Find the curvature of y^2 + x^3= 0 at (-1,1)

Answer 1

2yy' + 3x² = 0; 2y' + 3 = 0 y' = - 3/2 at (- 1, 1)

2y'² + 2yy'' + 6x = 0; at (- 1, 1): 9/2 + 2y'' - 6 = 0; y'' = 3/4.

Curvaturre k = Iy''I/(1 + y'²)^3/2 = 3/4/(1 + 9/4)^3/2 = 3/4/(13/4)^3/2 = 6/(13√13) = 0.128

Answer 2

We can either rewrite as y in terms of x or use a nice formula presented in https://en.wikipedia.org/wiki/Implicit_curve

κ = |(-F_y²F_xx + 2 F_xF_yF_xy -F_x²F_yy)/(F_x² + F_y²)^3/2|

F = y² + x³

F_y = 2y

F_x = 3x²

F_yy = 2

F_xx = 6x

F_xy = 0

Then,

F_y (-1,1)= 2y = 2*1 = 2

F_x (-1,1)= 3x² = 3*1² = 3

F_yy (-1,1)= 2

F_xx (-1,1)= 6x = 6 * -1 = -6

F_xy (-1,1)= 0

|(-F_y²F_xx + 2 F_xF_yF_xy -F_x²F_yy)/(F_x² + F_y²)^3/2| =

|(-2²*-6 + 2*2*3*0 -3²*2)/(F_x² + F_y²)^3/2| =

6/(3² + 2²)^3/2 = 6/(13√13) = 6√13/169