1
$\begingroup$

The rate of change of position with respect to time is velocity. But we measure two common velocities, average velocity over an interval of time and instantaneous velocity.

Now, I understand average velocity as the rate of change. I believe it is the slope of the secant line whose x values are the values across which we are computing the average.

My doubt is, how is the instantaneous velocity a rate of change? If you compute the rate of change, at an instant (i.e at an point), you'll see that the x axis is not changing.

While discussing this with another group, they told me to use the definition of limits. The instantaneous velocity is $\Delta{x}/\Delta{t}, \Delta{t} \rightarrow 0$. But that just defines the tangent (which is the derivative) at some point, t. I still don't see the relation between instantaneous velocity and rate of change.

Regarding my prerequisite knowledge, I have a Masters in Engineering. Therefore I understand the formulae of calculus quite well. I just never bothered to understand some of the underlying concepts.

I hope my question is sufficiently clear.

Edit 1 -

One of the answers here has the following text - "it is important to note that as we decrease the size of h , we are not caring about the total change in f from x0 to x0+h , but rather the convergent value of the ratio of these quantities. Even if f(x0+h) gets ever closer to f(x0) , because h is growing small itself, there is a good chance that the two 'balance each other out' in some meaningful way."

Perhaps my confusion stems from what $h$ is doing. I believe that, at the derivative, i.e the point where we are calculating the slope of the tangent, $h$ isn't tending to 0. Rather it is 0. Therefore, the ratio can either be some finite value divided by 0 or 0/0.

$\endgroup$
1
  • 1
    $\begingroup$ Let $x(t)$ be the position, then $\frac{x(t+h)-x(t)}{h}$ gives a rate of change of position in the time interval $(t,t+h)$. Taking the limit as $h\to 0^+$ gives the instantaneous rate of change, which is exactly $x'(t)$. Right? Also see: math.stackexchange.com/questions/3750972/… $\endgroup$ Commented Mar 9 at 22:49

2 Answers 2

3
$\begingroup$

It may be helpful to first clear up some confusion. The instantaneous velocity of a function $f(x)$, represented by its derivative $f'(x)$, at some point $x_0$ represents the slope of the tangent line at that point, not the tangent line itself.

When another group mentioned the limit definition of the derivative, namely

$$ f'(x_0) = \lim_{h \to 0}\frac{f(x_0+h)-f(x_0)}{h} $$

it is important to note that as we decrease the size of $h$, we are not caring about the total change in $f$ from $x_0$ to $x_0+h$, but rather the convergent value of the ratio of these quantities. Even if $f(x_0+h)$ gets ever closer to $f(x_0)$, because $h$ is growing small itself, there is a good chance that the two 'balance each other out' in some meaningful way.

Instantaneous velocity is therefore the limiting ratio, for $h \to 0$. If we include units in the problem more explicitly, the fact that this converges to a rate of change with respect to the variable $x$ can also be seen. I hope this answer addresses some of your concerns, but I am keen to make edits if I can get closer to the heart of your question.

START OF EDIT: In response to the edit that "$h$ isn't tending to $0$, $h$ is $0$ itself"

That is actually the key observation. The definition of the derivative is as a limit. You are correct for saying that if you evaluate the difference at the same point, you evaluate

$$\frac{f(x) - f(x)}{x-x} = \frac{0}{0}.$$

Many limiting quantities do not make sense if you actually 'plug in' the limiting values. For example, the function $\sin(x)/x$ has a limit of $1$ as $x \to 0$, but $\sin(0)/0$ is undefined. The role of $h$ in the definition of the derivative is as a "small timestep", as $f(x+h)$ is supposed to be "the value of $f$ after a small change in $x$". Our intuition is that as we take "smaller timesteps ($h\to 0$)" from our initial data, our approximation of the rate of change grows more accurate. If this limit as $h\to 0$ happens to converge, we call that value the derivative.

