Evolute and Involute
Solved Problems
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 \(4.\)
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 \(5\)).
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 \(6\)).
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 \(f\left( {\xi ,\eta } \right) = 0.\)
