BROCK UNIVERSITY MATHEMATICS MODULES
Module 12D1.4: Average Rates of Change
Author: Erin Kox
WWW
• What it is: For a function, an average rate of change is the change in the function’s value
divided by the change in value of its variable. Graphically, an average rate of change of
a function is the slope of a secant line joining two points on its graph.
• Why you need it: Rates of change are used everywhere in our lives, from the speed of
a car, to a city’s population growth, to pizza dough rising in the oven. If you know the
average rate of change of a quantity, you can make predictions about the future values
of the quantity.
• When to use it: Use it when you are interested in estimating the rate of change of a
quantity and you have experimental data, or when you have the graph of a function but
don’t know its formula.
PREREQUISITES
Before you tackle this module, make sure you have completed these modules:
Understanding Rates of Change D1.1, D1.2, D1.3
WARMUP
Before you tackle this module, make sure you can solve the following exercises. If you have
difficulties, please review the appropriate prerequisite modules.
(Answers below.)
Please refer to the following graph for the first three warm-up questions:
y
3
2
1
−3 −2 −1
1
1
2
3
x
1. For each interval, determine whether the function’s rate of change is positive or negative:
(a) −3 ≤ x ≤ 0 (b) 0 ≤ x ≤ 3
2. Without doing any calculations, determine at which x-value the function has a greater
rate of change.
(a) x = −2 (b) x = 2
3. Suppose this is a distance-time graph, and write a story that it could tell.
4. Find the slope between the following sets of points (this question does not refer to the
graph):
(a) (0, 0) and (1, 1) (b) (−5, 3) and (3, −3) (c) (4, −10) and (4, 10)
Answers: 1.(a) negative (b) positive 2.(b) The rate of change is greater at x = 2, because the slope of the
graph is positive at x = 2, and negative at x = −2. 3. Example: On Saturday, I wanted to go to the movies
so my older brother drove me. He was driving moderately fast. When we got to the movie, I realized I had
forgotten my wallet, so we immediately headed back home. Traffic was not very busy so we made it there pretty
quickly. (For this story, position y = 0 represents the movie theatre, and position y = 3 represents home; time
= 0 represents the moment of arrival at the movie theatre.) 4.(a) 1; (b) − 34 ; (c) 0
Introduction
The idea of rate of change is easiest to understand (and describe) when the independent variable
is time, and the dependent variable is any other quantity. In any situation where a quantity
changes in time, one of the most important questions is, “How fast is the change occurring?” In
such situations, the rate of change of the quantity tells us how fast the quantity is changing.
If the graph of the quantity is linear, the quantity’s rate of change is constant, and we can
calculate it just by determining the slope of the line. However, many important functions are
not linear; mathematical models frequently involve polynomials, exponentials, sinusoids, and many
other types of functions. How can we determine rate of change in these situations? The full answer
to this question involves learning some calculus concepts, which will be introduced in later modules.
In this module we’ll lay the foundation for calculating precise “instantaneous” rates of change
(that is, the rate of change at a certain moment of time) for nonlinear graphs by learning how to
calculate average rates of change over various time intervals. Graphically, an instantaneous rate of
change amounts to calculating the precise slope of a curve at a particular point (that’s the part
that involves calculus, which you’ll learn in subsequent modules). On the other hand, an average
rate of change amounts to determining the slope of a line (called a secant line) that joins two points
on a curve.
If time is not involved, the calculation procedures are identical, but the interpretation of rate
of change is a little more delicate. We can’t really say that rate of change represents “how fast a
quantity is changing” if the independent variable is not time. For example, imagine blowing up a
balloon, and plotting the volume of the balloon versus the pressure of the air inside it. If an average
rate of change (over a certain interval of pressure) is 0.5 mL/kPa, this means that for every change
in pressure of 1 kPa, the volume changes by 0.5 mL. For a “stretchier” balloon, the rate of change
would be greater, whereas for a stiffer balloon, the rate of change would be less.
Rates of change can be positive or negative, very large or very small. A function with a nonlinear
graph will have many different average rates of change, depending on which interval of values you
analyze. We will use tables of values, graphs, and equations to practice this concept.
The calculations of average rate of change are very similar to what you have done in Grades 9
2
and 10, but the key is to understand what an average rate of change means. You will interpret the
average rates of change that you calculate, make predictions, and create graphs of functions.
