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
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:
\(\ds \rho \sin \psi\) | \(=\) | \(\ds \dfrac {\d s} {\d \psi} \dfrac {\d y} {\d s}\) | ||||||||||||
\(\text {(3)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {\d y} {\d \psi}\) |
and:
\(\ds \rho \cos \psi\) | \(=\) | \(\ds \dfrac {\d s} {\d \psi} \dfrac {\d x} {\d s}\) | ||||||||||||
\(\text {(4)}: \quad\) | \(\ds \) | \(=\) | \(\ds \dfrac {\d x} {\d \psi}\) |
Differentiating $(2)$ with respect to $\psi$:
\(\ds \dfrac {\d X} {\d \psi}\) | \(=\) | \(\ds \dfrac {\d x} {\d \psi} - \rho \cos \psi - \dfrac {\d \rho} {\d \psi} \sin \psi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds - \dfrac {\d \rho} {\d \psi} \sin \psi\) | from $(3)$ |
and:
\(\ds \dfrac {\d Y} {\d \psi}\) | \(=\) | \(\ds \dfrac {\d y} {\d \psi} - \rho \sin \psi + \dfrac {\d \rho} {\d \psi} \cos \psi\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \dfrac {\d \rho} {\d \psi} \cos \psi\) | from $(4)$ |
By assumption, $\dfrac {\d \rho} {\d \psi} \ne 0$ on $C$.
Hence we have:
\(\ds \dfrac {\d Y} {\d X}\) | \(=\) | \(\ds -\dfrac {\cos \psi} {\sin \psi}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\dfrac 1 {\tan \psi}\) |
We have Slope of Normal is Minus Reciprocal of Tangent.
Thus the slope of the tangent to $E$ equals the slope of the normal to $C$.
The result follows.
Note the case when $\dfrac {\d \rho} {\d \psi} = 0$ 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.
$\blacksquare$
Sources
- 1992: George F. Simmons: Calculus Gems ... (previous) ... (next): Chapter $\text {B}.23$: Evolutes and Involutes. The Evolute of a Cycloid