6
$\begingroup$

According to this answer, instantaneous rates of change are more intuitive than they are rigorous.

I tend to agree with that answer because, in the Wikipedia article on differential calculus, they aren't defining the derivative to be the slope at a particular point. They define it as, "The derivative of a function at a chosen input value describes the rate of change of the function near that input value." Although this isn't wrong, the definition has been written rather safely, and I think that was intentional. They didn't define it as the slope of the graph at a particular point. It is only in the explanations section of the Derivative wiki article that they did that: "The derivative of a function y = f(x) of a variable x is a measure of the rate at which the value y of the function changes with respect to the change of the variable x. It is called the derivative of f with respect to x. If x and y are real numbers, and if the graph of f is plotted against x, derivative is the slope of this graph at each point."

So, are physicists using terms like "instantaneous velocity" merely from an intuitive standpoint? What is the physical significance of instantaneous rates of change?

$\endgroup$
9
  • 27
    $\begingroup$ I don't agree with that answer. The derivative gives precise meaning to "instantaneous rate of change." $\endgroup$
    – Randall
    Oct 7, 2021 at 15:31
  • 6
    $\begingroup$ As a general rule, don't rely on Wikipedia to give you definitive answers to questions like this. Articles get edited and re-edited and sometimes the exactly correct answer gets edited out. The information you're getting from Wikipedia in this case is not suitable for your purpose. $\endgroup$
    – David K
    Oct 8, 2021 at 0:12
  • 2
    $\begingroup$ What makes you think rigor is more important than intuition in physics? $\endgroup$
    – David K
    Oct 8, 2021 at 0:15
  • 6
    $\begingroup$ Are you willing to accept "instantaneous position"? Or even the idea of a single point in time? Because those seem be only slightly more rigorous than instantaneous rates of change. $\endgroup$ Oct 8, 2021 at 3:24
  • 2
    $\begingroup$ "instantaneous rate of change" is an informal statement of a formal concept (the derivative), that correctly applies to almost all physical circumstances (because physical reality tends to be almost everywhere continuous). Mathematicians use informal intuition to explore mathematical concepts all the time. Then we see if they are valid by trying to formalize them. $\endgroup$ Oct 8, 2021 at 11:49

5 Answers 5

17
$\begingroup$

It's perfectly fine to use intuition in applying mathematics - it's just that in mathematics itself we want rigorous definitions so we can actually prove stuff. We seek definitions that formalize our intuition about something.

Just consider the simple example of finding the derivative of $f(x) = x^2$. Using the "intuitive definition" it's not really clear that this should equal $2x$. You could of course look at a few examples and extrapolate from those that it should be true, but how can you really be sure? In contrast the "hard" definition (which can also be considered to be a rather intuitive one) directly allows you to construct the derivative.

The approach mathematicians often times take is to take some concept, state which properties it should have, try formalizing those and seeing if the resulting thing:

  • already "nails down" a concept well enough or if it's still too general
  • conforms to our intuition.

So we defined the mathematical concept of derivative the way we did, because it corresponds to our intuitive notion of rate of change and thus should be applicable in circumstances where the intuitive thing is asked for.

When applying math in physics, engineering etc. you of course always have to consider whether it makes sense to model some real life phenomenon via the mathematical idealized version: What assumptions go into a derivative? Are they compatible with the real world? Surely some notion of continuity is needed for a derivative. Is the real world continuous? We really don't know, and afaik (not a physicist) we can never find out. That's why physics is more than just building theories - we also need to do experiments to see if our theory corresponds with the real world up to some acceptable margin of error. And judging from experiments and how succesful we are in modelling the real world using differential calculus, it would seem that using the intuition behind derivatives in the real world isn't totally wrong.

$\endgroup$
9
$\begingroup$
  1. I disagree with the Answer cited in your first line: ‘instantaneous rate of change’ does have a formal meaning. Just because a concept is defined with reference to the details of its neighbourhood does not mean that the idea/concept is fuzzy or informal.

    Instantaneous rate of change $(R)$ at $q$ is coarsely analogous to the country of birth $(N)$ of $p:$ sure, the neighbourhoods of $q$ and $p$ exist independently of $q$ and $p,$ but this does not detract from the fact that the values of $R$ and $N$ at $q$ and $p,$ respectively, depend—in fact, are defined based—on the respective neighbourhoods of $q$ and $p.$ The definitions of $R$ and $N$ are rigorous, precise and unambiguous.

  2. You agree that for a car moving at a constant $70\mathrm{km/h}$ velocity, its instantaneous velocity at $t=100\mathrm s$ rigorously and precisely equals $70\mathrm{km/h}.$

    Presumably, you also agree that even if the car was travelling at a varying velocity, it continues to have some instantaneous velocity at $t=100\mathrm s.$

    If so, then your question becomes: is the formalisation of ‘instantaneous velocity’ (rate of change, derivative) somewhat arbitrary in caputuring the actual instantaneous velocity? I'd say, it really does “accurately”  and—at the risk of sounding circular—definitively reflect the actuality.

