Calculate an average rate
A falling rock
The curve shows how far a rock has fallen, in feet, \(t\) seconds after it is dropped. The two orange dots mark the start and the end of a stretch of time.
From \(t = 1\) to \(t = 3\), the rock falls \(144 - 16 = 128\) ft while \(3 - 1 = 2\) s pass. The green segment is the rise, the distance fallen, and the dashed segment is the run, the time that passes.
To cover 128 ft in 2 s at one steady speed, the rock would need \(128 \div 2 = 64\) ft/s. This speed is the average rate.
The orange line through the two dots is a secant line, a line through two points of a curve. Its slope is the rise over the run, so it equals the average rate.
Drag the end to the right. The rock speeds up as it falls, so the rise grows faster than the run and the secant becomes steeper.
Any quantity
The curve can show any quantity \(f\) against an input \(x\). Drag the end \(b\) to the right and to the left.
The rise is the change in the output, and the run is the change in the input. The rise is green when the output at \(b\) is higher than at \(a\), and red when it is lower.
Wherever \(b\) is, the average rate is the rise divided by the run, which is the slope of the secant line.
The average rate describes the whole stretch, not what the curve does between the dots.
Notation
The symbol \(\Delta\) means “change in”. The green rise is the change in the output, \(\cg{\Delta f}\), and the run is the change in the input, \(\Delta x\).
\[\begin{aligned} \cg{\Delta f} &= f(b) - f(a) \\ \Delta x &= b - a \end{aligned}\]The average rate of change is the change in the output divided by the change in the input.
\[\text{average rate} = \dfrac{\cg{f(b) - f(a)}}{b - a}\]The units are the output units per input unit, such as feet per second. A negative rate means that the quantity falls.
The inputs \(a\) and \(b\) must be different, because you cannot divide by a run of 0.
When it fails
When the end reaches the start, the rise and the run are both 0, and 0 divided by 0 has no value.
An average rate needs two different inputs. The next skills ask what happens when the two dots come close.
Example 1: the rock again
A rock falls \(d(t) = 16t^2\) feet in \(t\) seconds. Find its average rate from \(t = 2\) to \(t = 5\).
Find the distance at the start and at the end.
\[d(2) = 64, \quad d(5) = 400\]The rise is the distance fallen, and the run is the time that passes.
\[\cg{400 - 64 = 336 \text{ ft}}, \quad 5 - 2 = 3 \text{ s}\]Divide the rise by the run. The unit is feet per second.
\[\dfrac{336}{3} = \cy{112 \text{ ft/s}}\]
Example 2: a cooling drink
A drink cools. Its temperature is \(T(t) = 20 + 0.8(10 - t)^2\) degrees Celsius after \(t\) minutes. Find the average rate from \(t = 2\) to \(t = 5\).
The temperature is 71.2 °C at the start and 40 °C at the end. The drink cools, so the rise is negative.
\[\begin{aligned} &T(2) = 71.2, \quad T(5) = 40 \\ &\cr{40 - 71.2 = -31.2} \end{aligned}\]A tempting wrong move: Divide by the end time, 5.
\[\dfrac{-31.2}{5} = -6.24\]The red dashed segment on the axis has length 5 because it starts at t = 0, but the stretch starts at t = 2. The run of the triangle is the time that passes, 5 − 2 = 3 minutes.
Divide the rise by the run of 3 minutes. The rate is negative because the drink cools.
\[\dfrac{-31.2}{3} = \cr{-10.4 \text{ °C/min}}\]
Practice
The number of bacteria in a culture is \(N(t) = 100 + 4t^2\) thousand after \(t\) hours. Find the average rate from \(t = 1\) to \(t = 5\).
Drag the end to t = 5. The average rate is the rise of the secant line divided by its run.
Solution
Find the number at the start and at the end.
\[N(1) = 104, \quad N(5) = 200\]The rise is 96 thousand, and the run is 4 hours.
\[200 - 104 = 96, \quad 5 - 1 = 4\]Divide the rise by the run. The average rate is 24 thousand bacteria per hour.
\[\dfrac{96}{4} = 24\]
Questions
The graph shows the distance \(D(t)\) of a car from home, in kilometers, \(t\) hours after it leaves.
What is the average rate of the car from hour 0 to hour 2?
- 50 km/h
- 100 km/h
- 0.02 h/km
Answer
50 km/h
On which stretch is the average rate greatest?
- Hour 0 to hour 2
- Hour 2 to hour 4
- Hour 4 to hour 6
Answer
Hour 0 to hour 2
A town had 12,400 people in 2010 and 11,800 people in 2016. What is the average rate of change?
- −100 people per year
- 100 people per year
- −600 people per year
Answer
−100 people per year
Summary
You divide the change in the output by the change in the input, and you keep the sign and the units.
This works because the quotient is the rise over the run, which is the slope of the secant line.
Use the average rate when a problem asks how fast a quantity changed over a stretch of input.
This is one of four skills in topic 1.1 of Which & Why: AP Calculus.