Example: Computing an Average Rate of Change

Using the data in the table below, find the average rate of change of the price of gasoline between 2007 and 2009.

[latex]y[/latex]	2005	2006	2007	2008	2009	2010	2011	2012
[latex]C\left(y\right)[/latex]	2.31	2.62	2.84	3.30	2.41	2.84	3.58	3.68

Answer: In 2007, the price of gasoline was $2.84. In 2009, the cost was $2.41. The average rate of change is

[latex]\begin{array}{l}\frac{\Delta y}{\Delta x}=\frac{{y}_{2}-{y}_{1}}{{x}_{2}-{x}_{1}}\\\,\,\,\,\,\,\,\,\,\,=\frac{2.41-2.84}{2009 - 2007}\\\,\,\,\,\,\,\,\,\,\,=\frac{-0.43}{2\text{ years}}\\\,\,\,\,\,\,\,\,\,\,={-0.22}\text{ per year}\end{array}[/latex]

Analysis of the Solution

Note that a decrease is expressed by a negative change or "negative increase." A rate of change is negative when the output decreases as the input increases or when the output increases as the input decreases. The following video provides another example of how to find the average rate of change between two points from a table of values.

Example: Computing Average Rate of Change from a Graph

Given the function [latex]g\left(t\right)[/latex], find the average rate of change on the interval [latex]\left[-1,2\right][/latex].

Answer: At [latex]t=-1[/latex], the graph shows [latex]g\left(-1\right)=4[/latex]. At [latex]t=2[/latex], the graph shows [latex]g\left(2\right)=1[/latex]. The horizontal change [latex]\Delta t=3[/latex] is shown by the red arrow, and the vertical change [latex]\Delta g\left(t\right)=-3[/latex] is shown by the turquoise arrow. The output changes by –3 while the input changes by 3, giving an average rate of change of

[latex]\frac{1 - 4}{2-\left(-1\right)}=\frac{-3}{3}=-1[/latex]

Analysis of the Solution

Note that the order we choose is very important. If, for example, we use [latex]\frac{{y}_{2}-{y}_{1}}{{x}_{1}-{x}_{2}}[/latex], we will not get the correct answer. Decide which point will be 1 and which point will be 2, and keep the coordinates fixed as [latex]\left({x}_{1},{y}_{1}\right)[/latex] and [latex]\left({x}_{2},{y}_{2}\right)[/latex].

Example: Finding the Average Rate of Change of a Force

The electrostatic force [latex]F[/latex], measured in newtons, between two charged particles can be related to the distance between the particles [latex]d[/latex], in centimeters, by the formula [latex]F\left(d\right)=\frac{2}{{d}^{2}}[/latex]. Find the average rate of change of force if the distance between the particles is increased from 2 cm to 6 cm.

Answer: We are computing the average rate of change of [latex]F\left(d\right)=\frac{2}{{d}^{2}}[/latex] on the interval [latex]\left[2,6\right][/latex].

[latex]\begin{array}{l}\text{Average rate of change.}=\frac{F\left(6\right)-F\left(2\right)}{6 - 2}\\ \,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\frac{2}{{6}^{2}}-\frac{2}{{2}^{2}}}{6 - 2} \,\,\,\,\,\,\,\,\,\,\,\,\,\text{Simplify.} \\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\frac{2}{36}-\frac{2}{4}}{4}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{-\frac{16}{36}}{4}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Combine numerator terms.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=-\frac{1}{9}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Simplify.}\end{array}[/latex]

The average rate of change is [latex]-\frac{1}{9}[/latex] newton per centimeter.

Example: Finding an Average Rate of Change as an Expression

Find the average rate of change of [latex]g\left(t\right)={t}^{2}+3t+1[/latex] on the interval [latex]\left[0,a\right][/latex]. The answer will be an expression involving [latex]a[/latex].

Answer: We use the average rate of change formula.

[latex]\begin{array}{l}\text{Average rate of change}=\frac{g\left(a\right)-g\left(0\right)}{a - 0}\text{Evaluate.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{\left({a}^{2}+3a+1\right)-\left({0}^{2}+3\left(0\right)+1\right)}{a - 0}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Simplify.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{{a}^{2}+3a+1 - 1}{a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Simplify and factor.}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=\frac{a\left(a+3\right)}{a}\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\text{Divide by the common factor }a.\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,=a+3\end{array}[/latex]

This result tells us the average rate of change in terms of [latex]a[/latex] between [latex]t=0[/latex] and any other point [latex]t=a[/latex]. For example, on the interval [latex]\left[0,5\right][/latex], the average rate of change would be [latex]5+3=8[/latex].