# Limits confusion -- What is meant by "instantaneous rate of change"?

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In summary, an instant is a single point in time and the instantaneous rate of change at that instant is the slope of the tangent line to the curve at that point. To understand this concept, we use the definition of average rate of change which requires knowledge at two instants, but we can obtain the limiting value by plugging in the only value we are not allowed to plug.
Suppose a particle is falling under the pull of gravity, the distance it has fallen is given by s=16t^2.Suppose we wish to find the instantaneous speed at t=1.
Find the average speed between t=1 and t=1+h where h is any real number except 0.
Distance traveled/Time it takes to travel the distance =16(1+h)^2-16=16+32h+16h^2-16/h=32+16h

We obtained the correct answer for the wrong problem. I have never fully understood what the instantaneous rate of change of any quantity meant. What is an instant? From experience, we say that an instant is an extremely tiny interval, but tiny to whom? It's quite subjective. Also, in keeping the definition of average rate of change to understand the instantaneous rate of change we require the knowledge at two instants, How can we recourse to a definition that requires the measurement at two instants to understand what happens at one?

Accepting that an instant is an extremely time interval and that nothing drastic happens over that tiny interval we observe that the average rate of change serves as a better approximation as the time interval shrinks but that we loose all information if no time transpires (we wind up with 0/0).
32+16h ≅ 32 we sense that 16h approaches as h approaches 0 and 32+16h tends to 32.
Why do we say the limiting value is equal to 32 and not approximately 32?

It seems to me that at first we use h≠0 to steer away from 0/0 and then we obtain the limiting value by plugging in the only value we are not allowed to plug.

Suppose a particle is falling under the pull of gravity, the distance it has fallen is given by s=16t^2.
This is meaningless without units. I assume you mean ## s = 16 t^2 \ \mathrm{ft} ##. This comes from the equation ## s = ut + \frac 1 2 a t^2 ##, where ## u ## is the initial speed which is nil in this case and ## a \approx 32 \mathrm{ft/s^2}##.

Suppose we wish to find the instantaneous speed at t=1.
Then we would use a different equation: ## v = u + at ##, in this case ## v = 0 + 32 \times 1 = 32 \mathrm{ft/s}##.

Find the average speed between t=1 and t=1+h where h is any real number except 0.
Distance traveled/Time it takes to travel the distance =16(1+h)^2-16=16+32h+16h^2-16/h=32+16h
I don't understand any of that. Edit: If you travel 16 ft in 1 second then your average speed is 16 ft/s.

I have never fully understood what the instantaneous rate of change of any quantity meant.
At any particular time ## t ##, the instantaneous rate of change of distance is the speed at that time, the instantaneous rate of change of speed is acceleration at that time. However note that 'instantaneous speed' is exactly the same as 'speed'.

What is an instant? From experience, we say that an instant is an extremely tiny interval, but tiny to whom? It's quite subjective.
No, an instant is an interval of exactly zero length.

Edit: none of this has anything to do with limits.

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I have never fully understood what the instantaneous rate of change of any quantity meant. What is an instant? From experience, we say that an instant is an extremely tiny interval, but tiny to whom? It's quite subjective.
An instant is a single point in time, which in your example is ##t=1~\rm s##. It's not a tiny interval. It's a single point.

In your example, you can think of the instantaneous rate of change as the slope of the line tangent to the curve ##s(t) = (16~{\rm ft/s^2}) t^2## at the point ##(1~{\rm s},\ 16~{\rm ft})##. There's no interval involved here at all.

Also, in keeping the definition of average rate of change to understand the instantaneous rate of change we require the knowledge at two instants, How can we recourse to a definition that requires the measurement at two instants to understand what happens at one?
To put it mathematically, you're asking how we know
$$f'(t) = \lim_{h \to 0} \frac{f(t+h)-f(t)}{h}.$$

Why do we say the limiting value is equal to 32 and not approximately 32?
It seems to me that at first we use h≠0 to steer away from 0/0 and then we obtain the limiting value by plugging in the only value we are not allowed to plug.
You have two different functions here. First, you have
$$f_1(h) = \frac{16(1+h)^2 - 16}{h},$$ which is not defined for ##h=0##. The second function is ##f_2(h) = 32 + 16h,## which is defined for ##h=0##. The only difference between the two functions is that ##f_1## has a hole at ##h=0##.

