Thomas' Calculus. SOLUTION MANUAL

CHAPTER 1 PRELIMINARIES 1.1 REAL NUMBERS AND THE REAL LINE 1. Executing long division,

" 9

2. Executing long division,

" 11

œ 0.1,

2 9

œ 0.2,

œ 0.09,

2 11

3 9

œ 0.3,

œ 0.18,

3 11

8 9

œ 0.8,

œ 0.27,

9 11

9 9

œ 0.9

œ 0.81,

11 11

œ 0.99

3. NT = necessarily true, NNT = Not necessarily true. Given: 2 < x < 6. a) NNT. 5 is a counter example. b) NT. 2 < x < 6 Ê 2
2 < x
2 < 6
2 Ê 0 < x
2 < 2. c) NT. 2 < x < 6 Ê 2/2 < x/2 < 6/2 Ê 1 < x < 3. d) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2. e) NT. 2 < x < 6 Ê 1/2 > 1/x > 1/6 Ê 1/6 < 1/x < 1/2 Ê 6(1/6) < 6(1/x) < 6(1/2) Ê 1 < 6/x < 3. f) NT. 2 < x < 6 Ê x < 6 Ê (x
4) < 2 and 2 < x < 6 Ê x > 2 Ê
x <
2 Ê
x + 4 < 2 Ê
(x
4) < 2. The pair of inequalities (x
4) < 2 and
(x
4) < 2 Ê | x
4 | < 2. g) NT. 2 < x < 6 Ê
2 >
x >
6 Ê
6 <
x <
2. But
2 < 2. So
6 <
x <
2 < 2 or
6 <
x < 2. h) NT. 2 < x < 6 Ê
1(2) >
1(x) <
1(6) Ê
6 <
x <
2 4. NT = necessarily true, NNT = Not necessarily true. Given:
1 < y
5 < 1. a) NT.
1 < y
5 < 1 Ê
1 + 5 < y
5 + 5 < 1 + 5 Ê 4 < y < 6. b) NNT. y = 5 is a counter example. (Actually, never true given that 4  y  6) c) NT. From a),
1 < y
5 < 1, Ê 4 < y < 6 Ê y > 4. d) NT. From a),
1 < y
5 < 1, Ê 4 < y < 6 Ê y < 6. e) NT.
1 < y
5 < 1 Ê
1 + 1 < y
5 + 1 < 1 + 1 Ê 0 < y
4 < 2. f) NT.
1 < y
5 < 1 Ê (1/2)(
1 + 5) < (1/2)(y
5 + 5) < (1/2)(1 + 5) Ê 2 < y/2 < 3. g) NT. From a), 4 < y < 6 Ê 1/4 > 1/y > 1/6 Ê 1/6 < 1/y < 1/4. h) NT.
1 < y
5 < 1 Ê y
5 >
1 Ê y > 4 Ê
y <
4 Ê
y + 5 < 1 Ê
(y
5) < 1. Also,
1 < y
5 < 1 Ê y
5 < 1. The pair of inequalities
(y
5) < 1 and (y
5) < 1 Ê | y
5 | < 1. 5.
2x  4 Ê x 
2 6. 8
3x   5 Ê
3x  
3 Ê x Ÿ 1 7. 5x
$ Ÿ (
3x Ê 8x Ÿ 10 Ê x Ÿ

ïïïïïïïïïñqqqqqqqqp x 1 5 4

8. 3(2
x)  2(3  x) Ê 6
3x  6  2x Ê 0  5x Ê 0  x 9. 2x


10.

" #

Ê

" 5

6
x 4



  7x 

ˆ


10 ‰ 6

3x
4 2

7 6

Ê
"#


  x or


" 3

7 6

ïïïïïïïïïðqqqqqqqqp x 0

  5x

 x

Ê 12
2x  12x
16

Ê 28  14x Ê 2  x

qqqqqqqqqðïïïïïïïïî x 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 11.

Chapter 1 Preliminaries 4 5

" 3

(x
2) 

(x
6) Ê 12(x
2)  5(x
6)

Ê 12x
24  5x
30 Ê 7x 
6 or x 
67 12.
x2 5 Ÿ

123x 4

Ê
(4x  20) Ÿ 24  6x

Ê
44 Ÿ 10x Ê
22 5 Ÿ x

qqqqqqqqqñïïïïïïïïî x
22/5

13. y œ 3 or y œ
3 14. y
3 œ 7 or y
3 œ
7 Ê y œ 10 or y œ
4 15. 2t  5 œ 4 or 2t  & œ
4 Ê 2t œ
1 or 2t œ
9 Ê t œ
"# or t œ
9# 16. 1
t œ 1 or 1
t œ
1 Ê
t œ ! or
t œ
2 Ê t œ 0 or t œ 2 17. 8
3s œ 18.

s #

9 2

or 8
3s œ
#9 Ê
3s œ
7# or
3s œ
25 # Ê sœ


1 œ 1 or

s #


1 œ
1 Ê

s #

œ 2 or

s #

7 6

or s œ

25 6

œ ! Ê s œ 4 or s œ 0

19.
2  x  2; solution interval (
2ß 2) 20.
2 Ÿ x Ÿ 2; solution interval [
2ß 2]

qqqqñïïïïïïïïñqqqqp x
2 2

21.
3 Ÿ t
1 Ÿ 3 Ê
2 Ÿ t Ÿ 4; solution interval [
2ß 4] 22.
1  t  2  1 Ê
3  t 
1; solution interval (
3ß
1)

qqqqðïïïïïïïïðqqqqp t
3
1

23.
%  3y
7  4 Ê 3  3y  11 Ê 1  y  solution interval ˆ1ß

11 3

;

11 ‰ 3

24.
1  2y  5  " Ê
6  2y 
4 Ê
3  y 
2; solution interval (
3ß
2) 25.
1 Ÿ

z 5


1Ÿ1 Ê 0Ÿ

z 5

qqqqðïïïïïïïïðqqqqp y
3
2

Ÿ 2 Ê 0 Ÿ z Ÿ 10;

solution interval [0ß 10] 26.
2 Ÿ


1 Ÿ 2 Ê
1 Ÿ solution interval

23 ß 2‘ 3z #

27.
"#  3
Ê

2 7

28.
3 

" x



x 2 x

2 5

" #

2 7

Ÿ 3 Ê
32 Ÿ z Ÿ 2; qqqqñïïïïïïïïñqqqqp z 2
2/3

Ê
7# 
x" 
5# Ê

7 #



" x



5 #

; solution interval ˆ 27 ß 25 ‰


43 Ê 1

Ê 2x

3z #

Ê

2 7

2 x

( Ê 1

x #



" 7

 x  2; solution interval ˆ 27 ß 2‰

qqqqðïïïïïïïïðqqqqp x 2 2/7

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.1 Real Numbers and the Real Line 29. 2s   4 or
2s   4 Ê s   2 or s Ÿ
2; solution intervals (
_ß
2]  [2ß _) 30. s  3  

" #

or
(s  3)  

" #

Ê s  
5# or
s  

7 #

Ê s  
5# or s Ÿ
7# ; solution intervals ˆ
_ß
7# ‘ 

5# ß _‰

ïïïïïïñqqqqqqñïïïïïïî s
7/2
5/2

31. 1
x  1 or
("
x)  1 Ê
x  0 or x  2 Ê x  0 or x  2; solution intervals (
_ß !)  (2ß _) 32. 2
3x  5 or
(2
3x)  5 Ê
3x  3 or 3x  7 Ê x 
1 or x  73 ; solution intervals (
_ß
1)  ˆ 73 ß _‰ 33.

r" #

ïïïïïïðqqqqqqðïïïïïïî x
1 7/3

  1 or
ˆ r# 1 ‰   1 Ê r  1   2 or r  1 Ÿ
2

Ê r   1 or r Ÿ
3; solution intervals (
_ß
3]  [1ß _) 34.

3r 5


"

Ê

or
ˆ 3r5
"‰ 

2 5

or
3r5 
53 Ê r  37 or r  1 solution intervals (
_ß ")  ˆ 73 ß _‰ 3r 5



2 5 7 5

ïïïïïïðqqqqqqðïïïïïïî r 1 7/3

35. x#  # Ê kxk  È2 Ê
È2  x  È2 ; solution interval Š
È2ß È2‹

qqqqqqðïïïïïïðqqqqqqp x È#
È #

36. 4 Ÿ x# Ê 2 Ÿ kxk Ê x   2 or x Ÿ
2; solution interval (
_ß
2]  [2ß _)

ïïïïïïñqqqqqqñïïïïïïî r
2 2

37. 4  x#  9 Ê 2  kxk  3 Ê 2  x  3 or 2 
x  3 Ê 2  x  3 or
3  x 
2; solution intervals (
3ß
2)  (2ß 3) 38.

" 9

 x# 

Ê

x

" #

" 3

 kxk 

" #

Ê

" 3

x

or
#"  x 
3" ; solution intervals ˆ
"# ß
3" ‰  ˆ 3" ß #" ‰ Ê

" 3

" 4

" #

or

" 3


x 

39. (x
1)#  4 Ê kx
1k  2 Ê
2  x
1  2 Ê
1  x  3; solution interval (
"ß $)

qqqqðïïïïðqqqqðïïïïðqqqp x
3
2 2 3 " #

qqqqðïïïïðqqqqðïïïïðqqqp x
1/2
1/3 1/3 1/2

qqqqqqðïïïïïïïïðqqqqp x
1 3

40. (x  3)#  # Ê kx  3k  È2 Ê
È2  x  3  È2 or
3
È2  x 
3  È2 ; solution interval Š
3
È2ß
3  È2‹

qqqqqqðïïïïïïïïðqqqqp x
3
È #
3  È #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

3

4

Chapter 1 Preliminaries

41. x#
x  0 Ê x#
x +

1 4

<

1 4

2 Ê ˆx
12 ‰ <

1 4

ʹx


1 2

¹<

Ê
12 < x


1 2

1 2

<

1 2

Ê 0 < x < 1.

So the solution is the interval (0ß 1) 42. x#
x
2   0 Ê x#
x +

1 4

 

9 4

Ê ¹x


1 2

¹  

3 2

Ê x


1 2

 

3 2

or
ˆx
12 ‰  

3 2

Ê x   2 or x Ÿ
1.

The solution interval is (
_ß
1]  [2ß _) 43. True if a   0; False if a  0. 44. kx
1k œ 1
x Í k
(x
1)k œ 1
x Í 1
x   0 Í x Ÿ 1 45. (1) ka  bk œ (a  b) or ka  bk œ
(a  b); both squared equal (a  b)# (2) ab Ÿ kabk œ kak kbk (3) kak œ a or kak œ
a, so kak# œ a# ; likewise, kbk# œ b# (4) x# Ÿ y# implies Èx# Ÿ Èy# or x Ÿ y for all nonnegative real numbers x and y. Let x œ ka  bk and y œ kak  kbk so that ka  bk# Ÿ akak  kbkb# Ê ka  bk Ÿ kak  kbk . 46. If a   0 and b   0, then ab   0 and kabk œ ab œ kak kbk . If a  0 and b  0, then ab  0 and kabk œ ab œ (
a)(
b) œ kak kbk . If a   0 and b  0, then ab Ÿ 0 and kabk œ
(ab) œ (a)(
b) œ kak kbk . If a  0 and b   0, then ab Ÿ 0 and kabk œ
(ab) œ (
a)(b) œ kak kbk . 47.
3 Ÿ x Ÿ 3 and x 
"# Ê


" #

 x Ÿ 3.

48. Graph of kxk  kyk Ÿ 1 is the interior of “diamond-shaped" region.

49. Let $ be a real number > 0 and f(x) = 2x + 1. Suppose that | x
1 | < $ . Then | x
1 | < $ Ê 2| x
1 | < 2$ Ê | 2x
# | < 2$ Ê | (2x + 1)
3 | < 2$ Ê | f(x)
f(1) | < 2$ 50. Let % > 0 be any positive number and f(x) = 2x + 3. Suppose that | x
0 | < % /2. Then 2| x
0 | < % and | 2x + 3
3 | < %. But f(x) = 2x + 3 and f(0) = 3. Thus | f(x)
f(0) | < %. 51. Consider: i) a > 0; ii) a < 0; iii) a = 0. i) For a > 0, | a | œ a by definition. Now, a > 0 Ê
a < 0. Let
a = b. By definition, | b | œ
b. Since b =
a, |
a | œ
(
a) œ a and | a | œ |
a | œ a. ii) For a < 0, | a | œ
a. Now, a < 0 Ê
a > 0. Let
a œ b. By definition, | b | œ b and thus |
a| œ
a. So again | a | œ |
a|. iii) By definition | 0 | œ 0 and since
0 œ 0, |
0 | œ 0. Thus, by i), ii), and iii) | a | œ |
a | for any real number. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles and Parabolas Prove | x | > 0 Ê x > a or x <
a for any positive number, a. For x   0, | x | œ x. | x | > a Ê x > a. For x < 0, | x | œ
x. | x | > a Ê
x > a Ê x <
a. ii) Prove x > a or x <
a Ê | x | > 0 for any positive number, a. a > 0 and x > a Ê | x | œ x. So x > a Ê | x | > a. For a > 0,
a < 0 and x <
a Ê x < 0 Ê | x | œ
x. So x <
a Ê
x > a Ê | x | > a.

52. i)

53. a)

1=1 Ê |1|=1 ʹb

b)

lal lbl

œ ¹a

†

" b

¹ œ ¹ a¹

† "b ¹ œ

†¹

" b

l bl lbl

¹ œ ¹ a¹

Ê ¹ b¹

† l bl "

† ¹ b" ¹ œ

œ

lbl lbl

Ê

†

¹ b ¹ ¹ "b ¹ ¹ b¹

œ

¹ b¹

†

¹ b¹ ¹ b¹

Ê ¹ b" ¹ œ "

¹ b¹

lal lbl

54. Prove Sn œ kan k œ kakn for any real number a and any positive integer n. ka" k œ kak " œ a, so S" is true. Now, assume that Sk œ ¸ak ¸ œ kak k is true form some positive integer 5 . Since ka" k œ kak " and ¸ak ¸ œ kak k , we have ¸ak" ¸ œ ¸ak † a" ¸ œ ¸ak ¸ka" k œ kak k kak " œ kak k+" . Thus, Sk" œ ¸ak" ¸ œ kak k+" is also true. Thus by the Principle of Mathematical Induction, Sn œ l an l œ l a ln is true for all n positive integers. 1.2 LINES, CIRCLES, AND PARABOLAS 1. ?x œ
1
(
3) œ 2, ?y œ
2
2 œ
4; d œ È(?x)#  (?y)# œ È4  16 œ 2È5 2. ?x œ
$
(
1) œ
2, ?y œ 2
(
2) œ 4; d œ È(
2)#  4# œ 2È5 3. ?x œ
8.1
(
3.2) œ
4.9, ?y œ
2
(
2) œ 0; d œ È(
4.9)#  0# œ 4.9 #

4. ?x œ 0
È2 œ
È2, ?y œ 1.5
4 œ
2.5; d œ ÊŠ
È2‹  (
2.5)# œ È8.25 5. Circle with center (!ß !) and radius 1.

6. Circle with center (!ß !) and radius È2.

7. Disk (i.e., circle together with its interior points) with center (!ß !) and radius È3. 8. The origin (a single point). 9. m œ

?y ?x

œ


1
2
2
(
1)

œ3

perpendicular slope œ
"3

10. m œ

?y ?x

œ


#
" 2
(
2)

œ
34

perpendicular slope œ

4 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5

6

Chapter 1 Preliminaries

11. m œ

?y ?x

œ

3
3
1
2

œ0

12. m œ

14. (a) x œ È2

(b) y œ


#
0
#
(
#)

; no slope

15. (a) x œ 0

16. (a) x œ
1

(b) y œ
È2

(b) y œ
1.3

4 3

œ

perpendicular slope œ 0

perpendicular slope does not exist

13. (a) x œ
1

?y ?x

(b) y œ 0

17. P(
1ß 1), m œ
1 Ê y
1 œ
1ax
(
1)b Ê y œ
x 18. P(2ß
3), m œ

" #

Ê y
(
3) œ

19. P(3ß 4), Q(
2ß 5) Ê m œ

?y ?x

20. P(
8ß 0), Q(
1ß 3) Ê m œ

œ

?y ?x

" #

(x
2) Ê y œ

5
4
2
3

œ

" #

x
4

œ
"5 Ê y
4 œ
"5 (x
3) Ê y œ
"5 x 

3
0
1
(
8)

œ

3 7

Ê y
0œ

3 7

ax
(
8)b Ê y œ

3 7

23 5

x

21. m œ
54 , b œ 6 Ê y œ
54 x  6

22. m œ "# , b œ
3 Ê y œ

" #

23. m œ 0, P(
12ß
9) Ê y œ
9

24. No slope, P ˆ "3 ß %‰ Ê x œ

24 7

x
3 " 3

25. a œ
1, b œ 4 Ê (0ß 4) and (
"ß 0) are on the line Ê m œ

?y ?x

œ

0
4
1
0

œ 4 Ê y œ 4x  4

26. a œ 2, b œ
6 Ê (2ß 0) and (!ß
6) are on the line Ê m œ

?y ?x

œ


6
0 0
2

œ 3 Ê y œ 3x
6

27. P(5ß
1), L: 2x  5y œ 15 Ê mL œ
25 Ê parallel line is y
(
1) œ
25 (x
5) Ê y œ
25 x  1 È È È 28. P Š
È2ß 2‹ , L: È2x  5y œ È3 Ê mL œ
52 Ê parallel line is y
2 œ
52 Šx
Š
È2‹‹ Ê y œ
52 x 

8 5

29. P(4ß 10), L: 6x
3y œ 5 Ê mL œ 2 Ê m¼ œ
"# Ê perpendicular line is y
10 œ
"# (x
4) Ê y œ
"# x  12 30. P(!ß 1), L: 8x
13y œ 13 Ê mL œ

8 13

13 Ê m¼ œ
13 8 Ê perpendicular line is y œ
8 x  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles and Parabolas 31. x-intercept œ 4, y-intercept œ 3

32. x-intercept œ
4, y-intercept œ
2

33. x-intercept œ È3, y-intercept œ
È2

34. x-intercept œ
2, y-intercept œ 3

35. Ax  By œ C" Í y œ
AB x 

C" B

and Bx
Ay œ C# Í y œ

B A

x


C# A.

Since ˆ
AB ‰ ˆ AB ‰ œ
1 is the

product of the slopes, the lines are perpendicular. 36. Ax  By œ C" Í y œ
AB x  slope


AB ,

C" B

and Ax  By œ C# Í y œ
AB x 

C# B.

Since the lines have the same

they are parallel.

37. New position œ axold  ?xß yold  ?yb œ (
#  &ß 3  (
6)) œ ($ß
3). 38. New position œ axold  ?xß yold  ?yb œ (6  (
6)ß 0  0) œ (0ß 0). 39. ?x œ 5, ?y œ 6, B(3ß
3). Let A œ (xß y). Then ?x œ x#
x" Ê 5 œ 3
x Ê x œ
2 and ?y œ y#
y" Ê 6 œ
3
y Ê y œ
9. Therefore, A œ (
#ß
9). 40. ?x œ "
" œ !, ?y œ !
! œ !

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

7

8

Chapter 1 Preliminaries

41. C(!ß 2), a œ 2 Ê x#  (y
2)# œ 4

42. C(
$ß 0), a œ 3 Ê (x  3)#  y# œ 9

43. C(
1ß 5), a œ È10 Ê (x  1)#  (y
5)# œ 10

44. C("ß "), a œ È2 Ê (x
1)#  (y
1)# œ 2 x œ 0 Ê (0
1)#  (y
1)# œ 2 Ê (y
1)# œ 1 Ê y
1 œ „ 1 Ê y œ 0 or y œ 2. Similarly, y œ 0 Ê x œ 0 or x œ 2

#

45. C Š
È3ß
2‹ , a œ 2 Ê Šx  È3‹  (y  2)# œ 4, #

x œ 0 Ê Š0  È3‹  (y  2)# œ 4 Ê (y  2)# œ 1 Ê y  2 œ „ 1 Ê y œ
1 or y œ
3. Also, y œ 0 #

#

Ê Šx  È3‹  (0  2)# œ 4 Ê Šx  È3‹ œ 0 Ê x œ
È 3 # 46. C ˆ3ß "# ‰, a œ 5 Ê (x
3)#  ˆy
"# ‰ œ 25, so # x œ 0 Ê (0
3)#  ˆy
"# ‰ œ 25 # Ê ˆy
"# ‰ œ 16 Ê y


" #

œ „4 Ê yœ

9 #

# or y œ
7# . Also, y œ 0 Ê (x
3)#  ˆ0
"# ‰ œ 25

Ê (x
3)# œ Ê xœ3„

99 4 3È11 #

Ê x
3œ „

3È11 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles and Parabolas 47. x#  y#  4x
4y  % œ 0 Ê x#  %B  y#
4y œ
4 Ê x#  4x  4  y#
4y  4 œ 4 Ê (x  2)#  (y
2)# œ 4 Ê C œ (
2ß 2), a œ 2.

48. x#  y#
8x  4y  16 œ 0 Ê x#
8x  y#  4y œ
16 Ê x#
8x  16  y#  4y  4 œ 4 Ê (x
4)#  (y  2)# œ 4 Ê C œ (%ß
2), a œ 2.

49. x#  y#
3y
4 œ 0 Ê x#  y#
3y œ 4 Ê x#  y#
3y  94 œ 25 4 # Ê x#  ˆy
3# ‰ œ

25 4

Ê C œ ˆ0ß 3# ‰ ,

a œ 5# .

50. x#  y#
4x
#

9 4 #

œ0

Ê x
4x  y œ

9 4 #

Ê x#
4x  4  y œ Ê (x
2)#  y# œ

25 4

25 4

Ê C œ (2ß 0), a œ 5# .

51. x#  y#
4x  4y œ 0 Ê x#
4x  y#  4y œ 0 Ê x#
4x  4  y#  4y  4 œ 8 Ê (x
2)#  (y  2)# œ 8 Ê C(2ß
2), a œ È8.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

9

10

Chapter 1 Preliminaries

52. x#  y#  2x œ 3 Ê x#  2x  1  y# œ 4 Ê (x  1)#  y# œ 4 Ê C œ (
1ß 0), a œ 2.


2 53. x œ
#ba œ
2(1) œ1

Ê y œ (1)#
2(1)
3 œ
4 Ê V œ ("ß
4). If x œ 0 then y œ
3. Also, y œ 0 Ê x#
2x
3 œ 0 Ê (x
3)(x  1) œ 0 Ê x œ 3 or x œ
1. Axis of parabola is x œ 1.

4 54. x œ
#ba œ
2(1) œ
2

Ê y œ (
2)#  4(
2)  3 œ
1 Ê V œ (
2ß
1). If x œ 0 then y œ 3. Also, y œ 0 Ê x#  4x  3 œ 0 Ê (x  1)(x  3) œ 0 Ê x œ
1 or x œ
3. Axis of parabola is x œ
2.

55. x œ
#ba œ
2(
4 1) œ 2 Ê y œ
(2)#  4(2) œ 4 Ê V œ (2ß 4). If x œ 0 then y œ 0. Also, y œ 0 Ê
x#  4x œ 0 Ê
x(x
4) œ 0 Ê x œ 4 or x œ 0. Axis of parabola is x œ 2.

56. x œ
#ba œ
2(
4 1) œ 2 Ê y œ
(2)#  4(2)
5 œ
1 Ê V œ (2ß
1). If x œ 0 then y œ
5. Also, y œ 0 Ê
x#  4x
5 œ 0 Ê x#
4x  5 œ 0 Ê x œ

4 „È
4 #

Ê no x intercepts. Axis of parabola is x œ 2.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles and Parabolas 57. x œ
#ba œ
2(

61) œ
3 Ê y œ
(
3)#
6(
3)
5 œ 4 Ê V œ (
3ß %). If x œ 0 then y œ
5. Also, y œ 0 Ê
x#
6x
5 œ 0 Ê (x  5)(x  1) œ 0 Ê x œ
5 or x œ
1. Axis of parabola is x œ
3.


1 58. x œ
#ba œ
2(2) œ

" 4

#

Ê y œ 2 ˆ "4 ‰
4"  3 œ 23 8 ‰ Ê V œ ˆ "4 ß 23 . If x œ 0 then y œ 3. 8

Also, y œ 0 Ê 2x#
x  3 œ 0 Ê xœ

1„È
23 4

Ê no x intercepts.

Axis of parabola is x œ "4 .

1 59. x œ
#ba œ
2(1/2) œ
1 " #

(
1)#  (
1)  4 œ 72 Ê V œ ˆ
"ß 72 ‰ . If x œ 0 then y œ 4. Ê yœ

Also, y œ 0 Ê Ê xœ


1 „ È
7 1

" #

x#  x  4 œ 0 Ê no x intercepts.

Axis of parabola is x œ
1. 60. x œ
#ba œ
2(
21/4) œ 4 Ê y œ
"4 (4)#  2(4)  4 œ 8 Ê V œ (4ß 8) . If x œ 0 then y œ 4. Also, y œ 0 Ê
"4 x#  2x  4 œ 0 Ê xœ


2 „ È 8
1/2

œ 4 „ 4È2.

Axis of parabola is x œ 4.

61. The points that lie outside the circle with center (!ß 0) and radius È7. 62. The points that lie inside the circle with center (!ß 0) and radius È5. 63. The points that lie on or inside the circle with center ("ß 0) and radius 2. 64. The points lying on or outside the circle with center (!ß 2) and radius 2. 65. The points lying outside the circle with center (!ß 0) and radius 1, but inside the circle with center (!ß 0), and radius 2 (i.e., a washer).

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11

12

Chapter 1 Preliminaries

66. The points on or inside the circle centered at (!ß !) with radius 2 and on or inside the circle centered at (
2ß 0) with radius 2.

67. x#  y#  6y  0 Ê x#  (y  3)#  9. The interior points of the circle centered at (!ß
3) with radius 3, but above the line y œ
3.

68. x#  y#
4x  2y  4 Ê (x
2)#  (y  1)#  9. The points exterior to the circle centered at (2ß
1) with radius 3 and to the right of the line x œ 2.

69. (x  2)#  (y
1)#  6

70. (x  4)#  (y
2)#  16

71. x#  y# Ÿ 2, x   1

72. x#  y#  4, (x
1)#  (y
3)#  10

73. x#  y# œ 1 and y œ 2x Ê 1 œ x#  4x# œ 5x# Ê Šx œ

" È5

and y œ

2 È5 ‹

or Šx œ
È"5 and y œ
È25 ‹ .

Thus, A Š È"5 ß È25 ‹ , B Š
È"5 ß
È25 ‹ are the points of intersection.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles, and Parabolas 74. x  y œ 1 and (x
1)#  y# œ 1 Ê 1 œ (
y)#  y# œ 2y# Ê Šy œ

" È2

and x œ "


Šy œ
È"2 and x œ 1  A Š"


" È2

" È2 ‹

" È2 ‹ .

ß È"2 ‹ and B Š1 

or

Thus,

" È2

ß
È"2 ‹

are intersection points.

75. y
x œ 1 and y œ x# Ê x#
x œ 1 1 „È 5 . # 1
È 5 3
È 5 If x œ # , then y œ x  1 œ # . È È If x œ 1# 5 , then y œ x  1 œ 3# 5 . È È È È Thus, A Š 1
# 5 ß 3
# 5 ‹ and B Š 1# 5 ß 3# 5 ‹

Ê x#
x
1 œ 0 Ê x œ

are the intersection points.

76. y œ
x and C œ
(x
1)# Ê (x
1)# œ x 3 „È 5 . # È 5 3 3 È 5 x œ # , then y œ
x œ # . If È È x œ 3
# 5 , then y œ
x œ
3
# 5 . È È È Thus, A Š 3# 5 ß 5#3 ‹ and B Š 3
# 5

Ê x#
3x  " œ 0 Ê x œ

If

È

ß
3
# 5 ‹

are the intersection points.

77. y œ 2x#
1 œ
x# Ê 3x# œ 1 Ê x œ È"3 and y œ
3" or x œ
È"3 and y œ
3" . Thus, A Š È"3 ß
3" ‹ and B Š
È"3 ß
3" ‹ are the intersection points.

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13

14

Chapter 1 Preliminaries

78. y œ

x# 4

œ (x
1)# Ê 0 œ #

3x# 4


2x  1

Ê 0 œ 3x
8x  4 œ (3x
2)(x
2) Ê x œ 2 and y œ yœ

#

x 4

x# 4

œ 1, or x œ

œ 9" . Thus, A(2ß 1) and

2 3 and 2 B ˆ 3 ß 9" ‰

are the intersection points.

79. x#  y# œ 1 œ (x
1)#  y# Ê x# œ (x
1)# œ x#
2x  1 Ê 0 œ
2x  1 Ê x œ "# . Hence y# œ "
x # œ A Š "# ß

È3 # ‹

and

3 4

or y œ „

È3 #

È B Š "# ß
#3 ‹

. Thus,

are the

intersection points.

80. x#  y# œ 1 œ x#  y Ê y# œ y Ê y(y
1) œ 0 Ê y œ 0 or y œ 1. If y œ 1, then x# œ "
y# œ 0 or x œ 0. If y œ 0, then x# œ 1
y# œ 1 or x œ „ 1. Thus, A(0ß 1), B("ß 0), and C(
1ß 0) are the intersection points.

81. (a) A ¸ (69°ß 0 in), B ¸ (68°ß .4 in) Ê m œ (b) A ¸ (68°ß .4 in), B ¸ (10°ß 4 in) Ê m œ (c) A ¸ (10°ß 4 in), B ¸ (5°ß 4.6 in) Ê m œ 82. The time rate of heat transfer across a material, to the temperature gradient across the material, of the material.

?U ?>

œ

X -kA ? ?B

Ê

?U ÎA k =
??> X . ?B

68°  69° .4  0 ¸
2.5°/in. 10°  68° 4  .4 ¸
16.1°/in. 5°  10° 4.6  4 ¸
8.3°/in. ?U ?> , is directly ?X ?B (the slopes

Note that

?U ?>

proportional to the cross-sectional area, A, of the material, from the previous problem), and to a constant characteristic

and

?X ?B

are of opposite sign because heat flow is toward lower

temperature. So a small value of k corresponds to low heat flow through the material and thus the material is a good insulator.Since all three materials have the same cross section and the heat flow across each is the same (temperatures are X not changing), we may define another constant, K, characteristics of the material: K œ
?"X Þ Using the values of ? ?B from ?B

the prevous problem, fiberglass has the smallest K at 0.06 and thus is the best insulator. Likewise, the wallboard is the poorest insulator, with K œ 0.4. 83. p œ kd  1 and p œ 10.94 at d œ 100 Ê k œ

10.94" 100

œ 0.0994. Then p œ 0.0994d  1 is the diver's

pressure equation so that d œ 50 Ê p œ (0.0994)(50)  1 œ 5.97 atmospheres. 84. The line of incidence passes through (!ß 1) and ("ß 0) Ê The line of reflection passes through ("ß 0) and (#ß ") 0 Ê m œ 1# 1 œ 1 Ê y
0 œ 1(x
1) Ê y œ x
1 is the line of reflection.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.2 Lines, Circles, and Parabolas 85. C œ

5 9

(F
32) and C œ F Ê F œ

86. m œ

37.1 100

œ

14 ?x

Ê ?x œ

14 .371 .

5 9

F


160 9

Ê

4 9

15

F œ
160 9 or F œ
40° gives the same numerical reading.

#

14 ‰ Therefore, distance between first and last rows is É(14)#  ˆ .371 ¸ 40.25 ft.

87. length AB œ È(5
1)#  (5
2)# œ È16  9 œ 5 length AC œ È(4
1)#  (
#
#)# œ È9  16 œ 5 length BC œ È(4
5)#  (
#
5)# œ È1  49 œ È50 œ 5È2 Á 5 #

88. length AB œ Ê(1
0)#  ŠÈ3
0‹ œ È1  3 œ 2 length AC œ È(2
0)#  (0
0)# œ È4  0 œ 2 #

length BC œ Ê(2
1)#  Š0
È3‹ œ È1  3 œ 2 89. Length AB œ È(?x)#  (?y)# œ È1#  4# œ È17 and length BC œ È(?x)#  (?y)# œ È4#  1# œ È17. Also, slope AB œ 41 and slope BC œ "4 , so AB ¼ BC. Thus, the points are vertices of a square. The coordinate increments from the fourth vertex D(xß y) to A must equal the increments from C to B Ê 2
x œ ?x œ 4 and
1
y œ ?y œ " Ê x œ
2 and y œ
2. Thus D(
#ß
2) is the fourth vertex.

90. Let A œ (xß 2) and C œ (9ß y) Ê B œ (xß y). Then 9
x œ kADk and 2
y œ kDCk Ê 2(9
x)  2(2
y) œ 56 and 9
x œ 3(2
y) Ê 2(3(2
y))  2(2
y) œ 56 Ê y œ
5 Ê 9
x œ 3(2
(
5)) Ê x œ
12. Therefore, A œ (
12ß 2), C œ (9ß
5), and B œ (
12ß
5). 91. Let A(
"ß "), B(#ß $), and C(2ß !) denote the points. Since BC is vertical and has length kBCk œ 3, let D" (
"ß 4) be located vertically upward from A and D# (
"ß
2) be located vertically downward from A so that kBCk œ kAD" k œ kAD# k œ 3. Denote the point D$ (xß y). Since the slope of AB equals the slope of 3 " CD$ we have yx 2 œ
3 Ê 3y
9 œ
x  2 or

x  3y œ 11. Likewise, the slope of AC equals the slope 0 2 of BD$ so that yx   2 œ 3 Ê 3y œ 2x
4 or 2x
3y œ 4.

Solving the system of equations

x  3y œ "" we find x œ 5 and y œ 2 yielding the vertex D$ (5ß #). 2x
3y œ 4


92. Let ax, yb, x Á ! and/or y Á ! be a point on the coordinate plane. The slope, m, of the segment a!ß !b to ax, yb is yx . A 90‰

rotation gives a segment with slope mw œ
m" œ
xy . If this segment has length equal to the original segment, its endpoint will be a
y, xb or ay,
xb, the first of these corresponds to a counter-clockwise rotation, the latter to a clockwise rotation. (a) (
"ß 4); (b) (3ß
2); (c) (5ß 2); (d) (0ß x);

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16

Chapter 1 Preliminaries (e) (
yß 0);

(f) (
yß x);

(g) (3ß
10)

93. 2x  ky œ 3 has slope
2k and 4x  y œ 1 has slope
4. The lines are perpendicular when
2k (
4) œ
1 or k œ
8 and parallel when
2k œ
4 or k œ "# .

94. At the point of intersection, 2x  4y œ 6 and 2x
3y œ
1. Subtracting these equations we find 7y œ 7 or y œ 1. Substitution into either equation gives x œ 1 Ê (1ß 1) is the intersection point. The line through (1ß 1) and ("ß #) is vertical with equation x œ 1. 95. Let M(aß b) be the midpoint. Since the two triangles shown in the figure are congruent, the value a must lie midway between x" and x# , so a œ x"
#x# . Similarly, b œ

y "
y # # .

96. (a) L has slope 1 so M is the line through P(2ß 1) with slope
1; or the line y œ
x  3. At the intersection point, Q, we have equal y-values, y œ x  2 œ
x  3. Thus, 2x œ 1 or x œ "# . Hence Q has coordinates ˆ "# ß 5# ‰ . The distance from P to L œ the distance from P to Q œ Ɉ #3 ‰#  ˆ
3# ‰# œ É 18 4 œ (b) L has slope
43 so M has slope

3 4

3È 2 # .

and M has the equation 4y
3x œ 12. We can rewrite the equations of

84 the lines as L: x  y œ 3 and M:
B  43 y œ 4. Adding these we get 25 12 y œ 7 so y œ 25 . Substitution 12 ‰ ˆ 12 84 ‰ into either equation gives x œ 43 ˆ 84 25
4 œ 25 so that Q 25 ß 25 is the point of intersection. The distance 3 4

from P to L œ Ɉ4


12 ‰# 25

 ˆ6


84 ‰# 25

œ

22 5 .

(c) M is a horizontal line with equation y œ b. The intersection point of L and M is Q(
"ß b). Thus, the distance from P to L is È(a  1)#  0# œ ka  1k . (d) If B œ 0 and A Á 0, then the distance from P to L is ¸ AC
x! ¸ as in (c). Similarly, if A œ 0 and B Á 0, the distance is ¸ CB
y! ¸ . If both A and B are Á 0 then L has slope
AB so M has slope AB . Thus, L: Ax  By œ C and M:
Bx  Ay œ
Bx!  Ay! . Solving these equations simultaneously we find the point of intersection Q(xß y) with x œ

ACB aAy! Bx! b A#
B#

P to Q equals È(?x)#  (?y)# , where (?x)# œ œ

A# aAx!
By!
Cb# aA#
B# b#

#

#

BC
A aAy! Bx! b . A#
B# # # # #
ABy! B x! Š x! aA
B bAAC ‹ #
B#

and y œ

#

#

A y!
ABx! , and (?y)# œ Š y! aA
B bABC ‹ œ #
B# #

!
Cb Thus, È(?x)#  (?y)# œ É aAx!A
#By œ
B#

kAx!
By!
Ck ÈA#
B#

The distance from

B# aAx!
By!
Cb# . aA#
B# b#

.

1.3 FUNCTIONS AND THEIR GRAPHS 1. domain œ (
_ß _); range œ [1ß _) 3. domain œ (!ß _); y in range Ê y œ

2. domain œ [0ß _); range œ (
_ß 1] " Èt

, t  0 Ê y# œ

" t

and y  ! Ê y can be any positive real number

Ê range œ (!ß _).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.3 Functions and Their Graphs 4. domain œ [0ß _); y in range Ê y œ

" 1
È t

17

, t  0. If t œ 0, then y œ 1 and as t increases, y becomes a smaller

and smaller positive real number Ê range œ (0ß 1]. 5. 4
z# œ (2
z)(2  z)   0 Í z − [
2ß 2] œ domain. Largest value is g(0) œ È4 œ 2 and smallest value is g(
2) œ g(2) œ È0 œ 0 Ê range œ [0ß 2]. 6. domain œ (
2ß 2) from Exercise 5; smallest value is g(0) œ "# and as 0  z increases to 2, g(z) gets larger and larger (also true as z  0 decreases to
2) Ê range œ  "# ß _‰ . 7. (a) Not the graph of a function of x since it fails the vertical line test. (b) Is the graph of a function of x since any vertical line intersects the graph at most once. 8. (a) Not the graph of a function of x since it fails the vertical line test. (b) Not the graph of a function of x since it fails the vertical line test. 9. y œ Ɉ "x ‰
" Ê (a) No (x  !Ñ; (c) No; if x   ",

" x

" x


"   ! Ê x Ÿ 1 and x  !. So,

"Ê

" x

(b) No; division by ! undefined; (d) Ð!ß "Ó


"  !;

10. y œ É#
Èx Ê #
Èx   ! Ê Èx   ! and Èx Ÿ #. Èx   ! Ê x   ! and Èx Ÿ # Ê x Ÿ %Þ So, ! Ÿ x Ÿ %. (a) No; (b) No; (c) Ò!ß %Ó #

11. base œ x; (height)#  ˆ #x ‰ œ x# Ê height œ

È3 #

x; area is a(x) œ

" #

(base)(height) œ

" #

(x) Š

È3 # x‹

œ

È3 4

x# ;

perimeter is p(x) œ x  x  x œ 3x. 12. s œ side length Ê s#  s# œ d# Ê s œ

d È2

; and area is a œ s# Ê a œ

" #

d#

13. Let D œ diagonal of a face of the cube and j œ the length of an edge. Then j#  D# œ d# and (by Exercise 10) D# œ 2j# Ê 3j# œ d# Ê j œ

d È3

. The surface area is 6j# œ

6d# 3

14. The coordinates of P are ˆxß Èx‰ so the slope of the line joining P to the origin is m œ ˆx, Èx‰ œ ˆ m"# ,

#

œ 2d# and the volume is j$ œ Š d3 ‹ Èx x

œ

" Èx

"‰ m .

15. The domain is a
_ß _b.

16. The domain is a
_ß _b.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

$Î#

œ

d$ 3È 3

(x  0). Thus,

.

18

Chapter 1 Preliminaries

17. The domain is a
_ß _b.

18. The domain is Ð
_ß !Ó.

19. The domain is a
_ß !b  a!ß _b.

20. The domain is a
_ß !b  a!ß _b.

21. Neither graph passes the vertical line test (a)

(b)

22. Neither graph passes the vertical line test (a)

(b)

Ú xyœ" Þ Ú yœ1
x Þ or or kx  yk œ 1 Í Û Í Û ß ß Ü x  y œ
" à Ü y œ
"
x à

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.3 Functions and Their Graphs 23.

x y

0 0

25. y œ œ

1 1

2 0

24.

x y

0 1

1 0

2 0

" , x0 26. y œ œ x x, 0 Ÿ x

3
x, x Ÿ 1 2x, 1  x

27. (a) Line through a!ß !b and a"ß "b: y œ x Line through a"ß "b and a#ß !b: y œ
x  2 x, 0 Ÿ x Ÿ 1 f(x) œ œ
x  2, 1  x Ÿ 2 Ú Ý Ý 2, ! Ÿ x  " !ß " Ÿ x  # (b) f(x) œ Û Ý Ý 2ß # Ÿ x  $ Ü !ß $ Ÿ x Ÿ % 28. (a) Line through a!ß 2b and a#ß !b: y œ
x  2 " Line through a2ß "b and a&ß !b: m œ !&

# œ
x  #, 0  x Ÿ # f(x) œ œ "
$ x  &$ , #  x Ÿ &


" $


$
! !
Ð
"Ñ œ
"
$
% #
! œ #

(b) Line through a
"ß !b and a!ß
$b: m œ Line through a!ß $b and a#ß
"b: m œ f(x) œ œ

œ
"$ , so y œ
"$ ax
2b  " œ
"$ x 

& $


$, so y œ
$x
$ œ
#, so y œ
#x  $


$x
$,
"  x Ÿ !
#x  $, !  x Ÿ #

29. (a) Line through a
"ß "b and a!ß !b: y œ
x Line through a!ß "b and a"ß "b: y œ " Line through a"ß "b and a$ß !b: m œ !
" $
" œ Ú
x
" Ÿ x  ! " !xŸ" f(x) œ Û Ü
"# x  $# "x$


" #

œ
"# , so y œ
"# ax
"b  " œ
"# x 

$ #

(b) Line through a
#ß
"b and a!ß !b: y œ "# x

Line through a!ß #b and a"ß !b: y œ
#x  # Line through a"ß
"b and a$ß
"b: y œ
"

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19

20

Chapter 1 Preliminaries Ú

" #x

f(x) œ Û
#x  # Ü
"


# Ÿ x Ÿ ! !xŸ" "xŸ$

30. (a) Line through ˆ T# ß !‰ and aTß "b: m œ

Ú A, Ý Ý Ý
Aß f(x) œ Û Aß Ý Ý Ý Ü
Aß

! Ÿ x  T# T # Ÿ x T T Ÿ x  $#T $T # Ÿ x Ÿ #T x #

31. (a) From the graph, (b)

x #

1

x  0:

x #

x  0:

x 2

œ T# , so y œ T# ˆx
T# ‰  0 œ T# x
"

!, 0 Ÿ x Ÿ T# # T T x
", #  x Ÿ T

f(x) œ


(b)

"
! T
aTÎ#b

4 x


1


x #

Ê 4 x



1

4 x

Ê x − (
2ß 0)  (%ß _)


1
4x  0 # 2x
8 0 Ê x
2x

0 Ê

(x
4)(x2) #x

0

(x
4)(x2) #x

0

Ê x  4 since x is positive;
1


4 x

0 Ê

x#
2x
8 2x

0 Ê

Ê x 
2 since x is negative; sign of (x
4)(x  2)  
ïïïïïðïïïïïðïïïïî
2 % Solution interval: (
#ß 0)  (%ß _)

3 2 x
1  x  1 3 2 x
1  x  1

32. (a) From the graph, (b) Case x 
1:

Ê x − (
_ß
5)  (
1ß 1) Ê

3(x1) x
1

2

Ê 3x  3  2x
2 Ê x 
5. Thus, x − (
_ß
5) solves the inequality. Case
1  x  1:

3 x
1



2 x 1

Ê

3(x1) x
1

2

Ê 3x  3  2x
2 Ê x 
5 which is true if x 
1. Thus, x − (
1ß 1) solves the inequality. Case 1  x: x
3 1  x2 1 Ê 3x  3  2x
2 Ê x 
5 which is never true if 1  x, so no solution here. In conclusion, x − (
_ß
5)  (
1ß 1). 33. (a) ÚxÛ œ 0 for x − [0ß 1)

(b) ÜxÝ œ 0 for x − (
1ß 0]

34. ÚxÛ œ ÜxÝ only when x is an integer. 35. For any real number x, n Ÿ x Ÿ n  ", where n is an integer. Now: n Ÿ x Ÿ n  " Ê
Ðn  "Ñ Ÿ
x Ÿ
n. By definition: Ü
xÝ œ
n and ÚxÛ œ n Ê
ÚxÛ œ
n. So Ü
xÝ œ
ÚxÛ for all x − d .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.3 Functions and Their Graphs

21

36. To find f(x) you delete the decimal or fractional portion of x, leaving only the integer part.

37. v œ f(x) œ xÐ"%
2xÑÐ22
2xÑ œ %x$
72x#  $!)x; !  x  7Þ 38. (a) Let h œ height of the triangle. Since the triangle is isosceles, AB #  AB # œ 2# Ê AB œ È2Þ So, #

h#  "# œ ŠÈ2‹ Ê h œ " Ê B is at a!ß "b Ê slope of AB œ
" Ê The equation of AB is y œ f(x) œ
B  "; x − Ò!ß "Ó. (b) AÐxÑ œ 2x y œ 2xÐ
x  "Ñ œ
2x#  #x; x − Ò!ß "Ó. 39. (a) Because the circumference of the original circle was )1 and a piece of length x was removed. x x (b) r œ )1#
1 œ %
#1 (c) h œ È"'
r# œ É"'
ˆ%
#

x‰ (d) V œ "$ 1 r# h œ "$ 1ˆ )1#
† 1

x ‰# #1

œ É"'
ˆ16


È"'1x
x# #1

œ

4x 1



x# ‰ %1#

œ É 4x 1


x# %1#

œ É "'%11#x


x# %1#

œ

È"'1x
x# #1

a)1
xb# È"'1x
x# #%1#

40. (a) Note that 2 mi = 10,560 ft, so there are È)!!#  x# feet of river cable at $180 per foot and a"!ß &'!
xb feet of land cable at $100 per foot. The cost is Caxb œ ")!È)!!#  x#  "!!a"!ß &'!
xb. (b) Ca!b œ $"ß #!!ß !!! Ca&!!b ¸ $"ß "(&ß )"# Ca"!!!b ¸ $"ß ")'ß &"# Ca"&!!b ¸ $"ß #"#ß !!! Ca#!!!b ¸ $"ß #%$ß ($# Ca#&!!b ¸ $"ß #()ß %(* Ca$!!!b ¸ $"ß $"%ß )(! Values beyond this are all larger. It would appear that the least expensive location is less than 2000 feet from the point P. 41. A curve symmetric about the x-axis will not pass the vertical line test because the points ax, yb and ax,
yb lie on the same vertical line. The graph of the function y œ faxb œ ! is the x-axis, a horizontal line for which there is a single y-value, !, for any x. 42. Pick 11, for example: ""  & œ "' Ä # † "' œ $# Ä $#
' œ #' Ä faxb œ

#ax&b
' #

#' #

œ "$ Ä "$
# œ "", the original number.


# œ x, the number you started with.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

22

Chapter 1 Preliminaries

1.4 IDENTIFYING FUNCTIONS; MATHEMATICAL MODELS 1. (a) linear, polynomial of degree 1, algebraic. (c) rational, algebraic.

(b) power, algebraic. (d) exponential.

2. (a) polynomial of degree 4, algebraic. (c) algebraic.

(b) exponential. (d) power, algebraic.

3. (a) rational, algebraic. (c) trigonometric.

(b) algebraic. (d) logarithmic.

4. (a) logarithmic. (c) exponential.

(b) algebraic. (d) trigonometric.

5. (a) Graph h because it is an even function and rises less rapidly than does Graph g. (b) Graph f because it is an odd function. (c) Graph g because it is an even function and rises more rapidly than does Graph h. 6. (a) Graph f because it is linear. (b) Graph g because it contains a!ß "b. (c) Graph h because it is a nonlinear odd function. 7. Symmetric about the origin Dec:
_  x  _ Inc: nowhere

8. Symmetric about the y-axis Dec:
_  x  ! Inc: !  x  _

9. Symmetric about the origin Dec: nowhere Inc:
_  x  ! !x_

10. Symmetric about the y-axis Dec: !  x  _ Inc:
_  x  !

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.4 Identifying Functions; Mathematical Models 11. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _

12. No symmetry Dec:
_  x Ÿ ! Inc: nowhere

13. Symmetric about the origin Dec: nowhere Inc:
_  x  _

14. No symmetry Dec: ! Ÿ x  _ Inc: nowhere

15. No symmetry Dec: ! Ÿ x  _ Inc: nowhere

16. No symmetry Dec:
_  x Ÿ ! Inc: nowhere

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

23

24

Chapter 1 Preliminaries

17. Symmetric about the y-axis Dec:
_  x Ÿ ! Inc: !  x  _

18. Symmetric about the y-axis Dec: ! Ÿ x  _ Inc:
_  x  !

19. Since a horizontal line not through the origin is symmetric with respect to the y-axis, but not with respect to the origin, the function is even. 20. faxb œ x
& œ

" x&

and fa
xb œ a
xb
& œ

" a
x b&

œ
ˆ x"& ‰ œ
faxb. Thus the function is odd.

21. Since faxb œ x#  " œ a
xb#  " œ
faxb. The function is even. 22. Since Òfaxb œ x#  xÓ Á Òfa
xb œ a
xb#
xÓ and Òfaxb œ x#  xÓ Á Ò
faxb œ
axb#
xÓ the function is neither even nor odd. 23. Since gaxb œ x$  x, ga
xb œ
x$
x œ
ax$  xb œ
gaxb. So the function is odd. 24. gaxb œ x%  $x#  " œ a
xb%  $a
xb#
" œ ga
xbß thus the function is even. 25. gaxb œ

" x#
"

26. gaxb œ

x x#
" ;

27. hatb œ

" t
";

œ

" a
xb#
"

œ ga
xb. Thus the function is even.

ga
xb œ
x#x
" œ ga
xb. So the function is odd.

h a
t b œ

"
t
" ;


h at b œ

" "
t.

Since hatb Á
hatb and hatb Á ha
tb, the function is neither even nor odd.

28. Since l t$ | œ l a
tb$ |, hatb œ ha
tb and the function is even. 29. hatb œ 2t  ", ha
tb œ
2t  ". So hatb Á ha
tb.
hatb œ
2t
", so hatb Á
hatb. The function is neither even nor odd. 30. hatb œ 2l t l  " and ha
tb œ 2l
t l  " œ 2l t l  ". So hatb œ ha
tb and the function is even. 31. (a)

The graph supports the assumption that y is proportional to x. The constant of proportionality is estimated from the slope of the regression line, which is 0.166.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.4 Identifying Functions; Mathematical Models (b)

25

The graph supports the assumption that y is proportional to x"Î# . The constant of proportionality is estimated from the slope of the regression line, which is 2.03.

32. (a) Because of the wide range of values of the data, two graphs are needed to observe all of the points in relation to the regression line.

The graphs support the assumption that y is proportional to $x . The constant of proportionality is estimated from the slope of the regression line, which is 5.00. (b) The graph supports the assumption that y is proportional to ln x. The constant of proportionality is extimated from the slope of the regression line, which is 2.99.

33. (a) The scatterplot of y œ reaction distance versus x œ speed is

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 1.1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

26

Chapter 1 Preliminaries (b) Calculate x w œ speed squared. The scatterplot of x w versus y œ braking distance is:

Answers for the constant of proportionality may vary. The constant of proportionality is the slope of the line, which is approximately 0.059. 34. Kepler's 3rd Law is Tadaysb œ !Þ%"R$Î# , R in millions of miles. "Quaoar" is 4 ‚ "!* miles from Earth, or about 4 ‚ "!*  *$ ‚ "!' ¸ % ‚ "!* miles from the sun. Let R œ 4000 (millions of miles) and T œ a!Þ%"ba%!!!b$Î# days ¸ "!$ß (#$ days. 35. (a)

The hypothesis is reasonable. (b) The constant of proportionality is the slope of the line ¸ (c) y(in.) œ a!Þ)( in./unit massba"$ unit massb œ ""Þ$" in. 36. (a)

)Þ(%"
! "!
!

in./unit mass œ !Þ)(% in./unit mass.

(b)

Graph (b) suggests that y œ k x$ is the better model. This graph is more linear than is graph (a). 1.5 COMBINING FUNCTIONS; SHIFTING AND SCALING GRAPHS 1. Df :
_  x  _, Dg : x   1 Ê Dfbg œ Dfg : x   1. Rf :
_  y  _, Rg : y   0, Rfbg : y   1, Rfg : y   0 2. Df : x  1   0 Ê x  
1, Dg : x
1   0 Ê x   1. Therefore Dfbg œ Dfg : x   1. Rf œ Rg : y   0, Rfbg : y   È2, Rfg : y   0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs 3. Df :
_  x  _, Dg :
_  x  _ Ê DfÎg :
_  x  _ since g(x) Á 0 for any x; DgÎf :
_  x  _ since f(x) Á 0 for any x. Rf : y œ 2, Rg : y   1, RfÎg : 0  y Ÿ 2, RgÎf : y   "# 4. Df :
_  x  _, Dg : x   0 Ê DfÎg : x   0 since g(x) Á 0 for any x   0; DgÎf : x   0 since f(x) Á 0 for any x   0. Rf : y œ 1, Rg : y   1, RfÎg : 0  y Ÿ 1, RgÎf : y   " 5. (a) (b) (c) (d) (e) (f) (g) (h)

f(g(0)) œ f(
3) œ 2 g(f(0)) œ g(5) œ 22 f(g(x)) œ f(x#
3) œ x#
3  5 œ x#  2 g(f(x)) œ g(x  5) œ (x  5)#
3 œ x#  10x  22 f(f(
5)) œ f(0) œ 5 g(g(2)) œ g(1) œ
2 f(f(x)) œ f(x  5) œ (x  5)  5 œ x  10 g(g(x)) œ g(x#
3) œ (x#
3)#
3 œ x%
6x#  6

6. (a) f ˆg ˆ "# ‰‰ œ f ˆ 23 ‰ œ
3" (b) g ˆf ˆ "# ‰‰ œ g ˆ
"# ‰ œ 2 (c) f(g(x)) œ f ˆ x " 1 ‰ œ

" x 1


1œ

(d) g(f(x)) œ g(x
1) œ

" (x
1)  1

(e) f(f(2)) œ f(1) œ 0 (f) g(g(2)) œ g ˆ "3 ‰ œ

œ

" 4 3

œ


x x1 " x

3 4

(g) f(f(x)) œ f(x
1) œ (x
1)
1 œ x
2 " (h) g(g(x)) œ g ˆ x " 1 ‰ œ " " 1 œ xx  # (x Á
1 and x Á
2) x
1

# 7. (a) u(v(f(x))) œ u ˆv ˆ "x ‰‰ œ u ˆ x"# ‰ œ 4 ˆ x" ‰
5 œ x4#
5 (b) u(f(v(x))) œ u af ax# bb œ u ˆ x"# ‰ œ 4 ˆ x"# ‰
5 œ x4#
5 # (c) v(u(f(x))) œ v ˆu ˆ "x ‰‰ œ v ˆ4 ˆ x" ‰
5‰ œ ˆ 4x
5‰

(d) v(f(u(x))) œ v(f(4x
5)) œ v ˆ 4x "
5 ‰ œ ˆ 4x "
5 ‰ (e) f(u(v(x))) œ f au ax# bb œ f a4 ax# b
5b œ

" 4x#
5

(f) f(v(u(x))) œ f(v(4x
5)) œ f a(4x
5)# b œ 8. (a) h(g(f(x))) œ h ˆg ˆÈx‰‰ œ h Š

Èx 4 ‹

#

œ 4Š

" (4x
5)#

Èx 4 ‹


8 œ Èx
8

(b) h(f(g(x))) œ h ˆf ˆ x4 ‰‰ œ h ˆÈ x4 ‰ œ 4È x4
8 œ 2Èx
8 4È x
8 œ Èx
2 4 È4x
8 È œ 4 œ x#
2

(c) g(h(f(x))) œ g ˆh ˆÈx‰‰ œ g ˆ4Èx
8‰ œ (d) g(f(h(x))) œ g(f(4x
8)) œ g ŠÈ4x
8‹

(e) f(g(h(x))) œ f(g(4x
8)) œ f ˆ 4x 4
8 ‰ œ f(x
2) œ Èx
2 (f) f(h(g(x))) œ f ˆh ˆ x ‰‰ œ f ˆ4 ˆ x ‰
8‰ œ f(x
8) œ Èx
8 4

4

9. (a) y œ f(g(x)) (c) y œ g(g(x)) (e) y œ g(h(f(x)))

(b) y œ j(g(x)) (d) y œ j(j(x)) (f) y œ h(j(f(x)))

10. (a) y œ f(j(x)) (c) y œ h(h(x)) (e) y œ j(g(f(x)))

(b) y œ h(g(x)) œ g(h(x)) (d) y œ f(f(x)) (f) y œ g(f(h(x))) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

27

28

Chapter 1 Preliminaries g(x)

f(x)

(f ‰ g)(x)

(a)

x
7

Èx

Èx
7

(b)

x2

3x

3(x  2) œ 3x  6

(c)

x#

Èx
5

Èx#
5

(d)

x x
1

x x
1

" x
1 " x

1

11.

(e) (f)

x xc1 x xc1
1

" x

(b) af‰gbaxb œ

gaxb
" g ax b

œ

x x
(x
1)

œx

x

" x

12. (a) af‰gbaxb œ lgaxbl œ

œ

x

" lx
"l . x x"

Ê"


" g ax b

œ

x x"

Ê"


x x"

œ

" g ax b

Ê

" x"

œ

" gaxb ß so

gaxb œ x  ".

(c) Since af‰gbaxb œ Ègaxb œ lxl, gaxb œ x . (d) Since af‰gbaxb œ fˆÈx‰ œ l x l, faxb œ x# . (Note that the domain of the composite is Ò!ß _Ñ.) #

The completed table is shown. Note that the absolute value sign in part (d) is optional. gaxb faxb af‰gbaxb " " lxl x
" lx
"l x
" x

x"

Èx

x# Èx

x#

x x"

lxl lxl

13. (a) fagaxbb œ É 1x  1 œ É 1x x gafaxbb œ

1 Èx  1

(b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð
1, _Ñ (c) Range af‰gb: Ð1, _Ñ, range ag‰f b: Ð0, _Ñ 14. (a) fagaxbb œ 1
2Èx  x gafaxbb œ 1
kxk (b) Domain af‰gb: Ð0, _Ñ, domain ag‰f b: Ð0, _Ñ (c) Range af‰gb: Ð0, _Ñ, range ag‰f b: Ð
_, 1Ñ 15. (a) y œ
(x  7)#

(b) y œ
(x
4)#

16. (a) y œ x#  3

(b) y œ x#
5

17. (a) Position 4

(b) Position 1

(c) Position 2

(d) Position 3

18. (a) y œ
(x
1)#  4

(b) y œ
(x  2)#  3

(c) y œ
(x  4)#
1

(d) y œ
(x
2)#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs 19.

20.

21.

22.

23.

24.

25.

26.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

29

30

Chapter 1 Preliminaries

27.

28.

29.

30.

31.

32.

33.

34.

35.

36.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs 37.

38.

39.

40.

41.

42.

43.

44.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

31

32

Chapter 1 Preliminaries

45.

46.

47.

48.

49. (a) domain: [0ß 2]; range: [#ß $]

(b) domain: [0ß 2]; range: [
1ß 0]

(c) domain: [0ß 2]; range: [0ß 2]

(d) domain: [0ß 2]; range: [
1ß 0]

(e) domain: [
2ß 0]; range: [!ß 1]

(f) domain: [1ß 3]; range: [!ß "]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs (g) domain: [
2ß 0]; range: [!ß "]

(h) domain: [
1ß 1]; range: [!ß "]

50. (a) domain: [0ß 4]; range: [
3ß 0]

(b) domain: [
4ß 0]; range: [!ß $]

(c) domain: [
4ß 0]; range: [!ß $]

(d) domain: [
4ß 0]; range: ["ß %]

(e) domain: [#ß 4]; range: [
3ß 0]

(f) domain: [
2ß 2]; range: [
3ß 0]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

33

34

Chapter 1 Preliminaries (g) domain: ["ß 5]; range: [
3ß 0]

(h) domain: [0ß 4]; range: [0ß 3]

51. y œ 3x#
3 52. y œ a2xb#
1 œ %x#
1 53. y œ "# ˆ"  54. y œ 1 

"‰ x#

" axÎ$b#

œ

" #



œ1

" #x#

* x#

55. y œ È%x  1 56. y œ 3Èx  1 # 57. y œ É%
ˆ x# ‰ œ "# È16
x#

58. y œ "$ È%
x# 59. y œ "
a3xb$ œ "
27x$ $

60. y œ "
ˆ x# ‰ œ "


x$ )

"Î# "Î# 61. Let y œ
È#x  " œ faxb and let gaxb œ x"Î# , haxb œ ˆx  "# ‰ , iaxb œ È#ˆx  "# ‰ , and "Î# jaxb œ
’È#ˆx  "# ‰ “ œ faBb. The graph of haxb is the graph of gaxb shifted left

" #

unit; the graph of iaxb is the graph

of haxb stretched vertically by a factor of È#; and the graph of jaxb œ faxb is the graph of iaxb reflected across the x-axis.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs 62. Let y œ È"


x #

œ faxbÞ Let gaxb œ a
xb"Î# , haxb œ a
x  #b"Î# , and iaxb œ

" È # a
x

 #b"Î# œ È"


x #

35

œ faxbÞ

The graph of gaxb is the graph of y œ Èx reflected across the x-axis. The graph of haxb is the graph of gaxb shifted right two units. And the graph of iaxb is the graph of haxb compressed vertically by a factor of È#.

63. y œ faxb œ x$ . Shift faxb one unit right followed by a shift two units up to get gaxb œ ax
"b3  #.

64. y œ a"
Bb$  # œ
Òax
"b$  a
#bÓ œ faxb. Let gaxb œ x$ , haxb œ ax
"b$ , iaxb œ ax
"b$  a
#b, and jaxb œ
Òax
"b$  a
#bÓ. The graph of haxb is the graph of gaxb shifted right one unit; the graph of iaxb is the graph of haxb shifted down two units; and the graph of faxb is the graph of iaxb reflected across the x-axis.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

36

Chapter 1 Preliminaries

65. Compress the graph of faxb œ get haxb œ

" #x


".

66. Let faxb œ

" x#

and gaxb œ

# x#

" x

horizontally by a factor of 2 to get gaxb œ

"œ

" # Š B# ‹

"œ

"

#

ŠxÎÈ#‹

"œ

" # ’Š"ÎÈ#‹B“

" #x .

Then shift gaxb vertically down 1 unit to

 "Þ Since È# ¸ "Þ%, we see that the graph of

faxb stretched horizontally by a factor of 1.4 and shifted up 1 unit is the graph of gaxb.

$ $ 67. Reflect the graph of y œ faxb œ È x across the x-axis to get gaxb œ
È x.

68. y œ faxb œ a
#xb#Î$ œ Òa
"ba#bxÓ#Î$ œ a
"b#Î$ a#xb#Î$ œ a#xb#Î$ . So the graph of faxb is the graph of gaxb œ x#Î$ compressed horizontally by a factor of 2.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.5 Combining Functions; Shifting and Scaling Graphs 69.

70.

71. *x#
#&y# œ ##& Ê

x# &#

73. $x#
ay  #b# œ $ Ê




x# "#




75. $ax  "b#
#ay
#b# œ ' Ê

ax
" b # #

ŠÈ#‹





y
a
#b‘# #

ŠÈ$‹

y# $#

a y
#b # #

ŠÈ$‹

œ"

74. ax
"b#
#y# œ % Ê




y# %#


x
a
"b‘# ##

œ"




# # 76. 'ˆx
$# ‰
*ˆy  "# ‰ œ &% #

œ"

x# # È Š (‹

72. "'x#
(y# œ ""# Ê

œ"

Ê

’x
ˆ
$# ‰“ $#




ˆy
"# ‰# #

ŠÈ'‹

œ"

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

y# # ŠÈ#‹

œ"

37

38 77.

Chapter 1 Preliminaries x# "'




y# *

œ " has its center at a!ß !b. Shiftinig 4 units left and 3 units up gives the center at ah, kb œ a%ß $b. So the

equation is


x
a
4b‘# 4#




ay
3 b # 3#

œ"Ê

ax  %b # 4#




a y
$b # 3#

œ ". Center, C, is a%ß $b, and major axis, AB, is the segment

from a)ß $b to a!ß $b.

78. The ellipse

x# %




y# #&

œ " has center ah, kb œ a!ß !b. Shifting the ellipse 3 units right and 2 units down produces an ellipse

with center at ah, kb œ a$ß #b and an equation a$ß $b to a$ß (b is the major axis.

ax
3 b# %





y
a
#b‘# #&

œ ". Center, C, is a3ß #b, and AB, the segment from

79. (a) (fg)(x) œ f(x)g(x) œ f(x)(g(x)) œ (fg)(x), odd (b) Š gf ‹ (x) œ (c) ˆ gf ‰ (x) œ (d) (e) (f) (g) (h) (i)

f(
x) g(
x) g(
x) f(
x)

œ œ

f(x)
g(x)
g(x) f(x)

œ  Š gf ‹ (x), odd œ  ˆ gf ‰ (x), odd

f # (x) œ f(x)f(x) œ f(x)f(x) œ f # (x), even g# (x) œ (g(x))# œ (g(x))# œ g# (x), even (f ‰ g)(x) œ f(g(x)) œ f(g(x)) œ f(g(x)) œ (f ‰ g)(x), even (g ‰ f)(x) œ g(f(x)) œ g(f(x)) œ (g ‰ f)(x), even (f ‰ f)(x) œ f(f(x)) œ f(f(x)) œ (f ‰ f)(x), even (g ‰ g)(x) œ g(g(x)) œ g(g(x)) œ g(g(x)) œ (g ‰ g)(x), odd

80. Yes, f(x) œ 0 is both even and odd since f(x) œ 0 œ f(x) and f(x) œ 0 œ f(x).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.6 Trigonometric Functions 81. (a)

(b)

(c)

(d)

82.

1.6 TRIGONOMETRIC FUNCTIONS 1. (a) s œ r) œ (10) ˆ 451 ‰ œ 81 m radians and

51 4

1 ‰ 3. ) œ 80° Ê ) œ 80° ˆ 180° œ

41 9

2. ) œ

s r

œ

101 8

œ

51 4

1 ‰ (b) s œ r) œ (10)(110°) ˆ 180° œ

1101 18

œ

551 9

m

ˆ 180° ‰ œ 225° 1 Ê s œ (6) ˆ 491 ‰ œ 8.4 in. (since the diameter œ 12 in. Ê radius œ 6 in.)

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39

40

Chapter 1 Preliminaries

4. d œ 1 meter Ê r œ 50 cm Ê ) œ 5.

1

)

 231

0

1 #

s r

œ

30 50

31 4 " È2  È" 2

sin )

0

cos )

1

È  #3  "#

tan )

0

È3

0

und.

"

und.

" È3

und.

0

1

und.

È 2

cot )

1

#

und.

 È23

sec ) csc )

0

"

"

0

" und.

7. cos x œ  45 , tan x œ  34 9. sin x œ 

È8 3

, tan x œ È8

"

‰ ¸ 34° œ 0.6 rad or 0.6 ˆ 180° 1 6.

È2

 3#1

) sin )

"

cos )

!

" #

tan )

und.

È 3

cot )

!

 È"3

sec )

und.

#

csc )

"

 È23

2 È5

10. sin x œ

12 13

11. sin x œ  È"5 , cos x œ  È25

12. cos x œ 

13.

14.

15.

 1'

È  #3

8. sin x œ

period œ 1

 13

, cos x œ

" È2

&1 ' " # È  #3

 È"3

"

 È"3

È 3

"

È 3

2 È3

È2

 È23

#

È2

#

 "# È3 #

" È5

, tan x œ  12 5 È3 #

, tan x œ

" È3

period œ 41 16.

period œ 2 17.

period œ 4 18.

period œ 6

period œ 1

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1 % " È2

Section 1.6 Trigonometric Functions 19.

20.

period œ 21

period œ 21

21.

22.

period œ 21

period œ 21

23. period œ 1# , symmetric about the origin

24. period œ 1, symmetric about the origin

25. period œ 4, symmetric about the y-axis

26. period œ 41, symmetric about the origin

27. (a) Cos x and sec x are positive in QI and QIV and negative in QII and QIII. Sec x is undefined when cos x is 0. The range of sec x is (_ß 1]  ["ß _); the range of cos x is ["ß 1].

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

41

42

Chapter 1 Preliminaries (b) Sin x and csc x are positive in QI and QII and negative in QIII and QIV. Csc x is undefined when sin x is 0. The range of csc x is (_ß 1]  [1ß _); the range of sin x is ["ß "].

28. Since cot x œ

" tan x

, cot x is undefined when tan x œ 0

and is zero when tan x is undefined. As tan x approaches zero through positive values, cot x approaches infinity. Also, cot x approaches negative infinity as tan x approaches zero through negative values.

29. D: _  x  _; R: y œ 1, 0, 1

30. D: _  x  _; R: y œ 1, 0, 1

31. cos ˆx  1# ‰ œ cos x cos ˆ 1# ‰  sin x sin ˆ 1# ‰ œ (cos x)(0)  (sin x)(1) œ sin x 32. cos ˆx
1# ‰ œ cos x cos ˆ 1# ‰  sin x sin ˆ 1# ‰ œ (cos x)(0)  (sin x)(1) œ sin x 33. sin ˆx
1# ‰ œ sin x cos ˆ 1# ‰
cos x sin ˆ 1# ‰ œ (sin x)(0)
(cos x)(1) œ cos x 34. sin ˆx  1# ‰ œ sin x cos ˆ 1# ‰
cos x sin ˆ 1# ‰ œ (sin x)(0)
(cos x)(1) œ cos x 35. cos (A  B) œ cos (A
(B)) œ cos A cos (B)  sin A sin (B) œ cos A cos B  sin A (sin B) œ cos A cos B
sin A sin B 36. sin (A  B) œ sin (A
(B)) œ sin A cos (B)
cos A sin (B) œ sin A cos B
cos A (sin B) œ sin A cos B  cos A sin B 37. If B œ A, A  B œ 0 Ê cos (A  B) œ cos 0 œ 1. Also cos (A  B) œ cos (A  A) œ cos A cos A
sin A sin A œ cos# A
sin# A. Therefore, cos# A
sin# A œ 1. 38. If B œ 21, then cos (A
21) œ cos A cos 21  sin A sin 21 œ (cos A)(1)  (sin A)(0) œ cos A and sin (A
21) œ sin A cos 21
cos A sin 21 œ (sin A)(1)
(cos A)(0) œ sin A. The result agrees with the fact that the cosine and sine functions have period 21. 39. cos (1
x) œ cos 1 cos B  sin 1 sin x œ (1)(cos x)  (0)(sin x) œ cos x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.6 Trigonometric Functions 40. sin (21
x) œ sin 21 cos (
x)  cos (21) sin (
x) œ (0)(cos (
x))  (1)(sin (
x)) œ
sin x 41. sin ˆ 3#1
x‰ œ sin ˆ 3#1 ‰ cos (
x)  cos ˆ 3#1 ‰ sin (
x) œ (
1)(cos x)  (0)(sin (
x)) œ
cos x 42. cos ˆ 3#1  x‰ œ cos ˆ 3#1 ‰ cos x
sin ˆ 3#1 ‰ sin x œ (0)(cos x)
(
1)(sin x) œ sin x œ sin ˆ 14  13 ‰ œ sin

44. cos

111 1#

45. cos

1 12

œ cos ˆ 13
14 ‰ œ cos

46. sin

51 1#

œ sin ˆ 231
14 ‰ œ sin ˆ 231 ‰ cos ˆ
14 ‰  cos ˆ 231 ‰ sin ˆ
14 ‰ œ Š

21 ‰ 3

cos

œ cos

È

47. cos#

1 8

œ

1
cos ˆ 281 ‰ #

œ

1
# 2 #

49. sin#

1 1#

œ

1cos ˆ 211# ‰ #

œ

1 # 3 #

È

1 3

1 4

1 3

 cos

cos

21 3

1 4

1 3

È2 È3 # ‹Š # ‹

71 1#

œ cos ˆ 14 

1 4

È2 ˆ"‰ # ‹ #

43. sin

sin


sin

1 4

cos ˆ
14 ‰
sin

œŠ

sin 1 3

21 3

œŠ

Š

È2 ˆ "‰ # ‹
#

sin ˆ
14 ‰ œ ˆ "# ‰ Š

œ

2
È 2 4

48. cos#

1 1#

œ

2 È 3 4

50. sin#

1 8


Š

È2 # ‹

œ

È2 È3 # ‹Š # ‹


Š

1
cos ˆ 211# ‰ #

1cos ˆ 281 ‰ #

51. tan (A  B) œ

sin (A
B) cos (A
B)

œ

sin A cos B
cos A cos B cos A cos Bsin A sin B

œ

sin A cos B cos A sin B cos A cos B
cos A cos B sin A sin B cos A cos B  cos A cos B cos A cos B

œ

tan A
tan B 1tan A tan B

52. tan (A
B) œ

sin (AB) cos (AB)

œ

sin A cos Bcos A cos B cos A cos B
sin A sin B

œ

sin A cos B cos A sin B cos A cos B  cos A cos B sin A sin B cos A cos B
cos A cos B cos A cos B

œ

tan Atan B 1
tan A tan B

È 2
È 6 4

œ


È3 È2 # ‹ Š
# ‹

È3 È2 # ‹Š # ‹

œ

È 6
È 2 4

œ

 ˆ
"# ‰ Š


œ

œ

œ

È

1
# 3 #

È

1 # 2 #

œ œ

1
È 3 2È 2 È2 # ‹

œ

1
È 3 2È 2

2
È 3 4

2 È 2 4

53. According to the figure in the text, we have the following: By the law of cosines, c# œ a#  b#
2ab cos ) œ 1#  1#
2 cos (A
B) œ 2
2 cos (A
B). By distance formula, c# œ (cos A
cos B)#  (sin A
sin B)# œ cos# A
2 cos A cos B  cos# B  sin# A
2 sin A sin B  sin# B œ 2
2(cos A cos B  sin A sin B). Thus c# œ 2
2 cos (A
B) œ 2
2(cos A cos B  sin A sin B) Ê cos (A
B) œ cos A cos B  sin A sin B. 54. (a) cosaA
Bb œ cos A cos B  sin A sin B sin ) œ cosˆ 1#
)‰ and cos ) œ sinˆ 1#
)‰ Let ) œ A  B

sinaA  Bb œ cos’ 1#
aA  Bb“ œ cos’ˆ 1#
A‰
B“ œ cos ˆ 1#
A‰ cos B  sin ˆ 1#
A‰ sin B œ sin A cos B  cos A sin B (b) cosaA
Bb œ cos A cos B  sin A sin B cosaA
a
Bbb œ cos A cos a
Bb  sin A sin a
Bb Ê cosaA  Bb œ cos A cos a
Bb  sin A sin a
Bb œ cos A cos B  sin A a
sin Bb œ cos A cos B
sin A sin B Because the cosine function is even and the sine functions is odd. 55. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (60°) œ 4  9
12 cos (60°) œ 13
12 ˆ "# ‰ œ 7. Thus, c œ È7 ¸ 2.65. 56. c# œ a#  b#
2ab cos C œ 2#  3#
2(2)(3) cos (40°) œ 13
12 cos (40°). Thus, c œ È13
12 cos 40° ¸ 1.951.

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43

44

Chapter 1 Preliminaries

57. From the figures in the text, we see that sin B œ hc . If C is an acute angle, then sin C œ hb . On the other hand, if C is obtuse (as in the figure on the right), then sin C œ sin (1
C) œ hb . Thus, in either case, h œ b sin C œ c sin B Ê ah œ ab sin C œ ac sin B. a #
b # c # 2ab

By the law of cosines, cos C œ

and cos B œ

a #
c #  b # . 2ac

Moreover, since the sum of the

interior angles of a triangle is 1, we have sin A œ sin (1
(B  C)) œ sin (B  C) œ sin B cos C  cos B sin C #

#

#

#

#

#

b c c b ˆ h ‰ h ‰ œ ˆ hc ‰ ’ a
2ab a2a#  b#
c#  c#
b# b œ “  ’ a
2ac “ b œ ˆ 2abc

ah bc

Ê ah œ bc sin A.

Combining our results we have ah œ ab sin C, ah œ ac sin B, and ah œ bc sin A. Dividing by abc gives h sin A sin C sin B bc œ ðóóóóóóóñóóóóóóóò a œ c œ b . law of sines 58. By the law of sines, Thus sin B œ

3È 3 2È 7

sin A #

œ

sin B 3

œ

È3/2 c .

By Exercise 55 we know that c œ È7.

¶ 0.982.

59. From the figure at the right and the law of cosines, b# œ a#  2#
2(2a) cos B œ a#  4
4a ˆ "# ‰ œ a#
2a  4. Applying the law of sines to the figure, Ê

È2/2 a

œ

È3/2 b

sin A a

œ

sin B b

Ê b œ É 3# a. Thus, combining results,

a#
2a  4 œ b# œ

3 #

a# Ê 0 œ

" #

a#  2a
4

Ê 0 œ a#  4a
8. From the quadratic formula and the fact that a  0, we have aœ

4
È4# 4(1)(8) #

œ

4 È 3 4 #

¶ 1.464.

60. (a) The graphs of y œ sin x and y œ x nearly coincide when x is near the origin (when the calculator is in radians mode). (b) In degree mode, when x is near zero degrees the sine of x is much closer to zero than x itself. The curves look like intersecting straight lines near the origin when the calculator is in degree mode. 61. A œ 2, B œ 21, C œ
1, D œ
1

62. A œ "# , B œ 2, C œ 1, D œ

" #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.6 Trigonometric Functions 63. A œ
12 , B œ 4, C œ 0, D œ

64. A œ

L 21 ,

" 1

B œ L, C œ 0, D œ 0

65. (a) amplitude œ kAk œ 37 (c) right horizontal shift œ C œ 101

(b) period œ kBk œ 365 (d) upward vertical shift œ D œ 25

66. (a) It is highest when the value of the sine is 1 at f(101) œ 37 sin (0)  25 œ 62° F. The lowest mean daily temp is 37(
1)  25 œ
12° F. (b) The average of the highest and lowest mean daily temperatures œ The average of the sine function is its horizontal axis, y œ 25.

62°
(12)° #

œ 25° F.

67-70. Example CAS commands: Maple f := x -> A*sin((2*Pi/B)*(x-C))+D1; A:=3; C:=0; D1:=0; f_list := [seq( f(x), B=[1,3,2*Pi,5*Pi] )]; plot( f_list, x=-4*Pi..4*Pi, scaling=constrained, color=[red,blue,green,cyan], linestyle=[1,3,4,7], legend=["B=1","B=3","B=2*Pi","B=3*Pi"], title="#67 (Section 1.6)" ); Mathematica Clear[a, b, c, d, f, x] f[x_]:=a Sin[21/b (x
c)] + d Plot[f[x]/.{a Ä 3, b Ä 1, c Ä 0, d Ä 0}, {x,
41, 41 }] 67. (a) The graph stretches horizontally.

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45

46

Chapter 1 Preliminaries (b) The period remains the same: period œ l B l. The graph has a horizontal shift of

" #

period.

68. (a) The graph is shifted right C units.

(b) The graph is shifted left C units. (c) A shift of „ one period will produce no apparent shift. l C l œ ' 69. The graph shifts upwards l D lunits for D  ! and down l D lunits for D  !Þ

70. (a) The graph stretches l A l units.

(b) For A  !, the graph is inverted. 1.7 GRAPHING WITH CALCULATORS AND COMPUTERS 1-4.

The most appropriate viewing window displays the maxima, minima, intercepts, and end behavior of the graphs and has little unused space.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.7 Graphing with Calculators and Computers 1. d.

2. c.

3. d.

4. b.

5-30.

For any display there are many appropriate display widows. The graphs given as answers in Exercises 5
30 are not unique in appearance.

5. Ò
2ß 5Ó by Ò
15ß 40Ó

6. Ò
4ß 4Ó by Ò
4ß 4Ó

7. Ò
2ß 6Ó by Ò
250ß 50Ó

8. Ò
1ß 5Ó by Ò
5ß 30Ó

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47

48

Chapter 1 Preliminaries

9. Ò
4ß 4Ó by Ò
5ß 5Ó

10. Ò
2ß 2Ó by Ò
2ß 8Ó

11. Ò
2ß 6Ó by Ò
5ß 4Ó

12. Ò
4ß 4Ó by Ò
8ß 8Ó

13. Ò
"ß 'Ó by Ò
"ß %Ó

14. Ò
"ß 'Ó by Ò
"ß &Ó

15. Ò
3ß 3Ó by Ò!ß "!Ó

16. Ò
"ß #Ó by Ò!ß "Ó

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.7 Graphing with Calculators and Computers 17. Ò
&ß "Ó by Ò
&ß &Ó

18. Ò
&ß "Ó by Ò
#ß %Ó

19. Ò
%ß %Ó by Ò!ß $Ó

20. Ò
&ß &Ó by Ò
#ß #Ó

21. Ò
"!ß "!Ó by Ò
'ß 'Ó

22. Ò
&ß &Ó by Ò
#ß #Ó

23. Ò
'ß "!Ó by Ò
'ß 'Ó

24. Ò
$ß &Ó by Ò
#ß "!Ó

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49

50

Chapter 1 Preliminaries

25. Ò
!Þ!$ß !Þ!$Ó by Ò
"Þ#&ß "Þ#&Ó

26. Ò
!Þ"ß !Þ"Ó by Ò
$ß $Ó

27. Ò
$!!ß $!!Ó by Ò
"Þ#&ß "Þ#&Ó

28. Ò
&!ß &!Ó by Ò
!Þ"ß !Þ"Ó

29. Ò
!Þ#&ß !Þ#&Ó by Ò
!Þ$ß !Þ$Ó

30. Ò
!Þ"&ß !Þ"&Ó by Ò
!Þ!#ß !Þ!&Ó

31. x#  #x œ %  %y
y# Ê y œ # „ È
x#
#x  ). The lower half is produced by graphing y œ #
È
x#
#x  ).

32. y#
"'x# œ " Ê y œ „ È"  "'x# . The upper branch is produced by graphing y œ È"  "'x# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 1.7 Graphing with Calculators and Computers 33.

34.

35.

36.

37.

38Þ

39.

40.

41. (a) y œ "!&*Þ"%x
#!(%*(# (b) m œ "!&*Þ"% dollars/year, which is the yearly increase in compensation.

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51

52

Chapter 1 Preliminaries (c)

(d) Answers may vary slightly. y œ a"!&*Þ14ba#!"!b
#!(%*(# œ $&$ß 899 42. (a) Let C œ cost and x œ year. C œ a(*'!Þ("bx
"Þ' ‚ "!( (b) Slope represents increase in cost per year (c) C œ a#'$(Þ"%bx
&Þ# ‚ "!' (d) The median price is rising faster in the northeast (the slope is larger). 43. (a) Let x represent the speed in miles per hour and d the stopping distance in feet. The quadratic regression function is d œ !Þ!)''x#
"Þ*(x  &!Þ". (b)

(c) From the graph in part (b), the stopping distance is about $(! feet when the vehicle is (# mph and it is about &#& feet when the speed is )& mph. Algebraically: dquadratic a(#b œ !Þ!)''a(#b#
"Þ*(a(#b  &!Þ" œ $'(Þ' ft. dquadratic a)&b œ !Þ!)''a)&b#
"Þ*(a)&b  &!Þ" œ &##Þ) ft. (d) The linear regression function is d œ 'Þ)*x
"%!Þ% Ê dlinear a(#b œ 'Þ)*a(#b
"%!Þ% œ $&&Þ( ft and dlinear a)&b œ 'Þ)*a)&b
"%!Þ% œ %%&Þ# ft. The linear regression line is shown on the graph in part (b). The quadratic regression curve clearly gives the better fit.

44. (a) The power regression function is y œ %Þ%%'%(x!Þ&""%"% .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 1 Practice Exercises (b)

(c) 15Þ2 km/h (d) The linear regression function is y œ !Þ*"$'(&x  %Þ")**(' and it is shown on the graph in part (b). The linear regession function gives a speed of "%Þ# km/h when y œ "" m. The power regression curve in part (a) better fits the data. CHAPTER 1 PRACTICE EXERCISES 1. (  2x   $ Ê #x  
% Ê x  
# 2.


3x  "! Ê x 
"! $

3.

" & ax


"b  "% ax
#b Ê %ax
"b  &ax
#b

x
$ #

 
%$ x Ê $ax
$b  
#a%  xb

qqqqqqqqðïïïïïïïî x
"! $

Ê %x
%  &x
"! Ê '  x

4.

Ê $x
*  
)
#x Ê &x   " Ê x   5.

qqqqqqqqñïïïïïïïî x " &

" &

lx  " l œ ( Ê x  " œ ( or
ax  "b œ ( Ê x œ ' or x œ
)

6. ly
$ l  % Ê
%  y
$  % Ê
"  y  ( 7. ¹"
x# ¹ 

$ #

Ê "


x #


$# or "


x #



$ #

Ê
x# 
&# or
x# 

" #

Ê
x 
& or
x  "

Ê x  & or x 
" 8. ¹ #x$( ¹ Ÿ & Ê
& Ÿ

#x( $

Ÿ & Ê
1& Ÿ #x  ( Ÿ 1& Ê
22 Ÿ #x Ÿ 8 Ê
"" Ÿ x Ÿ %

9. Since the particle moved to the y-axis,
#  ?x œ ! Ê ?x œ 2. Since ?y œ 3?x œ 6, the new coordinates are (x  ?xß y  ?y) œ (
#  #ß &  ') œ (0ß 11). 10. (a)

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53

54

Chapter 1 Preliminaries (b)

line AB

slope 10
1 9 3 2
8 œ
6 œ
# 10
6 4 2 2
(
4) œ 6 œ 3 6
(
3) 9 3
%
2 œ
6 œ
# 1
(
3) 4 2 8
2 œ 6 œ 3 6
6 œ0
%
14 3

BC CD DA CE

BD is vertical and has no slope (c) Yes; A, B, C and D form a parallelogram. 3 ˆ 14 ‰ (d) Yes. The line AB has equation y
1 œ
3# (x
8). Replacing x by 14 3 gives y œ
# 3
8  " 3 ˆ 10 ‰ 14 œ
#
3  1 œ 5  1 œ 6. Thus, E ˆ 3 ß 6‰ lies on the line AB and the points A, B and E are collinear. (e) The line CD has equation y  3 œ
3# (x
2) or y œ
3# x. Thus the line passes through the origin. 11. The triangle ABC is neither an isosceles triangle nor is it a right triangle. The lengths of AB, BC and AC are È53, È72 and È65, respectively. The slopes of AB, BC and AC are 7 ,
1 and " , respectively. #

8

12. P(xß 3x  1) is a point on the line y œ 3x  1. If the distance from P to (!ß 0) equals the distance from P to (
$ß %), then x#  (3x  1)# œ (x  3)#  (3
3x)# Ê x#  9x#  6x  1 œ x#  6x  9  9
18x  9x# 23 ˆ 17 ‰ ˆ 17 23 ‰ Ê 18x œ 17 or x œ 17 18 Ê y œ 3x  1 œ 3 18  1 œ 6 . Thus the point is P 18 ß 6 . 13. y œ $ax
"b  a
'b Ê y œ $x
* 14. y œ
"# ax  "b  # Ê y œ
"# x 

$ #

15. x œ ! 16. m œ


#
' "
a
$b

œ


) %

œ
# Ê y œ
#ax  $b  ' Ê y œ
#x

17. y œ # 18. m œ

&
$
#
$

œ

#
&

œ
&# Ê y œ
&# ax
$b  $ Ê y œ
&# x 

#" &

19. y œ
$x  $ 20. Since #x
y œ
# is equivalent to y œ #x  #, the slope of the given line (and hence the slope of the desired line) is 2. y œ #a x
"b  " Ê y œ # x
& 21. Since %x  $y œ "# is equivalent to y œ
%$ x  %, the slope of the given line (and hence the slope of the desired line) is
%$ . y œ
%$ ax
4b
"2 Ê y œ
%$ x


#! $

22. Since $x
&y œ " is equivalent to y œ $& x
"& , the slope of the given line is
5$ .

yœ


5$ ax

 #b
$ Ê y œ


5$ x




"* $

$ &

and the slope of the perpendicular line is

23. Since "# x  "$ y œ " is equivalent to y œ
$# x  $, the slope of the given line is
$# and the slope of the perpendicular line is #$ . y œ #$ ax  "b  # Ê y œ #$ x 

) $

24. The line passes through a!ß
&b and a$ß !b. m œ

!
a
&b $
!

œ

& $

Ê y œ $& x
&

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 1 Practice Exercises 25. The area is A œ 1 r# and the circumference is C œ #1 r. Thus, r œ 26. The surface area is S œ %1 r# Ê r œ ˆ %S1 ‰ surface area gives S œ %1 r# œ %1 ˆ $%V1 ‰

"Î#

#Î$

C #1

#

Ê A œ 1ˆ #C1 ‰ œ

C# %1 .

$ $V . The volume is V œ %$ 1 r$ Ê r œ É %1 . Substitution into the formula for

.

27. The coordinates of a point on the parabola are axß x# b. The angle of inclination ) joining this point to the origin satisfies the equation tan ) œ 28. tan ) œ

rise run

œ

h &!!

x# x

œ x. Thus the point has coordinates axß x# b œ atan )ß tan# )b.

Ê h œ &!! tan ) ft.

29.

30.

Symmetric about the origin.

Symmetric about the y-axis.

31.

32.

Neither

Symmetric about the y-axis.

33. ya
xb œ a
xb#  " œ x#  " œ yaxb. Even. 34. ya
xb œ a
xb&
a
xb$
a
xb œ
x&  x$  x œ
yaxb. Odd. 35. ya
xb œ "
cosa
xb œ "
cos x œ yaxb. Even. 36. ya
xb œ seca
xb tana
xb œ 37. ya
xb œ

a
xb% " a
xb$
#a
xb

œ

x% "
x$ #x

sina
xb cos# a
xb

œ


sin x cos# x

œ
sec x tan x œ
yaxb. Odd.

%

" œ
xx$
# x œ
yaxb. Odd.

38. ya
xb œ "
sina
xb œ "  sin x. Neither even nor odd. 39. ya
xb œ
x  cosa
xb œ
x  cos x. Neither even nor odd. 40. ya
xb œ Éa
xb%
" œ Èx%
" œ yaxb. Even.

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55

56

Chapter 1 Preliminaries

41. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since l x l attains all nonnegative values, the range is Ò
#ß _Ñ. 42. (a) Since the square root requires "
x   !, the domain is Ð
_ß "Ó. (b) Since È"
x attains all nonnegative values, the range is Ò
#ß _Ñ. 43. (a) Since the square root requires "'
x#   !, the domain is Ò
%ß %Ó. (b) For values of x in the domain, ! Ÿ "'
x# Ÿ "', so ! Ÿ È"'
x# Ÿ %. The range is Ò!ß %Ó. 44. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since $#
x attains all positive values, the range is a"ß _b. 45. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) Since #e
x attains all positive values, the range is a
$ß _b. 46. (a) The function is equivalent to y œ tan #x, so we require #x Á

k1 #

for odd integers k. The domain is given by x Á

k1 %

for

odd integers k. (b) Since the tangent function attains all values, the range is a
_ß _b. 47. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The sine function attains values from
" to ", so
# Ÿ #sina$x  1b Ÿ # and hence
$ Ÿ #sina$x  1b
" Ÿ ". The range is Ò
3ß 1Ó. 48. (a) The function is defined for all values of x, so the domain is a
_ß _b. & (b) The function is equivalent to y œ È x# , which attains all nonnegative values. The range is Ò!ß _Ñ. 49. (a) The logarithm requires x
$  !, so the domain is a$ß _b. (b) The logarithm attains all real values, so the range is a
_ß _b. 50. (a) The function is defined for all values of x, so the domain is a
_ß _b. (b) The cube root attains all real values, so the range is a
_ß _b. 51. (a) The function is defined for
% Ÿ x Ÿ %, so the domain is Ò
%ß %Ó. (b) The function is equivalent to y œ Èl x l,
% Ÿ x Ÿ %, which attains values from ! to # for x in the domain. The range is Ò!ß #Ó. 52. (a) The function is defined for
# Ÿ x Ÿ #, so the domain is Ò
#ß #Ó. (b) The range is Ò
"ß "Ó. 53. First piece: Line through a!ß "b and a"ß !b. m œ Second piece: Line through a"ß "b and a#ß !b. m faxb œ œ

"
x, ! Ÿ x  " #
x, " Ÿ x Ÿ #

54. First piece: Line through a!ß !b and a2ß 5b. m œ Second piece: Line through a2ß 5b and a4ß !b. m faxb œ


10


5 2 x, 5x 2 ,

!
"
" "
! œ " œ "
" œ !#

" œ "


" Ê y œ
x  " œ "
x œ
" Ê y œ
ax
"b  " œ
x  # œ #
x

5
! 5 5 2
! œ 2 Ê y œ 2x
5
5 5 œ !4
2 œ 2 œ
2 Ê

y œ
52 ax
2b  5 œ
52 x  10 œ 10


!Ÿx2 (Note: x œ 2 can be included on either piece.) 2ŸxŸ4 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5x 2

Chapter 1 Practice Exercises 55. (a) af‰gba
"b œ faga
"bb œ fŠ È
""  # ‹ œ fa"b œ (b) ag‰f ba#b œ gafa#bb œ gˆ "2 ‰ œ (c) af‰f baxb œ fafaxbb œ fˆ "x ‰ œ

" É "#  #

" "Îx

œ

" È#Þ&

" "

œ"

or É &#

œ x, x Á !

(d) ag‰gbaxb œ gagaxbb œ gŠ Èx" # ‹ œ

" " É Èx
# #

œ

% x# È É "  #È x  #

$ 56. (a) af‰gba
"b œ faga
"bb œ fˆÈ
"  "‰ œ fa!b œ #
! œ # $ (b) ag‰f ba#b œ faga#bb œ ga#
#b œ ga!b œ È !"œ"

(c) af‰f baxb œ fafaxbb œ fa#
xb œ #
a#
xb œ x $ $ $ È (d) ag‰gbaxb œ gagaxbb œ gˆÈ x  "‰ œ É x""

#

57. (a) af‰gbaxb œ fagaxbb œ fˆÈx  #‰ œ #
ˆÈx  #‰ œ
x, x  
#. ag‰f baxb œ fagaxbb œ ga#
x# b œ Èa#
x# b  # œ È%
x# (b) Domain of f‰g: Ò
#ß _ÑÞ Domain of g‰f: Ò
#ß #ÓÞ (c) Range of f‰g: Ð
_ß #ÓÞ Range of g‰f: Ò!ß #ÓÞ % 58. (a) af‰gbaxb œ fagaxbb œ fŠÈ"
x‹ œ ÉÈ"
x œ È "
x.

ag‰f baxb œ fagaxbb œ gˆÈx‰ œ É"
Èx (b) Domain of f‰g: Ð
_ß "ÓÞ Domain of g‰f: Ò!ß "ÓÞ (c) Range of f‰g: Ò!ß _ÑÞ Range of g‰f: Ò!ß "ÓÞ 59.

60.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

57

58

Chapter 1 Preliminaries

61.

62.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis.

It does not change the graph.

63.

64.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis. 65.

66.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis. 67.

The graph of f# (x) œ f" akxkb is the same as the graph of f" (x) to the right of the y-axis. The graph of f# (x) to the left of the y-axis is the reflection of y œ f" (x), x   0 across the y-axis.

It does not change the graph.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 1 Practice Exercises

59

68.

Whenever g" (x) is positive, the graph of y œ g# (x) œ kg" (x)k is the same as the graph of y œ g" (x). When g" (x) is negative, the graph of y œ g# (x) is the reflection of the graph of y œ g" (x) across the x-axis.

69.

70.

period œ 1

period œ 41

71.

72.

period œ 2

period œ 4

73.

74.

period œ 21

period œ 21

75. (a) sin B œ sin

1 3

œ

b c

œ

b #

Ê b œ 2 sin

1 3

œ 2Š

È3 # ‹

œ È3. By the theorem of Pythagoras,

a#  b# œ c# Ê a œ Èc#
b# œ È4
3 œ 1. (b) sin B œ sin

1 3

œ

b c

œ

2 c

Ê cœ

2 sin 13

œ È23 œ Š ‹ #

4 È3

#

. Thus, a œ Èc#
b# œ ÊŠ È43 ‹
(2)# œ É 43 œ

76. (a) sin A œ

a c

Ê a œ c sin A

(b) tan A œ

a b

Ê a œ b tan A

77. (a) tan B œ

b a

Ê aœ

(b) sin A œ

a c

Ê cœ

b tan B

a sin A

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 È3

.

60

Chapter 1 Preliminaries

78. (a) sin A œ

(c) sin A œ

a c

a c

œ

È c #
b # c

79. Let h œ height of vertical pole, and let b and c denote the distances of points B and C from the base of the pole, measured along the flatground, respectively. Then, tan 50° œ hc , tan 35° œ hb , and b
c œ 10. Thus, h œ c tan 50° and h œ b tan 35° œ (c  10) tan 35° Ê c tan 50° œ (c  10) tan 35° Ê c (tan 50°
tan 35°) œ 10 tan 35° tan 35° Ê c œ tan10 50°
tan 35° Ê h œ c tan 50° œ

10 tan 35° tan 50° tan 50°
tan 35°

¸ 16.98 m.

80. Let h œ height of balloon above ground. From the figure at the right, tan 40° œ ha , tan 70° œ hb , and a  b œ 2. Thus, h œ b tan 70° Ê h œ (2
a) tan 70° and h œ a tan 40° Ê (2
a) tan 70° œ a tan 40° Ê a(tan 40°  tan 70°) 70° œ 2 tan 70° Ê a œ tan 240°tantan 70° Ê h œ a tan 40° œ

2 tan 70° tan 40° tan 40°tan 70°

¸ 1.3 km.

81. (a)

(b) The period appears to be 41. (c) f(x  41) œ sin (x  41)  cos ˆ x#41 ‰ œ sin (x  21)  cos ˆ x#  21‰ œ sin x  cos since the period of sine and cosine is 21. Thus, f(x) has period 41.

x #

82. (a)

(b) D œ (
_ß 0)  (!ß _); R œ [
1ß 1] (c) f is not periodic. For suppose f has period p. Then f ˆ #"1  kp‰ œ f ˆ #"1 ‰ œ sin 21 œ 0 for all integers k. Choose k so large that

" #1

 kp 

" 1

Ê 0

" (1/21)kp

 1. But then

f ˆ #"1  kp‰ œ sin Š (1/#1")kp ‹  0 which is a contradiction. Thus f has no period, as claimed.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 1 Additional and Advanced Exercises CHAPTER 1 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The given graph is reflected about the y-axis.

(b) The given graph is reflected about the x-axis.

(c) The given graph is shifted left 1 unit, stretched vertically by a factor of 2, reflected about the x-axis, and then shifted upward 1 unit.

2. (a)

(d) The given graph is shifted right 2 units, stretched vertically by a factor of 3, and then shifted downward 2 units.

(b)

3. There are (infinitely) many such function pairs. For example, f(x) œ 3x and g(x) œ 4x satisfy f(g(x)) œ f(4x) œ 3(4x) œ 12x œ 4(3x) œ g(3x) œ g(f(x)). 4. Yes, there are many such function pairs. For example, if g(x) œ (2x  3)$ and f(x) œ x"Î$ , then (f ‰ g)(x) œ f(g(x)) œ f a(2x  3)$ b œ a(2x  3)$ b

"Î$

œ 2x  3.

5. If f is odd and defined at x, then f(
x) œ
f(x). Thus g(
x) œ f(
x)
2 œ
f(x)
2 whereas
g(x) œ
(f(x)
2) œ
f(x)  2. Then g cannot be odd because g(
x) œ
g(x) Ê
f(x)
2 œ
f(x)  2 Ê 4 œ 0, which is a contradiction. Also, g(x) is not even unless f(x) œ 0 for all x. On the other hand, if f is even, then g(x) œ f(x)
2 is also even: g(
x) œ f(
x)
2 œ f(x)
2 œ g(x). 6. If g is odd and g(0) is defined, then g(0) œ g(
0) œ
g(0). Therefore, 2g(0) œ 0 Ê g(0) œ 0.

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61

62

Chapter 1 Preliminaries

7. For (xß y) in the 1st quadrant, kxk  kyk œ 1  x Í x  y œ 1  x Í y œ 1. For (xß y) in the 2nd quadrant, kxk  kyk œ x  1 Í
x  y œ x  1 Í y œ 2x  1. In the 3rd quadrant, kxk  kyk œ x  1 Í
x
y œ x  1 Í y œ
2x
1. In the 4th quadrant, kxk  kyk œ x  1 Í x  (
y) œ x  1 Í y œ
1. The graph is given at the right. 8. We use reasoning similar to Exercise 7. (1) 1st quadrant: y  kyk œ x  kxk Í 2y œ 2x Í y œ x. (2) 2nd quadrant: y  kyk œ x  kxk Í 2y œ x  (
x) œ 0 Í y œ 0. (3) 3rd quadrant: y  kyk œ x  kxk Í y  (
y) œ x  (
x) Í 0 œ 0 Ê all points in the 3rd quadrant satisfy the equation. (4) 4th quadrant: y  kyk œ x  kxk Í y  (
y) œ 2x Í 0 œ x. Combining these results we have the graph given at the right:

9. By the law of sines, 10. By the law of sines,

sin 13 È3

œ

sin 14 4

œ

sin A a

œ

sin A a

œ

sin B b

œ

sin B b

œ

sin 14 b

Ê bœ

sin B 3

Ê sin B œ

È3 sin (1/4) sin (1/3) 3 4

11. By the law of cosines, a# œ b#  c#
2bc cos A Ê cos A œ

sin

œ

È3 Š È2 ‹

È3

#

œ È2.

#

1 4

œ

b #  c #
a# 2bc

3 4

Š

œ

12. By the law of cosines, c# œ a#  b#
2ab cos C œ 2#  3#
(2)(2)(3) cos

È2 # ‹

œ

3È 2 8

2#  3#
2# 2(2)(3)

1 4

.

œ 34 .

œ 4  9
12 Š

È2 # ‹

œ 13
6È2 Ê c œ É13
6È2 , since c  0. # a # c #
b # 4 #
3 # œ 2(2)(2)(4) #ac È135 3È15 121 256 œ 16 œ 16 .

œ

416
9 16

# 4 #
5 # a # b #
c # œ 2(2)(2)(4) 2ab È231 25 256 œ 16 .

œ

416
25 16

13. By the law of cosines, b# œ a#  c#
2ac cos B Ê cos B œ œ

11 16 .

Since 0  B  1, sin B œ È1
cos# B œ É1


14. By the law of cosines, c# œ a#  b#
2ab cos C Ê cos C œ 5 œ
16 . Since 0  C  1, sin C œ È1
cos# C œ É1


15. (a) sin# x  cos# x œ 1 Ê sin# x œ 1
cos# x œ (1
cos x)(1  cos x) Ê (1
cos x) œ Ê

1
cos x sin x

œ

sin# x 1cos x

sin x 1cos x

(b) Using the definition of the tangent function and the double angle formulas, we have tan# ˆ x# ‰ œ

sin# ˆ x# ‰ cos# ˆ #x ‰

œ

"
cos Š2 Š #x ‹‹ #

"cos Š2 Š #x ‹‹ #

œ

1
cos x 1cos x

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 1 Additional and Advanced Exercises 16. The angles labeled # in the accompanying figure are equal since both angles subtend arc CD. Similarly, the two angles labeled ! are equal since they both subtend arc AB. Thus, triangles AED and BEC are similar which )
b implies a
b c œ 2a cos a c Ê (a
c)(a  c) œ b(2a cos )
b) Ê a#
c# œ 2ab cos )
b# Ê c# œ a#  b#
2ab cos ).

17. As in the proof of the law of sines of Section P.5, Exercise 57, ah œ bc sin A œ ab sin C œ ac sin B Ê the area of ABC œ "# (base)(height) œ "# ah œ "# bc sin A œ "# ab sin C œ "# ac sin B. 18. As in Section P.5, Exercise 57, (Area of ABC)# œ œ

" 4

(base)# (height)# œ

" 4

a# h # œ

" 4

a# b# sin# C

a# b# a"
cos# Cb . By the law of cosines, c# œ a#  b#
2ab cos C Ê cos C œ

Thus, (area of ABC)# œ œ

" 4

" 16

" 4

a# b# a"
cos# Cb œ #

Š4a# b#
aa#  b#
c# b ‹ œ

" 16

" 4

a# b# Œ"
Š a

#

 b #
c# ‹ #ab

#

œ

a# b# 4

a #  b #
c# 2ab

Š"


#

.

#

aa  b
c # b 4a# b#

#

‹

ca2ab  aa#  b#
c# bb a2ab
aa#  b#
c# bbd

" ca(a  b)#
c# b ac#
(a
b)# bd œ 16 c((a  b)  c)((a  b)
c)(c  (a
b))(c
(a
b))d a  b  c
a  b  c a
b  c a  b
c œ ˆ # ‰ ˆ # ‰ ˆ # ‰ ˆ # ‰‘ œ s(s
a)(s
b)(s
c), where s œ a#bc .

œ

" 16

Therefore, the area of ABC equals Ès(s
a)(s
b)(s
c) . 19. 1. 2. 3. 4. 5. 6.

b  c
(a  c) œ b
a, which is positive since a  b. Thus, a  c  b  c. b
c
(a
c) œ b
a, which is positive since a  b. Thus, a
c  b
c. c  0 and a  b Ê c
0 œ c and b
a are positive Ê (b
a)c œ bc
ac is positive Ê ac  bc. a  b and c  0 Ê b
a and
c are positive Ê (b
a)(
c) œ ac
bc is positive Ê bc  ac. Since a  0, a and "a are positive Ê "a  0. Since 0  a  b, both "a and b" are positive. By (3), a  b and "a  0 Ê a ˆ "a ‰  b ˆ "a ‰ or 1  ba Ê 1 ˆ "b ‰ 

7.

b a

b a

" a

" b

0 Ê

" b



" a . " a 

and b" are both negative, i.e., 0 and b"  0. By (4), a  b and  1 Ê 1 ˆ b" ‰  ba ˆ b" ‰ by (4) since b"  0 Ê b"  "a .

ab0 Ê Ê

ˆ b" ‰ by (3) since

" a

 0 Ê b ˆ "a ‰  a ˆ "a ‰

20. (a) If a œ 0, then 0 œ kak  kbk Í b Á 0 Í 0 œ kak#  kbk# . Since kak# œ kak kak œ ka# k œ a# and kbk# œ b# we obtain a#  b# . If a Á 0 then kak  0 and kak  kbk Ê a#  b# . On the other hand, if a#  b# then a# œ kak#  kbk# œ b# Ê 0  kbk#
kak# œ akbk
kakb akbk  kakb . Since akbk  kakb  0 and the product akbk
kakb akbk  kakb is positive, we must have akbk
kakb  0 Ê kbk  kak . Thus kak  kbk Í a#  b# . (b) ab Ÿ kabk Ê
ab  
2 kabk by Exercise 19(4) above Ê a#
2ab  b#   kak#
2 kak kbk  kbk# , since kak# œ a# and kbk# œ b# . Factoring both sides, (a
b)#   akak
kbkb# Ê ka
bk   kkak
kbkk , by part (a). 21. The fact that ka"  a#  á  an k Ÿ ka" k  ka# k  á  kan k holds for n œ 1 is obvious. It also holds for n œ 2 by the triangle inequality. We now show it holds for all positive integers n, by induction. Suppose it holds for n œ k   1: ka"  a#  á  ak k Ÿ ka" k  ka# k  á  kak k (this is the induction hypothesis). Then ka"  a#  á  ak  akb1 k œ kaa"  a#  á  ak b  akb1 k Ÿ ka"  a#  á  ak k  kakb1 k (by the triangle inequality) Ÿ ka" k  ka# k  á  kak k  kakb1 k (by the induction hypothesis) and the inequality holds for n œ k  1. Hence it holds for all n by induction. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

63

64

Chapter 1 Preliminaries

22. The fact that ka"  a#  á  an k   ka" k
ka# k
á
kan k holds for n œ 1 is obvious. It holds for n œ 2 by Exercise 21(b), since ka"  a# k œ ka"
(
a# )k   kka" k
k
a# kk œ kka" k
ka# kk   ka" k
ka# k . We now show it holds for all positive integers n by induction. Suppose the inequality holds for n œ k   1. Then ka"  a#  á  ak k   ka" k
ka# k
á
kak k (this is the induction hypothesis). Thus ka"  á  ak  akb1 k œ kaa"  á  ak b
a
akb1 bk   kkaa"  á  ak bk
k
akb1 kk (by Exercise 21(b)) œ kka"  á  ak k
kakb1 kk   ka"  á  ak k
kakb1 k   ka" k
ka# k
á
kak k
kakb1 k (by the induction hypothesis). Hence the inequality holds for all n by induction. 23. If f is even and odd, then f(
x) œ
f(x) and f(
x) œ f(x) Ê f(x) œ
f(x) for all x in the domain of f. Thus 2f(x) œ 0 Ê f(x) œ 0. f(
x)  f(
(
x)) œ f(x) #f(
x) œ E(x) Ê E # even function. Define O(x) œ f(x)
E(x) œ f(x)
f(x) #f(
x) œ f(x)
#f(
x) . Then O(
x) œ f(
x)
#f(
(
x)) œ f(
x)#
f(x) œ
Š f(x)
#f(
x) ‹ œ
O(x) Ê O is an odd function

24. (a) As suggested, let E(x) œ

f(x)  f(
x) #

Ê E(
x) œ

is an

Ê f(x) œ E(x)  O(x) is the sum of an even and an odd function. (b) Part (a) shows that f(x) œ E(x)  O(x) is the sum of an even and an odd function. If also f(x) œ E" (x)  O" (x), where E" is even and O" is odd, then f(x)
f(x) œ 0 œ aE" (x)  O" (x)b
(E(x)  O(x)). Thus, E(x)
E" (x) œ O" (x)
O(x) for all x in the domain of f (which is the same as the domain of E
E" and O
O" ). Now (E
E" )(
x) œ E(
x)
E" (
x) œ E(x)
E" (x) (since E and E" are even) œ (E
E" )(x) Ê E
E" is even. Likewise, (O"
O)(
x) œ O" (
x)
O(
x) œ
O" (x)
(
O(x)) (since O and O" are odd) œ
(O" (x)
O(x)) œ
(O"
O)(x) Ê O"
O is odd. Therefore, E
E" and O"
O are both even and odd so they must be zero at each x in the domain of f by Exercise 23. That is, E" œ E and O" œ O, so the decomposition of f found in part (a) is unique. 25. y œ ax#  bx  c œ a Šx#  ba x 

b# 4a# ‹




b# 4a

 c œ a ˆx 

b ‰# 2a




b# 4a

c

(a) If a  0 the graph is a parabola that opens upward. Increasing a causes a vertical stretching and a shift of the vertex toward the y-axis and upward. If a  0 the graph is a parabola that opens downward. Decreasing a causes a vertical stretching and a shift of the vertex toward the y-axis and downward. (b) If a  0 the graph is a parabola that opens upward. If also b  0, then increasing b causes a shift of the graph downward to the left; if b  0, then decreasing b causes a shift of the graph downward and to the right. If a  0 the graph is a parabola that opens downward. If b  0, increasing b shifts the graph upward to the right. If b  0, decreasing b shifts the graph upward to the left. (c) Changing c (for fixed a and b) by ?c shifts the graph upward ?c units if ?c  0, and downward
?c units if ?c  0. 26. (a) If a  0, the graph rises to the right of the vertical line x œ
b and falls to the left. If a  0, the graph falls to the right of the line x œ
b and rises to the left. If a œ 0, the graph reduces to the horizontal line y œ c. As kak increases, the slope at any given point x œ x! increases in magnitude and the graph becomes steeper. As kak decreases, the slope at x! decreases in magnitude and the graph rises or falls more gradually. (b) Increasing b shifts the graph to the left; decreasing b shifts it to the right. (c) Increasing c shifts the graph upward; decreasing c shifts it downward. 27. If m  0, the x-intercept of y œ mx  2 must be negative. If m  0, then the x-intercept exceeds Ê 0 œ mx  2 and x 

" #

Ê x œ
m2 

" #

Ê 0  m 
4.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #

Chapter 1 Additional and Advanced Exercises 28. Each of the triangles pictured has the same base b œ v?t œ v(1 sec). Moreover, the height of each triangle is the same value h. Thus "# (base)(height) œ

" #

bh

œ A" œ A# œ A$ œ á . In conclusion, the object sweeps out equal areas in each one second interval.

29. (a) By Exercise #95 of Section 1.2, the coordinates of P are ˆ a# 0 ß b# 0 ‰ œ ˆ #a ß b# ‰ . Thus the slope of OP œ

(b)

?y ?x

œ

b/2 a/2

œ

b a . b
0 The slope of AB œ 0
a œ
ba . The line # of their slopes is
" œ ˆ ba ‰ ˆ
ba ‰ œ
ba#

segments AB and OP are perpendicular when the product . Thus, b# œ a# Ê a œ b (since both are positive). Therefore, AB

is perpendicular to OP when a œ b.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

65

66

Chapter 1 Preliminaries

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 2 LIMITS AND CONTINUITY 2.1 RATES OF CHANGE AND LIMITS 1. (a) Does not exist. As x approaches 1 from the right, g(x) approaches 0. As x approaches 1 from the left, g(x) approaches 1. There is no single number L that all the values g(x) get arbitrarily close to as x Ä 1. (b) 1 (c) 0 2. (a) 0 (b)
1 (c) Does not exist. As t approaches 0 from the left, f(t) approaches
1. As t approaches 0 from the right, f(t) approaches 1. There is no single number L that f(t) gets arbitrarily close to as t Ä 0. 3. (a) True (d) False

(b) True (e) False

(c) False (f) True

4. (a) False (d) True

(b) False (e) True

(c) True

5.

x

lim x Ä 0 kx k x kx k

does not exist because

x kx k

œ

x x

œ 1 if x  0 and

approaches
1. As x approaches 0 from the right,

x kx k

x kxk

œ

x
x

œ
1 if x  0. As x approaches 0 from the left,

approaches 1. There is no single number L that all

the function values get arbitrarily close to as x Ä 0. 6. As x approaches 1 from the left, the values of

" x
1

become increasingly large and negative. As x approaches 1

from the right, the values become increasingly large and positive. There is no one number L that all the function values get arbitrarily close to as x Ä 1, so lim x
" 1 does not exist. xÄ1

7. Nothing can be said about f(x) because the existence of a limit as x Ä x! does not depend on how the function is defined at x! . In order for a limit to exist, f(x) must be arbitrarily close to a single real number L when x is close enough to x! . That is, the existence of a limit depends on the values of f(x) for x near x! , not on the definition of f(x) at x! itself. 8. Nothing can be said. In order for lim f(x) to exist, f(x) must close to a single value for x near 0 regardless of xÄ0

the value f(0) itself. 9. No, the definition does not require that f be defined at x œ 1 in order for a limiting value to exist there. If f(1) is defined, it can be any real number, so we can conclude nothing about f(1) from lim f(x) œ 5. xÄ1

10. No, because the existence of a limit depends on the values of f(x) when x is near 1, not on f(1) itself. If lim f(x) exists, its value may be some number other than f(1) œ 5. We can conclude nothing about lim f(x), xÄ1

xÄ1

whether it exists or what its value is if it does exist, from knowing the value of f(1) alone.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

68

Chapter 2 Limits and Continuity

11. (a) f(x) œ ax#
*b/(x  3) x
3.1 f(x)
6.1
2.9
5.9

x f(x)


3.01
6.01


3.001
6.001


3.0001
6.0001


3.00001
6.00001


3.000001
6.000001


2.99
5.99


2.999
5.999


2.9999
5.9999


2.99999
5.99999


2.999999
5.999999

The estimate is lim f(x) œ
6. x Ä
$

(b)

(c) f(x) œ

x#
9 x3

œ

(x  3)(x
3) x3

œ x
3 if x Á
3, and lim (x
3) œ
3
3 œ
6. x Ä
$

12. (a) g(x) œ ax#
#b/ Šx
È2‹ x g(x)

1.4 2.81421

1.41 2.82421

1.414 2.82821

1.4142 2.828413

1.41421 2.828423

1.414213 2.828426

(b)

(c) g(x) œ

x#
2 x
È2

œ

Šx  È2‹ Šx
È2‹ Šx
È2‹

œ x  È2 if x Á È2, and

13. (a) G(x) œ (x  6)/ ax#  4x
12b x
5.9
5.99 G(x)
.126582
.1251564 x G(x)


6.1
.123456


6.01
.124843


5.999
.1250156
6.001
.124984

lim

x Ä È#


5.9999
.1250015
6.0001
.124998

Šx  È2‹ œ È2  È2 œ 2È2.


5.99999
.1250001
6.00001
.124999


5.999999
.1250000
6.000001
.124999

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.1 Rates of Change and Limits (b)

(c) G(x) œ

x6 ax#  4x
12b

œ

x6 (x  6)(x
2)

œ

" x
#

14. (a) h(x) œ ax#
2x
3b / ax#
4x  3b x 2.9 2.99 h(x) 2.052631 2.005025 x h(x)

3.1 1.952380

3.01 1.995024

"

if x Á
6, and lim

x Ä
' x
2

œ

"
'
2

œ
"8 œ
0.125.

2.999 2.000500

2.9999 2.000050

2.99999 2.000005

2.999999 2.0000005

3.001 1.999500

3.0001 1.999950

3.00001 1.999995

3.000001 1.999999

(b)

(c) h(x) œ

x#
2x
3 x#
4x  3

œ

(x
3)(x  1) (x
3)(x
1)

œ

x1 x
1

15. (a) f(x) œ ax#
1b / akxk
1b x
1.1
1.01 f(x) 2.1 2.01 x f(x)


.9 1.9


.99 1.99

if x Á 3, and lim

x1

x Ä $ x
1

œ

31 3
1

œ

4 #

œ 2.


1.001 2.001


1.0001 2.0001


1.00001 2.00001


1.000001 2.000001


.999 1.999


.9999 1.9999


.99999 1.99999


.999999 1.999999

(b)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

69

70

Chapter 2 Limits and Continuity (c) f(x) œ

x#
" kx k
1

(x  1)(x
1)


1 œ
(x  x1)(x
1)
(x  1)

œ x  1, x   0 and x Á 1 , and lim (1
x) œ 1
(
1) œ 2. x Ä
1 œ 1
x, x  0 and x Á
1

16. (a) F(x) œ ax#  3x  2b / a2
kxkb x
2.1
2.01 F(x)
1.1
1.01
1.9
.9

x F(x)


1.99
.99


2.001
1.001


2.0001
1.0001


2.00001
1.00001


2.000001
1.000001


1.999
.999


1.9999
.9999


1.99999
.99999


1.999999
.999999

(b)

(c) F(x) œ

x#  3x  2 2
kx k

(x  2)(x  1)

œ
(x  2)(x#
x") 2x

17. (a) g()) œ (sin ))/) ) .1 g()) .998334

, x 0 , and lim (x  1) œ
2  1 œ
1. x Ä
# œ x  1, x  0 and x Á
2

.01 .999983

.001 .999999

.0001 .999999

.00001 .999999

.000001 .999999


.1 .998334


.01 .999983


.001 .999999


.0001 .999999


.00001 .999999


.000001 .999999

18. (a) G(t) œ (1
cos t)/t# t .1 G(t) .499583

.01 .499995

.001 .499999

.0001 .5

.00001 .5

.000001 .5


.1 .499583


.01 .499995


.001 .499999


.0001 .5


.00001 .5


.000001 .5

) g()) lim g()) œ 1

)Ä!

(b)

t G(t)

lim G(t) œ 0.5

tÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.1 Rates of Change and Limits (b)

Graph is NOT TO SCALE 19. (a) f(x) œ x"ÎÐ"
xÑ x .9 f(x) .348678 x f(x)

1.1 .385543

.99 .366032

.999 .367695

.9999 .367861

.99999 .367877

.999999 .367879

1.01 .369711

1.001 .368063

1.0001 .367897

1.00001 .367881

1.000001 .367878

lim f(x) ¸ 0.36788

xÄ1

(b)

Graph is NOT TO SCALE. Also the intersection of the axes is not the origin: the axes intersect at the point (1ß 2.71820). 20. (a) f(x) œ a3x
1b /x x .1 f(x) 1.161231

.01 1.104669

.001 1.099215

.0001 1.098672

.00001 1.098618

.000001 1.098612


.1 1.040415


.01 1.092599


.001 1.098009


.0001 1.098551


.00001 1.098606


.000001 1.098611

x f(x)

lim f(x) ¸ 1.0986

xÄ!

(b)

21. lim 2x œ 2(2) œ 4

22. lim 2x œ 2(0) œ 0

23. lim" (3x
1) œ 3 ˆ "3 ‰
1 œ 0

24. lim

xÄ#

xÄ

$

xÄ!


1

x Ä 1 3x
1

œ


" 3(1)
1

œ
#"

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

71

72 25.

Chapter 2 Limits and Continuity lim 3x(2x
1) œ 3(
1)(2(
1)
1) œ 9

26.

x Ä
"

1 #

27. lim1 x sin x œ xÄ

#

1 #

sin

œ

1 #

28. xlim Ä1

29. (a)

?f ?x

œ

f(3)
f(2) 3
#

30. (a)

?g ?x

œ

g(1)
g(
1) 1
(
1)

31. (a)

?h ?t

œ

h ˆ 341 ‰
h ˆ 14 ‰ 1 31 4
4

œ

?g ?t

œ

g(1)
g(0) 1
0

(2
1)
(2  1) 1
0

32. (a) 33.

?R ?)

œ

R(2)
R(0) 2
0

34.

?P ?)

œ

P(2)
P(1) 2
1

35. (a)

œ œ

28
9 1

œ

œ

œ

1
1 2

Q% (18ß 550)

œ

3
3

œ
1


" 1
1

œ

" 1
1

œ

f(1)
f(
") 1
(
1)

œ

2
0 #

œ1

œ0

(b)

?g ?x

œ

g(0)
g(
2) 0
(
2)

œ

0
4 #

œ
2

(b)

?h ?t

œ

h ˆ 1# ‰
h ˆ 16 ‰ 1
1 # 6

œ

?g ?t

œ

g(1)
g(
1) 1
(
1)

œ


1
1 1 #

œ
14

œ

650
225 20
10 650
375 20
14 650
475 20
16.5 650
550 20
18

Q$ (16.5ß 475)

cos 1 1
1

œ

?f ?x

œ
12

3
" #

(b)

0
È3 1 3

œ


3 È 3 1

(2
1)
(2
") #1

œ0

œ1 œ2
2œ0

Slope of PQ œ

Q# (14ß 375)

œ

3(
1)# 2(
1)
1

(b)

(8
16  10)
("
%  &) 1

Q" (10ß 225)

cos x 1
1

œ

œ 19

È 8 1
È 1 #

Q

3x#

lim x Ä
1 2x
1

?p ?t

œ 42.5 m/sec œ 45.83 m/sec œ 50.00 m/sec œ 50.00 m/sec

(b) At t œ 20, the Cobra was traveling approximately 50 m/sec or 180 km/h. 36. (a)

Slope of PQ œ

Q Q" (5ß 20) Q# (7ß 39) Q$ (8.5ß 58) Q% (9.5ß 72)

80
20 10
5 80
39 10
7 80
58 10
8.5 80
72 10
9.5

?p ?t

œ 12 m/sec œ 13.7 m/sec œ 14.7 m/sec œ 16 m/sec

(b) Approximately 16 m/sec 37. (a)

(b)

?p ?t

œ

174
62 1994
1992

œ

112 #

œ 56 thousand dollars per year

(c) The average rate of change from 1991 to 1992 is ??pt œ The average rate of change from 1992 to 1993

is ??pt

œ

62
27 1992
1991 111
62 1993
1992

œ 35 thousand dollars per year. œ 49 thousand dollars per year.

So, the rate at which profits were changing in 1992 is approximatley "# a35  49b œ 42 thousand dollars per year.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.1 Rates of Change and Limits 38. (a) F(x) œ (x  2)/(x
2) x 1.2 F(x)
4.0 ?F ?x ?F ?x ?F ?x

œ

?g ?x ?g ?x

œ

œ œ

1.1
3.4

1.01
3.04

1.001
3.004

1.0001
3.0004

1
3


4.0
(
3) œ
5.0; 1.2
1
3.04
(
3) œ
4.04; 1.01
1
3.!!!%
(
3) œ
4.!!!%; 1.0001
1

?F ?x ?F ?x

œ œ


3.4
(
3) œ
4.4; 1.1
1
3.004
(
3) œ
4.!!%; 1.001
1

È g(2)
g(1) œ #2

1" ¸ 0.414213 2
1 È1  h
" g(1  h)
g(1) (1  h)
1 œ h

?g ?x

œ

g(1.5)
g(1) 1.5
1

(b) The rate of change of F(x) at x œ 1 is
4. 39. (a)

œ

œ

È1.5
" 0.5

¸ 0.449489

(b) g(x) œ Èx 1h È1  h

1.1 1.04880

1.01 1.004987

1.001 1.0004998

1.0001 1.0000499

1.00001 1.000005

1.000001 1.0000005

ŠÈ1  h
1‹ /h

0.4880

0.4987

0.4998

0.499

0.5

0.5

(c) The rate of change of g(x) at x œ 1 is 0.5. (d) The calculator gives lim hÄ! 40. (a) i) ii) (b)

f(3)
f(2) 3
2 f(T)
f(2) T
#

œ œ

"
" 3 #

1 " " T
# T
#

T f(T) af(T)
f(2)b/aT
2b

œ œ


" 6

1

È1  h
" h

œ
6"


#TT T
#

2 #T

œ "# .

œ

2.1 0.476190
0.2381

2
T #T(T
2)

œ

2
T
#T(2
T)

2.01 0.497512
0.2488

œ
#"T , T Á 2

2.001 0.499750
0.2500

2.0001 0.4999750
0.2500

2.00001 0.499997
0.2500

2.000001 0.499999
0.2500

(c) The table indicates the rate of change is
0.25 at t œ 2. " ‰ (d) lim ˆ
#T œ
4" TÄ#

41-46. Example CAS commands: Maple: f := x -> (x^4
16)/(x
2); x0 := 2; plot( f(x), x œ x0-1..x0+1, color œ black, title œ "Section 2.1, #41(a)" ); limit( f(x), x œ x0 ); In Exercise 43, note that the standard cube root, x^(1/3), is not defined for x<0 in many CASs. This can be overcome in Maple by entering the function as f := x -> (surd(x+1, 3)
1)/x. Mathematica: (assigned function and values for x0 and h may vary) Clear[f, x] f[x_]:=(x3
x2
5x
3)/(x  1)2 x0=
1; h = 0.1; Plot[f[x],{x, x0
h, x0  h}] Limit[f[x], x Ä x0]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

73

74

Chapter 2 Limits and Continuity

2.2 CALCULATING LIMITS USING THE LIMIT LAWS 1. 3. 4. 5.

lim (2x  5) œ 2(
7)  5 œ
14  5 œ
9

lim a
x#  5x
2b œ
(2)#  5(2)
2 œ
4  10
2 œ 4 lim ax$
2x#  4x  8b œ (
2)$
2(
2)#  4(
2)  8 œ
8
8
8  8 œ
16

x Ä
#

lim 8(t
5)(t
7) œ 8(6
5)(6
7) œ
8 x3

œ

9.

lim y Ä
& 5
y

y#

23 26

# y Ä # y  5y  6

œ

5 8

(
5)# 5
(
5)

œ

y2

10. lim

13.

6.

tÄ'

lim x Ä # x6

12.

lim (10
3x) œ 10
3(12) œ 10
36 œ
26

x Ä 1#

xÄ#

7.

11.

2.

x Ä
(

œ

8. œ

25 10

œ

22 (2)#  5(#)  6

lim# 3s(2s
1) œ 3 ˆ 23 ‰ 2 ˆ 23 ‰
1‘ œ 2 ˆ 43
1‰ œ

sÄ

$

4

lim x Ä & x
7

œ

4 5
7

œ

4
#

œ
2

5 #

œ

4 4  10  6

œ

4 #0

œ

" 5

lim 3(2x
1)# œ 3(2(
1)
1)# œ 3(
3)# œ 27

x Ä
"

lim (x  3)"*)% œ (
4  3)"*)% œ (
1)"*)% œ 1

x Ä
%

%

lim (5
y)%Î$ œ [5
(
3)]%Î$ œ (8)%Î$ œ ˆ(8)"Î$ ‰ œ 2% œ 16

y Ä
$

14. lim (2z
8)"Î$ œ (2(0)
8)"Î$ œ (
8)"Î$ œ
2 zÄ!

15. lim

3

œ

3 È3(0)  1  1

œ

3 È1  1

œ

3 2

16. lim

5

œ

5 È5(0)  4  2

œ

5 È4  #

œ

5 4

17. lim

È3h  1
" h

h Ä ! È3h  1  1 h Ä ! È5h  4  2

hÄ0

œ

3 È"  "

œ

È5h  4
2 h hÄ0 5 È4  2

19. lim

œ

x
5

# x Ä & x
25

20. 21.

œ lim

a3h  "b
1

È5h  4
2 h hÄ0

†

È5h  4  2 È5h  4  2

œ lim

a5h  4b
4

h Ä 0 hŠÈ3h  1  "‹

œ lim

3h

œ lim

5h

h Ä 0 hŠÈ3h  1  "‹

œ lim

3

h Ä 0 È3h1"

h Ä 0 hŠÈ5h  4  2‹

h Ä 0 hŠÈ5h  4  2‹

œ lim

5 4 x
5

œ lim

x Ä & (x  5)(x
5)

x3

lim

È3h  1  1 È3h  1  1

œ lim

lim # x Ä
$ x  4x  3 x Ä
&

†

hÄ0

$ #

18. lim œ

È3h  1
" h

œ lim

x#  3x
"0 x5

œ lim

œ lim

x3

x Ä
$ (x  3)(x  1)

œ lim

x Ä
&

1

x Ä & x5

(x  5)(x
2) x5

œ

œ lim

" 55

œ

" 10

1

œ

"
3  1

x Ä
$ x  1

œ
"2

œ lim (x
2) œ
&
# œ
7 x Ä
&

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5

h Ä 0 È5h  4  2

2 3

Section 2.2 Calculating Limits Using the Limit Laws (x
5)(x
2) x
2

22. lim

x#
7x  "0 x
#

œ lim

23. lim

t#  t
2 t#
1

t Ä " (t
1)(t  1)

xÄ#

tÄ"

t#  3t  2

lim # t Ä
" t
t
2

25.

lim $ # x Ä
# x  2x


2x
4

5y$  8y#

u%
"

$ u Ä 1 u
1

œ lim

y# (5y  8)

œ lim

œ lim

4x
x#

œ lim

x Ä 1 Èx  3
2

lim

x Ä
"

œ lim

xÄ%

œ

x Ä
"

35.

x2 x Ä
2 È x #  5
3

œ lim

œ

lim

"

œ lim

x ˆ2  È x ‰ ˆ 2
È x ‰ 2
Èx

œ lim

xÄ1

x
1

œ lim

x2

x Ä 2 Èx#  12  4

œ lim

x Ä
2

(x  3) Š2  Èx#
5‹

9
x#

œ

lim

12 32

œ

3 8

" 6

œ lim x ˆ2  Èx‰ œ 4(2  2) œ 16 xÄ%

(x
1) ˆÈx  3  #‰ (x  3)
4


2 33

œ lim ŠÈx  3  #‹ xÄ1

œ
"3

ax#  12b
16 x Ä 2 (x
2) ŠÈx#  12  4‹

œ lim 4 È16  4

œ

œ lim

" 2

ax  2b ŠÈx#  5  3‹ ax #  5 b
9

x Ä
2

œ

Š2
Èx#
5‹ Š2  Èx#
5‹

x Ä
3

x Ä
3 (x  3) Š2  Èx#
5‹

È x#  5  3 x
2

œ

œ

4 3

ax #  8 b
* x Ä
1 (x  1) ŠÈx#  8  $‹

ax  2b ŠÈx#  5  3‹

œ lim

444 (4)(8)

œ

œ

œ lim

œ

œ

(1  1)(1  1) 111

" È9  3

œ

x Ä * Èx  3

x Ä
2 ŠÈx#  5
3‹ ŠÈx#  5  3‹

(x  2)(x
2)

œ

v#  2v  4 (v  2) av#  4b vÄ#

(x
2) ŠÈx#  12  4‹

œ lim

œ
#"

œ lim

ŠÈx#  12
4‹ ŠÈx#  12  4‹

xÄ2

œ
13

au#  "b (u  1) u#  u  1

x Ä
1 È x #  )  $

ax  2b ŠÈx#  5  3‹

2
È x#
5 x3 x Ä
3

lim

uÄ1

œ lim

(x
2)(x  2)

lim

x Ä
2

8
16

(x  1) ŠÈx#  8  $‹

x Ä 2 (x
2) ŠÈx#  12  4‹

lim

œ

ŠÈx#  8
$‹ ŠÈx#  8  $‹

lim

(x  1)(x
1)

Èx#  12
4 x
2 xÄ2

œ

5y  8

œ È4  2 œ 4

33. lim

34.

œ
21

x Ä 1 ˆÈ x  3
# ‰ ˆ È x  3  # ‰

x Ä
1 (x  1) ŠÈx#  )  $‹

œ lim


2 4

(x
1) ˆÈx  3  2‰

œ lim

È x#  8
3 x1

œ lim

x(4
x)

x Ä % 2
Èx

x
1

31. lim

Èx
3

x Ä * ˆÈ x
3 ‰ ˆ È x  3 ‰

x Ä % 2
Èx

œ


2

œ lim

(v
2) av#  2v  4b (v
2)(v  2) av#  4b vÄ#

3 #


1  2
1
2

# y Ä ! 3y
16

au#  "b (u  1)(u
1) au#  u  1b (u
1)

œ

œ

# x Ä
# x

# # y Ä ! y a3y
16b

Èx
3 x
9

t2

t Ä
" t
2

œ lim

uÄ1

12 11

œ

œ lim


2(x  2)

œ lim

30. lim

t2

œ lim

v$
8 %
16 v vÄ#

xÄ*

32.

t Ä
" (t
2)(t  1)

œ lim

28. lim

29. lim

xÄ#

t Ä " t1

(t  2)(t  1)

œ lim

œ lim (x
5) œ 2
5 œ
3

œ lim

# x Ä
# x (x  2)

% # y Ä 0 3y
16y

27. lim

(t  2)(t
1)

œ lim

24.

26. lim

xÄ#

È9  3
4

œ
23

œ lim

(3
x)(3  x)

4
ax #
5 b

x Ä
3 (x  3) Š2  Èx#
5‹

x Ä
3 (x  3) Š2  Èx#
5‹

œ lim

3
x

x Ä
3 2  È x #
5

œ

6 2  È4

œ

3 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

75

76

Chapter 2 Limits and Continuity 4
x x Ä 4 5
È x#  9

œ lim

a4
xb Š5  Èx#  9‹

œ lim

36. lim

x Ä 4 Š5
Èx#  9‹ Š5  Èx#  9‹

a4
xb Š5  Èx#  9‹

xÄ4

16
x#

œ lim

(4
x)(4  x)

xÄ4

25
ax#  9b

xÄ4

a4
xb Š5  Èx#  9‹

œ lim

a4
xb Š5  Èx#  9‹

5  È x#  9 4x

œ lim

xÄ4

œ

5  È25 8

œ

5 4

37. (a) quotient rule (b) difference and power rules (c) sum and constant multiple rules 38. (a) quotient rule (b) power and product rules (c) difference and constant multiple rules 39. (a) xlim f(x) g(x) œ ’xlim f(x)“ ’ x lim g(x)“ œ (5)(
2) œ
10 Äc Äc Äc (b) xlim 2f(x) g(x) œ 2 ’xlim f(x)“ ’ xlim g(x)“ œ 2(5)(
2) œ
20 Äc Äc Äc (c) xlim [f(x)  3g(x)] œ xlim f(x)  3 xlim g(x) œ 5  3(
2) œ
1 Äc Äc Äc lim f(x) f(x) 5 5 xÄc (d) xlim œ lim f(x)
lim g(x) œ 5
(
2) œ 7 Ä c f(x)
g(x) x

40. (a) (b) (c) (d) 41. (a) (b) (c) (d) 42. (a) (b) (c)

Äc

Äc

lim [g(x)  3] œ lim g(x)  lim 3 œ
$  $ œ !

xÄ%

xÄ%

xÄ%

lim xf(x) œ lim x † lim f(x) œ (4)(0) œ 0

xÄ%

xÄ%

#

xÄ%

#

lim [g(x)] œ ’ lim g(x)“ œ [
3]# œ 9

xÄ%

g(x) x Ä % f(x)
1

lim

xÄ%

œ

Ä%

lim g(x)

x

lim f(x)
lim 1

xÄ%

xÄ%

œ


3 0
1

œ3

lim [f(x)  g(x)] œ lim f(x)  lim g(x) œ 7  (
3) œ 4

xÄb

xÄb

xÄb

lim f(x) † g(x) œ ’ lim f(x)“ ’ lim g(x)“ œ (7)(
3) œ
21

xÄb

xÄb

xÄb

lim 4g(x) œ ’ lim 4“ ’ lim g(x)“ œ (4)(
3) œ
12

xÄb

xÄb

xÄb

lim f(x)/g(x) œ lim f(x)/ lim g(x) œ

xÄb

xÄb

xÄb

7
3

œ
73

lim [p(x)  r(x)  s(x)] œ lim p(x)  lim r(x)  lim s(x) œ 4  0  (
3) œ 1

x Ä
#

x Ä
#

x Ä
#

x Ä
#

lim p(x) † r(x) † s(x) œ ’ lim p(x)“ ’ lim r(x)“ ’ lim s(x)“ œ (4)(0)(
3) œ 0

x Ä
#

x Ä
#

(1  h)#
1# h hÄ!

œ lim

hÄ!

(
2  h)#
(
2)# h

45. lim

[3(2  h)
4]
[3(2)
4] h

hÄ!

x Ä
#

x Ä
#

44. lim

hÄ!

x Ä
#

lim [
4p(x)  5r(x)]/s(x) œ ’
4 lim p(x)  5 lim r(x)“ ‚ lim s(x) œ [
4(4)  5(0)]/
3 œ

x Ä
#

43. lim

"‰ ˆ
#" h ‰
ˆ
# h hÄ!

46. lim

x

1  2h  h#
1 h

œ lim

hÄ!

œ lim

hÄ!

4
4hh#
4 h

œ lim

hÄ!

œ lim

3h

hÄ! h


2
2  h
"
2h

œ lim

x Ä
#

h(2  h) h

hÄ!

x Ä
#

œ lim (2  h) œ 2

h(h
4) h

hÄ!

œ lim (h
4) œ
4 hÄ!

œ3

œ lim

hÄ!


2
(
2  h)
2h(
#  h)

œ lim


h

h Ä ! h(4
2h)

œ
"4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

"6 3

Section 2.2 Calculating Limits Using the Limit Laws È7  h
È7 h hÄ!

47. lim

œ lim

ŠÈ7  h
È7‹ ŠÈ7  h  È7‹

œ lim

h ŠÈ7  h  È7‹

hÄ!

h

h Ä ! h ŠÈ7hÈ7‹

È3(0  h)  1
È3(0)  1 h hÄ! 3h

h Ä ! h ŠÈ3h  1  "‹

œ

h Ä ! È 7 h  È 7

48. lim

œ lim

"

œ lim

œ lim

" #È 7

ŠÈ3h  1
"‹ ŠÈ3h  1  "‹

œ lim

h ŠÈ3h  1 "‹

hÄ!

œ lim

œ

3

h Ä ! È3h  1  1

(7  h)
7

h Ä ! h ŠÈ7  h  È7‹

(3h  1)
"

œ lim

h Ä ! h ŠÈ3h  1  1 ‹

3 #

49. lim È5
2x# œ È5
2(0)# œ È5 and lim È5
x# œ È5
(0)# œ È5; by the sandwich theorem, xÄ!

xÄ!

lim f(x) œ È5

xÄ!

50. lim a2
x# b œ 2
0 œ 2 and lim 2 cos x œ 2(1) œ 2; by the sandwich theorem, lim g(x) œ 2 xÄ!

51. (a)

xÄ!

lim Š1


xÄ!

x# 6‹

œ1


0 6

xÄ!

œ 1 and lim 1 œ 1; by the sandwich theorem, lim

(b) For x Á 0, y œ (x sin x)/(2
2 cos x) lies between the other two graphs in the figure, and the graphs converge as x Ä 0.

52. (a)

lim Š "#


xÄ!

lim

xÄ!

1
cos x x#

x# 24 ‹

œ lim

1

xÄ! #


lim

x#

x Ä ! #4

œ

" #

x sin x

x Ä ! 2
2 cos x

xÄ!


0œ

" #

and lim

"

xÄ! #

œ1

œ "# ; by the sandwich theorem,

œ "# .

(b) For all x Á 0, the graph of f(x) œ (1
cos x)/x# lies between the line y œ "# and the parabola yœ

" #


x# /24, and the graphs converge as x Ä 0.

53. xlim f(x) exists at those points c where xlim x% œ xlim x# . Thus, c% œ c# Ê c# a1
c# b œ 0 Äc Äc Äc Ê c œ 0, 1, or
1. Moreover, lim f(x) œ lim x# œ 0 and lim f(x) œ lim f(x) œ 1. xÄ!

xÄ!

x Ä
1

xÄ1

54. Nothing can be concluded about the values of f, g, and h at x œ 2. Yes, f(2) could be 0. Since the conditions of the sandwich theorem are satisfied, lim f(x) œ
5 Á 0. xÄ#

55. 1 œ lim

xÄ%

f(x)
5 x
2

lim f(x)
lim 5 % xÄ% œ xÄlim x
lim 2 œ xÄ%

xÄ%

lim f(x)
5

xÄ%

%
#

Ê lim f(x)
5 œ 2(1) Ê lim f(x) œ 2  5 œ 7. xÄ%

xÄ%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

77

78

Chapter 2 Limits and Continuity

56. (a) 1 œ lim

f(x) x#

lim f(x) lim f(x)
# xÄ
# œ xÄlim Ê % x# œ

(b) 1 œ lim

f(x) x#

œ ’ lim

x Ä
# x Ä
#

xÄ
#

x Ä
#

57. (a) 0 œ 3 † 0 œ ’ lim

xÄ#

lim f(x) œ 4.

x Ä
#

f(x) lim x" “ x “ ’x Ä
#

œ ’ lim

x Ä
#

f(x) ˆ " ‰ x “
#

Ê

lim

x Ä
#

f(x) x

œ
2.

f(x)
5 x
# “ ’xlim Ä#


5 (x
2)“ œ lim ’Š f(x) x
# ‹ (x
2)“ œ lim [f(x)
5] œ lim f(x)
5

f(x)
5 x
# “ ’xlim Ä#

(x
2)“ Ê lim f(x) œ 5 as in part (a).

xÄ#

Ê lim f(x) œ 5.

xÄ#

xÄ#

xÄ#

(b) 0 œ 4 † 0 œ ’ lim

xÄ#

58. (a) 0 œ 1 † 0 œ ’ lim

f(x) # “ ’ lim xÄ! x xÄ!

(b) 0 œ 1 † 0 œ 59. (a)

lim x sin

xÄ!

(b)
1 Ÿ sin

60. (a)

" x

’ lim f(x) # “ ’ lim xÄ! x xÄ!

" x

xÄ#

#

x“ œ ’ lim

f(x)

# xÄ! x

x“ œ

lim ’ f(x) x# xÄ!

# “ ’ lim x# “ œ lim ’ f(x) x# † x “ œ lim f(x). That is, lim f(x) œ 0.

xÄ!

† x“ œ

xÄ!

lim f(x) . xÄ! x

That is,

xÄ!

lim f(x) xÄ! x

œ 0.

œ0

Ÿ 1 for x Á 0:

x  0 Ê
x Ÿ x sin

" x

Ÿ x Ê lim x sin

" x

œ 0 by the sandwich theorem;

x  0 Ê
x   x sin

" x

  x Ê lim x sin

" x

œ 0 by the sandwich theorem.

xÄ! xÄ!

lim x# cos ˆ x"$ ‰ œ 0

xÄ!

(b)
1 Ÿ cos ˆ x"$ ‰ Ÿ 1 for x Á 0 Ê
x# Ÿ x# cos ˆ x"$ ‰ Ÿ x# Ê lim x# cos ˆ x"$ ‰ œ 0 by the sandwich theorem since lim x# œ 0.

xÄ!

xÄ!

2.3 PRECISE DEFINITION OF A LIMIT 1. Step 1: Step 2:

kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 $  5 œ 7 Ê $ œ 2, or
$  5 œ 1 Ê $ œ 4. The value of $ which assures kx
5k  $ Ê 1  x  7 is the smaller value, $ œ 2.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

xÄ!

Section 2.3 Precise Definition of a Limit 2. Step 1: Step 2:

kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  2
$  2 œ 1 Ê $ œ 1, or $  2 œ 7 Ê $ œ 5. The value of $ which assures kx
2k  $ Ê 1  x  7 is the smaller value, $ œ 1.

Step 1: Step 2:

kx
(
3)k  $ Ê
$  x  $  $ Ê
$
3  x  $
3
$
3 œ
7# Ê $ œ "# , or $
$ œ
"# Ê $ œ 5# .

3.

The value of $ which assures kx
(
3)k  $ Ê
7#  x 
"# is the smaller value, $ œ "# .

4.

Step 1:

¸x
ˆ
3# ‰¸  $ Ê
$  x 

Step 2:


$


Step 1:

¸x
"# ¸  $ Ê
$  x


Step 2:


$ 

œ


3 #

 $ Ê
$
" #

3 #

x$


3 #

Ê $ œ #, or $
œ
Ê $ œ 1. The value of $ which assures ¸x
ˆ
3# ‰¸  $ Ê
7#  x 
"# is the smaller value, $ œ ". 3 #

7 #

3 #

5. " #

 $ Ê
$ 

" #

x$

" or $  #" œ 47 Ê $ œ 14 . " 4 The value of $ which assures ¸x
# ¸  $ Ê 9  x 

œ

4 9

Ê $œ

" 18 ,

" #

4 7

" #

is the smaller value, $ œ

" 18 .

6.

Step 1: Step 2:

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3
$  $ œ 2.7591 Ê $ œ 0.2409, or $  $ œ 3.2391 Ê $ œ 0.2391. The value of $ which assures kx
3k  $ Ê 2.7591  x  3.2391 is the smaller value, $ œ 0.2391.

7. Step 1: Step 2:

kx
5k  $ Ê
$  x
5  $ Ê
$  5  x  $  5 From the graph,
$  5 œ 4.9 Ê $ œ 0.1, or $  5 œ 5.1 Ê $ œ 0.1; thus $ œ 0.1 in either case.

8. Step 1: Step 2:

kx
(
3)k  $ Ê
$  x  3  $ Ê
$
3  x  $
3 From the graph,
$
3 œ
3.1 Ê $ œ 0.1, or $
3 œ
2.9 Ê $ œ 0.1; thus $ œ 0.1.

9. Step 1: Step 2:

kx
1k  $ Ê
$  x
1  $ Ê
$  1  x  $  1 9 7 From the graph,
$  1 œ 16 Ê $ œ 16 , or $  1 œ 25 16 Ê $ œ

10. Step 1: Step 2:

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3 From the graph,
$  3 œ 2.61 Ê $ œ 0.39, or $  3 œ 3.41 Ê $ œ 0.41; thus $ œ 0.39.

11. Step 1:

kx
2k  $ Ê
$  x
2  $ Ê
$  2  x  $  2 From the graph,
$  2 œ È3 Ê $ œ 2
È3 ¸ 0.2679, or $  2 œ È5 Ê $ œ È5
2 ¸ 0.2361; thus $ œ È5
2.

Step 2:

9 16 ;

thus $ œ

7 16 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

79

80

Chapter 2 Limits and Continuity

12. Step 1: Step 2:

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 From the graph,
$
1 œ
thus $ œ

È5
2 # .

È5 #

Ê $œ

È5
2 #

¸ 0.1180, or $
1 œ


13. Step 1: Step 2:

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
1  x  $
1 7 16 From the graph,
$
1 œ
16 9 Ê $ œ 9 ¸ 0.77, or $
1 œ
25 Ê

14. Step 1:

¸x
"# ¸  $ Ê
$  x


Step 2:

From the graph,
$  thus $ œ 0.00248.

" #

œ

" #  1 2.01

$ Ê
$  Ê $œ

1 2




" #

x$

" #.01

" #

¸ 0.00248, or $ 

" #

œ

È3 #

9 25

Ê $œ

2
È3 #

œ 0.36; thus $ œ

1 1.99

Ê $œ

1 1.99




¸ 0.1340;

9 25

" #

œ 0.36.

¸ 0.00251;

15. Step 1: Step 2:

k(x  1)
5k  0.01 Ê kx
4k  0.01 Ê
0.01  x
4  0.01 Ê 3.99  x  4.01 kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4 Ê $ œ 0.01.

16. Step 1:

k(2x
2)
(
6)k  0.02 Ê k2x  4k  0.02 Ê
0.02  2x  4  0.02 Ê
4.02  2x 
3.98 Ê
2.01  x 
1.99 kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2 Ê $ œ 0.01.

Step 2: 17. Step 1: Step 2: 18. Step 1: Step 2:

¹Èx  1
"¹  0.1 Ê
0.1  Èx  1
"  0.1 Ê 0.9  Èx  1  1.1 Ê 0.81  x  1  1.21 Ê
0.19  x  0.21 kx
0k  $ Ê
$  x  $ . Then,
$ œ
!Þ"* Ê $ œ !Þ"* or $ œ !Þ#"; thus, $ œ 0.19. ¸Èx
"# ¸  0.1 Ê
0.1  Èx
"#  0.1 Ê 0.4  Èx  0.6 Ê 0.16  x  0.36 ¸x
"4 ¸  $ Ê
$  x
4"  $ Ê
$  4"  B  $  4" . Then,
$ 

19. Step 1: Step 2:

20. Step 1: Step 2:

21. Step 1: Step 2:

22. Step 1: Step 2:

" 4

œ 0.16 Ê $ œ 0.09 or $ 

" 4

œ 0.36 Ê $ œ 0.11; thus $ œ 0.09.

¹È19
x
$¹  " Ê
"  È19
x
$  1 Ê 2  È19
x  % Ê 4  19
x  16 Ê
%  x
19 
16 Ê 15  x  3 or 3  x  15 kx
10k  $ Ê
$  x
10  $ Ê
$  10  x  $  10. Then
$  10 œ 3 Ê $ œ 7, or $  10 œ 15 Ê $ œ 5; thus $ œ 5. ¹Èx
7
4¹  1 Ê
"  Èx
7
%  1 Ê 3  Èx
7  5 Ê 9  x
7  25 Ê 16  x  32 kx
23k  $ Ê
$  x
23  $ Ê
$  23  x  $  23. Then
$  23 œ 16 Ê $ œ 7, or $  23 œ 32 Ê $ œ 9; thus $ œ 7. ¸ "x
4" ¸  0.05 Ê
0.05 

" x




" 4

 0.05 Ê 0.2 

" x

 0.3 Ê

kx
4k  $ Ê
$  x
4  $ Ê
$  4  x  $  4. 2 2 Then
$  % œ 10 3 or $ œ 3 , or $  4 œ 5 or $ œ 1; thus $ œ 3 .

10 #

x

10 3

or

10 3

 x  5.

kx#
3k  !.1 Ê
0.1  x#
3  0.1 Ê 2.9  x#  3.1 Ê È2.9  x  È3.1 ¹x
È3¹  $ Ê
$  x
È3  $ Ê
$  È3  x  $  È3. Then
$  È3 œ È2.9 Ê $ œ È3
È2.9 ¸ 0.0291, or $  È3 œ È3.1 Ê $ œ È3.1
È3 ¸ 0.0286; thus $ œ 0.0286.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.3 Precise Definition of a Limit 23. Step 1: Step 2:

81

kx#
4k  0.5 Ê
0.5  x#
4  0.5 Ê 3.5  x#  4.5 Ê È3.5  kxk  È4.5 Ê
È4.5  x 
È3.5, for x near
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
#  x  $
2. Then
$
# œ
È4.5 Ê $ œ È4.5
# ¸ 0.1213, or $
# œ
È3.5 Ê $ œ #
È3.5 ¸ 0.1292; thus $ œ È4.5
2 ¸ 0.12.

24. Step 1: Step 2:

25. Step 1: Step 2:

¸ "x
(
1)¸  0.1 Ê
0.1 

" x

11  1  0.1 Ê
10 

" x

9 10 10 10 
10 Ê
10 11  x 
9 or
9  x 
11 .

kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
". " 10 " Then
$
" œ
10 9 Ê $ œ 9 , or $
" œ
11 Ê $ œ 11 ; thus $ œ

" 11 .

kax#
5b
11k  " Ê kx#
16k  1 Ê
"  x#
16  1 Ê 15  x#  17 Ê È15  x  È17. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  % œ È15 Ê $ œ %
È15 ¸ 0.1270, or $  % œ È17 Ê $ œ È17
% ¸ 0.1231; thus $ œ È17
4 ¸ 0.12.

26. Step 1: Step 2:

27. Step 1: Step 2:

28. Step 1: Step 2:

29. Step 1: Step 2:

¸ 120 ¸ x
5  " Ê
" 

Step 2:


&1 Ê 4

120 x

6 Ê

" 4



x 120



" 6

Ê 30  x  20 or 20  x  30.

kx
24k  $ Ê
$  x
24  $ Ê
$  24  x  $  24. Then
$  24 œ 20 Ê $ œ 4, or $  24 œ 30 Ê $ œ 6; thus Ê $ œ 4. kmx
2mk  0.03 Ê
0.03  mx
2m  0.03 Ê
0.03  2m  mx  0.03  2m Ê 0.03 2
0.03 m x2 m . kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  #. 0.03 0.03 Then
$  # œ #
0.03 m Ê $ œ m , or $  # œ #  m Ê $ œ

0.03 m .

In either case, $ œ

kmx
3mk  c Ê
c  mx
3m  c Ê
c  3m  mx  c  3m Ê 3
kx
3k  $ Ê
$  x
3  $ Ê
$  $  B  $  $. Then
$  $ œ $
mc Ê $ œ mc , or $  $ œ $  mc Ê $ œ ¸(mx  b)
ˆ m#  b‰¸  - Ê
c  mx
m#  c Ê
c  ¸x
"# ¸  $ Ê
$  x
"#  $ Ê
$  "#  x  $  "# . Then
$ 

30. Step 1:

120 x

" #

œ

" #




c m

Ê $œ

c m,

or $ 

" #

œ

" #



c m

c m. m #

Ê $œ

c m

 x 3

In either case, $ œ




c m.

In either case, $ œ

c m.

m #

Ê

c m

c m. " #

 mx  c 

0.03 m .

c m

x

" #



c m.

k(mx  b)
(m  b)k  0.05 Ê
0.05  mx
m  0.05 Ê
0.05  m  mx  0.05  m 0.05 Ê 1
0.05 m x" m . kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". 0.05 0.05 Then
$  " œ "
0.05 m Ê $ œ m , or $  " œ "  m Ê $ œ

0.05 m .

In either case, $ œ

0.05 m .

31. lim (3
2x) œ 3
2(3) œ
3 xÄ3

Step 1: Step 2:

32.

ka3
2xb
(
3)k  0.02 Ê
0.02  6
2x  0.02 Ê
6.02 
2x 
5.98 Ê 3.01  x  2.99 or 2.99  x  3.01. 0  k x
3k  $ Ê
$  x
3  $ Ê
$  $  x  $  $ . Then
$  $ œ 2.99 Ê $ œ 0.01, or $  $ œ 3.01 Ê $ œ 0.01; thus $ œ 0.01.

lim (
3x
#) œ (
3)(
1)
2 œ 1

x Ä
1

Step 1:

k(
3x
2)
1k  0.03 Ê
0.03 
3x
3  0.03 Ê 0.01  x  1 
0.01 Ê
1.01  x 
0.99.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

82

Chapter 2 Limits and Continuity kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
1. Then
$
" œ
1.01 Ê $ œ 0.01, or $
" œ
0.99 Ê $ œ 0.01; thus $ œ 0.01.

Step 2:

33. lim

x#
4

x Ä # x
#

34.

35.

œ lim

xÄ#

#

(x  2)(x
2) (x
2)

œ lim (x  2) œ #  # œ 4, x Á 2 xÄ#

(x  2)(x
2) (x
2)

Step 1:

¹Š xx

24 ‹

Step 2:

Ê 1.95  x  2.05, x Á 2. kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  2. Then
$  # œ 1.95 Ê $ œ 0.05, or $  # œ 2.05 Ê $ œ 0.05; thus $ œ 0.05.

lim

x Ä
&

x#  6x  5 x5


4¹  0.05 Ê
0.05 

œ lim

x Ä
&

(x  5)(x  1) (x  5)


%  0.05 Ê 3.95  x  2  4.05, x Á 2

œ lim (x  1) œ
4, x Á
5. x Ä
&

(x  5)(x  ") (x  5)

Step 1:

# ¹Š x x 6x5 5 ‹

Step 2:

Ê
5.05  x 
4.95, x Á
5. kx
(
5)k  $ Ê
$  x  5  $ Ê
$
&  x  $
&. Then
$
& œ
5.05 Ê $ œ 0.05, or $
& œ
4.95 Ê $ œ 0.05; thus $ œ 0.05.


(
4)¹  0.05 Ê
0.05 

 4  0.05 Ê
4.05  x  1 
3.95, x Á
5

lim È1
5x œ È1
5(
3) œ È16 œ 4

x Ä
$

Step 1:

¹È1
5x
4¹  0.5 Ê
0.5  È1
5x
4  0.5 Ê 3.5  È1
5x  4.5 Ê 12.25  1
5x  20.25

Step 2:

Ê 11.25 
5x  19.25 Ê
3.85  x 
2.25. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
$. Then
$
$ œ
3.85 Ê $ œ 0.85, or $
$ œ
2.25 Ê 0.75; thus $ œ 0.75.

36. lim

4

xÄ# x

œ

4 #

œ2

Step 1:

¸ 4x
2¸  0.4 Ê
0.4 

Step 2:

kx
2k  $ Ê
$  x
2  $ Ê
$  #  x  $  #. Then
$  # œ 53 Ê $ œ "3 , or $  # œ 5# Ê $ œ "# ; thus $ œ 3" .

4 x


2  0.4 Ê 1.6 

4 x

 2.4 Ê

10 16



x 4



10 24

Ê

10 4

x

10 6

or

5 3

 x  25 .

37. Step 1: Step 2:

k(9
x)
5k  % Ê
%  4
x  % Ê
%
4 
x  %
4 Ê %  %  x  4
% Ê %
%  x  4  %. kx
4k  $ Ê
$  x
4  $ Ê
$  %  x  $  %. Then
$  4 œ
%  4 Ê $ œ %, or $  % œ %  % Ê $ œ %. Thus choose $ œ %.

38. Step 1:

k(3x
7)
2k  % Ê
%  3x
9  % Ê 9
%  3x  *  % Ê 3


Step 2:

39. Step 1: Step 2:

40. Step 1:

% 3

 x  3  3% .

kx
3k  $ Ê
$  x
3  $ Ê
$  3  x  $  3. Then
$  3 œ $
3% Ê $ œ 3% , or $  3 œ 3  3% Ê $ œ 3% . Thus choose $ œ 3% . ¹Èx
5
2¹  % Ê
%  Èx
5
#  % Ê #
%  Èx
5  #  % Ê (#
%)#  x
5  (#  %)# Ê (#
%)#  &  x  (#  %)#  5. kx
9k  $ Ê
$  x
9  $ Ê
$  9  x  $  9. Then
$  * œ %#
%%  * Ê $ œ %%
%# , or $  * œ %#  %%  * Ê $ œ %%  %# . Thus choose the smaller distance, $ œ %%
%# . ¹È4
x
2¹  % Ê
%  È4
x
#  % Ê #
%  È4
x  #  % Ê (#
%)#  %
x  (#  %)# Ê
(#  %)#  x
4 
(#
%)# Ê
(#  %)#  %  x 
(#
%)#  %.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.3 Precise Definition of a Limit Step 2:

41. Step 1: Step 2:

42. Step 1: Step 2:

43. Step 1: Step 2:

kx
0k  $ Ê
$  x  $ . Then
$ œ
(#  %)#  4 œ
%#
%% Ê $ œ %%  %# , or $ œ
(#
%)#  4 œ 4%
%# . Thus choose the smaller distance, $ œ 4%
%# . For x Á 1, kx#
1k  % Ê
%  x#
"  % Ê "
%  x#  "  % Ê È1
%  kxk  È1  % Ê È"
%  x  È1  % near B œ ". kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". Then
$  " œ È1
% Ê $ œ "
È1
%, or $  1 œ È"  % Ê $ œ È"  %
1. Choose $ œ min š"
È1
%ß È1  %
"›, that is, the smaller of the two distances. For x Á
2, kx#
4k  % Ê
%  x#
4  % Ê 4
%  x#  4  % Ê È4
%  kxk  È4  % Ê
È4  %  x 
È4
% near B œ
2. kx
(
2)k  $ Ê
$  x  2  $ Ê
$
2  x  $
2. Then
$
2 œ
È%  % Ê $ œ È%  %
#, or $
# œ
È%
% Ê $ œ #
È%
%. Choose $ œ min šÈ%  %
#ß #
È%
%› . ¸ "x
1¸  % Ê
% 

" x


"% Ê "
% 

" x

"% Ê

" 1%

% "%,

" 1
%.

x

kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ " " % Ê $ œ "
" " % œ " % % , or "  $ œ "
" % Ê $ œ Choose $ œ

44. Step 1:

83

" "
%


"œ

% "
%.

the smaller of the two distances.

¸ x"#
"3 ¸  % Ê
% 

" x#




" 3

% Ê

" 3


% 

" x#



" 3

% Ê

1
3% 3



" x#



1  $% 3

Ê

3 "


%$ x# 

3 " 

%$3 È $. Ê É 1 3 $%  kxk  É "
3 $% , or É " 3 $%  x  É "
$ % for x near

Step 2:

¹x
È3¹  $ Ê
$  x
È3  $ Ê È3
$  x  È3  $ . Then È3
$ œ É " 3 $% Ê $ œ È3
É " 3 $% , or È3  $ œ É "
3 $% Ê $ œ É "
3 $%
È3. Choose $ œ min šÈ3
É " 3 $% ß É "
3 $%
È3›.

45. Step 1: Step 2:

46. Step 1: Step 2:

47. Step 1:

#

¹Š xx
*3 ‹
(
6)¹  % Ê
%  (x
3)  6  %, x Á
3 Ê
%  x  3  % Ê
%
$  x  %
$. kx
(
3)k  $ Ê
$  x  3  $ Ê
$
$  x  $
3. Then
$
$ œ
%
$ Ê $ œ %, or $
$ œ %
$ Ê $ œ %. Choose $ œ %. #

¹Š xx

11 ‹
2¹  % Ê
%  (x  1)
2  %, x Á 1 Ê "
%  x  "  %. kx
1k  $ Ê
$  x
1  $ Ê "
$  x  "  $ . Then "
$ œ "
% Ê $ œ %, or "  $ œ "  % Ê $ œ %. Choose $ œ %. x  1: l(4
2x)
2l  % Ê !  2
2x  % since x  1Þ Thus, 1


% #

 x  !;

x   1: l(6x
4)
2l  % Ê ! Ÿ 6x
6  % since x   1. Thus, " Ÿ x  1  6% . Step 2:

48. Step 1:

kx
1k  $ Ê
$  x
1  $ Ê "
$  x  1  $ . Then 1
$ œ "
#% Ê $ œ #% , or "  $ œ 1  6% Ê $ œ 6% . Choose $ œ 6% . x  !: k2x
0k  % Ê
%  2x  ! Ê
#%  x  0; x   0: ¸ x#
!¸  % Ê ! Ÿ x  #%.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

84

Chapter 2 Limits and Continuity Step 2:

kx
0k  $ Ê
$  x  $ . Then
$ œ
#% Ê $ œ #% , or $ œ #% Ê $ œ #%. Choose $ œ #% .

49. By the figure,
x Ÿ x sin

" x

Ÿ x for all x  0 and
x   x sin

then by the sandwich theorem, in either case, lim x sin xÄ!

50. By the figure,
x# Ÿ x# sin

" x

" x

" x

  x for x  0. Since lim (
x) œ lim x œ 0, xÄ!

œ 0.

xÄ!

Ÿ x# for all x except possibly at x œ 0. Since lim a
x# b œ lim x# œ 0, then

by the sandwich theorem, lim x# sin xÄ!

" x

xÄ!

œ 0.

xÄ!

51. As x approaches the value 0, the values of g(x) approach k. Thus for every number %  0, there exists a $  ! such that !  kx
0k  $ Ê kg(x)
kk  %. 52. Write x œ h  c. Then !  lx
cl  $ Í
$  x
c  $ , x Á c Í
$  ah  cb
c  $ , h  c Á c Í
$  h  $ , h Á ! Í !  lh
!l  $ . Thus, limfaxb œ L Í for any %  !, there exists $  ! such that lfaxb
Ll  % whenever !  lx
cl  $ xÄc

Í lfah  cb
Ll  % whenever !  lh
!l  $ Í limfah  cb œ L. hÄ!

53. Let f(x) œ x# . The function values do get closer to
1 as x approaches 0, but lim f(x) œ 0, not
1. The xÄ!

function f(x) œ x# never gets arbitrarily close to
1 for x near 0.

54. Let f(x) œ sin x, L œ "# , and x! œ 0. There exists a value of x (namely, x œ 16 ) for which ¸sin x
"# ¸  % for any given %  0. However, lim sin x œ 0, not "# . The wrong statement does not require x to be arbitrarily close to xÄ!

x! . As another example, let g(x) œ sin "x , L œ #" , and x! œ 0. We can choose infinitely many values of x near 0 such that sin

" x

œ

" #

as you can see from the accompanying figure. However, lim sin xÄ!

" x

fails to exist. The

wrong statement does not require all values of x arbitrarily close to x! œ 0 to lie within %  0 of L œ "# . Again you can see from the figure that there are also infinitely many values of x near 0 such that sin "x œ 0. If we choose %  4" we cannot satisfy the inequality ¸sin x"
#" ¸  % for all values of x sufficiently near x! œ 0.

#

55. kA
*k Ÿ 0.01 Ê
0.01 Ÿ 1 ˆ x# ‰
9 Ÿ 0.01 Ê 8.99 Ÿ Ê

2É 8.99 1

ŸxŸ

2É 9.01 1

1 x# 4

Ÿ 9.01 Ê

4 1

(8.99) Ÿ x# Ÿ

4 1

(9.01)

or 3.384 Ÿ x Ÿ 3.387. To be safe, the left endpoint was rounded up and the right

endpoint was rounded down. 56. V œ RI Ê

V R

œ I Ê ¸ VR
5¸ Ÿ 0.1 Ê
0.1 Ÿ

120 R


5 Ÿ 0.1 Ê 4.9 Ÿ

120 R

Ÿ 5.1 Ê

10 49

 

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

R 1#0

 

10 51

Ê

Section 2.3 Precise Definition of a Limit (120)(10) 51

ŸRŸ

(120)(10) 49

85

Ê 23.53 Ÿ R Ÿ 24.48.

To be safe, the left endpoint was rounded up and the right endpoint was rounded down. 57. (a)
$  x
1  0 Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
2k œ kx
2k œ 2
x  2
1 œ 1. That is, kf(x)
2k   1   "# no matter how small $ is taken when "
$  x  1 Ê lim f(x) Á 2. xÄ1

(b) 0  x
1  $ Ê "  x  "  $ Ê f(x) œ x  1. Then kf(x)
1k œ k(x  1)
1k œ kxk œ x  1. That is, kf(x)
1k   1 no matter how small $ is taken when "  x  "  $ Ê lim f(x) Á 1. xÄ1

(c)
$  x
1  ! Ê "
$  x  1 Ê f(x) œ x. Then kf(x)
1.5k œ kx
1.5k œ 1.5
x  1.5
1 œ 0.5. Also, !  x
1  $ Ê 1  x  "  $ Ê f(x) œ x  1. Then kf(x)
1.5k œ k(x  1)
1.5k œ kx
0.5k œ x
0.5  "
0.5 œ 0.5. Thus, no matter how small $ is taken, there exists a value of x such that
$  x
1  $ but kf(x)
1.5k   "# Ê lim f(x) Á 1.5. xÄ1

58. (a) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
4k œ 2. Thus for %  2, kh(x)
4k   % whenever 2  x  2  $ no matter how small we choose $  0 Ê lim h(x) Á 4. xÄ#

(b) For 2  x  2  $ Ê h(x) œ 2 Ê kh(x)
3k œ 1. Thus for %  1, kh(x)
3k   % whenever 2  x  2  $ no matter how small we choose $  0 Ê lim h(x) Á 3. xÄ#

(c) For 2
$  x  2 Ê h(x) œ x# so kh(x)
2k œ kx#
2k . No matter how small $  0 is chosen, x# is close to 4 when x is near 2 and to the left on the real line Ê kx#
2k will be close to 2. Thus if %  1, kh(x)
2k   % whenever 2
$  x  2 no mater how small we choose $  0 Ê lim h(x) Á 2. xÄ#

59. (a) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
4k   0.8. Thus for %  0.8, kf(x)
4k   % whenever 3
$  x  3 no matter how small we choose $  0 Ê lim f(x) Á 4. xÄ$

(b) For 3  x  3  $ Ê f(x)  3 Ê kf(x)
4.8k   1.8. Thus for %  1.8, kf(x)
4.8k   % whenever 3  x  3  $ no matter how small we choose $  0 Ê lim f(x) Á 4.8. xÄ$

(c) For 3
$  x  3 Ê f(x)  4.8 Ê kf(x)
3k   1.8. Again, for %  1.8, kf(x)
3k   % whenever $
$  x  3 no matter how small we choose $  0 Ê lim f(x) Á 3. xÄ$

60. (a) No matter how small we choose $  0, for x near
1 satisfying
"
$  x 
"  $ , the values of g(x) are near 1 Ê kg(x)
2k is near 1. Then, for % œ "# we have kg(x)
2k   "# for some x satisfying
"
$  x 
"  $ , or !  kx  1k  $ Ê

lim g(x) Á 2.

x Ä
1

(b) Yes, lim g(x) œ 1 because from the graph we can find a $  ! such that kg(x)
1k  % if !  kx
(
1)k  $ . x Ä
1

61-66. Example CAS commands (values of del may vary for a specified eps): Maple: f := x -> (x^4-81)/(x-3);x0 := 3; plot( f(x), x=x0-1..x0+1, color=black, # (a) title="Section 2.3, #61(a)" ); L := limit( f(x), x=x0 ); # (b) epsilon := 0.2; # (c) plot( [f(x),L-epsilon,L+epsilon], x=x0-0.01..x0+0.01, color=black, linestyle=[1,3,3], title="Section 2.3, #61(c)" ); q := fsolve( abs( f(x)-L ) = epsilon, x=x0-1..x0+1 ); # (d) delta := abs(x0-q); plot( [f(x),L-epsilon,L+epsilon], x=x0-delta..x0+delta, color=black, title="Section 2.3, #61(d)" ); for eps in [0.1, 0.005, 0.001 ] do # (e)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

86

Chapter 2 Limits and Continuity q := fsolve( abs( f(x)-L ) = eps, x=x0-1..x0+1 ); delta := abs(x0-q); head := sprintf("Section 2.3, #61(e)\n epsilon = %5f, delta = %5f\n", eps, delta ); print(plot( [f(x),L-eps,L+eps], x=x0-delta..x0+delta, color=black, linestyle=[1,3,3], title=head )); end do: Mathematica (assigned function and values for x0, eps and del may vary): Clear[f, x] y1: œ L
eps; y2: œ L  eps; x0 œ 1; f[x_]: œ (3x2
(7x  1)Sqrt[x]  5)/(x
1) Plot[f[x], {x, x0
0.2, x0  0.2}] L: œ Limit[f[x], x Ä x0] eps œ 0.1; del œ 0.2; Plot[{f[x], y1, y2},{x, x0
del, x0  del}, PlotRange Ä {L
2eps, L  2eps}]

2.4 ONE-SIDED LIMITS AND LIMITS AT INFINITY 1. (a) True (e) True (i) False

(b) True (f) True (j) False

(c) False (g) False (k) True

(d) True (h) False (l) False

2. (a) True (e) True (i) True

(b) False (f) True (j) False

(c) False (g) True (k) True

(d) True (h) True

3. (a)

lim f(x) œ

x Ä #b

2 #

 " œ #, lim c f(x) œ $
# œ " xÄ#

(b) No, lim f(x) does not exist because lim b f(x) Á lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 4#  1 œ 3, lim b f(x) œ 4#  " œ $ xÄ%

xÄ%

(d) Yes, lim f(x) œ 3 because 3 œ lim c f(x) œ lim b f(x) xÄ% xÄ% xÄ% 4. (a)

lim f(x) œ

x Ä #b

2 #

œ 1, lim c f(x) œ $
# œ ", f(2) œ 2 xÄ#

(b) Yes, lim f(x) œ 1 because " œ lim b f(x) œ lim c f(x) xÄ# xÄ# xÄ# (c) lim c f(x) œ 3
(
1) œ 4, lim b f(x) œ 3
(
1) œ 4 x Ä
"

x Ä
"

(d) Yes, lim f(x) œ 4 because 4 œ x Ä
"

lim

x Ä
"c

f(x) œ

lim

x Ä
"b

f(x)

5. (a) No, lim b f(x) does not exist since sin ˆ "x ‰ does not approach any single value as x approaches 0 xÄ! (b) lim c f(x) œ lim c 0 œ 0 xÄ!

(c)

xÄ!

lim f(x) does not exist because lim b f(x) does not exist xÄ! xÄ!

6. (a) Yes, lim b g(x) œ 0 by the sandwich theorem since
Èx Ÿ g(x) Ÿ Èx when x  0 xÄ! (b) No, lim c g(x) does not exist since Èx is not defined for x  0 xÄ!

(c) No, lim g(x) does not exist since lim c g(x) does not exist xÄ! xÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.4 One-Sided Limits and Limits at Infinity 7. (aÑ

lim f(x) œ " œ lim b f(x) xÄ1 (c) Yes, lim f(x) œ 1 since the right-hand and left-hand (b)

x Ä 1c

xÄ1

limits exist and equal 1

8. (a)

(b)

lim f(x) œ 0 œ lim c f(x) xÄ1

x Ä 1b

(c) Yes, lim f(x) œ 0 since the right-hand and left-hand xÄ1

limits exist and equal 0

9. (a) domain: 0 Ÿ x Ÿ 2 range: 0  y Ÿ 1 and y œ 2 (b) xlim f(x) exists for c belonging to Äc (0ß 1)  ("ß #) (c) x œ 2 (d) x œ 0

10. (a) domain:
_  x  _ range:
" Ÿ y Ÿ 1 (b) xlim f(x) exists for c belonging to Äc (
_ß
1)  (
"ß ")  ("ß _) (c) none (d) none

11.

x Ä
!Þ&c

lim

13.

x Ä
#b

14.

x Ä 1c

15.

lim b

lim

lim

hÄ!

2 0.5  2 È3 É 3/2 É xx  É

1 œ
0.5  1 œ 1/2 œ

12.

lim

x Ä 1b


" 1 È0 œ ! É "1
É xx  # œ # œ

5‰ ˆ x x 1 ‰ ˆ 2x ˆ
2 ‰ 2(
2)  5 ˆ"‰ x#  x œ
#  " Š (
#)#  (
2) ‹ œ (2) # œ 1

ˆ x " 1 ‰ ˆ x x 6 ‰ ˆ 3
7 x ‰ œ ˆ 1 " 1 ‰ ˆ 1 1 6 ‰ ˆ 3
7 1 ‰ œ ˆ "# ‰ ˆ 71 ‰ ˆ 27 ‰ œ 1 Èh#  4h  5
È5 h

œ lim b hÄ!

œ lim b Š hÄ!

ah#  4h  5b
5 h ŠÈh#  4h  5  È5‹

Èh#  4h  5
È5 È # 4h  5  È5 ‹ Š Èhh#  ‹ h  4h  5  È5

œ lim b hÄ!

h(h  4) h ŠÈh#  4h  5  È5‹

œ

04 È5  È5

œ

2 È5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

87

88 16.

Chapter 2 Limits and Continuity lim

h Ä !c

È6
È5h#  11h  6 h

6
a5h#  11h  6b

œ lim c hÄ! 17. (a)

19. (a)

) Ä $b

20. (a)

t Ä %b

(x2) (x#)

Ú) Û )

lim

akx  2k œ x  2 for x 
2b

(x  3) ’
(x(x#2) ) “

lim

x Ä
#c

akx  2k œ
(x  2) for x 
2b

(x  3)(
1) œ
(
2  3) œ
1

È2x (x
1) (x
1)

akx
1k œ x
1 for x  1b

œ lim b È2x œ È2 xÄ1

œ lim c xÄ1

È2x (x
1)
(x
1)

akx
1k œ
(x
1) for x  1b

œ

œ1

3 3

lim at
ÚtÛb œ 4
4 œ 0 sin È2) È 2)

22. lim

sin kt t

23. lim

sin 3y 4y

)Ä!

tÄ!

yÄ!

œ

26. lim

2t

t Ä ! tan t

27. lim

xÄ!

)Ä!

3 sin 3y " 4 ylim 3y Ä!

sin 2x ‰ ˆ cos 2x x

œ lim

xÄ!

œ 2 lim

t

sin t t Ä ! ˆ cos t ‰

x csc 2x cos 5x

œ

œ

œ lim

tÄ!

" ‰ cos 5x

29. lim

x  x cos x

) Ä $c

(b)

t Ä %c

Ú) Û )

lim

œ

2 3

lim at
ÚtÛb œ 4
3 œ 1

t cos t sin t

œ lim ˆ sin xxcos x  xÄ!

œk†1œk

3 sin ) 4 )lim Ä! )

œ

" 3

Œ

œ

" lim

)Ä!c

(where ) œ kt) (where ) œ 3y)

3 4

œ

sin ) )

"

œ Š lim

‹ Š lim x Ä ! cos 2x xÄ!

sin 2x

" 3

†1œ

2 sin 2x #x ‹

" 3

(where ) œ 3h)

œ1†2œ2

œ 2 Š lim cos t‹ Œ lim" sin t  œ 2 † " † " œ 2 tÄ!

œ Š #" lim

t

Ä!

t

"

‹ Š lim cos 5x ‹ x Ä ! sin 2x xÄ!

6x# cos x sin x sin 2x xÄ!

x Ä ! sin x cos x

œ

" " sin 3h 3 h lim Ä !c ˆ 3h ‰

28. lim 6x# (cot x)(csc 2x) œ lim xÄ!

)Ä!

x Ä ! x cos 2x

œ 2 lim

sin ) )

œ k lim

sin 3y 3 4 ylim Ä ! 3y

3h ‰ sin 3h

œ lim ˆ sinx2x † xÄ!

k sin ) )

œ lim

œ lim c ˆ "3 † hÄ!

h

tan 2x x

k sin kt kt

(b)

(where x œ È2))

œ1

sin x x

xÄ!

tÄ!

lim h Ä !c sin 3h

xÄ!

œ lim

œ lim

25. lim

œ
211 È6

œ lim c
È2x œ
È2 xÄ1

21. lim

24.

lim


(0  11) È6  È6

œ

(x  3) œ (
2)  3 œ 1

lim

x Ä
#b

x Ä
#c

œ lim b xÄ1

È2x (x
1) kx
1 k

lim

x Ä 1c

œ


h(5h  11)

h ŠÈ6  È5h#  11h  6‹

(x  3)

lim

x Ä
#b

œ

È2x (x
1) kx
1 k

lim

(b)

kx  2 k x2

(x  3)

lim

x Ä 1b

œ

œ

x Ä
#c

18. (a)

kx 2 k x 2

(x  3)

lim

È5h#  11h  6 È6
È5h#  11h  6 È ‹ Š È66  ‹ h  È5h#  11h  6

œ lim c hÄ!

h ŠÈ6  È5h#  11h  6‹

x Ä
#b

(b)

œ lim c Š hÄ!

2x

œ lim ˆ3 cos x † xÄ!

x cos x ‰ sin x cos x

œ lim ˆ sinx x † xÄ!

x sin x

†

" ‰ cos x

œ ˆ #" † 1‰ (1) œ

2x ‰ sin 2x

" #

œ3†"†1œ3

 lim

x

x Ä ! sin x

œ lim Š sin" x ‹ † lim ˆ cos" x ‰  lim Š sin" x ‹ œ (1)(1)  1 œ 2 xÄ!

30. lim

xÄ!

x

x#
x  sin x #x

xÄ!

œ lim ˆ #x
xÄ!

xÄ!

" #

x

 "# ˆ sinx x ‰‰ œ 0


" #

 "# (1) œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.4 One-Sided Limits and Limits at Infinity 31. lim

sin(1
cos t) 1
cos t

32. lim

sin (sin h) sin h

tÄ!

hÄ!

sin )

33. lim

) Ä ! sin 2)

34. lim

sin 5x

35. lim

tan 3x

œ

3 8 xlim Ä!

36. lim

yÄ!

œ

)Ä!

sin ) )

sin ) )

œ lim

)Ä!

œ 1 since ) œ 1
cos t Ä 0 as t Ä 0

œ 1 since ) œ sin h Ä 0 as h Ä 0

sin ) œ lim ˆ sin 2) †

2) ‰ #)

5x œ lim ˆ sin sin 4x †

4x 5x

sin 3x œ lim ˆ cos 3x †

" ‰ sin 8x

)Ä!

x Ä ! sin 4x

x Ä ! sin 8x

œ lim

xÄ!

xÄ!

" # )lim Ä!

œ

† 54 ‰ œ

œ lim

yÄ!

2) ‰ sin 2)

ˆ sin5x5x †

5 4 xlim Ä!

sin 3x œ lim ˆ cos 3x † 3 8

†1†1†1œ

4x ‰ sin 4x

†

8x 3x

†1†1œ œ

5 4

3 8

"

lim

xÄ „_

12 5

œ

12 5

œ 0 whenever

m n

 0. This result follows immediately from

ˆ xm"În ‰ œ

lim

xÄ „_

37. (a)
3

(b)
3

38. (a) 1

(b) 1

39. (a)

" #

(b)

" #

40. (a)

" 8

(b)

" 8

41. (a)
53

45.

lim

tÄ_

46. r Ä lim_

ˆ x" ‰mÎn œ Š

"

lim ‹ xÄ „_ x

mÎn

(b)
53

3 4

(b)

44.
3") Ÿ

5 4

† 83 ‰

œ1†1†1†1†

lim mÎn xÄ „_ x

Example 6 and the power rule in Theorem 8:

43.
"x Ÿ

†1†1œ

yÄ!

cos 5y ˆ 3†4 ‰ lim Š sin3y3y ‹ Š sin4y4y ‹ Š sin5y5y ‹ Š cos 4y ‹ 5 yÄ!

42. (a)

" #

sin 4y cos 5y 3†4†5y œ lim Š siny3y ‹ Š cos 4y ‹ Š sin 5y ‹ Š 3†4†5y ‹

sin 3y sin 4y cos 5y y cos 4y sin 5y

Note: In these exercises we use the result

" #

œ

" sin 8x

xÄ!

ˆ cos"3x ‰ ˆ sin3x3x ‰ ˆ sin8x8x ‰ œ

sin 3y cot 5y y cot 4y

ˆ sin) ) †

sin 2x x

Ÿ

" x

cos ) 3)

Ÿ

" 3)

2
t  sin t t  cos t

Ê x lim Ä_ Ê

47. (a) x lim Ä_

lim

) Ä
_

œ lim

2 t

tÄ_

r  sin r 2r  7
5 sin r

2x  3 5x  7

$

sin 2x x

œrÄ lim_

œ x lim Ä_

œ 0 by the Sandwich Theorem

cos ) 3)

œ 0 by the Sandwich Theorem


1  ˆ sint t ‰ 1  ˆ cost t ‰

œ

1  ˆ sinr r ‰ 2  7r
5 ˆ sinr r ‰ 2  3x 5  7x

2x  7 48. (a) x lim œ x lim Ä _ x$
x#  x  7 Ä_ (b) 2 (same process as part (a))

3 4

œ

0
10 10

œ
1

œrÄ lim_

2 5

2  Š x7$ ‹

1
"x  x"#  x7$

10 20
0

œ

(b)

" #

2 5

(same process as part (a))

œ2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 0mÎn œ 0.

89

90

Chapter 2 Limits and Continuity " x

 x"#

49. (a) x lim Ä_

x1 x#  3

œ x lim Ä_

1  x3#

50. (a) x lim Ä_

3x  7 x#
2

œ x lim Ä_

1
x2#

51. (a) x lim Ä_

7x$ x$
3x#  6x

52. (a) x lim Ä_

" x$
4x  1

&

3 x

 x7#

œ x lim Ä_

(b)

9 #

2x%

œ0

(b) 0 (same process as part (a))

" x$

œ x lim Ä_

10 x

œ x lim Ä_

œ(

 x"#  x31' 1

2

(b) 7 (same process as part (a))

œ!

1
x4#  x"$

%

9x%  x  5x#
x  6

(b) 0 (same process as part (a))

7 1
3x  x6#

10x  x  31 53. (a) x lim œ x lim x' Ä_ Ä_ (b) 0 (same process as part (a))

54. (a) x lim Ä_

œ0

(b) 0 (same process as part (a))

œ0

9  x"$

5 x#


x"$  x6%

œ

9 #

(same process as part (a))

55. (a) x lim Ä_


2x$
2x  3 3x$  3x#
5x

œ x lim Ä_


2
x2#  x3$ 3  3x
x5#

œ
23

(b)
23 (same process as part (a)) %


x 56. (a) x lim œ x lim Ä _ x%
7x$  7x#  9 Ä_ (b)
1 (same process as part (a))

57. x lim Ä_

2Èx  x
" 3x
7

59. x Ä lim
_

œ x lim Ä_

$ & x È x
È $ xÈ & x È

x
"  x
% x
#
x
$

61. x lim Ä_

2x&Î$
x"Î$  7 x)Î&  3x  Èx

62. x Ä lim
_

2 Š "Î# ‹  Š x"# ‹ x 3
7x

œxÄ lim
_

60. x lim Ä_

œ x lim Ä_

$ È x
5x  3 2x  x#Î$
4


" 1
7x  x7#  x9%

œ0

1
xÐ"Î&Ñ
Ð"Î$Ñ 1  xÐ"Î&Ñ
Ð"Î$Ñ

x  x"# 1
x"

œ x lim Ä_ œxÄ lim
_

œ
1

58. x lim Ä_

œxÄ lim
_

" ‹ 1
Š #Î"& x

" ‹ 1  Š #Î"&

2  Èx 2
Èx

œ x lim Ä_

2 Š "Î# ‹" x 2 Š "Î# ‹
1 x

œ
1

œ1

x

œ_

"  7 2x"Î"&
"*Î"& x x)Î& " 3 1  $Î&  ""Î"! x

" x#Î$

2


5  3x " x"Î$

œ_

x


4x

œ
5#

63. Yes. If lim b f(x) œ L œ lim c f(x), then xlim f(x) œ L. If lim b f(x) Á lim c f(x), then xlim f(x) does not exist. Äa Äa xÄa xÄa xÄa xÄa 64. Since xlim f(x) œ L if and only if lim b f(x) œ L and lim c f(x) œ L, then xlim f(x) can be found by calculating Äc Äc xÄc xÄc lim b f(x). xÄc

65. If f is an odd function of x, then f(
x) œ
f(x). Given lim b f(x) œ 3, then lim c f(x) œ
$. xÄ! xÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.4 One-Sided Limits and Limits at Infinity

91

66. If f is an even function of x, then f(
x) œ f(x). Given lim c f(x) œ 7 then lim b f(x) œ 7. However, nothing xÄ# x Ä
# can be said about 67. Yes. If x lim Ä_

lim

x Ä
#c

f(x) g(x)

f(x) because we don't know lim b f(x). xÄ#

œ 2 then the ratio of the polynomials' leading coefficients is 2, so x Ä lim
_

f(x) g(x)

œ 2 as well.

68. Yes, it can have a horizontal or oblique asymptote. 69. At most 1 horizontal asymptote: If x lim Ä_ f(x) lim x Ä
_ g(x)

f(x) g(x)

œ L, then the ratio of the polynomials' leading coefficients is L, so

œ L as well.

Èx#  x
Èx#
x œ lim ’Èx#  x
Èx#
x“ † ’ Èx#  x  Èx#
x “ œ lim 70. x lim È x#  x  È x#
x Ä_ xÄ_ xÄ_ 2x 2 2 œ x lim œ lim œ œ 1 È # 1 1 " " Ä_ È # xÄ_ x x

x
x

ax #  x b
a x #
x b È x#  x  È x#
x

É1  x  É1
x

71. For any %  0, take N œ 1. Then for all x  N we have that kf(x)
kk œ kk
kk œ 0  %. 72. For any %  0, take N œ 1. Then for all y 
N we have that kf(x)
kk œ kk
kk œ 0  %. 73. I œ (5ß 5  $ ) Ê 5  x  &  $ . Also, Èx
5  % Ê x
5  %# Ê x  &  %# . Choose $ œ %# Ê lim Èx
5 œ 0. x Ä &b

74. I œ (%
$ ß %) Ê %
$  x  4. Also, È%
x  % Ê %
x  %# Ê x  %
%# . Choose $ œ %# Ê lim È%
x œ 0. x Ä %c

75. As x Ä 0
the number x is always negative. Thus, ¹ kxxk
(
1)¹  % Ê ¸
xx  1¸  % Ê 0  % which is always true independent of the value of x. Hence we can choose any $  0 with
$  x  ! Ê

x

lim x Ä ! c kx k

œ
1.

2 ¸ x
2 ¸ 76. Since x Ä # we have x  2 and kx
2k œ x
2. Then, ¹ kxx

2 k
" ¹ œ x
2
"  % Ê 0  %

which is always true so long as x  #. Hence we can choose any $  !, and thus #  x  #  $ 2 Ê ¹ kxx

2k
"¹  % . Thus,

77. (a)

lim

x Ä %!!b

x
2

lim x Ä
#b kx
2k

œ 1.

ÚxÛ œ 400. Just observe that if 400  x  401, then ÚxÛ œ 400. Thus if we choose $ œ ", we have for any

number %  ! that 400  x  400  $ Ê lÚxÛ
400l œ l400
400l œ !  %. (b) lim c ÚxÛ œ 399. Just observe that if 399  x  400 then ÚxÛ œ 399. Thus if we choose $ œ ", we have for any x Ä %!!

number %  ! that 400
$  x  400 Ê lÚxÛ
399l œ l399
399l œ !  %. (c) Since lim b ÚxÛ Á lim c ÚxÛ we conclude that lim ÚxÛ does not exist. x Ä %!!

78. (a)

x Ä %!!

x Ä %!!

lim f(x) œ lim b Èx œ È0 œ 0; ¸Èx
0¸  % Ê
%  Èx  % Ê !  x  %# for x positive. Choose $ œ %# xÄ! Ê lim b f(x) œ 0.

x Ä !b

xÄ!

(b)

lim f(x) œ lim c x# sin ˆ x" ‰ œ 0 by the sandwich theorem since
x# Ÿ x# sin ˆ x" ‰ Ÿ x# for all x Á 0. x Ä !c xÄ! Since kx#
0k œ k
x#
0k œ x#  % whenever kxk  È%, we choose $ œ È% and obtain ¸x# sin ˆ "x ‰
0¸  % if
$  x  0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

92

Chapter 2 Limits and Continuity (c) The function f has limit 0 at x! œ 0 since both the right-hand and left-hand limits exist and equal 0.

79.

81.

82. 83. 84.

lim

" x

x sin

xÄ „_

3x  4 x Ä „ _ 2x
5

œ lim

œ

lim

"

)Ä0 )

sin ) œ 1, ˆ) œ x" ‰

3  4x 5 x Ä „ _ 2
x

lim

œ lim

3  4t

t Ä 0 2
5t

œ

80. 3 #

cos

lim

" x

x Ä
_ 1  x"

œ lim c )Ä!

cos ) 1)

œ

" 1

œ 1, ˆ) œ x" ‰

, ˆt œ "x ‰

"Îx lim ˆ "x ‰ œ lim b zz œ 1, ˆz œ x" ‰ zÄ!

xÄ_

ˆ3  2x ‰ ˆcos "x ‰ œ lim (3  2))(cos )) œ (3)(1) œ 3, ˆ) œ x" ‰

lim

xÄ „_

)Ä0

lim ˆ x3#
cos x" ‰ ˆ1  sin x" ‰ œ lim b a3)#
cos )b (1  sin )) œ (0
1)(1  0) œ
1, ˆ) œ x" ‰ )Ä!

xÄ_

2.5 INFINITE LIMITS AND VERTICAL ASYMPTOTES "

œ_

1.

lim x Ä !b 3x

3.

lim x Ä #c x
2

5.

lim x Ä
)b x8

7.

lim # x Ä ( (x
7)

3

2x

4

œ
_ œ
_ œ_

lim "Î$ x Ä !b 3x

10. (a)

lim "Î& x Ä !b x 4

11. lim

#Î& xÄ! x

13.

œ lim

4

# x Ä ! ax"Î& b

œ_

Š positive positive ‹

lim x Ä !c 2x

positive Š negative ‹

4.

lim x Ä $b x
3

Š negative positive ‹

6.

lim x Ä
&c 2x10

3x

œ_

positive Š positive ‹

8.

lim # x Ä ! x (x1)

œ
_

(b)

lim "Î$ x Ä !c 3x

(b)

lim "Î& x Ä !c x

œ_

2

positive Š negative ‹

2.

œ_

2

9. (a)

œ
_

Š positive positive ‹

œ_

5

"


"

2

2

12. lim

"

#Î$ xÄ! x

lim
tan x œ _

14.

x Ä ˆ 1# ‰

Š negative negative ‹ negative Š positive †positive ‹

œ
_ œ
_

œ lim

"

# x Ä ! ax"Î$ b

œ_

lim  sec x œ _

x Ä ˆ
#1 ‰

lim (1  csc )) œ
_

15.

) Ä !


16.

) Ä !b

lim (2
cot )) œ
_ and lim c (2
cot )) œ _, so the limit does not exist )Ä!

"

œ lim b xÄ#

" (x2)(x
2)

œ_

Š positive"†positive ‹

"

œ lim c xÄ#

" (x2)(x
2)

œ
_

Š positive†"negative ‹

17. (a)

lim # x Ä # b x
4

(b)

lim # x Ä # c x
4

(c)

lim # x Ä
#b x
4

(d)

lim # x Ä
#c x
4

"

œ

lim x Ä
#b (x2)(x
2)

"

œ
_

Š positive†"negative ‹

"

œ

lim x Ä
#c (x2)(x
2)

"

œ_

Š negative"†negative ‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.5 Infinite Limits and Vertical Asymptotes x

œ lim b xÄ"

x (x1)(x
1)

œ_

positive Š positive †positive ‹

x

œ lim c xÄ"

x (x1)(x
1)

œ
_

positive Š positive †negative ‹

18. (a)

lim # x Ä "b x
1

(b)

lim # x Ä "c x
1

(c)

lim # x Ä
"b x
1

(d)

lim # x Ä
"c x
1

x

œ

lim x Ä
"b (x1)(x
1)

x

œ_

negative Š positive †negative ‹

x

œ

lim x Ä
"c (x1)(x
1)

x

œ
_

negative Š negative †negative ‹

19. (a)

lim x Ä !b #

x#




" x

œ 0  lim b xÄ!

"
x

œ
_

" Š negative ‹

(b)

lim x Ä !c #

x#




" x

œ 0  lim c xÄ!

"
x

œ_

" Š positive ‹

(c)

lim # x Ä $È2

(d)

lim x Ä
1 #

20. (a)

x#

x#

lim x Ä
#b

" x





" x

œ

x#
1 2x  4 x#
1

2#Î$ #

œ

" #

(d)

lim x Ä !c 2x  4

œ

(b) (c) (d) (e) 22. (a)

x#
3x  2 x$
2x#

lim b

x#
3x  2 x$
2x# #

x
3x  2 x$
2x#

lim

x Ä #c

lim

xÄ#

#

x
3x  2 x$
2x# #

x
3x  2 x$
2x#

lim

xÄ!

lim x Ä
#b

(c)

x Ä 0c

œ lim b xÄ# œ lim c xÄ#

(d)

x Ä "b

(e)

lim x Ä !b x(x  #)

x#
1 2x  4

lim x Ä
#c

œ
_

positive Š negative ‹

œ0

(x
2)(x
1) x# (x
2)

œ
_

(x
2)(x
1) x# (x
2)

œ lim b xÄ#

(x
2)(x
1) x# (x
2)

œ lim c xÄ#

œ lim

œ lim

(x
2)(x
1) x# (x
2)

œ
_

xÄ!

x#
3x  2 x$
4x

lim

x#
3x  2 x$
4x x
"

2†0 #4

(x
2)(x
1) x# (x
2)

xÄ#

lim

and

œ lim b xÄ!

œ lim b xÄ#

x#
3x  2 x$
4x

(b)

œ

œ lim

x#
3x  2 x$
4x

lim

x Ä #b

(x  1)(x
1) 2x  4

(b)


" 4

lim b

xÄ#

3 #

Š positive positive ‹

œ lim b xÄ"

xÄ!

œ 2
"Î$
2
"Î$ œ 0

œ_

lim x Ä "b 2x  4

21. (a)

" #"Î$


ˆ
"1 ‰ œ

(c)

x#
1




œ

xÄ#

(x
2)(x
") x(x
#)(x  2)

(x
2)(x
")

œ lim c xÄ! œ lim b xÄ"

(x
2)(x
") x(x
#)(x  2) (x
2)(x
") x(x
#)(x  2)

œ

x
1 x#

œ

" 4

,xÁ2

x
1 x#

œ

" 4

,xÁ2

x
1 x#

œ

" 4

,xÁ2 †negative Š negative positive†negative ‹

œ lim b xÄ#

lim x Ä
#b x(x
#)(x  2)

†negative Š negative positive†negative ‹

(x
1) x(x  #)

œ

(x
1)

lim x Ä
#b x(x  #)

œ lim c xÄ! œ lim b xÄ"

œ_

(x
1) x(x  #)

œ

negative Š positive †positive ‹

x
"

negative Š negative †positive ‹

œ_

œ

" 8

œ_

(x
1) x(x  #)

œ
_

lim x Ä !c x(x  #)

" #(4)

0 (1)(3)

negative Š negative †positive ‹ negative Š negative †positive ‹

œ0

so the function has no limit as x Ä 0. lim
2


23. (a)

t Ä !b

24. (a)

t Ä !b

25. (a)

x Ä !b

(c)

x Ä "b

3 ‘ t"Î$

œ
_

" lim
t$Î&  7‘ œ _

lim


2


lim




lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(b)

t Ä !c

(b)

t Ä !c

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(b)

x Ä !c

lim

"  ’ x#Î$

2 “ (x
1)#Î$

œ_

(d)

x Ä "c

" t$Î&

3 ‘ t"Î$

œ_

 7‘ œ
_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

93

94

Chapter 2 Limits and Continuity

26. (a)

x Ä !b

(c)

x Ä "b

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ_

(b)

x Ä !c

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

(d)

x Ä "c

lim

"
’ x"Î$

1 “ (x
1)%Î$

œ
_

27. y œ

" x
1

28. y œ

" x1

29. y œ

" #x  4

30. y œ


3 x
3

31. y œ

x3 x2

32. y œ

2x x1

œ1

" x#

œ#


2 x1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.5 Infinite Limits and Vertical Asymptotes 33. y œ

x# x
"

35. y œ

x#
% x
"

œx"


37. y œ

x#
1 x

œx


œx1

" x
"

$ x
"

" x

39. Here is one possibility.

34. y œ

x#  " x
1

œx"

36. y œ

x2
" #x  %

œ #" x
" 

38. y œ

x$  1 x#

œx

# x
1

$ #x  %

" x#

40. Here is one possibility.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

95

96

Chapter 2 Limits and Continuity

41. Here is one possibility.

42. Here is one possibility.

43. Here is one possibility.

44. Here is one possibility.

45. Here is one possibility.

46. Here is one possibility.


" x#

47. For every real number
B  0, we must find a $  0 such that for all x, 0  kx
0k  $ Ê


" x#

Ê


B  ! Í
" x#

" x#

#

B0 Í x 

" B

" ÈB

Í kxk 

. Choose $ œ

" ÈB

, then 0  kxk  $ Ê kxk 

xÄ!

" B.

 B  ! Í lxl 

" B.

Choose $ œ

Then !  kx
0k  $ Ê lxl 

" B

Ê

" lx l

" lx l


2 (x
3)#


B  ! Í

2 (x
3)#

$ œ É B2 , then 0  kx
3k  $ Ê

B0 Í
2 (x
3)#

(x
3)# 2



" B

Í (x
3)# 


B  0 so that lim


2

# x Ä $ (x
3)

2 B

 B. Now,

x Ä ! lx l
2 (x
3)#

Now,

#

 B  ! Í (x  5) 

Ê kx  5k 

" ÈB

Ê

" (x  5)#

" B

Í kx  5k 

 B so that lim

"

" ÈB

# x Ä
& (x  5)

. Choose $ œ

œ _.


B.

Í !  kB
$k  É B2 . Choose

œ
_.

50. For every real number B  0, we must find a $  0 such that for all x, 0  kx
(
5)k  $ Ê 1 (x  5)#

"

 B so that lim

49. For every real number
B  0, we must find a $  0 such that for all x, 0  kx
3k  $ Ê Now,

" ÈB


B so that lim
x"# œ
_.

48. For every real number B  0, we must find a $  0 such that for all x, !  kx
0k  $ Ê " lx l


B. Now,

" ÈB

1 (x  5)#

 B.

. Then 0  kx
(
5)k  $

œ _.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.5 Infinite Limits and Vertical Asymptotes

97

51. (a) We say that f(x) approaches infinity as x approaches x! from the left, and write lim
f(x) œ _, if x Ä x!

for every positive number B, there exists a corresponding number $  0 such that for all x, x!
$  x  x! Ê f(x)  B. (b) We say that f(x) approaches minus infinity as x approaches x! from the right, and write lim
f(x) œ
_, x Ä x!

if for every positive number B (or negative number
B) there exists a corresponding number $  0 such that for all x, x!  x  x!  $ Ê f(x) 
B. (c) We say that f(x) approaches minus infinity as x approaches x! from the left, and write lim
f(x) œ
_, x Ä x!

if for every positive number B (or negative number
B) there exists a corresponding number $  0 such that for all x, x!
$  x  x! Ê f(x) 
B. 52. For B  0,

" x

 B  0 Í x  B" . Choose $ œ B" . Then !  x  $ Ê 0  x 

53. For B  0,

" x


B  0 Í
x"  B  0 Í
x 

Ê
B"  x Ê 54. For B  !,

" x
#

" x


B so that lim c xÄ!

" x

" B

Ê

" x
#

" x
#

Ê

" x
#

" 1
x#

" x

œ _.

" B

Í x
2 
B" Í x  2
B" . Choose $ œ B" . Then " x
#


B  0 so that lim c xÄ#

" x
#

œ
_.

œ _.

 B Í 1
x# 

" #B . Then "
$  x  " Ê " 1
x#  B for !  x  1 and " x

 B so that lim b xÄ!

 B Í !  x
2  B" . Choose $ œ B" . Then #  x  #  $ Ê !  x
#  $ Ê !  x
2 

 B  ! so that lim b xÄ#

57. y œ sec x 

" x

œ
_.


B Í
x
" #  B Í
(x
2) 

56. For B  0 and !  x  1, $

Ê

Í
B"  x. Choose $ œ B" . Then
$  x  !

2
$  x  2 Ê
$  x
2  ! Ê
B"  x
2  0 Ê 55. For B  0,

" B

" B

Í ("
x)("  x)  B" . Now


$  x
1  0 Ê "
x  $  x near 1 Ê

"

lim # x Ä "c "
x

" #B

1x #

 1 since x  1. Choose Ê ("
x)("  x)  B" ˆ 1 # x ‰  B"

œ _.

58. y œ sec x


" x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" B

98

Chapter 2 Limits and Continuity

59. y œ tan x 

61. y œ

" x#

x È 4
x#

63. y œ x#Î$ 

60. y œ

" x

62. y œ


" È 4
x#


tan x

64. y œ sin ˆ x# 1 1 ‰

" x"Î$

2.6 CONTINUITY 1. No, discontinuous at x œ 2, not defined at x œ 2 2. No, discontinuous at x œ 3, " œ lim c g(x) Á g(3) œ 1.5 xÄ$

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.6 Continuity 3. Continuous on [
1ß 3] 4. No, discontinuous at x œ 1, 1.5 œ lim c k(x) Á lim b k(x) œ ! xÄ" xÄ" 5. (a) Yes

(b) Yes,

(c) Yes

(d) Yes

6. (a) Yes, f(1) œ 1

lim

x Ä
"b

f(x) œ 0

(b) Yes, lim f(x) œ 2 xÄ1

(c) No

(d) No

7. (a) No

(b) No

8. [
"ß !)  (!ß ")  ("ß #)  (#ß $) 9. f(2) œ 0, since lim c f(x) œ
2(2)  4 œ 0 œ lim b f(x) xÄ# xÄ# 10. f(1) should be changed to 2 œ lim f(x) xÄ1

11. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( limc f(x) œ 1 and lim b f(x) œ 0). xÄ" xÄ1 xÄ"

Removable discontinuity at x œ 0 by assigning the number lim f(x) œ 0 to be the value of f(0) rather than xÄ!

f(0) œ 1.

12. Nonremovable discontinuity at x œ 1 because lim f(x) fails to exist ( lim c f(x) œ 2 and lim b f(x) œ 1). xÄ" xÄ1 xÄ" Removable discontinuity at x œ 2 by assigning the number lim f(x) œ 1 to be the value of f(2) rather than xÄ#

f(2) œ 2. 13. Discontinuous only when x
2 œ 0 Ê x œ 2

14. Discontinuous only when (x  2)# œ 0 Ê x œ
2

15. Discontinuous only when x#
%x  $ œ ! Ê (x
3)(x
1) œ 0 Ê x œ 3 or x œ 1 16. Discontinuous only when x#
3x
10 œ 0 Ê (x
5)(x  2) œ 0 Ê x œ 5 or x œ
2 17. Continuous everywhere. ( kx
1k  sin x defined for all x; limits exist and are equal to function values.) 18. Continuous everywhere. ( kxk  " Á 0 for all x; limits exist and are equal to function values.) 19. Discontinuous only at x œ 0 20. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
") 1# , n an integer, but continuous at all other x. 21. Discontinuous when 2x is an integer multiple of 1, i.e., 2x œ n1, n an integer Ê x œ

n1 # ,

n an integer, but

continuous at all other x. 22. Discontinuous when

1x #

is an odd integer multiple of 1# , i.e.,

1x #

œ (2n
1) 1# , n an integer Ê x œ 2n
1, n an

integer (i.e., x is an odd integer). Continuous everywhere else.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

99

100

Chapter 2 Limits and Continuity

23. Discontinuous at odd integer multiples of 1# , i.e., x = (2n
1) 1# , n an integer, but continuous at all other x. 24. Continuous everywhere since x%  1   1 and
" Ÿ sin x Ÿ 1 Ê 0 Ÿ sin# x Ÿ 1 Ê 1  sin# x   1; limits exist and are equal to the function values. 25. Discontinuous when 2x  3  0 or x 
3# Ê continuous on the interval

3# ß _‰ . 26. Discontinuous when 3x
1  0 or x 

" 3

Ê continuous on the interval
3" ß _‰ .

27. Continuous everywhere: (2x
1)"Î$ is defined for all x; limits exist and are equal to function values. 28. Continuous everywhere: (2
x)"Î& is defined for all x; limits exist and are equal to function values. 29. xlim sin (x
sin x) œ sin (1
sin 1) œ sin (1
0) œ sin 1 œ 0, and function continuous at x œ 1. Ä1 30. lim sin ˆ 1# cos (tan t)‰ œ sin ˆ 1# cos (tan (0))‰ œ sin ˆ 1# cos (0)‰ œ sin ˆ 1# ‰ œ 1, and function continuous at t œ !. tÄ!

31. lim sec ay sec# y
tan# y
1b œ lim sec ay sec# y
sec# yb œ lim sec a(y
1) sec# yb œ sec a("
") sec# 1b yÄ1

yÄ1

yÄ1

œ sec 0 œ 1, , and function continuous at y œ ". 32. lim tan
14 cos ˆsin x"Î$ ‰‘ œ tan
14 cos (sin(0))‘ œ tan ˆ 14 cos (0)‰ œ tan ˆ 14 ‰ œ 1, and function continuous at x œ !. xÄ!

33. lim cos ’ È19
13 sec 2t “ œ cos ’ È19
13 sec 0 “ œ cos tÄ!

1 È16

œ cos

1 4

œ

È2 # ,

and function continuous at t œ !.

34. lim1 Écsc# x  5È3 tan x œ Écsc# ˆ 16 ‰  5È3 tan ˆ 16 ‰ œ Ê4  5È3 Š È"3 ‹ œ È9 œ 3, and function continuous at xÄ

'

x œ 1' . 35. g(x) œ

x#
9 x
3

(x  3)(x
3) (x
3)

œ

36. h(t) œ

t#  3t
10 t
#

37. f(s) œ

s$
" s#
1

38. g(x) œ

œ

œ

œ x  3, x Á 3 Ê g(3) œ lim (x  3) œ 6

(t  5)(t
2) t
#

as#  s  1b (s
1) (s  1)(s
1)

x#
16 x#
3x
4

œ

xÄ$

œ t  5, t Á # Ê h(2) œ lim (t  5) œ 7 tÄ#

œ

(x  4)(x
4) (x
4)(x  1)

s#  s  " s1 ,

œ

x4 x1

s Á 1 Ê f(1) œ lim Š s sÄ1

#

s1 s1 ‹

4‰ , x Á 4 Ê g(4) œ lim ˆ xx  1 œ

xÄ%

œ

3 #

8 5

39. As defined, lim c f(x) œ (3)#
1 œ 8 and lim b (2a)(3) œ 6a. For f(x) to be continuous we must have xÄ$ xÄ$ 6a œ 8 Ê a œ 43 .

40. As defined,

lim

x Ä
#c

g(x) œ
2 and

4b œ
2 Ê b œ
"# .

lim

x Ä
#b

g(x) œ b(
2)# œ 4b. For g(x) to be continuous we must have

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.6 Continuity 41. The function can be extended: f(0) ¸ 2.3.

42. The function cannot be extended to be continuous at x œ 0. If f(0) ¸ 2.3, it will be continuous from the right. Or if f(0) ¸
2.3, it will be continuous from the left.

43. The function cannot be extended to be continuous at x œ 0. If f(0) œ 1, it will be continuous from the right. Or if f(0) œ
1, it will be continuous from the left.

44. The function can be extended: f(0) ¸ 7.39.

101

45. f(x) is continuous on [!ß "] and f(0)  0, f(1)  0 Ê by the Intermediate Value Theorem f(x) takes on every value between f(0) and f(1) Ê the equation f(x) œ 0 has at least one solution between x œ 0 and x œ 1.

46. cos x œ x Ê (cos x)
x œ 0. If x œ
1# , cos ˆ
1# ‰
ˆ
1# ‰  0. If x œ 1# , cos ˆ 1# ‰
for some x between


1 #

and

1 #

1 #

 0. Thus cos x
x œ 0

according to the Intermediate Value Theorem.

47. Let f(x) œ x$
15x  1 which is continuous on [
4ß 4]. Then f(
4) œ
3, f(
1) œ 15, f(1) œ
13, and f(4) œ 5. By the Intermediate Value Theorem, f(x) œ 0 for some x in each of the intervals
%  x 
1,
"  x  1, and "  x  4. That is, x$
15x  1 œ 0 has three solutions in [
%ß 4]. Since a polynomial of degree 3 can have at most 3 solutions, these are the only solutions. 48. Without loss of generality, assume that a  b. Then F(x) œ (x
a)# (x
b)#  x is continuous for all values of x, so it is continuous on the interval [aß b]. Moreover F(a) œ a and F(b) œ b. By the Intermediate Value Theorem, since a  a # b  b, there is a number c between a and b such that F(x) œ a # b .

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102

Chapter 2 Limits and Continuity

49. Answers may vary. Note that f is continuous for every value of x. (a) f(0) œ 10, f(1) œ 1$
8(1)  10 œ 3. Since $  1  10, by the Intermediate Value Theorem, there exists a c so that !  c  1 and f(c) œ 1. (b) f(0) œ 10, f(
4) œ (
4)$
8(
4)  10 œ
22. Since
22 
È3  10, by the Intermediate Value Theorem, there exists a c so that
4  c  0 and f(c) œ
È3. (c) f(0) œ 10, f(1000) œ (1000)$
8(1000)  10 œ 999,992,010. Since 10  5,000,000  999,992,010, by the Intermediate Value Theorem, there exists a c so that !  c  1000 and f(c) œ 5,000,000. 50. All five statements ask for the same information because of the intermediate value property of continuous functions. (a) A root of f(x) œ x$
3x
1 is a point c where f(c) œ 0. (b) The points where y œ x$ crosses y œ 3x  1 have the same y-coordinate, or y œ x$ œ 3x  1 Ê f(x) œ x$
3x
1 œ 0. (c) x$
3x œ 1 Ê x$
3x
1 œ 0. The solutions to the equation are the roots of f(x) œ x$
3x
1. (d) The points where y œ x$
3x crosses y œ 1 have common y-coordinates, or y œ x$
3x œ 1 Ê f(x) œ x$
3x
1 œ !. (e) The solutions of x$
3x
1 œ 0 are those points where f(x) œ x$
3x
1 has value 0. 51. Answers may vary. For example, f(x) œ

sin (x
2) x
2

is discontinuous at x œ 2 because it is not defined there.

However, the discontinuity can be removed because f has a limit (namely 1) as x Ä 2. 52. Answers may vary. For example, g(x) œ

" x1

has a discontinuity at x œ
1 because lim g(x) does not exist. x Ä
"

Š lim c g(x) œ
_ and lim b g(x) œ _.‹ x Ä
" x Ä
" 53. (a) Suppose x! is rational Ê f(x! ) œ 1. Choose % œ "# . For any $  0 there is an irrational number x (actually infinitely many) in the interval (x!
$ ß x!  $ ) Ê f(x) œ 0. Then 0  kx
x! k  $ but kf(x)
f(x! )k œ 1  "# œ %, so x lim f(x) fails to exist Ê f is discontinuous at x! rational. Äx !

On the other hand, x! irrational Ê f(x! ) œ 0 and there is a rational number x in (x!
$ ß x!  $ ) Ê f(x) œ 1. Again x lim f(x) fails to exist Ê f is discontinuous at x! irrational. That is, f is discontinuous at Äx !

every point. (b) f is neither right-continuous nor left-continuous at any point x! because in every interval (x!
$ ß x! ) or (x! ß x!  $ ) there exist both rational and irrational real numbers. Thus neither limits lim
f(x) and x Ä x!

lim  f(x) exist by the same arguments used in part (a).

x Ä x!

54. Yes. Both f(x) œ x and g(x) œ x
g ˆ "# ‰ œ 0 Ê

f(x) g(x)

" #

are continuous on [!ß "]. However

f(x) g(x)

is undefined at x œ

" #

since

is discontinuous at x œ "# .

55. No. For instance, if f(x) œ 0, g(x) œ ÜxÝ, then h(x) œ 0 aÜxÝb œ 0 is continuous at x œ 0 and g(x) is not. 56. Let f(x) œ œ

" x
1

" (x  1)
1

œ

and g(x) œ x  1. Both functions are continuous at x œ 0. The composition f ‰ g œ f(g(x)) " x

is discontinuous at x œ 0, since it is not defined there. Theorem 10 requires that f(x) be

continuous at g(0), which is not the case here since g(0) œ 1 and f is undefined at 1. 57. Yes, because of the Intermediate Value Theorem. If f(a) and f(b) did have different signs then f would have to equal zero at some point between a and b since f is continuous on [aß b].

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.7 Tangents and Derivatives 58. Let f(x) be the new position of point x and let d(x) œ f(x)
x. The displacement function d is negative if x is the left-hand point of the rubber band and positive if x is the right-hand point of the rubber band. By the Intermediate Value Theorem, d(x) œ 0 for some point in between. That is, f(x) œ x for some point x, which is then in its original position. 59. If f(0) œ 0 or f(1) œ 1, we are done (i.e., c œ 0 or c œ 1 in those cases). Then let f(0) œ a  0 and f(1) œ b  1 because 0 Ÿ f(x) Ÿ 1. Define g(x) œ f(x)
x Ê g is continuous on [0ß 1]. Moreover, g(0) œ f(0)
0 œ a  0 and g(1) œ f(1)
1 œ b
1  0 Ê by the Intermediate Value Theorem there is a number c in (!ß ") such that g(c) œ 0 Ê f(c)
c œ 0 or f(c) œ c. 60. Let % œ

kf(c)k #

 0. Since f is continuous at x œ c there is a $  0 such that kx
ck  $ Ê kf(x)
f(c)k  %

Ê f(c)
%  f(x)  f(c)  %. If f(c)  0, then % œ "# f(c) Ê " #

" #

If f(c)  0, then % œ
f(c) Ê

f(c)  f(x)  3 #

3 #

f(c)  f(x) 

f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ). " #

f(c) Ê f(x)  0 on the interval (c
$ ß c  $ ).

61. By Exercises 52 in Section 2.3, we have xlim faxb œ L Í lim fac  hb œ L. Äc hÄ0

Thus, faxb is continuous at x œ c Í xlim faxb œ facb Í lim fac  hb œ facb. Äc hÄ0

62. By Exercise 61, it suffices to show that lim sinac  hb œ sin c and lim cosac  hb œ cos c. hÄ0

hÄ0

Now lim sinac  hb œ lim
asin cbacos hb  acos cbasin hb‘ œ asin cbŠ lim cos h‹  acos cbŠ lim sin h‹ hÄ0

hÄ0

hÄ0

hÄ0

By Example 6 Section 2.2, lim cos h œ " and lim sin h œ !. So lim sinac  hb œ sin c and thus faxb œ sin x is hÄ0

continuous at x œ c. Similarly,

hÄ0

hÄ0

lim cosac  hb œ lim
acos cbacos hb
asin cbasin hb‘ œ acos cbŠ lim cos h‹
asin cbŠ lim sin h‹ œ cos c.

hÄ0

hÄ0

Thus, gaxb œ cos x is continuous at x œ c.

hÄ0

63. x ¸ 1.8794,
1.5321,
0.3473

64. x ¸ 1.4516,
0.8547, 0.4030

65. x ¸ 1.7549

66. x ¸ 1.5596

67. x ¸ 3.5156

68. x ¸
3.9058, 3.8392, 0.0667

69. x ¸ 0.7391

70. x ¸
1.8955, 0, 1.8955

hÄ0

2.7 TANGENTS AND DERIVATIVES 1. P" : m" œ 1, P# : m# œ 5

2. P" : m" œ
2, P# : m# œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

103

104

Chapter 2 Limits and Continuity

3. P" : m" œ 5# , P# : m# œ
"# 5. m œ lim

hÄ!

4. P" : m" œ 3, P# : m# œ
3

c4
(
"  h)# d
a4
(
1)# b h


a1
2h  h# b1 h hÄ!

œ lim

œ lim

hÄ!

h(#
h) h

œ 2;

at (
"ß $): y œ $  #(x
(
1)) Ê y œ 2x  5, tangent line

6. m œ lim

hÄ!

c(1  h
1)#  1d
c("
")#  1d h

h#

œ lim

hÄ! h

œ lim h œ 0; at ("ß "): y œ 1  0(x
1) Ê y œ 1, hÄ!

tangent line

È 2È 1  h
2È 1 œ lim 2 1 h h
2 h hÄ! hÄ! 4(1  h)
4 œ lim œ lim È1 2h  1 h Ä ! 2h ŠÈ1  h  1‹ hÄ!

7. m œ lim

†

2È 1  h  2 2È 1  h  #

œ 1;

at ("ß #): y œ 2  1(x
1) Ê y œ x  1, tangent line

8. m œ lim

hÄ!

"

(
1  h)#


(
"" )#

h


a
2h  h# b # h Ä ! h(
1  h)

œ lim

1
(
1  h)# # h Ä ! h(
1h) 2
h lim # œ 2; h Ä ! (
1  h)

œ lim œ

at (
"ß "): y œ 1  2(x
(
1)) Ê y œ 2x  3, tangent line

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.7 Tangents and Derivatives (
2  h)$
(
2)$ h

9. m œ lim

hÄ!


8  12h
6h#  h$  8 h

œ lim

hÄ!

œ lim a12
6h  h# b œ 12; hÄ!

at (
2ß
8): y œ
8  12(x
(
2)) Ê y œ 12x  16, tangent line

"

(
#  h)$

10. m œ lim

h

hÄ!

œ

"2 8(
8)

hÄ!

hÄ!

at ˆ
#ß
"8 ‰ : y œ
8"
Ê yœ


11. m œ lim

hÄ!

x


" #,

12
6h  h# 8(
2  h)$

œ lim

3 œ
16 ;

3 16


8
(
#  h)$
8h(
#  h)$

œ lim


a12h
6h#  h$ b
8h(
#  h)$

œ lim

hÄ!


(
#" )$

3 16 (x


(
2))

tangent line

c(2  h)#  1d
5 h

œ lim

hÄ!

a5  4h  h# b
5 h

hÄ!

at (2ß 5): y
5 œ 4(x
2), tangent line 12. m œ lim

hÄ!

c("  h)
2(1  h)# d
(
1) h

œ lim

hÄ!

h(4  h) h

œ lim

a1  h
2
4h
2h# b  1 h

hÄ!

3
h (3
h)  2


3

h

œ lim

hÄ!

(3  h)
3(h  1) h(h  1)

h Ä ! h(h  1)

at ($ß $): y
3 œ
2(x
3), tangent line 14. m œ lim

hÄ!

8 (2
h)#


2

h

hÄ!

hÄ!

(2  h)$
8 h

œ lim

hÄ!

œ lim

a8  12h  6h#  h$ b
8 h

hÄ!

at (2ß )): y
8 œ 12(t
2), tangent line 16. m œ lim

hÄ!

c(1  h)$  3(1  h)d
4 h

a1  3h  3h#  h$  3  3hb
4 h

œ lim

hÄ!

at ("ß %): y
4 œ 6(t
1), tangent line È4  h
2 h hÄ!

17. m œ lim

È4  h
2 h hÄ!

œ lim

œ "4 ; at (%ß #): y
2 œ 18. m œ lim

hÄ!

œ

" È9  3

È(8  h)  1
3 h

" 4

†

È4  h  2 È4  h  2

œ lim

hÄ!

h a12  6h  h# b h

œ lim

œ
3;

œ
2;

8
2 a4  4h  h# b h(2  h)# hÄ!

8
2(2  h)# # h Ä ! h(2  h)

œ lim

at (2ß 2): y
2 œ
2(x
2) 15. m œ lim


2h

œ lim

h(
3
2h) h

œ lim

at ("ß
"): y  1 œ
3(x
1), tangent line 13. m œ lim

œ %;

œ lim


2h(4  h) h(2  h)#

œ


8 4

œ
2;

œ 12;

œ lim

hÄ!

(4  h)
4

h Ä ! h ŠÈ4  h  #‹

h a6  3h  h# b h

œ lim

œ 6;

h

h Ä ! h ŠÈ4  h  #‹

œ

" È4  #

(x
4), tangent line

œ lim

hÄ!

È9  h
3 h

œ 6" ; at (8ß 3): y
3 œ

19. At x œ
1, y œ 5 Ê m œ lim

hÄ!

" 6

†

È9  h  3 È9  h  3

œ lim

(9  h)
9

h Ä ! h ŠÈ9  h  3‹

œ lim

h

h Ä ! h ŠÈ9  h  3‹

(x
8), tangent line

5(
"  h)#
5 h

œ lim

hÄ!

5 a1
2h  h# b
5 h

œ lim

hÄ!

5h(
2  h) h

œ
10, slope

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

105

106

Chapter 2 Limits and Continuity c1
(2  h)# d
(
3) h

20. At x œ 2, y œ
3 Ê m œ lim

hÄ!

21. At x œ 3, y œ

" #

"

(3
h) 1

Ê m œ lim


#"

22. At x œ 0, y œ
1 Ê m œ lim

hÄ!

h1 h
1

hÄ!

hÄ!


(
1) h

a1
4
4h
h# b  3 h

2
(2  h) 2h(2  h)

œ lim

h

hÄ!

œ lim

hÄ!

hÄ!


h

œ lim

h Ä ! 2h(2  h)

(h
1)  (h  ") h(h  1)

œ lim

œ lim

œ lim


h(4  h) h

œ
4, slope

œ
"4 , slope 2h

h Ä ! h(h  1)

œ 2, slope

c(x  h)#  4(x  h)
1d
ax#  4x
1b h hÄ! a2xh  h#  4hb lim œ lim (2x  h  4) œ 2x h hÄ! hÄ!

23. At a horizontal tangent the slope m œ 0 Ê 0 œ m œ lim ax#  2xh  h#  4x  4h
1b
ax#  4x
1b h hÄ!

œ lim

œ

 4;

2x  4 œ 0 Ê x œ
2. Then f(
2) œ 4
8
1 œ
5 Ê (
2ß
5) is the point on the graph where there is a horizontal tangent. 24. 0 œ m œ lim

hÄ!

c(x  h)$
3(x  h)d
ax$
3xb h

3x# h  3xh#  h$
3h h

œ lim

hÄ!

œ lim

hÄ!

ax$  3x# h  3xh#  h$
3x
3hb
ax$
3xb h

œ lim a3x#  3xh  h#
3b œ 3x#
3; 3x#
3 œ 0 Ê x œ
1 or x œ 1. Then hÄ!

f(
1) œ 2 and f(1) œ
2 Ê (
"ß 2) and ("ß
2) are the points on the graph where a horizontal tangent exists. "

(x
h)  1

25.
1 œ m œ lim


x " 1

h

hÄ!

(x
1)
(x  h
1) h(x
1)(x  h
1)

œ lim

hÄ!


h

œ lim

h Ä ! h(x
1)(x  h
1)

œ
(x
" 1)#

Ê (x
1)# œ 1 Ê x#
2x œ 0 Ê x(x
2) œ 0 Ê x œ 0 or x œ 2. If x œ 0, then y œ
1 and m œ
1 Ê y œ
1
(x
0) œ
(x  1). If x œ 2, then y œ 1 and m œ
1 Ê y œ 1
(x
2) œ
(x
3). 26.

" 4

œ m œ lim

Èx  h
Èx

œ lim

h

y œ 2  "4 (x
4) œ

hÄ!

f(2  h)
f(2) h

x 4

Èx  h
Èx h

hÄ!

h Ä ! h ŠÈx  h  Èx‹

27. lim

œ lim

h

hÄ!

œ

" #È x

. Thus,

" 4

œ

†

Èx  h  Èx Èx  h  Èx

" #Èx

(x  h)
x

œ lim

h Ä ! h ŠÈx  h  Èx‹

Ê Èx œ 2 Ê x œ 4 Ê y œ 2. The tangent line is

 1.

œ lim

hÄ!

a100
4.9(#  h)# b
a100
4.9(2)# b h


4.9 a4  4h  h# b  4.9(4) h

œ lim

hÄ!

œ lim (
19.6
4.9h) œ
19.6. The minus sign indicates the object is falling downward at a speed of hÄ!

19.6 m/sec. f(10  h)
f(10) h hÄ!

28. lim

3(10  h)#
3(10)# h hÄ!

œ lim

29. lim

f(3  h)
f(3) h

œ lim

30. lim

f(2  h)
f(2) h

œ lim

hÄ!

hÄ!

hÄ!

hÄ!

1(3  h)#
1(3)# h 41 3

œ lim

(2  h)$
431 (2)$ h

f(0  h)
f(0) h hÄ!

31. Slope at origin œ lim

3 a20h  h# b h hÄ!

œ lim

hÄ!

œ lim

1 c9  6h h#
9d h 41 3

hÄ!

h# sin ˆ "h ‰ h hÄ!

œ lim

œ 60 ft/sec. œ lim 1(6  h) œ 61

c12h  6h#  h$ d h

hÄ!

œ lim

hÄ!

41 3

c12  6h  h# d œ 161

œ lim h sin ˆ h" ‰ œ 0 Ê yes, f(x) does have a tangent at hÄ!

the origin with slope 0. 32. lim

hÄ!

g(0  h)
g(0) h

œ lim

hÄ!

h sin ˆ "h ‰ h

œ lim sin h" . Since lim sin hÄ!

hÄ!

" h

does not exist, f(x) has no tangent at

the origin.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.7 Tangents and Derivatives 33.

lim

h Ä !c

lim

hÄ!

34.

f(0  h)
f(0) h f(0  h)
f(0) h

œ lim c hÄ!


1
0 h

œ _, and lim b hÄ!

f(0  h)
f(0) h

1
0 h

œ lim b hÄ!

œ _ Ê yes, the graph of f has a vertical tangent at the origin.

œ _, and lim b U(0  h)h
U(0) œ lim b hÄ! hÄ! does not have a vertical tangent at (!ß ") because the limit does not exist. lim

h Ä !c

œ _. Therefore,

U(0  h)
U(0) h

œ lim c hÄ!

0
1 h

1
1 h

œ 0 Ê no, the graph of f

35. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

h Ä !c

f(0  h)
f(0) h

œ lim c hÄ!

h#Î&
0 h

œ lim c hÄ!

" h$Î&

œ
_ and lim b hÄ!

" h$Î&

œ _ Ê limit does not exist

Ê the graph of y œ x#Î& does not have a vertical tangent at x œ 0.

36. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

h Ä !c

f(0  h)
f(0) h

œ lim c hÄ!

h%Î&
0 h

œ lim c hÄ!

" h"Î&

œ
_ and lim b hÄ!

" h"Î&

œ _ Ê limit does not exist

Ê y œ x%Î& does not have a vertical tangent at x œ 0.

37. (a) The graph appears to have a vertical tangent at x œ !.

(b)

f(0  h)
f(0) h hÄ!

lim

h"Î&
0 h hÄ!

œ lim

œ lim

"

%Î& hÄ! h

œ _ Ê y œ x"Î& has a vertical tangent at x œ 0.

38. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim

hÄ!

f(0  h)
f(0) h

at x œ 0.

œ lim

hÄ!

h$Î&
0 h

œ lim

"

#Î& hÄ! h

œ _ Ê the graph of y œ x$Î& has a vertical tangent

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107

108

Chapter 2 Limits and Continuity

39. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

h Ä !c

f(0  h)
f(0) h

œ lim c hÄ!

4h#Î&
2h h

œ lim c hÄ!

4 h$Î&


2 œ
_ and lim b hÄ!

4 h$Î&


#œ_

Ê limit does not exist Ê the graph of y œ 4x#Î&
2x does not have a vertical tangent at x œ 0.

40. (a) The graph appears to have a cusp at x œ 0.

(b)

lim

hÄ!

f(0  h)
f(0) h

œ lim

hÄ!

h&Î$
5h#Î$ h

œ lim h#Î$
hÄ!

5 h"Î$

œ 0
lim

5

"Î$ hÄ! h

y œ x&Î$
5x#Î$ does not have a vertical tangent at x œ !.

does not exist Ê the graph of

41. (a) The graph appears to have a vertical tangent at x œ 1 and a cusp at x œ 0.

(b) x œ 1:

(1  h)#Î$
(1  h
1)"Î$
" h hÄ! #Î$ "Î$

lim

Ê yœx

x œ 0:


(x
1)

lim f(0  h)h
f(0) hÄ!

(1  h)#Î$
h"Î$
" h hÄ!

œ lim

has a vertical tangent at x œ 1;

h#Î$
(h
1)"Î$
(
1)"Î$ h hÄ! #Î$ "Î$

œ lim

does not exist Ê y œ x

œ
_


(x
1)

" œ lim ’ h"Î$


hÄ!

(h
")"Î$ h

 h" “

does not have a vertical tangent at x œ 0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 2.7 Tangents and Derivatives 42. (a) The graph appears to have vertical tangents at x œ 0 and x œ 1.

(b) x œ 0:

h"Î$  (h
1)"Î$
(
")"Î$ h hÄ!

f(0  h)
f(0) h hÄ!

œ lim

f(1  h)
f(1) h

œ lim

lim

œ _ Ê y œ x"Î$  (x
1)"Î$ has a

vertical tangent at x œ 0;

x œ 1:

lim

hÄ!

hÄ!

(1  h)"Î$  ("  h
1)"Î$
1 h

œ _ Ê y œ x"Î$  (x
1)"Î$ has a

vertical tangent at x œ ".

43. (a) The graph appears to have a vertical tangent at x œ 0.

(b)

lim

h Ä !b

f(0  h)
f(0) h

œ lim b xÄ!

Èh
0 h

œ lim


È kh k
0

f(0  h)
f(0) h

"

h Ä ! Èh

œ lim c œ lim c h hÄ! hÄ! Ê y has a vertical tangent at x œ 0. lim

h Ä !c

œ _;
È kh k
kh k

œ lim c hÄ!

" È kh k

œ_

44. (a) The graph appears to have a cusp at x œ 4.

(b)

lim

f(4  h)
f(4) h

œ lim b hÄ!

Èk4
(4  h)k
0 h

lim

f(4  h)
f(4) h

œ lim c hÄ!

Èk4
(4  h)k h

h Ä !b h Ä !c

œ lim b hÄ!

œ lim c hÄ!

È kh k h

È kh k
lhl

œ lim b hÄ!

œ lim c hÄ!

" Èh


" È kh k

œ _;

œ
_

Ê y œ È%
x does not have a vertical tangent at x œ 4. 45-48. Example CAS commands: Maple: f := x -> x^3 + 2*x;x0 := 0; plot( f(x), x=x0-1/2..x0+3, color=black, # part (a) title="Section 2.7, #45(a)" ); q := unapply( (f(x0+h)-f(x0))/h, h ); # part (b) L := limit( q(h), h=0 ); # part (c) sec_lines := seq( f(x0)+q(h)*(x-x0), h=1..3 ); # part (d) tan_line := f(x0) + L*(x-x0); plot( [f(x),tan_line,sec_lines], x=x0-1/2..x0+3, color=black, Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

109

110

Chapter 2 Limits and Continuity

linestyle=[1,2,5,6,7], title="Section 2.7, #45(d)", legend=["y=f(x)","Tangent line at x=0","Secant line (h=1)", "Secant line (h=2)","Secant line (h=3)"] ); Mathematica: (function and value for x0 may change) Clear[f, m, x, h] x0 œ p; f[x_]: œ Cos[x]  4Sin[2x] Plot[f[x], {x, x0
1, x0  3}] dq[h_]: œ (f[x0+h]
f[x0])/h m œ Limit[dq[h], h Ä 0] ytan: œ f[x0]  m(x
x0) y1: œ f[x0]  dq[1](x
x0) y2: œ f[x0]  dq[2](x
x0) y3: œ f[x0]  dq[3](x
x0) Plot[{f[x], ytan, y1, y2, y3}, {x, x0
1, x0  3}] CHAPTER 2 PRACTICE EXERCISES 1. At x œ
1: Ê

lim

x Ä
"c

f(x) œ

lim

x Ä
"b

f(x) œ 1

lim f(x) œ 1 œ f(
1)

x Ä
1

Ê f is continuous at x œ
1. At x œ 0: lim c f(x) œ lim b f(x) œ 0 Ê lim f(x) œ 0. xÄ!

xÄ!

xÄ!

But f(0) œ 1 Á lim f(x) xÄ!

Ê f is discontinuous at x œ 0. If we define fa!b œ !, then the discontinuity at x œ ! is removable. At x œ 1: lim c f(x) œ
1 and lim b f(x) œ 1 xÄ"

Ê lim f(x) does not exist

xÄ"

xÄ1

Ê f is discontinuous at x œ 1. 2. At x œ
1: Ê

lim

x Ä
"c

f(x) œ 0 and

lim

x Ä
"b

f(x) œ
1

lim f(x) does not exist

x Ä
"

Ê f is discontinuous at x œ
1. At x œ 0: lim c f(x) œ
_ and lim b f(x) œ _ xÄ!

Ê lim f(x) does not exist

xÄ!

xÄ!

Ê f is discontinuous at x œ 0. At x œ 1: lim c f(x) œ lim b f(x) œ 1 Ê lim f(x) œ 1. xÄ"

xÄ1

xÄ"

But f(1) œ 0 Á lim f(x) xÄ1

Ê f is discontinuous at x œ 1. If we define fa"b œ ", then the discontinuity at x œ " is removable. 3. (a) (b)

lim a3fatbb œ 3 lim fatb œ 3(
7) œ
21

t Ä t!

t Ä t!

#

#

lim afatbb œ Š lim fatb‹ œ a
(b# œ 49

t Ä t!

t Ä t!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 2 Practice Exercises (c) (d) (e) (f)

111

lim afatb † gatbb œ lim fatb † lim gatb œ (
7)(0) œ 0

t Ä t!

t Ä t!

lim fatb t Ä t! g(t)
7

Ät

t Ä t!

lim fatb

œ

Ät

t

t

Ät

t

!

Ät

t

!


7 0
7

œ

lim gatb
lim 7

t

!

Ät

lim fatb

œ

!

lim agatb
7b

!

œ1

lim cos agatbb œ cos Š lim gatb‹ œ cos ! œ 1

t Ä t!

t Ä t!

lim kfatbk œ ¹ lim fatb¹ œ k
7k œ 7

t Ä t!

t Ä t!

(g) lim afatb  gatbb œ lim fatb
lim gatb œ
7  0 œ
7 t Ä t!

(h)

4. (a) (b) (c) (d) (e) (f)

t Ä t!

lim Š " ‹ t Ä t! fatb

œ

" lim fatb

t

Ät

t Ä t!

"
7

œ

!

œ
71

lim
g(x) œ
lim g(x) œ
È2

xÄ!

xÄ!

lim ag(x) † f(x)b œ lim g(x) † lim f(x) œ ŠÈ2‹ ˆ "# ‰ œ

xÄ!

xÄ!

xÄ!

lim af(x)  g(x)b œ lim f(x)  lim g(x) œ

xÄ!

"

lim x Ä ! f(x)

œ

xÄ!

" lim f(x)

œ

xÄ!

" " #

xÄ!

œ2

" #

lim ax  f(x)b œ lim x  lim f(x) œ 0 

xÄ!

xÄ!

f(x)†cos x x
1 xÄ!

lim

xÄ!

lim f(x)† lim cos x

œ

xÄ!

xÄ!

lim x
lim 1

xÄ!

xÄ!

œ

ˆ "# ‰ (1) 0
1

" #

È2 #

 È2

œ

" #

œ
#"

5. Since lim x œ 0 we must have that lim (4
g(x)) œ 0. Otherwise, if lim (%
g(x)) is a finite positive xÄ!

xÄ!

xÄ!

’ 4
xg(x) “

’ 4
xg(x) “

œ
_ and lim b œ _ so the limit could not equal 1 as xÄ! x Ä 0. Similar reasoning holds if lim (4
g(x)) is a finite negative number. We conclude that lim g(x) œ 4. number, we would have lim c xÄ!

xÄ!

6. 2 œ lim

x Ä
%

xÄ!

’x lim g(x)“ œ lim x † lim xÄ!

x Ä
%

x Ä
%

’ lim g(x)“ œ
4 lim xÄ!

(since lim g(x) is a constant) Ê lim g(x) œ xÄ!

xÄ!

2
%

x Ä
%

œ
#" .

’ lim g(x)“ œ
4 lim g(x) xÄ!

xÄ!

7. (a) xlim faxb œ xlim x"Î$ œ c"Î$ œ facb for every real number c Ê f is continuous on a
_ß _b. Äc Äc (b) xlim gaxb œ xlim x$Î% œ c$Î% œ gacb for every nonnegative real number c Ê g is continuous on Ò!ß _Ñ. Äc Äc (c) xlim haxb œ xlim x
#Î$ œ Äc Äc (d) xlim kaxb œ xlim x
"Î' œ Äc Äc

" c#Î$ " c"Î'

œ hacb for every nonzero real number c Ê h is continuous on a
_ß !b and a
_ß _b. œ kacb for every positive real number c Ê k is continuous on a!ß _b

8. (a) - ˆˆn
"# ‰1ß ˆn  "# ‰1‰, where I œ the set of all integers. n−I (b) - an1ß an  1b1b, where I œ the set of all integers. n−I (c) a
_ß 1b  a1ß _b (d) a
_ß !b  a!ß _b 9.

(a)

(b) 10. (a)

(x
2)(x
2) x#
4x  4 $  5x#
14x œ lim x xÄ! x Ä ! x(x  7)(x
2) x
2 x
2 lim œ _ and lim b x(x  7) x Ä !c x(x  7)

œ lim

lim (x
2)(x
2) x Ä # x(x  7)(x
#)

œ lim

lim

#

x
4x  4

xÄ!

lim $ # x Ä # x  5x
14x x#  x

lim & % $ x Ä ! x  2x  x Now lim c xÄ!

œ

œ lim

x(x  1)

$ # x Ä ! x ax  2x  1b

1 x# (x  1)

œ _ and lim b xÄ!

x
2

, x Á 2; the limit does not exist because

x
2

, x Á 2, and lim

x Ä ! x(x  7)

œ
_

x Ä # x(x  7)

œ lim

x1

# x Ä ! x (x  1)(x  1)

1 x# (x  1)

x
2

x Ä # x(x  7)

œ lim

œ _ Ê lim

"

# x Ä 0 x (x  1) #

x x

& % $ x Ä ! x  2x  x

œ

0 2(9)

œ0

, x Á 0 and x Á
1.

œ _.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

112

Chapter 2 Limits and Continuity

(b)

x#  x

exist because

$ # x Ä
" x ax  2x  1b

"

1
Èx 1
x

œ lim

12. xlim Äa

x #
a# x %
a%

œ xlim Äa

13. lim

(x  h)#
x# h

œ lim

(x  h)#
x# h xÄ!

œ lim

hÄ!

"

15. lim

#
x
#

16. lim

(#  x)$
8 x

xÄ!

xÄ!

17.

18.

xÄ!

xÄ!

lim

xÄ1

" x  g(x)

3x#  1 g(x)

œ2 Ê

xÄ!


"

x Ä ! 4  #x

xÄ!

xÄ1

5
x#

œ0 Ê

" #

lim 4 g(x) œ 8, since 2$ œ 8. Then lim b g(x) œ 2. xÄ!

x Ä !b

Ê È5  lim

x Ä È5

g(x) œ

" #

Ê

lim

x Ä È5

g(x) œ


%x  ) $x $

x Ä
#

#! &!

ˆ"
œxÄ lim
_ $x

% $x#

"

#

#

# &

œ

" x# 24. x lim œ x lim œ Ä _ x #
(x  " Ä _ "
(x  x"#

#x  $ ##. x Ä lim œxÄ lim
_ &x#  (
_ &  ) ‰ $x$



! "
!!

$ x# ( x#

œ

#! &!

œ!
!!œ!

œ!

#

%

x
(x x
( 25. x Ä lim œxÄ lim œ
_
_ x  1
_ "  "x

$

x x x" #'. x lim œxÄ lim œ_ Ä _ "#x$  "#)
_ "#  "#) x$

lsin xl lsin xl " 27. x lim Ÿ x lim œ ! since int x Ä _ as x Ä _ Êx lim œ !. Ä _ gx h Ä _ gx h Ä _ gx h

lim

)Ä_

29. x lim Ä_

lcos )
"l )

Ÿ lim


"

l
#l

)Ä_ )

x  sin x  #Èx x  sin x

#Î$


È5

lim g(x) œ _ since lim a5
x# b œ 1

# $

28.

" #

xÄ1

x Ä
#

#x  $ x 21. x lim œ x lim œ Ä _ &x  ( Ä _ &  (x #

œ2 Ê

œ _ Ê lim g(x) œ 0 since lim a3x#  1b œ 4

lim x Ä
# Èg(x)

x 23. x Ä lim
_

œ
"4

œ lim ax#  6x  12b œ 12

(x  g(x)) œ

lim

" #a#

œ lim (2x  h) œ h

œ lim

x Ä È&

œ

" #

hÄ!

"Î$

x Ä È&

œ

œ _.

œ lim (2x  h) œ 2x

ax$  6x#  12x  8b
8 x

œ lim

" x #  a#

œ xlim Äa

ax#  2hx  h# b
x# h

2
(2  x) 2x(#  x)

œ lim

"

lim [4 g(x)]"Î$ œ 2 Ê ’ lim b 4 g(x)“ x Ä !b xÄ!

19. lim 20.

x

ax #
a # b ax #  a # b a x #
a # b

"

lim # x Ä
"b x (x  1)

x Ä 1 1  Èx

, x Á 0 and x Á
1. The limit does not

1

# x Ä
" x (x  1)

œ lim

ax#  2hx  h# b
x# h xÄ!

14. lim

"

"
Èx

x Ä 1 ˆ1
È x ‰ ˆ 1  È x ‰

hÄ!

œ lim

œ
_ and

lim # x Ä
"c x (x  1)

11. lim

xÄ1

x(x  1)

œ lim

lim & % $ x Ä
" x  2x  x

œ ! Ê lim

œ x lim Ä_

)Ä_

"  sinx x  È#x "  sinx x


&Î$

x x " x 30. x lim œ x lim #x  œ Ä _ x#Î$  cos# x Ä _Œ "  cos#Î$

lcos )
"l )

œ !.

"!! "!

œ"

œ

"! "!

œ"

x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

# &

Chapter 2 Practice Exercises 31. At x œ
1: œ

lim

x Ä
"c

lim

x Ä
"c

lim

x Ä
"b

f(x) œ

x ax #
1 b x#
1

f(x) œ

œ

lim

x ax #
1 b kx #
1 k

lim

x Ä
"c

lim

x Ä
"c

x Ä
"b

x œ
1, and

x ax #
1 b kx #
1 k

œ

x ax #
1 b

lim # x Ä
"b
ax
"b

œ lim (
x) œ
(
1) œ 1. Since x Ä
1

lim f(x) Á lim b f(x) x Ä
"c x Ä
" Ê

lim f(x) does not exist, the function f cannot be

x Ä
1

extended to a continuous function at x œ
1. At x œ 1:

lim f(x) œ lim c xÄ"

x Ä "c

#

x ax #
1 b kx #
1 k

œ lim c xÄ" #

x ax #
1 b
ax #
1 b

œ lim c (
x) œ
1, and xÄ"

lim f(x) œ lim b xkaxx#

11k b œ lim b x axx#

"1b œ lim b x œ 1. Again lim f(x) does not exist so f xÄ1 xÄ" xÄ" xÄ1 cannot be extended to a continuous function at x œ 1 either.

x Ä "b

32. The discontinuity at x œ 0 of f(x) œ sin ˆ "x ‰ is nonremovable because lim sin xÄ!

" x

does not exist.

33. Yes, f does have a continuous extension to a œ 1: " define f(1) œ lim xx

È œ 43 . % x xÄ1

34. Yes, g does have a continuous extension to a œ 1# : ) 5 g ˆ 1# ‰ œ lim1 45)cos
#1 œ
4 . )Ä #

35. From the graph we see that lim
h(t) Á lim h(t) tÄ! tÄ! so h cannot be extended to a continuous function at a œ 0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

113

114

Chapter 2 Limits and Continuity

36. From the graph we see that lim c k(x) Á lim b k(x) xÄ! xÄ! so k cannot be extended to a continuous function at a œ 0.

37. (a) f(
1) œ
1 and f(2) œ 5 Ê f has a root between
1 and 2 by the Intermediate Value Theorem. (b), (c) root is 1.32471795724 38. (a) f(
2) œ
2 and f(0) œ 2 Ê f has a root between
2 and 0 by the Intermediate Value Theorem. (b), (c) root is
1.76929235424 CHAPTER 2 ADDITIONAL AND ADVANCED EXERCISES 1. (a)

x xx

0.1 0.7943

0.01 0.9550

0.001 0.9931

10

100

1000

0.3679

0.3679

0.3679

0.0001 0.9991

0.00001 0.9999

Apparently, lim b xx œ 1 xÄ! (b)

2. (a)

x ˆ "x ‰"ÎÐln xÑ Apparently,

"ÎÐln xÑ lim ˆ " ‰ xÄ_ x

œ 0.3678 œ

" e

(b)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 2 Additional and Advanced Exercises 115 3.

lim L œ lim c L! É"
vc# œ L! É1
vÄcc
# œ L! É1
cc# œ 0 vÄc The left-hand limit was needed because the function L is undefined if v  c (the rocket cannot move faster than the speed of light). lim v#

#

v Ä cc

#

4. ¹

Èx #


1¹  0.2 Ê
0.2 

Èx #


1  0.2 Ê 0.8 

Èx #

 1.2 Ê 1.6  Èx  2.4 Ê 2.56  x  5.76.

¹

Èx #


1¹  0.1 Ê
0.1 

Èx #


1  0.1 Ê 0.9 

Èx #

 1.1 Ê 1.8  Èx  2.2 Ê 3.24  x  4.84.

5. k10  (t
70) ‚ 10
%
10k  0.0005 Ê k(t
70) ‚ 10
% k  0.0005 Ê
0.0005  (t
70) ‚ 10
%  0.0005 Ê
5  t
70  5 Ê 65°  t  75° Ê Within 5° F. 6. We want to know in what interval to hold values of h to make V satisfy the inequality lV
"!!!l œ l$'1h
"!!!l Ÿ "!. To find out, we solve the inequality: **! l$'1h
"!!!l Ÿ "! Ê
"! Ÿ $'1h
"!!! Ÿ "! Ê **! Ÿ $'1h Ÿ "!"! Ê $' 1 Ÿ hŸ

"!"! $'1

Ê )Þ) Ÿ h Ÿ )Þ*. where 8.8 was rounded up, to be safe, and 8.9 was rounded down, to be safe. The interval in which we should hold h is about )Þ*
)Þ) œ !Þ" cm wide (1 mm). With stripes 1 mm wide, we can expect to measure a liter of water with an accuracy of 1%, which is more than enough accuracy for cooking.

7. Show lim f(x) œ lim ax#
7b œ
' œ f(1). xÄ1

xÄ1

Step 1: kax#
7b  6k  % Ê
%  x#
1  % Ê 1
%  x#  1  % Ê È1
%  x  È1  %. Step 2: kx
1k  $ Ê
$  x
1  $ Ê
$  "  x  $  ". Then
$  " œ È1
% or $  " œ È1  %. Choose $ œ min š1
È1
%ß È1  %
1› , then 0  kx
1k  $ Ê kax#
(b
6k  % and lim f(x) œ
6. By the continuity test, f(x) is continuous at x œ 1. xÄ1

8. Show lim" g(x) œ lim" xÄ

xÄ

%

" 2x

œ 2 œ g ˆ "4 ‰ .

%

Step 1: ¸ #"x
2¸  % Ê
%  #"x
#  % Ê #
%  #"x  #  % Ê Step 2: ¸B
"4 ¸  $ Ê
$  x
4"  $ Ê
$  4"  x  $  4" . Then
$  Choose $ œ

" 4

œ

" 4  #%

% 4(#%)

Ê $œ

" 4




" 4  #%

œ

% 4(2  %)

, or $ 

" 4

œ

, the smaller of the two values. Then 0  ¸x

By the continuity test, g(x) is continuous at x œ

" 4

" 4
#% Ê
4" ¸  $

" 4
#%

x

" 4
#% ¸ #"x


" 4#%

.

" 4

% 4(2
%)

$œ




œ

Ê

2¸  % and lim"

.

xÄ

%

" #x

œ 2.

.

9. Show lim h(x) œ lim È2x
3 œ " œ h(2). xÄ#

xÄ#

Step 1: ¹È2x
3
1¹  % Ê
%  È2x
3
"  % Ê "
%  È2x
3  "  % Ê

(1
%)#  $ #

x

("  %)#  3 . #

Step 2: kx
2k  $ Ê
$  x
2  $ or
$  #  x  $  #. ("
% )#  $ Ê $œ # ("  % Ñ #  $ ("  %Ñ#
"
#œ # #

Then
$  # œ

#


Ê $œ

œ%

# ("
%)#  $ œ "
(1#
%) # #  %# . Choose $ œ %

œ%



%# #,

%# #

, or $  # œ

("  %)#  $ #

the smaller of the two values . Then,

!  kx
2k  $ Ê ¹È2x
3
"¹  %, so lim È2x
3 œ 1. By the continuity test, h(x) is continuous at x œ 2. xÄ#

10. Show lim F(x) œ lim È9
x œ # œ F(5). xÄ&

xÄ&

Step 1: ¹È9
x
2¹  % Ê
%  È9
x
#  % Ê 9
(2
%)#  x  *
(#  %)# . Step 2: 0  kx
5k  $ Ê
$  x
&  $ Ê
$  &  x  $  &. Then
$  & œ *
(#  %)# Ê $ œ (#  %)#
% œ %#  #%, or $  & œ *
(#
%)# Ê $ œ %
(#
%)# œ %#
#%.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

116

Chapter 2 Limits and Continuity

Choose $ œ %#
#%, the smaller of the two values. Then, !  kx
5k  $ Ê ¹È9
x
#¹  %, so lim È9
x œ #. By the continuity test, F(x) is continuous at x œ 5.

xÄ&

11. Suppose L" and L# are two different limits. Without loss of generality assume L#  L" . Let % œ

" 3

(L#
L" ).

Since x lim f(x) œ L" there is a $"  0 such that 0  kx
x! k  $" Ê kf(x)
L" k  % Ê
%  f(x)
L"  % Äx !

Ê
"3 (L#
L" )  L"  f(x) 

" 3

(L#
L" )  L" Ê 4L"
L#  3f(x)  2L"  L# . Likewise, x lim f(x) œ L# Ä x! so there is a $# such that 0  kx
x! k  $# Ê kf(x)
L# k  % Ê
%  f(x)
L#  % Ê
"3 (L#
L" )  L#  f(x)  3" (L#
L" )  L# Ê 2L#  L"  3f(x)  4L#
L" Ê L"
4L# 
3f(x) 
2L#
L" . If $ œ min e$" ß $# f both inequalities must hold for 0  kx
x! k  $ : 4L"
L#  3f(x)  2L"  L# Ê 5(L"
L# )  0  L"
L# . That is, L"
L#  0 and L"
L#  0, L"
%L# 
3f(x) 
2L#
L"  a contradiction. 12. Suppose xlim f(x) œ L. If k œ !, then xlim kf(x) œ xlim 0 œ ! œ ! † xlim f(x) and we are done. Äc Äc Äc Äc % If k Á 0, then given any %  !, there is a $  ! so that !  lx
cl  $ Ê lfaxb
Ll  l5l Ê lkllfaxb
Ll  % Ê lkafaxb
Lb|  % Ê lakfaxbb
akLbl  %. Thus, xlim kf(x) œ kL œ kŠxlim f(x)‹. Äc Äc 13. (a) Since x Ä 0 , 0  x$  x  1 Ê ax$
xb Ä 0
Ê

lim f ax$
xb œ lim c f(y) œ B where y œ x$
x. yÄ!

x Ä !b

(b) Since x Ä 0
,
1  x  x$  0 Ê ax$
xb Ä 0 Ê

(c) Since x Ä 0 , 0  x%  x#  1 Ê ax#
x% b Ä 0 Ê

lim f ax$
xb œ lim b f(y) œ A where y œ x$
x. yÄ!

x Ä !c

lim f ax#
x% b œ lim b f(y) œ A where y œ x#
x% . yÄ!

x Ä !b

(d) Since x Ä 0
,
1  x  0 Ê !  x%  x#  1 Ê ax#
x% b Ä 0 Ê

lim f ax#
x% b œ A as in part (c).

x Ä !b

14. (a) True, because if xlim (f(x)  g(x)) exists then xlim (f(x)  g(x))
xlim f(x) œ xlim [(f(x)  g(x))
f(x)] Äa Äa Äa Äa œ xlim g(x) exists, contrary to assumption. Äa

(b) False; for example take f(x) œ

" x

and g(x) œ
x" . Then neither lim f(x) nor lim g(x) exists, but

lim (f(x)  g(x)) œ lim ˆ "x
x" ‰ œ lim 0 œ 0 exists.

xÄ!

xÄ!

xÄ!

xÄ!

xÄ!

(c) True, because g(x) œ kxk is continuous Ê g(f(x)) œ kf(x)k is continuous (it is the composite of continuous functions).
1, x Ÿ 0 Ê f(x) is discontinuous at x œ 0. However kf(x)k œ 1 is (d) False; for example let f(x) œ œ 1, x  0 continuous at x œ 0. 15. Show lim f(x) œ lim x Ä
1

x#
"

x Ä
1 x  1

œ lim

x Ä
1

(x  1)(x
") (x  1)

Define the continuous extension of f(x) as F(x) œ œ

œ
#, x Á
1.

x#
1 x1 ,


2

x Á
" . We now prove the limit of f(x) as x Ä
1 , x œ
1

exists and has the correct value. #

Step 1: ¹ xx 
1"
(
#)¹  % Ê
% 

(x  1)(x
") (x  1)

 #  % Ê
%  (x
1)  #  %, x Á
" Ê
%
"  x  %
".

Step 2: kx
(
1)k  $ Ê
$  x  1  $ Ê
$
"  x  $
". Then
$
" œ
%
" Ê $ œ %, or $
" œ %
" Ê $ œ %. Choose $ œ %. Then !  kx
(
1)k  $ #

Ê ¹ xx 
1"
a
#b¹  % Ê

lim F(x) œ
2. Since the conditions of the continuity test are met by F(x), then f(x) has a

x Ä
1

continuous extension to F(x) at x œ
1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 2 Additional and Advanced Exercises 117 16. Show lim g(x) œ lim xÄ$

xÄ$

x#
2x
3 2x
6

œ lim

xÄ$

(x
3)(x  ") 2(x
3)

œ #, x Á 3. #

x
2x
3 2x
6 ,

Define the continuous extension of g(x) as G(x) œ œ

xÁ3 . We now prove the limit of g(x) as , xœ3

2

x Ä 3 exists and has the correct value. Step 1: ¹ x

#


2x
3 #x
6


2¹  % Ê
% 

(x
3)(x  ") 2(x
3)


#  % Ê
% 

x" #


#  % , x Á $ Ê $
#%  x  $  #% .

Step 2: kx
3k  $ Ê
$  x
3  $ Ê $
$  x  $  $. Then, $
$ œ $
#% Ê $ œ #%, or $  $ œ $  #% Ê $ œ #%. Choose $ œ #%. Then !  kx
3k  $ Ê ¹x

#


2x
3 2x
6


2¹  % Ê lim

xÄ$

(x
3)(x  ") #(x
3)

œ 2. Since the conditions of the continuity test hold for G(x),

g(x) can be continuously extended to G(x) at B œ 3. 17. (a) Let %  ! be given. If x is rational, then f(x) œ x Ê kf(x)
0k œ kx
0k  % Í kx
0k  %; i.e., choose $ œ %. Then kx
0k  $ Ê kf(x)
0k  % for x rational. If x is irrational, then f(x) œ 0 Ê kf(x)
0k  % Í !  % which is true no matter how close irrational x is to 0, so again we can choose $ œ %. In either case, given %  ! there is a $ œ %  ! such that !  kx
0k  $ Ê kf(x)
0k  %. Therefore, f is continuous at x œ 0. (b) Choose x œ c  !. Then within any interval (c
$ ß c  $ ) there are both rational and irrational numbers. If c is rational, pick % œ #c . No matter how small we choose $  ! there is an irrational number x in (c
$ ß c  $ ) Ê kf(x)
f(c)k œ k0
ck œ c 

c #

œ %. That is, f is not continuous at any rational c  0. On

the other hand, suppose c is irrational Ê f(c) œ 0. Again pick % œ #c . No matter how small we choose $  ! there is a rational number x in (c
$ ß c  $ ) with kx
ck  œ kxk 

c #

œ% Í

œ % Ê f is not continuous at any irrational c  0.

If x œ c  0, repeat the argument picking % œ nonzero value x œ c. 18. (a) Let c œ

c #

kc k #

œ


c # .

x

c #

Then kf(x)
f(c)k œ kx
0k

3c #.

Therefore f fails to be continuous at any

m n

be a rational number in [0ß 1] reduced to lowest terms Ê f(c) œ "n . Pick % œ

" #n

œ %. Therefore f is discontinuous at x œ c, a rational number.

" #n .

No matter how small $  ! is taken, there is an irrational number x in the interval (c
$ ß c  $ ) Ê kf(x)
f(c)k œ ¸0
"n ¸ œ

" n



(b) Now suppose c is an irrational number Ê f(c) œ 0. Let %  0 be given. Notice that number reduced to lowest terms with denominator 2 and belonging to [0ß 1]; denominator 3 belonging to [0ß 1];

" 4

and

[0ß 1]; etc. In general, choose N so that

" N

3 4

with denominator 4 in [0ß 1];

" 3

and

" 2 3 5, 5, 5

2 3

and

" #

is the only rational

the only rationals with 4 5

with denominator 5 in

 % Ê there exist only finitely many rationals in [!ß "] having

denominator Ÿ N, say r" , r# , á , rp . Let $ œ min ekc
ri k : i œ 1ß á ß pf . Then the interval (c
$ ß c  $ ) contains no rational numbers with denominator Ÿ N. Thus, 0  kx
ck  $ Ê kf(x)
f(c)k œ kf(x)
0k œ kf(x)k Ÿ N"  % Ê f is continuous at x œ c irrational.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

118

Chapter 2 Limits and Continuity

(c) The graph looks like the markings on a typical ruler when the points (xß f(x)) on the graph of f(x) are connected to the x-axis with vertical lines.

19. Yes. Let R be the radius of the equator (earth) and suppose at a fixed instant of time we label noon as the zero point, 0, on the equator Ê 0  1R represents the midnight point (at the same exact time). Suppose x" is a point on the equator “just after" noon Ê x"  1R is simultaneously “just after" midnight. It seems reasonable that the temperature T at a point just after noon is hotter than it would be at the diametrically opposite point just after midnight: That is, T(x" )
T(x"  1R)  0. At exactly the same moment in time pick x# to be a point just before midnight Ê x#  1R is just before noon. Then T(x# )
T(x#  1R)  0. Assuming the temperature function T is continuous along the equator (which is reasonable), the Intermediate Value Theorem says there is a point c between 0 (noon) and 1R (simultaneously midnight) such that T(c)
T(c  1R) œ 0; i.e., there is always a pair of antipodal points on the earth's equator where the temperatures are the same. #

#

# # " 20. xlim f(x)g(x) œ xlim af(x)  g(x)b‹
Šxlim af(x)
g(x)b‹ “ ’af(x)  g(x)b
af(x)
g(x)b “ œ "% ’Šxlim Äc Äc % Äc Äc œ "% ˆ$#
a
"b# ‰ œ #.

21. (a) At x œ 0: lim r (a) œ lim aÄ!

œ lim

1
("  a)

aÄ!

a Ä ! a ˆ
"
È1  a‰

At x œ
1: (b) At x œ 0:

lim

a Ä
"b

œ

r (a) œ


"  È1  a a
1
"
È1  0

œ lim c aÄ!

1
("  a) a ˆ
"  È1  a‰

œ

" #

aÄ!

1
(1  a)

lim

a Ä
"b a ˆ
1
È1  a‰

lim r
(a) œ lim c aÄ!

a Ä !c

È1  a

œ lim Š
"  a


"
È1  a a

œ lim c aÄ!


"
È1  a

‹ Š
"
È1  a ‹
a

œ lim

a Ä
1 a ˆ
"
È1  a‰ È1  a

œ lim c Š
"
a aÄ!


a a ˆ
1  È 1  a ‰

œ lim c aÄ!

œ


"
"
È0

œ1


"  È1  a

‹ Š
"  È1  a ‹


" œ _ (because the
"  È1  a
" œ
_ (because the
"  È1  a

denominator is always negative); lim b r
(a) œ lim b aÄ! aÄ! is always positive). Therefore, lim r
(a) does not exist. aÄ!

At x œ
1:

lim r
(a) œ lim b a Ä
"b a Ä
"


1
È 1  a a

œ

lim


"

a Ä
1b
"  È1  a

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

denominator

Chapter 2 Additional and Advanced Exercises 119 (c)

(d)

22. f(x) œ x  2 cos x Ê f(0) œ 0  2 cos 0 œ 2  0 and f(
1) œ
1  2 cos (
1) œ
1
#  0. Since f(x) is continuous on [
1ß !], by the Intermediate Value Theorem, f(x) must take on every value between [
1
#ß #]. Thus there is some number c in [
1ß !] such that f(c) œ 0; i.e., c is a solution to x  2 cos x œ 0. 23. (a) The function f is bounded on D if f(x)   M and f(x) Ÿ N for all x in D. This means M Ÿ f(x) Ÿ N for all x in D. Choose B to be max ekMk ß kNkf . Then kf(x)k Ÿ B. On the other hand, if kf(x)k Ÿ B, then
B Ÿ f(x) Ÿ B Ê f(x)  
B and f(x) Ÿ B Ê f(x) is bounded on D with N œ B an upper bound and M œ
B a lower bound. (b) Assume f(x) Ÿ N for all x and that L  N. Let % œ L
# N . Since x lim f(x) œ L there is a $  ! such that Äx !

0  kx
x! k  $ Ê kf(x)
Lk  % Í L
%  f(x)  L  % Í L
Í

LN #

 f(x) 

3L
N # .

But L  N Ê

LN #

Ê L


 f(x)  L 

M
L #

Í

3L
M #

 f(x)

M
L # . As in part (b), 0  kx
L  M  M, a contradiction. #

24. (a) If a   b, then a
b   0 Ê ka
bk œ a
b Ê max (aß b) œ

ab #



ka
b k #

If a Ÿ b, then a
b Ÿ 0 Ê ka
bk œ
(a
b) œ b
a Ê max (aß b) œ œ

2b #

 f(x)  L 

x! k  $

ab a
b 2a #  # œ # œ a. ka
b k ab œ a # b  b
# a #  #

œ

œ b.

(b) Let min (aß b) œ

ab #




ka
b k #

L
N #

 N Ê N  f(x) contrary to the boundedness assumption

f(x) Ÿ N. This contradiction proves L Ÿ N. (c) Assume M Ÿ f(x) for all x and that L  M. Let % œ M
L #

L
N #

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

120

Chapter 2 Limits and Continuity

25. lim œ xÄ0

sina"
cos xb x

œ lim

œ lim

xÄ0

†

sin x

xÄ0 x

26.

lim

sin x

x Ä 0b sin Èx

œ

sina"
cos xb "
cos x

sin x "  cos x

sin x

lim x Ä 0b B

†

†

"
cos x x

Èx sin Èx

†

†

œ lim

x Èx

œ lim

sinasin xb sin x

28. lim

sinax#  xb x

œ lim

sinax#  xb x#  x

† ax  "b œ lim

sinax#
%b x Ä 2 x
2

œ lim

sinax#
%b # x Ä 2 x
%

† ax  2b œ lim

xÄ0

29. lim

xÄ0

sinˆÈx
$‰ x
9 xÄ9

30. lim

sin x x

xÄ0

sinasin xb sin x

†

sina"
cos xb "
cos x

† lim

"
cos# x

x Ä 0 xa"  cos xb

" Èx  $

† lim

sin x

xÄ0 x

œ " † lim

œ " † " œ ".

sinax#  xb x#  x

† lim ax  "b œ " † " œ "

sinax#
%b # x Ä 2 x
%

† lim ax  2b œ " † % œ %

xÄ0

sinˆÈx
$‰ x Ä 9 Èx
$

œ lim

xÄ0

œ " † lim b sin"Èx † lim b Èx œ " † ! † ! œ !. x Ä 0 Š Èx ‹ x Ä 0

sinasin xb x

xÄ0

œ lim

œ " † ˆ #! ‰ œ !.

27. lim

xÄ0

"  cos x "  cos x

†

xÄ0

xÄ2

sinˆÈx
$‰ x Ä 9 Èx
$

œ lim

† lim

"

x Ä 9 Èx  $

œ"†

" '

œ

" '

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

sin# x

x Ä 0 xa"  cos xb

CHAPTER 3 DIFFERENTIATION 3.1 THE DERIVATIVE OF A FUNCTION 1. Step 1: f(x) œ 4
x# and f(x  h) œ 4
(x  h)# f(x
h)  f(x) h

Step 2:

œ

c4  (x
h)# d  a4  x# b h

œ

a4  x#  2xh  h# b  4
x# h

œ

2xh  h# h

œ

h(2x  h) h

œ
2x
h Step 3: f w (x) œ lim (
2x
h) œ
2x; f w (
$) œ 6, f w (0) œ 0, f w (1) œ
2 hÄ!

c(x
h  1)#
1d  c(x  1)#
1d h hÄ! 2xh
h#  2h lim œ lim (2x  h
2) h hÄ! hÄ!

2. F(x) œ (x
1)#  1 and F(x  h) œ (x  h
1)#  " Ê Fw (x) œ lim ax#
2xh
h#  2x  2h
1
1b  ax#  2x
1
1b h

œ lim

hÄ!

œ

œ 2(x
1); Fw (
1) œ
4, Fw (0) œ
2, Fw (2) œ 2 3. Step 1: g(t) œ

" t#

and g(t  h) œ "

"

#
# g(t
h)  g(t) œ (t
h)h t h 2t  h) 2t  h œ h( (t
h)# t# h œ (t
h)# t#

Step 2:

2t  h

Step 3: gw (t) œ lim

1 z #z

4. k(z) œ

and k(z  h) œ

Œ

œ

œ

# # h Ä ! (t
h) t

" (t
h)#

h

2t t# †t#

1  (z
h) 2(z
h)

œ

 ("  z)(z
h) lim (1  z  h)z #(z
h)zh hÄ!

œ

" 2z#

œ

t#  (t
h)# (t
h)# †t#


œ

2 t$

t#  at#
2th
h# b (t
h)# †t# †h

œ

œ

2th  h# (t
h)# t# h

2 ; gw (
1) œ 2, gw (2) œ
"4 , gw ŠÈ3‹ œ
3È 3

Ê kw (z) œ lim

Š

"
(z  h)
"
z ‹ #(z  h) #z h

hÄ!

#  z  h
z#
zh lim z  z  zh 2(z
h)zh hÄ!

h

œ lim

h Ä ! 2(z
h)zh

œ lim

"

h Ä ! #(z
h)z

; kw (
") œ
"# , kw (1) œ
"# , kw ŠÈ2‹ œ
"4

5. Step 1: p()) œ È3) and p()  h) œ È3()  h) Step 2:

p()
h)  p()) h

œ

œ

È3()
h)  È3) h

3h h ŠÈ3)
3h
È3)‹

Step 3: pw ()) œ lim

œ

ŠÈ3)
3h  È3)‹

œ

3 È3)
3h
È3)

3

œ

h Ä ! È3)
3h
È3)

†

h

3 È 3)
È 3)

œ

3 2È 3 )

ŠÈ3)
3h
È3)‹ ŠÈ3)
3h
È3)‹

; pw (1) œ

6. r(s) œ È2s  1 and r(s  h) œ È2(s  h)  1 Ê rw (s) œ lim

hÄ!

œ lim

hÄ!

œ lim

ŠÈ2s
h
1  È2s
1‹ h

†

2h

h Ä ! h ŠÈ2s
2h
1
È2s
1‹

œ

" È2s
1

; rw (0) œ 1, rw (1) œ

ŠÈ2s
2h
1
È2s
1‹ ŠÈ2s
2h
1
È2s
1‹

œ lim

" È3

6x# h
6xh#
2h$ h hÄ!

3 #È2

È2s
2h
1  È2s
1 h

œ

2 È2s
1
È2s
1

œ

2 2È2s
1

" È2

dy dx

h a6x#
6xh
2h# b h hÄ!

œ lim

, pw (3) œ "# , pw ˆ 32 ‰ œ

h Ä ! h ŠÈ2s
2h
1
È2s
1‹

2

7. y œ f(x) œ 2x$ and f(x  h) œ 2(x  h)$ Ê œ lim

(3)
3h)  3) h ŠÈ3)
3h
È3)‹

(2s
2h
1)  (2s
1)

œ lim

h Ä ! È2s
2h
1
È2s
1

, rw ˆ #" ‰ œ

3 2È 3

œ

2(x
h)$  2x$ h hÄ!

œ lim

2 ax$
3x# h
3xh#
h$ b  2x$ h hÄ!

œ lim

œ lim a6x#  6xh  2h# b œ 6x# hÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

122

Chapter 3 Differentiation

8. r œ œ

s$ #

1 Ê

”

œ lim

dr ds

(s
h)$

#

t 2t
1

Š

œ lim

(t b h)(2t b 1) c t(2t b 2h b 1) ‹ (2t b 2h b 1)(2t b 1)

h

œ

2t#
t
2ht
h  2t#  2ht  t (2t
2h
1)(2t
1)h hÄ! " " (2t
1)(2t
1) œ (2t
1)#

dv dt

œ lim

œ

"

t
h“ 

ˆt  "t ‰

# # t
h lim ht h(t

h h)t hÄ!

" Èq
1

11. p œ f(q) œ

œ

œ lim

h

"
" t

h

t
h

œ

t#
1 t#

" È(q
h)
1

Ê

h Ä ! (2t
2h
1)(2t
1)

Š

h(t
h)t  t
(t
h) ‹ (t
h)t

h

hÄ!

œ1

" t#

Š È(q
"h)
1 ‹  Š Èq"
1 ‹

œ lim

dp dq

"

œ lim

œ lim

h

hÄ!

h

hÄ!

Èq
1  Èq
h
1

œ lim

h Ä ! hÈ q
h
1 È q
1

h

hÄ!

3 # # s

h

hÄ!

h Ä ! (2t
2h
1)(2t
1)h

and f(q  h) œ

 3sh  h# b œ

t ‰ Š 2(t bt bh)hb 1 ‹  ˆ 2t b 1

œ lim

ds dt

œ lim

#
1 lim t (t

hth)t hÄ!

Èq b 1 c Èq b h b 1 Œ Èq b h b 1 Èq b 1


œ

(t
h)(2t
1)  t(2t
2h
1) (2t
2h
1)(2t
1)h

hÄ!

œ lim

h

hÄ!

Ê

œ lim

œ lim

’(t
h) 

œ

t
h 2(t
h)
1

and r(t  h) œ

hÄ!

c(s
h)$
2d  cs$
2d " h # hlim Ä! h c3s#
3sh
h# d " " œ # lim a3s# # hlim h Ä! hÄ!

h

hÄ!

" s$
3s# h
3sh#
h$
2  s$  2 # hlim h Ä!

9. s œ r(t) œ

10.

$


1•  ’ s#
1“

œ

ˆÈ q
1  È q
h
1 ‰ ˆ È q
1
È q
h
1 ‰ 1)  (q
h
1) † ˆÈq
1
Èq
h
1‰ œ lim hÈq
h
1(qÈ
q
1 ˆÈ q
1
È q
h
1 ‰ h Ä ! h Èq
h
1 Èq
1 hÄ! h " lim œ lim Èq
h
1 Èq
1 ˆÈq
1
Èq
h
1‰ h Ä ! h È q
h
1 È q
1 ˆÈ q
1
È q
h
1 ‰ hÄ! " " œ È q
1 È q
1 ˆÈ q
1
È q
1 ‰ 2(q
1) Èq
1

dz dw

œ lim

œ lim œ

12.

"
Š È3(w
h)  2 h

hÄ!

"

È3w  2 ‹

ŠÈ3w  2  È3w
3h  2‹

œ lim

hÈ3w
3h  2 È3w  2

hÄ!

È3w  2  È3w
3h  2

œ lim

h Ä ! hÈ3w
3h  2 È3w  2

†

ŠÈ3w2
È3w
3h2‹ ŠÈ3w  2
È3w
3h  2‹

œ lim

(3w  2)  (3w
3h  2)

œ lim

3

h Ä ! hÈ3w
3h  2 È3w  2 ŠÈ3w  2
È3w
3h  2‹ h Ä ! È3w
3h  2 È3w  2 ŠÈ3w  2
È3w
3h  2‹

œ

9 x

and f(x  h) œ (x  h) 

œ

x(x
h)#
9x  x# (x
h)  9(x
h) x(x
h)h

œ

h(x#
xh  9) x(x
h)h

14. k(x) œ

" #
x

œ lim

hÄ!

œ lim

(#
x)  (2
x
h) h(2
x)(2
x
h)

hÄ!

œ lim

hÄ!

œ

x#
xh  9 x(x
h)

œ

w

œ

9 (x
h)

Ê

f(x
h)  f(x) h

œ

’(x
h)


x$
2x# h
xh#
9x  x$  x# h  9x  9h x(x
h)h

; f (x) œ

and k(x  h) œ

" kw (2) œ
16

ds dt

3 È3w  2 È3w  2 ŠÈ3w  2
È3w  2‹

3 2(3w  2) È3w  2

13. f(x) œ x 

15.

œ

# lim x
xh  9 h Ä ! x(x
h)

œ

x#  9 x#

9 9 (x b h) “  ’x
x “

h

œ

œ1


x# h
xh#  9h x(x
h)h 9 x#

; m œ f w (
3) œ 0

Š #
"x
h
k(x
h)  k(x) œ lim h h hÄ! hÄ! h " " lim œ lim (2
x)(#
x
h) œ (2
x)# ; h Ä ! h(2
x)(2
x
h) hÄ! " 2
(x
h)

c(t
h)$  (t
h)# d  at$  t# b h

3t# h
3th#
h$  2th  h# h

Ê kw (x) œ lim

œ lim

hÄ!

œ lim

hÄ!

" #
x‹

at$
3t# h
3th#
h$ b  at#
2th
h# b  t$
t# h

h a3t#
3th
h#  2t  hb h

œ lim a3t#  3th  h#
2t
hb hÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.1 The Derivative of a Function œ 3t#
2t; m œ 16.

dy dx

ds ¸ dt tœ"

œ5

(x
h
1)$  (x
1)$ h

œ lim

hÄ!

(x
1)$
3(x
")# h
3(x
1)h#
h$ (x
1)$ h

œ lim

hÄ!

œ lim c3(x  1)#  3(x  1)h  h# d œ 3(x  1)# ; m œ hÄ!

17. f(x) œ œ

8 Èx  2

and f(x  h) œ

8 ŠÈx  2  Èx
h  2‹ hÈ x
h  2 È x  2

†

8 È(x
h)  2

f(x
h)  f(x) h

Ê

ŠÈx  2
Èx
h  2‹

œ

ŠÈx  2
Èx
h  2‹

œ

8h hÈx
h  2 Èx  2 ŠÈx  2
Èx
h  2‹

œ

8 Èx  2 Èx  2 ŠÈx  2
Èx  2‹

œ

œ3

dy dx ¹ x=
#

È(x b h) c 2
Èx c 2 8

œ

8

h

8[(x  2)  (x
h  2)] hÈx
h  2 Èx  2 ŠÈx  2
Èx
h  2‹ 8

Ê f w (x) œ lim

h Ä ! Èx
h  2 Èx  2 ŠÈx  2
Èx
h  2‹

4 (x  2)Èx  2

; m œ f w (6) œ

4 4È 4

œ
"# Ê the equation of the tangent

line at (6ß 4) is y
4 œ
"# (x
6) Ê y œ
"# x  $  % Ê y œ
"# x  (. 18. gw (z) œ lim

ˆ1
È4  (z
h)‰  Š1
È4  z‹ h

hÄ!

œ

h lim h Ä ! h ŠÈ4  z  h
È4  z‹

œ

†

h

hÄ!

(4  z  h)  (4  z) lim h Ä ! h ŠÈ4  z  h
È4  z‹ " œ
"# 2È 4  3
"# z  $#  # Ê w

ŠÈ4  z  h  È4  z‹

œ lim

ŠÈ4  z  h
È4  z‹ ŠÈ4  z  h
È4  z‹ "

œ lim

h Ä ! ŠÈ4  z  h
È4  z‹

œ

" 2È 4  z

m œ gw (3) œ

Ê the equation of the tangent line at ($ß #) is w
2 œ
"# (z
3)

Êwœ

œ
"# z  (# .

19. s œ f(t) œ 1
3t# and f(t  h) œ 1
3(t  h)# œ 1
3t#
6th
3h# Ê a1  3t#  6th  3h# b  a1  3t# b h hÄ!

œ lim

20. y œ f(x) œ "
œ lim

hÄ!

h h Ä ! x(x
h)h

hÄ!

2È 4  )  2È 4  )  h hÈ 4  ) È 4  )  h

œ

œ

" x
h

Ê

" lim h Ä ! x(x
h)

2 È4  ()
h)

4(%  ))  4(%  )  h)

œ lim

h Ä ! 2hÈ4  ) È4  )  h ŠÈ4  )
È4  )  h‹

œ

2 (4  )) Š2È4  )‹

œ

dy dx

œ

œ lim

" x#

" 3

Ê

" (4  ))È4  )

Ê

Šz
h
Èz
h‹  ˆz
Èz‰

hÄ!

h

œ 1  lim

(z
h)  z

h Ä ! h ŠÈz
h
Èz‹

hÄ!

Š1 

dr ¸ d) )œ!

Ê

dr d)

È œ

È4 c ) c h  È4 c ) 2

2

h

2

h Ä ! È4  ) È4  )  h ŠÈ4  )
È%  )  h‹

œ

" 8

h
Èz
h  Èz h hÄ!

œ lim

œ 1  lim

"

h

œ lim

œ lim

"

x
h ‹  Š1  x ‹

hÄ!

dy dx ¹x= 3

f(t
h)  f(t) h

œ6

f(x
h)  f(x) h hÄ!

œ lim

22. w œ f(z) œ z  Èz and f(z  h) œ (z  h)  Èz  h Ê œ lim

ds ¸ dt t=
"

œ lim

f()
h)  f()) œ lim h hÄ! hÄ! È È 2È4  )  #È%  )  h Š2 %  )
2 4  )  h‹ lim † È Š2 4  )
#È4  )  h‹ h Ä ! hÈ 4  ) È 4  )  h

and f()  h) œ

2 È4  )

hÄ!

and f(x  h) œ 1


" " x x
h œ lim h

21. r œ f()) œ œ lim

" x

œ lim (
6t
3h) œ
6t Ê

ds dt

;

"

h Ä ! Èz
h
Èz

dw dz

œ lim

hÄ!

œ lim –1  hÄ!

œ"

" 2È z

Ê

f(z
h)  f(z) h Èz
h  Èz

dw ¸ dz zœ4

h

œ

†

ŠÈz
h
Èz‹ ŠÈz
h
Èz‹ —

5 4

"  "

fazb  faxb a x
#b  a z
# b xz " z
# x
# 23. f w axb œ zlim œ zlim œ zlim œ zlim œ zlim Äx zx Ä x zx Ä x az  xbaz
#bax
#b Ä x az  xbaz
#bax
#b Ä x az
#bax
#b œ ax "
#b#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

123

124

Chapter 3 Differentiation "

"

#

#

#

#

Òax  "b  az  "bÓÒax  "b
az  "bÓ ax  "b  az"b fazb  faxb az
"b ax
"b 24. f w axb œ zlim œ zlim œ zlim œ zlim zx Ä x zx Äx Ä x az  xbaz  "b# ax  "b# Äx az  xbaz  "b# ax  "b# ax  zbax
z  2b "ax
z  2b œ zlim œ zlim œ Ä x az  xbaz  "b# ax  "b# Ä x a z  " b # a x  "b# z



"a#x  #b a x  "b %

œ

#ax  "b a x  "b %

œ

# a x  "b $

x

gazb  gaxb z a x  "b  x a z  " b z
x " zc" x
" 25. gw axb œ zlim œ zlim œ zlim œ zlim œ zlim Äx zx Äx zx Ä x az  xbaz  "bax  "b Ä x az  xbaz  "bax  "b Ä x az  "bax  "b œ ax "  "b# gazb  gaxb 26. gw axb œ zlim œ zlim Äx zx Äx " " œ zlim œ #È x Ä x Èz
Èx

ˆ"
Èz‰ˆ"
Èx‰ zx

œ zlim Äx

Èz  Èx zx

†

Èz
Èx Èz
Èx

zx œ zlim Ä x az  x bˆÈ z
È x ‰

27. Note that as x increases, the slope of the tangent line to the curve is first negative, then zero (when x œ 0), then positive Ê the slope is always increasing which matches (b). 28. Note that the slope of the tangent line is never negative. For x negative, f#w (x) is positive but decreasing as x increases. When x œ 0, the slope of the tangent line to x is 0. For x  0, f#w (x) is positive and increasing. This graph matches (a). 29. f$ (x) is an oscillating function like the cosine. Everywhere that the graph of f$ has a horizontal tangent we expect f$w to be zero, and (d) matches this condition. 30. The graph matches with (c). 31. (a) f w is not defined at x œ 0, 1, 4. At these points, the left-hand and right-hand derivatives do not agree. For example, lim c xÄ!

f(x)  f(0) x0

œ slope of line joining (
%ß 0) and (!ß #) œ

" #

but lim b xÄ!

line joining (0ß 2) and ("ß
2) œ
4. Since these values are not equal, f w (0) œ lim

xÄ!

f(x)  f(0) x0

f(x)  f(0) x0

(b)

32. (a)

(b) Shift the graph in (a) down 3 units

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ slope of

does not exist.

Section 3.1 The Derivative of a Function 33.

(b) The fastest is between the 20th and 30th days; slowest is between the 40th and 50th days.

34. (a)

35. Left-hand derivative: For h  0, f(0  h) œ f(h) œ h# (using y œ x# curve) Ê œ lim c hÄ!

h#  0 h

œ lim c h œ 0; hÄ!

Right-hand derivative: For h  0, f(0  h) œ f(h) œ h (using y œ x curve) Ê œ lim b hÄ! Then lim c hÄ!

h0 h

lim

h Ä !c

œ lim b 1 œ 1; hÄ!

f(0
h)  f(0) h

Á lim b hÄ!

f(0
h)  f(0) h

lim

h Ä !b

œ lim c 0 œ 0; hÄ!

f(1
h)  f(1) h

lim

h Ä !c

Right-hand derivative: When h  !, 1  h  1 Ê f(1  h) œ 2(1  h) œ 2  2h Ê

Then lim c hÄ!

(2
2h)2 h

œ lim b hÄ!

f(1
h)  f(1) h

2h h

È1
h  " h

œ lim c hÄ!

lim

h Ä !b

ŠÈ1
h  "‹ h

†

ŠÈ1
h
"‹ ŠÈ1
h
1‹

œ lim c hÄ!

lim

h Ä !c

Then lim c hÄ!

(2h
1)  " h f(1
h)  f(1) h

38. Left-hand derivative:

lim

h Ä !c

Right-hand derivative: œ lim b hÄ! Then lim c hÄ!

h h(1
h)

œ lim b 2 œ 2; hÄ! Á lim b hÄ!

lim b

hÄ!

œ lim b hÄ!

f(1
h)  f(1) h

f(1
h)  f(1) h

f(1
h)  f(") h

(1
h)  " h ŠÈ1
h
"‹

œ lim c hÄ!

Á lim b hÄ!

" È1
h
1

lim

h Ä !b

Ê the derivative f w (1) does not exist.

œ lim c

f(1
h)  f(") h " 1
h

f(1
h)  f(1) h

f("
h)  f(1) h

Right-hand derivative: When h  0, 1  h  1 Ê f(1  h) œ 2(1  h)
1 œ 2h  1 Ê œ lim b hÄ!

22 h

Ê the derivative f w (1) does not exist.

37. Left-hand derivative: When h  0, 1  h  1 Ê f(1  h) œ È1  h Ê œ lim c hÄ!

œ lim c hÄ!

œ lim b 2 œ 2; hÄ!

f(1
h)  f(1) h

Á lim b hÄ!

f(0
h)  f(0) h

Ê the derivative f w (0) does not exist.

36. Left-hand derivative: When h  !, 1  h  1 Ê f(1  h) œ 2 Ê

œ lim b hÄ!

f(0
h)  f(0) h

hÄ!

œ lim b hÄ!

(1
h)  " h

œ lim c 1 œ 1;

"  "‹ Š1
h h

hÄ!

œ lim b hÄ!

Š

1  (1
h) 1
h ‹

h

œ
1; f(1
h)  f(1) h

Ê the derivative f w (1) does not exist.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ #" ;

f("
h)f(1) h

125

126

Chapter 3 Differentiation

39. (a) The function is differentiable on its domain
$ Ÿ x Ÿ 2 (it is smooth) (b) none (c) none 40. (a) The function is differentiable on its domain
# Ÿ x Ÿ 3 (it is smooth) (b) none (c) none 41. (a) The function is differentiable on
$ Ÿ x  0 and !  x Ÿ 3 (b) none (c) The function is neither continuous nor differentiable at x œ 0 since lim c f(x) Á lim b f(x) xÄ! xÄ! 42. (a) f is differentiable on
# Ÿ x 
1,
"  x  0, 0  x  2, and 2  x Ÿ 3 (b) f is continuous but not differentiable at x œ
1: lim f(x) œ 0 exists but there is a corner at x œ
1 since x Ä
1

œ
3 and lim b f("
h)h  f(1) œ 3 Ê f w (
1) does not exist hÄ! hÄ! (c) f is neither continuous nor differentiable at x œ 0 and x œ 2: at x œ 0, lim c f(x) œ 3 but lim b f(x) œ 0 Ê lim f(x) does not exist; lim c

f(1
h)  f(") h

xÄ!

xÄ0

xÄ!

at x œ 2, lim f(x) exists but lim f(x) Á f(2) xÄ#

xÄ#

43. (a) f is differentiable on
" Ÿ x  0 and 0  x Ÿ 2 (b) f is continuous but not differentiable at x œ 0: lim f(x) œ 0 exists but there is a cusp at x œ 0, so f(0
h)  f(0) h hÄ!

f w (0) œ lim

xÄ!

does not exist

(c) none 44. (a) f is differentiable on
$ Ÿ x 
2,
2  x  2, and 2  x Ÿ 3 (b) f is continuous but not differentiable at x œ
2 and x œ 2: there are corners at those points (c) none 45. (a) f w (x) œ lim

hÄ!

f(x
h)  f(x) h

œ lim

hÄ!

(x
h)#  ax# b h

œ lim

hÄ!

x#  2xh  h#
x# h

œ lim (
2x
h) œ
2x hÄ!

(b)

(c) yw œ
2x is positive for x  0, yw is zero when x œ 0, yw is negative when x  0 (d) y œ
x# is increasing for
_  x  0 and decreasing for !  x  _; the function is increasing on intervals where yw  0 and decreasing on intervals where yw  0 f(x
h)  f(x) h hÄ!

46. (a) f w (x) œ lim

œ lim

hÄ!

Š xc" h  h


1 ‹ x

œ lim

hÄ!

x
(x
h) x(x
h)h

œ lim

"

h Ä ! x(x
h)

œ

" x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.1 The Derivative of a Function (b)

(c) yw is positive for all x Á 0, yw is never 0, yw is never negative (d) y œ
"x is increasing for
_  x  0 and !  x  _ w

47. (a) Using the alternate formula for calculating derivatives: f (x) œ œ

$ $ lim z  x z Ä x 3(z  x)

œ

az#
zx
x# b lim (z  x)3(z  x) zÄx

œ

# # lim z
zx3
x zÄx

f(x) lim f(z)z  x zÄx # w

$

Š z3 

œ zlim Äx

x$ 3 ‹

zx

œ x Ê f (x) œ x#

(b)

(c) yw is positive for all x Á 0, and yw œ 0 when x œ 0; yw is never negative (d) y œ

x$ 3

is increasing for all x Á 0 (the graph is horizontal at x œ 0) because y is increasing where yw  0; y is

never decreasing %

48. (a) Using the alternate œ zlim Äx

z%  x% 4(z  x)

œ

%

z x Œ4  4
f(z)  f(x) form for calculating derivatives: f (x) œ zlim œ lim zx zx Äx zÄx (z  x) az$
xz#
x# z
x$ b z$
xz#
x# z
x$ $ w lim œ zlim œ x Ê f (x) œ x$ 4(z  x) 4 zÄx Äx

w

(b)

(c) yw is positive for x  0, yw is zero for x œ 0, yw is negative for x  0 (d) y œ

x% 4

is increasing on 0  x  _ and decreasing on
_  x  0 #

#

(xc) ax
xc
c b f(x)  f(c) x c 49. yw œ xlim œ xlim œ xlim œ xlim ax#  xc  c# b œ 3c# . xc xc Äc Ä c xc Äc Äc The slope of the curve y œ x$ at x œ c is yw œ 3c# . Notice that 3c#   0 for all c Ê y œ x$ never has a negative slope. $

$

50. Horizontal tangents occur where yw œ 0. Thus, yw œ lim

hÄ!

œ lim

hÄ!

2 ŠÈx
h  Èx‹ h

†

ŠÈx
h
Èx‹ ŠÈx
h
Èx‹

œ lim

2È x
h  2È x h

2((x
h)  x))

h Ä ! h ŠÈx
h
Èx‹

œ lim

2

h Ä ! Èx
h
Èx

œ

" Èx

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

127

128

Chapter 3 Differentiation

Then yw œ 0 when 51. yw œ lim

hÄ!

œ lim

hÄ!

" Èx

œ 0 which is never true Ê the curve has no horizontal tangents.

a2(x
h)#  13(x
h)
5b  a2x#  13x
5b h

4xh
2h#  13h h

œ lim

hÄ!

2x#
4xh
2h#  13x  13h
5  2x#
13x  5 h

œ lim (4x  2h
13) œ 4x
13, slope at x. The slope is
1 when 4x
13 œ
" hÄ!

Ê 4x œ 12 Ê x œ 3 Ê y œ 2 † 3#
13 † 3  5 œ
16. Thus the tangent line is y  16 œ (
1)(x
3) Ê y œ
x
"$ and the point of tangency is (3ß
16). 52. For the curve y œ Èx, we have yw œ lim

ŠÈx
h  Èx‹

hÄ!

"

œ lim

h Ä ! Èx
h
Èx

œ

" #Èx

h

†

ŠÈx
h
Èx‹ ŠÈx
h
Èx‹

h Ä ! ŠÈx
h
Èx‹ h

. Suppose ˆ+ß Èa‰ is the point of tangency of such a line and (
"ß !) is the point

on the line where it crosses the x-axis. Then the slope of the line is " ; 2È a

(x
h)  x

œ lim

using the derivative formula at x œ a Ê

exist: its point of tangency is ("ß "), its slope is

Èa a
1

œ

" #È a

œ

Èa  0 a  (1)

œ

Èa a
1

which must also equal

" Ê 2a œ a  1 Ê a œ 1. #Èa " # ; and an equation of the line is

Thus such a line does y
1œ

" #

(x
1)

Ê y œ "# x  "# . 53. No. Derivatives of functions have the intermediate value property. The function f(x) œ ÚxÛ satisfies f(0) œ 0 and f(1) œ 1 but does not take on the value "# anywhere in [!ß "] Ê f does not have the intermediate value

property. Thus f cannot be the derivative of any function on [!ß "] Ê f cannot be the derivative of any function on (
_ß _).

54. The graphs are the same. So we know that for f(x) œ kxk , we have f w (x) œ

kx k x

.

55. Yes; the derivative of
f is
f w so that f w (x! ) exists Ê
f w (x! ) exists as well. 56. Yes; the derivative of 3g is 3gw so that gw (7) exists Ê 3gw (7) exists as well. 57. Yes, lim

g(t)

t Ä ! h(t)

can exist but it need not equal zero. For example, let g(t) œ mt and h(t) œ t. Then g(0) œ h(0)

œ 0, but lim

g(t)

t Ä ! h(t)

œ lim

tÄ!

mt t

œ lim m œ m, which need not be zero. tÄ!

58. (a) Suppose kf(x)k Ÿ x# for
" Ÿ x Ÿ 1. Then kf(0)k Ÿ 0# Ê f(0) œ 0. Then f w (0) œ lim œ lim

hÄ!

f(h)  0 h

œ lim

hÄ!

f(h) h .

For khk Ÿ 1,
h# Ÿ f(h) Ÿ h# Ê
h Ÿ

hÄ!

f(h) h

f(0
h)  f(0) h

Ÿ h Ê f w (0) œ lim

hÄ!

f(h) h

œ0

by the Sandwich Theorem for limits. (b) Note that for x Á 0, kf(x)k œ ¸x# sin "x ¸ œ kx# k ksin xk Ÿ kx# k † 1 œ x# (since
" Ÿ sin x Ÿ 1). By part (a), f is differentiable at x œ 0 and f w (0) œ 0.

59. The graphs are shown below for h œ 1, 0.5, 0.1. The function y œ y œ Èx so that

" #È x

œ lim

hÄ!

Èx
h  Èx h

" 2È x

. The graphs reveal that y œ

is the derivative of the function

Èx
h  Èx h

gets closer to y œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #È x

Section 3.1 The Derivative of a Function as h gets smaller and smaller.

60. The graphs are shown below for h œ 2, 1, 0.5. The function y œ 3x# is the derivative of the function y œ x$ so that 3x# œ lim

hÄ!

(x
h)$ x$ h

. The graphs reveal that y œ

(x
h)$ x$ h

gets closer to y œ 3x# as h

gets smaller and smaller.

61. Weierstrass's nowhere differentiable continuous function.

62-67. Example CAS commands: Maple: f := x -> x^3 + x^2 - x;

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

129

130

Chapter 3 Differentiation

x0 := 1; plot( f(x), x=x0-5..x0+2, color=black, title="Section 3_1, #62(a)" ); q := unapply( (f(x+h)-f(x))/h, (x,h) ); # (b) L := limit( q(x,h), h=0 ); # (c) m := eval( L, x=x0 ); tan_line := f(x0) + m*(x-x0); plot( [f(x),tan_line], x=x0-2..x0+3, color=black, linestyle=[1,7], title="Section 3.1 #62(d)", legend=["y=f(x)","Tangent line at x=1"] ); Xvals := sort( [ x0+2^(-k) $ k=0..5, x0-2^(-k) $ k=0..5 ] ): # (e) Yvals := map( f, Xvals ): evalf[4](< convert(Xvals,Matrix) , convert(Yvals,Matrix) >); plot( L, x=x0-5..x0+3, color=black, title="Section 3.1 #62(f)" ); Mathematica: (functions and x0 may vary) (see section 2.5 re. RealOnly ): <<Miscellaneous`RealOnly` Clear[f, m, x, y, h] x0= 1 /4; f[x_]:=x2 Cos[x] Plot[f[x], {x, x0
3, x0  3}] q[x_, h_]:=(f[x  h]
f[x])/h m[x_]:=Limit[q[x, h], h Ä 0] ytan:=f[x0]  m[x0] (x
x0) Plot[{f[x], ytan},{x, x0
3, x0  3}] m[x0
1]//N m[x0  1]//N Plot[{f[x], m[x]},{x, x0
3, x0  3}] 3.2 DIFFERENTIATION RULES 1. y œ
x#  3 Ê

dy dx

œ

2. y œ x#  x  8 Ê

dy dx

3. s œ 5t$
3t& Ê

œ

ds dt

a
x# b 

d dx

4 3

6. y œ

x$ 3

x$
x Ê x# #

dy dx

(3) œ
2x  0 œ
#B Ê

d dt

a5t$ b
dw dz

d dt

d# y dx# " 4

œ x#  x 

7. w œ 3z#
z" Ê

dw dz

œ
6z$  z# œ

x 4

8. s œ
2t"  4t# Ê

ds dt

Ê

œ 2t#
8t$ œ

9. y œ 6x#
10x
5x# Ê

dy dx

œ
2

d# s dt#

œ

d dt

a15t# b


d dt

a15t% b œ 30t
60t$

œ 126z&
42z  42

œ 8x

dy dx



d# w dz#

œ 21z'
21z#  42z Ê

œ 4x#
1 Ê

d# y dx#

œ#

a3t& b œ 15t#
15t% Ê

Ê



d# y dx#

œ 2x  1  0 œ 2x  1 Ê

4. w œ 3z(
7z$  21z# Ê 5. y œ

d dx

d# y dx# 6 z$ 2 t#

œ 2x  1  0 œ 2x  1 




8 t$

" z#

Ê

Ê

d# w dz#

d# s dt#

œ 18z%
2z$ œ

œ
4t$  24t% œ

œ 12x
10  10x$ œ 12x
10 

10 x$

Ê

d# y dx#

18 z%

4 t$






2 z$

24 t%

œ 12
0
30x% œ 12


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

30 x%

Section 3.2 Differentiation Rules 10. y œ 4
2x
x$ Ê " # 3 s

11. r œ

œ
#  3x% œ
# 

dy dx


5# s" Ê

œ
32 s$  5# s# œ

dr ds

12. r œ 12)"
4)$  )% Ê œ




24 )$

48 )&



dr d)

3 x%

2 3s$

Ê

d# y dx#

5 2s#

Ê



12 x&

œ 0
12x& œ d# r ds#

œ
12)#  12)%
4)& œ

œ 2s%
5s$ œ 12 )#



12 )%




4 )&




2 s%

d# r d) #

Ê

5 s$

œ 24)$
48)&  20)'

20 )'

13. (a) y œ a3
x# b ax$
x  1b Ê yw œ a3
x# b †

ax$
x  1b  ax$
x  1b †

d dx

d dx

a3
x# b

œ a3
x# b a3x#
1b  ax$
x  1b (
2x) œ
5x%  12x#
2x
3 (b) y œ
x&  4x$
x#
3x  3 Ê yw œ
5x%  12x#
2x
3 14. (a) y œ (x
1) ax#  x  1b Ê yw œ (x
1)(2x  1)  ax#  x  1b (") œ 3x# (b) y œ (x
1) ax#  x  1b œ x$
1 Ê yw œ 3x# 15. (a) y œ ax#  1b ˆx  5  "x ‰ Ê yw œ ax#  1b †

d dx

ˆx  5  "x ‰  ˆx  5  x" ‰ †

d dx

ax#  1b

œ ax#  1b a1
x# b  ax  5  x" b (2x) œ ax#
1  1
x# b  a2x#  10x  2b œ 3x#  10x  2
(b) y œ x$  5x#  2x  5  16. y œ ˆx  "x ‰ ˆx
w

" x

" x

Ê yw œ 3x#  10x  2


" x#

 1‰

"

(a) y œ ax  x b † a1  x# b  ax
x"  1b a1
x# b œ 2x  1
(b) y œ x#  x 

" x




" x#

Ê yw œ 2x  1


" x#



2x
5 3x  2 ; use the quotient rule: u œ 2x  5 and (3x  2)(2)  (2x
5)(3) 4  6x  15 œ 6x (3x œ (3x192)# (3x  2)#  #)#

17. y œ œ

18. z œ

2x
1 x#  1

Ê

dz dx

œ

ax#  1b (2)  (2x
1)(2x) ax #  1 b #

œ

t#  " t#
t  2

œ

at  "bat
"b at
#bat  "b

21. v œ (1
t) a1  t# b 22. w œ

x
5 2x  7

23. f(s) œ

24. u œ

Ê ww œ

Ès  " Ès
1

NOTE:

d ds

5x
" #È x

"

du dx

œ

1 t 1
t#

t
" t
2,

2x#  2  4x#  2x a x #  1 b#

Ê

dv dt

œ

" #È s

ˆ È s
"‰ Š

œ

2x  7  2x  10 (2x  7)#

"

"

ˆÈ s
1 ‰

#

2x#  2x  2 a x #  1 b#

at
#ba"b  at
"ba"b at
2 b 2

a1
t# b (")  (1  t)(2t) a1
t# b#

Ès ‹  ˆÈs  1‰ Š #Ès ‹

#

œ

œ

vuw  uvw v#

 2 a x #
x
1b ax#  1b#

v œ x  0.5 Ê uw œ 2x and vw œ 1 Ê gw (x) œ

t Á " Ê f w (t) œ

(2x  7)(1)  (x
5)(2) (2x  7)#

Ê f w (s) œ

ˆÈ s ‰ œ

Ê

œ

œ

2 x$

v œ 3x
2 Ê uw œ 2 and vw œ 3 Ê yw œ

x#  4 # x
0.5 ; use the quotient rule: u œ x
4 and # # (x
0.5)(2x)  ax#  4b (") x#
4 x
4 œ 2x (x

x 0.5) œ x(x
# (x
0.5)#
0.5)#

20. f(t) œ



2 x$

19. g(x) œ œ

" x#

œ

œ

œ

œ

t
#t" at
2 b 2

"  t#  2t
2t# a1
t# b#

œ

œ

" at
2 b 2

t#  2t  " a1
t# b#

17 (2x  7)#

ˆ È s
"‰  ˆ È s  1 ‰ 2 È s ˆÈ s
1 ‰

#

œ

" È s ˆÈ s
1 ‰#

from Example 2 in Section 2.1

ˆ2Èx‰ (5)  (5x
1) Š È" ‹ x 4x

œ

5x  1 4x$Î#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

vuw  uvw v#

" x#

131

132

Chapter 3 Differentiation

25. v œ

1
x  4È x x

Ê vw œ

x Š1 

È x ‹  ˆ1
x  4 È x ‰ 2

x#

26. r œ 2 Š È"  È)‹ Ê rw œ 2
) 27. y œ

" ax #  1 b a x #
x
1 b #

È)(0)  1 Š )

"

È) ‹

#

œ



2È x  " x#

" #È) 

" œ
)$Î# 

" )"Î#

; use the quotient rule: u œ 1 and v œ ax#
1b ax#  x  1b Ê uw œ 0 and

vw œ ax
1b (2x  1)  ax#  x  1b (2x) œ 2x$  x#
2x
1  2x$  2x#  2x œ 4x$  3x#
1 Ê 28. y œ œ

vuw  uvw v#

œ

dy dx

(x
1)(x
2) (x  1)(x  #)

œ

œ

0  1 a4x$
3x#  1b ax #  1 b # a x #
x
1 b#

x#
3x
2 x#  3x
2

4x$  3x#
1 ax#  1b# ax#
x
1b#

œ

ax#  3x
2b (2x
3)  ax#
3x
2b (2x  3) (x  1)# (x  2)#

Ê yw œ

 6 ax #  2 b (x  1)# (x  2)#

29. y œ

" #

30. y œ

" 1 #0

31. y œ

x $
7 x

32. s œ

t#
5t  1 œ 1  5t
t"# œ 1  5t" t# d# s ' $
6t% œ "! dt# œ 10t t$
t%

Ê

x%


3 #

x#
x Ê yw œ 2x$
3x
1 Ê yww œ 6x#
3 Ê ywww œ 12x Ê yÐ%Ñ œ 12 Ê yÐnÑ œ 0 for all n   5

34. u œ Ê

" #4

x& Ê yw œ

x% Ê yww œ

œ x#  7x" Ê

()  " ) a ) #
)
1 b )$

33. r œ

œ

)$  " )$

dy dx

" 6

" #

x$ Ê ywww œ

" )$

œ"



t# Ê

x# Ê yÐ%Ñ œ x Ê yÐ&Ñ œ 1 Ê yÐnÑ œ 0 for all n   6 ( x#

œ 2x
7x# œ #x


Ê

œ 1
)$ Ê

dr d)

" z#


"Ê

d w dz#

œ 2z$
0 œ 2z$ œ

" 3

1

#

%

Ê 38. p œ Ê

d# p dq#

œ

" 6


#" q%
5q' œ

q#
3 (q  1)$
(q
1)$ dp dq

q'  q#
3q%  3 12q%

œ

" 6




œ

" #q %

q#


" 1#

q# 

d# p dq#

& t#



Ê

d# r d) #


z Ê

dw dz

œ
z#  0
1 œ
z#
1

œ
12)& œ

(c)

d dx d dx

ˆ vu ‰ œ ˆ uv ‰ œ

vuw  uvw v#

œ q$ œ

w

uv  vu u#

w

Ê

d dx

ˆ vu ‰¸

x=0

œ

Ê

d dx

ˆ uv ‰¸

x=0

œ

"# )&

œ 1  x$

8 3

" 4

dw dz

œ 4z$
0 œ 4z$ Ê


4" q% Ê

dp dq

œ

" 6

d# w dz#

œ 12z#

q  6" q$  q& œ

œ

q#
3 2q$
6q

œ

q#
3 2q aq#
3b

œ

" #q

œ

" #

q"

" q$

39. u(0) œ 5, uw (0) œ
3, v(0) œ
1, vw (0) œ 2 d d (a) dx (uv) œ uvw  vuw Ê dx (uv)¸ x=0 œ u(0)vw (0)  v(0)uw (0) œ 5 † 2  (
1)(
3) œ 13 (b)

# t$

$ )%

& q6

q#
3 aq$  3q#
3q  1b
aq$
3q#
3q
1b

œ
"# q# œ
#"q# Ê

"% x$

# z$

" 1#




x x%

 3
z œ z" 

36. w œ (z  1)(z
1) az#  1b œ az#
1b az#  1b œ z%
1 Ê
3 37. p œ Š q12q ‹ Š q q$ 1 ‹ œ

œ 2  14x$ œ # 

œ 0  $)% œ $)% œ

# $ % ax #
x b a x #  x
1 b œ x(x
1) axx%  x
"b œ x axx%
1b œ x x
% x œ x% # du % œ
3x% œ $ Ê ddxu# œ 12x& œ "# dx œ 0
3x x& x%

#

d# y dx#

œ 0
5t#  2t$ œ
5t#  2t$ œ

ds dt

35. w œ ˆ 1
3z3z ‰ (3
z) œ ˆ "3 z"  1‰ (3
z) œ z"
œ

6x#
12 (x  1)# (x  2)#

œ

v(0)uw (0)  u(0)vw (0) (v(0))# u(0)vw (0)  v(0)uw (0) (u(0))#

œ œ

(")(3)  (5)(2) (1)# (5)(2)  (1)(3) (5)#

œ
7 œ

7 25

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 6

q

" 6q$



" q&

Section 3.2 Differentiation Rules (d)

d dx

(7v
2u) œ 7vw
2uw Ê

d dx

133

(7v
2u)¸ x=0 œ 7vw (0)
2uw (0) œ 7 † 2
2(
3) œ 20

40. u(1) œ 2, uw (1) œ 0, v(1) œ 5, vw (1) œ
1 d (a) dx (uv)¸ x=1 œ u(1)vw (1)  v(1)uw (1) œ 2 † (
1)  5 † 0 œ
2 (b) (c) (d)

v(1)uw (")u(1)vw (1) (v(1))# u(1)vw (")v(1)uw (1) d ˆ v ‰¸ dx u x=1 œ (u(1))# d w ¸ dx (7v
2u) x=1 œ 7v (1)
d dx

ˆ vu ‰¸

x=1

œ

œ œ

5†02†(1) (5)# 2†(1)5†0 (2)#

œ

2 25

œ
12

2uw (1) œ 7 † (
1)
2 † 0 œ
7

41. y œ x$
4x  1. Note that (#ß ") is on the curve: 1 œ 2$
4(2)  1 (a) Slope of the tangent at (xß y) is yw œ 3x#
4 Ê slope of the tangent at (#ß ") is yw (2) œ 3(2)#
4 œ 8. Thus the slope of the line perpendicular to the tangent at (#ß ") is
"8 Ê the equation of the line perpendicular to to the tangent line at (#ß ") is y
1 œ
"8 (x
2) or y œ
x8  54 .

(b) The slope of the curve at x is m œ 3x#
4 and the smallest value for m is
4 when x œ 0 and y œ 1. (c) We want the slope of the curve to be 8 Ê yw œ 8 Ê 3x#
4 œ 8 Ê 3x# œ 12 Ê x# œ 4 Ê x œ „ 2. When x œ 2, y œ 1 and the tangent line has equation y
1 œ 8(x
2) or y œ 8x
15; when x œ
2, y œ (
2)$
4(
2)  1 œ 1, and the tangent line has equation y
1 œ 8(x  2) or y œ 8x  17. 42. (a) y œ x$
3x
2 Ê yw œ 3x#
3. For the tangent to be horizontal, we need m œ yw œ 0 Ê 0 œ 3x#
3 Ê 3x# œ 3 Ê x œ „ 1. When x œ
1, y œ 0 Ê the tangent line has equation y œ 0. The line perpendicular to this line at (
"ß !) is x œ
1. When x œ 1, y œ
4 Ê the tangent line has equation y œ
4. The line perpendicular to this line at ("ß
%) is x œ 1. (b) The smallest value of yw is
3, and this occurs when x œ 0 and y œ
2. The tangent to the curve at (!ß
2) has slope
3 Ê the line perpendicular to the tangent at (!ß
2) has slope "3 Ê y  2 œ "3 (x
0) or yœ 43. y œ

4x x#
1

" 3

x
2 is an equation of the perpendicular line. Ê

dy dx

œ

ax#
1b(4)  (4x)(2x) ax #
1 b #

œ

4x#
4  8x# a x #
1 b#

œ

4 ax#
"b a x #
1 b#

. When x œ 0, y œ 0 and yw œ

4(0
1) 1

œ %, so the tangent to the curve at (!ß !) is the line y œ 4x. When x œ 1, y œ 2 Ê yw œ 0, so the tangent to the curve at ("ß 2) is the line y œ 2. 44. y œ

8 x#
4

Ê yw œ

ax#
4b(0)  8(2x) ax #
4 b #

œ

16x a x #
4 b#

. When x œ 2, y œ 1 and yw œ

16(2) a 2 #
4 b#

œ
"# , so the tangent

line to the curve at (2ß ") has the equation y
1 œ
"# (x
2), or y œ
x#  2. 45. y œ ax#  bx  c passes through (!ß !) Ê 0 œ a(0)  b(0)  c Ê c œ 0; y œ ax#  bx passes through ("ß #) Ê 2 œ a  b; yw œ 2ax  b and since the curve is tangent to y œ x at the origin, its slope is 1 at x œ 0 Ê yw œ 1 when x œ 0 Ê 1 œ 2a(0)  b Ê b œ 1. Then a  b œ 2 Ê a œ 1. In summary a œ b œ 1 and c œ 0 so the curve is y œ x#  x. 46. y œ cx
x# passes through ("ß !) Ê 0 œ c(1)
1 Ê c œ 1 Ê the curve is y œ x
x# . For this curve, yw œ 1
2x and x œ 1 Ê yw œ
1. Since y œ x
x# and y œ x#  ax  b have common tangents at x œ 0, y œ x#  ax  b must also have slope
1 at x œ 1. Thus yw œ 2x  a Ê
1 œ 2 † 1  a Ê a œ
3 Ê y œ x#
3x  b. Since this last curve passes through ("ß !), we have 0 œ 1
3  b Ê b œ 2. In summary, a œ
3, b œ 2 and c œ 1 so the curves are y œ x#
3x  2 and y œ x
x# . 47. (a) y œ x$
x Ê yw œ 3x#
1. When x œ
1, y œ 0 and yw œ 2 Ê the tangent line to the curve at (
"ß !) is y œ 2(x  1) or y œ 2x  2.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

134

Chapter 3 Differentiation

(b)

(c)

y œ x$
x Ê x$
x œ 2x  2 Ê x$
3x
2 œ (x
2)(x  1)# œ 0 Ê x œ 2 or x œ
1. Since y œ 2x  2  y œ 2a2b  2 œ 6; the other intersection point is (2ß 6)

48. (a) y œ x$
6x#  5x Ê yw œ 3x#
12x  5. When x œ 0, y œ 0 and yw œ 5 Ê the tangent line to the curve at (0ß 0) is y œ 5x. (b)

(c)

y œ x$
6x#  5x $ # $ # #  Ê x
6x  5x œ 5x Ê x
6x œ 0 Ê x (x
6) œ 0 Ê x œ 0 or x œ 6. y œ 5x Since y œ 5a6b œ $!, the other intersection point is (6ß 30).

49. Paxb œ an xn  an" xn"  â  a# x#  a" x  a! Ê P w axb œ nan xn"  an
"ban" xn#  â  #a# x  a" 50. R œ M# ˆ C#


M‰ 3

œ

C #

51. Let c be a constant Ê

M#
"3 M$ , where C is a constant Ê dc dx

œ0 Ê

d dx

(u † c) œ u †

dc dx

c†

dR dM

œ CM
M#

œu†0c

du dx

œc

du dx

du dx

. Thus when one of the

functions is a constant, the Product Rule is just the Constant Multiple Rule Ê the Constant Multiple Rule is a special case of the Product Rule. 52. (a) We use the Quotient rule to derive the Reciprocal Rule (with u œ 1): œ
v"# †

dv dx

d dx

ˆ "v ‰ œ

v†0  1† dv dx v#

d ˆ"‰ " du dx v  v † dx (Product Rule) dv v du dx  u dx , the Quotient Rule. v#

œu†

53. (a)

d dx

(uvw) œ w

"† dv dx v#

.

(b) Now, using the Reciprocal Rule and the Product Rule, we'll derive the Quotient Rule:

œ

œ

d dx w

œ u † ˆ v#1 ‰ dv dx 

((uv) † w) œ (uv) dw dx  w † w

d dx

(uv) œ uv

œ uvw  uv w  u vw d d % (b) dx au" u# u$ u% b œ dx aau" u# u$ b u% b œ au" u# u$ b du dx  u% du$ du# du" ‰ % ˆ œ u" u# u$ du dx  u% u" u# dx  u$ u" dx  u$ u# dx

" du v dx

dw dx d dx

(Reciprocal Rule) Ê

 w ˆu

dv dx

v

a u" u# u$ b Ê

d dx

du ‰ dx

d dx

œ uv

d dx

ˆ vu ‰ œ

dw dx

d ˆ "‰ dx u † v du u dv dx
v dx v#

ˆ uv ‰ œ

 wu

dv dx

 wv

a u" u# u$ u% b

(using (a) above)

du$ du# du" % Ê au" u# u$ u% b œ u" u# u$ du dx  u" u# u% dx  u" u$ u% dx  u# u$ u% dx œ u" u# u$ u%w  u" u# u$w u%  u" u#w u$ u%  u"w u# u$ u% d Generalizing (a) and (b) above, dx au" âun b œ u" u# âun" unw  u" u# âun# unw " un d dx

(c)

 á  u"w u# âun

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

du dx

Section 3.3 The Derivative as a Rate of Change

135

54. In this problem we don't know the Power Rule works with fractional powers so we can't use it. Remember d ˆÈ ‰ " x œ #È (from Example 2 in Section 2.1) dx x " #È x

Èx #

3È x #

(a)

d dx

ˆx$Î# ‰ œ

d dx

ˆx † x"Î# ‰ œ x †

(b)

d dx

ˆx&Î# ‰ œ

d dx

ˆx# † x"Î# ‰ œ x#

d dx

ˆÈx‰  Èx

d dx

" ax# b œ x# † Š #È ‹  Èx † 2x œ x

(c)

d dx

ˆx(Î# ‰ œ

d dx

ˆx$ † x"Î# ‰ œ x$

d dx

ˆÈx‰  Èx

d dx

" ax$ b œ x$ † Š #È ‹  Èx † 3x# œ x

" #

x&Î#  3x&Î# œ

d dx

ˆx(Î# ‰ œ

d dx

axnÎ# b œ

(d) We have

d dx

ˆx$Î# ‰ œ

3 #

x"Î# ,

d dx

d dx

ˆÈx‰  Èx

ˆx&Î# ‰ œ

5 #

d dx

x$Î# ,

(x) œ x †

whenever n is an odd positive integer   3. 55. p œ Ê

nRT an# Vnb
V# . We are holding T constant, and a, b, n, R are # # (V  nb)†0  (nRT)(1) dP 2an#
V (0) aVa#anb# b (2V) œ (VnRT dV œ (Vnb)# nb)#  V$

56. Aaqb œ

km q

 cm 

hq #

œ akmbq"  cm  ˆ h# ‰q Ê

dA dq

7 #

 Èx † 1 œ

 Èx œ

x&Î# so it appears that

" #

œ

3 #

x"Î#

x$Î#  2x$Î# œ

n #

5 #

x$Î# 7 #

x&Î#

xÐnÎ#Ñ"

also constant so their derivatives are zero

œ
akmbq#  ˆ #h ‰ œ
km q# 

h #

Ê

d# A dt#

œ #akmbq$ œ

#km q$

3.3 THE DERIVATIVE AS A RATE OF CHANGE 1. s œ t#
$t  #, 0 Ÿ t Ÿ # (a) displacement œ ?s œ s(#)
s(0) œ !m
#m œ
# m, vav œ (b) v œ aœ

ds dt œ #t d# s dt# œ #

?s ?t

œ

Ê a(0) œ # m/sec# and a(#) œ # m/sec#

changes direction at t œ

$ #.

2. s œ 't
t# , ! Ÿ t Ÿ ' (a) displacement œ ?s œ s(')
s(0) œ ! m, vav œ aœ

œ
" m/sec


$ Ê kv(0)k œ l
$l œ $ m/sec and kv(#)k œ 1 m/sec;

(c) v œ 0 Ê #t
$ œ 0 Ê t œ $# . v is negative in the interval !  t 

(b) v œ

# #

ds dt œ '
d# s dt# œ
#

?s ?t

œ

! '

$ #

and v is positive when

$ #

 t  # Ê the body

œ ! m/sec

#> Ê kv(0)k œ l 'l œ ' m/sec and kv(')k œ l
'l œ ' m/sec; Ê a(0) œ
# m/sec# and a(') œ
# m/sec#

(c) v œ 0 Ê '
#t œ 0 Ê t œ $. v is positive in the interval !  t  $ and v is negative when $  t  ' Ê the body changes direction at t œ $. 3. s œ
t$  3t#
3t, 0 Ÿ t Ÿ 3 (a) displacement œ ?s œ s(3)
s(0) œ
9 m, vav œ (b) v œ

ds dt

?s ?t

œ

9 3

œ
3 m/sec

œ
3t#  6t
3 Ê kv(0)k œ k
3k œ 3 m/sec and kv(3)k œ k
12k œ 12 m/sec; a œ #

#

d# s dt#

œ
6t  6

Ê a(0) œ 6 m/sec and a(3) œ
12 m/sec (c) v œ 0 Ê
3t#  6t
3 œ 0 Ê t#
2t  1 œ 0 Ê (t
1)# œ 0 Ê t œ 1. For all other values of t in the interval the velocity v is negative (the graph of v œ
3t#  6t
3 is a parabola with vertex at t œ 1 which opens downward Ê the body never changes direction). 4. s œ

t% 4


t$  t# , 0 Ÿ t Ÿ $

(a) ?s œ s($)
s(0) œ

* %

m, vav œ

?s ?t

œ

* %

$

œ

$ %

m/sec

(b) v œ t$
3t#  2t Ê kv(0)k œ 0 m/sec and kv($)k œ ' m/sec; a œ 3t#
6t  2 Ê a(0) œ 2 m/sec# and a($) œ "" m/sec# (c) v œ 0 Ê t$
3t#  2t œ 0 Ê t(t
2)(t
1) œ 0 Ê t œ 0, 1, 2 Ê v œ t(t
2)(t
1) is positive in the interval for 0  t  1 and v is negative for 1  t  2 and v is positive for #  t  $ Ê the body changes direction at

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136

Chapter 3 Differentiation t œ 1 and at t œ #.

5. s œ

25 t#


5t , 1 Ÿ t Ÿ 5

(a) ?s œ s(5)
s(1) œ
20 m, vav œ (b) v œ

50 t$



5 t#

m/sec#

(c) v œ 0 Ê

50
5t t$

6. s œ

25 t
5

œ
5 m/sec

Ê kv(1)k œ 45 m/sec and kv(5)k œ

4 25

a(5) œ

20 4

" 5

m/sec; a œ

150 t%




10 t$

Ê a(1) œ 140 m/sec# and

œ 0 Ê
50  5t œ 0 Ê t œ 10 Ê the body does not change direction in the interval

,
% Ÿ t Ÿ 0

(a) ?s œ s(0)
s(
4) œ
20 m, vav œ
20 4 œ
5 m/sec (b) v œ a(0) (c) v œ

25 (t
5)# Ê kv(
4)k œ œ 25 m/sec# 0 Ê (t
25 5)# œ 0 Ê

25 m/sec and kv(0)k œ " m/sec; a œ

50 (t
5)$

Ê a(
4) œ 50 m/sec# and

v is never 0 Ê the body never changes direction

7. s œ t$
6t#  9t and let the positive direction be to the right on the s-axis. (a) v œ 3t#
12t  9 so that v œ 0 Ê t#
4t  3 œ (t
3)(t
1) œ 0 Ê t œ 1 or 3; a œ 6t
12 Ê a(1) œ
6 m/sec# and a(3) œ 6 m/sec# . Thus the body is motionless but being accelerated left when t œ 1, and motionless but being accelerated right when t œ 3. (b) a œ 0 Ê 6t
12 œ 0 Ê t œ 2 with speed kv(2)k œ k12
24  9k œ 3 m/sec (c) The body moves to the right or forward on 0 Ÿ t  1, and to the left or backward on 1  t  2. The positions are s(0) œ 0, s(1) œ 4 and s(2) œ 2 Ê total distance œ ks(1)
s(0)k  ks(2)
s(1)k œ k4k  k
2k œ 6 m. 8. v œ t#
4t  3 Ê a œ 2t
4 (a) v œ 0 Ê t#
4t  3 œ 0 Ê t œ 1 or 3 Ê a(1) œ
2 m/sec# and a(3) œ 2 m/sec# (b) v  0 Ê (t
3)(t
1)  0 Ê 0 Ÿ t  1 or t  3 and the body is moving forward; v  0 Ê (t
3)(t
1)  0 Ê "  t  3 and the body is moving backward (c) velocity increasing Ê a  0 Ê 2t
4  0 Ê t  2; velocity decreasing Ê a  0 Ê 2t
4  0 Ê ! Ÿ t  2 9. sm œ 1.86t# Ê vm œ 3.72t and solving 3.72t œ 27.8 Ê t ¸ 7.5 sec on Mars; sj œ 11.44t# Ê vj œ 22.88t and solving 22.88t œ 27.8 Ê t ¸ 1.2 sec on Jupiter. 10. (a) v(t) œ sw (t) œ 24
1.6t m/sec, and a(t) œ vw (t) œ sw w (t) œ
1.6 m/sec# (b) Solve v(t) œ 0 Ê 24
1.6t œ 0 Ê t œ 15 sec (c) s(15) œ 24(15)
.8(15)# œ 180 m (d) Solve s(t) œ 90 Ê 24t
.8t# œ 90 Ê t œ

30„15È2 #

¸ 4.39 sec going up and 25.6 sec going down

(e) Twice the time it took to reach its highest point or 30 sec 11. s œ 15t
"# gs t# Ê v œ 15
gs t so that v œ 0 Ê 15
gs t œ 0 Ê gs œ

15 t

. Therefore gs œ

15 20

œ

3 4

œ 0.75 m/sec#

12. Solving sm œ 832t
2.6t# œ 0 Ê t(832
2.6t) œ 0 Ê t œ 0 or 320 Ê 320 sec on the moon; solving se œ 832t
16t# œ 0 Ê t(832
16t) œ 0 Ê t œ 0 or 52 Ê 52 sec on the earth. Also, vm œ 832
5.2t œ 0 Ê t œ 160 and sm (160) œ 66,560 ft, the height it reaches above the moon's surface; ve œ 832
32t œ 0 Ê t œ 26 and se (26) œ 10,816 ft, the height it reaches above the earth's surface. 13. (a) s œ 179
16t# Ê v œ
32t Ê speed œ kvk œ 32t ft/sec and a œ
32 ft/sec#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.3 The Derivative as a Rate of Change (b) s œ 0 Ê 179
16t# œ 0 Ê t œ É 179 16 ¸ 3.3 sec È É 179 (c) When t œ É 179 16 , v œ
32 16 œ
8 179 ¸
107.0 ft/sec 14. (a)

lim1 v œ lim1 9.8(sin ))t œ 9.8t so we expect v œ 9.8t m/sec in free fall

)Ä

(b) a œ

#

dv dt

)Ä

#

œ 9.8 m/sec# (b) between 3 and 6 seconds: $ Ÿ t Ÿ 6 (d)

15. (a) at 2 and 7 seconds (c)

16. (a) P is moving to the left when 2  t  3 or 5  t  6; P is moving to the right when 0  t  1; P is standing still when 1  t  2 or 3  t  5 (b)

17. (a) (c) (e) (f)

190 ft/sec at 8 sec, 0 ft/sec From t œ 8 until t œ 10.8 sec, a total of 2.8 sec Greatest acceleration happens 2 sec after launch

(g) From t œ 2 to t œ 10.8 sec; during this period, a œ

(b) 2 sec (d) 10.8 sec, 90 ft/sec

v(10.8)v(2) 10.82

¸
32 ft/sec#

18. (a) Forward: 0 Ÿ t  1 and 5  t  7; Backward: 1  t  5; Speeds up: 1  t  2 and 5  t  6; Slows down: 0 Ÿ t  1, 3  t  5, and 6  t  7 (b) Positive: 3  t  6; negative: 0 Ÿ t  2 and 6  t  7; zero: 2  t  3 and 7  t  9 (c) t œ 0 and 2 Ÿ t Ÿ 3 (d) 7 Ÿ t Ÿ 9 19. s œ 490t# Ê v œ 980t Ê a œ 980 (a) Solving 160 œ 490t# Ê t œ

4 7

sec. The average velocity was

s(4/7)s(0) 4/7

œ 280 cm/sec.

(b) At the 160 cm mark the balls are falling at v(4/7) œ 560 cm/sec. The acceleration at the 160 cm mark was 980 cm/sec# . 17 (c) The light was flashing at a rate of 4/7 œ 29.75 flashes per second.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

137

138

Chapter 3 Differentiation

20. (a)

(b)

21. C œ position, A œ velocity, and B œ acceleration. Neither A nor C can be the derivative of B because B's derivative is constant. Graph C cannot be the derivative of A either, because A has some negative slopes while C has only positive values. So, C (being the derivative of neither A nor B) must be the graph of position. Curve C has both positive and negative slopes, so its derivative, the velocity, must be A and not B. That leaves B for acceleration. 22. C œ position, B œ velocity, and A œ acceleration. Curve C cannot be the derivative of either A or B because C has only negative values while both A and B have some positive slopes. So, C represents position. Curve C has no positive slopes, so its derivative, the velocity, must be B. That leaves A for acceleration. Indeed, A is negative where B has negative slopes and positive where B has positive slopes. 23. (a) c(100) œ 11,000 Ê cav œ #

11,000 100

œ $110 w

(b) c(x) œ 2000  100x
.1x Ê c (x) œ 100
.2x. Marginal cost œ cw (x) Ê the marginal cost of producing 100 machines is cw (100) œ $80 (c) The cost of producing the 101st machine is c(101)
c(100) œ 100
201 10 œ $79.90 24. (a) r(x) œ 20000 ˆ1
"x ‰ Ê rw (x) œ (b) rw a"!!b œ

20000 100#

20000 x#

, which is marginal revenue.

œ $#Þ

(c) x lim rw (x) œ x lim Ä_ Ä_ will approach zero.

20000 x#

œ 0. The increase in revenue as the number of items increases without bound

25. b(t) œ 10'  10% t
10$ t# Ê bw (t) œ 10%
(2) a10$ tb œ 10$ (10
2t) (a) bw (0) œ 10% bacteria/hr (b) bw (5) œ 0 bacteria/hr w % (c) b (10) œ
10 bacteria/hr 26. Q(t) œ 200(30
t)# œ 200 a900
60t  t# b Ê Qw (t) œ 200(
60  2t) Ê Qw (10) œ
8,000 gallons/min is the rate the water is running at the end of 10 min. Then

Q(10)Q(0) 10

œ
10,000 gallons/min is the average rate the

water flows during the first 10 min. The negative signs indicate water is leaving the tank.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.3 The Derivative as a Rate of Change 27. (a) y œ 6 ˆ1


t ‰# 1#

œ 6 Š1


(b) The largest value of (c)

dy dt

t 6



t# 144 ‹

Ê

dy dt

œ

t 12

139


1

is 0 m/h when t œ 12 and the fluid level is falling the slowest at that time. The

smallest value of dy dt is
1 m/h, when t œ 0, and dy In this situation, dt Ÿ 0 Ê the graph of y is always decreasing. As dy dt increases in value,

the fluid level is falling the fastest at that time.

the slope of the graph of y increases from
1 to 0 over the interval 0 Ÿ t Ÿ 12.

28. (a) V œ

4 3

1 r$ Ê

(b) When r œ 2,

dV dr

dV dr

œ 41 r # Ê

dV ¸ dr r=2

œ 41(2)# œ 161 ft$ /ft

œ 161 so that when r changes by 1 unit, we expect V to change by approximately 161.

Therefore when r changes by 0.2 units V changes by approximately (161)(0.2) œ 3.21 ¸ 10.05 ft$ . Note that V(2.2)
V(2) ¸ 11.09 ft$ . 29. 200 km/hr œ 55 59 m/sec œ t œ 25, D œ

10 9

#

(25) œ

500 9 m/sec, 6250 9 m

and D œ

10 # 9 t

30. s œ v! t
16t# Ê v œ v!
32t; v œ 0 Ê t œ

v! 32

Ê Vœ

20 9

t. Thus V œ

500 9

Ê

; 1900 œ v! t
16t# so that t œ

Ê v! œ È(64)(1900) œ 80È19 ft/sec and, finally,

80È19 ft sec

†

60 sec 1 min

†

60 min 1 hr

†

1 mi 5280 ft

v! 32

20 9

tœ

500 9

Ê t œ 25 sec. When

Ê 1900 œ

v!# 3#




¸ 238 mph.

31.

v œ 0 when t œ 6.25 sec v  0 when 0 Ÿ t  6.25 Ê body moves up; v  0 when 6.25  t Ÿ 12.5 Ê body moves down body changes direction at t œ 6.25 sec body speeds up on (6.25ß 12.5] and slows down on [0ß 6.25) The body is moving fastest at the endpoints t œ 0 and t œ 12.5 when it is traveling 200 ft/sec. It's moving slowest at t œ 6.25 when the speed is 0. (f) When t œ 6.25 the body is s œ 625 m from the origin and farthest away. (a) (b) (c) (d) (e)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

v!# 64

140

Chapter 3 Differentiation

32.

(a) v œ 0 when t œ

3 #

sec

(b) v  0 when 0 Ÿ t  1.5 Ê body moves down; v  0 when 1.5  t Ÿ 5 Ê body moves up (c) body changes direction at t œ 3# sec (d) body speeds up on ˆ 3# ß &‘ and slows down on
!ß 3# ‰

(e) body is moving fastest at t œ 5 when the speed œ kv(5)k œ 7 units/sec; it is moving slowest at t œ 3# when the speed is 0 (f) When t œ 5 the body is s œ 12 units from the origin and farthest away.

33.

6 „ È15 3 6
È15 t 3

(a) v œ 0 when t œ

sec

(b) v  0 when



6  È15 3

Ê body moves left; v  0 when 0 Ÿ t 

6
È15 3

or

6  È15 3

tŸ4

Ê body moves right 6 „ È15 sec 3 È È Š 6
3 15 ß #‹  Š 6  3 15 ß %“

(c) body changes direction at t œ (d) body speeds up on

È15

and slows down on ’0ß 6
3

È15

‹  Š#ß 6  3

‹.

(e) The body is moving fastest at t œ 0 and t œ 4 when it is moving 7 units/sec and slowest at t œ (f) When t œ

6È15 3

the body is at position s ¸
6.303 units and farthest from the origin.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

6„È15 3

sec

Section 3.4 Derivatives of Trigonometric Functions

141

34.

6 „ È15 3 È È v  0 when 0 Ÿ t  6
3 15 or 6  3 15  t Ÿ 4 Ê body is moving left; v  0 when È 6
È15  t  6  3 15 Ê body is moving right 3 È body changes direction at t œ 6 „ 3 15 sec È È È È body speeds up on Š 6
3 15 ß #‹  Š 6  3 15 ß %“ and slows down on ’!ß 6
3 15 ‹  Š#ß 6  3 15 ‹

(a) v œ 0 when t œ (b)

(c) (d)

(e) The body is moving fastest at 7 units/sec when t œ 0 and t œ 4; it is moving slowest and stationary at tœ

6 „ È15 3

(f) When t œ

6  È15 3

the position is s ¸ 10.303 units and the body is farthest from the origin.

35. (a) It takes 135 seconds. (b) Average speed œ ??Ft œ

&
! ($
!

œ

& ($

¸ !Þ!') furlongs/sec.

(c) Using a symmetric difference quotient, the horse's speed is approximately

?F ?t

œ

%
# &*
$$

œ

# #'

¸ !Þ!(( furlongs/sec.

(d) The horse is running the fastest during the last furlong (between the 9th and 10th furlong markers). This furlong takes only 11 seconds to run, which is the least amount of time for a furlong. (e) The horse accelerates the fastest during the first furlong (between markers 0 and 1). 3.4 DERIVATIVES OF TRIGONOMETRIC FUNCTIONS 1. y œ
10x  3 cos x Ê

dy dx

œ
10  3

œ


3 x#

3. y œ csc x
4Èx  7 Ê

dy dx

2. y œ

3 x

 5 sin x Ê

4. y œ x# cot x


" x#

dy dx

Ê

dy dx

5

(cos x) œ
10
3 sin x

(sin x) œ


3 x#

œ
csc x cot x


œ x#

œ
x# csc# x  2x cot x 

d dx

d dx

d dx

 5 cos x

4 #È x

 0 œ
csc x cot x
ax# b 

2 x$

œ (sec x  tan x)

d dx

(cot x)  cot x †

d dx

2 Èx

œ
x# csc# x  (cot x)(2x) 

2 x$

2 x$

5. y œ (sec x  tan x)(sec x
tan x) Ê #

dy dx

(sec x
tan x)  (sec x
tan x) #

d dx

(sec x  tan x)

œ (sec x  tan x) asec x tan x
sec xb  (sec x
tan x) asec x tan x  sec xb œ asec# x tan x  sec x tan# x
sec$ x
sec# x tan xb  asec# x tan x
sec x tan# x  sec$ x
tan x sec# xb œ 0. ŠNote also that y œ sec# x
tan# x œ atan# x  1b
tan# x œ 1 Ê

dy dx

œ 0.‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

142

Chapter 3 Differentiation

6. y œ (sin x  cos x) sec x Ê

œ (sin x  cos x)

dy dx

d dx

(sec x)  sec x

œ (sin x  cos x)(sec x tan x)  (sec x)(cos x
sin x) œ œ

sin# x  cos x sin x  cos# x
cos x sin x cos# x

œ

" cos# x

d dx (sin x  cos x) (sin x  cos x) sin x x
sin x  cos cos cos# x x

œ sec# x

ŠNote also that y œ sin x sec x  cos x sec x œ tan x  1 Ê 7. y œ œ

(1  cot x)

œ

dy dx


csc# x
csc# x cot x  csc# x cot x (1  cot x)#

8. y œ œ

Ê

cot x 1  cot x

Ê

cos x 1  sin x


sin x
sin# x
cos# x (1  sin x)#

9. y œ

4 cos x



" tan x

10. y œ

cos x x



x cos x

œ

(cot x)
(cot x) (1  cot x)#

œ

(1  sin x)

œ

dy dx

d dx

œ

(1  cot x) a
csc# xb
(cot x) a
csc# xb (1  cot x)#

d (cos x)
(cos x) dx (1  sin x) b
(cos x) acos xb œ (1  sin x) a
(1sin xsin (1  sin x)# x)#
(1  sin x)
sin x
1
" (1  sin x)# œ (1  sin x)# œ 1  sin x

œ

dy dx

(1  cot x)

œ sec# x.‹


csc# x (1  cot x)#

d dx

œ 4 sec x  cot x Ê Ê

d dx

dy dx

œ 4 sec x tan x
csc# x

dy dx

x(
sin x)
(cos x)(1) x#

11. y œ x# sin x  2x cos x
2 sin x Ê #



(cos x)(1)
x(
sin x) cos# x

œ


x sin x
cos x x#



cos x  x sin x cos# x

dy dx

œ ax# cos x  (sin x)(2x)b  a(2x)(
sin x)  (cos x)(2)b
2 cos x

dy dx

œ ax# (
sin x)  (cos x)(2x)b
a2x cos x  (sin x)(2)b
2(
sin x)

œ x cos x  2x sin x
2x sin x  2 cos x
2 cos x œ x# cos x 12. y œ x# cos x
2x sin x
2 cos x Ê #

œ
x sin x  2x cos x
2x cos x
2 sin x  2 sin x œ
x# sin x 13. s œ tan t
t Ê

ds dt

œ

14. s œ t#
sec t  1 Ê 15. s œ œ 16. s œ œ

1  csc t 1
csc t

Ê

ds dt

œ

d dt

(tan t)
1 œ sec# t
1 œ tan# t

ds dt

œ 2t


d dt

(sec t) œ 2t
sec t tan t

(1
csc t)(
csc t cot t)
("  csc t)(csc t cot t) (1
csc t)#


csc t cot t  csc# t cot t
csc t cot t
csc# t cot t (1
csc t)# sin t 1
cos t " cos t
1

Ê

ds dt

œ

17. r œ 4
)# sin ) Ê

œ


2 csc t cot t (1
csc t)#

(1
cos t)(cos t)
(sin t)(sin t) (1
cos t)#

dr d)

18. r œ ) sin )  cos ) Ê

œ
ˆ) # dr d)

d d)

œ

cos t
cos# t
sin# t (1
cos t)#

œ

cos t
" (1
cos t)#

œ
1
"cos t

(sin ))  (sin ))(2))‰ œ
a)# cos )  2) sin )b œ
)() cos )  # sin ))

œ () cos )  (sin ))(1))
sin ) œ ) cos )

dr 19. r œ sec ) csc ) Ê d) œ (sec ))(
csc ) cot ))  (csc ))(sec ) tan ))
" "
" " cos ) sin ) ‰ # # œ ˆ cos ) ‰ ˆ sin ) ‰ ˆ sin ) ‰  ˆ sin" ) ‰ ˆ cos" ) ‰ ˆ cos ) œ sin# )  cos# ) œ sec )
csc )

20. r œ (1  sec )) sin ) Ê 21. p œ & 

" cot q

dr d)

œ ("  sec )) cos )  (sin ))(sec ) tan )) œ (cos )  ")  tan# ) œ cos )  sec# )

œ 5  tan q Ê

22. p œ (1  csc q) cos q Ê

dp dq

dp dq

œ sec# q

œ (1  csc q)(
sin q)  (cos q)(
csc q cot q) œ (
sin q
1)
cot# q œ
sin q
csc# q

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.4 Derivatives of Trigonometric Functions 23. p œ œ

sin q  cos q cos q

Ê

œ

dp dq

(cos q)(cos q
sin q)
(sin q  cos q)(
sin q) cos# q

cos# q
cos q sin q  sin# q  cos q sin q cos# q

24. p œ

tan q 1  tan q

Ê

dp dq

œ

œ

" cos# q

œ sec# q

(1  tan q) asec# qb
(tan q) asec# qb (1  tan q)#

œ

sec# q  tan q sec# q
tan q sec# q (1  tan q)#

œ

sec# q (1  tan q)#

25. (a) y œ csc x Ê yw œ
csc x cot x Ê yww œ
a(csc x) a
csc# xb  (cot x)(
csc x cot x)b œ csc$ x  csc x cot# x œ (csc x) acsc# x  cot# xb œ (csc x) acsc# x  csc# x
1b œ 2 csc$ x
csc x (b) y œ sec x Ê yw œ sec x tan x Ê yww œ (sec x) asec# xb  (tan x)(sec x tan x) œ sec$ x  sec x tan# x œ (sec x) asec# x  tan# xb œ (sec x) asec# x  sec# x
1b œ 2 sec$ x
sec x 26. (a) y œ
2 sin x Ê yw œ
2 cos x Ê yww œ
2(
sin x) œ 2 sin x Ê ywww œ 2 cos x Ê yÐ%Ñ œ
2 sin x (b) y œ 9 cos x Ê yw œ
9 sin x Ê yww œ
9 cos x Ê ywww œ
9(
sin x) œ 9 sin x Ê yÐ%Ñ œ 9 cos x 27. y œ sin x Ê yw œ cos x Ê slope of tangent at x œ
1 is yw (
1) œ cos (
1) œ
"; slope of tangent at x œ 0 is yw (0) œ cos (0) œ 1; and slope of tangent at x œ 3#1 is yw ˆ 3#1 ‰ œ cos 3#1

œ 0. The tangent at (
1ß !) is y
0 œ
1(x  1), or y œ
x
1; the tangent at (0ß 0) is y
0 œ 1(x
0), or y œ x; and the tangent at ˆ 31 ‰ # ß
1 is y œ
1.

28. y œ tan x Ê yw œ sec# x Ê slope of tangent at x œ
13 is sec# ˆ
13 ‰ œ 4; slope of tangent at x œ 0 is sec# (0) œ 1; and slope of tangent at x œ

1 3

is sec# ˆ 13 ‰ œ 4. The tangent

at ˆ
13 ß tanˆ
13 ‰‰ œ Š
13 ß
È3‹ is y  È3 œ 4ˆx  13 ‰ ; the tangent at (0ß 0) is y œ x; and the tangent at ˆ 13 ß tan ˆ 13 ‰‰ œ Š 13 ß È3‹ is y
È3 œ 4 ˆx
13 ‰ . 29. y œ sec x Ê yw œ sec x tan x Ê slope of tangent at x œ
13 is sec ˆ
13 ‰ tan ˆ
13 ‰ œ
2È3 ; slope of tangent is sec ˆ 14 ‰ tan ˆ 14 ‰ œ È2 . The tangent at the point ˆ
1 ß sec ˆ
1 ‰‰ œ ˆ
1 ß #‰ is y
2 œ
#È3 ˆx  1 ‰ ; at x œ

1 4

3

3

3

3

the tangent at the point ˆ 14 ß sec ˆ 14 ‰‰ œ Š 14 ß È2‹ is y
È2 œ È2 ˆx
14 ‰ .

30. y œ 1  cos x Ê yw œ
sin x Ê slope of tangent at È

x œ
13 is
sin ˆ
13 ‰ œ #3 ; slope of tangent at x œ ‰ œ 1. The tangent at the point is
sin ˆ 31 #

31 #

ˆ
13 ß "  cos ˆ
13 ‰‰ œ ˆ
13 ß 3# ‰ È

is y
3# œ #3 ˆx  13 ‰ ; the tangent at the point ˆ 3#1 ß "  cos ˆ 3#1 ‰‰ œ ˆ 3#1 ß 1‰ is y
1 œ x
3#1

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143

144

Chapter 3 Differentiation

31. Yes, y œ x  sin x Ê yw œ "  cos x; horizontal tangent occurs where 1  cos x œ 0 Ê cos x œ
1 Ê xœ1 32. No, y œ 2x  sin x Ê yw œ 2  cos x; horizontal tangent occurs where 2  cos x œ 0 Ê cos x œ
#. But there are no x-values for which cos x œ
#. 33. No, y œ x
cot x Ê yw œ 1  csc# x; horizontal tangent occurs where 1  csc# x œ 0 Ê csc# x œ
1. But there are no x-values for which csc# x œ
1. 34. Yes, y œ x  2 cos x Ê yw œ 1
2 sin x; horizontal tangent occurs where 1
2 sin x œ 0 Ê 1 œ 2 sin x Ê "# œ sin x Ê x œ 16 or x œ 561 35. We want all points on the curve where the tangent line has slope 2. Thus, y œ tan x Ê yw œ sec# x so that yw œ 2 Ê sec# x œ 2 Ê sec x œ „ È2 Ê x œ „ 14 . Then the tangent line at ˆ 14 ß "‰ has equation y
1 œ 2 ˆx
14 ‰ ; the tangent line at ˆ
14 ß
"‰ has equation y  1 œ 2 ˆx  14 ‰ .

36. We want all points on the curve y œ cot x where the tangent line has slope
1. Thus y œ cot x Ê yw œ
csc# x so that yw œ
1 Ê
csc# x œ
1 Ê csc# x œ 1 Ê csc x œ „ 1 Ê x œ 1# . The tangent line at ˆ 1# ß !‰ is y œ
x  12 .

2 cos x ‰ 37. y œ 4  cot x
2 csc x Ê yw œ
csc# x  2 csc x cot x œ
ˆ sin" x ‰ ˆ 1
sin x

(a) When x œ 1# , then yw œ
1; the tangent line is y œ
x  w

1 #

 2.

(b) To find the location of the horizontal tangent set y œ 0 Ê 1
2 cos x œ 0 Ê x œ then y œ %
È3 is the horizontal tangent. 38. y œ 1  È2 csc x  cot x Ê yw œ
È2 csc x cot x
csc# x œ
ˆ sin" x ‰ Š

1 3

È2 cos x  1 ‹ sin x

(a) If x œ 14 , then yw œ
4; the tangent line is y œ
4x  1  4. (b) To find the location of the horizontal tangent set yw œ 0 Ê È2 cos x  1 œ 0 Ê x œ xœ

31 4 ,

radians. When x œ 13 ,

31 4

radians. When

then y œ 2 is the horizontal tangent.

39. lim sin ˆ "x
#" ‰ œ sin ˆ #"
#" ‰ œ sin 0 œ 0 xÄ2

40.

lim

x Ä
16

È1  cos (1 csc x) œ É1  cos ˆ1 csc ˆ
1 ‰‰ œ È1  cos a1 † a
2bb œ È2 6

1 ‰ ‘
ˆ 1 ‰ ‘
ˆ1‰ ‘ 41. lim sec
cos x  1 tan ˆ 4 sec x
1 œ sec cos 0  1 tan 4 sec 0
1 œ sec 1  1 tan 4
1 œ sec 1 œ
1

xÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.4 Derivatives of Trigonometric Functions x ‰ ˆ 1tan 0 ‰ ˆ 1‰ 42. lim sin ˆ tan1x
tan 2 sec x œ sin tan 0
2 sec 0 œ sin
# œ
1

xÄ!

43. lim tan ˆ1
tÄ!

sin t ‰ t

œ tan Š1
lim

tÄ!

1) ‰ 44. lim cos ˆ sin ) œ cos Š1 lim

)

sin t t ‹

‹ ) Ä ! sin )

)Ä!

œ tan (1
1) œ 0 "

œ cos Œ1 †

lim

"  œ cos ˆ1 † 1 ‰ œ
1

sin )

)Ä!

)

dv da ˆ1‰ 45. s œ #
# sin t Ê v œ ds dt œ
2 cos t Ê a œ dt œ 2 sin t Ê j œ dt œ 2 cos t. Therefore, velocity œ v 4 œ
È2 m/sec; speed œ ¸v ˆ 1 ‰¸ œ È2 m/sec; acceleration œ a ˆ 1 ‰ œ È2 m/sec# ; jerk œ j ˆ 1 ‰ œ È2 m/sec$ . 4

46. s œ sin t  cos t Ê v œ

4

œ cos t
sin t Ê a œ

ds dt

v ˆ 14 ‰

velocity œ œ 0 m/sec; speed œ 1‰ ˆ jerk œ j 4 œ 0 m/sec$ . 47. lim f(x) œ lim xÄ!

Ê 9 œ c.

48.

xÄ!

sin# 3x x#

¸v ˆ 14 ‰¸

dv dt

4

œ
sin t
cos t Ê j œ

œ 0 m/sec; acceleration œ

a ˆ 14 ‰

œ
cos t  sin t. Therefore È œ
2 m/sec# ; da dt

œ lim 9 ˆ sin3x3x ‰ ˆ sin3x3x ‰ œ 9 so that f is continuous at x œ 0 Ê lim f(x) œ f(0) xÄ!

xÄ!

lim g(x) œ lim c (x  b) œ b and lim b g(x) œ lim b cos x œ 1 so that g is continuous at x œ 0 Ê lim c g(x) xÄ! xÄ! xÄ! xÄ! œ lim b g(x) Ê b œ 1. Now g is not differentiable at x œ 0: At x œ 0, the left-hand derivative is

x Ä !c

xÄ!

d dx

(x  b)¸ x=0 œ 1, but the right-hand derivative is

d dx

(cos x)¸ x=0 œ
sin 0 œ 0. The left- and right-hand

derivatives can never agree at x œ 0, so g is not differentiable at x œ 0 for any value of b (including b œ 1). 49.

d*** dx***

d% dx%

(cos x) œ sin x because

(cos x) œ cos x Ê the derivative of cos x any number of times that is a

multiple of 4 is cos x. Thus, dividing 999 by 4 gives 999 œ 249 † 4  3 Ê œ

d$ dx$

d#%*†% ’ dx #%*†%

(cos x)“ œ

50. (a) y œ sec x œ Ê

d dx

" cos x

d$ dx$

Ê

d*** dx***

(cos x)

(cos x) œ sin x.

dy dx

œ

(cos x)(0)
(1)(
sin x) (cos x)#

œ

(sin x)(0)
(1)(cos x) (sin x)#

œ

sin x cos# x

sin x ‰ œ ˆ cos" x ‰ ˆ cos x œ sec x tan x

(sec x) œ sec x tan x

(b) y œ csc x œ Ê

d dx

d dx

Ê

dy dx

œ


cos x sin# x


" ‰ ˆ cos x ‰ œ ˆ sin x sin x œ
csc x cot x

(csc x) œ
csc x cot x

(c) y œ cot x œ Ê

" sin x

cos x sin x

Ê

dy dx #

œ

(sin x)(
sin x)
(cos x)(cos x) (sin x)#

œ


sin# x
cos# x sin# x

œ


" sin# x

œ
csc# x

(cot x) œ
csc x

51.

As h takes on the values of 1, 0.5, 0.3 and 0.1 the corresponding dashed curves of y œ closer and closer to the black curve y œ cos x because

d dx

(sin x) œ

is true as h takes on the values of
1,
0.5,
0.3 and
0.1.


sin x lim sin (x  h) h hÄ!

sin (x  h)
sin x h

get

œ cos x. The same

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

145

146

Chapter 3 Differentiation

52.

cos (x  h)
cos x h
cos x lim cos (x  h) œ
sin x. h hÄ!

As h takes on the values of 1, 0.5, 0.3, and 0.1 the corresponding dashed curves of y œ

get

closer and closer to the black curve y œ
sin x because

The

d dx

(cos x) œ

same is true as h takes on the values of
1,
0.5,
0.3, and
0.1. 53. (a)

The dashed curves of y œ

sinax  hb
sinax
hb #h

are closer to the black curve y œ cos x than the corresponding dashed

curves in Exercise 51 illustrating that the centered difference quotient is a better approximation of the derivative of this function. (b)

The dashed curves of y œ

cosax  hb
cosax
hb #h

are closer to the black curve y œ
sin x than the corresponding dashed

curves in Exercise 52 illustrating that the centered difference quotient is a better approximation of the derivative of this function. 54. lim

hÄ!

k0  h k
k 0
h k 2h

œ lim

xÄ!

k h k
k hk 2h

œ lim 0 œ 0 Ê the limits of the centered difference quotient exists even hÄ!

though the derivative of f(x) œ kxk does not exist at x œ 0. 55. y œ tan x Ê yw œ sec# x, so the smallest value yw œ sec# x takes on is yw œ 1 when x œ 0; yw has no maximum value since sec# x has no largest value on ˆ
1# ß 1# ‰ ; yw is never negative since sec# x   1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.4 Derivatives of Trigonometric Functions 56. y œ cot x Ê yw œ
csc# x so yw has no smallest value since
csc# x has no minimum value on (!ß 1); the largest value of yw is
1, when x œ 1# ; the slope is never positive since the largest value yw œ
csc2 x takes on is
1.

57. y œ

sin x x appears to cross the y-axis at y œ 1, since lim sin x œ 1; y œ sinx2x appears to cross the y-axis xÄ! x

at y œ 2, since lim sinx2x œ 2; y œ sinx4x appears to xÄ! cross the y-axis at y œ 4, since lim sinx4x œ 4. xÄ!

However, none of these graphs actually cross the y-axis since x œ 0 is not in the domain of the functions. Also, lim

xÄ!

sin 5x x

sin (
3x) x

œ 5, lim

xÄ!

œ k Ê the graphs of y œ yœ

sin kx x

œ
3, and lim

sin kx x

xÄ!

yœ

sin 5x x ,

sin (
3x) , x

and

approach 5,
3, and k, respectively, as

x Ä 0. However, the graphs do not actually cross the y-axis. 58. (a)

sin h h

h 1 0.01 0.001 0.0001

ˆ sinh h ‰ ˆ 180 ‰ 1 .99994923 1 1 1

.017452406 .017453292 .017453292 .017453292 1 ‰ sin ˆh† 180 h xÄ!

œ lim

sin h° h hÄ!

lim

œ lim

1

180

hÄ!

1 ‰ sin ˆh† 180 1 †h 180

œ lim

1 sin )

180

)Ä!

)

œ

1 180

() œ h †

1 180 )

(converting to radians) (b)

cos h
1 h

h 1 0.01 0.001 0.0001 lim

hÄ!


0.0001523
0.0000015
0.0000001 0

cos h
1 h

(c) In degrees,

œ 0, whether h is measured in degrees or radians.

d dx

(sin x) œ lim

hÄ!

œ lim ˆsin x † hÄ!

cos h
1 ‰ h

sin (x  h)
sin x h

 lim ˆcos x † hÄ!

œ lim

hÄ!

sin h ‰ h

(sin x cos h  cos x sin h)
sin x h

œ (sin x) † lim ˆ cos hh
1 ‰  (cos x) † lim ˆ sinh h ‰ hÄ!

1 ‰ œ (sin x)(0)  (cos x) ˆ 180 œ

(d)

1 180 cos x
cos x d In degrees, dx (cos x) œ lim cos (x  h) œ lim (cos x cos h
sinh x sin h)
cos x h hÄ! hÄ! (cos x)(cos h
1)
sin x sin h ˆ œ lim œ lim cos x † cos hh
1 ‰
lim ˆsin x † sinh h ‰ h hÄ! hÄ! hÄ!

œ (cos x) lim ˆ hÄ!

(e)

d# dx# d# dx#

cos h
1 ‰ h

hÄ!

1 ‰ 1
(sin x) lim ˆ sinh h ‰ œ (cos x)(0)
(sin x) ˆ 180 œ
180 sin x

hÄ!

(sin x) œ

d dx

1 1 ‰# ˆ 180 cos x‰ œ
ˆ 180 sin x;

(cos x) œ

d dx

1 1 ‰# ˆ
180 sin x‰ œ
ˆ 180 cos x;

d$ dx$

(sin x) œ d$ dx$

d dx

(cos x) œ

#

$

1 ‰ 1 ‰ sin x‹ œ
ˆ 180 cos x; Š
ˆ 180 d dx

#

$

1 ‰ 1 ‰ cos x‹ œ ˆ 180 sin x Š
ˆ 180

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147

148

Chapter 3 Differentiation

3.5 THE CHAIN RULE AND PARAMETRIC EQUATIONS 1. f(u) œ 6u
9 Ê f w (u) œ 6 Ê f w (g(x)) œ 6; g(x) œ œ 6 † 2x$ œ 12x$

" #

x% Ê gw (x) œ 2x$ ; therefore

dy dx

œ f w (g(x))gw (x)

2. f(u) œ 2u$ Ê f w (u) œ 6u# Ê f w (g(x)) œ 6(8x
1)# ; g(x) œ 8x
1 Ê gw (x) œ 8; therefore œ 6(8x
1)# † 8 œ 48(8x
1)#

dy dx

œ f w (g(x))gw (x)

3. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (3x  1); g(x) œ 3x  1 Ê gw (x) œ 3; therefore

œ f w (g(x))gw (x)

dy dx

œ (cos (3x  1))(3) œ 3 cos (3x  1) 4. f(u) œ cos u Ê f w (u) œ
sin u Ê f w (g(x)) œ
sin ˆ
3x ‰ ; g(x) œ ‰ œ "3 sin ˆ
3x ‰ œ
sin ˆ
3x ‰ † ˆ
" 3


x 3

Ê gw (x) œ
"3 ; therefore

dy dx

œ f w (g(x))gw (x)

5. f(u) œ cos u Ê f w (u) œ
sin u Ê f w (g(x)) œ
sin (sin x); g(x) œ sin x Ê gw (x) œ cos x; therefore dy dx

œ f w (g(x))gw (x) œ
(sin (sin x)) cos x

6. f(u) œ sin u Ê f w (u) œ cos u Ê f w (g(x)) œ cos (x
cos x); g(x) œ x
cos x Ê gw (x) œ 1  sin x; therefore dy dx

œ f w (g(x))gw (x) œ (cos (x
cos x))(1  sin x)

7. f(u) œ tan u Ê f w (u) œ sec# u Ê f w (g(x)) œ sec# (10x
5); g(x) œ 10x
5 Ê gw (x) œ 10; therefore dy dx

œ f w (g(x))gw (x) œ asec# (10x
5)b (10) œ 10 sec# (10x
5)

8. f(u) œ
sec u Ê f w (u) œ
sec u tan u Ê f w (g(x)) œ
sec ax#  7xb tan ax#  7xb ; g(x) œ x#  7x Ê gw (x) œ 2x  7; therefore

œ f w (g(x))gw (x) œ
(2x  7) sec ax#  7xb tan ax#  7xb

dy dx

9. With u œ (2x  1), y œ u& :

dy dx

œ

dy du du dx

œ 5u% † 2 œ 10(2x  1)%

10. With u œ (4
3x), y œ u* :

dy dx

œ

dy du du dx

œ 9u) † (
3) œ
27(4
3x))

11. With u œ ˆ1
x7 ‰ , y œ ?
( :

œ

dy dx

12. With u œ ˆ x#
1‰ , y œ ?
"! :

dy dx

œ

#

13. With u œ Š x8  x
"x ‹ , y œ ?% : 14. With u œ ˆ x5 

" ‰ 5x ,

y œ ?& :

15. With u œ tan x, y œ sec u: 16. With u œ 1
"x , y œ cot u: 17. With u œ sin x, y œ u$ :

dy dx

dy dx dy dx

œ œ

dy du du dx

dy dx

œ

œ

dy du du dx

œ

dy du du dx

18. With u œ cos x, y œ 5u
% :

dy dx

œ

œ
10u
"" † ˆ "# ‰ œ
5 ˆ x#
1‰

dy du du dx

dy du du dx

dy du du dx

dy dx


) œ
7u
) † ˆ
"7 ‰ œ ˆ"
x7 ‰

dy du du dx

œ 4u$ † ˆ x4  1 

œ 5u% † ˆ 15


" ‰ 5x#

"‰ x#

œ ˆ x5 


""

$

#

œ 4 Š x8  x
x" ‹ ˆ x4  1  " ‰% 5x

ˆ1


"‰ x#

œ (sec u tan u) asec# xb œ (sec (tan x) tan (tan x)) sec# x œ a
csc# ub ˆ x"# ‰ œ
x"# csc# ˆ1
x" ‰

œ 3u# cos x œ 3 asin# xb (cos x)

dy du du dx

œ a
20u
& b (
sin x) œ 20 acos
& xb (sin x)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

"‰ x#

Section 3.5 The Chain Rule and Parametric Equations 19. p œ È3
t œ (3
t)"Î# Ê 20. q œ È2r
r# œ a2r
r# b 21. s œ œ

4 31 4 1

sin 3t 

4 51

dp dt

"Î#

Ê

cos 5t Ê

23. r œ (csc )  cot ))
" Ê

dr d)

dq dr

(3
t)
"Î# † œ

" #

d dt

a2r
r# b

(3
t) œ
"# (3
t)
"Î# œ


"Î#

†

d dr

ds dt

œ

cos 3t †

d dt

(3t) 

ds dt

œ cos ˆ 3#1t ‰ †

d dt

ˆ 3#1t ‰
sin ˆ 3#1t ‰ †

4 31

4 51

" #

a2r
r# b œ

(
sin 5t) †

a2r
r# b

(5t) œ

d dt


" 2È 3
t

4 1


"Î#

"
r È2r
r#

(2
2r) œ

cos 3t


4 1

sin 5t

d dt

ˆ 3#1t ‰ œ

31 2

cos ˆ 3#1t ‰


31 2

sin ˆ 3#1t ‰

œ
(csc )  cot ))
#

d d)

(csc )  cot )) œ

csc ) cot )  csc# ) (csc )  cot ))#

œ

csc ) (cot )  csc )) (csc )  cot ))#

œ (sec )  tan ))
#

d d)

(sec )  tan )) œ

sec ) tan )  sec# ) (sec )  tan ))#

œ

sec ) (tan )  sec )) (sec )  tan ))#

csc ) csc )  cot )

24. r œ
(sec )  tan ))
" Ê œ

" #

(cos 3t
sin 5t)

22. s œ sin ˆ 3#1t ‰  cos ˆ 3#1t ‰ Ê œ 321 ˆcos 3#1t
sin 3#1t ‰

œ

œ

dr d)

sec ) sec )  tan )

d d # d % % #
# 25. y œ x# sin% x  x cos
# x Ê dy xb  cos
# x † dx œ x dx asin xb  sin x † dx ax b  x dx acos d d œ x# ˆ4 sin$ x dx (sin x)‰  2x sin% x  x ˆ
2 cos
$ x † dx (cos x)‰  cos
# x

d dx

(x)

œ x# a4 sin$ x cos xb  2x sin% x  xa a
2 cos
$ xb (
sin x)b  cos
# x œ 4x# sin$ x cos x  2x sin% x  2x sin x cos
$ x  cos
# x d ˆ"‰ x d $ $ asin
& xb  sin
& x † dx x
3 dx acos xb
cos x † œ "x a
5 sin
' x cos xb  asin
& xb ˆ
x"# ‰
3x a a3 cos# xb (
sin x)b
acos$ xb ˆ 3" ‰

26. y œ

" x

sin
& x


x 3

cos$ x Ê yw œ

œ
5x sin
' x cos x


" x#

" d x dx

sin
& x  x cos# x sin x


" 3

ˆ x3 ‰

cos$ x

" " ‰
" 7 d ( ' ˆ Ê dy 21 (3x
2)  4
#x# dx œ 21 (3x
2) † dx (3x
2)  7 " ‰
# ˆ " ‰ " ' ' ˆ # 21 (3x
2) † 3  (
1) 4
#x# x$ œ (3x
2)
$ x Š4
"# ‹

27. y œ œ

d dx

(
1) ˆ4


" ‰
# #x #

†

d dx

ˆ4


" ‰ #x #

#x

% 28. y œ (5
2x)
$  "8 ˆ 2x  1‰ Ê

dy dx

$ $ œ
3(5
2x)
% (
2)  84 ˆ x2  1‰ ˆ
x2# ‰ œ 6(5
2x)
%
ˆ x"# ‰ ˆ 2x  1‰

$

œ

6 (5
2x)%




Š 2x  1‹ x#

29. y œ (4x  3)% (x  1)
$ Ê %

dy dx

œ (4x  3)% (
3)(x  1)
% †


%


$

d dx

(x  1)  (x  1)
$ (4)(4x  3)$ †

$

%

œ (4x  3) (
3)(x  1) (1)  (x  1) (4)(4x  3) (4) œ
3(4x  3) (x  1) œ

$

(4x  3) (x  1)%

c
3(4x  3)  16(x  1)d œ '

30. y œ (2x
5)
" ax#
5xb Ê &

œ 6 ax#
5xb


2 ax#
5xb (2x
5)#

dy dx


%

d dx $

(4x  3)

 16(4x  3) (x  1)
$

$

(4x  3) (4x  7) (x  1)% &

'

œ (2x
5)
" (6) ax#
5xb (2x
5)  ax#
5xb (
1)(2x
5)
# (2)

'

d ˆ d 31. h(x) œ x tan ˆ2Èx‰  7 Ê hw (x) œ x dx tan ˆ2x"Î# ‰‰  tan ˆ2x"Î# ‰ † dx (x)  0 " d ˆ "Î# ‰ # ˆ "Î# ‰ "Î# # † dx 2x œ x sec 2x  tan ˆ2x ‰ œ x sec ˆ2Èx‰ † È  tan ˆ2Èx‰ œ Èx sec# ˆ2Èx‰  tan ˆ2Èx‰ x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

149

150

Chapter 3 Differentiation

d ˆ d 32. k(x) œ x# sec ˆ "x ‰ Ê kw (x) œ x# dx sec x" ‰  sec ˆ x" ‰ † dx ax# b œ x# sec ˆ x" ‰ tan ˆ x" ‰ † œ x# sec ˆ "x ‰ tan ˆ x" ‰ † ˆ
x"# ‰  2x sec ˆ x" ‰ œ 2x sec ˆ x" ‰
sec ˆ x" ‰ tan ˆ x" ‰ #

33. f()) œ ˆ 1 sincos) ) ‰ Ê f w ()) œ 2 ˆ 1 sincos) ) ‰ † œ

(2 sin )) acos )  cos# )  sin# )b (1  cos ))$

34. g(t) œ ˆ 1 sincost t ‰ œ


"

(2 sin )) (cos )  1) (1  cos ))$

œ

Ê gw (t) œ
ˆ 1 sincost t ‰


a
sin# t
cos t
cos# tb (1  cos t)#


#

†

(1  cos ))(cos ))
(sin ))(
sin )) (1  cos ))#

2 sin ) (1  cos ))#

œ †

2 sin ) 1  cos )

d dt

#

ˆ 1 sincost t ‰ œ
(1 sincost t)# †

(sin t)(
sin t)
("  cos t)(cos t) (sin t)#

" 1  cos t

œ

35. r œ sin a)# b cos (2)) Ê

ˆ 1 sincos) ) ‰ œ

d d)

ˆ x" ‰  2x sec ˆ x" ‰

d dx

œ sin a)# b (
sin 2))

dr d)

(2))  cos (2)) acos a)# bb †

d d)

d d)

a) # b

œ sin a)# b (
sin 2))(2)  (cos 2)) acos a)# bb (2)) œ
2 sin a)# b sin (#))  2) cos (2)) cos a)# b 36. r œ Šsec È)‹ tan ˆ ") ‰ Ê

dr d)

œ
)"# sec È) sec# ˆ ") ‰  37. q œ sin Š Ètt 1 ‹ Ê œ cos Š Ètt 1 ‹ †

39. y œ sin# (1t
2) Ê

Èt
1 t

2

t1

dq dt

" #È )

tan ˆ ") ‰ sec È) tan È) œ Šsec È)‹ ”

œ cos Š Ètt 1 ‹ †

dq dt

Èt  1


38. q œ cot ˆ sint t ‰ Ê

œ Šsec È)‹ ˆ sec# ") ‰ ˆ
)"# ‰  tan ˆ ") ‰ Šsec È) tan È)‹ Š

d dt

Èt  1 (1)
t †

d dt




sec# ˆ )" ‰ )# •

ˆÈ t  1 ‰

ˆÈ t  1 ‰

#

 1)
t œ cos Š Ètt 1 ‹ Š 2(t ‹ œ Š 2(tt1)2$Î# ‹ cos Š Ètt 1 ‹ 2(t  1)$Î#

œ
csc# ˆ sint t ‰ †

d dt

ˆ sint t ‰ œ ˆ
csc# ˆ sint t ‰‰ ˆ t cos tt#
sin t ‰

œ 2 sin (1t
2) †

dy dt

Š Ètt 1 ‹ œ cos Š Ètt 1 ‹ †

tan È) tan ˆ ") ‰ #È )

" ‹ #È )

d dt

sin (1t
2) œ 2 sin (1t
2) † cos (1t
2) †

d dt

(1t
2)

œ 21 sin (1t
2) cos (1t
2) 40. y œ sec# 1t Ê

dy dt

œ (2 sec 1t) †

41. y œ (1  cos 2t)
% Ê 42. y œ ˆ1  cot ˆ #t ‰‰ œ


#

dy dt

Ê

(sec 1t) œ (2 sec 1t)(sec 1t tan 1t) †

œ
4(1  cos 2t)
& † dy dt

ˆ #t ‰ $ ˆ1  cot ˆ t ‰‰ # csc#

43. y œ sin acos (2t
5)b Ê

d dt

dy dt

œ
2 ˆ1  cot ˆ #t ‰‰

(1t) œ 21 sec# 1t tan 1t

(1  cos 2t) œ
4(1  cos 2t)
& (
sin 2t) †

d dt
$

œ cos (cos (2t
5)) †

d dt

 cot ˆ #t ‰‰ œ
2 ˆ1  cot ˆ #t ‰‰


$

d dt

(2t) œ

8 sin 2t (1  cos 2t)&

† ˆ
csc# ˆ #t ‰‰ †

†

dˆ dt 1

d dt

cos (2t
5) œ cos (cos (2t
5)) † (
sin (2t
5)) †

d dt

dˆt‰ dt #

(2t
5)

œ
2 cos (cos (2t
5))(sin (2t
5)) ˆ ˆ t ‰‰ † 44. y œ cos ˆ5 sin ˆ 3t ‰‰ Ê dy dt œ
sin 5 sin 3 5 t t œ
3 sin ˆ5 sin ˆ 3 ‰‰ ˆcos ˆ 3 ‰‰

d dt

$

dy % ˆ t ‰‘# d

† dt 1 dt œ 3 1  tan 1# # tan% ˆ 1t# ‰‘
tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰ † 1"# ‘ œ
1 

45. y œ
1  tan% ˆ 1t# ‰‘ Ê œ 12
1  46. y œ

" 6

ˆ5 sin ˆ 3t ‰‰ œ
sin ˆ5 sin ˆ 3t ‰‰ ˆ5 cos ˆ 3t ‰‰ †

$

c1  cos# (7t)d Ê

dy dt

œ

3 6

#

d dt

ˆ 3t ‰

#  tan% ˆ 1t# ‰‘ œ 3
1  tan% ˆ 1t# ‰‘
4 tan$ ˆ 1t# ‰ †

# tan% ˆ 1t# ‰‘
tan$ ˆ 1t# ‰ sec# ˆ 1t# ‰‘

#

d dt

tan ˆ 1t# ‰‘

c1  cos# (7t)d † 2 cos (7t)(
sin (7t))(7) œ
7 c1  cos# (7t)d (cos (7t) sin (7t))

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.5 The Chain Rule and Parametric Equations "Î#

Ê

œ
"# a1  cos at# bb


"Î#

47. y œ a1  cos at# bb

dy dt

2 cos ŒÉ1  Èt
É1  Èt†2Èt

œ

a1  cos at# bb


"Î#

†

d dt

a1  cos at# bb œ

#

" #

a1  cos at# bb

œ 4 cos ŒÉ1  Èt †

dy dt

" ‰# x

œ

6 x%

œ

" #x

51. y œ

" 9

œ


"

ˆ1
Èx‰
$ Š


ˆ1  x" ‰ 

" #

6 x$

ˆ1 

" ‰# x

œ

d dx 6 x$

ˆ1  x" ‰#
ˆ1  x" ‰# †

d dt

ˆ1  Èt‰

d dx

ˆ x3# ‰

ˆ1  x" ‰ ˆ x"  1  x" ‰

ˆ1
Èx‰
# x
"Î#

" #È x



" #

 1‹ œ

" #x

" #


$ x
" ˆ1
Èx‰

"# x
"Î# ˆ1
Èx‰  1‘

ˆ1
Èx‰
$ Š 3#


" ‹ #Èx

cot (3x
1) Ê yw œ
9" csc# (3x
1)(3) œ
3" csc# (3x
1) Ê yww œ ˆ
32 ‰ (csc (3x
1) † 2 3

†

# É 1 È t


#
$ ’ˆ1
Èx‰ ˆ
"# x
$Î# ‰  x
"Î# (
2) ˆ1
Èx‰ ˆ
"# x
"Î# ‰“


#
$
$Î# ˆ 1
Èx‰  x
" ˆ1
Èx‰ “ œ ’
" # x

" #

a t# b ‰

É t  Èt


"
# 50. y œ ˆ1
Èx‰ Ê yw œ
ˆ1
Èx‰ ˆ
"# x
"Î# ‰ œ

œ

d dt

cos ŒÉ1  Èt


œ ˆ
x3# ‰ ˆ2 ˆ1  x" ‰ ˆ
x"# ‰‰  ˆ x6$ ‰ ˆ1  œ x6$ ˆ1  x" ‰ ˆ1  2x ‰

" #

ˆ
sin at# b †

ŒÉ1  Èt œ 4 cos ŒÉ1  Èt †

d dt

$ # # 49. y œ ˆ1  "x ‰ Ê yw œ 3 ˆ1  x" ‰ ˆ
x"# ‰ œ
x3# ˆ1  x" ‰ Ê yww œ ˆ
x3# ‰ †

Ê yww œ


"Î#

at b asin at# bb † 2t œ
È1t sin  cos at# b

48. y œ 4 sin ŒÉ1  Èt Ê œ

" #

œ

151

csc (3x
1)(
csc (3x
1) cot (3x
1) †

d dx

csc (3x
1))

d dx

#

(3x
1)) œ 2 csc (3x
1) cot (3x
1)

52. y œ 9 tan ˆ x3 ‰ Ê yw œ 9 ˆsec# ˆ x3 ‰‰ ˆ "3 ‰ œ 3 sec# ˆ x3 ‰ Ê yww œ 3 † 2 sec ˆ x3 ‰ ˆsec ˆ x3 ‰ tan ˆ x3 ‰‰ ˆ "3 ‰ œ 2 sec# ˆ 3x ‰ tan ˆ 3x ‰ 53. g(x) œ Èx Ê gw (x) œ

" #È x

Ê g(1) œ 1 and gw (1) œ

therefore, (f ‰ g)w (1) œ f w (g(1)) † gw (1) œ 5 †

" #

œ

" u#


1 10
14

=

" (1
x)#

Ê g(
1) œ

" # w

and gw (
1) œ

" 4

; f(u) œ 1


Ê f w (g(
1)) œ f w ˆ #" ‰ œ 4; therefore, (f ‰ g)w (
1) œ f (g(
1))gw (
1) œ 4 †

55. g(x) œ 5Èx Ê gw (x) œ œ

; f(u) œ u&  1 Ê f w (u) œ 5u% Ê f w (g(1)) œ f w (1) œ 5;

5 #

54. g(x) œ (1
x)
" Ê gw (x) œ
(1
x)
# (
1) œ Ê f w (u) œ

" #

5 #Èx

Ê g(1) œ 5 and gw (1) œ

5 #

" 4

" u

œ1

1‰ ; f(u) œ cot ˆ 110u ‰ Ê f w (u) œ
csc# ˆ 110u ‰ ˆ 10

1 1 1 csc# ˆ 110u ‰ Ê f w (g(1)) œ f w (5) œ
10 csc# ˆ 1# ‰ œ
10 ; therefore, (f ‰ g)w (1) œ f w (g(1))gw (1) œ
10 †

5 #

56. g(x) œ 1x Ê gw (x) œ 1 Ê g ˆ "4 ‰ œ 14 and gw ˆ 4" ‰ œ 1; f(u) œ u  sec# u Ê f w (u) œ 1  2 sec u † sec u tan u œ 1  2 sec# u tan u Ê f w ˆg ˆ "4 ‰‰ œ f w ˆ 14 ‰ œ 1  2 sec# 14 tan 14 œ 5; therefore, (f ‰ g)w ˆ 4" ‰ œ f w ˆg ˆ 4" ‰‰ gw ˆ 4" ‰ œ 51 57. g(x) œ 10x#  x  1 Ê gw (x) œ 20x  1 Ê g(0) œ 1 and gw (0) œ 1; f(u) œ œ


2u#  2 au #  1 b #

Ê f w (u) œ

au#  1b(2)
(2u)(2u) au #  1 b #

Ê f w (g(0)) œ f w (1) œ 0; therefore, (f ‰ g)w (0) œ f w (g(0))gw (0) œ 0 † 1 œ 0

" 2 w x#
1 Ê g (x) œ
x$ Ê g(
1) œ 0 and 4(u
1) 1 ‰ (u  1)(1)
(u
1)(1) 2 ˆ uu
œ 2(u(u
1)(2) 1 † (u  1)# 1)$ œ (u  1)$ w w w

58. g(x) œ œ

2u u # 1

#

1‰ 1‰ gw (
1) œ 2; f(u) œ ˆ uu
Ê f w (u) œ 2 ˆ uu
1 1

Ê f w (g(
1)) œ f w (0) œ
4; therefore,

(f ‰ g) (
1) œ f (g(
1))g (
1) œ (
4)(2) œ
8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

d du

1‰ ˆ uu
1

152

Chapter 3 Differentiation

59. (a) y œ 2f(x) Ê

œ 2f w (x) Ê

dy dx

(b) y œ f(x)  g(x) Ê (c) y œ f(x) † g(x) Ê

œ 2f w (2) œ 2 ˆ "3 ‰ œ

dy dx ¹ x=2

œ f w (x)  gw (x) Ê

dy dx

dy dx

œ

g(x)f w (x)
f(x)gw (x) [g(x)]#

(e) y œ f(g(x)) Ê

dy dx

œ f w (g(x))gw (x) Ê

(d) y œ

f(x) g(x)

Ê

" #

Ê

dy dx ¹ x=2

œ

(g) y œ (g(x))
# Ê

dy dx

œ
2(g(x))
$ † gw (x) Ê

(h) y œ a(f(x))#  (g(x))# b dy dx ¹ x=2

œ

" #

(f(x))
"Î# † f w (x) œ

"Î#

Ê

" # # a(f(x))  #
"Î# w

a(f(2))#  (g(2)) b

Ê

dy dx ¹ x=3

œ

dy dx

(2) ˆ "3 ‰
(8)(
3) ##

œ

" 3

œ f w (g(2))gw (2) œ f w (2)(
3) œ

f w (x) #Èf(x)

dy dx

œ f(3)gw (3)  g(3)f w (3) œ 3 † 5  (
4)(21) œ 15
81

dy dx ¹ x=3

g(2)f w (2)
f(2)gw (2) [g(2)]#

œ

dy dx ¹ x=2

(f) y œ (f(x))"Î# Ê

Ê

œ f w (3)  gw (3) œ 21  5

dy dx ¹ x=3

œ f(x)gw (x)  g(x)f w (x) Ê

dy dx

2 3

dy dx ¹ x=2

œ

f w (2) #Èf(2)

ˆ "3 ‰

œ

œ

(
3) œ
1 œ

#È 8

37 6

" 6È 8

œ

" 1 #È 2

œ
2(g(3))
$ gw (3) œ
2(
4)
$ † 5 œ

(g(x))# b


"Î#

œ

È2 24

5 3#

a2f(x) † f w (x)  2g(x) † gw (x)b

a2f(2)f (2)  2g(2)gw (2)b œ

" #

a8#  2# b


"Î#

ˆ2 † 8 †

" 3

 2 † 2 † (
3)‰

œ
3È517 60. (a) y œ 5f(x)
g(x) Ê (b) y œ f(x)(g(x))$ Ê

œ 5f w (x)
gw (x) Ê

dy dx

dy dx ¹ x=1

œ 5f w (1)
gw (1) œ 5 ˆ
"3 ‰
ˆ
38 ‰ œ 1

œ f(x) a3(g(x))# gw (x)b  (g(x))$ f w (x) Ê

dy dx

dy dx ¹ x=0

œ $f(0)(g(0))# gw (0)  (g(0))$ f w (0)

œ 3(1)(1)# ˆ 3" ‰  (1)$ (5) œ 6 (c) y œ œ

f(x) g(x)  1

Ê

(g(x)  1)f w (x)
f(x) gw (x) (g(x)  1)#

œ

dy dx

(
4") ˆ
"3 ‰
(3) ˆ
83 ‰ (
41)#

Ê

dy dx ¹ x=1

œ

(g(1)  1)f w (1)
f(1)gw (1) (g(1)  1)#

œ1

(d) y œ f(g(x)) Ê

dy dx

œ f w (g(x))gw (x) Ê

dy dx ¹ x=0

œ f w (g(0))gw (0) œ f w (1) ˆ "3 ‰ œ ˆ
"3 ‰ ˆ 3" ‰ œ
9"

(e) y œ g(f(x)) Ê

dy dx

œ gw (f(x))f w (x) Ê

dy dx ¹ x=0

œ gw (f(0))f w (0) œ gw (1)(5) œ ˆ
83 ‰ (5) œ
40 3

(f) y œ ax""  f(x)b


#

Ê

œ
2 ax""  f(x)b

dy dx


$

a11x"!  f w (x)b Ê

dy dx ¹ x=1

œ
2(1  f(1))
$ a11  f w (1)b

" ‰ œ
2(1  3)
$ ˆ11
"3 ‰ œ ˆ
42$ ‰ ˆ 32 3 œ
3

(g) y œ f(x  g(x)) Ê

œ f w (x  g(x)) a1  gw (x)b Ê

dy dx

dy dx ¹ x=0

œ f w (0  g(0)) a1  gw (0)b œ f w (1) ˆ1  "3 ‰

œ ˆ
"3 ‰ ˆ 43 ‰ œ
49 61.

ds dt

œ

ds d)

†

d) dt :

s œ cos ) Ê

62.

dy dt

œ

dy dx

†

dx dt :

y œ x#  7x
5 Ê

63. With y œ x, we should get (a) y œ (b)

u 5

7 Ê

y œ 1  "u Ê
" œ
" u# † (x
1)#

œ

dy du dy du

œ

dy dx

" 5

dy du

dy dx

œ


" u#

ds ¸ d) )= 321

œ
sin ˆ 3#1 ‰ œ 1 so that

œ 2x  7 Ê

dy dx ¹ x=1

œ 9 so that

dy dt

œ

ds dt

dy dx

œ †

ds d)

dx dt

†

d) dt

œ9†

œ 1†5œ 5 " 3

œ3

œ 1 for both (a) and (b):

; u œ 5x
35 Ê ; u œ (x
1)


" a(x
1)
" b#

64. With y œ x$Î# , we should get (a) y œ u$ Ê

œ
sin ) Ê

ds d)

†

dy dx


" (x
1)#

œ

3 #


"

du dx

œ 5; therefore,

Ê

du dx

œ (x
1)# †

dy dy dx œ du
#

†

œ
(x
1) (1) œ

" (x
1)#

du " dx œ 5
" (x
1)# ;

† 5 œ 1, as expected therefore

dy dx

œ

dy du

†

du dx

œ 1, again as expected

x"Î# for both (a) and (b):

œ 3u# ; u œ Èx Ê

du dx

" #È x

; therefore,

dy dx

œ

dy du

†

du dx

œ 3u# †

œ 3x# ; therefore,

dy dx

œ

dy du

†

du dx

œ

œ

" #È x

#

œ 3 ˆÈx‰ †

" #Èx

œ

3 #

Èx,

as expected. (b) y œ Èu Ê

dy du

œ

" #È u

; u œ x$ Ê

du dx

" #Èu

† 3x# œ

" #È x $

again as expected. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

† 3x# œ

3 #

x"Î# ,

Section 3.5 The Chain Rule and Parametric Equations 65. y œ 2 tan ˆ 14x ‰ Ê (a)

dy dx ¹ x=1

œ

1 #

dy dx

œ ˆ2 sec#

1x ‰ ˆ 1 ‰ 4 4

œ

1 #

sec#

1x 4

sec# ˆ 14 ‰ œ 1 Ê slope of tangent is 2; thus, y(1) œ 2 tan ˆ 14 ‰ œ 2 and yw (1) œ 1 Ê tangent line is

given by y
2 œ 1(x
1) Ê y œ 1x  2
1 (b) yw œ 1# sec# ˆ 14x ‰ and the smallest value the secant function can have in
#  x  2 is 1 Ê the minimum value of yw is 1# and that occurs when 1# œ 1# sec# ˆ 14x ‰ Ê 1 œ sec# ˆ 14x ‰ Ê „ 1 œ sec ˆ 14x ‰ Ê x œ 0. 66. (a) y œ sin 2x Ê yw œ 2 cos 2x Ê yw (0) œ 2 cos (0) œ 2 Ê tangent to y œ sin 2x at the origin is y œ 2x; y œ
sin ˆ x# ‰ Ê yw œ
"# cos ˆ x# ‰ Ê yw (0) œ
"# cos 0 œ
"# Ê tangent to y œ
sin ˆ x# ‰ at the origin is y œ
"# x. The tangents are perpendicular to each other at the origin since the product of their slopes is


1. (b) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m cos 0 œ m; y œ
sin ˆ mx ‰ Ê yw œ
m" cos ˆ mx ‰ Ê yw (0) œ
m" cos (0) œ
m" . Since m † ˆ
m" ‰ œ
1, the tangent lines are perpendicular at the origin.

