Rate of change measures how fast a process is going when it moves from one point to another. It measures the change of, say, $Y$ when $X$ moves from $X$ to $X + \Delta X$. But my problem arises when rate of change is concerned at only a single point; the mathematical tool to solve this is to use derivative at that point provided the graph is continuous. I am unable to understand the rate of change concept at a single point; I have always connected the rate of change with two points. So what is the rate of change concept at a single point? Let a spherical balloon have a variable radius & the rate of change of volume w.r.t radius when radius is $7$ unit is $196\pi$. What is the meaning of this statement when it mentions the rate of change at a single point i.e. at $r=7$? Thanks for your help.

  • $\begingroup$ It is the instantaneous rate of change at that point itself. You can think of it as speed over an infinitely small interval. $\endgroup$ – user142299 Apr 20 '14 at 18:11
  • $\begingroup$ It's the derivative of $Y$ evaluated at $X_0$. $\endgroup$ – user122283 Apr 20 '14 at 18:12
  • $\begingroup$ I know to find it using derivative but my problem is the using of rate of change at a single point.It measures how fast a process goes when it moves from one point to other.But what does jt measure at a single point?I want to know this only;just read my question & give me the intuition. $\endgroup$ – user142971 Apr 20 '14 at 18:29
  • $\begingroup$ @NotNotLogical:what do u want to say by infinitely small interval?That means i want to know how much small u want to say? $\endgroup$ – user142971 Apr 28 '14 at 10:02
  • $\begingroup$ @user36790 I meant it is the limiting speed of an interval which is shrinking to zero length - it is the instantaneous speed at the point itself. See the second edit I added to my post. $\endgroup$ – user142299 Apr 28 '14 at 15:38

This is very good intuitive question to ask, since when we say "rate of change at a point", what we should really be saying is "rate of change around a point".

As I'm sure you know, in one variable the derivative of a function $f(x)$ can be defined as follows:

$f'(x)=\lim_{\Delta x\to 0}\frac{f(x+\Delta x)-f(x)}{\Delta x}$

But if we step back and consider the reality of a limit. If a limit exists we say:

$\lim_{x\to a}f(x)=\lim_{x\to a^+}f(x)=\lim_{x\to a^-}f(x)$

I.e. we are considering what is happening around a point, and not actually what is happening at the point.

The derivative is the same, and we are thinking about what is happening around $x$.

If we only knew what was happening at $x$, and nothing else, then we could not define the derivative, that is why a function can only be considered to be differentiable on open sets.

edit: to bring context to your question:

Let $V$ be the volume of the spherical balloon and $r$ the radius, the rate of change of the volume with respect to to the the radius (or namely the derivative) can be written as:

$\frac{dV}{dr}=\lim_{\Delta r\to 0}\frac{V(r+\Delta r)-V(r)}{\Delta r}$

Now say we want to know the rate of change "at" $r=7$, then we consider the behaviour around $r=7$:

$\frac{dV}{dr}(7)=\lim_{\Delta r\to 0}\frac{V(7+\Delta r)-V(7)}{\Delta r}$, again we see that in the limit we are considering what happens around $r=7$.

  • $\begingroup$ So, according to u,the instantaneous speed at a point gives the speed around that point,is not it?Can u explain my spherical balloon example with ur answer?Tttttttthhhhhaank you....... $\endgroup$ – user142971 Apr 20 '14 at 18:58
  • $\begingroup$ I've put in an edit, hope it helps! $\endgroup$ – Ellya Apr 20 '14 at 19:04
  • $\begingroup$ U r the best;when it is said instantaneous, it is describing around the point...thnk u.. $\endgroup$ – user142971 Apr 20 '14 at 19:21
  • $\begingroup$ No worries, it was a good question, not something to be taken for granted $\endgroup$ – Ellya Apr 20 '14 at 19:22
  • $\begingroup$ so when the statement says the instantaneous velocity at position p ,is it actually giving the information of the velocity around the position p ? $\endgroup$ – user142971 Apr 23 '14 at 7:33

The rate of change at a single point (i.e. the derivative at this point) is the slope of the best linear approximation the function has at this point. This means, if you have a function $f:D\rightarrow \mathbb R$ with the derivative $c$ at the point $x_0$, than the best linear approximation of the function at the point $x_0$ is given by $y\mapsto f(x_0) + c\cdot (y-x_0)$. For the error

$$r(y)=\underbrace{f(y)}_{\text{real value}} - \underbrace{(f(x_0)+c\cdot(y-x_0))}_{\text{approximated value}}$$

one has $\lim_{y\rightarrow x_0} \frac{r(y)}{|y-x_0|}=0$.

