# Evolute question

1. Aug 22, 2007

### John O' Meara

Prove that the evolute of the rectangular hyperbola $$x=a\cosh\theta$$, $$y=a\sinh\theta$$ is $$x^{\frac{2}{3}} - y^{\frac{2}{3}}= (2a)^{\frac{2}{3}}\\$$.
Let (X,Y) be a pair of coordinates on the center of curvature (the evolute) of the hyperbola. $$X =x -\frac{y^{'}(1 + (y^')^2)}{y^{''}} \\$$ and $$Y = y-\frac{(1 + (y^')^2)}{y^{''}} \\$$, where $$y^{'}=\frac{dy}{dx} \\$$.
Now $$\frac{dx}{d\theta} = a sinh\theta$$, $$\frac{dy}{d\theta}=a cosh\theta \\$$; therefore $$\frac{dy}{dx} = \coth\theta \\$$;
hence $$\frac{d^2y}{dx^2} = \frac{d \coth\theta}{d\theta}\frac{d\theta}{dx} =-\cosech^2\theta \frac{d\theta}{dx}\\$$.
Therefore $$y^{''} =-\frac{cosech^2 \theta}{a\sinh\theta}\\$$.
Therefore $$X = a\cosh\theta + a\cosh\theta(\sinh^2\theta + \cosh^2\theta)\\$$ and $$Y = a\sinh\theta + a\sinh\theta(\sinh^2\theta + \cosh^2\theta) \\$$.
Thest are the parametric equations of the evolute. How do I get the cartesian form of the equations for the evolute. I mean how do I eliminate $$\theta$$. Thanks for the help.

Last edited: Aug 23, 2007
2. Aug 31, 2007

### John O' Meara

Therefore I get $$x^{\frac{1}{3}} - y^{\frac{1}{3}} = (2a)^{\frac{1}{3}}((\cosh^3\theta)^{\frac{1}{3}} \\ - ( \sinh^{\frac{1}{3}}\theta + \sinh^3\theta)^{\frac{1}{3}}) \\$$ which gives $$x^{\frac{2}{3}} - y^{\frac{2}{3}} = (2a)^{\frac{2}{3}}( 1 - \sinh^{\frac{2}{3}}\theta)\\$$ Can anyone see where I went wrong? Thanks for the help.

3. Sep 3, 2007

### SanjeevGupta

John

There is a typo in the formula for Y above. It should be Y=y + etc. Then you get the required result.

Regards

4. Sep 3, 2007