Is this statement correct : $\frac{dv}{dt}\times dx=\frac{dx}{dt}\times dv$

Question

So I just came across a physics derivation where the treat the $dv$ and $dx$ operators like fractions while I have always heard it's a mistake. But so far what I came up with:

$$\begin{align*} \frac{dv}{dt}\times dx &= \lim_{h\to0} \frac{v(t+h)-v(t)}h \times (x(t+h)-x(t)) \\ &= \lim_{h\to0} \frac{x(t+h)-x(t)}h \times (v(t+h)-v(t)) \\ &= \frac{dx}{dt}\times dv \end{align*}$$

Is my reasoning correct? If not please explain what is the correct way to think about this.

Feel free to take a moment to visit the MathJax tutorial to familiarize yourself. — user170231, Mar 16 '23 at 18:56
I don't understand why this question is being closed. It is fine in my opinion. +1 — Kamal Saleh, Mar 16 '23 at 18:58
@KamalSaleh It wasn't fine initially. OP has been a member for more than one year. — Kurt G., Mar 16 '23 at 18:59
@KurtG. I understand. But notice that the OP has asked only two questions. — Kamal Saleh, Mar 16 '23 at 19:03
@KamalSaleh and presumably read many more, including the comments asking for MathJax formatting. — Kurt G., Mar 16 '23 at 19:05
Back to work: this contains a fairly rigorous version of what you were doing, @JustPassingby. Nothing really wrong with your approach from my point of view. It is the chain rule in disguise. — Kurt G., Mar 16 '23 at 19:06
Your limit statement unfortunately doesn't make sense since $dx$ is not the limit of $x(t+h)-x(t)$ which is $0$. The way to make sense of this statement is by the chain rule. See my answer here — Ninad Munshi, Mar 16 '23 at 19:06
The limits are certainly equal, because both limits are zero. But when $dv$ is used without a denominator, it doesn't really mean the limit definition of a derivative. — Thomas Andrews, Mar 16 '23 at 19:44

score 3 · Answer 1 · answered Mar 16 '23 at 21:00

Just before a differential disappears in the limit we are allowed to view them as small segments of length or small packets of velocity. Even after disappearance their ratio continues to make sense.

$$ \frac{\Delta y}{\Delta x} \to\frac{dy}{dx} =y' = \tan \phi$$ $$ \frac{\Delta x}{\Delta s} \to\frac{dx}{ds} =\cos \phi$$ $$ \frac{\Delta y}{\Delta s} \to\frac{dy}{ds} =\sin \phi. $$

Similarly packets of mass or velocity in a packet (interval) of time made.. even now makes sense in the Newtonian formulation of force :

$$ F =\frac{\Delta(m v)}{\Delta t }= \frac{v\Delta m}{\Delta t }+\frac{m\Delta v}{\Delta t }$$

$$\to v\frac{dm}{dt }+m \frac{ dv}{dt } $$

Even after the disappearance of finite differential form their product continues to make sense.

Joseph Robert Jepson · Answer 2 · 2023-03-17T02:37:36.380

As stated in the comments, taking the limit doesn't work because technically the limit goes to $0$ as $h\rightarrow0$. As stated in Narasimham's answer, you think of differentials as representing very small (infinitesimally small), but non-zero, changes.

The rigorous way to derive the statement you mentioned above is:

$$\cfrac{\text{d}x}{\text{d}t}\text{d}v=\cfrac{\text{d}x}{\text{d}t}\left(\cfrac{\text{d}v}{\text{d}x}\text{d}x\right)=\cfrac{\text{d}x}{\text{d}t}\left(\cfrac{\cfrac{\text{d}v}{\text{d}t}}{\cfrac{\text{d}x}{\text{d}t}}\text{d}x\right)=\cfrac{\text{d}v}{\text{d}t}\text{d}x\ ,$$ where I used the fact that $\cfrac{\text{d}v}{\text{d}t}=\cfrac{\text{d}v}{\text{d}x}\cfrac{\text{d}x}{\text{d}t}$, which is the chain rule.

I think it's good practice to do things more rigorously like this if you ever take (or have taken) multi-variable calculus. Because then differential cancellations are less straightforward.

This just hit the point point. Thanks for taking your time to help. I kinda got a feeling had to use the chain rule here. — JustPassingby, Mar 17 '23 at 03:16

Is this statement correct : $\frac{dv}{dt}\times dx=\frac{dx}{dt}\times dv$

2 Answers2