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Math Help - Can someone explain how this makes 2nd order diff. makes sense

  1. #1
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    Can someone explain how this makes 2nd order diff. makes sense

    Just reading through a proof which states that:

    dy/dt=(-2x^-3)(dx/dt)

    Then it states that:

    d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

    I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

    Many thanks,

    Bryn
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  2. #2
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by Bryn View Post
    Just reading through a proof which states that:

    dy/dt=(-2x^-3)(dx/dt)

    Then it states that:

    d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

    I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

    Many thanks,

    Bryn
    You are differentiating with respect to t..

    what is the derivative of -2x^{-3}? it is 6x^{-4} \frac{dx}{dt}

    that is where the second \frac{dx}{dt} came from..
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  3. #3
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    Quote Originally Posted by Bryn View Post
    Just reading through a proof which states that:

    dy/dt=(-2x^-3)(dx/dt)

    Then it states that:

    d2y/dt2= (-2x^-3)(d2x/dt2)+(6x^-4)(dx/dt)^2

    I can see that the chain rule is used, although I do not understand why the last dx/dt is squared - could somone explain how the first equation transforms to the next please,

    Many thanks,

    Bryn
    \frac{dy}{dt}=(-2x^{-3}) \frac{dx}{dt}

    Now taking the derivative with respect to t of both sides gives

    \frac{d}{dt} \left( \frac{dy}{dt} \right)= \frac{d}{dt} \left(-2x^{-3}\frac{dx}{dt} \right)

    Now using the product rule on the right side we get

    \frac{d^2y}{dt^2}=\left(\frac{d}{dt}(-2x^{-3}) \right)\cdot \frac{dx}{dt}+(-2x^{-3})\cdot \frac{d}{dt}\left( \frac{dx}{dt}\right)

    Now using the chain rule we get

    \frac{d^2y}{dt^2}=\left((6x^{-4})\cdot \frac{dx}{dt} \right)\cdot \frac{dx}{dt}+(-2x^{-3})\cdot \left( \frac{d^2x}{dt^2}\right)


    \frac{d^2y}{dt^2}=\left((6x^{-4})\cdot\left( \frac{dx}{dt} \right)^2\right)-(2x^{-3})\left( \frac{d^2x}{dt^2}\right)
    Last edited by TheEmptySet; June 15th 2008 at 06:38 PM. Reason: Thanks
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  4. #4
    MHF Contributor kalagota's Avatar
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    Quote Originally Posted by TheEmptySet View Post

    \frac{d^2y}{dt^2}=\left((6x^{-4})\cdot \frac{d^2x}{dt^2} \right)-(2x^{-3})\left( \frac{d^2x}{dt^2}\right)
    beware of these.. \frac{d^2x}{dt^2} \neq \left(\frac{dx}{dt}\right)^2
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