FOCUS QUESTION
To help you understand an important aspect of this lesson, focus your attention on this
question, which will be answered towards the end of the lesson.
Alice runs a landscaping business, so her earnings vary from day to day. At the
end of each day last week she made the following deposits in her bank account:
Monday $45, Tuesday $85, Wednesday $65, Thursday $30, and Friday $125. What
was the average rate of change in Alice’s bank balance last week?
Finding the average rate of change
As you have learned in previous grades, to determine the rate of change of a function whose
graph is a line, you just have to calculate its slope. You can find the slope, m, of any linear function
that includes the points (x1 , y1 ) and (x2 , y2 ) by calculating
m=
y2 − y1
rise
=
x2 − x1
run
EXAMPLE 1
For the function pictured below, determine the average rate of change over the intervals −3 ≤
x ≤ 0 and 0 ≤ x ≤ 3.
y
3
2
1
−3 −2 −1
1
2
3
x
SOLUTION
The endpoints of the graph over the first interval have coördinates (−3, 3) and (0, 0); for the
second interval, the coördinates are (0, 0) and (3, 3). Therefore,
Average rate of change over [−3, 0] =
y2 − y1
0−3
−3
=
=
= −1
x2 − x1
0 − (−3)
3
Average rate of change over [0, 3] =
y2 − y1
3−0
3
=
= =1
x2 − x1
3−0
3
The rate of change on the left part of the graph is −1; this means that as you walk along the
graph from x = −3 to x = 0, for every increase by 1 of the x-value, the y-value decreases by 1.
The rate of change on the right part of the graph is 1; this means that as you walk along the
graph from x = 0 to x = 3, for every increase by 1 of the x-value, the y-value increases by 1.
3
DEFINITION
An average rate of change for a function over an interval [x1 , x2 ] can be calculated by dividing
the net change in y-values by the net change in x-values.
Average Rate of Change =
y2 − y1
x2 − x1
Notice we are using the same formula you learned in Grade 9 for slope, but the difference is
that now we are applying it for any interval of x-values, whether the function has a constant or
changing slope over that interval.
EXAMPLE 2
y
3
2
1
−3 −2 −1
1
2
3
x
For the function in the graph, determine the average rate of change over the interval −3 ≤ x ≤ 3.
(Note that the function does not have a constant slope, but rather has two different slopes on
this interval.)
SOLUTION
The coördinates of the endpoints of the graph over this interval are (−3, 3) and (3, 3). Therefore,
Average rate of change over [−3, 3] =
y2 − y1
3−3
0
=
=
=0
x2 − x1
−3 − 3
−6
The average rate of change of the function over the interval [−3, 3] is zero because there is no
net change in y-values; both end points have the same y-value.
We can also find the average rate of change for the function in the previous example over any
other interval in its domain; say, −2 ≤ x ≤ 3:
Average rate of change over [−2, 3] =
y 2 − y1
3−2
1
=
=
x2 − x1
3 − (−2)
5
As you can see, we only care about the coordinates on the graph at the two endpoints of our
interval and disregard what happens in the middle.
Now let’s apply this understanding to find the average rate of change of a curve. A curve is
non-linear (not a straight line), which means that the slope is always changing for every value of
x. The average rate of change gives us an average value of the slope over the interval of interest.
4
Using a table of values, you can tell that a function is non-linear by noting that the first
differences are not constant. Consider the following table of values and the corresponding graph:
y
y = x2
4
3
2
1
−2 −1
Figure 1:
1
2
x
Figure 2:
EXAMPLE 3
Determine the average rate of change for the quadratic function in Figure 2 over the interval
0 ≤ x ≤ 2.
SOLUTION
Using the coördinates in the highlighted rows of the table,
Average rate of change over [0, 2] =
y2 − y1
4−0
4
=
= =2
x2 − x1
2−0
2
PRACTICE
(Answers below.)
1. Determine the average rate of change of the function in the previous example over the
intervals
(a) [−2, 1] (b) [−0.5, 1.5]
Answers:
1.(a) −1 (b) 1
5
Next we’ll determine an average rate of change for an exponential function.
EXAMPLE 4
Determine the average rate of change of the function y = 2x over the interval −3 ≤ x ≤ 5.
