What are "instantaneous" rates of change, really?

Question

Here's how I see it (please read the following if you can, because I address a lot of arguments people have already made):

Let's take instantaneous speed, for example. If it's truly instantaneous, then there is no change in $x$ (time), since there's no time interval.

Thus, in $\frac{f(x+h) - f(x)}{h}$, $h$ should actually be zero (not arbitrarily close to zero, since that would still be an interval) and therefore instantaneous speed is undefined.

If "instantaneous" is just a figure of speech for "very very very small", then I have two problems with it:

Firstly, well it's not instantaneous at all in the sense of "at a single moment".

Secondly, how is "very very very small" conceptually different from "small"? What's really the difference between considering $1$ second and $10^{-200}$ of a second?

I've heard some people talk about "infinitely small" quantities. This doesn't make any sense to me. In this case, what's the process by which a number goes from "not infinitely small" to "ok, now you're infinitely small"? Where's the dividing line in degree of smallness beyond which a number is infinitely small?

I understand $\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$ as the limit of an infinite sequence of ratios, I have no problem with that.

But I thought the point of a limit and infinity in general, is that you never get there. For example, when people say "the sum of an infinite geometric series", they really mean "the limit", since you can't possibly add infinitely many terms in the arithmetic sense of the word.

So again in this case, since you never get to the limit, $h$ is always some interval, and therefore the rate is not "instantaneous". Same problem with integrals actually; how do you add up infinitely many terms? Saying you can add up an infinity or terms implies that infinity is a fixed number.

Answer 1

In math, there's intuition and there's rigor. Saying $$f'(x)=\lim_{h\to0}\frac{f(x+h)-f(x)}h$$ is a rigorous statement. It's very formal. Saying "the derivative is the instantaneous rate of change" is intuitive. It has no formal meaning whatsovever. Many people find it helpful for informing their gut feelings about derivatives.

(Edit I should not understate the importance of gut feelings. You'll need to trust your gut if you ever want to prove hard things.)

That being said, here's no reason why you should find it helpful. If it's too fluffy to be useful for you that's fine. But you'll need some intuition on what derivatives are supposed to be describing. I like to think of it as "if I squinted my eyes so hard that $f$ became linear near some point, then $f$ would look like $f'$ near that point." Find something that works for you.

Answer 2

The idea behind $$\lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ is the slope of the graph $y=f(x)$ .

Now for a moment forget about your instantaneous velocity and think about your average velocity. What is average velocity? I think average velocity is $$\frac{x_f-x_i}{t_f-t_i}$$. Now if look at this carefully this is my slope but in a given interval of time.

So you might question what is the difference between instantaneous velocity and average velocity , Both of them talk about intervals .

No that not the idea over here. Now if it had been a linear graph it would have been very easy to calculate your instantaneous velocity , but in your you graph that's not the case. Here you see the particle (or object) is changing its velocity every moment of time and that becomes impossible to deal with . So to remove this element of doubt what we do is try to take the limit and try to get close to the frame of time and try finding the velocity and name it instantaneous velocity.

Answer 3

Instantaneous Rate of Change at a Point, not over an Interval

You can describe rate of change without any interval at all, at a specific point. In fact, to try and involve an interval, even a very small one, is somewhat counter to the concept. If you involve an interval in describing a rate of change, then you would be talking about an average rate of change. For example, when people describe the velocity(rate of position change) of a vehicle over an interval of 100ft, then you would usually be talking about the average speed, assuming it wasn't constant over the entire interval.

Therefore, a rate of change is always instantaneous. You can identify any point, which has no length/interval, and say what the rate of change at that point is. Instantaneous rate of change is analogous to a point.

This also helps understand why certain points have an undefined rate of change, and why the "ever smaller interval" concept is somewhat misleading.

