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 .