B.3 Generalizing Average Rate of Change

Calculus Home Introductory Ideas B. Average Rate of Change; Better estimate for df/dx at x = a B.3 Generalizing Average Rate of Change

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:

Average Rate of Change of a Function
\begin{align*} \text{average rate of change}_{[a,\,b]} &= \frac{f(b)\, -\, f(a)}{b\, -\, a}\end{align*}

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.

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.

Example 1: Water tank empties

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.

Find the average rate of change of the tank’s water volume between 11:00 am and 8:00 pm.
Explain the physical significance of the positive or negative sign of your answer.

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. $\, \cmark$

Example 2: Potato heats and cools

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.

You place the room-temperature potato in a hot oven at 5:00 pm, and take it out at 6:00 pm. Is the average rate of change of $f(t)$ positive or negative over the interval from 5:00 pm – 6:00 pm? Explain.
After taking the potato out at 6:00 pm, you leave it on the counter until 6:20 pm. Is the average rate of change of $f(t)$ positive or negative over the interval from 6:00 pm – 6:20 pm? Explain.
What are the units of the average rate of change of $f(t)$ in this scenario?

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.

Check Question 1: Compare average rate of change for two functions

Application to Chemistry: Average Reaction Rates

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.

Example 3: Decomposition rate of dinitrogen pentoxide

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.

Concentration of $\ce{N2O5}$ versus 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

Do you expect the average reaction rate for this decomposition to be positive or negative? Why?
Do you expect the average reaction rate from t = 0 s to t = 100 s to be greater than, less than, or equal to the average rate from t = 500 s to t = 600 s? Why? [Hints: (1) Physically, as the container holds less and less $\ce{N2O5},$ what do you think happens to the rate? (2) Looking at the graph, picture the relevant two secant lines, which reflect that physical evolution.]
Find the average reaction rate from t = 0 s to t = 100 s.
This calculator can help with your calculation.
Find the average reaction rate from t = 500 s to t = 600 s.
This calculator can help with your calculation.
Do your answers to (c) and (d) correspond to your answers to (a) and (b)?

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.

Practice Problem #1: Population Growth of Frisco, Texas

The table (broken into two parts for space reasons) shows population data for the town of Frisco, Texas. [ Ref. ] \[ \begin{array}{|c||c|c|c|c|c|} \hline t \text{ (Year)} & 2009 & 2010 & 2011 & 2012 & 2013 \\ \hline P \text{ (Population)} & 108{,}387 & 116{,}064 & 121{,}680 & 125{,}640 & 129{,}850 \\ \hline \end{array}\] \[ \begin{array}{|c||c|c|c|c|c|c|c|c|c|c|c|} \hline t \text{ (Year)} & 2014 & 2015 & 2016 & 2017 & 2018 & 2019 \\ \hline P \text{ (Population)} & 137{,}460 & 145{,}430 & 152{,}790 & 161{,}530 & 172{,}673 & 183{,}173\\ \hline \end{array}\]

The average population rate of change for 2009-2011 compared to 2017-2019 is \begin{array}{ll} \text{(A) }13{,}293 \, \tfrac{\text{people}}{\text{year}};\; 21{,}643\, \tfrac{\text{people}}{\text{year}} && \text{(B) }6{,}647\, \tfrac{\text{people}}{\text{year}};\; 10{,}822 \, \tfrac{\text{people}}{\text{year}} \end{array} \[\text{(C) }230{,}067\, \tfrac{\text{people}}{\text{year}};\; 644{,}703 \, \tfrac{\text{people}}{\text{year}}\] \begin{array}{ll} \text{(D) }115{,}034\, \tfrac{\text{people}}{\text{year}};\; 172{,}342 \, \tfrac{\text{people}}{\text{year}} && \text{(E) }5{,}364 \, \tfrac{\text{people}}{\text{year}};\; 9{,}143 \, \tfrac{\text{people}}{\text{year}} \end{array}

Show/Hide Solution

The average population rate of change from 2009-2011 is \begin{align*} \text{average population rate of change}_{[2009,2011]} &= \frac{P(2011) – P(2009)}{2011 – 2009} \\[8px] &= \frac{121{,}680 – 108{,}387}{2} \\[8px] &= \frac{13{,}293}{2} = 6{,}647 \, \tfrac{\text{people}}{\text{year}} \quad \blacktriangleleft \end{align*} The average population rate of change from 2019-2017 is \begin{align*} \text{average population rate of change}_{[2017,2019]} &= \frac{P(2019) – P(2017)}{2017 – 2019} \\[8px] &= \frac{183{,}173 – 161{,}530}{2} \\[8px] &= \frac{21{,}643}{2} = 10{,}822\, \tfrac{\text{people}}{\text{year}} \quad \blacktriangleleft \end{align*} So comparing 2009-2011 to 2017-2019 we have \[6{,}647 \, \tfrac{\text{people}}{\text{year}};\; 10{,}822\, \tfrac{\text{people}}{\text{year}} \implies \quad\text{ (B)} \quad \cmark\]

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Practice Problem #2: Bacterial Growth

Stock image of bacteria growing in petri dish, held by gloved hands.

