Normal to Curve is Tangent to Evolute

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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:

\(\ds \rho \sin \psi\) \(=\) \(\ds \dfrac {\d s} {\d \psi} \dfrac {\d y} {\d s}\)
\(\text {(3)}: \quad\) \(\ds \) \(=\) \(\ds \dfrac {\d y} {\d \psi}\)

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

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