(c) y œ sin (mx) Ê yw œ m cos (mx). The largest value cos (mx) can attain is 1 at x œ 0 Ê the largest value yw can attain is kmk because kyw k œ km cos (mx)k œ kmk kcos mxk Ÿ kmk † 1 œ kmk . Also, y œ
sin ˆ mx ‰ ˆ x ‰¸ Ÿ ¸ m" ¸ ¸cos ˆ mx ‰¸ Ÿ km" k Ê the largest value yw can attain is ¸ m" ¸ . Ê yw œ
m" cos ˆ mx ‰ Ê kyw k œ ¸
" m cos m (d) y œ sin (mx) Ê yw œ m cos (mx) Ê yw (0) œ m Ê slope of curve at the origin is m. Also, sin (mx) completes m periods on [0ß 21]. Therefore the slope of the curve y œ sin (mx) at the origin is the same as the number of periods it completes on [0ß 21]. In particular, for large m, we can think of “compressing" the graph of y œ sin x horizontally which gives more periods completed on [0ß 21], but also increases the slope of the graph at the origin. 67. x œ cos 2t, y œ sin 2t, 0 Ÿ t Ÿ 1 Ê cos# 2t  sin# 2t œ 1 Ê x#  y# œ 1

68. x œ cos (1
t), y œ sin (1
t), 0 Ÿ t Ÿ 1 Ê cos# (1
t)  sin# (1
t) œ 1 Ê x#  y# œ 1, y   !

69. x œ 4 cos t, y œ 2 sin t, 0 Ÿ t Ÿ 21

70. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21

Ê

16 cos# t 16



4 sin# t 4

œ1 Ê

x# 16



y# 4

œ1

Ê

16 sin# t 16



25 cos# t 25

œ1 Ê

x# 16



Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

y# #5

œ1

153

154

Chapter 3 Differentiation

71. x œ 3t, y œ 9t# ,
_  t  _ Ê y œ x#

72. x œ
Èt , y œ t, t   0 Ê x œ
Èy or y œ x# , x Ÿ 0

73. x œ 2t
5, y œ 4t
7,
_  t  _ Ê x  5 œ 2t Ê 2(x  5) œ 4t Ê y œ 2(x  5)
7 Ê y œ 2x  3

75. x œ t, y œ È1
t# ,
1 Ÿ t Ÿ 0 Ê y œ È1
x#

#

Ê y œ 2
23 x, ! Ÿ x Ÿ $

76. x œ Èt  1, y œ Èt, t   0 Ê y# œ t Ê x œ Èy#  1, y   0

77. x œ sec# t
1, y œ tan t,
1#  t  #

74. x œ 3
3t, y œ 2t, 0 Ÿ t Ÿ 1 Ê y# œ t Ê x œ 3
3 ˆ y# ‰ Ê 2x œ 6
3y

#

Ê sec t
1 œ tan t Ê x œ y

1 #

78. x œ
sec t, y œ tan t,
1#  t  #

#

#

1 # #

Ê sec t
tan t œ 1 Ê x
y œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.5 The Chain Rule and Parametric Equations 79. (a) x œ a cos t, y œ
a sin t, 0 Ÿ t Ÿ 21

80. (a) x œ a sin t, y œ b cos t,

(b) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 21 (c) x œ a cos t, y œ
a sin t, 0 Ÿ t Ÿ 41

1 #

ŸtŸ

155

51 #

(b) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 21 (c) x œ a sin t, y œ b cos t, 1# Ÿ t Ÿ 9#1

(d) x œ a cos t, y œ a sin t, 0 Ÿ t Ÿ 41

(d) x œ a cos t, y œ b sin t, 0 Ÿ t Ÿ 41

81. Using a
"ß
$b we create the parametric equations x œ
"  at and y œ
$  bt, representing a line which goes through a
"ß
$b at t œ !. We determine a and b so that the line goes through a%ß "b when t œ ". Since % œ
"  a Ê a œ &. Since " œ
$  b Ê b œ %. Therefore, one possible parameterization is x œ
"  &t, y œ
$
%t, 0 Ÿ t Ÿ ". 82. Using a
"ß $b we create the parametric equations x œ
"  at and y œ $  bt, representing a line which goes through a
"ß $b at t œ !. We determine a and b so that the line goes through a$ß
#b when t œ ". Since $ œ
"  a Ê a œ %. Since
# œ $  b Ê b œ
&. Therefore, one possible parameterization is x œ
"  %t, y œ
$
&t, 0 Ÿ t Ÿ ". 83. The lower half of the parabola is given by x œ y#  " for y Ÿ !. Substituting t for y, we obtain one possible parameterization x œ t#  ", y œ t, t Ÿ 0Þ 84. The vertex of the parabola is at a
"ß
"b, so the left half of the parabola is given by y œ x#  #x for x Ÿ
". Substituting t for x, we obtain one possible parametrization: x œ t, y œ t#  #t, t Ÿ
". 85. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a#ß $b for t œ ! and passes through a
"ß
"b at t œ ". Then x œ fatb, where fa!b œ # and fa"b œ
". Since slope œ ??xt œ
"
# "
! œ
$, x œ fatb œ
$t  # œ #
$t. Also, y œ gatb, where ga!b œ $ and ga"b œ
". Since slope œ

?y ?t


"
3 "
!

œ

œ
4. y œ gatb œ
%t  $ œ $
%t.

One possible parameterization is: x œ #
$t, y œ $
%t, t   !. 86. For simplicity, we assume that x and y are linear functions of t and that the pointax, yb starts at a
"ß #b for t œ ! and passes through a!ß !b at t œ ". Then x œ fatb, where fa!b œ
" and fa"b œ !. Since slope œ Since slope œ

?x ?t ?y ?t

!
a
"b "
! !
# "
! œ

œ œ

œ ", x œ fatb œ "t  a
"b œ
"  t. Also, y œ gatb, where ga!b œ # and ga"b œ !.
#. y œ gatb œ
#t  # œ #
#t.

One possible parameterization is: x œ
"  t, y œ #
#t, t   !. 87. t œ Ê

1 4 Ê dy dx ¹ tœ 1

x œ 2 cos

d# y dx#

dyw /dt dx/dt

œ
cot

1 4

1 4

œ È2, y œ 2 sin

1 4

œ È2;

dx dt

œ
2 sin t,

dy dt

œ 2 cos t Ê

dy dx

œ
1; tangent line is y
È2 œ
1 Šx
È2‹ or y œ
x 

œ

dy/dt dx/dt w È 2 2 ; dy dt

œ

2 cos t
2 sin t

œ
cot t

œ csc# t

4

Ê 88. t œ Ê

œ

21 3 Ê dy dx ¹ tœ 21

œ

csc# t
2 sin t

x œ cos

21 3

" œ
2 sin $t Ê

d# y dx# ¹ tœ 21

œ
È 2

4

È3 dx 21 3 œ
# ; dt È3 È 3
x
# ‹œ

œ
"# , y œ È3 cos

œ È3 ; tangent line is y
Š


3

Ê

d# y dx# ¹ tœ 1

œ
sin t,

dy dt

œ
È3 sin t Ê

ˆ
"# ‰‘ or y œ È3 x;

dy dx

œ


È3 sin t
sin t

œ È3

d# y dx#

œ

œ0

dyw dt

œ0 Ê

"

œ 1; tangent line is

0
sin t

œ0

3

89. t œ y


1 4 " #

Ê xœ

1 4

,yœ

" #

;

dx dt

œ 1,

dy dt

œ 1 † ˆx
4" ‰ or y œ x  4" ;

œ

dyw dt

" #Èt

Ê

dy dx

œ

œ
4" t
$Î# Ê

dy/dt dx/dt d# y dx#

œ œ

1 2È t dyw /dt dx/dt

Ê

dy dx ¹ tœ 1 4

œ

#É "4

œ
4" t
$Î# Ê

d# y dx# ¹ tœ 1

œ
2

4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

156

Chapter 3 Differentiation

90. t œ 3 Ê x œ
È3  1 œ
2, y œ È3(3) œ 3; È

œ
3 Èt3t1 œ dyw dt

Ê

œ

dy dx ¹ tœ3


3 È 3  1 È3(3)

œ

d# y dx# ¹ tœ3

#

3t

œ 4t,

dx dt

y
1 œ 1 † (x
5) or y œ x
4;

œ

1 3

1 3

Ê xœ Ê

sin t 1
cos t

93. t œ Ê

1 3


sin

dy dx ¹ tœ 1

œ

3

Ê y œ È3x
œ


1 (1
cos t)#

1 2 Ê dy dx ¹ tœ 1

1È3 3

œ

1 3




sin ˆ 13 ‰ 1
cos ˆ 13 ‰

 2;

d# y dx# ¹ tœ 1

Ê

œ

Ê

3 2tÈ3t Èt  1

d# y dx#

dyw dt

œ 4t$ Ê

dy dt

dyw dt

œ 2t Ê

dy dx

d# y dx#

œ

œ

(3t)
"Î# Ê

dy dx

ˆ 3# ‰ (3t)
"Î# ˆ
"# ‰ (t  1)
"Î#

œ

Š 2tÈ3t3Èt b 1 ‹

œ
tÈ33t

Š 2Èct1b 1 ‹

œ

dyw /dt dx/dt

4t$ 4t

œ

œ t# Ê

œ

2t 4t

" #

dy dx ¹ tœc1

d# y dx# ¹ tœc1

Ê

œ (
1)# œ 1; tangent line is

œ

dy , y œ 1
cos 13 œ 1
"# œ "# ; dx dt œ 1
cos t, dt œ sin t Ê È Š #3 ‹ È œ ˆ " ‰ œ È3 ; tangent line is y
"# œ È3 Šx
13  #3 ‹ #

œ

" #

È3 #

(1
cos t)(cos t)
(sin t)(sin t) (1
cos t)#

œ


1 1
cos t

d# y dx#

Ê

dyw /dt dx/dt

œ

œ

dy dx

œ

dy/dt dx/dt

1 ‰ ˆ 1

cos t 1
cos t

œ
4

x œ cos œ
cot

1 2

œ 0, y œ 1  sin

1 #

1 2

œ 2;

œ
sin t,

dx dt

œ 0; tangent line is y œ 2;

sec# t 2 sec# t tan t

œ

" 2 tan t

œ

" #

cot t Ê

w

dy dt

dy dx ¹ tœc 1 4

y
(
1) œ
"# (x
1) or y œ
"# x
"# ; Ê

œ

œ

dy/dt dx/dt

d# y dx# ¹ tœc 1

œ

dyw dt

œ

dy dt

œ cos t Ê

œ csc# t Ê

94. t œ
14 Ê x œ sec# ˆ
14 ‰
1 œ 1, y œ tan ˆ
14 ‰ œ
1; dy dx

3 #

3

2

Ê

œ

œ
"3

91. t œ
1 Ê x œ 5, y œ 1;

92. t œ

dy dt

œ
2; tangent line is y
3 œ
2[x
(
2)] or y œ
2x
1;

È3t 
3 (t  1)
"Î# ‘  3Èt  1  3 (3t)
"Î# ‘

#

œ
"# (t  1)
"Î# ,

dx dt

" #

dx dt

#

d y dx#

œ

dy dx #

csc t
sin t

œ

cos t
sin t

œ
cot t

œ
csc$ t Ê

d# y dx# ¹ tœ 1

œ
1

2

œ 2 sec# t tan t,

dy dt

œ sec# t

cot ˆ
14 ‰ œ
#" ; tangent line is

œ
"# csc# t Ê

d# y dx#

œ


"# csc# t 2 sec# t tan t

œ
"4 cot$ t

" 4

4

95. s œ A cos (21bt) Ê v œ

ds dt

œ
A sin (21bt)(21b) œ
21bA sin (21bt). If we replace b with 2b to double the

frequency, the velocity formula gives v œ
41bA sin (41bt) Ê doubling the frequency causes the velocity to # # double. Also v œ
#1bA sin (21bt) Ê a œ dv dt œ
41 b A cos (21bt). If we replace b with 2b in the

acceleration formula, we get a œ
161# b# A cos (41bt) Ê doubling the frequency causes the acceleration to $ $ quadruple. Finally, a œ
41# b# A cos (21bt) Ê j œ da dt œ 81 b A sin (21bt). If we replace b with 2b in the jerk formula, we get j œ 641$ b$ A sin (41bt) Ê doubling the frequency multiplies the jerk by a factor of 8. 21 21 21 ‰ 96. (a) y œ 37 sin
365 (x
101)‘  25 Ê yw œ 37 cos
365 (x
101)‘ ˆ 365 œ

741 365

21 cos
365 (x
101)‘ .

The temperature is increasing the fastest when yw is as large as possible. The largest value of 21 21 cos
365 (x
101)‘ is 1 and occurs when 365 (x
101) œ 0 Ê x œ 101 Ê on day 101 of the year

( µ April 11), the temperature is increasing the fastest. 1 741
21 ‘ 741 (b) yw (101) œ 74 365 cos 365 (101
101) œ 365 cos (0) œ 365 ¸ 0.64 °F/day 97. s œ ("  4t)"Î# Ê v œ

ds dt

œ

v œ 2("  4t)
"Î# Ê a œ

dv dt

" #

(1  4t)
"Î# (4) œ 2(1  4t)
"Î# Ê v(6) œ 2("  % † 6)
"Î# œ " #

2 5

m/sec;

œ
† 2(1  4t)
$Î# (4) œ
4(1  4t)
$Î# Ê a(6) œ
4(1  4 † 6)
$Î# œ
14#5 m/sec#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.5 The Chain Rule and Parametric Equations 98. We need to show a œ œ

k 2È s

† kÈs œ

99. v proportional to œ
2sk$Î# † dx dt

100. Let

k Ès

kT 2

k #

is constant: a œ

dv dt

œ

dv ds

†

ds dt

and

dv ds

œ

d ds

ˆkÈs‰ œ

Ê aœ

k 2È s

dv ds

†

†

ds dt

ds dt

œ

dv ds

†v

which is a constant.

" Ès

Ê vœ

k Ès

for some constant k Ê

#

dv ds

œ
2sk$Î# . Thus, a œ

œ
k# ˆ s"# ‰ Ê acceleration is a constant times

œ f(x). Then, a œ

101. T œ 21É Lg Ê œ

#

dv dt

dT dL

dv dt

œ 21 †

œ

" #É Lg

dv dx

†

dx dt

œ

†

" g

œ

1 gÉ Lg

dv dx

† f(x) œ œ

1 ÈgL

d dx

" s#

œ

dv ds

œ

dv ds

†v

so a is inversely proportional to s# .

ˆ dx ‰ dt † f(x) œ

. Therefore,

dv dt

dT du

œ

d dx

(f(x)) † f(x) œ f w (x)f(x), as required.

dT dL

†

dL du

œ

1 ÈgL

† kL œ

1 kÈ L Èg

œ

" #

† 21kÉ Lg

, as required.

102. No. The chain rule says that when g is differentiable at 0 and f is differentiable at g(0), then f ‰ g is differentiable at 0. But the chain rule says nothing about what happens when g is not differentiable at 0 so there is no contradiction. 103. The graph of y œ (f ‰ g)(x) has a horizontal tangent at x œ 1 provided that (f ‰ g)w (1) œ 0 Ê f w (g(1))gw (1) œ 0 Ê either f w (g(1)) œ 0 or gw (1) œ 0 (or both) Ê either the graph of f has a horizontal tangent at u œ g(1), or the graph of g has a horizontal tangent at x œ 1 (or both). 104. (f ‰ g)w (
5)  0 Ê f w (g(
5)) † gw (
5)  0 Ê f w (g(
5)) and gw (
5) are both nonzero and have opposite signs. That is, either cf w (g(
5))  0 and gw (
5)  0d or cf w (g(
5))  0 and gw (
5)  0d . 105. As h Ä 0, the graph of y œ

sin 2(xh)
sin 2x h

approaches the graph of y œ 2 cos 2x because lim

hÄ!

sin 2(xh)
sin 2x h

œ

d dx

(sin 2x) œ 2 cos 2x.

106. As h Ä 0, the graph of y œ

cos c(x  h)# d
cos ax# b h #

approaches the graph of y œ
2x sin ax b because lim

hÄ!

cos c(x  h)# d
cos ax# b h

œ

d dx

ccos ax# bd œ
2x sin ax# b.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

157

158 107.

Chapter 3 Differentiation dx dt

œ cos t and

œ 2 cos 2t Ê

dy dt

dy dx

Ê 2 cos# t
1 œ 0 Ê cos t œ „ y œ sin 2 ˆ 14 ‰ œ 1 Ê Š

È2 # ß 1‹

œ

dy/dt dx/dt

" È2

œ

œ

2 cos 2t cos t

Ê tœ

1 4

31 4

,

2 a2 cos# t
1b cos t 51 4

,

108.

dx dt

œ 2 cos 2t and

dy dx ¹ tœ0

3 ca2 cos# t
1b (cos t)
2 sin t cos t sin td 2 a2 cos# t
1b

œ dy dx

œ 3 cos 3t Ê

dy dt

(3 cos t) a4 cos# t
3b 2 a2 cos# t
1b

œ0 Ê

and y œ sin 3 ˆ 16 ‰ œ 1 Ê Š

œ

œ 2 Ê y œ 2x and

dy dx

œ

œ

dy/dt dx/dt

3 cos 3t 2 cos 2t

È3 #

È3 # ß 1‹

1 6

Ê tœ

,

51 6

,

71 6

œ

" #É È x

†

d dx

ˆÈx‰ œ

" #ÉÈx

†

" #Èx

œ

dy dx

œ

" #É x È x

3È x 4È x É È x

œ

3 4

†

d dx

ˆxÈx‰ Ê

dy dx

œ

dy dx

œ

110. From the power rule, with y œ x$Î% , we get Ê

œ

df dt

œ

(3 cos t) a4 cos# t
3b 2 a2 cos# t
1b

,

111 6

. In the 1st quadrant: t œ

1 6

1 #

,

31 #

and

and

Ê x œ sin 2 ˆ 16 ‰ œ

1 #

, 1,

31 #

dy dx ¹ tœ0

and t œ 0, œ

3 cos 0 2 cos 0

1 3

œ

,

21 3

3 #

, 1,

41 3

Ê yœ

,

51 3

3 #

È3 #

x, and

Ê t œ 0 and t œ 1 give dy dx ¹ tœ1

" 4

dy dx

" #ÉxÈx

œ

" 4

x
$Î% . From the chain rule, y œ ÉÈx

x
$Î% , in agreement.

œ

3 4

x
"Î% . From the chain rule, y œ ÉxÈx

† Šx †

" #È x

 È x‹ œ

" #ÉxÈx

† ˆ 3# Èx‰ œ

3È x 4É xÈ x

œ 1.27324 sin 2t  0.42444 sin 6t  0.2546 sin 10t  0.18186 sin 14t df dt

È2 #

; then

x
"Î% , in agreement.

(c) The curve of y œ

œ

3(cos 2t cos t
sin 2t sin t) 2 a2 cos# t
1b

111. (a)

(b)

1 4

1 give the tangent lines at

œ
3# Ê y œ
3# x

109. From the power rule, with y œ x"Î% , we get dy dx

Ê x œ sin

is the point where the graph has a horizontal tangent. At the origin: x œ 0

the tangent lines at the origin. Tangents at the origin:

Ê

1 31 # , 1, # ; thus t œ 0 and t œ dy dx ¹ tœ1 œ
2 Ê y œ
2x

(3 cos t) a2 cos# t
1
2 sin# tb 2 a2 cos# t
1b

and y œ 0 Ê sin 2t œ 0 and sin 3t œ 0 Ê t œ 0, 3 cos (31) 2 cos (21)

1 4

. In the 1st quadrant: t œ

œ 0 Ê 3 cos t œ 0 or 4 cos# t
3 œ 0: 3 cos t œ 0 Ê t œ

4 cos# t
3 œ 0 Ê cos t œ „

œ

71 4

œ0

is the point where the tangent line is horizontal. At the origin: x œ 0 and y œ 0

Ê sin t œ 0 Ê t œ 0 or t œ 1 and sin 2t œ 0 Ê t œ 0, the origin. Tangents at origin:

,

2 a2 cos# t
1b cos t

œ0 Ê

dy dx

; then

approximates y œ

dg dt

the best when t is not
1,
1# , 0, 1# , nor 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.5 The Chain Rule and Parametric Equations 112. (a)

(b)

dh dt

œ 2.5464 cos (2t)  2.5464 cos (6t)  2.5465 cos (10t)  2.54646 cos (14t)  2.54646 cos (18t)

(c)

111-116. Example CAS commands: Maple: f := t -> 0.78540 - 0.63662*cos(2*t) - 0.07074*cos(6*t) - 0.02546*cos(10*t) - 0.01299*cos(14*t); g := t -> piecewise( t<-Pi/2, t+Pi, t<0, -t, t
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

159

160

Chapter 3 Differentiation

3.6 IMPLICIT DIFFERENTIATION 1. y œ x*Î% Ê

dy dx

œ

x&Î%

9 4

3 3. y œ È 2x œ (2x)"Î$ Ê

2. y œ x
$Î& Ê " 3

œ

dy dx

(2x)
#Î$ † 2 œ

5. y œ 7Èx  6 œ 7(x  6)"Î# Ê

œ

dy dx

6. y œ
2Èx
1 œ
2(x
1)"Î# Ê 7. y œ (2x  5)
"Î# Ê 8. y œ ("
6x)#Î$ Ê 9. y œ x ax#  1b 10. y œ x ax#  1b

"Î#

dy dx

7 # 11. s œ È t œ t#Î( Ê

7 2È x  6

(1
6x)
"Î$ (
6) œ
4(1
6x)
"Î$

2 3

œ


"Î#

a#xb  ax#  1b
$Î#

dy dt

14. z œ cos ˆ("
6t)#Î$ ‰ Ê

dz dt

† " œ ax#  1b
"Î#


"Î#

ax#  x#  "b œ

† " œ ax#  1b

4 12. r œ È )
$ œ )
$Î% Ê


$Î#

dr d)

2x#  1 È x#  1

a
x#  x#  "b œ

œ
43 )
(Î%

œ
sin ˆ("
6t)#Î$ ‰ † 23 (1
6t)
"Î$ (
') œ 4(1
6t)
"Î$ sin ˆ(1
6t)#Î$ ‰ "Î#

Ê f w (x) œ

" #

ˆ1
x"Î# ‰
"Î# ˆ
#" x
"Î# ‰ œ

Ê gw (x) œ
23 ˆ2x
"Î#  1‰

3 17. h()) œ È 1  cos (2)) œ (1  cos 2))"Î$ Ê hw ()) œ

18. k()) œ (sin ()  5))&Î% Ê kw ()) œ

5 4

" 3


%Î$

† (
1)x
$Î# œ

2 3


" 4 ŠÉ1
Èx‹ Èx

œ


" 4 É x ˆ1
È x ‰

ˆ2x
"Î#  1‰
%Î$ x
$Î#

(1  cos 2))
#Î$ † (
sin 2)) † 2 œ
23 (sin 2))(1  cos 2))
#Î$

(sin ()  5))"Î% † cos ()  5) œ

5 4

cos ()  5)(sin ()  5))"Î%

19. x# y  xy# œ 6: Step 1:

Šx #

Step 2:

x#

dy dx

Step 3:

dy dx dy dx

ax#  2xyb œ
2xy
y#

Step 4:

dy dx

œ

 y † 2x‹  Šx † 2y

 2xy

dy dx

dy dx

 y# † 1‹ œ 0

œ
2xy
y#


2xy
y# x# 2xy

20. x$  y$ œ 18xy Ê 3x#  3y#

dy dx

œ 18y  18x

dy dx

Ê a3y#
18xb

dy dx

œ 18y
3x# Ê

dy dx

œ

21. 2xy  y# œ x  y: Step 1:

Š2x

dy dx

" ax# 1b$Î#

œ cos ˆ(2t  5)
#Î$ ‰ † ˆ
23 ‰ (2t  5)
&Î$ † 2 œ
43 (2t  5)
&Î$ cos ˆ(2t  5)
#Î$ ‰

15. f(x) œ É1
Èx œ ˆ1
x"Î# ‰
"Î$

"Î#

a#xb  ax#  1b

2
&Î( 7 t

13. y œ sin ˆ(2t  5)
#Î$ ‰ Ê

16. g(x) œ 2 ˆ2x
"Î#  1‰

5"Î% 4x$Î%

(5x)
$Î% † 5 œ

x
1

Ê yw œ x † ˆ
"# ‰ax#  1b ds dt

" 4

œ
1(x
1)
"Î# œ
È "

Ê yw œ x † "# ax#  1b


"Î#

œ

dy dx

œ
"# (2x  5)
$Î# † 2 œ
(2x  5)
$Î#

œ

dy dx

4 4. y œ È 5x œ (5x)"Î% Ê

(x  6)
"Î# œ

7 #

dy dx

2"Î$ 3x#Î$

œ
35 x
)Î&

dy dx

 2y‹  2y

dy dx

œ1

dy dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

6y
x# y#
6x

Section 3.6 Implicit Differentiation Step 2:

2x

Step 3:

dy dx dy dx

Step 4:

dy dx

 2y

dy dx




dy dx

161

œ 1
2y

(2x  2y
1) œ "
2y œ

1
2y 2x  2y
1

22. x$
xy  y$ œ 1 Ê 3x#
y
x

dy dx

 3y#

œ 0 Ê a3y#
xb

dy dx

dy dx

œ y
3x# Ê

dy dx

y
3x# 3y#
x

œ

23. x# (x
y)# œ x#
y# : Step 1:

x# ’2(x
y) Š1


Step 2:


2x# (x
y)

Step 3:

dy dx dy dx

Step 4:

dy dx

dy dx ‹“

 2y

 (x
y)# (2x) œ 2x
2y

dy dx

œ 2x
2x# (x
y)
2x(x
y)#

c
2x# (x
y)  2yd œ 2x c1
x(x
y)
(x
y)# d œ

œ

2x c1
x(x
y)
(x
y)# d
2x# (x
y)  2y

œ

dy dx

x c1
x(x
y)
(x
y)# d y
x# (x
y)

œ

x a1
x#  xy
x#  2xy
y# b x# y
x$  y

x
2x$  3x# y
xy# x# y
x$  y

24. (3xy  7)# œ 6y Ê 2(3xy  7) † Š3x Ê

dy dx

dy dx

 3y‹ œ 6

[6x(3xy  7)
6] œ
6y(3xy  7) Ê (x  1)
(x
1) (x  1)#

dy dx

dy dx

Ê 2(3xy  7)(3x)

y(3xy  7) œ
x(3xy  7)
1 œ

dy dx


6

dy dx

œ
6y(3xy  7)

#

3xy  7y 1
3x# y
7x

25. y# œ

x
" x1

Ê 2y

26. x# œ

x
y xy

Ê x$  x# y œ x
y Ê 3x#  2xy  x# yw œ 1
yw Ê ax#  1b yw œ 1
3x#
2xy Ê yw œ

dy dx

œ

27. x œ tan y Ê 1 œ asec# yb

Ê

dy dx

œ

dy dx

Ê

2 (x  1)#

œ

" sec# y

œ

dy dx

" y(x  1)# 1
3x#
2xy x#  1

œ cos# y

dy dy dy # # # 28. xy œ cot axyb Ê x dy dx  y œ
csc (xy)Šx dx  y‹ Ê x dx  x csc (xy) dx œ
y csc (xy)
y

Ê

dy
dx x

 x csc# (xy)‘ œ
y
csc# (xy)  "‘ Ê

29. x  tan (xy) œ ! Ê 1  csec# (xy)d Šy  x œ


1 x sec# (xy)




y x

œ


cos# (xy) x




y x

30. x  sin y œ xy Ê 1  (cos y)

œ dy dx

dy dx ‹

œyx

’
"y cos Š "y ‹  sin Š y" ‹  x“ œ
y Ê

’sin Š y" ‹  2y cos Š y" ‹
2“ œ 2 Ê

33. )"Î#  r"Î# œ 1 Ê

" #


y
csc# (xy)  "‘ x
"  csc# (xy)‘

œ
yx

œ 0 Ê x sec# (xy)

Ê (cos y
x)

dy dx

" y# dy dx

†

œ

)
"Î#  "# r
"Î# †

dy dx

dr d)

œ

dy dx

 sin Š y" ‹ †

dy dx “

32. y# cos Š "y ‹ œ 2x  2y Ê y# ’
sin Š y" ‹ † (
1) dy dx

œ

dy dx

œ
1
y sec# (xy) Ê


cos# (xy)
y x

31. y sin Š "y ‹ œ 1
xy Ê y ’cos Š y" ‹ † (
1) dy dx

dy dx

dy dx

œy
1 Ê œ
x


y


" y#

" y

†

cos Š "y ‹  sin Š "y ‹  x

dy dx “

œ

 cos Š y" ‹ † 2y

dy dx

dy dx

y
1 cos y
x

œ


y Ê
y #

y

sin Š "y ‹
cos Š "y ‹  xy

dy dx

œ22

dy dx

Ê

2 sin Š "y ‹  2y cos Š "y ‹
#

œ0 Ê

dr d)

" ’ #È “œ r


" #È )

Ê

dr d)

œ


2È r 2È )

Èr

œ
È

)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dy dx

œ


"
y sec# (xy) x sec# (xy)

162

Chapter 3 Differentiation

34. r
2È) œ " #

35. sin (r)) œ

)#Î$  43 )$Î% Ê

3 #


)
"Î# œ )
"Î$  )
"Î% Ê

dr d)

Ê [cos (r))] ˆr  )

dr ‰ d)

œ0 Ê

dr d)

œ )
"Î#  )
"Î$  )
"Î%

dr d)

[) cos (r))] œ
r cos (r)) Ê


r cos (r)) ) cos (r))

œ

dr d)

œ
)r ,

cos (r)) Á 0 36. cos r  cot ) œ r) Ê (
sin r)

dr d)


csc# ) œ r  )

37. x#  y# œ 1 Ê 2x  2yyw œ 0 Ê 2yyw œ
2x Ê y(
1)xyw y#

Ê yww œ


yx Š
xy ‹

œ

38. x#Î$  y#Î$ œ 1 Ê

y#

2 3

œ0 Ê

dy dx

dy dx d# y dx#

since yw œ
xy Ê

x
"Î$  23 y
"Î$

Ê

dr d)

dy dx

Ê

œ

" 3


#Î$
"Î$

x

y

2x  2 2y

39. y# œ x#  2x Ê 2yyw œ 2x  2 Ê yw œ d# y dx#

Ê

y#
(x 1)# y$

ww

œy œ

x1 y

œ

œ yw œ
xy ; now to find
y #
x # y$

d# y dx#


y#
a"
y# b y$

œ

d dx

, œ

#

 csc ) œ
rsin r)

dr d)

ayw b œ

d dx

Š
xy ‹


" y$

dy dx


"Î$

"Î$

œ
yx
"Î$ œ
ˆ yx ‰

"Î$

x#Î$

y
(x  1)yw y#

; then yww œ

" y1

œ

y
(x  1) Š x
y 1 ‹ y#

œ (y  1)
" ; then yww œ
(y  1)
# † yw


" (y  1)$

œ yww œ

41. 2Èy œ x
y Ê y
"Î# yw œ 1
yw Ê yw ˆy
"Î#  1‰ œ 1 Ê

dy dx

œ yw œ

" y
"Î#  1

Èy Èy  1

œ

; we can

differentiate the equation yw ˆy
"Î#  1‰ œ 1 again to find yww : yw ˆ
"# y
$Î# yw ‰  ˆy
"Î#  1‰ yww œ 0 Ê ˆy
"Î#  1‰ yww œ

" #

w #
$Î#

Ê

cy d y

d# y dx#

" #

œ yww œ

#

"
$Î# Œ y
"Î#  1  y

ay
"Î#  1b

œ

" $ 2y$Î# ay
"Î#  1b

œ

" $ # ˆ1  È y ‰

42. xy  y# œ 1 Ê xyw  y  2yyw œ 0 Ê xyw  2yyw œ
y Ê yw (x  2y) œ
y Ê yw œ œ


(x  2y)yw  y(1  2yw ) (x  2y)#

œ

2y(x  2y)
2y# (x  2y)$

œ

œ


y


y


(x  2y) ’ (x  2y) “  y ’1  2 Š (x  2y) ‹“ (x  2y)#

2y#  2xy (x  2y)$

œ

œ

"

(x  2y)


y (x2y)

;

d# y dx#

œ yww

cy(x  2y)  y(x  2y)
2y# d (x  2y)#

2y(x  y) (x  2y)$ #

43. x$  y$ œ 16 Ê 3x#  3y# yw œ 0 Ê 3y# yw œ
3x# Ê yw œ
xy# ; we differentiate y# yw œ
x# to find yww : # ww

w

w

w #

# ww

y y  y c2y † y d œ
2x Ê y y œ
2x
2y cy d œ


2xy$
2x% y&

Ê

d# y dx# ¹ (2ß2)

œ


32
32 32

ww

Ê y œ


2x
2y Š
y#

44. xy  y# œ 1 Ê xyw  y  2yyw œ 0 Ê yw (x  2y) œ
y Ê yw œ since yw k (0
1) œ
"# we obtain yww k (0ß
1) œ ß

#

x# ‹ y#

œ


2x
y#

2x% y$

œ
2

(
2) ˆ "# ‰
(1)(0) 4

45. y#  x# œ y%
2x at (
#ß ") and (
#ß
1) Ê 2y

dy dx

;

x"Î$ †ˆ
"3 y
#Î$ ‰ Œ
y"Î$   y"Î$ ˆ "3 x
#Î$ ‰ x

40. y#
2x œ 1
2y Ê 2y † yw
2 œ
2yw Ê yw (2y  2) œ 2 Ê yw œ œ
(y  1)
# (y  1)
" Ê

d# y dx#


23 y
"Î$ ‘ œ
23 x
"Î$ Ê yw œ

x"Î$ †ˆ
3" y
#Î$ ‰ yw  y"Î$ ˆ "3 x
#Î$ ‰ œ x#Î$ "Î$ y " "Î$
%Î$ " x œ 3x %Î$  3y"Î$ x#Î$ 3 y



[
sin r
)] œ r  csc# ) Ê

œ yww œ

Differentiating again, yww œ d# y dx#

dr d)


y (x2y)

Ê yww œ

(x  2y) a
yw b
(
y) a1  2yw b (x  2y)#

œ
4"  2x œ 4y$

dy dx


2 Ê 2y

dy dx


4y$

dy dx

œ
2
2x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

Section 3.6 Implicit Differentiation Ê

dy dx

a2y
4y$ b œ
2
2x Ê

dy dx

œ

x" #y $
y

Ê

dy dx ¹ (
2ß1)

œ
1 and

#

46. ax#  y# b œ (x
y)# at("ß !) and ("ß
1) Ê 2 ax#  y# b Š2x  2y Ê and

dy dx

c2y ax#  y# b  (x
y)d œ
2x ax#  y# b  (x
y) Ê

dy dx ¹ (1ßc1)

dy dx

dy dx ¹ (
2ß
1)

dy dx ‹

œ

œ1

œ 2(x
y) Š1



2x ax#  y# b  (x
y) 2y ax#  y# b  (x
y)

dy dx ‹

Ê

dy dx ¹ (1ß0)

œ
1

œ1

47. x#  xy
y# œ 1 Ê 2x  y  xyw
2yyw œ 0 Ê (x
2y)yw œ
2x
y Ê yw œ (a) the slope of the tangent line m œ yw k (2 3) œ ß

Ê the tangent line is y
3 œ

7 4

(b) the normal line is y
3 œ
47 (x
2) Ê y œ
47 x 

7 4

2x  y 2y
x

;

(x
2) Ê y œ

7 4

x


" #

29 7

48. x#  y# œ 25 Ê 2x  2yyw œ 0 Ê yw œ
xy ; (a) the slope of the tangent line m œ yw k (3
4) œ
xy ¹ ß

Ê yœ

3 4

x


(3ß
4)

œ

Ê the tangent line is y  4 œ

3 4

3 4

(x
3)

25 4

(b) the normal line is y  4 œ
43 (x
3) Ê y œ
43 x 49. x# y# œ 9 Ê 2xy#  2x# yyw œ 0 Ê x# yyw œ
xy# Ê yw œ
yx ; (a) the slope of the tangent line m œ yw k (
1ß3) œ
yx ¸ (
1ß3) œ 3 Ê the tangent line is y
3 œ 3(x  1) Ê y œ 3x  6 (b) the normal line is y
3 œ
"3 (x  1) Ê y œ
3" x 

8 3

50. y#
2x
4y
" œ ! Ê 2yyw
2
4yw œ 0 Ê 2(y
2)yw œ 2 Ê yw œ

" y
#

;

w

(a) the slope of the tangent line m œ y k (
2ß1) œ
1 Ê the tangent line is y
1 œ
1(x  2) Ê y œ
x
1 (b) the normal line is y
1 œ 1(x  2) Ê y œ x  3 51. 6x#  3xy  2y#  17y
6 œ 0 Ê 12x  3y  3xyw  4yyw  17yw œ 0 Ê yw (3x  4y  17) œ
12x
3y 12x
3y Ê yw œ 3x
 4y  17 ; (a) the slope of the tangent line m œ yw k (
1ß0) œ Ê yœ

6 7

x


"2x
3y 3x  4y  17 ¹ (
1ß0)

œ

6 7

Ê the tangent line is y
0 œ

6 7

(x  1)

6 7

(b) the normal line is y
0 œ
76 (x  1) Ê y œ
76 x


7 6

52. x#
È3xy  2y# œ 5 Ê 2x
È3xyw
È3y  4yyw œ 0 Ê yw Š4y
È3x‹ œ È3y
2x Ê yw œ (a) the slope of the tangent line m œ yw k ŠÈ3 2‹ œ ß

È3y
2x ¹ 4y
È3x ŠÈ3ß2‹

œ 0 Ê the tangent line is y œ 2

(b) the normal line is x œ È3 53. 2xy  1 sin y œ 21 Ê 2xyw  2y  1(cos y)yw œ 0 Ê yw (2x  1 cos y) œ
2y Ê yw œ (a) the slope of the tangent line m œ yw k ˆ1 12 ‰ œ ß


2y 2x  1 cos y ¹ ˆ1ß 1 ‰ 2

y


1 #


2y 2x  1 cos y

œ
1# Ê the tangent line is

œ
1# (x
1) Ê y œ
1# x  1

(b) the normal line is y


1 #

œ

2 1

(x
1) Ê y œ

2 1

x


2 1



1 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

È3y
2x 4y
È3x

;

163

164

Chapter 3 Differentiation

54. x sin 2y œ y cos 2x Ê x(cos 2y)2yw  sin 2y œ
2y sin 2x  yw cos 2x Ê yw (2x cos 2y
cos 2x) œ
sin 2y
2y sin 2x Ê yw œ

sin 2y  2y sin 2x cos 2x
2x cos 2y

(a) the slope of the tangent line m œ yw k ˆ 14

ß

1‰ 2

œ

; sin 2y  2y sin 2x cos 2x
2x cos 2y ¹ ˆ 1 ß 1 ‰

œ

4 2

1 #

y


œ 2 ˆx
14 ‰ Ê y œ 2x

(b) the normal line is y


1 #

œ
"# ˆx
14 ‰ Ê y œ
"# x 

1 1 #

œ 2 Ê the tangent line is

51 8

55. y œ 2 sin (1x
y) Ê yw œ 2 [cos (1x
y)] † a1
yw b Ê yw [1  2 cos (1x
y)] œ 21 cos (1x
y) Ê yw œ

21 cos (1x
y) 1  # cos (1x
y)

;

(a) the slope of the tangent line m œ yw k (1 0) œ ß

21 cos (1x
y) 1  2 cos (1x
y) ¹(1ß0)

y
0 œ 21(x
1) Ê y œ 21x
21 (b) the normal line is y
0 œ
#"1 (x
1) Ê y œ
2x1 

œ 21 Ê the tangent line is

" #1

56. x# cos# y
sin y œ 0 Ê x# (2 cos y)(
sin y)yw  2x cos# y
yw cos y œ 0 Ê yw c
2x# cos y sin y
cos yd 2x cos# y 2x# cos y sin y  cos y

œ
2x cos# y Ê yw œ

;

(a) the slope of the tangent line m œ yw k (0 1) œ ß

2x cos# y 2x# cos y sin y  cos y ¹ (0ß1)

œ 0 Ê the tangent line is y œ 1

(b) the normal line is x œ 0 57. Solving x#  xy  y# œ 7 and y œ 0 Ê x# œ 7 Ê x œ „ È7 Ê Š
È7ß !‹ and ŠÈ7ß !‹ are the points where the curve crosses the x-axis. Now x#  xy  y# œ 7 Ê 2x  y  xyw  2yyw œ 0 Ê (x  2y)yw œ
2x
y
2 È 7 y 2x  y È È Ê yw œ
2x x  2y Ê m œ
x  2y Ê the slope at Š
7ß !‹ is m œ

È7 œ
2 and the slope at Š 7ß !‹ is È

m œ
2È77 œ
2. Since the slope is
2 in each case, the corresponding tangents must be parallel. 58. x#  xy  y# œ 7 Ê 2x  y  x dy dx

(a) Solving

dy dx

 2y

dy dx

œ 0 Ê (x  2y)

dy dx

œ
2x
y Ê

dy dx

œ


2x
y x  2y

and

dx dy

œ

x  2y
2x
y

;

œ 0 Ê
2x
y œ 0 Ê y œ
2x and substitution into the original equation gives

x#  x(
2x)  (
2x)# œ 7 Ê 3x# œ 7 Ê x œ „ É 73 and y œ … 2É 73 when the tangents are parallel to the x-axis. dx dy

(b) Solving

#

œ 0 Ê x  2y œ 0 Ê y œ
x# and substitution gives x#  x ˆ
x# ‰  ˆ
x# ‰ œ 7 Ê

3x# 4

œ7

Ê x œ „ 2É 73 and y œ … É 73 when the tangents are parallel to the y-axis. 59. y% œ y#
x# Ê 4y$ yw œ 2yyw
2x Ê 2 a2y$
yb yw œ
2x Ê yw œ y
x2y$ ; the slope of the tangent line at È3 " È È È Š 43 ß #3 ‹ is y
x2y$ ¹ È3 È3 œ È3 4 6È3 œ "
4 3 œ #
" 3 œ
1; the slope of the tangent line at Š 43 ß #" ‹
4 # Œ

is

x y
2y$ ¹

Œ

È3 4

ß

1 2




œ

È3 " #

4


28

4

ß

2

œ




2È 3 4
2

#

8

œ È3

60. y# (2
x) œ x$ Ê 2yyw (2
x)  y# (
1) œ 3x# Ê yw œ mœ

#

#

y  3x 2y(2
x) ¹ (1ß1)

œ

4 #

y#  3x# 2y(2
x)

; the slope of the tangent line is

œ 2 Ê the tangent line is y
1 œ 2(x
1) Ê y œ 2x
1; the normal line is

y
1 œ
"# (x
1) Ê y œ
"# x 

3 #

61. y%
4y# œ x%
9x# Ê 4y$ yw
8yyw œ 4x$
18x Ê yw a4y$
8yb œ 4x$
18x Ê yw œ

4x$
18x 4y$
8y

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

2x$
9x 2y$
4y

Section 3.6 Implicit Differentiation x a2x#
9b y a2y#
4b

œ

œ m; (
3ß 2): m œ

(
3)(18
9) 2(8
4)

œ
27 8 ; (
$ß
#): m œ

; (3ß #): m œ

27 8

27 8

; (3ß
#): m œ
27 8

62. x$  y$
9xy œ 0 Ê 3x#  3y# yw
9xyw
9y œ 0 Ê yw a3y#
9xb œ 9y
3x# Ê yw œ (a) yw k (4 2) œ ß

and yw k (2 4) œ

5 4

ß

#

3y
x y#
3x

(b) yw œ 0 Ê

4 5

165

9y
3x# 3y#
9x

œ

3y
x# y#
3x

; x# 3

œ 0 Ê 3y
x# œ 0 Ê y œ

$

#

#

Ê x$  Š x3 ‹
9x Š x3 ‹ œ 0 Ê x'
54x$ œ 0

Ê x$ ax$
54b œ 0 Ê x œ 0 or x œ $È54 œ 3 $È2 Ê there is a horizontal tangent at x œ 3 $È2 . To find the corresponding y-value, we will use part (c). dx dy

(c)

œ0 Ê

$

y#
3x 3y
x#

œ 0 Ê y#
3x œ 0 Ê y œ „ È3x ; y œ È3x Ê x$  ŠÈ3x‹
9xÈ3x œ 0

Ê x$
6È3 x$Î# œ 0 Ê x$Î# Šx$Î#
6È3‹ œ 0 Ê x$Î# œ 0 or x$Î# œ 6È3 Ê x œ 0 or x œ $È108 œ 3 $È4 . Since the equation x$  y$
9xy œ 0 is symmetric in x and y, the graph is symmetric about the line y œ x. That is, if (aß b) is a point on the folium, then so is (bß a). Moreover, if yw k (a b) œ m, then yw k (b a) œ m" . ß

ß

Thus, if the folium has a horizontal tangent at (aß b), it has a vertical tangent at (bß a) so one might expect that with a horizontal tangent at x œ $È54 and a vertical tangent at x œ 3 $È4, the points of tangency are Š $È54ß 3 $È4‹ and Š3 $È4ß $È54‹, respectively. One can check that these points do satisfy the equation x$  y$
9xy œ 0. 63. x#
2tx  2t# œ 4 Ê 2x 2y$
3t# œ 4 Ê 6y# #

dx dt


2x
2t


6t œ 0 Ê

dy dt

#

dx dt

 4t œ 0 Ê (2x
2t) œ

dy dt

6t 6y#

œ

t y#

; thus

œ

dy dx

#

œ 2x
4t Ê

dx dt

dy/dt dx/dt

œ

#

Š yt# ‹ ˆ xx

2tt ‰

œ

dx dt

œ

t(x
t) y# (x
2t)

2x
4t 2x
2t

œ

x
2t x
t

;

;tœ2

Ê x
2(2)x  2(2) œ 4 Ê x
4x  4 œ 0 Ê (x
2) œ 0 Ê x œ 2; t œ 2 Ê 2y$
3(2)# œ 4 Ê 2y$ œ 16 Ê y$ œ 8 Ê y œ 2; therefore 64. x œ É5
Èt Ê Ê at
1b

dy dt

œ

" #È t

therefore,

dy dx ¹ tœ4

œ

dy dt

dy dt

œ


"Î#

" #Èt
y

at
1 b

œ

"
#y È t #tÈt
2Èt

œ

"
4 dx dt

 3x"Î#

dx dt

œ

dy t Èy ‹ dt

dy dx

œ 2t  1 Ê ˆ1  3x"Î# ‰ #

œ


y 2Èt1

Œ 2Èy (t b 1) b 2tÈt b 1
Š

"
"Î# ; y(t
1) œ Èt Ê y  (t
1) dy dt œ # t

œ

dy dt dx dt

"
#yÈt "
#yÈt 4Èt É5
Èt #tÈt
2Èt œ œ È †
"
" # tat
" b 4È t É 5
È t

14 3

#

œ

"

t œ 4 Ê x œ É5
È4 œ È3; t œ 4 Ê y(3) œ È4 œ 2


2Èy Ê

dy dt

œ

cyÈy c 4yÈt b 1

dy/dt dx/dt

œ0

4È t É 5
È t

; thus

Èt  1  y ˆ " ‰ (t  1)
"Î#  2Èy  2t ˆ " y
"Î# ‰

Ê ŠÈ t  1  dy dx

2(2
2) (2)# (2
2(2))

œ

ˆ
"# t
"Î# ‰ œ


2Š"
2a2bÈ4‹É&
È4

65. x  2x$Î# œ t#  t Ê Ê

ˆ5
Èt‰


y Ê

#ˆ"
#yÈt‰É&
Èt ; "
t

œ

" #

œ

dx dt

dy dx ¹ tœ2

2t b 1 ‹ 1 b 3x"Î#

dy dt

dx dt

œ 2t  1 Ê

œ0 Ê

Š 2Èct yb 1
2Èy‹ ŠÈt  1  Èt y ‹

œ

dy dt

dx dt

Èt  1 

œ

2t1 13x"Î#

y 2È t  1


yÈy
4yÈt  1 2Èy (t  1)  2tÈt  1

; yÈt  1  2tÈy œ 4

 2Èy  Š Èt y ‹

; thus

; t œ 0 Ê x  2x$Î# œ 0 Ê x ˆ1  2x"Î# ‰ œ 0 Ê x œ 0; t œ 0

Ê yÈ0  1  2(0)Èy œ 4 Ê y œ 4; therefore


4È4
4(4)È0  1

dy dx ¹ tœ0

œ

Œ 2È4(0  1)  2(0)È0  1  2(0)  1

Œ 1  3(0)"Î# 

œ
6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dy dt

œ0

166

Chapter 3 Differentiation

66. x sin t  2x œ t Ê

dx dt

sin t  x cos t  2

t sin t
2t œ y Ê sin t  t cos t
2 œ Ê xœ

1 #

; therefore

œ

dy dx ¹ tœ1

œ 1 Ê (sin t  2)

dx dt dy dt ;

thus

sin 1  1 cos 1
2

–

1
Š1 # ‹ cos 1 sin 1  2

œ

dy dx

œ


41
8 21

dx dt

œ 1
x cos t Ê

sin t  t cos t
2 c x cos t ‰ ˆ 1sin tb2

(c) (d)

;

œ
4

2x w 68. 2x#  3y# œ 5 Ê 4x  6yyw œ 0 Ê yw œ
2x 3y Ê y k (1 1) œ
3y ¹ ß

3x# 2y

1
x cos t sin t2

—

3 #Î$
3, then f w (x) œ x
"Î$ and f ww (x) œ
"3 x
%Î$ so the # x 9 &Î$ if f(x) œ 10 x
7, then f w (x) œ 3# x#Î$ and f ww (x) œ x
"Î$ is true f ww (x) œ x
"Î$ Ê f www (x) œ
"3 x
%Î$ is true if f w (x) œ #3 x#Î$  6, then f ww (x) œ x
"Î$ is true

also, y# œ x$ Ê 2yyw œ 3x# Ê yw œ

œ

; t œ 1 Ê x sin 1  2x œ 1

67. (a) if f(x) œ (b)

dx dt

Ê yw k (1 1) œ ß

3x# 2y ¹ (1ß1)

œ

(1ß1)

3 #

claim f ww (x) œ x
"Î$ is false

œ
23 and yw k (1
1) œ
2x 3y ¹ ß

and yw k (1
1) œ ß

3x# 2y ¹ (1ß
1)

(1ß
1)

œ

2 3

;

œ
#3 . Therefore

the tangents to the curves are perpendicular at (1ß 1) and (1ß
1) (i.e., the curves are orthogonal at these two points of intersection). 69. x#  2xy
3y# œ 0 Ê 2x  2xyw  2y
6yyw œ 0 Ê yw (2x
6y) œ
2x
2y Ê yw œ tangent line m œ yw k (1ß1) œ

xy 3y
x ¹ (1ß1)

xy 3y
x

Ê the slope of the

œ 1 Ê the equation of the normal line at (1ß 1) is y
1 œ
1(x
1)

Ê y œ
x  2. To find where the normal line intersects the curve we substitute into its equation: x#  2x(2
x)
3(2
x)# œ 0 Ê x#  4x
2x#
3 a4
4x  x# b œ 0 Ê
4x#  16x
12 œ 0 Ê x#
4x  3 œ 0 Ê (x
3)(x
1) œ 0 Ê x œ 3 and y œ
x  2 œ
1. Therefore, the normal to the curve at (1ß 1) intersects the curve at the point (3ß
1). Note that it also intersects the curve at (1ß 1). 70. xy  2x
y œ 0 Ê x

dy dx

y2


dy dx

œ0 Ê

dy dx

œ

y2 1
x

; the slope of the line 2x  y œ 0 is
2. In order to be

parallel, the normal lines must also have slope of
2. Since a normal is perpendicular to a tangent, the slope of 2 " the tangent is "# . Therefore, y1 
x œ # Ê 2y  4 œ 1
x Ê x œ
3
2y. Substituting in the original equation,

y(
3
2y)  2(
3
2y)
y œ 0 Ê y#  4y  3 œ 0 Ê y œ
3 or y œ
1. If y œ
3, then x œ 3 and y  3 œ
2(x
3) Ê y œ
2x  3. If y œ
1, then x œ
1 and y  1 œ
2(x  1) Ê y œ
2x
3. 71. y# œ x Ê

dy dx

œ

" #y

y"
0 x"
a

. If a normal is drawn from (aß 0) to (x" ß y" ) on the curve its slope satisfies

œ
2y"

Ê y" œ
2y" (x"
a) or a œ x"  "# . Since x"   0 on the curve, we must have that a   "# . By symmetry, the two points on the parabola are ˆx" ß Èx" ‰ and ˆx" ß
Èx" ‰ . For the normal to be perpendicular, Èx

Èx

" " Š x "
a ‹ Š a
x " ‹ œ
1 Ê

x" (a
x" )#

Therefore, ˆ "4 ß „ #" ‰ and a œ 72. Ex. 6b.) Ex. 7a.)

3 4

œ 1 Ê x" œ (a
x" )# Ê x" œ ˆx" 

" 4

and y" œ „ #" .

.

"Î%

y œ a1
x# b has a derivative only on (
"ß ") because the function is defined only on [
"ß "] and the slope of the tangent becomes vertical at both x œ
1 and x œ 1. dy dx ‹

 y$  x#

dy dx

 2xy œ 0 Ê

$

y  2xy $ # # $ œ
3xy #  x# ; also, xy  x y œ 6 Ê x a3y b  y dx dy

#


x" ‰ Ê x" œ

y œ x"Î# has no derivative at x œ 0 because the slope of the graph becomes vertical at x œ 0.

73. xy$  x# y œ 6 Ê x Š3y#

Ê

" #

#

#

x œ
3xy y$  2xy ; thus

dx dy

appears to equal

"

dy dx

dx dy

dy dx

a3xy#  x# b œ
y$
2xy Ê

 x#  y Š2x

dx dy ‹

œ0 Ê

dx dy

dy dx

œ


y$
2xy 3xy#  x#

ay$  2xyb œ
3xy#
x#

. The two different treatments view the graphs as functions

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.6 Implicit Differentiation symmetric across the line y œ x, so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a). 74. x$  y# œ sin# y Ê 3x#  2y œ

3x# 2 sin y cos y
2y

appears to equal

dy dx

œ (2 sin y)(cos y)

; also, x$  y# œ sin# y Ê 3x# "

dy dx

dx dy

dy dx

Ê

dy dx

(2y
2 sin y cos y) œ
3x# Ê

 2y œ 2 sin y cos y Ê

dx dy

œ

2 sin y cos y
2y 3x#

75. x%  4y# œ 1: (a)

% y œ 1
4x Ê dy dx œ

œ


3x# 2y
2 sin y cos y

; thus

dx dy

. The two different treatments view the graphs as functions symmetric across the line

y œ x so their slopes are reciprocals of one another at the corresponding points (aß b) and (bß a).

#

dy dx

(b) Ê yœ

„ "# È1 %
"Î#




x%

„x $ "Î# a1
x % b differentiating implicitly, we find, 4x$  8y dy dx dy
4x$
4x$ „x $ Ê dx œ 8y œ œ "Î# . a1
x % b 8 Š„ "# È1
x% ‹

„ "4 a1
x b

a
4x$ b œ

76. (x
2)#  y# œ 4: (a) y œ „ È4
(x
2)#

; œ0

(b)

dy " #
"Î# (
2(x
2)) dx œ „ # a4
(x
2) b „(x
2) œ ; differentiating implicitly, c4
(x
2)# d"Î# dy
2(x
2) 2(x
2)  2y dy dx œ 0 Ê dx œ 2y
(x
2)
(x
2) „(x
2) œ y œ œ . „c4
(x
2)# d "Î# c4
(x
#)# d "Î#

Ê

77-84. Example CAS commands: Maple: q1 := x^3-x*y+y^3 = 7; pt := [x=2,y=1]; p1 := implicitplot( q1, x=-3..3, y=-3..3 ): p1;

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

167

168

Chapter 3 Differentiation

eval( q1, pt ); q2 := implicitdiff( q1, y, x ); m := eval( q2, pt ); tan_line := y = 1 + m*(x-2); p2 := implicitplot( tan_line, x=-5..5, y=-5..5, color=green ): p3 := pointplot( eval([x,y],pt), color=blue ): display( [p1,p2,p3], ="Section 3.6 #77(c)" ); Mathematica: (functions and x0 may vary): Note use of double equal sign (logic statement) in definition of eqn and tanline. <
dA dt dS dt

œ 21r

dr dt

œ 81r

3. (a) V œ 1r# h Ê

dV dt dV dt

(c) V œ 1r# h Ê

dr dt

œ 1 r# œ

4. (a) V œ "3 1r# h Ê

dh dt # dh 1r dt

dV dt œ " # dh 2 3 1r dt  3 1rh

(b) V œ 1r# h Ê  21rh

dV dt

" # dh 3 1r dt dr dt

(b) V œ "3 1r# h Ê

dV dt

dV dt

œ

5. (a)

dV dt dV dt dR dt

" œ 1 volt/sec (b) dI dt œ
3 amp/sec " ˆ dV dI ‰ " ˆ dV V dI ‰ ‰ ˆ dR ‰ Ê dR œ R ˆ dI Ê dR dt  I dt dt œ I dt
R dt dt œ I dt
I dt ˆ " ‰‘ œ ˆ #" ‰ (3) œ 3# ohms/sec, R is increasing œ "#
1
12 #
3

(d)

6. (a) P œ RI# Ê

dP dt

œ I#

(b) P œ RI# Ê 0 œ

dP dt

dR dt

 2RI

œ I#

dR dt

dr dt

dr dt

(c)

(c)

œ 21rh

œ 32 1rh

dr dt

dI dt

 2RI

dI dt

Ê

dR dt

œ
2RI I#

7. (a) s œ Èx#  y# œ ax#  y# b

"Î#

Ê

ds dt

œ

x dx Èx#  y# dt

(b) s œ Èx#  y# œ ax#  y# b

"Î#

Ê

ds dt

œ

x dx Èx#  y# dt

(c) s œ Èx#  y# Ê s# œ x#  y# Ê 2s

ds dt

œ 2x



dx dt

8. (a) s œ Èx#  y#  z# Ê s# œ x#  y#  z# Ê 2s

dI dt

2 ˆ PI ‰ dI I# dt

œ
2P I$

dI dt

y dy Èx#  y# dt

 2y

ds dt

œ


dy dt

œ 2x

Ê 2s † 0 œ 2x dx dt

 2y

dy dt

 2z

dx dt

 2y

dy dt

Ê

dz dt

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dx dt

œ
yx

dy dt

Section 3.7 Related Rates Ê

ds dt

œ

x dx Èx#  y#  z# dt



y dy Èx#  y#  z# dt

(b) From part (a) with

dx dt

œ0 Ê

(c) From part (a) with

ds dt

œ 0 Ê 0 œ 2x

9. (a) A œ (c) A œ

" # " #

ab sin ) Ê ab sin ) Ê

10. Given A œ 1r# ,

dr dt

dA dt dA dt

œ œ

" # " #

œ

ds dt



z dz Èx#  y#  z# dt

y dy Èx#  y#  z# dt dx dt

d) dt d) dt

ab cos ) ab cos )

 2y

dy dt



z dz Èx#  y#  z# dt

 2z

Ê

dz dt

(b) A œ  "# b sin )

z dz x dt

" #

ab sin ) Ê

dA dt

 "# a sin )

da dt

œ 0.01 cm/sec, and r œ 50 cm. Since



dx dt

œ 21r

dA dt



y dy x dt

œ0 œ

" #

ab cos )

d) dt

 "# b sin )

db dt

dr dt

, then

dA ¸ dt r=50

" ‰ œ 21(50) ˆ 100

œ 1 cm# /min.

(a) (b) (c)

dj dt

dw dt œ 2 cm/sec, j œ 12 cm and w œ 5 cm. dj dA # A œ jw Ê œ j dw dt  w dt Ê dt œ 12(2)  5(
2) œ 14 cm /sec, increasing dP dj dw P œ 2j  2w Ê dt œ 2 dt  2 dt œ 2(
2)  2(2) œ 0 cm/sec, constant "Î# " dj ‰ # #
"Î# ˆ D œ Èw#  j# œ aw#  j# b Ê dD 2w dw Ê dD dt œ # aw  j b dt  2j dt dt

11. Given

œ
2 cm/sec, dA dt

œ

(5)(2) (12)(
2) È25  144

12. (a) V œ xyz Ê

dV dt

œ

dj w dw dt  j dt Èw#  j#

œ
14 13 cm/sec, decreasing œ yz

 xz

dx dt

 xy

dy dt

dz dt

Ê

dV ¸ dt (4ß3ß2)

œ (3)(2)(1)  (4)(2)(
2)  (4)(3)(1) œ 2 m$ /sec

dx (b) S œ 2xy  2xz  2yz Ê dS dt œ (2y  2z) dt  (2x  2z) ¸ Ê dS œ (10)(1)  (12)(
2)  (14)(1) œ 0 m# /sec dt

dy dt

 (2x  2y)

dz dt

(4ß3ß2)

(c) j œ Èx#  y#  z# œ ax#  y#  z# b Ê 13. Given:

dj ¸ dt (4ß3ß2)

dx dt

"Î#

dj dt

Ê

œ

x dx Èx#  y#  z# dt



y dy Èx#  y#  z# dt



z dz Èx#  y#  z# dt

œ Š È429 ‹ (1)  Š È329 ‹ (
2)  Š È229 ‹ (1) œ 0 m/sec

œ 5 ft/sec, the ladder is 13 ft long, and x œ 12, y œ 5 at the instant of time

(a) Since x#  y# œ 169 Ê

dy dt

œ
xy

dx dt

‰ œ
ˆ 12 5 (5) œ
12 ft/sec, the ladder is sliding down the wall

(b) The area of the triangle formed by the ladder and walls is A œ is changing at (c) cos ) œ

x 13

" #

xy Ê

dA dt

œ ˆ "# ‰ Šx

dy dt

y

dx dt ‹ .

The area

# [12(
12)  5(5)] œ
119 # œ
59.5 ft /sec.

Ê
sin )

14. s# œ y#  x# Ê 2s

" #

ds dt

d) dt

œ 2x

†

dx dt

Ê

d) dt

" œ
13 sin ) †

 2y

dy dt

Ê

ds dt

œ

œ

dx dt

" 13

" s

Šx

dx dt

dx dt

y

œ
ˆ 5" ‰ (5) œ
1 rad/sec

dy dt ‹

Ê

ds dt

œ

" È169

[5(
442)  12(
481)]

œ
614 knots 15. Let s represent the distance between the girl and the kite and x represents the horizontal distance between the girl and kite Ê s# œ (300)#  x# Ê

ds dt

œ

x dx s dt

œ

400(25) 500

œ 20 ft/sec.

" # 16. When the diameter is 3.8 in., the radius is 1.9 in. and dr dt œ 3000 in/min. Also V œ 61r Ê $ ˆ " ‰ Ê dV dt œ 121(1.9) 3000 œ 0.00761. The volume is changing at about 0.0239 in /min.

17. V œ (a) (b)

" " ˆ 4h ‰# 1 h$ 3 3r 4h # h œ 1627 3 1r h, h œ 8 (2r) œ 4 Ê r œ 3 Ê V œ 3 1 3 dh ¸ 90 ˆ 9 ‰ dt h=4 œ 1614# (10) œ 2561 ¸ 0.1119 m/sec œ 11.19 cm/sec dr 4 dh 4 ˆ 90 ‰ 15 r œ 4h 3 Ê dt œ 3 dt œ 3 2561 œ 321 ¸ 0.1492 m/sec œ 14.92

Ê

dV dt

œ

dV dt

œ 121r

161h# dh 9 dt

cm/sec

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dr dt

da dt

169

170

Chapter 3 Differentiation

18. (a) V œ

" 3

1r# h and r œ

Ê Vœ

15h #

" 3

#

‰ hœ 1 ˆ 15h #

751h$ 4

¸
0.0113 m/min œ
1.13 cm/min dr 15 dh dr ¸
8 ‰ ˆ 15 ‰ ˆ 225 (b) r œ 15h # Ê dt œ # dt Ê dt h=5 œ # 1 œ 19. (a) V œ

1 3

y# (3R
y) Ê

y œ 8 we have

" 1441 (
6)

œ

dy dt

1 3

œ

dV dt


4 151

c2y(3R
y)  y# (
1)d œ

œ

dV dt

2251h# dh 4 dt

Ê

œ

dh ¸ dt h=5

4(
50) 2251(5)#

œ


8 2251

¸
0.0849 m/sec œ
8.49 cm/sec dy dt

Ê

dy dt


" dV dt

œ
13 a6Ry
3y# b‘

Ê at R œ 13 and

m/min

(b) The hemisphere is on the circle r  (13
y)# œ 169 Ê r œ È26y
y# m (c) r œ a26y
y# b
5 2881

œ 20. If V œ

4 3

"Î#

Ê

" #

œ

dr dt

#


" 241

Ê

a26y
y# b


"Î#

(26
2y)

dy dt

Ê

dr dt

œ

13
y dy È26y
y# dt

Ê

dr ¸ dt y=8

œ

13
8 È26†8
64

ˆ #
" ‰ 41

m/min

1r$ , S œ 41r# , and

dV dt

œ kS œ 4k1r# , then

dV dt

œ 41r#

Ê 4k1r# œ 41r#

dr dt

dr dt

Ê

dr dt

œ k, a constant.

Therefore, the radius is increasing at a constant rate. 4 dV dV dr $ $ # dr 3 1r , r œ 5, and dt œ 1001 ft /min, then dt œ 41r dt Ê dt dr # dt œ 81(5)(1) œ 401 ft /min, the rate at which the surface area

21. If V œ œ 81r

œ 1 ft/min. Then S œ 41r# Ê

dS dt

is increasing.

22. Let s represent the length of the rope and x the horizontal distance of the boat from the dock. s ds s ds (a) We have s# œ x#  36 Ê dx dt œ x dt œ È # dt . Therefore, the boat is approaching the dock at s
36

dx ¸ dt s=10

œ

(b) cos ) œ d) dt

Ê

10 È10#
36

6 r

œ

(
2) œ
2.5 ft/sec.

Ê
sin ) 6 8 ‰ 10# ˆ 10

d) dt

œ
r6#

† (
2) œ


3 20

Ê

dr dt

d) dt

œ

6 dr r# sin ) dt

. Thus, r œ 10, x œ 8, and sin ) œ

8 10

rad/sec

23. Let s represent the distance between the bicycle and balloon, h the height of the balloon and x the horizontal distance between the balloon and the bicycle. The relationship between the variables is s# œ h#  x# " ˆ dh dx ‰ " Ê ds Ê ds dt œ s h dt  x dt dt œ 85 [68(1)  51(17)] œ 11 ft/sec. 24. (a) Let h be the height of the coffee in the pot. Since the radius of the pot is 3, the volume of the coffee is dh dh " dV 10 V œ 91h Ê dV dt œ 91 dt Ê the rate the coffee is rising is dt œ 91 dt œ 91 in/min. (b) Let h be the height of the coffee in the pot. From the figure, the radius of the filter r œ œ

1 h$ 1#

, the volume of the filter. The rate the coffee is falling is

25. y œ QD
" Ê 26. (a) (b)

dc dt dc dt dp dt

dy dt

œ D
"

dQ dt

œ a3x#
12x  15b #


QD
#

dD dt

œ

" 41

(0)


233 (41)#

(
2) œ

dh dt

466 1681

œ a3(2)#
12(2)  15b (0.1) œ 0.3,

dx dt
#

œ a3x
12x
45x b

dx dt

dr dt
#

#

œ

4 dV 1h# dt

œ

4 #5 1

Ê Vœ

h #

" 3

1r# h

(
10) œ
581 in/min.

L/min Ê increasing about 0.2772 L/min œ9

dx dt

œ 9(0.1) œ 0.9,

œ a3(1.5)
12(1.5)
45(1.5) b (0.05) =
1.5625,

dr dt

dp dt

œ 70

œ 0.9
0.3 œ 0.6 dx dt

œ 70(0.05) œ 3.5,

œ 3.5
(
1.5625) œ 5.0625

27. Let P(xß y) represent a point on the curve y œ x# and ) the angle of inclination of a line containing P and the origin. Consequently, tan ) œ #

and cos )kx=3 œ

x# y # x #

œ

28. y œ (
x)"Î# and tan ) œ

3# 9 # 3 #

y x

y x

œ

Ê tan ) œ " 10

, we have

Ê tan ) œ

(
x)"Î# x

x# d) # x œ x Ê sec ) dt d) ¸ dt x=3 œ 1 rad/sec.

Ê sec# )

d) dt

œ

œ

dx dt

Ê

d) dt

œ cos# )

dx dt

. Since

ˆ "# ‰ (
x)
"Î# (
1)x
(
x)"Î# (1) dx x# dt

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dx dt

œ 10 m/sec

Section 3.7 Related Rates d) dt

Ê d) dt

cx

È
È
x dx # œ Œ 2 cx x#  acos )b ˆ dt ‰ . Now, tan ) œ 4


2

œ Š 4 16 ‹ ˆ 45 ‰ (
8) œ

2 5

2
4

œ
#" Ê cos ) œ
È25 Ê cos# ) œ

" #

a x#  y # b


"Î#

Š2x

dx dt

. Then

rad/sec.

29. The distance from the origin is s œ Èx#  y# and we wish to find œ

4 5

 2y

dy dt ‹¹ (5ß12)

œ

(5)(
1)  (12)(
5) È25  144

ds ¸ dt (5ß12)

œ
5 m/sec

30. When s represents the length of the shadow and x the distance of the man from the streetlight, then s œ

3 5

x.

(a) If I represents the distance of the tip of the shadow from the streetlight, then I œ s  x Ê œ  dx dt 3 dx dx ¸ 8 ¸ ¸ dx ¸ 8 ¸ ¸ ¸ (which is velocity not speed) Ê ¸ dI œ  œ œ k
5 k œ 8 ft/sec, the speed the tip of the dt 5 dt dt 5 dt 5 dI dt

ds dt

shadow is moving along the ground. ds 3 dx 3 dt œ 5 dt œ 5 (
5) œ
3 ft/sec, so the length of the shadow is decreasing at a rate of 3 ft/sec.

(b)

31. Let s œ 16t# represent the distance the ball has fallen, h the distance between the ball and the ground, and I the distance between the shadow and the point directly beneath the ball. Accordingly, s  h œ 50 and since the triangle LOQ and triangle PRQ are similar we have Iœ œ

30h 50
h 1500 16t#


Ê h œ 50
16t# and I œ 30 Ê

dI dt

œ
1500 8t$ Ê

30 a50
16t# b 50
a50
16t# b

dI ¸ dt t= 12

œ
1500 ft/sec.

32. Let s œ distance of car from foot of perpendicular in the textbook diagram Ê tan ) œ Ê

d) dt

œ

#

cos ) ds 132 dt

;

ds dt

œ
264 and ) œ 0 Ê

d) dt

4 3

1r$
43 14$ Ê

thickness of the ice is decreasing at

5 721

Ê sec# )

d) dt

" ds 13# dt

œ

œ
2 rad/sec. A half second later the car has traveled 132 ft

right of the perpendicular Ê k)k œ 14 , cos# ) œ "# , and 33. The volume of the ice is V œ

s 13#

dV dt

ds dt

œ 264 (since s increases) Ê

œ 41r#

dr dt

Ê

dr ¸ dt r=6

œ


5 721

10 3

œ

in./min when

in/min. The surface area is S œ 41r# Ê

# œ
10 3 in /min, the outer surface area of the ice is decreasing at

d) dt

dS dt

œ 81r

dr dt

in# /min.

ˆ "# ‰ 132

(264) œ 1 rad/sec.

œ
10 in$ /min, the
5 ‰ ¸ œ 481 ˆ 72 Ê dS dt 1

dV dt

r=6

34. Let s represent the horizontal distance between the car and plane while r is the line-of-sight distance between r dr ds ¸ 5 the car and plane Ê 9  s# œ r# Ê ds dt œ È # dt Ê dt r=5 œ È16 (
160) œ
200 mph r
9

Ê speed of plane  speed of car œ 200 mph Ê the speed of the car is 80 mph. 35. When x represents the length of the shadow, then tan ) œ We are given that ¸ dx ¸ dt

œ

d) dt

œ 0.27° œ

# # ¹
x 80sec ) ddt) ¹¹ d) Š = dt

31 2000

31 #000

80 x

Ê sec# )

rad/min. At x œ 60, cos ) œ

and sec ) = 35 ‹

œ

31 16

3 5

d) dt

œ
80 x#

dx dt

Ê

dx dt

œ


x# sec# ) d) 80 dt

Ê

ft/min ¸ 0.589 ft/min ¸ 7.1 in./min.

36. Let A represent the side opposite ) and B represent the side adjacent ). tan ) œ AB Ê sec# ) ddt) œ B" dA dt
2 d) " ‰ 10 4‰
ˆ ˆ ‰ ‘ ˆ t Ê at A œ 10 m and B œ 20 m we have cos ) œ 20 œ and œ (
2)
(1) È È dt #0 400 5 10

œ ˆ
" 10


" ‰ ˆ4‰ 40 5

5

#

A dB B# dt

5

" œ
10 rad/sec œ
18° 1 /sec ¸
6°/sec

37. Let x represent distance of the player from second base and s the distance to third base. Then #

.

(a) s œ x  8100 Ê 2s

ds dt

œ 2x

dx dt

Ê

ds dt

œ

x dx s dt

dx dt

œ
16 ft/sec

. When the player is 30 ft from first base, x œ 60

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

171

172

Chapter 3 Differentiation Ê s œ 30È13 and

(b) cos )" œ Ê

(c)

d)" dt

œ


90 sx

d)" dt

œ

90 s

œ

œ

ds dt

Ê
sin )"

90 Š30È13‹ (60)

†

ds dt . 90 s# sin )"

60 30È13

d)" dt


32 È13

(
16) œ

œ
90 s# †


32 † ŠÈ ‹œ 13


8 65

ds dt

Ê

¸
8.875 ft/sec

d)" dt

œ

90 s# sin )"

rad/sec; sin )# œ

90 s

†

ds dt

œ

90 sx

†

Ê cos )#

. Therefore, x œ 60 and s œ 30È13

ds dt d)# dt

œ
90 s# †

8 Therefore, x œ 60 and s œ 30È13 Ê ddt)# œ 65 rad/sec. ds 90 x dx 90 dx 90 † dt œ ˆs# † x ‰ † ˆ s ‰ † ˆ dt ‰ œ ˆ s# ‰ ˆ dt ‰ œ ˆ x#  8100 ‰ dx dt Ê lim

xÄ!

s

œ lim ˆ x# 908100 ‰ (
15) œ
6" rad/sec; xÄ!


90 ‰ œ ˆ x#  8100

dx dt

d)# x Ä ! dt

Ê lim

œ

" 6

d)# dt

œ


90 s# cos )#

†

ds dt

ds dt

Ê

d)# dt

œ


90 s# cos )#

†

ds dt

d)" dt

90 ˆ x ‰ ˆ dx ‰ ˆ
s90 ‰ ˆ dx ‰ œ Š
# s# † x ‹ s dt œ dt s

rad/sec

38. Let a represent the distance between point O and ship A, b the distance between point O and ship B, and D the distance between the ships. By the Law of Cosines, D# œ a#  b#
2ab cos 120° "
da db db da ‘ da db dD 413 Ê dD dt œ #D 2a dt  2b dt  a dt  b dt . When a œ 5, dt œ 14, b œ 3, and dt œ 21, then dt œ 2D where D œ 7. The ships are moving

dD dt

œ 29.5 knots apart.

3.8 LINEARIZATION AND DIFFERENTIALS 1. f(x) œ x$
2x  3 Ê f w (x) œ 3x#
2 Ê L(x) œ f w (2)(x
2)  f(2) œ 10(x
2)  7 Ê L(x) œ 10x
13 at x œ 2 2. f(x) œ Èx#  9 œ ax#  9b

"Î#

Ê f w (x) œ ˆ "# ‰ ax#  9b

œ
45 (x  4)  5 Ê L(x) œ
45 x  3. f(x) œ x 

" x

9 5


"Î#

(2x) œ

x È x#  9

Ê L(x) œ f w (
4)(x  4)  f(
4)

at x œ
4

Ê f w (x) œ 1
x
# Ê L(x) œ f(1)  f w (1)(x
1) œ #  !(x
1) œ #

4. f(x) œ x"Î$ Ê f w (x) œ

" $x#Î$

Ê L(x) œ f w (
8)ax
a
8bb  fa
8b œ

" 1#

(x  8)
2 Ê L(x) œ

" 1#

x


4 3

5. f(x) œ x#  2x Ê f w (x) œ 2x  2 Ê L(x) œ f w (0)(x
0)  f(0) œ 2(x
0)  0 Ê L(x) œ 2x at x œ 0 6. f(x) œ x
" Ê f w (x) œ
x
# Ê L(x) œ f w (1)(x
1)  f(1) œ (
1)(x
1)  1 Ê L(x) œ
x  2 at x œ 1 7. f(x) œ 2x#  4x
3 Ê f w (x) œ 4x  4 Ê L(x) œ f w (
1)(x  1)  f(
1) œ 0(x  1)  (
5) Ê L(x) œ
5 at x œ
1 8. f(x) œ 1  x Ê f w (x) œ 1 Ê L(x) œ f w (8)(x
8)  f(8) œ 1(x
8)  9 Ê L(x) œ x  1 at x œ 8 9. f(x) œ $Èx œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x
#Î$ Ê L(x) œ f w (8)(x
8)  f(8) œ 10. f(x) œ

x x1

Ê L(x) œ

Ê f w (x) œ " 4

x

" 4

(1)(x  1)
(")(x) (x  1)#

œ

" (x  1)#

" 1#

(x
8)  2 Ê L(x) œ

Ê L(x) œ f w (1)(x
1)  f(1) œ

" 4

(x
1) 

" #

at x œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 1#

x

4 3

at x œ 8

Section 3.8 Linearization and Differentials 11. f(x) œ sin x Ê f w (x) œ cos x (a) L(x) œ f w (0)(x
0)  f(0) œ 1(x
0)  0 Ê L(x) œ x at x œ 0 (b) L(x) œ f w (1)(x
1)  f(1) œ (
1)(x
1)  0 Ê L(x) œ 1
x at x œ 1

12. f(x) œ cos x Ê f w (x) œ
sin x (a) L(x) œ f w (0)(x
0)  f(0) œ 0(x
0)  1 Ê L(x) œ 1 at x œ 0 (b) L(x) œ f w ˆ
1# ‰ ˆx  1# ‰  f ˆ
1# ‰ œ
(
1) ˆx  1# ‰  0 Ê L(x) œ x 

at x œ
1#

1 2

13. f(x) œ sec x Ê f w (x) œ sec x tan x (a) L(x) œ f w (0)(x
0)  f(0) œ 0(x
0)  1 Ê L(x) œ 1 at x œ 0 (b) L(x) œ f w ˆ
13 ‰ ˆx  13 ‰  f ˆ
13 ‰ œ
2È3 ˆx  13 ‰  2 Ê L(x) œ 2
2È3 ˆx  13 ‰ at x œ
13

14. f(x) œ tan x Ê f w (x) œ sec# x (a) L(x) œ f w (0)(x
0)  f(0) œ 1(x
0)  0 œ x Ê L(x) œ x at x œ 0 (b) L(x) œ f w ˆ 14 ‰ ˆx
14 ‰  f ˆ 14 ‰ œ 2 ˆx
14 ‰  1 Ê L(x) œ 1  2 ˆx
14 ‰ at x œ 14

15. f w axb œ ka"  xbk
" . We have fa!b œ " and f w a!b œ k. Laxb œ fa!b  f w a!bax
!b œ "  kax
!b œ "  kx '

16. (a) faxb œ a"
xb' œ
"  a
xb‘ ¸ "  'a
xb œ "
'x (b) faxb œ

# "
x


"

œ #
"  a
xb‘

(c) faxb œ a"  xb


"Î#

(d) faxb œ È"  x# œ

¸ #
"  a
"ba
xb‘ œ #  #x

¸ "  ˆ
"# ‰x œ "
x# "Î# È#Š"  x# ‹ ¸ È#Š" #

(e) faxb œ a%  $xb"Î$ œ %"Î$ ˆ" 

$x ‰"Î$ %



" x# # #‹

¸ %"Î$ ˆ" 

œ È#Š" 

" $x ‰ $ %

x# %‹

œ %"Î$ ˆ"  x% ‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

173

174

Chapter 3 Differentiation

(f) faxb œ ˆ"


" ‰2Î$ #x

2Î$

œ ’"  ˆ
# " x ‰“

¸ "  $# ˆ
# " x ‰ œ "


# '$x

17. (a) (1.0002)&! œ (1  0.0002)&! ¸ 1  50(0.0002) œ 1  .01 œ 1.01 (b) $È1.009 œ (1  0.009)"Î$ ¸ 1  ˆ " ‰ (0.009) œ 1  0.003 œ 1.003 3

18. f(x) œ Èx  1  sin x œ (x  1)"Î#  sin x Ê f w (x) œ ˆ "# ‰ (x  1)
"Î#  cos x Ê Lf (x) œ f w (0)(x
0)  f(0) œ 3 (x
0)  1 Ê Lf (x) œ 3 x  1, the linearization of f(x); g(x) œ Èx  1 œ (x  1)"Î# Ê gw (x) #

#

œ ˆ "# ‰ (x  1)
"Î# Ê Lg (x) œ gw (0)(x
0)  g(0) œ w

w

" #

(x
0)  1 Ê Lg (x) œ

" #

x  1, the linearization of g(x);

h(x) œ sin x Ê h (x) œ cos x Ê Lh (x) œ h (0)(x
0)  h(0) œ (1)(x
0)  0 Ê Lh (x) œ x, the linearization of h(x). Lf (x) œ Lg (x)  Lh (x) implies that the linearization of a sum is equal to the sum of the linearizations. 19. y œ x$
3Èx œ x$
3x"Î# Ê dy œ ˆ3x#
#3 x
"Î# ‰ dx Ê dy œ Š3x#


3 ‹ 2È x

dx

"Î# "Î#
"Î# 20. y œ xÈ1
x# œ x a1
x# b Ê dy œ ’(1) a1
x# b  (x) ˆ "# ‰ a1
x# b (
2x)“ dx

œ a1
x# b


"Î#

#

a1
2x# b È 1
x#

Ê dy œ Š (2) a1 a1xb x
# b(2x)(2x) ‹ dx œ #

21. y œ

2x 1 x #

22. y œ

2È x 3 ˆ1  È x ‰

Ê dy œ

ca1
x# b
x# d dx œ

œ

2x"Î# 3 a1  x"Î# b

" # 3 È x ˆ1  È x ‰

Ê dy œ Š

dx 2
2x# a 1  x # b#

dx

x
"Î# ˆ3 ˆ1  x"Î# ‰‰
2x"Î# ˆ #3 x
"Î# ‰ 9 a1  x"Î# b

#

‹ dx œ

3x
"Î#  3
3 # 9 a1  x"Î# b

dx

dx

23. 2y$Î#  xy
x œ 0 Ê 3y"Î# dy  y dx  x dy
dx œ 0 Ê ˆ3y"Î#  x‰ dy œ (1
y) dx Ê dy œ

1
y 3 È y x

24. xy#
4x$Î#
y œ 0 Ê y# dx  2xy dy
6x"Î# dx
dy œ 0 Ê (2xy
1) dy œ ˆ6x"Î#
y# ‰ dx Ê dy œ

6È x
y# 2xy
1

dx

25. y œ sin ˆ5Èx‰ œ sin ˆ5x"Î# ‰ Ê dy œ ˆcos ˆ5x"Î# ‰‰ ˆ 5# x
"Î# ‰ dx Ê dy œ

5 cos ˆ5Èx‰ 2È x

dx

26. y œ cos ax# b Ê dy œ c
sin ax# bd (2x) dx œ
2x sin ax# b dx $

$

$

27. y œ 4 tan Š x3 ‹ Ê dy œ 4 Šsec# Š x3 ‹‹ ax# b dx Ê dy œ 4x# sec# Š x3 ‹ dx 28. y œ sec ax#
1b Ê dy œ csec ax#
1b tan ax#
1bd (2x) dx œ 2x csec ax#
1b tan ax#
1bd dx 29. y œ 3 csc ˆ1
2Èx‰ œ 3 csc ˆ1
2x"Î# ‰ Ê dy œ 3 ˆ
csc ˆ1
2x"Î# ‰‰ cot ˆ1
2x"Î# ‰ ˆ
x
"Î# ‰ dx Ê dy œ È3 csc ˆ1
2Èx‰ cot ˆ1
2Èx‰ dx x

30. y œ 2 cot Š È"x ‹ œ 2 cot ˆx
"Î# ‰ Ê dy œ
2 csc# ˆx
"Î# ‰ ˆ
#" ‰ ˆx
$Î# ‰ dx Ê dy œ

" È x$

csc# Š È"x ‹ dx

31. f(x) œ x#  2x, x! œ 1, dx œ 0.1 Ê f w (x) œ 2x  2 (a) ?f œ f(x!  dx)
f(x! ) œ f(1.1)
f(1) œ 3.41
3 œ 0.41 (b) df œ f w (x! ) dx œ [2(1)  2](0.1) œ 0.4 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dx

Section 3.8 Linearization and Differentials

175

(c) k?f
df k œ k0.41
0.4k œ 0.01 32. f(x) œ 2x#  4x
3, x! œ
1, dx œ 0.1 Ê f w (x) œ 4x  4 (a) ?f œ f(x!  dx)
f(x! ) œ f(
.9)
f(
1) œ .02 (b) df œ f w (x! ) dx œ [4(
1)  4](.1) œ 0 (c) k?f
df k œ k.02
0k œ .02 33. f(x) œ x$
x, x! œ 1, dx œ 0.1 Ê f w (x) œ 3x#
1 (a) ?f œ f(x!  dx)
f(x! ) œ f(1.1)
f(1) œ .231 (b) df œ f w (x! ) dx œ [3(1)#
1](.1) œ .2 (c) k?f
df k œ k.231
.2k œ .031 34. f(x) œ x% , x! œ 1, dx œ 0.1 Ê f w (x) œ 4x$ (a) ?f œ f(x!  dx)
f(x! ) œ f(1.1)
f(1) œ .4641 (b) df œ f w (x! ) dx œ 4(1)$ (.1) œ .4 (c) k?f
df k œ k.4641
.4k œ .0641 35. f(x) œ x
" , x! œ 0.5, dx œ 0.1 Ê f w (x) œ
x
# (a) ?f œ f(x!  dx)
f(x! ) œ f(.6)
f(.5) œ
"3 " ‰ (b) df œ f w (x! ) dx œ (
4) ˆ 10 œ
25 (c) k?f
df k œ ¸
"3  25 ¸ œ

" 15

36. f(x) œ x$
2x  3, x! œ 2, dx œ 0.1 Ê f w (x) œ 3x#
2 (a) ?f œ f(x!  dx)
f(x! ) œ f(2.1)
f(2) œ 1.061 (b) df œ f w (x! ) dx œ (10)(0.10) œ 1 (c) k?f
df k œ k1.061
1k œ .061 37. V œ

4 3

1r$ Ê dV œ 41r!# dr

38. V œ x$ Ê dV œ 3x!# dx

39. S œ 6x# Ê dS œ 12x! dx 40. S œ 1rÈr#  h# œ 1r ar#  h# b Ê

dS dr

œ

1 ar#  h# b  1r# È r#  h #

"Î#

Ê dS œ

, h constant Ê

1 a2r#!  h# b Ér#!  h#

dS dr

œ 1 ar#  h# b

"Î#

 1r † r ar#  h# b


"Î#

dr, h constant

41. V œ 1r# h, height constant Ê dV œ 21r! h dr

42. S œ 21rh Ê dS œ 21r dh

43. Given r œ 2 m, dr œ .02 m (a) A œ 1r# Ê dA œ 21r dr œ 21(2)(.02) œ .081 m# 1‰ (b) ˆ .08 41 (100%) œ 2% 44. C œ 21r and dC œ 2 in. Ê dC œ 21 dr Ê dr œ œ 21(5) ˆ 1" ‰ œ 10 in.#

" 1

Ê the diameter grew about

45. The volume of a cylinder is V œ 1r# h. When h is held fixed, we have

dV dr

2 1

in.; A œ 1r# Ê dA œ 21r dr

œ #1rh, and so dV œ #1rh dr. For h œ $! in.,

r œ ' in., and dr œ !Þ& in., the volume of the material in the shell is approximately dV œ #1rh dr œ #1a'ba$!ba!Þ&b œ ")!1 ¸ &'&Þ& in$ .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

176

Chapter 3 Differentiation

46. Let ) œ angle of elevation and h œ height of building. Then h œ $!tan ), so dh œ $!sec# ) d). We want ldhl  !Þ!%h, &1 &1 sin ) which gives: l$!sec# ) d)l  !Þ!%l$!tan )l Ê cos"# ) ld)l  !Þ!% cos ) Ê ld)l  !Þ!%sin ) cos ) Ê ld)l  !Þ!%sin "# cos "# œ !Þ!" radian. The angle should be measured with an error of less than !Þ!" radian (or approximatley !Þ&( degrees), which is a percentage error of approximately !Þ('%.

47. V œ 1h$ Ê dV œ 31h# dh; recall that ?V ¸ dV. Then k?Vk Ÿ (1%)(V) œ (1) a1h$ b 100

Ê k31h# dhk Ÿ of h is

" 3

Ê kdhk Ÿ

" 300

(1) a1h$ b 100

Ê kdVk Ÿ

(1) a1h$ b 100

h œ ˆ "3 %‰ h. Therefore the greatest tolerated error in the measurement

%. #

48. (a) Let Di represent the inside diameter. Then V œ 1r# h œ 1 ˆ D#i ‰ h œ

1D#i h 4

dV œ 51Di dDi . Recall that ?V ¸ dV. We want k?Vk Ÿ (1%)(V) Ê kdVk Ÿ Ê 51Di dDi Ÿ

1D#i 40

Ê

dDi Di

51D#i # 1 D# œ 40i

and h œ 10 Ê V œ " ‰ 51D#i ˆ 100 Š # ‹

Ê

Ÿ 200. The inside diameter must be measured to within 0.5%.

(b) Let De represent the exterior diameter, h the height and S the area of the painted surface. S œ 1De h Ê dS œ 1hdDe dDe Ê dS S œ De . Thus for small changes in exterior diameter, the approximate percentage change in the exterior diameter is equal to the approximate percentage change in the area painted, and to estimate the amount of paint required to within 5%, the tanks's exterior diameter must be measured to within 5%. 49. V œ 1r# h, h is constant Ê dV œ 21rh dr; recall that ?V ¸ dV. We want k?Vk Ÿ Ê k21rh drk Ÿ

1r# h 1000

Ê kdrk Ÿ

r #000

" 1000

V Ê kdVk Ÿ

1 r# h 1000

œ (.05%)r Ê a .05% variation in the radius can be tolerated.

50. Volume œ (x  ?x)$ œ x$  3x# (?x)  3x(?x)#  (?x)$

51. W œ a 

b g

œ a  bg
" Ê dW œ
bg
# dg œ
bgdg Ê #

dWmoon dWearth

œ

b dg ‹ (5.2)# b dg Š
# ‹ (32)

Š


#

32 ‰ œ ˆ 5.2 œ 37.87, so a change of

gravity on the moon has about 38 times the effect that a change of the same magnitude has on Earth. 52. (a) T œ 21 Š Lg ‹

"Î#

Ê dT œ 21ÈL ˆ
"# g
$Î# ‰ dg œ
1ÈL g
$Î# dg

(b) If g increases, then dg  0 Ê dT  0. The period T decreases and the clock ticks more frequently. Both the pendulum speed and clock speed increase. (c) 0.001 œ
1È100 ˆ980
$Î# ‰ dg Ê dg ¸
0.977 cm/sec# Ê the new g ¸ 979 cm/sec# 53. The error in measurement dx œ (1%)(10) œ 0.1 cm; V œ x$ Ê dV œ 3x# dx œ 3(10)# (0.1) œ 30 cm$ Ê the 30 ‰ percentage error in the volume calculation is ˆ 1000 (100%) œ 3%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.8 Linearization and Differentials 54. A œ s# Ê dA œ 2s ds; recall that ?A ¸ dA. Then k?Ak Ÿ (2%)A œ Ê kdsk Ÿ

s# (2s)(50)

œ

s 100

œ” 56. V œ œ

4 3

1 D$ 200

# • a10 %b œ ”

1 r$ œ

4 3

10' 1 # 10' 1 6

$

1 ˆ D# ‰ œ

Ê kdVk Ÿ

1 D$ 200

œ

s# 50

Ê kdAk Ÿ

s# 50

Ê k2s dsk Ÿ

s# 50

œ (1%) s Ê the error must be no more than 1% of the true value.

55. Given D œ 100 cm, dD œ 1 cm, V œ 10% 1 # 10' 1 6

2s# 100

4 3

$

1 ˆ D# ‰ œ

1 D$ 6

Ê dV œ

1 #

D# dD œ

1 #

(100)# (1) œ

10% 1 #

. Then

dV V

(100%)

• % œ 3% 1 D$ 6

Ê dV œ #

Ê ¹ 1D# dD¹ Ÿ

1 D# #

3 ‰ 1D dD; recall that ?V ¸ dV. Then k?Vk Ÿ (3%)V œ ˆ 100 Š 6 ‹

$

1 D$ #00

Ê kdDk Ÿ

D 100

œ (1%) D Ê the allowable percentage error in

measuring the diameter is 1%. 57. A 5% error in measuring t Ê dt œ (5%)t œ

t 20

32t# 20

. Then s œ 16t# Ê ds œ 32t dt œ 32t ˆ 20t ‰ œ

œ

16t# 10

" ‰ œ ˆ 10 s

œ (10%)s Ê a 10% error in the calculation of s. 58. From Example 8 we have 59. lim

xÄ0

È1  x 1  x#

œ

È1  0 1  #0

dV V

œ4

dr r

. An increase of 12.5% in r will give a 50% increase in V.

œ1

60. lim

xÄ0

tan x x

œ lim ˆ sinx x ‰ ˆ cos" x ‰ œ (1)(1) œ 1 xÄ0

61. E(x) œ f(x)
g(x) Ê E(x) œ f(x)
m(x
a)
c. Then E(a) œ 0 Ê f(a)
m(a
a)
c œ 0 Ê c œ f(a). Next f(x)
m(x
a)
c f(a) œ 0 Ê xlim œ 0 Ê xlim ’ f(x)x
x
a
a
m“ œ 0 (since c œ f(a)) Äa Äa Ê f w (a)
m œ 0 Ê m œ f w (a). Therefore, g(x) œ m(x
a)  c œ f w (a)(x
a)  f(a) is the linear approximation, as claimed.

we calculate m: xlim Äa

E(x) x
a

62. (a) i. Qaab œ faab implies that b! œ faab. ii. Since Qw axb œ b"  #b# ax
ab, Qw aab œ f w aab implies that b" œ f w aab. iii. Since Qww axb œ #b# , Qww aab œ f ww aab implies that b2 œ In summary, b! œ faab, b" œ f w aab, and b2 œ (b) faxb œ a"
xb
"

ww

f aa b # .

ww

f aa b # .

f w axb œ
"a"
xb
# a
"b œ a"
xb
# f ww axb œ
#a"
xb
$ a
"b œ #a"
xb
$ Since fa!b œ ", f w a!b œ ", and f ww a!b œ #, the coefficients are b! œ ", b" œ ", b# œ #

approximation is Qaxb œ "  x  x . (c)

# #

œ ". The quadratic

As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.

(d) gaxb œ x
" gw axb œ
"x
# gww axb œ #x
$ Since ga"b œ ", gw a"b œ
", and gww a"b œ # , the coefficients are b! œ ", b" œ
", b# œ

# #

œ ". The quadratic

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

177

178

Chapter 3 Differentiation approximation is Qaxb œ "
ax
"b  ax
"b# . As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.

(e) haxb œ a"  xb"Î# hw axb œ "# a"  xb
"Î#

hww axb œ
"% a"  xb
$Î# Since ha!b œ ", hw a!b œ "# , and hww a!b œ
"% , the coefficients are b! œ ", b" œ "# , b# œ approximation is Qaxb œ " 

x #




#

x 8


"% 2

œ
"8 . The quadratic

. As one zooms in, the two graphs quickly become indistinguishable. They appear to be identical.

(f) The linearization of any differentiable function uaxb at x œ a is Laxb œ uaab  uw aabax
ab œ b!  b" ax
ab, where b! and b" are the coefficients of the constant and linear terms of the quadratic approximation. Thus, the linearization for faxb at x œ ! is "  x; the linearization for gaxb at x œ " is "
ax
"b or #
x; and the linearization for haxb at x œ ! is "  x# . 63. (a) x œ 1

(b) x œ 1; m œ 2.5, e1 ¸ 2.7

x œ 0; m œ 1, e0 œ 1

x œ
1; m œ 0.3, e
1 ¸ 0.4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 3.8 Linearization and Differentials

179

64. If f has a horizontal tangent at x œ a, then f w (a) œ 0 and the linearization of f at x œ a is L(x) œ f(a)  f w (a)(x
a) œ f(a)  0 † (x
a) œ f(a). The linearization is a constant. 65. Find lvl when m œ "Þ!"m! . m œ Ê lvl œ cÉ"
dv œ

"!"$ "!!$

c†m# Í ! Í m3! Í"
Ì

m#! m#

m!

# É"
v# c

Ê dv œ c † "# Š"


m# ! "!"# m# "!!# !

m! ‰ ˆ "!! œ

Ê mÉ"
m#! m# ‹

1!!! "!"$ Ê"


"!!# "!"#


"Î#

v# c#

œ m! Ê É"


v# c#

œ

m! m

Ê"


#m#

Š m$! ‹dm, dm œ !Þ!"m! Ê dv œ

v# c#

c m#! m$

Ê"


œ

m# ! m#

m!# m#

Ê v# œ c# Š"


m! ‰ ˆ "!! . mœ

m!# m# ‹

"!" "!! m! ,

¸ 0.69c. Body at rest Ê v! œ ! and v œ v!  dv

Ê v œ 0.69c. 66. (a) The successive square roots of 2 appear to converge to the number 1. For tenth roots the convergence is more rapid. (b) Successive square roots of 0.5 also converge to 1. In fact, successive square roots of any positive number converge to 1. A graph indicates what is going on:

Starting on the line y œ x, the successive square roots are found by moving to the graph of y œ Èx and then across to the line y œ x again. From any positive starting value x, the iterates converge to 1. 67-70. Example CAS commands: Maple: with(plots): a:= 1: f:=x -> x • 3  x • 2
2*x; plot(f(x), x=
1..2); diff(f(x),x); fp := unapply (ww ,x); L:=x -> f(a)  fp(a)*(x
a); plot({f(x), L(x)}, x=
1..2); err:=x -> abs(f(x)
L(x)); plot(err(x), x=
1..2, title = #absolute error function#); err(
1); Mathematica: (function, x1, x2, and a may vary): Clear[f, x] {x1, x2} = {
1, 2}; a = 1; f[x_]:=x3  x2
2x Plot[f[x], {x, x1, x2}] lin[x_]=f[a]  f'[a](x
a) Plot[{f[x], lin[x]}, {x, x1, x2}] err[x_]=Abs[f[x]
lin[x]] Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

180

Chapter 3 Differentiation

Plot[err[x], {x, x1,x 2}] err//N After reviewing the error function, plot the error function and epsilon for differing values of epsilon (eps) and delta (del) eps = 0.5; del = 0.4 Plot[{err[x], eps},{x, a
del, a  del}] CHAPTER 3 PRACTICE EXERCISES 1. y œ x&
0.125x#  0.25x Ê 2. y œ 3
0.7x$  0.3x( Ê

4. y œ x(  È7x


" 1 1

Ê

œ 5x%
0.25x  0.25

œ
2.1x#  2.1x'

dy dx

3. y œ x$
3 ax#  1# b Ê

dy dx

dy dx

œ 3x#
3(2x  0) œ 3x#
6x œ 3x(x
2)

dy dx

œ 7x'  È7

5. y œ (x  1)# ax#  2xb Ê

dy dx

œ (x  1)# (2x  2)  ax#  2xb (2(x  1)) œ 2(x  1) c(x  1)#  x(x  2)d

dy dx

œ (2x
5)(
1)(4
x)
# (
1)  (4
x)
" (2) œ (4
x)
# c(2x
5)  2(4
x)d

#

œ 2(x  1) a2x  4x  1b 6. y œ (2x
5)(4
x)
" Ê œ 3(4
x)


#

$

7. y œ a)#  sec )  1b Ê 8. y œ Š
1


csc ) #




)# 4‹

9. s œ

Èt 1  Èt

Ê

ds dt

œ

10. s œ

" Èt
1

Ê

ds dt

œ

#

Ê

ˆ1  Èt‰†

" sin# x




2 sin x

œ 2 Š
1


"

"

Èt
Èt Š #Èt ‹

#

ˆÈ t
1 ‰ dy dx

#

"

Èt ‹

#

œ

ds dt




)# ˆ csc ) cot ) 4‹ #

ˆ1  Èt‰
Èt 2Èt ˆ1  Èt‰

#

œ


#) ‰ œ Š
1


csc ) #




)# 4 ‹ (csc

) cot )
))

" # #Èt ˆ1  Èt‰


" # 2 È t ˆÈ t
1 ‰

dy dx

œ (2 csc x)(
csc x cot x)
2(
csc x cot x) œ (2 csc x cot x)(1
csc x)

œ 4 cos$ (1
2t)(
sin (1
2t))(
2) œ 8 cos$ (1
2t) sin (1
2t)

œ 3 cot# ˆ 2t ‰ ˆ
csc# ˆ 2t ‰‰ ˆ
t#2 ‰ œ

15. s œ (sec t  tan t)& Ê

ds dt

16. s œ csc& a1
t  3t# b Ê &

csc ) #

œ (4 tan x) asec# xb
(2 sec x)(sec x tan x) œ 2 sec# x tan x

œ csc# x
2 csc x Ê ds dt

œ

#

ˆÈt
1‰ (0)
1 Š

13. s œ cos% (1
2t) Ê 14. s œ cot$ ˆ 2t ‰ Ê

dy d)

ˆ1  Èt‰

11. y œ 2 tan# x
sec# x Ê 12. y œ

#

œ 3 a)#  sec )  1b (2)  sec ) tan ))

dy d)

6 t#

cot# ˆ 2t ‰ csc# ˆ 2t ‰

œ 5(sec t  tan t)% asec t tan t  sec# tb œ 5(sec t)(sec t  tan t)& ds dt

œ 5 csc% a1
t  3t# b a
csc a1
t  3t# b cot a1
t  3t# bb (
1  6t)

œ
5(6t
1) csc a1
t  3t# b cot a1
t  3t# b 17. r œ È2) sin ) œ (2) sin ))"Î# Ê

dr d)

œ

" #

(2) sin ))
"Î# (#) cos )  2 sin )) œ

) cos )  sin ) È2) sin )

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Practice Exercises 18. r œ 2)Ècos ) œ 2) (cos ))"Î# Ê

œ 2) ˆ "# ‰ (cos ))
"Î# (
sin ))  2(cos ))"Î# œ

dr d)


) sin ) Ècos )

181

 2Ècos )

2 cos )
) sin ) Ècos )

œ

19. r œ sin È2) œ sin (2))"Î# Ê 20. r œ sin Š)  È)  1‹ Ê

cos È2) È 2)

œ cos (2))"Î# ˆ "# (2))
"Î# (2)‰ œ

œ cos Š)  È)  1‹ Š1 

" ‹ 2È )  1

2È)"1 #È )  "

œ

œ

" #

22. y œ 2Èx sin Èx Ê

dy dx

" 2 œ 2Èx ˆcos Èx‰ Š 2È ‹  ˆsin Èx‰ Š 2È ‹ œ cos Èx  x x

x# csc

Ê

2 x

x# ˆ
csc

2 x

cot x2 ‰ ˆ
x#2 ‰  ˆcsc x2 ‰ ˆ "# † 2x‰ œ csc

cos Š)  È)  1‹

dy dx

21. y œ

" #

dr d)

dr d)

2 x

cot

2 x

 x csc

2 x

sin Èx Èx

dy
"Î# sec (2x)# tan (2x)# (2(2x) † 2)  sec (2x)# ˆ
"# x
$Î# ‰ dx œ x 8x"Î# sec (2x)# tan (2x)#
"# x
$Î# sec (2x)# œ "# x"Î# sec (2x)# c16 tan (2x)#
x
# d or #x"$Î# seca#xb2
16x# tana2xb#

23. y œ x
"Î# sec (2x)# Ê œ


"‘

24. y œ Èx csc (x  1)$ œ x"Î# csc (x  1)$ Ê

dy dx

œ x"Î# a
csc (x  1)$ cot (x  1)$ b a3(x  1)# b  csc (x  1)$ ˆ "# x
"Î# ‰

œ
3Èx (x  1)# csc (x  1)$ cot (x  1)$  or

" csc(x #È x

csc (x  1)$ 2È x

œ

" #

Èx csc (x  1)$
x"
6(x  1)# cot (x  1)$ ‘

 1)$ c1
6x(x  1)# cot (x  1)$ d

25. y œ 5 cot x# Ê 26. y œ x# cot 5x Ê

dy dx

œ 5 a
csc# x# b (2x) œ
10x csc# ax# b œ x# a
csc# 5xb (5)  (cot 5x)(2x) œ
5x# csc# 5x  2x cot 5x

dy dx

27. y œ x# sin# a2x# b Ê

dy dx

œ x# a2 sin a2x# bb acos a2x# bb (4x)  sin# a2x# b (2x) œ 8x$ sin a2x# b cos a2x# b  2x sin# a2x# b

28. y œ x
# sin# ax$ b Ê

dy dx

œ x
# a2 sin ax$ bb acos ax$ bb a3x# b  sin# ax$ b a
2x
$ b œ 6 sin ax$ b cos ax$ b
2x
$ sin# ax$ b

29. s œ ˆ t 4t 1 ‰ 30. s œ


#


" 15(15t
1)$

Èx

Ê

œ
2 ˆ t 4t 1 ‰


$


(4t)(1) Š (t  1)(4) ‹ œ
2 ˆ t 4t 1 ‰ (t  1)#

" œ
15 (15t
1)
$ Ê

#

31. y œ Š x  1 ‹ Ê 2È x

ds dt

#

dy dx

32. y œ Š 2Èx  1 ‹ Ê

dy dx

(x  1)#

2È x

œ 2 Š 2È x  1 ‹ 

# "Î# 33. y œ É x x# x œ ˆ1  "x ‰ Ê

dy dx

œ

34. y œ 4xÉx  Èx œ 4x ˆx  x"Î# ‰
"Î# œ ˆx  Èx‰ ’2x Š1 

" ‹ #È x

"Î#

" #

4 (t  1)#

" œ
15 (
3)(15t
1)
% (15) œ

ds dt

" (x  1) Š #È ‹
ˆÈx‰ (1) x

Èx

œ 2 Šx1‹ †


$

œ

(x  1)
2x (x  1)$

ˆ2Èx  1‰ Š È" ‹
ˆ2Èx‰ Š È" ‹ x x ˆ2 È x  1 ‰

#

ˆ1  "x ‰
"Î# ˆ
x"# ‰ œ


Ê

dy dx

œ

œ "


"Î#

3 (15t
1)%

1
x (x  1)$

#x # É 1 

œ 4x ˆ "# ‰ ˆx  x"Î# ‰

œ
(t 8t$1)

4Èx Š È"x ‹

ˆ2 È x  1 ‰$

œ

4 ˆ2 È x  1 ‰$

" x

ˆ1  "# x
"Î# ‰  ˆx  x"Î# ‰"Î# (4)


"Î# ˆ2x  Èx  4x  4Èx‰ œ  4 ˆx  Èx‰“ œ ˆx  Èx‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

6x  5Èx É x  Èx

182

Chapter 3 Differentiation #

35. r œ ˆ cossin)
) 1 ‰ Ê œ 2 ˆ cossin)
) " ‰ Š cos

#

#

1 ‰ 36. r œ ˆ 1sin
)cos Ê )

œ

2(sin )  ") (1
cos ))$

))
(sin ))(
sin )) œ 2 ˆ cossin)
) 1 ‰ ’ (cos )
1)(cos “ (cos )
1)#

dr d)

)
cos )  sin# ) ‹ (cos )
")#

œ

(2 sin )) (1
cos )) (cos )
1)$

 1 ‰ (1
cos ))(cos ))
(sin )  ")(sin )) œ 2 ˆ 1sin
)cos “ ) ’ (1
cos ))#

dr d)

2(sin )  1)(cos )
sin )
1) (1
c os ))$

acos )
cos# )
sin# )
sin )b œ

37. y œ (2x  1) È2x  1 œ (2x  1)$Î# Ê

œ

dy dx

3 #

(2x  1)"Î# (2) œ 3È2x  1

38. y œ 20(3x
4)"Î% (3x
4)
"Î& œ 20(3x
4)"Î#! Ê 39. y œ 3 a5x#  sin 2xb 40. y œ a3  cos$ 3xb


2 sin ) (cos )
")#

œ


$Î#


"Î$

Ê

Ê

" ‰ œ 20 ˆ 20 (3x
4)
"*Î#! (3) œ

œ 3 ˆ
3# ‰ a5x#  sin 2xb

dy dx

œ
"3 a3  cos$ 3xb

dy dx

dy dx


%Î$


&Î#

[10x  (cos 2x)(2)] œ

a3 cos# 3xb (
sin 3x)(3) œ

3 (3x
4)"*Î#!


9(5x  cos 2x) a5x#  sin 2xb&Î#

3 cos# 3x sin 3x a3  cos$ 3xb%Î$

2 41. xy  2x  3y œ 1 Ê axyw  yb  2  3yw œ 0 Ê xyw  3yw œ
2
y Ê yw (x  3) œ
2
y Ê yw œ
yx  3

42. x#  xy  y#
5x œ 2 Ê 2x  Šx œ 5
2x
y Ê

œ

dy dx

dy dx

dy dx

dy dx

ˆ4x
4y"Î$ ‰ œ 2
3x#
4y Ê

44. 5x%Î&  10y'Î& œ 15 Ê 4x
"Î&  12y"Î& " #

45. (xy)"Î# œ 1 Ê

(xy)
"Î# Šx

46. x# y# œ 1 Ê x# Š2y 47. y# œ

x x 1

Ê 2y

x‰ 48. y# œ ˆ 11 
x

"Î#

dy dx

dy dx ‹

œ

dp dq

œ

dp dq

dy dx

 2y

dy dx

œ 5
2x
y Ê

dy dx

(x  2y)

"x 1
x

Ê 4y$ dp dq

dy dx

œ 2 Ê 4x

œ

dy dx

œ 0 Ê 12y"Î& dy dx

dy dx


4y"Î$

œ
4x
"Î& Ê

œ
x
"Î# y"Î# Ê

œ
2xy# Ê

dy dx

œ
yx

Ê

dy dx

œ


6q œ 0 Ê 3p#

dp dq

dy dx

dy dx

œ 2
3x#
4y

dy dx

dy dx

" œ
"3 x
"Î& y
"Î& œ
3(xy) "Î&

œ
x
" y Ê

dy dx

œ
yx

" #y(x  1)#

dy dx

œ

œ

(1
x)(1)
(1  x)Ð
") ("
x)#

 4 Šp  q

dy dx

2
3x#
4y 4x
4y"Î$

dy dx

 y‹ œ 0 Ê x"Î# y
"Î#

Ê

dy dx

dp dq ‹

" 2y$ (1
x)#

 4q

dp dq

œ 6q
4p Ê

dp dq

a3p#  4qb œ 6q
4p

6q
4p 3p#  4q

50. q œ a5p#  2pb Ê


&œ! Ê x

 4y‹
4y"Î$

 y# (2x) œ 0 Ê 2x# y

49. p$  4pq
3q# œ 2 Ê 3p# Ê

dy dx

(x  1)(1)
(x)(1) (x  1)#

Ê y% œ

dy dx

5
2x
y x  2y

43. x$  4xy
3y%Î$ œ 2x Ê 3x#  Š4x Ê

 y‹  2y


$Î#

#

Ê 1 œ
3# a5p#  2pb


&Î#

Š10p

dp dq

2

dp dq ‹

Ê
23 a5p#  2pb

&Î#

œ

dp dq

(10p  2)

&Î#

œ
a5p3(5p 2p1)b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Practice Exercises dr ‰ 51. r cos 2s  sin# s œ 1 Ê r(
sin 2s)(2)  (cos 2s) ˆ ds  2 sin s cos s œ 0 Ê

Ê

dr ds

2r sin 2s
sin 2s cos 2s

œ

œ

(2r
1)(sin 2s) cos 2s

52. 2rs
r
s  s# œ
3 Ê 2 ˆr  s 53. (a) x$  y$ œ 1 Ê 3x#  3y# Ê

d# y dx#

(b) y# œ 1
Ê

d# y dx#

x# ‹ y#


2xy#  a2yx# b Š


œ

y%

Ê 2y

2 x


2xy
x# Š

œ

œ

dy dx

y# x%

" ‹ yx#

œ

54. (a) x#
y# œ 1 Ê 2x
2y (b)

dy dx

œ

x y

d# y dx#

Ê

œ

y(1)
x y#

dr ds

d# y dx#

œ

#

œ
xy# Ê

dy dx 2x% y

y%

œ

" yx#

œ

dy dx


1  2s œ 0 Ê

(2s
1) œ 1
2s
2r Ê y# (
2x)
a
x# b Š2y

dr ds

"
2s
2r 2s
1

œ

dy dx ‹

y%


2xy$
2x% y&

Ê

dy dx

œ ayx# b


"

Ê

dy dx

œ

d# y dx#

œ
ayx# b


#

’y(2x)  x#

dy dx “


2xy#
1 y$ x%

œ 0 Ê
2y

dy dx dy dx

dr ds


2xy#


Ê

2 x#




œ0 Ê

œ

(cos 2s) œ 2r sin 2s
2 sin s cos s

œ (2r
1)(tan 2s)

dr ‰ ds

dy dx

dr ds

y
x Š xy ‹

œ

œ

y#

dy dx

œ
2x Ê

y#
x# y$

œ


" y$

x y

(since y#
x# œ
1)

55. (a) Let h(x) œ 6f(x)
g(x) Ê hw (x) œ 6f w (x)
gw (x) Ê hw (1) œ 6f w (1)
gw (1) œ 6 ˆ "# ‰
a
%b œ (

(b) Let h(x) œ f(x)g# (x) Ê hw (x) œ f(x) a#g(x)b gw (x)  g# (x)f w (x) Ê hw (0) œ #f(0)g(0)gw (0)  g# (0)f w (0) œ #(1)(1) ˆ "# ‰  (1)# (
$) œ
# (c) Let h(x) œ œ

f(x) g(x)  1

(&  1) ˆ "# ‰
3 a
%b (&  1)#

(g(x)  1)f (x)
f(x)g (x) (g(x)  1)#

Ê hw (x) œ œ

w

w

Ê hw (1) œ

(g(1)  ")f (1)
f(1)g (1) (g(1)  1)# w

w

& "#

(d) Let h(x) œ f(g(x)) Ê hw (x) œ f w (g(x))gw (x) Ê hw (0) œ f w (g(0))gw (0) œ f w (1) ˆ "# ‰ œ ˆ "# ‰ ˆ "# ‰ œ

" %

(e) Let h(x) œ g(f(x)) Ê hw (x) œ gw (f(x))f w (x) Ê hw (0) œ gw (f(0))f w (0) œ gw (1)f w (0) œ a
%b (
$) œ "# (f) Let h(x) œ (x  f(x))$Î# Ê hw (x) œ 3# (x  f(x))"Î# a1  f w (x)b Ê hw (1) œ 3# (1  f(1))"Î# a1  f w (1)b œ 3# (1  3)"Î# ˆ1  "# ‰ œ *# (g) Let h(x) œ f(x  g(x)) Ê hw (x) œ f w (x  g(x)) a1  gw (x)b Ê hw (0) œ f w (g(0)) a1  gw (0)b œ f w (1) ˆ1  "# ‰ œ ˆ "# ‰ ˆ $# ‰ œ

%$56. (a) Let h(x) œ Èx f(x) Ê hw (x) œ Èx f w (x)  f(x) † (b) Let h(x) œ (f(x))"Î# Ê hw (x) œ

" #

" #È x

(f(x))
"Î# af w (x)b Ê hw (0) œ

(c) Let h(x) œ f ˆÈx‰ Ê hw (x) œ f w ˆÈx‰ †

" #È x

" œ 5"  (
3) ˆ #" ‰ #È 1 "
"Î# (
2) œ
3" # (9)

Ê hw (1) œ È1 f w (1)  f(1) † " #

(f(0))
"Î# f w (0) œ

Ê hw (1) œ f w ŠÈ1‹ †

" #È 1 w

œ

" 5

†

" #

œ

œ
13 10

" 10

(d) Let h(x) œ f(1
5 tan x) Ê hw (x) œ f w (1
5 tan x) a
5 sec# xb Ê h (0) œ f w (1
5 tan 0) a
5 sec# 0b œ f w (1)(
5) œ "5 (
5) œ
1 (2  cos x)f (x)
f(x)(
sin x)
f(0)(0) Ê hw (0) œ (2  1)f(2(0) œ 3(
9 2) œ (2  cos x)#  1)# h(x) œ 10 sin ˆ 1#x ‰ f # (x) Ê hw (x) œ 10 sin ˆ 1#x ‰ a2f(x)f w (x)b  f # (x) ˆ10 cos ˆ 1#x ‰‰ ˆ 1# ‰ hw (1) œ 10 sin ˆ 1# ‰ a2f(1)f w (1)b  f # (1) ˆ10 cos ˆ 1# ‰‰ ˆ 1# ‰ œ 20(
3) ˆ "5 ‰  ! œ
12

(e) Let h(x) œ (f) Let Ê

57. x œ t#  1 Ê dy dt

œ

dy dx

†

dx dt

f(x) 2  cos x

dx dt

Ê hw (x) œ

œ 2t; y œ 3 sin 2x Ê

œ 6 cos a2t# b † 2t Ê

58. t œ au#  2ub

"Î$

œ 2 au#  2ub

"Î$

w

Ê

dt du

œ

 5; thus

" 3

dy dt ¹ t=0

au#  2ub

ds du

œ

ds dt

†

dy dx


32

œ 3(cos 2x)(2) œ 6 cos 2x œ 6 cos a2t#  21b œ 6 cos a2t# b ; thus,

œ 6 cos (0) † 0 œ 0


#Î$

dt du

w

(2u  2) œ

2 # 3 au "Î$

œ ’2 au#  2ub

 2ub




#Î$

(u  1); s œ t#  5t Ê

5“ ˆ 32 ‰ au#

 2ub


#Î$

(u  1)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

ds dt

œ 2t  5

183

184

Chapter 3 Differentiation Ê

ds ¸ du u=2

œ ’2 a2#  2(2)b

59. r œ 8 sin ˆs  16 ‰ Ê œ

; thus,

2É8 sin ˆs  16 ‰

Ê

dw ¸ ds s=0

œ


#Î$

(2  1) œ 2 ˆ2 † 8"Î$  5‰ ˆ8
#Î$ ‰ œ 2(2 † 2  5) ˆ 4" ‰ œ

dw dr

†

œ

2É8 sin ˆ 16 ‰ d ) ‰‰ dt



d) dt

œ

dr ds

cos ŠÉ8 sin ˆs  16 ‰
2‹ # É8 sinˆ s  16 ‰

(cos 0)(8) Š

È3 ‹

2È4

œ0 Ê

d) dt

#

and

61. y$  y œ 2 cos x Ê 3y#
2 sin (0) 3 1

œ 0;

d# y dx# ¹ (0ß1)

Ê

œ

d# y dx#

d# y dx#

œ

 "3 8#Î$

œ

" 3

2 3

dy dx

œ
2 sin x Ê

a3y#  1b (
2 cos x)
(
2 sin x) Š6y

" 3

œ

" œ cos ˆÈr
2‰ Š #È ‹ r

†
8 cos ˆs  16 ‰‘

(2)t  1) œ
)# Ê

dy dx

d) dt

œ


) # 2)t1

; r œ a)#  7b

0 and )# t  ) œ 1 Ê ) œ 1 so that œ

ˆ 6" ‰ (
1)

œ


"Î$

d) ¸ dt t=0, )=1

œ


1 1

œ
1

" 6

a3y#  1b œ
2 sin x Ê

dy dx

œ


2 sin x 3y# 1

Ê

dy dx ¹ (0ß1)

dy dx ‹

a3y#  1b#

x
#Î$  3" y
#Î$

ˆx#Î$ ‰ Š
23 y
"Î$

4



(3  1)(
2 cos 0)
(
2 sin 0)(6†0) (3  1)#

62. x"Î$  y"Î$ œ 4 Ê Ê

œ

dy dx

dw dr

9 #

œ È3


#Î$
#Î$ dr " # (2)) œ 32 ) a)#  7b ; now t œ d) œ 3 a)  7b dr ¸ 2 dr ¸ dr ¸ "
#Î$ œ 6 Ê dt t=0 œ d) t=0 † ddt) ¸ t=0 d) )=1 œ 3 (1  7)

Ê

œ

œ

dw ds

cos ŠÉ8 sin ˆ 16 ‰
2‹†8 cos ˆ 16 ‰

60. )# t  ) œ 1 Ê ˆ)#  t ˆ2)

œ

 5“ ˆ 23 ‰ a2#  2(2)b

œ 8 cos ˆs  16 ‰ ; w œ sin ˆÈr
2‰ Ê

dr ds

cos É8 sin ˆs  16 ‰
2

"Î$

dy ˆ #Î$ ‰ ˆ 23 dx ‹

y

œ
#"

dy dx

x
"Î$ ‰

# ax#Î$ b

œ0 Ê Ê

dy dx

#Î$

œ
yx#Î$ Ê

d# y dx# ¹ (8ß8)

œ

dy dx ¹ (8ß8)

œ
1;

dy dx

œ


y#Î$ x#Î$

ˆ8#Î$ ‰

23 †8
"Î$ †(
1)‘  ˆ8#Î$ ‰ ˆ 23 †8
"Î$ ‰ 8%Î$

" 6

" " f(t  h)
f(t) #(t
h)
1
#t
1 œ 2t  1
(2t  2h  1) " " and f(t  h) œ Ê œ 2t  1 #(t  h)  1 h h (2t  2h  1)(2t  1)h f(t  h)
f(t)
2h
2
2 w œ Ê f (t) œ lim œ lim (2t  2h  1)(2t  1)h (2t  2h  1)(2t  1) h hÄ! h Ä ! (2t  2h  1)(#t  1)
# (2t  1)#

63. f(t) œ œ œ

g(x  h)
g(x) h lim g(x  h)h
g(x) œ lim hÄ! hÄ!

64. g(x) œ 2x#  1 and g(x  h) œ 2(x  h)#  1 œ 2x#  4xh  2h#  1 Ê œ

#

#

#

a2x  4xh  2h  1b
a2x  1b h

œ

4xh  2h# h

œ 4x  2h Ê gw (x) œ

œ 4x

(4x  2h)

65. (a)

lim f(x) œ lim c x# œ 0 and lim b f(x) œ lim b
x# œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0) it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c (2x) œ 0 and lim b f w (x) œ lim b (
2x) œ 0 Ê lim f w (x) œ 0. Since this limit exists, it (b)

x Ä !c

xÄ!

xÄ!

xÄ!

xÄ!

xÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Practice Exercises

185

follows that f is differentiable at x œ 0. 66. (a)

lim f(x) œ lim c x œ 0 and lim b f(x) œ lim b tan x œ 0 Ê lim f(x) œ 0. Since lim f(x) œ 0 œ f(0), it xÄ! xÄ! xÄ! xÄ! xÄ! follows that f is continuous at x œ 0. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b sec# x œ 1 Ê lim f w (x) œ 1. Since this limit exists it (b)

x Ä !c

xÄ!

xÄ!

xÄ!

xÄ!

xÄ!

follows that f is differentiable at x œ 0. 67. (a)

lim f(x) œ lim c x œ 1 and lim b f(x) œ lim b (2
x) œ 1 Ê lim f(x) œ 1. Since lim f(x) œ 1 œ f(1), it xÄ" xÄ" xÄ" xÄ" xÄ" follows that f is continuous at x œ 1. (c) lim c f w (x) œ lim c 1 œ 1 and lim b f w (x) œ lim b
1 œ
1 Ê lim c f w (x) Á lim b f w (x), so lim f w (x) does (b)

x Ä "c

xÄ"

xÄ"

xÄ"

not exist Ê f is not differentiable at x œ 1.

xÄ"

xÄ"

xÄ1

xÄ"

lim f(x) œ lim c sin 2x œ 0 and lim b f(x) œ lim b mx œ 0 Ê lim f(x) œ 0, independent of m; since xÄ! xÄ! xÄ! xÄ! f(0) œ 0 œ lim f(x) it follows that f is continuous at x œ 0 for all values of m.

68. (a)

x Ä !c

xÄ!

lim f w (x) œ lim c (sin 2x)w œ lim c 2 cos 2x œ 2 and lim b f w (x) œ lim b (mx)w œ lim b m œ m Ê f is x Ä !c xÄ! xÄ! xÄ! xÄ! xÄ! differentiable at x œ 0 provided that lim c f w (x) œ lim b f w (x) Ê m œ 2.

(b)

xÄ!

69. y œ œ

" #

x #



" #x
4

œ


2(2x
4)

" # x
#

 (2x
4)
" Ê

dy dx

œ

" #

xÄ!


2(2x
4)
# ; the slope of the tangent is
3# Ê
3#

Ê
2 œ
2(2x
4)
# Ê 1 œ

" (2x
4)#

Ê 4x#
16x  15 œ 0 Ê (2x
5)(2x
3) œ 0 Ê x œ

Ê (2x
4)# œ 1 Ê 4x#
16x  16 œ 1

5 #

or x œ

3 #

Ê ˆ 5# ß 94 ‰ and ˆ 3# ß
"4 ‰ are points on the

curve where the slope is
. 3 #

70. y œ x


" 2x

Ê xœ „

Ê " #

dy dx

œ1

2 (2x)#

Ê ˆ "# ß
"# ‰ and ˆ


71. y œ 2x$
3x#
12x  20 Ê #

#

" " #x# ; the slope of the tangent is 3 Ê 3 œ 1  #x# " "‰ # ß # are points on the curve where the slope is 3.

œ1

dy dx

Ê 2œ

œ 6x#
6x
12; the tangent is parallel to the x-axis when

dy dx

" #x #

Ê x# œ

" 4

œ0

Ê 6x
6x
12 œ 0 Ê x
x
2 œ 0 Ê (x
2)(x  1) œ 0 Ê x œ 2 or x œ
1 Ê (#ß !) and (
"ß #7) are points on the curve where the tangent is parallel to the x-axis. 72. y œ x$ Ê

dy dx

œ 3x# Ê

dy dx ¹ (
2ß
8)

œ 12; an equation of the tangent line at (
#ß
)) is y  8 œ 12(x  2)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

186

Chapter 3 Differentiation Ê y œ 12x  16; x-intercept: 0 œ 12x  16 Ê x œ
43 Ê ˆ
43 ß !‰ ; y-intercept: y œ 12(0)  16 œ 16 Ê (0ß 16)

73. y œ 2x$
3x#
12x  20 Ê

dy dx

œ 6x#
6x
12

(a) The tangent is perpendicular to the line y œ 1


x 24

when

dy dx

œ
Š
ˆ" " ‰ ‹ œ 24; 6x#
6x
12 œ 24 #4

Ê x#
x
2 œ 4 Ê x#
x
6 œ 0 Ê (x
3)(x  2) œ 0 Ê x œ
2 or x œ 3 Ê (
#ß 16) and ($ß 11) are x points where the tangent is perpendicular to y œ 1
24 . dy (b) The tangent is parallel to the line y œ È2
12x when dx œ
12 Ê 6x#
6x
12 œ
12 Ê x#
x œ 0 Ê x(x
1) œ 0 Ê x œ 0 or x œ 1 Ê (!ß 20) and ("ß () are points where the tangent is parallel to y œ È2
12x. 74. y œ

1 sin x x

Ê

dy dx

œ

x(1 cos x)
(1 sin x)(1) x#

Ê m" œ

dy dx ¹ x=1

œ


1 # 1#

œ
1 and m# œ

Since m" œ
m"# the tangents intersect at right angles. 75. y œ tan x,
1#  x 

1 #

Ê

dy dx

dy 1# dx ¹ x=c1 1#

œ 1.

œ sec# x; now the slope

of y œ
x# is
"# Ê the normal line is parallel to y œ
x# when #

Ê cos x œ

dy dx

" #

œ 2. Thus, sec# x œ 2 Ê

Ê cos x œ

for
1#  x 

1 #

„" È2

Ê xœ


1 4

" cos# x

œ2

and x œ

1 4

Ê ˆ
14 ß
1‰ and ˆ 14 ß "‰ are points

where the normal is parallel to y œ
x# .

76. y œ 1  cos x Ê

œ
sin x Ê

dy dx

dy dx ¹ ˆ 1 ß1‰

œ
1

2

Ê the tangent at ˆ 1# ß 1‰ is the line y
1 œ
ˆx
1# ‰ Ê y œ
x  1#  1; the normal at ˆ 1# ß 1‰ is

y
1 œ (1) ˆx
1# ‰ Ê y œ x


77. y œ x#  C Ê thus,

" #

œ

ˆ "# ‰#

78. y œ x$ Ê

dy dx

dy dx

1 #

1

œ 2x and y œ x Ê

C Ê Cœ œ 3x# Ê

dy dx

œ 1; the parabola is tangent to y œ x when 2x œ 1 Ê x œ

" #

Ê yœ

" #

;

" 4

dy dx ¹ x=a

œ 3a# Ê the tangent line at aaß a$ b is y
a$ œ 3a# (x
a). The tangent line

intersects y œ x$ when x$
a$ œ 3a# (x
a) Ê (x
a) ax#  xa  a# b œ 3a# (x
a) Ê (x
a) ax#  xa
2a# b œ 0 Ê (x
a)# (x  2a) œ 0 Ê x œ a or x œ
2a. Now

dy dx ¹ x=c2a

œ 3(
2a)# œ 12a# œ 4 a3a# b, so the slope at

x œ
2a is 4 times as large as the slope at aaß a$ b where x œ a. 79. The line through (!ß $) and (5ß
2) has slope m œ y œ
x  3; y œ

c x1

Ê

dy dx

œ


c (x  1)# ,

3
(
2) 0
5

œ
1 Ê the line through (!ß $) and (&ß
2) is

so the curve is tangent to y œ
x  3 Ê

Ê (x  1)# œ c, x Á
1. Moreover, y œ

c x1

intersects y œ
x  3 Ê #

c x 1

dy dx

œ
1 œ


c (x  1)#

œ
x  3, x Á
1

Ê c œ (x  1)(
x  3), x Á
1. Thus c œ c Ê (x  1) œ (x  1)(
x  3) Ê (x  1)[x  1
(
x  3)] œ !, x Á
1 Ê (x  1)(2x
2) œ 0 Ê x œ 1 (since x Á
1) Ê c œ 4.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Practice Exercises 80. Let Šbß „ Èa#
b# ‹ be a point on the circle x#  y# œ a# . Then x#  y# œ a# Ê 2x  2y Ê

dy dx ¹ x=b

œ


b „È a #
b #

y
Š „ Èa#
b# ‹ œ

Ê normal line through Šbß „ Èa#
b# ‹ has slope „È a #
b # b

(x
b) Ê y … Èa#
b# œ

„È a #
b # b

„È a #
b # b

œ0 Ê

dy dx

dy dx

œ
xy

Ê normal line is

x … Èa#
b# Ê y œ „

È a#
b # b

x

which passes through the origin. 81. x#  2y# œ 9 Ê 2x  4y œ
"4 x 

9 4

5 #

x œ
2y Ê

dy dx

œ
"4 Ê the tangent line is y œ 2
"4 (x
1)

dy dx ¹ (1ß2)

and the normal line is y œ 2  4(x
1) œ 4x
2.

82. x$  y# œ 2 Ê 3x#  2y œ
3# x 

œ0 Ê

dy dx

œ0 Ê

dy dx

œ

dy dx


3x# 2y

Ê

dy dx ¹ (1ß1)

and the normal line is y œ 1  23 (x
1) œ

83. xy  2x
5y œ 2 Ê Šx

 y‹  2
5

dy dx

œ0 Ê

dy dx

(x
5) œ
y
2 Ê

Ê the tangent line is y œ 2  2(x
3) œ 2x
4 and the normal line is y œ 2  84. (y
x)# œ 2x  4 Ê 2(y
x) Š dy dx
1‹ œ 2 Ê (y
x) Ê the tangent line is y œ 2  34 (x
6) œ 85. x  Èxy œ 6 Ê 1 

" #Èxy

dy dx

Šx

3 4

x


dy dx

œ 1  (y
x) Ê


y
2 x
5

dy dx

œ

Ê


1 #

(x
3) œ
"# x  7# .

dy dx

œ

1y
x y
x

3 2

x"Î#  3y"Î#

y œ 4
"4 (x
1) œ
4" x 

dy dx

dy dx ‹

dy dx

 y œ
2Èxy Ê

dy dx

œ

 y$ a3x# b“  2y

a3x$ y#  2y
1b œ 1
3x# y$ Ê

dy dx


2Èxy
y x

œ


x"Î# 2y"Î#

Ê

dy dx ¹ (1ß4)

dy dx

dy dx ¹ (6ß2)

œ

3 4

Ê

œ

dy dx ¹ (4ß1) 4 5

x



5 4

11 5

.

œ
"4 Ê the tangent line is

dy dx

œ

dy dx

œ1

1
3x# y$ 3x$ y#  2y
1

Ê

dy dx

Ê 3x$ y#

dy dx ¹ (1ß1)

dy dx

 2y

œ
24 , but

Therefore, the curve has slope
"# at ("ß ") but the slope is undefined at ("ß
1). 88. y œ sin (x
sin x) Ê

œ2

and the normal line is y œ 4  4(x
1) œ 4x.

17 4

87. x$ y$  y# œ x  y Ê ’x$ Š3y# Ê

œ0 Ê

dy dx

dy dx ¹ (3ß2)

Ê

Ê the tangent line is y œ 1
54 (x
4) =
54 x  6 and the normal line is y œ "  45 (x
4) œ 86. x$Î#  2y$Î# œ 17 Ê

(x
1)

and the normal line is y œ 2
43 (x
6) œ
43 x  10.

5 #

 y‹ œ 0 Ê x


3 #

x  "3 .

2 3

dy dx

œ
#3 Ê the tangent line is y œ 1 

dy dx




dy dx ¹ (1ßc1) is

dy dx

œ "
3x# y$

undefined.

œ [cos (x
sin x)](1
cos x); y œ 0 Ê sin (x
sin x) œ 0 Ê x
sin x œ k1,

k œ
2,
1, 0, 1, 2 (for our interval) Ê cos (x
sin x) œ cos (k1) œ „ 1. Therefore,

dy dx

œ 0 and y œ 0 when

1
cos x œ 0 and x œ k1. For
#1 Ÿ x Ÿ 21, these equations hold when k œ
2, 0, and 2 (since cos (
1) œ cos 1 œ
1). Thus the curve has horizontal tangents at the x-axis for the x-values
21, 0, and 21 (which are even integer multiples of 1) Ê the curve has an infinite number of horizontal tangents. 89. x œ

" #

tan t, y œ

Ê xœ

" #

tan

1 3

œ 2 cos$ ˆ 13 ‰ œ 90. x œ " 

" t#

" #

sec t Ê

œ

È3 #

dy dx

œ

dy/dt dx/dt

œ

" #

sec

1 3

and y œ

" #

sec t tan t " # # sec t

œ

tan t sec t

œ sin t Ê

œ1 Ê yœ

È3 #

x  4" ;

d# y dx#

" 4

,yœ"


3 t

Ê

dy dx

œ

dy/dt dx/dt

œ

Š t3# ‹ Š
t2$ ‹

œ
32 t Ê

dy dx ¹ tœ2

dy dx ¹ tœ1Î3

œ

dyw /dt dx/dt

œ sin

œ

" #

1 3

cos t sec# t

œ

È3 #

;tœ

1 3

œ 2 cos$ t Ê

d# y ¸ dx# tœ1Î3

œ
3# (2) œ
3; t œ 2 Ê x œ 1 

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" ##

œ

5 4

and

187

188

Chapter 3 Differentiation

yœ1


3 #

œ
"# Ê y œ
3x 

"3 4

d# y dx#

;

œ

dyw /dt dx/dt

œ

ˆ
#3 ‰ Š
t2$ ‹

œ

3 $ 4 t

d# y dx# ¹ tœ2

Ê

œ

3 4

(2)$ œ 6

91. B œ graph of f, A œ graph of f w . Curve B cannot be the derivative of A because A has only negative slopes while some of B's values are positive. 92. A œ graph of f, B œ graph of f w . Curve A cannot be the derivative of B because B has only negative slopes while A has positive values for x  0. 93.

94.

95. (a) 0, 0

(b) largest 1700, smallest about 1400

96. rabbits/day and foxes/day sin x

97. lim

# x Ä ! 2x
x

98. lim

3x
tan 7x #x

99. lim

sin r

xÄ!

r Ä ! tan 2r

100.

sin 7x ‰ 2x cos 7x

xÄ!

2r tan 2r

103.

104. 105.

)Ä!

xÄ!

4 tan# )  tan )  1 tan# )  &

œ

lim c ) Ä ˆ1‰

lim

tan x x

œ lim

)Ä!

2 sin# ˆ #) ‰ )# )Ä!

œ lim ˆ cos" x †

tan ) )

xÄ!

sin (sin )) sin )

Š"  tan5# ) ‹

Š5
cot7 )
cot8# ) ‹

œ

œ

‹œ

3 #


ˆ1 † 1 † 27 ‰ œ
2

" #

. Let x œ sin ). Then x Ä 0 as ) Ä 0

(4  0  0) (1  0)

(0
2) (5
0
0)

œ lim

x sin x # x x Ä ! 2 ˆ2 sin ˆ # ‰‰

œ4

œ
52 †

x x

œ lim ’ sin## ˆ# x ‰ † xÄ!

#

sin x x “

(1)(1)(1) œ 1

œ lim ’

sin x ‰ x

" ˆ 27 ‰

†

œ ˆ "# ‰ (1) ˆ 1" ‰ œ

cos 2r

Š4  tan" )  tan"# ) ‹

Š cot"# )
2‹

œ lim b )Ä!

œ lim

sin 7x 7x

œ1

sin x x

x sin x œ lim 2(1x
sincosx x) x Ä ! 2
2 cos x xÄ! ˆ x# ‰ ˆx‰ œ lim ’ sin ˆ x ‰ † sin #ˆ x ‰ † sinx x “ œ xÄ! # #

xÄ!

xÄ!

sin 2r r Ä ! ˆ 2r ‰

lim

1
cos ) )# )Ä!


lim Š cos"7x †

† "# ‰ œ ˆ "# ‰ (1) lim

2

lim

3 #

)Ä!

œ lim

1
2 cot# ) 5 cot# )
7 cot )
8

lim b

œ

)Ä!

2

102.

œ (1) ˆ
"1 ‰ œ
1

(sin )) ˆ sin ) ‰ œ lim Š sinsin œ lim ) ‹ )

sin (sin )) sin )

)Ä!

lim c ) Ä ˆ1‰

œ lim ˆ 3x 2x


rÄ!

Ê lim

101.

" (#x
1) “

œ lim ˆ sinr r †

sin (sin )) )

lim

)Ä!

xÄ!

œ lim ’ˆ sinx x ‰ †

)Ä!

sin ˆ #) ‰ ˆ #) ‰

†

sin ˆ #) ‰ ˆ #) ‰

† "# “ œ (1)(1) ˆ "# ‰ œ

" #

œ 1; let ) œ tan x Ê ) Ä 0 as x Ä 0 Ê lim g(x) œ lim xÄ!

xÄ!

œ 1. Therefore, to make g continuous at the origin, define g(0) œ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

tan (tan x) tan x

Chapter 3 Practice Exercises 106.

lim f(x) œ lim

xÄ!

(tan x) œ lim ’ tantan † x

tan (tan x)

x Ä ! sin (sin x)

sin x sin (sin x)

xÄ!

#105); let ) œ sin x Ê ) Ä 0 as x Ä 0 Ê

(b) S œ 21r#  21rh and r constant Ê (c) S œ 21r#  21rh Ê (d) S constant Ê

dh dt

(b) r constant Ê

dr dt

109. A œ 1r# Ê 110. V œ s$ Ê 111.

dR" dt

dV dt

œ0 Ê

dh ‰ ˆr dr dt  h dt È r#  h #

dr dt

œ 3s# †

ds dt

œ
1 ohm/sec,

dR# dt

ds dt

œ

œ

(using the result of

œ 1. Therefore, to make f

œ (41r  21h)

dr dt

Ê (2r 

dh dt

dr dt

dr dt  21r h) dr dt œ
r

dh dt dh dt

Ê

dr dt

œ


r dh 2rh dt

;

 1Èr#  h#

dr 1 r# Èr#  h# “ dt

; so r œ 10 and Ê

dr dt

lim ) ) Ä ! sin )

œ (41r  21h)

 21r

dr dt

sin x

x Ä ! sin (sin x)

œ ’1Èr#  h# 

dr dt

1 r# dr Èr#  h# “ dt

1rh dh Èr#  h# dt

œ

dS dt

dr ‰ dt

 1Èr#  h#

1r# dr dt È r#  h #

œ

dS dt

œ0 Ê

œ 21 r

dA dt

dr dt

œ ’1Èr#  h# 

dS dt

(c) In general,

œ 1r †

dS dt

(a) h constant Ê

œ 21r dh dt  #1 ˆr dh  h dt

œ 0 Ê 0 œ (41r  21h)

dS dt

108. S œ 1rÈr#  h# Ê

œ 41r

dS dt

œ 41r dr dt  21 h

dS dt dS dt

œ 1 † lim

sin x lim x Ä ! sin (sin x)

continuous at the origin, define f(0) œ 1. 107. (a) S œ 21r#  21rh and h constant Ê

" cos x “

†

dr dt

" dV 3s# dt



dh 1rh Èr#  h# dt

œ
12 m/sec Ê

; so s œ 20 and

œ 0.5 ohm/sec; and

" R

œ

" R"



" R#

dV dt

dA dt

œ (21)(10) ˆ
12 ‰ œ
40 m# /sec

œ 1200 cm$ /min Ê
" dR R# dt

Ê

œ


" dR" R"# dt




ds dt

œ

" dR# R## dt

" 3(20)#

(1200) œ 1 cm/min

. Also,

" " R" œ 75 ohms and R# œ 50 ohms Ê R" œ 75  50 Ê R œ 30 ohms. Therefore, from the derivative 9(625)
" dR
" " " " " ˆ ‰ Ê dR ˆ 5000
5625 ‰ (30)# dt œ (75)# (
1)
(50)# (0.5) œ 5625
5000 dt œ (
900) 5625†5000 œ 50(5625) œ 50

equation,

œ 0.02 ohm/sec. 112.

dR dt

œ 3 ohms/sec and

X œ 20 ohms Ê

dZ dt

dX dt

œ
2 ohms/sec; Z œ ÈR#  X# Ê

œ

(10)(3)(20)(
2) È10#  20#

113. Given

dx dt

œ 10 m/sec and

œ 2x

dx dt

 2y

&

dD dt

dy dt

Ê D


" È5

œ

dX R dR dt  X dt È R #  X#

so that R œ 10 ohms and

¸
0.45 ohm/sec.

œ 5 m/sec, let D be the distance from the origin Ê D# œ x#  y# Ê 2D

dy dt

dD dt

œ

dZ dt

œx

œ (5)(10)  (12)(5) Ê

dD dt

y

dx dt

œ

110 5

dy dt

dD dt

. When (xß y) œ ($ß
%), D œ É$#  a
%b# œ & and

œ 22. Therefore, the particle is moving away from the origin at 22 m/sec

(because the distance D is increasing). 114. Let D be the distance from the origin. We are given that œ x#  ˆ x

$Î# ‰#

œ x#  x$ Ê 2D

œ 2x

dD dt

 3x#

dx dt

dx dt

dD dt

œ 11 units/sec. Then D# œ x#  y#

œ x(2  3x)

dx dt

and substitution in the derivative equation gives (2)(6)(11) œ (3)(2  9) 115. (a) From the diagram we have (b) V œ

" 3

1 r# h œ

" 3

#

10 h

1 ˆ 25 h‰ h œ

œ

4 r 41 h$ 75

116. From the sketch in the text, s œ r) Ê Ê

ds dt

œr

d) dt

œ (1.2)

d) dt

. Therefore,

Ê rœ Ê

dV dt

2 5

œ

; x œ 3 Ê D œ È 3#  3$ œ 6 dx dt

Ê

dx dt

œ 4 units/sec.

h. 41h# dh 25 dt

ds d) dr dt œ r dt  ) dt . ds dt œ 6 ft/sec and r

œ
5 and h œ 6 Ê

dh dt

125 œ
144 1 ft/min.

Also r œ 1.2 is constant Ê

dr dt

œ0

, so

dV dt

œ 1.2 ft Ê

d) dt

œ 5 rad/sec

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

189

190

Chapter 3 Differentiation

117. (a) From the sketch in the text,

d) dt

point A, x œ 0 Ê ) œ 0 Ê

œ
0.6 rad/sec and x œ tan ). Also x œ tan ) Ê dx dt

dx dt

œ sec# )

d) dt ;

at

#

œ asec 0b (
0.6) œ
0.6. Therefore the speed of the light is 0.6 œ

3 5

km/sec

when it reaches point A. (3/5) rad sec

(b)

†

1 rev 21 rad

118. From the figure,

a r

†

60 sec min

œ

b BC

œ

18 1

Ê

a r

revs/min œ

b Èb#
r#

. We are given

that r is constant. Differentiation gives, " r

†

da dt

‰ ŠÈb#
r# ‹ ˆ db dt
(b) Š È

œ

b#
r#

b œ 2r and Ê œ

da dt

db dt

b ‰ ‹ ˆ db dt b#
r#

. Then,

œ
0.3r

Ô È(2r)#
r# (
0.3r)
(2r)  É2r(
#0.3r)#  × (2r)
r Ù œ rÖ (2r)#
r# Õ Ø

È3r# (
0.3r)  4r# (0.3r)

È3r#

3r

œ

a3r# b (
0.3r)  a4r# b (0.3r) 3 È 3 r#

œ

0.3r 3È 3

œ

r 10È3

m/sec. Since

da dt

is positive,

the distance OA is increasing when OB œ 2r, and B is moving toward O at the rate of 0.3r m/sec. 119. (a) If f(x) œ tan x and x œ
14 , then f w (x) œ sec# x, f ˆ
14 ‰ œ
1 and f w ˆ
14 ‰ œ 2. The linearization of f(x) is L(x) œ 2 ˆx  14 ‰  (
1) œ 2x 

1
2 #

.

(b) If f(x) œ sec x and x œ
14 , then f w (x) œ sec x tan x, f ˆ
1 ‰ œ È2 and f w ˆ
1 ‰ œ
È2. The linearization 4

4

of f(x) is L(x) œ
È2 ˆx  14 ‰  È2 œ
È2x 

120. f(x) œ

" 1  tan x

È2(%
1) . 4

Ê f w (x) œ


sec# x (1  tan x)#

. The linearization at x œ 0 is L(x) œ f w (0)(x
0)  f(0) œ 1
x.

121. f(x) œ Èx  1  sin x
0.5 œ (x  1)"Î#  sin x
0.5 Ê f w (x) œ ˆ "# ‰ (x  1)
"Î#  cos x

Ê L(x) œ f w (0)(x
0)  f(0) œ 1.5(x
0)  0.5 Ê L(x) œ 1.5x  0.5, the linearization of f(x).

122. f(x) œ œ

2 1
x

2 (1
x)#

 È1  x
3.1 œ 2(1
x)
"  (1  x)"Î#
3.1 Ê f w (x) œ
2(1
x)
# (
1)  "# (1  x)
"Î#



" 2È 1  x

Ê L(x) œ f w (0)(x
0)  f(0) œ 2.5x
0.1, the linearization of f(x).

123. S œ 1 rÈr#  h# , r constant Ê dS œ 1 r † "# ar#  h# b Ê dS œ


"Î#

#h dh œ

1rh Èr#  h# dh.

Height changes from h! to h!  dh

1 r h! adhb Ér#  h#!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Additional and Advanced Exercises 124. (a) S œ 6r# Ê dS œ 12r dr. We want kdSk Ÿ (2%) S Ê k12r drk Ÿ

12r# 100

Ê kdrk Ÿ

r 100

191

. The measurement of the

edge r must have an error less than 1%. #

3r dr ‰ (b) When V œ r$ , then dV œ 3r# dr. The accuracy of the volume is ˆ dV V (100%) œ Š r$ ‹ (100%) r ‰ œ ˆ 3r ‰ (dr)(100%) œ ˆ 3r ‰ ˆ 100 (100%) œ 3%

125. C œ 21r Ê r œ dV œ

#

C 21

, S œ 41 r # œ

C# 1

, and V œ

4 3

1 r$ œ

C$ 61 #

. It also follows that dr œ

" #1

dC, dS œ

2C 1

dC and

dC. Recall that C œ 10 cm and dC œ 0.4 cm. 0.2 ˆ drr ‰ (100%) œ ˆ 0.2 ‰ ˆ 2101 ‰ (100%) œ (.04)(100%) œ 4% (a) dr œ 0.4 21 œ 1 cm Ê 1 8 1 ‰ ˆ dS ‰ ˆ 8 ‰ ˆ 100 (b) dS œ 20 (100%) œ 8% 1 (0.4) œ 1 cm Ê S (100%) œ 1 C 21 #

10# 21 #

#

(0.4) œ

20 1#

‰ ˆ 20 ‰ 61 cm Ê ˆ dV V (100%) œ 1# Š 1000 ‹ (100%) œ 12%

126. Similar triangles yield

35 h

œ

(c) dV œ

Ê dh œ
120a
# da œ


15 6 120 a#

Ê h œ 14 ft. The same triangles imply that 20h a œ 6a Ê h œ 120a
"  6 " ‰ 2 ‰ ˆ „ 1"# ‰ œ ˆ
"#! ‰ˆ „ "# da œ ˆ
120 œ „ 45 ¸ „ .0444 ft œ „ 0.53 inches. a# "&#

CHAPTER 3 ADDITIONAL AND ADVANCED EXERCISES 1. (a) sin 2) œ 2 sin ) cos ) Ê #

#

d d)

Ê cos 2) œ cos )
sin ) (b) cos 2) œ cos# )
sin# ) Ê

(sin 2)) œ d d)

d d)

(cos 2)) œ

(2 sin ) cos )) Ê 2 cos 2) œ 2[(sin ))(
sin ))  (cos ))(cos ))] d d)

acos# )
sin# )b Ê
2 sin 2) œ (2 cos ))(
sin ))
(2 sin ))(cos ))

Ê sin 2) œ cos ) sin )  sin ) cos ) Ê sin 2) œ 2 sin ) cos ) 2. The derivative of sin (x  a) œ sin x cos a  cos x sin a with respect to x is cos (x  a) œ cos x cos a
sin x sin a, which is also an identity. This principle does not apply to the equation x#
2x
8 œ 0, since x#
2x
8 œ 0 is not an identity: it holds for 2 values of x (
2 and 4), but not for all x. 3. (a) f(x) œ cos x Ê f w (x) œ
sin x Ê f ww (x) œ
cos x, and g(x) œ a  bx  cx# Ê gw (x) œ b  2cx Ê gww (x) œ 2c; also, f(0) œ g(0) Ê cos (0) œ a Ê a œ 1; f w (0) œ gw (0) Ê
sin (0) œ b Ê b œ 0; f ww (0) œ gww (0) Ê
cos (0) œ 2c Ê c œ
"# . Therefore, g(x) œ 1
"# x# . (b) f(x) œ sin (x  a) Ê f w (x) œ cos (x  a), and g(x) œ b sin x  c cos x Ê gw (x) œ b cos x
c sin x; also, f(0) œ g(0) Ê sin (a) œ b sin (0)  c cos (0) Ê c œ sin a; f w (0) œ gw (0) Ê cos (a) œ b cos (0)
c sin (0) Ê b œ cos a. Therefore, g(x) œ sin x cos a  cos x sin a. (c) When f(x) œ cos x, f www (x) œ sin x and f Ð%Ñ (x) œ cos x; when g(x) œ 1
"# x# , gwww (x) œ 0 and gÐ%Ñ (x) œ 0. Thus f www (0) œ 0 œ gwww (0) so the third derivatives agree at x œ 0. However, the fourth derivatives do not agree since f Ð%Ñ (0) œ 1 but gÐ%Ñ (0) œ 0. In case (b), when f(x) œ sin (x  a) and g(x) œ sin x cos a  cos x sin a, notice that f(x) œ g(x) for all x, not just x œ 0. Since this is an identity, we have f ÐnÑ (x) œ gÐnÑ (x) for any x and any positive integer n.

4. (a) y œ sin x Ê yw œ cos x Ê yww œ
sin x Ê yww  y œ
sin x  sin x œ 0; y œ cos x Ê yw œ
sin x Ê yww œ
cos x Ê yww  y œ
cos x  cos x œ 0; y œ a cos x  b sin x Ê yw œ
a sin x  b cos x Ê yww œ
a cos x
b sin x Ê yww  y œ (
a cos x
b sin x)  (a cos x  b sin x) œ 0 (b) y œ sin (2x) Ê yw œ 2 cos (2x) Ê yww œ
4 sin (2x) Ê yww  4y œ
4 sin (2x)  4 sin (2x) œ 0. Similarly, y œ cos (2x) and y œ a cos (2x)  b sin (2x) satisfy the differential equation yw w  4y œ 0. In general, y œ cos (mx), y œ sin (mx) and y œ a cos (mx)  b sin (mx) satisfy the differential equation yww  m# y œ 0.

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192

Chapter 3 Differentiation

5. If the circle (x
h)#  (y
k)# œ a# and y œ x#  1 are tangent at ("ß #), then the slope of this tangent is m œ 2xk (1 2) œ 2 and the tangent line is y œ 2x. The line containing (hß k) and ("ß #) is perpendicular to ß

y œ 2x Ê

k
2 h
1

œ
"# Ê h œ 5
2k Ê the location of the center is (5
2kß k). Also, (x
h)#  (y
k)# œ a#

Ê x
h  (y
k)yw œ 0 Ê 1  ayw b#  (y
k)yw w œ 0 Ê yww œ w

1  ay b# k
y w

. At the point ("ß #) we know

ww

y œ 2 from the tangent line and that y œ 2 from the parabola. Since the second derivatives are equal at ("ß #) we obtain 2 œ

1  (2)# k
#

Ê kœ

9 #

# . Then h œ 5
2k œ
4 Ê the circle is (x  4)#  ˆy
9# ‰ œ a# . Since ("ß #)

lies on the circle we have that a œ

5È 5 2

.

6. The total revenue is the number of people times the price of the fare: r(x) œ xp œ x ˆ3


x ‰# , where 40 x ‰
ˆ x ‰
40 3
40

" ‰ dr x ‰# x ‰ˆ dr ‘ 0 Ÿ x Ÿ 60. The marginal revenue is dx œ ˆ3
40  2x ˆ3
40
40 Ê dx œ ˆ3
2x 40 x x dr œ 3 ˆ3
40 ‰ ˆ1
40 ‰ . Then dx œ 0 Ê x œ 40 (since x œ 120 does not belong to the domain). When 40 people

are on the bus the marginal revenue is zero and the fare is p(40) œ ˆ3
7. (a) y œ uv Ê

dy dt

œ

du dt

x ‰# 40 ¹ x=40

œ $4.00.

v  u dv dt œ (0.04u)v  u(0.05v) œ 0.09uv œ 0.09y Ê the rate of growth of the total production is

9% per year. (b) If

œ
0.02u and

du dt

dv dt

œ 0.03v, then

dy dt

œ (
0.02u)v  (0.03v)u œ 0.01uv œ 0.01y, increasing at 1% per

year. 8. When x#  y# œ 225, then yw œ
xy . The tangent line to the balloon at (12ß
9) is y  9 œ Ê yœ

4 3

4 3

(x
12)

x
25. The top of the gondola is 15  8

œ 23 ft below the center of the balloon. The intersection of y œ
23 and y œ 43 x
25 is at the far right edge of the gondola Ê
23 œ Ê xœ

3 #

4 3

x
25

. Thus the gondola is 2x œ 3 ft wide.

9. Answers will vary. Here is one possibility.

10. s(t) œ 10 cos ˆt  14 ‰ Ê v(t) œ 10 (a) s(0) œ 10 cos ˆ 14 ‰ œ È

ds dt

œ
10 sin ˆt  14 ‰ Ê a(t) œ

dv dt

œ

d# s dt#

œ
10 cos ˆt  14 ‰

2

(b) Left:
10, Right: 10 (c) Solving 10 cos ˆt  14 ‰ œ
10 Ê cos ˆt  14 ‰ œ
1 Ê t œ 341 when the particle is farthest to the left. Solving 10 cos ˆt  14 ‰ œ 10 Ê cos ˆt  14 ‰ œ 1 Ê t œ
14 , but t   0 Ê t œ 21 
41 œ 741 when the particle is farthest to the right. Thus, v ˆ 341 ‰ œ 0, v ˆ 741 ‰ œ 0, a ˆ 341 ‰ œ 10, and a ˆ 741 ‰ œ
10. (d) Solving 10 cos ˆt  14 ‰ œ 0 Ê t œ

1 4

Ê v ˆ 14 ‰ œ
10, ¸v ˆ 14 ‰¸ œ 10 and a ˆ 14 ‰ œ !.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Additional and Advanced Exercises 11. (a) s(t) œ 64t
16t# Ê v(t) œ

ds dt

193

œ 64
32t œ 32(2
t). The maximum height is reached when v(t) œ 0

Ê t œ 2 sec. The velocity when it leaves the hand is v(0) œ 64 ft/sec. (b) s(t) œ 64t
2.6t# Ê v(t) œ ds dt œ 64
5.2t. The maximum height is reached when v(t) œ 0 Ê t ¸ 12.31 sec. The maximum height is about s(12.31) œ 393.85 ft. 12. s" œ 3t$
12t#  18t  5 and s# œ
t$  9t#
12t Ê v" œ 9t#
24t  18 and v# œ
3t#  18t
12; v" œ v# Ê 9t#
24t  18 œ
3t#  18t
12 Ê 2t#
7t  5 œ 0 Ê (t
1)(2t
5) œ 0 Ê t œ 1 sec and t œ 2.5 sec. 13. m av#
v#! b œ k ax#!
x# b Ê m ˆ2v substituting

dx dt

œv Ê m

dv dt

dv ‰ dt

œ k ˆ
2x

dx ‰ dt

Ê m

dv dt

2x ‰ œ k ˆ
2v

dx dt

Ê m

dv dt

œ
kx ˆ "v ‰

dx dt

œ 2At  B Ê v ˆ t" # t# ‰ œ 2A ˆ t" # t# ‰  B œ A at"  t# b  B is the

instantaneous velocity at the midpoint. The average velocity over the time interval is vav œ œ

 Bt#  Cb
aAt#" t#
t"

. Then

œ
kx, as claimed.

14. (a) x œ At#  Bt  C on ct" ß t# d Ê v œ aAt##

dx dt

 Bt"  Cb

œ

at#
t" b cA at#  t" b  Bd t#
t" #

?x ?t

œ A at#  t" b  B.

(b) On the graph of the parabola x œ At  Bt  C, the slope of the curve at the midpoint of the interval ct" ß t# d is the same as the average slope of the curve over the interval. 15. (a) To be continuous at x œ 1 requires that lim c sin x œ lim b (mx  b) Ê 0 œ m1  b Ê m œ
1b ; xÄ1 xÄ1 (b) If yw œ œ

cos x, x  1 is differentiable at x œ 1, then lim c cos x œ m Ê m œ
1 and b œ 1. xÄ1 m, x   1

16. faxb is continuous at ! because lim

xÄ!

œ

x ‰ ˆ 1  cos x ‰ lim ˆ 1
xcos # 1  cos x xÄ!

œ

"
cos x x

œ ! œ fa!b. f w (0) œ lim

f(x)
f(0) x
0

ˆ 1  "cos x ‰

w

# lim ˆ sinx x ‰ xÄ!

xÄ!

œ

" #

œ lim

xÄ!

1 c cos x
0 x

x

. Therefore f (0) exists with value

" #

.

17. (a) For all a, b and for all x Á 2, f is differentiable at x. Next, f differentiable at x œ 2 Ê f continuous at x œ 2 Ê lim c f(x) œ f(2) Ê 2a œ 4a
2b  3 Ê 2a
2b  3 œ 0. Also, f differentiable at x Á 2 xÄ2

Ê f w (x) œ œ

a, x  2 . In order that f w (2) exist we must have a œ 2a(2)
b Ê a œ 4a
b Ê 3a œ b. 2ax
b, x  2

Then 2a
2b  3 œ 0 and 3a œ b Ê a œ

3 4

and b œ

9 4

.

(b) For x  #, the graph of f is a straight line having a slope of

$ %

and passing through the origin; for x   #, the graph of f

is a parabola. At x œ #, the value of the y-coordinate on the parabola is

$ #

which matches the y-coordinate of the point

on the straight line at x œ #. In addition, the slope of the parabola at the match up point is

$ %

which is equal to the

slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 18. (a) For any a, b and for any x Á
1, g is differentiable at x. Next, g differentiable at x œ
1 Ê g continuous at x œ
1 Ê lim b g(x) œ g(
1) Ê
a
1  2b œ
a  b Ê b œ 1. Also, g differentiable at x Á
1 x Ä
"

Ê gw (x) œ œ

a, x 
1 . In order that gw (
1) exist we must have a œ 3a(
1)#  1 Ê a œ 3a  1 3ax#  1, x 
1

Ê a œ
"# . (b) For x Ÿ
", the graph of f is a straight line having a slope of


" #

and a y-intercept of ". For x 
", the graph of f is

a parabola. At x œ
", the value of the y-coordinate on the parabola is

$ #

which matches the y-coordinate of the point

on the straight line at x œ
". In addition, the slope of the parabola at the match up point is
"# which is equal to the slope of the straight line. Therefore, since the graph is differentiable at the match up point, the graph is smooth there. 19. f odd Ê f(
x) œ
f(x) Ê

d dx

(f(
x)) œ

d dx

(
f(x)) Ê f w (
x)(
1) œ
f w (x) Ê f w (
x) œ f w (x) Ê f w is even.

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194

Chapter 3 Differentiation

20. f even Ê f(
x) œ f(x) Ê

d dx

(f(
x)) œ

d dx

(f(x)) Ê f w (
x)(
1) œ f w (x) Ê f w (
x) œ
f w (x) Ê f w is odd.

21. Let h(x) œ (fg)(x) œ f(x) g(x) Ê hw (x) œ x lim Äx œ x lim Äx œ

f(x) g(x)
f(x) g(x! )  f(x) g(x! )
f(x! ) g(x! ) x
x!

!

g(x! ) f(x! ) x lim ’ g(x)x

x! “ Ä x!

!

h(x)
h(x! ) x
x!

œ x lim Äx

!

f(x) g(x)
f(x! ) g(x! ) x
x!

f(x! ) !) œ x lim ’f(x) ’ g(x)x

xg(x ““  x lim ’g(x! ) ’ f(x)x

x! ““ Äx Äx ! !

w

 g(x! ) f (x! ) œ 0 †

g(x! ) lim ’ g(x)x

x! “ x Ä x!

!

w

 g(x! ) f (x! ) œ g(x! ) f w (x! ), if g is

continuous at x! . Therefore (fg)(x) is differentiable at x! if f(x! ) œ 0, and (fg)w (x! ) œ g(x! ) f w (x! ). 22. From Exercise 21 we have that fg is differentiable at 0 if f is differentiable at 0, f(0) œ 0 and g is continuous at 0. (a) If f(x) œ sin x and g(x) œ kxk , then kxk sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ kxk is continuous at x œ 0. (b) If f(x) œ sin x and g(x) œ x#Î$ , then x#Î$ sin x is differentiable because f w (0) œ cos (0) œ 1, f(0) œ sin (0) œ 0 and g(x) œ x#Î$ is continuous at x œ 0. (c) If f(x) œ 1
cos x and g(x) œ $Èx, then $Èx (1
cos x) is differentiable because f w (0) œ sin (0) œ 0, f(0) œ 1
cos (0) œ 0 and g(x) œ x"Î$ is continuous at x œ 0. (d) If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable because f w (0) œ 1, f(0) œ 0 and sin ˆ "x ‰

lim x sin ˆ "x ‰ œ lim

xÄ!

" x

xÄ!

œ lim

tÄ_

sin t t

œ 0 (so g is continuous at x œ 0).

23. If f(x) œ x and g(x) œ x sin ˆ "x ‰ , then x# sin ˆ x" ‰ is differentiable at x œ 0 because f w (0) œ 1, f(0) œ 0 and lim x sin ˆ "x ‰ œ lim

xÄ!

sin ˆ "x ‰ " x

xÄ!

œ lim

tÄ_

sin t t

œ 0 (so g is continuous at x œ 0). In fact, from Exercise 21,

h (0) œ g(0) f (0) œ 0. However, for x Á 0, hw (x) œ
x# cos ˆ "x ‰‘ ˆ
x"# ‰  2x sin ˆ x" ‰ . But lim hw (x) œ lim

cos ˆ "x ‰  2x sin ˆ x" ‰‘ does not exist because cos ˆ x" ‰ has no limit as x Ä 0. Therefore, w

w

xÄ!

xÄ!

the derivative is not continuous at x œ 0 because it has no limit there. 24. From the given conditions we have f(x  h) œ f(x) f(h), f(h)
1 œ hg(h) and lim g(h) œ 1. Therefore, hÄ!

w

f (x) œ

lim f(xh)h
f(x) hÄ! w

œ

lim f(x) f(h)h
f(x) hÄ!

œ

lim f(x) ’ f(h)h
1 “ hÄ!

œ f(x) ’ lim g(h)“ œ f(x) † 1 œ f(x)

Ê f w (x) œ f(x) and f axbexists at every value of x.

hÄ!

25. Step 1: The formula holds for n œ 2 (a single product) since y œ u" u# Ê

dy dx

œ

du" dx

u#  u"

du# dx

.

Step 2: Assume the formula holds for n œ k: y œ u" u# âuk Ê

du# duk dx u$ âuk  á  u" u# âuk-1 dx d(u" u# âuk ) If y œ u" u# âuk ukb1 œ au" u# âuk b ukb1 , then dy ukb1  u" u# âuk dudxkb1 dx œ dx dukb1 du# duk ‰ " œ ˆ du dx u# u$ âuk  u" dx u$ âuk  â  u" u# âukc1 dx ukb1  u" u# âuk dx dukb1 du# duk " œ du dx u# u$ âukb1  u" dx u$ â ukb1  â  u" u# âukc1 dx ukb1  u" u# âuk dx . dy dx

œ

du" dx

u# u$ âuk  u"

.

Thus the original formula holds for n œ (k1) whenever it holds for n œ k. 26. Recall ˆ mk ‰ œ œ

m! m! m! m! ˆm‰ ˆm‰ ˆ m ‰ k! (m
k)! . Then 1 œ 1! (m
1)! œ m and k  k  1 œ k! (m
k)!  (k  1)! (m
k
1)! m! (k  1)  m! (m
k) (m  1)! ˆm1‰ œ (k m!1)!(m(m
1)k)! œ (k  1)! ((m (k  1)! (m
k)!  1)
(k  1))! œ k  1 . Now, we prove

Leibniz's rule by mathematical induction. Step 1: If n œ 1, then

d(uv) dv du dx œ u dx  v dx . Assume that the statement is true for n œ k, that is:
" # k k
# k
" d (uv) du d u dv dk v ˆk‰ d u d v ˆ k ‰ du d v dxk œ dxk v  k dxk
" dx  2 dxk
# dx#  á  k
1 dv dxk
"  u dxk . kb" k k
" k (uv) d dk u dv dk" u d# v
ddxk
"u v  ddxuk dv ‘ If n œ k  1, then d dx(uv) œ dx Š d dx k
" k ‹ œ dx  ’k dxk dx  k dxk" dx# “ k

Step 2:

k

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 3 Additional and Advanced Exercises  ’ˆ k2 ‰ 
du dx  

dk
" u d# v dxk
" dx#

 ˆ k2 ‰

kb"

dk
# u d$ v dxk
# dx$ “

 á  ’ˆ k
k 1 ‰

k
"

d# u dk
" v dx# dxk
"

dv d u‘ d u d u dv dxk  u dxk
" œ dxk
" v  (k  1) dxk dx  k kb" k
" du d v d v d u
ˆ k
k 1 ‰  ˆ kk ‰‘ dx dxk  u dxk" œ dxk" v  (k  k kb" dv d v ˆ k k 1 ‰ du dx dxk  u dxk
" . k


ˆ k1 ‰

k

1)

 ˆ k
k 1 ‰

du dk u dx dxk

v“

k" #  ˆ k2 ‰‘ ddxk"u ddxv#  á dk u dv dk
" u d# v ˆ k 2 1 ‰ dx k
" dxk dx  dx#

á

Therefore the formula (c) holds for n œ (k  1) whenever it holds for n œ k. 27. (a) T# œ (b) T# œ

41 # L g #

41 L g

ÊLœ

T# g 41 #

ÊTœ

#1 È L; Èg

ÊLœ

a1 sec# ba32.2 ft/sec# b 41 #

dT œ

#1 Èg

†

" dL #È L

Ê L ¸ 0.8156 ft œ

1 ÈLg dL;

dT œ

1 Èa!Þ)"&' ftba32.2 ft/sec# b a!Þ!"

ftb ¸ 0.00613 sec.

(c) Since there are 86,400 sec in a day, we have a0.00613 secba86,400 sec/dayb ¸ 529.6 sec/day, or 8.83 min/day; the clock will lose about 8.83 min/day. 28. v œ s$ Ê

dv dt

# œ $s# ds dt œ
ka's b Ê

ds dt

œ
#k. If s! œ the initial length of the cube's side, then s" œ s!
#k

Ê #k œ s!
s" . Let t œ the time it will take the ice cube to melt. Now, t œ œ

" "Î$ "
ˆ $% ‰

s! #k

œ

s! s !
s "

œ

av! b"Î$ "Î$ av! b
ˆ $% v! ‰ "Î$

¸ "" hr.

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195

196

Chapter 3 Differentiation

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 4 APPLICATIONS OF DERIVATIVES 4.1 EXTREME VALUES OF FUNCTIONS 1. An absolute minimum at x œ c# , an absolute maximum at x œ b. Theorem 1 guarantees the existence of such extreme values because h is continuous on [aß b]. 2. An absolute minimum at x œ b, an absolute maximum at x œ c. Theorem 1 guarantees the existence of such extreme values because f is continuous on [aß b]. 3. No absolute minimum. An absolute maximum at x œ c. Since the function's domain is an open interval, the function does not satisfy the hypotheses of Theorem 1 and need not have absolute extreme values. 4. No absolute extrema. The function is neither continuous nor defined on a closed interval, so it need not fulfill the conclusions of Theorem 1. 5. An absolute minimum at x œ a and an absolute maximum at x œ c. Note that y œ g(x) is not continuous but still has extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 6. Absolute minimum at x œ c and an absolute maximum at x œ a. Note that y œ g(x) is not continuous but still has absolute extrema. When the hypothesis of Theorem 1 is satisfied then extrema are guaranteed, but when the hypothesis is not satisfied, absolute extrema may or may not occur. 7. Local minimum at a
"ß !b, local maximum at a"ß !b 8. Minima at a
#ß !b and a#ß !b, maximum at a!ß #b 9. Maximum at a!ß &b. Note that there is no minimum since the endpoint a#ß !b is excluded from the graph. 10. Local maximum at a
$ß !b, local minimum at a#ß !b, maximum at a"ß #b, minimum at a!ß
"b 11. Graph (c), since this the only graph that has positive slope at c. 12. Graph (b), since this is the only graph that represents a differentiable function at a and b and has negative slope at c. 13. Graph (d), since this is the only graph representing a funtion that is differentiable at b but not at a. 14. Graph (a), since this is the only graph that represents a function that is not differentiable at a or b.

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198

Chapter 4 Applications of Derivatives

15. f(x) œ

2 3

x
5 Ê f w (x) œ

f(
2) œ


19 3 ,

2 3

Ê no critical points;

f(3) œ
3 Ê the absolute maximum

is
3 at x œ 3 and the absolute minimum is
19 3 at x œ
2

16. f(x) œ
x
4 Ê f w (x) œ
1 Ê no critical points; f(
4) œ 0, f(1) œ
5 Ê the absolute maximum is 0 at x œ
4 and the absolute minimum is
5 at x œ "

17. f(x) œ x#
1 Ê f w (x) œ 2x Ê a critical point at x œ 0; f(
1) œ 0, f(0) œ
1, f(2) œ 3 Ê the absolute maximum is 3 at x œ 2 and the absolute minimum is
1 at x œ 0

18. f(x) œ %
x# Ê f w (x) œ
2x Ê a critical point at x œ 0; f(
3) œ
5, f(0) œ 4, f(1) œ 3 Ê the absolute maximum is 4 at x œ 0 and the absolute minimum is
5 at x œ
3

19. F(x) œ
x"# œ
x
# Ê Fw (x) œ 2x
$ œ

2 x$

, however

x œ 0 is not a critical point since 0 is not in the domain; F(0.5) œ
4, F(2) œ
0.25 Ê the absolute maximum is
0.25 at x œ 2 and the absolute minimum is
4 at x œ 0.5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.1 Extreme Values of Functions 20. F(x) œ
"x œ
x
" Ê Fw (x) œ x
# œ

" x#

, however

x œ 0 is not a critical point since 0 is not in the domain; F(
2) œ "# , F(
1) œ 1 Ê the absolute maximum is 1 at x œ
1 and the absolute minimum is

21. h(x) œ $Èx œ x"Î$ Ê hw (x) œ

" 3

" #

at x œ
2

x
#Î$ Ê a critical point

at x œ 0; h(
1) œ
1, h(0) œ 0, h(8) œ 2 Ê the absolute maximum is 2 at x œ 8 and the absolute minimum is
1 at x œ
1

22. h(x) œ
3x#Î$ Ê hw (x) œ
#x
"Î$ Ê a critical point at x œ 0; h(
1) œ
3, h(0) œ 0, h(1) œ
3 Ê the absolute maximum is 0 at x œ 0 and the absolute minimum is
3 at x œ 1 and at x œ
1

23. g(x) œ È4
x# œ a4
x# b Ê gw (x) œ

" #

a4
x# b


"Î#

"Î#

(
2x) œ


x È 4
x#

Ê critical points at x œ
2 and x œ 0, but not at x œ 2 because 2 is not in the domain; g(
2) œ 0, g(0) œ 2, g(1) œ È3 Ê the absolute maximum is 2 at x œ 0 and the absolute minimum is 0 at x œ
2 24. g(x) œ
È5
x# œ
a&
x# b a5
x# b (
2x) x "‰ w ˆ Ê g (x) œ
# œ È # Ê critical points at x œ
È5 "Î#


"Î#

&
x

and x œ 0, but not at x œ È5 because È5 is not in the domain; f Š
È5‹ œ 0, f(0) œ
È5 Ê the absolute maximum is 0 at x œ
È5 and the absolute minimum is
È5 at x œ 0

25. f()) œ sin ) Ê f w ()) œ cos ) Ê ) œ
1 #

1 #

is a critical point,

but ) œ is not a critical point because
#1 is not interior the domain; f ˆ
#1 ‰ œ
1, f ˆ 1# ‰ œ 1, f ˆ 561 ‰ œ "# Ê the absolute maximum is 1 at ) œ 1# and the absolute minimum is
1 at ) œ
#1

to

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199

200

Chapter 4 Applications of Derivatives

26. f()) œ tan ) Ê f w ()) œ sec# ) Ê f has no critical points in 1‰ ˆ
1 3 ß 4 . The extreme values therefore occur at the ‰ œ
È3 and f ˆ 14 ‰ œ 1 Ê the absolute endpoints: f ˆ
1 3 maximum is 1 at ) œ 14 and the absolute minimum is
È3 at ) œ
1 3

27. g(x) œ csc x Ê gw (x) œ
(csc x)(cot x) Ê a critical point at x œ 1# ; g ˆ 13 ‰ œ È23 , g ˆ 1# ‰ œ 1, g ˆ 231 ‰ œ È23 Ê the absolute maximum is

at x œ

2 È3

absolute minimum is 1 at x œ

1 3

and x œ

21 3 ,

and the

1 #

28. g(x) œ sec x Ê gw (x) œ (sec x)(tan x) Ê a critical point at x œ 0; g ˆ
13 ‰ œ 2, g(0) œ 1, g ˆ 16 ‰ œ È23 Ê the absolute maximum is 2 at x œ
13 and the absolute minimum is 1 at x œ 0

29. f(t) œ 2
ktk œ #
Èt# œ #
at# b Ê f w (t) œ
"# at# b


"Î#

"Î#

(2t) œ
Èt # œ
kttk t

Ê a critical point at t œ 0; f(
1) œ 1, f(0) œ 2, f(3) œ
1 Ê the absolute maximum is 2 at t œ 0 and the absolute minimum is
1 at t œ 3

30. f(t) œ kt
5k œ È(t
5)# œ a(t
5)# b œ

" #

a(t
5)# b


"Î#

(2(t
5)) œ

t
5 È(t
5)#

"Î#

œ

Ê f w (t) t
5 kt
5 k

Ê a critical point at t œ 5; f(4) œ 1, f(5) œ 0, f(7) œ 2 Ê the absolute maximum is 2 at t œ 7 and the absolute minimum is 0 at t œ 5 31. f(x) œ x%Î$ Ê f w (x) œ

4 3

x"Î$ Ê a critical point at x œ 0; f(
1) œ 1, f(0) œ 0, f(8) œ 16 Ê the absolute

maximum is 16 at x œ 8 and the absolute minimum is 0 at x œ 0 32. f(x) œ x&Î$ Ê f w (x) œ

5 3

x#Î$ Ê a critical point at x œ 0; f(
1) œ
1, f(0) œ 0, f(8) œ 32 Ê the absolute

maximum is 32 at x œ 8 and the absolute minimum is
1 at x œ
1 33. g()) œ )$Î& Ê gw ()) œ

3 5

)
#Î& Ê a critical point at ) œ 0; g(
32) œ
8, g(0) œ 0, g(1) œ 1 Ê the absolute

maximum is 1 at ) œ 1 and the absolute minimum is
8 at ) œ
32

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.1 Extreme Values of Functions 34. h()) œ 3)#Î$ Ê hw ()) œ 2)
"Î$ Ê a critical point at ) œ 0; h(
27) œ 27, h(0) œ 0, h(8) œ 12 Ê the absolute maximum is 27 at ) œ
27 and the absolute minimum is 0 at ) œ 0 35. Minimum value is 1 at x œ #.

36. To find the exact values, note that yw œ $x#
#, which is zero when x œ „ É #$ . Local maximum at Š
É #$ ß % 

%È ' * ‹

¸ a
!Þ)"'ß &Þ!)*b; local

minimum at ŠÉ #$ ß %


%È ' * ‹

¸ a!Þ)"'ß #Þ*""b

37. To find the exact values, note that that yw œ $x#  #x
) œ a$x
%bax  #b, which is zero when x œ
# or x œ %$ . ‰ Local maximum at a
#ß "(b; local minimum at ˆ %$ ß
%" #(

38. Note that yw œ $x#
'x  $ œ $ax
"b# , which is zero at x œ ". The graph shows that the function assumes lower values to the left and higher values to the right of this point, so the function has no local or global extreme values.

39. Minimum value is 0 when x œ
" or x œ ".

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201

202

Chapter 4 Applications of Derivatives

40. The minimum value is 1 at x œ !.

41. The actual graph of the function has asymptotes at x œ „ ", so there are no extrema near these values. (This is an example of grapher failure.) There is a local minimum at a!ß "b.

42. Maximum value is 2 at x œ "; minimum value is 0 at x œ
" and x œ $.

" # at x œ "à
"# as x œ
".

43. Maximum value is minimum value is

" # at x œ 0à
"# as x œ
2.

44. Maximum value is minimum value is

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.1 Extreme Values of Functions 45. yw œ x#Î$ a"b  #$ x
"Î$ ax  #b œ crit. pt. x œ
%& xœ!

derivative ! undefined

&x  % $ x $È

extremum local max local min

46. yw œ x#Î$ a#xb  #$ x
"Î$ ax#
%b œ crit. pt. x œ
" xœ! xœ"

derivative ! undefined !

" a
#xb  a"bÈ% #È %
x #
x#  a%
x# b %
#x # œÈ È %
x# %
x#

crit. pt. x œ
# x œ
È # x œ È# xœ#

)x#
) $ x $È

extremum minimum local max minimum

47. yw œ x œ

value "# "Î$ œ "Þ!$% #& "! 0

derivative undefined ! ! undefined

value
$ 0 $


x#

extremum local max minimum maximum local min

value !
# # !

48. yw œ x# #È$"
x a
1b  #xÈ$
x œ


x#  a%xba$
xb #È $
x

crit. pt. xœ0 x œ "# & xœ$

_5x#  "#x #È $
x

derivative ! ! undefined


#, 49. yw œ œ ", crit. pt. xœ"

œ

extremum minimum local max minimum

value ! "%% "Î# ¸ %Þ%'# "#& "& !

extremum minimum

value #

x" x"

derivative undefined

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203

204

Chapter 4 Applications of Derivatives


", x  ! 50. yw œ œ #
#x, x  ! crit. pt. xœ! xœ"

51. yw œ œ

derivative undefined !


2x
2,
2x  6,

crit. pt. x œ
1 xœ1 xœ3

extremum local min local max

value $ %

x1 x1

derivative ! undefined !

extremum maximum local min maximum

value 5 1 5


"% x#
"# x  "& % , xŸ" x$
'x#  )x, x" w w # if x  ", and limc f a"  hb œ
". Also, f axb œ $x
"#x  ) if x  ", and

52. We begin by determining whether f w axb is defined at x œ ", where faxb œ œ Clearly, f w axb œ
"# x


" #

hÄ!

limb f w a"  hb œ
". Since f is continuous at x œ ", we have that f w a"b œ
". Thus,

hÄ!

f w axb œ œ


"# x
"# , $x
"#x  ) , #

Note that
"# x
But #


#È $ $

crit. pt. x œ
" x ¸ $Þ"&&

" #

xŸ" x"

œ ! when x œ
", and $x#
"#x  ) œ ! when x œ

¸ !Þ)%&  ", so the critical points occur at x œ
" and x œ derivative ! !

extremum local max local min

È "# „ È"##
%a$ba)b œ "# „' %) #a$b È #  # $ $ ¸ $Þ"&&.

œ#„

#È$ $ .

value 4 ¸
$Þ!(*

53. (a) No, since f w axb œ #$ ax
#b
"Î$ , which is undefined at x œ #.

(b) The derivative is defined and nonzero for all x Á #. Also, fa#b œ ! and faxb  ! for all x Á #. (c) No, faxb need not have a global maximum because its domain is all real numbers. Any restriction of f to a closed interval of the form Òa, bÓ would have both a maximum value and minimum value on the interval. (d) The answers are the same as (a) and (b) with 2 replaced by a.
x$  *x, x Ÿ
$ or ! Ÿ x  $
$x$  *, x 
$ or !  x  $ . Therefore, f w axb œ œ . $ x
*x,
$  x  ! or x   $ $x$
*,
$  x  ! or x  $ (a) No, since the left- and right-hand derivatives at x œ !, are
* and *, respectively. (b) No, since the left- and right-hand derivatives at x œ $, are
") and "), respectively.

54. Note that faxb œ œ

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Section 4.1 Extreme Values of Functions

205

(c) No, since the left- and right-hand derivatives at x œ
$, are ") and
"), respectively. (d) The critical points occur when f w axb œ ! (at x œ „ È$) and when f w axb is undefined (at x œ ! and x œ „ $). The minimum value is ! at x œ
$, at x œ !, and at x œ $; local maxima occur at Š
È$ß 'È$‹ and ŠÈ$ß 'È$‹. 55.

(a) The construction cost is Caxb œ !Þ$È"'  x#  !Þ#a*
xb million dollars, where ! Ÿ x Ÿ * miles. The following is a graph of Caxb.

Solving Cw axb œ

!Þ$x È"'  x#


!Þ# œ ! gives x œ „

)È & &

¸ „ $Þ&) miles, but only x œ $Þ&) miles is a critical point is È

the specified domain. Evaluating the costs at the critical and endpoints gives Ca!b œ $3 million, CŠ ) & & ‹ ¸ $2.694 million, and Ca*b ¸ $2.955 million. Therefore, to minimize the cost of construction, the pipeline should be placed from the docking facility to point B, 3.58 miles along the shore from point A, and then along the shore from B to the refinery. (b) If the per mile cost of underwater construction is p, then Caxb œ pÈ"'  x#  !Þ#a*
xb and Cw axb œ È !Þ$x x#
!Þ# œ ! gives xc œ Èp#!Þ) , which minimizes the construction cost provided xc Ÿ *. The value
!Þ!% "' 

of p that gives xc œ * miles is !Þ#"))'%. Consequently, if the underwater construction costs $218,864 per mile or less, then running the pipeline along a straight line directly from the docking facility to the refinery will minimize the cost of construction. In theory, p would have to be infinite to justify running the pipe directly from the docking facility to point A (i.e., for xc to be zero). For all values of p  !Þ#"))'% there is always an xc − Ð!ß *Ñ that will give a minimum value for C. This is proved by looking at Cww axc b œ

"'p a"'  x#c b$Î#

which is always positive for p  !.

56. There are two options to consider. The first is to build a new road straight from Village A to Village B. The second is to build a new highway segment from Village A to the Old Road, reconstruct a segment of Old Road, and build a new highway segment from Old Road to Village B, as shown in the figure. The cost of the first option is C" œ !Þ&a"&!b million dollars œ 75 million dollars.

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206

Chapter 4 Applications of Derivatives

The construction cost for the second option is C# axb œ !Þ&Š#È#&!!  x# ‹  !Þ$a"&!
#xb million dollars for ! Ÿ x Ÿ (& miles. The following is a graph of C# axb.

Solving Cw# axb œ

x È#&!!  x#


!Þ' œ ! give x œ „ $(Þ& miles, but only x œ $(Þ& miles is in the specified domain. In

summary, C" œ $75 million, C# a!b œ $95 million, C# a$(Þ&b œ $85 million, and C# a(&b œ $90.139 million. Consequently, a new road straight from village A to village B is the least expensive option. 57.

The length of pipeline is Laxb œ È%  x#  É#&  a"!
xb# for ! Ÿ x Ÿ "!. The following is a graph of Laxb.

Setting the derivative of Laxb equal to zero gives Lw axb œ "!
x É#&  a"!
xb#

x È %  x#




a"!
xb É#&  a"!
xb#

œ !. Note that

x È %  x#

œ cos )A and

œ cos )B , therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACP is similar to ˜BDP. Use

simple proportions to determine x as follows:

x 2

œ

"!
x &

Êxœ

#! (

¸ #Þ)&( miles along the coast from town A to town B.

If the two towns were on opposite sides of the river, the obvious solution would be to place the pump station on a straight line (the shortest distance) between two towns, again forcing )A œ )B . The shortest length of pipe is the same regardless of whether the towns are on thee same or opposite sides of the river.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.1 Extreme Values of Functions

207

58.

(a) The length of guy wire is Laxb œ È*!!  x#  É#&!!  a"&!
xb# for ! Ÿ x Ÿ "&!. The following is a graph of Laxb.

Setting Lw axb equal to zero gives Lw axb œ a"&!
xb É#&!!  a"&!
xb#

x È*!!  x#




a"&!
xb É#&!!  a"&!
xb#

œ !. Note that

x È*!!  x#

œ cos )A and

œ cos )B . Therefore, Lw axb œ ! when cos )A œ cos )B , or )A œ )B and ˜ACE is similar to ˜ABD.

Use simple proportions to determine x:

x $!

œ

"&!
x &!

Êxœ

##& %

œ &'Þ#& feet.

(b) If the heights of the towers are hB and hC , and the horizontal distance between them is s, then Laxb œ Éh#C  x#  Éh#B  as
xb# and Lw axb œ as
x b É h B  as
x b #

x Éh#C  x#




as
x b É h B  as
x b #

. However,

x Éh#C  x#

œ cos )G and

œ cos )B . Therefore, Lw axb œ ! when cos )C œ cos )B , or )C œ )B and ˜ACE is similar to ˜ABD.

Simple proportions can again be used to determine the optimum x: hxc œ

s
x hB

Ê x œ Š hB hc hc ‹s.

59. (a) Vaxb œ "'!x
&#x#  %x$ Vw axb œ "'!
"!%x  "#x# œ %ax
#ba$x
#!b The only critical point in the interval a!ß &b is at x œ #. The maximum value of Vaxb is 144 at x œ #. (b) The largest possible volume of the box is 144 cubic units, and it occurs when x œ # units. 60. (a) Pw axb œ #
#!!x
# The only critical point in the interval a!ß _b is at x œ "!. The minimum value of Paxb is %! at x œ "!. (b) The smallest possible perimeter of the rectangel is 40 units and it occurs at x œ "! units which makes the rectangle a 10 by 10 square. 61. Let x represent the length of the base and È#&
x# the height of the triangle. The area of the triangle is represented by # Aaxb œ x È#&
x# where ! Ÿ x Ÿ &. Consequently, solving Aw axb œ ! Ê #&
#x œ ! Ê x œ & . Since #È#&
x#

#

Aa!b œ Aa&b œ !, Aaxb is maximized at x œ

& È# .

The largest possible area is AŠ È&# ‹ œ

È#

#& %

cm# .

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208

Chapter 4 Applications of Derivatives

62. (a) From the diagram the perimeter P œ #x  #1r œ %!! Ê x œ #!!
1r. The area A is 2rx Ê Aarb œ %!!r
#1r# where ! Ÿ r Ÿ #!! 1 . (b) Aw arb œ %!!
%1r so the only critical point is r œ

"!! 1 .

Since Aarb œ ! if r œ ! and x œ #!!
1r œ !, the values r œ "!! 1 ¸ 31.83 m and x œ "!! m maximize the area over the interval ! Ÿ r Ÿ 63. s œ
"# gt#  v! t  s! Ê

ds dt

#!! 1 .

œ
gt  v! œ ! Ê t œ

2

Thus sŠ vg! ‹ œ
"# gŠ vg! ‹  v0 Š vg! ‹  s0 œ 64.

Now satb œ s0 Í tˆ
gt2  v0 ‰ œ 0 Í t œ 0 or t œ

 s0  s0 is the maximum height over the interval 0 Ÿ t Ÿ

œ ! Ê tan t œ " Ê t œ never negative) Ê the peak current is #È# amps. dI dt

œ
#sin t  #cos t, solving

v!2 2g

v! g.

dI dt

65. Yes, since f(x) œ kxk œ Èx# œ ax# b

"Î#

Ê f w (x) œ

" #

ax# b

1 %

2v0 g . 2v0 g .

 n1 where n is a nonnegative integer (in this exercise t is


"Î#

(2x) œ

x ax# b"Î#

œ

x kx k

is not defined at x œ 0. Thus it

is not required that f w be zero at a local extreme point since f w may be undefined there. 66. If f(c) is a local maximum value of f, then f(x) Ÿ f(c) for all x in some open interval (aß b) containing c. Since f is even, f(
x) œ f(x) Ÿ f(c) œ f(
c) for all
x in the open interval (
bß
a) containing
c. That is, f assumes a local maximum at the point
c. This is also clear from the graph of f because the graph of an even function is symmetric about the y-axis. 67. If g(c) is a local minimum value of g, then g(x)   g(c) for all x in some open interval (aß b) containing c. Since g is odd, g(
x) œ
g(x) Ÿ
g(c) œ g(
c) for all
x in the open interval (
bß
a) containing
c. That is, g assumes a local maximum at the point
c. This is also clear from the graph of g because the graph of an odd function is symmetric about the origin. 68. If there are no boundary points or critical points the function will have no extreme values in its domain. Such functions do indeed exist, for example f(x) œ x for
_  x  _. (Any other linear function f(x) œ mx  b with m Á 0 will do as well.) 69. (a) f w axb œ $ax#  #bx  c is a quadratic, so it can have 0, 1, or 2 zeros, which would be the critical points of f. The function faxb œ x$
$x has two critical points at x œ
" and x œ ". The function faxb œ x$
" has one critical point at x œ !Þ The function faxb œ x$  x has no critical points.

(b) The function can have either two local extreme values or no extreme values. (If there is only one critical point, the cubic function has no extreme values.)

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Section 4.1 Extreme Values of Functions

209

70. (a)

fa!b œ ! is not a local extreme value because in any open interval containing x œ !, there are infinitely many points where faxb œ " and where faxb œ
". (b) One possible answer, on the interval Ò!ß "Ó: " a"
xbcos "
x , !Ÿx" faxb œ œ !, x œ " This function has no local extreme value at x œ ". Note that it is continuous on Ò!ß "Ó. 71. Maximum value is 11 at x œ &; minimum value is 5 on the interval Ò
$ß #Ó; local maximum at a
&ß *b

72. Maximum value is 4 on the interval Ò&ß (Ó; minimum value is
% on the interval Ò
#ß "Ó.

73. Maximum value is & on the interval Ò$ß _Ñ; minimum value is
& on the interval Ð
_ß
#Ó.

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210

Chapter 4 Applications of Derivatives

74. Minimum value is 4 on the interval Ò
"ß $Ó

75-80. Example CAS commands: Maple: with(student): f := x -> x^4 - 8*x^2 + 4*x + 2; domain := x=-20/25..64/25; plot( f(x), domain, color=black, title="Section 4.1 #75(a)" ); Df := D(f); plot( Df(x), domain, color=black, title="Section 4.1 # 75(b)" ) StatPt := fsolve( Df(x)=0, domain ) SingPt := NULL; EndPt := op(rhs(domain)); Pts :=evalf([EndPt,StatPt,SingPt]); Values := [seq( f(x), x=Pts )]; Maximum value is 2.7608 and occurs at x=2.56 (right endpoint). %

Minimum value $ is -6.2680 and occurs at x=1.86081 (singular point). Mathematica: (functions may vary) (see section 2.5 re. RealsOnly ): <<Miscellaneous `RealOnly` Clear[f,x] a =
1; b = 10/3; f[x_] =2  2x
3 x2/3 f'[x] Plot[{f[x], f'[x]}, {x, a, b}] NSolve[f'[x]==0, x] {f[a], f[0], f[x]/.%, f[b]//N In more complicated expressions, NSolve may not yield results. In this case, an approximate solution (say 1.1 here) is observed from the graph and the following command is used: FindRoot[f'[x]==0,{x, 1.1}] 4.2 THE MEAN VALUE THEOREM 1. When f(x) œ x#  2x
1 for 0 Ÿ x Ÿ 1, then 2. When f(x) œ x#Î$ for 0 Ÿ x Ÿ 1, then 3. When f(x) œ x 

" x

for

" #

f(1)
f(0) 1
0

Ÿ x Ÿ 2, then

f(1)
f(0) 1
0

œ f w (c) Ê 1 œ ˆ 32 ‰ c
"Î$ Ê c œ

f(2)
f(1/2) 2
1/2

4. When f(x) œ Èx
1 for 1 Ÿ x Ÿ 3, then

œ f w (c) Ê 3 œ 2c  2 Ê c œ #" .

f(3)
f(1) 3
1

œ f w (c) Ê 0 œ "
œ f w (c) Ê

È2 #

œ

" c#

8 #7 .

Ê c œ 1.

" #Èc
1

Ê c œ #3 .

5. Does not; f(x) is not differentiable at x œ 0 in (
"ß 8). Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.2 The Mean Value Theorem 6. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1). 7. Does; f(x) is continuous for every point of [0ß 1] and differentiable for every point in (0ß 1). 8. Does not; f(x) is not continuous at x œ 0 because lim c f(x) œ 1 Á 0 œ f(0). xÄ! 9. Since f(x) is not continuous on 0 Ÿ x Ÿ 1, Rolle's Theorem does not apply: Á 0 œ f(1).

lim f(x) œ lim c x œ 1

x Ä 1c

xÄ1

10. Since f(x) must be continuous at x œ 0 and x œ 1 we have lim b f(x) œ a œ f(0) Ê a œ 3 and xÄ! lim c f(x) œ lim b f(x) Ê
1  3  a œ m  b Ê 5 œ m  b. Since f(x) must also be differentiable at xÄ1

xÄ1

x œ 1 we have lim c f w (x) œ lim b f w (x) Ê
2x  3k x=1 œ mk x=1 Ê 1 œ m. Therefore, a œ 3, m œ 1 and b œ 4. xÄ1 xÄ1 11. (a) i ii iii iv (b) Let r" and r# be zeros of the polynomial P(x) œ xn  an-1 xn-1  á  a" x  a! , then P(r" ) œ P(r# ) œ 0. Since polynomials are everywhere continuous and differentiable, by Rolle's Theorem Pw (r) œ 0 for some r between r" and r# , where Pw (x) œ nxn-1  (n
1) an-1 xn-2  á  a" . 12. With f both differentiable and continuous on [aß b] and f(r" ) œ f(r# ) œ f(r$ ) œ 0 where r" , r# and r$ are in [aß b], then by Rolle's Theorem there exists a c" between r" and r# such that f w (c" ) œ 0 and a c# between r# and r$ such that f w (c# ) œ 0. Since f w is both differentiable and continuous on [aß b], Rolle's Theorem again applies and we have a c$ between c" and c# such that f w w (c$ ) œ 0. To generalize, if f has n1 zeros in [aß b] and f ÐnÑ is continuous on [aß b], then f ÐnÑ has at least one zero between a and b. 13. Since f ww exists throughout [aß b] the derivative function f w is continuous there. If f w has more than one zero in [aß b], say f w (r" ) œ f w (r# ) œ 0 for r" Á r# , then by Rolle's Theorem there is a c between r" and r# such that f ww (c) œ 0, contrary to f ww  0 throughout [aß b]. Therefore f w has at most one zero in [aß b]. The same argument holds if f ww  0 throughout [aß b]. 14. If f(x) is a cubic polynomial with four or more zeros, then by Rolle's Theorem f w (x) has three or more zeros, f ww (x) has 2 or more zeros and f www (x) has at least one zero. This is a contradiction since f www (x) is a non-zero constant when f(x) is a cubic polynomial. 15. With f(
2) œ 11  0 and f(
1) œ
1  0 we conclude from the Intermediate Value Theorem that f(x) œ x%  3x  1 has at least one zero between
2 and
1. Then
2  x 
1 Ê
)  x$ 
1 Ê
32  4x$ 
4 Ê
29  4x$  3 
1 Ê f w (x)  0 for
2  x 
1 Ê f(x) is decreasing on [
#ß
1] Ê f(x) œ 0 has exactly one solution in the interval (
#ß
1). 16. f(x) œ x$ 

4 x#

 7 Ê f w (x) œ 3x#


8 x$

 0 on (
_ß 0) Ê f(x) is increasing on (
_ß 0). Also, f(x)  0 if

x 
2 and f(x)  0 if
2  x  0 Ê f(x) has exactly one zero in (
_ß !). 17. g(t) œ Èt  Èt  1
4 Ê gw (t) œ

" #È t



" 2Èt1

 0 Ê g(t) is increasing for t in (!ß _); g(3) œ È3
2  0

and g(15) œ È15  0 Ê g(t) has exactly one zero in (!ß _)Þ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

211

212

Chapter 4 Applications of Derivatives

18. g(t) œ

" "
t


È1
t  3.1 Ê gw (t) œ

" ("
t)#




" 2 È 1 t

 0 Ê g(t) is increasing for t in (1ß 1);

g(0.99) œ 2.5 and g(0.99) œ 98.3 Ê g(t) has exactly one zero in (1ß 1). 19. r()) œ )
sin# ˆ 3) ‰  8 Ê rw ()) œ 1


sin ˆ 3) ‰ cos ˆ 3) ‰ œ 1
"3 sin ˆ 23) ‰  0 on (_ß _) Ê r()) is increasing on (_ß _); r(0) œ 8 and r(8) œ sin# ˆ 83 ‰  0 Ê r()) has exactly one zero in (_ß _). 2 3

20. r()) œ 2)  cos# )
È2 Ê rw ()) œ 2
2 sin ) cos ) œ 2
sin 2)  0 on (_ß _) Ê r()) is increasing on (_ß _); r(#1) œ 41  cos (#1)
È2 œ 41  1
È2  0 and r(21) œ 41  1
È2  0 Ê r()) has exactly one zero in (_ß _).  0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ!ß 1# ‰ ; r(0.1) ¸ 994 and r(1.57) ¸ 1260.5 Ê r()) has exactly one zero in ˆ!ß 1# ‰ .

21. r()) œ sec ) 

" )$


5 Ê rw ()) œ (sec ))(tan ))


3 )%

22. r()) œ tan )  cot )  ) Ê rw ()) œ sec# )
csc# )  1 œ sec# )
cot# )  0 on ˆ!ß 1# ‰ Ê r()) is increasing on ˆ0ß 1# ‰ ; r ˆ 14 ‰ œ  14  0 and r(1.57) ¸ 1254.2 Ê r()) has exactly one zero in ˆ!ß 1# ‰ . 23. By Corollary 1, f w (x) œ 0 for all x Ê f(x) œ C, where C is a constant. Since f(1) œ 3 we have C œ 3 Ê f(x) œ 3 for all x. 24. g(x) œ 2x
5 Ê gw (x) œ 2 œ f w (x) for all x. By Corollary 2, f(x) œ g(x)
C for some constant C. Then f(0) œ g(0)
C Ê 5 œ 5
C Ê C œ 0 Ê f(x) œ g(x) œ 2x
5 for all x. 25. g(x) œ x# Ê gw (x) œ 2x œ f w (x) for all x. By Corollary 2, f(x) œ g(x)
C. (a) f(0) œ 0 Ê 0 œ g(0)
C œ 0
C Ê C œ 0 Ê f(x) œ x# Ê f(2) œ 4 (b) f(1) œ 0 Ê 0 œ g(1)
C œ 1
C Ê C œ 1 Ê f(x) œ x#  1 Ê f(2) œ 3 (c) f(2) œ 3 Ê 3 œ g(2)
C Ê 3 œ 4
C Ê C œ 1 Ê f(x) œ x#  1 Ê f(2) œ 3 26. g(x) œ mx Ê gw (x) œ m, a constant. If f w (x) œ m, then by Corollary 2, f(x) œ g(x)
b œ mx
b where b is a constant. Therefore all functions whose derivatives are constant can be graphed as straight lines y œ mx
b. 27. (a) y œ

x# #


C

(b) y œ

28. (a) y œ x#
C

x$ 3


C

(b) y œ x#  x
C

29. (a) yw œ x
# Ê y œ

" x


C

(b) y œ x


" x

" #

31. (a) y œ  "# cos 2t
C (c) y œ 

" #

cos 2t
2 sin

32. (a) y œ tan )
C


C

(c) y œ 5x 

" x


C

(b) y œ 2Èx
C

(b) y œ 2 sin t #

x% 4

(c) y œ x$
x#  x
C


C

x
"Î# Ê y œ x"Î#
C Ê y œ Èx
C (c) y œ 2x#  2Èx
C

30. (a) yw œ

(c) y œ

t #


C


C (b) yw œ )"Î# Ê y œ

2 3

)$Î#
C

(c) y œ

33. f(x) œ x#  x
C; 0 œ f(0) œ 0#  0
C Ê C œ 0 Ê f(x) œ x#  x

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2 3

)$Î#  tan )
C

Section 4.2 The Mean Value Theorem

213

34. g(x) œ  "x
x#
C; 1 œ g(1) œ 
"1
(1)#
C Ê C œ 1 Ê g(x) œ  x"
x#  1 35. r()) œ 8)
cot )
C; 0 œ r ˆ 14 ‰ œ 8 ˆ 14 ‰
cot ˆ 14 ‰
C Ê 0 œ 21
1
C Ê C œ 21  1 Ê r()) œ 8)
cot )  21  1 36. r(t) œ sec t  t
C; 0 œ r(0) œ sec (0)  0
C Ê C œ 1 Ê r(t) œ sec t  t  1 37. v œ

ds dt

œ *Þ)t
& Ê s œ %Þ*t#
&t
C; at s œ "! and t œ ! we have C œ "! Ê s œ %Þ*t#
&t
"!

38. v œ

ds dt

œ $#t  # Ê s œ "'t#  #t
C; at s œ % and t œ

39. v œ

ds dt

œ sina1tb Ê s œ  1" cosa1tb
C; at s œ ! and t œ ! we have C œ

40. v œ

ds dt

œ 12 cosˆ #1t ‰ Ê s œ sinˆ #1t ‰
C; at s œ " and t œ 1# we have C œ " Ê s œ sinˆ #1t ‰
"

" #

we have C œ " Ê s œ 't#  #t
" " 1

Êsœ

"
cosa1tb 1

41. a œ $# Ê v œ $#t
C" ; at v œ #! and t œ ! we have C" œ #! Ê v œ $#t
#! Ê s œ "'t#
#!t
C# ; at s œ & and t œ ! we have C# œ & Ê s œ "'t#
#!t
& 42. a œ 9.8 Ê v œ 9.8t
C" ; at v œ $ and t œ ! we have C" œ $ Ê v œ *Þ)t  $ Ê s œ %Þ*t#  $t
C# ; at s œ ! and t œ ! we have C# œ ! Ê s œ %Þ*t#  $t 43. a œ %sina#tb Ê v œ #cosa#tb
C" ; at v œ # and t œ ! we have C" œ ! Ê v œ #cosa#tb Ê s œ sina#tb
C# ; at s œ $ and t œ ! we have C# œ $ Ê s œ sina#tb  $ Ê v œ 1$ sinˆ $1t ‰
C" ; at v œ ! and t œ ! we have C" œ ! Ê v œ 1$ sinˆ $1t ‰ Ê s œ cosˆ $1t ‰
C# ; at s œ " and t œ ! we have C# œ ! Ê s œ cosˆ $1t ‰

44. a œ

* ˆ $t ‰ 1# cos 1

45. If T(t) is the temperature of the thermometer at time t, then T(0) œ 19° C and T(14) œ 100° C. From the Mean Value Theorem there exists a 0  t!  14 such that

T(14)
T(0) 14
0

œ 8.5° C/sec œ Tw (t! ), the rate at which

the temperature was changing at t œ t! as measured by the rising mercury on the thermometer. 46. Because the trucker's average speed was 79.5 mph, by the Mean Value Theorem, the trucker must have been going that speed at least once during the trip. 47. Because its average speed was approximately 7.667 knots, and by the Mean Value Theorem, it must have been going that speed at least once during the trip. 48. The runner's average speed for the marathon was approximately 11.909 mph. Therefore, by the Mean Value Theorem, the runner must have been going that speed at least once during the marathon. Since the initial speed and final speed are both 0 mph and the runner's speed is continuous, by the Intermediate Value Theorem, the runner's speed must have been 11 mph at least twice. 49. Let d(t) represent the distance the automobile traveled in time t. The average speed over 0 Ÿ t Ÿ 2 is d(2)
d(0) #
0 .

The Mean Value Theorem says that for some 0  t!  2, dw (t! ) œ

d(2)
d(0) #
0 .

The value dw (t! ) is

the speed of the automobile at time t! (which is read on the speedometer). 50. aatb œ vw atb œ "Þ' Ê vatb œ "Þ't
C; at a!ß !b we have C œ ! Ê vatb œ "Þ't. When t œ $!, then va$!b œ %) m/sec.

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214

Chapter 4 Applications of Derivatives

51. The conclusion of the Mean Value Theorem yields 52. The conclusion of the Mean Value Theorem yields

" b


"a b
a

b #
a# b
a


b‰ œ  c"# Ê c# ˆ a ab œ a  b Ê c œ Èab.

œ 2c Ê c œ

a b # .

53. f w (x) œ [cos x sin (x
2)
sin x cos (x
2)]  2 sin (x
1) cos (x
1) œ sin (x
x
2)  sin 2(x
1) œ sin (2x
2)  sin (2x
2) œ 0. Therefore, the function has the constant value f(0) œ sin# 1 ¸ 0.7081 which explains why the graph is a horizontal line. 54. (a) faxb œ ax
#bax
"bxax  "bax  #b œ x&  &x$
%x is one possibility. (b) Graphing faxb œ x&  &x$
%x and f w axb œ &x%  "&x#
% on Ò$ß $Ó by Ò(ß (Ó we see that each x-intercept of f w axb lies between a pair of x-intercepts of faxb, as expected by Rolle's Theorem.

(c) Yes, since sin is continuous and differentiable on a  _ß _b. 55. faxb must be zero at least once between a and b by the Intermediate Value Theorem. Now suppose that faxb is zero twice between a and b. Then by the Mean Value Theorem, f w axb would have to be zero at least once between the two zeros of faxb, but this can't be true since we are given that f w axb Á ! on this interval. Therefore, faxb is zero once and only once between a and b. 56. Consider the function kaxb œ faxb  gaxb. kaxb is continuous and differentiable on Òa, bÓ, and since kaab œ faab  gaab and kabb œ fabb  gabb, by the Mean Value Theorem, there must be a point c in aa, bb where kw acb œ !. But since kw acb œ f w acb  gw acb, this means that f w acb œ gw acb, and c is a point where the graphs of f and g have tangent lines with the same slope, so these lines are either parallel or are the same line. 57. Yes. By Corollary 2 we have f(x) œ g(x)
c since f w (x) œ gw (x). If the graphs start at the same point x œ a, then f(a) œ g(a) Ê c œ 0 Ê f(x) œ g(x). 58. Let f(x) œ sin x for a Ÿ x Ÿ b. From the Mean Value Theorem there exists a c between a and b such that sin b
sin a sin a sin a ¸ œ cos c Ê 1 Ÿ sin bb
Ÿ 1 Ê ¸ sin bb
Ÿ 1 Ê ksin b  sin ak Ÿ kb  ak. b
a
a
a 59. By the Mean Value Theorem we have w

we have f(b)  f(a)  0 Ê f (c)  0.

f(b)
f(a) b
a

œ f w (c) for some point c between a and b. Since b  a  0 and f(b)  f(a),

60. The condition is that f w should be continuous over [aß b]. The Mean Value Theorem then guarantees the existence of a point c in (aß b) such that

f(b)
f(a) b
a w

œ f w (c). If f w is continuous, then it has a minimum and

maximum value on [aß b], and min f w Ÿ f (c) Ÿ max f w , as required.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.3 Monotonic Functions and the First Derivative Test 61. f w (x) œ a1
x% cos xb


"

Ê f ww (x) œ  a1
x% cos xb


#

a4x$ cos x  x% sin xb


#

œ x$ a1
x% cos xb (4 cos x  x sin x)  0 for 0 Ÿ x Ÿ 0.1 Ê f w (x) is decreasing when 0 Ÿ x Ÿ 0.1 f(0.1)
" 0.1

Ê min f w ¸ 0.9999 and max f w œ 1. Now we have 0.9999 Ÿ

Ÿ 1 Ê 0.09999 Ÿ f(0.1)  1 Ÿ 0.1

Ê 1.09999 Ÿ f(0.1) Ÿ 1.1. 62. f w (x) œ a1  x% b


"

Ê f ww (x) œ  a1  x% b


#

a4x$ b œ

4x$ $ a1
x % b

 0 for 0  x Ÿ 0.1 Ê f w (x) is increasing when

0 Ÿ x Ÿ 0.1 Ê min f w œ 1 and max f w œ 1.0001. Now we have 1 Ÿ

f(0.1)
2 0.1

Ÿ 1.0001

Ê 0.1 Ÿ f(0.1)  2 Ÿ 0.10001 Ê 2.1 Ÿ f(0.1) Ÿ 2.10001. 63. (a) Suppose x  1, then by the Mean Value Theorem (b)

f(1) Mean Value Theorem f(x)x
0
1 f(x)
f(1) Yes. From part (a), lim c x
1 xÄ1

f(x)
f(1) x
1

 0 Ê f(x)  f(1). Suppose x  1, then by the

Ê f(x)  f(1). Therefore f(x)   1 for all x since f(1) œ 1.

f(1) Ÿ 0 and lim b f(x)x
  0. Since f w (1) exists, these two one-sided
1 xÄ1 limits are equal and have the value f w (1) Ê f w (1) Ÿ 0 and f w (1)   0 Ê f w (1) œ 0.

64. From the Mean Value Theorem we have has only one solution c œ 

q #p .

f(b)
f(a) b
a

œ f w (c) where c is between a and b. But f w (c) œ 2pc
q œ 0

(Note: p Á 0 since f is a quadratic function.)

4.3 MONOTONIC FUNCTIONS AND THE FIRST DERIVATIVE TEST 1. (a) f w (x) œ x(x  1) Ê critical points at 0 and 1 (b) f w œ


±  ±


Ê increasing on (_ß !) and ("ß _), decreasing on (!ß ") ! " (c) Local maximum at x œ 0 and a local minimum at x œ 1 2. (a) f w (x) œ (x  1)(x
2) Ê critical points at 2 and 1 (b) f w œ


±  ±


Ê increasing on (_ß #) and ("ß _), decreasing on (2ß ") # " (c) Local maximum at x œ 2 and a local minimum at x œ 1 3. (a) f w (x) œ (x  1)# (x
2) Ê critical points at 2 and 1 (b) f w œ  ±


±


Ê increasing on (2ß 1) and ("ß _), decreasing on (_ß 2) # " (c) No local maximum and a local minimum at x œ 2 4. (a) f w (x) œ (x  1)# (x
2)# Ê critical points at 2 and 1 (b) f w œ


±


±


Ê increasing on (_ß 2)  (#ß ")  ("ß _), never decreasing # " (c) No local extrema 5. (a) f w (x) œ (x  1)(x
2)(x  3) Ê critical points at 2, 1 and 3 (b) f w œ  ±


±  ±


Ê increasing on (2ß 1) and ($ß _), decreasing on (_ß 2) and ("ß $) # " $ (c) Local maximum at x œ 1, local minima at x œ 2 and x œ 3 6. (a) f w (x) œ (x  7)(x
1)(x
5) Ê critical points at 5, 1 and 7 (b) f w œ  ±


±  ±


Ê increasing on (5ß 1) and (7ß _), decreasing on (_ß 5) and ("ß 7) & " ( (c) Local maximum at x œ 1, local minima at x œ 5 and x œ 7

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215

216

Chapter 4 Applications of Derivatives

7. (a) f w (x) œ x
"Î$ (x
2) Ê critical points at 2 and 0 (b) f w œ


±  )(


Ê increasing on (_ß 2) and (0ß _), decreasing on (2ß 0) ! # (c) Local maximum at x œ 2, local minimum at x œ 0 8. (a) f w (x) œ x
"Î# (x  3) Ê critical points at 0 and 3 (b) f w œ (  ±


Ê increasing on ($ß _), decreasing on (0ß 3) ! $ (c) No local maximum and a local minimum at x œ 3 9. (a) g(t) œ t#  3t
3 Ê gw (t) œ 2t  3 Ê a critical point at t œ  3# ; gw œ


± , increasing on $Î# ˆ_ß  3# ‰ , decreasing on ˆ 3# ß _‰ (b) local maximum value of g ˆ 3# ‰ œ (c) absolute maximum is

21 4

21 4

at t œ  3#

at t œ  3#

(d)

10. (a) g(t) œ 3t#
9t
5 Ê gw (t) œ 6t
9 Ê a critical point at t œ

3 #

; gw œ


± , increasing on $Î#

ˆ_ß 32 ‰ , decreasing on ˆ 3# ß _‰

(b) local maximum value of g ˆ 3# ‰ œ (c) absolute maximum is

47 4

at t œ

47 4

at t œ

3 #

3 #

(d)

11. (a) h(x) œ x$
2x# Ê hw (x) œ 3x#
4x œ x(4  3x) Ê critical points at x œ 0, 43 Ê hw œ  ±


± , increasing on ˆ0ß 43 ‰ , decreasing on (_ß !) and ˆ 43 ß _‰ ! %Î$ 4 (b) local maximum value of h ˆ 43 ‰ œ 32 27 at x œ 3 ; local minimum value of h(0) œ 0 at x œ 0 (c) no absolute extrema

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.3 Monotonic Functions and the First Derivative Test

217

(d)

12. (a) h(x) œ 2x$  18x Ê hw (x) œ 6x#  18 œ 6 Šx
È3‹ Šx  È3‹ Ê critical points at x œ „ È3 Ê hw œ


|  |


, increasing on Š_ß È3‹ and ŠÈ$ß _‹ , decreasing on ŠÈ$ß È3‹ È$ È $ (b) a local maximum is h ŠÈ3‹ œ 12È3 at x œ È3; local minimum is h ŠÈ3‹ œ 12È3 at x œ È3 (c) no absolute extrema (d)

13. (a) f()) œ 3)#  4)$ Ê f w ()) œ 6)  12)# œ 6)(1  2)) Ê critical points at ) œ 0,

" #

Ê f w œ  ±


± , ! "Î#

increasing on ˆ0ß "# ‰ , decreasing on (_ß !) and ˆ "# ß _‰ (b) a local maximum is f ˆ "# ‰ œ 4" at ) œ #" , a local minimum is f(0) œ 0 at ) œ 0 (c) no absolute extrema (d)

14. (a) f()) œ 6)  )$ Ê f w ()) œ 6  3)# œ 3 ŠÈ2  )‹ ŠÈ2
)‹ Ê critical points at ) œ „ È2 Ê f w œ  ±


± , increasing on ŠÈ2ß È2‹, decreasing on Š_ß È2‹ and ŠÈ2ß _‹ È# È # (b) a local maximum is f ŠÈ2‹ œ 4È2 at ) œ È2, a local minimum is f ŠÈ2‹ œ %È2 at ) œ È2 (c) no absolute extrema

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218

Chapter 4 Applications of Derivatives

(d)

15. (a) f(r) œ 3r$
16r Ê f w (r) œ 9r#
16 Ê no critical points Ê f w œ




, increasing on (_ß _), never decreasing (b) no local extrema (c) no absolute extrema (d)

16. (a) h(r) œ (r
7)$ Ê hw (r) œ 3(r
7)# Ê a critical point at r œ 7 Ê hw œ


±


, increasing on ( (_ß 7)  ((ß _), never decreasing (b) no local extrema (c) no absolute extrema (d)

17. (a) f(x) œ x%  8x#
16 Ê f w (x) œ 4x$  16x œ 4x(x
2)(x  2) Ê critical points at x œ 0 and x œ „ 2 Ê f w œ  ±


±  ±


, increasing on (#ß !) and (#ß _), decreasing on (_ß 2) and (!ß #) # ! # (b) a local maximum is f(0) œ 16 at x œ 0, local minima are f a „ 2b œ 0 at x œ „ 2 (c) no absolute maximum; absolute minimum is 0 at x œ „ 2 (d)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.3 Monotonic Functions and the First Derivative Test

219

18. (a) g(x) œ x%  4x$
4x# Ê gw (x) œ 4x$  12x#
)x œ 4x(x  2)(x  1) Ê critical points at x œ 0, 1, 2 Ê gw œ  ±


±  ±


, increasing on (0ß 1) and (#ß _), decreasing on (_ß 0) and (1ß #) ! " # (b) a local maximum is g(1) œ 1 at x œ 1, local minima are g(0) œ 0 at x œ 0 and g(2) œ 0 at x œ 2 (c) no absolute maximum; absolute minimum is 0 at x œ 0, 2 (d)

19. (a) H(t) œ

3 % # t

 t' Ê Hw (t) œ 6t$  6t& œ 6t$ (1
t)("  t) Ê critical points at t œ 0, „ 1

Ê Hw œ


±  ±


± , increasing on (_ß 1) and (0ß 1), decreasing on ("ß 0) and ("ß _) " ! " (b) the local maxima are H(1) œ "# at t œ 1 and H(1) œ "# at t œ 1, the local minimum is H(0) œ 0 at t œ 0 (c) absolute maximum is

" #

at t œ „ 1; no absolute minimum

(d)

20. (a) K(t) œ 15t$  t& Ê Kw (t) œ 45t#  5t% œ 5t# (3
t)(3  t) Ê critical points at t œ 0, „ 3 Ê Kw œ  ±


±


± , increasing on (3ß 0)  (0ß 3), decreasing on (_ß 3) and (3ß _) $ ! $ (b) a local maximum is K(3) œ 162 at t œ 3, a local minimum is K(3) œ 162 at t œ 3 (c) no absolute extrema (d)

21. (a) g(x) œ xÈ8  x# œ x a8  x# b

"Î#

Ê gw (x) œ a8  x# b

"Î#


x ˆ "# ‰ a)  x# b


"Î#

(2x) œ

2(2
x)(2  x) ÊŠ2È2
x‹ Š2È2  x‹

Ê critical points at x œ „ 2, „ 2È2 Ê gw œ (  ±


±  Ñ , increasing on (#ß #), decreasing on # # #È# #È # Š#È2ß #‹ and Š#ß #È2‹

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220

Chapter 4 Applications of Derivatives

(b) local maxima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2, local minima are g(2) œ 4 at x œ 2 and g Š2È2‹ œ 0 at x œ 2È2 (c) absolute maximum is 4 at x œ 2; absolute minimum is 4 at x œ 2 (d)

22. (a) g(x) œ x# È5  x œ x# (5  x)"Î# Ê gw (x) œ 2x(5  x)"Î#
x# ˆ "# ‰ (5  x)
"Î# (1) œ

5x(4
x) 2 È 5
x

Ê critical points at x œ 0, 4 and 5 Ê gw œ  ±


±  Ñ , increasing on (0ß 4), decreasing on (_ß !) & ! % and (%ß &) (b) a local maximum is g(4) œ 16 at x œ 4, a local minimum is 0 at x œ 0 and x œ 5 (c) no absolute maximum; absolute minimum is 0 at x œ 0, 5 (d)

23. (a) f(x) œ

x#
3 x
#

Ê f w (x) œ

2x(x
2)
ax#
3b (1) (x
2)#

œ

(x
3)(x
") (x
#)#

Ê critical points at x œ 1, 3

Ê f w œ


±  )(  ±


, increasing on (_ß 1) and ($ß _), decreasing on ("ß #) and (#ß $), # " $ discontinuous at x œ 2 (b) a local maximum is f(1) œ 2 at x œ 1, a local minimum is f(3) œ 6 at x œ 3 (c) no absolute extrema (d)

24. (a) f(x) œ

x$ 3x#  1

Ê f w (x) œ

3x# a3x#  1b
x$ (6x) a3x#  1b#

œ

3x# ax#  1b a3x#  1b#

Ê a critical point at x œ 0

w

Ê f œ


±


, increasing on (_ß !)  (!ß _), and never decreasing ! (b) no local extrema (c) no absolute extrema

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Section 4.3 Monotonic Functions and the First Derivative Test

221

(d)

25. (a) f(x) œ x"Î$ (x
8) œ x%Î$
8x"Î$ Ê f w (x) œ

4 3

x"Î$
83 x
#Î$ œ

w

4(x  2) 3x#Î$

Ê critical points at x œ 0, 2

Ê f œ  ±


)(


, increasing on (#ß !)  (!ß _), decreasing on (_ß 2) ! # (b) no local maximum, a local minimum is f(2) œ 6 $È2 ¸ 7.56 at x œ 2 (c) no absolute maximum; absolute minimum is 6 $È2 at x œ 2 (d)

26. (a) g(x) œ x#Î$ (x
5) œ x&Î$
5x#Î$ Ê gw (x) œ

5 3

x#Î$


10 3

x
"Î$ œ

5(x  2) $ 3È x

Ê critical points at x œ 2 and

x œ 0 Ê gw œ


±  )(


, increasing on (_ß 2) and (!ß _), decreasing on (2ß !) ! # (b) local maximum is g(2) œ 3 $È4 ¸ 4.762 at x œ 2, a local minimum is g(0) œ 0 at x œ 0 (c) no absolute extrema (d)

27. (a) h(x) œ x"Î$ ax#  4b œ x(Î$  4x"Î$ Ê hw (x) œ x œ 0,

„2 È7

7 3

x%Î$  43 x
#Î$ œ

ŠÈ7x  2‹ ŠÈ7x
#‹ $ # 3È x

Ê critical points at


2 Ê hw œ


±  )(  ±


, increasing on Š_ß È ‹ and Š È27 ß _‹ , decreasing on 7 ! È È #Î ( #Î (


2 ß !‹ and Š!ß È27 ‹ ŠÈ 7
2 (b) local maximum is h Š È ‹œ 7

$ 24 È 2 7(Î'

¸ 3.12 at x œ


2 È7 ,

the local minimum is h Š È27 ‹ œ 

$ 24 È 2 7(Î'

(c) no absolute extrema

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

¸ 3.12

222

Chapter 4 Applications of Derivatives

(d)

28. (a) k(x) œ x#Î$ ax#  4b œ x)Î$  4x#Î$ Ê kw (x) œ

8 3

x&Î$  83 x
"Î$ œ

8(x  1)(x
1) $ 3È x

Ê critical points at

x œ 0, „ 1 Ê kw œ  ±


)(  ±


, increasing on ("ß !) and ("ß _), decreasing on (_ß 1) ! " " and (!ß 1) (b) local maximum is k(0) œ 0 at x œ 0, local minima are k a „ 1b œ 3 at x œ „ 1 (c) no absolute maximum; absolute minimum is 3 at x œ „ 1 (d)

29. (a) f(x) œ 2x  x# Ê f w (x) œ 2  2x œ 2(1  x) Ê a critical point at x œ 1 Ê f w œ


±  ] and f(1) œ 1, # " f(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is 0 at x œ 2 (b) absolute maximum is 1 at x œ 1; no absolute minimum (c)

30. (a) f(x) œ (x
1)# Ê f w (x) œ 2(x
1) Ê a critical point at x œ 1 Ê f w œ  ±


] and f(1) œ 0, f(0) œ 1 ! " Ê a local maximum is 1 at x œ 0, a local minimum is 0 at x œ 1 (b) no absolute maximum; absolute minimum is 0 at x œ 1 (c)

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Section 4.3 Monotonic Functions and the First Derivative Test

223

31. (a) g(x) œ x#  4x
4 Ê gw (x) œ 2x  4 œ 2(x  2) Ê a critical point at x œ 2 Ê gw œ [  ±


and " # g(1) œ 1, g(2) œ 0 Ê a local maximum is 1 at x œ 1, a local minimum is g(2) œ 0 at x œ 2 (b) no absolute maximum; absolute minimum is 0 at x œ 2 (c)

32. (a) g(x) œ x#  6x  9 Ê gw (x) œ 2x  6 œ 2(x
3) Ê a critical point at x œ 3 Ê gw œ [


±  and % $ g(4) œ 1, g(3) œ 0 Ê a local maximum is 0 at x œ 3, a local minimum is 1 at x œ 4 (b) absolute maximum is 0 at x œ 3; no absolute minimum (c)

33. (a) f(t) œ 12t  t$ Ê f w (t) œ 12  3t# œ 3(2
t)(2  t) Ê critical points at t œ „ 2 Ê f w œ [  ±


±  $ # # and f(3) œ 9, f(2) œ 16, f(2) œ 16 Ê local maxima are 9 at t œ 3 and 16 at t œ 2, a local minimum is 16 at t œ 2 (b) absolute maximum is 16 at t œ 2; no absolute minimum (c)

34. (a) f(t) œ t$  3t# Ê f w (t) œ 3t#  6t œ 3t(t  2) Ê critical points at t œ 0 and t œ 2 Ê f w œ


±  ±


] and f(0) œ 0, f(2) œ 4, f(3) œ 0 Ê a local maximum is 0 at t œ 0 and t œ 3, a $ ! # local minimum is 4 at t œ 2 (b) absolute maximum is 0 at t œ 0, 3; no absolute minimum

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

224

Chapter 4 Applications of Derivatives

(c)

x$ 3


2x#  4x Ê hw (x) œ x#
4x  4 œ (x
2)# Ê a critical point at x œ 2 Ê hw œ [  ±  and ! # h(0) œ 0 Ê no local maximum, a local minimum is 0 at x œ 0 (b) no absolute maximum; absolute minimum is 0 at x œ 0 (c)

35. (a) h(x) œ

36. (a) k(x) œ x$  3x#  3x  1 Ê kw (x) œ 3x#  6x  3 œ 3(x  1)# Ê a critical point at x œ
1 Ê kw œ  ±  ] and k(
1) œ 0, k(0) œ 1 Ê a local maximum is 1 at x œ 0, no local minimum !
" (b) absolute maximum is 1 at x œ 0; no absolute minimum (c)

37. (a) f(x) œ

x #


2 sin ˆ x# ‰ Ê f w (x) œ

Ê f w œ [


±  ] ! #1 #1Î$


cos ˆ x# ‰ , f w (x) œ 0 Ê cos ˆ x# ‰ œ "# Ê a critical point at x œ 231 and f(0) œ 0, f ˆ 231 ‰ œ 13
È3, f(21) œ 1 Ê local maxima are 0 at x œ 0 and 1 " #

at x œ 21, a local minimum is 13
È3 at x œ 231 (b) The graph of f rises when f w  0, falls when f w  0, and has a local minimum value at the point where f w changes from negative to positive.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.3 Monotonic Functions and the First Derivative Test

225

38. (a) f(x) œ
2 cos x
cos# x Ê f w (x) œ 2 sin x  2 cos x sin x œ 2(sin x)(1  cos x) Ê critical points at x œ
1, 0, 1 Ê f w œ [


±  ] and f(
1) œ 1, f(0) œ
3, f(1) œ 1 Ê a local maximum is 1 at
1 1 ! x œ „ 1, a local minimum is
3 at x œ 0 (b) The graph of f rises when f w  0, falls when f w  0, and has local extreme values where f w œ 0. The function f has a local minimum value at x œ 0, where the values of f w change from negative to positive.

39. (a) f(x) œ csc# x
2 cot x Ê f w (x) œ 2(csc x)(
csc x)(cot x)
2 a
csc# xb œ
2 acsc# xb (cot x
1) Ê a critical point at x œ 14 Ê f w œ (


±  ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14 1 ! 1Î% w (b) The graph of f rises when f  0, falls when f w  0, and has a local minimum value at the point where f w œ 0 and the values of f w change from negative to positive. The graph of f steepens as f w (x) Ä „ _.

40. (a) f(x) œ sec# x
2 tan x Ê f w (x) œ 2(sec x)(sec x)(tan x)
2 sec# x œ a2 sec# xb (tan x
1) Ê a critical point at x œ 14 Ê f w œ (


±  ) and f ˆ 14 ‰ œ 0 Ê no local maximum, a local minimum is 0 at x œ 14 1Î#
1Î# 1Î% w (b) The graph of f rises when f  0, falls when f w  0, and has a local minimum value where f w œ 0 and the values of f w change from negative to positive.

41. h()) œ 3 cos ˆ #) ‰ Ê hw ()) œ
3# sin ˆ #) ‰ Ê hw œ [


] , (!ß $) and (#1ß
3) Ê a local maximum is 3 at ) œ 0, ! #1 a local minimum is
3 at ) œ 21 42. h()) œ 5 sin ˆ #) ‰ Ê hw ()) œ minimum is 0 at ) œ 0

5 #

cos ˆ #) ‰ Ê hw œ [  ] , (!ß 0) and (1ß 5) Ê a local maximum is 5 at ) œ 1, a local 1 !

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226

Chapter 4 Applications of Derivatives

43. (a)

(b)

44. (a)

(b)

(c)

(d)

(c)

45. (a)

(b)

46. (a)

(b)

(d)

47. f(x) œ x$
3x  2 Ê f w (x) œ 3x#
3 œ 3(x
1)(x  1) Ê f w œ  ±


±  Ê rising for x œ c œ # since
" " f w (x)  0 for x œ c œ 2. 48. f(x) œ ax#  bx  c œ a ˆx#  ba x‰  c œ a Šx#  ba x 

b# 4a# ‹




b# 4a

 c œ a ˆx 

b ‰# 2a




b#
4ac 4a

, a parabola whose

b vertex is at x œ
2a . Thus when a  0, f is increasing on ˆ
2ab ß _‰ and decreasing on ˆ
_ß
#ab ‰ ; when a  0, f is increasing on ˆ
_ß
#ab ‰ and decreasing on ˆ
#ab ß _‰ . Also note that f w (x) œ 2ax  b œ 2a ˆx  #ba ‰ Ê for

a  0, f w œ


|  ; for a  0, f w œ  ±


.
bÎ2a
bÎ2a 4.4 CONCAVITY AND CURVE SKETCHING x$ 3

x# #

Ê yw œ x#
x
2 œ (x
2)(x  1) Ê yww œ 2x
1 œ 2 ˆx
"# ‰ . The graph is rising on (
_ß
1) and (#ß _), falling on (
"ß #), concave up on ˆ "# ß _‰ and concave down on ˆ
_ß "# ‰ . Consequently, a local maximum is 3# at x œ
1, a local minimum is
3 at x œ 2, and ˆ "# ß
34 ‰ is a point of inflection.

1. y œ





2x 

" 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 2. y œ

227


2x#  4 Ê yw œ x$
4x œ x ax#
4b œ x(x  2)(x
2) Ê yww œ 3x#
4 œ ŠÈ3x  2‹ ŠÈ3x
2‹ . The

x% 4

graph is rising on (
2ß 0) and (#ß _), falling on (
_ß
#) and (!ß #), concave up on Š
_ß È23 ‹ and Š È23 ß _‹ and concave down on Š
È23 ß È23 ‹ . Consequently, a local maximum is 4 at x œ 0, local minima are 0 at 2 16 x œ „ 2, and Š
È23 ß 16 9 ‹ and Š È3 ß 9 ‹ are points of inflection.

3 4

ax#
1b

#Î$

Ê yw œ ˆ 34 ‰ ˆ 23 ‰ ax#
1b


"Î$

minima are 0 at x œ „ 1; yww œ ax#
1b


"Î$

(2x) œ x ax#
1b


"Î$

, yw œ


) (  ±


)( 
" " ! Ê the graph is rising on (
"ß !) and ("ß _), falling on (
_ß
") and (!ß ") Ê a local maximum is 34 at x œ 0, local

3. y œ

 (x) ˆ
3" ‰ ax#
1b


%Î$

(2x) œ

x #
3 $ 3É ax #
1 b %

,

yww œ  ±


) (


)(


±  Ê the graph is concave up on Š
_ß
È3‹ and ŠÈ3ß _‹, concave
" " È$
È $ $ È down on Š
È3ß È3‹ Ê points of inflection at Š „ È3ß $ % ‹ %

x
#Î$ ax#
1b, yw œ  ±


)(


±  !
" " Ê the graph is rising on (
_ß
1) and ("ß _), falling on (
1ß ") Ê a local maximum is 27 at x œ
1, a local 7

4. y œ

9 14

x"Î$ ax#
7b Ê yw œ

3 14

x
#Î$ ax#
7b 

9 14

x"Î$ (2x) œ

3 #

ww
&Î$ minimum is
27 ax#
1b  3x"Î$ œ 2x"Î$  x
&Î$ œ x
&Î$ a2x#  1b , 7 at x œ 1; y œ
x

yww œ


)(  Ê the graph is concave up on (!ß _), concave down on (
_ß !) Ê a point of inflection at ! (!ß !) 5. y œ x  sin 2x Ê yw œ 1  2 cos 2x, yw œ [


±  ±


] Ê the graph is rising on ˆ
13 ß 13 ‰ , falling
#1Î$ #1Î$
1Î$ 1Î$ on ˆ
#31 ß
13 ‰ and ˆ 13 ß 231 ‰ Ê local maxima are
231 
13


È3 #

at x œ
13 and

21 3

È3 #

È3 #

at x œ
231 and

1 3



È3 #

at x œ

1 3

, local minima are

21 3

; yww œ
4 sin 2x, yww œ [


±  ±


±  ] Ê the !
#1Î$ #1Î$
1Î# 1Î# graph is concave up on ˆ
1# ß !‰ and ˆ 1# ß 231 ‰ , concave down on ˆ
231 ß
1# ‰ and ˆ!ß 1# ‰ Ê points of inflection at


at x œ

ˆ
1# ß
1# ‰ , (!ß !), and ˆ 1# ß 1# ‰ 6. y œ tan x
4x Ê yw œ sec# x
4, yw œ (  ±


±  ) Ê the graph is rising on ˆ
12 ß
13 ‰ and 1Î#
1Î# 1Î$
1Î$ ˆ 13 ß 1# ‰ , falling on ˆ
13 ß 13 ‰ Ê a local maximum is
È3  431 at x œ
13 , a local minimum is È3
431 at x œ 13 ; yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ (


±  ) Ê the graph is concave up on ˆ0ß 1# ‰ , !
1Î# 1Î# concave down on ˆ
12 ß 0‰ Ê a point of inflection at (0ß !)

7. If x   0, sin kxk œ sin x and if x  0, sin kxk œ sin (
x) œ
sin x. From the sketch the graph is rising on ˆ
3#1 ß
1# ‰ , ˆ!ß 1# ‰ and ˆ 3#1 ß #1‰ , falling on ˆ
21ß
3#1 ‰ , ˆ
1# ß !‰ and ˆ 1# ß 3#1 ‰ ; local minima are
1 at x œ „ 3#1 and 0 at x œ !; local maxima are 1 at x œ „

1 #

and

0 at x œ „ #1; concave up on (
#1ß
1) and (1ß #1), and concavedown on (
1ß 0) and (!ß 1) Ê points of inflection are (
1ß !) and (1ß !)

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228

Chapter 4 Applications of Derivatives

8. y œ 2 cos x
È2 x Ê yw œ
2 sin x
È2, yw œ [


±  ±


±  ] Ê rising on
1 $1Î#
$1Î%
1Î% &1Î% ˆ
341 ß
14 ‰and ˆ 541 ß 3#1 ‰ , falling on ˆ
1ß
341 ‰ and ˆ
14 ß 541 ‰ Ê local maxima are
2  1È2 at x œ
1, È2  È2

at x œ
14 and
31#

at x œ

31 # ,

È at x œ
341 and
È2
514 2 at x œ 541 ; Ê concave up on ˆ
1ß
1# ‰ and ˆ 1# ß 3#1 ‰ , concave down on

and local minima are
È2 

yww œ
2 cos x, yww œ [  ±


±  ]
1 $1Î#
1Î# 1Î# ˆ
1# ß 1# ‰ Ê points of inflection at Š
1# ß

È 21 # ‹

and Š 1# ß


31 È 2 4

1È2 4

È 21 # ‹

9. When y œ x#
4x  3, then yw œ 2x
4 œ 2(x
2) and yww œ 2. The curve rises on (#ß _) and falls on (
_ß #). At x œ 2 there is a minimum. Since yw w  0, the curve is concave up for all x.

10. When y œ '
2x
x# , then yw œ
#
2x œ
2("  x) and yww œ
2. The curve rises on (
_ß
1) and falls on (
1ß _). At x œ
1 there is a maximum. Since yw w  0, the curve is concave down for all x.

11. When y œ x$
3x  3, then yw œ 3x#
3 œ 3(x
1)(x  1) and yww œ 6x. The curve rises on (
_ß
1)  ("ß _) and falls on (
1ß 1). At x œ
1 there is a local maximum and at x œ 1 a local minimum. The curve is concave down on (
_ß 0) and concave up on (!ß _). There is a point of inflection at x œ 0.

12. When y œ x(6
2x)# , then yw œ
4x(6
2x)  ('
2x)# œ 12(3
x)("
x) and yww œ
12(3
x)
12("
x) œ 24(x
2). The curve rises on (
_ß ")  ($ß _) and falls on ("ß $). The curve is concave down on (
_ß #) and concave up on (#ß _). At x œ 2 there is a point of inflection.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 13. When y œ
2x$  6x#
3, then yw œ
6x#  12x œ
6x(x
2) and yww œ
12x  12 œ
12(x
1). The curve rises on (!ß #) and falls on (
_ß 0) and (#ß _). At x œ 0 there is a local minimum and at x œ 2 a local maximum. The curve is concave up on (
_ß ") and concave down on ("ß _). At x œ 1 there is a point of inflection.

14. When y œ 1
9x
6x#
x$ , then yw œ
9
12x
3x# œ
$(x  3)(B  1) and yww œ
12
6x œ
6(x  2). The curve rises on (
$ß
") and falls on (
_ß
3) and (
"ß _). At x œ
1 there is a local maximum and at x œ
3 a local minimum. The curve is concave up on (
_ß
2) and concave down on (
#ß _). At x œ
2 there is a point of inflection.

15. When y œ (x
2)$  1, then yw œ 3(x
2)# and yww œ 6(x
2). The curve never falls and there are no local extrema. The curve is concave down on (
_ß #) and concave up on (#ß _). At x œ 2 there is a point of inflection.

16. When y œ 1
(x  1)$ , then yw œ
3(x  1)# and yww œ
6(x  1). The curve never rises and there are no local extrema. The curve is concave up on (
_ß
1) and concave down on (
"ß _). At x œ
1 there is a point of inflection.

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229

230

Chapter 4 Applications of Derivatives

17. When y œ x%
2x# , then yw œ 4x$
4x œ 4x(x  1)(x
1) and yww œ 12x#
4 œ 12 Šx 

" È 3 ‹ Šx




" È3 ‹ .

The curve

rises on (
"ß !) and ("ß _) and falls on (
_ß
1) and (!ß "). At x œ „ 1 there are local minima and at x œ 0 a local maximum. The curve is concave up on Š
_ß
È"3 ‹ and Š È"3 ß _‹ and concave down on Š
È"3 ß È"3 ‹ . At x œ

„" È3

there are points of inflection. 18. When y œ
x%  6x#
4, then yw œ
4x$  12x œ
4x Šx  È3‹ Šx
È3‹ and yww œ
12x#  12 œ
12(x  1)(x
1). The curve rises on Š
_ß
È3‹ and Š!ß È3‹ , and falls on Š
È3ß !‹ and ŠÈ3ß _‹ . At x œ „ È3there are local maxima and at x œ 0 a local minimum. The curve is concave up on (
"ß ") and concave down on (
_ß
1) and ("ß _). At x œ „ 1 there are points of inflection.

19. When y œ 4x$
x% , then yw œ 12x#
4x$ œ 4x# ($
x) and yww œ 24x
12x# œ 12x(2
x). The curve rises on a
_ß $b and falls on a$ß _b. At x œ 3 there is a local maximum, but there is no local minimum. The graph is concave up on a!ß #b and concave down on a
_ß !b and a#ß _b. There are inflection points at x œ 0 and x œ 2. 20. When y œ x%  2x$ , then yw œ 4x$  6x# œ 2x# (2x  3) and yww œ 12x#  12x œ 12x(x  1). The curve rises on ˆ
3# ß _‰ and falls on ˆ
_ß
32 ‰ . There is a local minimum at x œ
3# , but no local maximum. The curve is

concave up on (
_ß
1) and (!ß _), and concave down on (
1ß 0). At x œ
1 and x œ 0 there are points of inflection.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 21. When y œ x&
5x% , then yw œ 5x%
20x$ œ 5x$ (x
4) and yww œ 20x$
60x# œ 20x# (x
3). The curve rises on (
_ß !) and (%ß _), and falls on (!ß %). There is a local maximum at x œ 0, and a local minimum at x œ 4. The curve is concave down on (
_ß 3) and concave up on (3ß _). At x œ 3 there is a point of inflection. % % $ 22. When y œ x ˆ x#
5‰ , then yw œ ˆ x#
5‰  x(4)ˆ x#
5‰ ˆ "# ‰ $ ww ‰ ˆx ‰# ˆ "# ‰ ˆ 5x ‰ œ ˆ x#
5‰ ˆ 5x #
5 , and y œ 3 #
5 #
5

$ #  ˆ x#
5‰ ˆ 5# ‰ œ 5 ˆ x#
5‰ (x
4). The curve is rising

on (
_ß #) and (10ß _), and falling on (#ß 10). There is a local maximum at x œ 2 and a local minimum at x œ 10. The curve is concave down on (
_ß %) and concave up on (%ß _). At x œ 4 there is a point of inflection.

23. When y œ x  sin x, then yw œ "  cos x and yww œ
sin x. The curve rises on (!ß 21). At x œ 0 there is a local and absolute minimum and at x œ 21 there is a local and absolute maximum. The curve is concave down on (!ß 1) and concave up on (1ß #1). At x œ 1 there is a point of inflection.

24. When y œ x
sin x, then yw œ "
cos x and yww œ sin x. The curve rises on (!ß 21). At x œ 0 there is a local and absolute minimum and at x œ 21 there is a local and absolute maximum. The curve is concave up on (!ß 1) and concave down on (1ß #1). At x œ 1 there is a point of inflection.

25. When y œ x"Î& , then yw œ

" 5

4
*Î& x
%Î& and yww œ
25 x .

The curve rises on (
_ß _) and there are no extrema. The curve is concave up on (
_ß !) and concave down on (!ß _). At x œ 0 there is a point of inflection.

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231

232

Chapter 4 Applications of Derivatives

26. When y œ x$Î& , then yw œ

3 5

6
(Î& x
#Î& and yww œ
25 x .

The curve rises on (
_ß _) and there are no extrema. The curve is concave up on (
_ß !) and concave down on (!ß _). At x œ 0 there is a point of inflection.

27. When y œ x#Î& , then yw œ

2 5

6
)Î& x
$Î& and yww œ
25 x .

The curve is rising on (0ß _) and falling on (
_ß !). At x œ 0 there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (
_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0. 28. When y œ x%Î& , then yw œ

4 5

4
'Î& x
"Î& and yww œ
25 x .

The curve is rising on (0ß _) and falling on (
_ß !). At x œ 0 there is a local and absolute minimum. There is no local or absolute maximum. The curve is concave down on (
_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0.

29. When y œ 2x
3x#Î$ , then yw œ 2
2x
"Î$ and yww œ 23 x
%Î$ . The curve is rising on (
_ß !) and ("ß _), and falling on (!ß "). There is a local maximum at x œ 0 and a local minimum at x œ 1. The curve is concave up on (
_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0.

30. When y œ 5x#Î&
2x, then yw œ 2x
$Î&
2 œ 2 ˆx
$Î&
1‰ and yww œ
65 x
)Î& . The curve is rising on (0ß 1) and falling on (
_ß 0) and ("ß _). There is a local minimum at x œ 0 and a local maximum at x œ 1. The curve is concave down on (
_ß !) and (!ß _). There are no points of inflection, but a cusp exists at x œ 0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 31. When y œ x#Î$ ˆ #5
x‰ œ

5 #Î$
x&Î$ , then # x yw œ 53 x
"Î$
53 x#Î$ œ 53 x
"Î$ (1
x) and
"Î$ yww œ
59 x
%Î$
10 œ
95 x
%Î$ (1  2x). 9 x

The curve is rising on (!ß ") and falling on (
_ß !) and ("ß _). There is a local minimum at x œ 0 and a local maximum at x œ 1. The curve is concave up on ˆ
_ß
"# ‰ and concave down on ˆ
"# ß !‰ and (0ß _). There is a point of inflection at x œ
"# and a cusp at x œ 0. 32. When y œ x#Î$ (x
5) œ x&Î$
5x#Î$ , then
"Î$ yw œ 53 x#Î$
10 œ 35 x
"Î$ (x
2) and 3 x yww œ

10 9

x
"Î$ 

10 9

x
%Î$ œ

10 9

x
%Î$ (x  1). The curve

is rising on (
_ß !) and (#ß _), and falling on (!ß #). There is a local minimum at x œ 2 and a local maximum at x œ 0. The curve is concave up on (
"ß 0) and (!ß _), and concave down on (
_ß
1). There is a point of inflection at x œ
1 and a cusp at x œ 0. "Î# 33. When y œ xÈ8
x# œ x a8
x# b , then

yw œ a8
x# b

"Î#

#
"Î#

œ a8
x b

 (x) ˆ "# ‰ a8
x# b a8
2x# b œ

yww œ ˆ
"# ‰a8
x# b œ

2x ax#
12b É a8
x # b $


$#


"Î#

(
#x)

2(2
x)(2  x) ÊŠ2È2  x‹ Š2È2
x‹

and

(
2x)a8
2x# b  a8
x# b


"#

(
4x)

. The curve is rising on (
#ß #), and falling

on Š
2È2ß
2‹ and Š#ß 2È2‹ . There are local minima x œ
2 and x œ 2È2, and local maxima at x œ
2È2 and x œ 2. The curve is concave up on Š
2È2ß !‹ and concave down on Š!ß #È2‹ . There is a point of inflection at x œ 0. 34. When y œ a2
x# b

$Î#

, then yw œ ˆ 3# ‰ a2
x# b

"Î#

(
2x)

œ
3xÈ2
x# œ
3xÊŠÈ2
x‹ ŠÈ2  x‹ and yww œ (
3) a2
x# b œ

"Î#


6("
x)(1  x) ÊŠÈ2
x‹ ŠÈ2  x‹

 (
3x) ˆ "# ‰ a2
x# b


"Î#

(
2x)

. The curve is rising on

Š
È2ß !‹ and falling on Š!ß È2‹ . There is a local maximum at x œ 0, and local minima at x œ „ È2. The curve is concave down on (
"ß ") and concave up on Š
È2ß
"‹ and Š"ß È2‹ . There are points of inflection at x œ „ 1.

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233

234

Chapter 4 Applications of Derivatives x#
3 x
# , then (x
3)(x
1) and (x
2)#

35. When y œ œ

yww œ

yw œ

2x(x
2)
ax#
3b (") (x
2)#

(2x
4)(x
2)#
ax#
4x  3b# (x
2) (x
2)%

œ

2 (x
2)$

.

The curve is rising on (
_ß ") and ($ß _), and falling on ("ß #) and (#ß $). There is a local maximum at x œ 1 and a local minimum at x œ 3. The curve is concave down on (
_ß #) and concave up on (#ß _). There are no points of inflection because x œ 2 is not in the domain. 36. When y œ œ

3x# ax#  1b a3x#  1b#

then yw œ

3x# a3x#  1b
x$ (6x) a3x#  1b#

and

# a12x  6xb a3x#  "b
2 a3x#  "b(6x)ˆ3x%  3x# ‰ a3x#  1b% 6x(1
x)("  x) . The curve is rising on (
_ß _) a3x#  1b$

yww œ œ

x$ 3x#  1 ,

$

so

there are no local extrema. The curve is concave up on (
_ß
") and (!ß "), and concave down on (
"ß !) and ("ß _). There are points of inflection at x œ
", x œ 0, and x œ 1. x#
1, kxk   1 , then 1
x# , kxk  1 2x, kxk  1 2, kxk  " yw œ œ and yww œ œ . The
2x, kxk  "
#, kxk  "

37. When y œ kx#
1k œ œ

curve rises on (
"ß !) and ("ß _) and falls on (
_ß
1) and (0ß 1). There is a local maximum at x œ 0 and local minima at x œ „ 1. The curve is concave up on (
_ß
1) and ("ß _), and concave down on (
"ß "). There are no points of inflection because y is not differentiable at x œ „ 1 (so there is no tangent line at those points). Ú x#
2x, x  0 38. When y œ kx
2xk œ Û 2x
x# , 0 Ÿ x Ÿ 2 , then Ü x#
2x, x  2 Ú 2x
2, x  0 Ú 2, x  0 w ww y œ Û 2
2x, 0  x  2 , and y œ Û
2, 0  x  2 . Ü 2x
2, x  2 Ü 2, x  2 #

The curve is rising on (!ß 1) and (#ß _), and falling on (
_ß !) and ("ß #). There is a local maximum at x œ 1 and local minima at x œ 0 and x œ 2. The curve is concave up on (
_ß !) and (#ß _), and concave down on (!ß #). There are no points of inflection because y is not differentiable at x œ 0 and x œ 2 (so there is no tangent at those points).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 39. When y œ Èkxk œ


Èx , x   0 , then È
x , x  0

Ú

"
x
$Î# , x0 , x0 #È x y œ Û
" and yww œ

(
x)4
$Î# . , x0 Ü 2 È
x , x  0 4 w

Since lim c yw œ
_ and lim b yw œ _ there is a xÄ! xÄ!

cusp at x œ 0. There is a local minimum at x œ 0, but no local maximum. The curve is concave down on (
_ß !) and (!ß _). There are no points of inflection. 40. When y œ Èkx
4k œ
Ú

Èx
4 , x   4 , then È4
x , x  4

"
(x
4)
$Î# , x4 Èx
4 , x  4 y œ Û 2
" and yww œ

(4
4x)
$Î# . , x4 Ü #È 4
x , x  4 4 w

Since lim c yw œ
_ and lim b yw œ _ there is a cusp xÄ4 xÄ4

at x œ 4. There is a local minimum at x œ 4, but no local maximum. The curve is concave down on (
_ß %) and (%ß _). There are no points of inflection. 41. yw œ 2  x
x# œ (1  x)(#
x), yw œ


±  ±



" # Ê rising on (
"ß #), falling on (
_ß
1) and (#ß _) Ê there is a local maximum at x œ 2 and a local minimum at x œ
1; yww œ 1
2x, yww œ  ±


"Î# " Ê concave up on ˆ
_ß # ‰ , concave down on ˆ "# ß _‰ Ê a point of inflection at x œ

" #

42. yw œ x#
x
6 œ (x
3)(x  2), yw œ  ±


± 
# $ Ê rising on (
_ß
#) and (3ß _), falling on (
2ß 3) Ê there is a local maximum at x œ
2 and a local minimum at x œ 3; yww œ 2x
1, yww œ


±  "Î# " ˆ ‰ Ê concave up on # ß _ , concave down on ˆ
_ß "# ‰ Ê a point of inflection at x œ

" #

43. yw œ x(x
3)# , yw œ


±  ±  Ê rising on ! $ (!ß _), falling on (
_ß !) Ê no local maximum, but there is a local minimum at x œ 0; yww œ (x
3)#  x(2)(x
3) œ 3(x
3)(x
1), yww œ  ±


±  Ê concave " $ up on (
_ß ") and ($ß _), concave down on ("ß $) Ê points of inflection at x œ 1 and x œ 3

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Chapter 4 Applications of Derivatives

44. yw œ x# (2
x), yw œ  ±  ±


Ê rising on ! # (
_ß #), falling on (2ß _) Ê there is a local maximum at x œ 2, but no local minimum; yww œ 2x(2
x)  x# (
1) œ x(4
3x), yww œ


±  ±


Ê concave up ! %Î$ 4‰ ˆ on !ß 3 , concave down on (
_ß !) and ˆ 43 ß _‰ Ê points of inflection at x œ 0 and x œ

4 3

45. yw œ x ax#
12b œ x Šx
2È3‹ Šx  2È3‹ , yw œ


±  ±


±  Ê rising on !
#È$ #È $ Š
2È3ß !‹ and Š#È3ß _‹ , falling on Š
_ß
#È3‹ and Š!ß #È3‹ Ê a local maximum at x œ 0, local minima at x œ „ #È3 ; yww œ (1) ax#
12b  (x)(2x) œ 3(x
2)(x  2), yww œ  ±


± 
# # Ê concave up on (
_ß
#) and (#ß _), concave down on (
#ß #) Ê points of inflection at x œ „ 2 46. yw œ (x
1)# (2x  3), yw œ


±  ±  "
$Î# 3 3‰ ˆ ‰ ˆ Ê rising on
# ß _ , falling on
_ß
# Ê no local maximum, a local minimum at x œ
3# ;

yww œ 2(x
1)(2x  3)  (x
1)# (2) œ 2(x
1)(3x  2), yww œ  ±


±  Ê concave up on "
#Î$ 2 ˆ
_ß
3 ‰ and ("ß _), concave down on ˆ
23 ß "‰ Ê points of inflection at x œ
23 and x œ 1 47. yw œ a8x
5x# b (4
x)# œ x(8
5x)(%
x)# , yw œ


±  ±


±


Ê rising on ˆ!ß 85 ‰ , ! % )Î& 8 falling on (
_ß !) and ˆ 5 ß _‰ Ê a local maximum at xœ ww

8 5

, a local minimum at x œ 0;

y œ (8
10x)(4
x)#  a8x
5x# b (2)(%
x)(
1) œ 4(4
x) a5x#
16x  8b , yww œ  ±


±  ±


Ê concave up % )
#È' )#È' & È6

on Š
_ß 8
52 Š 8
52 xœ

&

‹ and Š 8  52

È 6 8  2È 6 ß 5 ‹

8 „ 2È 6 5

È6

ß %‹ , concave down on

and (4ß _) Ê points of inflection at

and x œ 4

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Section 4.4 Concavity and Curve Sketching 48. yw œ ax#
2xb (x
5)# œ x(x
2)(x
5)# , yw œ  ±


±  ±  Ê rising on (
_ß !) ! # & and (#ß _), falling on (!ß #) Ê a local maximum at x œ 0, a local minimum at x œ 2; yww œ (2x
2)(x
5)#  ax#
2xb (2)(x
5) œ 2(x
5) a2x#
8x  5b , yww œ


±  ±


±  Ê concave up & %
È' %È' on

# 4
È6 4  È6 Š # ß 2 ‹ È6

Š
_ß %
2 xœ

4 „È 6 2

#

and (5ß _), concave down on È6

‹ and Š 4 #

ß &‹ Ê points of inflection at

and x œ 5

49. yw œ sec# x, yw œ (  ) Ê rising on ˆ
1# ß 1# ‰ ,
1Î# 1Î# never falling Ê no local extrema; yww œ 2(sec x)(sec x)(tan x) œ 2 asec# xb (tan x), yww œ (


±  ) Ê concave up on ˆ!, 1# ‰, !
1Î# 1Î# 1 ˆ concave down on
# ß !‰, ! is a opoint of inflection. 50. yw œ tan x, yw œ (


±  ) Ê rising on ˆ0ß 1# ‰ , !
1Î# 1Î# 1 ˆ ‰ falling on
# ß ! Ê no local maximum, a local minimum at x œ 0; yww œ sec# x, yww œ (  ) Ê concave up 1Î#
1Î# 1 1‰ ˆ on
# ß # Ê no points of inflection

51. yw œ cot

) #

, yw œ (  ±


) Ê rising on (!ß 1), 1 ! #1 falling on (1ß #1) Ê a local maximum at ) œ 1, no local minimum; yww œ
"# csc# #) , yww œ (


) Ê never ! #1 concave up, concave down on (!ß #1) Ê no points of inflection

52. yw œ csc#

) #

, yw œ (  ) Ê rising on (!ß 21), never ! #1 falling Ê no local extrema; yww œ 2 ˆcsc #) ‰ ˆ
csc #) ‰ ˆcot #) ‰ ˆ "# ‰ œ
ˆcsc# #) ‰ ˆcot #) ‰, yww œ (


±  ) 1 ! #1 Ê concave up on (1ß #1), concave down on (!ß 1) Ê a point of inflection at ) œ 1

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237

238

Chapter 4 Applications of Derivatives

53. yw œ tan# )
1 œ (tan )
1)(tan )  1), yw œ (  |


±  ) Ê rising on
1Î#
1Î% 1Î# 1Î% ˆ
1# ß
14 ‰ and ˆ 14 ß 1# ‰ , falling on ˆ
14 ß 14 ‰

Ê a local maximum at ) œ
14 , a local minimum at ) œ 14 ;

yww œ 2 tan ) sec# ), yww œ (


±  ) !
1Î# 1Î# Ê concave up on ˆ!ß 1# ‰ , concave down on ˆ
1# ß !‰ Ê a point of inflection at ) œ 0

54. yw œ 1
cot# ) œ ("
cot ))(1  cot )), yw œ (


|  ±


) Ê rising on ˆ 14 ß 341 ‰ , 1 ! 1Î% $1Î% 1 3 1 falling on ˆ0ß 4 ‰ and ˆ 4 ß 1‰ Ê a local maximum at )œ ww

31 4 ,

a local minimum at ) œ #

ww

1 4

;

y œ
2(cot )) a
csc )b, y œ (  ±


) 1 ! 1Î# 1 1 Ê concave up on ˆ!ß # ‰ , concave down on ˆ # ß 1‰ Ê a point of inflection at ) œ

1 #

55. yw œ cos t, yw œ [  ±


±  ] Ê rising on ! #1 1Î# $1Î# 1 3 1 1 3 1 ˆ!ß # ‰ and ˆ # ß 21‰ , falling on ˆ # ß # ‰ Ê local maxima at tœ ww

1 #

and t œ 21, local minima at t œ 0 and t œ ww

y œ
sin t, y œ [


±  ] 1 ! #1 Ê concave up on (1ß #1), concave down on (!ß 1) Ê a point of inflection at t œ 1

31 #

;

56. yw œ sin t, yw œ [  ±


] Ê rising on (!ß 1), 1 ! #1 falling on (1ß 21) Ê a local maximum at t œ 1, local minima at t œ 0 and t œ 21; yww œ cos t, yww œ [  ±


± ] Ê concave up on ˆ!ß 1# ‰ ! #1 1Î# $1Î# 3 1 and ˆ # ß #1‰ , concave down on ˆ 1# ß 3#1 ‰ Ê points of inflection at t œ

1 #

and t œ

31 #

57. yw œ (x  1)
#Î$ , yw œ  ) (  Ê rising on
" (
_ß _), never falling Ê no local extrema; yww œ
23 (x  1)
&Î$ , yww œ  ) (



" Ê concave up on (
_ß
1), concave down on (
"ß _) Ê a point of inflection and vertical tangent at x œ
1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 58. yw œ (x
2)
"Î$ , yw œ


)(  Ê rising on (2ß _), # falling on (
_ß #) Ê no local maximum, but a local minimum at x œ 2; yww œ
13 (x
2)
%Î$ , yww œ


)(


Ê concave down on (
_ß 2) and # (#ß _) Ê no points of inflection, but there is a cusp at xœ2

59. yw œ x
#Î$ (x
1), yw œ


)(


±  Ê rising on ! " ("ß _), falling on (
_ß ") Ê no local maximum, but a local minimum at x œ 1; yww œ "3 x
#Î$  23 x
&Î$ " 3

x
&Î$ (x  2), yww œ  ±


)(  !
# Ê concave up on (
_ß
2) and (!ß _), concave down on (
#ß !) Ê points of inflection at x œ
2 and x œ 0, and a vertical tangent at x œ 0 œ

60. yw œ x
%Î& (x  1), yw œ


±  )(  Ê rising on !
" (
"ß 0) and (!ß _), falling on (
_ß
") Ê no local maximum, but a local minimum at x œ
1; yww œ "5 x
%Î&
45 x
*Î& œ "5 x
*Î& (x
4), yww œ  )(


±  Ê concave up on (
_ß 0) and ! % (4ß _), concave down on (0ß 4) Ê points of inflection at x œ 0 and x œ 4, and a vertical tangent at x œ 0

61. yw œ œ


#x, x Ÿ 0 w , y œ  ±  Ê rising on 2x, x  0 !

(
_ß _) Ê no local extrema; yww œ œ


2, x  0 , 2, x  0

yww œ


)(  Ê concave up on (!ß _), concave ! down on (
_ß !) Ê a point of inflection at x œ 0 62. yw œ œ


x# , x Ÿ 0 w , y œ


±  Ê rising on x# , x  0 !

(!ß _), falling on (
_ß !) Ê no local maximum, but a
2x, x Ÿ 0 local minimum at x œ 0; yww œ œ , 2x, x  0 yww œ  ±  Ê concave up on (
_ß _) ! Ê no point of inflection

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239

240

Chapter 4 Applications of Derivatives

63. The graph of y œ f ww (x) Ê the graph of y œ f(x) is concave up on (!ß _), concave down on (
_ß !) Ê a point of inflection at x œ 0; the graph of y œ f w (x) Ê yw œ  ±


±  Ê the graph y œ f(x) has both a local maximum and a local minimum

64. The graph of y œ f ww (x) Ê yww œ  ±


Ê the graph of y œ f(x) has a point of inflection, the graph of y œ f w (x) Ê yw œ


±  ±


Ê the graph of y œ f(x) has both a local maximum and a local minimum

65. The graph of y œ f ww (x) Ê yww œ


±  ±


Ê the graph of y œ f(x) has two points of inflection, the graph of y œ f w (x) Ê yw œ


±  Ê the graph of y œ f(x) has a local minimum

66. The graph of y œ f ww (x) Ê yww œ  ±


Ê the graph of y œ f(x) has a point of inflection; the graph of y œ f w (x) Ê yw œ


±  ±


Ê the graph of y œ f(x) has both a local maximum and a local minimum

67.

Point P Q R S T

yw
  !


yww  !




Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.4 Concavity and Curve Sketching 68.

241

69.

70.

71. Graphs printed in color can shift during a press run, so your values may differ somewhat from those given here. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 0  t  2 and 6  t  9.5; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 2  t  6 and 9.5  t  15 (b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero when t is approximately 2, 6, or 9.5 sec. (c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The acceleration is zero when t is approximately 4, 7.5, or 12.5 sec. (d) The acceleration is positive when the concavity is up, 4  t  7.5 and 12.5  t  15; the acceleration is negative when the concavity is down, 0  t  4 and 7.5  t  12.5 72. (a) The body is moving away from the origin when kdisplacementk is increasing as t increases, 1.5  t  4, 10  t  12 and 13.5  t  16; the body is moving toward the origin when kdisplacementk is decreasing as t increases, 0  t  1.5, 4  t  10 and 12  t  13.5 (b) The velocity will be zero when the slope of the tangent line for y œ s(t) is horizontal. The velocity is zero when t is approximately 0, 4, 12 or 16 sec. (c) The acceleration will be zero at those values of t where the curve y œ s(t) has points of inflection. The acceleration is zero when t is approximately 1.5, 6, 8, 10.5, or 13.5 sec. (d) The acceleration is positive when the concavity is up, 0  t  1.5, 6  t  8 and 10  t  13.5, the acceleration is negative when the concavity is down, 1.5  t  6, 8  t  10 and 13.5  t  16. 73. The marginal cost is

dc dx

which changes from decreasing to increasing when its derivative

d# c dx#

is zero. This is a

point of inflection of the cost curve and occurs when the production level x is approximately 60 thousand units. 74. The marginal revenue is

dy dx

d# y dx# is positive Ê the curve is concave up # when ddxy#  0 Ê the curve is concave down

and it is increasing when its derivative

Ê !  t  2 and 5  t  9; marginal revenue is decreasing Ê 2  t  5 and 9  t  12.

75. When yw œ (x
1)# (x
2), then yww œ 2(x
1)(x
2)  (x
1)# . The curve falls on (
_ß 2) and rises on (#ß _). At x œ 2 there is a local minimum. There is no local maximum. The curve is concave upward on (
_ß ") and Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

242

Chapter 4 Applications of Derivatives

ˆ 53 ß _‰ , and concave downward on ˆ"ß 53 ‰ . At x œ 1 or x œ

5 3

there are inflection points.

76. When yw œ (x
1)# (x
2)(x
4), then yww œ 2(x
1)(x
2)(x
4)  (x
1)# (x
4)  (x
1)# (x
2) œ (x
1) c2 ax#
6x  8b  ax#
5x  4b  ax#
3x  2bd œ 2(x
1) a2x#
10x  11b. The curve rises on (
_ß 2) and (%ß _) and falls on (#ß %). At x œ 2 there is a local maximum and at x œ 4 a local minimum. The È3 5  È3 ß # ‹

curve is concave downward on (
_ß ") and Š 5
2 È Š 5 # 3 ß _‹ .

At x œ 1,

5
È3 #

and

5  È3 #

È3

and concave upward on Š1ß 5
#

‹ and

there are inflection points.

77. The graph must be concave down for x  0 because f ww (x) œ
x"#  0.

78. The second derivative, being continuous and never zero, cannot change sign. Therefore the graph will always be concave up or concave down so it will have no inflection points and no cusps or corners. 79. The curve will have a point of inflection at x œ 1 if 1 is a solution of yww œ 0; y œ x$  bx#  cx  d Ê yw œ 3x#  2bx  c Ê yw w œ 6x  2b and 6(1)  2b œ 0 Ê b œ
3. 80. (a) True. If f(x) is a polynomial of even degree then f w is of odd degree. Every polynomial of odd degree has at least one real root Ê f w (x) œ 0 for some x œ r Ê f has a horizontal tangent at x œ r. (b) False. For example, f(x) œ x
1 is a polynomial of odd degree but f w (x) œ 1 is never 0. As another example, y œ "3 x$  x#  x is a polynomial of odd degree, but yw œ x#  2x  1 œ (x  1)#  0 for all x. 81. (a) f(x) œ ax#  bx  c œ a ˆx#  ba x‰  c œ a Šx#  ba B 

b# 4a# ‹




b# 4a

 c œ a ˆx 

b b whose vertex is at x œ
2a Ê the coordinates of the vertex are Š
2a ß
b

#

b ‰# #a




b#
4ac 4a

a parabola


4ac 4a ‹

(b) The second derivative, f ww (x) œ 2a, describes concavity Ê when a  0 the parabola is concave up and when a  0 the parabola is concave down. 82. No, f ww (x) could be decreasing to zero at x œ c and then increase again so it would be concave up on every interval even though f ww (x) œ 0. For example f(x) œ x% is always concave up even though f ww (0) œ 0. 83. A quadratic curve never has an inflection point. If y œ ax#  bx  c where a Á 0, then yw œ 2ax  b and yww œ 2a. Since 2a is a constant, it is not possible for yww to change signs. 84. A cubic curve always has exactly one inflection point. If y œ ax$  bx#  cx  d where a Á 0, then yw œ 3ax#  2bx  c and yww œ 6ax  2b. Since
3ab is a solution of yww œ 0, we have that yww changes its sign b b at x œ
3a and yw exists everywhere (so there is a tangent at x œ
3a ). Thus the curve has an inflection b point at x œ
3a . There are no other inflection points because yww changes sign only at this zero.

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Section 4.4 Concavity and Curve Sketching 85. If y œ x&
5x%
240, then yw œ 5x$ (x
4) and yww œ 20x# (x
3). The zeros of yw are extrema, and there is a point of inflection at x œ $Þ

86. If y œ x$
12x# , then yw œ 3x(x
8) and yww œ 6(x
4). The zeros of yw and yww are extrema and points of inflection, respectively.

87. If y œ ww

4 5

x&  16x#
25, then yw œ 4x ax$  8b and

y œ 16 ax$  2b . The zeros of yw and yww are extrema and points of inflection, respectively.

88. If y œ w

x% 4 $




x$ 3 #


4x#  12x  20, then

y œ x
x
)x  "# œ (x  3)(x
2)# Þ So y has a local minimum at x œ
$ as its only extreme value. Also yww œ $x#
#x
) œ (3x  4)(x
2) and there are inflection points at both zeros,
%$ and 2, of yww .

89. The graph of f falls where f w  0, rises where f w  0, and has horizontal tangents where f w œ 0. It has local minima at points where f w changes from negative to positive and local maxima where f w changes from positive to negative. The graph of f is concave down where f ww  0 and concave up where f ww  0. It has an inflection point each time f ww changes sign, provided a tangent line exists there.

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243

244

Chapter 4 Applications of Derivatives

90. The graph f is concave down where f ww  0, and concave up where f ww  0. It has an inflection point each time f ww changes sign, provided a tangent line exists there.

91. (a) It appears to control the number and magnitude of the local extrema. If k  0, there is a local maximum to the left of the origin and a local minimum to the right. The larger the magnitude of k (k  0), the greater the magnitude of the extrema. If k  0, the graph has only positive slopes and lies entirely in the first and third quadrants with no local extrema. The graph becomes increasingly steep and straight as k Ä _. (b) f w (x) œ 3x#  k Ê the discriminant 0#
4(3)(k) œ
12k is positive for k  0, zero for k œ 0, and negative for k  0; f w has two zeros x œ „ É
k3 when k  0, one zero x œ 0 when k œ 0 and no real zeros when k  0; the sign of k controls the number of local extrema. (c) As k Ä _, f w (x) Ä _ and the graph becomes increasingly steep and straight. As k Ä
_, the crest of the graph (local maximum) in the second quadrant becomes increasingly high and the trough (local minimum) in the fourth quadrant becomes increasingly deep.

92. (a) It appears to control the concavity and the number of local extrema.

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Section 4.5 Applied Optimization Problems

245

(b) f(x) œ x%  kx$  6x# Ê f w (x) œ 4x$  3kx#  12x Ê f ww (x) œ 12x#  6kx  12 Ê the discriminant is 36k#
4(12)(12) œ 36(k  4)(k
4), so the sign line of the discriminant is  ±


±  Ê the
% % discriminant is positive when kkk  4, zero when k œ „ 4, and negative when kkk  4; f ww (x) œ 0 has two zeros when kkk  4, one zero when k œ „ 4, and no real zeros for kkk  4; the value of k controls the number of possible points of inflection. 93. (a) If y œ x#Î$ ax#
2b , then yw œ yww œ

4 9

4 3

x
"Î$ a2x#
1b and

x
%Î$ a10x#  1b . The curve rises on

Š
È"2 ß 0‹ and Š È"2 ß _‹ and falls on Š
_ß
È"2 ‹ and Š!ß È"2 ‹ . The curve is concave up on (
_ß !) and (!ß _). (b) A cusp since lim c yw œ _ and lim b yw œ
_. xÄ! xÄ! 94. (a) If y œ 9x#Î$ (x
1), then yw œ

15 ˆx
25 ‰ x"Î$

and

10 ˆx  "5 ‰ x%Î$

yww œ . The curve rises on (
_ß !) and ˆ 25 ß _‰ and falls on ˆ!ß 25 ‰ . The curve is concave

down on ˆ
_ß
"5 ‰ and concave up on ˆ
5" ß !‰ and (!ß _). (b) A cusp since lim c yw œ _ and lim b yw œ
_. xÄ! xÄ! 95. Yes: y œ x#  3 sin 2x Ê yw œ 2x  6 cos 2x. The graph of yw is zero near
3 and this indicates a horizontal tangent near x œ
3.

4.5 APPLIED OPTIMIZATION PROBLEMS 1. Let j and w represent the length and width of the rectangle, respectively. With an area of 16 in.# , we have that (j)(w) œ 16 Ê w œ 16j
" Ê the perimeter is P œ 2j  2w œ 2j  32j
" and Pw (j) œ 2
w

Solving P (j) œ 0 Ê

2(j  4)(j
4) j#

32 j#

œ

2 aj#
16b j#

.

œ 0 Ê j œ
4, 4. Since j  0 for the length of a rectangle, j must be 4 and

w œ 4 Ê the perimeter is 16 in., a minimum since Pww (j) œ

16 j$

 0.

2. Let x represent the length of the rectangle in meters (!  x  %) Then the width is %
x and the area is Aaxb œ xa%
xb œ %x
x# . Since Aw axb œ %
#x, the critical point occurs at x œ #. Since, Aw axb  ! for !  x  # and Aw axb  ! for #  x  %, this critical point corresponds to the maximum area. The rectangle with the largest area measures # m by %
# œ # m, so it is a square.

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246

Chapter 4 Applications of Derivatives

Graphical Support:

3. (a) The line containing point P also contains the points (!ß ") and ("ß !) Ê the line containing P is y œ 1
x Ê a general point on that line is (xß 1
x). (b) The area A(x) œ 2x(1
x), where 0 Ÿ x Ÿ 1. (c) When A(x) œ 2x
2x# , then Aw (x) œ 0 Ê 2
4x œ 0 Ê x œ "# . Since A(0) œ 0 and A(1) œ 0, we conclude that A ˆ "# ‰ œ "# sq units is the largest area. The dimensions are " unit by "# unit. 4. The area of the rectangle is A œ 2xy œ 2x a12
x# b , where 0 Ÿ x Ÿ È12 . Solving Aw (x) œ 0 Ê 24
6x# œ 0 Ê x œ
2 or 2. Now
2 is not in the domain, and since A(0) œ 0 and A ŠÈ12‹ œ 0, we conclude that A(2) œ 32 square units is the maximum area. The dimensions are 4 units by 8 units. 5. The volume of the box is V(x) œ x(15
2x)(8
2x) œ 120x
46x#  4x$ , where 0 Ÿ x Ÿ 4. Solving Vw (x) œ 0 Ê 120
92x  12x# œ 4(6
x)(5
3x) œ 0 Ê x œ 53 or 6, but 6 is not in the domain. Since V(0) œ V(4) œ 0, $ V ˆ 53 ‰ œ #%&! #( ¸ *" in must be the maximum volume of the box with dimensions

14 3

‚

35 3

‚

5 3

inches.

6. The area of the triangle is A œ "# ba œ b# È400
b# , where b# 0 Ÿ b Ÿ 20. Then dA œ " È400
b#
db

œ

200
b# È400
b#

#

2È400
b#

œ 0 Ê the interior critical point is b œ 10È2.

When b œ 0 or 20, the area is zero Ê A Š10È2‹ is the maximum area. When a#  b# œ 400 and b œ 10È2, the value of a is also 10È2 Ê the maximum area occurs when a œ b. 7. The area is A(x) œ x(800
2x), where 0 Ÿ x Ÿ 400. Solving Aw (x) œ 800
4x œ 0 Ê x œ 200. With A(0) œ A(400) œ 0, the maximum area is A(200) œ 80,000 m# . The dimensions are 200 m by 400 m.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.5 Applied Optimization Problems 8. The area is 2xy œ 216 Ê y œ

108 x

247

. The amount of fence

needed is P œ 4x  3y œ 4x  324x
" , where !  x; dP 324 # dx œ 4
x# œ 0 Ê x
81 œ 0 Ê the critical points are 0 and „ 9, but 0 and
9 are not in the domain. Then Pww (9)  0 Ê at x œ 9 there is a minimum Ê the dimensions of the outer rectangle are 18 m by 12 m Ê 72 meters of fence will be needed. 9. (a) We minimize the weight œ tS where S is the surface area, and t is the thickness of the steel walls of the tank. The surace area is S œ x#  %xy where x is the length of a side of the square base of the tank, and y is its depth. The ˆ # #!!! ‰. Treating the volume of the tank must be &!!ft$ Ê y œ &!! x# . Therefore, the weight of the tank is waxb œ t x  x #!!! thickness as a constant gives ww axb œ tˆ#x
x# ‰ for xÞ!. The critical value is at x œ "!. Since www a"!b œ tˆ# 

%!!! ‰ "!$

 !, there is a minimum at x œ "!. Therefore, the optimum dimensions of the tank are "! ft on

the base edges and & ft deep. (b) Minimizing the surface area of the tank minimizes its weight for a given wall thickness. The thickness of the steel walls would likely be determined by other considerations such as structural requirements. 10. (a) The volume of the tank being ""#& ft$ , we have that yx# œ ""#& Ê y œ ""#& x# . The cost of building the tank is ""#& $$(&! # w caxb œ &x  $!xˆ x# ‰, where !  x. Then c axb œ "!x
x# œ ! Ê the critical points are ! and "&, but ! is not

in the domain. Thus, cww a"&b  ! Ê at x œ "& we have a minimum. The values of x œ "& ft and y œ & ft will minimize the cost. (b) The cost function c œ &ax#  %xyb  "!xy, can be separated into two items: (1) the cost of the materials and labor to fabricate the tank, and (2) the cost for the excavation. Since the area of the sides and bottom of the tanks is ax#  %xyb, it can be deduced that the unit cost to fabricate the tanks is $&/ft# . Normally, excavation costs are per unit volume of ‰ # excavated material. Consequently, the total excavation cost can be taken as "!xy œ ˆ "! x ax yb. This suggests that the unit cost of excavation is

$"!Îft# x

where x is the length of a side of the square base of the tank in feet. For the least

expensive tank, the unit cost for the excavation is

$"!Îft# "& ft

œ

$!Þ'( ft$

œ

$") yd$ .

The total cost of the least expensive tank is

$$$(&, which is the sum of $#'#& for fabrication and $(&! for the excavation. 11. The area of the printing is (y
4)(x
8) œ 50. ‰ Consequently, y œ ˆ x 50
8  4. The area of the paper is 50 A(x) œ x ˆ x
8  4‰ , where 8  x. Then 50 ‰ Aw (x) œ ˆ x 50
8  4
x Š (x
8)# ‹ œ

4(x
8)#
400 (x
8)#

œ0

Ê the critical points are
2 and 18, but
2 is not in the domain. Thus Aww (18)  0 Ê at x œ 18 we have a minimum. Therefore the dimensions 18 by 9 inches minimize the amount of paper. 12. The volume of the cone is V œ Thus, V(y) œ

1 3

" 3

1r# h, where r œ x œ È9
y# and h œ y  3 (from the figure in the text).

a9
y# b (y  3) œ

1 3

a27  9y
3y#
y$ b Ê Vw (y) œ ww

1 3

a9
6y
3y# b œ 1(1
y)(3  y).

The critical points are
3 and 1, but
3 is not in the domain. Thus V (1) œ we have a maximum volume of V(1) œ

1 3

(8)(4) œ

321 3

1 3

(
'
6(1))  0 Ê at y œ 1

cubic units.

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248

Chapter 4 Applications of Derivatives

13. The area of the triangle is A()) œ ab cos ) #

ab sin ) #

, where 0  )  1.

Solving A ()) œ 0 Ê œ 0 Ê ) œ 1# . Since Aww ()) ) œ
ab sin Ê Aww ˆ 1# ‰  0, there is a maximum at ) œ 1# . # w

14. A volume V œ 1r# h œ 1000 Ê h œ

1000 1 r#

. The amount of

material is the surface area given by the sides and bottom of # the can Ê S œ 21rh  1r# œ 2000 r  1r , 0  r. Then dS dr

œ
2000 r#  21r œ ! Ê

are 0 and #

d S dr#

rœ

œ

4000 r$

10 $ È 1

10 $ È 1

1r$
1000 r#

œ 0. The critical points

, but 0 is not in the domain. Since

 #1  0, we have a minimum surface area when

cm and h œ

1000 1 r#

œ

10 $ È 1

cm. Comparing this result to

the result found in Example 2, if we include both ends of the can, then we have a minimum surface area when the can is shorter-specifically, when the height of the can is the same as its diameter. 15. With a volume of 1000 cm and V œ 1r# h, then h œ A œ 8r#  21rh œ 8r# 

2000 r

. Then Aw (r) œ

but r œ 0 results in no can. Since Aww (r) œ 16 16. (a) The base measures "!
#x in. by

"&
#x #

1000 1r# . The amount of aluminum used per can is 8r$
1000 16r
2000 œ 0 Ê the critical points are 0 and 5, r# œ 0 Ê r# 1000  r$  0 we have a minimum at r œ 5 Ê h œ 40 1 and h:r œ 8:1.

in., so the volume formula is Vaxb œ

xa"!
#xba"&
#xb #

œ #x$
#&x#  (&x.

(b) We require x  !, #x  "!, and #x  "&. Combining these requirements, the domain is the interval a!ß &b.

(c) The maximum volume is approximately 66.02 in.$ when x ¸ "Þ*' in.

(d) Vw axb œ 'x#
&!x  (&. The critical point occurs when Vw axb œ !, at x œ œ ww

#& „ &È( , that '

&! „ Éa
&!b#
%a'ba(&b # a 'b

œ

&! „ È(!! "#

is, x ¸ "Þ*' or x ¸ 'Þ$(. We discard the larger value because it is not in the domain. Since

V axb œ "#x
&!, which is negative when x ¸ "Þ*' , the critical point corresponds to the maximum volume. The maximum volume occurs when x œ

#&
&È( '

¸ "Þ*', which comfimrs the result in (c).

17. (a) The" sides" of the suitcase will measure #%
#x in. by ")
#x in. and will be #x in. apart, so the volume formula is Vaxb œ #xa#%
#xba")
#xb œ )x$
"')x#  )'#x.

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Section 4.5 Applied Optimization Problems

249

(b) We require x
!, #x  "), and #x  #%. Combining these requirements, the domain is the interval a!ß *b.

(c) The maximum volume is approximately 1309.95 in.$ when x ¸ 3Þ3* in.

(d) Vw axb œ #%x#  $$'x  )'% œ #%ax#  "%x  $'b. The critical point is at x œ

"% „ Éa
"%b#
%a"ba$'b # a" b

œ

"% „ È&# #

œ ( „ È"$, that is, x ¸ $Þ$* or x ¸ "!Þ'". We discard the larger value because it is not in the domain. Since Vww axb œ #%a#x  "%b which is negative when x ¸ $Þ$*, the critical point corresponds to the maximum volume. The maximum value occurs at x œ (  È"$ ¸ $Þ$*, which confirms the results in (c). (e) )x$  "')x#  )'#x œ ""#! Ê 8ax$  #"x#  "!)x  "%!b œ ! Ê )ax  #bax  &bax  "%b œ !. Since "% is not in the fomain, the possible values of x are x œ # in. or x œ & in. (f) The dimensions of the resulting box are #x in., a#%  #xb in., and a")  #xb. Each of these measurements must be positive, so that gives the domain of a!ß *b. 18. If the upper right vertex of the rectangle is located at axß % cos !Þ& xb for !  x  1, then the rectangle has width #x and height % cos !Þ&x, so the area is Aaxb œ )x cos !Þ&x. Solving Aw axb œ ! graphically for !  x  1, we find that x ¸ #Þ#"%. Evaluating #x and % cos !Þ&x for x ¸ #Þ#"%, the dimensions of the rectangle are approximately %Þ%$ (width) by "Þ(* (height), and the maximum area is approximately (Þ*#$. 19. Let the radius of the cylinder be r cm, !  r  "!. Then the height is #È"!!  r# and the volume is Varb œ #1r# È"!!  r# cm$ . Then, Vw arb œ #1r# Š " ‹a#rb  Š#1È"!!  r# ‹a#rb È"!!
r#

œ

$

#


#1r  %1ra"!!
r b È"!!
r#

œ

#

#1ra#!!
$r b È"!!
r# .

É $# . Since Vw arb
! for The critical point for !  r  "! occurs at r œ É #!! $ œ "!

!  r  "!É #$ and Vw arb  ! for "!É #$  r  "!, the critical point corresponds to the maximum volume. The dimensions are r œ "!É #$ ¸ )Þ"' cm and h œ

#! È$

¸ ""Þ&& cm, and the volume is

%!!!1 $È$

¸ #%")Þ%! cm$ .

20. (a) From the diagram we have 4x  j œ 108 and V œ x# j. The volume of the box is V(x) œ x# (108  4x), where 0 Ÿ x  27. Then Vw (x) œ 216x  12x# œ 12x(18  x) œ 0 Ê the critical points are 0 and 18, but x œ 0 results in no box. Since Vww (x) œ 216  24x  0 at x œ 18 we have a maximum. The dimensions of the box are 18 ‚ 18 ‚ 36 in.

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250

Chapter 4 Applications of Derivatives #

(b) In terms of length, V(j) œ x# j œ ˆ 1084
j ‰ j. The graph indicates that the maximum volume occurs near j œ 36, which is consistent with the result of part (a).

21. (a) From the diagram we have 3h  2w œ 108 and V œ h# w Ê V(h) œ h# ˆ54  #3 h‰ œ 54h#  3# h$ . Then Vw (h) œ 108h  9# h# œ

9 #

h(24  h) œ 0

Ê h œ 0 or h œ 24, but h œ 0 results in no box. Since Vww (h) œ 108  9h  0 at h œ 24, we have a maximum volume at h œ 24 and w œ 54  3# h œ 18. (b)

22. From the diagram the perimeter is P œ 2r  2h  1r, where r is the radius of the semicircle and h is the height of the rectangle. The amount of light transmitted proportional to A œ 2rh  "4 1r# œ r(P  2r  1r)  4" 1r# œ rP  2r#  34 1r# . Then

dA dr

œ P  4r  3# 1r œ 0

(4  1)P 2P 4P 21 P 8  31 Ê 2h œ P  8  31  8  31 œ 8  31 . Therefore, 2rh œ 4 8 1 gives the proportions that admit the # most light since ddrA# œ 4  3# 1  0.

Ê rœ

23. The fixed volume is V œ 1r# h  23 1r$ Ê h œ

V 1 r#



2r 3

, where h is the height of the cylinder and r is the radius

of the hemisphere. To minimize the cost we must minimize surface area of the cylinder added to twice the 8 # surface area of the hemisphere. Thus, we minimize C œ 21rh  41r# œ 21r ˆ 1Vr#  2r3 ‰  41r# œ 2V r  3 1r . Then œ

dC dr

œ  2V r# 

4V"Î$ 1"Î$ †3#Î$



16 3

2†3"Î$ †V"Î$ 3†#†1"Î$

1r œ 0 Ê V œ œ

8 3

‰ 1r$ Ê r œ ˆ 3V 81

3"Î$ †2†4†V"Î$
2†3"Î$ †V"Î$ 3†#†1"Î$

‰ œ ˆ 3V 1

"Î$

"Î$

. From the volume equation, h œ

. Since

d# C dr#

œ

4V r$



16 3

V 1 r#



2r 3

1
0, these

dimensions do minimize the cost. 24. The volume of the trough is maximized when the area of the cross section is maximized. From the diagram the area of the cross section is A()) œ cos )  sin ) cos ), 0  )  1# . Then Aw ()) œ sin )  cos# )  sin# ) œ  a2 sin# )  sin )  1b œ (2 sin )  1)(sin )  1) so Aw ()) œ 0 Ê sin ) œ

sin ) Á 1 when 0  ) 

1 #

w

. Also, A ())
0 for 0  ) 

1 6

w

and A ())  0 for

1 6

" #

or sin ) œ 1 Ê ) œ

)

1 #

1 6

because

. Therefore, at ) œ

there is a maximum.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 6

Section 4.5 Applied Optimization Problems 25. (a) From the diagram we have: AP œ x, RA œ ÈL  x# , PB œ 8.5  x, CH œ DR œ 11  RA œ 11  ÈL  x# , QB œ Èx#  (8.5  x)# , HQ œ 11  CH  QB œ 11  ’11  ÈL  x#  Èx#  (8.5  x)# “ #

#

#

œ ÈL  x#  Èx#  (8.5  x)# , RQ œ RH  HQ #

œ (8.5)#  ŠÈL  x#  Èx#  (8.5  x)# ‹ . It #

#

#

#

follows that RP œ PQ  RQ Ê L# œ x#  ŠÈL#  x#  Èx#  (x  8.5)# ‹  (8.5)# Ê L# œ x#  L#  x#  2ÈL#  x# È17x  (8.5)#  17x  (8.5)#  (8.5)# Ê 17# x# œ 4 aL#  x# b a17x  (8.5)# b Ê L# œ x#  œ

17x$ ‰# 17x
ˆ 17 #

(b) If f(x) œ

4x$ 4x
17

œ

4x$ 4x
17

œ

(c) When x œ

51 8

. Thus L# is minimized when x œ

cylinder is formed, x œ 21r Ê r œ Then Vww (x)

17x$ 17x
(8.5)#

4x# (8x
51) (4x
17)#

Ê f w (x)  0 when x 

51 8

51 8 .

then L ¸ 11.0 in.

26. (a) From the figure in the text we have P œ 2x  2y Ê y œ Ê V(x) œ

œ

2x$ 2x
8.5 .

is minimized, then L# is minimized. Now f w (x) œ

and f w (x)
0 when x
51 8 ,

17# x# 4 c17x
(8.5)# d

#

$

18x
x 41 œ 13 ˆ3

x #1

P #

 x. If P œ 36, then y œ 18  x. When the

and h œ y Ê h œ 18  x. The volume of the cylinder is V œ 1r# h


x) . Solving Vw (x) œ 3x(12 œ 0 Ê x œ 0 or 12; but when x œ 0, there is no cylinder. 41 x‰ ww  # Ê V (12)  0 Ê there is a maximum at x œ 12. The values of x œ 12 cm and

y œ 6 cm give the largest volume. (b) In this case V(x) œ 1x# (18  x). Solving Vw (x) œ 31x(12  x) œ 0 Ê x œ 0 or 12; but x œ 0 would result in no cylinder. Then Vww (x) œ 61(6  x) Ê Vww (12)  0 Ê there is a maximum at x œ 12. The values of x œ 12 cm and y œ 6 cm give the largest volume. 27. Note that h#  r# œ $ and so r œ È$  h# . Then the volume is given by V œ 1$ r# h œ 1$ a$  h# bh œ 1h  1$ h$ for dV # # !  h  È$, and so dV dh œ 1  1r œ 1a"  r b. The critical point (for h
!) occurs at h œ ". Since dh
! for !  h  ", and dV  ! for "  h  È$, the critical point corresponds to the maximum volume. The cone of greatest dh

volume has radius È# m, height "m, and volume 28. (a) f(x) œ x#  (b) f(x) œ x# 

a x a x

#1 $

m$ .

Ê f w (x) œ x
# a2x$  ab , so that f w (x) œ 0 when x œ 2 implies a œ 16 Ê f ww (x) œ 2x
$ ax$  ab , so that f ww (x) œ 0 when x œ 1 implies a œ 1

29. If f(x) œ x#  xa , then f w (x) œ 2x  ax
# and f ww (x) œ 2  2ax
$ . The critical points are 0 and $È #a , but x Á 0. Now f ww ˆ $È #a ‰ œ 6
0 Ê at x œ $È #a there is a local minimum. However, no local maximum exists for any a. 30. If f(x) œ x$  ax#  bx, then f w (x) œ 3x#  2ax  b and f ww (x) œ 6x  2a. (a) A local maximum at x œ 1 and local minimum at x œ 3 Ê f w (1) œ 0 and f w (3) œ 0 Ê 3  2a  b œ 0 and 27  6a  b œ 0 Ê a œ 3 and b œ 9. (b) A local minimum at x œ 4 and a point of inflection at x œ 1 Ê f w (4) œ 0 and f ww (1) œ 0 Ê 48  8a  b œ 0

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251

252

Chapter 4 Applications of Derivatives and 6  2a œ 0 Ê a œ 3 and b œ 24.

31. (a) satb œ "'t#  *'t  ""# Ê vatb œ sw atb œ $#t  *'. At t œ !, the velocity is va!b œ *' ft/sec. (b) The maximum height ocurs when vatb œ !, when t œ $. The maximum height is sa$b œ #&' ft and it occurs at t œ $ sec. (c) Note that satb œ "'t#  *'t  ""# œ "'at  "bat  (b, so s œ ! at t œ " or t œ (. Choosing the positive value of t, the velocity when s œ ! is va(b œ "#) ft/sec. 32.

Let x be the distance from the point on the shoreline nearest Jane's boat to the point where she lands her boat. Then she needs to row È%  x# mi at 2 mph and walk '  x mi at 5 mph. The total amount of time to reach the village is faxb œ

È %  x# #



have: #È%x x# œ

'
x & " &

hours (! Ÿ x Ÿ '). Then f w axb œ

" " # #È%  x# a#xb



" &

œ

x #È%  x#

 &" . Solving f w axb œ !, we

Ê &x œ #È%  x# Ê #&x# œ %a%  x# b Ê #"x# œ "' Ê x œ „

% È#" .

We discard the negative

value of x because it is not in the domain. Checking the endpoints and critical point, we have fa!b œ #Þ#, fŠ È%#" ‹ ¸ #Þ"#, and fa'b ¸ $Þ"'. Jane should land her boat

% È#"

¸ !Þ)( miles donw the shoreline from the point

nearest her boat. 33.

) x

œ

h x  #(

Êhœ)

œ Ɉ) 

#"' ‰# x

#"' x

and Laxb œ Éh#  ax  #(b#

 ax  #(b# when x   !. Note that Laxb is

minimized when faxb œ ˆ) 

#"' ‰# x

 ax  #(b# is

minimized. If f w axb œ !, then ‰ˆ #"' ‰ #ˆ)  #"' x x#  #ax  #(b œ ! Ê ax  #(bˆ" 

"(#) ‰ x$

œ ! Ê x œ #( (not acceptable

since distance is never negative or x œ "#. Then La"#b œ È#"*( ¸ %'Þ)( ftÞ 34. (a) From the diagram we have d# œ 4r#  w# . The strength of the beam is S œ kwd# œ kw a4r#  w# b . When r œ 6, then S œ 144kw  kw$ . Also, Sw (w) œ 144k  3kw# œ 3k a48  w# b so Sw (w) œ 0 Ê w œ „ 4È3 ; Sww Š4È3‹  0 and 4È3 is not acceptable. Therefore S Š4È3‹ is the maximum strength. The dimensions of the strongest beam are 4È3 by 4È6 inches. (b)

(c)

Both graphs indicate the same maximum value and are consistent with each other. Changing k does not change the dimensions that give the strongest beam (i.e., do not change the values of w and d that produce the strongest beam).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.5 Applied Optimization Problems 35. (a) From the situation we have w# œ 144  d# . The stiffness of the beam is S œ kwd$ œ kd$ a144  d# b # # where 0 Ÿ d Ÿ 12. Also, Sw (d) œ 4kd a108
d b Ê critical points at 0, 12, and 6È3. Both d œ 0 and

"Î#

253

,

È144
d#

d œ 12 cause S œ 0. The maximum occurs at d œ 6È3. The dimensions are 6 by 6È3 inches. (b) (c)

Both graphs indicate the same maximum value and are consistent with each other. The changing of k has no effect. 36. (a) s" œ s# Ê sin t œ sin ˆt  13 ‰ Ê sin t œ sin t cos Ê tœ

1 3

or

1 3

 sin

1 3

cos t Ê sin t œ

41 3

(b) The distance between the particles is s(t) œ ks"  s# k œ ¸sin t  sin ˆt  13 ‰¸ œ Ê sw (t) œ are 0, 13 ,

Šsin t
È3 cos t‹ Šcos t  È3 sin t‹ 2 ¹sin t
È3 cos t¹

51 41 111 6 , 3 , 6 ,

21; then s(0) œ

since

È3 #

d dx

kxk œ

x kx k

" #

" #

sin t 

È3 #

cos t Ê tan t œ È3

¹sin t  È3 cos t¹

Ê critical times and endpoints

, s ˆ 13 ‰ œ 0, s ˆ 561 ‰ œ 1, s ˆ 431 ‰ œ 0, s ˆ 1161 ‰ œ 1, s(21) œ

È3 #

Ê the

greatest distance between the particles is 1. (c) Since sw (t) œ

Šsin t
È3 cos t‹ Šcos t  È3 sin t‹ 2 ¹sin t
È3 cos t¹

we can conclude that at t œ

1 3

and

41 w 3 , s (t)

has cusps and

the distance between the particles is changing the fastest near these points. 37. (a) s œ 10 cos (1t) Ê v œ 101 sin (1t) Ê speed œ k101 sin (1t)k œ 101 ksin (1t)k Ê the maximum speed is 101 ¸ 31.42 cm/sec since the maximum value of ksin (1t)k is 1; the cart is moving the fastest at t œ 0.5 sec, 1.5 sec, 2.5 sec and 3.5 sec when ksin (1t)k is 1. At these times the distance is s œ 10 cos ˆ 1# ‰ œ 0 cm and a œ 101# cos (1t) Ê kak œ 101# kcos (1t)k Ê kak œ 0 cm/sec# (b) kak œ 101# kcos (1t)k is greatest at t œ 0.0 sec, 1.0 sec, 2.0 sec, 3.0 sec and 4.0 sec, and at these times the magnitude of the cart's position is ksk œ 10 cm from the rest position and the speed is 0 cm/sec.

38. (a) 2 sin t œ sin 2t Ê 2 sin t  2 sin t cos t œ 0 Ê (2 sin t)(1  cos t) œ 0 Ê t œ k1 where k is a positive integer (b) The vertical distance between the masses is s(t) œ ks"  s# k œ ˆas"  s# b# ‰ Ê sw (t) œ ˆ "# ‰ a(sin 2t  2 sin t)# b 2(cos 2t
cos t)(sin 2t
2 sin t) ksin 2t
2 sin tk

œ 0,

21 3 ,

1,

41 3 ,

œ


"Î#

4(2 cos t  1)(cos t
")(sin t)(cos t
1) ksin 2t
2 sin tk

3È 3 2 ,

ds dt

œ

Ê

" #

a(12  12t)#  64t# b

ds ¸ dt t=0

Ê critical times at

œ 12 knots and


"Î#

, s(1) œ 0, s ˆ 431 ‰ 3È 3 #

at t œ

21 3

and

"Î#

[2(12  12t)(12)  128t] œ

ds ¸ dt t=1

3È 3 #

s(21) œ 0 Ê the greatest distance is

39. (a) s œ È(12  12t)#  (8t)# œ a(12  12t)#  64t# b (b)

œ a(sin 2t  2 sin t)# b

(2)(sin 2t  2 sin t)(2 cos 2t  2 cos t)

21; then s(0) œ 0, s ˆ 231 ‰ œ ¸sin ˆ 431 ‰  2 sin ˆ 231 ‰¸ œ

œ ¸sin ˆ 831 ‰  2 sin ˆ 431 ‰¸ œ

"Î#

208t
144 È(12
12t)#  64t#

œ 8 knots

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

41 3

"Î#

254

Chapter 4 Applications of Derivatives

(c) The graph indicates that the ships did not see each other because s(t)
5 for all values of t.

(d) The graph supports the conclusions in parts (b) and (c).

(e)

lim ds t Ä _ dt

œ

(208t
144)# É lim 144( "
t)#  64t# tÄ_

Š208


œ Ë lim

#

144 t ‹ #

t Ä _ 144 Š "
1‹ 64 t

#

œ É 144208 64 œ È208 œ 4È13

which equals the square root of the sums of the squares of the individual speeds. 40. The distance OT  TB is minimized when OB is a straight line. Hence n! œ n" Ê )" œ )# .

41. If v œ kax  kx# , then vw œ ka  2kx and vww œ 2k, so vw œ 0 Ê x œ vww ˆ #a ‰

œ 2k  0. The maximum value of v is

ka# 4

a #

. At x œ

a #

there is a maximum since

.

42. (a) According to the graph, yw a!b œ !. (b) According to the graph, yw aLb œ !. (c) ya!b œ !, so d œ !. Now yw axb œ $ax#  #bx  c, so yw a!b œ ! implies that c œ !. There fore, yaxb œ ax$  bx# and yw axb œ $ax#  #bx. Then yaLb œ aL$  bL# œ H and yw aLb œ $aL#  #bL œ !, so we have two linear equations in two unknowns a and b. The second equation gives b œ $aL # . Substituting into the first equation, we have aL$ 

$aL$ #

œ H, or

aL$ #

œ H, so a œ # LH$ . Therefore, b œ $ LH# and the equation for y is $

#

yaxb œ # LH$ x$  $ LH# x# , or yaxb œ H’#ˆ Lx ‰  $ˆ Lx ‰ “. 43. The profit is p œ nx  nc œ n(x  c) œ ca(x  c)
"  b(100  x)d (x  c) œ a  b(100  x)(x  c) œ a  (bc  100b)x  100bc  bx# . Then pw (x) œ bc  100b  2bx and pww (x) œ 2b. Solving pw (x) œ 0 Ê x œ #c  50. At x œ #c  50 there is a maximum profit since pww (x) œ 2b  0 for all x. 44. Let x represent the number of people over 50. The profit is p(x) œ (50  x)(200  2x)  32(50  x)  6000 œ 2x#  68x  2400. Then pw (x) œ 4x  68 and pww œ 4. Solving pw (x) œ 0 Ê x œ 17. At x œ 17 there is a maximum since pww (17)  0. It would take 67 people to maximize the profit.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.5 Applied Optimization Problems 45. (a) A(q) œ kmq
"  cm  h# q, where q
0 Ê Aw (q) œ kmq
# 

h #

œ

hq#
2km 2q#

255

and Aww (q) œ 2kmq
$ . The

ww É 2km É 2km É 2km critical points are É 2km h , 0, and h , but only h is in the domain. Then A Š h ‹
0 Ê at

q œ É 2km h there is a minimum average weekly cost. (b) A(q) œ

(kbq)m q

 cm  h# q œ kmq
"  bm  cm  h# q, where q
0 Ê Aw (q) œ 0 at q œ É 2km h as in (a).

Also Aww (q) œ 2kmq
$
0 so the most economical quantity to order is still q œ É 2km h which minimizes the average weekly cost. 46. We start with caxb œ the cost of producing x items, x
!, and

c ax b x

œ the average cost of producing x items, assumed

to be differentiable. If the average cost can be minimized, it will be at a production level at which Ê w

w

x c ax b
c a x b x#

d c ax b dx Š x ‹

œ ! (by the quotient rule) Ê x cw axb  caxb œ ! (multiply both sides by x# ) Ê cw axb œ

c ax b x

œ! where

c axb is the marginal cost. This concludes the proof. (Note: The theorem does not assure a production level that will give a minimum cost, but rather, it indicates where to look to see if there is one. Find the production levels where the average cost equals the marginal cost, then check to see if any of them give a mimimum.) 47. The profit p(x) œ r(x)  c(x) œ 6x  ax$  6x#  15xb œ x$  6x#  9x, where x   0. Then pw (x) œ 3x#  12x  9 œ 3(x  3)(x  1) and pw w (x) œ 6x  12. The critical points are 1 and 3. Thus pww (1) œ 6
0 Ê at x œ 1 there is a local minimum, and pww (3) œ '  0 Ê at x œ 3 there is a local maximum. But p(3) œ 0 Ê the best you can do is break even. 48. The average cost of producing x items is caxb œ

œ x#  #!x  #!ß !!! Ê c w axb œ #x  #! œ ! Ê x œ "!, the

c ax b x

only critical value. The average cost is ca"!b œ $"*ß *!! per item is a minimum cost because c ww a"!b œ #
!. 49. (a) The artisan should order px units of material in order to have enough until the next delivery. ˆ px ‰ (b) The average number of units in storage until the next delivery is px # and so the cost of storing then is s # per ‰ day, and the total cost for x days is ˆ px # sx. When added to the delivery cost, the total cost for delivery and storage for each cycle is: cost per cycle œ d 

px # sx.

ˆd


(c) The average cost per day for storage and delivery of materials is: average cost per day œ To minimize the average cost per day, set the derivative equal to zero.


" d dx Šdaxb



ps # ‰ #x

x

ps # x‹



œ  daxb

d x
#

 

ps # x. ps # œ

Ê x œ „ É #psd . Only the positive root makes sense in this context so that x‡ œ É #psd . To verify that x‡ gives a d ˆ minimum, check the second derivative ’ dx  daxb
# 

ps ‰ # “º

É #psd

œ

#d x$ º

É #psd

œ

#d #d

$

΃ ps



! Ê a minimum.

The amount to deliver is px‡ œ É #pd s . (d) The line and the hyperbola intersect when

d x

œ

ps # x.

Solving for x gives xintersection œ „ É #psd . For x
!,

xintersection œ É #psd œ x‡ . From this result, the average cost per day is minimized when the average daily cost of delivery is equal to the average daily cost of storage. 50. Average Cost:

c ax b x

c ax b d# dx# Š x ‹º

œ

xœ"!!

œ

#!!! x

%!!! "!!$

 *'  %x"Î# Ê

d c ax b dx Š x ‹


"Î# œ  #!!! œ ! Ê x œ "!!. Check for a minimum: x #  #x

 "!!
$Î# œ !Þ!!$
! Ê a minimum at x œ "!!. At a production level of "!!ß !!! units,

the average cost will be minimized at $"&' per unit.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

!

256

Chapter 4 Applications of Derivatives

51. We have

dR dM

œ CM  M# . Solving

d# R dM#

œ C  #M œ ! Ê M œ C# . Also,

d$ R dM$

œ  #  ! Ê at M œ

C #

there is a

maximum. 52. (a) If v œ cr! r#  cr$ , then vw œ 2cr! r  3cr# œ cr a2r!  3rb and vww œ 2cr!  6cr œ 2c ar!  3rb . The solution of vw œ 0 is r œ 0 or 2r3! , but 0 is not in the domain. Also, vw
0 for r  2r3! and vw  0 for r
2r3! Ê at rœ

2r! 3

there is a maximum.

(b) The graph confirms the findings in (a).

53. If x
0, then (x  1)#   0 Ê x#  1   2x Ê then Š a

#

#

#

#

1 b 1 c 1 d " a ‹Š b ‹Š c ‹Š d ‹

54. (a) f(x) œ

x È a#  x#

Ê f w (x) œ

aa #  x # b

"Î#


"Î#


x # aa #  x # b aa #  x # b

Ê gw (x) œ


ab#  (d
x)# b  (d
x)# ab#  (d
x)# b$Î#


ab#  (d
x)# b

œ

a#  x#
x# aa#  x# b$Î#

œ

a# aa#  x# b$Î#


0

"Î#
"Î#  (d
x)# ab#  (d
x)# b b#  (d
x)#


b #  0 Ê g(x) is a decreasing function of x ab#  (d
x)# b$Î# dt Since c" , c#
0, the derivative dx is an increasing function of x (from part (a)) minus a decreasing dt d# t " w " w w function of x (from part (b)): dx œ c"" f(x)  c"# g(x) Ê dx # œ c" f (x)  c# g (x)
0 since f (x)
dt gw (x)  0 Ê dx is an increasing function of x.

œ

(c)

d
x Èb#  (d
x)#

  2. In particular if a, b, c and d are positive integers,

  16.

Ê f(x) is an increasing function of x (b) g(x) œ

x#  1 x

œ

0 and

55. At x œ c, the tangents to the curves are parallel. Justification: The vertical distance between the curves is D(x) œ f(x)  g(x), so Dw (x) œ f w (x)  gw (x). The maximum value of D will occur at a point c where Dw œ 0. At such a point, f w (c)  gw (c) œ 0, or f w (c) œ gw (c). 56. (a) f(x) œ 3  4 cos x  cos 2x is a periodic function with period 21 (b) No, f(x) œ 3  4 cos x  cos 2x œ 3  4 cos x  a2 cos# x  1b œ 2 a1  2 cos x  cos# xb œ 2(1  cos x)#   0 Ê f(x) is never negative 57. (a) If y œ cot x  È2 csc x where 0  x  1, then yw œ (csc x) ŠÈ2 cot x  csc x‹. Solving yw œ 0 Ê cos x œ

" È2

Ê x œ 14 . For 0  x 

1 4

we have yw
0, and yw  0 when

1 4

 x  1. Therefore, at x œ

there is a maximum value of y œ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 4

Section 4.5 Applied Optimization Problems

257

(b)

The graph confirms the findings in (a). 58. (a) If y œ tan x  3 cot x where 0  x  Ê xœ „

1 3,

Therefore at x œ

1 #

, then yw œ sec# x  3 csc# x. Solving yw œ 0 Ê tan x œ „ È3

is not in the domain. Also, yww œ 2 sec# x tan x  3 csc# x cot x
0 for all 0  x  there is a minimum value of y œ 2È3.

but  1 3

1 3

1 2

.

(b)

The graph confirms the findings in (a). # # 59. (a) The square of the distance is Daxb œ ˆx  $# ‰  ˆÈx  !‰ œ x#  #x  *% , so Dw axb œ #x  # and the critical

point occurs at x œ ". Since Dw axb  ! for x  " and Dw axb
! for x
", the critical point corresponds to the minimum distance. The minimum distance is ÈDa"b œ

È& # .

(b)

The minimum distance is from the point ˆ $# ß !‰ to the point a"ß "b on the graph of y œ Èx, and this occurs at the

value x œ " where Daxb, the distance squared, has its minimum value. 60. (a) Calculus Method:

The square of the distance from the point Š"ß È$‹ to Šxß È"'  x# ‹ is given by #

Daxb œ ax  "b#  ŠÈ"'  x#  È$‹ œ x#  #x  "  "'  x#  #È%)  $x#  $ œ  #x  #!  #È%)  $x# . Then Dw axb œ  # 

# È%)
$x# a'xb

œ #

'x È%)
$x# .

Solving Dw axb œ ! we have: 'x œ #È%)  $x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

258

Chapter 4 Applications of Derivatives Ê $'x# œ %a%)  $x# b Ê *x# œ %)  $x# Ê "#x# œ %) Ê x œ „ #. We discard x œ # as an extraneous solution, leaving x œ #. Since Dw axb  ! for %  x  # and Dw axb
! for #  x  %, the critical point corresponds to the minimum distance. The minimum distance is ÈDa#b œ #. Geometry Method: The semicircle is centered at the origin and has radius %. The distance from the origin to Š"ß È$‹ is #

Ê"#  ŠÈ$‹ œ #. The shortest distance from the point to the semicircle is the distance along the radius containing the point Š"ß È$‹. That distance is %  # œ #. (b)

The minimum distance is from the point Š"ß È$‹ to the point Š#ß #È$‹ on the graph of y œ È"'  x# , and this occurs at the value x œ # where Daxb, the distance squared, has its minimum value. 61. (a) The base radius of the cone is r œ

#1 a
x #1

#

and so the height is h œ Èa#  r# œ Éa#  ˆ #1a#1
x ‰ . Therefore,

# # Vaxb œ 1$ r# h œ 1$ ˆ #1a#1
x ‰ Éa#  ˆ #1a#1
x ‰ .

(b) To simplify the calculations, we shall consider the volume as a function of r: volume œ farb œ 1$ r# Èa#  r# , where !  r  a. f w arb œ #

$

œ 1$ ” #Èa r#
$r# • œ a
r

1 d #È # a $ dr Šr 1ra#a#
$r# b . $Èa#
r#

h œ Èa#  r# œ Éa# 

#a # $

 r# ‹ œ 1$ ”r# †

" a#rb # È a #
r#

 ŠÈa#  r# ‹a#rb• œ 1$ ”
r

The critical point occurs when r# œ #

œ É a$ œ

aÈ $ $ .

Using r œ

aÈ ' $

#a # $ ,

and h œ

$

 #raa#
r# b • Èa#
r#

which gives r œ aÉ #$ œ

aÈ $ $ ,

aÈ ' $ .

Then

we may now find the values of r and h

for the given values of a. hœ

When a œ &: r œ

%È' $ , È & ' $ ,

When a œ ): r œ

)È' $ ,

hœ

)È$ $ à

aÈ $ $ ,

the relationship is

When a œ %: r œ

%È$ $ à È & $ $ à

hœ È When a œ ': r œ # ', h œ #È$à (c) Since r œ

aÈ ' $

and h œ

r h

œ È #.

62. (a) Let x! represent the fixed value of x at the point P, so that P has the coordinates ax! ß ab, and let m œ f w ax! b be the slope of the line RT. Then the equation of the line RT is y œ max  x! b  a. The y-intercept of this line is ma!  x! b  a œ a  mx! , and the x-intercept is the solution of max  x! b  a œ !, or x œ mx!m
a . Let O designate the origin. Then (Area of triangle RST) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.5 Applied Optimization Problems

259

œ #(Area of triangle ORT) œ # † "# (x-intercept of line RT)(y-intercept of line RT) œ # † "# ˆ mx!m
a ‰aa  mx! b œ mˆ mx!m
a ‰ˆ mx!m
a ‰ œ mˆ mx!m
a ‰ œ mˆx! 

#

a ‰# m

Substituting x for x! , f w axb for m, and faxb for a, we have Aaxb œ f w axb”x 

faxb f ax b • w

(b) The domain is the open interval a!ß "!b. To graph, let y" œ faxb œ &  &É"  y$ œ Aaxb œ y# Šx 

y" y# ‹

#

x# "!! ,

#

. y# œ f w axb œ NDERay" b, and

. The graph of the area function y$ œ Aaxb is shown below.

The vertical asymptotes at x œ ! and x œ "! correspond to horizontal or vertical tangent lines, which do not form triangles. (c) Using our expression for the y-intercept of the tangent line, the height of the triangle is
x x# a  mx œ faxb  f w axb † x œ &  " È"!!  x#  x œ &  " È"!!  x#  #

#

2È"!!
x#

2È"!!
x#

We may use graphing methods or the analytic method in part (d) to find that the minimum value of Aaxb occurs at x ¸ )Þ''. Substituting this value into the expression above, the height of the triangle is 15. This is 3 times the y-coordinate of the center of the ellipse. (d) Part (a) remains unchanged. Assuming C   B, the domain is a!ß Cb. To graph, note that faxb œ B  BÉ" 

x# C#

Aaxb œ f w axb”x 

faxb f ax b •

œ B  BC ÈC#  x# and f w axb œ #

œ

w

Bx Œx CÈC#
x#

" # ”Bx BCxÈC#  x#

œ

" ”BCŠC  ÈC# BCxÈC#  x#

A axb œ BC † œ œ œ œ

x# aC#  x# b BCŠC
ÈC#  x# ‹ x# aC#  x# b BCŠC
ÈC#  x# ‹ x# aC#  x# b x# aC#  x# b$Î#

È C#
x #

#


œ

 ŠBC  BÈC#  x# ‹ŠÈC#  x# ‹• œ #

 x# ‹• œ

œ

Bx . C È C #  x#

Bx x C È C #  x# 

" # ”Bx BCxÈC#  x#

Therefore we have #

ŠBC
BÈC#  x# ‹ŠÈC#  x# ‹ Bx



 BCÈC#  x#  BaC#  x# b•

#

#

BCŠC
ÈC#  x# ‹ xÈC#  x#


x ‹ C#
x #

#


ŠC
ÈC#  x# ‹ Šx È

x# aC#




x# b


x
ÈC# C#
x #

 x# a"b‹

#


x # È # È # # # –#x  ŠC  C  x ‹Š ÈC#
x#  C  x ‹— # ”#x 

Š È Cx #

#

C


BC# ŠC
ÈC# x# ‹


Bx

C

ŠxÈC#  x# ‹a#bŠC
ÈC#  x# ‹Š È

BCŠC
ÈC#  x# ‹

B
BC ÈC#  x#

#

œ

w



B " C #ÈC#  x# a#xb

x#

Cx# ÈC#
x#

 CÈC#  x#  x# aC#  x# b•

 CÈC#  x#  C# ‹ œ

BCŠC
ÈC#  x# ‹ x# aC#  x# b$Î#

’Cx#  CaC#  x# b  C# ÈC#  x# “

Š#x#  C#  CÈC#  x# ‹

To find the critical points for !  x  C, we solve: #x#  C# œ CÈC#  x# Ê %x%  %C# x#  C% œ C%  C# x# Ê %x%  $C# x# œ ! Ê x# a%x#  $C# b œ !. The minimum value of Aaxb for !  x  C occurs at the critical point Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

260

Chapter 4 Applications of Derivatives xœ

CÈ$ # ,

or x# œ

$C# % . w

The corresponding triangle height is

a  mx œ faxb  f axb † x œ B  B ÈC#  x#  C

Bx# É C C# 

$C# %

#

œB

BÈ # C C



œ B  BC ˆ C# ‰  œB



B #

$B #

x#



BŠ $C% ‹ $C# %

CÉ C# 

$BC# % C# #

œ $B This shows that the traingle has minimum arrea when its height is $B. ^ 4.6 INDETERMINATE FORMS AND L'HOPITAL'S RULE ^ 1. l'Hopital: lim

x
2

œ

" #x ¹xœ#

^ 2. l'Hopital: lim

sin 5x x

œ

5 cos 5x ¹ 1 xœ!

# x Ä 2 x
4

xÄ0

5x#
3x 7x#  1

^ 3. l'Hopital: lim xÄ_

œ x lim Ä_

x$
1 $ x Ä 1 4x
x
3 # 3 lim ax 2 x  "b œ 11 x Ä 1 a4x + 4x + 3b

^ 4. l'Hopital: lim œ

1
cos x x#

^ 5. l'Hopital: lim

xÄ! sin# x

œ lim

2 x Ä ! x a"  cos xb

7.

lim

tÄ0

sin t# t

tÄ0

2x
1 cos x

œ

8.

lim x Ä 1 Î2

9.

lim ) Ä 1 1
)

10.

lim x Ä 1Î2 1  cos 2x

sin )

œ

11.

sin x
cos x x
1% x Ä 1Î%

12.

cos x
1 x Ä 1Î$ x
$

lim

lim

" #

œ

xÄ0

"0x
3 14x

sin x 2x

œ x lim Ä_

x Ä 2 ax
#bax  #b

œ x lim Ä_ œ

3 11

œ lim

10 14

œ

cos )
"

œ

lim x Ä 1 Î2

œ

%x  3 $x #  "


"
"

œ

2
"

" #

or lim

xÄ!

œ x lim Ä_

% 'x

œ

" 4

œ5†1œ5

5x#
3x 7x# 1

or x lim Ä_

x $
1 $ x Ä 1 4x
x
3

œ

sin 5x 5x

"

x Ä 2 x#

œ x lim Ä_

5
7

3 x " x

œ

5 7

ax
"bax#  x  "b 2 a x Ä 1 x
"ba4x + 4x + 3b

œ lim

1
cos x x#

xb ˆ "  cos x ‰ œ lim ” a"
xcos 2 "  cos x • xÄ!

" #

œ ! or x lim Ä_

#x#  3x x$  x  1

œ x lim Ä_

# x

"

 " x#

3 x#



œ 2

œ"


cos x
2 sin 2x

œ

cos x  sin x " x Ä 1Î%

lim

lim

5 7

or lim

cos x 2

xÄ!

œ 5 lim 5x Ä 0

sin 5x x

œ lim

œ0 2

)Ä1

œ 5 or lim

x
2

œ lim

# x Ä 2 x
4

œ lim ”ˆ sinx x ‰ˆ sinx x ‰ˆ "  "cos x ‰• œ xÄ!

lim ) Ä 1Î2
sin x

œ lim

1
sin x

xÄ!

x
2

or lim

3x# # x Ä 1 12x
1

œ lim

2t cos t# 1

œ lim

" 4

œ lim

#x #  $ x x$  x  1

^ 6. l'Hopital: lim xÄ_

œ

x Ä 1Î$


sin x "

œ

lim x Ä 1 Î2 œ

È# #

sin x
4 cos 2x



È# #

œ

"
4(
1)

œ

" 4

œ È#

È$ #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" x$

œ

! "

œ!

^ Section 4.6 Indeterminate Forms and L'Hopital's Rule 13.

lim

x Ä 1 Î2

#x

14. lim


ˆx
1# ‰ sin x cos x x Ä 1 Î2

ˆx  1# ‰ tan x œ

lim

#

œ lim

#x #
a$ x  " b È x  # x
" xÄ"

x Ä ! 2Èx(

#x#
$x$Î#
x"Î#  # x
" xÄ"

" ax#  5b
"Î# a2xb #

16. lim

È x#  5
$ x#
%

17. lim

È a aa  x b
a x

18. lim

10(sin t
t) t$

œ lim

10(cos t
") $t#

19. lim

x(cos x
") sin x
x

œ lim


xsin x  cos x
" cos x
"

20. lim

sinaa  hb
sin a h

21. lim

rÄ"

a ar n
" b r
"

lim

Š "x 

xÄ0

tÄ0

xÄ0

hÄ0

22.

x Ä !b

œ lim

xÄ#

œ

a

x Ä 0 #Èa#  ax

tÄ0

xÄ0

œ lim

hÄ0

aan†rn
" b 1

œ lim

rÄ1

" Èx ‹

"

"!a
sin tb 't

œ lim

xÄ0

cosaa  hb
cos a "

"

œ

"

Èx

#

œ lim

tÄ0


xcos x
#sin x
sin x

" '


"! cos t '

œ lim

œ

xÄ0


10†" '


"

xcos x  #sin x sin x

$x
&

sin (x

x Ä 0 tan ""x

27. x lim Ä_

Èx

lim x Ä !b Èsin x

"
Èx x ‹

tanŠ "x ‹

29.

lim x Ä 1Î2c tan x

sec x

lim b

xÄ!

cot x csc x

" x

rule È x‰ † ˆ œ Š l'Hopital's does not apply ‹ œ lim b "  xÄ!

œ

3

œ!

(cosa(xb

(†" ""†"

œ

" lim

xÄ!b

lim

œ

9x  1 x1

sin x x

x Ä 1 Î2 c

œ lim b xÄ!

œ x lim Ä_

lim x Ä „ _ 4x
"

œ Éx lim Ä_

œÊ


xsin x  $ cos x cos x

rÄ1

# x Ä 0 ""sec a""xb

28.

30.

œ

œ lim

È9x  1 Èx  1

xÄ0

#

1 x

œ_

x #
ax #  x b x  È x#  x

œ x lim Ä_


xx x x

rule œ  "# Š l'Hopital's is unnecessary ‹

" x

lim # x Ä „ _ #x
x  #

26. lim

œ lim

œ an lim rn
" œ an, where n is a positive integer.

œ lim b Š xÄ!

24. x lim x tanˆ "x ‰ œ x lim Ä_ Ä_ 25.

œ  &$

œ!

È

"  É" 

œ"

œ "

x x 23. x lim Šx  Èx#  x‹ œ x lim Šx  Èx#  x‹Š xx  ‹ œ x lim Ä_ Ä_ Ä_  È x#  x

œ x lim Ä_


"
"

œ "# , where a
0.

a #Èa#

tÄ0

%x
*# x"Î#


xÄ1

x Ä # 2È x#  5

œ lim

œ

œ!

œ lim

œ lim

2x

œ lim

ˆ 1
x‰ cos x  sin xa
"b #
sin x x Ä 1 Î2

lim

%†! #†!(

œ

œ lim

15. lim

xÄ#

%È x

œ lim

( x Ä ! "  #Èx

x Ä ! x  (È x

œ

 x"#

œ x lim sec# ˆ x" ‰ œ sec# ! œ " Ä_

( ""

œ Éx lim Ä_

9 1

œ È9 œ 3

œ É 1" œ 1

x‰ ˆ cos" x ‰ ˆ cos sin x œ

ˆ cos x ‰ sin x

 x"# sec# ˆ x" ‰

Š sin" x ‹

"

lim x Ä 1Î2c sin x

œ1

œ lim b cos x œ 1 xÄ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley



È x#
x È x#

œ

$ "

œ$

261

262

Chapter 4 Applications of Derivatives

31. Part (b) is correct because part (a) is neither in the

0 0

nor

_ _

^ form and so l'Hopital's rule may not be used.

32. Answers may vary. (a) faxb œ $x  ";gaxb œ x lim x Ä _ g(x)

œ x lim Ä_

$x  " x #

lim x Ä _ g(x)

f(x)

œ x lim Ä_

x" x#

f(x)

œ x lim Ä_

x# x"

f(x)

(b) faxb œ x  ";gaxb œ x #

(c) faxb œ x ;gaxb œ x  " lim x Ä _ g(x)

$ "

œ$

œ x lim Ä_

" #x

œ!

œ x lim Ä_

#x "

œ_

œ x lim Ä_

33. If f(x) is to be continuous at x œ 0, then lim f(x) œ f(0) Ê c œ f(0) œ lim œ lim

xÄ0

27 sin 3x 30x

œ lim

34. (a) For x Á 0, f w (x) œ œ

x2 x1

œ

81 cos 3x 30

xÄ0

02 01

d dx

xÄ0

œ

27 10

xÄ0

9x
3 sin 3x 5x$

œ lim

xÄ0

9
9 cos 3x 15x#

.

(x  2) œ 1 and gw (x) œ

(x  1) œ 1. Therefore, lim

d dx

f w (x)

w x Ä 0 g (x)

œ 2.

œ

1 1

œ 1, while lim

f(x)

x Ä 0 g(x)

^ (b) This does not contradict l'Hopital's rule because neither f nor g is differentiable at x œ 0 ^ (as evidenced by the fact that neither is continuous at x œ 0), so l'Hopital's rule does not apply. 35. The graph indicates a limit near 1. The limit leads to the indeterminate form 00 : lim

xÄ1

2x#
3x$Î#
x"Î#  2 x
1 xÄ1 4
9#
#" œ 4
1 5 œ 1 1

œ lim œ

2x#
(3x  1) Èx  2 x
1 4x


œ lim

xÄ1

9 #

x"Î#
1

" #

x
"Î#

36. (a)

(b) The limit leads to the indeterminate form _  _: È # lim Šx  Èx#  x‹ œ x lim Šx  Èx#  x‹Š x  Èx# x ‹ œ x lim Š xÄ_ Ä_ Ä_ x

œ x lim Ä_


" "  É" 

37. Graphing faxb œ '

" x

"
cos x' x"#

œ


" "  È"  !

x #
ax #  x b ‹ x  È x#  x

œ x lim Ä_


x x  È x#  x

œ  "#

on th window Ò"ß "Ó by Ò!Þ&ß "Ó it appears that lim faxb œ !. However, we see that if we let "
cos u # uÄ0 u

u œ x , then lim faxb œ lim xÄ0

x x

38. (a) We seek c in a#ß !b so that

f ac b g ac b w

w

œ lim

sin u u Ä 0 #u

œ

œ lim

fa!b
fa
#b ga!b
ga
#b

cos u uÄ0 #

œ

!# !
%

œ "# .

xÄ0

œ  #" . Since f w acb œ " and gw acb œ #c we have that

Ê c œ ".

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #c

œ  #"

(b) We seek c in any open interval aaß bb so that (c) We seek c in a!ß $b so that (Note that c œ


"
È$( $

œ

w

fa$b
fa!b ga$b
ga!b

w

œ

œ


$
! *
!

fabb
faab g ab b
g a a b

œ

œ  "$ Ê

b
a b #
a# c#
% #c

œ

b
a ab
abab  ab

ba #

œ

" ba

Ê

" #c

œ

" ba

œ  "$ Ê $c#  #c  "# œ ! Ê c œ

"
x )

"
cos ) )
sin ) ,

œ

œ

CE EB

)Ä_

since the coordinates of C are acos )ß sin )b. Hence, "  x œ

)Ä_

”

)a"
cos )b )
sin )

 a"  cos )b• œ lim

)Ä_

As ) Ä _, a"  cos )b oscillates between ! and #, and so it is bounded. Since lim

)Ä_

a"  cos )b” )
)sin )  "•

ˆ )
)sin )  "‰ œ "  " œ !,

a"  cos )b” )
)sin )  "• œ !. Geometrically, this means that as ) Ä _, the distance between points P and D

lim

)Ä_

approaches 0. 40. Throughout this problem note that r# œ y#  ", r
y and that both r Ä _ and y Ä _ as ) Ä 1# Þ (b)


"  È$( . $

←→ ←→ ←→ where E is the point on AB such that CE ¼ AB :

a"  xb  a"  cos )b‘ œ lim

(c) We have that lim

(a)

Êcœ

)a"
cos )b )
sin ) . )a"
cos )b ) sin )  "
cos ) ) cos )  sin )  sin ) )  #sin ) lim a"  xb œ lim )
sin ) œ lim œ lim œ lim ) cos sin "
cos ) sin ) ) )Ä0 )Ä0 )Ä0 )Ä0 )Ä0 )a
sin )b  cos )  #cos )
) sin )  $cos ) !$ œ lim œ lim œ " œ$ cos ) cos ) )Ä0 )Ä0

Thus (b)

w

w

263

is not in the given interval a!ß $b.)

PA AB

39. (a) By similar triangles,

f ac b g ac b

f ac b g ac b

Section 4.7 Newton's Method

"

lim

ryœ

lim

r #  y# œ

) Ä 1 Î2 ) Ä 1 Î2

lim ) Ä 1 Î 2 r y lim

) Ä 1 Î2

œ!

"œ"

(c) We have that r$  y$ œ ar  ybar#  ry  y# b œ Since

lim

) Ä 1 Î2

$y †

y r

œ

lim

) Ä 1 Î2

r#  ry  y# ry




$sin ) † y œ _ we have that

y #  y †y  y # r $

lim

) Ä 1 Î2

œ

$y # r $

œ $y † yr .

r  y œ _.

4Þ7 NEWTON'S METHOD 1. y œ x#  x  1 Ê yw œ 2x  1 Ê xnb1 œ xn  Ê x# œ

2 3



4 2 9  3
1 4 3 1

Ê x# œ 2 

Ê x# œ

4
2
"
4  1

œ

5 3

2 3



4 6
9 129

œ

2 3



x#n
xn 1 # xn
1

; x! œ 1 Ê x " œ 1 

" #1

¸ .61905; x! œ 1 Ê x" œ 1 


#"7
11 " 3 3

13 21

œ

2 3 1
1
1
#1

¸ 1.66667

2. y œ x$  3x  1 Ê yw œ 3x#  3 Ê xnb1 œ xn  Ê x# œ  "3 

œ

1 1
1 #1

œ  3" 

" 90

x$n
3xn
1 3xn#
3

; x! œ 0 Ê x" œ 0 

" 3

œ  "3

œ  29 90 ¸ 0.32222

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 2

264

Chapter 4 Applications of Derivatives x%n
xn 3 4xn$
1

3. y œ x%  x  3 Ê yw œ 4x$  1 Ê xnb1 œ xn  Ê x# œ

6 5



1296 6 625  5
3 864 125 1

œ 2 Ê x# œ # 

œ

6 5

1296750
1875 4320625



16
2
3
3#1

œ 2 

œ

6 5



Ê x# œ  "# 



œ

¸ 2.41667

5 #



" 1#

5 4



113 2000

œ

2500
113 2000

œ

2387 #000

6. From Exercise 5, xnb1 œ xn  625
512
2000

œ  54 

œ  54 

œ

6 5

¸ 1.16542; x! œ 1 Ê x" œ 1 

"
1
3
4 1

2xn xn#
1 22xn

; x! œ 0 Ê x" œ 0 

x%n 2 4xn$ ; x!

œ 1 Ê x" œ 1 

"
2 4

œ

5 4

0
0" #
0

4
4" #
4

œ  "#

œ

5 #

Ê x# œ

5 4

Ê x# œ



625 256
2 125 16

œ

5 #



5 4

; x! œ  1 Ê x" œ  1 



"
2
4

œ 1 

" 4

œ  54 Ê x# œ  54 

625 256
2
125 16

f axn b f axn b w

gives x" œ x! Ê x# œ x! Ê xn œ x! for all n   0. That is, all of

the approximations in Newton's method will be the root of f(x) œ 0. 8. It does matter. If you start too far away from x œ

1 #

, the calculated values may approach some other root.

Starting with x! œ 0.5, for instance, leads to x œ  1# as the root, not x œ

œh

Èh Š

f(x! ) f w (x! )

if x! œ h  0 Ê x" œ x!  œ h 

Èh Š
" È‹ 2

œh

1 #

.

f(h) f w (h)

œ h  ŠÈh‹ Š2Èh‹ œ h;

"

Èh ‹

#

625
512 2000

¸ 1.1935

7. f(x! ) œ 0 and f w (x! ) Á ! Ê xnb1 œ xn 

9. If x! œ h
0 Ê x" œ x! 

5
25 4 1 #
5

¸ 1.1935

x%n 2 4xn$

113 2000

5763 4945

œ  15# ¸ .41667; x! œ 2 Ê x" œ 2 

5. y œ x%  2 Ê yw œ 4x$ Ê xnb1 œ xn  œ

œ

1 1
3 4 1

51 œ  31 ¸ 1.64516

11 31

4. y œ 2x  x#  1 Ê yw œ 2  2x Ê xnb1 œ xn 
1
"4 " œ  #" #1 20
254 29 œ #5  1"# œ 12
1 #

171 4945

; x! œ 1 Ê x " œ 1 

f(x! ) f w (x! )

œ h 

f(
h) f w (
h)

œ h  ŠÈh‹ Š2Èh‹ œ h.

h

"Î$

10. f(x) œ x"Î$ Ê f w (x) œ ˆ "3 ‰ x
#Î$ Ê xnb1 œ xn  ˆ " ‰xn
#Î$ xn 3

œ 2xn ; x! œ 1 Ê x" œ 2, x# œ 4, x$ œ 8, and x% œ 16 and so forth. Since kxn k œ 2lxnc1 l we may conclude that n Ä _ Ê kxn k Ä _.

11. i) is equivalent to solving x$  $x  " œ !. ii) is equivalent to solving x$  $x  " œ !. iii) is equivalent to solving x$  $x  " œ !. iv) is equivalent to solving x$  $x  " œ !. All four equations are equivalent. 12. f(x) œ x  1  0.5 sin x Ê f w (x) œ 1  0.5 cos x Ê xnb1 œ xn 

xn  1  0.5 sin xn 1  0.5 cos xn

; if x! œ 1.5, then

x" œ 1.49870

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.7 Newton's Method 13. For x! œ
!Þ$, the procedure converges to the root
!Þ$##")&$&ÞÞÞÞ (a)

(b)

(c)

(d) Values for x will vary. One possible choice is x! œ !Þ1.

(e) Values for x will vary. 14. (a) f(x) œ x$
3x
1 Ê f w (x) œ 3x#
3 Ê xnb1 œ xn


x$n
3xn
1 3xn#
3

Ê the two negative zeros are
1.53209

and
0.34730 (b) The estimated solutions of x$
3x
1 œ 0 are
1.53209,
0.34730, 1.87939.

(c) The estimated x-values where g(x) œ 0.25x%
1.5x#
x  5 has horizontal tangents are the roots of gw (x) œ x$
3x
1, and these are
1.53209,
0.34730, 1.87939.

15. f(x) œ tan x
2x Ê f w (x) œ sec# x
2 Ê xnb1 œ xn


tan axn b
2xn sec# axn b

; x! œ 1 Ê x" œ 12920445

Ê x# œ 1.155327774 Ê x16 œ x17 œ 1.165561185 16. f(x) œ x%
2x$
x#
2x  2 Ê f w (x) œ 4x$
6x#
2x
2 Ê xnb1 œ xn


x%n
2xn$
xn#
2xn  2 4xn$
6xn#
2xn
2

if x! œ 0.5, then x% œ 0.630115396; if x! œ 2.5, then x% œ 2.57327196

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

265

266

Chapter 4 Applications of Derivatives

17. (a) The graph of f(x) œ sin 3x
0.99  x# in the window
2 Ÿ x Ÿ 2,
2 Ÿ y Ÿ 3 suggests three roots. However, when you zoom in on the x-axis near x œ 1.2, you can see that the graph lies above the axis there. There are only two roots, one near x œ
1, the other near x œ 0.4. (b) f(x) œ sin 3x
0.99  x# Ê f w (x) œ 3 cos 3x  2x Ê xnb1 œ xn


sin (3xn )
0.99xn# 3 cos (3xn )  2xn

and the solutions

are approximately 0.35003501505249 and
1.0261731615301 18. (a) Yes, three times as indicted by the graphs (b) f(x) œ cos 3x
x Ê f w (x) œ
3 sin 3x
1 Ê xnb1 œ xn


cos a3xn b
xn
3 sin a3xn b
1

; at

approximately
0.979367,
0.887726, and 0.39004 we have cos 3x œ x 19. f(x) œ 2x%
4x#  1 Ê f w (x) œ 8x$
8x Ê xnb1 œ xn


2x%n
4xn# 1 8xn$
8xn

; if x! œ
2, then x' œ
1.30656296; if

x! œ
0.5, then x$ œ
0.5411961; the roots are approximately „ 0.5411961 and „ 1.30656296 because f(x) is an even function. 20. f(x) œ tan x Ê f w (x) œ sec# x Ê xnb1 œ xn


tan axn b sec# axn b

; x! œ 3 Ê x" œ 3.13971 Ê x# œ 3.14159 and we

approximate 1 to be 3.14159. 21. From the graph we let x! œ 0.5 and f(x) œ cos x
2x Ê xnb1 œ xn


cos axn b
2xn
sin axn b
2

Ê x" œ .45063

Ê x# œ .45018 Ê at x ¸ 0.45 we have cos x œ 2x.

22. From the graph we let x! œ
0.7 and f(x) œ cos x  x Ê xnb1 œ xn


xn  cos axn b 1
sin axn b

Ê x" œ
.73944

Ê x# œ
.73908 Ê at x ¸
0.74 we have cos x œ
x.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.7 Newton's Method 23. If f(x) œ x$  2x
4, then f(1) œ
1  0 and f(2) œ 8  0 Ê by the Intermediate Value Theorem the equation x$  2x
4 œ 0 has a solution between 1 and 2. Consequently, f w (x) œ 3x#  2 and xnb1 œ xn


x$n  2xn
4 3x#n  2

.

Then x! œ 1 Ê x" œ 1.2 Ê x# œ 1.17975 Ê x$ œ 1.179509 Ê x% œ 1.1795090 Ê the root is approximately 1.17951. 24. We wish to solve 8x%
14x$
9x#  11x
1 œ 0. Let f(x) œ 8x%
14x$
9x#  11x
1, then f w (x) œ 32x$
42x#
18x  11 Ê xnb1 œ xn
x!
1.0 0.1 0.6 2.0

8x%n
14xn$
9xn#  11xn
1 3#xn$
42xn#
18xn  11

.

approximation of corresponding root
0.976823589 0.100363332 0.642746671 1.983713587

25. f(x) œ 4x%
4x# Ê f w (x) œ 16x$
8x Ê xib1 œ xi


faxi b f w axi b

œ xi


xi$
xi . %x#i
#

Iterations are performed using the

procedure in problem 13 in this section. (a) For x! œ # or x! œ
!Þ), xi Ä
" as i gets large. (b) For x! œ
!Þ& or x! œ !Þ#&, xi Ä ! as i gets large. (c) For x! œ !Þ) or x! œ #, xi Ä " as i gets large. (d) (If your calculator has a CAS, put it in exact mode, otherwise approximate the radicals with a decimal value.) For x! œ
x! œ

È
721

È21 7

or x! œ


or x! œ

È
721

È21 7 ,

Newton's method does not converge. The values of xi alternate between

as i increases.

26. (a) The distance can be represented by # D(x) œ É(x
2)#  ˆx#  "# ‰ , where x   0. The

distance D(x) is minimized when # f(x) œ (x
2)#  ˆx#  "# ‰ is minimized. If

# f(x) œ (x
2)#  ˆx#  "# ‰ , then

f w (x) œ 4 ax$  x
1b and f w w (x) œ 4 a3x#  1b  0. Now f w (x) œ 0 Ê x$  x
1 œ 0 Ê x ax#  1b œ 1 Ê x œ x#"
1 . (b) Let g(x) œ

" x #
1

Ê xnb1 œ xn



x œ ax#  1b " Œ x#
1
xn
n

Î

2xn Ñ # Ï Šx#n
1‹  1 Ò

"

#


x Ê gw (x) œ
ax#  1b (2x)
1 œ

2x ax #
1 b #


1

; x! œ 1 Ê x% œ 0.68233 to five decimal places.

27. f(x) œ (x
1)%! Ê f w (x) œ 40(x
1)$* Ê xnb1 œ xn


axn
1b%! 40 axn
1b$*

œ

39xn  " 40

. With x! œ 2, our computer

gave x)( œ x)) œ x)* œ â œ x#!! œ 1.11051, coming within 0.11051 of the root x œ 1. 28. f(x) œ 4x%
4x# Ê f w (x) œ 16x$
8x œ 8x a2x#
1b Ê xnb1 œ xn


xn ax#n
"b 2 a2x#n
1b

; if x! œ .65, then

x"# ¸
.000004, if x! œ .7, then x"# œ
1.000004; if x! œ .8, then x' œ 1.000000. NOTE: 29. f(x) œ x$  3.6x#
36.4 Ê f w (x) œ 3x#  7.2x Ê xnb1 œ xn


x$n  3.6xn#
36.4 3xn#  7.2xn

È21 7

¸ .654654

; x! œ 2 Ê x" œ 2.5303

Ê x# œ 2.45418225 Ê x$ œ 2.45238021 Ê x% œ 2.45237921 which is 2.45 to two decimal places. Recall that

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

267

268

Chapter 4 Applications of Derivatives

x œ 10% cH$ O
d Ê cH$ O
d œ (x) a10% b œ (2.45) a10% b œ 0.000245 30. Newton's method yields the following: the initial value

2

i

the approached value

1


5.55931i

È3  i
29.5815
17.0789i

4.8 ANTIDERIVATIVES 1. (a) x#

(b)

x$ 3

(c)

x$ 3


x#  x

2. (a) 3x#

(b)

x) 8

(c)

x) 8


3x#  8x

3. (a) x$

(b)
x3

4. (a)
x#

(b)
x4 


$


$

(c)
x3  x#  3x


#

x$ 3

(c)

x
# #



x# 2


x

5. (a)

" x

(b)

5 x

(c) 2x 

6. (a)

" x#

(b)

" 4x#

(c)

x% 4

(c)

2 3

Èx$  2Èx

(c)

3 4

x%Î$  3# x#Î$



5 x " #x #

7. (a) Èx$

(b) Èx

8. (a) x%Î$

(b)

9. (a) x#Î$

(b) x"Î$

(c) x"Î$

10. (a) x"Î#

(b) x"Î#

(c) x$Î#

11. (a) cos (1x)

(b)
3 cos x

(c)

12. (a) sin (1x)

(b) sin ˆ 1#x ‰

(c) ˆ 12 ‰ sin ˆ 1#x ‰  1 sin x

13. (a) tan x

(b) 2 tan ˆ x3 ‰

‰ (c)
23 tan ˆ 3x #

14. (a)
cot x

‰ (b) cot ˆ 3x #

(c) x  4 cot (2x)

15. (a)
csc x

(b)

" 5

csc (5x)

(c) 2 csc ˆ 1#x ‰

16. (a) sec x

(b)

4 3

sec (3x)

(c)

17.

' (x  1) dx œ

19.

'

x# #

xC

ˆ3t#  #t ‰ dt œ t$ 

t# 4

C

" #

x#Î$

18.

' (5
6x) dx œ 5x
3x#  C

20.

'

#

Š t#  4t$ ‹ dt œ

t$ 6

cos (1x) 1

2 1

 t%  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

 cos (3x)

sec ˆ 1#x ‰

Section 4.8 Antiderivatives 21.

'

a2x$
5x  7b dx œ

23.

'

ˆ x"#
x#
3" ‰ dx œ ' ˆx#
x#
3" ‰ dx œ

24.

'

ˆ "5


25.

' x"Î$ dx œ

27.

'

ˆÈx  $Èx‰ dx œ ' ˆx"Î#  x"Î$ ‰ dx œ

28.

'

Š

29.

'

Š8y


30.

'

Èx #

2 x$



Š "7


" #

x%
5# x#  7x  C x
" 1

 2x‰ dx œ ' ˆ 5"
2x$  2x‰ dx œ x#Î$ 2 3

2 Èx ‹

2 ‹ y"Î%

1 ‹ y&Î%

Cœ

3 #

'

22. x$ 3





3" x  C œ
x"

#

" 5

x
Š 2x# ‹ 

x#Î$  C

dx œ ' ˆ "# x"Î#  2x"Î# ‰ dx œ dy œ ' ˆ8y
2y"Î% ‰ dy œ

dy œ ' ˆ "7
y&Î% ‰ dy œ

" 7



3 #

8y# #

" #

x%Î$

Cœ

4 3

2x# #

Cœ

' x&Î% dx œ

26. x$Î#

a1
x#
3x& b dx œ x
"3 x$
"# x'  C

2 3

5 x

x$ 3






x
"Î%  "4

" x#

x 3

C

 x#  C

Cœ

4 % x È

C

x$Î#  34 x%Î$  C

$Î#

"Î#

#

#

Šx3 ‹  2 Šx" ‹  C œ

" 3

x$Î#  4x"Î#  C

$Î%


2 Š y 3 ‹  C œ 4y#
83 y$Î%  C 4


"Î%

y
Š y 1 ‹  C œ 4

y 7



C

4 y"Î%

31.

'

2x a1
x$ b dx œ ' a2x
2x# b dx œ

32.

'

x$ (x  1) dx œ ' ax#  x$ b dx œ

33.

'

tÈt
Èt t#

34.

'

4
È t t$

35.

'
2 cos t dt œ
2 sin t  C

36.

'
5 sin t dt œ 5 cos t  C

37.

' 7 sin 3) d) œ
21 cos 3)  C

38.

' 3 cos 5) d) œ 35 sin 5)  C

39.

'
3 csc# x dx œ 3 cot x  C

40.

'
sec3# x dx œ
tan3 x  C

41.

'

42.

'

43.

' a4 sec x tan x
2 sec# xb dx œ 4 sec x
2 tan x  C

44.

'

45.

' asin 2x
csc# xb dx œ
"# cos 2x  cot x  C

46.

' (2 cos 2x
3 sin 3x) dx œ sin 2x  cos 3x  C

47.

'

1
cos 4t #

dt œ ' ˆ "# 

" #

cos 4t‰ dt œ

" #

t  "# ˆ sin4 4t ‰  C œ

t 2



sin 4t 8

C

48.

'

1  cos 6t #

dt œ ' ˆ "#


" #

cos 6t‰ dt œ

" #

t
"# ˆ sin6 6t ‰  C œ

t 2




sin 6t 12

C

$Î#

dt œ ' Š t4$ 

csc ) cot ) #

" 2

dt œ ' Š t t# 

t"Î# t# ‹

t"Î# t$ ‹

2x# #

x
" 1


"


2 Š x1 ‹  C œ x# 
#

 Š x# ‹  C œ
"x


dt œ ' ˆt"Î#  t$Î# ‰ dt œ

t"Î#

" #

2 x

C

" #x #

C


"Î#

 Š t " ‹  C œ 2Èt
#

t dt œ ' ˆ4t$  t&Î# ‰ dt œ 4 Š # ‹  Š t 3 ‹  C œ
t2#



#


$Î# #

d) œ
"# csc )  C

acsc# x
csc x cot xb dx œ
#" cot x 

" #

2 5

sec ) tan ) d) œ

2 5

2 Èt

C

2 3t$Î#

C

sec )  C

csc x  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

269

270

Chapter 4 Applications of Derivatives

49.

' a1  tan# )b d) œ ' sec# ) d) œ tan )  C

50.

' a2  tan# )b d) œ ' a1  1  tan# )b d) œ ' a1  sec# )b d) œ )  tan )  C

51.

' cot# x dx œ ' acsc# x
1b dx œ
cot x
x  C

52.

' a1
cot# xb dx œ ' a1
acsc# x
1bb dx œ ' a2
csc# xb dx œ 2x  cot x  C

53.

' cos ) (tan )  sec )) d) œ ' (sin )  1) d) œ
cos )  )  C

54.

'

csc ) csc )  sin )

55.

d dx

 2)  C‹ œ Š (7x28

56.

d dx

Š
(3 x
3 5)

57.

d dx

ˆ "5 tan (5x
1)  C‰ œ

58.

d dx

ˆ
3 cot ˆ x 3 " ‰  C‰ œ
3 ˆ
csc# ˆ x 3 " ‰‰ ˆ "3 ‰ œ csc# ˆ x 3 " ‰

59.

d dx

# ˆ x" ‰ œ
1  C œ (
1)(
1)(x  1)

) )‰ ' ‰ ˆ sin d) œ ' ˆ csc csc )  sin ) sin ) d) œ

%


"

4(7x  2)$ (7) 28

d) œ '


# (3)

asec# (5x
1)b (5) œ sec# (5x
1)

#

" (x
1)# x# #

Š x# sin x  C‹ œ

(b) Wrong:

d dx

(
x cos x  C) œ
cos x  x sin x Á x sin x

62. (a) Wrong: (b) Right: (c) Right:

2x #

d dx

ˆ x
x 1  C‰ œ

cos x œ x sin x 

x# #

$

Š sec3 )  C‹ œ

3 sec# ) 3

ˆ "# tan# )  C‰ œ "# (2 tan )) sec# ) œ tan ) sec# ) ˆ "# sec# )  C‰ œ "# (2 sec )) sec ) tan ) œ tan ) sec# ) $

3(2x
1)# (2) 3

Š (2x
3 1)  C‹ œ

(b) Wrong:

d dx

a(2x  1)$  Cb œ 3(2x  1)# (2) œ 6(2x  1)# Á 3(2x  1)#

d dx

ax#  x  Cb

(b) Wrong:

d dx

Šax#  xb

d dx

œ 2(2x  1)# Á (2x  1)#

a(2x  1)$  Cb œ 6(2x  1)#

64. (a) Wrong:

(c) Right:

" (x
1)#

(sec ) tan )) œ sec$ ) tan ) Á tan ) sec# )

d dx

d dx

œ

cos x Á x sin x

63. (a) Wrong:

(c) Right:

(x
1)(")  x(1) (x
1)#

(
x cos x  sin x  C) œ
cos x  x sin x  cos x œ x sin x

d d) d d) d d)

sin x 

60.

d dx

(c) Right:

d) œ ' sec# ) d) œ tan )  C

‹ œ (3x  5)#

61. (a) Wrong:

d dx

" cos# )

œ (7x
2)$

 C‹ œ
Š
(3x
35) " 5

" 1sin# )

"Î#

"Î#

œ

" #

ax#  x  Cb

 C‹ œ $

" #

ax#  xb

" Œ 3 ŠÈ2x  1‹  C
œ

d dx

"Î#

"Î#

(2x  1) œ

(2x  1) œ

2x
1 2 È x#
x
C

2x
1 2 È x#
x

ˆ 3" (2x  1)$Î#  C‰ œ

3 6

Á È2x  1

Á È2x  1

(2x  1)"Î# (2) œ È2x  1

65. Graph (b), because

dy dx

œ 2B Ê y œ x#  C. Then y(1) œ 4 Ê C œ 3.

66. Graph (b), because

dy dx

œ
B Ê y œ
"# x#  C. Then y(
1) œ 1 Ê C œ

3 #

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.8 Antiderivatives

271

67.

dy dx

œ 2x
7 Ê y œ x#
7x  C; at x œ 2 and y œ 0 we have 0 œ 2#
7(2)  C Ê C œ 10 Ê y œ x#
7x  10

68.

dy dx

œ 10
x Ê y œ 10x


Ê y œ 10x
69.

dy dx

œ

" x#

x# #

 C; at x œ 0 and y œ
1 we have
1 œ 10(0)


0# #

 C Ê C œ
1


1

 x œ x#  x Ê y œ
x" 

Ê y œ
x"  70.

x# #

#

x #




" #

or y œ
x" 

#

x #

x# #

 C; at x œ 2 and y œ 1 we have 1 œ
2" 




" #

2# #

 C Ê C œ
"#

dy dx

œ 9x#
4x  5 Ê y œ 3x$
2x#  5x  C; at x œ
1 and y œ 0 we have 0 œ 3(
1)$
2(
1)#  5(
1)  C

dy dx

œ $x#Î$ Ê y œ

Ê C œ 10 Ê y œ 3x$
2x#  5x  10

71.

$x"Î$ " $

 C œ *; at x œ 9x"Î$  C; at x œ
" and y œ
& we have
& œ *(
")"Î$  C Ê C œ %

Ê y œ 9x"Î$  % " #È x

œ

" #

x"Î# Ê y œ x"Î#  C; at x œ 4 and y œ 0 we have 0 œ 4"Î#  C Ê C œ
2 Ê y œ x"Î#
2

72.

dy dx

œ

73.

ds dt

œ 1  cos t Ê s œ t  sin t  C; at t œ 0 and s œ 4 we have 4 œ 0  sin 0  C Ê C œ 4 Ê s œ t  sin t  4

74.

ds dt

œ cos t  sin t Ê s œ sin t
cos t  C; at t œ 1 and s œ 1 we have 1 œ sin 1
cos 1  C Ê C œ 0

Ê s œ sin t
cos t 75.

dr d)

œ
1 sin 1) Ê r œ cos (1))  C; at r œ 0 and ) œ 0 we have 0 œ cos (10)  C Ê C œ
" Ê r œ cos (1))
1

76.

dr d)

œ cos 1) Ê r œ

77.

dv dt

œ

78.

dv dt

œ 8t  csc# t Ê v œ 4t#
cot t  C; at v œ
7 and t œ

" #

" 1

sin(1))  C; at r œ 1 and ) œ 0 we have 1 œ

sec t tan t Ê v œ

" #

sec t  C; at v œ 1 and t œ 0 we have 1 œ 1 #

Ê v œ 4t#
cot t
7
1#

79.

d# y dx#

Ê

œ 2
6x Ê dy dx

œ 2x
3x#  C" ; at

dy dx #

" 1

dy dx

sin (10)  C Ê C œ " Ê r œ " #

" #

sec (0)  C Ê C œ

" 1

Ê vœ

sin (1))  1 " #

sec t 

" #

#

we have
7 œ 4 ˆ 1# ‰
cot ˆ 1# ‰  C Ê C œ
7
1#

œ 4 and x œ 0 we have 4 œ 2(0)
3(0)#  C" Ê C" œ 4

œ 2x
3x  4 Ê y œ x#
x$  4x  C# ; at y œ 1 and x œ 0 we have 1 œ 0#
0$  4(0)  C# Ê C# œ 1

Ê y œ x#
x$  4x  1 80.

d# y dx#

œ0 Ê

dy dx

œ C" ; at

dy dx

œ 2 and x œ 0 we have C" œ 2 Ê

dy dx

œ 2 Ê y œ 2x  C# ; at y œ 0 and x œ 0 we

have 0 œ 2(0)  C# Ê C# œ 0 Ê y œ 2x 81.

d# r dt#

œ

d# s dt#

œ

2 t$

œ 2t$ Ê

dr dt

œ
t#  C" ; at

dr dt

œ 1 and t œ 1 we have 1 œ
(1)#  C" Ê C" œ 2 Ê

dr dt

œ
t#  2

Ê r œ t"  2t  C# ; at r œ 1 and t œ 1 we have 1 œ 1"  2(1)  C# Ê C# œ
2 Ê r œ t"  2t
2 or r œ "t  2t
2 82.

3t 8

Ê

ds dt

œ

3t# 16

 C" ; at

s œ 4 and t œ 4 we have 4 œ

ds dt $

4 16

œ 3 and t œ 4 we have 3 œ

 C# Ê C# œ 0 Ê s œ

3(4)# 16

 C" Ê C" œ 0 Ê

ds dt

œ

3t# 16

$

t 16

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ê sœ

t$ 16

 C# ; at

272 83.

Chapter 4 Applications of Derivatives d$ y dx$

œ6 Ê

Ê

dy dx

d# y dx# #

œ 6x  C" ; at

œ 3x
8x  C# ; at

d$ ) dt$

œ0 Ê

#

dy dx

œ
8 and x œ 0 we have
8 œ 6(0)  C" Ê C" œ
8 Ê

œ 0 and x œ 0 we have 0 œ 3(0)#
8(0)  C# Ê C# œ 0 Ê

d# y dx# œ 6x
8 dy # dx œ 3x
8x $ #

Ê y œ x
4x  C$ ; at y œ 5 and x œ 0 we have 5 œ 0$
4(0)#  C$ Ê C$ œ 5 Ê y œ x
4x  5 84.

$

d# y dx#

d# ) dt#

d# ) dt#

œ C" ; at

d# ) d) dt# œ
2 Ê dt " d) # dt œ
2t
# Ê ) œ
t ) œ
t#
"# t  È2

œ
2 and t œ 0 we have

have
"# œ
2(0)  C# Ê C# œ
"# Ê È2 œ
0#
" (0)  C$ Ê C$ œ È2 Ê #

œ
"# and t œ 0 we
"# t  C$ ; at ) œ È2 and t œ 0 we have œ
2t  C# ; at

d) dt

85. yÐ%Ñ œ
sin t  cos t Ê ywww œ cos t  sin t  C" ; at ywww œ 7 and t œ 0 we have 7 œ cos (0)  sin (0)  C" Ê C" œ 6 Ê ywww œ cos t  sin t  6 Ê yww œ sin t
cos t  6t  C# ; at yww œ
1 and t œ 0 we have
1 œ sin (0)
cos (0)  6(0)  C# Ê C# œ 0 Ê yww œ sin t
cos t  6t Ê yw œ
cos t
sin t  3t#  C$ ; at yw œ
1 and t œ 0 we have
1 œ
cos (0)
sin (0)  3(0)#  C$ Ê C$ œ 0 Ê yw œ
cos t
sin t  3t# Ê y œ
sin t  cos t  t$  C% ; at y œ 0 and t œ 0 we have 0 œ
sin (0)  cos (0)  0$  C% Ê C% œ
1 Ê y œ
sin t  cos t  t$
1 86. yÐ%Ñ œ
cos x  8 sin (2x) Ê ywww œ
sin x
4 cos (2x)  C" ; at ywww œ 0 and x œ 0 we have 0 œ
sin (0)
% cos (2(0))  C" Ê C" œ 4 Ê ywww œ
sin x
4 cos (2x)  4 Ê yww œ cos x
2 sin (2x)  4x  C# ; at yww œ 1 and x œ 0 we have 1 œ cos (0)
2 sin (2(0))  4(0)  C# Ê C# œ 0 Ê yww œ cos x
2 sin (2x)  4x Ê yw œ sin x  cos (2x)  2x#  C$ ; at yw œ 1 and x œ 0 we have 1 œ sin (0)  cos (2(0))  2(0)#  C$ Ê C$ œ 0 Ê yw œ sin x  cos (2x)  2x# Ê y œ
cos x  "# sin (2x)  23 x$  C% ; at y œ 3 and x œ 0 we have 3 œ
cos (0) 

" #

sin (2(0))  23 (0)$  C% Ê C% œ 4 Ê y œ
cos x 

" #

sin (2x)  23 x$  4

87. m œ yw œ 3Èx œ 3x"Î# Ê y œ 2x$Î#  C; at (*ß 4) we have 4 œ 2(9)$Î#  C Ê C œ
50 Ê y œ 2x$Î#
50 88. (a)

d# y dx#

œ 6x Ê

dy dx

œ 3x#  C" ; at yw œ 0 and x œ 0 we have 0 œ 3(0)#  C" Ê C" œ 0 Ê

dy dx

œ 3x#

Ê y œ x$  C# ; at y œ 1 and x œ 0 we have C# œ 1 Ê y œ x$  1 (b) One, because any other possible function would differ from x$  1 by a constant that must be zero because of the initial conditions 89.

dy dx

œ 1
34 x"Î$ Ê y œ ' ˆ1
34 x"Î$ ‰ dx œ x
x%Î$  C; at (1ß 0.5) on the curve we have 0.5 œ 1
1%Î$  C

Ê C œ 0.5 Ê y œ x
x%Î$  90.

dy dx

œ x
1 Ê y œ ' (x
1) dx œ

Ê C œ
"# Ê y œ 91.

dy dx

" #

#

x #


x


x# #


x  C; at (
1ß 1) on the curve we have 1 œ

(")# #


(
1)  C

" #

œ sin x
cos x Ê y œ ' (sin x
cos x) dx œ
cos x
sin x  C; at (
1ß
1) on the curve we have


" œ
cos (
1)
sin (
1)  C Ê C œ
2 Ê y œ
cos x
sin x
2 92.

dy dx

œ

" #Èx

 1 sin 1x œ

" #

x"Î#  1 sin 1x Ê y œ ' ˆ #" x"Î#  sin 1x‰ dx œ x"Î#
cos 1x  C; at (1ß #) on the

curve we have 2 œ 1"Î#
cos 1(1)  C Ê C œ 0 Ê y œ Èx
cos 1x 93. (a)

ds dt

œ 9.8t
3 Ê s œ 4.9t#
3t  C; (i) at s œ 5 and t œ 0 we have C œ 5 Ê s œ 4.9t#
3t  5;

displacement œ s(3)
s(1) œ ((4.9)(9)
9  5)
(4.9
3  5) œ 33.2 units; (ii) at s œ
2 and t œ 0 we have C œ
2 Ê s œ 4.9t#
3t
2; displacement œ s(3)
s(1) œ ((4.9)(9)
9
2)
(4.9
3
2) œ 33.2 units; (iii) at s œ s! and t œ 0 we have C œ s! Ê s œ 4.9t#
3t  s! ; displacement œ s(3)
s(1) œ ((4.9)(9)
9  s! )
(4.9
3  s! ) œ 33.2 units Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 4.8 Antiderivatives

273

(b) True. Given an antiderivative f(t) of the velocity function, we know that the body's position function is s œ f(t)  C for some constant C. Therefore, the displacement from t œ a to t œ b is (f(b)  C)
(f(a)  C) œ f(b)
f(a). Thus we can find the displacement from any antiderivative f as the numerical difference f(b)
f(a) without knowing the exact values of C and s. 94. a(t) œ vw (t) œ 20 Ê v(t) œ 20t  C; at (0ß !) we have C œ 0 Ê v(t) œ 20t. When t œ 60, then v(60) œ 20(60) œ 1200 m/sec. 95. Step 1:

d# s dt#

œ
k Ê

ds dt

œ
kt  C" ; at

ds dt

œ 88 and t œ 0 we have C" œ 88 Ê

#

ds dt

œ
kt  88 Ê

#

s œ
k Š t# ‹  88t  C# ; at s œ 0 and t œ 0 we have C# œ 0 Ê s œ
kt#  88t Step 2:

ds dt

œ 0 Ê 0 œ
kt  88 Ê t œ

Step 3: 242 œ 96.

d# s dt#

œ
k Ê

Ê

ds dt

‰ k ˆ 88 k #

#

#

‰ Ê 242 œ
(88)  88 ˆ 88 k 2k 

œ '
k dt œ
kt  C; at

ds dt

œ
kt  44 Ê s œ
#

Ê s œ
kt#  44t. Then Ê
968 k 

1936 k

88 k

œ 45 Ê

kt# #

ds dt

(88)# k

Ê 242 œ

(88)# 2k

Ê k œ 16

œ 44 when t œ 0 we have 44 œ
k(0)  C Ê C œ 44 #

 44t  C" ; at s œ 0 when t œ 0 we have 0 œ
k(0) #  44(0)  C" Ê C" œ 0

ds 44 dt œ 0 Ê
kt  44 œ 0 Ê t œ k 968 968 ft k œ 45 Ê k œ 45 ¸ 21.5 sec2 .

97. (a) v œ ' a dt œ ' ˆ15t"Î#
3t"Î# ‰ dt œ 10t$Î#
6t"Î#  C;

‰ and s ˆ 44 k œ


ds dt

‰ k ˆ 44 k #

#

‰  44 ˆ 44 k œ 45

(1) œ 4 Ê 4 œ 10(1)$Î#
6(1)"Î#  C Ê C œ 0

Ê v œ 10t$Î#
6t"Î#

(b) s œ ' v dt œ ' ˆ10t$Î#
6t"Î# ‰ dt œ 4t&Î#
4t$Î#  C; s(1) œ 0 Ê 0 œ 4(1)&Î#
4(1)$Î#  C Ê C œ 0 Ê s œ 4t&Î#
4t$Î# 98.

d# s dt#

œ
5.2 Ê

ds dt

œ
5.2t  C" ; at

ds dt

œ 0 and t œ 0 we have C" œ 0 Ê

ds dt

œ
5.2t Ê s œ
2.6t#  C# ; at s œ 4

4 and t œ 0 we have C# œ 4 Ê s œ
2.6t#  4. Then s œ 0 Ê 0 œ
2.6t#  4 Ê t œ É 2.6 ¸ 1.24 sec, since t  0

99.

d# s dt#

œa Ê

ds dt

œ ' a dt œ at  C;

when t œ 0 Ê s! œ

a(0)# #

ds dt

œ v! when t œ 0 Ê C œ v! Ê

 v! (0)  C" Ê C" œ s! Ê s œ

#

at #

100. The appropriate initial value problem is: Differential Equation: s œ s! when t œ 0. Thus, Ê

ds dt

ds dt

œ '
g dt œ
gt  C" ;

œ
gt  v! . Thus s œ ' a
gt  v! b dt œ


Thus s œ
"# gt#  v! t  s!.

ds dt

œ at  v! Ê s œ

at# #

 v! t  C" ; s œ s!

 v! t  s! d# s dt#

œ
g with Initial Conditions:

ds dt (0) œ v! Ê v! œ (
g)(0)  " " # # gt  v! t  C# ; s(0) œ s! œ
#

ds dt

œ v! and

C" Ê C" œ v! (g)(0)#  v! (0)  C# Ê C# œ s!

' f(x) dx œ 1
Èx  C" œ
Èx  C (b) ' g(x) dx œ x  2  C" œ x  C (c) '
f(x) dx œ
ˆ1
Èx‰  C" œ Èx  C (d) '
g(x) dx œ
(x  2)  C" œ
x  C (e) ' [f(x)  g(x)] dx œ ˆ1
Èx‰  (x  2)  C" œ x
Èx  C (f) ' [f(x)
g(x)] dx œ ˆ1
Èx‰
(x  2)  C" œ
x
Èx  C

101. (a)

102. Yes. If F(x) and G(x) both solve the initial value problem on an interval I then they both have the same first derivative. Therefore, by Corollary 2 of the Mean Value Theorem there is a constant C such that F(x) œ G(x)  C for all x. In particular, F(x! ) œ G(x! )  C, so C œ F(x! )
G(x! ) œ 0. Hence F(x) œ G(x) for all x.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

274

Chapter 4 Applications of Derivatives

103
106 Example CAS commands: Maple: with(student): f := x -> cos(x)^2 + sin(x); ic := [x=Pi,y=1]; F := unapply( int( f(x), x ) + C, x ); eq := eval( y=F(x), ic ); solnC := solve( eq, {C} ); Y := unapply( eval( F(x), solnC ), x ); DEplot( diff(y(x),x) = f(x), y(x), x=0..2*Pi, [[y(Pi)=1]], color=black, linecolor=black, stepsize=0.05, title="Section 4.8 #103" ); Mathematica: (functions and values may vary) The following commands use the definite integral and the Fundamental Theorem of calculus to construct the solution of the initial value problems for exercises 103 - 105. Clear[x, y, yprime] yprime[x_] = Cos[x]2  Sin[x]; initxvalue = 1; inityvalue = 1; y[x_] = Integrate[yprime[t], {t, initxvalue, x}]  inityvalue If the solution satisfies the differential equation and initial condition, the following yield True yprime[x]==D[y[x], x] //Simplify y[initxvalue]==inityvalue Since exercise 106 is a second order differential equation, two integrations will be required. Clear[x, y, yprime] y2prime[x_] = 3 Exp[x/2]  1; initxval = 0; inityval = 4; inityprimeval =
1; yprime[x_] = Integrate[y2prime[t],{t, initxval, x}]  inityprimeval y[x_] = Integrate[yprime[t], {t, initxval, x}]  inityval Verify that y[x] solves the differential equation and initial condition and plot the solution (red) and its derivative (blue). y2prime[x]==D[y[x], {x, 2}]//Simplify y[initxval]==inityval yprime[initxval]==inityprimeval Plot[{y[x], yprime[x]}, {x, initxval
3, initxval  3}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,0,1]}] CHAPTER 4 PRACTICE EXERCISES 1. No, since f(x) œ x$  2x  tan x Ê f w (x) œ 3x#  2  sec# x  0 Ê f(x) is always increasing on its domain cos x 2. No, since g(x) œ csc x  2 cot x Ê gw (x) œ
csc x cot x
2 csc# x œ
sin #x


2 sin# x

œ
sin"# x (cos x  2)  0

Ê g(x) is always decreasing on its domain 3. No absolute minimum because x lim (7  x)(11
3x)"Î$ œ
_. Next f w (x) œ Ä_ (11
3x)"Î$
(7  x)(11
3x)#Î$ œ w

(11  3x)  (7
x) (11  3x)#Î$

œ

4(1  x) (11  3x)#Î$

Ê x œ 1 and x œ

11 3

are critical points.

w

Since f  0 if x  1 and f  0 if x  1, f(1) œ 16 is the absolute maximum. 4. f(x) œ

ax
b x#  1

Ê f w (x) œ

We require also that f(3) w

#a$x  "bax  $b ax #  1 b #

# a ax#  1b  2x(ax
b)
ab œ aaxax#
2bx 1 b# ax #  1 b# œ 1. Thus " œ 3a8
b Ê 3a  b œ w

" ; f w (3) œ 0 Ê
'% (*a  'b  a) œ ! Ê &a  $b œ !.

). Solving both equations yields a œ 6 and b œ
10. Now,

so that f œ


±


±  ±  ±


. Thus f w changes sign at x œ $ from
1 1 3 1/3 positive to negative so there is a local maximum at x œ $ which has a value f(3) œ 1. f (x) œ

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Chapter 4 Practice Exercises

275

5. Yes, because at each point of [!ß "Ñ except x œ 0, the function's value is a local minimum value as well as a local maximum value. At x œ 0 the function's value, 0, is not a local minimum value because each open interval around x œ 0 on the x-axis contains points to the left of 0 where f equals
1. 6. (a) The first derivative of the function f(x) œ x$ is zero at x œ 0 even though f has no local extreme value at x œ 0. (b) Theorem 2 says only that if f is differentiable and f has a local extreme at x œ c then f w (c) œ 0. It does not assert the (false) reverse implication f w (c) œ 0 Ê f has a local extreme at x œ c. 7. No, because the interval 0  x  1 fails to be closed. The Extreme Value Theorem says that if the function is continuous throughout a finite closed interval a Ÿ x Ÿ b then the existence of absolute extrema is guaranteed on that interval. 8. The absolute maximum is k
1k œ 1 and the absolute minimum is k0k œ 0. This is not inconsistent with the Extreme Value Theorem for continuous functions, which says a continuous function on a closed interval attains its extreme values on that interval. The theorem says nothing about the behavior of a continuous function on an interval which is half open and half closed, such as Ò
"ß "Ñ, so there is nothing to contradict. 9. (a) There appear to be local minima at x œ
1.75 and 1.8. Points of inflection are indicated at approximately x œ 0 and x œ „ 1.

(b) f w (x) œ x(
3x&
5x%  15x# œ x# ax#
3b ax$
5b. The pattern yw œ


±  ±  ±


±  $È ! È$
È $ & $È indicates a local maximum at x œ 5 and local minima at x œ „ È3 . (c)

10. (a) The graph does not indicate any local extremum. Points of inflection are indicated at approximately x œ
$% and x œ ".

(b) f w (x) œ x(
2x%
5 

10 x$

œ x$ ax$
2b ax(
5b . The pattern f w œ


)(  ±


±  indicates (È $È ! & #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

276

Chapter 4 Applications of Derivatives a local maximum at x œ (È5 and a local minimum at x œ $È2 .

(c)

11. (a) g(t) œ sin# t
3t Ê gw (t) œ 2 sin t cos t
3 œ sin (2t)
3 Ê gw  0 Ê g(t) is always falling and hence must decrease on every interval in its domain. (b) One, since sin# t
3t
5 œ 0 and sin# t
3t œ 5 have the same solutions: f(t) œ sin# t
3t
5 has the same derivative as g(t) in part (a) and is always decreasing with f(
3)  0 and f(0)  0. The Intermediate Value Theorem guarantees the continuous function f has a root in [
$ß 0]. 12. (a) y œ tan ) Ê

dy d)

œ sec# )  0 Ê y œ tan ) is always rising on its domain Ê y œ tan ) increases on every

interval in its domain (b) The interval  14 ß 1‘ is not in the tangent's domain because tan ) is undefined at ) œ

1 #

. Thus the tangent

need not increase on this interval. 13. (a) f(x) œ x%  2x#
2 Ê f w (x) œ 4x$  4x. Since f(0) œ
2  0, f(1) œ 1  0 and f w (x)   0 for 0 Ÿ x Ÿ 1, we may conclude from the Intermediate Value Theorem that f(x) has exactly one solution when 0 Ÿ x Ÿ 1. È (b) x# œ 2 „ 4
8  0 Ê x# œ È3
1 and x   0 Ê x ¸ È.7320508076 ¸ .8555996772 #

14. (a) y œ

x x
1

Ê yw œ

" (x
1)#

 0, for all x in the domain of

x x
1

Ê yœ

x x
1

is increasing in every interval in

its domain (b) y œ x$  2x Ê yw œ 3x#  2  0 for all x Ê the graph of y œ x$  2x is always increasing and can never have a local maximum or minimum 15. Let V(t) represent the volume of the water in the reservoir at time t, in minutes, let V(0) œ a! be the initial amount and V(1440) œ a!  (1400)(43,560)(7.48) gallons be the amount of water contained in the reservoir after the rain, where 24 hr œ 1440 min. Assume that V(t) is continuous on [!ß 1440] and differentiable on (!ß 1440). The Mean Value Theorem says that for some t! in (!ß 1440) we have Vw (t! ) œ œ

a!
(1400)(43,560)(7.48)  a! 1440

œ

456,160,320 gal 1440 min

V(1440)  V(0) 1440  0

œ 316,778 gal/min. Therefore at t! the reservoir's volume

was increasing at a rate in excess of 225,000 gal/min. 16. Yes, all differentiable functions g(x) having 3 as a derivative differ by only a constant. Consequently, the d difference 3x
g(x) is a constant K because gw (x) œ 3 œ dx (3x). Thus g(x) œ 3x  K, the same form as F(x). x 1 x 1 x
1 œ 1  x
1 Ê x
1 differs from x
1 (x
1)  x(1) d ˆ x ‰ d ˆ " ‰ œ (x
" 1)# œ dx dx x
1 œ (x
1)# x
1 .

17. No,

18. f w (x) œ gw (x) œ

2x ax #
1 b #

by the constant 1. Both functions have the same derivative

Ê f(x)
g(x) œ C for some constant C Ê the graphs differ by a vertical shift.

19. The global minimum value of

" #

occurs at x œ #.

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Chapter 4 Practice Exercises

277

20. (a) The function is increasing on the intervals Ò
$ß
#Ó and Ò"ß #Ó. (b) The function is decreasing on the intervals Ò
#ß !Ñ and Ð!ß "Ó. (c) The local maximum values occur only at x œ
#, and at x œ #; local minimum values occur at x œ
$ and at x œ " provided f is continuous at x œ !. 21. (a) t œ 0, 6, 12

(b) t œ 3, 9

(c) 6  t  12

(d) 0  t  6, 12  t  14

22. (a) t œ 4

(b) at no time

(c) 0  t  4

(d) 4  t  8

23.

24.

25.

26.

27.

28.

29.

30.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

278

Chapter 4 Applications of Derivatives

31.

32.

33. (a) yw œ 16
x# Ê yw œ


±  ±


Ê the curve is rising on (
%ß %), falling on (
_ß
4) and (%ß _)
% % Ê a local maximum at x œ 4 and a local minimum at x œ
4; yww œ
2x Ê yww œ  ±


Ê the curve ! is concave up on (
_ß !), concave down on (!ß _) Ê a point of inflection at x œ 0 (b)

34. (a) yw œ x#
x
6 œ (x
$)(x  2) Ê yw œ  ±


±  Ê the curve is rising on (
_ß
2) and ($ß _),
# $ falling on (
#ß $) Ê local maximum at x œ
2 and a local minimum at x œ 3; yww œ 2x
1 Ê yww œ


±  Ê concave up on ˆ "# ß _‰ , concave down on ˆ
_ß "# ‰ Ê a point of inflection at x œ "# "Î# (b)

35. (a) yw œ 6x(x  1)(x
2) œ 6x$
6x#
12x Ê yw œ


±  ±


±  Ê the graph is rising on (
"ß !)
" ! # and (#ß _), falling on (
_ß
1) and (!ß #) Ê a local maximum at x œ 0, local minima at x œ
1 and x œ 2; yww œ 18x#
12x
12 œ 6 a3x#
2x
2b œ 6 Šx
yww œ  ± on




±

"È( $ È È Š 1 3 7 ß 1
3 7 ‹

"
È( $

1  È7 3 ‹ Šx

 Ê the curve is concave up on Š
_ß

Ê points of inflection at x œ




1
È7 3 ‹

1  È7 3 ‹

Ê È7

and Š 1
3

1 „ È7 3

(b)

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ß _‹ , concave down

Chapter 4 Practice Exercises

279

36. (a) yw œ x# (6
4x) œ 6x#
4x$ Ê yw œ  ±  ±


Ê the curve is rising on ˆ
_ß #3 ‰, falling on ˆ #3 ß _‰ ! $Î# 3 ww Ê a local maximum at x œ # ; y œ 12x
12x# œ 12x("
x) Ê yww œ


±  ±


Ê concave up on ! " (!ß "), concave down on (
_ß !) and ("ß _) Ê points of inflection at x œ 0 and x œ 1 (b)

37. (a) yw œ x%
2x# œ x# ax#
2b Ê yw œ  ±


±


±  Ê the curve is rising on Š
_ß
È2‹ and ! È#
È # ŠÈ2ß _‹ , falling on Š
È2ß È2‹ Ê a local maximum at x œ
È2 and a local minimum at x œ È2 ; yww œ 4x$
4x œ 4x(x
1)(x  1) Ê yww œ


±  ±


±  Ê concave up on (
"ß 0) and ("ß _),
" ! " concave down on (
_ß
1) and (0ß 1) Ê points of inflection at x œ 0 and x œ „ 1 (b)

38. (a) yw œ 4x#
x% œ x# a4
x# b Ê yw œ


±  ±  ±


Ê the curve is rising on (
2ß 0) and (0ß 2),
# ! # falling on (
_ß
2) and (#ß _) Ê a local maximum at x œ 2, a local minimum at x œ
2; yww œ 8x
4x$ œ 4x a2
x# b Ê yww œ  ±


±  ±


Ê concave up on Š
_ß
È2‹ and Š0ß È2‹ , concave ! È#
È # down on Š
È2ß 0‹ and ŠÈ2ß _‹ Ê points of inflection at x œ 0 and x œ „ È2 (b)

39. The values of the first derivative indicate that the curve is rising on (!ß _) and falling on (
_ß 0). The slope of the curve approaches
_ as x Ä ! , and approaches _ as x Ä 0
and x Ä 1. The curve should therefore have a cusp and local minimum at x œ 0, and a vertical tangent at x œ 1.

