Instantaneous Rate of Change: The Limit of Average Rates
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Direct link | https://openstax.org/books/calculus-volume-1/pages/3-1-defining-the-derivative |
| Supplementary | OpenStax Calculus Volume 1, Section 2.2: “The Limit of a Function” |
| Supplementary link | https://openstax.org/books/calculus-volume-1/pages/2-2-the-limit-of-a-function |
Both sources are free and openly licensed.
Try This First
A small object slides along a track. A sensor records its position $s$ (in meters) at each whole second.
| Time $t$ (s) | Position $s$ (m) |
|---|---|
| 2 | 14 |
| 3 | 19 |
| 4 | 26 |
| 5 | 35 |
Before reading further:
- Compute the average speed over the interval from $t = 3$ to $t = 5$. What is the unit of your answer?
- Compute the average speed over $[3, 4]$ and over $[4, 5]$ separately. Do they match? What does it mean that they do not?
- If you could narrow the interval to $[3.9, 4]$ or $[4, 4.1]$ -- data you do not have -- what would you be trying to find?
Hold these answers. The ideas below name what question (3) is really asking.
The Covariation Strand so Far
You have climbed four rungs of the covariation ladder:
- MA1: You identified which quantity depends on which.
- MA2: You tracked whether the output grows or shrinks as the input grows.
- MA3: You quantified how much the output changes over an interval ($\Delta y$).
- MA4: You divided that change by the interval width to find the average rate ($\Delta y / \Delta x$).
Each rung made the description more precise. The final rung -- MA5 -- asks about change at a single instant rather than across an interval.
The average rate of change answers “how fast, on average, over this stretch?” The instantaneous rate answers “how fast, exactly, at this moment?”
Key Idea
For the object above, the position follows the rule $s(t) = t^2 + 10$. Compute the average rate of change on a sequence of intervals, each one starting at $t = 4$ and shrinking toward it from the right:
| Interval $[4,\ 4+h]$ | $h$ | $s(4+h)$ | $\Delta s$ | Average rate $\frac{\Delta s}{h}$ |
|---|---|---|---|---|
| $[4,\ 5]$ | 1 | 35 | 9 | 9 |
| $[4,\ 4.5]$ | 0.5 | 30.25 | 4.25 | 8.5 |
| $[4,\ 4.1]$ | 0.1 | 26.81 | 0.81 | 8.1 |
| $[4,\ 4.01]$ | 0.01 | 26.0801 | 0.0801 | 8.01 |
| $[4,\ 4.001]$ | 0.001 | 26.008001 | 0.008001 | 8.001 |
Predict: what value does the average rate approach as $h \to 0$?
The average rates are closing in on 8. The sequence does not reach a stopping point at $h = 0$ -- it approaches a limit. That limit, 8 meters per second, is the instantaneous rate of change of $s$ at $t = 4$.
In symbols: $$s'(4) = \lim_{h \to 0} \frac{s(4 + h) - s(4)}{h}.$$
This is the derivative of $s$ at $t = 4$: the number that the average rates home in on as the interval collapses to a single moment.
Two Connected Representations
Numerically: the table above shows the convergence. Average rates on smaller and smaller intervals home in on a single value. That value is the instantaneous rate.
Graphically: on a graph of $s$ vs $t$, draw the secant line through $(4, 26)$ and $(4 + h,\ s(4+h))$. As $h \to 0$, that secant line rotates until it becomes the tangent line at $(4, 26)$. The slope of the tangent line is the instantaneous rate of change.
Slope of secant from $(4, 26)$ to $(5, 35)$: $\dfrac{35 - 26}{5 - 4} = 9$. (Steep.)
Slope of tangent at $(4, 26)$: 8. (The limit.)
Translation prompt: In the table, what happens to the average rate as $h$ shrinks? On the graph, what happens to the secant line as the second point slides back toward $(4, 26)$? Name one feature they share.
The Definition
Let $f$ be a function defined near $x = a$. The instantaneous rate of change of $f$ at $a$, written $f'(a)$, is:
$$f'(a) = \lim_{h \to 0} \frac{f(a + h) - f(a)}{h},$$
provided this limit exists.