SOLUTION
We could draw a graph or make a table of values to solve this problem, but we can also use the
definition of rate of change. Let x1 = −3 and x2 = 5; then the corresponding y-values are
y1 = 2−3 = 0.125 and y2 = 25 = 32
Using our formula,
Average rate of change =
y2 − y1
32 − 0.125
31.875
=
=
≈ 3.98
x2 − x1
5 − (−3)
8
So, what does it mean? Graphically this means that the slope of the line joining the two points
on the graph (−3, 2−3 ) and (5, 25 ) is about 3.98. Another way of saying this is that the average
rate of change of y with respect to x for the given interval is about 3.98. That is, for the given
interval, the average change in y is about 3.98 times the corresponding change in x.
DISCUSSION PROBLEM
The following problem is open in the sense that there may be no definitive solution. Unlike
typical textbook exercises, real-life problems rarely have cut-and-dried solutions. Discuss this
problem with classmates or friends, then do your best to come up with a reasonable solution,
and be prepared to identify and defend the assumptions you make.
Using what you know about averages, why do you think it is called an average
rate of change?
What happens if you calculate the average of the two slopes from Example 1? Do you get the
same answer as we did for the average rate of change for −3 ≤ x ≤ 3 in Example 2?
What does this mean to you? When will this work? When will this not work?
Although we will not take this idea any further in this module, by digging into this you will
explore a concept that is fundamental to calculus!
Interpreting average rate of change
We’ve calculated some average rates of change and understood them graphically in terms of
slope. Now let’s consider how such average rates of change are interpreted physically.
Let’s begin with the example of average velocity. If I tell you I drove 200 kilometers in 2 hours,
what does this mean to you? What is my average rate of change?
Asked in this way, the question is too vague to answer, because we haven’t specified the average
change of something with respect to something else. To be more specific, the average velocity is
6
the average rate of change of position1 with respect to time. For my two-hour drive, the average
velocity is
Average rate of change of position =
200 − 0
change in position
y2 − y1
200
=
=
=
= 100 km/h
change in time
x2 − x1
2−0
2
Thus, the average velocity for my trip was 100 km/h. This represents the average rate of change
of position for my trip; on average, my position changed by 100 km each hour.
Of course, there are many possible motions that could have the same average velocity. Being
an average, the concept of average rate of change ignores the details of the motion and focuses
attention on a single quantity that gives a sense of the overall motion. For instance, it is possible
that I drove a steady 100 km/h for the whole 2 hours, but it’s also possible that I got stuck in
traffic for a bit, or maybe took a break to grab a coffee. This means that at some points in my
trip, my actual speed2 (as shown on my speedometer) would have been less than 100 km/h, and at
other times my actual rate of change of position with respect to time may have been greater than
100 km/h.
Such details of my motion are ignored by the average rate of change. Similar to the quadratic
function we looked at previously, my car trip very likely has a non-constant rate of change, and
100 km/h is just the average of all of the actual values for the speed as shown on my speedometer.
This means the graph for my trip could look like any of the following graphs, for example:
Do you see any similarities between these four graphs?
Notice that the end points of the motion for each graph are the same, beginning at (0, 0) and
ending at (2, 200) for each graph. This is because we have specified that I drove 200 km in 2 hours.
However, you can see that the motions represented by the four graphs are quite different from each
other. (Describe them in words!3 ) Yet they all have the same average velocity, because they all
cover 200 km in 2 hours.
We can also use our interpretation to try to make predictions about the future, which is called
extrapolation. From our example, knowing my average velocity was 100 km/h, how long would it
take me to travel 400 km? Notice that in order to do this calculation, you have to assume that
the average speed will be the same for the 400-km journey as it was for the 200-km journey. In
practice, one would have to think carefully about whether this is a reasonable assumption; in some
cases it might be, and in others it might not be.
1
Some books use “displacement” in place of position, but it amounts to the same thing.
There is an important distinction between speed and velocity, but we shall ignore it for now. This is related to
the distinction between change in position and distance travelled. These distinctions are carefully explained in a later
module.
3
Also identify where the law was broken!
