# Calculus

## Applications of the Derivative # Evolute and Involute

Let a plane curve γ be given by the natural equation

$\mathbf{r} = \mathbf{r}\left( s \right),$

where the parameter s means the arc length of the curve. Suppose that at each point, the curvature of the curve is not zero: K(s) ≠ 0. Then at any point M we can define a finite radius of curvature:

$R = R\left( s \right) = \frac{1}{{K\left( s \right)}}.$

On the normal n we draw the segment MC equal to the radius of curvature R(s) at the point M (Figure 1).

The point $$C$$ is called the center of curvature of the curve $$\gamma$$ at point $$M.$$

If the radius vector of the center of curvature is denoted by $$\boldsymbol{\rho},$$ then

$\boldsymbol{\rho} = \mathbf{OM} + \mathbf{MC} = \mathbf{r} + R\mathbf{n}.$

The normal vector $$\mathbf{n}$$ is determined by the expression

$\mathbf{n} = \frac{1}{K}\frac{{d\boldsymbol{\tau} }}{{ds}} = \frac{1}{K}\frac{{{d^2}\mathbf{r}}}{{d{s^2}}} = R\frac{{{d^2}\mathbf{r}}}{{d{s^2}}},$

where $$\boldsymbol\tau$$ is the unit tangent vector. Consequently, the position of the center of curvature corresponding to the point $$M$$ is described by the formula

$\boldsymbol\rho = \mathbf{r} + R\mathbf{n} = \mathbf{r} + {R^2}\frac{{{d^2}\mathbf{r}}}{{d{s^2}}}.$

For each point of the curve (assuming $$K \ne 0$$), we can find the center of curvature. The set of all centers of curvature of the curve $$\gamma$$ is called the evolute of the curve.

If the curve $${\gamma_1}$$ is the evolute of the curve $$\gamma,$$ then the initial curve $$\gamma$$ is called the involute of the curve $${\gamma_1}.$$

We denote the center of curvature by the point $$C$$ with coordinates $$\left( {\xi ,\eta } \right).$$ If the curve $$\gamma$$ is given in parametric form

$x = x\left( t \right),\;\;y = y\left( t \right),\;\;\alpha \le t \le \beta ,$

the coordinates of the center of curvature $$\left( {\xi ,\eta } \right)$$ are calculated according to formulas

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

These formulas follow from the expression for the radius vector $$\boldsymbol\rho.$$

If the curve $$\gamma$$ is the graph of a function $$y = f\left( x \right),$$ the coordinates of the center of curvature are expressed in the form

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

Note that the condition of non-zero curvature at all points of the curve is rigid enough. As a result, certain curves, for example, with inflection points are excluded from analysis. Therefore, sometimes a more general case of arbitrary curvature is considered. If the curvature at a point is zero, the evolute at this point has a discontinuity. Such case is shown schematically in Figure $$2.$$

## Solved Problems

### Example 1.

Determine the evolute of the circle ${x^2} + {y^2} = {R^2}.$

Solution.

Write the equation of the circle in parametric form:

$x = R\cos t,\;\;\;y = R\sin t.$

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

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

The coordinates of the center of curvature $$\left( {\xi ,\eta } \right)$$ are calculated by the formulas

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

Substituting the expressions for the coordinates $$x, y$$ and their derivatives into this formula, we find

$\xi = x - y'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = R\cos t - R\cos t \cdot \frac{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}}{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}} = \cancel{R\cos t} - \cancel{R\cos t} \equiv 0;$
$\eta = y + x'\frac{{{{\left( {x'} \right)}^2 + {\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = R\sin t - R\sin t \cdot \frac{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}}{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}} = R\sin t - R\sin t \equiv 0.$

Thus, we have a trivial result: the evolute of the circle is only one single point - the center of the circle.

### Example 2.

Find the evolute of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1.$

Solution.