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280

Chapter 4 Applications of Derivatives

40. The values of the first derivative indicate that the curve is rising on ˆ!ß "# ‰ and ("ß _), and falling on (
_ß !) and ˆ "# ß "‰ . The derivative changes from positive to negative at x œ "# , indicating a local maximum there. The slope of the curve approaches
_ as x Ä 0
and x Ä 1
, and approaches _ as x Ä 0 and as x Ä 1 , indicating cusps and local minima at both x œ 0 and x œ 1.

41. The values of the first derivative indicate that the curve is always rising. The slope of the curve approaches _ as x Ä 0 and as x Ä 1, indicating vertical tangents at both x œ 0 and x œ 1.

È33

42. The graph of the first derivative indicates that the curve is rising on Š!ß 17
16 on (
_ß !) and xœ

17  È33 16

È È Š 17
16 33 ß 17 16 33 ‹

Ê a local maximum at x œ

17
È33 16

È33

‹ and Š 17 16

ß _‹ , falling

, a local minimum at

. The derivative approaches
_ as x Ä 0
and x Ä 1, and approaches _ as x Ä 0 ,

indicating a cusp and local minimum at x œ 0 and a vertical tangent at x œ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 4 Practice Exercises

43. y œ

x1 x
3

45. y œ

x#  1 x

œ1

œx

4 x
3

" x

44. y œ

2x x5

œ2


46. y œ

x#
x  1 x

10 x5

œx
1

" x

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281

282

Chapter 4 Applications of Derivatives

47. y œ

x$  2 #x

œ

49. y œ

x#
4 x#
3

œ1


51. lim

xÄ"

52. lim

54. lim

tan x x

œ

tan x

sin# x

sinamxb

x Ä 1Î#c

58.

xÄ"

axa
" b
" x Ä " bx

x Ä ! sinanxb

57. lim

œ lim

œ lim

# x Ä ! tanax b

56. lim

" x

" x#
3

xa
"

x Ä ! x  sin x

55. lim



x #  $x
% x
"

b x Ä " x
"

53. xlim Ä1

x# #

tan 1 1

#x  $ "

œ

x%
1 x#

œ x#


50. y œ

x# x#
4

œ1

" x#

4 x#
4

œ&

a b

œ!

œ lim

sec# x

x Ä ! "  cos x

œ lim

#sin x†cos x

# # x Ä ! #x sec ax b

œ lim

xÄ!

m cosamxb n cosanxb

œ

" ""

œ

" # sina#xb

œ lim

# # x Ä ! #x sec ax b

œ

cosa$xb

x Ä 1Î#c cosa(xb Èx cos x

59. lim acsc x
cot xb œ lim

xÄ!

#cosa#xb

œ lim

# # # # # x Ä ! #x a#sec ax btanax b†#xb  #sec ax b

œ

# !  #†"

œ"

m n

seca(xbcosa$xb œ lim

lim Èx sec x œ lim b x Ä !b xÄ! xÄ!

48. y œ

œ

! "


$sina$xb

œ lim

x Ä 1Î#c
(sina(xb

œ

$ (

œ!

"
cos x sin x

œ lim

sin x

x Ä ! cos x

œ

! "

œ!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 4 Practice Exercises 60. lim ˆ x"%
xÄ!

61.

#

œ lim Š "
x%x ‹ œ lim a"
x# b † xÄ!

xÄ!

" x%

œ lim a"
x# b œ lim

ŠÈx#  x  "
Èx#
x‹ œ lim ŠÈx#  x  "
Èx#
x‹ † xÄ_ œ lim È # #x  " È # xÄ_ x x" x
x Notice that x œ Èx# for x  ! so this is equivalent to lim

#x
" x # x x

"
É x #  x É # x x# $

œ lim xÄ_

$

lim Š x#x
"
x#x " ‹ œ lim xÄ_ xÄ_ "# " œ lim #% œ lim œ! x xÄ_ x Ä _ #x

"

% xÄ! x

xÄ!

xÄ_

œ lim xÄ_

62.

"‰ x#

œ"†_œ_

Èx#  x  "Èx#
x È x#  x  "  È x#
x

#
x" œ È"# È" œ " " É"
x
x"#
É"  x"

x $ a x #  "b
x $ a x #
" b ax#
"bax#  "b

œ lim xÄ_

#x $ x%
"

œ lim xÄ_

'x# %x $

œ lim xÄ_

"#x "#x#

63. (a) Maximize f(x) œ Èx
È36
x œ x"Î#
(36
x)"Î# where 0 Ÿ x Ÿ 36 Ê f w (x) œ

" #

x
"Î#
"# (36
x)
"Î# (
1) œ

È36
x  Èx #Èx È36
x

Ê derivative fails to exist at 0 and 36; f(0) œ
6,

and f(36) œ 6 Ê the numbers are 0 and 36 (b) Maximize g(x) œ Èx  È36
x œ x"Î#  (36
x)"Î# where 0 Ÿ x Ÿ 36 Ê gw (x) œ

" #

x
"Î#  "# (36
x)
"Î# (
1) œ

È36
x
Èx #Èx È36
x

Ê critical points at 0, 18 and 36; g(0) œ 6,

g(18) œ 2È18 œ 6È2 and g(36) œ 6 Ê the numbers are 18 and 18 64. (a) Maximize f(x) œ Èx (20
x) œ 20x"Î#
x$Î# where 0 Ÿ x Ÿ 20 Ê f w (x) œ 10x
"Î#
3# x"Î# œ

20
3x #È x

œ 0 Ê x œ 0 and x œ

œ

40È20 3È 3

Ê the numbers are

20 3

20 3

‰ É 20 ˆ are critical points; f(0) œ f(20) œ 0 and f ˆ 20 3 œ 3 20


and

40 3

.

(b) Maximize g(x) œ x  È20
x œ x  (20
x)"Î# where 0 Ÿ x Ÿ 20 Ê gw (x) œ Ê È20
x œ

" #

Ê xœ

the numbers must be 65. A(x) œ

" #

79 4

and

79 4 . " 4 .

The critical points are x œ

79 4

2È20
x
1 #È20
x

‰ and x œ 20. Since g ˆ 79 4 œ

(2x) a27
x# b for 0 Ÿ x Ÿ È27

Ê Aw (x) œ 3(3  x)(3
x) and Aw w (x) œ
6x. The critical points are
3 and 3, but
3 is not in the domain. Since Aw w (3) œ
18  0 and A ŠÈ27‹ œ 0, the maximum occurs at x œ 3 Ê the largest area is A(3) œ 54 sq units. 66. The volume is V œ x# h œ 32 Ê h œ 32 x# . The 32 ‰ # ˆ surface area is S(x) œ x  4x x# œ x#  128 x , where x  0 Ê Sw (x) œ

20 ‰ 3

2(x
4) ax#  4x  16b x#

Ê the critical points are 0 and 4, but 0 is not in the domain. Now Sw w (4) œ 2  256 4$  0 Ê at x œ 4 there is a minimum. The dimensions 4 ft by 4 ft by 2 ft minimize the surface area.

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81 4

œ0 and g(20) œ 20,

283

284

Chapter 4 Applications of Derivatives #

67. From the diagram we have ˆ h# ‰  r# œ ŠÈ3‹ Ê r# œ

12
h# 4

#

. The volume of the cylinder is #

V œ 1r# h œ 1 Š 12
4 h ‹ h œ

1 4

0 Ÿ h Ÿ 2È3 . Then Vw (h) œ

a12h
h$ b , where 31 4

(2  h)(2
h)

Ê the critical points are
2 and 2, but
2 is not in the domain. At h œ 2 there is a maximum since Vw w (2) œ
31  0. The dimensions of the largest cylinder are radius œ È2 and height œ 2. 68. From the diagram we have x œ radius and y œ height œ 12
2x and V(x) œ "3 1x# (12
2x), where

0 Ÿ x Ÿ 6 Ê Vw (x) œ 21x(4
x) and Vw w (4) œ
81. The critical points are 0 and 4; V(0) œ V(6) œ 0 Ê x œ 4 gives the maximum. Thus the values of r œ 4 and h œ 4 yield the largest volume for the smaller cone.

‰ , where p is the profit on grade B tires and 0 Ÿ x Ÿ 4. Thus 69. The profit P œ 2px  py œ 2px  p ˆ 405

10x x Pw (x) œ

2p (5
x)#

ax#
10x  20b Ê the critical points are Š5
È5‹, 5, and Š5  È5‹ , but only Š5
È5‹ is in

the domain. Now Pw (x)  0 for 0  x  Š5
È5‹ and Pw (x)  0 for Š5
È5‹  x  4 Ê at x œ Š5
È5‹ there is a local maximum. Also P(0) œ 8p, P Š5
È5‹ œ 4p Š5
È5‹ ¸ 11p, and P(4) œ 8p Ê at x œ Š5
È5‹ there is an absolute maximum. The maximum occurs when x œ Š5
È5‹ and y œ 2 Š5
È5‹ , the units are hundreds of tires, i.e., x ¸ 276 tires and y ¸ 553 tires. 70. (a) The distance between the particles is lfatbl where fatb œ
cos t  cosˆt  1% ‰. Then, f w atb œ sin t
sinˆt  1% ‰. Solving f w atb œ ! graphically, we obtain t ¸ "Þ"(), t ¸ %Þ$#!, and so on.

Alternatively, f w atb œ ! may be solved analytically as follows. f w atb œ sin’ˆt  1) ‰
1) “
sin’ˆt  1) ‰  1) “ œ ’sinˆt  1) ‰cos 1)
cosˆt  1) ‰sin 1) “
’sinˆt  1) ‰cos 1)  cosˆt  1) ‰sin 1) “ œ
#sin 1) cosˆt  1) ‰ so the critical points occur when cosˆt  1) ‰ œ !, or t œ

$1 )

 k1. At each of these values, fatb œ „ cos $)1

¸ „ !Þ('& units, so the maximum distance between the particles is !Þ('& units.

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Chapter 4 Practice Exercises

285

(b) Solving cos t œ cos ˆt  1% ‰ graphically, we obtain t ¸ #Þ(%*, t ¸ &Þ)*!, and so on.

Alternatively, this problem can be solved analytically as follows. cos t œ cos ˆt  1% ‰ cos’ˆt  1) ‰
1) “ œ cos’ˆt  1) ‰  1) “ cosˆt  1) ‰cos 1)  sinˆt  1) ‰sin 1) œ cosˆt  1) ‰cos 1)
sinˆt  1) ‰sin 1) #sin ˆt  1) ‰sin 1) œ ! sin ˆt  1) ‰ œ ! tœ The particles collide when t œ

(1 )

(1 )

 k1

¸ #Þ(%*. (plus multiples of 1 if they keep going.)

71. The dimensions will be x in. by "!
#x in. by "'
#x in., so Vaxb œ xa"!
#xba"'
#xb œ %x$
&#x#  "'!x for !  x  &. Then Vw axb œ "#x#
"!%x  "'! œ %ax
#ba$x
#!b , so the critical point in the correct domain is x œ #. This critical point corresponds to the maximum possible volume because Vw axb  ! for !  x  # and Vw axb  ! for 2  x  &. The box of largest volume has a height of 2 in. and a base measuring 6 in. by 12 in., and its volume is 144 in.$ Graphical support:

72. The length of the ladder is d"  d# œ 8 sec )  6 csc ). We wish to maximize I()) œ 8 sec )  6 csc ) Ê Iw ()) œ 8 sec ) tan )
6 csc ) cot ). Then Iw ()) œ 0 Ê 8 sin$ )
6 cos$ ) œ 0 Ê tan ) œ

$ È 6

#

Ê

d" œ 4 É4  $È36 and d# œ $È36 É4  $È36 Ê the length of the ladder is about Š4  $È36‹ É4  $È36 œ Š4  $È36‹

$Î#

¸ "*Þ( ft.

73. g(x) œ 3x
x$  4 Ê g(2) œ 2  0 and g(3) œ
14  0 Ê g(x) œ 0 in the interval [#ß 3] by the Intermediate Value Theorem. Then gw (x) œ 3
3x# Ê xnb1 œ xn


3xn  x$n
4 33xn#

; x! œ 2 Ê x" œ 2.22 Ê x# œ 2.196215, and

so forth to x& œ 2.195823345.

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286

Chapter 4 Applications of Derivatives

74. g(x) œ x%
x$
75 Ê g(3) œ
21  0 and g(4) œ 117  0 Ê g(x) œ 0 in the interval [$ß %] by the Intermediate Value Theorem. Then gw (x) œ 4x$
3x# Ê xnb1 œ xn


x%n  x$n  75 4xn$  3xn#

; x! œ 3 Ê x" œ 3.259259

Ê x# œ 3.229050, and so forth to x& œ 3.22857729. 75.

' ax$  5x
7b dx œ

76.

' Š8t$
t#  t‹ dt œ 8t4%
t6$  t##  C œ 2t%
t6$  t##  C

77.

' ˆ3Èt  t4# ‰ dt œ ' ˆ3t"Î#  4t
# ‰ dt œ 3t$Î#  4t

"1  C œ 2t$Î#
4t  C

78.

' Š #È" t
t3% ‹ dt œ ' ˆ #" t
"Î#
3t
% ‰ dt œ #" Œ t"Î#

(3t

$3)  C œ Èt  t"$  C

x% 4



5x# #


7x  C

#

Š 3# ‹

" #

79. Let u œ r  5 Ê du œ dr

' ar dr5b

œ'

#

du u#

u
"
1

œ ' u
# du œ

 C œ
u
"  C œ
ar " 5b  C

80. Let u œ r
È2 Ê du œ dr

'

6 dr

$

Šr
È2‹

œ 6'

dr

$

Šr
È2‹

œ 6'

du u$

81. Let u œ )#  1 Ê du œ 2) d) Ê

'

" #

3)È)#  1 d) œ ' Èu ˆ #3 du‰ œ

82. Let u œ 7  )2 Ê du œ 2) d) Ê

'È)

d) œ '

7 ) 2

" Èu

ˆ #" du‰ œ

" #

x$ a 1  x % b


"Î%

#

C

du œ ) d) 3 #

" #

3

Šr
È2‹

$Î# $Î# ' u"Î# du œ 3# Œ u$Î#  C œ a ) #  1b  C 3
 C œ u #

du œ ) d)

"Î# ' u
"Î# du œ #" Œ u"Î#  C œ È7  )2  C "
 C œ u #

83. Let u œ 1  x% Ê du œ 4x$ dx Ê

'


#

œ 6' u
$ du œ 6 Š u
# ‹  C œ
3u
#  C œ


" 4

du œ x$ dx

dx œ ' u
"Î% ˆ "4 du‰ œ

" 4

$Î% " $Î% ' u
"Î% du œ 4" Œ u$Î%  C œ 3" a1  x% b  C 3
 C œ 3 u 4

84. Let u œ 2
x Ê du œ
dx Ê
du œ dx

' (2
x)$Î& dx œ ' u$Î& (
du) œ
' u$Î& du œ
u

)Î&

Š 85 ‹

85. Let u œ

'

s 10

sec# 10s

Ê du œ

" 10

 C œ
58 u)Î&  C œ
58 (2
x))Î&  C

ds Ê 10 du œ ds

ds œ ' asec# ub (10 du) œ 10 ' sec# u du œ 10 tan u  C œ 10 tan

86. Let u œ 1s Ê du œ 1 ds Ê

" 1

s 10

C

du œ ds

' csc# 1s ds œ ' acsc# ub ˆ 1" du‰ œ 1" ' csc# u du œ
1" cot u  C œ
1" cot 1s  C

87. Let u œ È2 ) Ê du œ È2 d) Ê

' csc È2) cot È2) d) œ '

" È2

du œ d)

(csc u cot u) Š È"2 du‹ œ

" È2

(
csc u)  C œ
È"2 csc È2)  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 4 Additional and Advanced Exercises ) 3

88. Let u œ

'

sec

) 3

tan

89. Let u œ

'

Ê du œ

x 4

) 3

" 3

d) Ê 3 du œ d)

d) œ ' (sec u tan u)(3 du) œ 3 sec u  C œ 3 sec

Ê du œ

" 4

) 3

C

dx Ê 4 du œ dx

2u ‰ dx œ ' asin# ub (4 du) œ ' 4 ˆ 1
cos du œ 2' (1
cos 2u) du œ 2 ˆu
# œ 2u
sin 2u  C œ 2 ˆ x4 ‰
sin 2 ˆ x4 ‰  C œ x#
sin x#  C

sin#

x 4

90. Let u œ

'

cos#

œ

x #



91. y œ '

x # x # " #

Ê du œ

" #

2u ‰ dx œ ' acos# ub (2 du) œ ' 2 ˆ 1  cos du œ ' (1  cos 2u) du œ u  #

x#  " x#

dx œ ' a1  x
# b dx œ x
x
"  C œ x


y œ 1 when x œ 1 Ê

" x

œ ' Š15Èt 

3 Èt ‹

" 3

2


1 1

"‰ x#

Ê

" x

 C; y œ
1 when x œ 1 Ê 1


dx œ ' ax#  2  x
# b dx œ

 C œ 1 Ê C œ
3" Ê y œ

$

x 3

x$ 3

 2x
x
"  C œ

 2x


dt œ ' ˆ15t"Î#  3t
"Î# ‰ dt œ 10t$Î#  6t"Î#  C; dr dt œ &Î#

œ 4t&Î#  4t$Î#
8t  C; r œ 0 when t œ 1 Ê 4(1) r œ 4t&Î#  4t$Î#
8t d# r dt#

C

dr dt

" x




 C œ
1

x$ 3

 2x


" x

 C;

" 3

œ 8 when t œ 1

10t$Î#  6t"Î#
8 Ê r œ ' ˆ10t$Î#  6t"Î#
8‰ dt

 4(1)$Î#
8(1)  C" œ 0 Ê C" œ 0. Therefore,

œ '
cos t dt œ
sin t  C; rw w œ 0 when t œ 0 Ê
sin 0  C œ 0 Ê C œ 0. Thus, dr dt

1 1


1

Ê 10(1)$Î#  6(1)"Î#  C œ 8 Ê C œ
8. Thus

94.

sin 2u #

sin x  C

# 92. y œ ' ˆx  x" ‰ dx œ ' ˆx#  2 

dr dt

C

dx Ê 2 du œ dx

Ê C œ
1 Ê y œ x


93.

sin 2u ‰ #

œ '
sin t dt œ cos t  C" ; rw œ 0 when t œ 0 Ê 1  C" œ 0 Ê C" œ
1. Then

d# r dt# œ
sin t dr dt œ cos t


1

Ê r œ ' (cos t
1) dt œ sin t
t  C# ; r œ
1 when t œ 0 Ê 0
0  C# œ
1 Ê C# œ
1. Therefore, r œ sin t
t
1 CHAPTER 4 ADDITIONAL AND ADVANCED EXERCISES 1. If M and m are the maximum and minimum values, respectively, then m Ÿ f(x) Ÿ M for all x − I. If m œ M then f is constant on I. 3x  6,
2 Ÿ x  0 has an absolute minimum value of 0 at x œ
2 and an absolute 9
x# , 0 Ÿ x Ÿ 2 maximum value of 9 at x œ 0, but it is discontinuous at x œ 0.

2. No, the function f(x) œ œ

3. On an open interval the extreme values of a continuous function (if any) must occur at an interior critical point. On a half-open interval the extreme values of a continuous function may be at a critical point or at the closed endpoint. Extreme values occur only where f w œ 0, f w does not exist, or at the endpoints of the interval. Thus the extreme points will not be at the ends of an open interval. 4. The pattern f w œ  ±



±



±  ±  indicates a local maximum at x œ 1 and a local " # $ % minimum at x œ 3.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

287

288

Chapter 4 Applications of Derivatives

5. (a) If yw œ 6(x  1)(x
2)# , then yw  0 for x 
1 and yw  0 for x 
1. The sign pattern is f w œ


±  ±  Ê f has a local minimum at x œ
1. Also yww œ 6(x
2)#  12(x  1)(x
2)
" # œ 6(x
2)(3x) Ê yw w  0 for x  0 or x  2, while yww  0 for 0  x  2. Therefore f has points of inflection at x œ 0 and x œ 2. There is no local maximum. (b) If yw œ 6x(x  1)(x
2), then yw  0 for x 
1 and 0  x  2; yw  0 for
"  x  0 and x  2. The sign sign pattern is yw œ


±  ±


±  . Therefore f has a local maximum at x œ 0 and
" ! # È7

local minima at x œ
1 and x œ 2. Also, yww œ ") ’x
Š 1
$ 1
È 7 $

x

1 È 7 $

È7

‹“ ’x
Š 1 $

‹“ , so yww  0 for

and yww  0 for all other x Ê f has points of inflection at x œ

6. The Mean Value Theorem indicates that

f(6)
f(0) 6
0

1 „È 7 $

.

œ f w (c) Ÿ 2 for some c in (0ß 6). Then f(6)
f(0) Ÿ 12

indicates the most that f can increase is 12. 7. If f is continuous on [aß c) and f w (x) Ÿ 0 on [aß c), then by the Mean Value Theorem for all x − [aß c) we have f(c)
f(x) c
x

Ÿ 0 Ê f(c)
f(x) Ÿ 0 Ê f(x)   f(c). Also if f is continuous on (cß b] and f w (x)   0 on (cß b], then for f(x)
f(c) x
c

all x − (cß b] we have

  0 Ê f(x)
f(c)   0 Ê f(x)   f(c). Therefore f(x)   f(c) for all x − [aß b].

8. (a) For all x,
(x  1)# Ÿ 0 Ÿ (x
1)# Ê
a1  x# b Ÿ 2x Ÿ a1  x# b Ê
"# Ÿ (b) There exists c − (aß b) such that Ê kf(b)
f(a)k Ÿ

" #

c 1  c#

œ

f(b)
f(a) b
a


f(a) ¸ c ¸ Ê ¹ f(b)b
a ¹ œ 1  c# Ÿ

" #

x 1  x#

Ÿ

" #

.

, from part (a)

kb
ak .

9. No. Corollary 1 requires that f w (x) œ 0 for all x in some interval I, not f w (x) œ 0 at a single point in I. 10. (a) h(x) œ f(x)g(x) Ê hw (x) œ f w (x)g(x)  f(x)gw (x) which changes signs at x œ a since f w (x), gw (x)  0 when x  a, f w (x), gw (x)  0 when x  a and f(x), g(x)  0 for all x. Therefore h(x) does have a local maximum at x œ a. (b) No, let f(x) œ g(x) œ x$ which have points of inflection at x œ 0, but h(x) œ x' has no point of inflection (it has a local minimum at x œ 0). 11. From (ii), f(
1) œ lim

xÄ „_

f(x) œ

12.

dy dx

œ 0 Ê a œ 1; from (iii), either 1 œ x lim f(x) or 1 œ x Ä lim f(x). In either case, Ä_
_

x" # x Ä „ _ bx  cx  #

1
"x x
c
2x xÄ „_

lim


"  a b
c#

œ ! and if c œ 0,

œ 3x#  2kx  3 œ 0 Ê x œ

1
"x bx
c
2x xÄ „_ 1
x" then lim bx
2x xÄ „_

œ

lim

lim


2k „ È4k#
36 6

œ " Ê b œ 0 and c œ ". For if b œ ", then œ

lim

xÄ „_

1
x" 2 x

œ „ _. Thus a œ 1, b œ 0, and c œ 1.

Ê x has only one value when 4k#
36 œ 0 Ê k# œ 9 or

k œ „ 3. 13. The area of the ?ABC is A(x) œ w

where 0 Ÿ x Ÿ 1. Thus A (x) œ

" #

(2) È1
x# œ a1
x# b


x È 1
x#

"Î#

,

Ê 0 and „ 1 are

critical points. Also A a „ 1b œ 0 so A(0) œ 1 is the maximum. When x œ 0 the ?ABC is isosceles since AC œ BC œ È2 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 4 Additional and Advanced Exercises f (c  h)
f (c) œ f ww (c) Ê for % œ "# kf ww (c)k  0 h hÄ0 Ê ¹ f (ch)h
f (c)
f ww (c)¹  "# kf ww (c)k . Then f w (c) œ w

14. lim

w

w

w

3 #

f ww (c) 

0 Ê
"# kf ww (c)k 

f (c  h)  f ww (c)  "# kf ww (c)k . If f ww (c)  0, then h f (c  h)  "# f ww (c)  0; likewise if f ww (c)  0, then 0  "# h w

Ê f ww (c)
"# kf ww (c)k  Ê

there exists a $  0 such that 0  khk  $

w

w

f (c  h) h

" #


f ww (c) 

w

kf ww (c)k

kf ww (c)k œ
f ww (c) f ww (c) 

f (c  h) h w



3 #

f ww (c).

(a) If f ww (c)  0, then
$  h  0 Ê f (c  h)  0 and 0  h  $ Ê f w (c  h)  0. Therefore, f(c) is a local maximum. (b) If f ww (c)  0, then
$  h  0 Ê f w (c  h)  0 and 0  h  $ Ê f w (c  h)  0. Therefore, f(c) is a local minimum. 15. The time it would take the water to hit the ground from height y is É 2y g , where g is the acceleration of gravity. The product of time and exit velocity (rate) yields the distance the water travels: È64(h
y) œ 8 É 2 ahy
y# b D(y) œ É 2y g g are critical points. Now D(0) œ 0,

D ˆ h# ‰

16. From the figure in the text, tan ("  )) œ give

ba h

œ

tan "
1  ha tan " a h

œ

h tan "  a h
a tan "

œ

"Î#

, 0 Ÿ y Ÿ h Ê Dw (y) œ
4 É g2 ahy
y# b

8h Èg

ba h ;


"Î#

(h
2y) Ê 0,

and D(h) œ 0 Ê the best place to drill the hole is at y œ

tan ("  )) œ

tan "  tan ) 1
tan " tan )

. Solving for tan " gives tan " œ

; and tan ) œ

bh h#  a(b  a)

a h

h #

h #

and h

.

. These equations

or

#

ah
a(b  a)b tan " œ bh. Differentiating both sides with respect to h gives 2h tan "  ah#  a(b  a)b sec# "

d" dh

œ b. Then

d" dh

bh œ 0 Ê 2h tan " œ b Ê 2h Š h#  a(b  a) ‹ œ b

Ê 2bh# œ bh#  ab(b  a) Ê h# œ a(b  a) Ê h œ Èa(a  b) . 17. The surface area of the cylinder is S œ 21r#  21rh. From the diagram we have Rr œ H H
h Ê h œ RH R
rH and S(r) œ 21r(r  h) œ 21r ˆr  H
r HR ‰ œ 21 ˆ1
HR ‰ r#  21Hr, where 0 Ÿ r Ÿ R.

Case 1: H  R Ê S(r) is a quadratic equation containing the origin and concave upward Ê S(r) is maximum at r œ R. Case 2: H œ R Ê S(r) is a linear equation containing the origin with a positive slope Ê S(r) is maximum at r œ R. Case 3: H  R Ê S(r) is a quadratic equation containing the origin and concave downward. Then dS H‰ dS H‰ RH ˆ ˆ dr œ 41 1
R r  21H and dr œ 0 Ê 41 1
R r  21H œ 0 Ê r œ 2(H
R) . For simplification we let r‡ œ

RH 2(H
R)

.

(a) If R  H  2R, then 0  H
2R Ê H  2(H
R) Ê (b) If H œ 2R, then r‡ œ

#

2R 2R

RH 2(H
R)

 R which is impossible.

œ R Ê S(r) is maximum at r œ R.

(c) If H  2R, then 2R  H  2H Ê H  2(H
R) Ê S(r) is a maximum at r œ r‡ œ

RH 2(H
R)

H 2(H
R)

1 Ê

RH 2(H
R)

 R Ê r‡  R. Therefore,

.

Conclusion: If H − (0ß R] or H œ 2R, then the maximum surface area is at r œ R. If H − (Rß 2R), then r  R which is not possible. If H − (2Rß _), then the maximum is at r œ r‡ œ 2(HRH
R) . 18. f(x) œ mx
1 

" x

Ê f w (x) œ m


" x#

and f w w (x) œ

2 x$

 0 when x  0. Then f w (x) œ 0 Ê x œ

minimum. If f Š È"m ‹   0, then Èm
1  Èm œ 2Èm
1   0 Ê m  

" 4

" Èm

yields a

. Thus the smallest acceptable value

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

289

290

Chapter 4 Applications of Derivatives

for m is 19. (a) (b) (c)

" 4

.

lim

xÄ!

#sina&xb $x

œ lim

xÄ!

x x Ä ! sin# È#x

lim

x Ä 1/2

"

xÄ!

(g) (h)

cos$ x
#

sinax# b

lim x Ä ! xsin x lim

xÄ!

x$
)

20. (a) x lim Ä_ (b) x lim Ä_

"! $


$sina&xbsina$xb  &cosa&xbcosa$xb $cosa$xb

xÄ!

È #x

" œ lim #sinÈ#x cosÈ#x xÄ! È#x

xÄ!

sinŠ#È#x‹

œ

& $ "

È#x

œ lim

x Ä ! cosŠ#È#x‹ È# #x

"
sin x cos x

lim "
cos#x x Ä ! "
sec x

œ


cos x

œ lim

œ !Þ

x Ä 1/2
sin x

lim "
cos# x x Ä !
tan x

œ

lim cos x#
" x Ä ! tan x

œ lim


sin x

# x Ä ! # tan x sec x

œ lim

sin x

xÄ!

# sin x cos$ x

œ

œ
#" #x cosax# b

œ lim

x Ä ! xcos xsin x

sec x
" x#

lim # x Ä # x
%

†"œ

" #

œ

x Ä 1/2

œ

"! $

œ

œ lim

œ lim

asec x
tan xb œ lim

lim x
sin x x Ä ! x
tan x œ lim

(f)

sina&xbcosa$xb sina$xb

lim x csc# È#x œ lim x Ä ! cosŠ#È#x‹†#

(e)

xÄ! $

xÄ!

xÄ!

"! sina&xb a&xb

œ lim

lim sina&xbcota$xb œ lim

xÄ!

œ lim (d)

#sina&xb $ & a &x b

œ lim

xÄ!

œ lim

xÄ#

sec x tan x #x

xÄ!

ax
#bax#  #x  %b ax
#bax  #b

œ x lim Ä_

#

sec x  tan x sec x #

xÄ!

x #  #x  % x#

œ lim

xÄ#

œ x lim œ x lim Ä _ Èx
& Ä_

#x x  (È x

$

œ lim

Èx
& Èx Èx

Èx  & Èx  &


a#x# bsinax# b  #cosax# b
xsin x#cos x

œ lim

É"
x& "
È&x

#x x x
( x x

È œ x lim Ä_

œ

# "
(É x"

œ œ

" "

œ"

œ

# "!

œ

"! #

# #

œ"

œ

%%% %

" #

œ$

œ#

21. (a) The profit function is Paxb œ ac
exbx
aa  bxb œ
ex#  ac
bbx
a. Pw axb œ
#ex  c
b œ ! Ê x œ c#
eb . Pww axb œ
#e  ! if e  ! so that the profit function is maximized at x œ c #
e b . (b) The price therefore that corresponds to a production level yeilding a maximum profit is p¹

xœ c#
eb

œ c
eˆ c #
e b ‰ œ

c b #

dollars. #

(c) The weekly profit at this production level is Paxb œ
eˆ c #
e b ‰  ac
bbˆ c #
e b ‰
a œ

ac
b b #
%e #

a.

(d) The tax increases cost to the new profit function is Faxb œ ac
exbx
aa  bx  txb œ
ex  ac
b
tbx
a. b
c c
b
t ww Now Fw axb œ
#ex  c
b
t œ ! when x œ t 
# e œ #e . Since F axb œ
#e  ! if e  !, F is maximized when x œ c
#be
t units per week. Thus the price per unit is p œ c
eˆ c
#be
t ‰ œ c  #b  t dollars. Thus, such a tax increases the cost per unit by

cbt #

The x-intercept occurs when

" x




cb #

œ

t #

dollars if units are priced to maximize profit.

22. (a)


$œ!Ê

(b) By Newton's method, xn" œ xn
œ xn  xn


$x#n

œ #xn


$xn#

faxn b f ax n b . w

" x

œ $ Ê x œ $" .

Here f w axn b œ
x
# n œ

" x#n .

" $

So xn" œ xn
xn
" œ xn  Š x"n
$‹x#n x# n

œ xn a#
$xn b.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 4 Additional and Advanced Exercises 23. x" œ x!
and

a q
" x!

fax! b f w ax ! b

q

qx!q  xq!  a qxq!
"

x a œ x!
! q
" œ qx

!

with weights m! œ

In the case where x! œ

a xq!
"

q
" q

q

x aq  "b  a œ ! q
" œ x! Š q
q " ‹ 

a Š"‹ xq!
" q

qx!

and m" œ "q .

we have xq! œ a and x" œ

q
" a a " q
" Š q ‹  q
" Š q ‹ x! x!

œ

291

so that x" is a weighted average of x!

q
" a q
" Š q x!

 q" ‹ œ

a q
" . x! #

dy d y 24. We have that ax
hb#  ay
hb# œ r# and so #ax
hb  #ay
hb dy dx œ ! and #  # dx  #ay
hb dx# œ ! hold. dy x
y dx dy . "
dx

dy Thus #x  #y dy dx œ #h  #h dx , by the former. Solving for h, we obtain h œ #

d y equation yields #  # dy dx  #y dx#
#Œ

dy x
y dx dy
"
dx

œ !. Dividing by 2 results in " 

Substituting this into the second

dy dx

#

 y ddxy#
Œ

dy x
y dx dy
"
dx

œ !.

25. (a) aatb œ sww atb œ
k ak  !b Ê sw atb œ
kt  C" , where sw a!b œ )) Ê C" œ )) Ê sw atb œ
kt  )). So satb œ
kt# #


kt# #

 ))t  C# where sa!b œ ! Ê C# œ ! so satb œ

 ))t œ "!!. Solving for t we obtain t œ


kŠ )) 

È))#
#!!k ‹ k ))# #!!

so that k œ

 )) œ ! or
kŠ ))


)) „ È))#
#!!k . k

È))#
#!!k ‹ k


kt# #

 ))t. Now satb œ "!! when

At such t we want sw atb œ !, thus

 )) œ !. In either case we obtain ))#
#!!k œ !

¸ $)Þ(# ft/sec# .

(b) The initial condition that sw a!b œ %% ft/sec implies that sw atb œ
kt  %% and satb œ w

The car is stopped at a time t such that s atb œ
kt  %% œ ! Ê t œ ‰ sˆ %% k

œ


k ˆ %% ‰# # k



‰ %%ˆ %% k

œ

%%# #k

œ

*') k

œ

‰ *')ˆ #!! ))#

%% k .


kt# #

 %%t where k is as above.

At this time the car has traveled a distance

œ #& feet. Thus halving the initial velocity quarters

stopping distance. 26. haxb œ f # axb  g# axb Ê hw axb œ #faxbf w axb  #gaxbgw axb œ #faxbf w axb  gaxbgw axb‘ œ #faxbgaxb  gaxba
faxbb‘ œ # † ! œ !. Thus haxb œ c, a constant. Since ha!b œ &, haxb œ & for all x in the domain of h. Thus ha"!b œ &. 27. Yes. The curve y œ x satisfies all three conditions since

dy dx

œ " everywhere, when x œ !, y œ !, and

d# y dx#

œ ! everywhere.

28. yw œ $x#  # for all x Ê y œ x$  #x  C where
" œ "$  # † "  C Ê C œ
% Ê y œ x$  #x
%. 29. sww atb œ a œ
t# Ê v œ sw atb œ maximum for this t‡ . Since satb t‡ œ a$Cb"Î$ . So ÊCœ

a%bb$Î% $ .



a$Cb"Î$ ‘% 12


t$ w ‡ ‡ $  C. We seek v! œ s a!b œ C. We know that sat b œ b for some t and s is at a % % œ
12t  Ct  k and sa!b œ ! we have that satb œ
12t  Ct and also sw at‡ b œ ! so that

 Ca$Cb"Î$ œ b Ê a$Cb"Î$ ˆC


Thus v! œ sw a!b œ

a%bb$Î% $

œ

#È# $Î% . $ b

$C ‰ "#

œ b Ê a$Cb"Î$ ˆ $%C ‰ œ b Ê $"Î$ C%Î$ œ

30. (a) sww atb œ t"Î#
t
"Î# Ê vatb œ sw atb œ #$ t$Î#
#t"Î#  k where va!b œ k œ (b) satb œ

% &Î# "& t

%
%$ t$Î#  %$ t  k# where sa!b œ k# œ
"& . Thus satb œ

% # $Î#
#t"Î# $ Ê vatb œ $ t % &Î# %
%$ t$Î#  %$ t
"& . "& t

%b $

 %$ Þ

31. The graph of faxb œ ax#  bx  c with a  ! is a parabola opening upwards. Thus faxb   ! for all x if faxb œ ! for at most one real value of x. The solutions to faxb œ ! are, by the quadratic equation #


#b „ Éa#bb#
%ac . #a

Thus we require

#

a#bb
%ac Ÿ ! Ê b
ac Ÿ !. 32. (a) Clearly faxb œ aa" x  b" b#  Þ Þ Þ  aan x  bn b#   ! for all x. Expanding we see faxb œ aa#" x#  #a" b" x  b"# b  Þ Þ Þ  aan# x#  #an bn x  bn# b œ aa#"  a##  Þ Þ Þ  an# bx#  #aa" b"  a# b#  Þ Þ Þ  an bn bx  ab"#  b##  Þ Þ Þ  bn# b   !.

Thus aa" b"  a# b#  Þ Þ Þ  an bn b#
aa#"  a##  Þ Þ Þ  an# bab"#  b##  Þ Þ Þ  bn# b Ÿ ! by Exercise 31. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

292

Chapter 4 Applications of Derivatives

Thus aa" b"  a# b#  Þ Þ Þ  an bn b# Ÿ aa#"  a##  Þ Þ Þ  an# bab"#  b##  Þ Þ Þ  bn# b. (b) Referring to Exercise 31: It is clear that faxb œ ! for some real x Í b#
%ac œ !, by quadratic formula. Now notice that this implies that faxb œ aa" x  b" b#  Þ Þ Þ  aan x  bn b# œ aa#"  a##  Þ Þ Þ  an# bx#  #aa" b"  a# b#  Þ Þ Þ  an bn bx  ab"#  b##  Þ Þ Þ  bn# b œ ! Í aa" b"  a# b#  Þ Þ Þ  an bn b#
aa#"  a##  Þ Þ Þ  an# bab"#  b##  Þ Þ Þ  bn# b œ !

Í aa" b"  a# b#  Þ Þ Þ  an bn b# œ aa#"  a##  Þ Þ Þ  an# bab"#  b##  Þ Þ Þ  bn# b But now faxb œ ! Í ai x  bi œ ! for all i œ "ß #ß Þ Þ Þ ß n Í ai x œ
bi œ ! for all i œ "ß #ß Þ Þ Þ ß n.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 5 INTEGRATION 5.1 ESTIMATING WITH FINITE SUMS 1. faxb œ x#

Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.

"

#

iœ! $

" #

œ "# Š!#
ˆ "# ‰ ‹ œ

#

" 4

œ 4" Š!#
ˆ 4" ‰
ˆ #" ‰
ˆ 4$ ‰ ‹ œ

(a) ˜x œ

"
! #

œ

" #

and xi œ i˜x œ

i #

Ê a lower sum is !ˆ #i ‰ †

(b) ˜x œ

"
! %

œ

" %

and xi œ i˜x œ

i %

Ê a lower sum is !ˆ 4i ‰ †

(c) ˜x œ

"
! #

œ

" #

and xi œ i˜x œ

i #

Ê an upper sum is !ˆ #i ‰ †

(d) ˜x œ

"
! %

œ

" %

and xi œ i˜x œ

i %

Ê an upper sum is !ˆ 4i ‰ †

2. faxb œ x$

iœ! 2

iœ1 %

#

" )

#

#

#

#

" #

# œ "# Šˆ "# ‰ +1# ‹ œ

#

" 4

# # # œ 4" Šˆ 4" ‰
ˆ #" ‰
ˆ 4$ ‰ +1# ‹ œ

iœ"

" %

†

( )

" %

$! ‰ † ˆ "' œ

œ

( $#

& ) "& $#

Since f is increasing on Ò!ß "Ó, we use left endpoints to obtain lower sums and right endpoints to obtain upper sums.

"

$

iœ! $

" #

œ "# Š!$
ˆ "# ‰ ‹ œ

$

" 4

œ 4" Š!$
ˆ 4" ‰
ˆ #" ‰
ˆ 4$ ‰ ‹ œ

(a) ˜x œ

"
! #

œ

" #

and xi œ i˜x œ

i #

Ê a lower sum is !ˆ #i ‰ †

(b) ˜x œ

"
! %

œ

" %

and xi œ i˜x œ

i %

Ê a lower sum is !ˆ 4i ‰ †

(c) ˜x œ

"
! #

œ

" #

and xi œ i˜x œ

i #

Ê an upper sum is !ˆ #i ‰ †

(d) ˜x œ

"
! %

œ

" %

and xi œ i˜x œ

i %

Ê an upper sum is !ˆ 4i ‰ †

iœ! 2

iœ1 %

iœ"

$

" "'

$

$

$

$' #&'

$

" #

$ œ "# Šˆ "# ‰ +1$ ‹ œ

$

" 4

$ $ $ œ 4" Šˆ 4" ‰
ˆ #" ‰
ˆ 4$ ‰ +1$ ‹ œ œ

" #

†

* )

œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

* '%

"!! #&'

œ

* "' #& '%

294

Chapter 5 Integration

3. faxb œ

" x

Since f is decreasing on Ò!ß "Ó, we use left endpoints to obtain upper sums and right endpoints to obtain lower sums.

#

(a) ˜x œ

&
" #

œ # and xi œ "
i˜x œ "
#i Ê a lower sum is ! x"i † # œ #ˆ $"
&" ‰ œ

(b) ˜x œ

&
" %

œ 1 and xi œ "
i˜x œ "
i Ê a lower sum is

(c) ˜x œ

&
" #

œ # and xi œ "
i˜x œ "
#i Ê an upper sum is ! x"i † # œ #ˆ"
$" ‰ œ

(d) ˜x œ

&
" %

œ 1 and xi œ "
i˜x œ "
i Ê an upper sum is

4. faxb œ %  x#

iœ" % !" xi iœ" "

† " œ "ˆ #"


iœ! $ !" xi iœ!

" $

† " œ "ˆ"





" #

" %




"' "&


&" ‰ œ

" $

(( '!

) $


"% ‰ œ

#& "#

Since f is increasing on Ò#ß !Ó and decreasing on Ò!ß #Ó, we use left endpoints on Ò#ß !Ó and right endpoints on Ò!ß #Ó to obtain lower sums and use right endpoints on Ò#ß !Ó and left endpoints on Ò!ß #Ó to obtain upper sums.

(a) ˜x œ

#
a
#b #

œ # and xi œ #
i˜x œ #
#i Ê a lower sum is # † ˆ%  a#b# ‰
# † a%  ## b œ !

(b) ˜x œ

#
a
#b %

œ " and xi œ #
i˜x œ #
i Ê a lower sum is !ˆ%  axi b# ‰ † "
!ˆ%  axi b# ‰ † "

"

%

iœ!

iœ$

œ "ˆˆ%  a#b# ‰
ˆ%  a"b# ‰
a%  "# b
a%  ## b‰ œ ' (c) ˜x œ

#
a
#b #

œ # and xi œ #
i˜x œ #
#i Ê a upper sum is # † ˆ%  a!b# ‰
# † a%  !# b œ "'

(d) ˜x œ

#
a
#b %

œ " and xi œ #
i˜x œ #
i Ê a upper sum is !ˆ%  axi b# ‰ † "
!ˆ%  axi b# ‰ † "

#

$

iœ"

iœ#

œ "ˆˆ%  a"b# ‰
a%  !# b
a%  !# b
a%  "# b‰ œ "% 5. faxb œ x#

œ

" #

Using 4 rectangles Ê ˜x œ "
% ! œ Ê "% ˆfˆ ") ‰
fˆ $) ‰
fˆ &) ‰
fˆ () ‰‰

" %

Using 2 rectangles Ê ˜x œ #

#

œ "# Šˆ "% ‰
ˆ $% ‰ ‹ œ

#

#

"! $#

#

œ

"
! #

Ê "# ˆfˆ "% ‰
fˆ $% ‰‰

& "'

#

œ "% Šˆ ") ‰
ˆ $) ‰
ˆ &) ‰
ˆ () ‰ ‹ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

#" '%

Section 5.1 Estimating with Finite Sums 6. faxb œ x$

œ

" #

Using 4 rectangles Ê ˜x œ "
% ! œ Ê "% ˆfˆ ") ‰
fˆ $) ‰
fˆ &) ‰
fˆ () ‰‰

" %

$

$

œ "# Šˆ "% ‰
ˆ $% ‰ ‹ œ

$

$

$

$

( œ "% Š " $ )& ‹œ $

7. faxb œ

" x

"
! #

Using 2 rectangles Ê ˜x œ #) # † '%

%*' % † )$

œ

œ

Using 2 rectangles Ê ˜x œ œ #ˆ "#
"% ‰ œ $#

Ê "# ˆfˆ "% ‰
fˆ $% ‰‰

( $#

"#% )$

&
" #

œ

$" "#)

œ # Ê #afa#b
fa%bb

Using 4 rectangles Ê ˜x œ &
% " œ " Ê "ˆfˆ $# ‰
fˆ &# ‰
fˆ (# ‰
fˆ *# ‰‰ œ "ˆ #$


8. faxb œ %  x#

# &




# (


#* ‰ œ

"%)) $†&†(†*

Using 2 rectangles Ê ˜x œ œ #a$
$b œ "#

295

œ

#
a
#b #

%*' &†(†*

œ

%*' $"&

œ # Ê #afa"b
fa"bb

Using 4 rectangles Ê ˜x œ #
%a
#b œ " Ê "ˆfˆ $# ‰
fˆ "# ‰
fˆ "# ‰
fˆ $# ‰‰ # # # # œ "ŠŠ%  ˆ $# ‰ ‹
Š%  ˆ "# ‰ ‹
Š%  ˆ "# ‰ ‹
Š%  ˆ $# ‰ ‹‹ œ "'  ˆ *% † #
"% † #‰ œ "'  "! # œ ""

9. (a) D ¸ (0)(1)
(12)(1)
(22)(1)
(10)(1)
(5)(1)
(13)(1)
(11)(1)
(6)(1)
(2)(1)
(6)(1) œ 87 inches (b) D ¸ (12)(1)
(22)(1)
(10)(1)
(5)(1)
(13)(1)
(11)(1)
(6)(1)
(2)(1)
(6)(1)
(0)(1) œ 87 inches 10. (a) D ¸ (1)(300)
(1.2)(300)
(1.7)(300)
(2.0)(300)
(1.8)(300)
(1.6)(300)
(1.4)(300)
(1.2)(300)
(1.0)(300)
(1.8)(300)
(1.5)(300)
(1.2)(300) œ 5220 meters (NOTE: 5 minutes œ 300 seconds) (b) D ¸ (1.2)(300)
(1.7)(300)
(2.0)(300)
(1.8)(300)
(1.6)(300)
(1.4)(300)
(1.2)(300)
(1.0)(300)
(1.8)(300)
(1.5)(300)
(1.2)(300)
(0)(300) œ 4920 meters (NOTE: 5 minutes œ 300 seconds) 11. (a) D ¸ (0)(10)
(44)(10)
(15)(10)
(35)(10)
(30)(10)
(44)(10)
(35)(10)
(15)(10)
(22)(10)
(35)(10)
(44)(10)
(30)(10) œ 3490 feet ¸ 0.66 miles (b) D ¸ (44)(10)
(15)(10)
(35)(10)
(30)(10)
(44)(10)
(35)(10)
(15)(10)
(22)(10)
(35)(10)
(44)(10)
(30)(10)
(35)(10) œ 3840 feet ¸ 0.73 miles 12. (a) The distance traveled will be the area under the curve. We will use the approximate velocities at the midpoints of each time interval to approximate this area using rectangles. Thus, D ¸ (20)(0.001)
(50)(0.001)
(72)(0.001)
(90)(0.001)
(102)(0.001)
(112)(0.001)
(120)(0.001)
(128)(0.001)
(134)(0.001)
(139)(0.001) ¸ 0.967 miles (b) Roughly, after 0.0063 hours, the car would have gone 0.484 miles, where 0.0060 hours œ 22.7 sec. At 22.7 sec, the velocity was approximately 120 mi/hr.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

296

Chapter 5 Integration

13. (a) Because the acceleration is decreasing, an upper estimate is obtained using left end-points in summing acceleration † ?t. Thus, ?t œ 1 and speed ¸ [32.00
19.41
11.77
7.14
4.33](1) œ 74.65 ft/sec (b) Using right end-points we obtain a lower estimate: speed ¸ [19.41
11.77
7.14
4.33
2.63](1) œ 45.28 ft/sec (c) Upper estimates for the speed at each second are: t 0 1 2 3 4 5 v 0 32.00 51.41 63.18 70.32 74.65 Thus, the distance fallen when t œ 3 seconds is s ¸ [32.00
51.41
63.18](1) œ 146.59 ft. 14. (a) The speed is a decreasing function of time Ê right end-points give an lower estimate for the height (distance) attained. Also t 0 1 2 3 4 5 v 400 368 336 304 272 240 gives the time-velocity table by subtracting the constant g œ 32 from the speed at each time increment ?t œ 1 sec. Thus, the speed ¸ 240 ft/sec after 5 seconds. (b) A lower estimate for height attained is h ¸ [368
336
304
272
240](1) œ 1520 ft. 15. Partition [!ß #] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of these subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating 1 125 343 $ $ rectangles are f(m" ) œ (0.25)$ œ 64 , f(m# ) œ (0.75)$ œ 27 64 , f(m$ ) œ (1.25) œ 64 , and f(m% ) œ (1.75) œ 64 Notice that the average value is approximated by œ

" length of [!ß#]

†”

" #

$ $ $ $ ’ˆ 4" ‰ ˆ #" ‰
ˆ 34 ‰ ˆ #" ‰
ˆ 54 ‰ ˆ #" ‰
ˆ 74 ‰ ˆ #" ‰“ œ

$" "'

approximate area under • . We use this observation in solving the next several exercises. curve f(x) œ x$

16. Partition [1ß 9] into the four subintervals ["ß $], [3ß &], [&ß (], and [(ß *]. The midpoints of these subintervals are m" œ 2, m# œ 4, m$ œ 6, and m% œ 8. The heights of the four approximating rectangles are f(m" ) œ "# , f(m# ) œ "4 , f(m$ ) œ 6" , and f(m% ) œ 8" . The width of each rectangle is ?x œ 2. Thus, Area ¸ 2 ˆ "# ‰
2 ˆ 4" ‰
2 ˆ 6" ‰
2 ˆ 8" ‰ œ

Ê average value ¸

25 1#

area length of ["ß*]

œ

ˆ 25 ‰ 12 8

œ

25 96 .

17. Partition [0ß 2] into the four subintervals [0ß 0.5], [0.5ß 1], [1ß 1.5], and [1.5ß 2]. The midpoints of the subintervals are m" œ 0.25, m# œ 0.75, m$ œ 1.25, and m% œ 1.75. The heights of the four approximating rectangles are " #

f(m" ) œ œ

" #




" #


sin#

1 4

œ

" #




" #

œ 1, and f(m% ) œ

œ 1, f(m# ) œ

" 2


sin#

71 4

œ


sin#

" #


Š È"2 ‹ œ 1. The width of each rectangle is ?x œ #" . Thus,

31 4

œ

" #




" #

œ 1, f(m$ ) œ

" 2


sin#

51 4

œ

" #


Š È"2 ‹

#

" 2

#

Area ¸ (1
1
1
1) ˆ "# ‰ œ 2 Ê average value ¸

area length of [0ß2]

œ

2 #

œ 1.

18. Partition [0ß 4] into the four subintervals [0ß 1], [1ß 2ß ], [2ß 3], and [3ß 4]. The midpoints of the subintervals are m" œ "# , m# œ #3 , m$ œ 5# , and m% œ 7# . The heights of the four approximating rectangles are f(m" ) œ 1  Šcos Š

% 1 ˆ "# ‰ 4 ‹‹

œ 1  ˆcos ˆ 18 ‰‰ œ 0.27145 (to 5 decimal places),

f(m# ) œ 1  Šcos Š

% 1 ˆ 3# ‰ 4 ‹‹

œ 1  ˆcos ˆ 381 ‰‰ œ 0.97855, f(m3 ) œ 1  Šcos Š

%

œ 0.97855, and f(m% ) œ 1  Šcos Š

%

% 1 ˆ 7# ‰ 4 ‹‹

% 1 ˆ #5 ‰ 4 ‹‹

œ 1  ˆcos ˆ 581 ‰‰

%

%

œ 1  ˆcos ˆ 781 ‰‰ œ 0.27145. The width of each rectangle is

?x œ ". Thus, Area ¸ (0.27145)(1)
(0.97855)(1)
(0.97855)(1)
(0.27145)(1) œ 2.5 Ê average 2.5 5 value ¸ lengtharea of [0ß4] œ 4 œ 8 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.1 Estimating with Finite Sums

297

19. Since the leakage is increasing, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (70)(1)
(97)(1)
(136)(1)
(190)(1)
(265)(1) œ 758 gal, lower estimate œ (50)(1)
(70)(1)
(97)(1)
(136)(1)
(190)(1) œ 543 gal. (b) upper estimate œ (70
97
136
190
265
369
516
720) œ 2363 gal, lower estimate œ (50
70
97
136
190
265
369
516) œ 1693 gal. (c) worst case: 2363
720t œ 25,000 Ê t ¸ 31.4 hrs; best case: 1693
720t œ 25,000 Ê t ¸ 32.4 hrs 20. Since the pollutant release increases over time, an upper estimate uses right endpoints and a lower estimate uses left endpoints: (a) upper estimate œ (0.2)(30)
(0.25)(30)
(0.27)(30)
(0.34)(30)
(0.45)(30)
(0.52)(30) œ 60.9 tons lower estimate œ (0.05)(30)
(0.2)(30)
(0.25)(30)
(0.27)(30)
(0.34)(30)
(0.45)(30) œ 46.8 tons (b) Using the lower (best case) estimate: 46.8
(0.52)(30)
(0.63)(30)
(0.70)(30)
(0.81)(30) œ 126.6 tons, so near the end of September 125 tons of pollutants will have been released. #

21. (a) The diagonal of the square has length 2, so the side length is È#. Area œ ŠÈ#‹ œ # (b) Think of the octagon as a collection of 16 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 "' œ ) . Area œ "'ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ % sin 1 œ #È# ¸ #Þ)#) #

)

)

%

(c) Think of the 16-gon as a collection of 32 right triangles with a hypotenuse of length 1 and an acute angle measuring #1 1 $# œ "' . Area œ $#ˆ " ‰ˆsin 1 ‰ˆcos 1 ‰ œ ) sin 1 œ #È# ¸ $Þ!'" #

"'

"'

)

(d) Each area is less than the area of the circle, 1. As n increases, the area approaches 1. 22. (a) Each of the isosceles triangles is made up of two right triangles having hypotenuse 1 and an acute angle measuring #1 1 1 ‰ˆ ˆ " ‰ˆ cos 1n ‰ œ "# sin #n1 . #n œ n . The area of each isosceles triangle is AT œ # # sin n (b) The area of the polygon is AP œ nAT œ

n #

sin

#1 n ,

n nÄ_ #

so lim

sin

#1 n

œ lim 1 † nÄ_

sin #n1 ˆ #n1 ‰

œ1

(c) Multiply each area by r# . AT œ "# r# sin #n1 AP œ n# r# sin lim AP œ 1r

#

#1 n

nÄ_

23-26. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

298

Chapter 5 Integration

end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); Mathematica: (assigned function and values for a and b may vary): Symbols for 1, Ä , powers, roots, fractions, etc. are available in Palettes (under File). Never insert a space between the name of a function and its argument. Clear[x] f[x_]:=x Sin[1/x] {a,b}={1/4, 1} Plot[f[x],{x, a, b}] The following code computes the value of the function for each interval midpoint and then finds the average. Each sequence of commands for a different value of n (number of subdivisions) should be placed in a separate cell. n =100; dx = (b  a) /n; values = Table[N[f[x]], {x, a
dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =200; dx = (b  a) /n; values = Table[N[f[x]],{x, a + dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n n =1000; dx = (b  a) /n; values = Table[N[f[x]],{x, a
dx/2, b, dx}] average=Sum[values[[i]],{i, 1, Length[values]}] / n FindRoot[f[x] == average,{x, a}] 5.2 SIGMA NOTATION AND LIMITS OF FINITE SUMS 2

1. ! kœ1

3

2. ! kœ1

6k k1

œ

6(1) 11




6(2) 21

œ

6 2




k
1 k

œ

1
1 1




2
1 2




3
1 3

12 3

œ7

œ0


1 2




2 3

œ

7 6

4

3. ! cos k1 œ cos (11)
cos (21)
cos (31)
cos (41) œ 1
1  1
1 œ 0 kœ1

5

4. ! sin k1 œ sin (11)
sin (21)
sin (31)
sin (41)
sin (51) œ 0
0
0
0
0 œ 0 kœ1

3

5. ! (1)kb1 sin kœ1

1 k

œ (1)"" sin

1 1


(1)#" sin

1 #


(")$" sin

1 3

œ 01


È3 #

œ

È3
2 #

4

6. ! (1)k cos k1 œ (1)" cos (11)
(1)# cos (21)
(1)$ cos (31)
(1)% cos (41) kœ1

œ (1)
1  (1)
1 œ 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.2 Sigma Notation and Limits of Finite Sums 6

7. (a) ! 2kc1 œ 2"
"
2#
"
2$
"
2%
"
2&
"
2'
" œ 1
2
4
8
16
32 kœ1 5

(b) ! 2k œ 2!
2"
2#
2$
2%
2& œ 1
2
4
8
16
32 kœ0 4

(c) ! 2k1 œ 2
""
2!"
2""
2#"
2$"
2%" œ 1
2
4
8
16
32 kœ
"

All of them represent 1
2
4
8
16
32 6

8. (a) ! (2)k
1 œ (2)"
"
(2)#
"
(2)$
"
(2)%
"
(2)&
"
(2)'
" œ 1  2
4  8
16  32 kœ1 5

(b) ! (1)k 2k œ (1)! 2!
Ð")" 2"
(1)# 2#
(1)$ 2$
(1)% 2%
(1)& 2& œ 1  2
4  8
16  32 kœ0 3

(c) ! (1)k1 2k2 œ Ð")
#" 2
##
(")
"" 2
"#
(")!" 2!#
(1)"" 2"#
(")#" 2## kœ
2


(1)$" 2$# œ 1
2  4
8  16
32; (a) and (b) represent 1  2
4  8
16  32; (c) is not equivalent to the other two 4

9. (a) ! kœ2 2

(b) ! kœ0 1

(c) !

kœ
"

(
")k
" k
1

(
1)#
" 2
1

œ

(
")k k1

œ

(
1)! 01

(
")k k2

œ

(
1)
"
1  2







(
")$
" 3
1

(
")" 11







(
")! 02




(
")# 21




(
")%
" 4
1

œ 1


œ1

(
")" 12

" #




œ 1


" #

" #



" 3

" 3



" 3

(a) and (c) are equivalent; (b) is not equivalent to the other two. 4

10. (a) ! (k  1)# œ (1  1)#
(2  1)#
(3  1)#
(4  1)# œ 0
1
4
9 kœ1 3

(b) ! (k
1)# œ (1
1)#
(0
1)#
(1
1)#
(2
1)#
(3
1)# œ 0
1
4
9
16 kœ
1


"

(c) ! k# œ (3)#
(2)#
(1)# œ 9
4
1 kœ
3

(a) and (c) are equivalent to each other; (b) is not equivalent to the other two. 6

4

4

12. ! k#

11. ! k kœ1

13. !

kœ1

5

5

15. ! (1)k1

14. ! 2k kœ1

kœ1

n

kœ1

" k

5

16. ! (1)k kœ1

n

17. (a) ! 3ak œ 3 ! ak œ 3(5) œ 15 kœ1 n

(b) ! kœ1 n

bk 6

œ

" 6

kœ1 n

! bk œ kœ1

" 6

(6) œ 1

n

n

kœ1 n

kœ1 n

kœ1 n

kœ1 n

kœ1 n

kœ1

(c) ! (ak
bk ) œ ! ak
! bk œ 5
6 œ 1 (d) ! (ak  bk ) œ ! ak  ! bk œ 5  6 œ 11 n

(e) ! (bk  2ak ) œ ! bk  2 ! ak œ 6  2(5) œ 16 kœ1

kœ1

" #k

kœ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

k 5

299

300

Chapter 5 Integration n

n

kœ1 n

kœ1

n

18. (a) ! 8ak œ 8 ! ak œ 8(0) œ 0 n

n

kœ1

kœ1

(c) ! (ak
1) œ ! ak
! 1 œ 0
n œ n kœ1

10

19. (a) ! k œ kœ1

10(10  1) #

n

(b) ! 250bk œ 250 ! bk œ 250(1) œ 250 kœ1 n

n

kœ1

n

kœ1

kœ1

kœ1

(d) ! (bk  1) œ ! bk  ! 1 œ "  n 10

(b) ! k# œ

œ 55

kœ1

10(10  1)(2(10)  1) 6

œ 385

13(13  1)(2(13)  1) 6

œ 819

#

10

(c) ! k$ œ ’ 10(10#  1) “ œ 55# œ 3025 kœ1

13

20. (a) ! k œ kœ1

13(13  1) #

13

(b) ! k# œ

œ 91

kœ1

#

13

(c) ! k$ œ ’ 13(13#  1) “ œ 91# œ 8281 kœ1

7

7

kœ1

kœ1

6

6

6

kœ1

kœ1

kœ1

6

6

6

kœ1

kœ1

kœ1

5

21. ! 2k œ 2 ! k œ 2 Š 7(7 # ") ‹ œ 56

23. ! a3  k# b œ ! 3  ! k# œ 3(6) 

24. ! ak#  5b œ ! k#  ! 5 œ

22. ! kœ1

6(6  ")(2(6)  1) 6

6(6  ")(2(6)  1) 6

1k 15

œ

1 15

5

!kœ kœ1

1 15

Š 5(5 # 1) ‹ œ 1

œ 73

 5(6) œ 61

5

5

5

5

kœ1

kœ1

kœ1

kœ1

7

7

7

7

kœ1

kœ1

kœ1

kœ1

 1) 25. ! k(3k
5) œ ! a3k#
5kb œ 3 ! k#
5 ! k œ 3 Š 5(5  1)(2(5) ‹
5 Š 5(5 # 1) ‹ œ 240 6

 1) 26. ! k(2k
1) œ ! a2k#
kb œ 2 ! k#
! k œ 2 Š 7(7  1)(2(7) ‹
6

5

k$ 225

27. ! kœ1

7

kœ1

#

7

28. Œ! k
 ! kœ1

29. (a)

$

5


Œ! k
œ

kœ1

k$ 4

" 2 #5

7

5

5

kœ1

kœ1

#

œ Œ! k
 kœ1

$

! k $
Œ! k
œ

" 4

" #25

#

! k$ œ Š 7(7  1) ‹  # kœ1

(b)

œ 308

$

Š 5(5 # 1) ‹
Š 5(5 # 1) ‹ œ 3376 #

7

7(7  1) #

" 4

#

Š 7(7 # 1) ‹ œ 588 (c)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.2 Sigma Notation and Limits of Finite Sums 30. (a)

(b)

(c)

31. (a)

(b)

(c)

32. (a)

(b)

(c)

301

33. kx"  x! k œ k1.2  0k œ 1.2, kx#  x" k œ k1.5  1.2k œ 0.3, kx$  x# k œ k2.3  1.5k œ 0.8, kx%  x$ k œ k2.6  2.3k œ 0.3, and kx&  x% k œ k3  2.6k œ 0.4; the largest is lPl œ 1.2. 34. kx"  x! k œ k1.6  (2)k œ 0.4, kx#  x" k œ k0.5  (1.6)k œ 1.1, kx$  x# k œ k0  (0.5)k œ 0.5, kx%  x$ k œ k0.8  0k œ 0.8, and kx&  x% k œ k1  0.8k œ 0.2; the largest is lPl œ 1.1. 35. faxb œ "  x#

Since f is decreasing on Ò !, 1Ó we use left endpoints to obtain upper sums. ˜x œ "
n ! œ "n and xi œ i˜x œ ni . So an upper sum n
"

is ! a"  x#i b "n œ iœ!

œ

n$ n$



n "! # i n$ iœ!

" n

n
"

! Š"  ˆ i ‰# ‹ œ n

iœ!

œ"

an
" b n a# an
" b  " b 'n $

#
$n  n"# . Thus, ' n
" lim ! a"  x#i b "n œ lim Œ" nÄ_ iœ! nÄ_

" n$

n
"

! an#  i# b

iœ!

œ"

#n$
$n#  n 'n $

œ"



#
$n  n"# '

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley


œ"

" $

œ

# $

302

Chapter 5 Integration Since f is increasing on Ò !, $Ó we use right endpoints to obtain upper sums. ˜x œ $
n ! œ $n and xi œ i˜x œ $ni . So an upper

36. faxb œ #x

n

n

sum is !#xi ˆ n$ ‰ œ ! 'ni † iœ"

iœ"

n

Thus,

37. faxb œ x#
"

lim ! 'i nÄ_ iœ" n

†

$ n

œ

# œ lim *n n# *n nÄ_

$ n

n

") n#

") n#

!i œ

iœ"

n an  " b #

†

œ

*n#  *n n#

œ lim ˆ*
n* ‰ œ *. nÄ_

Since f is increasing on Ò !, $Ó we use right endpoints to obtain upper sums. ˜x œ $
n ! œ $n and xi œ i˜x œ $ni . So an upper n

n

# sum is !ax#i
"b $n œ !Šˆ $ni ‰
"‹ n$ œ

œ

iœ" n #( ! # i
n$ n iœ"

iœ"

†nœ

#( nan  "ba#n  "b ‹ n$ Š '

n

!Š *i##
"‹ n

$ n

iœ"


$

* ")  #( *a#n$  $n#  nb n  n#
$œ
$Þ Thus, #n $ # n #( ")  n  n*# lim !ax#i
"b $n œ lim Œ
$
œ # nÄ_ iœ" nÄ_

œ

38. faxb œ $x#

Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ "
n ! œ "n and xi œ i˜x œ ni . So an upper sum n

n

iœ"

iœ"

# is !$x#i ˆ "n ‰ œ !$ˆ ni ‰ ˆ n" ‰ œ

œ

#n$  $n#  n #n $

œ lim Œ nÄ_

39. faxb œ x
x# œ xa"
xb

œ

#  $n  n"# #

#  $n  n"# # # #


œ

n

$ n$

! i# œ

iœ"

$ n$

† Š nan  "ba' #n  "b ‹

n

. Thus, lim !$x#i ˆ "n ‰ nÄ_ iœ"

œ ".

Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ "
n ! œ "n and xi œ i˜x œ ni . So an upper sum n

n

# is !axi
xi# b n" œ !Š ni
ˆ ni ‰ ‹ n" œ iœ"

iœ"

œ

" n a n  "b ‹ n# Š #

œ

"  "n #


n"$ Š nan  "ba' #n  "b ‹ n # $  "




œ lim ”Š nÄ_

40. faxb œ $x
#x#

*
$ œ "#.

n

'

" #

" n

n#

n "! i n# iœ"




n "! # i n$ iœ"

n#  n #n#




#n $  $ n #  n 'n$

œ

. Thus, lim !axi
x#i b "n nÄ_ iœ"

‹
Œ

$ n

#   n"# '


• œ

" #




# '

œ &' .

Since f is increasing on Ò !, "Ó we use right endpoints to obtain upper sums. ˜x œ "
n ! œ "n and xi œ i˜x œ ni . So an upper sum n

n

# is !a$xi
#x#i b "n œ !Š $ni
#ˆ ni ‰ ‹ n" œ iœ"

iœ"

œ

$ n a n  "b ‹ n# Š #

œ

$  $n #




œ lim ”Š nÄ_


n#$ Š nan  "ba' #n  "b ‹ n # $  "

$ #

n

$

$ n

n#

œ

n $! i n# iœ"

$n#  $n #n#




. Thus, lim !a$xi
#x#i b "n

‹
Œ

nÄ_ iœ"

$ n

#   n"# $


• œ

$ #




Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

# $

œ

"$ ' .




n #! # i n$ iœ"

#n#  $n  " $n#

Section 5.3 The Definite Integral

303

5.3 THE DEFINITE INTEGRAL

1.

'02 x# dx

2.

'
"! 2x$ dx

3.

'
(& ax#  3xb dx

4.

'"% "x dx

5.

'#$ 1
" x dx

6.

'0" È4  x# dx

7.

'
! Î% (sec x) dx

8.

'0 Î% (tan x) dx 1

1

9. (a) (c) (e) (f)

10. (a) (b) (c) (d) (e) (f)

11. (a) (c)

12. (a) (c)

13. (a) (b)

14. (a) (b)

" & '#2 g(x) dx œ 0 (b) ' g(x) dx œ  ' g(x) dx œ 8 & " 2 2 & & 2 '" 3f(x) dx œ 3'" f(x) dx œ 3(4) œ 12 (d) ' f(x) dx œ ' f(x) dx  ' f(x) dx œ 6  (4) œ 10 # " " & & & '" [f(x)  g(x)] dx œ '" f(x) dx  '" g(x) dx œ 6  8 œ 2 '"& [4f(x)  g(x)] dx œ 4 '"& f(x) dx  '"& g(x) dx œ 4(6)  8 œ 16

'"* 2f(x) dx œ 2 '"* f(x) dx œ 2(1) œ 2 '(* [f(x)
h(x)] dx œ '(*f(x) dx
'(* h(x) dx œ 5
4 œ 9 '(* [2f(x)  3h(x)] dx œ 2 '(* f(x) dx  3 '(* h(x) dx œ 2(5)  3(4) œ 2 '*"f(x) dx œ  '"* f(x) dx œ (1) œ 1 '"( f(x) dx œ '"* f(x) dx  '(* f(x) dx œ 1  5 œ 6 '*( [h(x)  f(x)] dx œ '(* [f(x)  h(x)] dx œ '(* f(x) dx  '(* h(x) dx œ 5  4 œ 1 '"2 f(u) du œ '"2 f(x) dx œ 5 '#" f(t) dt œ  '"2 f(t) dt œ 5 '!
$ g(t) dt œ  '
$! g(t) dt œ È2 '
$! [g(x)] dx œ  '
$! g(x) dx œ È2

(b) (d)

(b) (d)

'"2 È3 f(z) dz œ È3 '"2 f(z) dz œ 5È3 '"2 [f(x)] dx œ  '"2 f(x) dx œ 5 '
$! g(u) du œ '
$! g(t) dt œ È2 '
$! Èg(r)2 dr œ È"2 '
$! g(t) dt œ Š È"2 ‹ ŠÈ2‹ œ 1

'$% f(z) dz œ '!% f(z) dz  '!$ f(z) dz œ 7  3 œ 4 '%$ f(t) dt œ  '$% f(t) dt œ 4 '"$ h(r) dr œ '
"$ h(r) dr  '
"" h(r) dr œ 6  0 œ 6 " $ $  ' h(u) du œ  Œ ' h(u) du
œ ' h(u) du œ 6 $ " "

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

304

Chapter 5 Integration

15. The area of the trapezoid is A œ œ

" #

(5
2)(6) œ 21 Ê

œ 21 square units

" #

(3
1)(1) œ 2 Ê

(B
b)h

" #

(B
b)h

'
# ˆ #x
3‰ dx %

16. The area of the trapezoid is A œ œ

" #

'"Î#

$Î#

(2x
4) dx

œ 2 square units

17. The area of the semicircle is A œ œ

9 #

1 Ê

" #

1r# œ

1(3)#

'
$$ È9  x# dx œ 9# 1 square units

18. The graph of the quarter circle is A œ œ 41 Ê

" #

" 4

1 r# œ

" 4

1(4)#

'
%! È16  x# dx œ 41 square units

19. The area of the triangle on the left is A œ

" #

bh œ

œ 2. The area of the triangle on the right is A œ œ Ê

" #

(1)(1) œ

" #.

" # " #

(2)(2) bh

Then, the total area is 2.5

'
# kxk dx œ 2.5 square units "

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.3 The Definite Integral 20. The area of the triangle is A œ Ê

" #

bh œ

'
" a1
kxkb dx œ 1 square unit "

21. The area of the triangular peak is A œ

" #

(2)(1) œ 1

" #

bh œ

" #

(2)(1) œ 1.

The area of the rectangular base is S œ jw œ (2)(1) œ 2. Then the total area is 3 Ê

'
"" a2
kxkb dx œ 3 square units

22. y œ 1  È1
x# Ê y
1 œ È1
x# Ê (y
1)# œ 1
x# Ê x#  (y
1)# œ 1, a circle with center (!ß ") and radius of 1 Ê y œ 1  È1
x# is the upper semicircle. The area of this semicircle is A œ "# 1r# œ "# 1(1)# œ 1# . The area of the rectangular base is A œ jw œ (2)(1) œ 2. Then the total area is 2  Ê

23.

'
"" Š1  È1
x# ‹ dx œ 2  1# square units

'!b x2 dx œ "# (b)( b2 ) œ b4

#

1 #

24.

'!b 4x dx œ "# b(4b) œ 2b#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

305

306

Chapter 5 Integration

25.

'ab 2s ds œ "# b(2b)
"# a(2a) œ b#
a#

27.

'"

29.

'1#1 ) d) œ (2#1)

31.

'0

33.

'!"Î# t# dt œ ˆ 3‰

œ

35.

'a#a x dx œ (2a)#




37.

'!

39.

'$" 7 dx œ 7(1
3) œ
14

41.

'!2 5x dx œ 5 '!2 x dx œ 5 ’ 2#

43.

'!2 (2t
3) dt œ 2 '"" t dt
'!2 3 dt œ 2 ’ 2#

44.

'!

45.

'#" ˆ1  #z ‰ dz œ '#" 1 dz  '#" #z dz œ '#" 1 dz
"# '"# z dz œ 1[1
2]
"# ’ 2#
1# “ œ
"
"# ˆ 3# ‰ œ
74

46.

'$! (2z
3) dz œ '$! 2z dz
'$! 3 dz œ
2 '!$ z dz
'$! 3 dz œ
2 ’ 3#

47.

'"# 3u# du œ 3 '"# u# du œ 3 ”'!# u# du
'!" u# du• œ 3 Š’ 23

È#

3 È 7

ŠÈ2‹ #

#







(1)# #

œ

1# #

œ

31 # #

" #

28.

'!Þ&#Þ& x dx œ (2.5)#

30.

'È& # # r dr œ Š5È#2‹

x dx œ

3 7‹ ŠÈ

œ

32.

'!!Þ$ s# ds œ (0.3)3

3

34.

'!1Î# )# d) œ ˆ 3‰

7 3

" 24

a# #

œ

36.

'a

$ b‹ ŠÈ

3

œ

b 3

38.

'!$b x# dx œ (3b)3

40.

'!
2 È2 dx œ È2 (
#
!) œ
2È2

#




0# #“

œ 10

42.

'$& 8x dx œ "8 '$& x dx œ 8" ’ 5#

#

Št
È2‹ dt œ

È2

'!

È2

t dt
' !




0# #“




(0.5)# #

#

$

1 $ #

È$a

3a# #

$

x# dx œ

#

È

$

#

#

È2

'ab 3t dt œ "# b(3b)
"# a(3a) œ 3# ab#
a# b

#

x dx œ

" $ #

$ È b

26.

œ3 #




ŠÈ2‹ #

œ 24

œ 0.009

œ

1$ #4

#

ŠÈ3a‹

x dx œ

#

$




a# #

œ a#

œ 9b$

#




3# #“

œ

16 16

œ1


3(2
0) œ 4
6 œ
2 #

È2 dt œ

ŠÈ2‹

–

#




0# #—


È2 ’È2
0“ œ 1
2 œ
1

#

#

$




0$ 3“

$





’ "3


0# #“

#


3[0
3] œ
9  9 œ 0

0$ 3 “‹

$

œ 3 ’ 23


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1$ 3“

œ 3 ˆ 73 ‰ œ 7

Section 5.3 The Definite Integral 48.

'"Î#" 24u# du œ 24 '"Î#"

49.

'!# a3x#  x
5b dx œ 3 '!# x# dx  '!# x dx
'!# 5 dx œ 3 ’ 23

50.

'"! a3x#  x
5b dx œ
'!" a3x#  x
5b dx œ
”3 '!" x# dx  '!" x dx
'!" 5 dx•

u# du œ 24 –

'!"

u# du


'!"Î#

$

u# du— œ 24 ” 13


$

$

0$ 3‹

b
0 n

œ

œ
’3 Š 13
51. Let ?x œ

#

 Š 1#


0# #‹


5(1
0)“ œ
ˆ 3#
5‰ œ




0$ 3“

ˆ "# ‰$ 3 •

#

 ’ 2#


œ 24 ’

0# #“

ˆ 78 ‰ 3

“œ7


5[2
0] œ (8  2)
10 œ 0

7 #

and let x! œ 0, x" œ ?x,

b n

x# œ 2?xß á ß xn
" œ (n
1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 3(?x)# ?x œ 3(?x)$ f(c# ) ?x œ f(2?x) ?x œ 3(2?x)# ?x œ 3(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 3(3?x)# ?x œ 3(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 3(n?x)# ?x œ 3(n)# (?x)$ n

n

kœ1 n

kœ1

Then Sn œ ! f(ck ) ?x œ ! 3k# (?x)$ $  1) œ 3(?x) ! k# œ 3 Š bn$ ‹ Š n(n  1)(2n ‹ 6

$

kœ1

œ

$

b #

ˆ2 

52. Let ?x œ



3 n

b
0 n

"‰ n#

œ

Ê

b n

'!b 3x# dx œ n lim Ä_

b$ #

ˆ2 

3 n



"‰ n#

œ b$ .

and let x! œ 0, x" œ ?x,

x# œ 2?xß á ß xn
" œ (n
1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 1(?x)# ?x œ 1(?x)$ f(c# ) ?x œ f(2?x) ?x œ 1(2?x)# ?x œ 1(2)# (?x)$ f(c$ ) ?x œ f(3?x) ?x œ 1(3?x)# ?x œ 1(3)# (?x)$ ã f(cn ) ?x œ f(n?x) ?x œ 1(n?x)# ?x œ 1(n)# (?x)$ n

n

Then Sn œ ! f(ck ) ?x œ ! 1k# (?x)$ kœ1 n

kœ1

 1) œ 1(?x)$ ! k# œ 1 Š bn$ ‹ Š n(n  1)(2n ‹ 6 $

kœ1

œ

1b 6

$

ˆ2 

3 n



"‰ n#

Ê

'!b 1x# dx œ n lim Ä_

1 b$ 6

ˆ2 

3 n



"‰ n#

œ

307

1 b$ 3 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

308

Chapter 5 Integration b
0 n

53. Let ?x œ

œ

b n

and let x! œ 0, x" œ ?x,

x# œ 2?xß á ß xn
" œ (n
1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: f(c" ) ?x œ f(?x) ?x œ 2(?x)(?x) œ 2(?x)# f(c# ) ?x œ f(2?x) ?x œ 2(2?x)(?x) œ 2(2)(?x)# f(c$ ) ?x œ f(3?x) ?x œ 2(3?x)(?x) œ 2(3)(?x)# ã f(cn ) ?x œ f(n?x) ?x œ 2(n?x)(?x) œ 2(n)(?x)# n

n

Then Sn œ ! f(ck ) ?x œ ! 2k(?x)# kœ1 n

kœ1

œ 2(?x)# ! k œ 2 Š bn# ‹ Š n(n 2 1) ‹ #

kœ1

œ b# ˆ1  "n ‰ Ê b
0 n

54. Let ?x œ

œ

'!b 2x dx œ n lim Ä_ b n

b# ˆ1  n" ‰ œ b# .

and let x! œ 0, x" œ ?x,

x# œ 2?xß á ß xn
" œ (n
1)?x, xn œ n?x œ b. Let the ck 's be the right end-points of the subintervals Ê c" œ x" , c# œ x# , and so on. The rectangles defined have areas: " # ‰ f(c" ) ?x œ f(?x) ?x œ ˆ ?x #  1 (?x) œ # (?x)  ?x 2 ? x " f(c# ) ?x œ f(2?x) ?x œ ˆ #  1‰ (?x) œ # (2)(?x)#  ?x f(c$ ) ?x œ f(3?x) ?x œ ˆ 3?# x  1‰ (?x) œ

" #

(3)(?x)#  ?x

f(cn ) ?x œ f(n?x) ?x œ ˆ n?# x  1‰ (?x) œ

" #

(n)(?x)#  ?x

ã

n

n

kœ1

kœ1

Then Sn œ ! f(ck ) ?x œ ! ˆ "# k(?x)#  ?x‰ œ œ

" 4

b# ˆ1  n1 ‰  b Ê

55. av(f) œ Š È3"
0 ‹ œ

'! ˆ x#  1‰ dx œ n lim Ä_ b

" È3

È$

'!

È$

'!

x# dx


" #

n

n

kœ1

kœ1

(?x)# ! k  ?x ! 1 œ

ˆ 4" b# ˆ1  n" ‰  b‰ œ

" 4

" #

Š bn# ‹ Š n(n 2 1) ‹  ˆ bn ‰ (n) #

b#  b.

ax#
1b dx È$

'!

" È3

1 dx

$

œ

" È3

ŠÈ3‹




3

56. av(f) œ ˆ 3
" 0 ‰ $




" È3

ŠÈ3
0‹ œ 1
1 œ 0.

'!$ Š
x# ‹ dx œ 3" ˆ
#" ‰ '!$ x# dx #

#

œ
"6 Š 33 ‹ œ
3# ;
x# œ
3# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.3 The Definite Integral

'!" a
3x#
1b dx œ " " œ
3 ' x# dx
' 1 dx œ
3 Š 13 ‹
(1
0) ! !

57. av(f) œ ˆ 1
" 0 ‰

$

œ
#.

'!" a3x#
3b dx œ " " œ 3 ' x# dx
' 3 dx œ 3 Š 13 ‹
3(1
0) ! !

58. av(f) œ ˆ 1
" 0 ‰

$

œ
#.

'!$ (t
1)# dt $ $ $ œ 3" ' t# dt
23 ' t dt  3" ' 1 dt ! ! !

59. av(f) œ ˆ 3
" 0 ‰

œ

" 3

$

#

Š 33 ‹
32 Š 3#


60. av(f) œ Š 1
1(
2) ‹

0# #‹

 3" (3
0) œ 1.

'
#" at#
tb dt

'
#" t# dt
3" '
#" t dt "
# œ "3 ' t# dt
3" ' t# dt
3" Š 1# ! ! œ

" 3

#

œ

" 3

$

Š 13 ‹
3" Š (
32) ‹  $

61. (a) av(g) œ Š 1
"(
1) ‹

" #

œ

3 #




(
2)# # ‹

.

'
"" akxk
1b dx

'
"! (
x
1) dx  "# '!" (x
1) dx ! ! " " œ
"# ' x dx
"# ' 1 dx  "# ' x dx
"# ' 1 dx
"
" ! ! œ

" #

#

œ
"# Š 0#


(
1)# # ‹

#


"# (0
(
1))  "# Š 1#


0# #‹


"# (1
0)

œ
"# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

309

310

Chapter 5 Integration

'"$ akxk
1b dx œ #" '"$ (x
1) dx $ $ œ "# ' x dx
"# ' 1 dx œ "# Š 3#
1# ‹
"# (3
1) " "

(b) av(g) œ ˆ 3
" 1 ‰

#

#

œ 1.

(c) av(g) œ Š 3
"(
1) ‹ œ

" 4 " 4

'
"$ akxk
1b dx

'
"" akxk
1b dx  4" '"$ akxk
1b dx " 4

(see parts (a) and (b) above).

62. (a) av(h) œ Š 0
"(
1) ‹

'
"0
kxk dx œ '
"0
(
x) dx

œ

œ

(
1  2) œ

'
"0 x dx œ 0#

#

(b) av(h) œ ˆ 1
" 0 ‰ #

œ
Š "#





(
1)# #

œ
"# .

'0"
kxk dx œ
'0" x dx

0# #‹

œ
"# .

(c) av(h) œ Š 1
"(
1) ‹

'
""
kxk dx

'
"0
kxk dx  '0"
kxk dx

œ

" #

Œ

œ

" #

ˆ
"#  ˆ
"# ‰‰ œ
"# (see parts (a) and (b)

above).

63. To find where x
x#   0, let x
x# œ 0 Ê x(1
x) œ 0 Ê x œ 0 or x œ 1. If 0  x  1, then 0  x
x# Ê a œ 0 and b œ 1 maximize the integral.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.3 The Definite Integral

311

64. To find where x%
2x# Ÿ 0, let x%
2x# œ 0 Ê x# ax#
2b œ 0 Ê x œ 0 or x œ „ È2. By the sign graph,  0

0

0  , we can see that x%
2x# Ÿ 0 on ’
È2ß È2“ Ê a œ
È2 and b œ È2 ! È#
È # minimize the integral. " 1 x #

65. f(x) œ

is decreasing on [0ß 1] Ê maximum value of f occurs at 0 Ê max f œ f(0) œ 1; minimum value of f

occurs at 1 Ê min f œ f(1) œ Ê

" #

Ÿ

'0" 1 " x

" 1  1#

œ

" #

. Therefore, (1
0) min f Ÿ

dx Ÿ 1. That is, an upper bound œ 1 and a lower bound œ

#

66. See Exercise 65 above. On [0ß 0.5], max f œ

'0

0.5

(0.5
0) min f Ÿ min f œ Then

" 4

" 1  1#



2 5

" 1  0#

'0

0.5

" 1  x#

dx 

'0.5 1 " x "

#

dx Ÿ

67.
1 Ÿ sin ax# b Ÿ 1 for all x Ê (1
0)(
1) Ÿ

" #

" 1  (0.5)#

œ 1, min f œ

f(x) dx Ÿ (0.5
0) max f Ê

Ÿ

2 5

'0

'0.5 1 " x "

œ 0.5. Therefore (1
0.5) min f Ÿ

Ÿ

'0" 1 " x



2 5

Ê

0.5

#

" 1  x#

#

" #

dx Ÿ (1
0) max f .

œ 0.8. Therefore

dx Ÿ

" #

. On [0.5ß 1], max f œ

dx Ÿ (1
0.5) max f Ê

13 20

Ÿ

'0 1 " x "

#

dx Ÿ

9 10

" 4

Ÿ

" 1  (0.5)#

'0.5 1 1 x "

#

dx Ÿ

œ 0.8 and 2 5

.

.

'0" sin ax# b dx Ÿ (1
0)(1) or '0"sin x# dx Ÿ 1

Ê

'0"sin x# dx cannot

equal 2. 68. f(x) œ Èx  8 is increasing on [!ß "] Ê max f œ f(1) œ È1  8 œ 3 and min f œ f(0) œ È0  8 œ 2È2 . Therefore, (1
0) min f Ÿ

'0" Èx  8 dx Ÿ (1
0) max f

Ê 2È 2 Ÿ

'0" Èx  8 dx Ÿ 3.

69. If f(x)   0 on [aß b], then min f   0 and max f   0 on [aß b]. Now, (b
a) min f Ÿ Then b   a Ê b
a   0 Ê (b
a) min f   0 Ê

'ab f(x) dx   0.

70. If f(x) Ÿ 0 on [aß b], then min f Ÿ 0 and max f Ÿ 0. Now, (b
a) min f Ÿ b   a Ê b
a   0 Ê (b
a) max f Ÿ 0 Ê

'ab f(x) dx Ÿ 0.

'ab f(x) dx Ÿ (b
a) max f.

'ab f(x) dx Ÿ (b
a) max f.

Then

'0" (sin x
x) dx Ÿ 0 (see Exercise 70) Ê '0" sin x dx
'0" x dx '0" sin x dx Ÿ Š 1#
0# ‹ Ê '0" sin x dx Ÿ "# . Thus an upper bound is "# .

71. sin x Ÿ x for x   0 Ê sin x
x Ÿ 0 for x   0 Ê Ÿ0 Ê

'0" sin x dx Ÿ '0" x dx

72. sec x   1 

x# #

Ê

#

on ˆ
1# ß 1# ‰ Ê sec x
Š1 

x# #‹

  0 on ˆ
1# ß 1# ‰ Ê

Exercise 69) since [0ß 1] is contained in ˆ
1# ß 1# ‰ Ê  

'0" Š1  x# ‹ dx #

Ê

#

'0" ’sec x
Š1  x# ‹“ dx   0 (see #

'0"sec x dx
'0" Š1  x# ‹ dx   0

'0" sec x dx   '0" 1 dx  "# '0" x# dx

#

Ê

Ê

'0" sec x dx

'0" sec x dx   (1
0)  "# Š 13 ‹ $

Ê

Thus a lower bound is 76 .

'ab f(x) dx is a constant K. Thus'ab av(f) dx œ 'ab K dx 'ab av(f) dx œ (b
a)K œ (b
a) † b
" a 'ab f(x) dx œ 'ab f(x) dx.

73. Yes, for the following reasons: av(f) œ œ K(b
a) Ê

" b
a

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

'0" sec x dx   76 .

312

Chapter 5 Integration

74. All three rules hold. The reasons: On any interval [aß b] on which f and g are integrable, we have: (a) av(f  g) œ

" b
a

'ab [f(x)  g(x)] dx œ b
" a ”'ab f(x) dx  'ab g(x) dx• œ b
" a 'ab f(x) dx  b
" a 'ab g(x) dx

œ av(f)  av(g) (b) av(kf) œ (c) av(f) œ

" b
a

" b
a

'ab kf(x) dx œ b
" a ”k 'ab f(x) dx• œ k ” b
" a 'ab f(x) dx• œ k av(f)

'ab f(x) dx Ÿ b
" a 'ab g(x) dx since f(x) Ÿ g(x) on [aß b], and b
" a 'ab g(x) dx œ av(g).

Therefore, av(f) Ÿ av(g). b
a n and let ck be the right n ab
a b × and ck œ a  kabn
ab . n

75. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ endpoint of each subinterval. So the partition is P œ Öa, a  n

n

kœ"

kœ" b

We get the Riemann sum ! fack b˜x œ ! c † this expression remains cab
ab. Thus,

b
a n

œ

b
a n , n

c ab
a b ! " n kœ"

a

œ

#a b
a b , n

c ab
a b n

...,a

† n œ cab
ab. As n Ä _ and mPm Ä !

'a c dx œ cab
ab.

76. Consider the partition P that subdivides the interval Òa, bÓ into n subintervals of width ˜x œ endpoint of each subinterval. So the partition is P œ n

n

We get the Riemann sum ! fack b˜x œ ! ck# ˆ b
n a ‰ œ

n b
a ! # a n Œ kœ"



kœ" n #a a b
a b ! k n kœ"

œ ab
aba#  aab
ab# †



n" n

kœ" n ab
a b # ! # k  n# kœ"



ab
a b $ '

†

œ

b
a n

b
a n

and let ck be the right

Öa, a  b
n a , a  #abn
ab , . . ., a  nabn
ab × and ck œ a  kabn
ab . n n # # # œ b
n a ! Ša  kabn
ab ‹ œ b
n a ! Ša#  #akabn
ab  k abn
# ab ‹ kœ" kœ" † na# 

an  "ba#n  "b n#

#a a b
a b# n#

†

n a n  "b #



ab
a b $ n$

†

nan  "ba#n  "b '

$ " "  n" ab
ab$ #  n  n# † "  ' " ab
a b $ †# ' b $ $ x# dx œ b$
a$ . a

œ ab
aba#  aab
ab# †

As n Ä _ and mPm Ä ! this expression has value ab
aba#  aab
ab# † "  œ ba#
a$  ab#
#a# b  a$  "$ ab$
$b# a  $ba#
a$ b œ

b$ $




a$ $.

Thus,

'

77. (a) U œ max" ?x  max# ?x  á  maxn ?x where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on [aß b]; L œ min" ?x  min# ?x  á  minn ?x where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U
L œ (max"
min" ) ?x  (max#
min# ) ?x  á  (maxn
minn ) ?x œ (f(x" )
f(x! )) ?x  (f(x# )
f(x" ))?x  á  (f(xn )
f(xnc1 )) ?x œ (f(xn )
f(x! )) ?x œ (f(b)
f(a)) ?x. (b) U œ max" ?x"  max# ?x#  á  maxn ?xn where max" œ f(x" ), max# œ f(x# ), á , maxn œ f(xn ) since f is increasing on[aß b]; L œ min" ?x"  min# ?x#  á  minn ?xn where min" œ f(x! ), min# œ f(x" ), á , minn œ f(xnc1 ) since f is increasing on [aß b]. Therefore U
L œ (max"
min" ) ?x"  (max#
min# ) ?x#  á  (maxn
minn ) ?xn œ (f(x" )
f(x! )) ?x"  (f(x# )
f(x" ))?x#  á  (f(xn )
f(xnc1 )) ?xn Ÿ (f(x" )
f(x! )) ?xmax  (f(x# )
f(x" )) ?xmax  á  (f(xn )
f(xnc1 )) ?xmax . Then U
L Ÿ (f(xn )
f(x! )) ?xmax œ (f(b)
f(a)) ?xmax œ kf(b)
f(a)k ?xmax since f(b)   f(a). Thus lim (U
L) œ lim (f(b)
f(a)) ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0

lPl Ä 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.3 The Definite Integral

313

78. (a) U œ max" ?x  max# ?x  á  maxn ?x where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xnc" ) since f is decreasing on [aß b]; L œ min" ?x  min# ?x  á  minn ?x where min" œ f(x" ), min# œ f(x# )ß á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U
L œ (max"
min" ) ?x  (max#
min# ) ?x  á  (maxn
minn ) ?x œ (f(x! )
f(x" )) ?x  (f(x" )
f(x# ))?x  á  (f(xn
" )
f(xn )) ?x œ (f(x! )
f(xn )) ?x œ (f(a)
f(b)) ?x. (b) U œ max" ?x"  max# ?x#  á  maxn ?xn where max" œ f(x! ), max# œ f(x" ), á , maxn œ f(xn
" ) since f is decreasing on[aß b]; L œ min" ?x"  min# ?x#  á  minn ?xn where min" œ f(x" ), min# œ f(x# ), á , minn œ f(xn ) since f is decreasing on [aß b]. Therefore U
L œ (max"
min" ) ?x"  (max#
min# ) ?x#  á  (maxn
minn ) ?xn œ (f(x! )
f(x" )) ?x"  (f(x" )
f(x# ))?x#  á  (f(xn
" )
f(xn )) ?xn Ÿ (f(x! )
f(xn )) ?xmax œ (f(a)
f(b) ?xmax œ kf(b)
f(a)k ?xmax since f(b) Ÿ f(a). Thus lim (U
L) œ lim kf(b)
f(a)k ?xmax œ 0, since ?xmax œ lPl . lPl Ä 0

lPl Ä 0

79. (a) Partition 0ß 1# ‘ into n subintervals, each of length ?x œ x# œ 2?x, á , xn œ n?x œ

1 #.

1 #n

with points x! œ 0, x" œ ?x,

Since sin x is increasing on 0ß 1# ‘ , the upper sum U is the sum of the areas

of the circumscribed rectangles of areas f(x" ) ?x œ (sin ?x)?x, f(x# ) ?x œ (sin 2?x) ?x, á , f(xn ) ?x œ (sin n?x) ?x. Then U œ (sin ?x  sin 2?x  á  sin n?x) ?x œ ” œ”

1
cos ˆˆn  " ‰ 1 ‰ cos 4n 1 # 2n 1 • ˆ #n ‰ # sin 4n

'!

1
cos ˆ 1  1 ‰‰ 1 ˆcos 4n # 4n 1 4n sin 4n

œ

1
cos ˆ 1  1 ‰ cos 4n # 4n sin 1

Š 14n ‹ 4n

1Î#

(b) The area is

œ

cos ?#x
cosˆ ˆn  #" ‰ ?x‰ • ?x # sin ?#x

sin x dx œ n lim Ä_

1
cos ˆ 1  1 ‰ cos 4n # 4n sin 1

Š 14n ‹

œ

1
cos 1# 1

œ 1.

4n

n

80. (a) The area of the shaded region is !˜xi † mi which is equal to L. iœ" n

(b) The area of the shaded region is !˜xi † Mi which is equal to U. iœ"

(c) The area of the shaded region is the difference in the areas of the shaded regions shown in the second part of the figure and the first part of the figure. Thus this area is U
L. n

n

iœ"

iœ"

81. By Exercise 80, U
L œ !˜xi † Mi
!˜xi † mi where Mi œ maxÖfaxb on the ith subinterval× and n

n

iœ"

iœ"

mi œ minÖfaxb on the ith subinterval×. Thus U
L œ !aMi
mi b˜xi  !% † ˜xi provided ˜xi  $ for each n

n

iœ"

iœ"

i œ "ß Þ Þ Þ , n. Since !% † ˜xi œ % !˜xi œ %ab
ab the result, U
L  %ab
ab follows.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

314

Chapter 5 Integration

82. The car drove the first 150 miles in 5 hours and the second 150 miles in 3 hours, which means it drove 300 miles in 8 hours, for an average of 300 8 mi/hr œ 37.5 mi/hr. In terms of average values of functions, the function whose average value we seek is 30, 0 Ÿ t Ÿ 5 v(t) œ œ , and the average value is 50, 5  1 Ÿ 8 (30)(5)  (50)(3) 8

œ 37.5.

83-88. Example CAS commands: Maple: with( plots ); with( Student[Calculus1] ); f := x -> 1-x; a := 0; b := 1; N :=[ 4, 10, 20, 50 ]; P := [seq( RiemannSum( f(x), x=a..b, partition=n, method=random, output=plot ), n=N )]: display( P, insequence=true ); 89-92. Example CAS commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi; plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); 83-92. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Sums of rectangles evaluated at left-hand endpoints can be represented and evaluated by this set of commands Clear[x, f, a, b, n]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.4 The Fundamental Theorem of Calculus {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a, b
dx, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals, xvals  dx, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N Sums of rectangles evaluated at right-hand endpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a  dx, b, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals
dx,xvals, yvals}]; Plot[f, {x, a, b}, Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1,Length[yvals]}]//N Sums of rectangles evaluated at midpoints can be represented and evaluated by this set of commands. Clear[x, f, a, b, n] {a, b}={0, 1}; n =10; dx = (b
a)/n; f = Sin[x]2 ; xvals =Table[N[x], {x, a  dx/2, b
dx/2, dx}]; yvals = f /.x Ä xvals; boxes = MapThread[Line[{{#1,0},{#1, #3},{#2, #3},{#2, 0}]&,{xvals
dx/2, xvals  dx/2, yvals}]; Plot[f, {x, a, b},Epilog Ä boxes]; Sum[yvals[[i]] dx, {i, 1, Length[yvals]}]//N 5.4 THE FUNDAMENTAL THEOREM OF CALCULUS 1.

'c (2x  5) dx œ cx#  5xd
#! œ a0#  5(0)b
a(
2)#  5(
2)b œ 6

2.

'c ˆ5
x# ‰ dx œ ’5x
x4 “ %

0

2

4

3.

#


$

3

'

4

0

Š3x


x$ 4‹

'c ax$
2x  3b dx œ ’ x4

%

'

6.

'

7.

'

1

#
#


Š5(
3)


4% 16 ‹

#


Š 3(0) #


(
3)# 4 ‹

(0)% 16 ‹

œ

133 4

œ8 %

œ Š 24
2#  3(2)‹
Š (
42)
(
2)#  3(
2)‹ œ 12 %

"

$

ˆx#  Èx‰ dx œ ’ x3  23 x$Î# “ œ ˆ "3  23 ‰
0 œ 1 !

0

0

#

4# 4‹

œ Š 3(4) #



x#  3x“

2

5.

5

&

x$Î# dx œ  25 x&Î# ‘ ! œ

32

1

(5)&Î#
0 œ 2(5)$Î# œ 10È5

$#

'cc x2 2

2 5

x
'Î& dx œ 
5x
"Î& ‘ " œ ˆ
#5 ‰
(
5) œ

1

8.

% x% 16 “ !

#

dx œ ’ 3x#


2

4.

œ Š5(4)


#

dx œ

5 #

'cc 2x
# dx œ c
2x
" d
" ˆ
2 ‰ ˆ
2 ‰
# œ
1

# œ 1 1

2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

315

316

Chapter 5 Integration

'

1

9.

'

1

10.

'

1Î3

11.

'

51Î6

12.

'

31Î4

13.

'

1Î3

14.

15.

'

0

0

sin x dx œ [
cos x]1! œ (
cos 1)
(
cos 0) œ
(
1)
(
1) œ 2 (1  cos x) dx œ [x  sin x]1! œ (1  sin 1)
(0  sin 0) œ 1

0

1Î$

1Î6

1Î4

0

œ ˆ2 tan ˆ 13 ‰‰
(2 tan 0) œ 2È3
0 œ 2È3

2 sec# x dx œ [2 tan x]!

&1Î' csc# x dx œ [
cot x]1Î' œ ˆ
cot ˆ 561 ‰‰
ˆ
cot ˆ 16 ‰‰ œ
Š
È3‹
Š
È3‹ œ 2È3

$1Î%

csc ) cot ) d) œ [
csc )]1Î% œ ˆ
csc ˆ 341 ‰‰
ˆ
csc ˆ 14 ‰‰ œ
È2
Š
È2‹ œ 0 1Î$

4 sec u tan u du œ [4 sec u]!

0

"  cos 2t # 1Î2

dt œ

'

0

ˆ" 

1Î2 #

" #

œ 4 sec ˆ 13 ‰
4 sec 0 œ 4(2)
4(1) œ 4

cos 2t‰ dt œ  "# t 

" 4

!

sin 2t‘ 1Î# œ ˆ "# (0) 

" 4

sin 2(0)‰
ˆ "# ˆ 1# ‰ 

" 4

sin 2 ˆ 1# ‰‰

œ
14 Î 1Î$ 2t 'c ÎÎ "
cos dt œ ' ˆ "#
"# cos 2t‰ dt œ  "# t
4" sin 2t‘
1Î$ #
Î 1 3

16.

1 3

1 3

1 3

œ ˆ "# ˆ 13 ‰


" 4

sin 2 ˆ 13 ‰‰
ˆ #" ˆ
13 ‰


" 4

sin 2 ˆ
13 ‰‰ œ

1 6




" 4

sin 231 

17.

'c

18.

'c
Î Î% ˆ4 sec# t  t1 ‰ dt œ '

Î Î% a4 sec# t  1t
# b dt œ 4 tan t
1t ‘

11Î%Î$

1Î#

1Î#

$

a8y#  sin yb dy œ ’ 8y3
cos y“

1Î#


1Î#

1

œŒ

8 ˆ 1# ‰ 3

$

8 ˆ
1# ‰ 3


cos 1# 
Œ

1 6 $

œ Š4 tan ˆ



cos ˆ
1# ‰ œ

1‰ 4





Š4 tan ˆ 13 ‰


1 ˆ
13 ‰ ‹




È3 4

21 $ 3

œ (4(
1)  4)
Š4 Š
È3‹  3‹ œ 4È3
3

'"
" (r  1)# dr œ '"
" ar#  2r  1b dr œ ’ r3  r#  r“
" œ Š (
31)

20.

'
È (t  1) at#  4b dt œ '
È at$  t#  4t  4b dt œ ’ t4  t3  2t#  4t“ÈÈ$

$

"

È3

È3

3

œ


%

$

$

 (
1)#  (
1)‹
Š 13  1#  1‹ œ
38

$




3

%

ŠÈ3‹

$

ŠÈ3‹



4

3

21.

'È" Š u#




" u& ‹

22.

' " ˆ v"




"‰ v%

23.

'

(

2

1

1 3

1 3

1 ˆ
14 ‰ ‹

19.

È2

sin ˆ
321 ‰ œ

1

#

1 3

1Î2

" 4



$

s#  È s s#

(

2

dv œ

ds œ

'

1


È 3 ‹

#

Š  2 ŠÈ3‹  4È3



" du œ 'È Š u#

%


&


u ‹ du œ

u) ’ 16



4

$



" " 4u% “È#

Š
È3‹

È2

ˆ1  s
$Î# ‰ ds œ ’s


#

)

œ

1) Š 16

$

È# 2 “ Ès "

#

 2 Š
È3‹  4 Š
È3‹ œ 10È3

3




" ' " av
$
v
% b dv œ  2v
1  3v" ‘ ""Î# œ Š 2(1) 1Î2

$

#

œ
È 2


" 4(1)% ‹







" 3(1)$ ‹

2 É È2 

ŠÈ2‹ 16



" % 4 ŠÈ2‹

" $ 3 ˆ "# ‰

œ
34


Œ


" # 2 ˆ "# ‰




Š1


2 È1 ‹

œ È2
2$Î%  1

œ È2
%È8  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ
56

Section 5.4 The Fundamental Theorem of Calculus 24.

'

4

1
Èu Èu 9

du œ

'

4

9

ˆu
"Î#
1‰ du œ 2Èu
u‘ % œ Š2È4
4‹
Š2È9
9‹ œ 3 *

'c% kxk dx œ '
%! kxk dx  '! 4

25.

4

kxk dx œ


'
%! x dx  '!

4

#

x dx œ ’
x# “

œ 16 26.

'

1

" ! #

acos x  kcos xk b dx œ 1 #

œ sin

27. (a)

!

d dx

28. (a)

'

(b)

d dx

29. (a)

'

(b)

d dt

30. (a)

'

!

" # (cos

x  cos x) dx 

'

1

" 1Î# #

#

 ’ x# “ œ Š
0#  !

(cos x
cos x) dx œ

'

1Î#

!

(
4)# # ‹

#

 Š 4#


Èx

cos t dt œ [sin t]! œ sin Èx
sin 0 œ sin Èx Ê

Èx

'

Œ

sin x

1

!

d dx

Œ

'

Èx

!

cos t dt œ

sin x

'

!

d ˆÈ ‰‰ cos t dt œ ˆcos Èx‰ ˆ dx x œ ˆcos Èx ‰ ˆ "# x
"Î# ‰ œ

3t# dt œ ct$ d "

Πt%

1Î#

d dx

ˆsin Èx‰ œ cos Èx ˆ "# x
"Î# ‰

sin x

Èu du œ t%

!

tan )

!

d dx

Œ

'

sin x

1

3t# dt œ

d dx

asin$ x
1b œ 3 sin# x cos x

d 3t# dt œ a3 sin# xb ˆ dx (sin x)‰ œ 3 sin# x cos x

1

Œ'

œ sin$ x
1 Ê

cos Èx 2È x

'

t%

!

t%

u"Î# du œ  23 u$Î# ‘ ! œ

2 3

at% b

$Î#


0œ

2 ' 3 t

Ê

d dt

Œ'

Œ

'

t%

!

Èu du œ

d dt

ˆ 23 t' ‰ œ 4t&

Èu du œ Èt% ˆ dtd at% b‰ œ t# a4t$ b œ 4t&

) sec# y dy œ [tan y]tan œ tan (tan ))
0 œ tan (tan )) Ê !

d d)

tan )

!

sec# y dy œ

d d)

(tan (tan )))

œ asec# (tan ))b sec# ) (b)

d d)

'

Œ

tan )

!

sec# y dy œ asec# (tan ))b ˆ dd) (tan ))‰ œ asec# (tan ))b sec# )

31. y œ

'

33. y œ

'È sin t# dt œ
'

34. y œ

'

35. y œ

'

36. y œ

'

x

!

È1  t# dt Ê

Èx

!

x

x#

cos Èt dt Ê

sin x

dt È1
t#

!

!

tan x

dt 1  t#

, kxk  Ê

dy dx

œ È1  x#

dy dx

!

!

0# #‹

cos x dx œ [sin x]!

cos Èx 2È x

œ (b)

1Î#

%

#


%


sin 0 œ 1

Èx

'

'

!

dy dx

1 #

sin t# dt Ê

32. y œ dy dx

'

1

x

" t

dt Ê

dy dx

œ

" x

,x0

#

d ˆÈ ‰‰ œ
Šsin ˆÈx‰ ‹ ˆ dx x œ
(sin x) ˆ "# x
"Î# ‰ œ
2sinÈxx

d œ Šcos Èx# ‹ ˆ dx ax# b‰ œ 2x cos kxk

Ê

dy dx

œ

" È1
sin# x

d ˆ dx (sin x)‰ œ

" Ècos# x

(cos x) œ

cos x kcos xk

œ

cos x cos x

œ 1 since kxk 

" d ‰ ˆ dx œ ˆ 1  tan (tan x)‰ œ ˆ sec"# x ‰ asec# xb œ 1 #x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 #

317

318

Chapter 5 Integration

37.
x#
2x œ 0 Ê
x(x  2) œ 0 Ê x œ 0 or x œ
2; Area œ


'
$
# a
x#
2xbdx  '
#! a
x#
2xbdx
'!# a
x#
2xbdx $

œ
’
x3
x# “ œ
ŠŠ


(
2)$ 3


#
$

$

 ’
x3
x# “ #


(
2) ‹
Š


!

$


#

(
3)$ 3


’
x3
x# “

#

!

#


(
3) ‹‹

$

 ŠŠ
03
0# ‹
Š
(
32)
(
2)# ‹‹ $

$

$


ŠŠ
23
2# ‹
Š
03
0# ‹‹ œ

28 3

38. 3x#
3 œ 0 Ê x# œ 1 Ê x œ „ 1; because of symmetry about the y-axis, Area œ 2 Œ


'!" a3x#
3bdx  '"# a3x#
3bdx

"

#

2 Š
cx$
3xd !  cx$
3xd " ‹ œ 2 c
aa1$
3(1)b
a0$
3(0)bb  aa2$
3(2)b
a1$
3(1)bd œ 2(6) œ 12

39. x$
3x#  2x œ 0 Ê x ax#
3x  2b œ 0 Ê x(x
2)(x
1) œ 0 Ê x œ 0, 1, or 2; Area œ

'!" ax$
3x#  2xbdx
'"# ax$
3x#  2xbdx "

%

%

œ ’ x4
x$  x# “
’ x4
x$  x# “ !

%

# "

%

œ Š 14
1$  1# ‹
Š 04
0$  0# ‹ %

%


’Š 24
2$  2# ‹
Š 14
1$  1# ‹“ œ

" #

40. x$
4x œ 0 Ê x ax#
4b œ 0 Ê x(x
2)(x  2) œ 0 Ê x œ 0, 2, or
2. Area œ œ

% ’ x4

#


2x “ %

!
#




% ’ x4

'c! ax$
4xbdx
'!# ax$
4xbdx 2

#

#

%


2x “ œ Š 04
2(0)# ‹ !


Š (
42)
2(
2)# ‹
’Š 24
2(2)# ‹
Š 04
2(0)# ‹“ œ 8 %

41. x"Î$ œ 0 Ê x œ 0; Area œ


%

'c! x"Î$ dx  '!) x"Î$ dx "

! ) œ 
34 x%Î$ ‘
"   34 x%Î$ ‘ ! œ ˆ
34 (0)%Î$ ‰
ˆ
34 (
1)%Î$ ‰  ˆ 34 (8)%Î$ ‰
ˆ 34 (0)%Î$ ‰

œ

51 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.4 The Fundamental Theorem of Calculus 42. x"Î$
x œ 0 Ê x"Î$ ˆ1
x#Î$ ‰ œ 0 Ê x"Î$ œ 0 or 1
x#Î$ œ 0 Ê x œ 0 or 1 œ x#Î$ Ê x œ 0 or 1 œ x# Ê x œ 0 or „ 1; Area œ
œ

'c! ˆx"Î$
x‰dx  '!" ˆx"Î$
x‰dx
'") ˆx"Î$
x‰dx "

%Î$


’ 34

x




! x# # “
"

œ
’Š 34 (0)%Î$


 ’ 34 x%Î$


0# #‹

" x# # “!


’ 43 x%Î$
(
1)# # ‹“


Š 34 (
1)%Î$


 ’Š 34 (1)%Î$


1# #‹


Š 34 (0)%Î$


0# # ‹“


’Š 34 (8)%Î$


8# #‹


Š 34 (1)%Î$


1# # ‹“

œ

" 4

" 4




ˆ
2!


$ 4

 #" ‰ œ

) x# # “"

83 4

43. The area of the rectangle bounded by the lines y œ 2, y œ 0, x œ 1, and x œ 0 is 21. The area under the curve y œ 1  cos x on [0ß 1] is

'!

1

(1  cos x) dx œ [x  sin x]!1 œ (1  sin 1)
(0  sin 0) œ 1. Therefore the area of

the shaded region is 21
1 œ 1. 44. The area of the rectangle bounded by the lines x œ 16 , x œ " #

51 6 ,

y œ sin

ˆ 561
16 ‰ œ 13 . The area under the curve y œ sin x on  16 ß 561 ‘ is

œ ˆ
cos

51 ‰ 6

È3 # ‹


ˆ
cos 16 ‰ œ
Š


È3 #



'

1 6

œ

51Î6

1Î6

" #

œ sin

51 6

, and y œ 0 is &1Î'

sin x dx œ [
cos x]1Î'

œ È3. Therefore the area of the shaded region is È3
13 .

45. On 
14 ß 0‘ : The area of the rectangle bounded by the lines y œ È2, y œ 0, ) œ 0, and ) œ
14 is È2 ˆ 14 ‰ œ

1È2 4

. The area between the curve y œ sec ) tan ) and y œ 0 is


'c

!

1Î4

sec ) tan ) d) œ [
sec )]!
1Î%

œ (
sec 0)
ˆ
sec ˆ
14 ‰‰ œ È2
1. Therefore the area of the shaded region on 
14 ß !‘ is

1È2 4

On 0ß 14 ‘ : The area of the rectangle bounded by ) œ 14 , ) œ 0, y œ È2, and y œ 0 is È2 ˆ 14 ‰ œ under the curve y œ sec ) tan ) is of the shaded region on !ß 14 ‘ is È

'

1Î4

!

1È2 4

1Î%

sec ) tan ) d) œ [sec )]!

œ sec

1 4

 ŠÈ2
1‹ .

1È2 4

. The area


sec 0 œ È2
1. Therefore the area


ŠÈ2
1‹ . Thus, the area of the total shaded region is

È

1È2 #

Š 1 4 2  È2
1‹  Š 1 4 2
È2  1‹ œ

.

46. The area of the rectangle bounded by the lines y œ 2, y œ 0, t œ
14 , and t œ 1 is 2 ˆ1
ˆ
14 ‰‰ œ 2  area under the curve y œ sec# t on 
14 ß !‘ is under the curve y œ 1
t# on [!ß "] is

'c

!

1Î4

"

$

œ

dt
3 œ 0
3 œ
3 Ê (d) is a solution to this problem.

dy dx

œ

48. y œ

'c sec t dt  4

Ê

dy dx

œ sec x and y(
1) œ

49. y œ

'

sec t dt  4 Ê

dy dx

œ sec x and y(0) œ

x

1

!

x

and y(1) œ

'

" t

dt
3 Ê

" x

1

1

Thus, the total

. Therefore the area of the shaded region is ˆ2  1# ‰


'

" 1 t

$

5 3

47. y œ

x

$

!

2 3

'

!

. The

! ˆ 1‰ sec# t dt œ [tan t]
1Î% œ tan 0
tan
4 œ 1. The area

'! a1
t# b dt œ ’t
t3 “ " œ Š1
13 ‹
Š0
03 ‹ œ 32 .

area under the curves on 
14 ß "‘ is 1 

1 #

'cc sec t dt  4 œ 0  4 œ 4 1

1

!

5 3

œ

" 3



1 #

.

Ê (c) is a solution to this problem.

sec t dt  4 œ 0  4 œ 4 Ê (b) is a solution to this problem.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

319

320

Chapter 5 Integration

50. y œ

'

51. y œ

'

53. s œ

'

x

" " t

"

" t

dt
3 œ 0
3 œ
3 Ê (a) is a solution to this problem.

'

'c ÎÎ

b 2 b 2

$


ˆ
bh # 

bh ‰ 6

Š2


b 
Œh ˆ
# ‰


2 (x  1)# ‹

bh ‰ 6

dx œ 2

'

$

!

œ

bh 3

t

t!

È1  t# dt
2 g(x) dx  v!

$

2 3

 bh

" (x  1)# ‹

Š1


"

bÎ2

4h ˆ
#b ‰ 3b#

œ bh


x

4hx$ 3b# “
bÎ2

ˆh
ˆ 4h ‰ # ‰ dx œ ’hx
b# x

œ ˆ bh #


!

"

54. v œ

4h ˆ b# ‰ 3b#

$

'

f(x) dx  s!

œ Œhˆ #b ‰


'

and y(1) œ

'

55. Area œ

56. r œ

" x

52. y œ

#

t!

œ

dy dx

sec t dt  3

x

t

dt
3 Ê

$

dx œ 2 x
ˆ x
11 ‰‘ ! œ 2 ’Š3 

" (3  1) ‹


Š0 

" (0  1) ‹“

œ 2 3 "4
1‘ œ 2 ˆ2 4" ‰ œ 4.5 or $4500 57.

dc dx

œ

" #È x

œ

" #

x
"Î# Ê c œ

'

x

!

"
"Î# dt # t

œ t"Î# ‘ 0 œ Èx x

c(100)
c(1) œ È100
È1 œ $9.00 58. By Exercise 57, c(400)
c(100) œ È400
È100 œ 20
10 œ $10.00 59. (a) v œ (b) a œ (c) s œ (d) (e) (f) (g)

ds dt df dt

'

!

œ

d dt

'

t

!

f(x) dx œ f(t) Ê v(5) œ f(5) œ 2 m/sec

is negative since the slope of the tangent line at t œ 5 is negative 3

f(x) dx œ

" #

(3)(3) œ

9 #

m since the integral is the area of the triangle formed by y œ f(x), the x-axis,

and x œ 3 t œ 6 since from t œ 6 to t œ 9, the region lies below the x-axis At t œ 4 and t œ 7, since there are horizontal tangents there Toward the origin between t œ 6 and t œ 9 since the velocity is negative on this interval. Away from the origin between t œ 0 and t œ 6 since the velocity is positive there. Right or positive side, because the integral of f from 0 to 9 is positive, there being more area above the x-axis than below it.

60. (a) v œ (b) a œ

dg dt df dt

œ

d dt

'

!

t

g(x) dx œ g(t) Ê v(3) œ g(3) œ 0 m/sec.

is positive, since the slope of the tangent line at t œ 3 is positive

(c) At t œ 3, the particle's position is

'

!

$

g(x) dx œ

" #

(3)(
6) œ
9

(d) The particle passes through the origin at t œ 6 because s(6) œ

'

!

'

g(x) dx œ 0

(e) At t œ 7, since there is a horizontal tangent there (f) The particle starts at the origin and moves away to the left for 0  t  3. It moves back toward the origin for 3  t  6, passes through the origin at t œ 6, and moves away to the right for t  6. (g) Right side, since its position at t œ 9 is positive, there being more area above the x-axis than below it at t œ *.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.4 The Fundamental Theorem of Calculus 61. k  0 Ê one arch of y œ sin kx will occur over the interval 0ß 1k ‘ Ê the area œ œ
"k cos ˆk ˆ 1k ‰‰
ˆ
k" cos (0)‰ œ 62. lim x"$ xÄ!

63.

'

64.

'

x

1

x

!

'

!

x

t%

t# dt "

œ lim

' x %t# dt ! t
" x$

xÄ!

x

66.

1

x#

" k

x#

d dx

xÄ!

'

d dx

'

!

1

x

x

f(t) dt œ

d dx

ax#
2x  1b œ 2x
2

f(t) dt œ cos 1x
1x sin 1x Ê f(4) œ cos 1(4)
1(4) sin 1(4) œ 1

Ê f w (x) œ
1  (x9  1) œ


9 x 2

Ê f w (1) œ
3; f(1) œ 2


'#"
" 1 9 t dt œ 2
0 œ 2;

sec (t
1) dt Ê gw (x) œ asec ax#
1bb (2x) œ 2x sec ax#
1b Ê gw (
1) œ 2(
1) sec a(
1)#
1b #

a
"b " œ
2; g(
1) œ 3  ' sec (t
1) dt œ 3  ' sec (t
1) dt œ 3  0 œ 3; L(x) œ
2(x
(
1))  g(
1) 1

1

œ
2(x  1)  3 œ
2x  1 67. (a) (b) (c) (d) (e) (f) (g)

True: since f is continuous, g is differentiable by Part 1 of the Fundamental Theorem of Calculus. True: g is continuous because it is differentiable. True, since gw (1) œ f(1) œ 0. False, since gww (1) œ f w (1)  0. True, since gw (1) œ 0 and gww (1) œ f w (1)  0. False: gww (x) œ f w (x)  0, so gww never changes sign. True, since gw (1) œ f(1) œ 0 and gw (x) œ f(x) is an increasing function of x (because f w (x)  0).

68. (a) True: by Part 1 of the Fundamental Theorem of Calculus, hw (x) œ f(x). Since f is differentiable for all x, h has a second derivative for all x. (b) True: they are continuous because they are differentiable. (c) True, since hw (1) œ f(1) œ 0. (d) True, since hw (1) œ 0 and hww (1) œ f w (1)  0. (e) False, since hww (1) œ f w (1)  0. (f) False, since hww (x) œ f w (x)  0 never changes sign. (g) True, since hw (1) œ f(1) œ 0 and hw (x) œ f(x) is a decreasing function of x (because f w (x)  0). 69.

1 Îk

cos kx‘ !

%

L(x) œ
3(x
1)  f(1) œ
3(x
1)  2 œ
3x  5 g(x) œ 3  '

sin kx dx œ 


2 k

xÄ!

f(t) dt œ x cos 1x Ê f(x) œ

'#
" 1 9 t dt

!

1Îk

œ lim x$x
#" œ lim $ax%" "b œ _.

f(t) dt œ x#
2x  1 Ê f(x) œ

65. f(x) œ 2


'

321

70. The limit is 3x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

322

Chapter 5 Integration

71-74. Example CAS commands: Maple: with( plots ); f := x -> x^3-4*x^2+3*x; a := 0; b := 4; F := unapply( int(f(t),t=a..x), x ); # (a) p1 := plot( [f(x),F(x)], x=a..b, legend=["y = f(x)","y = F(x)"], title="#71(a) (Section 5.4)" ): p1; dF := D(F); # (b) q1 := solve( dF(x)=0, x ); pts1 := [ seq( [x,f(x)], x=remove(has,evalf([q1]),I) ) ]; p2 := plot( pts1, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where F '(x)=0" ): display( [p1,p2], title="71(b) (Section 5.4)" ); incr := solve( dF(x)>0, x ); # (c) decr := solve( dF(x)<0, x ); df := D(f); # (d) p3 := plot( [df(x),F(x)], x=a..b, legend=["y = f '(x)","y = F(x)"], title="#71(d) (Section 5.4)" ): p3; q2 := solve( df(x)=0, x ); pts2 := [ seq( [x,F(x)], x=remove(has,evalf([q2]),I) ) ]; p4 := plot( pts2, style=point, color=blue, symbolsize=18, symbol=diamond, legend="(x,f(x)) where f '(x)=0" ): display( [p3,p4], title="71(d) (Section 5.4)" ); 75-78. Example CAS commands: Maple: a := 1; u := x -> x^2; f := x -> sqrt(1-x^2); F := unapply( int( f(t), t=a..u(x) ), x ); dF := D(F); # (b) cp := solve( dF(x)=0, x ); solve( dF(x)>0, x ); solve( dF(x)<0, x ); d2F := D(dF); # (c) solve( d2F(x)=0, x ); plot( F(x), x=-1..1, title="#75(d) (Section 5.4)" ); 79.

Example CAS commands: Maple: f := `f`; q1 := Diff( Int( f(t), t=a..u(x) ), x ); d1 := value( q1 );

80.

Example CAS commands: Maple: f := `f`; q2 := Diff( Int( f(t), t=a..u(x) ), x,x );

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.5 Indefinite Integrals and the Substitution Rule value( q2 ); 71-80. Example CAS commands: Mathematica: (assigned function and values for a, and b may vary) For transcendental functions the FindRoot is needed instead of the Solve command. The Map command executes FindRoot over a set of initial guesses Initial guesses will vary as the functions vary. Clear[x, f, F] {a, b}= {0, 21}; f[x_] = Sin[2x] Cos[x/3] F[x_] = Integrate[f[t], {t, a, x}] Plot[{f[x], F[x]},{x, a, b}] x/.Map[FindRoot[F'[x]==0, {x, #}] &,{2, 3, 5, 6}] x/.Map[FindRoot[f'[x]==0, {x, #}] &,{1, 2, 4, 5, 6}] Slightly alter above commands for 75 - 80. Clear[x, f, F, u] a=0; f[x_] = x2
2x
3 u[x_] = 1
x2 F[x_] = Integrate[f[t], {t, a, u(x)}] x/.Map[FindRoot[F'[x]==0,{x, #}] &,{1, 2, 3, 4}] x/.Map[FindRoot[F''[x]==0,{x,#}] &,{1, 2, 3, 4}] After determining an appropriate value for b, the following can be entered b = 4; Plot[{F[x], {x, a, b}] 5.5 INDEFINTE INTEGRALS AND THE SUBSTITUTION RULE " 3

1. Let u œ 3x Ê du œ 3 dx Ê

'

sin 3x dx œ '

" 3

du œ dx

sin u du œ
3" cos u  C œ
3" cos 3x  C " 4

2. Let u œ 2x# Ê du œ 4x dx Ê

'

x sin a2x b dx œ '

" 4

#

sin u du œ
4" cos u  C œ
4" cos 2x#  C " #

3. Let u œ 2t Ê du œ 2 dt Ê

'

sec 2t tan 2t dt œ '

" #

'

ˆ1
cos

du œ dt

sec u tan u du œ

4. Let u œ 1
cos 2t Ê du œ t ‰# #

" #

sin

28(7x
2)
& dx œ '

" 7

" 7

x$ ax%
1b dx œ ' #

" 4

2 3

" #

sec u  C œ

u$  C œ

t 2 2 3

sec 2t  C

dt

ˆ1
cos #t ‰$  C

du œ dx

(28)u
& du œ ' 4u
& du œ
u
%  C œ
(7x
2)
%  C

6. Let u œ x%
" Ê du œ 4x$ dx Ê

'

" #

dt Ê 2 du œ sin

t #

ˆsin #t ‰ dt œ ' 2u# du œ

5. Let u œ 7x
2 Ê du œ 7 dx Ê

'

du œ x dx

u# du œ

u$ 1#

" 4

du œ x$ dx

Cœ

" 1#

$

ax%
1b  C

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323

324

Chapter 5 Integration

7. Let u œ 1
r$ Ê du œ
3r# dr Ê
3 du œ 9r# dr

' È9r

# dr 1
r$

œ '
3u
"Î# du œ
3(2)u"Î#  C œ
6 a1
r$ b

"Î#

C

8. Let u œ y%  4y#  1 Ê du œ a4y$  8yb dy Ê 3 du œ 12 ay$  2yb dy

'

12 ay%  4y#  1b ay$  2yb dy œ ' 3u# du œ u$  C œ ay%  4y#  1b  C #

$

9. Let u œ x$Î#
1 Ê du œ

'

x"Î# dx Ê

Èx sin# ˆx$Î#
1‰ dx œ '

10. Let u œ
"x Ê du œ

'

3 #

" x#

2 3

2 3

du œ Èx dx

sin# u du œ

2 3

" 4

ˆ #u


sin 2u‰  C œ

" 3

ˆx$Î#
1‰


" 6

sin ˆ2x$Î#
2‰  C

dx

cos# ˆ x" ‰ dx œ ' cos# a
ub du œ " œ
2x
"4 sin ˆ 2x ‰  C " x#

'

cos# aub du œ ˆ u# 

" 4

" sin 2u‰  C œ
2x 

" 4

sin ˆ
x2 ‰  C

11. (a) Let u œ cot 2) Ê du œ
2 csc# 2) d) Ê
"# du œ csc# 2) d)

'

csc# 2) cot 2) d) œ
'

" #

#

#

u du œ
"# Š u# ‹  C œ
u4  C œ
4" cot# 2)  C

(b) Let u œ csc 2) Ê du œ
2 csc 2) cot 2) d) Ê
"# du œ csc 2) cot 2) d)

'

csc# 2) cot 2) d) œ '
"# u du œ
"# Š u# ‹  C œ
u4  C œ
4" csc# 2)  C #

" 5

12. (a) Let u œ 5x  8 Ê du œ 5 dx Ê

'

dx È5x8

œ'

" 5

Š È"u ‹ du œ " #

(b) Let u œ È5x  8 Ê du œ

'

dx È5x8

œ'

du œ

2 5

2 5

'

" 5

#

du œ dx

u
"Î# du œ

" 5

ˆ2u"Î# ‰  C œ

(5x  8)
"Î# (5) dx Ê

uCœ

2 5

2 5

du œ

2 5

u"Î#  C œ

2 5

È5x  8  C

dx È5x8

È5x  8  C

13. Let u œ 3
2s Ê du œ
2 ds Ê
"# du œ ds

'

È3
2s ds œ ' Èu ˆ
" du‰ œ
" ' u"Î# du œ ˆ
" ‰ ˆ 2 u$Î# ‰  C œ
" (3
2s)$Î#  C # # # 3 3 " #

14. Let u œ 2x  1 Ê du œ 2 dx Ê

'

(2x  1) dx œ ' u $

$

ˆ "#

du‰ œ

15. Let u œ 5s  4 Ê du œ 5 ds Ê

'

" È5s  4

ds œ '

" Èu

ˆ 5" du‰ œ

" 5

" #

'

" 5

'

du œ dx %

u$ du œ ˆ "# ‰ Š u4 ‹  C œ

" 8

(2x  1)%  C

du œ ds u
"Î# du œ ˆ 5" ‰ ˆ2u"Î# ‰  C œ

2 5

È5s  4  C

16. Let u œ 2
x Ê du œ
dx Ê
du œ dx

'

3 (2
x)#

dx œ '

3(
du) u#

œ
3 ' u
# du œ
3 Š u
1 ‹  C œ
"

3 2
x

C

17. Let u œ 1
)# Ê du œ
2) d) Ê
"# du œ ) d)

'

% &Î% ) È1
)# d) œ ' %Èu ˆ
"# du‰ œ
"# ' u"Î% du œ ˆ
"# ‰ ˆ 45 u&Î% ‰  C œ
25 a1
)# b  C

18. Let u œ )#
1 Ê du œ 2) d) Ê 4 du œ 8) d) ' 8) $È)#
1 d) œ ' $Èu (4 du) œ 4 ' u"Î$ du œ 4 ˆ 3 u%Î$ ‰  C œ 3 a)#
1b%Î$  C 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.5 Indefinite Integrals and the Substitution Rule 19. Let u œ 7
3y# Ê du œ
6y dy Ê
"# du œ 3y dy

'

$Î# 3yÈ7
3y# dy œ ' Èu ˆ
"# du‰ œ
"# ' u"Î# du œ ˆ
"# ‰ ˆ 23 u$Î# ‰  C œ
3" a7
3y# b  C

20. Let u œ 2y#  1 Ê du œ 4y dy

'

4y dy È2y#  1

œ'

" Èu

du œ ' u
"Î# du œ 2u"Î#  C œ 2È2y#  1  C " 2È x

21. Let u œ 1  Èx Ê du œ

'

dx œ '

" È x ˆ"  È x ‰ #

œ
2u  C œ

2 du u#

" 2È x

22. Let u œ 1  Èx Ê du œ

'

ˆ1  È x ‰ $ Èx

dx Ê 2 du œ

" 3

cos (3z  4) dz œ ' (cos

u) ˆ "3

sin (8z
5) dz œ ' (sin

u) ˆ "8

'

" 3

ˆ1  Èx‰%  C

" 3

' cos u du œ 3" sin u  C œ 3" sin (3z  4)  C

du œ dz

du‰ œ

25. Let u œ 3x  2 Ê du œ 3 dx Ê

" #

dx

du œ dz

du‰ œ " 8

dx

C

" Èx

dx œ ' u$ (2 du) œ 2 ˆ 4" u% ‰  C œ

24. Let u œ 8z
5 Ê du œ 8 dz Ê

'


2 1 È x

dx Ê 2 du œ

23. Let u œ 3z  4 Ê du œ 3 dz Ê

'

" Èx

" 8

'

" 8

sin u du œ

(
cos u)  C œ
8" cos (8z
5)  C

du œ dx

sec# (3x  2) dx œ ' asec# ub ˆ "3 du‰ œ

" 3

'

" 3

sec# u du œ

tan u  C œ

" 3

tan (3x  2)  C

26. Let u œ tan x Ê du œ sec# x dx

'

tan# x sec# x dx œ ' u# du œ

27. Let u œ sin ˆ x3 ‰ Ê du œ

'

r$ 18


1 Ê du œ

" #

r# 6

$

cos ˆ x3 ‰ dx Ê 3 du œ cos ˆ x3 ‰ dx " #

sin' ˆ x3 ‰  C

sec# ˆ x# ‰ dx Ê 2 du œ sec# ˆ x# ‰ dx " 4

tan) ˆ x# ‰  C

dr Ê 6 du œ r# dr

r % Š7


r& 10 $

r& 10 ‹

'

$

'

Ê du œ
"# r% dr Ê
2 du œ r% dr

dr œ ' u$ (
2 du) œ
2 ' u$ du œ
2 Š u4 ‹  C œ
"# Š7
%

31. Let u œ x$Î#  1 Ê du œ

'

tan$ x  C

r r r# Š 18
1‹ dr œ ' u& (6 du) œ 6 ' u& du œ 6 Š u6 ‹  C œ Š 18
1‹  C &

30. Let u œ 7


'

" 3

tan( ˆ x# ‰ sec# ˆ x# ‰ dx œ ' u( (2 du) œ 2 ˆ 8" u) ‰  C œ

29. Let u œ

'

u$  C œ

sin& ˆ x3 ‰ cos ˆ x3 ‰ dx œ ' u& (3 du) œ 3 ˆ 6" u' ‰  C œ

28. Let u œ tan ˆ x# ‰ Ê du œ

'

" 3

" 3

3 #

x"Î# dx Ê

2 3

r& 10 ‹

%

C

du œ x"Î# dx

x"Î# sin ˆx$Î#  1‰ dx œ ' (sin u) ˆ 23 du‰ œ

2 3

'

sin u du œ

2 3

(
cos u)  C œ
23 cos ˆx$Î#  1‰  C

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325

326

Chapter 5 Integration

32. Let u œ x%Î$
8 Ê du œ

'

4 3

x"Î$ dx Ê

3 4

du œ x"Î$ dx

x"Î$ sin ˆx%Î$
8‰ dx œ ' (sin u) ˆ 34 du‰ œ

3 4

'

sin u du œ

3 4

(
cos u)  C œ
34 cos ˆx%Î$
8‰  C

33. Let u œ sec ˆv  1# ‰ Ê du œ sec ˆv  1# ‰ tan ˆv  1# ‰ dv

'

sec ˆv  1# ‰ tan ˆv  1# ‰ dv œ ' du œ u  C œ sec ˆv  1# ‰  C

34. Let u œ csc ˆ v
# 1 ‰ Ê du œ
"# csc ˆ v
# 1 ‰ cot ˆ v
# 1 ‰ dv Ê
2 du œ csc ˆ v
# 1 ‰ cot ˆ v
# 1 ‰ dv

'

csc ˆ v
# 1 ‰ cot ˆ v
# 1 ‰ dv œ '
2 du œ
2u  C œ
2 csc ˆ v
# 1 ‰  C

35. Let u œ cos (2t  1) Ê du œ
2 sin (2t  1) dt Ê
"# du œ sin (2t  1) dt

'

sin (2t  1) cos# (2t  1)

dt œ '
#"

du u#

œ

" #u

Cœ

" # cos (2t  1)

C

36. Let u œ 2  sin t Ê du œ cos t dt

'

dt œ '

6 cos t (2  sin t)$

6 u$

du œ 6 ' u
$ du œ 6 Š u
# ‹  C œ
3(2  sin t)
#  C
#

37. Let u œ cot y Ê du œ
csc# y dy Ê
du œ csc# y dy

'

Ècot y csc# y dy œ ' Èu (
du) œ
' u"Î# du œ
23 u$Î#  C œ
23 (cot y)$Î#  C œ
23 acot$ yb"Î#  C

38. Let u œ sec z Ê du œ sec z tan z dz

'

sec z tan z Èsec z

39. Let u œ

'

" t#

" t

dz œ '

" Èu

du œ ' u
"Î# du œ 2u"Î#  C œ 2Èsec z  C

" Èt

" )#

"
"Î# # t

dt Ê 2 du œ

sin

" )

" )

cos

Ê du œ ˆcos ") ‰ ˆ
)"# ‰ d) Ê
du œ " )

cos È) È) sin# È)

" )#

cos

d) œ '
u du œ
#" u#  C œ
#" sin#

42. Let u œ csc È) Ê du œ Š
csc È) cot È)‹ Š

'

" Èt

dt

cos ˆÈt  3‰ dt œ ' (cos u)(2 du) œ 2 ' cos u du œ 2 sin u  C œ 2 sin ˆÈt  3‰  C

41. Let u œ sin

'

dt

cos ˆ "t
1‰ dt œ ' (cos u)(
du) œ
' cos u du œ
sin u  C œ
sin ˆ "t
1‰  C

40. Let u œ Èt  3 œ t"Î#  3 Ê du œ

'

" t#


1 œ t
"
1 Ê du œ
t
# dt Ê
du œ

d) œ '

" È)

" ‹ #È )

" )

" )

d)

C

d) Ê
2 du œ

" È)

cot È) csc È) d)

cot È) csc È) d) œ '
2 du œ
2u  C œ
2 csc È)  C œ


2 sin È)

C

43. Let u œ s$  2s#
5s  5 Ê du œ a3s#  4s
5b ds

' as$  2s#
5s  5b a3s#  4s
5b ds œ '

44.

u du œ

u# #

Let u œ )%
2)#  8)
2 Ê du œ a4)$
4)  8b d) Ê

'

as$  2s#
5s  5b #

Cœ

a)%
2)#  8)
2b a)$
)  2b d) œ ' u ˆ "4 du‰ œ

" 4

'

" 4

#

C

du œ a)$
)  2b d)

u du œ

" 4

#

Š u# ‹  C œ

ˆ) %
2 ) #  8 )
2 ‰ # 8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

C

Section 5.5 Indefinite Integrals and the Substitution Rule " 4

45. Let u œ 1  t% Ê du œ 4t$ dt Ê

'

t$ a1  t% b dt œ ' u$ ˆ "4 du‰ œ $

46. Let u œ 1


'

" x

Ê du œ

É x x
& 1 dx œ '

" x#

" x#

" 4

du œ t$ dt

ˆ 4" u% ‰  C œ

" 16

%

a 1  t% b  C

dx

É x
x 1 dx œ '

" x#

" x

É1


dx œ ' Èu du œ ' u"Î# du œ

2 3

u$Î#  C œ

2 3

ˆ1
"x ‰$Î#  C

47. Let u œ x#  ". Then du œ #xdx and "# du œ xdx and x# œ u
". Thus ' x$ Èx#  " dx œ ' au
"b "# Èu du œ

" #

' au$Î#
u"Î# bdu œ "# ’ #& u&Î#
#$ u$Î# “  C œ "& u&Î#
"$ u$Î#  C œ "& ax#  "b&Î#
"$ ax#  "b$Î#  C

48. Let u œ x$  " Ê du œ $x# dx and x$ œ u
". So ' $B& Èx$  " dx œ ' au
"bÈu du œ ' au$Î#
u"Î# bdu œ #& u&Î#
#$ u$Î#  C œ #& ax$  "b

&Î#


#$ ax$  "b

$Î#

C

49. (a) Let u œ tan x Ê du œ sec# x dx; v œ u$ Ê dv œ 3u# du Ê 6 dv œ 18u# du; w œ 2  v Ê dw œ dv

'

18 tan# x sec# x dx œ a2  tan$ xb# 6 œ
2  u$  C $

'

'

18u# du œ a 2  u $ b# 6
2  tan$ x  C # #

œ

6 dv (2  v)#

œ'

6 dw w#

œ 6 ' w
# dw œ
6w
"  C œ
# 6 v  C

(b) Let u œ tan x Ê du œ 3 tan x sec x dx Ê 6 du œ 18 tan# x sec# x dx; v œ 2  u Ê dv œ du

'

18 tan# x sec# x a2  tan$ xb#

dx œ '

œ'

6 du (2  u)#

6 dv v#

6 œ
v6  C œ
2 6 u  C œ
#  tan $x  C

(c) Let u œ 2  tan$ x Ê du œ 3 tan# x sec# x dx Ê 6 du œ 18 tan# x sec# x dx

'

18 tan# x sec# x a2  tan$ xb#

dx œ '

6 du u#

6 œ
u6  C œ
2  tan $x  C

50. (a) Let u œ x
1 Ê du œ dx; v œ sin u Ê dv œ cos u du; w œ 1  v# Ê dw œ 2v dv Ê

'

" #

Èw dw œ

" 3

w$Î#  C œ

" 3

a 1  v# b

$Î#

Cœ

" 3

a1  sin# ub

$Î#

#

Cœ

(b) Let u œ sin (x
1) Ê du œ cos (x
1) dx; v œ 1  u Ê dv œ 2u du Ê

'

È1  sin# (x
1) sin (x
1) cos (x
1) dx œ ' u È1  u# du œ ' œ ˆ "# ˆ 23 ‰ v$Î# ‰  C œ

" 3

v$Î#  C œ

" 3

a1  u # b

$Î#

Cœ

" 3

(c) Let u œ 1  sin (x
1) Ê du œ 2 sin (x
1) cos (x
1) dx Ê

'

È1  sin# (x
1) sin (x
1) cos (x
1) dx œ '

œ

" 3

a1  sin# (x
1)b

$Î#

" 6

dr œ ' Š

cos Èu ˆ" È u ‹ 1#

sin È) É) cos$ È)

d) œ

'

$Î#

dv œ u du

Èv dv œ ' $Î#

" #

v"Î# dv

C

" #

du œ sin (x
1) cos (x
1) dx

" #

u"Î# du œ

" #

ˆ 23 u$Î# ‰  C

" 1#

du œ (2r
1) dr; v œ Èu Ê dv œ

" #È u

du Ê

du‰ œ ' (cos v) ˆ 6" dv‰ œ

" 6

sin v  C œ

" 6

sin Èu  C

sin È3(2r
1)#  6  C

52. Let u œ cos È) Ê du œ Š
sin È)‹ Š

'

a1  sin# (x
1)b

du

(2r
1) cos È3(2r
1)#  6 È3(2r
1)#  6

œ

Èu du œ '

" 3

C

51. Let u œ 3(2r
1)#  6 Ê du œ 6(2r
1)(2) dr Ê " 1#Èu

" #

" #

" #

a1  sin# (x
1)b

#

'

dw œ v dv

È1  sin# (x
1) sin (x
1) cos (x
1) dx œ ' È1  sin# u sin u cos u du œ ' vÈ1  v# dv œ'

œ

" #

sin È) È) Écos$ È)

" ‹ #È )

d) œ '

d) Ê
2 du œ


2 du u$Î#

sin È) È)

d)

œ
2 ' u
$Î# du œ
2 ˆ
2u
"Î# ‰  C œ

4 Èu

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

C

" 6

dv

C

327

328

Chapter 5 Integration œ

4 Écos È)

C

53. Let u œ 3t#
1 Ê du œ 6t dt Ê 2 du œ 12t dt

s œ ' 12t a3t#
1b dt œ ' u$ (2 du) œ 2 ˆ "4 u% ‰  C œ $

s œ 3 when t œ 1 Ê 3 œ

" #

" #

u%  C œ

%

" #

a3t#
1b  C;

(3
1)%  C Ê 3 œ 8  C Ê C œ
5 Ê s œ

" #

%

a3t#
1b
5

54. Let u œ x#  8 Ê du œ 2x dx Ê 2 du œ 4x dx y œ ' 4x ax#  8b


"Î$

dx œ ' u
"Î$ (2 du) œ 2 ˆ 3# u#Î$ ‰  C œ 3u#Î$  C œ 3 ax#  8b

y œ 0 when x œ 0 Ê 0 œ 3(8)#Î$  C Ê C œ
12 Ê y œ 3 ax#  8b 55. Let u œ t 

1 1#

#Î$

#Î$

 C;


12

Ê du œ dt

s œ ' 8 sin# ˆt 

dt œ ' 8 sin# u du œ 8 ˆ u#
"4 sin 2u‰  C œ 4 ˆt  11# ‰
2 sin ˆ2t  16 ‰  C; s œ 8 when t œ 0 Ê 8 œ 4 ˆ 11# ‰
2 sin ˆ 16 ‰  C Ê C œ 8
13  1 œ 9
13 Ê s œ 4ˆt  11# ‰
2 sin ˆ2t  16 ‰  9
13 œ 4t
2 sin ˆ2t  16 ‰  9

56. Let u œ

1 4

1‰ 1#


) Ê
du œ d)

r œ ' 3 cos# ˆ 14
)‰ d) œ
' 3 cos# u du œ
3 ˆ u# 

sin 2u‰  C œ
3# ˆ 14
)‰
43 sin ˆ 1#
2)‰  C;  C Ê C œ 1#  43 Ê r œ
3# ˆ 14
)‰
43 sin ˆ 1#
2)‰  1# 

when ) œ 0 Ê 18 œ
381
43 sin 1# Ê r œ 3# )
34 sin ˆ 1#
2)‰  18  43 Ê r œ

rœ

1 8

57. Let u œ 2t
ds dt

1 #

3 2

)


3 4

" 4

cos 2) 

1 8



3 4

3 4

Ê du œ 2 dt Ê
2 du œ
4 dt

œ '
4 sin ˆ2t
1# ‰ dt œ ' (sin u)(
2 du) œ 2 cos u  C" œ 2 cos ˆ2t
1# ‰  C" ;

at t œ 0 and

ds dt

œ 100 we have 100 œ 2 cos ˆ
1# ‰  C" Ê C" œ 100 Ê

œ 2 cos ˆ2t
1# ‰  100

ds dt

Ê s œ ' ˆ2 cos ˆ2t
1# ‰  100‰ dt œ ' (cos u  50) du œ sin u  50u  C# œ sin ˆ2t
1# ‰  50 ˆ2t
1# ‰  C# ; at t œ 0 and s œ 0 we have 0 œ sin ˆ
1# ‰  50 ˆ
1# ‰  C# Ê C# œ 1  251 Ê s œ sin ˆ2t
1# ‰  100t
251  (1  251) Ê s œ sin ˆ2t
1# ‰  100t  1

58. Let u œ tan 2x Ê du œ 2 sec# 2x dx Ê 2 du œ 4 sec# 2x dx; v œ 2x Ê dv œ 2 dx Ê dy dx

œ ' 4 sec# 2x tan 2x dx œ ' u(2 du) œ u#  C" œ tan# 2x  C" ;

at x œ 0 and

dy dx

œ 4 we have 4 œ 0  C" Ê C" œ 4 Ê

Ê y œ ' asec# 2x  3b dx œ ' asec# v  3b ˆ "# dv‰ œ

at x œ 0 and y œ
1 we have
1 œ

" #

dy dx " #

" #

dv œ dx

œ tan# 2x  4 œ asec# 2x
1b  4 œ sec# 2x  3

tan v  3# v  C# œ

(0)  0  C# Ê C# œ
1 Ê y œ

" #

" #

tan 2x  3x  C# ;

tan 2x  3x
1

59. Let u œ 2t Ê du œ 2 dt Ê 3 du œ 6 dt

s œ ' 6 sin 2t dt œ ' (sin u)(3 du) œ
3 cos u  C œ
3 cos 2t  C; at t œ 0 and s œ 0 we have 0 œ
3 cos 0  C Ê C œ 3 Ê s œ 3
3 cos 2t Ê s ˆ 1# ‰ œ 3
3 cos (1) œ 6 m

60. Let u œ 1t Ê du œ 1 dt Ê 1 du œ 1# dt

v œ ' 1# cos 1t dt œ ' (cos u)(1 du) œ 1 sin u  C" œ 1 sin (1t)  C" ; at t œ 0 and v œ 8 we have 8 œ 1(0)  C" Ê C" œ 8 Ê v œ

ds dt

œ 1 sin (1t)  8 Ê s œ ' (1 sin (1t)  8) dt

œ ' sin u du  8t  C# œ
cos (1t)  8t  C# ; at t œ 0 and s œ 0 we have 0 œ
1  C# Ê C# œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.6 Substitution and Area Between Curves

329

Ê s œ 8t
cos (1t)  1 Ê s(1) œ 8
cos 1  1 œ 10 m 61. All three integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, sin# x  C" œ 1
cos# x  C" Ê C# œ 1  C" ; also
cos# x  C# œ
cos#2x
"#  C# Ê C$ œ C#
"# œ C"  "# . 62. Both integrations are correct. In each case, the derivative of the function on the right is the integrand on the left, and each formula has an arbitrary constant for generating the remaining antiderivatives. Moreover, # tan# x sec# x
1 ˆC
"# ‰  C œ sec# x  ðñò # Cœ # a constant 63. (a) Š

" 60

" ‹
0

'

0

1Î60

"Î'!

Vmax sin 1201t dt œ 60

Vmax ˆ 120" 1 ‰ cos (1201t)‘ !

œ
V#1max [1
1] œ 0 (b) Vmax œ È2 Vrms œ È2 (240) ¸ 339 volts (c)

'

1Î60

0

aVmax b# sin# 1201t dt œ aVmax b#

aVmax b #

œ

#

"Î'! 2401t‘ !


t
ˆ 240" 1 ‰ sin

'

1Î60

0

œ

ˆ 1
cos# 2401t ‰ dt œ

aVmax b #

#

aVmax b# #

œ
V#max 1 [cos 21
cos 0]

'

1Î60

0

(1
cos 2401t) dt

"
ˆ 60
ˆ 240" 1 ‰ sin (41)‰
ˆ0
ˆ #40" 1 ‰ sin (0)‰‘ œ

5.6 SUBSTITUTION AND AREA BETWEEN CURVES 1. (a) Let u œ y  1 Ê du œ dy; y œ 0 Ê u œ 1, y œ 3 Ê u œ 4

'

3

0

Èy  1 dy œ

'

4

1

%

u"Î# du œ
23 u$Î# ‘ " œ ˆ 23 ‰ (4)$Î#
ˆ 23 ‰ (1)$Î# œ ˆ 23 ‰ (8)
ˆ 23 ‰ (1) œ

14 3

(b) Use the same substitution for u as in part (a); y œ
1 Ê u œ 0, y œ 0 Ê u œ 1

'c Èy  1 dy œ ' 0

1

1

0

"

u"Î# du œ
23 u$Î# ‘ ! œ ˆ 23 ‰ (1)$Î#
0 œ

2 3

2. (a) Let u œ 1
r# Ê du œ
2r dr Ê
"# du œ r dr; r œ 0 Ê u œ 1, r œ 1 Ê u œ 0

'

1

0

r È1
r# dr œ

'

0

!


"# Èu du œ

"3 u$Î# ‘ " œ 0
ˆ
3" ‰ (1)$Î# œ

1

" 3

(b) Use the same substitution for u as in part (a); r œ
1 Ê u œ 0, r œ 1 Ê u œ 0

'c r È1
r# dr œ ' 1

1

0

0


"# Èu du œ 0

3. (a) Let u œ tan x Ê du œ sec# x dx; x œ 0 Ê u œ 0, x œ

'

1Î4

0

tan x sec# x dx œ

'

1

0

"

#

u du œ ’ u# “ œ !

1# #


0œ

1 4

Ê uœ1

" #

(b) Use the same substitution as in part (a); x œ
14 Ê u œ
1, x œ 0 Ê u œ 0

'c Î 0

1 4

tan x sec# x dx œ

'
u du œ ’ u# “ ! 0

#


"

1

œ0


" #

œ
"#

4. (a) Let u œ cos x Ê du œ
sin x dx Ê
du œ sin x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ
1

'

1

0

3 cos# x sin x dx œ

$ $ '

3u# du œ c
u$ d
" " œ
(
1)
a
(1) b œ 2 1

1

(b) Use the same substitution as in part (a); x œ 21 Ê u œ 1, x œ 31 Ê u œ
1

'

31

21

3 cos# x sin x dx œ

'
1

1


3u# du œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

aVmax b# 1#0

330

Chapter 5 Integration " 4

5. (a) u œ 1  t% Ê du œ 4t$ dt Ê

'

1

0

'

$

t$ a1  t% b dt œ

2

#

%

" 4

1

du œ t$ dt; t œ 0 Ê u œ 1, t œ 1 Ê u œ 2 2% 16

u u$ du œ ’ 16 “ œ "




1% 16

œ

15 16

(b) Use the same substitution as in part (a); t œ
1 Ê u œ 2, t œ 1 Ê u œ 2

'c

1

'

$

t$ a1  t% b dt œ

1

2

" 4

2

u$ du œ 0 " #

6. (a) Let u œ t#  1 Ê du œ 2t dt Ê

'

È7

t at#  1b

0

"Î$

'

dt œ

8

)

" #

1

du œ t dt; t œ 0 Ê u œ 1, t œ È7 Ê u œ 8

u"Î$ du œ
ˆ "# ‰ ˆ 34 ‰ u%Î$ ‘ " œ ˆ 38 ‰ (8)%Î$
ˆ 38 ‰ (1)%Î$ œ

45 8

(b) Use the same substitution as in part (a); t œ
È7 Ê u œ 8, t œ 0 Ê u œ 1

'cÈ 0

t at#  1b

7

"Î$

dt œ

'

1

" 8 #

" #

7. (a) Let u œ 4  r# Ê du œ 2r dr Ê

'
a4 5rr b 1

dr œ 5

# #

1

'

5

" #

5

'

u"Î$ du œ


8

1

" #

u"Î$ du œ
45 8

du œ r dr; r œ
1 Ê u œ 5, r œ 1 Ê u œ 5

u
# du œ 0

(b) Use the same substitution as in part (a); r œ 0 Ê u œ 4, r œ 1 Ê u œ 5

'

1

5r # 0 a4  r# b

dr œ 5

'

5

" #

4

&

u
# du œ 5

"# u
" ‘ % œ 5 ˆ
"# (5)
" ‰
5 ˆ
"# (4)
" ‰ œ

8. (a) Let u œ 1  v$Î# Ê du œ

'

1

10Èv a1  v$Î# b

0

#

dv œ

'

2

" u#

1

3 #

v"Î# dv Ê

ˆ 20 ‰ 3 du œ

'

20 3

20 3 2

1

" 8

du œ 10Èv dv; v œ 0 Ê u œ 1, v œ 1 Ê u œ 2

20
" 1‘
"‘# u
# du œ
20 3 u " œ
3 #
1 œ

10 3

(b) Use the same substitution as in part (a); v œ 1 Ê u œ 2, v œ 4 Ê u œ 1  4$Î# œ 9

'

4

10Èv

# 1 a1  v$Î# b

dv œ

'

9

" u#

2

20
" ‘ * 20 ˆ " 1‰ 20 ˆ 7 ‰ ˆ 20 ‰ 3 du œ
3 u # œ
3 9
2 œ
3
18 œ

70 #7

9. (a) Let u œ x#  1 Ê du œ 2x dx Ê 2 du œ 4x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4

'

È3

0

4x È x#  1

dx œ

'

4

2 1 Èu

'

du œ

4

1

%

2u
"Î# du œ
4u"Î# ‘ " œ 4(4)"Î#
4(1)"Î# œ 4

(b) Use the same substitution as in part (a); x œ
È3 Ê u œ 4, x œ È3 Ê u œ 4

È3

'cÈ

4x 3 È x#  1

dx œ

'

4

4

2 Èu

du œ 0

10. (a) Let u œ x%  9 Ê du œ 4x$ dx Ê

'

1

0

x$ È x%  9

dx œ

'

10

9

" 4

" 4

du œ x$ dx; x œ 0 Ê u œ 9, x œ 1 Ê u œ 10 "!

u
"Î# du œ
4" (2)u"Î# ‘ * œ

" #

(10)"Î#
#" (9)"Î# œ

È10
3 #

(b) Use the same substitution as in part (a); x œ
1 Ê u œ 10, x œ 0 Ê u œ 9

'c

0

x$ 1 È x%  9

dx œ

'

9

" 10 4

u
"Î# du œ


'

9

10

" 4

11. (a) Let u œ 1
cos 3t Ê du œ 3 sin 3t dt Ê

'

1Î6

0

(1
cos 3t) sin 3t dt œ

'

1

0

" 3

'

1Î6

(1
cos 3t) sin 3t dt œ

'

1

2

" 3

" 3

3
È10 #

du œ sin 3t dt; t œ 0 Ê u œ 0, t œ #

"

u du œ ’ 3" Š u# ‹ “ œ

(b) Use the same substitution as in part (a); t œ 1Î3

u
"Î# du œ

1 6

!

" 6

(1)#
6" (0)# œ

Ê u œ 1, t œ #

#

u du œ ’ 3" Š u# ‹ “ œ "

" 6

1 3

1 6

Ê u œ 1
cos

" 6

Ê u œ 1
cos 1 œ 2

(2)#
6" (1)# œ

" 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 #

œ1

Section 5.6 Substitution and Area Between Curves 12. (a) Let u œ 2  tan

'c

0

1Î2

t #

" #

Ê du œ

ˆ2  tan #t ‰ sec#

'

dt œ

t #

sec# 2

1

t #

dt Ê 2 du œ sec#

'c

1Î2

ˆ2  tan #t ‰ sec#

'

dt œ 2

t #

1

'

0

cos z È4  3 sin z

dz œ

'

4

" Èu

4


1 #

Ê u œ 2  tan ˆ
41 ‰ œ 1, t œ 0 Ê u œ 2

#

3


1 #

Ê u œ 1, t œ

1 #

Ê uœ3

$

u du œ cu# d " œ 3#
1# œ 8 " 3

13. (a) Let u œ 4  3 sin z Ê du œ 3 cos z dz Ê 21

dt; t œ

u (2 du) œ cu# d " œ 2#
1# œ 3

(b) Use the same substitution as in part (a); t œ 1Î2

t #

331

du œ cos z dz; z œ 0 Ê u œ 4, z œ 21 Ê u œ 4

ˆ 3" du‰ œ 0

(b) Use the same substitution as in part (a); z œ
1 Ê u œ 4  3 sin (
1) œ 4, z œ 1 Ê u œ 4

'c

1

cos z 1 È4  3 sin z

dz œ

'

4

" Èu

4

ˆ 3" du‰ œ 0

14. (a) Let u œ 3  2 cos w Ê du œ
2 sin w dw Ê
"# du œ sin w dw; w œ
1# Ê u œ 3, w œ 0 Ê u œ 5

'c

0

'

dw œ

sin w

# 1Î2 (3  2 cos w)

5

u
# ˆ
#" du‰ œ

3

" #

&

cu
" d $ œ

" #

" ˆ "5
"3 ‰ œ
15

(b) Use the same substitution as in part (a); w œ 0 Ê u œ 5, w œ

'

1Î2

!

sin w (3  2 cos w)#

'

dw œ

3

5

u
# ˆ
#" du‰ œ

" #

5

' u
# du œ

1 #

Ê uœ3

" 15

3

15. Let u œ t&  2t Ê du œ a5t%  2b dt; t œ 0 Ê u œ 0, t œ 1 Ê u œ 3

'

1

0

16. Let u œ 1  Èy Ê du œ

'

'

Èt&  2t a5t%  2b dt œ

4

dy # 1 2 È y ˆ1  È y ‰

œ

'

3

" # 2 u

3

0

$

u"Î# du œ
23 u$Î# ‘ ! œ

; y œ 1 Ê u œ 2, y œ 4 Ê u œ 3

dy 2È y

du œ

(3)$Î#
23 (0)$Î# œ 2È3

2 3

'

3

2

$

u
# du œ c
u
" d # œ ˆ
13 ‰
ˆ
12 ‰ œ

" 6

17. Let u œ cos 2) Ê du œ
2 sin 2) d) Ê
"# du œ sin 2) d); ) œ 0 Ê u œ 1, ) œ

'

1Î6

!

cos
$ 2) sin 2) d) œ

18. Let u œ tan ˆ 6) ‰ Ê du œ u œ tan

'

31Î2

1

1 4

'

1Î2

1

" 6

u
$ ˆ
"# du‰ œ
"#

!


#

u
$ du œ ’
2" Š u
# ‹“

"Î# "

œ

cot& ˆ 6) ‰ sec# ˆ 6) ‰ d) œ

'

" # 4 ˆ 1# ‰

1

!

" 4(1)#

" È3

œ

,)œ


%

"

"

u 3 3 3
& È3 u (6 du) œ ’6 Š
4 ‹“ "ÎÈ$ œ

2u% ‘ "ÎÈ$ œ
2(1)%

# Š "

" #

3 4

31 #

Ê

(1
sin 2t)$Î# cos 2t dt œ

%

È3 ‹

'

9

1

" 4

5u"Î% ˆ "4 du‰ œ

'

1

0

œ 12

du œ sin t dt; t œ 0 Ê u œ 5
4 cos 0 œ 1, t œ 1 Ê 5 4

'

1

9

*

u"Î% du œ
54 ˆ 45 u&Î% ‰‘ " œ 9&Î%
1 œ $&Î#
"

20. Let u œ 1
sin 2t Ê du œ
2 cos 2t dt Ê
"# du œ cos 2t dt; t œ 0 Ê u œ 1, t œ

'




1Î

5 (5
4 cos t)"Î% sin t dt œ

1Î4

Ê u œ cos 2 ˆ 16 ‰ œ

œ1

u œ 5
4 cos 1 œ 9 1

1

1 6

sec# ˆ 6) ‰ d) Ê 6 du œ sec# ˆ 6) ‰ d); ) œ 1 Ê u œ tan ˆ 16 ‰ œ

19. Let u œ 5
4 cos t Ê du œ 4 sin t dt Ê

'

'

1Î2

1 4

Ê uœ0

!


"# u$Î# du œ

"2 ˆ 25 u&Î# ‰‘ " œ ˆ
15 (0)&Î# ‰
ˆ
15 (1)&Î# ‰ œ

" 5

21. Let u œ 4y
y#  4y$  1 Ê du œ a4
2y  12y# b dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 4(1)
(1)#  4(1)$  1 œ 8

'

!

1

a4y
y#  4y$  1b


#Î$

a12y#
2y  4b dy œ

'

1

8

)

u
#Î$ du œ
3u"Î$ ‘ " œ 3(8)"Î$
3(1)"Î$ œ 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

332

Chapter 5 Integration " 3

22. Let u œ y$  6y#
12y  9 Ê du œ a3y#  12y
12b dy Ê Ê uœ4

'

1

!

ay$  6y#
12y  9b

œ


'

$

)"Î# d) Ê

È) cos# ˆ)$Î# ‰ d) œ

24. Let u œ 1 

'c
Î

1 2

1

" #

3 #

#

!

œ

ay#  4y
4b dy œ

'

4

9

%

" 3

u
"Î# du œ
3" ˆ2u"Î# ‰‘ * œ

2 3

(4)"Î#
32 (9)"Î# œ

2 3

(2
3)

2 3

23. Let u œ )$Î# Ê du œ

È1


"Î#

du œ ay#  4y
4b dy; y œ 0 Ê u œ 9, y œ 1

" t

'

" 4

!

du œ È) d); ) œ 0 Ê u œ 0, ) œ $È1# Ê u œ 1

cos# u ˆ 23 du‰ œ
23 ˆ #u 

" 4

1

sin 2u‰‘ ! œ

ˆ 1# 

2 3

" 4

sin 21‰
32 (0) œ

1 3

Ê du œ
t
# dt; t œ
1 Ê u œ 0, t œ
#" Ê u œ
1

t
# sin# ˆ1  "t ‰ dt œ




1

2 3

'!

sin# u du œ

ˆ u2
4" sin 2u‰‘
" œ
’ˆ
#"
4" sin (
2)‰
ˆ #0
4" sin 0‰“ ! 1

sin 2

25. Let u œ 4
x# Ê du œ
2x dx Ê
"# du œ x dx; x œ
2 Ê u œ 0, x œ 0 Ê u œ 4, x œ 2 Ê u œ 0 Aœ


'c

0 2

xÈ4
x# dx  %

œ
23 u$Î# ‘ ! œ

2 3

'

2

!

xÈ4
x# dx œ


(4)$Î#
23 (0)$Î# œ

'

4

!


"# u"Î# du 

'

0

4

'


"# u"Î# du œ 2

4

" ! #

u"Î# du œ

'

!

4

u"Î# du

16 3

26. Let u œ 1
cos x Ê du œ sin x dx; x œ 0 Ê u œ 0, x œ 1 Ê u œ 2

'

!

1

(1
cos x) sin x dx œ

'

2

!

#

#

u du œ ’ u2 “ œ !

2# #

0# #




œ2

27. Let u œ 1  cos x Ê du œ
sin x dx Ê
du œ sin x dx; x œ
1 Ê u œ 1  cos (
1) œ 0, x œ 0 Ê u œ 1  cos 0 œ 2 Aœ


'c

0

1

3 (sin x) È1  cos x dx œ


'

2

!

28. Let u œ 1  1 sin x Ê du œ 1 cos x dx Ê Because of symmetry about x œ
1# , A œ 2 œ

'

!

1

3u"Î# (
du) œ 3 " 1

'c

'

!

2

#

u"Î# du œ
2u$Î# ‘ ! œ 2(2)$Î#
2(0)$Î# œ 2&Î#

du œ cos x dx; x œ
1# Ê u œ 1  1 sin ˆ
1# ‰ œ 0, x œ 0 Ê u œ 1

0

1 1Î2 #

(cos x) (sin (1  1 sin x)) dx œ 2

'

1

("
cos 2x) # !

dx œ

" #

'

1

!

(1
cos 2x) dx œ

" #

30. For the sketch given, a œ
13 , b œ 13 ; f(t)
g(t) œ Aœ œ

" #

1

!

1 #

(sin u) ˆ 1" du‰

sin u du œ [
cos u]1! œ (
cos 1)
(
cos 0) œ 2

29. For the sketch given, a œ 0, b œ 1; f(x)
g(x) œ 1
cos# x œ sin# x œ Aœ

'


x
" #

sin 2x ‘ 1 # !

œ

" #

1
cos 2x ; #

[(1
0)
(0
0)] œ

sec# t
a
4 sin# tb œ

" #

1 #

sec# t  4 sin# t;

2t) 'c ÎÎ ˆ "# sec# t  4 sin# t‰ dt œ "# '
ÎÎ sec# t dt  4'
ÎÎ sin# t dt œ "# '
ÎÎ sec# t dt  4'
ÎÎ ("
cos dt # 1 3

1 3

1 3 1Î3

1Î3

1 3

1 3

1 3

1 3

'c Î sec# t dt  2'
Î (1
cos 2t) dt œ

1 3

" #

[tan

1Î$ t]
1Î$

 2[t


1 3

1 3

1 3

1Î$ sin 2t # ]
1Î$

1 3

œ È3  4 †

1 3


È3 œ

31. For the sketch given, a œ
2, b œ 2; f(x)
g(x) œ 2x#
ax%
2x# b œ 4x#
x% ; Aœ

'c

2 2

$

a4x#
x% b dx œ ’ 4x3


# x& 5 “
#

œ ˆ 32 3


32 ‰ 5

ˆ 32 ‰‘ œ


32 3

5

64 3




64 5

œ

320
192 15

œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

128 15

41 3

Section 5.6 Substitution and Area Between Curves

333

32. For the sketch given, c œ 0, d œ 1; f(y)
g(y) œ y#
y$ ; Aœ

'

1

!

ay#
y$ b dy œ

'

1

!

y# dy


'

!

1

$

"

%

"

("
0) 3

y$ dy œ ’ y3 “
’ y4 “ œ !

!

("
0) 4




" 3

œ




" 4

œ

" 1#

33. For the sketch given, c œ 0, d œ 1; f(y)
g(y) œ a12y#
12y$ b
a2y#
2yb œ 10y#
12y$  2y; Aœ

'

1

!

a10y#
12y$  2yb dy œ

'

‰ œ ˆ 10 3
0
(3
0)  (1
0) œ

!

1

10y# dy


'

1

!

12y$ dy 

'

1

!

"

"

"

$‘
12 % ‘
2 #‘ 2y dy œ
10 3 y !
4 y ! # y !

4 3

34. For the sketch given, a œ
1, b œ 1; f(x)
g(x) œ x#
a
2x% b œ x#  2x% ; Aœ

'c ax#  2x% b dx œ ’ x3 1

$

1



" 2x& 5 “
"

œ ˆ "3  25 ‰


3"  ˆ
52 ‰‘ œ

35. We want the area between the line y œ 1, 0 Ÿ x Ÿ 2, and the curve y œ

'

(formed by y œ x and y œ 1) with base 1 and height 1. Thus, A œ œ ˆ2


8 ‰ 1#




" #

œ2


2 3




" #

œ

!

2

2 3



x# 4,

4 5

œ

1012 15

œ

22 15

738?= the area of a triangle

Š1


x# 4‹

dx
"# (1)(1) œ ’x


# x$ 1# “ !




" #

5 6

36. We want the area between the x-axis and the curve y œ x# , 0 Ÿ x Ÿ 1 :6?= the area of a triangle (formed by x œ 1, x  y œ 2, and the x-axis) with base 1 and height 1. Thus, A œ

'

1

!

$

"

x# dx  "# (1)(1) œ ’ x3 “  !

" #

œ

" 3



" #

œ

5 6

37. AREA œ A1  A2 A1: For the sketch given, a œ
3 and we find b by solving the equations y œ x#
4 and y œ
x#
2x simultaneously for x: x#
4 œ
x#
2x Ê 2x#  2x
4 œ 0 Ê 2(x  2)(x
1) Ê x œ
2 or x œ 1 so b œ
2: f(x)
g(x) œ ax#
4b
a
x#
2xb œ 2x#  2x
4 Ê A1 œ $

œ ’ 2x3 

2x #

#


4x“


#
$

‰ œ ˆ
16 3  4  8
(
18  9  12) œ 9


16 3

œ

'cc a2x#  2x
4b dx 2

3

11 3 ;

A2: For the sketch given, a œ
2 and b œ 1: f(x)
g(x) œ a
x#
2xb
ax#
4b œ
2x#
2x  4

'c a2x#  2x
4b dx œ
’ 2x3 1

Ê A2 œ


$

2

œ
23
1  4


16 3

 x#
4x“

"
#

‰ œ
ˆ 23  1
4‰  ˆ
16 3 48

 4  8 œ 9;

Therefore, AREA œ A1  A2 œ

11 3

9œ

38 3

38. AREA œ A1  A2 A1: For the sketch given, a œ
2 and b œ 0: f(x)
g(x) œ a2x$
x#
5xb
a
x#  3xb œ 2x$
8x Ê A1 œ

'c a2x$
8xb dx œ ’ 2x4 0

%

2




! 8x# # “
#

œ 0
(8
16) œ 8;

A2: For the sketch given, a œ 0 and b œ 2: f(x)
g(x) œ a
x#  3xb
a2x$
x#
5xb œ 8x
2x$ Ê A2 œ

'

2

0

#

a8x
2x$ b dx œ ’ 8x2


Therefore, AREA œ A1  A2 œ 16

# 2x% 4 “!

œ (16
8) œ 8;

39. AREA œ A1  A2  A3 A1: For the sketch given, a œ
2 and b œ
1: f(x)
g(x) œ (
x  2)
a4
x# b œ x#
x
2 Ê A1 œ

'cc ax#
x
2b dx œ ’ x3 1

$

2




x# #


2x“


"
#

œ ˆ
"3


" #

 2‰
ˆ
83


4 2

 4‰ œ

7 3




" #

œ

14
3 6

œ

1" 6 ;

" #

œ 9# ;

A2: For the sketch given, a œ
1 and b œ 2: f(x)
g(x) œ a4
x# b
(
x  2) œ
ax#
x
2b Ê A2 œ


'c

2 1

$

ax#
x
2b dx œ
’ x3


x# #


2x“

#
"

œ
ˆ 83


4 #


4‰  ˆ
13


1 2

 2‰ œ
3  8


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

334

Chapter 5 Integration

A3: For the sketch given, a œ 2 and b œ 3: f(x)
g(x) œ (
x  2)
a4
x# b œ x#
x
2 Ê A3 œ

'

3

$

x# #

ax#
x
2b dx œ ’ x3


2

Therefore, AREA œ A1  A2  A3 œ

11 6



9 #

$


2x“ œ ˆ 27 3
# 9 #

 ˆ9


9 #


6‰
ˆ 38



83 ‰ œ 9


5 6

œ

4 2


4‰ œ 9


9 #


38 ;

49 6

40. AREA œ A1  A2  A3 $

A1: For the sketch given, a œ
2 and b œ 0: f(x)
g(x) œ Š x3
x‹
Ê A1 œ

" 3

'c

0 2

ax$
4xb dx œ

" 3

%

’ x4
2x# “

!

x 3

for x:


xœ

f(x)
g(x) œ " 3

œ

Ê

x 3

x$ 3


xœ0 Ê 4 3

x 3

" 3

ax$
4xb

x$ 3


x and y œ

x 3

$

'

0

2

ax$
4xb dx œ

" 3

'

0

2

a4x
x$ b œ

$

Ê A3 œ

" 3

'

3

2

ax$
4xb dx œ

" 3

Therefore, AREA œ A1  A2  A3 œ

$

%

’ x4
2x# “ œ #



4 3

4 3



25 12

œ

" 3

x 3

œ

" 3

ax$
4xb


ˆ 81 ‰ ˆ 16 ‰‘ œ 4
2†9
4
8

3225 1#

œ

" 3

ˆ 81 ‰ 4
14 œ

19 4

41. a œ
2, b œ 2; f(x)
g(x) œ 2
ax#
2b œ 4
x#

'c a4
x# bdx œ ’4x
x3 “ # 2

$


#

2

8‰ œ 2 † ˆ 24 3
3 œ

œ ˆ8
83 ‰
ˆ
8  83 ‰

32 3

42. a œ
1, b œ 3; f(x)
g(x) œ a2x
x# b
(
3) œ 2x
x#  3

'c a2x
x#  3b dx œ ’x#
x3 3

Ê Aœ œ ˆ9


$

1

27 3

 9‰
ˆ1 

1 3


3‰ œ 11


43. a œ 0, b œ 2; f(x)
g(x) œ 8x
x% Ê A œ #

œ ’ 8x2


# x& “ 5 !

œ 16


32 5

œ

'

2

0

" 3

 3x“ œ

$
"

32 3

a8x
x% b dx

80
32 5

œ

48 5

44. Limits of integration: x#
2x œ x Ê x# œ 3x Ê x(x
3) œ 0 Ê a œ 0 and b œ 3; f(x)
g(x) œ x
ax#
2xb œ 3x
x# Ê Aœ œ

27 #

" 3

(8
4) œ 43 ;

A3: For the sketch given, a œ 2 and b œ 3: f(x)
g(x) œ Š x3
x‹


Ê Aœ

simultaneously

(x
2)(x  2) œ 0 Ê x œ
2, x œ 0, or x œ 2 so b œ 2:


Š x3
x‹ œ
3" ax$
4xb Ê A2 œ
3"

x 3


43 x œ

œ 0
3" (4
8) œ 43 ;


#

A2: For the sketch given, a œ 0 and we find b by solving the equations y œ x$ 3

x$ 3

œ

'

0


9œ

3

#

a3x
x# b dx œ ’ 3x2
27
18 #

œ

$ x$ 3 “!

9 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

25 12 ;

’2x#


# x% 4 “!

Section 5.6 Substitution and Area Between Curves 45. Limits of integration: x# œ
x#  4x Ê 2x#
4x œ 0 Ê 2x(x
2) œ 0 Ê a œ 0 and b œ 2; f(x)
g(x) œ a
x#  4xb
x# œ
2x#  4x Ê Aœ

'

2

0

œ
16 3 

# 4x# 2 “!

$

a
2x#  4xb dx œ ’
2x 3  œ

16 #


32  48 6

œ

8 3

46. Limits of integration: 7
2x# œ x#  4 Ê 3x#
3 œ 0 Ê 3(x
1)(x  1) œ 0 Ê a œ
1 and b œ 1; f(x)
g(x) œ a7
2x# b
ax#  4b œ 3
3x# Ê Aœ

'c a3
3x# b dx œ 3 ’x
x3 “ " 1

$


"

1

œ 3
ˆ1
"3 ‰
ˆ
1  3" ‰‘ œ 6 ˆ 23 ‰ œ 4

47. Limits of integration: x%
4x#  4 œ x# Ê x%
5x#  4 œ 0 Ê ax#
4b ax#
1b œ 0 Ê (x  2)(x
2)(x  1)(x
1) œ ! Ê x œ
2,
1, 1, 2; f(x)
g(x) œ ax%
4x#  4b
x# œ x%
5x#  4 and g(x)
f(x) œ x#
ax%
4x#  4b œ
x%  5x#
4 Ê Aœ 

'

2

1

'cc a
x%  5x#
4bdx  'c ax%
5x#  4bdx 1

2

1

%

#

a
x  5x
4bdx &

œ ’
x5  œ ˆ "5
œ


1

5 3



60 5

5x$ 3


4x“


"

&

 ’ x5



#

 4‰
ˆ 32 5
60 3

œ

300
180 15

40 3

5x$ 3

 4x“

 8‰  ˆ 5"


5 3

"

&


"

 ’
5x 

5x$ 3

 4‰
ˆ
5" 


4x“ 5 3

# "


4‰  ˆ
32 5 

40 3


8‰
ˆ
5" 

œ8

48. Limits of integration: xÈa#
x# œ 0 Ê x œ 0 or Èa#
x# œ 0 Ê x œ 0 or a#
x# œ 0 Ê x œ
a, 0, a; Aœ œ œ

" # " 3

'c
xÈa#
x# dx  ' 0

a

0

’ 23 aa#
x# b # $Î#

aa b

a

$Î# !

“

" 3


a


"# ’ 23 aa#
x# b

# $Î#


’
aa b

xÈa#
x# dx

“œ

2a 3

$Î# a

“

!

$

49. Limits of integration: y œ Èkxk œ 

È
x, x Ÿ 0 and Èx, x   0

5y œ x  6 or y œ x5  65 ; for x Ÿ 0: È
x œ x5  65 Ê 5È
x œ x  6 Ê 25(
x) œ x#  12x  36 Ê x#  37x  36 œ 0 Ê (x  1)(x  36) œ 0 Ê x œ
1,
36 (but x œ
36 is not a solution); for x   0: 5Èx œ x  6 Ê 25x œ x#  12x  36 Ê x#
13x  36 œ 0 Ê (x
4)(x
9) œ 0 Ê x œ 4, 9; there are three intersection points and Aœ

'c ˆ x 5 6
È
x‰dx  ' 0

1

0

4

ˆ x 5 6
Èx‰dx 

'

4

9

ˆÈ x


x6‰ 5

dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5 3


4‰

335

336

Chapter 5 Integration #

œ ’ (x 106)  23 (
x)$Î# “ œ ˆ 36 10


25 10

!

%

#

 ’ (x 106)
23 x$Î# “  ’ 23 x$Î#



"


23 ‰  ˆ 100 10


2 3

† 4$Î#


!

36 10

 0‰  ˆ 32 † 9$Î#


* (x  6)# 10 “ % 225 10




2 3

† 4$Î# 

100 ‰ 10

œ
50 10 

20 3

50. Limits of integration: x#
4, x Ÿ
2 or x   2 y œ kx#
4k œ œ 4
x# ,
2 Ÿ x Ÿ 2 for x Ÿ
2 and x   2: x#
4 œ #

#

#

x# 2

4

Ê 2x
8 œ x  8 Ê x œ 16 Ê x œ „ 4; x# #

for
2 Ÿ x Ÿ 2: 4
x# œ #

 4 Ê 8
2x# œ x#  8

Ê x œ 0 Ê x œ 0; by symmetry of the graph,

Aœ2

'

2

0

#

’Š x2  4‹
a4
x# b“dx  2

œ 2 ˆ 8#
0‰  2 ˆ32


'

4

2

#

# !


16  68 ‰ œ 40


64 6

$

’Š x2  4‹
ax#
4b“dx œ 2 ’ x2 “  2 ’8x
56 3

œ

% x$ 6 “#

64 3

51. Limits of integration: c œ 0 and d œ 3; f(y)
g(y) œ 2y#
0 œ 2y#

'

Ê Aœ

0

3

$

$

2y# dy œ ’ 2y3 “ œ 2 † 9 œ 18 !

52. Limits of integration: y# œ y  2 Ê (y  1)(y
2) œ 0 Ê c œ
1 and d œ 2; f(y)
g(y) œ (y  2)
y# Ê Aœ

'c ay  2
y# b dy œ ’ y# 2

#

1

 2y


œ ˆ 4#  4
83 ‰
ˆ "#
2  3" ‰ œ 6


8 3




" #

# y$ 3 “
"

2


" 3

œ

9 #

53. Limits of integration: 4x œ y#
4 and 4x œ 16  y Ê y#
4 œ 16  y Ê y#
y
20 œ 0 Ê (y
5)(y  4) œ 0 Ê c œ
4 and d œ 5; #

f(y)
g(y) œ ˆ 164y ‰
Š y 4
4 ‹ œ Ê Aœ

" 4 y$ 3

'c a
y#  y  20b dy 5

4

y# #

œ

" 4

’


œ

" 4 " 4

ˆ
125 3  189 ˆ
3 

œ


y# y20 4



 20y“

&


% 25 ‰  100
4" ˆ 64 2 3 9 243 ‰  180 œ 2 8



16 #


80‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

5 3

Section 5.6 Substitution and Area Between Curves 54. Limits of integration: x œ y# and x œ 3
2y# Ê y# œ 3
2y# Ê 3y# œ 3 Ê 3(y
1)(y  1) œ 0 Ê c œ
1 and d œ 1; f(y)
g(y) œ a3
2y# b
y# œ 3
3y# œ 3 a1
y# b Ê A œ 3 œ 3 ’y


" y$ 3 “
"

œ 3 ˆ1


"‰ 3

'c a1
y# b dy 1

1


3 ˆ
1  3" ‰

œ 3 † 2 ˆ1
"3 ‰ œ 4 55. Limits of integration: x œ
y# and x œ 2
3y# Ê
y# œ 2
3y# Ê 2y#
2 œ 0 Ê 2(y
1)(y  1) œ 0 Ê c œ
1 and d œ 1; f(y)
g(y) œ a2
3y# b
a
y# b œ 2
2y# œ 2 a1
y# b Ê Aœ2

'c a1
y# b dy œ 2 ’y
y3 “ " 1

$


"

1

œ 2 ˆ1
"3 ‰
2 ˆ
1  3" ‰ œ 4 ˆ 23 ‰ œ

8 3

56. Limits of integration: x œ y#Î$ and x œ 2
y% Ê y#Î$ œ 2
y% Ê c œ
1 and d œ 1; f(y)
g(y) œ a2
y% b
y#Î$

'c ˆ2
y%
y#Î$ ‰ dy 1

Ê Aœ

1

œ ’2y


y& 5


35 y&Î$ “

"


"

œ ˆ2
"5
35 ‰
ˆ
2  œ 2 ˆ2
"5
35 ‰ œ 12 5

" 5

 35 ‰

57. Limits of integration: x œ y#
1 and x œ kyk È1
y# Ê y#
1 œ kyk È1
y# Ê y%
2y#  1 œ y# a1
y# b Ê y%
2y#  1 œ y#
y% Ê 2y%
3y#  1 œ 0 Ê a2y#
1b ay#
1b œ 0 Ê 2y#
1 œ 0 or y#
1 œ 0 Ê y# œ

" #

or y# œ 1 Ê y œ „ „È 2 #

È2 #

or y œ „ 1.

are not solutions Ê y œ „ 1; for
1 Ÿ y Ÿ 0, f(x)
g(x) œ
yÈ1
y#
ay#
1b Substitution shows that œ 1
y#
y a1
y# b

"Î#

, and by symmetry of the graph,

'c ’1
y#
y a1
y# b"Î# “ dy "Î# œ 2' a1
y# b dy
2 ' y a1
y# b dy c c 0

Aœ2

1

0

0

1

œ 2 ’y


1

$

y 3

“

!
"

# $Î#

 2 ˆ "# ‰ ” 2 a1
3y b •

!


"

œ 2
(!
0)
ˆ
1  3" ‰‘  ˆ 23
0‰ œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

337

338

Chapter 5 Integration

58. AREA œ A1  A2 Limits of integration: x œ 2y and x œ y$
y# Ê y$
y# œ 2y Ê y ay#
y
2b œ y(y  1)(y
2) œ 0 Ê y œ
1, 0, 2: for
1 Ÿ y Ÿ 0, f(y)
g(y) œ y$
y#
2y

'c

0

Ê A1 œ

1

œ 0
ˆ "4 

" 3

%

y$ 3

ay$
y#
2yb dy œ ’ y4

1‰ œ


y# “

!
"

5 12 ;

for 0 Ÿ y Ÿ 2, f(y)
g(y) œ 2y
y$  y#

'! a2y
y$  y# b dy œ ’y#
y4 2

Ê A2 œ Ê ˆ4


%

# y$ 3 “!



 38 ‰
0 œ 38 ;

16 4

Therefore, A1  A2 œ



5 12

8 3

œ

37 12

59. Limits of integration: y œ
4x#  4 and y œ x%
1 Ê x%
1 œ
4x#  4 Ê x%  4x#
5 œ 0 Ê ax#  5b (x
1)(x  1) œ 0 Ê a œ
1 and b œ 1; f(x)
g(x) œ
4x#  4
x%  1 œ
4x#
x%  5 Ê Aœ

'c a
4x#
x%  5b dx œ ’
4x3 1

$

1

œ ˆ
43


" 5

" 5

 5‰
ˆ 43 





5‰ œ 2 ˆ
43


"

x& 5

 5x“

" 5

 5‰ œ


" 104 15

60. Limits of integration: y œ x$ and y œ 3x#
4 Ê x$
3x#  4 œ 0 Ê ax#
x
2b (x
2) œ 0 Ê (x  1)(x
2)# œ 0 Ê a œ
1 and b œ 2; f(x)
g(x) œ x$
a3x#
4b œ x$
3x#  4 Ê Aœ

'c ax$
3x#  4b dx œ ’ x4

œ ˆ 16 4


24 3

2

%

1

 8‰
ˆ 41  "
4‰ œ

3x$ 3




 4x“

#
"

27 4

61. Limits of integration: x œ 4
4y# and x œ 1
y% Ê 4
4y# œ 1
y% Ê y%
4y#  3 œ 0 Ê Šy
È3‹ Šy  È3‹ (y
1)(y  1) œ 0 Ê c œ
1 and d œ 1 since x   0; f(y)
g(y) œ a4
4y# b
a1
y% b

'c a3
4y#  y% b dy 1

œ 3
4y#  y% Ê A œ œ ’3y


4y$ 3



" y& 5 “
"

1

œ 2ˆ3


4 3

 5" ‰ œ

56 15

#

62. Limits of integration: x œ 3
y# and x œ
y4 #

Ê 3
y# œ
y4 Ê

3y# 4


3œ0 Ê

3 4

(y
2)(y  2) œ 0 #

Ê c œ
2 and d œ 2; f(y)
g(y) œ a3
y# b
Š
y4 ‹ œ 3 Š1
œ 3
ˆ2


y# 4‹ 8 ‰ 12

'c Š1
y4 ‹ dy œ 3 ’y
1y# “ #

Ê Aœ3
ˆ
2 

2

#

$


#

2

8 ‰‘ 12

œ 3 ˆ4


16 ‰ 12

œ 12
4 œ 8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.6 Substitution and Area Between Curves 63. a œ 0, b œ 1; f(x)
g(x) œ 2 sin x
sin 2x Ê Aœ

'

1

0

(2 sin x
sin 2x) dx œ

2 cos x 

cos 2x ‘ 1 2 !

œ

2(
1)  "# ‘
ˆ
2 † 1  "# ‰ œ 4

64. a œ
13 , b œ 13 ; f(x)
g(x) œ 8 cos x
sec# x Ê Aœ œ Š8 †

'c

1Î3 1Î3

È3 #

1Î$

a8 cos x
sec# xb dx œ [8 sin x
tan x]
1Î$ È3 #


È3‹
Š
8 †

 È3‹ œ 6È3

‰ 65. a œ
1, b œ 1; f(x)
g(x) œ a1
x# b
cos ˆ 1x # Ê Aœ œ ˆ1


" 3

'c
1
x#
cos ˆ 1#x ‰‘ dx œ ’x
x3 1

$

1


12 ‰
ˆ
1 

" 3




 12 ‰ œ 2 ˆ 23
12 ‰ œ

2 1 4 3

sin ˆ 1#x ‰“


"
"

4 1

66. A œ A1  A2 a" œ
1, b" œ 0 and a# œ 0, b# œ 1; f" (x)
g" (x) œ x
sin ˆ 1#x ‰ and f# (x)
g# (x) œ sin ˆ 1#x ‰
x Ê by symmetry about the origin,

A"  A# œ 2A" Ê A œ 2 œ 2 ’
12 cos ˆ 1#x ‰


" x# # “!

œ 2 ˆ 12
"# ‰ œ 2 ˆ 42
11 ‰ œ

'

1

0


sin ˆ 1x ‰ ‘ #
x dx

œ 2
ˆ
12 † 0
"# ‰
ˆ
12 † 1
0‰‘ 4
1 1

67. a œ
14 , b œ 14 ; f(x)
g(x) œ sec# x
tan# x Ê Aœ œ

'c

1Î4 1Î4

'c

1Î4 1Î4

asec# x
tan# xb dx

csec# x
asec# x
1bd dx

1Î4

1Î% œ ' 1 † dx œ [x]
1Î% œ


1Î4

1 4


ˆ
14 ‰ œ

1 #

68. c œ
14 , d œ 14 ; f(y)
g(y) œ tan# y
a
tan# yb œ 2 tan# y œ 2 asec# y
1b Ê A œ

'c

1Î4 1Î4

2 asec# y
1b dy

1Î% œ 2[tan y
y]
1Î% œ 2
ˆ1
14 ‰
ˆ
1  14 ‰‘

œ 4 ˆ1
14 ‰ œ 4
1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

339

340

Chapter 5 Integration

69. c œ 0, d œ 1# ; f(y)
g(y) œ 3 sin yÈcos y
0 œ 3 sin yÈcos y Ê Aœ3

'

1Î2

0

1Î#

sin yÈcos y dy œ
3
23 (cos y)$Î# ‘ !

œ
2(0
1) œ 2

"Î$ ‰ 70. a œ
1, b œ 1; f(x)
g(x) œ sec# ˆ 1x 3
x

Ê Aœ œ

Š 13

'c
sec# ˆ 13x ‰
x"Î$ ‘ dx œ
13 tan ˆ 13x ‰
43 x%Î$ ‘ "
" 1

1

È3
3 ‹
’ 3 Š
È3‹
3 “ œ 4 1 4

6È 3 1

71. A œ A"  A# Limits of integration: x œ y$ and x œ y Ê y œ y$ Ê y$
y œ 0 Ê y(y
1)(y  1) œ 0 Ê c" œ
1, d" œ 0 and c# œ 0, d# œ 1; f" (y)
g" (y) œ y$
y and f# (y)
g# (y) œ y
y$ Ê by symmetry about the origin, A"  A# œ 2A# Ê A œ 2 œ 2 ˆ "#
4" ‰ œ

'

1

0

#

ay
y$ b dy œ 2 ’ y#


" #

" y% 4 “!

72. A œ A"  A# Limits of integration: y œ x$ and y œ x& Ê x$ œ x& Ê x&
x$ œ 0 Ê x$ (x
1)(x  1) œ 0 Ê a" œ
1, b" œ 0 and a# œ 0, b# œ 1; f" (x)
g" (x) œ x$
x& and f# (x)
g# (x) œ x&
x$ Ê by symmetry about the origin, A"  A# œ 2A# Ê A œ 2 œ 2 ˆ "4
6" ‰ œ

'

0

1

%

ax$
x& b dx œ 2 ’ x4


" 6

73. A œ A"  A# Limits of integration: y œ x and y œ $

" x#

Ê xœ

" x# ,

" x' 6 “!

xÁ0

Ê x œ 1 Ê x œ 1 , f" (x)
g" (x) œ x
0 œ x Ê A" œ

'

0

1

#

"

x dx œ ’ x2 “ œ "# ; f# (x)
g# (x) œ

œ x
# Ê A# œ A œ A"  A# œ

'

" #

!

2

1



" x#


0

#

" " ‘ x
# dx œ

" x " œ
#  1 œ #; " #

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.6 Substitution and Area Between Curves 74. Limits of integration: sin x œ cos x Ê x œ 1 4;

and b œ

œŠ

È2 #

1Î4

0

1Î%

(cos x
sin x) dx œ [sin x  cos x]!

È2 # ‹



Ê aœ0

f(x)
g(x) œ cos x
sin x

'

Ê Aœ

1 4

341


(0  1) œ È2
1

75. (a) The coordinates of the points of intersection of the line and parabola are c œ x# Ê x œ „ Èc and y œ c (b) f(y)
g(y) œ Èy
ˆ
Èy‰ œ 2Èy Ê the area of the lower section is, AL œ œ2

'

c

0

'

0

c

[f(y)
g(y)] dy

Èy dy œ 2
23 y$Î# ‘ ! œ c

c$Î# . The area of the

4 3

entire shaded region can be found by setting c œ 4: A œ ˆ 43 ‰ 4$Î# œ 43†8 œ 32 3 . Since we want c to divide the region 32 4 $Î# into subsections of equal area we have A œ 2AL Ê 3 œ 2 ˆ 3 c ‰ Ê c œ 4#Î$ (c) f(x)
g(x) œ c
x# Ê AL œ œ

4 3

Èc

Èc

c

c

'cÈ [f(x)
g(x)] dx œ 'cÈ ac
x# b dx œ ’cx
x3 “ È $

c

cÈc

œ 2 ’c$Î#


c$Î# . Again, the area of the whole shaded region can be found by setting c œ 4 Ê A œ

condition A œ 2AL , we get

4 3

$Î#

c

œ

#Î$

Ê cœ4

32 3

32 3 .

c$Î# 3 “

From the

as in part (b).

76. (a) Limits of integration: y œ 3
x# and y œ
1 Ê 3
x# œ
1 Ê x# œ 4 Ê a œ
2 and b œ 2; f(x)
g(x) œ a3
x# b
(
1) œ 4
x# Ê Aœ

'c a4
x# b dx œ ’4x
x3 “ # 2

$


# 32 3

1

œ ˆ8
83 ‰
ˆ
8  83 ‰ œ 16


16 3

œ

(b) Limits of integration: let x œ 0 in y œ 3
x# Ê y œ 3; f(y)
g(y) œ È3
y
ˆ
È3
y‰ œ 2(3
y)"Î# Ê Aœ2

'c (3
y)"Î# dy œ
2 'c (3
y)"Î# (
1) dy œ (
2) ’ 2(3
3y)

œ ˆ 43 ‰ (8) œ

3

3

1

1

“

$
"

œ ˆ
43 ‰
0
(3  1)$Î# ‘

32 3

77. Limits of integration: y œ 1  Èx and y œ Ê 1  Èx œ

$Î#

2 Èx

2 Èx

, x Á 0 Ê Èx  x œ 2 Ê x œ (2
x)#

Ê x œ 4
4x  x# Ê x#
5x  4 œ 0 Ê (x
4)(x
1) œ 0 Ê x œ 1, 4 (but x œ 4 does not satisfy the equation); y œ È2x and y œ x4 Ê È2x œ x4 Ê 8 œ xÈx Ê 64 œ x$ Ê x œ 4. Therefore, AREA œ A"  A# : f" (x)
g" (x) œ ˆ1  x"Î# ‰
Ê A" œ

'

œ ˆ1 


"8 ‰
0 œ

2 3

0

œ ˆ4 † 2


1

ˆ1  x"Î#
x4 ‰ dx œ ’x  23 x$Î#


16 ‰ 8

37 24 ; f# (x)

" x# 8 “!


g# (x) œ 2x
"Î#



ˆ4
"8 ‰ œ 4


15 8

œ

17 8 ;

x 4

x 4

Ê A# œ

'

1

4

ˆ2x
"Î#
4x ‰ dx œ ’4x"Î#


Therefore, AREA œ A"  A# œ

37 24



17 8

œ

3751 24

œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

88 24

% x# 8 “"

œ

11 3

342

Chapter 5 Integration

78. Limits of integration: (y
1)# œ 3
y Ê y#
2y  1 œ 3
y Ê y#
y
2 œ 0 Ê (y
2)(y  1) œ 0 Ê y œ 2 since y  0; also, 2Èy œ 3
y Ê 4y œ 9
6y  y# Ê y#
10y  9 œ 0 Ê (y
9)(y
1) œ 0 Ê y œ 1 since y œ 9 does not satisfy the equation; AREA œ A"  A# f" (y)
g" (y) œ 2Èy
0 œ 2y"Î# Ê A" œ 2 Ê A# œ

'

1

0

'

2

1

"

$Î#

y"Î# dy œ 2 ’ 2y3 “ œ 43 ; f# (y)
g# (y) œ (3
y)
(y
1)# !

# c3
y
(y
1) d dy œ
3y
"# y#
"3 (y
1)$ ‘ " œ ˆ6
2
3" ‰
ˆ3
#

Therefore, A"  A# œ

4 3



7 6

œ

œ

15 6

" #

80. A œ

'

b

a

2f(x) dx


'

a

b

'

0

a

'

b

a

a$ 3‹

aa#
x# b dx œ 2
a# x
"3 x$ ‘ ! œ 2 Ša$
a

a$ $ Š 4a3 ‹

(2a) aa# b œ a$ ; limit of ratio œ lim b aÄ!

f(x) dx œ 2

 0‰ œ 1


" 3



" #

œ 67 ;

5 2

79. Area between parabola and y œ a# : A œ 2 Area of triangle AOC:

" #

f(x) dx


'

b

a

f(x) dx œ

'

b

a

œ

3 4


0œ

4a$ 3 ;

which is independent of a.

f(x) dx œ 4

81. Neither one; they are both zero. Neither integral takes into account the changes in the formulas for the region's upper and lower bounding curves at x œ 0. The area of the shaded region is actually Aœ

'c [
x
(x)] dx  ' 0

1

0

1

[x
(
x)] dx œ

'c
2x dx  ' 0

1

0

1

2x dx œ #.

82. It is sometimes true. It is true if f(x)   g(x) for all x between a and b. Otherwise it is false. If the graph of f lies below the graph of g for a portion of the interval of integration, the integral over that portion will be negative and the integral over [aß b] will be less than the area between the curves (see Exercise 53). 83. Let u œ 2x Ê du œ 2 dx Ê

'

3

sin 2x x

1

dx œ

'

6

2

sin u ˆ u# ‰

" #

ˆ "# du‰ œ

du œ dx; x œ 1 Ê u œ 2, x œ 3 Ê u œ 6

'

6

du œ cF(u)d '# œ F(6)
F(2)

sin u u

2

84. Let u œ 1
x Ê du œ
dx Ê
du œ dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0

'

0

1

f(1
x) dx œ

'

1

0

f(u) (
du) œ


'

0

1

f(u) du œ

'

1

f(u) du œ

0

'

0

1

f(x) dx

85. (a) Let u œ
x Ê du œ
dx; x œ
1 Ê u œ 1, x œ 0 Ê u œ 0 f odd Ê f(
x) œ
f(x). Then

'c f(x) dx œ ' 0

1

0

1

f(
u) (
du) œ

'

1

œ
3 (b) Let u œ
x Ê du œ
dx; x œ
1 Ê u œ 1, x œ 0 Ê u œ 0 f even Ê f(
x) œ f(x). Then

'c f(x) dx œ ' 0

1

1

0

0

f(
u) (
du) œ


'


f(u) (
du) œ

1

0

f(u) du œ

'

0

1

'

1

0

f(u) du œ


'

0

1

f(u) du

f(u) du œ 3

'c f(x) dx when f is odd. Let u œ
x Ê du œ
dx Ê
du œ dx and x œ
a Ê u œ a and x œ ! Ê u œ !. Thus ' f(x) dx œ '
f(
u) du œ ' f(u) du œ
' f(u) du œ
' f(x) dx. c Thus ' f(x) dx œ ' f(x) dx  ' f(x) dx œ
' f(x) dx  ' f(x) dx œ !. c c 0

86. (a) Consider

a

0

0

a

0

a

a

0

a

a

a

a

0

a

0

a

0

a

0

a

0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 5.6 Substitution and Area Between Curves

'c

1/2

(b)

1/2

1Î#

sin x dx œ [
cos x]
1Î# œ
cos ˆ 1# ‰  cos ˆ
1# ‰ œ !  ! œ !.

87. Let u œ a
x Ê du œ
dx; x œ 0 Ê u œ a, x œ a Ê u œ 0 Iœ

'

' f(u)f(a
f(au)
duu) œ ' f(x)f(a
f(ax)
dxx) dx f(x)f(a
x) ' f(x)f(a
f(ax)
dxx) œ ' f(x) ' dx œ [x]! œ a
0 œ a. I  I œ ' f(x)f(x) f(a
x)  f(a
x) dx œ a

f(x) dx 0 f(x)f(a
x)

œ

'

0

f(a
u) a f(a
u)f(u)

a

Ê

a

0

0

a

a

a

a

0

0

Therefore, 2I œ a Ê I œ 88. Let u œ

'

a

(
du) œ

xy

x

" t

xy t

dt œ

a #

0

.

t Ê du œ
xy t# dt Ê
xy du œ

'

y

1


u" du œ


0

'

y

1

" u

du œ

'

y

1

" t

" u

dt Ê
u" du œ

du œ

'

1

y

" t

" t

dt; t œ x Ê u œ y, t œ xy Ê u œ 1. Therefore,

dt

89. Let u œ x  c Ê du œ dx; x œ a
c Ê u œ a, x œ b
c Ê u œ b

' cc

b c

a c

90. (a)

f(x  c) dx œ

'

a

b

f(u) du œ

'

a

b

f(x) dx (b)

(c)

91-94. Example CAS commands: Maple: f := x -> x^3/3-x^2/2-2*x+1/3; g := x -> x-1; plot( [f(x),g(x)], x=-5..5, legend=["y = f(x)","y = g(x)"], title="#91(a) (Section 5.6)" ); q1 := [ -5, -2, 1, 4 ]; # (b) q2 := [seq( fsolve( f(x)=g(x), x=q1[i]..q1[i+1] ), i=1..nops(q1)-1 )]; for i from 1 to nops(q2)-1 do # (c) area[i] := int( abs(f(x)-g(x)),x=q2[i]..q2[i+1] ); end do; add( area[i], i=1..nops(q2)-1 ); # (d) Mathematica: (assigned functions may vary) Clear[x, f, g] f[x_] = x2 Cos[x] g[x_] = x3
x Plot[{f[x], g[x]}, {x,
2, 2}] After examining the plots, the initial guesses for FindRoot can be determined. pts = x/.Map[FindRoot[f[x]==g[x],{x, #}]&, {
1, 0, 1}] i1=NIntegrate[f[x]
g[x], {x, pts[[1]], pts[[2]]}] i2=NIntegrate[f[x]
g[x], {x, pts[[2]], pts[[3]]}] i1  i2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

343

344

Chapter 5 Integration

CHAPTER 5 PRACTICE EXERCISES 1. (a) Each time subinterval is of length ?t œ 0.4 sec. The distance traveled over each subinterval, using the midpoint rule, is ?h œ "# avi
vib1 b ?t, where vi is the velocity at the left endpoint and vib1 the velocity at

the right endpoint of the subinterval. We then add ?h to the height attained so far at the left endpoint vi to arrive at the height associated with velocity vib1 at the right endpoint. Using this methodology we build the following table based on the figure in the text: t (sec) 0 0.4 0.8 1.2 1.6 2.0 2.4 2.8 3.2 3.6 4.0 4.4 4.8 5.2 5.6 6.0 v (fps) 0 10 25 55 100 190 180 165 150 140 130 115 105 90 76 65 h (ft) 0 2 9 25 56 114 188 257 320 378 432 481 525 564 592 620.2 t (sec) v (fps) h (ft)

6.4 50 643.2

6.8 37 660.6

7.2 25 672

7.6 12 679.4

8.0 0 681.8

NOTE: Your table values may vary slightly from ours depending on the v-values you read from the graph. Remember that some shifting of the graph occurs in the printing process. The total height attained is about 680 ft. (b) The graph is based on the table in part (a).

2. (a) Each time subinterval is of length ?t œ 1 sec. The distance traveled over each subinterval, using the midpoint rule, is ?s œ "# avi
vib1 b ?t, where vi is the velocity at the left, and vib1 the velocity at the

right, endpoint of the subinterval. We then add ?s to the distance attained so far at the left endpoint vi to arrive at the distance associated with velocity vib1 at the right endpoint. Using this methodology we build the table given below based on the figure in the text, obtaining approximately 26 m for the total distance traveled: t (sec) 0 1 2 3 4 5 6 7 8 9 10 v (m/sec) 0 0.5 1.2 2 3.4 4.5 4.8 4.5 3.5 2 0 s (m) 0 0.25 1.1 2.7 5.4 9.35 14 18.65 22.65 25.4 26.4

(b) The graph shows the distance traveled by the moving body as a function of time for 0 Ÿ t Ÿ 10.

3. (a) (c)

10

!

kœ1 10

ak 4

œ

" 4

10

! ak œ

kœ1

" 4

(2) œ  #"

(b)

10

10

10

kœ1

kœ1

kœ1

10

10

10

kœ1

kœ1

kœ1

! (bk  3ak ) œ ! bk  3 ! ak œ 25  3(2) œ 31

! (ak
bk  1) œ ! ak
! bk  ! " œ 2
25  (1)(10) œ 13

kœ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 5 Practice Exercises 10

10

kœ1

kœ1

! ˆ 5  bk ‰ œ ! #

(d)

20

5 #

10

 ! bk œ kœ1

20

kœ1 20

(b)

kœ1

! ˆ"  #

(c)

kœ1 20

2bk ‰ 7

20

œ !

kœ1 20

" #



20

! bk œ

2 7

kœ1 20

" #

20

20

20

kœ1

kœ1

kœ1

! (ak
bk ) œ ! ak
! bk œ 0
7 œ 7

(20)  27 (7) œ 8

! aak  2b œ ! ak  ! 2 œ 0  2(20) œ 40

(d)

kœ1

kœ1

kœ1 " #

5. Let u œ 2x  1 Ê du œ 2 dx Ê 5

1

'

(2x  1)
"Î# dx œ

9

1

3

1

x ax#  1b

7. Let u œ

"Î$

'

dx œ

8

0

du œ dx; x œ 1 Ê u œ 1, x œ 5 Ê u œ 9 *

u
"Î# ˆ "# du‰ œ
u"Î# ‘ " œ 3  1 œ 2

6. Let u œ x#  1 Ê du œ 2x dx Ê

'

(10)  25 œ 0

! 3ak œ 3 ! ak œ 3(0) œ 0

4. (a)

'

5 #

" #

du œ x dx; x œ 1 Ê u œ 0, x œ 3 Ê u œ 8 )

u"Î$ ˆ "# du‰ œ
38 u%Î$ ‘ ! œ

3 8

(16  0) œ 6

Ê 2 du œ dx; x œ 1 Ê u œ  1# , x œ 0 Ê u œ 0

x 2

'
cos ˆ x# ‰ dx œ '
Î (cos u)(2 du) œ [2 sin u]!
1Î# œ 2 sin 0  2 sin ˆ 1# ‰ œ 2(0  (1)) œ 2 0

0

1

1 2

8. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ

'

1Î2

0

(sin x)(cos x) dx œ

(e)

10. (a) (c) (e)

#

u du œ ’ u2 “ œ !

Ê uœ1

" #

2

2

5

2

2

(c)

0

"

'c f(x) dx œ "3 'c 3 f(x) dx œ 3" (12) œ 4 (b) ' f(x) dx œ ' f(x) dx  ' f(x) dx œ 6  4 œ 2 c c c ' g(x) dx œ  'c g(x) dx œ 2 (d) ' (1 g(x)) dx œ 1 ' g(x) dx œ 1(2) œ 21 c c 'c Š f(x) 5 g(x) ‹ dx œ 5" 'c f(x) dx
5" 'c g(x) dx œ 5" (6)
5" (2) œ 85 2

9. (a)

'

1

1 #

' ' '

2

5

5

2

5

5

5

2

2

2

2

0 0

2

' 7 g(x) dx œ "7 (7) œ 1 f(x) dx œ ' f(x) dx œ 1 [g(x)  3 f(x)] dx œ ' g(x) dx  3' g(x) dx œ

2

" 7

(b)

0 2

(d)

0

2

0

5

2

0

0

2

' '

2

2

2

5

5

2

2

2

1 2

0

g(x) dx œ

'

0

2

g(x) dx 

È2 f(x) dx œ È2

'

0

2

'

0

1

g(x) dx œ 1  2 œ 1

f(x) dx œ È2 (1) œ 1È2

f(x) dx œ 1  31

11. x#  4x
3 œ 0 Ê (x  3)(x  1) œ 0 Ê x œ 3 or x œ 1; Area œ

'

0

1

ax#  4x
3b dx  "

$

'

1

$

3

ax#  4x
3b dx

œ ’ x3  2x#
3x“  ’ x3  2x#
3x“ !

$ "

$

œ ’Š "3  2(1)#
3(1)‹  0“ $

$

 ’Š 33  2(3)#
3(3)‹  Š 13  2(1)#
3(1)‹“ œ ˆ "3
1‰ 
0  ˆ 3"
1‰‘ œ

8 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

345

346

Chapter 5 Integration

12. 1 

x# 4

œ 0 Ê 4  x#  0 Ê x œ „ 2;

Area œ

'c Š1  x4 ‹ dx  ' 2

2

œ ’x 

2

# x$ 12 “
# 2$ 12 ‹

œ ’Š2 

3

#

 ’x 

Š1 

x# 4‹

dx

$ x$ 12 “ #

 Š2 

(
2)$ 12 ‹“

œ
43  ˆ 43 ‰‘  ˆ 34  43 ‰ œ

 ’Š3 

3$ 12 ‹

2$ 12 ‹“

 Š2 

13 4

13. 5  5x#Î$ œ 0 Ê 1  x#Î$ œ 0 Ê x œ „ 1; Area œ

'c ˆ5  5x#Î$ ‰ dx  ' 1

1

1

8

ˆ5  5x#Î$ ‰ dx

" ) œ
5x  3x&Î$ ‘
" 
5x  3x&Î$ ‘ " œ
ˆ5(1)  3(1)&Î$ ‰  ˆ5(1)  3(1)&Î$ ‰‘


ˆ5(8)  3(8)&Î$ ‰  ˆ5(1)  3(1)&Î$ ‰‘ œ [2  (2)]  [(40  96)  2] œ 62 14. 1  Èx œ 0 Ê x œ 1; Area œ œ œ œ

'

0

1

ˆ1  Èx‰ dx 

4

1

ˆ1  Èx‰ dx


x  23 x$Î# ‘ " 
x  23 x$Î# ‘ % ! "
ˆ1  23 (1)$Î# ‰  0‘ 
ˆ4  23 "
ˆ4  16 ‰ "‘ 3  3  3 œ 2

15. f(x) œ x, g(x) œ œ

'

'

2

1

ˆx 

"‰ x#

œ

'

œ

Š 42

1

Šx 

a œ 1, b œ 2 Ê A œ

'

b

[f(x)  g(x)] dx

a

#

#

dx œ ’ x#
x" “ œ ˆ 4#
"# ‰  ˆ "#
1‰ œ 1

16. f(x) œ x, g(x) œ 2

" x# ,

(4)$Î# ‰  ˆ1  23 (1)$Î# ‰‘

"

" Èx

" Èx ‹dx

, a œ 1, b œ 2 Ê A œ # œ ’ x#  2Èx“

 2È2‹  ˆ "#  2‰ œ

'

a

b

[f(x)  g(x)] dx

#

" 7
4 È 2 #

'

# 17. f(x) œ ˆ1  Èx‰ , g(x) œ 0, a œ 0, b œ 1 Ê A œ

œ

'

0

1

ˆ1  2x"Î#
x‰ dx œ ’x  43 x$Î#
#

" x# # “!

x% #




" x( 7 “!

œ1

" #




" 7

œ

a

œ1

18. f(x) œ a1  x$ b , g(x) œ 0, a œ 0, b œ 1 Ê A œ œ ’x 

b

'

a

b

'

[f(x)  g(x)] dx œ 4 3




" #

œ

" 6

1

0

ˆ1  Èx‰# dx œ

(6  8
3) œ

[f(x)  g(x)] dx œ

'

0

1

'

0

1

ˆ1  2Èx
x‰ dx

" 6

#

a1  x$ b dx œ

'

0

9 14

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

a1  2x$
x' b dx

Chapter 5 Practice Exercises 19. f(y) œ 2y# , g(y) œ 0, c œ 0, d œ 3 Ê Aœ œ2

'

3

'

d

c

y# dy œ

0

'

[f(y)  g(y)] dy œ 2 3

3

0

a2y#  0b dy

$

cy$ d ! œ 18

20. f(y) œ 4  y# , g(y) œ 0, c œ 2, d œ 2 Ê Aœ

'

d

c

# y$ 3 “
#

œ ’4y 

'c a4  y# b dy 2

[f(y)  g(y)] dy œ œ 2 ˆ8 

8‰ 3

2

œ

32 3

y# 4

21. Let us find the intersection points:

y 2 4

œ

Ê y#  y  2 œ 0 Ê (y  2)(y
1) œ 0 Ê y œ 1 or y œ 2 Ê c œ 1, d œ 2; f(y) œ Ê Aœ

'

d

c

y 2 4

2

#

1

œ

" 4

'c ay
2  y# b dy œ 4" ’ y#

œ

" 4


ˆ 4#
4  83 ‰  ˆ "#  2
3" ‰‘ œ

#

1

y# 4

'c Š y4 2  y4 ‹ dy

[f(y)  g(y)] dy œ

2

, g(y) œ


2y  9 8

y#
4 4

22. Let us find the intersection points:

# y$ 3 “
"

œ

y  16 4

Ê y#  y  20 œ 0 Ê (y  5)(y
4) œ 0 Ê y œ 4 or y œ 5 Ê c œ 4, d œ 5; f(y) œ Ê Aœ

'

d

c

[f(y)  g(y)] dy œ

y  16 4

, g(y) œ

y#
4 4

'c Š y 416  y 4
4 ‹ dy 5

#

4

œ

" 4

'c ay
20  y# b dy œ "4 ’ y#

œ

" 4 " 4

125 ‰
ˆ 25 ‰‘  ˆ "#6  80
64 #
100  3 3 9 " 9 " ˆ #
180  63‰ œ 4 ˆ #
117‰ œ 8 (9
234) œ

œ

5

#


20y 

4

23. f(x) œ x, g(x) œ sin x, a œ 0, b œ Ê Aœ

'

b

a

#

[f(x)  g(x)] dx œ

œ ’ x#
cos x“

1Î% !

#

œ Š 31#


'

1Î4

(x  sin x) dx

1

24. f(x) œ 1, g(x) œ ksin xk , a œ  1# , b œ Ê Aœ œ

'c

œ2

0

'

a

b

[f(x)  g(x)] dx œ

(1
sin x) dx


1Î2 1Î2

'

0

'

0

1Î2

243 8

1 4

0

È2 # ‹

& y$ 3 “
%

'c

1Î2

1Î2

1 2

a1  ksin xkb dx

(1  sin x) dx 1Î#

(1  sin x) dx œ 2[x
cos x]!

œ 2 ˆ 1#  1‰ œ 1  2 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

347

348

Chapter 5 Integration

25. a œ 0, b œ 1, f(x)  g(x) œ 2 sin x  sin 2x

'

Ê Aœ

1

0

(2 sin x  sin 2x) dx œ
2 cos x


cos 2x ‘ 1 # !

œ
2 † (1)
"# ‘  ˆ2 † 1
"# ‰ œ 4

26. a œ  13 , b œ 13 , f(x)  g(x) œ 8 cos x  sec# x

'c

1Î3

Ê Aœ œ Š8 †

1Î3

È3 #

1Î$

a8 cos x  sec# xb dx œ [8 sin x  tan x]
1Î$

 È3‹  Š8 †

È3 #


È3‹ œ 6È3

27. f(y) œ Èy, g(y) œ 2  y, c œ 1, d œ 2

'

Ê Aœ œ

'

1

2

d

c

[f(y)  g(y)] dy œ

'

2

1


Èy  (2  y)‘ dy

ˆÈy  2
y‰ dy œ ’ 23 y$Î#  2y


œ Š 43 È2  4
2‹  ˆ 23  2
"# ‰ œ

4 3

# y# # “"

È2 

7 6

œ

8 È 2
7 6

28. f(y) œ 6  y, g(y) œ y# , c œ 1, d œ 2 Ê Aœ

'

œ ’6y 

y# #

œ4

c




7 3

d

[f(y)  g(y)] dy œ

 " #

# y$ 3 “"

œ

'

2

1

a6  y  y# b dy

œ ˆ12  2  83 ‰  ˆ6 

24
143 6

œ

" #

 3" ‰

13 6

29. f(x) œ x$  3x# œ x# (x  3) Ê f w (x) œ 3x#  6x œ 3x(x  2) Ê f w œ


±  ±


! # Ê f(0) œ 0 is a maximum and f(2) œ 4 is a minimum. A œ  ‰ œ  ˆ 81 4  27 œ 30. A œ

'

a

0

4 3

a# 6


"# ‰ œ

'

a

&Î$

A# œ

" y# # “!

œ

0

(6  8
3) œ

" 10

'

0

1

a

x# # “0

$ !

œ a#  34 Èa † aÈa


a# 6

ˆy#Î$  y‰ dy

; the area below the x-axis is

'c ˆy#Î$  y‰ dy œ ’ 3y5 0

%

ax$  3x# b dx œ  ’ x4  x$ “

ˆa  2Èa x"Î#
x‰ dx œ ’ax  43 Èa x$Î#


31. The area above the x-axis is A" œ œ ’ 3y5 

0

3

27 4

ˆa"Î#  x"Î# ‰# dx œ

œ a# ˆ1 

'

&Î$

1

Ê the total area is A"
A# œ



! y# # “
"

œ

11 10

6 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

a# #

Chapter 5 Practice Exercises

'

32. A œ


1Î4

0

'

31Î2

51Î4

(cos x  sin x) dx


'

51Î4

1Î4

(sin x  cos x) dx 1Î%

(cos x  sin x) dx œ [sin x
cos x]! &1Î%

$1Î#


[ cos x  sin x]1Î%
[sin x
cos x]&1Î% œ ’Š

È2 #

È2 # ‹




 (0
1)“
’Š


’(1
0)  Š

33. y œ x#
34. y œ

'

x

0

'

x

" 1 t

È2 #

dt Ê



dy dx

È2 # ‹“

xœ0 Ê yœ

'

0

0




È2 # ‹

È2 #

 Š



œ

8È 2 #

 2 œ 4È2  2

" x

Ê

d# y dx#

œ 2x


ˆ1
2Èsec t‰ dt Ê

È2 #

œ2

" x#

; y(1) œ 1


œ 1
2Èsec x Ê

dy dx

d# y dx#

ˆ1
2Èsec t‰ dt œ 0 and x œ 0 Ê

dy dx

36. y œ

'c È2  sin# t dt
2 so that dydx œ È2  sin# x; x œ 1

sin t t

dt  3 Ê

dy dx

œ

;xœ5 Ê yœ

sin x x

5

5

sin t t

1

1

" t

dt œ 1 and yw (1) œ 2
1 œ 3

œ 1
2Èsec 0 œ 3

'

x

'

œ 2 ˆ "# ‰ (sec x)
"Î# (sec x tan x) œ Èsec x (tan x);

35. y œ

5

'

È2 # ‹“

dt  3 œ 3

x

Ê yœ

1

'cc È2  sin# t dt
2 œ 2 1

1

37. Let u œ cos x Ê du œ sin x dx Ê  du œ sin x dx

' 2(cos x)
"Î# sin x dx œ ' 2u
"Î# ( du) œ 2 ' u
"Î# du œ 2 Š u"Î#" ‹
C œ 4u"Î#
C #

œ 4(cos x)"Î#
C 38. Let u œ tan x Ê du œ sec# x dx

' (tan x)
$Î# sec# x dx œ ' u
$Î# du œ ˆu

"Î#"‰
C œ 2u
"Î#
C œ (tan
x)2 "Î#
C #

39. Let u œ 2)
1 Ê du œ 2 d) Ê

" #

du œ d)

' [2)
1
2 cos (2)
1)] d) œ ' (u
2 cos u) ˆ "# du‰ œ u4

#

œ )#
)
sin (2)
1)
C, where C œ C"
40. Let u œ 2)  1 Ê du œ 2 d) Ê

'Š œ

41.

42.

" #

" È 2 )
1

" #


2 sec# (#)  1)‹ d) œ

" 4


sin u
C" œ

(2)1)# 4


sin (2)
1)
C"

is still an arbitrary constant

du œ d)

' Š È"u
2 sec# u‹ ˆ #" du‰ œ #" ' ˆu
"Î#
2 sec# u‰ du

"Î#

Š u " ‹
"# (2 tan u)
C œ u"Î#
tan u
C œ (2)  1)"Î#
tan (2)  1)
C #

' ˆt  2t ‰ ˆt
2t ‰ dt œ ' ˆt#  t4 ‰ dt œ ' at#  4t
# b dt œ t3$  4 Š
t
"1 ‹
C œ t3$
4t
C #

t
# ' (t1)t%#
1 dt œ ' t#t%2t dt œ ' ˆ t"#
t2$ ‰ dt œ ' at
#
2t
$ b dt œ (
t
"1)
2 Š
# ‹
C œ  "t  t"#
C

43. Let u œ #t$Î# Ê du œ $Èt dt Ê "$ du œ Èt dt

' Èt sin ˆ#t$Î# ‰dt œ "$ ' sin u du œ  "$ cos u
C œ  "$ cosˆ#t$Î# ‰
C Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

349

350

Chapter 5 Integration

44. Let u œ "
sec ) Ê du œ sec ) tan) d) Ê

' sec ) tan) È"
sec ) d) œ ' u"Î# du œ #$ u$Î#
C

œ #$ a"
sec )b$Î#
C

'c a3x#  4x
7b dx œ cx$  2x#
7xd "
" œ c1$  2(1)#
7(1)d  c(1)$  2(1)#
7(1)d œ 6  (10) œ 16 1

45.

1

46.

'

47.

'

48.

'

49.

'

1

0

"

a8s$  12s#
5b ds œ c2s%  4s$
5sd ! œ c2(1)%  4(1)$
5(1)d  0 œ 3

2

4 # 1 v 27

1

'

dv œ

2

#

4v
# dv œ c4v
" d " œ ˆ
#4 ‰  ˆ
14 ‰ œ 2

1

#(

x
%Î$ dx œ
3x
"Î$ ‘ " œ 3(27)
"Î$  ˆ3(1)
"Î$ ‰ œ 3 ˆ "3 ‰
3(1) œ 2

4

dt 1 tÈt

œ

'

4

œ

dt

t$Î#

1

'

4

1

%

50. Let x œ 1
Èu Ê dx œ

'

4

ˆ1  Èu‰"Î# Èu

1

du œ

'

3

2


2 È4

t
$Î# dt œ
2t
"Î# ‘ " œ " #

u
"Î# du Ê 2 dx œ

du Èu



(
2) È1

œ1

; u œ 1 Ê x œ 2, u œ 4 Ê x œ 3

$

x"Î# (2 dx) œ
2 ˆ 23 ‰ x$Î# ‘ # œ

4 3

ˆ3$Î# ‰  43 ˆ2$Î# ‰ œ 4È3  83 È2 œ

4 3

Š3È3  2È2‹

51. Let u œ 2x
1 Ê du œ 2 dx Ê 18 du œ 36 dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 3

'

1

36 dx $ 0 (2x1)

œ

'

3


# $

$



9 ‘ ˆ
9 ‰ ˆ
9 ‰ 18u
$ du œ ’ "8u
2 “ œ u # " œ 3 #  1 # œ 8 "

1

52. Let u œ 7  5r Ê du œ 5 dr Ê  "5 du œ dr; r œ 0 Ê u œ 7, r œ 1 Ê u œ 2

'

1

dr $ È (7
5r)#

0

œ

'

'

1

(7  5r)
#Î$ dr œ

0

2

7

#

u
#Î$ ˆ 5" du‰ œ  5"
3u"Î$ ‘ ( œ

53. Let u œ 1  x#Î$ Ê du œ  23 x
"Î$ dx Ê  3# du œ x
"Î$ dx; x œ x œ 1 Ê u œ 1  1#Î$ œ 0

'

1

1Î8

œ

x
"Î$ ˆ1  x#Î$ ‰

$Î#

'

dx œ

0

1Î2

0

x$ a1
9x% b

" ˆ 25 ‰ œ  18 16


"Î#


$Î#

dx œ

'

0

1

sin# 5r dr œ

56. Let u œ 4t 

'

1Î%

0

œ

1 8

1 4

$Î%

" 36

#Î$

œ

3 4

,

! &Î# œ
 35 u&Î# ‘ $Î% œ  35 (0)&Î#  ˆ 35 ‰ ˆ 34 ‰

'

51

0

" 16




" 16

" 5


"Î# #

1 8

'
Î acos 1 4

"

%

Ê u œ 1
9 ˆ #" ‰ œ

25 16

#&Î"'

"
"Î# ‘ œ
 18 u "

du œ dr; r œ 0 Ê u œ 0, r œ 1 Ê u œ 51

Ê du œ 4 dt Ê

œ

#&Î"'

" #

" 90

asin# ub ˆ "5 du‰ œ

31Î4

du œ x$ dx; x œ 0 Ê u œ 1, x œ

" " u
$Î# ˆ 36 du‰ œ ’ 36 Š u
" ‹“

"  ˆ 18 (1)
"Î# ‰ œ

cos# ˆ4t  14 ‰ dt œ 

25Î16

1

55. Let u œ 5r Ê du œ 5 dr Ê

'

!

#

Ê u œ 1  ˆ 8" ‰

27È3 160

54. Let u œ 1
9x% Ê du œ 36x$ dx Ê

'

&Î#

u$Î# ˆ #3 du‰ œ ’ˆ 32 ‰ Š u 5 ‹“

3Î4

" 8

Š $È7  $È2‹

3 5

" 4

#

" 5


u2 

sin 2u ‘ &1 4 !

œ ˆ 1# 

sin 101 ‰ #0

du œ dt; t œ 0 Ê u œ  14 , t œ ub ˆ "4 du‰ œ

" 4


u2


sin 2u ‘ $1Î% 4
1Î%

œ

 ˆ0  1 4 " 4

sin 0 ‰ 20

Ê uœ Š 381


œ

1 #

31 4

sin ˆ 3#1 ‰ ‹ 4

 4" Š 18


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

sin ˆ
1# ‰ ‹ 4

Chapter 5 Practice Exercises

'

1Î$

57.

'

31Î4

58.

0

1Î$

sec# ) d) œ [tan )]!

1Î4

31

1

1 3

 tan 0 œ È3

$1Î%

csc# x dx œ [cot x]1Î% œ ˆ cot

59. Let u œ

'

œ tan

cot#

" 6

Ê du œ

x 6

dx œ

x 6

'

1Î6

31 ‰ 4

 ˆ cot 14 ‰ œ 2

dx Ê 6 du œ dx; x œ 1 Ê u œ 16 , x œ 31 Ê u œ

1Î2

351

'

6 cot# u du œ 6

1Î2

1 # 1Î#

acsc# u  1b du œ [6(cot u  u)]1Î' œ 6 ˆ cot

1Î6

1 #

 1# ‰  6 ˆcot

1 6

 16 ‰

œ 6È3  21 60. Let u œ

'

1

0

tan#

) 3 ) 3

œ
3 tan

" 3

Ê du œ d) œ 1 3

'

0

1

d) Ê 3 du œ d); ) œ 0 Ê u œ 0, ) œ 1 Ê u œ ) 3

ˆsec#

 1‰ d) œ

'

1Î3

0

1 3 1Î$

3 asec# u  1b du œ [3 tan u  3u]!

 3 ˆ 13 ‰‘  (3 tan 0  0) œ 3È3  1

'c

sec x tan x dx œ [sec x]!
1Î$ œ sec 0  sec ˆ 13 ‰ œ 1  2 œ 1

'

csc z cot z dz œ [csc z]1Î% œ ˆ csc

0

61.

62.

1Î3

31Î4

1Î4

$1Î%

31 ‰ 4

 ˆ csc 14 ‰ œ È2
È2 œ 0

63. Let u œ sin x Ê du œ cos x dx; x œ 0 Ê u œ 0, x œ

'

1Î2

0

'

5(sin x)$Î# cos x dx œ

1

0

1 #

Ê uœ1

"

"

5u$Î# du œ
5 ˆ 25 ‰ u&Î# ‘ ! œ
2u&Î# ‘ ! œ 2(1)&Î#  2(0)&Î# œ 2

64. Let u œ 1  x# Ê du œ 2x dx Ê  du œ 2x dx; x œ 1 Ê u œ 0, x œ 1 Ê u œ 0

'c

1 1

'

2x sin a1  x# b dx œ

0

 sin u du œ 0

0

" 3

65. Let u œ sin 3x Ê du œ 3 cos 3x dx Ê œ 1

'c ÎÎ

1 2

15 sin% 3x cos 3x dx œ

1 2

du œ cos 3x dx; x œ  1# Ê u œ sin ˆ 3#1 ‰ œ 1, x œ

1 #

& & '
15u% ˆ "3 du‰ œ '
5u% du œ cu& d
" " œ (1)  (1) œ 2 1

1

1

1

66. Let u œ cos ˆ x# ‰ Ê du œ  "# sin ˆ x# ‰ dx Ê 2 du œ sin ˆ x# ‰ dx; x œ 0 Ê u œ cos ˆ 0# ‰ œ 1, x œ œ

'

" #

21Î3

cos
% ˆ x# ‰ sin ˆ x# ‰ dx œ

0

'

1

1Î2


$

u
% (2 du) œ ’2 Š u
3 ‹“

67. Let u œ 1
3 sin# x Ê du œ 6 sin x cos x dx Ê Ê u œ 1
3 sin#

'

1Î2

0

3 sin x cos x È1  3 sin# x

1 #

œ4

dx œ

'

4

" Èu

1

ˆ #" du‰ œ

'

4

1

" #

'

0

1Î4

1 4

Ê u œ 1
7 tan

sec# x (1  7 tan x)#Î$

dx œ

'

1

1 4 8

" #

"Î# "

œ

2 3

ˆ "# ‰
$  32 (1)
$ œ

2 3

(8  1) œ

du œ 3 sin x cos x dx; x œ 0 Ê u œ 1, x œ "Î#

%

#

"

21 3

21

Ê u œ cos Š #3 ‹

14 3

1 #

%

u
"Î# du œ ’ 2" Š u " ‹“ œ
u"Î# ‘ " œ 4"Î#  1"Î# œ 1 " 7

68. Let u œ 1
7 tan x Ê du œ 7 sec# x dx Ê xœ

Ê u œ sin ˆ 3#1 ‰

du œ sec# x dx; x œ 0 Ê u œ 1
7 tan 0 œ 1,

œ8 " u#Î$

ˆ 7" du‰ œ

'

1

8

" 7

"Î$

)

3

"

)

u
#Î$ du œ ’ 7" Š u " ‹“ œ
37 u"Î$ ‘ " œ

3 7

(8)"Î$  37 (1)"Î$ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

3 7

352

Chapter 5 Integration

69. Let u œ sec ) Ê du œ sec ) tan ) d); ) œ 0 Ê u œ sec 0 œ 1, ) œ

'

1Î3

0

tan ) È2 sec )

" È2

œ

'

d) œ


"Î#

#

#

"

'

1Î3

0

’ ˆu
" ‰ “ œ ’

1Î3

sec ) tan ) sec ) tan ) d) œ È2 (sec ))$Î# sec ) È2 sec ) 0 # 2 2 2 È2u “ œ  È2(2)  Š È2(1) ‹ œ " cos Èt 2È t

70. Let u œ sin Èt Ê du œ ˆcos Èt‰ ˆ "# t
"Î# ‰ dt œ #

1 4

tœ

'

1 Î4 #

Ê u œ sin

71. (a) av(f) œ

'

" 1
(
1)

'c

" k
(
k)

(b) av(f) œ " #k

œ

1

" 1Î2 Èu

1 1

'c

k k

(2 du) œ 2

73. favw œ

'

a

b

" È2 u$Î#

du œ

" È2

'

1

2

œ2 u
$Î# du

È2  1 dt Ê 2 du œ

cos Èt Èt

dt; t œ

1# 36

Ê u œ sin

1 6

œ

" #

,

"

"

" #k

’ mx2
bx“


"

#

’ mx2
bx“

(2bk) œ b " 3
0

1

1 3

u
"Î# du œ
4Èu‘ "Î# œ 4È1  4É #" œ 2 Š2  È2‹ #

(mx
b) dx œ

3

0

3

0

0

" b
a

1

1Î2

" #

a

(b) yav

'

(mx
b) dx œ

' È3x dx œ "3 ' œ a
" 0 ' Èax dx œ "a '

72. (a) yav œ

'

Ê u œ sec

œ1

dt œ

cos Èt Ét sin Èt

1# Î36

1 #

d) œ

2

1 3

È3 x"Î# dx œ

a

Èa x"Î# dx œ

0

" b
a

Èaxf w (x) dx œ

[f(x)]ab œ

" b
a

k

ck

#

#

m(
1)
b(1)‹“ œ ’Š m(1) 2
b(1)‹  Š #

œ

" #

œ

" #k

#

" #

(2b) œ b

#

m(
k)
b(k)‹“ ’Š m(k) 2
b(k)‹  Š #

È3 3


23 x$Î# ‘ $ œ !

È3 3


23 (3)$Î#  23 (0)$Î# ‘ œ

È3 3

Š2È3‹ œ 2

Èa a


23 x$Î# ‘ a œ !

Èa a

ˆ 23 (a)$Î#  23 (0)$Î# ‰ œ

Èa a

ˆ 32 aÈa‰ œ

[f(b)  f(a)] œ

f(b)
f(a) b
a

2 3

a

so the average value of f w over [aß b] is the

slope of the secant line joining the points (aß f(a)) and (bß f(b)), which is the average rate of change of f over [aß b]. 74. Yes, because the average value of f on [aß b] is and the average value of the function is

" #

'

a

" b
a

'

a

b

f(x) dx. If the length of the interval is 2, then b  a œ 2

b

f(x) dx.

75. We want to evaluate " $'&
!

'

$'&

!

f(x) dx œ

" $'&

'

$'&

!

#1 Œ$(sin” $'& ax  "!"b•
#&dx œ

#1 Notice that the period of y œ sin” $'& ax  "!"b• is

length 365. Thus the value of

76.

" '(&
#!

œ

'#

'(& !

$( $'&

'

$'&

!

" '&& Œ”)Þ#(a'(&b




#'a'(&b #†"!&



#1 $'&

"Þ)(a'(&b $†"!&

$

'

!

$'&

#1 sin” $'& ax  "!"b•dx


#& $'&

'

$'&

!

dx

œ $'& and that we are integrating this function over an iterval of

#1 ax  "!"b•dx
sin” $'&

a)Þ#(
"!
& a#'T  "Þ)(T# bbdT œ #

#1

$( $'&

" '&& ”)Þ#(T

•  ”)Þ#(a#!b





#& $'&

#'T# #†"!&

#'a#!b #†"!&

#

'

!

 

$'&

dx is

"Þ)(T$ $†"!& • "Þ)(a#!b $†"!&

$

$( $'&

†!


#& $'&

† $'& œ #&.

'(& #!

• ¸

" '&& a$(#%Þ%%

 "'&Þ%!b

œ &Þ%$ œ the average value of Cv on [20, 675]. To find the temperature T at which Cv œ &Þ%$, solve &Þ%$ œ )Þ#(
"!
& a#'T  "Þ)(T# b for T. We obtain "Þ)(T#  #'T  #)%!!! œ ! ÊTœ

#' „ Éa#'b#
%a"Þ)(ba
#)%!!!b È#"#%**' œ #' „ $Þ(% . #a"Þ)(b ‰

So T œ $)#Þ)# or T œ $*'Þ(#. Only T œ $*'Þ(# lies in the

interval [20, 675], so T œ $*'Þ(# C. 77.

dy dx

œ È#
cos$ x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 5 Practice Exercises 78.

dy dx

œ È#
cos$ a(x# b †

79.

dy dx

œ

d dx Œ

80.

dy dx

œ

d dx Œ

d # dx a(x b

œ "%xÈ#
cos$ a(x# b

' x $ ' t dt œ  $'x %

1

353

%

'sec# x t " " dt œ  dxd Œ'#sec x t " " dt œ  sec "x  " dxd asec xb œ  sec" xsectan xx #

#

#

#

81. Yes. The function f, being differentiable on [aß b], is then continuous on [aß b]. The Fundamental Theorem of Calculus says that every continuous function on [aß b] is the derivative of a function on [aß b]. 82. The second part of the Fundamental Theorem of Calculus states that if F(x) is an antiderivative of f(x) on [aß b], then

'

0

1

'

b

a

f(x) dx œ F(b)  F(a). In particular, if F(x) is an antiderivaitve of È1
x% on [0ß 1], then

È1
x% dx œ F(1)  F(0).

83. y œ

' x È1
t# dt œ ' x È1
t# dt

84. y œ

'

1

1

0

" # cos x 1
t

dt œ 

'

0

cos x

" 1
t#

dt Ê

Ê

dy dx

œ

d dx

dy dx

œ

d dx

”

' x È1
t# dt• œ  dxd ”' x È1
t# dt• œ È1
x#

”

'

1

cos x

0

" d ‰ ˆ dx œ  ˆ 1
cos (cos x)‰ œ  ˆ sin"# x ‰ ( sin x) œ #x

1

" 1
t#

" sin x

d dt• œ  dx ”

'

0

cos x

" 1
t#

dt•

œ csc x

85. We estimate the area A using midpoints of the vertical intervals, and we will estimate the width of the parking lot on each interval by averaging the widths at top and bottom. This gives the estimate A ¸ "& † ˆ ! # $'
$' # &%
&% # &"
&" #%*Þ&
%*Þ&# &%
&% #'%Þ%
'%Þ% # '(Þ&
'(Þ&# %# ‰

A ¸ &*'" ft# . The cost is Area † ($2.10/ft# ) ¸ a5961 ft# b a$2.10/ft# b œ $12,518.10 Ê the job cannot be done for $11,000.

86. (a) Before the chute opens for A, a œ 32 ft/sec# . Since the helicopter is hovering, v! œ 0 ft/sec

Ê v œ ' 32 dt œ 32t
v! œ 32t. Then s! œ 6400 ft Ê s œ ' 32t dt œ 16t#
s! œ 16t#
6400.

At t œ 4 sec, s œ 16(4)#
6400 œ 6144 ft when A's chute opens;

(b) For B, s! œ 7000 ft, v! œ 0, a œ 32 ft/sec# Ê v œ ' 32 dt œ 32t
v! œ 32t Ê s œ ' 32t dt œ 16t#
s! œ 16t#
7000. At t œ 13 sec, s œ 16(13)#
7000 œ 4296 ft when B's chute opens;

(c) After the chutes open, v œ 16 ft/sec Ê s œ ' 16 dt œ 16t
s! . For A, s! œ 6144 ft and for B,

s! œ 4296 ft. Therefore, for A, s œ 16t
6144 and for B, s œ 16t
4296. When they hit the ground, 4296 s œ 0 Ê for A, 0 œ 16t
6144 Ê t œ 6144 16 œ 384 seconds, and for B, 0 œ 16t
4296 Ê t œ 16 œ 268.5 seconds to hit the ground after the chutes open. Since B's chute opens 58 seconds after A's opens Ê B hits the ground first. 87. av(I) œ œ

" 30

" 30

'

30

0

(1200  40t) dt œ

" 30

$!

c1200t  20t# d! œ

" 30

ca(1200(30)  20(30)# b  a1200(0)  20(0)# bd

(18,000) œ 600; Average Daily Holding Cost œ (600)($0.03) œ $18

88. av(I) œ

" 14

'

0

14

(600
600t) dt œ

" 14

"%

c600t
300t# d! œ

" 14

c600(14)
300(14)#  0d œ 4800; Average Daily

Holding Cost œ (4800)($0.04) œ $192

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

354

Chapter 5 Integration

'

" 30

89. av(I) œ

30

#

0

$

" 30

Š450  t# ‹ dt œ

’450t  t6 “

œ (300)($0.02) œ $6

œ

" 60

'

" 60

90. av(I) œ

60

0

40È15 3

’600(60) 

(60)

$Î#

'

" 60

Š600  20È15t‹ dt œ

0

" 60

 0“ œ

60

$! !

" 30

œ

30$ 6

’450(30) 

 0“ œ 300; Average Daily Holding Cost

Š600  20È15 t"Î# ‹ dt œ

" 60

’600t  20È15 ˆ 23 ‰ t$Î# “

'! !

ˆ36,000  ˆ 320 ‰ 15# ‰ œ 200; Average Daily Holding Cost 3

œ (200)($0.005) œ $1.00 CHAPTER 5 ADDITIONAL AND ADVANCED EXERCISES

'

1. (a) Yes, because

1

0

(b) No. For example, 4È 2 3

œ

'

1

0

'

" 7

f(x) dx œ

1

0

" 7

7f(x) dx œ

(7) œ 1

'

"

8x dx œ c4x# d ! œ 4, but

1

0

"

È8x dx œ ’2È2 Š x$Î# œ 3 ‹“

2

2

5

5

5

2

2

2

2

2

œ4
3
2œ9

'c f(x) dx œ 4
3 œ 7  2 œ 'c g(x) dx

(c) False:

5

5

2

2

œ


'

x

0

sin ax a sin ax a

'

0

f(t) cos at dt 

'

d

Πdx

œ cos ax

x

0

'

x

0

f(x) dx


5

2

5

2

x

0

0

x

x

0

0

'

f(t) cos at dt


sin ax a

x

0

f(t) sin at dt 

(f(x) cos ax)
sin ax

x

cos ax a

'

x

0

d

Πdx

'

0

x

f(t) sin at dt

f(t) sin at dt 

cos ax a

(f(x) sin ax)

x

0 x

#

#

0

x

0

'

d
(sin ax) Πdx


a cos ax

'

0

x

0

x

0

f(t) sin at dt œ a sin ax

4. x œ

'

" #

x

0

y

" 0 È1  4t#

Ê 1œ œ

'

'

f(t) cos at dt 

dt Ê

" È14y#

a1
4y# b

'

0

x

d dx

Š dy dx ‹ Ê


"Î#

(x) œ dy dx

0

x

'

d dx

0

'

0

y

x

'

'

" È1  4t#

dy 4y Š dx ‹

È1  4y#

0

x

0

x

f(t) cos at dt
a cos ax

'

0

x

f(t) sin at dt
f(x).

f(t) cos at dt
f(x)

f(t) sin at dt œ f(x). Note also that yw (0) œ y(0) œ 0.

dt œ

œ È1
4y# . Then

(8y) Š dy dx ‹ œ

f(t) sin at dt

f(t) cos at dt
(cos ax)f(x) cos ax

f(t) sin at dt  a sin ax

cos ax a

x

0

f(t) sin at dt
(sin ax)f(x) sin ax œ a sin ax

Therefore, yww
a# y œ a cos ax
a# Πsinaax

2

2

x

" a

f(t) cos at dt
sin ax

œ cos ax

dy dx

5

5

' f(t) cos at dt
sin ax ' f(t) sin at dt. Next, d y ' f(t) cos at dt
(cos ax) Πdxd ' f(t) cos at dt
a cos ax ' dx œ a sin ax Ê

'c g(x) dx

' f(t) sin ax cos at dt  "a ' f(t) cos ax sin at dt cos ax ' ' f(t) cos at dt f(t) sin at dt Ê dy a dx œ cos ax Œ

f(t) sin a(x  t) dt œ x

2

5

'c [f(x)  g(x)] dx  0 Ê 'c [g(x)  f(x)] dx  0. Ê ' [g(x)  f(x)] dx   0 which is a contradiction. c

Ê

On the other hand, f(x) Ÿ g(x) Ê [g(x)  f(x)]   0 " a

ˆ1$Î#  0$Î# ‰

5

5

(b) True:

4È 2 3

Á È4

' f(x) dx œ  ' f(x) dx œ 3 'c [f(x)
g(x)] dx œ 'c f(x) dx
'c g(x) dx œ 'c f(x) dx
'

2. (a) True:

3. y œ

!

#

œ

'

d dy

”

d# y dx#

œ

4y ˆÈ1  4y# ‰ È1  4y#

y

0

" È1  4t#

d dx

ˆÈ1
4y# ‰ œ

œ 4y. Thus

dt• Š dy dx ‹ from the chain rule

d# y dx#

d dy

ˆÈ1
4y# ‰ Š dy dx ‹

œ 4y, and the constant of

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 5 Additional and Advanced Exercises

355

proportionality is 4.

'

5. (a)

x#

f(t) dt œ x cos 1x Ê

0

cos 1x
1x sin 1x . 2x

Ê f ax# b œ

'

(b)

f(x)

0

d dx

$

t# dt œ ’ t3 “

f(x)

!

œ

" 3

'

x#

f(t) dt œ cos 1x  1x sin 1x Ê f ax# b (2x) œ cos 1x  1x sin 1x

0

cos 21
21 sin 21 4

Thus, x œ 2 Ê f(4) œ " 3

(f(x))$ Ê

" 4

œ

(f(x))$ œ x cos 1x Ê (f(x))$ œ 3x cos 1x Ê f(x) œ $È3x cos 1x

Ê f(4) œ $È3(4) cos 41 œ $È12 6.

'

a

f(x) dx œ

0

a# #




a #

Ê f(a) œ Fw (a) œ a
7.

'

b

1

1 #

sin a
" #

cos a. Let F(a) œ

sin a


a #

1 #

cos a 

f(x) dx œ Èb#
1  È2 Ê f(b) œ

d db

'

a

0

f(t) dt Ê f(a) œ Fw (a). Now F(a) œ

sin a Ê f ˆ 1# ‰ œ

'

b

1

f(x) dx œ

" #

side of the equation is: œ

d dx

œ

'

0

'

”x

0

x

f(u) du• 

'

d dx

”

d dx

'

x

0

x

0

f(u)(x  u) du• œ

u f(u) du œ

'

x

0

' ”'

dy dx

d dx

0

'

x

0

0

u

" #

1 #

sin


"Î#

'

0

x

d dx




(2b) œ

f(t) dt• du• œ

f(u) x du 

d f(u) du
x ” dx

'

'

0

cos

1 #






1 #

a #

sin a


sin

1 #

Ê f(x) œ

b È b#  1

œ

1 #

1 #

cos a




" #



1 #

œ

" #

x È x#  1

x

f(t) dt; the derivative of the right

x

0

u f(u) du

f(u) du•  xf(x) œ

'

0

x

f(u) du
xf(x)  xf(x)

x

f(u) du. Since each side has the same derivative, they differ by a constant, and since both sides equal 0

when x œ 0, the constant must be 0. Therefore,

9.

”




ab#
1b

x

d dx

8. The derivative of the left side of the equation is:

1 #

ˆ 1# ‰ #

a# #

' ”' x

0

0

u

f(t) dt• du œ

'

0

x

f(u)(x  u) du.

œ 3x#
2 Ê y œ ' a3x#
2b dx œ x$
2x
C. Then (1ß 1) on the curve Ê 1$
2(1)
C œ 1 Ê C œ 4

Ê y œ x$
2x  4 10. The acceleration due to gravity downward is 32 ft/sec# Ê v œ ' 32 dt œ 32t
v! , where v! is the initial

velocity Ê v œ 32t
32 Ê s œ ' (32t
32) dt œ 16t#
32t
C. If the release point, at t œ !, is s œ 0, then C œ 0 Ê s œ 16t#
32t. Then s œ 17 Ê 17 œ 16t#
32t Ê 16t#  32t
17 œ 0. The discriminant of this quadratic equation is 64 which says there is no real time when s œ 17 ft. You had better duck.

11.

'c f(x) dx œ 'c x#Î$ dx
' œ œ œ

12.

3

0

8

8

3

4 dx

0


35 x&Î$ ‘ !
[4x]!$
) ˆ0  35 (8)&Î$ ‰
(4(3) 36 5

 0) œ

'c f(x) dx œ 'c Èx dx
' 3

0

4

4

!

3

0

$

œ
 23 (x)$Î# ‘
%
’ x3  4x“

96 5

 12

ax#  4b dx $ !

$

œ
0  ˆ 23 (4)$Î# ‰‘
’ Š 33  4(3)‹  0 “ œ

16 3

3œ

7 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

356 13.

Chapter 5 Integration

'

2

0

'

g(t) dt œ

1

t dt


0

"

#

'

2

1

sin 1t dt

#

œ ’ t2 “

 1" cos 1t‘ " !

œ ˆ "#  0‰

 1" cos 21  ˆ 1" cos 1‰‘ " #

œ

14.

'

2

0



2 1

h(z) dz œ

'

1

0

'

È1  z dz


1

2

(7z  6)
"Î$ dz

" # 3 œ
 23 (1  z)$Î# ‘ !

14 (7z  6)#Î$ ‘ " œ
 23 (1  1)$Î#  ˆ 23 (1  0)$Î# ‰‘ 3

14 (7(2)  6)#Î$  6 3 ‰ 55 œ
ˆ 7  14 œ 42

3 14

(7(1)  6)#Î$ ‘

2 3

'c f(x) dx œ 'cc dx
'c a1  x# b dx
' 2

15.

1

2

1

2

1

"

x$ 3 “
"

œ [x]
"
#
’x 

1$ 3‹

16.

 ˆ 23 ‰
4  2 œ

2 3

'c h(r) dr œ 'c r dr
' 2

0

1

1

#

œ ’ r2 “

!


"

2 3


Š Š1 


1œ

a1  r# b dr


" b
a

1$ 3‹

'

2

1

7 6

'

b

a

f(x) dx œ

" #
0

'

2

0

f(x) dx œ

#

’Š 1#  0‹
Š 2#  2‹  Š 1#  1‹“ œ

'

20. f(x) œ

'

x

" t

1/x

" b
a

sin x

" " t 1
t# " sin x




'

a

'ÈÈ sin t# dt

22. f(x) œ

'

y

x

xb3

f(x) dx œ

y

" x

" 3
0

'

3

0

" #

”

'

1

0

x dx


'

2

1

(x  1) dx• œ

" #

#

"

#

’ x2 “
#" ’ x2  x“ !

# "

" #

f(x) dx œ

" 3

”

'

dx ‰ d ˆ " ‰‰ ˆ dx  Š "" ‹ ˆ dx œ x x

0

" x

1

dx


'

1

2

0 dx


 x ˆ x"# ‰ œ

" x

'




3

2

dx• œ

" x

œ

#

#

" 3

[1  0
0
3  2] œ

2 3

2 x

" d " d ‰ ˆ dx ‰ ˆ dx dt Ê f w (x) œ ˆ 1
sin (sin x)‰  ˆ 1
cos (cos x)‰ œ #x #x

21. g(y) œ

2

b

dt Ê f w (x) œ

cos x

" cos x

dr

 0‹
a2  1b

#

18. Ave. value œ

œ

13 3

"

#

19. f(x) œ


’2(2)  2(1)“

!

(
1)# # ‹

17. Ave. value œ " #

(
1)$ 3 ‹•

 Š1 


’r  r3 “
[r]#"

œ  "#


œ

1

0

$

œ Š0 

2 dx


[2x]#"

œ a1  (2)b
”Š1  œ1


2

1

cos x cos# x




sin x sin# x

d ˆ d ˆ Èy‰‹ œ Ê gw (y) œ Šsin ˆ2Èy‰ ‹ Š dy 2Èy‰‹  Šsin ˆÈy‰ ‹ Š dy

sin 4y Èy



sin y 2È y

d ‰ t(5  t) dt Ê f w (x) œ (x
3)(&  (x
3)) ˆ dx (x
3)‰  x(5  x) ˆ dx dx œ (x
3)(2  x)  x(5  x)

œ 6  x  x#  5x
x# œ 6  6x. Thus f w (x) œ 0 Ê 6  6x œ 0 Ê x œ 1. Also, f ww (x) œ 6  ! Ê x œ 1 gives a

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 5 Additional and Advanced Exercises maximum. 23. Let f(x) œ x& on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ

1
0 n

œ "n . Then "n , n2 , á ,

_

n n

&

are the

right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for

_

œ

j 1

&

! Š j ‹ ˆ"‰ œ f(x) œ x& on [0ß 1] Ê n lim n n Ä_ jœ1 œ

'

1

'

"

x& dx œ ’ x6 “ œ !

0

& lim " ’ˆ n" ‰ nÄ_ n




ˆ n2 ‰&

&


á
ˆ nn ‰ “ œ n lim ’1 Ä_

&

 2&  á  n& “ n'

" 6

24. Let f(x) œ x$ on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ

1
0 n

œ "n . Then "n , n2 , á ,

_

n n

$

are the

right-hand endpoints of the subintervals. Since f is increasing on [0ß 1], U œ ! Š nj ‹ ˆ "n ‰ is the upper sum for

_

œ

j 1

$

! Š j ‹ ˆ " ‰ œ lim f(x) œ x$ on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1 œ

'

0

1

%

"

x$ dx œ ’ x4 “ œ !

" n

$ $ $ ’ˆ n" ‰
ˆ n2 ‰
á
ˆ nn ‰ “ œ n lim ’1 Ä_

$

 2$  á  n$ “ n%

" 4

25. Let y œ f(x) on [0ß 1]. Partition [0ß 1] into n subintervals with ?x œ

1
0 n

œ "n . Then "n , 2n , á ,

_

n n

are the

right-hand endpoints of the subintervals. Since f is continuous on [!ß 1], ! f Š nj ‹ ˆ "n ‰ is a Riemann sum of

œ

j 1

_

! f Š j ‹ ˆ " ‰ œ lim y œ f(x) on [0ß 1] Ê n lim n n Ä_ nÄ_ jœ1

" n

'

1

" 26. (a) n lim [2
4
6
á
2n] œ n lim Ä _ n# Ä_ on [0ß 1] (see Exercise 25)

" n


n2


" (b) n lim c1"&
2"&
á
n"& d œ n lim Ä _ n"' Ä_ "& f(x) œ x on [0ß 1] (see Exercise 25)

" n

"& "& "& ’ˆ 1n ‰
ˆ 2n ‰
á
ˆ nn ‰ “ œ

'

'


f ˆ n" ‰
f ˆ n2 ‰
á
f ˆ nn ‰‘ œ

4 n




6 n


á


œ

2n ‘ n

0

1

0

f(x) dx

"

2x dx œ cx# d! œ 1, where f(x) œ 2x

'

0

1

"'

"

" 16 ,

x"& dx œ ’ x16 “ œ !

where

1

" "
(c) n lim sin 1n
sin 2n1
á
sin nn1 ‘ œ sin n1 dx œ
 1" cos 1x‘ ! œ  1" cos 1  ˆ 1" cos 0‰ Ä_ n 0 œ 12 , where f(x) œ sin 1x on [0ß 1] (see Exercise 25) " (d) n lim c1"&
2"&
á
n"& d œ Šn lim Ä _ n"( Ä_ " ‰ ˆ œ 0 16 œ 0 (see part (b) above) " n"&

(e) n lim Ä_

c1"&
2"&
á
n"& d œ n lim Ä_

œ Šn lim n‹ Šn lim Ä_ Ä_

" n"'

" " n ‹ Šn lim Ä _ n"'

n n"'

c1"&
2"&
á
n"& d‹ œ Šn lim Ä_

" n‹

'

1

0

x"& dx

c1"&
2"&
á
n"& d

c1"&
2"&
á
n"& d‹ œ Šn lim n‹ Ä_

'

0

1

x"& dx œ _ (see part (b) above)

27. (a) Let the polygon be inscribed in a circle of radius r. If we draw a radius from the center of the circle (and the polygon) to each vertex of the polygon, we have n isosceles triangles formed (the equal sides are equal to r, the radius of the circle) and a vertex angle of )n where )n œ 2n1 . The area of each triangle is An œ

" # # r

sin )n Ê the area of the polygon is A œ nAn œ

(b) n lim A œ n lim Ä_ Ä_

nr# #

sin

21 n

œ n lim Ä_

n1r# 21

sin

21 n

nr# #

nr# 21 # sin n . sin ˆ 2n1 ‰ # ˆ 2n1 ‰ œ a1r b

sin )n œ

œ n lim a1 r # b Ä_

lim

2 1 În Ä 0

sin ˆ 2n1 ‰ ˆ 2n1 ‰

œ 1 r#

'x cos 2t dt
" œ sin x  ' x cos 2t dt
" Ê yw œ cos x  cosa2xb; when x œ 1 we have 1 yw œ cos 1  cosa21b œ "  " œ #. And yww œ sin x
2sina2xb; when x œ 1, y œ sin 1
' cos 2t dt
" x

28. y œ sin x


1

1

œ !
!
" œ ". Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

357

358

Chapter 5 Integration

' ga$b œ '

29. (a) ga"b œ (b)

1

1

$

1

(c) ga"b œ

fatb dt œ ! fatb dt œ  "# a#ba"b œ "

'
fatb dt œ '
fatb dt œ  "% a1 ## b œ 1 1

1

1

1

(d) gw axb œ faxb œ ! Ê x œ $, ", $ and the sign chart for gw axb œ faxb is relative maximum at x œ ". (e) gw a"b œ fa"b œ # is the slope and ga"b œ

±


±  ±


. So g has a 3 1 3

'
fatb dt œ 1, by (c). Thus the equation is y
1 œ #ax
"b 1

1

y œ #x
#  1 . (f) gww axb œ f w axb œ ! at x œ " and gww axb œ f w axb is negative on a$ß "b and positive on a"ß "b so there is an inflection point for g at x œ ". We notice that gww axb œ f w axb  ! for x on a"ß #b and gww axb œ f w axb  ! for x on a#ß %b, even though gww a#b does not exist, g has a tangent line at x œ #, so there is an inflection point at x œ #. (g) g is continuous on Ò$ß %Ó and so it attains its absolute maximum and minimum values on this interval. We saw in (d) that gw axb œ ! Ê x œ $, ", $. We have that ga$b œ

'
$ fatb dt œ '
$" fatb dt œ  1## 1

#

œ #1

' fatb dt œ ! $ ga$b œ ' fatb dt œ " % ga%b œ ' fatb dt œ "
"# † " † " œ  "#

ga"b œ

1

1

1

1

Thus, the absolute minimum is #1 and the absolute maximum is !. Thus, the range is Ò#1ß !Ó.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 5 Additional and Advanced Exercises NOTES:

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359

360

Chapter 5 Integration

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 6 APPLICATIONS OF DEFINITE INTEGRALS 6.1 VOLUMES BY SLICING AND ROTATION ABOUT AN AXIS 1. (a) A œ 1(radius)# and radius œ È1
x# Ê A(x) œ 1 a1
x# b (b) A œ width † height, width œ height œ 2È1
x# Ê A(x) œ 4 a1
x# b (diagonal)# ; #

(c) A œ (side)# and diagonal œ È2(side) Ê A œ (d) A œ

È3 4

diagonal œ 2È1
x# Ê A(x) œ 2 a1
x# b

(side)# and side œ 2È1
x# Ê A(x) œ È3 a1
x# b

2. (a) A œ 1(radius)# and radius œ Èx Ê A(x) œ 1x (b) A œ width † height, width œ height œ 2Èx Ê A(x) œ 4x (diagonal)# ; #

(c) A œ (side)# and diagonal œ È2(side) Ê A œ (d) A œ

È3 4

(side)# and side œ 2Èx Ê A(x) œ È3x

(diagonal)# #

3. A(x) œ

diagonal œ 2Èx Ê A(x) œ 2x

œ

ˆ È x
ˆ
È x ‰ ‰ # #

œ 2x (see Exercise 1c); a œ 0, b œ 4;

V œ 'a A(x) dx œ '0 2x dx œ cx# d ! œ 16 b

4

1(diameter)# 4

4. A(x) œ

œ

%

1 c a2
x # b
x # d 4

#

œ

1c2 a1
x# bd 4

#

œ 1 a1
2x#  x% b ; a œ
1, b œ 1;

V œ 'a A(x) dx œ 'c1 1 a1
2x#  x% b dx œ 1 ’x
23 x$  b

1

" x& 5 “
"

#

œ 21 ˆ1


2 3

 5" ‰ œ

161 15

#

5. A(x) œ (edge)# œ ’È1
x#
Š
È1
x# ‹“ œ Š2È1
x# ‹ œ 4 a1
x# b ; a œ
1, b œ 1; V œ 'a A(x) dx œ 'c1 4a1
x# b dx œ 4 ’x
b

1

#

(diagonal)# #

6. A(x) œ

œ

œ

#

Š2È1
x# ‹

V œ 'a A(x) dx œ 2'c1 a1
x# b dx œ 2 ’x
1

" #

7. (a) STEP 1) A(x) œ

#

" x$ 3 “
"

(side) † (side) † ˆsin 13 ‰ œ

STEP 2) a œ 0, b œ 1

œ 8 ˆ1
"3 ‰ œ

16 3

#

’È1
x#
Š
È1
x# ‹“

b

" x$ 3 “
"

" #

œ 2 a1
x# b (see Exercise 1c); a œ
1, b œ 1; œ 4 ˆ1
"3 ‰ œ

8 3

† Š2Èsin x‹ † Š2Èsin x‹ ˆsin 13 ‰ œ È3 sin x

STEP 3) V œ 'a A(x) dx œ È3 '0 sin x dx œ ’
È3 cos x“ œ È3(1  1) œ 2È3 1

1

b

!

#

(b) STEP 1) A(x) œ (side) œ Š2Èsin x‹ Š2Èsin x‹ œ 4 sin x STEP 2) a œ 0, b œ 1

STEP 3) V œ 'a A(x) dx œ '0 4 sin x dx œ c
4 cos xd 1! œ 8 1

b

#

8. (a) STEP 1) A(x) œ 1(diameter) œ 14 (sec x
tan x)# œ 4 sin x ‘ œ 14
sec# x  asec# x
1b
2 cos #x STEP 2) a œ
13 , b œ

asec# x  tan# x
2 sec x tan xb

1 3

STEP 3) V œ 'a A(x) dx œ 'c1Î3 b

1 4

1Î3

1 4

ˆ2 sec# x
1


2 sin x ‰ cos# x

dx œ

1 4


2 tan x
x  2 ˆ
cos" x ‰‘1Î$
1Î$

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362

Chapter 6 Applications of Definite Integrals œ

1 4

’2È3


1 3

 2 Š
ˆ "" ‰ ‹
Š
2È3  #

1 3

 2 Š
ˆ "" ‰ ‹‹“ œ #

(b) STEP 1) A(x) œ (edge)# œ (sec x
tan x)# œ ˆ2 sec# x
1
2 STEP 2) a œ
13 , b œ

1 3

STEP 3) V œ 'a A(x) dx œ 'c1Î3 ˆ2 sec# x
1
1Î3

b

1 4

9. A(y) œ

(diameter)# œ

1 4

#

ŠÈ5y#
0‹ œ

c œ 0, d œ 2; V œ 'c A(y) dy œ '0 d

#

&

œ ’ˆ 541 ‰ Š y5 ‹“ œ !

" #

10. A(y) œ

2

1 4

51 4

51 4

Š4È3


21 3 ‹

sin x ‰ cos# x

dx œ 2 Š2È3
13 ‹ œ 4È3


21 3

y% ;

y% dy

a2&
0b œ 81

" #

(leg)(leg) œ

#
È1
y#
ˆ
È1
y# ‰‘ œ

V œ 'c A(y) dy œ 'c1 2a1
y# b dy œ 2 ’y
d

2 sin x ‰ cos# x

1 4

1

" y$ 3 “
"

" #

#

ˆ2È1
y# ‰ œ 2 a1
y# b ; c œ
1, d œ 1;

œ 4 ˆ1
"3 ‰ œ

8 3

11. (a) It follows from Cavalieri's Principle that the volume of a column is the same as the volume of a right prism with a square base of side length s and altitude h. Thus, STEP 1) A(x) œ (side length)# œ s# ; STEP 2) a œ 0, b œ h; STEP 3) V œ 'a A(x) dx œ '0 s# dx œ s# h b

h

(b) From Cavalieri's Principle we conclude that the volume of the column is the same as the volume of the prism described above, regardless of the number of turns Ê V œ s# h 12. 1) The solid and the cone have the same altitude of 12. 2) The cross sections of the solid are disks of diameter x
ˆ x# ‰ œ x# . If we place the vertex of the cone at the origin of the coordinate system and make its axis of symmetry coincide with the x-axis then the cone's cross sections will be circular disks of diameter x ˆ x‰ x 4

4 œ # (see accompanying figure). 3) The solid and the cone have equal altitudes and identical parallel cross sections. From Cavalieri's Principle we conclude that the solid and the cone have the same volume. 13. R(x) œ y œ 1
œ 1 ˆ2


4 2



14. R(y) œ x œ

3y #

x #

8 ‰ 12

# Ê V œ '0 1[R(x)]# dx œ 1'0 ˆ1
x# ‰ dx œ 1'0 Š1
x  2

œ

2

2

x# 4‹

dx œ 1 ’x


x# #



21 3

‰ dy œ 1' Ê V œ '0 1[R(y)]# dy œ 1'0 ˆ 3y # 0 2

15. R(x) œ tan ˆ 14 y‰ ; u œ

1 4

y Ê du œ

2

1 4

#

2

9 4

#

y# dy œ 1
34 y$ ‘ ! œ 1 †

3 4

dy Ê 4 du œ 1 dy; y œ 0 Ê u œ 0, y œ 1 Ê u œ

# x$ 12 “ !

† 8 œ 61 1 4

;

1Î% V œ '0 1[R(y)]# dy œ 1'0
tan ˆ 14 y‰‘ dy œ 4 '0 tan# u du œ 4 '0 a
1  sec# ub du œ 4[
u  tan u]! 1

1

#

1Î4

1Î4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.1 Volumes by Slicing and Rotation About an Axis œ 4 ˆ
14  1
0‰ œ 4
1 1 #

16. R(x) œ sin x cos x; R(x) œ 0 Ê a œ 0 and b œ œ 1'0

1Î2

xœ

1 #

(sin x cos x)# dx œ 1 '0

1Î2

Ê u œ 1‘ Ä V œ 1'0

1

" 8

(sin 2x)# 4

are the limits of integration; V œ '0

1Î2

dx;
u œ 2x Ê du œ 2 dx Ê

sin# u du œ

1 8


#u




" 4

sin

1 2u‘ !

œ

1 8


ˆ 1#

du 8

œ

dx 4

1[R(x)]# dx

; x œ 0 Ê u œ 0,


0‰
0‘ œ

1# 16

17. R(x) œ x# Ê V œ '0 1[R(x)]# dx œ 1 '0 ax# b dx 2

2

œ 1 '0 x% dx œ 1 ’ x5 “ œ 2

#

&

#

321 5

!

18. R(x) œ x$ Ê V œ '0 1[R(x)]# dx œ 1'0 ax$ b dx 2

2

œ 1 '0 x' dx œ 1 ’ x7 “ œ 2

(

# !

#

1281 7

19. R(x) œ È9
x# Ê V œ 'c3 1[R(x)]# dx œ 1 'c3 a9
x# b dx 3

$ x$ 3 “
$

œ 1 ’9x


3

œ 21
9(3)


27 ‘ 3

œ 2 † 1 † 18 œ 361

20. R(x) œ x
x# Ê V œ '0 1[R(x)]# dx œ 1'0 ax
x# b dx 1

1

œ 1'0 ax#
2x$  x% b dx œ 1 ’ x3
1

œ 1 ˆ 13


$

" #

 5" ‰ œ

1 30

(10
15  6) œ

21. R(x) œ Ècos x Ê V œ '0

1Î2

1Î#

œ 1 csin xd !

2x% 4

 1 30

#

" x& 5 “!

1[R(x)]# dx œ 1'0 cos x dx 1Î2

œ 1(1
0) œ 1

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363

364

Chapter 6 Applications of Definite Integrals 1Î4

1Î4

22. R(x) œ sec x Ê V œ 'c1Î4 1[R(x)]# dx œ 1 '
1Î4 sec# x dx 1Î%

œ 1 ctan xd
1Î% œ 1[1
(
1)] œ 21

23. R(x) œ È2
sec x tan x Ê V œ œ1

'01Î4 1[R(x)]# dx

'01Î4 ŠÈ2
sec x tan x‹# dx

œ 1 '0 Š2
2È2 sec x tan x  sec# x tan# x‹ dx 1Î4

œ 1 Œ'0 2 dx
2È2 '0 sec x tan x dx  1Î4

1Î%

œ 1 Œ[2x]!

'01Î4 (tan x)# sec# x dx

1Î4

1Î%


2È2 [sec x]!

$

 ’ tan3 x “

1Î%



!

œ 1 ’ˆ 1#
0‰
2È2 ŠÈ2
1‹  "3 a1$
0b“ œ 1 Š 1#  2È2


11 3 ‹

24. R(x) œ 2
2 sin x œ 2(1
sin x) Ê V œ '0 1[R(x)]# dx 1Î2

œ 1 '0 4(1
sin x)# dx œ 41 '0 a1  sin# x
2 sin xb dx 1Î2

1Î2

œ 41'0
1  "# (1
cos 2x)
2 sin x‘ dx 1Î2

œ 41'0 ˆ 3#
1Î2


2 sin x‰

cos 2x 2

1Î# œ 41
3# x
sin42x  2 cos x‘ ! œ 41
ˆ 341
0  0‰
(0
0  2)‘ œ 1(31
8)

25. R(y) œ È5 † y# Ê V œ 'c1 1[R(y)]# dy œ 1 'c1 5y% dy 1

1

"

œ 1 cy& d
" œ 1[1
(
1)] œ 21

26. R(y) œ y$Î# Ê V œ '0 1[R(y)]# dy œ 1'0 y$ dy 2

%

2

#

œ 1 ’ y4 “ œ 41 !

27. R(y) œ È2 sin 2y Ê V œ '0 1[R(y)]# dy 1Î2

œ 1'0 2 sin 2y dy œ 1 c
cos 2yd ! 1Î2

1Î#

œ 1[1
(
1)] œ 21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.1 Volumes by Slicing and Rotation About an Axis 28. R(y) œ Écos

1y 4

Ê V œ 'c2 1[R(y)]# dy 0

œ 1 'c2 cos ˆ 14y ‰ dy œ 4
sin 0

29. R(y) œ

2 y1

1y ‘ ! 4
#

œ 4[0
(
1)] œ 4

Ê V œ '0 1[R(y)]# dy œ 41 '0 3

3

" (y  1)#

dy

$


" ‘ œ 41 ’ y
" 1 “ œ 41
4
(
1) œ 31 !

30. R(y) œ

È2y y # 1

Ê V œ '0 1[R(y)]# dy œ 1'0 2y ay#  1b 1

1


#

dy;

#

cu œ y  1 Ê du œ 2y dy; y œ 0 Ê u œ 1, y œ 1 Ê u œ 2d Ä V œ 1'1 u
# du œ 1

"u ‘ " œ 1

#"
(
1)‘ œ 2

#

1 #

31. For the sketch given, a œ
1# , b œ 1# ; R(x) œ 1, r(x) œ Ècos x; V œ 'a 1 a[R(x)]#
[r(x)]# b dx b

œ 'c1Î2 1(1
cos x) dx œ 21'0 (1
cos x) dx œ 21[x
sin x]! 1Î2

1Î2

1Î#

œ 21 ˆ 1#
1‰ œ 1#
21

32. For the sketch given, c œ 0, d œ 14 ; R(y) œ 1, r(y) œ tan y; V œ 'c 1 a[R(y)]#
[r(y)]# b dy d

œ 1'0 a1
tan# yb dy œ 1 '0 a2
sec# yb dy œ 1[2y
tan y]! 1Î4

1Î4

33. r(x) œ x and R(x) œ 1 Ê V œ œ '0 1 a1
x# b dx œ 1 ’x
1

1Î%

œ 1 ˆ 1#
1‰ œ

1# #


1

'01 1 a[R(x)]#
[r(x)]# b dx

" x$ 3 “!

œ 1
ˆ1
"3 ‰
0‘ œ

21 3

34. r(x) œ 2Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]#
[r(x)]# b dx 1

œ 1'0 (4
4x) dx œ 41’x
1

" x# # “!

œ 41 ˆ1
"# ‰ œ 21

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365

366

Chapter 6 Applications of Definite Integrals

35. r(x) œ x#  1 and R(x) œ x  3

Ê V œ 'c1 1 a[R(x)]#
[r(x)]# b dx 2

œ 1'c1 ’(x  3)#
ax#  1b “ dx 2

#

œ 1 'c1 cax#  6x  9b
ax%  2x#  1bd dx 2

œ 1 'c1 a
x%
x#  6x  8b dx 2

&

œ 1 ’
x5


x$ 3

œ 1
ˆ
32 5


8 3

#



6x# #

 8x“



24 #

 16‰
ˆ 5" 


"

" 3



6 #

‰ ˆ 5†305
33 ‰ œ
8‰‘ œ 1 ˆ
33 5
3  28
3  8 œ 1

36. r(x) œ 2
x and R(x) œ 4
x#

Ê V œ 'c1 1 a[R(x)]#
[r(x)]# b dx 2

œ 1'c1 ’a4
x# b
(2
x)# “ dx 2

#

œ 1 'c1 ca16
8x#  x% b
a4
4x  x# bd dx 2

œ 1'c1 a12  4x
9x#  x% b dx 2

œ 1 ’12x  2x#
3x$  œ 1
ˆ24  8
24 

# x& 5 “
"

32 ‰ 5


ˆ
12  2  3
"5 ‰‘ œ 1 ˆ15 

33 ‰ 5

œ

1081 5

37. r(x) œ sec x and R(x) œ È2

Ê V œ 'c1Î4 1 a[R(x)]#
[r(x)]# b dx 1Î4

œ 1 'c1Î4 a2
sec# xb dx œ 1[2x
tan x]
1Î% 1Î4

1Î%

œ 1
ˆ 1#
1‰
ˆ
1#  1‰‘ œ 1(1
2)

38. R(x) œ sec x and r(x) œ tan x

Ê V œ '0 1 a[R(x)]#
[r(x)]# b dx 1

œ 1 '0 asec# x
tan# xb dx œ 1 '0 1 dx œ 1[x]!" œ 1 1

1

39. r(y) œ 1 and R(y) œ 1  y

Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 1

œ 1'0 c(1  y)#
1d dy œ 1 '0 a1  2y  y#
1b dy 1

1

œ 1 '0 a2y  y# b dy œ 1 ’y#  1

" y$ 3 “!

œ 1 ˆ1  3" ‰ œ

41 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1171 5

Section 6.1 Volumes by Slicing and Rotation About an Axis 40. R(y) œ 1 and r(y) œ 1
y Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 1

œ 1'0 c1
(1
y)# d dy œ 1'0 c1
a1
2y  y# bd dy 1

1

œ 1'0 a2y
y# b dy œ 1 ’y#
1

" y$ 3 “!

œ 1 ˆ1
"3 ‰ œ

21 3

41. R(y) œ 2 and r(y) œ Èy Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 4

œ 1'0 (4
y) dy œ 1 ’4y
4

% y# 2 “!

œ 1(16
8) œ 81

42. R(y) œ È3 and r(y) œ È3
y#

È3

Ê V œ '0

È3

œ 1 '0

$

œ 1 ’ y3 “

1 a[R(y)]#
[r(y)]# b dy

È3

c3
a3
y# bd dy œ 1'0 È$ !

y# dy

œ 1È3

43. R(y) œ 2 and r(y) œ 1  Èy Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 1

œ 1'0 ’4
ˆ1  Èy‰ “ dy 1

#

œ 1 '0 ˆ4
1
2Èy
y‰ dy 1

œ 1 '0 ˆ3
2Èy
y‰ dy 1

œ 1 ’3y
43 y$Î#
œ 1 ˆ3


" y# # “!


"# ‰ œ 1 ˆ 18
68
3 ‰ œ

4 3

71 6

44. R(y) œ 2
y"Î$ and r(y) œ 1

Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 1

# œ 1'0 ’ˆ2
y"Î$ ‰
1“ dy 1

œ 1'0 ˆ4
4y"Î$  y#Î$
1‰ dy 1

œ 1 '0 ˆ3
4y"Î$  y#Î$ ‰ dy 1

œ 1 ’3y
3y%Î$ 

" 3y&Î$ 5 “!

œ 1 ˆ3
3  53 ‰ œ

31 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

367

368

Chapter 6 Applications of Definite Integrals

45. (a) r(x) œ Èx and R(x) œ 2 Ê V œ '0 1 a[R(x)]#
[r(x)]# b dx 4

œ 1'0 (4
x) dx œ 1 ’4x
4

(b) r(y) œ 0 and R(y) œ y#

% x# # “!

œ 1(16
8) œ 81

Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 2

œ 1'0 y% dy œ 1 ’ y5 “ œ 2

&

# !

321 5

# (c) r(x) œ 0 and R(x) œ 2
Èx Ê V œ '0 1 a[R(x)]#
[r(x)]# b dx œ 1'0 ˆ2
Èx‰ dx 4

œ 1'0 ˆ4
4Èx  x‰ dx œ 1 ’4x
4

4

8x$Î# 3



% x# # “!

œ 1 ˆ16


64 3



16 ‰ #

œ

81 3

(d) r(y) œ 4
y# and R(y) œ 4 Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy œ 1 '0 ’16
a4
y# b “ dy 2

2

œ 1 '0 a16
16  8y#
y% b dy œ 1 '0 a8y#
y% b dy œ 1 ’ 83 y$
2

2

46. (a) r(y) œ 0 and R(y) œ 1


# y& 5 “!

#

œ 1 ˆ 64 3


32 ‰ 5

œ

2241 15

y #

Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy 2

# œ 1'0 ˆ1
y# ‰ dy œ 1'0 Š1
y  2

œ 1 ’y


2

y# #



# y$ 12 “ !

œ 1 ˆ#


(b) r(y) œ 1 and R(y) œ 2


4 2



8 ‰ 12

y# 4‹

œ

dy

21 3

y #

# Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy œ 1 '0 ’ˆ2
y# ‰
1“ dy œ 1 '0 Š4
2y  2

2

œ 1'0 Š3
2y  2

y# 4‹

dy œ 1 ’3y
y# 

# y$ 12 “ !

2

œ 1 ˆ6
4 

8 ‰ 12

œ 1 ˆ2  23 ‰ œ

y# 4


1‹ dy

81 3

47. (a) r(x) œ 0 and R(x) œ 1
x#

Ê V œ 'c1 1 a[R(x)]#
[r(x)]# b dx 1

œ 1 'c1 a1
x# b dx œ 1 'c1 a1
2x#  x% b dx 1

œ 1 ’x


1

#

2x$ 3



" x& 5 “
"

103 ‰ œ 21 ˆ 15
15 œ

œ 21 ˆ1


2 3

 15 ‰

161 15

(b) r(x) œ 1 and R(x) œ 2
x# Ê V œ 'c1 1 a[R(x)]#
[r(x)]# b dx œ 1 'c1 ’a2
x# b
1“ dx 1

1

œ 1 'c1 a4
4x#  x%
1b dx œ 1'c1 a3
4x#  x% b dx œ 1 ’3x
43 x$  1

œ

21 15

1

(45
20  3) œ

561 15

#

" x& 5 “
"

œ 21 ˆ3


4 3

 15 ‰

2 3


15 ‰

(c) r(x) œ 1  x# and R(x) œ 2 Ê V œ 'c1 1 a[R(x)]#
[r(x)]# b dx œ 1 'c1 ’4
a1  x# b “ dx 1

1

œ 1 'c1 a4
1
2x#
x% b dx œ 1'c1 a3
2x#
x% b dx œ 1 ’3x
23 x$
1

œ

21 15

1

(45
10
3) œ

641 15

#

" x& 5 “
"

œ 21 ˆ3


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.1 Volumes by Slicing and Rotation About an Axis

369

48. (a) r(x) œ 0 and R(x) œ
hb x  h

Ê V œ '0 1 a[R(x)]#
[r(x)]# b dx b

# œ 1 '0 ˆ
hb x  h‰ dx b

œ 1'0 Š hb# x#
b

#

$

x œ 1h# ’ 3b #


x# b

2h# b

x  h# ‹ dx b

 x“ œ 1h# ˆ b3
b  b‰ œ !

1 h# b 3

# (b) r(y) œ 0 and R(y) œ b ˆ1
yh ‰ Ê V œ '0 1 a[R(y)]#
[r(y)]# b dy œ 1b# '0 ˆ1
yh ‰ dy h

œ 1b# '0 Š1
h

2y h



y# h# ‹

dy œ 1b# ’y


y# h

h



h

y$ 3h# “ !

1 b# h 3

œ 1b# ˆh
h  3h ‰ œ

49. R(y) œ b  Èa#
y# and r(y) œ b
Èa#
y# Ê V œ 'ca 1 a[R(y)]#
[r(y)]# b dy a

œ 1 'ca ’ˆb  Èa#
y# ‰
ˆb
Èa#
y# ‰ “ dy #

a

#

œ 1 'ca 4bÈa#
y# dy œ 4b1'ca Èa#
y# dy a

a

1a# #

œ 4b1 † area of semicircle of radius a œ 4b1 †

œ 2a# b1#

50. (a) A cross section has radius r œ È#y and area 1r# œ #1y. The volume is '0 #1ydy œ 1 cy# d ! œ #&1. &

(b) Vahb œ ' Aahbdh, so

dV dh

œ Aahb. Therefore

For h œ %, the area is #1a%b œ )1, so

dh dt

œ

dV dt

" )1

œ

dV dh

†

œ Aahb †

dh dt

$

$ )1

† $ units sec œ

hca

51. (a) R(y) œ Èa#
y# Ê V œ 1'ca aa#
y# b dy œ 1 ’a# y
œ 1 ’a# h
"3 ah$
3h# a  3ha#
a$ b
dV $ dt œ 0.2 m /sec dV # dh œ 101h
1h

(b) Given Ê

and a œ 5 m, find Ê

dV dt

œ

dV dh

†

dh dt

a$ 3“

œ 1 Ša# h


†

so

dh dt

œ

" A ah b

†

dV dt .

units$ sec . hca

y$ 3 “ ca

h$ 3

dh dt ,

&

œ 1 ’a# h
a$


 h# a
ha# ‹ œ

(h
a)$ 3


Š
a$ 

a$ 3 ‹“

1h# (3a
h) 3

#

From part (a), V(h) œ 1h (153
h) œ 51h#
13h dh ¸ 0.2 " " œ 1h(10
h) dh dt Ê dt hœ4 œ 41(10
4) œ (201)(6) œ 1#01 m/sec. dh ¸ dt hœ4 .

$

52. Suppose the solid is produced by revolving y œ 2
x about the y-axis. Cast a shadow of the solid on a plane parallel to the xy-plane. Use an approximation such as the Trapezoid Rule, to #

estimate 'a 1cRaybd# dy ¸ ! 1Œ #k  ˜y. b

n

d^

kœ"

53. The cross section of a solid right circular cylinder with a cone removed is a disk with radius R from which a disk of radius h has been removed. Thus its area is A" œ 1R#
1h# œ 1 aR#
h# b . The cross section of the hemisphere is a disk of #

radius ÈR#
h# . Therefore its area is A# œ 1 ŠÈR#
h# ‹ œ 1 aR#
h# b . We can see that A" œ A# . The altitudes of both solids are R. Applying Cavalieri's Principle we find Volume of Hemisphere œ (Volume of Cylinder)
(Volume of Cone) œ a1R# b R
"3 1 aR# b R œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 3

1 R$ .

370

Chapter 6 Applications of Definite Integrals

54. R(x) œ

rx h

Ê V œ '0 1[R(x)]# dx œ 1'0 h

h # # r x

dx œ

h#

1r# h#

h

$

#

$

’ x3 “ œ Š 1hr# ‹ Š h3 ‹ œ !

" 3

1r# h, the volume of

a cone of radius r and height h. c7

c7

55. R(y) œ È256
y# Ê V œ 'c16 1[R(y)]# dy œ 1'c16 a256
y# b dy œ 1 ’256y
œ 1 ’(256)(
7)  56. R(x) œ œ

1 144

x 1#

7$ 3


Š(256)(
16) 

16$ 3 ‹“

$

È36
x# Ê V œ ' 1[R(x)]# dx œ 1' 0 0 6

' x& 5 “!

’12x$


œ

1 144

6

Š12 † 6$


6& 5‹

œ

16$ 3 ‹

œ 1 Š 73  256(16
7)


1 †6 $ 144

x# 144

ˆ12


a36
x# b dx œ

36 ‰ 5

1 144


( y$ 3 “
"'

œ 10531 cm$ ¸ 3308 cm$

'06 a36x#
x% b dx

1 ‰ ˆ 60
36 ‰ œ ˆ 196 œ 144 5

361 5

cm$ . The plumb bob will

weigh about W œ (8.5) ˆ 3651 ‰ ¸ 192 gm, to the nearest gram. 57. (a) R(x) œ kc
sin xk , so V œ 1'0 [R(x)]# dx œ 1'0 (c
sin x)# dx œ 1'0 ac#
2c sin x  sin# xb dx 1

1

œ 1'0 ˆc#
2c sin x 

'1

1

(b)

1

1
cos 2x ‰ dx œ 1 0 ˆc#  "#
2c sin x
cos#2x ‰ dx # 1 œ 1
ˆc#  "# ‰ x  2c cos x
sin42x ‘ ! œ 1
ˆc# 1  1#
2c
0‰
(0  2c
0)‘ œ 1 ˆc# 1  1#
4c‰ . Let 2 V(c) œ 1 ˆc# 1  1#
4c‰ . We find the extreme values of V(c): dV dc œ 1(2c1
4) œ 0 Ê c œ 1 is a critical # # point, and V ˆ 12 ‰ œ 1 ˆ 14  1#
18 ‰ œ 1 ˆ 1#
14 ‰ œ 1#
4; Evaluate V at the endpoints: V(0) œ 1# and # # V(1) œ 1 ˆ 3# 1
4‰ œ 1#
(4
1)1. Now we see that the function's absolute minimum value is 1#
4, taken on at the critical point c œ 12 . (See also the accompanying graph.) # From the discussion in part (a) we conclude that the function's absolute maximum value is 1# , taken on at

the endpoint c œ 0. (c) The graph of the solid's volume as a function of c for 0 Ÿ c Ÿ 1 is given at the right. As c moves away from [!ß "] the volume of the solid increases without bound. If we approximate the solid as a set of solid disks, we can see that the radius of a typical disk increases without bounds as c moves away from [0ß 1]. 58. (a) R(x) œ 1


œ 1 'c4 Š1
4

œ 1 ’x


x$ 24

œ 21 ˆ4


Ê V œ 'c4 1[R(x)]# dx 4

x# 16

8 3

x# 16 ‹



#

dx œ 1'c4 Š1
4

%

x& 5†16# “
%

 45 ‰ œ

21 15

œ 21 Š4


x# 8



x% 16# ‹

4$ 24



4& 5†16# ‹

(60
40  12) œ

641 15

dx

ft$

(b) The helicopter will be able to fly ˆ 64151 ‰ (7.481)(2) ¸ 201 additional miles. 6.2 VOLUME BY CYLINDRICAL SHELLS 1. For the sketch given, a œ 0, b œ 2;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š1  b

2

x# 4‹

dx œ 21'0 Šx 

x# 4‹

dx œ 21'0 Š2x


2

x$ 4‹

œ 21 † 3 œ 61 2. For the sketch given, a œ 0, b œ 2;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š2
b

2

2

#

dx œ 21 ’ x# 

x$ 4‹

# x% 16 “ !

dx œ 21 ’x#


œ 21 ˆ 4# 

# x% 16 “ !

16 ‰ 16

œ 21(4
1) œ 61

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.2 Volume by Cylindrical Shells 3. For the sketch given, c œ 0, d œ È2;

È2

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 d

È2

21y † ay# b dy œ 21'0

%

y$ dy œ 21 ’ y4 “

4. For the sketch given, c œ 0, d œ È3;

È3

È3

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † c3
a3
y# bd dy œ 21 '0 d

È# !

œ 21

%

y$ dy œ 21 ’ y4 “

È3 !

œ

5. For the sketch given, a œ 0, b œ È3;

È3

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † ŠÈx#  1‹ dx; b

’u œ x#  1 Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ È3 Ê u œ 4“ Ä V œ 1'1 u"Î# du œ 1
23 u$Î# ‘ " œ %

4

21 3

ˆ4$Î#
1‰ œ ˆ 231 ‰ (8
1) œ

141 3

6. For the sketch given, a œ 0, b œ 3;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x Š Èx9x ‹ dx; $
9 b

3

cu œ x$  9 Ê du œ 3x# dx Ê 3 du œ 9x# dx; x œ 0 Ê u œ 9, x œ 3 Ê u œ 36d Ä V œ 21 '9 3u"Î# du œ 61
2u"Î# ‘ * œ 121 ŠÈ36
È9‹ œ 361 $'

36

7. a œ 0, b œ 2;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x
x
ˆ
x2 ‰‘ dx b

2

œ '0 21x# † 2

3 #

dx œ 1 '0 3x# dx œ 1 cx$ d ! œ 81 2

#

8. a œ 0, b œ 1;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2x
x2 ‰ dx b

1

œ 1 '0 2 Š 3x# ‹ dx œ 1 ' 3x# dx œ 1 cx$ d ! œ 1 1

1

#

"

0

9. a œ 0, b œ 1;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x c(2
x)
x# d dx b

1

œ 21'0 a2x
x#
x$ b dx œ 21 ’x#
1

œ 21 ˆ1


" 3


4" ‰ œ 21 ˆ 12 124  3 ‰ œ

x$ 3




101 12

œ

" x% 4 “!

51 6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

91 #

371

372

Chapter 6 Applications of Definite Integrals

10. a œ 0, b œ 1;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ca2
x# b
x# d dx b

1

œ 21'0 x a2
2x# b dx œ 41'0 ax
x$ b dx 1

1

" x% 4 “!

#

œ 41 ’ x#


œ 41 ˆ "2
4" ‰ œ 1

11. a œ 0, b œ 1;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x
Èx
(2x
1)‘ dx b

1

" œ 21'0 ˆx$Î#
2x#  x‰ dx œ 21
25 x&Î#
23 x$  "# x# ‘ ! 1

œ 21 ˆ 25


2 3


15 ‰  "# ‰ œ 21 ˆ 12  20 œ 30

71 15

12. a œ ", b œ 4;

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ˆ 32 x"Î# ‰ dx b

4

œ 31'1 x"Î# dx œ 31
23 x$Î# ‘ " œ 21 ˆ4$Î#
"‰ %

4

œ 21(8
1) œ 141

13. (a) xf(x) œ œ

xf(x) œ œ

x†

sin x, 0  x Ÿ 1 0xŸ1 Ê xf(x) œ œ ; since sin 0 œ 0 we have 0, x œ 0 x, x œ 0

sin x x ,

sin x, 0  x Ÿ 1 Ê xf(x) œ sin x, 0 Ÿ x Ÿ 1 sin x, x œ 0

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † f(x) dx and x † f(x) œ sin x, 0 Ÿ x Ÿ 1 by part (a) 1

b

Ê V œ 21'0 sin x dx œ 21[
cos x]1! œ 21(
cos 1  cos 0) œ 41 1

tan# x x ,

tan# x, 0  x Ÿ 1/4 0  x Ÿ 14 Ê xg(x) œ œ ; since tan 0 œ 0 we have 0, x œ 0 x † 0, x œ 0 tan# x, 0  x Ÿ 1/4 Ê xg(x) œ tan# x, 0 Ÿ x Ÿ 1/4 xg(x) œ œ tan# x, x œ 0

14. (a) xg(x) œ œ

x†

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x † g(x) dx and x † g(x) œ tan# x, 0 Ÿ x Ÿ 1/4 by part (a) 1Î4

b

Ê V œ 21'0 tan# x dx œ 21'0 asec# x
1b dx œ 21[tan x
x]! 1Î4

1Î4

1Î%

œ 21 ˆ1
14 ‰ œ

41  1 # #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.2 Volume by Cylindrical Shells 15. c œ 0, d œ 2;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y
Èy
(
y)‘ dy d

2

œ 21'0 ˆy$Î#  y# ‰ dy œ 21 ’ 2y5 2

&Î#

&

œ 21 ” 25 ŠÈ2‹  œ

161 15

2$ 3•



# y$ 3 “!

È

œ 21 Š 8 5 2  83 ‹ œ 161 Š

È2 5

 "3 ‹

Š3È2  5‹

16. c œ 0, d œ 2;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y cy#
(
y)ddy d

2

œ 21'0 ay$  y# b dy œ 21 ’ y4  2

%

œ 161 ˆ 56 ‰ œ

401 3

# y$ 3 “!

œ 161 ˆ 24  "3 ‰

17. c œ 0, d œ 2;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y
y# bdy d

2

œ 21'0 a2y#
y$ b dy œ 21 ’ 2y3
2

$

œ 321 ˆ "3
4" ‰ œ

321 12

œ

81 3

# y% 4 “!

œ 21 ˆ 16 3


"6 ‰ 4

18. c œ 0, d œ 1;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y a2y
y#
ybdy d

1

œ 21'0 y ay
y# b dy œ 21'0 ay#
y$ b dy 1

1

$

œ 21 ’ y3


" y% “ 4 !

œ 21 ˆ 13
"4 ‰ œ

1 6

19. c œ 0, d œ 1;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ 21'0 y[y
(
y)]dy d

1

œ 21'0 2y# dy œ 1

41 3

"

cy$ d ! œ

41 3

20. c œ 0, d œ 2;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 yˆy
y2 ‰dy d

2

œ 21 '0

2

y2 2 dy

1

œ 13 c y3 d ! œ

81 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

373

374

Chapter 6 Applications of Definite Integrals

21. c œ 0, d œ 2;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2  y)
y# d dy d

2

œ 21 '0 a2y  y#
y$ b dy œ 21 ’y#  2

œ 21 ˆ4 




8 3

16 ‰ 4

1 6

œ

y$ 3

(48  32
48) œ




# y% 4 “!

161 3

22. c œ 0, d œ 1;

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y c(2
y)
y# d dy d

1

œ 21'0 a2y
y#
y$ b dy œ 21 ’y#
1

œ 21 ˆ1



14 ‰ œ

1 3

1 6

(12
4
3) œ

y$ 3




51 6

" y% 4 “!

shell ‰ shell 23. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y † 12 ay#
y$ b dy œ 241 '0 ay$
y% b dy œ 241 ’ y4
d

1

œ 241 ˆ 14
15 ‰ œ

241 20

œ

1

" y& 5 “!

%

61 5

shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1
y) c12 ay#
y$ bd dy œ 241'0 (1
y) ay#
y$ b dy d

1

1

œ 241'0 ay#
2y$  y% b dy œ 241 ’ y3
1

$

y% 2



" y& 5 “!

œ 241 ˆ "3


1 2

" ‰  51 ‰ œ 241 ˆ 30 œ

41 5

shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆ 85
y‰ c12 ay#
y$ bd dy œ 241 '0 ˆ 85
y‰ ay#
y$ b dy d

1

œ 241'0 ˆ 85 y#
1

œ

241 12

13 5

1

8 $ y$  y% ‰ dy œ 241 ’ 15 y


13 20

y% 

" y& 5 “!

8 œ 241 ˆ 15


13 20

241 60

 15 ‰ œ

(32
39  12)

œ 21

shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy  25 ‰ c12 ay#
y$ bd dy œ 241'0 ˆy  25 ‰ ay#
y$ b dy d

1

1

2 $ œ 241'0 ˆy$
y%  25 y#
25 y$ ‰ dy œ 241'0 ˆ 25 y#  35 y$
y% ‰ dy œ 241 ’ 15 y  1

1

2 œ 241 ˆ 15 

3 20


15 ‰ œ

241 60

(8  9
12) œ

241 12

2

%

œ 21 ’ y4


#

y' 24 “ !

%

2' 24 ‹

œ 21 Š 24


#

%

œ 321 ˆ 4"


4 ‰ 24

dy œ '0 21y Šy#
2

y# # ‹“

2

œ 21 '0 Š2y#
2

y% 2


y$ 

y& 4‹

#

$

dy œ 21 ’ 2y3





y% 4



2

œ 21'0 Š5y#
54 y%
y$  2

y& 4‹

#

$

dy œ 21 ’ 5y3





y% 4

# y' #4 “ !

œ 21 ˆ 16 3


œ

2

œ 21'0 Šy$
2

y& 4

 58 y#


5 32

#

%

y% ‹ dy œ 21 ’ y4


%

y' #4



5y$ #4

2

y# # ‹“




16 4



64 ‰ 24

2

y' #4 “ ! y# # ‹“




32 10

dy œ '0 21(5
y) Šy#
#



shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21 ˆy  58 ‰ ’ y#
Š y4
d

81 3 y% 4‹

%

5y& 20

2

dy œ '0 21(2
y) Šy#


shell ‰ shell (c) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(5
y) ’ y#
Š y4
d

dy œ 21'0 Šy$


y# # ‹“

%

y& 10

y% 4‹

2 ‰ œ 321 ˆ 4"
6" ‰ œ 321 ˆ 24 œ

shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2
y) ’ y#
Š y4
d

" y& 5 “!

y%


œ 21

shell ‰ shell 24. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y ’ y#
Š y4
d

3 20

œ 21 ˆ 40 3


160 20




16 4



dy œ '0 21 ˆy  58 ‰ Šy#


# 5y& 160 “ !

2

œ 21 ˆ 16 4


64 24



40 24




Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

81 5

y% 4‹

64 ‰ 24

dy

dy

œ 81 y% 4‹

160 ‰ 160

dy

œ 41

y& 4‹

dy

Section 6.2 Volume by Cylindrical Shells

375

shell ‰ shell 25. (a) About x-axis: V œ 'c 21 ˆ radius Š height ‹dy d

œ '0 21yˆÈy
y‰dy œ 21'0 ˆy$Î#
y# ‰dy 1

1

" œ 21
#& y&Î#
"$ y$ ‘ ! œ 21ˆ #&
"$ ‰ œ

#1 "&

shell ‰ shell About y-axis: V œ 'a 21 ˆ radius Š height ‹dx b

œ '0 21xax
x# bdx œ 21'0 ax2
x3 bdx 1

1

$

œ 21’ x$


" x% % “!

œ 21ˆ "$
"% ‰ œ

1 '

(b) About x-axis: Raxb œ x and raxb œ x# Ê V œ 'a 1
Raxb#
raxb# ‘dx œ '0 1cx#
x% ddx b

$

œ 1’ x$


" x& & “!

œ 1ˆ "$
"& ‰ œ

1

#1 "&

About y-axis: Rayb œ Èy and rayb œ y Ê V œ 'c 1
Rayb#
rayb# ‘dy œ '0 1cy
y2 ddy d

#

œ 1’ y#


" y$ $ “!

œ 1ˆ "#
"$ ‰ œ

1

1 '

# 26. (a) V œ 'a 1
Raxb#
raxb# ‘dx œ 1'0 ’ˆ #x  #‰
x# “dx %

b

œ 1'0 ˆ
$% x#  #x  %‰dx œ 1’
x%  x#  %x“ %

$

œ 1a
"'  "'  "'b œ "'1

% !

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1xˆ x#  #
x‰dx %

b

œ '0 #1xˆ#
x# ‰dx œ #1'0 Š#x
%

%

œ #1’x#


% x$ ' “!

'% ‰ '

œ #1ˆ"'


œ

x# # ‹dx $#1 $

shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹dx œ '0 #1a%
xbˆ x#  #
x‰dx œ '0 #1a%
xbˆ#
x# ‰dx œ #1'0 Š)
%x  %

b

œ #1’)x
#x# 

% x$ “ ' !

œ #1ˆ$#
$# 

%

'% ‰ '

%

x# # ‹dx

'%1 $

œ

# (d) V œ 'a 1
Raxb#
raxb# ‘dx œ 1'0 ’a)
xb#
ˆ'
#x ‰ “dx œ 1'0 ’a'%
"'x  x# b
Š$'
'x x% ‹“dx %

b

%

#

1'0 ˆ $% x#
"!x  #)‰dx œ 1’ x%
&x#  #)x“ œ 1
"'
a&ba"'b  a(ba"'b‘ œ 1a$ba"'b œ %)1 %

%

$

!

shell ‰ shell 27. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21y(y
1) dy d

2

œ 21'1 ay#
yb dy œ 21 ’ y3
2

$

# y# # “"

œ 21
ˆ 83
42 ‰
ˆ "3
#" ‰‘ œ 21 ˆ 73
2  "# ‰ œ 13 (14
12  3) œ

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx

51 3

b

œ '1 21x(2
x) dx œ 21'1 a2x
x# b dx œ 21 ’x#
2

2

œ 21
ˆ 12 3 8 ‰
ˆ 3 3 " ‰‘ œ 21 ˆ 34
32 ‰ œ

41 3

# x$ 3 “"

œ 21
ˆ4
83 ‰
ˆ1
"3 ‰‘

shell ‰ shell ' ˆ 203
‰ (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21 ˆ 10 3
x (2
x) dx œ 21 1 b

2

#

" $‘ 8 #
ˆ 40 œ 21
20 3 x
3 x  3 x " œ 21 3


2

32 3

 38 ‰
ˆ 20 3


8 3

16 3

x  x# ‰ dx

 3" ‰‘ œ 21 ˆ 33 ‰ œ 21

shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '1 21(y
1)(y
1) dy œ 21'1 (y
1)# œ 21 ’ (y31) “ œ d

2

2

$

# "

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

21 3

376

Chapter 6 Applications of Definite Integrals

shell ‰ shell 28. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay#
0b dy d

2

œ 21'0 y$ dy œ 21 ’ y4 “ œ 21 Š 24 ‹ œ 81 2

%

#

%

!

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx b

œ '0 21x ˆ2
Èx‰ dx œ 21'0 ˆ2x
x$Î# ‰ dx 4

4

%

2 †2 & 5 ‹

œ 21
x#
25 x&Î# ‘ ! œ 21 Š16
œ 21 ˆ16


64 ‰ 5

21 5

œ

321 5

(80
64) œ

shell ‰ shell (c) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4
x) ˆ2
Èx‰ dx œ 21'0 ˆ8
4x"Î#
2x  x$Î# ‰ dx b

4

4

%

œ 21
8x
83 x$Î#
x#  25 x&Î# ‘ ! œ 21 ˆ32


64 3


16 

64 ‰ 5

œ

21 15

(240
320  192) œ

21 15

shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(2
y) ay# b dy œ 21 '0 a2y#
y$ b dy œ 21 ’ 23 y$
d

2

œ 21 ˆ 16 3


16 ‰ 4

œ

321 12

2

81 3

(4
3) œ

(112) œ

2241 15

# y% 4 “!

shell ‰ shell 29. (a) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21yay
y$ b dy d

1

œ '0 21 ay#
y% b dy œ 21 ’ y3
1

œ

$

41 15

" y& “ 5 !

œ 21 ˆ "3
"5 ‰

shell ‰ shell (b) V œ 'c 21 ˆ radius Š height ‹ dy d

œ '0 21(1
y) ay
y$ b dy 1

œ 21 '0 ay
y#
y$  y% b dy œ 21 ’ y#
1

#

y$ 3




y% 4



" y& 5 “!

œ 21 ˆ "#


" 3




" 4

 5" ‰ œ

21 60

(30
20
15  12) œ

71 30

shell ‰ shell 30. (a) V œ 'c 21 ˆ radius Š height ‹dy d

œ '0 21y c1
ay
y$ bddy 1

œ 21 '0 ay
y#  y% b dy œ 21 ’ y#
1

#

œ 21 ˆ "#
œ

" 3

 5" ‰ œ

21 30

y$ 3



" y& 5 “!

(15
10  6)

111 15

(b) Use the washer method: V œ 'c 1 cR# (y)
r# (y)d dy œ '0 1 ’1#
ay
y$ b “ dy œ 1 '0 a1
y#
y'  2y% b dy œ 1 ’y
d

1

œ 1 ˆ1


" 3




" 7

 25 ‰ œ

1 105

1

#

(105
35
15  42) œ

y$ 3




y( 7



971 105

" 2y& 5 “!

(c) Use the washer method: V œ 'c 1 cR# (y)
r# (y)d dy œ '0 1 ’c1
ay
y$ bd
0“ dy œ 1'0 ’1
2 ay
y$ b  ay
y$ b “ dy d

1

œ 1'0 a1  y#  y'
2y  2y$
2y% b dy œ 1 ’y  1

œ

1 210

(70  30  105
2 † 42) œ

1

#

y$ 3



y( 7


y# 

#

y% #




1211 210

" 2y& 5 “!

œ 1 ˆ1 

" 3



" 7


1

" #


25 ‰

shell ‰ shell (d) V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(1
y) c1
ay
y$ bd dy œ 21 '0 (1
y) a1
y  y$ b dy d

1

1

œ 21'0 a1
y  y$
y  y#
y% b dy œ 21'0 a1
2y  y#  y$
y% b dy œ 21 ’y
y#  1

œ 21 ˆ1
1 

1

" 3



" 4


5" ‰ œ

21 60

(20  15
12) œ

231 30

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

y$ 3



y% 4




" y& 5 “!

Section 6.2 Volume by Cylindrical Shells shell ‰ shell 31. (a) V œ 'c 21 ˆ radius Š height ‹dy œ '0 21y ˆÈ8y
y# ‰ dy d

2

œ 21'0 Š2È2 y$Î#
y$ ‹ dy œ 21 ’ 4 5 2 y&Î#
È

2

# y% 4 “!

&

œ 21 

4È2†ŠÈ2‹

2% 4




5

œ 21 † 4 ˆ 85
1‰ œ

81 5

$

œ 21 Š 4†52


(8
5) œ

4 †4 4 ‹

241 5

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ŠÈx
b

4

&

œ 21 Š 2†52


4% 3# ‹

'

œ 21 Š 25


2) 32 ‹

œ

1†2( 160

x# 8‹

dx œ 21'0 Šx$Î#


(32
20) œ

4

1†2* †3 160

œ

1†2% †3 5

œ

x$ 8‹

dx œ 21 ’ 25 x&Î#


481 5

shell ‰ shell 32. (a) V œ 'a 21 ˆ radius Š height ‹ dx b

œ '0 21x ca2x
x# b
xd dx 1

œ 21 '0 x ax
x# b dx œ 21'0 ax#
x$ b dx 1

$

1

œ 21 ’ x3


" x% 4 “!

œ 21 ˆ "3
4" ‰ œ

1 6

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(1
x) ca2x
x# b
xd dx œ 21'0 (1
x) ax
x# b dx b

1

1

œ 21 '0 ax
2x#  x$ b dx œ 21 ’ x2
23 x$  1

" x% 4 “!

#

œ 21 ˆ 12


2 3

 "4 ‰ œ

21 1#

(6
8  3) œ

1 6

33. (a) V œ 'a 1 cR# (x)
r# (x)d dx œ 1 '1Î16 ˆx"Î#
1‰ dx b

1

"

œ 1
2x"Î#
x‘"Î"' œ 1
(2
1)
ˆ2 † œ 1 ˆ1


7 ‰ 16

œ

" 4




" ‰‘ 16

91 16

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dy œ '1 21y Š y"%
b

2

œ 21'1 ˆy$
2

y ‰ 16

dy œ 21 ’
12 y#


œ 21
ˆ
"8
8" ‰
ˆ
#"
œ

21 32

(8  1) œ

91 16

" ‰‘ 3#

d

2

œ 1

"3 y$
œ

1 48

y ‘# 16 "

y# 32 “ "

" ‰ 32

" 16 ‹

dy

" œ 1
ˆ
24
8" ‰
ˆ
3"


(
2
6  16  3) œ

dy

#

œ 21 ˆ 4" 

34. (a) V œ 'c 1 cR# (y)
r# (y)d dy œ '1 1 Š y"%


" 16 ‹

" ‰‘ 16

111 48

shell ‰ shell (b) V œ 'a 21 ˆ radius Š height ‹ dx œ '1Î4 21x Š È"x
"‹ dx b

1

œ 21 '1Î4 ˆx"Î#
x‰ dx œ 21 ’ 23 x$Î#
1

œ 21
ˆ 23
"# ‰
ˆ 23 †

" 8




" ‰‘ 3#

" x# 2 “ "Î%

œ 1 ˆ 43
1


" 6



" ‰ 16

œ

1 48

(4 † 16
48
8  3) œ

111 48

35. (a) H3=k: V œ V"
V#

V" œ 'a 1[R" (x)]# dx and V# œ 'a 1[R# (x)]# with R" (x) œ É x
3 2 and R# (x) œ Èx, b"

b#

"

#

a" œ
2, b" œ 1; a# œ 0, b# œ 1 Ê two integrals are required (b) [ +=2/<: V œ V"
V#

V" œ 'a 1 a[R" (x)]#
[r" (x)]# b dx with R" (x) œ É x
3 2 and r" (x) œ 0; a" œ
2 and b" œ 0; b"

"

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

% x% 32 “ !

377

378

Chapter 6 Applications of Definite Integrals V# œ 'a 1 a[R# (x)]#
[r# (x)]# b dx with R# (x) œ É x
3 2 and r# (x) œ Èx; a# œ 0 and b# œ 1 b#

#

Ê two integrals are required

shell ‰ shell shell (c) W2/66: V œ 'c 21 ˆ radius Š height ‹ dy œ 'c 21y Š height ‹ dy where shell height œ y#
a3y#
2b œ 2
2y# ; d

d

c œ 0 and d œ 1. Only 98/ integral is required. It is, therefore preferable to use the =2/66 method. However, whichever method you use, you will get V œ 1. 36. (a) H3=k: V œ V"
V#
V$

Vi œ 'c 1[Ri (y)]# dy, i œ 1, 2, 3 with R" (y) œ 1 and c" œ
1, d" œ 1; R# (y) œ Èy and c# œ 0 and d# œ 1; di

i

R$ (y) œ (
y)"Î% and c$ œ
1, d$ œ 0 Ê three integrals are required (b) [ +=2/<: V œ V"  V#

Vi œ 'c 1a[Ri (y)]#
[ri (y)]# b dy, i œ 1, 2 with R" (y) œ 1, r" (y) œ Èy, c" œ 0 and d" œ 1; di

i

R# (y) œ 1, r# (y) œ (
y)"Î% , c# œ
1 and d# œ 0 Ê two integrals are required

shell ‰ shell shell (c) W2/66: V œ 'a 21 ˆ radius Š height ‹dx œ 'a 21xŠ height ‹dx, where shell height œ x#
a
x% b œ x#  x% , b

b

a œ 0 and b œ 1 Ê only one integral is required. It is, therefore preferable to use the =2/66 method. However, whichever method you use, you will get V œ 561 . 6.3 LENGTHS OF PLANE CURVES

1.

dx dt

œ
1 and

dy dt

#

#

È(
1)#  (3)# œ È10 ‰  Š dy œ 3 Ê Êˆ dx dt dt ‹ œ

Ê Length œ '2/3 È10 dt œ È10 ctd 12/3 œ È10
Š
23 È10‹ œ 1

2.

dx dt

œ
sin t and

dy dt

5È10 3

#

#

È(
sin t)#  (1  cos t)# œ È2  2 cos t ‰  Š dy œ 1  cos t Ê Êˆ dx dt dt ‹ œ

cos t ‰ È2 ' É sin# t dt Ê Length œ '! È2  2 cos t dt œ È2 '! Ɉ 11   cos t (1  cos t) dt œ 1  cos t ! 1

œ È2'!

1

sin t È1  cos t

1

1

dt (since sin t   0 on [0ß 1]); [u œ 1
cos t Ê du œ sin t dt; t œ 0 Ê u œ 0,

# t œ 1 Ê u œ 2] Ä È2 '! u"Î# du œ È2
2u"Î# ‘ ! œ 4 2

3.

dx dt

œ 3t# and

dy dt

#

#

‰  Š dy Éa3t# b#  (3t)# œ È9t%  9t# œ 3tÈt#  1 Šsince t   0 on ’0ß È3“‹ œ 3t Ê Êˆ dx dt dt ‹ œ

È3

Ê Length œ '! 3tÈt#  1 dt; ’u œ t#  1 Ê Ä

4.

dx dt

3 #

du œ 3t dt; t œ 0 Ê u œ 1, t œ È3 Ê u œ 4“

'14 3# u"Î# du œ
u$Î# ‘ %" œ (8
1) œ 7

œ t and

dy dt

#

#

Èt#  (2t  1) œ È(t  1)# œ kt  1k œ t  1 since 0 Ÿ t Ÿ 4 ‰  Š dy œ (2t  1)"Î# Ê Êˆ dx dt dt ‹ œ

Ê Length œ '0 (t  1) dt œ ’ t2  t“ œ (8  4) œ 12 4

%

#

!

5.

dx dt

œ (2t  3)"Î# and

dy dt

#

#

È(2t  3)  (1  t)# œ Èt#  4t  4 œ kt  2k œ t  2 ‰  Š dy œ 1  t Ê Êˆ dx dt dt ‹ œ

since 0 Ÿ t Ÿ 3 Ê Length œ '0 (t  2) dt œ ’ t2  2t“ œ $

#

$ !

21 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.3 Lengths of Plane Curves 6.

dx dt

œ 8t cos t and

dy dt

œ k8tk œ 8t since 0 Ÿ t Ÿ 7.

dy dx

œ

" 3

† 3# ax#  2b

#

#

È(8t cos t)#  (8t sin t)# œ È64t# cos# t  64t# sin# t ‰  Š dy œ 8t sin t Ê Êˆ dx dt dt ‹ œ

"Î#

Ê Length œ '0 8t dt œ c4t# d ! 1Î2

1 #

1Î#

œ 1#

† 2x œ Èax#  2b † x

Ê L œ '0 È1  ax#  2b x# dx œ '0 È1  2x#  x% dx $

3

œ '0 Éa1  x# b# dx œ '0 a1  x# b dx œ ’x  $

œ3

8.

dy dx

œ

3 #

œ 12

27 3

Èx Ê L œ ' É1  94 x dx;
u œ 1  94 x 0 4

Ê du œ

dx Ê

9 4

du œ dx; x œ 0 Ê u œ 1; x œ 4

4 9

Ê u œ 10d Ä L œ '1 u"Î# ˆ 49 du‰ œ 10

œ

9.

dx dy

8 27

œ y#


#

" 4y#

% Ê Š dx dy ‹ œ y


3

œ '1 Éy%  3

" #

œ '1 ÊŠy#  3

$

œ ’ y3
dx dy

œ

4 9


23 u$Î# ‘ "! "

Š10È10
1‹

Ê L œ '1 É1  y%


10.

$ x$ 3 “!

$

" #



" 16y%

" 4y# ‹

$ y
" 4 “"

#

" #

" 16y%



" #



" 16y%

dy

dy

dy œ '1 Šy#  3

" ‰ 1#

œ ˆ 27 3


" 4y# ‹

dy


ˆ 3"
4" ‰ œ 9
#

y"Î#
"# y"Î# Ê Š dx dy ‹ œ

" 4

" 1#




" 3



" 4

œ9

(1  4
3) 1#

œ9

(2) 1#

œ

53 6

Šy
2  y" ‹

Ê L œ '1 Ê1  "4 Šy
2  y" ‹ dy 9

œ '1 Ê "4 Šy  2  y" ‹ dy œ '1 9

œ

" #

9

" #

" Èy ‹

Ê ŠÈ y 

*

$Î#

$

"

dx dy

dy

'19 ˆy"Î#  y"Î# ‰ dy œ "#
23 y$Î#  2y"Î# ‘ *"

œ ’ y 3  y"Î# “ œ Š 33  3‹
ˆ "3  1‰ œ 11


11.

#

œ y$


" 4y$

#

' Ê Š dx dy ‹ œ y


Ê L œ '1 É1  y'
2

œ '1 Éy'  2

œ '1 Šy$  2

œ Š 16 4


" 2



y
$ 4 ‹

" (16)(2) ‹

" 16y'

" 2



" 16y'

" #



32 3

dy

2

%

œ

" 16y'

dy œ '1 ÊŠy$ 

dy œ ’ y4


" 3

y
$ 4 ‹

#

dy

# y
# 8 “"


ˆ "4
"8 ‰ œ 4


" 32




" 4



" 8

œ

12818
4 32

œ

123 32

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

379

380 12.

Chapter 6 Applications of Definite Integrals dx dy

œ

y# #




" #y #

#

Ê Š dx dy ‹ œ

" 4

ay%
2  y% b

Ê L œ '2 É1  "4 ay%
2  y% b dy 3

œ '2 É "4 ay%  2  y% b dy 3

13.

œ

" #

'23 Éay#  y# b# dy œ "# '23 ay#  y# b dy

œ

" #

’ y3
y" “ œ

dy dx

$

$

" #

#

"‰
ˆ 27 ˆ 8 " ‰‘ œ 3
3
3
# #

#Î$ œ x"Î$
"4 x"Î$ Ê Š dy
dx ‹ œ x

Ê L œ '1 É1  x#Î$
8

œ '1 Éx#Î$  8

" #



x
#Î$ 16

" #

x
#Î$ 16



" #



" #

ˆ 26 3


8 3

 #" ‰ œ

" #

ˆ6  #" ‰ œ

13 4

x
#Î$ 16

dx

dx

# œ '1 Ɉx"Î$  "4 x"Î$ ‰ dx œ '1 ˆx"Î$  "4 x"Î$ ‰ dx 8

8

)

œ
34 x%Î$  38 x#Î$ ‘ " œ œ 14.

dy dx

3 8


2x%Î$  x#Î$ ‘ ) "

3 8

ca2 † 2%  2# b
(2  1)d œ

œ x#  2x  1


œ (1  x)#


#

2

œ '0 É(1  x)% 

" #

œ '0 Ê’(1  x)#  2

œ '0 ’(1  x)#  2



" #

(1
x)
% 16

# (1
x)
# “ 4

(1
x)
# “ 4



(1
x)
% 16

99 8

" " 4 (1
x)#

% Ê Š dy dx ‹ œ (1  x)


Ê L œ '0 É1  (1  x)%
2

(32  4
3) œ

œ x#  2x  1


4 (4x
4)#

" " 4 (1
x)#

3 8

" #



" 16(1
x)%

dx

dx

dx

dx; cu œ 1  x Ê du œ dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 3d

Ä L œ '1 ˆu#  "4 u# ‰ du œ ’ u3
"4 u" “ œ ˆ9
3

$

$

"

15.

dx dy

" ‰ 1#


ˆ 3"
4" ‰ œ

10814
3 12

œ

106 12

œ

53 6

#

% œ Èsec% y
1 Ê Š dx dy ‹ œ sec y
1 1Î4

1Î4

Ê L œ '
1Î4 È1  asec% y
1b dy œ '
1Î4 sec# y dy 1Î%

œ ctan yd 1Î% œ 1
(
1) œ 2

16.

dy dx

#

% œ È3x%
1 Ê Š dy dx ‹ œ 3x
1

c1

c1

Ê L œ 'c2 È1  a3x%
1b dx œ 'c2 È3 x# dx $

œ È3 ’ x3 “

" #

œ

È3 3

c
1
(
2)$ d œ

È3 3

(
"  8) œ

7È 3 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.3 Lengths of Plane Curves 17. (a)

dy dx

#

(b)

# œ 2x Ê Š dy dx ‹ œ 4x

Ê L œ 'c1 Ê1
Š dy dx ‹ dx #

2

œ 'c1 È1
4x# dx 2

(c) L ¸ 6.13

18. (a)

dy dx

#

(b)

% œ sec# x Ê Š dy dx ‹ œ sec x

Ê L œ 'c1Î3 È1
sec% x dx 0

(c) L ¸ 2.06

19. (a)

dx dy

#

(b)

# œ cos y Ê Š dx dy ‹ œ cos y

Ê L œ '0 È1
cos# y dy 1

(c) L ¸ 3.82

20. (a)

dx dy

#

œ  È1y
y# Ê Š dx dy ‹ œ 1Î2

Ê L œ 'c1Î2 É1


œ 'c1Î2 a1  y# b 1Î2


"Î#

y# a1
y # b

y# 1
y#

(b) 1Î2

dy œ '
1Î2 É 1
" y# dy

dy

(c) L ¸ 1.05

21. (a) 2y
2 œ 2

dx dy

#

# Ê Š dx dy ‹ œ (y
1)

(b)

Ê L œ 'c1 È1
(y
1)# dy 3

(c) L ¸ 9.29

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

381

382

Chapter 6 Applications of Definite Integrals

22. (a)

dy dx

#

(b)

# # œ cos x - cos x + x sin x Ê Š dy dx ‹ œ x sin x

Ê L œ '0 È1
x# sin# x dx 1

(c) L ¸ 4.70

23. (a)

dy dx

#

(b)

# œ tan x Ê Š dy dx ‹ œ tan x

# x  cos# x Ê L œ '0 È1
tan# x dx œ '0 É sin cos dx #x

1Î6

œ '0

1Î6

1Î6

œ '0 sec x dx 1Î6

dx cos x

(c) L ¸ 0.55

24. (a)

dx dy

#

(b)

# œ Èsec# y  1 Ê Š dx dy ‹ œ sec y  1

Ê L œ 'c1Î3 È1
asec# y  1b dy 1Î4

1Î4

1Î4

œ 'c1Î3 ksec yk dy œ '
1Î3 sec y dy (c) L ¸ 2.20

È2 œ Ê1
Š dy ‹ Ê 25. È2 x œ '0 Ê1
Š dy dt ‹ dt, x   0 Ê dx #

x

#

dy dx

œ „ 1 Ê y œ f(x) œ „ x
C where C is any

real number. 26. (a) From the accompanying figure and definition of the differential (change along the tangent line) we see that dy œ f w (xkc1 ) ˜ xk Ê length of kth tangent fin is È( ˜ xk )#
(dy)# œ È( ˜ xk )#
[f w (xkc1 ) ˜ xk ]# .

n

n

! (length of kth tangent fin) œ lim ! È( ˜ xk )#
[f w (xk
1 ) ˜ xk ]# (b) Length of curve œ n lim Ä_ nÄ_ kœ1

! È1
[f w (xk
1 )]# ˜ xk œ ' È1
[f w (x)]# dx œ n lim Ä_ a n

kœ1

b

kœ1

#

" 27. (a) Š dy dx ‹ correspondes to 4x here, so take So y œ Èx from ("ß ") to (4ß 2).

dy dx

as

" . #È x

Then y œ Èx
C and since ("ß ") lies on the curve, C œ 0.

(b) Only one. We know the derivative of the function and the value of the function at one value of x. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.3 Lengths of Plane Curves #

28. (a) Š dx dy ‹ correspondes to So y œ

" y%

here, so take

dy dx

as

" y# .

Then x œ  y"
C and, since (!ß ") lies on the curve, C œ 1

" "
x.

(b) Only one. We know the derivative of the function and the value of the function at one value of x.

29. (a)

dx dt

œ 2 sin 2t and

dy dt

Ê Length œ '0 2 dt œ c2td ! 1Î2

(b)

dx dt

œ 1 cos 1t and

1Î#

dy dt

#

#

È(2 sin 2t)#
(2 cos 2t)# œ 2 ‰
Š dy œ 2 cos 2t Ê Êˆ dx dt dt ‹ œ œ1 #

#

È(1 cos 1t)#
(1 sin 1t)# œ 1 ‰
Š dy œ 1 sin 1t œ ʈ dx dt dt ‹ œ

Ê Length œ 'c1Î2 1 dt œ c1td
"Î# œ 1 1Î2

30. x œ a()  sin )) Ê Ê

dy d)

dx d)

"Î#

‰# œ a# a1  2 cos )
cos# )b and y œ a(1  cos )) œ a(1  cos )) Ê ˆ dx d)

# # ' ˆ dx ‰#
Š dyd) ‹ d) œ '0 È2a# (1  cos )) d) œ a sin ) Ê Š dy d) ‹ œ a sin ) Ê Length œ 0 Ê d) #

œ aÈ2'0 È2 É 1
#cos ) d) œ 2a '0 ¸sin #) ¸ d) œ 2a '0 sin 21

#

21

21

21

) #

21

#1 d) œ 4a
cos 2) ‘ ! œ 8a

31-36. Example CAS commands: Maple: with( plots ); with( Student[Calculus1] ); with( student ); f := x -> sqrt(1-x^2);a := -1; b := 1; N := [2, 4, 8 ]; for n in N do xx := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x,f(x)],x=xx)]; L := simplify(add( distance(pts[i+1],pts[i]), i=1..n )); T := sprintf("#31(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [f(x),pts], x=a..b, title=T ): end do: display( [seq(P[n],n=N)], insequence=true, scaling=constrained ); L := ArcLength( f(x), x=a..b, output=integral ): L = evalf( L );

# (b) # (a)

# (c)

37-40. Example CAS commands: Maple: with( plots ); with( student ); x := t -> t^3/3; y := t -> t^2/2; a := 0; b := 1; N := [2, 4, 8 ]; for n in N do tt := [seq( a+i*(b-a)/n, i=0..n )]; pts := [seq([x(t),y(t)],t=tt)]; Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

383

384

Chapter 6 Applications of Definite Integrals L := simplify(add( student[distance](pts[i+1],pts[i]), i=1..n )); T := sprintf("#37(a) (Section 6.3)\nn=%3d L=%8.5f\n", n, L ); P[n] := plot( [[x(t),y(t),t=a..b],pts], title=T ): end do: display( [seq(P[n],n=N)], insequence=true ); ds := t ->sqrt( simplify(D(x)(t)^2 + D(y)(t)^2) ): L := Int( ds(t), t=a..b ): L = evalf(L);

# (b) # (a)

# (c)

31-40. Example CAS commands: Mathematica: (assigned function and values for a, b, and n may vary) Clear[x, f] {a, b} = {1, 1}; f[x_] = Sqrt[1  x2 ] p1 = Plot[f[x], {x, a, b}] n = 8; pts = Table[{xn, f[xn]}, {xn, a, b, (b  a)/n}]/ / N Show[{p1,Graphics[{Line[pts]}]}] Sum[ Sqrt[ (pts[[i
1, 1]]  pts[[i, 1]])2
(pts[[i
1, 2]]  pts[[i, 2]])2 ], {i, 1, n}] NIntegrate[ Sqrt[ 1
f'[x]2 ],{x, a, b}] 6.4 MOMENTS AND CENTERS OF MASS 1. Because the children are balanced, the moment of the system about the origin must be equal to zero: 5 † 80 œ x † 100 Ê x œ 4 ft, the distance of the 100-lb child from the fulcrum. 2. Suppose the log has length 2a. Align the log along the x-axis so the 100-lb end is placed at x œ a and the 200-lb end at x œ a. Then the center of mass x satisfies x œ at a distance a  or

2 3

a 3

œ

2a 3

œ

" 3

(2a) which is

" 3

100(
a)  200(a) 300

Ê x œ 3a . That is, x is located

of the length of the log from the 200-lb (heavier) end (see figure)

of the way from the lighter end toward the heavier end. " 3

(2a)

èëëéëëê 100 lbs.  ñïïïïïïïïïïïïïïñïïïïñïïïïïïñ a 200 lbs a x œ a/3 ! 3. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point m masses located at the centers of the rods at coordinates ˆ L# ß !‰ and ˆ0ß L# ‰. Therefore x œ y m

œ

x" m"
x# m# m"
m#

œ

L # †m
0

m
m

œ

L 4

and y œ

mx m

œ

y" m#
y# m# m"
m#

œ

0
L2 †m m
m

œ

L 4

Ê

ˆ L4 ß L4 ‰

is the center of

mass location. 4. Let the rods have lengths x œ L and y œ 2L. The center of mass of each rod is in its center (see Example 1). The rod system is equivalent to two point masses located at the centers of the rods at coordinates ˆ L# ß !‰ and (!ß L). Therefore x œ

L # †m
0†2m

m
2m

œ

5. M! œ '0 x † 4 dx œ ’4 x# “ œ 4 † 2

#

# !

6. M! œ '1 x † 4 dx œ ’4 x# “ œ 3

#

$ "

4 #

!†m
L†2m m
2m

L 6

and y œ

œ

4 #

œ 8; M œ '0 4 dx œ [4x]#! œ 4 † 2 œ 8 Ê x œ 2

2L 3

‰ Ê ˆ L6 ß 2L 3 is the center of mass location. M! M

œ1

(9  1) œ 16; M œ '1 4 dx œ [4x]$" œ 12  4 œ 8 Ê x œ 3

M! M

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

16 8

œ2

Section 6.4 Moments and Centers of Mass 7. M! œ '0 x ˆ1
x3 ‰ dx œ '0 Šx
3

œ3


3

œ

9 6

9 #

Ê xœ

ˆ 15 ‰ 2 ˆ 92 ‰

œ

M! M

x# 3‹

œ

8. M! œ '0 x ˆ2  x4 ‰ dx œ '0 Š2x  4

4

M œ '0 ˆ2  x4 ‰ dx œ ’2x  4

9. M! œ '1 x Š1
4

" Èx ‹

% x# 8 “!

15 9

$ x$ 9 “!

#

dx œ ’ x#
œ

x# 4‹

œ

27 ‰ 9

M œ '0 ˆ1
3x ‰ dx œ ’x
3

15 # ;

16 8

% x$ 12 “ !

œ ˆ16 

64 ‰ 12

œ

œ

œ6 Ê xœ

dx œ '1 ˆx
x"Î# ‰ dx œ ’ x#
4

#

M! M

% 2x$Î# 3 “"

32 3 †6

œ ˆ8


œ 16 

16 ‰ 3

M! M

10. M! œ '1Î4 x † 3 ˆx
$Î#
x
&Î# ‰ dx œ 3'1Î4 ˆx
"Î#
x
$Î# ‰ dx œ 3
2x"Î#  1

1

œ 6
14 œ 20 Ê x œ

M! M

œ

2 ‘" 3x$Î# "Î%

œ

9 3

1

2

œ 3; M œ '0 (2  x) dx
'1 x dx œ ’2x  1

2

" x# # “!




#

# # ’ x# “ "

œ ˆ2 

12. M! œ '0 x(x
1) dx
'1 2x dx œ '0 ax#
xb dx
'1 2x dx œ ’ x3
œ3


œ

32 3 ;

ˆ 73 ‰ 6 5

œ

2 ‘" x"Î# "Î%

œ

15 #




14 3

2

1

1

2

2

œ

23 6 ;

Ê xœ

M! M

‰ ˆ 72 ‰ œ œ ˆ 23 6

œ

73 6

;

73 30

œ 3 ’(2  2)  Š2 † 16 ‰‘ 3

$

"‰ #

" x$ 3 “!

" #



2 ˆ "# ‰ ‹“

œ 3 ˆ2


14 ‰ 3




" x# 2 “!

ˆ 4#

#

$


’ x3 “ œ ˆ1  "3 ‰
ˆ 83  "3 ‰ "



"‰ #

œ3 Ê xœ

#

"

M! M

œ1

#
cx# d " œ ˆ "3
2" ‰
(4  1)

M œ '0 (x
1) dx
'1 2 dx œ ’ x#
x“
c2xd #" œ ˆ "#
1‰
(4  #) œ 2


5 6

4528 6

œ

9 20

2

1

2 3

œ 3
ˆ2  32 ‰  ˆ4 

11. M! œ '0 x(2  x) dx
'1 x † x dx œ '0 a2x  x# b dx
'1 x# dx œ ’ 2x#  1

œ 16 †

 ˆ "#
32 ‰ œ

4

œ 3(4  1) œ 9; M œ 3'1Î4 ˆx
$Î#
x
&Î# ‰ dx œ 3
x
"Î#2 

16 3

16 9

% M œ '1 ˆ1
x
"Î# ‰ dx œ
x
2x"Î# ‘ " œ (4
4)  (1
2) œ 5 Ê x œ 1

$ x# 6 “!

5 3

dx œ ’x# 

œ8

œ ˆ 92


385

!

3 #

œ

7 #

23 21

13. Since the plate is symmetric about the y-axis and its density is constant, the distribution of mass is symmetric about the y-axis and the center of mass lies on the y-axis. This means that x œ 0. It remains to find y œ MMx . We model the distribution of mass with @/<>3-+6 strips. The typical strip has center of mass: # (µ x ßµ y ) œ Šxß x  4 ‹ , length: 4  x# , width: dx, area: #

#

dA œ a4  x b dx, mass: dm œ $ dA œ $ a4  x# b dx. The moment of the strip about the x-axis is # µ C dm œ Š x #4 ‹ $ a4  x# b dx œ #$ a16  x% b dx. The moment of the plate about the x-axis is Mx œ ' µ C dm œ 'c2 #$ a16  x% b dx œ 2

$ #

’16x 

# x& 5 “
#

plate is M œ ' $ a4  x# b dx œ $ ’4x 

œ

$ #

’Š16 † 2 

# x$ 3 “
#

2& 5‹

œ 2$ ˆ8  83 ‰ œ

32$ 3 .

2& 5 ‹“

œ

$ †2 #

ˆ32 

Therefore y œ

Mx M

œ

 Š16 † 2


32 ‰ 5

$ Š 128 5 ‹

Š 323$ ‹

‰ mass is the point (xß y) œ ˆ!ß 12 5 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

œ

128$ 5 .

12 5 .

The mass of the

The plate's center of

386

Chapter 6 Applications of Definite Integrals

14. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. To find y œ MMx , we use the @/<>3-+6 strips technique. The typical strip has center of # mass: (µ x ßµ y ) œ Šxß 25
x ‹ , length: 25  x# , width: dx, #

#

area: dA œ a25  x bdx, mass: dm œ $ dA œ $ a25  x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š 25
# x ‹ $ a25  x# b dx œ

œ 'c5 #$ a25  x# b dx œ 5

#

œ $ † 625 ˆ5  œ 2$ Š5$ 

10 3

5$ 3‹

$ #

'c55

a25  x# b dx. The moment of the plate about the x-axis is Mx œ ' µ y dm #

$ #

$ #

a625  50x#
x% b dx œ

’625x 

50 3

x$


& x& 5 “
&

œ 2 † #$ Š625 † 5 

50 3

† 5$



1‰ œ $ † 625 † ˆ 38 ‰ . The mass of the plate is M œ ' dm œ 'c5 $ a25  x# b dx œ $ ’25x  5

œ

$ † 5$ . Therefore y œ

4 3

Mx M

œ

$ †5% † ˆ 83 ‰ $ †5$ †ˆ 43 ‰

5& 5‹ & x$ 3 “
&

œ 10. The plate's center of mass is the point (xß y) œ (!ß 10).

15. Intersection points: x  x# œ x Ê 2x  x# œ 0 Ê x(2  x) œ 0 Ê x œ 0 or x œ 2. The typical @/<>3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß ax
x b  (
x) ‹ #

œ Šxß 

x# # ‹,

#

length: ax  x b  (x) œ 2x  x# , width: dx,

area: dA œ a2x  x# b dx, mass: dm œ $ dA œ $ a2x  x# b dx. The moment of the strip about the x-axis is # µ y dm œ Š x# ‹ $ a2x  x# b dx; about the y-axis it is µ x dm œ x † $ a2x  x# b dx. Thus, Mx œ ' µ y dm œ  '0 ˆ #$ x# ‰ a2x  x# b dx œ  #$ '0 a2x$  x% b dx œ  #$ ’ x#  2

2

%

# x& 5 “!

œ  #$ Š2$ 

œ  45$ ; My œ ' µ x dm œ '0 x † $ a2x  x# b dx œ $ '0 a2x#  x$ b œ $ ’ 23 x$  2

2

M œ ' dm œ '0 $ a2x  x# b dx œ $ '0 a2x  x# b dx œ $ ’x#  2

2

œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1 and y œ

Mx M

# x$ 3 “!

# x% 4 “!

2& 5‹

œ  #$ † 2$ ˆ1  45 ‰

œ $ Š2 †

œ $ ˆ4  83 ‰ œ

4$ 3

2$ 3



2% 4‹

œ

. Therefore, x œ

$ †2% 1#

œ

My M

œ ˆ 45$ ‰ ˆ 43$ ‰ œ  35 Ê (xß y) œ ˆ1ß  35 ‰ is the center of mass.

16. Intersection points: x#  3 œ 2x# Ê 3x#  3 œ 0 Ê 3(x  1)(x
1) œ 0 Ê x œ 1 or x œ 1. Applying the symmetry argument analogous to the one in Exercise 13, we find x œ 0. The typical @/<>3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß
2x  ax
3b ‹ œ Šxß
x
3 ‹ , #

#

#

#

#

length: 2x  ax  3b œ 3 a1  x b, width: dx, area: dA œ 3 a1  x# b dx, mass: dm œ $ dA œ 3$ a1  x# b dx. The moment of the strip about the x-axis is µ y dm œ 3 $ ax#  3b a1  x# b dx œ 3 $ ax%
3x#  x#  3b dx œ #

œ

3 #

$ 'c1 ax%
2x#  3b dx œ 1

#

3 #

&

$ ’ x5


2x$ 3

M œ ' dm œ 3$ 'c1 a1  x# b dx œ 3$ ’x  1

 3x“ " x$ 3 “
"

"
"

œ

3 #

3 #

$ ax%
2x#  3b dx; Mx œ ' µ y dm

† $ † 2 ˆ 5"


2 3


45 ‰  3‰ œ 3$ ˆ 310 œ  325$ ; 15

œ 3$ † 2 ˆ1  3" ‰ œ 4$ . Therefore, y œ

Mx M

œ  5$††$32†4 œ  58

Ê (xß y) œ ˆ0ß  85 ‰ is the center of mass.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

4$ 3

;

Section 6.4 Moments and Centers of Mass

387

17. The typical 29<3D98>+6 strip has center of mass: $ (µ x ßµ y ) œ Š y
y ß y‹ , length: y  y$ , width: dy, #

area: dA œ ay  y$ b dy, mass: dm œ $ dA œ $ ay  y$ b dy. The moment of the strip about the y-axis is $ # µ x dm œ $ Š y
y ‹ ay  y$ b dy œ $ ay  y$ b dy #

œ

$ #

#

#

%

'

ay  2y
y b dy; the moment about the x-axis is

1 $ µ y dm œ $ y ay  y$ b dy œ $ ay#  y% b dy. Thus, Mx œ ' µ y dm œ $ '0 ay#  y% b dy œ $ ’ y3 

My œ ' µ x dm œ

$ #

'01 ay#  2y%
y' b dy œ #$ ’ y3

$

œ $ '0 ay  y$ b dy œ $ ’ y#  1

œ

#

" y% 4 “!

2y& 5



œ $ ˆ "#  4" ‰ œ

$ 4




" y( 7 “!

œ

$ #

ˆ "3 

. Therefore, x œ

2 5


7" ‰ œ

$ #

œ $ ˆ "3  "5 ‰ œ

 15 ‰ ˆ 35
3†42 œ 5†7

4$ ‰ ˆ 4 ‰ œ ˆ 105 $ œ

My M

" y& 5 “!

16 105

2$ 15

;

4$ 105

; M œ ' dm

Mx M

2$ ‰ ˆ 4 ‰ œ ˆ 15 $

and y œ

16 8 ‰ Ê (xß y) œ ˆ 105 ß 15 is the center of mass.

8 15

18. Intersection points: y œ y#  y Ê y#  2y œ 0 Ê y(y  2) œ 0 Ê y œ 0 or y œ 2. The typical 29<3D98>+6 strip has center of mass: # # (µ x ßµ y ) œ Š ay
yby ß y‹ œ Š y ß y‹ , #

2

#

length: y  ay  yb œ 2y  y# , width: dy, area: dA œ a2y  y# b dy, mass: dm œ $ dA œ $ a2y  y# b dy. The moment about the y-axis is µ x dm œ #$ † y# a2y  y# b dy œ #$ a2y$  y% b dy; the moment about the x-axis is µ y dm œ $ y a2y  y# b dy œ $ a2y#  y$ b dy. Thus, Mx œ ' µ y dm œ '0 $ a2y#  y$ b dy œ $ ’ 2y3  2

œ '0

2

$ #

$

a2y$  y% b dy œ

œ $ ’y# 

# y$ 3 “!

$ #

%

’ y2 

œ $ ˆ4  83 ‰ œ

# y& 5 “!

4$ 3

œ

$ #

ˆ8 

# y% 4 “!

16$ 1#

ˆ 40
5 32 ‰ œ

4$ 5

; M œ ' dm œ '0 $ a2y  y# b dy

œ

$ #

My M

œ ˆ 45$ ‰ ˆ 43$ ‰ œ

32 ‰ 5

. Therefore, x œ

œ

(4  3) œ

4$ 3

; My œ ' µ x dm

16 ‰ 4

œ $ ˆ 16 3 

2

3 5

and y œ

Mx M

œ ˆ 43$ ‰ ˆ 43$ ‰ œ 1

Ê (xß y) œ ˆ 35 ß "‰ is the center of mass. 19. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ ˆxß cos# x ‰ , length: cos x, width: dx,

area: dA œ cos x dx, mass: dm œ $ dA œ $ cos x dx. The moment of the strip about the x-axis is µ y dm œ $ † cos# x † cos x dx 2x ‰ œ #$ cos# x dx œ #$ ˆ 1  cos dx œ 4$ (1
cos 2x) dx; thus, # 1Î2

Mx œ ' µ y dm œ 'c1Î2 4$ (1
cos 2x) dx œ 1Î#

œ $ [sin x]
1Î# œ 2$ . Therefore, y œ

Mx M

œ

$ 4


x
$1 4 †# $

œ

sin 2x ‘ 1Î# #
1Î# 1 8

œ

$ 4


ˆ 1#
0‰  ˆ 1# ‰‘ œ

$1 4

Ê (xß y) œ ˆ!ß 18 ‰ is the center of mass.

20. Applying the symmetry argument analogous to the one used in Exercise 13, we find x œ 0. The typical vertical strip has # center of mass: (µ x ßµ y ) œ Šxß sec# x ‹ , length: sec# x, width: dx, area: dA œ sec# x dx, mass: dm œ $ dA œ $ sec# x dx. The # moment about the x-axis is µ y dm œ Š sec x ‹ a$ sec# xb dx œ

$ #

1Î4

sec% x dx. Mx œ 'c1Î4 µ y dm œ

#

$ #

1Î2

; M œ ' dm œ $ '
1Î2 cos x dx

'
11ÎÎ44 sec% x dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

388

Chapter 6 Applications of Definite Integrals œ

$ #

'c11ÎÎ44 atan# x
1b asec# xb dx œ #$ '
11ÎÎ44 (tan x)# asec# xb dx
#$ '
11ÎÎ44 sec# x dx œ 2$ ’ (tan3x) “ 1Î4

œ

$ 2


3"  ˆ 3" ‰‘
#$ [1  (1)] œ

$

1Î%


1 Î4

Therefore, y œ

Mx M

œ ˆ 43$ ‰ ˆ 2"$ ‰ œ

2 3

$ 3


$ œ

4$ 3


#$ [tan x]
1Î%

; M œ ' dm œ $ 'c1Î4 sec# x dx œ $ [tan x]
1Î4 œ $ [1  (1)] œ 2$ . 1Î4

1Î4

Ê (xß y) œ ˆ!ß 32 ‰ is the center of mass.

21. Since the plate is symmetric about the line x œ 1 and its density is constant, the distribution of mass is symmetric about this line and the center of mass lies on it. This means that x œ 1. The typical @/<>3-+6 strip has center of mass: # # # (µ x ßµ y ) œ Šxß a2x
x ba2x
4xb ‹ œ Šxß x
2x ‹ , #

#

#

#

#

length: a2x  x b  a2x  4xb œ 3x
6x œ 3 a2x  x# b , width: dx, area: dA œ 3 a2x  x# b dx, mass: dm œ $ dA œ 3$ a2x  x# b dx. The moment about the x-axis is # µ y dm œ 3 $ ax#  2xb a2x  x# b dx œ  3 $ ax#  2xb dx #

#

œ  3# $ ax%  4x$
4x# b dx. Thus, Mx œ ' µ y dm œ '0

2

œ $ 3 2

& Š 25

%

2


4 3

$

%

†2 ‹œ $†2 3 #

œ '0 3$ a2x  x# b dx œ 3$ ’x#  2

# x$ 3 “!

ˆ 25

1


2‰ 3

3 2

&

$ ax%  4x$
4x# b dx œ  32 $ ’ x5  x%
34 x$ “ %

œ $ †2 3 #

 10 ‰ ˆ 6
15 15

œ 3$ ˆ4  83 ‰ œ 4$ . Therefore, y œ

Mx M

œ

8$ 5

; M œ ' dm

# !

œ ˆ 85$ ‰ ˆ 4"$ ‰ œ  25

Ê (xß y) œ ˆ1ß  25 ‰ is the center of mass. 22. (a) Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/<>3-+6 strip has center of mass: È # (µ x ßµ y ) œ Šxß 9
x ‹ , length: È9  x# , width: dx, #

area: dA œ È9  x# dx, mass: dm œ $ dA œ $ È9  x# dx. The moment about the x-axis is È # µ y dm œ $ Š 9#
x ‹ È9  x# dx œ

$ #

a9  x# b dx. Thus, Mx œ ' µ y dm œ '0

3

$ #

a9  x# b dx œ

$ #

’9x 

$ x$ 3 “!

(27  9) œ 9$ ; M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a quarter of a circle of radius 3) œ $ ˆ 941 ‰ œ 4 ‰ Therefore, y œ MMx œ (9$ ) ˆ 91$ œ 14 Ê (xß y) œ ˆ 14 ß 14 ‰ is the center of mass. œ

$ #

(b) Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical vertical strip has the same parameters as in part (a). 3 Thus, M œ ' µ y dm œ ' $ a9  x# b dx x

œ #'0

3

$ #

c3 #

a9  x# b dx œ 2(9$ ) œ 18$ ;

M œ ' dm œ ' $ dA œ $ ' dA œ $ (Area of a semi-circle of radius 3) œ $ ˆ 921 ‰ œ 91$ 2 . Therefore, y œ 4 as in part (a) Ê (xß y) œ ˆ0ß 1 ‰ is the center of mass.

Mx M

2 ‰ œ (18$ ) ˆ 91$ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

4 1

, the same y

91$ 4

.

Section 6.4 Moments and Centers of Mass

389

23. Since the plate is symmetric about the line x œ y and its density is constant, the distribution of mass is symmetric about this line. This means that x œ y. The typical @/<>3-+6 strip has È # center of mass: (µ x ßµ y ) œ Šxß 3  9
x ‹ , #

length: 3  È9  x# , width: dx, area: dA œ Š3  È9  x# ‹ dx, mass: dm œ $ dA œ $ Š3  È9  x# ‹ dx. The moment about the x-axis is µ y dm œ $

Š3  È9
x# ‹ Š3
È9
x# ‹ #

dx œ

$ #

c9  a9  x# bd dx œ

$ x# #

dx. Thus, Mx œ '0

3

$ x# #

dx œ

$ 6

$

cx$ d ! œ #

9$ #

equals the area of a square with side length 3 minus one quarter the area of a disk with radius 3 Ê A œ 3  œ

9 4

9$ 4

(4  1) Ê M œ $ A œ

(4  1). Therefore, y œ

Mx M

œ ˆ 9#$ ‰ ’ 9$(44
1) “ œ

2 4
1

. The area 19 4

Ê (xß y) œ ˆ 4
2 1 ß 4
2 1 ‰ is the

center of mass. 24. Applying the symmetry argument analogous to the one used in Exercise 13, we find that y œ 0. The typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ Œxß " x$

length:

 ˆ x"$ ‰ œ

2 x$

" x$


x"$ #

 œ (xß 0),

, width: dx, area: dA œ

2 x$

dx,

2$ x$

mass: dm œ $ dA œ dx. The moment about the y-axis is a µ x dm œ x † 2x$$ dx œ 2x$# dx. Thus, My œ ' µ x dm œ '1 2x$# dx œ 2$
 x" ‘ " œ 2$ ˆ "a
1‰ œ a

xœ

My M

œ

’ 2$(aa
1) “

25. Mx œ ' µ y dm œ '1

2

# ’ $ aa#a
1b “

Š x2# ‹ #

2$ (a
1) a

œ

2a a1

; M œ ' dm œ '1

a

Ê (xß y) œ

2$ x$

ˆ a 2a ‰ 1ß 0 .

dx œ $
 x"# ‘ " œ $ ˆ a"#
1‰ œ a

$ aa#
1b a#

. Therefore,

Also, a lim x œ 2. Ä_

† $ † ˆ x2# ‰ dx

œ '1 ˆ x"# ‰ ax# b ˆ x2# ‰ dx œ '1 2

2

2 x#

dx œ 2'1 x
# dx 2

#

œ 2 cx
" d " œ 2
ˆ "# ‰  (1)‘ œ 2 ˆ "# ‰ œ 1;

My œ ' µ x dm œ '1 x † $ † ˆ x2# ‰ dx 2

œ '1 x ax# b ˆ x2# ‰ dx œ 2'1 x dx œ 2 ’ x# “ 2

2

#

# "

œ 2 ˆ2  "# ‰ œ 4  1 œ 3; M œ ' dm œ '1 $ ˆ x2# ‰ dx œ '1 x# ˆ x2# ‰ dx œ 2'1 dx œ 2[x]"# œ 2(2  1) œ 2. So xœ

My M

œ

3 #

and y œ

Mx M

œ

" #

2

2

2

Ê (xß y) œ ˆ 3# ß "# ‰ is the center of mass.

26. We use the @/<>3-+6 strip approach: 1 # M œ'µ y dm œ ' ax  x b ax  x# b † $ dx x

œ

0

" #

#

'0 ax#  x% b † 12x dx 1

œ 6'0 ax$  x& b dx œ 6 ’ x4  1

œ 6 ˆ "4  6" ‰ œ

%

6 4

1œ

" #

" x' 6 “!

;

My œ ' µ x dm œ '0 x ax  x# b † $ dx œ '0 ax#  x$ b † 12x dx œ 12'0 ax$  x% b dx œ 12 ’ x4  1

1

1

%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" x& 5 “!

œ 12 ˆ "4  5" ‰

390

Chapter 6 Applications of Definite Integrals œ

12 #0

xœ

œ

; M œ ' dm œ ' ax  x# b † $ dx œ 12'0 ax#  x$ b dx œ 12 ’ x3  1

3 5

1

$

0

My M

œ

3 5

and y œ

Mx M

œ

" #

" x% 4 “!

œ 12 ˆ "3  4" ‰ œ

12 12

œ 1. So

Ê ˆ 35 ß "# ‰ is the center of mass.

shell ‰ shell 27. (a) We use the shell method: V œ 'a 21 ˆ radius Š height ‹ dx œ '1 21x ’ È4x  Š È4x ‹“ dx b

œ 161'1

4

x Èx

4

% dx œ 161'1 x"Î# dx œ 161
32 x$Î# ‘ " œ 161 ˆ 32 † 8  32 ‰ œ 4

(b) Since the plate is symmetric about the x-axis and its density $ (x) œ

" x

321 3

(8  1) œ

2241 3

is a function of x alone, the

distribution of its mass is symmetric about the x-axis. This means that y œ 0. We use the vertical strip 4 4 4 approach to find x: My œ ' µ x dm œ '1 x † ’ È4x  Š È4x ‹“ † $ dx œ '1 x † È8x † x" dx œ 8'1 x
"Î# dx
4 œ 8
2x"Î# ‘ " œ 8(2 † 2  2) œ 16; M œ ' dm œ '1 ’ È4x  Š È ‹“ † $ dx œ 8'1 Š È"x ‹ ˆ "x ‰ dx œ 8'1 x
$Î# dx x %

4

%

œ 8
2x
"Î# ‘ " œ 8[1  (2)] œ 8. So x œ

My M

4

œ

4

œ 2 Ê (xß y) œ (2ß 0) is the center of mass.

16 8

(c)

28. (a) We use the disk method: V œ 'a 1R# (x) dx œ '1 1 ˆ x4# ‰ dx œ 41'1 x
# dx œ 41
 x" ‘ " b

4

4

%

‘ œ 41

" 4  (1) œ 1[1
4] œ 31

(b) We model the distribution of mass with vertical strips: Mx œ ' µ y dm œ '1

4


2 œ 2'1 x
$Î# dx œ 2 ’ È x dm œ '1 x † “ œ 2[1  (2)] œ 2; My œ ' µ x %

4

$Î#

4

"

%

2‘ œ 2 ’ 2x3 “ œ 2
16 3  3 œ "

œ 2(4  2) œ 4. So x œ

My M

28 3

œ

; M œ ' dm œ '1

4

ˆ 28 ‰ 3 4

œ

7 3

and y œ

2 x

Mx M

† $ dx œ 2'1

4

œ

2 4

œ

" #

Èx x

2 x

ˆ 2x ‰ 2

† ˆ 2x ‰ † $ dx œ '1

4

2 x#

† Èx dx

† $ dx œ 2'1 x"Î# dx 4

dx œ 2'1 x
"Î# dx œ 2
2x"Î# ‘ " %

4

Ê (xß y) œ ˆ 73 ß "# ‰ is the center of mass.

(c)

29. The mass of a horizontal strip is dm œ $ dA œ $ L dy, where L is the width of the triangle at a distance of y above its base on the x-axis as shown in the figure in the text. Also, by similar triangles we have Ê Lœ

b h

(h  y). Thus, Mx œ ' µ y dm œ '0 $ y ˆ bh ‰ (h  y) dy œ h

œ

$b h

Š h# 

$

h$ 3‹

œ

$b h

#

h# 2‹

Šh 

œ $ bh# ˆ "#  3" ‰ œ œ

$ bh 2

. So y œ

Mx M

$ bh# 6

œ

$b h

# ˆ 2 ‰ Š $bh 6 ‹ $ bh

œ

h 3

œ

h
y h

'0h ahy  y# b dy œ $hb ’ hy#

; M œ ' dm œ '0 $ ˆ hb ‰ (h  y) dy œ h

L b

$b h

#



h

y$ 3 “!

'0h ah  yb dy œ $hb ’hy  y2 “ h

Ê the center of mass lies above the base of the

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

#

!

Section 6.4 Moments and Centers of Mass triangle one-third of the way toward the opposite vertex. Similarly the other two sides of the triangle can be placed on the x-axis and the same results will occur. Therefore the centroid does lie at the intersection of the medians, as claimed. 30. From the symmetry about the y-axis it follows that x œ 0. It also follows that the line through the points (!ß !) and (!ß $) is a median Ê y œ "3 (3  0) œ 1 Ê (xß y) œ (!ß ").

31. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the points (!ß !) and ˆ "# ß "# ‰ is a median Ê y œ x œ 23 † ˆ "#  0‰ œ 3" Ê (xß y) œ ˆ "3 ß 3" ‰ . 32. From the symmetry about the line x œ y it follows that x œ y. It also follows that the line through the point (!ß !) and ˆ #a ß #a ‰ is a median Ê y œ x œ 32 ˆ #a  0‰ œ 3" a Ê (xß y) œ ˆ 3a ß 3a ‰ . 33. The point of intersection of the median from the vertex (0ß b) to the opposite side has coordinates ˆ!ß #a ‰ Ê y œ (b  0) † "3 œ b3 and x œ ˆ #a  !‰ † 32 œ 3a Ê (xß y) œ ˆ 3a ß b3 ‰ .

34. From the symmetry about the line x œ

a #

it follows that

xœ It also follows that the line through the points a ˆ # ß !‰ and ˆ #a ß b‰ is a median Ê y œ "3 (b  0) œ b3 a #.

Ê (xß y) œ ˆ #a ß b3 ‰ .

35. y œ x"Î# Ê dy œ

" #

x
"Î# dx

Ê ds œ È(dx)#
(dy)# œ É1
Mx œ $ '0 Èx É1
2

œ $ '0 Éx
2

dx œ

" ‰$Î# 4

œ

2$ 3

œ

2$ ˆ 9 ‰$Î# 3 ’ 4

’ˆ2


" 4



 ˆ 4" ‰

" 4x

" 4x

dx

2$ 3

$Î# ’ˆx
4" ‰ “

dx ;

# !

ˆ 4" ‰$Î# “ $Î#

“œ

2$ 3

"‰ ˆ 27 8  8 œ

13$ 6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

391

392

Chapter 6 Applications of Definite Integrals

36. y œ x$ Ê dy œ 3x# dx Ê dx œ É(dx)#
a3x# dxb# œ È1
9x% dx; Mx œ $ '0 x$ È1
9x% dx; 1

" 36

[u œ 1
9x% Ê du œ 36x$ dx Ê

du œ x$ dx;

x œ 0 Ê u œ 1, x œ 1 Ê u œ 10] Ä Mx œ $ '1

10

" 36

u"Î# du œ

$ 36


23 u$Î# ‘ "! œ "

$ 54

ˆ10$Î#  1‰

37. From Example 6 we have Mx œ '0 a(a sin ))(k sin )) d) œ a# k'0 sin# ) d) œ 1

œ

a# k #


) 

sin 2) ‘ 1 # !

œ

a# k 1 #

1

'01 (1  cos 2)) d)

; My œ '0 a(a cos ))(k sin )) d) œ a# k '0 sin ) cos ) d) œ 1

1

M œ '0 ak sin ) d) œ ak[ cos )]1! œ 2ak. Therefore, x œ 1

a# k #

My M

œ 0 and y œ

Mx M

a# k #

1

csin# )d ! œ 0;

#

" ‰ œ Š a 2k1 ‹ ˆ 2ak œ

a1 4

Ê ˆ!ß a41 ‰

is the center of mass. 38. Mx œ ' µ y dm œ '0 (a sin )) † $ † a d) 1

œ '0 aa# sin )b a1
k kcos )kb d) 1

œ a# '0 (sin ))(1
k cos )) d) 1Î2


a# '1Î2 (sin ))(1  k cos )) d) 1

œ a# '0 sin ) d)
a# k'0 sin ) cos ) d)
a# '1Î2 sin ) d)  a# k '1Î2 sin ) cos ) d) 1Î2

1Î2

1Î#

#

1


a# k ’ sin# ) “

œ a# [ cos )]!

1Î# !

1

#


a# [ cos )]11Î#  a# k ’ sin# ) “

1 1Î#

œ a [0  (1)]
a k ˆ "#  0‰
a# [(1)  0]  a# k ˆ0  "# ‰ œ a#
#

#

a# k #


a#


œ 2a#
a# k œ a# (2
k); 1 1 M œ'µ x dm œ ' (a cos )) † $ † a d) œ ' aa# cos )b a1
k kcos )kb d) y

0

0

œa

'0

œ a#

'01Î2 cos ) d)
a# k '

#

1Î2

#

(cos ))(1
k cos )) d)
a 1Î2

0

#

œ a [sin

1Î# ) ]!




œ a# (1  0)


a# k #

a# k #

a# k #


)


'1Î2 (cos ))(1  k cos )) d) 1

2) ‰ 2) ‰ ˆ 1  cos d)
a# '1Î2 cos ) d)  a# k'1Î2 ˆ 1  cos d) # #

sin 2) ‘ 1Î# # !

1


a# [sin )]11Î# 

1

a# k #


ˆ 1#  0‰  (!
0)‘
a# (0  1) 


)


a# k #

sin 2) ‘ 1 # 1Î#


(1
0)  ˆ 1#
0‰‘ œ a#


a# k 1 4

 a# 

M œ '0 $ † a d) œ a'0 (1
k kcos )k) d) œ a '0 (1
k cos )) d)
a'1Î2 (1  k cos )) d) 1

1

1Î#

œ a[)
k sin )]! œ

a1 #

1Î2

a# k 1 4

œ 0;

1


a[)  k sin )]11Î# œ a
ˆ 1#
k‰  0‘
a
(1
0)  ˆ 1#  k‰‘


ak
a ˆ 1#
k‰ œ a1
2ak œ a(1
2k). So x œ

My M

œ 0 and y œ

Mx M

œ

a# (2  k) a(1  #k)

œ

a(2  k) 1  #k

 ka ‰ Ê ˆ0ß 2a 1  #k is the center of mass.

39. Consider the curve as an infinite number of line segments joined together. From the derivation of arc length we have that the length of a particular segment is ds œ È(dx)#
(dy)# . This implies that Mx œ ' $ y ds, My œ ' $ x ds and M œ ' $ ds. If $ is constant, then x œ yœ

Mx M

' y ds

œ ' ds œ

' y ds length

My M

' x ds

œ ' ds œ

' x ds length

and

.

40. Applying the symmetry argument analogous to the one used in Exercise 13, we find that x œ 0. The typical

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus x#

a vertical strip has center of mass: (µ x ßµ y ) œ Œxß 2 4p  , length: a 

mass: dm œ $ dA œ $ Ša  œ

$ #

2

%

&

#

œ $ ’ax  8a$ Èpa 3

2

Èpa œ 2 † $ ’ax  È Mx M

œŠ

Èpa

2

Èpa

x$ 12p “ !

œ

3 5

Èpa

8a# $Èpa 5

2$ paÈpa 12p ‹

œ 2$ Š2aÈpa 

8a# $ Èpa 3 ‹ Š 8a$È 5 pa ‹

2

x& 80p# “ 0

œ 2 † #$ ’a# x 


"6 ‰ ‰ œ 2a# $ Èpa ˆ 8080 œ 2a# $ Èpa ˆ 64 80 œ

x$ 12p “ c2 pa

. So y œ

c2Èpa

#

16 ‰ 80

width: dx, area: dA œ Ša 

dx. Thus, Mx œ ' µ y dm œ 'c2Èpa "# Ša


#Èpa x x 'c22ÈÈpapa Ša#  16p ‹ dx œ #$ ’a# x  80p “

œ 2a# $ Èpa ˆ1 

œ

x# 4p ‹

x# 4p ,

x# 4p ‹ Ša



x# 4p ‹ $

œ 4a$ Èpa ˆ1 

dx,

dx 2& p# a# Èpa ‹ 80p#

œ $ Š2a# Èpa 

; M œ ' dm œ $

x# 4p ‹

2

Èpa

'

c2Èpa

4 ‰ 12

Ša 

x# 4p ‹

dx

œ 4a$ Èpa ˆ 121
#4 ‰

a, as claimed.

41. Since the density is constant, its value will not affect our answers, so we can set $ œ ". 1Î2  ! A generalization of Example 6 yields M œ ' µ y dm œ ' a# sin ) d) œ a# [ cos )]1Î2  ! 1Î2
!

1Î#
!

x

1Î2  !

œ a#
 cos ˆ 1#
!‰
cos ˆ 1#  !‰‘ œ a# (sin !
sin !) œ 2a# sin !; M œ ' dm œ '1Î#
! a d) œ a[)]11ÎÎ22 
!! œ a
ˆ 1#
!‰  ˆ 1#  !‰‘ œ 2a!. Thus, y œ Ê c œ 2a sin !. Then y œ

a(2a sin !) 2a!

œ

ac s ,

Mx M

œ

2a# sin ! 2a!

œ

a sin ! !

lim

! Ä !b

(b)

sin !
! cos ! !
! cos !

! f(!)

¸

d h

œ

a sin ! ! sin !
! cos ! !
! cos ! .

a(sin !
! cos !) a(!
! cos !)

œ

œ a cos !
d Ê d œ

0.4 0.664879

0.6 0.662615

0.8 0.659389

1.0 0.655145

6.5 AREAS OF SURFACES OF REVOLUTION AND THE THEOREMS OF PAPPUS 1. (a)

dy dx

#

% œ sec# x Ê Š dy dx ‹ œ sec x

Ê S œ 21'0

1Î4

a(sin !
! cos !) . !

The graphs below suggest that

2 3.

0.2 0.666222

c #

as claimed.

42. (a) First, we note that y œ (distance from origin to AB)
d Ê Moreover, h œ a  a cos ! Ê

. Now s œ a(2!) and a sin ! œ

(b)

(tan x) È1
sec% x dx

(c) S ¸ 3.84

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

393

394 2. (a)

Chapter 6 Applications of Definite Integrals dy dx

#

(b)

2 œ 2x Ê Š dy dx ‹ œ 4x

Ê S œ 21'0 x# È1
4x# dx 2

(c) S ¸ 53.23

3. (a) xy œ 1 Ê x œ Ê S œ 21'1

2

" y

" y

Ê

dx dy

#

œ  y"# Ê Š dx dy ‹ œ

" y%

(b)

È1
y
% dy

(c) S ¸ 5.02

4. (a)

dx dy

#

# œ cos y Ê Š dx dy ‹ œ cos y

(b)

Ê S œ 21'0 (sin y) È1
cos# y dy 1

(c) S ¸ 14.42

# 5. (a) x"Î#
y"Î# œ 3 Ê y œ ˆ3  x"Î# ‰ "Î# ‰ ˆ ˆ Ê dy  "# x
"Î# ‰ dx œ 2 3  x #


"Î# ‰ ˆ Ê Š dy dx ‹ œ 1  3x

(b)

#

# # Ê S œ 21'1 ˆ3  x"Î# ‰ É1
a1  3x
"Î# b dx 4

(c) S ¸ 63.37

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus dx dy

6. (a)

#


"Î# ‰ ˆ œ 1
y
"Î# Ê Š dx dy ‹ œ 1
y

#

395

(b)

# Ê S œ 21 '1 ˆy
2Èy‰ É1
a1
y
"Î# b dx 2

(c) S ¸ 51.33

dx dy

7. (a)

#

(b)

# œ tan y Ê Š dx dy ‹ œ tan y

Ê S œ 21'0 Š'0 tan t dt‹ È1
tan# y dy 1Î3

y

œ 21'0 Š'0 tan t dt‹ sec y dy 1Î3

y

(c) S ¸ 2.08

dy dx

8. (a)

#

(b)

# œ Èx#  1 Ê Š dy dx ‹ œ x  1

È5

Ê S œ 21'1 Š'1 Èt#  1 dt‹ È1
ax#  1b dx

È5

x

œ 21'1 Š'1 Èt#  1 dt‹ x dx x

(c) S ¸ 8.55

9. y œ œ

x #

1È5 #

Ê

dy dx

' ˆ x ‰ É1
œ "# ; S œ 'a 21y Ê1
Š dy dx ‹ dx Ê S œ 0 21 #

x #

4

" 4

dx œ

1È5 #

'04 x dx

%

# ’ x# “ œ 41È5; Geometry formula: base circumference œ 21(2), slant height œ È4#
2# œ 2È5

!

Ê Lateral surface area œ

10. y œ

#

b

Ê x œ 2y Ê

dx dy

" #

(41) Š2È5‹ œ 41È5 in agreement with the integral value

# È È ' È # ' œ 2; S œ 'c 21x Ê1
Š dx dy ‹ dy œ 0 21 † 2y 1
2 dy œ 41 5 0 y dy œ 21 5 cy d ! #

d

2

2

#

œ 21È5 † 4 œ 81È5; Geometry formula: base circumference œ 21(4), slant height œ È4#
2# œ 2È5 Ê Lateral surface area œ " (81) Š2È5‹ œ 81È5 in agreement with the integral value #

11.

dy dx

' œ "# ; S œ 'a 21yÊ1
Š dy dx ‹ dx œ 1 21 b

#

3

(x  1) #

É1
ˆ "# ‰# dx œ

1È5 #

'13 (x
1) dx

œ

1È5 #

1È5 #

#

’ x#
x“

$ "

È
ˆ 9#
3‰  ˆ "#
1‰‘ œ 1 # 5 (4
2) œ 31È5; Geometry formula: r" œ "#
"# œ 1, r# œ 3#
"# œ 2, œ slant height œ È(2  1)#
(3  1)# œ È5 Ê Frustum surface area œ 1(r"
r# ) ‚ slant height œ 1(1
2)È5

œ 31È5 in agreement with the integral value

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

396

Chapter 6 Applications of Definite Integrals

12. y œ

x #




" #

Ê x œ 2y  1 Ê

dx œ 2; S œ 'c 21x Ê1
Š dy ‹ dy œ '1 21(2y  1)È1
4 dy œ 21È5 '1 (2y  1) dy #

d

dx dy

2

#

2

œ 21È5 cy#  yd " œ 21È5 [(4  2)  (1  1)] œ 41È5; Geometry formula: r" œ 1, r# œ 3, slant height œ È(2  1)#
(3  1)# œ È5 Ê Frustum surface area œ 1(1
3)È5 œ 41È5 in agreement with the integral value 13.

dy dx

#

x# 3

œ

’u œ 1


x% 9

Ê S œ '0

2

x% 9

Ê Š dy dx ‹ œ Ê du œ

4 9

x œ 0 Ê u œ 1, x œ 2 Ê u œ Ä S œ 21 '1

25Î9

14.

œ

1 3

dy dx

œ

" 1
2 4 du œ # 3 1 ˆ 125
27 ‰ œ 98811 3 27 #

Ê S œ '3Î4 21Èx É1
œ 21'3Î4 Éx
15Î4

15.

œ

’ˆ 15 4

œ

41 3

(8  1) œ

dy dx




" (2
2x) # È2x
x#

œ



ˆ 43




dx;

dx;

#&Î*

u$Î# ‘ "

" 4x

dx

$Î# dx œ 21 ’ 23 ˆx
4" ‰ “

" 4

" ‰$Î# 4

41 3

x$ 9

du œ

x% 9

" 4x

x
"Î# Ê Š dy dx ‹ œ 15Î4

É1


25 ‘ 9

u"Î# †

ˆ 125 ‰ 27  1 œ " #

" 4

x$ dx Ê

21 x$ 9

" ‰$Î# “ 4

"&Î% $Î%

41 3

$ ’ˆ 24 ‰

Ê Š dy dx ‹ œ

(1
x)# 2x
x#

œ

 1“

281 3

œ

#

1
x È2x
x#

Ê S œ '0 5 21È2x  x# É1
1Þ5 Þ

œ 21'0 5 È2x  1Þ5 Þ

(1
x)# 2x
x#

È x#  1
2x  x# x# 2x
È 2x
x#

dx

dx

œ 21'0 5 dx œ 21[x]"Þ& !Þ& œ 21 1Þ5 Þ

16.

dy dx

" 2È x  1

œ

#

dy Ê Š dx ‹ œ

" 4(x  1)

Ê S œ '1 21Èx
1 É1
5

œ 21'1 É(x
1)
5

" 4

" 4(x  1)

dx

dx œ 21'1 Éx
5

&

$Î# œ 21 ’ 23 ˆx
54 ‰ “ œ

17.

œ

41 3

œ

1 6

dx dy

‰$Î#  ’ˆ 25 4

5 4

41 ˆ 5 ‰$Î#  3 ’ 5
4 " $ $ ˆ 94 ‰$Î# “ œ 431 Š 52$  32$ ‹

(125  27) œ

981 6

œ

dx ˆ1
45 ‰$Î# “

491 3

% ' œ y# Ê Š dx dy ‹ œ y Ê S œ 0 #

1


u œ 1
y% Ê du œ 4y$ dy Ê

" 4

21 y$ 3

È1
y% dy;

du œ y$ dy; y œ 0

Ê u œ 1, y œ 1 Ê u œ 2d Ä S œ '1 21 ˆ "3 ‰ u"Î# ˆ 4" du‰ 2

œ

1 6

'12 u"Î# du œ 16
32 u$Î# ‘ #" œ 19 ŠÈ8  1‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 18. x œ ˆ "3 y$Î#  y"Î# ‰ Ÿ 0, when 1 Ÿ y Ÿ 3. To get positive area, we take x œ  ˆ "3 y$Î#  y"Î# ‰ Ê

dx dy

#

œ  "# ˆy"Î#  y
"Î# ‰ Ê Š dx dy ‹ œ

" 4

ay  2
y
" b

Ê S œ  '1 21 ˆ "3 y$Î#  y"Î# ‰ É1
4" ay  2
y
" b dy 3

œ 21'1 ˆ 3" y$Î#  y"Î# ‰ É 4" ay
2
y
" b dy 3

Éay"Î#  y
"Î# b#

œ 21'1 ˆ "3 y$Î#  y"Î# ‰ 3

œ 1'1 ˆ "3 y#  3

2 3

dx dy

œ


" È 4
y

œ 41 '0

15Î4

œ

20.

dx dy

81 3

œ

3

$

y  1‰ dy œ 1 ’ y9 

œ  19 (18  1
3) œ 19.

dy œ 1'1 y"Î# ˆ 3" y  1‰ Šy"Î#


#

#

Ê Š dx dy ‹ œ

Ê S œ '0

15Î4

" 4
y

5È 5 8 ‹

œ

81 3

#

" È2y
1

Ê Š dx dy ‹ œ

Š 40

È 5
5 È 5 ‹ 8

œ

È

È5

1 12

‹œ

#

41 È 2 3

" 2y
1

$Î#

’1$Î#  ˆ 58 ‰

“œ

#


1 dy œ ÊŠy' 

" #




" 16y' ‹

2

È2

È2

dy dx

œ

" #

aa#  x# b

Ê S œ 21'ca Èa#  x# É1
a

$Î#

œ

21 r h

dy dx

#

#

É h h# r

25. y œ cos x Ê

œ

41 È 2 3

5È 5 ‹ 8È 8

Š1 

21 40

" #




" 16y' ‹

dy

dy œ 21'1 ˆy%
"4 y
# ‰ dy 2

" 4y$ ‹

(8 † 31
5) œ

2531 20

È2

x ax#
1b dx œ 21'0 ax$
xb dx œ 21 ’ x4



"Î#

x# aa #
x # b

(2x) œ


x È a#
x#

%

#

Ê Š dy dx ‹ œ

È# x# # “!

œ 21 ˆ 44
22 ‰ œ 41

x# aa #
x # b

dx œ 21'ca Èaa#  x# b
x# dx œ 21'ca a dx œ 21a[x]ca a a

a

r h

#

Ê Š dy dx ‹ œ

r# h#

Ê S œ 21 '0

h

r h

x É1


r# h#

dx œ 21'0

h

r h

#

#

x É h h# r dx

'0h x dx œ 2h1r Èh#
r# ’ x# “ h œ 2h1r Èh#
r# Š h# ‹ œ 1rÈh#
r# #

#

dy dx

 5$Î# “

1

œ 21a[a  (a)] œ (21a)(2a) œ 41a# x Ê


1‰

È2

23. y œ Èa#  x# Ê

r h

" 3

Ê dy œ xÈx#
2 dx Ê ds œ È1
a2x#
x% b dx Ê S œ 21'0 x È1
2x#
x% dx

$Î#

œ 21'0 xÉax#
1b# dx œ 21'0

24. y œ




dy œ 21'5Î8 È(2y  1)
1 dy

2

#

ax#
2b

" 9

È(4  y)
1 dy

 5$Î# “ œ  831 ’ˆ 45 ‰

dy; S œ '1 21y ds œ 21'1 y Šy$


" 4y$ ‹

"

" 3

 1‰‘ œ 1 ˆ3 


1 dy œ ÊŠy'


"‰ "‰ ˆ " " ‰‘ œ 21 ˆ 31 œ 21 ’ y5  4" y
" “ œ 21
ˆ 32 5  8  5  4 5
8 œ

22. y œ

15Î4

15 ‰$Î# 4

1

" 4y$ ‹

dy œ Šy$


&

dy œ 41'0

" 3

Š16È2  5È5‹

21. ds œ Èdx#
dy# œ ÊŠy$  " 4y$ ‹

3

351È5 3

"


5 Š 8†2 8†22È 2

dy œ 1 '1 ˆ 3" y  1‰ (y
1) dy

 3‰  ˆ "9 

9 3

" 4
y

21 † 2È4  y É1


Ê S œ '5Î8 21È2y  1 É1


" 2y
1

1

œ ÊŠy$


"

È5  y dy œ 41
23 (5  y)$Î# ‘ "&Î% œ  831 ’ˆ5  !

Š 5È 5 

41 È 2 3

 y“ œ 1
ˆ 27 9 

161 9

œ 21'5Î8 È2 y"Î# dy œ 21È2
23 y$Î# ‘ &Î) œ œ

$

y# 3

" ‹ y"Î#

#

0

#

# È1
sin# x dx ' œ  sin x Ê Š dy dx ‹ œ sin x Ê S œ 21 c1Î2 (cos x) #

1Î2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

397

398

Chapter 6 Applications of Definite Integrals

26. y œ ˆ1  x#Î$ ‰

$Î#

Ê

dy dx

œ

Ê S œ 2'0 21 ˆ1  x#Î$ ‰ 1

3 #

ˆ1  x#Î$ ‰"Î# ˆ 23 x
"Î$ ‰ œ 

$Î#

#

ˆ1
x#Î$ ‰"Î# x"Î$

Ê Š dy dx ‹ œ

1
x#Î$ x#Î$

œ

" x#Î$

1

$Î# " É1
ˆ x#Î$  1‰ dx œ 41'0 ˆ1  x#Î$ ‰ Èx
#Î$ dx 1

$Î# œ 41'0 ˆ1  x#Î$ ‰ x
"Î$ dx;
u œ 1  x#Î$ Ê du œ  23 x
"Î$ dx Ê  32 du œ x
"Î$ dx; 1

! x œ 0 Ê u œ 1, x œ 1 Ê u œ 0d Ä S œ 41'1 u$Î# ˆ 3# du‰ œ 61
25 u&Î# ‘ " œ 61 ˆ0  25 ‰ œ 0

121 5

# # # È16#  y# 27. The area of the surface of one wok is S œ 'c 21x Ê1
Š dx dy ‹ dy. Now, x
y œ 16 Ê x œ #

d

Ê

dx dy

œ c7

#


y È16#
y#

Ê Š dx dy ‹ œ

c7

; S œ 'c16 21È16#  y# É1


y# 16#
y#

y# 16#
y#

c7

dy œ 21'c16 Èa16#  y# b
y# dy

œ 21'c16 16 dy œ 321 † 9 œ 2881 ¸ 904.78 cm# . The enamel needed to cover one surface of one wok is

V œ S † 0.5 mm œ S † 0.05 cm œ (904.78)(0.05) cm$ œ 45.24 cm$ . For 5000 woks, we need 5000 † V œ 5000 † 45.24 cm$ œ (5)(45.24)L œ 226.2L Ê 226.2 liters of each color are needed. 28. y œ Èr#  x# Ê œ 21'a

abh

œ 21'a

2x È r#
x #

œ

Èar#  x# b
x# dx œ 21r' a

29. y œ ÈR#  x# Ê abh

œ  "#

dy dx

dy dx

œ  "#

2x È R #
x#


x Èr#
x#

abh

œ

abh

30. (a) x#
y# œ 45# Ê x œ È45#  y# Ê 45

x# r#
x # ;

S œ 21 'a

abh

Èr#  x# É1


x# r#
x#

dx

dx œ 21rh, which is independent of a. #


x È R #
x#

ÈaR#  x# b
x# dx œ 21R ' a

S œ 'c22Þ5 21 È45#  y# É1


#

Ê Š dx dy ‹ œ

y# 45#
y#

dx Ê Š dy ‹ œ

x# R #
x# ;

S œ 21'a

abh

ÈR#  x# É1


x# R #
x#

dx

dx œ 21Rh dx dy

œ


y È45#
y#

#

Ê Š dx dy ‹ œ

y# 45#
y#

;

dy œ 21 '
22Þ5 Èa45#  y# b
y# dy œ 21 † 45'
22Þ5 dy 45

45

œ (21)(45)(67.5) œ 60751 square feet (b) 19,085 square feet dy ' È1
1 dx œ 21 ' (x)È2 dx
21' xÈ2 dx 31. y œ x Ê Š dy dx ‹ œ 1 Ê Š dx ‹ œ 1 Ê S œ 21 c1 kxk c1 0 #

# œ 2È21 ’ x# “

32.

dy dx

œ

x# 3

!

2

#


"

!

#

4 9

0

#
2È21 ’ x# “ œ 2È21 ˆ0  "# ‰
2È21(2  0) œ 5È21

Ê Š dy dx ‹ œ

Ê du œ

2

x% 9

Ê by symmetry of the graph that S œ 2 'cÈ3 21 Š x9 ‹ É1
0

$

x% 9

dx; ’u œ 1


x$ dx Ê  "4 du œ  x9 dx; x œ È3 Ê u œ 2, x œ 0 Ê u œ 1“ Ä S œ 41'2 u"Î# ˆ "4 ‰ du 1

$

" œ 1'2 u"Î# du œ 1
23 u$Î# ‘ # œ 1 Š 23  23 È8‹ œ 1

È3

È3

21 3

ŠÈ8  1‹ . If the absolute value bars are dropped the

integral for S œ 'cÈ3 21f(x) ds will equal zero since 'cÈ3 21 Š x9 ‹ É1
$

x% 9

dx is the integral of an odd function

over the symmetric interval È3 Ÿ x Ÿ È3.

33.

dx dt

x% 9

œ  sin t and

dy dt

#

È( sin t)#
(cos t)# œ 1 Ê S œ ' 21y ds ‰
Š dy œ cos t Ê Êˆ dx dt dt ‹ œ #

œ '0 21(2
sin t)(1) dt œ 21 c2t  cos td #!1 œ 21[(41  1)  (0  1)] œ 81# 21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.5 Areas of Surfaces of Revolution and the Theorems of Pappus 34.

dx dt

œ t"Î# and

È3

dy dt

œ '0 21 ˆ 23 t$Î# ‰ É t

#

" t

dt œ

È3

'0

#

21 ˆ 23 t$Î# ‰ É t #

f(t) œ 21 ˆ 23 t$Î# ‰ É t Ê

35.

dx dt

È3

'0

F(t) dt œ

œ 1 and

È2

dy dt

#

281 9

" t

41 3

È3

'0

1 t

Ê S œ ' 21x ds

tÈt#
1 dt; cu œ t#
1 Ê du œ 2t dt; t œ 0 Ê u œ 1,

'14 231 Èu du œ
491 u$Î# ‘ %" œ 2891

’t œ È3 Ê u œ 4“ Ä Note:

#

#

Èt
t
" œ É t ‰
Š dy œ t
"Î# Ê Êˆ dx dt dt ‹ œ

1 t

dt is an improper integral but limb f(t) exists and is equal to 0, where tÄ!

. Thus the discontinuity is removable: define F(t) œ f(t) for t  0 and F(0) œ 0

. #

#

# È2‹ œ Ét#
2È2 t
3 Ê S œ ' 21x ds ‰
Š dy œ t
È2 Ê Êˆ dx dt dt ‹ œ Ê1
Št
#

œ 'cÈ2 21 Št
È2‹ Ét#
2È2 t
3 dt; ’u œ t#
2È2 t
3 Ê du œ Š2t
2È2‹ dt; t œ È2 Ê u œ 1,

* t œ È2 Ê u œ 9“ Ä '1 1Èu du œ
23 1u$Î# ‘ " œ 9

36.

dx dt

œ aa1  cos tb and

dy dt

21 3

(27  1) œ

521 3

#

#

‰
Š dy Éc aa1  cos tb d#
aa sin tb# œ a sin t Ê Êˆ dx dt dt ‹ œ

œ Èa2  2 a2 cos t
a2 cos2 t
a2 sin2 t œ È2a2  2a2 cos t œ aÈ2È1  cos t Ê S œ ' 21y ds œ '0 21 aa1  cos tb † aÈ2È1  cos t dt œ 2È2 1 a2 '0 a1  cos tb3/2 dt 21

37.

dx dt

œ 2 and

21

dy dt

È2#
1# œ È5 Ê S œ ' 21y ds œ ' 21(t
1)È5 dt ‰
Š dy œ 1 Ê Êˆ dx dt dt ‹ œ 0 #

#

1

"

#

œ 21È5 ’ t2
t“ œ 31È5. Check: slant height is È5 Ê Area is 1(1
2)È5 œ 31È5 . !

38.

dx dt

œ h and

dy dt

Èh#
r# Ê S œ ' 21y ds œ ' 21rtÈh#
r# dt ‰
Š dy œ r Ê Êˆ dx dt dt ‹ œ 0

œ 21rÈh#
r#

#

#

1

'01 t dt œ 21rÈh#
r# ’ t2 “ " œ 1rÈh#
r# . #

!

Check: slant height is Èh#
r# Ê Area is

1rÈh#
r# . 39. (a) An equation of the tangent line segment is (see figure) y œ f(mk )
f w (mk )(x  mk ). When x œ xkc1 we have r" œ f(mk )
f w (mk )(x5
1  mk ) œ f(mk )
f w (mk ) ˆ ?#xk ‰ œ f(mk )  f w (mk ) when x œ xk we have r# œ f(mk )
f w (mk )(x5  mk ) k œ f(mk )
f w (mk ) ?x # ;

(b) L#k œ (?xk )#
(r#  r" )#

;

#

 ˆf w (mk ) ?#xk ‰‘ œ (?xk )#
[f w (mk )?xk ]# Ê Lk œ È(?xk )#
[f w (mk )?xk ]# , as claimed (c) From geometry it is a fact that the lateral surface area of the frustum obtained by revolving the tangent œ (?xk )#

f w (mk )

?x k #

?x k #

line segment about the x-axis is given by ?Sk œ 1(r"
r# )Lk œ 1[2f(mk )] Éa?xk b#
[f w (mk )?xk ]# using parts (a) and (b) above. Thus, ?Sk œ 21f(mk ) È1
[f w (mk )]# ?xk .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

399

400

Chapter 6 Applications of Definite Integrals

! ?Sk œ lim ! 21f(mk ) È1
[f w (mk )]# ?xk œ ' 21f(x) È1
[f w (x)]# dx (d) S œ n lim Ä_ nÄ_ a kœ1 kœ1 n

n

È3

40. S œ 'a 21f(x) dx œ '0 b

21 †

x È3

b

dx œ

1 È3

È3

c x# d ! œ

31 È3

œ È31

41. The centroid of the square is located at (#ß #). The volume is V œ (21) ayb (A) œ (21)(2)(8) œ 321 and the surface area is S œ (21) ayb (L) œ (21)(2) Š4È8‹ œ 32È21 (where È8 is the length of a side). 42. The midpoint of the hypotenuse of the triangle is ˆ 3# ß 3‰ Ê y œ 2x is an equation of the median Ê the line y œ 2x contains the centroid. The point ˆ 3# ß $‰ is 3È 5 #

units from the origin Ê the x-coordinate of the

# centroid solves the equation Ɉx  3# ‰
(2x  3)#

œ

È5 #

Ê ˆx#  3x
94 ‰
a4x#  12x
9b œ

5 4

Ê 5x#  15x
9 œ 1 Ê x#  3x
2 œ (x  2)(x  1) œ 0 Ê x œ 1 since the centroid must lie inside the triangle Ê y œ 2. By the Theorem of Pappus, the volume is V œ (distance traveled by the centroid)(area of the region) œ 21 a5  xb
"# (3)(6)‘ œ (21)(4)(9) œ 721

43. The centroid is located at (#ß !) Ê V œ (21) axb (A) œ (21)(2)(1) œ 41# 44. We create the cone by revolving the triangle with vertices (0ß 0), (hß r) and (hß 0) about the x-axis (see the accompanying figure). Thus, the cone has height h and base radius r. By Theorem of Pappus, the lateral surface area swept out by the hypotenuse L is given by S œ 21yL œ 21 ˆ r ‰ Èh#
r# #

œ 1rÈr#
h# . To calculate the volume we need the position of the centroid of the triangle. From the diagram we see that the centroid lies on the line y œ œ

" 3

Éh#


r# 4

#

r 2h

#

# x. The x-coordinate of the centroid solves the equation É(x  h)#
ˆ 2hr x  #r ‰ #

#

Ê Š 4h4h# r ‹ x#  Š 4h 2h r ‹ x


inside the triangle Ê y œ

r 2h

47. V œ 21 yA Ê

4 3




2 ar#  4h# b 9

œ0 Ê xœ

2h 3

or

4h 3

Ê xœ

x œ 3r . By the Theorem of Pappus, V œ
21 ˆ 3r ‰‘ ˆ "# hr‰ œ

45. S œ 21 y L Ê 41a# œ a21yb (1a) Ê y œ 46. S œ 213 L Ê
21 ˆa 

r# 4

2a ‰‘ (1a) 1

2a 1,

since the centroid must lie

1r# h.

and by symmetry x œ 0

œ 21a# (1  2)

1ab# œ a21yb ˆ 1#ab ‰ Ê y œ

48. V œ 213A Ê V œ
21 ˆa


" 3

2h 3 ,

4a ‰‘ 1a# Š # ‹ 31

œ

4b 31

and by symmetry x œ 0

1a$ (31  4) 3

49. V œ 213 A œ (21)(area of the region) † (distance from the centroid to the line y œ x  a). We must find the 4a ‰ distance from ˆ0ß 31 to y œ x  a. The line containing the centroid and perpendicular to y œ x  a has slope 1 and contains the point ˆ!ß 34a1 ‰ . This line is y œ x
34a1 . The intersection of y œ x  a and y œ x
34a1 is the point ˆ 4a 613a1 ß 4a
613a1 ‰ . Thus, the distance from the centroid to the line y œ x  a is Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.6 Work Ɉ 4a 613a1 ‰#
ˆ 34a1 

4a 61




3a1 ‰# 61

œ

È2 (4a  3a1) 61

Ê V œ (21) Š

È2 (4a  3a1) # ‹ Š 1#a ‹ 61

œ

È2 1a$ (4  31) 6

‰ 50. The line perpendicular to y œ x  a and passing through the centroid ˆ!ß 2a 1 has equation y œ x
intersection of the two perpendicular lines occurs when x  a œ x


Ê xœ

2a 1

2a  a1 21

Ê yœ

2a ‰# #

a(21) È 21

#

the distance from the centroid to the line y œ x  a is Ɉ 2a 2 1a  0‰
ˆ 2a
2 1a 

œ

2a 1 . The 2a
a1 21 . Thus

.

1 ) Therefore, by the Theorem of Pappus the surface area is S œ 21 ’ a(2 “ (1a) œ È21a# (2
1). È 21

51. From Example 4 and Pappus's Theorem for Volumes we have the moment about the x-axis is Mx œ y M #

œ ˆ 34a1 ‰ Š 1#a ‹ œ

2a$ 3

.

6.6 WORK 1. The force required to stretch the spring from its natural length of 2 m to a length of 5 m is F(x) œ kx. The work done by F is W œ '0 F(x) dx œ k '0 x dx œ 3

3

k #

$

cx# d ! œ

9k # .

This work is equal to 1800 J Ê

k œ 1800

9 #

Ê k œ 400 N/m 2. (a) We find the force constant from Hooke's Law: F œ kx Ê k œ

Ê kœ

F x

800 4

œ 200 lb/in.

(b) The work done to stretch the spring 2 inches beyond its natural length is W œ '0 kx dx 2

œ 200 '0 x dx œ 200 ’ x# “ œ 200(2  0) œ 400 in † lb œ 33.3 ft † lb 2

#

# !

(c) We substitute F œ 1600 into the equation F œ 200x to find 1600 œ 200x Ê x œ 8 in. 3. We find the force constant from Hooke's law: F œ kx. A force of 2 N stretches the spring to 0.02 m N Ê 2 œ k † (0.02) Ê k œ 100 m . The force of 4 N will stretch the rubber band y m, where F œ ky Ê y œ Ê yœ

4N N 100 m

œ 100 '0

0Þ04

Ê y œ 0.04 m œ 4 cm. The work done to stretch the rubber band 0.04 m is W œ '0

F k

0Þ04

#

x dx œ 100 ’ x# “

!Þ!%

œ

!

(100)(0.04)# #

kx dx

œ 0.08 J

4. We find the force constant from Hooke's law: F œ kx Ê k œ

F x

Ê kœ

90 1

Ê k œ 90

N m. &

The work done to

‰ stretch the spring 5 m beyond its natural length is W œ '0 kx dx œ 90 '0 x dx œ 90 ’ x# “ œ (90) ˆ 25 # œ 1125 J 5

5

#

!

5. (a) We find the spring's constant from Hooke's law: F œ kx Ê k œ

F x

œ

21,714 8
5

œ

21,714 3

Ê k œ 7238

(b) The work done to compress the assembly the first half inch is W œ '0 kx dx œ 7238 '0 0Þ5

#

œ 7238 ’ x# “

!Þ& !

#

œ (7238) (0.5) # œ

(7238)(0.25) #

1Þ0

1Þ0

Þ

Þ

#

¸ 2714 in † lb 6. First, we find the force constant from Hooke's law: F œ kx Ê k œ compresses the scale x œ scale this far is W œ '0

1Î8

in, he/she must weigh F œ kx œ #

kx dx œ 2400 ’ x# “

"Î) !

x dx

¸ 905 in † lb. The work done to compress the assembly the

second half inch is: W œ '0 5 kx dx œ 7238 '0 5 x dx œ 7238 ’ x# “

" 8

lb in

0Þ5

œ

2400 2†64

F x

2,400 ˆ 8" ‰

"Þ! !Þ&

œ

œ

150 " ‰ ˆ 16

7238 #

c1  (0.5)# d œ

(7238)(0.75) #

œ 16 † 150 œ 2,400

lb in .

If someone

œ 300 lb. The work done to compress the

œ 18.75 lb † in. œ

25 16

ft † lb

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

401

402

Chapter 6 Applications of Definite Integrals

7. The force required to haul up the rope is equal to the rope's weight, which varies steadily and is proportional to x, the length of the rope still hanging: F(x) œ 0.624x. The work done is: W œ '0 F(x) dx œ '0 0.624x dx 50

#

œ 0.624 ’ x# “

&! !

50

œ 780 J

8. The weight of sand decreases steadily by 72 lb over the 18 ft, at 4 lb/ft. So the weight of sand when the bag is x ft off the ground is Faxb œ "%%  %x. The work done is: W œ 'a F(x) dx œ '0 a"%%  %xbdx œ c144x  2x# d ! œ 1944 ft † lb b

18

")

9. The force required to lift the cable is equal to the weight of the cable paid out: F(x) œ (4.5)(180  x) where x is the position of the car off the first floor. The work done is: W œ '0

180

œ 4.5 ’180x 

")! x# # “!

180# # ‹

œ 4.5 Š180# 

œ

4.5†180# #

F(x) dx œ 4.5'0

180

(180  x) dx

œ 72,900 ft † lb

10. Since the force is acting toward the origin, it acts opposite to the positive x-direction. Thus F(x) œ  xk# . The b work done is W œ 'a  xk# dx œ k 'a  x"# dx œ k
x" ‘ a œ k ˆ b"  "a ‰ œ b

b

k(a
b) ab

11. The force against the piston is F œ pA. If V œ Ax, where x is the height of the cylinder, then dV œ A dx Ê Work œ ' F dx œ ' pA dx œ 'ap

ap# ßV# b " ßV" b

p dV.

12. pV"Þ% œ c, a constant Ê p œ cV
"Þ% . If V" œ 243 in$ and p" œ 50 lb/in$ , then c œ (50)(243)"Þ% œ 109,350 lb. ‘ ˆ 3#"!Þ%  Thus W œ '243 109,350V
"Þ% dV œ
 109,350 œ  109,350 0.4 0.4V!Þ% #%$ $#

32

" ‰ #43!Þ%

ˆ 4"  9" ‰ œ  109,350 0.4

œ  (109,350)(5) (0.4)(36) œ 37,968.75 in † lb. Note that when a system is compressed, the work done by the system is negative. 13. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#!  xb, the distance the bucket is being raised. The leakage rate of the water is 0.8 lb/ft raised and the weight of the water in the bucket is F œ 0.8a#!  xb. So: W œ '0 0.8a#!  xb dx œ 0.8 ’20x  20

#! x# # “!

œ 160 ft † lb.

14. Let r œ the constant rate of leakage. Since the bucket is leaking at a constant rate and the bucket is rising at a constant rate, the amount of water in the bucket is proportional to a#!  xb, the distance the bucket is being raised. The leakage rate of the water is 2 lb/ft raised and the weight of the water in the bucket is F œ 2a#!  xb. So: W œ '0 2a#!  xb dx œ 2 ’20x  20

#! x# # “!

œ 400 ft † lb.

Note that since the force in Exercise 14 is 2.5 times the force in Exercise 13 at each elevation, the total work is also 2.5 times as great.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.6 Work

403

15. We will use the coordinate system given. (a) The typical slab between the planes at y and y
?y has a volume of ?V œ (10)(12) ?y œ 120 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 120 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance œ 62.4 † 120 † y † ?y ft † lb. The work it takes to lift all 20

the water is approximately W ¸ ! ?W 0 20

œ ! 62.4 † 120y † ?y ft † lb. This is a Riemann sum for 0

the function 62.4 † 120y over the interval 0 Ÿ y Ÿ 20. The work of pumping the tank empty is the limit of these sums: W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 20

#

#! !

‰ œ (62.4)(120)(200) œ 1,497,600 ft † lb œ (62.4)(120) ˆ 400 #

5 ‰ (b) The time t it takes to empty the full tank with ˆ 11 –hp motor is t œ

W †lb 250 ftsec

œ

1,497,600 ft†lb †lb 250 ftsec

œ 5990.4 sec

œ 1.664 hr Ê t ¸ 1 hr and 40 min (c) Following all the steps of part (a), we find that the work it takes to lower the water level 10 ft is W œ '0 62.4 † 120y dy œ (62.4)(120) ’ y# “ 10

#

œ 1497.6 sec œ 0.416 hr ¸ 25 min (d) In a location where water weighs 62.26

"! !

‰ œ 374,400 ft † lb and the time is t œ œ (62.4)(120) ˆ 100 #

W †lb 250 ftsec

lb ft$ :

a) W œ (62.26)(24,000) œ 1,494,240 ft † lb. b) t œ 1,494,240 œ 5976.96 sec ¸ 1.660 hr Ê t ¸ 1 hr and 40 min 250 In a location where water weighs 62.59

lb ft$

a) W œ (62.59)(24,000) œ 1,502,160 ft † lb b) t œ 1,502,160 œ 6008.64 sec ¸ 1.669 hr Ê t ¸ 1 hr and 40.1 min 250 16. We will use the coordinate system given. (a) The typical slab between the planes at y and y
?y has a volume of ?V œ (20)(12) ?y œ 240 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 62.4 ?V œ 62.4 † 240 ?y lb. The distance through which F must act is about y ft, so the work done lifting the slab is about ?W œ force ‚ distance 20

œ 62.4 † 240 † y † ?y ft † lb. The work it takes to lift all the water is approximately W ¸ ! ?W 10 20

œ ! 62.4 † 240y † ?y ft † lb. This is a Riemann sum for the function 62.4 † 240y over the interval 10

10 Ÿ y Ÿ 20. The work it takes to empty the cistern is the limit of these sums: W œ '10 62.4 † 240y dy 20

#

œ (62.4)(240) ’ y# “ (b) t œ

W †lb 275 ftsec

œ

#!

œ (62.4)(240)(200  50) œ (62.4)(240)(150) œ 2,246,400 ft † lb

"! 2,246,400 ft†lb 275

¸ 8168.73 sec ¸ 2.27 hours ¸ 2 hr and 16.1 min

(c) Following all the steps of part (a), we find that the work it takes to empty the tank halfway is W œ '10 62.4 † 240y dy œ (62.4)(240) ’ y# “ 15

#

Then the time is t œ

W †lb 275 ftsec

œ

936,000 #75

"& "!

œ (62.4)(240) ˆ 225 # 

100 ‰ #

‰ œ 936,000 ft. œ (62.4)(240) ˆ 125 #

¸ 3403.64 sec ¸ 56.7 min

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

404

Chapter 6 Applications of Definite Integrals lb ft$ :

(d) In a location where water weighs 62.26

a) W œ (62.26)(240)(150) œ 2,241,360 ft † lb. b) t œ 2,241,360 œ 8150.40 sec œ 2.264 hours ¸ 2 hr and 15.8 min 275 ‰ œ 933,900 ft † lb; t œ 933,900 c) W œ (62.26)(240) ˆ 125 # #75 œ 3396 sec ¸ 0.94 hours ¸ 56.6 min lb ft$

In a location where water weighs 62.59

a) W œ (62.59)(240)(150) œ 2,253,240 ft † lb. b) t œ 2,253,240 œ 8193.60 sec œ 2.276 hours ¸ 2 hr and 16.56 min 275 ‰ œ 938,850 ft † lb; t œ 938,850 c) W œ (62.59)(240) ˆ 125 # 275 ¸ 3414 sec ¸ 0.95 hours ¸ 56.9 min #

17. The slab is a disk of area 1x# œ 1ˆ y# ‰ , thickness ˜y, and height below the top of the tank a"!  yb. So the work to pump #

the oil in this slab, ˜W, is 57a"!  yb1ˆ y# ‰ . The work to pump all the oil to the top of the tank is W œ '0

10

571 # 4 a"!y

 y$ bdy œ

571 4

$

’ "!$y 

"! y% % “!

œ 11,8751 ft † lb ¸ 37,306 ft † lb. #

18. Each slab of oil is to be pumped to a height of 14 ft. So the work to pump a slab is a"%  yba1bˆ y# ‰ and since the tank is half full and the volume of the original cone is V œ "$ 1r# h œ "$ 1a&# ba"!b œ with half the volume the cone is filled to a height y, œ

571 "%y$ 4 ’ $



$ È &!! y% “ % !

#&!1 '

#&!1 $

ft3 , half the volume œ $ È &!!

$ œ $" 1 y% y Ê y œ È &!! ft. So W œ '0 #

#&!1 '

ft3 , and

571 # 4 a"%y

 y$ b dy

¸ 60,042 ft † lb. #

‰ ?y 19. The typical slab between the planes at y and and y
?y has a volume of ?V œ 1(radius)# (thickness) œ 1 ˆ 20 # œ 1 † 100 ?y ft$ . The force F required to lift the slab is equal to its weight: F œ 51.2 ?V œ 51.2 † 1001 ?y lb Ê F œ 51201 ?y lb. The distance through which F must act is about (30  y) ft. The work it takes to lift all the 30

30

kerosene is approximately W ¸ ! ?W œ ! 51201(30  y) ?y ft † lb which is a Riemann sum. The work to pump the 0

0

tank dry is the limit of these sums: W œ '0 51201(30  y) dy œ 51201 ’30y  30

¸ 7,238,229.48 ft † lb

$! y# # “!

‰ œ (5120)(4501) œ 51201 ˆ 900 #

20. (Alternate Solution) Each method must pump all of the water the 15 ft to the base of the tank. Pumping to the rim requires all the water to be pumped an additional 6 feet. Pumping into the bottom requires that the water be pumped an average of 3 additional feet. Thus pumping through the valve requires È$ fta%1b6 ft3 a'#Þ% lb/ft3 b ¸ 14,115 ft † lb less work and thus less time. 21. (a) Follow all the steps of Example 5 but make the substitution of 64.5 W œ '0

8

œ

64.51 4

64.51†8$ 3

(10  y)y# dy œ

64.51 4

$

’ 10y 3 

%

y 4

)

“ œ !

64.51 4

$

Š 103†8 

lb ft$

8% 4‹

for 57

lb ft$ .

Then,

1‰ ‰ œ ˆ 64.5 a8$ b ˆ 10 4 3 2

œ 21.51 † 8$ ¸ 34,582.65 ft † lb

(b) Exactly as done in Example 5 but change the distance through which F acts to distance ¸ (13  y) ft. Then W œ '0

8

571 4

(13  y)y# dy œ

571 4

œ (191) a8# b (7)(2) ¸ 53.482.5 ft † lb

$

’ 13y 3 

) y% 4 “!

œ

571 4

$

Š 133†8 

8% 4‹

‰ œ ˆ 5741 ‰ a8$ b ˆ 13 3 2 œ

571†8$ †7 3 †4

22. The typical slab between the planes of y and y
?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈy‰ ?y œ xy ?y m$ . The force F(y) is equal to the slab's weight: F(y) œ 10,000 mN$ † ?V œ 110,000y ?y N. The height of the tank is 4# œ 16 m. The distance through which F(y) must act to lift the slab to the level of the top of the tank is about (16  y) m, so the work done lifting the slab is about

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.6 Work

405

?W œ 10,0001y(16  y) ?y N † m. The work done lifting all the slabs from y œ 0 to y œ 16 to the top is 16

approximately W ¸ ! 10,0001y(16  y)?y. Taking the limit of these Riemann sums, we get 0

W œ '0 10,0001y(16  y) dy œ 10,0001'0 a16y  y# b dy œ 10,0001 ’ 16y #  16

œ

16

10,000†1†16$ 6

#

"' y$ 3 “!

$

œ 10,0001 Š 16# 

16$ 3 ‹

¸ 21,446,605.9 J

23. The typical slab between the planes at y and y
?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ25  y# ‰ ?y m$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 9800 † ?V #

œ 98001 ˆÈ25  y# ‰ ?y œ 98001 a25  y# b ?y N. The distance through which F(y) must act to lift the slab to the level of 4 m above the top of the reservoir is about (4  y) m, so the work done is approximately ?W ¸ 98001 a25  y# b (4  y) ?y N † m. The work done lifting all the slabs from y œ 5 m to y œ 0 m is 0

approximately W ¸ ! 98001 a25  y# b (4  y) ?y N † m. Taking the limit of these Riemann sums, we get c5

W œ 'c5 98001 a25  y# b (4  y) dy œ 98001 'c5 a100  25y  4y#
y$ b dy œ 98001 ’100C  0

0

25†25 #

œ 98001 ˆ500 




† 125


4 3

625 ‰ 4

25 #

y#  34 y$


¸ 15,073,099.75 J

24. The typical slab between the planes at y and y
?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆÈ100  y# ‰ ?y œ 1 a100  y# b ?y ft$ . The force is F(y) œ 56ft$lb † ?V œ 561 a100  y# b ?y lb. The distance through which F(y) must act to lift the slab to the level of 2 ft above the top of the tank is about (12  y) ft, so the work done is ?W ¸ 561 a100  y# b (12  y) ?y lb † ft. The work done lifting all the slabs 10

from y œ 0 ft to y œ 10 ft is approximately W ¸ ! 561 a100  y# b (12  y) ?y lb † ft. Taking the limit of these 0

Riemann sums, we get W œ '0 561 a100  y b (12  y) dy œ 561'0 a100  y# b (12  y) dy 10

10

#

œ 561'0 a1200  100y  12y#
y$ b dy œ 561 ’1200C  10

œ 561 ˆ12,000 

10,000 #

 4 † 1000


10,000 ‰ 4

100y# #



12y$ 3




"! y% 4 “!

œ (561) ˆ12  5  4
5# ‰ (1000) ¸ 967,611 ft † lb.

It would cost (0.5)(967,611) œ 483,805¢ œ $4838.05. Yes, you can afford to hire the firm. 25. F œ m œ

" #

dv dt

œ mv

by the chain rule Ê W œ 'x mv x#

dv dx #

"

m cv# (x# )  v (x" )d œ

26. weight œ 2 oz œ

" #

weight 32

œ

" 8

3#

œ

" #56

28. weight œ 1.6 oz œ 0.1 lb Ê m œ " 8

x# "

dv ‰ dx

dx œ m
"# v# (x)‘ x" x#

" slugs; W œ ˆ "# ‰ ˆ #56 slugs‰ (160 ft/sec)# ¸ 50 ft † lb

hr 1 min 5280 ft 27. 90 mph œ 901 hrmi † 601 min † 60 sec † 1 mi œ 132 ft/sec; m œ 0.3125 lb ‰ # W œ ˆ "# ‰ ˆ 32 ft/sec# (132 ft/sec) ¸ 85.1 ft † lb

29. weight œ 2 oz œ

dx œ m'x ˆv

mv##  "# mv"# , as claimed.

lb; mass œ

2 16

dv dx

lb Ê m œ

" 8

32

0.1 lb 32 ft/sec#

slugs œ

œ " #56

" 3 #0

0.3125 lb 32 ft/sec#

œ

0.3125 32

slugs;

slugs; W œ ˆ "# ‰ ˆ 3"#0 slugs‰ (280 ft/sec)# œ 122.5 ft † lb

slugs; 124 mph œ

(124)(5280) (60)(60)

¸ 181.87 ft/sec;

" W œ ˆ "# ‰ ˆ 256 slugs‰ (181.87 ft/sec)# ¸ 64.6 ft † lb

30. weight œ 14.5 oz œ 31. weight œ 6.5 oz œ

14.5 16

6.5 16

lb Ê m œ

lb Ê m œ

14.5 (16)(32)

6.5 (16)(32)

"4.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (88 ft/sec)# ¸ 109.7 ft † lb

6.5 slugs; W œ ˆ "# ‰ Š (16)(32) slugs‹ (132 ft/sec)# ¸ 110.6 ft † lb

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

! y% 4 “
&

406

Chapter 6 Applications of Definite Integrals

32. F œ (18 lb/ft)x Ê W œ '0 18x dx œ c9x# d ! 1Î6

mœ

" 8

32

œ

" #56

"Î6

slugs and v" œ 0 ft/sec. Thus,

1 4

œ

1 4

ft † lb. Now W œ È2 4

È2 4 ‹

 (16) Š

È2 4 ‹

1 4

ft † lb,

sec when the bearing is at the top of its path. È2 4

The height the bearing reaches is s œ 8È2 t  16t# Ê at t œ Š8È2‹ Š

mv#  "# mv"# , where W œ

" ft † lb. œ ˆ #" ‰ ˆ #56 slugs‰ v# Ê v œ 8È2 ft/sec. With v œ 0

at the top of the bearing's path and v œ 8È2  32t Ê t œ #

" #

the bearing reaches a height of

œ 2 ft

33. (a) From the diagram, rayb œ '!  x œ '!  É&!#  ay  325b# for 325 Ÿ y Ÿ 375 ft. (b) The volume of a horizontal slice of the funnel # is ˜V ¸ 1
rayb‘ ˜y #

œ 1”'!  É&!#  ay  325b# • ˜y (c) The work required to lift the single slice of water is ˜W ¸ 62.4˜Va$(&  yb #

œ 62.4a$(&  yb1”'!  É&!#  ay  325b# • ˜y. The total work to pump our the funnel is W #

œ '325 62.4a375  yb1”'!  É50#  ay  325b# • dy 375

¸ 6.3358 † 10( ft † lb. 34. (a) From the result in Example 6, the work to pump out the throat is 1,353,869,354 ft † lb. Therefor, the total work required to pump out the throat and the funnel is 1,353,869,354
63,358,000 œ 1,417227,354 ft † lb. (b) In horsepower-hours, the work required to pump out the glory hole is 1,417227,354 œ 715.8. Therefore, it would take 1.98†106 715.8 hp†h 1000 hp

œ 0.7158 hours ¸ 43 minutes.

35. We imagine the milkshake divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [!ß (]. The typical slab between the planes at y and y
?y has a volume of about #

17.5 ‰ ?V œ 1(radius)# (thickness) œ 1 ˆ y14 ?y in$ . The force F(y) required to lift this slab is equal to its

weight: F(y) œ

4 9

?V œ

41 9

#

17.5 ‰ ˆ y14 ?y oz. The distance through which F(y) must act to lift this slab to

the level of 1 inch above the top is about (8  y) in. The work done lifting the slab is about 17.5)# ?W œ ˆ 491 ‰ (y14 (8  y) ?y in † oz. The work done lifting all the slabs from y œ 0 to y œ 7 is # 7

approximately W œ ! 0

41 9†14#

(y
17.5)# (8  y) ?y in † oz which is a Riemann sum. The work is the limit of

these sums as the norm of the partition goes to zero: W œ '0

7

œ

41 9†14#

œ

41 9†14#

'07 a2450  26.25y  27y#  y$ b dy œ 9†4141 ’

7% 4

$

9†7 

26.25 #

41 9†14#

%

#

(y
17.5)# (8  y) dy

’ y4  9y$ 

26.25 #

y#
2450y“

( !

#

† 7
2450 † 7“ ¸ 91.32 in † oz

36. We fill the pipe and the tank. To find the work required to fill the tank follow Example 6 with radius œ 10 ft. Then ?V œ 1 † 100 ?y ft$ . The force required will be F œ 62.4 † ?V œ 62.4 † 1001 ?y œ 62401 ?y lb. The distance through which F must act is y so the work done lifting the slab is about ?W" œ 62401 † y † ?y lb † ft. The work it takes to

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 6.7 Fluid Pressures and Forces 385

385

360

360

lift all the water into the tank is: W" ¸ ! ?W" œ ! 62401 † y † ?y lb † ft. Taking the limit we end up with W" œ '360 62401y dy œ 62401 ’ y# “ 385

#

$)& $'!

62401 #

œ

c385#  360# d ¸ 182,557,949 ft † lb 4 #

To find the work required to fill the pipe, do as above, but take the radius to be Then ?V œ 1 †

" 36

$

?y ft and F œ 62.4 † ?V œ

integration: W# ¸ ! ?W# Ê W# œ '0 360

360

0

62.4 36

62.41 36

" 6

in œ

ft.

?y. Also take different limits of summation and

1y dy œ

62.41 36

#

$'!

#

1 ‰ 360 œ ˆ 62.4 Š # ‹ ¸ 352,864 ft † lb. 36

’ y# “

!

The total work is W œ W"
W# ¸ 182,557,949
352,864 ¸ 182,910,813 ft † lb. The time it takes to fill the W tank and the pipe is Time œ 1650 ¸ 182,910,813 ¸ 110,855 sec ¸ 31 hr 1650 37. Work œ '6 370 000

35ß780ß000 ß

1000 MG r#

ß

dr œ 1000 MG '6 370 000

35ß780ß000 ß

ß

" œ (1000) a5.975 † 10#% b a6.672 † 10
"" b Š 6,370,000 

$&ß()!ß!!!

œ 1000 MG
 "r ‘ 'ß$(!ß!!!

dr r#

" 35,780,000 ‹

¸ 5.144 ‚ 10"! J

38. (a) Let 3 be the x-coordinate of the second electron. Then r# œ (3  1)# Ê W œ 'c1 F(3) d3 0

œ 'c1 a23(3‚
101)# b d3 œ  ’ 233‚
10" “


#* !


#*

0


"

œ a23 ‚ 10
#* b ˆ1  #" ‰ œ 11.5 ‚ 10
#*

(b) W œ W"
W# where W" is the work done against the field of the first electron and W# is the work done against the field of the second electron. Let 3 be the x-coordinate of the third electron. Then r#" œ (3  1)# and r## œ (3
1)# Ê W" œ '3

5

œ a23 ‚ 10

b ˆ "4


#*

"‰ #



œ

23 4

23‚10
#* r#"

‚ 10

d3 œ '3


#*

5

23‚10
#* (3
")#

, and W# œ '

d3 œ 23 ‚ 10
#* ’ 3
" " “

23‚10
#* r## 3

&

œ 23 ‚ 10
#* ’ 3 " " “ œ a23 ‚ 10
#* b ˆ 6"  4" ‰ œ $


#* ‰
#* ‰ W œ W"
W# œ ˆ 23
ˆ 23 œ 4 ‚ 10 12 ‚ 10

23 3

5

23‚10
#* 12

d3 œ '

23‚10
#* # 3 (3  " ) 5

(3  2) œ

23 12

& $

d3

‚ 10
#* . Therefore

‚ 10
#* ¸ 7.67 ‚ 10
#* J

6.7 FLUID PRESSURES AND FORCES 1. To find the width of the plate at a typical depth y, we first find an equation for the line of the plate's right-hand edge: y œ x  5. If we let x denote the width of the right-hand half of the triangle at depth y, then x œ 5
y and the total width is L(y) œ 2x œ 2(5
y). The depth of the strip is (y). The force exerted by the c2

c2

water against one side of the plate is therefore F œ 'c5 w(y) † L(y) dy œ 'c5 62.4 † (y) † 2(5
y) dy c2

œ 124.8 'c5 a5y  y# b dy œ 124.8
 5# y#  "3 y$ ‘
& œ 124.8
ˆ 5# † 4

#

œ (124.8) ˆ 105 # 

117 ‰ 3

" 3

† 8‰  ˆ 5# † 25


" 3

† 125‰‘

œ (124.8) ˆ 315
6 234 ‰ œ 1684.8 lb

2. An equation for the line of the plate's right-hand edge is y œ x  3 Ê x œ y
3. Thus the total width is L(y) œ 2x œ 2(y
3). The depth of the strip is (2  y). The force exerted by the water is F œ 'c3 w(2  y)L(y) dy œ 'c3 62.4 † (2  y) † 2(3
y) dy œ 124.8'c3 a6  y  y# b dy œ 124.8 ’6y  0

0

œ (124.8) ˆ18 

9 #

0

y# #



! y$ 3 “
$

‰
9‰ œ (124.8) ˆ 27 # œ 1684.8 lb

3. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge is y œ x  3 Ê x œ y
3. Thus the total width is L(y) œ 2x œ 2(y
3). The depth of the strip changes to (4  y) Ê F œ 'c3 w(4  y)L(y) dy œ 'c3 62.4 † (4  y) † 2(y
3) dy œ 124.8'c3 a12
y  y# b dy 0

œ 124.8 ’12y


0

y# #



! y$ “ 3
$

œ (124.8) ˆ36


0

9 #

‰
9‰ œ (124.8) ˆ 45 # œ 2808 lb

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

407

408

Chapter 6 Applications of Definite Integrals

4. Using the coordinate system of Exercise 4, we see that the equation for the line of the plate's right-hand edge remains the same: y œ x  3 Ê x œ 3
y and L(y) œ 2x œ 2(y
3). The depth of the strip changes to (y) Ê F œ 'c3 w(y)L(y) dy œ 'c3 62.4 † (y) † 2(y
3) dy œ 124.8'c3 ay#  3yb dy œ 124.8 ’ y3  3# y# “ 0

0

œ (124.8) ˆ 27 3 

œ

27 ‰ #

0

(
124.8)(27)(2
3) 6

$

!
$

œ 561.6 lb

5. Using the coordinate system of Exercise 4, we find the equation for the line of the plate's right-hand edge to be y œ 2x  4 Ê x œ y # 4 and L(y) œ 2x œ y
4. The depth of the strip is (1  y). (a) F œ 'c4 w(1  y)L(y) dy œ 'c4 62.4 † (1  y)(y
4) dy œ 62.4 'c4 a4  3y  y# b dy œ 62.4 ’4y  0

0

œ (62.4) ’(4)(4) 

(3)(16) #

(b) F œ (64.0) ’(4)(4) 

0




(3)(16) #

64 3 “




œ (62.4) ˆ16  24


64 3 “

œ

(
64.0)(
120  64) 3

64 ‰ 3

œ

(
62.4)(
120  64) 3

3y# #



! y$ 3 “
%

œ 1164.8 lb

¸ 1194.7 lb

6. Using the coordinate system given, we find an equation for the line of the plate's right-hand edge to be y œ 2x
4 Ê x œ 4
#y and L(y) œ 2x œ 4  y. The depth of the strip is (1  y) Ê F œ '0 w(1  y)(4  y) dy 1

œ 62.4'0 ay#  5y
4b dy œ 62.4 ’ y3  1

$

œ (62.4) ˆ "3 

5 #

5y# #


4y“


4‰ œ (62.4) ˆ 2
156  24 ‰ œ

"

! (62.4)(11) 6

œ 114.4 lb

7. Using the coordinate system given in the accompanying figure, we see that the total width is L(y) œ 63 and the depth of the strip is (33.5  y) Ê F œ '0 w(33.5  y)L(y) dy 33

œ '0

33

64 1 #$

64 ‰ † (33.5  y) † 63 dy œ ˆ 12 (63)'0 (33.5  y) dy $ 33

$$ y# # “!

64 ‰ œ ˆ 12 (63) ’33.5y  $

œ

(64)(63)(33)(67
33) (#) a12$ b

‰ ’(33.5)(33)  œ ˆ 641#†63 $

33# # “

œ 1309 lb

8. (a) Use the coordinate system given in the accompanying ‰ figure. The depth of the strip is ˆ 11 6  y ft Ê F œ '0

11Î6

‰ w ˆ 11 6  y (width) dy

œ (62.4)(width)'0

11Î6

ˆ 11 ‰ 6  y dy

œ (62.4)(width) ’ 11 6 y

""Î' y# # “!

#

‰ † "# “ Ê Fend œ (62.4)(2) ˆ 121 ‰ ˆ "# ‰ ¸ 209.73 lb and Fside œ (62.4)(4) ˆ 121 ‰ ˆ "# ‰ ¸ 419.47 lb œ (62.4)(width) ’ˆ 11 6 36 36 (b) Use the coordinate system given in the accompanying figure. Find Y from the condition that the entire volume of the water is conserved (no spilling): 11 6 †2†4œ 2†2†Y 11 ‰ Ê Y œ 3 ft. The depth of a typical strip is ˆ 11 3  y ft and the total width is L(y) œ 2 ft. Thus, F œ '0

113

‰ w ˆ 11 3  y L(y) dy

11 ‰ œ '0 (62.4) ˆ 11 3  y † 2 dy œ (62.4)(2) ’ 3 y  113

""Î$ y# # “!

‰# “ œ œ (62.4)(2) ’ˆ "# ‰ ˆ 11 3

(62.4)(12") 9

force doubles.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

¸ 838.93 lb Ê the fluid

Section 6.7 Fluid Pressures and Forces 9. Using the coordinate system given in the accompanying figure, we see that the right-hand edge is x œ È1  y# so the total width is L(y) œ 2x œ 2È1  y# and the depth of the strip is (y). The force exerted by the water is therefore F œ 'c1 w † (y) † 2È1  y# dy 0

œ 62.4'c1 È1  y# d a1  y# b œ 62.4 ’ 23 a1  y# b 0

$Î# !

“


"

œ (62.4) ˆ 23 ‰ (1  0) œ 416 lb

10. Using the same coordinate system as in Exercise 15, the right-hand edge is x œ È3#  y# and the total width is L(y) œ 2x œ 2È9  y# . The depth of the strip is (y). The force exerted by the milk is therefore F œ 'c3 w † (y) † 2È9  y# dy œ 64.5'c3 È9  y# d a9  y# b œ 64.5 ’ 23 a9  y# b 0

0

$Î# !

“

œ (64.5)(18) œ 1161 lb


$

œ (64.5) ˆ 23 ‰ (27  0)

11. The coordinate system is given in the text. The right-hand edge is x œ Èy and the total width is L(y) œ 2x œ 2Èy. (a) The depth of the strip is (2  y) so the force exerted by the liquid on the gate is F œ '0 w(2  y)L(y) dy 1

" œ '0 50(2  y) † 2Èy dy œ 100 '0 (2  y)Èy dy œ 100'0 ˆ2y"Î#  y$Î# ‰ dy œ 100
43 y$Î#  25 y&Î# ‘ ! 1

1

1

‰ œ 100 ˆ 43  25 ‰ œ ˆ 100 15 (20  6) œ 93.33 lb

2‰ (b) We need to solve 160 œ '0 w(H  y) † 2Èy dy for h. 160 œ 100 ˆ 2H 3  5 Ê H œ 3 ftÞ 1

12. Use the coordinate system given in the accompanying figure. The total width is L(y) œ 1. (a) The depth of the strip is (3  1)  y œ (2  y) ft. The force exerted by the fluid in the window is F œ '0 w(2  y)L(y) dy œ 62.4 '0 (2  y) † 1 dy œ (62.4) ’2y  1

1

" y# # “!

œ (62.4) ˆ2  "# ‰ œ

(62.4)(3) #

œ 93.6 lb

(b) Suppose that H is the maximum height to which the tank can be filled without exceeding its design limitation. This means that the depth of a typical strip is (H  1)  y and the force is F œ '0 w[(H  1)  y]L(y) dy œ Fmax , where 1

Fmax œ 312 lb. Thus, Fmax œ w'0 [(H  1)  y] † 1 dy œ (62.4) ’(H  1)y 

" y# # “!

1

œ (62.4) ˆH  3# ‰

‰ (2H  3) œ 93.6
62.4H. Then Fmax œ 93.6
62.4H Ê 312 œ 93.6
62.4H Ê H œ œ ˆ 62.4 #

405.6 62.4

œ 6.5 ft

13. Suppose that h is the maximum height. Using the coordinate system given in the text, we find an equation for the line of the end plate's right-hand edge is y œ 5# x Ê x œ 25 y. The total width is L(y) œ 2x œ 45 y and the depth of the typical horizontal strip at level y is (h  y). Then the force is F œ '0 w(h  y)L(y) dy œ Fmax , h

where Fmax œ 6667 lb. Hence, Fmax œ w'0 (h  y) † 45 y dy œ (62.4) ˆ 45 ‰ ' ahy  y# b dy h

h

0

œ

# (62.4) ˆ 45 ‰ ’ hy#



h

y$ 3 “0

$

œ (62.4) ˆ 45 ‰ Š h# 

$

h 3

‰ œ $Ɉ 54 ‰ ˆ 6667 10.4 ¸ 9.288 ft. The volume of water which the tank can hold is V œ Height œ h and

" #

(Base) œ

2 5

$

max ‰ ‹ œ (62.4) ˆ 45 ‰ ˆ "6 ‰ h$ œ (10.4) ˆ 45 ‰ h$ Ê h œ Ɉ 54 ‰ ˆ F10.4

" #

(Base)(Height) † 30, where

h Ê V œ ˆ 25 h# ‰ (30) œ 12h# ¸ 12(9.288)# ¸ 1035 ft$ .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

409

410

Chapter 6 Applications of Definite Integrals

14. (a) After 9 hours of filling there are V œ 1000 † 9 œ 9000 cubic feet of water in the pool. The level of the water V is h œ Area , where Area œ 50 † 30 œ 1500 Ê h œ 9000 1500 œ 6 ft. The depth of the typical horizontal strip at level y is then (6  y) for the coordinate system given in the text. An equation for the drain plate's right-hand edge is y œ x Ê total width is L(y) œ 2x œ 2y. Thus the force against the drain plate is F œ '0 w(6  y)L(y) dy œ 62.4 '0 (6  y) † 2y dy œ (62.4)(2)'0 a6y  y# b œ (62.4)(2) ’ 6y#  1

1

1

#

œ (124.8) ˆ3  "3 ‰ œ (124.8) ˆ 83 ‰ œ 332.8 lb

" y$ 3 “!

(b) Suppose that h is the maximum height. Then, the depth of a typical strip is (h  y) and the force F œ '0 w(h  y)L(y) dy œ Fmax , where Fmax œ 520 lb. Hence, Fmax œ (62.4)'0 (h  y) † 2y dy 1

1

œ 124.8'0 ahy  y# b dy œ (124.8) ’ hy#  1

Êhœ

27 3

#

" y$ 3 “!

œ (124.8) ˆ h#  "3 ‰ œ (20.8)(3h  2) Ê

520 20.8

œ 3h  2

œ 9 ft

15. The pressure at level y is p(y) œ w † y Ê the average pressure is p œ œ

# ˆ wb ‰ Š b# ‹

œ

" b

'0b p(y) dy œ b" '0b w † y dy œ b" w ’ y# “ b #

0

wb #

. This is the pressure at level

b #

, which

is the pressure at the middle of the plate. 16. The force exerted by the fluid is F œ '0 w(depth)(length) dy œ '0 w † y † a dy œ (w † a)'0 y dy œ (w † a) ’ y# “ b

œ

# w Š ab# ‹

œ

ˆ wb ‰ # (ab)

b

b

#

b 0

œ p † Area, where p is the average value of the pressure (see Exercise 21).

17. When the water reaches the top of the tank the force on the movable side is 'c2 (62.4) ˆ2È4  y# ‰ (y) dy 0

œ (62.4)'c2 a4  y# b 0

"Î#

(2y) dy œ (62.4) ’ 23 a4  y# b

$Î# !

“


#

œ (62.4) ˆ 23 ‰ ˆ4$Î# ‰ œ 332.8 ft † lb. The force

compressing the spring is F œ 100x, so when the tank is full we have 332.8 œ 100x Ê x ¸ 3.33 ft. Therefore the movable end does not reach the required 5 ft to allow drainage Ê the tank will overflow. 18. (a) Using the given coordinate system we see that the total width is L(y) œ 3 and the depth of the strip is (3  y).

Thus, F œ '0 w(3  y)L(y) dy œ '0 (62.4)(3  y) † 3 dy 3

3

œ (62.4)(3)'0 (3  y) dy œ (62.4)(3) ’3y  3

$ y# # “!

œ (62.4)(3) ˆ9  9# ‰ œ (62.4)(3) ˆ 9# ‰ œ 842.4 lb

(b) Find a new water level Y such that FY œ (0.75)(842.4 lb) œ 631.8 lb. The new depth of the strip is (Y  y) and Y is the new upper limit of integration. Thus, FY œ '0 w(Y  y)L(y) dy Y

œ 62.4'0 (Y  y) † 3 dy œ (62.4)(3)'0 (Y  y) dy œ (62.4)(3) ’Yy  Y

Y

Y

y# # “0

œ (62.4)(3) ŠY# 

Y# # ‹

#

2FY È6.75 ¸ 2.598 ft. So, ?Y œ 3  Y œ (62.4)(3) Š Y# ‹ . Therefore, Y œ É (62.4)(3) œ É 1263.6 187.2 œ

¸ 3  2.598 ¸ 0.402 ft ¸ 4.8 in 19. Use a coordinate system with y œ 0 at the bottom of the carton and with L(y) œ 3.75 and the depth of a typical strip being (7.75  y). Then F œ '0

7Þ75

' ‰ w(7.75  y)L(y) dy œ ˆ 64.5 12$ (3.75) 0

7Þ75

‰ (7.75  y) dy œ ˆ 64.5 12$ (3.75) ’7.75y 

#

(7.75) ‰ œ ˆ 64.5 ¸ 4.2 lb 12$ (3.75) #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

(Þ(& y# # “!

Chapter 6 Practice Exercises

411

57 ‰ 20. The force against the base is Fbase œ pA œ whA œ w † h † (length)(width) œ ˆ 12 (10)(5.75)(3.5) ¸ 6.64 lb. $

To find the fluid force against each side, use a coordinate system with y œ 0 at the bottom of the can, so that the depth of a of ‰ of ‰ ˆ 57 ‰ ˆ width typical strip is (10  y): F œ '0 w(10  y) ˆ width the side dy œ 12$ the side ’10y  10

"! y# # “!

57 ‰ ˆ width of ‰ ˆ 100 ‰ 57 ‰ 57 ‰ œ ˆ 12 Ê Fend œ ˆ 12 (50)(3.5) ¸ 5.773 lb and Fside œ ˆ 12 (50)(5.75) ¸ 9.484 lb $ $ $ the side #

21. (a) An equation of the right-hand edge is y œ

x Ê xœ

3 #

2 3

y and L(y) œ 2x œ

4y 3

. The depth of the strip

is (3  y) Ê F œ '0 w(3  y)L(y) dy œ '0 (62.4)(3  y) ˆ 43 y‰ dy œ (62.4) † ˆ 43 ‰'0 a3y  y# b dy 3

3

$ y$ 3 “!

œ (62.4) ˆ 43 ‰ ’ 3# y# 

œ (62.4) ˆ 43 ‰
27 # 

3

27 ‘ 3

‰ œ (62.4) ˆ 34 ‰ ˆ 27 6 œ 374.4 lb

(b) We want to find a new water level Y such that FY œ

" #

(374.4) œ 187.2 lb. The new depth of the strip is

(Y  y), and Y is the new upper limit of integration. Thus, FY œ '0 w(Y  y)L(y) dy Y

œ 62.4'0 (Y  y) ˆ 43 y‰ dy œ (62.4) ˆ 43 ‰'0 aYy  y# b dy œ (62.4) ˆ 43 ‰ ’Y † Y

Y

œ (62.4) ˆ 29 ‰ Y$ . Therefore Y$ œ

9FY 2†(62.4)

œ

(9)(187.2) 124.8

y# #



Y

y$ 3 “!

$

œ (62.4) ˆ 34 ‰ Š Y2 

Y$ 3 ‹

$ $È Ê Y œ É (9)(187.2) 13.5 ¸ 2.3811 ft. So, 124.8 œ

?Y œ 3  Y ¸ 3  2.3811 ¸ 0.6189 ft ¸ 7.5 in. to the nearest half inch. (c) No, it does not matter how long the trough is. The fluid pressure and the resulting force depend only on depth of the water. ‰ 22. The area of a strip of the face of height ?y and parallel to the base is 100ˆ 26 24 † ?y, where the factor of

26 24

inclination of the face of the dam. With the origin at the bottom of the dam, the force on the face is then: ‰ F œ '0 w(24  y)a100bˆ 26 24 dy œ '('! ’#%y  24

#%

y# # “!

œ '('!Š#%# 

#%# # ‹

œ 1,946,880 lb.

CHAPTER 6 PRACTICE EXERCISES # 1. A(x) œ 14 (diameter)# œ 14 ˆÈx  x# ‰ œ 14 ˆx  2Èx † x#
x% ‰ ; a œ 0, b œ 1

Ê V œ 'a A(x) dx œ b

œ

1 4

œ

1 4†70

#

’ x#  74 x(Î#


È3 4

x& 5 “!

(35  40
14) œ

2. A(x) œ œ

1 4 "

" #

'01 ˆx  2x&Î#
x% ‰ dx 1 4

œ

ˆ "# 

4 7


5" ‰

91 280

(side)# ˆsin 13 ‰ œ

È3 4

ˆ2Èx  x‰#

ˆ4x  4xÈx
x# ‰ ; a œ 0, b œ 4

Ê V œ 'a A(x) dx œ

È3 4

b

œ

È3 4

œ

32È3 4

’2x#  85 x&Î#
ˆ1 

8 5

'04 ˆ4x  4x$Î#
x# ‰ dx

% x$ 3 “!


32 ‰ œ

8È 3 15

œ

È3 4

ˆ32 

8†32 5




(15  24
10) œ

64 ‰ 3

8È 3 15

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

accounts for the

412

Chapter 6 Applications of Definite Integrals

3. A(x) œ œ

1 4

1 4

(diameter)# œ

1 4

(2 sin x  2 cos x)#

† 4 asin# x  2 sin x cos x
cos# xb

œ 1(1  sin 2x); a œ

1 4

,bœ

51 4

Ê V œ 'a A(x) dx œ 1 '1Î4 (1  sin 2x) dx 51Î4

b

œ 1
x


cos 2x ‘ &1Î% # 1Î%

œ 1 ’Š 541


cos 5#1 #

cos 1# #

‹  Š 14 

‹“ œ 1 # #

#

%

4. A(x) œ (edge)# œ ŒŠÈ6  Èx‹  0 œ ŠÈ6  Èx‹ œ 36  24È6 Èx
36x  4È6 x$Î#
x# ; a œ 0, b œ 6 Ê V œ 'a A(x) dx œ '0 Š36  24È6 Èx
36x  4È6 x$Î#
x# ‹ dx b

6

œ ’36x  24È6 † 23 x$Î#
18x#  4È6 † 25 x&Î#
œ 216  576
648  5. A(x) œ

(diameter)# œ

1 4


72 œ 360 

Š2Èx 

x# 4‹

#

œ

1728 5

œ

1 4

œ 216  16 † È6 È6 † 6
18 † 6#  58 È6 È6 † 6#


1800
1728 5

Š4x  x&Î#


œ

6$ 3

72 5

x% 16 ‹ ;

a œ 0, b œ 4 Ê V œ 'a A(x) dx b

'04 Š4x  x&Î#
16x ‹ dx œ 14 ’2x#  27 x(Î#
5x†16 “ % œ 14 ˆ32  32 † 87
25 † 32‰ %

œ

1 4

œ

321 4

ˆ1 

6. A(x) œ œ

1 4

1728 5

' x$ 3 “!

È3 4

" #

8 7


25 ‰ œ

81 35

&

(35  40
14) œ

(edge)# sin ˆ 13 ‰ œ

È3 4

!

721 35


2Èx  ˆ2Èx‰‘#

ˆ4Èx‰# œ 4È3 x; a œ 0, b œ 1

Ê V œ 'a A(x) dx œ '0 4È3 x dx œ ’2È3 x# “ b

1

" !

œ 2È3

7. (a) .3=5 7/>29. :

V œ 'a 1R# (x) dx œ 'c1 1 a3x% b dx œ 1 'c1 9x) dx b

1

1

#

"

œ 1 cx* d
" œ 21

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x a3x% b dx œ 21 † 3'0 x& dx œ 21 † 3 ’ x6 “ œ 1 b

1

1

'

!

Note: The lower limit of integration is 0 rather than 1. (c) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'c1 (1  x) a3x% b dx œ 21 ’ 3x5  b

1

"

&

(d) A+=2/< 7/>29. :

" x' 2 “
"

œ 21
ˆ 35  "# ‰  ˆ 35  "# ‰‘ œ

R(x) œ 3, r(x) œ 3  3x% œ 3 a1  x% b Ê V œ 'a 1 cR# (x)  r# (x)d dx œ 'c1 1 ’9  9 a1  x% b “ dx b

1

œ 91 'c1 c1  a1  2x%
x) bd dx œ 91 'c1 a2x%  x) b dx œ 91 ’ 2x5  1

1

&

" x* 9 “
"

#

œ 181
25  9" ‘ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

21†13 5

œ

261 5

121 5

Chapter 6 Practice Exercises 8. (a) A+=2/< 7/>29. : R(x) œ

, r(x) œ

4 x$

" #

# # #
& Ê V œ 'a 1cR# (x)
r# (x)d dx œ '1 1 ’ˆ x4$ ‰
ˆ "# ‰ “ dx œ 1

16
x4 ‘ " 5 x b

2

"‰ " ˆ 16 " ‰‘ œ 1 ˆ
10 œ 1
ˆ 5
†16
32
#

5
4

(b) =2/66 7/>29. :

V œ 21'1 x ˆ x4$
"# ‰ dx œ 21 ’
4x
"
2

(c) =2/66 7/>29. :

" #

# x# 4 “"



 4" ‰ œ

16 5

1 20

(
2
10  64  5) œ

b

2

4 x


x

(d) A+=2/< 7/>29. :

# x# 4 “"

571 #0

œ 21
ˆ
4#
1‰
ˆ
4
4" ‰‘ œ 21 ˆ 54 ‰ œ

shell ‰ shell V œ 21'a ˆ radius Š height ‹ dx œ 21'1 (2
x) ˆ x4$
"# ‰ dx œ 21'1 ˆ x8$


œ 21 ’
x4# 

413

2

4 x#

51 #


1  x# ‰ dx

œ 21
(
1  2
2  1)
ˆ
4  4
1  4" ‰‘ œ

31 #

V œ 'a 1cR# (x)
r# (x)d dx b

# œ 1 '1 ’ˆ 7# ‰
ˆ4
2

dx

œ

491 4


161'1 a1
2x
$  x
' b dx

œ

491 4


161 ’x  x
#


œ

491 4 491 4 491 4


161
ˆ2  4"
5†"3# ‰
ˆ1  1
5" ‰‘ "
161 ˆ 4"
160  5" ‰

œ œ 9.

4 ‰# x$ “

2




161 160

# x
& 5 “"

(40
1  32) œ

491 4




711 10

1031 20

œ

(a) .3=5 7/>29. :

V œ 1 '1 ŠÈx
1‹ dx œ 1'1 (x
1) dx œ 1 ’ x#
x“ #

5

5

#

‰ ˆ" ‰‘ œ 1 ˆ 24 ‰ œ 1
ˆ 25 #
5
#
1 #
4 œ 81

& "

(b) A+=2/< 7/>29. :

R(y) œ 5, r(y) œ y#  1 Ê V œ 'c 1 cR# (y)
r# (y)d dy œ 1 'c2 ’25
ay#  1b “ dy d

2

œ 1'c2 a25
y%
2y#
1b dy œ 1 'c2 a24
y%
2y# b dy œ 1 ’24y
2

2

œ 321 ˆ3


2 5


"3 ‰ œ

321 15

(45
6
5) œ

10881 15

#

y& 5


23 y$ “

#
#

œ 21 ˆ24 † 2


(c) .3=5 7/>29. : R(y) œ 5
ay#  1b œ 4
y#

Ê V œ 'c 1R# (y) dy œ 'c2 1 a4
y# b dy d

2

#

œ 1 'c2 a16
8y#  y% b dy 2

œ 1 ’16y


8y$ 3

œ 641 ˆ1


2 3



# y& 5 “
#

 "5 ‰ œ

œ 21 ˆ32


641 15

64 3



(15
10  3) œ

32 ‰ 5 5121 15

10. (a) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21y Šy
d

4

œ 21'0 Šy#
4

œ

21 1#

† 64 œ

y$ 4‹

321 3

$

dy œ 21 ’ y3


%

y% 16 “ !

y# 4‹

dy

œ 21 ˆ 64 3


64 ‰ 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

32 5




2 3

† 8‰

414

Chapter 6 Applications of Definite Integrals

(b) =2/66 7/>29. : shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21x ˆ2Èx
x‰ dx œ 21'0 ˆ2x$Î#
x# ‰ dx œ 21 ’ 45 x&Î#
b

4

œ 21 ˆ 45 † 32


64 ‰ 3

œ

4

1281 15

% x$ 3 “!

(c) =2/66 7/>29. :

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21(4
x) ˆ2Èx
x‰ dx œ 21'0 ˆ8x"Î#
4x
2x$Î#  x# ‰ dx b

4

$Î# œ 21 ’ 16
2x#
54 x&Î#  3 x

œ 641 ˆ1
45 ‰ œ

641 5

4

% x$ 3 “!

œ 21 ˆ 16 3 † 8
32


(d) =2/66 7/>29. :

shell ‰ shell V œ 'c 21 ˆ radius Š height ‹ dy œ '0 21(4
y) Šy
d

4

œ 21'0 Š4y
2y#  4

y$ 4‹

y# 4‹ %

y% 16 “ !

dy œ 21 ’2y#
23 y$ 

4 5

† 32 

64 ‰ 3

œ 641 ˆ 34
1


dy œ 21'0 Š4y
y#
y#  4

œ 21 ˆ32


2 3

4 5

y$ 4‹

 32 ‰

dy

† 64  16‰ œ 321 ˆ2


321 3

 1‰ œ

8 3

11. .3=5 7/>29. : R(x) œ tan x, a œ 0, b œ

1 3

Ê V œ 1 '0 tan# x dx œ 1'0 asec# x
1b dx œ 1[tan x
x]! 1Î3

12. .3=5 7/>29. :

1Î3

1Î$

V œ 1'0 (2
sin x)# dx œ 1 '0 a4
4 sin x  sin# xb dx œ 1'0 ˆ4
4 sin x  1

œ 1
4x  4 cos x 

1

x #




sin 2x ‘ 1 4 !

1

œ 1
ˆ41
4 

1 #


0‰
(0  4  0
0)‘ œ

œ

1
cos 2x ‰ dx # 9 1 1 ˆ #
8‰ œ 1#

1 Š3È3
1‹ 3

(91
16)

13. (a) .3=5 7/>29. :

V œ 1'0 ax#
2xb dx œ 1'0 ax%
4x$  4x# b dx œ 1 ’ x5
x%  43 x$ “ œ 1 ˆ 32 5
16  2

œ

161 15

2

#

(6
15  10) œ

#

&

!

161 15

32 ‰ 3

(b) A+=2/< 7/>29. :

V œ '0 1’1#
ax#
2x  "b “ dx œ '0 1 dx  '0 1 ax
"b% dx œ #1
’1 2

2

#

2

(c) =2/66 7/>29. :

# ax
"b& & “!

œ #1
1 †

# &

œ

)1 &

shell ‰ shell V œ 'a 21 ˆ radius Š height ‹ dx œ 21'0 (2
x) c
ax#
2xbd dx œ 21'0 (2
x) a2x
x# b dx b

2

2

œ 21'0 a4x
2x#
2x#  x$ b dx œ 21'0 ax$
4x#  4xb dx œ 21 ’ x4
43 x$  2x# “ œ 21 ˆ4
2

œ

21 3

2

(36
32) œ

#

%

!

81 3

32 3

 8‰

(d) A+=2/< 7/>29. :

V œ 1 '0 c2
ax#
2xbd dx
1'0 2# dx œ 1'0 ’4
4 ax#
2xb  ax#
2xb “ dx
81 2

2

#

2

#

œ 1'0 a4
4x#  8x  x%
4x$  4x# b dx
81 œ 1'0 ax%
4x$  8x  4b dx
81 2

&

2

#

‰ œ 1 ’ x5
x%  4x#  4x“
81 œ 1 ˆ 32 5
16  16  8
81 œ !

1 5

(32  40)
81 œ

721 5

14. .3=5 7/>29. :

V œ 21'0 4 tan# x dx œ 81'0 asec# x
1b dx œ 81[tan x
x]! 1Î4

1Î4

1Î%

œ 21(4
1)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley




401 5

œ

321 5

Chapter 6 Practice Exercises 15. The material removed from the sphere consists of a cylinder and two "caps." From the diagram, the height of the cylinder #

is 2h, where h#  ŠÈ$‹ œ ## , i.e. h œ ". Thus #

Vcyl œ a#hb1ŠÈ$‹ œ '1 ft$ . To get the volume of a cap, use the disk method and x#  y# œ ## : Vcap œ '" 1x# dy 2

œ '" 1a%
y# bdy œ 1’%y
2

œ 1
ˆ8
83 ‰
ˆ%
"3 ‰‘ œ

# y3 3 “"

&1 3

ft$ . Therefore,

Vremoved œ Vcyl  #Vcap œ '1 

"!1 3

#)1 3

œ

ft$ .

16. We rotate the region enclosed by the curve y œ É12 ˆ1


4x# ‰ 121

and the x-axis around the x-axis. To find the

11Î2

volume we use the .3=5 method: V œ 'a 1R# (x) dx œ 'c11Î2 1 ŠÉ12 ˆ1
b

œ 121'c11Î2 Š1
11Î2

4x# 121 ‹

œ 1321 ˆ1
"3 ‰ œ 17. y œ x"Î#


x$Î# 3

Ê


ˆ4 

" #

œ

#

x
"Î#
"# x"Î# Ê Š dy dx ‹ œ

" 4

œ '1

8

dx dy

È9x#Î$  4 3x"Î$

œ

5 12

2 3

#

4

4

ˆ2 

14 ‰ 3

#

x
"Î$ Ê Š dx dy ‹ œ

œ

4x
#Î$ 9

" #

ˆx
"Î#  x"Î# ‰ dx œ

" #


2x"Î#  23 x$Î# ‘ % "

10 3 dx Ê L œ '1 Ê1  Š dy ‹ dy œ '1 É1  #

8

8

4 9x#Î$

dy

'1 È9x#Î$  4 ˆx
"Î$ ‰ dx;
u œ 9x#Î$  4 Ê du œ 6y
"Î$ dy; x œ 1 40 " ' "
2 $Î# ‘ %! Ä L œ 18 u"Î# du œ 18 œ #"7
40$Î#
13$Î# ‘ ¸ 7.634 3 u "$ 13

dx œ

x œ 8 Ê u œ 40d 19. y œ

" #

† 8‰
ˆ2  23 ‰‘ œ

18. x œ y#Î$ Ê

8

" 3

x'Î&
58 x%Î& Ê

dx

ˆ x"
2  x‰ Ê L œ ' É1  4" ˆ x"
2  x‰ dx 1

4

2 3

4x# 121 ‹

œ 881 ¸ 276 in$

4

" #

11Î2

dx œ 1 '
11Î2 12 Š1


4 ‰ 11 ˆ 4 ‰ ˆ 11 ‰ “ œ 1321 ’1
ˆ 363 œ 241 ’ 11 Š 4 ‹“ 2
363 #

# Ê L œ '1 É 4" ˆ x"  2  x‰ dx œ '1 É 4" ax
"Î#  x"Î# b dx œ '1

œ

#

$

4x$ 363 “
""Î#

dx œ 121 ’x


2641 3

dy dx

""Î#

4x# ‰ 121 ‹

#

" #

œ

dy dx

x"Î&
"# x
"Î& Ê Š dy dx ‹ œ

" 4

Ê u œ 13,

ˆx#Î&
2  x
#Î& ‰

# Ê L œ '1 É1  4" ax#Î&
2  x
#Î& b dx Ê L œ '1 É 4" ax#Î&  2  x
#Î& b dx œ ' É 4" ax"Î&  x
"Î& b dx 32

32

32

1

œ '1

32

œ

" 48

20. x œ

" #

ˆx

"Î&

x


"Î& ‰

(1260  450) œ

" 1#

y$ 

" y

Ê

" % œ '1 É 16 y  2

" #

dx œ œ

1710 48

dx dt



5 4

$# x%Î& ‘ "

#

œ

" 4



" y%

dy œ '1 ÊŠ 4" y# 

y#


" y#

x

'Î&

dx dy

8 œ ˆ 12
"# ‰
ˆ 1"#
1‰ œ

21.

"
5 # 6 285 8

dx Ê Š dy ‹ œ

2

7 1#

œ
5 sin t  5 sin 5t and



dy dt

" #

œ

" y# ‹

" 16 #

œ

" #


ˆ 65

y%


" #

'

†2 



" y%

5 4




ˆ 56



5 ‰‘ 4

œ

" #

ˆ 315 6 

" % Ê L œ '1 Ê1  Š 16 y


dy œ '1 Š 4" y#  2

†2

%‰

2

" y# ‹

dy œ ’ 1"# y$
y" “

" #



75 ‰ 4

" y% ‹

dy

# "

13 12

#

‰  Š dy œ 5 cos t
5 cos 5t Ê Êˆ dx dt dt ‹

#

œ Éa
5 sin t  5 sin 5tb#  a5 cos t
5 cos 5tb# œ 5Èsin# 5t
#sin t sin 5t  sin# t  cos# t
#cos t cos 5t  cos# 5t œ &È#
#asin t sin 5t  cos t cos 5 tb œ 5È#a"
cos %tb œ 5É%ˆ "# ‰a"
cos %tb œ "!Èsin# #t œ "!lsin #tl œ "!sin #t (since ! Ÿ t Ÿ 1# ) Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

415

416

Chapter 6 Applications of Definite Integrals Ê Length œ '!

1 Î2

22.

dx dt

œ 3t2
12t and

1Î#

"!sin #t dt œ c
5 cos #td ! dy dt

œ a
&ba
"b
a
&ba"b œ "! #

#

‰  Š dy Éa3t2
12tb#  a3t2  12tb# œ È288t#  "8t4 œ 3t2  12t Ê Êˆ dx dt dt ‹ œ

œ 3È2 ktkÈ16  t2 Ê Length œ '! 3È2 ktkÈ16  t2 dt œ 3È2'! t È16  t2 dt; ’u œ 16  t2 Ê du œ 2t dt "

"

3È 2 2

Ê "# du œ t dt; t œ 0 Ê u œ 16; t œ 1 Ê u œ 17“; œ

23.

dx d)

3È 2 2

† 23 Ša17b3/2
64‹ œ È2Ša17b3/2
64‹ ¸ 8.617.

œ
$ sin ) and

Ê Length œ '!

dy d)

$1Î2

#

24. x œ t and y œ œ'

'16"7 Èu du œ 3È2 2
23 u3/2 ‘1617 œ 3È2 2 Š 23 a17b3/2
23 a16b3/2 ‹

t$ 3

$ d) œ $'!

$1Î2

d) œ $ˆ $#1
!‰ œ


t,
È3 Ÿ t Ÿ È3 Ê

È3


È 3

#

#

‰  Š dy Éa
$ sin )b#  a$ cos )b# œ È$asin# )  cos# )b œ $ œ $ cos ) Ê Êˆ dx d) d) ‹ œ

Èt%  #t#  " dt œ

'

dx dt

*1 #

œ 2t and

dy dt

È3


È 3

Èt%  2t#  " dt œ

œt

'

#


" Ê Length œ '

È3


È 3

È3


È 3

Éat#  "b# dt œ

Éa2tb#  at#
"b# dt

È

'
È33 at#  "b dt œ ’ t3  t“ 3

È3
È 3

œ 4È3 25. Intersection points: 3
x# œ 2x# Ê 3x#
3 œ 0 Ê 3(x
1)(x  1) œ 0 Ê x œ
1 or x œ 1. Symmetry suggests that x œ 0. The typical @/<>3-+6 strip has # # # center of mass: (µ x ßµ y ) œ Šxß 2x  a3
x b ‹ œ Šxß x  3 ‹ , #

#

#

#

#

length: a3
x b
2x œ 3 a1
x b, width: dx, area: dA œ 3 a1
x# b dx, and mass: dm œ $ † dA œ 3$ a1
x# b dx Ê the moment about the x-axis is µ y dm œ œ

3 #

3 #

$ ax#  3b a1
x# b dx œ &

$ ’
x5


œ 3$ ’x


2x$ 3

" x$ 3 “
"

 3x“

"
"

3 #

$ a
x%
2x#  3b dx Ê Mx œ ' µ y dm œ

œ 3$ ˆ
5"


2 3

 3‰ œ

œ 6$ ˆ1
"3 ‰ œ 4$ Ê y œ

Mx M

œ

3$ 15

(
3
10  45) œ

32$ 5 †4 $

œ

8 5

32$ 5

3 #

$ 'c1 a
x%
2x#  3b dx "

; M œ ' dm œ 3$ 'c1 a1
x# b dx "

. Therefore, the centroid is (xß y) œ ˆ!ß 85 ‰ .

26. Symmetry suggests that x œ 0. The typical @/<>3-+6 # strip has center of mass: (µ x ßµ y ) œ Šxß x# ‹ , length: x# , width: dx, area: dA œ x# dx, mass: dm œ $ † dA œ $ x# dx Ê the moment about the x-axis is µ y dm œ #$ x# † x# dx x% dx Ê Mx œ ' µ y dm œ

œ

$ #

œ

2$ 10

a2& b œ

32$ 5

$ #

; M œ ' dm œ $

'c22 x% dx œ 10$ cx& d #
# 'c22 x# dx œ $ ’ x3 “ # $


#

œ

2$ 3

a2$ b œ

16$ 3

Ê yœ

Mx M

œ

32†$ †3 5†16†$

œ

centroid is(xß y) œ ˆ!ß 65 ‰ .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

6 5

. Therefore, the

Chapter 6 Practice Exercises

417

27. The typical @/<>3-+6 strip has: center of mass: (µ x ßµ y ) œ Œxß

#

4  x4 #

 , length: 4


area: dA œ Š4
œ $ Š4


x# 4‹

x# 4 ‹dx,

width: dx,

mass: dm œ $ † dA

dx Ê the moment about the x-axis is #

Š4  x4 ‹

µ y dm œ $ †

x# 4,

x# 4‹

Š4


#

$ #

dx œ

moment about the y-axis is µ x dm œ $ Š4
œ

$ 2

’16x


% x& 5†16 “ !

$ #

œ


64


#

œ

16†$ †3 32†$

œ

Mx M

and y œ

3 2

œ

dx. Thus, Mx œ ' µ y dm œ

4

4

My M

x$ 4‹

‹ † x dx œ $ Š4x


x 4

œ $ (32
16) œ 16$ ; M œ ' dm œ $ '0 Š4
Ê xœ

dx; the

; My œ ' µ x dm œ $ '0 Š4x


128$ 5

œ

64 ‘ 5

x% 16 ‹

Š16


128†$ †3 5†32†$

x# 4‹

œ

dx œ $ ’4x


12 5

% x$ 12 “ !

x$ 4‹

dx œ $ ’2x#


œ $ ˆ16


64 ‰ 1#

$ #

'04 Š16
16x ‹ dx %

% x% 16 “ !

32$ 3

œ

‰ . Therefore, the centroid is (xß y) œ ˆ 3# ß 12 5 .

28. A typical 29<3D98>+6 strip has: # center of mass: (µ x ßµ y ) œ Š y # 2y ß y‹ , length: 2y
y# , width: dy, area: dA œ a2y
y# b dy, mass: dm œ $ † dA œ $ a2y
y# b dy; the moment about the x-axis is µ y dm œ $ † y † a2y
y# b dy œ $ a2y#
y$ b ; the moment # about the y-axis is µ x dm œ $ † ay  2yb † a2y
y# b dy #

œ a4y
y b dy Ê Mx œ ' µ y dm œ $ '0 a2y#
y$ b dy $ #

#

œ $ ’ 23 y$
œ

$ #

ˆ 43†8

yœ

Mx M

2

%




œ

# y% 4 “!

32 ‰ 5

4†$ †3 3†4†$

œ

œ $ ˆ 23 † 8
32$ 15

16 ‰ 4

œ $ ˆ 16 3


16 ‰ 4

œ

$ †16 12

œ

4$ 3

; My œ ' µ x dm œ

$ #

'02 a4y#
y% b dy œ #$ ’ 34 y$
y5 “ # &

$ ; M œ ' dm œ $ '0 a2y
y# b dy œ $ ’y#
y3 “ œ $ ˆ4
83 ‰ œ

#

2

!

!

4$ 3

Ê xœ

My M

œ

$ †32†3 15†$ †4

œ

œ 1. Therefore, the centroid is (xß y) œ ˆ 85 ß 1‰ .

29. A typical horizontal strip has: center of mass: (µ x ßµ y ) œ Šy

#

 2y # ß y‹ ,

length: 2y
y# , width: dy,

area: dA œ a2y
y# b dy, mass: dm œ $ † dA œ (1  y) a2y
y# b dy Ê the moment about the x-axis is µ y dm œ y(1  y) a2y
y# b dy # œ a2y  2y$
y$
y% b dy œ a2y#  y$
y% b dy; the moment about the y-axis is # µ x dm œ Š y  2y ‹ (1  y) a2y
y# b dy œ " a4y#
y% b (1  y) dy œ #

Ê Mx œ ' µ y dm œ '0 a2y#  y$
y% b dy œ ’ 23 y$  2

œ

16 60

œ

" #

" #

#

(20  15
24) œ $

Š 4†32  2%


2& 5




4 15

(11) œ

2' 6‹

‰ ˆ 38 ‰ œ œ ˆ 44 15

44 40

œ

11 10

y$ 3




; My œ ' µ x dm œ '0

2

œ 4 ˆ 43  2


œ '0 a2y  y#
y$ b dy œ ’y#  2

44 15

y% 4




4 5

" #

# y& 5 “!

œ ˆ4 

8 3




16 ‰ 4

œ ˆ 16 3 

œ

8 3

24 5

area: dA œ

x$Î#




32 ‰ 5

œ 16 ˆ "3  " #

dx, mass: dm œ $ † dA œ $ †

x$Î#


25 ‰

’ 43 y$  y%


y& 5

2

Ê xœ

My M

‰ ˆ 83 ‰ œ œ ˆ 24 5

9 5

‰ . Therefore, the center of mass is (xß y) œ ˆ 95 ß 11 10 .

3

" 4

; M œ ' dm œ '0 (1  y) a2y
y# b dy

30. A typical vertical strip has: center of mass: (µ x ßµ y ) œ ˆxß 2x3$Î# ‰ , length: 3

16 4

a4y#  4y$
y%
y& b dy œ


86 ‰ œ 4 ˆ2
45 ‰ œ

# y% 4 “!

a4y#  4y$
y%
y& b dy

3 x$Î#

, width: dx,

dx Ê the moment about the x-axis is

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

and y œ

Mx M




# y' 6 “!

8 5

and

418

Chapter 6 Applications of Definite Integrals

µ y dm œ

9$ 2x$

3 † $ x$Î# dx œ

3 #x$Î#

(a) Mx œ $ '1

9

M œ $ '1

9

(b) Mx œ '1

9

" #

9$ #

ˆ x9$ ‰ dx œ


#

*

20$ 9

’
x# “ œ "

*

ˆ x9$ ‰ dx œ

9 #

*

dx.

*

My M

œ

12$ 4$

œ 3 and y œ

Mx M

œ

ˆ 209$ ‰ 4$

œ

5 9

* 3 ‰ 3 ‰

"x ‘ * œ 4; My œ ' x# ˆ $Î# dx œ
2x$Î# ‘ " œ 52; M œ '1 x ˆ x$Î# dx " x 1 9

œ 6
x"Î# ‘ " œ 12 Ê x œ 31. S œ 'a 21y Ê1  Š dy dx ‹ dx; #

b

3$ x"Î#

3 ‰ ; My œ $ '1 x ˆ x$Î# dx œ 3$
2x"Î# ‘ " œ 12$ ; 9

dx œ
6$
x
"Î# ‘ " œ 4$ Ê x œ

3 x$Î# x #

3 dx; the moment about the y-axis is µ x dm œ x † $ x$Î# dx œ

My M

dy dx

œ

œ

13 3

and y œ

Mx M

9

œ

" 3

#

" È2x  1

Ê Š dy dx ‹ œ

Ê S œ '0 21È2x  1 É1  3

" #x  1

" #x  1

dx

2 È ' Èx  1 dx œ 2È21
2 (x  1)$Î# ‘ $ œ 2È21 † 2 (8
1) œ œ 21'0 È2x  1 É 2x 2x1 dx œ 2 21 0 3 3 ! 3

3

32. S œ 'a 21y Ê1  Š dy dx ‹ dx; #

b

œ

1 6

dy dx

% ' œ x# Ê Š dy dx ‹ œ x Ê S œ 0 21 † #

1

x$ 3

È1  x% dx œ

1 6

281È2 3

'01 È1  x% a4x$ b dx

'01 È1  x% d a1  x% b œ 16 ’ 32 a1  x% b$Î# “ " œ 19 ’2È2
1“ !

33. S œ 'c 21x Ê1  Š dx dy ‹ dy; #

d

dx dy

œ

ˆ "# ‰ (4
2y) È4y
y#

2
y È4y
y#

œ

#

Ê 1  Š dx dy ‹ œ

4y
y#  4
4y  y# 4y
y#

œ

4 4y
y#

Ê S œ '1 21 È4y
y# É 4y
4 y# dy œ 41'1 dx œ 41 2

2

34. S œ 'c 21x Ê1  Š dx dy ‹ dy; #

d

œ 1'2 È4y  1 dy œ 6

35. x œ

t# #

1 4

dx dy

œ

*

36. x œ t# 

" 2t

761 3

" È2

1

œ 21 Š2


(125
27) œ

œ t and

ŸtŸ1 Ê

Ê Surface Area œ '1ÎÈ2 21 ˆt#  1

dx dt

1 6

dy dt

1 6

" 4y

œ

Ê S œ '2 21Èy † 6

4y  1 4y

(98) œ

È4y  1 È4y

dy

491 3

È5

œ 2 Ê Surface Area œ '0 21(2t)Èt#  4 dt œ '4 21u"Î# du 9

, where u œ t#  4 Ê du œ 2t dt; t œ 0 Ê u œ 4, t œ È5 Ê u œ 9

and y œ 4Èt ,

œ 21 '1ÎÈ2 ˆt# 

#

Ê 1  Š dx dy ‹ œ 1 


23 (4y  1)$Î# ‘ ' œ #

and y œ 2t, 0 Ÿ t Ÿ È5 Ê

œ 21
23 u$Î# ‘ % œ

1 2È y

" ‰ˆ 2t 2t

" ‰ 2t#



"‰ #t

dx dt

œ 2t


ʈ2t


" ‰# 2t#

dt œ 21 '1ÎÈ2 ˆ2t$  1

" 2t#

and

dy dt

œ

2 Èt

 Š È2 t ‹ dt œ 21 '1ÎÈ2 ˆt#  #

3 #

1

" ‰ Ɉ 2t #t



" ‰# #t#

dt

"

 4" t
$ ‰ dt œ 21
2" t%  3# t
8" t
# ‘ "ÎÈ#

3È 2 4 ‹

37. The equipment alone: the force required to lift the equipment is equal to its weight Ê F" (x) œ 100 N.

The work done is W" œ 'a F" (x) dx œ '0 100 dx œ [100x]%! ! œ 4000 J; the rope alone: the force required b

40

to lift the rope is equal to the weight of the rope paid out at elevation x Ê F# (x) œ 0.8(40
x). The work done is W# œ 'a F# (x) dx œ '0 0.8(40
x) dx œ 0.8 ’40x
b

40

the total work is W œ W"  W# œ 4000  640 œ 4640 J

%! x# # “!

œ 0.8 Š40#


40# # ‹

œ

(0.8)(1600) #

œ 640 J;

38. The force required to lift the water is equal to the water's weight, which varies steadily from 8 † 800 lb to 8 † 400 lb over the 4750 ft elevation. When the truck is x ft off the base of Mt. Washington, the water weight is
x‰ F(x) œ 8 † 800 † ˆ 2†24750 œ (6400) ˆ1
†4750

x ‰ 9500

lb. The work done is W œ 'a F(x) dx b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 6 Practice Exercises œ '0

4750

6400 ˆ1


x ‰ 9500

dx œ 6400 ’x


œ 22,800,000 ft † lb

%(&! x# 2†9500 “ !

œ 6400 Š4750


4750# 4†4750 ‹

419

œ ˆ 34 ‰ (6400)(4750)

39. Force constant: F œ kx Ê 20 œ k † 1 Ê k œ 20 lb/ft; the work to stretch the spring 1 ft is W œ '0 kx dx œ k'0 x dx œ ’20 x# “ œ 10 ft † lb; the work to stretch the spring an additional foot is 1

1

#

! # x# 20 ’ # “ "

W œ '1 kx dx œ k '1 x dx œ 2

"

2

œ 20 ˆ 4#
"# ‰ œ 20 ˆ 3# ‰ œ 30 ft † lb

40. Force constant: F œ kx Ê 200 œ k(0.8) Ê k œ 250 N/m; the 300 N force stretches the spring x œ œ

300 250

F k

œ 1.2 m; the work required to stretch the spring that far is then W œ '0 F(x) dx œ '0 250x dx 1Þ2

1Þ2

œ [125x# ]!"Þ# œ 125(1.2)# œ 180 J 41. We imagine the water divided into thin slabs by planes perpendicular to the y-axis at the points of a partition of the interval [0ß 8]. The typical slab between the planes at y and y  ?y has a volume of about ?V œ 1(radius)# (thickness) # œ 1 ˆ 54 y‰ ?y œ 25161 y# ?y ft$ . The force F(y) required to lift this slab is equal to its weight: F(y) œ 62.4 ?V œ

(62.4)(25) 16

1y# ?y lb. The distance through which F(y)

must act to lift this slab to the level 6 ft above the top is about (6  8
y) ft, so the work done lifting the slab is about ?W œ

(62.4)(25) 16

1y# (14
y) ?y ft † lb. The work done

lifting all the slabs from y œ 0 to y œ 8 to the level 6 ft above the top is approximately 8

W¸! !

(62.4)(25) 16

1y# (14
y) ?y ft † lb so the work to pump the water is the limit of these Riemann sums as the norm of

the partition goes to zero: W œ '0

8

œ

(62.4) ˆ 25161 ‰ Š 14 3

$

†8


8% 4‹

(62.4)(25) (16)

1y# (14
y) dy œ

(62.4)(25)1 16

'08 a14y#
y$ b dy œ (62.4) ˆ 25161 ‰ ’ 143 y$
y4 “ ) %

!

¸ 418,208.81 ft † lb

42. The same as in Exercise 41, but change the distance through which F(y) must act to (8
y) rather than (6  8
y). Also change the upper limit of integration from 8 to 5. The integral is: W œ '0

5

(62.4)(25)1 16

y# (8
y) dy œ (62.4) ˆ 25161 ‰'0 a8y#
y$ b dy œ (62.4) ˆ 25161 ‰ ’ 83 y$
5

œ (62.4) ˆ 25161 ‰ Š 38 † 5$


5% 4‹

& y% 4 “!

¸ 54,241.56 ft † lb

43. The tank's cross section looks like the figure in Exercise 41 with right edge given by x œ #

horizontal slab has volume ?V œ 1(radius)# (thickness) œ 1 ˆ #y ‰ ?y œ slab is its weight: F(y) œ 60 †

1 4

1 4

10

22,500 ft†lb 275 ft†lb/sec

y œ y# . A typical

y# ?y. The force required to lift this

y# ?y. The distance through which F(y) must act is (2  10
y) ft, so the

work to pump the liquid is W œ 60'0 1(12
y) Š y4 ‹ dy œ 151 ’ 12y 3
to empty the tank is

5 10

#

$

"! y% 4 “!

œ 22,5001 ft † lb; the time needed

¸ 257 sec

44. A typical horizontal slab has volume about ?V œ (20)(2x)?y œ (20) ˆ2È16
y# ‰ ?y and the force required to lift this slab is its weight F(y) œ (57)(20) ˆ2È16
y# ‰ ?y. The distance through which F(y) must act is (6  4
y) ft, so the work to pump the olive oil from the half-full tank is

W œ 57'c4 (10
y)(20) ˆ2È16
y# ‰ dy œ 2880 'c4 10È16
y# dy  1140'c4 a16
y# b 0

0

0

"Î#

(
2y) dy

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

420

Chapter 6 Applications of Definite Integrals œ 22,800 † (area of a quarter circle having radius 4)  23 (1140) ’a16
y# b

$Î# !

“

œ 335,153.25 ft † lb


%

œ (22,800)(41)  48,640

strip 45. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 2 '0 (62.4)(2
y)(2y) dy œ 249.6'0 a2y
y# b dy œ 249.6 ’y#
b

2

2

# y$ 3 “!

œ (249.6) ˆ4
83 ‰ œ (249.6) ˆ 43 ‰ œ 332.8 lb strip 46. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ '0 75 ˆ 56
y‰ (2y  4) dy œ 75'0 ˆ 53 y  5Î6

b

5Î6

10 3


2y#
4y‰ dy

7 7 # 2 $ ‘ &Î' 50 ‰ 25 ‰ 125 ‰‘ #‰
ˆ 18 œ 75 '0 ˆ 10 dy œ 75
10
ˆ 67 ‰ ˆ 36
ˆ 32 ‰ ˆ 216 3
3 y
2y 3 y
6 y
3 y ! œ (75) 5Î6

œ (75) ˆ 25 9


175 216




250 ‰ 3†#16

‰ œ ˆ 9†75 #16 (25 † 216
175 † 9
250 † 3) œ

strip 47. F œ 'a W † Š depth ‹ † L(y) dy Ê F œ 62.4'0 (9
y) Š2 † b

4

%

œ 62.4
6y$Î#
25 y&Î# ‘ ! œ (62.4) ˆ6 † 8


2 5

Èy 2 ‹

(75)(3075) 9†#16

¸ 118.63 lb.

dy œ 62.4'0 ˆ9y"Î#
3y$Î# ‰ dy 4

‰ (48 † 5
64) œ † 32‰ œ ˆ 62.4 5

(62.4)(176) 5

œ 2196.48 lb

strip 48. Place the origin at the bottom of the tank. Then F œ '0 W † Š depth ‹ † L(y) dy, h œ the height of the mercury column, h

strip depth œ h
y, L(y) œ 1 Ê F œ '0 849(h
y) " dy œ (849)'0 (h
y) dy œ 849’hy
h

œ

849 # 2 h .

Now solve

849 # 2 h

h

h

y# # “!

œ 849 Šh#


h# #‹

œ 40000 to get h ¸ 9.707 ft. The volume of the mercury is s2 h œ 12 † 9.707 œ 9.707 ft$ Þ

49. F œ w" '0 (8
y)(2)(6
y) dy  w# 'c6 (8
y)(2)(y  6) dy œ 2w" '0 a48
14y  y# b dy  2w# '
6 a48  2y
y# b dy 6

0

œ 2w" ’48y
7y# 

' y$ 3 “!

6

 2w# ’48y  y#


! y$ 3 “
'

0

œ 216w"  360w#

50. (a) F œ 62.4'0 (10
y)
ˆ8
y6 ‰
ˆ y6 ‰‘ dy 6

œ

6

62.4 3

' a240
34y  y# b dy 0

œ

62.4 3

’240y
17y# 

œ 18,720 lb.

' y$ 3 “!

œ

62.4 3

(1440
612  72)

(b) The centroid ˆ 72 ß 3‰ of the parallelogram is located at the intersection of y œ

6 7

x and y œ
65 x 

36 5 .

The centroid of

the triangle is located at (7ß 2). Therefore, F œ (62.4)(7)(36)  (62.4)(8)(6) œ (300)(62.4) œ 18,720 lb CHAPTER 6 ADDITIONAL AND ADVANCED EXERCISES 1. V œ 1 'a cf(x)d# dx œ b#
ab Ê 1'a cf(t)d# dt œ x#
ax for all x  a Ê 1 [f(x)]# œ 2x
a Ê f(x) œ „ É 2x1
a b

x

2. V œ 1 '0 [f(x)]# dx œ a#  a Ê 1 '0 [f(t)]# dt œ x#  x for all x  a Ê 1[f(x)]# œ 2x  1 Ê f(x) œ „ É 2x1 1 a

x

3. s(x) œ Cx Ê '0 È1  [f w (t)]# dt œ Cx Ê È1  [f w (x)]# œ C Ê f w (x) œ ÈC#
1 for C   1 x

Ê f(x) œ '0 ÈC#
1 dt  k. Then f(0) œ a Ê a œ 0  k Ê f(x) œ '0 ÈC#
1 dt  a Ê f(x) œ xÈC#
1  a, x

x

where C   1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 6 Additional and Advanced Exercises

421

4. (a) The graph of f(x) œ sin x traces out a path from (!ß !) to (!ß sin !) whose length is L œ '0 È1  cos# ) d). !

The line segment from (0ß 0) to (!ß sin !) has length È(!
0)#  (sin !
0)# œ È!#  sin# !. Since the shortest distance between two points is the length of the straight line segment joining them, we have ! immediately that ' È1  cos# ) d)  È!#  sin# ! if 0  ! Ÿ 1 . #

0

(b) In general, if y œ f(x) is continuously differentiable and f(0) œ 0, then '0 È1  [f w (t)]# dt  È!#  f # (!) !

for !  0. 5. From the symmetry of y œ 1
xn , n even, about the y-axis for
1 Ÿ x Ÿ 1, we have x œ 0. To find y œ MMx , we n use the vertical strips technique. The typical strip has center of mass: (µ x ßµ y ) œ ˆxß 1
2 x ‰ , length: 1
xn , width: dx, area: dA œ a1
xn b dx, mass: dm œ 1 † dA œ a1
xn b dx. The moment of the strip about the 1 1 n # n # " nb1 2n b 1 x-axis is µ y dm œ a1
x b dx Ê M œ ' a1
x b dx œ 2' " a1
2xn  x2n b dx œ
x
2x  x ‘ #

œ1


2 n1



" #n  1

œ

x c1 # (n  1)(2n  1)
2(2n  ")  (n  1) (n  1)(#n  1)

œ

0 # 2n#  3n  1
4n
2  n  1 (n  1)(#n  1)

Also, M œ 'c1 dA œ 'c1 a1
xn b dx œ 2 '0 a1
xn b dx œ 2
x
1

yœ

Mx M

œ

1

#

2n (n  1)(2n  1)

†

1

(n  1) 2n

œ

n 2n  1

xn b 1 ‘ " n1 !

n1

œ

2n# (n  1)(#n  1)

œ 2 ˆ1


" ‰ n1

#n  1 !

. œ

2n n1.

Therefore,

Ê ˆ!ß #n n 1 ‰ is the location of the centroid. As n Ä _, y Ä

" #

so

the limiting position of the centroid is ˆ!ß "# ‰ . 6. Align the telephone pole along the x-axis as shown in the accompanying figure. The slope of the top length of pole is 9 ‰ ˆ 14.5 " 81
81 œ 8"1 † 40 † (14.5
9) œ 815.5 †40 40 11 ‰ y œ 891  8111†80 x œ 8"1 ˆ9  80 x is an

œ

11 81†80 .

Thus,

equation of the

line representing the top of the pole. Then, My œ 'a x † 1y# dx œ 1 '0 x
8"1 ˆ9  b

40

11 80

# x‰‘ dx

b 11 ‰# '040 x ˆ9  80 x dx; M œ 'a 1y# dx 40 40 ‰‘# dx œ 64"1 ' ˆ9  11 ‰# dx. œ 1 '0
8"1 ˆ9  11 80 x 80 x 0

œ

" 641

My M

Thus, x œ

¸

129,700 5623.3

¸ 23.06 (using a calculator to compute

the integrals). By symmetry about the x-axis, y œ 0 so the center of mass is about 23 ft from the top of the pole. 7. (a) Consider a single vertical strip with center of mass (µ x ßµ y ). If the plate lies to the right of the line, then µ µ
b) $ dA Ê the plate's first moment the moment of this strip about the line x œ b is (x
b) dm œ (x about x œ b is the integral ' (x
b)$ dA œ ' $ x dA
' $ b dA œ My
b$ A. (b) If the plate lies to the left of the line, the moment of a vertical strip about the line x œ b is ab
µ x b dm œ ab
µ x b $ dA Ê the plate's first moment about x œ b is ' (b
x)$ dA œ ' b$ dA
' $ x dA œ b$ A
My . 8. (a) By symmetry of the plate about the x-axis, y œ 0. A typical vertical strip has center of mass: (µ x ßµ y ) œ (xß 0), length: 4Èax, width: dx, area: 4Èax dx, mass: dm œ $ dA œ kx † 4Èax dx, for some a proportionality constant k. The moment of the strip about the y-axis is M œ ' µ x dm œ ' 4kx# Èax dx y

œ 4kÈa'0 x&Î# dx œ 4kÈa
27 x(Î# ‘ 0 œ 4ka"Î# † 27 a(Î# œ a

a

œ 4kÈa'0 x$Î# dx œ 4kÈa
25 x&Î# ‘ 0 œ 4ka"Î# † 25 a&Î# œ a

a

8ka 7

%

8ka$ 5

0

. Also, M œ ' dm œ '0 4kxÈax dx a

. Thus, x œ

My M

œ

8ka% 7

†

5 8ka$

œ

5 7

a

‰ Ê (xß y) œ ˆ 5a 7 ß 0 is the center of mass. y#

# # a (b) A typical horizontal strip has center of mass: (µ x ßµ y ) œ Œ 4a # ß y œ Š y 8a4a ß y‹ , length: a


width: dy, area: Ša


y# 4a ‹

dy, mass: dm œ $ dA œ kyk Ša


y# 4a ‹

dy. Thus, Mx œ ' µ y dm

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

y# 4a

,

422

Chapter 6 Applications of Definite Integrals œ 'c2a y kyk Ša
2a

œ 'c2a Š
ay#  0

%

32a& 20a

œ
8a3 

y# 4a ‹

0

dy  '0 Šay#
2a

y% 4a ‹ 8a% 3



dy œ 'c2a
y# Ša





#

œ

%

&

'c2a a
16a y  y b dy 

œ

" 32a#

’8a% † 4a#


64a' 6 “

" 32a#

'0

2a

" 32a#

2a

#

c2a

" 3 #a #

a16a% y
y& b dy œ 64a' 6 “

" 16a#

œ

’
8a% y#  32a' 3 ‹

Š32a'


œ

œ2†

16a% 4 ‹

#

Š2a † 4a


" #a

œ

! y' 6 “
#a



1 3#a#

’8a% y#


† 32 a32a' b œ

4 3

#a y' 6 “!

a% ;

'c2a2a kyk a4a#
y# b dy

" 4a


#a

#

dy

" 16a#

%

" 4a

#a y& “ #0a !

y# 4a ‹

kyk Ša


'c02a a
4a# y  y$ b dy  4a" '02a a4a# y
y$ b dy œ 4a" ’
2a# y#  y4 “ !

" 4a

yœ

#

 4a# 8a ‹

 ’ 3a y$


' kyk a16a%
y% b dy

#

’8a% † 4a#


M œ ' dm œ 'c2a kyk Š 4a 4a
y ‹ dy œ

y& #0a “
#a

dy

2a

#

" 3 #a #

#

y# 4a ‹

!

dy œ ’
3a y$ 

'c2a2a kyk ay#  4a# b Š 4a 4a
y ‹ dy œ 32a" 

2a

2a

" 8a

0

dy  '0 y# Ša


œ 0; My œ ' µ x dm œ 'c2a Š y

32a& #0a

œ

œ

y% 4a ‹

y# 4a ‹

%

%

$

a8a
4a b œ 2a . Therefore, x œ



" 4a

’2a# y#


œ ˆ 34 a% ‰ ˆ 2a"$ ‰ œ

My M

#a y% 4 “!

2a 3

and

œ 0 is the center of mass.

Mx M

9. (a) On [0ß a] a typical @/<>3-+6 strip has center of mass: (µ x ßµ y ) œ Šx,

È b #
x #  È a#
x# ‹, #

length: Èb#
x#
Èa#
x# , width: dx, area: dA œ ŠÈb#
x#
Èa#
x# ‹ dx, mass: dm œ $ dA œ $ ŠÈb#
x#
Èa#
x# ‹ dx. On [aß b] a typical @/<>3-+6 strip has center of mass: È # # (µ x ßµ y ) œ Šxß b #
x ‹ , length: Èb#
x# , width: dx, area: dA œ Èb#
x# dx,

mass: dm œ $ dA œ $ Èb#
x# dx. Thus, Mx œ ' µ y dm œ '0

a

" #

ŠÈb#
x#  Èa#
x# ‹ $ ŠÈb#
x#
Èa#
x# ‹ dx  'a

b

" #

Èb#
x# $ Èb#
x# dx

œ

$ #

'0a cab#
x# b
aa#
x# bd dx  #$ 'ab ab#
x# b dx œ #$ '0a ab#
a# b dx  #$ 'ab ab#
x# b dx

œ

$ #

cab#
a# b xd !  #$ ’b# x


œ

$ #

aab#
a$ b  #$ Š 23 b$
ab# 

a

b

x$ 3 “a

œ a$ 3‹

$ #

b$ 3‹

cab#
a# b ad  #$ ’Šb$


œ

$ b$ 3




$ a$ 3

œ $ Šb

$


a$ 3 ‹;

a$ 3 ‹“


Š b# a


My œ ' µ x dm

œ '0 x$ ŠÈb#
x#
Èa#
x# ‹ dx  'a x$ Èb#
x# dx a

b

œ $ '0 x ab#
x# b a

œ


$ #

”

2 ab #
x # b 3

$Î#

dx
$ '0 x aa#
x# b a

"Î# a

$ 2 aa • #”

#

$Î#


x# b 3

# $Î#

#

œ
’ab
a b

# $Î#


ab b

a

dx  $ 'a x ab#
x# b

$ 2 ab •
#”

b

#

!

0

$ 3

"Î#

# $Î#

$ 3

“  ’0
aa b

• a

$ 3

“
’0
ab#
a# b #

#

$Î#

We calculate the mass geometrically: M œ $ A œ $ Š 14b ‹
$ Š 14a ‹ œ œ

$ ab $
a $ b 3

yœ (b) lim

œ

Mx M 4

b Ä a 31

†

4 $1 ab#
a# b

4 aa# abb# b 31(ab)

Ša

#

 ab  b# ‹ ab

œ

4 31

$

$


a Š bb#
a# ‹ œ

dx

b

$Î#


x# b 3

"Î#

4 (b
a) aa#  ab  b# b 31 (b
a)(b  a)

“œ

$1 4

$ b$ 3




$ a$ 3

œ

$ ab $
a $ b 3

ab#
a# b . Thus, x œ

œ Mx ; My M

œ

4 aa#  ab  b# b 31(a  b)

2a 1

2a ‰ Ê (xß y) œ ˆ 2a 1 ß 1 is the limiting

; likewise

. œ ˆ 341 ‰ Š a

#

 a#  a# ‹ aa

#

œ ˆ 341 ‰ Š 3a 2a ‹ œ

position of the centroid as b Ä a. This is the centroid of a circle of radius a (and we note the two circles coincide when b œ a).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 6 Additional and Advanced Exercises 10. Since the area of the traingle is 36, the diagram may be labeled as shown at the right. The centroid of the triangle is ˆ 3a , 24 ‰ a . The shaded portion is 144
36 œ 108. Write ax, yb for the centroid of the remaining region. The centroid of the whole square is obviously a6, 6b. Think of the square as a sheet of uniform density, so that the centroid of the square is the average of the centroids of the two regions, weighted by area: 'œ

$'ˆ 3a ‰
"!)axb "%%

and ' œ

‰ $'ˆ 24 a
"!)ayb "%%

which we solve to get x œ )


a *

and y œ

)a a
" b . a

Set

x œ 7 in. (Given). It follows that a œ *, whence y œ œ

7 "*

'% *

in. The distances of the centroid ax, yb from the other sides are easily computed. (Note that if we set y œ 7 in.

above, we will find x œ 7 "* .) 11. y œ 2Èx Ê ds œ É "x  1 dx Ê A œ '0 2Èx É "x  1 dx œ 3

4 3


(1  x)$Î# ‘ $ œ !

28 3

12. This surface is a triangle having a base of 21a and a height of 21ak. Therefore the surface area is " # # # (21a)(21ak) œ 21 a k. d# x dt#

13. F œ ma œ t# Ê

œaœ

t# m

Ê vœ

x œ 0 when t œ 0 Ê C" œ 0 Ê x œ W œ ' F dx œ '0

Ð12mhÑ"Î%

œ

(12mh)$Î# 18m

œ

F(t) †

12mh†È12mh 18m

œ

2h 3

dx dt

dx t$ dt œ 3m  C; v œ 0 when t œ 0 Ê t% "Î% . 12m . Then x œ h Ê t œ (12mh)

dt œ '0

Ð12mhÑ"Î%

† 2È3mh œ

14. Converting to pounds and feet, 2 lb/in œ

t# †

t$ 3m

dt œ

4h 3

È3mh

†

12 in 1 ft

2 lb 1 in

" 3m

'

’ t6 “

Ð12mh)"Î%

0

Cœ0 Ê

dx dt

œ

t$ 3m

Ê xœ

The work done is

" ‰ œ ˆ 18m (12mh)'Î%

œ 24 lb/ft. Thus, F œ 24x Ê W œ '0

1Î2

24x dx

"Î# " " ‰ œ c12x# d ! œ 3 ft † lb. Since W œ "# mv!#
"# mv"# , where W œ 3 ft † lb, m œ ˆ 10 lb‰ ˆ 3# ft/sec # " œ 320 slugs, and v" œ 0 ft/sec, we have 3 œ ˆ #" ‰ ˆ 3#"0 v#! ‰ Ê v!# œ 3 † 640. For the projectile height, s œ
16t#  v! t (since s œ 0 at t œ 0) Ê ds dt œ v œ
32t  v! . At the top of the ball's path, v œ 0 Ê #

and the height is s œ
16 ˆ 3v#! ‰  v! ˆ 3v#! ‰ œ

v!# 64

œ

3†640 64

œ 30 ft.

15. The submerged triangular plate is depicted in the figure at the right. The hypotenuse of the triangle has slope
1 Ê y
(
2) œ
(x
0) Ê x œ
(y  2) is an equation of the hypotenuse. Using a typical horizontal strip, the fluid strip strip pressure is F œ ' (62.4) † Š depth ‹ † Š length ‹ dy

c2

c2

œ 'c6 (62.4)(
y)[
(y  2)] dy œ 62.4 'c6 ay#  2yb dy $

œ 62.4 ’ y3  y# “


#
'

‰‘ œ (62.4)
ˆ
83  4‰
ˆ
216 3  36

‰ œ (62.4) ˆ 208 3
32 œ

(62.4)(112) 3

t% 12m

¸ 2329.6 lb

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

tœ

v! 3#

 C" ;

423

424

Chapter 6 Applications of Definite Integrals

16. Consider a rectangular plate of length j and width w. The length is parallel with the surface of the fluid of weight density =. The force on one side of the plate is F œ ='cw (
y)(j) dy œ
=j ’ y# “ 0

#

!

œ


w

=jw# #

. The = w

average force on one side of the plate is Fav œ œ

= w

#

’
y# “

!

=w #

œ


w

. Therefore the force

'c0w (
y)dy

=jw# #

‰ œ ˆ =w # (jw) œ (the average pressure up and down) † (the area of the plate). 17. (a) We establish a coordinate system as shown. A typical horizontal strip has: center of pressure: (µ x ßµ y ) b ˆ ‰ œ # ß y , length: L(y) œ b, width: dy, area: dA

œ b dy, pressure: dp œ = kyk dA œ =b kyk dy 0 0 Ê Fx œ ' µ y dp œ 'ch y † =b kyk dy œ
=b 'ch y# dy $

œ
=b ’ y3 “

!

œ
=b ’0
Š
3h ‹“ œ


=bh$ 3

'c0h = kyk L(y) dy œ
=b 'c0h

F œ ' dp œ œ

$


h

# !
=b ’ y# “
h

œ
=b ’0


h# #“

=bh# #

œ

;

y dy $

. Thus, y œ

œ

Fx F

Š
=3bh ‹

#

Š =bh # ‹

œ


2h 3

Ê the distance below the surface is

(b) A typical horizontal strip has length L(y). By similar triangles from the figure at the right,

L(y) b

œ


y
a h

Ê L(y) œ
bh (y  a). Thus, a typical strip has center of pressure: (µ x ßµ y ) œ (µ x ß y), length: L(y) œ
bh (y  a), width: dy, area: dA œ
bh (y  a) dy, pressure: dp œ = kyk dA œ =(
y) ˆ
bh ‰ (y  a) dy œ =b ay#  ayb dy Ê F œ ' µ y dp h

x

a

œ 'cÐa
hÑ y † %

=b h

#

ay  ayb dy œ

'Ðaa
hÑ

=b h


a ay$ 3 “ cÐa
hÑ

ay$  ay# b dy

œ

=b h

’ y4 

œ

=b h

’Š a4


œ œ

=b 12h =b 12h

œ


=bh 12

œ

=b h

’Š
3a 

œ

=b h

’a

œ

=b 6h

a6a# h  6ah#  2h$
6a# h
3ah# b œ

œ

%

a% 3‹

%


Š (a 4 h)


a(a  h)$ ‹“ 3

œ

=b h

’a

%


(a  h)% 4




a%
a(a  h)$ “ 3

c3 aa%
aa%  4a$ h  6a# h#  4ah$  h% bb
4 aa%
a aa$  3a# h  3ah#  h$ bbd a12a$ h  12a# h#  4ah$
12a$ h
18a# h#
12ah$
3h% b œ a6a#  8ah  3h# b ; F œ ' dp œ ' = kyk L(y) dy œ $

$

a$ #‹

$


Š
(a 3 h) 

 3a# h  3ah#  h$
a$ 3

ˆ
1=#bh ‰ a6a#  8ah  3h# b ˆ =6bh ‰ (3a  2h)

6a#  8ah  3h# 6a  4h



a(a  h)# ‹“ #

œ

=b h

a$
aa$  2a# h  ah# b “ #

‰ 6a œ ˆ
" # Š

#

=b 6h

 8ah  3h# ‹ 3a  2h

$

’ (a  h)3

œ

=b 6h


a$

a

=b 12h

a
6a# h#
8ah$
3h% b

=b h

'cÐa
hÑ



a$
a(a  h)# “ 2

ay#  ayb dy œ

=b h

$

’ y3 

c2 a3a# h  3ah#  h$ b
3 a2a# h  ah# bd

a3ah#  2h$ b œ

=bh 6

(3a  2h). Thus, y œ

Ê the distance below the surface is

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Fx F


a ay# 2 “ Ða
hÑ

2 3

h.

CHAPTER 7 TRANSCENDENTAL FUNCTIONS 7.1 INVERSE FUNCTIONS AND THEIR DERIVATIVES 1. Yes one-to-one, the graph passes the horizontal test. 2. Not one-to-one, the graph fails the horizontal test. 3. Not one-to-one since (for example) the horizontal line y œ # intersects the graph twice. 4. Not one-to-one, the graph fails the horizontal test. 5. Yes one-to-one, the graph passes the horizontal test 6. Yes one-to-one, the graph passes the horizontal test 7. Domain: 0
x Ÿ 1, Range: 0 Ÿ y

9. Domain: 1 Ÿ x Ÿ 1, Range:  1# Ÿ y Ÿ

8. Domain: x
1, Range: y  0

1 #

10. Domain: _
x
_, Range:  1#
y Ÿ

11. The graph is symmetric about y œ x.

(b) y œ È1  x# Ê y# œ 1  x# Ê x# œ 1  y# Ê x œ È1  y# Ê y œ È1  x# œ f
" (x)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 #

426

Chapter 7 Transcendental Functions

12. The graph is symmetric about y œ x.

yœ

" x

Ê xœ

" y

Ê yœ

" x

œ f
" (x)

13. Step 1: y œ x#  1 Ê x# œ y  1 Ê x œ Èy  1 Step 2: y œ Èx  1 œ f
" (x) 14. Step 1: y œ x# Ê x œ Èy, since x Ÿ !. Step 2: y œ Èx œ f
" (x) 15. Step 1: y œ x$  1 Ê x$ œ y  1 Ê x œ (y  1)"Î$ Step 2: y œ $Èx  1 œ f
" (x) 16. Step 1: y œ x#  2x  1 Ê y œ (x  1)# Ê Èy œ x  1, since x   1 Ê x œ 1  Èy Step 2: y œ 1  Èx œ f
" (x) 17. Step 1: y œ (x  1)# Ê Èy œ x  1, since x   1 Ê x œ Èy  1 Step 2: y œ Èx  1 œ f
" (x) 18. Step 1: y œ x#Î$ Ê x œ y$Î# Step 2: y œ x$Î# œ f
" (x) 19. Step 1: y œ x& Ê x œ y"Î& Step 2: y œ &Èx œ f
" (x); Domain and Range of f
" : all reals; &

f af
" (x)b œ ˆx"Î& ‰ œ x and f
" (f(x)) œ ax& b

"Î&

œx

"Î%

œx

20. Step 1: y œ x% Ê x œ y"Î% Step 2: y œ %Èx œ f
" (x); Domain of f
" : x   0, Range of f
" : y   0; %

f af
" (x)b œ ˆx"Î% ‰ œ x and f
" (f(x)) œ ax% b

21. Step 1: y œ x$  1 Ê x$ œ y  1 Ê x œ (y  1)"Î$ Step 2: y œ $Èx  1 œ f
" (x); Domain and Range of f
" : all reals; $

f af
" (x)b œ ˆ(x  1)"Î$ ‰  1 œ (x  1)  1 œ x and f
" (f(x)) œ aax$  1b  1b

"Î$

œ ax$ b

"Î$

œx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.1 Inverse Functions and Their Derivatives 22. Step 1: y œ

" #

x

" #

Ê

7 #


"

xœy

7 #

Ê x œ 2y  7

Step 2: y œ 2x  7 œ f (x); Domain and Range of f
" : all reals; f af
" (x)b œ "# (2x  7)  7# œ ˆx  7# ‰  23. Step 1: y œ Step 2: y œ

" x#

Ê x# œ

" y

" Èx

œ f
" (x)

Ê xœ

7 #

œ x and f
" (f(x)) œ 2 ˆ "# x  7# ‰  7 œ (x  7)  7 œ x

" Èy

Domain of f
" : x  0, Range of f
" : y  0; f af
" (x)b œ "" # œ "" œ x and f
" (f(x)) œ Š Èx ‹

24. Step 1: y œ

" x$

Ê x$ œ

" x"Î$
"

Step 2: y œ Domain of f f af
" (x)b œ

Šx‹

" y

Ê xœ

(c)

26. (a) y œ

" 5

" $ ax
"Î$ b

" x
"

œ

œ 2,

df
" dx ¹ xœ1

x7 Ê

df ¸ dx xœ
1

(c)

œ x since x  0

" y"Î$

œ x and f
" (f(x)) œ ˆ x"$ ‰

" 5

œ

" df
"

œ 5,

dx

¹

œ 4,

df
" dx ¹ xœ3

œ ˆ x" ‰


"

œx

(b) x #



3 #

xœy7

xœ$%Î&


"Î$

" #


"

(b)

(x) œ 5x  35

œ5

27. (a) y œ 5  4x Ê 4x œ 5  y Ê x œ 54  y4 Ê f
" (x) œ df ¸ dx xœ1Î#

" Š "x ‹

: x Á 0, Range of f
" : y Á 0;

Ê x œ 5y  35 Ê f (c)

œ

œ $É x" œ f
" (x);

25. (a) y œ 2x  3 Ê 2x œ y  3 Ê x œ y#  3# Ê f
" (x) œ df ¸ dx xœ
1

" É x"#

œ

(b) 5 4



x 4

" 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

427

428

Chapter 7 Transcendental Functions " #

28. (a) y œ 2x# Ê x# œ Ê xœ (c)

df ¸ dx xœ&

" È2

(b)

y

Èy Ê f


"

(x) œ

È x#

œ 4xk xœ5 œ 20,

df
" dx ¹ xœ&0

œ

" #È 2

x
"Î# ¹

xœ50

" #0

œ

$ $ 29. (a) f(g(x)) œ ˆ $Èx‰ œ x, g(f(x)) œ Èx$ œ x

w

#

w

(b)

w

(c) f (x) œ 3x Ê f (1) œ 3, f (1) œ 3; gw (x) œ 3" x
#Î$ Ê gw (1) œ 3" , gw (1) œ

" 3

(d) The line y œ 0 is tangent to f(x) œ x$ at (!ß !); the line x œ 0 is tangent to g(x) œ $Èx at (0ß 0)

30. (a) h(k(x)) œ

" 4

ˆ(4x)"Î$ ‰$ œ x,

k(h(x)) œ Š4 † (c) hw (x) œ w

k (x) œ

x$ 4‹

"Î$

(b)

œx

#

3x w w 4 Ê h (2) œ 3, h (2) 4
#Î$ Ê kw (2) œ "3 , 3 (4x)

œ 3; kw (2) œ

(d) The line y œ 0 is tangent to h(x) œ

x$ 4

" 3

at (!ß !);

the line x œ 0 is tangent to k(x) œ (4x)"Î$ at (!ß !)

œ 3x#  6x Ê

31.

df dx

33.

df
" dx ¹ x œ 4

df
" dx ¹ x œ f(3)

df
" dx ¹ x œ f(2)

œ

35. (a) y œ mx Ê x œ

" m

œ

(b) The graph of y œ f 36. y œ mx  b Ê x œ

y m

"

df dx

º

œ

xœ2

"

df dx

œ

º

xœ3

" ˆ 3" ‰

œ3

y Ê f
" (x) œ
"

" 9

œ

" m

œ 2x  4 Ê

32.

df dx

34.

dg
" dx ¹x œ 0

b m

dg
" dx ¹ x œ f(0)

œ

"

dg dx

º

œ

xœ0

"

df dx

º

œ

xœ5

œ

" 6

" 2

x

(x) is a line through the origin with slope



œ

df
" dx ¹ x œ f(5)

Ê f
" (x) œ

" m

x

b m;

" m.

the graph of f
" (x) is a line with slope

" m

and y-intercept  mb .

37. (a) y œ x  1 Ê x œ y  1 Ê f
" (x) œ x  1 (b) y œ x  b Ê x œ y  b Ê f
" (x) œ x  b (c) Their graphs will be parallel to one another and lie on opposite sides of the line y œ x equidistant from that line.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.1 Inverse Functions and Their Derivatives 38. (a) y œ x  1 Ê x œ y  1 Ê f
" (x) œ 1  x; the lines intersect at a right angle (b) y œ x  b Ê x œ y  b Ê f
" (x) œ b  x; the lines intersect at a right angle (c) Such a function is its own inverse.

39. Let x" Á x# be two numbers in the domain of an increasing function f. Then, either x"
x# or x"  x# which implies f(x" )
f(x# ) or f(x" )  f(x# ), since f(x) is increasing. In either case, f(x" ) Á f(x# ) and f is one-to-one. Similar arguments hold if f is decreasing. 40. f(x) is increasing since x#  x" Ê

" 3

x# 

5 6



" 3

x"  56 ;

df dx

œ

" 3

41. f(x) is increasing since x#  x" Ê 27x$#  27x"$ ; y œ 27x$ Ê x œ df dx

œ 81x# Ê

df
" dx

œ

" ¸ 81x# 13 x"Î$

œ

" 9x#Î$

œ

" 9

df
" dx

Ê " 3

œ

df dx

œ 24x# Ê

dx

œ

" ¸
24x# 12 Ð1
xÑ"Î$

œ

œ3

y"Î$ Ê f
" (x) œ

" 3

x"Î$ ;

x
#Î$

42. f(x) is decreasing since x#  x" Ê 1  8x$#
1  8x"$ ; y œ 1  8x$ Ê x œ df
"

" ˆ "3 ‰


" 6("
x)#Î$

" #

(1  y)"Î$ Ê f
" (x) œ

" #

(1  x)"Î$ ;

œ  "6 (1  x)
#Î$

43. f(x) is decreasing since x#  x" Ê (1  x# )$
(1  x" )$ ; y œ (1  x)$ Ê x œ 1  y"Î$ Ê f
" (x) œ 1  x"Î$ ; df dx

œ 3(1  x)# Ê

df
" dx

œ

"
3(1
x)# ¹ 1cx"Î$ &Î$

44. f(x) is increasing since x#  x" Ê x# df dx

œ

5 3

x#Î$ Ê

df
" dx

œ

5 3

" ¹ x#Î$ x$Î&

œ

3 5x#Î&

œ


" 3x#Î$

œ  "3 x
#Î$

&Î$

 x" ; y œ x&Î$ Ê x œ y$Î& Ê f
" (x) œ x$Î& ; œ

3 5

x
#Î&

45. The function g(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x" ) Á f(x# ) and therefore g(x" ) Á g(x# ). Therefore g(x) is one-to-one as well. 46. The function h(x) is also one-to-one. The reasoning: f(x) is one-to-one means that if x" Á x# then f(x" ) Á f(x# ), so f(x"" ) Á f(x"# ) , and therefore h(x" ) Á h(x# ). 47. The composite is one-to-one also. The reasoning: If x" Á x# then g(x" ) Á g(x# ) because g is one-to-one. Since g(x" ) Á g(x# ), we also have f(g(x" )) Á f(g(x# )) because f is one-to-one. Thus, f ‰ g is one-to-one because x" Á x# Ê f(g(x" )) Á f(g(x# )). 48. Yes, g must be one-to-one. If g were not one-to-one, there would exist numbers x" Á x# in the domain of g with g(x" ) œ g(x# ). For these numbers we would also have f(g(x" )) œ f(g(x# )), contradicting the assumption that f ‰ g is one-to-one.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

429

430

Chapter 7 Transcendental Functions

49. The first integral is the area between f(x) and the x-axis over a Ÿ x Ÿ b. The second integral is the area between f(x) and the y-axis for f(a) Ÿ y Ÿ f(b). The sum of the integrals is the area of the larger rectangle with corners at (0ß 0), (bß 0), (bß f(b)) and (0ß f(b)) minus the area of the smaller rectangle with vertices at (0ß 0), (aß 0), (aß f(a)) and (0ß f(a)). That is, the sum of the integrals is bf(b)  af(a).

50. f w axb œ

acx  dba
aax  bbc acx  db#

œ

ad
bc . acx  db#

Thus if ad  bc Á !, f w axb is either always positive or always negative. Hence faxb is

either always increasing or always decreasing. If follows that faxb is one-to-one if ad  bc Á !. 51. (g ‰ f)(x) œ x Ê g(f(x)) œ x Ê gw (f(x))f w (x) œ 1 52. W(a) œ 'f(a) 1 ’af
" (y)b  a# “ dy œ 0 œ 'a 21x[f(a)  f(x)] dx œ S(a); Ww (t) œ 1’af
" (f(t))b  a# “ f w (t) f(a)

a

#

#

œ 1 at#  a# b f w (t); also S(t) œ 21f(t)'a x dx  21'a xf(x) dx œ c1f(t)t#  1f(t)a# d  21'a xf(x) dx t

t

t

Ê Sw (t) œ 1t# f w (t)  21tf(t)  1a# f w (t)  21tf(t) œ 1 at#  a# b f w (t) Ê Ww (t) œ Sw (t). Therefore, W(t) œ S(t) for all t − [aß b]. 53-60. Example CAS commands: Maple: with( plots );#53 f := x -> sqrt(3*x-2); domain := 2/3 .. 4; x0 := 3; Df := D(f); # (a) plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#53(a) (Section 7.1)" ); q1 := solve( y=f(x), x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#53(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) If a function requires the odd root of a negative number, begin by loading the RealOnly package that allows Mathematica to do this. See section 2.5 for details. <<Miscellaneous `RealOnly` Clear[x, y] Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.1 Inverse Functions and Their Derivatives

431

{a,b} = {2, 1}; x0 = 1/2 ; f[x_] = (3x  2) / (2x  11) Plot[{f[x], f'[x]}, {x, a, b}] solx = Solve[y == f[x], x] g[y_] = x /. solx[[1]] y0 = f[x0] ftan[x_] = y0  f'[x0] (x-x0) gtan[y_] = x0  1/ f'[x0] (y  y0) Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b}, Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a,b},{a,b}}, AspectRatio Ä Automatic] 61-62. Example CAS commands: Maple: with( plots ); eq := cos(y) = x^(1/5); domain := 0 .. 1; x0 := 1/2; f := unapply( solve( eq, y ), x ); # (a) Df := D(f); plot( [f(x),Df(x)], x=domain, color=[red,blue], linestyle=[1,3], legend=["y=f(x)","y=f '(x)"], title="#62(a) (Section 7.1)" ); q1 := solve( eq, x ); # (b) g := unapply( q1, y ); m1 := Df(x0); # (c) t1 := f(x0)+m1*(x-x0); y=t1; m2 := 1/Df(x0); # (d) t2 := g(f(x0)) + m2*(x-f(x0)); y=t2; domaing := map(f,domain); # (e) p1 := plot( [f(x),x], x=domain, color=[pink,green], linestyle=[1,9], thickness=[3,0] ): p2 := plot( g(x), x=domaing, color=cyan, linestyle=3, thickness=4 ): p3 := plot( t1, x=x0-1..x0+1, color=red, linestyle=4, thickness=0 ): p4 := plot( t2, x=f(x0)-1..f(x0)+1, color=blue, linestyle=7, thickness=1 ): p5 := plot( [ [x0,f(x0)], [f(x0),x0] ], color=green ): display( [p1,p2,p3,p4,p5], scaling=constrained, title="#62(e) (Section 7.1)" ); Mathematica: (assigned function and values for a, b, and x0 may vary) For problems 61 and 62, the code is just slightly altered. At times, different "parts" of solutions need to be used, as in the definitions of f[x] and g[y] Clear[x, y] {a,b} = {0, 1}; x0 = 1/2 ; eqn = Cos[y] == x1/5 soly = Solve[eqn, y] f[x_] = y /. soly[[2]] Plot[{f[x], f'[x]}, {x, a, b}] solx = Solve[eqn, x] g[y_] = x /. solx[[1]] y0 = f[x0] ftan[x_] = y0  f'[x0] (x  x0)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

432

Chapter 7 Transcendental Functions gtan[y_] = x0  1/ f'[x0] (y  y0) Plot[{f[x], ftan[x], g[x], gtan[x], Identity[x]},{x, a, b}, Epilog Ä Line[{{x0, y0},{y0, x0}}], PlotRange Ä {{a, b}, {a, b}}, AspectRatio Ä Automatic]

7.2 NATURAL LOGARITHMS 1. (a) ln 0.75 œ ln (b) ln

4 9 " #

3 4

œ ln 3  ln 4 œ ln 3  ln 2# œ ln 3  2 ln 2

œ ln 4  ln 9 œ ln 2#  ln 3# œ 2 ln 2  2 ln 3

œ ln 1  ln 2 œ  ln 2 (d) ln $È9 œ "3 ln 9 œ (e) ln 3È2 œ ln 3  ln 2"Î# œ ln 3  "# ln 2 (f) ln È13.5 œ " ln 13.5 œ " ln 27 œ " aln 3$  ln 2b œ " (3 ln 3  ln 2) (c) ln

#

2. (a) ln

" 125

#

(e) ln 0.056 œ ln ln 35  ln ln 25

" 7

7 125

3 #

" #

2 3

ln 3

(b) ln 9.8 œ ln

49 5

œ ln 7#  ln 5 œ 2 ln 7  ln 5

(d) ln 1225 œ ln 35# œ 2 ln 35 œ 2 ln 5  2 ln 7

œ ln 7  ln 5$ œ ln 7  3 ln 5

ln 5  ln 7
ln 7 # ln 5

œ

ln 3# œ

#

ln 7 " #

œ

3. (a) ln sin )  ln ˆ sin5 ) ‰ œ ln
(c)

#

œ ln 1  3 ln 5 œ 3 ln 5

(c) ln 7È7 œ ln 7$Î# œ (f)

#

" 3

sin ) Š sin5 ) ‹ 

#

" ‰ (b) ln a3x#  9xb  ln ˆ 3x œ ln Š 3x 3x
9x ‹ œ ln (x  3)

œ ln 5

#

ln a4t% b  ln 2 œ ln È4t%  ln 2 œ ln 2t#  ln 2 œ ln Š 2t# ‹ œ ln at# b

4. (a) ln sec )  ln cos ) œ ln [(sec ))(cos ))] œ ln 1 œ 0 (b) ln (8x  4)  ln 2# œ ln (8x  4)  ln 4 œ ln ˆ 8x 4 4 ‰ œ ln (2x  1) $ "Î$
") (c) 3 ln Èt#  1  ln (t  1) œ 3 ln at#  1b  ln (t  1) œ 3 ˆ "3 ‰ ln at#  1b  ln (t  1) œ ln Š (t (t1)(t ‹  1)

œ ln (t  1) 1 ‰ 5. y œ ln 3x Ê yw œ ˆ 3x (3) œ

7. y œ ln at# b Ê 9. y œ ln

3 x

10. y œ ln

10 x

œ ˆ t"# ‰ (2t) œ

dy dt

œ ln 3x
" Ê

dy dx

œ ln 10x
" Ê

11. y œ ln ()  1) Ê 13. y œ ln x$ Ê

dy dx

15. y œ t(ln t)# Ê

dy d)

17. y œ

x% 4

ln x 

dy dt

8. y œ ln ˆt$Î# ‰ Ê

2 t

dy dx

dy dx

" ‰ ˆ 3 "Î# ‰ œ ˆ t$Î# œ # t

3 2t

œ ˆ 10x"
" ‰ a10x
# b œ  x" " )1

dy dt

12. y œ ln (2)  2) Ê 14. y œ (ln x)$ Ê

3 x

œ (ln t)#  2t(ln t) †

Ê

dy dt

œ ˆ 3x"
" ‰ a3x
# b œ  x"

œ ˆ x"$ ‰ a3x# b œ

" #(ln t)"Î# x% 16

" ‰ 6. y œ ln kx Ê yw œ ˆ kx (k) œ x

œ ˆ ) " 1 ‰ (1) œ

16. y œ tÈln t œ t(ln t)"Î# Ê œ (ln t)"Î# 

" x

d dt

(ln t) œ (ln t)# 

œ (ln t)"Î#  "# t(ln t)
"Î# †

œ x$ ln x 

x% 4

†

" x



4x$ 16

d dt

2t ln t t

dy dx

dy d)

œ ˆ #) " 2 ‰ (2) œ

œ 3(ln x)# †

d dx

œ (ln t)#  2 ln t

(ln t) œ (ln t)"Î# 

t(ln t)
"Î# #t

œ x$ ln x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" )1

(ln x) œ

3(ln x)# x

Section 7.2 Natural Logarithms 18. y œ

x$ 3

ln x 

x$ 9

19. y œ

ln t t

Ê

20. y œ

"  ln t t

Ê

21. y œ

ln x 1  ln x

Ê yw œ

(1  ln x) ˆ "x ‰
(ln x) ˆ x" ‰ (1  ln x)#

22. y œ

x ln x 1  ln x

Ê yw œ

(1ln x) ˆln x  x† "x ‰
(x ln x) ˆ x" ‰ (1ln x)#

dy dt

Ê

œ x# ln x 

dy dx

œ

t ˆ "t ‰
(ln t)(1) t#

dy dt

œ

23. y œ ln (ln x) Ê yw œ ˆ ln"x ‰ ˆ "x ‰ œ " ln (ln x)

†

25. y œ )[sin (ln ))  cos (ln ))] Ê

†

" x

3x# 9



œ x# ln x

1
ln t t#

œ

t ˆ "t ‰
("  ln t)(1) t#

24. y œ ln (ln (ln x)) Ê yw œ

x$ 3

"
1
ln t t#

œ

" x

œ

œ  lnt# t

 lnxx
lnxx (1  ln x)#

œ

œ

" x(1  ln x)#

("  ln x)#
ln x (1  ln x)#

œ1

ln x (1  ln x)#

" x ln x

d dx

(ln (ln x)) œ

" ln (ln x)

†

" ln x

†

(ln x) œ

d dx

" x (ln x) ln (ln x)

œ [sin (ln ))  cos (ln ))]  ) cos (ln )) †

dy d)

œ sin (ln ))  cos (ln ))  cos (ln ))  sin (ln )) œ 2 cos (ln )) 26. y œ ln (sec )  tan )) Ê

dy d)

" xÈ x  1

" #

27. y œ ln 28. y œ

" #

29. y œ

1  ln t 1
ln t

ln

1x 1
x

œ  ln x  œ

Ê

" #

dy dt

" #

(1
ln t)#

" t"Î#

31. y œ ln (sec (ln ))) Ê

œ

" #

Èsin ) cos ) 1  2 ln )

" #

œ

"Î#

Ê

œ

œ

dy dt

† #" t
"Î# œ dy d)

" #

" #

œ

" #

 1 " x  ˆ 1
" x ‰ (1)‘ œ

"
ln t  "  ln t t

t

t

ˆln t"Î# ‰
"Î# †

œ

t

(1
ln t)#

" #

1
x1x ’ (1  x)("
x) “ œ

2 t(1
ln t)#

ˆln t"Î# ‰ œ

" #

ˆln t"Î# ‰
"Î# †

sec (ln )) tan (ln )) sec (ln ))

†

d d)

d dt

" t"Î#

†

d dt

ˆt"Î# ‰

" sec (ln ))

†

d d)

(sec (ln ))) œ

dy d)

œ

" #

(ln )) œ

) ˆ cos sin ) 

tan (ln )) )

sin ) ‰ cos )



2 )

1  # ln )

" #

ln (1  x) Ê yw œ

5†2x x#  1

[5 ln (x  1)  20 ln (x  2)] Ê yw œ

" #

 #" ˆ 1
" x ‰ (1) œ

ˆ x 5 1 

20 ‰ x#

œ

5 #

10x x#  1



" #(1
x)


4(x  1) ’ (x(x2)1)(x  2) “

2 œ  5# ’ (x 3x1)(x  #) “

35. y œ 'x#Î2 ln Èt dt Ê x#

" 1
x#

4 )(1  2 ln )) “

&

34. y œ ln É (x(x2)1)#! œ

œ sec )

" 4tÉln Èt

x  1b 33. y œ ln Š aÈ ‹ œ 5 ln ax#  1b  1
x &

sec )(tan )  sec )) tan )  sec )

(ln sin )  ln cos ))  ln (1  2 ln )) Ê

’cot )  tan )  #

œ

 sin (ln )) † ") ‘

 1)  x 3x  2 ln (x  1) Ê yw œ  x"  #" ˆ x " 1 ‰ œ  2(x 2x(x  1) œ  2x(x  1)

‰ (1
ln t) ˆ "t ‰
(1  ln t) ˆ
" t

œ

ˆln t"Î# ‰
"Î# †

32. y œ ln

sec ) tan )  sec# ) sec )  tan )

cln (1  x)  ln (1  x)d Ê yw œ

30. y œ Éln Èt œ ˆln t"Î# ‰ œ

œ

" )

dy dx

œ Šln Èx# ‹ †

d dx

#

ax# b  Šln É x# ‹ †

d dx

#

Š x# ‹ œ 2x ln kxk  x ln

kx k È2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

433

434

Chapter 7 Transcendental Functions Èx

36. y œ 'Èx ln t dt Ê $

œ

$ ln È x

ˆ $Èx‰  ˆln Èx‰ †

d dx

d dx

ˆÈx‰ œ ˆln $Èx‰ ˆ "3 x
#Î$ ‰  ˆln Èx‰ ˆ #" x
"Î# ‰

ln Èx 2È x



$

3 È x#

œ ˆln $Èx‰ †

dy dx

37.

2 'cc32 x" dx œ cln kxkd
#
$ œ ln 2  ln 3 œ ln 3

38.

'c01 3x3
# dx œ cln k3x  2kd !
" œ ln 2  ln 5 œ ln 52

39.

' y 2y
25 dy œ ln ky#  25k  C

40.

' 4r8r
5 dr œ ln k4r#  5k  C

41.

42.

#

#

'01 2
sincost t dt œ cln k2  cos tkd !1 œ ln 3  ln 1 œ ln 3; or let u œ 2  cos t Ê du œ sin t dt with t œ 0 1 3 Ê u œ 1 and t œ 1 Ê u œ 3 Ê '0 2
sincost t dt œ '1 "u du œ cln kukd $" œ ln 3  ln 1 œ ln 3 '01Î3 1
4 4sincos) ) d) œ cln k1  4 cos )kd !1Î$ œ ln k1  2k œ  ln 3 œ ln "3 ; or let u œ 1  4 cos ) Ê du œ 4 sin ) d) 1Î3 c1 " with ) œ 0 Ê u œ 3 and ) œ 13 Ê u œ 1 Ê '0 1
4 4sincos) ) d) œ 'c3 u" du œ cln kukd
"
$ œ  ln 3 œ ln 3

43. Let u œ ln x Ê du œ

'1

2

2 ln x x

dx œ '0

ln 2

'2

dx x ln x

œ 'ln 2

ln 4

" u

ln 2

'2

dx x(ln x)#

'2

dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4; #

" x

dx; x œ 2 Ê u œ ln 2 and x œ 4 Ê u œ ln 4;

œ 'ln 2 u
# du œ  "u ‘ ln 2 œ  ln"4  ln 4

ln 4

46. Let u œ ln x Ê du œ 16

" x

4‰ ln 2 ˆ 2 ln 2 ‰ œ ln 2 du œ cln ud lnln 42 œ ln (ln 4)  ln (ln 2) œ ln ˆ ln ln 2 œ ln Š ln 2 ‹ œ ln ln 2

45. Let u œ ln x Ê du œ 4

dx; x œ 1 Ê u œ 0 and x œ 2 Ê u œ ln 2;

2u du œ cu# d 0 œ (ln 2)#

44. Let u œ ln x Ê du œ 4

" x

dx 2xÈln x

œ

" #

'ln 2

ln 16

" x

" ln #

œ  ln"## 

" ln 2

œ  2 ln" # 

" ln #

œ

" # ln 2

œ

" ln 4

dx; x œ 2 Ê u œ ln 2 and x œ 16 Ê u œ ln 16;

u
"Î# du œ u"Î# ‘ ln 2 œ Èln 16  Èln 2 œ È4 ln 2  Èln 2 œ 2Èln 2  Èln 2 œ Èln 2 ln 16

47. Let u œ 6  3 tan t Ê du œ 3 sec# t dt;

t ' 6 3sec ' duu œ ln kuk  C œ ln k6  3 tan tk  C 3 tan t dt œ #

48. Let u œ 2  sec y Ê du œ sec y tan y dy;

' sec# ysectanyy dy œ ' duu œ ln kuk  C œ ln k2  sec yk  C

49. Let u œ cos

x #

Ê du œ  "# sin

sin '01Î2 tan x# dx œ '01Î2 cos

x #

dx Ê 2 du œ sin 1Î

50. Let u œ sin t Ê du œ cos t dt; t œ

'11ÎÎ42 cot t dt œ '11ÎÎ42 51. Let u œ sin

) 3

cos t sin t

Ê du œ

'1Î2 2 cot 3) d) œ '1Î2 1

1 4

du u

dx; x œ 0 Ê u œ 1 and x œ

È2

œ c2 ln kukd 11Î

Ê uœ

" È2

and t œ

1 #

" dt œ '1ÎÈ2 du u œ cln kukd "ÎÈ# œ  ln 1

" 3

1 2 cos

sin

È2

dx œ 2 '1

x # x #

x #

)

3

cos )

3

) 3

d) Ê 6 du œ 2 cos

È3Î2

d) œ 6 '1Î2

du u

) 3

œ 2 ln

Ê uœ

" È2

œ 2 ln È2 œ ln 2

Ê u œ 1;

" È2

d) ; ) œ

È3Î2

" È2

1 #

œ ln È2 1 #

Ê uœ

œ 6 cln kukd 1Î2 œ 6 Šln

È3 #

" #

and ) œ 1 Ê u œ

È3 #

 ln "# ‹ œ 6 ln È3 œ ln 27

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

;

Section 7.2 Natural Logarithms 1 1#

52. Let u œ cos 3x Ê du œ 3 sin 3x dx Ê 2 du œ 6 sin 3x dx; x œ 0 Ê u œ 1 and x œ 1ÎÈ2

'01Î12 6 tan 3x dx œ '01Î12 6cossin3x3x dx œ 2 '1 53.

'

œ'

dx 2Èx  2x

dx ; 2 È x ˆ1  È x ‰

du u

È2

œ 2 cln kukd 11Î

let u œ 1  Èx Ê du œ

" #È x

œ 2 ln

" È2

" È2

;

 ln 1 œ 2 ln È2 œ ln 2

' 2Èx ˆdx1  Èx‰ œ '

dx;

Ê uœ

œ ln kuk  C

du u

œ ln ¸1  Èx¸  C œ ln ˆ1  Èx‰  C 54. Let u œ sec x  tan x Ê du œ asec x tan x  sec# xb dx œ (sec x)(tan x  sec x) dx Ê sec x dx œ

'

sec x dx Èln (sec x  tan x)

œ'

du uÈln u

œ ' (ln u)
"Î# † " #

55. y œ Èx(x  1) œ (x(x  1))"Î# Ê ln y œ Ê yw œ ˆ "# ‰ Èx(x  1) ˆ x" 

" ‰ x 1

œ " #

56. y œ Èax#  1b (x  1)# Ê ln y œ Ê yw œ Èax# +1b (x  1)# ˆ x# x 1  57. y œ É t t 1 œ ˆ t t 1 ‰ Ê

dy dt

œ

" #

"Î#

É t t 1 ˆ "t 

" #

Ê ln y œ " ‰ t1

œ

" #

dy dt

du œ 2(ln u)"Î#  C œ 2Èln (sec x  tan x)  C

ln (x(x  1)) Ê 2 ln y œ ln (x)  ln (x  1) Ê

Èx(x  1) (2x  1) 2x(x  1)

" ‰ x
1

dy d)

" dy y dt

dy d)

62. y œ Ê

dy dt

" #

œ

ˆ x#2x 1 

#

" x 1

ˆ "t 

a2x#
x  1b kx
1k Èx#  1 (x
1)

" ‰ t1

" dy y dt

œ  #" ˆ "t 

" ‰ t1

" #

ln ()  3)  ln (sin )) Ê

" dy y d)

œ

" #()  3)



#

) œ (tan )) È2)  1 Š sec tan ) 

" #)  1 ‹

œ asec# )b È2)  1 

œ t(t  1)(t+2) ˆ "t 

œ

" t1



" ‰ t#

cos ) sin )

 "t 

" t1



" ‘ t#

ln (2)  1) Ê

" dy y d)

œ

sec# ) tan )

)5 ) cos ) Ê ln y œ ln ()  5)  ln ) dy ˆ )5 ‰ ˆ ) " 5  ")  tan )‰ d) œ ) cos )

 ˆ #" ‰ ˆ #) 2 1 ‰

tan ) È 2)  1

" dy y dt

œ

" t



" t1



" t#

t(t  2)  t(t  1) œ t(t  1)(t  2) ’ (t  1)(t t(t2)1)(t “ œ 3t#  6t  2  2)

Ê ln y œ ln 1  ln t  ln (t  1)  ln (t  2) Ê

" t(t  1)(t  2)

" #

œ


" t(t  1)(t  #)

" dy y dt

œ  "t 

" t1



" t#

t(t  2)  t(t  1) ’ (t  1)(t t(t2)1)(t “  2)

#

Ê



2 ‰ x
1

œ  at$3t3t#6t2t2b# 63. y œ

" x

œ È)  3 (sin )) ’ 2() " 3)  cot )“

" t(t  1)(t  2) dy dt

œ

2t  1 2 at#  tb$Î#

61. y œ t(t  1)(t  2) Ê ln y œ ln t  ln (t  1)  ln (t  2) Ê Ê

w

" 2Èt (t  1)$Î#

60. y œ (tan )) È2)  1 œ (tan ))(2)  1)"Î# Ê ln y œ ln (tan ))  Ê

" #

œ

[ln t  ln (t  1)] Ê

59. y œ È)  3 (sin )) œ ()  3)"Î# sin ) Ê ln y œ Ê

w

y y

#

[ln t  ln (t  1)] Ê

" #

2y y

2x  " 2Èx(x  1)

œ Èax#  1b (x  1)# ’ axx#
x1b (xx
1)1 “ œ

É t t 1 ’ t(t " 1) “ œ

" œ  "# É t(t 1 1) ’ t(t2t  1) “œ

œ

cln ax#  1b  2 ln (x  1)d Ê

58. y œ É t(t 1 1) œ [t(t  1)]
"Î# Ê ln y œ Ê

" u

du u ;

 ln (cos )) Ê

" dy y d)

œ

" )5



" )



sin ) cos )

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

435

436

Chapter 7 Transcendental Functions

64. y œ

) sin ) Èsec ) Ê ln y dy ) sin ) ˆ " d) œ Èsec ) )

Ê 65. y œ Ê

" #

œ ln )  ln (sin ))   cot ) 

" #

"!

" #

"!

$

$


2) ˆ " É x(x x#  1 x 

" 3

Ê yw œ

" 3

1b 

2 3

ln (x  1) Ê

cln x  ln (x  2)  ln ax#  1bd Ê " x
#



$

" 3

$

 1)(x
2) ˆ " É ax(x x#  1b (2x  3) x 

" 3

cos ) sin )

(sec ))(tan )) 2 sec ) “



yw y

yw y

" x

œ

œ



5 x1

x x#  1





2 3(x  1)

5 2x  1

5 ‰ 2x  1

 1)(x
2) 68. y œ É ax(x x#  1b (2x  3) Ê ln y œ

Ê yw œ

œ ’ ") 

[10 ln (x  1)  5 ln (2x  1)] Ê

(x  1) ˆ 5 Ê yw œ É (2x x1   1)&


2) 67. y œ É x(x x#  1 Ê ln y œ

" dy y d)

tan )‰

xÈ x#  1 Ê ln y œ ln x  "# ln ax#  (x  1)#Î$ È # yw œ x(x x1)#Î$1 ’ "x  x# x 1  3(x 2 1) “

(x  1) 66. y œ É (2x  1)& Ê ln y œ

ln (sec )) Ê

yw y

œ

" 3

ˆ "x 

" x
#



2x ‰ x#  1

2x ‰ x#  1

cln x  ln (x  1)  ln (x  2)  ln ax#  1b  ln (2x  3)d " x1



" x
#



2x x#  1



2 ‰ 2x  3

sin x 1 w w 69. (a) f(x) œ ln (cos x) Ê f w (x) œ  cos x œ  tan x œ 0 Ê x œ 0; f (x)  0 for  4 Ÿ x
0 and f (x)
0 for 0
x Ÿ 13 Ê there is a relative maximum at x œ 0 with f(0) œ ln (cos 0) œ ln 1 œ 0; f ˆ 14 ‰ œ ln ˆcos ˆ 14 ‰‰

œ ln Š È"2 ‹ œ  #" ln 2 and f ˆ 13 ‰ œ ln ˆcos ˆ 13 ‰‰ œ ln xœ

1 3

" #

œ  ln 2. Therefore, the absolute minimum occurs at

with f ˆ 13 ‰ œ  ln 2 and the absolute maximum occurs at x œ 0 with f(0) œ 0.

(b) f(x) œ cos (ln x) Ê f w (x) œ


sin (ln x) x

œ 0 Ê x œ 1; f w (x)  0 for

" #

Ÿ x
1 and f w (x)
0 for 1
x Ÿ 2 Ê there is a relative maximum at x œ 1 with f(1) œ cos (ln 1) œ cos 0 œ 1; f ˆ "# ‰ œ cos ˆln ˆ "# ‰‰

œ cos ( ln 2) œ cos (ln 2) and f(2) œ cos (ln 2). Therefore, the absolute minimum occurs at x œ x œ 2 with f ˆ "# ‰ œ f(2) œ cos (ln 2), and the absolute maximum occurs at x œ 1 with f(1) œ 1.

" #

and

70. (a) f(x) œ x  ln x Ê f w (x) œ 1  "x ; if x  1, then f w (x)  0 which means that f(x) is increasing (b) f(1) œ 1  ln 1 œ 1 Ê f(x) œ x  ln x  0, if x  1 by part (a) Ê x  ln x if x  1 71.

'15 (ln 2x  ln x) dx œ '15 ( ln x  ln 2  ln x) dx œ (ln 2)'15 dx œ (ln 2)(5  1) œ ln 2% œ ln 16

72. A œ

'c01Î4  tan x dx  '01Î3 tan x dx œ '
01Î4
cossinxx dx  '01Î3
cossinxx dx œ cln kcos xkd !
1Î%  cln kcos xkd !1Î$

œ Šln 1  ln

" È2 ‹

 ˆln

" #

 ln 1‰ œ ln È2  ln 2 œ

73. V œ 1'0 Š Èy2 1 ‹ dy œ 41 '0 #

3

3

74. V œ 1 '1Î6 cot x dx œ 1 '1Î6 1Î2

1Î2

" y 1

cos x sin x

3 #

ln 2

dy œ 41 cln ky  1kd $! œ 41(ln 4  ln 1) œ 41 ln 4

1Î# dx œ 1 cln (sin x)d 1Î' œ 1 ˆln 1  ln "# ‰ œ 1 ln 2

75. V œ 21'1Î2 x ˆ x"# ‰ dx œ 21 '1Î2 x" dx œ 21 cln kxkd #"Î# œ 21 ˆln 2  ln #" ‰ œ 21(2 ln 2) œ 1 ln 2% œ 1 ln 16 2

2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.2 Natural Logarithms

437

76. V œ 1 '0 Š Èx9x ‹ dx œ 271'0 dx œ 271 cln ax$  9bd ! œ 271(ln 36  ln 9) $9 #

3

3

$

œ 271(ln 4  ln 9  ln 9) œ 271 ln 4 œ 541 ln 2 77. (a) y œ

x# 8

œ '4

#  ln x Ê 1  ayw b# œ 1  ˆ x4  x" ‰ œ 1  Š x 4x
4 ‹ œ Š x 4x 4 ‹ Ê L œ '4 É1  ayw b# dx

8 # x 4

(b) x œ

#

#

4x

ˆ y4 ‰#

#

#

8

dx œ '4 ˆ x4  "x ‰ dx œ ’ x8  ln kxk“ œ (8  ln 8)  (2  ln 4) œ 6  ln 2 8

)

#

%

2

ln ˆ y4 ‰

Ê

dx dy

œ

y 8



2 y

' Ê L œ '4 Ê1  Š dx dy ‹ dy œ 4 #

12

#

#

y y 2 Ê 1  Š dx dy ‹ œ 1  Š 8  y ‹ œ 1  Š

12 # y 16

8y

#

y dy œ '4 Š y8  2y ‹ dy œ ’ 16  2 ln y“ 12

#


16 8y ‹ "# %

#

œ Šy

#

 16 8y ‹

#

œ (9  2 ln 12)  (1  2 ln 4)

œ 8  2 ln 3 œ 8  ln 9 78. L œ '1 É1  2

" x#

dx Ê

dy dx

œ

" x

Ê y œ ln kxk  C œ ln x  C since x  0 Ê 0 œ ln 1  C Ê C œ 0 Ê y œ ln x

" ‰ ˆ"‰ 79. (a) My œ '1 x ˆ x" ‰ dx œ 1, Mx œ '1 ˆ 2x x dx œ 2

Ê xœ

2

My M

œ

" ln 2

¸ 1.44 and y œ

Mx M

œ

ˆ "4 ‰ ln 2

" #

'12 x"

" ‘ dx œ  2x œ 4" , M œ '1 "

2

#

#

dx œ cln kxkd #" œ ln 2

" x

¸ 0.36

(b)

80. (a) My œ '1 x Š È"x ‹ dx œ '1 x"Î# dx œ 16

œ

16

' cln kxkd "' " œ ln 4, M œ 1

16

" #

" Èx

2 3

" " x$Î# ‘ "' œ 42; Mx œ ' Š #È x ‹ Š Èx ‹ dx œ " 1 16

"'

dx œ 2x"Î# ‘ " œ 6 Ê x œ

My M

œ 7 and y œ

Mx M

" #

œ

'116 x" dx

ln 4 6

" " 4 '
$Î# dx (b) My œ '1 x Š È"x ‹ Š È4x ‹ dx œ 4'1 dx œ 60, Mx œ '1 Š #È x ‹ Š Èx ‹ Š Èx ‹ dx œ # 1 x 16

16

16

œ 4 x
"Î# ‘" œ 3, M œ '1 Š È"x ‹ Š È4x ‹ dx œ 4'1 "'

yœ 81.

dy dx

82.

d# y dx#

Mx M

œ1

" x

œ

16

16

" x

16

dx œ c4 ln kxkd "' " œ 4 ln 16 Ê x œ

My M

œ

15 ln 16

and

3 4 ln 16

at ("ß 3) Ê y œ x  ln kxk  C; y œ 3 at x œ 1 Ê C œ 2 Ê y œ x  ln kxk  2

œ sec# x Ê

dy dx

œ tan x  C and 1 œ tan 0  C Ê

dy dx

œ tan x  1 Ê y œ ' (tan x  1) dx

œ ln ksec xk  x  C" and 0 œ ln ksec 0k  0  C" Ê C" œ 0 Ê y œ ln ksec xk  x 83. (a) L(x) œ f(0)  f w (0) † x, and f(x) œ ln (1  x) Ê f w (x)k xœ0 œ ww

(b) Let faxb œ lnax  "b. Since f axb œ

"  ax" b#

" ¸ 1x xœ0

œ 1 Ê L(x) œ ln 1  1 † x Ê L(x) œ x


! on Ò!ß !Þ"Ó, the graph of f is concave down on this interval and the

largest error in the linear approximation will occur when x œ !Þ". This error is !Þ"  lna"Þ"b ¸ !Þ!!%'* to five decimal places.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

438

Chapter 7 Transcendental Functions

(c) The approximation y œ x for ln (1  x) is best for smaller positive values of x; in particular for 0 Ÿ x Ÿ 0.1 in the graph. As x increases, so does the error x  ln (1  x). From the graph an upper bound for the error is 0.5  ln (1  0.5) ¸ 0.095; i.e., kE(x)k Ÿ 0.095 for 0 Ÿ x Ÿ 0.5. Note from the graph that 0.1  ln (1  0.1) ¸ 0.00469 estimates the error in replacing ln (1  x) by x over 0 Ÿ x Ÿ 0.1. This is consistent with the estimate given in part (b) above. 84. For all positive values of x,

d ln a x dx c

dœ

1 a x

†  xa2 œ  1x and

d ln a dx c

 ln x d œ 0 

1 x

œ  1x . Since ln xa and ln a  ln x have

the same derivative, then ln xa œ ln a  ln x  C for some constant C. Since this equation holds for all positve values of x, it must be true for x œ 1 Ê ln 1x œ ln 1  ln x  C œ 0  ln x  C Ê ln 1x œ ln x  C. By part 3 we know that ln 1x œ ln x Ê C œ 0 Ê ln xa œ ln a  ln x. 85. y œ ln kx Ê y œ ln x  ln k; thus the graph of y œ ln kx is the graph of y œ ln x shifted vertically by ln k, k  0.

86. To turn the arches upside down we would use the formula y œ  ln ksin xk œ ln ksin" xk .

87. (a)

(b) yw œ

cos x asin x .

Since lsin xl and lcos xl are less than

or equal to 1, we have for a  "
" " w a
" Ÿ y Ÿ a
" for all x.

Thus, lim yw œ ! for all x Ê the graph of y looks aÄ_

more and more horizontal as a Ä  _.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.3 The Exponential Function 88. (a) The graph of y œ Èx  ln x appears to be concave upward for all x  0.

(b) y œ Èx  ln x Ê yw œ

" #È x



" x

Ê yww œ  4x"$Î# 

" x#

œ

" x#

Š

Èx 4

 1‹ œ 0 Ê Èx œ 4 Ê x œ 16.

Thus, yww  0 if 0
x
16 and yww
0 if x  16 so a point of inflection exists at x œ 16. The graph of y œ Èx  ln x closely resembles a straight line for x   10 and it is impossible to discuss the point of inflection visually from the graph. 7.3 THE EXPONENTIAL FUNCTION #

(b) e
ln x œ

1. (a) eln 7.2 œ 7.2 #

#

2. (a) eln ax y b œ x#  y#

(b) e
ln 0Þ3 œ

3. (a) 2 ln Èe œ 2 ln e"Î# œ (2) ˆ "# ‰ ln e œ 1 #

" eln x#

œ

" eln 0 3 Þ

" x#

œ

(c) eln x
ln y œ elnÐxÎyÑ œ

" 0.3

x y

(c) eln 1x
ln 2 œ elnÐ1xÎ2Ñ œ

1x #

(b) ln aln ee b œ ln (e ln e) œ ln e œ 1

#

(c) ln ea
x
y b œ ax#  y# b ln e œ x#  y# 4. (a) ln ˆesec ) ‰ œ (sec ))(ln e) œ sec )

x

(b) ln eae b œ aex b (ln e) œ ex

#

(c) ln ˆe2 ln x ‰ œ ln Šeln x ‹ œ ln x# œ 2 ln x 5. ln y œ 2t  4 Ê eln y œ e2t4 Ê y œ e2t4

6. ln y œ t  5 Ê eln y œ e
t5 Ê y œ e
t5

7. ln (y  40) œ 5t Ê eln Ðy
40) œ e5t Ê y  40 œ e5t Ê y œ e5t  40 " 8. ln (1  2y) œ t Ê eln Ð1
2y) œ et Ê 1  2y œ et Ê 2y œ et  1 Ê y œ  Š e
# ‹ t


1‰ 9. ln (y  1)  ln 2 œ x  ln x Ê ln (y  1)  ln 2  ln x œ x Ê ln ˆ y 2x œ x Ê eln ˆ

y
1‰ 2x

œ ex Ê

Ê y  1 œ 2xex Ê y œ 2xex  1 #

y
1 #x

œ ex

10. ln ay#  1b  ln (y  1) œ ln (sin x) Ê ln Š yy 
1" ‹ œ ln (sin x) Ê ln (y  1) œ ln (sin x) Ê eln Ðy
1Ñ œ eln Ðsin xÑ Ê y  1 œ sin x Ê y œ sin x  1 11. (a) e2k œ 4 Ê ln e2k œ ln 4 Ê 2k ln e œ ln 2# Ê 2k œ 2 ln 2 Ê k œ ln 2 (b) 100e10k œ 200 Ê e10k œ 2 Ê ln e10k œ ln 2 Ê 10k ln e œ ln 2 Ê 10k œ ln 2 Ê k œ (c) ekÎ1000 œ a Ê ln ekÎ1000 œ ln a Ê 12. (a) e5k œ

" 4

k 1000

ln e œ ln a Ê

k 1000

œ ln a Ê k œ 1000 ln a

Ê ln e5k œ ln 4
" Ê 5k ln e œ  ln 4 Ê 5k œ  ln 4 Ê k œ  ln54

(b) 80ek œ 1 Ê ek œ 80
" Ê ln ek œ ln 80
" Ê k ln e œ  ln 80 Ê k œ  ln 80 k (c) eÐln 0Þ8Ñk œ 0.8 Ê ˆeln 0Þ8 ‰ œ 0.8 Ê (0.8)k œ 0.8 Ê k œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

ln 2 10

439

440

Chapter 7 Transcendental Functions

13. (a) e
0Þ3t œ 27 Ê ln e
0Þ3t œ ln 3$ Ê (0.3t) ln e œ 3 ln 3 Ê 0.3t œ 3 ln 3 Ê t œ 10 ln 3 (b) ekt œ "# Ê ln ekt œ ln 2
" œ kt ln e œ  ln 2 Ê t œ  lnk2 t (c) eÐln 0Þ2Ñt œ 0.4 Ê ˆeln 0Þ2 ‰ œ 0.4 Ê 0.2t œ 0.4 Ê ln 0.2t œ ln 0.4 Ê t ln 0.2 œ ln 0.4 Ê t œ

ln 0.4 ln 0.2

14. (a) e
0Þ01t œ 1000 Ê ln e
0Þ01t œ ln 1000 Ê (0.01t) ln e œ ln 1000 Ê 0.01t œ ln 1000 Ê t œ 100 ln 1000 " (b) ekt œ 10 Ê ln ekt œ ln 10
" œ kt ln e œ  ln 10 Ê kt œ  ln 10 Ê t œ  lnk10 " #

(c) eÐln 2Ñt œ Èt

15. e

t Ê ˆeln 2 ‰ œ 2
" Ê 2t œ 2
" Ê t œ 1 Èt

œ x# Ê ln e

#

œ ln x# Ê Èt œ 2 ln x Ê t œ 4(ln x)# #

#

16. ex e2x1 œ et Ê ex 2x1 œ et Ê ln ex 2x1 œ ln et Ê t œ x#  2x  1 17. y œ e
5x Ê yw œ e
5x

d dx

18. y œ e2xÎ3 Ê yw œ e2xÎ3

(5x) Ê yw œ 5e
5x

d dx

19. y œ e5
7x Ê yw œ e5
7x #

d dx

ˆ 2x ‰ Ê yw œ 3

2 3

e2xÎ3

(5  7x) Ê yw œ 7e5
7x #

20. y œ eˆ4Èxx ‰ Ê yw œ eˆ4Èxx ‰

d dx

ˆ4Èx  x# ‰ Ê yw œ Š È2  2x‹ eˆ4Èxx# ‰ x

21. y œ xex  ex Ê yw œ aex  xex b  ex œ xex 22. y œ (1  2x) e
2x Ê yw œ 2e
2x  (1  2x)e
2x

d dx

(2x) Ê yw œ 2e
2x  2(1  2x) e
2x œ 4xe
2x

23. y œ ax#  2x  2b ex Ê yw œ (2x  2)ex  ax#  2x  2b ex œ x# ex 24. y œ a9x#  6x  2b e3x Ê yw œ (18x  6)e3x  a9x#  6x  2b e3x

d dx

(3x) Ê yw œ (18x  6)e3x  3 a9x#  6x  2b e3x

# 3x

œ 27x e

25. y œ e) (sin )  cos )) Ê yw œ e) (sin )  cos ))  e) (cos )  sin )) œ 2e) cos ) 26. y œ ln ˆ3)e
) ‰ œ ln 3  ln )  ln e
) œ ln 3  ln )  ) Ê #

27. y œ cos Še
) ‹ Ê

dy d)

28. y œ )$ e
#) cos 5) Ê

#

œ  sin Še
) ‹ dy d)

d d)

dy d)

#

œ

" )

#

#

Še
) ‹ œ Š sin Še
) ‹‹ Še
) ‹

œ a3)# b ˆe
#) cos 5)‰  a)$ cos 5)b e
#)

œ )# e
#) (3 cos 5)  2) cos 5)  5) sin 5)) 29. y œ ln a3te
t b œ ln 3  ln t  ln e
t œ ln 3  ln t  t Ê

dy dt

œ

" t

d d)

d d)

#

#

a)# b œ 2)e
) sin Še
) ‹

(2))  5(sin 5)) ˆ)$ e
#) ‰

1
t t

1œ

30. y œ ln a2e
t sin tb œ ln 2  ln e
t  ln sin t œ ln 2  t  ln sin t Ê œ

1

dy dt

œ 1  ˆ sin" t ‰

d dt

(sin t) œ 1 

cos t
sin t sin t

31. y œ ln

e) 1  e)

œ ln e)  ln ˆ1  e) ‰ œ )  ln ˆ1+e) ‰ Ê

dy d)

œ 1  ˆ 1 " e) ‰

d d)

ˆ 1  e) ‰ œ 1 

e) 1  e)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

" 1  e)

cos t sin t

Section 7.3 The Exponential Function 441 32. y œ ln

È) 1  È)

œ ln È)  ln Š1  È)‹ Ê

" " œ Š È" ‹ Š È ‹  Š 1 "È) ‹ Š #È ‹œ ) # ) )

33. y œ eÐcos tln tÑ œ ecos t eln t œ tecos t Ê 34. y œ esin t aln t#  1b Ê 35.

'0ln x sin et dt

dy dt

œ Š È" ‹ )

Š1  È)‹
È)

dy dt

2) Š1  È)‹

œ

d d)

ŠÈ)‹  Š 1 "È) ‹

" #) Š1  È)‹

œ ecos t  tecos t

d dt

œ

d d)

Š1  È ) ‹

" #) a1)"Î# b

(cos t) œ (1  t sin t) ecos t

œ esin t (cos t) aln t#  1b  2t esin t œ esin t aln t#  1b (cos t)  2t ‘

Ê yw œ ˆsin eln x ‰ †

d dx

36. y œ 'e4Èx ln t dt Ê yw œ aln e2x b † e2x

dy d)

(ln x) œ

d dx

sin x x

ae2x b  Šln e4Èx ‹ †

d dx

Še4Èx ‹ œ (2x) a2e2x b  ˆ4Èx‰ Še4Èx ‹ †

d dx

ˆ4Èx‰

œ 4xe2x  4Èx e4Èx Š È2x ‹ œ 4xe2x  8e4Èx 37. ln y œ ey sin x Ê Š y" ‹ yw œ ayw ey b (sin x)  ey cos x Ê yw Š y"  ey sin x‹ œ ey cos x Ê yw Š 1
yey sin x ‹ œ ey cos x Ê yw œ y

38. ln xy œ exy Ê ln x  ln y œ exy Ê Ê yw Š 1
ye y

xby

‹œ

xex b y
" x

Ê yw œ

yey cos x 1
yey sin x

" x

 Š y" ‹ yw œ a1  yw b exy Ê yw Š y"  exy ‹ œ exy 

y axex b y
"b x a1
yex b y b

39. e2x œ sin (x  3y) Ê 2e2x œ a1  3yw b cos (x  3y) Ê 1  3yw œ w

Ê y œ

" x

2e2x cos (x  3y)

Ê 3yw œ

2e2x cos (x  3y)

1

2e2x
cos (x  3y) 3 cos (x  3y)

40. tan y œ ex  ln x Ê asec# yb yw œ ex  41.

' ae3x  5e
x b dx œ e3

43.

" x

Ê yw œ

axex  "b cos# y x

42.

' a2ex  3e
2x b dx œ 2ex  #3 e
2x  C

'lnln23 ex dx œ cex d lnln 32 œ eln 3  eln 2 œ 3  2 œ 1

44.

'
lnln32 e
x dx œ ce
x d 0
ln 2 œ e!  eln 2 œ 1  2 œ 1

45.

' 8eÐx1Ñ dx œ 8eÐx1Ñ  C

46.

' 2eÐ2x
1Ñ dx œ eÐ2x
1Ñ  C

47.

'lnln49 exÎ2 dx œ 2exÎ2 ‘ lnln 94 œ 2 eÐln 9ÑÎ2  eÐln 4)Î2 ‘ œ 2 ˆeln 3  eln 2 ‰ œ 2(3  2) œ 2

48.

'0ln 16 exÎ4 dx œ 4exÎ4 ‘ ln0 16 œ 4 ˆeÐln 16ÑÎ4  e0 ‰ œ 4 ˆeln 2  1‰ œ 4(2  1) œ 4

3x

 5e
x  C

49. Let u œ r"Î# Ê du œ "# r
"Î# dr Ê 2 du œ r
"Î# dr; ' eÈÈrr dr œ ' er"Î# † r
"Î# dr œ 2 ' eu du œ 2eu  C œ 2er"Î#  C œ 2eÈr  C 50. Let u œ r"Î# Ê du œ  "# r
"Î# dr Ê 2 du œ r
"Î# dr; ' eÈ
Èrr dr œ ' e
r"Î# † r
"Î# dr œ 2 ' eu du œ 2e
r"Î#  C œ 2e
Èr  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

442

Chapter 7 Transcendental Functions

51. Let u œ t# Ê du œ 2t dt Ê du œ 2t dt;

' 2te
t

dt œ ' eu du œ eu  C œ e
t  C

#

#

" 4

52. Let u œ t% Ê du œ 4t$ dt Ê

'

%

t$ et dt œ " x

53. Let u œ

' ex#

" 4

du œ t$ dt;

' eu du œ 4" et%  C

Ê du œ  x"# dx Ê du œ

" x#

dx;

dx œ ' eu du œ eu  C œ e1Îx  C

1Îx

54. Let u œ x
# Ê du œ 2x
$ dx Ê

' e
x$Î #

dx œ ' e
x † x
$ dx œ
#

1 x

" #

" #

du œ x
$ dx;

' eu du œ "# eu  C œ "# e
x
#  C œ "# e
1Îx#  C

55. Let u œ tan ) Ê du œ sec# ) d); ) œ 0 Ê u œ 0, ) œ

'0

1 Î4

ˆ1  etan ) ‰ sec# ) d) œ '

1Î4

0

1 4

Ê u œ 1;

sec# ) d)  '0 eu du œ ctan )d 0 1

1 Î4

 ceu d "! œ tan ˆ 14 ‰  tan (0)‘  ae"  e! b

œ (1  0)  (e  1) œ e 56. Let u œ cot ) Ê du œ  csc# ) d); ) œ

'1Î4 ˆ1  ecot ) ‰ csc# ) d) œ '1Î4 1 Î2

1Î2

1 4

Ê u œ 1, ) œ

1 2

Ê u œ 0;

csc# ) d)  '1 eu du œ c cot )d 1Î4  ceu d !" œ  cot ˆ 12 ‰  cot ˆ 14 ‰‘  ae!  e" b 0

1Î2

œ (0  1)  (1  e) œ e 57. Let u œ sec 1t Ê du œ 1 sec 1t tan 1t dt Ê

' esec Ð1tÑ sec (1t) tan (1t) dt œ 1" ' eu du œ e1

u

du 1

œ sec 1t tan 1t dt;

Cœ

esec a1tb 1

C

58. Let u œ csc (1  t) Ê du œ  csc (1  t) cot (1  t) dt;

' ecsc Ð1tÑ csc (1  t) cot (1  t) dt œ ' eu du œ eu  C œ ecsc Ð1tÑ  C

59. Let u œ ev Ê du œ ev dv Ê 2 du œ 2ev dv; v œ ln

1 6

Ê u œ 16 , v œ ln

1 #

Ê u œ 1# ;

'lnlnÐÐ11ÎÎ62ÑÑ 2ev cos ev dv œ 2 '11ÎÎ62 cos u du œ c2 sin ud 11ÎÎ26 œ 2 sin ˆ 1# ‰  sin ˆ 16 ‰‘ œ 2 ˆ1  "# ‰ œ 1 #

#

60. Let u œ ex Ê du œ 2xex dx; x œ 0 Ê u œ 1, x œ Èln 1 Ê u œ eln 1 œ 1; Èln 1

'0

2xex cos Šex ‹ dx œ '1 cos u du œ csin ud 1" œ sin (1)  sin (1) œ  sin (1) ¸ 0.84147 #

1

#

61. Let u œ 1  er Ê du œ er dr;

' 1 e e

r

dr œ '

x

dx œ '

r

62.

' 1 " e


x

let u œ e

63.

" u

du œ ln kuk  C œ ln a1  er b  C

ecx ecx  1

dx;

 1 Ê du œ e
x dx Ê  du œ e
x dx;

'

ecx ecx  1

dx œ  '

dy dt

œ et sin aet  2b Ê y œ ' et sin aet  2b dt;

" u

du œ  ln kuk  C œ  ln ae
x  1b  C

let u œ et  2 Ê du œ et dt Ê y œ ' sin u du œ  cos u  C œ  cos aet  2b  C; y(ln 2) œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.3 The Exponential Function 443 Ê  cos ˆeln 2  2‰  C œ 0 Ê  cos (2  2)  C œ 0 Ê C œ cos 0 œ 1; thus, y œ 1  cos aet  2b 64.

dy dt

œ e
t sec# a1e
t b Ê y œ ' e
t sec# a1e
t b dt;

let u œ 1e
t Ê du œ 1e
t dt Ê  1" du œ e
t dt Ê y œ  1" ' sec# u du œ  1" tan u  C œ  1" tan a1e
t b  C; y(ln 4) œ 12 Ê  1" tan ˆ1e
ln 4 ‰  C œ 12 Ê  1" tan ˆ1 † 4" ‰  C œ 12 Ê  1" (1)  C œ

65.

d# y dx#

œ 2e
x Ê
x

Ê y œ 2e 66.

dy dx

2 1

Ê C œ 13 ; thus, y œ

œ 2e
x  C; x œ 0 and

3 1



dy dx

" 1

tan a1e
t b

œ 0 Ê 0 œ 2e!  C Ê C œ 2; thus

dy dx
x

!

 2x  1 œ 2 ae
x  xb  1

 2x  C" ; x œ 0 and y œ 1 Ê 1 œ 2e  C" Ê C" œ 1 Ê y œ 2e

d# y dy " 2t " # 2t Ê dy dt# œ 1  e dt œ t  # e  C; t œ 1 and dt œ 0 Ê 0 œ 1  # e  C Ê dy " 2t " # " # " 2t ˆ" # ‰ dt œ t  # e  # e  1 Ê y œ # t  4 e  # e  1 t  C" ; t œ 1 and y œ " " # " # " 2t " # Ê C" œ  #  4 e Ê y œ # t  4 e  ˆ # e  1‰ t  ˆ #"  4" e# ‰

Cœ

œ 2e
x  2

" #

e#  1; thus " #

1 Ê " œ

 4" e#  #" e#  1  C"

67. f(x) œ ex  2x Ê f w (x) œ ex  2; f w (x) œ 0 Ê ex œ 2 Ê x œ ln 2; f(0) œ 1, the absolute maximum; f(ln 2) œ 2  2 ln 2 ¸ 0.613706, the absolute minimum; f(1) œ e  2 ¸ 0.71828, a relative or local maximum since f ww (x) œ ex is always positive. 68. The function f(x) œ 2esin ÐxÎ2Ñ has a maximum whenever sin

x #

œ 1 and a minimum whenever sin

x #

œ 1.

Therefore the maximums occur at x œ 1  2k(21) and the minimums occur at x œ 31  2k(21), where k is any integer. The maximum is 2e ¸ 5.43656 and the minimum is 2e ¸ 0.73576. 69. f(x) œ x# ln

" x

Ê f w (x) œ 2x ln

" x

 x# Š "" ‹ ax
# b œ 2x ln x

" x

ln x œ  "# . Since x œ 0 is not in the domain of f, x œ e
"Î# œ f w (x)
0 for x 

" Èe

of f assumed at x œ

. Therefore, f Š È"e ‹ œ

" Èe

" e

ln Èe œ

" e

 x œ x(2 ln x  1); f w (x) œ 0 Ê x œ 0 or " Èe

. Also, f w (x)  0 for 0
x


ln e"Î# œ

" #e

ln e œ

" #e

" Èe

and

is the absolute maximum value

.

70. f(x) œ (x  3)# ex Ê f w (x) œ 2(x  3) ex  (x  3)# ex œ (x  3) ex (2  x  3) œ (x  1)(x  3) ex ; thus f w (x)  0 for x
1 or x  3, and f w (x)
0 for 1
x
3 Ê f(1) œ 4e ¸ 10.87 is a local maximum and f(3) œ 0 is a local minimum. Since f(x)   0 for all x, f(3) œ 0 is also an absolute minimum.

71.

'0ln 3 ae2x  ex b dx œ ’ e#

72.

'02 ln 2 ˆexÎ2  e
xÎ2 ‰ dx œ 2exÎ2  2e
xÎ2 ‘ 20 ln 2 œ ˆ2eln 2  2e
ln 2 ‰  a2e!  2e! b œ (4  1)  (2  2) œ 5  4 œ 1

2x

73. L œ '0 É1  1

ex 4

dx Ê

 ex “

dy dx

œ

ln 3 0

exÎ2 #

!

œ Š e #  eln 3 ‹  Š e#  e! ‹ œ ˆ 9#  3‰  ˆ "#  1‰ œ 2 ln 3

8 #

2œ2

Ê y œ exÎ2  C; y(0) œ 0 Ê 0 œ e0  C Ê C œ 1 Ê y œ exÎ2  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

444

Chapter 7 Transcendental Functions

74. S œ 21'0 ˆ e ln 2

œ 21'0

ln 2

œ œ 75. (a)

1 # 1 #

ˆe

y

y

É1  ˆ ey
#ecy ‰# dy œ 21 ' 0

 ecy ‰ #

 ecy ‰ #

ln 2

Ɉ ey #ecy ‰# dy œ 21 '0ln 2 ˆ e

y

ˆe

y

 e cy ‰ #

 e cy ‰ # #

É1  "4 ae2y  2  e
2y b dy

dy œ

1 #

'0ln 2 ae2y  2  e
2y bdy

 "# e2y  2y  "# e
2y ‘ ln 2 œ 1# ˆ "# e2 ln 2  2 ln 2  "# e
2 ln 2 ‰  ˆ "#  0  "# ‰‘ 0 ˆ "# † 4  2 ln 2  "# † 4" ‰ œ 1# ˆ2  8"  2 ln 2‰ œ 1 ˆ 15 ‰ 16  ln 2 " x

(x ln x  x  C) œ x †

d dx

(b) average value œ œ

" e
1

'1

e

" e
1

(e  e  1) œ

76. average value œ

" 2
1

 ln x  1  0 œ ln x

ln x dx œ

" e
1

cx ln x  xd e1 œ

" e
1

[(e ln e  e)  (1 ln 1  1)]

" e
1

'12 "x dx œ cln kxkd #" œ ln 2  ln 1 œ ln 2

77. (a) f(x) œ ex Ê f w (x) œ ex ; L(x) œ f(0)  f w (0)(x  0) Ê L(x) œ 1  x (b) f(0) œ 1 and L(0) œ 1 Ê error œ 0; f(0.2) œ e0 2 ¸ 1.22140 and L(0.2) œ 1.2 Ê error ¸ 0.02140 (c) Since yww œ ex  0, the tangent line approximation always lies below the curve y œ ex . Thus L(x) œ x  1 never overestimates ex . Þ

78. (a) ex e
x œ eÐx
xÑ œ e! œ 1 Ê e
x œ x" x#

(b) y œ ae b

" ex

ex" ex#

for all x;

œ ex" ˆ e"x# ‰ œ ex" e
x# œ ex"
x#

Ê ln y œ x# ln e œ x# x" œ x" x# Ê eln y œ ex" x# Ê y œ ex" x# Ê aex" bx# œ ex" x# x"

79. f(x) œ ln(x)  1 Ê f w (x) œ

" x

Ê x n  1 œ xn 

ln (xn )
1 ˆ x1 ‰ n

Ê xn1 œ xn c2  ln (xn )d . Then x" œ 2

Ê x# œ 2.61370564, x$ œ 2.71624393 and x& œ 2.71828183, where we have used Newton's method. 80. eln x œ x and ln aex b œ x for all x  0 81. Note that y œ ln x and ey œ x are the same curve; '1 ln x dx œ area under the curve between 1 and a; a

'0ln a ey dy œ area to the left of the curve between 0 and ln a. a ln a Ê '1 ln x dx  '0 ey dy œ a ln a.

The sum of these areas is equal to the area of the rectangle

82. (a) y œ ex Ê yww œ ex  0 for all x Ê the graph of y œ ex is always concave upward (b) area of the trapezoid ABCD
'ln a ex dx
area of the trapezoid AEFD Ê ln b


'ln a ex dx
Š e ln b

ln a

e #

ln b

‹ (ln b  ln a). Now

" #

" #

(AB  CD)(ln b  ln a)

(AB  CD) is the height of the midpoint

M œ eÐln a
ln bÑÎ2 since the curve containing the points B and C is linear Ê eÐln a
ln bÑÎ2 (ln b  ln a)
'ln a ex dx
Š e ln b

ln a

 eln b ‹ (ln #

b  ln a)

'ln a ex dx œ cex d lnln ba œ eln b  eln a œ b  a, so part (b) implies that ln b

(c)

eÐln aln bÑÎ2 (ln b  ln a)
b  a
Š e Ê eln aÎ2 † eln bÎ2


b
a ln b
ln a




ab #

ln a

 eln b ‹ (ln #

b  ln a) Ê eÐln aln bÑÎ2


Ê Èeln a Èeln b


b
a ln b
ln a




ab #

b
a ln b
ln a

Ê Èab





ab #

b
a ln b
ln a

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley




ab #

Section 7.4 ax and loga x 7.4 ax and loga x 1. (a) 5log5 7 œ 7

(b) 8log8

È2 œ È 2

#

Þ

(e) log3 È3 œ log3 3"Î# œ

(d) log4 16 œ log4 4 œ 2 log4 4 œ 2 † 1 œ 2 (f) log4 ˆ "4 ‰ œ log4 4
" œ 1 log4 4 œ 1 † 1 œ 1 2. (a) 2log2 3 œ 3

(c) 1.3log1 3 75 œ 75

(b) 10log10 Ð1Î2Ñ œ

(d) log11 121 œ log11 11# œ 2 log11 11 œ 2 † 1 œ 2 (e) log121 11 œ log121 121"Î# œ ˆ "# ‰ log121 121 œ ˆ "# ‰ † 1 œ (f) log3 ˆ "9 ‰ œ log3 3
# œ 2 log3 3 œ 2 † 1 œ 2

" #

log3 3 œ

" #

" #

" #

†1œ

œ 0.5

(c) 1log1 7 œ 7

" #

3. (a) Let z œ log4 x Ê 4z œ x Ê 22z œ x Ê a2z b# œ x Ê 2z œ Èx (b) Let z œ log3 x Ê 3z œ x Ê a3z b# œ x# Ê 32z œ x# Ê 9z œ x# (c) log2 aeÐln 2Ñ sin x b œ log2 2sin x œ sin x 4. (a) Let z œ log5 a3x# b Ê 5z œ 3x# Ê 25z œ 9x% (b) loge aex b œ x x x x x (c) log4 ˆ2e sin x ‰ œ log4 4ˆe sin x‰Î# œ e sin # 5. (a) (c) 6. (a) (b) (c)

log2 x ln x ln x ln x ln 3 ln 3 log3 x œ ln # ƒ ln 3 œ ln # † ln x œ ln 2 logx a ln a ln a ln a ln x# 2 ln x logx# a œ ln x ƒ ln x# œ ln x † ln a œ ln x

(b)

œ

ln b ln a

ƒ

ln a ln b

œ

ln b ln a

†

ln b ln a

b‰ œ ˆ ln ln a

œ

ln x ln #

ƒ

ln x ln 8

œ

ln x ln #

†

ln 8 ln x

œ

3 ln 2 ln 2

œ3

œ2

log9 x ln x ln x ln x ln 3 1 log3 x œ ln 9 ƒ ln 3 œ 2 ln 3 † ln x œ 2 logÈ10 x ˆ "# ‰ ln 2 ln x ln x ln x logÈ2 x œ ln È10 ƒ ln È2 œ ˆ "# ‰ ln 10 † ln x loga b logb a

log2 x log8 x

œ

ln 2 ln 10

#

7. 3log3 Ð7Ñ  2log2 Ð5Ñ œ 5log5 ÐxÑ Ê 7  5 œ x Ê x œ 12 8. 8log8 Ð3Ñ  eln 5 œ x#  7log7 Ð3xÑ Ê 3  5 œ x#  3x Ê 0 œ x#  3x  2 œ (x  1)(x  2) Ê x œ 1 or x œ 2 #

9. 3log3 ax b œ 5eln x  3 † 10log10 Ð2Ñ Ê x# œ 5x  6 Ê x#  5x  6 œ 0 Ê (x  2)(x  3) œ 0 Ê x œ 2 or x œ 3 10. ln e  4
2 log4 ÐxÑ œ #

" x

log10 100 Ê 1  4log4 ax

c# b

œ

#

" x

log10 10# Ê 1  x
# œ ˆ x" ‰ (2) Ê 1 

" x#



2 x

œ0

Ê x  2x  1 œ 0 Ê (x  1) œ 0 Ê x œ 1 12. y œ 3cx Ê yw œ 3cx (ln 3)(1) œ 3cx ln 3

11. y œ 2x Ê yw œ 2x ln 2 13. y œ 5Ès Ê #

14. y œ 2s Ê

dy ds

dy ds

œ 5Ès (ln 5) ˆ "# s
"Î# ‰ œ Š 2lnÈ5s ‹ 5Ès #

#

œ 2s (ln 2)2s œ aln 2# b Šs2s ‹ œ (ln 4)s2s

15. y œ x1 Ê yw œ 1xÐ1
1Ñ 17. y œ (cos ))

È2 Ê

dy d)

#

16. y œ t1
e Ê

dy dt

œ (1  e) t
e

È œ È2 (cos ))Š 2c1‹ (sin ))

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

445

446

Chapter 7 Transcendental Functions

18. y œ (ln ))1 Ê

œ 1(ln ))Ð1
1Ñ ˆ ") ‰ œ

dy d)

1(ln ))Ð1
1Ñ )

19. y œ 7sec ) ln 7 Ê

dy d)

œ a7sec ) ln 7b(ln 7)(sec ) tan )) œ 7sec ) (ln 7)# (sec ) tan ))

20. y œ 3tan ) ln 3 Ê

dy d)

œ a3tan ) ln 3b(ln 3) sec# ) œ 3tan ) (ln 3)# sec# )

21. y œ 2sin 3t Ê

dy dt

œ a2sin 3t ln 2b(cos 3t)(3) œ (3 cos 3t) a2sin 3t b (ln 2)

22. y œ 5c cos 2t Ê

dy dt

œ a5c cos 2t ln 5b(sin 2t)(2) œ (2 sin 2t) a5c cos 2t b (ln 5)

23. y œ log2 5) œ

ln 5) ln #

Ê

24. y œ log3 (1  ) ln 3) œ 25. y œ

ln x ln 4



26. y œ

x ln e ln #5

ln x# ln 4



œ

ln x 2 ln 5

œ

ln (1  ) ln 3) ln 3

2

ln x ln 4

œ ˆ ln"# ‰ ˆ 5") ‰ (5) œ

dy d)

x # ln 5

ln x ln 4



Ê

œ3

ln x 2 ln 5

" ) ln #

œ ˆ ln"3 ‰ ˆ 1  )" ln 3 ‰ (ln 3) œ

dy d)

Ê yw œ

ln x ln 4

" 1  ) ln 3

3 x ln 4

œ ˆ # ln" 5 ‰ (x  ln x) Ê yw œ ˆ # ln" 5 ‰ ˆ1  "x ‰ œ

x
1 2x ln 5

27. y œ log2 r † log4 r œ ˆ lnln #r ‰ ˆ lnln 4r ‰ œ

ln# r (ln 2)(ln 4)

Ê

dy dr

" ˆ"‰ œ ’ (ln 2)(ln 4) “ (2 ln r) r œ

2 ln r r(ln 2)(ln 4)

28. y œ log3 r † log9 r œ ˆ lnln 3r ‰ ˆ lnln 9r ‰ œ

ln# r (ln 3)(ln 9)

Ê

dy dr

" ˆ"‰ œ ’ (ln 3)(ln 9) “ (2 ln r) r œ

2 ln r r(ln 3)(ln 9)

1‰ 29. y œ log3 Šˆ xx  ‹œ
1 ln 3

Ê

dy dx

œ

" x1



" x
1

30. y œ log5 Ɉ 3x7x 2 ‰ œ

" #

ln 7x 

" #

œ

1 ln ˆ xx b c1‰

ln 3

œ

ln 3

Ðln 5ÑÎ2 œ log5 ˆ 3x7x 2 ‰ œ

ln 5

ln (3x  2) Ê

)‰ 32. y œ log7 ˆ sin e) #cos œ ) )

dy d)

œ

cos ) (sin ))(ln 7)

33. y œ log5 ex œ

ln ex ln 5



œ

# #

2 x ln 2



œ

dy dx

7 2†7x dy d)



ln ˆ 3x7x
2 ‰

Ðln 5ÑÎ2

ln 5

3 2†(3x  2)

œ

œ ˆ ln#5 ‰ ”

(3x  2)
3x 2x(3x  2)

œ

ln ˆ 3x7x
2 ‰ ln 5

dy dt

œ

4(x  1)
x 2x(x  1)(ln 2)

œ

" #

ln ˆ 3x7x # ‰

" x(3x  2) " ln 7

cos (log7 ))

2)

" ln 5

œ

œ

2 ln x  2
ln 2
"# ln (x  1) ln 2

3x  4 2x(x  1) ln #

œ c3Ðln tÑÎÐln 2Ñ (ln 3)d ˆ t ln" 2 ‰ œ

3 ln (log2 t) ln 8

•œ

ln ) ‰  ˆ ln ) ‰‘ ˆ ) ln" 7 ‰ œ sin (log7 ))  œ sin ˆ ln 7  ) cos ln 7

ln x#  ln e#
ln 2
ln Èx  1 ln 2

35. y œ 3log2 t œ 3Ðln tÑÎÐln 2Ñ Ê

œ

Ê yw œ

" # (ln 2)(x  1)

36. y œ 3 log8 (log2 t) œ

1‰ œ ln ˆ xx 
1 œ ln (x  1)  ln (x  1)

ln (sin ))  ln (cos ))
ln e)
ln 2) ))
)
) ln 2 œ ln (sin ))  ln (cos ln 7 ln 7 sin ) " ln 2 ˆ " ‰ (cos ))(ln 7)  ln 7  ln 7 œ ln 7 (cot )  tan )  1  ln

x ln 5

34. y œ log2 Š 2Èx xe 1 ‹ œ Ê yw œ

ln 3


2 (x  1)(x
1)

ln ) ‰ 31. y œ ) sin (log7 )) œ ) sin ˆ ln Ê 7

Ê

1 (ln 3) ln Š xx b c1‹

3 ln ˆ lnln 2t ‰ ln 8

Ê

dy dt

" t

alog2 3b 3log2 t

" ˆ " ‰ œ ˆ ln38 ‰ ’ (ln t)/(ln 2) “ t ln # œ

3 t(ln t)(ln 8)

" t(ln t)(ln #)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.4 ax and loga x ln 8  ln ˆtln 2 ‰ ln #

37. y œ log2 a8tln 2 b œ 38. y œ

t ln Šˆeln 3 ‰

sin t

‹

ln 3

t ln ˆ3sin t ‰ ln 3

œ

3 ln 2  (ln 2)(ln t) ln #

œ

œ

œ 3  ln t Ê

œ t sin t Ê

t(sin t)(ln 3) ln 3

40. y œ xÐx
"Ñ Ê ln y œ ln xÐx
"Ñ œ (x  1) ln x Ê Ê yw œ xÐx
1Ñ ˆ1  t

" x

w

dy dt

dy dt

Ê yw œ (x  1)x  x x 1  ln (x  1)‘

œ ln x  (x  1) ˆ x" ‰ œ ln x  1 

w

y y

" x

 ln x‰ " dy y dt

t

œ ˆ #" ‰ (ln t)  ˆ #t ‰ ˆ "t ‰ œ



ln t #

" #

t

œ ˆÈt‰ ˆ ln# t  "# ‰

42. y œ tÈt œ tˆt Ê

" (x  1)

œ ln (x  1)  x †

y y

41. y œ ˆÈt‰ œ ˆt"Î# ‰ œ ttÎ# Ê ln y œ ln ttÎ# œ ˆ #t ‰ ln t Ê Ê

" t

œ sin t  t cos t

dy dt

39. y œ (x  1)x Ê ln y œ ln (x  1)x œ x ln (x  1) Ê

œ

dy dt

"Î# ‰

Ê ln y œ ln tˆt

œ ˆt"Î# ‰ (ln t) Ê

"Î# ‰

" dy y dt

t2 œ Š ln2È ‹ tÈ t t

43. y œ (sin x)x Ê ln y œ ln (sin x)x œ x ln (sin x) Ê w

44. y œ xsin x Ê ln y œ ln xsin x œ (sin x)(ln x) Ê

y y

œ ˆ "# t
"Î# ‰ (ln t)  t"Î# ˆ "t ‰ œ

ln t2 2È t

x‰ œ ln (sin x)  x ˆ cos Ê yw œ (sin x)x cln (sin x)  x cot xd sin x

w

y y

œ (cos x)(ln x)  (sin x) ˆ x" ‰ œ

sin x  x (ln x)(cos x) x

Ê yw œ xsin x ’ sin x  x(lnx x)(cos x) “ 45. y œ xln x , x  0 Ê ln y œ (ln x)# Ê

w

y y

#

œ 2(ln x) ˆ "x ‰ Ê yw œ axln x b Š lnxx ‹ w

46. y œ (ln x)ln x Ê ln y œ (ln x) ln (ln x) Ê

y y

œ ˆ "x ‰ ln (ln x)  (ln x) ˆ ln"x ‰

d dx

(ln x) œ

ln (ln x) x



Ê yw œ Š ln (ln xx)  " ‹ (ln x)ln x 47.

' 5x dx œ ln5 5  C

49.

'01 2c

50.

x

)

'c02 5c

)

48.

d) œ '
2 ˆ "5 ‰ d) œ – 0

)

'1

52. Let u œ x"Î# Ê du œ

'14 È2Èx dx œ '14 2x x

"Î#

" #

!

" #

ln Š "# ‹



" ln Š "# ‹

!

)

ln Š 5" ‹ —

# x2ax b dx œ '1 ˆ "# ‰ 2u du œ

2

— œ

Š "5 ‹

51. Let u œ x# Ê du œ 2x dx Ê

È2

"

)

" #‹ ln Š "# ‹

1 Š ) d) œ '0 ˆ "# ‰ d) œ –

œ

' (1.3)x dx œ ln(1.3) (1.3)  C x

" #

œ

ln Š "# ‹


" 2(ln 1
ln 2)

œ

" # ln 2

c#

œ
#

" ln Š 5" ‹



Š 5" ‹

ln Š 5" ‹

œ

" ln Š 5" ‹

(1  25) œ


24 ln 1
ln 5

" #

du œ x dx; x œ 1 Ê u œ 1, x œ È2 Ê u œ 2;

" #

 ln2 # ‘ # œ ˆ 2 ln" 2 ‰ a2#  2" b œ " u

x
"Î# dx Ê 2 du œ

dx Èx

œ

24 ln 5

" ln #

; x œ 1 Ê u œ 1, x œ 4 Ê u œ 2;

† x
"Î# dx œ 2'1 2u du œ ’ 2ln # “ œ ˆ ln"# ‰ a2$  2# b œ 2

Ðu
1Ñ # "

447

4 ln #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" x

448

Chapter 7 Transcendental Functions

53. Let u œ cos t Ê du œ  sin t dt Ê  du œ sin t dt; t œ 0 Ê u œ 1, t œ

'0

1Î2

7cos t sin t dt œ '1 7u du œ  0

7 ‘! ln 7 " u

œ ˆ ln
"7 ‰ a7!  7b œ

54. Let u œ tan t Ê du œ sec# t dt; t œ 0 Ê u œ 0, t œ " 3‹ ln Š 3" ‹

'01Î4 ˆ 3" ‰tan t sec# t dt œ '01 ˆ 3" ‰u du œ – Š " du u dx

55. Let u œ x2x Ê ln u œ 2x ln x Ê

"

u

1 4

1 #

Ê u œ 0;

6 ln 7

Ê u œ 1; !

"

" " " — œ ˆ ln 3 ‰ ’ˆ 3 ‰  ˆ 3 ‰ “ œ

2 3 ln 3

!

œ 2 ln x  (2x) ˆ x" ‰ Ê

du dx

œ 2u(ln x  1) Ê

x œ 2 Ê u œ 2% œ 16, x œ 4 Ê u œ 4) œ 65,536;

" #

du œ x2x (1  ln x) dx;

'24 x2x (1  ln x) dx œ "# '1665 536 du œ "# cud 6516 536 œ "# (65,536  16) œ 65,520 œ 32,760 # ß

" x

56. Let u œ ln x Ê du œ

'1

2 ln x 2

x

dx œ '0

ln 2

È3 dx œ

ß

dx; x œ 1 Ê u œ 0, x œ 2 Ê u œ ln 2;

2u du œ  ln2 # ‘ 0 œ ˆ ln"# ‰ a2ln 2  2! b œ u

ln 2

2ln 2
" ln #

57.

'

C

58.

'

59.

'03 ŠÈ2  1‹ xÈ2 dx œ ’xŠÈ2
"‹ “ $ œ 3ŠÈ2
"‹

60.

'1e xÐln 2Ñ1 dx œ  lnx # ‘ e1 œ e

61.

'

3x

È

!

x ‰ ˆ"‰  dx œ ' ˆ lnln10 x dx; u œ ln x Ê du œ

log10 x x

" x

È xŠ 2c1‹ dx œ

È

x 2 È2

C

ln 2


1ln 2 ln 2

ln 2

#

'14 logx x dx œ '14 ˆ lnln #x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ x" dx; x œ 1 Ê u œ 0, x œ 4 Ê u œ ln 4‘ 4 x‰ ˆ"‰ ' ln 4 ˆ ln"# ‰ u du œ ˆ ln"# ‰  #" u# ‘ ln0 4 œ ˆ ln"# ‰  #" (ln 4)# ‘ œ (ln2 ln4)# œ (lnln4)4 œ ln 4 Ä '1 ˆ ln ln # x dx œ 0 2

2

" #

(2 ln 2)# œ 2(ln 2)#

64.

'1e 2 ln 10 x(log

65.

'02 logx (x # 2) dx œ ln"# '02 cln (x  2)d ˆ x " # ‰ dx œ ˆ ln"# ‰ ’ (ln (x # 2)) “ # œ ˆ ln"# ‰ ’ (ln#4)

10

x)

dx œ '1

e

(ln 10)(2 ln x) (ln 10)

ˆ x" ‰ dx œ c(ln x)# d e1 œ (ln e)#  (ln 1)# œ 1 #

2

#

'110Î10 log œ

10

(10x) x

dx œ

# ˆ ln"010 ‰ ’ 4(ln#010) “

(ln 2)# # “

"0 ln 10

œ

3 #



"! '110Î10 cln (10x)d ˆ 10x" ‰ dx œ ˆ ln"010 ‰ ’ (ln (10x)) “ #0 #

"Î"!

#

œ ˆ ln"010 ‰ ’ (ln #100)  0

(ln 1)# # “

œ # ln 10

'09 2 logx (x1 1) dx œ ln210 '09 ln (x  1) ˆ x " 1 ‰ dx œ ˆ ln210 ‰ ’ (ln (x# 1)) “ * œ ˆ ln210 ‰ ’ (ln 210) #

10

!

'23 2 logx
(x1
1) dx œ ln22 '23 ln (x  1) ˆ x
" 1 ‰ dx œ ˆ ln22 ‰ ’ (ln (x#
1)) “ $ œ ˆ ln22 ‰ ’ (ln22) 2

(ln 2)# # “

ln 2

œ ln 10 68.

#

!

œ ˆ ln"# ‰ ’ 4(ln# 2) 

67.

#

4 x " " x‰ #‘ % # # # '14 ln 2 log ' 4 ln x " dx œ '1 ˆ lnx2 ‰ ˆ ln x ln # dx œ 1 x dx œ # (ln x) " œ # c(ln 4)  (ln 1) d œ # (ln 4)

œ

66.

2
1 ln 2

dx‘

#

63.

œ

' ˆ lnln10x ‰ ˆ x" ‰ dx œ ln"10 ' u du œ ˆ ln"10 ‰ ˆ #" u# ‰  C œ 2(lnlnx)10  C

Ä 62.

3xŠ 3b1‹ È 3 1

#

#

#



#



(ln ")# # “

(ln ")# # “

œ ln 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

" ln #

Section 7.4 ax and loga x 69.

'

‰ ˆ x" ‰ dx œ (ln 10) ' ˆ ln"x ‰ ˆ x" ‰ dx; u œ ln x Ê du œ œ ' ˆ lnln10 x

dx x log10 x

Ä (ln 10) ' ˆ ln"x ‰ ˆ "x ‰ dx œ (ln 10) '

'

71.

'1ln x "t dt œ cln ktkd ln1 x œ ln kln xk  ln 1 œ ln (ln x), x  1

72.

'1e "t dt œ cln ktkd e1

73.

'11/x "t dt œ cln ktkd 1"Îx œ ln ¸ x" ¸  ln 1 œ aln 1  ln kxkb  ln 1 œ ln x, x  0

74.

" ln a

œ (ln 8)#

dx x ‰# x ˆ ln ln 8

'

(ln x)
# x

dx‘

du œ (ln 10) ln kuk  C œ (ln 10) ln kln xk  C

70.

dx x (log8 x)#

œ'

" u

" x

dx œ (ln 8)#

(ln x)
"
1

#

 C œ  (lnln 8)x  C

x

x

œ ln ex  ln 1 œ x ln e œ x

'1x "t dt œ  ln"a ln ktk‘ x1 œ lnln xa  lnln 1a œ loga x, x  0

75. A œ 'c2 1 2xx# dx œ 2'0 2

2

Ä A œ 2'1

5

76. A œ '
1 1

" u

2x 1  x#

dx; cu œ 1  x# Ê du œ 2x dx; x œ 0 Ê u œ 1, x œ 2 Ê u œ 5d

du œ 2 cln kukd &" œ 2(ln 5  ln 1) œ 2 ln 5

2Ð1
xÑ dx œ 2

'

x

1

ˆ " ‰x
1 #

dx œ 2 –

Š "# ‹

ln Š "# ‹ —

"

œ  ln2# ˆ "#  2‰ œ ˆ ln2# ‰ ˆ 3# ‰ œ

3 ln #


"

77. Let cH$ O d œ x and solve the equations 7.37 œ  log10 x and 7.44 œ  log10 x. The solutions of these equations are 10
(Þ$( and 10
(Þ%% . Consequently, the bounds for cH$ O d are c10
(Þ%% ß 10
(Þ$( d . 78. pH œ  log10 a4.8 ‚ 10
) b œ  (log10 4.8)  8 œ 7.32 79. Let O œ original sound level œ 10 log10 aI ‚ 10"# b db from Equation (6) in the text. Solving O  10 œ 10 log10 akI ‚ 10"# b for k Ê 10 log10 aI ‚ 10"# b  10 œ 10 log10 akI ‚ 10"# b Ê log10 aI ‚ 10"# b  1 k œ log10 akI ‚ 10"# b Ê log10 aI ‚ 10"# b  1 œ log10 k  log10 aI ‚ 10"# b Ê 1 œ log10 k Ê 1 œ lnln10 Ê ln k œ ln 10 Ê k œ 10 80. Sound level with 10I œ 10 log10 a10I ‚ 10"# b œ 10 clog10 10  log10 aI ‚ 10"# bd œ 10  10 log10 aI ‚ 10"# b œ original sound level  10 Ê an increase of 10 db 81. (a) If x œ cH$ O d and S  x œ cOH
d , then x(S  x) œ 10
"% Ê S œ x  and

d# S dx#

œ

2†10
"% x$

10c"% x

Ê

dS dx

œ1

œ

ln b ln a

10
"% x#

 0 Ê a minimum exists at x œ 10
(

(b) pH œ  log10 a10
( b œ 7 (c)

cOHc d cH $ O
d

œ

S
x x

œ

Šx 

10
"% x ‹
x

x

œ

10"% x#

Ê the ratio

82. Yes, it's true for all positive values of a and b: loga b œ

cOHc d c H $ O
d ln b ln a

equals 1 at x œ 10
(

and logb a œ

ln a ln b

Ê

" logb a

œ loga b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

449

450

Chapter 7 Transcendental Functions

83. From zooming in on the graph at the right, we estimate the third root to be x ¸ 0.76666

84. The functions f(x) œ xln 2 and g(x) œ 2ln x appear to have identical graphs for x  0. This is no accident, because xln 2 œ eln 2 ln x œ aeln 2 bln x œ 2ln x . †

85. (a) f(x) œ 2x Ê f w (x) œ 2x ln 2; L(x) œ a2! ln 2b x  2! œ x ln 2  1 ¸ 0.69x  1 (b)

86. (a) f(x) œ log3 x Ê f w (x) œ

" x ln 3

, and f(3) œ

ln 3 ln 3

Ê L(x) œ

" 3 ln 3

(x  3) 

ln 3 ln 3

œ

x 3 ln 3



" ln 3

1

¸ 0.30x  0.09 (b)

87. (a) log3 8 œ (c) log20 17

ln 8 ln 3 ¸ 17 œ ln ln #0

1.89279

(b) log7 0.5 œ

¸ 0.94575

(d) log0 5 7 œ Þ

(e) ln x œ (log10 x)(ln 10) œ 2.3 ln 10 ¸ 5.29595 (g) ln x œ (log2 x)(ln 2) œ 1.5 ln 2 ¸ 1.03972 88. (a) 89.

ln 10 ln #

d ˆ " # dx  # x

† log10 x œ

ln 10 ln #

†

 k‰ œ x and

Since x †

" x

ln x ln 10 d dx aln

œ

ln x ln #

œ log2 x

ln 0.5 ln 7 ¸ 0.35621 ln 7 ln 0.5 ¸ 2.80735

(f) ln x œ (log2 x)(ln 2) œ 1.4 ln 2 ¸ 0.97041 (h) ln x œ (log10 x)(ln 10) œ 0.7 ln 10 ¸ 1.61181 (b)

ln a ln b

† loga x œ

ln a ln b

†

ln x ln a

œ

ln x ln b

œ logb x

x  cb œ x" .

œ " for any x Á !, these two curves will have perpendicular tangent lines.

90. eln x œ x for x  ! and lnaex b œ x for all x Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.5 Exponential Growth and Decay " x

91. Using Newton's Method: faxb œ lnaxb  " Ê f w axb œ

Ê xn" œ xn 

lnaxn b
" " x8

451

Ê xn" œ xn ’#  lnaxn b“.

Then, x1 œ 2, x2 œ 2.61370564, x3 œ 2.71624393, and x& œ 2.71828183. Many other methods may be used. For example, graph y œ ln x  " and determine the zero of y. 92. (a) The point of tangency is apß ln pb and mtangent œ

" p

since

œ x" . The tangent line passes through a!ß !b Ê the

dy dx

equation of the tangent line is y œ "p x. The tangent line also passes throughapß ln pb Ê ln p œ "p p œ " Ê p œ e, and the tangent line equation is y œ "e x. (b)

d# y dx#

œ  x"# for x Á ! Ê y œ ln x is concave downward over its domain. Therefore, y œ ln x lies below the graph of

y œ "e x for all x  !, x Á e, and ln x


for x  !, x Á e.

x e

(c) Multiplying by e, e ln x
x or ln x
x. e (d) Exponentiating both sides of ln xe
x, we have eln x
ex , or xe
ex for all positive x Á e. (e) Let x œ 1 to see that 1e
e1 . Therefore, e1 is bigger. e

7.5 EXPONENTIAL GROWTH AND DECAY 1. (a) y œ y! ekt Ê 0.99y! œ y! e1000k Ê k œ

ln 0.99 1000

¸ 0.00001

(b) 0.9 œ eÐ
0Þ00001)t Ê (0.00001)t œ ln (0.9) Ê t œ (c) y œ y! eÐ20ß000Ñk ¸ y! e
0Þ2 œ y! (0.82) Ê 82% 2. (a)

dp dh

ln (0.9)
0.00001

œ kp Ê p œ p! ekh where p! œ 1013; 90 œ 1013e20k Ê k œ

(b) p œ 1013e
6Þ05 ¸ 2.389 millibars 900 ‰ (c) 900 œ 1013eÐ
0Þ121Ñh Ê 0.121h œ ln ˆ 1013 Ê hœ 3.

dy dt

¸ 10,536 years

ln (90)
ln (1013) 20

ln (1013)
ln (900) 0.121

¸ 0.121

¸ 0.977 km

œ 0.6y Ê y œ y! e
0Þ6t ; y! œ 100 Ê y œ 100e
0Þ6t Ê y œ 100e
0Þ6 ¸ 54.88 grams when t œ 1 hr

4. A œ A! ekt Ê 800 œ 1000e10k Ê k œ

ln (0.8) 10

Ê A œ 1000eÐln (0Þ8ÑÎ10Ñt , where A represents the amount of

sugar that remains after time t. Thus after another 14 hrs, A œ 1000eÐln Ð0Þ8ÑÎ10Ñ24 ¸ 585.35 kg

5. L(x) œ L! e
kx Ê

L! #

œ L! e
18k Ê ln

is one-tenth of the surface value,

L! 10

" #

œ 18k Ê k œ

ln 2 18

¸ 0.0385 Ê L(x) œ L! e
0Þ0385x ; when the intensity

œ L! ec0Þ0385x Ê ln 10 œ 0.0385x Ê x ¸ 59.8 ft

6. V(t) œ V! e
tÎ40 Ê 0.1V! œ V! e
tÎ40 when the voltage is 10% of its original value Ê t œ 40 ln (0.1) ¸ 92.1 sec 7. y œ y! ekt and y! œ 1 Ê y œ ekt Ê at y œ 2 and t œ 0.5 we have 2 œ e0Þ5k Ê ln 2 œ 0.5k Ê k œ Therefore, y œ eÐln 4Ñt Ê y œ e24 ln 4 œ 424 œ 2.81474978 ‚ 1014 at the end of 24 hrs

ln 2 0.5

œ ln 4.

8. y œ y! ekt and y(3) œ 10,000 Ê 10,000 œ y! e3k ; also y(5) œ 40,000 œ y! e5k . Therefore y! e5k œ 4y! e3k Ê e5k œ 4e3k Ê e2k œ 4 Ê k œ ln 2. Thus, y œ y! eÐln 2Ñt Ê 10,000 œ y! e3 ln 2 œ y! eln 8 Ê 10,000 œ 8y! Ê y! œ 10,000 œ 1250 8 9. (a) 10,000ekÐ1Ñ œ 7500 Ê ek œ 0.75 Ê k œ ln 0.75 and y œ 10,000eÐln 0Þ75Ñt . Now 1000 œ 10,000eÐln 0Þ75Ñt 0.1 Ê ln 0.1 œ (ln 0.75)t Ê t œ lnln0.75 ¸ 8.00 years (to the nearest hundredth of a year) (b) 1 œ 10,000eÐln 0Þ75Ñt Ê ln 0.0001 œ (ln 0.75)t Ê t œ

ln 0.0001 ln 0.75

¸ 32.02 years (to the nearest hundredth of a

year)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

452

Chapter 7 Transcendental Functions

10. (a) There are (60)(60)(24)(365) œ 31,536,000 seconds in a year. Thus, assuming exponential growth, P œ 257,313,431ekt and 257,313,432 œ 257,313,431eÐ14kÎ31ß536ß000Ñ Ê ln Š 257,313,432 257,313,431 ‹ œ

14k 31,536,000

Ê k ¸ 0.0087542 (b) P œ 257,313,431eÐ0.0087542Ña"&b ¸ 293,420,847 (to the nearest integer). Answers will vary considerably with the number of decimal places retained. 11. 0.9P! œ P! ek Ê k œ ln 0.9; when the well's output falls to one-fifth of its present value P œ 0.2P! 0.2 Ê 0.2P! œ P! eÐln 0Þ9Ñt Ê 0.2 œ eÐln 0Þ9Ñt Ê ln (0.2) œ (ln 0.9)t Ê t œ ln ln 0.9 ¸ 15.28 yr 12. (a)

dp dx

" œ
100 p Ê

dp p

" " œ
100 dx Ê ln p œ
100 x  C Ê p œ eÐ
0Þ01xCÑ œ eC e
0Þ01x œ C" e
0Þ01x ;

p(100) œ 20.09 Ê 20.09 œ C" eÐ
0Þ01ÑÐ100Ñ Ê C" œ 20.09e ¸ 54.61 Ê p(x) œ 54.61e
0Þ01x (in dollars) (b) p(10) œ 54.61eÐ
0Þ01ÑÐ10Ñ œ $49.41, and p(90) œ 54.61eÐ
0Þ01ÑÐ90Ñ œ $22.20 (c) r(x) œ xp(x) Ê rw (x) œ p(x)  xpw (x); pw (x) œ
.5461e
0Þ01x Ê rw (x) œ (54.61
.5461x)e
0Þ01x . Thus, rw (x) œ 0 Ê 54.61 œ .5461x Ê x œ 100. Since rw  0 for any x  100 and rw  0 for x  100, then r(x) must be a maximum at x œ 100. 13. (a) A! eÐ0Þ04Ñ5 œ A! e0Þ2 (b) 2A! œ A! eÐ0Þ04Ñt Ê ln 2 œ (0.04)t Ê t œ Ê tœ

ln 3 0.04

ln 2 0.04

¸ 17.33 years; 3A! œ A! eÐ0Þ04Ñt Ê ln 3 œ (0.04)t

¸ 27.47 years

14. (a) The amount of money invested A! after t years is A(t) œ A! et (b) If A(t) œ 3A! , then 3A! œ A! et Ê ln 3 œ t or t ¸ 1.099 years (c) At the beginning of a year the account balance is A! et , while at the end of the year the balance is A! eÐt1Ñ . The amount earned is A! eÐt1Ñ
A! et œ A! et (e
1) ¸ 1.7 times the beginning amount. 15. A(100) œ 90,000 Ê 90,000 œ 1000erÐ100Ñ Ê 90 œ e100r Ê ln 90 œ 100r Ê r œ 16. A(100) œ 131,000 Ê 131,000 œ 1000e100r Ê ln 131 œ 100r Ê r œ

ln 131 100

17. y œ y! e
0Þ18t represents the decay equation; solving (0.9)y! œ y! e
0Þ18t Ê t œ 18. A œ A! ekt and Ê tœ

" #

ln 0.05
0.00499

A! œ A! e139k Ê

20. (a) A œ A! e
kt Ê (b)

œ e139k Ê k œ

ln (0.5) 139

¸ 0.0450 or 4.50%

¸ 0.04875 or 4.875% ln (0.9)
0.18

¸ 0.585 days

¸
0.00499; then 0.05A! œ A! e
0Þ00499t

¸ 600 days

19. y œ y! e
kt œ y! e
ÐkÑÐ3ÎkÑ œ y! e
3 œ

" k

" #

ln 90 100

" #

y! e$



y! 20

œ e
2Þ645k Ê k œ

œ (0.05)(y! ) Ê after three mean lifetimes less than 5% remains

ln 2 #.645

¸ 0.262

¸ 3.816 years

ln 2 ‰ ln 2 ‰ (c) (0.05)A œ A exp ˆ
2.645 t Ê
ln 20 œ ˆ
2.645 t Ê tœ

2.645 ln 20 ln #

¸ 11.431 years

21. T
Ts œ (T!
Ts ) e
kt , T! œ 90°C, Ts œ 20°C, T œ 60°C Ê 60
20 œ 70e
10k Ê Ê kœ

ln ˆ 74 ‰ 10

4 7

œ e
10k

¸ 0.05596

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.6 Relative Rates of Growth (a) 35
20 œ 70e
0Þ05596t Ê t ¸ 27.5 min is the total time Ê it will take 27.5
10 œ 17.5 minutes longer to reach 35°C (b) T
Ts œ (T!
Ts ) e
kt , T! œ 90°C, Ts œ
15°C Ê 35  15 œ 105e
0Þ05596t Ê t ¸ 13.26 min 22. T
65° œ (T!
65°) e
kt Ê 35°
65° œ (T!
65°) e
10k and 50°
65° œ (T!
65°) e
20k . Solving
30° œ (T!
65°) e
10k and
15° œ (T!
65°) e
20k simultaneously Ê (T!
65°) e
10k œ 2(T!
65°) e
20k ln 2 Ê e10k œ 2 Ê k œ ln102 and
30° œ T!e
10k65° Ê
30°
e10 ˆ 10 ‰ ‘ œ T!
65° Ê T! œ 65°
30° ˆeln 2 ‰ œ 65°
60° œ 5° 23. T
Ts œ (T!
Ts ) eckt Ê 39
Ts œ (46
Ts ) ec10k and 33
Ts œ (46
Ts ) ec20k Ê 33
Ts 46
Ts

œ ec20k œ aec10k b# Ê

33
Ts 46
Ts

39
Ts 46
Ts

œ ec10k and

#


Ts # # œ Š 39 46
Ts ‹ Ê (33
Ts )(46
Ts ) œ (39
Ts ) Ê 1518
79Ts  Ts

œ 1521
78Ts  T#s Ê
Ts œ 3 Ê Ts œ
3°C 24. Let x represent how far above room temperature the silver will be 15 min from now, y how far above room temperature the silver will be 120 min from now, and t! the time the silver will be 10°C above room temperature. We then have the following time-temperature table: time in min. 0 20 (Now) 35 140 t! temperature Ts  70° Ts  60° Ts  x Ts  y Ts  10° " ‰ T
Ts œ (T!
Ts ) eckt Ê (60  Ts )
Ts œ c(70  Ts )
Ts d ec20k Ê 60 œ 70ec20k Ê k œ ˆ
20 ln ˆ 67 ‰ ¸ 0.00771 (a) T
Ts œ (T!
Ts ) ec0Þ00771t Ê (Ts  x)
Ts œ c(70  Ts )
Ts d e
Ð0Þ00771ÑÐ35Ñ Ê x œ 70e
0Þ26985 ¸ 53.44°C (b) T
Ts œ (T!
Ts ) e
0Þ00771t Ê (Ts  y)
Ts œ c(70  Ts )
Ts d e
Ð0Þ00771ÑÐ140Ñ Ê y œ 70e
1Þ0794 ¸ 23.79°C (c) T
Ts œ (T!
Ts ) e
0Þ00771t Ê (Ts  10)
Ts œ c(70  Ts )
Ts d e
Ð0Þ00771Ñ t! Ê 10 œ 70e
0Þ00771t! " ‰ ln ˆ "7 ‰ œ 252.39 Ê 252.39
20 ¸ 232 minutes from now the Ê ln ˆ "7 ‰ œ
0.00771t! Ê t! œ ˆ
0.00771 silver will be 10°C above room temperature 25. From Example 5, the half-life of carbon-14 is 5700 yr Ê Ê c œ c! e
0Þ0001216t Ê (0.445)c! œ c! e
0Þ0001216t Ê t œ

" ckÐ5700Ñ # c! œ c! e ln (0.445)
0.0001216 ¸ 6659

Ê kœ

ln 2 5700

¸ 0.0001216

years

26. From Exercise 25, k ¸ 0.0001216 for carbon-14. (a) c œ c! e
0Þ0001216t Ê (0.17)c! œ c! e
0Þ0001216t Ê t ¸ 14,571.44 years Ê 12,571 BC (b) (0.18)c! œ c! e
0Þ0001216t Ê t ¸ 14,101.41 years Ê 12,101 BC (c) (0.16)c! œ c! e
0Þ0001216t Ê t ¸ 15,069.98 years Ê 13,070 BC 27. From Exercise 25, k ¸ 0.0001216 for carbon-14. Thus, c œ c! e
0Þ0001216t Ê (0.995)c! œ c! e
0Þ0001216t Ê tœ

ln (0.995)
0.0001216

¸ 41 years old

7.6 RELATIVE RATES OF GROWTH lim x  3 œ lim e"x œ 0 x Ä _ ex xÄ_ $ #x # x cos x 2x (b) slower, lim x esin œ lim 3x  2 sin œ lim 6x  2excos 2x œ lim 6
4esin œ 0 by the x x ex xÄ_ xÄ_ xÄ_ xÄ_ Sandwich Theorem because e2x Ÿ 6
4exsin 2x Ÿ 10 lim e2x œ 0 œ lim "e0x ex for all reals and x Ä _ xÄ_

1. (a) slower,

"Î#

Èx

lim œ lim xex œ lim x Ä _ ex xÄ_ xÄ_ x ˆ 4e ‰x œ _ since (d) faster, lim 4ex œ lim xÄ_ xÄ_ (c) slower,

Š "# ‹ x
"Î# ex

4 e

œ

lim xÄ_

" #Èx ex

œ0

1

x

(e) slower,

lim xÄ_

Š 3# ‹ ex

œ

x lim ˆ 3 ‰ œ 0 since x Ä _ 2e

3 2e

1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

453

454

Chapter 7 Transcendental Functions

(f) slower, (g) same,

lim xÄ_

lim xÄ_

x Š e# ‹

ex

(h) slower,

lim xÄ_

2. (a) slower,

lim xÄ_

(b) slower, œ

lim xÄ_

lim xÄ_

Š "x ‹ ex

(c) slower,

exÎ2 ex

lim xÄ_ œ È0 œ 0

œ

lim xÄ_

" exÎ2

œ

lim xÄ_

" #

œ

log10 x ex

lim xÄ_

10x%  30x  1 ex x ln x
x ex

œ

œ

lim xÄ_

È 1  x% ex

œ

œ0

œ

ln x (ln 10) ex

lim xÄ_

œ

" x

lim xÄ_

40x$  30 ex

x (ln x
1) ex

lim xÄ_ " xex

" #

œ

œ

œ

(ln 10) ex

lim xÄ_

lim xÄ_

ln

lim xÄ_

"20x# ex

œ

" (ln 10)xex

240x ex

lim xÄ_

x
1  x Š "x ‹ ex

œ

œ0

lim xÄ_

œ

240 ex

lim xÄ_

ln x
1  1 ex

œ

œ0 ln x ex

lim xÄ_

œ0

œ É lim xÄ_

1  x% e2x

œ É lim xÄ_

4x$ 2e2x

œ É lim xÄ_

12x# 4e2x

œ É lim xÄ_

24x 8e2x

œ É lim xÄ_

x

Š 5# ‹

x

5 lim œ lim ˆ #5e ‰ œ 0 since 2e 1 x Ä _ ex xÄ_
x e " (e) slower, lim œ lim e2x œ 0 x Ä _ ex xÄ_ xex (f) faster, lim œ lim x œ _ x Ä _ ex xÄ_ (g) slower, since for all reals we have
1 Ÿ cos x Ÿ 1 Ê e
" Ÿ ecos x Ÿ e" Ê

(d) slower,

e
" ex

e" ex ,

e
" ex

Ÿ

ecos x ex

lim œ 0 œ lim so by the Sandwich Theorem we conclude that lim xÄ_ xÄ_ xÄ_ x
1 e " " " (h) same, lim œ lim eÐx
x  1Ñ œ lim e œ e x Ä _ ex xÄ_ xÄ_ 3. (a) same,

lim xÄ_ (b) faster, lim xÄ_ (c) same,

lim xÄ_

(d) same,

lim xÄ_

x#  4x x#

lim 2x  4 œ lim x Ä _ 2x xÄ_ $ œ lim ax
1b œ _ xÄ_

œ

x&
x# x#

È x%  x$ x# (x  3)# x#

œ É lim xÄ_

œ

lim xÄ_

x%  x$ x%

2(x  3) #x

œ

Ÿ ecos x ex

e" ex

and also

œ0

œ1

2 #

œ É lim ˆ1  "x ‰ œ È1 œ 1 xÄ_ lim xÄ_

2 #

œ1

Š "x ‹

lim x ln x œ lim lnxx œ lim œ0 x Ä _ x# xÄ_ xÄ_ 1 x # x x 2) 2 (f) faster, lim 2x# œ lim (ln2x œ lim (ln 2)# 2 œ _ xÄ_ xÄ_ xÄ_ $
x (g) slower, lim x xe# œ lim exx œ lim e"x œ 0 xÄ_ xÄ_ xÄ_ # (h) same, lim 8x œ lim 8 œ 8 # xÄ_ x xÄ_ (e) slower,

# È

x  x " ‰ lim œ lim ˆ1  x$Î# œ1 x Ä _ x# xÄ_ "0x# (b) same, lim œ lim 10 œ 10 x Ä _ x# xÄ_ #
x x e (c) slower, lim œ lim e"x œ 0 x Ä _ x# xÄ_

4. (a) same,

#

#

Š ln x ‹

ln 10 lim log10 x œ lim œ ln"10 lim x Ä _ x# x Ä _ x# xÄ_ x $
x # (e) faster, lim œ lim (x
1) œ _ x Ä _ x# xÄ_

(d) slower,

2 ln x x#

œ

2 ln 10

lim xÄ_

Š x" ‹ 2x

œ

" ln 10

x

" Š 10 ‹

lim œ lim 10"x x# œ 0 x Ä _ x# xÄ_ x (1.1)x (g) faster, lim œ lim (ln 1.1)(1.1) œ lim #x x Ä _ x# xÄ_ xÄ_ x#  100x 100 ˆ ‰ (h) same, lim œ lim 1  œ 1 x# x xÄ_ xÄ_ (f) slower,

(ln 1.1)# (1.1)x #

œ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

lim xÄ_

" x#

œ0

24 16e2x

Section 7.6 Relative Rates of Growth 5. (a) same,

lim xÄ_

log3 x ln x

(b) same,

lim xÄ_

ln 2x ln x

(c) same,

lim xÄ_

(d) faster,

lim xÄ_

(e) faster, (f) same,

œ œ

ln Èx ln x

x Š ln ln 3 ‹

lim xÄ_

ln x

ˆ #2x ‰ ˆ x" ‰

lim xÄ_

œ

œ

lim xÄ_

lim xÄ_

x ln x

œ

lim xÄ_

lim xÄ_

5 ln x ln x

œ

Š"‹

lim xÄ_

" ln 3

œ

" ln 3

" #

œ

" #

œ1

Š "# ‹ ln x

lim xÄ_

Èx ln x

œ

455

œ

ln x

lim xÄ_

Š "# ‹ x
"Î#

x"Î# ln x

œ

lim xÄ_

" Š "x ‹

œ

lim x œ _ xÄ_

Š x" ‹

œ

lim xÄ_

x #Èx

œ

Èx #

lim xÄ_

œ_

lim 5 œ 5 xÄ_

x lim œ lim x ln" x œ 0 x Ä _ ln x xÄ_ x x (h) faster, lim lne x œ lim ˆe" ‰ œ lim xex œ _ xÄ_ xÄ_ x xÄ_

(g) slower,

6. (a) same,

lim xÄ_

(b) same,

lim xÄ_

lim xÄ_

(d) slower,

lim xÄ_

lim xÄ_

(g) slower,

lim xÄ_

lim xÄ_

lim xÄ_

ln x

œ

lim xÄ_

ecx ln x

œ

œ

" ln #

œ

ln x

œ

ln x " ‹ x#

lim xÄ_

lim xÄ_

Š È"x ‹ Š

Š lnlnx2 ‹

œ

x
2 ln x ln x

lim xÄ_

(f) slower,

(h) same,

œ

log10 10x ln x

(c) slower,

(e) faster,

#

log2 x# ln x

Š lnln10x 10 ‹ ln x

œ

" ˆÈx‰ (ln x) " x# ln x

lim xÄ_

" ln 10

ln x# ln x

lim xÄ_

" ex ln x

ln (ln x) ln x

œ

lim xÄ_

ln (2x5) ln x

œ

lim xÄ_

" ln #

ln 10x ln x

œ

lim xÄ_ " ln 10

2 ln x ln x

lim xÄ_

œ

" ln #

"0 Š 10x ‹

Š x" ‹

lim 2 œ xÄ_ œ

" ln 10

2 ln #

lim 1 œ xÄ_

" ln 10

œ0

œ0

lim ˆ x
2‰ œ Š lim x Ä _ ln x xÄ_

lim xÄ_

œ

x ln x ‹


2 œ  lim xÄ_

" Š "x ‹ 


2 œ Š lim x‹
2 œ _ xÄ_

œ0

"/x Š ln x‹

Š x" ‹ Š 2x2
5 ‹ Š x" ‹

œ

lim xÄ_

œ

" ln x

œ0

2x 2x5

lim xÄ_

œ

lim xÄ_

2 #

œ

lim 1 œ 1 xÄ_ x

7.

e lim œ lim exÎ2 œ _ Ê ex grows faster than exÎ2 ; since for x  ee we have ln x  e and lim (lnexx) x Ä _ exÎ2 x Ä _ xÄ_ x x x œ lim ˆ lne x ‰ œ _ Ê (ln x)x grows faster than ex ; since x  ln x for all x  0 and lim (lnxx)x œ lim ˆ lnxx ‰ xÄ_ xÄ_ xÄ_ œ _ Ê xx grows faster than (ln x)x . Therefore, slowest to fastest are: exÎ2 , ex , (ln x)x , xx so the order is d, a, c, b

8.

2) lim (ln 2) œ lim (ln (ln 2))(ln œ lim (ln (ln 2))# (ln 2) œ (ln (ln# 2)) lim (ln 2)x œ 0 #x x Ä _ x# xÄ_ xÄ_ xÄ_ # Ê (ln 2)x grows slower than x# ; lim x2x œ lim (ln2x#)2x œ lim (ln 2)2 # #x œ 0 Ê x# grows slower than 2x ; xÄ_ xÄ_ xÄ_ x x lim 2ex œ lim ˆ 2e ‰ œ 0 Ê 2x grows slower than ex . Therefore, the slowest to the fastest is: (ln 2)x , x# , 2x xÄ_ xÄ_ and ex so the order is c, b, a, d

x

x

#

x

x

#

lim x œ 1 xÄ_ x (b) false; lim x x 5 œ 1" œ 1 xÄ_ (c) true; x  x  5 Ê xx 5  1 if x  1 (or sufficiently large)

9. (a) false;

(d) true; x  2x Ê (e) true;

lim xÄ_

(f) true;

x  ln x x

ex e2x

x 2x

 1 if x  1 (or sufficiently large)

œ lim xÄ0

œ1

ln x x

" ex

1

œ0 Èx x

œ1

" Èx

 2 if x  1 (or sufficiently large)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

456

Chapter 7 Transcendental Functions

(g) false;

È x#  5 x

(h) true;

" Šx
3‹

10. (a) true;

ln x ln 2x

lim xÄ_

Š x" ‹

 œ

œ

Š "x ‹

lim xÄ_

È(x  5)# x



œ

Š #2x ‹

x5 x

œ1

lim 1 œ 1 xÄ_ 5 x

 6 if x  1 (or sufficiently large)

 1 if x  1 (or sufficiently large)

x x3

Š "x


"‹ x# œ 1  "x  2 if x  1 (or sufficiently large) " Š ‹

(b) true;

x

(c) false;

lim xÄ_

Š "x 

" ‹ x#

Š "x ‹

(d) true; 2  cos x Ÿ 3 Ê e x ex x

(e) true;

œ1

(f) true;

lim xÄ_ (g) true; lnln(lnxx)  (h) false;

lim xÄ_

x ex

and

2  cos x # x ex Ä

Ÿ

3 #

if x is sufficiently large

0 as x Ä _ Ê 1  Š "x ‹

lim ln x œ lim xÄ_ x xÄ_ œ 1 if x is sufficiently large

x ln x x# ln x ln x

lim ˆ1
"x ‰ œ 1 xÄ_

œ

œ

ln x ln ax#  1b

œ

lim xÄ_

Š "x ‹ Š

2x

‹ x#
1

œ

1

x ex

 2 if x is sufficiently large

œ0

lim xÄ_

x#  " #x #

œ

lim ˆ "  xÄ_ #

lim xÄ_

f(x) g(x)

œLÁ0 Ê

lim xÄ_

f(x)
L¹  1 if x is sufficiently large Ê L
1  ¹ g(x)

f(x) g(x)

L1 Ê

f(x) g(x)

11. If f(x) and g(x) grow at the same rate, then

Ê f œ O(g). Similarly,

g(x) f(x)

" ‰ #x#

g(x) f(x)

œ

œ " L

" #

Á 0. Then

Ÿ kLk  1 if x is sufficiently large

Ÿ ¸ L" ¸  1 Ê g œ O(f).

12. When the degree of f is less than the degree of g since in that case

lim xÄ_

f(x) g(x)

œ 0.

f(x) lim œ 0 when the degree of f is smaller x Ä _ g(x) (the ratio of the leading coefficients) when the degrees are the same.

13. When the degree of f is less than or equal to the degree of g since than the degree of g, and

lim xÄ_

f(x) g(x)

œ

a b

14. Polynomials of a greater degree grow at a greater rate than polynomials of a lesser degree. Polynomials of the same degree grow at the same rate.

15.

lim xÄ_ œ

16.

17.

ln (x  ") ln x

lim xÄ_

lim xÄ_

œ

x x  999

ln (x  a) ln x

œ

lim xÄ_

" Šx
1‹

Š x" ‹

œ

lim xÄ_

x x 1

œ

lim xÄ_

œ

lim xÄ_

x x a

œ

lim xÄ_

ln (x  999) ln x

Š x
"999 ‹

" 1

œ 1 and

" 1

œ 1. Therefore, the relative rates are the same.

lim xÄ_

œ

lim xÄ_

Š x" ‹

œ1

lim xÄ_

" ‹ Šx
a Š x" ‹

È10x  1 È x 1 lim œ É lim "0xx 1 œ È10 and lim œ É lim x x 1 œ È1 œ 1. Since the growth rate Èx xÄ_ xÄ_ x Ä _ Èx xÄ_ is transitive, we conclude that È10x  1 and Èx  1 have the same growth rate ˆthat of Èx‰ . È x%  x x#

È

%

$

x
x œ É lim x x% x œ 1 and lim œ É lim x x
% x œ 1. Since the growth rate is x# xÄ_ xÄ_ xÄ_ È % % $ transitive, we conclude that Èx  x and x
x have the same growth rate athat of x# b .

18.

lim xÄ_

19.

lim xÄ_

xn ex

œ

%

lim xÄ_

nxnc1 ex

œá œ

lim xÄ_

n! ex

%

$

œ 0 Ê xn œ o aex b for any non-negative integer n

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.6 Relative Rates of Growth 20. If p(x) œ an xn  an
1 xn
1  á  a" x  a! , then

n

457

nc1

lim p(x) œ an lim xex  an
1 lim xex  á x Ä _ ex xÄ_ xÄ_  a" lim exx  a! lim e"x where each limit is zero (from Exercise 19). Therefore, lim p(x) œ0 xÄ_ xÄ_ x Ä _ ex x Ê e grows faster than any polynomial.

21. (a)

x1În ln x

lim xÄ_

œ

lim xÄ_

xÐ1
nÑÎn n ˆ "x ‰

œ ˆ "n ‰

lim x1În œ _ Ê ln x œ o ˆx1În ‰ for any positive integer n xÄ_ '

(b) ln ae17ß000ß000 b œ 17,000,000  Še"(‚"! ‹

1Î10'

œ e"( ¸ 24,154,952.75

(c) x ¸ 3.430631121 ‚ 10"& (d) In the interval c3.41 ‚ 10"& ß 3.45 ‚ 10"& d we have ln x œ 10 ln (ln x). The graphs cross at about 3.4306311 ‚ 10"& .

Š "x ‹

22.

lim xÄ_ œ

23. (a)

œ

ln x an xn  anc1 xn
1  á  a" x  a!

lim xÄ_

" aan b anxn b

lim nÄ_

lim xÄ_

lim Š ln x ‹ xÄ_ xn an
1 Ša n  x  á 

a" xn
1



a! xn

‹

œ

lim xÄ_ – nxn
1 — an

œ 0 Ê ln x grows slower than any non-constant polynomial (n   1) œ

n log2 n n alog2 nb#

slower than n (log2 n)# ; Š "n ‹

(b) œ 0 Ê n log2 n grows

n log2 n n$Î#

lim nÄ_

œ

lim nÄ_

n Š ln ln 2 ‹

n"Î#

" lim œ ln2# lim n"Î# œ0 n Ä _ ˆ "# ‰ n
"Î# nÄ_ $Î# Ê n log2 n grows slower than n . Therefore, n log2 n grows at the slowest rate Ê the algorithm that takes O(n log2 n) steps is the most efficient in the long run.

œ

" ln #

" log2 n

lim nÄ_

#

24. (a)

lim nÄ_

(log2 n)# n

œ

lim nÄ_

œ

2 (ln 2)#

œ

2(ln n) Š "n ‹ (ln 2)#

lim nÄ_ than n; lim nÄ_ œ

lim nÄ_

n Š ln ln 2 ‹

n"Î#

œ

Š "n ‹ 1

œ

lim nÄ_

lim nÄ_

(ln n)# n(ln 2)#

(b)

ln n n

œ 0 Ê (log2 n)# grows slower

" ln #

Š"‹

n

2 (ln 2)#

(log2 n)# Èn log2 n

œ

n Š ln ln 2 ‹

œ

lim nÄ_

lim nÄ_

log2 n Èn

ln n n"Î#

n " lim œ ln2# lim n"Î# œ 0 Ê (log2 n)# grows slower than Èn log2 n. Therefore (log2 n)# grows x Ä _ ˆ "# ‰ n
"Î# nÄ_ at the slowest rate Ê the algorithm that takes O a(log2 n)# b steps is the most efficient in the long run.

œ

" ln #

lim nÄ_

25. It could take one million steps for a sequential search, but at most 20 steps for a binary search because 2"* œ 524,288  1,000,000  1,048,576 œ 2#! . 26. It could take 450,000 steps for a sequential search, but at most 19 steps for a binary search because 2") œ 262,144  450,000  524,288 œ 2"* .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

458

Chapter 7 Transcendental Functions

7.7 INVERSE TRIGONOMETRIC FUNCTIONS 1. (a)

1 4

(b)
13

3. (a)
16

1 6

(c)

2. (a)
14

(b)

1 4

(c)
13

1 3

(b)

(c)
16

4. (a)

1 6

(b)
14

(c)

1 3

5. (a)

1 3

(b)

31 4

(c)

1 6

6. (a)

21 3

(b)

1 4

(c)

51 6

7. (a)

31 4

(b)

1 6

(c)

21 3

8. (a)

1 4

(b)

51 6

(c)

1 3

9. (a)

1 4

(b)
13

(c)

1 6

10. (a)
14

(b)

1 3

(c)
16

11. (a)

31 4

(b)

(c)

21 3

12. (a)

1 4

(b)

51 6

(c)

csc ! œ

13 5 ,

1 6

5 ‰ 13. ! œ sin
" ˆ 13 Ê cos ! œ

12 13 ,

tan ! œ

5 12 ,

sec ! œ

13 12 ,

and cot ! œ

14. ! œ tan
" ˆ 43 ‰ Ê sin ! œ 45 , cos ! œ 35 , sec ! œ 53 , csc ! œ 54 , and cot ! œ 15. ! œ sec
" Š
È5‹ Ê sin ! œ 16. ! œ sec
" Š
17. sin Šcos
"

È13 # ‹

È2 # ‹

Ê sin ! œ

œ sin ˆ 14 ‰ œ

2 È5 ,

cos ! œ
È15 , tan ! œ
2, csc ! œ

3 È13 ,

12 5

3 4

È5 2 ,

cos ! œ
È213 , tan ! œ
3# , csc ! œ

1 3

and cot ! œ
12

È13 3 ,

and cot ! œ
23

18. sec ˆcos
" #" ‰ œ sec ˆ 13 ‰ œ 2

" È2

19. tan ˆsin
" ˆ
"# ‰‰ œ tan ˆ
16 ‰ œ
È"3

20. cot Šsin
" Š


È3 # ‹‹

œ cot ˆ
13 ‰ œ
È"3

21. csc asec
" 2b  cos Štan
" Š
È3‹‹ œ csc ˆcos
" ˆ "# ‰‰  cos ˆ
13 ‰ œ csc ˆ 13 ‰  cos ˆ
13 ‰ œ

2 È3



" #

œ

4  È3 2È 3

22. tan asec
" 1b  sin acsc
" a
2bb œ tan ˆcos
" "1 ‰  sin ˆsin
" ˆ
12 ‰‰ œ tan (0)  sin ˆ
16 ‰ œ 0  ˆ
#" ‰ œ
#" 23. sin ˆsin
" ˆ
"# ‰  cos
" ˆ
"# ‰‰ œ sin ˆ
16 

21 ‰ 3

œ sin ˆ 1# ‰ œ 1

24. cot ˆsin
" ˆ
"# ‰
sec
" 2‰ œ cot ˆ
16
cos
" ˆ "# ‰‰ œ cot ˆ
16
13 ‰ œ cot ˆ
1# ‰ œ 0 25. sec atan
" 1  csc
" 1b œ sec ˆ 14  sin
" 11 ‰ œ sec ˆ 14  1# ‰ œ sec ˆ 341 ‰ œ
È2 26. sec Šcot
" È3  csc
" (
1)‹ œ sec ˆ 16  sin
" (
"1 )‰ œ sec ˆ 1#
27. sec
" ˆsec ˆ
16 ‰‰ œ sec
" Š È23 ‹ œ cos
" Š 28. cot
" ˆcot ˆ
14 ‰‰ œ cot
" (
1) œ

È3 # ‹

œ

1 3


12 ‰ œ sec ˆ
13 ‰ œ 2

1 6

31 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.7 Inverse Trigonometric Functions 29. ! œ tan
"

x #

indicates the diagram

30. ! œ tan
" 2x indicates the diagram

Ê sec ˆtan
" x# ‰ œ sec ! œ

È x#  4 #

Ê sec atan
" 2xb œ sec ! œ È4x#  1

31. ! œ sec
" 3y indicates the diagram

Ê tan asec
" 3yb œ tan ! œ È9y#
1

32. ! œ sec
"

Ê tan ˆsec
" y5 ‰ œ tan ! œ

y 5

indicates the diagram

33. ! œ sin
" x indicates the diagram

34. ! œ cos
" x indicates the diagram

Èy#
25 5

Ê cos asin
" xb œ cos ! œ È1
x#

Ê tan acos
" xb œ tan ! œ

35. ! œ tan
" Èx#
2x indicates the diagram

È 1
x# x

Ê sin Štan
" Èx#
2x‹

Èx#
2x x
1

œ sin ! œ

36. ! œ tan
"

x È x#  1

37. ! œ sin
"

2y 3

38. ! œ sin
"

y 5

indicates the diagram

39. ! œ sec
"

x 4

indicates the diagram

indicates the diagram

indicates the diagram

Ê sin Štan
"

Ê cos ˆsin
"

2y ‰ 3

x È x#  1 ‹

œ cos ! œ

Ê cos ˆsin
" y5 ‰ œ cos ! œ

œ sin ! œ

x È2x#  1

È9
4y# 3

È25
y# 5

Ê sin ˆsec
" x4 ‰ œ sin ! œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Èx#
16 x

459

460

Chapter 7 Transcendental Functions

40. ! œ sec
"

È x#  4 x

È x#  4 ‹ x

Ê sin Šsec
"

indicates the diagram

œ sin ! œ

sin
" x œ

1 #

42.

43. x lim tan
" x œ Ä_

1 #

44. x Ä lim tan
" x œ
1#
_

45. x lim sec
" x œ Ä_

1 #

46. x Ä lim sec
" x œ x Ä lim cos
" ˆ "x ‰ œ
_
_

41.

lim

x Ä 1c

47. x lim csc
" x œ x lim sin
" ˆ "x ‰ œ 0 Ä_ Ä_ 49. y œ cos
" ax# b Ê

dy dx

œ


51. y œ sin
" È2t Ê

dy dt

œ

2x É 1
ax # b#

È2 #

Ê1
ŠÈ2t‹

53. y œ sec
" (2s  1) Ê

dy ds

54. y œ sec
" 5s Ê

5 k5sk È(5s)#
1

dy ds

œ

55. y œ csc
" ax#  1b Ê

56. y œ csc
" ˆ x# ‰ Ê

dy dx

œ

œ


œ


2x È 1
x%

50. y œ cos
" ˆ "x ‰ œ sec
" x Ê

œ

È2 È1
2t#

52. y œ sin
" (1
t) Ê

#

57. y œ sec
" ˆ "t ‰ œ cos
" t Ê

œ

#

59. y œ cot
" Èt œ cot
" t"Î# Ê

œ

62. y œ tan
" (ln x) Ê

dy dx

œ

63. y œ csc
" aet b Ê

dy dt

œ


Š

" ‹ 1
x#

tanc" x

œ


œ


ˆ "x ‰

œ

1  (ln x)#

œ

œ

œ

" k2s  1k Ès#  s


2x ax#  1b Èx%  2x#


" kx k É x


2 #
4 œ kx k È x #
4 4

ˆ 2t3 ‰ #

#

#

¹ t3 ¹ ÊŠ t3 ‹
1

Š "# ‹ tc"Î# 1  at"Î# b

dy dt

#

œ

œ


œ


2t % t# É t 9
9

œ


6 t Èt%
9


" #Èt(1  t)

Š "# ‹ (t
1)c"Î# 1  c(t
1)"Î# d

#

œ


" 2Èt
1 (1  t
1)

œ


" 2tÈt
1

" atan
" xb a1x# b

œ

et ke t k É a e t b #
1

1 #

" x c1  (ln x)# d

œ

" kx k È x #
1


" È1
(1
t)#


" È1
t#

dy dt

dy dt

œ

60. y œ cot
" Èt
1 œ cot
" (t
1)"Î# Ê

dy dx

œ

2x

Ɉ x# ‰#
1

dy dt

2 k2s  1k È4s#  4s

kx #  1 k É a x #  1 b #
1

58. y œ sin
" ˆ t3# ‰ œ csc
" Š t3 ‹ Ê

61. y œ ln atan
" xb Ê

œ

dy dt

dy dx

" ksk È25s#
1

œ

Š "# ‹

¸x¸

cos
" x œ 1

48. x Ä lim csc
" x œ x Ä lim sin
" ˆ "x ‰ œ 0
_
_

2 k2s  1k È(2s  1)#
1

œ


dy dx

lim

x Ä
1 b

2 È x#  4


" Èe2t
1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ


" È2t
t#

Section 7.7 Inverse Trigonometric Functions 64. y œ cos
" ae
t b Ê

dy dt


e
t É1
ae
t b#

œ


œ

e
t È1
e
2t

"Î# 65. y œ sÈ1
s#  cos
" s œ s a1
s# b  cos
" s Ê s# È 1
s#

œ È1
s#


" È 1
s#




œ È1
s#


66. y œ Ès#
1
sec
" s œ as#
1b œ

"Î#

s#  1 È 1
s#


sec
" s Ê

dy ds

œ dy dx

œ a1
s# b

1
s #
s#
1 È 1
s#

 s ˆ "# ‰ a1
s# b


"Î#

(
2s)



"Î#

(2s)


" ks k È s #
1

œ

s È s#
1

" x È x#
1




" kx k È x #
1

œ sin
" x 

1 #


tan
" ax
" b
tan
" x Ê

x È 1
x#




x È 1
x#

dy dx

œ

Š "# ‹ ax#
1b


"Î#

(2x)

#

1  ’ax#
1b"Î# “




" kx k È x #
1

"Î#

Ê

dy dx

œ0



xc# 1  ax
" b#




œ sin
" x  x Š È

dy dx

" 1  x#

" ‹ 1
x#

œ

" x#  1




 ˆ #" ‰ a1
x# b

œ

dy dx

2x 4

x#


tan
" ˆ #x ‰
x –

Š "# ‹ 1  ˆ #x ‰

#

—œ

2x x#  4


tan
" ˆ #x ‰


œ
tan
" ˆ x# ‰ 71.

'È"

72.

'È

dx œ sin
" ˆ x3 ‰  C

9
x# " 1
4x#

" #

œ 73.

' 17 " x

74.

' 9 "3x

75.

'

sin
" u  C œ

#

dx œ '

#

dx œ

" È2

dx xÈ5x#
4

œ

'01 È4 ds

4
s#

" #

" 3

" #

" # ŠÈ17‹  x#

'

œ'

dx xÈ25x#
2

œ

' È1
2(2x)

" #

dx œ

dx œ

dx œ

du uÈ u#
4

' È du

1
u #

, where u œ 2x and du œ 2 dx

" È17

tan
"

" 3È 3

x È17

C

tan
" Š Èx3 ‹  C œ

È3 9

tan
" Š Èx3 ‹  C

, where u œ 5x and du œ 5 dx

sec
" ¹ Èu ¹  C œ 2

œ'

" #

sin
" (2x)  C

" # ŠÈ3‹  x#

du uÈ u#
2

dx œ

#

" È2

5x sec
" ¹ È ¹C 2

, where u œ È5x and du œ È5 dx

sec
" ¸ u# ¸  C œ

" #

" 1  x#

sec
" ¹

" œ
4 sin
" #s ‘ ! œ 4 ˆsin
"

È5x # ¹

" #

C


sin
" 0‰ œ 4 ˆ 16
0‰ œ

21 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ0


"Î#

œ sin
" x

70. y œ ln ax#  4b
x tan
" ˆ x# ‰ Ê

77.

" k sk È s #
1

œ 0, for x  1

69. y œ x sin
" x  È1
x# œ x sin
" x  a1
x# b

'




s ks k
1 ks k È s #
1

68. y œ cot
" ˆ x" ‰
tan
" x œ

76.

" È 1
s#


2s# È1
s#

œ

œ ˆ "# ‰ as#
1b

"Î# 67. y œ tan
" Èx#
1  csc
" x œ tan
" ax#
1b  csc
" x Ê

œ

"Î#

2x 4  x#

(
2x)

461

462 78.

'03

Chapter 7 Transcendental Functions È2Î4

œ

ds È9
4s#

" #

'03

3 œ
"# sin
" u3 ‘ 0

79.

80.

'02 8 dt2t

" È2

'02

œ ’ È"2 †

" È8

#

'c22

œ

œ ’ È"3 †

81.

È2Î2

'cc1

È2Î2

È

" #


È #

œ csec
" kukd
#

82.

È

È2

'c
2Î32Î3 yÈ9ydy
1 œ '

2 #


È #

' È1
34(rdr
1) œ

84.

3 #

85.

' 2  (xdx
1) œ

86.

œ 87.

#

" 3
"

œ

tan

" #

" #

†

#È$
#È$

3 #

du È 4
u#

u È2

1 8

œ

" 4

Štan
"

2È 2 È8


tan
" 0‹ œ

" 4

atan
" 1
tan
" 0b œ

" 4

ˆ 14
0‰ œ

œ

" #È3

’tan
" È3
tan
" Š
È3‹“ œ

" #È3


13
ˆ
13 ‰‘ œ

' 1 duu

#

, where u œ 2y and du œ 2 dy; y œ
1 Ê u œ
2, y œ
1 4




1 3

È2 #

œ
11#

, where u œ 3y and du œ 3 dy; y œ
23 Ê u œ
2, y œ
1 4




1 3

Ê u œ
È 2

È2 3

Ê u œ
È 2

œ
11#

sin
" 2(r
1)  C , where u œ r  1 and du œ dr


"

sec

¸ u# ¸

œ 2 '
1 1

" È2

tan
" Š xÈ
1 ‹  C 2

, where u œ 3x  1 and du œ 3 dx

uCœ



" 3

tan
" (3x  1)  C

du , where u œ 2x
uÈ u#
4 " C œ 4 sec
" ¸ 2x #
1 ¸  C

du , where u œ x uÈu#
25 C œ 5" sec
" ¸ x5 3 ¸  C

du 1  u#

1 16

1 3È 3

, where u œ 2(r
1) and du œ 2 dr

Cœ

sec
" ¸ u5 ¸ 

#

ˆ 14
0‰ œ

, where u œ x
1 and du œ dx

#

'c11ÎÎ22 12cos(sin) d)))

" #

3È 2 #

 C œ 6 sin
" ˆ r# 1 ‰  C

' (x  3)È(xdx 3)
25 œ ' " 5


sin
" 0‹ œ

Ê uœ

, where u œ È3t and du œ È3 dt; t œ
2 Ê u œ
2È3, t œ 2 Ê u œ 2È3

du È 1
u#

du 2  u#

tan
"

È2 #

œ sec
" ¹
È2¹
sec
" k
2k œ

#

œ 89.

œ'

!

du uÈ u#
1

' (2x
1)Èdx(2x
1)
4 œ "# ' œ

88.

sin
" u#

" È2

" 3

'

œ6'

#

#

' 1  (3xdx 1)

3 #

#È #

du uÈ u#
1

sin
" u  C œ

' È4
6 (rdr 1) œ6

œ

#

Šsin
"

œ sec
" ¹
È2¹
sec
" k
2k œ

œ csec
" kukd
# 83.

u È8 “

tan
" u# “
È2

" #

3È 2 4

, where u œ È2t and du œ È2 dt; t œ 0 Ê u œ 0, t œ 2 Ê u œ 2È2

du 4  u#

œ 'c2

dy yÈ4y#
1

œ

tan
"

, where u œ 2s and du œ 2 ds; s œ 0 Ê u œ 0, s œ

du È 9
u#

du 8  u#

'c22È33

" È3

œ

dt 4  3t#

È2

È2Î4

1 and du œ 2 dx

 3 and du œ dx

, where u œ sin ) and du œ cos ) d); ) œ
1# Ê u œ
", ) œ

1 #

Êuœ"

"

œ c2 tan
" ud
" œ 2 atan
" 1
tan
" (
1)b œ 2
14
ˆ
14 ‰‘ œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.7 Inverse Trigonometric Functions 90.

'11ÎÎ64

œ

91.

'0ln

œ
'È3 1

csc# x dx 1  (cot x)#

È3

" c
tan
" ud È$

È3

œ '1

ex dx 1  e2x

È$

'1e

1Î%

' Èy1 dy
y

" #

œ 94.

œ tan
" È3
tan
" 1 œ

1 3

1Î4

du 1  u# , where u œ ln t and 1Î% tan
" ud ! œ 4 ˆtan
" 14
tan
" 0‰

œ

%

" #

' È du

1
u#

#

#

1
u#

œ

Ê u œ È3 , x œ




1 4

du œ

œ " t

1 1#

dt; t œ 1 Ê u œ 0, t œ e1Î4 Ê u œ

œ 4 tan
"

1 4

1 4

" #

sin
" y#  C

, where u œ tan y and du œ sec# y dy

œ sin
" u  C œ sin
" (tan y)  C 95.

'È

96.

' È dx

97.

'
01

dx
x#  4x
3

2x
x#

œ'

œ'

6 dt È3
2t
t#

dx È1
ax#
4x  4b

dx È1
ax#
2x  1b

œ'

œ'

dx È1
(x
2)#

dx È1
(x
1)#

œ sin
" (x
2)  C

œ sin
" (x
1)  C

œ 6 '
1 È4
at#dt 2t  1b œ 6 '
1 È2#
dt(t  1)# œ 6
sin
" ˆ t # 1 ‰‘
" 0

0

!

œ 6
sin
" ˆ "# ‰
sin
" 0‘ œ 6 ˆ 16
0‰ œ 1 98.

'11Î2

6 dt È3  4t
4t#

dt 1 ‰‘ œ 3'1Î2 È4
a4t2#dt
4t  1b œ 3 '1Î2 È2#
2 (2t œ 3
sin
" ˆ 2t
# "Î#
1)# 1

1

œ 3
sin
" ˆ "# ‰
sin
" 0‘ œ 3 ˆ 16
0‰ œ 99.

#

#

"

1 #

' y
dy2y  5 œ ' 4  y dy
2y  1 œ ' #  (ydy
1) #

#

œ

" #

tan
" ˆ y
# 1 ‰  C

100.

' y  6ydy  10 œ ' 1  ay dy 6y  9b œ ' 1  (ydy 3)

101.

'12 x
8 2xdx 2 œ 8'12 1  ax dx
2x  1b œ 8'12 1  (xdx
1)

#

#

#

#

#

œ tan
" (y  3)  C

#

#

œ 8 ctan
" (x
1)d "

œ 8 atan
" 1
tan
" 0b œ 8 ˆ 14
0‰ œ 21 102.

'24

2 dx x#
6x  10
"

œ 2 ctan 103.

'

œ 2'2

4

1


dx 1  ax#
6x  9b tan
" (
1)d œ 2
14

dx œ (x  1)Èx#  2x œ ' Èdu# , u u
1
"

œ sec

'

1 4

Ê uœ1

1 1#

, where u œ y# and du œ 2y dy

sin
" u  C œ

' Èsec1
ytandy y œ ' È du

1 3

1 6

, where u œ ex and du œ ex dx; x œ 0 Ê u œ 1, x œ ln È3 Ê u œ È3

œ 4'0

4 dt t a1  ln# tb

œ c4 93.

where u œ cot x and du œ
csc# x dx; x œ

œ
tan
" 1  tan
" È3 œ
14 

du 1  u#

œ ctan
" ud " 92.

du 1  u# ,

œ 2'2

4

dx 1  (x
3)#

%

œ 2 ctan
" (x
3)d #


ˆ
14 ‰‘ œ 1

dx (x  1)Èx#  2x  1
1

œ'

dx (x  1)È(x  1)#
1

where u œ x  1 and du œ dx

kuk  C œ sec
" kx  1k  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

463

464 104.

Chapter 7 Transcendental Functions

'

dx (x
2)Èx#
4x  3

œ'

" uÈ u#
1
"

œ sec 105.

sinc" x

' Èe

œ'

kuk  C œ sec
" kx
2k  C

c" x

dx œ
' eu du, where u œ cos
" x and du œ

1
x#

' aÈsinc xb "

108.

u$ 3

c" x

C

dx œ ' u# du, where u œ sin
" x and du œ Cœ

asin
" xb 3

$

dx È 1
x#

C

2 3

u$Î#  C œ

2 3

atan
" xb

' atanc" yb"a1  y#b dy œ '

Š

" ‹ 1
y#

'

$Î#

Cœ

dy œ '

tan
" y
"

œ ln kuk  C œ ln ktan

110.


dx È 1
x#

' È1tan
"x# x dx œ ' u"Î# du, where u œ tan
" x and du œ 1 dxx# œ

109.

#

1
x#

œ

dx È 1
x#

C

œ
eu  C œ
ecos 107.

dx (x
2)È(x
2)#
1

dx œ ' eu du, where u œ sin
" x and du œ

1
x#

cos
" x

' Èe

œ'

du, where u œ x
2 and du œ dx

œ eu  C œ esin 106.

dx (x
2)Èx#
4x  4
1

" u

2 3

Éatan
" xb$  C

du, where u œ tan
" y and du œ

dy 1  y#

yk  C

"

" asinc" yb È1  y#

dy œ

'
Ésin
"
y#  dy œ ' u" du, where u œ sin
" y and du œ È1dy
y# 1

y

œ ln kuk  C œ ln ksin
" yk  C 111.

'È22

sec# asec
" xb xÈ x#
1

dx œ '1Î4 sec# u du, where u œ sec
" x and du œ 1Î3

1Î$

œ ctan ud 1Î4 œ tan 112.

'22ÎÈ3

cos asecc" xb xÈ x#
1

114.

sinc" 5x x xÄ0

lim b

xÄ1

œ lim

È x#
1 sec
" x

1 3


sin

ŠÈ

5 ‹ 1
25x#

xÄ0

œ lim b xÄ1

1

2 tanc" 3x# 7x# xÄ0

lim

œ lim

xÄ0

1 6

Š

14x

1 4

,xœ2 Ê uœ

1 3

dx xÈ x#
1

;xœ

1 6

,xœ2 Ê uœ

1 3

œ È3
1

œ

2 È3

Ê uœ

È3
" #

œ lim b xÄ1

tanc" a2x
" b x
"

12x ‹ 1
9x%

; x œ È2 Ê u œ

œ5

"Î#

ax #
1 b sec
" x

115. x lim x tan
" ˆ 2x ‰ œ x lim Ä_ Ä_ 116.

1 4

1Î3

1Î$

lim


tan

dx œ '1Î6 cos u du, where u œ sec
" x and du œ

œ csin ud 1Î' œ sin

113.

1 3

dx xÈ x#
1

Š "# ‹ ax#
1b Œ

œ x lim Ä_

œ lim

6


"Î#

"

x

k k

È

Š

% x Ä 0 7 a1  9x b

x#
1

(2x)




2x
# ‹ 1  4x
#
x
#

œ

œ lim b x kxk œ 1 xÄ1

œ x lim Ä_

2 14x
#

œ2

6 7

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.7 Inverse Trigonometric Functions " #

117. If y œ ln x
œ Š x"


x 1  x#

ln a1  x# b



" x a1  x # b

tanc" x x

tan
" x x# ‹



 C, then dy œ – x"
dx œ

x 1  x#




Š

x ‹
tan
" x 1
x#

x#

x a1  x# b
x$
x  atan
" xba1  x# b x # a1  x # b

dx œ

465

— dx

tan
" x x#

dx,

which verifies the formula 118. If y œ

x% 4

cos
" 5x 

5 4

'È

x% 1
25x#

%

dx, then dy œ ’x$ cos
" 5x  Š x4 ‹ Š È


5 ‹ 1
25x#

 45 Š È

x% ‹“ 1
25x#

dx

œ ax$ cos
" 5xb dx, which verifies the formula #

119. If y œ x asin
" xb
2x  2È1
x# sin
" x  C, then c" # dy œ ’asin
" xb  2x asin xb
2 
2x sin
" x  2È1
x# Š È 1
x#

È 1
x#

" È 1
x # ‹“

#

dx œ asin
" xb dx, which verifies

the formula 120. If y œ x ln aa#  x# b
2x  2a tan
" ˆ xa ‰  C, then dy œ –ln aa#  x# b  #

2x# a#  x#


2

2 # 1  Š x# ‹ —

dx

a

#

x # # œ ’ln aa#  x# b  2 Š aa#   x# ‹
2“ dx œ ln aa  x b dx, which verifies the formula

121.

dy dx

œ

" È 1
x#

122.

dy dx

œ

" x#  1

Ê dy œ

dx È 1
x#

Ê y œ sin
" x  C; x œ 0 and y œ 0 Ê 0 œ sin
" 0  C Ê C œ 0 Ê y œ sin
" x


1 Ê dy œ ˆ 1 " x#
1‰ dx Ê y œ tan
" (x)
x  C; x œ 0 and y œ 1 Ê 1 œ tan
" 0
0  C

Ê C œ 1 Ê y œ tan
" (x)
x  1 123.

dy dx

œ

œ1 124.

dy dx

œ

" Ê dy œ Èdx# Ê y œ sec
" xÈ x#
1 x x
1
13 œ 231 Ê y œ sec
" (x)  231 , x  1 " 1  x#




2 È 1
x#

Ê dy œ Š 1 " x#


kxk  C; x œ 2 and y œ 1 Ê 1 œ sec
" 2  C Ê C œ 1
sec
" 2

2 È 1
x# ‹

dx Ê y œ tan
" x
2 sin
" x  C; x œ 0 and y œ 2

Ê 2 œ tan
" 0
2 sin
" 0  C Ê C œ 2 Ê y œ tan
" x
2 sin
" x  2 125. The angle ! is the large angle between the wall and the right end of the blackboard minus the small angle x ‰ between the left end of the blackboard and the wall Ê ! œ cot
" ˆ 15
cot
" ˆ x3 ‰ . 126. V œ 1'0 c2#
(sec y)# d dy œ 1 c4y
tan yd ! 1Î3

1Î$

œ 1 Š 431
È3‹

127. V œ ˆ "3 ‰ 1r# h œ ˆ 3" ‰ 1(3 sin ))# (3 cos )) œ 91 acos )
cos$ )b, where 0 Ÿ ) Ÿ œ ! Ê sin ) œ 0 or cos ) œ „

" È3

1 #

Ê

dV d)

œ
91(sin )) a1
3 cos# )b

Ê the critical points are: 0, cos
" Š È"3 ‹ , and cos
" Š
È"3 ‹ ; but

cos
" Š
È"3 ‹ is not in the domain. When ) œ 0, we have a minimum and when ) œ cos
" Š È"3 ‹ ¸ 54.7°, we have a maximum volume. ‰ 128. 65°  (90°
" )  (90°
!) œ 180° Ê ! œ 65°
" œ 65°
tan
" ˆ 21 50 ¸ 65°
22.78° ¸ 42.22° 129. Take each square as a unit square. From the diagram we have the following: the smallest angle ! has a tangent of 1 Ê ! œ tan
" 1; the middle angle " has a tangent of 2 Ê " œ tan
" 2; and the largest angle # has a tangent of 3 Ê # œ tan
" 3. The sum of these three angles is 1 Ê !  "  # œ 1 Ê tan
" 1  tan
" 2  tan
" 3 œ 1. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

466

Chapter 7 Transcendental Functions

130. (a) From the symmetry of the diagram, we see that 1
sec
" x is the vertical distance from the graph of y œ sec
" x to the line y œ 1 and this distance is the same as the height of y œ sec
" x above the x-axis at
x; i.e., 1
sec
" x œ sec
" (
x). (b) cos
" (
x) œ 1
cos
" x, where
1 Ÿ x Ÿ 1 Ê cos
" ˆ
"x ‰ œ 1
cos
" ˆ "x ‰, where x   1 or x Ÿ
1 Ê sec
" (
x) œ 1
sec
" x

131. sin
" (1)  cos
" (1) œ

1 #

0œ

1 #

; sin
" (0)  cos
" (0) œ 0 

1 #

œ

1 #

; and sin
" (
1)  cos
" (
1) œ
1#  1 œ 1# .

If x − (
"ß 0) and x œ
a, then sin
" (x)  cos
" (x) œ sin
" (
a)  cos
" (
a) œ
sin
" a  a1
cos
" ab œ 1
asin
" a  cos
" ab œ 1
1# œ 1# from Equations (3) and (4) in the text.

" x

Ê tan ! œ x and tan " œ

132.

Ê

1 #

œ !  " œ tan
" x  tan
"

" x

.

133. (a) Defined; there is an angle whose tangent is 2. (b) Not defined; there is no angle whose cosine is 2. 134. (a) Not defined; there is no angle whose cosecant is "# . (b) Defined; there is an angle whose cosecant is 2. 135. (a) Not defined; there is no angle whose secant is 0. (b) Not defined; there is no angle whose sine is È2. 136. (a) Defined; there is an angle whose cotangent is
"# . (b) Not defined; there is no angle whose cosine is
5.

x ‰ 137. !(x) œ cot
" ˆ 15
cot
" ˆ x3 ‰ , x  0 Ê !w (x) œ


15 225  x#



3 9  x#

œ


15 a9  x# b  3 a225  x# b a225  x# b a9  x# b

; solving

!w (x) œ 0 Ê
135
15x#  675  3x# œ 0 Ê x œ 3È5 ; !w (x)  0 when 0  x  3È5 and !w (x)  0 for x  3È5 Ê there is a maximum at 3È5 ft from the front of the room 138. From the accompanying figure, !  "  ) œ 1, cot ! œ 2
x Ê ) œ 1
cot
" x
cot
" (2 1 1  (2
x)#
a1  x# b " " 1  x#
1  (2
x)# œ a1  x# b c1  (2
x)# d

and cot " œ Ê

œ

d) dx

œ

4
4x a1  x# b c1  (2
x)# d

; solving

d) dx

œ 0 Ê x œ 1;

Ê at x œ 1 there is a maximum ) œ 1
cot


"

139. Yes, sin
" x and
cos
" x differ by the constant

x 1


x)

d) dx

 0 for !  x  1 and
"

1
cot

(2
1) œ 1


d) dx  0 for 1 1 1 4
4 œ #

1 #

140. Yes, the derivatives of y œ
cos
" x  C and y œ cos
" (
x)  C are both 141. csc
" u œ

1 #


sec
" u Ê

d dx

acsc
" ub œ

d dx

x1

ˆ 1#
sec
" u‰ œ 0


du dx

ku k È u #
1

" È 1
x#

œ


du dx

ku k È u #
1

, ku k  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.7 Inverse Trigonometric Functions

467

142. y œ tan
" x Ê tan y œ x Ê Ê asec# yb œ

" 1  x#

dy dx

œ1 Ê

d d dx (tan y) œ dx (x) " " # sec# y œ È Š 1  x# ‹

œ

dy dx

, as indicated by the triangle

143. f(x) œ sec x Ê f w (x) œ sec x tan x Ê

df
" dx ¹xœb

œ

" df dx ¹x œ f
" abb

œ

" secasec
" bbtanasec
" bb

Since the slope of sec
" x is always positive, we the right sign by writing 144. cot
" u œ

1 #


tan
" u Ê

d dx

acot
" ub œ

d dx

du dx

ˆ 1#
tan
" u‰ œ 0


xœ

" . b Š„ È b #
" ‹ " . lx l È x #
"

du

1  u#

145. The functions f and g have the same derivative (for x   0), namely

d
" dx sec

œ

œ
1 dxu#

" Èx (x  1)

. The functions therefore differ

by a constant. To identify the constant we can set x equal to 0 in the equation f(x) œ g(x)  C, obtaining 1‰
" È sin
" (
1) œ 2 tan
" (0)  C Ê
1# œ 0  C Ê C œ
1# . For x   0, we have sin
" ˆ xx
x
 1 œ 2 tan 146. The functions f and g have the same derivative for x  0, namely


" 1  x#

. The functions therefore differ by a

constant for x  0. To identify the constant we can set x equal to 1 in the equation f(x) œ g(x)  C, obtaining sin
" Š È" ‹ œ tan
" 1  C Ê 2

È3

1 4

œ

1 4

 C Ê C œ 0. For x  0, we have sin
"

È3

147. V œ 1 'cÈ3Î3 Š È1" x# ‹ dx œ 1 '
È3Î3 #

œ 1
13
ˆ
16 ‰‘ œ

1Î2

" È 1
x#

149. (a) A(x) œ

1 4

œ tan
"

È3

dx œ 1 ctan
" xd
È3Î3 œ 1 ’tan
" È3
tan
" Š


" x

.

È3 3 ‹“

1# #

148. y œ È1
x# œ a1
x# b œ 2 '0

" 1  x#

" È x#  1

"Î#

Ê yw œ ˆ "# ‰ a1
x# b


"Î#

1Î2

dx œ 2 csin
" xd 0 œ 2 ˆ 16
0‰ œ

(diameter)# œ

1 4

#

"

" 1  x#

1# #


Š
È

# " ‹ “ # 1x

œ

" 1
x#

; L œ 'c1Î2 É1  ayw b# dx 1Î2

1 3

’ È1" x#
Š
È1" x# ‹“ œ

œ 1 ctan
" xd
" œ (1)(2) ˆ 14 ‰ œ (b) A(x) œ (edge)# œ ’ È

(
2x) Ê 1  ayw b# œ

4 1  x#

1 1  x#

Ê V œ 'a A(x) dx œ 'c1 b

1

1 dx 1  x#

Ê V œ 'a A(x) dx œ 'c1 14dxx# b

1

"

œ 4 ctan
" xd
" œ 4 ctan
" (1)
tan
" (
1)d œ 4
14
ˆ
14 ‰‘ œ 21 150. (a) A(x) œ

1 4

È2Î2

œ 'cÈ2Î2 (b) A(x) œ

(diameter)# œ 1 È 1
x#

(diagonal)# 2

1 4

Š

2 % È 1
x#

#


0‹ œ

1 4

È2Î2

ŠÈ

4 ‹ 1
x#

dx œ 1 csin
" xd
È2Î2 œ 1 ’sin
" Š œ

È2Î2

1 2

Š

2 % È 1
x#

#


0‹ œ

2 È 1
x#

È2 # ‹

œ

1 È 1
x#

Ê V œ 'a A(x) dx


sin
" Š


b

È2 # ‹“

œ 1
14
ˆ
14 ‰‘ œ

È2Î2

Ê V œ 'a A(x) dx œ 'cÈ2Î2 b

2 È 1
x#

1# #

dx

œ 2 csin
" xd
È2Î2 œ 2 ˆ 14 † 2‰ œ 1 151. (a) sec
" 1.5 œ cos
" (c) cot
" 2 œ

1 #


tan

" 1.5
"

¸ 0.84107

2 ¸ 0.46365

152. (a) sec
" (
3) œ cos
" ˆ
"3 ‰ ¸ 1.91063 (c) cot


"

(
2) œ

1 #

" ‰ (b) csc
" (
1.5) œ sin
" ˆ
1.5 ¸
0.72973


tan


"

" ‰ (b) csc
" 1.7 œ sin
" ˆ 1.7 ¸ 0.62887

(
2) ¸ 2.67795

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 #

.

468

Chapter 7 Transcendental Functions

153. (a) Domain: all real numbers except those having the form 1#  k1 where k is an integer. Range:
1#  y 

1 #

(b) Domain:
_  x  _; Range:
_  y  _ The graph of y œ tan
" (tan x) is periodic, the graph of y œ tan atan
" xb œ x for
_ Ÿ x  _.

154. (a) Domain:
_  x  _; Range:
1# Ÿ y Ÿ

1 #

(b) Domain:
" Ÿ x Ÿ 1; Range:
" Ÿ y Ÿ 1 The graph of y œ sin
" (sin x) is periodic; the graph of y œ sin asin
" xb œ x for
" Ÿ x Ÿ 1.

155. (a) Domain:
_  x  _; Range: 0 Ÿ y Ÿ 1

(b) Domain:
1 Ÿ x Ÿ 1; Range:
" Ÿ y Ÿ 1 The graph of y œ cos
" (cos x) is periodic; the graph of y œ cos acos
" xb œ x for
" Ÿ x Ÿ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.7 Inverse Trigonometric Functions 156. Since the domain of sec
" x is (
_ß
1]  ["ß _), we have sec asec
" xb œ x for kxk   1. The graph of y œ sec asec
" xb is the line y œ x with the open line segment from (
"ß
") to ("ß ") removed.

157. The graphs are identical for y œ 2 sin a2 tan
" xb œ 4 csin atan
" xbd ccos atan
" xbd œ 4 Š È œ

4x x#  1

x ‹ Š Èx#"  1 ‹ x#  1

from the triangle

158. The graphs are identical for y œ cos a2 sec
" xb œ cos# asec
" xb
sin# asec
" xb œ œ

2
x # x#

" x#




x#
1 x#

from the triangle

159. The values of f increase over the interval [
"ß 1] because f w  0, and the graph of f steepens as the values of f w increase towards the ends of the interval. The graph of f is concave down to the left of the origin where f ww  0, and concave up to the right of the origin where f ww  0. There is an inflection point at x œ 0 where f ww œ 0 and f w has a local minimum value.

160. The values of f increase throughout the interval (
_ß _) because f w  0, and they increase most rapidly near the origin where the values of f w are relatively large. The graph of f is concave up to the left of the origin where f ww  0, and concave down to the right of the origin where f ww  0. There is an inflection point at x œ 0 where f ww œ 0 and f w has a local maximum value.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

469

470

Chapter 7 Transcendental Functions

7.8 HYPERBOLIC FUNCTIONS #

1. sinh x œ
34 Ê cosh x œ È1  sinh# x œ É1  ˆ
34 ‰ œ É1  coth x œ 2. sinh x œ sech x œ 3. cosh x œ œ

8 17

" tanh x

coth x œ

œ

, and csch x œ

4 5

Ê cosh x œ È1  sinh# x œ É1 

4 3

" cosh x

œ

3 5

, and csch x œ

" sinh x

œ

16 9

" sin x

œ É 25 9 œ

25 œ É 16 œ

5 4

œ

17 8

, sech x œ

5 3

, tanh x œ

" tanh x

œ

5. 2 cosh (ln x) œ 2 Š e

6. sinh (2 ln x) œ

" cosh x

, sech x œ

13 12

" cosh x

ln x

 ecln x ‹ #

e2 ln x
ec2 ln x #

œ

œ

5 13

œ eln x 

" eln x

#


#

eln x
eln x #

œ

15 17

, and csch x œ

, and csch x œ

œ

œx Šx#
x"# ‹ #

7. cosh 5x  sinh 5x œ

e5x  e
5x #



e5x
e
5x #

œ e5x

8. cosh 3x
sinh 3x œ

e3x  e
3x #




e3x
e
3x #

œ e
3x


ecx #



e x  e cx ‰ % #

9. (sinh x  cosh x)% œ ˆ e

x

œ
35 ,

sinh x cosh x

œ

ˆ 43 ‰ ˆ 53 ‰

4 5

, coth x œ

" tanh x

œ

5 4

8 15

, tanh x œ

sinh x cosh x

œ

8 ‰ ˆ 15 ˆ 17 ‰ 15

œ

" sinh x

œ

" sinh x

œ

15 8

12 5

, tanh x œ

sinh x cosh x

œ

ˆ 12 ‰ 5 ˆ 13 ‰ 5

œ

5 12

" x

œ

x%
" #x#

œ aex b% œ e4x

10. ln (cosh x  sinh x)  ln (cosh x
sinh x) œ ln acosh# x
sinh# xb œ ln 1 œ 0 11. (a) sinh 2x œ sinh (x  x) œ sinh x cosh x  cosh x sinh x œ 2 sinh x cosh x (b) cosh 2x œ cosh (x  x) œ cosh x cosh x  sinh x sin x œ cosh# x  sinh# x  ecx ‰#
# " " ! 4 a4e b œ 4

12. cosh# x
sinh# x œ ˆ e œ

" 4

a2ex b a2e
x b œ

13. y œ 6 sinh 14. y œ

" #

x 3

Ê

dy dx

x

ˆe

x


e cx ‰ # #

16. y œ t# tanh

" t

" 4

dy dx

œ

" #

caex  e
x b  aex
e
x bd caex  e
x b
aex
e
x bd

(4) œ 1 x 3

[cosh (2x  1)](2) œ cosh (2x  1)

15. y œ 2Èt tanh Èt œ 2t"Î# tanh t"Î# Ê œ sech# Èt 

œ

œ 6 ˆcosh x3 ‰ ˆ "3 ‰ œ 2 cosh

sinh (2x  1) Ê

dy dt

œ
sech# ˆt"Î# ‰‘ ˆ "# t
"Î# ‰ ˆ2t"Î# ‰  ˆtanh t"Î# ‰ ˆt
"Î# ‰

tanh Èt Èt

œ t# tanh t
" Ê

17. y œ ln (sinh z) Ê

dy dz

œ

cosh z sinh z

dy dt

œ csech# at
" bd a
t
# b at# b  (2t) atanh t
" b œ
sech#

œ coth z

,

3 4

É 144 , x  0 Ê sinh x œ Ècosh# x
1 œ É 169 25
1 œ 25 œ

13 5

ˆ
34 ‰ ˆ 54 ‰

sinh x cosh x

œ
34

#

" tanh x

œ

, tanh x œ

17 ‰ 289 64 , x  0 Ê sinh x œ Ècosh# x
1 œ Ɉ 15
1 œ É 225
1 œ É 225 œ

17 15

, coth x œ

4. cosh x œ

" cosh x

œ
35 , sech x œ

9 16

18. y œ ln (cosh z) Ê

dy dz

œ

" t

sinh z cosh z

 2t tanh œ tanh z

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" t

12 13

,

Section 7.8 Hyperbolic Functions 471 19. y œ (sech ))(1
ln sech )) Ê

dy d)

) tanh ) ‰ œ ˆ

sech (sech ))  (
sech ) tanh ))(1
ln sech )) sech )

dy d)

) coth ) ‰ œ (csch )) ˆ

csch  (1
ln csch ))(
csch ) coth )) csch )

œ sech ) tanh )
(sech ) tanh ))(1
ln sech )) œ (sech ) tanh ))[1
(1
ln sech ))] œ (sech ) tanh ))(ln sech )) 20. y œ (csch ))(1
ln csch )) Ê

œ csch ) coth )
(1
ln csch ))(csch ) coth )) œ (csch ) coth ))(1
1  ln csch )) œ (csch ) coth ))(ln csch )) 21. y œ ln cosh v


" #

tanh# v Ê

" #

coth# v Ê

dy dv

œ

dy dv

œ

sinh v cosh v


ˆ "# ‰ (2 tanh v) asech# vb œ tanh v
(tanh v) asech# vb

cosh v sinh v


ˆ "# ‰ (2 coth v) a
csch# vb œ coth v  (coth v) acsch# vb

œ (tanh v) a1
sech# vb œ (tanh v) atanh# vb œ tanh$ v 22. y œ ln sinh v


œ (coth v) a1  csch# vb œ (coth v) acoth# vb œ coth$ v

23. y œ ax#  1b sech (ln x) œ ax#  1b ˆ eln x 2ec ln x ‰ œ ax#  1b ˆ x 2xc" ‰ œ ax#  1b ˆ x#2x1 ‰ œ 2x Ê

dy dx

œ2

2 4x # 24. y œ a4x#
1b csch (ln 2x) œ a4x#
1b ˆ eln 2x 2e
ln 2x ‰ œ a4x#
1b Š 2x
(2x)
" ‹ œ a4x
1b ˆ 4x#
1 ‰

œ 4x Ê

dy dx

œ4

25. y œ sinh
" Èx œ sinh
" ˆx"Î# ‰ Ê

dy dx

Š "# ‹ xc"Î#

œ

É1  ax"Î# b#

26. y œ cosh
" 2Èx  1 œ cosh
" ˆ2(x  1)"Î# ‰ Ê 27. y œ (1
)) tanh
" ) Ê

dy d)

) #  2)
) #
2 )

œ

" #È x È 1  x

dy d)

œ

" #Èx(1  x)

(2) Š "# ‹ (x  1)c"Î#

œ

Éc2(x  1)"Î# d#
1

œ (1
)) ˆ 1
" )# ‰  (
1) tanh
" ) œ

28. y œ a)#  2)b tanh
" ()  1) Ê œ

dy dx

œ

" 1)

" Èx  1 È4x  3

œ

" È4x#  7x  3


tanh
" )

œ a)#  2)b ’ 1
()" 1)# “  (2)  2) tanh
" ()  1)

 (2)  2) tanh
" ()  1) œ (2)  2) tanh
" ()  1)
1

29. y œ (1
t) coth
" Èt œ (1
t) coth
" ˆt"Î# ‰ Ê

30. y œ a1
t# b coth
" t Ê

dy dt

dy dt

œ (1
t) –

Š "# ‹ tc"Î# 1
at"Î# b

#


"

ˆt"Î# ‰ œ



" È 1
x#

—  (
1) coth

" #È t


coth
" Èt

œ a1
t# b ˆ 1
" t# ‰  (
2t) coth
" t œ 1
2t coth
" t

31. y œ cos
" x
x sech
" x Ê

œ

dy dx


" È 1
x#


’x Š


" ‹ xÈ 1
x#

 (1) sech
" x“ œ


" È 1
x#


sech
" x

œ
sech
" x 32. y œ ln x  È1
x# sech
" x œ ln x  a1
x# b œ

" x

 a1
x# b


"

33. y œ csch

"Î#

Š

ˆ "# ‰) Ê


" ‹ xÈ 1
x#

 ˆ "# ‰ a1
x# b

"Î#


"Î#

sech
" x Ê

dy dx

(
2x) sech
" x œ

" x

" x







x È 1
x#

sech
" x œ

)

dy d)

œ


’ln Š "# ‹“ Š "# ‹ )

Š "# ‹

) #

" Ë 1  ”Š # ‹

•

œ
ln (1)
ln (2) œ #) Ê1  Š "# ‹

ln 2 #)

Ê1  Š "# ‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley


x È 1
x#

sech
" x

472

Chapter 7 Transcendental Functions

34. y œ csch
" 2) Ê

œ


dy d)

35. y œ sinh
" (tan x) Ê

œ

dy dx

36. y œ cosh
" (sec x) Ê

(ln 2) 2) 2 ) É 1  a2 ) b

œ

dy dx


ln 2 È1  2#)

œ

sec# x Èsec# x

(sec x)(tan x) Èsec# x
1

œ

(sec x)(tan x) Ètan# x

(b) If y œ sin
" (tanh x)  C, then x# #

œ

sec# x È1  (tan x)#

37. (a) If y œ tan
" (sinh x)  C, then

38. If y œ

#

dy dx dy dx

œ œ

œ

sec# x ksec xk

œ

œ

ksec xk ksec xk ksec xk

(sec x)(tan x) ktan xk

œ ksec xk

œ sec x, 0  x 

1 #

cosh x cosh x 1  sinh# x œ cosh# x œ sech x, which verifies the formula # sech x sech# x È1
tanh# x œ sech x œ sech x, which verifies the formula

sech
" x
"# È1
x#  C, then

œ x sech
" x 

dy dx

x# #

Š


" ‹ xÈ 1
x#



2x 4È 1
x#

œ x sech
" x,

which verifies the formula x#
" #

39. If y œ

coth
" x 

x #

 C, then

dy dx

œ x coth
" x  Š x

#


" ˆ " ‰ # ‹ 1
x#



" #

œ x coth
" x, which verifies

the formula 40. If y œ x tanh
" x 

" #

ln a1
x# b  C, then

dy dx

œ tanh
" x  x ˆ 1
" x# ‰  #" ˆ 1

2xx# ‰ œ tanh
" x, which verifies

the formula 41.

'

sinh 2x dx œ œ

42.

'

sinh

x 5

cosh u #

" #

'

sinh u du, where u œ 2x and du œ 2 dx

Cœ

cosh 2x #

C

dx œ 5 ' sinh u du, where u œ

œ 5 cosh u  C œ 5 cosh 43.

'

44.

' '

" 5

and du œ

4 cosh (3x
ln 2) dx œ

tanh

x 7

4 3

sinh u  C œ

dx œ 7 '

sinh u cosh u

4 3

4 3

'

dx

C

6 cosh ˆ x#
ln 3‰ dx œ 12 ' cosh u du, where u œ œ 12 sinh u  C œ 12 sinh ˆ x#
ln 3‰  C

œ 45.

x 5

x 5

x #


ln 3 and du œ

" #

dx

cosh u du, where u œ 3x
ln 2 and du œ 3 dx

sinh (3x
ln 2)  C

du, where u œ

x 7

and du œ

" 7

dx

œ 7 ln kcosh uk  C" œ 7 ln ¸cosh x7 ¸  C" œ 7 ln ¹ e

xÎ7

 e
xÎ7 ¹ #

 C" œ 7 ln ¸exÎ7  e
xÎ7 ¸
7 ln 2  C"

œ 7 ln kexÎ7  e
xÎ7 k  C 46.

'

coth

) È3

d) œ È 3 '

cosh u sinh u

du, where u œ

œ È3 ln ksinh uk  C" œ È3 ln ¹sinh œ È3 ln ¹e)Î 47.

'

È$


e
)Î

È$

) È3

and du œ

) È3 ¹

d) È3

 C" œ È3 ln ¹ e

¹
È3 ln 2  C" œ È3 ln ¹e)Î

È3

È$
e
)ÎÈ$

)Î

#


e
)Î

È3

¹  C"

¹C

sech# ˆx
"# ‰ dx œ ' sech# u du, where u œ ˆx
"# ‰ and du œ dx œ tanh u  C œ tanh ˆx
"# ‰  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.8 Hyperbolic Functions 473 48.

'

csch# (5
x) dx œ
' csch# u du, where u œ (5
x) and du œ
dx œ
(
coth u)  C œ coth u  C œ coth (5
x)  C

49.

'

dt œ 2 ' sech u tanh u du, where u œ Èt œ t"Î# and du œ

sech Èt tanh Èt Èt

dt 2È t

œ 2(
sech u)  C œ
2 sech Èt  C 50.

'

csch (ln t) coth (ln t) t

dt œ ' csch u coth u du, where u œ ln t and du œ

dt t

œ
csch u  C œ
csch (ln t)  C 51.

x 'lnln24 coth x dx œ 'lnln24 cosh ' 15Î8 sinh x dx œ 3Î4

" u

"&Î) ¸ ¸3¸ ¸ 15 4 ¸ du œ cln kukd $Î% œ ln ¸ 15 8
ln 4 œ ln 8 † 3 œ ln

where u œ sinh x, du œ cosh x dx, the lower limit is sinh (ln 2) œ limit is sinh (ln 4) œ 52.

eln 4
ec ln 4 #

œ

4
Š "4 ‹ #

œ

17Î8 sinh 2x " ' '0ln 2 tanh 2x dx œ '0ln 2 cosh 2x dx œ # 1

eln 2
ec ln 2 #

œ

2
Š "# ‹

5 #

,

œ

3 4

and the upper


ln ˆ 17 ‰ ‘ 8
ln 1 œ

" #

ln

#

15 8

" u

du œ

" #

"(Î)

cln kukd "

" #

œ

17 8

, where

u œ cosh 2x, du œ 2 sinh (2x) dx, the lower limit is cosh 0 œ 1 and the upper limit is cosh (2 ln 2) œ cosh (ln 4) œ 53.

eln 4  ec ln 4 #

c2 ln 2 #

'0ln 2 4ec

)


ln 2‹
Š e

#

e
2 ln 2 # ‹


ec) ‹ #

c ln 2

d) œ 'c ln 4 ae2)  1b d) œ
e#  )‘ c ln 4

"
ln 4‹ œ ˆ 8"
ln 2‰
ˆ 32
ln 4‰ œ


Š0 

)

e0 # ‹“


ec) ‹ #


ln 2

2)

3 32


ln 2  2 ln 2 œ

d) œ 2 '0 a1
ec2) b d) œ 2 ’)  ln 2

œ 2 ˆln 2 

" 8


"# ‰ œ 2 ln 2 

" 4

3 3#

 ln 2

ec#) # “0

ln 2


1 œ ln 4


3 4

'c11ÎÎ44 cosh (tan )) sec# ) d) œ '
11 cosh u du œ csinh ud "
" œ sinh (1)
sinh (
1) œ Š e"
#e
" ‹
Š e
" #
e" ‹
e
" , where u œ tan ), du œ sec# ) d), the lower limit is tan ˆ
14 ‰ œ
1 and the upper

'01Î2 2 sinh (sin )) cos ) d) œ 2'01 sinh u du
"

œee

œ 2 ccosh ud "! œ 2(cosh 1
cosh 0) œ 2 Š e #e

c"


1‹


2, where u œ sin ), du œ cos ) d), the lower limit is sin 0 œ 0 and the upper limit is sin ˆ 1# ‰ œ 1

'12 cosht(ln t) dt œ '0ln 2 cosh u du œ csinh ud ln0 2 œ sinh (ln 2)
sinh (0) œ e u œ ln t, du œ

58.

c2 ln 4

ln 2

e
e
"
e
"  e œe # 1 limit is tan ˆ 4 ‰ œ 1

57.

17 8

sinh ) d) œ '0 4ec) Š e

œ

56.

œ

)

œ 2 ’Šln 2 

55.

#

'cclnln42 2e) cosh ) d) œ 'cclnln42 2e) Š e œ Še

54.

4  Š "4 ‹

œ

" t

ln 2


ec ln 2 #


0œ

2
#

" #

œ

, where

dt, the lower limit is ln 1 œ 0 and the upper limit is ln 2

2 Èx
#
" '14 8 cosh dx œ 16'1 cosh u du œ 16 csinh ud #" œ 16(sinh 2
sinh 1) œ 16 ’Š e
#e ‹
Š e
#e ‹“ Èx #

œ 8 ae#
e
#
e  e
" b , where u œ Èx œ x"Î# , du œ

" #

x
"Î# dx œ

dx 2È x

, the lower limit is È1 œ 1 and the upper

limit is È4 œ 2 59.

3 4

'c0ln 2 cosh# ˆ x# ‰ dx œ 'c0ln 2 cosh#x  " dx œ "# 'c0ln 2 (cosh x  1) dx œ "# csinh x  xd c0 ln 2 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

474

60.

Chapter 7 Transcendental Functions œ

" #

[(sinh 0  0)
(sinh (
ln 2)
ln 2)] œ

œ

" #

ˆ1


" 4

 ln 2‰ œ

3 8



" #

ln 2 œ

" #

’(0  0)
Š e

c ln 2
eln 2 #


ln 2‹“ œ

" #

–


Š "# ‹
2 #

 ln 2—

 ln È2

3 8

'0ln 10 4 sinh# ˆ x# ‰ dx œ '0ln 10 4 ˆ cosh#x
1 ‰ dx œ 2'0ln 10 (cosh x
1) dx œ 2 csinh x
xd ln0 10 œ 2[(sinh (ln 10)
ln 10)
(sinh 0
0)] œ eln 10
ec ln 10
2 ln 10 œ 10


5 25 61. sinh
" ˆ
1#5 ‰ œ ln Š
12  É 144  1‹ œ ln ˆ 32 ‰

63. tanh
" ˆ
"# ‰ œ

" #

65. sech
" ˆ 35 ‰ œ ln Š

67. (a)

'02

È3

dx È 4  x#

" 10


2 ln 10 œ 9.9
2 ln 10

62. cosh
" ˆ 53 ‰ œ ln Š 53  É 25 9
1‹ œ ln 3

(1/2) ln 3 ln Š 11
 (1/2) ‹ œ
#

64. coth
" ˆ 45 ‰ œ

1È1
(9/25) ‹ (3/5)

66. csch
" Š
È"3 ‹ œ ln 
È3 

2 œ
sinh
" x# ‘ 0

œ ln 3

È3

" #

ln Š (9/4) (1/4) ‹ œ

" #

ln 9 œ ln 3

È4/3 Š1/È3‹ 

œ ln Š
È3  2‹

œ sinh
" È3
sinh 0 œ sinh
" È3

(b) sinh
" È3 œ ln ŠÈ3  È3  1‹ œ ln ŠÈ3  2‹

68. (a)

'01Î3 È 6 dx

œ 2'0

1

1  9x#

dx È a#  u# ,

" sinh
" ud !

œ c2

where u œ 3x, du œ 3 dx, a œ 1

œ 2 asinh
" 1
sinh
" 0b œ 2 sinh
" 1

(b) 2 sinh
" 1 œ 2 ln Š1  È1#  1‹ œ 2 ln Š1  È2‹

69. (a)

'52Î4

#

" 1
x#

dx œ ccoth
" xd &Î% œ coth
" 2
coth
"

(b) coth
" 2
coth
" 70. (a)

'01Î2

71. (a)

'13ÎÎ513

" #

" #

œ


ln 3
ln ˆ 9/4 ‰‘ œ 1/4 "Î#

" 1
x #

(b) tanh
"

5 4

dx œ ctanh
" xd ! " #

œ

(1/2) ln Š 11 
(1/2) ‹ œ

dx xÈ1
16x#

œ '4Î5

12Î13

œ tanh
" " #

œ c
sech
" ud 4Î5 œ
sech
" (b)
sech
"

12 13

 sech
"

œ
ln Š 13 

È169
144 ‹ 1#

œ ln ˆ2 † 23 ‰ œ ln 72. (a) (b)

73. (a)

'12 " #

dx xÈ 4  x#

ˆcsch
"

'01

" 3


tanh
" 0 œ tanh
"

" #

where u œ 4x, du œ 4 dx, a œ 1

12 13

œ
ln Š

4 5

ln

ln 3

du , u È a#
u#

12Î13

" #

" #

5 4

 sech
"

4 5

1È1
(12/13)# ‹ (12/13)

 ln Š 5 

È25
16 ‹ 4

 ln Š

1È1
(4/5)# ‹ (4/5)

œ ln ˆ 5 4 3 ‰
ln ˆ 1312 5 ‰ œ ln 2
ln

4 3

# œ

"# csch
" ¸ x# ¸‘ " œ
"# ˆcsch
" 1
csch
" "# ‰ œ " #


csch
" 1‰ œ

cos x È1  sin# x

3 #

dx œ '0

0

" #

’ln Š2 

" È 1  u#

È5/4 (1/2) ‹


ln Š1  È2‹“ œ

" #

ˆcsch
"

" #

" #

5 ln Š 21  ‹  È2


csch
" 1‰

È

!

du œ csinh
" ud ! œ sinh
" 0
sinh
" 0 œ 0, where u œ sin x, du œ cos x dx

(b) sinh
" 0
sinh
" 0 œ ln Š0  È0  1‹
ln Š0  È0  1‹ œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 7.8 Hyperbolic Functions 475 74. (a)

'1e xÈ1 dx(ln x)

œ '0

1

#

du È a#  u#

, where u œ ln x, du œ

" x

dx, a œ 1

"

œ csinh
" ud ! œ sinh
" 1
sinh
" 0 œ sinh
" 1 (b) sinh
" 1
sinh
" 0 œ ln Š1  È1#  1‹
ln Š0  È0#  1‹ œ ln Š1  È2‹ f(x)
f(
x) . Then E(x)  O(x) œ f(x) #f(
x)  f(x)
#f(
x) # f(
x)  f(
(
x)) œ 2f(x) œ f(x) #f(
x) œ E(x) Ê E(x) is even, and # œ f(x). Also, E(
x) œ # O(
x) œ f(
x)
#f(
(
x)) œ
f(x)
#f(
x) œ
O(x) Ê O(x) is odd. Consequently, f(x) can

75. (a) Let E(x) œ

f(x)  f(
x) #

and O(x) œ

be written as

a sum of an even and an odd function. (b) f(x) œ

f(x)  f(
x) #

because

Thus, if f is even f(x) œ

f(x)
f(
x) œ # 2f(x) #  0 and

76. y œ sinh
" x Ê x œ sinh y Ê x œ Ê ey œ

2x „ È4x#  4 #

f(x)
f(
x) # 2f(x) #

0 if f is even and f(x) œ if f is odd, f(x) œ 0 

ey
ecy #

Ê 2x œ ey


" ey

because

f(x)  f(
x) #

œ 0 if f is odd.

Ê 2xey œ e2y
1 Ê e2y
2xey
1 œ 0

Ê ey œ x  Èx#  1 Ê sinh
" x œ y œ ln Šx  Èx#  1‹ . Since ey  0, we cannot

choose ey œ x
Èx#  1 because x
Èx#  1  0. É gk 77. (a) v œ É mg k tanhŒ m t Ê

dv dt

# É gk # É gk É gk œ É mg k ”sech Œ m t•Œ m  œ g sech Œ m t.

# É gk # É gk # Thus m dv dt œ mg sech Œ m t œ mgŒ"
tanh Œ m t œ mg
kv . Also, since tanh x œ ! when x œ !, v œ !

when t œ !. (b)

mg mg mg É kg lim v œ lim É mg lim tanh ŒÉ kg k tanh Œ m t œ É k t Ä m t œ É k (1) œ É k tÄ_ _

tÄ_

160 (c) É 0.005 œ É 160,000 œ 5

400 È5

œ 80È5 ¸ 178.89 ft/sec

78. (a) s(t) œ a cos kt  b sin kt Ê

ds dt

œ
ak sin kt  bk cos kt Ê

d# s dt#

œ
ak# cos kt
bk# sin kt

œ
k# (a cos kt  b sin kt) œ
k# s(t) Ê acceleration is proportional to s. The negative constant
k# implies that the acceleration is directed toward the origin. (b) s(t) œ a cosh kt  b sinh kt Ê

ds dt

œ ak sinh kt  bk cosh kt Ê

d# s dt#

œ ak# cosh kt  bk# sinh kt

œ k# (a cosh kt  b sinh kt) œ k# s(t) Ê acceleration is proportional to s. The positive constant k# implies that the acceleration is directed away from the origin. 79.

dy dx

œ


" xÈ 1
x#



x È 1
x#

Ê yœ'

dx  '


" xÈ 1
x#

x È 1
x#

dx Ê y œ sech
" (x)
È1
x#  C; x œ 1 and

y œ 0 Ê C œ 0 Ê y œ sech
" (x)
È1
x# 80. To find the length of the curve: y œ

" a

cosh ax Ê yw œ sinh ax Ê L œ '0 È1  (sinh ax)# dx b

Ê L œ '0 cosh ax dx œ
"a sinh ax‘ 0 œ b

b

œ
a"# sinh ax‘ 0 œ b

" a#

" a

sinh ab. Then the area under the curve is A œ '0

b

sinh ab œ ˆ "a ‰ ˆ "a sinh ab‰ which is the area of the rectangle of height

" a

" a

and length L

as claimed. 81. V œ 1'0 acosh# x
sinh# xb dx œ 1'0 1 dx œ 21 2

2

82. V œ 21 '0

ln

È3

sech# x dx œ 21 ctanh xd ln0

È3

È3
Š1/È3‹

œ 21 – È

3  Š1/È3‹ —

cosh ax dx

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

476

Chapter 7 Transcendental Functions

83. y œ

" #

cosh 2x Ê yw œ sinh 2x Ê L œ '0

ln

œ ’ "# Š e

2x


ec2x ‹“ # 0

ln

È5

" 4

œ

ˆ5
"5 ‰ œ

È5

È1  (sinh 2x)# dx œ ' 0

ln

È5

cosh 2x dx œ
"# sinh 2x‘ 0

ln

È5

6 5

84. (a) Let the point located at (cosh uß 0) be called T. Then A(u) œ area of the triangle ?OTP minus the area under the curve y œ Èx#
1 from A to T Ê A(u) œ " #

(b) A(u) œ œ

" #

cosh u sinh u
'1

cosh u

cosh# u 

(c) Aw (u) œ 85. y œ 4 cosh

x 4

" #

" #

Èx#
1 dx Ê Aw (u) œ

sinh# u
sinh# u œ

Ê A(u) œ

u #

" #

" #

'1cosh u Èx#
1 dx.

cosh u sinh u
" #

acosh# u  sinh# ub
ŠÈcosh# u
1‹ (sinh u)

acosh# u
sinh# ub œ ˆ "# ‰ (1) œ

" #

 C, and from part (a) we have A(0) œ 0 Ê C œ 0 Ê A(u) œ

u #

Ê u œ 2A

dy # ˆx‰ # ˆx‰ ' Ê 1  Š dy dx ‹ œ 1  sinh 4 œ cosh 4 ; the surface area is S œ c ln 16 21yÊ1  Š dx ‹ dx #

#

ln 81

ln 81 œ 81 'c ln 16 cosh# ˆ x4 ‰ dx œ 41 'c ln 16 ˆ1  cosh #x ‰ dx œ 41
x  2 sinh x# ‘ c ln 16 œ 41
ˆln 81  2 sinh ˆ ln#81 ‰‰
ˆ
ln 16  2 sinh ˆ
ln# 16 ‰‰‘ œ 41 cln (81 † 16)  2 sinh (ln 9)  2 sinh (ln 4)d 15 ‰ œ 41 cln (9 † 4)#  aeln 9
ec ln 9 b  aeln 4
ec ln 4 bd œ 41
2 ln 36  ˆ9
"9 ‰  ˆ4
"4 ‰‘ œ 41 ˆ4 ln 6  80 9  4 ln 81

œ 41 ˆ4 ln 6 

ln 81

320  135 ‰ 36

œ 161 ln 6 

4551 9

86. (a) y œ cosh x Ê ds œ È(dx)#  (dy)# œ È(dx)#  asinh# xb (dx)# œ cosh x dx; Mx œ 'c ln 2 y ds ln 2 ln 2 ln 2 ln 2 œ ' cosh x ds œ ' cosh# x dx œ ' (cosh 2x  1) dx œ
sinh# 2x  x‘ œ "4 aeln 4
ec ln 4 b  ln 2 ln 2

c ln 2

œ

15 16

c ln 2

0

0

ln 2 ln 2  ln 2; M œ 2 '0 È1  sinh# x dx œ 2 '0 cosh x dx œ 2 csinh xd ln0 2 œ eln 2
ec ln 2 œ 2
"# œ 3# .

Therefore, y œ

Mx M

œ

ˆ 15 ‰ 16  ln 2 ˆ #3 ‰

œ

5 8



ln 4 3

, and by symmetry x œ 0.

(b) x œ 0, y ¸ 1.09

87. (a) y œ

H w

‰ cosh ˆ w H x Ê tan 9 œ

dy dx

‰
w ˆ w ‰‘ œ sinh ˆ w ‰ œ ˆH w H sinh H x H x

(b) The tension at P is given by T cos 9 œ H Ê T œ H sec 9 œ HÈ1  tan# 9 œ HÉ1  ˆsinh

w H

# x‰

‰ ˆH‰ ˆw ‰ œ H cosh ˆ w H x œ w w cosh H x œ wy " a

sinh
" as; y œ

" a

cosh ax œ

89. (a) Since the cable is 32 ft long, s œ 16 and x œ 15. From Exercise 88, x œ

" a

sinh
" as Ê 15a œ sinh
" 16a

88. s œ œ

" a

sinh ax Ê sinh ax œ as Ê ax œ sinh
" as Ê x œ

Èsinh# ax  1 œ

" a

Èa# s#  1 œ És# 

" a

" a

Ècosh# ax

" a#

Ê sinh 15a œ 16a.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 7 Practice Exercises (b) The intersection is near (0.042ß 0.672).

(c) Newton's method indicates that at a ¸ 0.0417525 the curves y œ 16a and y œ sinh 15a intersect. " ‰ ¸ 47.90 lb (d) T œ wy ¸ (2 lb) ˆ 0.0417525 (e) The sag is "a cosha"&ab


" a

¸ 4.85 ft.

CHAPTER 7 PRACTICE EXERCISES 1. y œ 10ecxÎ5 Ê

dy dx

È2x Ê

2. y œ È2 e 3. y œ

" 4

" 16

xe4x


œ (10) ˆ
5" ‰ e
xÎ5 œ
2e
xÎ5

È2x œ 2eÈ2x

œ ŠÈ2‹ ŠÈ2‹ e

dy dx

dy dx

œ

" 4

Ê

dy dx

œ x# ca2x
# b e
2x d  e
2x (2x) œ (2  2x)e
2x œ 2e
2Îx (1  x)

e4x Ê

4. y œ x# ec2Îx œ x# e
2x


"

cx a4e4x b  e4x (1)d


dy d)

œ

2(sin ))(cos )) sin# )

6. y œ ln asec# )b Ê

dy d)

œ

2(sec ))(sec ) tan )) sec# )

#

#

ln Š x# ‹ ln #

8. y œ log5 (3x
7) œ 9. y œ 8ct Ê 11. y œ 5x3Þ6 Ê

dy dt

ln (3x
7) ln 5

dy dx

œ

Ê

œ

" ln #

dy dx

2 cos ) sin )


"


"

œ 2 cot )

œ 2 tan )

 Š x# ‹  œ x #

2 (ln 2)x

œ ˆ ln"5 ‰ ˆ 3x3
7 ‰ œ

œ 8ct (ln 8)(
1) œ
8ct (ln 8)

3 (ln 5)(3x
7)

10. y œ 92t Ê

dy dt

œ 92t (ln 9)(2) œ 92t (2 ln 9)

œ 5(3.6)x2Þ6 œ 18x2Þ6

dy dx

È2 Ê

12. y œ È2 x


Ê

a4e4x b œ xe4x  4" e4x
4" e4x œ xe4x


"

5. y œ ln asin# )b Ê

7. y œ log2 Š x# ‹ œ

" 16

dy dx

È È œ ŠÈ2‹ Š
È2‹ xŠ
2
1‹ œ
2xŠ
2
1‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

477

478

Chapter 7 Transcendental Functions

13. y œ (x
2)xb2 Ê ln y œ ln (x
2)xb2 œ (x
2) ln (x
2) Ê Ê

dy dx

w

y y

œ (x
2)xb2 cln (x
2)
1d

" ‰ œ (x
2) ˆ x
#
(1) ln (x
2)

w

14. y œ 2(ln x)xÎ2 Ê ln y œ ln c2(ln x)xÎ2 d œ ln (2)
ˆ x# ‰ ln (ln x) Ê Ê yw œ
# ln" x
ˆ "# ‰ ln (ln x)‘ 2 (ln x)xÎ2 œ (ln x)xÎ2
ln (ln x)
15. y œ sin" È1  u# œ sin" a1  u# b œ

u uÈ 1  u#

" È 1  u#

œ

Ê

dy du

œ

" #

a1  u # b


1 Î2

"Î#

#

u È 1  u # È 1  a1  u # b

œ

“

u ku k È 1  u #

œ

œ

dy dv



" v
$Î# #

É1  av
"Î# b#

" 2v$Î# È1  v
"

œ

œ

"

1 2v$Î# É v
v

œ

È v 2v$Î# Èv  1

" 2vÈv  1

17. y œ ln acos" xb Ê yw œ

Š È
"

1
x#

cos
" x

‹

œ

" È1  x# cos
" x

18. y œ z cos" z  È1  z# œ z cos" z  a1  z# b "

œ cos

z

z È 1  z#




"

œ cos

z È 1  z#

"Î#

Ê

œ cos" z 

dy dz

dy dt

20. y œ a1
t# b cot" 2t Ê

œ 2t cot" 2t
a1
t# b ˆ 1
24t# ‰

dy dt

œ tan" t
t ˆ 1
" t# ‰  ˆ #" ‰ ˆ "t ‰ œ tan" t


21. y œ z sec" z  Èz#  1 œ z sec" z  az#  1b z kz k È z #  1



"


sec

z È z#  1

z È 1  z#

zœ

 ˆ "# ‰ a1  z# b

"Î#

(2z)

z

19. y œ t tan" t  ˆ "# ‰ ln t Ê

œ

" ‘ ln x

(2u)

Ê1  ’a1  u# b

ˆ"‰

œ 0
ˆ x# ‰ ’ lnx x “
(ln (ln x)) ˆ "# ‰

,0u1

16. y œ sin" Š È"v ‹ œ sin" v"Î# Ê œ

"Î#

y y

1z È z#  1

"Î#

"


sec

Ê

dy dz

œ zŠ



t 1
t#

" ‹ kz k È z #  1

" 2t


asec" zb (1)  #" az#  1b

"Î#

(2z)

z, z  1

22. y œ 2Èx  1 sec" Èx œ 2(x  1)"Î# sec" ˆx"Î# ‰ Ê

dy dx

Š " ‹ x
"Î#

œ 2 –ˆ "# ‰ (x  1)"Î# sec" ˆx"Î# ‰
(x  1)"Î#  È# È x

23. y œ csc" (sec )) Ê 24. y œ a1
x# b etan 25. y œ

2 ax#
1b Ècos 2x


" x

dy d)

œ

 sec ) tan ) ksec )k Èsec# )  1

Ê yw œ 2xetan


" x

tan
" x


a1
x# b Š e1
x# ‹ œ 2xetan

#

2 ax #
1 b Ècos 2x

"! 3x
4 "! 3x
4 É 26. y œ É 2x  4 Ê ln y œ ln 2x  4 œ

" 10

) œ  ktan tan )k œ 1, 0  ) 

Ê ln y œ ln Š 2Èaxcos
2x1b ‹ œ ln (2)
ln ax#
1b 

Ê yw œ ˆ x#2x
1
tan 2x‰ y œ

Ê yw œ

x  1 —

ˆ 3x 3
4 

" ‰ x2 y

" #

œ 2Š

sec
" Èx 2È x  1




" #x ‹

œ

sec
" Èx Èx  1


" x


etan


" x

ln (cos 2x) Ê

cln (3x
4)  ln (2x  4)d Ê

"! 3x
4 ˆ " ‰ ˆ 3x 3
4  σ 2x  4 10

" x

1 #

yw y

œ0


2x x #
1

2 sin 2x)  ˆ #" ‰ (cos 2x

ˆ x#2x
1
tan 2x‰

" 10




yw y

œ

" 10

ˆ 3x 3
4 

2 ‰ 2x  4

" ‰ x2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 7 Practice Exercises &

1)(t  1) dy " 27. y œ ’ (t(t
 2)(t
3) “ Ê ln y œ 5 cln (t
1)
ln (t  1)  ln (t  2)  ln (t
3)d Ê Š y ‹ Š dt ‹

œ 5 ˆ t
" 1
28. y œ Ê

" t1

2u2u Èu#
1 Ê ln y dy 2u2u ˆ " du œ Èu#
1 u È)

29. y œ (sin )) Ê

" t#



dy d)



" ‰ t
3

Ê

dy dt

&


1)(t  1) ˆ t
" 1
œ 5 ’ (t(t  2)(t
3) “

œ ln 2
ln u
u ln 2 
ln 2 

" #

" t1

" t#



ln au#
1b Ê Š "y ‹ Š dy du ‹ œ

" u



" ‰ t
3


ln 2  #" ˆ u#2u
1 ‰

u ‰ u#
1

È) ˆ cos ) ‰
" )"Î# ln (sin )) Ê ln y œ È) ln (sin )) Ê Š "y ‹ Š dy d) ‹ œ sin ) #

œ (sin ))

È)

ŠÈ) cot )


ln (sin )) ‹ 2È )

30. y œ (ln x)1Îln x Ê ln y œ ˆ ln"x ‰ ln (ln x) Ê

w

y y

œ ˆ ln"x ‰ ˆ ln"x ‰ ˆ x" ‰
ln (ln x) ’ (ln"x)# “ ˆ x" ‰

ln (ln x) Ê yw œ (ln x)1Îln x ’ 1 x(ln x)# “

31.

' ex sin aex b dx œ ' sin u du, where u œ ex and du œ ex dx œ  cos u
C œ  cos aex b
C

32.

' et cos a3et  2b dt œ 3" ' cos u du, where u œ 3et  2 and du œ 3et dt œ

33.

" 3

sin u
C œ

" 3

sin a3et  2b
C

' ex sec# aex  7b dx œ ' sec# u du, where u œ ex  7 and du œ ex dx œ tan u
C œ tan aex  7b
C

34.

' ey csc aey
1b cot aey
1b dy œ ' csc u cot u du, where u œ ey
1 and du œ ey dy œ  csc u
C œ  csc aey
1b
C

35.

' asec# xb etan x dx œ ' eu du, where u œ tan x and du œ sec# x dx œ eu
C œ etan x
C

36.

' acsc# xb ecot x dx œ  ' eu du, where u œ cot x and du œ  csc# x dx œ eu
C œ ecot x
C

' c1

37.

'c11

38.

'1e Èlnx x dx œ '01 u"Î# du, where u œ ln x, du œ "x dx; x œ 1

" " " 3x  4 dx œ 3 c7 u du, where u œ 3x  4, du œ 3 dx; x œ 1 Ê " " ln 7 œ "3 cln kukd " ( œ 3 cln k1k  ln k7kd œ 3 [0  ln 7] œ  3

œ 39.


23 u$Î# ‘ " !

œ
23 1$Î#  23 0$Î# ‘ œ

u œ 7, x œ 1 Ê u œ 1

Ê u œ 0, x œ e Ê u œ 1

2 3

sin ˆ ‰ " '01 tan ˆ x3 ‰ dx œ '01 cos ' 1Î2 " ˆx‰ ˆx‰ ˆ ‰ dx œ 3 1 u du, where u œ cos 3 , du œ  3 sin 3 dx; x œ 0 x 3 x 3

Ê uœ œ 3 cln

"Î# k u kd "

œ 3
ln ¸ "# ¸  ln k1k‘ œ 3 ln

" #

" # $

œ ln 2 œ ln 8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ê u œ 1, x œ 1

479

480 40.

Chapter 7 Transcendental Functions

'11ÎÎ64

2 cot 1x dx œ 2'1Î6

1Î4

cos 1x sin 1x

dx œ

2 1

È2

'11ÎÎ2

" u

du, where u œ sin 1x, du œ 1 cos 1x dx; x œ Ê uœ

œ

41.

'04

È

cln kukd 11ÎÎ2 2 œ

2 1

c9

dt œ 'c25

2t t#  25

" u

2 1

’ln ¹ È"2 ¹  ln ¸ #" ¸“ œ

2 1


ln 1 

" #

"Î#

43.

'

" u


#" ln 2‘ œ

1 6

œ 
ln ¸ "# ¸  ln k2k‘ œ  ln 1
ln 2
ln 2 œ 2 ln 2 œ ln 4 sin u cos u

du, where u œ ln v and du œ

" v

dv

œ  ln kcos uk
C œ  ln kcos (ln v)k
C 44.

'

" v ln v

dv œ '

" u

du, where u œ ln v and du œ

" v

dv

œ ln kuk
C œ ln kln vk
C 45.

'

(ln x)
$ x

œ 46.

'

ln (x  5) x5

œ 47.

dx œ ' u$ du, where u œ ln x and du œ u
# #

'

" r

" #


C œ  (ln x)

#

" x

dx


C

dx œ ' u du, where u œ ln (x  5) and du œ #

u #


Cœ

cln (x5)d 2

#

" x 5

dx


C

csc# (1
ln r) dr œ ' csc# u du, where u œ 1
ln r and du œ

" r

dr

œ  cot u
C œ  cot (1
ln r)
C 48.

'

cos (1  ln v) v

dv œ  ' cos u du, where u œ 1  ln v and du œ  "v dv

œ  sin u
C œ  sin (1  ln v)
C 49.

'

#

œ

50.

'

" #

x3x dx œ " # ln 3

'

3u du, where u œ x# and du œ 2x dx

a3u b
C œ

" # ln 3

#

Š3x ‹
C

2tan x sec# x dx œ ' 2u du, where u œ tan x and du œ sec# x dx œ

" ln #

a2u b
C œ

dx œ 3'1

2tan x ln #


C

51.

'17

52.

" " È '132 5x" dx œ "5 '132 x" dx œ 5" cln kxkd $# " œ 5 aln 32  ln 1b œ 5 ln 32 œ ln Š 32‹ œ ln 2

3 x

7

" x

ln 2 1

9 25

du, where u œ 1  sin t, du œ  cos t dt; t œ  1# Ê u œ 2, t œ

dv œ ' tan u du œ '

tan (ln v) v

2 1

du, where u œ t#  25, du œ 2t dt; t œ 0 Ê u œ 25, t œ 4 Ê u œ 9

'c11ÎÎ62 1 cossint t dt œ '21Î2 œ  cln kukd #

Ê uœ

" È2

ln 2  ln 1
ln 2‘ œ

œ cln kukd * #& œ ln k9k  ln k25k œ ln 9  ln 25 œ ln 42.

" 6

dx œ 3 cln kxkd (" œ 3 aln 7  ln 1b œ 3 ln 7 &

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ê uœ

" #

" #

,xœ

" 4

Chapter 7 Practice Exercises 53.

15 '14 ˆ x8
#"x ‰ dx œ #" '14 ˆ 4" x
x" ‰ dx œ #"
8" x#
ln kxk‘ %" œ #"
ˆ 168
ln 4‰  ˆ 8"
ln 1‰‘ œ 16
#" ln 4

œ 54.


ln È4 œ

15 16

ˆln 8


'c0ln 2

e2w dw œ

'0ln 9 e

)

ˆln 8 

21 ‰ #

œ

2 3

(ln 8)  7 œ ln ˆ8#Î$ ‰  7 œ ln 4  7

" #

Ê u œ 1, x œ 1 Ê u œ 0

œ  ae!  e" b œ e  1

" #

'ln0Ð1Î4Ñ eu du, where u œ 2w, du œ 2 dw; w œ  ln 2 " #

ceu d 0ln Ð1Î4Ñ œ

ce!  eln Ð1Î4Ñ d œ

" #

ˆ1  4" ‰ œ

Ê u œ ln "4 , w œ 0 Ê u œ 0

3 8

2 3


u"Î# ‘ "' %

œ  ˆ16"Î#  4"Î# ‰ œ ˆ 2 3

2‰ ˆ" 3 4



"‰ #

œ ˆ

Ê u œ 4, r œ ln 5 Ê u œ 16

2‰ ˆ  4" ‰ 3

œ

8

2 3


u$Î# ‘ ) œ !

ˆ8$Î#  0$Î# ‰ œ

2 3

2 3

ˆ2*Î#  0‰ œ

2""Î# 3

œ

32È2 3

'1e "x (1
7 ln x)"Î$ dx œ 7" '18 u"Î$ du, where u œ 1
7 ln x, du œ 7x dx, x œ 1 'ee

#

" 6

ae)  1b"Î# d) œ '0 u"Î# du, where u œ e)  1, du œ e) d); ) œ 0 Ê u œ 0, ) œ ln 9 Ê u œ 8

œ 60.

2 3

'1ln 5 er a3er
1b$Î# dr œ "3 '416 u$Î# du, where u œ 3er
1, du œ 3er dr; r œ 0

œ 59.

 12‰ œ

 ceu d !"

œ 58.

3 #

'cc21 eÐx
1Ñ dx œ '10 eu du, where u œ (x
1), du œ  dx; x œ 2

œ 57.


ln 2

#

2 3

œ 56.

15 16

'18 ˆ 3x2  x8 ‰ dx œ 23 '18 ˆ "x  12x# ‰ dx œ 23 cln kxk
12x" d )" œ 23
ˆln 8
128 ‰  (ln 1
12)‘ œ

55.

481

" xÈln x

3
#Î$ ‘ ) 14 u "

œ

3 14

3 ‰ ˆ8#Î$  1#Î$ ‰ œ ˆ 14 (4  1) œ

dx œ 'e (ln x)"Î# e#

" x

Ê u œ 1, x œ e Ê u œ 8

9 14

dx œ '1 u"Î# du, where u œ ln x, du œ 2

" x

dx; x œ e Ê u œ 1, x œ e# Ê u œ 2

# œ 2
u"Î# ‘ " œ 2 ŠÈ2  1‹ œ 2È2  2

61.

'13 [ln (vv

11)]

#

dv œ '1 [ln (v
1)]# 3

" v
1

dv œ 'ln 2 u# du, where u œ ln (v
1), du œ ln 4

" v
1

dv;

v œ 1 Ê u œ ln 2, v œ 3 Ê u œ ln 4; œ 62.

" 3

$ ln 4

cu d ln 2 œ

" 3

$

$

c(ln 4)  (ln 2) d œ

" 3

$

$

c(2 ln 2)  (ln 2) d œ

(ln 2)$ 3

(8  1) œ

7 3

(ln 2)$

'24 (1
ln t)(t ln t) dt œ '24 (t ln t)(1
ln t) dt œ '24lnln24 u du, where u œ t ln t, du œ ˆ(t) ˆ "t ‰
(ln t)(1)‰ dt œ (1
ln t) dt; t œ 2 Ê u œ 2 ln 2, t œ 4 Ê u œ 4 ln 4 œ

63.

cu# d 2 ln 2 œ 4 ln 4

" #

c(4 ln 4)#  (2 ln 2)# d œ

" #

c(8 ln 2)#  (2 ln 2)# d œ

(2 ln 2)# #

(16  1) œ 30 (ln 2)#

'18 log) ) d) œ ln"4 '18 (ln )) ˆ ") ‰ d) œ ln"4 '0ln 8 u du, where u œ ln ), du œ ") d), ) œ 1 4

œ 64.

" #

'1e

" # ln 4

cu # d

8(ln 3)(log3 )) )

ln 8 !

" ln 16

œ

d) œ '1

c(ln 8)#  0# d œ

e

8(ln 3)(ln )) )(ln 3)

#

#

#

(3 ln 2) 4 ln 2

œ

Ê u œ 0, ) œ 8 Ê u œ ln 8

9 ln 2 4

d) œ 8 '1 (ln )) ˆ ") ‰ d) œ 8'0 u du, where u œ ln ), du œ e

1

" )

d) ;

) œ 1 Ê u œ 0, ) œ e Ê u œ 1 œ

" 4 cu # d !

œ 4 a1  0 b œ 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

482 65.

Chapter 7 Transcendental Functions

'c33ÎÎ44 È

6 9  4x#

3Î4

3Î2

dx œ 3 '
3Î4 È3# 2 (2x)# dx œ 3'
3Î2 È

" 3#  u#

du, where u œ 2x, du œ 2 dx; x œ  34 Ê u œ  3# , x œ

$Î#

Ê uœ

3 4

3 #

œ 3
sin" ˆ u3 ‰‘ $Î# œ 3
sin" ˆ "# ‰  sin" ˆ "# ‰‘ œ 3
16  ˆ 16 ‰‘ œ 3 ˆ 13 ‰ œ 1 66.

'c11ÎÎ55

6 È4  25x#

œ 67.

'c22

6 5

3 4
3t#

dx œ "


sin

'
11ÎÎ55

6 5

ˆ u2 ‰‘ " "

dt œ È3 'c2

œ

2

6 5

"


sin

69.

'

" 3
t#

dt œ 'È3 3

" yÈ4y#  1

œ sec

'

71.

'È2Î23Î3

24 yÈy#  16

#

È3

"

 sin

" ‰‘ #

ˆ

È3

" 2#
u#

È3 #

du, where u œ 5x, du œ 5 dx; œ

6 5


16

 ˆ

1 ‰‘ 6

œ

ˆ 13 ‰

6 5

È3 #

dy œ '

" uÈ u#  1

3

È3 œ

" È3

Štan" È3  tan" 1‹ œ

" yÈ y#  4#

3 'È2 dy œ 'È2Î3 k3yk È(3y) #  1 dy œ

" kyk È5y#  3

ˆ 13  14 ‰ œ

È 31 36

du, where u œ 2y and du œ 2 dy

dy œ 24 ˆ 4" sec" ¸ y4 ¸‰
C œ 6 sec" ¸ y4 ¸
C

2Î3

2

œ csec" ud È2 œ ’sec" 2  sec" È2“ œ

È È5

" È3

1 È3

k2yk
C

2

'cc2/È65Î

Ê uœ1

21 5


13  ˆ 13 ‰‘ œ

" ku k È u #  1

du, where u œ 3y, du œ 3 dy; yœ

72.

œ

1 5

du, where u œ È3t, du œ È3 dt;

’tan" ŠÈ3‹  tan" ŠÈ3‹“ œ

dt œ ’ È"3 tan" Š Èt 3 ‹“

kuk
C œ sec

" kyk È9y#  1

" 2#  u#

t œ 2 Ê u œ 2È3, t œ 2 Ê u œ 2È3

2 (2y)È(2y)#  1 "

dy œ 24 '

ˆ "# ‰

dt œ È3 'c2È3

#

2#
ŠÈ3t‹

ŠÈ3‹
t#

dy œ ' "

70.

"

'
11 È

2

È3

2

'È33

6 5

x œ  15 Ê u œ 1, x œ

œ È3
"# tan" ˆ u2 ‰‘ c2È3 œ 68.

dx œ

5 È2#  (5x)#


È6ÎÈ5

dy œ 'c2/È5

1 3

1 4



œ

#

#

È5y ÊŠÈ5y‹  ŠÈ3‹

Ê u œ È2, y œ

2 3

Ê uœ2

1 12


È6

È5

È2 3

dy œ 'c2

"

#

du,

uÊu#  ŠÈ3‹

È

œ ’ È"3 sec" ¹ Èu3 ¹“ 73.

'

dx œ '

" È2x  x#

cÈ6 c2

where u œ È5y, du œ È5 dy; y œ  È25 Ê u œ 2, y œ  È65 Ê u œ È6 œ

" È1  ax#
2x
1b

1 È3

’sec" È2  sec"

dx œ '

" È1  (x
1)#

2 È3 “

dx œ '

œ

" È3

" È 1  u#

ˆ 14  16 ‰ œ

" È3


3121 

21 ‘ 1#

œ

1 12È3

œ

È 3 1 36

du, where u œ x
1 and du œ dx

œ sin" u
C œ sin" (x
1)
C 74.

'È

" x#
4x  1

dx œ '

" È3  ax#  4x
4b

dx œ '

"

#

ÊŠÈ3‹  (x  2)#

dx œ '

"

#

du

ÊŠÈ3‹  u#

where u œ x  2 and du œ dx "

œ sin

Š Èu3 ‹

"


C œ sin

Š xÈ32 ‹


C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 7 Practice Exercises 75.

'cc21 v
4v2
5 dv œ 2 'cc21 1
av
" 4v
4b dv œ 2 'cc21 1
(v"
2) #

#

"

œ 2 ctan 76.

77.

'c11

3 4v#
4v
4

dv œ

"

œ 2 atan

3 4

'c11

2u ’ È23 tan" Š È ‹“ 3

œ

3 4

œ

È 31 4

'

" ud !

" (t
1)Èt#
2t  8

#

3 4

$Î# "Î#

dt œ '


Šv#

1  tan

"

"
v
4" ‹

dv œ

0b œ 3 4

dv œ 2'0

1

" 1
u#

483

du,

2 ˆ 14

where u œ v
2, du œ dv; v œ 2 Ê u œ 0, v œ 1 Ê u œ 1  0‰ œ 1#

'c11

È3

Œ

#

"

#



Šv


dv œ

#

" #‹

3 4

'c31ÎÎ22

Œ

È3 #

"

du

#



u#

" #

where u œ v
, du œ dv; v œ 1 Ê u œ  "# , v œ 1 Ê u œ œ

È3 #

È3 #

’tan" È3  tan" Š È"3 ‹“ œ

" (t
1)Èat#
2t
1b  9

dt œ '


13  ˆ 16 ‰‘ œ

dt œ '

" (t
1)È(t
1)#  3#

" uÈ u#  3#

È3 #

ˆ 261
16 ‰ œ

È3 #

†

3 #

1 #

du

where u œ t
1 and du œ dt œ 78.

'

" 3

"

sec

" (3t
1)È9t#
6t

¸ u3 ¸


Cœ

dt œ '

" 3

"

sec

¸ t
31 ¸


C

" (3t
1)Èa9t#
6t
1b  1

dt œ '

" (3t
1)È(3t
1)#  1#

dt œ

" 3

'

" uÈ u#  1

du

where u œ 3t
1 and du œ 3 dt œ

" 3

sec" kuk
C œ

" 3

sec" k3t
1k
C

79. 3y œ 2y
1 Ê ln 3y œ ln 2y
1 Ê y(ln 3) œ (y
1) ln 2 Ê (ln 3  ln 2)y œ ln 2 Ê ˆln 3# ‰ y œ ln 2 Ê y œ

ln 2 ln Š 3# ‹

80. 4y œ 3y
2 Ê ln 4y œ ln 3y
2 Ê y ln 4 œ (y
2) ln 3 Ê 2 ln 3 œ (ln 3
ln 4)y Ê (ln 12)y œ 2 ln 3 9 Ê y œ  lnln12 81. 9e2y œ x# Ê e2y œ

x# 9

#

#

Ê ln e2y œ ln Š x9 ‹ Ê 2y(ln e) œ ln Š x9 ‹ Ê y œ

82. 3y œ 3 ln x Ê ln 3y œ ln (3 ln x) Ê y ln 3 œ ln (3 ln x) Ê y œ

ln (3 ln x) ln 3

œ

" #

#

#

ln Š x9 ‹ œ ln É x9 œ ln ¸ x3 ¸ œ ln kxk  ln 3

ln 3
ln (ln x) ln 3

83. ln (y  1) œ x
ln y Ê eln Ðy1Ñ œ eÐx
ln yÑ œ ex eln y Ê y  1 œ yex Ê y  yex œ 1 Ê y a1  ex b œ 1 Ê y œ 1 " ex 84. ln (10 ln y) œ ln 5x Ê eln Ð10 ln yÑ œ eln 5x Ê 10 ln y œ 5x Ê ln y œ 85. The limit leads to the indeterminate form 00 : lim

10x  1 x

86. The limit leads to the indeterminate form 00 : lim

3)  1 )

87. The limit leads to the indeterminate form 00 : lim

2sin x  1 ex  1

xÄ0

)Ä0

xÄ0

)Ä0

2c sin x  " ex  1 xÄ0

89. The limit leads to the indeterminate form 00 : lim

(ln 3)3) 1

œ lim xÄ0

88. The limit leads to the indeterminate form 00 : lim

5  5 cos x

x x Ä 0 e x1

Ê eln y œ exÎ2 Ê y œ exÎ2

(ln 10)10x 1

œ lim xÄ0 œ lim

x #

œ lim

xÄ0

œ lim

xÄ0

œ ln 10

œ ln 3

2sin x (ln 2)(cos x) ex

œ ln 2

2c sin x (ln 2)( cos x) ex

5 sin x ex  1

œ lim xÄ0

œ  ln 2 5 cos x ex

œ5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

484

Chapter 7 Transcendental Functions 4  4ex xex

90. The limit leads to the indeterminate form 00 : lim

xÄ0

91. The limit leads to the indeterminate form 00 :

t  ln (1
2t) t#

lim


tÄ!

sin# (1x)

92. The limit leads to the indeterminate form 00 : lim

xc4 x Ä 4 e
3x

œ lim xÄ4

1 sin (21x) exc4  1

21# cos (21x) exc4

œ lim xÄ4

94. The limit leads to the indeterminate form yÄ!

eyc"

_ _:

indeterminate form

_ _:

tÄ!

lim
e1Îy ln y œ

yÄ!

ln a1
3xc" b x
"

lim b

Š

xÄ!

œ x lim Ä_

x Š ln ln 2 ‹ x Š ln ln 3 ‹


3x
# ‹ 1  3x
" xc#

œ x lim Ä_

x# x #
1

œ x lim Ä_

ln 3 ln 2

œ x lim Ä_

ˆ x ‰ xex ex 100 œ x lim xe
x œ x lim Ä _ 100x Ä _ 100 x tan
" x œ _ Ê faster

(e) x lim Ä_

csc
" x Š "x ‹

œ x lim Ä_

(f) x lim Ä_

sinh x ex

œxÄ lim_

$ # lim 10x e
x 2x xÄ_

(f) x lim Ä_

tanc" Š "x ‹ Š "x ‹

sinc" Š "x ‹ " ‹ x#

sech x ecx

yÄ!

œ

lim


yÄ!

sin
" ax
" b x
" aex ecx b #e x

2x #x

œ1 y
"
" ˆy
# ‰

e y

œ

# lim 30x ex
4x xÄ_

œ x lim Ä_ œ

tan
" ax
" b x
"


"
" lim sin x
#ax b xÄ_

œ x lim Ä_

Š ex b2e
x ‹ e cx

œ

the limit leads to the

œ0 Ê

ln a1
3xc" b ; x
"

the limit leads to the

ˆ1
x3 ‰x œ lim eln fÐxÑ œ e! œ 1 x Ä !b

lim

x Ä !b

Ê same rate

ln 3 ln #

œ x lim " œ 1 Ê same rate Ä_

œ _ Ê faster

œ x lim Ä_

œxÄ lim_

œ

ln a1
3x
" b ; x
"

lim ln f(x) œ lim
x Ä !
xÄ! 3x x
3

Š
x
# ‹

#

Ê1
Šx
" ‹

x
#

"ec2x #

3cx ˆ 23 ‰x œ 0 Ê slower 2cx œ x lim Ä_ ln 2x ln 2
ln x ˆ ln 2 œ x lim ln x# œ x lim Ä _ 2 (ln x) Ä _ # ln x

Š

ln y eyc"

lim


et 1

ˆ1
x3 ‰x œ lim eln fÐxÑ œ e$ œ 3 Ê x lim Ä_ xÄ_

3

œ lim b xÄ!

(c) x lim Ä_ (d) x lim Ä_

(e) x lim Ä_

tÄ!

Ê x lim ln f(x) œ x lim Ä_ Ä_

x Ä _ 1
3x

x x
Š "x ‹

(d) x lim Ä_

t

tÄ!

(b) x lim Ä_

(c)

œ _

21(sin 1x)(cos 1x) exc4  1

œ lim xÄ4

t

x 96. Let f(x) œ ˆ1
3x ‰ Ê ln f(x) œ x ln ˆ1
3x ‰ Ê

98. (a) x lim Ä_ (b) x lim Ä_

2t

lim
Š et  "t ‹ œ lim
Š e " t ‹ œ lim



# Š
3x
" ‹ 1  3x œ lim c x #

indeterminate form 00 : x lim Ä_

log2 x log3 x

Š1  1 b2 2t ‹

lim t Ä !


œ0

y

x 95. Let f(x) œ ˆ1
3x ‰ Ê ln f(x) œ

97. (a) x lim Ä_

œ

œ 4

œ 21 #

93. The limit leads to the indeterminate form 00 :

œ  lim


4ex ex
xex

œ lim xÄ0

œ

" #

x
#

"

œ 1 Ê same rate

Ê same rate

œ x lim Ä_


# Š
x
# ‹ 1x

"

Ê 1  Š x# ‹

Ê same rate


"# ‰ œ

lim 60xex
4 xÄ_

œ x lim Ä_

" #

œ x lim Ä_

60 ex

œ x lim Ä_

œ 0 Ê slower " 1
"#

œ 1 Ê same rate

x


x
#

œ x lim Ä_

œ x lim Ä_

 É1
ˆx
" ‰#  2x
$

2 e c x ae x
e c x b

œ x lim Ä_

x 2É1  x"#

œ _ Ê faster

ˆ 2 ‰ œ 2 Ê same rate œ x lim Ä _ 1
ec2x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 7 Practice Exercises 99. (a) (b)

"
"‹ x% œ 1
x"# Ÿ 2 for x sufficiently large Ê true Š "# ‹ x

Š

x#

Š

" x#

Š

" ‹ x%




" ‹ x%

œ x#
1  M for any positive integer M whenever x  ÈM Ê false

(c) x lim Ä_

x x
ln x

œ x lim Ä_

(d) x lim Ä_

ln (ln x) ln x

œ x lim Ä_

(e) (f)

tan
" x 1 cosh x ex

Š

100. (a)

Š "# x

Ÿ œ

Š

x

df dx

ˆ "x ‰

" %‹ " ‹ x%

œ 0 Ê grows slower Ê true

1) œ 1 if x  0 Ê true

ˆ " ‰ œ 0 Ê true œ x lim Ä _ x#
1

Š "x ‹ ln x œ0 Ê x
1 œ x lim Ä_ 1 ln 2 ln x
1 Ÿ 1
1 œ 2 if x  

secc" x œ 1 sinh x " ex œ #

" ln x

œ x lim Ä_

Ÿ 1 if x  0 Ê true

cos
" Š " ‹

(f) 101.

" x #
1

œ

x (b) x lim Ä _ Š "#


(e)

œ 1 Ê the same growth rate Ê false

" x " Šx‹ – ln x —

1 # for all x Ê true " c2x b Ÿ " (1
# a1
e #

" ‹ x%
"% ‹ x

(c) x lim Ä_ (d) lnln2x x œ

" 1


x

1

ˆ 1# ‰ 1

Ÿ

a1  ec2x b Ÿ

" #

102. y œ f(x) Ê y œ 1


x œ fÐln 2Ñ

" x

Ê

f af " (x)b œ 1


" " Šx
1‹

f w (x) œ  x"# Ê

df
" dx ¹ fÐxÑ

2 Ê true

if x  1 Ê true

if x  0 Ê true

c"

œ ex
1 Ê Š dfdx ‹

1 #

œ

true

" x

œ


"

"

Ê Š dfdx ‹

df Š dx ‹ x œ ln 2

œy1 Ê xœ

œ 1
(x  1) œ x; œ

" y 1

df c" dx ¹ fÐxÑ

œ

x œ fÐln 2Ñ

œ

" aex
1bx œ ln 2

Ê f " (x) œ " (x  1)# ¹ fÐxÑ

œ

" x 1

œ

" #
1

œ

; f " (f(x)) œ "

’Š1
x" ‹1“

#

" 3

" Š1
"x ‹1

œ x# ;

" f w (x)

2 ‰ 103. y œ x ln 2x  x Ê yw œ x ˆ 2x
ln (2x)  1 œ ln 2x;

solving yw œ 0 Ê x œ x and

" #

" #

; yw  0 for x 

Ê relative minimum of  "#

f ˆ #e ‰

" #

and yw  0 for at x œ "# ; f ˆ #"e ‰ œ  "e

œ 0 Ê absolute minimum is  "# at x œ

the absolute maximum is 0 at x œ

" #

and

e #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

" Š x" ‹

œ x and

485

486

Chapter 7 Transcendental Functions

104. y œ 10x(2  ln x) Ê yw œ 10(2  ln x)  10x ˆ "x ‰

œ 20  10 ln x  10 œ 10(1  ln x); solving yw œ 0 Ê x œ e; yw  0 for x  e and yw  0 for x  e Ê relative maximum at x œ e of 10e; y   ! on Ð!ß e# Ó and y ae# b œ 10e# (2  2 ln e) œ 0 Ê absolute minimum is 0 at x œ e# and the absolute maximum is 10e at x œ e

105. A œ '1

e

dx œ '0 2u du œ cu# d ! œ 1, where 1

2 ln x x

" x

u œ ln x and du œ 106. (a) A" œ '10

20

" x

(b) A" œ 'ka

kb

"

dx; x œ 1 Ê u œ 0, x œ e Ê u œ 1

dx œ cln kxkd #! "! œ ln 20  ln 10 œ ln

20 10

œ ln 2, and A# œ '1

kb dx œ cln kxkd ka œ ln kb  ln ka œ ln

kb ka

œ ln

" x

2

b a

" x

dx œ cln kxkd #" œ ln 2  ln 1 œ ln 2

œ ln b  ln a, and A# œ 'a

b

" x

dx œ cln kxkd ab

œ ln b  ln a 107. y œ ln x Ê

dy dx

108. y œ 9ecxÎ3 Ê Ê

dx ¸ dt xœ9

œ

dy dx

" x

Š "4 ‹ É9 

œ

Š

xœ

" È2

;

dA dx

dA dx

œ

Ê

dy dx dx dt

œ

dx dt " 4

œ

(dy/dt) (dy/dx)

œ ˆ "x ‰ Èx œ

Ê

dx dt

œ

" Èx

Ê

Š "4 ‹ È9  y 3e
xÎ3

œ

dy dt ¹ e#

" e

ln x x and dA dx

#

Ê

m/sec

; x œ 9 Ê y œ 9e$

Èe$ Èe$  1 ¸ 5 ft/sec #

" È2

and

units long by y œ e"Î# œ

 0 for x  e

dy dt

œ ecx
(x)(2x) ecx œ ecx a1  2x# b . Solving

 0 for x 

110. A œ xy œ x ˆ lnx#x ‰ œ dA dx

9 e$

3 ‹ e$

#

" È2

dy dt

œ 3e
xÎ3 ;

109. A œ xy œ xecx Ê Ê xœ

;

dA dx

œ

" Èe " x#

dA dx

#

 0 for 0  x 

" È2

dA dx

œ 0 Ê 1  2x# œ 0

Ê absolute maximum of

" È2

e"Î# œ

" È2e

at

units high. 

ln x x#

œ

1ln x x#

. Solving

dA dx

 0 for x  e Ê absolute maximum of

œ 0 Ê 1  ln x œ 0 Ê x œ e; ln e e

œ

" e

at x œ e units long and y œ

high. 111. K œ ln (5x)  ln (3x) œ ln 5
ln x  ln 3  ln x œ ln 5  ln 3 œ ln

5 3

112. (a) No, there are two intersections: one at x œ 2 and the other at x œ 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" e#

units

Chapter 7 Practice Exercises (b) Yes, because there is only one intersection

113.

log4 x log2 x

œ

x Š ln ln 4 ‹ x Š ln ln # ‹

114. (a) f(x) œ

œ

ln 2 ln x

ln x ln 4

†

ln 2 ln x

, g(x) œ

œ

ln 2 ln 4

œ

ln 2 2 ln 2

œ

" #

ln x ln #

(b) f is negative when g is negative, positive when g is positive, and undefined when g œ 0; the values of f decrease as those of g increase

115. (a) y œ

Ê yw œ

ln x Èx

" xÈ x



ln x 2x$Î#

œ

2  ln x 2xÈx

Ê yww œ  34 x&Î# (2  ln x)  "# x&Î# œ x&Î# ˆ 34 ln x  2‰ ;

solving yw œ 0 Ê ln x œ 2 Ê x œ e# ; yw  0 for x  e# and and yw  0 for x  e# Ê a maximum of 2e ; yww œ 0 Ê ln x œ 83 Ê x œ e)Î$ ; the curve is concave down on ˆ0ß e)Î$ ‰ and concave up on ˆe)Î$ ß _‰; so there is an inflection point at ˆe)Î$ ß #

) ‰ . $e%Î$ x #

(b) y œ ex Ê yw œ 2xe

#

Ê yww œ 2ex
4x# ex

#

#

œ a4x#  2bex ; solving yw œ 0 Ê x œ 0; yw  0 for x  0 and yw  0 for x  0 Ê a maximum at x œ 0 of e! œ 1; there are points of inflection at x œ „ È"2 ; the curve is concave down for  È"2  x 

" È2

and concave

up otherwise. (c) y œ (1
x) ecx Ê yw œ ecx  (1
x) ecx œ xecx Ê yww œ ecx
xecx œ (x  1) ecx ; solving yw œ 0 Ê xecx œ 0 Ê x œ 0; yw  0 for x  0 and yw  0 for x  0 Ê a maximum at x œ 0 of (1
0) e! œ 1; there is a point of inflection at x œ 1 and the curve is concave up for x  1 and concave down for x  1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

487

488

Chapter 7 Transcendental Functions

116. y œ x ln x Ê yw œ ln x
x ˆ "x ‰ œ ln x
1; solving yw œ 0

Ê ln x
1 œ 0 Ê ln x œ 1 Ê x œ e" ; yw  0 for x  e" and yw  0 for x  e" Ê a minimum of e" ln e" œ  "e at x œ e" . This minimum is an absolute minimum since yww œ

" x

is positive for all x  !.

117. Since the half life is 5700 years and A(t) œ A! ekt we have Ê kœ

ln (0.5) 5700

Ê ln (0.1) œ

A! #

" #

œ A! e5700k Ê

œ e5700k Ê ln (0.5) œ 5700k ln Ð0Þ5Ñ

. With 10% of the original carbon-14 remaining we have 0.1A! œ A! e 5700 ln (0.5) 5700

t Ê tœ

(5700) ln (0.1) ln (0.5)

t

ln Ð0Þ5Ñ

Ê 0.1 œ e 5700

t

¸ 18,935 years (rounded to the nearest year).

118. T  Ts œ (To  Ts ) eckt Ê 180  40 œ (220  40) eckÎ4 , time in hours, Ê k œ 4 ln ˆ 79 ‰ œ 4 ln ˆ 97 ‰ Ê 70  40 œ (220  40) ec4 ln Ð9Î7Ñ t Ê t œ

ln 6 4 ln ˆ 97 ‰

¸ 1.78 hr ¸ 107 min, the total time Ê the time it took to cool from

180° F to 70° F was 107  15 œ 92 min x ‰ 119. ) œ 1  cot" ˆ 60  cot" ˆ 53 

œ 30 ’ 60# 2
x# 

" 30#
(50  x)# “ ;

x ‰ 30 ,

solving

100
20È17 is not in the domain;

d) dx

0  x  50 Ê d) dx

d) dx

œ

1 Š 60 ‹ x ‰# 1
ˆ 60

Š




" 30 ‹

cx‹ 1
Š 5030

#

œ 0 Ê x#  200x
3200 œ 0 Ê x œ 100 „ 20È17, but d) dx

 0 for x  20 Š5  È17‹ and

 0 for 20 Š5  È17‹  x  50

Ê x œ 20 Š5  È17‹ ¸ 17.54 m maximizes ) 120. v œ x# ln ˆ "x ‰ œ x# (ln 1  ln x) œ x# ln x Ê Ê 2 ln x
1 œ 0 Ê ln x œ  maximum at x œ e
1Î2 ;

r h

dv dx

Ê x œ ec1Î2 ;

" #

œ 2x ln x  x# ˆ "x ‰ œ x(2 ln x
1); solving dv dx

 0 for x  e
1Î2 and

œ x and r œ 1 Ê h œ e1Î2 œ Èe ¸ 1.65 cm

dv dx

dv dx

œ0

 0 for x  e
1Î2 Ê a relative

CHAPTER 7 ADDITIONAL AND ADVANCED EXERCISES lim
'0

b

1.

bÄ1

2. x lim Ä_

" x

" È 1  x#

dx œ lim
csin" xd 0 œ lim
asin" b  sin" 0b œ lim
asin" b  0b œ lim
sin" b œ bÄ1 bÄ1 bÄ1 bÄ1 b

'0x tan" t dt œ x lim Ä_

œ x lim Ä_

tan
" x 1

œ

" x
"Î# sec# Èx # " x
"Î# #

4. y œ ax
ex b2Îx Ê ln y œ x 2Îx

Ê x lim ax
e b Ä_ ˆ "
5. x lim Ä _ n
1

" n
#

x

_ ˆ_ form‰

1 #

1Îx 3. y œ ˆcos Èx‰ Ê ln y œ

œ  "# lim b xÄ!

'0x tanc" t dt

" x

ln ˆcos Èx‰ x

ln ˆcos Èx‰ and lim b xÄ!

œ  "# Ê

lim

x Ä !b

ˆcos Èx‰1Îx œ e
1Î2 œ

2 ln ax
ex b x

Ê x lim ln y œ x lim Ä_ Ä_ y # œ x lim e œ e Ä_


á


" ‰ #n

œ lim b xÄ!

2 a1
e x b x
ex

 sin Èx 2Èx cos Èx

œ

" #

lim

x Ä !b

tan Èx Èx

" Èe

œ x lim Ä_

2ex 1
ex

œ x lim Ä_

2ex ex

œ2

1 ˆ"‰ œ x lim
ˆ n" ‰ – 1 " —
á
ˆ n" ‰ – 1 " — Ä _  n – 1
Š"‹ — 1
2Š ‹ 1
nŠ ‹ n

n

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

n

1 #

Chapter 7 Additional and Advanced Exercises which can be interpreted as a Riemann sum with partitioning ?x œ œ '0

1

6. x lim Ä_

ˆ "
Ê x lim Ä _ n
1

" n

Ê x lim Ä_

" n

t

(c) 8. (a)

1

t

lim A(t) œ lim a1  ect b œ 1

tÄ_

tÄ_

V(t) t Ä _ A(t) limb V(t) A(t)

lim

œ lim

tÄ_

œ limb tÄ!

tÄ!

1 #

ˆ1  ec2t ‰ 1  ect

1 #

ˆ1  ec2t ‰ 1  ect

œ

1 #

ln 2 ln a

œ 0;

lim loga 2 œ lim c aÄ1

ln 2 ln a

œ _;

lim loga 2 œ lim b aÄ1

ln 2 ln 1

œ _;

lim loga 2 œ a lim Ä_

ln 2 ln a

œ0

a Ä 1b aÄ_

9. A" œ '1

e

2 log2 x x

#

e

x) œ ’ (ln 2 ln # “ œ 1

dx œ

" # ln 2

2 ln 2



" 1
x#

#

a 1  e ct b a 1
e c t b a1  e ct b

œ limb tÄ!

1 #

1 #

a1  ec2t b

a1
ect b œ 1

(b)

'1e lnxx dx œ ’ (lnlnx)2 “ e œ ln"# ; A# œ '1e 2 log4 #

1

4

x

dx œ

2 ln 4

'1e lnxx dx

Ê A" : A# œ 2 : 1

10. y œ tan" x
tan" ˆ "x ‰ Ê yw œ " 1
x#

1

œ limb tÄ!

lim loga 2 œ lim b aÄ!

a Ä !b a Ä 1c

œ

" ‰ #n

ce1În
e2În
á
ed œ '0 ex dx œ cex d "! œ e  1

t 7. A(t) œ '0 ecx dx œ cecx d t0 œ 1  ect , V(t) œ 1'0 ec2x dx œ
 1# ec2x ‘ 0 œ

(b)


á



ˆ n" ‰ eÐ1ÎnÑ
ˆ n" ‰ e2Ð1ÎnÑ
á
ˆ n" ‰ enÐ1ÎnÑ ‘ which can be interpreted as a ce1În
e2În
á
ed œ x lim Ä_

Riemann sum with partitioning ?x œ

(a)

" n
#

dx œ cln (1
x)d "! œ ln 2

" 1
x " n

" n

489

" 1
x#




Š

" ‹ x#

Š1
x"# ‹

œ 0 Ê tan" x
tan" ˆ x" ‰ is a constant

and the constant is 1# for x  0; it is  1# for x  0 since tan" x
tan" ˆ "x ‰ is odd. Next the lim

x Ä !b


tan" x
tan" ˆ "x ‰‘ œ !


1 #

œ

1 #

and

lim

x Ä !c

ˆtan" x
tan" ˆ x" ‰‰ œ 0
ˆ 1# ‰ œ  1#

11. ln xax b œ xx ln x and ln axx bx œ x ln xx œ x# ln x; then, xx ln x œ x# ln x Ê axx  x# bln x œ ! Ê xx œ x# or ln x œ !Þ x ln x œ ! Ê x œ "; xx œ x# Ê x ln x œ 2 ln x Ê x œ 2. Therefore, xax b œ axx bx when x œ 2 or x œ ". x

12. In the interval 1  x  21 the function sin x  0 Ê (sin x)sin x is not defined for all values in that interval or its translation by 21.

13. f(x) œ egÐxÑ Ê f w (x) œ egÐxÑ gw (x), where gw (x) œ

x 1
x%

Ê f w (2) œ e! ˆ 1
2 16 ‰ œ

2 17

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

490

Chapter 7 Transcendental Functions

14. (a)

df dx

œ

2 ln ex ex

(b) f(0) œ '1

1

(c)

df dx

† ex œ 2x 2 ln t t

dt œ 0

œ 2x Ê f(x) œ x#
C; f(0) œ 0 Ê C œ 0 Ê f(x) œ x# Ê the graph of f(x) is a parabola

15. Triangle ABD is an isosceles right triangle with its right angle at B and an angle of measure 1 4

therefore have

tan" 3"

œ nDAB œ nDAE
nCAB œ

tan" #"




1 4

at A. We

.

16. (a) The figure shows that lne e  ln11 Ê 1 ln e  e ln 1 Ê ln e1  ln 1e Ê e1  1e (b) y œ lnxx Ê yw œ ˆ "x ‰ ˆ x" ‰  lnx#x Ê 1 x#ln x ; solving yw œ 0 Ê ln x œ 1 Ê x œ e; yw  0 for x  e and yw  0 for 0  x  e Ê an absolute maximum occurs at x œ e

17. The area of the shaded region is '0 sin" x dx œ '0 sin" y dy, which is the same as the area of the region to 1

1

the left of the curve y œ sin x (and part of the rectangle formed by the coordinate axes and dashed lines y œ 1, x œ 1# ) . The area of the rectangle is 1 #

1 #

œ '0 sin" x dx
'0 sin x dx Ê 1Î2

1

œ '0 sin" y dy
'0

1Î2

1

'0

1Î2

sin x dx œ

18. (a) slope of L$  slope of L#  slope of L" Ê

" b



1 #

sin x dx, so we have

 '0 sin" x dx. 1

ln b  ln a ba



" a

(b) area of small (shaded) rectangle  area under curve  area of large rectangle " b

Ê

(b  a)  'a

b

" x

dx 

" a

" b

(b  a) Ê



ln b  ln a ba



" a

19. (a) g(x)
h(x) œ 0 Ê g(x) œ h(x); also g(x)
h(x) œ 0 Ê g(x)
h(x) œ 0 Ê g(x)  h(x) œ 0 Ê g(x) œ h(x); therefore h(x) œ h(x) Ê h(x) œ 0 Ê g(x) œ 0 (b)

f(x)
f(x) # f(x)  f(x) #

œ œ

cfE (x)
fO (x)d b
fE (x)
fO (x)‘ œ fE (x)
fO (x)
# fE (x)  fO (x) œ fE (x); # cfE (x)
fO (x)d  cfE (x)
fO (x)d œ fE (x)
fO (x) # fE (x)
fO (x) œ fO (x) #

(c) Part b Ê such a decomposition is unique. 20. (a) g(0
0) œ

g(0)
g(0) 1  g(0) g(0) #

Ê c1  g# (0)d g(0) œ 2g(0) Ê g(0)  g$ (0) œ 2g(0) Ê g$ (0)
g(0) œ 0

Ê g(0) cg (0)
1d œ 0 Ê g(0) œ 0

g(x) b g(h)

’ “  g(x) g(x
h)  g(x)  g(x)
g# (x) g(h) œ lim 1 c g(x) g(h) œ lim g(x)
g(h) h h h c 1 g(x) g(h)d  hÄ0 hÄ0 hÄ0 g(h) 1
g# (x) # # lim ’ h “ ’ 1  g(x) g(h) “ œ 1 † c1
g (x)d œ 1
g (x) œ 1
[g(x)]# hÄ0

(b) gw (x) œ lim œ (c)

dy dx

œ 1
y# Ê

dy 1
y# "

Ê C œ 0 Ê tan

œ dx Ê tan" y œ x
C Ê tan" (g(x)) œ x
C; g(0) œ 0 Ê tan" 0 œ 0
C

(g(x)) œ x Ê g(x) œ tan x

21. M œ '0

1

œ

" 2 " xd ! œ 1# 1
x# dx œ 2 ctan ln 2 ln 4 ˆ 1 ‰ œ 1 ; y œ 0 by symmetry #

and My œ '0

1

2x 1
x#

"

dx œ cln a1
x# bd ! œ ln 2 Ê x œ

My M

" 22. (a) V œ 1 '1Î4 Š #È ‹ dx œ x

1 4

'14Î4 x" dx œ 14 cln kxkd %"Î% œ 14 ˆln 4  ln 4" ‰ œ 14 ln 16 œ 14 ln a2% b œ 1 ln 2

" (b) My œ '1Î4 x Š #È ‹ dx œ x

1 2

63 '14Î4 x"Î# dx œ
"3 x$Î# ‘ %"Î% œ ˆ 83  24" ‰ œ 64#4 1 œ 24 ;

4

4

#

" " Mx œ '1Î4 "# Š #È ‹ Š 2È ‹ dx œ x x 4

1 8

'14Î4

" x

%

dx œ
8" ln kxk‘ "Î% œ

" 8

ln 16 œ

" #

ln 2;

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 7 Additional and Advanced Exercises M œ '1Î4 4

yœ

Mx M

" #È x

dx œ '1Î4 "2 x"Î# dx œ
x"Î# ‘ "Î% œ 2  %

4

œ ˆ "# ln 2‰ ˆ 32 ‰ œ

" #

œ

3 #

; therefore, x œ

ds dt

œ ks Ê

ds s

63 ‰ ˆ 2 ‰ œ ˆ 24 3 œ

œ

21 1#

ln 2 3

23. A(t) œ A! ert ; A(t) œ 2A! Ê 2A! œ A! ert Ê ert œ 2 Ê rt œ ln 2 Ê t œ 24.

My M

ln 2 r

Ê t¸

.7 r

œ

70 100r

œ

70 (r%)

œ k dt Ê ln s œ kt
C Ê s œ s! ekt

Ê the 14th century model of free fall was exponential; note that the motion starts too slowly at first and then becomes too fast after about 7 seconds

25. (a) L œ k ˆ a  bR%cot )


b csc ) ‰ r%

Ê

dL d)

#

œ k Š b csc R%

)



b csc ) cot ) ‹; r%

solving

dL d)

œ0

Ê r% b csc# )  bR% csc ) cot ) œ 0 Ê (b csc )) ar% csc )  R% cot )b œ 0; but b csc ) Á 0 since )Á

1 #

Ê r% csc )  R% cot ) œ 0 Ê cos ) œ

r% R%

%

Ê ) œ cos" Š Rr % ‹ , the critical value of )

%

(b) ) œ cos" ˆ 56 ‰ ¸ cos" (0.48225) ¸ 61° 26. Two views of the graph of y œ 1000
1  (.99)x
"x ‘ are shown below.

(a) At about x œ 11 there is a minimum (b) There is no maximum; however, the curve is asymptotic to y œ 1000. The curve is near 1000 when x   643.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

7 4

and

491

492

Chapter 7 Transcendental Functions

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 8 TECHNIQUES OF INTEGRATION 8.1 BASIC INTEGRATION FORMULAS u œ 8x#
1 Ä du œ 16x dx •

1.

' È16x dx

2.

' È3 cos x dx

3.

'

4.

' cot$ y csc# y dy; ”

5.

'01 8x16x
dx2 ; Ô

6.

'11ÎÎ43

7.

8x#
1

;”

1
3 sin x

' Èduu œ 2Èu
C œ 2È1
3 sin x
C

u œ 1
3 sin x Ä du œ 3 cos x dx •

u œ sin v Ä du œ cos v dv •

'

3Èu du œ 3 † 23 u$Î#
C œ 2(sin v)$Î#
C

u œ cot y Ä du œ  csc# y dy •

'

u$ ( du) œ  u4
C œ

3Èsin v cos v dv; ”

#

'

;”

' Èduu œ 2Èu
C œ 2È8x#
1
C

%

 cot% y 4


C

u œ 8x#
2 × 10 "! Ä '2 du du œ 16x dx u œ cln kukd # œ ln 10  ln 2 œ ln 5 Õ x œ 0 Ê u œ 2, x œ 1 Ê u œ 10 Ø

sec# z dz tan z

Ô ;Ö

dx È x ˆÈ x
1 ‰

Õz œ

1 4

u œ tan z × Ù Ä du œ sec# z dz 1 È Ê u œ 1, z œ 3 Ê u œ 3 Ø

Ô u œ Èx
" × " Ö Ù ; Ö du œ #Èx dx Ù Ä dx Õ 2 du œ Èx Ø

8.

' x dxÈx œ '

9.

' cot (3  7x) dx; ” u œ 3  7x •

Ä  "7

10.

' csc (1x  1) dx; ” u œ 1x  1 •

Ä

du œ 1 dx

" u

È

du œ cln kukd 1 3 œ ln È3  ln 1 œ ln È3

' 2 udu œ 2 ln kuk
C œ 2 ln ˆÈx
1‰
C

Ô u œ Èx  1 × Ö du œ " dx Ù dx Ù Ä #È x È x ˆÈ x  1 ‰ ; Ö dx Õ 2 du œ Èx Ø

du œ 7 dx

È3

'1

' 2 udu œ 2 ln kuk
C œ 2 ln ¸Èx  1¸
C

' cot u du œ  7" ln ksin uk
C œ  7" ln ksin (3  7x)k
C

' csc u † du1 œ "1 ln kcsc u
cot uk
C

œ  1" ln kcsc (1x  1)
cot (1x  1)k
C 11.

' e) csc ˆe)
1‰ d); ” u œ e

12.

' cot (3 x
ln x) dx; ” u œ 3
dxln x •


1 Ä du œ e) d) • )

du œ

Ä

' csc u du œ  ln kcsc u
cot uk
C œ  ln ¸csc ˆe)
1‰
cot ˆe)
1‰¸
C

' cot u du œ ln ksin uk
C œ ln ksin (3
ln x)k
C

x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

494

Chapter 8 Techniques of Integration u œ 3t Ä du œ dt3 —

13.

' sec 3t dt; –

' 3 sec u du œ 3 ln ksec u
tan uk
C œ 3 ln ¸sec 3t
tan 3t ¸
C

14.

' x sec ax#  5b dx; ” u œ x

#

' "# sec u du œ "# ln ksec u
tan uk
C

5 Ä du œ 2x dx •

œ

" #

ln ksec ax#  5b
tan ax#  5bk
C

15.

' csc (s  1) ds; ” u œ s  1 •

16.

' )"

du œ ds

csc

#

Èln 2

" )

d); –

u œ ") Ä du œ )d# ) —

'  csc u du œ ln kcsc u
cot uk
C œ ln ¸csc ")
cot ") ¸
C

u œ x# × ln 2 du œ 2x dx Ä '0 eu du œ ceu d ln0 2 œ eln 2  e! œ 2  1 œ 1 2xe dx; Õ x œ 0 Ê u œ 0, x œ Èln 2 Ê u œ ln 2 Ø Ô

17.

'0

18.

'11Î2 sin (y) ecos y dy; Ô

19.

' etan v sec# v dv; ”

20.

' eÈÈtdt ; –

21.

' 3x
1 dx; ” u œ x
1 •

22.

' 2x

23.

' 2È#Èwdw ; –

24.

' 102) d); ”

25.

' 1 9
du9u

26.

4 dx ' 1
(2x
1)

27.

'01Î6 È dx

x#

Õy œ

t

u œ Èt — Ä du œ 2dt Èt

u œ ln x • Ä du œ dx x

dx; ”

w

#

;”

1 #

Ä

u œ 2) Ä du œ 2 d) •

1  9x#

;

' 3u du œ ˆ ln"3 ‰ 3u
C œ 3lnÐ
3 Ñ
C x 1

' 2u du œ ln2 #
C œ 2ln #
C

u œ Èw — Ä du œ #dw Èw

;”

' eu du œ eu
C œ etan v
C

' 2eu du œ 2eu
C œ 2eÈt
C

x œ 3u Ä dx œ 3 du •

#

u œ cos y × c1 0 du œ  sin y dy Ä '0 eu du œ 'c1 eu du œ ceu d !" œ 1  e" œ Ê u œ 0, y œ 1 Ê u œ 1 Ø

u œ tan v Ä du œ sec# v dv •

du œ dx

ln x

' csc u du œ  ln kcsc u
cot uk
C œ  ln kcsc (s  1)
cot (s  1)k
C

Ä

u

ln x

' 2u du œ ln2 #
C œ 2lnÈ# u

w


C

' "# 10u du œ # 10ln 10
C œ "# Š ln1010 ‹
C 2)

u

' 13
dxx

u œ 2x
1 Ä du œ 2 dx •

#

œ 3 tan" x
C œ 3 tan" 3u
C

' 12
duu

#

œ 2 tan" u
C œ 2 tan" (2x
1)
C

u œ 3x du œ 3 dx Õ x œ 0 Ê u œ 0, x œ " Ê u œ 6 Ô

× " #

Ø

Ä '0

1Î2

" du 3 È 1  u#

"Î#

œ
3" sin" u‘ !

œ

" 3

ˆ 16  0‰ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 18

e" e

Section 8.1 Basic Integration Formulas 28.

'01 È dt

29.

' È2s ds

30.

'

31.

'

32.

'

33.

' e
dxe


34.

' È dy

35.

'1e

" œ
sin" #t ‘ ! œ sin" ˆ "# ‰  0 œ

4  t#

;”

1  s%

u œ s# Ä du œ 2s ds •

2 dx xÈ1  4 ln# x

6 dx xÈ25x#  1

" 3

e2y  1

dx x cos (ln x)

œ ln ¸sec

ex dx e2x
1

œ'

1 Î3

1 3

œ

6 dx 5xÉx#  #"5

6 5

œ sin" u
C œ sin" s#
C

' È du

1  u#

œ sin" u
C œ sin" (2 ln x)
C

† 5 sec" k5xk
C œ 6 sec" k5xk
C

sec" ¸ 3r ¸
C

œ'

x

1  u#

u œ 2 ln x Ä du œ 2 xdx •

œ'

œ

dr r È r#  9

x

;”

' È du

1 6

;”

u œ ex Ä' du œ ex dx •

ey dy e y É ae y b #

1

;”

du u#
1

u œ ey Ä du œ ey dy •

'

œ tan" u
C œ tan" ex
C

du uÈ u#  1

u œ ln x du œ dx ; x Õ x œ 1 Ê u œ 0, x œ e1Î3 Ê u œ Ô

œ sec" kuk
C œ sec" ey
C

× 1 3

Ä '0

1Î3

Ø

du cos u

œ '0

1Î3

1Î$

sec u du œ cln ksec u
tan ukd !


tan 13 ¸  ln ksec 0
tan 0k œ ln Š2
È3‹  ln (1) œ ln Š2
È3‹ u œ ln# x Ä du œ 2x ln x dx •

36.

' x
ln4xx dxln x œ ' x a1ln
x4dxln xb ; ”

37.

'12

38.

'24 x 26xdx
10 œ 2 '24 (x  dx3)
1 ; Ô

39.

'È

40.

' È d)

41.

'

#

#

' #"

du 1
4u

œ

" 8

ln k1
4uk
C œ

" 8

ln a1
4 ln# xb
C

uœx1 × 1 " du œ dx Ä 8'0 1
duu# œ 8 ctan" ud ! Õ x œ 1 Ê u œ 0, x œ 2 Ê u œ " Ø œ 8 atan" 1  tan" 0b œ 8 ˆ 14  0‰ œ 21 x#

8 dx  2x
2

œ 8'1

2

dx 1
(x  1)#

;

Ô

uœx3 × 1 du œ dx Ä 2'c1 Õ x œ 2 Ê u œ 1, x œ 4 Ê u œ 1 Ø œ 2 ctan" 1  tan" (1)d œ 2
14  ˆ 14 ‰‘ œ 1 #

#

dt t#
4t  3

2)  ) #

œ'

œ'

dx (x
1)Èx#
2x

dt È1  (t  2)#

d) È1  ()  1)#

œ'

;”

;”

uœt2 Ä' du œ dt •

uœ)1 Ä' du œ d) •

dx (x
1)È(x
1)#  1

;”

du È 1  u#

du È 1  u#

du u#
1

"

œ 2 ctan" ud "

œ sin" u
C œ sin" (t  2)
C

œ sin" u
C œ sin" ()  1)
C

uœx
1 Ä' du œ dx •

du uÈ u#  1

œ sec" kuk
C œ sec" kx
1k
C,

kuk œ kx
1k  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

495

496 42.

Chapter 8 Techniques of Integration

'

dx (x  2)Èx#  4x
3

œ'

dx (x  2)È(x  2)# 1

;”

uœx2 Ä du œ dx •

'

du uÈ u#  1

œ sec" kuk
C

œ sec" kx  2k
C, kuk œ kx  2k  1 43.

' (sec x
cot x)# dx œ ' asec# x
2 sec x cot x
cot# xb dx œ ' sec# x dx
' 2 csc x dx
' acsc# x  1b dx œ tan x  2 ln kcsc x
cot xk  cot x  x
C

44.

' (csc x  tan x)# dx œ ' acsc# x  2 csc x tan x
tan# xb dx œ ' csc# x dx  ' 2 sec x dx
' asec# x  1b dx œ  cot x  2 ln ksec x
tan xk
tan x  x
C

45.

' csc x sin 3x dx œ ' (csc x)(sin 2x cos x
sin x cos 2x) dx œ ' (csc x) a2 sin x cos# x
sin x cos 2xb dx œ ' a2 cos# x
cos 2xb dx œ ' [(1
cos 2x)
cos 2x] dx œ ' (1
2 cos 2x) dx œ x
sin 2x
C

46.

' (sin 3x cos 2x  cos 3x sin 2x) dx œ ' sin (3x  2x) dx œ ' sin x dx œ  cos x
C

47.

' x
x 1 dx œ ' ˆ1  x
"1 ‰ dx œ x  ln kx
1k
C

48.

' x x
1 dx œ ' ˆ1  x "
1 ‰ dx œ x  tan" x
C

49.

'È32

50.

'c31 4x2x
37 dx œ 'c31
(2x  3)
2x 2
3 ‘ dx œ cx#  3x
ln k2x
3kd $" œ (9  9
ln 9)  (1
3
ln 1) œ ln 9  4

51.

' 4t t t

4 16t dt œ '
(4t  1)
t
4 4 ‘ dt œ 2t#  t
2 tan" ˆ #t ‰
C

52.

' 2) 2)7) 5
7) d) œ '
a)#  )
1b
2) 5 5 ‘ d) œ )3

53.

' È1  x

54.

' x
È2Èx  1 dx œ '

55.

'01Î4

56.

'01Î2 12
4x8x

#

#

#

dx œ 'È2 ˆ2x
3

2x$ x#  1

2x ‰ x#  1

dx œ cx#
ln kx#  1kd È2 œ (9
ln 8)  (2
ln 1) œ 7
ln 8 3

#

$

#

#

#

$

#

1  x#

2x

$

dx œ '

dx È 1  x#

x1

1
sin x cos# x

#

'

dx 2È x  1

dx x

5 #

dx œ '0 asec# x
sec x tan xb dx œ ctan x
sec xd ! 1Î4

dx œ '0 ˆ 1
24x#  1Î2

ln k2)  5k
C

1Î%

8x ‰ 1
4x#

œ Š1
È2‹  (!
1) œ È2 "Î#

dx œ ctan" (2x)  ln k1
4x# kd ! 1 4

 ln 2

 sin x) ' 1
dxsin x œ ' a(11 sinsin x)xb dx œ ' (1cos ' asec# x  sec x tan xb dx œ tan x  sec x
C x dx œ #

#

58. 1
cos x œ 1
cos ˆ2 † #x ‰ œ 2 cos#

59.


)


œ (x  1)"Î#
ln kxk
C

œ atan" 1  ln 2b  atan" 0  ln 1b œ 57.

)# #

œ sin" x
È1  x#
C

x dx È 1  x#


'



' sec )
" tan ) d) œ '

d); ”

x #

Ê

' 1
dxcos x œ ' 2 cosdx ˆ ‰ œ #" ' sec# ˆ #x ‰ dx œ tan x#
C

u œ 1
sin ) Ä du œ cos ) d) •

#

x #

' duu œ ln kuk
C œ ln k1
sin )k
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.1 Basic Integration Formulas 60.

61.

62.

' csc )
" cot ) d) œ ' 1
sincos) ) d); ”

#

#


1 ' 1  "csc x dx œ ' sinsinx x 1 dx œ ' ˆ1
sin x"  1 ‰ dx œ ' Š1
(sin x sin1)x(sin x
1) ‹ dx 1
sin x ‰ cos# x

dx œ ' ˆ1  sec# x 

'021 É 1  #cos x dx œ '021 ¸sin x# ¸ dx; ” œ (2)(2) œ 4

64.

' udu œ  ln kuk
C œ  ln k1
cos )k
C

cos x ‰ ' 1  "sec x dx œ ' coscosx x 1 dx œ ' ˆ1
cos x"  1 ‰ dx œ ' ˆ1  1
sincosx x ‰ dx œ ' ˆ1  csc# x  sin x dx œ ' a1  csc# x  csc x cot xb dx œ x
cot x
csc x
C

œ ' ˆ1 

63.

u œ 1
cos ) Ä du œ  sin ) d) •

sin x ‰ cos# x

dx œ ' a1  sec# x  sec x tan xb dx œ x  tan x  sec x
C

21 sin #x   0 #1 Ä '0 sin ˆ x# ‰ dx œ
2 cos x# ‘ ! œ 2(cos 1  cos 0) • x for 0 Ÿ # Ÿ 21

'01 È1  cos 2x dx œ '01 È2 ksin xk dx; ”

1 1 sin x   0 Ä È2 '0 sin x dx œ ’È2 cos x“ • for 0 Ÿ x Ÿ 1 !

œ È2 (cos 1  cos 0) œ 2È2 65.

'11Î2 È1
cos 2t dt œ '11Î2 È2 kcos tk dt; ” œ È2 ˆsin 1  sin

66.

1‰ #

cos t Ÿ 0 Ä for 1# Ÿ t Ÿ 1 •

'11Î2 È2 cos t dt œ ’È2 sin t“ 1

1Î#

œ È2

'c01 È1
cos t dt œ 'c01 È2 ¸cos #t ¸ dt; ”

0 cos #t   0 • Ä 'c1 È2 cos for 1 Ÿ t Ÿ 0

t #

dt œ ’2È2 sin #t “

! 1

œ 2È2
sin 0  sin ˆ 1# ‰‘ œ 2È2 67.

'c01 È1  cos# ) d) œ 'c01 ksin )k d); ”

0 sin ) Ÿ 0 Ä 'c1  sin ) d) œ ccos )d !1 œ cos 0  cos (1) • for 1 Ÿ ) Ÿ 0

œ 1  (1) œ 2 cos ) Ÿ 0 Ä for 1# Ÿ ) Ÿ 1 •

68.

'11Î2 È1  sin# ) d) œ '11Î2 kcos )k d); ”

69.

'c11ÎÎ44 Ètan# y
1 dy œ '
11ÎÎ44 ksec yk dy; ”

sec y   0 for  14 Ÿ y Ÿ

'11Î2  cos ) d) œ c sin )d 11Î# œ  sin 1
sin 1# œ 1 1Î4

1 4

• Ä '
1Î4 sec y dy œ cln ksec y
tan ykd 1Î% 1Î%

œ ln ¹È2
1¹  ln ¹È2  1¹

70.

'c01Î4 Èsec# y  1 dy œ '
01Î4 ktan yk dy; ”

tan y Ÿ 0 Ä for  14 Ÿ y Ÿ 0 •

'
01Î4  tan y dy œ cln kcos ykd !1Î% œ  ln Š È"

œ ln È2 71.

'13Î14Î4 (csc x  cot x)# dx œ '13Î14Î4 acsc# x  2 csc x cot x
cot# xb dx œ '13Î14Î4 a2 csc# x  1  2 csc x cot xb dx $1Î%

œ c2 cot x  x
2 csc xd 1Î% œ ˆ2 cot œ ’2(1) 

31 4


2 ŠÈ2‹“  ’2(1) 

31 4



31 4

1 4


2 ŠÈ 2‹ “ œ 4 


2 csc

31 ‰ 4

 ˆ2 cot

1 4



1 4


2 csc 14 ‰

1 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2

‹

497

498 72.

Chapter 8 Techniques of Integration

1Î% 2x ‰‘ '01Î4 (sec x
4 cos x)# dx œ '01Î4
sec# x
8
16 ˆ 1
cos dx œ ctan x
16x  4 sin 2xd ! #

œ ˆtan 73.

1 4


41  4 sin 1# ‰  (tan 0
0  4 sin 0) œ 5
41

' cos ) csc (sin )) d); ” u œ sin ) • Ä ' csc u du œ  ln kcsc u
cot uk
C du œ cos ) d) œ  ln kcsc (sin ))
cot (sin ))k
C

74.

' ˆ1
"x ‰ cot (x
ln x) dx; ” u œ x
ln" x • Ä ' cot u du œ ln ksin uk
C œ ln ksin (x
ln x)k
C du œ ˆ1
‰ dx x

75.

' (csc x  sec x)(sin x
cos x) dx œ ' (1
cot x  tan x  1) dx œ ' cot x dx  ' tan x dx œ ln ksin xk
ln kcos xk
C

76.

'

77.

' Èy6(1dy
y) ; –

78.

'

79.

'

3 sinhˆ x2
ln 5‰dx œ ”

dx xÈ4x#  1

u œ x2
ln 5 œ ' ' sinh u du œ 6 cosh u
C œ 6 coshˆ x#
ln 5‰
C 2 du œ dx •

u œ Èy " — Ä du œ 2È y dy

œ'

2 dx 2xÈ(2x)#  1

7 dx (x  1)Èx#  2x  48

œ'

;”

' 121
duu œ 12 tan" u
C œ 12 tan" Èy
C #

u œ 2x Ä' du œ 2 dx •

7 dx (x  1)È(x  1)#  49

;”

du uÈ u#  1

œ sec" kuk
C œ sec" k2xk
C

uœx1 Ä du œ dx •

'

7 du uÈu#  49

œ7†

" 7

sec" ¸ 7u ¸
C

œ sec" ¸ x 7 " ¸
C 80.

'

dx (2x
1)È4x#
4x

œ

" #

œ'

dx (2x
1)È(2x
1)#  1

;”

u œ 2x
1 Ä' du œ 2 dx •

du 2uÈu#  1

œ

" #

sec" kuk
C

sec" k2x
1k
C

81.

' sec# t tan (tan t) dt; ” u œ tan# t • Ä ' tan u du œ  ln kcos uk
C œ ln ksec uk
C œ ln ksec (tan t)k
C du œ sec t dt

82.

'

dx xÈ $
x#

83. (a) (b)

œ  "$ csch" ¹ Èx$ ¹
C

' cos$ ) d) œ ' (cos )) a1  sin# )b d); ” u œ sin ) • Ä ' a1  u# b du œ u  u3$
C œ sin )  "3 sin$ )
C du œ cos ) d) ' cos& ) d) œ ' (cos )) a1  sin# )b# d) œ ' a1  u# b# du œ ' a1  2u#
u% b du œ u  23 u$
u5&
C œ sin ) 

(c)

84. (a)

2 3

sin$ )


sin& )
C

' cos* ) d) œ ' acos) )b (cos )) d) œ ' a1  sin# )b% (cos )) d) ' sin$ ) d) œ ' a1  cos# )b (sin )) d); ” œ  cos )


(b)

" 5

" 3

cos$ )
C

2 3

cos$ ) 

u œ cos ) Ä ' a1  u# b ( du) œ du œ  sin ) d) •

u$ 3

u
C

' sin& ) d) œ ' a1  cos# )b# (sin )) d) œ ' a1  u# b# ( du) œ ' a1
2u#  u% b du œ  cos )


" 5

cos& )
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.1 Basic Integration Formulas ' sin( ) d) œ ' a1  u# b$ ( du) œ ' a1
3u#  3u%
u' b du œ  cos )
cos$ )  35 cos& )
cos7( )
C ' (d) ' sin"$ ) d) œ ' asin"# )b (sin )) d) œ ' a1  cos# )b (sin )) d) (c)

85. (a)

' tan$ ) d) œ ' asec# )  1b (tan )) d) œ ' sec# ) tan ) d)  ' tan ) d) œ "# tan# )  ' tan ) d) œ

" #

tan# )
ln kcos )k
C

' tan& ) d) œ ' asec# )  1b atan$ )b d) œ ' tan$ ) sec# ) d)  ' tan$ ) d) œ "4 tan% )  ' tan$ ) d) (c) ' tan( ) d) œ ' asec# )  1b atan& )b d) œ ' tan& ) sec# ) d)  ' tan& ) d) œ "6 tan' )  ' tan& ) d) (d) ' tan2kb1 ) d) œ ' asec# )  1b atan2kc1 )b d) œ ' tan2kc1 ) sec# ) d)  ' tan2kc1 ) d); (b)

u œ tan ) ” du œ sec# ) d) • Ä 86. (a)

' u2kc1 du  ' tan2kc1 ) d) œ

" 2k

u2k  ' tan2kc1 ) d) œ

tan2k )  ' tan2kc1 ) d)

" #k

' cot$ ) d) œ ' acsc# )  1b (cot )) d) œ ' cot ) csc# ) d)  ' cot ) d) œ  "# cot# )  ' cot ) d) œ  "# cot# )  ln ksin )k
C

' cot& ) d) œ ' acsc# )  1b acot$ )b d) œ ' cot$ ) csc# ) d)  ' cot$ ) d) œ  "4 cot% )  ' cot$ ) d) (c) ' cot( ) d) œ ' acsc# )  1b acot& )b d) œ ' cot& ) csc# ) d)  ' cot& ) d) œ  "6 cot' )  ' cot& ) d) (d) ' cot2kb1 ) d) œ ' acsc# )  1b acot2kc1 )b d) œ ' cot2kc1 ) csc# ) d)  ' cot2kc1 ) d); (b)

u œ cot ) 2kc1 du  ' cot2kc1 ) d) œ  #"k u2k  ' cot2kc1 ) d) ” du œ  csc# ) d) • Ä ' u " œ  2k cot2k )  ' cot2kc1 ) d)

87. A œ 'c1Î4 (2 cos x  sec x) dx œ c2 sin x  ln ksec x
tan xkd 1Î% 1Î4

1Î%

œ ’È2  ln ŠÈ2
1‹“  ’È2  ln ŠÈ2  1‹“ #

È 2
1‹

Š È œ 2È2  ln Š È2
" ‹ œ 2È2  ln  21

21



œ 2È2  ln Š3
2È2‹ 88. A œ '1Î6 (csc x  sin x) dx œ c ln kcsc x
cot xk
cos xd 1Î' 1Î2

1Î#

œ  ln k1
0k
ln ¹2
È3¹ 

1Î4

È3 #

œ ln Š2
È3‹ 

1Î4

È3 #

1Î4

1Î4

89. V œ 'c1Î4 1(2 cos x)# dx  '
1Î4 1 sec# x dx œ 41 '
1Î4 cos# x dx  1'
1Î4 sec# x dx œ 21 'c1Î4 (1
cos 2x) dx  1 ctan xd 1Î% œ 21
x
1Î4

1Î%

" #

1Î%

sin 2x‘ 1Î%  1[1  (1)]

œ 21
ˆ 14
"# ‰  ˆ 14  "# ‰‘  21 œ 21 ˆ 1#
1‰  21 œ 1# 90. V œ '1Î6 1 csc# x dx  '1Î6 1 sin# x dx œ 1 '1Î6 csc# x dx  1Î2

1Î2

1Î#

œ 1 c cot xd 1Î'  œ 1È3 

1 #

Š 261


1 #


x 

È3 4 ‹

" #

1Î2

1Î#

1 #

sin 2x‘ 1Î' œ 1 ’0  ŠÈ3‹“ 

'11ÎÎ62 (1  cos 2x) dx 1 #

’ˆ 1#  0‰  Š 16 

" #

†

È3 # ‹“

È

œ 1 Š 7 8 3  16 ‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

499

500

Chapter 8 Techniques of Integration

91. y œ ln (cos x) Ê

dy dy sin x # # ' œ  cos x Ê Š dx ‹ œ tan x œ sec x  1; L œ a Ê1
Š dx ‹ dx #

dy dx

#

b

1Î$ œ '0 È1
asec# x  1b dx œ '0 sec x dx œ cln ksec x
tan xkd ! œ ln ¹2
È3¹  ln k1
0k œ ln Š2
È3‹ 1Î3

1Î3

92. y œ ln (sec x) Ê

œ

dy dx

sec x tan x sec x

dy # # ' Ê Š dy dx ‹ œ tan x œ sec x  1; L œ a Ê1
Š dx ‹ dx #

œ '0 sec x dx œ cln ksec x
tan xkd ! 1Î4

1Î%

1Î4

93. Mx œ 'c1Î4 ˆ "# sec x‰ (sec x) dx œ œ

" #

1Î%

ctan xd 1Î% œ

" #

b

#

œ ln ¹È2
1¹  ln k1
0k œ ln ŠÈ2
1‹

'
11ÎÎ44 sec# x dx

" #

c1  (1)d œ 1;

M œ 'c1Î4 sec x dx œ cln ksec x
tan xkd 1Î% 1Î4

1Î%

È

œ ln ¹È2
1¹  ln ¹È2  1¹ œ ln Š È2
" ‹ 2 1

#

œ ln 

ŠÈ2
"‹ #1

 œ ln Š3
2

È2‹ ; x œ 0 by

symmetry of the region, and y œ

94. Mx œ '1Î6 ˆ "# csc x‰ (csc x) dx œ 51Î6

œ

" #

&1Î'

c cot xd 1Î' œ

" #

œ

Mx M

" #

" ln Š3
2È2‹

'15Î16Î6 csc# x dx

’ ŠÈ3‹  ŠÈ3‹“ œ È3;

M œ '1Î6 csc x dx œ c ln kcsc x
cot xkd 1Î' 51Î6

&1Î' È

3 œ  ln ¹2  È3¹  Š ln ¹2
È3¹‹ œ ln ¹ 22
¹  È3 #

œ ln 

Š2
È3‹ 43

 œ 2 ln Š2


of the region, and y œ

95.

Mx M

œ

È 3‹ ; x œ

1 #

by symmetry

È3 2 ln Š2
È3‹

x
cot x ‰ x cot x ' csc x dx œ ' (csc x)(1) dx œ ' (csc x) ˆ csc ' csccscx
x csc dx; csc x
cot x dx œ
cot x #

u œ csc x
cot x ” du œ a csc x cot x  csc# xb dx • Ä 96. cax#  1b (x
1)d

#Î$

1‰ œ (x
1)# ˆ xx 
1

' udu œ  ln kuk
C œ  ln kcsc x
cot xk
C

œ c(x  1)(x
1)# d

#Î$

#Î$

œ (x
1)# ˆ1 

œ (x  1)#Î$ (x
1)%Î$ œ (x
1)#
(x  1)#Î$ (x
1)#Î$ ‘

2 ‰#Î$ x
1

u œ x
" 1 du œ  (x
" 1)# dx —

(a)

' cax#  1b (x
1)d#Î$ dx œ '

(b)

' (1  2u)#Î$ du œ 3# (1  2u)"Î$
C œ 3# ˆ1  x
2 1 ‰"Î$
C œ #3 ˆ xx 
11 ‰"Î$
C ' cax#  1b (x
1)d#Î$ dx œ ' (x
1)# ˆ xx 
11 ‰#Î$ dx; u œ ˆ xx 
11 ‰k

(x
1)# ˆ1 

2 ‰#Î$ x
1

dx; –

Ä

Ê du œ k ˆ xx 
11 ‰ œ

(x
1)# 2k

œ

" #k

kc1

c(x
1)  (x  1)d (x
1)#

k c1

(x  1) dx œ 2k (x
1)kb1 dx; dx œ

1 ‰1ck 1 ‰#Î$ ˆ xx  du; then, ' ˆ xx 
1
1

" #k

 1 ‰1ck ˆ xx
du œ 1

(x
1)# #k " #k


1 ‰kc1 ˆ xx  du 1

' ˆ xx
 11 ‰Ð1Î3
kÑ du

' ˆ xx 
11 ‰kÐ1Î3k
1Ñ du œ #"k ' uÐ1Î3k
1Ñ du œ #"k (3k) u1Î3k
C œ #3 u1Î3k
C œ 3# ˆ xx 
11 ‰"Î$
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.2 Integration by Parts (c)

' cax#  1b (x
1)d#Î$ dx œ ' (x
1)# ˆ xx 
11 ‰#Î$ dx;

" Ô u œ tan x × x œ tan u Ä Õ dx œ du# Ø cos u

' (tan u"
1)

#

tan u  1 ‰#Î$ ˆ du ‰ ' ˆ tan u
1 cos# u œ

" (sin u
cos u)#

sin u
cos u œ sin u
sin ˆ 1#  u‰ œ 2 sin 14 cos ˆu  14 ‰ – sin u  cos u œ sin u  sin ˆ 1  u‰ œ 2 cos 1 sin ˆu  1 ‰ — Ä 4 4 #

u  cos u ‰#Î$ ˆ sin du; sin u
cos u

sin ˆu  1 ‰ #Î$ ' 2 cos ˆ"u  1 ‰ ’ cos du ˆu  1 ‰ “ 4

#

4

4

1

"Î$

œ

" #

u  tan 4 ' tan#Î$ ˆu  14 ‰ sec# ˆu  14 ‰ du œ 3# tan"Î$ ˆu  14 ‰
C œ 3# ’ 1tan
C
tan u tan 14 “

œ

3 #

1 ‰"Î$ ˆ xx 
C
1

(d) u œ tan" Èx Ê tan u œ Èx Ê tan# u œ x Ê dx œ 2 tan u ˆ cos"# u ‰ du œ sin# u  cos# u cos# u

x  1 œ tan# u  1 œ

1  2 cos# u cos# u

œ

; x
1 œ tan# u
1 œ


#Î$ ub ' (x  1)#Î$ (x
1)%Î$ dx œ ' a12 cos #
#Î$ † #

œ ' a1  2 cos# ub œ

a1  2 cos# ub

3 #

acos ub

#Î$

"Î$

(e) u œ tan" ˆ x # 1 ‰ Ê

'

(x  1)

#Î$


Cœ

%Î$

3 #

–

Š 1
2 cos #

#u

cos u

Š


Cœ

" ‹ — cos# u

dx œ ' (tan u)

#Î$

3 #

(tan u
1)

2d(cos u) cos$ u

1 ‰"Î$ ˆ xx 
C
1

%Î$

†2

#

œ 2d(tan u);

2 du cos# u

† 2 † d(tan u)

œ

" #

' ˆ1  tan u"
1 ‰#Î$ d ˆ1  tan u"
1 ‰ œ #3 ˆ1  tan u"
1 ‰"Î$
C œ #3 ˆ1  x
2 1 ‰"Î$
C

œ

3 #

1 ‰"Î$ ˆ xx 
C
1

œ'

du %Î$ (sin u)"Î$ ˆ2#Î$ cos u# ‰

œ

'

3 #

' cax œ

u #

'

œ  "#

du "Î$ &Î$ 2 ˆsin #u ‰ ˆcos #u ‰

œ  ' tan"Î$ ˆ u# ‰ d ˆtan u# ‰ œ  3# tan#Î$ œ

œ

sin u du $ Éacos# u  1b# (cos u
1)#


Cœ

u)  2d(cos cos$ u ;

' a1  2 cos# ub#Î$ † d a1  2 cos# ub

"Î$

‹

†

œ tan u Ê x
1 œ 2(tan u
1) Ê dx œ

" Ô u œ cos x × Ä ' (f) x œ cos u Õ dx œ  sin u du Ø

(g)

" #

† (2) † cos u † d(cos u) œ

x1 #

(x
1)

" acos# ub
%Î$

2 sin u cos$ u du œ cos# u
sin# u œ cos"# u ; cos# u

3 #

sin u du %Î$ asin%Î$ ub ˆ2#Î$ cos #u ‰

' Š cos sin

u # u #

‹

"Î$

du ˆcos# #u ‰

ˆ tan# u# ‰"Î$
C œ

3 #

u  1 ‰"Î$ ˆ cos
C cos u
1

3 #

1 ‰"Î$ ˆ xx 
C
1

" ‰"Î$ ˆ xx 
C
1 #

'

 1b (x
1)d

#Î$

sinh u du $ È asinh% ub (cosh u
1)#

" Ô u œ cosh x × dx; Ä x œ cosh u Õ dx œ sinh u Ø

'

œ

" #

'

du $ É (sinh u) ˆ4 cosh% #u ‰

"Î$ œ ' ˆtanh u# ‰ d ˆtanh u# ‰ œ

3 #

œ

ˆtanh u# ‰#Î$
C œ

sinh u du $ Éacosh# u1b# (cosh u
1)#

' 3 #

du $ É sinh ˆ #u ‰ cosh& ˆ #u ‰ u  1 ‰"Î$ ˆ cosh
Cœ cosh u
1

8.2 INTEGRATION BY PARTS 1. u œ x, du œ dx; dv œ sin

'

x sin

x #

dx œ 2x cos

x #

x #

dx, v œ 2 cos

) cos 1) d) œ

) 1

;

 ' ˆ2 cos x# ‰ dx œ 2x cos ˆ x# ‰
4 sin ˆ x# ‰
C

2. u œ ), du œ d); dv œ cos 1) d), v œ

'

x #

sin 1)  '

" 1

" 1

sin 1);

sin 1) d) œ

) 1

sin 1)


" 1#

cos 1)
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

501

502

Chapter 8 Techniques of Integration

3.

cos t Ð
Ñ t# ïïïïî ÐÑ 2t ïïïïî Ð
Ñ 2 ïïïïî

sin t cos t sin t

0 4.

'

t# cos t dt œ t# sin t
2t cos t  2 sin t
C

'

x# sin x dx œ x# cos x
2x sin x
2 cos x
C

sin x Ð
Ñ x# ïïïïî ÐÑ 2x ïïïïî Ð
Ñ 2 ïïïïî

cos x sin x cos x

0 5. u œ ln x, du œ

dv œ x dx, v œ

dx x ;

'1 x ln x dx œ ’ x# 2

6. u œ ln x, du œ

ln x“  '1 #

#

dx x ;

dx x

ln x“  '1 e

e % x

1

dy 1
y #

dx x

4

; #

#

œ 2 ln 2  ’ x4 “ œ 2 ln 2  "

x% 4

dv œ x$ dx, v œ

%

7. u œ tan" y, du œ

#

"

'1 x$ ln x dx œ ’ x4 e

2 # x

x# #

3 4

œ ln 4 

3 4

; œ

e% 4

%

e

x  ’ 16 “ œ 1

3e%
1 16

; dv œ dy, v œ y;

' tan" y dy œ y tan" y  ' a1y
dyy b œ y tan" y  #" ln a1
y# b
C œ y tan" y  ln È1
y#
C #

8. u œ sin" y, du œ

dy È 1  y#

; dv œ dy, v œ y;

' sin" y dy œ y sin" y  ' Èy1 dy y

#

œ y sin" y
È1  y#
C

9. u œ x, du œ dx; dv œ sec# x dx, v œ tan x;

' x sec# x dx œ x tan x  ' tan x dx œ x tan x
ln kcos xk
C

10.

' 4x sec# 2x dx; [y œ 2x]

' y sec# y dy œ y tan y  ' tan y dy œ y tan y  ln ksec yk
C

Ä

œ 2x tan 2x  ln ksec 2xk
C ex

11. Ð
Ñ x$ ïïïïî ÐÑ 3x# ïïïïî Ð
Ñ 6x ïïïïî ÐÑ 6 ïïïïî 0

ex ex ex ex

'

x$ ex dx œ x$ ex  3x# ex
6xex  6ex
C œ ax$  3x#
6x  6b ex
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.2 Integration by Parts ecp

12. Ð
Ñ ïïïïî ÐÑ 4p$ ïïïïî Ð
Ñ 12p# ïïïïî ÐÑ 24p ïïïïî Ð
Ñ 24 ïïïïî p%

ecp ecp ecp ecp ecp

'

0

p% ecp dp œ p% ecp  4p$ ecp  12p# ecp  24pecp  24ecp
C œ ap%  4p$  12p#  24p  24b ecp
C

ex

13.

Ð
Ñ x#  5x ïïïïî ex ÐÑ 2x  5 ïïïïî ex Ð
Ñ 2 ïïïïî ex

' ax#  5xb ex dx œ ax#  5xb ex  (2x  5)ex
2ex
C œ x# ex  7xex
7ex
C

0

œ ax#  7x
7b ex
C er

14.

Ð
Ñ r#
r
1 ïïïïî er ÐÑ 2r
1 ïïïïî er Ð
Ñ 2 ïïïïî er 0

' ar#
r
1b er dr œ ar#
r
1b er  (2r
1) er
2er
C œ car#
r
1b  (2r
1)
2d er
C œ ar#  r
2b er
C

ex

15. x& 5x% 20x$ 60x# 120x 120 0

Ð
Ñ ïïïïî ex ÐÑ ïïïïî ex Ð
Ñ ïïïïî ex ÐÑ ïïïïî ex Ð
Ñ ïïïïî ex ÐÑ ïïïïî ex

' x& ex dx œ x& ex  5x% ex
20x$ ex  60x# ex
120xex  120ex
C œ ax&  5x%
20x$  60x#
120x  120b ex
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

503

504

Chapter 8 Techniques of Integration e4t

16. Ð
Ñ t# ïïïïî ÐÑ 2t ïïïïî Ð
Ñ 2 ïïïïî

" 4

e4t

" 16

e4t

" 64

e4t

' t# e4t dt œ t4 e4t  162t e4t
642 e4t
C œ t4 e4t  8t e4t
3"# e4t
C #

0

#

œ Š t4  17.




" 4t 3# ‹ e


C

sin 2) Ð
Ñ )# ïïïïî  "2 cos 2) ÐÑ 2) ïïïïî  "4 sin 2) Ð
Ñ 2 ïïïïî "8 cos 2)

'01Î2 )# sin 2) d) œ ’ )#

#

0

œ ’ 18.

t 8

#

1# 8

† (1)


1 4

cos 2)


†0


" 4

) #

sin 2)


" 4

cos 2)“

† (1)“ 
0
0


" 4

1Î# !

† 1‘ œ

1# 8



" #

œ

1#  4 8

cos 2x Ð
Ñ x$ ïïïïî "2 sin 2x ÐÑ 3x# ïïïïî  "4 cos 2x Ð
Ñ 6x ïïïïî  "8 sin 2x ÐÑ " 6 ïïïïî 16 cos 2x

'01Î2 x$ cos 2x dx œ ’ x#

$

0

œ 19. u œ sec" t, du œ

dt tÈt# 1

'22ÎÈ3 t sec" t dt œ ’ t#

#

œ

51 9

 ’ "# Èt#  1“

20. u œ sin" ax# b , du œ

È2

'01Î

1$ ’ 16

†0


; dv œ t dt, v œ

sec" t“

# #ÎÈ$

œ

2x dx È 1 x %

#


’È1  x% “

!

#

dt tÈt#  1

œ

†0

1 1#

3 8

œ ˆ2 †

 "# ŠÈ3  É 43  1‹ œ

; dv œ 2x dx, v œ x# ;

"ÎÈ#

31 8

3x# 4

cos 2x 

3x 4

sin 2x 

3 8

cos 2x“

† (1)“ 
0
0  0 

3 8

1Î# ! #

1 † 1‘ œ  316


3 4

œ

3 a4  1 # b 16

;

"ÎÈ#

1 1#

† (1) 

 '2ÎÈ3 Š t# ‹

2x sin" ax# b dx œ cx# sin" ax# bd !

œ

t# # 2

#ÎÈ$

51 9

31 # 16

sin 2x


 '0


É 34  1 œ

È2

1Î

x# †

51 9

1 3



2 3

† 16 ‰  '2ÎÈ3

 "# ŠÈ3 

2x dx È 1  x%

2

È3 3 ‹

œ

œ ˆ "# ‰ ˆ 16 ‰
'0

t dt 2Èt#  1

51 9



È3 3

œ

51  3È 3 9

È 2 d ˆ1  x % ‰

1Î

2È 1  x%

1
6È312 1#

21. I œ ' e) sin ) d); cu œ sin ), du œ cos ) d); dv œ e) d), v œ e) d Ê I œ e) sin )  ' e) cos ) d); cu œ cos ), du œ  sin ) d); dv œ e) d), v œ e) d Ê I œ e) sin )  Še) cos )
' e) sin ) d)‹ œ e) sin )  e) cos )  I
Cw Ê 2I œ ae) sin )  e) cos )b
Cw Ê I œ

" #

ae) sin )  e) cos )b
C, where C œ

another arbitrary constant

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

w

C #

is

Section 8.2 Integration by Parts 22. I œ ' ecy cos y dy; cu œ cos y, du œ  sin y dy; dv œ ecy dy, v œ ecy d Ê I œ ecy cos y  ' aecy b (sin y) dy œ ecy cos y  ' ecy sin y dy; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecy d Ê I œ ecy cos y  Šecy sin y  ' aey b cos y dy‹ œ ecy cos y
ecy sin y  I
Cw Ê 2I œ ecy (sin y  cos y)
Cw Ê I œ

" #

aecy sin y  ecy cos yb
C, where C œ

23. I œ ' e2x cos 3x dx;
u œ cos 3x; du œ 3 sin 3x dx, dv œ e2x dx; v œ

is another arbitrary constant

e2x ‘

Ê

' e2x sin 3x dx;
u œ sin 3x, du œ 3 cos 3x, dv œ e2x dx; v œ "# e2x ‘ I œ "# e2x cos 3x
3# Š "# e2x sin 3x  3# ' e2x cos 3x dx‹ œ "# e2x cos 3x
34 e2x sin 3x  94 I
Cw

Ê

13 4

Ê Iœ

24.

" #

w

C #

" #

e2x cos 3x


Iœ

" #

3 #

e2x cos 3x
34 e2x sin 3x
Cw Ê

e2x 13

(3 sin 3x
2 cos 3x)
C, where C œ

4 13

Cw

' ec2x sin 2x dx; [y œ 2x] Ä "# ' ecy sin y dy œ I; cu œ sin y, du œ cos y dy; dv œ ecy dy, v œ ecyd Ê I œ "# Šecy sin y
' ecy cos y dy‹ cu œ cos y, du œ  sin y; dv œ ecy dy, v œ ecy d Ê I œ  "# ecy sin y
"# Šecy cos y  ' aecy b ( sin y) dy‹ œ  "# ecy (sin y
cos y)  I
Cw c2x

Ê 2I œ  "# ecy (sin y
cos y)
Cw Ê I œ  "4 ecy (sin y
cos y)
C œ  e 4 (sin 2x
cos 2x)
C, where Cœ

w

C #

#

25.

' eÈ3sb9 ds; ” 3s
92 œ x ds œ

2 3

' xex dx œ

2 3

3

x dx •

Ä

' ex † 23 x dx œ 23 ' xex dx; cu œ x, du œ dx; dv œ ex dx, v œ ex d ;

Šxex  ' ex dx‹ œ

2 3

axex  ex b
C œ

2 3

ŠÈ3s
9 eÈ3sb9  eÈ3sb9 ‹
C

26. u œ x, du œ dx; dv œ È1  x dx, v œ  23 È(1  x)$ ;

'01 xÈ1  x dx œ
 23 È(1  x)$ x‘ "!
23 '01 È(1  x)$ dx œ 23
 25 (1  x)&Î# ‘ "! œ 154

27. u œ x, du œ dx; dv œ tan# x dx, v œ ' tan# x dx œ ' œ tan x  x;'0

1Î3

œ

1 3

1Î$

x tan# x dx œ cx(tan x  x)d !

ŠÈ3  13 ‹
ln

28. u œ ln ax
x# b, du œ œ x ln ax
x# b  '

" #




1# 18

œ

(2x
1) dx x
x #

(2x
1) dx x
1

1È3 3

 ln 2 

sin# x cos# x

dx œ '

"  cos# x cos# x

 '0 (tan x  x) dx œ 1Î3

1 3

dx œ '

dx cos# x

 ' dx

ŠÈ3  13 ‹
’ln kcos xk


1# 18

; dv œ dx, v œ x; ' ln ax
x# b dx œ x ln ax
x# b  '

œ x ln ax
x# b  '

2(x
1)  " x
1

2x
" x(x
1)

† x dx

dx œ x ln ax
x# b  2x
ln kx
1k
C

u œ ln x

29.

1Î$ x# # “!

' sin (ln x) dx; Ô du œ "x dx ×

Ä ' (sin u) eu du. From Exercise 21, ' (sin u) eu du œ eu ˆ sin u # cos u ‰
C Õ dx œ eu du Ø œ "# cx cos (ln x)
x sin (ln x)d
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

505

506

Chapter 8 Techniques of Integration u œ ln z

30.

' z(ln z)# dz; Ô du œ "z dz × Õ dz œ eu du Ø e2u

Ð
Ñ u# ïïïïî ÐÑ 2u ïïïïî Ð
Ñ 2 ïïïïî

" 2

e2u

" 4

e2u

" 8

e2u

'

0

Ä

' eu † u# † eu du œ ' e2u † u# du;

u# e2u du œ œ

z# 4

u# # #

e2u  #u e2u
"4 e2u
C œ

e2u 4

c2u#  2u
1d
C

c2(ln z)  2 ln z
1d
C

31. (a) u œ x, du œ dx; dv œ sin x dx, v œ  cos x;

S" œ '0 x sin x dx œ [x cos x]!1
'0 cos x dx œ 1
[sin x]1! œ 1 1

1

(b) S# œ '1 x sin x dx œ ”[x cos x]#11
'1 cos x dx• œ  c31
[sin x]1#1 d œ 31 21

21

(c) S$ œ '21 x sin x dx œ [x cos x]$#11
'21 cos x dx œ 51
[sin x]$#11 œ 51 31

31

Ðn
1Ñ1

(d) S8
" œ (1)nb1 'n1

x sin x dx œ (1)nb1 c[x cos x]Ðnn1
1Ñ1
[sin x]Ðnn1
1Ñ1 d

œ (1)nb1 c(n
1)1(1)n
n1(1)nb1 d
0 œ (2n
1)1

32. (a) u œ x, du œ dx; dv œ cos x dx, v œ sin x; 31Î2 1‰ S" œ  '1Î2 x cos x dx œ  ”[x sin x]311Î2Î2  '1Î2 sin x dx• œ  ˆ 31 #  #  [cos x]1Î2 œ 21 31Î2

31Î2

(b) S# œ '31Î2 x cos x dx œ [x sin x]&$11ÎÎ22  '31Î2 sin x dx œ
5#1  ˆ 3#1 ‰‘  [cos x]&$11ÎÎ22 œ 41 51Î2

51Î2

(c) S$ œ  '51Î2 x cos x dx œ  ”[x sin x](&11ÎÎ22  '51Î2 sin x dx• œ  ˆ 7#1  71Î2

71Î2

Ð2n1Ñ1Î2

51 ‰ #

 [cos x](&11ÎÎ22 œ 61

Ð2n1Ñ1Î2

n1Ñ1Î2 n' (d) Sn œ (1)n 'Ð2n
1Ñ1Î2 x cos x dx œ (1)n ”[x sin x]Ð# sin x dx• Ð2n
1Ñ1Î2  Ð2n
1Ñ1Î2

œ (1)n ’ (2n
# 1)1 (1)n  33. V œ '0

ln 2

(2n1)1 #

n
1Ñ1Î2 (1)nc1 “  [cos x]Ð# Ð2n1Ñ1Î2 œ

" #

(2n1
1
2n1  1) œ 2n1

21(ln 2  x) ex dx œ 21 ln 2 '0 ex dx  21'0 xex dx ln 2

ln 2

œ (21 ln 2) cex d ln0 2  21 Œcxex d ln0 2  '0 ex dx ln 2

œ 21 ln 2  21 ˆ2 ln 2  cex d ln0 2 ‰ œ 21 ln 2
21 œ 21(1  ln 2)

34. (a) V œ '0 21xecx dx œ 21 Œcxecx d "!
'0 ecx dx 1

1

œ 21 Š "e
cecx d "! ‹ œ 21 ˆ "e  œ 21 

" e


1‰

41 e

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.2 Integration by Parts (b) V œ '0 21(1  x)ecx dx; u œ 1  x, du œ  dx; dv œ ecx dx, 1

v œ ecx ; V œ 21 ”c(1  x) aecx bd "!  '0 ecx dx• 1

œ 21 ’[0  1(1)]
cecx d "! “ œ 21 ˆ1


" e

35. (a) V œ '0 21x cos x dx œ 21 Œ[x sin x] ! 1Î2

1Î#

 1‰ œ

21 e

 '0 sin x dx 1Î2

1Î#

œ 21 Š 1#
[cos x] ! ‹ œ 21 ˆ 1#
0  1‰ œ 1(1  2)

(b) V œ '0 21 ˆ 1#  x‰ cos x dx; u œ 1Î2

V œ 21
ˆ 1#  x‰ sin

1Î# x‘ !

1 #

 x, du œ  dx; dv œ cos x dx, v œ sin x;


21'0 sin x dx œ 0
21[ cos x] ! 1Î2

1Î#

œ 21(0
1) œ 21

36. (a) V œ '0 21x(x sin x) dx; 1

sin x Ð
Ñ x# ïïïïî cos x ÐÑ 2x ïïïïî sin x Ð
Ñ 2 ïïïïî cos x 0

Ê V œ 21'0 x# sin x dx œ 21 cx# cos x
2x sin x
2 cos xd ! œ 21 a1#  4b 1

1

(b) V œ '0 21(1  x)x sin x dx œ 21# '0 x sin x dx  21 '0 x# sin x dx œ 21# [x cos x
sin x]1!  a21$  81b 1

1

1

œ 81 37. av(y) œ œ

" 1

" #1

'021 2ect cos t dt


ect ˆ sin t # cos t ‰‘ #1 !

(see Exercise 22) Ê av(y) œ

" #1

a1  ec21 b

'021 4ect (sin t  cos t) dt 21 21 œ 12 '0 ect sin t dt  12 '0 ect cos t dt

38. av(y) œ

œ œ

2 1 2 1

" #1


ect ˆ  sin t# cos t ‰  ect ˆ sin t # cos t ‰‘ #1 ! cect sin td #!1 œ 0

39. I œ ' xn cos x dx; cu œ xn , du œ nxn" dx; dv œ cos x dx, v œ sin xd Ê I œ xn sin x  ' nxn" sin x dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

507

508

Chapter 8 Techniques of Integration

40. I œ ' xn sin x dx; cu œ xn , du œ nxn" dx; dv œ sin x dx, v œ cos xd Ê I œ xn cos x
' nxn" cos x dx 41. I œ ' xn eax dx;
u œ xn , du œ nxn" dx; dv œ eax dx, v œ "a eax ‘ ÊIœ

xn eax ax a e



' xn" eax dx, a Á !

n a

42. I œ ' aln xbn dx; ’u œ aln xbn , du œ

naln xbn
" x

dx; dv œ " dx, v œ x“

Ê I œ xaln xbn  ' naln xbn" dx 43.

' sin" x dx œ x sin" x  ' sin y dy œ x sin" x
cos y
C œ x sin" x
cos asin" xb
C

44.

' tan" x dx œ x tan" x  ' tan y dy œ x tan" x
ln kcos yk
C œ x tan" x
ln kcos atan" xbk
C

45.

' sec" x dx œ x sec" x  ' sec y dy œ x sec" x  ln ksec y
tan yk
C œ x sec" x  ln ksec asec" xb
tan asec" xbk
C œ x sec" x  ln ¹x
Èx#  1¹
C

46.

' log2 x dx œ x log2 x  ' 2y dy œ x log2 x  ln2 #
C œ x log2 x  lnx#
C y

47. Yes, cos" x is the angle whose cosine is x which implies sin acos" xb œ È1  x# . 48. Yes, tan" x is the angle whose tangent is x which implies sec atan" xb œ È1
x# . 49. (a)

' sinh" x dx œ x sinh" x  ' sinh y dy œ x sinh" x  cosh y
C œ x sinh" x  cosh asinh" xb
C; check: d cx sinh" x  cosh asinh" xb
Cd œ ’sinh" x


x È 1
x#

œ sinh" x dx

(b)

' sinh" x dx œ x sinh" x  ' œ x sinh" x  a1
x# b "

check: d ’x sinh 50. (a)

"Î#

x ŠÈ

" ‹ 1
x#

dx œ x sinh" x 

" #

 sinh asinh" xb

dx

' a1
x# b"Î# 2x dx


C

x  a1
x# b

"Î#


C“ œ ’sinh" x


x È 1
x#



x È 1
x# “

dx œ sinh" x dx

' tanh" x dx œ x tanh" x  ' tanh y dy œ x tanh" x  ln kcosh yk
C œ x tanh" x  ln kcosh atanh" xbk
C; check: d cx tanh" x  ln kcosh atanh" xbk
Cd œ ’tanh" x
"

œ
tanh

(b)

" È 1
x# “

x


x 1  x#



x ‘ 1  x#

dx œ tanh

' tanh" x dx œ x tanh" x  ' 1 x x check: d
x tanh" x


" #

"

x 1  x#



sinh atanh
" xb " cosh atanh
" xb 1  x# “

dx

x dx

dx œ x tanh" x  #" ' 12xx# dx œ x tanh" x
#" ln k1  x# k
C ln k1  x# k
C‘ œ
tanh" x
1 x x#  1 x x# ‘ dx œ tanh" x dx #

8.3 INTEGRATION OF RATIONAL FUNCTIONS BY PARTIAL FRACTIONS 1.

5x  13 (x  3)(x  2)

Ê

œ

A x3




B x2

Ê 5x  13 œ A(x  2)
B(x  3) œ (A
B)x  (2A
3B)

A
Bœ5 Ê B œ (10  13) Ê B œ 3 Ê A œ 2; thus, 2A
3B œ 13 

5x  13 (x  3)(x  2)

œ

2 x3




3 x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.3 Integration of Rational Functions by Partial Fractions 2.

5x  7 x#  3x
2

Ê

3.

œ

A x2

x
4 (x
1)#

œ




A x
1

B x1

Ê 5x  7 œ A(x  1)
B(x  2) œ (A
B)x  (A
2B)

x
4 (x
1)#

2x
2 x#  2x
1

œ

1 x
1

2x
2 (x  1)#

œ


œ

z
1 z# (z  1)

5x  7 x#  3x
2

œ

3 x2




A x1

B (x  1)#

Ê 2x
2 œ A(x  1)
B œ Ax
(A
B) Ê

2x
2 x#  2x
1

œ

2 x1




Aœ2 A
B œ 2 

4 (x  1)#

œ




A z

B z#

" z#  z  6

œ

z z$  z#  6z

œ

" (z  3)(z
2)

œ

A z3




B z
#

z
" z# (z  1)

A
Bœ0 Ê 5B œ 1 Ê B œ  "5 Ê A œ 5" ; thus, 2A  3B œ 1 

t#
8 t#  5t
6

œ1


5t
2 t#  5t
6

(after long division);

Ê B œ 12 Ê A œ 17; thus, t%
9 t%
9t#

9t#
9 t%
9t#

œ1


œ1


t#
8 t#  5t
6

5t
2 t#  5t
6

9t#
9 t# at#
9b #

œ1


œ

2 z




" z#




2 z1

Ê 1 œ A(z
2)
B(z  3) œ (A
B)z
(2A  3B)

œ

17 t3




œ

z z$  z#  6z

5t
2 (t  3)(t  2)

Ê 5t
2 œ A(t  2)
B(t  3) œ (A
B)t
(2A  3B) Ê

8.

Aœ1 Ê A œ 1 and B œ 3; A
B œ 4


z C 1 Ê z
1 œ Az(z  1)
B(z  1)
Cz# Ê z
1 œ (A
C)z#
(A
B)z  B A
Cœ0 Þ

Ê

7.

2 x1

3 (x
1)#

Ê A
B œ 1 ß Ê B œ 1 Ê A œ 2 Ê C œ 2; thus, B œ 1 à 6.




Ê x
4 œ A(x
1)
B œ Ax
(A
B) Ê

B (x
1)#

Ê A œ 2 and B œ 4; thus, 5.




A
Bœ5 Ê B œ 2 Ê A œ 3; thus, A
2B œ 7 

thus,

4.

5x  7 (x  2)(x  1)

œ

œ

A t3




" 5

z3




 "5 z
2

B t2

A
Bœ5 Ê B œ (10
2) œ 12 2A  3B œ 2 

12 t2

(after long division);

9t#
9 t# at#
9b

œ

A t




B t#

Ct
D t#
9 #




Ê 9t#
9 œ At at#
9b
B at
9b
(Ct
D)t# œ (A
C)t$
(B
D)t
9At
9B A
Cœ0 Þ á á % B
D œ 9 Ê Ê A œ 0 Ê C œ 0; B œ 1 Ê D œ 10; thus, t%t

9t9# œ 1
t"#
t#
109 ß 9A œ 0 á á 9B œ 9 à 9.

" 1  x#

' 10.

œ

dx 1  x#

" x#
2x

A 1x

œ

A x

Ê 1 œ A(1
x)
B(1  x); x œ 1 Ê A œ

B 1
x

'

dx 1x







B x
2

Ê 1 œ A(x
2)
Bx; x œ 0 Ê A œ

" #

œ




" #

'

dx 1
x

œ

" #

" #

; x œ 1 Ê B œ

" #

;

cln k1
xk  ln k1  xkd
C

' x dx
2x œ #" ' dxx  #" ' x dx
2 œ #" cln kxk  ln kx
2kd
C

" #

; x œ 2 Ê B œ  #" ;

#

11.

x
4 x#
5x  6

'x 12.

#

x
4
5x  6

2x
1 x#  7x
12

'x

#

œ

A x
6




dx œ

2 7

œ

2x
1  7x
12

A x4

B x1

Ê x
4 œ A(x  1)
B(x
6); x œ 1 Ê B œ

; x œ 6 Ê A œ

5 7

2 7

œ

2 7

;

œ 7 ; x œ 4 Ê A œ

9 1

' x dx
6
75 ' x dx 1 œ 72 ln kx
6k
75 ln kx  1k
C œ 7" ln k(x
6)# (x  1)& k
C




B x3

dx œ 9 '

Ê 2x
1 œ A(x  3)
B(x  4); x œ 3 Ê B œ

dx x4

 7'

dx x3

œ 9 ln kx  4k  7 ln kx  3k
C œ ln

7 1

 4)* ¹ (x (x  3)( ¹


C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 9;

509

510 13.

Chapter 8 Techniques of Integration y y#  2y 3

'4

8

œ 14.

y
4 y#
y

'1Î2

œ

A y

'

" 8

" t$
t# 2t




8

" 16


ln

œ




B t
2

1 6

 3 '1Î2

dy y

œ ln ˆ 27 8 †

27 8




œ  "# ln ktk


" 6

x
3 2x$  8x

œ
43 ln ky  3k


)

" 4

B x
2

" 3

ln kt
2k


A x




Ê Bœ

" 16

;xœ2 Ê Cœ " 16




;'

" 3

" 8

† 16‰ œ ln

3x
2 (x
1)#

1

1

dx x
1

3x  2 (x  1)#

œ  #" '

dt t




" 6

;'

5 16

dx œ  83 '

x
3 2x$  8x

ln kx  2k
C œ

(after long division);

" 16

dx x




" 16

 '0

1

dx (x
1)#




A x
1

3x
2 (x
1)#

œ

A x
1




B (x
1)#

x œ 1 Ê C œ

3x  2 (x  1)#

œ

A x1

D (x  1)# " 4

;xœ1 Ê Dœ




B (x  1)#

œ 2  3 ln 2

Ê 1 œ A(x
1)(x  1)#
B(x  1)(x
1)#
C(x  1)#
D(x
1)# ;

; coefficient of x$ œ A
B Ê A
B œ 0; constant œ A  B
C
D ; thus, A œ

" 4

Ê B œ  4" ; '

' x dx
1  4" ' x dx 1
4" ' (x
dx1)

" 4

' (x dx1)

" 4


1¸ ln ¸ xx  1 

" 4

x# (x  1) ax#
2x
1b

œ

A x1




œ œ

" 4




B x
1

" 4

ln k(x  1)(x
1) k 4

1 (x
1) ax#
1b

œ




A x
1

" 2(x
1)







#

œ

dx ax #  1 b #

x 2 ax#  1b


C

Ê x# œ A(x
1)#
B(x  1)(x
1)
C(x  1); x œ 1

C (x
1)#

; coefficient of x# œ A
B Ê A
B œ 1 Ê B œ

' x dx 1
43 ' x dx
1  2" ' (x
dx1) $

#

Ê 3x  2 œ A(x  1)
B

x$ dx x#  2x
1 !  x " 1 “ "

" #

Ê C œ  2" ; x œ 1 Ê A œ

21.




C (x
1)#

" (#) ‹

Ê 3x
2 œ A(x
1)
B

" " x
1“!

Ê AB
C
Dœ1 Ê ABœ œ 20.

" 4




#

œ

" 4

ln kx  1k


3 4

ln kx
1k


" #(x
1)

3 4

;'

x# dx (x  1) ax#
2x
1b


C


C

Bx
C x#
1

; x œ 2

x$ dx x#
2x
1

#

 Š #"  2
3 ln 2 

3 8

&

#

0

" (1) ‹

B x1

 3 ln 3# ‰

ln ¹ (x  2)x'(x
2) ¹
C

œ ’ x#  2x
3 ln kx
1k


(after long division);

0

œ

" #

' x dx
2
165 ' x dx 2

x ' dx œ 'c1 (x
2) dx
3 'c1 x dx  1
c1 (x  1)# œ ’ #
2x
3 ln kx  1k

" ax #  1 b #

œ 3 ;

(x
3) œ A(x
2)(x  2)
Bx(x  2)
Cx(x
2); x œ 0 Ê A œ

œ Ax
(A
B) Ê A œ 3, A
B œ 2 Ê A œ 3, B œ 1; 'c1

œ Š0
0
3 ln 1 

ln 5‰

' t
dt2
3" ' tdt1

0

0

3 1

" 4

27 4

œ ˆ "#  2
3 ln 2
"# ‰  (1) œ 3 ln 2  2 œ (x
2)


;

ln 9‰  ˆ 43 ln 1


œ c4 ln kyk  3 ln ky
1kd ""Î# œ (4 ln 1  3 ln 2)  ˆ4 ln

dt t$
t# 2t

" #

5 16

ln kx
2k


œ '0 (x  2) dx
3 '0

19.

" 4

3 4

ln kt  1k
C

Ê

C x#

œ (x  2)


dy y
1

œ Ax
(A
B) Ê A œ 3, A
B œ 2 Ê A œ 3, B œ 1; '0

x$ x#  2x
1

;yœ3 Ê Aœ

ln ky
1k‘ % œ ˆ 43 ln 5


1

18.

" 4

Ê 1 œ A(t
2)(t  1)
Bt(t  1)
Ct(t
2); t œ 0 Ê A œ  "# ; t œ 2

C t1

œ

x$ x#
2x
1

dy y
1

1

;tœ1 Ê Cœ

œ  38 ln kxk


'4

8

œ

Ê y
4 œ A(y
1)
By; y œ 0 Ê A œ 4; y œ 1 Ê B œ

B y
1

 ln A t

1 4

Ê y œ A(y
1)
B(y  3); y œ 1 Ê B œ

B y
1

dy œ 4 '1Î2

Ê Bœ

17.




1

y
4 y#
y

œ ln

16.

A y3

y dy dy " 3 y#  2y  3 œ 4 4 y  3
4 " " ln 15 # ln 5
# ln 3 œ #

1

15.

œ

Ê 1 œ A ax#
1b
(Bx
C)(x
1); x œ 1 Ê A œ

œ A
B Ê A
B œ 0 Ê B œ  12 ; constant œ A
C Ê A
C œ 1 Ê C œ

" #

" #

; coefficient of x#

; '0

1

dx (x
1) ax#
1b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.3 Integration of Rational Functions by Partial Fractions

22.

'01 x dx
1
#" '01 (x x

11) dx œ
#" ln kx
1k  4" ln ax#
1b
#" tan" x‘ "!

œ

" 2

œ

ˆ "#

#

" 4

ln 2 

3t#
t
4 t$
t

œ

A t

" #

ln 2





Bt
C t#
1

tan" 1‰  ˆ #" ln 1 

" 4

ln 1


" #

tan" 0‰ œ

" 4

ln 2
#" ˆ 14 ‰ œ

È3

œ 4 '1

È3


'1

dt t

œ Š4 ln È3  " #

œ 2 ln 3  y#
2y
1 ay #
1 b # $

" #

È3

dt œ
4 ln ktk 

" 2

1 1#

œ ln Š È92 ‹


Cy
D a y #
1 b#




3t#
t
4 t$
1

dt

1 3




È$

ln at#
1b
tan" t‘ "

ln 4
tan" È3‹  ˆ4 ln 1 

Ay
B y#
1

œ

(t
1) t#
1

ln 2


(1
2 ln 2) 8

Ê 3t#
t
4 œ A at#
1b
(Bt
C)t; t œ 0 Ê A œ 4; coefficient of t#

œ A
B Ê A
B œ 3 Ê B œ 1; coefficient of t œ C Ê C œ 1; '1

23.

511

" #

ln 2
tan" 1‰ œ 2 ln 3  ln 2


" #

ln 2 

1 4

1 1#

Ê y#
2y
1 œ (Ay
B) ay#
1b
Cy
D

œ Ay
By#
(A
C)y
(B
D) Ê A œ 0, B œ 1; A
C œ 2 Ê C œ 2; B
D œ 1 Ê D œ 0;

'y 24.

dy œ '

#


2y
1 ay #
1 b #

8x#
8x
2 a4x#
1b# $

œ

Ax
B 4x#
1

" y#
1

dy
2 '

Cx
D a4x#
1b#




y a y #
1 b#

" y#
1

dy œ tan" y 


C

Ê 8x#
8x
2 œ (Ax
B) a4x#
1b
Cx
D

œ 4Ax
4Bx#
(A
C)x
(B
D); A œ 0, B œ 2; A
C œ 8 Ê C œ 8; B
D œ 2 Ê D œ 0;

' 8x

dx œ 2 '

2s
2 as#
1b (s  1)$

œ

#
8x
2 a4x#
1b#

25.

As
B s#
1 $

dx 4x#
1




C s1 #


8'


x dx a4x#
1b#

D (s  1)#




œ tan" 2x 

E (s  1)$ #

" 4x#
1


C

Ê 2s
2 #

œ (As
B)(s  1)
C as
1b (s  1)
D as
1b (s  1)
E as#
1b œ cAs%
(3A
B)s$
(3A  3B)s#
(A
3B)s  Bd
C as%  2s$
2s#  2s
1b
D as$  s#
s  1b
E as#
1b œ (A
C)s%
(3A
B  2C
D)s$
(3A  3B
2C  D
E)s#
(A
3B  2C
D)s
(B
C  D
E)
C œ0 Þ A á á á 3A
B  2C
D œ0á Ê 3A  3B
2C  D
E œ 0 ß summing all equations Ê 2E œ 4 Ê E œ 2; á A
3B  2C
D œ2á á á B
CD
Eœ2 à summing eqs (2) and (3) Ê 2B
2 œ 0 Ê B œ 1; summing eqs (3) and (4) Ê 2A
2 œ 2 Ê A œ 0; C œ 0 from eq (1); then 1
0  D
2 œ 2 from eq (5) Ê D œ 1;

' as

26.

#

2s
2
1b (s  1)$

s%
81 œ As s as #
9 b # %

ds œ '




Bs
C s#
9

ds s#
1




'

Ds
E a s #
9 b# %

ds (s  1)#


2'

ds (s  1)$

œ (s  1)#
(s  1)"
tan" s
C #

Ê s%
81 œ A as#
9b
(Bs
C)s as#
9b
(Ds
E)s

œ A as
18s#
81b
aBs
Cs$
9Bs#
9Csb
Ds#
Es œ (A
B)s%
Cs$
(18A
9B
D)s#
(9C
E)s
81A Ê 81A œ 81 or A œ 1; A
B œ 1 Ê B œ 0;

C œ 0; 9C
E œ 0 Ê E œ 0; 18A
9B
D œ 0 Ê D œ 18; ' œ ln ksk
27.

9 as #
9 b

2) $
5) #
8)
4 a) #
2 )
2 b# $

œ

s%
81 s as #
9 b #

ds œ '

ds s

 18 '

s ds a s #
9 b#


C A)
B ) #
#)
2




C)
D a ) #
2 )
2 b#

Ê 2)$
5)#
8)
4 œ (A)
B) a)#
2)
2b
C)
D

œ A)
(2A
B))#
(2A
2B
C))
(2B
D) Ê A œ 2; 2A
B œ 5 Ê B œ 1; 2A
2B
C œ 8 Ê C œ 2;

2B
D œ 4 Ê D œ 2; ' œ'

2)
2 ) #
2)
2

d)  '

2) $
5) #
8)
4 a) #
2 )
2 b#

d) ) #
2)
#


'

d) œ '

d a) #
2 )
2 b a) #
2 )
2 b #

2)
1 a) #
2 )
2 b

œ'

d)
'

d a) #
2 )
2 b ) #
2)
#

2)
2 a ) #
2 )
2 b#

'

d) ()
1)#
1

d) 

" ) #
2)
#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

512

Chapter 8 Techniques of Integration " ) #
2)
#

œ 28.


ln a)#
2)
2b  tan" ()
1)
C

) %  4) $
2) #  3)
1 a) #
1 b $

œ

A)
B )#
1




C)
D a ) #
1 b#

E)
F a ) #
1 b$




Ê ) %  4) $
2 ) #  3 )
1

#

œ (A)
B) a)#
1b
(C)
D) a)#
1b
E)
F œ (A)
B) a)%
2)#
1b
aC)$
D)#
C)
Db
E)
F œ aA)&
B)%
2A)$
2B)#
A)
Bb
aC)$
D)#
C)
Db
E)
F œ A)&
B)%
(2A
C))$
(2B
D))#
(A
C
E))
(B
D
F) Ê A œ 0; B œ 1; 2A
C œ 4 Ê C œ 4; 2B
D œ 2 Ê D œ 0; A
C
E œ 3 Ê E œ 1; B
D
F œ 1 Ê F œ 0;

') 29.

%

 4) $
2) #  3)
1 a) #
1 b $

2x$  2x#
1 x#  x

œ 2x


" x#  x

x œ 1 Ê B œ 1; ' 30.

x% x#  1

œ ax#
1b


d) œ '

œ 2x


$

#

2x  2x
1 x#  x " x#  1

 4'

d) )#
1

" x(x  1)

" 3

31.

x$
x 

9x$  3x
1 x$  x# #

" #

œ9


9x#  3x
" x# (x  1)

" #

" x(x  1)


'

œ

dx x

" #

;'

ln kx  1k


A x




Ê 1 œ A(x  1)
Bx; x œ 0 Ê A œ 1;

B x1

dx x1

x ax# x#  1 dx œ $ C œ x3
x
#" ln

(after long division);

9x#  3x
1 x# (x  1) #


C

œ x#  ln kxk
ln kx  1k
C œ x#
ln ¸ x x 1 ¸
C

'

%

 "4 a)#
1b

#

œ tan" )
2 a)#
1b

" (x
1)(x  1)

;

"

) d) a ) #
1 b$


'

" (x
1)(x  1)

œ ax#
1b


ln kx
1k


;

œ ' 2x dx  '

x œ 1 Ê A œ  "# ; x œ 1 Ê B œ œ

) d) a ) #
1 b#

œ

œ

A x
1




B x1

' x dx
1
#" ' x dx 1

" #


1b dx 

Ê 1 œ A(x  1)
B(x
1);

 1¸ ¸ xx
1
C




A x

B x#




C x1

Ê 9x  3x
1 œ Ax(x  1)
B(x  1)
Cx ; x œ 1 Ê C œ 7; x œ 0 Ê B œ 1; A
C œ 9 Ê A œ 2;

' 9x x3xx
1 dx œ ' 9 dx
2 ' dxx  ' dxx
7 ' xdx1 œ 9x
2 ln kxk
x"
7 ln kx  1k
C $

$

32.

#

16x$ 4x#  4x
1

#

œ (4x
4)


12x  4 4x#  4x
1

12x  4 (2x  1)#

;

œ

A 2x  1

Ê A œ 6; A
B œ 4 Ê B œ 2; ' œ 2(x
1)#
3 ln k2x  1k 

33.

y%
y#  1 y$
y

œy

" y ay #
1 b

" 2x  1

" y ay #
1 b

;

œ




16x$ 4x#  4x
1 C" œ 2x#


A y




By
C y#
1

Ê A œ 1; A
B œ 0 Ê B œ 1; C œ 0; ' y# #

œ 34.

 ln kyk


2y% y$  y#
y  1

" #




B (2x  1)#

Ê 12x  4 œ A(2x  1)
B

dx œ 4 ' (x
1) dx
6 '

dx 2x  1 "

4x
3 ln k2x  1k  (2x  1)


2'

dx (2x  1)#


C, where C œ 2
C"

Ê 1 œ A ay#
1b
(By
C)y œ (A
B)y#
Cy
A y%
y#  1 y$
y

dy œ ' y dy  '

dy y


'

y dy y#
1

ln a1
y# b
C

œ 2y
2


2 y$  y#
y  1

;

2 y$  y#
y  1 #

œ

2 ay#
1b (y  1) #

œ

A y1




By
C y#
1

Ê 2 œ A ay#
1b
(By
C)(y  1) œ aAy
Ab
aBy
Cy  By  Cb œ (A
B)y#
(B
C)y
(A  C) Ê A
B œ 0, B
C œ 0 or C œ B, A  C œ A  B œ 2 Ê A œ 1, B œ 1, C œ 1;

'y

$

2y%  y#
y  1 #

dy œ 2 ' (y
1) dy
'

œ (y
1)
ln ky  1k  where C œ C"
1 35.

'e

36.

'e œ

2t

4t

et dt
3et
2


2e2t  et e2t
1

y2 2




" #

œ cet œ yd' dt œ '

" #

dy y1

'

y y#
1

dy  '

dy y#
1

ln ay#
1b  tan" y
C" œ y#
2y
ln ky  1k 

dy y#
3y
2

e3t
2et  1 t e2t
1 e dt;

œ'

dy y
1

'

dy y
2

y œ et ” dy œ et dt • Ä

ln ay#
1b  tan" y
C œ

" 2

e2t


" #

" #

ln ay#
1b  tan" y
C,


1 e
1 œ ln ¹ yy
2 ¹
C œ ln Š et
# ‹
C

'y

t

3


2y  1 y#
1

dy œ ' Šy


y1 y#
1 ‹

dy œ

y2 2

ln ae2t
1b  tan" aet b
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley


'

y y#
1

dy  '

dy y#
1

Section 8.3 Integration of Rational Functions by Partial Fractions 37.

' sin ycos
ysindyy  6 ; [sin y œ t, cos y dy œ dt] #

œ

38.

39.

't

#

œ

" 3

dy
t6

œ

" 5

' ˆ t " 2  t
" 3 ‰ dt œ "5 ln ¸ tt
 23 ¸
C

y2 ln ¹ sin sin y
3 ¹
C

" 5

' cos )sin
)cosd))  2 ; ccos ) œ yd

Ä '

#

œ

Ä

" 3

dy y#
y  2

' y dy
2  3" '

dy y1

œ

" 3


2 ln ¹ yy  1¹
C œ

" 3

cos )
2 ¸ ln ¸ cos )1
C


cos ) ¸ " ¸ cos )  1 ¸ ln ¸ 21  cos )
C œ  3 ln cos )
2
C


" (2x)  12x$  3x ' (x  2)a4xtan#
"
1(2x) ' (x x 2)# dx dx œ ' tan b (x  2)# 4x#
1 dx  3 ' (x dx2) œ atan
"4 2xb#  3 ln kx  2k
x 6 2
C œ "# ' tan" (2x) d atan" (2x)b  3' x dx 2 6 #

#

40.


" (3x)
9x$
x
" (3x) ' (x
1)a9xtan ' (x
x 1)# dx dx œ ' tan #
1b (x
1)# 9x#
1 dx
' (x
dx1) œ atan
"6 3xb#
ln kx
1k
x
1 1
C œ "3 ' tan" (3x) d atan" (3x)b
' x dx
1  #

#

41. at#  3t
2b " #

Ê

dx dt

œC Ê

42. a3t%
4t#
1b

œ 1; x œ ' t2 t1

dx dt

t2  2¸ x œ ' t dt 2  ' t dt 1 œ ln ¸ tt  1
C; t  1 œ Ce ; t œ 3 and x œ 0 2 ‰¸ œ "# ex Ê x œ ln ¸2 ˆ tt  œ ln kt  2k  ln kt  1k
ln 2 1 dt t#  3t
2

œ 2È3; x œ 2È3'

dt 3t%
4t#
1

œ È 3'

œ 3 tan" ŠÈ3t‹  È3 tan" t
C; t œ 1 and x œ

1 È 3 4

dt t#


" 3

 È 3'

Ê 

È 31 4

dt t#
1

œ1

È3 4

1
C Ê C œ 1

Ê x œ 3 tan" ŠÈ3t‹  È3 tan" t  1 43. at#
2tb

dx dt

œ 2x
2;

" #

' x dx
1 œ ' t " 3

t œ 1 and x œ 1 Ê ln 2 œ ln Ê xœ 44. (t
1)

6t t
#

tan" 14

Ê Cœ

Ê

dt
2t

' x dx
1 œ ' t
dt 1 #

"

œ 1 Ê tan

45. V œ 1 '0 5 y# dx œ 1 '0 5 2Þ5

2Þ5

Þ

Þ

1

9 3x  x#

" A

dt t
2


"# ln ax#
1b‘ 0 œ x tan" x dx

È3

È3

1È3 3

 ln 2;

È

" #

È3

'0

œ

" A

3 " # " Œ
# x tan x‘ 0 

œ

" A

’ 1# 
"# ax  tan" xb‘ 0 “

œ

" A

Š 1# 

1 4

Ê tan"

Þ

'1

È3

'0

 #" '

2Þ5

2x ˆ " (x
1)(2  x) dx œ 41 0  3 " ln k2  xkb‘ ! œ 431 (ln 2)

È3

xœ

dt t

#Þ& dx œ 31 Œ'0 5 ˆ x " 3
x" ‰ dx œ
31 ln ¸ x x 3 ¸‘ !Þ& œ 31 ln 25

47. A œ '0 tan" x dx œ cx tan" xd 0  '0

È3

'

x œ ln kt
1k
1 Ê x œ tan (ln (t
1)
1), t  1

1

œ
 431 aln kx
1k
2

œ

" #

Ê tan" x œ ln kt
1k
C; t œ 0 and x œ

46. V œ 21 '0 xy dx œ 21'0

1È3 3

ln kx
1k œ

 1, t  0

œ x#
1 Ê

dx dt

" #

Ê ln kx
1k œ ln ¸ t
t # ¸
C;
C Ê C œ ln 2
ln 3 œ ln 6 Ê ln kx
1k œ ln 6 ¸ t
t 2 ¸ Ê x
1 œ t
6t # #

x 1
x#

x# 1
x#

dx

È3 # ‹

µ 1.10 œ

ˆ x
" 1 ‰
23 ˆ 2 " x ‰‰ dx

dx

È3

È3 #


16 ‹ œ

" A

Š 231 

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 4

œ ln k1k
C

513

514

Chapter 8 Techniques of Integration

48. A œ '3

5

xœ

49. (a)

4x#
13x  9 x$
2x#  3x

dx œ 3'3

5

" A

'3

dx dt

œ kx(N  x) Ê '

5

kœ

#

x a4x
13x  9b x$
2x#  3x

" 250 ,

dx x " A

dx œ

 '3

5

dx x
3

50.

dx dt

" #

Šc4xd $
3 '3 &

N œ 1000, t œ 0 and x œ 2 Ê

œ k(a  x)(b  x) Ê

Ê

' (a dxx)

" ax

(b) a Á b:

œ

#

œ ' k dt Ê

akt
" a

dx ' (a  x)(b ' k dt  x) œ

t œ 0 and x œ 0 Ê Ê xœ 51. (a) (b)

ab
1  eÐb
aÑkt ‘ a  beÐb
aÑkt

" ba

ln

b a

œ c3 ln kxk  ln kx
3k
2 ln kx  1kd &$ œ ln


2 '3

5

dx x  1‹

œ

" A

;

µ 3.90 (8
11 ln 2  3 ln 6) œ

' dxx
N" ' Ndx x œ ' k dt 2 ¸ ln ¸ 998 œC Ê

125 9

" 1000

Ê

" N

x ¸ ln ¸ N  x œ kt
C;

ln ¸ 1000x  x ¸ œ

t 250




" 1000

" ‰ ln ˆ 499

499x œ e4t (1000  x) Ê a499
e4t b x œ 1000e4t Ê x œ † 499
500e4t œ 1000e4t Ê e4t œ 499 Ê t œ

" 4

1000e4t 499
e4t

ln 499 ¸ 1.55 days

œ k dt " a x

Ê axœ

" N

" 1000

499x 4t Ê 1000  x œ e 4t 1000e œ 499
e4t Ê 500

dx (a  x)(bx)

dx x1

dx x
3

œ ' k dt Ê

dx x(N  x)

N œ 500 Ê 500

(a) a œ b:

5

5

499x ¸ Ê ln ¸ 1000  x œ 4t Ê

(b) x œ


2'3

œ kt
C; t œ 0 and x œ 0 Ê

a akt
1

Ê

" b a

Ê xœa

a akt
1

œ

" a

œC Ê

" ax

œ kt


" a

a# kt akt
1

' a dx x  b " a ' b dx x œ ' k dt

Ê

x¸ ˆb‰ Ê œ C Ê ln ¸ ba  x œ (b  a)kt
ln a

" ¸bx¸ ba ln a  x œ kt bx b ÐbaÑkt ax œ a e


C;

'01 x%x(x#
 11)% dx œ '01 ˆx'  4x&
5x%  4x#
4  x# 4
1 ‰ dx œ 227  1 22 7

1 1

† 100% µ œ 0.04%

(c) The area is less than 0.003

52. P(x) œ ax#
bx
c, P(0) œ c œ 1 and Pw (0) œ 0 Ê b œ 0 Ê P(x) œ ax#
1. Next, ax#
1 x$ (x  1)#

œ

A x




B x# #




C x$




D x1




E (x  1)# #

; for the integral to be a rational function, we must have A œ 0 and

D œ 0. Thus, ax
1 œ Bx(x  1)
C(x  1)#
Ex$ œ (B
E)x$
(C  2B)x#
(B  2C)x
C B
E œ 0Þ Ê C  2B œ a ß Ê E œ B; x œ 1 Ê a
1 œ E; therefore, 1  2B œ a Ê 1
2E œ a Ê 1
2(a
1) œ a C œ "à Ê a œ 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.4 Trigonometric Integrals

515

8.4 TRIGONOMETRIC INTEGRALS 1.

'01/2 sin5 x dx œ '01/2 asin# xb# sin x dx œ '01/2 a"  cos# xb# sin x dx œ '01/2 a"  2cos# x
cos4 xbsin x dx œ '0 sin x dx  '0 2cos# x sin x dx
'0 cos4 x sin x dx œ ’cos x
# cos3 x  1/2

1/2

# $

œ a!b  ˆ"
2.

1/2

 "& ‰ œ

3

) "&

'01 sin5 ˆ x2 ‰dx (using Exercise 1) œ '01 sinˆ x2 ‰dx  '01 2cos# ˆ x2 ‰sinˆ x2 ‰dx
'01 cos4 ˆ x2 ‰sinˆ x2 ‰dx 1

œ
#cos ˆ x2 ‰
%3 cos3 ˆ 2x ‰  5# cos5 ˆ 2x ‰‘ 0 œ a!b  ˆ#
3.

% $

 &# ‰ œ

"' "&

'11/2/2 cos3 x dx œ '11/2/2 acos# xbcos x dx œ '11/2/2 a"  sin# xbcos x dx œ '11/2/2 cos x dx  '11/2/2 sin# x cos x dx œ ’sin x 

4.

1/2 cos5 x 5 “0

1Î# sin3 x 3 “ 1Î#

œ ˆ"  "$ ‰  ˆ"
"$ ‰ œ

% $

'01/6 3cos5 3x dx œ '01/6 acos# 3xb# cos 3x † 3dx œ '01/6 a"  sin# 3xb# cos 3x † 3dx œ '01/6 a"  #sin# 3x
sin% 3xbcos 3x † 3dx œ '0 cos 3x † 3dx  #'0 sin# 3x cos 3x † 3dx
'0 sin% 3x cos 3x † 3dx œ ’sin 3x  # sin33x
1/6

œ ˆ"  5.

1/6

2 $


"& ‰  a!b œ

1/6

3

) "&

'01/2 sin7 y dy œ '01/2 sin6 y sin y dy œ '01/2 a"  cos2 yb$ sin y dy œ '01/2 sin y dy  $'01/2 cos2 y sin y dy
$'0 cos4 y sin y dy  '0 cos6 y sin y dy œ ’cos y
$ cos3 y  $ cos5 y
1/2

6.

1/2

&

3

8.

9.

1Î# sin( t ( “0

œ 7ˆ"  "


$ &

 "( ‰  7a!b œ

1Î# cos( y ( “0

œ a!b  ˆ"
" 

$ &


"( ‰ œ

"' $&

"' &

1 1 1 1 4x #x ‰ # '01 )sin4 x dx œ )'01 ˆ "  cos dx œ #'0 a"  #cos #x
cos# #xbdx œ #'0 dx  #'0 cos #x † #dx
#'0 "
cos dx # # 1 1 1 œ c#x  #sin #xd 10
'0 dx
'0 cos 4x dx œ #1

x
"# sin 4x‘ 0 œ #1
1 œ $1

'01 )cos4 21x dx œ )'01 ˆ "
cos# 41x ‰# dx œ #'01 a"
#cos 41x
cos# 41xbdx œ #'01 dx
%'01 cos 41x dx
#'01 "
cos# )1x dx 1 1 " " œ
#x
1" sin 41x‘ 0
'0 dx
'0 cos )1x dx œ #

x
)"1 sin )1x‘ 0 œ #
" œ $ 1/4 1/4 1/4 #x ‰ˆ "
cos #x ‰ 4x ‰ '11/4/4 16 sin# x cos# x dx œ 16'11/4/4 ˆ "  cos dx œ %'1/4 a"  cos# #xbdx œ %'1/4 dx  %'1/4 ˆ "
cos dx # # # 1/4 ' '
œ c4xd  1/4  # 1/4 dx  # 1/4 cos 4x dx œ 1
1  2x
1/4

10.

&

3

'01/2 7cos7 t dt (using Exercise 5) œ 7”'01/2 cos t dt  $'01/2 sin2 t cos t dt
$'01/2 sin4 t cos t dt  '01/2 sin6 t cos t dt• œ 7’sin t  $ sin3 t
$ sin5 t 

7.

1Î6 sin5 3x 5 “0

1/4

sin 4x ‘ 1/4 =21 2 1/4

 ˆ 1#  ˆ 1# ‰‰ œ 1

1 1 1 1 #y ‰# ˆ "
cos #y ‰ '01 8 sin4 y cos# y dy œ 8'01 ˆ "  cos dy œ '0 dy  '0 cos #y dy  '0 cos# #y dy
'0 cos$ #y dy # # 1 1 1 1 1 1 4y ‰ œ
y  2" sin 2y‘ 0  '0 ˆ "
cos dy
'0 a"  sin# #ybcos #y dy œ 1  "# '0 dy  "# '0 cos 4y dy
'0 cos #y dy # 1 1  '0 sin# #y cos #y dy œ 1
’ "# y  ") sin 4y
"# sin 2y  "# † sin32y “ œ 1  1# œ 1# 3

0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

516 11.

Chapter 8 Techniques of Integration

'01/2 35 sin4 x cos3 x dx œ '01/2 35 sin4 x a"  sin# xbcos x dx œ 35'01/2 sin4 x cos x dx  35'01/2 sin6 x cos x dx 5

7

œ ’35 sin5 x  35 sin7 x “ 12.

1/2 0

œ a7  5b  a!b œ #

'01 cos2 2x sin 2x dx œ ’ "# cos3 2x “ 1 œ  "'
"' œ ! 3

0

13.

'01/4 8cos3 2) sin 2) d) œ ’8ˆ "# ‰ cos4 2) “ 1/4 œ ccos4 2)d 10 /4 œ a!b  a"b œ " 4

0

14.

'01/2 sin2 2) cos3 2) d) œ '01/2 sin2 2)a"  sin2 2)bcos 2) d) œ '01/2 sin2 2) cos 2) d)  '01/2 sin4 2) cos 2) d) œ ’ "# †

sin3 2) 3



" #

†

1/2 sin5 2) 5 “0

œ!

15.

'021 É "  #cos x dx œ '021 ¹ sin x# ¹dx œ '021 sin x# dx œ
#cos x# ‘ 20 1 œ #
# œ %

16.

'01 È"  cos 2x dx œ '01 È# lsin 2x ldx œ '01 È# sin 2x dx œ ’È#cos 2x“ 1 œ È#
È# œ #È# 0

17.

'01 È"  sin# t dt œ '01 l cos t ldt œ '01/2 cos t dt  '11/2 cos t dt œ csin td 10 /2  csin td 11/2 œ "  !  !
" œ #

18.

'01 È"  cos# ) d) œ '01 l sin ) ld) œ '01 sin ) d) œ ccos )d 10 œ "
" œ #

19.

'11/4/4 È"
tan# x dx œ '11/4/4 l sec x ldx œ '11/4/4 sec x dx œ clnl sec x
tan x ld 1/41/4 œ lnŠÈ#
"‹  lnŠÈ#  "‹ È

" œ lnŠ È##
‹ œ # lnŠ"
È#‹ "

20.

'11/4/4 Èsec# x  " dx œ '11/4/4 l tan x ldx œ '!1/4 tan x dx
'!1/4 tan x dx œ clnl sec x ld !c1

/4


clnl sec x ld 1! /4

œ lna"b
lnÈ#
lnÈ#  lna"b œ # lnÈ# œ ln # 21.

'01/2 )È"  cos 2) d) œ '01/2 )È# l sin ) l d) œ È#'01/2 ) sin ) d) œ È# c)cos )
sin )d 10 /2 œ È#a"b œ È#

22.

'11 a"  cos# tb$Î# dt œ '11 asin# tb$Î# dt œ '11 ¸ sin$ t¸ dt œ '!1 sin$ t dt
'!1 sin$ t dt œ '!1 a"  cos# tbsin t dt
'! a"  cos# tbsin t dt œ '1 sin t dt
'1 cos# t sin t dt
'! sin t dt  '! cos# t sin t dt œ ’cos t  1


’cos t


23.

!

1 cos3 t 3 “!

œ ˆ" 

" $

!


"  "$ ‰
ˆ" 

1

" $


"  "$ ‰ œ

1

! cos3 t 3 “ 1

) $

'!1/3 2 sec$ x dx; u œ sec x, du œ sec x tan x dx, dv œ sec# x dx, v œ tan x;

'!1/3 2 sec$ x dx œ c2 sec x tan xd !1Î$  #'!1/3 sec x tan2 x dx œ # † " † !  # † # † È$  #'!1/3 sec x asec2 x  "bdx œ %È$  #'1/3 sec$ x dx
#'1/3 sec x dx; 2'1/3 2 sec$ x dx œ %È$
c#ln l sec x + tan xld !1Î$ !

!

!

2'1/3 2 sec$ x dx œ %È$
#ln l " + !l  #ln l #  È$ l œ %È$  # ln Š#  È$‹ !

'!1/3 2 sec$ x dx œ #È$  ln Š#

 È $‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.4 Trigonometric Integrals 24.

25.

' ex sec$ aex bdx; u œ secaex b, du œ secaex btanaex bex dx, dv œ sec# aex bex dx, v œ tanaex b. ' ex sec$ aex b dx œ secaex btanaex b  ' secaex btan# aex bex dx œ secaex btanaex b  ' secaex basec# aex b  "bex dx œ secaex btanaex b  ' sec$ aex bex dx
' secaex bex dx #' ex sec$ aex b dx œ secaex btanaex b
ln¸secaex b
tanaex b¸
C ' ex sec$ aex b dx œ "# ˆsecaex btanaex b
ln¸secaex b
tanaex b¸‰
C '!1/4 sec4 ) d) œ '!1/4 œ ˆ"
"$ ‰  a!b œ

26.

1/4

sec2 ) d)
'!

1/4

tan2 ) sec2 ) d) œ ’tan )


% $

1Î12 tan3 a3xb “ 3 !

œ ˆ"
"$ ‰  a!b œ

sec2 a3xb3dx
'!

1/12

1 Î4 tan3 ) 3 “!

tan2 a3xb sec2 a3xb3dx

% $

'11/4/2 csc4 ) d) œ '11/4/2 a"
cot# )bcsc# ) d) œ '11/4/2 csc# ) d)
'11/4/2 cot# ) csc# ) d) œ ’cot )  cot3 ) “ 1Î2 $

œ a!b  ˆ"  "$ ‰ œ 28.

a"
tan2 )bsec2 ) d) œ '!

'!1/12 3sec4 a3xb dx œ '!1/12 a"
tan2 a3xbbsec2 a3xb3dx œ '!1/ œ ’tan a3xb


27.

517

1/4

% $

'11/2 $csc4 #) d) œ $'11/2 ˆ"
cot# #) ‰csc# #) d) œ $'11/2 csc# #) d)
$'11/2 cot# #) csc# #) d) œ ’'cot #)  ' cot3

$) #

“

1 1 Î2

œ a' † !  # † !b  a' † "  # † "b œ ) 29.

'01/4 4 tan3 x dx œ 4'01/4 asec# x  "btan x dx œ 4'01/4 sec# x tan x dx  4'01/4 tan x dx œ ’% tan# x  % ln lsec xl“ 1Î4 #

!

œ 2a"b  %lnÈ#  # † !
%ln " œ #  #ln # 30.

'11/4/4 6 tan4 x dx œ 6'11/4/4

asec# x  "btan2 x dx œ 6'1/4 sec# x tan2 x dx  6'1/4 tan2 x dx 1/4

1/4

œ 6'1/4 sec2 x tan2 x dx  6'1/4 asec2 x  1bdx œ ’' tan$ x “ 1/4

1/4

$

1 Î4

1 Î4

1Î4

œ #a"  a"bb  c'tan xd 1Î4
c'xd 1Î4 œ %  'a"  a"bb
31.

$1 #




$1 #

1/4

œ $1  )

'11/6/3 cot3 x dx œ '11/6/3 acsc2 x  " bcot x dx œ '11/6/3 csc2 x cot x dx  '11/6/3 cot x dx œ ’ cot# x
ln l csc xl“ 1Î3 #

1Î6

œ

32.

 6'1/4 sec# x dx
6'1/4 dx 1/4

 1 Î4

 "# ˆ "$

 $‰


Šln È#$

 ln #‹ œ

% $

 lnÈ$

'11/4/2 8 cot4 t dt œ 8'11/4/2 acsc2 t  " bcot2 t dt œ 8'11/4/2 csc2 t cot2 t dt  8'11/4/2 cot2 t dt 3

œ 8’ cot3 t “

1 Î2 1 Î4

 8'1/4 acsc2 t  " bdt œ  3) a!  "b
c)cot td 1Î4
c)td 1Î4 œ 1/2

1 Î2

1Î2

) $


)a!  "b
%1  #1 œ #1 

33.

'!1 sin 3x cos 2x dx œ "# '!1 asin x
sin 5xb dx œ "#
cos x  "& cos 5x‘ !1 œ "# ˆ  "  "&  "  "& ‰ œ  '&

34.

'!1Î2 sin 2x cos 3x dx œ "# '!1Î2 asinaxb
sin 5xb dx œ "#
cosaxb  "& cos 5x‘ 1! Î2 œ "# a!b  "# ˆ"  "& ‰ œ  #&

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

"' $

518

Chapter 8 Techniques of Integration

35.

'11 sin 3x sin 3x dx œ "# '11 acos !  cos 6xb dx œ "# '11 dx  "# '11 cos 6x dx œ "#
x  "#" sin 6x‘ 11 œ 1#
1#  ! œ 1

36.

'!1Î2 sin x cos x dx œ "# '!1Î2 asin !
sin 2xb dx œ "# '!1Î2 sin 2x dx œ  "% ccos 2xd 1! Î# œ  "% a"  "b œ "#

37.

'!1 cos 3x cos 4x dx œ "# '!1 acosaxb
cos 7xb dx œ "2
sinaxb
(" sin 7x‘ 1! œ 2" a0b œ 0

38.

'11ÎÎ22 cos 7x cos x dx œ "# '11ÎÎ22 acos 6x
cos 8xb dx œ 2"
6" sin 6x
8" sin 8x‘ 1Î12Î2 œ 0 t# #

39. x œ t#Î$ Ê t# œ x$ ; y œ A œ '!

2

œ

#Î$

#1 ˆ #( ’ "

Êyœ

x$ # ;!

$

#1 Š x# ‹É"
*% x% dx; ” #Î$ ‰‰$Î#


* ˆ2

40. y œ lnacos xb; y w œ

Ÿ t Ÿ # Ê ! Ÿ x Ÿ 2#Î$ ;

u œ 9% x% •Ä du œ 9x$ dx

1 9

'!*Ð2

#Î$

Ñ

*Ð2#Î$ Ñ

È"
u du œ ’ 1 † # a"
ub$Î# “ 9 $

!

 "“

sin x cos x

œ tan x; ay w b# œ tan# x; '!

1Î3

È"
tan# x dx œ '

1Î3

È"
tan# x dx œ '

1Î4

!

lsec xl dx œ clnlsec x
tan xld !1/3

œ lnŠ2
È$‹  lna"
!b œ lnŠ2
È$‹ 41. y œ lnasec xb; y w œ

sec x tan x sec x

œ tan x;ay w b# œ tan# x; '!

1Î4

!

lsec xl dx œ clnlsec x
tan xld !1/4

œ lnŠÈ#
"‹  lna!
"b œ lnŠÈ#
"‹ " 42. M œ '1Î4 sec x dx œ clnlsec x
tan xld 1/41Î4 œ lnŠÈ#
"‹  ln lÈ#  "l œ ln È##
" 1 Î4

yœ

È

È ' ln È#
" 1Î4 #" 1Î4

"

sec# x #

Ê ax, yb œ Œ!ß Šln

dx œ

È#
" " È#" ‹

43. V œ 1'! sin# x dx œ 1'! 1

1

"

È

#ln È#
"

#"

1Î4

ctan xd 1Î4 œ

"

È

#ln È#
"

#"

œ

È# #

"  cos 2x 2

dx œ

1 #

1Î%



#

1Î%

$1Î%

È# #

csin 2xd 1Î%


È# #

csin 2xd 1$1Î% œ

45. (a) m# Á n# Ê m
n Á ! and m  n Á ! Ê 'k

k
#1

œ œ œ

#"

'!1 dx  1# '!1 cos 2x dx œ 1# cxd 1!  14 csin 2xd 1! œ 1# a1  !b  14 a!  !b œ 1#

1

csin 2xd !

È

ln È#
"



44. A œ '! È"
cos 4x dx œ '! È# lcos 2xldx œ È# '! 1

"

a"  a"bb œ

cos 2x dx  È# '1Î% cos 2x dx
È# '$1Î% cos 2x dx

È# #

$1Î%

a"  !b 

È# #

sin mx sin nx dx œ

" #

a"  "b


1

È# #

a !
"b œ È #
È # œ # È #

'kk
#1 ccosam  nbx  cosam
nbxddx

"
" " ‘ k
#1 # m  n sinam  nbx  m
n sinam
nbx k "ˆ " " " ‰ "ˆ " # m  n sinaam  nba k
#1bb  m
n sinaam
nba k
#1bb  # m  n sinaam  nbkb  m
n sinaam
" " " " #amnb sinaam  nbkb  #am
nb sinaam
nbkb  #am  nb sinaam  nbkb
#am
nb sinaam
nbkb œ !

nbkb‰

Ê sin mx and sin nx are orthogonal. (b) Same as part since

'k

k
#1

œ

" #

œ

" #am  nb sinaam " #am  nb sinaam

œ

" #

'kk
#1 cos ! dx œ 1.

m# Á n# Ê m
n Á ! and m  n Á ! Ê 'k



" #


m " n sinam  nbx


cos mx cos nx dx

k
#1

" ‘ m
n sinam
nbx k " " nba k
#1bb
#am
nb sinaam
nba k
#1bb  #am  nb sinaam  nbkb  #am"
nb sinaam
nbkb
#am"
nb sinaam
nbkb  #am" nb sinaam  nbkb  #am"
nb sinaam
nbkb œ !

ccosam  nbx
cosam
nbxddx œ 

k
#1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

nbkb

Section 8.5 Trigonometric Substitutions

519

Ê cos mx and cos nx are orthogonal. (c) Let m œ n Ê sin mx cos nx œ "# asin !
sinaam
nbxbb and

" #

'kk
#1 sin ! dx œ ! and "# 'kk
#1 sinaam
nbxb dx œ 0

Ê sin mx and cos nx are orthogonal if m œ n. Let m Á n.

'kk
#1 sin mx cos nx dx œ "# 'kk
#1 csinam  nbx
sinam
nbxddx œ "#
 m " n cosam  nbx  m
" n cosam
nbx‘ k
#1 k

œ  #am" nb cosaam  nba k
#1bb  œ  #am" nb cosaam  nbkb 

" #am
nb cosaam


" #am
nb cosaam


nbkb


" " #am  nb cosaam  nbkb
#am
nb cosaam
" " #am  nb cosaam  nbkb
#am
nb cosaam
nbkb œ !

nba k
#1bb


nbkb

Ê sin mx and cos nx are orthogonal. 46.

" 1

'11 faxbsin mx dx œ ! a1 '11 sin nx sin mx dx. Since 1" '11 sin nx sin mx dx œ œ ! N

n

1

nœ"

am 1

the sum on the right has only one nonzero term, namely

'11 sin mx sin mx dx œ am .

for m Á n , for m œ n

8.5 TRIGONOMETRIC SUBSTITUTIONS 1. y œ 3 tan ),  1#  )  1# , dy œ

3 d) cos# )

, 9
y# œ 9 a1
tan# )b œ

Ê

9 cos# )

" È 9
y#

kcos )k 3

œ

œ

cos ) 3

ˆbecause cos )  0 when  1#  )  1# ‰ ;

' È9dy
y

œ 3'

#

œ'

d) cos )

œ ln ksec )
tan )k
Cw œ ln ¹

È 9
y# 3


y3 ¹
Cw œ ln ¸È9
y#
y¸
C

; c3y œ xd Ä

'

' È dx

œ'

œ ln ksec t
tan tk
C œ ln ¹Èx#
1
x¹
C œ ln ¸È1
9y#
3y¸
C

3.

'22

# œ
#" tan" #x ‘ # œ

4.

'02 8
dx2x

5.

'03Î2 È dx

6.

'01Î2

2.

' È13
dy9y

cos ) d) 3 cos# )

#

1
x#

dx 4
x#

#

œ

9  x#

È2

" #

dt " cos# t Š cos t‹

'02 4
dxx

#

œ

" #

dx È 1
x#

" #

; x œ tan t,  1#  t 

tan" 1 


#" tan" #x ‘ # œ !

$Î# œ
sin" 3x ‘ ! œ sin"

2 dx È1  4x#

; ct œ 2xd Ä

'01Î2

È2

" #

" #

" #

, dx œ

ˆ #" tan" 1  1 6

" #

0œ

È2

1Î

œ csin" td 0

25 #

8. t œ

" 3

a)
sin ) cos )b
C œ

25 #

sin ),  1#  )  1# , dt œ

" 3

’sin" ˆ 5t ‰


" cos t

, È1
x# œ

;

1 4

tan" 0‰ œ ˆ #" ‰ ˆ #" ‰ ˆ 14 ‰  0 œ

1 16

1 6

œ sin"

7. t œ 5 sin ),  1#  )  1# , dt œ 5 cos ) d), È25  t# œ 5 cos ); ' È25  t# dt œ ' (5 cos ))(5 cos )) d) œ 25 ' cos# ) d) œ 25 ' œ

dt cos# t

tan" (1) œ ˆ #" ‰ ˆ 14 ‰  ˆ #" ‰ ˆ 14 ‰ œ

 sin" 0 œ

dt È1  t#

1 #

È ˆ 5t ‰ Š 255 t# ‹“

" È2

 sin" 0 œ

1
cos 2) #


Cœ

25 #

1 4

0œ

d) œ 25 ˆ #)


sin" ˆ 5t ‰


1 4

sin 2) ‰ 4

tÈ25  t# #


C


C

cos ) d), È1  9t# œ cos );

' È1  9t# dt œ "3 ' (cos ))(cos )) d) œ 3" ' cos# ) d) œ 6" a)
sin ) cos )b
C œ 6" ’sin" (3t)
3tÈ1  9t# “
C 9. x œ

'È

7 #

sec ), 0  )  1# , dx œ

dx 4x#  49

œ'

ˆ 7# sec ) tan )‰ d) 7 tan )

7 #

œ

sec ) tan ) d), È4x#  49 œ È49 sec# )  49 œ 7 tan ); " #

' sec ) d) œ "# ln ksec )
tan )k
C œ "# ln ¹ 2x7
È4x7#  49 ¹
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

520

Chapter 8 Techniques of Integration

10. x œ

3 5

sec ), 0  )  1# , dx œ

' È 5 dx

25x#  9

œ'

sec ) tan ) d), È25x#  9 œ È9 sec# )  9 œ 3 tan );

3 5

œ ' sec ) d) œ ln ksec )
tan )k
C œ ln ¹ 5x 3


5 ˆ 35 sec ) tan )‰ d) 3 tan )

È25x#  9 ¹ 3


C

11. y œ 7 sec ), 0  )  1# , dy œ 7 sec ) tan ) d), Èy#  49 œ 7 tan );

) tan )) d) ' Èy y 49 dy œ ' (7 tan ))(77 sec œ 7 ' tan# ) d) œ 7 ' asec# )  1b d) œ 7(tan )  ))
C sec ) #

œ 7’

Èy#  49 7

 sec" ˆ y7 ‰“
C

12. y œ 5 sec ), 0  )  1# , dy œ 5 sec ) tan ) d), Èy#  25 œ 5 tan ); sec ) tan )) d) ' Èyy 25 dy œ ' (5 tan ))(5 œ "5 ' 125 sec ) #

$

œ

" 10

$

a)  sin ) cos )b
C œ

" 10

tan# ) cos# ) d) œ

’sec" ˆ y5 ‰  Š

" 5

Èy#  25 ‹ Š 5y ‹“ y

'

sin# ) d) œ


Cœ’

sec
" 10

" 10

ˆ 5y ‰

' (1  cos 2)) d)



Èy#  25 “ #y #


C

13. x œ sec ), 0  )  1# , dx œ sec ) tan ) d), Èx#  1 œ tan );

'

dx x# È x#  1

œ'

sec ) tan ) d) sec# ) tan )

œ'

d) sec )

œ sin )
C œ

È x#  1 x


C

14. x œ sec ), 0  )  1# , dx œ sec ) tan ) d), Èx#  1 œ tan );

'

2 dx x$ È x#  1

œ'

2 tan ) sec ) d) sec$ ) tan )

2) ‰ œ 2 ' cos# ) d) œ 2 ' ˆ 1
cos d) œ )
sin ) cos )
C # #

œ )
tan ) cos# )
C œ sec" x
Èx#  1 ˆ "x ‰
C œ sec" x
15. x œ 2 tan ),  1#  )  1# , dx œ

' Èx

$ dx x#
4

œ'

ct œ cos )d Ä 8 ' œ 8 Œ

œ 8'

a8 tan$ )b (cos )) d) cos# )

È x#
4 #




#

t 1 t%

2 d) cos# )

$Î#


Cœ


C

, Èx#
4 œ

sin$ ) d) cos% )

œ 8'

2 cos ) ; acos# )  1b ( sin )) d) cos% )

dt œ 8' ˆ t"#  t"% ‰ dt œ 8 ˆ "t


ax #
4 b 8 †3

È x#  1 x#

" 3

ax#
4b

$Î#

" ‰ 3t$

;


C œ 8 Š sec )


sec$ ) 3 ‹


C

 4Èx#
4
C

16. x œ tan ),  1#  )  1# , dx œ sec# ) d), Èx#
1 œ sec );

'

dx x# È x#
1

œ'

sec# ) d) tan# ) sec )

œ'

cos ) d) sin# )

œ  sin" )
C œ

È x #
1 x


C

17. w œ 2 sin ),  1#  )  1# , dw œ 2 cos ) d), È4  w# œ 2 cos );

'

8 dw w # È 4  w#

œ'

8†2 cos ) d) 4 sin# )†2 cos )

œ 2'

d) sin# )

œ 2 cot )
C œ

2 È 4  w # w


C

18. w œ 3 sin ),  1#  )  1# , dw œ 3 cos ) d), È9  w# œ 3 cos );

' È9w w #

#

dw œ '

3 cos )†3 cos ) d) 9 sin# )

œ  cot )  )
C œ 

È 9  w# w

) ' acsc# )  1b d) œ ' cot# ) d) œ ' Š 1 sinsin # ) ‹ d) œ #

 sin" ˆ w3 ‰
C

19. x œ sin ), 0 Ÿ ) Ÿ 13 , dx œ cos ) d), a1  x# b

È3Î2

'0

4x# dx a1  x# b$Î#

œ '0

1Î3

1Î$

œ 4 ctan )  )d !

4 sin# ) cos ) d) cos$ )

œ 4È3 

$Î#

œ cos$ );

) '0 asec# )  1b d) œ 4 '0 Š 1 coscos # ) ‹ d) œ 4 1Î3

#

1Î3

41 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.5 Trigonometric Substitutions 20. x œ 2 sin ), 0 Ÿ ) Ÿ 16 , dx œ 2 cos ) d), a4  x# b

'0

1

œ '0

1Î6

dx a4  x# b$Î#

2 cos ) d) 8 cos$ )

œ

'0

1Î6

" 4

d) cos# )

œ

" 4

$Î#

1Î'

ctan )d !

21. x œ sec ), 0  )  12 , dx œ sec ) tan ) d), ax#  1b

'

œ'

dx ax#  1b$Î#

sec ) tan ) d) tan$ )

œ'

cos ) d) sin# )

œ'

x# dx ax#  1b&Î#

sec# )†sec ) tan ) d) tan& )

23. x œ sin ),  1#  ) 

' a1  xxb

# $Î# '

dx

œ'

' a1  xxb

dx

%

" #

25. x œ

' '

" 3

œ'

œ'

œ'

v# dv a1  v# b&Î#

# &Î# )

œ'

dr

"Î#

" #

œ cos$ );

t

e2t
9

œ 'tanc" Ð1Î3Ñ

œ ln ˆ 53
43 ‰  ln Š

È10 3

31.

et dt a1
e2t b$Î#

'11ÎÎ124

2 dt Èt
4tÈt

'11ÎÈ3

2 du 1
u#

" 3


C

4x a4x#
1b


C

sec# ) d), 9t#
1 œ sec# );

&Î#

3t a9t#
1b


C

œ cos& );

tan$ ) 3


Cœ

" 3

ŠÈ

v ‹ 1  v#

œ ' cot' ) csc# ) d) œ  cot7 )
C œ  "7 ’ (

3 tan )†sec# ) d) tan )†3 sec )

tan
" Ð4Î3Ñ

sec# ) tan )

$


C

È 1  r# ( “ r


C

d), Èe2t
9 œ È9 tan# )
9 œ 3 sec );

Ð4Î3Ñ œ 'tan
" Ð1Î3Ñ sec ) d) œ cln ksec )
tan )kd tan tanc" Ð1Î3Ñ


"


"3 ‹ œ ln 9  ln Š1
È10‹

tan
" Ð4Î3Ñ

(tan

#) )) Š sec tan ) ‹

sec$ )

; ’u œ 2Èt, du œ 1Î4

È 1  x# $ ‹ x

#

œ ' tan# ) sec# ) d) œ

œ 'tanc" Ð3Î4Ñ

œ '1Î6


C

sec# ) d), a4x#
1b œ sec% );

30. Let et œ tan ), t œ ln (tan )), tan" ˆ 34 ‰ Ÿ ) Ÿ tan" ˆ 43 ‰ , dt œ

'lnlnÐ3Ð4ÎÎ43Ñ Ñ

È 1  x# & ‹ x

œ cos );

29. Let et œ 3 tan ), t œ ln (3 tan )), tan" ˆ "$ ‰ Ÿ ) Ÿ tan" ˆ %$ ‰, dt œ

'0ln 4 Èe dt


C

œ 2 ' cos# ) d) œ )
sin ) cos )
C œ tan" 3t


cos& )†cos ) d) sin) )

tan
" Ð4Î3Ñ

x$ 3 ax#  1b$Î#

$

28. r œ sin ),  1#  )  1# ;

' a1  r rb

œ tan& );

œ 4 ' cos# ) d) œ 2()
sin ) cos ))
C œ 2 tan" 2x


8 ˆ "# sec# )‰ d) sec% )

sin# ) cos ) d) cos& )


C

x x#  1

œ ' cot# ) csc# ) d) œ  cot3 )
C œ  "3 Š

cos )†cos ) d) sin% )

27. v œ sin ),  1#  )  1# , dv œ cos ) d), a1  v# b

'

$Î#

" 4È 3

œ tan$ );

d) œ  3 sin" $ )
C œ 

, dx œ cos ) d), a1  x# b

6 ˆ "3 sec# )‰ d) sec% )

œ

&

tan ),  1#  )  1# , dt œ

6 dt a9t#
1b#

&Î#

È3 12

œ ' cot% ) csc# ) d) œ  cot5 )
C œ  5" Š

tan ),  1#  )  1# , dx œ

8 dx a4x#
1b#

26. t œ

œ'

1 #

cos ) sin% )

, dx œ cos ) d), a1  x# b

cos$ )†cos ) d) sin' )

24. x œ sin ),  1#  )  # "Î#

1 #

œ'

$Î#

œ

œ  sin" )
C œ  È

22. x œ sec ), 0  )  12 , dx œ sec ) tan ) d), ax#  1b

'

œ 8 cos$ );

2 sec# ) d) sec# )

" Èt

d)

sec# ) tan )

d), 1
e2t œ 1
tan# ) œ sec# );

tan
" Ð4Î3Ñ

Ð4Î3Ñ œ 'tan
" Ð3Î4Ñ cos ) d) œ csin )d tan œ tanc" Ð$Î%Ñ

dt“ Ä '1ÎÈ3 1

2 du 1
u#


"

; u œ tan ),

1Î% œ c2)d 1Î' œ 2 ˆ 14  16 ‰ œ

1 6

4 5



3 5

œ

" 5

Ÿ ) Ÿ 14 , du œ sec# ) d), 1
u# œ sec# );

1 6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

521

522

Chapter 8 Techniques of Integration

32. y œ etan ) , 0 Ÿ ) Ÿ 14 , dy œ etan ) sec# ) d), È1
(ln y)# œ È1
tan# ) œ sec );

'1e yÈ1
dy(ln y)

œ '0

1Î4

#

d) œ '0 sec ) d) œ cln ksec )
tan )kd ! 1Î4

etan ) sec# ) etan ) sec )

1Î%

œ ln Š1
È2‹

33. x œ sec ), 0  )  1# , dx œ sec ) tan ) d), Èx#  1 œ Èsec# )  1 œ tan );

'

dx xÈ x#  1

œ'

sec ) tan ) d) sec ) tan )

œ )
C œ sec" x
C

34. x œ tan ), dx œ sec# ) d), 1
x# œ sec# );

' x dx
1 œ ' secsec) )d) œ )
C œ tan" x
C #

#

#

35. x œ sec ), dx œ sec ) tan ) d), Èx#  1 œ Èsec# )  1 œ tan ); ' x dx œ ' sec )†sec ) tan ) d) œ ' sec# ) d) œ tan )
C œ Èx#  1
C tan )

È x#  1

36. x œ sin ), dx œ cos ) d),  1#  ) 

' È dx

œ'

1  x#

37. x

dy dx

cos ) d) cos )

œ 2’

38. Èx#  9

;yœ'

1

È x#  4 x

;yœ'

dx È x#  9

Ê y œ ln ¹ x3
dy dx

Ô x œ 3 sec ), 0  )  # × ; Ö dx œ 3 sec ) tan ) d) Ù Ä y œ ' Õ Èx#  9 œ 3 tan ) Ø

yœ'

3 dx x #
4 ; y œ 3 œ 3# tan" ˆ x# ‰

œ Èx#
1, dy œ

sec# ) d) sec$ )

3

A œ '0

È 9  x# 3

1Î2

3 sec ) tan ) d) 3 tan )


C; x œ 5 and y œ ln 3 Ê ln 3 œ ln 3
C Ê C œ 0

#



x È x#
1

dx ax#
1b$Î#

; x œ tan ), dx œ sec# ) d), ax#
1b

œ 3'0

tan ) sec )


Cœ

$Î#

3 #

tan" 1
C

œ sec$ );

x È x#
1


C; x œ 0 and y œ 1


1

dx; x œ 3 sin ), 0 Ÿ ) Ÿ

3 cos )†3 cos ) d) 3

Ê 0œ

31 8

œ ' cos ) d) œ sin )
C œ tan ) cos )
C œ

Ê 1œ0
C Ê yœ 41. A œ '0

È x#  9 ¹ 3

' x dx
4 œ #3 tan" #x
C; x œ 2 and y œ 0

œ 3, dy œ

# dy dx

 sec" x# “

È x#  9 ¹ 3

Ê C œ  381 Ê y 40. ax#
1b

È x#  4 #

1

dx È x#  9

œ ' sec ) d) œ ln ksec )
tan )k
C œ ln ¹ x3


39. ax#
4b

Ô x œ 2 sec ), 0  )  # × dx; Ö dx œ 2 sec ) tan ) d) Ù Õ Èx#  4 œ 2 tan ) Ø

 sec" ˆ x# ‰“
C; x œ 2 and y œ 0 Ê 0 œ 0
C Ê C œ 0 Ê y œ 2 ’

œ 1, dy œ

dy dx

dx x

œ 2' tan# ) d) œ 2 ' asec# )  1b d) œ 2(tan)  ))
C

(2 tan ))(2 sec ) tan )) d) 2 sec )

È x#  4 #

;

œ )
C œ sin" x
C

œ Èx#  4; dy œ Èx#  4

Ä yœ'

1 #

1Î2

1 #

, dx œ 3 cos ) d), È9  x# œ È9  9 sin# ) œ 3 cos );

cos# ) d) œ

3 #

1Î#

c)
sin ) cos )d !

œ

31 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.5 Trigonometric Substitutions 42. V œ '0 1 ˆ 1
2 x# ‰ dx œ 41 '0 1

1

#

dx ax #
1 b #

;

x œ tan ), dx œ sec# ) d), x#
1 œ sec# ); V œ 41 '0

1Î4

œ 21 '0

1Î4

sec# ) d) sec% )

œ 41 '0

1Î4

cos# ) d) sin 2) ‘ 1Î% # !

(1
cos 2)) d) œ 21
)


43.

' 1 dxsin x œ '

44.

' 1
sin dxx
cos x œ '

œ'

2 dz ‹ 1
z# 1  Š 2z # ‹ 1
z

Š

œ

2 dz (1  z)#

2 1z

œ'

2 dz ‹ 1
z# # 1
Š 2z #
1  z# ‹ 1
z 1
z

Š

œ 1 ˆ 1#
1‰


Cœ


C

2 1  tan ˆ #x ‰

2 dz 1
z#
2z
1  z#

œ'

œ ln k1
zk
C

dz 1
z

œ ln ¸tan ˆ x# ‰
1¸
C 45.

'01Î2 1
dxsin x œ '01

46.

'11ÎÎ32

47.

) '01Î2 2
dcos '1 ) œ 0

œ

48.

dx 1  cos x

1 3È 3

'12Î12Î3

œ

50.

" È2

51.

È$ z# 4 “"

2 dz (1
z)#

œ '1ÎÈ3 1

œ '0

Š

1

dz z#

" œ
 1z ‘ "ÎÈ$ œ È3  1

œ '0

1

2 dz 2
2z#
1  z#

È3

œ '1

#

Š 1
z# ‹ Š 1z

Ô 2z Š1
z# ‹

œ

2 dz z#
3

È3

2 dz ‹ 1  z#

× 2z #
Š 1  z# ‹Ø Õ Š1  z# ‹

œ '1

2 È3

’tan"

2 a1  z# b dz 2z  2z$
2z
2z$

œ Š "# ln È3  34 ‹  ˆ0  4" ‰ œ

Š 2 dz# ‹ 1
z # 2z  1  z# ‹ Š # 1
z 1
z

œ'

2 dz 2z  1
z#

z È3 “

" !

œ

2 È3

tan"

" È3

œ'

ln 3 4

1  z# 2z

œ

" 4

œ

" È2

1 2 ln ¹ zz
¹
C
1
È2

(ln 3  2) œ

" #

dz

" #



2 dz (z
1)#  2

È3

œ '1

Šln È3  1‹

È

tan ˆ t ‰
1  È2 #

a1  z# b dz a1
z # b z #

1
z# ‹ Š 2 dz# ‹ 1  z# 1 z # 1  Š 1
z# ‹ 1z

œ'

dz z # a1
z # b

œ' '

' sec ) d) œ ' cosd) ) œ ' Š

2 dz ‹ 1
z# 1  z# Š ‹ 1
z#

2 a1  z# b dz a1
z # b #  a 1
z # b a 1  z # b dz 1
z#

œ'

dz z#

œ'

2 dz 1  z#

œ'

œ ln k1
zk  ln k1  zk
C œ ln »

52.

"

œ 
1
2 z ‘ ! œ (1  2) œ 1

ln º tan ˆ #t ‰
1
È2 º
C

t dt ' 1cos cos 'Š t œ

œ'

2 dz ‹ 1
z# # 1  Š 1  z# ‹ 1
z

Š

2 dz ‹ 1
z# 1  z# 2
Š ‹ 1
z#

cos ) d) sin ) cos )
sin )

' sin t dt cos t œ ' œ

'11ÎÈ3

1

È 31 9

œ ’ "# ln z 

49.

œ

œ '0

2 dz ‹ 1
z# 1
Š 2z # ‹ 1
z

Š

' csc ) d) œ ' sind)) œ ' Š

2 dz ‹ 1
z# Š 2z # ‹ 1
z

1
tan Š )# ‹

2'

1  tan Š )# ‹ »

œ'

dz z

œ'

dz z#
1

2 dz (1
z)(1  z)

2 a1  z# b dz a1
z # b a 1
z #  1
z # b

œ  "z  2 tan" z
C œ  cot ˆ #t ‰  t
C œ'

dz 1
z


'

dz 1z


C

œ ln kzk
C œ ln ¸tan #) ¸
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

523

524

Chapter 8 Techniques of Integration

8.6 INTEGRAL TABLES AND COMPUTER ALGEBRA SYSTEMS 1.

'

dx xÈ x  3

œ

tan" É x 3 3
C

2 È3

(We used FORMULA 13(a) with a œ 1, b œ 3) 2.

' xÈdxx
4 œ È"

È

È

4 4 ln ¹ Èxx
¹
Cœ
4
È4

4

" #

È

42 ln ¹ Èxx
¹
C
4
2

(We used FORMULA 13(b) with a œ 1, b œ 4) 3.

' Èx dx

œ'

x2


2'

(x  2) dx Èx  2 $

œ ˆ 21 ‰

ŠÈx  2‹

œ ' ŠÈx  2‹ dx
2 ' ŠÈx  2‹ "

"

dx

"


2 ˆ 12 ‰

3

dx Èx  2

ŠÈx  2‹ 1

œ Èx  2 ’ 2(x 3 2)
4“
C

(We used FORMULA 11 with a œ 1, b œ 2, n œ 1 and a œ 1, b œ 2, n œ 1) 4.

' (2xx
dx3)

" #

œ

$Î#


3) dx ' (2x  3# ' (2x
dx3) (2x
3) $Î#

œ

$Î#

" #

' È2xdx
3  3# '

dx $ ˆÈ2x
3‰

' ŠÈ2x
3‹" dx  3# ' ŠÈ2x
3‹$ dx œ ˆ "# ‰ ˆ 2# ‰ ˆÈ2x1
3‰

œ

" #

œ

" #È2x
3

(x
3) È2x
3

(2x
3
3)
C œ

"

 ˆ 3# ‰ ˆ 2# ‰

ˆÈ2x
3‰ (1)


"


C


C

(We used FORMULA 11 with a œ 2, b œ 3, n œ 1 and a œ 2, b œ 3, n œ 3) 5.

' xÈ2x  3 dx œ "# ' (2x  3)È2x  3 dx
3# ' È2x  3 dx œ "# ' ŠÈ2x  3‹$ dx
3# ' ŠÈ2x  3‹" dx &

œ ˆ "# ‰ ˆ 2# ‰

ŠÈ2x  3‹ 5

$


ˆ 3# ‰ ˆ 2# ‰

ŠÈ2x  3‹


Cœ

3

(2x  3)$Î# #


2x 5 3
1‘
C œ

(2x  3)$Î# (x
1) 5


C

(We used FORMULA 11 with a œ 2, b œ 3, n œ 3 and a œ 2, b œ 3, n œ 1)

6.

' x(7x
5)$Î# dx œ "7 ' (7x
5)(7x
5)$Î# dx  57 ' (7x
5)$Î# dx œ 7" ' ˆÈ7x
5‰& dx  75 ' ˆÈ7x
5‰$ dx œ ˆ 7" ‰ ˆ 72 ‰

ˆÈ7x
5‰ 7

(

 ˆ 75 ‰ ˆ 72 ‰

ˆÈ7x
5‰ 5

&

&Î#


C œ ’ (7x
495) “ ’ 2(7x7
5)  2“
C

&Î#

œ ’ (7x
495) “ ˆ 14x7 4 ‰
C (We used FORMULA 11 with a œ 7, b œ 5, n œ 5 and a œ 7, b œ 5, n œ 3) 7.

' È9x 4x dx œ  È9x 4x
(#4) ' #

dx xÈ9  4x


C

(We used FORMULA 14 with a œ 4, b œ 9) œ

È9  4x x

È

È

4x  9  2 Š È"9 ‹ ln ¹ È99  ¹
C  4x
È9

(We used FORMULA 13(b) with a œ 4, b œ 9) œ

8.

'

È9  4x x

dx x# È4x  9



2 3

œ

È

4x  3 ln ¹ È99  ¹
C  4x
3 È4x  9 (9)x




4 18

'

dx xÈ4x  9


C

(We used FORMULA 15 with a œ 4, b œ 9) œ

È4x  9 9x


ˆ 92 ‰ Š È29 ‹ tan" É 4x 9 9
C

(We used FORMULA 13(a) with a œ 4, b œ 9) œ

È4x  9 9x




4 27

tan" É 4x 9 9
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems 9.

'

xÈ4x  x# dx œ ' xÈ2 † 2x  x# dx œ

œ

(x
2)(2x  6)È4x  x# 6


4 sin" ˆ x # 2 ‰
C

(We used FORMULA 51 with a œ 2) 10.

' Èxx x

dx œ '

#

É2† "# x  x# x

$ (x
2)(2x  3†2)È2†2†x  x#
2# sin" ˆ x # 2 ‰
6 È # œ (x
2)(x 33) 4x  x
4 sin" ˆ x # 2 ‰
C

dx œ É2 † "# x  x#


" #

sin" Š

x  "#

ˆWe used FORMULA 52 with a œ "# ‰

11.

'

dx xÈ 7
x#

œ'

dx #

xÊŠÈ7‹
x#

œ

" #

‹
C œ Èx  x#


C

" #

sin" (2x  1)
C

â â â È7
ÊŠÈ7‹#
x# â È 7
È 7
x# â â ln â ¹
C â
C œ  È"7 ln ¹ x x â â â â

" È7

ŠWe used FORMULA 26 with a œ È7‹

12.

'

dx xÈ 7  x#

œ'

dx #

xÊŠÈ7‹  x#

œ

" È7

â â â È7
ÊŠÈ7‹#  x# â È 7
È 7  x# â â ln â ¹
C â
C œ  È"7 ln ¹ x x â â â â

ŠWe used FORMULA 34 with a œ È7‹ 13.

' È4x x

#

dx œ '

È 2#  x# x

dx œ È2#  x#  2 ln ¹ 2


È 2#  x# ¹ x


C œ È4  x#  2 ln ¹ 2


È 4  x# ¹ x


C

(We used FORMULA 31 with a œ #) 14.

' Èxx  4 dx œ ' Èx x 2 #

#

#

dx œ Èx#  2#  2 sec" ¸ #x ¸
C œ Èx#  4  2 sec" ¸ x# ¸
C

(We used FORMULA 42 with a œ #) 15.

'

È25  p# dp œ ' È5#  p# dp œ

p #

È 5#  p#


5# #

sin"

p 5


Cœ

p #

È25  p#


25 #

sin"

p 5


C

(We used FORMULA 29 with a œ 5) 16.

' q# È25  q# dq œ ' q# È5#  q# dq œ 58

%

œ

625 8

sin" ˆ q5 ‰  "8 qÈ5#  q# a5#  2q# b
C

sin" ˆ q5 ‰  "8 qÈ25  q# a25  2q# b
C

(We used FORMULA 30 with a œ 5) 17.

' Èr

#

4  r#

dr œ '

r# È2#  r#

dr œ

2# #

sin" ˆ #r ‰  "# rÈ2#  r#
C œ 2 sin" ˆ #r ‰  "# rÈ4  r#
C

(We used FORMULA 33 with a œ 2)

18.

' È ds s#

2

œ'

ds #

Ês#  ŠÈ2‹

œ cosh"

s È2

#


C œ ln »s
Ês#  ŠÈ2‹ »
C œ ln ¹s
Ès#  2¹
C

ŠWe used FORMULA 36 with a œ È2‹ 19.

) ' 5
4dsin 2) œ

2 2È25  16

4 ˆ1 tan" ’É 55 
4 tan 4 

2) ‰ # “


C œ  "3 tan"
"3 tan ˆ 14  )‰‘
C

(We used FORMULA 70 with b œ 5, c œ 4, a œ 2) 20.

) ' 4
5dsin 2) œ

1 2È25  16

È


3 cos 2) ¸ ln ¹ 5
4 sin 24)

5 sin252) 16 cos 2) ¹
C œ  6" ln ¸ 5
4 4sin
25)sin
C 2)

(We used FORMULA 71 with a œ 2, b œ 4, c œ 5)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

525

526 21.

Chapter 8 Techniques of Integration

' e2t cos 3t dt œ 2 e
3 2t

#

e2t 13

(2 cos 3t
3 sin 3t)
C œ

#

(2 cos 3t
3 sin 3t)
C

(We used FORMULA 108 with a œ 2, b œ 3) 22.

' ec3t sin 4t dt œ (3)ec
4 3t

#

(3 sin 4t  4 cos 4t)
C œ

#

ec3t 25

($ sin 4t  4 cos 4t)
C

(We used FORMULA 107 with a œ 3, b œ 4) 23.

' x cos" x dx œ '

x1b1 1
1

x" cos" x dx œ

cos" x


" 1
1

' Èx b

1 1

(We used FORMULA 100 with a œ 1, n œ 1) # œ x cos" x
" ˆ " sin" x‰  " Š " xÈ1  x# ‹
C œ #

#

#

#

œ

dx 1  x#

x# #

#

x# #

cos" x


cos" x
" 4

" #

' Èx

# dx 1  x#

sin" x  4" xÈ1  x#
C

(We used FORMULA 33 with a œ 1) 24.

' x tan" x dx œ '

x1b1 1
1

x" tan" a"xb dx œ

tan" a"xb 

" 1
1

' 1 x
ba"bdxx 1 1

# #

œ

x# #

tan" x 

" #

' 1x
dxx #

(We used FORMULA 101 with a œ 1, n œ 1)

25.

œ

x# #

œ

x# #

'

' ˆ"  1
" x ‰dx (after long division) tan" x  "# ' dx
"# ' 1
" x dx œ x# tan" x  #" x
#" tan" x
C œ #" aax#
"btan" x  xb
C tan" x 

" #

#

#

#

ds a9  s # b #

œ'

ds a 3 $  s # b#

œ

s 2†3# †a3#  s# b

" 4†33 ln





3¸ ¸ ss  3
C

(We used FORMULA 19 with a œ 3) "
3¸ œ 18 a9s s# b
108 ln ¸ ss  3
C 26.

'

d) a2  ) # b #

œ'

d) #

”ŠÈ2‹  )# •

#

œ

#

)

#

2 ŠÈ2‹ ”ŠÈ2‹  )# •




È

"

3

4 ŠÈ2‹

ln ¹ )
È2 ¹
C )

2

ŠWe used FORMULA 19 with a œ È2‹ œ

27.

'

) 4 a2  ) # b

È4x
9 x#




" 8È 2

dx œ 

È

2 ln ¹ ))
¹
C  È2 È4x
9 x




4 2

' xÈ4xdx
9

(We used FORMULA 14 with a œ 4, b œ 9) œ

È4x
9 x

È

È

4x
9  9
2 Š È"9 ln ¹ È4x ¹‹
C œ 
9
È9

È4x
9 x




2 3

È

4x
9  3 ln ¹ È4x ¹
C
9
3

(We used FORMULA 13(b) with a œ 4, b œ 9) 28.

' È9xx  4 dx œ  È9xx  4
92 ' #

dx xÈ9x  4


C

(We used FORMULA 14 with a œ 9, b œ 4) œ

È9x  4 x


#9 Š È2 tan" É 9x 4 4 ‹
C œ  4

È9x  4 x




9 #

tan"

È9x  4 #


C

È3t  4 #


C

(We used FORMULA 13(a) with a œ 9, b œ 4) 29.

#

' È3tt  4 dt œ 2È3t  4
(4)'

dt tÈ3t  4

(We used FORMULA 12 with a œ 3, b œ 4) œ 2È3t  4  4 Š È24 tan" É 3t 4 4 ‹
C œ 2È3t  4  4 tan" (We used FORMULA 13(a) with a œ 3, b œ 4)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems 30.

' È3tt
9 dt œ 2È3t
9
9 ' tÈ3tdt
9 (We used FORMULA 12 with a œ 3, b œ 9) È
9  È9 È3t
9  3 œ 2È3t
9
9 Š È19 ln ¹ È3t ¹‹
C œ 2È3t
9
3 ln ¹ È3t
9
3 ¹
C 3t
9
È9

(We used FORMULA 13(b) with a œ 3, b œ 9) 31.

' x# tan" x dx œ 2x

1 tan" x  2
" 1 ' 1x

x# dx œ x3$ tan" x  3" ' 1
x$x# dx 2 1

2 1

(We used FORMULA 101 with a œ 1, n œ 2);

' 1
x$x# dx œ ' x$ 3

œ 32.

' '

tanc" x x#

x# 6




" 6

x dx 1
x#

œ

x# #



" #

' x# tan" x dx

ln a1
x# b
C Ê

ln a1
x# b
C

dx œ ' x# tan" x dx œ

xÐ
21Ñ (2
1)

tan" x 

" (2
1)

' 1xÐ

x#Ñ dx œ (x
"1) tan" x
' a1 x

"x#b dx 2 1

(We used FORMULA 101 with a œ 1, n œ 2); xc" dx 1
x #

Ê' 33.

tan" x 

x dx  '

œ'

tanc" x x#

dx x a1
x# b

œ'

'

dx x

x dx 1
x #

œ ln kxk 

dx œ  x" tan" x
ln kxk 

" #

" #

ln a1
x# b
C

ln a1
x# b
C

' sin 3x cos 2x dx œ  cos105x  cos# x
C (We used FORMULA 62(a) with a œ 3, b œ 2)

34.

' sin 2x cos 3x dx œ  cos105x
cos# x
C (We used FORMULA 62(a) with a œ 2, b œ 3)

35.

'

8 sin 4t sin

t #

dx œ

8 7

sin ˆ 7t# ‰ 

8 9

sin ˆ 9t# ‰
C œ 8 –

sin Š 7t #‹ 7



sin Š 9t #‹ 9

—
C

(We used FORMULA 62(b) with a œ 4, b œ "# ) 36.

' sin 3t sin 6t dt œ 3 sin ˆ 6t ‰  sin ˆ #t ‰
C

(We used FORMULA 62(b) with a œ "3 , b œ 6" )

37.

' cos 3) cos 4) d) œ 6 sin ˆ 12) ‰
67 sin ˆ 17#) ‰
C

(We used FORMULA 62(c) with a œ "3 , b œ 4" )

38.

13) # ‹

' cos 2) cos 7) d) œ 13" sin ˆ 132) ‰
151 sin ˆ 15#) ‰
C œ sin Š13 " 2,

(We used FORMULA 62(c) with a œ 39.

'x

$
x
1 ax #
1 b # œ "# lnax#

dx œ '
1b


x dx x#
1


'

x 2 a1
x # b




dx a x #
1 b# " " # tan

œ

" #




sin Š 15#) ‹ 15


C

b œ 7)

' d axx

11b
' #

#

dx a x #
1 b#

x
C

(For the second integral we used FORMULA 17 with a œ 1) 40.

'

x#
6x ax #
3 b #

dx œ '

dx x#
3


'

6x dx a x #
3 b#

'

3 dx a x #
3 b#

œ'

x#

dx #
ŠÈ3‹


3'

d ax #
3 b ax#
3b#

 3'

dx ”

x#

# #


ŠÈ3‹ •

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

527

528

Chapter 8 Techniques of Integration œ

" È3

tan" Š Èx3 ‹ 

3

3 ax #
3 b

Î

x # Ï 2 ŠÈ3‹ ŒŠÈ3‹
x#  #




" $ 2 ŠÈ3‹

tan" Š Èx3 ‹

Ñ Ò


C

ŠFor the first integral we used FORMULA 16 with a œ È3; for the third integral we used FORMULA 17 with a œ È3‹ œ

41.

'

" 2È 3

tan" Š Èx3 ‹ 

sin" Èx dx;

3 x #
3



x 2 ax #
3 b


C

Ô u œ Èx × 1b1 Ä 2 ' u" sin" u du œ 2 Š 1u
1 sin" u  x œ u# Õ dx œ 2u du Ø

œ u# sin" u  '

" 1
1

' Èu b

1 1

1  u#

du‹

u# du È 1  u#

(We used FORMULA 99 with a œ 1, n œ 1) œ u# sin" u  Š "# sin" u  "# uÈ1  u# ‹
C œ ˆu#  "# ‰ sin" u
"# uÈ1  u#
C (We used FORMULA 33 with a œ 1) œ ˆx  "# ‰ sin" Èx
"# Èx  x#
C u œ Èx ×
" 42. Ä ' cosu u † 2u du œ 2' cos" u du œ 2 Šu cos" u  1" È1  u# ‹
C x œ u# Õ dx œ 2u du Ø (We used FORMULA 97 with a œ 1) œ 2 ŠÈx cos" Èx  È1  x‹
C

' cosÈc"xÈx dx; Ô

43.

Ô u œ Èx × Ä x œ u# 1x Õ dx œ 2u du Ø œ sin" u  uÈ1  u#
C

' ÈÈx

dx;

' Èu†2u

1  u#

du œ 2 '

u# È 1  u#

du œ 2 Š "# sin" u  "# uÈ1  u# ‹
C

(We used FORMULA 33 with a œ 1) œ sin" Èx  Èx È1  x
C œ sin" Èx  Èx  x#
C

44.

u œ Èx × Ä x œ u# Õ dx œ 2u du Ø

' ÈÈ2 x x dx; Ô œ

2 – u#

' È2 u u

#

† 2u du œ 2 ' ÊŠÈ2‹  u# du #

#

#

ÊŠÈ2‹  u#


ŠÈ2‹ #

sin" Š Èu2 ‹—
C œ uÈ2  u#
2 sin" Š Èu2 ‹
C

ŠWe used FORMULA 29 with a œ È2‹ œ È2x  x#
2 sin" È x#
C 45.

' (cot t) È1  sin# t dt œ ' È1  sinsintt(cos t) dt ; ” #

œ È1  u#  ln ¹ 1


È 1  u# ¹ u

u œ sin t Ä du œ cos t dt •

' È1 u u

#

du


C

(We used FORMULA 31 with a œ 1) È # œ È1  sin# t  ln ¹ 1
1  sin t ¹
C sin t

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems 46.

'

œ'

dt (tan t) È4  sin# t

cos t dt (sin t) È4  sin# t

;”

u œ sin t Ä' du œ cos t dt •

du uÈ4  u#

œ  "# ln ¹ 2


È 4  u# ¹ u


C

(We used FORMULA 34 with a œ 2) œ  "# ln ¹ 2


47.

È4  sin# t ¹ sin t


C

Ô u œ ln y × y œ eu Ä # Õ dy œ eu du Ø œ ln ¸ln y
È3
(ln y)# ¸
C

' yÈ3
dy(ln y)

'

;

eu du eu È 3
u#

œ'

du È 3
u#

œ ln ¹u
È3
u# ¹
C

ŠWe used FORMULA 20 with a œ È3‹

48.

' Ècos ) d)

5
sin# )

;”

u œ sin ) Ä du œ cos ) d) •

' È du

5
u#

œ ln ¹u
È5
u# ¹
C œ ln ¹sin )
È5
sin# )¹
C

ŠWe used FORMULA 20 with a œ È5‹

49.

' È 3 dr

9r#  1

;”

' È du

u œ 3r Ä du œ 3 dr •

u#  1

œ ln ¹u
Èu#  1¹
C œ ln ¹3r
È9r#  1¹
C

(We used FORMULA 36 with a œ 1) 50.

' È13
dy9y

#

;”

u œ 3y Ä du œ 3 dy •

' È du

1
u#

œ ln ¹u
È1
u# ¹
C œ ln ¸3y
È1
9y# ¸
C

(We used FORMULA 20 with a œ 1) Ô t œ Èx × # 51. cos Ä 2 ' t cos" t dt œ 2 Š t# cos" t
x œ t# Õ dx œ 2t dt Ø (We used FORMULA 100 with a œ 1, n œ 1) œ t# cos" t
" sin" t  " tÈ1  t#
C

'

"

#

#

53.

#

t œ Èy × # Ä 2 ' t tan" t dt œ 2 ’ t# tan" t  y œ t# Õ dy œ 2t dt Ø (We used FORMULA 101 with n œ 1, a œ 1)

' tan" Èy dy; Ô œ t# tan" t  '

' Èt

#

1  t#

t#
1 t#
1

dt
'

dt 1
t#

" #

t# È1  t#

dt

" #

sin" Èx  "# Èx  x#
C

' 1
t t #

#

dt“ œ t# tan" t  '

t# 1
t#

dt

œ t# tan" t  t
tan" t
C œ y tan" Èy
tan" Èy  Èy
C

' sin& 2x dx œ  sin 2x5†#cos 2x
5 5 1 ' sin$ 2x dx œ  sin 2x10cos 2x
45 ’ sin 2x3†#cos 2x
3 3 1 ' sin 2x dx“ %

%

#

(We used FORMULA 60 with a œ 2, n œ 5 and a œ 2, n œ 3) % % 2 8 ˆ œ  sin 2x10cos 2x  15 sin# 2x cos 2x
15  "# ‰ cos 2x
C œ  sin 2x10cos 2x  54.

dt‹ œ t# cos" t
'

#

(We used FORMULA 33 with a œ 1) œ x cos" Èx
" sin" Èx  " ÈxÈ1  x
C œ x cos" Èx


52.

" #

Èx dx;

' sin& #) d) œ  sin 5† cos % ) #

" #

)

#




51 5

2 sin# 2x cos 2x 15

' sin$ #) d) œ  25 sin% #) cos #)
45 ’ sin 3† cos

ˆWe used FORMULA 60 with a œ "# , n œ 5 and a œ "# , n œ 3‰ 8 8 ˆ œ  25 sin% #) cos #)  15 sin# #) cos #)
15 2 cos #) ‰
C œ  25 sin%

# ) #

) #

cos

" #

) #



4 cos 2x 15


C

#




31 3

' sin #) d)“



8 15

sin#

) #

)

cos

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

) #



16 15

cos

) #


C

529

530 55.

Chapter 8 Techniques of Integration

'

8 cos% 21t dt œ 8 Š cos

$

21t sin 21t 4 †2 1




41 4

' cos# 21t dt‹

(We used FORMULA 61 with a œ 21, n œ 4) œ

cos$ 21t sin 21t 1


6 ’ #t


sin (2†21†t) 4†21 “


C

(We used FORMULA 59 with a œ 21) œ 56.

57.

'

cos$ 21t sin 21t 1


3t


3 sin 41t 41


Cœ

cos$ 21t sin 21t 1




3 cos 21t sin 21t 21

%

3y sin 3y 5 †3




51 5

' cos$ 3y dy‹

#

3y sin 3y 3†3




31 3

' cos 3y dy‹

3 cos& 3y dy œ 3 Š cos


3t
C

œ

cos% 3y sin 3y 5

œ

(We used FORMULA 61 with a œ 3, n œ 5 and a œ 3, n œ 3) " 4 8 % # 5 cos 3y sin 3y
15 cos 3y sin 3y
15 sin 3y
C




12 5

Š cos

' sin# 2) cos$ 2) d) œ sin #2(2)
cos3) 2)
33
 #1 ' sin# 2) cos 2) d) $

#

(We used FORMULA 69 with a œ 2, m œ 3, n œ 2) œ

58.

'

sin$ 2) cos# 2) 10




2 5

' sin# 2) cos 2) d) œ sin 2)10cos 2)
25 ’ "# ' sin# 2) d(sin 2))“ œ sin 2)10cos 2)
sin152)
C $

#

&Î#

9 sin$ ) cos$Î# ) d) œ 9 ’ sin3 )
cos ˆ3‰ #

)




#

31 3
ˆ #3 ‰

$

#

$

' sin ) cos$Î# ) d)“

œ 2 sin# ) cos&Î# )
4 ' cos$Î# ) sin ) d) ˆWe used FORMULA 68 with a œ 1, n œ 3, m œ 3# ‰

œ 2 sin# ) cos&Î# )  4 ' cos$Î# ) d(cos )) œ 2 sin# ) cos&Î# )  4 ˆ 25 cos&Î# )‰
C œ ˆ2 cos&Î# )‰ ˆsin# )
45 ‰
C 59.

'

2 sin# t sec% t dt œ ' 2 sin# t cos% t dt œ 2 Š sin2t cos 4

c$ t




21 24

' cos% t dt‹

(We used FORMULA 68 with a œ 1, n œ 2, m œ 4)

t œ sin t cos$ t  ' cos% t dt œ sin t cos$ t  ' sec% t dt œ sin t cos$ t  Š sec4 t tan 1
#

42 41

'

sec# t dt‹

(We used FORMULA 92 with a œ 1, n œ 4) œ sin t cos$ t  Š sec œ

2 3

#

t tan t ‹ 3



2 3

tan t
C œ

2 3

sec# t tan t 

2 3

tan t
C œ

2 3

tan t asec# t  1b
C

tan$ t
C

An easy way to find the integral using substitution:

' 2 sin# t cos% t dt œ '

60.

' œ

2 tan# t sec# t dt œ ' 2 tan# t d(tan t) œ

csc# y cos& y dy œ ' sin# y cos& y dy œ Š sin" y ‹ cos% y 3


43 

Š sin" y ‹ cos# y 32




31 3#

Š sin" y ‹ cos% y 5 2




51 5#

2 3

tan$ t
C

' sin# y cos$ y dy

' sin# y cos y dy

(We used FORMULA 69 with n œ 2, m œ 5, a œ 1 and n œ 2, m œ 3, a œ 1) œ

61.

Š sin" y ‹ cos% y 3


43 Š sin" y ‹ cos# y


8 3

y 8 ' sin# y d(sin y) œ 3cossin yy
43cos sin y  3 sin y
C %

#

' 4 tan$ 2x dx œ 4 Š tan2†#2x  ' tan 2x dx‹ œ tan# 2x  4' tan 2x dx #

(We used FORMULA 86 with n œ 3, a œ 2) œ tan# 2x  4# ln ksec 2xk
C œ tan# 2x  2 ln ksec 2xk
C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems 62.

' tan% ˆ x# ‰ dx œ

tan$ ˆ x# ‰ (4  1)

" #

 ' tan# ˆ x# ‰ dx œ

2 3

tan$ ˆ x# ‰  ' tan# ˆ x# ‰ dx

(We used FORMULA 86 with n œ 4, a œ "# ) œ

2 3

tan$

x #

 2 tan

x #


x
C

(We used FORMULA 84 with a œ "# ) 63.

' 8 cot% t dt œ 8 Š cot3 t  ' cot# t dt‹ $

(We used FORMULA 87 with a œ 1, n œ 4) œ 8 ˆ "3 cot$ t
cot t
t‰
C (We used FORMULA 85 with a œ 1) 64.

cot 2t ' 4 cot$ 2t dt œ 4 ’ 2(3 ' cot 2t dt“ œ  cot# 2t  4 ' cot 2t dt 1)  #

(We used FORMULA 87 with a œ 2, n œ 3) œ  cot# 2t  4# ln ksin 2tk
C œ  cot# 2t  2 ln ksin 2tk
C (We used FORMULA 83 with a œ 2)

65.

' 2 sec$ 1x dx œ 2 ’ sec11(3xtan1) 1x
33  12 ' sec 1x dx “ (We used FORMULA 92 with n œ 3, a œ 1) œ 1" sec 1x tan 1x
1" ln ksec 1x
tan 1xk
C (We used FORMULA 88 with a œ 1)

66.

' "# csc$ #x dx œ #" Š csc(3 cot1) " #

x #

x #




32 31

' csc #x dx‹

ˆWe used FORMULA 93 with a œ "# , n œ 3‰ œ "#
 csc x# cot x#  ln ¸csc x#
cot x# ¸‘
C œ  "# csc ˆWe used FORMULA 89 with a œ "# ‰ 67.

x #

cot

x #



" #

ln ¸csc

x #


cot x# ¸
C

' 3 sec% 3x dx œ 3 ’ sec3(43xtan1) 3x
44  12 ' sec# 3x dx“ #

(We used FORMULA 92 with n œ 4, a œ 3) œ

sec# 3x tan 3x 3




2 3

tan 3x
C

(We used FORMULA 90 with a œ 3) 68.

' csc% 3) d) œ  csc (4 cot1) " 3

# ) 3

)

3




42 41

'

csc#

) 3

d)

ˆWe used FORMULA 93 with n œ 4, a œ "3 ‰ œ  csc# 3) cot 3)  23 † 3 cot 3)
C œ  csc# ˆWe used FORMULA 91 with a œ "3 ‰ 69.

) 3

cot

) 3

 2 cot

) 3


C

' csc& x dx œ  csc5 xcot1 x
55  12 ' csc$ x dx œ  csc x4 cot x
34 Š csc3xcot1 x
33  21 ' csc x dx‹ $

$

(We used FORMULA 93 with n œ 5, a œ 1 and n œ 3, a œ 1) œ  "4 csc$ x cot x  38 csc x cot x  38 ln kcsc x
cot xk
C (We used FORMULA 89 with a œ 1) 70.

' sec& x dx œ sec5 xtan1 x
55  12 ' sec$ x dx œ sec x4 tan x
43 Š sec3xtan1 x
33  12 ' sec x dx‹ $

$

(We used FORMULA 92 with a œ 1, n œ 5 and a œ 1, n œ 3)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

531

532

Chapter 8 Techniques of Integration œ

" 4

sec$ x tan x


3 8

sec x tan x


3 8

ln ksec x
tan xk
C

(We used FORMULA 88 with a œ 1) 71.

' 16x$ (ln x)# dx œ 16 ’ x (ln4 x) %

#



'

2 4

%

#

%

x) x$ ln x dx“ œ 16 ’ x (ln4 x)  "# ’ x (ln  4

" 4

'

x$ dx““

(We used FORMULA 110 with a œ 1, n œ 3, m œ 2 and a œ 1, n œ 3, m œ 1) %

#

œ 16 Š x (ln4 x)  72.

'

(ln x)$ dx œ

x% (ln x) 8

x(ln x)$ 1






3 1

x% 3# ‹


C œ 4x% (ln x)#  2x% ln x


' (ln x)# dx œ x(ln x)$  3 ’ x(ln1 x)

#

x% #




C 2 1

'

ln x dx“ œ x(ln x)$  3x(ln x)#
'Š x ln1 x 

œ x(ln x)$  3x(ln x)#
6x ln x  6x
C (We used FORMULA 110 with n œ 0, a œ 1 and m œ 3, 2, 1) 73.

' xe3x dx œ e3

3x

(3x  1)
C œ

#

e3x 9

(3x  1)
C

(We used FORMULA 104 with a œ 3) 74.

'

xec2x dx œ

e
2x (2)#


2x

(2x  1)
C œ  e 4 (2x
1)
C

(We used FORMULA 104 with a œ 2) 75.

' x$ exÎ2 dx œ 2x$ exÎ2  3 † 2 '

x# exÎ2 dx œ 2x$ exÎ2  6 Š2x# exÎ2  2 † 2 ' xexÎ2 dx‹

œ 2x$ exÎ2  12x# exÎ2
24 † 4exÎ2 ˆ #x  1‰
C œ 2x$ exÎ2  12x# exÎ2
96exÎ2 ˆ x#  1‰
C ˆWe used FORMULA 105 with a œ "# twice and FORMULA 104 with a œ "# ‰ 76.

' x# e1x dx œ 1" x# e1x  12 ' xe1x dx (We used FORMULA 105 with n œ 2, a œ 1) 1x œ 1" x# e1x  1†21# † e1x (1x  1)
C œ 1" x# e1x  ˆ 2e1$ ‰ (1x  1)
C (We used FORMULA 104 with a œ 1)

77.

' x# 2x dx œ xln22

# x



2 ln 2

'

x2x dx œ

x# 2x ln #



2 ln 2

x

x2 Š ln 2 

" ln 2

' 2x dx‹ œ xln2#

# x



2 ln #

x

x2 ’ ln # 

2x (ln 2)# “


C

(We used FORMULA 106 with a œ 1, b œ 2, n œ 2, n œ 1) 78.

'

x# 2cx dx œ # cx

œ  xln2#


x# 2
x  ln 2

2 ln #




2 ln 2

cx

’ x2ln # 

'

x2cx dx œ

2cx (ln 2)# “

x # 2
x ln #




2 ln 2


x

Š x2 ln 2


" ln 2

'

2x dx‹


C

(We used FORMULA 106 with a œ 1, b œ 2, n œ 2, n œ 1) 79.

'

x1x dx œ

x1 x ln 1



" ln 1

'

x1 x ln 1

1x dx œ



" ln 1

ˆ ln1 1 ‰
C œ x

x1 x ln 1



1x (ln 1)#


C

(We used FORMULA 106 with n œ 1, b œ 1, a œ 1) 80.

'

x2

È2x

dx œ

È

x2 2x È2 ln 2



" È2 ln 2

'

È2x

2

dx œ

È

x2x 2 È2 ln 2



È

2x 2 #(ln 2)#


C

(We used FORMULA 106 with a œ È2, b œ 2, n œ 1)

' et sec$ aet  "b dt; ” x œ e

" x 2 ' Ä ' sec$ x dx œ sec3xtan
33 sec x dx 1 1 dx œ et dt • (We used FORMULA 92 with a œ 1, n œ 3) œ sec x#tan x
"# ln ksec x
tan xk
C œ "# csec aet  1b tan aet  1b
ln ksec aet  1b
tan aet  1bkd
C t

81.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" "

' dx‹

Section 8.6 Integral Tables and Computer Algebra Systems 82.

t œ È) × t cot t 32 ' csc t dt“ Ä 2 ' csc$ t dt œ 2 ’ csc3  ) œ t# 1
31 ) Õ d) œ 2t dt Ø (We used FORMULA 93 with a œ 1, n œ 3) œ 2
 csc t cot t  " ln kcsc t
cot tk‘
C œ  csc È) cot È)  ln ¹csc È)
cot È)¹
C

' cscÈÈ) d); Ô $

#

83.

#

'01 2Èx#
1 dx; cx œ tan td

Ä 2 '0

1Î4

sec t † sec# t dt œ 2'0

1Î4

1Î%

t‘ sec$ t dt œ 2 ”
sec3 t†tan 1 !




32 31

'01Î4 sec t dt•

(We used FORMULA 92 with n œ 3, a œ 1) 1Î% œ csec t † tan t
ln ksec t
tan tkd œ È2
ln ŠÈ2
1‹ !

84.

È3Î2

'0

dy a1  y# b&Î#

'01Î3

; cy œ sin xd Ä

cos x dx cos& x

x œ '0 sec% x dx œ ’ sec4 xtan 1 “ 1Î3

#

(We used FORMULA 92 with a œ 1, n œ 4) œ ’ sec

85.

#

'12 ar r1b #

œ




x tan x 3

$ ’ tan3 )

$Î#

2 3

tan x“

1Î$

dr; cr œ sec )d Ä

 tan )
)“

!

!




42 41

'01Î3 sec# x dx

œ ˆ 34 ‰ È3
ˆ 32 ‰ È3 œ 2È3

!

1Î$

1Î$

1Î$ 1Î3 ) ) '01Î3 tan ' 1Î3 tan% ) d) œ ’ tan  '0 tan# ) d) sec ) (sec ) tan )) d) œ 0 41 “ $

!

3È 3 3

œ

$

1 3

 È3


œ

1 3

(We used FORMULA 86 with a œ 1, n œ 4 and FORMULA 84 with a œ 1) 86.

È3

'01Î

dt at#
1b(Î#

œ ’ cos

%

œ ’ cos

%

; ct œ tan )d Ä '0

1Î6

1Î' ) sin ) “ 5 ! ) sin ) 5





45 ”’ cos

#

1Î' ) sin ) “ 3 !

cos# ) sin )


4 15

sec# ) d) sec( )

8 15

œ '0

1Î6


ˆ 3 3 1 ‰'0

cos& ) d) œ ’ cos

1Î6

sin )“

%

1Î' ) sin ) “ 5 !

1Î6

cos$ ) d)

cos ) d)•

1Î' !

(We used FORMULA 61 with a œ 1, n œ 5 and a œ 1, n œ 3) È % # Š #3 ‹ ˆ #" ‰ 4 ‰ È3 8 ‰ ˆ"‰ 9 " 4 3†9
48
32†4 œ
ˆ 15 œ Š # ‹ ˆ "# ‰
ˆ 15 5 # œ 160
10
15 œ 480 87.


ˆ 5 5 1 ‰ '0

203 480

' "8 sinh& 3x dx œ "8 Š sinh 3x5†3cosh 3x  5 5 1 ' sinh$ 3x dx‹ %

œ

sinh% 3x cosh 3x 120



" 10

3x  Š sinh 3x3†cosh 3

31 3

' sinh 3x dx‹

(We used FORMULA 117 with a œ 3, n œ 5 and a œ 3, n œ 3) % 3x 2 ˆ" ‰ œ sinh 3x1#cosh  sinh 3x90cosh 3x
30 0 3 cosh 3x
C œ

88.

sinh% 3x cosh 3x 

' coshÈxÈx dx; – %

œ

89.

" 1#0

cosh$ u sinh u #

" 90

sinh 3x cosh 3x


1 45

cosh 3x
C

u œ Èx cosh$ u sinh u ' %
— Ä 2 cosh u du œ 2 Š 4 du œ 2dx Èx

41 4

' cosh# u du‹


3# ˆ sinh4 2u
u# ‰
C

œ

(We used FORMULA 118 with a œ 1, n œ 4 and FORMULA 116 with a œ 1) " $È x sinh Èx
38 sinh 2Èx
34 Èx
C # cosh

'

x# cosh 3x dx œ

x# 3

sinh 3x 

2 3

' x sinh 3x dx œ x3

#

sinh 3x  32 Š 3x cosh 3x 

" 3

' cosh 3x dx‹

(We used FORMULA 122 with a œ 3, n œ 2 and FORMULA 121 with a œ 3, n œ 1)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

533

534

Chapter 8 Techniques of Integration œ

90.

'

x# 3

sinh 3x 

2x 9

x sinh 5x dx œ

cosh 3x


2 27

sinh 3x
C

cosh 5x 

" 25

sinh 5x
C

x 5

(We used FORMULA 119 with a œ 5) 91.

' sech( x tanh x dx œ  sech7 x
C (

(We used FORMULA 135 with a œ 1, n œ 7) 92.

' csch$ 2x coth 2x dx œ  csch3†#2x
C œ  csch6 2x
C $

$

(We used FORMULA 136 with a œ 2, n œ 3) ub a

93. u œ ax
b Ê x œ

' (axx
dxb)

#

œ'

(u  b) du au# a

œ

Ê dx œ " a#

du a

;

' ˆ "u  ub ‰ du œ a"
ln kuk
bu ‘
C œ a"
ln kax
bk
ax b
b ‘
C #

#

#

94. x œ a tan ) Ê a#
x# œ a# sec# ) Ê 2x dx œ 2a# sec# ) tan ) Ê dx œ a sec# ) d);

'

'

' secd) ) œ 2a" ' (1
cos 2)) d) œ 2a"

# dx ) œ aaa# sec d) œ a"$ aa #
x # b # sec# )b# œ #"a$ a)
sin ) cos )b
C œ 2a"$

œ

" 2a$

" x a

–tan




x # a Š1
xa# ‹ —

#

$

sin ) ‰ # ‘
)
ˆ cos ) cos )
C œ


Cœ

x 2a# aa#
x# b




" 2a$

tan"

x a

" 2a$

$

ˆ)


ˆ)


" #

sin 2)‰
C

tan ) ‰ 1
tan# )


C


C

95. x œ a sin ) Ê a#  x# œ a# cos# ) Ê 2x dx œ 2a# cos ) sin ) d) Ê dx œ a cos ) d); ' Èa#  x# dx œ ' a cos )(a cos )) d) œ a# ' cos# ) d) œ a# ' (1
cos 2)) d) œ a# ˆ)
sin 2) ‰
C #

œ

a# #

œ

#

a #

a)
cos ) sin )b
C œ sin"

x a

a# #

Š)
È1  sin# ) † sin )‹
C œ

a# #

Šsin" xa




# È a#  x# a

#

†

x a‹


C


x# Èa#  x#
C

96. x œ a sec ) Ê x#  a# œ a# tan# ) Ê 2x dx œ 2a# tan ) sec# ) d) Ê dx œ a sec ) tan ) d); ' dx œ ' a#tan )# sec ) d) œ ' # d) œ "# ' cos ) d) œ "# sin )
C œ "# È1  cos# )
C aa sec )b a tan )

x # È x #  a#

œ ˆ a"# ‰ 97.

'

É cos"# )  1 ˆ cos" ) ‰


C œ ˆ a"# ‰

a sec )

Èsec# )  1 sec )

a

a


C œ ˆ a"# ‰

#

É x#  " a ˆ xa ‰


Cœ

xn sin ax dx œ  ' xn ˆ "a ‰ d(cos ax) œ (cos ax) xn ˆ "a ‰
n

œ  xa cos ax


n a

'

a

" a

È x #  a# a# x


C

' cos ax † nxn1 dx

xn1 cos ax dx

ŠWe used integration by parts ' u dv œ uv  ' v du with u œ xn , v œ  "a cos ax‹ 98.

' xn (ln ax)m dx œ ' œ

xnb1 (ln ax)m n
1



m n
1

nb1

(ln ax)m d Š nx
1 ‹ œ

'

xnb1 (ln ax)m n
1

 ' Š xn
1 ‹ m(ln ax)m1 ˆ x" ‰ dx nb1

xn (ln ax)m1 dx, n Á "

ŠWe used integration by parts ' u dv œ uv  ' v du with u œ (ln ax)m , v œ 99.

' xn sin" ax dx œ ' œ

xn
1 n
1

sin" ax 

nb1

sin" ax d Š nx
1 ‹ œ

a n
1

' Èx


n 1 dx 1  a# x#

xnb1 n
1

xnb1 n
1‹

sin" ax  ' Š nx
1 ‹ È1 a (ax)# dx nb1

, n Á 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems ŠWe used integration by parts ' u dv œ uv  ' v du with u œ sin" ax, v œ 100.

' xn tan" ax dx œ ' œ

xn
1 n
1

tan" ax 

nb1

tan" ax d Š nx
1 ‹ œ

a n
1

xnb1 n
1‹

tan" ax  ' Š nx
1 ‹ 1
a(ax)# dx

xnb1 n
1

nb1

' 1x

a#dxx# , n Á 1 n 1

ŠWe used integration by parts ' u dv œ uv  ' v du with u œ tan" ax, v œ

È2

'0

101. S œ

xnb1 n
1 ‹

21yÈ1
(yw )# dx

È2

œ 21 '0

È x#
2 É 1


È2

œ 2È21 '0 œ 2È21 ’ x

x# x#
2

dx

Èx#
1 dx

È x#
1 #




" #

ln ¹x
Èx#
1¹“

È2 0

(We used FORMULA 21 with a œ 1) œ È21 ’È6
ln ŠÈ2
È3‹“ œ 21È3
1È2 ln ŠÈ2
È3‹

È3Î2

102. L œ '0

È3Î2

È1
(2x)# dx œ 2 ' 0

É "4
x# dx œ 2 ’ x# É 4"
x#
ˆ 4" ‰ ˆ #" ‰ ln Šx
É 4"
x# ‹“

ˆWe used FORMULA 2 with a œ "# ‰ " 4

œ ’ x# È1
4x#
œ

È3 4

(2)


103. A œ '0

3

" 4

dx È x
1

ln Š

ln Šx
#" È1
4x# ‹“

È3 #

" 4


1‹


ln 2 œ

$

œ ’2Èx
1“ œ 2; x œ !

œ

" A

'0 Èx
1 dx  '0

œ

" #

$ † 23
(x
1)$Î# ‘ !  1 œ

3

" A

3

È3 #

" A

È3Î2 0




" 4

œ

È3 4

É1
4 ˆ 43 ‰


" 4

ln Š

È3 #

ln ŠÈ3
2‹

'03 Èxxdx
1

;

(We used FORMULA 11 with a œ 1, b œ 1, n œ 1 and a œ 1, b œ 1, n œ 1) yœ

" #A

'03 x dx
1 œ 4" cln (x
1)d $! œ 4" ln 4 œ #" ln 2 œ ln È2

104. My œ '0 x ˆ 2x36
3 ‰ dx œ 18 '0 3

3

2x
3 2x
3

dx  54'0

3

dx 2x
3

œ c18x  27 ln k2x
3kd $!

œ 18 † 3  27 ln 9  (27 ln 3) œ 54  27 † 2 ln 3
27 ln 3 œ 54  27 ln 3 105. S œ 21 'c1 x# È1
4x# dx; 1

u œ 2x ” du œ 2 dx • Ä œ

1 4

1 4

'c22 u# È1
u# du

’ u8 a1
2u# bÈ1
u# 

" 8

ln Šu
È1
u# ‹“

(We used FORMULA 22 with a œ 1) œ 1 ’ 2 (1
2 † 4)È1
4  " ln Š2
È1
4‹ 4

8


(1
2 † 4)È1
4
œ

1 4

’ 9# È5 

# #

8

2 8

" 8

" 8

0


"# É1
4 ˆ 34 ‰‹ 

dx Èx
1 4 3

È3Î2

ln Š2
È1
4‹“

È

ln Š 22

È55 ‹“ ¸ 7.62

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 4

ln

" #

535

536

Chapter 8 Techniques of Integration

106. (a) The volume of the filled part equals the length of the tank times the area of the shaded region shown in the accompanying figure. Consider a layer of gasoline of thickness dy located at height y where r  y  r
d. The width of this layer is crbd

2Èr#  y# . Therefore, A œ 2 'cr crbd

and V œ L † A œ 2L 'cr

crbd

Èr#  y# dy

Èr#  y# dy

crbd

# yÈ r  y (b) 2L 'cr Èr#  y# dy œ 2L ’ #
r# sin" yr “ cr (We used FORMULA 29 with a œ r) # # # r) È ‰ È2rd  d#
Š r# ‹ ˆsin" ˆ d r r ‰
1# ‰“ œ 2L ’ (d  2rd  d#
r# sin" ˆ d r r ‰
r# ˆ 1# ‰“ œ 2L ’ˆ d-r # # #

#

107. The integrand f(x) œ Èx  x# is nonnegative, so the integral is maximized by integrating over the function's entire domain, which runs from x œ 0 to x œ 1 Ê

'01 Èx  x# dx œ '01 É2 † "# x  x# dx œ ” ˆx #

"‰ #

É2 † "# x  x#


ˆWe used FORMULA 48 with a œ "# ‰ œ’

ˆx  "# ‰ #

Èx  x#


" 8

"

" 8

sin" (2x  1)“ œ !

†

1 #

 8" ˆ 1# ‰ œ

ˆ "# ‰# #

sin" Š

x  "# " #

"

‹•

!

1 8

108. The integrand is maximized by integrating g(x) œ xÈ2x  x# over the largest domain on which g is nonnegative, namely [!ß 2] Ê

'02 xÈ2x  x# dx œ ’ (x
1)(2x 63)È2x  x

#




" #

sin" (x  1)“

(We used FORMULA 51 with a œ ") œ "# † 1#  "# ˆ 1# ‰ œ 1#

# !

CAS EXPLORATIONS 109.

Example CAS commands: Maple: q1 := Int( x*ln(x), x ); q1 = value( q1 ); q2 := Int( x^2*ln(x), x ); q2 = value( q2 ); q3 := Int( x^3*ln(x), x ); q3 = value( q3 ); q4 := Int( x^4*ln(x), x ); q4 = value( q4 ); q5 := Int( x^n*ln(x), x ); q6 := value( q5 ); q7 := simplify(q6) assuming n::integer; q5 = collect( factor(q7), ln(x) );

# (a) # (b) # (c) # (d) # (e)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.6 Integral Tables and Computer Algebra Systems 110.

111.

Example CAS commands: Maple: q1 := Int( ln(x)/x, x ); q1 = value( q1 ); q2 := Int( ln(x)/x^2, x ); q2 = value( q2 ); q3 := Int( ln(x)/x^3, x ); q3 = value( q3 ); q4 := Int( ln(x)/x^4, x ); q4 = value( q4 ); q5 := Int( ln(x)/x^n, x ); q6 := value( q5 ); q7 := simplify(q6) assuming n::integer; q5 = collect( factor(q7), ln(x) ); Example CAS commands: Maple: q := Int( sin(x)^n/(sin(x)^n+cos(x)^n), x=0..Pi/2 ); q = value( q ); q1 := eval( q, n=1 ): q1 = value( q1 ); for N in [1,2,3,5,7] do q1 := eval( q, n=N ); print( q1 = evalf(q1) ); end do: qq1 := PDEtools[dchange]( x=Pi/2-u, q, [u] ); qq2 := subs( u=x, qq1 ); qq3 := q + q = q + qq2; qq4 := combine( qq3 ); qq5 := value( qq4 ); simplify( qq5/2 );

537

# (a) # (b) # (c) # (d) # (e)

# (a) # (b)

# (c)

109-111.Example CAS commands: Mathematica: (functions may vary) In Mathematica, the natural log is denoted by Log rather than Ln, Log base 10 is Log[x,10] Mathematica does not include an arbitrary constant when computing an indefinite integral, Clear[x, f, n] f[x_]:=Log[x] / xn Integrate[f[x], x] For exercise 111, Mathematica cannot evaluate the integral with arbitrary n. It does evaluate the integral (value is 1/4 in each case) for small values of n, but for large values of n, it identifies this integral as Indeterminate 109. (e)

' xn ln x dx œ x nb
ln1 x  n
" 1 ' xn dx, n Á 1 n 1

œ 110. (e)

(We used FORMULA 110 with a œ 1, m œ 1) xnb1 ln x xnb1 xnb1 ˆ " ‰ n
1  (n
1)#
C œ n
1 ln x  n
1
C

' xn ln x dx œ xcnb
ln1 x  (n)"
1 ' xn dx, n Á 1 n 1

(We used FORMULA 110 with a œ 1, m œ 1, n œ n)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

538

Chapter 8 Techniques of Integration œ

x1cn ln x 1n



" 1 n

1cn

Š 1x n ‹
C œ

x1cn 1n

ˆln x 

" ‰ 1n


C

111. (a) Neither MAPLE nor MATHEMATICA can find this integral for arbitrary n. (b) MAPLE and MATHEMATICA get stuck at about n œ 5. (c) Let x œ 1#  u Ê dx œ  du; x œ 0 Ê u œ 1# , x œ 1# Ê u œ 0; I œ '0

1Î2

sinn x dx sinn x
cosn x

œ '1Î2 sinn ˆ 1  u‰ #
cosn ˆ 1  u‰ œ '0 #

x
cos x ‰ ' Ê I
I œ '0 ˆ sin sinn x
cosn x dx œ 0 1Î2

n

1Î2

 sinn ˆ 1  u‰ du

0

1Î2

n

#

1 #

dx œ

cosn u du cosn u
sinn u

Ê Iœ

œ '0

1Î2

cosn x dx cosn x
sinn x

1 4

8.7 NUMERICAL INTEGRATION 1.

'12 x dx I.

(a) For n œ 4, ?x œ b n a œ 2 4 1 œ "4 Ê ! mf(xi ) œ 12 Ê T œ " (12) œ 3 ; 8 #

?x #

" 8

œ

;

f(x) œ x Ê f w (x) œ 1 Ê f ww œ 0 Ê M œ 0 Ê kE T k œ 0

(b)

'12 x dx œ ’ x2 “ # œ 2  "# œ 3#

(c)

kET k True Value

#

"

(b)

'12 x dx œ 3#

(c)

kES k True Value

21 " ?x 4 œ 4 Ê 3 " 3 12 (18) œ # ;

Ê kES k œ '1 x dx  S œ 2

œ

" 12

3 #



3 #

œ0

" 2

Ê

?x #

" 4

œ

;

f(x) œ 2x  1 Ê f w (x) œ 2 Ê f ww œ 0 Ê M œ 0 Ê kE T k œ 0

(b)

'13 (2x  1) dx œ cx#  xd $" œ (9  3)  (1  1) œ 6

(c)

kET k True Value

" 2

Ê

?x 3

" 6

œ

'1 (2x  1) dx œ 6

'11 ax#
1b dx I.

m 1 4 2 4 1

mf(xi ) 1 5 3 7 2

x! x" x# x$ x%

xi 1 3/2 2 5/2 3

f(xi ) 1 2 3 4 5

m 1 2 2 2 1

mf(xi ) 1 4 6 8 5

x! x" x# x$ x%

xi 1 3/2 2 5/2 3

f(xi ) 1 2 3 4 5

m 1 4 2 4 1

mf(xi ) 1 8 6 16 5

x! x" x# x$ x%

xi 1 1/2 0 1/2 1

f(xi ) 2 5/4 1 5/4 2

m 1 2 2 2 1

mf(xi ) 2 5/2 2 5/2 2

;

Ê kES k œ '1 (2x  1) dx  S 3

œ66œ0

3.

f(xi ) 1 5/4 3/2 7/4 2

3

(x) œ 0 Ê M œ 0 Ê kES k œ 0

3

kES k True Value

xi 1 5/4 3/2 7/4 2

Ê kET k œ '1 (2x  1) dx  T œ 6  6 œ 0

6

(c)

x! x" x# x$ x%

‚ 100 œ 0%

II. (a) For n œ 4, ?x œ b n a œ 3 4 1 œ 24 œ ! mf(xi ) œ 36 Ê S œ " (36) œ 6 ; (b)

mf(xi ) 1 5/2 3 7/2 2

;

‚ 100 œ 0%

(a) For n œ 4, ?x œ b n a œ 3 4 1 œ 24 œ ! mf(xi ) œ 24 Ê T œ " (24) œ 6 ; 4

f

m 1 2 2 2 1

2

'13 (2x  1) dx

Ð%Ñ

f(xi ) 1 5/4 3/2 7/4 2

Ê kET k œ '1 x dx  T œ 0

f Ð%Ñ (x) œ 0 Ê M œ 0 Ê kES k œ 0

I.

xi 1 5/4 3/2 7/4 2

‚ 100 œ 0%

II. (a) For n œ 4, ?x œ b n a œ ! mf(xi ) œ 18 Ê S œ

2.

x! x" x# x$ x%

‚ 100 œ 0%

(a) For n œ 4, ?x œ b n a œ 1  4(1) œ 24 œ ! mf(xi ) œ 11 Ê T œ " (11) œ 2.75à 4

" #

Ê

?x #

œ

" 4

à

f(x) œ x#
1 Ê f w (x) œ 2x Ê f ww (x) œ 2 Ê M œ 2 1) ˆ " ‰# " Ê kET k Ÿ 11( # # (2) œ 1# or 0.08333

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.7 Numerical Integration (b)

'11 ax#
1b dx œ ’ x3

$


x“

" "

œ ˆ 13
1‰  ˆ 13  1‰ œ

8 3

Ê ET œ '1 ax#
1b dx  T œ 1

8 3



11 4

539

œ  1"#

Ê kET k œ ¸ 1"# ¸ ¸ 0.08333 (c)

kET k True Value

"

‚ 100 œ Š 18# ‹ ‚ 100 ¸ 3% 3

II. (a) For n œ 4, ?x œ b n a œ 1  4(1) œ 24 œ "# Ê ?3x œ "6 à ! mf(xi ) œ 16 Ê S œ " (16) œ 8 œ 2.66667 ; 6

3

f Ð$Ñ (x) œ 0 Ê f Ð%Ñ (x) œ 0 Ê M œ 0 Ê kES k œ 0 (b)

'11 ax#
1b dx œ ’ x3

$


x“

" "

œ

8 3

Ê kES k œ '1 ax#
1b dx  S œ 1

(c) 4.

kES k True Value

8 3



(a) For n œ 4, ?x œ b n a œ 0  4(2) œ ! mf(xi ) œ 3 Ê T œ " (3) œ 3 ; 4

" #

œ

2 4

?x #

Ê

œ

" 4

4

f(x) œ x#  1 Ê f w (x) œ 2x Ê f ww (x) œ 2 2) ˆ " ‰# " Ê M œ 2 Ê kET k Ÿ 01( # # (2) œ 1# œ 0.08333

(b)

'02 ax#  1b dx œ ’ x3 -  x“ ! $

kET k True Value

#

" 12

Ê kE T k œ (c)

œ 0  ˆ 83
2‰ œ

3

f

5.

6

6

(x) œ 0 Ê f

Ð%Ñ

;

(x) œ 0 Ê M œ 0 Ê kES k œ 0

'2 ax#  1b dx œ 23 2 2 3  3 œ0 kES k True Value ‚ 100

2 3

Ê kES k œ '2 ax#  1b dx  S 0

(a) For n œ 4, ?x œ b n a œ 2 4 0 œ 24 œ "2 " ! Ê ?x mf(ti ) œ 25 Ê T œ 4" (25) œ # œ 4 ;

25 4

f(t) œ t$
t Ê f w (t) œ 3t#
1 Ê f ww (t) œ 6t 0 ˆ " ‰# Ê M œ 12 œ f ww (2) Ê kET k Ÿ 212 # (12) œ

(b)

'02 at$
tb dt œ ’ t4
t# “ # œ Š 24

(c)

kET k True Value

%

‚ 100 œ

¸ "4 ¸ 6

#

%

!




2# #‹

; " #

(t) œ 6 Ê f

Ð%Ñ

'0 at$
tb dt œ 6 2

(b)

kES k True Value

m 1 2 2 2 1

mf(xi ) 3 5/2 0 3/2 1

2 3

Ê ET œ '2 ax#  1b dx  T œ 0

x! x" x# x$ x%

xi 2 3/2 1  1/2 0

t! t" t# t$ t%

ti 0 1/2 1 3/2 2

f(xi ) 3 5/4 0 3/4 1

2 3

m 1 4 2 4 1

f(ti ) 0 5/8 2 39/8 10

m 1 2 2 2 1

2

" 2

Ê

?x 3

œ

" 6

;

(t) œ 0 Ê M œ 0 Ê kES k œ 0

Ê kES k œ '0 at$
tb dt  S

œ66œ0 (c)

f(xi ) 3 5/4 0 3/4 1

 0 œ 6 Ê kET k œ '0 at$
tb dt  T œ 6 

6

f

xi 2 3/2 1  1/2 0



3 4

" œ  12

mf(xi ) 3 5 0 3 1

mf(ti ) 0 5/4 4 39/4 10 25 4

œ  4" Ê kET k œ

‚ 100 ¸ 4%

II. (a) For n œ 4, ?x œ b n a œ 2 4 0 œ 24 œ ! mf(ti ) œ 36 Ê S œ " (36) œ 6 ; Ð$Ñ

x! x" x# x$ x%

œ 0%

'02 at$
tb dt I.

mf(xi ) 2 5 2 5 2

3

œ

(c)

m 1 4 2 4 1

"

0

(b)

f(xi ) 2 5/4 1 5/4 2

‚ 100 œ Š 122 ‹ ‚ 100 ¸ 13%

II. (a) For n œ 4, ?x œ b n a œ 0  4(2) œ 24 œ "# Ê ?x œ " ; ! mf(xi ) œ 4 Ê S œ " (4) œ Ð$Ñ

xi 1 1/2 0 1/2 1

‚ 100 œ 0%

'02 ax#  1b dx I.

œ0

8 3

x! x" x# x$ x%

2

t! t" t# t$ t%

ti 0 1/2 1 3/2 2

f(ti ) 0 5/8 2 39/8 10

m 1 4 2 4 1

‚ 100 œ 0%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

mf(ti ) 0 5/2 4 39/2 10

" 4

540 6.

Chapter 8 Techniques of Integration

'11 at$
1b dt I.

t! t" t# t$ t%

(a) For n œ 4, ?x œ b n a œ 1  4(1) œ 24 œ "# " ! mf(ti ) œ 8 Ê T œ " (8) œ 2 ; Ê ?x # œ 4 ; 4

f(t) œ t$
1 Ê f w (t) œ 3t# Ê f ww (t) œ 6t 1) ˆ " ‰# Ê M œ 6 œ f ww (1) Ê kET k Ÿ 11( # # (6) œ

(b)

'11 at$
1b dt œ ’ t4
t“ "

(c)

kET k True Value

%

3

6

6

'11 at$
1b dt œ 2 œ22œ0

kES k True Value

7.

'12 s"

#

I.

Ê kES k œ '1 at$
1b dt  S 1

t! t" t# t$ t%

ti 1 1/2 0 1/2 1

f(ti ) 0 7/8 1 9/8 2

s! s" s# s$ s%

si 1 5/4 3/2 7/4 2

f(si ) 1 16/25 4/9 16/49 1/4

m 1 4 2 4 1

mf(ti ) 0 7/2 2 9/2 2

‚ 100 œ 0%

ds ba n

(a) For n œ 4, ?x œ ! mf(si ) œ

179,573 44,100

œ

21 4

Ê Tœ

œ

" 8

" 4

Ê

?x #

œ

Š 179,573 44,100 ‹ œ

" 8

;

179,573 352,800

" 2 w s# Ê f (s) œ  s$ 6 Ê f ww (s) œ s% Ê M œ 6 œ f ww (1) # Ê kET k Ÿ 2 1# " ˆ "4 ‰ (6) œ 3"# œ 0.03125 2 2 # " ds œ s# ds œ
 1s ‘ " œ  #"  ˆ 1" ‰ # 1 s 1

¸ 0.50899; f(s) œ

'

(b)

mf(ti ) 0 7/4 2 9/4 2

‚ 100 œ 0%

f Ð$Ñ (t) œ 6 Ê f Ð%Ñ (t) œ 0 Ê M œ 0 Ê kES k œ 0

(c)

m 1 2 2 2 1 1

%

II. (a) For n œ 4, ?x œ b n a œ 1  4(1) œ 24 œ "# Ê ?x œ " ; ! mf(ti ) œ 12 Ê S œ " (12) œ 2 ; (b)

f(ti ) 0 7/8 1 9/8 2

œ Š 14
1‹  Š (41)
(1)‹ œ 2 Ê kET k œ '1 at$
1b dt  T œ 2  2 œ 0 %

"

" 4

ti 1 1/2 0 1/2 1

'

œ

1 #

Ê ET œ '1

2

" s#

ds  T œ

m 1 2 2 2 1

" #

mf(si ) 1 32/25 8/9 32/49 1/4

 0.50899 œ 0.00899

Ê kET k œ 0.00899 kET k True Value

(c)

‚ 100 œ

II. (a) For n œ 4, ?x œ ! mf(si ) œ

0.00899 0.5 ‚ 100 ¸ ba 21 " n œ 4 œ 4

264,821 44,100

Ê Sœ

" 12

2% Ê

?x 3

Š 264,821 44,100 ‹

Ð%Ñ ¸ 0.50042; f Ð$Ñ (s) œ  24 (s) œ s& Ê f

" 12

œ œ

;

264,821 529,200

120 s'

%

"¸ ˆ"‰ Ê M œ 120 Ê kES k Ÿ ¸ 2180 4 (120)

œ

'1

" 384

¸ 0.00260

'2

2

" 1 " " s# ds œ # Ê ES œ 1 s# ds  S œ # kES k 0.0004 True Value ‚ 100 œ 0.5 ‚ 100 ¸ 0.08%

(b) (c) 8.

'24 (s "1) I.

#

ba n

œ

42 4

œ

" 2

Ê

?x #

œ

" 4

450

Ê Tœ

" 4

ˆ 1269 ‰ 450 œ

f(s) œ (s  1) Ê f ww (s) œ Ê kE T k Ÿ

'2

4

(b)

" (s  1)#

#

œ 0.70500;

Ê f (s) œ  (s 2 1)$

6 (s  1)% 42 1#

1269 1800 w

f(si ) 1 16/25 4/9 16/49 1/4

m 1 4 2 4 1

mf(si ) 1 64/25 8/9 64/49 1/4

;

s! s" s# s$ s%

si 2 5/2 3 7/2 4

f(si ) 1 4/9 1/4 4/25 1/9

m 1 2 2 2 1

mf(si ) 1 8/9 1/2 8/25 1/9

Ê Mœ6

ˆ "# ‰# (6) œ %

" 4

œ 0.25

ˆ " ‰ ˆ " ‰ ds œ ’ (s"  1) “ œ 4  1  #1 œ

Ê kET k ¸ 0.03833

si 1 5/4 3/2 7/4 2

 0.50042 œ 0.00042 Ê kES k œ 0.00042

ds

(a) For n œ 4, ?x œ ! mf(si ) œ 1269

s! s" s# s$ s%

#

2 3

Ê ET œ '2

4

" (s  1)#

ds  T œ

2 3

 0.705 ¸ 0.03833

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.7 Numerical Integration (c)

kET k True Value

‚ 100 œ

II. (a) For n œ 4, ?x œ ! mf(si ) œ 1813

0.03833 ˆ 23 ‰ ‚ 100 ¸ ba 42 " n œ 4 œ 2

6% Ê

?x 3

" 6

œ

450 ‰ 1813 Ê S œ ˆ 1813 450 œ 2700 ¸ 0.67148; Ð%Ñ f Ð$Ñ (s) œ (s24 (s) œ (s120 1)& Ê f 1)' Ê M œ % 2 ˆ"‰ " Ê kES k Ÿ 4180 # (120) œ 12 ¸ 0.08333

; s! s" s# s$ s%

" 6

9.

(b)

'24 (s "1)

(c)

kES k True Value

#

ds œ

Ê ES œ '2

4

2 3

‚ 100 œ

0.00481 ˆ 23 ‰

" (s  1)#

120

ds  S ¸

2 3

I.

1 8

10 4

1 4

Ê

?x #

œ

1 8

t! t" t# t$ t%

;

Š2
2È2‹ ¸ 1.89612;

f(t) œ sin t Ê f w (t) œ cos t Ê f ww (t) œ sin t # 1$ Ê M œ 1 Ê kET k Ÿ 11# 0 ˆ 14 ‰ (1) œ 19 # ¸ 0.16149

(b)

'01 sin t dt œ [cos t] 1! œ (cos 1)  (cos 0) œ 2

(c)

kET k True Value

‚ 100 œ

0.10388 ‚ 100 ¸ 2 ba 10 1 n œ 4 œ 4

II. (a) For n œ 4, ?x œ ! mf(ti ) œ 2
4È2 ¸ 7.6569 Ê Sœ

1 12

Ê

?x 3

œ

1 12

f Ð$Ñ (t) œ cos t Ê f Ð%Ñ (t) œ sin t  0 ˆ 1 ‰% Ê M œ 1 Ê kES k Ÿ 1180 4 (1) ¸ 0.00664

10.

'01 sin t dt œ 2

(c)

kES k True Value

‚ 100 œ

1 8

0.00456 2

m 1 2 2 2 1

Ê kET k œ '0 sin t dt  T ¸ 2  1.89612 œ 0.10388 1

; t! t" t# t$ t%

ti 0 1/4 1/2 31/4 1

f(ti ) 0 È2/2 1 È2/2 0

m 1 4 2 4 1

1 4

Ê

?x #

œ

1 8

;

Š2
2È2‹ ¸ 0.60355; f(t) œ sin 1t

t! t" t# t$ t%

ti 0 1/4 1/2 3/4 1

f(ti ) 0 È2/2 1 È2/2 0

'01 sin 1t dt œ [ 1" cos 1t] "! œ ˆ 1" cos 1‰  ˆ 1" cos 0‰ œ 12 ¸ 0.63662 2 1  0.60355 œ 0.03307 kET k 0.03307 True Value ‚ 100 œ ˆ 12 ‰ ‚ 100

m 1 2 2 2 1

mf(ti ) 0 È 2 2 2 È 2 2 0

mf(ti ) 0 È2 2 È2 0

Ê kET k œ '0 sin 1t dt  T

¸

(c)

mf(ti ) 0 È2 2 È2 0

‚ 100 ¸ 0%

Ê f w (t) œ 1 cos 1t Ê f ww (t) œ 1# sin 1t Ê M œ 1# # Ê kET k Ÿ 1 1# 0 ˆ 14 ‰ a1# b ¸ 0.05140

(b)

f(ti ) 0 È2/2 1 È2/2 0

1

(a) For n œ 4, ?x œ b n a œ 1 4 0 œ ! mf(ti ) œ 2
2È2 ¸ 4.828 Ê Tœ

ti 0 1/4 1/2 31/4 1

Ê ES œ '0 sin t dt  S ¸ 2  2.00456 œ 0.00456 Ê kES k ¸ 0.00456

'01 sin 1t dt I.

mf(si ) 1 16/9 1/2 16/25 1/9

5%

Š2
4È2‹ ¸ 2.00456;

(b)

m 1 4 2 4 1

‚ 100 ¸ 1%

(a) For n œ 4, ?x œ œ œ ! mf(ti ) œ 2
2È2 ¸ 4.8284 Ê Tœ

f(si ) 1 4/9 1/4 4/25 1/9

 0.67148 œ 0.00481 Ê kES k ¸ 0.00481

'01 sin t dt ba n

si 2 5/2 3 7/2 4

¸ 5%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

541

542

Chapter 8 Techniques of Integration

II. (a) For n œ 4, ?x œ b n a œ 1 4 0 œ 14 Ê ! mf(ti ) œ 2
4È2 ¸ 7.65685 Ê Sœ

1 12

?x 3

œ

1 12

t! t" t# t$ t%

Š2
4È2‹ ¸ 0.63807;

f Ð$Ñ (t) œ 1$ cos 1t Ê f Ð%Ñ (t) œ 1% sin 1t  0 ˆ 1 ‰% Ê M œ 1% Ê kES k Ÿ 1180 a1% b ¸ 0.00211 4 (b)

'01 sin 1t dt œ 12 ¸ 0.63662

(c)

kES k True Value

‚ 100 œ

0.00145 ˆ 12 ‰

ti 0 1/4 1/2 3/4 1

;

Ê ES œ '0 sin 1t dt  S ¸ 1

2 1

f(ti ) 0 È2/2 1 È2/2 0

m 1 4 2 4 1

mf(ti ) 0 È 2 2 2 È 2 2 0

 0.63807 œ 0.00145 Ê kES k ¸ 0.00145

‚ 100 ¸ 0%

" 11. (a) n œ 8 Ê ?x œ "8 Ê ?#x œ 16 ; ! mf(xi ) œ 1(0.0)
2(0.12402)
2(0.24206)
2(0.34763)
2(0.43301)
2(0.48789)
2(0.49608)


2(0.42361)
1(0) œ 5.1086 Ê T œ " 8

?x 3

" 16

(5.1086) œ 0.31929

" 24

(b) n œ 8 Ê ?x œ Ê œ ; ! mf(xi ) œ 1(0.0)
4(0.12402)
2(0.24206)
4(0.34763)
2(0.43301)
4(0.48789)
2(0.49608)
4(0.42361)
1(0) œ 7.8749 Ê S œ

" 24

(7.8749) œ 0.32812

(c) Let u œ 1  x# Ê du œ 2x dx Ê  "# du œ x dx; x œ 0 Ê u œ 1, x œ 1 Ê u œ 0

'01 xÈ1  x# dx œ '10 Èu ˆ "# du‰ œ "# '01 u"Î# du œ ’ 2" Š u ET œ '0 xÈ1  x# dx  T ¸ 1

" 3

$Î# 3 #

"

"

‹“ œ
3" u$Î# ‘ ! œ !

" 3

 0.31929 œ 0.01404; ES œ '0 xÈ1  x# dx  S ¸ 1

$

$

ŠÈ1‹  3" ŠÈ0‹ œ " 3

" 3

;

 0.32812 œ 0.00521

3 12. (a) n œ 8 Ê ?x œ 38 Ê ?#x œ 16 ; ! mf()i ) œ 1(0)
2(0.09334)
2(0.18429)
2(0.27075)
2(0.35112)
2(0.42443)
2(0.49026)


2(0.58466)
1(0.6) œ 5.3977 Ê T œ ?x 3

3 16

(5.3977) œ 1.01207

" 8

(b) n œ 8 Ê ?x œ Ê œ ; ! mf()i ) œ 1(0)
4(0.09334)
2(0.18429)
4(0.27075)
2(0.35112)
4(0.42443)
2(0.49026) 3 8


4(0.58466)
1(0.6) œ 8.14406 Ê S œ " #

#

(c) Let u œ 16
) Ê du œ 2) d) Ê

'03

) È16
)#

ET œ '0

3

d) œ '16

25

) È16
)#

" Èu

ˆ #" du‰ œ

" #

" 8

(8.14406) œ 1.01801

du œ ) d); ) œ 0 Ê u œ 16, ) œ 3 Ê u œ 16
3# œ 25

'1625 u"Î# du œ ’ 2" Š u

"Î# 1 #

‹“

d)  T ¸ 1  1.01207 œ 0.01207; ES œ '0

3

#& "'

œ È25  È16 œ 1;

) È16
)#

d)  S ¸ 1  1.01801 œ 0.01801

1 13. (a) n œ 8 Ê ?x œ 18 Ê ?#x œ 16 ; ! mf(ti ) œ 1(0.0)
2(0.99138)
2(1.26906)
2(1.05961)
2(0.75)
2(0.48821)
2(0.28946)
2(0.13429)


1(0) œ 9.96402 Ê T œ 1 8

?x 3

1 16

(9.96402) ¸ 1.95643

1 24

(b) n œ 8 Ê ?x œ Ê œ ; ! mf(ti ) œ 1(0.0)
4(0.99138)
2(1.26906)
4(1.05961)
2(0.75)
4(0.48821)
2(0.28946)
4(0.13429)
1(0) œ 15.311 Ê S ¸

1 24

(15.311) ¸ 2.00421

(c) Let u œ 2
sin t Ê du œ cos t dt; t œ  1# Ê u œ 2
sin ˆ 1# ‰ œ 1, t œ

'c1Î2 1Î2

3 cos t (2
sin t)#

ET œ '

1Î2

dt œ '1

3

3 u#

Ê u œ 2
sin

du œ 3 '1 u# du œ ’3 Š u1 ‹“ œ 3 ˆ "3 ‰  3 ˆ 11 ‰ œ 2; 3


"

$ "

1Î2

dt  T ¸ 2  1.95643 œ 0.04357; ES œ '
1Î2 ¸ 2  2.00421 œ 0.00421 3 cos t c1Î2 (2
sin t)#

1 #

3 cos t (2
sin t)#

dt  S

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 #

œ3

Section 8.7 Numerical Integration

543

1 1 14. (a) n œ 8 Ê ?x œ 32 Ê ?#x œ 64 ; ! mf(yi ) œ 1(2.0)
2(1.51606)
2(1.18237)
2(0.93998)
2(0.75402)
2(0.60145)
2(0.46364)


2(0.31688)
1(0) œ 13.5488 Ê T ¸ 1 32

?x 3

1 64

(13.5488) œ 0.66508

1 96

(b) n œ 8 Ê ?x œ Ê œ ; ! mf(yi ) œ 1(2.0)
4(1.51606)
2(1.18237)
4(0.93988)
2(0.75402)
4(0.60145)
2(0.46364)
4(0.31688)
1(0) œ 20.29734 Ê S ¸ (c) Let u œ cot y Ê du œ csc# y dy; y œ

'1Î4

1Î2

(20.29734) œ 0.66423

Ê u œ 1, y œ 1

ET œ '1Î4 acsc yb Ècot y dy  T ¸ 2 3

1 96

1 #

Ê uœ0

acsc# yb Ècot y dy œ '1 Èu ( du) œ '0 u"Î# du œ ’ u 3 “ œ 0

1Î2

¸

1 4

#

2 3

$Î#

"

2

!

$

$

ŠÈ1‹  23 ŠÈ0‹ œ 32 ;

2 3

 0.66508 œ 0.00159; ES œ '1Î4 acsc# yb Ècot y dy  S 1Î2

 0.66423 œ 0.00244

15. (a) M œ 0 (see Exercise 1): Then n œ 1 Ê ?x œ 1 Ê kET k œ

" 1#

(1)# (0) œ 0  10% " #

(b) M œ 0 (see Exercise 1): Then n œ 2 (n must be even) Ê ?x œ 16. (a) M œ 0 (see Exercise 2): Then n œ 1 Ê ?x œ 2 Ê kET k œ

2 1#

Ê kE S k œ

Ê kET k Ÿ

2 n

2 12

ˆ 2n ‰# (2) œ

4 3n#

Ê kET k Ÿ

2 n

2 12

ˆ 2n ‰# (2) œ

4 3n#

Ê kET k Ÿ

2 n

2 12

ˆ 2n ‰# (12) œ

8 n#

2 n

Ê kET k Ÿ

2 12

ˆ 2n ‰# (6) œ

4 n#

1 n

Ê kET k Ÿ

1 12

ˆ 1n ‰# (6) œ

1 2n#

2 180

a10% b Ê n  É 43 a10% b

(1)% (0) œ 0  10% 4 3

a10% b Ê n  É 43 a10% b

(1)% (0) œ 0  10%

2 180

(1)% (0) œ 0  10%

 10% Ê n#  4 a10% b Ê n  È4 a10% b

œ 200, so let n œ 201 (b) M œ 0 (see Exercise 6): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 21. (a) M œ 6 (see Exercise 7): Then ?x œ

2 180

4 3

 10% Ê n#  8 a10% b Ê n  È8 a10% b

Ê n  282.8, so let n œ 283 (b) M œ 0 (see Exercise 5): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 20. (a) M œ 6 (see Exercise 6): Then ?x œ

(1)% (0) œ 0  10%

 10% Ê n# 

Ê n  115.4, so let n œ 116 (b) M œ 0 (see Exercise 4): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 19. (a) M œ 12 (see Exercise 5): Then ?x œ

2 180

 10% Ê n# 

Ê n  115.4, so let n œ 116 (b) M œ 0 (see Exercise 3): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 18. (a) M œ 2 (see Exercise 4): Then ?x œ

ˆ #" ‰% (0) œ 0  10%

(2)# (0) œ 0  10%

(b) M œ 0 (see Exercise 2): Then n œ 2 (n must be even) Ê ?x œ " Ê kES k œ 17. (a) M œ 2 (see Exercise 3): Then ?x œ

" 180

2 180

(1)% (0) œ 0  10%

 10% Ê n# 

" #

a10% b Ê n  É "# a10% b

Ê n  70.7, so let n œ 71 (b) M œ 120 (see Exercise 7): Then ?x œ

" n

Ê kES k œ

1 180

ˆ "n ‰% (120) œ

2 3n%

 10% Ê n% 

2 3

a10% b

Ê n  %É 23 a10% b Ê n  9.04, so let n œ 10 (n must be even) 22. (a) M œ 6 (see Exercise 8): Then ?x œ

2 n

Ê kET k Ÿ

2 12

ˆ 2n ‰# (6) œ

4 n#

 10% Ê n#  4 a10% b Ê n  È4 a10% b

Ê n  200, so let n œ 201 (b) M œ 120 (see Exercise 8): Then ?x œ

2 n

Ê kES k Ÿ

2 180

ˆ 2n ‰% (120) œ

64 3n%

 10% Ê n% 

% Ê n  %É 64 3 a10 b Ê n  21.5, so let n œ 22 (n must be even)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

64 3

a10% b

544

Chapter 8 Techniques of Integration

23. (a) f(x) œ Èx
1 Ê f w (x) œ Then ?x œ

Ê kET k Ÿ

3 n

" #

3 12

(x
1)"Î# Ê f ww (x) œ  4" (x
1)$Î# œ 

ˆ 3n ‰# ˆ 4" ‰ œ

9 16n#

 10% Ê n# 

so let n œ 76 (Î# (b) f Ð$Ñ (x) œ 38 (x
1)&Î# Ê f Ð%Ñ (x) œ  15 œ 16 (x
1) Ê kE S k Ÿ

3 180

15 ‰ ˆ n3 ‰% ˆ 16 œ

3& (15) 16(180)n%

Ê Mœ

"

$

4 ŠÈ1‹

" 4

œ

.

9 a10% b Ê n  É 16 a10% b Ê n  75,

Ê Mœ

15 ( 16 ˆÈx
1‰

3& (15) ˆ10% ‰ 16(180)

 10% Ê n% 

9 16

" $ 4 ˆÈ x
1 ‰

%

15 ( 16 ŠÈ1‹

&

œ

. Then ?x œ

15 16

3 n

%

(15) a10 b Ê n  É 3 16(180) Ê n  10.6, so let

n œ 12 (n must be even) 24. (a) f(x) œ

Ê f w (x) œ  #" (x
1)$Î# Ê f w w (x) œ

" È x
1

Then ?x œ

Ê kET k Ÿ

3 n

3 12

ˆ 3n ‰# ˆ 34 ‰ œ

n œ 130 (Î# (b) f Ð$Ñ (x) œ  15 Ê f Ð%Ñ (x) œ 8 (x
1) Ê kE S k Ÿ

3 180

ˆ n3 ‰% ˆ 105 ‰ 16 œ

3& (105) 16(180)n%

3% 48n#

105 16

3 4

(x
1)&Î# œ

 10% Ê n# 

(x
1)*Î# œ

3% ˆ10% ‰ 48

105 * 16 ˆÈx
1‰

3& (105) ˆ10% ‰ 16(180)

 10% Ê n% 

3 & 4 ˆÈ x
1 ‰

Ê Mœ

Ê n  É3 Ê Mœ %

%

a10% b 48

105

&

&

œ

3 4

.

Ê n  129.9, so let

œ

*

16 ŠÈ1‹

3

4 ŠÈ1‹

105 16

. Then ?x œ

3 n

%

a10 b Ê n  É 3 (105) Ê n  17.25, so 16(180)

let n œ 18 (n must be even) 25. (a) f(x) œ sin (x
1) Ê f w (x) œ cos (x
1) Ê f ww (x) œ sin (x
1) Ê M œ 1. Then ?x œ œ

8 12n#

 10% Ê n# 

8 ˆ10% ‰ 12

32 ˆ10% ‰ 180

%

2 12

ˆ n2 ‰# (1) œ

27.

5 2 a6.0

32 ˆ10% ‰ 180

ˆ 2n ‰# (1)

Ê kES k Ÿ

2 n

2 180

ˆ 2n ‰% (1) œ

32 180n%

 10%

%

a10 b Ê n  É 32180 Ê n  6.49, so let n œ 8 (n must be even)

8 12n#

 10% Ê n# 

8 ˆ10% ‰ 12

%

2 n

%

Ê n  É 8 a110# b Ê n  81.6, so let n œ 82

(b) f Ð$Ñ (x) œ sin (x
1) Ê f Ð%Ñ (x) œ cos (x
1) Ê M œ 1. Then ?x œ Ê n% 

2 12

%

26. (a) f(x) œ cos (x
1) Ê f w (x) œ sin (x
1) Ê f ww (x) œ cos (x
1) Ê M œ 1. Then ?x œ Ê kE T k Ÿ

Ê k ET k Ÿ

Ê n  É 8 a110# b Ê n  81.6, so let n œ 82

(b) f Ð$Ñ (x) œ cos (x
1) Ê f Ð%Ñ (x) œ sin (x
1) Ê M œ 1. Then ?x œ Ê n% 

2 n

2 n

Ê kE S k Ÿ

2 180

ˆ 2n ‰% (1) œ

32 180n%

 10%

%

a10 b Ê n  É 32180 Ê n  6.49, so let n œ 8 (n must be even)


2a8.2b
2a9.1bÞ Þ Þ
2a12.7b
13.0ba30b œ 15,990 ft3 .

200 28. (a) Using Trapezoid Rule, ?x œ 200 Ê ?x 2 œ 2 œ 100; ! mf(xi ) œ 13,180 Ê Area ¸ 100 (13,180)

œ 1,318,000 ft# . Since the average depth œ 20 ft we obtain Volume ¸ 20 (Area) ¸ 26,360,000 ft$ . (b) Now, Number of fish œ Volume 1000 œ 26,360 (to the nearest fish) Ê Maximum to be caught œ 75% of 26,360 œ 19,770 Ê Number of licenses œ 19,770 #0 œ 988

x! x" x# x$ x% x5 x' x( x)

xi 0 200 400 600 800 1000 1200 1400 1600

f(xi ) 0 520 800 1000 1140 1160 1110 860 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

m 1 2 2 2 2 2 2 2 1

mf(xi ) 0 1040 1600 2000 2280 2320 2220 1720 0

Section 8.7 Numerical Integration

545

29. Use the conversion 30 mph œ 44 fps (ft per sec) since time is measured in seconds. The distance traveled as the car accelerates from, say, 40 mph œ 58.67 fps to 50 mph œ 73.33 fps in (4.5  3.2) œ 1.3 sec is the area of the trapezoid (see figure) associated with that time interval: "# (58.67
73.33)(1.3) œ 85.8 ft. The total distance traveled by the Ford Mustang Cobra is the sum of all these eleven trapezoids (using ?t # and the table below):

s œ (44)(1.1)
(102.67)(0.5)
(132)(0.65)
(161.33)(0.7)
(190.67)(0.95)
(220)(1.2)
(249.33)(1.25)
(278.67)(1.65)
(308)(2.3)
(337.33)(2.8)
(366.67)(5.45) œ 5166.346 ft ¸ 0.9785 mi v (mph) 0 30 40 50 60 70 80 90 100 110 120 v (fps) 0 44 58.67 73.33 88 102.67 117.33 132 146.67 161.33 176 t (sec) 0 2.2 3.2 4.5 5.9 7.8 10.2 12.7 16 20.6 26.2 ?t/2 0 1.1 0.5 0.65 0.7 0.95 1.2 1.25 1.65 2.3 2.8 30. Using Simpson's Rule, ?x œ b n a œ 24 6 0 œ 24 6 œ 4; 4 1400 ! myi œ 350 Ê S œ (350) œ ¸ 466.7 in.# 3

3

" 31. Using Simpson's Rule, ?x œ 1 Ê ?x 3 œ 3 ; ! myi œ 33.6 Ê Cross Section Area ¸ " (33.6) 3

œ 11.2 ft# . Let x be the length of the tank. Then the Volume V œ (Cross Sectional Area) x œ 11.2x. Now 5000 lb of gasoline at 42 lb/ft$ $ Ê V œ 5000 42 œ 119.05 ft Ê 119.05 œ 11.2x Ê x ¸ 10.63 ft

32.

24 0.019 2 c

x! x" x# x$ x% x5 x'

xi 0 4 8 12 16 20 24

yi 0 18.75 24 26 24 18.75 0

m 1 4 2 4 2 4 1

myi 0 75 48 104 48 75 0

x! x" x# x$ x% x5 x'

xi 0 1 2 3 4 5 6

yi 1.5 1.6 1.8 1.9 2.0 2.1 2.1

m 1 4 2 4 2 4 1

myi 1.5 6.4 3.6 7.6 4.0 8.4 2.1

130 190.67 37.1 5.45


2a0.020b
2a0.021b
Þ Þ Þ
2a0.031b
0.035 d œ 4.2 L

33. (a) kES k Ÿ

ba % 180 a?x b M; n 1 ?x 1 8 Ê 3 œ #4 ;

œ 4 Ê ?x œ

1 #

0 4

œ

1 8

; ¸f Ð%Ñ ¸ Ÿ 1 Ê M œ 1 Ê kES k Ÿ

(b) ?x œ ! mf(xi ) œ 10.47208705 Ê Sœ

1 #4

x! x" x# x$ x%

(10.47208705) ¸ 1.37079

xi 0 1/8 1/4 31/8 1/2

ˆ 1#  0‰ 180

ˆ 18 ‰% (1) ¸ 0.00021

f(xi ) 1 0.974495358 0.900316316 0.784213303 0.636619772

‰ (c) ¸ ˆ 0.00021 1.37079 ‚ 100 ¸ 0.015% 34. (a) ?x œ

ba n

2 ae0 30È1

(b) kES k Ÿ

10 10

œ

0.01


4e

10 180

œ 0.1 Ê erfa1b œ 0.04


2e

0.09


4e

2 ˆ 0.1 ‰ È3 3 ay0


4y1
2y2
4y3
Þ Þ Þ
4y9
y10 b

0.81


Þ Þ Þ
4e


e1 b ¸ 0.843

a0.1b4 a12b ¸ 6.7 ‚ 106

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

m 1 4 2 4 1

mf(x1i ) 1 3.897981432 1.800632632 3.136853212 0.636619772

546

Chapter 8 Techniques of Integration

0 1 35. (a) n œ 10 Ê ?x œ 110 œ 10 Ê ?#x œ #10 ; ! mf(xi ) œ 1(0)
2(0.09708)
2(0.36932)
2(0.76248)
2(1.19513)
2(1.57080)
2(1.79270)


2(1.77912)
2(1.47727)
2(0.87372)
1(0) œ 19.83524 Ê T œ (b) 1  3.11571 ¸ 0.02588 (c) With M œ 3.11, we get kET k Ÿ

1 1#

1 ‰# ˆ 10 (3.11) œ

1$ 1200

1 #0

(19.83524) œ 3.11571

(3.11)  0.08036

36. (a) f ww (x) œ 2 cos x  x sin x Ê f Ð$Ñ (x) œ 3 sin x  x cos x Ê f Ð%Ñ (x) œ 4 cos x
x sin x. From the graphs shown below, k4 cos x
x sin xk  4.8 for 0 Ÿ x Ÿ 1.

1 1 ˆ 1 ‰% (b) n œ 10 Ê ?x œ 10 Ê kES k Ÿ 180 10 (4.8) ¸ 0.00082 (c) ! mf(xi ) œ 1(0)
4(0.09708)
2(0.36932)
4(0.76248)
2(1.19513)
4(1.57080)
2(1.79270)


4(1.77912)
2(1.47727)
4(0.87372)
1(0) œ 30.0016 Ê S œ

1 30

(30.0016) œ 3.14176

(d) k1  3.14176k ¸ 0.00017 37. T œ Tœ

?x ba 2 ay0
2y1
2y2
2y3
Þ Þ Þ
2yn1
yn b where ?x œ n and f b  a ay0
y1
y1
y2
y2
Þ Þ Þ
ync1
ync1
yn b œ b n a Š fax0 b
2 fax1 b
fax1 b
2 fax2 b n 2

Since f is continuous on each interval [xk1 , xk ], and [xk1 , xk ] with fack b œ

faxkc1 b
faxk b ; #

faxkc1 b
faxk b #

is continuous on [a, b]. So
ÞÞÞ


faxnc1 b
faxn b ‹. 2

is always between faxk1 b and faxk b, there is a point ck in

this is a consequence of the Intermediate Value Theorem. Thus our sum is

n

n

k œ1

k œ1

! ˆ b  a ‰fack b which has the form ! ?xk fack b with ?xk œ n

ba n

for all k. This is a Riemann Sum for f on [a, b].

?x b a 3 ay0
4y1
2y2
4y3
Þ Þ Þ
2yn2
4yn1
yn b where n is even, ?x œ n and S œ b n a ˆ y0
4y3 1
y2
y2
4y33
y4
y4
4y3 5
y6
Þ Þ Þ
ync2
4y3 nc1
yn ‰ œ b n a Š fax0 b
4fa6x1 b
fax2 b
fax2 b
4fa6x3 b
fax4 b
fax4 b
4fa6x5 b
fax6 b
Þ Þ Þ
faxnc2 b
4fa6xnc1 b
faxn b ‹ 2

38. S œ

fax2k b
4fax2kb1 b
fax2kb2 b 6

f is continuous on [a, b]. So

is the average of the six values of the continuous function on the interval [x2k , x2k
2 ], so it is between

the minimum and maximum of f on this interval. By the Extreme Value Theorem for continuous functions, f takes on its maximum and minimum in this interval, so there are xa and xb with x2k Ÿ xa , xb Ÿ x2k
2 and faxa b Ÿ

fax2k b
4fax2kb1 b
fax2kb2 b 6

Ÿ faxb b. By the Intermediate Value Theorem, there is ck in [x2k , x2k
2 ] with

fack b œ

fax2k b
4fax2kb1 b
fax2kb2 b . 6

So our sum has the form ! ?xk fack b with ?xk œ

n/2

k œ1

Exercises 39-42 were done using a graphing calculator with n œ 50 39. 1.08943 40. 1.37076 41. 0.82812

ba an/2b ß

a Riemann sum for f on [a, b].

42. 51.05400

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.7 Numerical Integration 43. (a) T10 ¸ 1.983523538 T100 ¸ 1.999835504 T1000 ¸ 1.999998355 (b) n k ET k œ 2  Tn 10 0.016476462 œ 1.6476462 ‚ 102 100 1.64496 ‚ 104 1000 1.646 ‚ 106

44. (a) S10 ¸ 2.000109517 S100 ¸ 2.000000011 S1000 ¸ 2.000000000 (b) n k ES k œ 2  Sn 10 1.09517 ‚ 104 100 1.1 ‚ 108 1000 0

(c) k ET10n k ¸ 102 k ETn k (d) b  a œ 1, a?xb2 œ k ETn k Ÿ

1 1 12 Š n2 ‹ 2

k ET10n k Ÿ

13 12a10nb2

œ

(c) k ES10n k ¸ 104 k ESn k 12 n2 ,

(d) b  a œ 1, a?xb4 œ

Mœ1

1 12n2 3

k ESn k Ÿ

Ÿ 102 k ETn k

1 1 180 Š n4 ‹

k ES10n k Ÿ

4

15 180a10nb4

œ

14 n4 ,

Mœ1

1 180n4 5

Ÿ 104 k ESn k

45. (a) f w axb œ 2x cosax2 b, f ww axb œ 2x † a2xbsinax2 b
2cosax2 b œ 4x2 sinax2 b
2cosax2 b (b)

(c) The graph shows that 3 Ÿ f ww axb Ÿ 2 so k f ww axbk Ÿ 3 for 1 Ÿ x Ÿ 1. (d) k ET k Ÿ

1a1b 2 12 a?xb a3b

(e) For 0  ?x  0.1, k ET k (f) n  

1a1b ?x

 

2 0.1

a ?x b 2 # a? x b 2 Ÿ #

œ

Ÿ

0.12 2

œ 0.005  0.01

œ 20

46. (a) f ''' axb œ 4x2 † 2x cosax2 b  8x sinax2 b  4x sinax2 b œ 8x3 cosax2 b  12x sinax2 b f a4b axb œ 8x3 † 2x sinax2 b  24x2 cosax2 b  12x † 2x cosax2 b  12 sinax2 b œ a16x4  12bsinax2 b  48 x2 cosax2 b (b)

(c) The graph shows that 30 Ÿ f a4b axb Ÿ 0 so ¸ f a4b axb¸ Ÿ 30 for 1 Ÿ x Ÿ 1. (d) k ES k Ÿ

1  a1b 4 180 a?xb a30b

(e) For 0  ?x  0.4, k ES k Ÿ (f) n  

1a1b ?x

 

2 0.4

a ?x b 4 3 2 a? x b 4 Ÿ 0.43 3

œ

¸ 0.00853  0.01

œ5 2

47. (a) Using d œ C1 , and A œ 1ˆ d2 ‰ œ

C2 41

yields the following areas (in square inches, rounded to the nearest tenth):

2.3, 1.6, 1.5, 2.1, 3.2, 4.8, 7.0, 9.3, 10.7, 10.7, 9.3, 6.4, 3.2 (b) If Cayb is the circumference as a function of y, then the area of a cross section is 2

Aayb œ 1Š Cay2bÎ1 ‹ œ

C2 ayb 41 ,

and the volume is

1 41

'06 C2 ayb dy.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

547

548

Chapter 8 Techniques of Integration

(c)

'06 Aayb dy œ 411 '06 C2 ayb dy ¸

1 ˆ60‰ 2 41 24 c5.4 3


2a4.52
4.42
5.12
6.32
7.82
9.42
10.82
11.62
11.62
10.82
9.02 b
6.32 d

¸ 34.7 in (d) V œ

1 41

'06 C2 ayb dy ¸ 411 ˆ 6 36 0 ‰’5.42
4a4.52 b
2a4.42 b
4a5.12 b
2a6.32 b
4a7.82 b
2a9.42 b
%a10.82 b
2a11.62 b
%a11.62 b
2a10.82 b
%a9.02 b
6.32 “ œ 34.792 in3

by Simpson's Rule. The Simpson's Rule estimate should be more accurate than the trapezoid estimate. The error in the Simpson's estimate is proportional to a?yb4 œ 0.0625 whereas the error in the trapezoid estimate is proportional to a?yb2 œ 0.25, a larger number when ?y œ 0.5 in. 48. (a) Displacement Volume V ¸

?x 3 ay0


4y1
2y2
4y3
Þ Þ Þ
2yn2
4yn1
yn b, x0 œ 0, xn œ 10  ?x,

?x œ 2.54, n œ 10 Ê 'x Aaxb dx ¸ xn 0

2.54 3 ’0


4a14.00b
2a9.21b
4a3.24b
0“ œ


4a1.07b
2a3.84b
4a7.82b
2a12.20b
4a15.18b
2a16.14b

2.54 3 a248.02b

œ 209.99 ¸ 210 ft3 .

(b) The weigth of water displaced is approximately 64 † 120 œ 13,440 lb. (c) The volume of a prism œ a2.54ba16.14b œ 409.96 ¸ 410 ft3 . Thus, the prismatic coefficient is 49. (a) a œ 1, e œ

" #

Ê Length œ 4 '0 É1  1Î2

œ 2 '0 È4  cos# t dt œ '0 1Î2

1Î2

1 #0 .

cos# t dt

f(t) dt; use the

Trapezoid Rule with n œ 10 Ê ?t œ œ

" 4

ba n

œ

ˆ 1# ‰  0 10

'01Î2 È4  cos# t dt ¸ ! mf(xn ) œ 37.3686183

Ê Tœ

10

?t #

nœ0

(37.3686183) œ

1 40

(37.3686183)

œ 2.934924419 Ê Length œ 2(2.934924419) ¸ 5.870 (b) kf ww (t)k  1 Ê M œ 1 Ê kE T k Ÿ 50. ?x œ ÊS

ba 1#

a?t# Mb Ÿ

ˆ 1# ‰  0 1#

ˆ #10 ‰# 1 Ÿ 0.0032

10 1 ?x 1 ! mf(xi ) 8 œ 8 Ê 3 œ 24 ; 1 œ 24 a29.18480779b ¸ 3.82028

xi 0 1/20 1/10 31/20 1/5 1/4 31/10 71/20 21/5 91/20 1/2

f(xi ) 1.732050808 1.739100843 1.759400893 1.790560631 1.82906848 1.870828693 1.911676881 1.947791731 1.975982919 1.993872679 2

m 1 2 2 2 1 1 2 2 2 2 1

mf(xi ) 1.732050808 3.478201686 3.518801786 3.581121262 3.658136959 3.741657387 3.823353762 3.895583461 3.951965839 3.987745357 2

x! x" x# x$ x% x5 x' x( x)

xi 0 1/8 1/4 31/8 1/2 51/8 31/4 71/8 1

f(xi ) 1.414213562 1.361452677 1.224744871 1.070722471 1 1.070722471 1.224744871 1.361452677 1.414213562

m 1 4 2 4 2 4 2 4 1

mf(xi ) 1.414213562 5.445810706 2.449489743 4.282889883 2 4.282889883 2.449489743 5.445810706 1.414213562

œ 29.184807792

#

20

91 # 400

cos# ˆ 3#10 x‰ Ê L œ '0 É1
20

91 # 400

25

#

1# 4

sin# ˆ 150x ‰ Ê L œ

dy dx

œ

31 #0

cos ˆ 3#10 x‰ Ê Š dy dx ‹

cos# ˆ 3#10 x‰ dx. Using numerical integration we find L ¸ 21.07 in

52. First, we'll find the length of the cosine curve: L œ 'c25 Ê1
Š dy dx ‹ dx; Ê Š dy dx ‹ œ

¸ 0.51.

x! x" x# x$ x% x5 x' x( x) x* x"!

51. The length of the curve y œ sin ˆ 3#10 x‰ from 0 to 20 is: L œ '0 Ê1
Š dy dx ‹ dx; œ

210 ft3 410 ft3

'c2525 É1
14

#

#

dy dx

œ  25501 sin ˆ 150x ‰

sin# ˆ 150x ‰ dx. Using a numerical integrator we find

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

#

Section 8.8 Improper Integrals L ¸ 73.1848 ft. Surface area is: A œ length † width ¸ (73.1848)(300) œ 21,955.44 ft. Cost œ 1.75A œ (1.75)(21,955.44) œ $38,422.02. Answers may vary slightly, depending on the numerical integration used. 53. y œ sin x Ê

dy dx

# È1
cos# x dx; a numerical integration gives ' œ cos x Ê Š dy dx ‹ œ cos x Ê S œ 0 21(sin x)

œ

x #

#

1

S ¸ 14.4 54. y œ

x# 4

Ê

dy dx

55. y œ x
sin 2x Ê S œ 2'0

21Î3

56. y œ œ

2

#

x# 4

dx; a numerical integration gives S ¸ 5.28

#

21(x
sin 2x) È1
(1
2 cos 2x)# dx; a numerical integration gives S ¸ 54.9

" a362x# b 1# È36x# 6

Ê S œ '0 21 Š x4 ‹ É1


x# 4

# œ 1
2 cos 2x Ê Š dy dx ‹ œ (1
2 cos 2x) ; by symmetry of the graph we have that

dy dx

È36  x# Ê

x 1#

œ '0

#

Ê Š dy dx ‹ œ

œ

dy dx




x 12

#

a18  x# b 6È36  x#

œ

È36  x# 12

†

" (2x) # È36  x# #

a18  x# b 36 a36  x# b

dy Ê Š dx ‹ œ

È36  x# 12

œ

Ê S œ '0

6



21†x 12

x# 12È36  x#

œ

" a36  x#  x# b È36  x# 1#

È36  x# É1


a18  x# b# 36 a36  x# b

dx

Éa36  x# b
ˆ 18  x# ‰# dx; using numerical integration we get S ¸ 41.8 6

1x 6

57. A calculator or computer numerical integrator yields sin" 0.6 ¸ 0.643501109. 58. A calculator or computer numerical integrator yields 1 ¸ 3.1415929. 8.8 IMPROPER INTEGRALS 1.

'0_ x dx
1 œ

2.

'1_ xdx

3.

'01 Èdxx œ

bÄ!

4.

'04 È dx

œ lim c

5.

'c11

6.

'c18 xdx

7.

'01 È dx

8.

'01 r dr

#

lim

bÄ_

œ lim

1Þ001

bÄ_

4x

lim b

#

'1b xdx

1Þ001

lim ctan" xd 0 œ lim atan" b  tan" 0b œ b

bÄ_

bÄ_

bÄ_

lim

b Ä 0b

bÄ_

0œ

1 #

Þ

"

c2x1Î2 d b œ lim b Š2  2Èb‹ œ 2  0 œ 2 bÄ0

'0b a4  xb"Î# dx œ
'0

1 #

‰ œ 1000 œ lim c1000x
0Þ001 d "b œ lim ˆ b1000 0 001
1000

'b1 x"Î# dx œ

bÄ4

œ 'c1

'0b x dx
1 œ

lim

b Ä 4c

’2È4  b  Š2È4‹“ œ 0
4 œ 4

" b œ lim c
3x"Î$ ‘ c1
lim b
3x"Î$ ‘ c bÄ! cÄ! œ lim c
3b"Î$  3(1)"Î$ ‘
lim b
3(1)"Î$  3c"Î$ ‘ œ (0
3)
(3  0) œ 6 bÄ! cÄ! dx x#Î$

0

dx x#Î$

1

'0 œ 'c8 xdx "Î$


dx x#Î$

" b œ lim c
#3 x#Î$ ‘ c)
lim

3# x#Î$ ‘ c bÄ! cÄ! œ lim c
3# b#Î$  3# (8)#Î$ ‘
lim b
3# (1)#Î$  3# c#Î$ ‘ œ
0  3# (4)‘
ˆ 3#  0‰ œ  9# bÄ! cÄ! "Î$

0

dx x"Î$

œ lim c csin" xd 0 œ lim c asin" b  sin" 0b œ bÄ1 bÄ1 b

1  x#

0Þ999

1

1 #

0œ

1 #

œ lim b c1000r0 001 d 1b œ lim b a1000  1000b0 001 b œ 1000  0 œ 1000 bÄ! bÄ! Þ

Þ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

549

550 9.

Chapter 8 Techniques of Integration


2 2 dx 'c_ '
2 dx '
2 dx x  1 œ
_ x  1 
_ x
1 œ

œ

cln kx  1kd
b 2 

lim

#

b Ä
_

3¸ ˆln ¸  ¸ b  1 ¸‰ œ ln 3  ln Š lim 1  ln b
1

lim

b Ä
_

b"

10.

'c2_ x2
dx4 œ

b Ä
_

11.

'2_ v2 dvv œ

bÄ_

12.

'2_ t 2dt1 œ

bÄ_

13.

_ 'c_

lim

#


tan" #x ‘ 2 œ b

lim

b Ä
_

lim

b Ä
_

1 ¸‘
#
ln ¸ xx 
1 b

œ ln 3  ln 1 œ ln 3

‹ b Ä
_ b
1

lim

b Ä
_

cln kx
1kd
b 2 œ

ˆtan" 1  tan" b2 ‰ œ

1 4

 ˆ 1# ‰ œ

31 4

b lim
2 ln ¸ v v " ¸‘ # œ lim ˆ2 ln ¸ b b " ¸  2 ln ¸ 2 # " ¸‰ œ 2 ln (1)  2 ln ˆ "# ‰ œ 0
2 ln 2 œ ln 4

#

bÄ_

 " ¸‘ b  "¸ ˆln ¸ bb
¸ 2  " ¸‰ œ ln (1)  ln ˆ "3 ‰ œ 0
ln 3 œ ln 3 lim
ln ¸ tt
1 # œ lim 1  ln #
1

#

bÄ_

_

dx œ '
_ ax2x
'0 #
1 b# 0

2x dx ax #
1 b #

2x dx a x #
1 b#

;”

'_1 duu
'1_ duu œ

u œ x#
1 Ä du œ 2x dx •

#

1
 u" ‘ c lim
 u" ‘ b
c lim 1 Ä_

#

bÄ_


 "c  (1)‘ œ (1
0)
(0
1) œ 0 œ lim ˆ1
"b ‰
c lim Ä_ bÄ_

14.

_ 'c_

ax #

x dx
4b$Î#

œ '
_ 0

ax #

x dx
4b$Î#

_


'0

ax #

%

x dx
4b$Î#

;”

u œ x#
4 Ä du œ 2x dx •

c

œ lim ’ È"u “
c lim ’ È"u “ œ lim Š #"
Ä_ bÄ_ bÄ_ % b u œ ) #
2) Ä
2) du œ 2()
") d) • œ È3  0 œ È3

15.

'01 È)
"

16.

'02 Ès
" 4

s#

" #

'02 È2s ds


'0

du 2È u


lim c cÄ2

'0c È ds

ds œ

œ lim b 'b bÄ!

4

4

2

s#

ds È 4  s#

4  s#

'0_ (1
dxx)Èx ; –

;”

c #

bÄ!

_


'4

du 2u$Î#


c lim Š È"c
#" ‹ œ ˆ #"
0‰
ˆ0
#" ‰ œ 0 Ä_

'b3 2Èduu œ

lim

b Ä !b


Èu‘ 3 œ lim ŠÈ3  Èb‹ b b Ä !b

0 c u œ 4  s# du Ä  "# '4 È
lim c '0 È ds # u 4s cÄ2 du œ 2s ds •

 sin" 0‰ œ (2  0)
ˆ 1#  0‰ œ

'0_ u2
du1 œ

u œ Èx — Ä du œ 2dx Èx

lim b

$Î#

4 c œ lim b
Èu‘ b
lim c
sin" #s ‘ ! cÄ2 bÄ!

œ lim b Š2  Èb‹
lim c ˆsin" cÄ2 bÄ!

17.

'03 2Èduu œ

d); ”

)#

" Èb ‹

'_4 2udu

#

lim

bÄ_

'0b u2
du1 œ #

4
1 #

lim c2 tan" ud 0 b

bÄ_

œ lim a2 tan" b  2 tan" 0b œ 2 ˆ 1# ‰  2(0) œ 1 bÄ_

18.

'1_

dx xÈ x#  1

œ '1

2

dx xÈ x#  1

_


'2

dx xÈ x#  1

œ lim b 'b bÄ1

2

dx x È x#  1


c lim Ä_

'2c

dx x È x#  1

œ lim b csec" kxkd b
c lim csec" kxkd 2 œ lim b asec" 2  sec" bb
c lim asec" c  sec" 2b Ä_ Ä_ bÄ1 bÄ1 œ ˆ 13  0‰
ˆ 1#  13 ‰ œ 1# 2

19.

'0_ a1
v b adv1
tan
" vb œ #

c

lim cln k1
tan" vkd 0 œ lim cln k1
tan" bkd  ln k1
tan" 0k b

bÄ_

bÄ_

œ ln ˆ1
1# ‰  ln (1
0) œ ln ˆ1
1# ‰ 20.

c x '0_ 161tan dx œ
x "

#

# b

#

#

#

lim ’8 atan" xb “ œ lim ’8 atan" bb “  8 atan" 0b œ 8 ˆ 1# ‰  8(0) œ 21#

bÄ_

0

bÄ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.8 Improper Integrals 21.

'c0_ )e) d) œ œ 1 

lim

b Ä
_


)e)  e) ‘ 0 œ a0 † e!  e! b  b

ˆ e"cb ‰

lim

b Ä
_

^ (l'Hopital's rule for

_ _

lim

b Ä
_

cbeb  eb d œ 1 

lim

b Ä
_

551

ˆ becb" ‰

form)

œ 1  0 œ  1 22.

'0_ 2e) sin ) d) œ œ lim 2 bÄ_ bÄ_

0 'c_ e
x

24.

_ 'c_ 2xe
x

k k

œ

25.

( sin )  cos ))“

b

(FORMULA 107 with a œ 1, b œ 1)

0

2(sin 0
cos 0) 2e!




dx œ '
_ ex dx œ 0

lim

b Ä
_

œ0


cex d b! œ _

2(0
1) 2

œ1

lim

a1  eb b œ (1  0) œ 1

b Ä
_

dx œ '
_ 2xe
x dx
'0 2xe
x dx œ 0

#

#

#

lim

b Ä
_

ce
x d b
c lim ce
x d 0 Ä_ 0

#

#

c

c1  aecb bd
c lim cecc  (1)d œ (1  0)
(0
1) œ 0 Ä_ #

lim

b Ä
_

'01 x ln x dx œ

#

#

lim b ’ x# ln x 

bÄ!

Š "b ‹

œ  "4  lim b bÄ!

26.

bÄ_

2(sin b
cos b) 2eb

œ lim

23.

c) ’ 1e
1

'0b 2e) sin ) d)

lim

Š

'01 ( ln x) dx œ

4 ‹ b$

1

x# 4 “b

#

œ ˆ "# ln 1  4" ‰  lim b Š b# ln b  bÄ!

b# 4‹

œ  4"  lim b bÄ!

ln b Š b2# ‹

#

œ  4"
lim b Š b4 ‹ œ  4"
0 œ  4" bÄ!

lim cx  x ln xd b" œ c1  1 ln 1d  lim b cb  b ln bd œ 1  0
lim b b Ä !b bÄ! bÄ!

ln b Š "b ‹

œ 1  lim b b œ 1  0 œ 1 bÄ! 27.

'02 È ds

b œ lim c
sin" #s ‘ 0 œ lim c ˆsin" b# ‰  sin" 0 œ

28.

'01 È4r dr

œ lim c c2 sin" ar# bd 0 œ lim c c2 sin" ab# bd  2 sin" 0 œ 2 †

29.

'12

30.

'24 È dt

31.

'41 Èdxkxk œ

4  s#

1  r%

bÄ2

ds sÈ s#  1

t

t#  4

bÄ2

1 #

0œ

1 #

b

bÄ1

bÄ1

#

œ lim b csec" sd b œ sec" 2  lim b sec" b œ bÄ1 bÄ1

1 3

% œ lim b
"# sec" #t ‘ b œ lim b
ˆ "# sec" 4# ‰  bÄ2 bÄ2

sec" ˆ b# ‰‘ œ

lim c

bÄ!

'b1 Èdxx


lim b 'c

cÄ!

4

dx Èx

" #

0œ

1 #

0œ1

" #

ˆ 13 ‰ 

1 3

" #

†0œ

1 6

b 4 œ lim c
2Èx‘ c1
lim b
2Èx‘ c bÄ! cÄ!

œ lim c Š2Èb‹  ˆ2È(1)‰
2È4  lim b 2Èc œ 0
2
2 † 2  0 œ 6 bÄ! cÄ! 32.

'02 Èkxdx 1k œ '01 È dx


'1

2

1x

dx Èx  1

b

#

œ lim c ’2È1x“
lim
’2Èx  1“ bÄ1 ! cÄ1 c

œ lim  Š2È1b‹  Š2È1  0‹
2È2  1  lim
Š2Èc  1‹ œ 0
2
2  0 œ 4 bÄ1 cÄ1 33.

'c_1 )

#

d)
5)
6


2 ¸‘ b
2 ¸‘ "
2 ¸
ln ¸ bb
ˆ"‰ œ lim
ln ¸ ))
 ln ¸  3
1 œ lim 3 1
3 œ 0  ln # œ ln 2

bÄ_


0

bÄ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 1
lim b bÄ!

Š b" ‹ Š

" ‹ b#

552 34.

Chapter 8 Techniques of Integration

'0_ (x
1)dxax

#

œ lim
#" ln kx
1k 


1b

bÄ_

" œ lim ’ "# ln Š Èbb
‹
#
1

bÄ_

35.

'01Î2 tan ) d) œ

lim b Ä 12 c

" #

" 4

" #

ln ax#
1b


tan" b“  ’ "# ln

" È1




" #

1 tan" x‘ 0 œ lim ’ #" ln Š Èxx
‹
#
1 b

bÄ_

tan" 0“ œ

" #

ln 1


" #

†

1 #



" #

" #

ln 1 

" #

tan" x“

†0œ

b 0

1 4

c ln kcos )kd b0 œ lim1 c c ln kcos bkd
ln 1 œ lim1 c c ln kcos bkd œ
_, bÄ

bÄ

2

2

the integral diverges 36.

'01Î2 cot ) d) œ

cln ksin )kd 1b Î# œ ln 1  lim
cln ksin bkd œ  lim
cln ksin bkd œ
_, bÄ0 bÄ0

lim

b Ä 0


the integral diverges 37.

'01 Èsin ) d) ; c1  ) œ xd 1)

converges, then '0

38.

'c11ÎÎ22

1

cos ) d) (1  2))"Î$

;

sin x Èx

0

21

œ '0

1

sin x dx Èx

sin x dx Èx .

Since 0 Ÿ

'201  cos 2xˆ 1 c #

"Î$

converges, then '0

dx 2x"Î$

'0ln 2 x# ec1Îx dx;
"x œ y‘

Ä

21

x‰ #

dx

sin x# dx #x"Î$

œ '0

21

sin ˆ x# ‰ dx #x"Î$

'01 eÈcÈx

41.

'01 Èt
dtsin t .

Ÿ

" Èx

for all 0 Ÿ x Ÿ 1 and '0

1

dx Èx

'_1Îln 2 y e
y#dy œ '1_Îln 2 e
y dy œ #

. Since 0 Ÿ

sin x# 2x"Î$

Ÿ

" 2x"Î$

for all

converges by the Direct Comparison Test.

y

œ 0
ec1Îln 2 œ e
1Îln 2 , so the integral converges. 40.

sin x Èx

dx converges by the Direct Comparison Test.

Ô x œ 11  2x) × )œ ## Ä Õ d) œ  dx Ø #

0 Ÿ x Ÿ 21 and '0 39.

Ä '1

lim ce
y d b1Îln 2 œ lim ce
b d  ce
1Îln 2 d

bÄ_

bÄ_

dx;
y œ Èx‘ Ä 2'0 ecy dy œ 2  2e , so the integral converges. 1

x

Since for 0 Ÿ t Ÿ 1, 0 Ÿ

" Èt
sin t

Ÿ

" Èt

and '0

1

dt Èt

converges, then the original integral

converges as well by the Direct Comparison Test. 42.

'01 t dtsin t ; let f(t) œ t  "sin t and g(t) œ t" , then $

œ 6. Now,'0

f(t)

lim t Ä 0 g(t)

œ lim

t$

t Ä 0 t  sin t

œ lim

3t#

t Ä 0 1  cos t

œ lim

6t

t Ä 0 sin t

1 œ lim b
 #"t# ‘ b œ  "#  lim b
 #"b# ‘ œ
_, which diverges Ê '0 bÄ! bÄ! diverges by the Limit Comparison Test. 1

œ lim

6 t Ä 0 cos t

43.

'02 1 dxx

œ '0

1

#

diverges Ê 44.

'02

dx 1x

45.


'1

2

'02 1 dxx

œ '0

diverges Ê

dx 1  x#

1

dx 1x

'02

1

dt t$

dx 1  x#

and'0

1

dx 1  x#

dt t  sin t


x ¸‘ b
b ¸‘
" ln ¸ "1  œ lim c
#" ln ¸ 11   0 œ _, which x 0 œ b lim b bÄ1 Ä 1c #

diverges as well.

#


'1

2

dx 1x

dx 1x

and '0

1

dx 1x

œ lim c c ln (1  x)d b0 œ lim c c ln (1  b)d  0 œ _, which bÄ1 bÄ1

diverges as well.

'c11 ln kxk dx œ 'c01 ln (x) dx
'01 ln x dx; '01 ln x dx œ œ 1  0 œ 1; 'c1 ln (x) dx œ 1 Ê 0

lim

b Ä !b

cx ln x  xd 1b œ [1 † 0  1]  lim b [b ln b  b] bÄ!

'c11 ln kxk dx œ 2 converges.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.8 Improper Integrals 46.

'c11 ax ln kxk b dx œ 'c01 cx ln (x)d dx
'01 (x ln x) dx œ #

'1_ 1
dxx

$

;0Ÿ

" x$
1

#

_ dx

for 1 Ÿ x  _ and '1

" x$

Ÿ

x$

x# 4 “b

bÄ!

#

œ
"# ln 1  4" ‘  lim b ’ b# ln b  b4 “ 
"# ln 1  4" -‘
lim b ’ c# ln c  bÄ! cÄ! converges (see Exercise 25 for the limit calculations). 47.

1

#

lim b ’ x# ln x 

'1_ 1
dxx

converges Ê

c# 4“

#

 lim b ’ x# ln x  cÄ!

œ  4"  0


" 4

1

x# 4 “c


0 œ 0 Ê the integral

converges by the Direct

$

Comparison Test.

48.

'4_ Èxdx 1 ; x lim Ä_

Š Èx1
1 ‹

which diverges Ê

49.

'2_ È dv

_

'4

ŠÈ 1

_ d)

Ê '0 51.

e)

_

)

" e)

Ÿ

_ d)

_

œ '0

1

x'
1

dx È x'
1

_

bÄ_

'2_ È dx x#

1

; x lim Ä_

which diverges

53.

ŠÈ

d) 1
e )


'1

" ‹ x#
1

Š"‹ x

dx È x'
1

_

#

Œ

Èx x#

1

e)

dx È x'
1

œ lim cec) d b0 œ lim aecb
1b œ 1 bÄ_

_ dx


'1

54.

'2_ Èx dx





Èx Èx
1

œ x lim Ä_

x%

1

which diverges Ê

55.

x ‹ x%
1 x ‹ x%

ŠÈ

_

'2

œ x lim Ä_

x dx È x%  1

È x% È x%  1

œ x lim Ä_

_ 2
cos x x

Ê

#

'1_ 2xdx converges #

b œ lim
 2x" # ‘ 1

bÄ_

'2_ x" dx œ

lim cln bd b2 œ _,

bÄ_

" É1
x#

" É1 

" œ 1;

x

'1_ Èx x dx œ '1_ xdx #

$Î#

dx converges by the Limit Comparison Test.

" œ 1; x%

'2_ Èx dx

x%

_ dx

œ '2

x

œ lim cln xd b2 œ _, bÄ_

lim cln xd 1b œ _, which diverges

bÄ_

dx diverges by the Direct Comparison Test.

'1_ 1
xsin x dx; 0 Ÿ "
xsin x Ÿ x2 #

x$

diverges by the Limit Comparison Test.

'1_ 2
xcos x dx; 0  "x Ÿ 2
xcos x for x   1 and '1_ dxx œ Ê '1

56.

ŠÈ

" œ 1; x#

_ Èx
1

bÄ_

; x lim Ä_

_ dx

and '1

x$

" É1 

œ x lim Ä_

b 2 œ lim
2x"Î# ‘ 1 œ lim Š È
2‹ œ 2 Ê '1 b

bÄ_

bÄ_

converges by the Direct Comparison Test.

x È x#  1

Èx
1 œ x lim Ä_ x#

b lim
2Èv‘ # œ _,

bÄ_

diverges by the Limit Comparison Test.

x#  1

Œ

 '0

dx È x'
1

œ x lim Ä_

_ Ê '2 È dx

'1_ Èxx
1 dx; x lim Ä_

'2_ Èdvv œ

converges by the Direct Comparison Test.

Ê '0

" #

bÄ_

diverges by the Limit Comparison Test.

for 0 Ÿ )  _ and '0

œ lim ˆ #"b#
"# ‰ œ

52.

Èv

dv Èv  1

b lim
2Èx‘ 4 œ _,

diverges by the Limit Comparison Test.

‹

converges Ê '0

'0_ È dx

'4_ Èdxx œ

" œ v lim œ v lim œ È1" 0 œ 1 and Ä _ Èv  1 Ä _ É1  " Š È"v ‹ v

'0_ 1
d)e ; 0 Ÿ 1
"e )

dx Èx  1

v
1

; lim v1 v Ä _

which diverges Ê '2 50.

Èx

" œ x lim œ x lim œ 1" 0 œ 1 and Ä _ Èx  1 Ä _ 1  È"x Š È"x ‹

#

Ê

_

for x   1 and '1

2 x#

b dx œ lim
 x2 ‘ 1 œ lim ˆ b2
12 ‰ œ

bÄ_

bÄ_

'1_ 1
xsin x dx converges by the Direct Comparison Test. #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

553

2 1

554 57.

Chapter 8 Techniques of Integration

'4_ t 2 dt 1 ; $Î#

_

Ê '4

_ 2 dt

œ 1 and '4

t$Î#

lim t Ä _ t$Î#  1

2 dt t$Î#
1

b 4 œ lim
4t"Î# ‘ 4 œ lim Š È
2‹ œ 2 Ê b

t$Î#

bÄ_

'2_ lndxx ; 0  "x  ln"x for x  2 and '2_ dxx diverges

59.

'1_ ex

60.

'e_ ln (ln x) dx; cx œ eyd

" x

dx; 0 



ex x

_ dx

for x  1 and '1

e

Ä

x

'1_ Èedx x ; x lim Ä_

ŠÈ

_

Ê '2

diverges Ê

dx ln x

$Î#

diverges by the Direct Comparison Test.

'1_ e xdx diverges by the Direct Comparison Test. x

'e_ (ln y) ey dy; 0  ln y  (ln y) ey for y   e and 'e_ ln y dy œ

_

œ _, which diverges Ê 'e ln ey dy diverges Ê

61.

'4_ 2t dt converges

converges by the Limit Comparison Test.

58.

x

bÄ_

" ‹ ex
x

lim cy ln y  yd be

bÄ_

'e_ ln (ln x) dx diverges by the Direct Comparison Test. e

'_

'_

" œ x lim œ x lim œ È1" 0 œ 1; 1 Èdxex œ 1 e
xÎ2 dx Ä _ È ex  x Ä _ É1  xx Š È" x ‹ e e _ _ c xÎ2 b œ lim c2e d " œ lim a2e
bÎ2
2e
1Î2 b œ È2e Ê 1 e
xÎ2 dx converges Ê 1 Èedx converges xx bÄ_ bÄ_ È ex

x

'

'

by the Limit Comparison Test. 62.

'1_ e dx 2 x

ˆ ex
" 2x ‰ ˆ e"x ‰

; x lim Ä_

x

œ x lim Ä_

œ lim aecb
e" b œ

" e

bÄ_

63.

_ dx '
_ È

_ dx

'1 64.

x#

_

x%
1

œ 2 '0

dx È x%
1

_

'1

Ê _

; '0

œ x lim Ä_

ex e x 2 x dx ex

dx È x%
1

" x 1Š 2e ‹

converges Ê œ '0

1

dx È x%
1

b œ lim
 x" ‘ 1 œ lim ˆ b"
1‰ œ 1 Ê

bÄ_

'c__ e
dxec

x

x

bÄ_

_

œ 2'0

dx ex
ecx

" ex
ecx

;0



" ex

œ _

'1

" 1 0

dx e x 2 x

_


'1

_

x%
1

_

dx ex

ex

œ lim cecx d b1 bÄ_

converges by the Limit Comparison Test.

dx È x%
1

'
_ È dx

for x  0; '0

_ dx

œ 1 and '1

 '0

1

dx È x%
1

_


'1

dx x#

and

converges by the Direct Comparison Test.

_

converges Ê 2 '0

dx ex
ecx

converges by the

Direct Comparison Test. 65. (a)

(b)

'12 x(lndxx)

; ct œ ln xd Ä '0

ln 2

ln 2

œ lim b ’ p"
1 t1cp “ œ lim b b bÄ! bÄ! Ê the integral converges for p  1 and diverges for p   1 p

'2_ x(lndxx)

dt tp

_

p

; ct œ ln xd Ä 'ln 2

dt tp

b1cp p1




" 1p

(ln 2)1cp

and this integral is essentially the same as in Exercise 65(a): it converges

for p  1 and diverges for p Ÿ 1 66.

'0_ x2x
dx1 œ #

lim cln ax#
1bd 0 œ lim cln ab#
1bd  0 œ lim ln ab#
1b œ _ Ê the integral b

bÄ_

'c_ x2x
dx1 œ b

diverges. But lim

bÄ_

#

œ lim (ln 1) œ 0

bÄ_

bÄ_

" lim cln ax#
1bd
b œ lim cln ab#
1b  ln ab#
1bd œ lim ln Š bb#

1‹ b

bÄ_

bÄ_

_

67. A œ '0 ecx dx œ lim cecx d b0 œ lim aecb b  aec0 b bÄ_

#

#

bÄ_

œ0
1œ1

_ 2x '
_ x
1 dx

bÄ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

bÄ_

Section 8.8 Improper Integrals 68. x œ yœ

" A " 2A

'0_ xecx dx œ _

'0

lim cxecx  ecx d b0 œ lim abecb  ecb b  a0 † ec0  ec0 b œ 0
1 œ 1;

bÄ_

aecx b# dx œ

" #

_

'0

_

bÄ_

ec2x dx œ lim

"

bÄ_ #


 #" ec2x ‘ b œ lim 0

"

bÄ_ #

ˆ #" ec2b ‰  #" ˆ #" ec2 0 ‰ œ 0
†

" 4

œ

" 4

_

69. V œ '0 21xecx dx œ 21 '0 xecx dx œ 21 lim cxecx  ecx d b0 œ 21 ’ lim abecb  ecb b  1“ œ 21 bÄ_

_

bÄ_

_

b 70. V œ '0 1 aecx b# dx œ 1 '0 ec2x dx œ 1 lim
 "# ec2x ‘ 0 œ 1 lim ˆ "# ec2b
"# ‰ œ

bÄ_

71. A œ '0 (sec x  tan x) dx œ lim1 c cln ksec x
tan xk  ln ksec xkd b0 œ lim1 c ˆln ¸1
bÄ 2 bÄ 2 œ lim1 c ln k1
sin bk œ ln 2 1Î2

bÄ

1 #

bÄ_

tan b ¸ sec b

 ln k1
0k‰

2

72. (a) V œ '0

1Î2

1 sec# x dx  '0 1 tan# x dx œ 1 '0 asec# x  tan# xb dx œ '0 1Î2

œ 1 '0 dx œ 1Î2

1Î2

1Î2

1 csec# x  asec# x  1bd dx

1# #

(b) Souter œ '0 21 sec xÈ1
sec# x tan# x dx   '0 21 sec x(sec x tan x) dx œ 1 lim1 c ctan# xd 0 1Î2

1Î2

b

bÄ

2

œ 1 ” lim1 c ctan bd  0• œ 1 lim1 c atan bb œ _ Ê Souter diverges; Sinner œ '0 21 tan xÈ1
sec% x dx bÄ bÄ #

1Î2

#

2

2

  '0 21 tan x sec x dx œ 1 lim1 c ctan# xd 0 œ 1 ” lim1 c ctan# bd  0• œ 1 lim1 c atan# bb œ _ bÄ bÄ bÄ 1Î2

#

b

2

2

2

Ê Sinner diverges 73. (a)

'3_ ec3x dx œ

b lim
 "3 ec3x ‘ 3 œ lim ˆ 3" ec3b ‰  ˆ 3" ec3 3 ‰ œ 0
†

bÄ_

bÄ_

_

" 3

† ec9 œ

" 3

ec9

¸ 0.0000411  0.000042. Since ecx Ÿ ec3x for x  3, then '3 ecx dx  0.000042 and therefore #

(b)

'0_ ecx '03 ecx

#

#

#

dx can be replaced by '0 ecx dx without introducing an error greater than 0.000042. 3

#

dx µ = 0.88621

_

# b 74. (a) V œ '1 1 ˆ "x ‰ dx œ 1 lim
 x" ‘ 1 œ 1 ’ lim ˆ b" ‰  ˆ 1" ‰“ œ 1(0
1) œ 1

bÄ_

bÄ_

(b) When you take the limit to _, you are no longer modeling the real world which is finite. The comparison step in the modeling process discussed in Section 4.2 relating the mathematical world to the real world fails to hold. 75. (a)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

555

556

Chapter 8 Techniques of Integration

(b) > int((sin(t))/t, t=0..infinity); ˆanswer is 1# ‰

76. (a)

(b) > f:= 2*exp(t^2)/sqrt(Pi); > int(f, t=0..infinity); (answer is 1) 77. (a) faxb œ

1 x2 /2 È 21 e

f is increasing on (_, 0]. f is decreasing on [0, _). f has a local maximum at a0, fa0bb œ Š!ß (b) Maple commands: >f: œ exp(x^2/2)(sqrt(2*pi); >int(f, x œ 1..1); >int(f, x œ 2..2); >int(f, x œ 3..3);

1 È 21 ‹

¸ 0.683 ¸ 0.954 ¸ 0.997

(c) Part (b) suggests that as n increases, the integral approaches 1. We can take 'cn f(x) dx as close to 1 as we want by n

_


n

choosing n  1 large enough. Also, we can make 'n f(x) dx and '
_ f(x) dx as small as we want by choosing n large enough. This is because 0  faxb  ex/2 for x  1. (Likewise, 0  faxb  ex/2 for x  1.)

_

_

Thus, 'n f(x) dx  'n ex/2 dx.

'n_ ex/2 dx œ lim 'nc ex/2 dx œ lim c 2ex/2 dcn œ lim c 2ec/2
2en/2 d œ 2en/2 cÄ_

cÄ_

_

cÄ_

As n Ä _, 2en/2 Ä 0, for large enough n, 'n f(x) dx is as small as we want. Likewise for large enough n,


n '
_ f(x) dx is as small as we want.

78.

'3_ ˆ x " #  x" ‰ dx Á '3_ x dx #  '3_ dxx , since the left hand integral converges but both of the right hand integrals diverge. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 8.8 Improper Integrals _

_

79. (a) The statement is true since 'c_ f(x) dx œ '
_ f(x) dx
'a f(x) dx, 'b f(x) dx œ 'a f(x) dx  'a f(x) dx b

a

b

b

and 'a f(x) dx exists since f(x) is integrable on every interval [aß b]. b

(b)

80. (a)

(b)

81.

a a b b _ _ '
_ f(x) dx
'a f(x) dx œ '
_ f(x) dx
'a f(x) dx  'a f(x) dx
'a f(x) dx b a b _ _ œ '
_ f(x) dx
'b f(x) dx
'a f(x) dx œ '
_ f(x) dx
'b f(x) dx 0 0 _ _ _ '
_ f(x) dx œ '
_ f(x) dx
'0 f(x) dx œ  '_ f(u) du
'0 f(x) dx _ _ _ œ '0 f(u) du
'0 f(x) dx œ 2 '0 f(x) dx, where u œ x 0 0 _ _ _ '
_ f(x) dx œ '
_ f(x) dx
'0 f(x) dx œ '_ f(u) du
'0 f(x) dx _ _ _ _ œ '0 f(u) du
'0 f(x) dx œ '0 f(x) dx
'0 f(x) dx œ 0, where u œ x

_ dx '
_ È x#

1


1

È x#
1 x

œ x lim Ä_ 82.

_

œ '
_ Èxdx
'1 #
1

_ " '
_ È

dx È x#
1

É1
œ x lim Ä_

" x#

_

1
x'

dx converges, since '
_

_

; '1

dx È x#
1

diverges because x lim Ä_

_ dx

œ 1 and '1 " È 1
x'

x

_

diverges; therefore, '
_ _

dx œ 2 '0

" È 1
x'

Š "x ‹ ŠÈ

" ‹ x#
1

dx È x#
1

diverges

dx which was shown to converge in

Exercise 51 83.

_ dx " 'c_ ' _ e dx e e
ec œ
_ e
1 ; e
1 œ e
ec _ _ Ê 'c_ ee
dx1 œ 2 '0 e
dxec converges x

x

x

x

2x

2x

x

x



" ex

_

and '0

œ c lim ce
x d c0 œ c lim ae
c
1b œ 1 Ä_ Ä_

dx ex

x

2x

84.

x

x

_ c e dx e '
_ '
1 ec dx ' _ ec dx '
1 ec dx ' _ e du x
1 œ
_ x
1

1 x
1 ;
_ x
1 œ 1 1
u , where u œ x, and since 1
u _ ec dx '1_ duu diverges, the integral '1_ 1e
duu diverges Ê '
_ x
1 diverges x

x

#

x

#

x

#

u

#

u

#

u

86.

_ 'c_ e
x

k k

_

dx œ 2 '0 e
x dx œ 2 lim

bÄ_

_ dx 'c_ '
# (x
1) œ
_ b lim c 'c# (x
dx1) b Ä
1

87.

dx (x
1)#

#

#

œ


1


'
# lim

b Ä
1 c

dx (x
1)#

'0b e
x dx œ 2

_


'
1 (x
dx1)#
'2 2

dx (x
1)#

;

_ _ xk x 'c__ ksin xkxk k

kcos dx œ 2 '0 ksin xxk

k1cos xk dx   2'0 sin xx

cos dx œ 2 1 1 #

#

_ dx '
_ (x
1)

lim

bÄ_

#

diverges

'0b x dx
1 dx

_ ksin xk
kcos xk 'c_ dx diverges kx k
1

_ x '
_ ax
1b ax
2b dx œ 0 by Exercise 80(b) because the integrand is odd and the integral '0_ ax
1xbdxax
2b Ÿ '0_ dxx converges #

#

#

#

(u  1) and

lim ce
x d b0 œ 2, so the integral converges.

bÄ_


x
" 1 ‘ b œ _, which diverges Ê
#

bÄ_

89.

" u

#

œ 2 lim cln kx
1kd b0 œ _, which diverges Ê 88.



x

#

85.

#

$

Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := 0..exp(1); fn_list := [seq( f(x,p), p=-2..2 )]; Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

557

558

Chapter 8 Techniques of Integration plot( fn_list, x=domain, y=-50..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p= -2","p = -1","p = 0","p = 1","p = 2"], title="#89 (Section 8.8)" ); q1 := Int( f(x,p), x=domain ); q2 := value( q1 ); q3 := simplify( q2 ) assuming p>-1; q4 := simplify( q2 ) assuming p<-1; q5 := value( eval( q1, p=-1 ) ); i1 := q1 = piecewise( p<-1, q4, p=-1, q5, p>-1, q3 );

90.

Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := exp(1)..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=exp(1)..10, y=0..100, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#90 (Section 8.8)" ); q6 := Int( f(x,p), x=domain ); q7 := value( q6 ); q8 := simplify( q7 ) assuming p>-1; q9 := simplify( q7 ) assuming p<-1; q10 := value( eval( q6, p=-1 ) ); i2 := q6 = piecewise( p<-1, q9, p=-1, q10, p>-1, q8 );

91.

Example CAS commands: Maple: f := (x,p) -> x^p*ln(x); domain := 0..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=0..10, y=-50..50, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], thickness=[3,4,1,2,0], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#91 (Section 8.8)" ); q11 := Int( f(x,p), x=domain ): q11 = lhs(i1+i2); `` = rhs(i1+i2); `` = piecewise( p<-1, q4+q9, p=-1, q5+q10, p>-1, q3+q8 ); `` = piecewise( p<-1, -infinity, p=-1, undefined, p>-1, infinity );

92.

Example CAS commands: Maple: f := (x,p) -> x^p*ln(abs(x)); domain := -infinity..infinity; fn_list := [seq( f(x,p), p=-2..2 )]; plot( fn_list, x=-4..4, y=-20..10, color=[red,blue,green,cyan,pink], linestyle=[1,3,4,7,9], legend=["p = -2","p = -1","p = 0","p = 1","p = 2"], title="#92 (Section 8.8)" ); q12 := Int( f(x,p), x=domain ); q12p := Int( f(x,p), x=0..infinity ); q12n := Int( f(x,p), x=-infinity..0 ); q12 = q12p + q12n; `` = simplify( q12p+q12n );

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises 89-92. Example CAS commands: Mathematica: (functions and domains may vary) Clear[x, f, p] f[x_]:= xp Log[Abs[x]] int = Integrate[f[x], {x, e, 100)] int /. p Ä 2.5 In order to plot the function, a value for p must be selected. p = 3; Plot[f[x], {x, 2.72, 10}] CHAPTER 8 PRACTICE EXERCISES

1.

# ' xÈ4x#
9 dx; ” u œ 4x
9 • Ä du œ 8x dx

2.

# ' 6xÈ3x#  5 dx; ” u œ 3x  5 • Ä ' Èu du œ 23 u$Î#  C œ 23 a3x#  5b$Î#  C du œ 6x dx

3.

' x(2x  1)"Î# dx; ” u œ 2x  1 • Ä du œ 2 dx œ

4.

(2x  1)&Î# 10

' Èx

dx; ”

1
x

œ

2 3

(2x  1)$Î# 6

" #

" 1#

a4x#
9b

$Î#

C

' ˆ u
# 1 ‰ Èu du œ "4 Š' u$Î# du
' u"Î# du‹ œ "4 ˆ 25 u&Î#
23 u$Î# ‰  C

C

uœ1
x Ä
' du œ
dx •

(1
u) Èu

du œ ' ŠÈu


" Èu ‹

du œ

2 3

u$Î#
2u"Î#  C

(1
x)$Î#
2(1
x)"Î#  C

5.

' È x dx

6.

' È x dx

7.

' 25ydyy

8.

' 4ydyy ; ”

9.

' Èt

10.

'

11.




' Èu du œ "8 † 23 u$Î#  C œ

" 8

8x#

; 1 ”

;”

9
4x#

#

;”

$

%

$ dt 9
4t%

2t dt t%  1

" 16

' Èduu œ 16" † 2u"Î#  C œ È8x8  1  C

u œ 9
4x# Ä
"8 ' du œ
8x dx •

u œ 25  y# Ä du œ 2y dy •

u œ 4  y% • Ä du œ 4y$ dy

;”

;”

u œ 8x#  1 Ä du œ 16x dx •

" #

" 4

#

du Èu

È9
4x# 4

C

' duu œ #" ln kuk  C œ #" ln a25  y# b  C

' duu œ 4" ln kuk  C œ 4" ln a4  y% b  C

u œ 9
4t% " • Ä
16 ' du œ
16t$ dt

u œ t# Ä du œ 2t dt •

œ
8" † 2u"Î#  C œ


du Èu

" œ
16 † 2u"Î#  C œ


È9
4t% 8

C

' u du1 œ tan
" u  C œ tan
" t#  C #

&Î$ " ' z#Î$ ˆz&Î$  1‰#Î$ dz; ” u œ z5 #Î$ • Ä du œ z dz 3

3 5

' u#Î$ du œ 35 † 35 u&Î$  C œ

9 #5

ˆz&Î$  1‰&Î$  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

559

560 12.

Chapter 8 Techniques of Integration

z%Î& ' z
"Î& ˆ1  z%Î& ‰
"Î# dz; ” u œ 41 
"Î& • Ä du œ z dz 5

5 4

' u
"Î# du œ 54 † 2Èu  C œ #5 ˆ1  z%Î& ‰"Î#  C

13.

2) d) ' (1sin
cos 2) )

14.

) d) ' (1 cos  sin ))

15.

u œ 3  4 cos t dt ' 3 sin4tcos • t ;”

16.

' 1cos sin2t dt2t ; ”

17.

' (sin 2x) ecos 2x dx; ” u œ cos 2x • Ä
"# ' eu du œ
"# eu  C œ
"# ecos 2x  C du œ
2 sin 2x dx

18.

u œ sec x ' (sec x tan x) esec x dx; ” Ä ' eu du œ eu  C œ esec x  C du œ sec x tan x dx •

19.

' e) sin ae) b cos# ae) b d); ”

20.

) ' e) sec# ae) b d); ” u œ e) • Ä ' sec# u du œ tan u  C œ tan ae) b  C du œ e d)

21.

' 2xc1 dx œ

23.

' v dvln v ; ”

24.

' v(2 dvln v) ; ” u œ 2 " ln v •

#

;”

u œ 1
cos 2) Ä du œ # sin 2) d) •

"Î#

;”

u œ 1  sin ) Ä du œ cos ) d) •

u œ 1  sin 2t Ä du œ 2 cos 2t dt •

v

dv

œ
4" ln kuk  C œ
4" ln k3  4 cos tk  C

' duu œ 2" ln kuk  C œ 2" ln k1  sin 2tk  C

26.

' sinÈ
" x dx# ; – u œ sin dx x —

27.

' È 2 dx

;”

28.

' È dx

œ

du œ


"

du œ

'
u# du œ
"3 u$  C œ
"3 cos$ ae) b  C

È 1
x#

u œ 2x Ä du œ 2 dx • dx É1
ˆ x7 ‰#

;”

' 5xÈ2 dx œ

" È2

È2

x

Š 5ln 5 ‹  C

' duu œ ln kuk  C œ ln k2  ln vk  C

Ä


"

#

'

" 2

du u

' duu œ ln kuk  C œ ln kln vk  C

' ax 1b adx2tan
" xb ; ” u œ 2  tan dx

" 7

œ 2u"Î#  C œ 2È1  sin )  C

22.

25.

49
x#

"Î#

C

du œ

1
4x#

' udu

#

u œ cos ˆe) ‰ • Ä du œ
sin ˆe) ‰ † e) d)

u œ ln v Ä du œ v" dv •

1
x

' duu œ
2u"  C œ
#(1
"cos 2))  C

Ä
4" '

du œ
4 sin t dt

2xc1 ln 2

" #

x # 1

Ä

x

• Ä

' duu œ ln kuk  C œ ln k2  tan
" xk  C

' u du œ "# u#  C œ "# asin
" xb#  C

' È du

1
u#

œ sin
" u  C œ sin
" (2x)  C

u œ x7 Ä du œ "7 dx •

' È du

1
u#

œ sin
" u  C œ sin
" ˆ x7 ‰  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises 29.

30.

31.

'È

' È dt

9
4t#

' 9 dt t

#

" 4

œ

dt 16
9t#

œ

'

" 9

'

" 3

œ

'

dt #

Ê1
Š 3t 4‹

dt

;–

#

Ê1
Š 2t 3‹

dt # 1ˆt‰

;–

3

uœ du œ

;–

uœ du œ

uœ du œ " 3 " 3

3 4 3 4

2 3 2 3

t Ä dt —

t Ä dt —

t Ä dt —

" 2

34.

'

35.

' È dx

36.

'È

37.


2) "
" ˆ y
2 ‰ ' y
dy4y  8 œ ' (yd(y C
2)  4 œ # tan #

38.

 2)
" ' t  dt4t  5 œ ' (t d(t (t  2)  C  2)  1 œ tan

4 dx 5xÈ25x#
16

œ 3'

6 dx xÈ4x#
9

4x
x#

œ

œ'

dx 4x
x#
3

4 25

'

dx xÉx#
94

œ'

d(x
2) È1
(x
2)#

40.

'

" 5

dv (v  1)Èv#  2v

œ'

d(x
1) (x
1)È(x
1)#
1

œ sec
" kx
1k  C

œ'

d(v  1) (v  1)È(v  1)#
1

œ sec
" kv  1k  C

6x ' cos# 3x dx œ ' "  cos dx œ x#  sin126x  C #

43.

' sin$ #) d) œ ' ˆ1
cos# #) ‰ ˆsin #) ‰ d); –

tan
" (5t)  C

cos$

) #


2 cos

) #

u œ cos #) Ä
2 ' a1
u# b du œ du œ
"# sin #) d) —




u$ 3

Cœ

2u$ 3


2u  C

C

' sin$ ) cos# ) d) œ ' a1
cos# )b (sin )) acos# )b d); ” u& 5

tan
" ˆ 3t ‰  C

œ sin
" (x
2)  C

42.

œ

" 5

" 3

sin
" ˆ 2t3 ‰  C

œ sin
" ˆ x
# 2 ‰  C

2x ' sin# x dx œ ' 1
cos dx œ #x
sin42x  C #

44.

tan
" u  C œ

" 2

sin
" ˆ 3t4 ‰  C

¸ sec
" ¸ 5x 4 C

41.

2 3

" 3

sin
" u  C œ

" 3

#

dx (x
1)Èx#
2x

œ

œ

" 2

sin
" u  C œ

#

#

'

œ

œ

#

œ

" 3

¸ œ 2 sec
" ¸ 2x 3 C

d(x
2) È4
(x
2)#

#

39.

dx xÉx#
16 25

#

1
u#

tan
" u  C œ

'

" 5

' È du

" 5

33.

u œ 5t Ä du œ 5 dt •

œ

1
u#

' 1 duu

' 1 dt25t

;”

' È du

" 3

32.

#

' 1 duu

" 3

cos& ) 5




cos$ ) 3

u œ cos ) Ä
' a1
u# b u# du œ ' au%
u# b du du œ
sin ) d) •

C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

561

562 45.

Chapter 8 Techniques of Integration

' tan$ 2t dt œ ' Ä œ

46.

" 4

" #

(tan 2t) asec# 2t
1b dt œ ' tan 2t sec# 2t dt
' tan 2t dt; ”

u œ 2t du œ 2 dt •

' tan u sec# u du
"# ' tan u du œ 4" tan# u  #" ln kcos uk  C œ 4" tan# 2t  #" ln kcos 2tk  C

tan# 2t


" #

ln ksec 2tk  C

' 6 sec% t dt œ 6' atan# t  1b asec# tb dt; ”

u œ tan t Ä 6 ' au#  1b du œ 2u$  6u  C du œ sec# t dt •

œ 2 tan$ t  6 tan t  C 47.

' 2 sin dxx cos x œ ' sindx2x œ '

48.

' cos 2x
dxsin x œ ' cos2 dx2x ; ” #

#

csc 2x dx œ
"# ln kcsc 2x  cot 2xk  C

u œ 2x Ä du œ 2 dx •

'

du cos u

œ ' sec u du œ ln ksec u  tan uk  C

œ ln ksec 2x  tan 2xk  C 49.

'11ÎÎ42 Ècsc# y
1 dy œ '11ÎÎ42 cot y dy œ cln ksin ykd 11ÎÎ24 œ ln 1
ln È"

50.

'13Î14Î4 Ècot# t  1 dt œ '13Î14Î4 csc t dt œ c
ln kcsc t  cot tkd 311Î4Î4 œ
ln ¸csc 341  cot 341 ¸  ln ¸csc 14  cot 14 ¸ È

œ
ln ¹È2
1¹  ln ¹È2  1¹ œ ln ¹ È2  " ¹ œ ln »

œ ln È2

ŠÈ2  "‹ ŠÈ2  1‹ #
1

2
1

51.

2

» œ ln Š3  2

È2‹

'01 È1
cos# 2x dx œ '01 ksin 2xk dx œ '01Î2 sin 2x dx
'11Î2 sin 2x dx œ

cos#2x ‘ 10 Î2 
cos#2x ‘ 11Î2 œ
ˆ
"#
"# ‰ 
"#
ˆ
"# ‰‘ œ 2

52.

'021 È1
sin# x# dx œ '021 ¸cos x# ¸ dx œ '01 cos x# dx
'121 cos x# dx œ
2 sin x# ‘ 10

2 sin x# ‘ 1#1 œ (2
0)
(0
2) œ 4

53.

'
11ÎÎ22 È1
cos 2t dt œ È2 '
11ÎÎ22 ksin tk dt œ 2È2 '01Î2 sin t dt œ ’
2È2 cos t“ 1Î2 œ 2È2 [0
(
1)] œ 2È2 0

54.

'121 È1  cos 2t dt œ È2 '121 kcos tk dt œ
È2 '131Î2 cos t dt  È2 '3211Î2 cos t dt œ
È2 csin td $11Î2  È2 csin td #$11Î# œ
È2 (
1
0)  È2 [0
(
1)] œ 2È2

55.

' xx dx4 œ x
' x4 dx4 œ x
2 tan
" ˆ #x ‰  C

56.

' 9xdxx

57.

' 4x2x
13 dx œ '
(2x  1)  #x 4
1 ‘ dx œ x  x#  2 ln k2x
1k  C

58.

' 2xx
dx4 œ ' ˆ2  x
8 4 ‰ dx œ 2x  8 ln kx
4k  C

59.

' 2yy 
41 dy œ '

#

#

#

$

#

œ ' ’ x ax x# 9b9
9x “ dx œ ' ˆx
#

9x ‰ x#  9

dx œ

x# #




9 #

ln a9  x# b  C

#

#

2y dy y#  4


'

dy y#  4

œ ln ay#  4b


" #

tan
" ˆ #y ‰  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises 60.

' yy 41 dy œ ' yy dy1  4 ' y dy 1 œ #" ln ay#  1b  4 tan
" y  C

61.

' Èt  2

62.

' 2t È È1
t

63.

#

#

4
t#

dt œ '

#

#

1
t#

t

#

t dt È4
t#

dt œ '

2t dt È1
t#

'

dt t

œ
È4
t#  2 sin
" ˆ #t ‰  C

œ
2È1
t#  ln ktk  C #

#

#

#


cos x) dx cos x dx ' cotcotx x csc ' cos ' (cos1x)(1 dx œ ' cos x
sin1 x sin x dx x œ x1 œ
cos x x) ' sindx x  ' dx œ
sin" x  cot x  x  C œ x  cot x
csc x  C œ ' d(sin sin x
#

#

#

65.

dt È4
t#


sin x) dx ' tantanx x sec ' sinsin xxdx1 œ ' (sin1x)(1 dx œ ' sin x
cos1 xcos x dx x œ
sin x x) ' cosdx x  ' dx œ cos" x
tan x  x  C œ x
tan x  sec x  C œ
' d(cos cos x
#

64.

 2'

#

#

' sec (5
3x) dx; ” y œ 5
3x •

Ä

dy œ
3 dx

' sec y † Š
dy3 ‹ œ
"3 ' sec y dy œ
3" ln ksec y  tan yk  C

œ
"3 ln ksec (5
3x)  tan (5
3x)k  C 66.

' x csc ax#  3b dx œ "# ' csc ax#  3b d ax#  3b œ
"# ln kcsc ax#  3b  cot ax#  3bk  C

67.

' cot ˆ x4 ‰ dx œ 4 ' cot ˆ x4 ‰ d ˆ x4 ‰ œ 4 ln ¸sin ˆ x4 ‰¸  C

68.

' tan (2x
7) dx œ "# ' tan (2x
7) d(2x
7) œ
"# ln kcos (2x
7)k  C œ "# ln ksec (2x
7)k  C

69.

' xÈ1
x dx; ” u œ 1
x •

Ä
' (1
u)Èu du œ ' ˆu$Î#
u"Î# ‰ du œ

du œ
dx

$

œ

70.

72.

(1
x)

&Î#


(1
x) 2 3

$Î#

 C œ
2 –

' 3xÈ2x  1 dx; ” u œ 2x  1 • du œ 2 dx

œ 71.

2 5

3 10

œ

sec ) tan ) 3
1

œ

sin ) 2 cos# )

z œ tan ) Ä dz œ sec# ) d) • 



" #

3
2 3
1

&

ŠÈ1
x‹




3

3 ˆÈ2x  1‰ 10

ŠÈ1
x‹

—C

5

&




ˆÈ2x  1‰ #

$

C

' Ètan# )  1 † sec# ) d) œ ' sec$ ) d) (FORMULA 92)

ln ksec )  tan )k  C œ

' a16  z# b
$Î# dz; ” z 16 a16  z# b"Î#

' sec ) d)

u&Î#
23 u$Î#  C

' 3 ˆ u
# 1 ‰ Èu † "# du œ 34 ' ˆu$Î#
u"Î# ‰ du œ 34 † 25 u&Î#
34 † 23 u$Î#  C

(2x  1)&Î#
"# (2x  1)$Î#  C œ

' Èz#  1 dz; ”

œ

Ä

2 5

z œ 4 tan ) Ä dz œ 4 sec# ) d) •

zÈ z#  1 #



" #

ln ¹z  È1  z# ¹  C

) d) " ' " ' 644 sec cos ) d) œ 16 sin )  C œ sec ) d) œ 16 #

$

z 16È16  z#

C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

C

563

564 73.

Chapter 8 Techniques of Integration

' È25dy y

œ

#

" 5

'

dy É1ˆ 5y ‰#

œ'

du È 1  u#

u œ tan ) ,
u œ y5 ‘ ; ” Ä du œ sec# ) d) •

' Èsec ) d) #

1  tan# )

œ ' sec ) d)

#

œ ln ksec )  tan )k  C" œ ln ¹È1  u#  u¹  C" œ ln ºÉ1  ˆ y5 ‰  y5 º  C" œ ln ¹

È25  y#  y ¹ 5

 C"

œ ln ¸y  È25  y# ¸  C 74.

' È25dy 9y " 3

Ä 75.

'

76.

' Èx

#

œ

" 5

'

dy Ê1  Š

3y # 5 ‹

œ

" 3

' È du

dx x# È 1
x#

$ dx 1
x#

;”

;”

x œ sin ) Ä dx œ cos ) d) •

x œ sin ) Ä dx œ cos ) d) •

Note: Ans ´

' Èx

# dx 1
x#

œ 78.

" #

" 3

ln ¹È1  u#  u¹  C" from Exercise 73

ln ¸È25  9y#  3y¸  C

' sincos))cosd) ) œ ' csc# ) d) œ
cot )  C œ
È1x
x

)


;” " #


x

#È

1
x#

3

' sin )coscos) ) d) œ ' sin$ ) d) œ ' a1
cos# )b (sin )) d); $

u$ 3

 C œ
cos ) 

" 3

cos$ ) œ
È1
x#  3" a1
x# b

x#
9

sin ) cos ) œ

;”

sin
" x #




xÈ 1
x# #

x œ 2 sin ) Ä dx œ 2 cos ) d) •

C

' 2 cos ) † 2 cos ) d) œ 2 ' (1  cos 2)) d) œ 2 ˆ)  "# sin 2)‰  C

x œ 3 sec ) Ä' dx œ 3 sec ) tan ) d) •

3 sec ) tan ) d) È9 sec# )
9 #

œ'

3 sec ) tan ) d) 3 tan )

œ ln ksec )  tan )k  C" œ ln º x3  Ɉ x3 ‰
1º  C" œ ln ¹ x 

80.

'

12 dx ax#
1b$Î#

;”

C

#

#

' È dx

$Î#

2) ' sin )coscos) ) d) œ ' sin# ) d) œ ' 1
cos d) œ "# )
4" sin 2)  C #

œ 2)  2 sin ) cos )  C œ 2 sin
" ˆ x# ‰  xÉ1
ˆ x# ‰  C œ 2 sin
" ˆ x# ‰  79.

C


23 È1
x#  C by another method

x œ sin ) Ä dx œ cos ) d) •

' È4
x# dx; ”

#

#

cu œ cos )d Ä
' a1
u# b du œ
u 

77.

œ

1  u#

x œ sec ) Ä dx œ sec ) tan ) d) •

È x#
9 ¹ 3

#

C

œ ' sec ) d)

cos ) d) ' 12 sectan) tan) ) d) œ ' 12 sin ;” ) $

xÈ 4
x# 2

 C" œ ln ¹x  Èx#
9¹  C u œ sin ) Ä du œ cos ) d) •

' 12u du #

12 12 x œ
12 C u  C œ
sin )  C œ
È # x
1

81.

' Èww
1 dw; ”

w œ sec ) tan ) ‰ ' tan# ) d) œ ' asec# )
1b d) Ä ' ˆ sec ) † sec ) tan ) d) œ dw œ sec ) tan ) d) • œ tan )
)  C œ Èw#
1
sec
" w  C

82.

' Èz z
16 dz; ”

#

#

z œ 4 sec ) Ä dz œ 4 sec ) tan ) d) •

sec ) tan ) d) ' 4 tan )†44sec œ 4' )

tan# ) d) œ 4(tan )
))  C

œ Èz#
16
4 sec
" ˆ 4z ‰  C 83. u œ ln (x  1), du œ

dx x1

; dv œ dx, v œ x;

' ln (x  1) dx œ x ln (x  1)
' x x 1 dx œ x ln (x  1)
' dx  ' x dx 1 œ x ln (x  1)
x  ln (x  1)  C" œ (x  1) ln (x  1)
x  C" œ (x  1) ln (x  1)
(x  1)  C, where C œ C"  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises 84. u œ ln x, du œ

dx x

; dv œ x# dx, v œ

" 3

x$ ;

' x# ln x dx œ 3" x$ ln x
' 3" x$ ˆ x" ‰ dx œ x3

$

85. u œ tan
" 3x, du œ

3 dx 1  9x#

#

" 6

x$ 9

C

; dv œ dx, v œ x;

' tan
" 3x dx œ x tan
" 3x
' 13x 9xdx œ x tan
" (3x)


ln x


;”

y œ 1  9x# Ä x tan
" 3x
dy œ 18x dx •

" 6

' dyy

ln a1  9x# b  C

86. u œ cos
" ˆ x# ‰ , du œ


dx È 4
x#

; dv œ dx, v œ x;

' cos
" ˆ x# ‰ dx œ x cos
" ˆ x# ‰  ' Èx dx

4
x #

;”

y œ 4
x# Ä x cos
" ˆ x# ‰
dy œ
2x dx •

" #

' Èdyy

#

œ x cos
" ˆ x# ‰
È4
x#  C œ x cos
" ˆ x# ‰
2É1
ˆ x# ‰  C ex

87.

ÐÑ (x  1)# ïïïïî ex Ð
Ñ 2(x  1) ïïïïî ex ÐÑ 2 ïïïïî ex Ê

0

' (x  1)# ex dx œ c(x  1)#
2(x  1)  2d ex  C

sin (1
x)

88. ÐÑ x# ïïïïî Ð
Ñ 2x ïïïïî ÐÑ 2 ïïïïî

cos (1
x)
sin (1
x)
cos (1
x) Ê

0

' x# sin (1
x) dx œ x# cos (1
x)  2x sin (1
x)
2 cos (1
x)  C

89. u œ cos 2x, du œ
2 sin 2x dx; dv œ ex dx, v œ ex ; I œ ' ex cos 2x dx œ ex cos 2x  2 ' ex sin 2x dx; u œ sin 2x, du œ 2 cos 2x dx; dv œ ex dx, v œ ex ;

I œ ex cos 2x  2 ’ex sin 2x
2 ' ex cos 2x dx“ œ ex cos 2x  2ex sin 2x
4I Ê I œ

ex cos 2x 5



2ex sin 2x 5

90. u œ sin 3x, du œ 3 cos 3x dx; dv œ ec2x dx, v œ
"# ec2x ; I œ ' ec2x sin 3x dx œ
"# ec2x sin 3x 

3 2

' ec2x cos 3x dx;

u œ cos 3x, du œ
3 sin 3x dx; dv œ ec2x dx, v œ
"# ec2x ; I œ
"# ec2x sin 3x  3# ’
"# ec2x cos 3x
Ê Iœ

4 13

3 #

' ec2x sin 3x dx“ œ
"# ec2x sin 3x
34 ec2x cos 3x
94 I

2 c2x ˆ
"# ec2x sin 3x
34 ec2x cos 3x‰  C œ
13 e sin 3x


91.

' x
x 3xdx 2 œ ' x2
dx2
' x dx
1 œ 2 ln kx
2k
ln kx
1k  C

92.

' x x 4xdx 3 œ #3 ' xdx3
#" ' x dx 1 œ #3 ln kx  3k
#" ln kx  1k  C

3 13

ec2x cos 3x  C

#

#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

C

565

566

Chapter 8 Techniques of Integration

93.

' x(xdx 1)

œ ' Š "x


94.

' x x(x
11) dx œ ' ˆ x
2 1
2x
x" ‰ dx œ 2 ln ¸ x
" ¸  "x  C œ
2 ln kxk  "x  2 ln kx
1k  C x

95.

' cos )sin )cosd))
2 ; ccos ) œ yd

#


1 (x  1)# ‹



1 x1

#

Ä
'

" 3

œ
"3 '

dy y#  y
2

C

dy y
1



" 3

' y dy # œ 3" ln ¹ yy
 12 ¹  C

)2¸ " ¸ cos )
1 ¸ ln ¸ cos cos )
1  C œ
3 ln cos )  2  C

96.

) d) ' sin )cos sin )
6 ; csin ) œ xd

97.

' 3x x 4xx 4 dx œ '

98.

' x4xdx4x œ ' x4 dx4 œ 2 tan
" ˆ #x ‰  C

99.

' (v2v
3)8vdv œ #" '

#

4 x

$

$

'

Ä

#

dx
'

dx x#  x
6

x
4 x#  1

œ

" 5

)
2¸ ' xdx
2
5" ' xdx 3 œ 5" ln ¸ sin sin )  3  C

dx œ 4 ln kxk


" #

ln ax#  1b  4 tan
" x  C

#

$

œ

" x1

#

#

œ

dx œ ln kxk
ln kx  1k 

" 16

3  Š
4v

5 8(v
2)



" 8(v  #) ‹

dv œ
38 ln kvk 

5 16

ln kv
2k 

&


7) dv ' (v
(3v ' (
v 2)
1dv  ' 1)(v
2)(v
3) œ

101.

' t  4tdt  3 œ "# ' t dt 1
#" ' t dt 3 œ #" tan
" t


102.

' t
t tdt
2 œ "3 ' t t
dt2
3" ' t t dt1 œ 6" ln kt#
2k
6" ln at#  1b  C

103.

' x xx x
2 dx œ ' ˆx  x 2xx
2 ‰ dx œ ' x dx  32 ' x dx
1  34 ' x dx 2

%

#

#

%

#

#

$

dv v
2

'

dv v
3

#


3) œ ln ¹ (v
(v2)(v
1)# ¹  C

" #È 3

tan
" Š Èt 3 ‹  C œ

" #

tan
" t




ln kx  2k 

2 3

105.

' x x4x4x 3 dx œ ' ˆx
x 3x4x  3 ‰ dx œ '

$

107.

$

$

#

#

'

#

#

x #




9 #

ln kx  3k 

3 #

2x$  x#
21x  24 dx œ x#  2x
8 2 # x
3x  3 ln kx  4k

dx x ˆ3 È x  1 ‰

œ

" 3

x dx 

3 #

 ln kx
1k
ln kxk  C

' x dx 1
#9 ' xdx3

ln kx  1k  C

'

œ

C

ln kx
1k  C

$

'

t È3

#

4 3

' xx 
1x dx œ ' ˆ1  xx 
"x ‰ dx œ ' ’1  x(x "
1) “ dx œ ' dx  ' x dx
1
' dxx œ x

106.

tan
"

#

104.

œ

È3 6

#

#

œ

ln kv  2k  C

ln ¹ (v
2)v'(v  2) ¹  C

100.

x# #

" 16


(2x
3)  

" 3

x ‘ x#  2x
8

dx œ ' (2x
3) dx 

" 3

' x dx
#  32 ' x dx 4

ln kx
2k  C

Ô u œ Èx  1 × Ù Ä ; Ö du œ 2Èdx x1 Õ dx œ 2u du Ø

2 3

' au u
du1b u œ 3" ' u du
1
3" ' u du 1 œ 3" ln ku
1k
3" ln ku  1k  C #

È

1
" ln ¹ Èxx  ¹C 11

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises $ Ô u œ Èx × dx Ö du œ dx Ù Ä ' $ x‰ ; x ˆ1  È 3x#Î$ Õ dx œ 3u# du Ø

108.

'

109.

' e ds
1 ; Ô du œ es ds ×

u œ es
1

110.

s

'

Õ ds œ

È es  1
" ¹ È es  1  1 ¹

' È16y dy
y

111. (a)

' È16y dy
y

(b)

Ø

du u1

#

; cy œ 4 sin xd Ä 4 '

d a16
y# b È16
y#

œ

(b)

' Èx dx

; cx œ 2 tan yd Ä

(b)

#

#

#

C

' Èx dx

' 4x
dxx ' 4x
dxx

du ' u a2uu du
1b œ 2 ' (u  1)(u ' u du
1
' u du 1 œ ln ¸ uu 
"1 ¸  C
1) œ

u
1

112. (a)

113. (a)

$ x È

œ 3 ln ¸ u u 1 ¸  C œ 3 ln ¹ 1È $ x¹  C

s

œ
"# '

4  x#

du u(1  u)

s

#

4  x#

œ 3'

' u(udu 1) œ
' udu1  ' duu œ ln ¸ u u 1 ¸  C œ ln ¸ e e
" ¸  C œ ln k1
ecs k  C

È s Ôu œ es 1× e ds Ö ; du œ 2Èes 1 Ù Ä Õ ds œ 2u# du Ø

ds È es  1

œ ln

Ä

3u# du u$ (1  u)

" #

œ
È16
y#  C sin x cos x dx cos x

œ
4 cos x  C œ


4È16
y# 4

 C œ
È16
y#  C

' dÈa4  x b œ È4  x#  C #

4  x#

œ
#" '

d a4
x # b 4
x#

; cx œ 2 sin )d Ä

' 2 tan y2†2secsecy y dy œ 2 ' sec y tan y dy œ 2 sec y  C œ È4  x#  C #

œ
#" ln k4
x# k  C

È 2 cos ) d) ' 2 sin4)†cos œ ' tan ) d) œ
ln kcos )k  C œ
ln Š 4 2
x ‹  C ) #

#

œ
"# ln k4
x# k  C 114. (a)

' È t dt

œ

(b)

' È t dt

;
t œ

4t#
1

4t#
1

" 8

' dÈa4t

#
1b 4t#
1

" #

œ

" 4

sec )‘ Ä

È4t#
1  C

'

u œ 9
x# Ä
#" ' du œ
2x dx •

" #

sec ) tan )† "# sec ) d) tan )

œ

" 4

' sec# ) d) œ tan4 )  C œ È4t4
1  C #

115.

' 9x
dxx

116.

' x a9dx
x b œ 9" ' dxx  18" ' 3 dx
x
18" ' 3 dx x œ 9" ln kxk
18" ln k3
xk
18" ln k3  xk  C

#

;”

du u

œ
#" ln kuk  C œ ln

" Èu

 C œ ln

" È 9
x#

C

#

œ

" 9

ln kxk


" 18

ln k9
x# k  C

117.

' 9
dxx

118.

' È dx

119.

' sin3 x cos4 x dx œ ' cos4 xa1
cos2 xbsin x dx œ ' cos4 x sin x dx
' cos6 x sin x dx œ
cos5 x  cos7 x  C

120.

#

œ

9
x#

" 6

;”

' 3 dx
x  6" ' 3dxx œ
6" ln k3
xk  6" ln k3  xk  C œ 6" ln ¸ xx
 33 ¸  C x œ 3 sin ) Ä dx œ 3 cos ) d) •

) ' 33 cos ' d) œ )  C œ sin
" x  C cos ) d) œ 3

5

' cos5 x sin5 x dx œ ' sin5 x cos4 x cos x dx œ ' sin5 x a1
sin2 xb2 cos x dx œ ' sin5 x cos x dx
2' sin7 x cos x dx  ' sin9 x cos x dx œ sin6 x
2sin8 x  sin10 x  C 6

8

10

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

7

567

568

Chapter 8 Techniques of Integration

121.

' tan4 x sec2 x dx œ tan5 x  C

122.

' tan3 x sec3 x dx œ ' asec2 x
1b sec2 x † sec x † tan x dx œ ' sec4 x † sec x † tan x dx
' sec2 x † sec x † tan x dx

5

œ 123.

sec5 x 5




sec3 x 3

C

' sin 5) cos 6) d) œ "# ' asina
)b  sina11)bb d) œ "# ' sina
)b d)  "# ' sina11)b d) œ "# cosa
)b
##" cos 11)  C œ "# cos )


" ## cos

11)  C

124.

' cos 3) cos 3) d) œ "# ' acos 0  cos 6)b d) œ "# '

d) 

125.

' É1  cosˆ 2t ‰ dt œ ' È2¸ cos

t¸ 4

126.

' et Ètan2 et  1 dt œ ' k sec et k et dt œ lnk sec et  tan et k  C 3
" 180 Ð%Ñ

127. kEs k Ÿ

t¸ 4

dt œ 4È2 ¸ sin

3
" n

(˜x)% M where ˜x œ

œ

2 n

; f(x) œ

" x

" #

' cos 6) d) œ "# )  121 sin 6)  C

C

œ x
" Ê f w (x) œ
x
# Ê f ww (x) œ 2x
$ Ê f'''(x) œ
6x
%

(x) œ 24x
& which is decreasing on [1ß 3] Ê maximum of f Ð%Ñ (x) on [1ß 3] is f Ð%Ñ (1) œ 24 Ê M œ 24. Then
" ‰ ˆ 2 ‰% ˆ 768 ‰ ˆ n"% ‰ Ÿ 0.0001 Ê n"% Ÿ (0.0001) ˆ 180 ‰ Ê n%   10,000 ˆ 768 ‰ kEs k Ÿ 0.0001 Ê ˆ 3180 n (24) Ÿ 0.0001 Ê 180 768 180 Ê f

Ê n   14.37 Ê n   16 (n must be even) 1
0 12

128. kET k Ÿ Ê

2 3n#

129. ˜x œ

(˜x)# M where ˜x œ

Ÿ 10
$ Ê

b
a n

œ

1
0 6

3n# #

œ

1 6

1
0 n #

œ

  1000 Ê n   Ê

˜x #

œ

1 1#

" n ;0 2000 3

! mf(xi ) œ 12 Ê T œ ˆ 1 ‰ (12) œ 1 ; 12 iœ0

6

iœ0

˜x 3

1 18

Ê

2
" n

œ

œ

1‰ S œ ˆ 18 (18) œ 1 .

130. ¸f Ð%Ñ (x)¸ Ÿ 3 Ê M œ 3; ˜x œ

" n

ˆ "n ‰# (8) Ÿ 10
$

" 12

Ê n   25.82 Ê n   26

x! x" x# x$ x% x& x'

xi 0 1/6 1/3 1/2 21/3 51/6 1

f(xi ) 0 1/2 3/2 2 3/2 1/2 0

m 1 2 2 2 2 2 1

mf(xi ) 0 1 3 4 3 1 0

x! x" x# x$ x% x& x'

xi 0 1/6 1/3 1/2 21/3 51/6 1

f(xi ) 0 1/2 3/2 2 3/2 1/2 0

m 1 4 2 4 2 4 1

mf(xi ) 0 2 3 8 3 2 0

;

6

! mf(xi ) œ 18 and

Ÿ f ww (x) Ÿ 8 Ê M œ 8. Then kET k Ÿ 10
$ Ê

%


"‰ ˆ"‰
& . Hence kEs k Ÿ 10
& Ê ˆ 2180 Ê n (3) Ÿ 10

" 60n%

Ÿ 10
& Ê n%  

Ê n   6.38 Ê n   8 (n must be even) 131. yav œ œ œ

" 365
0

21 "
'0365
37 sin ˆ 365 ˆ 21 ‰ ‰‘ $'& (x
101)‰  25‘ dx œ 365
37 ˆ 365 21 cos 365 (x
101)  25x !

"
ˆ ‰
21 ‘ ‰ ˆ ˆ 365 ‰
21 ‘ ‰‘
37 ˆ 365 365 21 cos 365 (365
101)  25(365)

37 21 cos 365 (0
101)  25(0) 21 21 21 21
2371 cos ˆ 365 (264)‰  25  2371 cos ˆ 365 (
101)‰ œ
2371 ˆcos ˆ 365 (264)‰
cos ˆ 365 (
101)‰‰ 

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

25

10& 60

Chapter 8 Practice Exercises

569

37 ¸
#1 (0.16705
0.16705)  25 œ 25° F " 675
20

132. av(Cv ) œ ¸

" 655

"
"3 '20675 c8.27  10
& a26T
1.87T# bd dT œ 655 8.27T  10

&

T#


0.62333 10&

'(&

T$ ‘ #!

[(5582.25  59.23125
1917.03194)
(165.4  0.052
0.04987)] ¸ 5.434;

8.27  10
& a26T
1.87T# b œ 5.434 Ê 1.87T#
26T
283,600 œ 0 Ê T ¸

26È676  4(1.87)(283,600) #(1.87)

¸ 396.45° C 133. (a) Each interval is 5 min œ

1 12

hour.

 2a2.4b  2a2.3b  Þ Þ Þ  2a2.4b  2.3 d œ ‰ (b) a60 mphbˆ 12 29 hours/gal ¸ 24.83 mi/gal 1 2.5 24 c

29 12

¸ 2.42 gal

134. Using the Simpson's rule, ˜x œ 15 Ê ˜x 3 œ 5; ! mf(xi ) œ 1211.8 Ê Area ¸ a1211.8ba5b œ 6059 ft2 ;

x! x" x# x$ x% x5 x' x( x)

The cost is Area † a$2.10/ft2 b ¸ a6059 ft2 ba$2.10/ft2 b œ $12,723.90 Ê the job cannot be done for $11,000.

135.

'03 È dx

œ lim c '0 bÄ3

136.

'01 ln x dx œ

b

9
x#

xi 0 15 30 45 60 75 90 105 120

f(xi ) 0 36 54 51 49.5 54 64.4 67.5 42

b œ lim c
sin
" ˆ x3 ‰‘ 0 œ lim c sin
" ˆ 3b ‰
sin
" ˆ 30 ‰ œ bÄ3 bÄ3

dx È 9
x#

lim cx ln x
xd 1b œ (1 † ln 1
1)
lim b cb ln b
bd œ
1
lim b b Ä !b bÄ! bÄ!

m 1 4 2 4 2 4 2 4 1 1 #

mf(xi ) 0 144 108 204 99 216 128.8 270 42 1 #


0œ

ln b Š "b ‹

œ
1
lim b bÄ!

œ
1  0 œ
1 137.

'c11 ydy

138.

'c_2 () d1))

#Î$

œ 'c1 0

dy y#Î$

 '0


1

œ 'c2

$Î&

)$Î&

lim $Î& ) Ä _ ()1)

1

dy y#Î$

d) ()  1)$Î&

1

_

d) )$Î&

diverges Ê

'3_ u 2
du2u œ '3_ u du
2
'3_ duu œ

140.

'1_ 4v3v

v1

#

_

#

dv œ '1 ˆ "v 

1 œ 2 † 3 lim b
y"Î$ ‘ b œ 6 Š1
lim b b"Î$ ‹ œ 6 bÄ! bÄ!

2

139.

$

dy y#Î$

) '2  'c1 () d1) $Î& 

_

œ 1 and '2

œ 2 '0

" v#




d) ()  1)$Î&

'
2 () d1))

$Î&

diverges

b lim
ln ¸ u
u 2 ¸‘ 3 œ lim
ln ¸ b
b 2 ¸‘
ln ¸ 3
3 2 ¸ œ 0
ln ˆ "3 ‰ œ ln 3

bÄ_

4 ‰ 4v
1

bÄ_

dv œ lim
ln v
bÄ_

œ lim
ln ˆ 4b b
1 ‰
b" ‘
(ln 1
1
ln 3) œ ln bÄ_

141.

'0_ x# ecx dx œ

142.

'c0_ xe3x dx œ

143.

'c__ 4xdx 9 œ 2 '0_ 4xdx 9 œ #" '0_ x dx #

converges if each integral converges, but

_

" 4

" v

b


ln (4v
1)‘ 1

 1  ln 3 œ 1  ln

3 4

lim c
x# ecx
2xecx
2ecx d 0 œ lim a
b# ecb
2becb
2ecb b
(
2) œ 0  2 œ 2 b

bÄ_

lim

b Ä
_

#

bÄ_


x3 e3x
9" e3x ‘ 0 œ
9"
b

#

9 4

œ

lim

b Ä
_

" # b lim Ä_

ˆ b3 e3b
9" e3b ‰ œ
9"
0 œ
9"


32 tan
" ˆ 2x ‰‘ b œ 3 0

" # b lim Ä_


32 tan
" ˆ 2b ‰‘
3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 3

tan
" (0)

Š b" ‹ Š


" ‹ b#

570

Chapter 8 Techniques of Integration " #

œ

ˆ 23 † 1# ‰
0 œ

1 6

144.

'c__ x 4dx16 œ 2'0_ x 4dx16 œ 2

145.

) ) Ä _ È)#  1

#

#

b lim
tan
" ˆ x4 ‰‘ 0 œ 2 Š lim
tan
" ˆ b4 ‰‘
tan
" (0)‹ œ 2 ˆ 1# ‰
0 œ 1

bÄ_

_ d)

œ 1 and '6

lim

bÄ_

_

diverges Ê '6

)

d) È)#  1

_

diverges

_

_

146. I œ '0 ecu cos u du œ lim c
ecu cos ud b0
'0 ecu sin u du œ 1  lim cecu sin ud b0
'0 aecu b cos u du bÄ_

Ê I œ 1  0
I Ê 2I œ 1 Ê I œ 147.

bÄ_

" #

converges

'1_ lnz z dz œ '1e lnz z dz  'e_ lnz z dz œ ’ (ln2z) “ e  #

1

œ _ Ê diverges 148. 0 

ect Èt

_

Ÿ ect for t   1 and '1 ect dt converges Ê

149.

_ 2 dx '
_ e  ec

_

150.

_ dx 'c_ '
1 dx ' 0 dx '1 x a1  e b œ
_ x a1  e b 
1 x a1  e b  0

œ 2'0

x

x

#

Š x"# ‹

x Ä 0 ’ x# a1"
ex b “ #

' 1 x dxÈx ; – œ

œ lim

xÄ0

2x$Î# 3

converges Ê

4 dx ex

#

x

x # a1  e x b x#

x

#

#

bÄ_

e

'1_ Èect dt converges t

_ 2 dx '
_ e  ec x

dx x # a1  e x b

x

converges _

 '1

œ lim a1  ex b œ 2 and '0

'c__ x a1dx e b diverges

Ê

151.

_

 '0

#

x

lim

2 dx ex  ecx

b

#

lim ’ (ln#z) “ œ Š 12
0‹  lim ’ (ln2b)
"# “

bÄ_

1

xÄ0

dx x#

dx x # a1  e x b

;

diverges Ê

'01 x a1dx e b diverges #

x

x

u œ Èx — Ä du œ #dx Èx

' u 1†2u udu œ ' ˆ2u#
2u  2
1 2 u ‰ du œ 32 u$
u#  2u
2 ln k1  uk  C #


x  2Èx
2 ln ˆ1  Èx‰  C

152.

' x4
x2 dx œ
' ˆx  4xx
 42 ‰ dx œ
'

153.

'

$

#

#

dx x ax #  1 b #

;”

x œ tan ) Ä dx œ sec# ) d) •

œ ln ksin )k


154.

' cosÈÈx x dx; –

155.

'È

156.

' (tÈ
1) dt ' ”

157.

' È du

dx
2x
x#

t#
2t

1  u#

" #

3 #

' x dx 2
#5 ' x dx
2 œ
x#

#




3 #

ln kx  2k


) d) ' tansec) sec œ ' cossin))d) œ ' Š 1
sinsin) ) ‹ d(sin )) ) #

$

#

%

#

sin# )  C œ ln ¹ Èxx#  1 ¹
"# Š Èx#x  1 ‹  C

u œ Èx — Ä du œ #dx Èx

œ'

x dx


d(x  1) È1
(x  1)#

' cos uu†2u du œ 2' cos u du œ 2 sin u  C œ 2 sin Èx  C

œ sin
" (x  1)  C

u œ t#
2t Ä du œ (2t
2) dt œ 2(t
1) dt •

; cu œ tan )d Ä

" #

' Èduu œ Èu  C œ Èt#
2t  C

' secsec))d) œ ln ksec )  tan )k  C œ ln ¹È1  u#  u¹  C #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5 #

ln kx
2k  C

Chapter 8 Practice Exercises 158.

' et cos et dt œ sin et  C

159.

' 2
cossinx x sin x dx œ ' #

2 csc# x dx
'

cos x dx sin# x

 ' csc x dx œ
2 cot x 

" sin x


ln kcsc x  cot xk  C

œ
2 cot x  csc x
ln kcsc x  cot xk  C 160.

sin ) ' cos ' 1
coscos) ) d) œ ' sec# ) d)
' d) œ tan )
)  C ) d) œ

161.

' 819
dvv

162.

x)
" ' 1cos sinx dxx œ ' 1d(sin (sin x)  C  sin x œ tan

163.

#

#

#

#

%

œ

" #

' v dv 9  12" ' 3 dv
v  1"# ' 3 dv v œ 12" ln ¸ 33
 vv ¸  6" tan
" v3  C #

#

#

cos (2)  1) ÐÑ " ) ïïïïî # sin (2)  1) Ð
Ñ 1 ïïïïî
"4 cos (2)  1) Ê

0

' ) cos (2)  1) d) œ #) sin (2)  1)  "4 cos (2)  1)  C

164.

'2_ (x
dx1)

165.

' x
x 2xdx 1 œ ' ˆx  2  x 3x
2x
2 1 ‰ dx œ '

b œ lim
1
" x ‘ 2 œ lim
1
" b
(
1)‘ œ 0  1 œ 1

bÄ_

bÄ_

$

#

œ

166.

#

'

#

x# #

 2x  3 ln kx
1k


d) É 1  È)

" x
1

× Ù Ä 2 ) Õ d) œ 2(x
1) dx Ø

œ

Š1  È ) ‹

$Î#


4 Š1  È ) ‹

y œ Èx — Ä dy œ 2dx Èx

"Î#

Cœ4

Ô ŒÉ1  È)
3

Õ

' 2Èsinx secÈxÈdxx ; –

168.

' xx
dx16 œ ' ˆx  x 16x ‰ dx œ x#
16

169.

'

170.

' )
d2))  4 œ ' ()
d1))  3 œ È33 tan
" Š )È
" ‹  C

171.

tan x ' cos ' tan x sec# x dx œ ' tan x † d(tan x) œ "# tan# x  C x dx œ

172.

&

y†2y dy ' 2 siny sec œ ' 2 sin 2y dy œ
cos (2y)  C œ
cos ˆ2Èx‰  C y

#

%

dy sin y cos y

œ'

 ' ˆ x#2x
4


2x ‰ x#  4

dx œ

x# #

#


4  ln ¹ xx#  4¹  C

œ ' 2 csc (2y) dy œ
ln kcsc (2y)  cot (2y)k  C

2 dy sin 2y

#

dx (x
1)#

×
É1  È)  C Ø

167.

%

'

' 2(x
È1)x dx œ 2 ' Èx dx
2 ' Èdxx œ 34 x$Î#
4x"Î#  C $

4 3

dx x
1

C

x œ 1  È) d) dx œ È

Ô ;Ö

(x  2) dx  3 '

#

3

#

'

dr (r  1)Èr#  2r

œ'

d(r  1) (r  1)È(r  1)#
1

œ sec
" kr  1k  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

571

572

Chapter 8 Techniques of Integration

173.

' È(r  2) dr

œ'

174.

' 4ydyy

" #

175.

2) d) ' (1sin cos 2) )


r#

%


4r

œ

#

'

(r  2) dr È4
(r  2)#

d ay # b 4  ay # b #

œ

œ
#" '

u œ %
(r  2)# Ä
' du œ
2(r  2) dr •

du 2È u

œ
Èu  C œ
È4
(r  2)#  C

#

tan
" Š y# ‹  C

d(1  cos 2)) (1  cos 2))#

œ

" #(1  cos 2))

Cœ

" 4

sec# )  C

176.

'

177.

'11ÎÎ42 È1  cos 4x dx œ
È2 '11ÎÎ42 cos 2x dx œ ’
È#2 sin 2x“ 1Î2 œ È#2

178.

' (15)2x
1 dx œ "# ' (15)2x
1 d(2x  1) œ "# Š 15ln 15b ‹  C

179.

' Èx dx

dx ax #
1 b #

œ'

" 4

;”

œ

dx a 1
x # b#



x 2 a1
x # b

" 4

"¸ ln ¸ xx
1  C (FORMULA 19)

1Î4

2x 1

;”

2
x

yœ2
x Ä
' dy œ
dx •

(2
y) dy Èy

œ

y$Î#
4y"Î#  C œ

2 3

2 3

(2
x)$Î#
4(2
x)"Î#  C

$

ŠÈ2
x‹

œ 2–

180.

3

' È1v
v

#

#


2È2
x—  C

' cos )sin†cos)) d) œ ' a1
sinsin ))b d) œ ' csc# ) d)
' d) œ cot )
)  C #

dv; cv œ sin )d Ä

œ
sin
" v


È 1
v# v

#

#

C

181.


1)
" ' y
dy2y  2 œ ' (yd(y (y
1)  C
1)  1 œ tan

182.

' ln Èx
1 dx; – y œ

#

#

Èx
1

dy œ

Ê œ

" #

dx 2È x
1

— Ä

ln y † 2y dy; u œ ln y, du œ

dy y

; dv œ 2y dy, v œ y#

' 2y ln y dy œ y# ln y
' y dy œ y# ln y
"# y#  C œ (x
1) ln Èx
1
"# (x
1)  C" c(x
1) ln kx
1k
xd  ˆC"  "# ‰ œ

183.

'

184.

'È

185.

' z azz1 4b dz œ "4 ' ˆ "z  z"

186.

' x$ ex

187.

' È t dt

188.

'01Î10 È1  cos 5) d) œ È2 '01Î10

)# tan a)$ b d) œ

#

'

x dx 8
2x#
x%

œ

" #

" 3

'

d ax #  1 b É 9
ax #  1 b #

#

dx œ

9
4t#

" #

' x# ex

œ
"8 '

cx ln kx
1k
x
ln kx
1kd  C

' tan a)$ b d a)$ b œ 3" ln ksec )$ k  C

#

#

" #

#




d a x# b œ

d a9
4t# b È9
4t#

œ

" #

z1 ‰ z#  4

" #

sin
" Š x

dz œ

" 4

#

1 3 ‹

C

ln kzk


" 4z

ˆx# ex#
ex# ‰  C œ




" 8

ln az#  4b
#

ax #
1 b e x #

" 8

tan
"

z #

C

C

œ
4" È9
4t#  C cos ˆ 5#) ‰ d) œ

2È 2 5


sin ˆ 5#) ‰‘ 1Î"! œ !

2È 2 5

ˆsin

1 4


0‰ œ

2 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises 189.

) d) ' 1cot sin) d)) œ ' (sin )cos ) a1  sin )b ; ” #

#

' x a1dx x b œ ' dxx
' xx dx1

x œ sin ) Ä dx œ cos ) d) •

#

#

œ ln ksin )k
"# ln a1  sin# )b  C 190. u œ tan
" x, du œ

dx 1  x#

; dv œ

dx x#

, v œ
x" ;

' tan
"x#x dx œ
"x tan
" x  ' x a1dx x#b œ
x" tan
" x  ' dxx
' 1xdxx# œ
x" tan
" x  ln kxk
191.

' tan2ÈÈyy dy ;
Èy œ x‘

192.

'e œ

193.


" x

" #

ln a1  x# b  C œ
tanx

' tan x2x†2x dx œ ln ksec xk  C œ ln ¸sec Èy¸  C

Ä

'

et dt dx t (x  1)(x  2)  3et  2 ; ce œ xd Ä t "¸ ˆe "‰ ln ¸ xx   #  C œ ln et  #  C 2t

' 4)
d))

œ ' ˆ
1 

#

#

œ
)  ln

2¸ ¸ ))
2

4 ‰ 4
)#

 ln kxk
ln È1  x#  C

œ'

dx x1

d) œ
' d )
'


'

d) )
#

'

dx x#

œ ln kx  1k
ln kx  2k  C

d) )#

œ
)
ln k)
2k  ln k)  2k  C

C

194.

2x ' 11
 cos ' cos 2x dx œ

tan# x dx œ ' asec# x
1b dx œ tan x
x  C

195.

' cos Èasin
" x#b dx ; – u œ sin dx x —

196.

' sincosx
x dxsin x œ
' (sin x)cosa1x
dxsin xb œ
' (sincosx) axcosdx xb œ
' sin2 dx2x œ
2' csc 2x dx


"

du œ

1
x

È 1
x#

$

Ä

' cos u du œ sin u  C œ sin asin
" xb  C œ x  C

#

#

œ ln kcsc (2x)  cot (2x)k  C 197.

'

198.

' x
x  2 dx œ ' x dx 2
'

sin

x #

cos

x #

dx œ '

" #

sin ˆ x#  x# ‰ dx œ

#

ax #  2 b#

œ

" È2

#

tan
" Š Èx2 ‹ 

199.

'

200.

' tan$ t dt œ '

201.

'1_ ln yy dy ; Ö

et dt 1  et

x dx a x #  2 b#

" # ax #  2 b

œ

" È2

" #

' sin x dx œ
"# cos x  C

tan
" Š Èx2 ‹  "# ax#  2b


"

C

C

œ ln a1  et b  C (tan t) asec# t
1b dt œ

Ô x œ ln y × Ù Ä dx œ dy y x Õ dy œ e dx Ø

$


b œ lim ˆ 2e 2b


bÄ_

" ‰ 4e2b

'0_ xe†e

x

3x


ˆ0
"4 ‰ œ

tan# t #


' tan t dt œ


ln ksec tk  C

_

b dx œ '0 xec2x dx œ lim

x# ec2x
4" ec2x ‘ 0

bÄ_

" 4

202.

' 3  sectanxx sin x dx œ 3 ' cot x dx  ' sectanxxdx  '

203.

u œ ln (sin v) v dv ' lncot(sinv dvv) œ ' (sincos cos v dv • v) ln (sin v) ; ”

#

#

du œ

tan# t #

sin v

Ä

cos x dx œ 3 ln ksin xk  ln ktan xk  sin x  C

' duu œ ln kuk  C œ ln kln (sin v)k  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

573

574

Chapter 8 Techniques of Integration

'

204.

œ'

dx (2x
1)Èx#
x

œ'

2 dx (2x
1)È4x#
4x

2 dx (2x
1)È(2x
1)#
1

;”

u œ 2x
1 Ä du œ 2 dx •

'

du uÈ u#
1

œ sec
" kuk  C œ sec
" k2x
1k  C 205.

' eln Èx dx œ ' Èx dx œ 23 x$Î#  C

206.

' e) È3  4e) d); ”

207.

' 1 sin(cos5t dt5t)

208.

' È dv

209.

'

e2v
1

#

;”

;”

5x% 20x$ 60x# 120x 120

'

x œ ev Ä dx œ ev dv • " 3

' È3  u du œ 4" † 23 (3  u)$Î#  C œ 6" a3  4e) b$Î#  C

dx xÈ x#
1

du 1  u#

œ
5" tan
" u  C œ
5" tan
" (cos 5t)  C

œ sec
" x  C œ sec
" aev b  C

' (27)3)
1 d(3)  1) œ

sin x ÐÑ ïïïïî
cos x Ð
Ñ ïïïïî
sin x ÐÑ ïïïïî cos x Ð
Ñ ïïïïî sin x ÐÑ ïïïïî
cos x Ð
Ñ ïïïïî
sin x

x&

" 4

u œ cos 5t Ä
"5 ' du œ
5 sin 5t dt •

(27)3)
1 d) œ

210.

u œ 4e) Ä du œ 4e) d) •

" 3 ln 27

(27)3)
"  C œ

" 3

3)b1

Š 27 ln 27 ‹  C

Ê ' x& sin x dx œ
x& cos x  5x% sin x  20x$ cos x
60x# sin x
120x cos x

0

 120 sin x  C u œ Èr dr — Ä du œ 2È r

211.

' 1 drÈr ; –

212.

d ˆx
10x  9‰
20x ' x 4x
10x dx œ ' x
10x  9 œ ln kx%
10x#  9k  C 9

213.

' y (y8 dy 2) œ ' dyy
' 2ydy  ' 4ydy
' (y dy 2) œ ln ¹ y y 2 ¹  2y
y2

214.

'

215.

'

216.

' t(1  ln t)Èdt(ln t)(2  ln t) ; ” u œ lndtt •

%

$

%

' 2u1 duu œ ' ˆ2
1 2 u ‰ du œ 2u
2 ln k1  uk  C œ 2Èr
2 ln ˆ1  Èr‰  C

#

%

$

(t  1) dt at#  2tb#Î$

#

#

#

;”

8 dm mÈ49m#
4

$

#

u œ t#  2t Ä du œ 2(t  1) dt •

œ

8 7

'

' udu

#Î$

œ

" #

† 3u"Î$  C œ

3 #

at#  2tb

"Î$

C

¸C œ 4 sec
" ¸ 7m #

dm # mÉm#
ˆ 27 ‰

du œ

" #

C

t

Ä

' (1  u)Èduu(2  u) œ ' (u  1)Èdu (u  1)
1 #

œ sec
" ku  1k  C œ sec
" kln t  1k  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Practice Exercises

'0x È1  (t
1)% dt and dv œ 3(x
1)# dx, then du œ È1  (x
1)% dx, and v œ (x
1)$ so integration

217. If u œ

by parts Ê '0 3(x
1)# ’'0 È1  (t
1)% dt“ dx œ ’(x
1)$ 1

'0x È1  (t
1)% dt“ "

x

1 È $Î# "
'0 (x
1)$ È1  (x
1)% dx œ ’
"6 a1  (x
1)% b “ œ 86
"

!

!

4v$  v
1 v# (v
1) av#  1b

218.

œ

A v #



B v#



C v
1



Dv  E v#  1 #

Ê 4v$  v
1

œ Av(v
1) av  1b  B(v
1) av  1b  Cv# av#  1b  (Dv  E) av# b (v
1) v œ 0:
1 œ
B Ê B œ 1; v œ 1: 4 œ 2C Ê C œ 2; coefficient of v% : 0 œ A  C  D Ê A  D œ
2; coefficient of v$ : 4 œ
A  B  E
D coefficient of v# : 0 œ A
B  C
E Ê C
D œ 4 Ê D œ
2 (summing with previous equation); coefficient of v: 1 œ
A  B Ê A œ 0; in summary: A œ 0, B œ 1, C œ 2, D œ
2 and E œ 1

'2_ v (v4v
1) vav
1 1b dv œ $

Ê

#

#

œ lim
ln (v
1)#
bÄ_

#

" v

1) œ lim ’ln Š (b1
 b# ‹


œ

bÄ_ 1 #

 ln (5) 

" #

lim

bÄ_

'2b ˆ v
2 1  v
#  1 " v

#




2v ‰ 1  v#

dv

 tan
" v
ln a1  v# b‘ 2 b

" b

 tan
" b“
ˆln 1


" #

 tan
" 2
ln 5‰ œ ˆ0
0  1# ‰
ˆ0


" #

 tan
" 2
ln 5‰

1 #


tan
" a;


tan
" 2

219. u œ f(x), du œ f w (x) dx; dv œ dx, v œ x;

'13Î12Î2 f(x) dx œ cx f(x)d 311Î2Î2
'13Î12Î2 xf w (x) dx œ
3#1 f ˆ 3#1 ‰
1# f ˆ 1# ‰‘
'13Î12Î2 cos x dx œ ˆ 31#b


'0a 1dxx

220.

1a ‰ #


csin xd 311Î2Î2 œ 1# a3b
ab
c(
1)
1d œ 1# a3b
ab  2

œ ctan
" xd 0 œ tan
" a; 'a a

#

therefore, tan
" a œ

1 #

_

dx 1 x #

œ lim ctan
" xd a œ lim atan
" b
tan
" ab œ b

bÄ_


tan
" a Ê tan
" a œ

1 4

bÄ_

Ê a œ 1 since a  0.

CHAPTER 8 ADDITIONAL AND ADVANCED EXERCISES #

1. u œ asin
" xb , du œ

'

#


'

2.

" x

; dv œ dx, v œ x;

asin
" xb dx œ x asin
" xb
' #

u œ sin
" x, du œ

'

2 sin
" x dx È 1
x#

2x sin
" x dx È 1
x#

dx È 1
x#

2x sin
" x dx È 1
x#

;

; dv œ
È2x dx # , v œ 2È1
x# ; 1
x

œ 2 asin
" xb È1
x#
' 2 dx œ 2 asin
" xb È1
x#
2x  C; therefore

#

#

asin
" xb dx œ x asin
" xb  2 asin
" xb È1
x#
2x  C œ

" x

,

" " " x(x  1) œ x
x  1 , " " " " x(x  1)(x  2) œ 2x
x  1  #(x  2) , " " " " " x(x  1)(x  2)(x  3) œ 6x
#(x  1)  #(x  2)
6(x  3) , " " " " " x(x  1)(x  2)(x  3)(x  4) œ #4x
6(x  1)  4(x  #)
6(x  3) " x(x  1)(x  2) â (x  m)

m

œ! kœ0

(
") (k!)(m
k)!(x  k) ; k

therefore '



" 24(x  4)

Ê the following pattern:

dx x(x  1)(x  2) â (x  m)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

575

576

Chapter 8 Techniques of Integration m

(
") œ ! ’ (k!)(m
k)! ln kx  kk“  C k

kœ0

3. u œ sin
" x, du œ

dx È 1
x#

' x sin
" x dx œ x#

#

4.

œ

x# #

sin
" x


œ

x# #

sin
" x 

sin
" x
'

'

z œ Èy

— Ä

dy 2È y

x# #

;

x œ sin ) Ä dx œ cos ) d) •

;”

x
"# ˆ #)


sin 2) ‰ 4

'

Cœ

x sin
" x dx œ

x# #

sin
" x 



C


" Èy

C

) ' 1
dtan) ) œ ' cos cos ' 12coscos#2)) d) œ #" ' (sec 2)  1) d) œ ln ksec 2) 4tan 2)k  2)  C )
sin ) d) œ #

#

#

#

x  "# œ "# sec ) – dx œ " sec ) tan ) d) — Ä # tan )
ln ksec )  tan )k #

œ

" ‹ 2È 1  x

'

Cœ

" 4

œ

dx 2È x È 1  x

dt t
È1
t#

Õ d) œ

' u du
1
#" ' u du 1
#" ' uu du1 œ #" ln ¹ Èu
1

œ

" #

ln Št
È1
t# ‹


3e2x


ex b dx
6ex
1

u
1œ

Ô

Õ du œ

2 È3

#

' È(2u
1) du

2 È3

sec )

' Š È4

" È3

œ

2 3

È3u#
6u
1 

œ

" È3

" 16

Ä

3u#
6u
1

" È3

3

" È3 " 3

" #

tan
" u  C œ

" #

œ

" È3

'

(2u
1) du É(u
1)#
43

Ä

' (u
1)duau  1b #

ln ¸ tansec)
) " ¸
#" )  C

sec )  1‹ (sec )) d) œ

4 3

;

' sec# ) d)  È" ' sec ) d) 3

† É 34 (u
1)#
1 

" È3

ln ¹

ln ¹u
1  É(u
1)#
43 ¹  ŠC" 

" È3

ln

ln ksec )  tan )k  C" œ

’2Ée2x
2ex


' x " 4 dx œ ' œ

×

sec ) tan ) d) Ø

tan ) 

¹


Ø

du u#  1

C

sin
" t  C

u œ ex Ä du œ ex dx •

4 3

%

u#  1

;”

œ

4

u œ tan )

" #

2x

Èx#  x
ln ¹2x  1  2Èx#  x¹

d) ' sincos)
) cos ' tan d))
1 ; Ô du œ sec# ) d) × ) œ

œ

' Èa2e

;

C

#

t œ sin ) Ä ;” dt œ cos ) d) •

" #

x dx Ɉx  "# ‰#
"4

" #

2Èx#  x
ln ¹2x  1  2Èx#  x¹

#

; dv œ dx, v œ x;

sec ) tan ) d) ' (sec )
ˆ1)†tan œ #" ' asec# )
sec )b d) )‰

' ln ŠÈx  È1  x‹ dx œ x ln ŠÈx  È1  x‹
2

Ê

9.

sin# ) cos ) d) 2 cos )

C

' ln ŠÈx  È1  x‹ dx œ x ln ŠÈx  È1  x‹
"# ' ÈxxÈdx1  x ; "# '

8.

sin
" x
'

sin ) cos )
) 4

' sin
" Èy dy œ y sin
" Èy  Èy È1
y#
sin

" 6. u œ ln ŠÈx  È1  x‹ , du œ Š Èx dx È1  x ‹ Š #Èx 

7.

x# #

' 2z sin
" z dz; from Exercise 3, ' z sin
" z dz

zÈ1
z#
sin
" z C Ê 4
" Èy # È sin
y y sin
" Èy  #
 #

z# sin
" z #

œy

x# dx 2È 1
x#

# " sin# ) d) œ x# sin
" # xÈ1
x#
sin
" x C 4

' sin
" Èy dy; – dz œ œ

5.

; dv œ x dx, v œ

4 3

È3 #

(u
1)  É 34 (u
1)#
1¹  C"

È3 # ‹

 ln ¹ex
1  Ée2x
2ex
3" ¹“  C

" ax#  2b#
4x#

dx œ '

" ax#  2x  2b ax#
2x  2b

dx

' ’ x 2x 2x 2 2  (x  1)2  1
x 2x
2x
2#  (x
1)2  1 “ dx #

#

#

#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Additional and Advanced Exercises

10.

#

" 16

œ

 2x  2 "
" ln ¹ xx#
(x  1)  tan
" (x
1)d  C 2x  # ¹  8 ctan

' x "
1 dx œ 6" '

ˆ x
" 1


'

" x1

x
2 x#
x  1



x2 ‰ x#  x  1




dx

œ

" 6


1¸ ln ¸ xx  1 

" 1#

' ” x 2x

x " 1
ˆ

œ

" 6


"¸ ln ¸ xx  1 

" 1#


x"
" 2x
1 È ’ln ¹ xx#  Š È3 ‹
2È3 tan
" Š 2xÈ3 1 ‹“  C x  1 ¹
2 3 tan

#

3 # x
"# ‰  34




2x  1 x#  x  1




3 ˆx  "# ‰#  34 •

dx

#

11. x lim Ä_

'cxx sin t dt œ x lim c
cos td xcx œ x lim c
cos x  cos (
x)d œ x lim a
cos x  cos xb œ x lim 0œ0 Ä_ Ä_ Ä_ Ä_

12.

'x1 cost t dt;

lim b

xÄ!

#

lim b x 'x

1

xÄ!

lim b

xÄ!

cos t t#

limb

tÄ!

œ '0

1

" È n#
k#


'1

cos t t#

k n

dt

" x

xÄ!

lim b 'x

1

œ1 Ê

xÄ!

cos t t#

dt diverges since '0

1

œ lim b xÄ!


Š cos# x ‹ Š


dt t#

diverges; thus

œ lim b cos x œ 1 xÄ!

x " ‹ x#

! ln ˆ1  k ˆ " ‰‰ ˆ " ‰ œ ' ln (1  x) dx; ” œ n lim n n Ä_ 0 n

1

kœ1

nc1

! ŠÈ œ n lim Ä_ kœ0

"

" È 1
x #

dx œ csin
" xd ! œ

n ‹ ˆ n" ‰ n#
k#

nc1

! œ n lim Ä_ kœ0

Î

"

u œ 1  x, du œ dx x œ 0 Ê u œ 1, x œ 1 Ê u œ 2 •

œ

Ê 1  Š dy dx ‹ œ 1Î2

#

#

a1
x# b  4x# a1
x # b #

x ' ˆ
1  œ '0 Š "1
x# ‹ dx œ 0 1Î2

1Î4

Ê1  ŠÈcos 2t‹ dt œ È2 '0 #

1Î4

œ1 #


2x 1
x#

ˆ n" ‰

1 #

#

1Î%

Ñ #

Ï Ê1
’k Š "n ‹“ Ò

# ' œ Ècos 2x Ê 1  Š dy dx ‹ œ 1  cos 2x œ 2 cos x; L œ 0

œ È2 csin td ! dy dx

" cos t

'12 ln u du œ cu ln u
ud #" œ (2 ln 2
2)
(ln 1
1) œ 2 ln 2
1 œ ln 4
1 nc1

16.

lim b

#

! 14. n lim Ä_ kœ0

dy dx

œ limb tÄ!

x

1 t x 'x cos dt œ t

n

15.

t Š cos# t ‹ t

^ dt is an indeterminate 0 † _ form and we apply l'Hopital's rule:

! ln nÉ1  13. n lim Ä_ kœ1 Ä

Š "# ‹

2 ‰ 1
x#

œ ˆ
"#  ln 3‰
(0  ln 1) œ ln 3


œ

1  2x#  x% a 1
x # b#

x ' œ Š 11 
x# ‹ ; L œ 0

dx œ '0 ˆ
1  1Î2

#

" 1x

#



1Î2

" ‰ 1
x

#

Ê1  Š dy dx ‹ dx "Î#

 x ¸‘ dx œ

x  ln ¸ 11
x !

" #

shell ‰ shell 17. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xy dx b

1

œ 61 '0 x# È1
x dx; 1

Ô uœ1
x × du œ
dx Õ x# œ (1
u)# Ø

Ä
61 '1 (1
u)# Èu du 0

œ
61 '1 ˆu"Î#
2u$Î#  u&Î# ‰ du 0

!

œ
61
23 u$Î#
45 u&Î#  27 u(Î# ‘ " œ 61 ˆ 23
84  30 ‰ 16 ‰ œ 61 ˆ 70
105 œ 61 ˆ 105 œ 32351

4 5

 27 ‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ècos# t dt

577

578

Chapter 8 Techniques of Integration

18. V œ 'a 1y# dx œ 1 '1 b

4

œ 1 '1 ˆ dx x  4

5 dx x#



25 dx x# (5
x)

dx ‰ 5
x

% œ 1
ln ¸ 5
x x ¸
5x ‘ " œ 1 ˆln 4
54 ‰
1 ˆln

œ

151 4

" 4


5‰

 21 ln 4

shell ‰ shell 19. V œ 'a 21 ˆ radius Š height ‹ dx œ '0 21xex dx b

1

œ 21 cxex
ex d "! œ 21

20. V œ '0

ln 2

21(ln 2
x) aex
1b dx

œ 21 '0

ln 2

c(ln 2) ex
ln 2
xex  xd dx ln 2

œ 21 ’(ln 2) ex
(ln 2)x
xex  ex 

x# # “0

œ 21 ’2 ln 2
(ln 2)#
2 ln 2  2 

(ln 2)# # “


21(ln 2  1)

#

œ 21 ’
(ln#2)
ln 2  1“ 21. (a) V œ '1 1 c1
(ln x)# d dx e

œ 1 cx
x(ln x)# d 1  21'1 ln x dx e

e

(FORMULA 110) e œ 1 cx
x(ln x)#  2(x ln x
x)d 1 e œ 1 c
x
x(ln x)#  2x ln xd 1 œ 1 c
e
e  2e
(
1)d œ 1

(b) V œ '1 1(1
ln x)# dx œ 1'1 c1
2 ln x  (ln x)# d dx e

e

œ 1 cx
2(x ln x
x)  x(ln x)# d 1
21'1 ln x dx e

e

œ 1 cx
2(x ln x
x)  x(ln x)#
2(x ln x
x)d 1 e œ 1 c5x
4x ln x  x(ln x)# d 1 œ 1 c(5e
4e  e)
(5)d œ 1(2e
5) e

22. (a) V œ 1 '0
aey b#
1‘ dy œ 1'0 ae2y
1b dy œ 1
e#
y‘ ! œ 1 ’ e#
1
ˆ "# ‰“ œ 1

1

"

2y

#

1 ae #
3 b #

(b) V œ 1'0 aey
1b# dy œ 1'0 ae2y
2ey  1b dy œ 1
e#
2ey  y‘ ! œ 1 ’Š e#
2e  1‹
ˆ "#
2‰“ 1

1

#

œ 1 Š e#
2e  5# ‹ œ 23. (a)

lim

x Ä !b

x ln x œ 0 Ê

2y

"

#

1 ae#
4e  5b #

lim

x Ä !b

f(x) œ 0 œ f(0) Ê f is continuous

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Additional and Advanced Exercises # Ô u œ (ln x) × Ö du œ (2 ln x) dx Ù # 2 2 x # # Ö Ù Ä 1 Œ lim ’ x$ (ln x)# “
' Š x$ ‹ (2 ln x) (b) V œ '0 1x (ln x) dx; Ö # Ù 3 3 b 0 dv œ x dx b bÄ! $ x Õ Ø vœ

579

dx x 

3

$

œ 1 ”ˆ 83 ‰ (ln 2)#
ˆ 23 ‰ lim b ’ x3 ln x
bÄ!

2

x$ 9 “ b•

#

œ 1 ’ 8(ln3 2)


16(ln 2) 9



16 27 “

24. V œ '0 1(
ln x)# dx 1

œ 1 Œ lim b cx(ln x)# d b
2'0 ln x dx bÄ! 1

1

œ
21 lim b cx ln x
xd 1b œ 21 bÄ!

25. M œ '1 ln x dx œ cx ln x
xd e1 œ (e
e)
(0
1) œ 1; e

Mx œ '1 (ln x) ˆ ln#x ‰ dx œ e

œ

" #

e

e

My œ '1 x ln x dx œ ’ x " #

'1e (ln x)# dx

Šcx(ln x)# d 1
2 '1 ln x dx‹ œ e

œ

" #

’x# ln x


My M

therefore, x œ 26. M œ '0

1

e

x# # “1

2 dx È 1
x#

œ

œ

#

e

ln x # “1

" #




’Še#


e#  1 4

" #

" #

'1

e# #‹

and y œ

(e
2);

e

x dx  "# “ œ

Mx M

œ

" 4

ae#  1b ;

e
2 #

"

œ 2 csin
" xd ! œ 1;

My œ '0

" 2x dx È # È1
x# œ 2 ’
1
x “ œ 2; ! M therefore, x œ My œ 12 and y œ 0 by symmetry 1

27. L œ '1 É1  e

tan
" e

œ '1Î4

" x#

dx œ '1

e

(sec )) atan# )  1b tan )

œ ŠÈ1  e#
ln ¹

È x#  1 x

tan
" e

d) œ '1Î4

È 1  e# e

dx; ”

tan
" e x œ tan ) ' Ä L œ 1Î4 dx œ sec# ) d) •

sec )†sec# ) d) tan )
" e

(tan ) sec )  csc )) d) œ csec )
ln kcsc )  cot )kd 1tanÎ4

# È  "e ¹‹
’È2
ln Š1  È2‹“ œ È1  e#
ln Š 1e e  "e ‹
È2  ln Š1  È2‹

# # ' È ' yÈ1  e2y dy; ” 28. y œ ln x Ê 1  Š dx dy ‹ œ 1  x Ê S œ 21 c x 1  x dy Ê S œ 21 0 e #

d

1

tan
" e

u œ tan ) Ä S œ 21'1 È1  u# du; ” Ä 21 '1Î4 du œ sec# ) d) • e


" e

œ 21 ˆ "# ‰ csec ) tan )  ln ksec )  tan )kd tan 1Î%

u œ ey du œ ey dy •

sec ) † sec# ) d)

œ 1 ’ŠÈ1  e# ‹ e  ln ¹È1  e#  e¹“
1 ’È2 † 1  ln ŠÈ2  1‹“

È 1  e#  e œ 1 ’eÈ1  e#  ln Š È ‹
È 2“ 21

#Î$ 29. L œ 4 '0 Ê1  Š dy  y#Î$ œ 1 Ê y œ ˆ1
x#Î$ ‰ dx ‹ dx; x 1

#

$Î#

Ê

dy dx

œ
3# ˆ1
x#Î$ ‰

"Î#

ˆx
"Î$ ‰ ˆ 23 ‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

580

Chapter 8 Techniques of Integration #

Ê Š dy dx ‹ œ

x Ê L œ 4 '0 Ê1  Š 1
x#Î$ ‹ dx œ 4'0 1

1
x#Î$ x#Î$

1

#Î$

30. S œ 21 'c1 f(x) É1  cf w (x)d# dx; f(x) œ ˆ1
x#Î$ ‰ 1

$Î# " ‰ dx; – œ 41'0 ˆ1
x#Î$ ‰ ˆ x"Î$ 1

" œ
61 † 25
(1
u)&Î# ‘ ! œ #

31. Š dy dx ‹ œ

" 4x

Ê

dy dx

„" #È x

œ

$Î#

"

œ 6
x#Î$ ‘ ! œ 6

dx x"Î$

Ê cf w (x)d#  1 œ

Ê S œ 21 'c1 ˆ1
x#Î$ ‰ 1

" x#Î$

$Î#

†

dx Èx#Î$

1 u œ x#Î$ 3 $Î# '1 du œ
61'0 (1
u)$Î# d(1
u) 2 dx — Ä 4 † # 1 0 (1
u) du œ 3 x"Î$

121 5

Ê y œ Èx or y œ
Èx, 0 Ÿ x Ÿ 4

32. The integral 'c1 È1
x# dx is the area enclosed by the x-axis and the semicircle y œ È1
x# . This area is half 1

the circle's area, or

1 #

and multiplying by 2 gives 1. The length of the circular arc y œ È1
x# from x œ
1 to


x ' x œ 1 is L œ 'c1 Ê1  Š dy ‹ dx œ 'c1 È dx # dx ‹ dx œ c1 Ê1  Š È #

1

#

1

1

1
x#

1
x

circle's circumference. In conclusion, 2 'c1 È1
x# dx œ 'c1 È dx 1

_ 'c_ e x
e a

x

b

_

dx œ '
_ ea
e b ex dx

'a0 e ce

œaÄ lim
_

a

x

b

(21) œ 1 since L is half the

.

(a)

x

ex dx 

" #

1

1
x#

33. (b)

œ

lim

b Ä _

'0b e ce a

x

b

ex dx;

u œ ex ” du œ ex dx • Ä lim ' ecu du  a Ä
_ ea

'1e

b

1

lim

b Ä _

œaÄ lim c
ecu d 1ea 
_

ecu du

c
ecu d e1

b

lim

b Ä
_



"e  ec aea b ‘  œaÄ lim
_



ec ˆeb ‰  "e ‘

lim

b Ä _

œ ˆ
"e  e! ‰  ˆ0  "e ‰ œ 1 34. u œ

" 1y

lim nÄ_

, du œ
(1 dyy)# ; dv œ nync1 dy, v œ yn ;

'01 ny1 
y

n 1

dy œ n lim ’ y “  '0 Ä _ Œ 1y 1

" #

0œ

1

!

' Ê 0 Ÿ n lim Ä_ 0 œ

"

n

yn 1  y#

dy Ÿ n lim Ä_

36.

1 6

uÐn
2ÑÎ2 n#

œ

" È2

" #

" #

 n lim Ä_

ˆÈu‰ n
2 n#

ŠÈx#
a# ‹

Cœ

n#

" œ
sin
" x# ‘ ! œ '0

1

È2


sin
" u# ‘ 0

œ

'01

yn 1  y#

n 1

!

" #

Cœ

œ sin
"

dy œ

b " '01 yn dy œ n lim ’ y “ œ n lim Ä _ n1 Ä_

35. u œ x#
a# Ê du œ 2x dx; ' x ŠÈx#
a# ‹ n dx œ "# ' ˆÈu‰ n du œ œ

yn 1  y#

" # n
2

" n 1

dy. Now, 0 Ÿ

yn 1  y#

' œ 0 Ê n lim Ä_ 0

1

' unÎ2 du œ "# Š u Î
1 ‹  C, n Á
2 n 2 n

1

#

C

'1

dx dx È 4
x#  0 È 4
x#
x$ 1È2 " " ˆ1‰
" È2 È2 sin # œ È2 4 œ 8

 '0

1

dx È4
2x#

œ

" È2

È2

'0

du È 4
u#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ÿ yn nync1 1y

dy

Chapter 8 Additional and Advanced Exercises 37.

'1_ ˆ x ax 1
#"x ‰ dx œ

lim

#

" bÄ_ #

œ lim

ab #  1 b b

’ln

bÄ_ a

integral diverges if a  " #

œ Ê

_

'1

ˆln 1


" #

lim ln

bÄ_

ˆ x#ax 1


bÄ_

ab #  1 b b bÄ_


ln 2a “ ; lim " #

ab #  1 b b

a

_ dx xp

a

" #

p " #

bÄ_

b2a bÄ_ b

" b#

" #

'0b ecxt dt œ

bÄ_

" #

" #

" bÄ_ #

lim

ax #  1 b x

b

a

“

1

Ê the improper

”ln

ab #  1 b b

"Î#


ln 2"Î# •

œ lim (b  1)2ac1 œ 0 bÄ_

; in summary, the improper integral

and has the value
ln42 cxb

b lim

"x ecxt ‘ 0 œ lim Š 1
xe ‹ œ

bÄ_

bÄ_

œ1 Ê

(b1)2a b 1

 lim

œ
_ Ê the improper integral diverges if a 

dx converges only when a œ

b

bÄ_

bÄ_ a

ln x‘ 1 œ lim ’ #" ln

1

œ lim É1 

ab #  1 b b bÄ_

" #

œ lim b2 ˆa c 2 ‰ œ _ if a 

 lim

: 0 Ÿ lim

bÄ_

1
0 x

œ

" x

if x  0 Ê xG(x) œ x ˆ "x ‰

converges if p  1 and diverges if p Ÿ 1. Thus, p Ÿ 1 for infinite area. The volume of the solid of revolution

_

_ dx

about the x-axis is V œ '1 1 ˆ x"p ‰ dx œ 1 '1 " #

lim
#a ln ax#  1b


È b#  1 " # : b lim b Ä_

; for a œ

œ 1 if x  0 39. A œ '1

#

ln 2‰ œ
ln42 ; if a 

" ‰ #x

38. G(x) œ lim

'1b ˆ x ax 1
#"x ‰ dx œ

#

x2p

which converges if 2p  1 and diverges if 2p Ÿ 1. Thus we want

for finite volume. In conclusion, the curve y œ xcp gives infinite area and finite volume for values of p satisfying

 p Ÿ 1.

40. The area is given by the integral A œ '0

1

dx xp

;

p œ 1: A œ lim b cln xd œ
lim b ln b œ _, diverges; bÄ! bÄ! 1 b

p  1: A œ lim b cx1cp d 1b œ 1
lim b b1cp œ
_, diverges; bÄ! bÄ!

p  1: A œ lim b cx1cp d 1b œ "
lim b b1cp œ 1
0, converges; thus, p   1 for infinite area. bÄ! bÄ! The volume of the solid of revolution about the x-axis is Vx œ 1'0

1

p

" #

" #

, and diverges if p  

dx x2p

which converges if 2p  1 or

. Thus, Vx is infinite whenever the area is infinite (p   1).

_

_

The volume of the solid of revolution about the y-axis is Vy œ 1 '1 cR(y)d# dy œ 1'1 converges if

2 p

dy y2Îp

which

 1 Í p  2 (see Exercise 39). In conclusion, the curve y œ xcp gives infinite area and finite

volume for values of p satisfying 1 Ÿ p  2, as described above. ÐÑ

cos 3x

2e2x

Ð
Ñ

1 3 sin

4e2x

ÐÑ


19 cos 3x

41. e2x

Iœ

e2x 3

581

sin 3x 

2e2x 9

cos 3x
49 I Ê

ÐÑ

42. e3x

3x

Ð
Ñ


4" cos 4x

9e3x

ÐÑ

"
16 sin 4x

3x

Iœ

e2x 9

(3 sin 3x  2 cos 3x) Ê I œ

e2x 13

(3 sin 3x  2 cos 3x)  C

sin 4x

3e3x

I œ
e4 cos 4x 

13 9

3e3x 16

sin 4x


9 16

I Ê

25 16

Iœ

e3x 16

(3 sin 4x
4 cos 4x) Ê I œ

e3x 25

(3 sin 4x
4 cos 4x)  C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

582

Chapter 8 Techniques of Integration sin 3x

ÐÑ

sin x

3 cos 3x

Ð
Ñ


cos x


9 sin 3x

ÐÑ


sin x

43.

I œ
sin 3x cos x  3 cos 3x sin x  9I Ê
8I œ
sin 3x cos x  3 cos 3x sin x Ê I œ sin 3x cos x
83 cos 3x sin x  C cos 5x

ÐÑ

sin 4x


sin 5x

Ð
Ñ


"4 cos 4x


25cos 5x

ÐÑ

"
16 sin 4

44.

I œ
"4 cos 5x cos 4x
Ê Iœ

" 9

Ð
Ñ


"b cos bx

a# eax

ÐÑ


b"# sin bx

ax

I œ
eb cos bx  eax a#  b#

aeax b#

sin bx


a# b#

I Ê Ša

Ð
Ñ

" b

a# eax

ÐÑ


b"# cos bx

eax b

sin bx 

Ê Iœ

sin 5x sin 4x

eax a#  b#

#

 b# b# ‹ I

œ

eax b#

(a sin bx
b cos bx)

cos bx

aeax

Iœ

5 16

(a sin bx
b cos bx)  C

ÐÑ

46. eax

9 I Ê
16 I œ
"4 cos 5x cos 4x


sin bx

aeax

Ê Iœ

25 16

(4 cos 5x cos 4x  5 sin 5x sin 4x)  C ÐÑ

45. eax

sin 5x sin 4x 

5 16

aeax b#

sin bx

cos bx


a# b#

I Ê Ša

#

 b# b# ‹ I

œ

eax b#

(a cos bx  b sin bx)

(a cos bx  b sin bx)  C

47. ln (ax)

ÐÑ

1

" x

ÐÑ

x

I œ x ln (ax)
' ˆ "x ‰ x dx œ x ln (ax)
x  C 48. ln (ax)

ÐÑ

x#

" x

ÐÑ

" 3

Iœ

" 3

x$

x$ ln (ax)
' ˆ "x ‰ Š x3 ‹ dx œ $

_

49. (a) >(1) œ '0 ect dt œ lim

bÄ_

" 3

x$ ln (ax)
9" x$  C

'0b ect dt œ

lim c
ect d b0 œ lim

e"b
(
1)‘ œ 0  1 œ 1

bÄ_

bÄ_

(b) u œ tx , du œ xtxc1 dt; dv œ ect dt, v œ
ect ; x œ fixed positive real _

_

Ê >(x  1) œ '0 tx e
t dt œ lim c
tx e
t d b0  x '0 tx
1 e
t dt œ lim ˆ
beb  0x e! ‰  x>(x) œ x>(x) bÄ_

x

bÄ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 8 Additional and Advanced Exercises (c) >(n  1) œ n>(n) œ n!: n œ 0: >(0  1) œ >(1) œ 0!; n œ k: Assume >(k  1) œ k! n œ k  1: >(k  1  1) œ (k  1) >(k  1) œ (k  1)k! œ (k  1)! Thus, >(n  1) œ n>(n) œ n! for every positive integer n.

for some k  0; from part (b) induction hypothesis definition of factorial

x n n 50. (a) >(x) ¸ ˆ xe ‰ É 2x1 and n>(n) œ n! Ê n! ¸ n ˆ ne ‰ É 2n1 œ ˆ ne ‰ È2n1

(b)

n 10 20 30 40 50 60

ˆ ne ‰n È2n1 3598695.619 2.4227868 ‚ 10") 2.6451710 ‚ 10$# 8.1421726 ‚ 10%( 3.0363446 ‚ 10'% 8.3094383 ‚ 10)"

calculator 3628800 2.432902 ‚ 10") 2.652528 ‚ 10$# 8.1591528 ‚ 10%( 3.0414093 ‚ 10'% 8.3209871 ‚ 10)"

(c)

n 10

ˆ ne ‰n È2n1 3598695.619

ˆ ne ‰n È2n1 e1Î12n 3628810.051

calculator 3628800

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

583

584

Chapter 8 Techniques of Integration

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 9 FURTHER APPLICATIONS OF INTEGRATION 9.1 SLOPE FIELDS AND SEPARABLE DIFFERENTIAL EQUATIONS 1. (a) y œ e
x Ê y w œ
e
x Ê 2y w  3y œ 2 a
e
x b  3e
x œ e
x (b) y œ e
x  e
3xÎ2 Ê y w œ
e
x
3# e
3xÎ2 Ê 2y w  3y œ 2 ˆ
e
x
3# e
3xÎ2 ‰  3 ae
x  e
3xÎ2 b œ e
x (c) y œ e
x  Ce
3xÎ2 Ê y w œ
e
x
3# Ce
3xÎ2 Ê 2y w  3y œ 2 ˆ
e
x
3# Ce
3xÎ2 ‰  3 ae
x  Ce
3xÎ2 b œ e
x 2. (a) y œ
"x Ê y w œ

" x#

(b) y œ
x
" 3 Ê y w œ (c) y œ 3. y œ

" x
C

'1x

" x

et t

Ê yw œ

#

œ ˆ
x" ‰ œ y# " (x
3)#

" (x
C)#

#

œ ’
(x
" 3) “ œ y# #

œ

x
" C ‘ œ y#

dt Ê yw œ
x"# '1

x t e

t

dt  ˆ x" ‰ ˆ ex ‰ Ê x# y w œ
'1

x t e

x

t

dt  ex œ
x Š x"

'1x et

t

dt‹  ex œ
xy  ex

Ê x# y w  xy œ ex 4. y œ

" È 1
x%

'1x È1  t% dt $

Ê y w œ Š 1
2xx% ‹ Š È

" 1
x%

Ê y w œ
#" –

' x È1  t% dt  È "

4x$ $— 1 È Š 1
x% ‹

'1x È1  t% dt‹  1

1
x%

ŠÈ 1  x% ‹

$

Ê y w œ Š 1
2xx% ‹ y  1 Ê y w 

2x$ 1
x%

†yœ1

5. y œ ecx tan" a2ex b Ê y w œ
ecx tan" a2ex b  ecx ’ 1
a"2ex b# “ a2ex b œ
ecx tan" a2ex b  Ê y w œ
y 

2 1
4e2x

Ê yw  y œ

2 1
4e2x

2 1
4e2x

; y(
ln 2) œ ecÐ
ln 2Ñ tan" a2e
ln 2 b œ 2 tan" 1 œ 2 ˆ 14 ‰ œ

1 #

# # # # # 6. y œ (x
2) e
x Ê y w œ e
x  ˆ
2xe
x ‰ (x
2) Ê y w œ e
x
2xy; y(2) œ (2
2) e
2 œ 0

7. y œ

Ê xy  y œ 8. y œ

x ln x

dy dx "Î#

9. 2Èxy œ 2x 10.

dy dx

x sin x  cos x Ê y w œ
sinx x x# 1/2)
sin x; y ˆ 1# ‰ œ cos(1(/2) œ0

Ê yw œ

cos x x w

Ê yw œ

ln x  x Š "x ‹ (ln x)#

Ê yw œ

" ln x





"x ˆ cosx x ‰ Ê y w œ
sinx x


" (ln x)#

Ê x# y w œ

œ 1 Ê 2x"Î# y"Î# dy œ dx Ê 2y"Î# dy œ x"Î# dx Ê  C" Ê

2 3

y$Î#
x"Î# œ C, where C œ

" #

"Î#

" 3




x# (ln x)#

Ê xy w œ
sin x
y

Ê x# y w œ xy
y# ; y(e) œ

' 2y"Î# dy œ ' x"Î# dx

e ln e

Ê 2 ˆ 23 y$Î# ‰

C"

œ x# Èy Ê dy œ x# y"Î# dx Ê y"Î# dy œ x# dx Ê

Ê 2y

x# ln x

y x

' y"Î# dy œ ' x# dx

Ê 2y"Î# œ

x$ 3

C

$


x œC

' ey dy œ ' ex dx

11.

dy dx

œ excy Ê dy œ ex ecy dx Ê ey dy œ ex dx Ê

12.

dy dx

œ 3x# ey Ê dy œ 3x# ey dx Ê ey dy œ 3x# dx Ê ' ey dy œ ' 3x# dx Ê ey œ x3  C Ê ey
x3 œ C

Ê ey œ ex  C Ê ey
ex œ C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ e.

586 13.

Chapter 9 Further Applications of Integration dy dx

sec# Èy Èy dy

œ Èy cos# Èy Ê dy œ ˆÈy cos# Èy‰ dx Ê

side, substitute u œ Èy Ê du œ

1 2Èy dy

Ê 2 du œ

1 Èy dy,

œ dx Ê '

sec# Èy Èy dy

œ ' dx. In the integral on the left-hand

and we have ' sec# u du œ ' dx Ê 2 tan u œ x  C

Ê
x  2 tan Èy œ C 14. È2xy dy dx œ 1 Ê dy œ Ê È2 15. Èx

dy dx

y3/2 3 2

dy œ

x1/2 " #

œ eyÈx Ê

1 È2xy

dx Ê È2Èydy œ

1 Èx

dx Ê È2 y1/2 dy œ x1/2 dx Ê È2 ' y1/2 dy œ ' x1/2 dx

3  C1 Ê È2 y3/2 œ 3Èx  32 C1 Ê È2 ˆÈy‰
3Èx œ C, where C œ 32 C1

dy dx

œ

ey eÈ x Èx

Ê dy œ

ey eÈ x Èx dx

right-hand side, substitute u œ Èx Ê du œ

" #È x

eÈ x Ê ecy dy œ È dx Ê x

dx Ê 2 du œ

Ê
ecy œ 2eu  C1 Ê
ecy œ 2eÈx  C, where C œ
C1

16. asec xb

dy dx

œ eysin x Ê

dy dx

" Èx

' ecy dy œ ' ÈeÈx dxÞ In the integral on the

dx, and we have

x

' ecy dy œ 2 ' eu du

œ eysin x cos x Ê dy œ aey esin x cos xbdx Ê e
y dy œ esin x cos x dx

Ê ' ecy dy œ ' esin x cos x dx Ê
ecy œ esin x  C1 Ê ecy  esin x œ C, where C œ
C1

17.

dy dx

œ 2xÈ1
y2 Ê dy œ 2xÈ1
y2 dx Ê

dy È 1  y2

œ 2x dx Ê '

dy È 1  y2

œ ' 2x dx Ê sin" y œ x#  C since kyk  "

Ê y œ sinax2  Cb 18.

dy dx

œ

e2xcy exby 2y

Ê dy œ

e2xcy exby dx

Ê dy œ

e2x ecy ex ey dx

œ

ex e2y dx

Ê e2y dy œ ex dx Ê ' e2y dy œ ' ex dx Ê

Ê e
2ex œ C where C œ 2C1 19. y w œ x  y Ê slope of 0 for the line y œ
x. For x, y  0, y w œ x  y Ê slope  0 in Quadrant I. For x, y  0, y w œ x  y Ê slope  0 in Quadrant III. For kyk  kxk, y  0, x  0, y w œ x  y Ê slope  0 in Quadrant II above y œ
x. For kyk  kxk, y  0, x  0, y w œ x  y Ê slope  0 in Quadrant II below y œ
x. For kyk  kxk, x  0, y  0, y w œ x  y Ê slope  0 in Quadrant IV above y œ
x. For kyk  kxk, x  0, y  0, y w œ x  y Ê slope  0 in Quadrant IV below y œ
x. All of the conditions are seen in slope field (d). 20. y w œ y  1 Ê slope is constant for a given value of y, slope is 0 for y œ
1, slope is positive for y  1 and negative for y 
1. These characteristics are evident in slope field (c).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

e2y #

œ ex  C1

Section 9.1 Slope Fields and Separable Differential Equations 21. y w œ
xy Ê slope œ 1 on y œ
x and
1 on y œ x. y w œ
xy Ê slope œ 0 on the y-axis, excluding a0, 0b, and is undefined on the x-axis. Slopes are positive for x  0, y  0 and x  0, y  0 (Quadrants II and IV), otherwise negative. Field (a) is consistent with these conditions.

22. y w œ y2
x2 Ê slope is 0 for y œ x and for y œ
x. For kyk  kxk slope is positive and for kyk  kxk slope is negative. Field (b) has these characteristics.

23.

24.

25-36. Example CAS commands: Maple: ode := diff( y(x), x ) = y(x); icA := [0, 1]; icB := [0, 2]; icC := [0,-1]; DEplot( ode, y(x), x=0..2, [icA,icB,icC], arrows=slim, linecolor=blue, title="#25 (Section 9.1)" ); Mathematica: To plot vector fields, you must begin by loading a graphics package. <
587

588

Chapter 9 Further Applications of Integration

solnset = Map[f, vals] ps = Plot[Evaluate[solnset, {x,
5, 5}]; Show[pv, ps, PlotRange Ä {
4, 6}]; The code for problems such as 33 & 34 is similar for the direction field, but the analytical solutions involve complicated inverse functions, so the numerical solver NDSolve is used. Note that a domain interval is specified. equation = y'[x] == Cos[2x
y[x]] ; initcond = y[0] == 2; sol = NDSolve[{equation, initcond}, y[x], {x, 0, 5}] ps = Plot[Evaluate[y[x]/.sol, {x, 0, 5}]; N[y[x] /. sol/.x Ä 2] Show[pv, ps, PlotRange Ä {0, 5}]; Solutions for 35 can be found one at a time and plots named and shown together. No direction fields here. For 36, the direction field code is similar, but the solution is found implicitly using integrations. The plot requires loading another special graphics package. <
26.

27.

28.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.2 First-Order Linear Differential Equations 29.

30.

9.2 FIRST-ORDER LINEAR DIFFERENTIAL EQUATIONS 1. x

dy dx

 y œ ex Ê

dy dx

 ˆ x" ‰ y œ

ex x

, P(x) œ

" x

, Q(x) œ

ex x

' P(x) dx œ ' "x dx œ ln kxk œ ln x, x  0 Ê v(x) œ e' PÐxÑ dx œ eln x œ x " ' C y œ v(x) v(x) Q(x) dx œ "x ' x ˆ ex ‰ dx œ "x aex  Cb œ e
x ,x0 x

x

2. ex

dy dx

 2ex y œ 1 Ê

dy dx

 2y œ ex , P(x) œ 2, Q(x) œ ex

' P(x) dx œ ' 2 dx œ 2x Ê v(x) œ e' PÐxÑ dx œ e2x y œ e" ' e2x † ecx dx œ e" ' ex dx œ e" aex  Cb œ ecx  Cec2x 2x

2x

3. xyw  3y œ

sin x x#

,x0Ê

2x

dy dx

 ˆ x3 ‰ y œ

, P(x) œ

sin x x$

3 x

, Q(x) œ

sin x x$

' 3x dx œ 3 ln kxk œ ln x$ , x  0 Ê v(x) œ eln x œ x$ y œ x" ' x$ ˆ sinx x ‰ dx œ x" ' sin x dx œ x" (
cos x  C) œ Cxcos x , x  0 $

$

$

$

$

4. yw  (tan x) y œ cos# x,
1#  x 

1 #

Ê

dy dx

$

 (tan x) y œ cos# x, P(x) œ tan x, Q(x) œ cos# x

sin x 1 1 " ln Ðcos xÑ
" ' tan x dx œ ' cos œ (cos x)" x dx œ
ln kcos xk œ ln (cos x) ,
#  x  # Ê v(x) œ e y œ (cos"x)
" ' (cos x)" † cos# x dx œ (cos x)' cos x dx œ (cos x)(sin x  C) œ sin x cos x  C cos x

5. x

dy dx

 2y œ 1
"x , x  0 Ê

dy dx

" x

 ˆ 2x ‰ y œ




" x#

, P(x) œ

2 x

, Q(x) œ

" x

" x#




' 2x dx œ 2 ln kxk œ ln x# , x  0 Ê v(x) œ eln x œ x# y œ x" ' x# ˆ x"
x" ‰ dx œ x" ' (x
1) dx œ x" Š x#
x  C‹ œ #"
x"  xC , x  0 #

#

#

#

#

6. (1  x) yw  y œ Èx Ê

dy dx

#

 ˆ 1
" x ‰ y œ

Èx 1
x

, P(x) œ

#

" 1
x

, Q(x) œ

Èx 1
x

' 1
" x dx œ ln (1  x), since x  0 Ê v(x) œ eln Ð1
xÑ œ 1 Èx 2x C y œ 1
" x ' (1  x) Š 1
x ‹ dx œ 1
" x ' Èx dx œ ˆ 1
" x ‰ ˆ 32 x$Î#  C‰ œ 3(1
x)  1
x $Î#

7.

dy dx


"# y œ

Ê yœ 8.

dy dx

" #

" e
xÎ2

exÎ2 Ê P(x) œ
"# , Q(x) œ

" #

exÎ2 Ê ' P(x) dx œ
"# x Ê v(x) œ e
xÎ2

' ecxÎ2 ˆ #" exÎ2 ‰ dx œ exÎ2 ' #" dx œ exÎ2 ˆ #" x  C‰ œ #" xexÎ2  CexÎ2

 2y œ 2xec2x Ê P(x) œ 2, Q(x) œ 2xec2x Ê ' P(x) dx œ ' 2 dx œ 2x Ê v(x) œ e2x

Ê yœ

" e2x

' e2x a2xec2x b dx œ e" ' 2x dx œ ec2x ax#  Cb œ x# ec2x  Cec2x 2x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

589

590 9.

Chapter 9 Further Applications of Integration dy dx


ˆ x" ‰ y œ 2 ln x Ê P(x) œ
x" , Q(x) œ 2 ln x Ê ' P(x) dx œ
'

Ê v(x) œ ec ln x œ 10.

dy dx

 ˆ 2x ‰ y œ

" x

" x#

Ê v(x) œ eln x œ x# Ê y œ #

11.

ds dt

 ˆ t 4 1 ‰ s œ

t
1 (t  1)$

2 x

Ê P(t) œ

12. (t  1)

dr d)

dx œ 2 ln kxk œ ln x# , x  0

4 t1

t$ 3(t  1)%

 2s œ 3(t  1) 




" (t
1)#

#

t
1 (t  1)$

sin x
C x#

Ê

$

t (t  1)%



Ê

 ˆ t
2 1 ‰ s œ 3 

ds dt

(sin x  C) œ

' P(t) dt œ ' t41 dt œ 4 ln kt
1k œ ln (t
1)% ' (t
1)% ’ (tt
1)" “ dt œ (t "1) ' at#
1b dt

, Q(t) œ " (t1)%

#

%

C (t  1)% " (t
1)$

Ê P(t) œ

2 t
1

, Q(t) œ 3  (t  1)$

#

Ê

#

" (t
1)#

œ

2 x

' P(t) dt œ ' t
2 1 dt œ 2 ln kt  1k œ ln (t  1)# Ê v(t) œ eln Ðt1Ñ œ (t  1)# s œ (t
"1) ' (t  1)# c3  (t  1)$ d dt œ (t
"1) ' c3(t  1)#  (t  1)" d dt

Ê

13.

ds dt

$

Š t3
t  C‹ œ

Ê ' P(x) dx œ '

cos x x#

#

%

" (t  1)%

, Q(x) œ

' x# ˆ cosx x ‰ dx œ x" ' cos x dx œ x"

Ê v(t) œ eln Ðt
1Ñ œ (t
1)% Ê s œ œ

dx œ
ln x, x  0

Ê y œ x' ˆ x" ‰ (2 ln x) dx œ x c(ln x)#  Cd œ x (ln x)#  Cx

, x  0 Ê P(x) œ

cos x x#

" x

#

$

c(t  1)  ln kt  1k  Cd œ (t  1)  (t  1)# ln (t  1) 

, t 
1

C (t
1)#

' P()) d) œ ' cot ) d) œ ln ksin )k Ê v()) œ eln sin r œ sin" ) ' (sin ))(sec )) d) œ sin" ) ' tan ) d) œ sin" ) aln ksec )k  Cb

 (cot )) r œ sec ) Ê P()) œ cot ), Q()) œ sec ) Ê 1 #

œ sin ) because 0  ) 

Ê

k

)k

œ (csc )) aln ksec )k  Cb 14. tan ) Ê

 r œ sin# ) Ê



r tan )

œ

sin# ) tan )

Ê

dr d)

 (cot )) r œ sin ) cos ) Ê P()) œ cot ), Q()) œ sin ) cos )

' P()) d) œ ' cot ) d) œ ln ksin )k œ ln (sin )) since 0  )  1# Ê v()) œ eln Ðsin )Ñ œ sin ) r œ sin" ) ' (sin )) (sin ) cos )) d) œ sin" ) ' sin# ) cos ) d) œ ˆ sin" ) ‰ Š sin3 )  C‹ œ sin3 )  sinC ) #

 2y œ 3 Ê P(t) œ 2, Q(t) œ 3 Ê ' P(t) dt œ ' 2 dt œ 2t Ê v(t) œ e2t Ê y œ œ e"2t ˆ 3# e2t  C‰ ; y(0) œ 1 Ê 3#  C œ 1 Ê C œ
"# Ê y œ 3#
"# ec2t

15.

dy dt

16.

dy dt



œ

" t#

dy d)

2y t

œ t# Ê P(t) œ &

 ˆ ") ‰ y œ " k) k

$

Ê P()) œ

" )

, Q()) œ

sin ) )

Ê

8 5



; y ˆ 1# ‰ œ 1 Ê C œ

1 #

C 4

œ 1 Ê C œ
12 5 Ê yœ

" )
#

t$ 5




' at# b at# b dt 12 5t#

k k

" ) 1 2)

Ê yœ

Ê y œ
") cos ) 

' sin ) d) œ ") (
cos )  C)


ˆ 2) ‰ y œ )# sec ) tan ) Ê P()) œ
2) , Q()) œ )# sec ) tan ) Ê

œ )# Ê y œ

" t#

Ê ' P()) d) œ ln k)k Ê v()) œ eln ) œ k)k

' k)k ˆ sin) ) ‰ d) œ ") ' ) ˆ sin) ) ‰ d) for ) Á 0 C )

' 3e2t dt

#

#

sin ) )

" e2t

, Q(t) œ t# Ê ' P(t) dt œ 2 ln ktk Ê v(t) œ eln t œ t# Ê y œ

#

œ
") cos )  dy d)

2 t

' t% dt œ t" Š t5  C‹ œ t5  tC ; y(2) œ 1

Ê yœ

18.

dr d)

$

Ê

17.

dr d)

' P()) d) œ
2 ln k)k

Ê v()) œ ec2 ln )

' a)# b a)# sec ) tan )b d) œ )# ' sec ) tan ) d) œ )# (sec )  C) œ )# sec )  C)# ; #

#

y ˆ 13 ‰ œ 2 Ê 2 œ Š 19 ‹ (2)  C Š 19 ‹ Ê C œ

18 1#

‰ #
2 Ê y œ )# sec )  ˆ 18 1#
2 )

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

k k

Section 9.2 First-Order Linear Differential Equations 19. (x  1) Q(x) œ œ ex

#

dy dx

x#

' (x
" 1)

dx œ ex ’ (x
1) 1 #

#

#

dy dx

dy dt

" ex# Î2

#

Šex Î2  C‹ œ 1 

dy dx


1)
2 ’ x(x x
1 “ y œ


"

#

ex (x
1)#

Ê #

Ê v(x) œ ex Ê y œ

x#

#

ex (x
1)#


2xy œ

dy dx

" e
x#

Ê P(x) œ
2x,

' e
x# ’ (x e
1) “ dx x#

#

#

 C“ œ
xe
1  Cex ; y(0) œ 5 Ê
0
1 1  C œ 5 Ê
"  C œ 5

#

ex x
1

C ex# Î2

x# #

#

Ê v(x) œ ex Î2 Ê y œ

; y(0) œ
6 Ê 1  C œ
6 Ê C œ
7 Ê y œ 1


" ex# Î2

' ex Î2 † x dx #

7 ex# Î2


ky œ 0 Ê P(t) œ
k, Q(t) œ 0 Ê ' P(t) dt œ '
k dt œ
kt Ê v(t) œ ekt

Ê yœ 22. (a)

dv dt



' ˆekt ‰ (0) dt œ ekt (0  C) œ Cekt ; y(0) œ y!

" eckt

k m

v œ 0 Ê P(t) œ " ektÎm

Ê yœ (b)

dv dt

23. x ' " cos x

" x

k m

, Q(t) œ 0 Ê

' ektÎm † 0 dt œ e C

œ
mk v Ê

Then v œ

24.

Ê

 xy œ x Ê P(x) œ x, Q(x) œ x Ê ' P(x) dx œ ' x dx œ

œ

21.

ex x
1

' P(x) dx œ '
2x dx œ
x#

Ê

e (x
1)#

Ê C œ 6 Ê y œ 6ex
20.

#


2 ax#  xb y œ

1 eÐkÎmÑt

ktÎm

dv v

Ê C œ y! Ê y œ y! ekt

' P(t) dt œ ' mk dt œ mk t œ mkt

; v(0) œ v! Ê

C ekÐ0ÑÎm

Ê v(t) œ ektÎm

œ v! Ê C œ v! Ê v œ v! eÐkÎmÑt

œ
mk dt Ê ln v œ
mk t  C Ê v œ eakÎmbt
C Ê v œ eÐkÎmÑt † eC . Let eC œ C1 Þ

† C1 and va0b œ v0 œ

1 eÐkÎmÑa0b

† C1 œ C1 . So v œ v0 eÐkÎmÑt

dx œ x aln kxk  Cb œ x ln kxk  Cx Ê (b) is correct

' cos x dx œ cos" x (sin x  C) œ tan x  cosC x

Ê (b) is correct

25. Let y(t) œ the amount of salt in the container and V(t) œ the total volume of liquid in the tank at time t. Then, the departure rate is (a) Rate entering œ

2 lb gal

†

y(t) V(t) 5 gal min

(the outflow rate). œ 10 lb/min

(b) Volume œ V(t) œ 100 gal  (5t gal
4t gal) œ (100  t) gal (c) The volume at time t is (100  t) gal. The amount of salt in the tank at time t is y lbs. So the concentration at any time t is 100y
t lbs/gal. Then, the rate leaving œ 100y
t (lbs/gal) † 4 (gal/min) œ (d)

dy dt

4y 100
t

lbs/min

œ 10


4y 100
t

Ê

dy dt

 ˆ 1004
t ‰ y œ 10 Ê P(t) œ

4 100
t

œ 4 ln (100  t) Ê v(t) œ e4 ln Ð100
tÑ œ (100  t)% Ê y œ œ

10 (100
t)%

&

Š (1005
t)  C‹ œ 2(100  t) 

C (100
t)%

Ê C œ
(150)(100)% Ê y œ 2(100  t)
(e) y(25) œ 2(100  25)
26. (a)

dV dt

(150)(100)% (100
25)%

, Q(t) œ 10 Ê " (100
t)%

' P(t) dt œ ' 1004
t dt

' (100  t)% (10 dt)

; y(0) œ 50 Ê 2(100  0)  %

(150)(100) (100
t)%

Ê y œ 2(100  t)


¸ 188.56 lbs Ê concentration œ

y(25) volume

¸

C (100
0)%

œ 50

150 t ‰% ˆ1
100

188.6 125

¸ 1.5 lb/gal

œ a5
3b œ 2 Ê V œ 100  2t

The tank is full when V œ 200 œ 100  2t Ê t œ 50 min (b) Let yatb be the amount of concentrate in the tank at time t. dy dt

gal gal lb lb œ Š "# gal ‹Š5 min ‹
Š 100y
2t gal ‹Š3 min ‹Ê

Q(t) œ 52 ; P(t) œ 32 ˆ 501
t ‰ Ê

dy dt

œ

5 2


32 ˆ 50y
t ‰ Ê

dy dt



3 2at
50b y

œ

5 2

' P(t) dt œ 32 ' t
150 dt œ 32 ln at  50b since t  50  0

vatb œ e' P(t) dt œ e 2 ln at
50b œ at  50b3/2 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

591

592

Chapter 9 Further Applications of Integration yatb œ

' 25 at  50b3/2 dt œ at  50b3/2
at  50b5/2  C ‘ Ê yatb œ t  50 

1 at
50b3/2

C at
50b3/2

Apply the initial condition (i.e., distilled water in the tank at t œ 0): ya0b œ 0 œ 50 

C 503/2

ya50b œ 100


5/2

Ê C œ
505/2 Ê yatb œ t  50


505/2 . at
50b3/2

When the tank is full at t œ 50,

¸ 83.22 pounds of concentrate.

50 1003/2

27. Let y be the amount of fertilizer in the tank at time t. Then rate entering œ 1

lb gal

†1

gal min

œ1

lb min

and the

volume in the tank at time t is V(t) œ 100 (gal)  [1 (gal/min)
3 (gal/min)]t min œ (100
2t) gal. Hence 3y ‰ dy ˆ ˆ 3 ‰ rate out œ ˆ 100y 2t ‰ 3 œ 1003y2t lbs/min Ê dy dt œ 1
100  2t lbs/min Ê dt  100  2t y œ 1 Ê P(t) œ

3 1002t

2t) ' P(t) dt œ ' 10032t dt œ 3 ln (100 Ê v(t) œ eÐ
3 ln Ð100
2tÑÑ2 #  2t)
"Î# y œ (100 "2t)c$Î# ' (100
2t)$Î# dt œ (100
2t)$Î# ’ 2(100#  C“

, Q(t) œ 1 Ê

œ (100
2t)$Î# Ê

œ (100
2t)  C(100
2t)$Î# ; y(0) œ 0 Ê [100
2(0)]  C[100
2(0)]$Î# Ê C(100)$Î# œ
100 " Ê C œ
(100)"Î# œ
10 Ê y œ (100
2t)


œ
2 

3È100  2t 10

(100  2t)$Î# 10

. Let

dy dt

œ0 Ê

dy dt

œ
2


ˆ #3 ‰ (100  2t)"Î# (2) 10

œ 0 Ê 20 œ 3È100
2t Ê 400 œ 9(100
2t) Ê 400 œ 900
18t Ê
500 œ
18t

Ê t ¸ 27.8 min, the time to reach the maximum. The maximum amount is then [100  2(27.8)]$Î# 10

y(27.8) œ [100
2(27.8)]


¸ 14.8 lb

28. Let y œ y(t) be the amount of carbon monoxide (CO) in the room at time t. The amount of CO entering the y ‰ˆ 3 ‰ y 4 3 ‰ 12 $ room is ˆ 100 ‚ 10 œ 1000 ft$ /min, and the amount of CO leaving the room is ˆ 4500 10 œ 15,000 ft /min. Thus,

œ

dy dt

y 15,000

Ê

dy dt

12 ' 1000 etÎ15ß000 dt

" etÎ15ß000

Ê yœ




12 1000



" 15,000

yœ

12 1000

" 15,000

Ê P(t) œ

, Q(t) œ

12 1000

Ê v(t) œ etÎ15ß000

15,000 tÎ15,000 Ê y œ ectÎ15ß000 ˆ 12†1000 e  C‰ œ e
tÎ15ß000 a180etÎ15ß000  Cb ;

y(0) œ 0 Ê 0 œ 1(180  C) Ê C œ
180 Ê y œ 180
180ectÎ15ß000 . When the concentration of CO is 0.01% y .01 in the room, the amount of CO satisfies 4500 œ 100 Ê y œ 0.45 ft$ . When the room contains this amount we
tÎ15ß000 ‰ ¸ 37.55 min. have 0.45 œ 180
180e
tÎ15ß000 Ê 179.55 Ê t œ
15,000 ln ˆ 179.55 180 œ e 180 29. Steady State œ Ê ln 30. (a) (b)

di dt

" #



R L

and we want i œ Ê
" i

iœ0 Ê

L R

" #

ln

31. (a) t œ (b) t œ di dt

L R

Ê iœIe Ê iœ

3L R 2L R



R L

Ê iœ iœ

Ê iœ

V L

" eRtÎL

(b) i(0) œ 0 Ê (c) i œ

V R

Ê

di dt

V R V R

" #

" #

ˆ VR ‰ Ê

œt Ê tœ

L R

ˆ VR ‰ œ

V R

a1
e
RtÎL b Ê

" #

œ 1
e
RtÎL Ê
"# œ
e
RtÎL

ln 2 sec

C" cRtÎL di œ
RL dt Ê ln i œ
Rt œ Ce
RtÎL ; i(0) œ I Ê I œ C L  C" Ê i œ e e

Ê i œ Ie
RtÎL amp œ I e
RtÎL Ê e
RtÎL œ

" #I

(c) t œ

32. (a)

œ


V R Rt L

" #

Ð
RtÎLÑÐLÎRÑ

Ê
Rt L œ ln

œ I ect amp

a1
eÐ
R/LÑÐ3LÎRÑ b œ a1
e

Ð
R/LÑÐ2LÎRÑ

Ê P(t) œ

R L

bœ

, Q(t) œ

' eRtÎL ˆ VL ‰ dt œ e "

RtÎL

V L

V R V R

" #

œ
ln 2 Ê t œ

a1
ec3 b ¸ 0.9502

V R V R

a1
ec2 b ¸ 0.8647

Ê

'

P(t) dt œ '


RL eRtÎL ˆ VL ‰  C‘ œ

V R

R L

L R

ln 2 sec

amp, or about 95% of the steady state value amp, or about 86% of the steady state value

dt œ

Rt L

 Ce
ÐRÎLÑt

 C œ 0 Ê C œ
VR Ê i œ VR
VR ecRtÎL R ˆ R ‰ ˆ VR ‰ œ VL Ê i œ œ 0 Ê di dt  L i œ 0  L

V R

V R

is a solution of Eq. (11); i œ Ce
ÐRÎLÑt

33. y w
y œ
y2 ; we have n œ 2, so let u œ y12 œ y1 . Then y œ u1 and 1 Ê
u2 du œ
u 2 Ê dx
u

du dx

Ê v(t) œ eRtÎL

du dx

œ
1y2 dy dx Ê

dy dx

 u œ 1. With e' dx œ ex as the integrating factor, we have

œ
y2 du dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.3 Euler's Method ‰ ex ˆ du dx  u œ

d x dx ae ub

œ ex . Integrating, we get ex u œ ex  C Ê u œ 1 

34. y w
y œ xy2 ; we have n œ 2, so let u œ y1 . Then y œ u1 and 1 Substituting:
u2 du œ xu2 Ê dx
u

‰ ex ˆ du dx  u œ

d x dx ae ub

du dx

du dx

C ex

œ

1 y

œ
y2 dy dx Ê

Êyœ dy dx

1 1
eCx

œ

593

ex ex
C

2 du œ
y2 du dx œ
u dx .

 u œ
x. Using e' dx œ ex as an integrating factor:

œ
x ex Ê ex u œ ex a1
xb  C Ê u œ

e x a1  x b
C ex

Ê y œ u 1 œ

ex ex  xex
C

35. xy w  y œ y2 Ê y w  ˆ 1x ‰y œ ˆ 1x ‰y2 . Let u œ y1a2b œ y3 Ê y œ u1/3 and y2 œ u2/3 . dy du 2 dy w ˆ 1 ‰ˆ du ‰ 2 ˆ 1 ‰ˆ du ‰ˆ 2/3 ‰. Thus we have dx œ 3y dx Ê y œ dx œ 3 dx ay b œ 3 dx u 1 du 1 1 du 3 ˆ 3 ‰ˆ dx ‰ˆu2/3 ‰  ˆ x ‰u1/3 œ ˆ x ‰u2/3 Ê dx  ˆ x ‰u œ ˆ 3x ‰1. The integrating factor, vaxb, 3 '3 d e x dx œ e3ln x œ eln x œ x3 . Thus dx ax3 ub œ ˆ 3x ‰x3 œ 3x2 Ê x3 u œ x3  C Ê u œ 1  xC3 1/3 Ê y œ ˆ1  xC3 ‰

is œ y3

36. x2 y w  2xy œ y3 Ê y w  ˆ 2x ‰y œ ˆ x12 ‰y3 . Paxb œ ˆ 2x ‰, Qaxb œ ˆ x12 ‰, n œ 3. Let u œ y13 œ y2 . du ˆ2‰ ˆ1‰ ˆ x4 ‰u œ x22 . Let the integrating factor, vaxb, be Substituting gives du dx  a
2b x u œ
2 x2 Ê dx  c4

e' a x bdx œ eln x

c4

œ x4 . Thus

2 Ê y œ ˆ 5x Cx4 ‰

d 4 dx ax

ub œ
2x6 Ê x4 u œ 25 x5  C Ê u œ

2 4 5x Cx

œ y2

1/2

9.3 EULER'S METHOD 1 ‰ # (.5)

1. y" œ y!  Š1


y! x! ‹

dx œ
1  ˆ1


y# œ y "  Š 1


y" x" ‹

dx œ
0.25  ˆ1


y$ œ y #  Š 1


y# x# ‹

dx œ 0.3  ˆ1


dy dx

 ˆ x" ‰ y œ 1 Ê P(x) œ

Ê yœ

" x

" x

' x † 1 dx œ x" Š x#

Ê y(3.5) œ

#

3.5 #




4 3.5

œ

4.25 7

œ
0.25,

0.25 ‰ (.5) 2.5

0.3 ‰ 3 (.5)

, Q(x) œ 1 Ê

œ 0.3,

œ 0.75;

' P(x) dx œ ' x" dx œ ln kxk œ ln x, x  0 Ê v(x) œ eln x œ x

 C‹ ; x œ 2, y œ
1 Ê
1 œ 1 

C 2

Ê C œ
4 Ê y œ

x #




4 x

¸ 0.6071

2. y" œ y!  x! (1
y! ) dx œ 0  1(1
0)(.2) œ .2, y# œ y"  x" (1
y" ) dx œ .2  1.2(1
.2)(.2) œ .392, y$ œ y#  x# (1
y# ) dx œ .392  1.4(1
.392)(.2) œ .5622; dy 1 y

œ x dx Ê
ln k1
yk œ

x# #

 C; x œ 1, y œ 0 Ê
ln 1 œ

" #

#

 C Ê C œ
#" Ê ln k1
yk œ
x# 

" #

#

Ê y œ 1
ea1
x bÎ2 Ê y(1.6) ¸ .5416 3. y" œ y!  (2x! y!  2y! ) dx œ 3  [2(0)(3)  2(3)](.2) œ 4.2, y# œ y"  (2x" y"  2y" ) dx œ 4.2  [2(.2)(4.2)  2(4.2)](.2) œ 6.216, y$ œ y#  (2x# y#  2y# ) dx œ 6.216  [2(.4)(6.216)  2(6.216)](.2) œ 9.6969; dy dx

œ 2y(x  1) Ê

dy y

œ 2(x  1) dx Ê ln kyk œ (x  1)#  C; x œ 0, y œ 3 Ê ln 3 œ 1  C Ê C œ ln 3
1

Ê ln y œ (x  1)#  ln 3
1 Ê y œ eÐx1Ñ ln 3
1 œ eln 3 ex 2x œ 3exÐx2Ñ Ê y(.6) ¸ 14.2765 #

#

4. y" œ y!  y#! (1  2x! ) dx œ 1  1# [1  2(
1)](.5) œ .5, y# œ y"  y#" (1  2x" ) dx œ .5  (.5)# [1  2(
.5)](.5) œ .5, y$ œ y#  y## (1  2x# ) dx œ .5  (.5)# [1  2(0)](.5) œ .625; dy y#

œ (1  2x) dx Ê
y" œ x  x#  C; x œ
1, y œ 1 Ê
1 œ
1  (
1)#  C Ê C œ
1 Ê

Ê yœ

" 1  x  x#

Ê y(.5) œ

" 1  .5  (.5)#

œ4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" y

œ 1
x
x#

594

Chapter 9 Further Applications of Integration #

5. y" œ y!  2x! ex! dx œ 2  2(0)(.1) œ 2, # # y# œ y"  2x" ex" dx œ 2  2(.1) eÞ1 (.1) œ 2.0202, # # y$ œ y#  2x# ex# dx œ 2.0202  2(.2) eÞ2 (.1) œ 2.0618, # # # # dy œ 2xex dx Ê y œ ex  C; y(0) œ 2 Ê 2 œ 1  C Ê C œ 1 Ê y œ ex  1 Ê y(.3) œ eÞ3  1 ¸ 2.0942 6. y" œ y!  ay!  ex!
2b dx œ 2  a2  e!
2b (.5) œ 2.5, y# œ y"  ay"  ex"
2b dx œ 2.5  a2.5  eÞ5
2b (.5) œ 3.5744, y$ œ y#  ay#  ex#
2b dx œ 3.5744  a3.5744  e"
2b (.5) œ 5.7207; dy dx


y œ ex
2 Ê P(x) œ
1, Q(x) œ ex
2 Ê

' P(x) dx œ
x

" ecx

Ê v(x) œ ex Ê y œ

' ecx aex
2b dx

œ ex ax  2ecx  Cb ; y(0) œ 2 Ê 2 œ 2  C Ê C œ 0 Ê y œ xex  2 Ê y(1.5) œ 1.5e1 5  2 ¸ 8.7225 Þ

7. y" y# y$ y% y& dy y

œ 1  1(.2) œ 1.2, œ 1.2  (1.2)(.2) œ 1.44, œ 1.44  (1.44)(.2) œ 1.728, œ 1.728  (1.728)(.2) œ 2.0736, œ 2.0736  (2.0736)(.2) œ 2.48832; œ dx Ê ln y œ x  C" Ê y œ Cex ; y(0) œ 1 Ê 1 œ Ce! Ê C œ 1 Ê y œ ex Ê y(1) œ e ¸ 2.7183

8. y" œ 2  ˆ 21 ‰ (.2) œ 2.4, ‰ y# œ 2.4  ˆ 2.4 1.2 (.2) œ 2.8, ‰ y$ œ 2.8  ˆ 2.8 1.4 (.2) œ 3.2, 3.2 y% œ 3.2  ˆ 1.6 ‰ (.2) œ 3.6, ‰ y& œ 3.6  ˆ 3.6 1.8 (.2) œ 4; dy y

œ

dx x

Ê ln y œ ln x  C Ê y œ kx; y(1) œ 2 Ê 2 œ k Ê y œ 2x Ê y(2) œ 4 #

9. y" œ
1  ’ (È1) “ (.5) œ
.5, 1 #

.5) y# œ
.5  ’ (È “ (.5) œ
.39794, 1.5 #

y$ œ
.39794  ’ (.39794) “ (.5) œ
.34195, È2 #

y% œ
.34195  ’ (.34195) È2.5 “ (.5) œ
.30497, y& œ
.27812, y' œ
.25745, y( œ
.24088, y) œ
.2272; dy dx " È y# œ Èx Ê
y œ 2 x  C; y(1) œ
1 Ê 1 œ 2  C Ê C œ
1 Ê y œ

" 1  #È x

Ê y(5) œ

" 1  #È 5

¸
.2880

10. y" œ 1  a1
e! b ˆ "3 ‰ œ 1, y# œ 1  ˆ1
e#Î$ ‰ ˆ "3 ‰ œ 0.68408, y$ œ 0.68408  ˆ0.68408
e%Î$ ‰ ˆ "3 ‰ œ
0.35245, y% œ
0.35245  ˆ
0.35245
e'Î$ ‰ ˆ "3 ‰ œ
2.93295, y& œ
2.93295  ˆ
2.93295
e)Î$ ‰ ˆ "3 ‰ œ
8.70790, y' œ
8.7079  ˆ
8.7079
e"!Î$ ‰ ˆ "3 ‰ œ
20.95441; yw
y œ
e2x Ê P(x) œ
1, Q(x) œ
e2x Ê

' P(x) dx œ
x

Ê v(x) œ ex Ê y œ

" ecx %

' ex a
e2x b dx

œ ex a
ex  Cb ; y(0) œ 1 Ê 1 œ
1  C Ê C œ 2 Ê y œ
e2x  2ex Ê y(2) œ
e  2e# ¸
39.8200 11. Let zn œ yn1  2yn1 axn1  1bdx and yn œ yn1  ayn1 axn1  1b  zn axn  1bbdx with x0 œ 0, y0 œ 3, and dx œ 0.2. The exact solution is y œ 3exax
2b . Using a programmable calculator or a spreadsheet (I used a spreadsheet) gives the values in the following table. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.3 Euler's Method x 0 0.2 0.4 0.6

z --4.2 6.81984 11.89262

y-approx 3 4.608 7.623475 13.56369

y-exact 3 4.658122 7.835089 14.27646

595

Error 0 0.050122 0.211614 0.712777

12. Let zn œ yn1  xn1 a1
yn1 bdx and yn œ yn1  Š xnc1 a1  ync12b
xn a1  zn b ‹dx with x0 œ 1, y0 œ 0, and dx œ 0.2. The exact solution is y œ 1
eˆ1
x ‰Î2 . Using a programmable calculator or a spreadsheet (I used a spreadsheet) gives the values in the following table. x z y-approx y-exact Error 1 --0 0 0 1.2 0.2 1.196 0.197481 0.001481 1.4 0.38896 0.378026 0.381217 0.003191 1.6 0.552178 0.536753 0.541594 0.004841 2

13.

dy dx

œ 2xex , ya0b œ 2 Ê yn
1 œ yn  2xn exn dx œ yn  2xn exn a0.1b œ yn  0.2xn exn 2

2

2

2

On a TI-92 Plus calculator home screen, type the following commands: 2 STO > y: 0 STO > x: y (enter) y  0.2*x*e^(x^2) STO > y: x  0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 3.45835 2 2 2 The exact solution: dy œ 2xex dx Ê y œ ex  C; ya0b œ 2 œ e0  C Ê C œ 1 Ê y œ 1  ex Ê yexact a1b œ 1  e ¸ 3.71828 14.

dy dx

œ y  ex
2, ya0b œ 2 Ê yn
1 œ yn  ayn  exn
2bdx œ yn  0.5ayn  exn
2b

On a TI-92 Plus calculator home screen, type the following commands: 2 STO > y: 0 STO > x: y (enter) y  0.5*(y  ex
2) STO > y: x  0.5 STO > x: y (enter, 4 times) The last value displayed gives yEuler a2b ¸ 9.82187 The exact solution: Êyœ

dy dx


y œ ex
2 Ê Paxb œ 1, Qaxb œ ex
2 Ê ' P(x) dx œ
x Ê vaxb œ ex

' ex aex
2bdx œ ex ax  2ex  Cb;ya0b œ 2 Ê 2 œ 2  C Ê C œ 0

1 ecx x

Ê y œ xe  2 Ê yexact a2b œ 2e2  2 ¸ 16.7781 15.

dy dx

œ

Èx y ,y

 0ß ya0b œ " Ê yn
1 œ yn 

È xn yn dx

œ yn 

È xn yn a0.1b

œ yn  0."

È xn yn

On a TI-92 Plus calculator home screen, type the following commands: 1 STO > y: 0 STO > x: y (enter) y  0.1*(Èx /y) STO > y: x  0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.5000 The exact solution: dy œ Ê 16.

dy dx

y2 2

œ 23 x3/2 

1 2

Èx y dx

Ê y dy œ Èx dx Ê

y2 2

2

œ 32 x3/2  C; aya20bb œ

12 2

œ

1 2

œ 23 a0b3/2  C Ê C œ

1 2

Ê y œ É 43 x3/2  1 Ê yexact a"b œ É 43 a1b3/2  1 ¸ 1.5275

œ 1  y2 , ya0b œ 0 Ê yn
1 œ yn  a1  y2n bdx œ yn  a1  y2n ba0.1b œ yn  0.1a1  y2n b

On a TI-92 Plus calculator home screen, type the following commands: 0 STO > y: 0 STO > x: y (enter) y  0.1*(1  y2 ) STO > y: x  0.1 STO > x: y (enter, 10 times) The last value displayed gives yEuler a1b ¸ 1.3964 The exact solution: dy œ a1  y2 bdx Ê

dy 1
y2

œ dx Ê tan1 y œ x  C; tan1 ya0b œ tan1 0 œ 0 œ 0  C Ê C œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

596

Chapter 9 Further Applications of Integration Ê tan1 y œ x Ê y œ tan x Ê yexact a"b œ tan 1 ¸ 1.5574

17. (a)

dy dx

œ 2y2 ax
1b Ê

dy y2

œ 2ax
1bdx Ê ' y2 dy œ ' a2x
2bdx Ê
y" œ x2
2x  C

Initial value: ya2b œ
"# Ê 2 œ 22
2a2b  C Ê C œ 2 1 Solution:
y" œ x2
2x  2 or y œ
x2  2x
2

ya3b œ
32  21a3b
2 œ
15 œ
0.2 (b) To find the approximation, set y1 œ 2y2 ax
1b and use EULERT with initial values x œ 2 and y œ
"# and step size

0.2 for 5 Points. This gives ya3b ¸
0.1851; error ¸ 0.0149. (c) Use step size 0.1 for 10 points. This gives ya3b ¸
0.1929; error ¸ 0.0071. (d) Use step size 0.05 for 20 points. This gives ya3b ¸
0.1965; error ¸ 0.0035. 18. (a)

dy dx

œy
1Ê'

dy y1

œ ' dx Ê ln ky
1k œ x  C Ê ky
1k œ ex
C Ê y
1 œ „ eC ex Ê y œ Aex  1

Initial value: ya0b œ 3 Ê 3 œ Ae0  1 Ê A œ 2 Solution: y œ 2ex  1 ya1b œ 2e  1 ¸ 6.4366 (b) To find the approximation, set y1 œ y
1 and use a graphing calculator or CAS with initial values x œ 0 and y œ 3 and step size 0.2 for 5 Points. This gives ya1b ¸ 5.9766; error ¸ 0.4599 (c) Use step size 0.1 for 10 points. This gives ya1b ¸ 6.1875; error ¸ 0.2491. (d) Use step size 0.05 for 20 points. This gives ya1b ¸ 6.3066; error ¸ 0.1300. 1 2 x2  2x
2 , so ya3b œ
0.2. To find the approximation, let zn œ yn1  2yn1 axn1
1bdx ay2n1 axn1
1b  z2n ax2n
1bbdx with initial values x0 œ 2 and y0 œ
"# . Use a spreadsheet, graphing

19. The exact solution is y œ yn œ y n  1 

and

calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya3b ¸
0.2024 Ê error ¸ 0.0024. (b) Use dx œ 0.1 with 10 steps to obtain ya3b ¸
0.2005 Ê error ¸ 0.0005. (c) Use dx œ 0.05 with 20 steps to obtain ya3b ¸
0.2001 Ê error ¸ 0.0001. (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 20. The exact solution is y œ 2ex  1, so ya1b œ 2e  1 ¸ 6.4366. To find the approximation, let zn œ yn1  ayn1
1bdx and yn œ yn1  ˆ ync1
2zn  2 ‰dx with initial value yn œ 3. Use a spreadsheet, graphing calculator, or CAS as indicated in parts (a) through (d). (a) Use dx œ 0.2 with 5 steps to obtain ya1b ¸ 6.4054 Ê error ¸ 0.0311. (b) Use dx œ 0.1 with 10 steps to obtain ya1b ¸ 6.4282 Ê error ¸ 0.0084 (c) Use dx œ 0.05 with 20 steps to obtain ya1b ¸ 6.4344 Ê error ¸ 0.0022 (d) Each time the step size is cut in half, the error is reduced to approximately one-fourth of what it was for the larger step size. 13-16. Example CAS commands: Maple: ode := diff( y(x), x ) = 2*x*exp(x^2);ic := y(0)=2; xstar := 1; dx := 0.1; approx := dsolve( {ode,ic}, y(x), numeric, method=classical[foreuler], stepsize=dx ): approx(xstar); exact := dsolve( {ode,ic}, y(x) ); eval( exact, x=xstar ); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.3 Euler's Method evalf( % ); 17.

Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; exact := dsolve( {ode,ic}, y(x) ); # (a) eval( exact, x=xstar ); evalf( % ); approx1 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[foreuler], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[foreuler], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (d) numeric, method=classical[foreuler], stepsize=0.05 ): approx3(xstar);

19.

Example CAS commands: Maple: ode := diff( y(x), x ) = 2*y(x)*(x-1);ic := y(2)=-1/2; xstar := 3; approx1 := dsolve( {ode,ic}, y(x), # (a) numeric, method=classical[heunform], stepsize=0.2 ): approx1(xstar); approx2 := dsolve( {ode,ic}, y(x), # (b) numeric, method=classical[heunform], stepsize=0.1 ): approx2(xstar); approx3 := dsolve( {ode,ic}, y(x), # (c) numeric, method=classical[heunform], stepsize=0.05 ): approx3(xstar);

21.

Example CAS commands: Maple: ode := diff( y(x), x ) = x + y(x);ic := y(0)=-7/10; x0 := -4;x1 := 4;y0 := -4; y1 := 4; b := 1; P1 := DEplot( ode, y(x), x=x0..x1, y=y0..y1, arrows=thin, title="#21(a) (Section 9.3)" ): P1; Ygen := unapply( rhs(dsolve( ode, y(x) )), x,_C1 ); # (b) P2 := seq( plot( Ygen(x,c), x=x0..x1, y=y0..y1, color=blue ), c=-2..2 ): # (c) display( [P1,P2], title="#21(c) (Section 9.3)" ); CC := solve( Ygen(0,C)=rhs(ic), C ); # (d) Ypart := Ygen(x,CC); P3 := plot( Ypart, x=0..b, title="#21(d) (Section 9.3)" ): P3; euler4 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/4 ): # (e) P4 := odeplot( euler4, [x,y(x)], x=0..b, numpoints=4, color=blue ):

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

597

598

Chapter 9 Further Applications of Integration display( [P3,P4], title="#21(e) (Section 9.3)" ); euler8 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/8 ): # (f) P5 := odeplot( euler8, [x,y(x)], x=0..b, numpoints=8, color=green ): euler16 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/16 ): P6 := odeplot( euler16, [x,y(x)], x=0..b, numpoints=16, color=pink ): euler32 := dsolve( {ode,ic}, numeric, method=classical[foreuler], stepsize=(x1-x0)/32 ): P7 := odeplot( euler32, [x,y(x)], x=0..b, numpoints=32, color=cyan ): display( [P3,P4,P5,P6,P7], title="#21(f) (Section 9.3)" ); << N|h | `percent error` >, # (g) < 4 | (x1-x0)/ 4 | evalf[5]( abs(1-eval(y(x),euler4(b))/eval(Ypart,x=b))*100 ) >, < 8 | (x1-x0)/ 8 | evalf[5]( abs(1-eval(y(x),euler8(b))/eval(Ypart,x=b))*100 ) >, < 16 | (x1-x0)/16 | evalf[5]( abs(1-eval(y(x),euler16(b))/eval(Ypart,x=b))*100 ) >, < 32 | (x1-x0)/32 | evalf[5]( abs(1-eval(y(x),euler32(b))/eval(Ypart,x=b))*100 ) > >;

13-24. Example CAS commands: Mathematica: (assigned functions, step sizes, and values for initial conditions may vary) For exercises 13 - 20, find the exact solution as follows. Set up two error lists. Clear[x, y, f] f[x_,y_]:= 2 y2 (x
1) a = 2; b =
1/2; xstar = 3; desol=DSolve[{y'[x] == f[x, y[x]], y[a] == b}, y[x], x] //Simplify actual[x_] = desol[[1, 1, 2]]; {xstar, actual[xstar]} errorlisteuler = { }; errorlisteulerimp = { }; pa = Plot[actual[x], {x, a, xstar}] Euler's method with error at x*. The Do command is used with a sequence of commands that are repeated n times. a = 2; b = -1/2; dx = 0.2; xstar = 3; n = (xstar
a) /dx; solnslist = {{a,b}}; Do[ {new = b + f[a,b] dx, a = a + dx, b = new, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar]
solnslist[[n, 2]] relativeerror= error / actual[xstar] AppendTo[errorlisteuler, error] pe = ListPlot[solnslist, PlotStyle Ä {Hue[.4], PointSize[0.02]}] Show[pa, pe] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteuler Improved Euler's method. with error at x* a = 2; b =
1/2; dx = 0.2; xstar = 3; n = (xstar
a) /dx; solnslist = {{a,b}}; Do[{new1 = b  f[a,b] dx, new2 = b + (f[a, b]  f[a+dx, new1])/2 dx, a = a  dx, b = new2, AppendTo[solnslist, {a,b}]},{n}] solnslist error= actual[xstar]
solnslist[[n, 2] Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.4 Graphical Solutions of Autonomous Differential Equations relativeerror= error / actual[xstar] AppendTo[errorlisteulerimp, error] peimp = ListPlot[solnslist, PlotStyle Ä {Hue[.8], PointSize[0.02]}] Show[pa, peimp] Rerun with different values for dx, starting from largest to smallest. After doing this, observe what happens to the error as the step size decreases by entering the input command: errorlisteulerimp You can also type Show[pa, pe, peimp]. This would be appropriate for a fixed value of dx with each method. You can also make a list of relative errors. Problems 21 - 24 involve use of code from section 9.1 together with the above code for Euler's method. 9.4 GRAPHICAL SOUTIONS OF AUTONOMOUS DIFFERENTIAL EQUATIONS 1. y w œ ay  2bay
3b (a) y œ
2 is a stable equilibrium value and y œ 3 is an unstable equilibrium. (b) yww œ a2y
1by w œ 2ay  2bˆy
12 ‰ay
3b

(c)

2. y w œ ay  2bay
2b (a) y œ
2 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ 2yy w œ 2ay  2byay
2b

(c)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

599

600

Chapter 9 Further Applications of Integration

3. y w œ y3
y œ ay  1byay
1b (a) y œ
1 and y œ 1 is an unstable equilibrium and y œ 0 is a stable equilibrium value. (b) yww œ a3y2
1by w œ 3ay  1bŠy 

1 È3 ‹yŠy




1 È3 ‹ay


1b

(c)

4. y w œ yay
2b (a) y œ 0 is a stable equilibrium value and y œ 2 is an unstable equilibrium. (b) yww œ a2y
2by w œ 2yay
1bay
2b

(c)

5. y w œ Èy, y  0 (a) There are no equilibrium values. 1 1 w Èy œ "# (b) yww œ 2È y y œ 2È y

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.4 Graphical Solutions of Autonomous Differential Equations (c)

6. y w œ y
Èy, y  0 (a) y œ 1 is an unstable equilibrium. (b) yww œ Š1


1 2È y ‹

y w œ Š1


1 ˆ 2È y ‹ y


Èy‰ œ ˆÈy
"# ‰ˆÈy
1‰

(c)

7. y w œ ay
1bay
2bay
3b (a) y œ 1 and y œ 3 is an unstable equilibrium and y œ 2 is a stable equilibrium value. (b) yww œ a3y2
12y  11bay
1bay
2bay
3b œ 3ay
1bŠy


6  È3 ‹ay 3


2bŠy


6
È3 3 ‹ay

(c)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley


3b

601

602

Chapter 9 Further Applications of Integration

8. y w œ y3
y2 œ y2 ay
1b (a) y œ 0 and y œ 1 is an unstable equilibrium. (b) yww œ a3y2
2ybay3
y2 b œ y3 a3y
2bay
1b

(c)

9.

10.

dP dt

œ 1
2P has a stable equilibrium at P œ "# .

dP dt œ Pa1
2Pb has an unstable equilibrium d2 P dP dt2 œ a1
4Pb dt œ Pa1
4Pba1
2Pb

d2 P dt2

œ
2 dP dt œ
2a1
2Pb

at P œ 0 and a stable equilibrium at P œ "# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.4 Graphical Solutions of Autonomous Differential Equations

11.

12.

dP dt œ 2PaP
3b has a d2 P dP dt2 œ 2a2P
3b dt œ

dP dt d2 P dt2

stable equilibrium at P œ 0 and an unstable equilibrium at P œ 3. 4Pa2P
3baP
3b

œ 3Pa1
PbˆP
"# ‰ has a stable equilibria at P œ 0 and P œ 1 an unstable equilibrium at P œ "# . 3 œ
#3 a6P2
6P+1b dP dt œ # PŠP


3
È3 ˆ ‹ P 6


"# ‰ŠP


3  È3 ‹aP 6


1b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

603

604

Chapter 9 Further Applications of Integration

13.

Before the catastrophe, the population exhibits logistic growth and Patb Ä M0 , the stable equilibrium. After the catastrophe, the population declines logistically and Patb Ä M1 , the new stable equilibrium. 14.

dP dt

œ rPaM
PbaP
mb, r, M, m  0

The model has 3 equilibrium points. The rest point P œ 0, P œ M are asymptotically stable while P œ m is unstable. For initial populations greater than m, the model predicts P approaches M for large t. For initial populations less than m, the model predicts extinction. Points of inflection occur at P œ a and P œ b where a œ "3
M  m
ÈM2
mM  m2 ‘ and b œ "3
M  m  ÈM2
mM  m2 ‘. (a) The model is reasonable in the sense that if P  m, then P Ä 0 as t Ä _; if m  P  M, then P Ä M as t Ä _; if P  M, then P Ä M as t Ä _. (b) It is different if the population falls below m, for then P Ä 0 as t Ä _ (extinction). If is probably a more realistic model for that reason because we know some populations have become extinct after the population level became too low. (c) For P  M we see that dP dt œ rPaM
PbaP
mb is negative. Thus the curve is everywhere decreasing. Moreover, P ´ M is a solution to the differential equation. Since the equation satisfies the existence and uniqueness conditions, solution trajectories cannot cross. Thus, P Ä M as t Ä _. (d) See the initial discussion above. (e) See the initial discussion above. 15.

dv dt

œg


k 2 mv ,

Equilibrium: Concavity:

g, k, m  0 and vatb   0

dv dt

d2 v dt2

œg


k 2 mv

œ 0 Ê v œ É mg k

ˆ k ‰ˆ œ
2ˆ mk v‰ dv dt œ
2 m v g


k 2‰ mv

(a)

(b)

160 (c) vterminal œ É 0.005 œ 178.9

ft s

œ 122 mph

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.4 Graphical Solutions of Autonomous Differential Equations

605

16. F œ Fp
Fr ma œ mg
kÈv dv kÈ v, va0b œ v0 dt œ g
m Thus,

dv dt

‰ , the terminal velocity. If v0  ˆ mg ‰ , the object will fall faster and faster, approaching the œ 0 implies v œ ˆ mg k k 2

2

‰ , the object will slow down to the terminal velocity. terminal velocity; if v0  ˆ mg k 2

17. F œ Fp
Fr ma œ 50
5kvk dv 1 dt œ m a50
5kvkb The maximum velocity occurs when

dv dt

œ 0 or v œ 10

ft sec .

18. (a) The model seems reasonable because the rate of spread of a piece of information, an innovation, or a cultural fad is proportional to the product of the number of individuals who have it (X) and those who do not (N
X). When X is small, there are only a few individuals to spread the item so the rate of spread is slow. On the other hand, when (N
X) is small the rate of spread will be slow because there are only a few indiciduals who can receive it during the interval of time. The rate of spread will be fastest when both X and (N
X) are large because then there are a lot of individuals to spread the item and a lot of individuals to receive it. (b) There is a stable equilibrium at X œ N and an unstable equilibrium at X œ 0. d2 X dt2

dX 2 œ k dX dt aN
Xb
kX dt œ k XaN
XbaN
2Xb Ê inflection points at X œ 0, X œ

N 2,

(c)

(d) The spread rate is most rapid when x œ

Eventually all of the people will receive the item.

œ VL
RL i œ RL ˆ VR
i‰, V, L, R  0 œ RL ˆ VR
i‰ œ 0 Ê i œ VR

19. L di dt  Ri œ V Ê

di dt d i 2 dt œ

Equilibrium: Concavity:

N 2.

2

di dt

ˆ R ‰2 ˆ VR
i‰
ˆ RL ‰ di dt œ
L

Phase Line:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

and X œ N.

606

Chapter 9 Further Applications of Integration

If the switch is closed at t œ 0, then ia0b œ 0, and the graph of the solution looks like this:

As t Ä _, it Ä isteady state œ

V R.

(In the steady state condition, the self-inductance acts like a simple wire connector and, as

a result, the current throught the resistor can be calculated using the familiar version of Ohm's Law.) 20. (a) Free body diagram of he pearl:

(b) Use Newton's Second Law, summing forces in the direction of the acceleration: ˆ m m
P ‰g
mk v. mg
Pg
kv œ ma Ê dv dt œ (c) Equilibrium: Ê vterminal œ Concavity:

dv dt

œ

k am
Pbg mŠ k


v‹ œ 0

am
Pbg k

d2 v dt2

ˆ k ‰ am
k Pbg
v‹ œ
mk dv dt œ
m Š 2

(d)

(e) The terminal velocity of the pearl is

am
Pbg . k

9.5 APPLICATIONS OF FIRST ORDER DIFFERENTIAL EQUATIONS 1. Note that the total mass is 66  7 œ 73 kg, therefore, v œ v0 e
akÎmbt Ê v œ 9e
3.9tÎ73
3.9tÎ73 (a) satb œ ' 9e
3.9tÎ73 dt œ
2190 C 13 e

Since sa0b œ 0 we have C œ

2190 13


3.9tÎ73 ‰ ˆ and lim satb œ lim 2190 œ 13 1
e tÄ_

tÄ_

2190 13

¸ 168.5

The cyclist will coast about 168.5 meters. 73 ln 9 (b) 1 œ 9e
3.9tÎ73 Ê 3.9t 73 œ ln 9 Ê t œ 3.9 ¸ 41.13 sec It will take about 41.13 seconds. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.5 Applications of First-Order Differential Equations 2. v œ v0 e
akÎmbt Ê v œ 9e
a59,000Î51,000,000bt Ê v œ 9e
59tÎ51,000 (a) satb œ ' 9e
59tÎ51,000 dt œ
459,0000 e
59tÎ51,000  C 59 Since sa0b œ 0 we have C œ

459,0000 59

ˆ1
e
59tÎ51,000 ‰ œ and lim satb œ lim 459,0000 59 tÄ_

tÄ_

459,0000 59

¸ 7780 m

The ship will coast about 7780 m, or 7.78 km. ln 9 59t (b) 1 œ 9e
59tÎ51,000 Ê 51,000 œ ln 9 Ê t œ 51,000 ¸ 1899.3 sec 59 It will take about 31.65 minutes. 3. The total distance traveled œ v0km Ê a2.75bak39.92b œ 4.91 Ê k œ 22.36. Therefore, the distance traveled is given by the function satb œ 4.91ˆ1
e
a22.36/39.92bt ‰. The graph shows satb and the data points.

4.

a0.80ba49.90b œ 1.32 Ê k œ 998 k 33 v0 m k 998 We know that k œ 1.32 and m œ 33a49.9b œ 20 . 33 v0 m ˆ
ak/mbt ‰ Using Equation 3, we have: satb œ k 1
e œ 1.32ˆ1 v0 m k

5. (a)

œ coasting distance Ê

dP dt

œ 0.0015Pa150
Pb œ

0.255 150 Pa150


Pb œ

Thus, k œ 0.255 and M œ 150, and P œ Initial condition: Pa0b œ 6 Ê 6 œ Formula: P œ

150
0.255t 1  24ec0.255t Ê 1  24e ln 48 Ê t œ 0.255 ¸ 17.21 weeks 150 125 œ 1  24ec0.255t Ê 1  24e
0.255t 120 Ê t œ ln0.255 ¸ 21.28

(b) 100 œ

M 1  Aeckt

150 1  Ae0

150 1  24ec0.255t

k M PaM

œ


e
20t/33 ‰ ¸ 1.32a1
e
0.606t b


Pb

150 1  Aec0.255t

Ê 1  A œ 25 Ê A œ 24

œ

3 2

Ê 24e
0.255t œ

" #

Ê e
0.255t œ

" 48

œ

6 5

Ê 24e
0.255t œ

" 5

Ê e
0.255t œ

" 120

Ê
0.255t œ
ln 48 Ê
0.255t œ
ln 120

It will take about 17.21 weeks to reach 100 guppies, and about 21.28 weeks to reach 125 guppies. 6. (a)

dP dt

œ 0.0004Pa250
Pb œ

0.1 250 Pa150


Pb œ

Thus, k œ 0.1 and M œ 250, and P œ

k M PaM
Pb M œ 1 250 Aec0.1t 1  Aeckt

Initial condition: Pa0b œ 28, where t œ 0 represents the year 1970 250 111 28 œ 1 250 Ae0 Ê 28a1  Ab œ 250 Ê A œ 28
1 œ 14 ¸ 7.9286 Formula: P œ

250 c0.1t 1  111 14 e

or approximately P œ

250 1  7.9286ec0.1t

(b) The population Patb will round to 250 when Patb   249.5 Ê 249.5 œ Ê

a249.5bˆ111ec0.1t ‰ 14

œ 0.5 Ê e
0.1t œ

14 55,389

Ê
0.1t œ ln

14 55,389

250 c0.1t 1  111 14 e

Ê 249.5ˆ1 

Ê t œ 10 aln 55,389
ln 14b ¸ 82.8.

It will take about 83 years. 7. (a) Using the general solution form Example 2, part (c), dy dt

œ a0.08875 ‚ 10
7 ba8 ‚ 107
yby Ê yatb œ

M 1  AecrMt

œ

111
0.1t ‰ 14 e

8‚107 1  Aeca!Þ!))(&ba)bt

œ

8‚107 1  Aec0.71t

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 250

607

608

Chapter 9 Further Applications of Integration Apply the initial condition: ya0b œ 1.6 ‚ 107 œ

(b) yatb œ 4 ‚ 107 œ

8‚107 1A

Ê

8‚107 1  4ec0.71t

8 1.6

8‚107 1  4ec0.71 ¸ 2.69671 ‚ lnˆ 1 ‰
0.714 ¸ 1.95253 years.


1 œ 4 Ê ya1b œ

Ê 4e
0.71t œ 1 Ê t œ

107 kg.

