Covers All 21 Chapters
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Chapter 1: FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS
2. Chapter 2: KEY CONCEPT: THE DERIVATIVE
3. Chapter 3: SHORT-CUTS TO DIFFERENTIATION
4. Chapter 4: USING THE DERIVATIVE
5. Chapter 5: KEY CONCEPT: THE DEFINITE INTEGRAL
6. Chapter 6: CONSTRUCTING ANTIDERIVATIVES
7. Chapter 7: INTEGRATION
8. Chapter 8: USING THE DEFINITE INTEGRAL
9. Chapter 9: SEQUENCES AND SERIES
10. Chapter 10: APPROXIMATING FUNCTIONS USING SERIES
11. Chapter 11: DIFFERENTIAL EQUATIONS
12. Chapter 12: FUNCTIONS OF SEVERAL VARIABLES
13. Chapter 13: A FUNDAMENTAL TOOL: VECTORS
14. Chapter 14: DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES
15. Chapter 15: OPTIMIZATION: LOCAL AND GLOBAL EXTREMA
16. Chapter 16: INTEGRATING FUNCTIONS OF SEVERAL VARIABLES
17. Chapter 17: PARAMETERIZATION AND VECTOR FIELDS
18. Chapter 18: LINE INTEGRALS
19. Chapter 19: FLUX INTEGRALS AND DIVERGENCE
20. Chapter 20: THE CURL AND STOKES’ THEOREM
21. Chapter 21: PARAMETERS, COORDINATES, AND INTEGRALS
, 1.1 SOLUTIONS 11
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since t represents the number of years since 1970, we see that f (35) represents the population of the city in
2005. In 2005, the city’s population was 12 million.
2. Since T = f ( P ), we see that f (200) is the value of T when P = 200; that is, the thickness of pelican eggs
when the concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of
workers increases, productivity also increases. As a result, the curve goes up initially. At a certain point the
curve reaches its highest level, after which it goes downward; in other words, as the number of workers
increases beyond that point, productivity decreases. This might, for example, be due either to the inefficiency
inherent in large organizations or simply to workers getting in each other’s way as too many are crammed on
the same line. Many other reasons are possible.
4. The slope is (1 − 0)/(1 − 0) = 1. So the equation of the line is y = x.
5. The slope is (3 − 2)/(2 − 0) = 1/2. So the equation of the line is y = (1/2)x + 2.
6. Using the points (−2, 1) and (2, 3), we have
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Now we know that y = (1/2)x + b. Using the point (−2, 1), we have 1 = −2/2 + b, which yields b = 2. Thus, the
equation of the line is y = (1/2)x + 2.
6−0
7. Slope = = 2 so the equation is y − 6 = 2(x − 2) or y = 2x + 2.
2 − (−1)
5 5
8. Rewriting the equation as y = − x + 4 shows that the slope is − and the vertical intercept is 4.
2 2
9. Rewriting the equation as
12 2
y =− x+
7 7
shows that the line has slope −12/7 and vertical intercept 2/7.
10. Rewriting the equation of the line as
−2
−y = x−2
14
y = x + 2,
2
we see the line has slope 1/2 and vertical intercept 2.
11. Rewriting the equation of the line as
12 4
y= x−
6 6
2
y = 2x − ,
3
we see that the line has slope 2 and vertical intercept −2/3.
12. (a) is (V), because slope is positive, vertical intercept is negative
(b) is (IV), because slope is negative, vertical intercept is positive
(c) is (I), because slope is 0, vertical intercept is positive
(d) is (VI), because slope and vertical intercept are both negative
(e) is (II), because slope and vertical intercept are both positive
(f) is (III), because slope is positive, vertical intercept is 0
,2 Chapter One /SOLUTIONS
13. (a) is (V), because slope is negative, vertical intercept is 0
(b) is (VI), because slope and vertical intercept are both
positive
(c) is (I), because slope is negative, vertical intercept is
positive
(d) is (IV), because slope is positive, vertical intercept is
negative
(e) is (III), because slope and vertical intercept are both 2
negative =− .
(f) is (II), because slope is positive, vertical intercept is 0
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
−3 6
Slope = =−
7.5 15 5
The y-intercept is at (0, 3), so a possible equation for the line is
2
y = − x + 3.
5
(Answers may
vary.)
15. y − c = m(x − a)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
29.2 − 27.8 1.4
Slope = = = 14.
5.3 − 5.2 0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
y − 27.8 = 14(x − 5.2)
which is equivalent
to y = 14x − 45.
