What is the time derivative of the linear velocity vector $\vec{v}\,(t)$? If $\vec{v}\,(t)$ denotes linear velocity, we can then write $\vec{v}\,(t)$ as $|v(t)|\hat{v}$. My question is what is $\displaystyle\frac{d\vec{v}\,(t)}{dt}?$
The answer I have seen to this question says that $$\displaystyle\frac{d\vec{v}\,(t)}{dt} = \hat{v}\frac{d|v(t)|}{dt}.$$ If you agree with this answer why is $\hat{v}$ treated like a constant that does not depend on $t$?
 A: In general, there is no reason to assume that the direction of $\vec{v}$ is constant. Therefore, the equation you quote cannot be explained without further context. The time derivative of velocity is more commonly called the acceleration $\vec{a}$:
$$\frac{\text{d}\vec{v}}{\text{d}t}= \vec{a}$$
A: First I apologise to the students who came to answer my question since I had not given the context of the question.
The background of this question came from the work kinetic energy theorem. The Professor while explaining wrote that $\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = d\vec{v}\cdot\vec{v}$ by telling us that we can cancel the differential $dt$ in the left hand side of the equation. Not happy with that explanation I asked the TA for further clarifications who then explained to me that $\displaystyle\frac{d\vec{v}}{dt} = \hat{v}\frac{dv}{dt}$ and then proceeding accordingly I will be able to show that $\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = d\vec{v}\cdot\vec{v}$
But now I realise how to show that $\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = d\vec{v}\cdot\vec{v}$ and here are my steps.
Linear velocity $\vec{v} = v\hat{v}$ where $v$ is the magnitude of $\vec{v}$.
$\displaystyle\frac{d\vec{v}}{dt} = v\frac{d\hat{v}}{dt}+\hat{v}\frac{dv}{dt}$.
$\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = (v\frac{d\hat{v}}{dt}+\hat{v}\frac{dv}{dt})\cdot \vec{v}dt$
The dot product of $v\frac{d\hat{v}}{dt}$ with $\vec{v}dt$ is zero because $\frac{d\hat{v}}{dt}$ is perpendicular to $\vec{v}$ (Reference Kleppner 1.10)
Therefore $\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt =\hat{v}\frac{dv}{dt}\cdot \vec{v}dt$
Now we can write $dt$ on the right hand side of the above equation as $dt = \displaystyle\frac{dv}{\frac{dv}{dt}}$
So we have
$\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt =\hat{v}\frac{dv}{dt}\cdot \vec{v}\frac{dv}{\frac{dv}{dt}}$
In the above equation we can cancel the $\frac{dv}{dt}$ terms as these are scalar terms and hence we get
$\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = \hat{v}\cdot \vec{v}dv= \hat{v}dv\cdot \vec{v}=d\vec{v}\cdot\vec{v}$.
End
