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$$ \color{darkcyan}{\frac{dy}{dx}} = \lim_{h \to 0} \frac{f(x+h) - f(x)}{h}$$ $$ \color{darkcyan}{m} = \lim_{x \to a} \frac{f(x) - f(a)}{x-a} $$ Text source: https://i.sstatic.net/Kn3Bm.png

I think I have a fairly solid understanding of the derivative, but I don't get how it helps us find instantaneous velocity at a point. It only gives us the velocity that we can get infinitely close to, but that's not the velocity at the point. The velocity at the point is undefined as x-x in the denominator = 0.

I get the following about limits and derivatives:

  1. That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.

  2. That the derivative is the limit of the slope of x and a, as a is moved infinitely closer to a. It is the slope that is being approached, as a gets infinitely close to x.

But while this lets us know what the velocity is between two points as they get infinitely close to each other, that still doesn't give the actual instantaneous velocity at that point, because to find the actual velocity at that single instant, you have to do f(x)-f(x)/x-x= 0/0 = undefined. So how does the concept of the derivative give us instantaneous velocity?

How can this be explained without epsilon delta proofs, at the level of someone learning Khan Academy calculus?

Peter Mortensen
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Ethan Chan
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  • Maybe you can see the Khan's video on Instantaneous speed. – Mauro ALLEGRANZA Sep 03 '18 at 14:23
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    The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative. – Mauro ALLEGRANZA Sep 03 '18 at 14:24
  • @MauroALLEGRANZA So there is no such thing as instantaneous speed? I get the notation, just don't get how it solves the problem of instantaneous speed. – Ethan Chan Sep 03 '18 at 14:25
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    There's nothing like "infinitely close to". Velocity over a time interval is the ratio between the space displacement $\Delta s$ and the time length $\Delta t$. Instantaneous velocity at a point is by definition the limit of that ratio for $\Delta t\to0$. – Andrea Mori Sep 03 '18 at 14:26
  • Think of it as a 'best' straight line approximation to a curve at a point. – copper.hat Sep 03 '18 at 14:27
  • @AndreaMori Exactly! So there is no such thing as instantaneous velocity right? It's only that we measure the velocity over a time period that is getting infinitely close to 0. – Ethan Chan Sep 03 '18 at 14:27
  • The velocity is, by definition, an 'instantaneous' quantity. – copper.hat Sep 03 '18 at 14:28
  • @copper.hat I'm sorry, I don't read you. Velocity is distance/time, so by definition, it can't be in an instant, because then time=0? – Ethan Chan Sep 03 '18 at 14:30
  • @EthanChan: of course there is a thing "instantaneous velocity". As I said it's by definition the limit of average velocity for $\Delta t\to 0$ – Andrea Mori Sep 03 '18 at 14:30
  • It is the limit of quotients not the quotient of limits. If my position is $x(t) = t^2$, then my velocity at time $t=1$ is $2$. – copper.hat Sep 03 '18 at 14:31
  • @AndreaMori But doesn't that mean not when Δt=0, but when Δt is approaching 0? And if it's over a period of time, it's not in an instant right? – Ethan Chan Sep 03 '18 at 14:31
  • Draw a picture, it is the limiting slope of the secant approximations to the curve at a point. – copper.hat Sep 03 '18 at 14:32
  • @copper.hat "It is the limit of quotients not the quotient of limits. If my position is x(t)=t^2, then my velocity at time t=1 is 2" But isn't that false? It's not that you velocity is 2, is that it's approaching 2 as it gets infinitely closer to 2. You can't find a velocity with t=0 as I was saying! – Ethan Chan Sep 03 '18 at 14:33
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    @EthanChan: you started your post by declaring "I think I have a fairly solid understanding of the derivative". Well if you keep asking this question, maybe your undertanding is not that solid. – Andrea Mori Sep 03 '18 at 14:33
  • @AndreaMori That's what I'm trying to figure out here. But can you try and help me see where I'm wrong? Because I don't. – Ethan Chan Sep 03 '18 at 14:34
  • The velocity at time $t$ is the limit of ${x(s)-x(t) \over s-t}$ as $s \to t$. You can explicitly compute the formula and show that the limit gives $x'(t) = 2t$. – copper.hat Sep 03 '18 at 14:35
  • @copper.hat I don't disagree. But it's lim x->t! Which means it's approaching that value, not at that value. The whole point of limit notation is to denote approaching a value right? – Ethan Chan Sep 03 '18 at 14:36
  • @EthanChan : HINT: the limit of a function $f(x)$ for $x\to a$ has nothing to do with the value $f(a)$ that may as well be not defined (as the case for the incremental ratio $\Delta f/\Delta x$: that's why the derivative is defined as a limit and not as some ratio). – Andrea Mori Sep 03 '18 at 14:37
  • @EthanChan: It is a definition. I don't understand your last comment. The definition coincides with our intuitive notion of velocity. – copper.hat Sep 03 '18 at 14:37
  • @copper.hat The derivative of 0.9999999999999999 is infinitely close to 2. The derivative of 1.000000000000001 is infinitely close to 2. That's because delta t here is infinitely close to 0. But when we actually get delta t to be equal to 0, we get 0/0, which is undefined. That is what it means. – Ethan Chan Sep 03 '18 at 14:41
  • @EthanChan: You have lost me. What do you mean by the derivative of a constant (which must be zero)??? The value of the derivative at a point (if it exists) is defined in terms of the value of the function in a neighbourhood of a point. – copper.hat Sep 03 '18 at 14:42
  • Related, possibly helpful https://math.stackexchange.com/questions/2136779/calculus-why-do-we-define-rate-of-change-as-dy-dx/2136831#2136831 – Ethan Bolker Sep 03 '18 at 14:43
  • @EthanChan: I strongly suggest you to go back to the textbook, study carefully the definition and work out some examples of limits – Andrea Mori Sep 03 '18 at 14:43
  • @EthanChan: The definition of velocity at a point is in terms of the limit of the quotients around that point. That is, in essence, the utility of the derivative - it gives a local approximation to the function. – copper.hat Sep 03 '18 at 14:46
  • @EthanChan: Not to confuse things, but the limit does not necessarily exist. For example, the function $x(t) = |t|$ does not have a velocity at $t=0$ because the limit does not exist. – copper.hat Sep 03 '18 at 14:47

