Answer to Question #145441 in Differential Geometry | Topology for Dolly

Question

Find the centre of curvature of astroid
x^2/3+y^2/3=a^2/3

Answer 1

$Solution$

Center of curvature is denoted by $P$

Lets determine the coordinates of the center of curvature at the point $P$ on the astroid

Let $\theta = \frac{\pi}{6}$

$\rho = \frac{\delta s}{\delta \psi}=12\ sin\ \psi\ cos\ \psi = 6\ sin\ \psi\\ At\ \theta_p=\frac{\pi}{6},\\ \psi_p=-\frac{\pi}{6},\\ \rho_p=6\ sin\ (-\frac{\pi}{3})=6\ (-\frac{\sqrt{3}}{2})=-3 \sqrt{3}$

Now

$(X,Y)=(x_p, y_p) +\rho n$

So written

$\theta=\frac{\pi}{6},\ \psi= -\frac{\pi}{6}, x_p=\frac{3 \sqrt{3}}{2}, y_p=\frac12, \rho_p=-3 \sqrt{3}\\ n_p=(-sin(-\frac{\pi}{6})cos(-\frac{\pi}{6}))=(sin(\frac{\pi}{6})cos(\frac{\pi}{6}))= (\frac12, \frac{\sqrt{3}}{2})\\ Thus;\\ \implies(X,Y)=(\frac{3 \sqrt{3}}{2}, \frac12)+|-3 \sqrt{3}|(\frac12,\frac{\sqrt{3}}{2})\\ \implies(X,Y)=(\frac{3 \sqrt{3}}{2}, \frac12)+(-\frac{3 \sqrt{3}}{2},\frac{9}{2})\\ \therefore\ (X,Y)=(3 \sqrt {3},\ 5)$

Hence, the center of curvature of astroid is $(3 \sqrt {3},\ 5)$