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Math Help - Differentiation

  1. #1
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    Differentiation

    A curve y=x^2e^{-kx}, where k is a positive constant has stationary points of the curve are at (0,0) and (8,a).

    Determine the exact values of the constants k and a.
    Last edited by BabyMilo; September 9th 2010 at 12:24 PM.
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  2. #2
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    May be the first derivative at stationary point is zero.
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  3. #3
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    Quote Originally Posted by zzzoak View Post
    May be the first derivative at stationary point is zero.
    \frac{dy}{dx}= x^2e^{-kx}-2xe^{-kx}=0

    cant finish
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  4. #4
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    Product rule? Just in case a picture helps...



    Or, zooming in on the chain rule...



    ... where (key in spoiler) ...

    Spoiler:


    ... is the product rule. Straight lines differentiate downwards (anti-diff up).



    ... is the chain rule. Straight continuous lines still differentiate downwards with respect to x, and the straight dashed line similarly but with respect to the dashed balloon expression (the inner function of the composite which is subject to the chain rule).


    Spoiler:


    _________________________________________

    Don't integrate - balloontegrate!

    Balloon Calculus; standard integrals, derivatives and methods

    Balloon Calculus Drawing with LaTeX and Asymptote!
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  5. #5
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    Quote Originally Posted by tom@ballooncalculus View Post
    Product rule? Just in case a picture helps...


    Spoiler:

    that's what i've done. how did you get k=\frac{2}{x} ?
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  6. #6
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    Factorised?
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  7. #7
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    cant see it. please show me.
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  8. #8
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    Quote Originally Posted by BabyMilo View Post
    \frac{dy}{dx}= x^2e^{-kx}-2xe^{-kx}=0
    You didn't differentiate it correctly -- that's why you can't finish off.
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  9. #9
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    From Tom's differentiation we have [tex]f'(x) = 2xe^{-kx}-kx^2e^{-kx} = 0 \Rightarrow e^{-kx}\left(2x-kx^2\right) = 0[/Math].
    Since e^{-kx} \ne 0 for any real number k, we have 2x-kx^2 = 0. We know that f'(8) = 0,
    so 16-64k = 0 \Rightarrow k = \frac{1}{4}. You can find a easily now as it's just f(8).
    Last edited by TheCoffeeMachine; September 10th 2010 at 03:05 AM.
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