4 Answers4

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Instantaneous rate of change of a differentiable function at a point is by definition the change in value of the function when the point is infinitesimally perturbed. The definition of instantaneous velocity at any point itself is the rate of change of position at that point, and is the velocity "at that point".

If $t$ denotes time and $f(t)$ denotes position at time $t$, then the velocity at time $t_0$ is defined as $$v_0 = f'(t_0)=\lim_{t\to t_0} \frac{f(t) - f(t_0)}{t-t_0}$$

As you mention, the limit is an "actual value", and may be defined based on $f$. When it is defined, it gives the exact velocity at time $t_0$. It is true that when you substitute $t_0$ in the limit, you get $\frac{0}{0}$, which is undefined. But the fact that a function is undefined at a point does not mean that the limit on approaching the point is undefined (take, for example, $\lim_\limits{x\to 2}\frac{x^2-4}{x-2}$). Moreover, substituting $t_0$ represents no change in time, whereas the velocity is defined for an infinitesimal change in time, for which the change in position is given by the limit, and is often well defined.

The key is that taking the limit allows you to exactly compute the change of the position for an infinitesimal change in time. Substituting $t_0$ represents no change in time, and substituting any non-infinitesimal change in $t_0$, say, $t_0+t_1$ gives the average velocity over the time period $t_1$.

