Please Subscribe here, thank you!!!The graph of mathx^2(y\sqrt3{x^2})^2=1/math is very interesting and is show below using desmos3 The Valentine equation (x2 y2 1)3 x2y3 = 0 contains the point (1;1) Near (1;1), we have y= y(x) so that (x2 y(x)2 1)3 x2y(x)3 = 0 Find y0 at the point x= 1 Solution Take the derivative 0 = 3(x 2 y 1)2(2x 2yy 0) 2xy3 x3y2y(x) and solve for y0 = 3(x2 y2 1)2x 2xy3 3(x2 y2 21)2y 3xy2 For x= 1;y= 1, we get 4=3
The Curve Below Is The Graph Of X 2 Y 2 1 3 X 2 Chegg Com