$\endgroup$
8
  • $\begingroup$ I see. Why did the Wikipedia articles define a derivative so safely? Why didn't they just directly say that the derivative is the slope of the curve at a particular point? $\endgroup$ Oct 7, 2021 at 16:48
  • 2
    $\begingroup$ @tryingtobeastoic This is the full quotation from the article (all the quoted sentences are actually technically consistent with one another):$\quad$"The derivative of a function at a chosen input value describes the rate of change of the function near that input value. The process of finding a derivative is called differentiation. Geometrically, the derivative at a point is the slope of the tangent line to the graph of the function at that point, provided that the derivative exists and is defined at that point. " $\endgroup$
    – ryang
    Oct 7, 2021 at 16:53
  • 3
    $\begingroup$ @tryingtobeastoic - the issue is that the slope of a curve at a particular point is a fuzzy concept. It relies on the idea of the function being able to be drawn on a graph, and the function being well-behaved. It doesn't generalise very nicely to alternative contexts, and itself relies on intuition. On the other hand, behaviour in the neighbourhood is a layman way of referring to limit behaviour, and thus becomes formal when expressed mathematically. $\endgroup$
    – Glen O
    Oct 8, 2021 at 1:28
  • $\begingroup$ @GlenO I see. Could you please help me out a bit? Could you please see if there is a difference between these two definitions: 1. The derivative is the slope of the curve at a particular point. 2. The derivative is the slope of the tangent line to the curve at a particular point. Is there any important difference between 1 & 2? Does 2 suffer from the same issue as you just mentioned? $\endgroup$ Oct 8, 2021 at 3:14
  • 1
    $\begingroup$ @tryingtobeastoic - they both have the same fundamental issue - they are geometric plotting definitions, which mean they cannot be easily generalised to situations where geometric reasoning starts to fail. It's not that they are bad definitions, just a bit fuzzy. $\endgroup$
    – Glen O
    Oct 8, 2021 at 4:16
7
$\begingroup$

I think you failed to realize that "slope of $f$ at $x$" is meaningless until you define it, but how can you define it precisely? That is the real problem.

You cannot just say "tangent", either, because that just swaps the problem for a worse problem. Not only is it difficult to define "tangent", it fails on cases like $f : ℝ→ℝ$ defined by $f(0) = 0$ and $f(x) = x^2·\cos(1/x)$ for every $x∈ℝ_{≠0}$, if you try to define "tangent" in terms of a line that touches only once in a small open disk around the point.

Ultimately, one of the best ways to define "slope/gradient of $f$ at $x$" is via the standard definition. You might ask "Why that definition?", and the answer is:

Because that definition gives nice properties, and even corresponds with intuition!

See, suppose you have a smooth hillside, and you happen to know a function $f$ that captures its elevation for every coordinate along a straight path up that hill. Then the rigorous definition of $f'$ actually gives you something that coincides with the intuitive notion of slope! The reason is that slope is intuitively "how fast it goes up", and you cannot measure it at a single point but you can measure it near a point. If you zoom in on that point on the hillside and it gradually looks more and more straight as you zoom in, then it is (rigorously provable to be) differentiable there and indeed $f'$ at that point is exactly the slope of the line that you approach as you zoom in.

Similarly, the instantaneous speed of a vehicle cannot really be measured in real life, but what the speedometer measures is how many rounds the wheel turns over a small time interval. In fact, you should notice that if you jerk the accelerator pedal on and off rapidly, the speedometer actually fails to show you the rapidly changing speed, precisely because it does not measure instantaneous speed and its time resolution is not that good. But if you travel on a relatively smooth journey, then the mathematically rigorous derivative matches closely with what the speedometer shows, because each small part of the speed-time graph is close to linear.

$\endgroup$
7
$\begingroup$

It is not that they aren't rigorous, it's that calculus books, as usual, don't necessarily take the care to make the relevant distinctions to make it fully rigorous. They can be made rigorous.

One thing I'd argue is that the "instantaneous rate of change" is something which can be defined as formally equivalent to, but conceptually distinct from, the derivative, with the derivative being more general. A derivative, in the case of a function of a real variable, is a certain quantity that characterizes the local behavior of such a function around an input point and how it responds to small changes in that input or, better, how its output differs when considering input values slightly different from a particular input and compared against the value it attains at that particular input.

The reason I say this is because the concept of a "rate of change" implicitly presumes a flow of time, and not all derivatives involve time.

The derivative of $f$ at $a$ is defined by

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

as you already know.

But now for the instantaneous rate of change. Parsing that term, we'd ideally want to say that, to make the intuition in it rigorous, we should define both what a "rate of change" is and, moreover, what it means for that rate to be "instantaneous".