GoodDeeds
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  • So does that mean that you can't actually find instantaneous velocity? You can only find the velocity of such a short time period that the value approaches instantaneous velocity? – Ethan Chan Sep 03 '18 at 14:42
  • The value approached by change in position in an infinitesimally short time is instantaneous velocity. – GoodDeeds Sep 03 '18 at 14:43
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    There are no actual infinitesimal numbers. The limit IS the instantaneous velocity. – A. Thomas Yerger Sep 03 '18 at 14:43
  • @GoodDeeds So instantaneous velocity isn't velocity over time=0. It is velocity over an arbitrarily short period of time, such that time approaches 0. – Ethan Chan Sep 03 '18 at 14:44
  • It is the value that the change in the position approaches when the change in time approaches 0. This value can be computed exactly by the derivative of the position with respect to time, if it exists. – GoodDeeds Sep 03 '18 at 14:46
  • @GoodDeeds I'm sorry, that's what I meant. So the time elapsed isn't actually 0, but arbitrarily close to 0. So like 0.00000001 seconds after we first started recording. So instantaneous velocity is at are arbitrarily close time elapsed, and not at 0 time elapsed? Am I right? – Ethan Chan Sep 03 '18 at 14:48
  • @GoodDeeds Finally someone understands what I'm saying! Thank you! – Ethan Chan Sep 03 '18 at 14:50
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    Do note that $0.0000001$ is not really "infinitesimally close". You can get closer than that. A continuous function $f(x)$ has a limit $L$ at $x=x_0$ if you can get $f(x)$ as close to $L$ as you wish (arbitrarily close), by making $x$ sufficiently close to $x_0$. So, if the instantaneous velocity is given by $v$ at time $t$, you can make the average velocity arbitrarily close to $v$ by choosing sufficiently small change in $t$. – GoodDeeds Sep 03 '18 at 14:55
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    @EthanChan: That's wrong. Instantaneous velocity is the value approximated by the ratio for arbitrarily close elapsed times. In the same way that the derivative is the value approximated by taking the difference quotient. –  Sep 03 '18 at 14:56
  • For $h \neq 0$, we have a function $h \mapsto {f(x+h)-f(x) \over h}$. If this function has a limit as $h\to 0$ then we say that $f$ is differentiable at $x$ and the value of the derivative is the limit. It is the limit not some nearby approximation. – copper.hat Sep 03 '18 at 14:57
  • @Hurkyl I'm pretty sure that's what I meant. I said it was the velocity that is approached when x and x+delta x get arbitrarily close to each other. That 0.0000001 was just a number I used to make me point. – Ethan Chan Sep 03 '18 at 14:58
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    I have to agree here with @Hurkyl comment. The instantaneous velocity is not the velocity when the time interval is $0$,but rather it is a specific value to which the velocity approaches in the mathematical sense of a limit as the time interval tends to $0$. – Paramanand Singh Sep 03 '18 at 15:00
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    @Winther I seem to have misread the original comment. What I mean to say is, assuming your position is differentiable w.r.t. time, the change in position can be made arbitrarily close to the instantaneous velocity by a sufficiently small change in time. Please see my second comment. – GoodDeeds Sep 03 '18 at 15:02
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    @Hurkyl I'm sorry if I misunderstood, but does that contradict with the answer? – GoodDeeds Sep 03 '18 at 15:04
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It is a relatively common circumstance in mathematics that there is something that is difficult to define 'directly', but we can nonetheless easily write down approximations of this thing.

In this case, the notion of "instantaneous velocity" is hard to define directly, but it would seem that we could easily approximate what the value should be by the ratio

$$ m \approx \frac{f(x)-f(a)}{x-a} \qquad \qquad \text{when }x \approx a $$

The general pattern for dealing with this situation is to use the approximations to determine what the thing we care about actually is; in cases like this one, the notion of limit is precisely the tool we need to identify what value these approximations are approximating:

$$ m = \lim_{x \to a} \frac{f(x)-f(a)}{x-a} $$

Once we have the definition, we do some theory development to decide if this definition is useful for whatever purpose we were ultimately trying to achieve.

There is, incidentally, an alternative visualization of what the limit does that may be appealing.

Suppose we define the quantity $u(x)$ to mean "the average velocity over the interval from $a$ to $x$. For $x \neq a$, this is clearly given by

$$ u(x) = \frac{f(x) - f(a)}{x - a}$$

Of course, this formula doesn't give us $u(a)$, which ought to be the instantaneous velocity, whatever that may mean.

