Rates of Change
Textbook Reference
| Primary source | OpenStax Calculus Volume 1, Section 3.1: “Defining the Derivative” |
| Book URL | https://openstax.org/details/books/calculus-volume-1 |
Freely available and openly licensed.
Key idea
A rate of change answers the question: “How fast is one quantity changing relative to another?” Average rate of change compares total change over an interval. Instantaneous rate of change captures what is happening at a single moment.
You already know average rate of change: it is $\Delta y / \Delta x$, the slope of the secant line. Instantaneous rate of change is the limit of the average rate as the interval shrinks to zero -- it is the slope of the tangent line, and it is the derivative. This connection between “rate of change” and “derivative” is not a separate idea; it is the definition.
Prerequisite Check
Before this lesson, make sure you can do all of the following:
Quick Reference
Average rate of change of $f$ over $[a, b]$: \[ \frac{f(b) - f(a)}{b - a} = \frac{\Delta y}{\Delta x}. \] This is the slope of the secant line through $(a, f(a))$ and $(b, f(b))$.
Instantaneous rate of change of $f$ at $a$: \[ \lim_{h \to 0} \frac{f(a+h) - f(a)}{h} = f'(a). \] This is the slope of the tangent line at $(a, f(a))$, which is the derivative at $a$.
Units. The rate of change has units of $\dfrac{\text{units of output}}{\text{units of input}}$.
Key Concepts
1. Average Rate of Change
The average rate of change of $f$ from $x = a$ to $x = b$ is: \[ \frac{f(b) - f(a)}{b - a}. \]
This is the slope of the secant line connecting the two points $(a, f(a))$ and $(b, f(b))$ on the graph. It measures the net change in output per unit change in input over the interval.
Example 1. A car’s position in kilometers at time $t$ hours is $s(t) = 5t^2 + 10t$. What is the average velocity from $t = 1$ to $t = 3$?
$s(1) = 5 + 10 = 15$ km. $s(3) = 45 + 30 = 75$ km.
\[ \text{Average velocity} = \frac{75 - 15}{3 - 1} = \frac{60}{2} = 30 \text{ km/h}. \]
The car traveled the net distance of 60 km over 2 hours, averaging 30 km/h. The car may have been faster or slower at any given instant.
Example 2. For $f(x) = x^2 - 3x$, find the average rate of change from $x = 2$ to $x = 5$.
$f(2) = 4 - 6 = -2$. $f(5) = 25 - 15 = 10$.
\[ \frac{10 - (-2)}{5 - 2} = \frac{12}{3} = 4. \]
The output of $f$ increased at an average rate of 4 units per unit of $x$ over this interval.
2. Instantaneous Rate of Change
The instantaneous rate of change at $x = a$ is the limit of the average rate as the interval $[a, a+h]$ shrinks: \[ f'(a) = \lim_{h \to 0} \frac{f(a+h) - f(a)}{h}. \]
Example 3. For $s(t) = 5t^2 + 10t$, find the instantaneous velocity at $t = 1$.
$s(1+h) = 5(1+h)^2 + 10(1+h) = 5 + 10h + 5h^2 + 10 + 10h = 15 + 20h + 5h^2$.
$s(1+h) - s(1) = 20h + 5h^2 = h(20 + 5h)$.
$\dfrac{s(1+h) - s(1)}{h} = 20 + 5h$.
$s'(1) = \lim_{h \to 0}(20 + 5h) = 20$ km/h.
At the instant $t = 1$, the car is moving at exactly 20 km/h.
3. Units and Interpretation
The units of a rate of change are always output-units per input-units.
| $f$ represents | $x$ represents | $f'(a)$ is | Units |
|---|---|---|---|
| Position (m) | Time (s) | Velocity | m/s |
| Position (km) | Time (h) | Velocity | km/h |
| Cost ($) | Quantity produced | Marginal cost | $/unit | |||
| Population | Time (yr) | Growth rate | people/yr |
| Temperature (K) | Distance (m) | Temperature gradient | K/m |
Example 4. If $C(q)$ is the cost in dollars of producing $q$ items, and $C'(50) = 8$ dollars/item, then producing one additional item beyond the 50th costs approximately 8 dollars. This is the marginal cost at $q = 50$.
4. Comparing Average and Instantaneous
These quantities answer different questions:
| Question | Use |
|---|---|
| “What was the average speed over the whole trip?” | Average rate of change |
| “How fast was the car going at the moment it passed the exit?” | Derivative (instantaneous) |
| “By how much did the cost increase per unit from 20 to 30 units?” | Average rate of change |
| “Is the cost increasing right now, at 25 units?” | Sign of the derivative |
The sign of the instantaneous rate tells you the direction:
- $f'(a) > 0$: $f$ is increasing at $a$.
- $f'(a) < 0$: $f$ is decreasing at $a$.
- $f'(a) = 0$: $f$ has a horizontal tangent at $a$ (possible maximum, minimum, or inflection).
the average rate of change over an interval equals the instantaneous rate at the midpoint.
This is the rate-as-fixed-number error. The average rate $\frac{f(b)-f(a)}{b-a}$ is one number summarizing the whole interval $[a, b]$. It is not the instantaneous rate at any particular input unless the function happens to be linear. For $f(t) = t^2$: the average rate over $[1, 3]$ is $\frac{9-1}{3-1} = 4$. The instantaneous rate at the midpoint $t = 2$ is $f'(2) = 4$ -- a coincidence for quadratics only. For $f(t) = t^3$, the average over $[1, 3]$ is $\frac{27-1}{2} = 13$, while $f'(2) = 12$ and the midpoint of the interval in a sense closer to the mean value occurs at $t = \sqrt{13/3} \approx 2.08$, not at $t=2$. The average and instantaneous rates are different objects that happen to coincide in special cases.
if $f'(a) = 0$, then $f(a) = 0$ too.
