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Dot product of a vector and a derivative of that vector

  1. Aug 5, 2014 #1
    I'm reading through Douglas Gregory's Classical Mechanics, and at the start of chapter 6 he says that [itex]m \vec{v} \cdot \frac{d\vec{v}}{dt} = \frac{d}{dt}\left(\frac12 m \vec{v} \cdot \vec{v}\right)[/itex], but I'm not sure how to get the right hand side from the left hand side.

    If someone could point me in the direction of where to look for this I would be grateful.

    Thank you.
     
  2. jcsd
  3. Aug 5, 2014 #2

    mathman

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    [itex]\vec{v} \cdot \vec{v}=v_x^2 + v_y^2 + v_z^2 [/itex]
    Just use elementary calculus to get derivative of the right hand side. Assumes m is constant.
     
    Last edited: Aug 5, 2014
  4. Aug 5, 2014 #3
    It's the product rule, ##\frac{d}{dt}[\mathbf{u}\cdot\mathbf{v}]=\frac{d\mathbf{u}}{dt}\cdot \mathbf{v}+\mathbf{u}\cdot\frac{d\mathbf{v}}{dt}##, that gives this result. It's probably easiest to see this if you work right to left.
     
  5. Aug 5, 2014 #4
    Yes, of course it is. Thank you.
     
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