# Activity A1 – Average Rate of Change ```ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
Page 1 of 3
Activity A1 – Average Rate of Change
The table and scatter plot at right shows the average gas prices in the United States for
various days in June 2004. Suppose we want to find the average change in gas prices
from the 1st day in June to the 27th day of June. The following calculation gives the
average change in gas prices from 6/1/04 to 6/27/04:
change in dollars per gallon 1.80  1.90 0.1


 0.00385
change in days
27  1
26
This tells us gas prices in the United States dropped at an average rate of 0.00385 dollars
per gallon per day from June 1st to June 27th. This average rate of change(RS2) is
represented by the straight line that connects the data point for June 1st to the data point
for June 27th . Now suppose that we want to calculate the average change in gas from the
21st day of June to the 27th day of June.
change in dollars per gallon 1.80  1.75 0.5


 0.0833
change in days
27  21
6
Thus, during this 6 day period, gas prices in the Unites States increased at an average rate
of 0.0833 dollars per gallon per day. This average rate of change(RS3) is represented by
the straight line that connects the data point for June 21st to the data point for June 27th.
Notice in the examples above, the formula for the average rate of change(RS4) is the
same formula for calculating the slope of a line.
1. a) Use the data set(RS5) to calculate the average rate of change in gas prices between
the 4th day of June and the 19th day of June.
b) Which of the following correctly interprets the answer from 1a)?
a) The gas prices in the United States increased at an average rate of 0.006
dollars per gallon per day from the 4th to the 19th day of June.
b) The gas prices in the United States decreased at an average rate of 0.006
dollars per gallon per day from the 4th to the 19th day of June.
c) The gas prices in the United States increased by 0.006 dollar per gallon
from the 4th day of June to the 19th day of June.
d) The gas prices in the United States decreased by 0.006 dollar per gallon
from the 4th day of June to the 19th day of June.
e) The gas prices in the Unites States didn’t change much from the 4th day of
June to the 19th day of June.
2. a) Use the data set(RS6) given at right to calculate the average rate of change in the
number of AIDS deaths from 1995 to 1997.
ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
Page 2 of 3
b) Which of the following correctly interprets the answer from 2a)?
a) The number of deaths from AIDS increased at an average rate of 14,000
deaths per year from 1995 to 1997.
b) The number of deaths from AIDS decreased at an average rate of 14,000
deaths per year from 1995 to 1997.
c) The number of deaths from AIDS decreased by 14 deaths from 1995 to
1997.
d) The number of deaths from AIDS increased by 14 deaths from 1995 to
1997.
e) The number of deaths from AIDS decreased at an average rate of 14 per
year from 1995 to 1997.
Suppose we want to predict the average rate of change in gas price from the 16th day of
June to the 24th day of June. Since these data points are not given in the chart, we can use
regression to find the quartic polynomial that fits the data(RS7). The quartic curve that
fits the data is g ( x)  0.000004 x 4  0.0002 x3  0.0028 x 2  0.019 x  1.9183 . We can use
g ( x ) to predict the average rate of change of gasoline prices from the 16th day of June to
the 24th day of June by calculating the slope of the line that contains (RS8) the points
16, g (16)  and  24, g (24)  .
m
g (24)  g (16) 1.6374  1.774 0.1366


 0.0171
24  16
8
8
So, from the 16th to the 24th day of June, gasoline prices decreased at an average rate of
0.0171 dollars per gallon per day.
3. a) Use the cubic curve, g ( x)  0.00003x3  0.0013x 2  0.0071x  1.8852 , which
represents gasoline prices in the Unites States in June 2004, to find the average rate
of change in gasoline prices from the 5th day of June to the 10th day of June.
b) Which of the following correctly interprets the answer from 3a)?
a) The gas prices in the United States increased at an average rate of dollars
per gallon per day from the 5th to the 10th day of June.
b) The gas prices in the United States decreased at an average rate of dollars
per gallon per day from the 5th to the 10th day of June.
c) The gas prices in the United States increased by dollar per gallon from
the 5th day of June to the 10th day of June.
d) The gas prices in the United States decreased by dollar per gallon from
the 5th day of June to the 10th day of June.
e) The gas prices in the Unites States didn’t change much from the 5th day of
June to the 10th day of June.
ACOW RATES OF CHANGE AND DERIVATIVES MODULE
Updated 5/28/2016
Page 3 of 3
Given a curve, the line that contains two points on the curve is known as the secant
line(RS9). Thus, a secant line is a geometric representation for the average rate of
change. We can use the secant/tangent applet(RS10) to help us make the connection
between the slope of the secant line and the average rate of change. Suppose the function
R( x)  0.23x3  1.5 x 2 for 0  x  5 represents a company’s revenue equation, in
thousands of dollars and x is the number of items sold (in hundreds). Using the applet at
right, type the company’s revenue equation in the f(x) cell and then click on graph. (Hint:
After the function has been graphed, click and drag the origin of the axes to the bottom
left portion of the display window.) Suppose we want to find the company’s average rate
of change in revenue from the sale of 200 to 400 items. Thus, we need to calculate the
slope of the secant line from x = 2 to x = 4. To have the applet calculate this, fill the cells
as shown below and then click on graph.
The applet shows the slope of the secant line is approximately 2.5599. This means that
when the company sales increase from 200 items to 400 items, the revenue increases at
an average rate of 2.5599 thousand dollars per hundred of items sold or \$25.60 dollars
per item. The calculation below finds the slope of the secant line using the average rate
of change formula. By rounding the answer the applet gave, we see that both answers are
the same.
R(4)  R(2) 9.28  4.16 5.12
m


 2.56
42
2
2
4. Use the secant/tangent applet to calculate the average rate of change in the company’s
revenue from the sale of 400 items to 550 items.
5. Which of the following expressions correctly calculates the average rate of change in
the company’s revenue from the sale of 400 items to 550 items?
7.1088  9.28
a) m 
5.5  4
7.1088  9.28
b) m 
4  5.5
7.1088  9.28
c) m 
550  400
7.1088  9.28
d) m 
400  550
5.5  4
e) m 
7.1088  9.28
```