Let's consider a biological application. Sam is growing a sample of baceteria in a lab, and records in the table below how many bacteria are present every fifteen minutes. The graph shows a hypothetical growth curve for the number present as time passes.

Data points from table and hypothetical (idealized) growth curve

Time (minutes)	Number of bacteria
0	187
15	252
30	340
45	460
60	620
75	838
90	1131

The average growth rate for the first thirty minutes of the experiment is \begin{array}{lllll} \text{(A) }-5.1 \, \tfrac{\text{bacteria}}{\text{min}} && \text{(B) }10.5 \, \tfrac{\text{bacteria}}{\text{min}} && \text{(C) }4.4 \, \tfrac{\text{bacteria}}{\text{min}} && \text{(D) }5.1 \, \tfrac{\text{bacteria}}{\text{min}} && \text{(E) none of these} \end{array}

Show/Hide Solution

\begin{align*} \text{average rate of change}_{[0, \, 30 \, \text{min}]} &= \frac{\text{change in number over first 30 min}}{\text{30 min}} \\[8px] &= \frac{N(30) – N(0)}{30 – 0} \\[8px] &= \frac{340 – 187}{30} \\[8px] &= 5.1 \, \tfrac{\text{bacteria}}{\text{min}} \implies \quad\text{ (D)} \quad \cmark \end{align*} This average rate is equal to the secant line that passes through the points of interest on the bacterial growth curve, as shown.

Same graph, now with a secant line passing through the points (0, N(0)) and (30, N(30)).

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Practice Problem #3: Sphere Expands

The surface area of a sphere as a function of its radius is $A(r) = 4\pi r^2.$
Consider a sphere whose radius increases from $r_1$ to $r_2.$
Find the average rate of change of the sphere's surface area on the interval $[r_1, r_2].$

[Hint: This problem may suddenly seem more abstract. Don't let that throw you! Use the same approach as before, computing $A(r_1)$ and $A(r_2),$ and see where that takes you.]
\begin{array}{lllll} \text{(A) }\dfrac{r_1 + r_2}{2} && \text{(B) }4\pi \left(r_2 - r_1 \right) && \text{(C) }2\pi \left(r_1^2 + r_2^2 \right) && \text{(D) }4\pi \left(r_2 + r_1 \right) && \text{(E) none of these} \end{array}

Show/Hide Solution

\begin{align*} \text{average rate of change}_{[a,\,b]} &= \frac{f(b)\, -\, f(a)}{b\, -\, a} \\[8px] \text{average rate of change}_{[r_1,\,r_2]} &= \frac{A(r_2)\, – \,A(r_1)}{r_2\, -\, r_1} \\[8px] &= \frac{4\pi r_2^2\, -\, 4\pi r_1^2}{r_2\, -\, r_1} \\[8px] &= \frac{4\pi \left(r_2^2\, -\, r_1^2 \right)}{r_2\, -\, r_1} \end{align*} The preceding line completes the Calculus portion of this problem, so make sure you understand how to develop it for yourself. The remainder of the problem is algebraic simplification. Since $r_2^2\, -\, r_1^2 = \left(r_2 + r_1 \right)\left(r_2\, -\, r_1 \right)$, we can rewrite the right-hand side of the equation as: \begin{align*} \phantom{ \text{average rate of change}_{[r_1,\,r_2]}}&= \frac{4\pi \left(r_2 + r_1 \right)\cancel{\left(r_2\, -\, r_1 \right)}}{\cancel{r_2\, -\, r_1}} \\[8px] &= 4\pi \left(r_2 + r_1 \right) \implies \quad\text{ (D)} \quad \cmark \end{align*} Graphically, the average rate of change equals the slope of the secant line that passes through the two points of interest as shown below.

The curve A = 4*pi*r^2, and the secant line that passes through the two points of interest.

