SOLUTION: Please help me solve this problem: Where on the curve x^2 - 2xy + y^3 = 1 is the slope of the curve vertical? Thanks

Question 457965: Please help me solve this problem:
Where on the curve x^2 - 2xy + y^3 = 1 is the slope of the curve vertical?

Thanks
Answer by richard1234(7193) (Show Source): You can put this solution on YOUR website!
Suppose we take d/dx of both sides:

Applying the basic derivation rules such as the power rule, chain rule, and product rule,

Solving for dy/dx yields

The slope is vertical when the denominator of dy/dx is zero and the numerator is non-zero. It must be noted that all (x,y) that satisfy 3y^2 - 2x = 0 must also satisfy the original curve.

I did not want to attempt to solve the system of two equations (it would be quite difficult as well since it involves cubics), so I used Wolfram to find the points on the graph where dy/dx has denominator zero. WolframAlpha lists (0.653, -0.660) and (2.046, 1.168) as the only such points. The graph of the curve and the parabola 3y^2 - 2x = 0 can be found by going to the URL below (the intersection of the curves show the desired points):

http://www.wolframalpha.com/input/?i=x^2+-+2xy+%2B+y^3+%3D+1+and+3y^2+-+2x+%3D+0

FYI: Another alternative might be to use polar coordinates and then differentiate. However, this solution might be longer since you will have to solve for r first.
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