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Classical physics Time dependent vector calculation

  1. Jul 9, 2014 #1
    1. The problem statement, all variables and given/known data
    If A is a time dependent vector, calculate

    [itex] \int_{t1}^{t2} dtA(t) \times \frac{d^2A}{dt^2} [\itex]

    2. Relevant equations



    3. The attempt at a solution

    I think we should somehow relate it with something's derivative.

    \int_{t1}^{t2}A(t)\frac{d^2A(t)}{dt^2}dt=
    A(t)\int_{t1}^{t2}\frac{d^2A(t)}{dt^2}dt-\int_{t1}^{t2}\frac{dA(t)}{dt}\left ( \int \frac{d^2A(t)}{dt^2} \right )dt=
    A(t)\frac{dA(t)}{dt}\left.\right|_a^b\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt=
    A(t2)\frac{dA(t)}{dt}\left.\right|_t_{2}-A(t1)\frac{dA(t)}{dt}\left.\right|_t_{1}-\int_{t1}^{t2}\left ( \frac{dA(t)}{dt} \right )^2dt
     
    Last edited: Jul 9, 2014
  2. jcsd
  3. Jul 9, 2014 #2

    SteamKing

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    Your Latex seems to have blown up. Would you like to edit the OP and try asking your question again?

    Hint: before posting, use the Preview Button to check the appearance of your post and correct as required.
     
  4. Jul 9, 2014 #3

    Orodruin

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    Your notation is a bit horrible, you really should not write ##A(t)## outside of the integrals and you are missing the ##\times## from the original expression - making it unclear whether or not it is a cross product or a scalar product. The approach with using partial integration does make sense (and in case of a cross product, what is ##\dot A \times \dot A##?). However, it is not necessary to use partial integration as you can rewrite ##A\times \ddot A## directly as the total derivative of something. What total derivative would have the given term as one of its terms? What would be the value of the other term?
     
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