ is a constant.