# Evolute and Involute

## Solved Problems

Click or tap a problem to see the solution.

### Example 3

Find the evolute of the parabola $y = {x^2}.$

### Example 4

Find the evolute of the logarithmic spiral $r = {e^\theta }.$

### Example 5

Determine the evolute of the cycloid $x = t - \sin t, y = 1 - \cos t.$

### Example 6

Prove that the curve given by the equations

$x = R\left( {\cos t + t\sin t} \right), y = R\left( {\sin t - t\cos t} \right)$

is the involute of the circle of radius $$R$$ centered at the origin.

### Example 7

Find an equation of the evolute of the hyperbola $y = \frac{1}{x}.$

### Example 3.

Find the evolute of the parabola $y = {x^2}.$

Solution.

For a curve given by an explicit equation, the coordinates of the center of curvature are determined by the formulas

$\xi = x - \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}}y',\;\;\;\eta = y + \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}}.$

Substituting the given function, we get:

$\xi = x - \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}}y' = x - \frac{{1 + {{\left( {2x} \right)}^2}}}{2} \cdot 2x = x - x\left( {1 + 4{x^2}} \right) = - 4{x^3};$
$\eta = y + \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}} = {x^2} + \frac{{1 + {{\left( {2x} \right)}^2}}}{2} = {x^2} + \frac{{1 + 4{x^2}}}{2} = 3{x^2} + \frac{1}{2}.$

Eliminating the variable $$x$$ from these expressions, we represent the equation of the evolute as a function $$\xi \left( \eta \right).$$ This yields:

$\eta = 3{x^2} + \frac{1}{2},\;\; \Rightarrow \eta - \frac{1}{2} = 3{x^2},\;\; \Rightarrow {x^2} = \frac{\eta }{3} - \frac{1}{6},\;\; \Rightarrow x = \pm {\left( {\frac{\eta }{3} - \frac{1}{6}} \right)^{\frac{1}{2}}}.$

Hence,

$\xi = - 4{x^3} = - 4 \cdot {\left[ { \pm {{\left( {\frac{\eta }{3} - \frac{1}{6}} \right)}^{\frac{1}{2}}}} \right]^3} = \pm 4{\left( {\frac{\eta }{3} - \frac{1}{6}} \right)^{\frac{3}{2}}},$

where $$\eta \ge \frac{1}{2}.$$

The parabola and its evolute are sketched in Figure $$4.$$

The evolute found above is shaped like a dovetail and its equation is a semicubic parabola.

### Example 4.

Find the evolute of the logarithmic spiral $r = {e^\theta }.$

Solution.

This curve in Cartesian coordinates is described by the following system of equations:

$x = r\cos \theta = {e^\theta }\cos \theta ,\;\;\;y = r\sin \theta = {e^\theta }\sin \theta .$

This expression is the equation of the curve in parametric form, where the angle $$\theta$$ plays the role of a parameter. Then the coordinates of the center of curvature can be found by the formulas

$\xi = x - y'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}},\;\;\;\eta = y + x'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}}.$

Compute the derivatives:

$x' = \left( {{e^\theta }\cos \theta } \right)^\prime = {e^\theta }\cos \theta - {e^\theta }\sin\theta = {e^\theta }\left( {\cos \theta - \sin\theta } \right);$
$x^{\prime\prime} = \left[ {{e^\theta }\left( {\cos \theta - \sin\theta } \right)} \right]^\prime = {e^\theta }\left( {\cos \theta - \sin\theta } \right) + {e^\theta }\left( { - \sin\theta - \cos \theta } \right) = {e^\theta }\left( {\cancel{\cos \theta} - \sin\theta - \sin\theta - \cancel{\cos \theta} } \right) = - 2{e^\theta }\sin\theta ;$
$y' = \left( {{e^\theta }\sin \theta } \right)^\prime = {e^\theta }\sin \theta + {e^\theta }\cos\theta = {e^\theta }\left( {\sin \theta + \cos\theta } \right);$
$y^{\prime\prime} = \left[ {{e^\theta }\left( {\sin \theta + \cos\theta } \right)} \right]^\prime = {e^\theta }\left( {\sin \theta + \cos\theta } \right) + {e^\theta }\left( {\cos\theta - \sin \theta } \right) = {e^\theta }\left( {\cancel{\sin \theta} + \cos\theta + \cos\theta - \cancel{\sin \theta} } \right) = 2{e^\theta }\cos\theta .$

