Why is $\vec{v}\cdot d \vec{v} = v dv$? Can someone help me understand why is this true: 
$$\vec{v} \cdot d \vec{v} = v \cdot dv$$ 
where $v$ is speed? I found somewhere that $\vec{v} \cdot d \vec{v}=|\vec{v}||d \vec{v}| \cos \phi$. And I don't know from where it follows that $|d \vec{v}| \cdot \cos \phi = dv$.
 A: Let $v=|\mathbf v|$. Then by Leibniz rule
$$\text d v^2 = 2v\text dv$$
Now $v^2 = \mathbf v\cdot\mathbf v$ by definition, so we also have
$$\text dv^2 = \text d(\mathbf v\cdot\mathbf v)=2\mathbf v\cdot\text d\mathbf v$$
whence
$$v\text d v = \mathbf v\cdot\text d\mathbf v.$$
A: 
I don't know from where it follows that $|d \vec{v}| \cdot cos \phi = dv$.

Not strict, yet more intuitive answer:
Imagine $d \vec{v}$ as a sum of two components:


*

*along $\vec{v}$

*and perpendicular to it.


The former changes only the magnitude of $\vec{v}$ (i.e. $v$) so it acts as $dv$; and the latter changes only the direction of $\vec{v}$.
Then geometrically, if $\phi$ is the angle between $\vec{v}$ and $d \vec{v}$, the former component magnitude is $|d \vec{v}| \cdot cos \phi$.
A: The usual case is in 3-dimensional. In this case we have
\begin{equation}
\vec v \cdot d \vec v = v_x dv_x + v_y dv_y+ v_z dv_z \;. \tag{1}
\end{equation}
On the other hand the magnitude of a vector $v$ is given by 
\begin{equation}
v =\sqrt{v_x^2 +v_y^2+v_z^2} \;, \tag{2}
\end{equation}
or  \begin{equation}
v^2 =v_x^2 +v_y^2+v_z^2 \;. \tag{3}
\end{equation}
Differentiate on eqn (3) gives us
\begin{equation}
2v \,dv = 2v_x dv_x + 2v_y dv_y+ 2v_z dv_z \;. \tag{4}
\end{equation}
From eqn(1) and eqn(4) we have
$$\vec v \cdot d \vec v = v dv$$
A: Update  2 after @DavidZ said I had not answered the question and quite correctly deleted my answer.
I rewrote my answer but could not undelete it as it was deleted by a moderator so this is the reason for a second answer.
This is an attempt to illustrate the two interpretations of the change in velocity, $d\vec v$ and $|d \vec v|$,  which is possibly the reason for the confusion.    
Here is a diagram which shows how under the action of a force $\vec F$ for a short interval of time the initial velocity $\vec v$ changes to a final velocity $\vec v + d \vec v$
 
$\vec v \cdot \Delta \vec v = v \, |\Delta \vec v| \cos \phi$  where $|\Delta \vec v|$ is the magnitude of the change in velocity $\Delta \vec v$
As the time interval over which the acceleration took place tends to zero this equation becomes
$\vec v \cdot d \vec v = v \, |d \vec v| \cos \phi$
The magnitude of the final velocity  $|\vec v + \Delta \vec v| = v +\Delta v = v\cos \Delta \phi+ |\Delta \vec v|\cos(\phi - \Delta \phi)$  
As the time interval over which the force acts tends to zero  $\Delta \phi \rightarrow 0$  
$v\cos \Delta \phi \rightarrow v$ and $|\Delta \vec v|\cos(\phi - \Delta \phi) \rightarrow |d \vec v|\cos\phi $ which is $dv$.
Finally the desired result $\vec v \cdot d \vec v = v \, |d \vec v| \cos \phi = v \, dv$
