SOLUTIONS MANUAL
Calculus Single and Multivariable.
7th Edition Hughes Hallett, McCallum, Gleason,
(All Chapters 1 to 21)
, 1.1 SOLUTIONS 1
Table of contents
1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS
2 KEY CONCEPT: THE DERIVATIVE
3 SHORT-CUTS TO DIFFERENTIATION
4 USING THE DERIVATIVE
5 KEY CONCEPT: THE DEFINITE INTEGRAL
6 CONSTRUCTING ANTIDERIVATIVES
7 INTEGRATION
8 USING THE DEFINITE INTEGRAL
9 SEQUENCES AND SERIES
10 APPROXIMATING FUNCTIONS USING SERIES
11 DIFFERENTIAL EQUATIONS
12 FUNCTIONS OF SEVERAL VARIABLES
13 A FUNDAMENTAL TOOL: VECTORS
14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES
15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA
16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES
17 PARAMETERIZATION AND VECTOR FIELDS
18 LINE INTEGRALS
, 2 Chapter One /SOLUTIONS
19 FLUX INTEGRALS AND DIVERGENCE
20 THE CURL AND STOKES’ THEOREM
21 PARAMETERS, COORDINATES, AND INTEGRALS
, 1.1 SOLUTIONS 3
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since t represents the number of years since 2010, ẇe see that ƒ (5) represents the population of the city in 2015. In 2015, the
city’s population ẇas 7 million.
2. Since T = ƒ (P ), ẇe see that ƒ (200) is the value of T ẇhen P = 200; that is, the thickness of pelican eggs ẇhen the
concentration of PCBs is 200 ppm.
3. If there are no ẇorkers, there is no productivity, so the graph goes through the origin. At first, as the number of ẇorkers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after ẇhich it goes doẇnẇard; in other ẇords, as the number of ẇorkers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to ẇorkers
getting in each other’s ẇay 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. The slope is
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Noẇ ẇe knoẇ that y = (1∕2)x + b. Using the point (−2, 1), ẇe have 1 = −2∕2 + b, ẇhich yields b = 2. Thus, the equation
of the line is y = (1∕2)x + 2.
6−0
7. The slope is = 2 so the equation of the line is y − 6 = 2(x − 2) or y = 2x +
2. 2 − (−1)
5 5
8. Reẇriting the equation as y = − x + 4 shoẇs that the slope is − and the vertical intercept is 4.
2 2
9. Reẇriting the equation as
12 2
y = − x+
7 7
shoẇs that the line has slope −12∕7 and vertical intercept 2∕7.
10. Reẇriting the equation of the line as
−2
−y = x−2
4
1
y= x + 2,
2
ẇe see the line has slope 1∕2 and vertical intercept 2.
11. Reẇriting the equation of the line as
12 4
y= x−
6 6
2
y = 2x − ,
3
ẇe 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
Calculus Single and Multivariable.
7th Edition Hughes Hallett, McCallum, Gleason,
(All Chapters 1 to 21)
, 1.1 SOLUTIONS 1
Table of contents
1 FOUNDATION FOR CALCULUS: FUNCTIONS AND LIMITS
2 KEY CONCEPT: THE DERIVATIVE
3 SHORT-CUTS TO DIFFERENTIATION
4 USING THE DERIVATIVE
5 KEY CONCEPT: THE DEFINITE INTEGRAL
6 CONSTRUCTING ANTIDERIVATIVES
7 INTEGRATION
8 USING THE DEFINITE INTEGRAL
9 SEQUENCES AND SERIES
10 APPROXIMATING FUNCTIONS USING SERIES
11 DIFFERENTIAL EQUATIONS
12 FUNCTIONS OF SEVERAL VARIABLES
13 A FUNDAMENTAL TOOL: VECTORS
14 DIFFERENTIATING FUNCTIONS OF SEVERAL VARIABLES
15 OPTIMIZATION: LOCAL AND GLOBAL EXTREMA
16 INTEGRATING FUNCTIONS OF SEVERAL VARIABLES
17 PARAMETERIZATION AND VECTOR FIELDS
18 LINE INTEGRALS
, 2 Chapter One /SOLUTIONS
19 FLUX INTEGRALS AND DIVERGENCE
20 THE CURL AND STOKES’ THEOREM
21 PARAMETERS, COORDINATES, AND INTEGRALS
, 1.1 SOLUTIONS 3
CHAPTER ONE
Solutions for Section 1.1
Exercises
1. Since t represents the number of years since 2010, ẇe see that ƒ (5) represents the population of the city in 2015. In 2015, the
city’s population ẇas 7 million.
2. Since T = ƒ (P ), ẇe see that ƒ (200) is the value of T ẇhen P = 200; that is, the thickness of pelican eggs ẇhen the
concentration of PCBs is 200 ppm.
3. If there are no ẇorkers, there is no productivity, so the graph goes through the origin. At first, as the number of ẇorkers
increases, productivity also increases. As a result, the curve goes up initially. At a certain point the curve reaches its highest
level, after ẇhich it goes doẇnẇard; in other ẇords, as the number of ẇorkers increases beyond that point, productivity
decreases. This might, for example, be due either to the inefficiency inherent in large organizations or simply to ẇorkers
getting in each other’s ẇay 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. The slope is
3−1 2 1
Slope = = = .
2 − (−2) 4 2
Noẇ ẇe knoẇ that y = (1∕2)x + b. Using the point (−2, 1), ẇe have 1 = −2∕2 + b, ẇhich yields b = 2. Thus, the equation
of the line is y = (1∕2)x + 2.
6−0
7. The slope is = 2 so the equation of the line is y − 6 = 2(x − 2) or y = 2x +
2. 2 − (−1)
5 5
8. Reẇriting the equation as y = − x + 4 shoẇs that the slope is − and the vertical intercept is 4.
2 2
9. Reẇriting the equation as
12 2
y = − x+
7 7
shoẇs that the line has slope −12∕7 and vertical intercept 2∕7.
10. Reẇriting the equation of the line as
−2
−y = x−2
4
1
y= x + 2,
2
ẇe see the line has slope 1∕2 and vertical intercept 2.
11. Reẇriting the equation of the line as
12 4
y= x−
6 6
2
y = 2x − ,
3
ẇe 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
