Let’s now extend the idea of average rate of change to apply to quantities other than position.
So far we’ve used motion and everyday ideas to develop the idea of average rate of change of an object’s position over the interval $[t_1, t_2]$, which is its average velocity:
\begin{align*}
\text{average velocity}_{[t_1,\, t_2]} &= \frac{\text{change in position}}{\text{change in time}} \\[8px]
&= \frac{s(t_2)\,- s(t_1)}{t_2 – t_1}
\end{align*}
But we can generalize the definition of “average rate of change” to apply to any function:
\[\text{average rate of change} = \frac{\text{change in the function’s value}}{\text{interval over which the change occurs}}\]
Consider the function $f(x)$ on the interval $a \le x \le b$, which we denote $[a, b].$ The function’s initial output value for the interval is then $f(a),$ while its final output value is $f(b).$
The function’s average rate of change over that interval is then:
Graphically, if we plot $y = f(x)$ versus $x$, as shown, then we have
\[ \text{average rate of change}_{[a,\,b]} = \frac{f(b)\, -\, f(a)}{b\, -\, a} = \frac{\Delta y}{\Delta x}\]
and we can associate the average rate of change with the slope of the line that passes through the endpoints of interval, as shown below.
![Title text: Computing Average Rate of Change. Graph of y = f(x) versus x, with the points (a, f(a)) and (b, f(b)) labelled, and a secant line passes through them. Text says: average rate of change = f(b) - f(a), divided by b - a, which equals the slope of the line that passes through the interval's initial and final points. Text at the bottom says: Graphically, finding the average rate of change on the interval [a, b] means computing the slope of the secant line that passes through the interval's initial and final points.](/wp-content/uploads/Deriv_AvgRateofChangeSecant.png)
The units of a function’s average rate of change will always be
\[\frac{\text{units of the function’s output}}{\text{units of the input variable}}\]
For instance, average velocity is the average rate of change of the position-versus-time function. In our initial examples about average velocity, the position function was stated in “miles,” and the independent variable time t had units of “hours.” Hence the average velocity (average rate of change of the position function) had units of $\dfrac{\text{miles}}{\text{hour}}$. In a different scenario, the position function had units of “meters,” and the independent variable time t had units of “seconds.” Hence the average velocity in this scenario had units of $\dfrac{\text{meters}}{\text{second}}.$
Let’s consider a few examples to see how we apply the idea of “average rate of change” to various situations.
The volume of water V stored in a large tank is given by $V = g(t),$ where V is in gallons. A graph of the function is shown.
Solution.
(a)
As shown on the graph, the relevant points are (11:00am, 2500 gallons) and (8:00pm, 1000 gallons). Then from the definition of average rate of change:
\begin{align*}
\text{average rate of change}_{[a,\,b]} &= \frac{g(b)\, -\, g(a)}{b\, -\, a} \\[5px]
\text{average rate of change}_{[\text{11:00 am}, \,\text{8:00 pm}]} &= \frac{g(\text{8:00 pm}) \,-\, g(\text{11:00 am})}{\text{8:00 pm}\, -\, \text{11:00 am}} \\[5px]
&= \frac{1000 \text{ gallons}\, -\, 2500 \text{ gallons} }{9 \text{ hours}} \\[5px]
&= \frac{-1500 \text{ gallons} }{9 \text{ hours}} \\[5px]
&= -167 \text{ gallons/hour} \quad \cmark
\end{align*}
(b) The negative value indicates that the volume of water in the tank decreases over the interval, as shown on the graph: the amount of water in the tank at the end of the interval is less than it was at the beginning, and so the slope of the line connecting the two points is negative. $\, \cmark$
The temperature T of a potato at time t is given by $T = f(t)$, where T is in degrees Fahrenheit, and t is in minutes.
Solution.
(a) The room-temperature potato goes into the oven at 5:00pm, and comes out an hour later hotter than it was, so its temperature has increased over the interval. Hence its average rate of temperature change is positive.$\quad \cmark$
(b) By contrast, as soon as the potato comes out of the oven at 6:00pm its temperature starts to decrease, and so its temperature at the end of the interval is less than it was at the start. Hence its average rate of temperature change over this period is negative.$\quad \cmark$
(c) The temperate function $T = f(t)$ outputs a value in “degrees Fahrenheit,” while and the independent variable time t is in units of “minutes.” Hence the average rate of change has units of $\dfrac{\text{degrees Fahrenheit}}{\text{minute}}. \quad \cmark$
Let’s do a quick Check Question.
As you know from everyday experience, chemical transformations occur at different rates depending on the substances involved, their temperature, and the amounts present, among other factors. For example, a piece of iron sitting outside reacts with oxygen in the presence of moisture to form rust, a process that happens very slowly. (Too slowly to watch continuously!)
By contrast, steel wool (primarily iron) burns quite dramatically when lit. This reaction, also between iron and oxygen, happens at a much faster rate, making for dramatic pictures.
