Can you use $\frac{\Delta v}{\Delta t}$ instead of $\frac{dv}{dt}$ to find instantaneous acceleration?
$\begingroup$
$\endgroup$
7
-
$\begingroup$ By "using $\Delta$", do you mean taking actual differences instead of taking a derivative? Why would you think you can do that? What's the context here? $\endgroup$– ACuriousMind ♦Commented Apr 10, 2020 at 15:28
-
$\begingroup$ Related question. $\endgroup$– CharlieCommented Apr 10, 2020 at 15:42
-
1$\begingroup$ Yes, if you take the limit $\Delta t\to 0$. $\endgroup$– G. SmithCommented Apr 10, 2020 at 16:45
-
$\begingroup$ @G. Smith, without the limit, i mean why can't we find the instantaneous acceleration if we are using the same equation but without the limit? For example like a=0.0000000000000002/0.0000000000000001=2 $\endgroup$– ZheerCommented Apr 10, 2020 at 16:51
-
1$\begingroup$ why can't we find the instantaneous acceleration if we are using the same equation but without the limit? Because if you don’t take the limit there is nothing instantaneous about it. It’s the average acceleration over a finite time interval, and in general you will get a different value for different choices of $\Delta t$. It may, however, be an excellent approximation to the instantaneous acceleration for sufficiently small $\Delta t$. $\endgroup$– G. SmithCommented Apr 10, 2020 at 17:23
|
Show 2 more comments
3 Answers
$\begingroup$
$\endgroup$
4
$d$ is used to denote an infinitesimal, so in the case of velocity $\frac{dv}{dt}\bigr |_{t_0}$ would represent the instantaneous rate of change in velocity at $t_0$, which is the acceleration at $t_0$. $\Delta$ is usually used to represent larger change, they are equal if $\frac{dv}{dt}$ is constant, but otherwise they are generally not.
-
$\begingroup$ "otherwise they are generally not" do you mean that we can't use Δ in a=Δv/Δt for infinitesimal change to find instantaneous acceleration? $\endgroup$– ZheerCommented Apr 10, 2020 at 15:38
-
$\begingroup$ You can think of $\frac{dv}{dt}$ as being $\frac{\Delta v}{\Delta t}$ when the "change" represented by $\Delta$ is infinitesimally small. If you look at any graph that is a curve you can't generally find the gradient using $\frac{\Delta y}{\Delta x}$, but you can find the instantaneous gradient at a point and this is written $\frac{dy}{dx}$ and has to be calculated at a point. $\endgroup$– CharlieCommented Apr 10, 2020 at 15:40
-
$\begingroup$ "you can't generally find the gradient using Δy/Δx" why can't we generally find the gradient if we can always have infinitesimally small change? $\endgroup$– ZheerCommented Apr 10, 2020 at 16:10
-
$\begingroup$ Because $\Delta y/\Delta x$ does not typical represent an infinitesimal change. Think of $\frac{dy}{dx}$ as being $\Delta y/\Delta x$ in the special case where the change represented by $\Delta$ is infinitesimally small. $\endgroup$– CharlieCommented Apr 10, 2020 at 16:24
$\begingroup$
$\endgroup$
2
The answer to the title is no. The first equation only gives you the average acceleration for the time interval. The only situation where it equals the instantaneous acceleration is if velocity is a linear function of time.
Hope this helps
-
$\begingroup$ According to my knowledge the change in Δy/Δx can be infinitesimally small that it equals the same number as the slope so why is the answer is no? $\endgroup$– ZheerCommented Apr 10, 2020 at 16:48
-
$\begingroup$ @Zheer Sure, but you didn't say that. Besides, when you make it infinitely small it becomes the derivative, the second equation. That's essentially the definition of the derivative! $\endgroup$– Bob DCommented Apr 10, 2020 at 17:37
$\begingroup$
$\endgroup$
Actually you do use it, implicitly, in the definition of the derivative. First replace the changes with small changes $${\Delta v\over \Delta t} \rightarrow {\delta v\over \delta t}$$ Then the derivative is defined by $${dv\over dt} = \lim\limits_{\delta t \to 0} {\delta v\over \delta t}$$
And recalling the $\epsilon-\delta$ definition of a limit, this really just means that $\delta t$ is a finite amount of time so small that taking it any smaller does not make any practical difference to the result.
answered Apr 10, 2020 at 18:28