2
7
EXAMPLE 5
For the graph in the figure, determine the average rate of change for each interval, and compare
the two: (a) −1 ≤ x ≤ 2, and (b) 1 ≤ x ≤ 2. (This is the same graph as was used in Example
3.)
y
y = x2
4
3
2
1
−2 −1
1
Figure 3:
x
2
Figure 4:
SOLUTION
For the time interval −1 ≤ x ≤ 2, the average rate of change is
Average rate of change =
rise
y2 − y1
4−1
3
=
=
= =1
run
x2 − x1
2 − (−1)
3
For the time interval 1 ≤ x ≤ 2, the average rate of change is
Average rate of change =
rise
y2 − y 1
4−1
3
=
=
= =3
run
x2 − x1
2−1
1
For the interval −1 ≤ x ≤ 2, the average rate of change is 1, whereas for the interval 1 ≤ x ≤ 2,
the average rate of change is 3. You can see that this is so by studying the slopes of the two
secant lines shown in the following graph.
y
y = x2
4
3
2
1
−2 −1
Figure 5:
1
2
x
Figure 6:
Can you explain why in the previous example the average rate of change is less for the time
8
interval −1 ≤ x ≤ 2 than it is for the time interval 1 ≤ x ≤ 2? It is important to notice that,
although the changes in y are the same, the changes in x differ. Relating this to velocity, we can
understand that in the longer time interval, it took more time to achieve the same net change in
position, resulting in a lower average velocity.
RECAP OF FOCUS QUESTION
Recall the focus question, which was asked earlier in the lesson.
Alice runs a landscaping business, so her earnings vary from day to day. At the
end of each day last week she made the following deposits in her bank account:
Monday $45, Tuesday $85, Wednesday $65, Thursday $30, and Friday $125. What
was the average rate of change in Alice’s bank balance last week?
SOLUTION
You can think of rate of change as a measure of net change divided by the time in which the
change occurred; this is the same as “rise-over-run.” Using this, we get
45 + 85 + 65 + 30 + 125
5
350
=
5
= 70
average rate of change of bank balance for the week =
This means that the average rate of change of Alice’s bank balance was $70 per day.
PRACTICE
(Answers below.)
2. Tell a story about a quantity that has (a) a positive average rate of change; (b) a negative
average rate of change; (c) a zero average rate of change.
3. Calculate the average rate of change for the graph on each of the indicated intervals.
a) −2 ≤ x ≤ 0 b) −1 ≤ x ≤ 3 c) −1 ≤ x ≤ 4 d) 0 ≤ x ≤ 4
4. Create a distance-time graph that displays the following story.
When I took my dog for a walk we went at total of 5 kilometers in 30 minutes. For the
first 15 minutes, Rufus had so much energy he ran the whole time. But that made me
tired so we stopped for a 5 minute break. After that, I kept him on a leash at a brisk
walking pace, so we went a further 1 kilometres in 10 minutes.
9
5. In the previous exercise, suppose that Rufus and I continue at the same walking pace for
an additional hour. How far do we travel in the additional hour at this walking pace?
6. An ocean’s water level y (as measured by a scale on the side of a pier) is described by
the function
1
y = x3 − 3x2 − x + 10
5
where x is the number of days after today (today is x = 0) and the water level is
measured in centimeters. What is the average rate of change in water level for (a) the
first five days, and (b) the first ten days? (c) Compare the results of Parts (a) and (b).
7. Explain why the average rate of change for a straight-line function is the same no matter
which interval is chosen.
Answers:
2. (a) House prices in St. Catharines have increased during the years 1990 to 2000.
(b) The
prices for electronic devices often decreases in the years after they are first marketed. (c) My car has been
sitting in the driveway for the past few days; it’s average rate of change of position is zero.
3. a) −2 b)
− 12 c) 0 d) 34 4. One possible answer is included here; your answer might be different and still satisfy the
conditions told in the story.
5. A further 6 km. 6.(a) 9.8 cm/day (b) 69.8 cm/day (c) Since the slope of the secant in Part (b) is so
greater than the slope of the secant in Part (a), the average rate of change is greater over the ten-day period
than it is over the first five-day period. 7. The average rate of change for a straight-line function is equal to
the slope of the line, which is the same for all parts of the line.
WWW
• What we did: In this module we learned how to calculate and interpret the average rate
of change for any curve given a table of values, graph, or equation of a function. You
should also be able to create your own graphs and make predictions about the future
when given an average.
• Why we did it: Average rates of change can be used in many aspects of our everyday
lives, as seen in the lesson. It is also a very important building block for calculus.
• What’s next: To build on this concept, towards beginning to learn calculus, study instantaneous rate of change, which amounts to the slope of a tangent to a graph.
10