Answer to Question #281785 in Differential Geometry | Topology for chaitu

Question

find the radius of the curvature at (acostheta^3,asintheta^3) on the curve x^2/3+y^2/3=a^2/3

Answer 1

radius of the curvature:

$R=\frac{(1+(dy/dx)^2)^{3/2}}{|d^2y/dx^2|}$

$2x/3+2yy'/3=0$

$y'=-\frac{x}{y}$

$y''=-\frac{y-xy'}{y^2}$

at point $(acos \theta^3 ,asin \theta ^3):$

$y'=-cot\theta^3$

$y''=-\frac{sin\theta^3-cos^2\theta^3/sin\theta^3}{sin^2\theta ^3}$

$R=\frac{sin^2\theta ^3(1+cot^2\theta^3)^{3/2}}{sin\theta^3-cos^2\theta^3/sin\theta^3}$