1
17. y = 5x − 3. Since the slope of this line is 5, we want a line with slope −
5 passing through the point (2, 1). The
equation
is (y − 1) = − 1 (x − 2), or y = − 1 x + 7 .
5 5 5
18. The line y + 4x = 7 has slope −4. Therefore the parallel line has slope −4 and equation y − 5 = −4(x − 1) or
y−1= −4x + 9. The perpendicular line has slope = 1 and equation y − 5 = 1 (x − 1) or y = 0.25x + 4.75.
(−4) 4 4
19. The line parallel to y = mx + c also has slope m, so its equation is
y = m(x − a) + b.
The line perpendicular to y = mx + c has slope −1/m, so its equation will be
1
y = − (x − a) + b.
m
20. Since the function goes from x = 0 to x = 4 and between y = 0 and y = 2, the domain is 0 ≤ x ≤ 4 and the range
is
0 ≤ y ≤ 2.
21. Since x goes from 1 to 5 and y goes from 1 to 6, the domain is 1 ≤ x ≤ 5 and the range is 1 ≤ y ≤ 6.
22. Since the function goes from x = −2 to x = 2 and from y = −2 to y = 2, the domain is −2 ≤ x ≤ 2 and the range is
−2 ≤ y ≤ 2.
23. Since the function goes from x = 0 to x = 5 and between y = 0 and y = 4, the domain is 0 ≤ x ≤ 5 and the range
is
0 ≤ y ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since x2 ≥ 0 for all x.
1
25. The domain is all x-values, as the denominator is never zero. The range is 0 < y .
≤ 2
26. The value of f (t) is real provided t 2 − 16 ≥ 0 or t2 ≥ 16. This occurs when either t ≥ 4, or t ≤ −4. Solving f (t) = 3,
we have
p
t2 − 16 = 3
t2 − 16 = 9
t2 = 25
, 1.1 SOLUTIONS 31
so
t = ±5.
4
27. We have V = kr3 . You may know that V =
πr 3.
3
d
28. If distance is d, then v = .
t
29. For some constant k, we have S = kh2 .
30. We know that E is proportional to v3 , so E = kv3 , for some constant k.
31. We know that N is proportional to 1/l2 , so
k
N= , for some constant
k. l2
Problems
32. The year 1983 was 25 years before 2008 so 1983 corresponds to t = 25. Thus, an expression that represents the
statement is:
f(25) = 7.019
33. The year 2008 was 0 years before 2008 so 2008 corresponds to t = 0. Thus, an expression that represents the
statement is:
f (0) meters.
34. The year 1965 was 2008 − 1865 = 143 years before 2008 so 1965 corresponds to t = 143. Similarly, we see
that the year 1911 corresponds to t = 97. Thus, an expression that represents the statement is:
f(143) = f(97)
35. Since t = 1 means one year before 2008, then t = 1 corresponds to the year 2007. Similarly, t = 0
corresponds to the year 2008. Thus, f (1) and f (0) are the average annual sea level values, in meters, in
2007 and 2008, respectively. Because 1 millimeter is the same as 0.001 meters, an expression that
represents the statement is:
f (0) = f (1) + 0.001.
Note that there are other possible equivalent expressions, such as: f (1) − f (0) = 0.001.
36. (a) Each date, t, has a unique daily snowfall, S, associated with it. So snowfall is a function of date.
(b) On December 12, the snowfall was approximately 5 inches.
(c) On December 11, the snowfall was above 10 inches.
(d) Looking at the graph we see that the largest increase in the snowfall was between December 10 to
December 11.
37. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in
Figure 1.1:
V (thousand dollars)
(5, 6)
a (years
Figure 1.1
(c) The vertical intercept is the value of V when a = 0, or the value of the car when it is new. The horizontal
intercept is the value of a when V = 0, or the age of the car when it is worth nothing.
,4 Chapter One /SOLUTIONS
38. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home.
(b) The story in (b) matches Graph (II), the flat tire story. Note the long period of time during which the
distance from home did not change (the horizontal part).
(c) The story in (c) matches Graph (III), in which the person started calmly but sped up later.
The first graph (I) does not match any of the given stories. In this picture, the person keeps going away
from home, but his speed decreases as time passes. So a story for this might be: I started walking to school at
a good pace, but since I stayed up all night studying calculus, I got more and more tired the farther I walked.
39. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30
minutes was 10◦ C. So, 30 minutes after the object was placed outside, it had cooled to 10 ◦ C.
(b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put
outside, it had a temperature of a◦C. The intercept b measures the value of t when f (t) = 0. In other
words, at time b the object’s temperature is 0 ◦C.
40. (a) The height of the rock decreases as time passes, so the graph falls as you move from left to right. One
possibility is shown in Figure 1.2.
s (meters)
t (sec)
Figure 1.2
(b) The statement f (7) = 12 tells us that 7 seconds after the rock is dropped, it is 12 meters above the ground.
(c) The vertical intercept is the value of s when t = 0; that is, the height from which the rock is dropped. The
horizontal intercept is the value of t when s = 0; that is, the time it takes for the rock to hit the ground.
41. (a) We find the slope m and intercept b in the linear equation C = b + mw. To find the slope m, we use
∆C 12.32 − 8
m= = = 0.12 dollars per gallon.
∆w 68 − 32
We substitute to find
b:
C = b + mw
8 = b + (0.12)(32)
b = 4.16 dollars.
The linear formula is C = 4.16 + 0.12w.
(b) The slope is 0.12 dollars per gallon. Each additional gallon of waste collected costs 12 cents.
(c) The intercept is $4.16. The flat monthly fee to subscribe to the waste collection service is $4.16. This is the
amount charged even if there is no waste.
42. We are looking for a linear function y = f (x) that, given a time x in years, gives a value y in dollars for the value
of the refrigerator. We know that when x = 0, that is, when the refrigerator is new, y = 950, and when x = 7,
the refrigerator is worthless, so y = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope
is then given by
950
m=
−7
It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the
point (7, 0) and use the point-slope formula:
y − y1 = m(x − x1).
So,
950
y − 0 = − 7 (x − 7)
950
y=− x + 950.
7
, 1.1 SOLUTIONS 51
43. (a) The first company’s price for a day’s rental with m miles on it is C1(m) = 40 + 0.15m. Its competitor’s price
for a day’s rental with m miles on it is C2 (m) = 50 + 0.10m.
(b) See Figure 1.3.
C (cost in dollars)
150 C 1(m) = 40 + 0.15m
100
50 C 2(m) = 50 + 0.10m
0 m (miles)
200 400 600 800
Figure 1.3
(c) To find which company is cheaper, we need to determine where the two lines intersect. We let C1 = C2, and
thus
40 + 0.15m = 50 + 0.10m
0.05m = 10
m = 200.
If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles
a day, the first company is cheaper.
∆$
44. (a) Charge per cubic foot = 55 − 40 = $0.025/cu. ft.
∆ cu. ft. = 1600 −1000
Alternatively, if we let c = cost, w = cubic feet of water, b = fixed charge, and m = cost/cubic feet, we
obtain
c = b + mw. Substituting the information given in the problem, we have
40 = b + 1000m
55 = b + 1600m.
Subtracting the first equation from the second yields 15 = 600m, so m = 0.025.
(b) The equation is c = b + 0.025w, so 40 = b + 0.025(1000), which yields b = 15. Thus the equation is c
= 15+0.025w.
(c) We need to solve the equation 100 = 15 + 0.025w, which yields w = 3400. It costs $100 to use 3400 cubic
feet of water.
45. See Figure 1.4.
driving speed
time
Figure 1.4
46. See Figure
1.5. distance driven
time
Figure 1.5
,6 Chapter One /SOLUTIONS
47. See Figure 1.6.
distance from exit
time
Figure 1.6
48. See Figure 1.7.
distance between cars
distance driven
Figure 1.7
49. (a) (i) f (1985) = 13
(ii) f (1990) = 99
(b) The average yearly increase is the rate of change.
f(1990) − f(1985) 99 − 13
Yearly increase = = = 17.2 billionaires per year.
1990 −1985 5
(c) Since we assume the rate of increase remains constant, we use a linear function with slope 17.2
billionaires per year. The equation is
f (t) = b + 17.2t
where f (1985) = 13, so
13 = b + 17.2(1985)
b = −34,129.
Thus, f(t) = 17.2t − 34,129.
50. (a) The largest time interval was 2008–2009 since the percentage growth rate increased from −11.7 to 7.3 from
2008 to 2009. This means the US consumption of biofuels grew relatively more from 2008 to 2009 than
from 2007 to 2008. (Note that the percentage growth rate was a decreasing function of time over 2005–
2007.)