8. (a) If a part of the population leaves or is removed from the environment (e.g., a preserve or a region) each year, then c would represent the rate of reduction of the population due to this removal and/or migration. When grizzly bears become a nuisance (e.g., feeding on livestock) or threaten human safety, they are often relocated to other areas or even eliminated, but only after relocation efforts fail. In addition, bears are killed, sometimes accidently and sometimes maliciously. For an environment that has a capacity of about 100 bears, a realistic value for c would probably be between 0 and 4. (b)

Equilibrium solutions:

dP dt

œ 0 œ 0.001a100
PbP
1 Ê P2
100P  1000 œ 0 Ê Peq ¸ 11.27 (unstable) and

Peq ¸ 88.73 (stable) (c)

For 0  Pa0b Ÿ 11, the bear population will eventually disappear, for 12 Ÿ Pa0b Ÿ 88, the population will grow to about 89, the population will remain at about 89, and for Pa0b  89, the population will decrease to about 89 bears. 9. (a)

dy dt

œ 1  y Ê dy œ a1  ybdt Ê

dy 1y

œ dt Ê ln k1  yk œ t  C1 Ê eln k1yk œ etC1 Ê k1  yk œ et eC1

1  y œ „ C2 et Ê y œ Cet
1, where C2 œ eC1 and C œ „ C2 . Apply the initial condition: ya0b œ 1 œ Ce0
1 Ê C œ 2 Ê y œ 2et
1. (b)

dy dt

œ 0.5a400
yby Ê dy œ 0.5a400
yby dt Ê

Example 2, part (c), we obtain Ê ' Š 1y  ln¹ y cy400 ¹

Êe Ê

y y
400

Êyœ

1 400
y ‹dy

1 1 400 Š y



1 400
y ‹dy

dy a400
yby

œ 0.5 dt. Using the partial fraction decomposition in

œ 0.5 dt Ê Š 1y 

1 400
y ‹dy

œ 200 dt

œ ' 200 dt Ê lnkyk
lnky
400k œ 200t  C1 Ê ln¹ y
y400 ¹ œ 200t  C1

œ e200tC1 œ e200t eC1 Ê ¹ y
y400 ¹ œ C2 e200t (where C2 œ eC1 ) Ê

y y
400

œ „ C2 e200t

œ Ce200t (where C œ „ C2 ) Ê y œ Ce200t y
400 Ce200t Ê a1
Ce200t by œ
400 Ce200t

400 Ce200t Ce200t
1

ya0b œ 2 œ

Êyœ

400 1  Ae0

400 1
C1 ec200t

œ

400 1  Aec200t ,

Ê A œ 199 Ê yatb œ

where A œ
C1 . Apply the initial condition:

400 1  199ec200t

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.5 Applications of First-Order Differential Equations 10.

dP dt

œ raM
PbP Ê dP œ raM
PbP dt Ê

we obtain

1 ˆ1 M P

1 ‰ M
P dP



dP aM
PbP

œ r dt. Using the partial fraction decomposition in Example 6, part (c),

'

œ r dt Ê ˆ P1 

Ê lnkPk
lnkP
Mk œ arMb Ê ¸ P
P M ¸ œ C2 earMbt (where

'

1 ‰ ˆ P1
P
1 M ‰dP œ rM dt M
P dP œ rM dt Ê P t  C1 Ê ln¸ P
P M ¸ œ arMb t  C1 Ê eln¸ P c M ¸ œ earMbtC1 œ earMbt eC1 C2 œ eC1 ) Ê P
PM œ „ C2 earMbt Ê P
PM œ CearMbt (where C œ „ C2 )

Ê P œ CearMbt P
M CearMbt Ê ˆ1
CearMbt ‰P œ
M CearMbt Ê P œ ÊPœ 11. (a)

dP dt

M , 1
AecarMbt

œ kP2 Ê ' P
2 dP œ ' k dt Ê
P
" œ kt  C Ê P œ

Initial condition: Pa0b œ P0 Ê P0 œ
C1 Ê C œ

dP dt

œ raM
PbaP
mb Ê

Ê

ÊPœ

M 1
C1 ecarMbt

aP
100ba1200
Pb dP a1200
PbaP
100b dt


1 P0


" kt  C

P0 1
kP0 t

(b) There is a vertical asymptote at t œ 12. (a)

M CearMbt CearMbt
1

where A œ
C1 .

Solution: P œ
kt
a11/P0 b œ

dP dt

1 kpO

œ ra1200
PbaP
100b Ê

œ 1100 r Ê

ˆ 12001
P



1 ‰ dP P
100 dt

1 dP a1200
PbaP
100b dt

œ 1100 r Ê

œrÊ

ˆ 12001
P



1100 dP a1200
PbaP
100b dt

1 ‰ P
100 dP

P
100 1200
P 1100 r t

where C œ „ eC1 Ê P
100 œ 1200Ce1100 r t
CPe1100 r t Ê Pa1  Ce1100 r t b œ 1200Ce ÊPœ

1200Ce  100 Ce1100 r t  1

œ

c1100 r t 1200  100 C e 1  C1 ec1100 r t

(b) Apply the initial condition: 300 œ

1200  100Aec1100 r t 1  Aec1100 r t

ÊPœ

1200  100A 1A

œ 1100 r

œ 1100 r dt

Ê ' ˆ 12001
P  P
1100 ‰dP œ ' 1100 r dt Ê
ln a1200
Pb  ln aP
100b œ 1100 r t  C1 P
100 ¸ P
100 C1 1100 r t ¸ P
100 ¸ Ê ln ¸ 1200 Ê
P œ 1100 r t  C1 Ê ln 1200
P œ 1100 r t  C1 Ê 1200
P œ „ e e 1100 r t

œ Ce1100 r t

 100

where A œ C1 .

Ê 300  300A œ 1200  100A Ê A œ

9 2

ÊPœ

2400  900Aec1100 r t Þ 2  9ec1100 r t

(Note that P Ä 1200 as t Ä _.) (c)

dP dt

œ raM
PbaP
mb Ê

Ê ˆ M 1
P 

1 ‰ dP P
m dt

1 dP aM
PbaP
mb dt

œrÊ

œ raM
mb Ê ' ˆ M 1
P 

Ê
ln aM
Pb  ln aP
mb œ aM
mb r t  Ê

P
m M
P

M
m dP aM
PbaP
mb dt

Ê Pˆ1  Ce

aM
mb r t

œ MCe

œ raM
mb Ê

'

1 ‰ raM
mbdt P
m dP œ P
m ¸ ¸ C1 Ê ln M
P œ aM
mb r t  aM
mb r t aM
mb r t

œ CeaM
mb r t where C œ „ eC1 Ê P
m œ MCe aM
mb r t ‰

mÊPœ

aP
mb  aM
Pb dP aM
PbaP
mb dt

C1 Ê

P
m M
P

Apply the initial condition Pa0b œ P0 MmA 1 A

Ê P0  P0 A œ M  mA Ê A œ

M
P0 P0
m

ÊPœ

ÊPœ

caMcmb r t M m Ce 1  C1 ecaMcmb r t

ÊPœ

MaP0
mb  maM
P0 becaMcmb r t aP0
mb  aM
P0 becaMcmb r t

(Note that P Ä M as t Ä _ provided P0  m.) 13. y œ mx Ê

y x

orthogonals:

œmÊ dy dx

xy
y x2 w

œ 0 Ê y w œ yx . So for

œ
xy Ê y dy œ
x dx Ê

y2 2



x2 2

œC

Ê x  y œ C1 2

œ raM
mb

œ „ eC1 eaM
mb r t


CPe

MCeaMcmb r t  m CeaMcmb r t  1

A œ C1 . P0 œ

609

2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

M  mAecaMcmb r t 1  AecaMcmb r t

610

Chapter 9 Further Applications of Integration

14. y œ cx2 Ê Ê yw œ

y x2

2y x .

œcÊ

x2 y
2xy x4

œ 0 Ê x2 y w œ 2xy

w

So for the orthogonals:

dy dx

x œ
2y

2

Ê 2ydy œ
xdx Ê y2 œ
x2  C Ê y œ „ É x2  C, 2

C0

15. kx2  y2 œ 1 Ê 1
y2 œ kx2 Ê

1
y2 x2

œk

x a2yby
ˆ1
y2 ‰2x Ê œ 0 Ê
2yx2 y w œ a1
y2 ba2xb x% ˆ1
y2 ‰a2xb ˆ1
y 2 ‰ Ê y w œ
2xy2 œ
xy . So for the orthogonals: 2 ˆ1
y 2 ‰ 2 dy xy dy œ x dx Ê ln y
y2 œ x2  C dx œ 1
y2 Ê y 2

w

2x 16. 2x2  y2 œ c2 Ê 4x  2yy w œ 0 Ê y w œ
4x 2y œ
y . For

orthogonals:

dy dx

œ

y 2x

Ê

œ

dy y

dx 2x

Ê ln y œ "# ln x  C

Ê ln y œ ln x1/2  ln C1 Ê y œ C1 kxk1/2

17. y œ ce
x Ê

y ecx

œcÊ

e
x y w c yae
x bac1b ae
x b2

œ!

Ê e
x y w œ
ye
x Ê y w œ
y. So for the orthogonals: dy dx

œ

1 y 2

Ê y dy œ dx Ê

y2 2

œxC

Ê y œ 2x  C1 Ê y œ „ È2x  C1

xŠ 1y ‹y c ln y w

18. y œ ekx Ê ln y œ kx Ê

ln y x

œkÊ

Ê Š xy ‹ y w
ln y œ 0 Ê y w œ dy dx

y ln y x .


x y ln y Ê y ln y dy œ
x dx 1 2 " 2 ˆ " 2‰  # y ln y
4 ay b œ
# x 2 y2 ln y
y2 œ
x2  C1

x2

œ0

So for the orthogonals:

œ

Ê Ê

C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 9.5 Applications of First-Order Differential Equations 2 w 19. 2x2  3y2 œ 5 and y2 œ x3 intersect at a1, 1b. Also, 2x2  3y2 œ 5 Ê 4x  6y y w œ 0 Ê y w œ
4x 6y Ê y a1, 1b œ
3

y21 œ x3 Ê 2y1 y1w œ 3x2 Ê y1w œ x2 2

20. (a) x dx  y dy œ 0 Ê



y2 2 w

3x2 2y1

Ê y1w a1, 1b œ 32 . Since y w † y1w œ ˆ
23 ‰ˆ 32 ‰ œ
1, the curves are orthogonal.

œ C is the general equation

of the family with slope y œ
xy . For the orthogonals: yw œ

y x

Ê

dy y

œ

dx x

Ê ln y œ ln x  C or y œ C1 x

(where C1 œ e Ñ is the general equation of the orthogonals. C

(b) x dy
2y dx œ 0 Ê 2y dx œ x dy Ê Ê "# Š dy y ‹œ

dx x

dy 2y

œ

dx x

Ê "# ln y œ ln x  C Ê y œ C1 x2 is

the equation for the solution family. " # ln

y
ln x œ C Ê

"y # y

w




Ê slope of orthogonals is

1 x dy dx

œ 0 Ê yw œ

2y x

x œ
2y 2

Ê 2y dy œ
x dx Ê y2 œ
x2  C is the general equation of the orthogonals.

2". y2 œ 4a2
4ax and y2 œ 4b2  4bx Ê (at intersection) 4a2
4ax œ 4b2  4bx Ê a2
b2 œ xaa  bb Ê aa  bbaa
bb œ aa  bbx Ê x œ a
b. Now, y2 œ 4a2
4aaa
bb œ 4a2
4a2  4ab œ 4ab Ê y œ „ 2Èab. 4a Thus the intersections are at Ša
b, „ 2Èab‹. So, y2 œ 4a2
4ax Ê y1w œ
2y which are equal to
4a and È 2Š2

4a
2Š
2Èab‹ 4b 2Š
2Èab‹

œ


È ba

and

È ba

at the intersections. Also, y œ 4b  4bx Ê 2

2

y2w

œ

4b 2y

which are equal to

œ
É ba and É ba at the intersections. ay1w b † ay2w b œ
". Thus the curves are orthogonal.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

ab‹

4b 2Š2Èab‹

and

611

612

Chapter 9 Further Applications of Integration

CHAPTER 9 PRACTICE EXERCISES ".

dy dx

œ Èy cos2 Èy Ê

2. y w œ

3yax1b2 y
1

Ê

œ dx Ê 2tanÈy œ x  C Ê y œ ˆtan
1 ˆ x 2 C ‰‰

dy Èy cos2 Èy

ay
1 b y dy

3. yy w œ secay2 bsec2 x Ê

œ 3ax  1b2 dx Ê y
ln y œ ax  1b3  C

y dy secay2 b

sinˆy2 ‰ 2

œ sec2 x dx Ê

œ tan x  C Ê sinay2 b œ 2tan x  C1

sin x 4. y cos2 axb dy  sin x dx œ 0 Ê y dy œ
cos 2 axb dx Ê

y2 2

œ
cos1axb  C Ê y œ „ É cos
a2xb  C1

2ax
2b3/2 a3x  4b  15
2ax
2b3/2 a3x  4b
C“ 15

5. y w œ xey Èx
2 Ê e
y dy œ xÈx
2 dx Ê
e
y œ
2ax
2b3/2 a3x  4b 15

Ê
y œ ln’ 6. y w œ xyex Ê 2

dy y

2


C“ Ê y œ
ln’

C Ê e
y œ


2ax
2b3/2 a3x  4b 15


C

œ ex x dx Ê ln y œ "# ex  C 2

2

7. sec x dy  x cos2 y dx œ 0 Ê

dy cos2 y

x dx œ
sec x Ê tan y œ
cos x
x sin x  C

8. 2x2 dx
3Èy csc x dy œ 0 Ê 3Èy dy œ

2x2 csc x dx

Ê 2y3/2 œ 2a2
x2 bcos x  4x sin x  C

Ê y3/2 œ a2
x2 bcos x  2x sin x  C1 9. y w œ

ey xy

Ê ye
y dy œ

Ê ay  1be
y œ
ln kxk  C

dx x

10. y w œ xex
y csc y Ê y w œ

x ex ey csc

yÊ

ey csc y dy

œ x ex dx Ê

11. xax
1bdy
y dx œ 0 Ê xax
1bdy œ y dx Ê

dy y

œ

ey 2 asin

dx x ax
1 b

y
cos yb œ ax
1bex  C

Ê ln y œ lnax
1b
lnaxb  C

Ê ln y œ lnax
1b
lnaxb  ln C1 Ê ln y œ lnŠ C1 axx
1b ‹ Ê y œ

12. y w œ ay2
1bax
1 b Ê

dy

y 2
1

œ

Ê

dx x

1 lnŠ yy c b1‹

2

C1 ax
1b x


1 œ ln x  C Ê lnŠ yy  1 ‹ œ 2ln x  ln C1 Ê

y
1 y1

œ C1 x2

13. 2y w
y œ xex/2 Ê y w
"# y œ x2 ex/2 .

' ˆ "‰ paxb œ
"# , vaxb œ e c # dx œ e
x/2 .

e
x/2 y w
"# e
x/2 y œ ˆe
x/2 ‰ˆ x2 ‰ˆex/2 ‰ œ 14.

w

y 2

x 2

Ê

d ˆ
x/2 dx e

y‰ œ

x 2

Ê e
x/2 y œ

x2 4

2

 C Ê y œ ex/2 Š x4  C‹

 y œ e
x sin x Ê y w  2y œ 2e
x sin x.

paxb œ 2, vaxb œ e' 2dx œ e2x . e2x y w  2e2x y œ 2e2x e
x sin x œ 2ex sin x Ê
x

d 2x dx ae

yb œ 2ex sin x Ê e2x y œ ex asin x
cos xb  C


2x

Ê y œ e asin x
cos xb  Ce

15. xy w  2y œ 1
x
1 Ê y w  ˆ 2x ‰y œ vaxb œ e2'

dx x

1 x




1 x2 .

2

œ e2ln x œ eln x œ x2 .

x2 y w  2xy œ x
1 Ê

d 2 dx ax yb

œ x
1 Ê x2 y œ

x2 2


xCÊyœ

" #




1 x



C x2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 9 Practice Exercises

613

16. xy w
y œ 2x ln x Ê y w
ˆ 1x ‰y œ 2 ln x. vaxb œ e
d ˆ1 dx x

' dxx

2 œ e
ln x œ 1x . ˆ 1x ‰y w
ˆ 1x ‰ y œ 2x ln x Ê

† y‰ œ 2x ln x Ê

† y œ c ln x d2  C Ê y œ xc ln x d2  Cx

1 x

17. a1  ex bdy  ayex  e
x bdx œ 0 Ê a1  ex by w  ex y œ
e
x Ê y w œ ' a exdx b

vaxb œ e 1 b ex œ elnae 1b œ ex  1. x cx aex  1by w  aex  1bˆ 1 e ex ‰y œ a1
e ex b aex  1b Ê Êyœ 18.

dx dy

ecx  C ex  1

ex 1  ex y

œ


ecx a1  e x b .

x

œ

e cx  C

 1by d œ
e
x Ê aex  1by œ e
x  C

d aex dx c

1  ex

 x
4yey œ 0 Ê x w  x œ 4yey . Let vayb œ e

' dy

œ ey . Then ey x w  xey œ 4ye2y Ê

d y dy axe b

œ 4ye2y

Ê xey œ a2y
1be2y  C Ê x œ a2y
1bey  Ce
y 19. ax  3y2 b dy  y dx œ 0 Ê x dy  y dx œ
3y2 dy Ê

d dx axyb

œ
3y2 dy Ê xy œ
y3  C '

20. y dx  a3x
y
2 cos yb dx œ 0 Ê x w  Š 3y ‹x œ y
3 cos y. Let vayb œ e

3dy y

3

œ e3ln y œ eln y œ y3 .

Then y3 x w  3y2 x œ cos y and y3 x œ ' cos y dy œ sin y  C. So x œ y
3 asin y  Cb 21.

œ e
x
y
2 Ê ey dy œ e
ax2b dx Ê ey œ
e
ax2b  C. We have ya0b œ
2, so e
2 œ
e
2  C Ê C œ 2e
2 and e œ
e
ax2b  2e
2 Ê y œ lnˆ
e
ax2b  2e
2 ‰

22.

dy dx

dy dx y

œ

y ln y 1  x2

Ê etan

Ê

c1 a0bC

dy y ln y

œ

dx 1  x2

Ê lnaln yb œ tan
1 axb  C Ê y œ ee

So y w ax  1b2  Ê yax  1b2 œ

x x1.

' x b2 1 dx

Let vaxb œ e

2 ax  1 b a x

 1b2 y œ

x3 3

 C Ê y œ ax  1b
2 Š x3 



y œ ax  1b
2 Š x3  3

x2 2 2

x 2

x ax  1 b a x

 1b2 Ê

d
dx yax

3

x2 2

dy dx

tanc1 axbbln 2

2

œ e2lnax1b œ elnax1b œ ax  1b2 .

2  1b2 ‘ œ xax  1b Ê yax  1b œ ' xax  1bdx

 1‹ ' ˆ 2 ‰dx

25.

tanc1 a0bbC

 C‹. We have ya0b œ 1 Ê 1 œ C. So

1 2 w ˆ2‰ 24. x dy dx  2y œ x  1 Ê y  x y œ x  x . Let vaxb œ e

So

. We have ya0b œ e2 Ê e2 œ ee

œ 2 Ê tan
1 a0b  C œ ln 2 Ê 0  C œ ln 2 Ê C œ ln 2 Ê y œ ee

w ˆ 2 ‰ 23. ax  1b dy dx  2y œ x Ê y  x  1 y œ

Ê

tanc1 axbbC

d x4 2 3 2 dx ax yb œ x  x Ê x y œ 4 2 4 2x2  1 y œ x4  4x1 2  "# œ x 4x 2



x2 2

CÊyœ

x2 4

x



C x2

œ eln x œ x2 . So x2 y w  2xy œ x3  x 2

 "# . We have ya1b œ 1 Ê 1 œ

 3x2 y œ x2 . Let vaxb œ e' 3x dx œ ex . So ex y w  3x2 ex y œ x2 ex Ê 2

3

3

3

3

We have ya0b œ
1 Ê e0 a
1b œ 13 e0  C Ê
1 œ

3

1 3

d dx axyb

dy ˆy È y ‰

3

" #

3

Ê C œ 14 .

3

œ x2 ex Ê ex y œ 13 ex  C.

3

cos x x .

C

3

4 3

Êyœ

1 3


43 e
x

3

' 1 dx x œ eln x œ x.

Let vaxb œ e

œ cos x Ê xy œ ' cos x dx Ê xy œ sin x  C. We have yˆ 12 ‰ œ 0 Ê ˆ 12 ‰0 œ 1  C

Ê C œ
1. So xy œ
1  sin x Ê y œ 27. x dy
ˆy  Èy‰dx œ 0 Ê

d x3 dx Še y‹

 C Ê C œ
43 and ex y œ 13 ex


26. xdy  ay
cos xbdx œ 0 Ê xy w  y
cos x œ 0 Ê y w  ˆ 1x ‰y œ So xy w  xˆ 1x ‰y œ cos x Ê

3

1 4

œ

dx x


1  sin x x

Ê 2lnˆÈy  1‰ œ ln x  C. We have ya1b œ 1 Ê 2 lnŠÈ1  1‹ œ ln 1  C

Ê 2 ln 2 œ C œ ln 22 œ ln 4. So 2 lnˆÈy  1‰ œ ln x  ln 4 œ lna4xb Ê lnˆÈy  1‰ œ "# lna4xb œ lna4xb1/2 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

614

Chapter 9 Further Applications of Integration Ê eln

ˆÈ y 1 ‰

28. y
2 dx dy œ So

y3 3

1/2

œ elna4xb Ê Èy  1 œ 2Èx Ê y œ ˆ2Èx
1‰

ex e2x  1

Ê

e2x  1 ex dx

œ ex
e
x 

1 3

œ

dy yc2

Ê

y3 3

2

œ ex
e
x  C. We have ya0b œ 1 Ê

a1 b 3 3

œ e0
e0  C Ê C œ 13 .

Ê y3 œ 3aex
e
x b  1 Ê y œ c3aex
e
x b  1 d1/3 xc2

29. xy w  ax
2by œ 3x3 e
x Ê y w  ˆ x
x 2 ‰y œ 3x2 e
x . Let vaxb œ e' ˆ x ‰dx œ ex
2ln x œ xe2 . So x x ex w ex ˆ x
2 ‰ d ˆ y œ 3 Ê dx y † xe2 ‰ œ 3 Ê y † xe2 œ 3x  C. We have ya1b œ 0 Ê 0 œ 3a1b  C Ê C œ
3 x2 y  x2 x Êy†

ex x2

œ 3x
3 Ê y œ x2 e
x a3x
3b

30. y dx  a3x
xy  2bdy œ 0 Ê Payb œ

3 y

x

dx dy



3x
xy  2 y

œ0Ê

dx dy



3x y


x œ
2y Ê

dx dy

 Š 3y
1‹x œ
2y .


1 Ê ' Paybdy œ 3ln y
y Ê vayb œ e3ln y
y œ y3 e
y

y3 e
y x w  y3 e
y Š 3y
1‹x œ
2y2 e
y Ê y3 e
y x œ '
2y2 e
y dy œ 2e
y ay2  2y  2b  C Ê y3 œ Ê y3 œ

2ˆy2  2y  2‰  Cey . x 2 yb1 ˆ ‰ 2 y  2y  2
4e x

We have ya2b œ
1 Ê
1 œ

2a1
2  2b  Cec1 2

Ê C œ
4e and

31. To find the approximate values let yn œ yn
1  ayn
1  cos xn
1 ba0.1b with x0 œ 0, y0 œ 0, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y 1.1 1.6241 0 0 1.2 1.8319 0.1 0.1000 1.3 2.0513 0.2 0.2095 1.4 2.2832 0.3 0.3285 1.5 2.5285 0.4 0.4568 1.6 2.7884 0.5 0.5946 1.7 3.0643 0.6 0.7418 1.8 3.3579 0.7 0.8986 1.9 3.6709 0.8 1.0649 2.0 4.0057 0.9 1.2411 1.0 1.4273 32. To find the approximate values let zn œ yn
1  aa2
yn
1 ba2 xn
1  3bba0.1b and yn œ yn
1  Š a2
ync1 ba2 xnc1  32b  a2
zn ba2 xn  3b ‹a0.1b with initial values x0 œ
3, y0 œ 1, and 20 steps. Use a spreadsheet, graphing calculator, or CAS to obtain the values in the following table. x y x y
1.9
5.9686
3 1
1.8
6.5456
2.9 0.6680
1.7
6.9831
2.8 0.2599
1.6
7.2562
2.7
0.2294
1.5
7.3488
2.6
0.8011
1.4
7.2553
2.5
1.4509
1.3
6.9813
2.4
2.1687
1.2
6.5430
2.3
2.9374
1.1
5.9655
2.2
3.7333
1.0
5.2805
2.1
4.5268
2.0
5.2840

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 9 Practice Exercises 2ync1 " xnc1
2ync1 33. To estimate ya3b, let zn œ yn
1  Š xncx1nc
1  1 ‹a0.05b and yn œ yn
1  # Š xnc1  1 

xn
2zn xn  1 ‹a0.05b

615

with initial values

x0 œ 0, y0 œ 1, and 60 steps. Use a spreadsheet, graphing calculator, or CAS to obtain ya3b ¸ 0.9063. 34. To estimate ya4b, let zn œ yn
1  Š

x2nc1
2ync1  1 ‹a0.05b xnc1

with initial values x0 œ 1, y0 œ 1, and 60 steps. Use a

spreadsheet, graphing calculator, or CAS to obtain ya4b ¸ 4.4974. 35. Let yn œ yn
1  ˆ exnc1 b1ync1 b 2 ‰adxb with starting values x0 œ 0 and y0 œ 2, and steps of 0.1 and
0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a)

(b) Note that we choose a small interval of x-values because the y-values decrease very rapidly and our calculator cannot handle the calculations for x Ÿ
1. (This occurs because the analytic solution is y œ
2  lna2
e
x b, which has an asymptote at x œ
ln 2 ¸ 0.69. Obviously, the Euler approximations are misleading for x Ÿ
0.7.)

y

y

36. Let zn œ yn
1
Š eynncc11  xnncc11 ‹adxb and yn œ yn
1  #" Š eynncc11  xnncc11  x2

x2

xn2  zn ezn  xn ‹adxb

with starting values x0 œ 0 and y0 œ 0,

and steps of 0.1 and
0.1. Use a spreadsheet, programmable calculator, or CAS to generate the following graphs. (a) (b)

37.

x y dy dx

1 1.2 1.4 1.6
1
0.8
0.56
0.28

œ x Ê dy œ x dx Ê y œ

Ê
1 œ

" #

Ê ya2b œ

CÊCœ 2

2 2




3 2

œ

" #


32

x2 2

1.8 0.04

2.0 0.4

 C; x œ 1 and y œ
1

Ê yaexactb œ

x2 2




3 2

is the exact value.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

616

Chapter 9 Further Applications of Integration

38.

x y œ

dy dx

1 1.2 1.4 1.6 1.8 2.0
1
0.8
0.6333
0.4904
0.3654
0.2544 1 x

Ê dy œ x1 dx Ê y œ lnkxk  C; x œ 1 and y œ
1

Ê
1 œ ln 1  C Ê C œ
1 Ê yaexactb œ lnkxk
1 Ê ya2b œ ln 2
1 ¸
0.3069 is the exact value.

39.

x y

1 1.2 1.4 1.6 1.8 2.0
1
1.2
0.488
1.9046
2.5141
3.4192

œ xy Ê

dy dx

Êyœe

dy y

x2 2 C

œ x dx Ê lnkyk œ x2 2

x2 2

C

x2 2

œ e † eC œ C1 e ; x œ 1 and y œ
1 x2

Ê
1 œ C1 e1/2 Ê C1 œ
e1/2 yaexactb œ
e1/2 † e 2 œ
eˆx
1‰/2 Ê ya2b œ
e3/2 ¸
4.4817 is the exact value. 2

40.

x y

1 1.2 1.4 1.6 1.8 2.0
1
1.2
1.3667
1.5130
1.6452
1.7688

dy y2 1 dx œ y Ê y dy œ dx Ê 2 œ x " " 2 # œ 1  C Ê C œ
# Ê y œ

 C; x œ 1 and y œ
1

2x
1 Ê yaexactb œ È2x
1 Ê ya2b œ
È3 ¸
1.7321 is the exact value.

41.

dy dx

œ y2
1 Ê y w œ ay  1bay
1b. We have y w œ 0 Ê ay  1b œ 0, ay
1b œ 0 Ê y œ
1, 1.

(a) Equilibrium points are
1 (stable) and 1 (unstable) (b) y w œ y2
1 Ê y ww œ 2yy w Ê y ww œ 2yay2
1b œ 2yay  1bay
1b. So y ww œ 0 Ê y œ 0, y œ
1, y œ 1.

(c)

42.

dy dx

œ y
y2 Ê y w œ ya1
yb. We have y w œ 0 Ê ya1
yb œ 0 Ê y œ 0, 1
y œ 0 Ê y œ 0, 1.

(a) The equilibrium points are 0 and 1. So, 0 is unstable and 1 is stable. (b) Let ïî œ increasing, íï œ decreasing. yw  ! yw  ! yw  ! qqíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqpy 0 1 y w œ y
y2 Ê y ww œ y w
2yy w Ê y ww œ ay
y2 b
2yay
y2 b œ y
y2
2y2  2y3 Ê y ww œ 2y3
3y2  y œ ya2y2
3y  1b Ê y ww œ ya2y
1bay
1b. So, y ww œ 0 Ê y œ 0, 2y
1 œ 0, y
1 œ 0 Ê y œ 0, y œ "# , Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 9 Additional and Advanced Exercises

617

y œ 1. Let ïî œ concave up, íï œ concave down. y ww  ! y ww  ! y ww  ! y ww  ! qíïïïïïqqñqqïïïïïîqqñqqíïïïïïqqñqqïïïïïîqpy 0 1 1/2 (c)

43. (a) Force œ Mass times Acceleration (Newton's Second Law) or F œ ma. Let a œ

dv dt

œ

dv ds

†

ds dt

œ v dv ds . Then

2
2 ma œ
mgR2 s
2 Ê a œ
gR2 s
2 Ê v dv Ê v dv œ
gR2 s
2 ds Ê ' v dv œ '
gR2 s
2 ds ds œ
gR s

Ê

v2 2

œ

ÊCœ

gR2 s  C1 2 v0
2gR

Ê v2 œ Êv œ 2

(b) If v0 œ È2gR, then v2 œ

2gR2 s 2gR2 s

2gR s

2

 2C1 œ 

v20

2gR2 s

 C. When t œ 0, v œ v0 and s œ R Ê v20 œ

2gR2 R

C


2gR 2

È2gR. Then Ê v œ É 2gR s , since v   0 if v0  

ds dt

œ

È2gR2 Ès

Ê Ès ds œ È2gR2 dt

Ê ' s1/2 ds œ ' È2gR2 dt Ê 23 s3/2 œ È2gR2 t  C1 Ê s3/2 œ ˆ 32 È2gR2 ‰t  C; t œ 0 and s œ R Ê R3/2 œ ˆ 32 È2gR2 ‰a0b  C Ê C œ R3/2 Ê s3/2 œ ˆ 32 È2gR2 ‰t  R3/2 œ ˆ 32 RÈ2g‰t  R3/2 3 œ R3/2
ˆ 32 R
1/2 È2g‰t  1 ‘ œ R3/2 ’ Š

44.

v0 m k

a0.86ba30.84b k
0.8866t

œ coasting distance Ê

Ê satb œ 0.97a1
e

È2gR 2R ‹t

2/3 0‰ 0‰ ‘ ‘ Ê s œ R
1  ˆ 3v  1 “ œ R3/2
ˆ 3v 2R t  1 2R t

œ 0.97 Ê k ¸ 27.343. satb œ

v0 m ˆ k 1


e
ak/mbt ‰ Ê satb œ 0.97ˆ1
e
a27.343/30.84bt ‰

b. A graph of the model is shown superimposed on a graph of the data.

CHAPTER 9 ADDITIONAL AND ADVANCED EXERCISES 1. (a)

dy dt

A œ kA V ac
yb Ê dy œ
k V ay
cbdt Ê

dy y
c

' œ
k A V dt Ê

dy y
c

A œ
' k A V dt Ê lnky
ck œ
k V t  C1

Ê y
c œ „ eC1 e
k V t . Apply the initial condition, ya0b œ y0 Ê y0 œ c  C Ê C œ y0
c A

Ê y œ c  ay0
cbe
k V t . A (b) Steady state solution: y_ œ lim yatb œ lim
c  ay0
cbe
k V t ‘ œ c  ay0
cba0b œ c A

tÄ_

tÄ_

2. Measure the amounts of oxygen involved in mL. Then the inflow of oxygen is 1000 mL/min (Assumed: it will take 5 minutes to deliver the 5L œ 5000mL); the amount of oxygen at t œ 0 is 210 mL; letting A œ the amount of oxygen in the flask, the concentration at time t is A mL/L; the outflow rate of oxygen is A mL/L (lb/sec). The rate of change in A, dA dt , equals the rate of gain (1000 mL/min) minus rate of loss (A mL/min). Thus: dA dA
t dt œ 1000
A Ê 1000
A œ dt Ê lnaA
1000b œ
t  C Ê A
1000 œ Ce . At t œ 0, A œ 210, so C œ
790 and A œ 1000
790e
t . Thus, Aa5b œ 1000
790e
5 ¸ 994.7 mL. The concentration is

994.7 mL 1000 mL

œ 99.47%.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

618

Chapter 9 Further Applications of Integration

3. The amount of CO2 in the room at time t is Aatb. The rate of change in the amount of CO2 ,

dA dt

is the rate of internal

production (R1 ) plus the inflow rate (R2 ) minus the outflow rate (R3 ). R1 œ ˆ20

breaths/min ‰ a30 student

R2 œ Š1000

ft3 CO2 ft3 min ‹Š0.0004 min ‹

A R3 œ Š 10,000 ‹1000 œ 0.1A dA dt

3

100 2 studentsbˆ 1728 ft3 ‰Š0.04 ft ftCO ‹ ¸ 1.39 3

œ 0.4

ft3 CO2 min

ft3 CO2 min

ft3 CO2 min

œ 1.39  0.4
0.1A œ 1.79
0.1A Ê Aw  0.1A œ 1.79. Let vatb œ e

' 0.1dt

. We have

' 0.1dt d ‹ dt ŠAe

' 0.1dt

œ 1.79e

Ê Ae0.1t œ ' 1.79e0.1t dt œ 17.9e0.1t  C. At t œ 0, A œ a10,000ba0.0004b œ 4 ft3 CO2 Ê C œ
13.9 Ê A œ 17.9
13.9e
0.1t . So Aa60b œ 17.9
13.9e
0.1a60b ¸ 17.87 ft3 of CO2 in the 10,000 ft3 room. The percent of 17.87 CO2 is 10,000 ‚ 100 œ 0.18% 4.

damvb damvb dm dm dv dm dm dm dv dm dt œ F  av  ub dt Ê F œ dt
av  ub dt Ê F œ m dt  v dt
v dt
u dt Ê F œ m dt
u dt . dm dt œ
b Ê m œ
kbkt  C. At t œ 0, m œ m0 , so C œ m0 and m œ m0
kbkt. u kb k m0
kbkt dv Thus, F œ am0
kbktb dv dt
ukbk œ
am0
kbktbkgk Ê dt œ
g  m0
kbkt Ê v œ
gt
u lnŠ m0 ‹  C1

v œ 0 at t œ 0 Ê C1 œ 0. So v œ
gt
u lnŠ m0 m
0kbkt ‹ œ t œ 0 Ê y œ
"# gt2  c’ t  Š

m0
kbkt m0
kbkt kbk ‹ lnŠ m0 ‹

dy dt

Ê y œ ' ’
gt
u lnŠ

m0
kbkt m0 ‹

“dt and u œ c, y œ 0 at

“

' 5. (a) Let y be any function such that vaxby œ ' vaxbQaxb dx  C, vaxb œ e Paxb dx . Then ' Paxb dx ' d w w Ê v w axb œ œ e Paxb dx Paxb œ vaxbPaxb. dx avaxb † yb œ vaxb † y  y † v axb œ vaxbQaxb. We have vaxb œ e

Thus vaxb † y w  y † vaxb Paxb œ vaxbQaxb Ê y w  y Paxb œ Qaxb Ê the given y is a solution.

(b) If v and Q are continuous on c a, b d and x − aa, bb, then Ê

d dx ’

'xx vatbQatb dt“ œ vaxbQaxb 0

'xx vatbQatb dt œ ' vaxbQaxb dx. So C œ y0 vax0 b
' vaxbQaxb dx. From part (a), vaxby œ ' vaxbQaxb dx  C. 0

Substituting for C: vaxby œ ' vaxbQaxb dx  y0 vax0 b
' vaxbQaxb dx Ê vaxby œ y0 vax0 b when x œ x0 . 6. (a) y w  Paxby œ 0, yax0 b œ 0. Use vaxb œ e' Paxb dx as an integrating factor. Then

d dx avaxbyb

œ 0 Ê vaxby œ C

Ê y œ Ce
' Paxb dx and y1 œ C1 e
' Paxb dx , y2 œ C# e
' Paxb dx , y1 ax0 b œ y2 ax0 b œ 0, y1
y2 œ aC1
C2 be
' Paxb dx œ C3 e
(b)

' Paxb dx

d y axb dx avaxbc 1

and y1
y2 œ 0
0 œ 0. So y1
y2 is a solution to y w  Paxby œ 0 with yax0 b œ 0.


y2 axb db œ

' Paxb dx

' Paxb dx d e aC1 dx Še


C2 b ‘‹ œ

d dx aC1


C2 b œ

d dx aC3 b

œ !.

' dxd avaxbc y1 axb
y2 axb dbdx œ avaxbc y1 axb
y2 axb db œ ' ! dx œ C ' ' ' ' (c) y1 œ C1 e
Paxb dx , y2 œ C# e
Paxb dx , y œ y1
y2 . So yax0 b œ 0 Ê C1 e
Paxb dx
C# e
Paxb dx œ ! Ê C1
C2 œ 0 Ê C1 œ C2 Ê y1 axb œ y2 axb for a  x  b.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 10 CONIC SECTIONS AND POLAR COORDINATES 10.1 CONIC SECTIONS AND QUADRATIC EQUATIONS 1. x œ

y# 8

Ê 4p œ 8 Ê p œ 2; focus is (2ß 0), directrix is x œ
2 #

2. x œ
y4 Ê 4p œ 4 Ê p œ 1; focus is (
1ß 0), directrix is x œ 1 #

3. y œ
x6 Ê 4p œ 6 Ê p œ 4. y œ

x# 2

Ê 4p œ 2 Ê p œ

1 #

3 #

; focus is ˆ!ß
3# ‰ , directrix is y œ

3 #

; focus is ˆ!ß 1# ‰ , directrix is y œ
1#

5.

x# 4




y# 9

œ 1 Ê c œ È4  9 œ È13 Ê foci are Š „ È13ß !‹ ; vertices are a „ 2ß 0b ; asymptotes are y œ „ 3# x

6.

x# 4



y# 9

œ 1 Ê c œ È9
4 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a0ß „ 3b

7.

x# 2

 y# œ 1 Ê c œ È2
1 œ 1 Ê foci are a „ 1ß 0b ; vertices are Š „ È2ß !‹

8.

y# 4


x# œ 1 Ê c œ È4  1 œ È5 Ê foci are Š0ß „ È5‹ ; vertices are a!ß „ 2b ; asymptotes are y œ „ 2x

9. y# œ 12x Ê x œ

y# 1#

Ê 4p œ 12 Ê p œ 3;

focus is ($ß !), directrix is x œ
3

11. x# œ
8y Ê y œ

x#
8

Ê 4p œ 8 Ê p œ 2;

focus is (!ß
2), directrix is y œ 2

#

10. x# œ 6y Ê y œ x6 Ê 4p œ 6 Ê p œ focus is ˆ!ß 3# ‰ , directrix is y œ
3#

#

3 #

;

y 12. y# œ
2x Ê x œ
# Ê 4p œ 2 Ê p œ " focus is ˆ
# ß !‰ , directrix is x œ "#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #

;

620

Chapter 10 Conic Sections and Polar Coordinates

13. y œ 4x# Ê y œ

x# ˆ "4 ‰

" 4

Ê 4p œ

Ê pœ

" 16

#

14. y œ
8x# Ê y œ
ˆx" ‰ Ê 4p œ

;

8

" ‰ " focus is ˆ!ß 16 , directrix is y œ
16

#

15. x œ
3y# Ê x œ
ˆy" ‰ Ê 4p œ 3

focus is ˆ
1"# ß !‰ , directrix is x œ

#

#

" 3

" 1#

y 17. 16x#  25y# œ 400 Ê #x5  16 œ1 Ê c œ Èa#
b# œ È25
16 œ 3

#

19. 2x#  y# œ 2 Ê x#  y# œ 1 Ê c œ Èa#
b# œ È2
1 œ 1

" ‰ focus is ˆ!ß
32 , directrix is y œ

Ê pœ

" 1#

;

16. x œ 2y# Ê x œ

y# ˆ "# ‰

Ê 4p œ

" #

" 8

" 3#

Ê pœ

focus is ˆ 8" ß !‰ , directrix is x œ
8"

#

#

x 18. 7x#  16y# œ 112 Ê 16  y7 œ 1 Ê c œ Èa#
b# œ È16
7 œ 3

#

#

20. 2x#  y# œ 4 Ê x#  y4 œ 1 Ê c œ Èa#
b# œ È4
2 œ È2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Ê pœ

" 8

;

" 32

;

Section 10.1 Conic Sections and Quadratic Equations #

#

21. 3x#  2y# œ 6 Ê x#  y3 œ 1 Ê c œ Èa#
b# œ È3
2 œ 1

#

#

23. 6x#  9y# œ 54 Ê x9  y6 œ 1 Ê c œ Èa#
b# œ È9
6 œ È3

#

#

x 22. 9x#  10y# œ 90 Ê 10  y9 œ 1 Ê c œ Èa#
b# œ È10
9 œ 1

#

#

y x 24. 169x#  25y# œ 4225 Ê 25  169 œ1 Ê c œ Èa#
b# œ È169
25 œ 12

#

25. Foci: Š „ È2ß !‹ , Vertices: a „ 2ß 0b Ê a œ 2, c œ È2 Ê b# œ a#
c# œ 4
ŠÈ2‹ œ 2 Ê 26. Foci: a!ß „ 4b , Vertices: a0ß „ 5b Ê a œ 5, c œ 4 Ê b# œ 25
16 œ 9 Ê 27. x#
y# œ 1 Ê c œ Èa#  b# œ È1  1 œ È2 ; asymptotes are y œ „ x

x# 9

 #

y# #5

œ1 #

x 28. 9x#
16y# œ 144 Ê 16
y9 œ 1 Ê c œ Èa#  b# œ È16  9 œ 5; asymptotes are y œ „ 34 x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x# 4



y# #

œ1

621

622

Chapter 10 Conic Sections and Polar Coordinates

29. y#
x# œ 8 Ê y8
x8 œ 1 Ê c œ Èa#  b# œ È8  8 œ 4; asymptotes are y œ „ x

# # 30. y#
x# œ 4 Ê y4
x4 œ 1 Ê c œ Èa#  b# œ È4  4 œ 2È2; asymptotes are y œ „ x

31. 8x#
2y# œ 16 Ê x#
y8 œ 1 Ê c œ Èa#  b# œ È2  8 œ È10 ; asymptotes are y œ „ 2x

32. y#
3x# œ 3 Ê y3
x# œ 1 Ê c œ Èa#  b# œ È3  1 œ 2; asymptotes are y œ „ È3x

# # 33. 8y#
2x# œ 16 Ê y#
x8 œ 1 Ê c œ Èa#  b# œ È2  8 œ È10 ; asymptotes are y œ „ x

y x 34. 64x#
36y# œ 2304 Ê 36
64 œ 1 Ê c œ Èa#  b# œ È36  64 œ 10; asymptotes are y œ „ 4

#

#

#

#

#

#

#

#

3

35. Foci: Š!ß „ È2‹ , Asymptotes: y œ „ x Ê c œ È2 and

a b

œ 1 Ê a œ b Ê c# œ a#  b# œ 2a# Ê 2 œ 2a#

Ê a œ 1 Ê b œ 1 Ê y#
x# œ 1 36. Foci: a „ 2ß !b , Asymptotes: y œ „ Ê 4œ

4a# 3

" È3

x Ê c œ 2 and

Ê a# œ 3 Ê a œ È3 Ê b œ 1 Ê

x# 3

b a

œ

" È3

Ê bœ

a È3

4 3

Ê c# œ a#  b# œ a# 


y# œ 1

37. Vertices: a „ 3ß 0b , Asymptotes: y œ „ 43 x Ê a œ 3 and

b a

œ

4 3

Ê bœ

(3) œ 4 Ê

38. Vertices: a!ß „ 2b , Asymptotes: y œ „ 12 x Ê a œ 2 and

a b

œ

1 2

Ê b œ 2(2) œ 4 Ê

x# 9 y# 4





Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

y# 16 x# 16

œ1 œ1

a# 3

œ

4a# 3

Section 10.1 Conic Sections and Quadratic Equations 39. (a) y# œ 8x Ê 4p œ 8 Ê p œ 2 Ê directrix is x œ
2, focus is (#ß !), and vertex is (!ß 0); therefore the new directrix is x œ
1, the new focus is (3ß
2), and the new vertex is (1ß
2)

40. (a) x# œ
4y Ê 4p œ 4 Ê p œ 1 Ê directrix is y œ 1, focus is (!ß
1), and vertex is (!ß 0); therefore the new directrix is y œ 4, the new focus is (
1ß 2), and the new vertex is (
1ß 3)

41. (a)

x# 16



y# 9

œ 1 Ê center is (!ß 0), vertices are (
4ß 0)

and (%ß !); c œ Èa#
b# œ È7 Ê foci are ŠÈ7ß 0‹ and Š
È7ß !‹ ; therefore the new center is (%ß $), the new vertices are (!ß 3) and (8ß 3), and the new foci are Š4 „ È7ß $‹

42. (a)

x# 9



y# 25

œ 1 Ê center is (!ß 0), vertices are (0ß 5) and (0ß
5); c œ Èa#
b# œ È16 œ 4 Ê foci are (!ß 4) and (!ß
4) ; therefore the new center is (
3ß
2), the new vertices are (
3ß 3) and (
3ß
7), and the new foci are (
3ß 2) and (
3ß
6)

43. (a)

x# 16




y# 9

œ 1 Ê center is (!ß 0), vertices are (
4ß 0)

and (4ß 0), and the asymptotes are x4 œ „ y3 or Èa#  b# œ È25 œ 5 Ê foci are y œ „ 3x 4 ;cœ (
5ß 0) and (5ß 0) ; therefore the new center is (2ß 0), the new vertices are (
2ß 0) and (6ß 0), the new foci are (
3ß 0) and (7ß 0), and the new asymptotes are yœ „

3(x
2) 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

623

624

Chapter 10 Conic Sections and Polar Coordinates

44. (a)

y# 4




x# 5

œ 1 Ê center is (!ß 0), vertices are (0ß
2)

and (0ß 2), and the asymptotes are yœ „

2x È5

y 2

œ „

x È5

or

; c œ Èa#  b# œ È9 œ 3 Ê foci are

(0ß 3) and (0ß
3) ; therefore the new center is (0ß
2), the new vertices are (0ß
4) and (0ß 0), the new foci are (0ß 1) and (0ß
5), and the new asymptotes are 2x y2œ „ È 5 45. y# œ 4x Ê 4p œ 4 Ê p œ 1 Ê focus is ("ß 0), directrix is x œ
1, and vertex is (0ß 0); therefore the new vertex is (
2ß
3), the new focus is (
1ß
3), and the new directrix is x œ
3; the new equation is (y  3)# œ 4(x  2) 46. y# œ
12x Ê 4p œ 12 Ê p œ 3 Ê focus is (
3ß 0), directrix is x œ 3, and vertex is (0ß 0); therefore the new vertex is (4ß 3), the new focus is (1ß 3), and the new directrix is x œ 7; the new equation is (y
3)# œ
12(x
4) 47. x# œ 8y Ê 4p œ 8 Ê p œ 2 Ê focus is (0ß 2), directrix is y œ
2, and vertex is (0ß 0); therefore the new vertex is (1ß
7), the new focus is (1ß
5), and the new directrix is y œ
9; the new equation is (x
1)# œ 8(y  7) Ê focus is ˆ!ß #3 ‰ , directrix is y œ
3# , and vertex is (0ß 0); therefore the new vertex is (
3ß
2), the new focus is ˆ
3ß
"# ‰ , and the new directrix is y œ
7# ; the new equation is

48. x# œ 6y Ê 4p œ 6 Ê p œ

3 #

(x  3)# œ 6(y  2) 49.

x# 6



y# 9

œ 1 Ê center is (!ß 0), vertices are (0ß 3) and (!ß
3); c œ Èa#
b# œ È9
6 œ È3 Ê foci are Š!ß È3‹

and Š!ß
È3‹ ; therefore the new center is (
#ß
1), the new vertices are (
2ß 2) and (
#ß
4), and the new foci are Š
#ß
1 „ È3‹ ; the new equation is 50.

x# 2

(x  2)# 6

(y  1)# 9



œ1

 y# œ 1 Ê center is (!ß 0), vertices are ŠÈ2ß !‹ and Š
È2ß !‹ ; c œ Èa#
b# œ È2
1 œ 1 Ê foci are

(
1ß 0) and ("ß !); therefore the new center is (3ß 4), the new vertices are Š3 „ È2ß 4‹ , and the new foci are (2ß 4) and (4ß 4); the new equation is 51.

x# 3



y# #

(x
3)# #

 (y
4)# œ 1

œ 1 Ê center is (!ß 0), vertices are ŠÈ3ß !‹ and Š
È3ß !‹ ; c œ Èa#
b# œ È3
2 œ 1 Ê foci are

(
1ß 0) and ("ß !); therefore the new center is (2ß 3), the new vertices are Š2 „ È3ß 3‹ , and the new foci are (1ß 3) and (3ß 3); the new equation is 52.

x# 16



y# #5

(x
2)# 3



(y
3)# #

œ1

œ 1 Ê center is (!ß 0), vertices are (!ß &) and (!ß
5); c œ Èa#
b# œ È25
16 œ 3 Ê foci are

(0ß 3) and (0ß
3); therefore the new center is (
4ß
5), the new vertices are (
4ß 0) and (
4ß
10), and the new foci are (
4ß
2) and (
4ß
8); the new equation is 53.

x# 4




y# 5

(x  4)# 16



(y  5)# #5

œ1

œ 1 Ê center is (!ß 0), vertices are (2ß 0) and (
2ß 0); c œ Èa#  b# œ È4  5 œ 3 Ê foci are ($ß !) and

(
3ß 0); the asymptotes are „

x #

œ

y È5

Ê yœ „

È5x #

; therefore the new center is (2ß 2), the new vertices are

(4ß 2) and (0ß 2), and the new foci are (5ß 2) and (
1ß 2); the new asymptotes are y
2 œ „

È5 (x
2) #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

; the new

Section 10.1 Conic Sections and Quadratic Equations equation is 54.

x# 16




y# 9

(x
2)# 4

(y
2)# 5




625

œ1

œ 1 Ê center is (!ß 0), vertices are (4ß 0) and (
4ß 0); c œ Èa#  b# œ È16  9 œ 5 Ê foci are (
5ß !)

and (5ß 0); the asymptotes are „

x 4

œ

Ê yœ „

y 3

3x 4

; therefore the new center is (
5ß
1), the new vertices are

(
1ß
1) and (
9ß
1), and the new foci are (
10ß
1) and (0ß
1); the new asymptotes are y  1 œ „ the new equation is

(x  5)# 16




(y  1)# 9

3(x  5) 4

;

œ1

55. y#
x# œ 1 Ê center is (!ß 0), vertices are (0ß 1) and (0ß
1); c œ Èa#  b# œ È1  1 œ È2 Ê foci are Š!ß „ È2‹ ; the asymptotes are y œ „ x; therefore the new center is (
1ß
1), the new vertices are (
1ß 0) and (
1ß
2), and the new foci are Š
1ß
1 „ È2‹ ; the new asymptotes are y  1 œ „ (x  1); the new equation is (y  1)#
(x  1)# œ 1 56.

y# 3


x# œ 1 Ê center is (!ß 0), vertices are Š0ß È3‹ and Š!ß
È3‹ ; c œ Èa#  b# œ È3  1 œ 2 Ê foci are (!ß #)

and (!ß
2); the asymptotes are „ x œ

y È3

Ê y œ „ È3x; therefore the new center is (1ß 3), the new vertices

are Š"ß $ „ È3‹ , and the new foci are ("ß &) and (1ß 1); the new asymptotes are y
3 œ „ È3 (x
1); the new equation is

(y
3)# 3


(x
1)# œ 1

57. x#  4x  y# œ 12 Ê x#  4x  4  y# œ 12  4 Ê (x  2)#  y# œ 16; this is a circle: center at C(
2ß 0), a œ 4 58. 2x#  2y#
28x  12y  114 œ 0 Ê x#
14x  49  y#  6y  9 œ
57  49  9 Ê (x
7)#  (y  3)# œ 1; this is a circle: center at C(7ß
3), a œ 1 59. x#  2x  4y
3 œ 0 Ê x#  2x  1 œ
4y  3  1 Ê (x  1)# œ
4(y
1); this is a parabola: V(
1ß 1), F(
1ß 0) 60. y#
4y
8x
12 œ 0 Ê y#
4y  4 œ 8x  12  4 Ê (y
2)# œ 8(x  2); this is a parabola: V(
#ß 2), F(!ß #) 61. x#  5y#  4x œ 1 Ê x#  4x  4  5y# œ 5 Ê (x  2)#  5y# œ 5 Ê

(x  2)# 5

 y# œ 1; this is an ellipse: the

center is (
2ß 0), the vertices are Š
2 „ È5ß 0‹ ; c œ Èa#
b# œ È5
1 œ 2 Ê the foci are (
4ß 0) and (!ß 0) #

62. 9x#  6y#  36y œ 0 Ê 9x#  6 ay#  6y  9b œ 54 Ê 9x#  6(y  3)# œ 54 Ê x6  (y 9 3) œ 1; this is an ellipse: the center is (0ß
3), the vertices are (!ß 0) and (!ß
6); c œ Èa#
b# œ È9
6 œ È3 Ê the foci are Š0ß
3 „ È3‹ #

63. x#  2y#
2x
4y œ
1 Ê x#
2x  1  2 ay#
2y  1b œ 2 Ê (x
1)#  2(y
1)# œ 2 # Ê (x
1)  (y
1)# œ 1; this is an ellipse: the center is (1ß 1), the vertices are Š" „ È2ß "‹ ; 2

c œ Èa#
b# œ È2
1 œ 1 Ê the foci are (2ß 1) and (0ß 1) 64. 4x#  y#  8x
2y œ
1 Ê 4 ax#  2x  1b  y#
2y  1 œ 4 Ê 4(x  1)#  (y
1)# œ 4 Ê (x  1)# 

(y
1)# 4

œ 1; this is an ellipse: the center is (
1ß 1), the vertices are (
1ß 3) and

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

626

Chapter 10 Conic Sections and Polar Coordinates

(
1ß
1); c œ Èa#
b# œ È4
1 œ È3 Ê the foci are Š
1ß " „ È3‹ 65. x#
y#
2x  4y œ 4 Ê x#
2x  1
ay#
4y  4b œ 1 Ê (x
1)#
(y
2)# œ 1; this is a hyperbola: the center is (1ß 2), the vertices are (2ß 2) and (!ß 2); c œ Èa#  b# œ È1  1 œ È2 Ê the foci are Š1 „ È2ß #‹ ; the asymptotes are y
2 œ „ (x
1) 66. x#
y#  4x
6y œ 6 Ê x#  4x  4
ay#  6y  9b œ 1 Ê (x  2)#
(y  3)# œ 1; this is a hyperbola: the center is (
2ß
3), the vertices are (
1ß
3) and (
3ß
3); c œ Èa#  b# œ È1  1 œ È2 Ê the foci are Š
2 „ È2ß
3‹ ; the asymptotes are y  3 œ „ (x  2) 67. 2x#
y#  6y œ 3 Ê 2x#
ay#
6y  9b œ
6 Ê

(y
3)# 6




x# 3

œ 1; this is a hyperbola: the center is (!ß $),

the vertices are Š!ß 3 „ È6‹ ; c œ Èa#  b# œ È6  3 œ 3 Ê the foci are (0ß 6) and (!ß 0); the asymptotes are y
3 È6

œ „

x È3

Ê y œ „ È2x  3

68. y#
4x#  16x œ 24 Ê y#
4 ax#
4x  4b œ 8 Ê

y# 8




(x
2)# 2

œ 1; this is a hyperbola: the center is (2ß 0),

the vertices are Š2ß „ È8‹ ; c œ Èa#  b# œ È8  2 œ È10 Ê the foci are Š2ß „ È10‹ ; the asymptotes are y È8

œ „

x
2 È2

Ê y œ „ 2(x
2)

69.

70.

71.

72.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.1 Conic Sections and Quadratic Equations

627

74. kx#
y# k Ÿ 1 Ê
1 Ÿ x#
y# Ÿ 1 Ê
1 Ÿ x#
y# and x#
y# Ÿ 1 Ê 1   y#
x# and x#
y# Ÿ 1

73.

75. Volume of the Parabolic Solid: V" œ '0 21x ˆh
bÎ2

œ

1hb# 8

76. y œ '

; Volume of the Cone: V# œ w H

x dx œ

w H

#

Š x# ‹  C œ

wx# 2H

" 3

#

1 ˆ b# ‰ h œ

" 3

4h b#

x# ‰ dx œ 21h '0 Šx
bÎ2

#

1 Š b4 ‹ h œ

1hb# 12

4x$ b# ‹

; therefore V" œ

 C; y œ 0 when x œ 0 Ê 0 œ

w(0)# 2H

#

dx œ 21h ’ x2
3 #

bÎ2

x% b# “ !

V#

 C Ê C œ 0; therefore y œ

wx# 2H

is the

equation of the cable's curve 77. A general equation of the circle is x#  y#  ax  by  c œ 0, so we will substitute the three given points into a  c œ
1 Þ b  c œ
1 ß Ê c œ 43 and a œ b œ
73 ; therefore this equation and solve the resulting system: 2a  2b  c œ
8 à 3x#  3y#
7x
7y  4 œ 0 represents the circle 78. A general equation of the circle is x#  y#  ax  by  c œ 0, so we will substitute each of the three given points 2a  3b  c œ
13 Þ into this equation and solve the resulting system:

3a  2b  c œ
13 ß Ê a œ 2, b œ 2, and c œ
23;
4a  3b  c œ
25 à

therefore x#  y#  2x  2y
23 œ 0 represents the circle 79. r# œ (
2
1)#  (1
3)# œ 13 Ê (x  2)#  (y
1)# œ 13 is an equation of the circle; the distance from the center to (1.1ß 2.8) is È(
#
1.1)#  (1
2.8)# œ È12.85  È13 , the radius Ê the point is inside the circle 80. (x
2)#  (y
1)# œ 5 Ê 2(x
2)  2(y
1)

dy dx

œ0 Ê

dy dx

2 # # œ
yx

1 ; y œ 0 Ê (x
2)  (0
1) œ 5

Ê (x
2)# œ 4 Ê x œ 4 or x œ 0 Ê the circle crosses the x-axis at (4ß 0) and (!ß 0); x œ 0 Ê (0
2)#  (y
1)# œ 5 Ê (y
1)# œ 1 Ê y œ 2 or y œ 0 Ê the circle crosses the y-axis at (!ß 2) and (!ß !). At (4ß 0): At (!ß !): At (!ß #):

dy dx dy dx dy dx

2 œ
40

1 œ 2 Ê the tangent line is y œ 2(x
4) or y œ 2x
8 2 œ
00

1 œ
2 Ê the tangent line is y œ
2x

2 œ
02

1 œ 2 Ê the tangent line is y
2 œ 2x or y œ 2x  2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

628

Chapter 10 Conic Sections and Polar Coordinates

81. (a) y# œ kx Ê x œ

y# k

; the volume of the solid formed by

Èkx

revolving R" about the y-axis is V" œ '0 œ

1 k#

Èkx

'0

y% dy œ

1x# Èkx 5

#

#

1 Š yk ‹ dy

; the volume of the right

circular cylinder formed by revolving PQ about the y-axis is V# œ 1x# Èkx Ê the volume of the solid formed by revolving R# about the y-axis is V$ œ V#
V" œ

41x# Èkx 5

. Therefore we can see the

ratio of V$ to V" is 4:1.

(b) The volume of the solid formed by revolving R# about the x-axis is V" œ '0 1 ŠÈkt‹ dt œ 1k'0 t dt #

x

œ

1kx# #

x

. The volume of the right circular cylinder formed by revolving PS about the x-axis is #

V# œ 1 ŠÈkx‹ x œ 1kx# Ê the volume of the solid formed by revolving R" about the x-axis is 1kx# #

V$ œ V#
V" œ 1kx#


œ

1kx# #

. Therefore the ratio of V$ to V" is 1:1.

82. Let P" (
pß y" ) be any point on x œ
p, and let P(xß y) be a point where a tangent intersects y# œ 4px. Now y# œ 4px Ê 2y

dy dx

œ 4p Ê

dy dx

œ

2p y

Ê y#
yy" œ 2px  2p# . Since x œ Ê

" #

tangents from P" are m" œ

y# 4p

2p y"  Èy#"  4p#

œ

dy dx

œ

#

y , we have y#
yy" œ 2p Š 4p ‹  2p# Ê y#
yy" œ

2y" „ È4y#"  16p# #

y#
yy"
2p# œ 0 Ê y œ

y
y" x
(
p)

; then the slope of a tangent line from P" is

and m# œ

" #

2p y

y#  2p#

œ y" „ Èy#"  4p# . Therefore the slopes of the two 2p y"
Èy#"  4p#

Ê m" m# œ

4p# y#"
ay#"  4p# b

œ
1

Ê the lines are perpendicular 83. Let y œ É1


x# 4

on the interval 0 Ÿ x Ÿ 2. The area of the inscribed rectangle is given by

A(x) œ 2x Š2É1
Ê Aw (x) œ 4É1


x# 4‹ x# 4

œ 4xÉ1





x# É1
x4#

x# 4

(since the length is 2x and the height is 2y)

. Thus Aw (x) œ 0 Ê 4É1


x# 4




x# É1
x4#

œ 0 Ê 4 Š1


x# 4‹


x# œ 0 Ê x# œ 2

Ê x œ È2 (only the positive square root lies in the interval). Since A(0) œ A(2) œ 0 we have that A ŠÈ2‹ œ 4 is the maximum area when the length is 2È2 and the height is È2. 84. (a) Around the x-axis: 9x#  4y# œ 36 Ê y# œ 9
94 x# Ê y œ „ É9
94 x# and we use the positive root #

Ê V œ 2 '0 1 ŠÉ9
94 x# ‹ dx œ 2 '0 1 ˆ9
94 x# ‰ dx œ 21
9x
34 x$ ‘ ! œ 241 2

2

#

(b) Around the y-axis: 9x#  4y# œ 36 Ê x# œ 4
49 y# Ê x œ „ É4
49 y# and we use the positive root #

Ê V œ 2'0 1 ŠÉ4
49 y# ‹ dy œ 2 '0 1 ˆ4
49 y# ‰ dy œ 21
4y
3

85. 9x#
4y# œ 36 Ê y# œ œ

91 4

9x#
36 4

'24 ax#
4b dx œ 941 ’ x3

$

3

4 27

$

y$ ‘ ! œ 161

Ê y œ „ #3 Èx#
4 on the interval 2 Ÿ x Ÿ 4 Ê V œ '2 1 Š #3 Èx#
4‹ dx #

4

%


4x“ œ #

91 4


ˆ 64 ‰ ˆ8 ‰‘ œ 3
16
3
8

91 4

ˆ 56 ‰ 3
8 œ

31 4

(56
24) œ 241

86. x#
y# œ 1 Ê x œ „ È1  y# on the interval
3 Ÿ y Ÿ 3 Ê V œ 'c3 1 ˆÈ1  y# ‰ dy œ 2'0 1 ˆÈ1  y# ‰ dy 3

œ 21'0 a1  y# b dy œ 21 ’y  3

$ y$ 3 “!

#

œ 241

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

3

#

Section 10.1 Conic Sections and Quadratic Equations 87. Let y œ É16


x# on the interval
3 Ÿ x Ÿ 3. Since the plate is symmetric about the y-axis, x œ 0. For a

16 9

É16
aµ x ßµ y b œ xß #

vertical strip:

Ê mass œ dm œ $ dA œ $É16
# É16
16 9 x

µ y dm œ

#

Š$ É16


16 9

16 9

16 9

x#

 , length œ É16


16 9

x# , width œ dx Ê area œ dA œ É16


16 9

x# dx

x# ‹ dx œ $ ˆ8
98 x# ‰ dx so the moment of the plate about the x-axis is

3

3

16 9

x# dx. Moment of the strip about the x-axis:

Mx œ ' µ y dm œ 'c3 $ ˆ8
89 x# ‰ dx œ $
8x
M œ 'c3 $ É16


629

8 27

$

x$ ‘
$ œ 32$ ; also the mass of the plate is

# x# dx œ 'c3 4$ É1
ˆ "3 x‰ dx œ 4$ 'c1 3È1
u# du where u œ 3

1

x 3

Ê 3 du œ dx; x œ
3

Ê u œ
1 and x œ 3 Ê u œ 1. Hence, 4$ 'c1 3È1
u# du œ 12$ 'c1 È1
u# du 1

œ 12$ ’ "2 ŠuÈ1
u#  sin
" u‹“ 88. y œ Èx#  1 Ê

dy dx

" #

œ

È2

1 ' œ É 2x x#  1 Ê S œ 0 #

–

89.

u œ È2x — Ä du œ È2 dx

drA dt

œ

drB dt

Ê

d dt

21 È2

ax#  1b

"


"

1

œ 61$ Ê y œ


"Î#

(2x) œ

x È x#  1

Mx M

œ

32$ 61$

#

œ

Ê Š dy dx ‹ œ

È2

16 31

. Therefore the center of mass is ˆ!ß 3161 ‰ .

x# x # 1

#

dy Ê Ê1  Š dx ‹ œ É1 

È2

dy 1 È # ' 21yÊ1  Š dx ‹ dx œ '0 21Èx#  1 É 2x x#  1 dx œ 0 21 2x  1 dx ; #

'02 Èu#  1 du œ È21

#

#

’ " ŠuÈu#  1  ln Šu  Èu#  1‹‹“ œ 2 2 !

1 È2

90. (a) tan " œ mL Ê tan " œ f w (x! ) where f(x) œ È4px ; œ

2p y!

" #

(4px)
"Î# (4p) œ

(b) tan 9 œ mFP œ

œ

2p È4px

Ê f w (x! ) œ

2p È4px!

Ê tan " œ

(c) tan ! œ

’2È5  ln Š2  È5‹“

(rA
rB ) œ 0 Ê rA
rB œ C, a constant Ê the points P(t) lie on a hyperbola with foci at A

and B

f w (x) œ

x# x#  1

2p y! . y!
0 y! x!
p œ x!
p

tan 9
tan " 1  tan 9 tan "

y#!
2p(x!
p) y! (x!
p  2p)

œ

œ

y! 2p Šx
p c y ‹

!

!

y! 2p 1 b Šx
p‹ Šy ‹

!

4px!
2px!  2p# y! (x!  p)

!

œ

2p(x!  p) y! (x!  p)

œ

2p y!

91. PF will always equal PB because the string has constant length AB œ FP  PA œ AP  PB. 92. (a) In the labeling of the accompanying figure we have y 1 œ tan t so the coordinates of A are (1ß tan t). The coordinates of P are therefore (1  rß tan t). Since 1#  y# œ (OA)# , we have 1#  tan# t œ (1  r)# Ê 1  r œ È1  tan# t œ sec t Ê r œ sec t
1. The coordinates of P are therefore (xß y) œ (sec tß tan t) Ê x#
y# œ sec# t
tan# t œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

630

Chapter 10 Conic Sections and Polar Coordinates

(b) In the labeling of the accompany figure the coordinates of A are (cos tß sin t), the coordinates of C are (1ß tan t), and the coordinates of P are (1  dß tan t). By similar triangles,

d AB

œ

Ê

OC OA

d 1
cos t

œ

È1  tan# t 1

Ê d œ (1
cos t)(sec t) œ sec t
1. The coordinates of P are therefore (sec tß tan t) and P moves on the hyperbola x#
y# œ 1 as in part (a).

93. x# œ 4py and y œ p Ê x# œ 4p# Ê x œ „ 2p. Therefore the line y œ p cuts the parabola at points (
2pß p) and (2pß p), and these points are È[2p
(
2p)]#  (p
p)# œ 4p units apart. 94. x lim Š b x
ba Èx#
a# ‹ œ Ä_ a œ

b a x lim Ä_

’

x #
ax #
a # b “ x  È x #
a#

œ

b a x lim Ä_

b a x lim Ä_

’

Šx
Èx#
a# ‹ œ

a# “ x  È x #
a#

b a x lim Ä_

–

Šx
Èx#
a# ‹ Šx  Èx#
a# ‹ x  È x #
a#

œ0

10.2 CLASSIFYING CONIC SECTIONS BY ECCENTRICITY # y# 1. 16x#  25y# œ 400 Ê #x5  16 œ 1 Ê c œ Èa#
b# œ È25
16 œ 3 Ê e œ ca œ 35 ; F a „ 3ß 0b ;

directrices are x œ 0 „

œ „

a e

5 ˆ 35 ‰

œ „

25 3

# x# 2. 7x#  16y# œ 112 Ê 16  y7 œ 1 Ê c œ Èa#
b# œ È16
7 œ 3 Ê e œ ca œ 34 ; F a „ 3ß 0b ;

directrices are x œ 0 „

œ „

a e

4 ˆ 34 ‰

œ „

16 3

3. 2x#  y# œ 2 Ê x#  y2 œ 1 Ê c œ Èa#
b# œ È2
1 œ 1 Ê e œ ca œ È12 ; F a0ß „ 1b ; #

directrices are y œ 0 „

a e

œ „

È2 Š È12 ‹

œ „2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

—

Section 10.2 Classifying Conic Sections by Eccentricity 4. 2x#  y# œ 4 Ê

x# #



œ 1 Ê c œ Èa#
b#

y# 4

œ È4
2 œ È2 Ê e œ directrices are y œ 0 „

a e

c a

œ

È2 2

; F Š0ß „ È2‹ ;

œ „ È22 œ „ 2È2 Š ‹ 2

# # 5. 3x#  2y# œ 6 Ê x#  y3 œ 1 Ê c œ Èa#
b# œ È3
2 œ 1 Ê e œ ca œ È13 ; F a0ß „ 1b ;

directrices are y œ 0 „

a e

œ „

È3

œ „3

Š È13 ‹

# x# 6. 9x#  10y# œ 90 Ê 10  y9 œ 1 Ê c œ Èa#
b# œ È10
9 œ 1 Ê e œ ca œ È110 ; F a „ 1ß 0b ;

directrices are x œ 0 „

7. 6x#  9y# œ 54 Ê

x# 9

a e

œ „



y# 6

œ È9
6 œ È3 Ê e œ directrices are x œ 0 „

a e

È10 Š È110 ‹

œ „ 10

œ 1 Ê c œ Èa#
b# c a

œ

È3 3

; F Š „ È3ß 0‹ ;

œ „ È33 œ „ 3È3 Š ‹ 3

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631

632

Chapter 10 Conic Sections and Polar Coordinates

y# x# 8. 169x#  25y# œ 4225 Ê 25  169 œ 1 Ê c œ Èa#
b# œ È169
25 œ 12 Ê e œ c œ 12 ; F a0ß „ 12b ; a

directrices are y œ 0 „

a e

œ „

13

13 ˆ 12 ‰ 13

œ „

169 12

x# #7

y# 36

9. Foci: a0ß „ 3b , e œ 0.5 Ê c œ 3 and a œ

c e

œ

3 0.5

œ 6 Ê b# œ 36
9 œ 27 Ê

10. Foci: a „ 8ß 0b , e œ 0.2 Ê c œ 8 and a œ

c e

œ

8 0.#

œ 40 Ê b# œ 1600
64 œ 1536 Ê



œ1 x# 1600



y# 1536

11. Vertices: a0ß „ 70b , e œ 0.1 Ê a œ 70 and c œ ae œ 70(0.1) œ 7 Ê b# œ 4900
49 œ 4851 Ê

œ1

x# 4851

y# 4900

œ1

Šx


9 È5 ‹



12. Vertices: a „ 10ß 0b , e œ 0.24 Ê a œ 10 and c œ ae œ 10(0.24) œ 2.4 Ê b# œ 100
5.76 œ 94.24 x# 100

Ê



y# 94.24

œ1

13. Focus: ŠÈ5ß !‹ , Directrix: x œ Ê eœ

È5 3

. Then PF œ

PF œ

Ê È(x
x

256 ‰ 9

Ê

a e

œ

5 9

Šx#


16 3

x

Ê

81 5 ‹

Ê c œ ae œ 4 and 0)#

œ

x#  y# œ

16 3

 (y


" 4

18 È5

È3 #

¸x


Ê

x# ˆ 64 ‰ 3

4 9

a e

È5 3



16 3

Ê

œ

ae e#

#

œ

ae e#

¹x




Ê

16 3 #

Ê (x
4)  y œ y# ˆ 16 ‰ 3

9 È5

9 È5 ¹

x# 9

x#  y# œ 4 Ê

œ

16 ¸ 3

Ê

9 È5

PD Ê ÊŠx
È5‹  (y
0)# œ

14. Focus: (%ß 0), Directrix: x œ 4)#

Ê c œ ae œ È5 and #

È5 3

Ê x#
2È5 x  5  y# œ

È œ #3 PD 3 ˆ # 32 4 x
3

9 È5

y# 4

4 e# 3 4

Ê

È5 e#

œ

Ê e# œ

9 È5 #

Ê Šx
È5‹  y# œ

5 9

5 9

œ1 œ

16 3

ˆx


Ê e# œ

16 ‰# 3

Ê eœ

3 4

È3 #

. Then

#

Ê x
8x  16  y#

œ1

4 " # 15. Focus: (
%ß 0), Directrix: x œ
16 Ê c œ ae œ 4 and ae œ 16 Ê ae e# œ 16 Ê e# œ 16 Ê e œ 4 Ê e œ PF œ 1 PD Ê È(x  4)#  (y
0)# œ 1 kx  16k Ê (x  4)#  y# œ 1 (x  16)# Ê x#  8x  16  y# #

œ

1 4

#

#

ax  32x  256b Ê

3 4

#

#

x  y œ 48 Ê

œ

" #

1 È2

. Then PF œ #

1 È2



y# 48

. Then

œ1

#

PD Ê ÊŠx  È2‹  (y
0)# œ

Šx  2È2‹ Ê x#  2È2 x  2  y# œ

1 #

4

x# 64

16. Focus: Š
È2ß !‹ , Directrix: x œ
2È2 Ê c œ ae œ È2 and Ê eœ

#

" #

a e

œ 2È 2 Ê

1 È2

ae e#

œ 2È 2 Ê

È2 e#

œ 2 È 2 Ê e# œ

#

¹x  2È2¹ Ê Šx  È2‹  y#

Šx#  4È2 x  8‹ Ê

" #

x#  y# œ 2 Ê

x# 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley



y# #

œ1

" #

Section 10.2 Classifying Conic Sections by Eccentricity 17. e œ

Ê take c œ 4 and a œ 5; c# œ a#
b#

4 5

Ê 16 œ 25
b# Ê b# œ 9 Ê b œ 3; therefore x# #5



y# 9

œ1

18. The eccentricity e for Pluto is 0.25 Ê e œ

c a

œ 0.25 œ

" 4 #

Ê take c œ 1 and a œ 4; c# œ a#
b# Ê 1 œ 16
b # # Ê b# œ 15 Ê b œ È15 ; therefore, x  y œ 1 is a 16

15

model of Pluto's orbit.

19. One axis is from A("ß ") to B("ß 7) and is 6 units long; the other axis is from C($ß %) to D(
1ß 4) and is 4 units long. Therefore a œ 3, b œ 2 and the major axis is vertical. The center is the point C("ß 4) and the ellipse is given by (x
1)# 4



(y
4)# 9

œ 1; c# œ a#
b# œ 3#
2# œ 5

Ê c œ È5 ; therefore the foci are F Š1ß 4 „ È5‹ , the eccentricity is e œ yœ4„

a e

œ

c a

È5 3

, and the directrices are

œ 4 „ È5 œ 4 „ Š ‹ 3

9È 5 5

.

3

20. Using PF œ e † PD, we have È(x
4)#  y# œ œ

4 9

ax#
18x  81b Ê

5 9

2 3

kx
9k Ê (x
4)#  y# œ

x#  y# œ 20 Ê 5x#  9y# œ 180 or

x# 36



#

y 20

4 9

(x
9)# Ê x#
8x  16  y#

œ 1.

21. The ellipse must pass through (!ß 0) Ê c œ 0; the point (
1ß 2) lies on the ellipse Ê
a  2b œ
8. The ellipse is tangent to the x-axis Ê its center is on the y-axis, so a œ 0 and b œ
4 Ê the equation is 4x#  y#
4y œ 0. Next, 4x#  y#
4y  4 œ 4 Ê 4x#  (y
24)# œ 4 Ê x# 

(y
2)# 4

standard symbols) Ê c# œ a#
b# œ 4
1 œ 3 Ê c œ È3 Ê e œ

œ 1 Ê a œ 2 and b œ 1 (now using the c a

œ

È3 #

.

22. We first prove a result which we will use: let m" , and m# be two nonparallel, nonperpendicular lines. Let ! be the acute angle between the lines. Then tan ! œ 1m" m
"mm## . To see this result, let )" be the angle of inclination of the line with slope m" , and )# be the angle of inclination of the line with slope m# . Assume m"  m# . Then )"  )# and we have ! œ )"
)# . Then tan ! œ tan ()"
)# ) )"
tan )# m"
m# œ 1tan  tan )" tan )# œ 1  m" m# , since m" œ tan )" and and m# œ tan )# .

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633

634

Chapter 10 Conic Sections and Polar Coordinates

Now we prove the reflective property of ellipses (see the x# a#

accompanying figure): If # #

# #

# #

b x  a y œ a b and y œ

 b a

y# b#

œ 1, then Èa#
x# Ê yw œ


bx aÈ a#
x#

.

Let P(x! ß y! ) be any point on the ellipse Ê yw (x! ) œ


bx!

œ

aÉa#
x#!


b # x ! a# y!

be the foci. Then mPF" œ

. Let F" (cß 0) and F# (
cß 0)

y! x!
c

and mPF# œ

y! x!  c

. Let ! and

" be the angles between the tangent line and PF" and PF# , respectively. Then b# x

tan ! œ

!
Œc a# y! c x!
c ! y

b# x y Š1 c a# y (x!
! c) ‹ ! !

Similarly, tan " œ

b# cy!

œ


b# x#!  b# x! c
a# y#! a # y ! x !
a# y! c
b# x! y!

œ

b# x! c
ab# x#!  a# y#! b
a # y ! c  aa #
b # b x ! y!

È2 1

b # x! c
a# b #
a # y ! c  c # x ! y !

œ

b# cy!

.

. Since tan ! œ tan " , and ! and " are both less than 90°, we have ! œ " .

23. x#
y# œ 1 Ê c œ Èa#  b# œ È1  1 œ È2 Ê e œ œ

œ

c a

œ È2 ; asymptotes are y œ „ x; F Š „ È2 ß !‹ ;

directrices are x œ 0 „

a e

œ „

" È2

# x# 24. 9x#
16y# œ 144 Ê 16
y9 œ 1 Ê c œ Èa#  b# œ È16  9 œ 5 Ê e œ ca œ 54 ; asymptotes are

y œ „ 34 x; F a „ 5ß !b ; directrices are x œ 0 „ œ „

a e

"6 5

# # 25. y#
x# œ 8 Ê y8
x8 œ 1 Ê c œ Èa#  b# œ È8  8 œ 4 Ê e œ ca œ È48 œ È2 ; asymptotes are

y œ „ x; F a0ß „ 4b ; directrices are y œ 0 „ œ „

È8 È2

a e

œ „2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.2 Classifying Conic Sections by Eccentricity y# 4

26. y#
x# œ 4 Ê




x# 4

œ 1 Ê c œ Èa#  b#

œ È4  4 œ 2È2 Ê e œ

c a

œ

2È 2 2

œ È2 ; asymptotes

are y œ „ x; F Š0ß „ 2È2‹ ; directrices are y œ 0 „ œ „

2 È2

a e

œ „ È2

27. 8x#
2y# œ 16 Ê

x# 2




y# 8

œ È2  8 œ È10 Ê e œ

œ 1 Ê c œ Èa#  b# c a

œ

È10 È2

œ È5 ; asymptotes

are y œ „ 2x; F Š „ È10ß !‹ ; directrices are x œ 0 „ œ „

È2 È5

635

œ „

a e

2 È10

# 28. y#
3x# œ 3 Ê y3
x# œ 1 Ê c œ Èa#  b# œ È3  1 œ 2 Ê e œ ca œ È23 ; asymptotes are

y œ „ È3 x; F a0ß „ 2b ; directrices are y œ 0 „ œ „

È3 Š È23 ‹

œ „

a e

3 #

29. 8y#
2x# œ 16 Ê

y# 2




x# 8

œ È2  8 œ È10 Ê e œ

œ 1 Ê c œ Èa#  b# c a

œ

È10 È2

œ È5 ; asymptotes

are y œ „ x# ; F Š0ß „ È10‹ ; directrices are y œ 0 „ œ „

È2 È5

œ „

a e

2 È10

y# x# 30. 64x#
36y# œ 2304 Ê 36
64 œ 1 Ê c œ Èa#  b# 5 œ È36  64 œ 10 Ê e œ ca œ 10 6 œ 3 ; asymptotes are

y œ „ 43 x; F a „ 10ß !b ; directrices are x œ 0 „ œ „

6 ˆ 53 ‰

œ „

a e

18 5

31. Vertices a!ß „ 1b and e œ 3 Ê a œ 1 and e œ

c a

œ 3 Ê c œ 3a œ 3 Ê b# œ c#
a# œ 9
1 œ 8 Ê y#


Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x# 8

œ1

636

Chapter 10 Conic Sections and Polar Coordinates


y# 1#

œ1

œ 3 Ê c œ 3a Ê a œ 1 Ê b# œ c#
a# œ 9
1 œ 8 Ê x#


y# 8

œ1

32. Vertices a „ 2ß !b and e œ 2 Ê a œ 2 and e œ 33. Foci a „ 3ß !b and e œ 3 Ê c œ 3 and e œ

c a

34. Foci a!ß „ 5b and e œ 1.25 Ê c œ 5 and e œ œ 25
16 œ 9 Ê

#

y 16

#




x 9

œ 2 Ê c œ 2a œ 4 Ê b# œ c#
a# œ 16
4 œ 12 Ê

c a

c a

œ 1.25 œ

5 4

Ê cœ

a Ê 5œ

5 4

5 4

x# 4

a Ê a œ 4 Ê b# œ c#
a#

œ1

4 # È2 . Then 35. Focus (4ß 0) and Directrix x œ 2 Ê c œ ae œ 4 and ae œ 2 Ê ae e# œ 2 Ê e# œ 2 Ê e œ # Ê e œ PF œ È2 PD Ê È(x
4)#  (y
0)# œ È2 kx
2k Ê (x
4)#  y# œ 2(x
2)# Ê x#
8x  16  y#

œ 2 ax#
4x  4b Ê
x#  y# œ
8 Ê

x# 8




y# 8

œ1

36. Focus ŠÈ10ß !‹ and Directrix x œ È2 Ê c œ ae œ È10 and

a e

œ È2 Ê

ae e#

œ È2 Ê

È10 e#

œ È 2 Ê e# œ È 5

#

#

Ê e œ %È5 . Then PF œ %È5 PD Ê ÊŠx
È10‹  (y
0)# œ %È5 ¹x
È2¹ Ê Šx
È10‹  y# #

œ È5 Šx
È2‹ Ê x#
2È10 x  10  y# œ È5 Šx#
2È2 x  2‹ Ê Š1
È5‹ x#  y# œ 2È5
10 Ê

Š1
È5‹ x# #È5
10



y# 2È5
10

œ1 Ê

x# 2È 5




y# 10
2È5

œ1

37. Focus (
2ß 0) and Directrix x œ
"# Ê c œ ae œ 2 and

a e

œ

" # #

Ê

ae e#

œ

" #

Ê

PF œ 2PD Ê È(x  2)#  (y
0)# œ 2 ¸x  "# ¸ Ê (x  2)  y# œ 4 ˆx  œ 4 ˆx#  x  "4 ‰ Ê
3x#  y# œ
3 Ê x#


#

y 3

2 e# " ‰# #

œ

" #

Ê e# œ 4 Ê e œ 2. Then

Ê x#  4x  4  y#

œ1

6 # È3. Then 38. Focus (
6ß 0) and Directrix x œ
# Ê c œ ae œ 6 and ae œ # Ê ae e# œ # Ê e# œ # Ê e œ 3 Ê e œ PF œ È3 PD Ê È(x  6)#  (y
0)# œ È3 kx  2k Ê (x  6)#  y# œ 3(x  2)# Ê x#  12x  36  y# x# 1#

œ 3 ax#  4x  4b Ê
2x#  y# œ
24 Ê 39. È(x
1)#  (y  3)# œ

3 #




y# 24

œ1

ky
2k Ê x#
2x  1  y#  6y  9 œ

Ê 4 ax#
2x  1b
5 ay#
12y  36b œ
4  4
180 Ê 40. c# œ a#  b# Ê b# œ c#
a# ; e œ x# a#




#

y b#

œ" Ê

x# a#




#

y a # ae #
1 b

c a

(y
6)# 36

9 4

ay#
4y  4b Ê 4x#
5y#
8x  60y  4 œ 0




(x
1)# 45

œ1

Ê c œ ea Ê c# œ e# a# Ê b# œ e# a#
a# œ a# ae#
1b ; thus,

œ 1; the asymptotes of this hyperbola are y œ „ ae#
1bx Ê as e increases, the

absolute values of the slopes of the asymptotes increase and the hyperbola approaches a straight line. 41. To prove the reflective property for hyperbolas: x# a#




y# b#

œ 1 Ê a# y# œ b# x#
a# b# and

dy dx

œ

xb# ya#

.

Let P(x! ß y! ) be a point of tangency (see the accompanying figure). The slope from P to F(
cß 0) is x!y! c and from P to F# (cß 0) it is

y! x !
c

. Let the tangent through P meet

the x-axis in point A, and define the angles nF" PA œ ! and nF# PA œ " . We will show that tan ! œ tan " . From the preliminary result in Exercise 22,

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.3 Quadratic Equations and Rotations x b#

tan ! œ

!
Πy! a# c x!
c !

x b# y! 1 b Š y! a# ‹ Š x
! c‹ ! y

tan " œ

y

!
Πx!
c

x! b# y! a#


y! x! b# 1  Šx
c ‹ Š y a# ‹

!

!

œ

œ

x#! b#  x! b# c
y#! a# x ! y ! a #  y ! a # c  x! y! b #

b# y! c

œ

a# b#  x! b# c x ! y ! c #  y! a# c

œ

b# y! c

. In a similar manner,

. Since tan ! œ tan " , and ! and " are acute angles, we have ! œ " .

42. From the accompanying figure, a ray of light emanating from the focus A that met the parabola at P would be reflected from the hyperbola as if it came directly from B (Exercise 41). The same light ray would be reflected off the ellipse to pass through B. Thus BPC is a straight line. Let " be the angle of incidence of the light ray on the hyperbola. Let ! be the angle of incidence of the light ray on the ellipse. Note that !  " is the angle between the tangent lines to the ellipse and hyperbola at P. Since BPC is a straight line, 2!  2" œ 180°. Thus !  " œ 90°. 10.3 QUADRATIC EQUATIONS AND ROTATIONS 1. x#
3xy  y#
x œ 0 Ê B#
4AC œ (
3)#
4(1)(1) œ 5  0 Ê Hyperbola 2. 3x#
18xy  27y#
5x  7y œ
4 Ê B#
4AC œ (
18)#
4(3)(27) œ 0 Ê Parabola 3. 3x#
7xy  È17y# œ 1 Ê B#
4AC œ (
7)#
4(3) È17 ¸
0.477  0 Ê Ellipse #

4. 2x#
È15 xy  2y#  x  y œ 0 Ê B#
4AC œ Š
È15‹
4(2)(2) œ
1  0 Ê Ellipse 5. x#  2xy  y#  2x
y  2 œ 0 Ê B#
4AC œ 2#
4(1)(1) œ 0 Ê Parabola 6. 2x#
y#  4xy
2x  3y œ 6 Ê B#
4AC œ 4#
4(2)(
1) œ 24  0 Ê Hyperbola 7. x#  4xy  4y#
3x œ 6 Ê B#
4AC œ 4#
4(1)(4) œ 0 Ê Parabola 8. x#  y#  3x
2y œ 10 Ê B#
4AC œ 0#
4(1)(1) œ
4  0 Ê Ellipse (circle) 9. xy  y#
3x œ 5 Ê B#
4AC œ 1#
4(0)(1) œ 1  0 Ê Hyperbola 10. 3x#  6xy  3y#
4x  5y œ 12 Ê B#
4AC œ 6#
4(3)(3) œ 0 Ê Parabola 11. 3x#
5xy  2y#
7x
14y œ
1 Ê B#
4AC œ (
5)#
4(3)(2) œ 1  0 Ê Hyperbola 12. 2x#
4.9xy  3y#
4x œ 7 Ê B#
4AC œ (
4.9)#
4(2)(3) œ 0.01  0 Ê Hyperbola 13. x#
3xy  3y#  6y œ 7 Ê B#
4AC œ (
3)#
4(1)(3) œ
3  0 Ê Ellipse 14. 25x#  21xy  4y#
350x œ 0 Ê B#
4AC œ 21#
4(25)(4) œ 41  0 Ê Hyperbola 15. 6x#  3xy  2y#  17y  2 œ 0 Ê B#
4AC œ 3#
4(6)(2) œ
39  0 Ê Ellipse

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

637

638

Chapter 10 Conic Sections and Polar Coordinates

16. 3x#  12xy  12y#  435x
9y  72 œ 0 Ê B#
4AC œ 12#
4(3)(12) œ 0 Ê Parabola 1 1 # Ê !œ 4 ; È È xw sin !  yw cos ! Ê x œ xw #2
yw #2 , y È È È È Š #2 xw
#2 yw ‹ Š #2 xw  #2 yw ‹ œ 2 Ê "#

17. cot 2! œ yœ Ê

18. cot 2! œ

A
C B

A
C B

œ

1
1 1

œ

therefore x œ xw cos !
yw sin !,

œ 0 Ê 2! œ

0 1

œ 0 Ê 2! œ

1 #

Ê !œ

È2 #

œ xw

xw #
"# yw # œ 2 Ê xw #
yw # œ 4 Ê Hyperbola 1 4

; therefore x œ xw cos !
yw sin !,

È2 w È2 w È2 #
y # ,yœ x #  È È È È Š #2 xw  #2 yw ‹ Š #2 xw
#2 yw ‹

y œ xw sin !  yw cos ! Ê x œ xw Ê Š Ê

" #

È2 #

xw


#

È2 #

yw ‹ 

È2 #

 yw

yw 

È2 # È Š #2

xw 

xw #
xw yw  "# yw #  "# xw #
"# yw #  "# xw #  xw yw  "# yw # œ 1 Ê

19. cot 2! œ

A
C B

3
1 2È 3

œ

œ

" È3

1 3

Ê 2! œ

1 6

Ê !œ

È2 #

#

yw ‹ œ 1

xw #  "# yw # œ 1 Ê 3xw #  yw # œ 2 Ê Ellipse

3 #

; therefore x œ xw cos !
yw sin !,

È3 w È3 w " w 1 w # x
# y,yœ # x  # y È È Š #3 xw  1# yw ‹ Š 1# xw  #3 yw ‹ 

y œ xw sin !  yw cos ! Ê x œ Ê 3Š

È3 #

#

xw
1# yw ‹  2È3

 8È3 Š "# xw  20. cot 2! œ

A
C B

È3 #

œ

È3 #

È3 #

#

yw ‹
8 Š

È3 #

xw
"# yw ‹

yw ‹ œ 0 Ê 4xw #  16yw œ 0 Ê Parabola

1
2
È 3

œ

" È3

#

1 3

Ê 2! œ

y œ xw sin !  yw cos ! Ê x œ Ê Š

Š 1# xw 

xw
1# yw ‹
È3 Š

È3 #

È3 #

1 6

Ê !œ

xw
#1 yw , y œ

" #

; therefore x œ xw cos !
yw sin !,

xw  È3 #

xw
1# yw ‹ Š 1# xw 

È3 #

yw

yw ‹  2 Š 1# xw 

È3 #

#

yw ‹ œ 1 Ê

" #

xw #  5# yw # œ 1

Ê xw #  5yw # œ 2 Ê Ellipse 21. cot 2! œ

A
C B

œ

1
1
2

œ 0 Ê 2! œ

y œ xw sin !  yw cos ! Ê x œ Ê Š

È2 #

xw


#

È2 #

yw ‹
2 Š

È2 #

È2 #

1 2

Ê !œ È2 #

; therefore x œ xw cos !
yw sin !,

È2 w È2 w # x  # y È2 w È2 w È2 w È2 # y ‹Š # x  # y ‹  Š #

xw


xw


1 4

yw , y œ

xw 

È2 #

#

yw ‹ œ 2 Ê yw # œ 1

Ê Parallel horizontal lines 22. cot 2! œ

A
C B

œ

3
1
2 È 3

œ
È"3 Ê 2! œ

y œ xw sin !  yw cos ! Ê x œ Ê 3 Š 1# xw


È3 #

#

1 #

xw


yw ‹
2È3 Š 1# xw


È3 # È3 #

21 3

Ê !œ

yw , y œ yw ‹ Š

È3 #

È3 #

1 3

; therefore x œ xw cos !
yw sin !,

xw  1# yw

xw  1# yw ‹  Š

È3 #

#

xw  1# yw ‹ œ 1 Ê 4yw # œ 1

Ê Parallel horizontal lines 23. cot 2! œ

A
C B

œ

È2
È2 2È 2

y œ xw sin !  yw cos ! Ê x œ È Ê È 2 Š # 2 xw


È2 #

Ê !œ

#

È2 #

xw


È2 #

yw , y œ

È2 #

xw


È2 #

yw ‹ œ 0 Ê 2È2xw #  8È2 yw œ 0 Ê Parabola

yw ‹  2È2 Š

È2 #

xw


È2 #

yw ‹  8 Š

24. cot 2! œ

A
C B

œ

0
0 1


8Š

1 2

œ 0 Ê 2! œ

È2 #

xw 

œ 0 Ê 2! œ

y œ xw sin !  yw cos ! Ê x œ

È2 #

1 2

xw


; therefore x œ xw cos !
yw sin !,

È2 w È2 w # x  # y È2 w È2 w È2 w # y ‹Š # x  # y ‹

Ê !œ È2 #

1 4

1 4

 È2 Š

È2 #

xw 

È2 #

yw ‹

#

; therefore x œ xw cos !
yw sin !,

yw , y œ

È2 #

xw 

È2 #

yw

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.3 Quadratic Equations and Rotations Ê Š

È2 #

È2 #

xw


yw ‹ Š

È2 #

xw 

È2 #

yw ‹
Š

È2 #

xw 

È2 #

yw ‹
Š

È2 #

xw


È2 #

639

yw ‹  1 œ 0 Ê xw #
yw #
2È2 xw  2

œ 0 Ê Hyperbola 25. cot 2! œ

A
C B

œ

3
3 2

œ 0 Ê 2! œ

y œ xw sin !  yw cos ! Ê x œ Ê 3Š

È2 #

xw


È2 #

#

yw ‹  2 Š

È2 #

È2 #

1 2

xw


xw


1 4

Ê !œ È2 #

È2 #

; therefore x œ xw cos !
yw sin !, È2 #

yw , y œ

yw ‹ Š

È2 #

xw +

xw  È2 #

È2 #

yw

yw ‹  3 Š

È2 #

xw 

È2 #

#

yw ‹ œ 19 Ê 4xw #  2yw # œ 19

Ê Ellipse 26. cot 2! œ

A
C B

œ

3
(
1) 4È 3

œ

" È3

Ê 2! œ

È3 # È3 È 4 3Š #

y œ xw sin !  yw cos ! Ê x œ Ê 3Š

È3 #

#

xw
1# yw ‹ 

1 3

Ê !œ

xw
#1 yw , y œ

1 #

1 6

; therefore x œ xw cos !
yw sin !,

xw 

xw
1# yw ‹ Š 1# xw 

È3 #

È3 #

yw

yw ‹
Š 1# xw 

È3 #

#

yw ‹ œ 7 Ê 5xw #
3yw # œ 7

Ê Hyperbola 27. cot 2! œ

14
2 16

œ

3 4

Ê cos 2! œ

2! and cos ! œ É 1  cos œÉ #

28. cot 2! œ œÉ

A
C B

1
ˆ
35 ‰ #

29. tan 2! œ

œ

œ


" 1
3 w

4
"
4

2 È5

1ˆ 35 ‰ #

3 5

2! (if we choose 2! in Quadrant I); thus sin ! œ É 1
cos σ 2

œ

2 È5

(or sin ! œ

2 È5

and cos ! œ

1
ˆ 35 ‰ #

œ

" È5


" È5 )

2! œ
34 Ê cos 2! œ
35 (if we choose 2! in Quadrant II); thus sin ! œ É 1
cos 2

2! and cos ! œ É 1  cos œÉ #

1  ˆ
35 ‰ #

œ

1 È5

(or sin ! œ

1 È5

and cos ! œ


2 È5 )

" #

Ê 2! ¸ 26.57° Ê ! ¸ 13.28° Ê sin ! ¸ 0.23, cos ! ¸ 0.97; then Aw ¸ 0.9, Bw ¸ 0.0,

œ

" 5

œ

Cw ¸ 3.1, D ¸ 0.7, Ew ¸
1.2, and Fw œ
3 Ê 0.9 xw #  3.1 yw #  0.7xw
1.2yw
3 œ 0, an ellipse 30. tan 2! œ

" 2
(
3)

Ê 2! ¸ 11.31° Ê ! ¸ 5.65° Ê sin ! ¸ 0.10, cos ! ¸ 1.00; then Aw ¸ 2.1, Bw ¸ 0.0,

Cw ¸
3.1, Dw ¸ 3.0, Ew ¸
0.3, and Fw œ
7 Ê 2.1 xw #
3.1 yw #  3.0xw
0.3yw
7 œ 0, a hyperbola 31. tan 2! œ


4 1
4 w

œ

Ê 2! ¸ 53.13° Ê ! ¸ 26.5(° Ê sin ! ¸ 0.45, cos ! ¸ 0.89; then Aw ¸ 0.0, Bw ¸ 0.0,

4 3

Cw ¸ 5.0, D ¸ 0, Ew ¸ 0, and Fw œ
5 Ê 5.0 yw #
5 œ 0 or yw œ „ 1.0, parallel lines 32. tan 2! œ


12 2
18 w

œ

3 4

Ê 2! ¸ 36.87° Ê ! ¸ 18.43° Ê sin ! ¸ 0.32, cos ! ¸ 0.95; then Aw ¸ 0.0, Bw ¸ 0.0,

Cw ¸ 20.1, D ¸ 0, Ew ¸ 0, and Fw œ
49 Ê 20.1 yw #
49 œ 0, parallel lines 33. tan 2! œ

œ 5 Ê 2! ¸ 78.69° Ê ! ¸ 39.35° Ê sin ! ¸ 0.63, cos ! ¸ 0.77; then Aw ¸ 5.0, Bw ¸ 0.0,

5 3
2

Cw ¸
0.05, Dw ¸
5.0, Ew ¸
6.2, and Fw œ
1 Ê 5.0 xw #
0.05 yw #
5.0xw
6.2yw
1 œ 0, a hyperbola 34. tan 2! œ

7 2
9 w

œ
1 Ê 2! ¸
45.00° Ê ! ¸
22.5° Ê sin ! ¸
0.38, cos ! ¸ 0.92; then Aw ¸ 0.5, Bw ¸ 0.0,

Cw ¸ 10.4, D ¸ 18.4, Ew ¸ 7.6, and Fw œ
86 Ê 0.5 xw #  10.4ayw b#  18.4xw  7.6yw
86 œ 0, an ellipse 35. ! œ 90° Ê x œ xw cos 90°
yw sin 90° œ
yw and y œ xw sin 90°  yw cos 90° œ xw (a)

xw # b#



yw # a#

œ1

(b)

yw # a#




xw # b#

œ1

(c) xw #  yw # œ a#

(d) y œ mx Ê y
mx œ 0 Ê D œ
m and E œ 1; ! œ 90° Ê Dw œ 1 and Ew œ m Ê myw  xw œ 0 Ê yw œ
m" xw (e) y œ mx  b Ê y
mx
b œ 0 Ê D œ
m and E œ 1; ! œ 90° Ê Dw œ 1, Ew œ m and Fw œ
b Ê myw  xw
b œ 0 Ê yw œ
m" xw  mb Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

640

Chapter 10 Conic Sections and Polar Coordinates

36. ! œ 180° Ê x œ xw cos 180°
yw sin 180° œ
xw and y œ xw sin 180°  yw cos 180° œ
yw (a)

xw # a#

yw # b#



œ1

(b)

xw # a#




yw # b#

(c) xw #  yw # œ a#

œ1

(d) y œ mx Ê y
mx œ 0 Ê D œ
m and E œ 1; ! œ 180° Ê Dw œ m and Ew œ
1 Ê
yw  mxw œ 0 Ê yw œ mxw (e) y œ mx  b Ê y
mx
b œ 0 Ê D œ
m and E œ 1; ! œ 180° Ê Dw œ m, Ew œ
1 and Fw œ
b Ê
yw  mxw
b œ 0 Ê yw œ mxw
b 37. (a) Aw œ cos 45° sin 45° œ Š " w# " # x
# Aw œ "# , Cw œ

Ê

(b)

c a

œ

2È 2 #

œ

" #

, Bw œ 0, Cw œ
cos 45° sin 45° œ
"# , Fw œ
1

yw # œ 1 Ê xw #
yw # œ 2
"# (see part (a) above), Dw œ Ew œ Bw œ 0, Fw œ
a Ê

38. xy œ 2 Ê xw #
yw # œ 4 Ê Ê eœ

È2 È2 # ‹Š # ‹

xw # 4




yw # 4

" #

xw #
"# yw # œ a Ê xw #
yw # œ 2a

œ 1 (see Exercise 37(b)) Ê a œ 2 and b œ 2 Ê c œ È4  4 œ 2È2

œ È2

39. Yes, the graph is a hyperbola: with AC  0 we have
4AC  0 and B#
4AC  0. 40. The one curve that meets all three of the stated criteria is the ellipse x#  4xy  5y#
1 œ 0. The reasoning: The symmetry about the origin means that (
xß
y) lies on the graph whenever (xß y) does. Adding Ax#  Bxy  Cy#  Dx  Ey  F œ 0 and A(
x)#  B(
x)(
y)  C(
y)#  D(
x)  E(
y)  F œ 0 and dividing the result by 2 produces the equivalent equation Ax#  Bxy  Cy#  F œ 0. Substituting x œ 1, y œ 0 (because the point (1ß 0) lies on the curve) shows further that A œ
F. Then
Fx#  Bxy  Cy#  F œ 0. By implicit differentiation,
2Fx  By  Bxyw  2Cyyw œ 0, so substituting x œ
2, y œ 1, and yw œ 0 (from Property 3) gives 4F  B œ 0 Ê B œ
4F Ê the conic is
Fx#
4Fxy  Cy#  F œ 0. Now substituting x œ
2 and y œ 1 again gives
4F  8F  C  F œ 0 Ê C œ
5F Ê the equation is now
Fx#
4Fxy
5Fy#  F œ 0. Finally, dividing through by
F gives the equation x#  4xy  5y#
1 œ 0. 41. Let ! be any angle. Then Aw œ cos# !  sin# ! œ 1, Bw œ 0, Cw œ sin# !  cos# ! œ 1, Dw œ Ew œ 0 and Fw œ
a# Ê xw #  yw # œ a# . 42. If A œ C, then Bw œ B cos 2!  (C
A) sin 2! œ B cos 2!. Then ! œ

1 4

Ê 2! œ

1 #

Ê Bw œ B cos

1 #

œ 0 so the

xy-term is eliminated. 43. (a) B#
4AC œ 4#
4(1)(4) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of x#  4xy  4y#  6x  12y  9 œ 0 factors as a perfect square: (x  2y  3)# œ 0 Ê x  2y  3 œ 0 Ê 2y œ
x
3; thus the curve is a degenerate parabola (i.e., a straight line). 44. (a) B#
4AC œ 6#
4(9)(1) œ 0, so the discriminant indicates this conic is a parabola (b) The left-hand side of 9x#  6xy  y#
12x
4y  4 œ 0 factors as a perfect square: (3x  y
2)# œ 0 Ê 3x  y
2 œ 0 Ê y œ
3x  2; thus the curve is a degenerate parabola (i.e., a straight line).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.3 Quadratic Equations and Rotations 45. (a) B#
4AC œ 1
4(0)(0) œ 1 Ê hyperbola (b) xy  2x
y œ 0 Ê y(x
1) œ
2x Ê y œ (c) y œ


2x x
1

Ê

dy dx

œ

2 (x
1)#

and we want

the slope of y œ
2x Ê
2 œ
(x
#1)


1 dy Š dx ‹


2x x
1

œ
2,

#

Ê (x
1)# œ 4 Ê x œ 3 or x œ
1; x œ 3 Ê y œ
3 Ê (3ß
3) is a point on the hyperbola where the line with slope m œ
2 is normal Ê the line is y  3 œ
2(x
3) or y œ
2x  3; x œ
1 Ê y œ
1 Ê (
1ß
1) is a point on the hyperbola where the line with slope m œ
2 is normal Ê the line is y  1 œ
2(x  1) or y œ
2x
3 46. (a) False: let A œ C œ 1, B œ 2 Ê B#
4AC œ 0 Ê parabola (b) False: see part (a) above (c) True: AC  0 Ê
4AC  0 Ê B#
4AC  0 Ê hyperbola 47. Assume the ellipse has been rotated to eliminate the xy-term Ê the new equation is Aw xw #  Cw yw # œ 1 Ê the semi-axes are É A" and É C" Ê the area is 1 ŠÉ A" ‹ ŠÉ C" ‹ œ w

w

w

w

1 ÈA C

œ Bw #
4Aw Cw œ
4Aw Cw (because Bw œ 0) we find that the area is

w

w

œ

21 È4A C

21 È4AC
B#

w

w

. Since B#
4AC

as claimed.

48. (a) Aw  Cw œ aA cos# !  B cos ! sin !  C sin# !b  aA sin# !
B cos ! sin !  C sin# !b œ A acos# !  sin# !b  C asin# !  cos# !b œ A  C (b) Dw #  Ew # œ (D cos !  E sin !)#  (
D sin !  E cos !)# œ D# cos# !  2DE cos ! sin !  E# sin# !  D# sin# !
2DE sin ! cos !  E# cos# ! œ D# acos# !  sin# !b  E# asin# !  cos# !b œ D#  E# 49. Bw #
4Aw Cw œ aB cos 2!  (C
A) sin 2!b#
4 aA cos# !  B cos ! sin !  C sin# !b aA sin# !
B cos ! sin !  C cos# !b œ B# cos# 2!  2B(C
A) sin 2! cos 2!  (C
A)# sin# 2!
4A# cos# ! sin# !  4AB cos$ ! sin !
4AC cos% !
4AB cos ! sin$ !  4B# cos# ! sin# !
4BC cos$ ! sin !
4AC sin% !  4BC cos ! sin$ !
4C# cos# ! sin# ! # œ B cos# 2!  2BC sin 2! cos 2!
2AB sin 2! cos 2!  C# sin# 2!
2AC sin# 2!  A# sin# 2!
4A# cos# ! sin# !  4AB cos$ ! sin !
4AC cos% !
4AB cos ! sin$ !  B# sin# 2!
4BC cos$ ! sin !
4AC sin% !  4BC cos ! sin$ !
4C# cos# ! sin# ! # œ B  2BC(2 sin ! cos !) acos# !
sin# !b
2AB(2 sin ! cos !) acos# !
sin# !b  C# a4 sin# ! cos# !b
2AC a4 sin# ! cos# !b  A# a4 sin# ! cos# !b
4A# cos# ! sin# !  4AB cos$ ! sin !
4AC cos% !
4AB cos ! sin$ !
4BC cos$ ! sin !
4AC sin% !  4BC cos ! sin$ !
4C# cos# ! sin# ! # œ B
8AC sin# ! cos# !
4AC cos% !
4AC sin% ! œ B#
4AC acos% !  2 sin# ! cos# !  sin% !b œ B#
4AC acos# !  sin# !b œ B#
4AC

#

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641

642

Chapter 10 Conic Sections and Polar Coordinates

10.4 CONICS AND PARAMETRIC EQUATIONS; THE CYCLOID 1. x œ cos t, y œ sin t, 0 Ÿ t Ÿ 1 Ê cos# t  sin# t œ 1 Ê x#  y# œ 1

2. x œ sin (21(1
t)), y œ cos (21(1
t)), 0 Ÿ t Ÿ 1 Ê sin# (21(1
t))  cos# (21(1
t)) œ 1 Ê x#  y# œ 1

3. x œ 4 cos t, y œ 5 sin t, 0 Ÿ t Ÿ 1

4. x œ 4 sin t, y œ 5 cos t, 0 Ÿ t Ÿ 21

Ê

16 cos# t 16



25 sin# t 25

x# 16

œ1 Ê



y# 25

œ1

5. x œ t, y œ Èt, t   0 Ê y œ Èx

Ê

16 sin# t 16



25 cos# t 25

œ1 Ê

x# 16



6. x œ sec# t
1, y œ tan t,
1#  t  Ê sec# t
1 œ tan# t Ê x œ y#

7. x œ
sec t, y œ tan t,
1#  t  #

#

#

1 # #

Ê sec t
tan t œ 1 Ê x
y œ 1

y# #5

œ1

1 #

8. x œ csc t, y œ cot t, 0  t  1 Ê 1  cot# t œ csc# t Ê 1  y# œ x# Ê x#
y# œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.4 Conics and Parametric Equations; The Cycloid 9. x œ t, y œ È4
t# , 0 Ÿ t Ÿ 2 Ê y œ È4
x#

10. x œ t# , y œ Èt%  1, t   0 Ê y œ Èx#  1, x   0

11. x œ
cosh t, y œ sinh t,
_  1  _ Ê cosh# t
sinh# t œ 1 Ê x#
y# œ 1

12. x œ 2 sinh t, y œ 2 cosh t,
_  t  _ Ê 4 cosh# t
4 sinh# t œ 4 Ê y#
x# œ 4

13. Arc PF œ Arc AF since each is the distance rolled and Arc PF œ nFCP Ê Arc PF œ b(nFCP); ArcaAF œ ) b Ê Arc AF œ a) Ê a) œ b(nFCP) Ê nFCP œ nOCG œ

1 #

a b

);


); nOCG œ nOCP  nPCE œ nOCP  ˆ 1#
!‰ . Now nOCP œ 1
nFCP œ 1
ba ). Thus nOCG œ 1
ba )  œ 1
ba ) 

1 #

1 #


! Ê


! Ê ! œ 1
ba )  ) œ 1


1 #
) ˆ a
b b )‰ .

Then x œ OG
BG œ OG
PE œ (a
b) cos )
b cos ! œ (a
b) cos )
b cos ˆ1
œ (a
b) cos )  b cos ˆ a
b b )‰ . Also y œ EG œ CG
CE œ (a
b) sin )
b sin !

a
b b

)‰

œ (a
b) sin )
b sin ˆ1
a
b b )‰ œ (a
b) sin )
b sin ˆ a
b b )‰ . Therefore x œ (a
b) cos )  b cos ˆ a
b b )‰ and y œ (a
b) sin )
b sin ˆ a
b b )‰ . If b œ 4a , then x œ ˆa
4a ‰ cos )  œ œ œ œ

3a 4 3a 4 3a 4 3a 4

cos ) 

œ œ œ

3a 4 3a 4 3a 4 3a 4

cos 3) œ

3a 4

cos Š

a
ˆ 4a ‰ ˆ 4a ‰

)‹

cos )  4a (cos ) cos 2)
sin ) sin 2))

cos )  a(cos )) acos# )
sin# )b
(sin ))(2 sin ) cos ))b a 2a # # 4 cos ) sin )
4 sin ) cos ) # $ )  cos$ )
3a 4 (cos )) a1
cos )b œ a cos ); a
ˆ 4a ‰ a‰ a 3a a 3a 4 sin )
4 sin Š ˆ 4a ‰ )‹ œ 4 sin )
4 sin 3) œ 4

cos )  cos

y œ ˆa
œ

a 4 a 4 a 4 a 4

a 4

cos$ )


sin )
4a (sin ) cos 2)  cos ) sin 2))

sin )
4a a(sin )) acos# )
sin# )b  (cos ))(2 sin ) cos ))b sin )
sin )
sin )


a 4 3a 4 3a 4

sin ) cos# )  sin ) cos# ) 

a 4 a 4 #

sin$ )


2a 4

cos# ) sin )

sin$ )

(sin )) a1
sin )b 

a 4

sin$ ) œ a sin$ ).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

643

644

Chapter 10 Conic Sections and Polar Coordinates

14. P traces a hypocycloid where the larger radius is 2a and the smaller is a Ê x œ (2a
a) cos )  a cos ˆ 2a a
a )‰ œ 2a cos ), 0 Ÿ ) Ÿ 21, and y œ (2a
a) sin )
a sin ˆ 2a a
a )‰ œ a sin )
a sin ) œ 0. Therefore P traces the diameter of the circle back and forth as ) goes from 0 to 21. 15. Draw line AM in the figure and note that nAMO is a right angle since it is an inscribed angle which spans the diameter of a circle. Then AN# œ MN#  AM# . Now, OA œ a, AN AM a œ tan t, and a œ sin t. Next MN œ OP Ê OP# œ AN#
AM# œ a# tan# t
a# sin# t Ê OP œ Èa# tan# t
a# sin# t œ (a sin t)Èsec# t
1 œ a sin$ t cos t œ #

x œ OP sin t œ

a sin# t cos t #

. In triangle BPO,

a sin t tan t and

y œ OP cos t œ a sin t Ê x œ a sin# t tan t and y œ a sin# t. 16. Let the x-axis be the line the wheel rolls along with the y-axis through a low point of the trochoid (see the accompanying figure).

Let ) denote the angle through which the wheel turns. Then h œ a) and k œ a. Next introduce xw yw -axes parallel to the xy-axes and having their origin at the center C of the wheel. Then xw œ b cos ! and yw œ b sin !, where ! œ 3#1
). It follows that xw œ b cos ˆ 3#1
)‰ œ
b sin ) and yw œ b sin ˆ 3#1
)‰

œ
b cos ) Ê x œ h  xw œ a)
b sin ) and y œ k  yw œ a
b cos ) are parametric equations of the trochoid.

# # # 17. D œ É(x
2)#  ˆy
"# ‰ Ê D# œ (x
2)#  ˆy
"# ‰ œ (t
2)#  ˆt#
"# ‰ Ê D# œ t%
4t 

Ê

d aD # b dt

17 4

œ 4t$
4 œ 0 Ê t œ 1. The second derivative is always positive for t Á 0 Ê t œ 1 gives a local

minimum for D# (and hence D) which is an absolute minimum since it is the only extremum Ê the closest point on the parabola is (1ß 1). # # 18. D œ Ɉ2 cos t
34 ‰  (sin t
0)# Ê D# œ ˆ2 cos t
34 ‰  sin# t Ê

d aD # b dt

œ 2 ˆ2 cos t
34 ‰ (
2 sin t)  2 sin t cos t œ (
2 sin t) ˆ3 cos t
3# ‰ œ 0 Ê
2 sin t œ 0 or 3 cos t
Ê t œ 0, 1 or t œ #

#

1 3

,

51 3

. Now

#

#

d aD b dt#

œ
6 cos# t  3 cos t  6 sin# t so that #

#

#

d aD b dt#

3 #

œ0

(0) œ
3 Ê relative

#

maximum, d dtaD# b (1) œ
9 Ê relative maximum, d dtaD# b ˆ 13 ‰ œ 92 Ê relative minimum, and d # aD # b ˆ 5 1 ‰ œ 9# Ê relative minimum. Therefore both t œ 13 and t œ 531 give points on the ellipse dt# 3 È È the point ˆ 34 ß !‰ Ê Š1ß #3 ‹ and Š1ß
#3 ‹ are the desired points.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

closest to

Section 10.4 Conics and Parametric Equations; The Cycloid 19. (a)

(b)

(c)

20. (a)

(b)

(c)

(b)

(c)

21.

22. (a)

23. (a)

(b)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

645

646

Chapter 10 Conic Sections and Polar Coordinates

24. (a)

25. (a)

(b)

(b)

(c)

26. (a)

(b)

(c)

(d)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.5 Polar Coordinates 10.5 POLAR COORDINATES 1. a, e; b, g; c, h; d, f

2. a, f; b, h; c, g; d, e

3. (a) ˆ2ß 1#
2n1‰ and ˆ2ß 1#
(2n
1)1‰ , n an integer

(b) (#ß 2n1) and (#ß (2n
1)1), n an integer (c) ˆ2ß 3#1
2n1‰ and ˆ2ß 3#1
(2n
1)1‰ , n an integer

(d) (#ß (2n
1)1) and (#ß 2n1), n an integer

4. (a) ˆ3ß 14
2n1‰ and ˆ3ß 541
2n1‰ , n an integer (b) ˆ3ß 14
2n1‰ and ˆ3ß 541
2n1‰ , n an integer (c) ˆ3ß  14
2n1‰ and ˆ3ß 341
2n1‰ , n an integer (d) ˆ3ß  14
2n1‰ and ˆ3ß 341
2n1‰ , n an integer

5. (a) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (b) x œ r cos ) œ 3 cos 0 œ 3, y œ r sin ) œ 3 sin 0 œ 0 Ê Cartesian coordinates are ($ß 0) (c) x œ r cos ) œ 2 cos 21 œ 1, y œ r sin ) œ 2 sin 21 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3

(d) x œ r cos ) œ 2 cos

71 3

3

œ 1, y œ r sin ) œ 2 sin

71 3

œ È3 Ê Cartesian coordinates are Š1ß È3‹

(e) x œ r cos ) œ 3 cos 1 œ 3, y œ r sin ) œ 3 sin 1 œ 0 Ê Cartesian coordinates are (3ß 0) (f) x œ r cos ) œ 2 cos 1 œ 1, y œ r sin ) œ 2 sin 1 œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3

3

(g) x œ r cos ) œ 3 cos 21 œ 3, y œ r sin ) œ 3 sin 21 œ 0 Ê Cartesian coordinates are (3ß 0) (h) x œ r cos ) œ 2 cos ˆ 1 ‰ œ 1, y œ r sin ) œ 2 sin ˆ 1 ‰ œ È3 Ê Cartesian coordinates are Š1ß È3‹ 3

6. (a) x œ È2 cos

1 4

œ 1, y œ È2 sin

3

1 4

œ 1 Ê Cartesian coordinates are (1ß 1)

(b) x œ 1 cos 0 œ 1, y œ 1 sin 0 œ 0 Ê Cartesian coordinates are (1ß 0) (c) x œ 0 cos 1# œ 0, y œ 0 sin 1# œ 0 Ê Cartesian coordinates are (!ß 0) (d) x œ È2 cos ˆ 1 ‰ œ 1, y œ È2 sin ˆ 1 ‰ œ 1 Ê Cartesian coordinates are (1ß 1) 4

(e) x œ 3 cos

51 6

œ

4

3È 3 2

, y œ 3 sin

51 6

È

œ  3# Ê Cartesian coordinates are Š 3 # 3 ß  3# ‹

(f) x œ 5 cos ˆtan
" 43 ‰ œ 3, y œ 5 sin ˆtan
" 43 ‰ œ 4 Ê Cartesian coordinates are ($ß 4) (g) x œ 1 cos 71 œ 1, y œ 1 sin 71 œ 0 Ê Cartesian coordinates are (1ß 0) (h) x œ 2È3 cos 231 œ È3, y œ 2È3 sin 231 œ 3 Ê Cartesian coordinates are ŠÈ3ß 3‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

647

648

Chapter 10 Conic Sections and Polar Coordinates

7.

8.

9.

10.

11.

12.

13.

14.

15.

16.

17.

18.

19.

20.

21.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.5 Polar Coordinates

649

22.

23. r cos ) œ 2 Ê x œ 2, vertical line through (#ß 0)

24. r sin ) œ 1 Ê y œ 1, horizontal line through (0ß 1)

25. r sin ) œ 0 Ê y œ 0, the x-axis

26. r cos ) œ 0 Ê x œ 0, the y-axis

27. r œ 4 csc ) Ê r œ

4 sin )

28. r œ 3 sec ) Ê r œ

Ê r sin ) œ 4 Ê y œ 4, a horizontal line through (0ß 4)


3 cos )

Ê r cos ) œ 3 Ê x œ 3, a vertical line through (3ß 0)

29. r cos )
r sin ) œ 1 Ê x
y œ 1, line with slope m œ 1 and intercept b œ 1 30. r sin ) œ r cos ) Ê y œ x, line with slope m œ 1 and intercept b œ 0 31. r# œ 1 Ê x#
y# œ 1, circle with center C œ (!ß 0) and radius 1 32. r# œ 4r sin ) Ê x#
y# œ 4y Ê x#
y#  4y
4 œ 4 Ê x#
(y  2)# œ 4, circle with center C œ (0ß 2) and radius 2 33. r œ

5 sin )
2 cos )

Ê r sin )  2r cos ) œ 5 Ê y  2x œ 5, line with slope m œ 2 and intercept b œ 5

34. r# sin 2) œ 2 Ê 2r# sin ) cos ) œ 2 Ê (r sin ))(r cos )) œ 1 Ê xy œ 1, hyperbola with focal axis y œ x )‰ˆ " ‰ 35. r œ cot ) csc ) œ ˆ cos Ê r sin# ) œ cos ) Ê r# sin# ) œ r cos ) Ê y# œ x, parabola with vertex (0ß 0) sin ) sin )

which opens to the right sin ) ‰ 36. r œ 4 tan ) sec ) Ê r œ 4 ˆ cos Ê r cos# ) œ 4 sin ) Ê r# cos# ) œ 4r sin ) Ê x# œ 4y, parabola with #)

vertex œ (!ß 0) which opens upward

37. r œ (csc )) er cos ) Ê r sin ) œ er cos ) Ê y œ ex , graph of the natural exponential function 38. r sin ) œ ln r
ln cos ) œ ln (r cos )) Ê y œ ln x, graph of the natural logarithm function 39. r#
2r# cos ) sin ) œ 1 Ê x#
y#
2xy œ 1 Ê x#
2xy
y# œ 1 Ê (x
y)# œ 1 Ê x
y œ „ 1, two parallel straight lines of slope 1 and y-intercepts b œ „ 1 40. cos# ) œ sin# ) Ê r# cos# ) œ r# sin# ) Ê x# œ y# Ê kxk œ kyk Ê „ x œ y, two perpendicular lines through the origin with slopes 1 and 1, respectively. 41. r# œ 4r cos ) Ê x#
y# œ 4x Ê x#
4x
y# œ 0 Ê x#
4x
4
y# œ 4 Ê (x
2)#
y# œ 4, a circle with center C(2ß 0) and radius 2

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650

Chapter 10 Conic Sections and Polar Coordinates

42. r# œ 6r sin ) Ê x#
y# œ 6y Ê x#
y#
6y œ 0 Ê x#
y#
6y
9 œ 9 Ê x#
(y
3)# œ 9, a circle with center C(0ß 3) and radius 3 43. r œ 8 sin ) Ê r# œ 8r sin ) Ê x#
y# œ 8y Ê x#
y#  8y œ 0 Ê x#
y#  8y
16 œ 16 Ê x#
(y  4)# œ 16, a circle with center C(0ß 4) and radius 4 44. r œ 3 cos ) Ê r# œ 3r cos ) Ê x#
y# œ 3x Ê x#
y#  3x œ 0 Ê x#  3x
# Ê ˆx  3# ‰
y# œ

9 4

, a circle with center C ˆ 3# ß !‰ and radius

9 4


y# œ

9 4

3 #

45. r œ 2 cos )
2 sin ) Ê r# œ 2r cos )
2r sin ) Ê x#
y# œ 2x
2y Ê x#  2x
y#  2y œ 0 Ê (x  1)#
(y  1)# œ 2, a circle with center C(1ß 1) and radius È2 46. r œ 2 cos )  sin ) Ê r# œ 2r cos )  r sin ) Ê x#
y# œ 2x  y Ê x#  2x
y#
y œ 0 # Ê (x  1)#
ˆy
"# ‰ œ 54 , a circle with center C ˆ1ß  "# ‰ and radius

È5 #

È

47. r sin ˆ)
16 ‰ œ 2 Ê r ˆsin ) cos 16
cos ) sin 16 ‰ œ 2 Ê #3 r sin )
"# r cos ) œ 2 Ê Ê È3 y
x œ 4, line with slope m œ  " and intercept b œ 4 È3

È3 #

È3

È

48. r sin ˆ 231  )‰ œ 5 Ê r ˆsin 231 cos )  cos 231 sin )‰ œ 5 Ê #3 r cos )
"# r sin ) œ 5 Ê Ê È3 x
y œ 10, line with slope m œ È3 and intercept b œ 10 49. x œ 7 Ê r cos ) œ 7 51. x œ y Ê r cos ) œ r sin ) Ê ) œ

y
"# x œ 2

È3 #

x
"# y œ 5

50. y œ 1 Ê r sin ) œ 1 1 4

52. x  y œ 3 Ê r cos )  r sin ) œ 3

53. x#
y# œ 4 Ê r# œ 4 Ê r œ 2 or r œ 2 54. x#  y# œ 1 Ê r# cos# )  r# sin# ) œ 1 Ê r# acos# )  sin# )b œ 1 Ê r# cos 2) œ 1 55.

x# 9




y# 4

œ 1 Ê 4x#
9y# œ 36 Ê 4r# cos# )
9r# sin# ) œ 36

56. xy œ 2 Ê (r cos ))(r sin )) œ 2 Ê r# cos ) sin ) œ 2 Ê 2r# cos ) sin ) œ 4 Ê r# sin 2) œ 4 57. y# œ 4x Ê r# sin# ) œ 4r cos ) Ê r sin# ) œ 4 cos ) 58. x#
xy
y# œ 1 Ê x#
y#
xy œ 1 Ê r#
r# sin ) cos ) œ 1 Ê r# (1
sin ) cos )) œ 1 59. x#
(y  2)# œ 4 Ê x#
y#  4y
4 œ 4 Ê x#
y# œ 4y Ê r# œ 4r sin ) Ê r œ 4 sin ) 60. (x  5)#
y# œ 25 Ê x#  10x
25
y# œ 25 Ê x#
y# œ 10x Ê r# œ 10r cos ) Ê r œ 10 cos ) 61. (x  3)#
(y
1)# œ 4 Ê x#  6x
9
y#
2y
1 œ 4 Ê x#
y# œ 6x  2y  6 Ê r# œ 6r cos )  2r sin )  6 62. (x
2)#
(y  5)# œ 16 Ê x#
4x
4
y#  10y
25 œ 16 Ê x#
y# œ 4x
10y  13 Ê r# œ 4r cos )
10r sin )  13 63. (!ß )) where ) is any angle Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.6 Graphing in Polar Coordinates 64. (a) x œ a Ê r cos ) œ a Ê r œ (b) y œ b Ê r sin ) œ b Ê r œ

a cos ) b sin )

Ê r œ a sec ) Ê r œ b csc )

10.6 GRAPHING IN POLAR COORDINATES 1. 1
cos ()) œ 1
cos ) œ r Ê symmetric about the x-axis; 1
cos ()) Á r and 1
cos (1  )) œ 1  cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin

2. 2  2 cos ()) œ 2  2 cos ) œ r Ê symmetric about the x-axis; 2  # cos ()) Á r and 2  2 cos (1  )) œ 2
2 cos ) Á r Ê not symmetric about the y-axis; therefore not symmetric about the origin

3. 1  sin ()) œ 1
sin ) Á r and 1  sin (1  )) œ 1  sin ) Á r Ê not symmetric about the x-axis; 1  sin (1  )) œ 1  sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin

4. 1
sin ()) œ 1  sin ) Á r and 1
sin (1  )) œ 1
sin ) Á r Ê not symmetric about the x-axis; 1
sin (1  )) œ 1
sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin

5. 2
sin ()) œ 2  sin ) Á r and 2
sin (1  )) œ 2
sin ) Á r Ê not symmetric about the x-axis; 2
sin (1  )) œ 2
sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

651

652

Chapter 10 Conic Sections and Polar Coordinates

6. 1
2 sin ()) œ 1  2 sin ) Á r and 1
2 sin (1  )) œ 1
2 sin ) Á r Ê not symmetric about the x-axis; 1
2 sin (1  )) œ 1
2 sin ) œ r Ê symmetric about the y-axis; therefore not symmetric about the origin

7. sin ˆ #) ‰ œ  sin ˆ #) ‰ œ r Ê symmetric about the y-axis; sin ˆ 21#
) ‰ œ sin ˆ 2) ‰ , so the graph is symmetric about the x-axis, and hence the origin.

8. cos ˆ #) ‰ œ cos ˆ #) ‰ œ r Ê symmetric about the x-axis; cos ˆ 21#
) ‰ œ cos ˆ 2) ‰ , so the graph is symmetric about the y-axis, and hence the origin.

9. cos ()) œ cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin

10. sin (1  )) œ sin ) œ r# Ê (rß 1  )) and (rß 1  )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.6 Graphing in Polar Coordinates 11.  sin (1  )) œ  sin ) œ r# Ê (rß 1  )) and (rß 1  )) are on the graph when (rß )) is on the graph Ê symmetric about the y-axis and the x-axis; therefore symmetric about the origin

12.  cos ()) œ  cos ) œ r# Ê (rß )) and (rß )) are on the graph when (rß )) is on the graph Ê symmetric about the x-axis and the y-axis; therefore symmetric about the origin

13. Since a „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ 4 cos 2( )) Ê r# œ 4 cos 2)‰ , the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin

14. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ 4 sin 2) Ê r# œ 4 sin 2)‰ , the graph is symmetric about the origin. But 4 sin 2()) œ 4 sin 2) Á r# and 4 sin 2(1  )) œ 4 sin (21  2)) œ 4 sin (2)) œ 4 sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 15. Since (rß )) on the graph Ê (rß )) is on the graph ˆa „ rb# œ  sin 2) Ê r# œ  sin 2)‰ , the graph is symmetric about the origin. But  sin 2()) œ ( sin 2)) sin 2) Á r# and  sin 2(1  )) œ  sin (21  2)) œ  sin (2)) œ ( sin 2)) œ sin 2) Á r# Ê the graph is not symmetric about the x-axis; therefore the graph is not symmetric about the y-axis 16. Sincea „ rß )b are on the graph when (rß )) is on the graph ˆa „ rb# œ  cos 2()) Ê r# œ  cos 2)‰, the graph is symmetric about the x-axis and the y-axis Ê the graph is symmetric about the origin.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

653

654

Chapter 10 Conic Sections and Polar Coordinates Ê r œ 1 Ê ˆ1ß 1# ‰ , and ) œ  1# Ê r œ 1 w )r cos ) Ê ˆ1ß  1# ‰ ; rw œ ddr) œ  sin ); Slope œ rrw sin cos )
r sin )

17. ) œ

1 #


sin# )r cos )
sin ) cos )
r sin )
sin# ˆ 1# ‰(
1) cos 1#
sin 1# cos 1#
(
1) sin 1#

œ

Ê Slope at ˆ1ß 1# ‰ is

œ 1; Slope at ˆ1ß  1# ‰ is


sin# ˆ
1# ‰(
1) cos ˆ
1# ‰
sin ˆ
1# ‰ cos ˆ
1# ‰
(
1) sin ˆ
1# ‰

œ1

18. ) œ 0 Ê r œ 1 Ê ("ß 0), and ) œ 1 Ê r œ 1 dr Ê ("ß 1); rw œ d) œ cos );

rw sin )r cos ) cos ) sin )r cos ) rw cos )
r sin ) œ cos ) cos )
r sin ) 0 sin 0(
1) cos 0 cos ) sin )r cos ) Ê Slope at ("ß 0) is coscos # 0
(
1) sin 0 cos# )
r sin ) cos 1 sin 1(
1) cos 1 1; Slope at ("ß 1) is cos# 1
(
1) sin 1 œ 1

Slope œ œ œ

Ê r œ 1 Ê ˆ"ß 14 ‰ ; ) œ  14 Ê r œ 1 Ê ˆ1ß  14 ‰ ; ) œ 341 Ê r œ 1 Ê ˆ"ß 341 ‰ ; ) œ  341 Ê r œ 1 Ê ˆ1ß  341 ‰ ;

19. ) œ

rw œ

1 4

dr d)

œ 2 cos 2);

Slope œ

r sin )r cos ) r cos )
r sin ) w w

Ê Slope at ˆ1ß 14 ‰ is Slope at ˆ1ß  14 ‰ is Slope at ˆ1ß 341 ‰ is Slope at ˆ1ß  341 ‰ is

2 cos 2) sin )r cos ) 2 cos 2) cos )
r sin ) 2 cos ˆ 1# ‰ sin ˆ 14 ‰(1) cos ˆ 14 ‰ 2 cos ˆ 1 ‰ cos ˆ 1 ‰
(1) sin ˆ 1 ‰

œ

#

4

4

œ 1;

2 cos ˆ
1# ‰ sin ˆ
14 ‰(
1) cos ˆ
14 ‰ 2 cos ˆ
1# ‰ cos ˆ
14 ‰
(
1) sin ˆ
14 ‰

2 cos Š 3#1 ‹ sin Š 341 ‹(
1) cos Š 341 ‹ 2 cos Š 3#1 ‹ cos Š 341 ‹
(
1) sin Š 341 ‹

œ 1;

œ 1;

2 cos Š
3#1 ‹ sin Š
341 ‹(1) cos Š
341 ‹ 2 cos Š
3#1 ‹ cos Š
341 ‹
(1) sin Š
341 ‹

œ 1

20. ) œ 0 Ê r œ 1 Ê (1ß 0); ) œ 12 Ê r œ 1 Ê ˆ1ß 12 ‰ ; ) œ  1# Ê r œ 1 Ê ˆ"ß  12 ‰ ; ) œ 1 Ê r œ 1 Ê (1ß 1); rw œ

dr d) œ 2 sin 2); )r cos )
2 sin 2) sin )r cos ) Slope œ rr sin cos )
r sin ) œ
2 sin 2) cos )
r sin ) 2 sin 0 sin 0cos 0 Ê Slope at (1ß 0) is

2 sin 0 cos 0
sin 0 , which is undefined;
2 sin 2 ˆ 1 ‰ sin ˆ 1 ‰(
1) cos ˆ 1 ‰ Slope at ˆ1ß 12 ‰ is
2 sin 2 ˆ 12 ‰ cos ˆ21 ‰
(
1) sin ˆ 21 ‰ œ 0; w w

2

Slope at ˆ1ß  12 ‰ is Slope at ("ß 1) is

2

2


2 sin 2 ˆ
1# ‰ sin ˆ
1# ‰(
1) cos ˆ
1# ‰
2 sin 2 ˆ
1 ‰ cos ˆ
1 ‰
(
1) sin ˆ
1 ‰ #


2 sin 21 sin 1cos 1
2 sin 21 cos 1
sin 1

#

#

œ 0;

, which is undefined

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.6 Graphing in Polar Coordinates 21. (a)

(b)

22. (a)

(b)

23. (a)

(b)

24. (a)

(b)

25.

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655

656

Chapter 10 Conic Sections and Polar Coordinates

26. r œ 2 sec ) Ê r œ

2 cos )

Ê r cos ) œ 2 Ê x œ 2

27.

28.

29. ˆ#ß 341 ‰ is the same point as ˆ
2ß
14 ‰ ; r œ 2 sin 2 ˆ
14 ‰ œ 2 sin ˆ
1# ‰ œ
2 Ê ˆ
2ß
14 ‰ is on the graph Ê ˆ#ß 341 ‰ is on the graph 30. ˆ "# ß 321 ‰ is the same point as ˆ
"# ß 12 ‰ ; r œ
sin Š

ˆ 1# ‰ 3 ‹

œ
sin

1 6

œ
"# Ê ˆ
"# ß 1# ‰ is on the graph Ê ˆ "# ß 3#1 ‰

is on the graph 31. 1  cos ) œ 1
cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê r œ 1; points of intersection are ˆ"ß 1# ‰ and ˆ"ß 3#1 ‰ . The point of intersection (!ß 0) is found by graphing.

32. 1  sin ) œ 1
sin ) Ê sin ) œ 0 Ê ) œ 0, 1 Ê r œ 1; points of intersection are (1ß 0) and (1ß 1). The point of intersection (!ß 0) is found by graphing.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.6 Graphing in Polar Coordinates 33. 2 sin ) œ 2 sin 2) Ê sin ) œ sin 2) Ê sin ) œ 2 sin ) cos ) Ê sin )
2 sin ) cos ) œ 0 Ê (sin ))(1
2 cos )) œ 0 Ê sin ) œ 0 or cos ) œ

, or
13 ; ) œ 0 or 1 Ê r œ 0, Ê r œ È3 , and ) œ
1 Ê r œ
È3 ; points of

Ê ) œ 0, 1, )œ

1 3

1 3

" #

3

intersection are (!ß 0), ŠÈ3 ß 13 ‹, and Š
È3 ß
13 ‹

34. cos ) œ 1
cos ) Ê 2 cos ) œ 1 Ê cos ) œ

" #

Ê ) œ 13 ,
13 Ê r œ "# ; points of intersection are ˆ "# ß 13 ‰ and ˆ "# ,
13 ‰ . The point (0ß 0) is found by graphing.

#

35. ŠÈ2‹ œ 4 sin ) Ê

" #

œ sin ) Ê ) œ

1 6

,

51 6

; points

of intersection are ŠÈ2ß 16 ‹ and ŠÈ2ß 561 ‹ . The points ŠÈ2ß
16 ‹ and ŠÈ2ß
561 ‹ are found by graphing.

36. È2 sin ) œ È2 cos ) Ê sin ) œ cos ) Ê ) œ )œ

1 4

#

Ê r œ 1 Ê r œ „ 1 and ) œ

51 4

1 4

,

51 4

;

#

Ê r œ
1

Ê no solution for r; points of intersection are ˆ „ 1ß 14 ‰ . The points (!ß 0) and ˆ „ 1ß 341 ‰ are found by graphing.

37. 1 œ 2 sin 2) Ê sin 2) œ

" #

Ê 2) œ

1 6

,

51 6

,

131 6

,

171 6

1 Ê ) œ 12 , 5121 , 13121 , 17121 ; points of intersection are 1‰ ˆ ˆ1ß 12 , 1ß 5121 ‰ , ˆ1ß 131#1 ‰, and ˆ1ß 171#1 ‰ . No other

points are found by graphing.

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657

658

Chapter 10 Conic Sections and Polar Coordinates

38. È2 cos 2) œ È2 sin 2) Ê cos 2) œ sin 2) Ê 2) œ 14 , 541 , 941 , 1341 Ê ) œ 18 , 581 , 981 , )œ

1 91 8 , 8 #

#

Ê r œ 1 Ê r œ „ 1; ) œ

51 8

,

131 8 131 8

;

Ê r œ
1 Ê no solution for r; points of intersection are ˆ1ß 18 ‰ and ˆ1ß 981 ‰ . The point of intersection (!ß 0) is found by graphing.

39. r# œ sin 2) and r# œ cos 2) are generated completely for 0 Ÿ ) Ÿ 1# . Then sin 2) œ cos 2) Ê 2) œ 14 is the only solution on that interval Ê ) œ 18 Ê r# œ sin 2 ˆ 18 ‰ œ È" Ê rœ „

" % È 2

2

" ß 1‹. % È 2 8

; points of intersection are Š „

The point of intersection (!ß 0) is found by graphing.

40. 1
sin

) #

31 #

Ê )œ )œ

71 #

œ 1  cos ,

71 #

) #

;)œ

Ê
sin 31 #

Ê r œ 1  cos

intersection are Š"


) #

œ cos

Ê r œ 1  cos 71 4

œ1

È 2 31 # ß # ‹

È2 #

) #

Ê

) #

31 4

œ1


31 71 4 , 4 È2 # ;

œ

; points of

and Š1 

È 2 71 # ß # ‹.

three points of intersection (0ß 0) and Š1 „

È2 #

The

ß 1# ‹ are

found by graphing and symmetry.

41. 1 œ 2 sin 2) Ê sin 2) œ

" #

Ê 2) œ

1 6

,

51 6

,

131 6

,

171 6

Ê ) œ 11# , 511# , 131#1 , 171#1 ; points of intersection are ˆ"ß 11# ‰ , ˆ"ß 511# ‰ , ˆ1ß 131#1 ‰ , and ˆ"ß 17121 ‰ . The points of intersection ˆ1ß 711# ‰ , ˆ"ß 111#1 ‰ , ˆ"ß 191#1 ‰ and ˆ"ß 231#1 ‰ are found by graphing and symmetry.

42. r# œ 2 sin 2) is completely generated on 0 Ÿ ) Ÿ " #

1 6

1 #

so

1 that 1 œ 2 sin 2) Ê sin 2) œ Ê 2) œ , 561 Ê ) œ 12 51 1 5 1 ˆ ‰ ˆ ‰ 1# ; points of intersection are 1ß 1# and "ß 1# . The 1 5 1 points of intersection ˆ
"ß 1# ‰ and ˆ
1ß 1# ‰ are found

,

by graphing.

43. Note that (rß )) and (
rß )  1) describe the same point in the plane. Then r œ 1
cos ) Í
1
cos ()  1) œ
1
(cos ) cos 1
sin ) sin 1) œ
1  cos ) œ
(1
cos )) œ
r; therefore (rß )) is on the graph of r œ 1
cos ) Í (
rß )  1) is on the graph of r œ
1
cos ) Ê the answer is (a).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.6 Graphing in Polar Coordinates

44. Note that (rß )) and (
rß )  1) describe the same point in the plane. Then r œ cos 2) Í
sin ˆ2()  1))  1# ‰ œ
sin ˆ2)  5#1 ‰ œ
sin (2)) cos ˆ 5#1 ‰
cos (2)) sin ˆ 5#1 ‰ œ
cos 2) œ
r; therefore (rß )) is on the graph of r œ
sin ˆ2)  1# ‰ Ê the answer is (a).

45.

47. (a)

46.

(b)

(c)

(d)

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659

660

Chapter 10 Conic Sections and Polar Coordinates

48. (a)

(b)

(d)

(c)

(e)

#

#

49. (a) r# œ
4 cos ) Ê cos ) œ
r4 ; r œ 1
cos ) Ê r œ 1
Š
r4 ‹ Ê 0 œ r#
4r  4 Ê (r
2)# œ 0 #

Ê r œ 2; therefore cos ) œ
24 œ
1 Ê ) œ 1 Ê (2ß 1) is a point of intersection (b) r œ 0 Ê 0# œ 4 cos ) Ê cos ) œ 0 Ê ) œ 1# , 3#1 Ê ˆ!ß 1# ‰ or ˆ!ß 3#1 ‰ is on the graph; r œ 0 Ê 0 œ 1
cos ) Ê cos ) œ 1 Ê ) œ 0 Ê (0ß 0) is on the graph. Since (!ß 0) œ ˆ!ß 1# ‰ for polar coordinates, the graphs intersect at the origin. 50. (a) Let r œ f()) be symmetric about the x-axis and the y-axis. Then (rß )) on the graph Ê (rß
)) is on the graph because of symmetry about the x-axis. Then (
rß
(
))) œ (
rß )) is on the graph because of symmetry about the y-axis. Therefore r œ f()) is symmetric about the origin. (b) Let r œ f()) be symmetric about the x-axis and the origin. Then (rß )) on the graph Ê (rß
)) is on the graph because of symmetry about the x-axis. Then (
rß
)) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the y-axis. (c) Let r œ f()) be symmetric about the y-axis and the origin. Then (rß )) on the graph Ê (
rß
)) is on the graph because of symmetry about the y-axis. Then (
(
r)ß
)) œ (rß
)) is on the graph because of symmetry about the origin. Therefore r œ f()) is symmetric about the x-axis. 51. The maximum width of the petal of the rose which lies along the x-axis is twice the largest y value of the curve on the interval 0 Ÿ ) Ÿ 14 . So we wish to maximize 2y œ 2r sin ) œ 2 cos 2) sin ) on 0 Ÿ ) Ÿ 14 . Let f()) œ 2 cos 2) sin ) œ 2 a1
2 sin# )b (sin )) œ 2 sin )
4 sin$ ) Ê f w ()) œ 2 cos )
12 sin# ) cos ). Then f w ()) œ 0 Ê 2 cos )
12 sin# ) cos ) œ 0 Ê (cos )) a1
6 sin# )b œ 0 Ê cos ) œ 0 or 1
6 sin# ) œ 0 Ê ) œ sin ) œ

„1 È6 .

Since we want 0 Ÿ ) Ÿ

œ 2 Š È"6 ‹
4 † interval 0 Ÿ ) Ÿ is

2È 6 9

" œ . We 6È 6 1 4 . Therefore the

1 4


"

, we choose ) œ sin

Š È"6 ‹

$

Ê f()) œ 2 sin )
4 sin )

can see from the graph of r œ cos 2) that a maximum does occur in the maximum width occurs at ) œ sin
" Š È"6 ‹ , and the maximum width

2È 6 9 .

52. We wish to maximize y œ r sin ) œ 2(1  cos ))(sin )) œ 2 sin )  2 sin ) cos ). Then dy # # # d) œ 2 cos )  2(sin ))(
sin ))  2 cos ) cos ) œ 2 cos )
2 sin )  2 cos ) œ 2 cos )  4 cos )
2; thus dy # # d) œ 0 Ê 4 cos )  2 cos )
2 = 0 Ê 2 cos )  cos )
1 œ 0 Ê (2 cos )
1)(cos )  1) œ 0 Ê cos ) œ or cos ) œ
1 Ê ) œ 13 , 531 , 1. From the graph, we can see that the maximum occurs in the first quadrant so È

we choose ) œ 13 . Then y œ 2 sin 13  2 sin 13 cos 13 œ 3 # 3 . The x-coordinate of this point is x œ r cos È œ 2 ˆ1  cos 13 ‰ ˆcos 13 ‰ œ 3# . Thus the maximum height is h œ 3 # 3 occurring at x œ 3# . Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 3

" #

1 #

or

Section 10.7 Area and Lengths in Polar Coordinates 10.7 AREA AND LENGTHS IN POLAR COORDINATES 1. A œ '0

21

" #

(4  2 cos ))# d) œ '0

21

" #

2) ‰‘ a16  16 cos )  4 cos# )b d) œ '0
8  8 cos )  2 ˆ 1  cos d) # 21

œ '0 (9  8 cos )  cos 2)) d) œ
9)  8 sin )  21

2. A œ '0

21

œ

" #

a#

" #

[a(1  cos ))]# d) œ '0

21

" #

#1

" 2

sin 2)‘ ! œ 181

a# a1  2 cos )  cos# )b d) œ

" #

a#

2) ‰ '021 ˆ1  2 cos )  1  cos d) #

'021 ˆ 3#  2 cos )  "# cos 2)‰ d) œ "# a#
3# )  2 sin )  4" sin 2)‘ #!1 œ 3# 1a#

3. A œ 2 '0

1Î4

" #

cos# 2) d) œ '0

" #

a2a# cos 2)b d) œ 2a# '
1Î4 cos 2) d) œ 2a#
sin22) ‘
1Î% œ 2a#

1Î4

4. A œ 2 '
1Î4 5. A œ '0

1Î2

" #

1Î4

1  cos 4) #

d) œ

" #


) 

sin 4) ‘ 1Î% 4 !

œ

1Î4

(4 sin 2)) d) œ '0

1Î2

6. A œ (6)(2)'0

1Î6

1 8 1Î%

1Î#

2 sin 2) d) œ c
cos 2)d !

œ2

(2 sin 3)) d) œ 12 '0 sin 3) d) œ 12

cos3 3) ‘ ! 1Î6

" #

1Î'

œ4

7. r œ 2 cos ) and r œ 2 sin ) Ê 2 cos ) œ 2 sin ) Ê cos ) œ sin ) Ê ) œ 14 ; therefore A œ 2 '0

1Î4

œ '0

1Î4

(2 sin ))# d) œ '0

1Î4

" #

2) ‰ 4 ˆ 1
cos d) œ '0 #

1Î4

œ c2)
sin

1Î% 2) d !

œ

1 #

4 sin# ) d)

(2
2 cos 2)) d)


1

8. r œ 1 and r œ 2 sin ) Ê 2 sin ) œ 1 Ê sin ) œ Ê )œ

1 6

or

51 6

" #

; therefore

A œ 1(1)#
'1Î6

51Î6

" #

c(2 sin ))#
1# d d)

œ 1
'1Î6 ˆ2 sin# )
"# ‰ d) 51Î6

œ 1
'1Î6 ˆ1
cos 2)
"# ‰ d) 51Î6

œ 1
'1Î6 ˆ "#
cos 2)‰ d) œ 1

"2 )


sin 2) ‘ &1Î' # 1Î'

œ 1
ˆ 511#


41
3È 3 6

51Î6

" #

sin

51 ‰ 3

1  ˆ 12


" #

sin 13 ‰ œ

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661

662

Chapter 10 Conic Sections and Polar Coordinates

9. r œ 2 and r œ 2(1
cos )) Ê 2 œ 2(1
cos )) Ê cos ) œ 0 Ê ) œ „ 1# ; therefore A œ 2 '0

1Î2

œ '0

1Î2

œ '0

1Î2

œ '0

1Î2

" #

[2(1
cos ))]# d)  "# area of the circle

4 a1
2 cos )  cos# )b d)  ˆ "# 1‰ (2)# 4 ˆ1
2 cos ) 

1  cos 2) ‰ #

d)  2 1

(4
8 cos )  2  2 cos 2)) d)  21 1Î#

œ c6)
8 sin )  sin 2)d !

 21 œ 51
8

10. r œ 2(1
cos )) and r œ 2(1  cos )) Ê 1
cos ) œ 1  cos ) Ê cos ) œ 0 Ê ) œ 1# or 3#1 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0

1Î2

" #

[2(1
cos ))]# d)  2 '1Î2 "# [2(1  cos ))]# d) 1

œ '0

4 a1
2 cos )  cos# )b d)

œ '0

4 ˆ1
2 cos ) 

œ '0

(6
8 cos )  2 cos 2)) d)  '1Î2 (6  8 cos )  2 cos 2)) d)

1Î2

 '1Î2 4 a1  2 cos )  cos# )b d) 1

1Î2 1Î2

1  cos 2) ‰ #

d)  '1Î2 4 ˆ1  2 cos )  1

1  cos 2) ‰ #

d)

1

1Î#

œ c6)
8 sin )  sin 2)d !

 c6)  8 sin )  sin 2)d 11Î# œ 61
16

11. r œ È3 and r# œ 6 cos 2) Ê 3 œ 6 cos 2) Ê cos 2) œ 1 6

Ê )œ

" #

(in the 1st quadrant); we use symmetry of the

graph to find the area, so A œ 4 '0 ” "# (6 cos 2))
"# ŠÈ3‹ • d) #

1Î6

œ 2 '0 (6 cos 2)
3) d) œ 2 c3 sin 2)
3)d ! 1Î6

1Î'

œ 3È3
1 12. r œ 3a cos ) and r œ a(1  cos )) Ê 3a cos ) œ a(1  cos )) Ê 3 cos ) œ 1  cos ) Ê cos ) œ "# Ê ) œ 13 or
13 ; the graph also gives the point of intersection (0ß 0); therefore A œ 2 '0

1Î3

" #

c(3a cos ))#
a# (1  cos ))# d d)

œ '0 a9a# cos# )
a#
2a# cos )
a# cos# )b d) 1Î3

œ '0

1Î3

a8a# cos# )
2a# cos )
a# b d)

œ '0 c4a# (1  cos 2))
2a# cos )
a# d d) 1Î3

œ '0 a3a#  4a# cos 2)
2a# cos )b d) 1Î3

1Î$

œ c3a# )  2a# sin 2)
2a# sin )d !

œ 1a#  2a# ˆ "# ‰
2a# Š

È3 # ‹

œ a# Š1  1
È3‹

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Section 10.7 Area and Lengths in Polar Coordinates

663

13. r œ 1 and r œ
2 cos ) Ê 1 œ
2 cos ) Ê cos ) œ
"# Ê )œ

A œ 2'

1

21 3

in quadrant II; therefore c(
2 cos ))#
1# d d) œ '21Î3 a4 cos# )
1b d) 1

" 21Î3 #

œ '21Î3 [2(1  cos 2))
1] d) œ '21Î3 (1  2 cos 2)) d) 1

1

œ c)  sin 2)d 1#1Î$ œ 14. (a) A œ 2 '0

21Î3

œ '0

21Î3

" #

1 3



È3 #

(2 cos )  1)# d) œ '0

21Î3

a4 cos# )  4 cos )  1b d) œ '0

21Î3

#1Î$

(3  2 cos 2)  4 cos )) d) œ c3)  sin 2)  4 sin )d ! 3È 3 # ‹

(b) A œ Š21 


Š1


3È 3 # ‹

1 6

; therefore A œ '1Î6

51Î6

51 6

or

œ '1Î6 ˆ18
51Î6

9 #

È3 #



4È 3 #

œ 21 

3È 3 #

œ 1  3È3 (from 14(a) above and Example 2 in the text) " #

15. r œ 6 and r œ 3 csc ) Ê 6 sin ) œ 3 Ê sin ) œ Ê )œ

œ 21


[2(1  cos 2))  4 cos )  1] d)

csc# )‰ d) œ
18) 

" #

a6#
9 csc# )b d)

9 #

cot )‘ 1Î'

&1Î'

œ Š151
9# È3‹
Š31  9# È3‹ œ 121
9È3

16. r# œ 6 cos 2) and r œ Ê

sec ) Ê

3 # #

9 4 %

sec# ) œ 6 cos 2) Ê

œ 2 cos% )
cos ) Ê 2 cos )
cos# )


3 8

1 6

(in the first quadrant); thus A œ 2 '0

œ
3 sin 2)


9 4

tan )‘ !

1Î6

1Î'

17. (a) r œ tan ) and r œ Š

œ 3Š

È2 # ‹

È3 # ‹




9 4È 3

œ

3È 3 #

csc ) Ê tan ) œ Š

" ˆ # 6 cos 2) 3È 3 3È 3 4 œ 4




È2 # ‹

È2 # ‹

cos ) Ê 1
cos# ) œ Š

Ê cos# )  Š

È2 # ‹

cos )
1 œ 0 Ê cos ) œ
È2 or 1 4

œ acos# )b a2 cos# )
1b




9 4

È3 #

(the second equation has no real

sec# )‰ d) œ '0 ˆ6 cos 2)
1Î6

9 4

sec# )‰ d)

csc )

Ê sin# ) œ Š

(use the quadratic formula) Ê ) œ

3 8

or cos# ) œ
"4 Ê cos ) œ „

3 4

roots) Ê ) œ

È2 #

œ cos# ) cos 2) Ê

œ 0 Ê 16 cos% )
8 cos# )
3 œ 0

3 8

Ê a4 cos# )  1ba4 cos# )
3b œ 0 Ê cos# ) œ

9 24

È2 # ‹

cos )

(the solution

in the first quadrant); therefore the area of R" is A" œ '0

1Î4

AO œ Š

" #

È2 # ‹

tan# ) d) œ 1 #

csc

œ

È2 #

" #

and OB œ Š

Ê the area of R# is A# œ 2 ˆ "#


1 8

 4" ‰ œ

3 #




1 4

'01Î4 asec# )
1b d)

" #

Š

rœ (b)

lim

lim

œ

" 4

1 4

1Î%

ctan )
)d !

œ

" #

ˆtan

œ 1 Ê AB œ Ê1#
Š

1 4


14 ‰ œ

È2 # ‹

#

œ

) Ä 1 Î2 c

r œ sec ) as ) Ä

" ‰ cos ) 1c #




1 8

;

; therefore the area of the region shaded in the text is

1 4

generates the arc OB of r œ tan ) but does not generate the segment AB of the line

tan ) œ _ and the line x œ 1 is r œ sec ) in polar coordinates; then sin ) ˆ cos )


" #

È2 #

csc ). Instead the interval generates the half-line from B to _ on the line r œ

) Ä 1 Î2


=

È2 È2 # ‹Š # ‹

csc

" #

. Note: The area must be found this way since no common interval generates the region. For

example, the interval 0 Ÿ ) Ÿ È2 #

È2 # ‹

œ

œ

lim

) Ä 1 Î2 c

ˆ sincos)
) 1 ‰ œ

lim

) Ä 1 Î2 c

lim

) Ä 1 Î2 c

È2 #

csc ).

(tan )
sec ))

) ‰ ˆ
cos sin ) œ 0 Ê r œ tan ) approaches

Ê r œ sec ) (or x œ 1) is a vertical asymptote of r œ tan ). Similarly, r œ
sec )

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

664

Chapter 10 Conic Sections and Polar Coordinates (or x œ
1) is a vertical asymptote of r œ tan ).

18. It is not because the circle is generated twice from ) œ 0 to 21. The area of the cardioid is A œ 2 '0

1

" #

2) (cos )  1)# d) œ '0 acos# )  2 cos )  1b d) œ '0 ˆ 1  cos  2 cos )  1‰ d) # 1

œ
32)  31 #

sin 2) 4 1 51 œ 4 4




1

 2 sin )‘ ! œ

19. r œ )# , 0 Ÿ ) Ÿ È5 Ê

#

È5

È5

œ '0 k)k È)#  4 d) œ (since )   0) '0

) œ È5 Ê u œ 9“ Ä '4

9

20. r œ

e) È2

,0Ÿ)Ÿ1 Ê

dr d)

" #

œ

1 4

. The area of the circle is A œ 1 ˆ "# ‰ œ

œ 2); therefore Length œ '0

dr d)

È5

31 #

1

Èu du œ

e) È2

" #

Ê the area requested is actually

Éa)# b#  (2))# d) œ '

0

) È ) #  4 d ) ;
u œ ) #  4 Ê


23 u$Î# ‘ * œ %

È5

" #

È ) %  4) # d)

du œ ) d); ) œ 0 Ê u œ 4,

19 3

; therefore Length œ '0 ÊŠ Èe 2 ‹  Š Èe 2 ‹ d) œ '0 Ê2 Š e# ‹ d) 1

#

)

#

)

1

2)

œ '0 e) d) œ
e) ‘ ! œ e1
1 1

1

21. r œ 1  cos ) Ê

dr d)

œ
sin ); therefore Length œ '0 È(1  cos ))#  (
sin ))# d) 21

1 œ 2 '0 È2  2 cos ) d) œ 2'0 É 4(1 #cos )) d) œ 4 '0 É 1  #cos ) d) œ 4 '0 cos ˆ #) ‰ d) œ 4
2 sin 2) ‘ ! œ 8 1

1

22. r œ a sin#

) #

1

, 0 Ÿ ) Ÿ 1, a  0 Ê

œ '0 Éa# sin% 1

) #

 a# sin#

) #

dr d) ) #

cos#

œ a sin

) #

cos

) #

1

# ; therefore Length œ '0 Ɉa sin# #) ‰  ˆa sin 1

d) œ '0 a ¸sin #) ¸ Ésin# 1

) #

) #

 cos#

1

6 1  cos )

œ '0

1Î2

,0Ÿ)Ÿ

1 #

É (1  36 cos ))# 

œ ˆsince

" 1  cos )

Ê

dr d)

œ

; therefore Length œ '0

1Î2

6 sin ) (1  cos ))#

d) œ 6 '0

1Î2

36 sin# ) a1  cos )b%

" ¸ 1cos ¸ ) É1 

0

1Î2

1Î2

1Î2

1Î2

) #

d)

cos# )  sin# )  0 on 0 Ÿ ) Ÿ 1# ‰ 6 '0 ˆ 1  "cos ) ‰ É 1  2 cos(1)cos d) ) )#

cos ) È ' œ 6 '0 ˆ 1  "cos ) ‰ É (12 2cos ) )# d) œ 6 2 0

œ 3'0 sec$

#

#

6 sin ) ʈ 1  6cos ) ‰  Š (1  cos ))# ‹ d)

sin# ) (1  cos ))#

d) œ 6'0

1Î4

d) (1  cos ))$Î#

œ 6È2 '0

1Î2

1Î% sec$ u du œ (use tables) 6 Œ
sec u2tan u ‘ ! 

d) ˆ2 cos# #) ‰$Î# " #

'01Î4

œ 3'0

1Î2

¸sec$ #) ¸ d)

sec u du

1Î% œ 6 Š È"2 
2" ln ksec u  tan uk‘ ! ‹ œ 3 ’È2  ln Š1  È2‹“

24. r œ

2 1
cos )

,

1 #

Ÿ)Ÿ1 Ê

4 œ '1Î2 Ê (1
cos ) ) # Š1  1

dr d)

œ


2 sin ) (1
cos ))#

sin# ) ‹ a1
cos )b#

œ ˆsince 1
cos )   0 on

1 #

sin ) ; therefore Length œ '1Î2 ʈ 1
2cos ) ‰  Š (1

2cos ))# ‹ d) 1

1

#

1

œ '1Î2 csc$ ˆ #) ‰ d) œ ˆsince csc 1

) #

cos ) d) È ' È ' œ 2 '1Î2 ˆ 1
"cos ) ‰ É (12

2cos ))# d) œ 2 2 1Î2 (1
cos ))$Î# œ 2 2 1Î2 1

#

d)

cos )  sin Ÿ ) Ÿ 1‰ 2 '1Î2 ˆ 1
"cos ) ‰ É 1
2 cos(1)
cos ) )# 1

) #

  0 on

1 #

1

#

#

)  sin d) œ '1Î2 ¸ 1
2cos ) ¸ É (1
(1cos
)cos ) )#

#

d) ˆ2 sin# )# ‰$Î#

)

d)

œ '1Î2 ¸csc$ #) ¸ d) 1

Ÿ ) Ÿ 1‰ 2 '1Î4 csc$ u du œ (use tables) 1Î2

# cos #) ‰ d)

d) œ (since 0 Ÿ ) Ÿ 1) a ' sin ˆ #) ‰ d)

1 œ

2a cos 2) ‘ ! œ 2a

23. r œ

) #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.7 Area and Lengths in Polar Coordinates 1Î# 2Œ

csc u2cot u ‘ 1Î% 

'11ÎÎ42

" #

1Î#

csc u du œ 2 Š È"2

2" ln kcsc u  cot uk‘ 1Î% ‹ œ 2 ’ È"2 

" #

ln ŠÈ2  1‹“

œ È2  ln Š1  È2‹ ) 3

25. r œ cos$ œ '0

Ê

dr d)

) 3

œ
sin

; therefore Length œ '0

1Î4

) 3

cos#

Écos' ˆ 3) ‰  sin# ˆ 3) ‰ cos% ˆ 3) ‰ d) œ '

1Î4

1Î4 1cos ˆ 2) ‰ 3

#

d) œ

" #


) 

3 2

2) ‘ 1Î% 3 !

sin

26. r œ È1  sin 2) , 0 Ÿ ) Ÿ 1È2 Ê 1È 2

Length œ '0 œ '0

È

1 2

É(1  sin 2)) 

sin 2) ' É 212sin 2) d) œ 0

27. r œ È1  cos 2) Ê 1È 2

œ '0

È

1 2

œ

dr d)

" #

cos# 2) (1  sin 2))



È

d) œ '0

1 2

1È 2

d) œ '0

2)

1È# !

#

sin 2)  cos É 1  2 sin 2)1   sin 2)

#

2)

d)

œ 21

È

1È 2

cos 2) ' É 212cos 2 ) d) œ 0

21

É(1  cos 2)) 

È2 d) œ ’È2 )“

œ 0; Length œ '0 Èa#  0# d) œ '0 kak d) œ ca)d #!1 œ 21a

1È# !

sin# 2) (1  cos 2))

21

œ
a sin ); Length œ '0 È(a cos ))#  (
a sin ))# d) œ '0 Èa# acos# )  sin# )b d)

dr d)

œ a cos ); Length œ '0 È(a cos ))#  (a sin ))# d) œ '0 Èa# acos# )  sin# )b d)

1

œ '0 kak d) œ ca)d 1! œ 1a 1

d)

œ 21

dr d)

(b) r œ a cos ) Ê

(c) r œ a sin ) Ê

cos# ˆ 3) ‰ d)

(1  sin 2))
"Î# (2 cos 2)) œ (cos 2))(1  sin 2))
"Î# ; therefore

È2 d) œ ’È2 )“

#

1Î4

3 8

1 2

#

dr d)

" #

œ

# cos# 3) ‰ d) 0

(1  cos 2))
"Î# (
2 sin 2)); therefore Length œ '0

cos 2)  sin É 1  2 cos 21) cos 2)

28. (a) r œ a Ê

dr d)

1 8

œ

) 3

ˆcos# 3) ‰ Écos# ˆ 3) ‰  sin# ˆ 3) ‰ d) œ '

1Î4

0

œ '0

Ɉcos$ 3) ‰#  ˆ
sin

1

1

1

œ '0 kak d) œ ca)d 1! œ 1a 1

29. r œ Ècos 2) , 0 Ÿ ) Ÿ œ '0

1Î4

1 4

Ê

dr d)

œ

" #

(cos 2))
"Î# (
sin 2))(2) œ


sin 2) Ècos 2)

; therefore Surface Area


sin 2) (21r cos )) ÊŠÈcos 2)‹  Š È ‹ d) œ '0 Š21Ècos 2)‹ (cos ))Écos 2)  cos 2) #

#

œ '0 Š21Ècos 2)‹ (cos ))É cos" 2) d) œ '0 1Î4

1Î4

30. r œ È2e)Î2 , 0 Ÿ ) Ÿ

1 #

Ê

dr d)

œ È2 ˆ "# ‰ e)Î2 œ

œ '0 Š21È2 e)Î2 ‹ (sin )) ÊŠÈ2 e)Î2 ‹  Š #

1Î2

1Î4

1Î%

21 cos ) d) œ c21 sin )d ! È2 #

È2 #

sin# 2) cos 2)

d)

œ 1È2

e)Î2 ; therefore Surface Area

e)Î2 ‹ d) œ '0 Š21È2 e)Î2 ‹ (sin )) É2e)  "# e) d) #

1Î2

œ '0 Š21È2 e)Î2 ‹ (sin )) É 5# e) d) œ '0 Š21È2 e)Î2 ‹ (sin )) Š È52 e)Î2 ‹ d) œ 21È5 '0 e) sin ) d) 1Î2

1Î2

)

1Î#

œ 21È5
e2 (sin )
cos ))‘ !

œ 1È5 ae1Î2  1b where we integrated by parts

31. r# œ cos 2) Ê r œ „ Ècos 2) ; use r œ Ècos 2) on
0ß 14 ‘ Ê therefore Surface Area œ 2 '0 Š21Ècos 2)‹ (sin )) Écos 2)  1Î4

œ 41 '0 sin ) d) œ 41 c
cos )d ! 1Î4

1Î2

È

1Î%

œ 41 ’


È2 #

dr d)

œ

sin# 2) cos 2)

" #

(cos 2))
"Î# (
sin 2))(2) œ

d) œ 41 '0

1Î4


sin 2) Ècos 2)

Ècos 2) (sin )) É


(
1)“ œ 21 Š2
È2‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

" cos 2)

d)

665

666

Chapter 10 Conic Sections and Polar Coordinates

32. r œ 2a cos ) Ê

dr d)

œ
2a sin ); therefore Surface Area œ '0 21(2a cos ))(cos ))È(2a cos ))#  (
2a sin ))# d) 1

œ 4a1 '0 acos# )b È4a# acos# )  sin# )b d) œ 8a1 '0 acos# )b kak d) œ 8a# 1 '0 cos# ) d) 1

1

1

2) ‰ œ 8a# 1 '0 ˆ 1  cos d) œ 4a# 1 '0 (1  cos 2)) d) œ 4a# 1
)  # 1

1

33. Let r œ f()). Then x œ f()) cos ) Ê

" 2

1

sin 2)‘ ! œ 4a# 1#

‰# œ cf w ()) cos )
f()) sin )d# œ f w ()) cos )
f()) sin ) Ê ˆ dx d)

dx d)

œ cf w ())d# cos# )
2f w ()) f()) sin ) cos )  [f())]# sin# ); y œ f()) sin ) Ê

dy d)

#

œ f w ()) sin )  f()) cos )

# # w w # w # # Ê Š dy d) ‹ œ cf ()) sin )  f()) cos )d œ cf ())d sin )  2f ())f()) sin ) cos )  [f())] cos ). Therefore #

# # w # # # # # w # # ˆ dx ‰#  Š dy ˆ dr ‰# d) d) ‹ œ cf ())d acos )  sin )b  [f())] acos )  sin )b œ cf ())d  [f())] œ r  d) .

' Ér#  ˆ ddr) ‰# d). ‰#  Š dy Thus, L œ '! ʈ dx d) d) ‹ d) œ ! "

"

#

'021 a(1
cos )) d) œ 2a1 c)
sin )d #!1 œ a 21 rav œ 21"
0 '0 a d) œ #"1 ca)d #!1 œ a 1Î2 1Î# rav œ ˆ 1 ‰
"ˆ
1 ‰ 'c1Î2 a cos ) d) œ 1" ca sin )d
1Î# œ 2a 1

34. (a) rav œ (b) (c)

" 2 1
0

#

#

35. r œ 2f()), ! Ÿ ) Ÿ " Ê

dr d)

œ 2f w ()) Ê r#  ˆ ddr) ‰ œ [2f())]#  c2f w ())d# Ê Length œ '! É4[f())]#  4 cf w ())d# d) "

#

œ 2 '! É[f())]#  cf w ())d# d) which is twice the length of the curve r œ f()) for ! Ÿ ) Ÿ " . "

36. Again r œ 2f()) Ê r#  ˆ ddr) ‰ œ [2f())]2  c2f w ())d# Ê Surface Area œ '! 21[2f()) sin )] É4[f())]#  4 cf w ())d# d) "

#

œ 4 '! 21[f()) sin )] É[f())]#  cf w ())d# d) which is four times the area of the surface generated by revolving "

r œ f()) about the x-axis for ! Ÿ ) Ÿ " . '021 r$ cos ) d) 37. x œ œ '021 r# d) 2 3

œ

2 3

2 3

'021 [a(1  cos ))]$ (cos )) d) œ '021 [a(1  cos ))]# d)

2 3

a$

'021 a1  3 cos )  3 cos# )  cos$ )b (cos )) d) 21 a# '0 a1  2 cos )  cos# )b d) #

2) 1
cos 2) # a '0 ’cos )  3 ˆ 1
cos # ‰  3 a1
sin )b (cos ))  ˆ # ‰ “ d) 21

'0

21

2)
1  2 cos )  ˆ 1
cos # ‰‘ d )

œ (After considerable algebra using

" 4 # ' 15 8 1  cos 2A ‰ a 0 ˆ 12  3 cos )  3 cos 2)
2 cos ) sin )  12 cos 4)‰ d) 21 # " 3 ' ˆ #  2 cos )  # cos 2)‰ d) 21

the identity cos# A œ

0

œ

#1

" 8 2 2 $ ‘ a
15 12 )  3 sin )  3 sin 2)
3 sin )  48 sin 4) !
#3 )  2 sin )  "4 sin 2)‘ #1

yœ

!

2 3

2 3

'2a2a
"a u$ du 31

œ

0 31

'01 r# d) œ '01 a# d) œ ca# )d !1 œ a# 1; x œ yœ

2 3

œ

5 6

a;

21 2' $ '021 r$ sin ) d) 3 0 [a(1  cos ))] (sin )) d) œ ;
u œ a(1  cos )) Ê
"a du œ sin ) d); ) œ 0 Ê u œ 2a; 21 31 '0 r # d )

) œ 21 Ê u œ 2ad Ä 38.

œ

‰ a ˆ 15 6 1 31

2' $ '0 r$ sin ) d) 3 0 a sin ) d) œ a# 1 '01 r# d) œ 1

1

2 3

œ 0. Therefore the centroid is aBß yb œ ˆ 56 aß 0‰ 2 3

'01 r$ cos ) d) œ '01 r# d)

a$ c
cos )d 1! a# 1

œ

ˆ 43 ‰ a$ a# 1

œ

2 3

'01 a$ cos ) d)

4a 31 .

a# 1

œ

2 3

a$ c sin )d 1! a# 1

œ

0 a# 1

œ 0;

Therefore the centroid is axß yb œ ˆ0ß 34a1 ‰ .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.8 Conic Sections in Polar Coordinates

667

10.8 CONIC SECTIONS IN POLAR COORDINATES 1. r cos ˆ)
16 ‰ œ 5 Ê r ˆcos ) cos œ 10 Ê y œ
È3 x  10 2. r cos ˆ)
Ê


È2 #

3. r cos ˆ)
Ê
1# x

31 ‰ 4

œ 2 Ê r ˆcos ) cos

x

È2 #

1 6

 sin ) sin 16 ‰ œ 5 Ê

31 4

31 ‰ 4

 sin ) sin

È3 #

œ2 Ê


È2 #

41 ‰ œ3 Ê 3 È3
# yœ3

r cos )


È2 #

r cos ) 

È2 #

È3 #

r ˆcos ) cos

41 3

41 ‰ 3

 sin ) sin

œ 3 Ê
#1 r cos )
È3 3

Ê x  È 3 y œ
6 Ê y œ


È2 #

È2 #

r sin ) œ 4 Ê

x


È2 #

r sin ) œ 2

È2

È2

È3 #

r sin ) œ 3

x
2È3 1 4


sin ) sin 14 ‰ œ 4

y œ 4 Ê È2 x
È2 y œ 8 Ê y œ x
4È2

5. r cos ˆ)
14 ‰ œ È2 Ê r ˆcos ) cos 14  sin ) sin 14 ‰ œ È2 Ê " r cos )  " r sin ) œ È2 Ê " x  È2

" È2

y

œ È2 Ê x  y œ 2 Ê y œ 2
x

6. r cos ˆ)  Ê


31 ‰ 4

œ 1 Ê r ˆcos ) cos

È2 2

r cos )
Ê y œ
x
È 2

7. r cos ˆ)


21 ‰ 3

È2 2

È3 2

31 4


sin ) sin

31 ‰ 4

œ1

r sin ) œ 1 Ê x  y œ
È2

œ 3 Ê r ˆcos ) cos

Ê
r cos )  1 2

x  "# y œ 5 Ê È3 x  y

y œ 2 Ê
È2 x  È2 y œ 4 Ê y œ x  2È2

4. r cos ˆ)
ˆ
14 ‰‰ œ 4 Ê r cos ˆ)  14 ‰ œ 4 Ê r ˆcos ) cos Ê

r cos )  "# r sin ) œ 5 Ê

21 3

 sin ) sin " #

r sin ) œ 3 Ê
x 

Ê
x  È 3 y œ 6 Ê y œ

È3 3

È3 #

21 ‰ 3

œ3

yœ3

x  2È 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

668

Chapter 10 Conic Sections and Polar Coordinates

8. r cos ˆ)  13 ‰ œ 2 Ê r ˆcos ) cos Ê

1 2

r cos )


È3 2

r sin ) œ 2 Ê

Ê x
È3 y œ 4 Ê y œ

È3 3

1 3


sin ) sin 13 ‰ œ 2

" #

x


x


È3 #

yœ2

4È 3 3

È 9. È2 x  È2 y œ 6 Ê È2 r cos )  È2 r sin ) œ 6 Ê r Š #2 cos ) 

È2 #

sin )‹ œ 3 Ê r ˆcos

1 4

cos )  sin

œ 3 Ê r cos ˆ)
14 ‰ œ 3 È 10. È3 x
y œ 1 Ê È3 r cos )
r sin ) œ 1 Ê r Š #3 cos )


œ

" #

Ê r cos ˆ)  16 ‰ œ

1 #

sin )‹ œ

" #

Ê r ˆcos

1 6

cos )
sin

1 6

sin )‰

" #

11. y œ
5 Ê r sin ) œ
5 Ê
r sin ) œ 5 Ê r sin (
)) œ 5 Ê r cos ˆ 1#
(
))‰ œ 5 Ê r cos ˆ)  1# ‰ œ 5 12. x œ
4 Ê r cos ) œ
4 Ê
r cos ) œ 4 Ê r cos ()
1) œ 4 13. r œ 2(4) cos ) œ 8 cos )

14. r œ
2(1) sin ) œ
2 sin )

15. r œ 2È2 sin )

16. r œ
2 ˆ "# ‰ cos ) œ
cos )

17.

18.

19.

20.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 4

sin )‰

Section 10.8 Conic Sections in Polar Coordinates 21. (x
6)#  y# œ 36 Ê C œ (6ß 0), a œ 6 Ê r œ 12 cos ) is the polar equation

22. (x  2)#  y# œ 4 Ê C œ (
2ß 0), a œ 2 Ê r œ
4 cos ) is the polar equation

23. x#  (y
5)# œ 25 Ê C œ (!ß 5), a œ 5 Ê r œ 10 sin ) is the polar equation

24. x#  (y  7)# œ 49 Ê C œ (!ß
7), a œ 7 Ê r œ
14 sin ) is the polar equation

25. x#  2x  y# œ 0 Ê (x  1)#  y# œ 1 Ê C œ (
1ß 0), a œ 1 Ê r œ
2 cos ) is the polar equation

26. x#
16x  y# œ 0 Ê (x
8)#  y# œ 64 Ê C œ (8ß 0), a œ 8 Ê r œ 16 cos ) is the polar equation

# 27. x#  y#  y œ 0 Ê x#  ˆy  "# ‰ œ 4" Ê C œ ˆ!ß
"# ‰ , a œ "# Ê r œ
sin ) is the

# 28. x#  y#
43 y œ 0 Ê x#  ˆy
23 ‰ œ 49 Ê C œ ˆ0ß 23 ‰ , a œ 23 Ê r œ 43 sin ) is the

polar equation

polar equation

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669

670

Chapter 10 Conic Sections and Polar Coordinates

29. e œ 1, x œ 2 Ê k œ 2 Ê r œ

2(1) 1  (1) cos )

œ

2 1cos )

30. e œ 1, y œ 2 Ê k œ 2 Ê r œ

2(1) 1  (1) sin )

œ

2 1sin )

31. e œ 5, y œ
6 Ê k œ 6 Ê r œ

6(5) 1
5 sin )

32. e œ 2, x œ 4 Ê k œ 4 Ê r œ

4(2) 1  2 cos )

33. e œ "# , x œ 1 Ê k œ 1 Ê r œ

ˆ "# ‰ (1) 1  ˆ "# ‰ cos )

35. e œ "5 , x œ
10 Ê k œ 10 Ê r œ

37. r œ

" 1  cos )

38. r œ

6 2  cos )

œ

30 1
5 sin )

8 12 cos )

œ

ˆ "4 ‰ (2) 1
ˆ "4 ‰ cos )

34. e œ 4" , x œ
2 Ê k œ 2 Ê r œ

36. e œ "3 , y œ 6 Ê k œ 6 Ê r œ

œ

1 2cos )

œ

ˆ "5 ‰ (10) 1
ˆ "5 ‰ sin )

ˆ "3 ‰ (6) 1  ˆ "3 ‰ sin )

œ

2 4
cos )

œ

10 5
sin )

6 3sin )

Ê e œ 1, k œ 1 Ê x œ 1

œ

3 1  ˆ "# ‰ cos )

Ê eœ

" #

, k œ 6 Ê x œ 6;

#

a a1
e# b œ ke Ê a ’1
ˆ "# ‰ “ œ 3 Ê

3 4

aœ3

Ê a œ 4 Ê ea œ 2

39. r œ

25 10
5 cos )

Ê eœ

" #

Ê rœ

œ

ˆ #5 ‰

1
ˆ "# ‰ cos )

, k œ 5 Ê x œ
5; a a1
e# b œ ke #

Ê a ’1
ˆ "# ‰ “ œ

40. r œ

ˆ 25 ‰ 10

5 ‰ 1
ˆ 10 cos )

4 2
2 cos )

Ê rœ

5 #

Ê

2 1
cos )

3 4

aœ

5 #

Ê aœ

10 3

Ê ea œ

5 3

Ê e œ 1, k œ 2 Ê x œ
2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.8 Conic Sections in Polar Coordinates 41. r œ eœ

400 16
8 sin ) " #

Ê rœ

ˆ 400 ‰ 16

8 ‰ 1
ˆ 16 sin )

25 1
ˆ "# ‰ sin )

, k œ 50 Ê y œ 50; a a1
e# b œ ke #

Ê a ’1
ˆ "# ‰ “ œ 25 Ê Ê ea œ

42. r œ

Ê rœ

a œ 25 Ê a œ

3 4

100 3

50 3

12 3
3 sin )

Ê rœ

4 1
sin )

Ê e œ 1,

43. r œ

kœ4 Ê yœ4

44. r œ

4 2  sin )

Ê rœ

8 2  2 sin )

Ê rœ

4 1  sin )

Ê e œ 1,

k œ 4 Ê y œ
4

2 1  ˆ "# ‰ sin )

Ê eœ

" #

,kœ4 #

Ê y œ
4; a a1
e# b œ ke Ê a ’1
ˆ "# ‰ “ œ 2 Ê

3 4

aœ2 Ê aœ

8 3

Ê ea œ

4 3

45.

46.

47.

48.

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671

672

Chapter 10 Conic Sections and Polar Coordinates

49.

50.

51.

52.

53.

54.

55.

56.

57. (a) Perihelion œ a
ae œ a(1
e), Aphelion œ ea  a œ a(1  e) (b) Planet Perihelion Aphelion Mercury 0.3075 AU 0.4667 AU Venus 0.7184 AU 0.7282 AU Earth 0.9833 AU 1.0167 AU Mars 1.3817 AU 1.6663 AU Jupiter 4.9512 AU 5.4548 AU Saturn 9.0210 AU 10.0570 AU Uranus 18.2977 AU 20.0623 AU Neptune 29.8135 AU 30.3065 AU Pluto 29.6549 AU 49.2251 AU

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 10.8 Conic Sections in Polar Coordinates (0.3871) a1  0.2056# b 0.3707 œ 1
0.2056 1
0.2056 cos ) cos ) (0.7233) a1  0.0068# b 0.7233 Venus: r œ 1
0.0068 cos ) œ 1
0.0068 cos )  0.0167# b 0.9997 Earth: r œ 11
a10.0167 cos ) œ 1
0.0617 cos ) a1  0.0934# b 1.511 Mars: r œ (1.524) œ 1
0.0934 1
0.0934 cos ) cos ) (5.203) a1  0.0484# b 5.191 Jupiter: r œ 1
0.0484 cos ) œ 1
0.0484 cos ) a1  0.0543# b 9.511 Saturn: r œ (9.539) œ 1
0.0543 1
0.0543 cos ) cos ) (19.18) a1  0.0460# b 19.14 Uranus: r œ 1
0.0460 cos ) œ 1
0.0460 cos ) a1  0.0082# b 30.06 Neptune: r œ (30.06) œ 1
0.0082 1
0.0082 cos ) cos )

58. Mercury: r œ

59. (a) r œ 4 sin ) Ê r# œ 4r sin ) Ê x#  y# œ 4y; È r œ È3 sec ) Ê r œ cos3) Ê r cos ) œ È3

(b)

#

Ê x œ È3 ; x œ È3 Ê ŠÈ3‹  y# œ 4y Ê y#
4y  3 œ 0 Ê (y
3)(y
1) œ 0 Ê y œ 3 or y œ 1. Therefore in Cartesian coordinates, the points of intersection are ŠÈ3ß 3‹ and ŠÈ3ß 1‹. In polar coordinates, 4 sin ) œ È3 sec ) Ê 4 sin ) cos ) œ È3 Ê 2 sin ) cos ) œ 21 3

Ê )œ

1 6

1 3

or

È3 #

;)œ

Ê sin 2) œ 1 6

È3 #

Ê 2) œ

Ê r œ 2, and ) œ

1 3

or

1 3

Ê r œ 2È3 Ê ˆ2ß 16 ‰ and Š2È3ß 13 ‹ are the points of intersection in polar coordinates. 60. (a) r œ 8 cos ) Ê r# œ 8r cos ) Ê x#  y# œ 8x Ê x#
8x  y# œ 0 Ê (x
4)#  y# œ 16; r œ 2 sec ) Ê r œ cos2 ) Ê r cos ) œ 2

(b)

Ê x œ 2; x œ 2 Ê 2#
8(2)  y# œ 0 Ê y# œ 12 Ê y œ „ 2È3. Therefore Š2ß „ 2È3‹

are the points of intersection in Cartesian coordinates. In polar coordinates, 8 cos ) œ 2 sec ) Ê 8 cos# ) œ 2 Ê cos# ) œ "4 Ê cos ) œ „ #" Ê ) œ 13 , 231 , 431 , or 51 3

Ê r œ 4, and ) œ 231 and 431 Ê r œ
4 Ê ˆ4ß 13 ‰ and ˆ4ß 531 ‰ are the points of intersection in polar coordinates. The points ˆ
4ß 231 ‰ and ˆ
4ß 431 ‰ are the same points. ;)œ

1 3

and

51 3

61. r cos ) œ 4 Ê x œ 4 Ê k œ 4: parabola Ê e œ 1 Ê r œ 62. r cos ˆ)
1# ‰ œ 2 Ê r ˆcos ) cos Ê rœ

1 #

4 1
cos )

 sin ) sin 1# ‰ œ 2 Ê r sin ) œ 2 Ê y œ 2 Ê k œ 2: parabola Ê e œ 1

2 1
sin )

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

673

674

Chapter 10 Conic Sections and Polar Coordinates

63. (a) Let the ellipse be the orbit, with the Sun at one focus.  rmin Then rmax œ a  c and rmin œ a
c Ê rrmax max
rmin œ

(a
c)  (a  c) (a
c)
(a  c)

œ

2c 2a

œ

c a

œe

(b) Let F" , F# be the foci. Then PF"  PF# œ 10 where P is any point on the ellipse. If P is a vertex, then PF" œ a  c and PF# œ a
c Ê (a  c)  (a
c) œ 10 Ê 2a œ 10 Ê a œ 5. Since e œ ca we have 0.2 œ

c 5

Ê c œ 1.0 Ê the pins should be 2 inches apart. 64. e œ 0.97, Major axis œ 36.18 AU Ê a œ 18.09, Minor axis œ 9.12 AU Ê b œ 4.56 (1 AU ¸ 1.49 ‚ 10) km) (a) r œ

ke 1
e cos )

œ

(b) ) œ 0 Ê r œ (c) ) œ 1 Ê r œ

(18.09) c1(0.97)# d a a1  e # b 1.07 œ 1
0.97 1
e cos ) œ 1
0.97 cos ) cos ) AU 1.07 ( 1
0.97 ¸ 0.5431 AU ¸ 8.09 ‚ 10 km 1.07 * 10.97 ¸ 35.7 AU ¸ 5.32 ‚ 10 km

65. x#  y#
2ay œ 0 Ê (r cos ))#  (r sin ))#
2ar sin ) œ 0 Ê r# cos# )  r# sin# )
2ar sin ) œ 0 Ê r# œ 2ar sin ) Ê r œ 2a sin )

66. y# œ 4ax  4a# Ê (r sin ))# œ 4ar cos )  4a# Ê r# sin# ) œ 4ar cos )  4a# Ê r# a1
cos# )b œ 4ar cos )  4a# Ê r#
r# cos# ) œ 4ar cos )  4a# Ê r# œ r# cos# )  4ar cos )  4a# Ê r# œ (r cos )  2a)# Ê r œ „ (r cos )  2a) Ê r
r cos ) œ 2a or 2a r  r cos ) œ
2a Ê r œ 12a cos ) or r œ 1
cos ) ; the equations have the same graph, which is a parabola opening to the right 67. x cos !  y sin ! œ p Ê r cos ) cos !  r sin ) sin ! œ p Ê r(cos ) cos !  sin ) sin !) œ p Ê r cos ()
!) œ p

#

68. ax#  y# b  2ax ax#  y# b
a# y# œ 0 Ê Ê Ê Ê Ê Ê Ê

#

ar# b  2a(r cos )) ar# b
a# (r sin ))# œ 0 r%  2ar$ cos )
a# r# sin# ) œ 0 r# cr#  2ar cos )
a# a1
cos# )bd œ 0 (assume r Á 0) r#  2ar cos )
a#  a# cos# ) œ 0 ar#  2ar cos )  a# cos# )b
a# œ 0 (r  a cos ))# œ a# Ê r  a cos ) œ „ a r œ a(1
cos )) or r œ
a(1  cos ));

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Practice Exercises the equations have the same graph, which is a cardioid 69 - 70. Example CAS commands: Maple: with( plots );#69 f := (r,k,e) -> k*e/(1+e*cos(theta)); elist := [3/4,1,5/4]; # (a) P1 := seq( plot( f(r,-2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P1], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=-2" ); P2 := seq( plot( f(r,2,e), theta=-Pi..Pi, coords=polar ), e=elist ): display( [P2], insequence=true, view=[-20..20,-20..20], title="#69(a) (Section 10.8)\nk=2" ); elist2 := [7/6,5/4,4/3,3/2,2,3,5,10,20]; # (b) P3 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist2 ): display( [P3], insequence=true, view=[-20..20,-20..20], title="#69(b) (Section 10.8)\nk=-1, e>1" ); elist3 := [1/2,1/3,1/4,1/10,1/20]; P4 := seq( plot( f(r,-1,e), theta=-Pi..Pi, coords=polar ), e=elist3 ): display( [P4], insequence=true, title="#69(b) (Section 10.8)\nk=-1, e<1" ); klist := -5..-1; # (c) P5 := seq( plot( f(r,k,1/2), theta=-Pi..Pi, coords=polar ), k=klist ): display( [P5], insequence=true, title="#69(c) (Section 10.8)\ne=1/2, k<0" ); P6 := seq( plot( f(r,k,1), theta=-Pi..Pi, coords=polar ), k=klist ): display( [P6], insequence=true, view=[-4..50,-50..50], title="#69(c) (Section 10.8)\ne=1, k<0" ); P7 := seq( plot( f(r,k,2), theta=-Pi..Pi, coords=polar ), k=klist ): display( [P5], insequence=true, title="#69(c) (Section 10.8)\ne=2, k<0" ); Mathematica: (assigned function and values for parameters and bounds may vary): To do polar plots in Mathematica, it is necessary to first load a graphics package In the PolarPlot command, it is assumed that the variable r is given as a function of the variable ). <
1. x# œ
4y Ê y œ
x4 Ê 4p œ 4 Ê p œ 1; therefore Focus is (0ß
1), Directrix is y œ 1

x# #

œ y Ê 4p œ 2 Ê p œ "# ; therefore Focus is ˆ!ß "# ‰; Directrix is y œ
"#

2. x# œ 2y Ê

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675

676

Chapter 10 Conic Sections and Polar Coordinates

3. y# œ 3x Ê x œ

y# 3

Ê 4p œ 3 Ê p œ

3 4

#

4. y# œ
83 x Ê x œ
ˆy8 ‰ Ê 4p œ

;

3

therefore Focus is ˆ 34 ß 0‰ , Directrix is x œ
34

x# 7

5. 16x#  7y# œ 112 Ê



y# 16

7. 3x#
y# œ 3 Ê x#
Ê c œ 2; e œ

c a

œ

2 1

y# 3

x# 4



c a

œ

8. 5y#
4x# œ 20 Ê

y# 4 3 # ;

6. x#  2y# œ 4 Ê c a

œ

Ê pœ

therefore Focus is ˆ
23 ß !‰ , Directrix is x œ

œ1

Ê c# œ 16
7 œ 9 Ê c œ 3; e œ

8 3

Ê c œ È2 ; e œ

3 4

œ 1 Ê c# œ 1  3 œ 4

œ 2; the asymptotes are

Ê c œ 3, e œ

c a

œ

y# # œ È2 #




x# 5

2 3

;

2 3

1 Ê c# œ 4
2 œ 2

œ 1 Ê c# œ 4  5 œ 9

the asymptotes are y œ „

2 È5

x

y œ „ È3 x

#

9. x# œ
12y Ê
1x# œ y Ê 4p œ 12 Ê p œ 3 Ê focus is (!ß
3), directrix is y œ 3, vertex is (0ß 0); therefore new vertex is (2ß 3), new focus is (2ß 0), new directrix is y œ 6, and the new equation is (x
2)# œ
12(y
3) #

y 10. y# œ 10x Ê 10 œ x Ê 4p œ 10 Ê p œ #5 Ê focus is ˆ #5 ß 0‰ , directrix is x œ
5# , vertex is (0ß 0); therefore new vertex is ˆ
"# ß
1‰ , new focus is (2ß
1), new directrix is x œ
3, and the new equation is (y  1)# œ 10 ˆx  "# ‰

11.

x# 9



y# #5

œ 1 Ê a œ 5 and b œ 3 Ê c œ È25
9 œ 4 Ê foci are a!ß „ 4b , vertices are a!ß „ 5b , center is

(0ß 0); therefore the new center is (
$ß
5), new foci are (
3ß
1) and (
3ß
9), new vertices are (
$ß
10) and (
$ß 0), and the new equation is 12.

x# 169



y# 144

(x
3)# 9



(y
5)# #5

œ1

œ 1 Ê a œ 13 and b œ 12 Ê c œ È169
144 œ 5 Ê foci are a „ 5ß 0b , vertices are a „ 13ß 0b , center

is (0ß 0); therefore the new center is (5ß 12), new foci are (10ß 12) and (0ß 12), new vertices are (18ß 12) and

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Practice Exercises (
8ß 12), and the new equation is 13.

y# 8




x# 2

(x  5)# 169

(y  12)# 144



677

œ1

œ 1 Ê a œ 2È2 and b œ È2 Ê c œ È8  2 œ È10 Ê foci are Š0ß „ È10‹ , vertices are

Š0ß „ 2È2‹ , center is (0ß 0), and the asymptotes are y œ „ 2x; therefore the new center is Š2ß 2È2‹, new foci are Š2ß 2È2 „ È10‹ , new vertices are Š2ß 4È2‹ and (#ß 0), the new asymptotes are y œ 2x
4  2È2 and #

y œ
2x  4  2È2; the new equation is 14.

x# 36




y# 64

Šy  2È2‹ 8




(x  2)# #

œ1

œ 1 Ê a œ 6 and b œ 8 Ê c œ È36  64 œ 10 Ê foci are a „ 10ß 0b , vertices are a „ 6ß 0b , the center

is (0ß 0) and the asymptotes are

y 8

œ „

x 6

or y œ „ 34 x; therefore the new center is (
10ß
3), the new foci are

(
20ß
3) and (0ß
3), the new vertices are (
16ß
3) and (
4ß
3), the new asymptotes are y œ y œ
43 x


49 3

; the new equation is

(x
10) 36

#




(y
3) 64

#

4 3

x

31 3

and

œ1

15. x#
4x
4y# œ 0 Ê x#
4x  4
4y# œ 4 Ê (x
2)#
4y# œ 4 Ê

(x  2)# 4


y# œ 1, a hyperbola; a œ 2 and

b œ 1 Ê c œ È1  4 œ È5 ; the center is (2ß 0), the vertices are (!ß 0) and (4ß 0); the foci are Š2 „ È5 ß 0‹ and the asymptotes are y œ „

x 2 #

16. 4x#
y#  4y œ 8 Ê 4x#
y#  4y
4 œ 4 Ê 4x#
(y
2)# œ 4 Ê x#


(y  2)# 4

œ 1, a hyperbola; a œ 1 and

b œ 2 Ê c œ È1  4 œ È5 ; the center is (!ß 2), the vertices are (1ß 2) and (
"ß 2), the foci are Š „ È5ß 2‹ and the asymptotes are y œ „ 2x  2 17. y#
2y  16x œ
49 Ê y#
2y  1 œ
16x
48 Ê (y
1)# œ
16(x  3), a parabola; the vertex is (
$ß 1); 4p œ 16 Ê p œ 4 Ê the focus is (
7ß 1) and the directrix is x œ 1 18. x#
2x  8y œ
17 Ê x#
2x  1 œ
8y
16 Ê (x
1)# œ
8(y  2), a parabola; the vertex is (1ß
2); 4p œ 8 Ê p œ 2 Ê the focus is (1ß
4) and the directrix is y œ 0 19. 9x#  16y#  54x
64y œ
1 Ê 9 ax#  6xb  16 ay#
4yb œ
1 Ê 9 ax#  6x  9b  16 ay#
4y  4b œ 144 Ê 9(x  3)#  16(y
2)# œ 144 Ê

(x
3)# 16



(y  2)# 9

œ 1, an ellipse; the center is (
3ß 2); a œ 4 and b œ 3

Ê c œ È16
9 œ È7 ; the foci are Š
$ „ È7ß 2‹ ; the vertices are (1ß 2) and (
7ß 2) 20. 25x#  9y#
100x  54y œ 44 Ê 25 ax#
4xb  9 ay#  6yb œ 44 Ê 25 ax#
4x  4b  9 ay#  6y  9b œ 225 # # Ê (x  2)  (y
3) œ 1, an ellipse; the center is (2ß
3); a œ 5 and b œ 3 Ê c œ È25
9 œ 4; the foci are 9

25

(2ß 1) and (2ß
7); the vertices are (2ß 2) and (2ß
8) 21. x#  y#
2x
2y œ 0 Ê x#
2x  1  y#
2y  1 œ 2 Ê (x
1)#  (y
1)# œ 2, a circle with center (1ß 1) and radius œ È2 22. x#  y#  4x  2y œ 1 Ê x#  4x  4  y#  2y  1 œ 6 Ê (x  2)#  (y  1)# œ 6, a circle with center (
2ß
1) and radius œ È6 23. B#
4AC œ 1
4(1)(1) œ
3  0 Ê ellipse

24. B#
4AC œ 4#
4(1)(4) œ 0 Ê parabola

25. B#
4AC œ 3#
4(1)(2) œ 1  0 Ê hyperbola

26. B#
4AC œ 2#
4(1)(
2) œ 12  0 Ê hyperbola

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678

Chapter 10 Conic Sections and Polar Coordinates

27. x#
2xy  y# œ 0 Ê (x
y)# œ 0 Ê x
y œ 0 or y œ x, a straight line 28. B#
4AC œ (
3)#
4(1)(4) œ
7  0 Ê ellipse AC B œ 0 È2 w È2 w # y ‹Š # x

29. B#
4AC œ 1#
4(2)(2) œ
15  0 Ê ellipse; cot 2! œ yœ

È2 #

xw 

È2 #

yw Ê 2 Š

È2 #

xw


È2 #

#

yw ‹  Š

È2 #

xw


1 #

Ê 2! œ 

È2 #

Ê !œ

yw ‹  2 Š

È2 #

1 4

xw 

;xœ È2 #

È2 # #

xw


È2 #

xw


È2 #

yw and

yw ‹
15 œ 0

Ê 5xw #  3yw # œ 30 AC B

30. B#
4AC œ 2#
4(3)(3) œ
32  0 Ê ellipse; cot 2! œ yœ

È2 #

xw 

È2 #

yw Ê 3 Š

È2 #

xw


È2 #

#

yw ‹  2 Š

È2 #

xw


È2 #

1 #

œ 0 Ê 2! œ

yw ‹ Š

È2 #

xw 

È2 #

Ê !œ

yw ‹  3 Š

È2 #

Ê 2! œ

1 3

1 4

;xœ

xw 

È2 #

#

È2 #

yw and

yw ‹ œ 19

Ê 4xw #  2yw # œ 19 #

œ

" È3

xw
1# yw ‹ Š 1# xw 

È3 #

31. B#
4AC œ Š2È3‹
4(1)(
1) œ 16 Ê hyperbola; cot 2! œ and y œ

1 #

xw 

È3 #

yw Ê Š

È3 #

#

xw
1# yw ‹  2È3 Š

È3 #

AC B

Ê !œ

yw ‹
Š 1# xw 

È3 #

1 6

;xœ

È3 #

xw
1# yw

#

yw ‹ œ 4

Ê 2xw #
2yw # œ
4 Ê yw #
xw # œ 2 32. B#
4AC œ (
3)#
4(1)(1) œ 5  0 Ê hyperbola; cot 2! œ and y œ Ê

5 #

È2 #

È2 #

yw Ê Š

È2 #

xw


È2 #

#

yw ‹
3 Š

È2 #

xw


È2 #

yw ‹ Š

œ 0 Ê 2! œ È2 #

xw 

È2 #

1 #

Ê !œ

yw ‹  Š

È2 #

1 4

xw 

;xœ È2 #

È2 # #

xw


yw ‹ œ 5

yw #
"# xw # œ 5 or 5yw #
xw # œ 10

" # tan t and y# œ "4 # #

33. x œ

xw 

AC B

and y œ

" #

sec t Ê x# œ

" 4

tan# t

sec# t Ê 4x# œ tan# t and

34. x œ
2 cos t and y œ 2 sin t Ê x# œ 4 cos# t and y# œ 4 sin# t Ê x#  y# œ 4

4y œ sec t Ê 4x#  1 œ 4y# Ê 4y#
4x# œ 1

35. x œ
cos t and y œ cos# t Ê y œ (
x)# œ x#

35. x œ 4 cos t and y œ 9 sin t Ê x# œ 16 cos# t and y# œ 81 sin# t Ê

x# 16



y# 81

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

È2 #

yw

Chapter 10 Practice Exercises 37.

38.

39. d

40.

e

41.

l

42.

f

43. k

44.

h

45.

i

46.

j

47. r œ sin ) and r œ 1  sin ) Ê sin ) œ 1  sin ) Ê 0 œ 1 so no solutions exist. There are no points of intersection found by solving the system. The point of intersection (0ß 0) is found by graphing.

48. r œ cos ) and r œ 1
cos ) Ê cos ) œ 1
cos ) Ê cos ) œ "# Ê ) œ 13 ,
13 ; ) œ 13 Ê r œ "# ; ) œ
13 Ê r œ "# . The points of intersection are ˆ "# ß 13 ‰ and ˆ "# ß
13 ‰ . The point of intersection (0ß 0) is found by graphing.

49. r œ 1  cos ) and r œ 1
cos ) Ê 1  cos ) œ 1
cos ) Ê 2 cos ) œ 0 Ê cos ) œ 0 Ê ) œ 1# , 3#1 ; ) œ 1# or 3#1 Ê r œ 1. The points of intersection are ˆ1ß 1# ‰ and ˆ1ß 3#1 ‰ . The point of intersection (0ß 0) is found by graphing.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

679

680

Chapter 10 Conic Sections and Polar Coordinates

50. r œ 1
sin ) and r œ 1  sin ) Ê 1
sin ) œ 1  sin ) Ê 2 sin ) œ 0 Ê sin ) œ 0 Ê ) œ 0, 1; ) œ 0 or 1 Ê r œ 1. The points of intersection are (1ß 0) and (1ß 1). The point of intersection (!ß 0) is found by graphing.

51. r œ 1
sin ) and r œ 1
sin ) intersect at all points of r œ 1
sin ) because the graphs coincide. This can be seen by graphing them.

52. r œ 1
cos ) and r œ 1
cos ) intersect at all points of r œ 1
cos ) because the graphs coincide. This can be seen by graphing them.

53. r œ sec ) and r œ 2 sin ) Ê sec ) œ 2 sin ) Ê 1 œ 2 sin ) cos ) Ê 1 œ sin 2) Ê 2) œ 1# Ê ) œ Ê r œ 2 sin 1 œ È2 Ê the point of intersection is

1 4

4

ŠÈ2ß 14 ‹ . No other points of intersection exist.

54. r œ 2 csc ) and r œ 4 cos ) Ê 2 csc ) œ 4 cos ) Ê 1 œ 2 sin ) cos ) Ê 1 œ sin 2) Ê 2) œ 1# , 5#1 Ê ) œ 1 , 51 ; ) œ 1 Ê r œ 4 cos 1 œ 2È2 ; 4

)œ

51 4

4

4

Ê r œ 4 cos

4

51 4

œ 2È2 . The point of

intersection is Š2È2ß 541 ‹ and the point Š2È2ß 14 ‹ is the same point.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Practice Exercises 55. r cos ˆ)
13 ‰ œ 2È3 Ê r ˆcos ) cos

1 3

 sin ) sin 13 ‰

È r cos )  #3 r sin ) œ 2È3 Ê r cos )  È3 r sin ) œ 4È3 Ê x  È3 y œ 4È3 " #

œ 2È3 Ê Ê yœ

È3 3

56. r cos ˆ)  œ

È2 #

x4

31 ‰ 4

Ê 

œ È2 #

È2 #

Ê r ˆcos ) cos

r cos )


Ê yœx
1

57. r œ 2 sec ) Ê r œ

2 cos )

È2 #

31 4

r sin ) œ


sin ) sin È2 #

31 ‰ 4

Ê x
y œ 1

Ê r cos ) œ 2 Ê x œ 2

58. r œ È2 sec ) Ê r cos ) œ È2 Ê x œ È2

59. r œ  3# csc ) Ê r sin ) œ  3# Ê y œ  3#

60. r œ 3È3 csc ) Ê r sin ) œ 3È3 Ê y œ 3È3

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681

682

Chapter 10 Conic Sections and Polar Coordinates

61. r œ 4 sin ) Ê r# œ 4r sin ) Ê x#
y#
4y œ 0 Ê x#
(y
2)# œ 4; circle with center (!ß 2) and radius 2.

62. r œ 3È3 sin ) Ê r# œ 3È3 r sin ) Ê x#
y#  3È3 y œ 0 Ê x#
Šy  circle with center Š!ß

3È 3 # ‹

and radius

3È 3 # ‹

#

œ

27 4

;

3È 3 #

63. r œ 2È2 cos ) Ê r# œ 2È2 r cos ) #

Ê x#
y#  2È2 x œ 0 Ê Šx  È2‹
y# œ 2; circle with center ŠÈ2ß 0‹ and radius È2

64. r œ 6 cos ) Ê r# œ 6r cos ) Ê x#
y#
6x œ 0 Ê (x
3)#
y# œ 9; circle with center (3ß 0) and radius 3

# 65. x#
y#
5y œ 0 Ê x#
ˆy
#5 ‰ œ

and a œ

5 #

25 4

Ê C œ ˆ!ß  5# ‰

; r#
5r sin ) œ 0 Ê r œ 5 sin )

66. x#
y#  2y œ 0 Ê x#
(y  1)# œ 1 Ê C œ (!ß 1) and a œ 1; r#  2r sin ) œ 0 Ê r œ 2 sin )

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Chapter 10 Practice Exercises # 67. x#
y#  3x œ 0 Ê ˆx  #3 ‰
y# œ

and a œ

3 #

9 4

Ê C œ ˆ 3# ß !‰

; r#  3r cos ) œ 0 Ê r œ 3 cos )

68. x#
y#
4x œ 0 Ê (x
2)#
y# œ 4 Ê C œ (2ß 0) and a œ 2; r#
4r cos ) œ 0 Ê r œ 4 cos )

69. r œ

2 1
cos )

Ê e œ 1 Ê parabola with vertex at (1ß 0)

70. r œ

8 2
cos )

Ê rœ

ke œ 4 Ê

" #

4 1
ˆ "# ‰ cos )

Ê eœ

k œ 4 Ê k œ 8; k œ

a e

" #

Ê ellipse;

 ea Ê 8 œ

a ˆ "# ‰

 "# a

ˆ " ‰ ˆ 16 ‰ 8 Ê a œ 16 3 Ê ea œ # 3 œ 3 ; therefore the center is ˆ 83 ß 1‰ ; vertices are ()ß 1) and ˆ 83 ß 0‰

71. r œ

6 1  2 cos )

Ê e œ 2 Ê hyperbola; ke œ 6 Ê 2k œ 6

Ê k œ 3 Ê vertices are (2ß 1) and (6ß 1)

72. r œ Ê

12 3
sin ) " 3

Ê rœ

4 1
ˆ "3 ‰ sin )

Ê eœ

" 3

; ke œ 4 #

k œ 4 Ê k œ 12; a a1  e# b œ 4 Ê a ’1  ˆ 3" ‰ “

œ 4 Ê a œ 9# Ê ea œ ˆ "3 ‰ ˆ 9# ‰ œ 3# ; therefore the center is ˆ 3# ß 3#1 ‰ ; vertices are ˆ3ß 1# ‰ and ˆ6ß 3#1 ‰

73. e œ 2 and r cos ) œ 2 Ê x œ 2 is directrix Ê k œ 2; the conic is a hyperbola; r œ Ê rœ

ke 1
e cos )

Ê rœ

4 1
# cos )

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(2)(2) 1
# cos )

683

684

Chapter 10 Conic Sections and Polar Coordinates

74. e œ 1 and r cos ) œ 4 Ê x œ 4 is directrix Ê k œ 4; the conic is a parabola; r œ Ê rœ 75. e œ

" #

and r sin ) œ 2 Ê y œ 2 is directrix Ê k œ 2; the conic is an ellipse; r œ

" 3

Ê rœ

(4)(1) 1  cos )

4 1  cos )

Ê rœ 76. e œ

ke 1  e cos )

Ê rœ

ke 1
e sin )

(2) ˆ "# ‰ 1
ˆ "# ‰ sin )

2 2
sin )

and r sin ) œ 6 Ê y œ 6 is directrix Ê k œ 6; the conic is an ellipse; r œ

Ê rœ

ke 1  e sin )

Ê rœ

(6) ˆ "3 ‰ 1  ˆ "3 ‰ sin )

6 3  sin )

77. A œ 2'0

1

" # # r

d) œ '0 (2  cos ))# d) œ '0 a4  4 cos )
cos# )b d) œ '0 ˆ4  4 cos )
1

œ '0 ˆ 9#  4 cos )
1

78. A œ '0

1Î3

" #

1

cos 2) ‰ #

1

sin 2) ‘ 1 4 !

d) œ
92 )  4 sin )


asin# 3)b d) œ '0

1Î3

6) ‰ ˆ 1  cos d) œ #

" 4


) 

" 6

œ

9 #

A œ 4'0

" #

1Î$

sin 6)‘ !

œ

1 12 1 #

1 4

Ê )œ

c(1
cos 2))#  1# d d) œ 2 '0 a1
2 cos 2)
cos# 2)  1b d) 1Î4

œ 2'0 ˆ2 cos 2)
1Î4

" #




cos 4) ‰ #

d) œ 2
sin 2)


" 2

)


d)

1

79. r œ 1
cos 2) and r œ 1 Ê 1 œ 1
cos 2) Ê 0 œ cos 2) Ê 2) œ 1Î4

1
cos 2) ‰ #

sin 4) ‘ 1Î% 8 !

œ 2 ˆ1


1 8

; therefore


0‰ œ 2


1 4

80. The circle lies interior to the cardioid (see the graphs in Exercises 61 and 63). Thus, 1Î2

A œ 2 '
1Î2 1Î2

" #

[2(1
sin ))]# d)  1 (the integral is the area of the cardioid minus the area of the circle) 1Î2

œ '
1Î2 4 a1
2 sin )
sin# )b d)  1 œ '
1Î2 (6
8 sin )  2 cos 2)) d)  1 œ c6)  8 cos )  sin 2)d 1Î#  1 1Î#

œ c31  (31)d  1 œ 51 81. r œ 1
cos ) Ê

dr d)

œ  sin ); Length œ '0 È(1
cos ))#
( sin ))# d) œ '0 È2  2 cos ) d) 21

œ '0 É 4(1 #cos )) d) œ '0 2 sin 21

21

82. r œ 2 sin )
2 cos ), 0 Ÿ ) Ÿ

1 #

83. r œ 8 sin$ ˆ 3) ‰ , 0 Ÿ ) Ÿ

1 4

Ê

dr d)

#1 d) œ
4 cos 2) ‘ ! œ (4)(1)  (4)(1) œ 8

) #

Ê

œ 8 asin# )
cos# )b œ 8 Ê L œ

21

dr d)

#

œ 2 cos )  2 sin ); r#
ˆ ddr) ‰ œ (2 sin )
2 cos ))#
(2 cos )  2 sin ))#

'01Î2 È8 d) œ ’2È2 )“ 1Î# œ 2È2 ˆ 1# ‰ œ 1È2 !

#

œ 64 sin% ˆ 3) ‰ Ê L œ '0 É64 sin% ˆ 3) ‰ d) œ '0 1Î4

#

œ 8 sin# ˆ 3) ‰ cos ˆ 3) ‰ ; r#
ˆ ddr) ‰ œ
8 sin$ ˆ 3) ‰‘

8 sin# ˆ 3) ‰ cos ˆ 3) ‰‘ 1Î4

8 sin# ˆ 3) ‰ d) œ '0 8 ’ 1Î4

1cos ˆ 23) ‰ “ #

d)

1Î% œ '0
4  4 cos ˆ 23) ‰‘ d) œ
4)  6 sin ˆ 23) ‰‘ ! œ 4 ˆ 14 ‰  6 sin ˆ 16 ‰  0 œ 1  3 1Î4

84. r œ È1
cos 2) Ê

dr d)

œ

" #

(1
cos 2))"Î# (2 sin 2)) œ

#

Ê r#
ˆ ddr) ‰ œ 1
cos 2)
œ

2
2 cos 2) 1
cos 2)

1Î2

sin# 2) 1
cos 2)

œ

(1
cos 2))#
sin# 2) 1
cos 2)

 sin 2) È1
cos 2)

œ

# Ê ˆ ddr) ‰ œ

sin# 2) 1
cos 2)

1
2 cos 2)
cos# 2)
sin# 2) 1
cos 2)

œ 2 Ê L œ '
1Î2 È2 d) œ È2
1#  ˆ 1# ‰‘ œ È2 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

#

Chapter 10 Practice Exercises 85. r œ Ècos 2) Ê œ '0

1Î4

dr d)

1Î%

r# œ sin 2) Ê 2r

œ 21 Š1  dr d)

1Î4

1Î2

1Î#

œ 2 c21 sin )d !

È2 # ‹

sin# 2) cos 2)

dr ‰# dt

#

d) œ '0 21Ècos 2) (sin )) É cos" 2) d) œ '0 1Î4

1Î4

21 sin ) d)

œ Š2  È2‹ 1

œ 2 cos 2) Ê r

œ 2'0 21(cos )) Ér%
ˆr

87.

; Surface Area œ '0 21(r sin )) Ér#
ˆ ddr) ‰ d)

21Ècos 2) (sin )) Écos 2)


œ c21( cos ))d ! 86.

 sin 2) Ècos 2)

œ

œ cos 2); Surface Area œ 2'0

1Î2

dr d)

d) œ 2 '0

1Î2

#

21(r cos )) Ér#
ˆ ddr) ‰ d)

21(cos )) È(sin 2))#
(cos 2))# d) œ 2 '0 21 cos ) d) 1Î2

œ 41

(a) Around the x-axis: 9x#
4y# œ 36 Ê y# œ 9  94 x# Ê y œ „ É9  94 x# and we use the positive root: #

V œ 2 '0 1 ŠÉ9  94 x# ‹ dx œ 2 '0 1 ˆ9  94 x# ‰ dx œ 21
9x  34 x$ ‘ ! œ 241 2

2

#

(b) Around the y-axis: 9x#
4y# œ 36 Ê x# œ 4  49 y# Ê x œ „ É4  49 y# and we use the positive root: #

V œ 2 '0 1 ŠÉ4  49 y# ‹ dy œ 2'0 1 ˆ4  49 y# ‰ dy œ 21
4y  3

88.

9x#  4y# œ 36, x œ 4 Ê y# œ œ

91 4

$

%

’ x3  4x“ œ #

91 4

3

9x#  36 4

Ê yœ


ˆ 64 ‰ ˆ8 ‰‘ œ 3  16  3  8

3 #

91 4

4 27

$

y$ ‘ ! œ 161

Èx#  4 ; V œ ' 1 Š 3 Èx#  4‹ dx œ # 2 4

ˆ 56 3 

24 ‰ 3

œ

31 4

#

91 4

'24 ax#  4b dx

(32) œ 241

89.

Each portion of the wave front reflects to the other focus, and since the wave front travels at a constant speed as it expands, the different portions of the wave arrive at the second focus simultaneously, from all directions, causing a spurt at the second focus.

90.

The velocity of the signals is v œ 980 ft/ms. Let t" be the time it takes for the signal to go from A to S. Then d" œ 980t" and d# œ 980(t"
1400) Ê d#  d" œ 980(1400) œ 1.372 ‚ 10' ft or 259.8 miles. The ship is 259.8 miles closer to A than to B. The difference of the distances is always constant (259.8 miles) so the ship is traveling along a branch of a hyperbola with foci at the two towers. The branch is the one having tower A as its focus.

91.

The time for the bullet to hit the target remains constant, say t œ t! . Let the time it takes for sound to travel from the target to the listener be t# . Since the listener hears the sounds simultaneously, t" œ t!
t# where t" is the time for the sound to travel from the rifle to the listener. If v is the velocity of sound, then vt" œ vt!
vt# or vt"  vt# œ vt! . Now vt" is the distance from the rifle to the listener and vt# is the distance from the target to the listener. Therefore the difference of the distances is constant since vt! is constant so the listener is on a branch of a hyperbola with foci at the rifle and the target. The branch is the one with the target as focus.

92.

Let (r" ß )" ) be a point on the graph where r" œ a)" . Let (r# ß )# ) be on the graph where r# œ a)# and )# œ )"
21. Then r" and r# lie on the same ray on consecutive turns of the spiral and the distance between the two points is r#  r" œ a)#  a)" œ a()#  )" ) œ 21a, which is constant.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

685

686 93.

Chapter 10 Conic Sections and Polar Coordinates (a) r œ

k 1
e cos ) #

Ê r
er cos ) œ k Ê Èx#
y#
ex œ k Ê Èx#
y# œ k  ex Ê x#
y#

œ k  2kex
e# x# Ê x#  e# x#
y#
2kex  k# œ 0 Ê a1  e# b x#
y#
2kex  k# œ 0 (b) e œ 0 Ê x#
y#  k# œ 0 Ê x#
y# œ k# Ê circle; 0  e  1 Ê e#  1 Ê e#  1  0 Ê B#  4AC œ 0#  4 a1  e# b (1) œ 4 ae#  1b  0 Ê ellipse; e œ 1 Ê B#  4AC œ 0#  4(0)(1) œ 0 Ê parabola; e  1 Ê e#  1 Ê B#  4AC œ 0#  4 a1  e# b (1) œ 4e#  4  0 Ê hyperbola 94.

(a) The length of the major axis is 300 miles
8000 miles
1000 miles œ 2a Ê a œ 4650 miles. If the center of the earth is one focus and the distance from the center of the earth to the satellite's low point is 4300 miles (half the diameter plus the distance above the North Pole), then the distance from the center of the ellipse to the focus (center of the earth) is 4650 miles  4300 miles œ 350 miles œ c. Therefore 350 miles 7 e œ ca œ 4650 miles œ 93 . a a1  e # b 1
e cos )

(b) r œ

#

Ê rœ

7 ‰ 4650 ’1  ˆ 93 “ 7 ˆ1
93 cos )‰

œ

430,000 93
7 cos )

mile

CHAPTER 10 ADDITIONAL AND ADVANCED EXERCISES 1. Directrix x œ 3 and focus (4ß 0) Ê vertex is ˆ 7# ß !‰ Ê pœ

" #

Ê the equation is x 

7 #

œ

y# #

2. x#  6x  12y
9 œ 0 Ê x#  6x
9 œ 12y Ê

(x3)# 12

œ y Ê vertex is (3ß 0) and p œ 3 Ê focus is (3ß 3) and the

directrix is y œ 3 3. x# œ 4y Ê vertex is (!ß 0) and p œ 1 Ê focus is (!ß 1); thus the distance from P(xß y) to the vertex is Èx#
y# and the distance from P to the focus is Èx#
(y  1)# Ê Èx#
y# œ 2Èx#
(y  1)# Ê x#
y# œ 4 cx#
(y  1)# d Ê x#
y# œ 4x#
4y#  8y
4 Ê 3x#
3y#  8y
4 œ 0, which is a circle 4. Let the segment a
b intersect the y-axis in point A and intersect the x-axis in point B so that PB œ b and PA œ a (see figure). Draw the horizontal line through P and let it intersect the y-axis in point C. Let nPBO œ ) Ê nAPC œ ). Then sin ) œ yb and cos ) œ xa Ê

x# a#




y# b#

œ cos# )
sin# ) œ 1.

5. Vertices are a!ß „ 2b Ê a œ 2; e œ

c a

Ê 0.5 œ

c #

Ê c œ 1 Ê foci are a0ß „ 1b

6. Let the center of the ellipse be (xß 0); directrix x œ 2, focus (4ß 0), and e œ 23 Ê ae  c œ 2 Ê ae œ 2
c a Ê a œ 23 (2
c). Also c œ ae œ 23 a Ê a œ 23 ˆ2
23 a‰ Ê a œ 43
49 a Ê 59 a œ 43 Ê a œ 12 5 ;x2œ e 28 28 8 # # # ‰ ˆ #3 ‰ œ 18 ˆ 28 ‰ Ê x  2 œ ˆ 12 5 5 Ê x œ 5 Ê the center is 5 ß 0 ; x  4 œ c Ê c œ 5  4 œ 5 so that c œ a  b #

#

‰  ˆ 58 ‰ œ œ ˆ 12 5

80 25

; therefore the equation is

ˆx  28 ‰# 5 144 ˆ 25 ‰




y#

80 ‰ ˆ 25

œ 1 or

‰ 25 ˆx  28 5 144

#




5y# 16

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Additional and Advanced Exercises 7. Let the center of the hyperbola be (0ß y). (a) Directrix y œ 1, focus (0ß 7) and e œ 2 Ê c  ae œ 6 Ê Ê a œ 2(2a)  12 Ê a œ 4

Ê cœ œ

625 16



25 4 25 16

a e

œ6 Ê

; y  (1) œ œ

75 #

a e a e

œ c  6 Ê a œ 2c  12. Also c œ ae œ 2a

Ê c œ 8; y  (1) œ œ 4# œ 2 # 48; therefore the equation is (y161) a e

Ê b# œ c#  a# œ 64  16 œ (b) e œ 5 Ê c 

a e

Ê y œ 1 Ê the center is (0ß 1); c# œ a#
b# 

x# 48

œ1

œ c  6 Ê a œ 5c  30. Also, c œ ae œ 5a Ê a œ 5(5a)  30 Ê 24a œ 30 Ê a œ œ

ˆ 54 ‰ 5

œ

" 4

Ê y œ  43 Ê the center is ˆ!ß  43 ‰ ; c# œ a#
b# Ê b# œ c#  a#

; therefore the equation is

ˆy
34 ‰# ˆ 25 ‰ 16

x# ˆ 75 ‰ #



œ 1 or

16 ˆy
34 ‰ 25

#



Ê



144 a4  a # b #

2x# 75

y# a#

8. The center is (0ß 0) and c œ 2 Ê 4 œ a#
b# Ê b# œ 4  a# . The equation is 49 a#

687

œ1



x# b#

œ1 Ê

49 a#



144 b#

œ1

œ 1 Ê 49 a4  a# b  144a# œ a# a4  a# b Ê 196  49a#  144a# œ 4a#  a% Ê a%  197a#
196

œ 0 Ê aa  196b aa#  1b œ 0 Ê a œ 14 or a œ 1; a œ 14 Ê b# œ 4  (14)#  0 which is impossible; a œ 1 Ê b# œ 4  1 œ 3; therefore the equation is y#  9. (a) b# x#
a# y# œ a# b# Ê

dy dx

x# 3

œ1 #

#

œ  ba# yx ; at (x" ß y" ) the tangent line is y  y" œ Š ba# yx"" ‹ (x  x" )

Ê a# yy"
b# xx" œ b# x"#
a# y"# œ a# b# Ê b# xx"
a# yy"  a# b# œ 0 (b) b# x#  a# y# œ a# b# Ê

dy dx

œ

b# x a# y

#

; at (x" ß y" ) the tangent line is y  y" œ Š ba# yx"" ‹ (x  x" )

Ê b# xx"  a# yy" œ b# x"#  a# y"# œ a# b# Ê b# xx"  a# yy"  a# b# œ 0 10. Ax#
Bxy
Cy#
Dx
Ey
F œ 0 has the derivative

dy dx

œ

2Ax  By  D Bx
2Cy
E

; at (x" ß y" ) the tangent line is

2Ax"  By"  D y  y" œ Š Bx ‹ (x  x" ) Ê Byx"
2Cyy"
Ey  By" x"  2Cy#"  Ey" "
2Cy"
E

œ 2Axx"  Bxy"  Dx
2Ax#"
Bx" y"
Dx" Ê 2Axx"
B(yx"
xy" )
2Cyy"
Dx  Dx"
Ey  Ey" œ 2Ax#"
2Bx" y"
2Cy"# . Now add 2Dx"
2Ey" to both sides of this last equation, divide the result by 2, and represent the constant value on the right by F to get: Axx"
B ˆ yx"
# xy" ‰
Cyy"
D ˆ x
# x" ‰
E ˆ y
# y" ‰ œ F 11.

12.

13.

14.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

5 4

688

Chapter 10 Conic Sections and Polar Coordinates

15. a9x#
4y#  36b a4x#
9y#  16b Ÿ 0 Ê 9x#
4y#  36 Ÿ 0 and 4x#
9y#  16   0 or 9x#
4y#  36   0 and 4x#
9y#  16 Ÿ 0

16. a9x#
4y#  36b a4x#
9y#  16b  0, which is the complement of the set in Exercise 15

#

17. x%  ay#  9b œ 0 Ê x#  ay#  9b œ 0 or x#
ay#  9b œ 0 Ê y#  x# œ 9 or x#
y# œ 9

18. x#
xy
y#  3 Ê tan 2! œ

" 1 1 w

which is undefined

Ê 2! œ 90° Ê ! œ 45° Ê A œ cos# 45°
cos 45° sin 45°
sin# 45° œ

3 #

, Bw œ 0,

Cw œ sin# 45°  sin 45° cos 45°
cos# 45° œ Ê

3 #

" #

xw #
"# yw #  3 which is the interior of a

rotated ellipse

19. Arc PF œ Arc AF since each is the distance rolled; nPCF œ Arcb PF Ê Arc PF œ b(nPCF); ) œ ArcaAF Ê Arc AF œ a) Ê a) œ b(nPCF) Ê nPCF œ ˆ ba ‰ ); nOCB œ

1 #

 ) and nOCB œ nPCF  nPCE œ nPCF  ˆ 1#  !‰ œ ˆ ba ‰ )  ˆ 1#  !‰ Ê 1#  ) œ ˆ ba ‰ )  ˆ 1#  !‰ Ê 1#  ) œ ˆ ba ‰ )  1#
! Ê ! œ 1  )  ˆ ba ‰ ) Ê ! œ 1  ˆ a
b b ‰ ).

Now x œ OB
BD œ OB
EP œ (a
b) cos )
b cos ! œ (a
b) cos )
b cos ˆ1  ˆ a
b b ‰ )‰ œ (a
b) cos )
b cos 1 cos ˆˆ a
b b ‰ )‰
b sin 1 sin ˆˆ a
b b ‰ )‰ œ (a
b) cos )  b cos ˆˆ a
b b ‰ )‰ and y œ PD œ CB  CE œ (a
b) sin )  b sin ! œ (a
b) sin )  b sin ˆˆ a
b b ‰ )‰ œ (a
b) sin )  b sin 1 cos ˆˆ a
b b ‰ )‰
b cos 1 sin ˆˆ a
b b ‰ )‰ œ (a
b) sin )  b sin ˆˆ a
b b ‰ )‰ ; therefore x œ (a
b) cos )  b cos ˆˆ a
b b ‰ )‰ and y œ (a
b) sin )  b sin ˆˆ a
b b ‰ )‰

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Additional and Advanced Exercises 20. (a) x œ a(t  sin t) Ê

dx dt

‰ œ a(1  cos t) and let $ œ 1 Ê dm œ dA œ y dx œ y ˆ dx dt dt

œ a(1  cos t) a (1  cos t) dt œ a# (1  cos t)# dt; then A œ '0 a# (1  cos t)# dt 21

'021 a1  2 cos t
cos# tb dt œ a# '021 ˆ1  2 cos t
"#
"# cos 2t‰ dt œ a#
32 t  2 sin t
sin4 2t ‘ #!1 œ 31a# ; µ x = x œ a(t  sin t) and µ y = "# y œ "# a(1  cos t) Ê Mx œ ' µ y dm œ ' µ y $ dA 21 21 21 œ '0 "# a(1  cos t) a# (1  cos t)# dt œ "# a$ '0 (1  cos t)$ dt œ a# '0 a1  3 cos t
3 cos# t  cos$ tb dt œ a#

$

œ œ

'021
1  3 cos t
3#
3 cos# 2t  a1  sin# tb (cos t)‘ dt œ a# ’ 52 t  3 sin t
3 sin4 2t  sin t
sin3 t “ #1

a$ #

$

51 a #

$

!

$

. Therefore y œ

Mx M

œ

$

Š 51#a ‹ 31 a#

œ

5 6

a. Also, My œ ' µ x dm œ ' µ x $ dA

œ '0 a(t  sin t) a# (1  cos t)# dt œ a$ '0 at  2t cos t
t cos# t  sin t
2 sin t cos t  sin t cos# tb dt 21

21

#

œ a$ ’ t2  2 cos t  2t sin t
"4 t#
My 31 # a$ 5 ˆ M œ 31a# œ 1a Ê 1aß 6 2 $Î# "Î# Ê dx and y œ 3 t dt œ t

xœ (b) x œ

È3 È3

œ '0

È$

2t$Î# dt œ
45 t&Î# ‘ ! 4 &Î# 3 t

œ

È$

8 (Î# ‘ dt œ
21 t !

cos 2t


t 4

sin 2t
cos t
sin# t


#1 cos$ t 3 “!

œ 31# a$ . Thus

a‰ is the center of mass. 2t"Î# Ê

Ê ˆ2t"Î# ‰ ˆt"Î# ‰ dt œ 2t dt; µ x =xœ œ '0

" 8

2 $Î# 3 t

‰ œ t"Î# ; let $ œ 1 Ê dm œ dA œ y dx œ y ˆ dx dt dt

and µ y =

y #

È3

œ t"Î# Ê Mx œ ' µ y dm œ '0

t"Î# (2t dt)

È3 4 È 3. Also, My œ ' µ x dm œ ' µ x dA œ '0 23 t$Î# (2t dt)

12 5

œ

dy dt

8 7

4 È 27 .

sin t 21. (a) x œ e2t cos t and y œ e2t sin t Ê x#
y# œ e4t cos# t
e4t sin# t œ e4t . Also yx œ ee2t cos t œ tan t
" y " ˆ ‰ # # % tan ayÎxb # # Ê t œ tan Ê x
y œe is the Cartesian equation. Since r œ x
y# and x 2t

) œ tan" ˆ yx ‰ , the polar equation is r# œ e4) or r œ e2) for r  0 (b) ds# œ r# d)#
dr# ; r œ e2) Ê dr œ 2e2) d) # # Ê ds# œ r# d)#
ˆ2e2) d)‰ œ ˆe2) ‰ d)#
4e4) d)# œ 5e4) d)# Ê ds œ È5 e2) d) Ê L œ '0 È5 e2) d) 21

œ’

È5 e2) #1 2 “!

œ

È5 #

ae41  1b

# # 22. r œ 2 sin$ ˆ 3) ‰ Ê dr œ 2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d) Ê ds# œ r# d)#
dr# œ
2 sin$ ˆ 3) ‰‘ d)#

2 sin# ˆ 3) ‰ cos ˆ 3) ‰ d)‘ œ 4 sin' ˆ 3) ‰ d)#
4 sin% ˆ 3) ‰ cos# ˆ 3) ‰ d)# œ
4 sin% ˆ 3) ‰‘
sin# ˆ 3) ‰
cos# ˆ 3) ‰‘ d)# œ 4 sin% ˆ 3) ‰ d)#

Ê ds œ 2 sin# ˆ 3) ‰ d). Then L œ '0 2 sin# ˆ 3) ‰ d) œ '0
1  cos ˆ 23) ‰‘ d) œ
)  31

31

3 2

$1

sin ˆ 23) ‰‘ ! œ 31

23. r œ 1
cos ) and S œ ' 213 ds, where 3 œ y œ r sin ); ds œ Èr# d)#
dr# œ È(1
cos ))# d)#
sin# ) d)# È1
2 cos )
cos# )
sin# ) d) œ È2
2 cos ) d) œ É4 cos# ˆ #) ‰ d) œ 2 cos ˆ #) ‰ d) since 0 Ÿ ) Ÿ

1 #

. Then S œ '0 21(r sin )) † 2 cos ˆ #) ‰ d) œ 1Î2

'01Î2 41(1
cos )) † sin ) cos ˆ #) ‰ d)

œ '0 41
2 cos# ˆ #) ‰‘
2 sin ˆ #) ‰ cos ˆ 2) ‰ cos ˆ #) ‰‘ d) œ '0 161 cos% ˆ #) ‰ sin ˆ #) ‰ d) œ ’ 1Î2

œ

1Î2

(321) Š 5

È2 ‹& #

 ˆ 3251 ‰ œ

1Î# 321 cos& ˆ #) ‰ “ 5 !

321  41È2 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

689

690

Chapter 10 Conic Sections and Polar Coordinates

24. The region in question is the figure eight in the middle. The arc of r œ 2a sin# ˆ #) ‰ in the first quadrant gives " 4

of that region. Therefore the area is A œ 4 '0

1Î2

œ 4 '0

1Î2


2a sin# ˆ #) ‰‘# d) œ 8a# ' 0

1Î2

" #

" # # r

d)

sin% ˆ #) ‰ d)

'01Î2 sin# ˆ #) ‰
1  cos# ˆ #) ‰‘ d) 1Î2 œ 8a# '0
sin# ˆ #) ‰  sin# ˆ #) ‰ cos# ˆ #) ‰‘ d) 1Î2 ) œ 8a# '0 Š 1cos  sin4 ) ‹ d) # œ 8a#

#

œ 2a#

1Î2 1Î# 2) ‰ '01Î2 ˆ2  2 cos )  1  cos d) œ a# '0 (3  4 cos )
cos 2)) d) œ a#
3)  4 sin )
"2 sin 2)‘ ! #

‰ œ a# ˆ 31 # 4 25. e œ 2 and r cos ) œ 2 Ê x œ 2 is the directrix Ê k œ 2; the conic is a hyperbola with r œ Ê rœ

(2)(2) 1
2 cos )

œ

ke 1
e cos )

4 1
2 cos )

26. e œ 1 and r cos ) œ 4 Ê x œ 4 is the directrix Ê k œ 4; the conic is a parabola with r œ Ê rœ 27. e œ

" #

" 3

œ

4 1  cos )

and r sin ) œ 2 Ê y œ 2 is the directrix Ê k œ 2; the conic is an ellipse with r œ

Ê rœ 28. e œ

(4)(1) 1  cos )

2 ˆ "# ‰ 1
ˆ "# ‰ sin

)

œ

ke 1
e sin )

2 2
sin )

and r sin ) œ 6 Ê y œ 6 is the directrix Ê k œ 6; the conic is an ellipse with r œ

Ê rœ

6 ˆ "3 ‰ 1  ˆ "3 ‰ sin

)

ke 1  e cos )

œ

ke 1  e sin )

6 3  sin )

29. The length of the rope is L œ 2x
2c
y   8c. (a) The angle A (nBED) occurs when the distance CF œ j is maximized. Now j œ Èx#  c#
y Ê j œ Èx#  c#
L  2x  2c dj dx

Ê Thus

œ

dj dx

" #

ax#  c# b

œ0 Ê

"Î#

x È x #  c#

(2x)  2 œ

x È x #  c#

 2 œ 0 Ê x œ 2Èx#  c#

Ê x# œ 4x#  4c# Ê 3x# œ 4c# Ê Ê Ê

c x A #

œ

È3 #

. Since

c x

 2.

œ sin

A #

c# x#

we have sin

œ

3 4

A #

œ

È3 #

œ 60° Ê A œ 120°

(b) If the ring is fixed at E (i.e., y is held constant) and E is moved to the right, for example, the rope will slip around the pegs so that BE lengthens and DE becomes shorter Ê BE
ED is always 2x œ L  y  2c, which is constant Ê the point E lies on an ellipse with the pegs as foci. (c) Minimal potential energy occurs when the weight is at its lowest point Ê E is at the intersection of the ellipse and its minor axis.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Additional and Advanced Exercises 30.

d" c




d# c

œ

30 c

Ê d"
d# œ 30;

d$ c




d% c

œ

691

30 c

Ê d$
d% œ 30. Therefore P and Q lie on an ellipse with F" and F# as foci. Now 2a œ d"
d# œ 30 Ê a œ 15 and the focal distance is 10 Ê b# œ 15#  10# œ 125 Ê an equation of the ellipse is x# œ x "
v ! t œ x "


v! Š 10 v! ‹

x# 225




y# 125

œ 1. Next

œ x"
10.

If the plane is flying level, then P and Q must be symmetric to the y-axis Ê x" œ x# Ê x# œ x#
10 #

y# 5# 1000 # 225
125 œ 1 Ê y# œ 9 10È10 3 ‹ where the origin (0ß 0)

Ê x# œ 5 Ê the plane is Š&ß

10È10 3

Ê y# œ

since y# must be positive. Therefore the position of

is located midway between the two stations.

31. If the vertex is (!ß 0), then the focus is (pß 0). Let P(xß y) be the present position of the comet. Then È(x  p)#
y# œ 4 ‚ 10( . Since y# œ 4px we have È(x  p)#
4px œ 4 ‚ 10( Ê (x  p)#
4px œ 16 ‚ 10"% . #

Also, x  p œ 4 ‚ 10( cos 60° œ 2 ‚ 10( Ê x œ p
2 ‚ 10( . Therefore a2 ‚ 10( b
4p ap
2 ‚ 10( b œ 16 ‚ 10"% Ê 4 ‚ 10"%
4p#
8p ‚ 10( œ 16 ‚ 10"% Ê 4p#
8p ‚ 10(  12 ‚ 10"% œ 0 Ê p#
2p ‚ 10(  3 ‚ 10"% œ 0 Ê ap
3 ‚ 10( b ap  10( b œ 0 Ê p œ 3 ‚ 10( or p œ 10( . Since p is positive we obtain p œ 10( miles. 32. x œ 2t and y œ t# Ê y œ œ

" 4

dD dx

œ

x# 4

#

#

; let D œ É(x  0)#
ˆ x4  3‰ œ Éx#


x% 16

%

x  3# x#
9 œ É 16  "# x#
9

Èx%  8x#
144 be the distance from any point on the parabola to (0ß 3). We want to minimize D. Then " 8

ax%  8x#
144b

"Î#

a4x$  16xb œ

ˆ "# ‰ x$  2x Èx%  8x#
144

œ0 Ê

" #

x$  2x œ 0 Ê x$  4x œ 0 Ê x œ 0 or

x œ „ 2. Now x œ 0 Ê y œ 0 and x œ „ 2 Ê y œ 1. The distance from (0ß 0) to (0ß 3) is D œ 3. The distance from a „ 2ß 1b to (!ß 3) is D œ Éa „ 2b#
(1  3)# œ 2È2 which is less than 3. Therefore the points closest to (0ß 3) are a „ 2ß 1b . 33. cot 2! œ

AC B

œ 0 Ê ! œ 45° is the angle of rotation Ê Aw œ cos# 45°
cos 45° sin 45°
sin# 45° œ

and Cw œ sin# 45°  sin 45° cos 45°
cos# 45° œ œ2

2 3

œ

4 3

Ê cœ

2 È3

" #

Ê

3 #

3 #

, Bw œ 0,

xw #
"# yw # œ 1 Ê b œ É 23 and a œ È2 Ê c# œ a#  b#

. Therefore the eccentricity is e œ

c a

œ

Š È23 ‹ È2

œ É 32 ¸ 0.82. w#

w#

34. The angle of rotation is ! œ 14 Ê Aw œ sin 14 cos 14 œ "# , Bw œ 0, and Cw œ  sin 14 cos 14 œ  "# Ê x#  y# œ 1 Ê a œ È2 and b œ È2 Ê c# œ a#
b# œ 4 Ê c œ 2. Therefore the eccentricity is e œ ca œ È22 œ È2 . 35. Èx
Èy œ 1 Ê x
2Èxy
y œ 1 Ê 2Èxy œ 1  (x
y) Ê 4xy œ 1  2(x
y)
(x
y)# Ê 4xy œ x#
2xy
y#  2x  2y
1 Ê x#  2xy
y#  2x  2y
1 œ 0 Ê B#  4AC œ (2)#  4(1)(1) œ 0 Ê the curve is part of a parabola 36. ! œ

1 4

Ê Aw œ 2 sin

1 4

cos

1 4

œ 1, Bw œ 0, Cw œ 2 sin

1 4

cos

1 4

œ 1, Dw œ È2 sin

1 4

œ 1, Ew œ È2 cos

1 4

œ 1, Fw œ 2 Ê xw #  yw #  xw  yw
2 œ 0 Ê Šxw #  xw ‹  Šyw #
yw ‹ œ 2 Ê Šxw #  xw
"4 ‹  Šyw #
yw
"4 ‹ ˆyw
"# ‰# #

œ 2 Ê yœ

" #

sin

1 4



" #

ˆxw  "# ‰# #



cos

1 4

œ 1. The center is (xw ß yw ) œ ˆ "# ß  "# ‰ Ê x œ

œ 0 or the center is (xß y) œ

È Š #2 ß 0‹ .

(xw ß yw ) œ Š "# ß È2  "# ‹ and Š "# ß È2  "# ‹ Ê x œ yœ

" #

sin

1 4


ŠÈ2  "# ‹ cos

1 4

œ 1 or (xß y) œ Š

È2 #

" #

cos

1 4

" #

cos

1 4

 ˆ "# ‰ sin

1 4

œ

È2 #

and

Next a œ È2 Ê the vertices are  ŠÈ2  "# ‹ sin

1 4

œ

 1ß 1‹ is one vertex, and x œ

È2 # " #

 1 and

cos

1 4

 ŠÈ2  "# ‹ sin

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 4

692

Chapter 10 Conic Sections and Polar Coordinates È2 #

œ

" #


1 and y œ

sin

1 4


ŠÈ2  "# ‹ sin

1 4

œ 1 or (xß y) œ Š

È2 #


1ß 1‹ is the other vertex. Also

c# œ 2
2 œ 4 Ê c œ 2 Ê the foci are (xw ß yw ) œ ˆ "# ß #3 ‰ and ˆ "# ß  5# ‰ Ê x œ yœ

" #

1 4

sin

3 #

1 4

cos

È Š3 # 2

" #

sin

yw


" #

œ „ ˆxw  "# ‰ in the rotated system. Since x œ

Ê

È2 #

È2 #

x


x




5 #

È2 #

È2 #

1 4

œ È2 or (xß y) œ Š

yœ



1 4




È2 È 2‹ # ß

cos

œ È2 or (xß y) œ

" #

œ „Š

È2 #

x


È2 #

The yw -axis is the line through

" È2

È2 #

xw  È2 #

x


" È2

yw and y œ

1 4




5 #



3 #

sin

sin

1 4

œ

È2 # ß 0‹

" È2

xw


" È2

yw Ê x
y œ

" È2

and

2 È2

xw

and È2 y œ 0 or

with a slope of 1 ˆrecall that ! œ 14 ‰ Ê y œ x 

with a slope of 1 Ê y œ x


2a 1
cos ˆ)  14 ‰

È2 1 4 œ # 3È 2 # and

y œ yw ; the asymptotes are

37. (a) The equation of a parabola with focus (!ß 0) and vertex (aß 0) is r œ through ! œ 45° gives r œ

cos

1 4

y  "# ‹ Ê the asymptotes are È2 x
1 œ 0 or x œ

y œ 0. Finally, the xw -axis is the line through Š È Š # 2 ß 0‹

" #

cos

ß È2‹ is the other focus. The asymptotes are

y œ xw and x  y œ  È22 yw Ê 

y


is one focus, and x œ

" #

È2 #

È2 #

.

.

2a 1
cos )

and rotating this parabola

.

(b) Foci at (!ß 0) and (2ß 0) Ê the center is (1ß 0) Ê a œ 3 and c œ 1 since one vertex is at (4ß 0). Then e œ œ

" 3

œ

3 ˆ1  "9 ‰ 1  ˆ 3" ‰ cos )

œ

(c) Center at ˆ#ß

1‰ #

œ

2 1

. For ellipses with one focus at the origin and major axis along the x-axis we have r œ 8 3  cos )

c a

a a1  e # b 1  e cos )

.

and focus at (!ß 0) Ê c œ 2; center at ˆ2ß 1# ‰ and vertex at ˆ"ß 1# ‰ Ê a œ 1. Then e œ

œ 2. Also k œ ae 

a e

œ (1)(2) 

" #

œ

3 #

. Therefore r œ

ke 1
e sin )

œ

ˆ 3# ‰ (2) 1
2 sin )

œ

3 1
2 sin )

c a

.

38. Let (d" ß )" ) and (d# ß )# ) be the polar coordinates of P" and P# , respectively. Then )# œ )"
1, and we have 3 3 " 2
cos )" and d# œ 2
cos ()"
1) . Therefore d" 4
cos )"
cos )" cos 1  sin )" sin 1 œ 43 . 3

d" œ œ




" d#

œ

2
cos )" 3




2
cos ()"
1) 3

39. Arc PT œ Arc TO since each is the same distance rolled. Now Arc PT œ a(nTAP) and Arc TO œ a(nTBO) Ê nTAP œ nTBO. Since AP œ a œ BO we have that ?ADP is congruent to ?BCO Ê CO œ DP Ê OP is parallel to AB Ê nTBO œ nTAP œ ). Then OPDC is a square Ê r œ CD œ AB  AD  CB œ AB  2CB Ê r œ 2a  2a cos ) œ 2a(1  cos )), which is the polar equation of a cardioid. 40. Note first that the point P traces out a circular arc as the door closes until the second door panel PQ is tangent to the circle. This happens when P is located at Š È"2 ß È"2 ‹ , since nOPQ is 90° at that time. Thus the curve is the circle x#
y# œ 1 for 0 Ÿ x Ÿ È"2 . When x   È"2 , the second door panel is tangent to the curve at P. Now let t represent nPOQ so that as t runs from 1# to 0, the door closes. The coordinates of P

are given by (cos tß sin t), and the coordinates of Q by (2 cos tß 0) (since triangle POQ is isosceles). Therefore at a fixed instant of time t, the slope of the line formed by the second panel PQ is m œ

?y ?x

œ

sin t  0 cos t  2 cos t

œ  tan t Ê the tangent line PQ is

y  0 œ ( tan t)(x  2 cos t) Ê y œ ( tan t) x
2 sin t. Now, to find an equation of the curve for " 1 È2 Ÿ x Ÿ 1, we want to find, for fixed x, the largest value of y as t ranges over the interval 0 Ÿ t Ÿ 4 . We solve d# y dt#

dy dt

œ 0 Ê a sec# tb x
2 cos t œ 0 Ê a sec# tb x œ 2 cos t Ê x œ 2 cos$ t. (Note that

œ a2 sec# t tan tb x  2 sin t  0 on 0 Ÿ t Ÿ

1 #

, so a maximum occurs for y.) Now x œ 2 cos$ t Ê the

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 10 Additional and Advanced Exercises

693

corresponding y value is y œ ( tan t) a2 cos$ tb
2 sin t œ 2 sin t cos# t
2 sin t œ (2 sin t) a cos# t
1b œ 2 sin$ t. Therefore parametric equations for the path of the curve are given by x œ 2 cos$ t and y œ 2 sin$ t for 0 Ÿ t Ÿ

1 4 #

. In Cartesian coordinates, we have the curve x#Î$
y#Î$ œ a2 cos$ tb

#Î$


a2 sin$ tb

#Î$

œ 2#Î$ acos t
sin# tb œ 2#Î$ Ê the curve traced out by the door is given by Þ x#
y # œ 1 for 0 Ÿ x Ÿ È" 2 ß x#Î$
y#Î$ œ 2#Î$ for È" Ÿ x Ÿ 1 à 2 41.

" œ <#  <" Ê tan " œ tan (<#  <" ) œ

tan <#  tan <" 1
tan <# tan <"

;

the curves will be orthogonal when tan " is undefined, or when tan <# œ tan"<" Ê g (r)) œ " r w

’ f ()) “ w

#

w

w

Ê r œ f ( ) ) g ( ) )

42.

r œ sin% ˆ 4) ‰ Ê

43.

r œ 2a sin 3) Ê Ê <œ

44.

œ sin$ ˆ 4) ‰ cos ˆ 4) ‰ Ê tan < œ

dr d)

dr d)

œ 6a cos 3) Ê tan < œ

1 #

r ˆ ddr) ‰

œ

sin% ˆ )4 ‰ sin$ ˆ 4) ‰ cos ˆ )4 ‰ 2a sin 3) 6a cos 3)

œ

œ tan ˆ 4) ‰

" 3

tan 3); when ) œ

1 6

, tan < œ

(b) r) œ 1 Ê r œ )" Ê

(a)

œ

)
" )
#

œ ) Ê

Ê < Ä

1 #

dr d)

" 3

tan

1 #

œ )# Ê tan
lim tan < œ _

)Ä_

from the right as the spiral winds in

around the origin.

45.

tan <" œ

È3 cos ) È3 sin )

œ  cot ) is  È"3 at ) œ

1 3

; tan <# œ

sin ) cos )

œ tan ) is È3 at ) œ

1 3

; since the product of

these slopes is 1, the tangents are perpendicular 46.

a(1
cos )) œ 3a cos ) Ê 1 œ 2 cos ) Ê cos ) œ ) œ 1 ; tan <1 œ a(1
cos )) is È3 at ) œ 1 ; 3

tan <2 œ tan " œ

3a cos ) 3a sin )

a sin )

1
Š È ‹ŠÈ3‹ 3 "

or

3

is  È"3 at ) œ

 È"  ŠÈ3‹ 3

" #

œ

1 3

. Then

 È"
È3 3

#

œ

" È3

Ê "œ

1 6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

694 47.

Chapter 10 Conic Sections and Polar Coordinates r" œ

" 1  cos )

Ê

œ  (1 sincos) ))# ; r# œ

dr" d)

Ê

3 1
cos )

dr# d)

œ

3 sin ) (1
cos ))#

" #

Ê )œ

Ê 1
cos ) œ 3  3 cos ) Ê 4 cos ) œ 2 Ê cos ) œ "

1 3

tan <" k )œ
1Î$ œ 

3

at ) œ  13 Ê " œ 48.

3 1
cos )

r" œ r# œ 2 Ê the curves intersect at the 3

; therefore tan " is undefined at ) œ 1  cos ˆ 13 ‰ sin ˆ 1 ‰

œ

ˆ ‰ œ  1 sincos) ) is  È"3 at ) œ 13 ; tan <# œ 13sincos)) œ 1
sincos) ) is
sin ) ’ (1
cos ))# “ ’ (1  cos ))# “ ˆ 1
cos ) ‰

points ˆ#ß „ 13 ‰ ; tan <" œ È3 at ) œ

" 1  cos ) „ 13 Ê

;

œ

" È3

1 3

since 1
tan <" tan <# œ 1
Š È"3 ‹ ŠÈ3‹ œ 0 Ê " œ 1
cos ˆ 13 ‰ sin ˆ 1 ‰

and tan <# k )œ1Î$ œ

3

1 #

œ È3 Ê tan " is also undefined

1 #

(a) We need <
) œ 1, so that tan < œ tan (1  )) œ  tan ). Now tan < œ

r ˆ ddr) ‰

œ

a(1
cos )) a sin )

sin ) # # œ  tan ) œ  cos ) Ê cos )
cos ) œ sin )

Ê cos )
cos# ) œ 1  cos# ) Ê 2 cos# )
cos )  1 œ 0 Ê cos ) œ "# or cos ) œ 1; cos ) œ Ê rœ

3a #

" #

Ê )œ „

1 3

; cos ) œ 1 Ê ) œ 1 Ê r œ 0.

Therefore the points where the tangent line 1‰ is horizontal are ˆ 3a # ß „ 3 and (!ß 1). 1 #

(b) We need <
) œ cos ) sin )

œ

so that tan < œ tan ˆ 1#  )‰ œ cot ). Thus tan < œ

r ˆ ddr) ‰

œ

a(1
cos )) a sin )

œ cot )

Ê sin )
sin ) cos ) œ  sin ) cos ) Ê cos ) œ  "# or sin ) œ 0; cos ) œ  "# Ê ) œ „

r" œ

a 1
cos )

Ê

21 3

; sin ) œ 0 Ê ) œ 0 (not 1, see part (a)) Ê r œ 2a. Therefore the points where the tangent line is vertical are ˆ #a ß „ 231 ‰ and (2aß 0). 49.

Ê rœ

a #

a ˆ 1
cos ‰ )

tan <" œ

œ

dr" d)

) ’ (1
a sin “ cos ))#

œ

a sin ) (1
cos ))# 1
cos ) sin )

and r# œ

b 1  cos )

and tan <# œ

cos ) ‰ œ 1
ˆ 1
sincos) ) ‰ ˆ 1sin œ1 )

1  cos# ) sin# )

Ê

ˆ 1  bcos ) ‰

sin ) ’ (1bcos “ ))#

dr# d)

œ

b sin ) œ  (1  cos ))# ; then

1cos )  sin )

Ê 1
tan <" tan <#

œ 0 Ê " is undefined Ê the parabolas are orthogonal at each

point of intersection tan < œ

51.

r œ 3 sec ) Ê r œ

r ˆ ddr) ‰

Ê cos ) œ œ  1
sincos) )

œ

a(1  cos )) a sin )

50.

3 cos )

is 1 at ) œ ;

3 cos )  3#

1 #

Ê <œ

1 4

œ 4
4 cos ) Ê 3 œ 4 cos )
4 cos# ) Ê (2 cos )
3)(2 cos )  1) œ 0

cos )) or cos ) œ Ê ) œ 13 or 531 (the second equation has no solutions); tan <# œ 4(14
sin ) 3 sec ) " 1 is È3 at 13 and tan <" œ 3 sec œ cot is at . Then tan is undefined since ) " È3 ) tan ) 3

" #

1
tan <" tan <# œ 1
Š È"3 ‹ ŠÈ3‹ œ 0 Ê " œ

1 #

. Also, tan <# k 51Î3 œ È3 and tan <" k 51Î3 œ  È"3

Ê 1
tan <" tan <# œ 1
Š È"3 ‹ ŠÈ3‹ œ 0 Ê tan " is also undefined Ê " œ 52.

tan < œ

53.

" 1  cos )

a tan ˆ )# ‰ sec# ˆ #) ‰

a #

œ

tan <# œ

œ 1 at ) œ

" 1  sin )

ˆ 1
"sin ) ‰

1 #

Ê <œ

1 4

; mtan œ tan ()
<) œ tan

31 4

Ê 1  cos ) œ 1  sin ) Ê cos ) œ sin ) Ê ) œ

) ’ (1
cos “ sin ))#

œ

1  sin ) cos )

. Thus at ) œ

1 4

, tan <" œ

1  cos ˆ 14 ‰  sin ˆ 14 ‰

1 #

.

œ 1 1 4

; tan <" œ

ˆ 1
"cos ) ‰


sin ) “ ’ (1
cos ))#

œ 1  È2 and

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

1  cos )  sin )

;

;

Chapter 10 Additional and Advanced Exercises tan <# œ

54.

1  sin ˆ 14 ‰ cos ˆ 14 ‰

œ È2  1. Then tan " œ

ŠÈ2  1‹  Š1  È2‹ 1
ŠÈ2  1‹ Š1  È2‹

œ

2È 2  2 2È 2  2

1 4

œ1 Ê "œ

(b) r# œ 2 csc 2) œ

(a)

695

2 sin 2)

œ

2 2 sin ) cos )

Ê r# sin ) cos ) œ 1 Ê xy œ 1, a hyperbola (c) At ) œ

1 4

,xœyœ1 Ê

œ mtan Ê 9 œ

55.

(a) tan ! œ Aœ œ

" #

r ˆ ddr) ‰

'))

tan ! 4

(b) tan ! œ

#

"

Ê

dr r

œ

d) tan !

Ê ln r œ #

B# e2) Î atan !b d) œ ’ B

) tan !

31 4

dy dx

Ê < œ9) œ


C (by integration) Ê r œ Be)Î atan !b for some constant B;

)# (tan !) e2)Î atan !b “ 4 ) "

tan ! 4

œ

cB# e2)#Î atan !b  B# e2)"Î atan !b d

ar##  r"# b since r## œ B# e2)#Î atan !b and r"# œ B# e2)"Î atan !b ; constant of proportionality K œ r ˆ ddr) ‰

Ê

dr d)

œ

r tan !

# Ê ˆ ddr) ‰ œ

r# tan# !

#

Ê r#
ˆ ddr) ‰ œ r#


! )Î atan !b ' ˆ sec ! ‰ ' œ r# Š sec † tan# ! ‹ Ê Length œ ) r tan ! d) œ ) Be #

œ  x"# œ 1

)#

)#

"

"

sec ! tan !

r# tan# !

tan ! 4

#

1 œ r# Š tantan!#
! ‹ )

d) œ cB (sec !) e)Î atan !b d )#"

œ (sec !) cBe)#Î atan !b  Be)" atan !b d œ K (r#  r" ) where K œ sec ! is the constant of proportionality 56.

r# sin 2) œ 2a# Ê r# sin ) cos ) œ a# Ê xy œ a# and dy dx

#

œ  xa# . If P(x" ß y" ) is a point on the curve, the tangent #

line is y  y" œ  xa# (x  x" ), so the tangent line crosses "

#

the x-axis when y œ 0 Ê y" œ  xa# (x  x" ) "

x#" y" a# œ x  since xa" y# " œ 1.

Ê

x" Ê x œ

x#" y" a#


x" œ x"
x" œ 2x"

Let Q be (2x" ß 0). Then È PQ œ (2x"  x" )#
(0  y" )# œ Èx"#
y"# and OP œ r œ È(x"  0)#
(y"  0)# œ Èx"#
y"# Ê OP œ PQ and the triangle is isosceles.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

31 4



1 4

œ

1 #

696

Chapter 10 Conic Sections and Polar Coordinates

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 11 INFINITE SEQUENCES AND SERIES 11.1 SEQUENCES 1. a" œ

1
1 1#

2. a" œ

1 1!

3.

a" œ

"
2 ##

œ 0, a# œ

œ 1, a# œ (
1)# #
1

" #!

œ

œ 1, a# œ

œ
"4 , a$ œ

1
3 3#

" 2

1 6

, a$ œ

(
")$ 4
1

œ

1 3!

, a% œ

œ
3" , a$ œ

1
4 4#

œ
92 , a% œ œ

1 4!

(
1)% 6
1

œ

" 5

3 œ
16

1 24 (
1)& 8
1

, a% œ

œ
7"

4. a" œ 2  (
1)" œ 1, a# œ 2  (
1)# œ 3, a$ œ 2  (
1)$ œ 1, a% œ 2  (
1)% œ 3 5. a" œ

2 ##

œ

6. a" œ

2
" #

" #

, a# œ

œ

" #

a( œ

, a) œ

8. a" œ 1, a# œ a* œ

" 362,880

œ

" # 255 128

" #

" #

œ

3 #

œ

511 256

, a$ œ 3 #



" #

œ

" ##

, a"! œ

ˆ #" ‰ " 3 œ 6 " 3,628,800

, a$ œ

, a"! œ

3 4

, a$ œ

, a* œ

2$ #%

, a$ œ

2#
1 2#

, a# œ

7. a" œ 1, a# œ 1  127 64

2# 2$

, a% œ

, a% œ

2$
1 2$

œ

7 4

œ

2% 2& 7 8

œ

" #

, a% œ

, a% œ

7 4



2%
" 2%

" #$

œ

a' œ

,

15 8

ˆ "6 ‰ 4

œ

" #4

, a& œ

ˆ #"4 ‰ 5

œ

$ (
1)% ˆ
"# ‰ (
1)# (2) œ 1, a$ œ (
1)2 (1) œ
"# , a% œ # # " " a( œ
3"# , a) œ
64 , a* œ 1#"8 , a"! œ 256

1†(
2) œ
1, a$ œ 2†(3
1) œ
23 , a% # a) œ
"4 , a* œ
29 , a"! œ
"5

10. a" œ
2, a# œ a( œ
27 ,

15 16

, a& œ

15 8



" #%

œ

œ

31 16 , a'

63 32

,

1023 512

9. a" œ 2, a# œ " 16

œ

œ

3†ˆ
23 ‰ 4

" 1#0

, a' œ

" 7#0

œ
4" , a& œ

œ
"# , a& œ

, a( œ

" 5040

(
1)& ˆ
4" ‰ #

4†ˆ
"# ‰ 5

, a) œ

œ

" 8

" 40,320

,

,

œ
25 , a' œ
3" ,

11. a" œ 1, a# œ 1, a$ œ 1  1 œ 2, a% œ 2  1 œ 3, a& œ 3  2 œ 5, a' œ 8, a( œ 13, a) œ 21, a* œ 34, a"! œ 55 12. a" œ 2, a# œ
1, a$ œ
"# , a% œ

ˆ
"# ‰
1

œ

" #

, a& œ

ˆ "# ‰ ˆ
"# ‰

œ
1, a' œ
2, a( œ 2, a) œ
1, a* œ
"# , a"! œ

13. an œ (
1)n1 , n œ 1, 2, á

14. an œ (
1)n , n œ 1, 2, á

15. an œ (
1)n1 n# , n œ 1, 2, á

16. an œ

17. an œ n#
1, n œ 1, 2, á

18. an œ n
4 , n œ 1, 2, á

19. an œ 4n
3, n œ 1, 2, á

20. an œ 4n
2 , n œ 1, 2, á

21. an œ

1  (
1)n
1 , #

n œ 1, 2, á

23. n lim 2  (0.1)n œ 2 Ê converges Ä_

22. an œ

(
")n
1 n#

, n œ 1, 2, á

n
"#  (
1)n ˆ "# ‰ #

œ Ú n# Û, n œ 1, 2, á

(Theorem 5, #4)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #

698

Chapter 11 Infinite Sequences and Series

24. n lim Ä_

n  (
")n n

25. n lim Ä_

"
2n 1  #n

26. n lim Ä_

2n  " 1
3È n

œ n lim Ä_

27. n lim Ä_

"
5n% n%  8n$

œ n lim Ä_

28. n lim Ä_

n3 n#  5n  6

œ n lim Ä_

n3 (n  3)(n  2)

œ n lim Ä_

29. n lim Ä_

n#
2n  1 n
1

œ n lim Ä_

(n
1)(n
1) n
1

œ n lim (n
1) œ _ Ê diverges Ä_

30 n lim Ä_

"
n$ 70
4n#

œ n lim 1 Ä_ ˆ "n ‰
2 ˆ "n ‰  2

œ n lim Ä_

(
1)n n

œ 1 Ê converges

2Èn  Š È"n ‹

Š n"% ‹
5 1  ˆ 8n ‰

" ‹
n n# 70 Š #‹
4 n

Š

œ
1 Ê converges

œ
_ Ê diverges

Š È"n
3‹

œ n lim Ä_


2 #

œ n lim Ä_

œ
5 Ê converges " n#

œ 0 Ê converges

œ _ Ê diverges 32. n lim (
1)n ˆ1
"n ‰ does not exist Ê diverges Ä_

31. n lim a1  (
1)n b does not exist Ê diverges Ä_ ˆ n #n " ‰ ˆ1
"n ‰ œ lim ˆ "#  33. n lim Ä_ nÄ_ ˆ2
34. n lim Ä_

" ‰ˆ 3 #n



"‰ #n

ˆ
"# ‰n œ lim 36. n lim Ä_ nÄ_

" (0.9)n


"n ‰ œ

" #

œ 6 Ê converges

(
")n #n

É n 2n 37. n lim  1 œ É n lim Ä_ Ä_ 38. n lim Ä_

" ‰ˆ 1 #n

Ê converges 35. n lim Ä_

(
")nb1 #n
1

œ 0 Ê converges

œ 0 Ê converges

2n n1

œ Ên lim Š 2 ‹ œ È2 Ê converges Ä _ 1
" n

ˆ "0 ‰n œ _ Ê diverges œ n lim Ä_ 9

ˆ 1  n" ‰‹ œ sin 39. n lim sin ˆ 1#  "n ‰ œ sin Šn lim Ä_ Ä_ #

1 #

œ 1 Ê converges

40. n lim n1 cos (n1) œ n lim (n1)(
1)n does not exist Ê diverges Ä_ Ä_ œ 0 because
n" Ÿ

" n

Ê converges by the Sandwich Theorem for sequences

sin n n

42. n lim Ä_

sin# n #n

43. n lim Ä_

n #n

œ n lim Ä_

" #n ln 2

^ œ 0 Ê converges (using l'Hopital's rule)

44. n lim Ä_

3n n$

œ n lim Ä_

3n ln 3 3n#

œ n lim Ä_

45. n lim Ä_

ln (n  ") Èn

œ 0 because 0 Ÿ

œ n lim Ä_

sin n n

Ÿ

41. n lim Ä_

sin# n #n

ˆ n
" 1 ‰

" ‹ Š #È n

Ÿ

" #n

Ê converges by the Sandwich Theorem for sequences

3n (ln 3)# 6n

œ n lim Ä_

œ n lim Ä_

2È n n1

3n (ln 3)$ 6

œ n lim Ä_

^ œ _ Ê diverges (using l'Hopital's rule)

Š È2n ‹

1  Š n" ‹

œ 0 Ê converges

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.1 Sequences 46. n lim Ä_

œ n lim Ä_

ln n ln 2n

ˆ "n ‰ 2 ‰ ˆ 2n

œ 1 Ê converges

47. n lim 81În œ 1 Ê converges Ä_

(Theorem 5, #3)

48. n lim (0.03)1În œ 1 Ê converges Ä_

(Theorem 5, #3)

ˆ1  7n ‰n œ e( Ê converges 49. n lim Ä_ ˆ1
"n ‰n œ lim ’1  50. n lim Ä_ nÄ_

(
") n “

(Theorem 5, #5) n

œ e
" Ê converges

(Theorem 5, #5)

n È 51. n lim 10n œ n lim 101În † n1În œ 1 † 1 œ 1 Ê converges Ä_ Ä_

# n n È ˆÈ 52. n lim n# œ n lim n‰ œ 1# œ 1 Ê converges Ä_ Ä_

ˆ 3 ‰1În œ nÄ_ 1În œ 53. n lim lim n Ä_ n nÄ_ lim 31În

" 1

œ 1 Ê converges

(Theorem 5, #3 and #2)

(Theorem 5, #2)

(Theorem 5, #3 and #2)

54. n lim (n  4)1ÎÐn4Ñ œ x lim x1Îx œ 1 Ê converges; (let x œ n  4, then use Theorem 5, #2) Ä_ Ä_ 55. n lim Ä_

ln n n1În

lim Ä_ ln1Înn œ œ nlim n n

Ä_

_ 1

œ _ Ê diverges

(Theorem 5, #2)

56. n lim cln n
ln (n  1)d œ n lim ln ˆ n n 1 ‰ œ ln Šn lim Ä_ Ä_ Ä_ n n È 57. n lim 4n n œ n lim 4È n œ 4 † 1 œ 4 Ê converges Ä_ Ä_

n n1‹

œ ln 1 œ 0 Ê converges

(Theorem 5, #2)

n È 58. n lim 32n1 œ n lim 32 a1Înb œ n lim 3# † 31În œ 9 † 1 œ 9 Ê converges Ä_ Ä_ Ä_

œ n lim Ä_

"†2†3â(n
1)(n) n†n†nân†n

59. n lim Ä_

n! nn

60. n lim Ä_

(
4)n n!

61. n lim Ä_

n! 106n

œ n lim Ä_

" 'n Š (10n! ) ‹

62. n lim Ä_

n! 2n 3n

œ n lim Ä_

" ˆ 6n!n ‰

œ 0 Ê converges

ˆ " ‰ œ 0 and Ÿ n lim Ä_ n

n! nn

  0 Ê n lim Ä_

n! nn

(Theorem 5, #3)

œ 0 Ê converges

(Theorem 5, #6)

œ _ Ê diverges

œ _ Ê diverges

(Theorem 5, #6)

(Theorem 5, #6)

ˆ " ‰1ÎÐln nÑ œ lim exp ˆ ln"n ln ˆ n" ‰‰ œ lim exp ˆ ln 1ln
nln n ‰ œ e
" Ê converges 63. n lim Ä_ n nÄ_ nÄ_ n ˆ1  n" ‰n ‹ œ ln e œ 1 Ê converges 64. n lim ln ˆ1  "n ‰ œ ln Šn lim Ä_ Ä_

(Theorem 5, #5)

 " ‰‰ ˆ 3n  " ‰n œ lim exp ˆn ln ˆ 3n 65. n lim œ n lim exp Š ln (3n  1)
" ln (3n
1) ‹ 3n
1 Ä _ 3n
1 nÄ_ Ä_ n

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

699

700

Chapter 11 Infinite Sequences and Series 3



3

6n #Î$ ˆ6‰ œ n lim exp
3n
1 "3n  1  œ n lim exp Š (3n  1)(3n Ê converges
1) ‹ œ exp 9 œ e Ä_ Ä_ Š ‹ #

n#

"

"



ˆ n ‰n œ lim exp ˆn ln ˆ n n 1 ‰‰ œ lim exp Š ln n
ln" (n  1) ‹ œ lim exp
n n
1  66. n lim ˆn‰ Ä _ n1 nÄ_ nÄ_ nÄ_ Š "# ‹ n

œ n lim exp Š
Ä_

n# n(n  1) ‹


"

œe

Ê converges

 1) ˆ x ‰1În œ lim x ˆ #n " 1 ‰1În œ x lim exp ˆ n" ln ˆ #n " 1 ‰‰ œ x lim exp Š
ln (2n 67. n lim ‹ n Ä _ 2n  1 nÄ_ nÄ_ nÄ_
2 ! œ x n lim exp ˆ 2n1 ‰ œ xe œ x, x  0 Ê converges Ä_ n

ˆ1
68. n lim Ä_

" ‰n n#

œ n lim exp ˆn ln ˆ1
Ä_

" ‰‰ n#

œ n lim exp
Ä_

ln Š1  n"# ‹

exp –  œ n lim Ä_

ˆ n" ‰

Š n2$ ‹‚Š1  n"# ‹ Š n"# ‹

—

œ n lim exp ˆ n
#
2n1 ‰ œ e! œ 1 Ê converges Ä_ 69. n lim Ä_

3 n †6 n 2cn †n!

œ n lim Ä_

36n n!

œ 0 Ê converges

ˆ 10 ‰n

11 70. n lim œ n lim n ‰n Ä _ ˆ 109 ‰  ˆ 11 Ä_ 12 (Theorem 5, #4)

71. n lim tanh n œ n lim Ä_ Ä_

73. n lim Ä_

n# sin ˆ "n ‰ 2n
1

ˆ 12 ‰n ˆ 10 ‰n 11 11 n 9 n 12 ‰n ˆ 11 ‰n ˆ 12 ‰ ˆ ‰ ˆ  11 10 11 12

en
e
n en  e
n

œ n lim Ä_

œ n lim Ä_

eln n
e
ln n 2

72. n lim sinh (ln n) œ n lim Ä_ Ä_

sin ˆ "n ‰ Š 2n



" ‹ n#

74. n lim n ˆ1
cos "n ‰ œ n lim Ä_ Ä_ 75. n lim tan
" n œ Ä_ ˆ " ‰n  77. n lim Ä_ 3

1 #

" È 2n

(Theorem 5, #6)

e2n
" e2n  1

œ n lim Ä_

œ n lim Ä_

ˆ"  cos "n ‰ ˆ n" ‰

œ n lim Ä_

œ n lim Ä_ n
ˆ "n ‰ #

ˆ 120 ‰n 121 n ˆ 108 ‰ 1 110

2e2n 2e2n

Š

œ n lim Ä_




œ n lim " œ 1 Ê converges Ä_

œ _ Ê diverges

 ˆcos ˆ n" ‰‰ Š n"# ‹ 2 n#

œ 0 Ê converges

2 ‹ n$

œ n lim Ä_


sin ˆ n" ‰‘ Š "# ‹ n Š n"# ‹

Ê converges

 cos ˆ n" ‰ #
ˆ 2n ‰

œ

" #

Ê converges

sin ˆ "n ‰ œ 0 Ê converges œ n lim Ä_

76. n lim Ä_

" Èn

tan
" n œ 0 †

1 #

œ 0 Ê converges

n

n œ n lim Šˆ 3" ‰  Š È"2 ‹ ‹ œ 0 Ê converges Ä_

(Theorem 5, #4)

#

n 1‰ ! È 78. n lim n#  n œ n lim exp ’ ln ann  nb “ œ n lim exp ˆ 2n n#  n œ e œ 1 Ê converges Ä_ Ä_ Ä_

79. n lim Ä_

(ln n)#!! n

œ n lim Ä_

200 (ln n)"** n

œ n lim Ä_

200†199 (ln n)"*) n

œ á œ n lim Ä_

200! n

œ 0 Ê converges

%

80.

& lim (ln n) n Ä _ Èn

œ n lim Ä_ –

Š 5(lnnn) ‹ " Š #È ‹ n

— œ n lim Ä_

10(ln n)% Èn

œ n lim Ä_

80(ln n)$ Èn

œ á œ n lim Ä_

3840 Èn

œ 0 Ê converges

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.1 Sequences È

#

81. n lim Šn
Èn#
n‹ œ n lim Šn
Èn#
n‹ Š n  Èn#
n ‹ œ n lim Ä_ Ä_ Ä_ n n
n œ

" #

n n  È n#
n

œ n lim Ä_

" 1  É1


" n

Ê converges

82. n lim Ä_

" È n#
1
È n#  n

É1
n"#  É1  "n

œ n lim Ä_ 83. n lim Ä_

œ n lim Š Ä_ È

ˆ
"n
1‰

' 84. n lim Ä_ 1

n

" xp

œ n lim Ä_

È n#
1  È n#  n
1
n

œ
2 Ê converges

'1n x" dx œ n lim Ä_

" n

È # È # " ‹ Š Èn#
1  Èn#  n ‹ n#
1
È n#  n n
1 n n

œ n lim Ä_

ln n n

dx œ n lim ’ " Ä _ 1
p

n

" xpc1 “ 1

" n

œ 0 Ê converges

œ n lim Ä_

" 1
p

(Theorem 5, #1)

ˆ np"c1
1‰ œ

" p
1

if p  1 Ê converges

85. 1, 1, 2, 4, 8, 16, 32, á œ 1, 2! , 2" , 2# , 2$ , 2% , 2& , á Ê x" œ 1 and xn œ 2nc2 for n   2 86. (a) 1#
2(1)# œ
1, 3#
2(2)# œ 1; let f(aß b) œ (a  2b)#
2(a  b)# œ a#  4ab  4b#
2a#
4ab
2b# œ 2b#
a# ; a#
2b# œ
1 Ê f(aß b) œ 2b#
a# œ 1; a#
2b# œ 1 Ê f(aß b) œ 2b#
a# œ
1 #

‰
2œ (b) r#n
2 œ ˆ aa2b b

a#  4ab  4b#
2a#
4ab
2b# (a  b)#

In the first and second fractions, yn   n. Let

a b

œ


aa#
2b# b (a  b)#

œ

„" y#n

Ê rn œ Ê2 „ Š y"n ‹

represent the (n
1)th fraction where

for n a positive integer   3. Now the nth fraction is lim rn œ È2.

a  2b ab

a b

#

  1 and b   n
1

and a  b   2b   2n
2   n Ê yn   n. Thus,

nÄ_

87. (a) f(x) œ x#
2; the sequence converges to 1.414213562 ¸ È2 (b) f(x) œ tan (x)
1; the sequence converges to 0.7853981635 ¸

1 4

(c) f(x) œ ex ; the sequence 1, 0,
1,
2,
3,
4,
5, á diverges 88. (a) n lim nf ˆ "n ‰ œ lim b f(??xx) œ lim b f(0??x)x
f(0) œ f w (0), where ?x œ Ä_ ?x Ä ! ?x Ä ! "
" ˆ " ‰ w
" (b) n lim n tan œ f (0) œ x # n 1  0 œ 1, f(x) œ tan Ä_

" n

(c) n lim n ae1În
1b œ f w (0) œ e! œ 1, f(x) œ ex
1 Ä_ (d) n lim n ln ˆ1  2n ‰ œ f w (0) œ 1 22(0) œ 2, f(x) œ ln (1  2x) Ä_ #

89. (a) If a œ 2n  1, then b œ Ú a# Û œ Ú 4n

#

 4n  1 Û # #

#

œ Ú2n#  2n  "# Û œ 2n#  2n, c œ Ü a# Ý œ Ü2n#  2n  "# Ý #

œ 2n#  2n  1 and a#  b# œ (2n  1)  a2n#  2nb œ 4n#  4n  1  4n%  8n$  4n# #

œ 4n%  8n$  8n#  4n  1 œ a2n#  2n  1b œ c# . #

(b) a lim Ä_

Ú a# Û # Ü a# Ý

œ a lim Ä_

2n#  2n 2n#  2n  1

œ 1 or a lim Ä_

#

Ú a# Û # Ü a# Ý

œ a lim sin ) œ Ä_

2n1 ‰ 90. (a) n lim (2n1)1Î a2nb œ n lim exp ˆ ln2n œ n lim exp
Ä_ Ä_ Ä_

21 Š 2n 1‹

#

lim

) Ä 1 Î2

exp ˆ #"n ‰ œ e! œ 1;  œ n lim Ä_

n n n! ¸ ˆ ne ‰ È 2n1 , Stirlings approximation Ê È n! ¸ ˆ ne ‰ (2n1)1Î a2nb ¸

(b)

n 40 50 60

Èn! n

15.76852702 19.48325423 23.19189561

sin ) œ 1

n e

for large values of n

n e

14.71517765 18.39397206 22.07276647

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

701

702

Chapter 11 Infinite Sequences and Series ˆ"‰

ln n " n 91. (a) n lim œ n lim œ n lim œ0 Ä _ nc Ä _ cncc1 Ä _ cnc
Ðln %ÑÎc (b) For all %  0, there exists an N œ e such that n  e
Ðln %ÑÎc Ê ln n 
lnc % Ê ln nc  ln ˆ "% ‰ Ê nc  "% Ê n"c  % Ê ¸ n"c
0¸  % Ê lim n"c œ 0 nÄ_

92. Let {an } and {bn } be sequences both converging to L. Define {cn } by c2n œ bn and c2nc1 œ an , where n œ 1, 2, 3, á . For all %  0 there exists N" such that when n  N" then kan
Lk  % and there exists N# such that when n  N# then kbn
Lk  %. If n  1  2max{N" ß N# }, then kcn
Lk  %, so {cn } converges to L. 93. n lim n1În œ n lim exp ˆ "n ln n‰ œ n lim exp ˆ n" ‰ œ e! œ 1 Ä_ Ä_ Ä_ 94. n lim x1În œ n lim exp ˆ "n ln x‰ œ e! œ 1, because x remains fixed while n gets large Ä_ Ä_ 95. Assume the hypotheses of the theorem and let % be a positive number. For all % there exists a N" such that when n  N" then kan
Lk  % Ê
%  an
L  % Ê L
%  an , and there exists a N# such that when n  N# then kcn
Lk  % Ê
%  cn
L  % Ê cn  L  %. If n  max{N" ß N# }, then L
%  an Ÿ bn Ÿ cn  L  % Ê kbn
Lk  % Ê n lim b œ L. Ä_ n 96. Let %  !. We have f continuous at L Ê there exists $ so that kx
Lk  $ Ê kf(x)
f(L)k  %. Also, an Ä L Ê there exists N so that for n  N kan
Lk  $ . Thus for n  N, kf(an )
f(L)k  % Ê f(an ) Ä f(L). 97. an1   an Ê

3(n  1)  1 (n  1)  1



3n  1 n1

Ê

3n  4 n#



3n  1 n1

Ê 3n#  3n  4n  4  3n#  6n  n  2

Ê 4  2; the steps are reversible so the sequence is nondecreasing;

3n  " n1

 3 Ê 3n  1  3n  3

Ê 1  3; the steps are reversible so the sequence is bounded above by 3 98. an1   an Ê

(2(n  1)  3)! ((n  1)  1)!



(2n  3)! (n  1)!

Ê

(2n  5)! (n  2)!



(2n  3)! (n  1)!

Ê

(2n  5)! (2n  3)!



(n  2)! (n  1)!

Ê (2n  5)(2n  4)  n  2; the steps are reversible so the sequence is nondecreasing; the sequence is not bounded since 99. an1 Ÿ an Ê

(2n  3)! (n  1)!

œ (2n  3)(2n  2)â(n  2) can become as large as we please

2nb1 3nb1 (n  1)!

Ÿ

2n 3n n!

2nb1 3nb1 2n 3n

Ê

(n  1)! n!

Ÿ

Ê 2 † 3 Ÿ n  1 which is true for n   5; the steps are

reversible so the sequence is decreasing after a& , but it is not nondecreasing for all its terms; a" œ 6, a# œ 18, a$ œ 36, a% œ 54, a& œ 324 5 œ 64.8 Ê the sequence is bounded from above by 64.8 100. an1   an Ê 2


2 n 1




" #nb1

 2


2 n




" #n

Ê

reversible so the sequence is nondecreasing; 2
101. an œ 1


" n

converges because

102. an œ n


" n

diverges because n Ä _ and

103. an œ

2 n
1 2n

œ1


" #n

and 0 

" #n

" n

2 n 2 n





2 " n1   #nb1 " #n Ÿ 2 Ê




" #n

Ê

2 n(n  1)

 
#n"b1 ; the steps are

the sequence is bounded from above

Ä 0 by Example 1; also it is a nondecreasing sequence bounded above by 1



" n

; since

" n " n

Ä 0 by Example 1, so the sequence is unbounded Ä 0 (by Example 1) Ê

" #n

Ä 0, the sequence converges; also it is

a nondecreasing sequence bounded above by 1 104. an œ

2 n
1 3n

n

œ ˆ 23 ‰


" 3n

; the sequence converges to ! by Theorem 5, #4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.1 Sequences 105. an œ a(
1)n  1b ˆ nn 1 ‰ diverges because an œ 0 for n odd, while for n even an œ 2 ˆ1  n" ‰ converges to 2; it diverges by definition of divergence 106. xn œ max {cos 1ß cos 2ß cos 3ß á ß cos n} and xn1 œ max {cos 1ß cos 2ß cos 3ß á ß cos (n  1)}   xn with xn Ÿ 1 so the sequence is nondecreasing and bounded above by 1 Ê the sequence converges. 107. If {an } is nonincreasing with lower bound M, then {
an } is a nondecreasing sequence with upper bound
M. By Theorem 1, {
an } converges and hence {an } converges. If {an } has no lower bound, then {
an } has no upper bound and therefore diverges. Hence, {an } also diverges. 108. an   an1 Í

n1 n

 

(n  1)  " n1

Í n#  2n  1   n#  2n Í 1   0 and

n1 n

  1; thus the sequence is

nonincreasing and bounded below by 1 Ê it converges 109. an   an1 Í and

1  È2n Èn

1  È2n Èn

 

"  È2(n  1) Èn  1

Í Èn  1  È2n#  2n   Èn  È2n#  2n Í Èn  1   Èn

  È2 ; thus the sequence is nonincreasing and bounded below by È2 Ê it converges

110. an   an1 Í

1
4n 2n n

 

1
4nb1 #nb1 n 1

Í 2n1
2n1 4n   2n
2n 4n1 Í 2n1
2n   2n1 4n
2n 4n1

Í 2
1 2†4
4 Í 1   4n (2
4) Í 1   (
2) † 4n ; thus the sequence is nonincreasing. However, 4n " " n an œ #n
2n œ #n
2 which is not bounded below so the sequence diverges 111.

4nb1  3n œ4 4n n 3 4  ˆ 4 ‰   4;

ˆ 43 ‰n so an   an1 Í 4  ˆ 43 ‰n   4  ˆ 43 ‰n" Í ˆ 43 ‰n   ˆ 43 ‰n1 Í 1  

3 4

and

thus the sequence is nonincreasing and bounded below by 4 Ê it converges

112. a" œ 1, a# œ 2
3, a$ œ 2(2
3)
3 œ 2#
a22
"b † 3, a% œ 2 a2#
a22
"b † 3b
3 œ 2$
a2$
1b 3, a& œ 2 c2$
a2$
1b 3d
3 œ 2%
a2%
1b 3, á , an œ 2n
"
a2n
"
1b 3 œ 2n
"
3 † 2n
1  3 œ 2n
1 (1
3)  3 œ
2n  3; an   an1 Í
2n  3  
2n1  3 Í
2n  
2n1 Í 1 Ÿ 2 so the sequence is nonincreasing but not bounded below and therefore diverges 113. Let 0  M  1 and let N be an integer greater than Ê n  M  nM Ê n  M(n  1) Ê

n n1

M 1
M

. Then n  N Ê n 

 M.

M 1
M

Ê n
nM  M

114. Since M" is a least upper bound and M# is an upper bound, M" Ÿ M# . Since M# is a least upper bound and M" is an upper bound, M# Ÿ M" . We conclude that M" œ M# so the least upper bound is unique. 115. The sequence an œ 1 

(
")n #

is the sequence

" #

,

3 #

,

" #

,

3 #

, á . This sequence is bounded above by

3 #

,

but it clearly does not converge, by definition of convergence. 116. Let L be the limit of the convergent sequence {an }. Then by definition of convergence, for corresponds an N such that for all m and n, m  N Ê kam
Lk  kam
an k œ kam
L  L
an k Ÿ kam
Lk  kL
an k 

% #



% #

% #

% #

there

and n  N Ê kan
Lk  #% . Now

œ % whenever m  N and n  N.

117. Given an %  0, by definition of convergence there corresponds an N such that for all n  N, kL"
an k  % and kL#
an k  %. Now kL#
L" k œ kL#
an  an
L" k Ÿ kL#
an k  kan
L" k  %  % œ 2%. kL#
L" k  2% says that the difference between two fixed values is smaller than any positive number 2%. The only nonnegative number smaller than every positive number is 0, so kL"
L# k œ 0 or L" œ L# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

703

704

Chapter 11 Infinite Sequences and Series

118. Let k(n) and i(n) be two order-preserving functions whose domains are the set of positive integers and whose ranges are a subset of the positive integers. Consider the two subsequences akÐnÑ and aiÐnÑ , where akÐnÑ Ä L" , aiÐnÑ Ä L# and L" Á L# . Thus ¸akÐnÑ
aiÐnÑ ¸ Ä kL"
L# k  0. So there does not exist N such that for all m, n  N Ê kam
an k  %. So by Exercise 116, the sequence Öan × is not convergent and hence diverges. 119. a2k Ä L Í given an %  0 there corresponds an N" such that c2k  N" Ê ka2k
Lk  %d . Similarly, a2k1 Ä L Í c2k  1  N# Ê ka2k1
Lk  %d . Let N œ max{N" ß N# }. Then n  N Ê kan
Lk  % whether n is even or odd, and hence an Ä L. 120. Assume an Ä 0. This implies that given an %  0 there corresponds an N such that n  N Ê kan
0k  % Ê kan k  % Ê kkan kk  % Ê kkan k
0k  % Ê kan k Ä 0. On the other hand, assume kan k Ä 0. This implies that given an %  0 there corresponds an N such that for n  N, kkan k
0k  % Ê kkan kk  % Ê kan k  % Ê kan
0k  % Ê an Ä 0. n " 121. ¹È 0.5
1¹  10
$ Ê
1000  ˆ #" ‰

Ê N œ 692; an œ ˆ "# ‰

1 În

and lim

1 În

nÄ_


1

" 1000

" 1000

123. (0.9)n  10
$ Ê n ln (0.9) 
3 ln 10 Ê n  2n n!

" #

n

‰ Ê n  ˆ 1001 1000

ln ˆ "# ‰ 999 ‰ ln ˆ 1000

Ê n  692.8

an œ 1

" n 122. ¸È n
1¸  10
$ Ê
1000  n1În
1  n an œ È n œ n1În and n Ä lim_ an œ 1

124.

999 ‰n Ê ˆ 1000 

999 ‰n ‰n Ê n  9123 Ê N œ 9123; Ê ˆ 1000  n  ˆ 1001 1000


3 ln 10 ln (0.9)

n

9 ‰ ¸ 65.54 Ê N œ 65; an œ ˆ 10 and n Ä lim_ an œ 0

 10
( Ê n!  2n 10( and by calculator experimentation, n  14 Ê N œ 14; an œ

125. (a) f(x) œ x#
a Ê f w (x) œ 2x Ê xn1 œ xn


x#n
a #xn

Ê xn1 œ

2x#n
ax#n
ab 2xn

œ

x#n  a 2xn

2n n!

and lim

œ

ˆxn  xa ‰

nÄ_

#

an œ 0

n

(b) x" œ 2, x# œ 1.75, x$ œ 1.732142857, x% œ 1.73205081, x& œ 1.732050808; we are finding the positive number where x#
3 œ 0; that is, where x# œ 3, x  0, or where x œ È3 . 126. x" œ 1.5, x# œ 1.416666667, x$ œ 1.414215686, x% œ 1.414213562, x& œ 1.414213562; we are finding the positive number x#
2 œ 0; that is, where x# œ 2, x  0, or where x œ È2 . 127. x" œ 1, x# œ 1  cos (1) œ 1.540302306, x$ œ 1.540302306  cos (1  cos (1)) œ 1.570791601, x% œ 1.570791601  cos (1.570791601) œ 1.570796327 œ 1# to 9 decimal places. After a few steps, the arc axnc1 b and line segment cos axnc1 b are nearly the same as the quarter circle. 128. (a) S" œ 6.815, S# œ 6.4061, S$ œ 6.021734, S% œ 5.66042996, S& œ 5.320804162, S' œ 5.001555913, S( œ 4.701462558, S) œ 4.419374804, S* œ 4.154212316, S"! œ 3.904959577, S"" œ 3.670662003, S"# œ 3.450422282 so it will take Ford about 12 years to catch up (b) x ¸ 11.8 129-140. Example CAS Commands: Maple: with( Student[Calculus1] ); f := x -> sin(x); a := 0; b := Pi;

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.1 Sequences

705

plot( f(x), x=a..b, title="#23(a) (Section 5.1)" ); N := [ 100, 200, 1000 ]; # (b) for n in N do Xlist := [ a+1.*(b-a)/n*i $ i=0..n ]; Ylist := map( f, Xlist ); end do: for n in N do # (c) Avg[n] := evalf(add(y,y=Ylist)/nops(Ylist)); end do; avg := FunctionAverage( f(x), x=a..b, output=value ); evalf( avg ); FunctionAverage(f(x),x=a..b,output=plot); # (d) fsolve( f(x)=avg, x=0.5 ); fsolve( f(x)=avg, x=2.5 ); fsolve( f(x)=Avg[1000], x=0.5 ); fsolve( f(x)=Avg[1000], x=2.5 ); Mathematica: (sequence functions may vary): Clear[a, n] a[n_]; = n1 / n first25= Table[N[a[n]],{n, 1, 25}] Limit[a[n], n Ä 8] The last command (Limit) will not always work in Mathematica. You could also explore the limit by enlarging your table to more than the first 25 values. If you know the limit (1 in the above example), to determine how far to go to have all further terms within 0.01 of the limit, do the following. Clear[minN, lim] lim= 1 Do[{diff=Abs[a[n]
lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN For sequences that are given recursively, the following code is suggested. The portion of the command a[n_]:=a[n] stores the elements of the sequence and helps to streamline computation. Clear[a, n] a[1]= 1; a[n_]; = a[n]= a[n
1]  (1/5)(n
1) first25= Table[N[a[n]], {n, 1, 25}] The limit command does not work in this case, but the limit can be observed as 1.25. Clear[minN, lim] lim= 1.25 Do[{diff=Abs[a[n]
lim], If[diff < .01, {minN= n, Abort[]}]}, {n, 2, 1000}] minN 141. Example CAS Commands: Maple: with( Student[Calculus1] ); A := n->(1+r/m)*A(n-1) + b; A(0) := A0; A(0) := 1000; r := 0.02015; m := 12; b := 50; pts1 := [seq( [n,A(n)], n=0..99 )]: plot( pts1, style=point, title="#141(a) (Section 11.1)");

# (a)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

706

Chapter 11 Infinite Sequences and Series

A(60); The sequence { A[n] } is not unbounded; limit( A[n], n=infinity ) = infinity. A(0) := 5000; r := 0.0589; m := 12; b := -50; # (b) pts1 := [seq( [n,A(n)], n=0..99 )]: plot( pts1, style=point, title="#141(b) (Section 11.1)"); A(60); pts1 := [seq( [n,A(n)], n=0..199 )]: plot( pts1, style=point, title="#141(b) (Section 11.1)"); # This sequence is not bounded, and diverges to -infinity: limit( A[n], n=infinity ) = -infinity. A(0) := 5000; r := 0.045; m := 4; b := 0; # (c) for n from 1 while A(n)<20000 do end do; n; It takes 31 years (124 quarters) for the investment to grow to $20,000 when the interest rate is 4.5%, compounded quarterly. r := 0.0625; for n from 1 while A(n)<20000 do end do; n; When the interest rate increases to 6.25% (compounded quarterly), it takes only 22.5 years for the balance to reach $20,000. B := k -> (1+r/m)^k * (A(0)+m*b/r) - m*b/r; # (d) A(0) := 1000.; r := 0.02015; m := 12; b := 50; for k from 0 to 49 do printf( "%5d %9.2f %9.2f %9.2f\n", k, A(k), B(k), B(k)-A(k) ); end do; A(0) := 'A(0)'; r := 'r'; m := 'm'; b := 'b'; n := 'n'; eval( AA(n+1) - ((1+r/m)*AA(n) + b), AA=B ); simplify( % ); 142. Example CAS Commands: Maple: r := 3/4.; # (a) for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 99 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,k/10] := [L]; end do: plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title="#142(a) (Section 11.1)" ); R1 := [1.1, 1.2, 1.5, 2.5, 2.8, 2.9]; # (b) for r in R1 do for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 99 do A := r*A*(1-A); L := L, [n,A];

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.1 Sequences end do; pt[r,k/10] := [L]; end do: t := sprintf("#142(b) (Section 11.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R1)], insequence=true ); R2 := [3.05, 3.1, 3.2, 3.3, 3.35, 3.4]; # (c) for r in R2 do for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 99 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,k/10] := [L]; end do: t := sprintf("#142(c) (Section 11.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R2)], insequence=true ); R3 := [3.46, 3.47, 3.48, 3.49, 3.5, 3.51, 3.52, 3.53, 3.542, 3.544, 3.546, 3.548]; for r in R3 do for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 199 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,k/10] := [L]; end do: t := sprintf("#142(d) (Section 11.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R3)], insequence=true ); R4 := [3.5695]; # (e) for r in R4 do for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 299 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,k/10] := [L]; end do: t := sprintf("#142(e) (Section 11.1)\nr = %f", r);

# (d)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

707

708

Chapter 11 Infinite Sequences and Series P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R4)], insequence=true ); R5 := [3.65]; # (f) for r in R5 do for k in $1..9 do A := k/10.; L := [0,A]; for n from 1 to 299 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,k/10] := [L]; end do: t := sprintf("#142(f) (Section 11.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[($1..9)/10] )], style=point, title=t ); end do: display( [seq(P[r], r=R5)], insequence=true ); R6 := [3.65, 3.75]; # (g) for r in R6 do for a in [0.300, 0.301, 0.600, 0.601 ] do A := a; L := [0,a]; for n from 1 to 299 do A := r*A*(1-A); L := L, [n,A]; end do; pt[r,a] := [L]; end do: t := sprintf("#142(g) (Section 11.1)\nr = %f", r); P[r] := plot( [seq( pt[r,a], a=[0.300, 0.301, 0.600, 0.601] )], style=point, title=t ); end do: display( [seq(P[r], r=R6)], insequence=true );

11.2 INFINITE SERIES 1. sn œ

a a1
r n b (1
r)

œ

n 2 ˆ1
ˆ "3 ‰ ‰ " 1
ˆ3‰

2. sn œ

a a1
r n b (1
r)

œ

9 ‰ˆ " ‰n ‰ ˆ 100 1
ˆ 100 " 1
ˆ 100 ‰

3. sn œ

a a1
r n b (1
r)

œ

1
ˆ
"# ‰ 1
ˆ
"# ‰

4. sn œ

1
(
2)n 1
(
2)

, a geometric series where krk  1 Ê divergence

5.

" (n  1)(n  #)

œ

" n1

n




Ê n lim s œ Ä_ n

Ê n lim s œ Ä_ n

Ê n lim s œ Ä_ n

" n#

2 1
ˆ "3 ‰

" ˆ #3 ‰

œ3 9 ‰ ˆ 100

" ‰ 1
ˆ 100

œ

œ

" 11

2 3

Ê sn œ ˆ #"
3" ‰  ˆ 3"
4" ‰  á  ˆ n " 1


" ‰ n#

œ

" #




" n#

Ê n lim s œ Ä_ n

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #

Section 11.2 Infinite Series 6.

5 n(n  1)

œ




5 n

5 n1

Ê sn œ ˆ5
52 ‰  ˆ 52
53 ‰  ˆ 53
54 ‰  á  ˆ n
5 1
n5 ‰  ˆ n5


5 ‰ n1

œ 5


709

5 n1

Ê n lim s œ5 Ä_ n 7. 1


8.

" 16

9.

7 4

" 4





10. 5


" 16

 " 64

7 16

5 4

" 256










7 64

5 16

" 64

 á , the sum of this geometric series is






" 1
ˆ "3 ‰

œ 10 

3 #

œ




" 1
ˆ "3 ‰

œ 10


3 #

œ

13. (1  1)  ˆ 1#
"5 ‰  ˆ 14  1 1
ˆ "# ‰

14. 2  15.

16.

4 5

 

" 1  ˆ "5 ‰

8 25



œ2

5 6

œ

" 1  ˆ "4 ‰

œ

A 2n
1



" ‰ 25 œ 17 6

B 2n  1

4 5

" 1#

7 3

5 1
ˆ
"4 ‰

œ4

" ‰ #7

 á , is the sum of two geometric series; the sum is

" ‰ #7

 á , is the difference of two geometric series; the sum is

 ˆ 18


2 5



" ‰ 1#5

4 25



 á , is the sum of two geometric series; the sum is

8 125

 á ‰ ; the sum of this geometric series is 2 Š 1
"ˆ 2 ‰ ‹ œ 5

4 " " "‰ " ‰ ˆ ˆ" "‰ ˆ" (4n
3)(4n  1) œ 4n
3
4n  1 Ê sn œ 1
5  5
9  9
13  ˆ1
4n " 1 ‰ œ 1  ˆ 4n "
3
4n " 1 ‰ œ 1
4n " 1 Ê n lim s œ n lim Ä_ n Ä_ 6 (2n
1)(2n  1)

œ

17 #

 á œ 2 ˆ1 

16 125

œ

œ

23 #

12. (5
1)  ˆ 5#
"3 ‰  ˆ 54
9" ‰  ˆ 85
5 1
ˆ "# ‰

ˆ 74 ‰

1
ˆ "4 ‰

 á , the sum of this geometric series is

5 64

11. (5  1)  ˆ 5#  "3 ‰  ˆ 54  9" ‰  ˆ 85  5 1
ˆ "# ‰

" ‰ ˆ 16 1
ˆ 4" ‰

 á , the sum of this geometric series is

 á , the sum of this geometric series is

" 1
ˆ
"4 ‰

œ

A(2n  1)  B(2n
1) (2n
1)(2n  1)

á  ˆ 4n "
7


10 3

" ‰ 4n
3

Ê A(2n  1)  B(2n
1) œ 6

2A  2B œ 0 ABœ0 Ê (2A  2B)n  (A
B) œ 6 Ê œ Ê œ Ê 2A œ 6 Ê A œ 3 and B œ
3. Hence, A
Bœ6 A
Bœ6 k

! nœ1

œ 3 ˆ1
17.

nœ1

" ‰ #k  1

40n (2n
1)# (2n  1)#

œ

k

œ 3 ! ˆ #n "
1


6 (2n
1)(2n  1)

œ

œ 3 Š 1"


Ê the sum is lim 3 ˆ1
kÄ_

A (2n
1) #

" ‰ #n  1



B (2n
1)# #



C (2n  1)



" 3

" ‰ #k  1



" 3




" 5



" 5




" 7

á


" #(k
1)  1



" 2k
1




" #k  1 ‹

œ3

D (2n  1)#

#

A(2n
1)(2n  1)  B(2n  1)  C(2n  1)(2n
1)  D(2n
1)# (2n
1)# (2n  1)# # #

Ê A(2n
1)(2n  1)  B(2n  1)  C(2n  1)(2n
1)#  D(2n
1)# œ 40n Ê A a8n$  4n#
2n
1b  B a4n#  4n  1b  C a8n$
4n#
2n  1b œ D a4n#
4n  1b œ 40n Ê (8A  8C)n$  (4A  4B
4C  4D)n#  (
2A  4B
2C
4D)n  (
A  B  C  D) œ 40n Ú Ú 8A  8C œ 0 8A  8C œ 0 Ý Ý Ý Ý 4A  4B
4C  4D œ 0 A B
C Dœ 0 B Dœ 0 Ê Û Ê Û Ê œ Ê 4B œ 20 Ê B œ 5


œ
 2A 4B 2C 4D 40 A 2

œ
2D œ 20 B C 2D 20 2B Ý Ý Ý Ý Ü
A  B  C  D œ 0 Ü
A  B  C  D œ 0 k ACœ0 Ê C œ 0 and A œ 0. Hence, ! ’ (#n
1)40n and D œ
5 Ê œ # (2n  1)# “
A  5  C
5 œ 0 nœ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

710

Chapter 11 Infinite Sequences and Series k

œ 5 ! ’ (#n
" 1)#
nœ1

18.

" (#n  1)# “

œ 5 Š1


" (2k  1)# ‹

2n  1 n# (n  1)#

" n#

œ

" È2 ‹

 Š È"
2

" ‰ #"Î#

"  ˆ #"Î#


Ê n lim s œ Ä_ n 21. sn œ ˆ ln"3
œ
ln"# 

" ‰ ln #

" #




" 1

œ

" #5






" #5

 Š È"3


" (2k  1)# ‹



" (#k
1)#




" (#k  1)# ‹

œ5 " ‰ 16

 á  ’ (n
" 1)#


" n# “

 ’ n"#


" (n  1)# “

" È4 ‹

 á  Š È "
 n

1

" Èn ‹

 Š È"n


" Èn  1 ‹

œ1


" Èn  1

œ1

"  ˆ #"Î$


" ‰ ln 3

" (2(k
1)  1)#


á


œ1

" Èn  1 ‹

" ‰ #"Î$
#"

 ˆ ln"4


" ln (n  2)

" (n  1)# “

" È3 ‹

Ê n lim s œ n lim Š1
Ä_ n Ä_ 20. sn œ ˆ "#


" 9



Ê sn œ ˆ1
4" ‰  ˆ 4"
9" ‰  ˆ 9"


Ê n lim s œ n lim ’1
Ä_ n Ä_ 19. sn œ Š1


" 9

Ê the sum is n lim 5 Š1
Ä_

" (n  1)#




œ 5 Š 1"


" ‰ #"Î%

 ˆ ln"5


 á  ˆ #1ÎÐ"n
1Ñ


" ‰ ln 4

" ‰ #1În

 á  Š ln (n" 1)


 ˆ #1"În


" ln n ‹

" ‰ #1ÎÐn1Ñ

 Š ln (n" 2)


œ

" #




" #1ÎÐn1Ñ

" ln (n  1) ‹

Ê n lim s œ
ln"# Ä_ n

22. sn œ ctan
" (1)
tan
" (2)d  ctan
" (2)
tan
" (3)d  á  ctan
" (n
1)
tan
" (n)d  ctan
" (n)
tan
" (n  1)d œ tan
" (1)
tan
" (n  1) Ê n lim s œ tan
" (1)
Ä_ n 23. convergent geometric series with sum

" 1
Š È" ‹

œ

2

È2 È 2
1

1 #

œ

1 4




1 #

œ
14

œ 2  È2

Š 3# ‹

24. divergent geometric series with krk œ È2  1

25. convergent geometric series with sum

26. n lim (
1)n1 n Á 0 Ê diverges Ä_

27. n lim cos (n1) œ n lim (
1)n Á 0 Ê diverges Ä_ Ä_

28. cos (n1) œ (
1)n Ê convergent geometric series with sum

29. convergent geometric series with sum

30. n lim ln Ä_

" n

" 1
Š

" ‹ e#

œ

" 1
Š "5 ‹

œ

5 6

e# e #
1

œ
_ Á 0 Ê diverges

31. convergent geometric series with sum

2 " 1
Š 10 ‹

32. convergent geometric series with sum

" 1
Š "x ‹

33. difference of two geometric series with sum ˆ1
"n ‰n œ lim ˆ1  34. n lim Ä_ nÄ_


" ‰n n


2œ œ

20 9




18 9

œ

2 9

x x
1

" 1
Š 23 ‹




" 1
Š 3" ‹

œ3


3 #

œ

3 #

œ e
" Á 0 Ê diverges

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1
Š "# ‹

œ1

Section 11.2 Infinite Series 35. n lim Ä_

n! 1000n

œ _ Á 0 Ê diverges

_

_

nœ1

nœ1

nn n!

36. n lim Ä_

œ n lim Ä_

n†nân 1†#ân

 n lim n œ _ Ê diverges Ä_

37. ! ln ˆ n n 1 ‰ œ ! cln (n)
ln (n  1)d Ê sn œ cln (1)
ln (2)d  cln (2)
ln (3)d  cln (3)
ln (4)d  á  cln (n
1)
ln (n)d  cln (n)
ln (n  1)d œ ln (1)
ln (n  1) œ
ln (n  1) Ê n lim s œ
_, Ê diverges Ä_ n 38. n lim a œ n lim ln ˆ 2n n 1 ‰ œ ln ˆ #" ‰ Á 0 Ê diverges Ä_ n Ä_ " 1
ˆ 1e ‰

39. convergent geometric series with sum 40. divergent geometric series with krk œ _

_

nœ0

nœ0

e1 1e

¸

1 1
e

œ

23.141 22.459

1

41. ! (
1)n xn œ ! (
x)n ; a œ 1, r œ
x; converges to _

_

nœ0

nœ0

" 1
(
x)

n 42. ! (
1)n x2n œ ! a
x# b ; a œ 1, r œ
x# ; converges to

43. a œ 3, r œ _

44. ! nœ0

œ

x
1 #

_

(
1)n #

ˆ 3  "sin x ‰n œ !

nœ0

3  sin x 2(4  sin x)

3  sin x 8  2 sin x

œ

3 " 1
Šx
# ‹

; converges to

" #

6 3
x

ˆ 3 
"sin x ‰n ; a œ

" 1
2x

46. a œ 1, r œ
x"# ; converges to

,rœ

for k2xk  1 or kxk 

" #

œ

Ÿ

" 3  sin x

x# x#  1

" 1  x#

x
" #


" 3  sin x " #

" 1
Š
" #‹

" 4

" #

for
1 

Ÿ

for all x ˆsince

45. a œ 1, r œ 2x; converges to

œ

œ

" 1x

for kxk  1

for kxk  1

 1 or
1  x  3

; converges to

ˆ "# ‰

1
Š 3 
" sin x ‹

for all x‰

for ¸ x1# ¸  1 or kxk  1.

x

47. a œ 1, r œ
(x  1)n ; converges to 48. a œ 1, r œ

3
x #

" x 1
Š3
# ‹

; converges to

" 1
sin x

49. a œ 1, r œ sin x; converges to 50. a œ 1, r œ ln x; converges to _

51. 0.23 œ !

nœ0

_

53. 0.7 œ !

nœ0

23 100

7 10

ˆ 10" # ‰n œ

" ‰n ˆ 10 œ

" 1
ln x

23 Š 100 ‹

" ‰ 1
ˆ 100

7 Š 10 ‹

1


" Š 10 ‹

" 1  (x  1)

œ

œ

2 x
1

for kx  1k  1 or
2  x  0

for ¸ 3
# x ¸  1 or 1  x  5

for x Á (2k  1) 1# , k an integer

for kln xk  1 or e
"  x  e _

52. 0.234 œ !

23 99

nœ0

_

54. 0.d œ !

7 9

_

6 Š 100 ‹

nœ0

" 1
Š 10 ‹

1 ‰ ˆ 6 ‰ ˆ " ‰n 55. 0.06 œ ! ˆ 10 œ 10 10

œ

" #x

œ

nœ0

œ

6 90

œ

d 10

234 1000

ˆ 10" $ ‰n œ

" ‰n ˆ 10 œ

234 Š 1000 ‹

" 1
Š 1000 ‹

d Š 10 ‹

" 1
Š 10 ‹

œ

d 9

" 15

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

234 999

711

712

Chapter 11 Infinite Sequences and Series _

56. 1.414 œ 1  !

nœ0

57. 1.24123 œ

124 100

414 1000

_

!

nœ0

_

58. 3.142857 œ 3  !

nœ0

_

59. (a) ! nœ
2 _

60. (a) ! nœ
1

414 Š 1000 ‹

ˆ 10" $ ‰n œ 1 

123 10&

ˆ 10" $ ‰n œ

142,857 10'

œ1

" 1
Š 1000 ‹

124 100



ˆ 10" ' ‰n œ 3 

Š 123& ‹ 10

1
Š

" ‹ 10$

Š 142,857 ' ‹ 10

1
Š

" ‹ 10'

414 999

œ

_

(b) !

5 (n  2)(n  3)

(b) !

nœ0 _

nœ3

" #



" 4

(b) one example is
3#
(c) one example is 1


" #



" 8



" 16

á œ

3 4




3 8




3 16




" 4




" 8




Š "# ‹ 1
Š "# ‹


á œ " 16



œ3

" (n  4)(n  5)

61. (a) one example is

124 100

"413 999

œ

œ

123 10&
10#

œ

142,857 10'
1

124 100

3,142,854 999,999



123 99,900

œ

œ

123,999 99,900

œ

41,333 33,300

116,402 37,037

_

" (n  2)(n  3)

(c) !

5 (n
2)(n
1)

(c) !

nœ5 _

nœ20

" (n
3)(n
#)

5 (n
19)(n
18)

œ1

Š
3# ‹ 1
Š "# ‹

œ
3


á ; the series

k #



k 4



k 8

á œ

Š k# ‹ 1
Š "# ‹

œ k where k is any positive or

negative number. _

Š k# ‹

nœ0

1
Š "# ‹

n 1 62. The series ! kˆ 12 ‰ is a geometric series whose sum is

œ k where k can be any positive or negative number.

_

_

_

_

_

nœ1

nœ1

nœ1

nœ1

nœ1

_

_

_

_

_

nœ1

nœ1

nœ1

nœ1

nœ1

n n 63. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! Š bann ‹ œ ! (1) diverges.

n n n 64. Let an œ bn œ ˆ "# ‰ . Then ! an œ ! bn œ ! ˆ "# ‰ œ 1, while ! aan bn b œ ! ˆ 4" ‰ œ

n

n

_

65. Let an œ ˆ "4 ‰ and bn œ ˆ #" ‰ . Then A œ ! an œ nœ1

" 3

_

_

_

nœ1

nœ1

nœ1

" 3

Á AB.

n , B œ ! bn œ 1 and ! Š bann ‹ œ ! ˆ #" ‰ œ 1 Á

66. Yes: ! Š a"n ‹ diverges. The reasoning: ! an converges Ê an Ä 0 Ê

" an

A B

.

Ä _ Ê ! Š a"n ‹ diverges by the

nth-Term Test. 67. Since the sum of a finite number of terms is finite, adding or subtracting a finite number of terms from a series that diverges does not change the divergence of the series. 68. Let An œ a"  a#  á  an and n lim A œ A. Assume ! aan  bn b converges to S. Let Ä_ n Sn œ (a"  b" )  (a#  b# )  á  (an  bn ) Ê Sn œ (a"  a#  á  an )  (b"  b#  á  bn ) Ê b"  b#  á  bn œ Sn
An Ê n lim ab"  b#  á  bn b œ S
A Ê ! bn converges. This Ä_ contradicts the assumption that ! bn diverges; therefore, ! aan  bn b diverges.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.3 The Integral Test 69. (a) (b)

œ5 Ê

2 1
r Š 13 2 ‹ 1
r

œ1
r Ê rœ

2 5

œ5 Ê

13 10

70. 1  eb  e2b  á œ

3 5

#

; 2  2 ˆ 35 ‰  2 ˆ 35 ‰  á

3 œ 1
r Ê r œ
10 ; " 1
e b

" 9

œ9 Ê

13 2




13 #

3 ‰ ˆ 10 

œ 1
eb Ê eb œ

3 ‰# ˆ 10


13 #

3 ‰$ ˆ 10 á

13 #

Ê b œ ln ˆ 89 ‰

8 9

71. sn œ 1  2r  r#  2r$  r%  2r&  á  r2n  2r2n1 , n œ 0, 1, á Ê sn œ a1  r#  r%  á  r2n b  a2r  2r$  2r&  á  2r2n1 b Ê n lim s œ Ä_ n 1  2r œ 1
r# , if kr# k  1 or krk  1 72. L
sn œ

a 1
r




a a1
r n b 1
r

" 1
r#



2r 1
r#

arn 1
r

œ

#

73. distance œ 4  2 ’(4) ˆ 34 ‰  (4) ˆ 34 ‰  á “ œ 4  2


3 1
Š 34 ‹ 

#

œ 28 m

$

#

4 4 ‰ ˆ3‰ 4 ‰ ˆ3‰ 4 ‰ ˆ3‰ 4 4 3 3 Ɉ 4.9 74. time œ É 4.9  2Ɉ 4.9  2Ɉ 4.9  á œ É 4.9  2É 4.9 ”É 4  É ˆ 4 ‰  á • 4 2 4 4

œ

2 È4.9

 Š È44.9 ‹ –

É 34 1
É 34

—œ

2 È4.9

#

Š4
2È3‹  4È3

È

3  Š È44.9 ‹ Š 2
È ‹œ 3

#

75. area œ 2#  ŠÈ2‹  (1)#  Š È" ‹  á œ 4  2  1  2 #

76. area œ 2 –

1 Š "# ‹ #

#

—  4–

1 Š 4" ‹ #

œ

È4.9 Š2
È3‹

" #

á œ

4 1


—  8–

" —  á œ 1 ˆ4 

#

#

nc1

" 8



 በœ 1


1 16

È3 ˆ " ‰2 4 ‹ 3

A% œ A$  3a4b2 Š An œ lim

È3 4

nÄ_

œ

È3 ˆ " ‰2 4 ‹ 33

n

 ! 3a4bk
2 Š kœ2

An œ

È3 4

_

nœ1

" n#

È3 1#

, A$ œ A#  3a4bŠ

, A5 œ A4  3a4b3 Š

È3 ˆ " ‰ k
1 4 ‹ 32

È3 lim nÄ_ Œ 4

78. Each term of the series !



œ

n

 3È3Œ! kœ2

È3 4

È3 ˆ " ‰2 4 ‹ 32

È3 ˆ " ‰2 4 ‹ 34 ,

œ

kœ2

œ

nc1

œ

1 #

œ_ A" œ

È3 4 ,

...,

n

È3 4

1
Š "# ‹ 

È3 2 4 s , we see that È3 È3 È3 4  12  #7 ,

k
1  ! 3È3a4bk
$ ˆ 9" ‰ œ

4kc$ 9k
1 

Š 4" ‹

Ê n lim L œ n lim 3 ˆ 43 ‰ Ä_ n Ä_

(b) Using the fact that the area of an equilateral triangle of side length s is

¸ 12.58 sec

œ 8 m#

" #

#

1 Š 8" ‹

77. (a) L" œ 3, L# œ 3 ˆ 43 ‰ , L$ œ 3 ˆ 43 ‰ , á , Ln œ 3 ˆ 43 ‰ A# œ A"  3Š

4  2È 3 È4.9 Š2
È3‹

1 36

 3È 3 Œ 1
4  œ 9

n

4kc$ . 9k
1 

È3 4

 3È3Œ!

È3 4

1 ‰  3È3ˆ 20 œ

kœ2

2È 3 5

represents the area of one of the squares shown in the figure, and all of the

_

n squares lie inside the rectangle of width 1 and length ! ˆ "# ‰ œ

œ

n 0

" 1
_

rectangle completely, and the area of the rectangle is 2, we have ! nœ1

" #

" n#

œ 2. Since the squares do not fill the  2.

11.3 THE INTEGRAL TEST 1. converges; a geometric series with r œ

" 10

1

2. converges; a geometric series with r œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" e

1

713

714

Chapter 11 Infinite Sequences and Series

3. diverges; by the nth-Term Test for Divergence, n lim Ä_ 4. diverges by the Integral Test; '1

n

_

5. diverges; ! nœ1

3 Èn

_

6. converges; ! nœ1

_

" Èn

œ3!

nœ1


2 nÈ n

_

dx œ 5 ln (n  1)
5 ln 2 Ê '1

5 x1

_

nœ1


8 n

_

œ
2 !

nœ1

_

œ
8 !

nœ1

5 x1

dx Ä _

, which is a divergent p-series (p œ #" ) " n$Î#

, which is a convergent p-series (p œ 3# )

7. converges; a geometric series with r œ 8. diverges; !

œ1Á0

n n1

" 8

1 _

and since !

1 n

nœ1

9. diverges by the Integral Test:

" n

_

diverges,
8 !

nœ1

'2n lnxx dx œ "# aln# n
ln 2b

Ê

t œ ln x × dt œ dx Ä 10. diverges by the Integral Test: x Õ dx œ et dt Ø œ lim 2ebÎ2 (b
2)
2eÐln 2ÑÎ2 (ln 2
2)‘ œ _

'2_ lnÈxx dx; Ô

1 n

diverges

'2_ lnxx dx

Ä _

'ln_2 tetÎ2 dt œ

b

lim 2tetÎ2
4etÎ2 ‘ ln 2

bÄ_

bÄ_

11. converges; a geometric series with r œ 12. diverges; n lim Ä_ _

13. diverges; ! nœ0


2 n 1

5n 4n  3

_

œ
2 !

nœ0

" n1

1

ˆ ln 5 ‰ ˆ 54 ‰n œ _ Á 0 œ n lim Ä _ ln 4

5n ln 5 4n ln 4

œ n lim Ä_

2 3

, which diverges by the Integral Test

14. diverges by the Integral Test:

'1n 2xdx
1 œ #" ln (2n
1)

15. diverges; n lim a œ n lim Ä_ n Ä_

2n n1

16. diverges by the Integral Test:

'1n Èx ˆÈdxx  1‰ ; – u œ

œ n lim Ä_

2n ln 2 1

Ä _ as n Ä _

œ_Á0 Èx  "

du œ

dx Èx

— Ä

Ènb1 du

'2

u

œ ln ˆÈn  1‰
ln 2

Ä _ as n Ä _

17. diverges; n lim Ä_

Èn ln n

œ n lim Ä_

" Š 2È ‹ n

Š "n ‹

œ n lim Ä_

Èn #

œ_Á0

ˆ1  n" ‰n œ e Á 0 18. diverges; n lim a œ n lim Ä_ n Ä_ 19. diverges; a geometric series with r œ

" ln #

20. converges; a geometric series with r œ

¸ 1.44  1

" ln 3

¸ 0.91  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.3 The Integral Test 21. converges by the Integral Test:

'3_ (ln x) ÈŠ(ln‹x)
1 dx; ” " x

#

u œ ln x Ä du œ x" dx •

'ln_3

" uÈ u#
1

du

b œ lim csec
" kukd ln 3 œ lim csec
" b
sec
" (ln 3)d œ lim cos
" ˆ "b ‰
sec
" (ln 3)‘

bÄ_

bÄ_

œ cos
" (0)
sec
" (ln 3) œ

1 #

22. converges by the Integral Test:

'1_ x a1 "ln xb dx œ '1_ 1 Š(ln‹x) " x

#

œ lim ctan
" ud 0 œ lim atan
" b
tan
" 0b œ b

bÄ_

bÄ_


sec
" (ln 3) ¸ 1.1439

bÄ_

1 #


0œ

dx; ”

#

u œ ln x Ä du œ "x dx •

'0_ 1"u

#

du

1 #

23. diverges by the nth-Term Test for divergence; n lim n sin ˆ "n ‰ œ n lim Ä_ Ä_

sin ˆ "n ‰ ˆ "n ‰

œ lim

24. diverges by the nth-Term Test for divergence; n lim n tan ˆ "n ‰ œ n lim Ä_ Ä_

tan ˆ "n ‰ ˆ "n ‰

œ n lim Ä_

xÄ0

œ1Á0

sin x x

Š
n"# ‹ sec# ˆ n" ‰ Š
n"# ‹

œ n lim sec# ˆ "n ‰ œ sec# 0 œ 1 Á 0 Ä_ 25. converges by the Integral Test: œ lim atan
" b
tan
" eb œ bÄ_

26. converges by the Integral Test: œ lim 2 ln bÄ_

u ‘b u1 e

"

1 #

dx; ”

2x

'e_

u œ ex Ä du œ ex dx •

" 1  u#

ctan
" ud e du œ n lim Ä_

b


tan
" e ¸ 0.35

_

'1

u œ ex × _ _ dx; du œ ex dx Ä 'e u(1 2 u) du œ 'e ˆ 2u
Õ dx œ " du Ø u Ô

2 1  ex

2 ‰ u1

du

bÄ_

28. diverges by the Integral Test: bÄ_ #

x

œ lim 2 ln ˆ b b 1 ‰
2 ln ˆ e e 1 ‰ œ 2 ln 1
2 ln ˆ e e 1 ‰ œ
2 ln ˆ e e 1 ‰ ¸ 0.63

27. converges by the Integral Test:

œ lim

'1_ 1 e e

'1_ 81tancx x dx; ” u œ tan dx x •
"

"

#

du œ

1  x#

'1_ x x1 dx; ” u œ x

#

#

1 Ä du œ 2x dx •

Ä

'11ÎÎ42 8u du œ c4u# d 11ÎÎ24 œ 4 Š 14

#

'2_ du4 œ

" #




1# 16 ‹

œ

31 # 4

b lim  #" ln u‘ 2

bÄ_

(ln b
ln 2) œ _

29. converges by the Integral Test:

'1_ sech x dx œ 2

'1b 1 eae b x

lim

x #

bÄ_

dx œ 2 lim ctan
" ex d 1 b

bÄ_

œ 2 lim atan
" eb
tan
" eb œ 1
2 tan
" e ¸ 0.71 bÄ_

30. converges by the Integral Test:

'1_ sech# x dx œ

œ 1
tanh 1 ¸ 0.76 31.

'1_ ˆ x a 2
x " 4 ‰ dx œ lim

bÄ_

(b  2)a b4

lim

bÄ_

'1b sech# x dx œ

lim ca ln kx  2k
ln kx  4kd 1 œ lim ln b

bÄ_

œ a lim (b  2)a
1 œ œ bÄ_

bÄ_

lim ctanh xd b1 œ lim (tanh b
tanh 1)

bÄ_

(b  2)a b4

bÄ_


ln ˆ 35 ‰ ; a

_, a  1 Ê the series converges to ln ˆ 53 ‰ if a œ 1 and diverges to _ if 1, a œ 1

a  1. If a  1, the terms of the series eventually become negative and the Integral Test does not apply. From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

715

716 32.

Chapter 11 Infinite Sequences and Series

'3_ ˆ x
" 1
x 2a 1 ‰ dx œ " 2ac1 b Ä _ #a(b  1)

œ lim if a 

" #

" #

. If a 

b

lim ’ln ¹ (xx
1)12a ¹“ œ lim ln

bÄ_

œ

3

bÄ_

b
1 (b  1)2a

b
"


ln ˆ 422a ‰ ; lim

2a b Ä _ (b  1)

1, a œ "# Ê the series converges to ln ˆ #4 ‰ œ ln 2 if a œ _, a  "#

" #

and diverges to _ if

, the terms of the series eventually become negative and the Integral Test does not apply.

From that point on, however, the series behaves like a negative multiple of the harmonic series, and so it diverges. 33. (a)

(b) There are (13)(365)(24)(60)(60) a10* b seconds in 13 billion years; by part (a) sn Ÿ 1  ln n where n œ (13)(365)(24)(60)(60) a10* b Ê sn Ÿ 1  ln a(13)(365)(24)(60)(60) a10* bb œ 1  ln (13)  ln (365)  ln (24)  2 ln (60)  9 ln (10) ¸ 41.55 _

34. No, because ! nœ1

" nx

œ

" x

_

" n

! nœ1

_

and ! nœ1

" n

diverges

_

_

_

nœ1

nœ1

nœ1

35. Yes. If ! an is a divergent series of positive numbers, then ˆ "# ‰ ! an œ ! ˆ a#n ‰ also diverges and

an #

 an .

_

There is no “smallest" divergent series of positive numbers: for any divergent series ! an of positive nœ1

_

numbers ! nœ1

ˆ a#n ‰

has smaller terms and still diverges.

_

_

_

nœ1

nœ1

nœ1

36. No, if ! an is a convergent series of positive numbers, then 2 ! an œ ! 2an also converges, and 2an   an . There is no “largest" convergent series of positive numbers. n

n

kœ1

kœ1

37. Let An œ ! ak and Bn œ ! 2k aa2k b , where {ak } is a nonincreasing sequence of positive terms converging to 0. Note that {An } and {Bn } are nondecreasing sequences of positive terms. Now, Bn œ 2a#  4a%  8a)  á  2n aa2n b œ 2a#  a2a%  2a% b  a2a)  2a)  2a)  2a) b  á ˆ2aa2n b  2aa2n b  á  2aa2n b ‰ Ÿ 2a"  2a#  a2a$  2a% b  a2a&  2a'  2a(  2a) b  á  ðóóóóóóóóóóóóóóñóóóóóóóóóóóóóóò 2n
1 terms _

 ˆ2aa2nc1 b  2aa2nc1 1b  á  2aa2n b ‰ œ 2Aa2n b Ÿ 2 ! ak . Therefore if ! ak converges, kœ1

then {Bn } is bounded above Ê ! 2k aa2k b converges. Conversely, _

An œ a"  aa#  a$ b  aa%  a&  a'  a( b  á  an  a"  2a#  4a%  á  2n aa2n b œ a"  Bn  a"  ! 2k aa2k b . kœ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.4 Comparison Tests

717

_

Therefore, if ! 2k aa2k b converges, then {An } is bounded above and hence converges. kœ1

38. (a) aa2n b œ

" 2n ln a2n b

_

" n ln n

Ê !

nœ2

(b) aa2n b œ

39. (a)

" #np

œ

" 2n †n(ln 2)

nœ2

_

_

"

nœ1

nœ1

u œ ln x • Ä du œ dx x

;”

" #n †n(ln 2)

œ

_

" ln #

" n

! nœ2

, which diverges

_

" #np

" a2n bpc1

œ!

nœ1

_

n œ ! ˆ #p"c1 ‰ , a geometric series that nœ1

 1 or p  1, but diverges if p Ÿ 1.

'2_ x(lndxx) œœ

nœ2

Ê ! 2 n a a2 n b œ ! 2 n † #pc1

(ln 2)cpb1 , p  1

" p
1

_

diverges.

converges if

p

_

Ê ! 2 n a a2 n b œ ! 2 n

_, p  "

if p  1. For p œ 1:

'ln_2 ucp du œ

cpb1

lim ’
up  1 “

bÄ_

b

œ lim Š 1
" p ‹ cb
p1
(ln 2)
p1 d bÄ_

ln 2

Ê the improper integral converges if p  1 and diverges

'2_ x dxln x œ

integral diverges if p œ 1.

lim cln (ln x)d b2 œ lim cln (ln b)
ln (ln 2)d œ _, so the improper

bÄ_

bÄ_

_

(b) Since the series and the integral converge or diverge together, ! nœ2

" n(ln n)p

converges if and only if p  1.

40. (a) p œ 1 Ê the series diverges (b) p œ 1.01 Ê the series converges _

(c) ! nœ2

" n aln n$ b

œ

" 3

_

! nœ2

" n(ln n)

; p œ 1 Ê the series diverges

(d) p œ 3 Ê the series converges 41. (a) From Fig. 11.8 in the text with f(x) œ

" x

Ÿ 1  '1 f(x) dx Ê ln (n  1) Ÿ 1  n

Ÿ ˆ1 

" #



" 3

á 

"‰ n

and ak œ " #



" 3

" k

á 


cln (n  1)
ln nd œ ˆ1

If we define an œ 1 

" #

œ

dx Ÿ 1 

" 3



" n

" #



" 3

á 

" n

Ÿ 1  ln n Ê 0 Ÿ ln (n  1)
ln n " #



" 3

 á  n" ‰
ln n™ is bounded above

' nb1

" " " x , n1  n x dx œ ln (n  1)
ln n  #"  3"  á  n" 1
ln (n  1)‰
ˆ1  #"

(b) From the graph in Fig. 11.8(a) with f(x) œ " n1

" n

" x


ln n Ÿ 1. Therefore the sequence ˜ˆ1 

by 1 and below by 0. Ê 0

nb1

, we have '1



" 3

á 

" n


ln n‰ .


ln n, then 0  an1
an Ê an1  an Ê {an } is a decreasing sequence of

nonnegative terms.

_

_

# # b 42. e
x Ÿ e
x for x   1, and '1 ecx dx œ lim c
e
x d " œ lim ˆ
e
b  e
1 ‰ œ ec1 Ê '1 ecx dx converges by

bÄ_

_

bÄ_


n #

the Comparison Test for improper integrals Ê ! e nœ0

_

#

œ 1  ! e
n converges by the Integral Test. nœ1

11.4 COMPARISON TESTS _

1. diverges by the Limit Comparison Test (part 1) when compared with ! nœ1

" Èn

, a divergent p-series:

"

lim nÄ_

Œ #Èn
È $ n Š È"n ‹

œ n lim Ä_

Èn $ n 2È n  È

ˆ " ‰œ œ n lim Ä _ #  n
1Î6

" #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

718

Chapter 11 Infinite Sequences and Series

2. diverges by the Direct Comparison Test since n  n  n  n  Èn  0 Ê

_

" n

term of the divergent series ! nœ1

3 n  Èn



" n

, which is the nth

" n

or use Limit Comparison Test with bn œ

3. converges by the Direct Comparison Test;

sin# n 2n

Ÿ

" #n

, which is the nth term of a convergent geometric series

4. converges by the Direct Comparison Test;

1  cos n n#

Ÿ

2 n#

5. diverges since n lim Ä_

2n 3n
1

œ

2 3

lim

È

" n#

converges

Á0

6. converges by the Limit Comparison Test (part 1) with Š nn#
"n ‹

and the p-series !

" n$Î#

, the nth term of a convergent p-series:

ˆ n n " ‰ œ 1 œ n lim Ä_

nÄ_ Š " ‹ n$Î#

n

n

n

n ‰ 7. converges by the Direct Comparison Test; ˆ 3n n 1 ‰  ˆ 3n œ ˆ "3 ‰ , the nth term of a convergent geometric

series 8. converges by the Limit Comparison Test (part 1) with "

Š $Î# ‹ n

lim nÄ_ Š " È$ n


2

$

‹

É n n$ 2 œ lim É1  œ n lim Ä_ nÄ_

" n$Î#

, the nth term of a convergent p-series:

œ1

2 n$

9. diverges by the Direct Comparison Test; n  ln n Ê ln n  ln ln n Ê

" n



" ln n



" ln (ln n)

_

and ! nœ3

" n

diverges _

10. diverges by the Limit Comparison Test (part 3) when compared with ! nœ2

lim nÄ_

Š (ln"n)# ‹ ˆ n" ‰

œ n lim Ä_

n (ln n)#

œ n lim Ä_

" #(ln n) Š n" ‹

œ

" n # n lim Ä _ ln n

œ

" n

" " # n lim Ä _ Š"‹ n _

11. converges by the Limit Comparison Test (part 2) when compared with ! nœ1 #

lim nÄ_

’ (lnn$n) “ Š n"# ‹

œ n lim Ä_

(ln n)# n

œ n lim Ä_

2(ln n) Š n" ‹ 1

œ 2 n lim Ä_

ln n n

_

nœ1 $

lim nÄ_

Š n"# ‹

œ n lim Ä_

(ln n)$ n

œ n lim Ä_

3(ln n)# Š n" ‹ 1

œ 3 n lim Ä_

œ

" # n lim Ä_

nœ_

" n#

, a convergent p-series:

" n#

, a convergent p-series:

œ0

12. converges by the Limit Comparison Test (part 2) when compared with ! ’ (lnn$n) “

, a divergent p-series:

(ln n)# n

œ 3 n lim Ä_

2(ln n) Š n" ‹ 1

œ 6 n lim Ä_

œ6†0œ0 13. diverges by the Limit Comparison Test (part 3) with lim

nÄ_

’È

1 “ n ln n ˆ n" ‰

œ n lim Ä_

Èn ln n

" n

, the nth term of the divergent harmonic series:

"

œ n lim Ä_

Š 2È n ‹ ˆ n" ‰

œ n lim Ä_

Èn 2

œ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

ln n n

Section 11.4 Comparison Tests " n&Î%

14. converges by the Limit Comparison Test (part 2) with lim

n)# ’ (ln$Î# “ n

nÄ_ Š

" ‹ n&Î%

(ln n)# n"Î%

œ n lim Ä_

œ n lim Ä_

ˆ 2 lnn n ‰

œ 8 n lim Ä_

" Š $Î% ‹ 4n

15. diverges by the Limit Comparison Test (part 3) with lim

nÄ_

ˆ 1
"ln n ‰ ˆ "n ‰

œ n lim Ä_

" Š n" ‹

œ n lim Ä_

n 1  ln n

lim

nÄ_

ˆ n" ‰

œ n lim Ä_

n (1  ln n)#

17. diverges by the Integral Test:

lim

nÄ_

ˆ "n ‰

œ n lim Ä_

" n

" ˆ 2 lnn n ‰

œ n lim Ä_

ˆ n" ‰ Š

" ‹ 4n$Î%

œ 32 n lÄ im_

" n"Î%

œ 32 † 0 œ 0

, the nth term of the divergent harmonic series: œ n lim Ä_

'2_ lnx(x11) dx œ 'ln_3 u du œ

n 1  ln# n

œ 8 n lim Ä_

, the nth term of the divergent harmonic series:

" ’ 2(1
n ln n) “

œ n lim Ä_

18. diverges by the Limit Comparison Test (part 3) with Š 1
"ln# n ‹

" n

ln n n"Î%

œ n lim nœ_ Ä_

16. diverges by the Limit Comparison Test (part 3) with Š (1
"ln n)# ‹

, the nth term of a convergent p-series:

" n

œ n lim Ä_

n #(1  ln n)

" Š 2n ‹

b lim  2" u# ‘ ln 3 œ lim

"

bÄ_ #

bÄ_

œ n lim Ä_

n #

œ_

ab#
ln# 3b œ _

, the nth term of the divergent harmonic series:

œ n lim Ä_

n # ln n

œ n lim Ä_

" Š 2n ‹

œ n lim Ä_

n #

œ_

" 19. converges by the Direct Comparison Test with n$Î# , the nth term of a convergent p-series: n#
1  n for " " n   2 Ê n# an#
1b  n$ Ê nÈn#
1  n$Î# Ê $Î#  or use Limit Comparison Test with nÈ n#
1

n

" n$Î# Èn n#  1

20. converges by the Direct Comparison Test with n#  1 Èn

Ê n#  1  Ènn$Î# Ê _

21. converges because ! nœ1 _

! nœ1

" n2n

"
n n2n

 n$Î# Ê

_

œ!

nœ1

" n2n

_

!

nœ1


" #n

1 n# .

, the nth term of a convergent p-series: n#  1  n# 

" n$Î#

or use Limit Comparison Test with

" . n$Î#

which is the sum of two convergent series:

converges by the Direct Comparison Test since

" n #n



" #n

_

, and ! nœ1


" 2n

is a convergent geometric

series _

22. converges by the Direct Comparison Test: ! nœ1

n  2n n# 2n

_

œ ! ˆ n2" n  nœ1

"‰ n#

and

" n2n



" n#

Ÿ

" #n



" n#

, the sum of

the nth terms of a convergent geometric series and a convergent p-series 23. converges by the Direct Comparison Test:

" 3nc1  1



" 3nc1

, which is the nth term of a convergent geometric

series 24. diverges; n lim Š3 Ä_

nc1

" 3n ‹

ˆ"  œ n lim Ä_ 3

"‰ 3n

œ

" 3

Á0

25. diverges by the Limit Comparison Test (part 1) with "‰ n

ˆsin lim n Ä _ ˆ "n ‰

œ lim

xÄ0

sin x x

" n

, the nth term of the divergent harmonic series:

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

719

720

Chapter 11 Infinite Sequences and Series

26. diverges by the Limit Comparison Test (part 1) with ˆtan "n ‰ lim n Ä _ ˆ "n ‰

œ n lim Š " ‹ Ä _ cos " n

ˆsin n" ‰ ˆ n" ‰

" n

, the nth term of the divergent harmonic series:

œ lim ˆ cos" x ‰ ˆ sinx x ‰ œ 1 † 1 œ 1 xÄ0

" n#

27. converges by the Limit Comparison Test (part 1) with lim

nÄ_


" Š n(n 10n
1)(n
2) ‹ Š n"# ‹

œ n lim Ä_

10n#  n n#  3n  2

20n  1 2n  3

œ n lim Ä_

lim

5n$
3n  n# (n
2) Šn#  5‹ 

Š n"# ‹

nÄ_

œ n lim Ä_

5n$
3n n$
2n#  5n
10

œ n lim Ä_

tanc" n n1Þ1

29. converges by the Direct Comparison Test:

œ n lim Ä_

" n#

28. converges by the Limit Comparison Test (part 1) with



, the nth term of a convergent p-series:

, the nth term of a convergent p-series: 15n#
3 3n#
4n  5

_

1 #

n1Þ1

œ 10

20 2

œ n lim Ä_

1

and !

œ

#

n1Þ1

nœ1

1 #

_

! nœ1

30n 6n
4

" n1Þ1

œ5

is the product of a

convergent p-series and a nonzero constant 30. converges by the Direct Comparison Test: sec
" n 

1 #

Ê

secc" n n1 3 Þ



ˆ 1# ‰ n1 3 Þ

_

and ! nœ1

ˆ 1# ‰ n1 3 Þ

œ

1 #

_

! nœ1

" n1 3 Þ

is the

product of a convergent p-series and a nonzero constant

31. converges by the Limit Comparison Test (part 1) with œ n lim Ä_

"  ec2n 1
ec2n

"
ec2n 1  ec2n

: n lim Ä_

" n#

: n lim Ä_

34. converges by the Limit Comparison Test (part 1) with " 123án

lim nÄ_ 36.

" ˆ n(n #
1) ‰

œ

Š nan 2b 1b ‹ Š n"# ‹

" 1  2#  3#  á  n#

Š n"# ‹

œ n lim coth n œ n lim Ä_ Ä_

en  ecn en
ecn

n Š tanh ‹ n#

Š n"# ‹

œ n lim tanh n œ n lim Ä_ Ä_

en
e
n en  e
n

œ1

33. diverges by the Limit Comparison Test (part 1) with 1n : n lim Ä_

35.

n Š coth ‹ n#

œ1

32. converges by the Limit Comparison Test (part 1) with œ n lim Ä_

" n#

œ

œ n lim Ä_ œ

"

2 n(n  1) .

2n# n#  n

n(n b 1)(2n b 1) 6

œ

1 Š nÈ n n‹

ˆ 1n ‰

" n# : n lim Ä_

Š

œ n lim Ä_

Èn n ‹ n#

Š n"# ‹

1 n n È

œ 1.

n È œ n lim nœ1 Ä_

The series converges by the Limit Comparison Test (part 1) with

œ n lim Ä_

4n 2n  1

6 n(n  1)(2n  1)

Ÿ

œ n lim Ä_ 6 n$

4 2

" n# :

œ 2.

Ê the series converges by the Direct

Comparison Test an 37. (a) If n lim œ 0, then there exists an integer N such that for all n  N, ¹ bann
0¹  1 Ê
1  Ä _ bn Ê an  bn . Thus, if ! bn converges, then ! an converges by the Direct Comparison Test.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

an bn

1

Section 11.4 Comparison Tests an (b) If n lim œ _, then there exists an integer N such that for all n  N, Ä _ bn ! bn diverges, then ! an diverges by the Direct Comparison Test. _

38. Yes, ! nœ1

an n

converges by the Direct Comparison Test because

an n

an bn

 1 Ê an  bn . Thus, if

 an

an 39. n lim œ _ Ê there exists an integer N such that for all n  N, Ä _ bn then ! bn converges by the Direct Comparison Test

an bn

 1 Ê an  bn . If ! an converges,

40. ! an converges Ê n lim a œ 0 Ê there exists an integer N such that for all n  N, 0 Ÿ an  1 Ê an#  an Ä_ n Ê ! a#n converges by the Direct Comparison Test 41. Example CAS commands: Maple: a := n -> 1./n^3/sin(n)^2; s := k -> sum( a(n), n=1..k ); # (a)] limit( s(k), k=infinity ); pts := [seq( [k,s(k)], k=1..100 )]: # (b) plot( pts, style=point, title="#41(b) (Section 11.4)" ); pts := [seq( [k,s(k)], k=1..200 )]: # (c) plot( pts, style=point, title="#41(c) (Section 11.4)" ); pts := [seq( [k,s(k)], k=1..400 )]: # (d) plot( pts, style=point, title="#41(d) (Section 11.4)" ); evalf( 355/113 ); Mathematica: Clear[a, n, s, k, p] a[n_]:= 1 / ( n3 Sin[n]2 ) s[k_]= Sum[ a[n], {n, 1, k}] points[p_]:= Table[{k, N[s[k]]}, {k, 1, p}] points[100] ListPlot[points[100]] points[200] ListPlot[points[200] points[400] ListPlot[points[400], PlotRange Ä All] To investigate what is happening around k = 355, you could do the following. N[355/113] N[1
355/113] Sin[355]//N a[355]//N N[s[354]] N[s[355]] N[s[356]]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

721

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Chapter 11 Infinite Sequences and Series

11.5 THE RATIO AND ROOT TESTS

1. converges by the Ratio Test: ˆ1  "n ‰ œ n lim Ä_

È2

ˆ #" ‰ œ

lim anb1 n Ä _ an " #

”

œ n lim Ä_

È

(n b 1) 2 2nb1 •

”

È n 2 #n

œ n lim Ä_

•

È2

(n  1) #nb1

n

† 2È2 n

1 Š (nenbb1)1 ‹ 2

2. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

3. diverges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

4. diverges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

œ n lim Ä_

#

Š nen ‹

Š (nenbb1)! 1 ‹ ˆ en!n ‰

b 1)! ‹ Š (n 10nb1 ˆ 10n!n ‰

œ

" 10

anb1 an

(n  ")! enb1

†

en n!

œ n lim Ä_

(n  ")! 10nb1

†

10n n!

Š (n10bn1)
1 ‹

œ n lim Ä_

en lim n2 œ n Ä _

†

œ n lim Ä_

"!

5. converges by the Ratio Test: n lim Ä_

(n  1)2 enb1

œ n lim Ä_

"! Š n10n ‹

(n  ")"! 10n
1

œ n lim Ä_

†

œ n lim Ä_

10n n"!

ˆ1  n" ‰# ˆ "e ‰ œ

n" e

n 10

" e

1

œ_

œ_

" ‰ ˆ1  n" ‰"! ˆ 10 œ n lim Ä_

1

ˆ n
n 2 ‰n œ lim ˆ1  6. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ 7. converges by the Direct Comparison Test:

2(
1)n (1.25)n


2 ‰ n n

œ e
# Á 0 n

n

œ ˆ 45 ‰ c2  (
1)n d Ÿ ˆ 45 ‰ (3) which is the nth term of a convergent

geometric series 8. converges; a geometric series with krk œ ¸
23 ¸  1 ˆ1
3n ‰n œ lim ˆ1  9. diverges; n lim a œ n lim Ä_ n Ä_ nÄ_ ˆ1
10. diverges; n lim a œ n lim Ä_ n Ä_

" ‰n 3n


3 ‰ n n

œ n lim 1 Ä_


11. converges by the Direct Comparison Test:

ln n n$



n n$

œ

œ e
$ ¸ 0.05 Á 0

Š
"3 ‹ n

" n#

n
"Î$ ¸ 0.72 Á 0  œe

for n   2, the nth term of a convergent p-series. n

n (ln n) n È É 12. converges by the nth-Root Test: n lim an œ n lim nn œ n lim Ä_ Ä_ Ä_

œ n lim Ä_

Š "n ‹ 1

œ n lim Ä_

ln n n

œ01

13. diverges by the Direct Comparison Test: with "n .

a(ln n)n b1În ann b1În

" n




" n#

œ

n
1 n#



" #

ˆ "n ‰ for n  2 or by the Limit Comparison Test (part 1)

n n ˆ n"
È É 14. converges by the nth-Root Test: n lim an œ n lim Ä_ Ä_ ˆ "
n"# ‰ œ 0  1 œ n lim Ä_ n

" ‰n n#

ˆˆ n"
œ n lim Ä_

" ‰n ‰1În n#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.5 The Ratio and Root Tests 15. diverges by the Direct Comparison Test:

ln n n



" n

for n   3

16. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

(n  1) ln (n  1) #nb1

†

17. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

(n  2)(n  3) (n  1)!

†

n! (n  1)(n  2)

18. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

(n  1)$ en
1

œ

19. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

(n  4)! 3! (n  1)! 3nb1

anb1 20. converges by the Ratio Test: n lim œ n lim Ä _ an Ä_ n  1 2 n  2 2 ˆ n ‰ ˆ3‰ ˆn1‰ œ 3  1 œ n lim Ä_

†

anb1 an

œ n lim Ä_

(n  1)! (2n  3)!

22. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

(n  1)! (n  1)nb1

"

ˆ1  "n ‰n

œ

" e

†

" e

†

(2n  1)! n!

† †

nn n!

" #

œ

1

œ01

1

3! n! 3n (n  3)!

(n  1)2nb1 (n  2)! 3nb1 (n  1)!

21. converges by the Ratio Test: n lim Ä_

œ n lim Ä_

en n$

2n n ln (n)

œ n lim Ä_

n4 3(n  1)

" 3

œ

1

3n n! n2n (n  1)!

n" (2n  3)(2n  2)

œ n lim Ä_

œ01

ˆ n ‰n œ lim œ n lim Ä _ n1 nÄ_

"

ˆ n bn " ‰n

1 n n È ln n

n n n È 23. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É (ln n)n Ä_

n n n È 24. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É (ln n)nÎ2 Ä_

" ln n

œ n lim Ä_

n n È Èln n

œ

œ01

Ä_ Ä_

n n lim È n Èln n lim n

œ01

n È n œ 1‹ Šn lim Ä_

25. converges by the Direct Comparison Test:

œ

n! ln n n(n  2)!

ln n n(n  1)(n  2)



n n(n  1)(n  2)

œ

" (n  1)(n  #)



" n#

which is the nth-term of a convergent p-series 26. diverges by the Ratio Test: n lim Ä_

an
1 an

27. converges by the Ratio Test: n lim Ä_ 28. converges by the Ratio Test: approaches 1 

1 #

œ n lim Ä_

anb1 an

lim anb1 n Ä _ an

3n
1 (n  1)$ 2n
1

œ n lim Ä_ œ n lim Ä_

ˆ 1 b nsin n ‰ an an Š 1 b tan n


" n

an

while the denominator tends to _

29. diverges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

†

c1‰ ˆ 3n 2n b 1 an an

n$ 2n 3n

œ n lim Ä_

n3 (n  1)3

ˆ #3 ‰ œ

3 #

1

œ01

‹ an

œ n lim Ä_

œ n lim Ä_

3n
1 2n  1

"  tan
" n n

œ

3 #

œ 0 since the numerator

1


2 ‰ 30. diverges; an1 œ n n 1 an Ê an1 œ ˆ n n 1 ‰ ˆ n
n 1 an
1 ‰ Ê an1 œ ˆ n n 1 ‰ ˆ n
n 1 ‰ ˆ nn
1 an
2 a" 2‰ 3 ˆ"‰ Ê an1 œ ˆ n n 1 ‰ ˆ n
n 1 ‰ ˆ nn

1 â # a" Ê an1 œ n  1 Ê an1 œ n  1 , which is a constant times the

general term of the diverging harmonic series

31. converges by the Ratio Test: n lim Ä_

anb1 an

œ n lim Ä_

Š 2n ‹ an an

œ n lim Ä_

2 n

œ01

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

723

724

Chapter 11 Infinite Sequences and Series

32. converges by the Ratio Test:

lim anb1 n Ä _ an

œ n lim Ä_

anb1 an

œ n lim Ä_

33. converges by the Ratio Test: n lim Ä_ 34.

n  ln n n  10

 0 and a" œ

Ê an1 œ

n  ln n n  10

" #

Œ

Èn n #


an

an

Š 1 bnln n ‹ an an

n n È

œ n lim Ä_

n

œ

"ln n n

œ n lim Ä_

" #

1

œ n lim Ä_

an  an ; thus an1  an  

" 3

35. diverges by the nth-Term Test: a" œ

œ01

n  ln n n  10

Ê an  0; ln n  10 for n  e"! Ê n  ln n  n  10 Ê " #

" n

1

Ê n lim a Á 0, so the series diverges by the nth-Term Test Ä_ n

3 3 6 " %! " 2 " 2 " 2 " É É É É , a# œ É 3 , a$ œ Ê 3 œ 3 , a% œ ËÊ 3 œ 3 ,á ,

%

n! " n! " n " É an œ É a œ 1 because šÉ 3 Ê n lim 3 › is a subsequence of š 3 › whose limit is 1 by Table 8.1 Ä_ n

36. converges by the Direct Comparison Test: a" œ n!

" #

# $

#

'

' %

#%

, a# œ ˆ "# ‰ , a$ œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , a% œ Šˆ "# ‰ ‹ œ ˆ "# ‰ , á

n

Ê an œ ˆ "# ‰  ˆ "# ‰ which is the nth-term of a convergent geometric series anb1 an

37. converges by the Ratio Test: n lim Ä_ n" " œ n lim œ  1 # Ä _ 2n  1

œ n lim Ä_

2nb1 (n  1)! (n  1)! (2n  2)!

†

(2n)! 2n n! n!

œ n lim Ä_

2(n  1)(n  1) (2n  #)(2n  1)

(3n  3)! 1)! (n  2)! anb1 38. diverges by the Ratio Test: n lim œ n lim † n! (n (3n)! Ä _ an Ä _ (n  1)! (n  2)! (n  3)! (3n  3)(3  2)(3n  1)  2 ‰ ˆ 3n  1 ‰ œ n lim œ n lim 3 ˆ 3n n# n  3 œ 3 † 3 † 3 œ 27  1 Ä _ (n  1)(n  2)(n  3) Ä_ n

n (n!) n È 39. diverges by the Root Test: n lim an ´ n lim œ n lim Ä_ Ä _ É an n b # Ä_

n

œ_1

n! n#

n

n (n!) n (n!) É 40. converges by the Root Test: n lim œ n lim œ n lim É an n b n Ä_ Ä_ Ä_ nn# " Ÿ n lim œ 0  1 Ä_ n

n! nn

ˆ " ‰ ˆ 2n ‰ ˆ 3n ‰ â ˆ n
n 1 ‰ ˆ nn ‰ œ n lim Ä_ n

n n n È 41. converges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É 2n# Ä_

n #n

œ n lim Ä_

n n n È 42. diverges by the Root Test: n lim an œ n lim œ n lim Ä_ Ä _ É a#n b # Ä_

n 4

œ_1

n

n

anb1 an

43. converges by the Ratio Test: n lim Ä_ 2n  " " œ n lim œ  1 4 Ä _ (4†#)(n  1) 44. converges by the Ratio Test: an œ Ê n lim Ä_

(2n  2)! c2nb1 (n  1)!d# a3nb1  1b #

œ n lim Š 4n  6n  2 ‹ Ä _ 4n#  8n  4 45. Ratio: n lim Ä_

anb1 an

†

a1  3cn b a3  3cn b

œ n lim Ä_

œ n lim Ä_

1†3 â (2n
1) (2†4 â #n) a3n  1b

a2n n!b# a3n  1b (2n)!

œ1†

" (n  1)p

†

" 3

np 1

œ

" 3

1†3† â †(2n
1)(2n  1) 4nb1 2nb1 (n  1)!

œ

1†2†3†4 â (2n
1)(2n) (2†4 â 2n)# a3n  1b

œ n lim Ä_

†

" #n ln 2

œ01

4n 2n n! 1†3† â †(2n
1)

œ

(2n)! a2n n!b# a3n  1b

(2n  ")(2n  2) a3n  1b 2# (n  1)# a3n
1  1b

1

ˆ n ‰p œ 1p œ 1 Ê no conclusion œ n lim Ä_ n1

n " n È É Root: n lim an œ n lim np œ n lim Ä_ Ä_ Ä_

" n n ‰p ˆÈ

œ

" (1)p

œ 1 Ê no conclusion

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.6 Alternating Series, Absolute and Conditional Convergence 46. Ratio: n lim Ä_

anb1 an

œ n lim Ä_

" (ln (n  1))p

†

(ln n)p 1

œ ’n lim Ä_

ln n ln (n  1) “

p

ˆ "n ‰ ˆ n b" 1 ‰ •

œ ”n lim Ä_

p

œ Šn lim Ä_

n" n ‹

725

p

œ (1)p œ 1 Ê no conclusion " n n È Root: n lim an œ n lim É (ln n)p œ Ä_ Ä_

"

p

lim (ln n)1În ‹ ŠnÄ_ ˆ

"

; let f(n) œ (ln n)1În , then ln f(n) œ

‰

" n ln n

ln (ln n) n ln n Ê n lim ln f(n) œ n lim œ n lim œ n lim n 1 Ä_ Ä_ Ä_ Ä_ " ln fÐnÑ ! n È an œ œ n lim e œ e œ 1; therefore lim Ä_ nÄ_

p

lim (ln n)1În ‹ ŠnÄ_

47. an Ÿ

n 2n

_

for every n and the series !

_

nœ1

n #n

ln (ln n) n

œ 0 Ê n lim (ln n)1În Ä_ œ (1)" p œ 1 Ê no conclusion

converges by the Ratio Test since n lim Ä_

(n  ") 2nb1

†

2n n

œ

" #

" n#

which is a

1

Ê ! an converges by the Direct Comparison Test nœ1

11.6 ALTERNATING SERIES, ABSOLUTE AND CONDITIONAL CONVERGENCE 1.

_

_

nœ1

nœ1

_

_

nœ1

nœ1

converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ ! convergent p-series

2. converges absolutely Ê converges by the Absolute Convergence Test since ! kan k œ !

" n$Î#

which is a

convergent p-series 3. diverges by the nth-Term Test since for n  10 Ê

4. diverges by the nth-Term Test since n lim Ä_

"0 n n"!

n 10

_

n ‰n ˆ n ‰n Á 0 Ê ! (
1)n1 ˆ 10  1 Ê n lim diverges Ä _ 10 nœ1

œ n lim Ä_

"0n (ln 10)"! 10!

^ œ _ (after 10 applications of L'Hopital's

rule) 5. converges by the Alternating Series Test because f(x) œ ln x is an increasing function of x Ê Ê un   un1 for n   1; also un   0 for n   1 and

" lim n Ä _ ln n

6. converges by the Alternating Series Test since f(x) œ

ln x x

ln n ln n#

œ n lim Ä_

Ê f w (x) œ

ln n 2 ln n

is decreasing

œ0 1
ln x x#

decreasing Ê un   un1 ; also un   0 for n   1 and n lim u œ n lim Ä_ n Ä_ 7. diverges by the nth-Term Test since n lim Ä_

" ln x

œ n lim Ä_

ln n n " #

 0 when x  e Ê f(x) is œ n lim Ä_

œ

8. converges by the Alternating Series Test since f(x) œ ln a1  x
" b Ê f w (x) œ

" #

Š "n ‹ 1

œ0

Á0


" x(x  1)

 0 for x  0 Ê f(x) is

ˆ1  n" ‰‹ œ ln 1 œ 0 decreasing Ê un   un1 ; also un   0 for n   1 and n lim u œ n lim ln ˆ1  "n ‰ œ ln Šn lim Ä_ n Ä_ Ä_ 9. converges by the Alternating Series Test since f(x) œ Ê un   unb1 ; also un   0 for n   1 and n lim u œ Ä_ n 10. diverges by the nth-Term Test since n lim Ä_

3È n  1 Èn  1

Èx  " x1

Ê f w (x) œ

Èn  " lim n Ä _ n1

œ n lim Ä_

3É 1  "

1
x
2È x 2Èx (x  1)#

 0 Ê f(x) is decreasing

œ0 " n

1  Š Èn ‹

œ3Á0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

726

Chapter 11 Infinite Sequences and Series _

_

nœ1

nœ1

" ‰n 11. converges absolutely since ! kan k œ ! ˆ 10 a convergent geometric series

12. converges absolutely by the Direct Comparison Test since ¹ (
1)

nb1

(0.1)n

n

¹œ

" (10)n n

n

" ‰  ˆ 10 which is the nth

term of a convergent geometric series 13. converges conditionally since

" Èn



" Èn  1

" Èn

 0 and n lim Ä_

_

_

nœ1

nœ1

œ 0 Ê convergence; but ! kan k œ !

" n"Î#

is a divergent p-series 14. converges conditionally since _

_

! kan k œ !

nœ1

nœ1

" 1  Èn

" 1  Èn



" 1  Èn  1

is a divergent series since _

_

15. converges absolutely since ! kan k œ ! nœ1

n n $ 1

nœ1

17. converges conditionally since _

œ!

nœ1

" n3

diverges because

" n3 " n3

  

" 1 È n

and

" (n  1)  3

" #È n

n n $ 1



and !

" n#

 0 and n lim Ä_

_

" 4n

 

_

nœ1

" n"Î#

œ 0 Ê convergence; but is a divergent p-series

which is the nth-term of a converging p-series

œ_

n! #n

16. diverges by the nth-Term Test since n lim Ä_

" 1  Èn

 0 and n lim Ä_

" n

and ! nœ1

" n 3

_

œ 0 Ê convergence; but ! kan k nœ1

is a divergent series

_

18. converges absolutely because the series ! ¸ sinn# n ¸ converges by the Direct Comparison Test since ¸ sinn# n ¸ Ÿ nœ1

3n 5n

19. diverges by the nth-Term Test since n lim Ä_

œ1Á0 " 3 ln x

20. converges conditionally since f(x) œ ln x is an increasing function of x Ê Ê

" 3 ln n _

œ!

nœ2



" 3 ln n

" 3 ln (n1)

 0 for n   2 and n lim Ä_

diverges because

" 3 ln n



21. converges conditionally since f(x) œ

" 3n

" x#

" 3 ln n

_

and ! nœ2



" x

" n

" ln ax$ b

œ

_

nœ2

nœ2

œ 0 Ê convergence; but ! kan k œ !

_

nœ1

" n#

_

!

nœ1

" n

" ln an$ b

diverges

Ê f w (x) œ
ˆ x2$ 

"‰ x#

 0 Ê f(x) is decreasing and hence _

_

nœ1

nœ1

ˆ "  n" ‰ œ 0 Ê convergence; but ! kan k œ ! un  unb1  0 for n   1 and n lim Ä _ n# œ!

is decreasing

_

1 n n#

is the sum of a convergent and divergent series, and hence diverges nb1

22. converges absolutely by the Direct Comparison Test since ¹ (
n2)5n ¹ œ

2nb1 n 5 n

n

 2 ˆ 25 ‰ which is the nth term

of a convergent geometric series 23. converges absolutely by the Ratio Test: n lim Š uunbn 1 ‹ œ n lim Ä_ Ä_ ”

(n")# ˆ 23 ‰ n n# ˆ 23 ‰

n
1

•œ

2 3

1

24. diverges by the nth-Term Test since n lim a œ n lim 101În œ 1 Á 0 Ä_ n Ä_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" n#

Section 11.6 Alternating Series, Absolute and Conditional Convergence _


" xb#

25. converges absolutely by the Integral Test since '1 atan
" xb ˆ 1 " x# ‰ dx œ lim ’ atan # bÄ_

#

#

" #

œ lim ’atan
" bb
atan
" 1b “ œ bÄ_

26. converges conditionally since f(x) œ

lim

" x

bÄ_

_

_

nœ1

nœ1

Ê ! kan k œ !

'2b
Šln x‹  dx œ " n ln n

" n ln n

lim cln (ln x)d b2 œ lim cln (ln b)
ln (ln 2)d œ _

œ

(x
ln x)#

œ n lim Ä_

Š "n ‹ 1
Š n" ‹

_

_

nœ1

nœ1

! kan k œ !

bÄ_

diverges

28. converges conditionally since f(x) œ x  Š lnxx ‹

œ 0 Ê convergence; but by the Integral Test,

bÄ_

n n1

27. diverges by the nth-Term Test since n lim Ä_

1
Š lnxx ‹
ln

œ

1
ln x (x
ln x)#

ln x x
ln x

œ1Á0

Ê f w (x) œ

Š "x ‹ (x
ln x)
(ln x) Š1
x" ‹ (x
ln x)#

 0 Ê un   un1  0 when n  e and n lim Ä_

œ 0 Ê convergence; but n
ln n  n Ê

ln n n
ln n

1

31 # 32

 1d Ê f w (x) œ
cln(x(x) ln x)#  0 Ê f(x) is decreasing

" x ln x

Ê un  unb1  0 for n   2 and n lim Ä_

'2_ x dxln x œ

# # ’ˆ 1# ‰
ˆ 14 ‰ “ œ

b

“

" n
ln n

" n



Ê

ln n n
ln n

ln n n
ln n



" n

so that

diverges by the Direct Comparison Test

29. converges absolutely by the Ratio Test: n lim Š uunbn 1 ‹ œ n lim Ä_ Ä_ _

_

nœ1

nœ1

("00)nb1 (n1)!

†

œ n lim Ä_

n! (100)n

"00 n1

œ01

n 30. converges absolutely since ! kan k œ ! ˆ 5" ‰ is a convergent geometric series

_

_

nœ1

nœ1

31. converges absolutely by the Direct Comparison Test since ! kan k œ ! " n#  2n  1



" n#

" n#  2n  1

and

which is the nth-term of a convergent p-series _

_

_

_

nœ1

nœ1

nœ1

nœ1

_

_

nœ1

nœ1

n n n 32. converges absolutely since ! kan k œ ! ˆ lnln nn# ‰ œ ! ˆ 2lnlnnn ‰ œ ! ˆ "# ‰ is a convergent

geometric series _

33. converges absolutely since ! kan k œ ! ¹ (n
È1)n ¹ œ ! _

34. converges conditionally since ! nœ1 _

_

nœ1

nœ1

! kan k œ !

" n

cos n1 n

n

nœ1

_

œ!

nœ1

(
1)n n

" n$Î#

is a convergent p-series

is the convergent alternating harmonic series, but

diverges

 1) n È kan k œ n lim 35. converges absolutely by the Root Test: n lim Š (n(2n) n ‹ Ä_ Ä_ n

36. converges absolutely by the Ratio Test: n lim ¹ anb1 ¹ œ n lim Ä _ an Ä_

a(n  1)!b# ((2n  2)!)

†

1În

œ n lim Ä_

(2n)! (n!)#

œ n lim Ä_

n" #n

œ

" #

1

(n  1)# (2n  2)(2n  1)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

" 4

1

727

728

Chapter 11 Infinite Sequences and Series

37. diverges by the nth-Term Test since n lim kan k œ n lim Ä_ Ä_

ˆ n # 1 ‰n
1 œ _ Á 0  n lim Ä_

(n  1)(n  2)â(n  (n
1)) #nc1

œ n lim Ä_

(2n)! 2n n! n

(n  1)! (n  1)! 3nb1 (2n  3)!

38. converges absolutely by the Ratio Test: n lim ¹ anabn 1 ¹ œ n lim Ä_ Ä_ (n  1)# 3 (2n  2)(2n  3)

œ n lim Ä_

œ

Èn  1
Èn 1

†

Èn  1  Èn Èn  1  Èn

œ

" Èn  1  Èn _

(
")n Èn  1  Èn

nœ1

_

nœ1

nœ1

" Èn  1  Èn

diverges by the Limit Comparison Test (part 1) with

"

Èn
1
Èn œ lim " Èn  n Ä _

lim nÄ_


(2n  1)! n! n! 3n

and š Èn  1"  Èn › is a

decreasing sequence of positive terms which converges to 0 Ê ! _

†

1

3 4

39. converges conditionally since

! kan k œ !

(n  ")(n  2)â(2n) 2n n

œ n lim Ä_

Èn Èn  1  Èn

œ n lim Ä_

1 É1  1n 1

œ

converges; but " Èn ;

a divergent p-series:

" #

È

#

n n 40. diverges by the nth-Term Test since n lim ŠÈn#  n
n‹ œ n lim ŠÈn#  n
n‹ † Š Ènn#  ‹ Ä_ Ä_ n  n

œ n lim Ä_

n È n # n  n

œ n lim Ä_

" É1 "n 1

œ

" #

Á0

É n  Èn  Èn

41. diverges by the nth-Term Test since n lim ŠÉn  Èn
Èn‹ œ n lim ŠÉn  Èn
Èn‹
Ä_ Ä_ – É n  È n  È n — Èn

œ n lim Ä_

É n  Èn  Èn

œ n lim Ä_

" É1 

"

Èn  1

œ

" #

Á0

42. converges conditionally since š Èn  "Èn  1 › is a decreasing sequence of positive terms converging to 0 _

Ê !

nœ1

(
")n Èn  Èn  1 _

so that ! nœ1

converges; but n lim Ä_

" Èn  Èn  1

Š Èn
"Èn
1 ‹ Š È"n ‹

Èn È n È n 1

œ n lim Ä_

_

" Èn

diverges by the Limit Comparison Test with ! nœ1

43. converges absolutely by the Direct Comparison Test since sech (n) œ

œ n lim Ä_

" 1É1 "n

" #

œ

which is a divergent p-series

2 en  ecn

œ

2en e2n  1



2en e2n

œ

2 en

nth term of a convergent geometric series _

_

nœ1

nœ1

44. converges absolutely by the Limit Comparison Test (part 1): ! kan k œ ! Apply the Limit Comparison Test with lim

nÄ_

Œ

2 en c ecn 1 en

2en en
ecn

 œ n lim Ä_

49.

" (2n)!



5 10'

(0.01)& 5 ¹

Ê (2n)! 

œ 2 ‚ 10
"" 10' 5

the n-th term of a convergent geometric series:

œ n lim Ä_

45. kerrork  ¸(
1)' ˆ "5 ‰¸ œ 0.2 47. kerrork  ¹(
1)'

1 en ,

2 en
ecn

2 1
ec2n

œ2 46. kerrork  ¸(
1)' ˆ 10" & ‰¸ œ 0.00001 48. kerrork  k(
1)% t% k œ t%  1

œ 200,000 Ê n   5 Ê 1


" #!



" 4!




" 6!



" 8!

¸ 0.54030

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

which is the

Section 11.6 Alternating Series, Absolute and Conditional Convergence 50.

" n!



10' 5

Ê

5 10'

" #!

 n! Ê n   9 Ê 1
1 

51. (a) an   an1 fails since _

_

nœ1

nœ1

" 3



" #




" 3!



" 4!




_

_

nœ1

nœ1

" 5!

" 6!






" 7!



" 8!

¸ 0.367881944

n n n n (b) Since ! kan k œ ! ˆ 3" ‰  ˆ #" ‰ ‘ œ ! ˆ 3" ‰  ! ˆ #" ‰ is the sum of two absolutely convergent

series, we can rearrange the terms of the original series to find its sum: ˆ 3" 

" 9



" 27

52. s#! œ 1


" #



" 3

 á ‰
ˆ #" 


" 4

á 

" 19

" 4






" 20

" 8

 በœ

ˆ "3 ‰

1
ˆ "3 ‰




ˆ "# ‰

1
ˆ "# ‰

œ

" #


1 œ
#"

" #

†

" #1

¸ 0.6687714032 Ê s#! 

¸ 0.692580927

_

53. The unused terms are ! (
1)j
1 aj œ (
1)n
1 aan
1
an
2 b  (
1)n
3 aan
3
an
4 b  á jœn
1

œ (
1)n
1 caan
1
an
2 b  aan
3
an
4 b  á d . Each grouped term is positive, so the remainder has the same sign as (
1)n
1 , which is the sign of the first unused term. 54. sn œ

" 1 †2



" #†3



" 3†4

" n(n  1)

á 

n

œ!

kœ1

" k(k  1)

n

œ ! ˆ k"
kœ1

œ ˆ1
"# ‰  ˆ "#
3" ‰  ˆ 3"
4" ‰  ˆ 4"
5" ‰  á  ˆ n"


" ‰ k1

" ‰ n1

which are the first 2n terms

of the first series, hence the two series are the same. Yes, for n

sn œ ! ˆ k"
kœ1

" ‰ k 1

œ ˆ1
#" ‰  ˆ #"
3" ‰  ˆ 3"
4" ‰  ˆ 4"
5" ‰  á  ˆ n
" 1
n" ‰  ˆ n"


" ‰ n1

œ 1


" n1

ˆ1
n " 1 ‰ œ 1 Ê both series converge to 1. The sum of the first 2n  1 terms of the first Ê n lim s œ n lim Ä_ n Ä_ ˆ1
n " 1 ‰ œ 1. series is ˆ1
n " 1 ‰  n " 1 œ 1. Their sum is n lim s œ n lim Ä_ n Ä_ _

_

_

_

nœ1

nœ1

nœ1

nœ1

55. Theorem 16 states that ! kan k converges Ê ! an converges. But this is equivalent to ! an diverges Ê ! kan k diverges. _

_

nœ1

nœ1

56. ka"  a#  á  an k Ÿ ka" k  ka# k  á  kan k for all n; then ! kan k converges Ê ! an converges and these _

_

nœ1

nœ1

imply that º ! an º Ÿ ! kan k _

57. (a) ! kan  bn k converges by the Direct Comparison Test since kan  bn k Ÿ kan k  kbn k and hence nœ1 _

! aan  bn b converges absolutely

nœ1 _

_

_

(b) ! kbn k converges Ê !
bn converges absolutely; since ! an converges absolutely and nœ1 _

nœ1

nœ1 _

_

!
bn converges absolutely, we have ! can  (
bn )d œ ! aan
bn b converges absolutely by part (a)

nœ1 _

nœ1

_

_

nœ1 _

nœ1

nœ1 _

(c) ! kan k converges Ê kkk ! kan k œ ! kkan k converges Ê ! kan converges absolutely nœ1

58. If an œ bn œ (
1)n

" Èn

, then ! (
1)n nœ1

59. s" œ
"# , s# œ
"#  1 œ " #

s$ œ
 1


" 4




" 6




" 8

" #

" Èn

nœ1

_

_

nœ1

nœ1

converges, but ! an bn œ !

" n

diverges

,




" 10




" 1#




" 14




" 16




" 18




" #0




" 2#

¸
0.5099,

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

729

730

Chapter 11 Infinite Sequences and Series

s% œ s$  s& œ s%
s' œ s&  s( œ s'


" 3 ¸
0.1766, " " " #4
#6
#8
" 5 ¸
0.312, " " " 46
48
50


" 30




" 3#




" 34




" 36




" 38




" 40




" 42




" 44

¸
0.512,

" 52




" 54




" 56




" 58




" 60




" 62




" 64




" 66

¸
0.51106

N"
1

60. (a) Since ! kan k converges, say to M, for %  0 there is an integer N" such that º ! kan k
Mº  nœ1

N"
1

N"
1

_

nœ1

nœ1

nœN"

_

% #

Í » ! kan k

! kan k  ! kan k » 

Í »
! kan k»  nœN"

_

% #

Í ! kan k  nœN"

% #

% #

. Also, ! an

converges to L Í for %  0 there is an integer N# (which we can choose greater than or equal to N" ) such that ksN#
Lk 

% #

_

. Therefore, ! kan k  nœN"

% #

% #

and ksN#
Lk 

.

_

k

nœ1

nœ1

(b) The series ! kan k converges absolutely, say to M. Thus, there exists N" such that º ! kan k
Mº  % whenever k  N" . Now all of the terms in the sequence ekbn kf appear in ekan kf. Sum together all of the N terms in ekbn kf, in order, until you include all of the terms ekan kf nœ" 1 , and let N# be the largest index in the N#

N#

_

nœ1

nœ1

nœ1

sum ! kbn k so obtained. Then º ! kbn k
Mº  % as well Ê ! kbn k converges to M. _

_

61. (a) If ! kan k converges, then ! an converges and nœ1

nœ1

converges where bn œ

a n  ka n k #

_

_

nœ1

nœ1

a n
ka n k #

_

! an 

nœ1

" #

_

_

nœ1

nœ1

! kan k œ !

a n  ka n k #

a , if an   0 œœ n . 0, if an  0

(b) If ! kan k converges, then ! an converges and converges where cn œ

" #

œœ

" #

_

! an


nœ1

" #

_

_

nœ1

nœ1

! kan k œ !

a n
ka n k #

0, if an   0 . an , if an  0

62. The terms in this conditionally convergent series were not added in the order given. 63. Here is an example figure when N œ 5. Notice that u$  u#  u" and u$  u&  u% , but un   un1 for n   5.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.7 Power Series

731

11.7 POWER SERIES _

nb1

1. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ xxn ¹  1 Ê kxk  1 Ê
1  x  1; when x œ
1 we have ! (
1)n , a divergent Ä_ Ä_ nœ1 _

series; when x œ 1 we have ! 1, a divergent series nœ1

(a) the radius is 1; the interval of convergence is
1  x  1 (b) the interval of absolute convergence is
1  x  1 (c) there are no values for which the series converges conditionally nb1

2. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x(x5)5)n ¹  1 Ê kx  5k  1 Ê
6  x 
4; when x œ
6 we have Ä_ Ä_ _

_

nœ1

nœ1

! (
1)n , a divergent series; when x œ
4 we have ! 1, a divergent series

(a) the radius is 1; the interval of convergence is
6  x 
4 (b) the interval of absolute convergence is
6  x 
4 (c) there are no values for which the series converges conditionally nb1

 1) " " 3. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (4x (4x  1)n ¹  1 Ê k4x  1k  1 Ê
1  4x  1  1 Ê
#  x  0; when x œ
# we Ä_ Ä_ _

_

_

_

nœ1

nœ1

nœ1

have ! (
1)n (
1)n œ ! (
1)2n œ ! 1n , a divergent series; when x œ 0 we have ! (
1)n (1)n nœ1 _

œ ! (
1)n , a divergent series nœ1

(a) the radius is "4 ; the interval of convergence is
#"  x  0 (b) the interval of absolute convergence is
"#  x  0

(c) there are no values for which the series converges conditionally nb1

4. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (3xn
2)1 Ä_ Ä_ Ê
1  3x
2  1 Ê

" 3

†

n (3x
2)n ¹

ˆ n ‰  1 Ê k3x
2k  1  1 Ê k3x
2k n lim Ä _ n1

 x  1; when x œ

" 3

_

nœ1

(b) the interval of absolute convergence is

" 3

(c) the series converges conditionally at x œ

nœ1

" n

conditionally convergent; when x œ 1 we have ! (a) the radius is "3 ; the interval of convergence is

_

we have !

" 3

(
")n n

which is the alternating harmonic series and is

, the divergent harmonic series

Ÿx1

x1 " 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

732

Chapter 11 Infinite Sequences and Series nb1


2) 5. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ (x 10 nb1 Ä_ Ä_

10n (x
2)n ¹

1 Ê

kx
2 k 10

 1 Ê kx
2k  10 Ê
10  x
2  10

_

_

nœ1

nœ1

Ê
8  x  12; when x œ
8 we have ! (
")n , a divergent series; when x œ 12 we have ! 1, a divergent series (a) the radius is "0; the interval of convergence is
8  x  12 (b) the interval of absolute convergence is
8  x  12 (c) there are no values for which the series converges conditionally nb1

6. n lim k2xk  1 Ê k2xk  1 Ê
"#  x  ¹ uunbn 1 ¹  1 Ê n lim ¹ (2x) (2x)n ¹  1 Ê n lim Ä_ Ä_ Ä_ _

! (
")n , a divergent series; when x œ

nœ1

" #

" #

; when x œ
"# we have

_

we have ! 1, a divergent series nœ1

(a) the radius is "# ; the interval of convergence is
"#  x  (b) the interval of absolute convergence is
"#  x 

" #

" #

(c) there are no values for which the series converges conditionally nb1

 1)x 7. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (n (n  3) † Ä_ Ä_

(n  2) nxn ¹

 1 Ê kxk n lim Ä_

_

Ê
1  x  1; when x œ
1 we have ! (
")n nœ1

_

have ! nœ1

n n#,

n n#

(n  1)(n  2) (n  3)(n)

 1 Ê kxk  1

, a divergent series by the nth-term Test; when x œ " we

a divergent series

(a) the radius is "; the interval of convergence is
"  x  " (b) the interval of absolute convergence is
"  x  " (c) there are no values for which the series converges conditionally nb1

8. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x n2)1 Ä_ Ä_

†

n (x  2)n ¹

ˆ  1 Ê kx  2k n lim Ä_ _

Ê
1  x  2  1 Ê
3  x 
1; when x œ
3 we have !

nœ1

_

! nœ1

(
1)n n ,

" n,

n ‰ n1

 1 Ê kx  2k  1

a divergent series; when x œ
" we have

a convergent series

(a) the radius is "; the interval of convergence is
3  x Ÿ
" (b) the interval of absolute convergence is
3  x 
" (c) the series converges conditionally at x œ
1 nb1

x 9. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä _ (n  1)Èn  1 3nb1

Ê

kx k 3

nÈ n 3n xn ¹

1 Ê

kx k 3

n n ‹ n  1 ‹ ŠÉ n lim Ä _ n1

Šn lim Ä_ _

(1)(1)  1 Ê kxk  3 Ê
3  x  3; when x œ
3 we have !

nœ1

_

when x œ 3 we have !

nœ1

1 , n$Î#

(
")n , n$Î#

1

an absolutely convergent series;

a convergent p-series

(a) the radius is 3; the interval of convergence is
3 Ÿ x Ÿ 3 (b) the interval of absolute convergence is
3 Ÿ x Ÿ 3 (c) there are no values for which the series converges conditionally nb1

10. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (xÈ
n1) 1 † Ä_ Ä_

Èn (x
1)n ¹

 1 Ê kx
1k Én lim Ä_ _

Ê
1  x
1  1 Ê 0  x  2; when x œ 0 we have !

nœ1

(
")n , n"Î#

n n1

 1 Ê kx
1k  1

a conditionally convergent series; when x œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.7 Power Series _

we have ! nœ1

1 , n"Î#

a divergent series

(a) the radius is 1; the interval of convergence is 0 Ÿ x  2 (b) the interval of absolute convergence is 0  x  2 (c) the series converges conditionally at x œ 0 nb1

ˆ " ‰  1 for all x 11. n lim † n! ¹  1 Ê kxk n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ x Ä_ Ä _ (n  1)! xn Ä _ n1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1

nb1

ˆ " ‰  1 for all x 12. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ 3 x † 3nn!xn ¹  1 Ê 3 kxk n lim Ä_ Ä _ (n  1)! Ä _ n1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 2nb3

ˆ " ‰  1 for all x 13. n lim † n! ¹  1 Ê x# n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ x Ä_ Ä _ (n  1)! x2nb1 Ä _ n1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 2nb3

ˆ " ‰  1 for all x 14. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (2x(n3)1)! † (2x n!3)2nb1 ¹  1 Ê (2x  3)# n lim Ä_ Ä_ Ä _ n 1 (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1

x 15. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä _ È(n  1)#  3

È n#  3 ¹ xn

_

Ê
1  x  1; when x œ
1 we have !

nœ1

_

! nœ1

" È n#  3

 1 Ê kxk Én lim Ä_

(
")n È n#  3

n#  3 n#  2n  4

 " Ê kxk  1

, a conditionally convergent series; when x œ 1 we have

, a divergent series

(a) the radius is 1; the interval of convergence is
1 Ÿ x  1 (b) the interval of absolute convergence is
1  x  1 (c) the series converges conditionally at x œ
1 n
1

x 16. n lim † ¹ uun
n 1 ¹  1 Ê n lim ¹ Ä_ Ä _ È(n  1)#  3

È n#  3 ¹ xn

_

Ê
1  x  1; when x œ
1 we have !

nœ1

 1 Ê kxk Én lim Ä_

" È n#  3

n#  3 n#  2n  4

 " Ê kxk  1 _

, a divergent series; when x œ 1 we have !

nœ1

(
")n È n#  3

a conditionally convergent series (a) the radius is 1; the interval of convergence is
1  x Ÿ 1 (b) the interval of absolute convergence is
1  x  1 (c) the series converges conditionally at x œ 1  3) 17. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (n  1)(x 5nb1 Ä_ Ä_

nb1

†

5n n(x  3)n ¹

1 Ê

kx  3 k lim 5 nÄ_

ˆ n n " ‰  1 Ê

kx  3 k 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

,

733

734

Chapter 11 Infinite Sequences and Series _

Ê kx  3k  5 Ê
5  x  3  5 Ê
8  x  2; when x œ
8 we have !

nœ1

_

series; when x œ 2 we have !

nœ1

n

n5 5n

n(
5)n 5n

_

œ ! (
1)n n, a divergent nœ1

_

œ ! n, a divergent series nœ1

(a) the radius is 5; the interval of convergence is
8  x  2 (b) the interval of absolute convergence is
8  x  2 (c) there are no values for which the series converges conditionally nb1

18. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ (n  1)x Ä_ Ä _ 4nb1 an#  2n  2b _

Ê
4  x  4; when x œ
4 we have !

nœ1

_

! nœ1

n n#  1

4 n an #  1 b ¹ nxn n(
1)n n#  1

1 Ê

kx k 4 n lim Ä_

#

(n 1) n 1 ¹ n an# a2n  2bb ¹  1 Ê kxk  4

, a conditionally convergent series; when x œ 4 we have

, a divergent series

(a) the radius is 4; the interval of convergence is
4 Ÿ x  4 (b) the interval of absolute convergence is
4  x  4 (c) the series converges conditionally at x œ
4 19. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä_

Èn  1 xnb1 3nb1

3n È n xn ¹

†

1 Ê

kx k 3

ˆ n n 1 ‰  1 Ê Én lim Ä_

kx k 3

 1 Ê kxk  3

_

Ê
3  x  3; when x œ
3 we have ! (
1)n Èn , a divergent series; when x œ 3 we have nœ1

_

! Èn, a divergent series nœ1

(a) the radius is 3; the interval of convergence is
3  x  3 (b) the interval of absolute convergence is
3  x  3 (c) there are no values for which the series converges conditionally 20. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä_

nbÈ 1

n  1 (2x5)nb1 ¹ n n (2x5)n È

 1 Ê k2x  5k n lim Š Ä_

nbÈ 1

n1 ‹ n n È

1

t È

lim t Ä_ Ê k2x  5k Œ tlim n n   1 Ê k2x  5k  1 Ê
1  2x  5  1 Ê
3  x 
2; when x œ
3 we have È n

_

Ä_

_

n ! (
1) nÈn, a divergent series since lim nÈn œ 1; when x œ
2 we have ! È n, a divergent series nÄ_ nœ1 nœ1

(a) the radius is "# ; the interval of convergence is
3  x 
2

(b) the interval of absolute convergence is
3  x 
2 (c) there are no values for which the series converges conditionally

21.

lim ¹ uunbn 1 ¹ nÄ_

 1 Ê n lim Ä_ »

Š1 

nb1

" n
1‹

xnb1

Š1  "n ‹ xn n

"

t

lim Š1  t ‹ e Ä_ »  1 Ê kxk
lim Š1  " ‹n   1 Ê kxk ˆ e ‰  1 Ê kxk  1 n nÄ_ t

_

n Ê
1  x  1; when x œ
1 we have ! (
1)n ˆ1  "n ‰ , a divergent series by the nth-Term Test since nœ1

lim ˆ1  nÄ_

" ‰n n

_

n œ e Á 0; when x œ 1 we have ! ˆ1  n" ‰ , a divergent series nœ1

(a) the radius is "; the interval of convergence is
1  x  1 (b) the interval of absolute convergence is
1  x  1 (c) there are no values for which the series converges conditionally 22. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ ln (nxnln1)xn Ä_ Ä_

nb1

ˆn
1‰ ˆ n ‰  1 Ê kxk  1 ¹  1 Ê kxk n lim º ˆ " ‰ º  1 Ê kxk n lim Ä_ Ä _ n1 "

n

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.7 Power Series

735

_

Ê
1  x  1; when x œ
1 we have ! (
1)n ln n, a divergent series by the nth-Term Test since nœ1

_

lim ln n Á 0; when x œ 1 we have ! ln n, a divergent series

nÄ_

nœ1

(a) the radius is 1; the interval of convergence is
1  x  1 (b) the interval of absolute convergence is
1  x  1 (c) there are no values for which the series converges conditionally nb1 nb1

x ˆ1  n" ‰n ‹ Š lim (n  1)‹  1 23. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (n  1) ¹  1 Ê kxk Šn lim nn xn Ä_ Ä_ Ä_ nÄ_ Ê e kxk n lim (n  1)  1 Ê only x œ 0 satisfies this inequality Ä_

(a) the radius is 0; the series converges only for x œ 0 (b) the series converges absolutely only for x œ 0 (c) there are no values for which the series converges conditionally nb1

24. n lim (n  1)  1 Ê only x œ 4 satisfies this ¹ uunbn 1 ¹  1 Ê n lim ¹ (n n!1)!(x(x

4)4)n ¹  1 Ê kx
4k n lim Ä_ Ä_ Ä_ inequality (a) the radius is 0; the series converges only for x œ 4 (b) the series converges absolutely only for x œ 4 (c) there are no values for which the series converges conditionally nb1

25. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ (x  2) Ä_ Ä _ (n  1) 2nb1

n2n (x  2)n ¹

1 Ê

kx  2 k lim # nÄ_

ˆ n n 1 ‰  1 Ê

kx  2 k #

 1 Ê kx  2k  2

_

Ê
2  x  2  2 Ê
4  x  0; when x œ
4 we have !
" n , a divergent series; when x œ 0 we have nœ1

_

! (
1) nœ1

nb1

, the alternating harmonic series which converges conditionally

n

(a) the radius is 2; the interval of convergence is
4  x Ÿ 0 (b) the interval of absolute convergence is
4  x  0 (c) the series converges conditionally at x œ 0 nb1

nb1

(n  2)(x
1) ˆ n  2 ‰  1 Ê 2 kx
1k  1 26. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (
(2)
2)n (n  1)(x
1)n ¹  1 Ê 2 kx
1k n lim Ä_ Ä_ Ä _ n1

Ê kx
1k 

" #

Ê
"#  x
1 

" #

Ê

" #

 x  3# ; when x œ

" #

_

we have ! (n  1) , a divergent series; when x œ nœ1

_

we have ! (
1)n (n  1), a divergent series n œ1

(a) the radius is "# ; the interval of convergence is (b) the interval of absolute convergence is

" #

" #

x

x

3 #

3 #

(c) there are no values for which the series converges conditionally nb1

x 27. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä _ (n  1) aln (n  1)b#

Ê kxk (1) Œn lim Ä_ _

! nœ1

(
1)n n(ln n)#

#

ˆ "n ‰ ˆ nb" 1 ‰ 

n(ln n)# xn ¹

 1 Ê kxk Šn lim Ä_

 1 Ê kxk Šn lim Ä_

n1 n ‹

#

n ln n ‹ n  1 ‹ Šn lim Ä _ ln (n  1)

#

1

 1 Ê kxk  1 Ê
1  x  1; when x œ
1 we have _

which converges absolutely; when x œ 1 we have !

nœ1

" n(ln n)#

which converges

(a) the radius is "; the interval of convergence is
1 Ÿ x Ÿ 1 (b) the interval of absolute convergence is
1 Ÿ x Ÿ 1 (c) there are no values for which the series converges conditionally

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

3 #

736

Chapter 11 Infinite Sequences and Series nb1

x 28. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ Ä_ Ä _ (n  1) ln (n  1)

n ln (n) xn ¹

 1 Ê kxk Šn lim Ä_

ln (n) n ‹ n  1 ‹ Šn lim Ä _ ln (n  1) _

(
1)n n ln n

Ê kxk (1)(1)  1 Ê kxk  1 Ê
1  x  1; when x œ
1 we have !

nœ2

_

" n ln n

when x œ 1 we have !

nœ2

1

, a convergent alternating series;

which diverges by Exercise 38, Section 11.3

(a) the radius is "; the interval of convergence is
1 Ÿ x  1 (b) the interval of absolute convergence is
1  x  1 (c) the series converges conditionally at x œ
1 2nb3


5) 29. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ (4x (n  1)$Î# Ä_ Ä_

n$Î# (4x
5)2n
1 ¹

 1 Ê (4x
5)# Šn lim Ä_

Ê k4x
5k  1 Ê
1  4x
5  1 Ê 1  x  absolutely convergent; when x œ

3 #

_

we have ! nœ1

(")2nb1 n$Î#

_

; when x œ 1 we have !

