# Dot product question

1. Feb 16, 2010

### 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.

2. Feb 16, 2010

### mgb_phys

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

3. Feb 17, 2010

### torquil

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 $$t=0.5t^2$$.

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