Find the point on the curve y = 1/x^2 ,( x > 0 ), where the curvature is maximum.

Answer 1

k = |y"|/(1+y'^2)^(3/2) = (6/x^4)/(1 + 4/x^6)^(3/2)
so, maximum k is where
24(5-x^6)/(x^6+4)^(5/2) = 0
That is, at x = 5^(1/6)
That would make the point (5^(1/6),5^(-1/3))