(b) The largest time interval was 2005–2007 since the percentage growth rates were positive for each of
these three consecutive years. This means that the amount of biofuels consumed in the US steadily
increased during the three year span from 2005 to 2007, then decreased in 2008.
51. (a) The largest time interval was 2005–2007 since the percentage growth rate decreased from −1.9 in 2005 to
−45.4 in
2007. This means that from 2005 to 2007 the US consumption of hydroelectric power shrunk relatively more
with
each successive year.
(b) The largest time interval was 2004–2007 since the percentage growth rates were negative for each of
these four consecutive years. This means that the amount of hydroelectric power consumed by the US
industrial sector steadily decreased during the four year span from 2004 to 2007, then increased in 2008.
52. (a) The largest time interval was 2004–2006 since the percentage growth rate increased from −5.7 in 2004 to
9.7 in 2006. This means that from 2004 to 2006 the US price per watt of a solar panel grew relatively
more with each successive year.
(b) The largest time interval was 2005–2006 since the percentage growth rates were positive for each of
these two consecutive years. This means that the US price per watt of a solar panel steadily increased
during the two year span from 2005 to 2006, then decreased in 2007.
, 1.1 SOLUTIONS 71
53. (a) Since 2008 corresponds to t = 0, the average annual sea level in Aberdeen in 2008 was 7.094 meters.
(b) Looking at the table, we see that the average annual sea level was 7.019 fifty years before 2008, or in the
year 1958. Similar reasoning shows that the average sea level was 6.957 meters 125 years before 2008, or
in 1883.
(c) Because 125 years before 2008 the year was 1883, we see that the sea level value corresponding to the
year 1883 is
6.957 (this is the sea level value corresponding to t = 125). Similar reasoning yields the table:
Year 1883 1908 1933 1958 1983 2008
S 6.957 6.938 6.965 6.992 7.019 7.094
54. (a) We find the slope m and intercept b in the linear equation S = b + mt. To find the slope m, we use
∆S 66 − 113
m= = = −0.94.
∆t 50 − 0
When t = 0, we have S = 113, so the intercept b is 113. The linear formula is
S = 113 − 0.94t.
(b) We use the formula S = 113 − 0.94t. When S = 20, we have 20 = 113 − 0.94t and so t = 98.9. If this
linear model were correct, the average male sperm count would drop below the fertility level during the
year 2038.
55. (a) This could be a linear function because w increases by 5 as h increases by 1.
(b) We find the slope m and the intercept b in the linear equation w = b + mh. We first find the slope m using the
first two points in the table. Since we want w to be a function of h, we take
∆w 171 − 166
m= = = 5.
∆h 69 − 68
Substituting the first point and the slope m = 5 into the linear equation w = b + mh, we have 166 = b +
(5)(68), so b = −174. The linear function is
w = 5h − 174.
The slope, m = 5, is in units of pounds per
inch.
(c) We find the slope and intercept in the linear function h = b + mw using m = ∆h/∆w to obtain the linear
function
h = 0.2w + 34.8.
Alternatively, we could solve the linear equation found in part (b) for h. The slope, m = 0.2, has units
inches per pound.
56. We will let
T = amount of fuel for take-
off, L = amount of fuel for
landing,
P = amount of fuel per mile in the
air, m = the length of the trip in miles.
Then Q, the total amount of fuel needed, is given by
Q(m ) = T + L + P m.
57. (a) The variable costs for x acres are $200x, or 0.2x thousand dollars. The total cost, C (again in thousands of
dollars), of planting x acres is:
C = f(x) = 10 + 0.2x.
This is a linear function. See Figure 1.8. Since C = f (x) increases with x, f is an increasing function of x.
Look at the values of C shown in the table; you will see that each time x increases by 1, C increases by