If you tried to use the very very tiny interval concept to identify the rate of change across point p and point q by sampling a point just to the left and right of p to define an interval, then you'd see a very large change. Perhaps point p is at x coordinate 1.00, and we define an interval from .99 to 1.01 and sample values of .99,2 and 1.01,10. This would be a rate of change of 400 y per x. If you make your interval smaller and smaller, your rate of change appears to approach infinity, which again is very misleading, because the instantaneous rate of change at p is undefined.

I'm not trying to disregard the intervals concept as useless, but to emphasize that very small intervals are a concept that can't be applied rigorously. They certainly help understand the concept of a derivatives, because if you were to imagine that you don't know what the instantaneous rate of change at a particular point is, then you'd have to observe how much change occurred over an interval in order to estimate the rate of change. It helps to understand how these functions were "designed".

There is no observed change, but we know the RATE of change

Let's take instantaneous speed, for example. If it's truly instantaneous, then there is no change in x (time), since there's no time interval'

You are absolutely right! There is no change in position as there is no defined interval. However, at a specific point, we can still describe what the instantaneous rate of change is, which is a way of saying how much it will change if an interval were defined. Imagine we freeze a frame of a movie where someone tells you a bullet just exited a barrel at an instantaneous velocity of 300 meters/s. That's not hard to understand right? If someone asked, disregarding air resistance and gravity, where will that bullet be two seconds from then? Now you've introduced an interval, and using two things, the initial position, and the instantaneous rate of change at that position, you can confidently draw a line along the bullet's trajectory and conclude it will be 600 meters away in that two seconds. Based on that instantaneous information, we can model for any interval how that bullet's position will change. We can apply the instantaneous velocity across the entire interval only because we assumed it was constant across the interval.

Where calculus comes to the rescue is that we often deal with non-constant rates of change. We know in reality that the bullet will slow down due to air resistance A(), and it's trajectory will be an arc instead of straight line due to the constant acceleration of gravity G() (which results in a non-constant velocity). If we have functions that describe air resistance and gravity's influence on our velocity, then we can describe the non-constant instantaneous velocities over any point on the entire interval. Since those are instantaneous points and are non-constant, they are somewhat useless. We could use the "very small interval" approach of sampling changes in position to sum up across the entire interval, but since the velocity is constantly changing, we'd be accumulating alot of error. E.g. if the initial velocity is 300 m/s but slows to 250 m/s after half a second, then how far did the bullet travel in half a second? You could try to average, but that assumes linear change in speed which might not be the case. Just about any iterative approach you take that involves small intervals will introduce error. However, with the help of calculus we can directly and accurately incorporate the influence of A() and G() on our non-constant velocity to directly calculate the change in position.

More than just a Concept

In addition to being an aid in understanding concepts, the "very small interval" concept is often applied in computational models where direct methods are not feasible, perhaps because a computer algorithm cannot determine a direct method. This is exactly how physics are simulated in many computer games. Each iteration it may know that .03 seconds has passed, and based on various rates of changes(usually determined based first on what forces are in affect) it will determine what has changed across that interval. In certain situations this can be extremely inaccurate, but in other cases with small enough intervals it may be sufficiently accurate to meet desired requirements.

Answer 4

Think of a mountain: far away it's nearly flat; near the top it's more steep.

If you ask "How steep is it where I am standing right now?", you're asking exactly the same thing as "What is the rate of change of my elevation with respect to my position right here?"

I feel that should be pretty intuitive; that's what instantaneous rate of change means.

(Side note: in this case, you can move in two independent directions: north/south and east/west. That means there is a separate rate of change of elevation for each of those directions at every possible position on the ground. But in your case, you're only dealing with time, which can only go forward/backward, and hence there's only one number to worry about at each instant, not two.)

Answer 5

As the accepted answer stated, the English language statement is informal and exists to provide intuition only.

That does not mean it might not be finessed.

If one rewords to

Rate of change applying at a specific instant

then I think this is consistent with:

1. A rate of change being defined only on intervals
2. Those intervals being arbitrarily small
3. We have a unique value meaningful for a given instant, provided of course that the limit exists.