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Practice Problem #4: $f(x) = 30 - x^3$

Consider the function $f(x) = 30 - x^3.$

Find the average rate of change on the interval $1 \le x \le 3.$
The answer to (a) is a negative value. What does this tell you about the function's behavior on this interval?

Show/Hide Solution

(a) \begin{align*} \text{average rate of change}_{[1,\,3]} &= \frac{f(3)\, -\, f(1)}{3\, -\,1} \\[8px] &= \frac{\left(30\, -\, 3^3 \right)\, -\, \left(30\, -\, 1^3 \right)}{2} \\[8px] &= \frac{(30\, -\, 27)\, -\, (30\,-\,1)}{2} \\[8px] &= \frac{3\,-\, 29}{2} \\[8px] &=\frac{-26}{2} = -13 \quad \cmark \end{align*}

Graph of y = f(x) versus x, and the secant line that passes through the points of interest.

(b) The negative value indicates that the function’s value at the end of the interval is less than its value at the start: $f(3) \lt f(1).$ That is, the function decreases over this interval. The slope of the secant line connecting the two points it thus negative, as shown on the graph.

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Let’s consider a new physical situation to complete this topic. We’ll return to this scenario several times over the coming chapters. The following is an extended problem, with multiple parts—an extended investigation. As best you can, picture the physical situation in your mind, and connect the images with the graph and static pictures shown.

Practice Problem #5: Cone Fills

The cone to the right is being filled slowly with water. The function that relates the volume of water V this particular cone holds, and the height h of that water in the cone, is
\[V(h) = \frac{\pi}{12} h^3 \]
The cone has total height H = 20 cm.

Since the water's volume is a function of its height in the cone, we can think about the average rate of change of volume V with respect to the water's height h.

(a) First, let's consider how the average rate of change for the first centimeter of height, raising the water's level from the bottom at h = 0 cm to h = 1 cm, is greater than, less than, or equal to the average rate of change from the final centimeter, from h = 19 cm to h = 20 cm? To decide, imagine the amount of water you have to add add to get that increase at the bottom. Then, imagine how much water you have to add to get that final centimeter at the top. Which do you think requires the larger amount of water, $\Delta V,$ to be added to get the same 1-cm increase in height?

(b) Compute the average rate of change for the water's volume with respect to its height in the cone for the change from h = 0 to h = 1 cm.

(c) Now compute the average rate of change for the water's volume with respect to its height in the cone for the change from h = 19 cm to h = 20 cm.

(d) Are your results of (b) and (c) consistent with your reasoning from part (a)?

Show/Hide Solution

Solution SummarySolution (a) DetailSolution (b) DetailSolution (c) DetailSolution (d) Detail

(a) $\text{average rate of change}_{[0 \, \text{cm, } 1 \, \text{cm}]}

(b) $\frac{\pi}{12} \, \text{cm}^2 \approx 0.262 \, \text{cm}^2$

(c) $\frac{\pi}{12}(1{,}141) \, \text{cm}^2 \approx 298.7 \, \text{cm}^2$

(d) See solution detail.

We’re reasoning conceptually here. The question is asking to compare \[ \text{average rate of change}_{[0 \, \text{cm, } 1 \, \text{cm}]} = \frac{\Delta V_{0 \rightarrow 1 \, \text{cm} }}{1 \, \text{cm} – 0 \, \text{cm}} = \frac{\Delta V_{0 \rightarrow 1 \, \text{cm} }}{1 \, \text{cm}} \] and \[ \text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} = \frac{\Delta V_{19 \rightarrow 20 \, \text{cm} }}{20 \, \text{cm} – 19 \, \text{cm}} = \frac{\Delta V_{19 \rightarrow 20 \, \text{cm} }}{1 \, \text{cm}} \] So a different way to ask the question is: is $\Delta V_{0 \rightarrow 1 \, \text{cm}}$ greater than, less than, or equal to $\Delta V_{19 \rightarrow 20 \, \text{cm}}$?
To answer that, we can think about the physical situation. As you can imagine, and as the figure below shows (open to view), it takes only a tiny bit of water to increase the water’s height by the first centimeter, $0 \rightarrow 1$ cm, since the cone is so narrow there. By contrast, at the top the cone is quite wide, and so it takes a LOT more water to get that final centimeter in.

Open/close figure

The figure shows the graph of $V(h) = \frac{\pi}{12} h^3,$ along with the physical situations depicted for the first and final centimeters of water height-gain.