0.2. Because C increases at a constant rate as x increases, the graph of C against x is a line.
, 8 Chapter One /SOLUTIONS
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
SOLUTIONS MANUAL
,TABLE OF CONTENTS
1. Chapter 1: FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS
2. Chapter 2: KEY CONCEPT: THE DERIVATIVE
3. Chapter 3: SHORT-CUTS TO DIFFERENTIATION
4. Chapter 4: USING THE DERIVATIVE
5. Chapter 5: KEY CONCEPT: THE DEFINITE INTEGRAL
6. Chapter 6: CONSTRUCTING ANTIDERIVATIVES
7. Chapter 7: INTEGRATION
8. Chapter 8: USING THE DEFINITE INTEGRAL
9. Chapter 9: SEQUENCES AND SERIES
10. Chapter 10: APPROXIMATING FUNCTIONS USING SERIES
11. Chapter 11: DIFFERENTIAL EQUATIONS
12. Chapter 12: FUNCTIONS OF SEVERAL VARIABLES
13. Chapter 13: A FUNDAMENTAL TOOL: VECTORS
14. Chapter 14: DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES
15. Chapter 15: OPTIMIZATION: LOCAL AND GLOBAL EXTREMA
16. Chapter 16: INTEGRATING FUNCTIONS OF SEVERAL VARIABLES
17. Chapter 17: PARAMETERIZATION AND VECTOR FIELDS
18. Chapter 18: LINE INTEGRALS
19. Chapter 19: FLUX INTEGRALS AND DIVERGENCE
20. Chapter 20: THE CURL AND STOKES’ THEOREM
21. Chapter 21: PARAMETERS, COORDINATES, AND INTEGRALS
, 1.1 SOLUTIONS 11
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since t represents the number of years since 1970, we see that f (35) represents the population of the city in
2005. In 2005, the city’s population was 12 million.
2. Since T = f ( P ), we see that f (200) is the value of T when P = 200; that is, the thickness of pelican eggs
when the concentration of PCBs is 200 ppm.
3. If there are no workers, there is no productivity, so the graph goes through the origin. At first, as the number of
workers increases, productivity also increases. As a result, the curve goes up initially. At a certain point the
curve reaches its highest level, after which it goes downward; in other words, as the number of workers
increases beyond that point, productivity decreases. This might, for example, be due either to the inefficiency
inherent in large organizations or simply to workers getting in each other’s way as too many are crammed on
the same line. Many other reasons are possible.
4. The slope is (1 − 0)/(1 − 0) = 1. So the equation of the line is y = x.
5. The slope is (3 − 2)/(2 − 0) = 1/2. So the equation of the line is y = (1/2)x + 2.
6. Using the points (−2, 1) and (2, 3), we have
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Now we know that y = (1/2)x + b. Using the point (−2, 1), we have 1 = −2/2 + b, which yields b = 2. Thus, the
equation of the line is y = (1/2)x + 2.
6−0
7. Slope = = 2 so the equation is y − 6 = 2(x − 2) or y = 2x + 2.
2 − (−1)
5 5
8. Rewriting the equation as y = − x + 4 shows that the slope is − and the vertical intercept is 4.
2 2
9. Rewriting the equation as
12 2
y =− x+
7 7
shows that the line has slope −12/7 and vertical intercept 2/7.
10. Rewriting the equation of the line as
−2
−y = x−2
14
y = x + 2,
2
we see the line has slope 1/2 and vertical intercept 2.
11. Rewriting the equation of the line as
12 4
y= x−
6 6
2
y = 2x − ,
3
we see that the line has slope 2 and vertical intercept −2/3.
12. (a) is (V), because slope is positive, vertical intercept is negative
(b) is (IV), because slope is negative, vertical intercept is positive
(c) is (I), because slope is 0, vertical intercept is positive
(d) is (VI), because slope and vertical intercept are both negative
(e) is (II), because slope and vertical intercept are both positive
(f) is (III), because slope is positive, vertical intercept is 0
,2 Chapter One /SOLUTIONS
13. (a) is (V), because slope is negative, vertical intercept is 0
(b) is (VI), because slope and vertical intercept are both
positive
(c) is (I), because slope is negative, vertical intercept is
positive
(d) is (IV), because slope is positive, vertical intercept is
negative
(e) is (III), because slope and vertical intercept are both 2
negative =− .
(f) is (II), because slope is positive, vertical intercept is 0
14. The intercepts appear to be (0, 3) and (7.5, 0), giving
−3 6
Slope = =−
7.5 15 5
The y-intercept is at (0, 3), so a possible equation for the line is
2
y = − x + 3.
5
(Answers may
vary.)
15. y − c = m(x − a)
16. Given that the function is linear, choose any two points, for example (5.2, 27.8) and (5.3, 29.2). Then
29.2 − 27.8 1.4
Slope = = = 14.
5.3 − 5.2 0.1
Using the point-slope formula, with the point (5.2, 27.8), we get the equation
y − 27.8 = 14(x − 5.2)
which is equivalent
to y = 14x − 45.
1
17. y = 5x − 3. Since the slope of this line is 5, we want a line with slope −
5 passing through the point (2, 1). The
equation
is (y − 1) = − 1 (x − 2), or y = − 1 x + 7 .
