Find the abscissa of the center of curvature of the curve x^3+y^3=4xy at (2,2)

Answer 1

Derivatives: 3x² + 3y²y' = 4y + 4xy', y' = (3x² - 4y)/(4x - 3y²); y'(2,2) = ((12 - 8)/(8 - 12) = - 1;

6x + 3y²y'' + 6yy'²= 8y' + 4xy''; 12 + 12y'' + 12 = - 8 + 8y''; 4y'' = - 32; y'' = - 8

So curvache: K = |y''|/(1 + y'²)^3/2 = 8/2^3/2 = 2^3/2 = 2√2.

Let find equation of normal line at point (2,2): slope m = -1/y' = -1/(-1) = 1; y - 2 = 1(x - 2); y = x is equation of normal line at point (2, 2).

Let (a,a) is center of curvature. Then (a - 2)² + (a - 2)² = (2√2)²; 2(a - 2)² = 8; (a - 2)² = 4; a - 2 = ±2

a = 4 or a = 0. Because at (2, 2) y '' = - 8 < 0 we have to take a = 0. So center of curvature K located at origin (0, 0) and x = 0 is x-coordinate of this point.