Parametric Equations for Evolute/Formulation 1

Theorem
Let $C$ be a curve expressed as the locus of an equation $f \left({x, y}\right) = 0$.

The parametric equations for the evolute of $C$ can be expressed as:


 * $\begin{cases}

X = x - \dfrac {y' \left({1 + y'^2}\right)} {y''} \\ Y = y + \dfrac {1 + y'^2} {y''} \end{cases}$

where:
 * $\left({x, y}\right)$ denotes the Cartesian coordinates of a general point on $C$
 * $\left({X, Y}\right)$ denotes the Cartesian coordinates of a general point on the evolute of $C$
 * $y'$ and $y''$ denote the derivative and second derivative respectively of $y$ $x$.

Proof

 * CenterOfCurvature.png

Let $P = \left({x, y}\right)$ be a general point on $C$.

Let $Q = \left({X, Y}\right)$ be the center of curvature of $C$ at $P$.

From the above diagram:
 * $x - X = \pm \rho \sin \psi$
 * $Y - y = \pm \rho \cos \psi$

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.

By definition of radius of curvature:
 * $(1): \quad \begin{cases} x - X = \dfrac 1 k \sin \psi \\

Y - y = \dfrac 1 k \cos \psi \end{cases}$

where $k$ is the curvature of $C$ at $P$, given by:
 * $k = \dfrac {y''} {\left({1 + y'^2}\right)^{3/2} }$

We have that:
 * $\sin \psi = \dfrac {\mathrm d y} {\mathrm d s} = \dfrac {y'} {\sqrt {1 + y'^2} }$


 * $\cos \psi = \dfrac {\mathrm d x} {\mathrm d s} = \dfrac 1 {\sqrt {1 + y'^2} }$

Substituting for $k$ and $\psi$ in $(1)$ gives:

and: