# Mathematics

Chapter 3, “Introduction to Graphing” from Elementary Algebra Textbook, Second Edition, by the Department of Mathematics, College of the Redwoods, is available under a Creative Commons Attribution-Non Commercial 3.0 Unported license. © 2011, Department of Mathematics, College of the Redwoods.

http://mathrev.redwoods.edu/ElemAlgText/ElementaryAlgebra.pdf

3.3. RATES AND SLOPE 183

3.3 Rates and Slope

Let’s open this section with an application of the concept of rate.

You Try It!

EXAMPLE 1. An object is dropped from rest, then begins to pick up speed Starting from rest, an automobile picks up speed at a constant rate of 5 miles per hour every second (5 (mi/hr)/s). Sketch the graph of the speed of the object versus time.

at a constant rate of 10 meters per second every second (10 (m/s)/s or 10m/s2). Sketch the graph of the speed of the object versus time.

Solution: In this example, the speed of the object depends on the time. This makes the speed the dependent variable and time the independent variable.

Independent versus dependent. It is traditional to place the independent variable on the horizontal axis and the dependent variable on the vertical axis.

Following this guideline, we place the time on the horizontal axis and the speed on the vertical axis. In Figure 3.41, note that we’ve labeled each axis with the dependent and independent variables (v and t), and we’ve included the units (m/s and s) in our labels.

Next, we need to scale each axis. In determining a scale for each axis, keep two thoughts in mind:

1. Pick a scale that makes it convenient to plot the given data.

2. Pick a scale that allows all of the given data to fit on the graph.

In this example, we want a scale that makes it convenient to show that the speed is increasing at a rate of 10 meters per second (10 m/s) every second (s). One possible approach is to make each tick mark on the horizontal axis equal to 1 s and each tick mark on the vertical axis equal to 10 m/s.

t(s)

v(m/s)

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

Figure 3.41: Label and scale each axis. Include units with labels.

t(s)

v(m/s)

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

Figure 3.42: Start at (0, 0), then continuously move 1 right and 10 up.

184 CHAPTER 3. INTRODUCTION TO GRAPHING

Next, at time t = 0 s, the speed is v = 0 m/s. This is the point (t, v) = (0, 0) plotted in Figure 3.42. Secondly, the rate at which the speed is increasing is 10 m/s per second. This means that every time you move 1 second to the right, the speed increases by 10 m/s.

In Figure 3.42, start at (0, 0), then move 1 s to the right and 10 m/s up. This places you at the point (1, 10), which says that after 1 second, the speed of the particle is 10 m/s. Continue in this manner, continuously moving 1 s to the right and 10 m/s upward. This produces the sequence of points shown in Figure 3.42. Note that this constant rate of 10 (m/s)/s forces the graph of the speed versus time to be a line, as depicted in Figure 3.43.

t(s)

v(m/s)

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

Figure 3.43: The constant rate forces the graph to be a line.

t(s)

v(mi/hr)

0 1 2 3 4 5 6 7 8 910 0 5

10 15 20 25 30 35 40 45 50

Measuring the Change in a Variable

To calculate the change in some quantity, we take a difference. For example, suppose that the temperature in the morning is 40◦ F, then in the afternoon the temperature measures 60◦ F (F stands for Fahrenheit temperature). Then the change in temperature is found by taking a difference.

Change in temperature = Afternoon temperature−Morning temperature = 60◦ F− 40◦ F = 20◦ F

Therefore, there was a twenty degree increase in temperature from morning to afternoon.

3.3. RATES AND SLOPE 185

Now, suppose that the evening temperature measures 50◦ F. To calculate the change in temperature from the afternoon to the evening, we again subtract.

Change in temperature = Evening temperature−Afternoon temperature = 50◦ F− 60◦ F = −10◦F

There was a ten degree decrease in temperature from afternoon to evening.

Calculating the Change in a Quantity. To calculate the change in a quantity, subtract the earlier measurement from the later measurement.

Let T represent the temperature. Mathematicians like to use the symbolism ΔT to represent the change in temperature. For the change in temperature from morning to afternoon, we would write ΔT = 20◦ F. For the afternoon to evening change, we would write ΔT = −10◦ F.

Mathematicians and scientists make frequent use of the Greek alphabet, the first few letters of which are:

α, β, γ, δ, . . . (Greek alphabet, lower case)

A,B,Γ,Δ, . . . (Greek alphabet, upper case)

a, b, c, d, . . . (English alphabet)

Thus, the Greek letter Δ, the upper case form of δ, correlates with the letter ‘d’ in the English alphabet. Why did mathematicians make this choice of letter to represent the change in a quantity? Because to find the change in a quantity, we take a difference, and the word “difference” starts with the letter ‘d.’ Thus, ΔT is also pronounced “the difference in T .”

Important Pronunciations. Two ways to pronounce the symbolism ΔT .

1. ΔT is pronounced “the change in T .”

2. ΔT is also pronounced “the difference in T .”

Slope as Rate

Here is the definition of the slope of a line.

Slope. The slope of a line is the rate at which the dependent variable is chang- ing with respect to the independent variable. For example, if the dependent

186 CHAPTER 3. INTRODUCTION TO GRAPHING

variable is y and the independent variable is x, then the slope of the line is:

Slope = Δy

Δx

You Try It!

EXAMPLE 2. In Example 1, an object released from rest saw that its speedStarting from rest, an automobile picks up speed at a constant rate of 5 miles per hour every second (5 (mi/hr)/s). The constant rate forces the graph of the speed of the object versus time to be a line. Calculate the slope of this line.

increased at a constant rate of 10 meters per second per second (10 (m/s)/s or 10m/s2). This constant rate forced the graph of the speed versus time to be a line, shown in Figure 3.43. Calculate the slope of this line.

Solution: Start by selecting two points P (2, 20) and Q(8, 80) on the line, as shown in Figure 3.44. To find the slope of this line, the definition requires that we find the rate at which the dependent variable v changes with respect to the independent variable t. That is, the slope is the change in v divided by the change in t. In symbols:

Slope = Δv

Δt

t(s)

v(m/s)

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

P (2, 20)

Q(8, 80)

Figure 3.44: Pick two points to compute the slope.

t(s)

v(m/s)

0 1 2 3 4 5 6 7 8 9 10 0

10 20 30 40 50 60 70 80 90

100

P (3, 30)

Q(7, 70)

Figure 3.45: The slope does not de- pend on the points we select on the line.

Now, as we move from point P (2, 20) to point Q(8, 80), the speed changes from 20 m/s to 80 m/s. Thus, the change in the speed is:

Δv = 80m/s− 20m/s = 60m/s

3.3. RATES AND SLOPE 187

Similarly, as we move from point P (2, 20) to point Q(8, 80), the time changes from 2 seconds to 8 seconds. Thus, the change in time is:

Δt = 8 s− 2 s = 6 s

Now that we have both the change in the dependent and independent variables, we can calculate the slope.

Slope = Δv

Δt

= 60m/s

6 s

= 10 m/s

s

Therefore, the slope of the line is 10 meters per second per second (10 (m/s)/s or 10m/s2).

The slope of a line does not depend upon the points you select. Let’s try the slope calculation again, using two different points and a more compact presentation of the required calculations. Pick points P (3, 30) and Q(7, 70) as shown in Figure 3.45. Using these two new points, the slope is the rate at which the dependent variable v changes with respect to the independent variable t.

Slope = Δv

Δt

= 70m/s− 30m/s

7 s− 3 s =

40m/s

4 s

= 10 m/s

s

Again, the slope of the line is 10 (m/s)/s. Answer: 5 (mi/hr)/s

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