What is the slope of the line tangent to the curve y^3-xy^2+x^3=5 at the point (1,2)?

Question

Options are as follows:

A. 1/10

B. 1/8

C. 5/12

D. 11/4

Answer 1

The slope of the tangent line at any point is basically the derivative at that point. So the first step is to take the derivative.

y^3 - xy^2 +x^3 = 5 ------> 3y^2 (y') - y^2 - 2xy (y') + 3x^2 = 0

Then you solve so that y' is on its own side of the equation

3y^2 (y') - y^2 - 2xy (y') + 3x^2 = 0 ----------> (y') ( 3y^2 - 2xy) = - 3x^2 + y^2 -----> y' = (y^2 - 3x^2)/ ( 3y^2 - 2xy)

You then plug in x = 1 and y = 2 to get that y' = (2^2 - 3*1^2)/ ( 3* 2^2 - 2*1*2) = (4-3) / (12-4) = 1/8