5 5 5
18. The line y + 4x = 7 has slope −4. Therefore the parallel line has slope −4 and equation y − 5 = −4(x − 1) or
y−1= −4x + 9. The perpendicular line has slope = 1 and equation y − 5 = 1 (x − 1) or y = 0.25x + 4.75.
(−4) 4 4
19. The line parallel to y = mx + c also has slope m, so its equation is
y = m(x − a) + b.
The line perpendicular to y = mx + c has slope −1/m, so its equation will be
1
y = − (x − a) + b.
m
20. Since the function goes from x = 0 to x = 4 and between y = 0 and y = 2, the domain is 0 ≤ x ≤ 4 and the range
is
0 ≤ y ≤ 2.
21. Since x goes from 1 to 5 and y goes from 1 to 6, the domain is 1 ≤ x ≤ 5 and the range is 1 ≤ y ≤ 6.
22. Since the function goes from x = −2 to x = 2 and from y = −2 to y = 2, the domain is −2 ≤ x ≤ 2 and the range is
−2 ≤ y ≤ 2.
23. Since the function goes from x = 0 to x = 5 and between y = 0 and y = 4, the domain is 0 ≤ x ≤ 5 and the range
is
0 ≤ y ≤ 4.
24. The domain is all numbers. The range is all numbers ≥ 2, since x2 ≥ 0 for all x.
1
25. The domain is all x-values, as the denominator is never zero. The range is 0 < y .
≤ 2
26. The value of f (t) is real provided t 2 − 16 ≥ 0 or t2 ≥ 16. This occurs when either t ≥ 4, or t ≤ −4. Solving f (t) = 3,
we have
p
t2 − 16 = 3
t2 − 16 = 9
t2 = 25
, 1.1 SOLUTIONS 31
so
t = ±5.
4
27. We have V = kr3 . You may know that V =
πr 3.
3
d
28. If distance is d, then v = .
t
29. For some constant k, we have S = kh2 .
30. We know that E is proportional to v3 , so E = kv3 , for some constant k.
31. We know that N is proportional to 1/l2 , so
k
N= , for some constant
k. l2
Problems
32. The year 1983 was 25 years before 2008 so 1983 corresponds to t = 25. Thus, an expression that represents the
statement is:
f(25) = 7.019
33. The year 2008 was 0 years before 2008 so 2008 corresponds to t = 0. Thus, an expression that represents the
statement is:
f (0) meters.
34. The year 1965 was 2008 − 1865 = 143 years before 2008 so 1965 corresponds to t = 143. Similarly, we see
that the year 1911 corresponds to t = 97. Thus, an expression that represents the statement is:
f(143) = f(97)
35. Since t = 1 means one year before 2008, then t = 1 corresponds to the year 2007. Similarly, t = 0
corresponds to the year 2008. Thus, f (1) and f (0) are the average annual sea level values, in meters, in
2007 and 2008, respectively. Because 1 millimeter is the same as 0.001 meters, an expression that
represents the statement is:
f (0) = f (1) + 0.001.
Note that there are other possible equivalent expressions, such as: f (1) − f (0) = 0.001.
36. (a) Each date, t, has a unique daily snowfall, S, associated with it. So snowfall is a function of date.
(b) On December 12, the snowfall was approximately 5 inches.
(c) On December 11, the snowfall was above 10 inches.
(d) Looking at the graph we see that the largest increase in the snowfall was between December 10 to
December 11.
37. (a) When the car is 5 years old, it is worth $6000.
(b) Since the value of the car decreases as the car gets older, this is a decreasing function. A possible graph is in
Figure 1.1:
V (thousand dollars)
(5, 6)
a (years
Figure 1.1
(c) The vertical intercept is the value of V when a = 0, or the value of the car when it is new. The horizontal
intercept is the value of a when V = 0, or the age of the car when it is worth nothing.
,4 Chapter One /SOLUTIONS
38. (a) The story in (a) matches Graph (IV), in which the person forgot her books and had to return home.
(b) The story in (b) matches Graph (II), the flat tire story. Note the long period of time during which the
distance from home did not change (the horizontal part).
(c) The story in (c) matches Graph (III), in which the person started calmly but sped up later.
The first graph (I) does not match any of the given stories. In this picture, the person keeps going away
from home, but his speed decreases as time passes. So a story for this might be: I started walking to school at
a good pace, but since I stayed up all night studying calculus, I got more and more tired the farther I walked.
