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This lesson will introduce a way to compare and quantify how fast one quantity changes with respect to another. This will allow a more in-depth study of the relationship between two variables and the classification of those variables according to their type of dependency. The ideas covered in this lesson are the foundation of many areas of modern mathematics.

Catch-Up and Review

Here are a few recommended readings before getting started with this lesson.

Variables
Table of Values
Linear Equations in Two Variables
Solution of an Equation
Coordinate Pair
Coordinate Plane
Line

Here are a few practice exercises before getting started with this lesson.

a Which of the following equations is a linear equation in two variables?

b Consider the linear equation in two variables

y = - 3 x + 5 .

Which of the following coordinate pairs is a solution to the equation?

c Take a look at the tables of values shown below.

Which of the tables represents the linear equation in two variables

y = 2 x + 4 ?

Challenge

Analyzing Dominika’s Speeds on Her Way to and From School

Dominika is curious about the amount of time she spends going to and from school. One day, she decides to measure the time it takes her to get to school on foot and compare it to the amount of time it takes her to arrive home by bus. She drew the following graph for her trip to school.

Graph representing Dominika's walk to school

a What was Dominika's average walking speed on the way to school?

b At

2 : 00 PM,

the school day was over and Dominika was waiting for the bus. After

9

minutes, the bus finally arrived, and she arrived back home at

2 : 12 PM .

Draw a graph similar to the one given in Part A, showing the distance from the school this time, and find Dominika's average speed when returning home.

c Was Dominika faster than the school bus? Explain.

Example

Exploring the Rate of Change in Different Scenarios

Diego is taking the bus back to his hometown. During the trip, he sees that there are signs by the side of the road that indicate the distance from the city.

He keeps track of some of the signs and notes the time taken to reach them. He puts the information together using a table of values.

Time (minutes)	Distance (kilometers)
$0$	$0$
$10$	$15$
$20$	$30$
$30$	$45$
$40$	$60$
$50$	$75$
$60$	$90$

a Considering the variables used in this context, what is the interpretation of the rate of change?

What is the rate of change between

10

and

20

minutes?

What is the rate of change between

0

and

60

minutes?

b When he arrived at his hometown, Diego found a magazine containing statistics for different cars. The following graph shows information about the performance of a new electric car, indicating the speed reached during the first few seconds of the car accelerating from zero.

Graph of a line showing three points: (0,0),(2,50), and (4,100)

Find the rate of change between

(0, 0)

and

(2, 50) .

Find the rate of change between

(2, 50)

and

(4, 100) .

c The magazine also included a linear equation in two variables describing the performance of a hybrid car. This time it refers to the amount of kilometers the car can travel depending on the amount of gallons of gas.

y = 50 x + 30

The

y -

variable represents the distance in kilometers that the car can travel and the

x -

variable the gas in gallons. Find the rate of change between

x = 1

and

x = 3 .

Find the rate of change between

x = 4

and

x = 7 .

Find the rate of change between

x = 1

and

x = 7 .

Hint

a In this case, the independent variable is the time, which is measured in seconds. On the other hand, the dependent variable is the distance, which is measured in meters.

b In a coordinate pair, the first coordinate represents the value of the independent variable, while the second coordinate represents the value of the dependent variable.

c A linear equation can be represented by a line on the coordinate plane. When the

x -

coordinate of a point on the line is known, its

y -

coordinate can be found by substituting the

x -

value into the equation of the line.

Solution

a First, the units of the rate of change will be determined and interpreted. Then, the rate of change will be calculated for the given interval.

Interpreting the Rate of Change

The rate of change is defined as the ratio of the change in the dependent variable to the change of the independent variable.

Rate of Change = \frac{Δ y}{Δ x}

Note that the distance covered by the car depends on the time elapsed. On the other hand, time passes by without relation to the the car and can take any non-negative value without restrictions. Therefore, the dependent variable in this case is the distance, and time is independent. Let

t

represent the time and

d

the distance.

Rate of Change = \frac{Δ d}{Δ t} = \frac{d _{2} - d _{1}}{t _{2} - t _{1}}

Recall that the unit of the rate of change is the ratio of the unit of the dependent variable to the unit of the independent variable.

Rate of Change Units = \frac{kilometers}{minute}

Therefore, the unit of the rate of change is

\frac{km}{min},

which quantifies kilometers traveled per minute. The rate of change can be interpreted as the speed of the bus.

\underline{Interpretation for the Rate of Change} The speed of the car during a specific period of time

Finding the Rate of Change Between Different Times

Now, by using the rate of change formula together with the information from the table, the rate of change of the car between

10

and

20

minutes will be calculated. Notice that the corresponding distances are

15

and

30

kilometers, respectively.