3 #

nœ1

$Î#

 1 Ê (4x
5)#  1

(
1)2nb1 n$Î#

_

œ!

nœ1


" n$Î#

which is

, a convergent p-series

(a) the radius is "4 ; the interval of convergence is 1 Ÿ x Ÿ (b) the interval of absolute convergence is 1 Ÿ x Ÿ

n n1‹

3 #

3 #

(c) there are no values for which the series converges conditionally nb2

30. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (3x2n1)4 Ä_ Ä_

2n  2 (3x  1)nb1 ¹

†

ˆ 2n  2 ‰  1 Ê k3x  1k  1  1 Ê k3x  1k n lim Ä _ 2n  4 _

Ê
1  3x  1  1 Ê
23  x  0; when x œ
23 we have !

nœ1

_

(")nb1 2n  1

when x œ 0 we have !

nœ1

_

" #n  1

œ!

nœ1

(
1)nb1 2n  1

, a conditionally convergent series;

, a divergent series

(a) the radius is "3 ; the interval of convergence is
23 Ÿ x  0 (b) the interval of absolute convergence is
23  x  0 (c) the series converges conditionally at x œ
23 nb1

31. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (xÈn1) 1 † Ä_ Ä_

Èn (x  1)n ¹

n  1 Ê kx  1k n lim ¹ ¹ 1 Ä _ Én1

ˆ n ‰  1 Ê kx  1 k  1 Ê
1  x  1  1 Ê
1
1  x  1
1 ; Ê kx  1k Én lim Ä _ n1 _

(
1)n Èn

when x œ
1
1 we have !

nœ1

_

! nœ1

"n Èn

_

œ!

nœ1

" n"Î#

_

œ!

nœ1

(
")n n"Î#

, a conditionally convergent series; when x œ 1
1 we have

, a divergent p-series

(a) the radius is "; the interval of convergence is (
1
1) Ÿ x  (1
1) (b) the interval of absolute convergence is
1
1  x  1
1 (c) the series converges conditionally at x œ
1
1 2nb3

32.

lim ¹ uunbn 1 ¹ nÄ_

 1 Ê n lim Ä_ »

Šx
È2‹

#

Ê

Šx
È2‹ #

2nb1

nœ1

nœ1

1 Ê

#

lim

nÄ_

k1k  1

#

_ Š
È2‹

we have !

†

2n
1

Šx
È2‹

 1 Ê Šx
È2‹  2 Ê ¹x
È2¹  È2 Ê
È2  x
È2  È2 Ê 0  x  2È2 ; when

x œ 0 we have ! _

#

2n 2n
1 » È Šx
2‹

2nb1

ŠÈ2‹ 2n

2n

_

œ
!

nœ1

_

œ!

nœ1

2

nb1Î2

2n

2nb1Î2 #n

_

œ
! È2 which diverges since n lim a Á 0; when x œ 2È2 Ä_ n nœ1

_

œ ! È2, a divergent series nœ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.7 Power Series (a) the radius is È2; the interval of convergence is 0  x  2È2 (b) the interval of absolute convergence is 0  x  2È2 (c) there are no values for which the series converges conditionally 2nb2

33. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x
4n1)
1 Ä_ Ä_

4n (x
1)2n ¹

†

1 Ê

(x
1)# lim 4 nÄ_ _

Ê
2  x
1  2 Ê
1  x  3; at x œ
1 we have !

nœ0

_

we have ! nœ0 _

! nœ0

(x
")2n 4n "

#

n

œ!

2 4n

4 4n

nœ0

_

œ!

nœ0

œ

" 1
Šxc # ‹

_

2n

4
(x
")# “ 4

œ

n

4 4
x#  2x
1

œ

4 3  2x
x#

9n (x  1)2n ¹

†

1 Ê

(x  1)# lim 9 nÄ_ _

nœ0

! !

nœ0

k1k  1 Ê (x  1)#  9 Ê kx  1k  3

(
3)2n 9n

_

œ ! 1 which diverges; at x œ 2 we have nœ0

_

œ ! " which also diverges; the interval of convergence is
4  x  2; the series

(x  1) 9n "

nœ0

is a convergent geometric series when
1  x  3 and the sum is

Ê
3  x  1  3 Ê
4  x  2; when x œ
4 we have !

nœ0 _

nœ0

nœ0

2nb2

32n 9n

_

œ ! 1, a divergent series; the interval of convergence is
1  x  3; the series

34. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x 9n1)
1 Ä_ Ä_

_

_

n œ ! 44n œ ! 1, which diverges; at x œ 3

_

# Šˆ x
# 1 ‰ ‹ "

’

(
2)2n 4n

k1k  1 Ê (x
1)#  4 Ê kx
1k  2

nœ0

2n

_

n

# œ ! Šˆ x3 1 ‰ ‹ is a convergent geometric series when
4  x  2 and the sum is nœ0

1 1
Šxb 3 ‹

#

œ

"

’

9
(x  1)# “ 9

œ

9 9
x#
2x
1

35. n lim ¹ uunbn 1 ¹  1 Ê n lim Ä_ Ä_ º

œ

9 8
2x
x#

ˆÈx
2‰nb1 2nb1

2n ˆÈ x
2 ‰ n º

†

 1 Ê ¸È x
2 ¸  2 Ê
2  È x
2  2 Ê 0  È x  4

_

_

Ê 0  x  16; when x œ 0 we have ! (
1)n , a divergent series; when x œ 16 we have ! (1)n , a divergent nœ0

nœ0

_

series; the interval of convergence is 0  x  16; the series !

nœ0

0  x  16 and its sum is

1
Œ

" Èx c 2 œ

#


Œ

2c

" Èx
2 œ

#




Èx
2 n Š # ‹

is a convergent geometric series when

2 4
Èx

nb1

36. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (ln(lnx)x)n ¹  1 Ê kln xk  1 Ê
1  ln x  1 Ê e
"  x  e; when x œ e
" or e we Ä_ Ä_ _

_

nœ0

nœ0

obtain the series ! 1n and ! (
1)n which both diverge; the interval of convergence is e
"  x  e; _

! (ln x)n œ

nœ0

" 1
ln x

when e
"  x  e

37. n lim ¹ uunbn 1 ¹  1 Ê n lim Šx Ä_ Ä_ º

#

1 3 ‹

n 1

n † ˆ x# 3 1 ‰ º  1 Ê

ax #  1 b lim 3 nÄ_

k1k  1 Ê

x#  " 3

 1 Ê x#  2

_

Ê kxk  È2 Ê
È2  x  È2 ; at x œ „ È2 we have ! (1)n which diverges; the interval of convergence is nœ0

_


È2  x  È2 ; the series !

nœ0

" # 1
Š x 3b 1 ‹

œ

" # Š 3 c x3 c 1 ‹

œ

# Š x 3 1 ‹

n

is a convergent geometric series when
È2  x  È2 and its sum is

3 #
x#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

737

738

Chapter 11 Infinite Sequences and Series

a 38. n lim ¹ uun
n 1 ¹  1 Ê n lim ¹x Ä_ Ä_

#
1bn
1 2n
1

2n ¹ ax #  1 b n

†

 1 Ê kx#
1k  2 Ê
È3  x  È3 ; when x œ „ È3 we

_

_

nœ0

nœ0

have ! 1n , a divergent series; the interval of convergence is
È3  x  È3 ; the series ! Š x " # 1
Š x 2
1 ‹

convergent geometric series when
È3  x  È3 and its sum is

nb1

39. n lim ¹ (x
#n3) b1 Ä_

2n (x
3)n ¹

†

"

œ 

2
Šx#
1 ‹

#

œ

#


1 2 ‹

n

is a

2 3
x#



_

 1 Ê kx
3k  2 Ê 1  x  5; when x œ 1 we have ! (1)n which diverges; nœ1

_

when x œ 5 we have ! (
1) which also diverges; the interval of convergence is 1  x  5; the sum of this n

nœ1

" 3 1  Šxc # ‹

convergent geometric series is œ

2 x
1

œ

2 x
1

n

. If f(x) œ 1
#" (x
3)  4" (x
3)#  á  ˆ
#" ‰ (x
3)n  á n

then f w (x) œ
#"  #" (x
3)  á  ˆ
#" ‰ n(x
3)n
1  á is convergent when 1  x  5, and diverges
2 (x
1)#

when x œ 1 or 5. The sum for f w (x) is

, the derivative of

2 x
1

.

n

40. If f(x) œ 1
"# (x
3)  4" (x
3)#  á  ˆ
#" ‰ (x
3)n  á œ œx


(x
3)# 4



(x
3)$ 12

_

the series

n ! (
1) 2 n 1

 á  ˆ
"# ‰

n (x
3)n
1 n 1

2 ln kx
1k  (3
ln 4), since '

œ

_

 á . At x œ 1 the series ! n
21 diverges; at x œ 5 nœ1

dx œ 2 ln kx
1k  C, where C œ 3
ln 4 when x œ 3.

2 x
1

41. (a) Differentiate the series for sin x to get cos x œ 1
x% 4!

then ' f(x) dx

converges. Therefore the interval of convergence is 1  x Ÿ 5 and the sum is

nœ1

x# #!

2 x
1

x' 6!

) x"! 1

 x8!
10! á . a#nb! x2nb2 2 lim ¹ † x#8 ¹ œ x n lim n Ä _ (2n  2)! Ä_

(b) sin 2x œ 2x


2$ x$ 3!

2& x& 5!

2( x( 7!

" 6!

œ 2x
42. (a) (b)

d x






5x% 5!




7x' 7!



9x) 9!




11x"! 11!

á

The series converges for all values of x since Š a2n  1ba" 2n  2b ‹ œ 0  1 for all x.

1†00†

" 4!

$ $

( (

2 x 3!

& &



2 x 5!




2x 2!



3x# 3!

aex b œ 1 

0† 

2 x 7!



4x$ 4!

' ex dx œ ex  C œ x  x#

#

#

(c) e
x œ 1
x  x#!
 ˆ1 † 3!"
1 † #"!   ˆ1 † 5!"
1 † 4!" 

x$ 3! " #! " #!



2* x* 9!

2"" x"" 11!

" 3!

* *

2 x 9!



" #

0†

"" ""

á

0†


5x% 5!

2 x 11!

œ

x #



%

x 1#



'

x 45

)



&

(

*

""

(b) sec# x œ when


d(tan x) dx 1 #

œ

x

d dx

" 5!

 0 † 1‰ x'  á ‘ œ 2 ’x


x# #!



x$ 3!



x% 4!

4x$ 3!



16x& 5!


á“

 á œ ex ; thus the derivative of ex is ex itself

x$ x% x& x 3!  4!  5!  á  C, which is the general antiderivative of e % &  x4!
x5!  á ; ecx † ex œ 1 † 1  (1 † 1
1 † 1)x  ˆ1 † #"!
1 † 1  #"! † 1
3!" † 1‰ x$  ˆ1 † 4!"
1 † 3!"  #"! † #"!
3!" † 1  4!" † 1‰ x% † 3!"
3!" † #"!  4!" † 1
5!" † 1‰ x&  á œ 1  0  0  0  0  0  á

17x 2520 1 # 

converges when


 á œ 2x


á œ1x



43. (a) ln ksec xk  C œ ' tan x dx œ ' Šx  #




8x$ 3!

128x 512x 2048x  32x 5!
7!  9!
11!  á " "‰ $ ‰ # ˆ (c) 2 sin x cos x œ 2 (0 † 1)  (0 † 0  1 † 1)x  ˆ0 †
" #  1 † 0  0 † 1 x  0 † 0
1 † #  0 † 0
1 † 3! x  ˆ0 † 4!"  1 † 0
0 † #"
0 † 3!"  0 † 1‰ x%  ˆ0 † 0  1 † 4!"  0 † 0  #" † 3!"  0 † 0  1 † 5!" ‰ x&

 ˆ0 †



3x# 3!

 x

x$ 3



2x& 15



17x( 315



62x* 2835

 á ‹ dx

"!

 á  C; x œ 0 Ê C œ 0 Ê ln ksec xk œ



2x& 15

31x 14,175  1#

Šx 

x$ 3



17x( 315



62x* 2835

† 1‰ x#

 á ‹ œ 1  x# 

2x% 3



x# #

17x' 45

 

x% 12

62x) 315



x' 45



17x) 2520



31x"! 14,175

 á , converges

1 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

á ,

Section 11.8 Taylor and Maclaurin Series x# #

(c) sec# x œ (sec x)(sec x) œ Š1  œ1

ˆ "# #

5  "# ‰ x#  ˆ 24  4" % ' 62x)  2x3  17x 45  315

œ1x





5x% 24

5 ‰ % 24 x

61x' 720



 1 #

á ,


61 ˆ 720

 á ‹ Š1  

x

44. (a) ln ksec x  tan xk  C œ ' sec x dx œ ' Š1  $

œx

x 6

œx

$

x 6

&



x 24



&

x 24

(b) sec x tan x œ when


1 #

(



61x 5040



(

œ

x# #

1 #

x

61x' 720

61x' 720

 á‹

á

 á ‹ dx

 á  C; x œ 0 Ê C œ 0 Ê ln ksec x  tan xk  á , converges when
1#  x 

Š1 



2 œ x  ˆ "3  #" ‰ x$  ˆ 15 






61 ‰ ' 720 x



x# #



5x% 24



61x' 720

5x$ 6

 á‹ œ x 

1 #



61x& 120



277x( 1008

 á , converges

1 #

(c) (sec x)(tan x) œ Š1  1 #

5x% 24





5x% 24

*

277x 72,576

d dx

x# 2

5 48



*

277x 72,576



61x 5040

d(sec x) dx

x





5 48 1 #

x# #

5x% 24 " 6





_

61x' 720

 á ‹ Šx 

5 ‰ & 24 x

17  ˆ 315 

" 15

x$ 3

2x& 15





5 72





17x( 315

61 ‰ ( 720 x

 á‹ á œ x

5x$ 6



61x& 120



277x( 1008

á ,

_

45. (a) If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n
1)(n
2)â(n
(k
1)) an xn
k and f ÐkÑ (0) œ k!ak nœ0

Ê ak œ

f

ÐkÑ

(0) k!

nœk _

; likewise if f(x) œ ! bn xn , then bk œ nœ0

f ÐkÑ (0) k!

Ê ak œ bk for every nonnegative integer k

_

(b) If f(x) œ ! an xn œ 0 for all x, then f ÐkÑ (x) œ 0 for all x Ê from part (a) that ak œ 0 for every nœ0

nonnegative integer k 46.

" 1
x

œ 1  x  x#  x$  x%  á Ê x ’ (1
" x)# “ œ x a1  2x  3x#  4x$  á b Ê

œ x  2x#  3x$  4x%  á Ê x ’ (11
x)x $ “ œ x a1  4x  9x#  16x$  á b Ê œ x  4x#  9x$  16x%  á Ê _

ˆ "#  4" ‰ ˆ "8 ‰

œ

" #



4 4



9 8



16 16

_

á Ê ! nœ1

n# 2n

x (1
x)#

x  x# (1
x)$

œ6

47. The series ! xn converges conditionally at the left-hand endpoint of its interval of convergence [
1ß 1Ñ; the n

nœ1

_

series ! nœ1

xn an # b

converges absolutely at the left-hand endpoint of its interval of convergence [
1ß 1]

48. Answers will vary. For instance: _

_

n (a) ! ˆ x3 ‰

(b) ! (x  1)n

nœ1

nœ1

_

n (c) ! ˆ x
# 3 ‰ nœ1

11.8 TAYLOR AND MACLAURIN SERIES 1. f(x) œ ln x, f w (x) œ

" x

, f ww (x) œ
x"# , f www (x) œ

2 x$ ;

f(1) œ ln 1 œ 0, f w (1) œ 1, f ww (1) œ
1, f www (1) œ 2 Ê P! (x) œ 0,

P" (x) œ (x
1), P# (x) œ (x
1)
"# (x
1)# , P$ (x) œ (x
1)
"# (x
1)#  "3 (x
1)$ 2. f(x) œ ln (1  x), f w (x) œ f w (0) œ œx


1 1 x# #

" 1x œ
#

œ 1, f ww (0) œ
(1) 

(1  x)
" , f ww (x) œ
(1  x)
# , f www (x) œ 2(1  x)
$ ; f(0) œ ln 1 œ 0,

œ
1, f www (0) œ 2(1)
$ œ 2 Ê P! (x) œ 0, P" (x) œ x, P# (x) œ x


x$ 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x# #,

P$ (x)

739

740

Chapter 11 Infinite Sequences and Series

3. f(x) œ

" x

œ x
" , f w (x) œ
x
# , f ww (x) œ 2x
$ , f www (x) œ
6x
% ; f(2) œ

Ê P! (x) œ " #

P$ (x) œ




" # , P" (x) œ " 4 (x
2) 

" # " 8


4" (x
2), P# (x) œ " 16

(x
2)#


(x
2)

" # $

" #

, f w (2) œ
4" , f ww (2) œ 4" , f www (x) œ
38


4" (x
2)  8" (x
2)# ,

4. f(x) œ (x  2)
" , f w (x) œ
(x  2)
# , f ww (x) œ 2(x  2)
$ , f www (x) œ
6(x  2)
% ; f(0) œ (2)
" œ œ
4" , f ww (0) œ 2(2)
$ œ " #

P$ (x) œ




x 4

x# 8






" 4

, f www (0) œ
6(2)
% œ
38 Ê P! (x) œ

x$ 16

" #

f ww ˆ 14 ‰ œ
sin È2 #

P# (x) œ



1 4 œ
È2 ˆx
#

È , f www ˆ 14 ‰ œ
cos 14 œ
#2 Ê È2 È 1‰ ˆx
14 ‰# , P$ (x) œ #2  4
4

6. f(x) œ cos x, f w (x) œ
sin x, f ww (x) œ
cos x, f www (x) œ sin x; f ˆ 14 ‰ œ cos f w ˆ 14 ‰ œ
sin

1 4

œ
È"2 , f ww ˆ 14 ‰ œ
cos

P" (x) œ

" È2




" È2

ˆx
14 ‰ , P# (x) œ

P$ (x) œ

" È2




" È2

ˆx
14 ‰


" #È 2

" È2

1 4




œ
È"2 , f www ˆ 14 ‰ œ sin

" È2

ˆx
14 ‰# 

" #È 2 1 ‰$ 4

ˆx
14 ‰
" 6È 2

ˆx



x4 , P# (x) œ

" #

, f w (0) œ
(2)
#




x 4



È2 È2 1 w ˆ1‰ 4 œ cos 4 œ # # ,f È È È P! œ #2 , P" (x) œ #2  #2 ˆx
14 ‰ , È2 È È ˆx
14 ‰
42 ˆx
14 ‰#
1#2 ˆx
14 ‰$ #

5. f(x) œ sin x, f w (x) œ cos x, f ww (x) œ
sin x, f www (x) œ
cos x; f ˆ 14 ‰ œ sin È2 #

" #

, P" (x) œ

" #

ˆx


1 4

œ

1 4

œ

1 4

œ

" È2

" È2

x# 8

,

,

,

Ê P! (x) œ

" È2

,

1 ‰# 4 ,

7. f(x) œ Èx œ x"Î# , f w (x) œ ˆ "# ‰ x
"Î# , f ww (x) œ ˆ
4" ‰ x
$Î# , f www (x) œ ˆ 38 ‰ x
&Î# ; f(4) œ È4 œ 2, " 3 f w (4) œ ˆ "# ‰ 4
"Î# œ 4" , f ww (4) œ ˆ
4" ‰ 4
$Î# œ
32 ,f www (4) œ ˆ 38 ‰ 4
&Î# œ 256 Ê P! (x) œ 2, P" (x) œ 2  "4 (x
4), P# (x) œ 2  4" (x
4)


" 64

(x
4)# , P$ (x) œ 2  4" (x
4)


" 64

(x
4)# 

" 51#

(x
4)$

8. f(x) œ (x  4)"Î# , f w (x) œ ˆ "# ‰ (x  4)
"Î# , f ww (x) œ ˆ
4" ‰ (x  4)
$Î# , f www (x) œ ˆ 38 ‰ (x  4)
&Î# ; f(0) œ (4)"Î# œ 2, 3 " f w (0) œ ˆ "# ‰ (4)
"Î# œ 4" , f ww (0) œ ˆ
4" ‰ (4)
$Î# œ
32 , f www (0) œ ˆ 38 ‰ (4)
&Î# œ 256 Ê P! (x) œ 2, P" (x) œ 2  4" x, P# (x) œ 2  4" x
_

xn n!

9. ex œ !

nœ0 _

xn n!

10. ex œ !

nœ0

_

Ê e
x œ !

nœ0

(
x)n n!

_ ˆ x ‰n

Ê exÎ2 œ !

nœ0

2

n!

" 64

x# , P$ (x) œ 2  4" x
x# #!




x$ 3!

x# 4†2!



x$ 2$ †3!

œ1
x

œ1

x #



x% 4!





" 64

x# 

" 512

x$


á

x% 2% †4!

á

11. f(x) œ (1  x)
" Ê f w (x) œ
(1  x)
# , f ww (x) œ 2(1  x)
$ , f www (x) œ
3!(1  x)
% Ê á f ÐkÑ (x) œ (
1)k k!(1  x)
k
1 ; f(0) œ 1, f w (0) œ
1, f ww (0) œ 2, f www (0) œ
3!, á ß f ÐkÑ (0) œ (
1)k k! Ê

" 1x

_

_

nœ0

nœ0

œ 1
x  x#
x$  á œ ! (
x)n œ ! (
1)n xn

12. f(x) œ (1
x)
" Ê f w (x) œ (1
x)
# , f ww (x) œ 2(1
x)
$ , f www (x) œ 3!(1
x)
% Ê á f ÐkÑ (x) œ k!(1
x)
k
1 ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3!, á ß f ÐkÑ (0) œ k! Ê

" 1
x

_

œ 1  x  x#  x$  á œ ! xn

_

13. sin x œ !

nœ0

nœ0

(
")n x2nb1 (#n1)!

_

Ê sin 3x œ !

nœ0

(
")n (3x)2nb1 (#n1)!

_

œ!

nœ0

(
")n 32nb1 x2nb1 (#n1)!

œ 3x


3$ x$ 3!



3& x& 5!


á

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.8 Taylor and Maclaurin Series _

14. sin x œ !

nœ0

(
")n x2nb1 (#n1)!

Ê sin _

15. 7 cos (
x) œ 7 cos x œ 7 !

nœ0

_

16. cos x œ !

nœ0

17. cosh x œ _

œ!

nœ0

_

nœ0

(
")n ˆ #x ‰ (#n1)!

nœ0

(
")n x2n (2n)!

œ7


Ê 5 cos 1x œ 5 !

nœ0

œ

" #

’Š1  x# 

œ

" #

’Š1  x 

x# #!



x$ 3!

_

œ!

nœ0

7x# #!



7x% 4!

(
1)n (1x)2n (#n)!



x% 4!

(
")n x2nb1 #2n
1 (2n1)!




7x' 6!

œ5


œ

x$ 2$ †3!




x #



x& 2& †5!

á

 á , since the cosine is an even function

51 # x# 2!

51 % x% 4!



 á ‹  Š1
x 

x# #!

51 ' x' 6!







x$ 3!



á

x% 4!


á ‹“ œ 1 

x# #!



x% 4!



x' 6!

á

x2n (2n)!

18. sinh x œ œ!

2n
1

_

œ!

_

(
1)n x2n (2n)!

ex  ecx #

x #

741

ex
ecx #

x# #!



x$ 3!



x% 4!

 á ‹
Š1
x 

x# #!




x$ 3!



x% 4!


á ‹“ œ x 

x$ 3!



x& 5!



x' 6!

á

x2n
1 (2n  1)!

19. f(x) œ x%
2x$
5x  4 Ê f w (x) œ 4x$
6x#
5, f ww (x) œ 12x#
12x, f www (x) œ 24x
12, f Ð4Ñ (x) œ 24 Ê f ÐnÑ (x) œ 0 if n   5; f(0) œ 4, f w (0) œ
5, f ww (0) œ 0, f www (0) œ
12, f Ð4Ñ (0) œ 24, f ÐnÑ (0) œ 0 if n   5 24 % $ % $ Ê x%
2x$
5x  4 œ 4
5x
12 3! x  4! x œ x
2x
5x  4 itself 20. f(x) œ (x  1)# Ê f w (x) œ 2(x  1); f ww (x) œ 2 Ê f ÐnÑ (x) œ 0 if n   3; f(0) œ 1, f w (0) œ 2, f ww (0) œ 2, f ÐnÑ (0) œ 0 if n   3 Ê (x  1)# œ 1  2x  #2! x# œ 1  2x  x# 21. f(x) œ x$
2x  4 Ê f w (x) œ 3x#
2, f ww (x) œ 6x, f www (x) œ 6 Ê f ÐnÑ (x) œ 0 if n   4; f(2) œ 8, f w (2) œ 10, 6 # $ f ww (2) œ 12, f www (2) œ 6, f ÐnÑ (2) œ 0 if n   4 Ê x$
2x  4 œ 8  10(x
2)  12 2! (x
2)  3! (x
2) œ 8  10(x
2)  6(x
2)#  (x
2)$

22. f(x) œ 2x$  x#  3x
8 Ê f w (x) œ 6x#  2x  3, f ww (x) œ 12x  2, f www (x) œ 12 Ê f ÐnÑ (x) œ 0 if n   4; f(1) œ
2, f w (1) œ 11, f ww (1) œ 14, f www (1) œ 12, f ÐnÑ (1) œ 0 if n   4 Ê 2x$  x#  3x
8 12 # $ # $ œ
2  11(x
1)  14 2! (x
1)  3! (x
1) œ
2  11(x
1)  7(x
1)  2(x
1) 23. f(x) œ x%  x#  1 Ê f w (x) œ 4x$  2x, f ww (x) œ 12x#  2, f www (x) œ 24x, f Ð4Ñ (x) œ 24, f ÐnÑ (x) œ 0 if n   5; f(
2) œ 21, f w (
2) œ
36, f ww (
2) œ 50, f www (
2) œ
48, f Ð4Ñ (
2) œ 24, f ÐnÑ (
2) œ 0 if n   5 Ê x%  x#  1 48 24 # $ % # $ % œ 21
36(x  2)  50 2! (x  2)
3! (x  2)  4! (x  2) œ 21
36(x  2)  25(x  2)
8(x  2)  (x  2) 24. f(x) œ 3x&
x%  2x$  x#
2 Ê f w (x) œ 15x%
4x$  6x#  2x, f ww (x) œ 60x$
12x#  12x  2, f www (x) œ 180x#
24x  12, f Ð4Ñ (x) œ 360x
24, f Ð5Ñ (x) œ 360, f ÐnÑ (x) œ 0 if n   6; f(
1) œ
7, f w (
1) œ 23, f ww (
1) œ
82, f www (
1) œ 216, f Ð4Ñ (
1) œ
384, f Ð5Ñ (
1) œ 360, f ÐnÑ (
1) œ 0 if n   6 216 384 360 # $ % & Ê 3x&
x%  2x$  x#
2 œ
7  23(x  1)
82 2! (x  1)  3! (x  1)
4! (x  1)  5! (x  1) œ
7  23(x  1)
41(x  1)#  36(x  1)$
16(x  1)%  3(x  1)& 25. f(x) œ x
# Ê f w (x) œ
2x
$ , f ww (x) œ 3! x
% , f www (x) œ
4! x
& Ê f ÐnÑ (x) œ (
1)n (n  1)! x
n
2 ; f(1) œ 1, f w (1) œ
2, f ww (1) œ 3!, f www (1) œ
4!, f ÐnÑ (1) œ (
1)n (n  1)! Ê x"# _

œ 1
2(x
1)  3(x
1)#
4(x
1)$  á œ ! (
1)n (n  1)(x
1)n nœ0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

742

Chapter 11 Infinite Sequences and Series

26. f(x) œ

Ê f w (x) œ (1
x)
# , f ww (x) œ 2(1
x)
$ , f www (x) œ 3! (1
x)
% Ê f ÐnÑ (x) œ n! (1
x)
n
1 ;

x 1
x

f(0) œ 0, f w (0) œ 1, f ww (0) œ 2, f www (0) œ 3! Ê

x 1
x

_

œ x  x#  x$  á œ ! xn1 nœ0

27. f(x) œ ex Ê f w (x) œ ex , f ww (x) œ ex Ê f ÐnÑ (x) œ ex ; f(2) œ e# , f w (2) œ e# , á f ÐnÑ (2) œ e# Ê ex œ e#  e# (x
2) 

e# #

(x
2)# 

e$ 3!

_

(x
2)$  á œ !

nœ0

e# n!

(x
2)n

28. f(x) œ 2x Ê f w (x) œ 2x ln 2, f ww (x) œ 2x (ln 2)# , f www (x) œ 2x (ln 2)$ Ê f ÐnÑ (x) œ 2x (ln 2)n ; f(1) œ 2, f w (1) œ 2 ln 2, f ww (1) œ 2(ln 2)# , f www (1) œ 2(ln 2)$ , á , f ÐnÑ (1) œ 2(ln 2)n Ê 2x œ 2  (2 ln 2)(x
1)  _

29. If ex œ !

nœ0

f ÐnÑ (a) n!

2(ln 2)# #

(x
1)# 

2(ln 2)$ 3!

_

(x
1)$  á œ !

nœ0

2(ln 2)n (x
1)n n!

(x
a)n and f(x) œ ex , we have f ÐnÑ (a) œ ea f or all n œ 0, 1, 2, 3, á !

Ê ex œ ea ’ (x
0!a) 

(x
a)" 1!



(x
a)# 2!

 á “ œ ea ’1  (x
a) 

(x
a)# 2!

 á “ at x œ a

30. f(x) œ ex Ê f ÐnÑ (x) œ ex for all n Ê f ÐnÑ (1) œ e for all n œ 0, 1, 2, á Ê ex œ e  e(x
1) 

e #!

(x
1)# 

e 3!

(x
1)$  á œ e ’1  (x
1) 

f ww (a) f www (a) # $ w # (x
a)  3! (x
a)  á Ê f (x) www œ f w (a)  f ww (a)(x
a)  f 3!(a) 3(x
a)#  á Ê f ww (x) œ f ww (a)  f www (a)(x Ðn
2Ñ Ê f ÐnÑ (x) œ f ÐnÑ (a)  f Ðn1Ñ (a)(x
a)  f # (a) (x
a)#  á w w Ðn Ñ Ðn Ñ

(x
1)# 2!

31. f(x) œ f(a)  f w (a)(x
a) 

Ê f(a) œ f(a)  0, f (a) œ f (a)  0, á , f

(a) œ f


a) 



(x
1)$ 3!

f Ð4Ñ (a) 4!

 á“

4 † 3(x
a)#  á

(a)  0

32. E(x) œ f(x)
b!
b" (x
a)
b# (x
a)#
b$ (x
a)$
á
bn (x
a)n Ê 0 œ E(a) œ f(a)
b! Ê b! œ f(a); from condition (b), lim

xÄa

Ê Ê Ê

f(x)
f(a)
b" (x
a)
b# (x
a)#
b$ (x
a)$
á
bn (x
a)n (x
a)n

œ0

w a)#
á
nbn (x
a)n
1 lim f (x)
b"
2b# (x
a)
n(x3b
$ (xa)
œ0 n
1 xÄa ww f (x) 2b 3! b (x a) n(n



á

")bn (x
a)n
2 # $ b" œ f w (a) Ê xlim œ n(n
1)(x
a)n
2 Äa f www (x)
3! b$
á
n(n
1)(n
2)bn (x
a)n
3 " ww b# œ # f (a) Ê xlim œ0 n(n
1)(n
#)(x
a)n
3 Äa

0

ÐnÑ

f (x)
n! bn f www (a) Ê xlim œ 0 Ê bn œ n!" f ÐnÑ (a); therefore, n! Äa ww ÐnÑ g(x) œ f(a)  f w (a)(x
a)  f 2!(a) (x
a)#  á  f n!(a) (x
a)n œ Pn (x)

œ b$ œ

" 3!

33. f(x) œ ln (cos x) Ê f w (x) œ
tan x and f ww (x) œ
sec# x; f(0) œ 0, f w (0) œ 0, f ww (0) œ
1 #

Ê L(x) œ 0 and Q(x) œ
x2

34. f(x) œ esin x Ê f w (x) œ (cos x)esin x and f ww (x) œ (
sin x)esin x  (cos x)# esin x ; f(0) œ 1, f w (0) œ 1, f ww (0) œ 1 Ê L(x) œ 1  x and Q(x) œ 1  x  35. f(x) œ a1
x# b w

ww


"Î#

Ê f w (x) œ x a1
x# b


$Î#

x# #

and f ww (x) œ a1
x# b

f (0) œ 0, f (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 


$Î#

 3x# a1
x# b


&Î#

; f(0) œ 1,

#

x #

36. f(x) œ cosh x Ê f w (x) œ sinh x and f ww (x) œ cosh x; f(0) œ 1, f w (0) œ 0, f ww (0) œ 1 Ê L(x) œ 1 and Q(x) œ 1 

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x# #

Section 11.9 Convergence of Taylor Series; Error Estimates 37. f(x) œ sin x Ê f w (x) œ cos x and f ww (x) œ
sin x; f(0) œ 0, f w (0) œ 1, f ww (0) œ 0 Ê L(x) œ x and Q(x) œ x 38. f(x) œ tan x Ê f w (x) œ sec# x and f ww (x) œ 2 sec# x tan x; f(0) œ 0, f w (0) œ 1, f ww œ 0 Ê L(x) œ x and Q(x) œ x 11.9 CONVERGENCE OF TAYLOR SERIES; ERROR ESTIMATES _

1. ex œ 1  x 

x# #!

á œ !

2. ex œ 1  x 

x# #!

á œ !

_

œ!

nœ0

xn n!

Ê e
5x œ 1  (
5x) 

(
5x)# #!

 á œ 1
5x 

xn n!

Ê e
xÎ2 œ 1  ˆ
#x ‰ 

ˆ
#x ‰# #!

á œ1


_

(
1)n x2n
1 (#n1)!

Ê 5 sin (
x) œ 5 ’(
x)


(
1)n x2n
1 (#n1)!

Ê sin

nœ0 _

nœ0

x# 2# #!



x #

5# x# #!

x$ 2$ 3!




_

5$ x$ 3!




á œ!

nœ0

(
1)n 5n xn n!

á

(
1)n xn 2n n!

3. sin x œ x
_

x$ 3!



x& 5!


á œ!

x& 5!


á œ!

nœ0

(
x)$ 3!

(
x)& 5!




á“

nb1 2nb1

x œ ! 5(
(1) #n1)! nœ0

4. sin x œ x
_

x$ 3!



_

nœ0

1x #

1x #

œ

ˆ 1#x ‰$ 3!




ˆ 1#x ‰& 5!






ˆ 1#x ‰( 7!

á

n 2nb1 2nb1

1 x œ ! (
21) 2nb1 (#n1)! nœ0

2n

_

nœ0

_

6. cos x œ !

(
1)n x2n (2n)!

x$ 2†2!

x' 2# †4!

nœ0

œ1
_

nœ0

_

8. sin x œ !

nœ0 _

9. cos x œ !

nœ0

œ

x% 4!



xn n!

7. ex œ !




Ê cos Èx  1 œ !

Ê

_

n œ0

(
1)n x2nb1 (2n1)!

(
1)n x2n (2n)!

_

11. cos x œ !

nœ0

x& 5!







(
1)n x2n (2n)!

xn n! 

_

xnb1 n!

œ!

nœ0 _

Ê x# sin x œ x# Œ !

nœ0

Ê

(
1)n x2nb1 (2n1)!



x) 8!

œ

$ cos ŒŠ x# ‹

x"! 10!

x# #


1  cos x œ

"Î#

$

(
1)n ŒŠ x# ‹

_

œ!



ax  1 b# 4!




a x  1 b$ 6!

_

_

(
1)n x3n 2n (2n)!

œ!

n œ0

x# #

x$ #!

œ x  x# 

(
1)n x2nb1 (#n1)! 

_


1!

nœ0

_

œ!

nœ0



x% 3!

x& 4!



(
1)n x2nb3 (2n1)!

(
1)n x2n (#n)!

œ

x# #

á

œ x$



11


x& 3!



x# 2

x( 5!

x* 7!




á



x% 4!

_

(
1)n x2n
1 (2n1)!




x' 6!



x) 8!




n 2n

nœ2



á

2n




(#n)!

nœ0

x 1 #!

œ1


x  á œ ! (
(1) #n)!

Ê sin x
x  x( 7!

"Î#

nœ0

(
1)n ax1bn (2n)!

á

Ê xex œ x Œ !

_

x$ 3!

x* 2$ †6!






nœ0

$Î# cos Š xÈ ‹ 2

_

œ!

(2n)!

nœ0

x' 6!

10. sin x œ ! œ Šx


_ (
1)n ’ ax1b"Î# “

(
1)n x2n (2n)!

5. cos x œ !

x* 9!




x"" 11!

x$ 3!

_

œ Œ!

nœ0

 á‹
x  _

Ê x cos 1x œ x !

nœ0

(
1)n x2nb1 (#n1)!  x$ 3!

(
1)n (1x)2n (#n)!

œ

x& 5!

_

œ!

nœ0





x

x( 7!



x* 9!

x$ 3!




(
")n 12n x2nb1 (#n)!

x"" 11!

á œ!

œx


nœ2

1 # x$ 2!



1 % x& 4!




1 ' x( 6!

á

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x"! 10!

á

743

744

Chapter 11 Infinite Sequences and Series _

nœ0

13. cos# x œ œ1


" #

_

(
1)n x2n (2n)!

12. cos x œ !



(2x)# 2†2!

Ê x# cos ax# b œ x# !

nœ0

" #

(2x)% 2†4!




(2x)' 2†6!



" #

" #



2x ‰ 14. sin# x œ ˆ 1
cos œ # _

œ!

nœ1

15.

x# 1
2x

(
1)nb1 (2x)2n #†(2n)!

_

n

2n

nœ0

(2x)) 2†8!

" #

_

nœ1

" #

_

œ!

nœ0

 "# ’1



á œ1!

cos 2x œ

2n


"# Š1


(2x)# 2!

(
1)n (2x)2n 2†(2n)!

(2x)# #!



(
")n x4n
2 (#n)!



(2x)% 4!

œ x#
(2x)' 6!


_

œ1!

nœ1

(2x)% 4!




(2x)' 6!



x' 2!



(2x)) 8!

x"! 4!




x"% 6!

á


á“

(
1)n 22n
1 x2n (2n)!

(2x)# 2†2!

 á‹ œ




(2x)% 2†4!



(2x)' 2†6!


á

(
1)n 22n
1 x2n (2n)!

nœ1

_

_

nœ0

nœ0

œ x# ˆ 1
"2x ‰ œ x# ! (2x)n œ ! 2n xn2 œ x#  2x$  2# x%  2$ x&  á _

nœ1

" 1
x




œ!

16. x ln (1  2x) œ x !

17.

_

œ



" #

! (
1) (2x) œ (2n)!

cos 2x #

(
1)n ax# b (#n)!

(
1)nc1 (2x)n n

_

œ!

nœ1

(
1)nc1 2n xn
1 n

_

œ ! xn œ 1  x  x#  x$  á Ê _

nœ0

d dx

œ 2x#


ˆ 1
" x ‰ œ

2# x$ #

" (1
x)#



2$ x% 4




2% x& 5

á _

œ 1  2x  3x#  á œ ! nxn
1 nœ1

œ ! (n  1)xn n œ0

18.

2 a1
x b $

œ

d# dx#

ˆ 1
" x ‰ œ

d dx

_

Š (1
"x)# ‹ œ

d dx

_

a1  2x  3x#  á b œ 2  6x  12x#  á œ ! n(n
1)xn
2 nœ2

œ ! (n  2)(n  1)x

n

nœ0

19. By the Alternating Series Estimation Theorem, the error is less than & Ê kxk&  600 ‚ 10
% Ê kxk  È6 ‚ 10
# ¸ 0.56968 20. If cos x œ 1


x# #

kx k & 5!

Ê kxk&  a5!b a5 ‚ 10
% b

%

and kxk  0.5, then the error is less than ¹ (.5) 24 ¹ œ 0.0026, by Alternating Series Estimation Theorem;

since the next term in the series is positive, the approximation 1


x# #

is too small, by the Alternating Series Estimation

Theorem 21. If sin x œ x and kxk  10
$ , then the error is less than

a10c$ b 3!

$

¸ 1.67 ‚ 10
10 , by Alternating Series Estimation Theorem; $

The Alternating Series Estimation Theorem says R# (x) has the same sign as
x3! . Moreover, x  sin x Ê 0  sin x
x œ R# (x) Ê x  0 Ê
10
$  x  0.

22. È1  x œ 1 

x #




x# 8



x$ 16

#


á . By the Alternating Series Estimation Theorem the kerrork  ¹
8x ¹ 

œ 1.25 ‚ 10
& c $

3Ð0Þ1Ñ (0.1)$ 3!

c $

(0.1)$ 3!

23. kR# (x)k œ ¹ e3!x ¹  24. kR# (x)k œ ¹ e3!x ¹ 

 1.87 ‚ 10
4 , where c is between 0 and x

œ 1.67 ‚ 10
% , where c is between 0 and x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

(0.01)# 8

Section 11.9 Convergence of Taylor Series; Error Estimates c &¸ e e 25. kR% (x)k  ¸ cosh 5! x œ ¹ # c

cc x&

5! ¹



" 1.65 1.65 #

†

(0.5)& 5!

&

œ (1.13) (0.5) 5! ¸ 0.000294 c #

26. If we approximate eh with 1  h and 0 Ÿ h Ÿ 0.01, then kerrork  ¹ e #h ¹ Ÿ

e0Þ01 h†h #

Ÿ Še

0Þ01

(0.0") ‹h #

œ 0.00505h  0.006h œ (0.6%)h, where c is between 0 and h. 27. kR" k œ ¹ (1"c)#

x# #! ¹

28. tan
" x œ x


x$ 3

 

x# # x& 5

œ ¸ #x ¸ kxk  .01 kxk œ (1%) kxk Ê ¸ #x ¸  .01 Ê 0  kxk  .02


x( 7

á Ê

1 4

œ tan
" 1 œ 1


" 3



" 5

x% 5!




x' 7!




" 7

 á ; kerrork 

" #n  1

 .01

Ê 2n  1  100 Ê n  49 x$ 3!

29. (a) sin x œ x




x& 5!

series representing L
s# œ

sin x x




graph of y œ 1


á Ê

sin x x

x# 3!

œ1




#

x 6

‹  0. Therefore 1


#



x 6

sin x x

x# 6 ; if sin x x
1

 á , s" œ 1 and s# œ 1


then by the Alternating Series Estimation Theorem, L
s" œ

sin x x ,


Š1


(b) The graph of y œ

x( 7!

L is the sum of the  0 and

1

sin x x , x Á 0, is bounded below by the x# 6 and above by the graph of y œ 1 as

derived in part (a).

30. (a) cos x œ 1


x# #!



x% 4!




x' 6!

 á Ê 1
cos x œ

if L is the sum of the series representing L
s" œ (b)

1
cos x x#




" #

 0 and

1
cos x x#

1
cos x x#


Š "#


x The graph of y œ 1
xcos is bounded below by # " x# the graph of y œ #
24 and above by the graph y œ "# as indicated in part (a).

x# #!




x% 4!



x' 6!




x) 8!

á Ê

1
cos x x#

œ

" #




1 4

33. tan
" x when x œ

; the sum is cos ˆ 14 ‰ œ 1 3

" È2



x% 6!

, then by the Alternating Series Estimation Theorem x# 4! ‹

 0. Therefore

" #




x# #4



1
cos x x#



" #

of

31. sin x when x œ 0.1; the sum is sin (0.1) ¸ 0.099833417 32. cos x when x œ

x# 4!

¸ 0.707106781

; the sum is tan
" ˆ 13 ‰ ¸ 0.808448

34. ln (1  x) when x œ 1; the sum is ln (1  1) ¸ 1.421080

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

.




x' 8!

á ;

745

746

Chapter 11 Infinite Sequences and Series

35. ex sin x œ 0  x  x#  x$ ˆ
3!"  " 3

#

$

œxx  x


" 30

" 90

&

x


'

" 6

$

%

œ1x
x
x
2x ‰ 37. sin# x œ ˆ 1
cos œ #

Ê

d dx

asin# xb œ

œ 2x


(2x)$ 3!






" #

#

Š 2x 2!


d dx

(2x)& 5!

" #

" 30




(2x)( 7!

&

"‰ #!

œ1




2$ x% 4!




2& x' 6!

"‰ 3!

 x& ˆ 5!"


" " #! 3!



"‰ 4!

 x' ˆ 5!"


" " 3! 3!



"‰ 5!

á

 x$ ˆ
#"! 

"‰ 3!

 x% ˆ 4!"


" " #! 2!



"‰ 4!

 x& ˆ 4!"


" " 2! 3!

"‰ 5!



á

x á " #

cos 2x œ 2$ x% 4!



2& x' 6!


"# Š1


(2x)# 2!


á ‹ œ 2x




(2x)% 4!

(2x)$ 3!






(2x)' 6!

(2x)& 5!

2x# #!

 á‹ œ (2x)( 7!







2$ x% 4!



2& x' 6!


á

 á Ê 2 sin x cos x

 á œ sin 2x, which checks

38. cos# x œ cos 2x  sin# x œ Š1
2x# #!

 x% ˆ
3!" 

x á

36. ex cos x œ 1  x  x# ˆ
#"!  " 3

"‰ #!

(2x)# #!



(2x)% 4!




(2x)' 6!

" 3

 á œ 1
x#  x%


2 45



(2x)) 8!

x' 

#

 á ‹  Š 2x #!


" 315

2$ x% 4!



2& x' 6!




2( x) 8!

 á‹

x)
á

39. A special case of Taylor's Theorem is f(b) œ f(a)  f w (c)(b
a), where c is between a and b Ê f(b)
f(a) œ f w (c)(b
a), the Mean Value Theorem. 40. If f(x) is twice differentiable and at x œ a there is a point of inflection, then f ww (a) œ 0. Therefore, L(x) œ Q(x) œ f(a)  f w (a)(x
a). 41. (a) f ww Ÿ 0, f w (a) œ 0 and x œ a interior to the interval I Ê f(x)
f(a) œ Ê f(x) Ÿ f(a) throughout I Ê f has a local maximum at x œ a (b) similar reasoning gives f(x)
f(a) œ local minimum at x œ a

f ww (c# ) #

f ww (c# ) #

(x
a)# Ÿ 0 throughout I

(x
a)#   0 throughout I Ê f(x)   f(a) throughout I Ê f has a

42. f(x) œ (1
x)
" Ê f w (x) œ (1
x)
# Ê f ww (x) œ 2(1
x)
$ Ê f Ð3Ñ (x) œ 6(1
x)
% Ê f Ð4Ñ (x) œ 24(1
x)
& ; therefore

" 1
x

¸ 1  x  x#  x$ . kxk  0.1 Ê

&

%

Ð4Ñ

10 11



" 1
x



10 9

‰ Ê ¹ (1
"x)& ¹  ˆ 10 9

&

%

‰ Ê the error e$ Ÿ ¹ max f 4! (x) x ¹  (0.1)% ˆ 10 ‰ œ 0.00016935  0.00017, since ¹ f Ê ¹ (1
x x)& ¹  x% ˆ 10 9 9

&

Ð4Ñ

(x) 4! ¹

œ ¹ (1
"x)& ¹ .

43. (a) f(x) œ (1  x)k Ê f w (x) œ k(1  x)k
1 Ê f ww (x) œ k(k
1)(1  x)k
2 ; f(0) œ 1, f w (0) œ k, and f ww (0) œ k(k
1) Ê Q(x) œ 1  kx  k(k #
") x# " (b) kR# (x)k œ ¸ 3†3!2†" x$ ¸  100 Ê kx$ k 

" 100

Ê 0x

" 100"Î$

or 0  x  .21544

44. (a) Let P œ x  1 Ê kxk œ kP
1k  .5 ‚ 10
n since P approximates 1 accurate to n decimals. Then, P  sin P œ (1  x)  sin (1  x) œ (1  x)
sin x œ 1  (x
sin x) Ê k(P  sin P)
1k œ ksin x
xk Ÿ

kx k $ 3!



0.125 3!

‚ 10
3n  .5 ‚ 10
3n Ê P  sin P gives an approximation to 1 correct to 3n

decimals. _

_

nœ0

nœk

45. If f(x) œ ! an xn , then f ÐkÑ (x) œ ! n(n
1)(n
2)â(n
k  1)an xn
k and f ÐkÑ (0) œ k! ak Ê ak œ

f ÐkÑ (0) k!

for k a nonnegative integer. Therefore, the coefficients of f(x) are identical with the

corresponding coefficients in the Maclaurin series of f(x) and the statement follows. 46. Note: f even Ê f(
x) œ f(x) Ê
f w (
x) œ f w (x) Ê f w (
x) œ
f w (x) Ê f w odd; f odd Ê f(
x) œ
f(x) Ê
f w (
x) œ
f w (x) Ê f w (
x) œ f w (x) Ê f w even; also, f odd Ê f(
0) œ f(0) Ê 2f(0) œ 0 Ê f(0) œ 0 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.9 Convergence of Taylor Series; Error Estimates (a) If f(x) is even, then any odd-order derivative is odd and equal to 0 at x œ 0. Therefore, a" œ a$ œ a& œ á œ 0; that is, the Maclaurin series for f contains only even powers. (b) If f(x) is odd, then any even-order derivative is odd and equal to 0 at x œ 0. Therefore, a! œ a# œ a% œ á œ 0; that is, the Maclaurin series for f contains only odd powers. 47. (a) Suppose f(x) is a continuous periodic function with period p. Let x! be an arbitrary real number. Then f assumes a minimum m" and a maximum m# in the interval [x! ß x!  p]; i.e., m" Ÿ f(x) Ÿ m# for all x in [x! ß x!  p]. Since f is periodic it has exactly the same values on all other intervals [x!  pß x!  2p], [x!  2pß x!  3p], á , and [x!
pß x! ], [x!
2pß x!
p], á , and so forth. That is, for all real numbers
_  x  _ we have m" Ÿ f(x) Ÿ m# . Now choose M œ max ekm" k , km# kf . Then
M Ÿ
km" k Ÿ m" Ÿ f(x) Ÿ m# Ÿ km# k Ÿ M Ê kf(x)k Ÿ M for all x. (b) The dominate term in the nth order Taylor polynomial generated by cos x about x œ a is

sin (a) n!

causing the graph of Pn (x) to move away from cos x. 48. (b) tan
" x œ x
" 3

œ




x# 5

x$ 3



x& 5



 á ; from the Alternating Series

x
tan
" x x$ " 3




x
tan
" x x$


á Ê x
tan
" x x$

Estimation Theorem, Ê

x# 5

x#


Š "3


5

" 3




0

‹0 Ê

; therefore, the lim

xÄ0

" 3



x
tan
" x x$

x
tan
" x x$

œ

" 3

49. (a) e
i1 œ cos (
1)  i sin (
1) œ
1  i(0) œ
1 (b) ei1Î4 œ cos ˆ 14 ‰  i sin ˆ 14 ‰ œ

" È2



i È2

œ Š È" ‹ (1  i) 2

(c) e
i1Î2 œ cos ˆ
1# ‰  i sin ˆ
1# ‰ œ 0  i(
1) œ
i 50. ei) œ cos )  i sin ) Ê e
i) œ ei(
)) œ cos (
))  i sin (
)) œ cos )
i sin ); ei)  eci) ; # ei)
eci) œ #i

ei)  e
i) œ cos )  i sin )  cos )
i sin ) œ 2 cos ) Ê cos ) œ ei)
e
i) œ cos )  i sin )
(cos )
i sin )) œ 2i sin ) Ê sin ) 51. ex œ 1  x 

x# #!

e
i) œ 1
i)  Ê œ

ei)  eci) œ # # % 1
)#!  )4!

ei)
eci) #i

œ)


œ )$ 3!

(x
a)n or

(x
a)n . In both cases, as kxk increases the absolute value of these dominate terms tends to _,

cos (a) n!

x$ 3! (
i))# 2!



)& 5!



Š1  i) 




)' 6!

Š1  i) 








(i))# #!



(i))$ 3!



(i))% 4!

œ 1  i) 

(i))# 2!

á œ 1
i) 

 á‹  Š1
i) 

(i))# #!

 (i))# #!




(i))$ 3!






(i))$ 3!

(i))$ 3!



(i))% 4!

(i))% 4!



 á and (i))% 4!


á


á‹

#

 á œ cos );

(i))# #!

)( 7!

x% i) 4!  á Ê e (
i))$ (
i))% 3!  4! 

$

%

#

$

%

)) )) )) ))  (i3!  (i4!  á‹
Š1
i)  (i#)!)
(i3!  (i4!
á‹

#i

 á œ sin )

52. ei) œ cos )  i sin ) Ê e
i) œ eiÐ
)Ñ œ cos (
))  i sin (
)) œ cos )
i sin ) (a) ei)  e
i) œ (cos )  i sin ))  (cos )
i sin )) œ 2 cos ) Ê cos ) œ

ei)  eci) #

œ cosh i)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

747

748

Chapter 11 Infinite Sequences and Series

(b) ei)
e
i) œ (cos )  i sin ))
(cos )
i sin )) œ 2i sin ) Ê i sin ) œ 53. ex sin x œ Š1  x 

x# #!



x$ 3!



x% 4!

 á ‹ Šx


x$ 3!



x& 5!




x( 7!

œ (1)x  (1)x#  ˆ
6"  #" ‰ x$  ˆ
6"  6" ‰ x%  ˆ 1#"0


ei)
eci) 2

œ sinh i)

 á‹

" 1#



" ‰ & #4 x x

 á œ x  x#  3" x$


" 30

x&  á ;

ex † eix œ eÐ1iÑx œ ex (cos x  i sin x) œ ex cos x  i aex sin xb Ê e sin x is the series of the imaginary part _

of eÐ1iÑx which we calculate next; eÐ1iÑx œ !

nœ0

œ 1  x  ix  Ð1iÑx

of e

is x 

(xix)n n!

œ 1  (x  ix) 

(x  ix)# #!



(x  ix)$ 3!



(x  ix)% 4!

á

" " " " " # $ $ % & & ' #! a2ix b  3! a2ix
2x b  4! a
4x b  5! a
4x
4ix b  6! a
8ix b  á Ê the imaginary 2 # 2 $ 4 & 8 ' " $ " & " ' # #! x  3! x
5! x
6! x  á œ x  x  3 x
30 x
90 x  á in agreement with our x

part

product calculation. The series for e sin x converges for all values of x. 54.

d dx

ˆeÐaibÑ ‰ œ

d dx

ceax (cos bx  i sin bx)d œ aeax (cos bx  i sin bx)  eax (
b sin bx  bi cos bx)

œ aeax (cos bx  i sin bx)  bieax (cos bx  i sin bx) œ aeÐaibÑx  ibeÐaibÑx œ (a  ib)eÐaibÑx 55. (a) ei)" ei)# œ (cos )"  i sin )" )(cos )#  i sin )# ) œ (cos )" cos )#
sin )" sin )# )  i(sin )" cos )#  sin )# cos )" ) œ cos()"  )# )  i sin()"  )# ) œ eiÐ)" )# Ñ " " )  i sin ) ‰ (b) e
i) œ cos(
))  i sin(
)) œ cos )
i sin ) œ (cos )
i sin )) ˆ cos cos )  i sin ) œ cos )  i sin ) œ ei) 56.

a
bi ÐabiÑx  C"  iC# œ ˆ aa#
bib# ‰ eax (cos bx  i sin bx)  C"  iC# a#  b# e ax œ a# e b# (a cos bx  ia sin bx
ib cos bx  b sin bx)  C"  iC# ax œ a# e b# [(a cos bx  b sin bx)  (a sin bx
b cos bx)i]  C"  iC# ax ax œ e (a cosa#bxb#b sin bx)  C"  ie (a sina#bx
b#b cos bx)  iC# ; ÐabiÑx ax ibx ax ax ax

e

'e

œe e

ÐabiÑx

dx œ

œ e (cos bx  i sin bx) œ e cos bx  ie sin bx, so that given

a
bi a#  b#

eÐabiÑx  C"  iC# we conclude that ' eax cos bx dx œ

and ' eax sin bx dx œ

e (a sin bx
b cos bx) a#  b# ax

eax (a cos bx  b sin bx) a#  b#

 C"

 C#

57-62. Example CAS commands: Maple: f := x -> 1/sqrt(1+x); x0 := -3/4; x1 := 3/4; # Step 1: plot( f(x), x=x0..x1, title="Step 1: #57 (Section 11.9)" ); # Step 2: P1 := unapply( TaylorApproximation(f(x), x = 0, order=1), x ); P2 := unapply( TaylorApproximation(f(x), x = 0, order=2), x ); P3 := unapply( TaylorApproximation(f(x), x = 0, order=3), x ); # Step 3: D2f := D(D(f)); D3f := D(D(D(f))); D4f := D(D(D(D(f)))); plot( [D2f(x),D3f(x),D4f(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 3: #57 (Section 11.9)" ); c1 := x0; M1 := abs( D2f(c1) ); c2 := x0; M2 := abs( D3f(c2) );

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.9 Convergence of Taylor Series; Error Estimates

749

c3 := x0; M3 := abs( D4f(c3) ); # Step 4: R1 := unapply( abs(M1/2!*(x-0)^2), x ); R2 := unapply( abs(M2/3!*(x-0)^3), x ); R3 := unapply( abs(M3/4!*(x-0)^4), x ); plot( [R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], title="Step 4: #57 (Section 11.9)" ); # Step 5: E1 := unapply( abs(f(x)-P1(x)), x ); E2 := unapply( abs(f(x)-P2(x)), x ); E3 := unapply( abs(f(x)-P3(x)), x ); plot( [E1(x),E2(x),E3(x),R1(x),R2(x),R3(x)], x=x0..x1, thickness=[0,2,4], color=[red,blue,green], linestyle=[1,1,1,3,3,3], title="Step 5: #57 (Section 11.9)" ); # Step 6: TaylorApproximation( f(x), view=[x0..x1,DEFAULT], x=0, output=animation, order=1..3 ); L1 := fsolve( abs(f(x)-P1(x))=0.01, x=x0/2 ); # (a) R1 := fsolve( abs(f(x)-P1(x))=0.01, x=x1/2 ); L2 := fsolve( abs(f(x)-P2(x))=0.01, x=x0/2 ); R2 := fsolve( abs(f(x)-P2(x))=0.01, x=x1/2 ); L3 := fsolve( abs(f(x)-P3(x))=0.01, x=x0/2 ); R3 := fsolve( abs(f(x)-P3(x))=0.01, x=x1/2 ); plot( [E1(x),E2(x),E3(x),0.01], x=min(L1,L2,L3)..max(R1,R2,R3), thickness=[0,2,4,0], linestyle=[0,0,0,2], color=[red,blue,green,black], view=[DEFAULT,0..0.01], title="#57(a) (Section 11.9)" ); abs(`f(x)`-`P`[1](x) ) <= evalf( E1(x0) ); # (b) abs(`f(x)`-`P`[2](x) ) <= evalf( E2(x0) ); abs(`f(x)`-`P`[3](x) ) <= evalf( E3(x0) ); Mathematica: (assigned function and values for a, b, c, and n may vary) Clear[x, f, c] f[x_]= (1  x)3/2 {a, b}= {
1/2, 2}; pf=Plot[ f[x], {x, a, b}]; poly1[x_]=Series[f[x], {x,0,1}]//Normal poly2[x_]=Series[f[x], {x,0,2}]//Normal poly3[x_]=Series[f[x], {x,0,3}]//Normal Plot[{f[x], poly1[x], poly2[x], poly3[x]}, {x, a, b}, PlotStyle Ä {RGBColor[1,0,0], RGBColor[0,1,0], RGBColor[0,0,1], RGBColor[0,.5,.5]}]; The above defines the approximations. The following analyzes the derivatives to determine their maximum values. f''[c] Plot[f''[x], {x, a, b}]; f'''[c] Plot[f'''[x], {x, a, b}]; f''''[c] Plot[f''''[x], {x, a, b}]; Noting the upper bound for each of the above derivatives occurs at x = a, the upper bounds m1, m2, and m3 can be defined and bounds for remainders viewed as functions of x. m1=f''[a] m2=-f'''[a] m3=f''''[a] r1[x_]=m1 x2 /2!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

750

Chapter 11 Infinite Sequences and Series

Plot[r1[x], {x, a, b}]; r2[x_]=m2 x3 /3! Plot[r2[x], {x, a, b}]; r3[x_]=m3 x4 /4! Plot[r3[x], {x, a, b}]; A three dimensional look at the error functions, allowing both c and x to vary can also be viewed. Recall that c must be a value between 0 and x, so some points on the surfaces where c is not in that interval are meaningless. Plot3D[f''[c] x2 /2!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f'''[c] x3 /3!, {x, a, b}, {c, a, b}, PlotRange Ä All] Plot3D[f''''[c] x4 /4!, {x, a, b}, {c, a, b}, PlotRange Ä All] 11.10 APPLICATIONS OF POWER SERIES 1. (1  x)"Î# œ 1  "# x 

ˆ "# ‰ ˆ
"# ‰ x#

2. (1  x)"Î$ œ 1  "3 x 

ˆ "3 ‰ ˆ
23 ‰ x#

8. a1  x# b


"Î$

œ 1
3" x# 

"Î# 9. ˆ1  1x ‰ œ 1  "# ˆ 1x ‰ 

$

(4)(3)x# #!

12. a1  x# b œ 1  3x# 

ˆ 3" ‰ ˆ
32 ‰ ˆ
53 ‰ x$

 á œ 1  3" x
9" x# 

5 81

x$
á



ˆ "# ‰ ˆ
"# ‰ (
2x)# #!

#

#!

(
2)(
3) ˆ
x# ‰

#

ˆ
"3 ‰ ˆ
43 ‰ ax# b#

(3)(2) ax# b #!

#



(3)(2)(
2x)# #!



% 14. ˆ1
#x ‰ œ 1  4 ˆ
x# ‰ 

(4)(3) ˆ
x# ‰

#

#!

3!

3!

$

 á œ 1
"3 x#  29 x%


á œ1

" #x




á œ1


1 8x#



" 16x$

2 3x




4 9x#

x*  á

5 16

14 81

x'  á

á




40 81x$

dy dx


á

œ 1  4x  6x#  4x$  x%

œ 1  3x#  3x%  x' œ 1
6x  12x#
8x$

(4)(3)(2) ˆ
x# ‰ 3!

$



(4)(3)(2)(1) ˆ
x# ‰ 4!

%

œ 1
2x  32 x#
"# x$ 

15. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

x$  á

 á œ 1
x
12 x#
12 x$
á

 á œ 1
"# x$  38 x'


3!

(3)(2)(1)(
2x)$ 3!



5 16

 á œ 1  x  34 x#  "# x$  á

ˆ 3" ‰ ˆ
32 ‰ ˆ
35 ‰ ˆ
x2 ‰$

(4)(3)(2)x% 4!

(3)(2)(1) ax# b 3!

13. (1
2x)$ œ 1  3(
2x) 

$

ˆ
"# ‰ ˆ
3# ‰ ˆ
5# ‰ ax$ b$



 á œ 1  "# x  38 x# 

 á œ 1
x  34 x#
"# x$

3!

3!

#!



$

ˆ "# ‰ ˆ
"# ‰ ˆ
3# ‰ ˆ 1x ‰$



ˆ "3 ‰ ˆ
23 ‰ ˆ
2x ‰#

(4)(3)(2)x$ 3!

3!

ˆ
3" ‰ ˆ
43 ‰ ˆ
73 ‰ ax# b$



#!



Š "# ‹ Š
"# ‹ Š
3# ‹ (
2x)$

(
2)(
3)(
4) ˆ
x# ‰

 

#!

#!

3!

3!

#!

ˆ "# ‰ ˆ
"# ‰ ˆ 1x ‰#



ˆ
"# ‰ ˆ
3# ‰ ˆ
5# ‰ (
x)$

(
2)(
3)(
4) ˆ x# ‰



ˆ
"# ‰ ˆ
3# ‰ ax$ b#

"Î$ 10. ˆ1
2x ‰ œ 1  "3 ˆ
2x ‰ 

11. (1  x)% œ 1  4x 

x$
á

3!

(
2)(
3) ˆ x# ‰

œ 1
"# x$ 

" 16

#!


# 6. ˆ1
x# ‰ œ 1
# ˆ
x# ‰ 
"Î#

 á œ 1  "# x
8" x# 

ˆ
"# ‰ ˆ
3# ‰ (
x)#

4. (1
2x)"Î# œ 1  "# (
2x) 
# 5. ˆ1  x# ‰ œ 1
# ˆ x# ‰ 

ˆ "# ‰ ˆ
"# ‰ ˆ
3# ‰ x$ 3!



#!

3. (1
x)
"Î# œ 1
"# (
x) 

7. a1  x$ b



#!

œ a"  2a# x  á  nan xn
1  á Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 16

x%

Section 11.10 Applications of Power Series Ê

dy dx

 y œ (a"  a! )  (2a#  a" )x  (3a$  a# )x#  á  (nan  an
1 )xn
1  á œ !

Ê a"  a! œ 0, 2a#  a" œ 0, 3a$  a# œ 0 and in general nan  an
1 œ 0. Since y œ 1 when x œ 0 we have a! œ 1. Therefore a" œ
1, a# œ Ê y œ 1
x  "# x#


" 3!


a " 2 †1

œ

x$  á 

" #

, a$ œ

(
1)n n!


a # 3

œ
3"†# , á , an œ _

(
")n xn n!

xn  á œ !

nœ0


an
1 n

œ

(
1)n n!

œ e
x

16. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê Ê

dy dx dy dx

œ a"  2a# x  á  nan xn
1  á
2y œ (a"
2a! )  (2a#
2a" )x  (3a$
2a# )x#  á  (nan
2an
1 )xn
1  á œ !

Ê a"
2a! œ 0, 2a#
2a" œ 0, 3a$
2a# œ 0 and in general nan
2an
1 œ 0. Since y œ 1 when x œ 0 we have a! œ 1. Therefore a" œ 2a! œ 2(1) œ 2, a# œ nc1

an œ ˆ 2n ‰ an
1 œ ˆ 2n ‰ Š n2
1 ‹ an
2 œ œ 1  (2x) 

(2x)# 2!



(2x)$ 3!

á 

2n n!

(2x)n n!

2 #

a" œ

2 #

(2) œ

Ê y œ 1  2x  _

(2x)n n!

á œ !

nœ0

2# #

2# #

, a$ œ

x# 

2$ 3!

2 3

a# œ

2 3

x$  á 

#

Š 2# ‹ œ 2n n!

2$ 3 †#

,á ,

xn  á

œ e2x

17. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê Ê

dy dx dy dx

œ a"  2a# x  á  nan xn
1  á
y œ (a"
a! )  (2a#
a" )x  (3a$
a# )x#  á  (nan
an
1 )xn
1  á œ 1

Ê a"
a! œ 1, 2a#
a" œ 0, 3a$
a# œ 0 and in general nan
an
1 œ 0. Since y œ 0 when x œ 0 we have a! œ 0. Therefore a" œ 1, a# œ a#" œ "# , a$ œ a3# œ 3"†# , a% œ a4$ œ 4†3"†# , á , an œ ann
1 œ n!" Ê y œ 0  1x  "# x#  œ ˆ1  1x  "# x# 

" 3 †#

" 3 †#

x$ 

x$ 

" 4†3†2

" 4†3†#

x%  á 

x%  á 

" n!

" n!

xn  á _

xn  á ‰
1 œ !

nœ0

xn n!


1 œ ex
1

18. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê Ê

dy dx dy dx

œ a"  2a# x  á  nan xn
1  á  y œ (a"  a! )  (2a#  a" )x  (3a$  a# )x#  á  (nan  an
1 )xn
1  á œ 1

Ê a"  a! œ 1, 2a#  a" œ 0, 3a$  a# œ 0 and in general nan  an
1 œ 0. Since y œ 2 when x œ 0 we have a! œ 2. Therefore a" œ 1
a! œ
1, a# œ Ê y œ 2
x  "# x#
_

œ1!

nœ0

(
1)n xn n!

" 3 †#

x$  á 

(
")n n!


a " # †1

œ

" #

, a$ œ


a # 3

œ
3"†# , á , an œ

xn  á œ 1  Š1
x  "# x#


" 3 †#


an
1 n

œ

x$  á 

(
")n n! (
")n n!

xn  á ‹

œ 1  e
x

19. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê Ê

dy dx dy dx

œ a"  2a# x  á  nan xn
1  á
y œ (a"
a! )  (2a#
a" )x  (3a$
a# )x#  á  (nan
an
1 )xn
1  á œ x

Ê a"
a! œ 0, 2a#
a" œ 1, 3a$
a# œ 0 and in general nan
an
1 œ 0. Since y œ 0 when x œ 0 we have a! œ 0. Therefore a" œ 0, a# œ 1 # a" œ "# , a$ œ a3# œ 3"†# , a% œ a4$ œ 4†3"†# , á , an œ ann
1 œ n!" Ê y œ 0  0x  "# x#  œ ˆ1  1x  "# x# 

" 3 †#

" 3 †#

x$ 

x$ 

" 4†3†#

" 4†3†#

x%   á 

x%  á 

" n!

" n!

xn  á _

xn  á ‰
1
x œ !

nœ0

xn n!


1
x œ ex
x
1

20. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê Ê

dy dx dy dx

œ a"  2a# x  á  nan xn
1  á  y œ (a"  a! )  (2a#  a" )x  (3a$  a# )x#  á  (nan  an
1 )xn
1  á œ 2x

Ê a"  a! œ 0, 2a#  a" œ 2, 3a$  a# œ 0 and in general nan  an
1 œ 0. Since y œ
1 when x œ 0 we have Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

751

752

Chapter 11 Infinite Sequences and Series

a! œ
1. Therefore a" œ 1, a# œ Ê y œ
1  1x  "# x#
œ Š1
1x  "# x#


" 3 †#

" 3 †#

2
a " #

œ

x$  á 

x$  á 

(
")n n!


a # " # , a$ œ 3 œ n (
") n n! x  á


an
1 n


3"†# , á , an œ _

xn  á ‹
2  2x œ !

nœ0

(
1)n xn n!

(
")n n!

œ


2  2x œ e
x  2x
2

21. yw
xy œ a"  (2a#
a! )x  (3a$
a" )x  á  (nan
an
2 )xn
1  á œ 0 Ê a" œ 0, 2a#
a! œ 0, 3a$
a" œ 0, 4a%
a# œ 0 and in general nan
an
2 œ 0. Since y œ 1 when x œ 0, we have a! œ 1. Therefore a# œ a#! œ "# , a$ œ

a# 4

œ

Ê y œ 1  "# x# 

" #†4

a" 3

œ 0, a% œ

" #†4

, a& œ

x% 

" #†4†6

a$ 5

œ 0, á , a2n œ

" #†4†6â2n

" #†4†6â2n

x'  á 

and a2n1 œ 0 _

x2n  á œ !

nœ0

_

x2n 2n n!

œ!

nœ0

n

#

Š x# ‹ n!

#

œ ex Î2

22. yw
x# y œ a"  2a# x  (3a$
a! )x#  (4a%
a" )x$  á  (nan
an
3 )xn
1  á œ 0 Ê a" œ 0, a# œ 0, 3a$
a! œ 0, 4a%
a" œ 0 and in general nan
an
3 œ 0. Since y œ 1 when x œ 0, we have a! œ 1. Therefore a$ œ a3! œ "3 , a% œ a4" œ 0, a& œ a5# œ 0, a' œ a6$ œ 3"†6 , á , a3n œ 3†6†9"â3n , a3n1 œ 0 and a3n2 œ 0 " 3

" 3 †6

$

Ê yœ1 x 

'

x 

" 3†6†9

*

x á 

" 3†6†9â3n

_

x á œ ! 3n

nœ0

n

x3n 3n n!

_ Š x$ ‹ 3

œ!

nœ0

n!

$

œ ex Î3

23. (1
x)yw
y œ (a"
a! )  (2a#
a"
a" )x  (3a$
2a#
a# )x#  (4a%
3a$
a$ )x$  á  (nan
(n
1)an
1
an
1 )xn
1  á œ 0 Ê a"
a! œ 0, 2a#
2a" œ 0, 3a$
3a# œ 0 and in general (nan
nan
1 ) œ 0. Since y œ 2 when x œ 0, we have a! œ 2. Therefore _

a" œ 2, a# œ 2, á , an œ 2 Ê y œ 2  2x  2x#  á œ ! 2xn œ nœ0

2 1
x

24. a1  x# b yw  2xy œ a"  (2a#  2a! )x  (3a$  2a"  a" )x#  (4a%  2a#  2a# )x$  á  (nan  nan
2 )xn
1  á œ 0 Ê a" œ 0, 2a#  2a! œ 0, 3a$  3a" œ 0, 4a%  4a# œ 0 and in general nan  nan
2 œ 0. Since y œ 3 when x œ 0, we have a! œ 3. Therefore a# œ
3, a$ œ 0, a% œ 3, á , a2n1 œ 0, a2n œ (
1)n 3 _

_

nœ0

nœ0

n Ê y œ 3
3x#  3x%
á œ ! 3(
1)n x2n œ ! 3 a
x# b œ

3 1  x#

25. y œ a!  a" x  a# x#  á  an xn  á Ê yww œ 2a#  3 † 2a$ x  á  n(n
1)an xn
2  á Ê yww
y œ (2a#
a! )  (3 † 2a$
a" )x  (4 † 3a%
a# )x#  á  (n(n
1)an
an
2 )xn
2  á œ 0 Ê 2a#
a! œ 0, 3 † 2a$
a" œ 0, 4 † 3a%
a# œ 0 and in general n(n
1)an
an
2 œ 0. Since yw œ 1 and y œ 0 when x œ 0, " we have a! œ 0 and a" œ 1. Therefore a# œ 0, a$ œ 3"†# , a% œ 0, a& œ 5†4"†3†# , á , a2n1 œ (#n  1)! and a2n œ 0 Ê y œ x 

" 3!

x$ 

" 5!

_

x&  á œ !

nœ0

x2nb1 (2n  1)!

œ sinh x

26. y œ a!  a" x  a# x#  á  an xn  á Ê yww œ 2a#  3 † 2a$ x  á  n(n
1)an xn
2  á Ê yww  y œ (2a#  a! )  (3 † 2a$  a" )x  (4 † 3a%  a# )x#  á  (n(n
1)an  an
2 )xn
2  á œ 0 Ê 2a#  a! œ 0, 3 † 2a$  a" œ 0, 4 † 3a%  a# œ 0 and in general n(n
1)an  an
2 œ 0. Since yw œ 0 and y œ 1 when x œ 0, we have a! œ 1 and a" œ 0. Therefore a# œ
"# , a$ œ 0, a% œ Ê y œ 1
"2 x# 

" 4!

_

x%
á œ !

nœ0

(
1)n x2n (2n)!

" 4 † 3 †#

, a& œ 0, á , a2n1 œ 0 and a2n œ

(
")n (#n)!

œ cos x

27. y œ a!  a" x  a# x#  á  an xn  á Ê yww œ 2a#  3 † 2a$ x  á  n(n
1)an xn
2  á Ê yww  y œ (2a#  a! )  (3 † 2a$  a" )x  (4 † 3a%  a# )x#  á  (n(n
1)an  an
2 )xn
2  á œ x Ê 2a#  a! œ 0, 3 † 2a$  a" œ 1, 4 † 3a%  a# œ 0 and in general n(n
1)an  an
2 œ 0. Since yw œ 1 and y œ 2 when x œ 0, we have a! œ 2 and a" œ 1. Therefore a# œ
", a$ œ 0, a% œ

" 4†3

, a& œ 0, á , a2n œ
2 †

(
1)nb1 (#n)!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

and

Section 11.10 Applications of Power Series a2n1 œ 0 Ê y œ 2  x
x#  2 †

x% 4!

_

(
1)n
1 x2n (2n)!

á œ2x
2!

nœ1

753

œ x  cos 2x

28. y œ a!  a" x  a# x#  á  an xn  á Ê yww œ 2a#  3 † 2a$ x  á  n(n
1)an xn
2  á Ê yww
y œ (2a#
a! )  (3 † 2a$
a" )x  (4 † 3a%
a# )x#  á  (n(n
1)an
an
2 )xn
2  á œ x Ê 2a#
a! œ 0, 3 † 2a$
a" œ 1, 4 † 3a%
a# œ 0 and in general n(n
1)an
an
2 œ 0. Since yw œ 2 and y œ
1 when x œ 0, "
" "
1 3 we have a! œ
1 and a" œ 2. Therefore a# œ
" # , a$ œ # , a% œ 2†3†4 , a& œ 5†4†# œ 5! , á , a2n œ (#n)! and a2n1 œ

Ê y œ
1  2x
#" x# 

3 (2n1)!

3 3!

_

x$
á œ
1  2x
!

x2n (2n)!

nœ1

_

!

nœ1

3x2nb1 (2n1)!

29. y œ a!  a" ax
2b  a# ax
2b#  á  an ax
2bn  á Ê yww œ 2a#  3 † 2a$ ax
2b  á  n(n
1)an ax
2bn
2  á Ê yww
y œ (2a#
a! )  (3 † 2a$
a" )(x
2)  (4 † 3a%
a# )(x
2)#  á  (n(n
1)an
an
2 )(x
2)n
2  á œ
x œ
ax
2b
2 Ê 2a#
a! œ
2, 3 † 2a$
a" œ
1, and n(n
1)an
an
2 œ 0 for n  3. Since y œ 0 when x œ 2, 2 we have a! œ 0, and since yw œ
2 when x œ 2, we have a" œ
2. Therefore a# œ
1, a$ œ
"# , a% œ 41†3 a
1b œ 4†
3 †2 †1 ,
2
3 a& œ 51†4 ˆ
#" ‰ œ 5†4
†33†2†1 , . . . , a2n œ a2n b! , and a2n1 œ (2n1)! . Since a" œ
2, we have a" ax
2b œ a
2bax
2b and a
2bax
2b œ a
3  1bax
2b œ a
3bax
2b  a1bax
2b œ x
2
3ax
2b. Ê y œ x
2
3ax
2b



2b5
. . .

Êyœx
2





_

Ê y œ x
2!

nœ0




2 2! ax

2 3 4 2 3 2 3 2! ax
2b
3! ax
2b
4! ax
2b
5! ax 2b2
4!2 ax
2b4
. . .
3ax
2bx
3!3 ax
2b3 _

(x
2)2n (2n)!


3!

nœ0

3 5! ax


2b5
. . .

(x
2)2nb1 (2n1)!

30. yww
x# y œ 2a#  6a$ x  (4 † 3a%
a! )x#  á  (n(n
1)an
an
4 )xn
2  á œ 0 Ê 2a# œ 0, 6a$ œ 0, 4 † 3a%
a! œ 0, 5 † 4a&
a" œ 0, and in general n(n
1)an
an
4 œ 0. Since yw œ b and y œ a when x œ 0, we have a! œ a, a" œ b, a# œ 0, a$ œ 0, a% œ 3a†4 , a& œ 4b†5 , a' œ 0, a( œ 0, a) œ 3†4a†7†8 , a* œ 4†5b†8†9 Ê y œ a  bx 

a 3†4

x% 

b 4†5

x& 

a 3†4†7†8

x) 

b 4†5†8†9

x*  á

31. yww  x# y œ 2a#  6a$ x  (4 † 3a%  a! )x#  á  (n(n
1)an  an
4 )xn
2  á œ x Ê 2a# œ 0, 6a$ œ 1, 4 † 3a%  a! œ 0, 5 † 4a&  a" œ 0, and in general n(n
1)an  an
4 œ 0. Since yw œ b and y œ a when x œ 0, we have a! œ a and a" œ b. Therefore a# œ 0, a$ œ #"†3 , a% œ
3a†4 , a& œ
4b†5 , a' œ 0, a( œ #†3
" †6†7 Ê y œ a  bx 

1 2†3

x$


a 3†4

x%


b 4†5

x&


1 2†3†6†7

ax) 3†4†7†8

x( 



bx* 4†5†8†9

á

32. yww
2yw  y œ (2a#
2a"  a! Ñ  (2 † 3a$
4a#  a" )x  (3 † 4a%
2 † 3a$  a# )x#  á  ((n
1)nan
2(n
1)an
"  an
2 )xn
2  á œ 0 Ê 2a#
2a"  a! œ 0, 2 † 3a$
4a#  a" œ 0, 3 † 4a%
2 † 3a$  a# œ 0 and in general (n
1)nan
2(n
1)an
"  an
2 œ 0. Since yw œ 1 and y œ 0 when 1 when x œ 0, we have a! œ 0 and a" œ 1. Therefore a# œ 1, a$ œ "# , a% œ 16 , a& œ 24 and an œ (n
"1)! Ê y œ x  x#  "# x$  6" x% 

'00 2 sin x# dx œ '00 2 Šx#
x3!  x5! Þ

33.

kE k Ÿ

Þ

(

(.2) 7†3!

Þ

x

œ ’
x 

'

"!

_

x&  á œ !

nœ1

xn (n
1)!

$


á ‹ dx œ ’ x3


_

œ!

x( 7†3!

nœ0

_

xnb1 n!

 á“

œx!

nœ0

!Þ# !

$

xn n!

¸ ’ x3 “

œ xex

!Þ# !

¸ 0.00267 with error

¸ 0.0000003

'00 2 ec x
" dx œ '00 2 Þ

34.

" #4

x# 4




x$ 18

" x

Š1
x 

 á“

!Þ# !

x# #!




x$ 3!



x% 4!


á
1‹ dx œ '0 Š
1 

¸
0.19044 with error kEk Ÿ

0Þ2

(0.2)% 96

x #




x# 6



x$ 24


á ‹ dx

¸ 0.00002

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

754

Chapter 11 Infinite Sequences and Series

'00 1 È "

1  x%

dx œ '0 Š1


kE k Ÿ

(0.1)& 10

œ 0.000001

'00 25

$È

0Þ1

Þ

35.

Þ

36.

kE k Ÿ

1  x# dx œ '0

0Þ25

&

(0.25) 45

Þ

#

x# 3




x% 9

 á“

x$ 9

 á ‹ dx œ ’x 

!Þ"

¸ [x]!Þ" ! ¸ 0.1 with error

!

x& 45




!Þ#&

 á“

¸ ’x 

!

!Þ#& x$ 9 “!

¸ 0.25174 with error

%

'

$

&

(

$

&

!

(0.1)7 7†7!

¸ 2.8 ‚ 10

Þ

%

(0.1)9 216

¸ 0.0996676643, kEk Ÿ 39. a1  x% b  Ê

"Î#

ˆ "# ‰ ˆ
"# ‰ ˆ
3# ‰ ˆ
5# ‰ 4!

'0

0Þ1

Š "# ‹

œ (1)"Î# 

Š1 

%

x #

)



x 8

x 16



x& 10

x( 42



 á“

!Þ" !

¸ ’x


x$ 3



x& 10




¸ 4.6 ‚ 10

ˆ "# ‰ ˆ
"# ‰ #!

%

"#

$


12

"'




5x 128

x% #

#

(1)
$Î# ax% b 


x) 8

 á ‹ dx ¸ ’x 

&

(1)
(Î# ax% b  á œ 1 

)




'

(1)
"Î# ax% b 

1

!


12

'00 1 exp a
x# b dx œ '00 1 Š1
x#  x2!
x3!  x4!
á ‹ dx œ ’x
x3 Þ

40.

Š1 

x& 10


á ‹ dx œ ’x


'00 1 sinx x dx œ '00 1 Š1
x3!  x5!
x7!  á ‹ dx œ ’x
3x†3!  5x†5!
7x†7!  á “ !Þ" ¸ ’x
3x†3!  5x†5! “ !Þ" ¸ 0.0999444611, kEk Ÿ

38.

3x) 8



¸ 0.0000217

Þ

37.

x% 2

x"# 16 !Þ"



ˆ "# ‰ ˆ
"# ‰ ˆ
3# ‰ 3!

5x"' 128




$

á (0.1)9 72

¸ 0.100001, kEk Ÿ

x 10 “ !

(1)
&Î# ax% b

¸ 1.39 ‚ 10
11

" x '01 ˆ 1
xcos x ‰ dx œ '01 Š "#
x4!  x6!
x8!  10!
á ‹ dx ¸ ’ x#
3x†4!  5x†6!
7x†8!  9†x10! “ #

%

'

)

$

&

(

*

#

¸ 0.4863853764, kEk Ÿ

!

¸ 1.9 ‚ 10
10

1 11†12!

41.

'01 cos t# dt œ '01 Š1
t#  4!t
t6!  á ‹ dt œ ’t
10t  9t†4!
13t †6!  á “ "

42.

'01 cos Èt dt œ '01 Š1
#t  4!t
6!t  8!t
á ‹ dt œ ’t
t4  3t†4!
4t†6!  5t†8!
á “ "

%

)

"#

&

*

"$

Ê kerrork 

!

#

Ê kerrork  x

Ê kerrork 

t' 3!

" 15†7!

¸






x( 7†2!




x

" 33†34

Ê kerrork  (b) kerrork 




t"% 7!

$

 á ‹ dt œ ’ t3


t' 2!

x* 9†3! t$ 3




t) 3!



x"" 11†4!



t& 5






t( 7

t"! 4!




t"# 5!

t( 7†3!

t"" 11†5!



t 2



t# 3




t$ 4

$




Ê kerrork 

" 13†5!

%

&

t"& 15†7!

 á“

x

¸

!

x$ 3




x( 7†3!



x"" 11†5!

#

#

x #




%

x 3†4

 á ‹ dt œ ’t
F(x) ¸ x


x# ##





t( 7†2!




t* 9†3!



t"" 11†4!




t"$ 13†5!

 á“

x

!

¸ 0.00064

 á ‹ dt œ ’ t2


(0.5)' 6# ¸ .00043 " 32# ¸ .00097 when

t& 5

 á ‹ dt œ ’ t3


¸ .00089 when F(x) ¸

46. (a) F(x) œ '0 Š1
x

$

¸ 0.000013

45. (a) F(x) œ '0 Št
(b) kerrork 

t"! 5!



x

x& 5

#

¸ 0.000004960

44. F(x) œ '0 Št#
t%  x$ 3

%

¸ .00011

!

" 5†8!

43. F(x) œ '0 Št#


$

" 13†6!

t% 1#





'

x

t' 30

x 5†6


á“ ¸




t# 2†2



t$ 3†3

x$ 3#




x% 4#

)

!

x 7†8




x# #




x% 1#

Ê kerrork 

 á  (
1)"&

t% 4†4



t& 5†5

x

 á  (
1)$"

¸ .00052

$#

x 31†32


á“ ¸ x
!

(0.5)' 30

x# ##



x$ 3#

x$" 31#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley




x% 4#



x& 5#

!Þ" x( 42 “ !

Section 11.10 Applications of Power Series 47.

" x#

aex
(1  x)b œ

" x

" x

’Š1  x 

aex
e
x b œ 2x#

" t%



3!

2x%

" )&

Š
) 

)$ 6

)Ä0

) 7!

)%

x$ 3!

 á ‹
1
x‹ œ

" #



x 3!



x# 4!

 á Ê lim



x% 4!

 á ‹
Š1
x 

x# #!




x$ 3!



x% 4!


á ‹“ œ

e
e
x x

œ x lim Š2  Ä_

x


Š1


ex
(1  x) x#

xÄ0

t# #

t% 4!



2x# 3!

2x%





5!

t' 6!

 á ‹“ œ
4!" 

)& 5!


á‹ œ




2x'

t# 6!

Š2x 

2x$ 3!



2x& 5!



2x( 7!

 á‹



y% 7


á‹

x& 5

á‹

 á‹ œ 2

7!




" x

#

t% 8!

 á Ê lim

"
cos t
Š t# ‹ t%

tÄ0

"  á ‹ œ
24

" )&

)$ 6

Š
) 


á‹ œ

9!

" y$

x$ 3!

" #



t# #

’1


t% 8!






ay
tan
" yb œ



" 5!




)# 7!

$

)% 9!




á Ê lim

sin )
)  Š )6 ‹ )&

)Ä0

" 1 #0

y$ 3

’y
Š y


)$ 3!

)


y& 5



" 3


á ‹“ œ

y# 5






y% 7

y
tan
" y y$


á Ê lim

yÄ0

œ lim Š 3"
yÄ0

y# 5

" 3

œ

52.

" t%

 sin )‹ œ #



xÄ0

#

t# 6!

x# #!

x# #

 á Ê lim

7!

Š1
cos t
t# ‹ œ

œ lim Š 5!"
" y$

2x'



5!

tÄ0

51.

 á‹ œ



œ lim Š
4!" 

50.

x# 4!

x 3!

œ2

49.

ŠŠ 1  x 

œ lim Š "#  xÄ0

48.

" x#

755

tanc" y
sin y y$ cos y

Ê lim

yÄ0

œ

Œy


y$ 3



y& 5


á

Œy
y$

tanc" y
sin y y$ cos y

Œ


œ lim

" 6

" 6x%




y& 5!


á


23y# 5!



" x#


á




" #x %

55.

56.

x#
4 ln (x
1)

œ

œ

x% #

1
Š1




x# #!

x' 3




á


x% 4!


á‹

œ

Œ1
Š #"!

(x
2)(x  2) ’(x
2)


(x c 2)#

#

x 2

œ lim



(x c 2)$ 3

#

x Ä 2 ’1
x c# 2  (x
32)
á“

" 3!(x  1)$


á“

x# #






x# 4!

œ

x% 3

" n10n



" 10)

x# #



x$ 3




" 6x'



23y& 5!

y$

cos y


á


 á ‰ œ
1 

x% 4

" 5!(x  1)&



" 3!(x  1)#


á


œ

" 6

Œ




23y# 5!


á


cos y

" #x #




’1





á‹ œ 1
" 5!(x  1)%

#

" 6x%

x# Še
1Îx
1‹  á Ê x lim Ä_

x2

(x c 2)# 3

x
2

#



x# #




á“

" 3!(x  1)#



" 5!(x  1)%


á


á‹ œ 1

a1  x # b lim ln1
cos x xÄ0

Ê

 á‹

Œ1


œ lim

Ê lim

Š #"!

xÄ0

x# #






x# 4!

x% 3


á


 á‹

œ 2! œ 2

x#
4

x Ä 2 ln (x
1)

œ4

x‰ 57. ln ˆ 11 
x œ ln (1  x)
ln (1
x) œ Šx


58. ln (1  x) œ x


y$ 6

œ
"6




Ê x lim (x  1) sin ˆ x " 1 ‰ œ x lim Š1
Ä_ Ä_ # Œx


Œ


 á ‰ œ
1

54. (x  1) sin ˆ x " 1 ‰ œ (x  1) Š x " 1


ln a1  x# b 1
cos x

œ

cos y

yÄ0

#

" #x#



cos y

53. x# Š
1  e
1Îx ‹ œ x# ˆ
1  1
ˆ
1  œ x lim Ä_

y$ 3!

á 

(
1)n
1 xn n

x$ 3




x% 4

 á ‹
Š
x


x# #

n
1 n

 á Ê kerrork œ ¹ (
")n

x




¹œ

x$ 3




" n10n

x% 4


á ‹ œ 2 Šx 

when x œ 0.1;

Ê n10n  10) when n   8 Ê 7 terms

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x$ 3



756

Chapter 11 Infinite Sequences and Series

59. tan
" x œ x
" #n
1



" 10$

x$ 3



x& 5

Ê n

60. tan
" x œ x


x$ 3








x* 9


á 

(
")n
1 x2n
1 2n
1

n
1 2n
1

 á Ê kerrork œ ¹ (
1)2n
x1




x( 7



x* 9


á 

(
1)n1 x2n1 2n
1

2n
1

(
1)n 2n
1

nœ1

(
1)nc1 2n
1

nœ1

2n
1 x2n1 ¹

 á and n lim † ¹x Ä _ 2n  1 _

_

" #n
1

when x œ 1;

st

Ê tan
" x converges for kxk  1; when x œ
1 we have ! we have !

¹œ

œ 500.5 Ê the first term not used is the 501 Ê we must use 500 terms

1001 # x& 5

x( 7

¸ 2n
1 ¸ œ x# œ x# n lim Ä _ #n  1

which is a convergent series; when x œ 1

which is a convergent series Ê the series representing tan
" x diverges for kxk  1

(
1)n
1 x2n
1 x$ x& x( x*  á and when the series representing 48 3  5
7  9
á  2n
1 "
' error less than 3 † 10 , then the series representing the sum " ‰ " ‰ " ‰ 48 tan
" ˆ 18  32 tan
" ˆ 57
20 tan
" ˆ #39 also has an error of magnitude less than 10
' ;

" ‰ tan
" ˆ 18 has an

61. tan
" x œ x


thus

2nc1

kerrork œ 48

" Š 18 ‹



#n
1

" 3†10'

Ê n   4 using a calculator Ê 4 terms

62. ln (sec x) œ '0 tan t dt œ '0 Št  x


"Î#

x

t$ 3

x# 3x% #  8  2nb3 lim ¹ 1†3†5â(2n
1)(2n  1)x † n Ä _ 2†4†6â(2n)(2n  2)(2n  3)

63. (a) a1
x# b

¸1



2t& 15

 á ‹ dt ¸

x# #



x% 12

$ 5x'
" x ¸ x  x6 16 Ê sin 2†4†6â(2n)(2n  ") 1†3†5â(2n
1)x2nb1 ¹  1 Ê





x' 45

3x& 40

á



5x( 112

; Using the Ratio Test: (2n  1)(2n1)

x# n lim ¹ ¹1 Ä _ (2n  2)(2n  3)

Ê kxk  1 Ê the radius of convergence is 1. See Exercise 69. d dx

(b)

acos
" xb œ
a1
x# b

64. (a) a1  t# b œ1



"Î# t# #



65.


" 1x

5x( 112

Ê cos
" x œ

¸ (1)
"Î#  ˆ
"# ‰ (1)
$Î# at# b  3t% 2# †2!

(b) sinh
" ˆ 4" ‰ ¸ term,


"Î#

" 4







3†5t' 2$ †3!

" 384



1 #


sin
" x ¸

x

, evaluated at t œ

" 4


Šx 

ˆ
"# ‰ ˆ
#3 ‰ (1)
&Î# at# b# #!

Ê sinh
" x ¸ '0 Š1


3 40,960

1 #

t# #



3t% 8




5t' 16 ‹



x$ 6



3x& 40



5x( 112 ‹

¸

1 #


x


x$ 6




3x& 40

ˆ
#" ‰ ˆ
#3 ‰ ˆ
#5 ‰ (1)
(Î# at# b$ 3!

dt œ x


x$ 6



3x& 40




5x( 112

œ 0.24746908; the error is less than the absolute value of the first unused

since the series is alternating Ê kerrork 

œ
1
"(
x) œ
1  x
x#  x$
á Ê

d dx

ˆ 1
1x ‰ œ

" 1  x#

œ

d dx

5 ˆ "4 ‰ 112

(

¸ 2.725 ‚ 10
'

a
1  x
x#  x$
á b

œ 1
2x  3x#
4x$  á 66.

" 1
x#

œ 1  x#  x%  x'  á Ê

d dx

ˆ 1
" x# ‰ œ

2x a1
x # b #

œ

d dx

a1  x#  x%  x'  á b œ 2x  4x$  6x&  á

8â(2n
2)†(2n) 67. Wallis' formula gives the approximation 1 ¸ 4 ’ 3†23††45††45††67††67†â (2n
1)†(2n
1) “ to produce the table

n 10 20 30 80 90 93 94 95 100

µ1 3.221088998 3.181104886 3.167880758 3.151425420 3.150331383 3.150049112 3.149959030 3.149870848 3.149456425

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley




5x( 112

Section 11.11 Fourier Series At n œ 1929 we obtain the first approximation accurate to 3 decimals: 3.141999845. At n œ 30,000 we still do not obtain accuracy to 4 decimals: 3.141617732, so the convergence to 1 is very slow. Here is a Maple CAS procedure to produce these approximations: pie := proc(n) local i,j; a(2) := evalf(8/9); for i from 3 to n do a(i) := evalf(2*(2*i
2)*i/(2*i
1)^2*a(i
1)) od; [[j,4*a(j)] $ (j = n
5 .. n)] end

68. ln 1 œ 0; ln 2 œ ln ¸ ln 2 



&

Š "5 ‹



3

Š 3" ‹

¸ 2
"3 

1
Š "3 ‹ $

2
"5

$

1  Š "3 ‹

3

&



Š 3" ‹ 5

(



Š 3" ‹ 7

 ¸ 0.69314; ln 3 œ ln 2  ln ˆ # ‰ œ ln 2  ln 3

1  Š 5" ‹ 1
Š 5" ‹

(

Š "5 ‹



5

Š "5 ‹ 7

 ¸ 1.09861; ln 4 œ 2 ln 2 ¸ 1.38628; ln 5 œ ln 4  ln ˆ 4 ‰ œ ln 4  ln 5

¸ 1.60943; ln 6 œ ln 2  ln 3 ¸ 1.79175; ln 7 œ ln 6  ln ˆ 76 ‰ œ ln 6  ln

" 1  Š 13 ‹

1  Š 9" ‹ 1
Š 9" ‹

¸ 1.94591; ln 8 œ 3 ln 2

" 1
Š 13 ‹

¸ 2.07944; ln 9 œ 2 ln 3 ¸ 2.19722; ln 10 œ ln 2  ln 5 ¸ 2.30258 69. a1
x# b 


"Î#

œ a1  a
x# bb


"Î#

ˆ
"# ‰ ˆ
3# ‰ ˆ
5# ‰ (1)
(Î# a
x# b$ 3!

á œ1

Ê sin
" x œ '0 a1
t b x

œ (1)
"Î#  ˆ
"# ‰ (1)
$Î# a
x# b  #

x #



dt œ '0 Œ1  ! _

x

#
"Î#

nœ1

%

1†3x 2# †#!



1†3†5x 2$ †3!

'

ˆ
"# ‰ ˆ
3# ‰ (1)
&Î# a
x# b# #!

_

1†3†5â(2n
1)x2n #n †n!

á œ1!

nœ1

1†3†5â(2n
1)x2n  #n †n!

_

1†3†5â(2n
1)x2nb1 #†4â(2n)(2n  1)

dt œ x  !

nœ1

,

where kxk  1 _

70. ctan
" td x œ

_

œ 'x ˆ t"#



tan
" x œ 'x

" t%



Ê tan
" x œ œ

_

1 #

1 #

" t'




" x


"t 

lim

b Ä
_

" t)


" 3t$

_

t#

bÄ_




_

Š 1# ‹

œ 'x – t " — dt œ 'x 1Š ‹

 á ‰ dt œ lim " 3x$



dt 1  t#


"t 

" 3t$




" 5t&



" t#

" 7t(

ˆ1


" t#



" t%




" t'

" x




" 3x$

b


á ‘x œ

"
" x td c_ œ tan
" x  1# 5x&  á , x  1; ctan " " " " " " ‘x 5t&  7t(
á b œ
x  3x$
5x&  7x(
á




œ

 á ‰ dt 

" 5x&




'
_ 1 dt t

" 7x(

á

x

#

Ê tan
" x œ
1#


" x



" 3x$




x 
1

71. (a) tan atan
" (n  1)
tan
" (n
1)b œ N

tan atanc" (n  1)b
tan atan
" (n
1)b 1  tan atan
" (n  1)b tan atan
" (n
1)b

œ

(n  1)
(n
1) 1  (n  1)(n
1)

œ

2 n#

N

(b) ! tan
" ˆ n2# ‰ œ ! ctan
" (n  1)
tan
" (n
1)d œ atan
" 2
tan
" 0b  atan
" 3
tan
" 1b nœ1

nœ1


"

 atan _

4
tan

(c) ! tan
" ˆ n2# ‰ œ nœ1


"

2b  á  atan
" (N  1)
tan
" (N
1)b œ tan
" (N  1)  tan
" N


lim tan
" (N  1)  tan
" N
14 ‘ œ nÄ_

1 #



1 #




1 4

œ

1 4

31 4

11.11 FOURIER SERIES 1. a0 œ

1 21

'021 1 dx œ 1, ak œ 11 '021 cos kx dx œ 11  sinkkx ‘201 œ 0, bk œ 11 '021 sin kx dx œ 11 
coskkx ‘201 œ 0.

Thus, the Fourier series for faxb is 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 5x&

á ,

757

758

Chapter 11 Infinite Sequences and Series

2. a0 œ

' 1 1 dx  '121
1 dx • œ 0, ak œ 11 ”'01 cos kx dx
'121 cos kx dx • œ 11 ” sinkkx ¹1
sinkkx ¹21 • œ 0,

1 21 ” 0

bk œ 11 ”'0 sin kx dx
'1 sin kx dx • œ 11 ”
cosk kx ¹  21

1 0

4 , k odd
2 cos k1b œ œ k1 . 0, k even Thus, the Fourier series for faxb is 14  sin x 

œ

1

0

1

21

cos kx k ¹1

•œ

1 a
cos k1 c

k1  1b  acos 21k
cos 1kb d

1 k 1 a2

3. a0 œ

sin 3x 3



sin 5x 5

 . . . ‘.

' 1 x dx  '121 ax
21b dx • œ 211  "# 12  "# a412
12 b
212 ‘ œ 0.

1 21 ” 0

Note,

'121 ax
21bcos kx dx œ
'01 u cos ku du (Let u œ 21
x). So ak œ 11 ”'01 x cos kx dx  '121 ax
21b cos kx dx • œ 0. Note, '1 ax
21bsin kx dx œ '0 u sin ku du (Let u œ 21
x). So bk œ 11 ”'0 x sin kx dx  '1 ax
21b sin kx dx • 21

œ

2 1

1

1

21

'01 x sin kx dx œ 12 
xk cos kx  k1 sin kx ‘01 œ
2k cos k1 œ 2k a
1bk1 . 2

_

Thus, the Fourier series for faxb is ! a
1bk1 2 sink kx . k œ1

4. a0 œ

1 21

'021 faxb dx œ 211 '01 x2 dx œ 16 12 , 2

œ 11 ’ Š xk

'021 faxb cos kx dx œ 11 '01 x2 cos kx dx 1 1 21
k2 ‹sin kx  k# x cos kx “ œ k# cos k1 œ a
1bk ˆ k# ‰, bk œ 11 '0 faxb sin kx dx œ 11 '0 x2 sin kx dx œ

œ 11 ’ Š k23


3

2

x k

2

‹cos kx 

# k2 x

0 1

ak œ

1 1

2

sin kx “ œ 11 ’ Š k23
0

2

1 k

2

k ‹a
1b


# k3

k k “ œ 11 ’ Ša
1b
1‹ k#3 “
1k a
1b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 11.11 Fourier Series œœ


14k3  1k , k odd .
1k , k even

Thus, the Fourier series for faxb is 16 12
2 cos x  Š 1

5. a0 œ bk œ

1 21 1 1

'021 ex dx œ 211 ae21
1b,

'0

21

ex sin kx dx

Thus, the Fourier series

6. a0 œ

1 21

ak œ

1 1


4 1 ‹sin

x  "# cos 2x


1 2

sin 2x
29 cos 3x  Š 9127
1 4 ‹sin 3x  . . . 2

'021 ex cos kx dx œ 11  1 e k acos kx  k sin kxb ‘201 œ 1ea11
k1 b , x

2

2

2

21 kˆ1
e21 ‰ œ 11  1e k2 asin kx
k cos kxb ‘0 œ 1a1  k2 b . _ 21 kx k sin kx ‰ for faxb is 211 ae21
1b  e 1
1 ! ˆ cos 1  k2
1  k2 . k œ1 x

x

2

1 k  1 1a1k2 b e a
1b

œ

1  ex asin 1 1  k2


1‘ œ 


a1  e 1 b 1 a1  k 2 b , e1
1 1 a1  k 2 b ,

1 kxb ‘0

kx
k cos

œ

k odd k even

. bk œ


k  1 k 1a1  k2 b e a
1b

Thus, the Fourier series for faxb is

7. a0 œ

2

'021 faxb dx œ 211 '01 ex dx œ e12
1 1 , ak œ 11 '021 faxb cos kx dx œ 11 '01 ex cos kx dx œ 11  1 e k acos kx  k sin kxb ‘01

œ

e1
1 21

759




a1  e 1 b 21 cos

1 21

'0

21

x

faxb dx œ

a1  e 1 b 21 sin

1 21

'0

21

x

e1
1 51 cos

2x 

cos x dx œ 0, ak œ

1 1

1 1

'021 faxb sin kx dx œ 11 '01 ex sin kx dx


1‘ œ 

2 a1
e 1 b sin 51

'0

21

k a1  e 1 b 1 a1  k 2 b , 1
e1 1 a1  k 2 b ,

2x


k odd k even

a1  e 1 b 101 cos

Ú cos x cos kx dx œ Û Ü

3x 

.

3 a1  e1 b 101 sin

1 sinak
1bx 1 ’ 2 ak
1 b



3x  . . .

sinak  1bx 2 ak  1 b

1 " 1 #x



Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

“ ,

0 1 " sin 2x ‘0 , 4

kÁ1 kœ1

760

Chapter 11 Infinite Sequences and Series !, œœ" #,

bk œ

1 1

kÁ1 . kœ1

'0

21

Ú Ý
1 ’ cosak
1bx  2 ak
1 b 1 cos x sin kx dx œ Û Ý Ü

Thus, the Fourier series for faxb is

8. a0 œ

1 21

" # cos

x! k even

1

cosak  1bx 2 ak  1 b

0


411 cos

2k 1ak2
1b sin

œ 11 ”'0 2 cos kx dx  '1
x cos kx dx • œ
11  cosk2kx  bk œ œ

1 1

21

1

2x¹ , k œ 1

!,

œ

k odd . k even

2k 1ak2
1b ,

0

kx.

'021 faxb dx œ 211 ”'01 2 dx  '121
x dx • œ 1
43 1, 1

kÁ1

“ ,

ak œ

x sin kx k

1 1

'021 faxb cos kx dx

‘21 œ 1


1  a
1bk 1 k2

œœ


12k2 , k odd . 0, k even

'021 faxb sin kx dx œ 11 ”'01 2 sin kx dx  '121
x sin kx dx • œ 11 ”
2k cos kx¹1  ˆ x cosk kx
sink kx ‰¹21 •

1ˆ4 k 1

2

0

 3‰, 1 k,

1

k odd . k even

Thus, the Fourier series for faxb is 1
34 1
12 cos x  ˆ 14  3‰sin x  "# sin 2x


2 91 cos

3x  13 ˆ 14  3‰sin 3x  . . . .

9.

'021 cos px dx œ 1p sin px¹21 œ 0 if p Á 0.

10.

'021 sin px dx œ
1p cos px¹21 œ
1p c 1
1 d œ 0 if p Á 0.

11.

'021 cos px cos qx dx œ '021 "# c cos ap  qbx  cos ap
qbx ddx œ "#  p 1 q sin ap  qbx  p
1 q sin ap
qbx ‘201 œ 0 if p Á q.

0

0

If p œ q then '0 cos px cos qx dx œ '0 cos2 px dx œ '0 21

12.

21

21

" # a1

 cos 2pxb dx œ "# Šx 

1 2p sin

2px‹¹

21 0

œ 1.

'021 sin px sin qx dx œ '021 "# c cos ap
qbx
cos ap  qbx ddx œ "#  p
1 q sin ap
qbx
p 1 q sin ap  qbx ‘201 œ 0 if p Á q.

If p œ q then '0 sin px sin qx dx œ '0 sin2 px dx œ '0 21

21

21

" # a1


cos 2pxb dx œ "# Šx


1 2p sin

2px‹¹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

21 0

œ 1.

Chapter 11 Practice Exercises 13.

761

'021 sin px cos qx dx œ '021 "# c sin ap  qbx  sin ap
qbx ddx œ
"#  p 1 q cos ap  qbx  p
1 q cos ap
qbx ‘201 21 21 21 œ
"#  a1
1b p 1 q  a1
1b p
1 q ‘ œ 0. If p œ q then '0 sin px cos qx dx œ '0 sin px cos px dx œ '0 "# sin 2px dx œ
411 cos 2px¹

21 0

œ
411 a1
1b œ 0.

14. Yes. Note that if f is continuous at c, then the expression

facb b  facc b 2

œ facb since fac b œ limb faxb œ facb and xÄc

fac
b œ limc faxb œ facb. Now since the sum of two piecewise continuous functions on Ò0, 21Ó is also continuous on Ò0, 21Ó, x Äc

the function f  g satisfies the hypothesis of Theorem 24, and so its Fourier series converges to

af  gbacb b  af  gbacc b 2

for 0  c  21. Let sf axb denote the Fourier series for faxb. Then for any c in the interval a0, 21b b c sfg acb œ af  gbac b  af  gbac b œ " ’ lim+ af  gbaxb  limc af  gbaxb “ œ " ’ limb faxb  limb gaxb  limc faxb  limc gaxb “ #

2

"  # c afac b

œ



x Äc

x Äc




#

x Äc

x Äc

x Äc

x Äc




 gac bb  afac b  gac bb d œ sf acb  sg acb, since f and g satisfy the hypothesis of Theorem 24.

15. (a) faxb is piecewise continuous on Ò0, 21Ó and f w axb œ 1 for all x Á 1 Ê f w axb is piecewise continuous on Ò0, 21Ó. Then by Theorem 24, the Fourier series for faxb converges to faxb for all x Á 1 and converges to "# afa1 b  fa1
bb œ "# a
1  1b œ 0 at x œ 1.

_

(b) The Fourier series for faxb is ! a
1bk1 2 sink kx . If we differentiate this series term by term we get the series k œ1

_

! a
1b

k 1

k œ1

2 cos kx, which diverges by the nth term test for divergence for any x since lim a
1bk1 2 cos kx Á 0. kÄ_

16. Since the Fourier series in discontinuous at x œ 1, by Theorem 24, the Fourier series will converge to at x œ 1 we have

fa1b b  fa1c b 2

œ

1 2 61

Ê

0  12 2

œ 16 12
2 cos 1  Š 1

Ê

0  12 2

œ 16 12  2 

12 2




12 6

_

œ 2!

n œ1

1 n2

Ê

12 3

" #



2 9

2

2 Š 1 1
4 ‹sin


2 cos x 


4 1 ‹sin

1  "# cos 21


 . . . œ 16 12  2ˆ1 

_

œ 2!

n œ1

1 n2

Ê

12 6

_

œ!

n œ1

1 4

1 2



x

" # cos

2x


1 2

sin 2x


2 9 cos

3x 

facb b  facc b . 2

2 Š 9127
1 4 ‹sin

sin 21
29 cos 31  Š 9127
1 4 ‹sin 31  . . . 2

1 9

_

 . . . ‰ œ 16 12  2! n12 Ê n œ1

12 2

œ

12 6

_

 2! n12 n œ1

1 n2 .

CHAPTER 11 PRACTICE EXERCISES 1. converges to 1, since n lim a œ n lim Š1  Ä_ n Ä_ 2. converges to 0, since 0 Ÿ an Ÿ

2 Èn

(
1)n n ‹

œ1

, n lim 0 œ 0, n lim Ä_ Ä_

2 Èn

œ 0 using the Sandwich Theorem for Sequences

ˆ 1
2n2 ‰ œ lim ˆ #"n
1‰ œ
1 3. converges to
1, since n lim a œ n lim Ä_ n Ä_ nÄ_ n

4. converges to 1, since n lim a œ n lim c1  (0.9)n d œ 1  0 œ 1 Ä_ n Ä_ 5. diverges, since ˜sin

n1 ™ #

œ e0ß 1ß 0ß
1ß 0ß 1ß á f

6. converges to 0, since {sin n1} œ {0ß 0ß 0ß á } 7. converges to 0, since n lim a œ n lim Ä_ n Ä_

ln n# n

œ 2 n lim Ä_

Š "n ‹ 1

œ0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Thus,

3x  . . . .

762

Chapter 11 Infinite Sequences and Series ln (2n") n

8. converges to 0, since n lim a œ n lim Ä_ n Ä_

Š 2n 2b 1 ‹

œ n lim Ä_

ˆ n nln n ‰ œ lim 9. converges to 1, since n lim a œ n lim Ä_ n Ä_ nÄ_ 10. converges to 0, since n lim a œ Ä_ n

$ lim ln a2nn  1b nÄ_

œ0

1 1Š "n ‹

œ1

1

Š

œ n lim Ä_

6n# ‹ 2n$
1

1

ˆ n
n 5 ‰n œ lim Š1  11. converges to ec5 , since n lim a œ n lim Ä_ n Ä_ nÄ_ ˆ1  n" ‰cn œ lim , since n lim a œ n lim Ä_ n Ä_ nÄ_

" e

12. converges to

œ n lim Ä_

(
5) n ‹

" ˆ1  "n ‰n

n

œ

12n 6n#

œ n lim Ä_

œ ec5 by Theorem 5 " e

by Theorem 5

ˆ 3 ‰1În œ lim 13. converges to 3, since n lim a œ n lim Ä_ n Ä_ n nÄ_

3 n1În

œ

3 1

œ 3 by Theorem 5

ˆ 3 ‰1În œ lim 14. converges to 1, since n lim a œ n lim Ä_ n Ä_ n nÄ_

31În n1În

œ

1 1

œ 1 by Theorem 5

n

15. converges to ln 2, since n lim a œ n lim n a21În
1b œ n lim Ä_ n Ä_ Ä_

21În
1 Š "n ‹

œ0

2 n

œ n lim Ä_

–

Š
21În ln 2‹ n#

—

œ n lim 21În ln 2 Ä_

Š
" ‹ n#

œ 2! † ln 2 œ ln 2 2

n È 16. converges to 1, since n lim a œ n lim 2n  1 œ n lim exp Š ln (2nn  1) ‹ œ n lim exp Œ 2n1b 1  œ e! œ 1 Ä_ n Ä_ Ä_ Ä_

17. diverges, since n lim a œ n lim Ä_ n Ä_

(n  1)! n!

œ n lim (n  1) œ _ Ä_

18. converges to 0, since n lim a œ n lim Ä_ n Ä_

(
4)n n!

Š "# ‹

Š "# ‹

19.

" (2n
3)(2n
1)

œ

#n
3




Š "# ‹

Ê sn œ –

2n
1

Š "# ‹

" Ê n lim s œ n lim
Ä_ n Ä _ –6

20.


2 n(n  1)

œ


2 n



2 n1

ˆ
1  œ n lim Ä_ 21.

22.

9 (3n
1)(3n  2) œ 3#
3n3#

œ

_

_

nœ0

nœ0

5

—–

Š "# ‹ 5




Š "# ‹ 7




Š "‹

# —  á  – #n
3


Š "# ‹

2n
1 —

œ

Š "# ‹

3 3n  2

3

" 6

Ê sn œ ˆ #3
35 ‰  ˆ 35
38 ‰  ˆ 38


ˆ3
Ê n lim s œ n lim Ä_ n Ä_ #

" en

Š "# ‹

2 ‰ n1

œ
#2 

2 n1

Ê n lim s Ä_ n

œ
1


8
2 2 (4n
3)(4n  1) œ 4n
3  4n  1 œ
29  4n21 Ê n lim s Ä_ n

23. ! e
n œ !

œ




Ê sn œ ˆ
#2  23 ‰  ˆ
32  42 ‰  á  ˆ
n2 

2 ‰ n1

3 3n
1

2n
1 —

3

œ 0 by Theorem 5

3 ‰ 3n  2

œ

 á  ˆ 3n 3
1


3 ‰ 3n  2

3 #

Ê sn œ ˆ
92  ˆ
œ n lim Ä_

3 ‰ 11

2 ‰ 2 ‰ ˆ
2 13  13  17 2 2 ‰ 2 9  4n1 œ
9

, a convergent geometric series with r œ

" e

 ˆ
172 

2 ‰ 21

 á  ˆ 4n

2 3 

and a œ 1 Ê the sum is

" 1
Š "e ‹

œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 ‰ 4n  1

e e
1




Š "# ‹

2n
1

Chapter 11 Practice Exercises _

24. ! (
1)n

3 4n

nœ1

ˆ
34 ‰ 1
ˆ c4" ‰

_

‰n a convergent geometric series with r œ
"4 and a œ œ ! ˆ
43 ‰ ˆ
" 4

_

nœ1

Ê the sum is

œ
35

25. diverges, a p-series with p œ 26. !


3 4

nœ0


5 n

_

œ
5 !

nœ1

27. Since f(x) œ _

" x"Î#

nœ1

diverges since it is a nonzero multiple of the divergent harmonic series

Ê f w (x) œ
#x"$Î#  0 Ê f(x) is decreasing Ê an1  an , and _

(
")n Èn

series !

" n,

" #

converges by the Alternating Series Test. Since ! nœ1

" Èn

lim a œ lim nÄ_ n nÄ_

1 Èn

œ 0, the

diverges, the given series converges

conditionally. 28. converges absolutely by the Direct Comparison Test since

" #n$



" n$

for n   1, which is the nth term of a

convergent p-series 29. The given series does not converge absolutely by the Direct Comparison Test since the nth term of decreasing Ê

" ln (n  1)



" n1

, which is

" a divergent series. Since f(x) œ ln (x" 1) Ê f w (x) œ
(ln (x  1)) # (x  1)  0 Ê f(x) is " an1  an , and n lim a œ n lim œ 0, the given series converges conditionally Ä_ n Ä _ ln (n  1)

by the

Alternating Series Test. 30.

'2_ x(ln" x)

#

dx œ lim

bÄ_

'2b

" x(ln x)#

b dx œ lim c
(ln x)
" d 2 œ
lim ˆ ln"b


bÄ_

bÄ_

" ‰ ln 2

œ

" ln #

Ê the series

converges absolutely by the Integral Test 31. converges absolutely by the Direct Comparison Test since

ln n n$



n n$

œ

" n#

, the nth term of a convergent p-series

n n 32. diverges by the Direct Comparison Test for en  n Ê ln ˆen ‰  ln n Ê nn  ln n Ê ln nn  ln (ln n)

Ê n ln n  ln (ln n) Ê

33. n lim Ä_

Š

ln n ln (ln n)

"

n

È n#
1 ‹ Š n"# ‹

34. Since f(x) œ

œ Én lim Ä_

3x# x$  1



n# n#  1

Ê f w (x) œ

" n

, the nth term of the divergent harmonic series

œ È1 œ 1 Ê converges absolutely by the Limit Comparison Test

3x a2
x$ b ax $  1 b#

 0 when x   2 Ê an1  an for n   2 and n lim Ä_

3n# n$  1

œ 0, the

series converges by the Alternating Series Test. The series does not converge absolutely: By the Limit #

Comparison Test, n lim Ä_

Š n$3n
1 ‹ ˆ n" ‰

œ n lim Ä_

3n$ n$  1

œ 3. Therefore the convergence is conditional.

35. converges absolutely by the Ratio Test since n lim ’ n 2 † Ä _ (n  1)! 36. diverges since n lim a œ n lim Ä_ n Ä_

(
")n an#  1b 2n#  n
1

n! n1“

œ n lim Ä_

n2 (n  1)#

œ01

does not exist nb1

37. converges absolutely by the Ratio Test since n lim † ’ 3 Ä _ (n  1)!

n! 3n “

œ n lim Ä_

3 n1

œ01

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

763

764

Chapter 11 Infinite Sequences and Series

n 2 3 n È É 38. converges absolutely by the Root Test since n lim an œ n lim nn œ n lim Ä_ Ä_ Ä_ n n

39. converges absolutely by the Limit Comparison Test since n lim Ä_

40. converges absolutely by the Limit Comparison Test since n lim Ä_ nb1

41. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x  4) † Ä_ Ä _ (n  1)3nb1

n3n (x  4)n ¹

1 Ê

" Š $Î# ‹ n

Š Èn(n
"1)(n
2) ‹

6 n

œ01

Š n"# ‹

Š

n # an #
1 b n%

" ‹ œ Én lim Ä_ nÈ n#
1

kx  4 k lim 3 nÄ_

n(n  1)(n  2) n$

œ Én lim Ä_

kx  4 k 3

ˆ n n 1 ‰  1 Ê _

(
1)n 3n n3n

Ê kx  4k  3 Ê
3  x  4  3 Ê
7  x 
1; at x œ
7 we have !

nœ1 _

alternating harmonic series, which converges conditionally; at x œ
1 we have !

nœ1

3n n3n

œ1

œ1

1

_

œ ! (
"n ) , the n

nœ1

_

œ!

nœ1

" n

, the divergent

harmonic series (a) the radius is 3; the interval of convergence is
7 Ÿ x 
1 (b) the interval of absolute convergence is
7  x 
1 (c) the series converges conditionally at x œ
7 42. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x
1) † (2n
1)! ¹  1 Ê (x
1)# n lim Ä_ Ä _ (2n1)! (x
1)2nc2 Ä_ all x (a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally 2n

nb1

43. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (3x(n
1)1)# † Ä_ Ä_

n# (3x
1)n ¹

 1 Ê k3x
1k n lim Ä_

Ê
1  3x
1  1 Ê 0  3x  2 Ê 0  x  _

" n#

œ
!

nœ1 _

have ! nœ1

" (#n)(2n1)

2 3

n# (n  1)# _

œ 0  1, which holds for

 1 Ê k3x
1k  1 nc1

_

2nc1

; at x œ 0 we have ! (
1) n# (
1) œ ! (
"n)# n

nœ1

nœ1

, a nonzero constant multiple of a convergent p-series, which is absolutely convergent; at x œ _

(
1)n
1 (1)n n#

(a) the radius is

n
1

œ ! (
"n)#

2 3

we

, which converges absolutely

nœ1

" 3

; the interval of convergence is 0 Ÿ x Ÿ

(b) the interval of absolute convergence is 0 Ÿ x Ÿ

2 3

2 3

(c) there are no values for which the series converges conditionally 44. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ n2 † Ä_ Ä _ 2n  3 Ê _

! nœ1

k2x  1k #

n1 2n  1

†

†

2n  1 n1

†

2n (2x  1)n ¹

1 Ê

k2x  1k lim 2 nÄ_

n2 ¸ 2n 3 †

2n  " ¸ n1

(1)  1 Ê k2x  1k  2 Ê
2  2x  1  2 Ê
3  2x  1 Ê
3#  x 

(
2)n #n

_

œ!

n1 ‰ lim ˆ 2n 1 œ

nÄ_

(2x  1)nb1 2nb1

nœ1

" #

(
")n (n1) 2n  1

which diverges by the nth-Term Test for Divergence since

Á 0; at x œ

" #

_

n1 we have ! 2n 1 † nœ1

2n #n

_

n" œ ! 2n  1 , which diverges by the nthnœ1

Term Test (a) the radius is 1; the interval of convergence is
3#  x  (b) the interval of absolute convergence is
3#  x 

" #

" #

(c) there are no values for which the series converges conditionally

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" #

1

; at x œ
3# we have

Chapter 11 Practice Exercises nb1

45. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ x Ä_ Ä _ (n  1)nb1 Ê

kx k e

¸ˆ n ‰n ˆ n " 1 ‰¸  1 Ê  1 Ê kxk n lim Ä _ n1

nn xn ¹

kx k e n lim Ä_

ˆ n " 1 ‰  1

† 0  1, which holds for all x

(a) the radius is _; the series converges for all x (b) the series converges absolutely for all x (c) there are no values for which the series converges conditionally nb1

46. n lim † ¹ uunbn 1 ¹  1 Ê n lim ¹ x Ä_ Ä _ Èn  1 _

Èn xn ¹

n  1 Ê kxk n lim  1 Ê kxk  1; when x œ
1 we have Ä _ Én1

_

! (
È1) , which converges by the Alternating Series Test; when x œ 1 we have ! n n

nœ1

nœ1

" Èn

, a divergent

p-series (a) the radius is 1; the interval of convergence is
1 Ÿ x  1 (b) the interval of absolute convergence is
1  x  1 (c) the series converges conditionally at x œ
1 2nb1

47. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (n 32)x nb1 Ä_ Ä_ _

_

nœ1

nœ1

the series !
nÈ31 and !

n1 È3

†

3n (n  1)x2n
1 ¹

1 Ê

x# 3 n lim Ä_

2‰ È È3; ˆ nn  1 1 Ê
3x

, obtained with x œ „ È3, both diverge

(a) the radius is È3; the interval of convergence is
È3  x  È3 (b) the interval of absolute convergence is
È3  x  È3 (c) there are no values for which the series converges conditionally 2nb3

48. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ (x
2n1)x 3 Ä_ Ä_

†

2n  1 (x
1)2nb1 ¹

ˆ 2n  1 ‰  1 Ê (x
1)# (1)  1  1 Ê (x
1)# n lim Ä _ 2n  3 _

Ê (x
1)#  1 Ê kx
1k  1 Ê
1  x
1  1 Ê 0  x  2; at x œ 0 we have ! (
1)#n(
1)1 n

2nb1

nœ1

_

œ!

nœ1 _

(
1)3nb1 2n  1

that ! nœ1

" 2n  1

_

œ!

nœ1

(
1)nc1 2n  1

which converges conditionally by the Alternating Series Test and the fact _

diverges; at x œ 2 we have !

nœ1

(
1)n (1)2nb1 2n  1

_

œ!

nœ1

(
1)n 2n  1

, which also converges

conditionally (a) the radius is 1; the interval of convergence is 0 Ÿ x Ÿ 2 (b) the interval of absolute convergence is 0  x  2 (c) the series converges conditionally at x œ 0 and x œ 2 nb1

(n  1)x 49. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹ cschcsch (n)xn Ä_ Ä_

c"


2n
1

Ê kxk n lim ¹ e1

e
e 2n
2 ¹  1 Ê Ä_

kx k e

¹  1 Ê kxk n lim Ä_ »

Š enb1 c2ecnc1 ‹ ˆ en c2ecn ‰

»1

_

 1 Ê
e  x  e; the series ! a „ ebn csch n, obtained with x œ „ e, nœ1

both diverge since n lim a „ e)n csch n Á 0 Ä_

(a) the radius is e; the interval of convergence is
e  x  e (b) the interval of absolute convergence is
e  x  e (c) there are no values for which the series converges conditionally 50. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹x Ä_ Ä_

nb1

coth (n  1) xn coth (n) ¹

c2nc2

 1 Ê kxk n lim † ¹1e Ä _ 1
ec2nc2

1
ec2n 1  ec2n ¹

 1 Ê kxk  1

_

Ê
1  x  1; the series ! a „ 1bn coth n, obtained with x œ „ 1, both diverge since n lim a „ 1bn coth n Á 0 Ä_ nœ1

(a) the radius is 1; the interval of convergence is
1  x  1 (b) the interval of absolute convergence is
1  x  1 Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

765

766

Chapter 11 Infinite Sequences and Series

(c) there are no values for which the series converges conditionally 51. The given series has the form 1
x  x#
x$  á  (
x)n  á œ 52. The given series has the form x
ln ˆ 53 ‰ ¸ 0.510825624

x# #



x$ 3


á  (
1)n
1

53. The given series has the form x


x$ 3!



x& 5!


á  (
1)n

x2n
1 (2n  1)!

x# 2!



x% 4!


á  (
1)n

x2n (2n)!

xn n

" 1x

, where x œ

" 4

; the sum is

 á œ ln (1  x), where x œ

2 3

" 1  ˆ "4 ‰

œ

4 5

œ

" #

; the sum is

 á œ sin x, where x œ 1; the sum is

sin 1 œ 0 54. The given series has the form 1


55. The given series has the form 1  x  56. The given series has the form x
tan
" Š È"3 ‹ œ 57. Consider

" 1
2x

x$ 3

x# 2!





x& 5

x# 3!

x2n
1 (2n
1)


á  (
1)n

 á œ tan
" x, where x œ

as the sum of a convergent geometric series with a œ 1 and r œ 2x Ê _

nœ0

nœ0

œ 1  (2x)  (2x)#  (2x)$  á œ ! (2x)n œ ! 2n xn where k2xk  1 Ê kxk 

58. Consider

" 1  x$

" È3

; the sum is

1 6

_

" 1
2x

" #

as the sum of a convergent geometric series with a œ 1 and r œ
x$ Ê

" 1  x$

œ

" 1
a
x$ b

_

# $ œ 1  a
x$ b  a
x$ b  a
x$ b  á œ ! (
1)n x3n where k
x$ k  1 Ê kx$ k  1 Ê kxk  1 nœ0

_

59. sin x œ !

nœ0

_

60. sin x œ !

nœ0 _

61. cos x œ !

nœ0 _

62. cos x œ !

nœ0

_

63. ex œ !

nœ0 _

64. ex œ !

nœ0

_

(
1)n x2nb1 (2n  1)!

Ê sin 1x œ !

nœ0

(
1)n x2nb1 (2n  1)!

_

nœ0

Ê sin

_ (
1)n Š 2x ‹ 3

œ!

2x 3

_

_

nœ0

(
1)n ˆx&Î# ‰ (2n)!

2n

_

(
1)n x2n (2n)!

Ê cos ˆx&Î# ‰ œ !

(
1)n x2n (2n)!

Ê cos È5x œ cos ˆ(5x)"Î# ‰ œ !

nœ0

_ ˆ 1 x ‰n

xn n!

# Ê e
x œ !

nœ0

_



3 8

œ

9 32

#

n!

a
x # b n!

nœ0

Ê f www (x) œ 3x$ a3  x# b f (
1) œ


nœ0

nœ0

Ê eÐ1xÎ2Ñ œ !

3 32

œ! _

xn n!

"Î#


&Î#

_

œ!

n œ0

n

_

œ!

nœ0

(
1)n 22nb1 x2nb1 32nb1 (#n  1)!

œ!

(2n  1)!

nœ0

(
1)n 12nb1 x2nb1 (#n  1)!

œ!

2nb1

65. f(x) œ È3  x# œ a3  x# b www

(
1)n (1x)2nb1 (2n  1)!

(
1)n x5n (#n)!

(
1)n ˆ(5x)"Î# ‰ (2n)!

2n

_

œ!

nœ0

(
1)n 5n xn (#n)!

1 n xn #n n!

(
1)n x2n n!

Ê f w (x) œ x a3  x# b


"Î#

Ê f ww (x) œ
x# a3  x# b


$Î#


$Î#

 a3  x# b


3x a3  x# b ; f(
1) œ 2, f w (
1) œ
"# , f ww (
1) œ
"8  # $ Ê È3  x# œ 2
(x  1)  3(x$ 1)  9(x& 1)  á 2†1!

1 3

 á œ ex , where x œ ln 2; the sum is eln Ð2Ñ œ 2

xn n!

á 

 á œ cos x, where x œ 13 ; the sum is cos

2 †2!

" #

2 †3!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ


"Î#

3 8

,

Chapter 11 Practice Exercises 66. f(x) œ

" 1
x

œ (1
x)
" Ê f w (x) œ (1
x)
# Ê f ww (x) œ 2(1
x)
$ Ê f www (x) œ 6(1
x)
% ; f(2) œ
1, f w (2) œ 1, " 1
x

f ww (2) œ
2, f www (2) œ 6 Ê " x1 œ œ
4"# ,

67. f(x) œ w

f (3)

68. f(x) œ

" x

f www (a) œ

œ
1  (x
2)
(x
2)#  (x
2)$
á

(x  1)
" Ê f w (x) œ
(x  1)
# Ê f ww (x) œ 2(x  1)
$ Ê f www (x) œ
6(x  1)
% ; f(3) œ ww

f (3) œ

www

, f (2) œ

2 4$


6 4%

Ê

" x 1

œ

" 4

" 4#




(x
3) 

" 4$

#

(x
3)


œ x
" Ê f w (x) œ
x
# Ê f ww (x) œ 2x
$ Ê f www (x) œ
6x
% ; f(a) œ
6 a%

Ê

" x

" a

œ




" a#

(x
a) 

" a$

(x
a)#


" a%

" 4% " a

" 4

,

2 a$

,

$

(x
3)  á , f w (a) œ
a"# , f ww (a) œ

(x
a)$  á

69. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê

dy dx

y

œ aa"  a! b  a2a#  a" bx  (3a$  a# )x#  á  (nan  an
1 )xn
1  á œ 0 Ê a"  a! œ 0, 2a#  a" œ 0, 3a$  a# œ 0 and in general nan  an
1 œ 0. Since y œ
1 when x œ 0 we have a! œ
1. Therefore a" œ 1, a# œ


a " 2 †1


a # 3

œ
"# , a$ œ


a $ 4

œ

" 3†2

" 3 †#

x$
á 

Ê y œ
1  x
"# x# 

, a% œ

œ
4†3"†# , á , an œ

(
1)n1 n!


an
1 n

_

xn  á œ
!

nœ0

œ

(
1)n xn n!


1 (
1)n n (n
1)!

œ

(
1)n1 n!

œ
e
x

70. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê

dy dx


y

#

œ aa"
a! b  a2a#
a" bx  (3a$
a# )x  á  (nan
an
1 )xn
1  á œ 0 Ê a"
a! œ 0, 2a#
a" œ 0, 3a$
a# œ 0 and in general nan
an
1 œ 0. Since y œ
3 when x œ 0 we have a! œ
3. Therefore a" œ
3, a# œ a2" œ
#3 , a$ œ a3# œ 3
†32 , an œ ann
1 œ
n!3 Ê y œ
3
3x
23†1 x#
33†# x$
á

n!3 xn  á œ
3 Š1  x 

x# #!



x$ 3!

á 

xn n!

_

 á ‹ œ
3 !

nœ0

xn n!

œ
3ex

71. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê

dy dx

 2y

#

œ aa"  2a! b  a2a#  2a" bx  (3a$  2a# )x  á  (nan  2an
1 )xn
1  á œ 0. Since y œ 3 when x œ 0 we #

have a! œ 3. Therefore a" œ
2a! œ
2(3) œ
3(2), a# œ
#2 a" œ
#2 (
2 † 3) œ 3 Š 2# ‹ , a$ œ
23 a# n
1 n
1

$

2 2 œ
23 ’3 Š 2# ‹“ œ
3 Š 32†# ‹ , á , an œ ˆ
n2 ‰ an
1 œ ˆ
n2 ‰ Š3 Š (
(n1)
1)! ‹‹ œ 3 Š (
1) n! ‹ #

#

$

n n

(
1) 2 $ Ê y œ 3
3(2x)  3 (2)# x#
3 (2) xn  á 3 †# x  á  3 n!

œ 3 ’1
(2x) 

(2x)# 2!




(2x)$ 3!

á 

n n

(
1)n (2x)n n!

_

 á “ œ 3!

nœ0

(
1)n (2x)n n!

œ 3e
2x

72. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê

dy dx

y

œ aa"  a! b  a2a#  a" bx  (3a$  a# )x#  á  (nan  an
1 )xn
1  á œ 1 Ê a"  a! œ 1, 2a#  a" œ 0, 3a$  a# œ 0 and in general nan  an
1 œ 0 for n  1. Since y œ 0 when x œ 0 we have a! œ 0. Therefore a" œ 1
a! œ 1, a# œ
2†a1" œ
"# , a$ œ
3a# œ 3"†2 , a% œ
4a$ œ
4†3"†# , á , an œ


an
1 n

œ ˆ
n1 ‰ (n(

1)1)! œ n

œ
1 ’1
x  "# x#


" 3 †#

(
1)n1 n!

Ê y œ 0  x
"# x# 

x$
á 

(
1)n n!

" 3†#

x$
á  _

xn  á “  1 œ
!

nœ0

(
1)n xn n!

(
1)n1 n!

xn  á

 1 œ 1
e
x

73. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

767

œ a"  2a# x  á  nan xn
1  á Ê

dy dx


y

#

œ aa"
a! b  a2a#
a" bx  (3a$
a# )x  á  (nan
an
1 )xn
1  á œ 3x Ê a"
a! œ 0, 2a#
a" œ 3, Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

768

Chapter 11 Infinite Sequences and Series

3a$
a# œ 0 and in general nan
an
1 œ 0 for n  2. Since y œ
1 when x œ 0 we have a! œ
1. Therefore a" œ
1, a# œ 3 2 a" œ #2 , a$ œ a3# œ 32†2 , a% œ a4$ œ 4†32†# , á , an œ ann
1 œ n!2 Ê y œ
1
x  ˆ 2# ‰ x#  33†# x$  4†23†# x%  á  n!2 xn  á œ 2 ˆ1  x  "# x# 

" 3 †#

x$ 

" 4†3†#

_

" n!

x%  á 

xn  á ‰
3
3x œ 2 !

nœ0

xn n!


3
3x œ 2ex
3x
3

74. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

œ a"  2a# x  á  nan xn
1  á Ê

dy dx

y

dy dx

#

œ aa"  a! b  a2a#  a" bx  (3a$  a# )x  á  (nan  an
1 )xn
1  á œ x Ê a"  a! œ 0, 2a#  a" œ 1, 3a$  a# œ 0 and in general nan  an
1 œ 0 for n  2. Since y œ 0 when x œ 0 we have a! œ 0. Therefore a" œ 0, a# œ

"
a" 2

" #

œ

Ê y œ 0
0x  "# x#
_

œ!

nœ0

(
1)n xn n!


a # 3

, a$ œ " 3 †#

œ
3"†2 , á , an œ

x$  á 

(
1)n n!


an
1 n

œ

(
1)n n!

xn  á œ Š1
x  #" x#


" 3 †#

x$  á 

(
1)n n!

xn  á ‹
1  x


1  x œ e
x  x
1

75. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê


y

dy dx

#

œ aa"
a! b  a2a#
a" bx  (3a$
a# )x  á  (nan
an
1 )xn
1  á œ x Ê a"
a! œ 0, 2a#
a" œ 1, 3a$
a# œ 0 and in general nan
an
1 œ 0 for n  2. Since y œ 1 when x œ 0 we have a! œ 1. Therefore a" œ 1, a# œ 1 2 a" œ #2 , a$ œ a3# œ 32†2 , a% œ a4$ œ 4†32†# , á , an œ ann
1 œ n!2 Ê y œ 1  x  ˆ #2 ‰ x#  32†# x$  4†22†# x%  á  n!2 xn  á œ 2 ˆ1  x  "# x# 

" 3 †#

x$ 

" 4†3†#

x%  á 

_

" n!

xn n!

xn  á ‰
1
x œ 2 !

nœ0


1
x œ 2ex
x
1

76. Assume the solution has the form y œ a!  a" x  a# x#  á  an
1 xn
1  an xn  á Ê

dy dx

œ a"  2a# x  á  nan xn
1  á Ê


y

dy dx

#

œ aa"
a! b  a2a#
a" bx  (3a$
a# )x  á  (nan
an
1 )xn
1  á œ
x Ê a"
a! œ 0, 2a#
a" œ
1, 3a$
a# œ 0 and in general nan
an
1 œ 0 for n  2. Since y œ 2 when x œ 0 we have a! œ 2. Therefore a" œ 2, a# œ
1 2 a" œ 1# , a$ œ a3# œ 31†2 , a% œ a4$ œ 4†31†# , á , an œ ann
1 œ n!1 Ê y œ 2  2x  1# x# 

œ ˆ1  x  "# x# 

77.

" 4†3†#

x%  á 

1 4 †3 † #

x%  á 

'

" #




" #% † 4

" n!



" #( †7†2!




" 2"! †10†3!

" 2"$ †13†4!

 *

&

'11Î2 ¸

" #

x"" 11†3!

tanc" x x




" 9†2$



x"( 17†5!




x#$ 23†7!

dx œ '1 Š1
1Î2



"

5# †#&

¸ 0.4872223583




"

7# †#(

x# 3





 "

"&

x#* 29†9!

x% 5

9# †2*




_

xn n!

nœ0

"#

 á ‹ dx œ ’x


" 2"' †16†5!

x#" 7!

xn  á



x#( 9!

x% 4

 1  x œ ex  x  1



x( 7†2!




x"! 10†3!

x"! 3!



x"' 5!




x## 7!

x& 25




x( 49



x* 81



21# †##"



x"$ 13†4!


á“

"Î# !

¸ 0.484917143  á ‹ dx œ '0 Šx%
1



x#) 9!


á ‹ dx

"


á “ ¸ 0.185330149 !










1 n!

xn  á ‰  1  x œ !

*

'01 x sin ax$ b dx œ '01 x Šx$
x3!  x5! œ ’ x5


79.

x$ 

x$ 

'01Î2 exp a
x$ b dx œ '01Î2 Š1
x$  x2!
x3!  x4! ¸

78.

" 3 †#

1 3 †#

x' 7 "



11# †2""

x) 9






x"! 11 "

13# †2"$

 á ‹ dx œ ’x



"

15# †2"&



"

17# †#"(

x$ 9




 "

19# †#"*




x"" 121

"

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

 á“

"Î# !

Chapter 11 Practice Exercises 80.

'01Î64

dx œ '0

1Î64

tanc" x Èx

œ  23 x$Î#


2 #1

x(Î# 

" Èx 2 55

x$ 3

Šx


x""Î#


2 105

$

7 sin x 2x x Ä 0 e
1

81. lim

82.

x Ä 0 Š2x 

œ lim

5! #

 á ‹ dx œ '0

1Î64

"Î'%

&

2$ x $ 3!



œ ˆ 3†28$
#

 á‹

x Ä 0 Š2 

2# x #!

 á‹

83.

œ lim

tÄ0

"‰ t#




# " 2 Š 4!
t6!  á‹ # Š1
2t4!  á‹

Š sinh h ‹
cos h

84. lim

œ

#



h# Π#!

$

œ lim






h% 5!




# lim t
# 2  2 cos t t Ä 0 2t (1
cos t)

"
cos# z z Ä 0 ln (1
z)  sin z

85. lim

2 105†8"&

 á ‰ ¸ 0.0013020379

œ

7 #

$

&

$

&

2 Š )3!  )5!  á‹

œ lim

$

&

Š )3!
)5!  á‹

)Ä0

h% 4!

%

t#
2  2 Œ1
t#  t4!
á


œ lim

t Ä 0 2t# Š1
1 

t# #




t% 4!

 á‹

%

t' 6!

2 Πt4!


œ lim

tÄ0

Št%


2t' 4!

 á
 á‹

" 1#

œ

Œ1


h# 3!

%

#



%

 h5!
á

Œ1
h#!  h4!
á
h#

h' 6!

h' 7!




 á


h#

hÄ0




2 55†8""

œ2

hÄ0

h# 3!

 á‹

)
Š)
)3!  )5!
á‹

œ lim

h#

hÄ0



2 21†8(

%

2$ x # 3!

#

lim ˆ " t Ä 0 #
2 cos t

ˆx"Î#
3" x&Î#  5" x*Î#
7" x"$Î#  á ‰ dx

7 Š1
x3!  x5!
á‹

œ lim

$

)Ä0

Š 3!"
)5!  á‹

)Ä0

2# x # 2!

x( 7

Š1  )  )#!  )3!  á‹
Š1
)  )#!
)3!  á‹
2)

œ lim )#




x"&Î#  á ‘ !

#

) c) lim e )

e sin
)2) )Ä0

x& 5

7 Šx
x3!  x5!
á‹

œ lim

2 Š 3!" 



œ lim

hÄ0

1
Š1
z# 

œ lim

#

z% 3

Š #"!


" 3!



h# 5!

h# 4!



h% 6!




h% 7!

 á‹ œ

" 3

%


á‹

$




$

&

z Ä 0 Š
z
z#
z3
á‹  Šz
z3!  z5!
á‹

Šz#
z3  á‹

œ lim

#

$

%

z Ä 0 Š
z#
2z3
z4
á‹

#

Š1
z3  á‹

œ lim

#

z z Ä 0 Š
"#
2z 3
4
á‹

y#
cos y cosh y yÄ0

86. lim

y#

œ lim

y Ä 0 Œ1


"

œ lim

œ
2

y Ä 0 Œ
1
2y%
á
6!

y# #



y% 4!




y' 6!

 á

Œ1 

xÄ0

Ê

r x#



3 x#

88. (a) csc x ¸

r x#

 s‰ œ lim – xÄ0

œ 0 and s
" x



x 6



y% 4!



y' 6!

 á


y#

œ lim

y Ä 0 Œ


2y# #

'


2y6!
á


œ
1

$

87. lim ˆ sinx$3x 

y# #!

9 #

&

(3x) Š3x
(3x) 6  120
á‹

x$



œ 0 Ê r œ
3 and s œ

Ê csc x ¸

(b) The approximation sin x ¸

6  x# 6x 6x 6  x#

Ê sin x ¸

r x#

 s— œ lim Š x3#
xÄ0

9 #



81x# 40

á 

9 #

6x 6  x#

is better than

sin x ¸ x.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

r x#

 s‹ œ 0

769

770

Chapter 11 Infinite Sequences and Series _

" 2n

89. (a) ! ˆsin nœ1

" ‰ #n  1


sin

_

 á œ ! (
1)n sin nœ2

" n

lim sin nÄ_

_

" 2n

nœ1

Ê tan

" n


sin 3" ‰  ˆsin

; f(x) œ sin

_

œ 0 Ê ! (
1)n sin nœ2

" ¸ 42

(b) kerrork  ¸sin 90. (a) ! ˆtan

" #

œ ˆsin

" ‰ #n  1

 tan

Test (b) kerrork  ¸tan

Ê f w (x) œ


sin 5" ‰  ˆsin


cos Š "x ‹ x#

" 6


sin 7" ‰  á  ˆsin

 0 if x   2 Ê sin

" n 1

" ¸ 42

" n

_

" n

œ ! (
1)n tan nœ2

, and n lim tan Ä_

(see Exercise 89); f(x) œ tan

_

" n

œ 0 Ê ! (
1)n tan nœ2

" n

" x

Ê f w (x) œ


sec# ˆ x" ‰ x#

0

2 3

nb1

kœ2

, and

¸ 0.02382 and the sum is an underestimate because the remainder is positive

(2n1)(2n3)(x
1) ¸ 2n  3 ¸  1 Ê kxk  92. n lim † 34††59††714ââ(2n(5n
1)x1)n ¹  1 Ê kxk n lim ¹ 3†5†47†â 9†14â(5n
1)(5n4) Ä_ Ä _ 5n  4 Ê the radius of convergence is 52

"‰ k#

" ‰ #n  1

converges by the Alternating Series

nb1

n

" n


sin

converges by the Alternating Series Test

(3n
1)(3n  2)x â(2n) ¸ 3n  2 ¸  1 Ê kxk  91. n lim † #†52†8†4â†6(3n ¹ 2†5†8#â †4†6â(2n)(2n  2)
1)xn ¹  1 Ê kxk n lim Ä_ Ä _ 2n  2 Ê the radius of convergence is 23

93. ! ln ˆ1


 sin

" #n

¸ 0.02381 and the sum is an underestimate because the remainder is positive


tan

" n1

" n

" x

" 4

n

n

kœ2

kœ2

5 2

œ ! ln ˆ1  k" ‰  ln ˆ1
k" ‰‘ œ ! cln (k  1)
ln k  ln (k
1)
ln kd

œ cln 3
ln 2  ln 1
ln 2d  cln 4
ln 3  ln 2
ln 3d  cln 5
ln 4  ln 3
ln 4d  cln 6
ln 5  ln 4
ln 5d  á  cln (n  1)
ln n  ln (n
1)
ln nd œ cln 1
ln 2d  cln (n  1)
ln nd after cancellation n

"‰ k#

Ê ! ln ˆ1
kœ2

n

94. ! kœ2

" k#
1 -

 ˆ n
" 1


_

Ê !

kœ2

" #

œ

n

!ˆ kœ2

" ‰‘ n1

" k #
1

_

1 ‰ œ ln ˆ n2n Ê ! ln ˆ1


"‰ k#

kœ2

" k
1

œ

" #




" ‰ k1

ˆ 1" 

" #

"ˆ3 œ n lim
Ä_ # 2

œ


1 n

" n




" #

1‰ œ n lim ln ˆ n 2n œ ln Ä_




" ‰ n1

1 ‰ n1

œ

œ

" #

ˆ #3


" n




" ‰ n1

œ

" #


2(n  1)
2n ’ 3n(n  1)2n(n “œ  1)

1†4†7â(3n
2) (3n)!

nœ1 _

dy dx

_

œ!

nœ1 _

1†4†7â(3n
2) (3n
1)!

x3nc1

1†4†7â(3n
5) (3n
3)!

x3nc2

(3n  ") (3n  1)(3n  2)(3n  3)

1†4†7â(3n
2) (3n
2)!

x3nc2 œ x  !

œ x Œ1  !

1†4†7â(3n
2) (3n)!

x3n  œ xy  0 Ê a œ 1 and b œ 0

x# 1x

œ x#  x# (
x)  x# (
x)#  x# (
x)$  á œ x#
x$  x%
x&  á œ ! (
1)n xn which

Ê

d# y dx#

œ!

nœ1 _

nœ1

96. (a)

x3n Ê

3n#
n
2 4n(n  1)

3 4

3nb3

_

is the sum

ˆ 11
3" ‰  ˆ #"
4" ‰  ˆ 3"
5" ‰  ˆ 4"
6" ‰  á  ˆ n
" #
n" ‰

1)x $ 95. (a) n lim † 1†4†7â(3n)! ¹ 1†4†7â(3n(3n
2)(3n 3)! (3n
2)x3n ¹  1 Ê kx k n lim Ä_ Ä_ œ kx$ k † 0  1 Ê the radius of convergence is _

(b) y œ 1  !

" #

œ

x# 1
(
x)

nœ2

_

nœ2

converges absolutely for kxk  1 _

_

nœ2

nœ2

(b) x œ 1 Ê ! (
1)n xn œ ! (
1)n which diverges

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 11 Practice Exercises _

771

_

97. Yes, the series ! an bn converges as we now show. Since ! an converges it follows that an Ä 0 Ê an  1 nœ1

nœ1

_

_

nœ1

nœ1

for n  some index N Ê an bn  bn for n  N Ê ! an bn converges by the Direct Comparison Test with ! bn _

98. No, the series ! an bn might diverge (as it would if an and bn both equaled n) or it might converge (as it nœ1

would if an and bn both equaled "n ). _

_

nœ1

kœ1

!(xk1
xk ) œ lim (xn1
x" ) œ lim (xn1 )
x" Ê both the series and 99. ! (xn1
xn ) œ n lim Ä_ nÄ_ nÄ_ sequence must either converge or diverge. Š 1 banan ‹

100. It converges by the Limit Comparison Test since n lim Ä_

an

œ n lim Ä_

_

" 1 a n

œ 1 because ! an converges nœ1

and so an Ä 0. 101. Newton's method gives xn1 œ xn
Lœ _

102. ! nœ1

an n

œ a" 

103. an œ

" ln n

" ln #

104. (a) T œ

xn 

" 40

, and if the sequence {xn } has the limit L, then

w



a# #



" 11

á 

a$ 3



a% 4

" #

ˆ "# ‰ #



" 3

" ‰ 16 a"'

á  

" #

" ln #



" ln 4

_

nœ2

#

Š0  2 ˆ "# ‰ e"Î#  e‹ œ x# #

1

" 8

" 6



" 7

 8" ‰ a)

(a#  a%  a)  a"'  á ) which is a divergent series

 á ‰ which diverges so that 1  !

(b) x# ex œ x# Š1  x 

39 40

 á   a"  ˆ "# ‰ a#  ˆ 3"  4" ‰ a%  ˆ 5" 

for n   2 Ê a#   a$   a%   á , and

ˆ1 

39 40

ww

L

" 10

œ

Ê L œ 1 and Öxn × converges since ¹ faÒfxbafxbaÓx2 b ¹ œ

39 40

 ˆ 9" 

œ

" 40

axn
1b%! 40 axn
1b$*



" n ln n

" ln 8

á œ

" ln #



" # ln 2



" 3 ln 2

á

diverges by the Integral Test.

e"Î#  4" e ¸ 0.885660616

 á ‹ œ x#  x$ 

x% #

á Ê

'01 Šx#  x$  x# ‹ dx œ ’ x3 %

$



x% 4



" x& 10 “ !

œ

41 60

(c) If the second derivative is positive, the curve is concave upward and the polygonal line segments used in the trapezoidal rule lie above the curve. The trapezoidal approximation is therefore greater than the actual area under the graph. (d) All terms in the Maclaurin series are positive. If we truncate the series, we are omitting positive terms and hence the estimate is too small. (e)

'01 x# ex dx œ cx# ex
2xex  2ex d "! œ e
2e  2e
2 œ e
2 ¸ 0.7182818285

105. a0 œ bk œ

1 21 1 1

'021 faxb dx œ 211 '121 1 dx œ "# ß ak œ 11 '021 faxb cos kx dx œ 11 '121 cos kx dx œ 0.

'021 faxb sin kx dx œ 11 '121 sin kx dx œ
cos1kkx ¹21 œ
11k Š1
a
1bk ‹ œ œ
1k , 2

1

Thus, the Fourier series of faxb is

" #


! k odd

2 1k sin

k odd . 0, k even

kx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 0.68333

772

Chapter 11 Infinite Sequences and Series

106. a0 œ œ

' 1 x dx  '121 1 dx • œ "#  4" 1, ak œ 11 ”'01 x cos kx dx  '121 cos kx dx • œ 11  cosk kx  x sink kx ‘01

1 21 ” 0

k 1 1k2 Ša
1b

2


1‹ œ œ


12k2 , 0,

k odd . k even

bk œ 11 ”'0 x sin kx dx  '1 sin kx dx • œ 11  sink2kx
1

1ˆ

œk

21

‘1
0

21 cos kx 1k ¹1

œ

a
1bkb1 k




1 1 k Š1


a
1bk ‹

1
12 ‰, k odd .
1k , k even

Thus, the Fourier series of faxb is

107. a0 œ

x cos kx k

" #

 4" 1
12 cos x  ˆ1
12 ‰sin x


" #

sin 2x


2 91 cos

3x  13 ˆ1
12 ‰ sin 3x  . . .

' 1 a1
xb dx  '121 ax
21b dx • œ 211 ’'01 a1
xb dx
'01 a1
ub du “ œ 0 where we used the

1 21 ” 0

substitution u œ x
1 in the second integral. We haveak œ 11 ”'0 a1
xb cos kx dx  '1 ax
21b cos kx dx •. Using 1

21

the substitution u œ x
1 in the second integral gives '1 ax
21b cos kx dx œ '0
a1
ub cosaku  k1b du Ú '01 a1
ub cos ku du, k odd œÛ 1 . Ü '0
a1
ub cos ku du, k even 21

#

Thus, ak œ  1

'01 a1
xb cos kx dx, 0,

Now, since k is odd, letting v œ 1
x Ê Exercise 106). So, ak œ œ

4 1 k2 ,

0,

1

k odd . k even # 1

'01 a1
xb cos kx dx œ
1# '01 v cos kv dv œ
1# ˆ
k2 ‰ œ 14k , k odd. (See 2

k odd . k even #

Using similar techniques we see that bk œ  1

'01 a1
ub sin ku du, 0,

2 , k odd k odd . œœk 0, k even k even

Thus, the Fourier series of faxb is ! ˆ 14k2 cos kx  2k sin kx‰. k odd

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2

Chapter 11 Additional and Advanced Exercises

'021 ksin xk dx œ 11 '01 sin x dx œ 12 . We have ak œ 11 '021 ksin xk cos kx dx 1 #1 œ 11 ’'0 sin x cos kx dx
'1 sin x cos kx dx “. Using techniques similar to those used in Exercise 107, we find

108. a0 œ

1 21

ak œ  bk œ

1 1

2 1

'01 sin x cos kx dx,

0,

k odd k even

œ œ

0,

k odd . k even


4 ak 2
1 b 1 ,

'021 ksin xk sin kx dx œ 11 ’'01 sin x sin kx dx
'1#1 sin x sin kx dx “ œ  2 ' 1 1

0

0, sin x sin kx dx,

k odd k even

for all k. 2 1

Thus, the Fourier series of faxb is

 ! Š ak2

41b1 cos kx‹. k even

CHAPTER 11 ADDITIONAL AND ADVANCED EXERCISES 1. converges since

" (3n
#)Ð2n
1ÑÎ2



" (3n
2)$Î#

" Š $Î# ‹ n

lim nÄ_

Š

ˆ 3n n
2 ‰ œ n lim " Ä_ ‹ (3n
2)$Î#

2. converges by the Integral Test: $

1 œ Š 24


1$ 192 ‹

œ

$Î#

_

and ! nœ1

" (3n
2)$Î#

converges by the Limit Comparison Test:

œ 3$Î#

'1_ atan
" xb# x dx1 œ #


" xb$

lim ’ atan 3

bÄ_


" bb$

b

tan “ œ lim ’ a 3 "

bÄ_




1$ 192 “

71 $ 192

c2n

e 3. diverges by the nth-Term Test since n lim a œ n lim (
1)n tanh n œ lim (
1)n Š 11
(
1)n  ec2n ‹ œ n lim Ä_ n Ä_ Ä_ bÄ_ does not exist

4. converges by the Direct Comparison Test: n!  nn Ê ln (n!)  n ln (n) Ê Ê logn (n!)  n Ê

logn (n!) n$



" n#

12 (3)(5)(4)#

n

, which is the nth-term of a convergent p-series

5. converges by the Direct Comparison Test: a" œ 1 œ œ

ln (n!) ln (n)

, a% œ ˆ 53††64 ‰ ˆ 42††53 ‰ ˆ 31††42 ‰ œ

"2 (4)(6)(5)#

12 (1)(3)(2)#

, a# œ

_

,á Ê 1!

nœ1

1 †2 3 †4

œ

12 (2)(4)(3)#

12 (n  1)(n  3)(n  2)#

, a$ œ ˆ 42††53 ‰ ˆ 31††42 ‰

represents the

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ0

773

774

Chapter 11 Infinite Sequences and Series 

12 (n  1)(n  3)(n  2)#

given series and

12 n%

, which is the nth-term of a convergent p-series

anb1 an

6. converges by the Ratio Test: n lim Ä_

œ n lim Ä_

n (n
1)(n  1)

œ01 " 1L

7. diverges by the nth-Term Test since if an Ä L as n Ä _, then L œ

Ê L#  L
1 œ 0 Ê L œ


1 „ È 5 #

Á0 _

_

" 32nb1

8. Split the given series into ! nœ1

and ! nœ1

2n 32n

; the first subseries is a convergent geometric series and the n n n È 2È

n 2n É second converges by the Root Test: n lim 32n œ n lim Ä_ Ä_

1 3

9. f(x) œ cos x with a œ " #

cos x œ

È3 #




9

È3 ww # ,f 1 ‰$ á 3

Ê f ˆ 13 ‰ œ 0.5, f w ˆ 13 ‰ œ


ˆx
13 ‰
4" ˆx
13 ‰# 

È3 1#

ˆx


œ

" †1 9

" 9

œ

1

ˆ 13 ‰ œ
0.5, f www ˆ 13 ‰ œ

È3 #

, f Ð4Ñ ˆ 13 ‰ œ 0.5;

10. f(x) œ sin x with a œ 21 Ê f(21) œ 0, f w (21) œ 1, f ww (21) œ 0, f www (21) œ
1, f Ð4Ñ (21) œ 0, f Ð5Ñ (21) œ 1, f Ð6Ñ (21) œ 0, f Ð7Ñ (21) œ
1; sin x œ (x
21)
x# #!

11. ex œ 1  x 



x$ 3!

(x
21)$ 3!



(x
21)& 5!




(x
21)( 7!

á

 á with a œ 0

12. f(x) œ ln x with a œ 1 Ê f(1) œ 0, f w (1) œ 1, f ww (1) œ
1,f www (1) œ 2, f Ð4Ñ (1) œ
6; ln x œ (x
1)


(x
1)# #



(x
1)$ 3




(x
1)% 4

á

13. f(x) œ cos x with a œ 221 Ê f(221) œ 1, f w (221) œ 0, f ww (221) œ
1, f www (221) œ 0, f Ð4Ñ (221) œ ", f Ð5Ñ (221) œ 0, f Ð6Ñ (221) œ
1; cos x œ 1
"# (x
221)#  4!" (x
221)%
6!" (x
221)'  á 14. f(x) œ tan
" x with a œ 1 Ê f(1) œ 1 4

tan
" x œ

(x
1) 2






(x
1)# 4

1 4

(x
1)$ 12



, f w (1) œ

" #

, f ww (1) œ
"# , f www (1) œ

ˆ ba ‰n ln ˆ ba ‰ a n nÄ_ ˆ b ‰  1

16. 1 

2 10



3 10#

_

œ1!

nœ0

œ1

200 999

n
1

17. sn œ !

kœ0



7 10$

'kkb1

2 10%

_

2

103n1





30 999

!

nœ0



dx 1  x#

0†ln ˆ ba ‰ 01

œ ln b 

7 999

 3



7 10'

_

103n2

œ

3 10&

!

999237 999

Ê sn œ '0

1

_

á œ1!

nœ1

7

œ

n

lnˆˆ ba ‰ 1‰ n nÄ_

Ê n lim c œ ln b  lim Ä_ n

œ1

2 ‰ ˆ 10

$ 1
ˆ " ‰ 10

2

_

103n
2



!

nœ1

Š 103# ‹

$ 1
ˆ " ‰ 10

3

_

103n
1



!

nœ1

Š 107$ ‹

"‰ 1
ˆ 10

7 103n

$

412 333

dx 1  x#

 '1

2

dx 1  x#

Ê n lim s œ n lim atan
" n
tan
" 0b œ Ä_ n Ä_ nb1

1În

œ ln b since 0  a  b. Thus, n lim c œ eln b œ b. Ä_ n

103n3

nœ0

;

á

n 15. Yes, the sequence converges: cn œ aan  bn b1În Ê cn œ b ˆˆ ba ‰  1‰

œ ln b  lim

" #

 á  'nc1 1 dxx# Ê sn œ '0 n

n

dx 1  x#

1 #

#

 1) 18. n lim † (n  1)(2x ¹ uunbn 1 ¹ œ n lim ¹ (n  1)x ¹ œ n lim ¹ x † (n  1) ¹ œ ¸ 2x x 1 ¸  1 nxn Ä_ Ä _ (n  2)(2x  1)n
1 Ä _ 2x  1 n(n  2) Ê kxk  k2x  1k ; if x  0, kxk  k2x  1k Ê x  2x  1 Ê x 
1; if
"#  x  0, kxk  k2x  1k n

Ê
x  2x  1 Ê 3x 
1 Ê x 
"3 ; if x 
#" , kxk  k2x  1k Ê
x 
2x
1 Ê x 
1. Therefore, Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 11 Additional and Advanced Exercises the series converges absolutely for x 
1 and x 
"3 . 19. (a) Each Anb1 fits into the corresponding upper triangular region, whose vertices are: (nß f(n)
f(n1)), (n1ß f(n1)) and (nß f(n)) along the line whose slope is f(n  1)
f(n).

(b) If Ak œ n
1

! Ak œ

kœ1

œ

kb1


'k

f(k  1)  f(k) #

œ

n 1

f(1)
f(2) #


'1 f(x) dx
'2 f(x) dx
á
'nc1 f(x) dx 2

3

 ! f(k)
'1 f(x) dx Ê ! Ak œ ! f(k)
n
1

Ê ! An 

f(x) dx, then

f(1)  f(2)  f(2)  f(3)  f(3)  á  f(n
1)  f(n) #

f(1)  f(n) #

_

f(1)
f(2) 2

All the An 's fit into the first upper triangular region whose area is

n

kœ2

n
1

n

kœ1

kœ1

n

f(1)  f(n) #




'1n f(x) dx  f(1)
# f(2) , from

n
1

part (a). The sequence œ ! Ak  is bounded above and increasing, so it converges and the limit in kœ1

question must exist. ! f(k)
' f(x) dx
" afa1b  fanbb•, which exists by part (b). Since f is positive and (c) Let L œ n lim # Ä_ ” 1 kœ1 n

n

! f(k)
' f(x) dx• œ L  " afa1b  Mb. decreasing lim fanb œ M   0 exists. Thus n lim # Ä_ ” 1 nÄ_ n

n

kœ1

20. The number of triangles removed at stage n is 3n
1 ; the side length at stage n is at stage n is È3 4

(a)

È3 4

b#  3

b 2nc1

; the area of a triangle

#

ˆ 2nbc1 ‰ . È3 4

#

Š b2# ‹  3#

È3 4

(b) a geometric series with sum

#

Š b2% ‹  3$ Š

È 3 b# ‹ 4

1
ˆ 34 ‰

È3 4

#

Š b2' ‹  á œ

È3 4

_

b# ! nœ0

3n 22n

œ

È3 4

_

n b# ! ˆ 43 ‰ nœ0

œ È3b#

(c) No; for instance,the three vertices of the original triangle are not removed. However the total area removed is È3b# which equals the area of the original triangle. Thus the set of points not removed has area 0. 21. (a) No, the limit does not appear to depend on the value of the constant a (b) Yes, the limit depends on the value of b (c) s œ Š1
œ n lim Ä_

cos ˆ na ‰ n n ‹ a n

lim Š1


nÄ_

Ê ln s œ

sin ˆ na ‰
cos ˆ na ‰ 1


cos ˆ na ‰ n

cos ˆ na ‰ n bn ‹

œ

ln Œ1


cos ˆ na ‰
n

ˆ "n ‰ 0
1 1
0

Ê n lim ln s œ Ä_


na sin ˆ na ‰  cos ˆ na ‰
n# " Š
# ‹ n

" cos ˆ na ‰  Œ n

œ e
1Îb 1În

_

" #

1


œ
1 Ê n lim s œ e
" ¸ 0.3678794412; similarly, Ä_

a n ‰n 22. ! an converges Ê n lim a œ 0; n lim ’ˆ 1  sin “ # Ä_ n Ä_ nœ1

œ



an ‰ ˆ 1sin œ n lim œ # Ä_

Ä_

1sin Šnlim an ‹ #

œ

1sin 0 #

Ê the series converges by the nth-Root Test nb1 nb1

23. n lim ¹ uunbn 1 ¹  1 Ê n lim ¹b x † Ä_ Ä _ ln (n  1)

ln n bn xn ¹

 1 Ê kbxk  1 Ê
"b  x 

" b

œ5 Ê bœ „

" 5

24. A polynomial has only a finite number of nonzero terms in its Taylor series, but the functions sin x, ln x and ex have infinitely many nonzero terms in their Taylor expansions.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

775

776

Chapter 11 Infinite Sequences and Series

sin (ax)
sin x
x x$ xÄ0

xÄ0

26.

lim cos#axx#
b xÄ0

" 3!



(b)

28.

un unb1

œ

(n  1)# n#

un unb1

œ

n1 n

œ

Ê Cœ

3 #

" 3!

œ

1 n

2 n

 á‹
x

a# x# #


á

b

œ
1 Ê lim Š "2x
#b
xÄ0

_

" n#

Ê C œ 2  1 and !

nœ1 _

" n

nœ1

œ1

Š 64 ‹



n

5n# 4n#
4n  1

 1 and kf(n)k œ

a% x% 4!



Ê C œ 1 Ÿ 1 and !

0 n#

4n#  2n 4n#
4n  1

 á “ is finite if a
2 œ 0 Ê a œ 2;

#x#

" n#





x$ 3!

œ
76

xÄ0

œ1

œ1

" # 5! ‹ x

Œ1


œ
1 Ê lim

un unb1

2n(2n  1) (2n
1)#

&


Š a5!


œ
23! 

Ê b œ 1 and a œ „ 2

27. (a)

 á‹
Šx
x$

$

sin 2x
sin x
x x$

lim

xÄ0

a$ x$ 3!

xÄ0

a$ 3!

œ lim ’ a x
# 2


Šax


œ lim

25. lim

œ

Š4


5 

4 n

_

_

_

nœ1

nœ1

nœ1



a# x# 48


á ‹ œ
1

converges

diverges

œ1

5 4n#
4n  1

a# 4

Š 3# ‹ n

5n#



– Š4n# c 4n b 1‹ — n#

after long division

_

" ‹ n#

Ÿ 5 Ê ! un converges by Raabe's Test nœ1

29. (a) ! an œ L Ê an# Ÿ an ! an œ an L Ê ! an# converges by the Direct Comparison Test an Š1c an ‹

(b) converges by the Limit Comparison Test: n lim Ä_

an

œ n lim Ä_

" 1
an

_

œ 1 since ! an converges and nœ1

therefore x lim a œ0 Ä_ n 30. If 0  an  1 then kln (1
an )k œ
ln (1
an ) œ an 

a#n #



a$n 3

 á  an  an#  an$  á œ

an 1
an

,

a positive term of a convergent series, by the Limit Comparison Test and Exercise 29b _

31. (1
x)
" œ 1  ! xn where kxk  1 Ê nœ1

#

" (1
x)#

$

4 œ 1  2 ˆ "# ‰  3 ˆ "# ‰  4 ˆ "# ‰  á  n ˆ "# ‰ _

32. (a) ! xn1 œ nœ1

_

Ê !

nœ1 _

(b) x œ !

nœ1

x# 1
x

n(n  1) xn n(n  ") xn

_

Ê ! (n  1)xn œ nœ1

œ

2 x

$

Š1
"x ‹

Ê xœ

œ

2x# (x
1)$

2x# (x
1)$

2x
x# (1
x)#

œ

n
1

d dx

_

(1
x)
" œ ! nxn
1 and when x œ nœ1

" #

we have

á _

Ê ! n(n  1)xn
1 œ nœ1

2 (1
x)$

_

Ê ! n(n  1)xn œ nœ1

2x (1
x)$

, kxk  1

Ê x$
3x#  x
1 œ 0 Ê x œ 1  Š1 

È57 "Î$ 9 ‹

 Š1


È57 "Î$ 9 ‹

¸ 2.769292, using a CAS or calculator 33. The sequence {xn } converges to

1 #

from below so %n œ

Estimation Theorem %n1 ¸ 3!" (%n )$ with overestimate Ê 0  %n1  "6 (%n )$ . _

kerrork 

" 5!

1 #


xn  0 for each n. By the Alternating Series &

(%n ) , and since the remainder is negative this is an

34. Yes, the series ! ln (1  an ) converges by the Direct Comparison Test: 1  an  1  an  nœ1

a#n #!



Ê 1  an  ean Ê ln (1  an )  an

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

a$n 3!

á

Chapter 11 Additional and Advanced Exercises " (1
x)#

35. (a)

œ

d dx

ˆ 1
" x ‰ œ

d dx

_

a1  x  x#  x$  á b œ 1  2x  3x#  4x$  á œ ! nxn
1 nœ1

_

! n ˆ 5 ‰ n
1 6

(b) from part (a) we have

nœ1 _

ˆ "6 ‰ œ ˆ "6 ‰ ’

(c) from part (a) we have ! npn
1 q œ

q (1
p)#

nœ1

_

_

ˆ "# ‰

36. (a) ! pk œ ! 2
k œ kœ1

1
ˆ "# ‰

kœ1

œ

q q#

2

" “ 1
ˆ 56 ‰

œ

œ6

" q

_

_

kœ1

kœ1

œ 1 and E(x) œ ! kpk œ ! k2
k œ

" #

_

! k21
k œ ˆ " ‰ #

kœ1

" 1
ˆ "# ‰‘#

œ2

by Exercise 35(a) _

_

kœ1

kœ1

(b) ! pk œ ! œ ˆ "6 ‰

" 1
ˆ 56 ‰‘#

_

_

(c) ! pk œ ! kœ1

kœ1

_

œ!

kœ1

5kc1 6k

" k1

_

5

_

_

kœ1

kœ1

! ˆ 5 ‰k œ ˆ " ‰ ’ ˆ 6 ‰5 “ œ 1 and E(x) œ ! kpk œ ! k 6 5 1
ˆ ‰

" 5

œ

kœ1

6

5kc1 6k

" 6

œ

_

" k(k1)

œ ! ˆ k"
kœ1

" ‰ k1

œ

lim ˆ1
kÄ_

" ‰ k1

_

_

kœ1

kœ1

œ 1 and E(x) œ ! kpk œ ! k Š k(k " 1) ‹

, a divergent series so that E(x) does not exist C! e
kt! ˆ1
e
nkt! ‰ 1
e
kt!

Ê Rœ

lim R œ nÄ_ n ¸ 0.58195028;

C! e
kt! 1
e
kt!


" e
1 a1
e
n b e
"! b Ê R" œ e
" ¸ 0.36787944 and R"! œ e 1a1

e
" 1
e
" R œ e
" 1 ¸ 0.58197671; R
R"! ¸ 0.00002643 Ê R
RR"!  0.0001
Þ1n e
Þ1 ˆ1
e
Þ1n ‰ R Rn œ , # œ #" ˆ eÞ1 "
1 ‰ ¸ 4.7541659; Rn  R# Ê 1e
Þ1 e
1  ˆ #" ‰ ˆ eÞ1 "
1 ‰ 1
e
Þ1 n n Ê 1
e
nÎ10  "# Ê e
nÎ10  "# Ê
10  ln ˆ "# ‰ Ê 10 
ln ˆ "# ‰ Ê n  6.93

(b) Rn œ

38. (a) R œ (b) t! œ

C! ekt!
1 " 0.05

kœ1

œ6

37. (a) Rn œ C! e
kt!  C! e
2kt!  á  C! e
nkt! œ

(c)

_

! k ˆ 5 ‰ k
1 6

Ê Rekt! œ R  C! œ CH Ê ekt! œ

CH CL

Ê t! œ

" k

C! ekt!
1

œ

Ê nœ7

ln Š CCHL ‹

ln e œ 20 hrs

(c) Give an initial dose that produces a concentration of 2 mg/ml followed every t! œ

" 0.0#

2 ‰ ln ˆ 0.5 ¸ 69.31 hrs

by a dose that raises the concentration by 1.5 mg/ml " 0.1 ‰ ‰ (d) t! œ 0.2 ln ˆ 0.03 œ 5 ln ˆ 10 3 ¸ 6 hrs _

39. The convergence of ! kan k implies that nœ1

ka n k 1
ka n k

Ê

lim ka k œ 0. Let N  0 be such that kan k  nÄ_ n

 2 kan k for all n  N. Now kln a1  an bk œ ¹an


 kan k  kan k #  kan k $  kan k %  á œ

ka n k 1
ka n k

a#n #



a$n 3




a%n 4

" #

Ê 1
kan k  #

$

 á ¹ Ÿ kan k  ¹ a#n ¹  ¹ a3n ¹  ¹ a4n ¹  á

_

 2 kan k . Therefore ! ln (1  an ) converges by the Direct nœ1

_

Comparison Test since ! kan k converges. nœ1

_

40. ! nœ3

" n ln n(ln (ln n))p

lim

bÄ_

converges if p  1 and diverges otherwise by the Integral Test: when p œ 1 we have

'3b x ln x(lndx(ln x)) œ cpb1

œ lim ’ (ln (ln1
x))p bÄ_

b

lim cln (ln (ln x))d b$ œ _; when p Á 1 we have lim

bÄ_

“ œ $

(ln (ln 3))cpb1 1
p

bÄ_

, _,

%

" #

'3b x ln x(lndx(ln x))

if p  1 if p  1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

p

777

778

Chapter 11 Infinite Sequences and Series

41. (a) s2n1 œ

c" 1



c# #



c$ 3

c2n
1 2n1

á 

œ

t" 1



t#
t" 2

" œ t" ˆ1
"# ‰  t# ˆ "#
3" ‰  á  t2n ˆ 2n
_

(b) ecn f œ e(
1)n f Ê !

nœ1

(
1)n n



t$
t# 3

" ‰ #n  1



t2n
1
t2n 2n1

á  t2n
1 2n1

2n

œ!

kœ1

tk k(k1)



t2n
1 2n1

.

" 3



" 5

" 6

converges

(c) ecn f œ e1ß
1ß
1ß 1ß 1ß
1ß
1ß 1ß 1ß á f Ê the series 1


" #






" 4







" 7

 á converges

42. (a) a1
t  t#
t$  á  (
1)n tn b (1  t) œ 1
t  t#
t$  á  (
1)n tn  t
t#  t$
t%  á  (
1)n tn1 œ 1  (
1)n tn1 Ê 1
t  t#
t$  á  (
1)n tn


'0

x

(b)

dt œ '0 ’1
t  t#  á  (
1)n tn  x

" 1t

œ ’t


t# 2



t$ 3

á 

œx


x# #



x$ 3


á 

(
1)n tn
1 n1 “! x

(c) x  0 and Rn1 œ (
1)n1

'0

x

tn
1 1t x

x

ktk nb1 1
kx k

dx œ

kxk nb2 a1
kxkb (n  2)

dt Ê ln k1  xk

dt Ê kRn1 k œ '0

x

tnb1 1t

" 1t

tnb1 1 t

(
1)n
1 tn
1 n1

dt

dt Ÿ '0 tn1 dt œ x

dt Ê kRn1 k œ ¹'0

x

tnb1 1t

xn
2 n#

dt¹ Ÿ '0 ¹ 1t  t ¹ dt x

nb1

since k1  tk   1
kxk

(e) From part (d) we have kRn1 k Ÿ kxk nb2 lim n Ä _ a1
kxkb (n  2)

(
1)n
1 tn
1 n1

œ

dt Ê cln k1  tkd !x

x

(d)
1  x  0 and Rnb1 œ (
1)nb1 '0 Ÿ '0

(
1)nb1 tnb1 “ 1t

 Rn1 , where Rn1 œ '0

(
1)n xn
1 n1 x

 '0

(
1)n tnb1 1t

kxk nb2 a1
kxkb (n2)

Ê the given series converges since

œ 0 Ê kRn1 k Ä 0 when kxk  1. If x œ 1, by part (c) kRn1 k Ÿ

kxknb2 n2

Thus the given series converges to lna1  xb for
1  x Ÿ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

1 n2

Ä 0.

CHAPTER 12 VECTORS AND THE GEOMETRY OF SPACE 12.1 THREE-DIMENSIONAL COORDINATE SYSTEMS 1. The line through the point (#ß $ß !) parallel to the z-axis 2. The line through the point (
1ß 0ß !) parallel to the y-axis 3. The x-axis 4. The line through the point (1ß !ß !) parallel to the z-axis 5. The circle x#  y# œ 4 in the xy-plane 6. The circle x#  y# œ 4 in the plane z =
2 7. The circle x#  z# œ 4 in the xz-plane 8. The circle y#  z# œ 1 in the yz-plane 9. The circle y#  z# œ 1 in the yz-plane 10. The circle x#  z# œ 9 in the plane y œ
4 11. The circle x#  y# œ 16 in the xy-plane 12. The circle x#  z# œ 3 in the xz-plane 13. (a) The first quadrant of the xy-plane

(b) The fourth quadrant of the xy-plane

14. (a) The slab bounded by the planes x œ 0 and x œ 1 (b) The square column bounded by the planes x œ 0, x œ 1, y œ 0, y œ 1 (c) The unit cube in the first octant having one vertex at the origin 15. (a) The solid ball of radius 1 centered at the origin (b) The exterior of the sphere of radius 1 centered at the origin 16. (a) The circumference and interior of the circle x#  y# œ 1 in the xy-plane (b) The circumference and interior of the circle x#  y# œ 1 in the plane z œ 3 (c) A solid cylindrical column of radius 1 whose axis is the z-axis 17. (a) The closed upper hemisphere of radius 1 centered at the origin (b) The solid upper hemisphere of radius 1 centered at the origin 18. (a) The line y œ x in the xy-plane (b) The plane y œ x consisting of all points of the form (xß xß z)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

780

Chapter 12 Vectors and the Geometry of Space

19. (a) x œ 3

(b) y œ
1

(c) z œ
2

20. (a) x œ 3

(b) y œ
1

(c) z œ 2

21. (a) z œ 1

(b) x œ 3

(c) y œ
1

22. (a) x#  y# œ 4, z œ 0

(b) y#  z# œ 4, x œ 0

(c) x#  z# œ 4, y œ 0

23. (a) x#  (y
2)# œ 4, z œ 0

(b) (y
2)#  z# œ 4, x œ 0

(c) x#  z# œ 4, y œ 2

24. (a) (x  3)#  (y
4)# œ 1, z œ 1 (c) (x  3)#  (z
1)# œ 1, y œ 4

(b) (y
4)#  (z
1)# œ 1, x œ
3

25. (a) y œ 3, z œ
1

(b) x œ 1, z œ
1

(c) x œ 1, y œ 3

26. Èx#  y#  z# œ Èx#  (y
2)#  z# Ê x#  y#  z# œ x#  (y
2)#  z# Ê y# œ y#
4y  4 Ê y œ 1 27. x#  y#  z# œ 25, z œ 3 Ê x#  y# œ 16 in the plane z œ 3 28. x#  y#  (z
1)# œ 4 and x#  y#  (z  1)# œ 4 Ê x#  y#  (z
1)# œ x#  y#  (z  1)# Ê z œ 0, x#  y# œ 3 29. 0 Ÿ z Ÿ 1

30. 0 Ÿ x Ÿ 2, 0 Ÿ y Ÿ 2, 0 Ÿ z Ÿ 2

31. z Ÿ 0

32. z œ È1
x#
y#

33. (a) (x
1)#  (y
1)#  (z
1)#  1

(b) (x
1)#  (y
1)#  (z
1)#  1

34. 1 Ÿ x#  y#  z# Ÿ 4 35. kP" P# k œ Éa3
1b#  a3
1b#  a0
1b# œ È9 œ 3 36. kP" P# k œ Éa2  1b#  a5
1b#  a0
5b# œ È50 œ 5È2 37. kP" P# k œ Éa4
1b#  a
2
4b#  a7
5b# œ È49 œ 7 38. kP" P# k œ Éa2
3b#  a3
4b#  a4
5b# œ È3 39. kP" P# k œ Éa2
0b#  a
2
0b#  a
2
0b# œ È3 † 4 œ 2È3 40. kP" P# k œ Éa0
5b#  a0
3b#  a0  2b# œ È38 41. center (
2ß 0ß 2), radius 2È2

42. center ˆ
"# ß
"# ß
"# ‰ , radius

43. center ŠÈ2ß È2ß
È2‹ , radius È2

44. center ˆ!ß
"3 ß 3" ‰ , radius

È21 #

È29 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.2 Vectors 45. (x
1)#  (y
2)#  (D
3)# œ 14

46. x#  (y  1)#  (z
5)# œ 4

47. (x  2)#  y#  z# œ 3

48. x#  (y  7)#  z# œ 49

781

49. x#  y#  z#  4x
4z œ 0 Ê ax#  4x  4b  y#  az#
4z  4b œ 4  4 #

Ê (x  2)#  (y
0)#  (z
2)# œ ŠÈ8‹ Ê the center is at (
2ß 0ß 2) and the radius is È8 50. x#  y#  z#
6y  8z œ 0 Ê x#  ay#
6y  9b  az#  8z  16b œ 9  16 Ê (x
0)#  (y
3)#  (z  4)# œ 5# Ê the center is at (0ß 3ß
4) and the radius is 5 51. 2x#  2y#  2z#  x  y  z œ 9 Ê x#  "# x  y#  "# y  z#  "# z œ Ê ˆx#  "# x 

" ‰ 16

 ˆy#  "# y 

" ‰ 16

 ˆz#  "# z 

Ê the center is at ˆ
"4 ß
4" ß
4" ‰ and the radius is

" ‰ 16

œ

9 #



3 16

9 # È

5È 3 4

52. 3x#  3y#  3z#  2y
2z œ 9 Ê x#  y#  23 y  z#
23 z œ 3 Ê x#  ˆy#  23 y  "9 ‰  ˆz#
23 z  "9 ‰ œ 3  # # Ê (x
0)#  ˆy  "3 ‰  ˆz
3" ‰ œ Š

È29 # 3 ‹

Ê the center is at ˆ0ß
3" ß 3" ‰ and the radius is

È29 3

53. (a) the distance between (xß yß z) and (xß 0ß 0) is Èy#  z# (b) the distance between (xß yß z) and (0ß yß 0) is Èx#  z#

(c) the distance between (xß yß z) and (0ß 0ß z) is Èx#  y#

54. (a) the distance between (xß yß z) and (xß yß 0) is z (b) the distance between (xß yß z) and (0ß yß z) is x (c) the distance between (xß yß z) and (xß 0ß z) is y 55. kABk œ Éa1
a
1bb#  a
1
2b#  a3
1b# œ È4  9  4 œ È17 kBCk œ Éa3
1b#  a4
a
1bb#  a5
3b# œ È4  25  4 œ È33 kCAk œ Éa
1
3b#  a2
4b#  a1
5b# œ È16  4  16 œ È36 œ 6 Thus the perimeter of triangle ABC is È17  È33  6. 56. kPAk œ Éa2
3b#  a
1
1b#  a3
2b# œ È1  4  1 œ È6 kPBk œ Éa4
3b#  a3
1b#  a1
2b# œ È1  4  1 œ È6 Thus P is equidistant from A and B. 12.2 VECTORS 1. (a)  3a3b, 3a
2b¡ œ  9,
6¡ (b) É92  a
6b2 œ È117 œ 3È13 3. (a)  3  a
2b,
2  5¡ œ  1, 3¡ (b) È12  32 œ È10

#

# # # Ê ˆx  4" ‰  ˆy  4" ‰  ˆz  4" ‰ œ Š 5 4 3 ‹

2. (a)  
2a
2b,
2a5b¡ œ  4,
10¡ (b) É42  a
10b2 œ È116 œ 2È29 4. (a)  3
a
2b,
2
5¡ œ  5,
7¡ (b) É52  a
7b2 œ È74

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 9

782

Chapter 12 Vectors and the Geometry of Space

5. (a) 2u œ  2a3b, 2a
2b¡ œ  6,
4¡ 3v œ  3a
2b, 3a5b¡ œ  
6, 15¡

6. (a)
2u œ  
2a3b,
2a
2b¡ œ  
6, 4¡ 5v œ  5a
2b, 5a5b¡ œ  
10, 25¡

2u
3v œ  6
a
4b,
4
15¡ œ  12,
19¡


2u  5v œ  
6  a
10b, 4  25¡ œ  
16, 29¡

(b) É122  a
19b2 œ È505 7. (a)

(b) Éa
16b2  292 œ È1097

3 5u

œ ¢ 53 a3b, 53 a
2b£ œ ¢ 59 ,
56 £

4 5v

œ ¢ 45 a
2b, 45 a5b£ œ ¢
85 , 4£

3 5u

 45 v œ ¢ 95  ˆ
85 ‰,
65  4£ œ ¢ 15 ,

2 ‰2 œ (b) Ɉ 15 ‰  ˆ 14 5

5 5 5 8. (a)
13 u œ ¢
13 a3b,
13 a
2b£ œ ¢
15 13 , 12 13 v

œ ¢ 12 13 a
2b,

5
13 u

14 5 £

È"97 5

12 13 v

12 13 a5b£

ˆ 24 ‰ œ ¢
15 13 
13 ,

70 ‰2 (b) Éa
3b2  ˆ 13 œ

10. ¢ 2
a#4b
0,

9.  2
1,
1
3¡ œ  1,
4¡

œ ¢
24 13 ,

"
3 #

10 13

10 13 £

60 13 £



60 13 £

œ ¢
3,

È6421 13


0£ œ  
1, 1¡

11.  0
2, 0
3¡ œ  
2,
3¡ Ä Ä Ä Ä 12. AB œ  2
1, 0
a
1b¡ œ  1, 1¡, CD œ  
2
a
1b, 2
3¡ œ  
1,
1¡, AB  CD œ  0, 0¡ 13. ¢cos

21 3 ,

sin

21 3 £

œ ¢
"# ,

È3 2 £

14. ¢cos ˆ
341 ‰, sin ˆ
341 ‰£ œ ¢
È"2 ,
È"2 £

15. This is the unit vector which makes an angle of 120‰  90‰ œ 210‰ with the positive x-axis; È3 2 ,


"# £

16.  cos 135‰ , sin 135‰ ¡ œ ¢
È" ,

" È2 £

 cos 210‰ , sin 210‰ ¡ œ ¢


2

Ä 17. P" P# œ a2
5bi  a9
7bj  a
2
a
1bbk œ
3i  2j
k Ä 18. P" P# œ a
3
1bi  a0
2bj  a5
0bk œ
4i
2j  5k Ä 19. AB œ a
10
a
7bbi  a8
a
8bbj  a1
1bk œ
3i  16j Ä 20. AB œ a
1
1bi  a4
0bj  a5
3bk œ
2i  4j  2k 21. 5u
v œ 5 1, 1,
1¡
 2, 0, 3¡ œ  5, 5,
5¡
 2, 0, 3¡ œ  5
2, 5
0,
5
3¡ œ  3, 5,
8¡ œ 3i  5j
8k 22.
2u  3v œ
2 
1, 0, 2¡  3 1, 1, 1¡ œ  2, 0,
4¡   3, 3, 3¡ œ  5, 3,
1¡ œ 5i  3j
k

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

70 13 £

Section 12.2 Vectors 23. The vector v is horizontal and 1 in. long. The vectors u and w are the horizontal. All vectors must be drawn to scale. (a)

(b)

(c)

(d)

"1 16

783

in. long. w is vertical and u makes a 45‰ angle with

24. The angle between the vectors is 120‰ and vector u is horizontal. They are all 1 in. long. Draw to scale. (a) (b)

(c)

(d)

25. length œ k2i  j
2kk œ È2#  1#  (
2)# œ 3, the direction is 26. length œ k9i
2j  6kk œ È81  4  36 œ 11, the direction is 9 2 6 ‰ œ 11 ˆ 11 i
11 j  11 k

2 3

9 11

i  "3 j
32 k Ê 2i  j
2k œ 3 ˆ 32 i  3" j
32 k‰

i


2 11

j

6 11

k Ê 9i
2 j  6k

27. length œ k5kk œ È25 œ 5, the direction is k Ê 5k œ 5(k) 9 28. length œ ¸ 35 i  45 k¸ œ É 25 

29. length œ ¹ È16 i
Ê

1 È6

i


1 È6

1 È6

j


j


" È6

" È6

16 25

œ 1, the direction is

3 5

#

i  45 k Ê

3 5

k¹ œ Ê3 Š È"6 ‹ œ É #" , the direction is

k œ É #" Š È13 i


1 È3

j


" È3

i  45 k œ 1 ˆ 35 i  45 k‰

1 È3

i


1 È3

j


" È3

k

k‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

784

Chapter 12 Vectors and the Geometry of Space

30. length œ ¹ È13 i  Ê

1 È3

i

1 È3

1 È3

j

j

" È3

" È3

#

k¹ œ Ê3 Š È"3 ‹ œ 1, the direction is

k œ 1 Š È13 i 

1 È3

j

31. (a) 2i

(b)
È3k

32. (a)
7j

(b)
3 5 2 i


È

33. kvk œ È12#  5# œ È169 œ 13; 34. kvk œ É 4" 

" 4



" 4

œ

È3 v # ; kv k

œ

œ

v kv k

" 13

" È3

4È 2 5

vœ

1 È3

i


1 È3

j


3 35. (a) 3i  4j
5k œ 5È2 Š 5È i 2

4 5È 2

j


" È2

" 13

1 È3

i

1 È3

j

" È3

k

k‹

k

(c)

3 10

(c)

1 4

j  25 k

(d) 6i
2j  3k

i
13 j
k

(12i
5k) Ê the desired vector is

1 È3

a È2

(d) 7 13

i

a È3

j


a È6

k

(12i
5k)

k Ê the desired vector is
3 Š È"3 i


" È3

j


" È3

k‹

œ
È3i  È3j  È3k

(b) the midpoint is

k‹ Ê the direction is

i

4 5È 2

j


" È2

k

ˆ "# ß 3ß 5# ‰

36. (a) 3i
6j  2k œ 7 ˆ 37 i
67 j  27 k‰ Ê the direction is (b) the midpoint is ˆ 5# ß 1ß 6‰ 37. (a)
i
j
k œ È3 Š
È13 i
(b) the midpoint is

3 5È 2

1 È3

j


1 È3

3 7

i
67 j  27 k

k‹ Ê the direction is
È13 i


1 È3

j


1 È3

k

ˆ 5# ß 7# ß 9# ‰

38. (a) 2i
2j
2k œ 2È3 Š È13 i


1 È3

j


1 È3

k‹ Ê the direction is

1 È3

i


1 È3

j


1 È3

k

(b) the midpoint is ("ß
"ß
1) Ä 39. AB œ (5
a)i  (1
b)j  (3
c)k œ i  4j
2k Ê 5
a œ 1, 1
b œ 4, and 3
c œ
2 Ê a œ 4, b œ
3, and c œ 5 Ê A is the point (4ß
3ß 5) Ä 40. AB œ (a  2)i  (b  3)j  (c
6)k œ
7i  3j  8k Ê a  2 œ
7, b  3 œ 3, and c
6 œ 8 Ê a œ
9, b œ 0, and c œ 14 Ê B is the point (
9ß 0ß 14) 41. 2i  j œ a(i  j)  b(i
j) œ (a  b)i  (a
b)j Ê a  b œ 2 and a
b œ 1 Ê 2a œ 3 Ê a œ bœa
"œ

" #

3 #

and

42. i
2j œ a(2i  3j)  b(i  j) œ (2a  b)i  (3a  b)j Ê 2a  b œ 1 and 3a  b œ
2 Ê a œ
3 and b œ 1
#a œ 7 Ê u" œ a(2i  3j) œ
6i
9j and u# œ b(i  j) œ 7i  7j 43. If kxk is the magnitude of the x-component, then cos 30° œ

kx k kFk

Ê kxk œ kFk cos 30° œ (10) Š

È3 # ‹

œ 5È3 lb

Ê Fx œ 5È3 i; if kyk is the magnitude of the y-component, then sin 30° œ

ky k kFk

Ê kyk œ kFk sin 30° œ (10) ˆ 1# ‰ œ 5 lb Ê Fy œ 5 j.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.2 Vectors 44. If kxk is the magnitude of the x-component, then cos 45° œ

kx k kFk

Ê kxk œ kFk cos 45° œ (12) Š

È2 # ‹

œ 6È2 lb

Ê Fx œ
6È2 i (the negative sign is indicated by the diagram) if kyk is the magnitude of the y-component, then sin 45° œ

ky k kFk

Ê kyk œ kFk sin 45° œ (12) Š

È2 # ‹

œ 6È2 lb

Ê Fy œ
6È2 j (the negative sign is indicated by the diagram) 45. 25‰ west of north is 90‰  25‰ œ 115‰ north of east. 800 cos 155‰ , sin 115‰ ¡ ¸  
338.095, 725.046¡ 46. 10‰ east of south is 270‰  10‰ œ 280‰ ”north” of east. 600 cos 280‰ , sin 280‰ ¡ ¸  104.189,
590.885¡ Ä È 47. (a) The tree is located at the tip of the vector OP œ (5 cos 60°)i  (5 sin 60°)j œ 5# i  5 # 3 j Ê P œ Š 5# ß Ä Ä (b) The telephone pole is located at the point Q, which is the tip of the vector OP  PQ œ Š 52 i 

5È 3 #

j‹  (10 cos 315°)i  (10 sin 315°)j œ Š 5# 

Ê Q œ Š 5
10 # 48. Let t œ

q p
q

and s œ

È2

ß

10È2 # ‹i

È

 Š5 # 3


5È 3 # ‹

10È2 # ‹j

5È3  10È2 ‹ #

p p
q.

Choose T on OP1 so that TQ is

parallel to OP2 , so that ˜TP1 Q is similar to ˜OP1 P2 . Then Ä Ä kOTk kOP1 k œ t Ê OT œ t OP1 so that T œ at x1 , t y1 , t z1 b. Ä Ä TQk Also, kkOP œ s Ê TQ œ s OP2 œ s x2 , y2 , z2 ¡. 2k Letting Q œ ax, y, zb, we have that Ä TQ œ  x
t x1 , y
t y1 , z
t z1 ¡ œ s x2 , y2 , z2 ¡ Thus x œ t x1  s x2 , y œ t y1  s y2 , z œ t z1  s z2 . (Note that if Q is the midpoint, then qp œ 1 and t œ s œ so that x œ

" #

x1  "# x2 œ

x1
x2 2 ,

yœ

y1
y2 2 ,

zœ

z1
z2 2

" #

so that this result agress with the midpoint formula.)

Ä 49. (a) the midpoint of AB is M ˆ 5# ß 5# ß 0‰ and CM œ ˆ 5#
1‰ i  ˆ 5#
1‰ j  (!
3)k œ Ä (b) the desired vector is ˆ 23 ‰ CM œ 23 ˆ 3# i  3# j
3k‰ œ i  j
2k

3 #

i  3# j
3k

(c) the vector whose sum is the vector from the origin to C and the result of part (b) will terminate at the center of mass Ê the terminal point of (i  j  3k)  (i  j
2k) œ 2i  2j  k is the point (2ß 2ß 1), which is the location of the center of mass Ä 50. The midpoint of AB is M ˆ 3# ß 0ß 5# ‰ and ˆ 32 ‰ CM œ

ˆ 3#  1‰ i  (0
2)j  ˆ 5#  1‰ k‘ œ 32 ˆ 5# i
2j  7# k‰ Ä œ 53 i
43 j  73 k. The terminal point of ˆ 53 i
43 j  73 k‰  OC œ ˆ 53 i
43 j  73 k‰  (
i  2j
k) œ 23 i  23 j  43 k is the point ˆ 23 ß 23 ß 43 ‰ which is the location of the intersection of the medians. 2 3

51. Without loss of generality we identify the vertices of the quadrilateral such that A(0ß 0ß 0), B(xb ß 0ß 0), C(xc ß yc ß 0) and D(xd ß yd ß zd ) Ê the midpoint of AB is MAB ˆ x#b ß 0ß 0‰ , the midpoint of BC is MBC ˆ xb
# xc ß y#c ß 0‰ , the midpoint of CD is MCD ˆ xc
# xd ß yc
# yd ß z#d ‰ and the midpoint of AD is xb

MAD ˆ x#d ß y#d ß z#d ‰ Ê the midpoint of MAB MCD is Œ # xb
xc

of MAD MBC œ Œ #

#

x

b #d

ß yc
4 yd ,

b

xc
xd

#

#

ß yc
4 yd ,

zd 4

which is the same as the midpoint

zd 4 .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

785

786

Chapter 12 Vectors and the Geometry of Space

52. Let V" , V# , V$ , á , Vn be the vertices of a regular n-sided polygon and vi denote the vector from the center to n

Vi for i œ 1, 2, 3, á , n. If S œ ! vi and the polygon is rotated through an angle of

i(21) n

iœ1

where

i œ 1, 2, 3, á , n, then S would remain the same. Since the vector S does not change with these rotations we conclude that S œ 0. 53. Without loss of generality we can coordinatize the vertices of the triangle such that A(0ß 0), B(bß 0) and Ä C(xc ß yc ) Ê a is located at ˆ b
# xc ß y#c ‰ , b is at ˆ x#c ß y#c ‰ and c is at ˆ b# ß 0‰ . Therefore, Aa œ ˆ b#  x#c ‰ i  ˆ y#c ‰ j , Ä Ä Ä Ä Ä Bb œ ˆ x#c
b‰ i  ˆ y#c ‰ j , and Cc œ ˆ b#
xc ‰ i  (
yc ) j Ê Aa  Bb  Cc œ 0 . 54. Let u be any unit vector in the plane. If u is positioned so that its initial point is at the origin and terminal point is at ax, yb, then u makes an angle ) with i, measured in the counter-clockwise direction. Since kuk œ 1, we have that x œ cos ) and y œ sin ). Thus u œ cos ) i  sin ) j. Since u was assumed to be any unit vector in the plane, this holds for every unit vector in the plane. 12.3 THE DOT PRODUCT ) NOTE: In Exercises 1-8 below we calculate projv u as the vector Š kuk kcos vk ‹ v , so the scalar multiplier of v is

the number in column 5 divided by the number in column 2. v†u

kvk

kuk

cos )

kuk cos )

projv u

1.
25

5

5


1


5


2i  4j
È 5k

2.

3

1

13

3 13

3

3 ˆ 35 i  45 k‰

3.

25

15

5

" 3

5 3

" 9

4.

13

15

3

13 45

13 15

13 225

5.

2

È34

È3

2 È3 È34

2 È34

" 17

6. È3
È2

È2

3

È3  È2 3È 2

È3  È2 È2

È3  È2 #

(
i  j )

7. 10  È17

È26

È21

10
È17 È546

10
È17 È26

10
È17 È26

(
5i  j)

È30 6

È30 6

" 5

" È30

" " " 5 ¢ È# , È3 £

8.

" 6


(1)(2)
(0)(1) 9. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È2#(2)(1) 1 # 0# È 1# 2# (


10. ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È

11. ) œ cos" Š kuuk†kvvk ‹ œ cos"

Î








1)#

(5j
3k)

œ cos" Š È910È25 ‹ œ cos" ˆ 23 ‰ ¸ 0.84 rad

ŠÈ3‹ ŠÈ3‹
(7)(1)
(0)(2)

Ï ÊŠÈ3‹

(2i  10j
11k)

‹ œ cos" Š È54È6 ‹ œ cos" Š È430 ‹ ¸ 0.75 rad

(2)(3)
(2)(0)
(1)(4) ‹ 2#
(2)#
1# È3#
0#
4#

#

(10i  11j
2k)

#

Ñ


(7)#
0# ÊŠÈ3‹
(1)#
(2)# Ò

7 œ cos" Š È352È ‹ 8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.3 The Dot Product

787

1 œ cos" Š È26 ‹ ¸ 1.77 rad

12. ) œ cos" Š kuuk†kvvk ‹ œ cos"

Î

Ñ

(1)(1)
ŠÈ2‹ (1)
ŠÈ2‹ (1) #

Ï Ê(1)#
ŠÈ2‹

#


ŠÈ2‹ È(1)#
(1)#
(1)# Ò

1 œ cos" Š È ‹ 5 È3

1 œ cos" Š È15 ‹ ¸ 1.83 rad

Ä Ä Ä Ä Ä Ä 13. AB œ  3, 1¡, BC œ  
1,
3¡, and AC œ  2,
2¡. BA œ  
3,
1¡, CB œ  1, 3¡, CA œ  
2, 2¡. Ä Ä Ä Ä Ä Ä ¹ AB ¹ œ ¹ BA ¹ œ È10, ¹ BC ¹ œ ¹ CB ¹ œ È10, ¹ AC ¹ œ ¹ CA ¹ œ 2È2, Ä Ä AB†AC

Ä Ä  œ cos" Angle at A œ cos" Œ ¹AB  ¹ ¹AC¹

3 a2 b
1 a 2 b ŠÈ10‹Š2È2‹ 

œ cos" Š È15 ‹ ¸ 63.435‰

Ä Ä BC†BA

a1ba3b
a3ba1b " 3 ‰ Ä Ä  œ cos" Angle at B œ cos" Œ ¹BC   œ cos ˆ 5 ‰ ¸ 53.130 , and ¹ ¹BA¹ È È Š

Ä Ä CB†CA

Ä Ä  œ cos" Angle at C œ cos" Œ ¹CB  ¹ ¹CA¹

10‹Š

10‹

1 a 2 b
3 a 2 b ŠÈ10‹Š2È2‹ 

œ cos" Š È15 ‹ ¸ 63.435‰

Ä Ä Ä Ä 14. AC œ  2, 4¡ and BD œ  4,
2¡. AC † BD œ 2a4b  4a
2b œ 0, so the angle measures are all 90‰ . 15. (a) cos ! œ

i†v kik kvk

œ

a kvk

, cos " œ

j †v kjk kvk

œ

b k vk

, cos # œ

#

#

k†v kkk kvk #

œ

cos# !  cos# "  cos# # œ Š kvak ‹  Š kbvk ‹  Š kvck ‹ œ (b) kvk œ 1 Ê cos ! œ

a kv k

œ a, cos " œ

b kv k

c kvk

and

a #
b #
c# k v k k vk

œ b and cos # œ

c kvk

œ

kv k k v k kvk kvk

œ1

œ c are the direction cosines of v

16. u œ 10i  2k is parallel to the pipe in the north direction and v œ 10j  k is parallel to the pipe in the east direction. The angle between the two pipes is ) œ cos" Š kuuk†kvvk ‹ œ cos" Š È1042È101 ‹ ¸ 1.55 rad ¸ 88.88°. 17. u œ ˆ vv††uv v‰  ˆu


v†u v†v

v‰ œ

3 #

(i  j)  (3j  4k)
3# (i  j)‘ œ ˆ 3# i  3# j‰  ˆ
3# i  3# j  4k‰ , where

v‰ œ

" #

v  ˆu
"# v‰ œ

v † u œ 3 and v † v œ 2 18. u œ ˆ vv††uv v‰  ˆu


v†u v†v

1 #

(i  j)  (j  k)
1# (i  j)‘ œ ˆ 1# i  1# j‰  ˆ
1# i  1# j  k‰ ,

where v † u œ 1 and v † v œ 2 28  ˆ 14 19. u œ ˆ vv††uv v‰  ˆu
vv††uv v‰ œ 14 3 (i  2j
k)  (8i  4j
12k)
3 i  3 j
28 14 ‰ 16 22 ‰ ˆ 10 œ ˆ 14 3 i  3 j
3 k  3 i
3 j
3 k , where v † u œ 28 and v † v œ 6

20. u œ ˆ vv††uv v‰  ˆu


v†u v†v

v‰ œ

" 1

14 3

k‰‘

(A)   (i  j  k)
ˆ 1" ‰ A‘ œ (i)  (j  k), where v † u œ 1 and v † v œ 1; yes

21. The sum of two vectors of equal length is always orthogonal to their difference, as we can see from the equation ( v"  v# ) † ( v "
v # ) œ v " † v "  v # † v "
v " † v #
v # † v # œ k v " k #
k v # k # œ 0 Ä Ä 22. CA † CB œ (
v  (
u)) † (
v  u) œ v † v
v † u  u † v
u † u œ kvk #
kuk # œ 0 because kuk œ kvk since both equal Ä Ä the radius of the circle. Therefore, CA and CB are orthogonal.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

788

Chapter 12 Vectors and the Geometry of Space

23. Let u and v be the sides of a rhombus Ê the diagonals are d" œ u  v and d# œ
u  v

Ê d" † d# œ (u  v) † (
u  v) œ
u † u  u † v
v † u  v † v œ kvk #
kuk # œ 0 because kuk œ kvk , since a rhombus has equal sides.

24. Suppose the diagonals of a rectangle are perpendicular, and let u and v be the sides of a rectangle Ê the diagonals are d" œ u  v and d# œ
u  v. Since the diagonals are perpendicular we have d" † d# œ 0

Í (u  v) † (
u  v) œ
u † u  u † v
v † u  v † v œ 0 Í kvk #
kuk # œ 0 Í akvk  kukb akvk
kukb œ 0 Í akvk  kukb œ 0 which is not possible, or akvk
kukb œ 0 which is equivalent to kvk œ kuk Ê the rectangle is a square.

25. Clearly the diagonals of a rectangle are equal in length. What is not as obvious is the statement that equal diagonals happen only in a rectangle. We show this is true by letting the adjacent sides of a parallelogram be the vectors (v" i  v# j) and (u" i  u# j). The equal diagonals of the parallelogram are d" œ (v" i  v# j)  (u" i  u# j) and d# œ (v" i  v# j)
(u" i  u# j). Hence kd" k œ kd# k œ k(v" i  v# j)  (u" i  u# j)k œ k(v" i  v# j)
(u" i  u# j)k Ê k(v"  u" )i  (v#  u# )jk œ k(v"
u" )i  (v#
u# )jk Ê È(v"  u" )#  (v#  u# )# œ È(v"
u" )#  (v#
u# )# Ê v#1  2v" u"  u1#  v##  2v# u#  u##

œ v#1
2v" u"  u1#  v##
2v# u#  u## Ê 2(v" u"  v# u# ) œ
2(v" u"  v# u# ) Ê v" u"  v# u# œ 0 Ê (v" i  v# j) † (u" i  u# j) œ 0 Ê the vectors (v" i  v# j) and (u" i  u# j) are perpendicular and the parallelogram must be a rectangle.

26. If kuk œ kvk and u  v is the indicated diagonal, then (u  v) † u œ u † u  v † u œ kuk #  v † u œ u † v  kvk # œ u † v  v † v œ (u  v) † v Ê the angle cos" Š k(uu

vvk )k†uuk ‹ between the diagonal and u and the angle cos" Š k(uu

vvk )k†vvk ‹ between the diagonal and v are equal because the inverse cosine function is one-to-one. Therefore, the diagonal bisects the angle between u and v. 27. horizontal component: 1200 cosa8‰ b ¸ 1188 ft/s; vertical component: 1200 sina8‰ b ¸ 167 ft/s 28. kwkcosa33‰
15‰ b œ 2.5 lb, so kwk œ

2.5 lb cos 18‰ .

Then w œ

2.5 lb   cos 18‰ cos

33‰ , sin 33‰ ¡ ¸  2.205, 1.432¡

29. (a) Since kcos )k Ÿ 1, we have ku † vk œ kuk kvk kcos )k Ÿ kuk kvk (1) œ kuk kvk . (b) We have equality precisely when kcos )k œ 1 or when one or both of u and v is 0. In the case of nonzero vectors, we have equality when ) œ 0 or 1, i.e., when the vectors are parallel. 30. (xi  yj) † v œ kxi  yjk kvk cos ) Ÿ 0 when

1 #

Ÿ ) Ÿ 1. This

means (xß y) has to be a point whose position vector makes an angle with v that is a right angle or bigger.

31. v † u" œ (au"  bu# ) † u" œ au" † u"  bu# † u" œ a ku" k #  b(u# † u" ) œ a(1)#  b(0) œ a 32. No, v" need not equal v# . For example, i  j Á i  2j but i † (i  j) œ i † i  i † j œ 1  0 œ 1 and i † (i  2j) œ i † i  2i † j œ 1  2 † 0 œ 1.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.3 The Dot Product

789

33. P(x" ß y" ) œ P ˆx" , bc
ba x" ‰ and Q(x# ß y# ) œ Q ˆx# ß bc
ba x# ‰ are any two points P and Q on the line with b Á 0 Ä Ä Ê PQ œ (x#
x" )i  ba (x"
x# )j Ê PQ † v œ (x#
x" )i  ba (x"
x# )j‘ † (ai  bj) œ a(x#
x" )  b ˆ ba ‰ (x"
x# ) Ä œ 0 Ê v is perpendicular to PQ for b Á 0. If b œ 0, then v œ ai is perpendicular to the vertical line ax œ c. Alternatively, the slope of v is ba and the slope of the line ax  by œ c is
ba , so the slopes are negative reciprocals Ê the vector v and the line are perpendicular. 34. The slope of v is

b a

and the slope of bx
ay œ c is ba , provided that a Á 0. If a œ 0, then v œ bj is parallel to

the vertical line bx œ c. In either case, the vector v is parallel to the line ax
by œ c. 35. v œ i  2j is perpendicular to the line x  2y œ c; P(2ß 1) on the line Ê 2  2 œ c Ê x  2y œ 4

36. v œ
2i
j is perpendicular to the line
2x
y œ c; P(
1ß 2) on the line Ê (
2)(
1)
2 œ c Ê
2x
y œ 0

37. v œ
2i  j is perpendicular to the line
2x  y œ c; P(
2ß
7) on the line Ê (
2)(
2)
7 œ c Ê
2x  y œ
3

38. v œ 2i
3j is perpendicular to the line 2x
3y œ c; P(11ß 10) on the line Ê (2)(11)
(3)(10) œ c Ê 2x
3y œ
8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

790

Chapter 12 Vectors and the Geometry of Space

39. v œ i
j is parallel to the line
x
y œ c; P(
2ß 1) on the line Ê
a
2b
1 œ c Ê
x
y œ 1 or x  y œ
1.

40. v œ 2i  3j is parallel to the line 3x
2y œ c; P(0ß
2) on the line Ê 0
2(
2) œ c Ê 3x
2y œ 4

41. v œ
i
2j is parallel to the line
2x  y œ c; P(1ß 2) on the line Ê
2(1)  2 œ c Ê
2x
y œ 0 or 2x
y œ 0.

42. v œ 3i
2j is parallel to the line
2x
3y œ c; P(1ß 3) on the line Ê (
2)(1)
(3)(3) œ c Ê
2x
3y œ
11 or 2x  3y œ 11

Ä Ä 43. P(0ß 0), Q(1ß 1) and F œ 5j Ê PQ œ i  j and W œ F † PQ œ (5j) † (i  j) œ 5 N † m œ 5 J 44. W œ kFk (distance) cos ) œ (602,148 N)(605 km)(cos 0) œ 364,299,540 N † km œ (364,299,540)(1000) N † m œ 3.6429954 ‚ 10"" J Ä 45. W œ kFk ¹ PQ ¹ cos ) œ (200)(20)(cos 30°) œ 2000È3 œ 3464.10 N † m œ 3464.10 J

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.3 The Dot Product Ä 46. W œ kFk ¹ PQ ¹ cos ) œ (1000)(5280)(cos 60°) œ 2,640,000 ft † lb In Exercises 47-52 we use the fact that n œ ai  bj is normal to the line ax  by œ c. 1 47. n" œ 3i  j and n# œ 2i
j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È610È ‹ œ cos" Š È"2 ‹ œ 5

1 4

3
1 " ˆ 48. n" œ
È3i  j and n# œ È3i  j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È
"2 ‰ œ È ‹ œ cos 4

49. n" œ È3i
j and n# œ i
È3j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š

21 3

4

È3
È3 È4 È4 ‹

œ cos" Š

È3 2 ‹

1 6

œ

50. n" œ i  È3j and n# œ Š1
È3‹ i  Š1  È3‹ j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" 

1  È3
È3
3 È 1
3 É 1  2È 3
3
1
2È 3
3 

4 œ cos" Š 2È ‹ œ cos" Š È"2 ‹ œ 8

1 4

4 7 51. n" œ 3i
4j and n# œ i
j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È3
È ‹ œ cos" Š 5È ‹ ¸ 0.14 rad 25 2 2

 10 " 52. n" œ 12i  5j and n# œ 2i
2j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È24169 Š È8 ‹ œ cos

14 ‹ 26È2

¸ 1.18 rad

53. The angle between the corresponding normals is equal to the angle between the corresponding tangents. The points of intersection are Š
y


3 4

œ f w Š


È3 # ‹ Šx

È3 3 # ß 4‹


Š


f(x) œ ˆ 3# ‰
x# is y


3 4

È3 # ‹‹

œ f w Š


and Š

È3 3 # ß 4‹ .

At Š


Ê y œ
È 3 Šx 

È3 # ‹ Šx


Š


È3 # ‹‹

È3 3 # ß 4‹

È3 # ‹



y œ È 3 Šx 

È3 # ‹



3 4

œ È3x 

9 4

21 3 ,

the angle is

1 3

3 4

Ê y œ
È3x
34 , and the tangent line for

Ê y œ È 3 Šx 

normals are n" œ È3i  j and n# œ
È3i  j. The angle at Š
œ cos" Š È43
È14 ‹ œ cos" ˆ
"# ‰ œ

the tangent line for f(x) œ x# is

and

21 3 .

È3 3 # ß 4‹

At Š

and the tangent line for f(x) œ

È3 # ‹

3 4

œ È3x  94 . The corresponding

is ) œ cos" Š knn""k†knn## k ‹

È3 3 # ß 4‹

3 #



the tangent line for f(x) œ x# is


x# is y œ
È3 Šx 

È3 # ‹



œ
È3x
34 . The corresponding normals are n" œ
È3i  j and n# œ È3i  j. The angle at Š ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È3
È1 ‹ œ cos" ˆ
"# ‰ œ 4 4 54. The points of intersection are Š!ß tangent line at Š!ß curve x œ y#


3 4

È3 # ‹

is y


È3 #

È3 # ‹

È3 # ‹.

" #y

œ

the angle is

The curve x œ

œ
È"3 (x
0) Ê n" œ

dy dx

has derivative

and Š!ß


21 3 ,

" È3

3 4

1 3

and

3 4

È3 3 # ß 4‹

is

21 3 .


y# has derivative

dy dx

œ
#"y Ê the

i  j is normal to the curve at that point. The

Ê the tangent line at Š!ß

È3 # ‹

is y


È3 #

œ

" È3

(x
0)

Ê n# œ
È"3 i  j is normal to the curve. The angle between the curves is ) œ cos" Š knn""k†knn## k ‹ œ cos" 

 "3
1 É 3"


1

É "3

"

œ cos
1

ˆ2‰

Š ˆ 34 ‰ ‹ œ cos" ˆ "# ‰ œ 3

1 3

and

21 3 .

Because of symmetry the angles between

the curves at the two points of intersection are the same. $

55. The curves intersect when y œ x$ œ ay# b œ y' Ê y œ 0 or y œ 1. The points of intersection are (!ß !) and ("ß "). Note that y   0 since y œ y' . At (!ß 0) the tangent line for y œ x$ is y œ 0 and the tangent line for Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

791

792

Chapter 12 Vectors and the Geometry of Space

y œ Èx is x œ 0. Therefore, the angle of intersection at (0ß 0) is 1# . At (1ß 1) the tangent line for y œ x$ is y œ 3x
2 and the tangent line for y œ Èx is y œ "# x  "# . The corresponding normal vectors are n" œ
3i  j and n# œ
"# i  j Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È" ‹ œ 14 , the angle is 2

1 4

and

31 4 .

56. The points of intersection for the curves y œ
x# and y œ $Èx are (!ß 0) and (
1ß
1). At (!ß 0) the tangent line for y œ
x# is y œ 0 and the tangent line for y œ $Èx is x œ 0. Therefore, the angle of intersection at (!ß 0) is 1# . At (
1ß
1) the tangent line for y œ
x# is y œ 2x  1 and the tangent line for y œ $Èx is y œ The corresponding normal vectors are n" œ 2i
j and n# œ œ cos"  È

2 3

5


"

É 9"

"

œ cos
1

ˆ5‰

"

Œ È5 3È10  œ cos 3

" 3

i
j Ê ) œ cos" Š knn""k†knn## k ‹

Š È"2 ‹ œ 14 , the angle is

1 4

and

31 4 .

12.4 THE CROSS PRODUCT â âi â 1. u ‚ v œ â 2 â â"

â j k â â
2
" â œ 3 ˆ 23 i  "3 j  32 k‰ Ê length œ 3 and the direction is 32 i  3" j  32 k; â 0
" â v ‚ u œ
(u ‚ v) œ
3 ˆ 23 i  "3 j  32 k‰ Ê length œ 3 and the direction is
32 i
3" j
32 k

â â i â 2. u ‚ v œ â 2 â â
"

â j kâ â 3 0 â œ 5(k) Ê length œ 5 and the direction is k â 1 0â

v ‚ u œ
(u ‚ v) œ
5(k) Ê length œ 5 and the direction is
k â â i â 3. u ‚ v œ â 2 â â
"

â j k â â
2 4 â œ 0 Ê length œ 0 and has no direction â 1
2 â

v ‚ u œ
(u ‚ v) œ 0 Ê length œ 0 and has no direction â âi â 4. u ‚ v œ â 1 â â0

j 1 0

â k â â
1 â œ 0 Ê length œ 0 and has no direction â 0 â

v ‚ u œ
(u ‚ v) œ 0 Ê length œ 0 and has no direction â âi â 5. u ‚ v œ â 2 â â0

j 0
3

â kâ â 0 â œ
6(k) Ê length œ 6 and the direction is
k â 0â

v ‚ u œ
(u ‚ v) œ 6(k) Ê length œ 6 and the direction is k â âi â 6. u ‚ v œ (i ‚ j) ‚ (j ‚ k) œ k ‚ i œ â 0 â â1

â j kâ â 0 1 â œ j Ê length œ 1 and the direction is j â 0 0â

v ‚ u œ
(u ‚ v) œ
j Ê length œ 1 and the direction is
j â â j k â â i â â 7. u ‚ v œ â
8
2
4 ⠜ 6i
12k Ê length œ 6È5 and the direction is È"5 i
È25 k â â 2 1 â â 2 v ‚ u œ
(u ‚ v) œ
(6i
12k) Ê length œ 6È5 and the direction is
" i  2 k È5

È5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 3

x
23 .

Section 12.4 The Cross Product âi â â 8. u ‚ v œ â 3# â â1

k ââ â 1 â œ
2i
2j  2k Ê length œ 2È3 and the direction is
È"3 i
È13 j  È"3 k â 2â v ‚ u œ
(u ‚ v) œ
(
2i
2j  2k) Ê length œ 2È3 and the direction is " i  1 j
" k j
"# 1

È3

â âi â 9. u ‚ v œ â 1 â â0

j 0 1

â kâ â 0â œ k â 0â

â âi â 10. u ‚ v œ â 1 â â0

â âi â 11. u ‚ v œ â 1 â â0

j 0 1

â k â â
1 ⠜ i
j  k â 1 â

â j âi â 12. u ‚ v œ â 2
1 â â1 2

â âi â 13. u ‚ v œ â 1 â â1

j 1
1

â kâ â 0 â œ
2k â 0â

â âi j â 14. u ‚ v œ â 0 1 â â1 0

â â j k â â i Ä Ä â â 15. (a) PQ ‚ PR œ â 1 1
3 â œ 8i  4j  4k Ê Area œ â â â
1 3
1 â (b) u œ „

Ä Ä PQ‚PR Ä Ä ¹PQ‚PR¹

œ „

" È6

j 0 1

" #

Ä Ä ¹ PQ ‚ PR ¹ œ

È3

È3

â k â â
1 â œ i  k â 0 â

â kâ â 0 â œ 5k â 0â

â kâ â 2 â œ 2j
k â 0â

" #

È64  16  16 œ 2È6

(2i  j  k)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

793

794

Chapter 12 Vectors and the Geometry of Space

â â j kâ âi Ä Ä â â 16. (a) PQ ‚ PR œ â 1 0 2 â œ 4i  4j
2k Ê Area œ â â â 2
2 0 â (b) u œ „

Ä Ä PQ‚PR Ä Ä ¹PQ‚PR¹

Ä Ä PQ‚PR Ä Ä ¹PQ‚PR¹

œ „

" È2

(
i  j ) œ „

" È2

" #

Ä Ä ¹ PQ ‚ PR ¹ œ

Ä Ä PQ‚PR Ä Ä ¹PQ‚PR¹

" #

È16  16  4 œ 3

œ „

" È14

" #

È1  1 œ

È2 #

(i
j )

â â j k â âi Ä Ä â â 18. (a) PQ ‚ PR œ â 2
1
1 â œ 2i  3j  k Ê Area œ â â â 1 0
2 â (b) u œ „

Ä Ä ¹ PQ ‚ PR ¹ œ

œ „ 3" (2i  2j
k)

â â â i j kâ Ä Ä â â 17. (a) PQ ‚ PR œ â 1 1 1 â œ
i  j Ê Area œ â â â1 1 0â (b) u œ „

" #

" #

Ä Ä ¹ PQ ‚ PR ¹ œ

" #

È4  9  1 œ

È14 #

(2i  3j  k)

â â â a" a# a$ â â â 19. If u œ a" i  a# j  a$ k, v œ b" i  b# j  b$ k, and w œ c" i  c# j  c$ k, then u † (v ‚ w) œ â b" b# b$ â , â â â c" c# c$ â â â â â â b" b # b $ â â c" c# c$ â â â â â v † (w ‚ u) œ â c" c# c$ â and w † (u ‚ v) œ â a" a# a$ â which all have the same value, since the â â â â â a" a# a$ â â b" b# b $ â interchanging of two pair of rows in a determinant does not change its value Ê the volume is â â â2 0 0â â â k(u ‚ v) † wk œ abs â 0 2 0 â œ 8 â â â0 0 2â â â 1 â 20. k(u ‚ v) † wk œ abs â 2 â â
1

â
1 1 â â 1
2 â œ 4 (for details about verification, see Exercise 19) â 2
1 â

â â2 1 â 21. k(u ‚ v) † wk œ abs â 2
1 â â1 0

â 0â â 1 â œ k
7k œ 7 (for details about verification, see Exercise 19) â 2â

â â â 1 1
2 â â â 22. k(u ‚ v) † wk œ abs â
1 0
1 â œ 8 (for details about verification, see Exercise 19) â â â 2 4
2 â 23. (a) u † v œ
6, u † w œ
81, v † w œ 18 Ê none â â â â â â j k â j k â j k â âi â i â i â â â â â â
1 1 â œ 0, v ‚ w œ â 0 1
5 â Á 0 (b) u ‚ v œ â 5
1 1 â Á 0, u ‚ w œ â 5 â â â â â â â 0 1
5 â â
15 3
3 â â
15 3
3 â Ê u and w are parallel 24. (a) u † v œ 0, u ‚ w œ 0, u † r œ
31, v † w œ 0, v † r œ 0, w † r œ 0 Ê u ¼ v, u ¼ w, v ¼ w, v ¼ r and w ¼ r

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.4 The Cross Product

795

â â â â â â j k â j k â â i âi j k â â i â â â â â â 2
1 â œ 0 (b) u ‚ v œ â 1 2
1 â Á 0, u ‚ w œ â 1 2
1 â Á 0, u ‚ r œ â 1 â â â â â
1
1 1 â â
1 1 1 â â1 0 1 â â # # â â â â â â â j kâ j kâ j kâ â i â i â i â â â â â â 1 1 â Á 0, w ‚ r œ â 1 0 1âÁ0 v ‚ w œ â
1 1 1 â Á 0, v ‚ r œ â
1 â â â
1
1 1 â â
1
1 1 â â 1 0 1â â # â # # â # â Ê u and r are parallel Ä Ä 25. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (60°) œ

2 3

Ä Ä 26. ¹ PQ ‚ F¹ œ ¹ PQ ¹ kFk sin (135°) œ

† 30 † 2 3

È3 #

† 30 †

ft † lb œ 10È3 ft † lb

È2 #

ft † lb œ 10È2 ft † lb

27. (a) true, kuk œ Èa#1  a##  a3# œ Èu † u (b) not always true, u † u œ kuk # â â â â j kâ j kâ â i â i â â â â (c) true, u ‚ 0 œ â a" a# a$ â œ 0i  0j  0k œ 0 and 0 ‚ u œ â 0 0 0 â œ 0i  0j  0k œ 0 â â â â â0 0 0â â a" a# a$ â â â j k â â i â â a# a$ â œ (
a# a$  a# a$ )i
(
a" a$  a" a$ )j  (
a" a#  a" a# )k œ 0 (d) true, u ‚ (
u) œ â a" â â â
a"
a#
a$ â (e) (f) (g) (h)

not always true, i ‚ j œ k Á
k œ j ‚ i for example true, distributive property of the cross product true, (u ‚ v) † v œ u † (v ‚ v) œ u † 0 œ 0 true, the volume of a parallelpiped with u, v, and w along the three edges is (u ‚ v) † w œ (v ‚ w) † u œ u † (v ‚ w), since the dot product is commutative.

28. (a) true, u † v œ a" b"  a# b#  a$ b$ œ b" a"  b# a#  b$ a$ œ v † u â â â â j kâ j kâ â i â i â â â â (b) true, u ‚ v œ â a" a# a$ â œ
â b" b# b$ â œ
(v ‚ u) â â â â â b" b# b$ â â a" a# a$ â â â â â j k â j kâ â i â i â â â â (c) true, (
u) ‚ v œ â
a"
a#
a$ ⠜
â a" a# a$ â œ
(u ‚ v) â â â â b# b$ â â b" â b" b # b $ â (d) true, (cu) † v œ (ca" )b"  (ca# )b#  (ca$ )b$ œ a" (cb" )  a# (cb# )  a$ (cb$ ) œ u † (cv) œ c(a" b"  a# b#  a$ b$ ) œ c(u † v) â â â â â â j kâ j k â j k â â i â i â i â â â â â â a# a$ â œ u ‚ (cv) (e) true, c(u ‚ v) œ c â a" a# a$ â œ â ca" ca# ca$ â œ (cu) ‚ v œ â a" â â â â â â â b" b# b$ â â b" b# b$ â â cb" cb# cb$ â #

(f) true, u † u œ a#1  a##  a3# œ ˆÈa#1  a##  a3# ‰ œ kuk # (g) true, (u ‚ u) † u œ 0 † u œ 0 (h) true, u ‚ v ¼ u and u ‚ v ¼ v Ê (u ‚ v) † u œ v † (u ‚ v) œ 0 29. (a) projv u œ Š kvukk†vvk ‹ v

(b) „ (u ‚ v)

(c) „ a(u ‚ v) ‚ wb

(d) k(u ‚ v) † wk

30. (a) (u ‚ v) ‚ (u ‚ w) (b) (u  v) ‚ (u
v) œ (u  v) ‚ u
(u  v) ‚ v œ u ‚ u  v ‚ u
u ‚ v
v ‚ v œ 0  v ‚ u
u ‚ v
0 œ 2(v ‚ u), or simply u ‚ v (c) kuk kvvk (d) ku ‚ wk Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

796

Chapter 12 Vectors and the Geometry of Space

31. (a) yes, u ‚ v and w are both vectors (c) yes, u and u ‚ w are both vectors

(b) no, u is a vector but v † w is a scalar (d) no, u is a vector but v † w is a scalar

32. (u ‚ v) ‚ w is perpendicular to u ‚ v, and u ‚ v is perpendicular to both u and v Ê (u ‚ v) ‚ w is parallel to a vector in the plane of u and v which means it lies in the plane determined by u and v. The situation is degenerate if u and v are parallel so u ‚ v œ 0 and the vectors do not determine a plane. Similar reasoning shows that u ‚ (v ‚ w) lies in the plane of v and w provided v and w are nonparallel. 33. No, v need not equal w. For example, i  j Á
i  j, but i ‚ (i  j) œ i ‚ i  i ‚ j œ 0  k œ k and i ‚ (
i  j ) œ
i ‚ i  i ‚ j œ 0  k œ k. 34. Yes. If u ‚ v œ u ‚ w and u † v œ u † w, then u ‚ (v
w) œ 0 and u † (v
w) œ 0. Suppose now that v Á w. Then u ‚ (v
w) œ 0 implies that v
w œ ku for some real number k Á 0. This in turn implies that u † (v
w) œ u † (ku) œ k kuk # œ 0, which implies that u œ 0. Since u Á 0, it cannot be true that v Á w, so v œ w. â j â i Ä Ä Ä Ä â 35. AB œ
i  j and AD œ
i
j Ê AB ‚ AD œ â
1 1 â â
1
1 â âi Ä Ä Ä Ä â 36. AB œ 7i  3j and AD œ 2i  5j Ê AB ‚ AD œ â 7 â â2 â âi Ä Ä Ä Ä â 37. AB œ 3i
2j and AD œ 5i  j Ê AB ‚ AD œ â 3 â â5

â kâ Ä Ä â 0 â œ 2k Ê area œ ¹ AB ‚ AD ¹ œ 2 â 0â

â j kâ Ä Ä â 3 0 â œ 29k Ê area œ ¹ AB ‚ AD ¹ œ 29 â 5 0â â kâ Ä Ä â 0 â œ 13k Ê area œ ¹ AB ‚ AD ¹ œ 13 â 0â

j
2 1

â âi Ä Ä Ä Ä â 38. AB œ 7i
4j and AD œ 2i  5j Ê AB ‚ AD œ â 7 â â2

j
4 5

â â i Ä Ä Ä Ä â 39. AB œ
2i  3j and AC œ 3i  j Ê AB ‚ AC œ â
2 â â 3 â âi Ä Ä Ä Ä â 40. AB œ 4i  4j and AC œ 3i  2j Ê AB ‚ AC œ â 4 â â3

j 3 1

â kâ Ä Ä â 0 â œ 43k Ê area œ ¹ AB ‚ AD ¹ œ 43 â 0â â kâ â 0 â œ
11k Ê area œ â 0â

â j kâ â 4 0 â œ
4k Ê area œ â 2 0â

" #

" #

Ä Ä ¹ AB ‚ AC ¹ œ

11 #

Ä Ä ¹ AB ‚ AC ¹ œ 2

â â i Ä Ä Ä Ä â 41. AB œ 6i
5j and AC œ 11i
5j Ê AB ‚ AC œ â 6 â â 11

j
5
5

â kâ â 0 â œ 25k Ê area œ â 0â

" #

Ä Ä ¹ AB ‚ AC ¹ œ

â â i Ä Ä Ä Ä â 42. AB œ 16i
5j and AC œ 4i  4j Ê AB ‚ AC œ â 16 â â 4

j
5 4

â kâ â 0 â œ 84k Ê area œ â 0â

" #

Ä Ä ¹ AB ‚ AC ¹ œ 42

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

25 #

Section 12.5 Lines and Planes in Space â â i â 43. If A œ a" i  a# j and B œ b" i  b# j, then A ‚ B œ â a" â â b" " #

kA ‚ Bk œ „

" #

a" ºb

"

j a# b#

â kâ a â 0â œ º " b" â 0â

a# k and the triangle's area is b# º

a# . The applicable sign is () if the acute angle from A to B runs counterclockwise b# º

in the xy-plane, and (
) if it runs clockwise, because the area must be a nonnegative number. Ä Ä 44. If A œ a" i  a# j, B œ b" i  b# j, and C œ c" i  c# j, then the area of the triangle is "# ¹ AB ‚ AC ¹ . Now, â â i j kâ â Ä Ä Ä Ä b
a" b#
a# â â k Ê "# ¹ AB ‚ AC ¹ AB ‚ AC œ â b"
a" b#
a# 0 â œ º " c"
a" c#
a# º â â â c"
a" c#
a# 0 â k(b"
a" )(c#
a# )
(c"
a" )(b#
a# )k œ "# ka" (b#
c# )  a# (c"
b" )  (b" c#
c" b# )k â â â a" a# 1 â â " â œ „ # â b" b# 1 â . The applicable sign ensures the area formula gives a nonnegative number. â â â c" c# 1 â

œ

" #

12.5 LINES AND PLANES IN SPACE 1. The direction i  j  k and P(3ß
4ß
1) Ê x œ 3  t, y œ
4  t, z œ
1  t Ä 2. The direction PQ œ
2i
2j  2k and P(1ß 2ß
1) Ê x œ 1
2t, y œ 2
2t, z œ
1  2t Ä 3. The direction PQ œ 5i  5j
5k and P(
2ß 0ß 3) Ê x œ
2  5t, y œ 5t, z œ 3
5t Ä 4. The direction PQ œ
j
k and P(1ß 2ß 0) Ê x œ 1, y œ 2
t, z œ
t 5. The direction 2j  k and P(!ß !ß !) Ê x œ 0, y œ 2t, z œ t 6. The direction 2i
j  3k and P(3ß
2ß 1) Ê x œ 3  2t, y œ
2
t, z œ 1  3t 7. The direction k and P(1ß 1ß 1) Ê x œ 1, y œ 1, z œ 1  t 8. The direction 3i  7j
5k and P(2ß 4ß 5) Ê x œ 2  3t, y œ 4  7t, z œ 5
5t 9. The direction i  2j  2k and P(0ß
7ß 0) Ê x œ t, y œ
7  2t, z œ 2t â â â i j kâ â â 10. The direction is A ‚ B œ â 1 2 3 â œ
2i  4j
2k and P(#ß $ß 0) Ê x œ 2
2t, y œ 3  4t, z œ
2t â â â3 4 5â 11. The direction i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0 12. The direction k and P(0ß 0ß 0) Ê x œ 0, y œ 0, z œ t

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797

798

Chapter 12 Vectors and the Geometry of Space

Ä 13. The direction PQ œ i  j  3# k and P(0ß 0ß 0) Ê x œ t, y œ t, z œ

3 #

t, where 0 Ÿ t Ÿ 1

Ä 14. The direction PQ œ i and P(0ß 0ß 0) Ê x œ t, y œ 0, z œ 0, where 0 Ÿ t Ÿ 1

Ä 15. The direction PQ œ j and P(1ß 1ß 0) Ê x œ 1, y œ 1  t, z œ 0, where
1 Ÿ t Ÿ 0

Ä 16. The direction PQ œ k and P(1ß 1ß 0) Ê x œ 1, y œ 1, z œ t, where 0 Ÿ t Ÿ 1

Ä 17. The direction PQ œ
2j and P(0ß 1ß 1) Ê x œ 0, y œ 1
2t, z œ 1, where 0 Ÿ t Ÿ 1

Ä 18. The direction PQ œ $i
2j and P(0ß 2ß 0) Ê x œ 3t, y œ 2
2t, z œ 0, where 0 Ÿ t Ÿ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.5 Lines and Planes in Space

799

Ä 19. The direction PQ œ
2i  2j
2k and P(2ß 0ß 2) Ê x œ 2
2t, y œ 2t, z œ 2
2t, where 0 Ÿ t Ÿ 1

Ä 20. The direction PQ œ
i  3j  k and P(1ß 0ß
1) Ê x œ 1
t, y œ 3t, z œ
1  t, where 0 Ÿ t Ÿ 1

21. 3(x
0)  (
2)(y
2)  (
1)(z  1) œ 0 Ê 3x
2y
z œ
3 22. 3(x
1)  (1)(y  1)  (1)(z
3) œ 0 Ê 3x  y  z œ 5 â â j kâ â i Ä Ä Ä Ä â â 23. PQ œ i
j  3k, PS œ
i
3j  2k Ê PQ ‚ PS œ â 1
1 3 ⠜ 7i
5j
4k is normal to the plane â â â
1
3 2 â Ê 7(x
2)  (
5)(y
0)  (
4)(z
2) œ 0 Ê 7x
5y
4z œ 6 â â i Ä Ä Ä Ä â 24. PQ œ
i  j  2k, PS œ
3i  2j  3k Ê PQ ‚ PS œ â
1 â â
3

j 1 2

â kâ â 2 â œ
i
3j  k is normal to the plane â 3â

Ê (
1)(x
1)  (
3)(y
5)  (1)(z
7) œ 0 Ê x  3y
z œ 9 25. n œ i  3j  4k, P(2ß 4ß 5) œ (1)(x
2)  (3)(y
4)  (4)(z
5) œ 0 Ê x  3y  4z œ 34 26. n œ i
2j  k, P(1ß
2ß 1) œ (1)(x
1)  (
2)(y  2)  (1)(z
1) œ 0 Ê x
2y  z œ 6 x œ 2t  1 œ s  2 2t
s œ 1 4t
2s œ 2 Ê œ Ê œ Ê t œ 0 and s œ
1; then z œ 4t  3 œ
4s
1 27. œ y œ 3t  2 œ 2s  4 3t
2s œ 2 3t
2s œ 2 Ê 4(0)  3 œ (
4)(
1)
1 is satisfied Ê the lines do intersect when t œ 0 and s œ
1 Ê the point of intersection is x œ 1, y œ 2, and z œ 3 or P(1ß 2ß 3). A vector normal to the plane determined by these lines is â â âi j k â â â n" ‚ n# œ â 2 3 4 â œ
20i  12j  k, where n" and n# are directions of the lines Ê the plane â â â 1 2
4 â containing the lines is represented by(
20)(x
1)  (12)(y
2)  (1)(z
3) œ 0 Ê
20x  12y  z œ 7. xœ t œ 2s  2 t
2s œ 2 Ê œ Ê s œ
1 and t œ 0; then z œ t  1 œ 5s  6 Ê 0  1 œ 5(
1)  6 28. œ y œ
t  2 œ s  3
t
s œ 1 is satisfied Ê the lines do intersect when s œ
1 and t œ 0 Ê the point of intersection is x œ 0, y œ 2 and z œ 1 â â j kâ âi â â or P(0ß 2ß 1). A vector normal to the plane determined by these lines is n" ‚ n# œ â 1
" 1 â â â â2 1 5â

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

800

Chapter 12 Vectors and the Geometry of Space

œ
6i
3j  3k, where n" and n# are directions of the lines Ê the plane containing the lines is represented by (
6)(x
0)  (
3)(y
2)  (3)(z
1) œ 0 Ê 6x  3y
3z œ 3. 29. The cross product of i  j
k and
4i  2j
2k has the same direction as the normal to the plane â â j k â â i â â Ê n œ â 1 "
1 ⠜ 6j  6k. Select a point on either line, such as P(
1ß 2ß 1). Since the lines are given â â â
4 2
2 â to intersect, the desired plane is 0(x  1)  6(y
2)  6(z
1) œ 0 Ê 6y  6z œ 18 Ê y  z œ 3. 30. The cross product of i
3j
k and i  j  k has the same direction as the normal to the plane â â j k â âi â â n œ â 1
3
1 ⠜
2i
2j  4k. Select a point on either line, such as P(0ß 3ß
2). Since the lines are â â 1 â â1 1 given to intersect, the desired plane is (
2)(x
0)  (
2)(y
3)  (4)(z  2) œ 0 Ê
2x
2y  4z œ
14 Ê x  y
2z œ 7. â âi j â 31. n" ‚ n# œ â 2 1 â â1 2

â k â â
1 ⠜ 3i
3j  3k is a vector in the direction of the line of intersection of the planes â 1 â Ê 3(x
2)  (
3)(y
1)  3(z  1) œ 0 Ê 3x
3y  3z œ 0 Ê x
y  z œ 0 is the desired plane containing P! (2ß 1ß
1)

â â j k â âi Ä â â 32. A vector normal to the desired plane is P" P# ‚ n œ â 2 0
2 ⠜
2i
12j
2k; choosing P" (1ß 2ß 3) as a â â â 4
1 2 â point on the plane Ê (
2)(x
1)  (
12)(y
2)  (
2)(z
3) œ 0 Ê
2x
12y
2z œ
32 Ê x  6y  z œ 16 is the desired plane â âi Ä â 33. S(0ß 0ß 12), P(0ß 0ß 0) and v œ 4i
2j  2k Ê PS ‚ v œ â 0 â â4 Ê dœ

Ä ¹PS‚v¹ kv k

œ

24È1
4 È16
4
4

œ

24È5 È24

â kâ â 12 â œ 24i  48j œ 24(i  2j) â 2 â

j 0
2

œ È5 † 24 œ 2È30 is the distance from S to the line

â j â i Ä â 34. S(0ß 0ß 0), P(5ß 5ß
3) and v œ 3i  4j
5k Ê PS ‚ v œ â
5
5 â 4 â 3 Ê dœ

Ä ¹PS‚v¹ kv k

œ

È169
256
25 È9
16
25

œ

È450 È50

â k â â 3 â œ 13i
16j
5k â
5 â

œ È9 œ 3 is the distance from S to the line

Ä 35. S(2ß 1ß 3), P(2ß 1ß 3) and v œ 2i  6j Ê PS ‚ v œ 0 Ê d œ

Ä ¹PS‚v¹ kv k

œ

0 È40

œ 0 is the distance from S to the line

(i.e., the point S lies on the line) â âi Ä â ß ß
ß ß œ   Ê ‚ œ 36. S(2 1 1), P(0 1 0) and v 2i 2j 2k PS v â 2 â â2 Ê dœ

Ä ¹PS‚v¹ kv k

œ

È4
36
16 È4
4
4

œ

È56 È12

j 0 2

â k â â
1 ⠜ 2 i
6 j  4 k â 2 â

œ É 14 3 is the distance from S to the line

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.5 Lines and Planes in Space â j â i Ä â 37. S(3ß
1ß 4), P(4ß 3ß
5) and v œ
i  2j  3k Ê PS ‚ v œ â
1
4 â â
1 2 Ê dœ

Ä ¹PS‚v¹ kv k

œ

È900
36
36 È1
4
9

œ

È972 È14

œ

È486 È7

œ

È81†6 È7

œ

9È42 7

â â i Ä â 38. S(
1ß 4ß 3), P(10ß
3ß 0) and v œ 4i  4k Ê PS ‚ v œ â
11 â â 4 Ê dœ

Ä ¹PS‚v¹ kv k

œ

28È1
4
1 4È 1
1

â kâ â 9 â œ
30i
6j
6k â 3â

is the distance from S to the line j 7 0

â kâ â 3 â œ 28i  56j
28k œ 28(i  2j
k) â 4â

œ 7È3 is the distance from S to the line

Ä 39. S(2ß
3ß 4), x  2y  2z œ 13 and P(13ß 0ß 0) is on the plane Ê PS œ
11i
3j  4k and n œ i  2j  2k Ä 9 Ê d œ ¹ PS † knnk ¹ œ ¹ È111
46

48 ¹ œ ¹ È ¹œ3 9 Ä 40. S(0ß 0ß 0), 3x  2y  6z œ 6 and P(2ß 0ß 0) is on the plane Ê PS œ
2i and n œ 3i  2j  6k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È9
46
36 ¹ œ È649 œ 67 Ä 41. S(0ß 1ß 1), 4y  3z œ
12 and P(0ß
3ß 0) is on the plane Ê PS œ 4j  k and n œ 4j  3k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È1616

39 ¹ œ 19 5 Ä 42. S(2ß 2ß 3), 2x  y  2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ 2j  3k and n œ 2i  j  2k Ä Ê d œ ¹ PS † knnk ¹ œ ¹ È42

16
4 ¹ œ 83 Ä 43. S(0ß
1ß 0), 2x  y  2z œ 4 and P(2ß 0ß 0) is on the plane Ê PS œ
2i
j and n œ 2i  j  2k Ä 4  1
0 Ê d œ ¹ PS † knnk ¹ œ ¹ È ¹ œ 53 4
1
4 Ä 44. S(1ß 0ß
1),
4x  y  z œ 4 and P(
1ß 0ß 0) is on the plane Ê PS œ 2i
k and n œ
4i  j  k Ä È Ê d œ ¹ PS † knnk ¹ œ ¹ È168
11
1 ¹ œ È918 œ 3 # 2 Ä 45. The point P(1ß 0ß 0) is on the first plane and S(10ß 0ß 0) is a point on the second plane Ê PS œ 9i, and Ä n œ i  2j  6k is normal to the first plane Ê the distance from S to the first plane is d œ ¹ PS † knnk ¹ œ ¹ È1
94
36 ¹ œ

9 È41

, which is also the distance between the planes.

46. The line is parallel to the plane since v † n œ ˆi  j
"# k‰ † (i  2j  6k) œ 1  2
3 œ 0. Also the point Ä S(1ß 0ß 0) when t œ
1 lies on the line, and the point P(10ß 0ß 0) lies on the plane Ê PS œ
9i. The distance Ä from S to the plane is d œ ¹ PS † knnk ¹ œ ¹ È1
49
36 ¹ œ È9 , which is also the distance from the line to the 41

plane. 47. n" œ i  j and n# œ 2i  j
2k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È22
È1 9 ‹ œ cos" Š È"2 ‹ œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 4

801

802

Chapter 12 Vectors and the Geometry of Space

523 48. n" œ 5i  j
k and n# œ i
2j  3k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È ‹ œ cos" (0) œ 27 È14

1 #

442 49. n" œ 2i  2j  2k and n# œ 2i
2j
k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È ‹ œ cos" Š 3" È3 ‹ ¸ 1.76 rad 12 È9

50. n" œ i  j  k and n# œ k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È 1È ‹ ¸ 0.96 rad 3 1 5 51. n" œ 2i  2j
k and n# œ i  2j  k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š 2È
94È61 ‹ œ cos" Š 3È ‹ ¸ 0.82 rad 6

18 26 ‰ 52. n" œ 4j  3k and n# œ 3i  2j  6k Ê ) œ cos" Š knn""k†knn## k ‹ œ cos" Š È825
È ¸ 0.73 rad ‹ œ cos" ˆ 35 49

53. 2x
y  3z œ 6 Ê 2(1
t)
(3t)  3(1  t) œ 6 Ê
2t  5 œ 6 Ê t œ
"# Ê x œ 3# , y œ
3# and z œ Ê ˆ 3# ß
3# ß "# ‰ is the point

" #

54. 6x  3y
4z œ
12 Ê 6(2)  3(3  2t)
4(
2
2t) œ
12 Ê 14t  29 œ
12 Ê t œ
41 14 Ê x œ 2, y œ 3
41 20 27 and z œ
2  7 Ê ˆ2ß
7 ß 7 ‰ is the point 55. x  y  z œ 2 Ê (1  2t)  (1  5t)  (3t) œ 2 Ê 10t  2 œ 2 Ê t œ 0 Ê x œ 1, y œ 1 and z œ 0 Ê (1ß 1ß 0) is the point 56. 2x
3z œ 7 Ê 2(
1  3t)
3(5t) œ 7 Ê
9t
2 œ 7 Ê t œ
1 Ê x œ
1
3, y œ
2 and z œ
5 Ê (
4ß
2ß
5) is the point â â â i j kâ â â 57. n" œ i  j  k and n# œ i  j Ê n" ‚ n# œ â 1 1 1 â œ
i  j, the direction of the desired line; (1ß 1ß
1) â â â1 1 0â is on both planes Ê the desired line is x œ 1
t, y œ 1  t, z œ
1 â âi â 58. n" œ 3i
6j
2k and n# œ 2i  j
2k Ê n" ‚ n# œ â 3 â â2

â j k â â
6
2 â œ 14i  2j  15k, the direction of the â 1
2 â desired line; (1ß 0ß 0) is on both planes Ê the desired line is x œ 1  14t, y œ 2t, z œ 15t

â âi â 59. n" œ i
2j  4k and n# œ i  j
2k Ê n" ‚ n# œ â 1 â â1

j
2 1

â k â â 4 â œ 6j  3k, the direction of the â
2 â

desired line; (4ß 3ß 1) is on both planes Ê the desired line is x œ 4, y œ 3  6t, z œ 1  3t â â j k â âi â â 60. n" œ 5i
2j and n# œ 4j
5k Ê n" ‚ n# œ â 5
2 0 â œ 10i  25j  20k, the direction of the â â â 0 4
5 â desired line; (1ß
3ß 1) is on both planes Ê the desired line is x œ 1  10t, y œ
3  25t, z œ 1  20t 2t
4s œ
2 2t
4s œ
2 Ê œ 61. L1 & L2: x œ 3  2t œ 1  4s and y œ
1  4t œ 1  2s Ê œ 4t
2s œ 2 2t
s œ 1 Ê
3s œ
3 Ê s œ 1 and t œ 1 Ê on L1, z œ 1 and on L2, z œ 1 Ê L1 and L2 intersect at (5ß 3ß 1).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

41 7 ,

Section 12.5 Lines and Planes in Space L2 & L3: The direction of L2 is " 3

" 6

(4i  2j  4k) œ

" 3

803

(2i  j  2k) which is the same as the direction

(2i  j  2k) of L3; hence L2 and L3 are parallel.

2t
2r œ 0 t
rœ0 Ê œ Ê 3t œ 3 L1 & L3: x œ 3  2t œ 3  2r and y œ
1  4t œ 2  r Ê œ 4t
r œ 3 4t
r œ 3 Ê t œ 1 and r œ 1 Ê on L1, z œ 2 while on L3, z œ 0 Ê L1 and L2 do not intersect. The direction of L1 is È"21 (2i  4j
k) while the direction of L3 is 3" (2i  j  2k) and neither is a multiple of the other; hence L1 and L3 are skew. 2t  s œ 1 Ê
5s œ 3 Ê s œ
35 and t œ 62. L1 & L2: x œ 1  2t œ 2
s and y œ
1
t œ 3s Ê œ
t
3s œ 1 zœ

12 5

while on L2, z œ 1


while the direction of L2 is

3 5

œ

" È11

2 5

Ê L1 and L2 do not intersect. The direction of L1 is

" È14

4 5

Ê on L1,

(2i
j  3k)

(
i  3j  k) and neither is a multiple of the other; hence, L1 and L2 are

skew.
s
2r œ 3 L2 & L3: x œ 2
s œ 5  2r and y œ 3s œ 1
r Ê œ Ê 5s œ 5 Ê s œ 1 and r œ
2 Ê on L2, 3s  r œ 1 z œ 2 and on L3, z œ 2 Ê L2 and L3 intersect at (1ß 3ß 2). L1 & L3: L1 and L3 have the same direction È"14 (2i
j  3k); hence L1 and L3 are parallel. 63. x œ 2  2t, y œ
4
t, z œ 7  3t; x œ
2
t, y œ
2  "# t, z œ 1
3# t 64. 1(x
4)
2(y
1)  1(z
5) œ 0 Ê x
4
2y  2  z
5 œ 0 Ê x
2y  z œ 7;
È2 (x
3)  2È2 (y  2)
È2 (z
0) œ 0 Ê
È2x  2È2y
È2z œ
7È2 65. x œ 0 Ê t œ
"# , y œ
"# , z œ
3# Ê ˆ!ß
"# ß
3# ‰ ; y œ 0 Ê t œ
1, x œ
1, z œ
3 Ê (
1ß 0ß
3); z œ 0 Ê t œ 0, x œ 1, y œ
1 Ê (1ß
1ß 0)

66. The line contains (0ß 0ß 3) and ŠÈ3ß 1ß 3‹ because the projection of the line onto the xy-plane contains the origin and intersects the positive x-axis at a 30° angle. The direction of the line is È3i  j  0k Ê the line in question is x œ È3t, y œ t, z œ 3. 67. With substitution of the line into the plane we have 2(1
2t)  (2  5t)
(
3t) œ 8 Ê 2
4t  2  5t  3t œ 8 Ê 4t  4 œ 8 Ê t œ 1 Ê the point (
1ß 7ß
3) is contained in both the line and plane, so they are not parallel. 68. The planes are parallel when either vector A" i  B" j  C" k or A# i  B# j  C# k is a multiple of the other or when k(A" i  B" j  C" k) ‚ (A# i  B# j  C# kk œ 0. The planes are perpendicular when their normals are perpendicular, or(A" i  B" j  C" k) † (A# i  B# j  C# k) œ 0. 69. There are many possible answers. One is found as follows: eliminate t to get t œ x
1 œ 2
y œ Ê x
1 œ 2
y and 2
y œ

z3 #

z3 #

Ê x  y œ 3 and 2y  z œ 7 are two such planes.

70. Since the plane passes through the origin, its general equation is of the form Ax  By  Cz œ 0. Since it meets the plane M at a right angle, their normal vectors are perpendicular Ê 2A  3B  C œ 0. One choice satisfying this equation is A œ 1, B œ
1 and C œ 1 Ê x
y  z œ 0. Any plane Ax  By  Cz œ 0 with 2A  3B  C œ 0 will pass through the origin and be perpendicular to M.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

804

Chapter 12 Vectors and the Geometry of Space

71. The points (aß 0ß 0), (0ß bß 0) and (0ß 0ß c) are the x, y, and z intercepts of the plane. Since a, b, and c are all nonzero, the plane must intersect all three coordinate axes and cannot pass through the origin. Thus, y x z a  b  c œ 1 describes all planes except those through the origin or parallel to a coordinate axis. 72. Yes. If v" and v# are nonzero vectors parallel to the lines, then v" ‚ v# Á 0 is perpendicular to the lines. Ä Ä 73. (a) EP œ cEP" Ê
x! i  yj  zk œ c c(x"
x! )i  y" j  z" kd Ê
x! œ c(x"
x! ), y œ cy" and z œ cz" , where c is a positive real number x ! (b) At x" œ 0 Ê c œ 1 Ê y œ y" and z œ z" ; at x" œ x! Ê x! œ 0, y œ 0, z œ 0; x lim c œ x lim Ä_ Ä _ x"  x! œ x lim Ä_ !

" 1

!

!

œ 1 Ê c Ä 1 so that y Ä y" and z Ä z"

74. The plane which contains the triangular plane is x  y  z œ 2. The line containing the endpoints of the line segment is x œ 1
t, y œ 2t, z œ 2t. The plane and the line intersect at ˆ 23 ß 23 ß 23 ‰ . The visible section of the line #

#

#

segment is Ɉ "3 ‰  ˆ 23 ‰  ˆ 23 ‰ œ 1 unit in length. The length of the line segment is È1#  2#  2# œ 3 Ê the line segment is hidden from view. 12.6 CYLINDERS AND QUADRIC SURFACES 1. d, ellipsoid

2. i, hyperboloid

3. a, cylinder

4. g, cone

5. l, hyperbolic paraboloid

6. e, paraboloid

7. b, cylinder

8. j, hyperboloid

9. k, hyperbolic paraboloid

10. f, paraboloid

11. h, cone

12. c, ellipsoid

13. x#  y# œ 4

14. x#  z# œ 4

15. z œ y#
1

16. x œ y#

17. x#  4z# œ 16

18. 4x#  y# œ 36

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2 3

of

Section 12.6 Cylinders and Quadric Surfaces 19. z#
y# œ 1

20. yz œ 1

21. 9x#  y#  z# œ 9

22. 4x#  4y#  z# œ 16

23. 4x#  9y#  4z# œ 36

24. 9x#  4y#  36z# œ 36

25. x#  4y# œ z

26. z œ x#  9y#

27. z œ 8
x#
y#

28. z œ 18
x#
9y#

29. x œ 4
4y#
z#

30. y œ 1
x#
z#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

805

806

Chapter 12 Vectors and the Geometry of Space

31. x#  y# œ z#

32. y#  z# œ x#

33. 4x#  9z# œ 9y#

34. 9x#  4y# œ 36z#

35. x#  y#
z# œ 1

36. y#  z#
x# œ 1

y# 4

37.

y# 4



z# 9




40.

y# 4




x# 4


z# œ 1

x# 4

œ1




z# 9

œ1

39. z#
x#
y# œ 1

41. x#
y#


z# 4

œ1

42.

38.

x# 4



x# 4


y#


z# 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ1

Section 12.6 Cylinders and Quadric Surfaces 43. y#
x# œ z

44. x# œ y# œ z

45. x#  y#  z# œ 4

46. 4x#  4y# œ z#

47. z œ 1  y#
x#

48. y#
z# œ 4

49. y œ
ax#  z# b

50. z#
4x#
4y# œ 4

51. 16x#  4y# œ 1

52. z œ x#  y#  1

53. x#  y#
z# œ 4

54. x œ 4
y#

55. x#  z# œ y

56. z#


x# 4

57. x#  z# œ 1


y# œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

807

808

Chapter 12 Vectors and the Geometry of Space

58. 4x#
4y#
z# œ 4

59. 16y#
9z# œ 4x#

60. z œ x#  y#  1

61. 9x#
4y#
z# œ 36

62. 4x#
9z# œ y#

63. x#
y#  16z# œ 16

64. z#
4y# œ 9

65. z œ  ax#
y# b

66. y#  x#  z# œ 1

67. x#  4y# œ 1

68. z œ 4x#
y#  4

69. 4y#
z#  4x# œ 4

70. z œ 1  x#

71. x#
y# œ z

72.

x# 4


y#  z# œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.6 Cylinders and Quadric Surfaces 74. 36x#
9y#
4z# œ 36

73. yz œ 1

75. 9x#
16y# œ 4z#

76. 4z#  x#  y# œ 4

77. (a) If x#
œ 1Š

y# 4

z# 9




œ 1 and z œ c, then x#


È 9
c# È # ‹ Š 2 93
c ‹ 3

œ

x# a#

(c)

Ê

x# # Š 9
9 c ‹

4 ˆ9
c# ‰ “ 9

2 1 a9
z # b 9

œ 1 Ê A œ ab1

, where 3 Ÿ z Ÿ 3. Thus V œ 2 '0

3

!

y# b#

z# c#




x#

œ1 Ê –

Ê V œ 2 '0 then V œ

41 r 3

$

1ab c#

a# Šc#
z# ‹ c#

y#


—

ac#  z# b dz œ

21 9

a9  z# b dz

– 21ab c#

b# Šc#
z# ‹ c#

’c# z 

œ 1 Ê A œ 1 Ša

È c#
z# È # # ‹ Š b cc
z ‹ c

— c

z$ 3 “!

œ

21ab c#

ˆ 23 c$ ‰ œ

41abc 3

. Note that if r œ a œ b œ c,

, which is the volume of a sphere.

78. The ellipsoid has the form of the barrel. Thus,

#

r R#

x# R# #




V œ 1 'ch y# dz. Now, h

y R#

y# R#




z# c# #

œ 1. To determine c# we note that the point (0ß rß h) lies on the surface

œ1 Ê c œ

h c# #







#

z c#

h# R # R#
r#

. We calculate the volume by the disk method:

œ 1 Ê y# œ R# Š1 

z# c# ‹

œ R # ’1 

Ê V œ 1 'ch ’R#  Š R h
# r ‹ z# “ dz œ 1 ’R# z  "3 Š R h
# r ‹ z$ “ h

4 3

y#

’

$

c

œ




'03 a9  z# b dz œ 491 ’9z  z3 “ $ œ 491 (27  9) œ 81

41 9




9
c # 9

œ

2 1 a9
c # b 9

(b) From part (a), each slice has the area œ

y# 4

#

#

#

#

h

ch

z# aR#
r# b “ h# R #

#

#

œ R#  Š R h
# r ‹ z# #

œ 21
R# h  "3 aR#  r# b h‘ œ 21 Š 2R3 h


1R# h
23 1r# h, the volume of the barrel. If r œ R, then V œ 21R# h which is the volume of a cylinder of

radius R and height 2h. If r œ 0 and h œ R, then V œ

4 3

1R$ which is the volume of a sphere.

79. We calculate the volume by the slicing method, taking slices parallel to the xy-plane. For fixed z, gives the ellipse

x# # Š zac ‹




y#

#

Š zbc ‹

œ 1. The area of this ellipse is 1 ˆaÈ cz ‰ ˆbÈ cz ‰ œ

the volume is given by V œ '0

h

Aœ

r# h 3 ‹

1abh c ,

1abz c

#

h

dz œ ’ 1abz 2c “ œ

as determined previously. Thus, V œ

! 1abh# c

1abh# c .

œ

" #

1abz c

x# a#




y# b#

œ

(see Exercise 77a). Hence

Now the area of the elliptic base when z œ h is

ˆ 1abh ‰h œ c

" #

z c

(base)(altitude), as claimed.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

809

810

Chapter 12 Vectors and the Geometry of Space

80. (a) For each fixed value of z, the hyperboloid x

–

#

a# Šc#
z# ‹ c#




y

—

–

#

b# Šc#
z# ‹ c#

h

h

1abh 3

Š2
1


(c) Am œ A ˆ h# ‰ œ œ œ

h# c# ‹

1ab c#

h
1ab # 6 1ab
c# a4c 1abh # # 3c# a3c
h b œ

81. y œ y" Ê

z c

œ

y#1 b#

œ

1ab # c # Šc

x# a#



y# b#



z# c#

œ 1 results in a cross-sectional ellipse

œ 1. The area of the cross-sectional ellipse (see Exercise 77a) is

ac#
z# b dz œ 1ab c#

(b) A! œ A(0) œ 1ab and Ah œ A(h) œ œ




—

A(z) œ 1 Š ac Èc#
z# ‹ Š bc Èc#
z# ‹ œ V œ '0 A(z) dz œ '0

x# a#




1abh 3 #

h 4

Š2


‹œ


h# b


1ab 4c#

1ab c#

1ab c#

ac#
z# b . The volume of the solid by the method of slices is

1ab c# #


c# z
"3 z$ ‘ h œ !

c#  h# c# ‹

œ

h 3


21ab


a4c#
h# b Ê

ac#
h# b‘ œ

1abh 6c#

h 6

1ab c#

ac#
h# b‘ œ

1abh 3c# h 3

1abh 3c#

a3c#
h# b

a3c#
h# b

(2A!
Ah )

(A!
4Am
Ah )

ac#
4c#
h#
c#
h# b œ

1abh 6c#

a6c#
2h# b

V from part (a)

, a parabola in the plane y œ y" Ê vertex when

#

#

ˆc# h
3" h$ ‰ œ

ac
h# b , from part (a) Ê V œ

#

a # 1 Ê Vertex Š0ß y" ß cy b# ‹ ; writing the parabola as x œ  c z
1 Ê Focus Š0ß y" ß cy b# 

1ab c#

a# y#1 b#

dz dx

œ 0 or c

dz dx

œ  2x a# œ 0 Ê x œ 0 #

#

a we see that 4p œ  ac Ê p œ  4c

a# 4c ‹

82. The curve has the general form Ax#
By#
Dxy
Gx
Hy
K œ 0 which is the same form as Eq. (1) in Section 10.3 for a conic section (including the degenerate cases) in the xy-plane. 83. No, it is not mere coincidence. A plane parallel to one of the coordinate planes will set one of the variables x, y, or z equal to a constant in the general equation Ax#
By#
Cz#
Dxy
Eyz
Fxz
Gx
Hy
Jz
K œ 0 for a quadric surface. The resulting equation then has the general form for a conic in that parallel plane. For example, setting y œ y" results in the equation Ax#
Cz#
Dw x
Ew z
Fxz
Gx
Jz
Kw œ 0 where Dw œ Dy" , Ew œ Ey" , and Kw œ K
By#1
Hy" , which is the general form of a conic section in the plane y œ y" by Section 10.3. 84. The trace will be a conic section. To see why, solve the plane's equation Ax
By
Cz œ 0 for one of the variables in terms of the other two and substitute into the equation Ax#
By#
Cz#
á
K œ 0. The result will be a second degree equation in the remaining two variables. By Section 10.3, this equation will represent a conic section. (See also the discussion in Exercises 82 and 83.) 85. z œ y#

86. z œ 1  y#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 12.6 Cylinders and Quadric Surfaces

811

87. z œ x#
y#

88. z œ x#
2y# (a)

(c)

(b)

(d)

89-94. Example CAS commands: Maple: with( plots ); eq := x^2/9 + y^2/36 = 1 - z^2/25; implicitplot3d( eq, x=-3..3, y=-6..6, z=-5..5, scaling=constrained, shading=zhue, axes=boxed, title="#89 (Section 12.6)" ); Mathematica: (functions and domains may vary): In the following chapter, you will consider contours or level curves for surfaces in three dimensions. For the purposes of plotting the functions of two variables expressed implicitly in this section, we will call upon the function ContourPlot3D.

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812

Chapter 12 Vectors and the Geometry of Space

To insert the stated function, write all terms on the same side of the equal sign and the default contour equating that expression to zero will be plotted. This built-in function requires the loading of a special graphics package. <
3. (a)  2a3b, 2a4b¡ œ  6, 8¡

(b) Éa1b2
a1b2 œ È2

(b) É62
a8b2 œ 10

4. (a)  5a2b, 5a5b¡ œ  10, 25¡ (b) É102
a25b2 œ È725 œ 5È29 5.

1 6

6. ¢

radians below the negative x-axis: ¢

È3 # ,

 "# £ [assuming counterclockwise].

È3 " # , #£

7. 2Š È

1 ‹a4i 42  12

 jb œ Š È817 i 

8. 5 3 21 4 2 ˆ 53 i
54 j‰ œ a3i  4jb Ɉ ‰  ˆ ‰

2 È17 j‹

5

9. length œ ¹È2i
È2j¹ œ È2
2 œ 2, È2i
È2j œ 2 Š È"2 i
10. length œ ki  jk œ È1
1 œ È2, i  j œ È2 Š È"2 i  11. t œ

1 2

" È2

" È2

5

j‹ Ê the direction is

" È2

i


j‹ Ê the direction is  È"2 i 

" È2

" È2

j

j

Ê v œ (2 sin 12 )i
ˆ2 cos 12 ‰ j œ 2i; length œ k2ik œ È4
0 œ 2; 2i œ 2aib Ê the direction is i

12. t œ ln 2 Ê v œ ˆeln 2 cosaln 2b  eln 2 sinaln 2b‰ i
ˆeln 2 sinaln 2b
eln 2 cosaln 2b‰ j œ a2 cosaln 2b  2 sinaln 2bb i
a2 sinaln 2b
2 cosaln 2bb j œ 2c acosaln 2b  sinaln 2bb i
asinaln 2b
cosaln 2bb j d length œ k2c acosaln 2b  sinaln 2bb i
asinaln 2b
cosaln 2bb j dk œ 2Éacosaln 2b  sinaln 2bb2
acosaln 2b
sinaln 2bb2 œ 2È2cos2 aln 2b
2sin2 aln 2b œ 2È2; 2c acosaln 2b  sinaln 2bb i
asinaln 2b
cosaln 2bb j d œ 2È2Š acosaln 2b
sinaln 2bb iÈ2asinaln 2b  cosaln 2bb j ‹ Ê direction œ

acosaln 2b
sinaln 2bb È2

i


asinaln 2b  cosaln 2bb È2

j

13. length œ k2i  3j
6kk œ È4
9
36 œ 7, 2i  3j
6k œ 7 ˆ 27 i  37 j
67 k‰ Ê the direction is

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 7

i  37 j
67 k

Chapter 12 Practice Exercises 14. length œ ki
2j  kk œ È1
4
1 œ È6, i
2j  k œ È6 Š È16 i


j

1 È6

â âi â 17. kvk œ È1
1 œ È2, kuk œ È4
1
4 œ 3, v † u œ 3, u † v œ 3, v ‚ u œ â 1 â â2

j 1 1

1 È6

15. 2

i


v kv k

16. 5

2 È6

œ2† v kv k

j

1 È6

k‹ Ê the direction is

k

4i
j  4k È4#  (
1)#  4#

œ 5 †

2 È6

œ2†

ˆ 35 ‰ i  ˆ 45 ‰ k Ɉ 35 ‰#

#  ˆ 45 ‰

4i
j  4k È33

œ 5 †

œ

8 È33

ˆ 35 ‰ iˆ 45 ‰k 9 16 É 25  25

i

2 È33

j


8 È33

k

œ 3i  4k â k â â 0 â œ  2i
2j  k , â 2 â

u ‚ v œ (v ‚ u) œ 2i  2j
k, kv ‚ uk œ È4
4
1 œ 3, ) œ cos
" Š kvvk†kuuk ‹ œ cos
" Š È"2 ‹ œ kuk cos ) œ

3 È2

813

, projv u œ Š kvvkk†uvk ‹ v œ

3 2

1 4

,

( i
j)

18. kvk œ È1#
1#
2# œ È6, kuk œ È(1)#
(1)# œ È2, v † u œ (1)(1)
(1)(0)
(2)(1) œ 3, â â j k â â i â â u † v œ 3, v ‚ u œ â 1 1 2 â œ i  j
k , u ‚ v œ (v ‚ u) œ i
j  k, â â â 1 0  1 â 3 3 kv ‚ uk œ È(1)#
(1)#
1# œ È3, ) œ cos
" Š kvvk†kuuk ‹ œ cos
" Š È
‹ œ cos
" Š È
12 ‹ 6 È2

œ cos
" Š

È3 # ‹

œ

51 6

È

, kuk cos ) œ È2 † Š
2 3 ‹ œ 

19. u œ Š kvvkk†uvk ‹ v
’u  Š kvvkk†uvk ‹ v“ œ

4 3

È6 2

, projv u œ Š kvvkk†uvk ‹ v œ


3 6

(i
j
2k) œ  "# (i
j
k)

(2i
j  k)

(i
j  5k)  43 (2i
j  k)‘ œ

4 3

(2i
j  k)  "3 (5i
j
11k),

where v † u œ 8 and v † v œ 6 20. u œ Š kvvkk†uvk ‹ v
’u  Š kvvkk†uvk ‹ v“ œ  13 (i  2j)

(i
j
k)  ˆ
31 ‰ (i  2j)‘ œ  13 (i  2j)
ˆ 43 i
53 j
k‰ , where v † u œ 1 and v † v œ 3 â âi â 21. u ‚ v œ â 1 â â1

j 0 1

â j âi â 22. u ‚ v œ â 1 1 â â1 1

â kâ â 0â œ k â 0â

â kâ â 0 â œ 2k â 0â

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

814

Chapter 12 Vectors and the Geometry of Space

23. Let v œ v" i
v# j
v$ k and w œ w" i
w# j
w$ k. Then kv  2wk # œ k(v" i
v# j
v$ k)  2(w" i
w# j
w$ k)k # œ k(v"  2w" )i
(v#  2w# )j
(v$  2w$ )kk # œ ˆÈ(v"  2w" )#
(v#  2w# )#
(v$  2w$ )# ‰

#

œ av#"
v##
v3# b  4(v" w"
v# w#
v$ w$ )
4 aw1#
w##
w3# b œ kvk #  4v † w
4 kwk # œ kvk #  4 kvk kwk cos )
4 kwk # œ 4  4(2)(3) ˆcos 13 ‰
36 œ 40  24 ˆ "# ‰ œ 40  12 œ 28 Ê kv  2wk œ È28 œ 2È7 â j â i â 4 24. u and v are parallel when u ‚ v œ 0 Ê â 2 â â 4  8 Ê 4a  40 œ 0 and 20  2a œ 0 Ê a œ 10

â k â â 5 â œ 0 Ê (4a  40)i
(20  2a)j
(0)k œ 0 â a â

â âi â 25. (a) area œ ku ‚ vk œ abs â 1 â â2

â j k â â 1 1 â œ k2i  3j  kk œ È4
9
1 œ È14 â 1 1 â â â 1 1 â â 1 â â 1 1 â œ 1(3
2)
1(1  6)  1(4
1) œ 1 (b) volume œ u † (v ‚ w) œ â 2 â â â 1  2 3 â

â âi â 26. (a) area œ ku ‚ vk œ abs â 1 â â0

â j kâ â 1 0 â œ kk k œ 1 â 1 0â â â â1 1 0â â â (b) volume œ u † (v ‚ w) œ â 0 1 0 â œ 1(1  0)
1(0  0)
0 œ 1 â â â1 1 1â

27. The desired vector is n ‚ v or v ‚ n since n ‚ v is perpendicular to both n and v and, therefore, also parallel to the plane. 28. If a œ 0 and b Á 0, then the line by œ c and i are parallel. If a Á 0 and b œ 0, then the line ax œ c and j are parallel. If a and b are both Á 0, then ax
by œ c contains the points ˆ ca ß !‰ and ˆ0ß bc ‰ Ê the vector ab ˆ ca i  bc j‰ œ c(bi  aj) and the line are parallel. Therefore, the vector bi  aj is parallel to the line ax
by œ c in every case. Ä 29. The line L passes through the point P(0ß 0ß 1) parallel to v œ i
j
k. With PS œ 2i
2j
k and â â j kâ â i Ä â â PS ‚ v œ â 2 2 1 â œ (2  1)i
(1  2)j
(2
2)k œ i  3j
4k, we find the distance â â â 1 1 1 â dœ

Ä ¹PS‚v¹ kv k

œ

È1  9  16 È1  1  1

œ

È26 È3

œ

È78 3

.

Ä 30. The line L passes through the point P(2ß 2ß 0) parallel to v œ i
j
k. With PS œ 2i
2j
k and â â j kâ â i Ä â â PS ‚ v œ â 2 2 1 â œ (2  1)i
(1
2)j
(2  2)k œ i
3j  4k, we find the distance â â â 1 1 1â dœ

Ä ¹PS‚v¹ kv k

œ

È1  9  16 È1  1  1

œ

È26 È3

œ

È78 3

.

31. Parametric equations for the line are x œ 1  3t, y œ 2, z œ 3
7t.

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Chapter 12 Practice Exercises Ä 32. The line is parallel to PQ œ 0i
j  k and contains the point P(1ß 2ß 0) Ê parametric equations are x œ 1, y œ 2
t, z œ t for 0 Ÿ t Ÿ 1. Ä 33. The point P(4ß 0ß 0) lies on the plane x  y œ 4, and PS œ (6  4)i
0j
(6
0)k œ 2i  6k with n œ i  j Ê dœ

Ä ¹n†PS¹ knk

œ ¹ È210100 ¹ œ

2 È2

œ È2.

Ä 34. The point P(0ß 0ß 2) lies on the plane 2x
3y
z œ 2, and PS œ (3  0)i
(0  0)j
(10
2)k œ 3i
8k with n œ 2i
3j
k Ê d œ

Ä ¹n†PS¹ knk

œ ¹ È640981 ¹ œ

14 È14

œ È14.

35. P(3ß 2ß 1) and n œ 2i
j  k Ê (2)(x  3)
(1)(y  (2))
(1)(z  1) œ 0 Ê 2x
y
z œ 5 36. P(1ß 6ß 0) and n œ i  2j
3k Ê (1)(x  (1))
(2)(y  6)
(3)(z  0) œ 0 Ê x  2y
3z œ 13 Ä Ä Ä Ä 37. P(1ß 1ß 2), Q(2ß 1ß 3) and R(1ß 2ß 1) Ê PQ œ i
2j
k, PR œ 2i
3j  3k and PQ ‚ PR â â j k â â i â â œ â 1 2 1 â œ 9i
j
7k is normal to the plane Ê (9)(x  1)
(1)(y
1)
(7)(z  2) œ 0 â â â 2 3  3 â Ê 9x
y
7z œ 4 Ä Ä Ä Ä 38. P(1ß 0ß 0), Q(0ß 1ß 0) and R(0ß 0ß 1) Ê PQ œ i
j, PR œ i
k and PQ ‚ PR â â j kâ â i â â œ â 1 1 0 â œ i
j
k is normal to the plane Ê (1)(x  1)
(1)(y  0)
(1)(z  0) œ 0 â â â 1 0 1 â Ê x
y
zœ1 39. ˆ0ß  "# ß  3# ‰ , since t œ  "# , y œ  "# and z œ  3# when x œ 0; ("ß 0ß 3), since t œ 1, x œ 1 and z œ 3 when y œ 0; (1ß 1ß 0), since t œ 0, x œ 1 and y œ 1 when z œ 0

40. x œ 2t, y œ t, z œ t represents a line containing the origin and perpendicular to the plane 2x  y  z œ 4; this line intersects the plane 3x  5y
2z œ 6 when t is the solution of 3(2t)  5(t)
2(t) œ 6 Ê t œ 23 Ê ˆ 43 ß  23 ß  23 ‰ is the point of intersection 41. n" œ i and n# œ i
j
È2k Ê the desired angle is cos
" Š knn""k†knn## k ‹ œ cos
" ˆ "# ‰ œ 42. n" œ i
j and n# œ j
k Ê the desired angle is cos
" Š knn""k†knn## k ‹ œ cos
" ˆ "# ‰ œ

1 3

1 3

â â j kâ âi â â 43. The direction of the line is n" ‚ n# œ â 1 2 1 â œ 5i  j  3k. Since the point (5ß 3ß 0) is on â â â 1 1 2 â both planes, the desired line is x œ 5
5t, y œ 3  t, z œ 3t. â âi â 44. The direction of the intersection is n" ‚ n# œ â 1 â â5 same as the direction of the given line.

â j k â â 2 2 â œ 6i  9j  12k œ 3(2i
3j
4k) and is the â 2  1 â

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815

816

Chapter 12 Vectors and the Geometry of Space

45. (a) The corresponding normals are n" œ 3i
'k and n# œ 2i
2j  k and since n" † n# œ (3)(2)
(0)(2)
(6)(1) œ 6
0  6 œ 0, we have that the planes are orthogonal â â âi j k â â â (b) The line of intersection is parallel to n" ‚ n# œ â 3 0 6 â œ 12i
15j
6k. Now to find a point in â â â 2 2 1 â the intersection, solve œ

3x
6z œ 1 3x
6z œ 1 Ê œ Ê 15x
12y œ 19 Ê x œ 0 and y œ 2x
2y  z œ 3 12x
12y  6z œ 18

"‰ Ê ˆ!ß 19 1# ß 6 is a point on the line we seek. Therefore, the line is x œ 12t, y œ

19 12


15t and z œ

" 6


6t.

â â j kâ âi â â 46. A vector in the direction of the plane's normal is n œ u ‚ v œ â 2 3 1 â œ 7i  3j  5k and P("ß 2ß 3) on â â â 1 1 2 â the plane Ê 7(x  1)  3(y  2)  5(z  3) œ 0 Ê 7x  3y  5z œ 14. 47. Yes; v † n œ (2i  4j
k) † (2i
j
0k) œ 2 † 2  4 † 1
1 † 0 œ 0 Ê the vector is orthogonal to the plane's normal Ê v is parallel to the plane Ä 48. n † PP!  0 represents the half-space of points lying on one side of the plane in the direction which the normal n points â âi Ä Ä â 49. A normal to the plane is n œ AB ‚ AC œ â 2 â â2

j 0 1

â k â Ä â 1 â œ i  2j  2k Ê the distance is d œ ¹ APn†n ¹ â 0 â

j)†(
i
2j
#k) œ ¹ (i  4È ¹ œ ¸
1
38  0 ¸ œ 3 144

Ä 50. P(0ß 0ß 0) lies on the plane 2x
3y
5z œ 0, and PS œ 2i
2j
3k with n œ 2i
3j
5k Ê Ä

4  6  15 d œ ¹ nk†PS nk ¹ œ ¹ È4  9  25 ¹ œ

25 È38

â âi â 51. n œ 2i  j  k is normal to the plane Ê n ‚ v œ â 2 â â1

â j k â â 1 1 â œ 0i  3j
3k œ 3j
3k is orthogonal â 1 1 â

to v and parallel to the plane 52. The vector B ‚ C is normal to the plane of B and C Ê A ‚ (B ‚ C) is orthogonal to A and parallel to the plane of B and C: â â â â j k â âi j k â â i â â â â B ‚ C œ â 1 2 1 â œ 5i
3j  k and A ‚ (B ‚ C) œ â 2 1 1 â œ 2i  3j
k â â â â â 1 1 2 â â 5 3  1 â Ê kA ‚ (B ‚ C)k œ È4
9
1 œ È14 and u œ " (2i  3j
k) is the desired unit vector. È14

â âi â 53. A vector parallel to the line of intersection is v œ n" ‚ n# œ â 1 â â1 Ê kvk œ È25
1
9 œ È35 Ê 2 Š kvvk ‹ œ

2 È35

j 2 1

â kâ â 1 â œ 5i  j  3k â 2â

(5i  j  3k) is the desired vector.

54. The line containing (0ß 0ß 0) normal to the plane is represented by x œ 2t, y œ t, and z œ t. This line intersects the plane 3x  5y
2z œ 6 when 3(2t)  5(t)
2(t) œ 6 Ê t œ 23 Ê the point is ˆ 43 ß  23 ß  23 ‰ . Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

19 12

Chapter 12 Practice Exercises 55. The line is represented by x œ 3
2t, y œ 2  t, and z œ 1
2t. It meets the plane 2x  y
2z œ 2 when 26 7‰ 2(3
2t)  (2  t)
2("
2t) œ 2 Ê t œ  89 Ê the point is ˆ 11 9 ß 9 ß 9 . â âi â 56. The direction of the intersection is v œ n" ‚ n# œ â 2 â â1

j 1 1

â k â â 1 â œ 3i  5j
k Ê ) œ cos
" Š kvvk†kiik ‹ â 2 â

œ cos
" Š È335 ‹ ¸ 59.5° 57. The intersection occurs when (3
2t)
3(2t)  t œ 4 Ê t œ 1 Ê the point is (1ß 2ß 1). The required line â â âi j k â â â must be perpendicular to both the given line and to the normal, and hence is parallel to â 2 2 1 â â â â 1 3 1 â œ 5i
3j
4k Ê the line is represented by x œ 1  5t, y œ 2
3t, and z œ 1
4t. 58. If P(aß bß c) is a point on the line of intersection, then P lies in both planes Ê a  2b
c
3 œ 0 and 2a  b  c
1 œ 0 Ê (a  2b
c
3)
k(2a  b  c
1) œ 0 for all k. â i j k ââ â Ä Ä â 3 2 4 ââ œ 26 59. The vector AB ‚ CD œ â 5 (2i
7j
2k) is normal to the plane and A(2ß 0ß 3) lies on the â 26 â 0  26 â 5 â 5 plane Ê 2(x
2)
7(y  0)
2(z  (3)) œ 0 Ê 2x
7y
2z
10 œ 0 is an equation of the plane. 60. Yes; the line's direction vector is 2i
3j  5k which is parallel to the line and also parallel to the normal 4i  6j
10k to the plane Ê the line is orthogonal to the plane. â â j kâ â i Ä Ä â â 61. The vector PQ ‚ PR œ â 2 1 3 â œ i  11j  3k is normal to the plane. â â â 3 0 1 â Ä Ä (a) No, the plane is not orthogonal to PQ ‚ PR . (b) No, these equations represent a line, not a plane.

Ä Ä (c) No, the plane (x
2)
11(y  1)  3z œ 0 has normal i
11j  3k which is not parallel to PQ ‚ PR . (d) No, this vector equation is equivalent to the equations 3y
3z œ 3, 3x  2z œ 6, and 3x
2y œ 4 Ê x œ  43  23 t, y œ t, z œ 1  t, which represents a line, not a plane. Ä Ä (e) Yes, this is a plane containing the point R(2ß 1ß 0) with normal PQ ‚ PR .

62. (a) The line through A and B is x œ 1
t, y œ t, z œ 1
5t; the line through C and D must be parallel and is L" : x œ 1
t, y œ 2  t, z œ 3
5t. The line through B and C is x œ 1, y œ 2
2s, z œ 3
4s; the line through A and D must be parallel and is L# : x œ 2, y œ 1
2s, z œ 4
4s. The lines L" and L# intersect at D(2ß 1ß 8) where t œ 1 and s œ 1. (b) cos ) œ

(2j  4k)†(i
j  5k) È20 È27

Ä Ä Ä †BC (c) Š BA Ä Ä ‹ BC œ BC†BC

18 20

œ

Ä BC œ

9 5

3 È15

Ä Ä (j
2k) where BA œ i  j
5k and BC œ 2j
4k

(d) area œ k(2j
4k) ‚ (i  j
5k)k œ k14i
4j  2kk œ 6È6 (e) From part (d), n œ 14i
4j  2k is normal to the plane Ê 14(x  1)
4(y  0)  2(z
1) œ 0 Ê 7x
2y  z œ 8. (f) From part (d), n œ 14i
4j  2k Ê the area of the projection on the yz-plane is kn † ik œ 14; the area of the projection on the xy-plane is kn † jk œ 4; and the area of the projection on the xy-plane is kn † kk œ 2.

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817

818

Chapter 12 Vectors and the Geometry of Space

â â i Ä Ä Ä â 63. AB œ 2i
j
k, CD œ i
4j  k, and AC œ 2i
j Ê n œ â 2 â â 1 †(
5i
j
9k) d œ ¹ (2i Èj)25 ¹œ  1  81

â j k â â 1 1 â œ 5i  j  9k Ê the distance is â 4 1 â

11 È107

â â i Ä Ä Ä â 64. AB œ 2i
4j  k, CD œ i  j
2k, and AC œ 3i
3j Ê n œ â 2 â â 1 j)†(7i  3j
2k) is d œ ¹ (
3i È349 ¹œ 94

â j k â â 4 1 â œ 7i
3j  2k Ê the distance â 1 2 â

12 È62

65. x#
y#
z# œ 4

66. x#
(y  1)#
z# œ 1

67. 4x#
4y#
z# œ 4

68. 36x#
9y#
4z# œ 36

69. z œ  ax#
y# b

70. y œ  ax#
z# b

71. x#
y# œ z#

72. x#
z# œ y#

73. x#
y#  z# œ 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 12 Additional and Advanced Exercises 74. 4y#
z#  4x# œ 4

75. y#  x#  z# œ 1

819

76. z#  x#  y# œ 1

CHAPTER 12 ADDITIONAL AND ADVANCED EXERCISES 1. Information from ship A indicates the submarine is now on the line L" : x œ 4
2t, y œ 3t, z œ  "3 t; information from ship B indicates the submarine is now on the line L# : x œ 18s, y œ 5  6s, z œ s. The current position of the sub is ˆ6ß 3ß  "3 ‰ and occurs when the lines intersect at t œ 1 and s œ 3" . The straight line path of the submarine contains both points P ˆ2ß 1ß  "3 ‰ and Q ˆ6ß 3ß  3" ‰ ; the line representing this path

is L: x œ 2
4t, y œ 1
4t, z œ  "3 . The submarine traveled the distance between P and Q in 4 minutes Ê Ä ¹PQ¹

È32 4

œ È2 thousand ft/min. In 20 minutes the submarine will move 20È2 thousand ft from Q along the line L Ê 20È2 œ È(2
4t  6)#
(1
4t  3)#
0# Ê 800 œ 16(t  1)#
16(t  1)# œ 32(t  1)# "‰ ˆ Ê (t  1)# œ 800 32 œ 25 Ê t œ 6 Ê the submarine will be located at 26ß 23ß  3 in 20 minutes. a speed of

4

œ

2. H# stops its flight when 6
110t œ 446 Ê t œ 4 hours. After 6 hours, H" is at P(246ß 57ß 9) while H# is at (446ß 13ß 0). The distance between P and Q is È(246  446)#
(57  13)#
(9  0)# ¸ 204.98 miles. At 150 mph, it would take about 1.37 hours for H" to reach H# . Ä Ä 3. Torque œ ¹ PQ ‚ F¹ Ê 15 ft-lb œ ¹ PQ ¹ kFk sin

1 #

œ

3 4

ft † kFk Ê kFk œ 20 lb

4. Let a œ i
j
k be the vector from O to A and b œ i
3j
2k be the vector from O to B. The vector v orthogonal to a and b Ê v is parallel to b ‚ a (since the rotation is closkwise). Now b ‚ a œ i
j  2k; proja b œ ˆ aa††ba ‰a œ 2i
2j
2k Ê a2, 2, 2b is the center of the circular path a1, 3, 2b takes Ê radius œ É12
a1b2
02 œ È2 Ê arc length per second covered by the point is 3# È2 units/sec œ kvk (velocity is constant). A unit vector in the direction of v is œ

" È6 i




" È6 j



2 È6 k

‚a 3È Ê v œ kvkŠ kbb‚ 2Š È"6 i
ak ‹ œ #

" È6 j



2 È 6 k‹

œ

È3 2 i




È3 2 j

b‚a kb‚ak

 È 3k

5. (a) If P(xß yß z) is a point in the plane determined by the three points P" (x" ß y" ß z" ), P# (x# ß y# ß z# ) and Ä Ä Ä Ä Ä Ä P$ (x$ ß y$ ß z$ ), then the vectors PP" , PP# and PP$ all lie in the plane. Thus PP" † (PP# ‚ PP$ ) œ 0 â â â x "  x y "  y z"  z â â â Ê â x#  x y#  y z#  z â œ 0 by the determinant formula for the triple scalar product in Section 10.4. â â â x $  x y $  y z$  z â (b) Subtract row 1 from rows 2, 3, and 4 and evaluate the resulting determinant (which has the same value as the given determinant) by cofactor expansion about column 4. This expansion is exactly the determinant in part (a) so we have all points P(xß yß z) in the plane determined by P" (x" ß y" ß z" ), P# (x# ß y# ß z# ), and P$ (x$ ß y$ ß z$ ).

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820

Chapter 12 Vectors and the Geometry of Space

6. Let L" : x œ a" s
b" , y œ a# s
b# , z œ a$ s
b$ and L# : x œ c" t
d" , y œ c# t
d# , z œ c$ t
d$ . If L" ² L# , â â â â â a" c" b"  d" â â kc" c" b"  d" â â â â â then for some k, ai œ kci , i œ 1, 2, 3 and the determinant â a# c# b#  d# â œ â kc# c# b#  d# â œ 0, â â â â â a$ c$ b$  d$ â â kc$ c$ b$  d$ â since the first column is a multiple of the second column. The lines L" and L# intersect if and only if the Ú a s  c t
(b  d ) œ 0 " " " " system Û a# s  c# t
(b#  d# ) œ 0 has a nontrivial solution Í the determinant of the coefficients is zero. Ü a$ s  c$ t
(b$  d$ ) œ 0 Ä Ä Ä 7. (a) BD œ AD  AB Ä Ä Ä Ä Ä (b) AP œ AB
"# BD œ "# Š AB
AD ‹ Ä Ä Ä Ä (c) AC œ AB
AD, so by part (b), AP œ

" #

Ä AC

Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä Ä 8. Extend CD to CG so that CD œ DG. Then CG œ t CF œ CB
BG and t CF œ 3 CE
CA, since ACBG is a Ä Ä Ä parallelogram. If t CF  3 CE  CA œ 0, then t  3  1 œ 0 Ê t œ 4, since F, E, and A are collinear. Ä Ä Ä Ä Therefore, CG œ 4 CF Ê CD œ 2 CF Ê F is the midpoint of CD. Ä 9. If Q(xß y) is a point on the line ax
by œ c, then P" Q œ (x  x" )i
(y  y" )j , and n œ ai
bj is normal to the Ä x" )  b(y
y" )k line. The distance is ¹projn P" Q¹ œ ¹ [(x
x" )i È(y#
y"#)j]†(ai  bj) ¹ œ ka(x
È # # a b

œ

kax"  by"
ck È a#  b#

a b

, since c œ ax
by.

Ä 10. (a) Let Q(xß yß z) be any point on Ax
By
Cz  D œ 0. Let QP" œ (x  x" )i
(y  y" )j
(z  z" )k, and Ä Bj  Ck Bj  Ck n œ ÈAi  . The distance is ¹projn QP" ¹ œ ¹((x  x" )i
(y  y" )j
(z  z" )k) † Š ÈAi  ‹¹ # # # # # # œ

A B C kAx"  By"  Cz"
(Ax  By  Cz)k ÈA#  B#  C#

A B C

œ

kAx"  By"  Cz"
Dk ÈA#  B#  C#

.

(b) Since both tangent planes are parallel, one-half of the distance between them is equal to the radius of the sphere, i.e., r œ " k3
9k œ È3 (see also Exercise 17a). Clearly, the points (1ß 2ß 3) and (1ß 2ß 3) # È1  1  1

are on the line containing the sphere's center. Hence, the line containing the center is x œ 1
2t, y œ 2
4t, z œ 3
6t. The distance from the plane x
y
z  3 œ 0 to the center is È3 Ê

k(1  2t)  (2  4t)  (3  6t)
3k È1  1  1

œ È3 from part (a) Ê t œ 0 Ê the center is at (1ß 2ß 3). Therefore

an equation of the sphere is (x  1)#
(y  2)#
(z  3)# œ 3. 11. (a) If (x" ß y" ß z" ) is on the plane Ax
By
Cz œ D" , then the distance d between the planes is kAx"  By"  Cz"
D# k D# k œ kAikD"B
j  ÈA#  B#  C# Ckk , since Ax"
By"
Cz" œ D" , by d œ Èk412
9 6k 1 œ È614 k2(3)  (
1)(2)  2(
1)  4k  2(
1)
Dk œ k2(3)  (
1)(2) Ê D œ 8 or 4 Ê the È14 È14

dœ (b) (c)

Exercise 10(a).

desired plane is

2x  y
2x œ 8 (d) Choose the point (2ß 0ß 1) on the plane. Then

k3
D k È6

œ 5 Ê D œ 3 „ 5È6 Ê the desired planes are

x  2y
z œ 3
5È6 and x  2y
z œ 3  5È6. Ä Ä 12. Let n œ AB ‚ BC and D(xß yß z) be any point in the plane determined by A, B and C. Then the point D lies in Ä Ä Ä Ä this plane if and only if AD † n œ 0 Í AD † (AB ‚ BC ) œ 0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 12 Additional and Advanced Exercises â âi â 13. n œ i
2j
6k is normal to the plane x
2y
6z œ 6; v ‚ n œ â 1 â â1

821

â kâ â 1 â œ 4i  5j
k is parallel to the â 6â â â j kâ âi â â plane and perpendicular to the plane of v and n Ê w œ n ‚ (v ‚ n) œ â 1 2 6 â œ 32i
23j  13k is a â â â 4 5 1 â j 1 2

vector parallel to the plane x
2y
6z œ 6 in the direction of the projection vector projP v. Therefore, projP v œ projw v œ Šv †

w w kwk ‹ kwk

23
13 ‰ œ Š kvw†wk # ‹ w œ ˆ 3232# 23 wœ #  13#

42 1722

wœ

" 41

wœ

32 41

i


23 41

j

13 41

k

14. projz w œ projz v and w  projz w œ v
projz v Ê w œ (w  projz w)
projz w œ (v  projz v)
projz w œ v  2 projz v œ v  2 Š kvz†kz# ‹ z 15. (a) u ‚ v œ 2i ‚ 2j œ 4k Ê (u ‚ v) ‚ C œ 0 ; (u † w)v  (v † w)u œ 0v  0u œ 0; v ‚ w œ 4i Ê u ‚ (v ‚ w) œ 0; (u † w)v  (u † v)w œ 0v  0w œ 0 â â â â j k â j k â âi â i â â â â (b) u ‚ v œ â 1 1 1 â œ i
4j
3k Ê (u ‚ v) ‚ w œ â 1 4 3 â œ 10i  2j
6k ; â â â â â 2 1 2 â â 1 2  1 â (u † w)v  (v † w)u œ 4(2i
j  2k)  2(i  j
k) œ 10i  2j
6k; â â â â j k â j kâ â i âi â â â â v ‚ w œ â 2 1  2 â œ 3i
4j
5k Ê u ‚ ( v ‚ w ) œ â 1  1 1 â œ  9i  2j
7k ; â â â â â 1 2  1 â â3 4 5â (u † w)v  (u † v)w œ 4(2i
j  2k)  (1)(i
2j  k) œ 9i  2j
7k â â â â j kâ j k â âi âi â â â â (c) u ‚ v œ â 2 1 0 â œ i  2j  4k Ê (u ‚ v) ‚ w œ â 1 2 4 â œ 4i  6j
2k ; â â â â 2 â â 2 1 1 â â1 0 (u † w)v  (v † w)u œ 2(2i  j
k)  4(2i
j) œ 4i  6j
2k; â â â j kâ j âi â i â â â 1 v ‚ w œ â 2  1 1 â œ  2i  3j
k Ê u ‚ ( v ‚ w ) œ â 2 â â â â1 0 2â â 2  3

â kâ â 0 â œ i  2j  4k ; â 1â

(u † w)v  (u † v)w œ 2(2i  j
k)  3(i
2k) œ i  2j  4k â â â â j k â j k â â i â i â â â â (d) u ‚ v œ â 1 1 2 â œ i
3j
k Ê (u ‚ v) ‚ w œ â 1 3 1 â œ 10i  10k ; â â â â â 1 0  1 â â 2 4 2 â (u † w)v  (v † w)u œ 10(i  k)  0(i
j  2k) œ 10i  10k; â â â â j k â j k â â i âi â â â â v ‚ w œ â 1 0 1 â œ 4i  4j  4k Ê u ‚ (v ‚ w) œ â 1 1 2 â œ 12i  4j  8k; â â â â â 2 4 2 â â 4 4  4 â (u † w)v  (u † v)w œ 10(i  k)  1(2i
4j  2k) œ 12i  4j  8k 16. (a) u ‚ (v ‚ w)
v ‚ (w ‚ u)
w ‚ (u ‚ v) œ (u † w)v  (u † v)w
(v † u)w  (v † w)u
(w † v)u  (w † u)v œ 0 (b) [u † (v ‚ i)]i
[(u † (v ‚ j)]j
[(u † (v ‚ k)]k œ [(u ‚ v) † i]i
[(u ‚ v) † j]j
[(u ‚ v) † k]k œ u ‚ v (c) (u ‚ v) † (w ‚ r) œ u † [v ‚ (w ‚ r)] œ u † [(v † r)w  (v † w)r] œ (u † w)(v † r)  (u † r)(v † w) u†w v†w œº u†r v†r º 17. The formula is always true; u ‚ [u ‚ (u ‚ v)] † w œ u ‚ [(u † v)u  (u † u)v] † w œ [(u † v)u ‚ u  (u † u)u ‚ v] † w œ  kuk # u ‚ v † w œ  kuk # u † v ‚ w

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

822

Chapter 12 Vectors and the Geometry of Space

18. If u œ (cos !)i
(sin !)j and v œ (cos " )i
(sin " )j , where "  !, then u ‚ v œ ckuk kvk sin ("  !)d k â â j kâ â i â â œ â cos ! sin ! 0 â œ (cos ! sin "  sin ! cos " )k Ê sin ("  !) œ cos ! sin "  sin ! cos " , since â â â cos " sin " 0 â kuk œ 1 and kvk œ 1. 19. If u œ ai
bj and v œ ci
dj , then u † v œ kuk kvk cos ) Ê ac
bd œ Èa#
b# Èc#
d# cos ) Ê (ac
bd)# œ aa#
b# b ac#
d# b cos# ) Ê (ac
bd)# Ÿ aa#
b# b ac#
d# b , since cos# ) Ÿ 1. 20. w œ projv u œ Š kvuk†kvvk ‹ v and r œ u  w œ u  Š kvuk†kvvk ‹ v 21. ku
vk # œ (u
v) † (u
v) œ u † u
2u † v
v † v Ÿ kuk #
2 kuk kvk
kvk # œ akuk
kvkb# Ê ku
vk Ÿ kuk
kvk 22. Let ! denote the angle between w and u, and " the angle between w and v. Let a œ kuk and b œ kvk . Then cos ! œ

w †u kwk kuk

œ

and likewise, cos

# (av  bu)†u œ (av†kuwk kubuk †u) œ (av†kuwk kubuk †u) œ aav†kuwk aba b kwk kuk " œ u†vkwk ba . Since the angle between u and v is

œ

v†u  ba k wk

,

always Ÿ

1 #

and cos ! œ cos " , we have

that ! œ " Ê w bisects the angle between u and v . 23. (av
bu) † (bu  av) œ av † bu
bu † bu  av † av  bu † av œ bu † av
b# u † u  a# v † v  bu † av œ b# a#  a# b# œ 0, where a œ kuk and b œ kvk 24. If u œ ai
bj
ck, then u † u œ a#
b#
c#   0 and u † u œ 0 iff a œ b œ c œ 0. 25. (a) The vector from (0ß d) to (kdß 0) is rk œ kdi  dj Ê

œ

1 kr k k $

" d$ ak#  1b$Î#

Ê

rk kr k k $

œ

ki
j d# ak#  1b$Î#

. The

total force on the mass (0ß d) due to the masses Qk for k œ n, n
1, á , n  1, n is Fœ

GMm d#

( j )





GMm 5d#

GMm 2d#

Š iÈ
2j ‹


GMm 5d#

2i
j Š
È ‹
á
5

i
j Š 2È ‹
á
5

GMm an #  1 b d #

GMm an #  1 b d #

Š Ènni #
j 1 ‹


GMm 2d#

Š
Èi
2 j ‹


n i
j ŠÈ ‹ n#  1

The i components cancel, giving Fœ

GMm d#

kF k œ

Š 1 

GMm d#

2 2È 2 n

Œ1
!



á 

2 5È 5

2 ‹j an#  1b an#  1b"Î#

Ê the magnitude of the force is

2 . ai#  1b$Î#

iœ1

(b) Yes, it is finite: n lim kF k œ Ä_

GMm d#

_

Œ1
!

iœ1

ai#

2   1b$Î#

_

is finite since !

2 #  1b$Î# i a iœ1

converges.

t œ (xt † y)x t t  (xt † x)y t t œ (xt † x)y. t t This means that 26. (a) If xt † yt œ 0, then xt ‚ (xt ‚ y) xt Š yt œ xt
yt


" c#

†

" 1  É1 c

xt†xt c#

#

t b yt œ xt
1  a(xt † x)

orthogonal, then kxt Š ytk œ kxtk
1  kxtk
1  1 

c#  Éc%
c# kxtk #

c#  Éc%
c# kxtk

#

c#  Éc%
c# kxtk #

#  kytk . A calculation will show that

# # # #  c œ c . Since kytk  c, then kytk  c so

#

kxtk #

t Since xt and yt are  y.

#

kxtk #

#


c# kxtk #

#

kxtk #

#

kxtk # c # É c %

#  kytk  1 

kxtk # c#  Éc%
c# kxtk #

#  c . This means that

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 12 Additional and Advanced Exercises #

kxtk #

#

kxt Š ytk œ kxtk
1 

c#  Éc%
c# kxtk #

# #

 kytk  kxtk
1 

#

kxtk #

#

c#  Éc%
c# kxtk #

We now have kxt Š ytk #  c# , so kxt Š ytk  c. t This gives xt Š yt œ t œ 0. (b) If xt and yt are parallel, then xt ‚ (xt ‚ y)

xt  yt xt†yt 1 #

# #  c œc .

.

c

(i) If xt and yt have the same direction, then xt Š yt œ Since kytk  c, kxtk  c, we have kytk Š1  Ê kxtk
kytk  c


kxtk kytk c

œ c Š1


kxtk c

†

kxtk c ‹ kytk c ‹

xt  yt k tk

x c

1

†

k tk

 c Š1  Ê

y c

kxtk c ‹

kxtk  kytk 1

k tk

x c

†

and kxt Š ytk œ

k tk

y c

Ê kytk 

Ê kxtk


kxtk kytk c

kxtk
kytk 1


k tk k tk

x y c#

kytk kxtk c

k tk

x c

kxtk kytk c

k tk

y c

.

 c  kxtk

xtyt 1


k tk k tk

x y c#

. Since kxtk  c, we have kxtk Š1


 c
kytk Ê kxtk  kytk  c 

†

 c. This means that kxt Š ytk  c.

(ii) If xt and yt have opposite directions, then xt † yt œ  kxtk kytk and xt Š yt œ Assume kxtk   kytk , then kxt Š ytk œ

kxtk  kytk 1

œ c Š1 

kxtk kytk c# ‹

Ê

.

kytk c ‹

kxtk
kytk 1


k tk k tk

x y c#

 c Š1
 c.

This means that kxt Š ytk  c. A similar argument holds if kxtk  kytk . t (c) c lim xt Š yt œ xt
y. Ä_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

kytk c ‹

823

824

Chapter 12 Vectors and the Geometry of Space

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 13 VECTOR-VALUED FUNCTIONS AND MOTION IN SPACE 13.1 VECTOR FUNCTIONS 1. x œ t
1 and y œ t#  1 Ê y œ (x  1)#  1 œ x#  2x; v œ

dr dt

œ i
2tj Ê a œ

dv dt

œ 2j Ê v œ i
2j and a œ 2j

at t œ 1 #

2. x œ t#
1 and y œ 2t  1 Ê x œ ˆ y
# 1 ‰
" Ê x œ Ê v œ i
2j and a œ 2i at t œ 3. x œ et and y œ

2 9

e2t Ê y œ

2 9

" #

x# ; v œ

dr dt

" 4

(y
1)#
1; v œ

dr dt

œ 2ti
2j Ê a œ

dr dt

œ (2 sin 2t)i
(6 cos 2t)j Ê a œ

dv dt

œ (4 cos 2t)i
(12 sin 2t)j Ê v œ 6j and a œ 4i at t œ 0 dr dt

œ (cos t)i  (sin t)j and a œ

Ê for t œ 14 , v ˆ 14 ‰ œ a ˆ 14 ‰ œ  a ˆ 1# ‰ œ i

È2 #

i

È2 #

È2 #

i

È2 #

dv dt

œ (sin t)i  (cos t)j

j and

j ; for t œ 1# , v ˆ 1# ‰ œ j and

6. v œ ddtr œ ˆ2 sin #t ‰ i
ˆ2 cos #t ‰ j and a œ ddtv œ ˆ cos #t ‰ i
ˆ sin #t ‰ j Ê for t œ 1, v(1) œ 2i and 31 # , È2 # j

a(1) œ j ; for t œ a ˆ 3#1 ‰ œ

7. v œ

dr dt

È2 #

i

v ˆ 3#1 ‰ œ È2 i  È2 j and

œ (1  cos t)i
(sin t)j and a œ

dv dt

œ (sin t)i
(cos t)j Ê for t œ 1, v(1) œ 2i and a(1) œ j ; for t œ 3#1 , v ˆ 3#1 ‰ œ i  j and a ˆ 3#1 ‰ œ i

8. v œ

dr dt

œ i
2tj and a œ

dv dt

œ 2i

œ et i
49 e2t j Ê a œ et i
89 e2t j Ê v œ 3i
4j and a œ 3i
8j at t œ ln 3

4. x œ cos 2t and y œ 3 sin 2t Ê x#
"9 y# œ 1; v œ

5. v œ

dv dt

œ 2j Ê for t œ 1,

v(1) œ i  2j and a(1) œ 2j ; for t œ 0, v(0) œ i and a(0) œ 2j ; for t œ 1, v(1) œ i
2j and a(1) œ 2j

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

826

Chapter 13 Vector-Valued Functions and Motion in Space

9. r œ (t
1)i
at#  1b j
2tk Ê v œ

t# È2

10. r œ (1
t)i


i
2(1)j
2k 3

œ

v(1) kv(1)k

Direction:

j


t$ 3

œ

" 3

i
32 j
32 k Ê v(1) œ 3 ˆ 3" i

k Ê vœ

dr dt

œi


#

# # œ Ê1#
Š 2(1) È2 ‹
(1 ) œ 2; Direction:

œ 2 Š "# i


" È2

d# r dt# œ 2j ; Speed:
32 j
32 k‰

2t È2

j
t# k Ê a œ

v(1) kv(1)k

i


œ

È

2(1) 2

j
(1# )k #

d# r dt#

œ

" #

œ

Speed: ¸v ˆ 1# ‰¸ œ Ɉ2 sin

1 ‰# #

12. r œ (sec t)i
(tan t)j
43 tk Ê v œ v ˆ 16 ‰ ¸v ˆ 1 ‰¸

œ

6

1 ‰# 2

dr dt

4 #È5

k œ  È"5 i


2 È5

t# 2

k Ê vœ

dr dt

4 3

k

œ

" 3

1 È6

k Ê v ˆ 1# ‰ œ 2È5 Š È"5 i


14. r œ aet b i
(2 cos 3t)j
(2 sin 3t)k Ê v œ

dr dt

2 È5

k‹

d# r dt#

# # # tan 16 ‰
ˆsec# 16 ‰
ˆ 43 ‰ œ 2;

1 6

œ ˆ t
2 1 ‰ i
2tj
tk Ê a œ

k Ê v(1) œ È6 Š È"6 i


œ (2 cos t)i  (3 sin t)j ;

i
23 j
23 k Ê v ˆ 16 ‰ œ 2 ˆ 3" i
32 j
32 k‰

#

j


d# r dt#

v ˆ 1# ‰ ¸v ˆ 1 ‰¸ #


4# œ 2È5; Direction:

Speed: kv(1)k œ Ɉ 1
2 1 ‰
(2(1))#
1# œ È6; Direction: 2 È6

j
"# k Ê v(1)

œ (sec t tan t)i
asec# tb j
43 k Ê a œ

ˆsec 1 tan 1 ‰ i
ˆsec# 1 ‰ j
6 6 6 #

13. r œ (2 ln (t
1))i
t# j


i


" È2

i


œ asec t tan# t
sec$ tb i
a2 sec# t tan tb j ; Speed: ¸v ˆ 16 ‰¸ œ Ɉsec

" È6

j
2tk ; Speed: kv(1)k

œ (2 sin t)i
(3 cos t)j
4k Ê a œ

dr dt


ˆ3 cos

2 3 œ Š #È sin 1# ‹ i
Š #È cos 1# ‹ j
5 5

œ

2 È2

j
#" k‹

11. r œ (2 cos t)i
(3 sin t)j
4tk Ê v œ

Direction:

kv(1)k œ È1#
(2(1))#
2# œ 3;

œ i
2tj
2k Ê a œ

dr dt

2 È6

j


" È6

v(1) kv(1)k

d# r dt#

œ ’ (t
21)# “ i
2j
k ;

2

Š ‹ i
2(1)j
(1)k œ "
1 È6

k‹

œ aet b i  (6 sin 3t)j
(6 cos 3t)k Ê a œ

d# r dt#

œ aet b i  (18 cos 3t)j  (18 sin 3t)k ; Speed: kv(0)k œ Éae! b#
[6 sin 3(0)]#
[6 cos 3(0)]# œ È37; Direction:

v(0) kv(0)k

œ

ae! b i  6 sin 3(0)j
6 cos 3(0)k È37

œ  È"37 i


6 È37

k Ê v(0) œ È37 Š È"37 i


6 È37

k‹ #

15. v œ 3i
È3 j
2tk and a œ 2k Ê v(0) œ 3i
È3 j and a(0) œ 2k Ê kv(0)k œ Ê3#
ŠÈ3‹
0# œ È12 and ka(0)k œ È2# œ 2; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 16. v œ

È2 #

i
Š

È2 #

 32t‹ j and a œ 32j Ê v(0) œ

œ 1 and ka(0)k œ È(32)# œ 32; v(0) † a(0) œ Š 1 ‰ # 17. v œ ˆ t# 2t
1 ‰ i
ˆ t#
1 j
tat
1b

"Î#

È2 #

1 #

i


È2 #

È2 # ‹ (32)

#

j and a(0) œ 32j Ê kv(0)k œ ÊŠ

œ 16È2 Ê cos ) œ

2t
2 k and a œ ’  “ i  ’ at#
2t 1b# “ j
’ at#
1b#

16È2 1(32)

" “k at#
1b$Î#

œ

È2 #

2 3

(1
t)"Î# i  23 (1  t)"Î# j
3" k and a œ

a(0) œ

" 3

#

#

" 3

(1
t)"Î# i
3" (1  t)"Î# j Ê v(0) œ #

#

#

i
3" j Ê kv(0)k œ Ɉ 23 ‰
ˆ 23 ‰
ˆ 3" ‰ œ 1 and ka(0)k œ Ɉ 3" ‰
ˆ 3" ‰ œ

œ 0 Ê cos ) œ 0 Ê ) œ


Š

È2 # # ‹

Ê )œ

31 4

Ê v(0) œ j and

a(0) œ 2i
k Ê kv(0)k œ 1 and ka(0)k œ È2#
1# œ È5; v(0) † a(0) œ 0 Ê cos ) œ 0 Ê ) œ 18. v œ

È2 # # ‹

2 3 i È2 3

1 2

 32 j
3" k and

1 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

; v(0) † a(0) œ

2 9



2 9

Section 13.1 Vector Functions 19. v œ (1  cos t)i
(sin t)j and a œ (sin t)i
(cos t)j Ê v † a œ (sin t)(1  cos t)
(sin t)(cos t) œ sin t. Thus, v † a œ 0 Ê sin t œ 0 Ê t œ 0, 1, or 21 20. v œ (cos t)i
j  (sin t)k and a œ ( sin t)i  (cos t)k Ê v † a œ  sin t cos t
sin t cos t œ 0 for all t   0 21.

'01 ct$ i
7j
(t
1)kd dt œ ’ t4 “ " i
[7t] "! j
’ t2
t“ " k œ 4" i
7j
3# k

22.

'12
(6  6t)i
3Èt j
ˆ t4 ‰ k‘ dt œ c6t  3t# d #" i

2t$Î# ‘ #" j
c4t" d #" k œ 3i
Š4È2  2‹ j
2k

23.

'
11Î%Î% c(sin t)i
(1
cos t)j
asec# tb kd dt œ c cos td 1Î%1Î% i
ct
sin td 1Î%1Î% j
ctan td 1Î%1Î% k

%

!

#

œ Š 1
#2 24.

#

!

È2

‹ j
2k

'01Î3 c(sec t tan t)i
(tan t)j
(2 sin t cos t) kd dt œ '01Î3 [(sec t tan t)i
(tan t)j
(sin 2t)k] dt 1Î$

1Î$

1Î$

œ csec td ! i
c ln (cos t)d ! j

 "# cos 2t‘ ! k œ i
(ln 2)j
34 k 25.

'14 ˆ "t i
5 " t j
#"t k‰ dt œ

26.

'01 Š È 2

1  t#

i


È3 1
t#

% œ cln td %" i
c ln (5  t)d %" j

#" ln t‘ " k œ (ln 4)i
(ln 4)j
(ln 2)k "

"

k‹ dt œ c2 sin" td ! i
’È3 tan" t“ k œ 1i
!

27. r œ ' (ti  tj  tk) dt œ  t# i  #

#

t# #

j

#

t# #

1È3 4

k

k
C ; r(0) œ 0i  0j  0k
C œ i
2j
3k Ê C œ i
2j
3k #

Ê r œ Š t#
1‹ i
Š t#
2‹ j
Š t#
3‹ k $‰ # #
28. r œ ' c(180t)i
a180t  16t# b jd dt œ 90t# i
ˆ90t#  16 3 t j
C ; r(0) œ 90(0) i
90(0)  $ ‰ œ 100j Ê C œ 100j Ê r œ 90t# i
ˆ90t#  16 3 t
100 j

16 3

(0)$ ‘ j
C

29. r œ '
ˆ 3# (t
1)"Î# ‰ i
e
t j
ˆ t
" 1 ‰ k‘ dt œ (t
1)$Î# i  e
t j
ln (t
1)k
C ; r(0) œ (0
1)$Î# i  e
! j
ln (0
1)k
C œ k Ê C œ i
j
k Ê r œ
(t
1)$Î#  1‘ i
a1  e
t b j
[1
ln (t
1)]k

30. r œ ' cat$
4tb i
tj
2t# kd dt œ Š t4
2t# ‹ i
%

%

t# 2

j


2t$ 3

%

k
C ; r(0) œ ’ 04
2(0)# “ i


#

œ i
j Ê C œ i
j Ê r œ Š t4
2t#
1‹ i
Š t#
1‹ j
31.

0# 2

j


2(0)$ 3

k
C

2t$ 3 k

œ ' (32k) dt œ 32tk
C" ; ddtr (0) œ 8i
8j Ê 32(0)k
C" œ 8i
8j Ê C" œ 8i
8j Ê dr œ 8i
8j  32tk ; r œ ' (8i
8j  32tk) dt œ 8ti
8tj  16t# k
C# ; r(0) œ 100k

dr dt

dt

Ê 8(0)i
8(0)j  16(0)# k
C# œ 100k Ê C# œ 100k Ê r œ 8ti
8tj
a100  16t# b k 32.

dr dt

œ ' (i
j
k) dt œ (ti
tj
tk)
C" ;

Ê

dr dt

dr dt

(0) œ 0 Ê (0i
0j
0k)
C" œ 0 Ê C" œ 0 #

œ (ti
tj
tk) ; r œ ' (ti
tj
tk) dt œ  Š t# i
#

Ê  Š 0# i


0# #

j


0# #

t# #

j


t# #

k‹
C# ; r(0) œ 10i
10j
10k

k‹
C# œ 10i
10j
10k Ê C# œ 10i
10j
10k

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

827

828

Chapter 13 Vector-Valued Functions and Motion in Space #

#

#

Ê r œ Š t#
10‹ i
Š t#
10‹ j
Š t#
10‹ k 33. r(t) œ (sin t)i
at#  cos tb j
et k Ê v(t) œ (cos t)i
(2t
sin t)j
et k ; t! œ 0 Ê v(t0 ) œ i
k and r(t0 ) œ P! œ (0ß 1ß 1) Ê x œ 0
t œ t, y œ 1, and z œ 1
t are parametric equations of the tangent line 34. r(t) œ (2 sin t)i
a2 cos tb j
5tk Ê v(t) œ (2 cos t)i  (2 sin t)j
5k ; t! œ 41 Ê v(t0 ) œ 2i
5k and r(t0 ) œ P! œ (0ß 2ß 201) Ê x œ 0
2t œ 2t, y œ 2, and z œ 201
5t are parametric equations of the tangent line 35. r(t) œ (a sin t)i
aa cos tb j
btk Ê v(t) œ (a cos t)i  (a sin t)j
bk ; t! œ 21 Ê v(t0 ) œ ai
bk and r(t0 ) œ P! œ (0ß aß 2b1) Ê x œ 0
at œ at, y œ a, and z œ 21b
bt are parametric equations of the tangent line 36. r(t) œ (cos t)i
asin tb j
(sin 2t)k Ê v(t) œ ( sin t)i
(cos t)j
(2 cos 2t)k ; t! œ

1 #

Ê v(t0 ) œ i  2k and

r(t0 ) œ P! œ (0ß 1ß 0) Ê x œ 0  t œ t, y œ 1, and z œ 0  2t œ 2t are parametric equations of the tangent line 37. (a) v(t) œ (sin t)i
(cos t)j Ê a(t) œ (cos t)i  (sin t)j ; (i) kv(t)k œ È(sin t)#
(cos t)# œ 1 Ê constant speed; (ii) v † a œ (sin t)(cos t)  (cos t)(sin t) œ 0 Ê yes, orthogonal; (iii) counterclockwise movement; (iv) yes, r(0) œ i
0j (b) v(t) œ (2 sin 2t)i
(2 cos 2t)j Ê a(t) œ (4 cos 2t)i  (4 sin 2t)j; (i) kv(t)k œ È4 sin# 2t
4 cos# 2t œ 2 Ê constant speed; (ii) v † a œ 8 sin 2t cos 2t  8 cos 2t sin 2t œ 0 Ê yes, orthogonal; (iii) counterclockwise movement; (iv) yes, r(0) œ i
0j (c) v(t) œ  sin ˆt  1# ‰ i
cos ˆt  1# ‰ j Ê a(t) œ  cos ˆt  1# ‰ i  sin ˆt  1# ‰ j ; (i)

kv(t)k œ Ésin# ˆt  1# ‰
cos# ˆt  1# ‰ œ 1 Ê constant speed;

(ii)

v † a œ sin ˆt  1# ‰ cos ˆt  1# ‰  cos ˆt  1# ‰ sin ˆt  1# ‰ œ 0 Ê yes, orthogonal;

(iii) counterclockwise movement; (iv) no, r(0) œ 0i  j instead of i
0j (d) v(t) œ (sin t)i  (cos t)j Ê a(t) œ (cos t)i
(sin t)j ; (i) kv(t)k œ È(sin t)#
( cos t)# œ 1 Ê constant speed; (ii) v † a œ (sin t)(cos t)  (cos t)(sin t) œ 0 Ê yes, orthogonal; (iii) clockwise movement; (iv) yes, r(0) œ i  0j (e) v(t) œ (2t sin t)i
(2t cos t)j Ê a(t) œ (2 sin t
2t cos t)i
(2 cos t  2t sin t)j ; (i)

kv(t)k œ Éc a2t sin tb d#
(2t cos t)# œ È4t# asin# t
cos# tb œ 2ktk œ 2t, t   0

Ê variable speed; (ii) v † a œ 4 at sin# t
t# sin t cos tb
4 at cos# t  t# cos t sin tb œ 4t Á 0 in general Ê not orthogonal in general; (iii) counterclockwise movement; (iv) yes, r(0) œ i
0j 38. Let p œ 2i
2j
k denote the position vector of the point a2, 2, 1b and let, u œ

" È2

i

" È2

j and v œ

" È3

i


" È3

Then r(t) œ p
(cos t)u
(sin t)v. Note that (2ß 2ß 1) is a point on the plane and n œ i
j  2k is normal to the plane. Moreover, u and v are orthogonal unit vectors with u † n œ v † n œ 0 Ê u and v are parallel to the plane. Therefore, r(t) identifies a point that lies in the plane for each t. Also, for each t, (cos t)u
(sin t)v

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

j


" È3

k.

Section 13.1 Vector Functions is a unit vector. Starting at the point Š2


2

1 È2 ,

1 È2 ,

829

1‹ the vector ratb traces out a circle of radius 1 and

center (2ß 2ß 1) in the plane x
y  2z œ 2. 39.

dv dt

œ a œ 3i  j
k Ê v(t) œ 3ti  tj
tk
C" ; the particle travels in the direction of the vector

(4  1)i
(1  2)j
(4  3)k œ 3i  j
k (since it travels in a straight line), and at time t œ 0 it has speed 2 Ê v(0) œ

(3i  j
k) œ C" Ê

Ê r(t) œ Š 3# t#


6 È11

t‹ i  Š "# t#


Ê r(t) œ Š 3# t#


6 È11

t
1‹ i  Š "# t#


œ Š "# t#
40.

2 È9
1
1

dv dt

2 È11

2 È11

dr dt

œ v(t) œ Š3t


t‹ j
Š "# t#
2 È11

2 È11

6 È11 ‹ i

 Št


2 È11 ‹ j


Št


2 È11 ‹ k

t‹ k
C# ; r(0) œ i
2j
3k œ C#

t  2‹ j
Š "# t#


2 È11

t
3‹ k

t‹ (3i  j
k)
(i
2j
3k)

œ a œ 2i
j
k Ê v(t) œ 2ti
tj
tk
C" ; the particle travels in the direction of the vector

(3  1)i
(0  (1))j
(3  2)k œ 2i
j
k (since it travels in a straight line), and at time t œ 0 it has speed 2 Ê v(0) œ

2 È4
1
1

(2i
j
k) œ C" Ê

Ê r(t) œ Št#


4 È6

t‹ i
Š "# t#


Ê r(t) œ Št#


4 È6

t
1‹ i
Š "# t#


2 È6

dr dt

œ v(t) œ Š2t


t‹ j
Š "# t#
2 È6

2 È6

4 È6 ‹ i


Št


2 È6 ‹ j


Št


2 È6 ‹ k

t‹ k
C# ; r(0) œ i  j
2k œ C#

t  1‹ j
Š "# t#


2 È6

t
2‹ k œ Š "# t#


2 È6

t‹ (2i
j
k)
(i  j
2k)

41. The velocity vector is tangent to the graph of y# œ 2x at the point (#ß #), has length 5, and a positive i component. Now, y# œ 2x Ê 2y

dy dx

œ2 Ê

vector i
"# j Ê the velocity vector is v œ

dy dx ¹ Ð#ß#Ñ

5 É1


" 4

œ

2 # †2

œ

ˆi
"# j‰ œ

" #

Œ

Ê the tangent vector lies in the direction of the

5 " È È È 5 ˆ i
# j‰ œ 2 5 i
5 j #




42. (a)

(b) v œ (1  cos t)i
(sin t)j and a œ (sin t)i
(cos t)j ; kvk# œ (1  cos t)#
sin# t œ 2  2 cos t Ê kvk# is at a max when cos t œ 1 Ê t œ 1, 31, 51, etc., and at these values of t, kvk# œ 4 Ê max kvk œ È4 œ 2; kvk# is at a min when cos t œ 1 Ê t œ 0, 21, 41, etc., and at these values of t, kvk# œ 0 Ê min kvk œ 0; kak# œ sin# t
cos# t œ 1 for every t Ê max kak œ min kak œ È1 œ 1 43. v œ (3 sin t)j
(2 cos t)k and a œ (3 cos t)j  (2 sin t)k ; kvk# œ 9 sin# t
4 cos# t Ê dtd ˆkvk# ‰ œ 18 sin t cos t  8 cos t sin t œ 10 sin t cos t; dtd ˆkvk# ‰ œ 0 Ê 10 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0 Ê t œ 0, 1 or t œ 1 , 31 . When t œ 0, 1, kvk# œ 4 Ê kvk œ È4 œ 2; when t œ 1 , 31 , kvk œ È9 œ 3. #

#

#

#

Therefore max kvk is 3 when t œ 1# , 3#1 , and min kvk œ 2 when t œ 0, 1. Next, kak# œ 9 cos# t
4 sin# t Ê dtd ˆkak# ‰ œ 18 cos t sin t
8 sin t cos t œ 10 sin t cos t; dtd ˆkak# ‰ œ 0 Ê 10 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0 Ê t œ 0, 1 or t œ 1# ,

31 # .

When t œ 0, 1, kak# œ 9 Ê kak œ 3; when t œ 1# ,

Therefore, max kak œ 3 when t œ 0, 1, and min kak œ 2 when t œ 1# ,

31 # .

31 # ,

kak# œ 4 Ê kak œ 2.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

830

Chapter 13 Vector-Valued Functions and Motion in Space

44. (a) r(t) œ (r! cos ))i
(r! sin ))j , and the distance traveled along the circle in time t is vt (rate times time) which equals the circular arc length r! ) Ê ) œ (b) v(t) œ

œ Šv sin

dr dt

vt r! ‹ i

#

œ  vr# ’Šr! cos !

vt r! ‹ i

(c) F œ ma Ê Š GmM ‹ r# !

r r!


Šv cos

vt r! ‹ j


Šr! sin

Ê r(t) œ Šr! cos

vt r!

Ê a(t) œ

vt r! ‹ j“

#

œ Š vr! cos

dv dt


Šr! sin

vt r! ‹ i

vt r! ‹ i

vt r! ‹ j

#


Š vr! sin

vt r! ‹ j

#

œ  vr# r(t) !

#

#

œ m Š vr# ‹ r Ê  GmM œ  mr#v Ê v# œ r$ !

!

!

GM r!

(d) T is the time for the satellite to complete one full orbit Ê vT œ circumference of circle Ê vT œ 21r! (e) Substitute v œ

2 1 r! T

into v# œ

dv dt

† v œ 2v †

41# r#! T#

Ê

GM r!

œ

41# r$! GM

Ê T# œ

GM r!

Ê T# is proportional to r!$ since

41 # GM

is a

constant 45.

d dt

(v † v) œ v †

46. (a)

d dt

dv dt




(u † v ‚ w ) œ

œ

du dt

du dt

dv dt

œ 2 † 0 œ 0 Ê v † v is a constant Ê kvk œ Èv † v is constant

† (v ‚ w )
u †

† (v ‚ w )
u †

dv dt

d dt

(v ‚ w ) œ

‚w
u†v‚

du dt

† (v ‚ w)
u † ˆ ddtv ‚ w
v ‚

dw ‰ dt

dw dt

(b) Each of the determinants is equivalent to each expression in Eq. 7 in part (a) because of the formual in Section 12.4 expressing the triple scalar product as a determinant. 47.

d dt

’r † Š ddtr ‚

d# r dt# ‹“

œ

dr dt

† Š ddtr ‚

d# r dt# ‹

#


r † Š ddt#r ‚

d# r dt# ‹


r † Š ddtr ‚

d$ r dt$ ‹

d$ r dt$ ‹ ,

œ r † Š ddtr ‚

since A † (A ‚ B) œ 0

and A † (B ‚ B) œ 0 for any vectors A and B 48. u œ C œ ai
bj
ck with a, b, c real constants Ê

du dt

œ

da dt

i


49. (a) u œ f(t)i
g(t)j
h(t)k Ê cu œ cf(t)i
cg(t)j
ch(t)k Ê œ

c Š df dt

i


dg dt

j


dh dt

k‹ œ c

df dt

dc dt

k œ 0 i
0 j
0k œ 0

(cu) œ c

d dt

df dt

i
c

dg dt

j
c

dh dt

k

du dt

(b) f u œ f f(t)i
f g(t)j
f h(t)k Ê œ

j


db dt

d dt

(f u) œ ’ ddtf f(t)
f

[f(t)i
g(t)j
h(t)k]
f ’ df dt i


dg dt

j


dh dt

k“ œ

df dt “ i df dt


’ ddtf g(t)
f

u
f

dg dt “ j


’ ddtf h(t)
f

dh dt “ k

du dt

50. Let u œ f" (t)i
f# (t)j
f$ (t)k and v œ g" (t)i
g# (t)j
g$ (t)k. Then u
v œ [f" (t)
g" (t)]i
[f# (t)
g# (t)]j
[f$ (t)
g$ (t)]k Ê dtd (u
v) œ [f"w (t)
gw" (t)]i
[f#w (t)
gw# (t)]j
[f$w (t)
gw$ (t)]k œ [f"w (t)i
f#w (t)j
f$w (t)k]
[gw" (t)i
gw# (t)j
gw$ (t)k] œ

du dt




dv dt

;

u  v œ [f" (t)  g" (t)]i
[f# (t)  g# (t)]j
[f$ (t)  g$ (t)]k Ê dtd (u  v) œ [f"w (t)  gw" (t)]i
[f#w (t)  gw# (t)]j
[f$w (t)  gw$ (t)]k œ [f"w (t)i
f#w (t)j
f$w (t)k]  [gw" (t)i
gw# (t)j
gw$ (t)k] œ

du dt



dv dt

51. Suppose r is continuous at t œ t! . Then lim r(t) œ r(t! ) Í lim [f(t)i
g(t)j
h(t)k] t Ä t!

t Ä t!

œ f(t! )i
g(t! )j
h(t! )k Í lim f(t) œ f(t! ), lim g(t) œ g(t! ), and lim h(t) œ h(t! ) Í f, g, and h are t Ä t!

t Ä t!

t Ä t!

continuous at t œ t! . â â i â 52. lim [r" (t) ‚ r# (t)] œ lim â f" (t) t Ä t! t Ä t! â â g" (t)

j f# (t) g# (t)

â â i k â ââ â â lim f" (t) f$ (t) â œ â t Ä t! â g$ (t) â ââ lim g" (t) tÄt !

j lim f# (t)

t Ä t!

lim g# (t)

t Ä t!

k lim

t Ä t!

lim

t Ä t!

â â f$ (t) ââ â g$ (t) ââ

œ lim r" (t) ‚ lim r# (t) œ A ‚ B t Ä t!

t Ä t!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.1 Vector Functions 53. rw (t! ) exists Ê f w (t! )i
gw (t! )j
hw (t! )k exists Ê f w (t! ), gw (t! ), hw (t! ) all exist Ê f, g, and h are continuous at t œ t! Ê r(t) is continuous at t œ t! 54. (a)

'ab kr(t) dt œ 'ab [kf(t)i
kg(t)j
kh(t)k] dt œ 'ab [kf(t)] dt i


'a [kg(t)] dt j
'a [kh(t)] dt k b

b

œ k Œ'a f(t) dt i
'a g(t) dt j
'a h(t) dt k œ k 'a r(t) dt b

(b)

b

b

b

'ab [r" (t) „ r# (t)] dt œ 'ab acf" (t)i
g" (t)j
h" (t)kd „ cf# (t)i
g# (t)j
h# (t)kdb dt b œ 'a acf" (t) „ f# (t)d i
[g" (t) „ g# (tb] j
[h" (t) „ h# (t)] k) dt b b b œ 'a cf" (t) „ f# (t)d dt i
'a cg" (t) „ g# (t)d dt j
'a ch" (t) „ h# (t)d dt k œ ”'a f" (t) dt i „ 'a f# (t) dt i•
”'a g" (t) dt j „ 'a g# (t) dt j •
”'a h" (t) dt k „ 'a h# (t) dt k• b

b

b

b

b

b

œ 'a r" (t) dt „ 'a r# (t) dt b

b

(c) Let C œ c" i
c# j
c$ k. Then 'a C † r(t) dt œ 'a cc" f(t)
c# g(t)
c$ h(t)d dt b

b

'ab f(t) dt
c# 'ab g(t) dt
c$ 'ab h(t) dt = C † 'ab r(t) dt; 'ab C ‚ r(t) dt œ 'ab cc# h(t)  c$ g(t)d i
cc$ f(t)  c" h(t)d j
cc" g(t)  c# f(t)d k dt œ c"

œ ”c#

'ab

h(t) dt  c$

'ab

g(t) dt• i
”c$

'ab

f(t) dt  c"

'ab h(t) dt• j
”c" 'ab g(t) dt  c# 'ab f(t) dt• k

œ C ‚ 'a r(t) dt b

55. (a) Let u and r be continuous on [aß b]. Then lim u(t)r(t) œ lim [u(t)f(t)i
u(t)g(t)j
u(t)h(t)k] t Ä t!

t Ä t!

œ u(t! )f(t! )i
u(t! )g(t! )j
u(t! )h(t! )k œ u(t! )r(t! ) Ê ur is continuous for every t! in [aß b]. (b) Let u and r be differentiable. Then dtd (ur) œ dtd [u(t)f(t)i
u(t)g(t)j
u(t)h(t)k] dg df ‰ du dh ‰ ˆ du œ ˆ du dt f(t)
u(t) dt i
Š dt g(t)
u(t) dt ‹j
dt h(t)
u(t) dt k df œ [f(t)i
g(t)j
h(t)k] du dt
u(t) Š dt i


dg dt

j


dh dt

dr k‹ œ r du dt
u dt

56. (a) If R" (t) and R# (t) have identical derivatives on I, then œ

d R# dt

Ê

df" dt

œ

df# dt

,

dg" dt

œ

dg# dt

,

dh" dt

œ

dh# dt

d R" dt

œ

df" dt

i


dg" dt

j


dh" dt

kœ

df# dt

i


dg# dt

j


dh# dt

k

Ê f" (t) œ f# (t)
c" , g" (t) œ g# (t)
c# , h" (t) œ h# (t)
c$

Ê f" (t)i
g" (t)j
h" (t)k œ [f# (t)
c" ]i
[g# (t)
c# ]j
[h# (t)
c$ ]k Ê R" (t) œ R# (t)
C, where C œ c" i
c# j
c$ k. (b) Let R(t) be an antiderivative of r(t) on I. Then Rw (t) œ r(t). If U(t) is an antiderivative of r(t) on I, then Uw (t) œ r(t). Thus Uw (t) œ Rw (t) on I Ê U(t) œ R(t)
C. 57.

'at r(7 ) d7 œ dtd 'at [f(7 )i
g(7 )j
h(7 )k] d7 œ dtd 'at f(7 ) d7 i
dtd 'at g(7 ) d7 j
dtd 'at h(7 ) d7 k t t œ f(t)i
g(t)j
h(t)k œ r(t). Since dtd 'a r(7 ) d7 œ r(t), we have that 'a r(7 ) d7 is an antiderivative of t a r. If R is any antiderivative of r, then R(t) œ 'a r(7 ) d7
C by Exercise 56(b). Then R(a) œ 'a r(7 ) d7
C t b œ 0
C Ê C œ R(a) Ê 'a r(7 ) d7 œ R(t)  C œ R(t)  R(a) Ê 'a r(7 ) d7 œ R(b)  R(a). d dt

58-61. Example CAS commands: Maple: > with( plots ); r := t -> [sin(t)-t*cos(t),cos(t)+t*sin(t),t^2];

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

831

832

Chapter 13 Vector-Valued Functions and Motion in Space t0 := 3*Pi/2; lo := 0; hi := 6*Pi; P1 := spacecurve( r(t), t=lo..hi, axes=boxed, thickness=3 ): display( P1, title="#58(a) (Section 13.1)" ); Dr := unapply( diff(r(t),t), t ); # (b) Dr(t0); # (c) q1 := expand( r(t0) + Dr(t0)*(t-t0) ); T := unapply( q1, t ); P2 := spacecurve( T(t), t=lo..hi, axes=boxed, thickness=3, color=black ): display( [P1,P2], title="#58(d) (Section 13.1)" );

62-63. Example CAS commands: Maple: a := 'a'; b := 'b'; r := (a,b,t) -> [cos(a*t),sin(a*t),b*t]; Dr := unapply( diff(r(a,b,t),t), (a,b,t) ); t0 := 3*Pi/2; q1 := expand( r(a,b,t0) + Dr(a,b,t0)*(t-t0) ); T := unapply( q1, (a,b,t) ); lo := 0; hi := 4*Pi; P := NULL: for a in [ 1, 2, 4, 6 ] do P1 := spacecurve( r(a,1,t), t=lo..hi, thickness=3 ): P2 := spacecurve( T(a,1,t), t=lo..hi, thickness=3, color=black ): P := P, display( [P1,P2], axes=boxed, title=sprintf("#62 (Section 13.1)\n a=%a",a) ); end do: display( [P], insequence=true ); 58-63. Example CAS commands: Mathematica: (assigned functions, parameters, and intervals will vary) The x-y-z components for the curve are entered as a list of functions of t. The unit vectors i, j, k are not inserted. If a graph is too small, highlight it and drag out a corner or side to make it larger. Only the components of r[t] and values for t0, tmin, and tmax require alteration for each problem. Clear[r, v, t, x, y, z] r[t_]={ Sin[t]  t Cos[t], Cos[t]
t Sin[t], t2} t0= 31 / 2; tmin= 0; tmax= 61; ParametricPlot3D[Evaluate[r[t]], {t, tmin, tmax}, AxesLabel Ä {x, y, z}]; v[t_]= r'[t] tanline[t_]= v[t0] t
r[t0] ParametricPlot3D[Evaluate[{r[t], tanline[t]}], {t, tmin, tmax}, AxesLabel Ä {x, y, z}]; For 62 and 63, the curve can be defined as a function of t, a, and b. Leave a space between a and t and b and t. Clear[r, v, t, x, y, z, a, b] r[t_,a_,b_]:={Cos[a t], Sin[a t], b t} t0= 31 / 2; tmin= 0; tmax= 41; v[t_,a_,b_]= D[r[t, a, b], t] tanline[t_,a_,b_]=v[t0, a, b] t
r[t0, a, b] pa1=ParametricPlot3D[Evaluate[{r[t, 1, 1], tanline[t, 1, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}];

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.2 Modeling Projectile Motion

833

pa2=ParametricPlot3D[Evaluate[{r[t, 2, 1], tanline[t, 2, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; pa4=ParametricPlot3D[Evaluate[{r[t, 4, 1], tanline[t, 4, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; pa6=ParametricPlot3D[Evaluate[{r[t, 6, 1], tanline[t, 6, 1]}], {t,tmin, tmax}, AxesLabel Ä {x, y, z}]; Show[GraphicsArray[{pa1, pa2, pa4, pa6}]] 13.2 MODELING PROJECTILE MOTION m‰ 1. x œ (v! cos !)t Ê (21 km)ˆ 1000 œ (840 m/s)(cos 60°)t Ê t œ 1 km v#! g

2. R œ

21,000 m (840 m/s)(cos 60°)

œ 50 seconds

v#

! sin 2! and maximum R occurs when ! œ 45° Ê 24.5 km œ Š 9.8 m/s # ‹ (sin 90°)

Ê v! œ È(9.8)(24,500) m# /s# œ 490 m/s

(b)

(c)

v!# g

(500 m/s)# 9.8 m/s# (sin 90°) ¸ 25,510.2 m 5000 m x œ (v! cos !)t Ê 5000 m œ (500 m/s)(cos 45°)t Ê t œ (500 m/s)(cos 45°) ¸ 14.14 s; thus, " " # y œ (v! sin !)t  # gt Ê y ¸ (500 m/s)(sin 45°)(14.14 s)  # a9.8 m/s# b (14.14 s)# ¸ 4020 m !)# 45°))# ymax œ (v! sin œ ((5002 m/s)(sin ¸ 6378 m 2g a9.8 m/s# b

3. (a) t œ

2v! sin ! g

œ

2(500 m/s)(sin 45°) 9.8 m/s#

¸ 72.2 seconds; R œ

sin 2! œ

4. y œ y!
(v! sin !)t  "# gt# Ê y œ 32 ft
(32 ft/sec)(sin 30°)t  "# a32 ft/sec# b t# Ê y œ 32
16t  16t# ;

the ball hits the ground when y œ 0 Ê 0 œ 32
16t  16t# Ê t œ 1 or t œ 2 Ê t œ 2 sec since t  0; thus,

x œ (v! cos !) t Ê x œ (32 ft/sec)(cos 30°)t œ 32 Š

È3 # ‹ (2)

¸ 55.4 ft

5. x œ x!
(v! cos !)t œ 0
(44 cos 45°)t œ 22È2t and y œ y!
(v! sin !)t  "# gt# œ 6.5
(44 sin 45°)t  16t# œ 6.5
22È2t  16t# ; the shot lands when y œ 0 Ê t œ

22È2 „ È968
416 3#

¸ 2.135 sec since t  0; thus

x œ 22È2t ¸ Š22È2‹ (2.135) ¸ 66.43 ft 6. x œ 0
(44 cos 40°)t ¸ 33.706t and y œ 6.5
(44 sin 40°)t  16t# ¸ 6.5
28.283t  16t# ; y œ 0 Ê t¸

28.283
È(28.283)#
416 3#

¸ 1.9735 sec since t  0; thus x ¸ (33.706)(1.9735) ¸ 66.52 ft Ê the

difference in distances is about 66.52  66.43 œ 0.09 ft or about 1 inch 7. (a) R œ

v#! g

(b) 6m ¸

v#

# # # ! sin 2! Ê 10 m œ Š 9.8 m/s Ê v! ¸ 9.9 m/s; # ‹ (sin 90°) Ê v! œ 98 m s

(9.9 m/s)# 9.8 m/s#

(sin 2!) Ê sin 2! ¸ 0.59999 Ê 2! ¸ 36.87° or 143.12° Ê ! ¸ 18.4° or 71.6°

8. v! œ 5 ‚ 10' m/s and x œ 40 cm œ 0.4 m; thus x œ (v! cos !)t Ê 0.4m œ a5 ‚ 10' m/sb (cos 0°)t Ê t œ 0.08 ‚ 10' s œ 8 ‚ 10) s; also, y œ y!
(v! sin !)t  "# gt# #

Ê y œ a5 ‚ 10' m/sb (sin 0°) a8 ‚ 10) sb  "# a9.8 m/s# b a8 ‚ 10) sb œ 3.136 ‚ 10"% m or

3.136 ‚ 10"# cm. Therefore, it drops 3.136 ‚ 10"# cm. 9. R œ

10. v! œ

v#! g

v#

# # # ! sin 2! Ê 3(248.8) ft œ Š 32 ft/sec Ê v! ¸ 278.02 ft/sec ¸ 190 mph # ‹ (sin 18°) Ê v! ¸ 77,292.84 ft /sec

80È10 3

ft/sec and R œ 200 ft Ê 200 œ

È

#

Š 80 3 10 ‹

2! ¸ 115.8° Ê ! ¸ 57.9°; If ! ¸ 32.1‰ , ymax

(sin 2!) Ê sin 2! œ 0.9 Ê 2! ¸ 64.2° Ê ! ¸ 32.1°; or # È ’Š 80 3 10 ‹ (sin 32.1°)“ œ ¸ 31.4 ft. If ! ¸ 57.9‰ , ymax ¸ 79.7 ft  75 ft. In 2(32)

32

order to reach the cushion, the angle of elevation will need to be about 32.1°. At this angle, the circus performer will go

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

834

Chapter 13 Vector-Valued Functions and Motion in Space

31.4 ft into the air at maximum height and will not strike the 75 ft high ceiling. 11. x œ (v! cos !)t Ê 135 ft œ (90 ft/sec)(cos 30°)t Ê t ¸ 1.732 sec; y œ (v! sin !)t  "# gt#

Ê y ¸ (90 ft/sec)(sin 30°)(1.732 sec)  "# a32 ft/sec# b (1.732 sec)# Ê y ¸ 29.94 ft Ê the golf ball will clip

the leaves at the top 12. v! œ 116 ft/sec, ! œ 45°, and x œ (v! cos !)t Ê 369 œ (116 cos 45°)t Ê t ¸ 4.50 sec; also y œ (v! sin !)t  "# gt# Ê y œ (116 sin 45°)(4.50)  "# (32)(4.50)#

¸ 45.11 ft. It will take the ball 4.50 sec to travel 369 ft. At that time the ball will be 45.11 ft in the air and will hit the green past the pin. 13. We do part b first. " 315 # (b) x œ (v! cos !)t Ê 315 ft œ (v! cos 20°)t Ê v! œ t cos 20° ; also y œ (v! sin !)t  # gt 315 ‰ " # # # # Ê 34 ft œ ˆ t cos 20° (t sin 20°)  # (32)t Ê 34 œ 315 tan 20°  16t Ê t ¸ 5.04 sec Ê t ¸ 2.25 sec (a) v! œ

315 (2.25)(cos 20°) v#! g

¸ 149 ft/sec

14. R œ

v#! g

sin 2! œ

15. R œ

v#! g

sin 2! Ê 16,000 m œ

(2 sin ! cos !) œ

v#! g

(400 m/s)# 9.8 m/s#

[2 cos (90°  !) sin (90°  !)] œ

v#! g

[sin 2(90°  !)]

sin 2! Ê sin 2! œ 0.98 Ê 2! ¸ 78.5° or 2! ¸ 101.5° Ê ! ¸ 39.3°

or 50.7° 16. (a) R œ

(2v! )# g

sin 2! œ

4v#! g

v#

sin 2! œ 4 Š g! sin !‹ or 4 times the original range.

(b) Now, let the initial range be R œ Ê

(pv! )# g

v#! g

sin 2!. Then we want the factor p so that pv! will double the range

v#!

sin 2! œ 2 Š g sin 2!‹ Ê p# œ 2 Ê p œ È2 or about 141%. The same percentage will approximately

double the height:

apv0 sin !b2 2g

œ

2av0 sin !b2 2g

Ê p# œ 2 Ê p œ È2.

17. x œ x!
(v! cos !)t œ 0
(v! cos 40°)t ¸ 0.766 v! t and y œ y!
(v! sin !)t  "# gt# œ 6.5
(v! sin 40°)t  16t# ¸ 6.5
0.643 v! t  16t# ; now the shot went 73.833 ft Ê 73.833 œ 0.766 v! t Ê t ¸ #

when y œ 0 Ê 0 œ 6.5
(0.643)(96.383)  16 Š 96.383 v! ‹ Ê 0 ¸ 68.474 

148,635 v#!

96.383 v!

sec; the shot lands

Ê v! ¸ É 148,635 68.474

¸ 46.6 ft/sec, the shot's initial speed 18. ymax œ

(v! sin !)# 2g

Ê

3 4

ymax œ

3(v! sin !)# and 8g # #

y œ (v! sin !)t  "# gt# Ê

3(v! sin !)# 8g

œ (v! sin !)t  "# gt#

Ê 3(v! sin !)# œ (8gv! sin !)t  4g t Ê 4g# t#  (8gv! sin !)t
3(v! sin !)# œ 0 Ê 2gt  3v! sin ! œ 0 or sin ! ! 2gt  v! sin ! œ 0 Ê t œ 3v!2gsin ! or t œ v! 2g . Since the time it takes to reach ymax is tmax œ v! sin , g then the time it takes the projectile to reach

3 4

of ymax is the shorter time t œ

v! sin ! 2g

or half the time it takes

to reach the maximum height. 19.

dr dt

œ ' (gj) dt œ gtj
C" and

dr dt

(0) œ (v! cos !)i
(v! sin !)j Ê g(0)j
C" œ (v! cos !)i
(v! sin !)j

Ê C" œ (v! cos !)i
(v! sin !)j Ê ddtr œ (v! cos !)i
(v! sin !  gt)j ; r œ ' [(v! cos !)i
(v! sin !  gt)j] dt œ (v! t cos !)i
ˆv! t sin !  "# gt# ‰ j
C# and r(0) œ x! i
y! j Ê [v! (0) cos !]i

v! (0) sin !  "# g(0)# ‘ j
C# œ x! i
y! j Ê C# œ x! i
y! j Ê r œ (x!
v! t cos !)i
ˆy!
v! t sin !  "# gt# ‰ j Ê x œ x!
v! t cos ! and Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.2 Modeling Projectile Motion y œ y!
v! t sin !  "# gt# 20. From Example 3(b) in the text, v! sin ! œ È(68)(64) Ê v! sin 56.5° ¸ 65.97 Ê v! ¸ 79 ft/sec 21. The horizontal distance from Rebollo to the center of the cauldron is 90 ft Ê the horizontal distance to the nearest rim is x œ 90  "# (12) œ 84 Ê 84 œ x!
(v! cos !)t ¸ 0
Š v!90g sin ! ‹ t Ê 84 œ " #

(90)(32) È(68)(64)

t

#

Ê t œ 1.92 sec. The vertical distance at this time is y œ y!
(v! sin !)t  gt ¸ 6
È(68)(64) (1.92)  16(1.92)# ¸ 73.7 ft Ê the arrow clears the rim by 3.7 ft 22. The projectile rises straight up and then falls straight down, returning to the firing point. 23. Flight time œ 1 sec and the measure of the angle of elevation is about 64° (using a protractor) so that tœ Rœ

2v! sin ! g v#! g

Ê 1œ

Ê v! ¸ 17.80 ft/sec. Then ymax œ

2v! sin 64° 32

sin 2! Ê R œ

#

(17.80) 32

(17.80 sin 64°)# 2(32)

¸ 4.00 ft and

sin 128° ¸ 7.80 ft Ê the engine traveled about 7.80 ft in 1 sec Ê the engine

velocity was about 7.80 ft/sec 24. When marble A is located R units downrange, we have x œ (v! cos !)t Ê R œ (v! cos !)t Ê t œ

R v! cos ! #

. At

" R R that time the height of marble A is y œ y!
(v! sin !)t  "# gt# œ (v! sin !) Š v! cos ! ‹  # g Š v! cos ! ‹ #

R Ê y œ R tan !  "# g Š v# cos # ! ‹ . The height of marble B at the same time t œ !

R v! cos !

seconds is

#

R h œ R tan !  "# gt# œ R tan !  "# g Š v# cos # ! ‹ . Since the heights are the same, the marbles collide regardless !

of the initial velocity v! . 25. (a) At the time t when the projectile hits the line OR we have tan " œ yx ; x œ [v! cos (!  " )]t and y œ [v! sin (!  " )]t  "# gt#  0 since R is

below level ground. Therefore let kyk œ "# gt#  [v! sin (!  " )]t  0

 "# gt# (v! sin (!  " ))t‘  " gt  v sin (!  " )‘ œ # v! cos! (!  ") [v! cos (!  " )]t v! cos (!  " ) tan " œ "# gt  v! sin (!  " ) t œ 2v! sin (!  ")
2vg ! cos (!  ") tan " , which is the time

so that tan " œ Ê Ê

when the projectile hits the downhill slope. Therefore, x œ [v! cos (!  " )] ’ 2v! sin (!  ")
2vg ! cos (!  ") tan " “ œ maximized, then OR is maximized:

dx d!

œ

2v#! g

2v#! g

ccos# (!  " ) tan "
sin (!  " ) cos (!  " )d . If x is

[ sin 2(!  " ) tan "
cos 2(!  " )] œ 0

Ê  sin 2(!  " ) tan "
cos 2(!  " ) œ 0 Ê tan " œ cot 2(!  " ) Ê 2(!  " ) œ 90°  " Ê !  " œ "# (90°  " ) Ê ! œ "# (90°
" ) œ "# of nAOR. (b) At the time t when the projectile hits OR we have tan " œ yx ; x œ [v! cos (!
" )]t and y œ [v! sin (!
" )]t  "# gt#

v! sin (!
" )  "# gt‘ v! cos (!
" ) v! cos (!
" ) tan " œ v! sin (!
" )  "# gt t œ 2v! sin (!
")  2vg ! cos (!
") tan " , which is the

Ê tan " œ Ê Ê

cv! sin (!
" )d t  "# gt# [v! cos (!
" )]t

œ

time

when the projectile hits the uphill slope. Therefore,

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

835

836

Chapter 13 Vector-Valued Functions and Motion in Space x œ [v! cos (!
" )] ’ 2v! sin (!
")  2vg ! cos (!
") tan " “ œ maximized, then OR is maximized: ddx! œ

2v#! g

2v#! g

csin (!
" ) cos (!
" )  cos# (!
" ) tan " d . If x is

[cos 2(!
" )
sin 2(!
" ) tan " ] œ 0

Ê cos 2(!
" )
sin 2(!
" ) tan " œ 0 Ê cot 2(!
" )
tan " œ ! Ê cot 2(!
" ) œ  tan " œ tan (" ) Ê 2(!
" ) œ 90°  (" ) œ 90°
" Ê ! œ "# (90°  " ) œ "# of nAOR. Therefore v! would bisect

nAOR for maximum range uphill.

26. (a) ratb œ axatbbi
ayatbbj; where xatb œ a145 cos 23‰  14bt and yatb œ 2.5
a145 sin 23‰ bt  16t2 . (b) ymax œ

av0 sin !b2 2g


2.5 œ

a145sin 23‰ b2 64

v0 sin ! g


2.5 ¸ 52.655 feet, which is reached at t œ ‰

œ

145sin 23‰ 32

¸ 1.771 seconds.

(c) For the time, solve y œ 2.5
a145 sin 23 bt  16t œ 0 for t, using the quadratic formula tœ

145 sin 23‰
Éa145 sin 23‰ b2
160 32

2

¸ 3.585 sec. Then the range at t ¸ 3.585 is about x œ a145 cos 23‰  14ba3.585b

¸ 428.311 feet. (d) For the time, solve y œ 2.5
a145 sin 23‰ bt  16t2 œ 20 for t, using the quadratic formula tœ

145 sin 23‰
Éa145 sin 23‰ b2  1120 32 ‰

¸ 0.342 and 3.199 seconds. At those times the ball is about

xa0.342b œ a145 cos 23  14ba0.342b ¸ 40.860 feet from home plate and xa3.199b œ a145 cos 23‰  14ba3.199b ¸ 382.195 feet from home plate. (e) Yes. According to part (d), the ball is still 20 feet above the ground when it is 382 feet from home plate. 27. (a) (Assuming that "x" is zero at the point of impact:) ratb œ axatbbi
ayatbbj; where xatb œ a35 cos 27‰ bt and yatb œ 4
a35 sin 27‰ bt  16t2 . (b) ymax œ

av0 sin !b2 2g


4 œ

a35sin 27‰ b2 64


4 ¸ 7.945 feet, which is reached at t œ ‰

v0 sin ! g

œ

35sin 27‰ 32

¸ 0.497 seconds.

(c) For the time, solve y œ 4
a35 sin 27 bt  16t œ 0 for t, using the quadratic formula tœ

35 sin 27‰
Éa35 sin 27‰ b2
256 32

2

¸ 1.201 sec. Then the range is about xa1.201b œ a35 cos 27‰ ba1.201b

¸ 37.453 feet. (d) For the time, solve y œ 4
a35 sin 27‰ bt  16t2 œ 7 for t, using the quadratic formula tœ

35 sin 27‰
Éa35 sin 27‰ b2  192 32 ‰

¸ 0.254 and 0.740 seconds. At those times the ball is about

xa0.254b œ a35 cos 27 ba0.254b ¸ 7.921 feet and xa0.740b œ a35 cos 27‰ ba0.740b ¸ 23.077 feet the impact point, or about 37.453  7.921 ¸ 29.532 feet and 37.453  23.077 ¸ 14.376 feet from the landing spot. (e) Yes. It changes things because the ball won't clear the net (ymax ¸ 7.945). 28. The maximum height is y œ

(v! sin !)# #g

and this occurs for x œ

v#! #g

sin 2! œ

v#! sin ! cos ! g

. These equations describe

parametrically the points on a curve in the xy-plane associated with the maximum heights on the parabolic trajectories in terms of the parameter (launch angle) !. Eliminating the parameter !, we have x# œ œ

v%! sin# ! g#



v%! sin% ! g#

Ê x#
4 Šy  29.

d2 r dt2

œ

v#! 4g ‹

#

v#! g

œ

(2y)  (2y)# Ê x#
4y#  Š v!% 4g#

2v#! g ‹y

v%! sin# ! cos# ! g# v#

œ 0 Ê x#
4 ’y#  Š 2g! ‹ y


œ

ˆv%! sin# !‰ a1  sin# !b g#

v%! 16g# “

œ

v%! 4g#

, where x   0.


k ddtr œ gj Ê Patb œ k and Qatb œ gj Ê ' Patb dt œ kt Ê vatb œ e' Patb dt œ ekt Ê

dr dt

œ

1 vatb

' vatb Qatb dt

œ gekt ' ekt j dt œ gekt
ek j
C1 ‘ œ  gk j
Cekt , where C œ gC1 ; apply the initial condition: kt

dr dt ¹tœ0

œ av0 cos !bi
av0 sin !bj œ  gk j
C Ê C œ av0 cos !bi
ˆ gk
v0 sin !‰j

Ê

œ ˆv0 ekt cos !‰i
ˆ gk
ekt ˆ gk
v0 sin !‰‰j, r œ ' c ˆv0 ekt cos !‰i
ˆ gk
ekt ˆ gk
v0 sin !‰‰j ddt

dr dt

gt v œ ˆ k0 ekt cos !‰i
Š k 

eckt ˆ g k k


v0 sin !‰‹j
C2 ; apply the initial condition:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.3 Arc Length and the Unit Tangent Vector T v0 sin ! ‰ j
C2 Ê C2 œ ˆ vk0 cos !‰i
ˆ kg2 k ˆ vk0 ˆ1  ekt ‰sin !
kg2 ˆ1  kt  ekt ‰‰j

ra0b œ 0 œ ˆ vk0 cos !‰i
ˆ kg2  Ê ratb œ ˆ vk0 ˆ1  ekt ‰cos !‰i





837

v0 sin ! ‰ j k

152 ‰ 30. (a) ratb œ axatbbi
ayatbbj; where xatb œ ˆ 0.12 a1  e0.12t bacos 20‰ b and 152 ‰ 32 ‰ 0.12t yatb œ 3
ˆ 0.12 a1  e0.12t basin 20‰ b
ˆ 0.12 b 2 a1  0.12t  e

(b) Solve graphically using a calculator or CAS: At t ¸ 1.484 seconds the ball reaches a maximum height of about 40.435 feet. (c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.126 seconds. The range is 152 ‰ˆ about xa3.126b œ ˆ 0.12 1  e0.12a3.126b ‰acos 20‰ b ¸ 372.311 feet. (d) Use a graphing calculator or CAS to find that y œ 30 for t ¸ 0.689 and 2.305 seconds, at which times the ball is about xa0.689b ¸ 94.454 feet and xa2.305b ¸ 287.621 feet from home plate. (e) Yes, the batter has hit a home run since a graph of the trajectory shows that the ball is more than 14 feet above the ground when it passes over the fence. 1 ‰ 31. (a) ratb œ axatbbi
ayatbbj; where xatb œ ˆ 0.08 a1  e0.08t ba152 cos 20‰  17.6b and 152 ‰ 32 ‰ 0.08t yatb œ 3
ˆ 0.08 a1  e0.08t basin 20‰ b
ˆ 0.08 b 2 a1  0.08t  e

(b) Solve graphically using a calculator or CAS: At t ¸ 1.527 seconds the ball reaches a maximum height of about 41.893 feet. (c) Use a graphing calculator or CAS to find that y œ 0 when the ball has traveled for ¸ 3.181 seconds. The range is 1 ‰ˆ about xa3.181b œ ˆ 0.08 1  e0.08a3.181b ‰a152 cos 20‰  17.6b ¸ 351.734 feet. (d) Use a graphing calculator or CAS to find that y œ 35 for t ¸ 0.877 and 2.190 seconds, at which times the ball is about xa0.877b ¸ 106.028 feet and xa2.190b ¸ 251.530 feet from home plate. (e) No; the range is less than 380 feet. To find the wind needed for a home run, first use the method of part (d) to find that 1 ‰ˆ y œ 20 at t ¸ 0.376 and 2.716 seconds. Then define xawb œ ˆ 0.08 1  e0.08a2.716b ‰a152 cos 20‰
wb, and solve xawb œ 380 to find w ¸ 12.846 ft/sec. 13.3 ARC LENGTH AND THE UNIT TANGENT VECTOR T 1. r œ (2 cos t)i
(2 sin t)j
È5tk Ê v œ (2 sin t)i
(2 cos t)j
È5k #

Ê kvk œ Ê(2 sin t)#
(2 cos t)#
ŠÈ5‹ œ È4 sin# t
4 cos# t
5 œ 3; T œ œ ˆ 23 sin t‰ i
ˆ 23 cos t‰ j


È5 3

1

v kv k

1

k and Length œ '0 kvk dt œ '0 3 dt œ c3td 1! œ 31

2. r œ (6 sin 2t)i
(6 cos 2t)j
5tk Ê v œ (12 cos 2t)i
(12 sin 2t)j
5k Ê kvk œ È(12 cos 2t)#
(12 sin 2t)#
5# œ È144 cos# 2t
144 sin# 2t
25 œ 13; T œ ‰ ˆ 12 ‰ œ ˆ 12 13 cos 2t i  13 sin 2t j


5 13

1

1

v kv k

k and Length œ '0 kvk dt œ '0 13 dt œ c13td 1! œ 131 #

3. r œ ti
23 t$Î# k Ê v œ i
t"Î# k Ê kvk œ É1#
at"Î# b œ È1
t ; T œ ) and Length œ '0 È1
t dt œ
23 (1
t)$Î# ‘ ! œ 8

v kv k

œ

" È1
t

i


Èt È1
t

52 3

4. r œ (2
t)i  (t
1)j
tk Ê v œ i  j
k Ê kvk œ È1#
(1)#
1# œ È3 ; T œ

v kv k

œ

$

and Length œ '0 È3 dt œ ’È3t“ œ 3È3 3

k

!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" È3

i

1 È3

j


" È3

k

838

Chapter 13 Vector-Valued Functions and Motion in Space

5. r œ acos$ tb j
asin$ tb k Ê v œ a3 cos# t sin tb j
a3 sin# t cos tb k Ê kvk œ Éa3 cos# t sin tb#
a3 sin# t cos tb# œ Èa9 cos# t sin# tb acos# t
sin# tb œ 3 kcos t sin tk ; Tœ

v kv k

œ

3 cos# t sin t 3 kcos t sin tk 1Î2

Length œ '0

j


3 sin# t cos t 3 kcos t sin tk

k œ ( cos t)j
(sin t)k , if 0 Ÿ t Ÿ 1Î2

1Î2

3 kcos t sin tk dt œ '0 3 cos t sin t dt œ '0

3 #

1 #

, and 1Î#

sin 2t dt œ
 34 cos 2t‘ !

œ

3 #

6. r œ 6t$ i  2t$ j  3t$ k Ê v œ 18t# i  6t# j  9t# k Ê kvk œ Éa18t# b#
a6t# b#
a9t# b# œ È441t% œ 21t# ; Tœ

v kv k

œ

"8t# 21t#

i

6t# 21t#

j

7. r œ (t cos t)i
(t sin t)j


9t# 21t#

kœ

6 7

2È2 $Î# k 3 t

i  27 j  37 k and Length œ '1 21t# dt œ c7t$ d " œ 49 2

#

Ê v œ (cos t  t sin t)i
(sin t
t cos t)j
ŠÈ2 t"Î# ‹ k #

Ê kvk œ Ê(cos t  t sin t)#
(sin t
t cos t)#
ŠÈ2 t‹ œ È1
t#
2t œ È(t
1)# œ kt
1k œ t
1, if t   0; Tœ

v kv k

œ ˆ cos tt 
t1sin t ‰ i
ˆ sin tt

t1cos t ‰ j
Š

È2 t"Î# t
1 ‹ k

1

1

and Length œ '0 (t
1) dt œ ’ t2
t“ œ #

!

1# 2


1

8. r œ (t sin t
cos t)i
(t cos t  sin t)j Ê v œ (sin t
t cos t  sin t)i
(cos t  t sin t  cos t)j œ (t cos t)i  (t sin t)j Ê kvk œ È(t cos t)#
(t sin t)# œ Èt# œ ktk œ t if È2 Ÿ t Ÿ 2; T œ v

kv k

t‰ t‰ œ ˆ t cos i  ˆ t sin j œ (cos t)i  (sin t)j and Length œ 'È2 t dt œ ’ t2 “ t t 2

#

# È#

œ1

9. Let P(t! ) denote the point. Then v œ (5 cos t)i  (5 sin t)j
12k and 261 œ '0 È25 cos# t
25 sin# t
144 dt t!

œ '0 13 dt œ 13t! Ê t! œ 21, and the point is P(21) œ (5 sin 21ß 5 cos 21ß 241) œ (0ß 5ß 241) t!

10. Let P(t! ) denote the point. Then v œ (12 cos t)i
(12 sin t)j
5k and 131 œ '0 È144 cos# t
144 sin# t
25 dt œ '0 13 dt œ 13t! Ê t! œ 1, and the point is t!

t!

P(1) œ (12 sin (1)ß 12 cos (1)ß 51) œ (0ß 12ß 51) 11. r œ (4 cos t)i
(4 sin t)j
3tk Ê v œ (4 sin t)i
(4 cos t)j
3k Ê kvk œ È(4 sin t)#
(4 cos t)#
3# œ È25 œ 5 Ê s(t) œ '0 5 d7 œ 5t Ê Length œ s ˆ 1# ‰ œ t

51 #

12. r œ (cos t
t sin t)i
(sin t  t cos t)j Ê v œ (sin t
sin t
t cos t)i
(cos t  cos t
t sin t)j t œ (t cos t)i
(t sin t)j Ê kvk œ È(t cos t)#
(t cos t)# œ œ Èt# œ t, since 1# Ÿ t Ÿ 1 Ê s(t) œ '0 7 d7 œ Ê Length œ s(1)  s ˆ 1# ‰ œ

1# #



ˆ 1# ‰# #

œ

t# #

31 # 8

13. r œ aet cos tb i
aet sin tb j
et k Ê v œ aet cos t  et sin tb i
aet sin t
et cos tb j
et k Ê kvk œ Éaet cos t  et sin tb#
aet sin t
et cos tb#
aet b# œ œ È3e2t œ È3 et Ê s(t) œ '0 È3 e7 d7 t

œ È3 et  È3 Ê Length œ s(0)  s( ln 4) œ 0  ŠÈ3 e ln 4  È3‹ œ

3È 3 4

14. r œ (1
2t)i
(1
3t)j
(6  6t)k Ê v œ 2i
3j  6k Ê kvk œ È2#
3#
(6)# œ 7 Ê s(t) œ '0 7 d7 œ 7t t

Ê Length œ s(0)  s(1) œ 0  (7) œ 7

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.3 Arc Length and the Unit Tangent Vector T #

839

#

15. r œ ŠÈ2t‹ i
ŠÈ2t‹ j
a1  t# b k Ê v œ È2i
È2j  2tk Ê kvk œ ÊŠÈ2‹
ŠÈ2‹
(2t)# œ È4
4t# œ 2È1
t# Ê Length œ '0 2È1
t# dt œ ’2 Š 2t È1
t#
1

" #

"

ln Št
È1
t# ‹‹“ œ È2
ln Š1
È2‹ !

16. Let the helix make one complete turn from t œ 0 to t œ 21. Note that the radius of the cylinder is 1 Ê the circumference of the base is 21. When t œ 21, the point P is (cos 21ß sin 21ß 21) œ (1ß 0ß 21) Ê the cylinder is 21 units high. Cut the cylinder along PQ and flatten. The resulting rectangle has a width equal to the circumference of the cylinder œ 21 and a height equal to 21, the height of the cylinder. Therefore, the rectangle is a square and the portion of the helix from t œ 0 to t œ 21 is its diagonal.

17. (a) r œ (cos t)i
(sin t)j
("  cos t)k, 0 Ÿ t Ÿ 21 Ê x œ cos t, y œ sin t, z œ 1  cos t Ê x#
y# œ cos# t
sin# t œ 1, a right circular cylinder with the z-axis as the axis and radius œ 1. Therefore P(cos tß sin tß 1  cos t) lies on the cylinder x#
y# œ 1; t œ 0 Ê P(1ß 0ß 0) is on the curve; t œ 1# Ê Q(!ß 1ß 1) Ä Ä is on the curve; t œ 1 Ê R(1ß 0ß 2) is on the curve. Then PQ œ i
j
k and PR œ 2i
2k j k× Ô i Ä Ä Ê PQ ‚ PR œ 1 " " œ 2i
2k is a vector normal to the plane of P, Q, and R. Then the Õ 2 0 2 Ø plane containing P, Q, and R has an equation 2x
2z œ 2(1)
2(0) or x
z œ 1. Any point on the curve will satisfy this equation since x
z œ cos t
(1  cos t) œ 1. Therefore, any point on the curve lies on the intersection of the cylinder x#
y# œ 1 and the plane x
z œ 1 Ê the curve is an ellipse. (b) v œ ( sin t)i
(cos t)j
(sin t)k Ê kvk œ Èsin# t
cos# t
sin# t œ È1
sin# t Ê T œ kvvk œ

( sin t)i
(cos t)j
(sin t)k È1
sin# t

Ê T(0) œ j , T ˆ 1# ‰ œ

i
k È2

, T(1) œ j , T ˆ 3#1 ‰ œ

ik È2

(c) a œ ( cos t)i  (sin t)j
(cos t)k ; n œ i
k is normal to the plane x
z œ 1 Ê n † a œ  cos t
cos t œ 0 Ê a is orthogonal to n Ê a is parallel to the plane; a(0) œ i
k , a ˆ 1# ‰ œ j , a a1b œ i  k , ‰œj a ˆ 31 #

21

(d) kvk œ È1
sin# t (See part (b) Ê L œ '0 È1
sin# t dt (e) L ¸ 7.64 (by Mathematica) 18. (a) r œ (cos 4t)i
(sin 4t)j
4tk Ê v œ (4 sin 4t)i
(4 cos 4t)j
4k Ê kvk œ È(4 sin 4t)#
(4 cos 4t)#
4# 1Î2

œ È32 œ 4È2 Ê Length œ '0 4È2 dt œ ’4È2 t“

1Î# !

œ 21 È 2

(b) r œ ˆcos #t ‰ i
ˆsin #t ‰ j
#t k Ê v œ ˆ "# sin #t ‰ i
ˆ "# cos #t ‰ j
"# k # # # Ê kvk œ Ɉ "# sin #t ‰
ˆ "# cos #t ‰
ˆ "# ‰ œ É 4"


" 4

œ

È2 #

41

Ê Length œ '0

È2 #

dt œ ’

È2 2

t“

%1 !

œ 21È2

(c) r œ (cos t)i  (sin t)j  tk Ê v œ ( sin t)i  (cos t)j  k Ê kvk œ È( sin t)#
( cos t)#
(1)# œ È1
1 œ È2 Ê Length œ 'c21 È2 dt œ ’È2 t“ 0

! #1

œ 21 È 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

840

Chapter 13 Vector-Valued Functions and Motion in Space

19. nPQB œ nQOB œ t and PQ œ arc (AQ) œ t since PQ œ length of the unwound string œ length of arc (AQ); thus x œ OB
BC œ OB
DP œ cos t
t sin t, and y œ PC œ QB  QD œ sin t  t cos t

20. r œ acos t
t sin tbi
asin t
t cos tbj Ê v œ asin t
t cos t
sin tbi
acos t  atasin tb
cos tbbj œ at cos tbi
at sin tbj Ê kvk œ Éat cos tb2
at sin tb2 œ Èt2 œ ktk œ t, t   0 Ê T œ

v kv k

œ

t cos t t i




t sin t t j

œ cos t i
sin t j 13.4 CURVATURE AND THE UNIT NORMAL VECTOR N sin t ‰ È1#
( tan t)# œ Èsec# t œ ksec tk œ sec t, since 1. r œ ti
ln (cos t)j Ê v œ i
ˆ cos t j œ i  (tan t)j Ê kvk œ 1 1 v " tan t  #  t  # Ê T œ kvk œ ˆ sec t ‰ i  ˆ sec t ‰ j œ (cos t)i  (sin t)j ; ddtT œ ( sin t)i  (cos t)j

Ê ¸ ddtT ¸ œ È( sin t)#
( cos t)# œ 1 Ê N œ ,œ

1 kv k

†

¸ ddtT ¸

" sec t

œ

ˆ ddtT ‰ ¸ ddtT ¸

œ ( sin t)i  (cos t)j ;

† 1 œ cos t.

t tan t ‰ 2. r œ ln (sec t)i
tj Ê v œ ˆ secsec i
j œ (tan t)i
j Ê kvk œ È( tan t)#
1# œ Èsec# t œ ksec tk œ sec t, t 1 1 v tan since  #  t  # Ê T œ kvk œ ˆ sec tt ‰ i  ˆ sec1 t ‰ j œ (sin t)i
(cos t)j ; ddtT œ (cos t)i  (sin t)j

Ê ¸ ddtT ¸ œ È(cos t)#
( sin t)# œ 1 Ê N œ ,œ

1 kv k

† ¸ ddtT ¸ œ

" sec t

ˆ ddtT ‰ ¸ ddtT ¸

œ (cos t)i  (sin t)j ;

† 1 œ cos t.

3. r œ (2t
3)i
a5  t# b j Ê v œ 2i  2tj Ê kvk œ È2#
(2t)# œ 2È1
t# Ê T œ kvvk œ È 2 # i
2 1
t Í # # Í Í t " t " dT ¸ ¸ œ È "
# i  È t
# j ; ddtT œ i  j Ê œ
 $ $ $  ŠÈ1
t# ‹$  dt 1 t 1 t ŠÈ1
t# ‹ ŠÈ1
t# ‹ Ì  ŠÈ1
t# ‹  œ É a1
"t# b# œ ,œ

1 kv k

" 1
t#

† ¸ ddtT ¸ œ

Ê Nœ

" #È1
t#

†

ˆ ddtT ‰ ¸ ddtT ¸

" 1
t#

œ

œ

t È1
t#

i

" È1
t#

2t 2È1
t#

j;

" # a1
t# b3/2

4. r œ (cos t
t sin t)i
(sin t  t cos t)j Ê v œ (t cos t)i
(t sin t)j Ê kvk œ È( t cos t)#
(t sin t)# œ Èt# œ ktk œ t, since t  0 Ê T œ

v kv k

œ

(t cos t)i
(t sin t)j t

œ (cos t)i
(sin t)j ;

Ê ¸ ddtT ¸ œ È( sin t)#
(cos t)# œ 1 Ê N œ 5. (a) ,axb œ

1 kvaxbk

ˆ ddtT ‰ ¸ ddtT ¸

dT dt

œ ( sin t)i
(cos t)j

œ ( sin t)i
(cos t)j ; , œ

1 kv k

† ¹ dTdtaxb ¹. Now, v œ i
f w axbj Ê kvaxbk œ É1
c f w axb d2 Ê T œ
1Î2

œ Š1
c f w axb d2 ‹


1Î2

i
f w axbŠ1
c f w axb d2 ‹

j. Thus

dT dt axb

Í 2 Í 2  f ax b f a x b cf Í dTaxb f ax b 3 2 Ê ¹ dt ¹ œ ” Š1
c f axb d2 ‹ •
 3 2 œ Ë 2 Ì Š 1
c f ax b d ‹ w

ww

ww

ww

œ

w

ww

Î

w

2

Š1
c f a x b d ‹ w

axb d2 Š1
c f axb d2 ‹ w

2

Š 1
c f ax b d ‹ w

3

" t

" t

†1œ

v kv k

 f ax b f a x b

Î

w

† ¸ ddtT ¸ œ

3Î2

œ

i


f ax b ww

3Î2

2 Š1
c f axb d ‹ w

kf axbk ww

2 ¹ 1
c f ax b d ¹ w

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

j

j

Section 13.4 Curvature and the Unit Normal Vector N Thus ,axb œ

1 a1
Òf axbÓ2 b1

(b) y œ ln (cos x) Ê œ

" sec x

†

2

Î

w

kf axbk k1
Òf axbÓ2 k ww

w

œ

kf axbk ww

3Î2

2 Š1
c f axb d ‹ w

œ ˆ cos" x ‰ ( sin x) œ  tan x Ê

dy dx

œ cos x, since  1#  x 

d# y dx#

œ  sec# x Ê , œ

k sec# xk c1
( tan x)# d$Î#

œ

sec# x ksec$ xk

1 #

(c) Note that f ww (x) œ 0 at an inflection point. Þ Þ Þ Þ 6. (a) r œ f(t)i
g(t)j œ xi
yj Ê v œ xi
yj Ê kvk œ Èx#
y# Ê T œ dT dt

œ

Þ Þ ÞÞ Þ ÞÞ Þ Þ ÞÞ Þ ÞÞ yay x  x yb xax y  y x b dT Þ # Þ # 3/2 i
Þ # Þ # 3/2 j Ê ¸ dt ¸ ax
y b ax
y b Þ ÞÞ Þ ÞÞ ky x  x yk 1 1 dT Þ Þ ; , œ kvk † ¸ dt ¸ œ È Þ # Þ # kx#
y# k x
y

œ

Þ Þ ÞÞ Þ ÞÞ yay x  x yb 2 Þ Þ 3/2 “
’ ax#
y# b Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ k y x  x yk k y x  x yk Þ # Þ # œ Þ # Þ # 3/2 . kx
y k ax
y b

œ Ê’ †

v kv k

Þ Þ ÞÞ Þ ÞÞ xax y  y xb 2 Þ Þ 3/2 “ ax#
y# b

œ

Þ

Þ y x ÈxÞ #
yÞ # i
ÈxÞ #
yÞ # j

œÊ

Þ Þ Þ ÞÞ Þ ÞÞ ay#
x# bay x  x yb2 Þ Þ 3 ax#
y# b

Þ ÞÞ Þ (b) r(t) œ ti
ln (sin t)j , 0  t  1 Ê x œ t and y œ ln (sin t) Ê x œ 1, x œ 0; y œ Ê ,œ

k csc# t  0k a1
cot# t)b$Î# "

œ

csc# t csc$ t

cos t sin t

ÞÞ œ cot t, y œ  csc# t

œ sin t

Þ t " (sinh t)i
ln (cosh t)j Ê x œ tan" (sinh t) and y œ ln (cosh t) Ê x œ 1
cosh sinh# t œ cosh t $ # ÞÞ Þ ÞÞ ksech t
sech t tanh tk sinh t # œ sech t, x œ  sech t tanh t; y œ cosh œ ksech tk asech# t
tanh# tb t œ tanh t, y œ sech t Ê , œ

(c) r(t) œ tan œ sech t

7. (a) r(t) œ f(t)i
g(t)j Ê v œ f w (t)i
gw (t)j is tangent to the curve at the point (f(t)ß g(t)); n † v œ c gw (t)i
f w (t)jd † cf w (t)i
gw (t)jd œ gw (t)f w (t)
f w (t)gw (t) œ 0; n † v œ (n † v) œ 0; thus, n and n are both normal to the curve at the point (b) r(t) œ ti
e2t j Ê v œ i
2e2t j Ê n œ 2e2t i
j points toward the concave side of the curve; N œ knk œ È4e4t
1 Ê N œ

2e È1
4e4t 2t

(c) r(t) œ È4  t# i
tj Ê v œ Nœ

n knk

and knk œ É1


t# 4  t#

i


t È4  t#

œ

" È1
4e4t

and

j

i
j Ê n œ i  Ê Nœ

2 È4  t#

n knk

" #

t È4  t#

j points toward the concave side of the curve;

ŠÈ4  t# i
tj‹

8. (a) r(t) œ ti
"3 t$ j Ê v œ i
t# j Ê n œ t# i  j points toward the concave side of the curve when t  0 and n œ t# i
j points toward the concave side when t  0 Ê N œ Nœ

" È1
t%

2ktk ; 1
t%

at# i  jb for t  0 and

at# i
jb for t  0

(b) From part (a), kvk œ È1
t% Ê T œ œ

" È1
t%

Nœ

ˆ ddtT ‰ ¸ ddtT ¸

œ

2t$ 1
t% 2ktk Š a1
t% b$Î# i

N does not exist at t œ 0, dt ¸ œ ¸ ddtT † ds œ 0 at t œ 0

" È1
t% i







t# È1
t% j

dT dt

t$ i ktkÈ1
t%

œ

2t$ $Î# i a1
t% b




2t $Î# j a1
t% b

6 2 Ê ¸ ddtT ¸ œ É a4t1

t%4tb$

t j; t Á 0 ktkÈ1
t% where the curve has a point of inflection; ddtT ¸ tœ0 œ 0 so the curvature , œ ¸ ddsT ¸ Ê N œ ," ddsT is undefined. Since x œ t and y œ 3" t$ Ê y œ 3" x$ , the curve is 2t $Î# j‹ a1
t% b

œ

Ê




the

cubic power curve which is concave down for x œ t  0 and concave up for x œ t  0.

