Prove that the evolute of the rectangular hyperbola [tex] x=a\cosh\theta [/tex], [tex] y=a\sinh\theta [/tex] is [tex] x^{\frac{2}{3}} - y^{\frac{2}{3}}= (2a)^{\frac{2}{3}}\\ [/tex].(adsbygoogle = window.adsbygoogle || []).push({});

Let (X,Y) be a pair of coordinates on the center of curvature (the evolute) of the hyperbola. [tex] X =x -\frac{y^{'}(1 + (y^')^2)}{y^{''}} \\ [/tex] and [tex] Y = y-\frac{(1 + (y^')^2)}{y^{''}} \\ [/tex], where [tex] y^{'}=\frac{dy}{dx} \\[/tex].

Now [tex] \frac{dx}{d\theta} = a sinh\theta [/tex], [tex] \frac{dy}{d\theta}=a cosh\theta \\ [/tex]; therefore [tex] \frac{dy}{dx} = \coth\theta \\ [/tex];

hence [tex] \frac{d^2y}{dx^2} = \frac{d \coth\theta}{d\theta}\frac{d\theta}{dx} =-\cosech^2\theta \frac{d\theta}{dx}\\ [/tex].

Therefore [tex] y^{''} =-\frac{cosech^2 \theta}{a\sinh\theta}\\ [/tex].

Therefore [tex] X = a\cosh\theta + a\cosh\theta(\sinh^2\theta + \cosh^2\theta)\\[/tex] and [tex] Y = a\sinh\theta + a\sinh\theta(\sinh^2\theta + \cosh^2\theta) \\[/tex].

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 [tex] \theta [/tex]. Thanks for the help.

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# Homework Help: Evolute question

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