Calculate an average rate from a formula or table
A hiker's distance, hour by hour
A hiker walks a trail, and a table lists her distance from the start. It is 0 km at hour 0, 4 km at hour 1, 10 km at hour 3 and 16 km at hour 6. Each chalk dot is one row of the table.
Press Hour 1 to 3. The green leg is the rise, her gain in distance: \(10 - 4 = 6\) km. Green marks a positive change. The faint leg is the run, the time that passed: \(3 - 1 = 2\) hours.
The rise divided by the run is \(6 \div 2 = 3\), which is the slope of the orange line through the two dots. On average, she walked 3 km each hour.
Press Hour 1 to 6. The run is 5 hours and the rise is 12 km, so the rate is 2.4 km per hour. A different interval gives a different rate.
Any two inputs
The curve is now any function \(f\), with two inputs \(a < b\). The orange line through the points of the curve above them is a secant line.
The run is \(b - a\), the change in input. The rise is \(f(b) - f(a)\), the change in output. If the curve is lower at \(b\) than at \(a\), the rise is red, because the output decreased.
Move \(b\). The secant turns, but its slope is always the rise divided by the run. This slope is the average rate of change of \(f\) from \(a\) to \(b\).
Notation
The rise is the change in output, and the run is the change in input.
\[\begin{aligned} \text{rise} &= \cg{f(b)-f(a)} \\ \text{run} &= \cq{b-a} \end{aligned}\]The average rate of change from \(a\) to \(b\) is the rise divided by the run, which is the slope of the secant.
\[\cy{\frac{f(b)-f(a)}{b-a}}\]Its unit is the output unit per input unit, such as kilometers per hour.
With a formula, substitute \(a\) and \(b\) into \(f\). With a table, read \(f(a)\) and \(f(b)\) from it. Then divide by \(b - a\), not by the number of rows.
The interval must have a positive length, so \(b \neq a\).
When it fails
Press Move b onto a. The rise and the run both become 0, and \(0 \div 0\) has no value, so an average rate needs an interval of positive length.
Press Move b away. The slope changes with \(b\), so an average rate belongs to one named interval.
Example 1: from a formula
After \(t\) minutes, an oil spill has area \(A(t) = t^2 + 3t\) square meters. Find the average rate of change of the area from \(t = 1\) to \(t = 4\).
Substitute each end of the interval into the formula.
\[\begin{aligned} A(1)&=1^2+3\cdot 1=4 \\ A(4)&=4^2+3\cdot 4=28 \end{aligned}\]The rise is the change in area, and the run is the length of the interval.
\[\cg{28-4=24}, \qquad \cq{4-1=3}\]Divide the rise by the run. The area grows by 8 square meters per minute on average.
\[\cy{\frac{24}{3}=8}\]
Example 2: from a table with uneven gaps
A town's population, in thousands, was 40 in 2010, 49 in 2013, 58 in 2018 and 66 in 2020. Find the average rate of change from 2013 to 2020.
Subtract the 2013 population from the 2020 population.
\[\cg{66-49=17}\]A tempting wrong move: The table goes from 2013 to 2018 and then to 2020, which is two steps. Divide the rise by 2.
\[\cr{\frac{17}{2}=8.5}\]The orange secant spans the 7 years from 2013 to 2020. The red line rises 17 thousand in only 2 years, so it reaches 66 by 2015 and is much too steep.
Divide by the gap between the inputs, 2020 − 2013 = 7 years.
\[\cy{\frac{17}{7}\approx 2.43}\]On average, the population grows by about 2.43 thousand per year.
Practice
A bacteria count is \(N(t) = 3t^2 - t\) thousand cells after \(t\) hours. Find the average rate of change from \(t = 2\) to \(t = 5\).
Drag the second dot to \(t = 5\). The green leg is the change in \(N\), and the faint leg is the length of the interval.
Solution
The count is 10 at t = 2 and 70 at t = 5.
\[\begin{aligned} N(2)&=3\cdot 4-2=10 \\ N(5)&=3\cdot 25-5=70 \end{aligned}\]The rise is 60 thousand cells and the run is 3 hours.
\[\cg{60}, \qquad \cq{3}\]The average rate is 20 thousand cells per hour.
\[\cy{\frac{60}{3}=20}\]
Questions
Questions 1 and 2 use the graph. Question 3 needs a calculation.
The secant joins the points at \(x = 1\) and \(x = 4\). What is the average rate of change of \(f\) from \(x = 1\) to \(x = 4\)?
- 1
- 3
- −1
Answer
1
The secant now joins the points at \(x = 0\) and \(x = 2\). What is the average rate of change of \(f\) from \(x = 0\) to \(x = 2\)?
- −2
- 2
- −4
Answer
−2
Find the average rate of change of \(g(x) = \sqrt{x}\) from \(x = 1\) to \(x = 9\).
- \(\tfrac14\)
- 2
- \(\tfrac13\)
Answer
\(\tfrac14\)
Summary
You find the output at both ends and subtract.
You divide by the gap between the inputs, because the average rate is the slope of the secant.
The result tells you how fast a quantity changes on average over a whole interval.
This is one of four skills in topic 2.1 of Which & Why: AP Calculus.