Answer 6

I will answer your question with a question.

You are driving a car. 60 miles per hour is 88 feet per second (which is pretty cool in and of itself. You look down at a text for 1/2 second? You moved 44 feet.) Now, when I ask where you were in a given second, you could have been anywhere within +/- 44 feet. So you say "which 1/100th second are you asking about?" But 44 feet is 528 inches, so even if I say at time = 6.31 seconds past 2:30, you only know within a 10 inch range.

Can you ever tell me precisely where you are if you are in motion?

Instantaneous rate is a calculus concept, but it should be understandable to you as the slope of a tangent line.

The line tangent to the parabola has a slope of 2. Even though the "slope" is just a point.

Answer 7

I think the answer of @ZachStone already expresses the essence of your question.

I would add a more Physical/Engineering point of view, because you call in the intuition, and in applied maths, as usually found in engineering, you need a lot of intuition.

As a concrete example, let's consider the speed of a car (or of any point mass). There is the average speed which is trivially and very understandably defined as:

\[ v_m = \dfrac{\Delta s} {\Delta t} \]

You define the instantaneous speed as the limit of that ratio as the time interval tends to zero:

\[ v = \lim_{\Delta t \to 0} \dfrac{\Delta s} {\Delta t} \]

Of course if you knew a mathematical expression for s(t) you could do all your usual math, and that's ok. But what does it mean that limit in practical applications? It means that you look for a value of the "average speed" over a "very small" time interval, where "very small" means "as small as possible in relation to the application". That usually means that $\Delta t$ is so small that s can be deemed constant over that time interval, because the allowable measurement error is larger that the variation of s over that interval.

In a sense, you approximate your motion as you go ahead: at any time instant you say to yourself "if I had to approximate now the motion with a constant speed motion, which speed would the car have?". The answer is "the instantaneous speed" (which is, as I said before, approximately computed as an average speed over a "small enough" time interval).

Answer 8

I think this question should belong into meta, as it is a question between finitism and platonism, but I try to answer you as best as I can:

The dividing line where a number $\epsilon$ is small enough is when someone gives you a number $\delta>|x-c|>0$, there exists $\epsilon>|f(x)-L|$, where $c$ is a point near $x$, $L$ is the limit. So you need not think $h$ as an interval, but as a number.

In arithmetic sense, you can add infinitely many terms if they follow a rule which can be mathematically constructed: the geometric series is a good example. The existence of integral is therefore quite natural.

Addressing your first question in a philosophical manner:

Say that there exists events that happen instantaneously. If they do exist, they need to be observed. For the observation to happen, time needs to pass, so we can think "instantaneous" as an event which occurs in the least time possible for the event to actually happen.

Answer 9

This was stated in another form in a different answer, but I thought I'd say it in a way that might be conceptually easier to grasp:

The instantaneous rate of change at any point in a function is the slope of the line that is tangent to the function at that point.

If you imagine a line that intersects two points on an arbitrary function that are some arbitrary distance apart, and then slide those points together, you will see that the slope of the line changes depending on the distance between them. When the two points are so close together that they become one, then the slope of the line is equal to the instantaneous rate of change of that function.

Answer 10

Why don't we just say "approximately instant" to appease our intuition, and then define our velocity as an arbitrarily close approximation to an average one?

As a bit of an aside, 'closeness' is tied to so many higher levels of Mathematics. So many things hang on it. Topology, Analysis, and the like to be sure. So, as mathematicians, we've become comfortable with the idea of the limit as a definite value to make so many other things work out.

I always have in the back if my mind that that denominator IS NOT zero but gets really close to it. And for me the limit is hanging onto that closeness and rejecting the idea that it can actually be zero - because it can't be.