This number has two faces:
- Rate of change: how fast the output is changing per unit of input, at the single moment $x = a$.
- Slope: the slope of the tangent line to the graph of $f$ at the point $(a, f(a))$.
These are the same number, read two ways. A positive $f'(a)$ means the graph is rising at $a$ and the output is increasing. A negative $f'(a)$ means the graph is falling and the output is decreasing. A zero $f'(a)$ means the graph has a horizontal tangent -- an instant when the output is momentarily neither increasing nor decreasing.
Worked Example 1: Verifying the Table Result
Confirm $s'(4) = 8$ using the definition, where $s(t) = t^2 + 10$.
$$\frac{s(4+h) - s(4)}{h} = \frac{(4+h)^2 + 10 - (16 + 10)}{h} = \frac{16 + 8h + h^2 + 10 - 26}{h} = \frac{8h + h^2}{h} = \frac{h(8 + h)}{h} = 8 + h.$$
As $h \to 0$: $\lim_{h \to 0}(8 + h) = 8$.
Confirm: the table’s limiting value and the algebraic limit agree. Both give $s'(4) = 8$ m/s.
Check your prediction: does this match what the table was converging toward?
Worked Example 2: A General Point
Find $s'(a)$ for $s(t) = t^2 + 10$ at a general time $a$.
$$\frac{s(a+h) - s(a)}{h} = \frac{(a+h)^2 + 10 - a^2 - 10}{h} = \frac{2ah + h^2}{h} = 2a + h.$$
$$s'(a) = \lim_{h \to 0}(2a + h) = 2a.$$
So the instantaneous speed at time $a$ is $2a$ m/s. At $t = 4$: $s'(4) = 8$. At $t = 1$: $s'(1) = 2$. At $t = 0$: $s'(0) = 0$ -- the object starts from rest.
Translation prompt: At which time is the object moving fastest in the interval $[0, 5]$? At which time is it slowest? How does the formula $s'(a) = 2a$ show this without computing each value separately?
the limit “never reaches” $h = 0$, so the instantaneous rate is only approximate.
This is the limit-as-unreachable-barrier error. The limit $\lim_{h \to 0}(8 + h)$ is not an approximation; it is exactly 8. The process of $h \to 0$ does not mean $h$ is stuck at some small positive number. A limit is a precise mathematical statement about what value the expression homes in on. The expression $8 + h$ can be made as close to 8 as desired by choosing $h$ small enough, so the limit is exactly 8, not “almost 8”. The derivative is an exact number, not an estimate.
a high position means a high speed.
This is the height-vs-slope error. In Worked Example 1, the object is at position $s(4) = 26$ meters. Its speed at that moment is $s'(4) = 8$ m/s. The position (height on the graph) and the speed (slope of the graph) are different quantities with different units. A large output value says nothing about how steeply the function is changing. A tall graph can be flat; a low graph can be steep. Speed is slope, not height.
the instantaneous rate is a fixed property of the function, not dependent on the point.
This is the rate-as-fixed-number error carried from MA4. At MA4, students learned that the average rate depends on the interval. At MA5, the same principle applies to instantaneous rates: $s'(a) = 2a$ changes with $a$. The instantaneous rate at $t = 1$ (which is 2 m/s) is different from the instantaneous rate at $t = 4$ (which is 8 m/s). The derivative is a quantity that varies from point to point; when you ask “what is the derivative?”, the answer depends on where you are on the function.
Predict-Then-Check Practice
Problem 1 (Low floor): Fill in the missing entry in the table below for $f(x) = x^2$ and the interval $[3,\ 3+h]$.
| $h$ | Average rate $\frac{f(3+h)-f(3)}{h}$ |
|---|---|
| 1 | 7 |
| 0.5 | ? |
| 0.1 | 6.1 |
| 0.01 | 6.01 |
Predict first: the values are decreasing toward 6. What should the missing entry be?
