Average Rate of Change

A rate of change tells you how fast one quantity changes compared with another: kilometres per hour, dollars per year, degrees per minute. For a straight line the rate is just the slope, and it never changes. For curves the rate keeps changing, so we measure an average over an interval. This idea is the starting point for calculus.

Key ideas

Rate of change compares two changes

A rate of change compares the change in the dependent variable ($y$) with the change in the independent variable ($x$):

$$\text{rate of change} = \frac{\Delta y}{\Delta x} = \frac{\text{change in } y}{\text{change in } x}$$

The symbol $\Delta$ (delta) means “change in”. The units are “$y$-units per $x$-unit”, for example metres per second (m/s) or dollars per year.

Zero, constant, or changing

Rate of change	Graph	Example
zero	horizontal line	the distance of a parked car from home
constant	straight line	the distance travelled by a plane cruising at $850$ km/h
changing	curve	the area of a circle as the radius grows; money growing with compound interest

A positive rate means $y$ is increasing; a negative rate means $y$ is decreasing.

Average rate of change over an interval

The average rate of change of $f$ from $x = a$ to $x = b$ is

$$\frac{\Delta y}{\Delta x} = \frac{f(b) - f(a)}{b - a}$$

It’s the constant rate that would produce the same overall change. For example, if you drive $240$ km in $3$ hours, your average speed is $\frac{240}{3} = 80$ km/h, even if you sometimes went faster or slower.

The slope of a secant

A secant is a line through two points on a curve. The average rate of change from $a$ to $b$ is exactly the slope of the secant through $\big(a, f(a)\big)$ and $\big(b, f(b)\big)$.

The average rate of change of $h(t) = -5t^2 + 30t$ from $t = 1$ to $t = 3$ is the slope of the secant: $\frac{20}{2} = 10$ m/s.

Three ways to find it

• From a table: read the two values and divide the differences.
• From a graph: read two points and find the slope of the secant joining them.
• From an equation: substitute both $x$-values, then divide.

Graphs from stories

You can sketch a graph from a description by thinking about the rate in each part of the story. On a distance–time graph, the slope is the speed: steady speed means a straight segment, stopped means horizontal, speeding up means the graph curves upward and gets steeper, slowing down means it levels off. On a speed–time graph, the height is the speed, so steady speed is a horizontal segment and stopping means dropping to $0$.

Worked examples

Example 1: From a table

A cup of hot chocolate cools on the counter. Its temperature is recorded every $5$ minutes.

Time (min)	$0$	$5$	$10$	$15$	$20$
Temperature ($^\circ\text{C}$)	$90$	$70$	$56$	$46$	$39$

Find the average rate of change over the whole $20$ minutes, over the first $5$ minutes, and over the last $5$ minutes. What do the results tell you?

Solution.

$$\begin{aligned} 0 \text{ to } 20\text{: } \quad \frac{39 - 90}{20 - 0} &= \frac{-51}{20} = -2.55\ ^\circ\text{C/min} \\ 0 \text{ to } 5\text{: } \quad \frac{70 - 90}{5 - 0} &= \frac{-20}{5} = -4\ ^\circ\text{C/min} \\ 15 \text{ to } 20\text{: } \quad \frac{39 - 46}{20 - 15} &= \frac{-7}{5} = -1.4\ ^\circ\text{C/min} \end{aligned}$$

All three rates are negative because the temperature is falling. The rate is changing: the hot chocolate cools quickly at first ($4$ degrees per minute) and much more slowly later ($1.4$ degrees per minute). The overall average of $-2.55\ ^\circ\text{C/min}$ hides that difference.

Example 2: From an equation (quadratic)

A ball is kicked straight up. Its height in metres after $t$ seconds is $h(t) = -5t^2 + 30t$. Find the average rate of change of height from $t = 1$ to $t = 3$, from $t = 3$ to $t = 5$, and from $t = 1$ to $t = 5$. Interpret each one.

Solution. First find the heights: $h(1) = -5 + 30 = 25$, $h(3) = -45 + 90 = 45$, $h(5) = -125 + 150 = 25$.

$$\begin{aligned} \frac{h(3) - h(1)}{3 - 1} &= \frac{45 - 25}{2} = 10 \text{ m/s} \\ \frac{h(5) - h(3)}{5 - 3} &= \frac{25 - 45}{2} = -10 \text{ m/s} \\ \frac{h(5) - h(1)}{5 - 1} &= \frac{25 - 25}{4} = 0 \text{ m/s} \end{aligned}$$

From $1$ s to $3$ s the ball rises at an average of $10$ m/s. From $3$ s to $5$ s it falls at an average of $10$ m/s (that’s what the negative sign means). From $1$ s to $5$ s the average rate is $0$: the ball is at the same height at both times. It certainly wasn’t standing still, which shows that an average can hide a lot of what happens in between.

