Answer to Question #94987 in Differential Geometry | Topology for sushmita

Question

find the radius of curvature at the point (1,1) of the folium x^3+y^3=2yx

Answer 1

$y'=\frac{2y-3x^2}{3y^2-2x},\;\;y'(1,1)=-1.$

$y''=\frac{(18xy-6y^2-4x)y'-18xy+4y+6x^2}{(3y^2-2x)^2},\;\;y''(1,1)=-16.$

Radius of curvature:

$R=|\frac{1+y'^2)^{\frac{3}{2}}}{y''}|=\frac{\sqrt{2}}{8}.$