BC Calc problem: 2y^3 + 6x^2(y) -12x^2 +6y = 1

[jagged]

I have a math problem and am sort of bewildered as how to do it. The question is:

Consider the curve defined by 2y^3 + 6x^2(y) -12x^2 +6y = 1

a. Find dy/dx

Done- i think. I got dy/dx = (4x-2xy)/(x^2 + y^2 + 1) Check perhaps?

b. Write an equation of each horizontal tangent line to the curve.

Okay, so dy/dx=0. I solved for that, and by i got that either x=0 or y=2. Plugging in y=2 into the original equation yields no solutions for x. However, plugging in x=0 i got a cubic equation with a single, irrational solution that i could only find using the cubic equation: 2y^3 + 6y = 1 (Can i solve it otherwise?)

On the basis that this problem seems to be sort of well-made, i don't see how it could come up to be such a messy answer. Am i doing anything wrong?

Thanks


Edit: does anybody have a f(x,y) graphing applet that show what this looks like?

 

[galactus]

There are applets about. Do a search. Here's one:

http://cs.jsu.edu/mcis/faculty/leathrum ... -main.html

 

[stapel]

I get the same results you do. Yes, there is a zero of dy/dx for x = 0, but no, you can't find the zero neatly. :shock:

My guess? There's a typo in the assignment.

Try taking your work, showing that you tried and got something unreasonably messy (so you clearly did try), and asking your instructor for advice. :idea:

Good luck!

Eliz.