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

Question #94987
find the radius of curvature at the point (1,1) of the folium x^3+y^3=2yx
1
Expert's answer
2019-09-23T09:31:54-0400

"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}."


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