Evolutes

The locus of the centers of curvature of a given curve is called the evolute of that curve. Consider the circle of curvature corresponding to a point $ P$ on a curve. If $ P$ moves along the given curve, we may suppose the corresponding circle of curvature to roll along the curve with it, its radius varying so as to be always equal to the radius of curvature of the curve at the point $ P$. The curve $ CC_7$ in Figure 14.5 described by the center of the circle is the evolute of $ PP_7$.

Figure 14.5: Geometric visualization of an evolute.
\includegraphics[height=5cm,width=6cm]{evolutes.eps}

It is instructive to make an approximate construction of the evolute of a curve by estimating (from the shape of the curve) the lengths of the radii of curvature at different points on the curve and then drawing them in and drawing the locus of the centers of curvature.

Formula (14.7) gives the coordinates of any point $ (\alpha,\beta)$ on the evolute expressed in terms of the coordinates of the corresponding point $ (x, y)$ of the given curve. But $ y$ is a function of $ x$; therefore

$\displaystyle \alpha
= x - \frac{ \left[ 1 + \left( \frac{dy}{dx} \right)^2 \r...
...\beta = y + \frac{ 1 + \left( \frac{dy}{dx} \right)^2 }{ \frac{d^2 y}{dx^2} }
$

give us at once the parametric equations of the evolute in terms of the parameter x.

To find the ordinary rectangular equation of the evolute we eliminate $ x$ between the two expressions. No general process of elimination can be given that will apply in all cases, the method to be adopted depending on the form of the given equation. In a large number of cases, however, the student can find the rectangular equation of the evolute by taking the following steps:

General directions for finding the equation of the evolute in rectangular coordinates.

Example 14.4.1   Find the equation of the evolute of the parabola $ y^2 = 4px$.

Figure 14.6: Evolute of a parabola.
\includegraphics[height=4cm,width=6cm]{evolute-of-parabola.eps}

Solution. $ \frac{dy}{dx} =\ \frac{2p}{y}, \frac{d^2 y}{dx^2} = -\frac{4p^2}{y^3}$.

First step. $ \alpha= 3x + 2p$, $ \beta = -\frac{y^3}{4p^2}$.

Second step. $ x= \frac{\alpha - 2p}{3}$, $ y = -(4p^2 \beta)^{\frac{1}{3}}$.

Third step $ (4p^2 \beta)^{\frac{2}{3}} = 4p \left( \frac{\alpha - 2p}{3} \right)$; or, $ p\beta^2\frac{4}{27} (\alpha - 2p)^3$.

Remembering that $ \alpha$ denotes the abscissa and $ \beta$ the ordinate of a rectangular system of coordinates, we see that the evolute of the parabola $ AOB$ is the semi-cubical parabola $ DC'E$; the centers of curvature for $ O$, $ P$, $ P_1$, $ P_2$ being at $ C'$, $ C$, $ C_1$, $ C_2$ respectively.

Example 14.4.2   Find the equation of the evolute of the parabola $ b^2x^2 + a^2y^2 = a^2b^2$.

Figure 14.7: Evolute of an ellipse.
\includegraphics[height=4cm,width=6cm]{evolute-of-ellipse.eps}

Solution. $ \frac{dy}{dx} =\ -\frac{b^2x}{a^2y}$, $ \frac{d^2 y}{dx^2} = -\frac{b^4}{a^2 y^3}$.

First step. $ \alpha = \frac{(a^2 - b^2)x^3}{a^4}$, $ \beta = -\frac{(a^2 - b^2)y^3}{b^4}$.

Second step. $ x = \left( \frac{a^4 \alpha}{a^2 - b^2} \right)^{frac{1}{3}}$, $ y = -\left( \frac{b^4 \beta}{a^2 - b^2} \right)^{\frac{1}{3}}$.

Third step. $ (a\alpha)^{\frac{2}{3}} + (b\beta)^{\frac{2}{3}}
= (a^2 - b^2)^{\frac{2}{3}}$, the equation of the evolute $ EHE'H'$ of the ellipse $ ABA'B'$, $ E$, $ E'$, $ H'$, $ H$ are the centers of curvature corresponding to the points $ A$, $ A'$, $ B$, $ B'$, on the curve, and $ C$, $ C'$, $ C''$ correspond to the points $ P$, $ P'$, $ P''$.