Intuitively, the limit as ##h \to 0## tells us what value a function is approaching as ##h \to 0##, not what the value of the function is when ##h=0##. Since ##f_1## and ##f_2## are identical everywhere except at ##h=0##, they have the same limit as ##h \to 0##. It doesn't matter that ##f_1## is not defined for ##h=0##. Clearly, we can find the limit of the second function as ##h \to 0## by simply setting ##h## to 0, and that gives us the limit of ##f_1## as well.

pbuk said:
This is meaningless without units. I assume you mean ## s = 16 t^2 \ \mathrm{ft} ##. This comes from the equation ## s = ut + \frac 1 2 a t^2 ##, where ## u ## is the initial speed which is nil in this case and ## a \approx 32 \mathrm{ft/s^2}##.Then we would use a different equation: ## v = u + at ##, in this case ## v = 0 + 32 \times 1 = 32 \mathrm{ft/s}##.I don't understand any of that. Edit: If you travel 16 ft in 1 second then your average speed is 16 ft/s.At any particular time ## t ##, the instantaneous rate of change of distance is the speed at that time, the instantaneous rate of change of speed is acceleration at that time. However note that 'instantaneous speed' is exactly the same as 'speed'.No, an instant is an interval of exactly zero length.

Edit: none of this has anything to do with limits.
I wanted to illustrate the concept of limit in terms of an example. I am discussing how we arrive at the expression v in the first place. Returning to the problem of finding the instantaneous speed at t=1.
To get around it, we instead compute the average rate of change and let the time interval get as close to zero as we wish but never equal to zero, had we let it equal 0 we would wind with a meaningless 0/0. The average rate of change of any quantity invokes a non zero interval, though the instantaneous speed involves an interval of exactly zero length. I am puzzled on why we equate them..
vela said:
An instant is a single point in time, which in your example is ##t=1~\rm s##. It's not a tiny interval. It's a single point.

In your example, you can think of the instantaneous rate of change as the slope of the line tangent to the curve ##s(t) = (16~{\rm ft/s^2}) t^2## at the point ##(1~{\rm s},\ 16~{\rm ft})##. There's no interval involved here at all.To put it mathematically, you're asking how we know
$$f'(t) = \lim_{h \to 0} \frac{f(t+h)-f(t)}{h}.$$You have two different functions here. First, you have
$$f_1(h) = \frac{16(1+h)^2 - 16}{h},$$ which is not defined for ##h=0##. The second function is ##f_2(h) = 32 + 16h,## which is defined for ##h=0##. The only difference between the two functions is that ##f_1## has a hole at ##h=0##.

Intuitively, the limit as ##h \to 0## tells us what value a function is approaching as ##h \to 0##, not what the value of the function is when ##h=0##. Since ##f_1## and ##f_2## are identical everywhere except at ##h=0##, they have the same limit as ##h \to 0##. It doesn't matter that ##f_1## is not defined for ##h=0##. Clearly, we can find the limit of the second function as ##h \to 0## by simply setting ##h## to 0, and that gives us the limit of ##f_1## as well.
What is a tangent line? Had we thought about it geometrically, How would we construct it and distinguish between a line which crosses the curve at the point we wish to find the instantaneous speed and a line which touches the curve at the same point? Also the approach of sliding in a ruler on the curve at the point is quite subjective.

But wasn't our goal to find the instantaneous rate of change at 1,namely f1(h=0?).

The average rate of change of any quantity invokes a non zero interval, though the instantaneous speed involves an interval of exactly zero length. I am puzzled on why we equate them..
We don't. ##v_{ave} \ne v_{inst}##.
The velocity (i.e., instantaneous rate of change of position relative to time) is the limit of the average velocities as the time interval decreases to zero.
What is a tangent line? Had we thought about it geometrically, How would we construct it and distinguish between a line which crosses the curve at the point we wish to find the instantaneous speed and a line which touches the curve at the same point?
Hasn't your textbook discussed tangent lines? A tangent line is different from a line that merely crosses the curve; a tangent line touches the curve at a single point.
Also the approach of sliding in a ruler on the curve at the point is quite subjective.
I disagree. You use some fixed point on the ruler and slide this point along the curve with the ruler staying on one side of the curve, at least locally.
But wasn't our goal to find the instantaneous rate of change at 1,namely f1(h=0?).
The goal was to find the instantaneous rate of change of f at 1, yes, but f1(0) is undefined. f2(0), however, is defined, and gives the instantaneous rate of change of f at 1.