So how do we do that? Analyzing the term further, we see we need to define "change" and "rate". The change - before we get to "rate of" - of a temporally-varying quantity from time $t_1$ to time $t_2$, given as a function $f$ of time, is thus defined by

$$\text{Change in $f(t)$ from time $t_1$ to time $t_2$} := f(t_2) - f(t_1)$$

i.e. change is just subtraction (difference). The rate of change then, is the ratio of two changes (note that the "change in time" can be understood as the change of the identity function of time, so we don't need another definition):

$$\text{Rate of change of $f(t)$ from time $t_1$ to time $t_2$} := \frac{\text{Change in $f(t)$ from time $t_1$ to time $t_2$}}{\text{Change in time from $t_1$ to $t_2$}}$$

from which we can see that

$$\text{Rate of change of $f(t)$ from time $t_1$ to time $t_2$} = \frac{f(t_2) - f(t_1)}{t_2 - t_1}$$

So what then is the instantaneous rate of change? Logically, it is the rate of change at a single instant, i.e. when $t_1 = t_2 = t_a$ at a particular instant $t_a$. However, we cannot achieve that with the above definition because we get a division-by-0 error. Instead, what we must do is use a limit to fill it in - in particular, we should take the following two-dimensional limit:

$$\text{Instantaneous rate of change of $f$ at $t_a$} := \lim_{(t_1, t_2) \rightarrow (t_a, t_a)} \frac{f(t_2) - f(t_1)}{t_2 - t_1}$$

where that we only consider points $t_1$ and $t_2$ such that both $t_1 \le t_a \le t_2$, i.e. the intervals of change "bracket" our desired point $t_a$, and $t_1 \ne t_2$. Then we have

Theorem: If the IRoC of $f$ exists at $a$, then it equals $f'(a)$.

$\endgroup$
4
  • $\begingroup$ You have a slight problem. You must require $t_1 ≠ t_2$ in your double-limit, otherwise it is ill-defined. That said, with that error fixed, it's a fine way to think of "instantaneous rate of change", though I don't quite believe that it is an intuitive way. In fact, I never had the idea that it was a double-limit until after I read your post! Haha.. $\endgroup$
    – user21820
    Oct 8, 2021 at 18:38
  • $\begingroup$ It's good now, thanks! By the way, the difference between your current version (i.e. non-zero interval straddling the point) and my version is that mine is actually stronger than differentiability (I think it is continuous differentiability, which is actually what we have in real life), whereas yours (now) is truly equivalent to standard differentiability. $\endgroup$
    – user21820
    Oct 9, 2021 at 9:23
  • $\begingroup$ @user21820: Thanks, yeah. Comments removed to clean up the comment section. ADD: Oh shit, I deleted the last one too, I didn't see that disclaimer :( Fortunately I screenshotted so I can repost it if it's useful for archive purposes $\endgroup$ Oct 9, 2021 at 20:28
  • $\begingroup$ It doesn't matter. Thank you! =) $\endgroup$
    – user21820
    Oct 10, 2021 at 3:01
1
$\begingroup$

In Mathematics, definitions and systems are not written once and baked into clay.

Instead, they develop over time. For derivatives, you can go back to Newton and Leibniz who developed ways to describe the slopes of equations using different syntax and different definitions. By todays standards, neither definitions where "formal".

Over time, the syntax we use to talk about derivatives, the terms we use, and the formal definitions have evolved.

The most common formal definition we use for derivatives of one dimensional functions is that epsion-delta based one. There are others which can be shown to give the same results on the set of common functions we collectively and intuitively agree on what "slope" means.

On edge cases, two different ways of talking about derivative may describe things differently; those edge cases tend to be pretty weird.

And then you go up and start talking about abstractions of the derivative. What happens when instead of functions from $\mathbb{R}$ to $\mathbb{R}$, we are talking about $\mathbb{Z}$ to $\mathbb{Z}$ or complex numbers or quaternions or polynomials or vectors or more exotic group structures or Lie algebras.

In those contexts, you'll find derivative-like structures that map back to important features of the derivative, be they "this is like a slope" or "it has similar algebraic properties as the differential operator".

But what makes it the derivative is that when you have a reasonably well behaved drawing of a line that doesn't fold over, the value it assigns to each point along the line is the slope of the line you drew. That is the core concept that unifies the derivative of Newton to modern epsilon-deltas or infinitesimal based derivative definitions.

Working with that directly is like cooking food when your definition of food is "stuff critters eat". Any technical definition of food has to be consistent with that. Any gastronomic definition of food has to be consistent with that. But that isn't a useful working definition for almost any practical purpose.

If tomorrow someone came up with and shared a better definition of derivative that encompassed the idea of instantaneous rate of change but turned out to be much more useful than existing ones, over a period of decades I'd expect the current epsilon-delta based definition to fall in prominence. The new definition would still be "the derivative" (or would eventually be it), so saying that the epsilon-delta definition is what the derivative "is" is misleading over the medium or long term.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.