Check your answer
$f(3 + 0.5) = (3.5)^2 = 12.25$. $f(3) = 9$. Average rate: $\dfrac{12.25 - 9}{0.5} = \dfrac{3.25}{0.5} = 6.5$.
Did your prediction fall between 7 and 6.1? It should: 6.5 is between those values and the sequence is converging to 6.
Problem 2: The height of a ball thrown straight up is given by $h(t) = 60t - 5t^2$ (meters, $t$ in seconds). Find the instantaneous velocity at $t = 3$ using the definition.
Predict first: between $t = 2$ and $t = 4$, is the ball rising or falling? At exactly $t = 3$, do you expect positive or negative velocity?
Check your answer
$h(3+k) = 60(3+k) - 5(3+k)^2 = 180 + 60k - 5(9 + 6k + k^2) = 180 + 60k - 45 - 30k - 5k^2 = 135 + 30k - 5k^2$.
$h(3) = 60(3) - 5(9) = 180 - 45 = 135$.
$\dfrac{h(3+k) - h(3)}{k} = \dfrac{30k - 5k^2}{k} = 30 - 5k$.
$h'(3) = \lim_{k \to 0}(30 - 5k) = 30$ m/s.
Positive velocity confirms the ball is still rising at $t = 3$. (It reaches its peak when $h'(t) = 60 - 10t = 0$, i.e., at $t = 6$, so at $t = 3$ it is only halfway to the top.)
Problem 3 (High ceiling): For a general function $f(x) = x^3$, find $f'(a)$ using the definition. Then find the specific point where $f'(a) = 12$ and give a geometric interpretation.
Check your answer
$\dfrac{(a+h)^3 - a^3}{h} = \dfrac{3a^2h + 3ah^2 + h^3}{h} = 3a^2 + 3ah + h^2$.
$f'(a) = \lim_{h \to 0}(3a^2 + 3ah + h^2) = 3a^2$.
Set $3a^2 = 12$: $a^2 = 4$, so $a = 2$ or $a = -2$.
Geometric interpretation: the tangent line to $y = x^3$ has slope 12 at two points -- $(2, 8)$ and $(-2, -8)$. At both points, the graph is equally steep (in the sense of slope magnitude), though the curves run in the same upward direction at both since $3a^2$ is always non-negative.
One way to see this: $y = x^3$ is symmetric about the origin (an odd function), so if the tangent is steep at a positive input, the graph is equally steep at the corresponding negative input.
Another way to see it: the formula $f'(a) = 3a^2$ depends on $a^2$, which equals $(−a)^2$, so the derivative is the same at $a$ and $-a$ for any $a$.
Which interpretation do you find more satisfying, and why?
Ask Why
One way to see the definition of the derivative: it formalizes “zoom in on a point.” If you could magnify the graph of a smooth function near $x = a$, the graph would look like a straight line with slope $f'(a)$. Why would zooming in produce a line rather than a curve?
Another way to see it: the derivative is the best linear approximation to $f$ near $a$. If you know only $f(a)$ and $f'(a)$, the line $y = f(a) + f'(a)(x - a)$ is the closest a straight line can come to agreeing with $f$ near $a$.
Why might both descriptions be equivalent? Why does the slope that the secant lines converge to match the slope of the line that best approximates $f$ near $a$? (You do not need to prove this now; just think about whether the two descriptions are saying the same thing in different language.)
Mastery Checklist
Strand Connection: MA1 to MA5 in One View
| Level | Question answered | Key idea |
|---|---|---|
| MA1 | Which quantity depends on which? | Identify input and output |
| MA2 | Does the output increase or decrease as the input grows? | Direction of covariation |
| MA3 | By how much does the output change? | $\Delta y = f(b) - f(a)$ |
| MA4 | How fast does the output change, on average? | $\Delta y / \Delta x$ over $[a, b]$ |
| MA5 | How fast is the output changing at a single instant? | $\lim_{h \to 0} \Delta y / h$ at $a$ |
Each step sharpens the description of how two quantities change together. MA5 is where the idea of rate becomes fully dynamic: not just across a stretch, but at a point.
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