When the equations of the curve are given in parametric form, we proceed to find $ \frac{dy}{dx}$ and $ \frac{d^2 y}{dx^2}$, as in §12.5, from

$\displaystyle \frac{dy}{dx} = \frac{ \frac{dy}{dt} }{ \frac{dx}{dt} }, \ \ \ \ ...
...y}{dt^2} - \frac{dy}{dt} \frac{d^2 x}{dt^2} }{ \left( \frac{dx}{dt} \right)^3 }$ (14.12)

and then substitute the results in formulas (14.9). This gives the parametric equations of the evolute in terms of the same parameter that occurs in the given equations.

Example 14.4.3   The parametric equations of a curve are

$\displaystyle x = \frac{t^2 + 1}{4}, \ \ \ \ \ \ y = \frac{t^3}{6}.$ (14.13)

Find the equation of the evolute in parametric form, plot the curve and the evolute, find the radius of curvature at the point where $ t = 1$, and draw the corresponding circle of curvature.

Solution. $ \frac{dx}{dt} = \frac{t}{2}$, $ \frac{d^2 x}{dt^2} = \frac{1}{2}$, $ \frac{dy}{dt} = \frac{t^2}{2}$, $ \frac{d^2 y}{dt^2} = t$. Substituting in above formulas (14.12) and then in (14.9), gives

$\displaystyle \alpha = \frac{1 - t^2 - 2t^4}{4}, \ \ \ \ \ \ \beta = \frac{4t^3 + 3t}{6},$ (14.14)

the parametric equations of the evolute. Assuming values of the parameter $ t$, we calculate $ x$, $ y$; $ \alpha,\beta$ from (14.13) and (14.14). Now plot the curve and its evolute.

Figure 14.8: Evolute of an parametric curve.
% latex2html id marker 58002
\includegraphics[height=4cm,width=6cm]{evolute-parametric-example.eps}

The point $ (\frac{1}{4}, 0)$ is common to the given curve and its evolute. The given curve (a semi-cubical parabola) lies entirely to the right and the evolute entirely to the left of $ x = \frac{1}{4}$.

The circle of curvature at $ A= (\frac{1}{2}, \frac{1}{6})$, where $ t = 1$, will have its center at $ A'= (-\frac{1}{2}, -\frac{7}{6})$ on the evolute and radius $ = AA'$. To verify our work, find radius of curvature at $ A$. From (12.5), we get

$\displaystyle R = \frac{t(1 + t^2)^{\frac{3}{2}}}{2} = \sqrt{2},
$

when $ t = 1$. This should equal the distance

$\displaystyle AA'
= \sqrt{(\frac{1}{2} + \frac{1}{2})^2 +
(\frac{1}{6} - \frac{7}{6})^2} = \sqrt{2}.
$

Example 14.4.4   Find the parametric equations of the evolute of the cycloid,

\begin{displaymath}\begin{cases}x = a(t - \sin t) \\ y = a(1 - \cos t). \end{cases}\end{displaymath} (14.15)

Figure 14.9: Evolute of a cycloid.
\includegraphics[height=4cm,width=6cm]{evolute-of-cycloid.eps}

Solution. As in Example 12.5.2, we get

$\displaystyle \frac{dy}{dx} = \frac{\sin t}{1 - \cos t},
\ \ \ \ \
\frac{d^2 y}{dx^2} = -\frac{1}{\alpha (1 - \cos t)^2}.
$

Substituting these results in formulas (14.9), we get the answer:

\begin{displaymath}\begin{cases}\alpha = a(t + \sin t), \\ \beta = -a(1 - \cos t). \end{cases}\end{displaymath} (14.16)

Remark 14.4.1   In the previous example, if we eliminate $ t$ between equations (14.16), there results the rectangular equation of the evolute $ OO'Q^v$ referred to the axes $ O'\alpha$ and $ O'\beta$. The coordinates of $ O$ with respect to these axes are $ (-\pi a, -2a)$. Let us transform equations (14.16) to the new set of axes $ OX$ and $ OY$. Then

$\displaystyle \alpha = x -\pi a,\ \ \ \ \beta = y -2a,\ \ \ \ t = t' -\pi.
$

Substituting in (14.16) and reducing, the equations of the evolute become

\begin{displaymath}\begin{cases}x = a(t' - \sin t'), \\ y = a(1 - \cos t'). \end{cases}\end{displaymath} (14.17)

Since (14.17) and (14.14) are identical in form, we have: The evolute of a cycloid is itself a cycloid whose generating circle equals that of the given cycloid.

david joyner 2008-11-22