However, if we consider the graph of $\frac{f(x)-f(a)}{x-a}$, while there is a discontinuity at $x=a$, it's a removable discontinuity: there is a value that the graph clearly wants to pass through to continue from the $x<a$ domain to the $x > a$ domain.

So, $u(a)$ ought to be the value that fills in the removable discontinuity to make a smooth graph. The limit is precisely the operation that gives the value that would fill in the removable discontinuity:

$$ u(a) = \lim_{x \to a} u(x) $$

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But while this lets us know what the velocity is between 2 points as they get infinitely close to each other, that still doesn't give the actual instantaneous velocity at that point, because to find the actual velocity at that single instant, you have to do f(x)-f(x)/x-x= 0/0 = undefined.

Clearly you don't have to do that, unless you're out to prove that instantaneous velocity is always undefined, which makes "instantaneous velocity" a useless concept. Since we generally prefer to work with concepts that tell us something useful, we don't define instantaneous velocity that way. Since defining instantaneous velocity as 0/0 doesn't tell us anything we want to know, we find it useful instead to define instantaneous velocity as the limit of the velocity as the change in time goes to zero.

You say that you understand the concept of a limit, and that it's "not an approximation", it's a value that's actually approached as you near some point, and that you can get approach the limit arbitrarily closely. This is good, but it's worth thinking more about what that means. For one, it means that a limit doesn't always exist. Some functions aren't "neat" enough at some points to have a limiting value that you can get arbitrarily close to. But in cases where the limit of a difference quotient does exist, then it corresponds what most people would agree "instantaneous velocity" should mean, and in cases where it doesn't, then it's not clear what "instantaneous velocity" should be either. This is a clue that maybe the limit of the difference quotient is a useful concept after all.

hobbs
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I think you may be worrying too much about the rather philosophical question of whether reality is discrete or continuous; in particular, about whether time is indeed continuous (so that representing it with a line would be justified). One may argue endlessly about this, and still get no further in understanding what's going on.

Thus, let us forget about this and abstract the mathematical essence of the derivative of a function $f$ at some point $p$ in an interval on which $f$ is defined; which is the fact that the number $f'(p)$ is well-defined, that is, it is uniquely specified by definition. Why is this? Because the derivative is just a limit of some (properly defined) function related to $f$ at $p,$ namely the quotient of differences. And the way we have defined what we mean by such a limit is exactly to make sure that limits, whenever they exist, are uniquely specified. Thus, we can speak of the limit of some function at a point embedded in its domain.

It is then clear that we can speak of the derivative of a function at a point (whenever it exists) without any ambiguity at all. It follows that velocity (if we choose to model it after the derivative) is also well-defined.

How is this definition done? Well, usually when we define some object, we use a finite list of conditions. This is why it usually seems strange to define some object by infinitely many conditions (which we do all the time in the study of the real line and related continua) when one comes to study analysis. However, we have little choice about it really if we want to have the set $\mathrm R.$ Otherwise, we are stuck with incomplete systems like $\mathrm Q.$ Furthermore, there's nothing logically spurious about this device (so long as we are careful to be consistent and not to make unwarranted assumptions). I hope you find this illuminating.

Allawonder
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  • The problem of $\mathbb {Q} $ as compared to $\mathbb{R} $ is not related to "infinitely many". Definition of limit applies in $\mathbb{Q} $ also without any change at all, but frankly speaking there are not many useful functions to deal with (we are just left with polynomials and their ratios). – Paramanand Singh Sep 04 '18 at 01:21
  • @ParamanandSingh Perhaps I was not clear enough, but by mentioning $\mathrm Q$ I only wanted to emphasise the fact that it is the completeness of $\mathrm R$ only which assures us that we can maximally enjoy our definition of limit; or something like that. In particular, I had the axiom of least upper bound in mind. – Allawonder Sep 04 '18 at 03:33