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Calc I problem

  1. Sep 12, 2008 #1
    1. The problem statement, all variables and given/known data
    Given 2y^3 +(6x^2)y-12x^2+6y = 1

    A. Show that dy/dx = 4x -2xy/x^2 + y^2 + 1
    B. Write an equation of each horizontal tangent line to the curve
    C. The line through the origin with slope -1 is tangent to the curve at point P. Find the x-y coordinates of point P.


    3. The attempt at a solution

    A. 6y^2(dy/dx) + 12xy + 6x^2(dy/dx) - 24x + 6(dy/dx) = 0

    6y^2(dy/dx) + 6x^2(dy/dx) + 6(dy/dx) = 24x - 12xy

    (dy/dx) (6y^2 + 6x^2 +6 ) = 24x - 12xy

    (dy/dx) = 24x - 12xy/ (6y^2 + 6x^2 +6 )

    (dy/dx) = 4x - 2xy/ x^2 + y^2 + 1 <--- (factoring out a 6)


    B. 4x - 2xy = 0

    2x( 2 - y )= 0

    x=0 , y=2

    This part confuses me a bit. Plugging x=0 back in the original equation gives me a nonsensical answer ( 28=1) and plugging y=2 in the original nets me a cubic equation.
    Is y=2 the equation for the tangent or what?

    C. 4x -2xy/x^2 + y^2 + 1 = -1

    4x -2xy = -x^2 - y^2 - 1

    Since the line goes through the origin with slope of -1, y= -x

    4x - 2x(-x) = -x^2 - (-x)^2 -1

    Rearranging gives 4x^2 + 4x +1 = 0 , Factoring that gives x= -1/2

    And again, since y= -x, then the coords are (-1/2, 1/2)


    **I may be doing this totally wrong. I haven't been in a calculus class for 2 years and I'm trying to brush up on it a bit.**

    Sorry for any confusion on my format. This is my first time posting here.

    Thanks!!!
     
  2. jcsd
  3. Sep 12, 2008 #2
    Part A looks good. Isn't implicit differentiation fun!?

    For Part B, you are solving for y, in terms of x (if necessary). Since the lines are horzontal, the solution is just y = 2.

    Part C looks good also.
     
  4. Sep 12, 2008 #3

    Dick

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    I can't see anything wrong with what you done so far. For B) since y=2 gives you an equation with no roots, the x=0 is the only possibility, Which tells you the tangent point is (0,A) where A is the root of 2y^3+6y-1=0. The equation of the tangent is y=A. But the cubic doesn't factor or anything nice. About all you could do is solve it numerically or use the dreaded 'cubic equation'.
     
  5. Sep 12, 2008 #4

    Dick

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    That would be right, except the poster got the two conditions backwards, y=2 doesn't give you any solutions to the original equation. And the x=0 case gives you the funny cubic. The person who composed the problem may not have thought that part through correctly.
     
  6. Sep 12, 2008 #5
    Bah, you're right. I guess I just mixed those 2 up when I was writing up the question.

    Thanks for all your input everyone!
     
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