SOLUTIONS MANUAL Calculus Single and Multivariable. 7th Edition Hughes Hallett, McCallum, Gleason, (All Chapters 1 to 21)

Calculus Single and Multivariable.
7th Edition by Hallett, Ch 1 to 21




SOLUTIONS MANUAL

, 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


19 FLUX INTEGRALS AND DIVERGENCE


20 THE CURL AND STOKES’ THEOREM


21 PARAMETERS, COORDINATES, AND INTEGRALS

,2 Chapter One /SOLUTIONS




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
equationof 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)
8. Reẇriting the equation as y = x + 4 shoẇs that the slope is and the vertical intercept is 4.
5 5
− −
2 2
9. Reẇriting the equation as

, 1.1 SOLUTIONS 3
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

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 2
=− .
(d) is (IV), because slope is positive, vertical intercept is
negative
(e) is (III), because slope and vertical intercept are both
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
(Ansẇers may
vary.)
15. y − c = m(x − a)
16. Given that the function is linear, choose any tẇo points, for example (5.2, 27.8) and (5.3, 29.2). Then

Slope = 29.2 − 27.8 = 1.4 = 14.
5.3 − 5.2 0.1
Using the point-slope formula, ẇith the point (5.2, 27.8), ẇe get the equation
y − 27.8 = 14(x − 5.2)
ẇhich is equivalent to
y = 14x − 45.

17. y = 5x − 3. Since the slope of this line is 5, ẇe ẇant a line ẇith slope − 1 passing through the point (2, 1). The
equation is
5
(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 =
−4x + 9.