Consider the closed curve in the day plane:

Question

2x^2-2xy+y^3=14

a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k

Answer 1

Consider the closed curve in the xy plane:
2x^2-2xy+y^3=14

a) show that dy/dx=2y-4x/3y^2-2x (I got this part)
b) find equation lines to the curve when y=2
c) if the point (2.5, k) is on the curve, use part b to find the best approximation of the value of k

Answer 2

Your dy/dx should be
(2y-4x)/(3y^2 - 2x)
(the way you have it, only the -4x is divided by 3y^2 )

so when y = 2 , in the original,
2x^2 - 4x + 8 = 14
2x^2 - 4x - 6 = 0
x^2 - 2x - 3 = 0
(x-3)(x+1) = 0
x = 3 or x = -1

b) you probably meant to say:
Find the equation of the tangent lines when y = 2

so the points of contact of the two tangents are
(3,2) and (-1,2)

at (3,2), slope = dy/dx = (4 - 8)/(12 - 6) = -4/6 = -2/3
equation of tangent:
y-2 = (-2/3)(x-3)
3y - 6 = -2x + 6
2x + 3y = 12 or y = (-2/3)x + 4

follow the same steps for the tangent to the other point

c) Not really sure what they want for that part.
Have you studied differentials?
If the point (2.5 , k) is supposed to be "close" to (3,2), that's quite a spread even if we use differentials
Just subbing in x = 2.5 yields the cubic
y^3 - 5y = -1.5

An appr just taking a few values of y gave me
y = appr 2.4 (y = 2.3732...)