Write the equation of the ellipse in parametric form:

$x = a\cos t,\;\;\;y = b\sin t.$

The derivatives of $$x$$ and $$y$$ with respect to $$t$$ are written as

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

To calculate the coordinates of the center of curvature $$\left( {\xi ,\eta } \right),$$ we use the formulas:

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

Substituting the expressions for $$x, y$$ and their derivatives, we obtain:

$\xi = x - y'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = {a\cos t - b\cos t} \cdot {\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{{ab\,{{\sin }^2}t + ab\,{{\cos }^2}t}} } = {a\cos t - \cos t} \cdot {\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{a} } = \frac{{{a^2}\cos t - {a^2}{{\sin }^2}t\cos t - {b^2}{{\cos }^3}t}}{a} = \frac{{{a^2}\cos t\left( {1 - {{\sin }^2}t} \right) - {b^2}{{\cos }^3}t}}{a} = \frac{1}{a}\left( {{a^2}{{\cos }^3}t - {b^2}{{\cos }^3}t} \right) = \frac{{{a^2} - {b^2}}}{a}{\cos ^3}t;$
$\eta = y + x'\frac{{{{\left( {x'} \right)}^2} + {{\left( {y'} \right)}^2}}}{{x'y^{\prime\prime} - x^{\prime\prime}y'}} = {b\sin t - a\sin t} \cdot {\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{{ab\,{{\sin }^2}t + ab\,{{\cos }^2}t}} } = {b\sin t - \sin t}\cdot {\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{b} } = \frac{{{b^2}\sin t - {a^2}{\sin^3}t - {b^2}{\cos^2}t\sin t}}{b} = \frac{{{b^2}\sin t\left( {1 - {{\cos }^2}t} \right) - {a^2}{\sin^3}t}}{b} = \frac{1}{b}\left( {{b^2}{\sin^3}t - {a^2}{\sin^3}t} \right) = \frac{{{b^2} - {a^2}}}{b}{\sin^3}t.$

Consequently, the evolute of the ellipse is described by the following parametric equations:

$\xi = \frac{{{a^2} - {b^2}}}{a}{\cos^3}t = \left( {a - \frac{{b^2}}{a}} \right){\cos ^3}t,\;\;\;\eta = \frac{{{b^2} - {a^2}}}{b}{\sin^3}t = \left( {b - \frac{{a^2}}{b}} \right){\sin ^3}t.$

Eliminating the parameter $$t,$$ we can write the equation of the evolute in implicit form:

$\xi = \frac{{{a^2} - {b^2}}}{a}{\cos^3}t,\;\;\Rightarrow a\xi = \left( {{a^2} - {b^2}} \right){\cos^3}t, \;\;\Rightarrow \frac{{a\xi }}{{{a^2} - {b^2}}} = {\cos^3}t,\;\;\Rightarrow \frac{{{{\left( {a\xi } \right)}^{\frac{2}{3}}}}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{2}{3}}}}} = {\cos^2}t;$
$\eta = \frac{{{b^2} - {a^2}}}{b}{\sin^3}t,\;\;\Rightarrow b\eta = \left( {{b^2} - {a^2}} \right){\sin^3}t, \;\;\Rightarrow \frac{{b\eta }}{{\left[ { - \left( {{a^2} - {b^2}} \right)} \right]}} = {\sin ^3}t,\;\;\Rightarrow \frac{{{{\left( {b\eta } \right)}^{\frac{2}{3}}}}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{2}{3}}}}} = {\sin ^2}t.$

Adding the squares of the cosine and sine, we get

$\frac{{{{\left( {a\xi } \right)}^{\frac{2}{3}}}}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{2}{3}}}}} + \frac{{{{\left( {b\eta } \right)}^{\frac{2}{3}}}}}{{{{\left( {{a^2} - {b^2}} \right)}^{\frac{2}{3}}}}} = 1,\;\;\Rightarrow \left( {a\xi } \right)^{\frac{2}{3}} + \left( {b\eta } \right)^{\frac{2}{3}} = \left( {{a^2} - {b^2}} \right)^{\frac{2}{3}}.$

We denote $$a\xi = X,$$ $$b\eta = Y,$$ $${a^2} - {b^2} = A.$$ Then the equation of the evolute may be represented as

${X^{\frac{2}{3}}} + {Y^{\frac{2}{3}}} = {A^{\frac{2}{3}}}.$

As can be seen, the evolute of the ellipse is a curve, which is quite similar to the astroid. In contrast to the "right" astroid, the given curve is elongated along one axis (Figure $$3$$).