Silly vector-related question

Question

It's probably a silly question but, I don't understand this :

How can this be always true when

$\vec{v}.d\vec{v} = v.dv.cos(\alpha) = v.dv \Leftrightarrow \alpha = [2\pi]$

It seems contradictory...

Answer 1

The physical context of the text most likely specifies or implies that $\vec{v}$ and $d\vec{v}$ are collinear, i.e. the angle between them is $\alpha = 0$, thus $\vec{v} \cdot d\vec{v} = v \cdot dv \cdot \cos(0) = v \cdot dv$.