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Finding and Interpreting Average Rates of Change

4.8.3 Finding and Interpreting Average Rates of Change

4.8.3 • Finding and Interpreting Average Rates of Change

Activity

The graphs show the populations of California and Texas over time.

Line graph showing the population growth of California and Texas from 1910 to 2010; Californias population rises more steeply, surpassing 35 million, while Texas grows steadily to just over 25 million.

1. Estimate the average rate of change in the population in California between 1970 and 2010. Be prepared to show your reasoning.

Compare your answer:

The population of California was about 20 million in 1970 and 37 million in 2010. The average rate of change is: 37 − 20 2010 − 1970 = 17 40 = 0.425 37 − 20 2010 − 1970 = 17 40 = 0.425 million per year.

2. In this situation, what does the rate of change mean for California?

Compare your answer:

On average, the population of California was growing by 425,000 people per year.

3. Estimate the average rate of change in the population in Texas between 1970 and 2010. Be prepared to show your reasoning.

Compare your answer:

The population of Texas was about 11 million in 1970 and 25 million in 2010. The average rate of change is 25 − 11 2010 − 1970 = 14 40 = 0.35 25 − 11 2010 − 1970 = 14 40 = 0.35 million per year.

4. In this situation, what does the rate of change mean for Texas?

Compare your answer:

On average, the population of Texas was growing by 350,000 people per year.

5. Which state's population grew more quickly between 1900 and 2000? Be prepared to show your reasoning.

Compare your answer:

California is growing more quickly. For example: If we connect each pair of points representing the population of each state in 1900 and 2000, the segment for California has a greater slope, so it has a greater average rate of change.

Video: Interpreting Average Rate of Change

Watch the following video to learn more about interpreting the average rate of change.

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Self Check

Brooke was racing bikes with a friend on the way to the park. A table of her distance traveled over time is shown.

Time (minutes) Distance (miles)
0 0
5 1.2
10 2.4
15 3.6
20 4.8
25 6.0
Brooke

What was Brooke’s average rate of change between 5 minutes and 20 minutes?

  1. The average rate of change is 4.17 minutes per mile.
  2. The average rate of change is 0.24 mile per minute.
  3. The average rate of change is 15 minutes per mile.
  4. The average rate of change is 3.6 miles per minute.

Additional Resources

Interpreting Slope

A Mars rover collected the following temperature data over 1.6 Martian days. A Martian day is called a Sol. Use the graph to answer the following questions.

Line graph titled GROUND AND AIR TEMPERATURE SENSOR showing ground and air temperature variations over time, with labeled lines for each and temperature in °C on the y-axis and Sol on the x-axis. The y-axis is from negative one hundred to twenty in increments of twenty. The x-axis is from ten to eleven point six in increments of point two. The air temperature and the ground temperature follow similar graph pattern. They remain steady from zero to 10.2 sol, increated from 10.2 to 10.6. They decrease from 10.6 to 11, they remain constand from 11 to 11.2, then increase from 11.2 to 11.6. At 11 sol the ground temperature is just under negative 60, the air temperature is just under negative 60.

Which had a greater average rate of change between 10.2 and 10.4 Sol, ground temperature or air temperature? What is the meaning of each rate change?

Since ground temperature has a greater slant or slope, it has a greater rate of change. Slope is another word for the rate of change of a graph.

To find the average rate of change, first you need to find Sol on the graph (which is what a day is called on Mars). Find the values of 10.2 and 10.4 on the x x -axis. Find the corresponding y y -values of the points for both the ground temperature and the air temperature graphs. By finding the average rate of change, you will be able to determine which one was greater.

Ground Temperature: ( 10.2 , − 90 ) ( 10.2 , − 90 ) ( 10.4 , − 20 ) ( 10.4 , − 20 )

Air Temperature: ( 10.2 , − 70 ) ( 10.2 , − 70 ) ( 10.4 , − 40 ) ( 10.4 , − 40 )

Substitute each set of points into the average rate of change formula:

Ground Temperature = − 20 − ( − 90 ) 10.4 − 10.2 = 70 0.2 = 350 − 20 − ( − 90 ) 10.4 − 10.2 = 70 0.2 = 350

Air Temperature = − 40 − ( − 70 ) 10.4 − 10.2 = 30 0.2 = 150 − 40 − ( − 70 ) 10.4 − 10.2 = 30 0.2 = 150

Remember that to interpret the meaning of the rate of change, you need to compare the axes. It is the change in the y y -axis units divided by the change in the x x -axis units, or c h a n g e i n t e m p e r a t u r e i n d e g r e e s c e l s i u s c h a n g e i n S o l c h a n g e i n t e m p e r a t u r e i n d e g r e e s c e l s i u s c h a n g e i n S o l . That means that the average rate of change of each can be interpreted as:

Ground Temperature’s average rate of change is 350 degrees celsius per Sol (or day on Mars).

Air Temperature’s average rate of change is 150 degrees celsius per Sol (or day on Mars).

Try it

Interpreting Slope

A Mars rover collected the following temperature data over 1.6 Martian days. A Martian day is called a Sol. Use the graph to answer the following questions.

Line graph titled GROUND AND AIR TEMPERATURE SENSOR showing ground and air temperature variations over time, with labeled lines for each and temperature in °C on the y-axis and Sol on the x-axis. The y-axis is from negative one hundred to twenty in increments of twenty. The x-axis is from ten to eleven point six in increments of point two. The air temperature and the ground temperature follow similar graph pattern. They remain steady from zero to 10.2 sol, increated from 10.2 to 10.6. They decrease from 10.6 to 11, they remain constand from 11 to 11.2, then increase from 11.2 to 11.6. At 11 sol the ground temperature is just under negative 60, the air temperature is just under negative 60.

Which had a greater average rate of change between 10.6 and 11 Sol, ground temperature or air temperature? What is the meaning of each rate change?

Here is how to find the average rate of change:

To find the average rate of change, first you need to find Sol on the graph (which is what a day is called on Mars). Find the values of 10.6 and 11 on the x x -axis. Find the corresponding y y -values of the points for both the ground temperature and air temperature graphs. By finding the average rate of change, you will be able to determine which one was greater.

Ground Temperature: ( 10.6 , − 3 ) ( 10.6 , − 3 ) ( 11 , − 85 ) ( 11 , − 85 )

Air Temperature: ( 10.6 , − 5 ) ( 10.6 , − 5 ) ( 11 , − 65 ) ( 11 , − 65 )

Substitute each set of points into the average rate of change formula:

Ground Temperature = − 85 − ( − 3 ) 11 − 10.6 = − 82 0.4 = − 205 − 85 − ( − 3 ) 11 − 10.6 = − 82 0.4 = − 205

Air Temperature = − 65 − ( − 5 ) 11 − 10.6 = − 60 0.4 = − 150 − 65 − ( − 5 ) 11 − 10.6 = − 60 0.4 = − 150

Remember that to interpret the meaning of the rate of change, you need to compare the axes. It is the change in the y y -axis units divided by the change in the x x -axis units, or c h a n g e i n t e m p e r a t u r e i n d e g r e e s c e l s i u s c h a n g e i n S o l c h a n g e i n t e m p e r a t u r e i n d e g r e e s c e l s i u s c h a n g e i n S o l . That means that each average rate of change of each can be interpreted as:

Ground temperature’s average rate of change is –205 degrees celsius per Sol (or day on Mars).

Air temperature’s average rate of change is –150 degrees celsius per Sol (or day on Mars).

The ground temperature had a greater rate of decrease per Sol.