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Implicit differentiation

  1. Feb 20, 2012 #1
    1. The problem statement, all variables and given/known data

    Find the coordinates of the stationary points on the curve:
    x^3 + (3x^2)(y) -2y^3=16


    2. Relevant equations
    Stationary points occur when the first derivative of y with respect to x is equal to zero



    3. The attempt at a solution
    I implicitly differentiated the equation and got
    dy/dx = (x^2 + 2xy) / (2y^2 - x^2)

    I know I have to make this equal to zero but then I'm not sure how to find the x and y coordinates of the stationary point.

    Help would be greatly appreciated :)
     
  2. jcsd
  3. Feb 20, 2012 #2

    ehild

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    Homework Helper
    Gold Member

    Hi aanandpatel,

    Find y in terms of x from the condition dy/dx=0. Substitute back into the original equation.

    ehild
     
  4. Feb 20, 2012 #3

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have two equations,
    [tex]x^3 + (3x^2)(y) -2y^3=16[/tex]
    and
    [tex](x^2 + 2xy) / (2y^2 - x^2)= 0[/tex]
    to solve for x and y. The second equation can easily be solved for y in terms of x since a fraction is equal to 0 if and only if the numerator is 0.
     
  5. Feb 21, 2012 #4
    Thanks guys - helped a lot! :)
     
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