Graph of V(h) versus h, which is a cubic. From h=0 to 1, the curve barely rises at all; an accompanying figure shows that only a tiny tiny bit of water is added for this first centimeter, so delta-V from 0 to 1 in tiny. By contrast, the change in volume from h=19 to 20 is much larger, both on the graph and in the accompanying figure showing that 'disk' of added water. Delta-V from 19 to 20 is BIG.

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That is, \[ \Delta V_{0 \rightarrow 1 \, \text{cm}}

Let’s compute the average rate of change for the water’s volume with respect to its height from h = 0 to h = 1 cm: \[\text{average rate of change}_{[0 \, \text{cm, } 1 \, \text{cm}]} = \frac{V(1 \, \text{cm}) – V(0 \, \text{cm})}{1 \, \text{cm} – 0 \, \text{cm}}\] We’ll first compute the quantities in the numerator. Recall that $V(h) = \frac{\pi}{12} h^3:$ \[V(1 \, \text{cm}) =\frac{\pi}{12} (1 \, \text{cm})^3 = \frac{\pi}{12} \, \text{cm}^3 \quad \blacktriangleleft\] \[V(0 \, \text{cm}) =\frac{\pi}{12} (0 \, \text{cm})^3 = 0 \, \text{cm}^3 \quad \blacktriangleleft \] \begin{align*} \text{average rate of change}_{[0 \, \text{cm, } 1 \, \text{cm}]} &= \frac{V(1 \, \text{cm}) – V(0 \, \text{cm})}{1 \, \text{cm} – 0 \, \text{cm}} \\[8px] &= \frac{\frac{\pi}{12} \, \text{cm}^3 – 0}{1 \, \text{cm}} \\[8px] &= \frac{\pi}{12} \, \text{cm}^2 \approx 0.262 \, \text{cm}^2 \quad \cmark \end{align*} We can represent the value of this average rate of change as the slope of the secant line passing through the points that mark the beginning and end of the interval, $(0, V(0))$ and $(1, V(1)),$ as shown in the figure.

Now the average rate of change from h = 19 cm to h = 20 cm: \[\text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} = \frac{V(20 \, \text{cm}) – V(19 \, \text{cm})}{20 \, \text{cm} – 19 \, \text{cm}}\] Let’s compute the quantities in the numerator. Recall that $V(h) = \frac{\pi}{12} h^3:$ \[V(20 \, \text{cm}) =\frac{\pi}{12} (20 \, \text{cm})^3 = \frac{\pi}{12}8{,}000 \, \text{cm}^3 \quad \blacktriangleleft\] \[V(19 \, \text{cm}) =\frac{\pi}{12} (19 \, \text{cm})^3 = \frac{\pi}{12}6{,}859 \, \text{cm}^3 \quad \blacktriangleleft \] \begin{align*} \text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} &= \frac{V(20 \, \text{cm}) – V(19 \, \text{cm})}{20 \, \text{cm} – 19 \, \text{cm}} \\[8px] &= \frac{\frac{\pi}{12} (8{,}000 \, \text{cm}^3 – 6{,}859\, \text{cm}^3)}{1 \, \text{cm}} \\[8px] &= \frac{\pi}{12}(1{,}141) \, \text{cm}^2 \approx 298.7 \, \text{cm}^2 \quad \cmark \end{align*} We can represent the value of this average rate of change as the slope of the secant line passing through the points that mark the beginning and end of the interval, $(19, V(19))$ and $(20, V(20)),$ as shown in the figure.

Graph of V(h) versus h, with the points (19, V(19)) and (20, V(20)) highlighted. The secant line that passes through them has slope equal to the average rate of change for the interval, 298.7 cm^2.

We have found \[\text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} \approx 0.262 \, \text{cm}^2\] and \[ \text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} \approx 298.7 \, \text{cm}^2 \] which is consistent with our reasoning of part (a): \[ \text{average rate of change}_{[0 \, \text{cm, } 1 \, \text{cm}]} < \text{average rate of change}_{[19 \, \text{cm, } 20 \, \text{cm}]} \quad \cmark\]

(much less, actually)

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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.

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The Upshot

The average rate of change for a function on the interval $[a, b]$ is defined
\[ \text{average rate of change}_{[a,\,b]} = \frac{f(b)\, -\, f(a)}{b\, -\, a} = \frac{\Delta y}{\Delta x}\]
On a graph of the function $y=f(x)$ versus $x,$ the average rate of change equals the slope of the secant line that passes through the points $(a, f(a))$ and $(b, f(b)).$

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