  • $\begingroup$ I think this is the best answer for someone who is thinking physically. "Instantaneous velocity" is not really a terribly physical concept: you can't measure instantaneous velocity, you can only directly measure average velocity over an interval. You can, on the other hand, ask about the best linear approximation at a particular time. So if "instantaneous velocity" is intuitively unsatisfying you should just think of it as a code phrase for "slope of best linear approximation". $\endgroup$ – Guest Apr 20 '14 at 18:46
  • $\begingroup$ @Guest: I guess you have commented the wrong answer... ;-) $\endgroup$ – Stephan Kulla Apr 20 '14 at 18:49
  • $\begingroup$ Nope, yours is the one I was aiming for. $\endgroup$ – Guest Apr 20 '14 at 18:50
  • $\begingroup$ Sir,can u explain what is linear approximation?An advanced thank u. $\endgroup$ – user142971 Apr 20 '14 at 19:02
  • 1
    $\begingroup$ @user36790: Of course: A linear approximation is an approximation of a given function with a linear function... $\endgroup$ – Stephan Kulla Apr 20 '14 at 19:08

Let's say the function is $f(t)$ which gives distance, so that the derivative is $f'(t)$ or instantaneous speed. You can think of it as follows: what speed is the car (for example) travelling at that precise moment in time. Or: if the car was to continue at the same speed, what would it be going. $f'(t)$ gives the speed at time t exactly. That is the intuition behind what a rate of change at one point means.

EDIT: So say we want to know the speed right at time $t$. We could go a little ways out to time $t+\Delta t$ where $\Delta t$ is a very small number, and see how much our distance has changed over that time interval: so take $f(t+\Delta t)-f(t)$. We have $$\text{speed}=\frac{\text{change in distance}}{\text{change in time}}$$ so the speed is $$\text{speed}=\frac{f(t+\Delta t)-f(t)}{\Delta t}$$ approximately. We can closer and closer to the exact speed at $t$ itself by making the change $\Delta t$ smaller and smaller. So the real speed is the limiting value $$\text{speed}=\lim_{\Delta t\to 0}{\frac{f(t+\Delta t)-f(t)}{\Delta t}}=f'(t)$$ Remember that limits don't tell us what the function actually reaches when $\Delta t=0$; we can make $\Delta t$ as small as we like, but it can never equal zero. The limit tells us what the function seems to be approaching as we make $\Delta t$ tend to $0$ - we infer that the speed at the point itself, if we could find it, would be that limiting value (which is, as you should notice, the derivative by definition).

EDIT2: How can the derivative be exact?

Consider the following graph: sine You are looking at the function $$f(x)=\frac{\sin x}{x}$$ This function approaches $1$ as $x$ goes to zero because $$\lim_{x\to 0}{\sin x\over x}=1$$ However, the function is undefined at $x=0$ itself.

Finding a derivative is a little like asking: if there was a point at $x=0$, what would it be? We can make $x$ get closer and closer to $0$, and the function $f$ will get closer and closer to $1$. But it will never get there exactly.

However, if there was, hypothetically speaking, a point at $x=0$ like the blue one I have drawn in, what would its y-coordinate be? Saying that "the limit of f is 1" does not mean that "the function equals 1". It means that "the function never equals 1, BUT... it gets closer and closer, and if there was a value there, it would have to be 1".

So when taking a derivative, you are working your way around a value that "could" exist, but that just doesn't work at the point itself. However, you can infer what the value must be by looking at what the function gets closer and closer to - in other words, by taking the limit. The exact limit.

  • $\begingroup$ You are awesome;but can you elaborate ur explanation?Again,thank you. $\endgroup$ – user142971 Apr 20 '14 at 18:48
  • $\begingroup$ @user36790 I added some things in edit, post any more questions you have please. $\endgroup$ – user142299 Apr 20 '14 at 18:55
  • $\begingroup$ Ur simple explanation enlightened me very much.So, instantaneous rate is an approximation.Thanks... $\endgroup$ – user142971 Apr 20 '14 at 19:26
  • $\begingroup$ @user36790 It's exact, it's not an approximation. The "approximations" get better and better as the limit goes to $0$. So the final answer, $f'(t)$, is the exact speed of the object at the time $t$. No error at all. $\endgroup$ – user142299 Apr 20 '14 at 19:29
  • $\begingroup$ How can it be exact?Delta t can be very,very small but cannot be equal to 0.If delta t = 0,then we could not find the derivative.So,the instantaneous rate is approximate;there is always an error,however small it might be.So why can it be exact? $\endgroup$ – user142971 Apr 21 '14 at 3:53

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