The expressions for $$\xi$$ and $$\eta$$ contain a common fraction equal to

$F = \frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = \frac{1}{2} \cdot \frac{2}{1} = 1.$

Consequently, the coordinates of the center of curvature $$\xi$$ and $$\eta$$ are defined by

$\xi = x - y'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = x - y'F = {e^\theta }\cos \theta - {e^\theta }\left( {\sin\theta + \cos \theta } \right) = - {e^\theta }\sin\theta = - y;$
$\eta = y + x'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = y + x'F = {e^\theta }\sin \theta + {e^\theta }\left( {\cos\theta - \sin \theta } \right) = {e^\theta }\cos\theta = x.$

Thus, if we take the initial coordinate system $${xOy}$$ and rotate it counterclockwise by $$\frac{\pi }{2},$$ we get the coordinate system $${\xi O \eta}.$$ The negative $$\left( { - y} \right)$$-semiaxis is mapped into the positive horizontal $$\xi$$-axis, and the positive $$x$$-semiaxis is mapped into the positive vertical $$\eta$$-axis. In other words, the evolute of the logarithmic spiral $$r = {e^\theta }$$ is again the same curve rotated counterclockwise by the angle $$\frac{\pi }{2}.$$

### Example 5.

Determine the evolute of the cycloid $x = t - \sin t, y = 1 - \cos t.$

Solution.

Find the derivatives of the given curve:

$x' = \left( {t - \sin t} \right)^\prime = 1 - \cos t,\;\;\;y' = \left( {1 - \cos t} \right)^\prime = \sin t.$
$x^{\prime\prime} = \left( {1 - \cos t} \right)^\prime = \sin t,\;\;\;y^{\prime\prime} = \left( {\sin t} \right)^\prime = \cos t.$

Calculate the coordinates of the center of curvature:

$\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = \frac{{{{\left( {1 - \cos t} \right)}^2} + {{\sin }^2}t}}{{\left( {1 - \cos t} \right)\cos t - \sin t\sin t}} = \frac{{1 - 2\cos t + {{\cos }^2}t + {{\sin }^2}t}}{{\cos t - {{\cos }^2}t - {{\sin }^2}t}} = \frac{{2\left( {1 - \cos t} \right)}}{{\cos t - 1}} = - 2;$
$\xi = x - y'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = t - \sin t - \sin t \cdot \left( { - 2} \right)= t + \sin t;$
$\eta = y + x'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = 1 - \cos t + \left( {1 - \cos t} \right) \cdot \left( { - 2} \right) = \cos t - 1.$

Next, we represent the parameter $$t$$ as $$t = \tau + \pi .$$ Here the variable $$\tau$$ plays the role of a "delayed" parameter, which lags by $$\pi$$ units behind $$t$$ (i.e. half of the arc of the cycloid).

Now we express the coordinates of the center of curvature $$\xi$$ and $$\eta$$ in terms of the variable $$\tau:$$

$\xi = t + \sin t = \tau + \pi + \sin \left( {\tau + \pi } \right) = \left[ {\tau - \sin \tau } \right] + \pi ;$
$\eta = \cos t - 1 = \cos \left( {\tau + \pi } \right) - 1 = - \cos \tau + 1 - 2 = \left[ {1 - \cos \tau } \right] - 2.$

So, the evolute of the cycloid is also a cycloid. Its position with respect to the initial curve is shifted by the vector $$\left( {\pi , - 2} \right).$$ In addition, due to replacement $$t \to \tau,$$ the evolute of the cycloid begins from the middle of the arc.

### Example 6.

Prove that the curve given by the equations

$x = R\left( {\cos t + t\sin t} \right), y = R\left( {\sin t - t\cos t} \right)$

is the involute of the circle of radius $$R$$ centered at the origin.

Solution.

