how to find the evolute of a curve when the curve is given in implicit equation?

Question

The parametric equations of the evolute of the curve are given by: $$X = x − \frac{y′((x′)^2+(y′)^2)}{x′y′′ − x′′y′}$$ $$Y = y + \frac{x′((x′)^2+(y′)^2)}{x′y′′ − x′′y′}$$ But what happens when the curve is given in implicit equation $F(x,y)=0,$ how do we find the evolute?

Answer 1

Assume that $\dfrac{\partial F}{\partial y}\ne 0$ so that locally you may write the curve $F(x,y)=0$ in the form $y=f(x)$. Then use $x=t$, $y=f(t)$: Substitute $x'=1$ and $y'=-\dfrac{\partial F/\partial x}{\partial F/\partial y}$ in your formulas. So $x''=0$ is easy enough, but you will need to use the chain rule and the quotient rule to find $y''$. Recall that $$\frac d{dx} g(x,f(x)) = \frac{\partial g}{\partial x}(x,f(x)) + \frac{\partial g}{\partial y}(x,f(x))\cdot f'(x).$$ Using this, we have \begin{align*} y'' = \frac {dy'}{dx} &= \frac d{dx}\left({-}\frac{\partial F/\partial x}{\partial F/\partial y}(x,f(x))\right) \\ &= \frac{\frac{\partial F}{\partial x} \cdot\frac{\partial^2 F}{\partial x\partial y} - \frac{\partial F}{\partial y} \cdot\frac{\partial^2 F}{\partial x^2}}{\left(\frac{\partial F}{\partial y}\right)^2} + \frac{\frac{\partial F}{\partial x} \cdot\frac{\partial^2 F}{\partial y^2} - \frac{\partial F}{\partial y} \cdot\frac{\partial^2 F}{\partial y\partial x}}{\left(\frac{\partial F}{\partial y}\right)^2} y' \\ &= \frac{\frac{\partial F}{\partial x} \cdot\frac{\partial^2 F}{\partial x\partial y} - \frac{\partial F}{\partial y} \cdot\frac{\partial^2 F}{\partial x^2}}{\left(\frac{\partial F}{\partial y}\right)^2} + \frac{\frac{\partial F}{\partial x} \cdot\frac{\partial^2 F}{\partial y^2} - \frac{\partial F}{\partial y} \cdot\frac{\partial^2 F}{\partial y\partial x}}{\left(\frac{\partial F}{\partial y}\right)^2}\left(-\frac{\frac{\partial F}{\partial x}}{\frac{\partial F}{\partial y}}\right). \end{align*} Have fun doing all the algebra.