The average rate of change of any quantity invokes a non zero interval, though the instantaneous speed involves an interval of exactly zero length. I am puzzled on why we equate them.
It's called calculus. It's a very useful branch of mathematics and worth learning.
What is a tangent line?
That's a concept from plane geometry. That's useful too - and also worth learning.
Also the approach of sliding in a ruler on the curve at the point is quite subjective.
Mathematics may appear subjective to those who don't take the time to learn it.

@PeroK @Mark44 I am taking my time to learn , I am posting my doubts because I find limits a difficult concept to grasp. The concept of tangent line is simple for straight lines and circle. But for any arbitrarily curve why don't we consider a line which crosses the curve and still touches the curve at a single point to be a tangent line.

But for any arbitrarily curve why don't we consider a line which crosses the curve and still touches the curve at a single point to be a tangent line.
We do. A tangent line at a point on a curve does not cross the curve locally (i.e. in some possibly small neighbourhood of the point), but it might cross the curve further away. See, for example:

https://en.wikipedia.org/wiki/Tangent

PeroK said:
We do. A tangent line at a point on a curve does not cross the curve locally
If the point is a point of inflection then the tangent does cross at that point (I prefer to think of this as the curve crossing the tangent).

For example the x-axis is tangent to, and crosses, the curve ## y = x^3 ## at ## (0, 0) ##.

Actually I think discussion of tangents is taking us a little off topic and in danger of getting stuck going round in circles . Let's get back to the heart of the problem.
...we instead compute the average rate of change and let the time interval get as close to zero as we wish but never equal to zero, had we let it equal 0 we would wind with a meaningless 0/0. The average rate of change of any quantity invokes a non zero interval, though the instantaneous speed involves an interval of exactly zero length. I am puzzled on why we equate them..
...
But wasn't our goal to find the instantaneous rate of change at 1,namely f1(h=0?).
No, we cannot find ## f_1(h = 0) ## because as you point out this would involve division by zero. Instead we look for the limit ## \lim_{h \to 0} f_1(h) ##.

In this case, as we take smaller and smaller values of ## h ## we get closer and closer to ## f_1(h) = 32 ## (assuming the approximation for acceleration) and so we say ## \lim_{h \to 0} f_1(h) = 32 ##. How do we know that? Well if you assert that we never get any closer than, say, 32.001 then I can find a value of h that proves you wrong: let's try ## h = \frac 1 {32000} ## which gives ## f_1( \frac 1 {32000}) = 32.0005 ## - that will do.

This is known as the epsilon-delta definition of a limit (in the example above ## \varepsilon = 0.001 ## and although I have not found ## \delta ## I have shown that it is greater than ## \frac 1 {32000} ##).

@PeroK @Mark44 I am taking my time to learn , I am posting my doubts because I find limits a difficult concept to grasp.
Your confusion might be arising from thinking of taking a limit as a process where you're getting closer and closer to some value. I don't really have time to go into this right now, but I think @fresh_42 discussed this misconception in another thread.

The concept of tangent line is simple for straight lines and circle. But for any arbitrarily curve why don't we consider a line which crosses the curve and still touches the curve at a single point to be a tangent line.
Because that's not the definition of a tangent line. Anyway, I agree with @pbuk that the talk of tangent lines is not useful here. My original point was that when we talk about a tangent to a curve, the line is tangent at a single point and is not based on some tiny interval.

Having read about the epsilon delta method and the comments above. I think (what I understood) the reason as to why we conjecture lim h->0 f1(h)=32 is that for smaller values of h the average velocities get closer to 32. If we assume a number other than 32 to be the limiting value then we can find a point x1≠1 in the deleted neighborhood surrounding x=1 such that f(x1)<L . Hence it cannot be the limiting value. Generalizing the idea ,as long as there is space in the deleted neighborhood and if the function is always rising or falling then no f(x) for any x in the deleted neighbourhood can be the limiting value.

