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How does the concept of a derivative solve the problem of instantaneous velocity?
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$$ colordarkcyanfracdydx
= lim_h to 0 fracf(x+h) - f(x)h$$
$$ colordarkcyanm
= lim_x to a fracf(x) - f(a)x-a $$
Text source: https://i.stack.imgur.com/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:
That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.
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?
calculus limits derivatives
edited Sep 3 at 21:08
Peter Mortensen
537310
asked Sep 3 at 14:21
Ethan Chan
1
 |Â
show 20 more comments
up vote
3
down vote
favorite
$$ colordarkcyanfracdydx
= lim_h to 0 fracf(x+h) - f(x)h$$
$$ colordarkcyanm
= lim_x to a fracf(x) - f(a)x-a $$
Text source: https://i.stack.imgur.com/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:
That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.
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?
calculus limits derivatives
edited Sep 3 at 21:08
Peter Mortensen
537310
asked Sep 3 at 14:21
Ethan Chan
1
Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
4
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 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 3 at 14:25
3
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
1
@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 3 at 14:33
 |Â
show 20 more comments
up vote
3
down vote
favorite
up vote
3
down vote
favorite
$$ colordarkcyanfracdydx
= lim_h to 0 fracf(x+h) - f(x)h$$
$$ colordarkcyanm
= lim_x to a fracf(x) - f(a)x-a $$
Text source: https://i.stack.imgur.com/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:
That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.
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?
calculus limits derivatives
edited Sep 3 at 21:08
Peter Mortensen
537310
asked Sep 3 at 14:21
Ethan Chan
1
$$ colordarkcyanfracdydx
= lim_h to 0 fracf(x+h) - f(x)h$$
$$ colordarkcyanm
= lim_x to a fracf(x) - f(a)x-a $$
Text source: https://i.stack.imgur.com/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:
That the limit is an actual value, not an approximation. The limit is the actual value that we are getting infinitely closer to.
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?
calculus limits derivatives
edited Sep 3 at 21:08
Peter Mortensen
537310
asked Sep 3 at 14:21
Ethan Chan
1
edited Sep 3 at 21:08
Peter Mortensen
537310
edited Sep 3 at 21:08
Peter Mortensen
537310
edited Sep 3 at 21:08
Peter Mortensen
537310
537310
asked Sep 3 at 14:21
Ethan Chan
1
asked Sep 3 at 14:21
Ethan Chan
1
asked Sep 3 at 14:21
Ethan Chan
1
1
Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
4
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 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 3 at 14:25
3
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
1
@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 3 at 14:33
 |Â
show 20 more comments
Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
4
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 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 3 at 14:25
3
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
1
@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 3 at 14:33
Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
4
4
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 at 14:24
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 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 3 at 14:25
@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 3 at 14:25
3
3
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
1
1
@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 3 at 14:33
@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 3 at 14:33
 |Â
show 20 more comments
4 Answers
4
active
oldest
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up vote
6
down vote
accepted
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
$$v_0 = f'(t_0)=lim_tto t_0 fracf(t) - f(t_0)t-t_0$$
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 $frac00$, 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, $lim_limitsxto 2fracx^2-4x-2$). 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 $t_1$.
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
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 3 at 14:55
1
@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.
– Hurkyl
Sep 3 at 14:56
2
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 3 at 15:00
1
@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 3 at 15:02
 |Â
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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
$$ m approx fracf(x)-f(a)x-a qquad qquad textwhen x approx a $$
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:
$$ m = lim_x to a fracf(x)-f(a)x-a $$
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
$$ u(x) = fracf(x) - f(a)x - a$$
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 $fracf(x)-f(a)x-a$, 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:
$$ u(a) = lim_x to a u(x) $$
answered Sep 3 at 15:03
Hurkyl
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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.
answered Sep 3 at 15:55
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.
answered Sep 3 at 16:19
Allawonder
1,852516
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
6
down vote
accepted
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
$$v_0 = f'(t_0)=lim_tto t_0 fracf(t) - f(t_0)t-t_0$$
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 $frac00$, 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, $lim_limitsxto 2fracx^2-4x-2$). 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 $t_1$.
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
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 3 at 14:55
1
@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.
– Hurkyl
Sep 3 at 14:56
2
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 3 at 15:00
1
@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 3 at 15:02
 |Â
show 9 more comments
up vote
6
down vote
accepted
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
$$v_0 = f'(t_0)=lim_tto t_0 fracf(t) - f(t_0)t-t_0$$
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 $frac00$, 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, $lim_limitsxto 2fracx^2-4x-2$). 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 $t_1$.
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
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 3 at 14:55
1
@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.
– Hurkyl
Sep 3 at 14:56
2
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 3 at 15:00
1
@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 3 at 15:02
 |Â
show 9 more comments
up vote
6
down vote
accepted
up vote
6
down vote
accepted
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
$$v_0 = f'(t_0)=lim_tto t_0 fracf(t) - f(t_0)t-t_0$$
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 $frac00$, 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, $lim_limitsxto 2fracx^2-4x-2$). 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 $t_1$.
