Calculus

Applications of the Derivative

Evolute and Involute

Page 1
Problems 1-3
Page 2
Problems 4-7

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 \(M.\)

Definition of the evolute of a curve

Figure 1.

An example of a curve with an inflection point and discontinuous evolute

Figure 2.

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 on problem description to see 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:

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

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 \(3\)).

The evolute of an ellipse is a curve similar to the astroid

Figure 3.

The evolute of the basic parabola is a semicubic parabola

Figure 4.

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 \(4.\) The evolute found above is shaped like a dovetail and its equation is a semicubic parabola.

Page 1
Problems 1-3
Page 2
Problems 4-7