39. (a) f (30) = 10 means that the value of f at t = 30 was 10. In other words, the temperature at time t = 30
minutes was 10◦ C. So, 30 minutes after the object was placed outside, it had cooled to 10 ◦ C.
(b) The intercept a measures the value of f (t) when t = 0. In other words, when the object was initially put
outside, it had a temperature of a◦C. The intercept b measures the value of t when f (t) = 0. In other
words, at time b the object’s temperature is 0 ◦C.
40. (a) The height of the rock decreases as time passes, so the graph falls as you move from left to right. One
possibility is shown in Figure 1.2.
s (meters)
t (sec)
Figure 1.2
(b) The statement f (7) = 12 tells us that 7 seconds after the rock is dropped, it is 12 meters above the ground.
(c) The vertical intercept is the value of s when t = 0; that is, the height from which the rock is dropped. The
horizontal intercept is the value of t when s = 0; that is, the time it takes for the rock to hit the ground.
41. (a) We find the slope m and intercept b in the linear equation C = b + mw. To find the slope m, we use
∆C 12.32 − 8
m= = = 0.12 dollars per gallon.
∆w 68 − 32
We substitute to find
b:
C = b + mw
8 = b + (0.12)(32)
b = 4.16 dollars.
The linear formula is C = 4.16 + 0.12w.
(b) The slope is 0.12 dollars per gallon. Each additional gallon of waste collected costs 12 cents.
(c) The intercept is $4.16. The flat monthly fee to subscribe to the waste collection service is $4.16. This is the
amount charged even if there is no waste.
42. We are looking for a linear function y = f (x) that, given a time x in years, gives a value y in dollars for the value
of the refrigerator. We know that when x = 0, that is, when the refrigerator is new, y = 950, and when x = 7,
the refrigerator is worthless, so y = 0. Thus (0, 950) and (7, 0) are on the line that we are looking for. The slope
is then given by
950
m=
−7
It is negative, indicating that the value decreases as time passes. Having found the slope, we can take the
point (7, 0) and use the point-slope formula:
y − y1 = m(x − x1).
So,
950
y − 0 = − 7 (x − 7)
950
y=− x + 950.
7
, 1.1 SOLUTIONS 51
43. (a) The first company’s price for a day’s rental with m miles on it is C1(m) = 40 + 0.15m. Its competitor’s price
for a day’s rental with m miles on it is C2 (m) = 50 + 0.10m.
(b) See Figure 1.3.
C (cost in dollars)
150 C 1(m) = 40 + 0.15m
100
50 C 2(m) = 50 + 0.10m
0 m (miles)
200 400 600 800
Figure 1.3
(c) To find which company is cheaper, we need to determine where the two lines intersect. We let C1 = C2, and
thus
40 + 0.15m = 50 + 0.10m
0.05m = 10
m = 200.
If you are going more than 200 miles a day, the competitor is cheaper. If you are going less than 200 miles
a day, the first company is cheaper.
∆$
44. (a) Charge per cubic foot = 55 − 40 = $0.025/cu. ft.
∆ cu. ft. = 1600 −1000
Alternatively, if we let c = cost, w = cubic feet of water, b = fixed charge, and m = cost/cubic feet, we
obtain
c = b + mw. Substituting the information given in the problem, we have
40 = b + 1000m
55 = b + 1600m.
Subtracting the first equation from the second yields 15 = 600m, so m = 0.025.
(b) The equation is c = b + 0.025w, so 40 = b + 0.025(1000), which yields b = 15. Thus the equation is c
= 15+0.025w.
(c) We need to solve the equation 100 = 15 + 0.025w, which yields w = 3400. It costs $100 to use 3400 cubic
feet of water.
45. See Figure 1.4.
driving speed
time
Figure 1.4
46. See Figure
1.5. distance driven
time
Figure 1.5
,6 Chapter One /SOLUTIONS
47. See Figure 1.6.
distance from exit
time
Figure 1.6
48. See Figure 1.7.
distance between cars
distance driven
Figure 1.7
49. (a) (i) f (1985) = 13
(ii) f (1990) = 99
(b) The average yearly increase is the rate of change.
f(1990) − f(1985) 99 − 13
Yearly increase = = = 17.2 billionaires per year.
1990 −1985 5
(c) Since we assume the rate of increase remains constant, we use a linear function with slope 17.2
billionaires per year. The equation is
f (t) = b + 17.2t
where f (1985) = 13, so
13 = b + 17.2(1985)
b = −34,129.
