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Math Help - Differentiating twice

  1. #1
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    Differentiating twice

    Alright, the question is:

    transform 2x\frac{d^2x}{dt^2} -6(\frac{dx}{dt})^2=x^2-3x^4

    into \frac{d^2y}{dt^2}+y=3 by using the substitution y=\frac{1}{x^2}



    I'm stuck on differentiatin twice

    Help, please?
    Last edited by Silver; June 14th 2008 at 12:37 PM.
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  2. #2
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    Pretty please?
    No??
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  3. #3
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    Question

    In the result you're trying to get, are the derivatives of y supposed to be with respect to t, or with respect to x (as you have it written)? I'm pretty sure it should be the former.

    --Kevin C.
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  4. #4
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    Quote Originally Posted by TwistedOne151 View Post
    In the result you're trying to get, are the derivatives of y supposed to be with respect to t, or with respect to x (as you have it written)? I'm pretty sure it should be the former.

    --Kevin C.
    woops, it was supposed to be with respect to t, not x, sorry
    edited it now

    thanks for pointing it out
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  5. #5
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    Chain rule

    Apply the chain rule. You should have gotten from y=\frac{1}{x^2} that
    \Large{\begin{array}{rcl}\frac{dy}{dt}&=&\frac{dy}  {dx}\frac{dx}{dt}\\&=&-\frac{2}{x^3}\frac{dx}{dt}\end{array}},
    as you said it was the second derivative you had trouble with.
    The key is to differentiale both sides of the above result with respect to t:
    \frac{d^2y}{dt^2}=\frac{d}{dt}\left(-\frac{2}{x^3}\frac{dx}{dt}\right).
    Use the product rule:
    \frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d  x}{dt}-\frac{2}{x^3}\frac{d}{dt}\left(\frac{dx}{dt}\right  )
    \frac{d^2y}{dt^2}=-\frac{d}{dt}\left(\frac{2}{x^3}\right)\cdot\frac{d  x}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}.
    Now apply the chain rule to the remaining derivative:
    \frac{d^2y}{dt^2}=-\frac{d}{dx}\left(\frac{2}{x^3}\right)\cdot\frac{d  x}{dt}\cdot\frac{dx}{dt}-\frac{2}{x^3}\frac{d^2x}{dt^2}
    \frac{d^2y}{dt^2}=-\left(-\frac{6}{x^4}\right)\cdot\left(\frac{dx}{dt}\right  )^2-\frac{2}{x^3}\frac{d^2x}{dt^2}
    \frac{d^2y}{dt^2}=\frac{6}{x^4}\left(\frac{dx}{dt}  \right)^2-\frac{2}{x^3}\frac{d^2x}{dt^2}
    And you should be able to proceed from here.

    --Kevin C.
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  6. #6
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    Hello, Silver!

    Is there a typo? . . . Is there really a square in the equation?
    . . Without the square, I can do it (backwards).


    Transform . 2x\frac{d^2x}{dt^2} -6 \frac{dx}{dt} \;=\;x^2-3x^4\;\text{ into }\;\frac{d^2y}{dt^2}+y\;=\;3
    by using the substitution y=\frac{1}{x^2}

    We have: . y \;=\;x^{-2}\quad\Rightarrow\quad \frac{dy}{dt} \:=\:-2x^{-3}\frac{dx}{dt}

    Then: . \frac{d^2y}{dx^2} \;=\;-2x^{-3}\frac{d^2x}{dt^2} + 6x^{-4}\frac{dx}{dt}\quad\hdots\quad\text{ product rule}


    Substitute into: . \underbrace{\frac{d^2y}{dt^2} }+ \underbrace{y} \;=\;3

    . . . . \overbrace{-2x^{-3}\frac{d^2}{dt^2} + 6x^{-4}\frac{dx}{dt}} + \;\overbrace{x^{-2}}\;=\;3

    . . -\frac{2}{x^3}\frac{dx^2}{dt^2} + \frac{6}{x^4}\frac{dx}{dt} \;=\;-\frac{1}{x^2}+3


    Multiply by -x^4\!:\;\;2x\frac{dx^2x}{dt^2} - 6\frac{dx}{dt}  \;=\;x^2-3x^4

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  7. #7
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    there is definitely a square, but thanks anyway
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