Let’s consider a particular, simpler process to think quantitatively about the rate at which a chemical reaction occurs: the gas dinitrogen pentoxide, $\ce{N2O5},$ decomposes at room temperature into $\ce{NO2}$ (very toxic) and $\ce{O2}$:
\[\ce{2N2O5(g) -> 4NO2(g) + O2(g)}\]
We can measure the concentration of $\ce{N2O5}$ we have present in a container as a function of time. Following tradition in Chemistry, we denote concentration by putting square brackets around the quantity: $\ce{[N2O5]}$ indicates the concentration in moles per liter (mol/L). Then the average rate of decomposition between initial time $t_i$ and final time $t_f$ is
\begin{align*}
\text{average reaction rate} &= \frac{\text{change in concentration}}{\text{change in time}} \\[8px]
&= \frac{\Delta \ce{[N2O5]}}{\Delta t} = \frac{\ce{[N2O5]}_f\, -\, \ce{[N2O5]}_i}{t_f\, -\, t_i}
\end{align*}
In the following Example, we’ll work through some calculations in this scenario.
An experiment measures the concentration of $\ce{N2O5}$ present in a container. The table shows the collected data. The plot shows the data as points, and also a curve that shows the concentration over a longer period of time.
| time (seconds, s) | Concentration $\ce{[N2O5]_}$ (moles per liter, mol/L) |
|---|---|
| 0 | 1.000 |
| 100 | 0.979 |
| 200 | 0.958 |
| 300 | 0.938 |
| 400 | 0.918 |
| 500 | 0.899 |
| 600 | 0.880 |
Solution.
(a) Since the concentration of $\ce{N2O5}$ decreases as time passes, we expect the reaction rate to be negative. $\quad \cmark$
(b) As there is less as less reactant in the container, the decomposition process will go slower and slower. Hence we expect the rate from t = 0 s to t = 100 s to be greater than the rate from t = 500 s to t = 600 s. $\cmark$
The greater rate is shown by the greater slope of the secant line in the graph for (c) below.
(c)
\begin{align*}
\text{average reaction rate}_{[0 \, \text{s}, \, 100 \, \text{s}]} &= \frac{\text{change in concentration from $t$
= 100 s to $t$ = 0 s}}{\text{change in time}} \\[8px]
&= \frac{\ce{[N2O5]_{t= 100 \, \text{s}}}\, -\, \ce{[N2O5]_{t = 0 \, \text{s}}}}{0 \, \text{s}\, -\, 100 \, \text{s}}
\\[8px]
&= \frac{0.979 \, \text{mol/L}\, -\, 1.000 \, \text{mol/L}}{100 \, \text{s}} \\[8px]
&= \frac{-0.021 \, \text{mol/L}}{100 \, \text{s}} = -0.00021 \, \tfrac{\text{mol/L}}{\text{s}} \quad \cmark
\end{align*}
This value equals the slope of the secant line that passes through the points of interest, as shown on the graph at the bottom of this solution.
(d)
\begin{align*}
\text{average reaction rate}_{[500 \, \text{s}, \, 600 \, \text{s}]} &= \frac{\text{change in concentration from
$t$ = 500 s to $t$ = 600 s}}{\text{change in time}} \\[8px]
&= \frac{\ce{[N2O5]_{t= 600 \, \text{s}}}\, -\, \ce{[N2O5]_{t = 500 \, \text{s}}}}{600 \, \text{s}\, -\, 500 \, \text{s}}
\\[8px]
&= \frac{0.0.880 \, \text{mol/L}\, -\, 0.899 \, \text{mol/L}}{100 \, \text{s}} \\[8px]
&= \frac{-0.019 \, \text{mol/L}}{100 \, \text{s}} = -0.00019 \, \tfrac{\text{mol/L}}{\text{s}} \quad \cmark
\end{align*}
This value equals the slope of the secant line that passes through the points of interest, as shown on the graph at the bottom of this solution.
(e) In both cases the rate is negative, as we expected from (a). And the rate we calculated in (c) is greater than that from the later time period in (d), so the rate is becoming slower as time passes as we expected from (b). The graphs below shows the concentration versus time. The average reaction rate for the two time intervals are shown, and equal the slope of the secant line that passes through the initial and final moments of interest.
Let’s consider some Practice Problems, some based on previous University exam questions.
This completes our examination of the average rate of change of any function.
With this knowledge in place, in the next Topic we’ll return to estimating a function’s instantaneous rate of change, but now with a more systematic approach than before, building off of the idea of average rate of change we developed above. It’s going to be super-interactive — so much so we’re calling it a lab. Really, this is all laying the groundwork for the concept of “the derivative,” one of the two important pillars of Calculus.
Matheno®
Berkeley, California
AP® is a trademark registered by the College Board, which is not affiliated with, and does not endorse, this site.
© 2014–2022 Matheno, Inc.
