Calc question please help?

Question

Find the radius of curvature and the coordinates of the center of curvature of the curve y=x^3-x-1 at the points for which x= -2, x=0, and x=1
thanks

Answer 1

first find the first and second derivative.

y' = 3x^2 - 1
y" = 6x

Radius of Curvature:
r(x) = |1 + (y')^2|^(3/2) / |y''|
= |1 + (3x^2 -1)^2|^(3/2) / |6x|
= |1 + 9x^4 - 6x^2 + 1|^(3/2)/|6x|
= |9x^4 - 6x^2 + 2|^(3/2)/|6x|
r(-2) = |9*16 - 24 +2|^(3/2)/12 = 61*sqrt(122)/6
r(0) = undefined
r(-1) = 5/6

I'm not sure how to find the center of curvature.

Answer 2

The center of curvature has coordinates
Xc =x-y´[(1+y´^2)/y´´] and Yc= y+[(1+y´^2)/y´´] at the point (x,y)