Example 3: From an equation (exponential)

The amount of caffeine in your body $t$ hours after drinking a large coffee is modelled by $A(t) = 200(0.5)^{t/5}$ milligrams. Compare the average rate of change over the first $5$ hours with the rate over the next $5$ hours.

Solution. $A(0) = 200$, $A(5) = 200(0.5) = 100$, and $A(10) = 200(0.25) = 50$.

$$\frac{A(5) - A(0)}{5 - 0} = \frac{100 - 200}{5} = -20 \text{ mg/h}, \qquad \frac{A(10) - A(5)}{10 - 5} = \frac{50 - 100}{5} = -10 \text{ mg/h}$$

The caffeine level drops by an average of $20$ mg per hour at first, but only $10$ mg per hour over the next $5$ hours. With exponential decay, the rate of change shrinks as the amount shrinks.

Example 4: Sketching graphs from a story

Amira walks from home to the bus stop at a steady pace, waits a few minutes, then rides the bus. The bus speeds up, travels at a steady speed, then slows down and stops at her school. Sketch graphs of her distance from home and her speed against time.

Solution. Go through the story one part at a time, thinking about the rate of change of distance (which is her speed).

Part of the trip	Distance–time graph	Speed–time graph
walking steadily	straight, gentle upward slope	low horizontal segment
waiting	horizontal (distance not changing)	at $0$
bus speeding up	curving upward, getting steeper	rising
bus at steady speed	straight, steep upward slope	high horizontal segment
bus slowing to a stop	curving, levelling off	falling to $0$
at school	horizontal	at $0$

The slope of the distance graph (left) is the height of the speed graph (right).

Notice that the distance graph never goes down: Amira never moves back toward home.

Common mistakes

Subtracting in different orders. Keep the same order on the top and bottom: $\dfrac{f(b) - f(a)}{b - a}$. Mixing them up, like $\dfrac{f(b) - f(a)}{a - b}$, flips the sign.

Forgetting the units. A rate of change always has units: the $y$-units per the $x$-units. "$-20$" alone doesn’t tell you anything; ”$-20$ mg/h” does.

Dividing by the wrong change. The change in the independent variable goes on the bottom. For heights over time, divide the change in height by the change in time, not the other way round.

Thinking an average tells the whole story. An average rate of $0$ m/s from $t = 1$ to $t = 5$ in Example 2 doesn’t mean the ball stopped. It only compares the start and the end of the interval.

Using the height of the graph instead of its slope. On a distance–time graph, speed is the slope, not the height. A high horizontal segment means “far from home but not moving”.

Practice

1. (Warm-up) Is the rate of change zero, constant, or changing?

• (a) The distance travelled by a plane cruising at a steady $850$ km/h, against time.
• (b) The height of a book sitting on a shelf, against time.
• (c) The area of a circle, against its radius.
• (d) The balance of a savings account earning compound interest, against time.

Solution

(a) Constant: the distance goes up by $850$ km every hour.

(b) Zero: the height doesn’t change.

(c) Changing: $A = \pi r^2$ is a curve; the area grows faster as the radius gets larger.

(d) Changing: compound interest is exponential growth, so the balance grows faster over time.

2. (Warm-up) Find the average rate of change of $f(x) = 3x^2 - 2$ from $x = 1$ to $x = 3$.

Solution

$f(1) = 3 - 2 = 1$ and $f(3) = 27 - 2 = 25$.

$$\frac{f(3) - f(1)}{3 - 1} = \frac{25 - 1}{2} = 12$$

3. (Core) The population of a city, in thousands, is shown below.

Year	$2000$	$2005$	$2010$	$2015$	$2020$
Population (thousands)	$520$	$548$	$590$	$612$	$655$

• (a) Find the average rate of change from 2000 to 2020.
• (b) In which five-year period did the population grow fastest?

Solution

(a) $\dfrac{655 - 520}{2020 - 2000} = \dfrac{135}{20} = 6.75$ thousand people per year (about $6750$ people per year).

(b) The rates for each five-year period are $\frac{28}{5} = 5.6$, $\frac{42}{5} = 8.4$, $\frac{22}{5} = 4.4$, and $\frac{43}{5} = 8.6$ thousand per year. The fastest growth was from 2015 to 2020.

4. (Core) An invasive plant covers $A(t) = 40(1.5)^t$ square metres of a pond $t$ years after it was first noticed. Find the average rate of change over the first $2$ years and over the next $2$ years. Is the rate constant?