Thus, f(t) = 17.2t − 34,129.
50. (a) The largest time interval was 2008–2009 since the percentage growth rate increased from −11.7 to 7.3 from
2008 to 2009. This means the US consumption of biofuels grew relatively more from 2008 to 2009 than
from 2007 to 2008. (Note that the percentage growth rate was a decreasing function of time over 2005–
2007.)
(b) The largest time interval was 2005–2007 since the percentage growth rates were positive for each of
these three consecutive years. This means that the amount of biofuels consumed in the US steadily
increased during the three year span from 2005 to 2007, then decreased in 2008.
51. (a) The largest time interval was 2005–2007 since the percentage growth rate decreased from −1.9 in 2005 to
−45.4 in
2007. This means that from 2005 to 2007 the US consumption of hydroelectric power shrunk relatively more
with
each successive year.
(b) The largest time interval was 2004–2007 since the percentage growth rates were negative for each of
these four consecutive years. This means that the amount of hydroelectric power consumed by the US
industrial sector steadily decreased during the four year span from 2004 to 2007, then increased in 2008.
52. (a) The largest time interval was 2004–2006 since the percentage growth rate increased from −5.7 in 2004 to
9.7 in 2006. This means that from 2004 to 2006 the US price per watt of a solar panel grew relatively
more with each successive year.
(b) The largest time interval was 2005–2006 since the percentage growth rates were positive for each of
these two consecutive years. This means that the US price per watt of a solar panel steadily increased
during the two year span from 2005 to 2006, then decreased in 2007.
, 1.1 SOLUTIONS 71
53. (a) Since 2008 corresponds to t = 0, the average annual sea level in Aberdeen in 2008 was 7.094 meters.
(b) Looking at the table, we see that the average annual sea level was 7.019 fifty years before 2008, or in the
year 1958. Similar reasoning shows that the average sea level was 6.957 meters 125 years before 2008, or
in 1883.
(c) Because 125 years before 2008 the year was 1883, we see that the sea level value corresponding to the
year 1883 is
6.957 (this is the sea level value corresponding to t = 125). Similar reasoning yields the table:
Year 1883 1908 1933 1958 1983 2008
S 6.957 6.938 6.965 6.992 7.019 7.094
54. (a) We find the slope m and intercept b in the linear equation S = b + mt. To find the slope m, we use
∆S 66 − 113
m= = = −0.94.
∆t 50 − 0
When t = 0, we have S = 113, so the intercept b is 113. The linear formula is
S = 113 − 0.94t.
(b) We use the formula S = 113 − 0.94t. When S = 20, we have 20 = 113 − 0.94t and so t = 98.9. If this
linear model were correct, the average male sperm count would drop below the fertility level during the
year 2038.
55. (a) This could be a linear function because w increases by 5 as h increases by 1.
(b) We find the slope m and the intercept b in the linear equation w = b + mh. We first find the slope m using the
first two points in the table. Since we want w to be a function of h, we take
∆w 171 − 166
m= = = 5.
∆h 69 − 68
Substituting the first point and the slope m = 5 into the linear equation w = b + mh, we have 166 = b +
(5)(68), so b = −174. The linear function is
w = 5h − 174.
The slope, m = 5, is in units of pounds per
inch.
(c) We find the slope and intercept in the linear function h = b + mw using m = ∆h/∆w to obtain the linear
function
h = 0.2w + 34.8.
Alternatively, we could solve the linear equation found in part (b) for h. The slope, m = 0.2, has units
inches per pound.
56. We will let
T = amount of fuel for take-
off, L = amount of fuel for
landing,
P = amount of fuel per mile in the
air, m = the length of the trip in miles.
Then Q, the total amount of fuel needed, is given by
Q(m ) = T + L + P m.
57. (a) The variable costs for x acres are $200x, or 0.2x thousand dollars. The total cost, C (again in thousands of
dollars), of planting x acres is:
C = f(x) = 10 + 0.2x.
This is a linear function. See Figure 1.8. Since C = f (x) increases with x, f is an increasing function of x.
Look at the values of C shown in the table; you will see that each time x increases by 1, C increases by
0.2. Because C increases at a constant rate as x increases, the graph of C against x is a line.
, 8 Chapter One /SOLUTIONS
THOSE WERE PREVIEW PAGES
TO DOWNLOAD THE FULL PDF
CLICK ON THE L.I.N.K
ON THE NEXT PAGE