Answer 11

The expression $\dfrac{x^3-8}{x-2}$ is undefined at $x = 2$. When $x = 2$ you get $\dfrac 0 0$, which is undefined. So suppose we decide to define the value of $\dfrac{x^3-8}{x-2}$ when $x = 2$. What number makes the most sense and why? Well, when $x \ne 2$ we can simplify

$$\dfrac{x^3-8}{x-2} = \dfrac{(x-2)(x^2 + 2x + 4)}{x-2}=x^2 + 2x + 4$$

and $x^2 + 2x + 4 = 12$ when $x = 2$.

If you graphed $y = \dfrac{x^3-8}{x-2}$ you would get the graph of $y = x^2 + 2x + 4$. Depending on how good your graphics software is, it may or may not indicate that there is something wrong at $x = 2$, but the point $(2, 12)$ is actually missing. We call points like this "removable discontinuities".

A rate of change of a function $f(t)$ between a variable time $t$ and a fixed time $a$ looks like $\dfrac{f(t) - f(a)}{t-a}$ and you notice that, if you let $t=a$, you get $\dfrac 0 0$.

A whole lot of the time, $ y = \dfrac{f(t) - f(a)}{t-a}$ has a removable discontinuity at $t = a$. If the missing point is (a, A), then it turns out that $\displaystyle lim_{t \to a}\frac{f(t)-f(a)}{t - a} = A$. The number A is called "the derivative of $f(t)$ at $t = a$" and it is described as "the instantaneous rate of change of $f(t)$ with respect to $t$ at $t=a$.

Answer 12

What makes Calculus "useful" in the "real world" isn't the formalism about limits or infinitesmials¹. It is the fact that most functions in the "real world" are relatively smooth and well behaved.

Such functions act linearly if you zoom in "enough" -- they act like little linear functions (well, affine functions -- linear with an offset).

The derivative associates each point on a function's domain (valid input) with a linear function. This describes the "slope" at that point.

If you are "close enough" to the point, then the linear function that the derivative gives you is a good approximation to the original function.

As linear functions are ridiculously easier to work with than other functions, this makes understanding the original function easier. Even if we are forced to understand an infinite number of such linear functions to understand a single non-linear function (the family of derivatives).

If a function goes from $\mathbb{R}$ to $\mathbb{R}$, then the linear maps have a natural bijection with elements of $\mathbb{R}$ -- so instead of the derivative returning a function, we denote it as returning a single real value.

If $f'(x) = g(x)$, then the affine map that approximates $f$ near $x_0$ is $h(x) = f(x_0) + g(x_0)*(x-x_0)$. Most of that is boring, so we just talk about $g(x_0)$.

In higher dimensions, the value returned from "the derivative" is a higher dimension linear function, which is represented by a matrix.

These linear functions are "guaranteed" to be "useful" if you are close enough to the point in question. How close can vary with the function in question (see the error bounds on prefixes of the Taylor series).

This is the "slope at the point" because talking about the interval around the point where it is half-decent is hard to talk about, but we do know that if we make the area small enough it is going to be good.

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¹ There are multiple formulation of calculus that acts on infinitesmials -- the smooth analytic analysis or non-standard analysis.

In it, you actually talk about points $x+h$ infinitesimally close to $x$, and evaluate $\frac{f(x+h)-f(x)}{(x+h)-x}$. The result is the derivative of $f$.

The delta -- $h$ -- is smaller than every classical real number. In a sense, $x+h$ is "still at $x$" (as far as classical real numbers are concerned), but in the non-standard analysis it isn't the same value.

Under this formalism, "at the point" could refer to "at the point, and infinitesimally close to the point".

Answer 13

To talk about instantaneous rate of change, you have to first have a sufficiently nontrivial idea of an instant. The main point of differential calculus is to flesh out the idea of an instant, how to do calculations in instants, and how to use integrals to "add" them all up.

In a more precise sense, the central idea is that at a point, you can have numbers that encode what happens near that point.