Solution

$A(0) = 40$, $A(2) = 40(2.25) = 90$, and $A(4) = 40(5.0625) = 202.5$.

$$\frac{90 - 40}{2} = 25 \text{ m}^2\text{/year}, \qquad \frac{202.5 - 90}{2} = 56.25 \text{ m}^2\text{/year}$$

The rate isn’t constant: it more than doubles. With exponential growth, the bigger the patch gets, the faster it spreads.

5. (Core) The depth of water in a harbour $t$ hours after high tide is modelled by $d(t) = 3\cos\left(\dfrac{\pi}{6}t\right) + 5$ metres, where the angle is in radians. Find the average rate of change of the depth from $t = 0$ to $t = 1$ and from $t = 2$ to $t = 3$. Round to three decimal places, and compare.

Solution

$d(0) = 3\cos 0 + 5 = 8$, $d(1) = 3\cos\frac{\pi}{6} + 5 = \frac{3\sqrt{3}}{2} + 5 \approx 7.598$, $d(2) = 3\cos\frac{\pi}{3} + 5 = 6.5$, and $d(3) = 3\cos\frac{\pi}{2} + 5 = 5$.

$$\frac{d(1) - d(0)}{1} \approx 7.598 - 8 = -0.402 \text{ m/h}, \qquad \frac{d(3) - d(2)}{1} = 5 - 6.5 = -1.5 \text{ m/h}$$

Both are negative because the tide is going out. Just after high tide the water level drops slowly (about $0.4$ m/h), but by $2$ to $3$ hours later it’s dropping much faster ($1.5$ m/h).

6. (Core) A ball is thrown upward from the top of a $10$ m cliff. Its height above the ground is $h(t) = -5t^2 + 15t + 10$ metres after $t$ seconds. Find the average rate of change of height over $[0, 1]$, $[1, 2]$, and $[2, 3]$, and interpret each result.

Solution

$h(0) = 10$, $h(1) = 20$, $h(2) = 20$, and $h(3) = 10$.

• $[0, 1]$: $\frac{20 - 10}{1} = 10$ m/s. The ball rises $10$ m in that second.
• $[1, 2]$: $\frac{20 - 20}{1} = 0$ m/s. It’s at the same height at both times: it went up, reached its peak, and came back down.
• $[2, 3]$: $\frac{10 - 20}{1} = -10$ m/s. It falls $10$ m in that second.

7. (Core) A skier rides a chairlift up a hill at a steady speed, waits at the top for two minutes, then skis down, speeding up as she goes, and finally slows to a stop at the bottom. Describe a graph of her height against time.

Solution

• Chairlift: a straight segment rising at a steady slope (constant positive rate of change of height).
• Waiting at the top: a horizontal segment (rate $0$).
• Skiing down and speeding up: the graph falls and gets steeper, curving downward (the rate is negative and getting more negative).
• Slowing to a stop: the graph keeps falling but levels off, ending in a horizontal segment at the bottom (height $0$ relative to the base).

8. (Challenge) The area of a circle is $A = \pi r^2$, with $r$ in centimetres.

• (a) Find the average rate of change of the area as $r$ goes from $1$ to $2$, from $2$ to $3$, and from $3$ to $4$. Leave your answers in terms of $\pi$.
• (b) Show that the average rate of change from $r$ to $r + 1$ is $\pi(2r + 1)$.

Solution

(a) Each interval has $\Delta r = 1$:

$$\frac{4\pi - \pi}{1} = 3\pi, \qquad \frac{9\pi - 4\pi}{1} = 5\pi, \qquad \frac{16\pi - 9\pi}{1} = 7\pi \quad \text{cm}^2\text{/cm}$$

The rate grows by $2\pi$ each time: the bigger the circle, the faster its area grows.

(b)

$$\frac{\pi(r + 1)^2 - \pi r^2}{(r + 1) - r} = \pi(r^2 + 2r + 1 - r^2) = \pi(2r + 1)$$

Check with $r = 2$: $\pi(5) = 5\pi$ ✓

9. (Challenge) For $f(x) = x^2 + kx$, the average rate of change from $x = 1$ to $x = 4$ is $9$. Find $k$.

Solution

$f(4) = 16 + 4k$ and $f(1) = 1 + k$.

$$\frac{(16 + 4k) - (1 + k)}{4 - 1} = \frac{15 + 3k}{3} = 5 + k$$

Set $5 + k = 9$, so $k = 4$.

Check: $f(x) = x^2 + 4x$ gives $f(4) = 32$ and $f(1) = 5$, and $\frac{32 - 5}{3} = 9$ ✓