# Calculus

## Applications of the Derivative # Evolute and Involute

Let a plane curve $$\gamma$$ 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: $$K\left( s \right) \ne 0.$$ 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 $$\mathbf{n}$$ we draw the segment $$\mathbf{MC}$$ equal to the radius of curvature $$R\left( s \right)$$ 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 $$\require{AMSmath.js}\boldsymbol{\rho},$$ 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

${x = x\left( t \right),}\;\;\; {y = y\left( t \right),}\;\;\;\kern-0.3pt {\alpha \le t \le \beta ,}$

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

${\xi = x – y’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}},}\;\;\;\kern-0.3pt {\eta = y + x’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}}.}$

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

${\xi = x – \frac{{1 + {{\left( {y’} \right)}^2}}}{{y^{\prime\prime}}}y’,}\;\;\;\kern-0.3pt {\eta = y + \frac{{1 + {{\left( {y’} \right)}^2}}}{{y^{\prime\prime}}}.}$

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

Click or tap a problem to see the solution.

### Example 1

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

### Example 2

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

### 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
${x = t – \sin t,}\;\;\;\kern-0.3pt {y = 1 – \cos t.}$

### Example 6

Prove that the curve given by the equations
${x = R\left( {\cos t + t\sin t} \right),}\;\;\;\kern-0.3pt {y = R\left( {\sin t – t\cos t} \right)}$
is the involute of the circle of radius $$R$$ centered at the origin.

### Example 7

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

### 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 $$t:$$

${x’ = {\left( {R\cos t} \right)^\prime } = – R\sin t,}\;\;\;\kern-0.3pt {y’ = {\left( {R\sin t} \right)^\prime } = R\cos t,}$

${x^{\prime\prime} = {\left( { – R\sin t} \right)^\prime } = – R\cos t,}\;\;\;\kern-0.3pt {y^{\prime\prime} = {\left( {R\cos t} \right)^\prime } = – R\sin t.}$

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

${\xi = x – y’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}},}\;\;\;\kern-0.3pt {\eta = y + x’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}}.}$

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

$\require{cancel} {\xi = x – y’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}} } = {{R\cos t – R\cos t \cdot}\kern0pt {\frac{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}}{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}} }} = {\cancel{R\cos t} – \cancel{R\cos t} \equiv 0;}$

${\eta = y + x’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}} } = {{R\sin t – R\sin t} \cdot\kern0pt{\frac{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}}{\cancel{{{R^2}{{\sin }^2}t + {R^2}{{\cos }^2}t}}} }} = {R\sin t – R\sin t \equiv 0.}$

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

${x’ = {\left( {a\cos t} \right)^\prime } = – a\sin t,}\;\;\;\kern-0.3pt {y’ = {\left( {b\sin t} \right)^\prime } = b\cos t,}$

${x^{\prime\prime} = {\left( { – a\sin t} \right)^\prime } = – a\cos t,}\;\;\;\kern-0.3pt {y^{\prime\prime} = {\left( {b\cos t} \right)^\prime } = – b\sin t.}$

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

${\xi = x – y’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}},}\;\;\;\kern-0.3pt {\eta = y + x’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}}.}$

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

${\xi = x – y’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}} } = {{a\cos t – b\cos t}\cdot\kern0pt{\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{{ab\,{{\sin }^2}t + ab\,{{\cos }^2}t}} }} = {{a\cos t – \cos t}\cdot\kern0pt{\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{a} }} = {\frac{{{a^2}\cos t – {a^2}{{\sin }^2}t\cos t – {b^2}{{\cos }^3}t}}{a} = \frac{{{a^2}\cos t\left( {1 – {{\sin }^2}t} \right) – {b^2}{{\cos }^3}t}}{a} } = {\frac{1}{a}\left( {{a^2}{{\cos }^3}t – {b^2}{{\cos }^3}t} \right) = \frac{{{a^2} – {b^2}}}{a}{\cos ^3}t;}$

${\eta = y + x’\frac{{{{\left( {x’} \right)}^2} + {{\left( {y’} \right)}^2}}}{{x’y^{\prime\prime} – x^{\prime\prime}y’}} } = {{b\sin t – a\sin t}\cdot\kern0pt{\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{{ab\,{{\sin }^2}t + ab\,{{\cos }^2}t}} }} = {{b\sin t – \sin t}\cdot\kern0pt{\frac{{{a^2}{{\sin }^2}t + {b^2}{{\cos }^2}t}}{b} }} = {\frac{{{b^2}\sin t – {a^2}{\sin^3}t – {b^2}{\cos^2}t\sin t}}{b} } = {\frac{{{b^2}\sin t\left( {1 – {{\cos }^2}t} \right) – {a^2}{\sin^3}t}}{b} } = {\frac{1}{b}\left( {{b^2}{\sin^3}t – {a^2}{\sin^3}t} \right) = \frac{{{b^2} – {a^2}}}{b}{\sin^3}t.}$

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

${\xi = \frac{{{a^2} – {b^2}}}{a}{\cos^3}t = \left( {a – \frac{{b^2}}{a}} \right){\cos ^3}t,}\;\;\;\kern-0.3pt {\eta = \frac{{{b^2} – {a^2}}}{b}{\sin^3}t = \left( {b – \frac{{a^2}}{b}} \right){\sin ^3}t.}$

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

${\xi = \frac{{{a^2} – {b^2}}}{a}{\cos^3}t,}\;\;\Rightarrow {a\xi = \left( {{a^2} – {b^2}} \right){\cos^3}t,}\;\;\Rightarrow {\frac{{a\xi }}{{{a^2} – {b^2}}} = {\cos^3}t,}\;\;\Rightarrow {\frac{{{{\left( {a\xi } \right)}^{\large\frac{2}{3}\normalsize}}}}{{{{\left( {{a^2} – {b^2}} \right)}^{\large\frac{2}{3}\normalsize}}}} = {\cos^2}t;}$

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

Adding the squares of the cosine and sine, we get

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

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

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

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 $$3$$).

### 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

${\xi = x – \frac{{1 + {{\left( {y’} \right)}^2}}}{{y^{\prime\prime}}}y’,}\;\;\;\kern-0.3pt {\eta = y + \frac{{1 + {{\left( {y’} \right)}^2}}}{{y^{\prime\prime}}}.}$

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:

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

Hence,

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

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

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