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
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
$$v_0 = f'(t_0)=lim_tto t_0 fracf(t) - f(t_0)t-t_0$$
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 $frac00$, 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, $lim_limitsxto 2fracx^2-4x-2$). 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 $t_1$.
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
answered Sep 3 at 14:38
GoodDeeds
10.2k21335
10.2k21335
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
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 3 at 14:55
1
@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.
– Hurkyl
Sep 3 at 14:56
2
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 3 at 15:00
1
@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 3 at 15:02
 |Â
show 9 more comments
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
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 3 at 14:55
1
@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.
– Hurkyl
Sep 3 at 14:56
2
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 3 at 15:00
1
@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 3 at 15:02
2
2
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
There are no actual infinitesimal numbers. The limit IS the instantaneous velocity.
– Alfred Yerger
Sep 3 at 14:43
1
1
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 3 at 14:55
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 3 at 14:55
1
1
@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.
– Hurkyl
Sep 3 at 14:56
@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.
– Hurkyl
Sep 3 at 14:56
2
2
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 3 at 15:00
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 3 at 15:00
1
1
@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 3 at 15:02
@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 3 at 15:02
 |Â
show 9 more comments
up vote
1
down vote
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
$$ m approx fracf(x)-f(a)x-a qquad qquad textwhen x approx a $$
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:
$$ m = lim_x to a fracf(x)-f(a)x-a $$
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
$$ u(x) = fracf(x) - f(a)x - a$$
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 $fracf(x)-f(a)x-a$, 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:
$$ u(a) = lim_x to a u(x) $$
answered Sep 3 at 15:03
Hurkyl
109k9113254
add a comment |Â
up vote
1
down vote
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
$$ m approx fracf(x)-f(a)x-a qquad qquad textwhen x approx a $$
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:
$$ m = lim_x to a fracf(x)-f(a)x-a $$
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
$$ u(x) = fracf(x) - f(a)x - a$$
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 $fracf(x)-f(a)x-a$, 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:
$$ u(a) = lim_x to a u(x) $$
answered Sep 3 at 15:03
Hurkyl
109k9113254
add a comment |Â
up vote
1
down vote
up vote
1
down vote
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
$$ m approx fracf(x)-f(a)x-a qquad qquad textwhen x approx a $$
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:
$$ m = lim_x to a fracf(x)-f(a)x-a $$
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
$$ u(x) = fracf(x) - f(a)x - a$$
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 $fracf(x)-f(a)x-a$, 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:
$$ u(a) = lim_x to a u(x) $$
answered Sep 3 at 15:03
Hurkyl
109k9113254
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
$$ m approx fracf(x)-f(a)x-a qquad qquad textwhen x approx a $$
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:
$$ m = lim_x to a fracf(x)-f(a)x-a $$
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
$$ u(x) = fracf(x) - f(a)x - a$$
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 $fracf(x)-f(a)x-a$, 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:
$$ u(a) = lim_x to a u(x) $$
answered Sep 3 at 15:03
Hurkyl
109k9113254
edited Sep 3 at 15:16
answered Sep 3 at 15:03
Hurkyl
109k9113254
answered Sep 3 at 15:03
Hurkyl
109k9113254
answered Sep 3 at 15:03
Hurkyl
109k9113254
109k9113254
add a comment |Â
add a comment |Â
up vote
0
down vote
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.
answered Sep 3 at 15:55
hobbs
21828
add a comment |Â
up vote
0
down vote
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.
answered Sep 3 at 15:55
hobbs
21828
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered Sep 3 at 15:55
hobbs
21828
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.
answered Sep 3 at 15:55
hobbs
21828
edited Sep 3 at 16:02
answered Sep 3 at 15:55
hobbs
21828
answered Sep 3 at 15:55
hobbs
21828
answered Sep 3 at 15:55
hobbs
21828
21828
add a comment |Â
add a comment |Â
up vote
0
down vote
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.
answered Sep 3 at 16:19
Allawonder
1,852516
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
add a comment |Â
up vote
0
down vote
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.
answered Sep 3 at 16:19
Allawonder
1,852516
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
add a comment |Â
up vote
0
down vote
up vote
0
down vote
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.
answered Sep 3 at 16:19
Allawonder
1,852516
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.
answered Sep 3 at 16:19
Allawonder
1,852516
edited Sep 4 at 3:38
answered Sep 3 at 16:19
Allawonder
1,852516
answered Sep 3 at 16:19
Allawonder
1,852516
answered Sep 3 at 16:19
Allawonder
1,852516
1,852516
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
add a comment |Â
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
The problem of $mathbb Q $ as compared to $mathbbR $ is not related to "infinitely many". Definition of limit applies in $mathbbQ $ 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 4 at 1:21
@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 4 at 3:33
@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 4 at 3:33
add a comment |Â
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Maybe you can see the Khan's video on Instantaneous speed.
– Mauro ALLEGRANZA
Sep 3 at 14:23
4
The intuitive concept of instantaneous speed is replaced by the mathematical notion of derivative.
– Mauro ALLEGRANZA
Sep 3 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 3 at 14:25
3
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 $Delta tto0$.
– Andrea Mori
Sep 3 at 14:26
1
@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 3 at 14:33