Implicit+Differentiation+Free+Response

Mimi: a) take the derivative with implicit differentiation: 2x + 8y(dy/dx) = 3x(dy/dx) + 3y 3y - 2x = 8y(dy/dx) - 3x(dy/dx) so dy/dx = 3y - 2x/ 8y - 3x b) If the slope of the tangent line is 0, the derivative is 0. so, 3y - 2x/8y - 3x = 0 3y - 2x = 0 when x = 3, 3y = 6, so y = 2 when plugging in those values, 3^2 + 4(2^2) = 25 and 7 + (3)(3)(2) = 25 so P = (3,2) is on the curve and slope = 0 at this point. c) d^2y/dx^2 = (8y - 3x)(3y' - 2) - (3y - 2x)(8y' - 3)/(8y - 3x)^2 at P = (3,2), d^2y/dx^2 = (16 - 9)(-2)/(16 - 9)^2 = -2/7 y' = 0 and y'' < 0, there is therefore a local maximum at P.

yutong a) use the implicit differentiation to find the derivative 2y(dy/dx)=x(dy/dx)+y 2y(dy/dx)-x(dy/dx)=y (2y-x)*(dy/dx) = y dy/dx=y/(2y-x) b) set the derivative to 1/2 to find all the points y/(2y-x)=0 x has to be 0 in this case and y = squre root (2) c) the tangent is horizontal only happens when the derivative of the function is 0 the limite as x approch 0 is 1/2 so therefore the function is always increasing therefore the derivative can never be 0 d) dy/dt= 6 so plug in 3 for y in the derivative function and set it equal to 6 to solve for x 6= 3/(2*3-x) x= 6/33 .....? not sure about D

Zielonka: I got 33/6 which simplifies to 11/2 for x. But I'm not sure where to go from there either...