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Simple differential questions

  1. Feb 29, 2008 #1
    A little thing that may be stupid but I am confused about it. Say we take any equation like f=ma
    If we take the derivative of both sides then we could either have

    df=m da
    or
    df =a dm

    Are both of these valid computations. If I am looking to change the integrating variable can I use this any way I want? And would this work with any equation relating 3 or more variables?

    thanks in advanced
     
  2. jcsd
  3. Feb 29, 2008 #2

    Andy Resnick

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    You are basically correct: the "rocket equation" comes from F = dp/dt = d(mv)dt = m dv/dt + v dm/dt.

    For what you wrote, f = ma, df = m da/dt + a dm/dt.
     
  4. Feb 29, 2008 #3

    pam

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    If f=abc, df=ab dc+ ac db + bc da, and so on for an number of variables.
    You just differentiate one at a time.
    Of course if any factor is a constant, then its differential is zero.
     
  5. Feb 29, 2008 #4

    pam

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    The "rocket equation" is a bit different, because the exit velocity of the gas enters instead of just v in the dm/dt term.
     
  6. Feb 29, 2008 #5
    thanks this clears some up. But as I said before could I take df=mda to subsitute df with da to integrate with respect to a, and in the same problem could I take df=adm to integrate with respect to m.

    so you could take s=rTheta and make it ds=rdtheta?

    I'm just seeing how flexible I can be when substituting vaiables to integrate or differentiate with.

    Edit, I just realized that what you did there was the product rule which makes sense. Is what I just said above wrong then?>
     
    Last edited: Feb 29, 2008
  7. Feb 29, 2008 #6

    pam

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    A derivative of all product of N variables will have N terms, each term being differentilated once. So d(r theta)=r dtheta+theta dr.
     
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