Answer to Question #145446 in Differential Geometry | Topology for Dolly

Question

Find the evolute of the four cusped hypocycloid
x=a cos^3θ y=a sin^3θ

Answer 1

Find the derivatives of $x$ and $y$ with respect to the parameter $\theta :$

$$x'=-3a\sin \theta\cos^2\theta,$$

$$y'=3a\sin^2 \theta\cos\theta,$$

$$x''=-3a\cos^3\theta+6a\sin^2 \theta\cos\theta$$

$$y''=-3a\sin^3 \theta+6a\sin \theta\cos^2\theta$$

To calculate the coordinates of the center of curvature $(\xi,\eta),$  we use the formulas:

$$\xi=x-y'\dfrac{(x')^2+(y')^2}{x'y''-x''y'}$$

$$\eta=y+x'\dfrac{(x')^2+(y')^2}{x'y''-x''y'}$$

$$(x')^2+(y')^2=9a^2 \sin^2\theta\cos^4\theta+9a^2 \sin^4\theta\cos^2\theta=$$

$$=9a^2 \sin^2\theta\cos^2\theta$$

$$x'y''-x''y'=9a^2\sin^4 \theta\cos^2\theta-18a^2\sin^2 \theta\cos^4\theta$$

$$+9a^2\sin^2 \theta\cos^4\theta-18a^2\sin^4 \theta\cos^2\theta=$$

$$=9a^2\sin^2 \theta\cos^2\theta-18a^2\sin^2 \theta\cos^2\theta$$

$$=-9a^2\sin^2 \theta\cos^2\theta$$

$$\dfrac{(x')^2+(y')^2}{x'y''-x''y'}=-1$$

$$\xi=a\cos^3\theta+3a\sin^2\theta \cos\theta$$

$$\eta=a\sin^3\theta+3a\sin\theta \cos^2\theta$$

Consequently, the evolute of the four cusped hypocycloid is described by the following parametric equations:

$$\xi=a\cos^3\theta+3a\sin^2\theta \cos\theta$$

$$\eta=a\sin^3\theta+3a\sin\theta \cos^2\theta$$

Thus the evolute of an astroid is an astroid .