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Dot product question

  1. Feb 16, 2010 #1
    OK, this has been bugging me for a while. Why is it that

    [tex]\mathbf{v} \cdot \frac{d \mathbf{v}}{d t} = 1/2 v^2 [/tex]

    where regular v is just the magnitude of bold v or more specifically where does the 1/2 coefficient turn up.
  2. jcsd
  3. Feb 16, 2010 #2


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    Essentially because it's the average of the v at the start of the dt and the end
  4. Feb 17, 2010 #3
    It's not true. Just use [tex]\vec{v} = t\vec{x}[/tex] where [tex]\vec{x}[/tex] is the unit vector in the x-direction. It then implies that [tex]t=0.5t^2[/tex].

    Maybe you really wanted to ask about this?:

    \frac{d}{dt}(\frac{1}{2}v^2) = \vec{v}\cdot\frac{d\vec{v}}{dt}

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