Find the derivatives of $$x$$ and $$y$$ with respect to the parameter $$t:$$

$x' = \left[ {R\left( {\cos t + t\sin t} \right)} \right]^\prime = R\left( { - \cancel{\sin t} + \cancel{\sin t} + t\cos t} \right) = Rt\cos t,$
$x^{\prime\prime} = \left( {Rt\cos t} \right)^\prime = R\left( {\cos t - t\sin t} \right),$
$y' = \left[ {R\left( {\sin t - t\cos t} \right)} \right]^\prime = R\left( {\cancel{\cos t} - \cancel{\cos t} + t\sin t} \right) = Rt\sin t,$
$y^{\prime\prime} = \left( {Rt\sin t} \right)^\prime = R\left( {\sin t + t\cos t} \right).$

Determine the coordinates of the center of curvature for the given curve. We first calculate the fractional expression $$D:$$

$D = \frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = 1.$

Consequently, the coordinates of the center are

$\xi = x - y'D = R\left( {\cos t + t\sin t} \right) - Rt\sin t \cdot 1 = R\cos t + \cancel{Rt\sin t} - \cancel{Rt\sin t} = R\cos t;$
$\eta = y + x'D = R\left( {\sin t - t\cos t} \right) + Rt\sin t \cdot 1 = R\sin t - \cancel{Rt\cos t} + \cancel{Rt\cos t} = R\sin t.$

So we have the equations of the evolute in the form

$\xi = R\cos t,\;\;\;\eta = R\sin t.$

These are parametric equations of the circle of radius $$R$$ centered at the origin. In Cartesian coordinates $$\left( {\xi ,\eta } \right),$$ the equation of the evolute is written as

${\xi ^2} + {\eta ^2} = {R^2}.$

The involute of the circle is shaped like a spiral (Figure $$5$$).

It describes the trajectory of any point of a straight line, which rolls along the circumference. The involute form is used, in particular, in designing gears (Figure $$6$$).

### Example 7.

Find an equation of the evolute of the hyperbola $y = \frac{1}{x}.$

Solution.

The first and second derivatives of the hyperbolic function can be written as

$y' = \left( {\frac{1}{x}} \right)^\prime = - \frac{1}{{{x^2}}},\;\;\;y^{\prime\prime} = \left( { - \frac{1}{{{x^2}}}} \right)^\prime = \frac{2}{{{x^3}}}.$

Then the coordinates of the center of curvature of the hyperbola are given by

$\xi = x - \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}}y' = x - \frac{{1 + {{\left( { - \frac{1}{{{x^2}}}} \right)}^2}}}{{\frac{2}{{{x^3}}}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = x - \frac{{1 + \frac{1}{{{x^4}}}}}{{\frac{2}{{{x^3}}}}} \cdot \left( { - \frac{1}{{{x^2}}}} \right) = x + \frac{{\left( {{x^4} + 1} \right){x^3}}}{{2{x^4}{x^2}}} = x + \frac{{{x^4} + 1}}{{2{x^3}}} = \frac{{2{x^4} + {x^4} + 1}}{{2{x^3}}} = \frac{{3{x^4} + 1}}{{2{x^3}}};$
$\eta = y + \frac{{1 + {{\left( {y'} \right)}^2}}}{{y^{\prime\prime}}} = \frac{1}{x} + \frac{{1 + {{\left( { - \frac{1}{{{x^2}}}} \right)}^2}}}{{\frac{2}{{{x^3}}}}} = \frac{1}{x} + \frac{{1 + \frac{1}{{{x^4}}}}}{{\frac{2}{{{x^3}}}}} = \frac{1}{x} + \frac{{\left( {{x^4} + 1} \right){x^3}}}{{2{x^4}}} = \frac{1}{x} + \frac{{{x^4} + 1}}{{2x}} = \frac{{{x^4} + 3}}{{2x}}.$

Using algebraic manipulations, we eliminate the variable $$x$$ (which is a parameter in the formulas for $$\xi$$ and $$\eta$$). First we find the expressions for the sum and difference of the coordinates $$\xi$$ and $$\eta:$$