$\endgroup$
5
  • $\begingroup$ Firstly, thank you for your detailed answer. I have tried to clarify my question by editing it. Let me know if something is not clear. $\endgroup$ Commented Mar 9 at 23:18
  • $\begingroup$ An edit has been made that may address your question regarding the role of $h$. $\endgroup$ Commented Mar 9 at 23:33
  • $\begingroup$ Thank you for the edit. What does this statement mean, "If this limit as h→0 happens to converge" $\endgroup$ Commented Mar 10 at 1:14
  • $\begingroup$ Not all limits exist. For example, taking $\lim_{x\to 0}x^{-1}$, this limit does not exist. For some functions, the derivative may also not exist at some points. $\endgroup$ Commented Mar 10 at 1:48
  • $\begingroup$ Do you agree that at the derivative, $h=0$? $\endgroup$ Commented Mar 10 at 2:54
1
$\begingroup$

My doubt is, how is the instantaneous velocity a rate of change? If you compute the rate of change, at an instant (i.e at an point), you'll see that the x axis is not changing.

From what I can gather, it's the use of instantaneous (instantaneous velocity) that's giving you a hard time. You are taking instantaneous to mean "at one instant", say, $t = 5 \text{s}$. And this is correct.

However, velocity, by definition, has to do with change. And a change requires a non-zero interval of time.

So, in an attempt to resolve this conflict, let me give you an example. Say you are driving in one direction along a perfectly straight line. Your speed is not constant though. Now, let's consider your motion from $t = 0 \text{ s}\;$ to $\;t = 60 \text{ s}$. We'll assume you cover $1$ mile in this duration.

Now, what's your average velocity (magnitude only)? $60$ miles per hour, right? It's easy.

But what was your velocity at $t = 5 \text{ s}$? Now, we are taking of instantaneous velocity. Was it the same as your average velocity, 60 mph? We don't know. You could be moving faster, slower. You may have been stationary, for all we know. The $30 \text{s}$ interval is too large for us to assume that your velocity didn't change within that interval.

What if we knew you covered $0.05$ miles in the interval from $4$ to $6$ seconds? This gives a velocity of $90$ mph. Now because our interval is shorter (only $2$ seconds), one might think it gives a more accurate representation of our velocity at $5$ s. But still it's the average velocity for those $2$ s. You could have accelerated and/or decelerated during those $2$ seconds. So, your velocities at $4$ s, $5$ s, and $6$ s could have different values. And we can't say for sure what your velocity was at $5$ s.

However, as we continue to shrink this interval further around $t = 5$ s, the average velocity for that interval starts to give an increasingly accurate picture of the instantaneous velocity at $5$ s. Just intuitively, one can understand that if we knew the displacement between $4.9999$ s and $5.0001$ s, the average velocity for that duration would almost be the same as your velocity at any instant during those $0.0002$ s. Won't you agree with that?

This is what we do when we take the derivative (say at $t_0 = 5$ s). We are shrinking the interval such that it approaches $0$. "Approaches" is the key word here. The interval must not become $0$ (or there won't be any velocity to talk about).

$$f'(t_0) = \lim_{\Delta t \to 0} \frac{f(t_0 + \Delta t) - f(t_0)}{\Delta t}$$

TL;DR - Again, for us to calculate the instantaneous velocity, the interval must not be $0$. It must, however, approach $0$ so that we can be confident that the velocity didn't change over that interval.

$\endgroup$
2
  • $\begingroup$ I follow everything you say. It cleared up major doubts; thank you! My followup question would be, while taking the derivative (at $t_0 = 5$), although we shrink the interval while computing the limit, the derivative still is being computed only at a point (5). The number 5 isn't an interval. $\endgroup$ Commented Mar 10 at 12:30
  • 1
    $\begingroup$ I am glad you found the answer helpful. When calculating $f'(t_0)$, we use both $t_0$ and $\Delta t$. You seem to be focusing only on $t_0$. $t_0$ gives us the point at which to calculate the derivative, while $\Delta t$ gives us the interval over which the change is obtained. And that interval is non-zero. Don't forget that $\lim_{\Delta t \to 0}$ is not the same as $\delta t = 0$. Hope this helps. $\endgroup$
    – Haris
    Commented Mar 10 at 12:53

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .