Parametric Equations for Evolute/Formulation 1

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Theorem

Let $C$ be a curve expressed as the locus of an equation $\map f {x, y} = 0$.

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

$\begin{cases}

X = x - \dfrac {y' \paren {1 + y'^2} } {y} \\ Y = y + \dfrac {1 + y'^2} {y} \end{cases}$

where:

$\tuple {x, y}$ denotes the Cartesian coordinates of a general point on $C$
$\tuple {X, Y}$ denotes the Cartesian coordinates of a general point on the evolute of $C$
$y'$ and $y$ denote the derivative and second derivative respectively of $y$ with respect to $x$.


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:

$x - X = \pm \rho \sin \psi$
$Y - y = \pm \rho \cos \psi$

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:

$k = \dfrac {y} {\paren {1 + y'^2}^{3/2} }$


We have that:

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


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

\(\ds x - X\) \(=\) \(\ds \dfrac {\paren {1 + y'^2}^{3/2} } {y} \dfrac {y'} {\sqrt {1 + y'^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {y' \paren {1 + y'^2} } {y}\)
\(\ds \leadsto \ \ \) \(\ds X\) \(=\) \(\ds x - \dfrac {y' \paren {1 + y'^2} } {y}\)

and:

\(\ds Y - y\) \(=\) \(\ds \dfrac {\paren {1 + y'^2}^{3/2} } {y} \dfrac 1 {\sqrt {1 + y'^2} }\)
\(\ds \) \(=\) \(\ds \dfrac {1 + y'^2} {y}\)
\(\ds \leadsto \ \ \) \(\ds Y\) \(=\) \(\ds y + \dfrac {1 + y'^2} {y}\)

$\blacksquare$


Sources

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