Question

Find radius of curvature at (x,y) for a2y =x3-a3

Answer 1

Solution:

let, R be the radius of curvature at (x, y) for a^2y =x3-a^3.

now, differentiating w.r.t x both side of this equation we get

a^2dy/dx =3x^2

d y/dx = 3x^2/a^2   ------1

again, differentiating w.r.t x both side of this equation we get

d^2y/dx^2 = 6x/a^2     -------2

now, we know that

R = [{1+(d y/dx)^2}^3/2]/d^2y/dx^2  ( from equation 1 and 2)

R = [{1+ (3x^2/a^2)^2}^3/2]/6x/a^2

R = [(1+3x^4/a^4)^3/2]^3/2/6x/a^2

R = [(a^3+3x^4)/a^4]^3/2/6x/a^2

R = [(a^3+3x^4)^3/2]/a^6*a^2/6x

R =  [(a^3+3x^4)^3/2] / [6xa^4]

therefore, radius of curvature at (x, y) for a2y =x3-a3 is  [(a^3+3x^4)^3/2] / [6xa^4]