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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 $t$?

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    $\begingroup$ Where is this claimed? $\endgroup$
    – Filippo
    Commented Aug 12, 2021 at 14:36
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    $\begingroup$ Where did you see that answer? It is surely incorrect as $d{\bf v}/dt$ is not parallel to ${\bf v}$ unless the motion is along a straight line. Perhaps that is what you mean by "linear" velocity? It's not a standard term. In that case $\hat v$ is a constant if the motion does not reverse. $\endgroup$
    – mike stone
    Commented Aug 12, 2021 at 14:40
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    $\begingroup$ As a rule of thumb, if you'd like to ask about something you've seen somewhere, then you should explain where you saw it; this helps users craft answers which you are likely to find helpful or relevant. Statements which are incorrect in general (like this one) may be correct in a limited context which may have been omitted from your question, so a reference to the source could be the difference between "that claim is simply wrong" and "as per the author, that claim is true if [...]." $\endgroup$
    – J. Murray
    Commented Aug 12, 2021 at 15:50

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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 $\vec{a}$: $$\frac{\text{d}\vec{v}}{\text{d}t}= \vec{a}$$

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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

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    $\begingroup$ My God that is confusing. Are you trying to prove the work-energy theorem? If so, just project ${\bf F} = m {\bf a}$ along ${\bf v}$: ${\bf F} \cdot {\bf v} = m {\bf a} \cdot {\bf v}$. Now we note that $\frac{d}{dt} ({\bf v} \cdot {\bf v}) = 2 {\bf v} \cdot {\bf a}$. Therefore, ${\bf F} \cdot {\bf v} = \frac{d}{dt} (\frac{1}{2} m {\bf v} \cdot {\bf v})$. In other words, $P = \dot{T}$, where $T$ is the kinetic energy and $P$ is the power supplied to the particle. $\endgroup$
    – Evan
    Commented Aug 12, 2021 at 17:58
  • $\begingroup$ Well actually I was trying to show $\displaystyle\frac{d\vec{v}}{dt}\cdot \vec{v}\,dt = d\vec{v}\cdot\vec{v}$. $\endgroup$
    – ADN
    Commented Aug 13, 2021 at 3:44
  • $\begingroup$ @ADN That is a mathematical fact that works for all smooth functions, because by definition of the derivative, we have $\text{d}f = (\text{d}f/\text{d}t) \text{d}t$ for $f=f(t)$. It seems to me that what you did is circular reasoning. $\endgroup$
    – Vincent Thacker
    Commented Aug 13, 2021 at 4:15
  • $\begingroup$ @VincentThacker what you wrote above is for scalar functions but what about vector functions? Unfortunately I had not come across a definition for vector functions plus the physics Prof. explained the above fact by asking us to imagine that the $dt$'s cancel each other. $\endgroup$
    – ADN
    Commented Aug 13, 2021 at 11:31
  • $\begingroup$ @ADN The derivative is defined the same way for vectors. The fundamental way of defining any derivative is $\lim \limits_{\Delta x \to 0}(f(x+\Delta x)-f(x))/\Delta x$. To say that the $\text{d}t$'s cancel each other is correct but not mathematically rigorous. A full mathematical treatment involves differential forms and so on. $\endgroup$
    – Vincent Thacker
    Commented Aug 13, 2021 at 11:49

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