Having read about the epsilon delta method and the comments above. I think (what I understood) the reason as to why we conjecture lim h->0 f1(h)=32 is that for smaller values of h the average velocities get closer to 32. If we assume a number other than 32 to be the limiting value then we can find a point x1≠1 in the deleted neighborhood surrounding x=1 such that f(x1)<L . Hence it cannot be the limiting value.
Not quite sure what you are saying here but in general it is true that if a limit exists then it is unique.

Generalizing the idea ,as long as there is space in the deleted neighborhood and if the function is always rising or falling then no f(x) for any x in the deleted neighbourhood can be the limiting value.
Again not quite sure what you are saying, but it is not always the case that a limit exists.

f1(h)=16(1+h)^2-16/h f2(h)=32+16h where h=x-1
f1(x-1)=f2(x-1) for all x-1 not equal to 0.
Lim h->0 f1(h)=Lim h->0=L f2(h) because h->0 implies that we get as close as we wish to 0 but not equal to 0. If L≠32 this means for values of h near 0 we can never get any closer to L. If we find the h for which f1(h)=L and f2(h)=L it turns out that x≠1, x=1 does not satisfy the criteria and so the implication is never met even no matter how big we choose ε to be.
0< |x-1|<δ -> |f1(h)-L|<ε

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f1(h)=16(1+h)^2-16/h f2(h)=32+16h where h=x-1
f1(x-1)=f2(x-1) for all x-1 not equal to 0.
You don't really need x - 1 here. Also, to be clear, you need parentheses in what you wrote for f1().
16(1 + h)^2 - 16/h means ##16(1 + h)^2 - \frac {16} h##, which is surely not what you meant. If you don't use LaTeX, write this expression as (16(1 + h)^2 - 16)/h.

To simplify what you wrote above,
##f_1(h) = \frac{16(1 + h)^2 - 16}h##
##f_2(h) = 32 + 16h##
The two functions have identical graphs except at the single point (0, 32), where ##f_1## has a point discontinuity, a "hole."

Lim h->0 f1(h)=Lim h->0=L f2(h) because h->0 implies that we get as close as we wish to 0 but not equal to 0. If L≠32 this means for values of h near 0 we can never get any closer to L. If we find the h for which f1(h)=L and f2(h)=L it turns out that x≠1, x=1 does not satisfy the criteria and so the implication is never met even no matter how big we choose ε to be.
The average velocity between the times t = 1 and t = 1 + h is ##\frac{16(1 + h)^2 - 16}h = \frac{16 + 32h + 16h^2 - 16}h##.

The instantaneous velocity at time t = 1, or v(1), is
##\lim_{h \to 0}\frac{16(1 + h)^2 - 16}h = \lim_{h \to 0}\frac{16 + 32h + 16h^2 - 16}h##
##= \lim_{h \to 0}\frac{32h + 16h^2}h = \lim_{h \to 0}\frac h h \lim_{h \to 0}(32 + 16h)##
Regarding the first limit of the last pair of limits, for any nonzero value of h, h/h will always be 1, so we say that ##\lim_{h \to 0}\frac h h = 1##.
For the second of these limits, the expression 32 + 16h can be made as close to 32 as anyone care, so we say that ##\lim_{h \to 0}(32 + 16h) = 32##.
From this we can conclude that v(1) = 32, the instantaneous velocity at time t = 1.
0< |x-1|<δ -> |f1(h)-L|<ε
Better would be for any choice of ##\epsilon > 0##, there exists a ##\delta > 0## such that ##|h| < \delta \Rightarrow |f_1(h) - 32| < \epsilon##. IOW, if you tell me how close you want ##f_1(h)## to be to 32, I'll tell you how small h needs to be.
The work I showed uses the properties of limits, all of which rely on the ##\delta - \epsilon## definition of the limit. After you get used to working with limits, then it might be more useful to dig into the details of more rigorous limit proofs.

I will attempt to articulate what I have been trying to say
Let the value of h get as close to 0 as we wish but that h≠0. Making a chart of sorts:
Time interval: Average speed
t=1 to t=2 48
t=1 to t=1.5 40
t=1 to t=1.1 33.6
t=1 to t=1.01 32.16
To emphasize that h need not be positive
t=0 to t=1 16
t=0.5 to t=1 24
t=0.9 to t=1 30.4
t=0.99 to t=1 31.84
f2(h) and f1(h) (which are equal functions except when h≠0) graph is made of continuous points therefore it seems that they converge to 32. We are not interested in what is happening at h=0 instead in what f1(h) and f2(h) approach for values of x in the neighborhood of h as it zeros in towards h=0.

