Dot Product: Explaining 1/2 Coefficient

[lordWilhelm]

OK, this has been bugging me for a while. Why is it that

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

where regular v is just the magnitude of bold v or more specifically where does the 1/2 coefficient turn up.

 

[mgb_phys]

Essentially because it's the average of the v at the start of the dt and the end

 

[torquil]

OK, this has been bugging me for a while. Why is it that

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

where regular v is just the magnitude of bold v or more specifically where does the 1/2 coefficient turn up.

It's not true. Just use \vec{v} = t\vec{x} where \vec{x} is the unit vector in the x-direction. It then implies that t=0.5t^2.

Maybe you really wanted to ask about this?:

<br /> \frac{d}{dt}(\frac{1}{2}v^2) = \vec{v}\cdot\frac{d\vec{v}}{dt}<br />

Torquil