Rate of Change = \frac{d _{2} - d _{1}}{t _{2} - t _{1}}

SubstitutePoints

Substitute $(10, 15)$ & $(20, 30)$

Rate of Change = \frac{3 0 - 1 5}{2 0 - 1 0}

Evaluate

SubTerms

Subtract terms

Rate of Change = \frac{1 5}{1 0}

CalcQuot

Calculate quotient

Rate of Change = 1.5

The rate of change between

10

and

20

minutes is

1.5 \frac{km}{min} .

By following the same procedure it is possible to find the rate of change between the other points of interest. The following table summarizes these results for the rate of change corresponding to the required time intervals.

Rate of Change $= \frac{d _{2} - d _{1}}{t _{2} - t _{1}}$
Times	Substitute	Evaluate
$10 - 20$ min	$\frac{3 0 km - 1 5 km}{2 0 min - 1 0 min}$	$1.5 \frac{km}{min}$
$0 - 60$ min	$\frac{9 0 km - 0 km}{6 0 min - 0 min}$	$1.5 \frac{km}{min}$

b Similarly to Part A, the rate of change formula will be used to find the rate of change of the graph. In this case, the variables are the speed

s

and the time

t .

Rate of Change = \frac{Δ s}{Δ t}

Notice that in a coordinate pair, the first coordinate represents the value of the independent variable, while the second coordinate corresponds to the dependent variable. Using this information, the rate of change between the points

(0, 0)

and

(2, 50)

will be calculated for the given graph.

Rate of Change = \frac{s _{2} - s _{1}}{t _{2} - t _{1}}

SubstitutePoints

Substitute $(0, 0)$ & $(2, 50)$

Rate of Change = \frac{5 0 - 0}{2 - 0}

Evaluate

SubTerm

Subtract term

Rate of Change = \frac{5 0}{2}

CalcQuot

Calculate quotient

Rate of Change = 25

The acceleration in the first

2

seconds is

25 \frac{km}{hr \cdot s} .

By following the same procedure, the rate of change for the other pair of points can be found. The following table summarizes the results for the rate of change of the graph between the specified pair of points.

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(0, 0), (2, 50)$	$\frac{5 0 - 0}{2 - 0}$	$25$
$(2, 50), (4, 100)$	$\frac{1 0 0 - 5 0}{4 - 2}$	$25$

c In this case, before using the rate of change formula, the corresponding

x -

and

y -

values must first be found. This can be done by substituting the

x -

value of interest into the linear equation and evaluating the resulting expression on the right-hand side. This will be done for

x = 1

first to illustrate the procedure.

y = 50 x + 30

Substitute

$x = 1$

y = 50 (1) + 30

Evaluate

Multiply

y = 50 + 30

AddTerms

Add terms

y = 80

It was found that the

y -

value corresponding to

x = 1

y = 80 .

Therefore, the coordinate pair

(1, 80)

is a point on the graph of the line representing the given linear equation. By repeating this process, the specific points of interest can be found.

Linear Equation: $y = 2 x + 3$
$x$	Substitute	Point
$x = 1$	$y y = 50 (1) + 30 = 80$	$(1, 80)$
$x = 3$	$y y = 50 (3) + 30 = 180$	$(3, 180)$
$x = 4$	$y y = 50 (4) + 30 = 230$	$(4, 230)$
$x = 7$	$y y = 50 (7) + 30 = 380$	$(7, 380)$

Now that the points are known, all that is left to do is to repeat the same procedure as in Part B. The following table summarizes the results for the pair of points of interest.

Rate of Change $= \frac{y _{2} - y _{1}}{x _{2} - x _{1}}$
Points	Substitute	Evaluate
$(1, 80), (3, 180)$	$\frac{1 8 0 - 8 0}{3 - 1}$	$50$
$(4, 230), (7, 380)$	$\frac{3 8 0 - 2 3 0}{7 - 4}$	$50$
$(1, 80), (7, 380)$	$\frac{3 8 0 - 8 0}{7 - 1}$	$50$

It is interesting to note that the rate of change is the same between all the pairs of points used. This occurred in Parts A and B as well.

Pop Quiz

Practice Calculating the Rate of Change

The following applet shows different representations for the data of two variables. Practice finding the rate of change between the indicated values. If the answer is not an integer number, round it to two decimal places.

Interactive graph showing a table/ expression, or a graph

Discussion

Linear Relations and Constant Rates of Change

It is possible to use the information presented above to set a criterion for identifying data that follows a linear relation. Data sets that have a constant rate of change between consecutive points can be described by linear relations, and the rate of change is the slope of the line.