$\xi + \eta = \frac{{3{x^4} + 1}}{{2{x^3}}} + \frac{{{x^4} + 3}}{{2x}} = \frac{{3{x^4} + 1 + {x^6} + 3{x^2}}}{{2{x^3}}} = \frac{{{x^6} + 3{x^4} + 3{x^2} + 1}}{{2{x^3}}} = \frac{{{{\left( {{x^2} + 1} \right)}^3}}}{{2{x^3}}} = \frac{1}{2}{\left( {\frac{{{x^2} + 1}}{x}} \right)^3} = \frac{1}{2}{\left( {x + \frac{1}{x}} \right)^3};$
$\xi - \eta = \frac{{3{x^4} + 1}}{{2{x^3}}} - \frac{{{x^4} + 3}}{{2x}} = \frac{{3{x^4} + 1 - {x^6} - 3{x^2}}}{{2{x^3}}} = - \frac{{{x^6} - 3{x^4} + 3{x^2} - 1}}{{2{x^3}}} = - \frac{{{{\left( {{x^2} - 1} \right)}^3}}}{{2{x^3}}} = - \frac{1}{2}{\left( {\frac{{{x^2} - 1}}{x}} \right)^3} = - \frac{1}{2}{\left( {x - \frac{1}{x}} \right)^3}.$

Extracting cube roots, we obtain the following relationships:

$\left( {\xi + \eta } \right)^{\frac{1}{3}} = \frac{1}{{\sqrt[3]{2}}}\left( {x + \frac{1}{x}} \right),\;\;\;{\left( {\xi - \eta } \right)^{\frac{1}{3}}} = - \frac{1}{{\sqrt[3]{2}}}\left( {x - \frac{1}{x}} \right).$

Now we square each equation:

$\left( {\xi + \eta } \right)^{\frac{2}{3}} = \frac{1}{{\sqrt[3]{4}}}{\left( {x + \frac{1}{x}} \right)^2} = \frac{1}{{\sqrt[3]{4}}}\left( {{x^2} + 2 + \frac{1}{{{x^2}}}} \right),$
$\left( {\xi - \eta } \right)^{\frac{2}{3}} = \frac{1}{{\sqrt[3]{4}}}{\left( {x - \frac{1}{x}} \right)^2} = \frac{1}{{\sqrt[3]{4}}}\left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right).$
$\sqrt[3]{4}{\left( {\xi + \eta } \right)^{\frac{2}{3}}} = {x^2} + 2 + \frac{1}{{{x^2}}},\;\;\;\sqrt[3]{4}{\left( {\xi - \eta } \right)^{\frac{2}{3}}} = {x^2} - 2 + \frac{1}{{{x^2}}}$

and subtract the second equation from the first to obtain:

$\sqrt[3]{4}\left[ {{{\left( {\xi + \eta } \right)}^{\frac{2}{3}}} - {{\left( {\xi - \eta } \right)}^{\frac{2}{3}}}} \right] = {x^2} + 2 + \frac{1}{{{x^2}}} - \left( {{x^2} - 2 + \frac{1}{{{x^2}}}} \right),\;\; \Rightarrow \sqrt[3]{4}\left[ {{{\left( {\xi + \eta } \right)}^{\frac{2}{3}}} - {{\left( {\xi - \eta } \right)}^{\frac{2}{3}}}} \right] = 4,\;\; \Rightarrow \sqrt[3]{4}\left[ {{{\left( {\xi + \eta } \right)}^{\frac{2}{3}}} - {{\left( {\xi - \eta } \right)}^{\frac{2}{3}}}} \right] = 4,\;\; \Rightarrow {\left( {\xi + \eta } \right)^{\frac{2}{3}}} - {\left( {\xi - \eta } \right)^{\frac{2}{3}}} = \frac{4}{{\sqrt[3]{4}}},\;\; \Rightarrow {\left( {\xi + \eta } \right)^{\frac{2}{3}}} - {\left( {\xi - \eta } \right)^{\frac{2}{3}}} = {4^{\frac{2}{3}}}.$

As a result, we have an equation of the evolute of the hyperbola in implicit form $$f\left( {\xi ,\eta } \right) = 0.$$