Normal to Curve is Tangent to Evolute

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

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

Let $P$ be a point on $C$.

Let $Q$ be the center of curvature of $C$ at $P$.

The normal to $C$ at $P$ is tangent to the evolute $E$ of $C$ at $Q$.

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:

By assumption, $\dfrac {\d \rho} {\d \psi} \ne 0$ on $C$.

Hence we have:

We have Slope of Normal is Minus Reciprocal of Tangent.

Thus the slope of the tangent to $E$ equals the slope of the normal.

The result follows.

Note the case when $\dfrac {\d \rho} {\d \psi} \ne 0$ on $C$.

In this case $C$ is a circle.

By Evolute of Circle is its Center its evolute is a single point and so has no tangent.