The simplest example of this is the meaning of continuity: if you know $f(a) = b$, then you also know that $f(c) \approx b$ whenever $c \approx a$. So when you're working with continuous functions, knowing what happens at a particular point means you have some idea about what happens near that point too. Another example of something you know is that $\lim_{x \to a} f(x) = b$.

The derivative is another example of this, and there is a lot of circumstantial evidence that links it to being a rate of change. For example:

• The mean value theorem, differential approximation, and similar things allow you to use a derivative to estimate a rate of change. e.g. $$y - y_0 \approx (x - x_0) \frac{\mathrm{d}y}{\mathrm{d}y}$$
• I can use an integral to "accumulate" the derivatives to produce an actual change: $$\int_a^b f'(x) \mathrm{d}x = f(b) - f(a)$$
• If you plot the graph of the rate of change (i.e. the difference quotient) of a differentiable function between $x_0$ and $x$, there is hole that really ought not be there (in a very similar sense how one would like to say that $t/t = 1$). The derivative is the value that fills that hole.
• In many circumstances (all of them, when suitably treated), one can compute a derivative formally by the procedure of writing down $(f(x) - f(x_0))/(x - x_0)$, doing simplifications to cancel out the denominator, then making the substitution $x \to x_0$.

Rather than being stuck on the fact that pre-calculus tools fail to express the idea of a rate of change in the situation where the derivative applies, you should instead be looking to see how the method of using derivatives allows you to study the idea of rates of change in situations where the pre-calculus tools were unable to help.

Answer 14

I had come to accept the slope intuition of derivatives a long time ago, when I first learned calculus. I never really understood it completely, I must confess, but since it proved useful in so many different applications (particularly in physics), I began to accept that interpretation without having to think about it at all after a certain point.

It was when I studied calculus slightly more formally that I gained a better understanding of the concept of limit and derivatives as a whole. The breakthrough came when I understood how sequences/series of numbers and their limit points work. Consider the following infinite sum:

\begin{equation} 1 + \frac{1}{2} + \frac{1}{4} + \frac{1}{8} + ... \end{equation}

The 'limit' of this sum is equal to $2$. Now obviously, you can't go on adding infinite terms to get exactly $2$, yet we say (and we say this as a proper mathematical statement - not as an ambiguous expression of out intuition) that the limit of the sequence is $2$.

The idea behind the limit (in the case of sequences/series) is that if you kept on evaluating the sequence with more and more terms, you would come closer and closer to $2$. You wouldn't be 'converging' to $\textbf{any other number}$ other than $2$. We call $2$ a limit point of the sequence.

In a similar manner, when we talk about 'instantaneous' rates, what we are saying is that if we calculate the rate with smaller and smaller intervals, we get closer and closer to the value we call the 'instantaneous' rate. We never actually attain that value, but we get closer and closer to it.

In a sense, it's kind of like a 'sacred' value that your expression is always striving to reach. You can't ever really be sure of it even if you study the behaviour of the expression for very long. As an example, you might conclude that the sequence I showed above has a limit of $2.000000000001$, just by computing the value with more and more terms. But if you went even further, you would discover that the sum's distance from $2.000000000001$ begins to increase after some time (suppose you hit the midpoint of $2.000000000001$ and $2$, the true limit; continuing with the evaluation of the sum from there on out would lead you closer to $2$, and thus farther away from $2.000000000001$).

So you see, the thing about the limiting value of an expression is that it is not an easy task to determine it just by studying the way the expression behaves numerically (which is what the intuitive explanation of 'instantaneous' may lead you to believe). This is why the mathematics of limits is so profound. You can compute the $\textbf{exact}$ value that the expression converges to, regardless of 'practical artefacts' like numeric precision. It is actually quite an elusive value, the limit of an expression...but it has a lot of meaning to it.

If you continued forever in the direction of the limit, you would continuously get closer to the limiting value (limit point), but you would never reach it. If we could measure the rate in an intervals of ever decreasing length, we would approach the instantaneous value - but we never get an exact match.