This is the height-vs-slope error. $f'(a) = 0$ means the tangent line is horizontal at $x = a$ -- the slope is zero. It says nothing about the height $f(a)$. For $f(x) = x^2 + 5$: $f'(x) = 2x = 0$ at $x = 0$, but $f(0) = 5 \neq 0$. The function has a horizontal tangent at height $5$. A zero slope and a zero height are independent conditions, and confusing them leads to misidentifying both critical numbers and x-intercepts.
Common Errors
| Error | Example | Correction |
|---|---|---|
| Confusing average and instantaneous | “The velocity at $t = 3$ is $\frac{s(3)-s(0)}{3}$” | That is average velocity over $[0,3]$; instantaneous requires a limit |
| Omitting units | “Rate is 20” | Include units: “20 km/h” |
| Using the wrong endpoints | Computing $\frac{f(3)-f(1)}{2}$ when asked for $f'(2)$ | $f'(2)$ requires the limit definition, not an average over $[1,3]$ |
Leveled Practice
Level 1 -- Average Rate of Change
Problem 1. Find the average rate of change of $f(x) = 2x^2 + 1$ from $x = 1$ to $x = 4$.
Show answer
$f(1) = 3$, $f(4) = 33$. Rate $= \dfrac{33-3}{3} = 10$.
Problem 2. A particle moves so that $s(t) = t^3 - 3t$ meters at time $t$ seconds. Find the average velocity from $t = 0$ to $t = 2$.
Show answer
$s(0) = 0$, $s(2) = 8 - 6 = 2$. Average velocity $= \dfrac{2}{2} = 1$ m/s.
Level 2 -- Instantaneous Rate of Change
Problem 3. Find $f'(2)$ for $f(x) = 3x^2$ using the limit definition.
Show answer
$f(2+h) = 3(4 + 4h + h^2) = 12 + 12h + 3h^2$.
$f(2+h) - f(2) = 12h + 3h^2 = h(12 + 3h)$.
$\lim_{h \to 0}(12 + 3h) = 12$.
$f'(2) = 12$.
Problem 4. A ball’s height in meters at time $t$ seconds is $h(t) = 20t - 5t^2$. Find the instantaneous velocity at $t = 2$ and interpret the result.
Show answer
$h(2+s) = 20(2+s) - 5(2+s)^2 = 40 + 20s - 5(4 + 4s + s^2) = 40 + 20s - 20 - 20s - 5s^2 = 20 - 5s^2$.
$h(2) = 40 - 20 = 20$ m.
$h(2+s) - h(2) = -5s^2$.
$\dfrac{-5s^2}{s} = -5s \to 0$ as $s \to 0$.
$h'(2) = 0$ m/s.
At $t = 2$ seconds, the ball is momentarily at rest -- it is at the peak of its trajectory. The velocity is zero; the ball is about to start falling.
Level 3 -- Interpretation and Context
Problem 5. The number of bacteria is $N(t) = 100 \cdot 2^t$ at time $t$ hours. Using the known result $\lim_{h \to 0} \frac{2^h - 1}{h} = \ln 2$, find $N'(0)$ and state its units and meaning.
Show answer
\[ N'(0) = \lim_{h \to 0} \frac{100 \cdot 2^h - 100}{h} = 100 \ln 2 \approx 69.3 \text{ bacteria per hour}. \]
At the initial moment, the culture is growing at approximately 69 bacteria per hour.
Problem 6. The revenue $R(p)$ from selling at price $p$ dollars is $R(p) = 1000p - 5p^2$. Find and interpret $R'(80)$.
Show answer
$R(80+h) = 1000(80+h) - 5(80+h)^2 = 80000 + 1000h - 5(6400 + 160h + h^2)$.
$= 80000 + 1000h - 32000 - 800h - 5h^2 = 48000 + 200h - 5h^2$.
$R(80) = 1000(80) - 5(6400) = 80000 - 32000 = 48000$.
$R(80+h) - R(80) = 200h - 5h^2 = h(200 - 5h)$.
$R'(80) = \lim_{h \to 0}(200 - 5h) = 200$ dollars per dollar of price.
When the price is 80 dollars, revenue increases at a rate of \$200 per unit increase in price. (This means raising the price by one dollar increases revenue by approximately 200 dollars at this price level.)
Mastery Checklist
You have mastered this skill when you can do all of the following without referring to notes:
Mental Model
Average rate of change is a compass bearing: the overall direction of travel from start to finish, regardless of what happened in between. Instantaneous rate of change is the speedometer: what is happening at a single instant.
A car averaging 60 mph over an hour may have been stopped at a light for five minutes and then driven faster the rest of the time. The speedometer never shows “60 mph” as the instantaneous reading during those five minutes. Average and instantaneous are genuinely different quantities, even when they happen to agree for special functions.
The derivative is the mathematical speedometer: it reports the instantaneous rate, not the average.
Connections
Within Calculus I (MATH161)
- Derivative at a point: $f'(a)$ is both the instantaneous rate of change at $a$ and the slope of the tangent to the graph at $a$. Both descriptions refer to the same number.
- Increasing and decreasing functions (Ch. 3): The sign of $f'$ throughout an interval tells you whether $f$ is increasing or decreasing on that interval.
- Optimization: Where the instantaneous rate of change is zero ($f'(a) = 0$), the function may have a maximum or minimum. Rate of change is the key tool for finding optimal values.