Rates and Slope

3.3. RATES AND SLOPE
3.3
149
Rates and Slope
1. Start by setting up a Cartesian Coordinate System, labeling and scaling each
axis. Plot the point (0, 10). This represents the fact that the initial velocity
is v = 10 m/s at time t = 0 s. Because the object accelerates at a rate of 5
m/s each second, start at the point (0,10), then every time you move 1 second
to the right, move 5 m/s upward. Do this for 5 consecutive points. Finally, as
evidenced from the initial points plotted on the left, the constant acceleration
of 5 m/s/s guarantees that the graph of the object’s velocity versus time will
be a line, shown on the right.
v(m/s)
v(m/s)
50
50
40
40
30
30
20
20
10
0
10
(0, 10)
0 1 2 3 4 5 6 7 8 9 10
t(s)0
(0, 10)
0 1 2 3 4 5 6 7 8 9 10
We know that every time the time increases by 1 second (Δt = 1 s), the object’s
speed increases by 5 m/s (Δv = 5 m/s). Thus, the slope of the line is:
Δv
Δt
5 m/s
=
1s
= 5 m/s/s
Slope =
Note that the slope of the line is identical to the rate at which the objects’s
velocity is increasing with respect to time.
3. Start by setting up a Cartesian Coordinate System, labeling and scaling
each axis. Plot the point (0, 90). This represents the fact that the David’s
initial distance from his brother is d = 90 feet at time t = 0 seconds. Because
David’s distance from his brother decreases at a constant rate of 10 feet per
second (10 ft/s), start at the point (0,90), then every time you move 1 second
to the right, move 10 ft downward. Do this for 5 consecutive points. Finally,
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t(s)
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as evidenced from the initial points plotted on the left, the decrease of David’s
distance from his brother at a constant rate of 10ft/s guarantees that the graph
of the David’s distance from his brother versus time will be a line, shown on
the right.
d(ft)
d(ft)
100 (0, 90)
100 (0, 90)
90
90
80
80
70
70
60
60
50
50
40
40
30
30
20
20
10
10
t(s)0
t(s)
0
0 1 2 3 4 5 6 7 8 9 10
0 1 2 3 4 5 6 7 8 9 10
We know that every time the time increases by 1 second (Δt = 1 s), David’s
distance from his brother decreases by 10 ft (Δd = −10 ft). Thus, the slope of
the line is:
Δd
Δt
−10 ft
=
1s
= −10 ft/s
Slope =
Note that the slope of the line is identical to the rate at which the David’s
distance from his brother is decreasing with respect to time. Note that the
minus sign indicates the the David’s distance from his brother is decreasing
with respect to time.
5. To calculate the slope, we’ll subtract the coordinates of point P (9, 0) from
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3.3. RATES AND SLOPE
151
the point Q(−9, 15).
Slope =
=
Δy
Δx
15 − 0
−9 − 9
15
−18
5
=−
6
=
Slope is the change in y
divided by the change in x.
Divide the difference in y
by the difference in x.
Simplify numerator and denominator.
Reduce to lowest terms.
7. To calculate the slope, we’ll subtract the coordinates of point P (0, 11) from
the point Q(16, −11).
Slope =
=
Δy
Δx
−11 − 11
16 − 0
−22
16
11
=−
8
=
Slope is the change in y
divided by the change in x.
Divide the difference in y
by the difference in x.
Simplify numerator and denominator.
Reduce to lowest terms.
9. To calculate the slope, we’ll subtract the coordinates of point P (11, 1) from
the point Q(−1, −1).
Slope =
=
Δy
Δx
−1 − 1
−1 − 11
−2
−12
1
=
6
=
Slope is the change in y
divided by the change in x.
Divide the difference in y
by the difference in x.
Simplify numerator and denominator.
Reduce to lowest terms.
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152
11. To calculate the slope, we’ll subtract the coordinates of point P (−18, 8)
from the point Q(3, −10).
Slope =
=
Δy
Δx
Slope is the change in y
divided by the change in x.
