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Derivative of velocity squared

  1. Mar 27, 2009 #1
    1. The problem statement, all variables and given/known data
    This really isn't a specific homework problem. It is just something that I've never actually known the reason for. An example is in the derivation of:

    [tex]\frac{dT}{dt} = F\dot v [\tex]

    In order to arrive at it, I replace T with [tex]1/2mv^2[\tex] and assume m is constant and then I have to do the product rule on v^2 and I'm never sure why I can't use the power rule. Any explanation is appreciated.

    On a side note, this makes sense to me when I substitute momentum in for one of the velocities and then do the product rule. I just don't understand why I can't leave it as velocity and use the power rule. Thanks!
     
  2. jcsd
  3. Mar 27, 2009 #2

    CompuChip

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    Well, v = v(t) is a function of t, and you are differentiating with respect to t.
    So when you differentiate v(t)^2 you use the chain rule with u = v(t) and get
    [tex]\frac{d(v(t))^2}{dt} = \frac{du}{dv} \frac{dv(t)}{dt} = (2v) (v'(t))[/tex]
    where du/dv = d(v^2)/dv = 2 v, just like differentiating
    [tex](x^2 + 6)^2[/tex]
    with respect to x first gives you 2(x^2 + 6) by the power rule and then another factor of 2x because of the chain rule.

    So, using v'(t) = a(t), and F = m v(t)
    [tex]\frac{dT}{dt} = \frac12 m (2 v a) = m v a = F v[/tex]
    [If you want to treat v(t) as a vector, you need to be a little more careful as you will get some dot products]
     
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