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Math Help - horizontal tangent on a curve

  1. #1
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    horizontal tangent on a curve

    The curve C in the xy-plane is determined by the equation:
    3x5 + 30xy + 12y3

    (a) Find an expression in x and y for dy/dx.

    (b) At what points different from (0,0) does C have a horizontal tangent?

    I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.
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  2. #2
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    Quote Originally Posted by jarrod573 View Post
    The curve C in the xy-plane is determined by the equation:
    3x5 + 30xy + 12y3

    (a) Find an expression in x and y for dy/dx.

    (b) At what points different from (0,0) does C have a horizontal tangent?

    I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.
    Put the value of dy/dx you got in part a as 0(zero) and all the values which satisfy the equality is you your answer

    dy/dx=0
    and all x and y ae your answer (other than (0,0))
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  3. #3
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    Quote Originally Posted by jarrod573 View Post
    The curve C in the xy-plane is determined by the equation:
    3x5 + 30xy + 12y3

    (a) Find an expression in x and y for dy/dx.

    (b) At what points different from (0,0) does C have a horizontal tangent?

    I was able to get part (a) but don't know how to get part (b). Any help is greatly appreciated.
    a) I've got:

    \dfrac{dy}{dx}=-\dfrac{5(x^4  + 2y)}{2(5x + 8y^2 )}

    b) \dfrac{dy}{dx} = 0 ~\implies~x^4+2y=0~\implies~x= \pm \sqrt[4]{-2y} which will yield real results if y < 0.

    Plug in this term instead of x into the original equation to calculate y. I've got: y = -\sqrt[7]{2^5}~\vee~y = 0

    Re-substitute this value into the term of x. I've got:

    x = -\sqrt[7]{2^3}

    The tangent is drawn in blue.
    Attached Thumbnails Attached Thumbnails horizontal tangent on a curve-tang_ankurve.png  
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