−10 − 8
3 − (−18)
−18
21
6
=−
7
=
Divide the difference in y
by the difference in x.
Simplify numerator and denominator.
Reduce to lowest terms.
13. To calculate the slope, we’ll subtract the coordinates of point P (−18, 10)
from the point Q(−9, 7).
Slope =
=
Δy
Δx
Slope is the change in y
divided by the change in x.
7 − 10
−9 − (−18)
−3
9
1
=−
3
=
Divide the difference in y
by the difference in x.
Simplify numerator and denominator.
Reduce to lowest terms.
15. Because each gridline on the x-axis represents 1 unit, Δx = 8. Because
each gridline on the y-axis represents 2 units, Δy = 12.
y
20
Δy = 12
0
Δx = 8
0
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x
10
3.3. RATES AND SLOPE
153
Therefore, the slope is:
Δy
Δx
12
=
8
3
=
2
Slope =
17. Because each gridline on the x-axis represents 1 unit, Δx = 6. Because
each gridline on the y-axis represents 2 units, Δy = −10.
y
10
Δx = 6
x
5
Δy = −10
−5
−10
Therefore, the slope is:
Δy
Δx
−10
=
6
5
=−
3
Slope =
19. First, sketch each of the lines passing through the given points.
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154
y
5
(0, 0)
(1, 3)
(1, 2)
(1, 1)
x
−5
5
−5
Next, calculate the slope of each line.
Slope through
and (1, 1).
(0, 0)
Δy
Δx
1−0
=
1−0
=1
Slope through
and (1, 2).
(0, 0)
Slope through
and (1, 3).
Δy
Δx
2−0
=
1−0
=2
m1 =
Δy
Δx
3−0
=
1−0
=3
m1 =
m1 =
Label each line with its slope.
y
5
m3 = 3
m2 = 2
m1 = 1
x
−5
5
−5
As the slope gets larger, the line gets steeper.
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(0, 0)
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155
21. First, set up a coordinate system on a sheet of graph paper. Label and scale
each axis, then plot the point P (−4, 0). Because the slope is Δy/Δx = −3/7,
start at the point P (−4, 0), then move 3 units downward and 7 units to the
right, reaching the point Q(3, −3). Draw the line through the points P and Q.
y
5
P (−4, 0)
−5
Δy = −3
5
x
Q(3, −3)
Δx = 7
−5
23. First, set up a coordinate system on a sheet of graph paper. Label and scale
each axis, then plot the point P (−3, 0). Because the slope is Δy/Δx = 3/7,
start at the point P (−3, 0), then move 3 units upward and 7 units to the right,
reaching the point Q(4, 3). Draw the line through the points P and Q.
y
5
Δx = 7
Q(4, 3)
Δy = 3
−5
P (−3, 0)
5
x
−5
25. First, set up a coordinate system on a sheet of graph paper. Label and scale
each axis, then plot the point P (−3, −3). Because the slope is Δy/Δx = 3/7,
start at the point P (−3, −3), then move 3 units upward and 7 units to the
right, reaching the point Q(4, 0). Draw the line through the points P and Q.
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CHAPTER 3. INTRODUCTION TO GRAPHING
156
y
5
Δx = 7
x
Q(4,
5 0)
−5
Δy = 3
P (−3, −3)
−5
27. First, set up a coordinate system on a sheet of graph paper. Label and scale
each axis, then plot the point P (−4, 3). Because the slope is Δy/Δx = −3/5,
start at the point P (−4, 3), then move 3 units downward and 5 units to the
right, reaching the point Q(1, 0). Draw the line through the points P and Q.
y
5
P (−4, 3)
Δy = −3
Q(1, 0)
−5
Δx = 5
5
x
−5
29. First, set up a coordinate system on a sheet of graph paper. Label and scale
each axis, then plot the point P (−1, 0). Because the slope is Δy/Δx = −3/4,
start at the point P (−1, 0), then move 3 units downward and 4 units to the
right, reaching the point Q(3, −3). Draw the line through the points P and Q.
Second Edition: 2012-2013