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Math Help - implicit differentiation easy question that i can't do...

  1. #1
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    Lightbulb implicit differentiation easy question that i can't do...

    Find the points on the graph  x^3 + y^3 = xy
    that are parallel to the x axis (gradient = 0)

    so I have to differentiate it
    I had
     3x^2 + 3y^2 (dy/dx) = dy/dx

    then what?

    Thank you
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  2. #2
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    Quote Originally Posted by differentiate View Post
    Find the points on the graph  x^3 + y^3 = xy
    that are parallel to the x axis (gradient = 0)

    so I have to differentiate it
    I had
     3x^2 + 3y^2 (dy/dx) = dy/dx

    correction, product rule on the right hand side ...

    \textcolor{red}{3x^2 + 3y^2 \frac{dy}{dx} = x \frac{dy}{dx} + y}


    then what? solve for \textcolor{red}{\frac{dy}{dx}} and set the derivative equal to 0. use that equation to find a relationship between y and x to find the points on the curve where the gradient is 0.
    ...
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  3. #3
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    thank you
    so then,

     3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y
     3y^2(dy/dx) - x(dy/dx) = y - 3x^2
     (dy/dx)(3y^2 - x) = y - 3x^2
     dy/dx = (y-3x^2)/(3y^2 - x)
    dy/dx = 0
     y - 3x^2 = 0
     y = 3x^2

    do I put that into the original equation?

     x^3 + (3x^2)^3 = x(3x^2)
     x^3 + 27x^6 = 3x^3
     27x^6 - 2x^3 = 0
     x^3(27x^3 - 2) = 0

    I'm stuck
    and I don't have answers to this question
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  4. #4
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    Quote Originally Posted by differentiate View Post
    thank you
    so then,

     3x^2 + 3y^2 (dy/dx) = x.(dy/dx) + y
     3y^2(dy/dx) - x(dy/dx) = y - 3x^2
     (dy/dx)(3y^2 - x) = y - 3x^2
     dy/dx = (y-3x^2)/(3y^2 - x)
    dy/dx = 0
     y - 3x^2 = 0
     y = 3x^2

    do I put that into the original equation?

     x^3 + (3x^2)^3 = x(3x^2)
     x^3 + 27x^6 = 3x^3
     27x^6 - 2x^3 = 0
     x^3(27x^3 - 2) = 0

    I'm stuck
    and I don't have answers to this question
    then the points are
    x = 0 \rightarrow y = 0
    x = \frac{\sqrt[3]2}{3} \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2
    Last edited by dedust; January 3rd 2010 at 08:16 PM.
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  5. #5
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    Smile

    Quote Originally Posted by dedust View Post
    then the points are
    x = 0 \rightarrow y = 0
    x = \frac{\sqrt[3]2}{3} \rightarrow y = \left(\frac{\sqrt[3]2}{3}\right)^2
    You mean x = \frac{\sqrt[3]2}{3} \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2
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  6. #6
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    Smile

    Quote Originally Posted by pickslides View Post
    You mean x = \frac{\sqrt[3]2}{3} \rightarrow y = 3\left(\frac{\sqrt[3]2}{3}\right)^2
    oops,..yeap
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