How is the epsilon delta method helping us in finding the Limiting value which is 32 in this case? In the epsilon delta method we say that we are content if we could be sure that the difference between f1(h)=f2(h) and L is ϵ. We would in turn be interested in knowing how much x would deviate from c to be and still have f(x) in the required range. In short, it seems that the the methods goal is to solve an inequality to meet a certain implication.

How is the epsilon delta method helping us in finding the Limiting value which is 32 in this case? In the epsilon delta method we say that we are content if we could be sure that the difference between f1(h)=f2(h) and L is ϵ.
##\epsilon## can be any positive number. It's not just one "small" value.

In the epsilon delta method we say that we are content if we could be sure that the difference between f1(h)=f2(h) and L is ϵ.
No, we are only content when we can find ## \delta ## so that the difference is less than ## \varepsilon ##. And because as @PeroK says we can make ## \varepsilon ## as small as we like, we are saying that we can make the difference smaller than any value we choose.

The only non-negative number that is smaller than every positive number is 0 and so the difference betwen the slope and 32 is 0.

(Edit: this is not really the way the maths works when you look at it rigorously, but I don't think that visualizing it this way is harmful).

I have never fully understood what the instantaneous rate of change of any quantity meant. What is an instant?
Forget the numbers and the context initially. Numbers can often give you the 'can't see the wood for the trees' problem. You are actually asking about the fundamental idea behind Differential Calculus tadaaah! Most of the basic problems we come across in Science (certainly in classical Science) deal with continuously variable values (i.e. smooth curves with no discontinuities). No diagram needed for this: imagine a road up a hill that starts horizontally at the bottom, climbs and then levels out at the top. You could estimate the slope at a point where you stop the car by using two poles and some surveyor's equipment to measure the rise from your start pole over, say 100m to the next pole. Do the same over 20m, with an extra pole and you won't necessarily get exactly the same answer because the in between pole won't exactly touch the line between the tops of the two original poles. Now do the same over 1m, 0.1m and 0.01m. The answers will still be a bit different but they will approach a 'limit' as the spacing approaches zero. (The differences will get less and less.) That limit is what you'd call the instantaneous slope - you'd call that a Tangent to the road surface.
Instantaneous velocity for an accelerating car (at a certain point in its journey) will be the limit of a small change in position (δx) divided by the small time (δt). That's the limit of δx/δt as δt approaches zero and it's renamed dx/dt. You don't need to look at the 'precise instant'; a bit before or a bit afterwards will still give you a (near enough) answer.
You can do this for any pair of related variables that you can measure as long as they vary 'smoothly' over the smallest scale of measurement you use.
Instantaneous dy/dx is undefined where a curve is not continuous.

## 1. What is the definition of "instantaneous rate of change"?

The instantaneous rate of change is the rate at which a function is changing at a specific point, or instant, on its graph. It represents the slope of the tangent line to the function at that point.

## 2. How is "instantaneous rate of change" different from "average rate of change"?

The average rate of change is the overall rate at which a function changes over a given interval. It is calculated by finding the slope of the secant line between two points on the function's graph. In contrast, the instantaneous rate of change focuses on the rate of change at a single point on the graph.

## 3. Can the instantaneous rate of change be negative?

Yes, the instantaneous rate of change can be either positive or negative. A positive instantaneous rate of change means that the function is increasing at that point, while a negative instantaneous rate of change indicates that the function is decreasing at that point.

## 4. How is the instantaneous rate of change related to the derivative?

The derivative of a function is the mathematical expression that represents the instantaneous rate of change at any point on the function's graph. In other words, the derivative is the function that gives the slope of the tangent line at any given point on the original function.

## 5. Why is understanding the concept of "instantaneous rate of change" important?

Understanding the concept of instantaneous rate of change is important because it allows us to analyze the behavior of a function at a specific point. This information is crucial in many real-world applications, such as calculating the velocity of an object at a certain moment or determining the rate of change of a chemical reaction at a particular time.

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