I heard some one talked about the instantaneous and average velocities.

He was using $\Delta t$ to denote a time frame, $dt$ denote a time point.

average velocities $\bar{v} = \dfrac{\Delta s}{\Delta t}$

the $\Delta t$ part is indeed common. my concern is about the $dt$ part

wiki use the notation

instantaneous velocities $v = \dfrac{ds}{dt}$

Is it reasonable and common to interpret this way?

  • $\begingroup$ Possible duplicates: physics.stackexchange.com/q/65724/2451 and links therein. $\endgroup$ – Qmechanic Aug 26 '19 at 11:02
  • $\begingroup$ I think this is off-topic: it's a question about how differentials are defined, and that's a question about analysis, not about physics, and would therefore be better asked on math SE, especially as physicists are generally fairly vague on this. $\endgroup$ – tfb Aug 26 '19 at 11:22
  • $\begingroup$ See also physics.stackexchange.com/q/153791/25301 $\endgroup$ – Kyle Kanos Aug 27 '19 at 11:59

$\text dt$ is not a point in time. It is an infinitesimal time interval. Physically, you could think of it as a time interval that is much much smaller than the relevant time scale of the system. Mathematically, it is the limit of $\Delta t$ as it approaches $0$ (not equal to $0$).

This just comes from the definition of the limit: $$v=\lim_{\Delta t\to0}\frac{x(t+\Delta t)-x(t)}{\Delta t}$$

Limits are not the same thing as equality. Plugging in $\Delta t=0$ makes the above definition undefined.

  • $\begingroup$ Thanks for your answer. Would you please give a concrete example about the relevant time scale of the system, such as a car, plane, rocket? $\endgroup$ – whnlp Aug 26 '19 at 11:21
  • $\begingroup$ Simply saying "it's undefined" doesn't really answer the question. Why is it undefined? Well, Suppose I say $x=0/0$, and I ask you to solve for $x$. What I'm asking is for the unique $x$ such that $0x=0$. It should be obvious that there is no unique $x$. But if $f(c)=0$ and $g(c)=0$, and I ask for the limit of $f(x)/g(x)$ as $x$ approaches $c$, then there often is a unique answer. $\endgroup$ – Solomon Slow Aug 26 '19 at 11:39
  • 2
    $\begingroup$ @SolomonSlow The question isn't "why is $c/0$ undefined". It's not the reason why $\text dt$ isn't $0$. It's just something to point out. It might not answer something you want answered, but that isn't the OP's question. If the OP wanted to know why we can't divide by $0$ then I would explain that. $\endgroup$ – Aaron Stevens Aug 26 '19 at 12:01

dt is indeed an infinitesimally small amount of time. It is so small, infact, that no matter in what directions and manner a body is undergoing motion, it's motion is always straight line for time dt.


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