Question

Find the evolute of the four cuspecl hypocycloid
x=acos^3θ y=asin^3θ

Answer 1

The equation of the evolute is

$$\vec{r}(t)=x(t)\cdot\vec{i}+y(t)\cdot\vec{j}\to\\[0.3cm] \left\{\begin{array}{l} X(t)=x(t)-\displaystyle\frac{y'(t)\cdot\left(x'(t)^2+y'(t)^2\right)}{x'(t)y''(t)-x''(t)y'(t)}\\[0.5cm] Y(t)=y(t)+\displaystyle\frac{x'(t)\cdot\left(x'(t)^2+y'(t)^2\right)}{x'(t)y''(t)-x''(t)y'(t)} \end{array}\right.$$

(more infotmation : https://en.wikipedia.org/wiki/Evolute )

In our case,

$$\left\{\begin{array}{c} x(\theta)=a\cos^3(\theta)\\[0.3cm] y(\theta)=a\sin^3(\theta) \end{array}\right.\to\\[0.3cm] \left\{\begin{array}{c} x'(\theta)=-3a\cos^2(\theta)\cdot\sin(\theta)\\[0.3cm] \boxed{x'(\theta)=\displaystyle\frac{-3a}{2}\cos(\theta)\sin(2\theta)}\\[0.5cm] x''(\theta)=\displaystyle\frac{-3a}{2}(-\sin(\theta)\sin(2\theta)+2\cos(\theta)\cos(2\theta))\\[0.5cm] y'(\theta)=3a\sin^2(\theta)\cos(\theta)\\[0.3cm] \boxed{y'(\theta)=\displaystyle\frac{3a}{2}\sin(\theta)\sin(2\theta)}\\[0.5cm] y''(\theta)=\displaystyle\frac{3a}{2}(\cos(\theta)\sin(2\theta)+2\sin(\theta)\cos(2\theta)) \end{array}\right.$$

Now we transform the expressions that enter the equations for the evolute :

$$x'(\theta)^2+y'(\theta)^2=\left(\frac{-3a}{2}\cos(\theta)\sin(2\theta)\right)^2+\left(\frac{3a}{2}\sin(\theta)\sin(2\theta)\right)^2\\[0.3cm] =\frac{9a^2}{4}\cdot\sin^2(2\theta)\cdot\left(\cos^2(\theta)+\sin^2(\theta)\right)\equiv\frac{9a^2}{4}\cdot\sin^2(2\theta)\\[0.3cm] \boxed{x'(\theta)^2+y'(\theta)^2=\frac{9a^2}{4}\cdot\sin^2(2\theta)}\\[0.3cm] x'(\theta)y''(\theta)-x''(\theta)y'(\theta)=\\[0.3cm] =\frac{-3a}{2}\cos(\theta)\sin(2\theta)\cdot\frac{3a}{2}(\cos(\theta)\sin(2\theta)+2\sin(\theta)\cos(2\theta))-\\[0.3cm] -\frac{-3a}{2}(-\sin(\theta)\sin(2\theta)+2\cos(\theta)\cos(2\theta))\cdot\frac{3a}{2}\sin(\theta)\sin(2\theta)=\\[0.3cm] =\frac{-9a^2}{4}\sin(2\theta)\cdot\left(\cos^2(\theta)\sin(2\theta)+2\sin(\theta)\cos(\theta)\cos(2\theta)+\right.\\[0.3cm] \left.+\sin^2(\theta)\sin(2\theta)-2\sin(\theta)\cos(\theta)\cos(2\theta)\right)=\\[0.3cm] =\frac{-9a^2}{4}\sin(2\theta)\cdot\left(\sin(2\theta)\cdot\left(\sin^2(\theta)+\cos^2(\theta)\right)\right)=\\[0.3cm] =\frac{-9a^2}{4}\sin^2(2\theta)\\[0.3cm] \boxed{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)=\frac{-9a^2}{4}\sin^2(2\theta)}$$

Substitute the found expressions into the equations for the evolute, which were indicated at the very beginning :

$$X(\theta)=x(\theta)-\frac{y'(\theta)\cdot\left(x'(\theta)^2+y'(\theta)^2\right)}{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}=\\[0.3cm] =a\cos^3(\theta)-\frac{\displaystyle\frac{3a}{2}\sin(\theta)\sin(2\theta)\cdot\displaystyle\frac{9a^2}{4}\sin^2(2\theta)}{\displaystyle\frac{-9a^2}{4}\sin^2(2\theta)}=\\[0.3cm] =a\cos^3(\theta)+\frac{3a}{2}\sin(\theta)\sin(2\theta)=a\cos^3(\theta)+3a\sin^2(\theta)\cos(\theta)=\\[0.3cm] =a\cos(\theta)\left(\cos^2(\theta)+3\sin^2(\theta)\right)\equiv a\cos(\theta)\left(1+2\sin^2(\theta)\right)\\[0.3cm] \boxed{X(\theta)=a\cos(\theta)\left(1+2\sin^2(\theta)\right)}\\[0.3cm] Y(\theta)=y(\theta)+\frac{x'(\theta)\cdot\left(x'(\theta)^2+y'(\theta)^2\right)}{x'(\theta)y''(\theta)-x''(\theta)y'(\theta)}=\\[0.3cm] =a\sin^3(\theta)+\frac{\displaystyle\frac{-3a}{2}\cos(\theta)\sin(2\theta)\cdot\displaystyle\frac{9a^2}{4}\sin^2(2\theta)}{\displaystyle\frac{-9a^2}{4}\sin^2(2\theta)}=\\[0.3cm] =a\sin^3(\theta)+\frac{3a}{2}\cos(\theta)\sin(2\theta)=a\sin^3(\theta)+3a\cos^2(\theta)\sin(\theta)=\\[0.3cm] =a\sin(\theta)\left(\sin^2(\theta)+3\cos^2(\theta)\right)\equiv a\sin(\theta)\left(1+2\cos^2(\theta)\right)\\[0.3cm] \boxed{Y(\theta)=a\sin(\theta)\left(1+2\cos^2(\theta)\right)}\\[0.3cm]$$

ANSWER

$$\left\{\begin{array}{c} x(\theta)=a\cos^3(\theta)\\[0.3cm] y(\theta)=a\sin^3(\theta) \end{array}\right.\to \left\{\begin{array}{c} X(\theta)=a\cos(\theta)\left(1+2\sin^2(\theta)\right)\\[0.3cm] Y(\theta)=a\sin(\theta)\left(1+2\cos^2(\theta)\right) \end{array}\right.$$