9. r œ (3 sin t)i
(3 cos t)j
4tk Ê v œ (3 cos t)i
(3 sin t)j
4k Ê kvk œ È(3 cos t)#
(3 sin t)#
4# œ È25 œ 5 Ê T œ kvvk œ ˆ 35 cos t‰ i  ˆ 35 sin t‰ j
45 k Ê ddtT œ ˆ 35 sin t‰ i  ˆ 35 cos t‰ j # # Ê ¸ ddtT ¸ œ Ɉ 35 sin t‰
ˆ 35 cos t‰ œ

3 5

Ê Nœ

ˆ ddtT ‰ ¸ ddtT ¸

œ ( sin t)i  (cos t)j ; , œ

1 5

†

3 5

œ

3 25

10. r œ (cos t
t sin t)i
(sin t  t cos t)j
3k Ê v œ (t cos t)i
(t sin t)j Ê kvk œ È(t cos t)#
(t sin t)# œ Èt# œ ktk œ t, if t  0 Ê T œ

v kv k

œ (cos t)i  (sin t)j , t  0 Ê

Ê ¸ ddtT ¸ œ È( sin t)#
(cos t)# œ 1 Ê N œ

ˆ ddtT ‰ ¸ ddtT ¸

dT dt

œ ( sin t)i
(cos t)j

œ ( sin t)i
(cos t)j ; , œ

" t

†1œ

" t

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

841

842

Chapter 13 Vector-Valued Functions and Motion in Space

11. r œ aet cos tb i
aet sin tb j
2k Ê v œ aet cos t  et sin tb i
aet sin t
et cos tb j Ê kvk œ Éaet cos t  et sin tb#
aet sin t
et cos tb# œ È2e2t œ et È2 ; Tœ

v kv k

t  sin t
cos t œ Š cos È ‹ i
Š sin tÈ ‹j Ê 2 2 #

dT dt

t  sin t œ Š  sinÈt 2 cos t ‹ i
Š cos È ‹j 2

#

t  sin t Ê ¸ ddtT ¸ œ ÊŠ  sinÈt 2 cos t ‹
Š cos È ‹ œ1 Ê Nœ 2

,œ

1 kv k

† ¸ ddtT ¸ œ

†1œ

1

et È2

ˆ ddtT ‰ ¸ ddtT ¸

œ Š  cosÈt 2 sin t ‹ i
Š  sinÈt
2 cos t ‹ j ;

1

et È2

12. r œ (6 sin 2t)i
(6 cos 2t)j
5tk Ê v œ (12 cos 2t)i  (12 sin 2t)j
5k Ê kvk œ È(12 cos 2t)#
(12 sin 2t)#
5# œ È169 œ 13 Ê T œ v kv k

‰ ˆ 12 ‰ œ ˆ 12 13 cos 2t i  13 sin 2t j


5 13

k Ê

‰ ˆ 24 ‰ œ ˆ 24 13 sin 2t i  13 cos 2t j

dT dt

‰#
ˆ 24 ‰# œ Ê ¸ ddtT ¸ œ Ɉ 24 13 sin 2t 13 cos 2t ,œ

1 kv k

† ¸ ddtT ¸ œ

$

#

1 13

†

24 13

œ

24 13

ˆ ddtT ‰ ¸ ddtT ¸

Ê Nœ

œ ( sin 2t)i  (cos 2t)j ;

24 169 .

13. r œ Š t3 ‹ i
Š t# ‹ j , t  0 Ê v œ t# i
tj Ê kvk œ Èt%
t# œ tÈt#
1, since t  0 Ê T œ œ

t Èt#
t

i


1 Èt#
1

#


t œ É at1#
œ 1 b$

j Ê

" t#
1

œ

dT dt

Ê Nœ

i

1 at#
1b$Î#

ˆ ddtT ‰ ¸ ddtT ¸

14. r œ acos$ tb i
asin$ tb j , 0  t 

œ 1 #

1 Èt#
1

t at#
1b$Î#

i

j Ê ¸ ddtT ¸ œ ÊŠ

t Èt#
1

j; , œ

1 kv k

" ‹ at#
1b$Î#

† ¸ ddtT ¸ œ

#


Š

1 tÈt#
1

†

t ‹ at#
1b$Î#

1 t#
1

œ

v kv k

#

" . t at#
1b$Î#

Ê v œ a3 cos# t sin tb i
a3 sin# t cos tb j

Ê kvk œ Éa3 cos# t sin tb#
a3 sin# t cos tb# œ È9 cos% t sin# t
9 sin% t cos# t œ 3 cos t sin t, since 0  t  Ê Tœ

v kv k

œ ( cos t)i
(sin t)j Ê

œ (sin t)i
(cos t)j; , œ

1 kv k

† ¸ ddtT ¸ œ

dT dt

œ (sin t)i
(cos t)j Ê ¸ ddtT ¸ œ Èsin# t
cos# t œ 1 Ê N œ

1 3 cos t sin t

†1œ

v kv k

œ ˆsech at ‰ i
ˆtanh at ‰ j Ê

Ê ¸ ddtT ¸ œ É a"# sech# ˆ at ‰ tanh# ˆ at ‰
,œ

1 kv k

† ¸ ddtT ¸ œ

1 cosh

t a

†

" a

sech ˆ at ‰ œ

" a

" a#

dT dt

ˆ ddtT ‰ ¸ ddtT ¸

1 3 cos t sin t .

15. r œ ti
ˆa cosh at ‰ j , a  0 Ê v œ i
ˆsinh at ‰ j Ê kvk œ É1
sinh# ˆ at ‰ œ Écosh# ˆ at ‰ œ cosh Ê Tœ

1 #

œ ˆ "a sech

sech% ˆ at ‰ œ

" a

t a

t a

tanh at ‰ i
ˆ "a sech# at ‰ j

sech ˆ at ‰ Ê N œ

ˆ ddtT ‰ ¸ ddtT ¸

œ ˆ tanh at ‰ i
ˆsech at ‰ j ;

sech# ˆ at ‰.

16. r œ (cosh t)i  (sinh t)j
tk Ê v œ (sinh t)i  (cosh t)j
k Ê kvk œ Èsinh# t
( cosh t)#
1 œ È2 cosh t Ê Tœ

v kv k

œ Š È"2 tanh t‹ i 

Ê ¸ ddtT ¸ œ É "# sech% t
,œ

1 kv k

† ¸ ddtT ¸ œ

1 È2 cosh t

†

" #

" È2

j
Š È"2 sech t‹ k Ê

sech# t tanh# t œ

" È2

sech t œ

" #

" È2

dT dt

œ Š È"2 sech# t‹ i  Š È"2 sech t tanh t‹ k

sech t Ê N œ

&Î#

œ (sech t)i  (tanh t)k ;

sech# t.

17. y œ ax# Ê yw œ 2ax Ê yww œ 2a; from Exercise 5(a), ,(x) œ Ê ,w (x) œ  3# k2ak a1
4a# x# b

ˆ ddtT ‰ ¸ ddtT ¸

k2ak a1
4a# x# b$Î#

œ k2ak a1
4a# x# b

$Î#

a8a# xb ; thus, ,w (x) œ 0 Ê x œ 0. Now, ,w (x)  0 for x  0 and ,w (x)  0 for

x  0 so that ,(x) has an absolute maximum at x œ 0 which is the vertex of the parabola. Since x œ 0 is the only critical point for ,(x), the curvature has no minimum value.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.4 Curvature and the Unit Normal Vector N 18. r œ (a cos t)i
(b sin t)j â i j â â œ â  a sin t b cos t â â  a cos t  b sin t

Ê v œ (a sin t)i
(b cos t)j Ê a œ (a cos t)i  (b sin t)j Ê v ‚ a â kâ â 0 â œ abk Ê kv ‚ ak œ kabk œ ab, since a  b  0; , (t) œ kvkv‚k$ak â 0â

œ ab aa# sin# t
b# cos# tb

$Î#

; ,w (t) œ  #3 (ab) aa# sin# t
b# cos# tb

œ  3# (ab) aa#  b# b (sin 2t) aa# sin# t
b# cos# tb points on the major axis, or t œ 1 #

0t

and for 1  t 

31 # ;

1 31 #, # w

&Î#

&Î#

a2a# sin t cos t  2b# sin t cos tb

; thus, ,w (t) œ 0 Ê sin 2t œ 0 Ê t œ 0, 1 identifying

identifying points on the minor axis. Furthermore, ,w (t)  0 for

, (t)  0 for

1 #

 t  1 and

31 #

 t  21. Therefore, the points associated

with t œ 0 and t œ 1 on the major axis give absolute maximum curvature and the points associated with t œ and t œ 19. , œ

31 #

1 #

on the minor axis give absolute minimum curvature. d, da

Ê

a a#
b#

a  b and

843

d, da

a #
b # aa #
b # b #

œ

;

d, da

œ 0 Ê a#
b# œ 0 Ê a œ „ b Ê a œ b since a, b   0. Now,

 0 if a  b Ê , is at a maximum for a œ b and , (b) œ

b b#
b#

œ

" 2b

d, da

 0 if

is the maximum value of , .

20. (a) From Example 5, the curvature of the helix r(t) œ (a cos t)i
(a sin t)j
btk, a, b   0 is , œ a#
a b# ; also kvk œ Èa#
b# . For the helix r(t) œ (3 cos t)i
(3 sin t)j
tk, 0 Ÿ t Ÿ 41, a œ 3 and b œ 1 Ê , œ 3#
3 1# œ and kvk œ È10 Ê K œ '0

41

3 10

È10 dt œ ’

3 È10

t“

%1 !

3 10

121 È10

œ

(b) y œ x# Ê x œ t and y œ t# , _  t  _ Ê r(t) œ ti
t# j Ê v œ i
2tj Ê kvk œ È1
4t# ; Tœ ,œ

1 È1
4t# i 1 È1
4t#

œaÄ lim
_




†

2t dT È1
4t# j; dt

2 1
4t#

'a0

œ

2 1
4t#

œ

4t i a1
4t# b3/2

2 3. ŠÈ1
4t# ‹

dt
lim

bÄ_

œaÄ lim a tan" 2ab
lim
_




_

Then K œ 'c_

'0b 1
24t

bÄ_

2 j; a1
4t# b3/2

#

¸ ddtT ¸ œ É

2 $ ŠÈ1
4t# ‹

16t2
4 a1
4t# b3

œ

2 1
4t# .

ŠÈ1
4t# ‹ dt œ '
_ 1
24t# dt !

dt œ a Ä lim ctan" 2td a
lim
_

a tan" 2bb œ

1 #




1 #

Thus _

bÄ_

œ1

ctan" 2td 0

b

21. r œ ti
(sin t)j Ê v œ i
(cos t)j Ê kvk œ È1#
(cos t)# œ È1
cos# t Ê ¸v ˆ 1# ‰¸ œ É1
cos# ˆ 1# ‰ œ 1; T œ œ

i
cos t j È1
cos2 t

Ê 3œ

" 1

Ê

dT dt

œ

sin t cos t i a1
cos2 tb3/2




sin t j a1
cos2 tb3/2

Ê ¸ ddtT ¸ œ

ksin tk 1
cos2 t ;

¸ ddtT ¸

tœ 12

œ

¸sin 12 ¸ 1
cos2 ˆ 12 ‰

œ

1 1

œ 1. Thus ,ˆ 12 ‰ œ

1 1

†1œ1

# œ 1 and the center is ˆ 1# ß 0‰ Ê ˆx  1# ‰
y# œ 1

2 22. r œ (2 ln t)i  ˆt
"t ‰ j Ê v œ ˆ 2t ‰ i  ˆ1  t"# ‰ j Ê kvk œ É t42
ˆ1  t12 ‰ œ dT dt

œ

2ˆt2  1‰ i at2
1b2

œ

" #

Ê 3œ

" ,



at2

4t j
1 b2

2

Ê ¸ ddtT ¸ œ Ê 4at

 1b2
16t2 at2
1b4

œ

2 t2
1 .

Thus , œ

1 kv k

t2
1 t2

† ¸ ddtT ¸ œ

ÊTœ

t2 t2
1

†

2 t2
1

v kv k

œ

œ

2t2 at2
1b2

2t t2
1 i



t2  1 t2
1 j;

Ê ,a1b œ

2 22

œ 2. The circle of curvature is tangent to the curve at P(0ß 2) Ê circle has same tangent as the curve

Ê v(1) œ 2i is tangent to the circle Ê the center lies on the y-axis. If t Á 1 (t  0), then (t  1)#  0 # Ê t#  2t
1  0 Ê t#
1  2t Ê t
t 1  2 since t  0 Ê t
"t  2 Ê  ˆt
"t ‰  2 Ê y  2 on both

sides of (0ß 2) Ê the curve is concave down Ê center of circle of curvature is (0ß 4) Ê x#
(y
4)# œ 4 is an equation of the circle of curvature

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

v kv k

844

Chapter 13 Vector-Valued Functions and Motion in Space

23. y œ x# Ê f w (x) œ 2x and f ww (x) œ 2 Ê ,œ

24. y œ

x% 4

k2 k a1
(2x)# b$Î#

œ

2 a1
4x# b$Î#

Ê f w (x) œ x$ and f ww (x) œ 3x#

Ê ,œ

k3x# k

$Î#

# Š1
ax$ b ‹

œ

3x# a1
x' b$Î#

25. y œ sin x Ê f w (x) œ cos x and f ww (x) œ  sin x Ê ,œ

k sin xk a1
cos# xb$Î#

œ

ksin xk a1
cos# xb$Î#

26. y œ ex Ê f w (x) œ ex and f ww (x) œ ex Ê ,œ

ke x k

# $Î# Š1
aex b ‹

œ

ex ˆ1
e2x ‰$Î#

27-34. Example CAS commands: Maple: with( plots ); r := t -> [3*cos(t),5*sin(t)]; lo := 0; hi := 2*Pi; t0 := Pi/4; P1 := plot( [r(t)[], t=lo..hi] ): display( P1, scaling=constrained, title="#27(a) (Section 13.4)" ); CURVATURE := (x,y,t) ->simplify(abs(diff(x,t)*diff(y,t,t)-diff(y,t)*diff(x,t,t))/(diff(x,t)^2+diff(y,t)^2)^(3/2)); kappa := eval(CURVATURE(r(t)[],t),t=t0); UnitNormal := (x,y,t) ->expand( [-diff(y,t),diff(x,t)]/sqrt(diff(x,t)^2+diff(y,t)^2) ); N := eval( UnitNormal(r(t)[],t), t=t0 ); C := expand( r(t0) + N/kappa ); OscCircle := (x-C[1])^2+(y-C[2])^2 = 1/kappa^2; evalf( OscCircle ); P2 := implicitplot( (x-C[1])^2+(y-C[2])^2 = 1/kappa^2, x=-7..4, y=-4..6, color=blue ): Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.5 Torsion and the Unit Binormal Vector B

845

display( [P1,P2], scaling=constrained, title="#27(e) (Section 13.4)" ); Mathematica: (assigned functions and parameters may vary) In Mathematica, the dot product can be applied either with a period "." or with the word, "Dot". Similarly, the cross product can be applied either with a very small "x" (in the palette next to the arrow) or with the word, "Cross". However, the Cross command assumes the vectors are in three dimensions For the purposes of applying the cross product command, we will define the position vector r as a three dimensional vector with zero for its z-component. For graphing, we will use only the first two components. Clear[r, t, x, y] r[t_]={3 Cos[t], 5 Sin[t] } t0= 1 /4; tmin= 0; tmax= 21; r2[t_]= {r[t][[1]], r[t][[2]]} pp=ParametricPlot[r2[t], {t, tmin, tmax}]; mag[v_]=Sqrt[v.v] vel[t_]= r'[t] speed[t_]=mag[vel[t]] acc[t_]= vel'[t] curv[t_]= mag[Cross[vel[t],acc[t]]]/speed[t]3 //Simplify unittan[t_]= vel[t]/speed[t]//Simplify unitnorm[t_]= unittan'[t] / mag[unittan'[t]] ctr= r[t0] + (1 / curv[t0]) unitnorm[t0] //Simplify {a,b}= {ctr[[1]], ctr[[2]]} To plot the osculating circle, load a graphics package and then plot it, and show it together with the original curve. <
â â 3 cos t â â 3 sin t â â 3 cos t

â 3 sin t 4 â â 3 sin t 0 â â 3 sin t 0 â 225

œ

4†a9 sin# t9 cos# tb 225

œ

36 225

4 œ  25

2. By Exercise 10 in Section 13.4, T œ (cos t)i
(sin t)j and N œ ( sin t)i
(cos t)j ; thus B œ T ‚ N â â j kâ â i â â œ â cos t sin t 0 â œ acos# t
sin# tb k œ k. Also v œ (t cos t)i
(t sin t)j â â â  sin t cos t 0 â Ê a œ atasin tb
cos tbi
at cos t
sin tbj Ê

œ at cos t  sin t  sin tbi
at sin t
cos t
cos tbj â â i j kâ â â â t cos t t sin t 0â œ at cos t  2 sin tbi
a2 cos t  t sin tbj. Thus v ‚ a œ â â â â at sin t
cos tb at cos t
sin tb 0 â da dt

#

œ [(t cos t)(t cos t
sin t)  (t sin t)(t sin t
cos t)]k œ t# k Ê kv ‚ ak2 œ at# b œ t4 . Thus Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

846

Chapter 13 Vector-Valued Functions and Motion in Space

7œ

â â t cos t t sin t 0â â â â sin t
t cos t 0 â â cos t  t sin t â â â 2 sin t  t cos t 2 cos t  t sin t 0 â t4

œ

0 t4

œ0

t  sin t
cos t 3. By Exercise 11 in Section 13.4, T œ Š cos È ‹ i
Š sin tÈ ‹ j and N œ Š  cosÈt 2 sin t ‹ i
Š  sinÈt
2 cos t ‹ j ; Thus 2 2 â i j k ââ â â cos t  sin t sin t
cos t 0 ââ œ ’Š cos# t  2 cos t sin t
sin# t ‹
Š sin# t
2 sin t cos t
cos# t ‹“ k È2 È2 B œ T ‚ N œ ââ 2 2 â â  cos t  sin t  sin t
cos t 0 â â â È2 È2 a2tb a2tb œ ’Š 1  sin ‹
Š 1
sin ‹“ k œ k . Also, v œ aet cos t  et sin tb i
aet sin t
et cos tb j 2 2

Ê a œ c et asin t  cos tb
et acos t  sin tb d i
c et acos t  sin tb
et asin t
cos tb d j=a2et sin tb i
a2et cos tb j â â i j kâ â â â Ê ddta œ 2et acos t
sin tb i
2et asin t
cos tbj. Thus v ‚ a œ â et acos t  sin tb et asin t
cos tb 0 â œ 2e2t k â â 2et cos t 0â â 2et sin t #

Ê kv ‚ ak2 œ a2e2t b œ 4e4t . Thus 7 œ

â t et asin t
cos tb â e acos t  sin tb â â 2et sin t 2et cos t â â 2et acos t
sin tb 2et asin t
cos tb 4e4t

â 0â â 0â â 0â

œ0

5 ‰ ˆ 12 ‰ 4. By Exercise 12 in Section 13.4, T œ ˆ 12 13 cos 2t i  13 sin 2t j
13 k and N œ ( sin 2t)i  (cos 2t)j so â â i j kâ â â ˆ 12 12 5 â 12 ‰ ˆ ‰ ˆ5 ‰ ˆ5 ‰ B œ T ‚ N œ â 13 cos 2t  13 sin 2t 13 ââ œ 13 cos 2t i  13 sin 2t j  13 k . Also, â â a sin 2tb a cos 2tb 0â

v œ (12 cos 2t)i  (12 sin 2t)j
5k â i j â â v ‚ a œ â 12 cos 2t 12 sin 2t â â 24 sin 2t 24 cos 2t

Ê a œ (24 sin 2t)i  (24 cos 2t)j and ddta œ (48 cos 2t)i
(48 sin 2t)j â kâ â 5 â œ (120 cos 2t)i  (120 sin 2t)j  288k Ê kv ‚ ak2 â 0â œ (120 cos 2t)#
(120 sin 2t)#
(288)# œ 120# acos# 2t
sin# 2tb
288# œ 97344. Thus

7œ

â â â 12 cos 2t 12 sin 2t 5 â â â â 24 sin 2t 24 cos 2t 0 â â â â 48 cos 2t 48 sin 2t 0 â 97344

œ

5†a24†48b 97344

10 œ  169

5. By Exercise 13 in Section 13.4, T œ at#
t1b1/2 i
at#
11b1/2 j and N œ Èt#1
1 i  Èt#t
1 j so that B œ T ‚ N â i â # â j k ââ â ât t 0â â t â 1 â 0 d a # â œ k. Also, v œ t i
tj Ê a œ 2ti
j Ê œ 2i so that â 2t " 0 ââ œ 0 Ê 7 œ 0 œ ââ Èt#
1 Èt#
1 dt â â â t â " â â 2 0 0â â Èt#
1 Èt#
1 0 â 6. By Exercise 14 in Section 13.4, T œ ( cos t)i
(sin t)j and N œ (sin t)i
(cos t)j so that B œ T ‚ N â â j kâ â i â â œ â  cos t sin t 0 â œ k . Also, v œ a3 cos# t sin tb i
a3 sin# t cos tb j â â cos t 0 â â sin t Ê a œ dtd a3 cos# t sin tb i
dtd a3 sin# t cos tb j Ê ddta œ dtd ˆ dtd a3 cos# t sin tb‰ i
dtd ˆ dtd a3 sin# t cos tb‰ j â â â 3 cos# t sin t 3 sin# t cos t 0â â d â d # 0 ââ œ 0 Ê 7 œ 0 Ê ââ dt a3 cos# t sin tb dt a3 sin t cos tb â d ˆ d a3 cos# t sin tb‰ d ˆ d a3 sin# t cos tb‰ 0 â â dt dt â dt dt 7. By Exercise 15 in Section 13.4, T œ kvvk œ ˆsech at ‰ i
ˆtanh at ‰ j and N œ ˆ tanh at ‰ i
ˆsech at ‰ j so that B œ T ‚ N â i j k ââ â â sech ˆ t ‰ tanh ˆ at ‰ 0 ââ œ k. Also, v œ i
ˆsinh at ‰ j Ê a œ ˆ "a cosh at ‰ j Ê ddta œ a"# sinh ˆ at ‰ j so that œâ a â â â  tanh ˆ at ‰ sech ˆ at ‰ 0 â Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.5 Torsion and the Unit Binormal Vector B â â1 â â0 â â0 â

sinh ˆ at ‰ " ˆt‰ a cosh a " ˆt‰ a# sinh a

847

â 0â â 0 ââ œ 0 Ê 7 œ 0 0 ââ

8. By Exercise 16 in Section 13.4, T œ Š È"2 tanh t‹ i  È"2 j
Š È"2 sech t‹ k and N œ (sech t)i  (tanh t)k so that â â â â i j k â " â " " B œ T ‚ N œ ââ È2 tanh t È2 È2 sech t ââ œ Š È"2 tanh t‹ i
È"2 j
Š È"2 sech t‹ k. Also, v œ (sinh t)i  (cosh t)j
k â sech t 0  tanh t ââ â â â j kâ â i â â da a œ (cosh t)i  (sinh t)j Ê dt œ (sinh t)i  (cosh t)j and v ‚ a œ â sinh t  cosh t 1 â â â â cosh t  sinh t 0 â œ (sinh t)i
(cosh t)j
acosh2 t  sinh2 tbk œ (sinh t)i
(cosh t)j
k Ê kv ‚ ak# œ sinh# t
cosh# t
1. Thus 7œ

â â â sinh t  cosh t 1 â â â â cosh t  sinh t 0 â â â â sinh t  cosh t 0 â sinh# t
cosh# t
1

œ

" sinh# t
cosh# t
1

œ

" # cosh# t .

9. r œ (a cos t)i
(a sin t)j
btk Ê v œ (a sin t)i
(a cos t)j
bk Ê kvk œ È(a sin t)#
(a cos t)#
b# œ Èa#
b# Ê aT œ d kvk œ 0; a œ (a cos t)i
(a sin t)j Ê kak œ È(a cos t)#
(a sin t)# œ Èa# œ kak dt

#

Ê aN œ Ékak 

a# T

œ Ékak#  0# œ kak œ kak Ê a œ (0)T
kak N œ kak N

10. r œ (1
3t)i
(t  2)j  3tk Ê v œ 3i
j  3k Ê kvk œ È3#
1#
(3)# œ È19 Ê aT œ

d dt

kvk œ 0; a œ 0

Ê aN œ Ékak#  a#T œ 0 Ê a œ (0)T
(0)N œ 0 11. r œ (t
1)i
2tj
t# k Ê v œ i
2j
2tk Ê kvk œ È1#
2#
(2t)# œ È5
4t# Ê aT œ œ 4t a5
4t# b œ É 20 9 œ

"Î#

2È 5 3

Ê aT (1) œ

Ê a(1) œ

4 3

4 È9

T


œ

2È 5 3

4 3

" #

a5
4t# b

"Î#

(8t)

; a œ 2k Ê a(1) œ 2k Ê ka(1)k œ 2 Ê aN œ Ékak#  a#T œ É2#  ˆ 43 ‰

#

N

12. r œ (t cos t)i
(t sin t)j
t# k Ê v œ (cos t  t sin t)i
(sin t
t cos t)j
2tk "Î# Ê kvk œ È(cos t  t sin t)#
(sin t
t cos t)#
(2t)# œ È5t#
1 Ê aT œ " a5t#
1b (10t) #

œ

5t È5t#
1

Ê aT (0) œ 0; a œ (2 sin t  t cos t)i
(2 cos t  t sin t)j
2k Ê a(0) œ 2j
2k Ê ka(0)k #

œ È2#
2# œ 2È2 Ê aN œ Ékak#  aT# œ ÊŠ2È2‹  0# œ 2È2 Ê a(0) œ (0)T
2È2N œ 2È2N 13. r œ t# i
ˆt
"3 t$ ‰ j
ˆt  3" t$ ‰ k Ê v œ 2ti
a1
t# b j
a1  t# b k Ê kvk œ É(2t)#
a1
t# b#
a1  t# b# œ È2 at%
2t#
1b œ È2 a1
t# b Ê aT œ 2tÈ2 Ê aT (0) œ 0; a œ 2i
2tj  2tk Ê a(0) œ 2i Ê ka(0)k œ 2 Ê aN œ Ékak#  a#T œ È2#  0# œ 2 Ê a(0) œ (0)T
2N œ 2N 14. r œ aet cos tb i
aet sin tb j
È2et k Ê v œ aet cos t  et sin tb i
aet sin t
et cos tb j
È2et k #

Ê kvk œ Êaet cos t  et sin tb#
aet sin t
et cos tb#
ŠÈ2et ‹ œ È4e2t œ 2et Ê aT œ 2et Ê aT (0) œ 2; a œ aet cos t  et sin t  et sin t  et cos tb i
aet sin t
et cos t
et cos t  et sin tb j
È2et k #

œ a2et sin tb i
a2et cos tb j
È2et k Ê a(0) œ 2j
È2k Ê ka(0)k œ Ê2#
ŠÈ2‹ œ È6

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

848

Chapter 13 Vector-Valued Functions and Motion in Space #

Ê aN œ Ékak#  a#T œ ÊŠÈ6‹  2# œ È2 Ê a(0) œ 2T
È2N 15. r œ (cos t)i
(sin t)j  k Ê v œ ( sin t)i
(cos t)j Ê kvk œ È( sin t)#
(cos t)# œ 1 Ê T œ È2 #

œ ( sin t)i
(cos t)j Ê T ˆ 14 ‰ œ  œ1 Ê Nœ

ˆ ddtT ‰ ¸ ddtT ¸

È2 #

v kv k

œ ( cos t)i  (sin t)j Ê ¸ ddtT ¸ œ È( cos t)#
( sin t)# â â j kâ â i È2 È2 â â 1 œ ( cos t)i  (sin t)j Ê N ˆ 4 ‰ œ  # i  # j ; B œ T ‚ N œ â  sin t cos t 0 â œ k â â â  cos t  sin t 0 â i


j;

dT dt

Ê B ˆ 14 ‰ œ k , the normal to the osculating plane; r ˆ 14 ‰ œ È2 # ‹

osculating plane Ê 0 Šx  to the normal plane Ê Š


0 Šy 

È2 # ‹ Šx

È2 # ‹



È2 # ‹


Š

È2 #

i


È2 #

jk Ê PœŠ

È2 È2 # ß # ß 1‹

lies on the


(z  (1)) œ 0 Ê z œ 1 is the osculating plane; T is normal

È2 # ‹ Šy



È2 # ‹


0(z  (1)) œ 0 Ê 

È2 #

x


È2 #

yœ0

Ê x
y œ 0 is the normal plane; N is normal to the rectifying plane Ê Š

È2 # ‹ Šx



È2 # ‹


Š

È2 # ‹ Šy



È2 # ‹


0(z  (1)) œ 0 Ê 

È2 #

x

È2 #

y œ 1 Ê x
y œ È2 is the

rectifying plane 16. r œ (cos t)i
(sin t)j
tk Ê v œ ( sin t)i
(cos t)j
k Ê kvk œ Èsin# t
cos# t
1 œ È2 Ê T œ œ Š È"2 sin t‹ i
Š È"2 cos t‹ j


" È2

k Ê

dT dt

v kv k

œ Š È"2 cos t‹ i
Š È"2 sin t‹ j Ê ¸ ddtT ¸

ˆ dT ‰

œ É "# cos# t
"# sin# t œ È"2 Ê N œ ¸ ddtT ¸ œ ( cos t)i  (sin t)j ; thus T(0) œ È"2 j
È"2 k and N(0) œ i dt â â â i j k â â " " ââ " " Ê B(0) œ ââ 0 È2 È2 â œ  È2 j
È2 k , the normal to the osculating plane; r(0) œ i Ê P(1ß 0ß 0) lies on â 1 0 â 0 â â the osculating plane Ê 0(x  1)  to the normal plane Ê 0(x  1)


" È2 " È2

(y  0)
(y  0)


" È2 " È2

(z  0) œ 0 Ê y  z œ 0 is the osculating plane; T is normal (z  0) œ 0 Ê y
z œ 0 is the normal plane; N is normal to

the rectifying plane Ê 1(x  1)
0(y  0)
0(z  0) œ 0 Ê x œ 1 is the rectifying plane 17. Yes. If the car is moving along a curved path, then , Á 0 and aN œ , kvk# Á 0 Ê a œ aT T
aN N Á 0 . 18. kvk constant Ê aT œ

d dt

kvk œ 0 Ê a œ aN N is orthogonal to T Ê the acceleration is normal to the path

19. a ¼ v Ê a ¼ T Ê aT œ 0 Ê

d dt

20. a(t) œ aT T
aN N , where aT œ

kvk œ

d dt

kvk œ 0 Ê kvk is constant d dt

(10) œ 0 and aN œ , kvk# œ 100, Ê a œ 0T
100,N. Now, from

Exercise 5(a) Section 13.4, we find for y œ f(x) œ x# that , œ

kf ww (x)k

$Î#

1
af w (x)b# ‘

œ

2 c1
(2x)# d$Î#

œ

2 a1
4x# b$Î#

; also,

r(t) œ ti
t# j is the position vector of the moving mass Ê v œ i
2tj Ê kvk œ È1
4t# Ê T œ È1 "
4t# (i
2tj). At (0ß 0): T(0) œ i, N(0) œ j and ,(0) œ 2 Ê F œ ma œ m(100,)N œ 200m j ; At ŠÈ2ß 2‹ : T ŠÈ2‹ œ

" 3

Š i
2 È 2j ‹ œ

‰ œ m(100,)N œ ˆ 200 27 m Š

2È 2 3

" 3

i


i
"3 j‹ œ 

2È 2 3

400È2 81

È

j , N ŠÈ2‹ œ  2 3 2 i
3" j , and , ŠÈ2‹ œ

mi


200 81

2 27

Ê F œ ma

mj

21. a œ aT T
aN N , where aT œ dtd kvk œ dtd (constant) œ 0 and aN œ , kvk# Ê F œ ma œ m, kvk# N Ê kFk œ m, kvk# œ ˆm kvk# ‰ ,, a constant multiple of the curvature , of the trajectory

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.5 Torsion and the Unit Binormal Vector B

849

22. aN œ 0 Ê , kvk# œ 0 Ê , œ 0 (since the particle is moving, we cannot have zero speed) Ê the curvature is zero so the particle is moving along a straight line 23. From Example 1, kvk œ t and aN œ t so that aN œ , kvk# Ê , œ

aN kv k #

œ

t t#

œ

" t

,tÁ0 Ê 3œ

" ,

œt

24. r œ (x!
At)i
(y!
Bt)j
(z!
Ct)k Ê v œ Ai
Bj
Ck Ê a œ 0 Ê v ‚ a œ 0 Ê , œ 0. Since the curve is a plane curve, 7 œ 0. 25. If a plane curve is sufficiently differentiable the torsion is zero as the following argument shows: r œ f(t)i
g(t)j Ê v œ f w (t)i
gw (t)j Ê a œ f w w (t)i
gw w (t)j Ê Ê 7œ

â w â f (t) gw (t) â ww â f (t) gww (t) â www â f (t) gwww (t) kv ‚a k #

â 0â â 0â â 0â

26. From Example 2, 7 œ

da dt

w

w

œ f w w (t)i
gw w (t)j

œ0

b a#
b#

Ê 7 w (b) œ

a#  b# aa #
b # b # w

; 7 w (b) œ 0 Ê

a#  b# a a #
b # b# w

œ 0 Ê a#  b# œ 0 Ê b œ „ a

Ê b œ a since a, b  0. Also b  a Ê 7  0 and b  a Ê 7  0 so 7max occurs when b œ a Ê 7max œ œ

" 2a

a a#
a#

27. r(t) œ f(t)i
g(t)j
h(t)k Ê v œ f w (t)i
gw (t)j
hw (t)k; v † k œ 0 Ê hw (t) œ 0 Ê h(t) œ C Ê r(t) œ f(t)i
g(t)j
Ck and r(a) œ f(a)i
g(a)j
Ck œ 0 Ê f(a) œ 0, g(a) œ 0 and C œ 0 Ê h(t) œ 0. 28. From Example 2, v œ (a sin t)i
(a cos t)j
bk Ê kvk œ Èa#
b# Ê T œ œ

" È a#
b#

œ

b sin t È a#
b#

v kv k

c(a sin t)i
(a cos t)j
bkd ; ddtT œ Èa#"
b# c(a cos t)i  (a sin t)jd Ê N œ â â â i j k â â â a sin t a cos t b œ (cos t)i  (sin t)j ; B œ T ‚ N œ ââ  Èa#
b# Èa#
b# Èa#
b# ââ â  cos t  sin t 0 ââ â i

Ê 7 œ

" kv k

b cos t È a#
b#

ˆ ddtB

j


a È a#
b #

† N‰ œ Š

k Ê

dB dt

œ

" È a#
b #

" b Èa#
b# ‹ Š Èa#
b# ‹

c(b cos t)i
(b sin t)jd Ê

œ

b a#
b#

dB dt

ˆ ddtT ‰ ¸ ddtT ¸

† N œ  Èa#b


b#

, which is consistent with the result in

Example 2. 29-32. Example CAS commands: Maple: with( LinearAlgebra ); r := < t*cos(t) | t*sin(t) | t >; t0 := sqrt(3); rr := eval( r, t=t0 ); v := map( diff, r, t ); vv := eval( v, t=t0 ); a := map( diff, v, t ); aa := eval( a, t=t0 ); s := simplify(Norm( v, 2 )) assuming t::real; ss := eval( s, t=t0 ); T := v/s; TT := vv/ss ; q1 := map( diff, simplify(T), t ): NN := simplify(eval( q1/Norm(q1,2), t=t0 )); Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

850

Chapter 13 Vector-Valued Functions and Motion in Space

BB := CrossProduct( TT, NN ); kappa := Norm(CrossProduct(vv,aa),2)/ss^3; tau := simplify( Determinant(< vv, aa, eval(map(diff,a,t),t=t0) >)/Norm(CrossProduct(vv,aa),2)^3 ); a_t := eval( diff( s, t ), t=t0 ); a_n := evalf[4]( kappa*ss^2 ); Mathematica: (assigned functions and value for t0 will vary) Clear[t, v, a, t] mag[vector_]:=Sqrt[vector.vector] Print["The position vector is ", r[t_]={t Cos[t], t Sin[t], t}] Print["The velocity vector is ", v[t_]= r'[t]] Print["The acceleration vector is ", a[t_]= v'[t]] Print["The speed is ", speed[t_]= mag[v[t]]//Simplify] Print["The unit tangent vector is ", utan[t_]= v[t]/speed[t] //Simplify] Print["The curvature is ", curv[t_]= mag[Cross[v[t],a[t]]] / speed[t]3 //Simplify] Print["The torsion is ", torsion[t_]= Det[{v[t], a[t], a'[t]}] / mag[Cross[v[t],a[t]]]2 //Simplify] Print["The unit normal vector is ", unorm[t_]= utan'[t] / mag[utan'[t]] //Simplify] Print["The unit binormal vector is ", ubinorm[t_]= Cross[utan[t],unorm[t]] //Simplify] Print["The tangential component of the acceleration is ", at[t_]=a[t].utan[t] //Simplify] Print["The normal component of the acceleration is ", an[t_]=a[t].unorm[t] //Simplify] You can evaluate any of these functions at a specified value of t. t0= Sqrt[3] {utan[t0], unorm[t0], ubinorm[t0]} N[{utan[t0], unorm[t0], ubinorm[t0]}] {curv[t0], torsion[t0]} N[{curv[t0], torsion[t0]}] {at[t0], an[t0]} N[{at[t0], an[t0]}] To verify that the tangential and normal components of the acceleration agree with the formulas in the book: at[t]== speed'[t] //Simplify an[t]==curv [t] speed[t]2 //Simplify 13.6 PLANETARY MOTION AND SATELLITES 1.

T# a$

œ

41 # GM

Ê T# œ

41 # GM

a$ Ê T# œ

41 # a6.6726‚10
"" Nm# kg
# b a5.975‚10#% kgb

(6,808,000 m)$

¸ 3.125 ‚ 10( sec# Ê T ¸ È3125 ‚ 10% sec# ¸ 55.90 ‚ 10# sec ¸ 93.2 min 2. e œ 0.0167 and perihelion distance œ 149,577,000 km and e œ Ê 0.0167 œ

(149,577,000,000 m)v#! a6.6726‚10
"" Nm# kg
# b a1.99‚10$! kgb

r! v#! GM

1

 1 Ê v#! ¸ 9.03 ‚ 10) m# /sec#

Ê v! ¸ È9.03 ‚ 10) m# /sec# ¸ 3.00 ‚ 10% m/sec T# 41 # GM # $ a$ œ GM Ê a œ 41# T c"" #
# #% $ a6.6726‚10 Nm kg b ˆ5.975‚10 kg‰ Ê a$ œ (5535 sec)# œ 3.094 ‚ 10#! m$ Ê a ¸ È3.094 41 # œ 6.764 ‚ 10' m ¸ 6764 km. Note that 6764 km ¸ "# a12,757 km
183 km
589 kmb.

3. 92.25 min œ 5535 sec and

‚ 10#! m$

# 4. T œ 1639 min œ 98,340 sec and mass of Mars œ 6.418 ‚ 10#$ kg Ê a$ œ GM 41# T
"" #
# #$ # $ ‚10 kgb (98,340 sec) œ a6.6726‚10 Nm kg b 4a6.418 ¸ 1.049 ‚ 10## m$ Ê a ¸ È1.049 ‚ 10## m$ 1#

œ 2.19 ‚ 10( m œ 21,900 km

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 13.6 Planetary Motion and Satellites 5. 2a œ diameter of Mars
perigee height
apogee height œ D
1499 km
35,800 km Ê 2(21,900) km œ D
37,299 km Ê D œ 6501 km 6. a œ 22,030 km œ 2.203 ‚ 10( m and T# œ Ê T# œ

#

41 a6.6720‚10
"" Nm# kg
# b a6.418‚10#$ kgb

41 # GM

a$

(2.203 ‚ 10( m)$ ¸ 9.856 ‚ 10* sec#

Ê T ¸ È9.856 ‚ 10) sec# ¸ 9.928 ‚ 10% sec ¸ 1655 min 7. (a) Period of the satellite œ rotational period of the Earth Ê period of the satellite œ 1436.1 min œ 86,166 sec; a$ œ

a6.6726‚10
"" Nm# kg
# b ˆ5.975‚10#% kg‰ (86,166 sec)#

GMT# 41 # $

Ê a$ œ 41 # $ ## È #" $ ¸ 7.4980 ‚ 10 m Ê a ¸ 74.980 ‚ 10 m ¸ 4.2168 ‚ 10( m œ 42,168 km (b) The radius of the Earth is approximately 6379 km Ê the height of the orbit is 42,168  6379 œ 35,789 km (c) Symcom 3, GOES 4, and Intelsat 5 GMT# 41 # a6.6726‚10c"" Nm# kg
# b a6.418‚10#$ kgb (88,644 sec)# œ 41 # (

8. T œ 1477.4 min œ 88,644 sec Ê a$ œ œ

$ 8.524 ‚ 10#" m$ Ê a ¸ È 8.524 ‚ 10#" m$

¸ 2.043 ‚ 10 m œ 20,430 km

9. Period of the Moon œ 2.36055 ‚ 10' sec Ê a$ œ œ

a6.6726‚10c"" Nm# kg
# b ˆ5.975‚10#% kg‰ (2.36055‚10' sec)# 41 # )

GMT# 41 #

$ ¸ 5.627 ‚ 10#& m$ Ê a ¸ È 5.627 ‚ 10#& m$

¸ 3.832 ‚ 10 m œ 383,200 km from the center of the Earth. 10. r œ

Ê v# œ

GM v#

11. Solar System:

T# a$

œ

É a6.6726‚10 Ê kvk œ É GM r œ


"" Nm# kg
# b a5.975‚10#% kgb

41 # a6.6726‚10
"" Nm# kg
# b a1.99‚10$! kgb

¸ 2.97 ‚ 10"* sec# /m$ ;

#

r

¸ 1.9967 ‚ 10( r"Î# m/sec

Earth:

#

T a$

œ

41 a6.6726‚10
"" Nm# kg
# b a5.975‚10#% kgb

¸ 9.902 ‚ 10"% sec# /m$ ;

Moon:

T# a$

œ

41 # a6.6726‚10
"" Nm# kg
# b a7.354‚10## kgb

¸ 8.045 ‚ 10"# sec# /m$ ;

r! v#! GM

12. e œ

GM r

 1 Ê v#! œ

GM(e
1) r!

Ê v! œ É GM(er!
1) ;

Circle: e œ 0 Ê v! œ É GM r! 2GM Ellipse: 0  e  1 Ê É GM r!  v!  É r!

Parabola: e œ 1 Ê v! œ É 2GM r! Hyperbola: e  1 Ê v!  É 2GM r! 13. r œ

Ê v# œ

GM v#

14. ?A œ

" #

GM r

Ê v œ É GM r which is constant since G, M, and r (the radius of orbit) are constant

kr(t
?t) ‚ r(t)k Ê

œ

" #

¹ r(t
??t)t  r(t) ‚ r(t)


œ

" #

¸ ddtr ‚ r(t)¸ œ

" #

?A ?t

œ

" #

?t) ‚ r(t)¹ œ ¹ r(t
?t

" #

 r(t)
r(t) ‚ r(t)¹ ¹ r(t
?t) ? t

r(t) ‚ r(t)¹ œ #" ¹ r(t
??t)t  r(t) ‚ r(t)¹ Ê ¸r(t) ‚ ddtr ¸ œ "# kr ‚ rÞ k " ?t

dA dt

œ lim

"

?t Ä 0 #

¹ r(t
??t)t  r(t) ‚ r(t)¹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

851

852

Chapter 13 Vector-Valued Functions and Motion in Space #

# %

r v#

# %

#

! ! 15. T œ Š 2r!1va! ‹ È1  e# Ê T# œ Š 4r1# va# ‹ a1  e# b œ Š 4r1# va# ‹ ”1  Š GM  1‹ • (from Equation 32) ! ! ! !

r# v %

# %

r v#

# %

! ! œ Š 4r1# va# ‹ ’ G!# M!#
2 Š GM ‹“ œ Š 4r1# va# ‹ ’ !

!

# %

œ a41 a

!

2GM  r v# 2 ‰ b Š 2r! GM! ! ‹ ˆ GM

!

2GMr! v#!  r#! v%! “ G# M#

œ

ˆ41# a% ‰ a2GM  r! v#! b r! G# M#

" ‰ˆ 2 ‰ # œ a41 a b ˆ 2a GM (from Equation 35) Ê T œ # %

4 1 # a$ GM

Ê

T# a$

œ

41 # GM

16. Let rAB (t) denote the vector from planet A to planet B at time t. Then rAB (t) œ rB (t)  rA (t) œ [3 cos (1t)  2 cos (21t)]i
[3 sin (1t)  2 sin (21t)]j œ c3 cos (1t)  2 acos# (1t)  sin# (1t)bd i
[3 sin (1t)  4 sin (1t) cos (1t)]j œ c3 cos (1t)  4 cos# (1t)
2d i
[(3  4 cos (1t)) sin (1t)]j Ê parametric equations for the path are x(t) œ 2
[3  4 cos (1t)] cos (1t) and y(t) œ [3  4 cos (1t)] sin (1t) 17. The graph of the path of planet B is the limacon ¸ at the right.

18. (i) (ii) (iii) (iv) (v)

Perihelion is the time t such that kr(t)k is a minimum. Aphelion is the time t such that kr(t)k is a maximum. Equinox is the time t such that r(t) † w œ 0 . Summer solstice is the time t such that the angle between r(t) and w is a maximum. Winter solstice is the time t such that the angle between r(t) and w is a minimum.

CHAPTER 13 PRACTICE EXERCISES 1. r(t) œ (4 cos t)i
ŠÈ2 sin t‹ j Ê x œ 4 cos t and y œ È2 sin t Ê

x# 16




y# #

œ 1;

v œ (4 sin t)i
ŠÈ2 cos t‹ j and a œ (4 cos t)i  ŠÈ2 sin t‹ j ; r(0) œ 4 i , v(0) œ È2j , a(0) œ 4i ; r ˆ 14 ‰ œ 2È2i
j , v ˆ 14 ‰ œ 2È2i
j , a ˆ 1 ‰ œ 2È2i  j ; kvk œ È16 sin# t
2 cos# t 4

Ê aT œ

d dt

kvk œ

at t œ 14 : aT œ

14 sin t cos t È16 sin# t
2 cos# t

7 È 8
1

œ

7 3

; at t œ 0: aT œ 0, aN œ Ékak#  0 œ 4, , œ

, aN œ É9 

49 9

œ

4È 2 3

,,œ

aN kv k #

œ

aN kv k #

œ

4 2

œ 2;

4È 2 27

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 13 Practice Exercises 2. r(t) œ ŠÈ3 sec t‹ i
ŠÈ3 tan t‹ j Ê x œ È3 sec t and y œ È3 tan t Ê

x# 3



y# 3

853

œ sec# t  tan# t œ 1;

Ê x#  y# œ 3; v œ ŠÈ3 sec t tan t‹ i
ŠÈ3 sec# t‹ j and a œ ŠÈ3 sec t tan# t
È3 sec$ t‹ i  Š2È3 sec# t tan t‹ j ; r(0) œ È3i , v(0) œ È3j , a(0) œ È3i ; kvk œ È3 sec# t tan# t
3 sec% t Ê aT œ

d dt

kvk œ

6 sec# t tan$ t
18 sec% t tan t 2È3 sec# t tan# t
3 sec% t

;

at t œ 0: aT œ 0, aN œ Ékak#  0 œ È3, ,œ 3. r œ

œ

È3 3

œ

" È1
t#

i


t È1
t#

aN kv k #

" È3

j Ê v œ  t a1
t# b

$Î#

i
a 1
t# b

#

#

Ê kvk œ Ê’t a1
t# b$Î# “
’a1
t# b$Î# “ œ d kv k dt

œ0 Ê

2t a1
t# b#

œ 0 Ê t œ 0. For t  0,

" 1
t#

2t a1
t# b#

$Î#

j

. We want to maximize kvk :

 0; for t  0,

2t a1
t# b#

d kv k dt

œ

2t a1
t# b#

and

 0 Ê kvk max occurs when

t œ 0 Ê kvk max œ 1 4. r œ aet cos tb i
aet sin tb j Ê v œ aet cos t  et sin tb i
aet sin t
et cos tb j Ê a œ aet cos t  et sin t  et sin t  et cos tb i
aet sin t
et cos t
et cos t  et sin tb j œ a2et sin tb i
a2et cos tb j . Let ) be the angle between r and a . Then ) œ cos" Š krrk†kaak ‹ œ cos" 


2e2t sin t cos t2e2t sin t cos t Éaet cos tb# aet sin tb# Éa
2et sin tb# a2et cos tb# 

â âi â 5. v œ 3i
4j and a œ 5i
15j Ê v ‚ a œ â 3 â â5 Ê ,œ 6. , œ

kv ‚ a k kv k $

kyww k $Î# 1
ayw b# ‘

œ ex a1
e2x b d, dx

œ

25 5$

œ

$Î#

&Î#

œ 0 Ê a1  2e2x b œ 0 Ê e2x œ

maximum at the point Š ln È2ß

x#
y# œ 1, 2x

dx dt

dx dt

i



2y

dy dt

dy dt

d, dx

Ê

 3e3x a1
e2x b

7. r œ xi
yj Ê v œ

for all t

â kâ â 0 â œ 25k Ê kv ‚ ak œ 25; kvk œ È3#
4# œ 5 â 0â

j 4 "5

" 5

œ ex a1
e2x b $Î#

1 #

œ cos" Š 2e02t ‹ œ cos" 0 œ

" #

œ ex a1
e2x b

œ ex a1
e2x b

$Î#


ex ’ #3 a1
e2x b

&Î#

&Î#

a2e2x b“ &Î#

ca1
e2x b  3e2x d œ ex a1
e2x b a1  2e2x b ; Ê 2x œ  ln 2 Ê x œ  "# ln 2 œ  ln È2 Ê y œ È"2 ; therefore , is at a

" È2 ‹

j and v † i œ y Ê

œ0 Ê

dy dt

œ

x dx y dt

dx dt

œ y. Since the particle moves around the unit circle

Ê

dy dt

œ  yx (y) œ x. Since

dx dt

œ y and

dy dt

œ x, we have

v œ yi  xj Ê at (1ß 0), v œ j and the motion is clockwise. dy dy " # dx # dx dt œ 3x dt Ê dt œ 3 x dt . If r œ xi
yj , where x and y are differentiable functions of dy dy dx " # dx " # then v œ dx dt i
dt j. Hence v † i œ 4 Ê dt œ 4 and v † j œ dt œ 3 x dt œ 3 (3) (4) œ 12 at (3ß 3). Also, # # # # # ‰#
ˆ 3" x# ‰ ddt#x . Hence a † i œ 2 Ê ddt#x œ 2 and a œ ddt#x i
ddt#y j and ddt#y œ ˆ 23 x‰ ˆ dx dt # a † j œ ddt#y œ 23 (3)(4)#
"3 (3)# (2) œ 26 at the point (xß y) œ (3ß 3).

8. 9y œ x$ Ê 9

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

t,

854 9.

Chapter 13 Vector-Valued Functions and Motion in Space dr dt

orthogonal to r Ê 0 œ

†rœ

dr dt

" dr # dt

† r
"# r †

œ

dr dt

#

" d # dt #

(r † r) Ê r † r œ K, a constant. If r œ xi
yj , where

x and y are differentiable functions of t, then r † r œ x
y Ê x#
y# œ K, which is the equation of a circle centered at the origin. 10. (b) v œ (1  1 cos 1t)i
(1 sin 1t)j Ê a œ a1# sin 1tb i
a1# cos 1tb j ; v(0) œ 0 and a(0) œ 1# j ; v(1) œ 21i and a(1) œ 1# j ; v(2) œ 0 and a(2) œ 1# j ; v(3) œ 21i and a(3) œ 1# j

(c) Forward speed at the topmost point is kv(1)k œ kv(3)k œ 21 ft/sec; since the circle makes

" #

revolution per

second, the center moves 1 ft parallel to the x-axis each second Ê the forward speed of C is 1 ft/sec. 11. y œ y!
(v! sin !)t  "# gt# Ê y œ 6.5
(44 ft/sec)(sin 45°)(3 sec)  "# a32 ft/sec# b (3 sec)# œ 6.5
66È2  144 ¸ 44.16 ft Ê the shot put is on the ground. Now, y œ 0 Ê 6.5
22È2t  16t# œ 0 Ê t ¸ 2.13 sec (the positive root) Ê x ¸ (44 ft/sec)(cos 45°)(2.13 sec) ¸ 66.27 ft or about 66 ft, 3 in. from the stopboard 12. ymax œ y!


(v! sin !)# #g

œ 7 ft


[(80 ft/sec)(sin 45°)]# (2) a32 ft/sec# b

¸ 57 ft (v! sin !)t  "# gt# (v sin !)  " gt y œ ! v! cos ! # x œ (v! cos !)t 2v! sin !  2v! cos ! tan 9 , which is the time when g

13. x œ (v! cos !)t and y œ (v! sin !)t  "# gt# Ê tan 9 œ Ê v! cos ! tan 9 œ v! sin !  "# gt Ê t œ

the golf ball

hits the upward slope. At this time x œ (v! cos !) Š 2v! sin !  2vg ! cos ! tan 9 ‹ œ Š 2g ‹ av#! sin ! cos !  v#! cos# ! tan 9b . Now OR œ

x cos 9

Ê OR œ Š g2 ‹ Š

v#! sin ! cos !  v#! cos# ! tan 9 ‹ cos 9

œŠ

2v#! cos ! sin ! ‹ Š cos g 9

œŠ

2v#! cos ! 9  cos ! sin 9 ‹ Š sin ! cos cos ‹ #9 g

œŠ

2v#! cos ! g cos# 9 ‹ [sin (!

cos ! tan 9 cos 9 ‹



 9)]. The distance OR is maximized

when x is maximized:

dx d!

œŠ

2v#! g ‹(cos

Ê cot 2! œ tan (9) Ê 2! œ 14. R œ

v#! g

1 #

2!
sin 2! tan 9) œ 0 Ê (cos 2!
sin 2! tan 9) œ 0 Ê cot 2!
tan 9 œ 0


9 Ê !œ

9 #




1 4 #

ft) a32 ft/sec b sin 2! Ê v! œ É sinRg2! ; for 4325 yards: 4325 yards œ 12,975 ft Ê v! œ É (12,975(sin 90°) #

ft) a32 ft/sec b ¸ 644 ft/sec; for 4752 yards: 4752 yards œ 14,256 ft Ê v! œ É (14,256(sin ¸ 675 ft/sec 90°)

15. (a) R œ

v#! g

v#

# # # ! sin 2! Ê 109.5 ft œ Š 32 ft/sec Ê v! œ È3504 ft# /sec# # ‹ (sin 90°) Ê v! œ 3504 ft /sec

¸ 59.19 ft/sec (b) x œ (v! cos !)t and y œ 4
(v! sin !)t  "# gt# ; when the cork hits the ground, x œ 177.75 ft and y œ 0 Ê 177.75 œ Šv! Ê v! œ

(177.75)È2 t

" È2 ‹ t

œ

and 0 œ 4
Šv!

4(177.75)È2 È181.75

" È2 ‹ t

 16t# Ê 16t# œ 4
177.75 Ê t œ

È181.75 4

¸ 74.58 ft/sec

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 13 Practice Exercises

855

5 16. (a) x œ v! (cos 40°)t and y œ 6.5
v! (sin 40°)t  "# gt# œ 6.5
v! (sin 40°)t  16t# ; x œ 262 12 ft and y œ 0 ft 5 Ê 262 12 œ v! (cos 40°)t or v! œ

262.4167 # # and 0 œ 6.5
’ (cos 40°)t “ (sin 40°)t  16t Ê t œ 14.1684

262.4167 (cos 40°)t

Ê t ¸ 3.764 sec. Therefore, 262.4167 ¸ v! (cos 40°)(3.764 sec) Ê v! ¸ #

(v! sin !) 2g

(b) ymax œ y!


a(91)(sin 40°)b (2)(32)

¸ 6.5


2

262.4167 (cos 40°)(3.764 sec)

Ê v! ¸ 91 ft/sec

¸ 60 ft

#

#

17. x# œ av!# cos# !b t# and ˆy
"# gt# ‰ œ av!# sin# !b t# Ê x#
ˆy
"# gt# ‰ œ v!# t# Þ ÞÞ Þ ÞÞ ÞÞ# ÞÞ# ÞÞ# ÞÞ# ÞÞ# axÞ ÞÞx  yÞ ÞÞyb# x x y y Ê x
y  s œ x
y  xÞ #  yÞ # Þ Þ # # Èx  y ÞÞ# ÞÞ# Þ # Þ # Þ # ÞÞ# Þ ÞÞ Þ ÞÞ Þ # ÞÞ# Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ Þ ÞÞ ax  y b ax  y b
ax x  2x x y y  y y b ax y
y x b# x# y#  y# x#
2x x y y œ œ œ Þ# Þ# Þ# Þ# Þ Þ x y x y x#  y# Þ ÞÞ Þ ÞÞ Þ # Þ # $Î# Þ Þ # # kx y
y x k ax  y b ÞÞ ÞÞ ÞÞ x y Ê È x#
y#  s # œ È Þ # Þ # Ê ÈÞÞ# ÞÞ# ÞÞ# œ kxÞ ÞÞy
yÞ ÞÞxk œ ," œ 3 x y x y
s

ÞÞ 18. s œ

d dt

ÈxÞ #
yÞ # œ

19. r(t) œ ’'0 cos ˆ "# 1)# ‰ d)“ i
’'0 sin ˆ "# 1)# ‰ d)“ j Ê v(t) œ cos Š 1#t ‹ i
sin Š 1#t â i j â â # 1t# # # â sin Š 1#t ‹ a(t) œ 1t sin Š 1#t ‹ i
1t cos Š 1#t ‹ j Ê v ‚ a œ â cos Š # ‹ â â 1t sin Š 1t# ‹ 1t cos Š 1t# ‹ â # # t

t

œ 1 tk Ê , œ 20. s œ a) Ê ) œ

kv ‚ a k kv k $ s a

#

œ 1t; kv(t)k œ

Ê 9œ

s a




1 #

ds dt

Ê

#

‹ j Ê kvk œ 1; k ââ â 0 ââ â 0 ââ

œ 1 Ê s œ t
C; r(0) œ 0 Ê s(0) œ 0 Ê C œ 0 Ê , œ 1s d9 ds

œ

" a

Ê , œ ¸ "a ¸ œ

" a

since a  0

21. r œ (2 cos t)i
(2 sin t)j
t# k Ê v œ (2 sin t)i
(2 cos t)j
2tk Ê kvk œ È(2 sin t)#
(2 cos t)#
(2t)# œ 2È1
t# Ê Length œ '0 2È1
t# dt œ ’tÈ1
t#
ln ¹t
È1
t# ¹“ 1Î4

1Î% !

œ

1 4

É1


1# 16


ln Š 14
É1


22. r œ (3 cos t)i
(3 sin t)j
2t$Î# k Ê v œ (3 sin t)i
(3 cos t)j
3t"Î# k

1# 16 ‹

$ # Ê kvk œ É(3 sin t)#
(3 cos t)#
a3t"Î# b œ È9
9t œ 3È1
t Ê Length œ '0 3È1
t dt œ
2(1
t)$Î# ‘ ! 3

œ 14 23. r œ

4 9

(1
t)$Î# i
49 (1  t)$Î# j
3" tk Ê v œ #

2 3

(1
t)"Î# i  32 (1  t)"Î# j
3" k

#

#

Ê kvk œ É
23 (1
t)"Î# ‘

 23 (1  t)"Î# ‘
ˆ 3" ‰ œ 1 Ê T œ i  23 j
3" k ;

" 3

(1
t)"Î# i
â â â Ê N(0) œ È" i
È" j ; B(0) œ T(0) ‚ N(0) œ ââ 2 2 â â Ê T(0) œ

2 3

dT dt

œ

2 3

(1
t)"Î# i  23 (1  t)"Î# j
3" k

" 3

(1  t)"Î# j Ê ddtT (0) œ 3" i
3" j Ê ¸ ddtT (0)¸ œ â i j kâ 2 " â " " 4 â  23 3 3 â œ  È i
È j
È k; 3 2 3 2 3 2 " " 0 ââ È2 È2

a œ "3 (1
t)"Î# i
3" (1  t)"Î# j Ê a(0) œ 3" i
3" j and v(0) œ 32 i  32 j
3" k Ê v(0) ‚ a(0) â i j k ââ È â Š 32 ‹ È2 È2 â 2 2 "â kv ‚a k " " 4 i j k k v a k œâ 3 œ 

Ê ‚ œ Ê , (0) œ œ â 3 3 9 9 9 3 1$ œ 3 ; kv k $ â " â " â 3 0â 3 Þ Þ a œ  "6 (1
t)$Î# i
"6 (1  t)$Î# j Ê a(0) œ  6" i
6" j Ê 7 (0) œ

â 2 â 3 â â " â 3 â " â 6

 23 " 3 " 6

kv ‚a k #

â â â 0 ââ 0 ââ " 3

œ

2 ‰ ˆ "3 ‰ ˆ 18

Š

È2 ‹# œ 3

t œ 0 Ê ˆ 49 ß 49 ß 0‰ is the point on the curve

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 6

;

È2 3

856

Chapter 13 Vector-Valued Functions and Motion in Space

24. r œ aet sin 2tb i
aet cos 2tb j
2et k Ê v œ aet sin 2t
2et cos 2tb i
aet cos 2t  2et sin 2tb j
2et k Ê kvk œ Éaet sin 2t
2et cos 2tb#
aet cos 2t  2et sin 2tb#
a2et b# œ 3et Ê T œ œ ˆ "3 sin 2t
dT dt

2 3

cos 2t‰ i
ˆ 3" cos 2t 

œ ˆ 23 cos 2t 

Ê N(0) œ

2 3

sin 2t‰ j
23 k Ê T(0) œ

sin 2t‰ i
ˆ 23 sin 2t 

4 3

4 3

cos 2t‰ j Ê

dT dt

È

Š2 3 5‹

i
3" j
32 k ;

i  43 j Ê ¸ ddtT (0)¸ œ 23 È5 â j kâ 1 2 â 4 2 5 â 3 3 â œ È i
È j  È k; 3 5 3 5 3 5 2 â  È5 0 â

(0) œ

â â i â 2 2 " œ È5 i  È5 j ; B(0) œ T(0) ‚ N(0) œ ââ 3 â " â È5

ˆ 23 i 43 j‰

2 3 2 3

v kv k

a œ a4et cos 2t  3et sin 2tb i
a3et cos 2t  4et sin 2tb j
2et k Ê a(0) œ 4i  3j
2k and v(0) œ 2i
j
2k â â j kâ âi â â Ê v(0) ‚ a(0) œ â 2 " 2 â œ 8i
4j  10k Ê kv ‚ ak œ È64
16
100 œ 6È5 and kv(0)k œ 3 â â â 4 3 2 â Ê ,(0) œ

6È 5 3$

œ

2È 5 9 t

â â 2 â â 4 â â 2

1 3 11

; Þ t a œ a4e cos 2t  8e sin 2t  3et sin 2t  6et cos 2tb i
a3et cos 2t
6et sin 2t  4et sin 2t  8et cos 2tb j
2et k Þ œ a2et cos 2t  11et sin 2tb i
a11 et cos 2t
2et sin 2tb j
2et k Ê a(0) œ 2i  11j
2k Ê 7 (0) œ

kv ‚ a k #

â 2â â 2â â 2â

œ

80 180

œ  94 ; t œ 0 Ê (!ß "ß 2) is on the curve

25. r œ ti
"# e2t j Ê v œ i
e2t j Ê kvk œ È1
e4t Ê T œ dT dt

œ

2 e ˆ1
e4t ‰$Î# 4t

i


2t

2e ˆ1
e4t ‰$Î#

j Ê

dT dt

(ln 2) œ

32 17È17

i


8 17È17

" È1
e4t

i


e2t È1
e4t

j Ê T (ln 2) œ

j Ê N (ln 2) œ  È417 i


" È17

" È17

i


4 È17

j;

j;

â i j k ââ â â " 4 0 ââ œ k ; a œ 2e2t j Ê a(ln 2) œ 8j and v(ln 2) œ i
4j È17 B (ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ È17 â " â 4 â â È17 È17 0 â â â â i j kâ â â Þ 8 Ê v(ln 2) ‚ a(ln 2) œ â " 4 0 â œ 8k Ê kv ‚ ak œ 8 and kv(ln 2)k œ È17 Ê ,(ln 2) œ 17È ; a œ 4e2t j 17 â â â0 8 0â Þ Ê a(ln 2) œ 16j Ê 7 (ln 2) œ

â â1 â â0 â â0

4 8 16 kv ‚a k #

â 0â â 0â â 0â

œ 0; t œ ln 2 Ê (ln 2ß 2ß 0) is on the curve

26. r œ (3 cosh 2t)i
(3 sinh 2t)j
6tk Ê v œ (6 sinh 2t)i
(6 cosh 2t)j
6k Ê kvk œ È36 sinh# 2t
36 cosh# 2t
36 œ 6È2 cosh 2t Ê T œ kvvk œ Š È" tanh 2t‹ i
2

Ê T(ln 2) œ #

15 17È2

i


" È2

j


8 17È2

k;

8 ‰ 8 ‰ ˆ 15 ‰ œ Š È22 ‹ ˆ 17 i  Š È22 ‹ˆ 17 17 k œ

dT dt

œ

Š È22

sech 2t‹ i 

Š È22

sech 2t tanh 2t‹ k Ê

j
Š È" sech 2t‹ k

dT dt

(ln 2)

2

#

#

È

8 2 128 240 k Ê ¸ ddtT (ln 2)¸ œ ÊŠ 289 È2 ‹
Š 289È2 ‹ œ 17 â â j k â â i â â " 15 8 " 8 15 8 Ê N(ln 2) œ 17 i  17 k ; B(ln 2) œ T(ln 2) ‚ N(ln 2) œ ââ 17È2 È2 17È2 ââ œ  1715 È2 i
È2 j  17È2 k ; â 8 â 0  15 â 17 17 â 17 ‰ 15 ‰ 51 ˆ ˆ a œ (12 cosh 2t)i
(12 sinh 2t)j Ê a(ln 2) œ 12 8 i
12 8 j œ # i
45 # j and â i j k ââ â â 45 51 45 51 ‰ ˆ 17 ‰ 6 ââ v(ln 2) œ 6 ˆ 15 4 8 i
6 8 j
6k œ 4 i
4 j
6k Ê v(ln 2) ‚ a(ln 2) œ ââ 4 â 45 â 51 0â 2 # 128 289È2

i

#

" È2

240 289È2

œ 135i
153j  72k Ê kv ‚ ak œ 153È2 and kv(ln 2)k œ

51 4

È2 Ê ,(ln 2) œ

153È2 $ Š 51 È2‹

œ

4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

32 867

;

Chapter 13 Practice Exercises

Þ Þ a œ (24 sinh 2t)i
(24 cosh 2t)j Ê a(ln 2) œ 45i
51j Ê 7 (ln 2) œ

â 45 â 4 â 5" â 2 â â 45

51 4 45 2

51 kv ‚a k #

â 6â â 0â â 0â

œ

32 867

857

;

45 ‰ t œ ln 2 Ê ˆ 51 8 ß 8 ß 6 ln 2 is on the curve

27. r œ a2
3t
3t# b i
a4t
4t# b j  (6 cos t)k Ê v œ (3
6t)i
(4
8t)j
(6 sin t)k Ê kvk œ È(3
6t)#
(4
8t)#
(6 sin t)# œ È25
100t
100t#
36 sin# t "Î#

" #

a25
100t
100t#
36 sin# tb (100
200t
72 sin t cos t) Ê aT (0) œ ddtkvk (0) œ 10; a œ 6i
8j
(6 cos t)k Ê kak œ È6#
8#
(6 cos t)# œ È100
36 cos# t Ê ka(0)k œ È136 Ê

d kv k dt

œ

Ê aN œ Ékak#  a#T œ È136  10# œ È36 œ 6 Ê a(0) œ 10T
6N 28. r œ (2
t)i
at
2t# b j
a1
t# b k Ê v œ i
(1
4t)j
2tk Ê kvk œ È1#
(1
4t)#
(2t)# "Î# œ È2
8t
20t# Ê ddtkvk œ "# a2
8t
20t# b (8
40t) Ê aT œ ddtkvk (0) œ 2È2; a œ 4j
2k #

Ê kak œ È4#
2# œ È20 Ê aN œ Ékak#  aT# œ Ê20  Š2È2‹ œ È12 œ 2È3 Ê a(0) œ 2È2T
2È3N 29. r œ (sin t)i
ŠÈ2 cos t‹ j
(sin t)k Ê v œ (cos t)i  ŠÈ2 sin t‹ j
(cos t)k #

Ê kvk œ Ê(cos t)#
ŠÈ2 sin t‹
(cos t)# œ È2 Ê T œ

v kv k

œ Š È"2 cos t‹ i  (sin t)j
Š È"2 cos t‹ k ; #

#

œ Š È"2 sin t‹ i  (cos t)j  Š È"2 sin t‹ k Ê ¸ ddtT ¸ œ ÊŠ È"2 sin t‹
( cos t)#
Š È"2 sin t‹ œ 1 â â i j k â â â " cos t  sin t â " ˆ ddtT ‰ cos t " " â â È2 Ê N œ ¸ dT ¸ œ Š È2 sin t‹ i  (cos t)j  Š È2 sin t‹ k ; B œ T ‚ N œ â È2 â dt â  " sin t  cos t  " sin t â â È2 â È2 â i â j k â â â â œ È"2 i  È"2 k ; a œ ( sin t)i  ŠÈ2 cos t‹ j  (sin t)k Ê v ‚ a œ â cos t È2 sin t cos t â â â â  sin t È2 cos t  sin t â Þ œ È2 i  È2 k Ê kv ‚ ak œ È4 œ 2 Ê , œ kv‚$ak œ 2 $ œ " ; a œ ( cos t)i
ŠÈ2 sin t‹ j  (cos t)k

dT dt

kv k

Ê 7œ

â â cos t â â  sin t â â â  cos t

â cos t ââ È2 sin t È2 cos t  sin t ââ È2 sin t  cos t ââ kv ‚a k #

œ

ŠÈ2‹

È2

(cos t) ŠÈ2‹  ŠÈ2 sin t‹ (0)
(cos t) ŠÈ2‹ 4

œ0

30. r œ i
(5 cos t)j
(3 sin t)k Ê v œ (5 sin t)j
(3 cos t)k Ê a œ (5 cos t)j  (3 sin t)k Ê v † a œ 25 sin t cos t  9 sin t cos t œ 16 sin t cos t; v † a œ 0 Ê 16 sin t cos t œ 0 Ê sin t œ 0 or cos t œ 0 Ê t œ 0, 1# or 1 31. r œ 2i
ˆ4 sin #t ‰ j
ˆ3  1t ‰ k Ê 0 œ r † (i  j) œ 2(1)
ˆ4 sin #t ‰ (1) Ê 0 œ 2  4 sin Ê tœ

1 3

t #

Ê sin

t #

œ

" #

Ê

t #

œ

(for the first time)

32. r(t) œ ti
t# j
t$ k Ê v œ i
2tj
3t# k Ê kvk œ È1
4t#
9t% Ê kv(1)k œ È14 Ê T(1) œ È"14 i
È214 j
È314 k , which is normal to the normal plane Ê

" È14

(x  1)


2 È14

(y  1)


3 È14

(z  1) œ 0 or x
2y
3z œ 6 is an equation of the normal plane. Next we

calculate N(1) which is normal to the rectifying plane. Now, a œ 2j
6tk Ê a(1) œ 2j
6k Ê v(1) ‚ a(1)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1 6

858

Chapter 13 Vector-Valued Functions and Motion in Space â âi â œ â" â â0 œ

" #

œ

22 È14

â j kâ â 2 3 â œ 6i  6j
2k Ê kv(1) ‚ a(1)k œ È76 Ê ,(1) œ â 2 6â

a1
4t#
9t% b

"Î#

2j
3k Š i
È ‹
14

a8t
36t$ b¹

È19 7È14

#

œ

tœ1

22 È14

, so a œ

ŠÈ14‹ N Ê N œ

È14 2È19

d# s dt#

È76 $ È Š 14‹

œ

È19 7È14

;

ds dt

œ kv(t)k Ê

d# s dt# ¹ tœ1

#

‰ N Ê 2j
6k T
, ˆ ds dt

8 9 ‰ 11 8 9 ˆ 11 7 i  7 j
7 k Ê  7 (x  1)  7 (y  1)
7 (z  1)

œ 0 or 11x
8y  9z œ 10 is an equation of the rectifying plane. Finally, B(1) œ T(1) ‚ N(1) â â j kâ â i È14 â â 2 3 â œ È" (3i  3j
k) Ê 3(x  1)  3(y  1)
(z  1) œ 0 or 3x  3y
z œ Š 2È19 ‹ Š È" ‹ ˆ 7" ‰ â " 19 14 â â â 11 8 9 â œ 1 is an equation of the osculating plane. " ‰ 33. r œ et i
(sin t)j
ln (1  t)k Ê v œ et i
(cos t)j  ˆ 1  t k Ê v(0) œ i
j  k ; r(0) œ i Ê (1ß 0ß 0) is on the line

Ê x œ 1
t, y œ t, and z œ t are parametric equations of the line

34. r œ ŠÈ2 cos t‹ i
ŠÈ2 sin t‹ j
tk Ê v œ ŠÈ2 sin t‹ i
ŠÈ2 cos t‹ j
k Ê v ˆ 14 ‰ œ ŠÈ2 sin 14 ‹ i
ŠÈ2 cos 14 ‹ j
k œ i
j
k is a vector tangent to the helix when t œ is parallel to v ˆ 14 ‰ ; also r ˆ 14 ‰ œ ŠÈ2 cos 14 ‹ i
ŠÈ2 sin 14 ‹ j
Ê x œ 1  t, y œ 1
t, and z œ 35. (a) ?SOT ¸ ?TOD Ê Ê y! œ

6380# 6817

DO OT

œ

1 4

1 4

1 4

Ê the tangent line

k Ê the point ˆ1ß 1ß 14 ‰ is on the line


t are parametric equations of the line

OT SO

Ê

y! 6380

œ

6380 6380
437

Ê y! ¸ 5971 km;

(b) VA œ '5971 21x Ê1
Š dx dy ‹ dy #

6380

6380 œ 21'5971 È6380#  y# Š È6380 #  y# ‹ dy 6817

œ 21 '5971 6380 dy œ 21 c6380yd ')"( &*(" 6817

œ 16,395,469 km# ¸ 1.639 ‚ 10( km# ; (c) percentage visible ¸

16,395,469 km# 41(6380 km)#

¸ 3.21%

CHAPTER 13 ADDITIONAL AND ADVANCED EXERCISES 1. (a) The velocity of the boat at (xß y) relative to land is the sum of the velocity due to the rower and the " velocity of the river, or v œ
 250 (y  50)#
10‘ i  20j . Now, dy dt œ 20 Ê y œ 20t
c; y(0) œ 100 " Ê c œ 100 Ê y œ 20t
100 Ê v œ
 250 (20t
50)#
10‘ i  20j œ ˆ 85 t#
8t‰ i  20j 8 $ Ê r(t) œ ˆ 15 t
4t# ‰ i  20tj
C" ; r(0) œ 0i
100j Ê 100j œ C" Ê r(t) 8 $ œ ˆ 15 t
4t# ‰ i
(100  20t) j

(b) The boat reaches the shore when y œ 0 Ê 0 œ 20t
100 from part (a) Ê t œ 5 8 100 ‰ Ê r(5) œ ˆ 15 † 125
4 † 25‰ i
(100  20 † 5)j œ ˆ 200 3
100 i œ 3 i ; the distance downstream is therefore

100 3

m

2. (a) Let ai
bj be the velocity of the boat. The velocity of the boat relative to an observer on the bank of the river is v œ ai
’b 

3x(20  x) “j. 100

x œ at Ê v œ ai
’b 

The distance x of the boat as it crosses the river is related to time by

3at(20  at) “j 100

œ ai
Šb


3a# t#  60at ‹j 100

Ê r(t) œ ati
Šbt


a# t$ 100

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley



30at# 100 ‹ j


C;

Chapter 13 Additional and Advanced Exercises a# t$  30at# ‹j. 100

r(0) œ 0i
0j Ê C œ 0 Ê r(t) œ ati
Šbt
Ê 20 œ at Ê t œ œ

2000b
8000  12,000 100a

#

ˆ 20 ‰$ a

The boat reaches the shore when x œ 20

‰  30a ˆ 20 a 100

#

$

#

(20) 30(20) œ 20b a
100a # # È È Ê b œ 2; the speed of the boat is 20 œ kvk œ a
b œ Èa#
4 Ê a# œ 16 20 a

‰ and y œ 0 Ê 0 œ b ˆ 20 a


a

Ê a œ 4; thus, v œ 4i
2j is the velocity of the boat a# t$  30at# ‹j 100

(b) r(t) œ ati
Šbt


$

(c) x œ 4t and y œ 2t
œ œ

16t$ 100

œ 4ti
Š2t


120t# 100 ‹ j



by part (a), where 0 Ÿ t Ÿ 5

#

16t 120t 100  100 4 $ 6 # 2 # 25 t  5 t
2t œ 25 t a2t  15t
25b 2 25 t(2t  5)(t  5), which is the graph of

the cubic displayed here

3. (a) r()) œ (a cos ))i
(a sin ))j
b)k Ê œ Èa#
b# (b)

d) dt

d) dt

Ê

) œ É a#2gb
b# Ê

d) dt

d) È)

dr dt

œ [(a sin ))i
(a cos ))j
bk]

) É a#2gb œ É a#2gz
b# œ
b# Ê

gbt# 2 aa #
b # b

; z œ b) Ê z œ

œ [(a sin ))i
(a cos ))j
bk]

d) dt

; kvk œ È2gz œ ¸ ddtr ¸

œ É a#41
gbb# œ 2É a#1
gbb#

gb# t# 2 aa #
b # b

œ [(a sin ))i
(a cos ))j
bk] Š a# gbt
b# ‹ , from part (b)

i
(a cos ))j
bk Ê v(t) œ ’ (a sin ))È “ Š Èagbt ‹œ #
b# a#
b# d# r dt#

d) ¸ dt )œ#1

d) dt

"Î# œ É a#2gb œ É a#2gb
b# dt Ê 2)
b# t
C; t œ 0 Ê ) œ 0 Ê C œ 0

Ê 2)"Î# œ É a#2gb
b# t Ê ) œ (c) v(t) œ

dr dt

gbt È a#
b #

T;

#

œ [(a cos ))i  (a sin ))j] ˆ ddt) ‰
[(a sin ))i
(a cos ))j
bk] #

d# ) dt#

gb œ Š a# gbt
b# ‹ [(a cos ))i  (a sin ))j]
[(a sin ))i
(a cos ))j
bk] Š a#
b# ‹ #

i
(a cos ))j
bk œ ’ (a sin ))È “ Š Èa#gb
b# ‹
a Š a# gbt
b# ‹ [( cos ))i  (sin ))j] a#
b#

œ

gb È a#
b#

#

T
a Š a# gbt
b# ‹ N (there is no component in the direction of B).

4. (a) r()) œ (a) cos ))i
(a) sin ))j
b)k Ê

dr dt

# "Î#

kvk œ È2gz œ ¸ ddtr ¸ œ aa#
a# )#
b b (b) s œ '0 kvk dt œ '0 aa#
a# )#
b# b t

t

œ '0 aÉ a )

#


b# a#

(1
e)r! 1
e cos )

Ê

dr d)

ˆ ddt) ‰ Ê

d) dt

œ

œ

c# #

)

ln ¹u
Èc#
u# ¹“ œ

(1
e)r! (e sin )) (1
e cos ))#

!

;

dr d)

œ0 Ê

Ê sin ) œ 0 Ê ) œ 0 or 1. Note that

dr d)

a #

;

È a#
a# ) #
b#

dt œ '0 aa#
a# )#
b# b


u# du œ a '0 Èc#
u# du, where c œ

d) dt

È2gb)

t

)

Ê s œ a ’ u# Èc#
u#
5. r œ

"Î# d) dt

œ [(a cos )  a) sin ))i
(a sin )
a) cos ))j
bk]

"Î#

d) œ '0 aa#
a# u#
b# b )

"Î#

du

È a#
b# ka k

Š)Èc#
)#
c# ln ¹)
Èc#
)# ¹  c# ln c‹

(1
e)r! (e sin )) (1
e cos ))#

œ 0 Ê (1
e)r! (e sin )) œ 0

 0 when sin )  0 and

dr d)

 0 when sin )  0. Since sin )  0 on

1  )  0 and sin )  0 on 0  )  1, r is a minimum when ) œ 0 and r(0) œ

(1
e)r! 1
e cos 0

œ r!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

859

860

Chapter 13 Vector-Valued Functions and Motion in Space

6. (a) f(x) œ x  1 

" #

sin x œ 0 Ê f(0) œ 1 and f(2) œ 2  1 

" #

sin 2  

" #

since ksin 2k Ÿ 1; since f is continuous

on [0ß 2], the Intermediate Value Theorem implies there is a root between 0 and 2 (b) Root ¸ 1.4987011335179 7. (a) v œ

dx dt

i


v†iœ Ê

j and v œ

dr dt

œ

dr dt

d) ˆ dr ‰ ˆ d) ‰ dt u) œ dt [(cos ))i
(sin ))j]
r dt [( sin ))i
(cos ))j] Ê dy dx dr d) dr d) dt œ dt cos )  r dt sin ); v † j œ dt and v † j œ dt sin )
r dt cos )

ur
r

cos )  r ddt) sin ) Ê

dr dt

dy dt

dy dt

v†i œ

dx dt

and

sin )
r ddt) cos )

dy (b) ur œ (cos ))i
(sin ))j Ê v † ur œ dx dt cos )
dt sin ) d) d) ‰ ˆ dr ‰ œ ˆ dr dt cos )  r dt sin ) (cos ) )
dt sin )
r dt cos ) (sin ) ) by part (a),

Ê v † ur œ

dr dt

; therefore,

œ

dr dt

dx dt

cos )


dy dt

sin );

u) œ (sin ))i
(cos ))j Ê v † u) œ  sin )
dy dt cos ) dr d) dr d) ˆ ‰ ˆ œ dt cos )  r dt sin ) ( sin ))
dt sin )
r dt cos )‰ (cos )) by part (a) Ê v † u) œ r ddt) ; dx dt

therefore, r ddt) œ  dx dt sin )
8. r œ f()) Ê

dr dt

œ f w () )

d) dt

Ê

d# r dt#

dy dt

cos ) #

œ f ww ()) ˆ ddt) ‰
f w ())

d# ) dt#

;vœ

dr dt

ur
r

d) dt

u) "Î#

"Î#

# d) ‰  r sin ) ddt) ‰ i
ˆsin ) dr
r# ˆ ddt) ‰ “ œ ’af w b#
f # “ dt
r cos ) dt j Ê kvk œ Þ ÞÞ Þ ÞÞ d) dr kv ‚ ak œ kx y  y xk , where x œ r cos ) and y œ r sin ). Then dx dt œ (r sin )) dt
(cos )) dt

œ ˆcos )

‰# ’ˆ dr dt

dr dt

# # d) dr d) ˆ d) ‰#  (r sin )) ddt#)
(cos )) ddt#r ; dy dt dt  (r cos )) dt dt œ (r cos )) dt
(sin # # ˆ d) ‰#
(r cos )) ddt#)
(sin )) ddt#r . Then kv ‚ ak Ê œ (2 cos )) ddt) dr dt  (r sin )) dt # $ $ d) d# r d) ˆ dr ‰# œ (after much algebra) r# ˆ ddt) ‰
r ddt#) dr œ ˆ ddt) ‰ Šf 2  f † f ww
2af w b2 ‹ dt  r dt dt#
2 dt dt

Ê

d# x dt# d# y dt#

Ê ,œ

œ (2 sin ))

kv ‚a k kv k

œ

dr dt

dr dt

f 2  f†f ww
2af w b2 $Î# af w b#
f # ‘

9. (a) Let r œ 2  t and ) œ 3t Ê vœ

))

ˆ ddt) ‰ ;

dr dt

œ 1 and

d) dt

d# r dt#

œ3 Ê #

ur
r ddt) u) Ê v(1) œ ur
3u) ; a œ ’ ddt#r 

œ

d# ) dt#

# r ˆ ddt) ‰ “ ur

œ 0. The halfway point is (1ß 3) Ê t œ 1; #


’r ddt#)
2 dr dt

d) dt “ u)

Ê a(1) œ 9ur  6u)

(b) It takes the beetle 2 min to crawl to the origin Ê the rod has revolved 6 radians # # Ê L œ '0 É[f())]#
cf w ())d# d) œ '0 Ɉ2  3) ‰
ˆ "3 ‰ d) œ '0 É4  6

6

œ '0 É 37  129 )
) d) œ 6

#

œ È37 

" 6

dL dt

4) 3




)# 9




" 9

'06 È()  6)#
1 d) œ "3 ’ ()#6) È()  6)#
1
"# ln ¸)  6
È()  6)#
1¸“ ' !

dL dt

œ ˆ ddtr ‚ mv‰
Šr ‚ m

d# r dt# ‹

Ê

dL dt

œ (v ‚ mv)
(r ‚ ma) œ r ‚ ma ; F œ ma Ê  krck$ r

œ r ‚ ma œ r ‚ Š krck$ r‹ œ  krck$ (r ‚ r) œ 0 Ê L œ constant vector

â â i â 11. (a) ur ‚ u) œ â cos ) â â  sin )

j sin ) cos )

â kâ â 0 â œ k Ê a right-handed frame of unit vectors â 0â

œ ( sin ))i
(cos ))j œ u) and ddu)) œ ( cos ))i  (sin ))j œ ur Þ ÞÞ Þ Þ Þ ÞÞ ÞÞ ÞÞ Þ Þ ÞÞ (c) From Eq. (7), v œ rur
r)u)
zk Ê a œ v œ a r ur
r ur b
ˆr )u)
r) u)
r) u) ‰
z k Þ# ÞÞ ÞÞ ÞÞ ÞÞ œ Š r  r) ‹ ur
ˆr)
2r )‰ u)
z k (b)

d)

ln ŠÈ37  6‹ ¸ 6.5 in.

10. L(t) œ r(t) ‚ mv(t) Ê œ ma Ê

" 3

6

dur d)

12. (a) x œ r cos ) Ê dx œ cos ) dr  r sin ) d); y œ r sin ) Ê dy œ sin ) dr
r cos ) d); thus dx# œ cos# ) dr#  2r sin ) cos ) dr d)
r# sin# ) d)# and Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 13 Additional and Advanced Exercises dy# œ sin# ) dr#
2r sin ) cos ) dr d)
r# cos# ) d)# Ê dx#
dy#
dz# œ dr#
r# d)#
dz# (b) (c) r œ e) Ê dr œ e) d) Ê L œ '0

ln 8

Èdr#
r# d)#
dz#

œ '0 Èe#)
e#)
e#) d) ln 8

œ '0 È3e) d) œ ’È3 e) “ ln 8

ln 8 0

œ 8È3  È3 œ 7È3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

861

862

Chapter 13 Vector-Valued Functions and Motion in Space

NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

CHAPTER 14 PARTIAL DERIVATIVES 14.1 FUNCTIONS OF SEVERAL VARIABLES 1. (a) (b) (c) (d) (e) (f)

Domain: all points in the xy-plane Range: all real numbers level curves are straight lines y
x œ c parallel to the line y œ x no boundary points both open and closed unbounded

2. (a) (b) (c) (d)

Domain: set of all (xß y) so that y
x   0 Ê y   x Range: z   0 level curves are straight lines of the form y
x œ c where c   0 boundary is Èy
x œ 0 Ê y œ x, a straight line

(e) closed (f) unbounded 3. (a) Domain: all points in the xy-plane (b) Range: z   0 (c) level curves: for f(xß y) œ 0, the origin; for f(xß y) œ c  0, ellipses with center (!ß 0) and major and minor axes along the x- and y-axes, respectively (d) no boundary points (e) both open and closed (f) unbounded 4. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves: for f(xß y) œ 0, the union of the lines y œ „ x; for f(xß y) œ c Á 0, hyperbolas centered at (0ß 0) with foci on the x-axis if c  0 and on the y-axis if c  0 (d) no boundary points (e) both open and closed (f) unbounded 5. (a) Domain: all points in the xy-plane (b) Range: all real numbers (c) level curves are hyperbolas with the x- and y-axes as asymptotes when f(xß y) Á 0, and the x- and y-axes when f(xß y) œ 0 (d) no boundary points (e) both open and closed (f) unbounded 6. (a) Domain: all (xß y) Á (0ß y) (b) Range: all real numbers (c) level curves: for f(xß y) œ 0, the x-axis minus the origin; for f(xß y) œ c Á 0, the parabolas y œ cx# minus the origin (d) boundary is the line x œ 0

864

Chapter 14 Partial Derivatives

(e) open (f) unbounded 7. (a) Domain: all (xß y) satisfying x#  y#  16 (b) Range: z   "4 (c) (d) (e) (f)

level curves are circles centered at the origin with radii r  4 boundary is the circle x#  y# œ 16 open bounded

8. (a) (b) (c) (d) (e) (f)

Domain: all (xß y) satisfying x#  y# Ÿ 9 Range: 0 Ÿ z Ÿ 3 level curves are circles centered at the origin with radii r Ÿ 3 boundary is the circle x#  y# œ 9 closed bounded

9. (a) (b) (c) (d) (e) (f)

Domain: (xß y) Á (0ß 0) Range: all real numbers level curves are circles with center (!ß 0) and radii r  0 boundary is the single point (0ß 0) open unbounded

10. (a) (b) (c) (d) (e) (f)

Domain: all points in the xy-plane Range: 0  z Ÿ 1 level curves are the origin itself and the circles with center (0ß 0) and radii r  0 no boundary points both open and closed unbounded

11. (a) Domain: all (xß y) satisfying
1 Ÿ y
x Ÿ 1 (b) Range:
1# Ÿ z Ÿ 1# (c) (d) (e) (f)

level curves are straight lines of the form y
x œ c where
1 Ÿ c Ÿ 1 boundary is the two straight lines y œ 1  x and y œ
1  x closed unbounded

12. (a) Domain: all (xß y), B Á 0 (b) Range:
1#  z  1# (c) (d) (e) (f)

level curves are the straight lines of the form y œ cx, c any real number and x Á 0 boundary is the line x œ 0 open unbounded

13. f

14. e

15. a

16. c

17. d

18. b

Section 14.1 Functions of Several Variables 19. (a)

(b)

20. (a)

(b)

21. (a)

(b)

22. (a)

(b)

865

866

Chapter 14 Partial Derivatives

23. (a)

(b)

24. (a)

(b)

25. (a)

(b)

Section 14.1 Functions of Several Variables 26. (a)

(b)

27. (a)

(b)

28. (a)

(b)

#

#

867

29. f(xß y) œ 16
x#
y# and Š2È2ß È2‹ Ê z œ 16
Š2È2‹
ŠÈ2‹ œ 6 Ê 6 œ 16
x#
y# Ê x#  y# œ 10 30. f(xß y) œ Èx#
1 and (1ß 0) Ê D œ È1#
1 œ 0 Ê x#
1 œ 0 Ê x œ 1 or x œ
1 31. f(xß y) œ 'x

y

1 1
t#

dt at Š
È2ß È2‹ Ê z œ tan" y
tan" x; at Š
È2ß È2‹ Ê z œ tan" È2
tan" Š
È2‹

œ 2 tan" È2 Ê tan" y
tan" x œ 2 tan" È2

_

n

32. f(xß y) œ ! Š xy ‹ at (1ß 2) Ê z œ

œ

n 0

Ê y œ 2x

" 1  Š xy ‹

œ

y yx

; at (1ß 2) Ê z œ

2 #1

œ2 Ê 2œ

y yx

Ê 2y
2x œ y

868

Chapter 14 Partial Derivatives

33.

34.

35.

36.

37.

38.

39.

40.

41. f(xß yß z) œ Èx
y
ln z at (3ß
1ß 1) Ê w œ Èx
y
ln z; at (3ß
1ß 1) Ê w œ È3
(
1)
ln 1 œ 2 Ê Èx
y
ln z œ 2

Section 14.1 Functions of Several Variables

869

42. f(xß yß z) œ ln ax#  y  z# b at (
"ß #ß ") Ê w œ ln ax#  y  z# b ; at (
"ß #ß ") Ê w œ ln (1  2  1) œ ln 4 Ê ln 4 œ ln ax#  y  z# b Ê x#  y  z# œ 4 _

(x b y)n n! zn

43. g(xß yß z) œ !

nœ0

_

at (ln 2ß ln 4ß 3) Ê w œ !

œ eÐln 8ÑÎ3 œ eln 2 œ 2 Ê 2 œ eÐx
yÑÎz Ê 44. g(xß yß z) œ 'x

y

d) È1  )#

 'È2 z

dt tÈt#  1

x
y z

nœ0

(x b y)n n! zn

œ eÐx
yÑÎz ; at (ln 2ß ln 4ß 3) Ê w œ eÐln 2
ln 4ÑÎ3

œ ln 2

at ˆ0ß "# ß 2‰ Ê w œ csin" )d x  csec" td È2 y

z

œ sin" y
sin" x  sec" z
sec" ŠÈ2‹ Ê w œ sin" y
sin" x  sec" z
Ê w œ sin"

" #


sin" 0  sec" 2


1 4

œ

1 4

Ê

1 #

1 4

; at ˆ0ß "# ß 2‰

œ sin" y
sin" x  sec" z

45. f(xß yß z) œ xyz and x œ 20
t, y œ t, z œ 20 Ê w œ (20
t)(t)(20) along the line Ê w œ 400t
20t# Ê

dw dt

œ 400
40t;

dw dt

œ 0 Ê 400
40t œ 0 Ê t œ 10 and

d# w dt#

œ
40 for all t Ê yes, maximum at t œ 10

Ê x œ 20
10 œ 10, y œ 10, z œ 20 Ê maximum of f along the line is f(10ß 10ß 20) œ (10)(10)(20) œ 2000 46. f(xß yß z) œ xy
z and x œ t
1, y œ t
2, z œ t  7 Ê w œ (t
1)(t
2)
(t  7) œ t#
4t
5 along the line Ê

dw dt

œ 2t
4;

dw dt

œ 0 Ê 2t
4 œ 0 Ê t œ 2 and

d# w dt#

œ 2 for all t Ê yes, minimum at t œ 2 Ê x œ 2
1 œ 1,

y œ 2
2 œ 0, and z œ 2  7 œ 9 Ê minimum of f along the line is f(1ß 0ß 9) œ (1)(0)
9 œ
9 ‰ 47. w œ 4 ˆ Th d

"Î#

"Î#

km) œ 4 ’ (290 5K)(16.8 “ K/km

¸ 124.86 km Ê must be

" #

(124.86) ¸ 63 km south of Nantucket

48. The graph of f(x" ß x# ß x$ ß x% ) is a set in a five-dimensional space. It is the set of points (x" ß x# ß x$ ß x% ß f(x" ß x# ß x$ ß x% )) for (x" ß x# ß x$ ß x% ) in the domain of f. The graph of f(x" ß x# ß x$ ß á ß xn ) is a set in an (n  1)-dimensional space. It is the set of points (x" ß x# ß x$ ß á ß xn ß f(x" ß x# ß x$ ß á ß xn )) for (x" ß x# ß x$ ß á ß xn ) in the domain of f. 49-52. Example CAS commands: Maple: with( plots ); f := (x,y) -> x*sin(y/2) + y*sin(2*x); xdomain := x=0..5*Pi; ydomain := y=0..5*Pi; x0,y0 := 3*Pi,3*Pi; plot3d( f(x,y), xdomain, ydomain, axes=boxed, style=patch, shading=zhue, title="#49(a) (Section 14.1)" ); plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, orientation=[-90,0], title="#49(b) (Section 14.1)" ); # (b) L := evalf( f(x0,y0) ); # (c) plot3d( f(x,y), xdomain, ydomain, grid=[50,50], axes=boxed, shading=zhue, style=patchcontour, contours=[L], orientation=[-90,0], title="#49(c) (Section 14.1)" ); 53-56. Example CAS commands: Maple: eq := 4*ln(x^2+y^2+z^2)=1; implicitplot3d( eq, x=-2..2, y=-2..2, z=-2..2, grid=[30,30,30], axes=boxed, title="#53 (Section 14.1)" );

870

Chapter 14 Partial Derivatives

57-60. Example CAS commands: Maple: x := (u,v) -> u*cos(v); y := (u,v) ->u*sin(v); z := (u,v) -> u; plot3d( [x(u,v),y(u,v),z(u,v)], u=0..2, v=0..2*Pi, axes=boxed, style=patchcontour, contours=[($0..4)/2], shading=zhue, title="#57 (Section 14.1)" ); 49-60. Example CAS commands: Mathematica: (assigned functions and bounds will vary) For 49 - 52, the command ContourPlot draws 2-dimensional contours that are z-level curves of surfaces z = f(x,y). Clear[x, y, f] f[x_, y_]:= x Sin[y/2]  y Sin[2x] xmin= 0; xmax= 51; ymin= 0; ymax= 51; {x0, y0}={31, 31}; cp= ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False]; cp0= ContourPlot[[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, Contours Ä {f[x0,y0]}, ContourShading Ä False, PlotStyle Ä {RGBColor[1,0,0]}]; Show[cp, cp0] For 53 - 56, the command ContourPlot3D will be used and requires loading a package. Write the function f[x, y, z] so that when it is equated to zero, it represents the level surface given. For 53, the problem associated with Log[0] can be avoided by rewriting the function as x2 + y2 +z2 - e1/4 <
2.

3.

4.

5.

3x#  y#
5 # # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x
y
2

lim

x

lim Ðx ß y Ñ Ä Ð 0 ß 4 Ñ È y lim

Ðx ß y Ñ Ä Ð 3 ß 4 Ñ

œ

0 È4

œ

3(0)#  0#
5 0#
0#
2

œ

5 #

œ0

Èx#  y#
1 œ È3#  4#
1 œ È24 œ 2È6 #

#

#

lim

Š "x  y" ‹ œ
#"  ˆ "3 ‰‘ œ ˆ 6" ‰ œ

lim

sec x tan y œ (sec 0) ˆtan 14 ‰ œ (1)(1) œ 1

Ðx ß y Ñ Ä Ð 2 ß
3 Ñ

Ðxß yÑ Ä ˆ0ß 14 ‰

" 36

Section 14.2 Limits and Continuity 6.

7.

8.

9.

10.

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

#

lim

œ

lim

Ðx ß y Ñ Ä Ð 0 ß 0Ñ

x sin y

œ

lim # Ðx ß y Ñ Ä Ð 1 ß 0 Ñ x
1

12.

lim y  sin x Ðxß yÑ Ä ˆ 12 ß 0‰

cos y
1

1†sin 0 1#
1

x#  2xy
y# xy Ðxß yÑ Ä Ð1ß 1Ñ

œ

lim

x#  y# Ðxß yÑ Ä Ð1ß 1Ñ x  y

lim

œ

16.

17.

lim

Ðxß yÑ Ä Ð1ß 1Ñ xÁ1

xÄ0

œ0

0 2

œ

œ 2

(x  y)# Ð x ß y Ñ Ä Ð 1 ß 1Ñ x  y

œ

(x
y)(x  y) xy

œ

lim

xy  y  2x
2 x1

1
1 1

lim

Ðx ß y Ñ Ä Ð 1 ß 1Ñ

xÁy

15.

œ

(cos 0)
" 0  sin ˆ 1# ‰

œ

xÁy

14.

aey b ˆ sinx x ‰ œ e! † lim ˆ sinx x ‰ œ 1 † 1 œ 1

cos ˆ $Èkxyk  1‰ œ cos ˆ $È(1)(1)  1‰ œ cos 0 œ 1

11.

13.

" #

ln k1
x# y# k œ ln k1
(1)# (1)# k œ ln 2

ey sin x x Ðxß yÑ Ä Ð0ß 0Ñ

lim

$

1

lim

Ðx ß y Ñ Ä Ð 1 ß 1 Ñ

#

exy œ e0  ln 2 œ eln ˆ 2 ‰ œ

lim

Ðxß yÑ Ä Ð0ß ln 2Ñ

Ðx ß y Ñ Ä Ð 1 ß 1 Ñ

$

cos Š xx

y
y 1 ‹ œ cos Š 00

0
0 1 ‹ œ cos 0 œ 1

œ

lim

Ðx ß y Ñ Ä Ð 1 ß 1Ñ xÁ1

y
4

œ

lim # # Ðxß yÑ Ä Ð2ß
4Ñ x y  xy
4x  4x # y Á
4, x Á x x  y
2È x  2È y Èx  Èy Ðxß yÑ Ä Ð0ß 0Ñ

lim

xÁy

œ

lim

(x  y) œ ("  1) œ 0

lim

(x
y) œ (1
1) œ 2

Ðxß yÑ Ä Ð1ß 1Ñ

Ðxß yÑ Ä Ð1ß 1Ñ

(x  1)(y  2) x1

œ

lim

Ðxß yÑ Ä Ð1ß 1Ñ

y
4

lim Ðxß yÑ Ä Ð2ß
4Ñ x(x  1)(y
4) y Á
4, x Á x# lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ xÁy

œ

(y  2) œ (1  2) œ 1

1

lim Ðxß yÑ Ä Ð2ß
4Ñ x(x  1) x Á x#

ˆÈ x  È y ‰ ˆ È x
È y
2 ‰ Èx  Èy

œ

lim

Ðxß yÑ Ä Ð0ß 0Ñ

œ

" #(2  1)

Note: (xß y) must approach (0ß 0) through the first quadrant only with x Á y. lim

Ðxß yÑ Ä Ð2ß 2Ñ xyÁ4

x
y4 Èx
y  2

œ

lim

Ðx ß y Ñ Ä Ð 2 ß 2Ñ xyÁ4

ˆÈ x
y
2 ‰ ˆ È x
y  2 ‰ Èx
y  2

œ

lim

Ðxß yÑ Ä Ð2ß 2Ñ xyÁ4

ˆÈ x
y
2‰

œ ŠÈ2
2
2‹ œ 2
2 œ 4 19.

lim

Ðx ß y Ñ Ä Ð 2 ß 0 Ñ 2x
y Á 4

œ 20.

È2x  y  2 2x  y  4

" È(2)(2)  0
#

lim

Ðxß yÑ Ä Ð4ß 3Ñ x
yÁ1

œ

" 2
2

È x  È y
1 xy1

œ œ œ

lim

È2x  y  2 ˆÈ2x  y
2‰ ˆÈ2x  y  2‰

lim

Èx  Èy
1 ˆÈ x
È y
1 ‰ ˆ È x  È y
1 ‰

Ð x ß y Ñ Ä Ð 2 ß 0Ñ 2x
y Á 4

œ

"

lim Ðxß yÑ Ä Ð2ß 0Ñ È2x  y
#

" 4

Ð x ß y Ñ Ä Ð 4 ß 3Ñ x
yÁ1

œ

" #

ˆÈ x
È y
2‰

œ ŠÈ0
È0
2‹ œ 2

18.

œ

"

lim Ðxß yÑ Ä Ð4ß 3Ñ Èx
Èy
1

871

872

Chapter 14 Partial Derivatives œ

21.

22.

23.

24.

25.

26.

" È4
È3
1

lim

T Ä Ð1 ß 3 ß 4 Ñ

œ

Š "x


lim

lim

lim

lim

" 4

œ


"z ‹ œ œ

" 1




" 3




" 4

12
4
3 12

œ

2(1)(1)
(1)(1) 1#
(1)#

œ

2
" 1
1

œ

19 12

œ  #"

asin# x
cos# y
sec# zb œ asin# 3
cos# 3b
sec# 0 œ 1
1# œ 2

T Ä ˆ
14 ß 12 ß 2‰

T Ä Ð1 ß 0 ß 3 Ñ

" y

2xy
yz x #
z#

lim

T Ä Ð1 ß
1 ß
1 Ñ

T Ä Ð3 ß 3 ß 0 Ñ

" 2
2

tan" (xyz) œ tan" ˆ "4 †

1 #

† 2‰ œ tan" ˆ 14 ‰

ze
2y cos 2x œ 3e
2Ð0Ñ cos 21 œ (3)(1)(1) œ 3

T Ä Ð0 ß
2 ß 0 Ñ

ln Èx#
y#
z# œ ln È0#
(2)#
0# œ ln È4 œ ln 2

27. (a) All (xß y) (b) All (xß y) except (0ß 0) 28. (a) All (xß y) so that x Á y (b) All (xß y) 29. (a) All (xß y) except where x œ 0 or y œ 0 (b) All (xß y) 30. (a) All (xß y) so that x#  3x
2 Á 0 Ê (x  2)(B  1) Á 0 Ê x Á 2 and x Á 1 (b) All (xß y) so that y Á x# 31. (a) All (xß yß z) (b) All (xß yß z) except the interior of the cylinder x#
y# œ 1 32. (a) All (xß yß z) so that xyz  0 (b) All (xß yß z) 33. (a) All (xß yß z) with z Á 0 (b) All (xß yß z) with x#
z# Á 1 34. (a) All (xß yß z) except (xß 0ß 0) (b) All (xß yß z) except (!ß yß 0) or (xß 0ß 0) 35.

lim



x È x#
y#

œ lim b  Èx#x
x# œ lim b  È2x kxk œ lim b  Èx2 x œ lim b  È"2 œ  È"2 ; xÄ0 xÄ0 xÄ0 xÄ0

lim



x È x#
y#

œ lim c  È2x kxk œ lim c  È2(xx) œ lim c xÄ0 xÄ0 xÄ0

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ x x0 Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ x x0

" È2

œ

" È2

Section 14.2 Limits and Continuity 36.

37.

38.

39.

40.

41.

42.

lim

x% x%
y#

œ lim

lim

x%  y# x%
y#

œ lim

lim

xy kxyk

lim

xy x
y

œ lim

x  kx

œ

1k 1
k

Ê different limits for different values of k, k Á 1

lim

x
y xy

œ lim

x
kx

œ

1
k 1k

Ê different limits for different values of k, k Á 1

lim

x#
y y

œ lim

x#
kx# kx#

œ

1
k k

Ê different limits for different values of k, k Á 0

lim

x# x#  y

œ lim

x#

œ

1 1k

Ê different limits for different values of k, k Á 1

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ 0

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ kx#

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ kx kÁ0

Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx k Á
1

Ðxß yÑ Ä Ð0ß 0Ñ along y œ kx kÁ1

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ kx# kÁ0

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ along y œ kx# kÁ1

x%

œ 1;

% # x Ä 0 x
0

#

x%  akx# b # x Ä 0 x%
akx# b

œ lim

kx#

# x Ä 0 kkx k

x Ä 0 x
kx

x Ä 0 x  kx

xÄ0

x%  k# x% % # % x Ä 0 x
k x

œ lim

œ lim

x(kx)

x Ä 0 kx(kx)k

# # x Ä 0 x  kx

x% x%
y#

lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ along y œ x#

œ lim

# x Ä 0 x %
ax # b

1  k# 1
k#

œ

x%

œ lim

œ lim

x%

% x Ä 0 2x

œ

873

" #

Ê different limits for different values of k

; if k  0, the limit is 1; but if k  0, the limit is 1

k

x Ä 0 kkk

43. No, the limit depends only on the values f(xß y) has when (xß y) Á (x! ß y! ) 44. If f is continuous at (x! ß y! ), then

lim

Ðx ß y Ñ Ä Ð x ! ß y ! Ñ

f(xß y) must equal f(x! ß y! ) œ 3. If f is not continuous at

(x! ß y! ), the limit could have any value different from 3, and need not even exist. 45.

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

Š1 

x# y# 3 ‹

œ 1 and

lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ

1œ1 Ê

# #

46. If xy  0,

2 kxyk  Š x 6y ‹

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

kxyk

tan
" xy xy

lim

Ðxß yÑ Ä Ð0ß 0Ñ

œ 1, by the Sandwich Theorem

# #

œ

2xy  Š x 6y ‹

lim

xy

Ð x ß y Ñ Ä Ð 0 ß 0Ñ

œ

lim

Ðxß yÑ Ä Ð0ß 0Ñ

ˆ2 

xy ‰ 6

œ 2 and

# #

2 kxyk Ðxß yÑ Ä Ð0ß 0Ñ kxyk

lim

œ

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

œ

lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ

ˆ2


xy ‰ 6

œ 2 and

2 œ 2; if xy  0, lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ

lim

2 kxyk  Š x 6y ‹ kxyk

Ðxß yÑ Ä Ð0ß 0Ñ

2 kxyk kxyk

œ2 Ê

lim

Ðxß yÑ Ä Ð0ß 0Ñ

# #

œ

lim

2xy  Š x 6y ‹

Ðxß yÑ Ä Ð0ß 0Ñ

4  4 cos Èkxyk kxyk

xy

œ 2, by the Sandwich

Theorem 47. The limit is 0 since ¸sin ˆ "x ‰¸ Ÿ 1 Ê 1 Ÿ sin ˆ x" ‰ Ÿ 1 Ê y Ÿ y sin ˆ x" ‰ Ÿ y for y   0, and y   y sin ˆ x" ‰   y for y Ÿ 0. Thus as (xß y) Ä (!ß !), both y and y approach 0 Ê y sin ˆ "x ‰ Ä 0, by the Sandwich Theorem. 48. The limit is 0 since ¹cos Š "y ‹¹ Ÿ 1 Ê 1 Ÿ cos Š y" ‹ Ÿ 1 Ê x Ÿ x cos Š y" ‹ Ÿ x for x   0, and x   x cos Š y" ‹   x for x Ÿ 0. Thus as (xß y) Ä (!ß !), both x and x approach 0 Ê x cos Š "y ‹ Ä 0, by the Sandwich Theorem.

874

Chapter 14 Partial Derivatives

49. (a) f(xß y)k yœmx œ

2m 1
m#

2 tan ) 1
tan# )

œ

œ sin 2). The value of f(xß y) œ sin 2) varies with ), which is the line's

angle of inclination. (b) Since f(xß y)k yœmx œ sin 2) and since 1 Ÿ sin 2) Ÿ 1 for every ),

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

f(xß y) varies from 1 to 1

along y œ mx. 50. kxy ax#  y# bk œ kxyk kx#  y# k Ÿ kxk kyk kx#
y# k œ Èx# Èy# kx#
y# k Ÿ Èx#
y# Èx#
y# kx#
y# k #

#

#

œ ax#
y# b Ê ¹ xyxa#x
y#y b ¹ Ÿ Ê

lim

xy

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

x#  y# x#
y#

ax #
y # b x#
y#

#

œ x#
y# Ê  ax#
y# b Ÿ

œ 0 by the Sandwich Theorem, since

xy ax#  y# b x#
y#

lim

Ð x ß y Ñ Ä Ð 0 ß 0Ñ

Ÿ a x#
y # b

„ ax#
y# b œ 0; thus, define

f(0ß 0) œ 0 51.

52.

53.

r$ cos$ )  (r cos )) ar# sin# )b r# cos# )
r# sin# ) rÄ0

x$  xy# # # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x
y

œ lim

lim

$

lim

Ðxß yÑ Ä Ð0ß 0Ñ

$

r acos$ )  cos ) sin# )b 1 rÄ0

œ lim

y r cos )  r sin ) cos Š xx# 
y# ‹ œ lim cos Š r# cos# )
r# sin# ) ‹ œ lim cos ’ $

$

$

$

rÄ0

y#

r# sin# ) r#

œ lim

lim # # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x
y

rÄ0

rÄ0

œ0

r acos$ )  sin$ )b “ 1

œ cos 0 œ 1

œ lim asin# )b œ sin# ); the limit does not exist since sin# ) is between rÄ0

0 and 1 depending on ) 54.

55.

2x

lim # # Ðxß yÑ Ä Ð0ß 0Ñ x
x
y lim

Ðxß yÑ Ä Ð0ß 0Ñ

2r cos )

œ lim

# r Ä 0 r
r cos )

œ lim

2 cos )

r Ä 0 r
cos )

2 cos ) cos )

; the limit does not exist for cos ) œ 0

ky k krk akcos )k
ksin )kb " kr cos )k
kr sin )k tan" ’ kxx#k
’ “ œ lim tan" ’ “;
y# “ œ lim tan r# r#

rÄ0




"

rÄ0

krk akcos )k
ksin )kb ’ “ r#

"

’

lim tan" ’ krk akcos )rk#
ksin )kb “ œ lim c tan" Š kcos )k
r ksin )k ‹ œ rÄ!

1 #

if r Ä 0 , then lim b tan rÄ!

œ lim b tan rÄ!

r Ä !c

56.

œ

x#  y#

œ lim

lim # # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x
y

rÄ0

r# cos# )  r# sin# ) r#

kcos )k
ksin )k “ r

œ

1 #

Ê the limit is

; if r Ä 0 , then 1 #

œ lim acos# )  sin# )b œ lim (cos 2)) which ranges between rÄ0

rÄ0

1 and 1 depending on ) Ê the limit does not exist 57.

lim

Ðx ß y Ñ Ä Ð 0 ß 0 Ñ

ln Š 3x

#

 x# y#
3y# ‹ x#
y#

œ lim ln Š 3r rÄ0

#

cos# )  r% cos# ) sin# )
3r# sin# ) ‹ r#

œ lim ln a3  r# cos# ) sin# )b œ ln 3 Ê define f(0ß 0) œ ln 3 rÄ0

58.

lim

Ðxß yÑ Ä Ð0ß 0Ñ

2xy# x #
y #

œ lim

rÄ0

(2r cos )) ar# sin# )b r#

œ lim 2r cos ) sin# ) œ 0 Ê define f(0ß 0) œ 0 rÄ0

59. Let $ œ 0.1. Then Èx#
y#  $ Ê Èx#
y#  0.1 Ê x#
y#  0.01 Ê kx#
y#  0k  0.01 Ê kf(xß y)  f(!ß !)k  0.01 œ %. 60. Let $ œ 0.05. Then kxk  $ and kyk  $ Ê kf(xß y)  f(0ß 0)k œ ¸ x# y
1  0¸ œ ¸ x# y
1 ¸ Ÿ kyk  0.05 œ %. 61. Let $ œ 0.005. Then kxk  $ and kyk  $ Ê kf(xß y)  f(0ß 0)k œ ¸ xx#

y1  0¸ œ ¸ xx#

y1 ¸ Ÿ kx
yk  kxk
kyk  0.005
0.005 œ 0.01 œ %.

Section 14.3 Partial Derivatives kx
y k " 62. Let $ œ 0.01. Since 1 Ÿ cos x Ÿ 1 Ê 1 Ÿ 2
cos x Ÿ 3 Ê "3 Ÿ #
cos Ÿ ¸ 2
x
cosy x ¸ Ÿ kx
yk x Ÿ 1 Ê 3 Ÿ kxk
kyk . Then kxk  $ and kyk  $ Ê kf(xß y)  f(0ß 0)k œ ¸ 2
x
cosy x  0¸ œ ¸ 2
x
cosy x ¸ Ÿ kxk
kyk  0.01
0.01

œ 0.02 œ %. 63. Let $ œ È0.015. Then Èx#
y#
z#  $ Ê kf(xß yß z)  f(!ß 0ß 0)k œ kx#
y#
z#  0k œ kx#
y#
z# k #

#

œ ŠÈx#
t#
x# ‹  ŠÈ0.015‹ œ 0.015 œ %. 64. Let $ œ 0.2. Then kxk  $ , kyk  $ , and kzk  $ Ê kf(xß yß z)  f(!ß 0ß 0)k œ kxyz  0k œ kxyzk œ kxk kyk kzk  (0.2)$ œ 0.008 œ %. 65. Let $ œ 0.005. Then kxk  $ , kyk  $ , and kzk  $ Ê kf(xß yß z)  f(!ß 0ß 0)k œ ¹ x#
x y
# y

z#z
1  0¹ œ ¹ x#
x y
# y

z#z
1 ¹ Ÿ kx
y
zk Ÿ kxk
kyk
kzk  0.005
0.005
0.005 œ 0.015 œ %. 66. Let $ œ tan" (0.1). Then kxk  $ , kyk  $ , and kzk  $ Ê kf(xß yß z)  f(!ß 0ß 0)k œ ktan# x
tan# y
tan# zk Ÿ ktan# xk
ktan# yk
ktan# zk œ tan# x
tan# y
tan# z  tan# $
tan# $
tan# $ œ 0.01
0.01
0.01 œ 0.03 œ %. 67.

f(xß yß z) œ

lim

Ðx ß y ß z Ñ Ä Ð x ! ß y ! ß z ! Ñ

lim

(x
y
z) œ x!
y!
z! œ f(x! ß y! ß z! ) Ê f is continuous at

lim

ax#
y#
z# b œ x!#
y!#
z!# œ f(x! ß y! ß z! ) Ê f is continuous at

Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ

every (x! ß y! ß z! ) 68.

f(xß yß z) œ

lim

Ðx ß y ß z Ñ Ä Ð x ! ß y ! ß z ! Ñ

Ðxß yß zÑ Ä Ðx! ß y! ß z! Ñ

every point (x! ß y! ß z! ) 14.3 PARTIAL DERIVATIVES 1.

`f `x

œ 4x,

`f `y

3.

`f `x

œ 2x(y
2),

5.

`f `x

œ 2y(xy  1),

7.

`f `x

œ

9.

`f `x

œ  (x
" y)# †

10.

`f `x

œ

11.

`f `x

œ œ

12.

`f `x

œ

13.

`f `x

œ eÐx
y
1Ñ †

œ 3 `f `y

x `f È x#
y# , ` y

œ x#  1

`f `y

œ

` `x

œ 2x(xy  1) y È x#
y#

(x
y) œ  (x
" y)# ,

ax#
y# b (1) x(2x) ax #
y # b #

œ

y#  x# a x #
y # b#

(xy  1)(1)  (x
y)(y) (xy  1)#

" # 1
ˆ xy ‰

†

` `x

ˆ xy ‰ œ 

` `x

œ

,

`f `y

œ

 y#  1 (xy  1)# y #

x# ’1
ˆ xy ‰ “

,

`f `y

œ  (x
" y)# †

ax#
y# b (0)  x(2y) a x #
y # b# `f `y

œ

`f `x

œ 2x  y,

4.

`f `x

œ 5y  14x
3,

6.

`f `x

œ 6(2x  3y)# ,

8.

`f `x

œ

` `y

`f `y

`f `y

$

`f `y

2x#
ˆ #y ‰

É x$

,

œ x
2y

`f `y

`f `y

œ 5x  2y  6

`f `y

œ 9(2x  3y)#

œ

" $ $ 3É x
ˆ #y ‰

(x
y) œ  (x
" y)#

œ  ax# 2xy
y# b#

(xy  ")(1)  (x
y)(x) (xy  1)#

œ  x#
y y# ,

(x
y
1) œ eÐx
y
1Ñ ,

2.

œ

œ

" # 1
ˆ xy ‰

œ eÐx
y
1Ñ †

` `y

†

x #  " (xy  1)# ` `y

ˆ xy ‰ œ

1 # x ’1
ˆ xy ‰ “

(x
y
1) œ eÐx
y
1Ñ

œ

x x#
y#

875

876

Chapter 14 Partial Derivatives

14.

`f `x

œ ex sin (x
y)
ex cos (x
y),

15.

`f `x

œ

16.

`f `x

œ exy †

`f `x `f `y

œ 2 sin (x  3y) †

`f `x

œ 2 cos a3x  y# b †

17.

18.

" x
y

†

` `x

` `x

(x
y) œ

" x
y

,

`f `y

(xy) † ln y œ yexy ln y, ` `x ` `y

œ 2 sin (x  3y) †

" x
y

œ

`f `y

`f `y

†

œ ex cos (x
y)

` `y

" x
y

(x
y) œ

œ exy †

` `y

(xy) † ln y
exy †

sin (x  3y) œ 2 sin (x  3y) cos (x  3y) † sin (x  3y) œ 2 sin (x  3y) cos (x  3y) †

` `x

" y

œ xexy ln y


` `x ` `y

(x  3y) œ 2 sin (x  3y) cos (x  3y),

exy y

(x  3y) œ 6 sin (x  3y) cos (x  3y)

cos a3x  y# b œ 2 cos a3x  y# b sin a3x  y# b †

œ 6 cos a3x  y# b sin a3x  y# b , `f ` # # # # ` y œ 2 cos a3x  y b † ` y cos a3x  y b œ 2 cos a3x  y b sin a3x  y b †

` `x

a3x  y# b

` `y

a3x  y# b

œ 4y cos a3x  y# b sin a3x  y# b 19.

`f `x

œ yxyc1 ,

21.

`f `x

œ g(x),

`f `y

œ xy ln x

`f `y

20. f(xß y) œ

Ê

`f `x

" x ln y

œ

and

`f `y

œ

 ln x y(ln y)#

œ g(y)

_

22. f(xß y) œ ! (xy)n , kxyk  1 Ê f(xß y) œ nœ0

`f `y

ln x ln y

œ  (1 "xy)# †

` `y

(1  xy) œ

" 1  xy

Ê

`f `x

œ  (1 "xy)# †

` `x

(1  xy) œ

y (1  xy)#

and

x (1  xy)#

23. fx œ 1
y# , fy œ 2xy, fz œ 4z

24. fx œ y
z, fy œ x
z, fz œ y
x

25. fx œ 1, fy œ  Èy#y
z# , fz œ  Èy#z
z# 26. fx œ x ax#
y#
z# b 27. fx œ

yz È 1  x # y # z#

28. fx œ

" kx
yzk È(x
yz)# 1

29. fx œ

" x
2y
3z

30. fx œ yz †

" xy

†

, fy œ

, fy œ ` `x

xz È 1  x # y # z#

, fy œ

(xy) œ

#

, fy œ y ax#
y#
z# b

#

` `z

, fz œ

(yz)(y) xy

, fz œ œ

yz x

$Î#

, fz œ z ax#
y#
z# b

$Î#

xy È1  x# y# z#

z kx
yzk È(x
yz)#  1

2 x
2y
3z

fz œ y ln (xy)
yz † #

$Î#

, fz œ

y kx
yzk È(x
yz)#  1

3 x
2y
3z

, fy œ z ln (xy)
yz †

` `y

ln (xy) œ z ln (xy)


ln (xy) œ y ln (xy) #

#

#

#

#

#

31. fx œ 2xe ax
y
z b , fy œ 2ye ax
y
z b , fz œ 2ze ax
y
z b 32. fx œ yzexyz , fy œ xzexyz , fz œ xyexyz 33. fx œ sech# (x
2y
3z), fy œ 2 sech# (x
2y
3z), fz œ 3 sech# (x
2y
3z) 34. fx œ y cosh axy  z# b , fy œ x cosh axy  z# b , fz œ 2z cosh axy  z# b

yz xy

†

` `y

(xy) œ z ln (xy)
z,

Section 14.3 Partial Derivatives 35.

`f `t

œ 21 sin (21t  !),

36.

`g `u

œ v# eÐ2uÎvÑ †

37.

`h `3

œ sin 9 cos ),

38.

`g `r

œ 1  cos ),

` `u

`g `)

$ v# 2g ,

`A `c

œ m,

`A `h

œ

q #

41.

`f `x

œ 1
y,

`f `y

œ 1
x,

42.

`f `x

œ y cos xy,

43.

`g `x

œ 2xy
y cos x,

44.

`h `x

œ ey ,

45.

`r `x

œ

46.

`s `x

œ”

` #s ` x#

œ

` #s ` y` x

œ

" #• 1
ˆ xy ‰

` #s ` x` y

œ

œ

Vv# 2g

m `A q , `m

œ

` #f ` x#

†

`g `y

` `x

` #h ` x#

2xy a x #
y # b#

,

œ xey , ` #r ` y#

œ

x(2y) a x #
y # b#

6 (2x
3y)#

`w `x

œ y#
2xy$
3x# y% ,

œ

3 2x
3y `w `y

,

œ

x y

`w `y

œ


ln x,

` #w ` y` x

` #g ` x#

` #h ` y` x

œ

h #

` #f ` x` y

` #f ` y#

œ1

œ x# sin xy,

œ 2y  y sin x,

œ

` #h ` x` y

y x#
y#

` #g ` y#

œ

` #f ` x` y

œ cos y,

œ cos xy  xy sin xy ` #g ` y` x

œ

` #g ` x` y

œ 2x
cos x

œ ey

œ

,

` #f ` y` x

` #r ` x` y

`s `y

œ”

œ

" (x
y)#

" #• 1
ˆ xy ‰

†

` `y




" x

ˆ xy ‰ œ ˆ x1 ‰ ”

" #• 1
ˆ xy ‰

œ

x x#
y#

,

œ  ax# 2xy ,
y# b#

œ

`w `y

, Wg œ  V#$gv#

œ  km q#


` #f ` y` x

" #• 1
ˆ xy ‰

#

V$ v g

" ` #r (x
y)# , ` y` x

` #w ` y` x

49.

, and

œ œ

" y

` #w ` x` y

6 (2x
3y)#

œ


"x , and

œ 2xy
3x# y#
4x$ y$ ,

` #w ` y` x

` #w ` x` y

œ

" y

œ 2y
6xy#
12x# y$ , and

œ 2y
6xy#
12x# y$

œ sin y
y cos x
y,

` w ` x` y

œ

œ

œ y# sin xy,

y#  x# a x #
y # b#

œ ex
ln y
xy ,

#

œ 0,

œ

`w `x

`w `x

` #f ` y#

2V$ v 2g

`A `q

ax#
y# b (1)
y(2y) ax #
y # b #

48.

50.


c,

k q

` #h ` y#

" (x
y)#

` #s ` y#

,

Ð2uÎvÑ ˆ 2u ‰  2ueÐ2uÎvÑ v œ 2ve

œ 3 sin 9 sin )

, Wv œ

ˆ xy ‰ œ ˆ xy# ‰ ”

œ

` #w ` x` y

` #f ` x#

œ 0,

œ

`w `x

,

`h `)

œ x#  sin y
sin x,

47.

2 2x
3y

œ 0,

œ x cos xy,

` `v

œ 2veÐ2uÎvÑ
v# eÐ2uÎvÑ †

œ 1

W$ œ

" ` #r x
y , ` x #

œ

y(2x) ax #
y # b #

œ

`A `k

œ xey
1,

" `r x
y , ` y

`g `z

œ r sin ),

40.

`h `y

`g `v

œ 3 cos 9 cos ),

39. Wp œ V, Wv œ P


`f `y

œ sin (21t  !)

Ð2uÎvÑ ˆ 2u ‰ , v œ 2ve

`h `9

,

`f `!

`w `y

œ x cos y
sin x
x,

` #w ` y` x

œ cos y
cos x
1, and

œ cos y
cos x
1

51. (a) x first

(b) y first

52. (a) y first three times

(c) x first (b) y first three times

(d) x first (c) y first twice

(e) y first

(f) y first (d) x first twice

877

878

Chapter 14 Partial Derivatives f(1
hß 2)  f(1ß 2) h

53. fx (1ß 2) œ lim

hÄ0

œ lim

hÄ0

13h  6h# h

œ lim

hÄ0

c1  (1
h)
2  6(1
h)# d  (2  6) h

œ lim

hÄ0

h  6 a1
2h
h# b
6 h

œ lim (13  6h) œ 13, hÄ0

f(1ß 2
h)  f(1ß 2) h hÄ0

fy (1ß 2) œ lim

œ lim

hÄ0

œ lim (2) œ 2

c1  1
(2
h)  3(2
h)d  (2  6) h

(2  6  2h)  (2  6) h

œ lim

hÄ0

hÄ0

54. fx (2ß 1) œ lim

hÄ0

œ lim

hÄ0

f(2
hß 1)  f(2ß 1) h

(2h  1  h)
1 h

c4
2(2
h)  3  (2
h)d  (3
2) h

œ lim

hÄ0

œ lim 1 œ 1, hÄ0


h# )d  (3
2) f(2ß 1
h)  f(2ß 1) œ lim c4  4  3(1
h)
2(1 h h hÄ0 hÄ0 # # lim a3  3h
2
h 4h
2h b
1 œ lim h
h2h œ lim (1
2h) œ 1 hÄ0 hÄ0 hÄ0

fy (2ß 1) œ lim œ

55. fz (x! ß y! ß z! ) œ lim

hÄ0

fz (1ß 2ß 3) œ lim

hÄ0

f(x! ß y! ß z!
h)  f(x! , y! ß z! ) h

f(1ß 2ß 3
h)  f(1, 2ß 3) h

56. fy (x! ß y! ß z! ) œ lim

hÄ0

œ lim

hÄ0

f(x! ß y!
hß z! )  f(x! , y! ß z! ) h

`z `x

`z ‰ `x x


z$  2y

`z `x

hÄ0

œ lim

"2h
2h# h

œ lim (12
2h) œ 12 hÄ0

œ lim (2h
9) œ 9 hÄ0

œ 0 Ê a3xz#  2yb `` xz œ y  z$ Ê at (1ß 1ß 1) we have (3  2) `` xz œ 1  1 or

œ 2

œ

y x

" 6

 2x‰ `` xz œ x Ê at (1ß 1ß 3) we have (3  1  2) `` xz œ 1 or

59. a# œ b#
c#  2bc cos A Ê 2a œ (2bc sin A) ``Aa Ê Ê 2c cos A  2b œ (2bc sin A) ``Ab Ê

60.

œ lim

;

58. ˆ `` xz ‰ z
x
ˆ yx ‰ `` xz  2x `` xz œ 0 Ê ˆz
`x `z

2(3
h)#  2(9) h

a2h#
9hb  0 h hÄ0

f(1ß hß 3)  f(1, 0ß 3) h hÄ0

fy (1ß 0ß 3) œ lim 57. y
ˆ3z#

;

a b sin A œ sin B ˆ sin" A ‰ ``Ba œ

Ê

(sin A) ``Aa  a cos A sin# A

`A `b

œ

`A `a c cos A  b bc sin A

œ

a bc sin A

; also 0 œ 2b  2c cos A
(2bc sin A) ``Ab

œ 0 Ê (sin A) `` xa  a cos A œ 0 Ê

b( csc B cot B) Ê

`a `B

`a `A

œ

a cos A sin A

; also

œ b csc B cot B sin A

61. Differentiating each equation implicitly gives 1 œ vx ln u
ˆ vu ‰ ux and 0 œ ux ln v
ˆ uv ‰ vx or (ln u) vx
ˆ vu ‰ ux œ 1 Ê vx œ ˆ uv ‰ vx
(ln v) ux œ 0 Ÿ

v " º 0 lnu v º v ln u u º u ln vº v

œ

ln v (ln u)(ln v)  1

62. Differentiating each equation implicitly gives 1 œ (2x)xu  (2y)yu and 0 œ (2x)xu  yu or (2x)xu  (2y)yu œ 1 Ê xu œ (2x)xu  yu œ 0
yu œ

" 0º #x
4xy 2x º 2x

œ

2x 2x
4xy

œ

`f `x

œ 2x,

`f `y

œ 2y,

`f `z

2y 1 º 2x 2y º 2x 1 º

2x 2x  4xy

œ 2x Š 2x " 4xy ‹
2y Š 1 " 2y ‹ œ 63.

" º0

œ 4z Ê

œ

1 1  2y

" 1  #y ` #f ` x#

œ




1 2x
4xy

1 2x  4xy

and

; next s œ x#
y# Ê

2y 1  2y

œ 2,

œ

` #f ` y#

œ

`s `u

œ 2x

`x `u


2y

` #f ` x#




` #f ` y#




` #f ` z#

`y `u

1
2y 1  2y

œ 2,

` #f ` z#

œ 4 Ê

œ 2
2
(4) œ 0

Section 14.3 Partial Derivatives 64.

`f `x

`f `y

œ 6xz,

`f `z

œ 6yz,

œ 6z#  3 ax#
y# b ,

` #f ` x#

` #f ` y#

œ 6z,

œ 6z,

` #f ` z#

` #f ` x#

œ 12z Ê




` #f ` y#




879

` #f ` z#

œ 6z  6z
12z œ 0 `f `x

œ 2ec2y sin 2x,

66.

`f `x

œ

67.

`f `x

œ  "# ax#
y#
z# b

65.

`f `y

œ 2ec2y cos 2x,

œ 4ec2y cos 2x
4ec2y cos 2x œ 0 x x#
y#

,

`f `y

œ

y x#
y#

` #f ` x#

,

$Î#

œ

y#  x# ax #
y # b #

œ 4ec2y cos 2x,

` #f ` x#

` #f ` y#

,

œ

x#  y# a x #
y # b#

` #f ` y#

` #f ` x#

Ê

$Î# ` f , `y $Î# #

(2x) œ x ax#
y#
z# b

œ 4ec2y cos 2x Ê




` #f ` y#

œ

y#  x# a x #
y # b#




œ  "# ax#
y#
z# b

` #f ` x#

x#  y# a x #
y # b#

$Î#

$Î# ` f $Î# œ y ax#
y#
z# b , ` z œ  "# ax#
y#
z b (2z) œ z ax#
y#
z# b ; ` #f # # # $Î# # # # # &Î# ` # f # # # $Î#
3x ax
y
z b , ` y # œ  ax
y
z b
3y# ` x # œ  ax
y
z b # # # &Î# ` #f # # # $Î#
3z# ax#
y#
z# b Ê `` xf#
`` yf#
`` zf# ` z # œ  ax
y
z b $Î#


3x# ax#
y#
z# b

$Î#


3z# ax#
y#
z# b

œ ’ ax#
y#
z# b
’ ax#
y#
z# b

&Î#

&Î#

` #f ` y#

œ0

(2y) ax#
y#
z# b &Î#

&Î#

,

$Î#


3y# ax#
y#
z# b

$Î#


a3x#
3y#
3z# b ax#
y#
z# b

“
’ ax#
y#
z# b

“ œ 3 ax#
y#
z# b




“

œ0 68.

69.

`f `x

œ 3e3x
4y cos 5z,

œ 25e3x
4y cos 5z Ê

`w `x

œ cos (x
ct),

`w `x

Ê 71.

72. 73.

74.

75.

œ 4e3x
4y cos 5z,

` #f ` z#

œ c# 70.

`f `y

`w `t

#

` #f ` x#




` #f ` y#




`f `z

` #f ` z#

` #w ` x#

œ c cos (x
ct);

œ 5e3x
4y sin 5z;

` #f ` x#

œ 9e3x
4y cos 5z,

` #w ` t#

œ  sin (x
ct),

œ c# sin (x
ct) Ê

` #w ` t#

œ c# [ sin (x
ct)]

` w ` x#

œ 2 sin (2x
2ct), #

` w ` t#

œ

œ 16e3x
4y cos 5z,

œ 9e3x
4y cos 5z
16e3x
4y cos 5z  25e3x
4y cos 5z œ 0

`w `t

œ 2c sin (2x
2ct);

œ c# [4 cos (2x
2ct)] œ c#

#

` #w ` x#

œ 4 cos (2x
2ct),

" x
ct

,

`w `t

œ

c x
ct

;

` #w ` x#

œ

1 (x
ct)#

` #w ` t#

œ 4c# cos (2x
2ct)

` w ` x#

`w `w ` x œ cos (x
ct)  2 sin (2x
2ct), ` t œ c cos (x
ct)  2c sin (2x ` #w ` #w # # ` x# œ  sin (x
ct)  4 cos (2x
2ct), ` t# œ c sin (x
ct)  4c # # Ê `` tw# œ c# [ sin (x
ct)  4 cos (2x
2ct)] œ c# `` xw# `w `x

` #f ` y#

,

` #w ` t#

œ

c # (x
ct)#

Ê

` #w ` t#


2ct); cos (2x
2ct)

" # œ c# ’ (x
ct)# “ œ c

` #w ` x#

`w `w ` #w # # # ` x œ 2 sec (2x  2ct), ` t œ 2c sec (2x  2ct); ` x# œ 8 sec (2x  2ct) tan (2x  2ct), ` #w ` #w # # # # # ` #w ` t# œ 8c sec (2x  2ct) tan (2x  2ct) Ê ` t# œ c [8 sec (2x  2ct) tan (2x  2ct)] œ c ` x# # `w xbct ` w , ` t œ 15c sin (3x
3ct)
cexbct ; `` xw# œ 45 cos (3x
` x œ 15 sin (3x
3ct)
e # # ` #w # # xbct Ê `` tw# œ c# c45 cos (3x
3ct)
exbct d œ c# `` xw# ` t# œ 45c cos (3x
3ct)
c e

`w `t

œ

`f `u `u `t

œ

` #f ` u#

Ê

œ a#

`f `u

(ac) Ê

` #w ` t#

œ a# c#

` #w ` t#

œ (ac) Š `` uf# ‹ (ac) œ a# c#

#

` #f ` u#

œ c# Ša#

` #f ` u# ‹

œ c#

` #w ` x#

` #f ` u#

;

`w `x

œ

`f `u `u `x

œ

`f `u

†a Ê

3ct)
exbct ,

` #w ` x#

#

œ Ša `` uf# ‹ † a

&Î#

880

Chapter 14 Partial Derivatives

76. If the first partial derivatives are continuous throughout an open region R, then by Theorem 3 in this section of the text, f(xß y) œ f(x! ß y! )
fx (x! ß y! ) ?x
fy (x! ß y! ) ?y
%" ?x
%# ?y, where %" , %# Ä 0 as ?x, ?y Ä 0. Then as (xß y) Ä (x! ß y! ), ?x Ä 0 and ?y Ä 0 Ê lim f(xß y) œ f(x! ß y! ) Ê f is continuous at every point Ðx ß y Ñ Ä Ð x ! ß y ! Ñ

(x! ß y! ) in R. 77. Yes, since fxx , fyy , fxy , and fyx are all continuous on R, use the same reasoning as in Exercise 76 with fx (xß y) œ fx (x! ß y! )
fxx (x! ß y! ) ?x
fxy (x! ß y! ) ?y
%" ?x
%# ?y and fy (xß y) œ fy (x! ß y! )
fyx (x! ß y! ) ?x
fyy (x! ß y! ) ?y
s%" ?x
s%# ?y. Then lim fx (xß y) œ fx (x! ß y! ) Ðx ß y Ñ Ä Ð x ! ß y ! Ñ

and

lim

Ðx ß y Ñ Ä Ð x ! ß y ! Ñ

fy (xß y) œ fy (x! ß y! ).

14.4 THE CHAIN RULE 1. (a)

`w `x

(b)

dw dt

(1 ) œ 0

2. (a)

`w `x

œ 2x,

œ 2x,

`w `y œ #

2y,

œ  sin t,

dx dt

dy dt #

œ cos t Ê

œ 0; w œ x
y# œ cos# t
sin t œ 1 Ê

`w `y

œ 2y,

œ  sin t
cos t,

dx dt

dy dt

dw dt

œ 2x sin t
2y cos t œ 2 cos t sin t
2 sin t cos t

dw dt

œ0

œ  sin t  cos t Ê

dw dt

œ (2x)( sin t
cos t)
(2y)( sin t  cos t) œ 2(cos t
sin t)(cos t  sin t)  2(cos t  sin t)(sin t
cos t) œ a2 cos# t  2 sin# tb  a2 cos# t  2 sin# tb œ 0; w œ x#
y# œ (cos t
sin t)#
(cos t  sin t)# œ 2 cos# t
2 sin# t œ 2 Ê dw dt œ 0 (b)

dw dt

(0) œ 0

3. (a)

`w `x

œ

Ê

" z

dw dt

,

`w `y

œ

" z

,

`w `z

œ

(x
y) z#

œ  2z cos t sin t


2 z

dx dt

,

œ 2 cos t sin t,

sin t cos t


x
y z# t#

dy dt

œ 2 sin t cos t,

cos# t
sin# t Š "# ‹ at# b

œ

œ 1; w œ

x z

dz dt

œ  t"#




y z

œ

t

(b)

dw dt

(3) œ 1

4. (a)

`w `x

œ

2x x #
y #
z#

,

`w `y

œ

2y x #
y #
z#

2y cos t 2x sin t dw dt œ x#
y#
z#
x#
y#
z# œ 1
1616t ; w œ ln ax#
y#
z# b dw 16 dt (3) œ 49

Ê

(b) 5. (a)

`w `x

œ 2yex ,

`w `y

œ 2ex ,

`w `z

,

`w `z




œ

2z x #
y #
z#

4zt
"Î# x#
y#
z# #

œ

œ ln acos t


œ  "z ,

dx dt

œ

2t t#
1

,

,

dx dt

œ  sin t,

6. (a)

dy dt

œ

" t#
1

,

dz dt

œ et Ê

# t (4t) atan
" tb at#
1b
2 at#t

11b  eet œ 4t tan" t
1; w œ 2yex  ln t#
1 " ˆ 2 ‰ # Ê dw tb (2t)  1 œ 4t tan" t
1 dt œ t#
1 at
1b
a2 tan dw ˆ1‰ dt (1) œ (4)(1) 4
1 œ 1
1 `w `x

œ y cos xy,

`w `y

œ x cos xy,

œ (ln t)[cos (t ln t)]  tc1

œe (b)

dw dt

 sin (t ln t) Ê

`w `z

œ 1,

dx dt

œ cos t,

dz dt

œ 1,

dy dt

œ

" t

,

dz dt

dw dt

œ




sin# t Š "t ‹

œt Ê

4ytex t#
1

œ

œ1




16 1
16t

2ex t#
1



et z

z œ a2 tan" tb at#
1b  t

œ etc1 Ê

dw dt

œ y cos xy 

t cos (t ln t)
etc1 œ (ln t)[cos (t ln t)]  cos (t ln t)
etc1 ; w œ z  sin t dw tc1  [cos (t ln t)] ln t
t ˆ "t ‰‘ œ etc1  (1
ln t) cos (t ln t) dt œ e

(1) œ 1  (1
0)(1) œ 0

dw dt

œ 2t"Î#

2 cos t sin t
2 sin t cos t
4 ˆ4t"Î# ‰ t
"Î# cos# t
sin# t
16t # sin t
16tb œ ln (1
16t) Ê dw dt

œ

(b)

dy dt

cos# t Š "t ‹

x cos xy t

xy


etc1

Section 14.4 The Chain Rule 7. (a)

`z `u

œ

`z `x `x `u




`z `y `y `u

x

v ‰ 4e œ a4ex ln yb ˆ ucos cos v
Š y ‹ (sin v) œ

4ex ln y u




4ex sin v y

œ

4(u cos v) ln (u sin v) v)(sin v)
4(u cos œ (4 cos v) ln (u sin v)
4 cos v; u u sin v `z `z `x `z `y 4ex x x ˆ u sin v ‰ ` v œ ` x ` v
` y ` v œ a4e ln yb u cos v
Š y ‹ (u cos v) œ  a4e

ln yb (tan v)


4ex u cos v y

4(u cos v)(u cos v) cos# v œ (4u sin v) ln (u sin v)
4usin u sin v v ; `z sin x z œ 4e ln y œ 4(u cos v) ln (u sin v) Ê ` u œ (4 cos v) ln (u sin v)
4(u cos v) ˆ u sinvv ‰ v‰ œ (4 cos v) ln (u sin v)
4 cos v; also `` vz œ (4u sin v) ln (u sin v)
4(u cos v) ˆ uu cos sin v # cos v œ (4u sin v) ln (u sin v)
4usin v At ˆ2ß 14 ‰ : `` uz œ 4 cos 14 ln ˆ2 sin 14 ‰
4 cos 14 œ 2È2 ln È2
2È2 œ È2 (ln 2
2); (4)(2) ˆcos# 14 ‰ `z 1 1‰ ˆ œ 4È2 ln È2
4È2 œ 2È2 ln 2
4È2 ˆsin 1 ‰ ` v œ (4)(2) sin 4 ln 2 sin 4


œ [4(u cos v) ln (u sin v)](tan v)


(b)

4

8. (a)

`z `u `z `v

œ– œ–

Š
#x ‹

Š "y ‹ #

Š xy ‹

— cos v
– Š x ‹#
1 — sin v œ
1 y

Š xy ‹



x sin v x #
y #

œ

(u sin v)(cos v)  (u cos v)(sin v) u#

Š
#x ‹

— (u sin v)
– Š x ‹#
1 — u cos v œ 
1 y

yu sin v x#
y#



(b) At 9. (a)

`w `u

" sin# v
cos# v œ 1 ˆ1.3ß 16 ‰ : `` uz œ 0

œ

xu cos v x#
y#

œ

(u sin v)(u sin v)  (u cos v)(u cos v) u#

y

œ sin# v  cos# v œ 1; z œ tan" Š xy ‹ œ tan" (cot v) Ê œ

œ 0;

y

Š "y ‹ #

y cos v x#
y#

`w `x `x `u




and

`w `y `y `u




`z `v

`z `u

œ 0 and

`z `v

" # ‰ œ ˆ 1
cot # v a csc vb

œ 1

`w `z `z `u

œ (y
z)(1)
(x
z)(1)
(y
x)(v) œ x
y
2z
v(y
x)

œ (u
v)
(u  v)
2uv
v(2u) œ 2u
4uv;

`w `v

œ

`w `x `x `v




`w `y `y `v




`w `z `z `v

œ (y
z)(1)
(x
z)(1)
(y
x)(u) œ y  x
(y
x)u œ 2v
(2u)u œ 2v
2u# ; w œ xy
yz
xz œ au#  v# b
au# v  uv# b
au# v
uv# b œ u#  v#
2u# v Ê ``wu œ 2u
4uv and `w `v

œ 2v
2u#

(b) At ˆ "# ß 1‰ : 10. (a)

`w `u

`w `u

œ 2 ˆ "# ‰
4 ˆ #" ‰ (1) œ 3 and

`w `v

#

œ 2(1)
2 ˆ "# ‰ œ  #3

2y 2z v v v v v œ Š x#
2x y#
z# ‹ ae sin u
ue cos ub
Š x#
y#
z# ‹ ae cos u  ue sin ub
Š x#
y#
z# ‹ ae b v

u ‰ aev sin u
uev cos ub œ ˆ u# e2v sin# u
2ueu# esin 2v cos# u
u# e2v v cos u v ‰ v
ˆ u# e2v sin# u 2ue
u# e2v cos# u
u# e2v ae cos u  ue sin ub v

‰ aev b œ 2u ;
ˆ u# e2v sin# u
u2ue # e2v cos# u
u# e2v `w `v

2y 2z v v v œ Š x#
2x y#
z# ‹ aue sin ub
Š x#
y#
z# ‹ aue cos ub
Š x#
y#
z# ‹ aue b v

u ‰ auev sin ub œ ˆ u# e2v sin# u
2ueu# esin 2v cos# u
u# e2v v cos u ‰ v
ˆ u# e2v sin# u 2ue
u# e2v cos# u
u# e2v aue cos ub

‰ auev b œ 2; w œ ln au# e2v sin# u
u# e2v cos# u
u# e2v b œ ln a2u# e2v b
ˆ u# e2v sin# u
u2ue # e2v cos# u
u# e2v v

œ ln 2
2 ln u
2v Ê (b) At (#ß !): 11. (a)

`w `u

œ

2 #

`w `u

œ

œ 1 and

2 `w u and ` v `w `v œ 2

œ2

rp pq qr
rp
pq `u `p `u `q `u `r " œ 0; ` p ` x
` q ` x
` r ` x œ q  r
(q  r)#
(q  r)# œ (q  r)# rp pq qrr
p
pq 2p  2r `u `u `p `u `q `u `r " œ (q ` y œ ` p ` y
` q ` y
` r ` y œ q  r  (q  r)#
(q  r)# œ (q  r)#  r)# (2x
2y
2z)  (2x
2y  2z) z `u `u `p `u `q `u `r œ œ (z  y)# ; ` z œ ` p ` z
` q ` z
` r ` z (2z  2y)# rp pq qr
rpp
q  2p 4y y " œ q œ 2q r
(q  r)#  (q  r)# œ (q  r)# (q  r)# œ (2z  2y)# œ  (z  y)# ; `u `x

œ

881

882

Chapter 14 Partial Derivatives pq 2y q  r œ 2z  2y  (zyy)#

uœ œ

(b) At ŠÈ3ß 2ß 1‹ : 12. (a)

`u `x

œ

eqr È 1  p#

œ

`u `x

Ê

y zy

œ 0,

`u `y

œ

`u `x

œ 0,

" (1  2)#

`u `y

œ

(z  y)  y(1) (z  y)#

œ 1, and

`u `z

œ

œ

z (z  y)#

2 (1  2)#

œ 2

(cos x)
areqr sin" pb (0)
aqeqr sin" pb (0) œ

eqr cos x È 1  p#

œ

z re sin
" p #

#

`u `z

, and

qr

(z  y)(0)  y(1) (z  y)#

œ

ez ln y cos x È1  sin# x

z# ˆ " ‰ y z x

œ yz if  1#  x 

`u `y

œ

e È 1  p#

(0)
areqr sin" pb Š zy ‹
aqeqr sin" pb (0) œ

`u `z

œ

eqr È 1  p#

(0)
areqr sin" pb (2z ln y)
aqeqr sin" pb ˆ z"# ‰ œ a2zreqr sin" pb (ln y) 

qr

œ (2z) ˆ "z ‰ ayz x ln yb  `u `y

œ xzyz1 , and

(b) At ˆ 14 ß "# ß  "# ‰ : œ1 œ

` z dx ` x dt

`u `z `u `x

az# ln yb ayz b x z# z

y

œ

z

y

;

œ xzyz1 ;

œ xyz ln y; u œ ez ln y sin" (sin x) œ xyz if  1# Ÿ x Ÿ

qeqr sin
" p z# 1 #

Ê

œ ˆ 14 ‰ ˆ "# ‰

"Î#

`u `x

œ yz ,

œ œ xy ln y from direct calculations œ ˆ "# ‰

"Î#

œ È2,

`u `y

œ ˆ 14 ‰ ˆ "# ‰ ˆ "# ‰

Ð"Î#Ñ"

È

œ  14 2 ,

`u `z

È2 ln 2 4




1 #

` z dy ` y dt

` z du ` u dt




` z dv ` v dt




` x dw ` w dt

dz dt

15.

`w `u

œ

`w `x `x `u




`w `y `y `u




`w `z `z `u

`w `v

œ

`w `x `x `v




`w `y `y `v




`w `z `z `v

16.

`w `x

œ

`w `r `r `x




`w `s `s `x




`w `t `t `x

`w `y

œ

`w `r `r `y




`w `s `s `y




`w `t `t `y

14.

dz dt

œ

13.

ln ˆ "# ‰

Section 14.4 The Chain Rule 17.

`w `u

œ

`w `x `x `u




`w `y `y `u

`w `v

œ

`w `x `x `v




`w `y `y `v

18.

`w `x

œ

`w `u `u `x




`w `v `v `x

`w `y

œ

`w `u `u `y




`w `v `v `y

19.

`z `t

œ

`z `x `x `t

`z `s

œ

`z `x `x `s

20.

`y `r

œ

dy ` u du ` r

`w `s

œ

dw ` u du ` s

22.

`w `p

œ

`w `x `x `p




`z `y `y `t

21.




`w `y `y `p




`w `z `z `p




`w `v `v `p




`z `y `y `s

`w `t

œ

dw ` u du ` t

883

884

Chapter 14 Partial Derivatives

23.

`w `r

œ

` w dx ` x dr




` w dy ` y dr

24.

`w `s

œ

`w `x `x `s




`w `y `y `s

œ

` w dx ` x dr

since

dy dr

`w `s

œ0

œ

` w dx ` x ds




` w dy ` y ds

œ

and Fy (xß y) œ 4y
x Ê dy dx

(1ß 1) œ

dy dx

œ0

dy dx

#

œ  FFxy œ  (3x4y

yx)

3 œ  FFxy œ  xy
2y

dy dx

(1ß 1) œ 2

27. Let F(xß y) œ x#
xy
y#  7 œ 0 Ê Fx (xß y) œ 2x
y and Fy (xß y) œ x
2y Ê Ê

dx ds

4 3

26. Let F(xß y) œ xy
y#  3x  3 œ 0 Ê Fx (xß y) œ y  3 and Fy (xß y) œ x
2y Ê dy dx

since

25. Let F(xß y) œ x$  2y#
xy œ 0 Ê Jx (xß y) œ 3x#
y Ê

Ê

` w dy ` y ds

dy dx

œ  FFxy œ  x2x

2yy

(1ß 2) œ  45

28. Let F(xß y) œ xey
sin xy
y  ln 2 œ 0 Ê Fx (xß y) œ ey
y cos xy and Fy (xß y) œ xey
x sin xy
1 Ê

dy dx

œ  FFxy œ  xeye

xysincosxyxy
1 Ê y

dy dx

(!ß ln 2) œ (2
ln 2)

29. Let F(xß yß z) œ z$  xy
yz
y$  2 œ 0 Ê Fx (xß yß z) œ y, Fy (xß yß z) œ x
z
3y# , Fz (xß yß z) œ 3z#
y Ê Ê

Fx `z ` x œ  Fz `z ` y (1ß 1ß 1)

30. Let F(xß yß z) œ Ê

`z `x

œ  3z#
y y œ

y 3z#
y

Ê

`z `x

(1ß 1ß 1) œ

" 4

;

`z `y

#

œ  Fyz œ  x3z
#z

y3y œ F

x  z  3y# 3z#
y

œ  43 " x




" y

œ  FFxz œ 




" z

 1 œ 0 Ê Fx (xß yß z) œ  x"# , Fy (xß yß z) œ  y"# , Fz (xß yß z) œ  z"#

Š x"# ‹ Š z"# ‹

#

œ  xz# Ê

`z `x

(2ß 3ß 6) œ 9;

`z `y

F

œ  Fyz œ 

Š y"# ‹ Š z"# ‹

#

œ  yz# Ê

`z `y

(2ß 3ß 6) œ 4

31. Let F(xß yß z) œ sin (x
y)
sin (y
z)
sin (x
z) œ 0 Ê Fx (xß yß z) œ cos (x
y)
cos (x
z), Fy (xß yß z) œ cos (x
y)
cos (y
z), Fz (xß yß z) œ cos (y
z)
cos (x
z) Ê `` xz œ  FFxz (x
y)
cos (x
z) œ  cos cos (y
z)
cos (x
z) Ê

`z `x

(1ß 1ß 1) œ 1;

`z `y

(x
y)
cos (y
z) œ  Fyz œ  cos cos (y
z)
cos (x
z) Ê F

`z `y

(1 ß 1 ß 1 ) œ  1

32. Let F(xß yß z) œ xey
yez
2 ln x  2  3 ln 2 œ 0 Ê Fx (xß yß z) œ ey
2x , Fy (xß yß z) œ xey
ez , Fz (xß yß z) œ yez Ê 33.

`w `r

`z `x

œ

œ  FFxz œ 

`w `x `x `r




`w `y `y `r

ˆey
2x ‰ yez




Ê

`w `z `z `r

`z `x

(1ß ln 2ß ln 3) œ  3 ln4 2 ;

`z `y

œ  Fyz œ  xeye
z e Ê F

y

z

`z `y

(1ß ln 2ß ln 3) œ  3 ln5 2

œ 2(x
y
z)(1)
2(x
y
z)[ sin (r
s)]
2(x
y
z)[cos (r
s)]

œ 2(x
y
z)[1  sin (r
s)
cos (r
s)] œ 2[r  s
cos (r
s)
sin (r
s)][1  sin (r
s)
cos (r
s)]

Section 14.4 The Chain Rule `w ¸ ` r rœ1ßsœ
1

Ê 34.

`w `v

œ 2(3)(2) œ 12

`w `x `x `v




`w `y `y `v




`w `x `x `v




`w `y `y `v

œ ˆ2x 

`w ¸ ` v uœ0ßvœ0

œ 7

œ

`w `z `z `v

‰ ˆ"‰ ˆ 2v ‰ œ y ˆ 2v u
x(1)
z (0) œ (u
v) u


v# u

Ê

`w ¸ ` v uœ
1ßvœ2

œ (1) ˆ 41 ‰
ˆ 41 ‰

œ 8 35.

`w `v

œ

Ê 36.

`z `u

œ

`z `x `x `u




`z `y `y `u

y‰ x# (2)


ˆ "x ‰ (1) œ ’2(u  2v
1) 

2u
v  2 (u  2v
1)# “ (2)




" u  2v
1

œ (y cos xy
sin y)(2u)
(x cos xy
x cos y)(v)

$

œ cuv cos au v
uv b
sin uvd (2u)
cau#
v# b cos au$ v
uv$ b
au#
v# b cos uvd (v) Ê `` uz ¸ uœ0ßvœ1 œ 0
(cos 0
cos 0)(1) œ 2 37.

38.

$

`z `u

œ

dz ` x dx ` u

œ ˆ 1
5 x# ‰ eu œ ’ 1
aeu 5
ln vb# “ eu Ê

`z `v

œ

dz ` x dx ` v

œ ˆ 1
5 x# ‰ ˆ "v ‰ œ ’ 1
aeu 5
ln vb# “ ˆ "v ‰ Ê

`z `u

œ

dz ` q dq ` u

œ Š "q ‹ Š

Ê

`z ¸ ` u uœ1ßvœ
2

Èv
3 1
u# ‹

œ Š Èv
3"tan
" u ‹ Š

" atan
" 1b a1
1# b

œ


"

u œ Š Èv
3"tan
" u ‹ Š 2tan Èv
3 ‹ œ `V `I

39. V œ IR Ê

œ (600 ohms) 40. V œ abc Ê ¸ Ê dV dt

`V `R

œ R and

œ

dV dt

œ

2 1

;

" #(v
3)

œ I;

dV dt

`z `v

œ

` V dI ` I dt




aœ1ßbœ2ßcœ3

` V da ` a dt




` V db ` b dt




` V dc ` c dt

œ

œ ’ 1
5(2)# “ (1) œ 1

" atan
" ub a1
u# b


"

u œ Š "q ‹ Š 2tan Èv
3 ‹

`z ¸ ` v uœ1ßvœ
2

Ê

œ ’ 1
5(2)# “ (2) œ 2;

`z ¸ ` v uœln 2ßvœ1

Èv
3 1
u# ‹

dz ` q dq ` v

œ


(0.04 amps)(0.5 ohms/sec)

dI dt

`z ¸ ` u uœln 2ßvœ1

œ

" #

` V dR dI dR ` R dt œ R dt
I dt Ê 0.01 Ê dI dt œ 0.00005 amps/sec

volts/sec

db dc œ (bc) da dt
(ac) dt
(ab) dt

œ (2 m)(3 m)(1 m/sec)
(1 m)(3 m)(1 m/sec)
(1 m)(2 m)(3 m/sec) œ 3 m$ /sec

and the volume is increasing; S œ 2ab
2ac
2bc Ê db dc dS ¸ œ 2(b
c) da dt
2(a
c) dt
2(a
b) dt Ê dt

œ

dS dt

` S da ` a dt

` S db ` b dt







` S dc ` c dt

aœ1ßbœ2ßcœ3

œ 2(5 m)(1 m/sec)
2(4 m)(1 m/sec)
2(3 m)(3 m/sec) œ 0 m# /sec and the surface area is not changing; " ˆa da
b db
c dc ‰ Ê dD ¸ D œ Èa#
b#
c# Ê dD œ ` D da
` D db
` D dc œ dt

œ

" Š È14 ‹ [(1 m

` a dt

` b dt

` c dt

È a #
b #
c#

dt

m)(1 m/sec)
(2 m)(1 m/sec)
(3 m)(3 m/sec)] œ 

dt

6 È14

dt

dt

aœ1ßbœ2ßcœ3

m/sec  0 Ê the diagonals are

decreasing in length 41.

`f `x `f `y `f `z

42. (a) (b)

œ œ œ

`f `u `f `u `f `u

`w `r `w `r

`u `x `u `y `u `z






`f `v `f `v `f `v







fy

Ê fy œ (sin )) œ

`f `w `f `w `f `w

`w `x `w `y `w `z

œ œ œ

`f `u `f `u `f `u

`f ` w (1) (1)
`` vf (1)
``wf (0) (0)
`` vf (1)
``wf (1)

(1)


`f `v

(0)


œ

`f `u

œ œ

`f `w

 `f `u `f `v





,

`f `v , `f `w

`y `r

and Ê

`f `x




`f `y

œ fx cos )
fy sin ) and ``w) œ fx (r sin ))
fy (r cos )) Ê sin ) œ fx sin ) cos )
fy sin# ) and ˆ cosr ) ‰ ``w) œ fx sin ) cos )
fy cos# ) œ fx

`x `r

`v `x `v `y `v `z

`w `r #

 asin# )b

`w `r `w `r




`f `z

" `w r `)

œ0 œ fx sin )
fy cos )


ˆ cosr ) ‰ ``w) ; then ``wr œ fx cos )
(sin )) ``wr
ˆ cosr ) ‰ ``w) ‘ (sin )) Ê fx cos )  ˆ sin ) rcos ) ‰ ``w) œ a1  sin# )b ``wr  ˆ sin ) rcos ) ‰ ``w) Ê fx œ (cos )) ``wr  ˆ sinr ) ‰ #

`w ‰ `)


Š sinr# ) ‹ ˆ ``w) ‰ and

#

`w ‰ `)


Š cosr# ) ‹ ˆ ``w) ‰ Ê afx b#
afy b# œ ˆ ``wr ‰


(c) afx b œ acos# )b ˆ ``wr ‰  ˆ 2 sin )r cos ) ‰ ˆ ``wr afy b# œ asin# )b ˆ ``wr ‰
ˆ 2 sin )r cos ) ‰ ˆ ``wr

#

#

#

#

#

" r#

ˆ ``w) ‰#

`w `)

885

886

Chapter 14 Partial Derivatives `w `x

43. wx œ

œ

`w `u `u `x #

œ

`w `u


x Š `` uw#

œ

`w `u


x#

` #w ` u#

`w `v `v `x




`u `x

` #w ` v ` v` u ` x ‹





2xy

` #w ` v` u #

Ê wyy œ  ``wu  y Š `` uw#

`w `u

œx

#

`w `v

#


y Š ``u`wv ` #w ` v#


y# `u `y


y

`u `x

; wy œ

` #w ` v ` v` u ` y ‹







` #w ` v ` v# ` x ‹

`w `y

œ #

`w `u `u `y


x Š ``u`wv

#

`w `u

Ê wxx œ

#

`u `y


x

œ





`w `u

` `x

ˆ ``wu ‰
y

wxx
wyy œ ax#
y# b 44.

` w ` u#


ax#
y# b

#

` w ` v#

ˆ ``wv ‰

#

#


x Šx `` uw#
y ``v`wu ‹
y Šx

`w `v `v `y

œ y

`w `u


x

` #w ` u` v

#


y `` vw# ‹

`w `v

` #w ` v ` v# ` y ‹

#

œ  ``wu  y Šy `` uw#
x ``v`wu ‹
x Šy ``u`wv
x `` vw# ‹ œ  ``wu
y# #

` `x

` #w ` u#

 2xy

` #w ` v` u


x#

` #w ` v#

; thus

œ ax#
y# b (wuu
wvv ) œ 0, since wuu
wvv œ 0

œ f w (u)(1)
gw (v)(1) œ f w (u)
gw (v) Ê wxx œ f ww (u)(1)
gw w (v)(1) œ f ww (u)
gww (v);

`w `x `w `y

œ f w (u)(i)
gw (v)(i) Ê wyy œ f ww (u) ai# b
gww (v) ai# b œ f ww (u)  gww (v) Ê wxx
wyy œ 0

45. fx (xß yß z) œ cos t, fy (xß yß z) œ sin t, and fz (xß yß z) œ t#
t  2 Ê œ (cos t)( sin t)
(sin t)(cos t)
at#
t  2b(1) œ t#
t  2;

df dt

df dt

` f dx ` x dt #

œ




` f dy ` y dt




` f dz ` z dt

œ 0 Ê t
t  2 œ 0 Ê t œ 2

or t œ 1; t œ 2 Ê x œ cos (2), y œ sin (2), z œ 2 for the point (cos (2)ß sin (2)ß 2); t œ 1 Ê x œ cos 1, y œ sin 1, z œ 1 for the point (cos 1ß sin 1ß 1) 46.

dw dt

` w dx ` x dt

œ




` w dy ` y dt




` w dz ` z dt

" ‰ œ a2xe2y cos 3zb ( sin t)
a2x# e2y cos 3zb ˆ t
#
a3x# e2y sin 3zb (1)

2x# e2y cos 3z  3x# e2y t
# 2(1)# (4)(1) 0œ4 #

œ 2xe2y cos 3z sin t
Ê 47. (a)

dw ¸ dt Ð1ßln 2ß0Ñ `T `x

œ0


œ 8x  4y and

`T `y

œ 8y  4x Ê

dT dt

sin 3z; at the point on the curve z œ 0 Ê t œ z œ 0

œ

` T dx ` x dt




` T dy ` y dt

œ (8x  4y)( sin t)
(8y  4x)(cos t)

œ (8 cos t  4 sin t)( sin t)
(8 sin t  4 cos t)(cos t) œ 4 sin# t  4 cos# t Ê dT dt

d# T dt#

œ 16 sin t cos t;

œ 0 Ê 4 sin t  4 cos t œ 0 Ê sin t œ cos t Ê sin t œ cos t or sin t œ  cos t Ê t œ 14 , #

#

#

#

51 31 71 4 , 4 , 4

on

the interval 0 Ÿ t Ÿ 21; d# T dt# ¹ tœ 1

œ 16 sin

1 4

1 4

cos

 0 Ê T has a minimum at (xß y) œ Š

4

È2 #

ß

È2 # ‹;

d# T dt# ¹ tœ 31

œ 16 sin

31 4

cos

31 4

 0 Ê T has a maximum at (xß y) œ Š

È2 #

ß

d# T dt# ¹ tœ 51

œ 16 sin

51 4

cos

51 4

 0 Ê T has a minimum at (xß y) œ Š

È2 #

ß

d# T dt# ¹ tœ 71

œ 16 sin

71 4

cos

71 4

 0 Ê T has a maximum at (xß y) œ Š

4

4

4

`T `T ` x œ 8x  4y, and ` y œ 8y È2 È2 È2 È2 # ß # ‹ œ TŠ # ß # ‹ œ

(b) T œ 4x#  4xy
4y# Ê found in part (a): T Š TŠ 48. (a)

`T `x

È2 #

ß

È2 # ‹

œ y and

œ T Š

`T `y

È2 #

œx Ê

ß

dT dt

È2 # ‹

œ

È2 #

ß

È2 # ‹; È2 # ‹;

È2 # ‹

 4x so the extreme values occur at the four points 4 ˆ "# ‰  4 ˆ "# ‰
4 ˆ "# ‰ œ 6, the maximum and

œ 4 ˆ "# ‰  4 ˆ "# ‰
4 ˆ "# ‰ œ 2, the minimum

` T dx ` x dt




` T dy ` y dt

œ y Š2È2 sin t‹
x ŠÈ2 cos t‹

œ ŠÈ2 sin t‹ Š2È2 sin t‹
Š2È2 cos t‹ ŠÈ2 cos t‹ œ 4 sin# t
4 cos# t œ 4 sin# t
4 a1  sin# tb œ 4  8 sin# t Ê 31 51 71 4 , 4 , 4 #

d T dt# ¹ tœ 1

d# T dt#

œ 16 sin t cost t;

dT dt

œ 0 Ê 4
8 sin# t œ 0 Ê sin# t œ

on the interval 0 Ÿ t Ÿ 21;

œ 8 sin 2 ˆ 14 ‰ œ 8 Ê T has a maximum at (xß y) œ (2ß 1);

4

d# T dt# ¹ tœ 31 4

œ 8 sin 2 ˆ 341 ‰ œ 8 Ê T has a minimum at (xß y) œ (2ß 1);

" #

Ê sin t œ „

" È2

Ê t œ 14 ,

Section 14.5 Directional Derivatives and Gradient Vectors d# T dt# ¹ tœ 51

œ 8 sin 2 ˆ 541 ‰ œ 8 Ê T has a maximum at (xß y) œ (2ß 1);

d# T dt# ¹ tœ 71

œ 8 sin 2 ˆ 741 ‰ œ 8 Ê T has a minimum at (xß y) œ (2ß 1)

4

4

(b) T œ xy  2 Ê

`T `x

`T `y

œ y and

œ x so the extreme values occur at the four points found in part (a):

T(2ß 1) œ T(2ß 1) œ 0, the maximum and T(2ß 1) œ T(2ß 1) œ 4, the minimum 49. G(uß x) œ 'a g(tß x) dt where u œ f(x) Ê u

dG dx

œ

` G du ` u dx




` G dx ` x dx

F(x) œ '0 Èt%
x$ dt Ê Fw (x) œ Éax# b%
x$ (2x)
'0 x#

x#

` `x

œ g(uß x)f w (x)
'a gx (tß x) dt; thus u

Èt%
x$ dt œ 2xÈx)
x$
'

x#

0

3x# 2Èt%
x$

50. Using the result in Exercise 49, F(x) œ 'x# Èt$
x# dt œ  '1 Èt$
x# dt Ê Fw (x) x#

1

œ ’ Éax# b$
x# x#  '

Èt$
x# dt “ œ x# Èx'
x#
' # È $x # dt x t
x

x#

1

` 1 `x

14.5 DIRECTIONAL DERIVATIVES AND GRADIENT VECTORS 1.

`f `x

œ 1,

`f `y

œ 1 Ê ™ f œ i
j ; f(2ß 1) œ 1

Ê 1 œ y  x is the level curve

2.

`f `x

œ

Ê

2y 2x `f `f x#
y# Ê ` x ("ß ") œ 1; ` y œ x#
y# `f ` y ("ß ") œ 1 Ê ™ f œ i
j ; f(1ß 1) # # # #

œ ln 2 Ê ln 2

œ ln ax
y b Ê 2 œ x
y is the level curve

3.

`g `x

`g `x

œ 2x Ê

(1ß 0) œ 2;

`g `y

œ1

Ê ™ g œ 2i
j ; g(1ß 0) œ 1 Ê 1 œ y  x# is the level curve

4.

`g `x

œx Ê

Ê

`g `y

`g `x

ŠÈ2ß "‹ œ È2;

`g `y

œ y

ŠÈ2ß "‹ œ 1 Ê ™ g œ È2 i  j ;

g ŠÈ2ß "‹ œ

" #

Ê

" #

œ

x# #



y# #

or 1 œ x#  y# is the level

curve 5.

`f `x

œ 2x


z x

Ê

`f `x

(1ß 1ß 1) œ 3;

thus ™ f œ 3i
2j  4k

`f `y

œ 2y Ê

`f `y

("ß "ß ") œ 2;

`f `z

œ 4z
ln x Ê

`f `z

("ß "ß ") œ 4;

dt

887

888 6.

Chapter 14 Partial Derivatives `f `x

œ 6xz


Ê 7.

8.

z x # z#
1 `f " ` z ("ß "ß ") œ # ;

`f `x

œ

`f `z

œ

`f `x

œ ex
y cos z


`f `z

œ ex
y sin z Ê

9. u œ

v kv k

x ax#
y#
z# b$Î# z ax#
y#
z# b$Î#

œ

4i
3j È 4#
3#

`f `x

Ê

`f `y

œ 6yz Ê

("ß "ß ") œ 6;

`f `z

œ 6z#  3 ax#
y# b


x x # z#
1

" thus ™ f œ  11 # i  6j
# k




" x

Ê

`f `x

(1ß 2ß 2) œ  26 27 ;




" z

Ê

`f `z

(1ß 2ß 2) œ  23 54 ; thus ™

y
1 È 1  x#

œ

`f `y

(1ß "ß ") œ  11 # ;

`f `z

Ê

`f `x

ˆ!ß !ß 16 ‰ œ

È3 #


1;

`f `y

œ

`f `y

ˆ!ß !ß 16 ‰ œ  "# ; thus ™ f œ Š

y
"y Ê `` yf ax#
y#
z# b$Î# 23 23 f œ  26 27 i
54 j  54 k

œ ex
y cos z
sin" x Ê È 3
2 # ‹i

È3 #




(1ß 2ß 2) œ

`f `y

23 54

È3 #

ˆ0ß 0ß 16 ‰ œ

;

;

j  "# k

i
35 j ; fx (xß y) œ 2y Ê fx (5ß 5) œ 10; fy (xß y) œ 2x  6y Ê fy (5ß 5) œ 20

4 5

Ê ™ f œ 10i  20j Ê (Du f)P! œ ™ f † u œ 10 ˆ 45 ‰  20 ˆ 35 ‰ œ 4 10. u œ

v kv k

œ

3i  4j È3#
(4)#

œ

3 5

i  45 j ; fx (xß y) œ 4x Ê fx (1ß 1) œ 4; fy (xß y) œ 2y Ê fy (1ß 1) œ 2

Ê ™ f œ 4i
2j Ê (Du f)P! œ ™ f † u œ  12 5  11. u œ

œ

12. u œ

12i
5j 12 5 È12#
5# œ 13 i
13 j ; gx (xß y) œ 1 2xÈ3 2y x
2xyÈ4x# y#  1 Ê gy (1ß 1) œ 1 Ê

v kv k

v kv k

œ

œ

hy (xß y) œ

3i  2j È3#
(2)# ˆ "x ‰

ˆ xy ‰#
1




œ

i

3 È13

ˆ x# ‰ È3 Ê1  Š

x# y# 4 ‹

2 È13







2yÈ3 2xyÈ4x# y#  1

Ê gx (1ß 1) œ 3; gy (xß y)

™ g œ 3i  j Ê (Du g)P! œ ™ g † u œ

j ; hx (xß y) œ

Ê hy (1ß 1) œ

y# x#

œ 4

8 5

3 #

Š
x#y ‹

ˆ xy ‰#
1




ˆ #y ‰ È3 Ê1  Š

Ê ™hœ

" #

36 13



5 13

œ

31 13

Ê hx (1ß 1) œ "# ;

x# y# 4 ‹

i
3# j Ê (Du h)P! œ ™ h † u œ

3 2È13



6 2È13

œ  2È313 13. u œ

v kv k

œ

3 i
6 j  #k È3#
6#
(2)#

œ

3 7

i
67 j  27 k ; fx (xß yß z) œ y
z Ê fx (1ß 1ß 2) œ 1; fy (xß yß z) œ x
z

Ê fy (1ß 1ß 2) œ 3; fz (xß yß z) œ y
x Ê fz (1ß 1ß 2) œ 0 Ê ™ f œ i
3j Ê (Du f)P! œ ™ f † u œ 14. u œ

v kv k

œ

i
j
k È 1#
1#
1#

œ

1 È3

i


1 È3

j


1 È3

3 7




18 7

œ3

k ; fx (xß yß z) œ 2x Ê fx (1ß 1ß 1) œ 2; fy (xß yß z) œ 4y

Ê fy (1ß 1ß 1) œ 4; fz (xß yß z) œ 6z Ê fz (1ß 1ß 1) œ 6 Ê ™ f œ 2i
4j  6k Ê (Du f)P! œ ™ f † u œ 2 Š È"3 ‹
4 Š È"3 ‹  6 Š È"3 ‹ œ 0 15. u œ

v kv k

œ

2i
j  2k È2#
1#
(2)#

œ

2 3

i
13 j  23 k ; gx (xß yß z) œ 3ex cos yz Ê gx (0ß 0ß 0) œ 3; gy (xß yß z) œ 3zex sin yz

Ê gy (0ß 0ß 0) œ 0; gz (xß yß z) œ 3yex sin yz Ê gz (0ß 0ß 0) œ 0 Ê ™ g œ 3i Ê (Du g)P! œ ™ g † u œ 2 16. u œ

v kv k

œ

i
2j
2k È 1#
2#
2#

œ

1 3

i
23 j
23 k ; hx (xß yß z) œ y sin xy


" x

Ê hx ˆ1ß 0ß #" ‰ œ 1;

hy (xß yß z) œ x sin xy
zeyz Ê hy ˆ"ß !ß "# ‰ œ "# ; hz (xß yß z) œ yeyz
Ê (Du h)P! œ ™ h † u œ

" 3




" 3




4 3

" z

Ê hz ˆ"ß !ß #" ‰ œ 2 Ê ™ h œ i
#" j
2k

œ2

17. ™ f œ (2x
y) i
(x
2y) j Ê ™ f(1ß 1) œ i
j Ê u œ most rapidly in the direction u œ  È"2 i


" È2

™f k™f k

œ

i
j È(1)#
1#

œ  È"2 i


" È2

j and decreases most rapidly in the direction u œ

(Du f)P! œ ™ f † u œ k ™ f k œ È2 and (Du f)P! œ È2

j ; f increases

" È2

i

" È2

j;

Section 14.5 Directional Derivatives and Gradient Vectors ™f k™f k

18. ™ f œ a2xy
yexy sin yb i
ax#
xexy sin y
exy cos yb j Ê ™ f(1ß 0) œ 2j Ê u œ

œ j ; f increases most

rapidly in the direction u œ j and decreases most rapidly in the direction u œ j ; (Du f)P! œ ™ f † u œ k ™ f k œ 2 and (Du f)P! œ 2 19. ™ f œ œ

" 3È 3

" y

i  Š yx#
z‹ j  yk Ê ™ f(4ß "ß ") œ i  5j  k Ê u œ

i

5 3È 3

" 3È 3

j

™f k ™f k

k ; f increases most rapidly in the direction of u œ

" most rapidly in the direction u œ  3È i
3

5 3È 3

j


" 3È 3

i  5j  k È1#
(5)#
(1)# " 5 " i  3È j  3È 3È 3 3 3

œ

k and decreases

k ; (Du f)P! œ ™ f † u œ k ™ f k œ 3È3 and

(Du f)P! œ 3È3 20. ™ g œ ey i
xey j
2zk Ê ™ g ˆ1ß ln 2ß "# ‰ œ 2i
2j
k Ê u œ g increases most rapidly in the direction u œ

2 3

™g k ™g k

œ

2i
2j
k È 2#
2#
1#

œ

2 3

i
32 j
3" k ;

i
23 j
3" k and decreases most rapidly in the direction

u œ  23 i  23 j  3" k ; (Du g)P! œ ™ g † u œ k ™ gk œ 3 and (Du g)P! œ 3 21. ™ f œ ˆ "x
x" ‰ i
Š y"
y" ‹ j
ˆ "z
"z ‰ k Ê ™ f("ß "ß ") œ 2i
2j
2k Ê u œ f increases most rapidly in the direction u œ u œ  È"3 i


" È3

j


" È3

" È3

i


" È3

j


" È3

™f k ™f k

2 7

" È3

i


" È3

j


" È3

k;

k and decreases most rapidly in the direction

k; (Du f)P! œ ™ f † u œ k ™ f k œ 2È3 and (Du f)P! œ 2È3

2y 22. ™ h œ Š x#
2x y#  1 ‹ i
Š x#
y#  1
1‹ j
6k Ê ™ h("ß "ß 0) œ 2i
3j
6k Ê u œ

œ

œ

i
37 j
67 k ; h increases most rapidly in the direction u œ

2 7

Ê Tangent line: 2È2 Šx  È2‹
2È2 Šy  È2‹ œ 0 Ê È2x
È2y œ 4

24. ™ f œ 2xi  j Ê ™ f ŠÈ2ß 1‹ œ 2È2 i  j Ê Tangent line: 2È2 Šx  È2‹  (y  1) œ 0 Ê y œ 2È2x  3

25. ™ f œ yi
xj Ê ™ f(2ß 2) œ 2i
2j Ê Tangent line: 2(x  2)
2(y
2) œ 0 Ê yœx4

œ

2i
3j
6k È 2#
3#
6#

i
37 j
67 k and decreases most rapidly in the

direction u œ  27 i  37 j  67 k ; (Du h)P! œ ™ h † u œ k ™ hk œ 7 and (Du h)P! œ 7 23. ™ f œ 2xi
2yj Ê ™ f ŠÈ2ß È2‹ œ 2È2 i
2È2 j

™h k ™h k

889

890

Chapter 14 Partial Derivatives

26. ™ f œ (2x  y)i
(2y  x)j Ê ™ f(1ß 2) œ 4i
5j Ê Tangent line: 4(x
1)
5(y  2) œ 0 Ê 4x
5y  14 œ 0

27. ™ f œ yi
(x
2y)j Ê ™ f(3ß 2) œ 2i
7j ; a vector orthogonal to ™ f is v œ 7i  2j Ê u œ œ

7 È53

28. ™ f œ

i

4xy# ax #
y # b #

Ê uœ

j and u œ  È753 i


2 È53

v kv k

4x# y j Ê a x #
y # b# i
j 1 È 1#
1# œ È 2 i

i

œ

2 È53

v kv k

œ

7i  2j È7#
(2)#

j are the directions where the derivative is zero

™ f("ß ") œ i  j ; a vector orthogonal to ™ f is v œ i
j


1 È2

j and u œ  È12 i 

1 È2

j are the directions where the derivative is zero

29. ™ f œ (2x  3y)i
(3x
8y)j Ê ™ f(1ß 2) œ 4i
13j Ê k ™ f(1ß 2)k œ È(4)#
(13)# œ È185 ; no, the maximum rate of change is È185  14 30. ™ T œ 2yi
(2x  z)j  yk Ê ™ T(1ß 1ß 1) œ 2i
j
k Ê k ™ T(1ß 1ß 1)k œ È(2)#
1#
1# œ È6 ; no, the minimum rate of change is È6  3 i
j È 1#
1#

31. ™ f œ fx ("ß #)i
fy ("ß #)j and u" œ

œ

" È2

i


" È2

j Ê (Du" f)(1ß 2) œ fx (1ß 2) Š È" ‹
fy (1ß 2) Š È" ‹ 2 2

œ 2È2 Ê fx (1ß 2)
fy (1ß 2) œ 4; u# œ j Ê (Du# f)(1ß 2) œ fx (1ß 2)(0)
fy (1ß 2)(1) œ 3 Ê fy (1ß 2) œ 3 Ê fy (1ß 2) œ 3; then fx (1ß 2)
3 œ 4 Ê fx (1ß 2) œ 1; thus ™ f(1ß 2) œ i
3j and u œ œ  È15 i 

2 È5

j Ê (Du f)P! œ ™ f † u œ  È"5 

32. (a) (Du f)P œ 2È3 Ê k ™ f k œ 2È3; u œ

v kv k

œ

v kv k

œ

i
j È 1#
1#

œ

" È2

i


" È2

" È3

i  2 j È(1)#
(2)#

œ

œ  È75

i
jk È1#
1#
(1)#

Ê ™ f œ k ™ f k u Ê ™ f œ 2È3 Š È"3 i
(b) v œ i
j Ê u œ

6 È5

v kv k

j

" È3

œ

1 È3

i


1 È3

j

" È3

k; thus u œ

™f k ™f k

k‹ œ 2i
2j  2k

j Ê (Du f)P! œ ™ f † u œ 2 Š È"2 ‹
2 Š È"2 ‹  2(0) œ 2È2

33. The directional derivative is the scalar component. With ™ f evaluated at P! , the scalar component of ™ f in the direction of u is ™ f † u œ (Du f)P! . 34. Di f œ ™ f † i œ (fx i
fy j
fz k) † i œ fx ; similarly, Dj f œ ™ f † j œ fy and Dk f œ ™ f † k œ fz 35. If (xß y) is a point on the line, then T(xß y) œ (x  x! )i
(y  y! )j is a vector parallel to the line Ê T † N œ 0 Ê A(x  x! )
B(y  y! ) œ 0, as claimed. 36. (a) ™ (kf) œ

` (kf) `x

(b) ™ (f
g) œ œ

`f `x

i


`g `x

i


` (kf) `y

` (f
g) `x

i


`f `y

i


j


j


` (kf) `z

` (f
g) `y `g `y

j


k œ k ˆ `` xf ‰ i
k Š `` yf ‹ j
k ˆ `` zf ‰ k œ k Š `` xf i


j
`f `z

` (f
g) `z

k


`g `z

k œ Š `` xf
k œ Š `` xf i


`g `x ‹ i `f `y


Š `` yf


j


`f `z

`g `y ‹ j

j



Š `` zf


k‹
Š `` gx i


(c) ™ (f  g) œ ™ f  ™ g (Substitute g for g in part (b) above)

`f `y

`g `y

j


`f `z

k‹ œ k ™ f

`g `z ‹ k `g `z

k‹ œ ™ f
™ g

Section 14.6 Tangent Planes and Differentials ` (fg) `x

(d) ™ (fg) œ

i


` (fg) `y

j


` (fg) `z

k œ Š `` xf g


`g `x

f ‹ i
Š `` yf g


`g `y

f ‹ j
Š `` zf g


`g `z

891

f‹ k

œ ˆ `` xf g‰ i
Š `` gx f ‹ i
Š `` yf g‹ j
Š `` gy f ‹ j
ˆ `` zf g‰ k
Š `` gz f ‹ k œ f Š `` gx i
(e) ™ Š gf ‹ œ œŒ œ

`g `y

j


` Š gf ‹ `x

`g `z

i


k‹
g Š `` xf i


` Š gf ‹ `y

g ``xf i
g ``yf j
g `` fz k  g#

g ™f g#



f™g g#

œ

j


Œ

` Š gf ‹ `z

`f `y

j


kœŠ

`f `z

k‹ œ f ™ g
g ™ f

g ``xf  f `` xg ‹i g#

f `` xg i
f `` gy j
f ``gz k  g#

œ


Œ

g ``yf  f `` gy j g#

g Š ``xf i
``yf j
`` fz k‹ g#




Š

g `` zf  f ``gz ‹k g#

f Š `` xg i
`` gy j
``gz k‹ g#

g™f  f™g g#

14.6 TANTGENT PLANES AND DIFFERENTIALS 1. (a) ™ f œ 2xi
2yj
2zk Ê ™ f(1ß 1ß 1) œ 2i
2j
2k Ê Tangent plane: 2(x  1)
2(y  1)
2(z  1) œ 0 Ê x
y
z œ 3; (b) Normal line: x œ 1
2t, y œ 1
2t, z œ 1
2t 2. (a) ™ f œ 2xi
2yj  2zk Ê ™ f(3ß 5ß 4) œ 6i
10j
8k Ê Tangent plane: 6(x  3)
10(y  5)
8(z
4) œ 0 Ê 3x
5y
4z œ 18; (b) Normal line: x œ 3
6t, y œ 5
10t, z œ 4
8t 3. (a) ™ f œ 2xi
2k Ê ™ f(2ß 0ß 2) œ 4i
2k Ê Tangent plane: 4(x  2)
2(z  2) œ 0 Ê 4x
2z
4 œ 0 Ê 2x
z
2 œ 0; (b) Normal line: x œ 2  4t, y œ 0, z œ 2
2t 4. (a) ™ f œ (2x
2y)i
(2x  2y)j
2zk Ê ™ f(1ß 1ß 3) œ 4j
6k Ê Tangent plane: 4(y
1)
6(z  3) œ 0 Ê 2y
3z œ 7; (b) Normal line: x œ 1, y œ 1
4t, z œ 3
6t 5. (a) ™ f œ a1 sin 1x  2xy
zexz b i
ax#
zb j
axexz
yb k Ê ™ f(0ß 1ß 2) œ 2i
2j
k Ê Tangent plane: 2(x  0)
2(y  1)
1(z  2) œ 0 Ê 2x
2y
z  4 œ 0; (b) Normal line: x œ 2t, y œ 1
2t, z œ 2
t 6. (a) ™ f œ (2x  y)i  (x
2y)j  k Ê ™ f(1ß 1ß 1) œ i  3j  k Ê Tangent plane: 1(x  1)  3(y  1)  1(z
1) œ 0 Ê x  3y  z œ 1; (b) Normal line: x œ 1
t, y œ 1  3t, z œ 1  t 7. (a) ™ f œ i
j
k for all points Ê ™ f(0ß 1ß 0) œ i
j
k Ê Tangent plane: 1(x  0)
1(y  1)
1(z  0) œ 0 Ê x
y
z  1 œ 0; (b) Normal line: x œ t, y œ 1
t, z œ t 8. (a) ™ f œ (2x  2y  1)i
(2y  2x
3)j  k Ê ™ f(2ß 3ß 18) œ 9i  7j  k Ê Tangent plane: 9(x  2)  7(y
3)  1(z  18) œ 0 Ê 9x  7y  z œ 21; (b) Normal line: x œ 2
9t, y œ 3  7t, z œ 18  t 9. z œ f(xß y) œ ln ax#
y# b Ê fx (xß y) œ

2x x#
y#

and fy (xß y) œ

2y x#
y#

Ê fx (1ß 0) œ 2 and fy (1ß 0) œ 0 Ê from

Eq. (4) the tangent plane at (1ß 0ß 0) is 2(x  1)  z œ 0 or 2x  z  2 œ 0

892

Chapter 14 Partial Derivatives #

#

#

#

#

#

10. z œ f(xß y) œ e ax
y b Ê fx (xß y) œ 2xe ax
y b and fy (xß y) œ 2ye ax
y b Ê fx (0ß 0) œ 0 and fy (!ß !) œ 0 Ê from Eq. (4) the tangent plane at (0ß 0ß 1) is z  1 œ 0 or z œ 1 11. z œ f(Bß y) œ Èy  x Ê fx (xß y) œ  "# (y  x)"Î# and fy (xß y) œ

" #

(y  x)"Î# Ê fx (1ß 2) œ  "# and fy ("ß #) œ

Ê from Eq. (4) the tangent plane at (1ß 2ß 1) is  "# (x  1)
"# (y  2)  (z  1) œ 0 Ê x  y
2z  1 œ 0

" #

12. z œ f(Bß y) œ 4x#
y# Ê fx (xß y) œ 8x and fy (xß y) œ #y Ê fx (1ß 1) œ 8 and fy ("ß 1) œ # Ê from Eq. (4) the tangent plane at (1ß 1ß 5) is 8(x  1)
2(y  1)  (z  5) œ 0 or 8x
2y  z  5 œ 0 13. ™ f œ i
2yj
2k Ê ™ f(1ß 1ß 1) œ i
2j
2k and ™ g œ i for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â " 2 2 â œ 2j  2k Ê Tangent line: x œ 1, y œ 1
2t, z œ 1  2t â â â" 0 0â 14. ™ f œ yzi
xzj
xyk Ê ™ f(1ß 1ß 1) œ i
j
k; ™ g œ 2xi
4yj
6zk Ê ™ g(1ß 1ß 1) œ 2i
4j
6k ; â â â i j kâ â â Ê v œ ™ f ‚ ™ g Ê â " 1 1 â œ 2i  4j
2k Ê Tangent line: x œ 1
2t, y œ 1  4t, z œ 1
2t â â â2 4 6â 15. ™ f œ 2xi
2j
2k Ê ™ f ˆ1ß 1ß "# ‰ œ 2i
2j
2k and ™ g œ j for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â 2 2 2 â œ 2i
2k Ê Tangent line: x œ 1  2t, y œ 1, z œ "#
2t â â â0 1 0â 16. ™ f œ i
2yj
k Ê ™ f ˆ "# ß 1ß "# ‰ œ i
2j
k and ™ g œ j for all points; v œ ™ f ‚ ™ g â â â i j kâ â â Ê v œ â 1 2 1 â œ i
k Ê Tangent line: x œ "#  t, y œ 1, z œ "#
t â â â0 1 0â 17. ™ f œ a3x#
6xy#
4yb i
a6x# y
3y#
4xb j  2zk Ê ™ f(1ß 1ß 3) œ 13i
13j  6k ; ™ g œ 2xi
2yj
2zk â â j k â â i â â Ê ™ g("ß "ß $) œ 2i
2j
6k ; v œ ™ f ‚ ™ g Ê v œ â "3 13 6 â œ 90i  90j Ê Tangent line: â â 2 6 â â 2 x œ 1
90t, y œ 1  90t, z œ 3 18. ™ f œ 2xi
2yj Ê ™ f ŠÈ2ß È2ß 4‹ œ 2È2 i
2È2 j ; ™ g œ 2xi
2yj  k Ê ™ g ŠÈ2ß È2ß 4‹ â i j k ââ â â â œ 2È2i
2È2j  k ; v œ ™ f ‚ ™ g Ê v œ â 2È2 2È2 0 â œ 2È2 i
2È2 j Ê Tangent line: â â â 2È2 2È2 1 â x œ È2  2È2 t, y œ È2
2È2 t, z œ 4 19. ™ f œ Š x#
yx#
z# ‹ i
Š x#
yy#
z# ‹ j
Š x#
yz#
z# ‹ k Ê ™ f(3ß 4ß 12) œ uœ

v kv k

œ

3i
6j  2k È3#
6#
(2)#

œ

3 7

i
j k Ê ™f†uœ 6 7

2 7

9 1183

3 169

1 È3

i


1 È3

j

1 È3

k Ê ™f†uœ

1 È3

4 169

j


and df œ ( ™ f † u) ds œ

20. ™ f œ aex cos yzb i  azex sin yzb j  ayex sin yzb k Ê ™ f(0ß 0ß 0) œ i ; u œ œ

i


and df œ ( ™ f † u) ds œ

1 È3

v kv k

œ

12 169

k;

9 ‰ ˆ 1183 (0.1)

2i
2j  2k È2#
2#
(2)#

(0.1) ¸ 0.0577

¸ 0.0008

Section 14.6 Tangent Planes and Differentials

893

Ä 21. ™ g œ (1
cos z)i
(1  sin z)j
(x sin z  y cos z)k Ê ™ g(2ß 1ß 0) œ 2i
j
k; A œ P! P" œ 2i
2j
2k i
2j
2k 1 1 1 Ê u œ kvvk œ È(22) ™ g † u œ 0 and dg œ ( ™ g † u) ds œ (0)(0.2) œ 0 #
2#
2# œ  È 3 i
È 3 j
È 3 k Ê 22. ™ h œ c1y sin (1xy)
z# d i  c1x sin (1xy)d j
2xzk Ê ™ h(1ß 1ß 1) œ (1 sin 1
1)i
(1 sin 1)j
2k Ä œ i
2k ; v œ P! P" œ i
j
k where P" œ (!ß !ß !) Ê u œ kvvk œ È i#
j
# k # œ È13 i
È13 j
È13 k 1
1
1

Ê ™h†uœ

œ È3 and dh œ ( ™ h † u) ds œ È3(0.1) ¸ 0.1732

3 È3

23. (a) The unit tangent vector at Š "# ß

È3 # ‹

in the direction of motion is u œ

™ T œ (sin 2y)i
(2x cos 2y)j Ê ™ T Š "# ß È3 #

œ

sin È3 

" #

È3 # ‹

È3 #

i  "# j ;

œ Šsin È3‹ i
Šcos È3‹ j Ê Du T Š "# ß

œ ™T†vœŠ™T† dT dt

œŠ

È3 #

œ ™T†u

cos È3 ¸ 0.935° C/ft ` T dx ` x dt




` T dy ` y dt

we have u œ

È3 #

i  "# j from part (a)

(b) r(t) œ (sin 2t)i
(cos 2t)j Ê v(t) œ (2 cos 2t)i  (2 sin 2t)j and kvk œ 2;

Ê

È3 # ‹

v kv k ‹

sin È3 

" #

kvk œ (Du T) kvk , where u œ

v kv k

; at Š "# ß

È3 # ‹

dT dt

œ

cos È3‹ † 2 œ È3 sin È3  cos È3 ¸ 1.87° C/sec

24. (a) ™ T œ (4x  yz)i  xzj  xyk Ê ™ T(8ß 6ß 4) œ 56i
32j  48k ; r(t) œ 2t# i
3tj  t# k Ê the particle is at the point P()ß 6ß 4) when t œ 2; v(t) œ 4ti
3j  2tk Ê v(2) œ 8i
3j  4k Ê u œ kvvk

(b)

œ

8 È89

dT dt

œ

i


` T dx ` x dt

3 È89




j

` T dy ` y dt

4 È89

k Ê Du T(8ß 6ß 4) œ ™ T † u œ

" È89

œ ™ T † v œ ( ™ T † u) kvk Ê at t œ 2,

[56 † 8
32 † 3  48 † (4)] œ

dT dt

736 È89

° C/m

736 œ Du T¸ tœ2 v(2) œ Š È ‹ È89 œ 736° C/sec 89

25. (a) f(!ß 0) œ 1, fx (xß y) œ 2x Ê fx (0ß 0) œ 0, fy (xß y) œ 2y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1
0(x  0)
0(y  0) œ 1 (b) f(1ß 1) œ 3, fx (1ß 1) œ 2, fy (1ß 1) œ 2 Ê L(xß y) œ 3
2(x  1)
2(y  1) œ 2x
2y  1 26. (a) f(!ß 0) œ 4, fx (xß y) œ 2(x
y
2) Ê fx (0ß 0) œ 4, fy (xß y) œ 2(x
y
2) Ê fy (0ß 0) œ 4 Ê L(xß y) œ 4
4(x  0)
4(y  0) œ 4x
4y
4 (b) f(1ß 2) œ 25, fx (1ß 2) œ 10, fy (1ß 2) œ 10 Ê L(xß y) œ 25
10(x  1)
10(y  2) œ 10x
10y  5 27. (a) f(0ß 0) œ 5, fx (xß y) œ 3 for all (xß y), fy (xß y) œ 4 for all (xß y) Ê L(xß y) œ 5
3(x  0)  4(y  0) œ 3x  4y
5 (b) f(1ß 1) œ 4, fx (1ß 1) œ 3, fy (1ß 1) œ 4 Ê L(xß y) œ 4
3(x  1)  4(y  1) œ 3x  4y
5 28. (a) f(1ß 1) œ 1, fx (xß y) œ 3x# y% Ê fx (1ß 1) œ 3, fy (xß y) œ 4x$ y$ Ê fy (1ß 1) œ 4 Ê L(xß y) œ 1
3(x  1)
4(y  1) œ 3x
4y  6 (b) f(0ß 0) œ 0, fx (!ß 0) œ 0, fy (0ß 0) œ 0 Ê L(xß y) œ 0 29. (a) f(0ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1
1(x  0)
0(y  0) œ x
1 (b) f ˆ0ß 1# ‰ œ 0, fx ˆ0ß 1# ‰ œ 0, fy ˆ0ß 1# ‰ œ 1 Ê L(xß y) œ 0
0(x  0)  1 ˆy  1# ‰ œ y
30. (a) f(0ß 0) œ 1, fx (xß y) œ e2yx Ê fx (!ß !) œ 1, fy (xß y) œ 2e2yx Ê fy (0ß 0) œ 2 Ê L(xß y) œ 1  1(x  0)
2(y  0) œ x
2y
1 (b) f(1ß 2) œ e$ , fx (1ß 2) œ e$ , fy (1ß 2) œ 2e$ Ê L(xß y) œ e$  e$ (x  1)
2e$ (y  2) œ e$ x
2e$ y  2e$

1 #

894

Chapter 14 Partial Derivatives

31. f(2ß 1) œ 3, fx (xß y) œ 2x  3y Ê fx (2ß 1) œ 1, fy (xß y) œ 3x Ê fy (2ß 1) œ 6 Ê L(xß y) œ 3
1(x  2)  6(y  1) œ 7
x  6y; fxx (xß y) œ 2, fyy (xß y) œ 0, fxy (xß y) œ 3 Ê M œ 3; thus kE(xß y)k Ÿ ˆ "# ‰ (3) akx  2k
ky  1kb# Ÿ ˆ 3# ‰ (0.1
0.1)# œ 0.06

32. f(2ß 2) œ 11, fx (xß y) œ x
y
3 Ê fx (2ß 2) œ 7, fy (xß y) œ x


y #

 3 Ê fy (2ß 2) œ 0

Ê L(xß y) œ 11
7(x  2)
0(y  2) œ 7x  3; fxx (xß y) œ 1, fyy (xß y) œ "# , fxy (xß y) œ 1 Ê M œ 1; thus kE(xß y)k Ÿ ˆ "# ‰ (1) akx  2k
ky  2kb# Ÿ ˆ 1# ‰ (0.1
0.1)# œ 0.02 33. f(0ß 0) œ 1, fx (xß y) œ cos y Ê fx (0ß 0) œ 1, fy (xß y) œ 1  x sin y Ê fy (0ß 0) œ 1 Ê L(xß y) œ 1
1(x  0)
1(y  0) œ x
y
1; fxx (xß y) œ 0, fyy (xß y) œ x cos y, fxy (xß y) œ  sin y Ê Q œ 1; thus kE(xß y)k Ÿ ˆ "# ‰ (1) akxk
kykb# Ÿ ˆ 1# ‰ (0.2
0.2)# œ 0.08 34. f("ß #) œ 6, fx (xß y) œ y#  y sin (x  1) Ê fx (1ß 2) œ 4, fy (xß y) œ 2xy
cos (x  1) Ê fy (1ß 2) œ 5 Ê L(xß y) œ 6
4(x  1)
5(y  2) œ 4x
5y  8; fxx (xß y) œ y cos (x  1), fyy (xß y) œ 2x, fxy (xß y) œ 2y  sin (x  1); kx  1k Ÿ 0.1 Ê 0.9 Ÿ x Ÿ 1.1 and ky  2k Ÿ 0.1 Ê 1.9 Ÿ y Ÿ 2.1; thus the max of kfxx (xß y)k on R is 2.1, the max of kfyy (xß y)k on R is 2.2, and the max of kfxy (xß y)k on R is 2(2.1)  sin (0.9  1) Ÿ 4.3 Ê M œ 4.3; thus kE(xß y)k Ÿ ˆ "# ‰ (4.3) akx  1k
ky  2kb# Ÿ (2.15)(0.1
0.1)# œ 0.086 35. f(!ß 0) œ 1, fx (xß y) œ ex cos y Ê fx (0ß 0) œ 1, fy (xß y) œ ex sin y Ê fy (0ß 0) œ 0 Ê L(xß y) œ 1
1(x  0)
0(y  0) œ 1
x; fxx (xß y) œ ex cos y, fyy (xß y) œ ex cos y, fxy (xß y) œ ex sin y; kxk Ÿ 0.1 Ê 0.1 Ÿ x Ÿ 0.1 and kyk Ÿ 0.1 Ê 0.1 Ÿ y Ÿ 0.1; thus the max of kfxx (xß y)k on R is e0Þ1 cos (0.1) Ÿ 1.11, the max of kfyy (xß y)k on R is e0Þ1 cos (0.1) Ÿ 1.11, and the max of kfxy (xß y)k on R is e0Þ1 sin (0.1) Ÿ 0.12 Ê M œ 1.11; thus kE(xß y)k Ÿ ˆ "# ‰ (1.11) akxk
kykb# Ÿ (0.555)(0.1
0.1)# œ 0.0222 36. f(1ß 1) œ 0, fx (xß y) œ

" x

Ê fx (1ß 1) œ 1, fy (xß y) œ

" y

Ê fy (1ß 1) œ 1 Ê L(xß y) œ 0
1(x  1)
1(y  1)

œ x
y  2; fxx (xß y) œ  x"# , fyy (xß y) œ  y"# , fxy (xß y) œ 0; kx  1k Ÿ 0.2 Ê 0.98 Ÿ x Ÿ 1.2 so the max of kfxx (xß y)k on R is " (0.98)#

" (0.98)#

Ÿ 1.04; ky  1k Ÿ 0.2 Ê 0.98 Ÿ y Ÿ 1.2 so the max of kfyy (xß y)k on R is

Ÿ 1.04 Ê M œ 1.04; thus kE(xß y)k Ÿ ˆ #" ‰ (1.04) akx  1k
ky  1kb# Ÿ (0.52)(0.2
0.2)# œ 0.0832

37. (a) f("ß "ß ") œ 3, fx (1ß 1ß 1) œ y
zkÐ1ß1ß1Ñ œ 2, fy (1ß 1ß 1) œ x
zkÐ1ß1ß1Ñ œ 2, fz (1ß 1ß 1) œ y
xkÐ1ß1ß1Ñ œ 2 Ê L(xß yß z) œ 3
2(x  1)
2(y  1)
2(z  1) œ 2x
2y
2z  3 (b) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ 0, fy (1ß 0ß 0) œ 1, fz (1ß 0ß 0) œ 1 Ê L(xß yß z) œ 0
0(x  1)
(y  0)
(z  0) œy
z (c) f(0ß 0ß 0) œ 0, fx (0ß 0ß 0) œ 0, fy (0ß 0ß 0) œ 0, fz (0ß 0ß 0) œ 0 Ê L(xß yß z) œ 0 38. (a) f(1ß 1ß 1) œ 3, fx (1ß 1ß 1) œ 2xkÐ"ß"ß"Ñ œ 2, fy (1ß 1ß 1) œ 2ykÐ"ß"ß"Ñ œ 2, fz (1ß 1ß 1) œ 2zkÐ"ß"ß"Ñ œ 2 Ê L(xß yß z) œ 3
2(x  1)
2(y  1)
2(z  1) œ 2x
2y
2z  3 (b) f(0ß 1ß 0) œ 1, fx (0ß 1ß 0) œ 0, fy (!ß 1ß 0) œ 2, fz (0ß 1ß 0) œ 0 Ê L(xß yß z) œ 1
0(x  0)
2(y  1)
0(z  0) œ 2y  1 (c) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ 2, fy (1ß 0ß 0) œ 0, fz (1ß 0ß 0) œ 0 Ê L(xß yß z) œ 1
2(x  1)
0(y  0)
0(z  0) œ 2x  1 39. (a) f(1ß 0ß 0) œ 1, fx (1ß 0ß 0) œ fz (1ß 0ß 0) œ

z È x #
y #
z# ¹

x È x #
y #
z# ¹

Ð1ß0ß0Ñ

Ð1ß0ß0Ñ

œ 1, fy (1ß 0ß 0) œ

y È x #
y #
z# ¹

Ð1ß0ß0Ñ

œ 0,

œ 0 Ê L(xß yß z) œ 1
1(x  1)
0(y  0)
0(z  0) œ x

Section 14.6 Tangent Planes and Differentials (b) f(1ß 1ß 0) œ È2, fx (1ß 1ß 0) œ Ê L(xß yß z) œ È2


" È2

" È2

, fy (1ß 1ß 0) œ

(x  1)


" È2

" È2

, fz (1ß 1ß 0) œ 0

(y  1)
0(z  0) œ

(c) f(1ß 2ß 2) œ 3, fx (1ß 2ß 2) œ 3" , fy (1ß 2ß 2) œ 32 , fz (1ß 2ß 2) œ œ

" 3

895

2 3

" È2

x


" È2

y

Ê L(xß yß z) œ 3
3" (x  1)
32 (y  2)
23 (z  2)

x
23 y
23 z

40. (a) f ˆ 12 ß 1ß 1‰ œ 1, fx ˆ 1# ß 1ß 1‰ œ fz ˆ 1# ß 1ß 1‰ œ

 sin xy z# ¹ ˆ 1 ß"ß"‰

y cos xy ¸ ˆ 1# ß"ß"‰ z

œ 0, fy ˆ 1# ß 1ß 1‰ œ

x cos xy ¸ ˆ 1# ß"ß"‰ z

œ 0,

œ 1 Ê L(xß yß z) œ 1
0 ˆx  1# ‰
0(y  1)  1(z  1) œ 2  z

#

(b) f(2ß 0ß 1) œ 0, fx (2ß 0ß 1) œ 0, fy (2ß 0ß 1) œ 2, fz (2ß 0ß 1) œ 0 Ê L(xß yß z) œ 0
0(x  2)
2(y  0)
0(z  1) œ 2y 41. (a) f(0ß 0ß 0) œ 2, fx (0ß 0ß 0) œ ex k Ð!ß!ß!Ñ œ 1, fy (0ß 0ß 0) œ  sin (y
z)k Ð!ß!ß!Ñ œ 0, fz (0ß 0ß 0) œ  sin (y
z)k Ð!ß!ß!Ñ œ 0 Ê L(xß yß z) œ 2
1(x  0)
0(y  0)
0(z  0) œ 2
x (b) f ˆ0ß 1# ß 0‰ œ 1, fx ˆ0ß 1# ß 0‰ œ 1, fy ˆ0ß 1# ß 0‰ œ 1, fz ˆ0ß 1# ß 0‰ œ 1 Ê L(xß yß z) œ 1
1(x  0)  1 ˆy  12 ‰  1(z  0) œ x  y  z
1#
1

(c) f ˆ0ß 14 ß 14 ‰ œ 1, fx ˆ0ß 14 ß 14 ‰ œ 1, fy ˆ0ß 14 ß 14 ‰ œ 1, fz ˆ0ß 14 ß 14 ‰ œ 1 Ê L(xß yß z) œ 1
1(x  0)  1 ˆy  14 ‰  1 ˆz  14 ‰ œ x  y  z
1#
1 42. (a) f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ fz (1ß 0ß 0) œ

xy (xyz)#
1 ¹ Ð"ß!ß!Ñ

yz (xyz)#
1 ¹ Ð"ß!ß!Ñ

œ 0, fy (1ß 0ß 0) œ

xz (xyz)#
1 ¹ Ð"ß!ß!Ñ

œ 0,

œ 0 Ê L(xß yß z) œ 0

(b) f(1ß 1ß 0) œ 0, fx (1ß 1ß 0) œ 0, fy (1ß 1ß 0) œ 0, fz (1ß 1ß 0) œ 1 Ê L(xß yß z) œ 0
0(x  1)
0(y  1)
1(z  0) œ z (c) f(1ß 1ß 1) œ 14 , fx (1ß 1ß 1) œ "# , fy (1ß 1ß 1) œ "# , fz (1ß 1ß 1) œ "# Ê L(xß yß z) œ 14
"# (x  1)
"# (y  1)
"# (z  1) œ

" #

x
"# y
"# z


1 4



3 #

43. f(xß yß z) œ xz  3yz
2 at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 2; fx œ z, fy œ 3z, fz œ x  3y Ê L(xß yß z) œ 2
2(x  1)  6(y  1)  2(z  2) œ 2x  6y  2z
6; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 0, fyz œ 3 Ê M œ 3; thus, kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01
0.01
0.02)# œ 0.0024 44. f(xß yß z) œ x#
xy
yz
"4 z# at P! (1ß 1ß 2) Ê f(1ß 1ß 2) œ 5; fx œ 2x
y, fy œ x
z, fz œ y
"# z

Ê L(xß yß z) œ 5
3(x  1)
3(y  1)
2(z  2) œ 3x
3y
2z  5; fxx œ 2, fyy œ 0, fzz œ "# , fxy œ 1, fxz œ 0, fyz œ 1 Ê M œ 2; thus kE(xß yß z)k Ÿ ˆ "# ‰ (2)(0.01
0.01
0.08)# œ 0.01

45. f(xß yß z) œ xy
2yz  3xz at P! (1ß 1ß 0) Ê f(1ß 1ß 0) œ 1; fx œ y  3z, fy œ x
2z, fz œ 2y  3x Ê L(xß yß z) œ 1
(x  1)
(y  1)  (z  0) œ x
y  z  1; fxx œ 0, fyy œ 0, fzz œ 0, fxy œ 1, fxz œ 3, fyz œ 2 Ê M œ 3; thus kE(xß yß z)k Ÿ ˆ "# ‰ (3)(0.01
0.01
0.01)# œ 0.00135 46. f(xß yß z) œ È2 cos x sin (y
z) at P! ˆ0ß 0ß 14 ‰ Ê f ˆ0ß 0ß 14 ‰ œ 1; fx œ È2 sin x sin (y
z), fy œ È2 cos x cos (y
z), fz œ È2 cos x cos (y
z) Ê L(xß yß z) œ 1  0(x  0)
(y  0)
ˆz  14 ‰ œ y
z  14
1; fxx œ È2 cos x sin (y
z), fyy œ È2 cos x sin (y
z), fzz œ È2 cos x sin (y
z), fxy œ È2 sin x cos (y
z), fxz œ È2 sin x cos (y
z), fyz œ È2 cos x sin (y
z). The absolute value of each of these second partial derivatives is bounded above by È2 Ê M œ È2; thus kE(xß yß z)k Ÿ ˆ " ‰ ŠÈ2‹ (0.01
0.01
0.01)# œ 0.000636. #

896

Chapter 14 Partial Derivatives

47. Tx (xß y) œ ey
ey and Ty (xß y) œ x aey  ey b Ê dT œ Tx (xß y) dx
Ty (xß y) dy œ aey
ey b dx
x aey  ey b dy Ê dTkÐ#ßln 2Ñ œ 2.5 dx
3.0 dy. If kdxk Ÿ 0.1 and kdyk Ÿ 0.02, then the maximum possible error in the computed value of T is (2.5)(0.1)
(3.0)(0.02) œ 0.31 in magnitude. #

21rh dr
1r dh 48. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr
Vh dh Ê dV œ 2r dr
"h dh; now ¸ drr † 100¸ Ÿ 1 and V œ 1 r# h ¸ dh ¸ ¸ dV ¸ ¸ˆ2 drr ‰ (100)
ˆ dh ‰ ¸ ¸ dr ¸ ¸ dh ¸ h † 100 Ÿ 1 Ê V † 100 Ÿ h (100) Ÿ 2 r † 100
h † 100 Ÿ 2(1)
1 œ 3 Ê 3%

49. Vr œ 21rh and Vh œ 1r# Ê dV œ Vr dr
Vh dh Ê dV œ 21rh dr
1r# dh Ê dVkÐ5ß12Ñ œ 1201 dr
251 dh; kdrk Ÿ 0.1 cm and kdhk Ÿ 0.1 cm Ê dV Ÿ (1201)(0.1)
(251)(0.1) œ 14.51 cm$ ; V(5ß 12) œ 3001 cm$ 1 Ê maximum percentage error is „ 14.5 3001 ‚ 100 œ „ 4.83% 50. (a)

" R

œ

" R"

(b) dR œ R


#

" R#

Ê  R"# dR œ  R"# dR"  "

’Š R"# ‹ "

dR"


Š R"# ‹ #

" R##

#

#

dR# Ê dR œ Š RR" ‹ dR"
Š RR# ‹ dR#

" dR# “ Ê dRk Ð100 400Ñ œ R# ’ (100) # dR"
ß

sensitive to a variation in R" since

" (100)#

#



" (400)#

dR# “ Ê R will be more

" (400)#

#

(c) From part (a), dR œ Š RR" ‹ dR"
Š RR# ‹ dR# so that R" changing from 20 to 20.1 ohms Ê dR" œ 0.1 ohm and R# changing from 25 to 24.9 ohms Ê dR# œ 0.1 ohms; Ê dRk Ð20 25Ñ œ ß

œ

0.011 ˆ 100 ‰ 9

ˆ 100 ‰# 9 (20)#

(0.1)


ˆ 100 ‰# 9 (25)#

" R

œ

" R"




" R#

Ê Rœ

(0.1) ¸ 0.011 ohms Ê percentage change is

100 9

ohms

dR ¸ R Ð20ß25Ñ

‚ 100

‚ 100 ¸ 0.1%

51. A œ xy Ê dA œ x dy
y dx; if x  y then a 1-unit change in y gives a greater change in dA than a 1-unit change in x. Thus, pay more attention to y which is the smaller of the two dimensions. 52. (a) fx (xß y) œ 2x(y
1) Ê fx (1ß 0) œ 2 and fy (xß y) œ x# Ê fy (1ß 0) œ 1 Ê df œ 2 dx
1 dy Ê df is more sensitive to changes in x dx " (b) df œ 0 Ê 2 dx
dy œ 0 Ê 2 dx dy
1 œ 0 Ê dy œ  # 53. (a) r# œ x#
y# Ê 2r dr œ 2x dx
2y dy Ê dr œ œ „ œ

0.07 5

y y#
x#

œ „ 0.014 Ê ¸ drr ‚ 100¸ œ ¸ „

dx


x y#
x#

0.014 5

x r

dx


y r

dy Ê dr|Ð$ß%Ñ œ ˆ 53 ‰ a „ 0.01b
ˆ 54 ‰ a „ 0.01b

‚ 100¸ œ 0.28%; d) œ

3 ‰ dy Ê d)|Ð$ß%Ñ œ ˆ 254 ‰ a „ 0.01b
ˆ 25 a „ 0.01b œ

y ‹ x# # y ˆ ‰
1 x

Š

…0.04 25

dx





Š x" ‹ y ˆ ‰#
1 x

dy

„0.03 #5

Ê maximum change in d) occurs when dx and dy have opposite signs (dx œ 0.01 and dy œ 0.01 or vice „0.0028 " ˆ 4 ‰ ¸ d)) ‚ 100¸ œ ¸ 0.927255218 versa) Ê d) œ „#0.07 ‚ 100¸ 5 ¸ „ 0.0028; ) œ tan 3 ¸ 0.927255218 Ê

¸ 0.30% (b) the radius r is more sensitive to changes in y, and the angle ) is more sensitive to changes in x

54. (a) V œ 1r# h Ê dV œ 21rh dr
1r# dh Ê at r œ 1 and h œ 5 we have dV œ 101 dr
1 dh Ê the volume is about 10 times more sensitive to a change in r " (b) dV œ 0 Ê 0 œ 21rh dr
1r# dh œ 2h dr
r dh œ 10 dr
dh Ê dr œ  10 dh; choose dh œ 1.5 Ê dr œ 0.15 Ê h œ 6.5 in. and r œ 0.85 in. is one solution for ?V ¸ dV œ 0 55. f(aß bß cß d) œ º

a b œ ad  bc Ê fa œ d, fb œ c, fc œ b, fd œ a Ê df œ d da  c db  b dc
a dd; since c dº

kak is much greater than kbk , kck , and kdk , the function f is most sensitive to a change in d.

Section 14.7 Extreme Values and Saddle Points 56. ux œ ey , uy œ xey
sin z, uz œ y cos z Ê du œ ey dx
axey
sin zb dy
(y cos z) dz Ê duk ˆ2ßln 3ß 12 ‰ œ 3 dx
7 dy
0 dz œ 3 dx
7 dy Ê magnitude of the maximum possible error Ÿ 3(0.2)
7(0.6) œ 4.8 57. QK œ

" #

ˆ 2KM ‰"Î# ˆ 2M ‰ , QM œ h h

" #

ˆ 2KM ‰"Î# ˆ 2K ‰ h h , and Qh œ

" #

ˆ 2KM ‰"Î# ˆ 2KM ‰ h h#

" ˆ 2KM ‰"Î# ˆ 2M ‰ ‰"Î# ˆ 2K ‰ dM
"# ˆ 2KM ‰"Î# ˆ 2KM ‰ dh dK
"# ˆ 2KM # h h h h h h# 2K 2KM ˆ 2KM ‰"Î#  2M ‘ h h dK
h dM  h# dh Ê dQk Ð2ß20ß0Þ0.05Ñ "Î# (2)(2) (2)(2)(20) ’ (2)(2)(20) ’ (2)(20) 0.05 “ 0.05 dK
0.05 dM  (0.05)# dh“ œ (0.0125)(800 dK
80 dM

Ê dQ œ œ

" #

œ

" #

 32,000 dh)

Ê Q is most sensitive to changes in h ab sin C Ê Aa œ "# b sin C, Ab œ "# a sin C, Ac œ "# ab cos C Ê dA œ ˆ "# b sin C‰ da
ˆ "# a sin C‰ db
ˆ "# ab cos C‰ dC; dC œ k2°k œ k0.0349k radians, da œ k0.5k ft,

58. A œ

" #

db œ k0.5k ft; at a œ 150 ft, b œ 200 ft, and C œ 60°, we see that the change is approximately dA œ "# (200)(sin 60°) k0.5k
"# (150)(sin 60°) k0.5k
"# (200)(150)(cos 60°) k0.0349k œ „ 338 ft#

59. z œ f(xß y) Ê g(xß yß z) œ f(xß y)  z œ 0 Ê gx (xß yß z) œ fx (xß y), gy (xß yß z) œ fy (xß y) and gz (xß yß z) œ 1 Ê gx (x! ß y! ß f(x! ß y! )) œ fx (x! ß y! ), gy (x! ß y! ß f(x! ß y! )) œ fy (x! ß y! ) and gz (x! ß y! ß f(x! ß y! )) œ 1 Ê the tangent plane at the point P! is fx (x! ß y! )(x  x! )
fy (x! ß y! )(y  y! )  [z  f(x! ß y! )] œ 0 or z œ fx (x! ß y! )(x  x! )
fy (x! ß y! )(y  y! )
f(x! ß y! ) 60. ™ f œ 2xi
2yj œ 2(cos t
t sin t)i
2(sin t  t cos t)j and v œ (t cos t)i
(t sin t)j Ê u œ œ

(t cos t)i
(t sin t)j È(t cos t)#
(t sin t)#

v kv k

œ (cos t)i
(sin t)j since t  0 Ê (Du f)P! œ ™ f † u

œ 2(cos t
t sin t)(cos t)
2(sin t  t cos t)(sin t) œ 2 61. ™ f œ 2xi
2yj
2zk œ (2 cos t)i
(2 sin t)j
2tk and v œ ( sin t)i
(cos t)j
k Ê u œ œ

( sin t)i
(cos t)j
k È(sin t)#
(cos t)#
1#

t œ Š Èsin2 t ‹ i
Š cos È2 ‹ j


" È2

k Ê (Du f)P! œ ™ f † u

t " œ (2 cos t) Š Èsin2 t ‹
(2 sin t) Š cos È2 ‹
(2t) Š È2 ‹ œ

(Du f) ˆ 14 ‰ œ

" #

" "Î# i
#" t"Î# j # t # #

" "Î# i # t # #

63. r œ Èti
Ètj
(2t  1)k Ê v œ " #

Ê (Du f) ˆ 41 ‰ œ

1 2È 2

, (Du f)(0) œ 0 and

 4" k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1)

i
"# j  4" k ; f(xß yß z) œ x
y  z  3 œ 0 Ê ™ f œ 2xi
2yj  k

Ê ™ f(1ß 1ß 1) œ 2i
2j  k ; therefore v œ

v(1) œ

2t È2

1 2È 2

62. r œ Èti
Ètj  4" (t
3)k Ê v œ and v(1) œ

v kv k

" 4

( ™ f) Ê the curve is normal to the surface


"# t"Î# j
2k ; t œ 1 Ê x œ 1, y œ 1, z œ 1 Ê P! œ (1ß 1ß 1) and

i
"# j
2k ; f(xß yß z) œ x
y  z  1 œ 0 Ê ™ f œ 2xi
2yj  k Ê ™ f(1ß 1ß 1) œ 2i
2j  k ;

now va1b † ™ fa1ß 1ß 1b œ 0, thus the curve is tangent to the surface when t œ 1 14.7 EXTREME VALUES AND SADDLE POINTS 1. fx (xß y) œ 2x
y
3 œ 0 and fy (xß y) œ x
2y  3 œ 0 Ê x œ 3 and y œ 3 Ê critical point is (3ß 3); # œ 3  0 and fxx  0 Ê local minimum of fxx (3ß 3) œ 2, fyy (3ß 3) œ 2, fxy (3ß 3) œ 1 Ê fxx fyy  fxy f(3ß 3) œ 5

897

898

Chapter 14 Partial Derivatives

2. fx (xß y) œ 2x
3y  6 œ 0 and fy (xß y) œ 3x
6y
3 œ 0 Ê x œ 15 and y œ 8 Ê critical point is (15ß 8); # œ 3  0 and fxx  0 Ê local minimum of fxx (15ß 8) œ 2, fyy (15ß 8) œ 6, fxy (15ß 8) œ 3 Ê fxx fyy  fxy f(15ß 8) œ 63 3. fx (xß y) œ 2y  10x
4 œ 0 and fy (xß y) œ 2x  4y
4 œ 0 Ê x œ 23 and y œ 43 Ê critical point is ˆ 23 ß 43 ‰ ; # fxx ˆ 23 ß 43 ‰ œ 10, fyy ˆ 23 ß 43 ‰ œ 4, fxy ˆ 23 ß 43 ‰ œ 2 Ê fxx fyy  fxy œ 36  0 and fxx  0 Ê local maximum of f ˆ 23 ß 43 ‰ œ 0 4. fx (xß y) œ 2y  10x
4 œ 0 and fy (xß y) œ 2x  4y œ 0 Ê x œ 49 and y œ 29 Ê critical point is ˆ 49 ß 29 ‰ ; # fxx ˆ 49 ß 29 ‰ œ 10, fyy ˆ 49 ß 29 ‰ œ 4, fxy ˆ 49 ß 29 ‰ œ 2 Ê fxx fyy  fxy œ 36  0 and fxx  0 Ê local maximum of f ˆ 49 ß 29 ‰ œ  28 9 5. fx (xß y) œ 2x
y
3 œ 0 and fy (xß y) œ x
2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # œ 1  0 Ê saddle point fxx (2ß 1) œ 2, fyy (2ß 1) œ 0, fxy (2ß 1) œ 1 Ê fxx fyy  fxy 6. fx (xß y) œ y  2 œ 0 and fy (xß y) œ 2y
x  2 œ 0 Ê x œ 2 and y œ 2 Ê critical point is (2ß 2); # fxx (2ß 2) œ 0, fyy (2ß 2) œ 2, fxy (2ß 2) œ 1 Ê fxx fyy  fxy œ 1  0 Ê saddle point ˆ 6 69 ‰ 7. fx (xß y) œ 5y  14x
3 œ 0 and fy (xß y) œ 5x  6 œ 0 Ê x œ 65 and y œ 69 #5 Ê critical point is 5 ß 25 ; # ‰ ˆ 6 69 ‰ ˆ 6 69 ‰ fxx ˆ 65 ß 69 25 œ 14, fyy 5 ß 25 œ 0, fxy 5 ß 25 œ 5 Ê fxx fyy  fxy œ 25  0 Ê saddle point 8. fx (xß y) œ 2y  2x
3 œ 0 and fy (xß y) œ 2x  4y œ 0 Ê x œ 3 and y œ 3# Ê critical point is ˆ3ß 32 ‰ ; # fxx ˆ3ß 32 ‰ œ 2, fyy ˆ3ß 32 ‰ œ 4, fxy ˆ3ß 32 ‰ œ 2 Ê fxx fyy  fxy œ 4  0 and fxx  0 Ê local maximum of f ˆ3ß 3# ‰ œ

17 #

9. fx (xß y) œ 2x  4y œ 0 and fy (xß y) œ 4x
2y
6 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # fxx (2ß 1) œ 2, fyy (2ß 1) œ 2, fxy (2ß 1) œ 4 Ê fxx fyy  fxy œ 12  0 Ê saddle point 3 3‰ ˆ 13 10. fx (xß y) œ 6x
6y  2 œ 0 and fy (xß y) œ 6x
14y
4 œ 0 Ê x œ 13 1# and y œ  4 Ê critical point is 12 ß  4 ; 3‰ 3‰ 3‰ # ˆ 13 ˆ 13 fxx ˆ 13 1# ß  4 œ 6, fyy 1# ß  4 œ 14, fxy 1# ß  4 œ 6 Ê fxx fyy  fxy œ 48  0 and fxx  0 Ê local minimum of 3‰ 31 f ˆ 13 12 ß  4 œ  1#

11. fx (xß y) œ 4x
3y  5 œ 0 and fy (xß y) œ 3x
8y
2 œ 0 Ê x œ 2 and y œ 1 Ê critical point is (2ß 1); # œ 23  0 and fxx  0 Ê local minimum of fxx (2ß 1) œ 4, fyy (2ß 1) œ 8, fxy (2ß 1) œ 3 Ê fxx fyy  fxy f(2ß 1) œ 6 12. fx (xß y) œ 8x  6y  20 œ 0 and fy (xß y) œ 6x
10y
26 œ 0 Ê x œ 1 and y œ 2 Ê critical point is (1ß 2); # fxx (1ß 2) œ 8, fyy (1ß 2) œ 10, fxy (1ß 2) œ 6 Ê fxx fyy  fxy œ 44  0 and fxx  0 Ê local minimum of f(1ß 2) œ 36 13. fx (xß y) œ 2x  2 œ 0 and fy (xß y) œ 2y
4 œ 0 Ê x œ 1 and y œ 2 Ê critical point is (1ß 2); fxx (1ß 2) œ 2, # œ 4  0 Ê saddle point fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy  fxy 14. fx (xß y) œ 2x  2y  2 œ 0 and fy (xß y) œ 2x
4y
2 œ 0 Ê x œ 1 and y œ 0 Ê critical point is (1ß 0); # œ 4  0 and fxx  0 Ê local minimum of fxx (1ß 0) œ 2, fyy (1ß 0) œ 4, fxy (1ß 0) œ 2 Ê fxx fyy  fxy f(1ß 0) œ 0

Section 14.7 Extreme Values and Saddle Points

899

15. fx (xß y) œ 2x
2y œ 0 and fy (xß y) œ 2x œ 0 Ê x œ 0 and y œ 0 Ê critical point is (0ß 0); fxx (0ß 0) œ 2, # œ 4  0 Ê saddle point fyy (0ß 0) œ 0, fxy (0ß 0) œ 2 Ê fxx fyy  fxy 16. fx (xß y) œ 2  4x  2y œ 0 and fy (xß y) œ 2  2x  2y œ 0 Ê x œ 0 and y œ 1 Ê critical point is (0ß 1); # œ 4  0 and fxx  0 Ê local maximum of f(0ß 1) œ 4 fxx (0ß 1) œ 4, fyy (0ß 1) œ 2, fxy (0ß 1) œ 2 Ê fxx fyy  fxy 17. fx (xß y) œ 3x#  2y œ 0 and fy (xß y) œ 3y#  2x œ 0 Ê x œ 0 and y œ 0, or x œ  23 and y œ 23 Ê critical points are (0ß 0) and ˆ 23 ß 23 ‰ ; for (0ß 0): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 2 # Ê fxx fyy  fxy œ 4  0 Ê saddle point; for ˆ 23 ß 23 ‰ : fxx ˆ 23 ß 23 ‰ œ 4, fyy ˆ 23 ß 23 ‰ œ 4, fxy ˆ 23 ß 23 ‰ œ 2 # Ê fxx fyy  fxy œ 12  0 and fxx  0 Ê local maximum of f ˆ 23 ß 23 ‰ œ 170 27

18. fx (xß y) œ 3x#
3y œ 0 and fy (xß y) œ 3x
3y# œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical points # are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 6xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 6yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 3 Ê fxx fyy  fxy # œ 9  0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3 Ê fxx fyy  fxy

œ 27  0 and fxx  0 Ê local maximum of f(1ß 1) œ 1 19. fx (xß y) œ 12x  6x#
6y œ 0 and fy (xß y) œ 6y
6x œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 Ê critical # points are (0ß 0) and (1ß 1); for (!ß !): fxx (0ß 0) œ 12  12xk Ð0ß0Ñ œ 12, fyy (0ß 0) œ 6, fxy (0ß 0) œ 6 Ê fxx fyy  fxy œ 36  0 and fxx  0 Ê local minimum of f(0ß 0) œ 0; for (1ß 1): fxx (1ß 1) œ 0, fyy (1ß 1) œ 6, # fxy (1ß 1) œ 6 Ê fxx fyy  fxy œ 36  0 Ê saddle point 20. fx (xß y) œ 6x
6y œ 0 Ê x œ y; fy (xß y) œ 6y  6y#
6x œ 0 Ê 12y  6y# œ 0 Ê 6y(2  y) œ 0 Ê y œ 0 or y œ 2 Ê (0ß 0) and (2ß 2) are the critical points; fxx (xß y) œ 6, fyy (xß y) œ 6  12y, fxy (xß y) œ 6; for (0ß 0): # fxx (0ß 0) œ 6, fyy (0ß 0) œ 6, fxy (0ß 0) œ 6 Ê fxx fyy  fxy œ 72  0 Ê saddle point; for (2ß 2): fxx (2ß 2) œ 6, # fyy (2ß 2) œ 18, fxy (2ß 2) œ 6 Ê fxx fyy  fxy œ 72  0 and fxx  0 Ê local maximum of f(2ß 2) œ 8

21. fx (xß y) œ 27x#  4y œ 0 and fy (xß y) œ y#  4x œ 0 Ê x œ 0 and y œ 0, or x œ 49 and y œ 43 Ê critical points are # (0ß 0) and ˆ 49 ß 43 ‰ ; for (!ß !): fxx (0ß 0) œ 54xk Ð0ß0Ñ œ 0, fyy (0ß 0) œ 2yk Ð0ß0Ñ œ 0, fxy (0ß 0) œ 4 Ê fxx fyy  fxy # œ 16  0 Ê saddle point; for ˆ 49 ß 43 ‰ : fxx ˆ 49 ß 43 ‰ œ 24, fyy ˆ 49 ß 43 ‰ œ 83 , fxy ˆ 49 ß 43 ‰ œ 4 Ê fxx fyy  fxy œ 48  0 4 4‰ 64 ˆ and fxx  0 Ê local minimum of f 9 ß 3 œ  81 #

22. fx (xß y) œ 24x#
6y œ 0 Ê y œ 4x# ; fy (xß y) œ 3y#
6x œ 0 Ê 3 a4x# b
6x œ 0 Ê 16x%
2x œ 0 Ê 2x a8x$
1b œ 0 Ê x œ 0 or x œ  "# Ê (!ß 0) and ˆ "# ß 1‰ are the critical points; fxx (xß y) œ 48x,

# fyy (xß y) œ 6y, and fxy (xß y) œ 6; for (0ß 0): fxx (0ß 0) œ 0, fyy (0ß 0) œ 0, fxy (0ß 0) œ 6 Ê fxx fyy  fxy œ 36  0 1 1 1 1 Ê saddle point; for ˆ 2 ß 1‰ : fxx ˆ 2 ß 1‰ œ 24, fyy ˆ 2 ß 1‰ œ 6, fxy ˆ 2 ß 1‰ œ 6 # Ê fxx fyy  fxy œ 108  0 and fxx  0 Ê local maximum of f ˆ 12 ß 1‰ œ 1

23. fx (xß y) œ 3x#
6x œ 0 Ê x œ 0 or x œ 2; fy (xß y) œ 3y#  6y œ 0 Ê y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2); for (!ß !): fxx (0ß 0) œ 6x
6k Ð0ß0Ñ œ 6, fyy (0ß 0) œ 6y  6k Ð0ß0Ñ œ 6, # fxy (0ß 0) œ 0 Ê fxx fyy  fxy œ 36  0 Ê saddle point; for (0ß 2): fxx (0ß 2) œ 6, fyy (0ß 2) œ 6, fxy (0ß 2) œ 0 # Ê fxx fyy  fxy œ 36  0 and fxx  0 Ê local minimum of f(0ß 2) œ 12; for (2ß 0): fxx (2ß 0) œ 6, # fyy (2ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy  fxy œ 36  0 and fxx  0 Ê local maximum of f(2ß 0) œ 4; # for (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0 Ê fxx fyy  fxy œ 36  0 Ê saddle point

900

Chapter 14 Partial Derivatives

24. fx (xß y) œ 6x#  18x œ 0 Ê 6x(x  3) œ 0 Ê x œ 0 or x œ 3; fy (xß y) œ 6y#
6y  12 œ 0 Ê 6(y
2)(y  1) œ 0 Ê y œ 2 or y œ 1 Ê the critical points are (0ß 2), (0ß 1), (3ß 2), and (3ß 1); fxx (xß y) œ 12x  18, fyy (xß y) œ 12y
6, and fxy (xß y) œ 0; for (!ß 2): fxx (0ß 2) œ 18, fyy (0ß 2) œ 18, fxy (0ß 2) œ 0 # Ê fxx fyy  fxy œ 324  0 and fxx  0 Ê local maximum of f(0ß 2) œ 20; for (0ß 1): fxx (0ß 1) œ 18, # fyy (0ß 1) œ 18, fxy (0ß 1) œ 0 Ê fxx fyy  fxy œ 324  0 Ê saddle point; for (3ß 2): fxx (3ß 2) œ 18, # fyy (3ß 2) œ 18, fxy (3ß 2) œ 0 Ê fxx fyy  fxy œ 324  0 Ê saddle point; for (3ß 1): fxx (3ß 1) œ 18, # fyy (3ß 1) œ 18, fxy (3ß 1) œ 0 Ê fxx fyy  fxy œ 324  0 and fxx  0 Ê local minimum of f(3ß 1) œ 34

25. fx (xß y) œ 4y  4x$ œ 0 and fy (xß y) œ 4x  4y$ œ 0 Ê x œ y Ê x a1  x# b œ 0 Ê x œ 0, 1, 1 Ê the critical points are (0ß 0), (1ß 1), and (1ß 1); for (!ß !): fxx (0ß 0) œ 12x# k Ð0ß0Ñ œ 0, fyy (0ß 0) œ 12y# k Ð0ß0Ñ œ 0, # fxy (0ß 0) œ 4 Ê fxx fyy  fxy œ 16  0 Ê saddle point; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 # Ê fxx fyy  fxy œ 128  0 and fxx  0 Ê local maximum of f(1ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12, # fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy  fxy œ 128  0 and fxx  0 Ê local maximum of f(1ß 1) œ 2

26. fx (xß y) œ 4x$
4y œ 0 and fy (xß y) œ 4y$
4x œ 0 Ê x œ y Ê x$
x œ 0 Ê x a1  x# b œ 0 Ê x œ 0, 1, 1 Ê the critical points are (0ß 0), (1ß 1), and (1ß 1); fxx (xß y) œ 12x# , fyy (xß y) œ 12y# , and fxy (xß y) œ 4; # for (!ß 0): fxx (0ß 0) œ 0, fyy (0ß 0) œ 0, fxy (0ß 0) œ 4 Ê fxx fyy  fxy œ 16  0 Ê saddle point; for (1ß 1): # fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy  fxy œ 128  0 and fxx  0 Ê local minimum of # f("ß 1) œ 2; for (1ß 1): fxx (1ß 1) œ 12, fyy (1ß 1) œ 12, fxy (1ß 1) œ 4 Ê fxx fyy  fxy œ 128  0 and

fxx  0 Ê local minimum of f(1ß 1) œ 2 27. fx (xß y) œ #

œ 0 and fy (xß y) œ

#

#

2y a x #
y #  1 b#

#

œ 0 Ê x œ 0 and y œ 0 Ê the critical point is (!ß 0);

4x  2y
2 , fyy œ ax2x#

y#4y 1
b$2 , fxy œ ax#
8xy ; fxx (!ß !) œ 2, fyy (0ß 0) ax #
y #  1 b $ y #  1 b$ # fxx fyy  fxy œ 4  0 and fxx  0 Ê local maximum of f(0ß 0) œ 1

fxx œ Ê

2x ax #
y #  1 b #

28. fx (xß y) œ  x1#
y œ 0 and fy (xß y) œ x  fxx œ

2 x$

, fyy œ

2 y$

1 y#

œ 2, fxy (0ß 0) œ 0

œ 0 Ê x œ 1 and y œ 1 Ê the critical point is (1ß 1);

# , fxy œ 1; fxx (1ß 1) œ 2, fyy (1ß 1) œ 2, fxy (1ß 1) œ 1 Ê fxx fyy  fxy œ 3  0 and fxx  2 Ê local

minimum of f(1ß 1) œ 3 29. fx (xß y) œ y cos x œ 0 and fy (xß y) œ sin x œ 0 Ê x œ n1, n an integer, and y œ 0 Ê the critical points are (n1ß 0), n an integer (Note: cos x and sin x cannot both be 0 for the same x, so sin x must be 0 and y œ 0); fxx œ y sin x, fyy œ 0, fxy œ cos x; fxx (n1ß 0) œ 0, fyy (n1ß 0) œ 0, fxy (n1ß 0) œ 1 if n is even and fxy (n1ß 0) œ 1 # if n is odd Ê fxx fyy  fxy œ 1  0 Ê saddle point. 30. fx (xß y) œ 2e2x cos y œ 0 and fy (xß y) œ e2x sin y œ 0 Ê no solution since e2x Á 0 for any x and the functions cos y and sin y cannot equal 0 for the same y Ê no critical points Ê no extrema and no saddle points On OA, f(xß y) œ f(0ß y) œ y#  4y
1 on 0 Ÿ y Ÿ 2; f w (0ß y) œ 2y  4 œ 0 Ê y œ 2; f(0ß 0) œ 1 and f(!ß #) œ 3 (ii) On AB, f(xß y) œ f(xß 2) œ 2x#  4x  3 on 0 Ÿ x Ÿ 1; f w (xß 2) œ 4x  4 œ 0 Ê x œ 1; f(0ß 2) œ 3 and f(1ß #) œ 5 (iii) On OB, f(xß y) œ f(xß 2x) œ 6x#  12x
1 on 0 Ÿ x Ÿ 1; endpoint values have been found above; f w (xß 2x) œ 12x  12 œ 0 Ê x œ 1 and y œ 2, but ("ß #) is not an interior point of OB

31. (i)

Section 14.7 Extreme Values and Saddle Points

901

(iv) For interior points of the triangular region, fx (xß y) œ 4x  4 œ 0 and fy (xß y) œ 2y  4 œ 0 Ê x œ 1 and y œ 2, but (1ß 2) is not an interior point of the region. Therefore, the absolute maximum is 1 at (0ß 0) and the absolute minimum is 5 at ("ß #). On OA, D(xß y) œ D(0ß y) œ y#
1 on 0 Ÿ y Ÿ 4; Dw (0ß y) œ 2y œ 0 Ê y œ 0; D(!ß !) œ 1 and D(!ß %) œ 17 (ii) On AB, D(xß y) œ D(xß 4) œ x#  4x
17 on 0 Ÿ x Ÿ 4; Dw (xß 4) œ 2x  4 œ 0 Ê x œ 2 and (2ß 4) is an interior point of AB; D(#ß %) œ 13 and D(%ß %) œ D(!ß %) œ 17 (iii) On OB, D(xß y) œ D(xß x) œ x#
1 on 0 Ÿ x Ÿ 4; Dw (xß x) œ 2x œ 0 Ê x œ 0 and y œ 0, which is not an interior point of OB; endpoint values have been found above (iv) For interior points of the triangular region, fx (xß y) œ 2x  y œ 0 and fy (xß y) œ x
2y œ 0 Ê x œ 0 and y œ 0, which is not an interior point of the region. Therefore, the absolute maximum is 17 at (!ß %) and (%ß %), and the absolute minimum is 1 at (0ß 0).

32. (i)

On OA, f(xß y) œ f(!ß y) œ y# on 0 Ÿ y Ÿ 2; f w (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; f(0ß 0) œ 0 and f(0ß #) œ 4 (ii) On OB, f(xß y) œ f(xß 0) œ x# on 0 Ÿ x Ÿ 1; f w (xß 0) œ 2x œ 0 Ê x œ 0 and y œ 0; f(0ß 0) œ 0 and f(1ß 0) œ 1 (iii) On AB, f(xß y) œ f(xß 2x
2) œ 5x#  8x
4 on 0 Ÿ x Ÿ 1; f w (xß 2x
2) œ 10x  8 œ 0 Ê x œ 45 and y œ 25 ; f ˆ 45 ß 25 ‰ œ 45 ; endpoint values have been found above.

33. (i)

(iv) For interior points of the triangular region, fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0, but (!ß 0) is not an interior point of the region. Therefore the absolute maximum is 4 at (0ß 2) and the absolute minimum is 0 at (0ß 0). 34. (i)

(ii)

On AB, T(xß y) œ T(!ß y) œ y# on 3 Ÿ y Ÿ 3; Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0; T(0ß 0) œ 0, T(!ß 3) œ 9, and T(!ß 3) œ 9 On BC, T(xß y) œ T(xß 3) œ x#  3x
9 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x  3 œ 0 Ê x œ #3 and y œ 3; T ˆ 3# ß 3‰ œ 27 4 and T(5ß 3) œ 19

(iii) On CD, T(xß y) œ T(5ß y) œ y#
5y  5 on 3 Ÿ y Ÿ 3;Tw (5ß y) œ 2y
5 œ 0 Ê y œ  5# and x œ 5;T ˆ5ß  5# ‰ œ  45 4 , T(&ß 3) œ 11 and T(5ß 3) œ 19

(iv) On AD, T(xß y) œ T(xß 3) œ x#  9x
9 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x  9 œ 0 Ê x œ T ˆ 9# ß 3‰ œ  45 4 , T(!ß 3) œ 9 and T(&ß 3) œ 11

(v)

9 #

and y œ 3;

For interior points of the rectangular region, Tx (xß y) œ 2x
y  6 œ 0 and Ty (xß y) œ x
2y œ 0 Ê x œ 4 and y œ 2 Ê (4ß 2) is an interior critical point with T(4ß 2) œ 12. Therefore the absolute maximum is 19 at (5ß 3) and the absolute minimum is 12 at (4ß 2).

902 35. (i)

(ii)

Chapter 14 Partial Derivatives On OC, T(xß y) œ T(xß 0) œ x#  6x
2 on 0 Ÿ x Ÿ 5; Tw (xß 0) œ 2x  6 œ 0 Ê x œ 3 and y œ 0; T(3ß 0) œ 7, T(0ß 0) œ 2, and T(5ß 0) œ 3 On CB, T(xß y) œ T(5ß y) œ y#
5y  3 on 3 Ÿ y Ÿ 0; Tw (5ß y) œ 2y
5 œ 0 Ê y œ  5# and x œ 5; T ˆ5ß  5# ‰ œ  37 4 and T(5ß 3) œ 9

(iii) On AB, T(xß y) œ T(xß 3) œ x#  9x
11 on 0 Ÿ x Ÿ 5; Tw (xß 3) œ 2x  9 œ 0 Ê x œ 9# and y œ 3; T ˆ 9# ß 3‰ œ  37 4 and T(!ß 3) œ 11

(iv) On AO, T(xß y) œ T(!ß y) œ y#
2 on 3 Ÿ y Ÿ 0; Tw (0ß y) œ 2y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is not an interior point of AO (v) For interior points of the rectangular region, Tx (xß y) œ 2x
y  6 œ 0 and Ty (xß y) œ x
2y œ 0 Ê x œ 4 and y œ 2, an interior critical point with T(%ß 2) œ 10. Therefore the absolute maximum is 11 at (!ß 3) and the absolute minimum is 10 at (4ß 2). 36. (i)

(ii)

On OA, f(xß y) œ f(!ß y) œ 24y# on 0 Ÿ y Ÿ 1; f w (0ß y) œ 48y œ 0 Ê y œ 0 and x œ 0, but (0ß 0) is not an interior point of OA; f(!ß 0) œ 0 and f(!ß 1) œ 24 On AB, f(xß y) œ f(xß 1) œ 48x  32x$  24 on 0 Ÿ x Ÿ 1; f w (xß 1) œ 48  96x# œ 0 Ê x œ È"2 and y œ 1, or x œ  È"2 and y œ 1, but Š È"2 ß 1‹ is not in the interior of AB; f Š È"2 ß 1‹ œ 16È2  24 and f(1ß 1) œ 8

(iii) On BC, f(xß y) œ f("ß y) œ 48y  32  24y# on 0 Ÿ y Ÿ 1; f w ("ß y) œ 48  48y œ 0 Ê y œ 1 and x œ 1, but ("ß ") is not an interior point of BC; f("ß 0) œ 32 and f("ß ") œ 8 (iv) On OC, f(xß y) œ f(xß 0) œ 32x$ on 0 Ÿ x Ÿ 1; f w (xß 0) œ 96x# œ 0 Ê x œ 0 and y œ 0, but (0ß 0) is not an interior point of OC; f(!ß 0) œ 0 and f("ß 0) œ 32 (v) For interior points of the rectangular region, fx (xß y) œ 48y  96x# œ 0 and fy (xß y) œ 48x  48y œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ "# , but (0ß 0) is not an interior point of the region; f ˆ "# ß "# ‰ œ 2. Therefore the absolute maximum is 2 at ˆ "# ß "# ‰ and the absolute minimum is 32 at (1ß 0). 37. (i)

On AB, f(xß y) œ f(1ß y) œ 3 cos y on  14 Ÿ y Ÿ w

1 4

;

1 4

;

f (1ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 1; f("ß 0) œ 3, f ˆ1ß  14 ‰ œ

(ii)

3È 2 #

, and f ˆ1ß 14 ‰ œ

3È 2 #

On CD, f(xß y) œ f($ß y) œ 3 cos y on  14 Ÿ y Ÿ f w (3ß y) œ 3 sin y œ 0 Ê y œ 0 and x œ 3;

È 3È 2 ˆ 1‰ 3 2 # and f 3ß 4 œ # È2 1‰ # 4 œ # a4x  x b on

f(3ß 0) œ 3, f ˆ3ß  14 ‰ œ (iii) On BC, f(xß y) œ f ˆxß

1 Ÿ x Ÿ 3; f w ˆxß 14 ‰ œ È2(2  x) œ 0 Ê x œ 2 and y œ f ˆ3ß 14 ‰ œ

; f ˆ2ß 14 ‰ œ 2È2, f ˆ1ß 14 ‰ œ

3È 2 #

, and

3È 2 # È2 # w # a4x  x b on 1 Ÿ x Ÿ 3; f È È œ 3 # 2 , and f ˆ3ß  14 ‰ œ 3 # 2

(iv) On AD, f(xß y) œ f ˆxß  14 ‰ œ f ˆ2ß  14 ‰ œ 2È2, f ˆ1ß  14 ‰

(v)

1 4

ˆxß  14 ‰ œ È2(2  x) œ 0 Ê x œ 2 and y œ  14 ;

For interior points of the region, fx (xß y) œ (4  2x) cos y œ 0 and fy (xß y) œ  a4x  x# b sin y œ 0 Ê x œ 2 and y œ 0, which is an interior critical point with f(2ß 0) œ 4. Therefore the absolute maximum is 4 at

Section 14.7 Extreme Values and Saddle Points (2ß 0) and the absolute minimum is

3È 2 #

903

at ˆ3ß  14 ‰ , ˆ3ß 14 ‰ , ˆ1ß  14 ‰ , and ˆ1ß 14 ‰ .

On OA, f(xß y) œ f(!ß y) œ 2y
1 on 0 Ÿ y Ÿ 1; f w (0ß y) œ 2 Ê no interior critical points; f(0ß 0) œ 1 and f(0ß 1) œ 3 (ii) On OB, f(xß y) œ f(xß 0) œ 4x
1 on 0 Ÿ x Ÿ 1; f w (xß 0) œ 4 Ê no interior critical points; f(1ß 0) œ 5 (iii) On AB, f(xß y) œ f(xß x
1) œ 8x#  6x
3 on 0 Ÿ x Ÿ 1; f w (xß x
1) œ 16x  6 œ 0 Ê x œ 38 and y œ 58 ; f ˆ 38 ß 58 ‰ œ 15 8 , f(0ß 1) œ 3, and f("ß 0) œ 5

38. (i)

(iv) For interior points of the triangular region, fx (xß y) œ 4  8y œ 0 and fy (xß y) œ 8x
2 œ 0 Ê y œ "# and x œ 4" which is an interior critical point with f ˆ 4" ß #" ‰ œ 2. Therefore the absolute maximum is 5 at (1ß 0) and the absolute minimum is 1 at (!ß 0).

39. Let F(aß b) œ 'a a6  x  x# b dx where a Ÿ b. The boundary of the domain of F is the line a œ b in the b

ab-plane, and F(aß a) œ 0, so F is identically 0 on the boundary of its domain. For interior critical points we have: ``Fa œ  a6  a  a# b œ 0 Ê a œ 3, 2 and `` Fb œ a6  b  b# b œ 0 Ê b œ 3, 2. Since a Ÿ b, there is only

one interior critical point (3ß 2) and F(3ß 2) œ 'c3 a6  x  x# b dx gives the area under the parabola 2

y œ 6  x  x# that is above the x-axis. Therefore, a œ 3 and b œ 2.

40. Let F(aß b) œ 'a a24  2x  x# b b

"Î$

dx where a Ÿ b. The boundary of the domain of F is the line a œ b and

on this line F is identically 0. For interior critical points we have: and

`F `b

# "Î$

œ a24  2b  b b

`F `a

œ  a24  2a  a# b

"Î$

œ 0 Ê a œ 4, 6

œ 0 Ê b œ 4, 6. Since a Ÿ b, there is only one critical point (6ß 4) and

F(6ß 4) œ 'c6 a24  2x  x# b dx gives the area under the curve y œ a24  2x  x# b 4

"Î$

that is above the x-axis.

Therefore, a œ 6 and b œ 4.

41. Tx (xß y) œ 2x  1 œ 0 and Ty (xß y) œ 4y œ 0 Ê x œ

" #

and y œ 0 with T ˆ "# ß 0‰ œ  4" ; on the boundary

x#
y# œ 1: T(xß y) œ x#  x
2 for 1 Ÿ x Ÿ 1 Ê Tw (xß y) œ 2x  1 œ 0 Ê x œ  "# and y œ „ T Š

" È3 #ß # ‹

Š "# ß 

œ

È3 # ‹;

9 4

, T Š

œ 2  ln

" #

œ

9 4

" 4

, T(1ß 0) œ 2, and T("ß 0) œ 0 Ê the hottest is 2 ° at Š

" È3 #ß # ‹

;

and

the coldest is  "4 ° at ˆ "# ß 0‰ .

42. fx (xß y) œ y
2  fyy ˆ "# ß 2‰ œ

È3 " #ß  # ‹

È3 #

2 x

" y # ¹ ˆ 1 ß2 ‰

œ 0 and fy (xß y) œ x  œ

2

" 4

" y

œ0 Ê xœ

" #

and y œ 2; fxx ˆ #" ß 2‰ œ

2¸ x# ˆ 12 ß2‰

œ 8,

# , fxy ˆ #" ß 2‰ œ 1 Ê fxx fyy  fxy œ 1  0 and fxx  0 Ê a local minimum of f ˆ "# ß 2‰

œ 2
ln 2

43. (a) fx (xß y) œ 2x  4y œ 0 and fy (xß y) œ 2y  4x œ 0 Ê x œ 0 and y œ 0; fxx (0ß 0) œ 2, fyy (0ß 0) œ 2, # fxy (0ß 0) œ 4 Ê fxx fyy  fxy œ 12  0 Ê saddle point at (0ß 0) (b) fx (xß y) œ 2x  2 œ 0 and fy (xß y) œ 2y  4 œ 0 Ê x œ 1 and y œ 2; fxx (1ß 2) œ 2, fyy (1ß 2) œ 2, # œ 4  0 and fxx  0 Ê local minimum at ("ß #) fxy (1ß 2) œ 0 Ê fxx fyy  fxy (c) fx (xß y) œ 9x#  9 œ 0 and fy (xß y) œ 2y
4 œ 0 Ê x œ „ 1 and y œ 2; fxx (1ß 2) œ 18xk Ð1ß2Ñ œ 18, # œ 36  0 and fxx  0 Ê local minimum at ("ß #); fyy (1ß 2) œ 2, fxy (1ß 2) œ 0 Ê fxx fyy  fxy # fxx (1ß 2) œ 18, fyy ("ß 2) œ 2, fxy ("ß 2) œ 0 Ê fxx fyy  fxy œ 36  0 Ê saddle point at ("ß 2)

904

Chapter 14 Partial Derivatives

44. (a) (b) (c) (d) (e) (f)

Minimum at (0ß 0) since f(xß y)  0 for all other (xß y) Maximum of 1 at (!ß !) since f(xß y)  1 for all other (xß y) Neither since f(xß y)  0 for x  0 and f(xß y)  0 for x  0 Neither since f(xß y)  0 for x  0 and f(xß y)  0 for x  0 Neither since f(xß y)  0 for x  0 and y  0, but f(xß y)  0 for x  0 and y  0 Minimum at (0ß 0) since f(xß y)  0 for all other (xß y)

45. If k œ 0, then f(xß y) œ x#
y# Ê fx (xß y) œ 2x œ 0 and fy (xß y) œ 2y œ 0 Ê x œ 0 and y œ 0 Ê (!ß 0) is the only critical point. If k Á 0, fx (xß y) œ 2x
ky œ 0 Ê y œ  2k x; fy (xß y) œ kx
2y œ 0 Ê kx
2 ˆ 2k x‰ œ 0 4‰ ˆ ˆ 2‰ Ê kx  4x k œ 0 Ê k  k x œ 0 Ê x œ 0 or k œ „ 2 Ê y œ  k (0) œ 0 or y œ „ x; in any case (0ß 0) is a critical point. # œ 4  k# ; f will have a 46. (See Exercise 45 above): fxx (xß y) œ 2, fyy (xß y) œ 2, and fxy (xß y) œ k Ê fxx fyy  fxy

saddle point at (0ß 0) if 4  k#  0 Ê k  2 or k  2; f will have a local minimum at (0ß 0) if 4  k#  0 Ê 2  k  2; the test is inconclusive if 4  k# œ 0 Ê k œ „ 2. 47. No; for example f(xß y) œ xy has a saddle point at (aß b) œ (0ß 0) where fx œ fy œ 0. # 48. If fxx (aß b) and fyy (aß b) differ in sign, then fxx (aß b) fyy (aß b)  0 so fxx fyy  fxy  0. The surface must therefore have a

saddle point at (aß b) by the second derivative test. 49. We want the point on z œ 10  x#  y# where the tangent plane is parallel to the plane x
2y
3z œ 0. To find a normal vector to z œ 10  x#  y# let w œ z
x#
y#  10. Then ™ w œ 2xi
2yj
k is normal to z œ 10  x#  y# at (xß y). The vector ™ w is parallel to i
2j
3k which is normal to the plane x
2y
3z " ‰ œ 0 if 6xi
6yj
3k œ i
2j
3k or x œ "6 and y œ 3" . Thus the point is ˆ 6" ß 3" ß 10  36  9" ‰ or ˆ 6" ß 3" ß 355 36 . 50. We want the point on z œ x#
y#
10 where the tangent plane is parallel to the plane x
2y  z œ 0. Let w œ z  x#  y#  10, then ™ w œ 2xi  2yj
k is normal to z œ x#
y#
10 at (xß y). The vector ™ w is parallel to i
2j  k which is normal to the plane if x œ "# and y œ 1. Thus the point ˆ "# ß 1ß 4"
1
10‰ # # ‰ or ˆ "# ß 1ß 45 4 is the point on the surface z œ x
y
10 nearest the plane x
2y  z œ 0. 51. No, because the domain x   0 and y   0 is unbounded since x and y can be as large as we please. Absolute extrema are guaranteed for continuous functions defined over closed and bounded domains in the plane. Since the domain is unbounded, the continuous function f(xß y) œ x
y need not have an absolute maximum (although, in this case, it does have an absolute minimum value of f(0ß 0) œ 0). 52. (a) (i) On x œ 0, f(xß y) œ f(0ß y) œ y#  y
1 for 0 Ÿ y Ÿ 1; f w (0ß y) œ 2y  1 œ 0 Ê y œ f ˆ0ß "# ‰ œ 34 , f(0ß 0) œ 1, and f(0ß 1) œ 1 (ii)

" #

and x œ 0;

On y œ 1, f(xß y) œ f(xß 1) œ x#
x
1 for 0 Ÿ x Ÿ 1; f w (xß 1) œ 2x
1 œ 0 Ê x œ  "# and y œ 1, but ˆ "# ß 1‰ is outside the domain; f(0ß 1) œ 1 and f("ß ") œ 3

(iii) On x œ 1, f(xß y) œ f("ß y) œ y#
y
1 for 0 Ÿ y Ÿ 1; f w (1ß y) œ 2y
1 œ 0 Ê y œ  "# and x œ 1, but ˆ1ß  "# ‰ is outside the domain; f(1ß 0) œ 1 and f("ß ") œ 3 (iv) On y œ 0, f(xß y) œ f(xß 0) œ x#  x
1 for 0 Ÿ x Ÿ 1; f w (xß 0) œ 2x  1 œ 0 Ê x œ f ˆ "# ß 0‰ œ 34 ; f(0ß 0) œ 1, and f("ß 0) œ 1 (v)

" #

and y œ 0;

On the interior of the square, fx (xß y) œ 2x
2y  1 œ 0 and fy (xß y) œ 2y
2x  1 œ 0 Ê 2x
2y œ 1 Ê (x
y) œ "# . Then f(xß y) œ x#
y#
2xy  x  y
1 œ (x
y)#  (x
y)
1 œ 34 is the absolute minimum value when 2x
2y œ 1.

Section 14.7 Extreme Values and Saddle Points (b) The absolute maximum is f("ß ") œ 3. 53. (a)

df dt

œ

` f dx ` x dt




` f dy ` y dt

œ




dx dt

dy dt

œ 2 sin t
2 cos t œ 0 Ê cos t œ sin t Ê x œ y

On the semicircle x#
y# œ 4, y   0, we have t œ

(i)

1 4

and x œ y œ È2 Ê f ŠÈ2ß È2‹ œ 2È2. At the

endpoints, f(2ß 0) œ 2 and f(#ß !) œ 2. Therefore the absolute minimum is f(2ß 0) œ 2 when t œ 1; the absolute maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ 1 . 4

On the quartercircle x#
y# œ 4, x   0 and y   0, the endpoints give f(!ß 2) œ 2 and f(#ß 0) œ 2. Therefore the absolute minimum is f(2ß 0) œ 2 and f(!ß 2) œ 2 when t œ 0, 1# respectively; the absolute

(ii)

maximum is f ŠÈ2ß È2‹ œ 2È2 when t œ (b)

dg dt

œ

` g dx ` x dt




` g dy ` y dt

œy

dx dt


x

dy dt

1 4

.

œ 4 sin# t
4 cos# t œ 0 Ê cos t œ „ sin t Ê x œ „ y.

tœ

31 4

and x œ È2, y œ È2 at

. Then g ŠÈ2ß È2‹ œ 2 and g ŠÈ2ß È2‹ œ 2. At the endpoints, g(2ß 0) œ g(#ß 0) œ 0.

Therefore the absolute minimum is g ŠÈ2ß È2‹ œ 2 when t œ g ŠÈ2 ß È2‹ œ 2 when t œ

1 4

31 4

; the absolute maximum is

.

On the quartercircle x#
y# œ 4, x   0 and y   0, the endpoints give g(!ß 2) œ 0 and g(#ß 0) œ 0. Therefore the absolute minimum is g(2ß 0) œ 0 and g(!ß 2) œ 0 when t œ 0, 1# respectively; the absolute

(ii)

maximum is g ŠÈ2ß È2‹ œ 2 when t œ (c)

1 4

On the semicircle x#
y# œ 4, y   0, we obtain x œ y œ È2 at t œ

(i)

dh dt

œ

` h dx ` x dt




` h dy ` y dt

1 4

.

dy œ 4x dx dt
2y dt œ (8 cos t)(2 sin t)
(4 sin t)(2 cos t) œ 8 cos t sin t œ 0

Ê t œ 0, 1# , 1 yielding the points (2ß 0), (0ß 2) for 0 Ÿ t Ÿ 1.

On the semicircle x#
y# œ 4, y   0 we have h(2ß 0) œ 8, h(0ß 2) œ 4, and h(2ß 0) œ 8. Therefore, the absolute minimum is h(!ß 2) œ 4 when t œ 1# ; the absolute maximum is h(2ß 0) œ 8 and h(2ß 0) œ 8

(i)

when t œ 0, 1 respectively. On the quartercircle x#
y# œ 4, x   0 and y   0 the absolute minimum is h(0ß 2) œ 4 when t œ

(ii)

absolute maximum is h(2ß 0) œ 8 when t œ 0. 54. (a)

df dt

(i)

œ

` f dx ` x dt




` f dy ` y dt

1 4

x# 9




y# 4

œ 1, y   0, f(xß y) œ 2x
3y œ 6 cos t
6 sin t œ 1 4

.

On the quarter ellipse, at the endpoints f(0ß 2) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6 È and f(0ß 2) œ 6 when t œ 0, 1 respectively; the absolute maximum is f Š 3 2 ß È2‹ œ 6È2 when t œ 1 .

(ii)

#

(i)

` g dy dx œ `` xg dx dt
` y dt œ y dt Ê t œ 14 , 341 for 0 Ÿ t Ÿ

dg dt


x

dy dt

#

4

#

#

œ (2 sin t)(3 sin t)
(3 cos t)(2 cos t) œ 6 acos t  sin tb œ 6 cos 2t œ 0

1. È

On the semi-ellipse, g(xß y) œ xy œ 6 sin t cos t. Then g Š 3 # 2 ß È2‹ œ 3 when t œ

1 4

, and

È

31 4

. At the endpoints, g(3ß 0) œ g($ß 0) œ 0. The absolute minimum is

È

31 4

; the absolute maximum is g Š 3 # 2 ß È2‹ œ 3 when t œ

g Š 3 # 2 ß È2‹ œ 3 when t œ g Š 3 # 2 ß È2‹ œ 3 when t œ (ii)

œ 6È2

. At the endpoints, f(3ß 0) œ 6 and f(3ß 0) œ 6. The absolute minimum is f(3ß 0) œ 6 when

È t œ 1; the absolute maximum is f Š 3 # 2 ß È2‹ œ 6È2 when t œ

(b)

1 4 for 0 Ÿ t Ÿ 1. È È 6 Š #2 ‹
6 Š #2 ‹

; the

dy œ 2 dx dt
3 dt œ 6 sin t
6 cos t œ 0 Ê sin t œ cos t Ê t œ

On the semi-ellipse, at t œ

1 #

È

1 4

.

On the quarter ellipse, at the endpoints g(!ß 2) œ 0 and g($ß 0) œ 0. The absolute minimum is g(3ß 0) œ 0 È and g(0ß 2) œ 0 at t œ 0, 1 respectively; the absolute maximum is g Š 3 2 ß È2‹ œ 3 when t œ 1 . #

#

4

905

906

Chapter 14 Partial Derivatives dh dt

(c)

œ

` h dx ` x dt

Ê t œ 0,

dy œ 2x dx dt
6y dt œ (6 cos t)(3 sin t)
(12 sin t)(2 cos t) œ 6 sin t cos t œ 0

, 1 for 0 Ÿ t Ÿ 1, yielding the points (3ß 0), (0ß 2), and (3ß 0).

On the quarter ellipse, the absolute minimum is h(3ß 0) œ 9 when t œ 0; the absolute maximum is h(!ß 2) œ 12 when t œ 1# .

(ii)

df dt

1 #

` h dy ` y dt

On the semi-ellipse, y   0 so that h(3ß 0) œ 9, h(0ß 2) œ 12, and h(3ß 0) œ 9. The absolute minimum is h(3ß 0) œ 9 and h(3ß 0) œ 9 when t œ 0, 1 respectively; the absolute maximum is h(!ß 2) œ 12 when t œ 1# .

(i)

55.




œ

` f dx ` x dt




` f dy ` y dt

dy œ y dx dt
x dt

" x œ 2t and y œ t
1 Ê df dt œ (t
1)(2)
(2t)(1) œ 4t
2 œ 0 Ê t œ  # Ê x œ 1 and y œ f ˆ1ß "# ‰ œ  "# . The absolute minimum is f ˆ1ß "# ‰ œ  "# when t œ  "# ; there is no absolute

(i)

" #

with

maximum. For the endpoints: t œ 1 Ê x œ 2 and y œ 0 with f(2ß 0) œ 0; t œ 0 Ê x œ 0 and y œ 1 with f(!ß 1) œ 0. The absolute minimum is f ˆ1ß "# ‰ œ  "# when t œ  "# ; the absolute maximum is f(0ß 1) œ 0

(ii)

and f(#ß 0) œ 0 when t œ 1, 0 respectively. (iii) There are no interior critical points. For the endpoints: t œ 0 Ê x œ 0 and y œ 1 with f(0ß 1) œ 0; t œ 1 Ê x œ 2 and y œ 2 with f(2ß 2) œ 4. The absolute minimum is f(0ß 1) œ 0 when t œ 0; the absolute maximum is f(2ß 2) œ 4 when t œ 1. 56. (a)

df dt

œ

` f dx ` x dt




` f dy ` y dt

dy œ 2x dx dt
2y dt

4 4 x œ t and y œ 2  2t Ê df dt œ (2t)(1)
2(2  2t)(2) œ 10t  8 œ 0 Ê t œ 5 Ê x œ 5 and y œ 4 f ˆ 45 ß 25 ‰ œ "#65
25 œ 45 . The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 when t œ 45 ; there is no absolute

(i)

2 5

with

maximum along the line. For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with f(0ß 2) œ 4; t œ 1 Ê x œ 1 and y œ 0 with f(1ß 0) œ 1. The absolute minimum is f ˆ 45 ß 25 ‰ œ 45 at the interior critical point when t œ 45 ; the absolute maximum is

(ii)

f(0ß 2) œ 4 at the endpoint when t œ 0. (b) (i)

dg dt

œ

` g dx ` x dt




` g dy ` y dt

œ ’ ax#
2xy# b# “

dx dt


’ ax#
2yy# b# “

x œ t and y œ 2  2t Ê x#
y# œ 5t#  8t
4 Ê #

œ  a5t#  8t
4b (10t
8) œ 0 Ê t œ maximum is (ii)

g ˆ 45 ß 25 ‰

œ

5 4

when t œ

bœ Ê

58. m œ bœ Ê

4 5

4 5

dg dt

#

œ  a5t#  8t
4b [(2t)(1)
(2)(2  2t)(2)]

Ê xœ

4 5

and y œ

(2)(1)  3("4) œ  20 (2)#  3(10) 13 and "  20 9 ˆ ‰ ‘ 3 1   13 (2) œ 13 9 ¸ y œ  20 13 x
13 ; y xœ4 œ

(0)(5)  3(6) 3 (0)#  3(8) œ 4 and "  3 ‘ 5 3 5  4 (0) œ 3 3 5 y œ 4 x
3 ; y¸ xœ4

œ

14 3

 71 13

" 4

2 5

with g ˆ 45 ß 25 ‰ œ

" ˆ 45 ‰

œ

5 4

. The absolute

; there is no absolute minimum along the line since x and y can be

as large as we please. For the endpoints: t œ 0 Ê x œ 0 and y œ 2 with g(0ß 2) œ The absolute minimum is g(0ß 2) œ

57. m œ

dy dt

" 4

; t œ 1 Ê x œ 1 and y œ 0 with g(1ß 0) œ 1. when t œ 0; the absolute maximum is g ˆ 45 ß 52 ‰ œ 45 when t œ 54 . k 1 2 3 D

xk 1 0 3 2

yk 2 1 4 1

x#k 1 0 9 10

xk yk 2 0 12 14

k 1 2 3 D

xk 2 0 2 0

yk 0 2 3 5

x#k 4 0 4 8

xk yk 0 0 6 6

Section 14.7 Extreme Values and Saddle Points 59. m œ bœ Ê

60. m œ bœ Ê

61. m œ bœ

(3)(5)  3(8) 3 (3)#  3(5) œ 2 and "  3 ‘ 1 3 5  2 (3) œ 6 y œ 32 x
16 ; y¸ xœ4

œ

(5)(5)  3(10) 5 (5)#  3(13) œ 14 and "  5 ‘ 15 3 5  14 (5) œ 14 5 15 y œ 14 x
14 ; y¸ xœ4

37 6

œ

35 14

œ

5 #

(162)(41.32)  6(1192.8) ¸ 0.122 and (162)#  6(5004) " 6 c41.32  (0.122)(162)d ¸ 3.59

Ê y œ 0.122x
3.59

(0.001863)(91)4(0.065852) (0.001863)# 4(0.000001323) ¸ 51,545 b œ "4 (91  51,545(0.001863)) ¸ F œ 51,545 D"#  1.26

62. m œ and Ê

1.26

k 1 2 3 4 D

ˆ D"# ‰

k 1 2 3 D

xk 0 1 2 3

yk 0 2 3 5

x#k 0 1 4 5

xk yk 0 2 6 8

k 1 2 3 D

xk 0 2 3 5

yk 1 2 2 5

x#k 0 4 9 13

xk yk 0 4 6 10

k 1 2 3 4 5 6 D

xk

k

0.001 0.0005 0.00024 0.000123 0.001863

12 18 24 30 36 42 162

Fk 51 22 14 4 91

x#k 144 324 576 900 1296 1764 5004

yk 5.27 5.68 6.25 7.21 8.20 8.71 41.32

ˆ D"# ‰# k

0.000001 0.00000025 0.0000000576 0.0000000153 0.000001323

xk yk 63.24 102.24 150 216.3 295.2 365.82 1192.8

ˆ D"# ‰ Fk k 0.051 0.011 0.00336 0.000492 0.065852

907

908

Chapter 14 Partial Derivatives (3201)(17,785)  10(5,710,292) (3201)#  10(1,430,389) " 0.0427 and b œ 10 [17,785

63. (b) m œ ¸

 (0.0427)(3201)]

¸ 1764.8 Ê y œ 0.0427K
1764.8

yk 1761 1771 1772 1775 1777 1780 1783 1786 1789 1791 17,785

K#

k 1 2 3 4 5 6 7 8 9 10 D

Kk 1 75 155 219 271 351 425 503 575 626 3201

k 1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 D

xk

yk

xk#

3 2 4 2 5 5 9 12 8 13 14 3 4 13 10 16 123

3 2 6 3 4 3 11 9 10 16 13 5 6 19 15 15 140

9 4 16 4 25 25 81 144 64 169 196 9 16 169 100 256 1287

1 5625 24,025 47,961 73,441 123,201 180,625 253,009 330,625 391,876 1,430,389

Kk yk 1761 132,825 274,660 388,725 481,567 624,780 757,775 898,358 1,028,675 1,121,166 5,710,292

(c) K œ 364 Ê y œ (0.0427)(364) Ê y œ (0.0427)(364)
1764.8 ¸ 1780 64. m œ bœ

(123)(140)  16(1431) (123)#  16(1287) ¸ 1.04 " 16 [140  (1.04)(123)] ¸

Ê y œ 1.04x
0.755

and 0.755

xk yk 9 4 24 6 20 15 99 108 80 208 182 15 24 247 150 240 1431

65-70. Example CAS commands: Maple: f := (x,y) -> x^2+y^3-3*x*y; x0,x1 := -5,5; y0,y1 := -5,5; plot3d( f(x,y), x=x0..x1, y=y0..y1, axes=boxed, shading=zhue, title="#65(a) (Section 14.7)" ); plot3d( f(x,y), x=x0..x1, y=y0..y1, grid=[40,40], axes=boxed, shading=zhue, style=patchcontour, title="#65(b) (Section 14.7)" ); fx := D[1](f); # (c) fy := D[2](f); crit_pts := solve( {fx(x,y)=0,fy(x,y)=0}, {x,y} ); fxx := D[1](fx); # (d) fxy := D[2](fx); fyy := D[2](fy); discr := unapply( fxx(x,y)*fyy(x,y)-fxy(x,y)^2, (x,y) ); for CP in {crit_pts} do # (e) eval( [x,y,fxx(x,y),discr(x,y)], CP );

Section 14.8 Lagrange Multipliers

909

end do; # (0,0) is a saddle point # ( 9/4, 3/2) is a local minimum Mathematica: (assigned functions and bounds will vary) Clear[x,y,f] f[x_,y_]:= x2
y3  3x y xmin= 5; xmax= 5; ymin= 5; ymax= 5; Plot3D[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, AxesLabel Ä {x, y, z}] ContourPlot[f[x,y], {x, xmin, xmax}, {y, ymin, ymax}, ContourShading Ä False, Contours Ä 40] fx= D[f[x,y], x]; fy= D[f[x,y], y]; critical=Solve[{fx==0, fy==0},{x, y}] fxx= D[fx, x]; fxy= D[fx, y]; fyy= D[fy, y]; discriminant= fxx fyy  fxy2 {{x, y}, f[x, y], discriminant, fxx} /.critical 14.8 LAGRANGE MULTIPLIERS 1.

™ f œ yi
xj and ™ g œ 2xi
4yj so that ™ f œ - ™ g Ê yi
xj œ -(2xi
4yj) Ê y œ 2x- and x œ 4yÊ x œ 8x-# Ê - œ „

È2 4

or x œ 0.

CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the ellipse so x Á 0. CASE 2: x Á 0 Ê - œ „

È2 4

Therefore f takes on its extreme values at Š „ are „ 2.

È2 #

#

Ê x œ „ È2y Ê Š „ È2y‹
2y# œ 1 Ê y œ „ "# . È2 " 2 ß #‹

and Š „

È2 " 2 ß  #‹ .

The extreme values of f on the ellipse

.

™ f œ yi
xj and ™ g œ 2xi
2yj so that ™ f œ - ™ g Ê yi
xj œ -(2xi
2yj) Ê y œ 2x- and x œ 2yÊ x œ 4x-# Ê x œ 0 or - œ „ 12 .

CASE 1: If x œ 0, then y œ 0. But (0ß 0) is not on the circle x#
y#  10 œ 0 so x Á 0. CASE 2: x Á 0 Ê - œ „ 12 Ê y œ 2x ˆ „ "# ‰ œ „ x Ê x#
a „ xb#  10 œ 0 Ê x œ „ È5 Ê y œ „ È5. Therefore f takes on its extreme values at Š „ È5ß È5‹ and Š „ È5ß È5‹ . The extreme values of f on the circle are 5 and 5. 3.

™ f œ 2xi  2yj and ™ g œ i
3j so that ™ f œ - ™ g Ê 2xi  2yj œ -(i
3j) Ê x œ  -# and y œ  3#Ê ˆ -# ‰
3 ˆ 3#- ‰ œ 10 Ê - œ 2 Ê x œ 1 and y œ 3 Ê f takes on its extreme value at (1ß 3) on the line.

The extreme value is f("ß $) œ 49  1  9 œ 39. 4.

™ f œ 2xyi
x# j and ™ g œ i
j so that ™ f œ - ™ g Ê 2xyi
x# j œ -(i
j) Ê 2xy œ - and x# œ Ê 2xy œ x# Ê x œ 0 or 2y œ x. CASE 1: If x œ 0, then x
y œ 3 Ê y œ 3. CASE 2: If x Á 0, then 2y œ x so that x
y œ 3 Ê 2y
y œ 3 Ê y œ 1 Ê x œ 2. Therefore f takes on its extreme values at (!ß 3) and (#ß "). The extreme values of f are f(0ß 3) œ 0 and f(#ß 1) œ 4.

910

Chapter 14 Partial Derivatives

5. We optimize f(xß y) œ x#
y# , the square of the distance to the origin, subject to the constraint g(xß y) œ xy#  54 œ 0. Thus ™ f œ 2xi
2yj and ™ g œ y# i
2xyj so that ™ f œ - ™ g Ê 2xi
2yj œ - ay# i
2xyjb Ê 2x œ -y# and 2y œ 2-xy. CASE 1: If y œ 0, then x œ 0. But (0ß 0) does not satisfy the constraint xy# œ 54 so y Á 0. CASE 2: If y Á 0, then 2 œ 2-x Ê x œ -" Ê 2 ˆ -" ‰ œ -y# Ê y# œ -2# . Then xy# œ 54 Ê ˆ -" ‰ ˆ -2# ‰ œ 54 Ê -$ œ " Ê - œ " Ê x œ 3 and y# œ 18 Ê x œ 3 and y œ „ 3È2. 27

3

Therefore Š$ß „ 3È2‹ are the points on the curve xy# œ 54 nearest the origin (since xy# œ 54 has points increasingly far away as y gets close to 0, no points are farthest away). 6. We optimize f(xß y) œ x#
y# , the square of the distance to the origin subject to the constraint g(xß y) œ x# y  2 œ 0. Thus ™ f œ 2xi
2yj and ™ g œ 2xyi
x# j so that ™ f œ - ™ g Ê 2x œ 2xy- and 2y œ x# ˆ 2y ‰ Ê x# œ 2y# Ê - œ 2y x# , since x œ 0 Ê y œ 0 (but g(0ß 0) Á 0). Thus x Á 0 and 2x œ 2xy x# Ê a2y# b y  2 œ 0 Ê y œ 1 (since y  0) Ê x œ „ È2 . Therefore Š „ È2ß 1‹ are the points on the curve x# y œ 2 nearest the origin (since x# y œ 2 has points increasingly far away as x gets close to 0, no points are farthest away). 7. (a) ™ f œ i
j and ™ g œ yi
xj so that ™ f œ - ™ g Ê i
j œ -(yi
xj) Ê 1 œ -y and 1 œ -x Ê y œ xœ

" -

Ê

" -#

" 4

œ 16 Ê - œ „ . Use - œ

" 4

" -

and

since x  0 and y  0. Then x œ 4 and y œ 4 Ê the minimum

value is 8 at the point (4ß 4). Now, xy œ 16, x  0, y  0 is a branch of a hyperbola in the first quadrant with the x-and y-axes as asymptotes. The equations x
y œ c give a family of parallel lines with m œ 1. As these lines move away from the origin, the number c increases. Thus the minimum value of c occurs where x
y œ c is tangent to the hyperbola's branch. (b) ™ f œ yi
xj and ™ g œ i
j so that ™ f œ - ™ g Ê yi
xj œ -(i
j) Ê y œ - œ x y
y œ 16 Ê y œ 8 Ê x œ 8 Ê f()ß )) œ 64 is the maximum value. The equations xy œ c (x  0 and y  0 or x  0 and y  0 to get a maximum value) give a family of hyperbolas in the first and third quadrants with the x- and yaxes as asymptotes. The maximum value of c occurs where the hyperbola xy œ c is tangent to the line x
y œ 16. 8. Let f(xß y) œ x#
y# be the square of the distance from the origin. Then ™ f œ 2xi
2yj and ™ g œ (2x
y)i
(2y
x)j so that ™ f œ - ™ g Ê 2x œ -(2x
y) and 2y œ -(2y
x) Ê

2y 2y
x

œ-

Ê 2x œ Š 2y2y
x ‹ (2x
y) Ê x(2y
x) œ y(2x
y) Ê x# œ y# Ê y œ „ x. CASE 1: y œ x Ê x#
x(x)
x#  1 œ 0 Ê x œ „

" È3

and y œ x.

CASE 2: y œ x Ê x#
x(x)
(x)#  1 œ 0 Ê x œ „ 1 and y œ x. Thus f Š È"3 ß È"3 ‹ œ

2 3

œ f Š È"3 ß  È"3 ‹ and f(1ß 1) œ 2 œ f(1ß 1). Therefore the points (1ß 1) and (1ß 1) are the farthest away; Š È"3 ß È"3 ‹ and Š È"3 ß  È"3 ‹ are the closest points to the origin. 9. V œ 1r# h Ê 161 œ 1r# h Ê 16 œ r# h Ê g(rß h) œ r# h  16; S œ 21rh
21r# Ê ™ S œ (21h
41r)i
21rj and ™ g œ 2rhi
r# j so that ™ S œ - ™ g Ê (21rh
41r)i
21rj œ - a2rhi
r# jb Ê 21rh
41r œ 2rh- and 21r œ -r# Ê r œ 0 or - œ 2r1 . But r œ 0 gives no physical can, so r Á 0 Ê - œ 2r1 Ê 21h
41r œ 2rh ˆ 2r1 ‰ Ê 2r œ h Ê 16 œ r# (2r) Ê r œ 2 Ê h œ 4; thus r œ 2 cm and h œ 4 cm give the only extreme surface area of 241 cm# . Since r œ 4 cm and h œ 1 cm Ê V œ 161 cm$ and S œ 401 cm# , which is a larger surface area, then 241 cm# must be the minimum surface area.

Section 14.8 Lagrange Multipliers

911

10. For a cylinder of radius r and height h we want to maximize the surface area S œ 21rh subject to the constraint # g(rß h) œ r#
ˆ h# ‰  a# œ 0. Thus ™ S œ 21hi
21rj and ™ g œ 2ri
h# j so that ™ S œ - ™ g Ê 21h œ 2-r and 21r œ

-h #

Ê

1h r

4r# 4

œ - and 21r œ ˆ 1rh ‰ ˆ h# ‰ Ê 4r# œ h# Ê h œ 2r Ê r#


œ a# Ê 2r# œ a# Ê r œ

a È2

Ê h œ aÈ2 Ê S œ 21 Š Èa2 ‹ ŠaÈ2‹ œ 21a# . #

#

x 11. A œ (2x)(2y) œ 4xy subject to g(xß y) œ 16
y9  1 œ 0; ™ A œ 4yi
4xj and ™ g œ 8x i
2y 9 j so that ™ A 2y 2y 32y x x ‰ ˆ 32y ‰ œ - ™ g Ê 4yi
4xj œ - ˆ 8 i
9 j‰ Ê 4y œ ˆ 8 ‰ - and 4x œ ˆ 9 ‰ - Ê - œ x and 4x œ ˆ 2y 9 x

Ê y œ „ 34 x Ê Then y œ

3 4

x# 16

Š2È2‹ œ

ˆ „43 x‰# œ 1 Ê x# 9 3È 2 # , so the length is




x# a#

12. P œ 4x
4y subject to g(xß y) œ




y# b#

and height œ 2y œ

2b# È a#
b#

2x œ 4È2 and the width is 2y œ 3È2.

 1 œ 0; ™ P œ 4i
4j and ™ g œ

‰ ˆ 2y ‰ Ê 4 œ ˆ 2x a# - and 4 œ b# - Ê - œ œ 1 Ê aa#
b# b x# œ a% Ê x œ

œ 8 Ê x œ „ 2È2 . We use x œ 2È2 since x represents distance.

2a# x

a# È a#
b#

#

2x a#

i


#

b ‰ 2a and 4 œ ˆ 2y b # Š x ‹ Ê y œ Š a# ‹ x Ê #

, since x  0 Ê y œ Š ba# ‹ x œ

Ê perimeter is P œ 4x
4y œ

4a#
4b# È a#
b#

b# È a#
b#

2y b# x# a#

j so that ™ P œ - ™ g #




#

Š ba# ‹ x# b#

œ1 Ê

Ê width œ 2x œ

x# a#




b# x# a%

2a# È a#
b #

œ 4Èa#
b#

13. ™ f œ 2xi
2yj and ™ g œ (2x  2)i
(2y  4)j so that ™ f œ - ™ g œ 2xi
2yj œ -[(2x  2)i
(2y  4)j] 2# # Ê 2x œ -(2x  2) and 2y œ -(2y  4) Ê x œ - 1 and y œ -1 , - Á 1 Ê y œ 2x Ê x  2x
(2x)  4(2x) œ 0 Ê x œ 0 and y œ 0, or x œ 2 and y œ 4. Therefore f(0ß 0) œ 0 is the minimum value and f(2ß 4) œ 20 is the maximum value. (Note that - œ 1 gives 2x œ 2x  2 or ! œ 2, which is impossible.)

14. ™ f œ 3i  j and ™ g œ 2xi
2yj so that ™ f œ - ™ g Ê 3 œ 2-x and 1 œ 2-y Ê - œ #

Ê y œ  x3 Ê x#
ˆ x3 ‰ œ 4 Ê 10x# œ 36 Ê x œ „ yœ

2 È10 .

Therefore f Š È610 ß  È210 ‹ œ

20 È10

6 È10

Ê xœ

6 È10

3 2x

3 ‰ and 1 œ 2 ˆ 2x y

and y œ  È210 , or x œ  È610 and


6 œ 2È10
6 ¸ 12.325 is the maximum value, and

f Š È610 ß È210 ‹ œ 2È10
6 ¸ 0.325 is the minimum value. 15. ™ T œ (8x  4y)i
(4x
2y)j and g(xß y) œ x#
y#  25 œ 0 Ê ™ g œ 2xi
2yj so that ™ T œ - ™ g Ê (8x  4y)i
(4x
2y)j œ -(2xi
2yj) Ê 8x  4y œ 2-x and 4x
2y œ 2-y Ê y œ -2x1 , - Á 1 Ê 8x  4 ˆ -2x1 ‰ œ 2-x Ê x œ 0, or - œ 0, or - œ 5. CASE 1: x œ 0 Ê y œ 0; but (0ß 0) is not on x#
y# œ 25 so x Á 0. CASE 2: - œ 0 Ê y œ 2x Ê x#
(2x)# œ 25 Ê x œ „ È5 and y œ 2x. CASE 3: - œ 5 Ê y œ and y œ È5 .

2x 4

#

œ  #x Ê x#
ˆ #x ‰ œ 25 Ê x œ „ 2È5 Ê x œ 2È5 and y œ È5, or x œ 2È5

Therefore T ŠÈ5ß 2È5‹ œ 0° œ T ŠÈ5ß 2È5‹ is the minimum value and T Š2È5ß È5‹ œ 125° œ T Š2È5ß È5‹ is the maximum value. (Note: - œ 1 Ê x œ 0 from the equation 4x
2y œ 2-y; but we found x Á 0 in CASE 1.) 16. The surface area is given by S œ 41r#
21rh subject to the constraint V(rß h) œ

4 3

1r$
1r# h œ 8000. Thus

™ S œ (81r
21h)i
21rj and ™ V œ a41r#
21rhb i
1r# j so that ™ S œ - ™ V œ (81r
21h)i
21rj œ - ca41r#
21rhb i
1r# jd Ê 81r
21h œ - a41r#
21rhb and 21r œ -1r# Ê r œ 0 or 2 œ r-. But r Á 0 so 2 œ r- Ê - œ 2r Ê 4r
h œ 2r a2r#
rhb Ê h œ 0 Ê the tank is a sphere (there is no cylindrical part) and 4 3

1r$ œ 8000 Ê r œ 10 ˆ 16 ‰

"Î$

.

912

Chapter 14 Partial Derivatives

17. Let f(xß yß z) œ (x  1)#
(y  1)#
(z  1)# be the square of the distance from (1ß 1ß 1). Then ™ f œ 2(x  1)i
2(y  1)j
2(z  1)k and ™ g œ i
2j
3k so that ™ f œ - ™ g Ê 2(x  1)i
2(y  1)j
2(z  1)k œ -(i
2j
3k) Ê 2(x  1) œ -, 2(y  1) œ 2-, 2(z  1) œ 3Ê 2(y  1) œ 2[2(x  1)] and 2(z  1) œ 3[2(x  1)] Ê x œ y
# 1 Ê z
2 œ 3 ˆ y
# 1 ‰ or z œ 3y # 1 ; thus y
1 ˆ 3y # 1 ‰  13 œ 0 Ê y œ 2 Ê x œ 3# and z œ 5# . Therefore the point ˆ 3# ß 2ß 5# ‰ is closest (since no #
2y
3 point on the plane is farthest from the point (1ß 1ß 1)).

18. Let f(xß yß z) œ (x  1)#
(y
1)#
(z  1)# be the square of the distance from (1ß 1ß 1). Then ™ f œ 2(x  1)i
2(y
1)j
2(z  1)k and ™ g œ 2xi
2yj
2zk so that ™ f œ - ™ g Ê x  1 œ -x, y
1 œ -y # ‰#
ˆ 1 " - ‰# œ 4 and z  1 œ -z Ê x œ 1 " - , y œ  1 " - , and z œ 1" - for - Á 1 Ê ˆ 1 " - ‰
ˆ 1" Ê

" "-

œ „

2 È3

Ê xœ

2 È3

, y œ  È23 , z œ

2 È3

or x œ  È23 , y œ

2 È3

, z œ  È23 . The largest value of f

occurs where x  0, y  0, and z  0 or at the point Š È23 ß È23 ß  È23 ‹ on the sphere. 19. Let f(xß yß z) œ x#
y#
z# be the square of the distance from the origin. Then ™ f œ 2xi
2yj
2zk and ™ g œ 2xi  2yj  2zk so that ™ f œ - ™ g Ê 2xi
2yj
2zk œ -(2xi  2yj  2zk) Ê 2x œ 2x-, 2y œ 2y-, and 2z œ 2z- Ê x œ 0 or - œ 1. CASE 1: - œ 1 Ê 2y œ 2y Ê y œ 0; 2z œ 2z Ê z œ 0 Ê x#  1 œ 0 Ê x œ „ 1 and y œ z œ 0. CASE 2: x œ 0 Ê y#  z# œ 1, which has no solution. Therefore the points on the unit circle x#
y# œ 1, are the points on the surface x#
y#  z# œ 1 closest to the originÞ The minimum distance is 1. 20. Let f(xß yß z) œ x#
y#
z# be the square of the distance to the origin. Then ™ f œ 2xi
2yj
2zk and ™ g œ yi
xj  k so that ™ f œ - ™ g Ê 2xi
2yj
2zk œ -(yi
xj  k) Ê 2x œ -y, 2y œ -x, and 2z œ Ê xœ

-y #

Ê 2y œ - Š -#y ‹ Ê y œ 0 or - œ „ 2.

CASE 1: y œ 0 Ê x œ 0 Ê z
1 œ 0 Ê z œ 1. CASE 2: - œ 2 Ê x œ y and z œ 1 Ê x#  (1)
1 œ 0 Ê x#
2 œ 0, so no solution. CASE 3: - œ 2 Ê x œ y and z œ 1 Ê (y)y  1
1 œ 0 Ê y œ 0, again. Therefore (0ß 0ß 1) is the point on the surface closest to the origin since this point gives the only extreme value and there is no maximum distance from the surface to the origin. 21. Let f(xß yß z) œ x#
y#
z# be the square of the distance to the origin. Then ™ f œ 2xi
2yj
2zk and ™ g œ yi  xj
2zk so that ™ f œ - ™ g Ê 2xi
2yj
2zk œ -(yi  xj
2zk) Ê 2x œ y-, 2y œ x-, and 2z œ 2z- Ê - œ 1 or z œ 0. CASE 1: - œ 1 Ê 2x œ y and 2y œ x Ê y œ 0 and x œ 0 Ê z#  4 œ 0 Ê z œ „ 2 and x œ y œ 0. CASE 2: z œ 0 Ê xy  4 œ 0 Ê y œ  4x . Then 2x œ

4 x

- Ê -œ

x# #

#

, and  8x œ x- Ê  8x œ x Š x# ‹

Ê x% œ 16 Ê x œ „ 2. Thus, x œ 2 and y œ 2, or x = 2 and y œ 2. Therefore we get four points: (#ß 2ß 0), (2ß 2ß 0), (0ß 0ß 2) and (!ß 0ß 2). But the points (!ß 0ß 2) and (!ß !ß 2) are closest to the origin since they are 2 units away and the others are 2È2 units away. 22. Let f(xß yß z) œ x#
y#
z# be the square of the distance to the origin. Then ™ f œ 2xi
2yj
2zk and ™ g œ yzi
xzj
xyk so that ™ f œ - ™ g Ê 2x œ -yz, 2y œ -xz, and 2z œ -xy Ê 2x# œ -xyz and 2y# œ -yxz Ê x# œ y# Ê y œ „ x Ê z œ „ x Ê x a „ xb a „ xb œ 1 Ê x œ „ 1 Ê the points are (1ß 1ß 1), ("ß 1ß 1), ("ß "ß "), and (1ß 1, 1). 23. ™ f œ i  2j
5k and ™ g œ 2xi
2yj
2zk so that ™ f œ - ™ g Ê i  2j
5k œ -(2xi
2yj
2zk) Ê 1 œ 2x-, 2 œ 2y-, and 5 œ 2z- Ê x œ #"- , y œ  -" œ 2x, and z œ #5- œ 5x Ê x#
(2x)#
(5x)# œ 30 Ê x œ „ 1.

Section 14.8 Lagrange Multipliers

913

Thus, x œ 1, y œ 2, z œ 5 or x œ 1, y œ 2, z œ 5. Therefore f(1ß 2ß 5) œ 30 is the maximum value and f(1ß 2ß 5) œ 30 is the minimum value. 24. ™ f œ i
2j
3k and ™ g œ 2xi
2yj
2zk so that ™ f œ - ™ g Ê i
2j
3k œ -(2xi
2yj
2zk) Ê 1 œ 2x-, 2 œ 2y-, and 3 œ 2z- Ê x œ #"- , y œ -" œ 2x, and z œ #3- œ 3x Ê x#
(2x)#
(3x)# œ 25 Ê x œ „ È514 . Thus, x œ

5 È14

,yœ

10 È14

,zœ

15 È14

or x œ  È514 , y œ  È1014 , z œ  È1514 . Therefore f Š È514 ß È1014 ß È1514 ‹

œ 5È14 is the maximum value and f Š È514 ß  È1014 , È1514 ‹ œ 5È14 is the minimum value. 25. f(xß yß z) œ x#
y#
z# and g(xß yß z) œ x
y
z  9 œ 0 Ê ™ f œ 2xi
2yj
2zk and ™ g œ i
j
k so that ™ f œ - ™ g Ê 2xi
2yj
2zk œ -(i
j
k) Ê 2x œ -, 2y œ -, and 2z œ - Ê x œ y œ z Ê x
x
x  9 œ 0 Ê x œ 3, y œ 3, and z œ 3. 26. f(xß yß z) œ xyz and g(xß yß z) œ x
y
z#  16 œ 0 Ê ™ f œ yzi
xzj
xyk and ™ g œ i
j
2zk so that ™ f œ - ™ g Ê yzi
xzj
xyk œ -(i
j
2zk) Ê yz œ -, xz œ -, and xy œ 2z- Ê yz œ xz Ê z œ 0 or y œ x. But z  0 so that y œ x Ê x# œ 2z- and xz œ -. Then x# œ 2z(xz) Ê x œ 0 or x œ 2z# . But x  0 so that 32 x œ 2z# Ê y œ 2z# Ê 2z#
2z#
z# œ 16 Ê z œ „ È45 . We use z œ È45 since z  0. Then x œ 32 5 and y œ 5 32 4 which yields f Š 32 5 ß 5 ß È5 ‹ œ

4096 25È5

.

27. V œ 6xyz and g(xß yß z) œ x#
y#
z#  1 œ 0 Ê ™ V œ 6yzi
6xzj
6xyk and ™ g œ 2xi
2yj
2zk so that ™ V œ - ™ g Ê 3yz œ -x, 3xz œ -y, and 3xy œ -z Ê 3xyz œ -x# and 3xyz œ -y# Ê y œ „ x Ê z œ „ x Ê x#
x#
x# œ 1 Ê x œ È"3 since x  0 Ê the dimensions of the box are È23 by È23 by È23 for maximum volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.) 28. V œ xyz with xß yß z all positive and

x a




y b




z c

œ 1; thus V œ xyz and g(xß yß z) œ bcx
acy
abz  abc œ 0

Ê ™ V œ yzi
xzj
xyk and ™ g œ bci
acj
abk so that ™ V œ - ™ g Ê yz œ -bc, xz œ -ac, and xy œ -ab Ê xyz œ -bcx, xyz œ -acy, and xyz œ -abz Ê - Á 0. Also, -bcx œ -acy œ -abz Ê bx œ ay, cy œ bz, and a cx œ az Ê y œ ba x and z œ ca x. Then xa
by
cz œ 1 Ê xa
"b ˆ ba x‰
"c ˆ ca x‰ œ 1 Ê 3x a œ 1 Ê xœ 3 Ê y œ ˆ ba ‰ ˆ 3a ‰ œ b3 and z œ ˆ ca ‰ ˆ 3a ‰ œ 3c Ê V œ xyz œ ˆ 3a ‰ ˆ 3b ‰ ˆ 3c ‰ œ abc 27 is the maximum volume. (Note that there is no minimum volume since the box could be made arbitrarily thin.) 29. ™ T œ 16xi
4zj
(4y  16)k and ™ g œ 8xi
2yj
8zk so that ™ T œ - ™ g Ê 16xi
4zj
(4y  16)k œ -(8xi
2yj
8zk) Ê 16x œ 8x-, 4z œ 2y-, and 4y  16 œ 8z- Ê - œ 2 or x œ 0. CASE 1: - œ 2 Ê 4z œ 2y(2) Ê z œ y. Then 4z  16 œ 16z Ê z œ  43 Ê y œ  43 . Then #

#

4x#
ˆ 43 ‰
4 ˆ 43 ‰ œ 16 Ê x œ „ 43 . CASE 2: x œ 0 Ê - œ

2z y

# # # # # Ê 4y  16 œ 8z Š 2z y ‹ Ê y  4y œ 4z Ê 4(0)
y
ay  4yb  16 œ 0

Ê y#  2y  8 œ 0 Ê (y  4)(y
2) œ 0 Ê y œ 4 or y œ 2. Now y œ 4 Ê 4z# œ 4#  4(4) Ê z œ 0 and y œ 2 Ê 4z# œ (2)#  4(2) Ê z œ „ È3. °

°

The temperatures are T ˆ „ 43 ß  43 ß  43 ‰ œ 642 23 , T(0ß 4ß 0) œ 600°, T Š0ß 2ß È3‹ œ Š600  24È3‹ , and °

T Š0ß 2ß È3‹ œ Š600
24È3‹ ¸ 641.6°. Therefore ˆ „ 43 ß  43 ß  43 ‰ are the hottest points on the space probe. 30. ™ T œ 400yz# i
400xz# j
800xyzk and ™ g œ 2xi
2yj
2zk so that ™ T œ - ™ g Ê 400yz# i
400xz# j
800xyzk œ -(2xi
2yj
2zk) Ê 400yz# œ 2x-, 400xz# œ 2y-, and 800xyz œ 2z-. Solving this system yields the points a!ß „ 1ß 0b , a „ 1ß 0ß 0b , and Š „ "# ß „ "# ß „

È2 # ‹.

The corresponding

914

Chapter 14 Partial Derivatives

temperatures are T a!ß „ 1ß 0b œ 0, T a „ 1ß 0ß 0b œ 0, and T Š „ "# ß „ "# ß „ maximum temperature at Š "# ß "# ß „ Š "# ß  "# ß „

È2 # ‹

and Š "# ß "# ß „

È2 # ‹

and Š "# ß  "# ß „

È2 # ‹;

È2 # ‹

œ „ 50. Therefore 50 is the

50 is the minimum temperature at

È2 # ‹.

31. ™ U œ (y
2)i
xj and ™ g œ 2i
j so that ™ U œ - ™ g Ê (y
2)i
xj œ -(2i
j) Ê y
# œ 2- and x œ - Ê y
2 œ 2x Ê y œ 2x  2 Ê 2x
(2x  2) œ 30 Ê x œ 8 and y œ 14. Therefore U(8ß 14) œ $128 is the maximum value of U under the constraint. 32. ™ M œ (6
z)i  2yj
xk and ™ g œ 2xi
2yj
2zk so that ™ M œ - ™ g Ê (6
z)i  2yj
xk œ -(2xi
2yj
2zk) Ê 6
z œ 2x-, 2y œ 2y-, x œ 2z- Ê - œ 1 or y œ 0. CASE 1: - œ 1 Ê 6
z œ 2x and x œ 2z Ê 6
z œ 2(2z) Ê z œ 2 and x œ 4. Then (4)#
y#
2#  36 œ 0 Ê y œ „ 4. x x ‰ CASE 2: y œ 0, 6
z œ 2x-, and x œ 2z- Ê - œ 2z Ê 6
z œ 2x ˆ 2z Ê 6z
z# œ x#

Ê a6z
z# b
0#
z# œ 36 Ê z œ 6 or z œ 3. Now z œ 6 Ê x# œ 0 Ê x œ 0; z œ 3 Ê x# œ 27 Ê x œ „ 3È3.

Therefore we have the points Š „ 3È3ß 0ß 3‹ , (0ß 0ß 6), and a4ß „ 4ß 2b . Then M Š3È3ß 0ß 3‹ œ 27È3
60 ¸ 106.8, M Š3È3ß 0ß 3‹ œ 60  27È3 ¸ 13.2, M(0ß 0ß 6) œ 60, and M(4ß 4ß 2) œ 12 œ M(4ß 4ß 2). Therefore, the weakest field is at a4ß „ 4ß 2b . 33. Let g" (xß yß z) œ 2x  y œ 0 and g# (xß yß z) œ y
z œ 0 Ê ™ g" œ 2i  j , ™ g# œ j
k , and ™ f œ 2xi
2j  2zk so that ™ f œ - ™ g"
. ™ g# Ê 2xi
2j  2zk œ -(2i  j)
.(j
k) Ê 2xi
2j  2zk œ 2-i
(.  -)j
.k Ê 2x œ 2-, 2 œ .  -, and 2z œ . Ê x œ -. Then 2 œ 2z  x Ê x œ 2z  2 so that 2x  y œ 0 Ê 2(2z  2)  y œ 0 Ê 4z  4  y œ 0. This equation coupled with y
z œ 0 implies z œ  43 and y œ 43 . Then x œ œ

4 3

2 3

#

so that ˆ 23 ß 43 ß  43 ‰ is the point that gives the maximum value f ˆ 23 ß 43 ß  43 ‰ œ ˆ 23 ‰
2 ˆ 43 ‰  ˆ 43 ‰

#

.

34. Let g" (xß yß z) œ x
2y
3z  6 œ 0 and g# (xß yß z) œ x
3y
9z  9 œ 0 Ê ™ g" œ i
2j
3k , ™ g# œ i
3j
9k , and ™ f œ 2xi
2yj
2zk so that ™ f œ - ™ g"
. ™ g# Ê 2xi
2yj
2zk œ -(i
2j
3k)
.(i
3j
9k) Ê 2x œ -
., 2y œ 2-
3., and 2z œ 3-
9.. Then 0 œ x
2y
3z  6 ‰ œ "# (-
.)
(2-
3.)
ˆ 9# -
27 # .  6 Ê 7-
17. œ 6; 0 œ x
3y
9z  9 " 9 27 81 Ê # (-
.)
ˆ3-
# .‰
ˆ # -
# .‰  9 Ê 34-
91. œ 18. Solving these two equations for - and . gives -
. 2-
3. 3-
9. 78 81 9 - œ 240 œ 123 œ 59 . The minimum value is 59 and . œ  59 Ê x œ # œ 59 , y œ # 59 , and z œ # 21,771 81 123 9 369 f ˆ 59 ß 59 ß 59 ‰ œ 59# œ 59 . (Note that there is no maximum value of f subject to the constraints because

at least one of the variables x, y, or z can be made arbitrary and assume a value as large as we please.) 35. Let f(xß yß z) œ x#
y#
z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject to the constraints g" (xß yß z) œ y
2z  12 œ 0 and g# (xß yß z) œ x
y  6 œ 0. Thus ™ f œ 2xi
2yj
2zk , ™ g" œ j
2k , and ™ g# œ i
j so that ™ f œ - ™ g"
. ™ g# Ê 2x œ ., 2y œ -
., and 2z œ 2-. Then 0 œ y
2z  12 œ ˆ -#
.# ‰
2-  12 Ê 5# -
"# . œ 12 Ê 5-
. œ 24; 0 œ x
y  6 œ .#
ˆ -#
.# ‰  6 Ê yœ

" # -
. œ 6 Ê -
#. œ 12. Solving these two equations for - and . gives - œ 4 and . œ -
. # œ 4, and z œ - œ 4. The point (2ß 4ß 4) on the line of intersection is closest to the origin.

maximum distance from the origin since points on the line can be arbitrarily far away.) 36. The maximum value is f ˆ 23 ß 43 ß  43 ‰ œ

4 3

from Exercise 33 above.

4 Ê xœ

. #

(There is no

œ 2,

Section 14.8 Lagrange Multipliers

915

37. Let g" (xß yß z) œ z  1 œ 0 and g# (xß yß z) œ x#
y#
z#  10 œ 0 Ê ™ g" œ k , ™ g# œ 2xi
2yj
2zk , and ™ f œ 2xyzi
x# zj
x# yk so that ™ f œ - ™ g"
. ™ g# Ê 2xyzi
x# zj
x# yk œ -(k)
.(2xi
2yj
2zk) Ê 2xyz œ 2x., x# z œ 2y., and x# y œ 2z.
- Ê xyz œ x. Ê x œ 0 or yz œ . Ê . œ y since z œ 1. CASE 1: x œ 0 and z œ 1 Ê y#  9 œ 0 (from g# ) Ê y œ „ 3 yielding the points a0ß „ 3ß 1b. CASE 2: . œ y Ê x# z œ 2y# Ê x# œ 2y# (since z œ 1) Ê 2y#
y#
1  10 œ 0 (from g# ) Ê 3y#  9 œ 0 #

Ê y œ „ È3 Ê x# œ 2 Š „ È3‹ Ê x œ „ È6 yielding the points Š „ È6ß „ È3ß "‹ . Now f a!ß „ 3ß 1b œ 1 and f Š „ È6ß „ È3ß "‹ œ 6 Š „ È3‹
1 œ 1 „ 6È3. Therefore the maximum of f is 1
6È3 at Š „ È6ß È3ß 1‹, and the minimum of f is 1  6È3 at Š „ È6ß È3ß "‹ . 38. (a) Let g" (xß yß z) œ x
y
z  40 œ 0 and g# (xß yß z) œ x
y  z œ 0 Ê ™ g" œ i
j
k , ™ g# œ i
j  k , and ™ w œ yzi
xzj
xyk so that ™ w œ - ™ g"
. ™ g# Ê yzi
xzj
xyk œ -(i
j
k)
.(i
j  k) Ê yz œ -
., xz œ -
., and xy œ -  . Ê yz œ xz Ê z œ 0 or y œ x. CASE 1: z œ 0 Ê x
y œ 40 and x
y œ 0 Ê no solution. CASE 2: x œ y Ê 2x
z  40 œ 0 and 2x  z œ 0 Ê z œ 20 Ê x œ 10 and y œ 10 Ê w œ (10)(10)(20) œ 2000 â â âi j k â â â " â œ 2i
2j is parallel to the line of intersection Ê the line is x œ 2t
10, (b) n œ â " " â â â " " " â y œ 2t
10, z œ 20. Since z œ 20, we see that w œ xyz œ (2t
10)(2t
10)(20) œ a4t#
100b (20) which has its maximum when t œ 0 Ê x œ 10, y œ 10, and z œ 20. 39. Let g" (Bß yß z) œ y  x œ 0 and g# (xß yß z) œ x#
y#
z#  4 œ 0. Then ™ f œ yi
xj
2zk , ™ g" œ i
j , and ™ g# œ 2xi
2yj
2zk so that ™ f œ - ™ g"
. ™ g# Ê yi
xj
2zk œ -(i
j)
.(2xi
2yj
2zk) Ê y œ -
2x., x œ -
2y., and 2z œ 2z. Ê z œ 0 or . œ 1. CASE 1: z œ 0 Ê x#
y#  4 œ 0 Ê 2x#  4 œ 0 (since x œ y) Ê x œ „ È2 and y œ „ È2 yielding the points Š „ È2ß „ È2ß !‹ . CASE 2: . œ 1 Ê y œ -
2x and x œ -
2y Ê x
y œ 2(x
y) Ê 2x œ 2(2x) since x œ y Ê x œ 0 Ê y œ 0 Ê z#  4 œ 0 Ê z œ „ 2 yielding the points a!ß !ß „ 2b . Now, f a!ß !ß „ 2b œ 4 and f Š „ È2ß „ È2ß !‹ œ 2. Therefore the maximum value of f is 4 at a!ß !ß „ 2b and the minimum value of f is 2 at Š „ È2ß „ È2ß !‹ . 40. Let f(xß yß z) œ x#
y#
z# be the square of the distance from the origin. We want to minimize f(xß yß z) subject to the constraints g" (xß yß z) œ 2y
4z  5 œ 0 and g# (xß yß z) œ 4x#
4y#  z# œ 0. Thus ™ f œ 2xi
2yj
2zk , ™ g" œ 2j
4k , and ™ g# œ 8xi
8yj  2zk so that ™ f œ - ™ g"
. ™ g# Ê 2xi
2yj
2zk œ -(2j
4k)
.(8xi
8yj  2zk) Ê 2x œ 8x., 2y œ 2-
8y., and 2z œ 4-  2z. Ê x œ 0 or . œ "4 . CASE 1: x œ 0 Ê 4(0)#
4y#  z# œ 0 Ê z œ „ 2y Ê 2y
4(2y)  5 œ 0 Ê y œ Ê y œ  56 yielding the points ˆ!ß "# ß "‰ and ˆ!ß  56 ß 53 ‰ . CASE 2: . œ #

" 4

" #

, or 2y
4(2y)  5 œ 0

Ê y œ -
y Ê - œ 0 Ê 2z œ 4(0)  2z ˆ 4" ‰ Ê z œ 0 Ê 2y
4(0) œ 5 Ê y œ #

# 4 ˆ #5 ‰

(0) œ 4x
Ê no solution. " 5 " Then f ˆ!ß # ß 1‰ œ 4 and f ˆ!ß  56 ß 53 ‰ œ 25 ˆ 36
"9 ‰ œ

125 36

5 #

and

Ê the point ˆ!ß #" ß 1‰ is closest to the origin.

41. ™ f œ i
j and ™ g œ yi
xj so that ™ f œ - ™ g Ê i
j œ -(yi
xj) Ê 1 œ y- and 1 œ x- Ê y œ x Ê y# œ 16 Ê y œ „ 4 Ê (4ß 4) and (%ß 4) are candidates for the location of extreme values. But as x Ä _, y Ä _ and f(xß y) Ä _; as x Ä _, y Ä 0 and f(xß y) Ä _. Therefore no maximum or minimum value exists subject to the constraint.

916

Chapter 14 Partial Derivatives 4

42. Let f(Aß Bß C) œ ! (Axk
Byk
C  zk )# œ C#
(B
C  1)#
(A
B
C  1)#
(A
C
1)# . We want kœ1

to minimize f. Then fA (Aß Bß C) œ 4A
2B
4C, fB (Aß Bß C) œ 2A
4B
4C  4, and fC (Aß Bß C) œ 4A
4B
8C  2. Set each partial derivative equal to 0 and solve the system to get A œ  "# , B œ 3# , and C œ  "4 or the critical point of f is ˆ "# ß 3# ß  4" ‰ . 43. (a) Maximize f(aß bß c) œ a# b# c# subject to a#
b#
c# œ r# . Thus ™ f œ 2ab# c# i
2a# bc# j
2a# b# ck and ™ g œ 2ai
2bj
2ck so that ™ f œ - ™ g Ê 2ab# c# œ 2a-, 2a# bc# œ 2b-, and 2a# b# c œ 2cÊ 2a# b# c# œ 2a# - œ 2b# - œ 2c# - Ê - œ 0 or a# œ b# œ c# . CASE 1: - œ 0 Ê a# b# c# œ 0. #

$

CASE 2: a# œ b# œ c# Ê f(aß bß c) œ a# a# a# and 3a# œ r# Ê f(aß bß c) œ Š r3 ‹ is the maximum value. (b) The point ŠÈaß Èbß Èc‹ is on the sphere if a
b
c œ r# . Moreover, by part (a), abc œ f ŠÈaß Èbß Èc‹ #

$

Ÿ Š r3 ‹ Ê (abc)"Î$ Ÿ

r# 3

œ

a
b
c 3

, as claimed.

n

44. Let f(x" ß x# ß á ß xn ) œ ! ai xi œ a" x"
a# x#
á
an xn and g(x" ß x# ß á ß xn ) œ x#"
x##
á
x#n  1. Then we iœ1

want ™ f œ - ™ g Ê a" œ -(2x" ), a# œ -(2x# ), á , an œ -(2xn ), - Á 0 Ê xi œ n

n

iœ1

iœ1

"Î#

Ê 4-# œ ! ai# Ê 2- œ Œ! ai# 

n

n

iœ1

iœ1

ai 2-

Ê f(x" ß x# ß á ß xn ) œ ! ai xi œ ! ai ˆ #a-i ‰ œ

Ê " #-

a#" 4- #




a## 4- #

a#n 4- # "Î#


á


n

n

iœ1

iœ1

! a#i œ Œ! a#i 

the maximum value. 45-50. Example CAS commands: Maple: f := (x,y,z) -> x*y+y*z; g1 := (x,y,z) -> x^2+y^2-2; g2 := (x,y,z) -> x^2+z^2-2; h := unapply( f(x,y,z)-lambda[1]*g1(x,y,z)-lambda[2]*g2(x,y,z), (x,y,z,lambda[1],lambda[2]) ); hx := diff( h(x,y,z,lambda[1],lambda[2]), x ); hy := diff( h(x,y,z,lambda[1],lambda[2]), y ); hz := diff( h(x,y,z,lambda[1],lambda[2]), z ); hl1 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[1] ); hl2 := diff( h(x,y,z,lambda[1],lambda[2]), lambda[2] ); sys := { hx=0, hy=0, hz=0, hl1=0, hl2=0 }; q1 := solve( sys, {x,y,z,lambda[1],lambda[2]} ); q2 := map(allvalues,{q1}); for p in q2 do eval( [x,y,z,f(x,y,z)], p ); ``=evalf(eval( [x,y,z,f(x,y,z)], p )); end do; Mathematica: (assigned functions will vary) Clear[x, y, z, lambda1, lambda2] f[x_,y_,z_]:= x y
y z g1[x_,y_,z_]:= x2
y2  2 g2[x_,y_,z_]:= x2
z2  2 h = f[x, y, z]  lambda1 g1[x, y, z]  lambda2 g2[x, y, z]; hx= D[h, x]; hy= D[h, y]; hz= D[h,z]; hL1=D[h, lambda1]; hL2= D[h, lambda2]; critical=Solve[{hx==0, hy==0, hz==0, hL1==0, hL2==0, g1[x,y,z]==0, g2[x,y,z]==0},

# (a) #(b)

# (c) # (d)

is

œ1

Section 14.9 Partial Derivatives with Constrained Variables {x, y, z, lambda1, lambda2}]//N {{x, y, z}, f[x, y, z]}/.critical 14.9 PARTIAL DERIVATIVES WITH CONSTRAINED VARIABLES 1. w œ x#
y#
z# and z œ x#
y# : Î x œ x(yß z) Ñ y yœy (a) Œ  Ä Ä w Ê Š ``wy ‹ œ z z Ï zœz Ò œ 2x `` xy
2y Ê 0 œ 2x `` xy
2y Ê

`x `y

`y `z

œ

" #y

`x `z

œ

1 2x

`w `x `x `z

`z `y

œ 0 and

œ 2x `` yx
2y `` yy




`w `y `y `z




`w `z `x `z `z ; `z

œ 0 and

`z `z

œ 2x `` xz
2y `` yz

`z `z

œ 2x `` xz
2y `` yz

" Ê ˆ ``wz ‰x œ (2x)(0)
(2y) Š 2y ‹
(2z)(1) œ 1
2z `w `x `x `z




`w `y `y `z




`w `z `y `z `z ; `z

œ 0 and

Ê ˆ ``wz ‰y œ (2x) ˆ #"x ‰
(2y)(0)
(2z)(1) œ 1
2z

2. w œ x#
y  z
sin t and x
y œ t: Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê Š ``wy ‹ œ (a) Ó zœz xz ÏzÒ Ït œ x
yÒ ß

`t `y

`w `z `z `z `y ; `y




z

Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ (c) Œ  Ä z Ï zœz Ò Ê 1 œ 2x `` xz Ê

`w `y `y `y




œ  yx Ê Š ``wy ‹ œ (2x) ˆ xy ‰
(2y)(1)
(2z)(0) œ 2y
2y œ 0

Î xœx Ñ x y œ y(xß z) Ä w Ê ˆ ``wz ‰x œ (b) Œ  Ä z Ï zœz Ò Ê 1 œ 2y `` yz Ê

`w `x `x `y

`w `x `x `y




`w `y `y `y




`w `z `z `y




`w `t `x `t `y ; `y

œ 0,

`z `y

œ 0, and

œ 1 Ê Š ``wy ‹ œ (2x)(0)
(1)(1)
(1)(0)
(cos t)(1) œ 1
cos t œ 1
cos (x
y) xßt

Îx œ t  yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê Š ``wy ‹ œ (b) Ó zœz zt ÏtÒ Ï tœt Ò ß

Ê

`x `y

œ

`t `y



`y `y

`w `x `x `y




`w `y `y `y




`w `z `z `y




`w `t `z `t `y ; `y

œ 0 and

`t `y

œ0

œ 1 Ê Š ``wy ‹ œ (2x)(1)
(1)(1)
(1)(0)
(cos t)(0) œ 1  2at  yb œ 1
2y  2t zßt

Î xœx Ñ ÎxÑ Ð yœy Ó y Ä Ð Ä w Ê ˆ ``wz ‰x y œ ``wx `` xz
``wy (c) Ó œ z z ÏzÒ Ït œ x
yÒ Ê ˆ ``wz ‰ œ (2x)(0)
(1)(0)
(1)(1)
(cos t)(0) œ 1

`y `z




`w `z `z `z




`w `t `x `t `z ; `z

œ 0 and

`y `z

œ0

`w `y `y `z




`w `z `z `z




`w `t `y `t `z ; `z

œ 0 and

`t `z

œ0




`w `z `z `t




`w `t `x `t `t ; `t

œ 0 and

`z `t

œ0

ß

xßy

Îx œ t  yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wz ‰y t œ (d) Ó zœz ÏtÒ Ï tœt Ò ß

`w `x `x `z




Ê ˆ ``wz ‰y t œ (2x)(0)
(1)(0)
(1)(1)
(cos t)(0) œ 1 ß

Î xœx Ñ ÎxÑ Ð y œ t  xÓ z Ä Ð Ä w Ê ˆ ``wt ‰x z œ (e) Ó zœz ÏtÒ Ï tœt Ò ß

`w `x `x `t




`w `y `y `t

Ê ˆ ``wt ‰x z œ (2x)(0)
(1)(1)
(1)(0)
(cos t)(1) œ 1
cos t ß

917

918

Chapter 14 Partial Derivatives

Îx œ t  yÑ ÎyÑ Ð yœy Ó z Ä Ð Ä w Ê ˆ ``wt ‰y z œ (f) Ó z œ z ÏtÒ Ï tœt Ò ß

`w `x `x `t




`w `y `y `t




`w `z `z `t




`w `t `y `t `t ; `t

œ 0 and

`z `t

œ0

Ê ˆ ``wt ‰y z œ (2x)(1)
(1)(0)
(1)(0)
(cos t)(1) œ cos t
2x œ cos t
2(t  y) ß

3. U œ f(Pß Vß T) and PV œ nRT Î PœP Ñ P VœV Ä U Ê ˆ ``UP ‰V œ (a) Œ  Ä V Ï T œ PV Ò nR V ‰ œ ``UP
ˆ ``UT ‰ ˆ nR

`U `P `P `P

nRT ÎP œ V Ñ V Ä U Ê ˆ ``UT ‰V œ (b) Œ  Ä VœV T Ï TœT Ò ‰ `U œ ˆ ``UP ‰ ˆ nR V
`T

`U `P `P `T

4. w œ x#
y#
z# and y sin z
z sin x œ 0 Î xœx Ñ x yœy Ä w Ê ˆ ``wx ‰y œ (a) Œ  Ä y Ï z œ z(xß y) Ò (y cos z) Ê

`z `x


(sin x)

ˆ ``wx ‰ yk (0ß1ß1)

`z `x


z cos x œ 0 Ê

`z `x

`x `z

`U `V `V `T




`U `T `T `P







`w `y `y `x

z cos x y cos z
sin x . #

œ

`U `T `T `T

`U `P

‰ ˆ `U ‰ œ ˆ ``UP ‰ ˆ nR V
` V (0)


`w `z `y `z `x ; `x




V ‰ ‰ ˆ ` U ‰ ˆ nR
ˆ `` U V (0)
` T

`U `T

œ 0 and

`z `x

œ

1 1

œ1

œ (2x)

`x `z


(2y)(0)
(2z)(1)

At (0ß 1ß 1),

œ (2x)(1)
(2y)(0)
(2z)(1)k Ð0ß1ß1Ñ œ 21

Î x œ x(yß z) Ñ y yœy Ä w Ê ˆ ``wz ‰y œ (b) Œ  Ä z Ï zœz Ò œ (2x)




`w `x `x `x

œ

`U `V `V `P




`y `z
y x) `` xz œ


2z. Now (sin z)

Ê y cos z
sin x
(z cos

`w `x `x `z




`w `y `y `z

cos z
sin x
(z cos x) 0 Ê

`x `z

œ

y cos z  sin x . z cos x

`w `z `z `z




`x `z

`y `z œ 0 (!ß "ß 1), `` xz œ (11)(1)0

œ 0 and

At

œ

" 1

Ê ˆ ``wz ‰Ck (!,"ß1Ñ œ 2(0) ˆ 1" ‰
21 œ 21 5. w œ x# y#
yz  z$ and x#
y#
z# œ 6 Î xœx Ñ x yœy Ä w Ê Š ``wy ‹ œ (a) Œ  Ä y x Ï z œ z(xß y) Ò œ a2xy# b (0)
a2x# y
zb (1)
ay  3z# b `x `y

œ 0 Ê 2y
(2z)

`z `y

œ0 Ê

#

`z `y

`z `y

`z `y




`w `y `y `y




`w `z `z `y

œ 2x# y
z
ay  3z# b

`z `y .

Now (2x)

œ  yz . At (wß xß yß z) œ (4ß 2ß 1ß 1),

`z `y

`x `y


2y
(2z)

`x `y

`w `x `x `y





a2x# y
zb (1)
ay  3z# b (0) œ a2x# yb

œ 0 Ê (2x)

`x `y


2y œ 0 Ê

`x `y

`w `y `y `y `x `y




œv

`u `y


u Š uv

`u `y


u

`u `y ‹

`v `y ;

`w `z `z `y


2x# y
z. Now (2x)

œ  yx . At (wß xß yß z) œ (4ß 2ß 1ß 1),

œ Šv

#

x œ u#
v# and u v

#

‹

`u `y

Ê

`u `y

`x `y

œ

œ 0 Ê 0 œ 2u

v v#  u# .

`u `y

œ 0 and

x (4ß2ß1ßc1)

`x `y

`x `y


2y
(2z)

œ  "2 Ê Š ``wy ‹ ¹ z

œ (2)(2)(1)# ˆ "# ‰
(2)(2)# (1)
(1) œ 5 6. y œ uv Ê 1 œ v

`z `y

œ  "1 œ 1 Ê Š ``wy ‹ ¹

#

œ c(2)(2) (1)
(1)d
c1  3(1) d (1) œ 5 Î x œ x(yß z) Ñ y yœy Ä w Ê Š ``wy ‹ œ (b) Œ  Ä z z Ï zœz Ò œ a2xy# b

`w `x `x `y


2v

`v `y

At (uß v) œ ŠÈ2ß 1‹ ,

Ê `u `y

`v `y

œ

œ ˆ uv ‰

" # 1#  ŠÈ2‹

`u `y

`z `y

(4ß2ß1ßc1)

Ê 1

œ 1

œ 0 and

Section 14.10 Taylor's Formula for Two Variables Ê Š `` uy ‹ œ 1 x

r x œ r cos ) 7. Œ  Ä Œ Ê ˆ ``xr ‰) œ cos ); x#
y# œ r# Ê 2x
2y y œ r sin )  ) Ê ``xr œ xr Ê ˆ ``xr ‰ œ È #x #

`y `x

`r `x

œ 2r

and

`y `x

8. If x, y, and z are independent, then ˆ ``wx ‰y z œ ß

`w `x `x `x




`w `y `y `x

`w `z `z `x







`w `t `t `x

œ (2x)(1)
(2y)(0)
(4)(0)
(1) ˆ ``xt ‰ œ 2x
``xt . Thus x
2z
t œ 25 Ê 1
0
Ê ˆ ``wx ‰ œ 2x  1. On the other hand, if x, y, and t are independent, then ˆ ``wx ‰ yßz

œ

Ê 1


`r `x

x
y

y

`w `x `x `x

œ 0 Ê 2x œ 2r

`t `x

œ0 Ê

`t `x

œ 1

yßt


``wy `` xy
``wz `` xz
``wt ``xt œ (2x)(1)
(2y)(0)
4 `` xz
(1)(0) œ 2 `` xz
0 œ 0 Ê `` xz œ  "# Ê ˆ ``wx ‰yßt œ 2x
4 ˆ "# ‰ œ 2x  2.

9. If x is a differentiable function of y and z, then f(xß yß z) œ 0 Ê

`f `x `x `x




`z `x .

2x
4

`f `y `y `x




`f `z `z `x

Thus, x
2z
t œ 25

œ0 Ê

`f `x




`f `y `y `x

œ0

Ê Š `` xy ‹ œ  `` f/f/`` yz . Similarly, if y is a differentiable function of x and z, Š `` yz ‹ œ  `` f/f/`` xz and if z is a z

x

differentiable function of x and y, ˆ `` xz ‰y œ  `` f/f/`` xy . Then Š `` xy ‹ Š `` yz ‹ ˆ `` xz ‰y z

œ Š

` f/` y ˆ ` f/` z ‰ ` f/` x ` f/` z ‹  ` f/` x Š ` f/` y ‹

10. z œ z
f(u) and u œ xy Ê œ x ˆ1
y

df ‰ du

 y ˆx

df ‰ du

`z `x

x

œ 1.

œ1


df ` u du ` x

œ1
y

df du ;

also

`z `y

œ0


df ` u du ` y

œx

df du

so that x

`z `x

y

œ 0 and

`x `y

œ0 Ê

`g `y




œx

11. If x and y are independent, then g(xß yß z) œ 0 Ê

`g `x `x `y

`g `y `y `y




`g `z `z `y




`y Ê Š `` yz ‹ œ  `` g/ g/` z , as claimed. x

12. Let x and y be independent. Then f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and Ê `` xf `` xx
`` yf `` xy
`` zf `` xz
``wf ``wx œ `` xf
`` zf `` xz
``wf ``wx `g `x `g `y `g `z `g `w `g `g `z `g `w `x `x
`y `x
`z `x
`w `x œ `x
`z `x
`w `x œ 0 

`f `z `g `z

`z `x `z `x





`f `w `g `w

`w `x `w `x

œ  `` xf œ

`g `x

Ê ˆ `` xz ‰y œ

c ``xf » c `g `x `f `z » `g `z

`f `w `g » `w `f `w `g » `w

œ

`f `y




`f `z `z `y `z `z `y

 `g

`f `z `z `y





`f `w `g `w


`w `y `w `y

`f `w `w `y

œ 0 and (similarly)

œ  `` yf œ

`g `y

Ê Š ``wy ‹ œ x

`f `z » `g `z `f `z » `g `z

c ``yf c `` gy » `f `w `g » `w

œ

œ0

imply

`g `g `f `w
`x `w `g `f `f `g  `z `w `z `w

`x `y œ 0 `g `g `z `y
`z `y

`y `x

œ 0 and

 ``xf

Likewise, f(xß yß zß w) œ 0, g(xß yß zß w) œ 0 and œ

`z `y

œ

`f `x `f `z

`g `w `g `w

 ``wf

 ``wf

`g `x `g `z

`f `x `f `y `f `z `x `y
`y `y
`z `y `g `w ` w ` y œ 0 imply

Ê


`g `g `f `y
`z `y `g `f `f `g `z `w  `z `w

 `` fz

œ

`f `z `f `z

`g `y `g `w

`g `z `f `g `w `z

 ``yf 

, as claimed.


`f `w `w `y

, as claimed.

14.10 TAYLOR'S FORMULA FOR TWO VARIABLES 1. f(xß y) œ xey Ê fx œ ey , fy œ xey , fxx œ 0, fxy œ ey , fyy œ xey Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d œ 0
x † 1
y † 0
"# ax# † 0
2xy † 1
y# † 0b œ x
xy quadratic approximation;

fxxx œ 0, fxxy œ 0, fxyy œ ey , fyyy œ xey

`g `z `z `y

œ0

919

920

Chapter 14 Partial Derivatives Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (!ß !)
3x# yfxxy (0ß 0)
3xy# fxyy (!ß !)
y$ fyyy (0ß 0)d

œ x
xy
"6 ax$ † 0
3x# y † 0
3xy# † 1
y$ † 0b œ x
xy
#" xy# , cubic approximation 2. f(xß y) œ ex cos y Ê fx œ ex cos y, fy œ ex sin y, fxx œ ex cos y, fxy œ ex sin y, fyy œ ex cos y Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (!ß 0)
"# cx# fxx (!ß !)
2xyfxy (!ß !)
y# fyy (0ß 0)d

œ 1
x † 1
y † 0
"# cx# † 1
2xy † 0
y# † (1)d œ 1
x
"# ax#  y# b , quadratic approximation;

fxxx œ ex cos y, fxxy œ ex sin y, fxyy œ ex cos y, fyyy œ ex sin y Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d œ 1
x
"# ax#  y# b
6" cx$ † 1
3x# y † 0
3xy# † (1)
y$ † 0d œ 1
x
"# ax#  y# b
6" ax$  3xy# b , cubic approximation

3. f(xß y) œ y sin x Ê fx œ y cos x, fy œ sin x, fxx œ y sin x, fxy œ cos x, fyy œ 0 Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (!ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d œ 0
x † 0
y † 0
"# ax# † 0
2xy † 1
y# † 0b œ xy, quadratic approximation;

fxxx œ y cos x, fxxy œ  sin x, fxyy œ 0, fyyy œ 0 Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d œ xy
"6 ax$ † 0
3x# y † 0
3xy# † 0
y$ † 0b œ xy, cubic approximation

4. f(xß y) œ sin x cos y Ê fx œ cos x cos y, fy œ  sin x sin y, fxx œ  sin x cos y, fxy œ  cos x sin y, fyy œ  sin x cos y Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d œ 0
x † 1
y † 0
"# ax# † 0
2xy † 0
y# † 0b œ x, quadratic approximation;

fxxx œ  cos x cos y, fxxy œ sin x sin y, fxyy œ  cos x cos y, fyyy œ sin x sin y Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d

œ x
"6 cx$ † (1)
3x# y † 0
3xy# † (1)
y$ † 0d œ x  6" ax$
3xy# b, cubic approximation

5. f(xß y) œ ex ln (1
y) Ê fx œ ex ln (1
y), fy œ

ex 1
y

, fxx œ ex ln (1
y), fxy œ

ex 1
y

x

, fyy œ  (1
e y)#

Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d

œ 0
x † 0
y † 1
"# cx# † 0
2xy † 1
y# † (1)d œ y
"# a2xy  y# b , quadratic approximation; fxxx œ ex ln (1
y), fxxy œ Ê f(xß y) ¸ quadratic


ex ex 2ex 1
y , fxyy œ  (1
y)# , fyyy œ (1
y)$ " $ # # 6 cx fxxx (0ß 0)
3x yfxxy (!ß 0)
3xy fxyy (0ß 0) $ # # $


y$ fyyy (0ß 0)d

œ y
"2 a2xy  y# b
6" cx † 0
3x y † 1
3xy † (1)
y † 2d

œ y
"# a2xy  y# b
6" a3x# y  3xy#
2y$ b , cubic approximation 4 2 (2x
y
1)# , fxy œ (2x
y
1)# , " # # fyy œ (2x
" y
1)# Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
# cx fxx (0ß 0)
2xyfxy (0ß 0)
y fyy (0ß 0)d œ 0
x † 2
y † 1
"# cx# † (4)
2xy † (2)
y# † (1)d œ 2x
y
"# a4x#  4xy  y# b œ (2x
y)  "# (2x
y)# , quadratic approximation; fxxx œ (2x
16y
1)$ , fxxy œ (2x
8y
1)$ , fxyy œ (2x
4y
1)$ , fyyy œ (2x
2y
1)$ Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d œ (2x
y)  "# (2x
y)#
6" ax$ † 16
3x# y † 8
3xy# † 4
y$ † 2b œ (2x
y)  "# (2x
y)#
3" a8x$
12x# y
6xy#
y# b œ (2x
y)  "# (2x
y)#
3" (2x
y)$ , cubic approximation

6. f(xß y) œ ln (2x
y
1) Ê fx œ

2 2x
y
1

, fy œ

" #x
y
1

, fxx œ

7. f(xß y) œ sin ax#
y# b Ê fx œ 2x cos ax#
y# b , fy œ 2y cos ax#
y# b , fxx œ 2 cos ax#
y# b  4x# sin ax#
y# b , fxy œ 4xy sin ax#
y# b , fyy œ 2 cos ax#
y# b  4y# sin ax#
y# b

Section 14.10 Taylor's Formula for Two Variables Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d

œ 0
x † 0
y † 0
"# ax# † 2
2xy † 0
y# † 2b œ x#
y# , quadratic approximation;

fxxx œ 12x sin ax#
y# b  8x$ cos ax#
y# b , fxxy œ 4y sin ax#
y# b  8x# y cos ax#
y# b , fxyy œ 4x sin ax#
y# b  8xy# cos ax#
y# b , fyyy œ 12y sin ax#
y# b  8y$ cos ax#
y# b Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d œ x#
y#
"6 ax$ † 0
3x# y † 0
3xy# † 0
y$ † 0b œ x#
y# , cubic approximation

8. f(xß y) œ cos ax#
y# b Ê fx œ 2x sin ax#
y# b , fy œ 2y sin ax#
y# b , fxx œ 2 sin ax#
y# b  4x# cos ax#
y# b , fxy œ 4xy cos ax#
y# b , fyy œ 2 sin ax#
y# b  4y# cos ax#
y# b Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d œ 1
x † 0
y † 0
"# cx# † 0
2xy † 0
y# † 0d œ 1, quadratic approximation;

fxxx œ 12x cos ax#
y# b
8x$ sin ax#
y# b , fxxy œ 4y cos ax#
y# b
8x# y sin ax#
y# b , fxyy œ 4x cos ax#
y# b
8xy# sin ax#
y# b , fyyy œ 12y cos ax#
y# b
8y$ sin ax#
y# b Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0)
y$ fyyy (0ß 0)d œ 1
"6 ax$ † 0
3x# y † 0
3xy# † 0
y$ † 0b œ 1, cubic approximation

9. f(xß y) œ

" 1xy

Ê fx œ

" (1  x  y)#

œ fy , fxx œ

Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)


2 (1  x  y)$ " # # cx fxx (0ß 0)

œ fxy œ fyy
2xyfxy (0ß 0)
y# fyy (0ß 0)d

œ 1
x † 1
y † 1
"# ax# † 2
2xy † 2
y# † 2b œ 1
(x
y)
ax#
2xy
y# b œ 1
(x
y)
(x
y)# , quadratic approximation; fxxx œ Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)


6 œ fxxy œ fxyy œ fyyy (1  x  y)% # 3xy fxyy (0ß 0)
y$ fyyy (0ß 0)d $

œ 1
(x
y)
(x
y)#
"6 ax$ † 6
3x# y † 6
3xy# † 6
y † 6b

œ 1
(x
y)
(x
y)#
ax$
3x# y
3xy#
y$ b œ 1
(x
y)
(x
y)#
(x
y)$ , cubic approximation 10. f(xß y) œ fxy œ

" 1  x  y
xy

1 ("  x  y
xy)#

Ê fx œ

, fyy œ

1y (1  x  y
xy)#

, fy œ

1x ("  x  y
xy)#

, fxx œ

2(1  y)# (1  x  y
xy)$

,

#

2("  x) (1  x  y
xy)$

Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d

œ 1
x † 1
y † 1
"# ax# † 2
2xy † 1
y# † 2b œ 1
x
y
x#
xy
y# , quadratic approximation; fxxx œ

6(1  y)$ (1  x  y
xy)%

, fxxy œ

[4(1  x  y
xy)
6(1  y)(1  x)](1  y) (1  x  y
xy)%

,

$

 x) [4(1  x  y
xy)
6(1  x)(1  y)](1  x) , fyyy œ (1 6(1 (1  x  y
xy)% x  y
xy)% Ê f(xß y) ¸ quadratic
"6 cx$ fxxx (0ß 0)
3x# yfxxy (!ß 0)
3xy# fxyy (0ß 0) œ 1
x
y
x#
xy
y#
"6 ax$ † 6
3x# y † 2
3xy# † 2
y$ † 6b # # $ # # $

fxyy œ


y$ fyyy (0ß 0)d

œ 1
x
y
x
xy
y
x
x y
xy
y , cubic approximation 11. f(xß y) œ cos x cos y Ê fx œ  sin x cos y, fy œ  cos x sin y, fxx œ  cos x cos y, fxy œ sin x sin y, fyy œ  cos x cos y Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d œ 1
x † 0
y † 0
"# cx# † (1)
2xy † 0
y# † (1)d œ 1 

x# #



y# #

, quadratic approximation. Since all partial

derivatives of f are products of sines and cosines, the absolute value of these derivatives is less than or equal to 1 Ê E(xß y) Ÿ "6 c(0.1)$
3(0.1)$
3(0.1)$
0.1)$ d Ÿ 0.00134. 12. f(xß y) œ ex sin y Ê fx œ ex sin y, fy œ ex cos y, fxx œ ex sin y, fxy œ ex cos y, fyy œ ex sin y Ê f(xß y) ¸ f(0ß 0)
xfx (0ß 0)
yfy (0ß 0)
"# cx# fxx (0ß 0)
2xyfxy (0ß 0)
y# fyy (0ß 0)d

œ 0
x † 0
y † 1
"# ax# † 0
2xy † 1
y# † 0b œ y
xy , quadratic approximation. Now, fxxx œ ex sin y,

fxxy œ ex cos y, fxyy œ ex sin y, and fyyy œ ex cos y. Since kxk Ÿ 0.1, kex sin yk Ÿ ke0Þ1 sin 0.1k ¸ 0.11 and kex cos yk Ÿ ke0Þ1 cos 0.1k ¸ 1.11. Therefore,

921

922

Chapter 14 Partial Derivatives

E(xß y) Ÿ

" 6

c(0.11)(0.1)$
3(1.11)(0.1)$
3(0.11)(0.1)$
(1.11)(0.1)$ d Ÿ 0.000814.

CHAPTER 14 PRACTICE EXERCISES 1. Domain: All points in the xy-plane Range: z   0 Level curves are ellipses with major axis along the y-axis and minor axis along the x-axis.

2. Domain: All points in the xy-plane Range: 0  z  _ Level curves are the straight lines x
y œ ln z with slope 1, and z  0.

3. Domain: All (xß y) such that x Á 0 and y Á 0 Range: z Á 0 Level curves are hyperbolas with the x- and y-axes as asymptotes.

4. Domain: All (xß y) so that x#  y   0 Range: z   0 Level curves are the parabolas y œ x#  c, c   0.

5. Domain: All points (xß yß z) in space Range: All real numbers Level surfaces are paraboloids of revolution with the z-axis as axis.

Chapter 14 Practice Exercises 6. Domain: All points (xß yß z) in space Range: Nonnegative real numbers Level surfaces are ellipsoids with center (0ß 0ß 0).

7. Domain: All (xß yß z) such that (xß yß z) Á (0ß !ß 0) Range: Positive real numbers Level surfaces are spheres with center (0ß 0ß 0) and radius r
0.

8. Domain: All points (xß yß z) in space Range: (0ß 1] Level surfaces are spheres with center (0ß 0ß 0) and radius r
0.

9.

lim

Ðxß yÑ Ä Ð1ß ln 2Ñ

ey cos x œ eln 2 cos 1 œ (2)(1) œ 2 2
y

10.

lim Ðxß yÑ Ä Ð0ß 0Ñ x
cos y

11.

lim # # Ðx ß y Ñ Ä Ð 1 ß 1 Ñ x  y xÁ „y

12.

13.

14.

xy

œ

œ

2
0 0
cos 0

œ2 xy

lim Ðxß yÑ Ä Ð1ß 1Ñ (x  y)(x
y) xÁ „y

œ

1

lim Ðxß yÑ Ä Ð1ß 1Ñ x
y

(xy  1) ax# y#
xy
1b xy  1

lim

x$ y$  1 xy  1

lim

ln kx  y  zk œ ln k1  (1)  ek œ ln e œ 1

Ðx ß y Ñ Ä Ð 1 ß 1 Ñ

P Ä Ð1 ß
1 ß e Ñ

lim

P Ä Ð1 ß
1 ß
1 Ñ

œ

lim

Ð x ß y Ñ Ä Ð 1 ß 1Ñ

œ

œ

lim

Ðxß yÑ Ä Ð1ß 1Ñ

" 1
1

œ

" #

ax# y#  xy  1b œ 1# † 1#  1 † 1  1 œ 3

tan" (x  y  z) œ tan" (1  (1)  (1)) œ tan" (1) œ  14

15. Let y œ kx# , k Á 1. Then

y

lim # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x  y # yÁx

œ

kx#

lim # # axß kx# b Ä Ð0ß 0Ñ x  kx

œ

k 1  k#

which gives different limits for

œ

1
k# k

which gives different limits for

different values of k Ê the limit does not exist. 16. Let y œ kx, k Á 0. Then

lim

Ðxß yÑ Ä Ð0ß 0Ñ xy Á 0

x#
y# xy

œ

lim

(xß kxÑ Ä Ð0ß 0Ñ

x#
(kx)# x(kx)

923

924

Chapter 14 Partial Derivatives

different values of k Ê the limit does not exist. 17. Let y œ kx. Then

x#  y#

œ

lim # # Ðx ß y Ñ Ä Ð 0 ß 0 Ñ x
y

x#  k# x# x #
k# x#

1  k# 1
k#

œ

which gives different limits for different values

of k Ê the limit does not exist so f(0ß 0) cannot be defined in a way that makes f continuous at the origin. sin (x  y) Ðx ß y Ñ Ä Ð 0 ß 0 Ñ k x
y k

18. Along the x-axis, y œ 0 and

œ lim

lim

sin x kx k

xÄ0

œœ

1, x
0 , so the limit fails to exist ", x  0

Ê f is not continuous at (0ß 0). 19.

`g `r

œ cos )  sin ),

20.

`f `x

œ

" #

Š x# 2x
y# ‹ 

`f `y

œ

" #

Š x# 2y
y# ‹ 

21.

`f ` R"

œ  R"# , "

`f ` R#

`g `)

œ r sin )  r cos )

y ‹ x# y # 1
ˆx‰

Š

Š 1x ‹ y #

1
ˆx‰

œ  R"# ,

œ

x x#
y#



y x#
y#

œ

xy x#
y#

œ

y x#
y#



x x#
y#

œ

x
y x#
y#

`f ` R$

#

,

œ  R"# $

22. hx (xß yß z) œ 21 cos (21x  y  3z), hy (xß yß z) œ cos (21x  y  3z), hz (xß yß z) œ 3 cos (21x  y  3z) 23.

`P `n

œ

RT V

,

`P `R

œ

nT V

`P `T

,

œ

nR V

,

`P `V

œ  nRT V#

24. fr (rß jß Tß w) œ  2r"# j É 1Tw , fj (rß jß Tß w) œ  #r"j# É

25.

œ

" 4rj

É T1"w œ

`g `x

œ

" y

`g `y

,

" 4rjT

œ1

, fT (rß jß Tß w) œ ˆ #"rj ‰ Š È"1w ‹ Š 2È" T ‹

T 1w

É 1Tw , fw (rß jß Tß w) œ ˆ #"rj ‰ É T1 ˆ "# w$Î# ‰ œ  4r"jw É 1Tw

x y#

Ê

` #g ` x#

œ 0,

` #g ` y#

œ

2x y$

,

` #g ` y` x

œ

` #g ` x` y

œ  y"#

26. gx (xß y) œ ex  y cos x, gy (xß y) œ sin x Ê gxx (xß y) œ ex  y sin x, gyy (xß y) œ 0, gxy (xß y) œ gyx (xß y) œ cos x 27.

`f `x

œ 1  y  15x# 

2x x#
1

,

`f `y

œx Ê

` #f ` x#

œ 30x 

22x# ax #
1 b #

,

` #f ` y#

œ 0,

` #f ` y` x

œ

` #f ` x` y

œ1

28. fx (xß y) œ 3y, fy (xß y) œ 2y  3x  sin y  7ey Ê fxx (xß y) œ 0, fyy (xß y) œ 2  cos y  7ey , fxy (xß y) œ fyx (xß y) œ 3 29.

`w `x

Ê Ê 30.

`w `x

Ê Ê 31.

`w `x

Ê

œ y cos (xy  1),

`w `y

œ x cos (xy  1),

dx dt

œ et ,

dy dt

dw t ˆ " ‰ dt œ [y cos (xy  1)]e  [x cos (xy  1)] t
1 ; dw ¸ ˆ " ‰ dt tœ0 œ 0 † 1  [1 † (1)] 0
1 œ 1

œ ey ,

`w `y

œ xey  sin z,

dw y "Î#  axey  dt œ e t dw ¸ dt tœ1 œ 1 † 1  (2 † 1

œ 2 cos (2x  y), `w `r

`w `y

`w `z

œ y cos z  sin z,

sin zb ˆ1 

"‰ t

dx dt

œ

" t
1

t œ 0 Ê x œ 1 and y œ 0

œ t"Î# ,

dy dt

œ 1  "t ,

dz dt

œ1

 (y cos z  sin z)1; t œ 1 Ê x œ 2, y œ 0, and z œ 1

 0)(2)  (0  0)1 œ 5

œ  cos (2x  y),

`x `r

œ 1,

`x `s

œ cos s,

`y `r

œ s,

`y `s

œr

œ [2 cos (2x  y)](1)  [ cos (2x  y)](s); r œ 1 and s œ 0 Ê x œ 1 and y œ 0

Chapter 14 Practice Exercises `w ¸ ` r Ð1ß0Ñ `w ¸ ` s Ð1ß0Ñ

Ê Ê 32.

33.

`w `u `w `v

œ

`f `x

œ y  z,

œ

Ê

`w `x

œ

`w `s

œ [2 cos (2x  y)](cos s)  [ cos (2x  y)](r)

œ (2 cos 21)(cos 0)  (cos 21)(1) œ 2  1

`x `u `x `v

œ ˆ 1
x x# 

" ‰ u x#
1 a2e cos vb ; u œ v œ 0 Ê " ‰ `w ¸ u x#
1 a2e sin vb Ê ` v Ð0ß0Ñ œ

`f `y

`f `z

œ ˆ 1
x x# 

œ x  z,

œ y  x,

dx dt

df dt œ (y  z)(sin t)  (x  z)(cos df ¸ dt tœ1 œ (sin 1  cos 2)(sin 1) 

Ê 34.

dw dx dw dx

œ (2 cos 21)  (cos 21)(0) œ 2;

dw ` s ds ` x

œ (5)

dw ds

and

`w `y

œ

dw ` s ds ` y

œ  sin t,

œ ˆ 52  "5 ‰ (2) œ

1
y cos xy 2y  x cos xy

dz dt

œ 2 sin 2t

(cos 1  cos 2)(cos 1)  2(sin 1  cos 1)(sin 2)

œ (1)

dw ds

œ

`w `x

Ê

dw ds

5

`w `y

œ5

œ

dy dx ¹ Ð0ß1Ñ

1
" 2

dw ds

5

dw ds

dy dx ¹ Ð0ßln 2Ñ

1  y cos xy œ  FFxy œ  2y  x cos xy

dy dx

xby

2y
e œ  FFxy œ  2x
exby

ß

1‰ 4

#

œ

i

j Ê f increases most rapidly in the direction u œ 

È2 #

i

u œ  È12 i 

1 È2

1 È2

i

1 È2

" È2

œ

È2 #

and decreases most

38. ™ f œ 2xec2y i  2x# ec2y j Ê ™ f k Ð1ß0Ñ œ #i  #j Ê k ™ f k œ È2#  (2)# œ 2È2; u œ Ê f increases most rapidly in the direction u œ

#

œ  "# i  "# j Ê k ™ f k œ Ɉ "# ‰  ˆ "# ‰ œ

È2 # j È È È È rapidly in the direction u œ #2 i  #2 j ; (Du f)P! œ k ™ f k œ #2 and (Dcu f)P! œ  #2 ; 7 u" œ kvvk œ È33i #
4j4# œ 35 i  45 j Ê (Du" f)P! œ ™ f † u" œ ˆ "# ‰ ˆ 35 ‰  ˆ "# ‰ ˆ 45 ‰ œ  10
™f k™f k

È2 #

dy dx

2 œ  2 ln0
2
2 œ (ln 2  1)

37. ™ f œ ( sin x cos y)i  (cos x sin y)j Ê ™ f k ˆ 14 È2 #

œ0

œ 1

36. F(xß y) œ 2xy  ex
y  2 Ê Fx œ 2y  ex
y and Fy œ 2x  ex
y Ê

uœ

;

t)  2(y  x)(sin 2t); t œ 1 Ê x œ cos 1, y œ sin 1, and z œ cos 2

Ê at (xß y) œ (!ß 1) we have

Ê at (xß y) œ (!ß ln 2) we have

2 5

ˆ 52  "5 ‰ (0) œ 0

œ cos t,

dy dt

`w ¸ ` u Ð0ß0Ñ

xœ2 Ê

35. F(xß y) œ 1  x  y#  sin xy Ê Fx œ 1  y cos xy and Fy œ 2y  x cos xy Ê œ

™f k™f k

œ

1 È2

i

1 È2

j

j and decreases most rapidly in the direction

j ; (Du f)P! œ k ™ f k œ 2È2 and (Dcu f)P! œ 2È2 ; u" œ

v kv k

œ

i
j È 1#
1#

œ

1 È2

i

1 È2

j

Ê (Du" f)P! œ ™ f † u" œ (2) Š È" ‹  (2) Š È" ‹ œ 0 2 2 2 3 6 39. ™ f œ Š 2x
3y
6z ‹ i  Š 2x
3y
6z ‹ j  Š 2x
3y
6z ‹ k Ê ™ f k Ð
1ß
1ß1Ñ œ 2i  3j  6k ;

uœ

™f k™f k

œ

2i
3j
6k È 2#
3#
6#

œ

2 7

i  37 j  67 k Ê f increases most rapidly in the direction u œ

2 7

i  37 j  67 k and

decreases most rapidly in the direction u œ  27 i  37 j  67 k ; (Du f)P! œ k ™ f k œ 7, (Du f)P! œ 7; u" œ

v kv k

œ

925

2 7

i  37 j  67 k Ê (Du" f)P! œ (Du f)P! œ 7

40. ™ f œ (2x  3y)i  (3x  2)j  (1  2z)k Ê ™ f k Ð0ß0ß0Ñ œ 2j  k ; u œ rapidly in the direction u œ

2 È5

j

" È5

™f k™f k

œ

2 È5

j

" È5

k Ê f increases most

k and decreases most rapidly in the direction u œ  È25 j 

(Du f)P! œ k ™ f k œ È5 and (Du f)P! œ È5 ; u" œ

v kv k

œ

i
j
k È 1#
1#
1#

Ê (Du" f)P! œ ™ f † u" œ (0) Š È"3 ‹  (2) Š È"3 ‹  (1) Š È"3 ‹ œ

3 È3

œ

" È3

œ È3

i

" È3

j

" È3

k

" È5

k;

;

926

Chapter 14 Partial Derivatives

41. r œ (cos 3t)i  (sin 3t)j  3tk Ê v(t) œ (3 sin 3t)i  (3 cos 3t)j  3k Ê v ˆ 13 ‰ œ 3j  3k Ê u œ  È"2 j 

" È2

k ; f(xß yß z) œ xyz Ê ™ f œ yzi  xzj  xyk ; t œ

Ê ™ f k Ð
1ß0ß1Ñ œ 1j Ê ™ f † u œ (1j) † Š È"2 j 

" È2

k‹ œ

1 3

yields the point on the helix (1ß 0ß 1)

1 È2

42. f(xß yß z) œ xyz Ê ™ f œ yzi  xzj  xyk ; at (1ß 1ß 1) we get ™ f œ i  j  k Ê the maximum value of Du f k œ k ™ f k œ È3 Ð1ß1ß1Ñ

43. (a) Let ™ f œ ai  bj at (1ß 2). The direction toward (2ß 2) is determined by v" œ (2  1)i  (2  2)j œ i œ u so that ™ f † u œ 2 Ê a œ 2. The direction toward (1ß 1) is determined by v# œ (1  1)i  (1  2)j œ j œ u so that ™ f † u œ 2 Ê b œ 2 Ê b œ 2. Therefore ™ f œ 2i  2j ; fx a1, 2b œ fy a1, 2b œ 2. (b) The direction toward (4ß 6) is determined by v$ œ (4  1)i  (6  2)j œ 3i  4j Ê u œ 35 i  45 j Ê ™f†uœ

14 5

.

44. (a) True

(b) False

(c) True

(d) True

45. ™ f œ 2xi  j  2zk Ê ™ f k Ð0ß
1ß
1Ñ œ j  2k , ™ f k Ð0ß0ß0Ñ œ j , ™ f k Ð0ß
1ß1Ñ œ j  2k

46. ™ f œ 2yj  2zk Ê ™ f k Ð2ß2ß0Ñ œ 4j , ™ f k Ð2ß
2ß0Ñ œ 4j , ™ f k Ð2ß0ß2Ñ œ 4k , ™ f k Ð2ß0ß
2Ñ œ 4k

47. ™ f œ 2xi  j  5k Ê ™ f k Ð2ß
1ß1Ñ œ 4i  j  5k Ê Tangent Plane: 4(x  2)  (y  1)  5(z  1) œ 0 Ê 4x  y  5z œ 4; Normal Line: x œ 2  4t, y œ 1  t, z œ 1  5t 48. ™ f œ 2xi  2yj  k Ê ™ f k Ð1ß1ß2Ñ œ 2i  2j  k Ê Tangent Plane: 2(x  1)  2(y  1)  (z  2) œ 0 Ê 2x  2y  z  6 œ 0; Normal Line: x œ 1  2t, y œ 1  2t, z œ 2  t 49.

`z `x

œ

2x x#
y#

Ê

`z ¸ ` x Ð0ß1ß0Ñ

œ 0 and

`z `y

œ

2y x#
y#

2(y  1)  (z  0) œ 0 or 2y  z  2 œ 0

Ê

`z ` y ¹ Ð0ß1ß0Ñ

œ 2; thus the tangent plane is

Chapter 14 Practice Exercises 50.

`z `x

œ 2x ax#  y# b

#

`z ¸ ` x ˆ1ß1ß 12 ‰

Ê

œ  #" and

`z `y

œ 2y ax#  y# b

#

Ê

`z ` y ¹ ˆ1ß1ß 1 ‰ 2

927

œ  #" ; thus the tangent

plane is  "# (x  1)  "# (y  1)  ˆz  "# ‰ œ 0 or x  y  2z  3 œ 0 51. ™ f œ ( cos x)i  j Ê ™ f k Ð1ß1Ñ œ i  j Ê the tangent line is (x  1)  (y  1) œ 0 Ê x  y œ 1  1; the normal line is y  1 œ 1(x  1) Ê y œ x  1  1

52. ™ f œ xi  yj Ê ™ f k Ð1ß2Ñ œ i  2j Ê the tangent line is (x  1)  2(y  2) œ 0 Ê y œ

" #

x  3# ; the normal

line is y  2 œ 2(x  1) Ê y œ 2x  4

53. Let f(xß yß z) œ x#  2y  2z  4 and g(xß yß z) œ y  1. Then ™ f œ 2xi  2j  2kk a1 1 12 b œ 2i  2j  2k â â â i j kâ â â and ™ g œ j Ê ™ f ‚ ™ g œ â 2 2 2 â œ 2i  2k Ê the line is x œ 1  2t, y œ 1, z œ "#  2t â â â0 " 0â ß ß

54. Let f(xß yß z) œ x  y#  z  2 and g(xß yß z) œ y  1. Then ™ f œ i  2yj  kk a 12 1 12 b œ i  2j  k and â â â i j kâ â â ™ g œ j Ê ™ f ‚ ™ g œ â 1 2 1 â œ i  k Ê the line is x œ "#  t, y œ 1, z œ "#  t â â â0 " 0â ß ß

55. f ˆ 14 ß 14 ‰ œ

" #

, fx ˆ 14 ß 14 ‰ œ cos x cos yk Ð1Î4ß1Î4Ñ œ

Ê L(xß y) œ

" #

 "# ˆx  14 ‰  "# ˆy  14 ‰ œ

" #

" # " #

, fy ˆ 14 ß 14 ‰ œ  sin x sin yk Ð1Î4ß1Î4Ñ œ  "#

 x  "# y; fxx (xß y) œ  sin x cos y, fyy (xß y) œ  sin x cos y, and

fxy (xß y) œ  cos x sin y. Thus an upper bound for E depends on the bound M used for kfxx k , kfxy k , and kfyy k . With M œ

È2 #

we have kE(xß y)k Ÿ

with M œ 1, kE(xß y)k Ÿ

" #

" #

Š

È2 ˆ¸ # ‹ x

#  14 ¸  ¸y  14 ¸‰ Ÿ

# (1) ˆ¸x  14 ¸  ¸y  14 ¸‰ œ

" #

È2 4

(0.2)# Ÿ 0.0142;

(0.2)# œ 0.02.

56. f(1ß 1) œ 0, fx (1ß 1) œ yk Ð1ß1Ñ œ 1, fy (1ß 1) œ x  6yk Ð1ß1Ñ œ 5 Ê L(xß y) œ (x  1)  5(y  1) œ x  5y  4; fxx (xß y) œ 0, fyy (xß y) œ 6, and fxy (xß y) œ 1 Ê maximum of kfxx k , kfyy k , and kfxy k is 6 Ê M œ 6 Ê kE(xß y)k Ÿ

" #

(6) akx  1k  ky  1kb# œ

" #

(6)(0.1  0.2)# œ 0.27

57. f(1ß 0ß 0) œ 0, fx (1ß 0ß 0) œ y  3zk Ð1ß0ß0Ñ œ 0, fy (1ß 0ß 0) œ x  2zk Ð1ß0ß0Ñ œ 1, fz (1ß 0ß 0) œ 2y  3xk Ð1ß0ß0Ñ œ 3 Ê L(xß yß z) œ 0(x  1)  (y  0)  3(z  0) œ y  3z; f(1ß 1ß 0) œ 1, fx (1ß 1ß 0) œ 1, fy (1ß 1ß 0) œ 1, fz ("ß "ß !) œ 1 Ê L(xß yß z) œ 1  (x  1)  (y  1)  1(z  0) œ x  y  z  1 58. f ˆ0ß !ß 14 ‰ œ 1, fx ˆ!ß 0ß 14 ‰ œ È2 sin x sin (y  z)¹

ˆ0ß0ß 1 ‰

œ 0, fy ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y  z)¹

4

fz ˆ!ß 0ß 14 ‰ œ È2 cos x cos (y  z)¹

ˆ0ß0ß 1 ‰

È2 #

œ 1 Ê L(xß yß z) œ 1  1(y  0)  1 ˆz  14 ‰ œ 1  y  z 

Ê L(xß yß z) œ

È2 È2 È2 ˆ1 1 ‰ ˆ1 1 ‰ # , fy 4 ß 4 ß 0 œ # , fz 4 ß 4 ß 0 œ # È È È È È  #2 ˆy  14 ‰  #2 (z  0) œ #2  #2 x  #2

, fx ˆ 14 ß 14 ß 0‰ œ  È2 #



È2 #

ˆx  14 ‰

œ 1,

4

4

f ˆ 14 ß 14 ß 0‰ œ

ˆ0ß0ß 1 ‰

y

È2 #

z

1 4

;

928

Chapter 14 Partial Derivatives

59. V œ 1r# h Ê dV œ 21rh dr  1r# dh Ê dVk Ð1Þ5ß5280Ñ œ 21(1.5)(5280) dr  1(1.5)# dh œ 15,8401 dr  2.251 dh. You should be more careful with the diameter since it has a greater effect on dV. 60. df œ (2x  y) dx  (x  2y) dy Ê df k Ð1ß2Ñ œ 3 dy Ê f is more sensitive to changes in y; in fact, near the point (1ß 2) a change in x does not change f. 61. dI œ

" R

dV 

V R#

" 100

dR Ê dI¸ Ð24ß100Ñ œ

dV 

24 100#

dR Ê dI¸ dVœ
1ßdRœ
20 œ 0.01  (480)(.0001) œ 0.038,

" ‰ 20 ‰ or increases by 0.038 amps; % change in V œ (100) ˆ 24 ¸ 4.17%; % change in R œ ˆ 100 (100) œ 20%;

Iœ

24 100

œ 0.24 Ê estimated % change in I œ

dI I

‚ 100 œ

0.038 0.24

‚ 100 ¸ 15.83% Ê more sensitive to voltage change.

62. A œ 1ab Ê dA œ 1b da  1a db Ê dAk Ð10ß16Ñ œ 161 da  101 db; da œ „ 0.1 and db œ „ 0.1 ¸ ¸ 2.61 ¸ Ê dA œ „ 261(0.1) œ „ 2.61 and A œ 1(10)(16) œ 1601 Ê ¸ dA A ‚ 100 œ 1601 ‚ 100 ¸ 1.625% 63. (a) y œ uv Ê dy œ v du  u dv; percentage change in u Ÿ 2% Ê kduk Ÿ 0.02, and percentage change in v Ÿ 3% Ê kdvk Ÿ 0.03;

dy y

Ÿ 2%  3% œ 5% (b) z œ u  v Ê dzz œ

œ

v du
u dv uv

du
dv u
v

œ

œ

du u
v

Ê ¸ dzz ‚ 100¸ Ÿ ¸ du u ‚ 100  64. C œ Ê

dv v



du u



dv u
v

Þ

Þ

Þ

Þ

Þ

Þ

Ÿ

¸ du Ê ¹ dy y ‚ 100¹ œ u ‚ 100  du u



dv v

dv v

¸ ¸ dv ¸ ‚ 100¸ Ÿ ¸ du u ‚ 100  v ‚ 100

(since u
0, v
0)

‚ 100¸ œ ¹ dy y ‚ 100¹

(0.425)(7) 7 71.84w0 425 h0 725 Ê Cw œ 71.84w1 425 h0 725 2.975 5.075 dC œ 71.84w 1 425 h0 725 dw  71.84w0 425 h1 725 Þ

dv v

Þ

and Ch œ

(0.725)(7) 71.84w0 425 h1 725 Þ

Þ

dh; thus when w œ 70 and h œ 180 we have

dCk Ð70ß180Ñ ¸ (0.00000225) dw  (0.00000149) dh Ê 1 kg error in weight has more effect 65. fx (xß y) œ 2x  y  2 œ 0 and fy (xß y) œ x  2y  2 œ 0 Ê x œ 2 and y œ 2 Ê (2ß 2) is the critical point; # fxx (2ß 2) œ 2, fyy (#ß 2) œ 2, fxy (#ß 2) œ 1 Ê fxx fyy  fxy œ 3
0 and fxx
0 Ê local minimum value of f(#ß 2) œ 8 66. fx (xß y) œ 10x  4y  4 œ 0 and fy (xß y) œ 4x  4y  4 œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1) is the critical point; # fxx (0ß 1) œ 10, fyy (0ß 1) œ 4, fxy (0ß 1) œ 4 Ê fxx fyy  fxy œ 56  0 Ê saddle point with f(0ß 1) œ 2 67. fx (xß y) œ 6x#  3y œ 0 and fy (xß y) œ 3x  6y# œ 0 Ê y œ 2x# and 3x  6 a4x% b œ 0 Ê x a1  8x$ b œ 0 Ê x œ 0 and y œ 0, or x œ  "# and y œ  "# Ê the critical points are (0ß 0) and ˆ "# ß  "# ‰ . For (!ß !):

# œ 9  0 Ê saddle point with fxx (!ß !) œ 12xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 12yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 Ê fxx fyy  fxy # f(0ß 0) œ 0. For ˆ "# ß  "# ‰: fxx œ 6, fyy œ 6, fxy œ 3 Ê fxx fyy  fxy œ 27
0 and fxx  0 Ê local maximum " "‰ " ˆ value of f  # ß  # œ 4

68. fx (xß y) œ 3x#  3y œ 0 and fy (xß y) œ 3y#  3x œ 0 Ê y œ x# and x%  x œ 0 Ê x ax$  1b œ 0 Ê the critical points are (0ß 0) and (1ß 1) . For (!ß !): fxx (!ß !) œ 6xk Ð0ß0Ñ œ 0, fyy (!ß !) œ 6yk Ð0ß0Ñ œ 0, fxy (!ß 0) œ 3 # Ê fxx fyy  fxy œ 9  0 Ê saddle point with f(0ß 0) œ 15. For (1ß 1): fxx (1ß 1) œ 6, fyy (1ß 1) œ 6, fxy (1ß 1) œ 3 # Ê fxx fyy  fxy œ 27
0 and fxx
0 Ê local minimum value of f(1ß 1) œ 14

69. fx (xß y) œ 3x#  6x œ 0 and fy (xß y) œ 3y#  6y œ 0 Ê x(x  2) œ 0 and y(y  2) œ 0 Ê x œ 0 or x œ 2 and y œ 0 or y œ 2 Ê the critical points are (0ß 0), (0ß 2), (2ß 0), and (2ß 2) . For (!ß !): fxx (!ß !) œ 6x  6k Ð0ß0Ñ # œ 6, fyy (!ß !) œ 6y  6k Ð0ß0Ñ œ 6, fxy (!ß 0) œ 0 Ê fxx fyy  fxy œ 36  0 Ê saddle point with f(0ß 0) œ 0. For # (0ß 2): fxx (!ß 2) œ 6, fyy (0ß #) œ 6, fxy (!ß 2) œ 0 Ê fxx fyy  fxy œ 36
0 and fxx
0 Ê local minimum value of

Chapter 14 Practice Exercises

929

# f(!ß 2) œ 4. For (#ß 0): fxx (2ß 0) œ 6, fyy (#ß 0) œ 6, fxy (2ß 0) œ 0 Ê fxx fyy  fxy œ 36
0 and fxx  0

Ê local maximum value of f(2ß 0) œ 4. For (2ß 2): fxx (2ß 2) œ 6, fyy (2ß 2) œ 6, fxy (2ß 2) œ 0 # Ê fxx fyy  fxy œ 36  0 Ê saddle point with f(2ß 2) œ 0. 70. fx (xß y) œ 4x$  16x œ 0 Ê 4x ax#  4b œ 0 Ê x œ 0, 2, 2; fy (xß y) œ 6y  6 œ 0 Ê y œ 1. Therefore the critical points are (0ß 1), (2ß 1), and (2ß 1). For (!ß 1): fxx (!ß 1) œ 12x#  16k Ð0ß1Ñ œ 16, fyy (!ß 1) œ 6, fxy (!ß 1) œ 0 # Ê fxx fyy  fxy œ 96  0 Ê saddle point with f(0ß 1) œ 3. For (2ß 1): fxx (2ß 1) œ 32, fyy (2ß 1) œ 6, # fxy (2ß 1) œ 0 Ê fxx fyy  fxy œ 192
0 and fxx
0 Ê local minimum value of f(2ß 1) œ 19. For (#ß 1): # fxx (2ß 1) œ 32, fyy (#ß 1) œ 6, fxy (2ß 1) œ 0 Ê fxx fyy  fxy œ 192
0 and fxx
0 Ê local minimum value of

f(2ß 1) œ 19. 71. (i)

On OA, f(xß y) œ f(0ß y) œ y#  3y for 0 Ÿ y Ÿ 4 Ê f w (!ß y) œ 2y  3 œ 0 Ê y œ  3# . But ˆ!ß  3# ‰

is not in the region. Endpoints: f(0ß 0) œ 0 and f(0ß 4) œ 28. (ii) On AB, f(xß y) œ f(xß x  4) œ x#  10x  28 for 0 Ÿ x Ÿ 4 Ê f w (xß x  4) œ 2x  10 œ 0 Ê x œ 5, y œ 1. But (5ß 1) is not in the region. Endpoints: f(4ß 0) œ 4 and f(!ß 4) œ 28. (iii) On OB, f(xß y) œ f(xß 0) œ x#  3x for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x  3 Ê x œ critical point with f ˆ 3# ß !‰ œ  94 .

3 #

and y œ 0 Ê ˆ 3# ß 0‰ is a

Endpoints: f(0ß 0) œ 0 and f(%ß 0) œ 4. (iv) For the interior of the triangular region, fx (xß y) œ 2x  y  3 œ 0 and fy (xß y) œ x  2y  3 œ 0 Ê x œ 3 and y œ 3. But (3ß 3) is not in the region. Therefore the absolute maximum is 28 at (0ß 4) and the absolute minimum is  94 at ˆ 3# ß !‰ .

On OA, f(xß y) œ f(0ß y) œ y#  4y  1 for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2y  4 œ 0 Ê y œ 2 and x œ 0. But (0ß 2) is not in the interior of OA. Endpoints: f(0ß 0) œ 1 and f(0ß 2) œ 5. (ii) On AB, f(xß y) œ f(xß 2) œ x#  2x  5 for 0 Ÿ x Ÿ 4 Ê f w (xß 2) œ 2x  2 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 4. Endpoints: f(4ß 2) œ 13 and f(!ß 2) œ 5. (iii) On BC, f(xß y) œ f(4ß y) œ y#  4y  9 for 0 Ÿ y Ÿ 2 Ê f w (4ß y) œ 2y  4 œ 0 Ê y œ # and x œ 4. But (4ß 2) is not in the interior of BC. Endpoints: f(4ß 0) œ 9 and f(%ß 2) œ 13. (iv) On OC, f(xß y) œ f(xß 0) œ x#  2x  1 for 0 Ÿ x Ÿ 4 Ê f w (xß 0) œ 2x  2 œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 0. Endpoints: f(0ß 0) œ 1 and f(4ß 0) œ 9. (v) For the interior of the rectangular region, fx (xß y) œ 2x  2 œ 0 and fy (xß y) œ 2y  4 œ 0 Ê x œ 1 and y œ 2. But (1ß 2) is not in the interior of the region. Therefore the absolute maximum is 13 at (4ß 2) and the absolute minimum is 0 at (1ß 0).

72. (i)

930 73. (i)

Chapter 14 Partial Derivatives On AB, f(xß y) œ f(2ß y) œ y#  y  4 for 2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y  1 Ê y œ "# and x œ 2 Ê ˆ2ß "# ‰ is an interior critical point in AB

with f ˆ2ß "# ‰ œ  17 4 . Endpoints: f(2ß 2) œ 2 and

f(2ß 2) œ 2. On BC, f(xß y) œ f(xß 2) œ 2 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 0 Ê no critical points in the interior of BC. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 2. (iii) On CD, f(xß y) œ f(2ß y) œ y#  5y  4 for 2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2y  5 œ 0 Ê y œ 5# and x œ 2. But ˆ#ß 5# ‰ is not in the region. (ii)

Endpoints: f(2ß 2) œ 18 and f(2ß 2) œ 2. (iv) On AD, f(xß y) œ f(xß 2) œ 4x  10 for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 4 Ê no critical points in the interior of AD. Endpoints: f(2ß 2) œ 2 and f(2ß 2) œ 18. (v) For the interior of the square, fx (xß y) œ y  2 œ 0 and fy (xß y) œ 2y  x  3 œ 0 Ê y œ 2 and x œ 1 Ê (1ß 2) is an interior critical point of the square with f(1ß 2) œ 2. Therefore the absolute maximum "‰ ˆ is 18 at (2ß 2) and the absolute minimum is  17 4 at #ß # . On OA, f(xß y) œ f(0ß y) œ 2y  y# for 0 Ÿ y Ÿ 2 Ê f w (!ß y) œ 2  2y œ 0 Ê y œ 1 and x œ 0 Ê (!ß 1) is an interior critical point of OA with f(0ß 1) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (ii) On AB, f(xß y) œ f(xß 2) œ 2x  x# for 0 Ÿ x Ÿ 2 Ê f w (xß 2) œ 2  2x œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of AB with f(1ß 2) œ 1. Endpoints: f(0ß 2) œ 0 and f(2ß 2) œ 0. (iii) On BC, f(xß y) œ f(2ß y) œ 2y  y# for 0 Ÿ y Ÿ 2 Ê f w (2ß y) œ 2  2y œ 0 Ê y œ 1 and x œ 2 Ê (2ß 1) is an interior critical point of BC with f(2ß 1) œ 1. Endpoints: f(2ß 0) œ 0 and f(2ß 2) œ 0. (iv) On OC, f(xß y) œ f(xß 0) œ 2x  x# for 0 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2  2x œ 0 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of OC with f(1ß 0) œ 1. Endpoints: f(0ß 0) œ 0 and f(0ß 2) œ 0. (v) For the interior of the rectangular region, fx (xß y) œ 2  2x œ 0 and fy (xß y) œ 2  2y œ 0 Ê x œ 1 and y œ 1 Ê (1ß 1) is an interior critical point of the square with f(1ß 1) œ 2. Therefore the absolute maximum is 2 at (1ß 1) and the absolute minimum is 0 at the four corners (0ß 0), (0ß 2), (2ß 2), and (2ß 0).

74. (i)

On AB, f(xß y) œ f(xß x  2) œ 2x  4 for 2 Ÿ x Ÿ 2 Ê f w (xß x  2) œ 2 œ 0 Ê no critical points in the interior of AB. Endpoints: f(2ß 0) œ 8 and f(2ß 4) œ 0. (ii) On BC, f(xß y) œ f(2ß y) œ y#  4y for 0 Ÿ y Ÿ 4 Ê f w (2ß y) œ 2y  4 œ 0 Ê y œ 2 and x œ 2 Ê (2ß 2) is an interior critical point of BC with f(2ß 2) œ 4. Endpoints: f(2ß 0) œ 0 and f(2ß 4) œ 0. (iii) On AC, f(xß y) œ f(xß 0) œ x#  2x for 2 Ÿ x Ÿ 2 Ê f w (xß 0) œ 2x  2 Ê x œ 1 and y œ 0 Ê (1ß 0) is an interior critical point of AC with f(1ß 0) œ 1. Endpoints: f(2ß 0) œ 8 and f(2ß 0) œ 0. (iv) For the interior of the triangular region, fx (xß y) œ 2x  2 œ 0 and fy (xß y) œ 2y  4 œ 0 Ê x œ 1 and y œ 2 Ê (1ß 2) is an interior critical point of the region with f(1ß 2) œ 3. Therefore the absolute maximum is 8 at (2ß 0) and the absolute minimum is 1 at (1ß 0).

75. (i)

Chapter 14 Practice Exercises 76. (i)

(ii)

931

On AB, f(xß y) œ f(xß x) œ 4x#  2x%  16 for 2 Ÿ x Ÿ 2 Ê f w (xß x) œ 8x  8x$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1, or x œ 1 and y œ 1 Ê (0ß 0), ("ß 1), (1ß 1) are all interior points of AB with f(0ß 0) œ 16, f(1ß 1) œ 18, and f(1ß 1) œ 18. Endpoints: f(2ß 2) œ 0 and f(2ß 2) œ 0. On BC, f(xß y) œ f(2ß y) œ 8y  y% for 2 Ÿ y Ÿ 2 Ê f w (2ß y) œ 8  4y$ œ 0 Ê y œ $È2 and x œ 2 Ê Š2ß $È2‹ is an interior critical point of BC with f Š2ß $È2‹ œ 6 $È2. Endpoints: f(2ß 2) œ 32 and f(2ß 2) œ 0.

(iii) On AC, f(xß y) œ f(xß 2) œ 8x  x% for 2 Ÿ x Ÿ 2 Ê f w (xß 2) œ 8  4x$ œ 0 Ê x œ $È2 and y œ 2 Ê Š $È2ß 2‹ is an interior critical point of AC with f Š $È2ß 2‹ œ 6 $È#. Endpoints: f(2ß 2) œ 0 and f(2ß 2) œ 32. (iv) For the interior of the triangular region, fx (xß y) œ 4y  4x$ œ 0 and fy (xß y) œ 4x  4y$ œ 0 Ê x œ 0 and y œ 0, or x œ 1 and y œ 1 or x œ 1 and y œ 1. But neither of the points (0ß 0) and (1ß 1), or (1ß 1) are interior to the region. Therefore the absolute maximum is 18 at (1ß 1) and (1ß 1), and the absolute minimum is 32 at (2ß 2). On AB, f(xß y) œ f(1ß y) œ y$  3y#  2 for 1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y#  6y œ 0 Ê y œ 0 and x œ 1, or y œ 2 and x œ 1 Ê (1ß 0) is an interior critical point of AB with f(1ß 0) œ 2; (1ß 2) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f(1ß 1) œ 0. (ii) On BC, f(xß y) œ f(xß 1) œ x$  3x#  2 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  6x œ 0 Ê x œ 0 and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an interior critical point of BC with f(!ß 1) œ 2; (2ß 1) is outside the boundary. Endpoints: f("ß 1) œ 0 and f("ß 1) œ 2. (iii) On CD, f(xß y) œ f("ß y) œ y$  3y#  4 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y#  6y œ 0 Ê y œ 0 and x œ 1, or y œ 2 and x œ 1 Ê ("ß 0) is an interior critical point of CD with f("ß 0) œ 4; (1ß 2) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0. (iv) On AD, f(xß y) œ f(xß 1) œ x$  3x#  4 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  6x œ 0 Ê x œ 0 and y œ 1, or x œ 2 and y œ 1 Ê (0ß 1) is an interior point of AD with f(0ß 1) œ 4; (#ß 1) is outside the boundary. Endpoints: f(1ß 1) œ 2 and f("ß 1) œ 0. (v) For the interior of the square, fx (xß y) œ 3x#  6x œ 0 and fy (xß y) œ 3y#  6y œ 0 Ê x œ 0 or x œ 2, and y œ 0 or y œ 2 Ê (0ß 0) is an interior critical point of the square region with f(!ß 0) œ 0; the points (0ß 2), (2ß 0), and (2ß 2) are outside the region. Therefore the absolute maximum is 4 at (1ß 0) and the absolute minimum is 4 at (0ß 1).

77. (i)

932

Chapter 14 Partial Derivatives

On AB, f(xß y) œ f(1ß y) œ y$  3y for 1 Ÿ y Ÿ 1 Ê f w (1ß y) œ 3y#  3 œ 0 Ê y œ „ 1 and x œ 1 yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2. (ii) On BC, f(xß y) œ f(xß 1) œ x$  3x  2 for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  3 œ 0 Ê no solution. Endpoints: f("ß 1) œ 2 and f("ß 1) œ 6. (iii) On CD, f(xß y) œ f("ß y) œ y$  3y  2 for 1 Ÿ y Ÿ 1 Ê f w ("ß y) œ 3y#  3 œ 0 Ê no solution. Endpoints: f(1ß 1) œ 6 and f("ß 1) œ 2. (iv) On AD, f(xß y) œ f(xß 1) œ x$  3x for 1 Ÿ x Ÿ 1 Ê f w (xß 1) œ 3x#  3 œ 0 Ê x œ „ 1 and y œ 1 yielding the corner points (1ß 1) and (1ß 1) with f(1ß 1) œ 2 and f(1ß 1) œ 2 (v) For the interior of the square, fx (xß y) œ 3x#  3y œ 0 and fy (xß y) œ 3y#  3x œ 0 Ê y œ x# and x%  x œ 0 Ê x œ 0 or x œ 1 Ê y œ 0 or y œ 1 Ê (!ß 0) is an interior critical point of the square region with f(0ß 0) œ 1; (1ß 1) is on the boundary. Therefore the absolute maximum is 6 at ("ß 1) and the absolute minimum is 2 at (1ß 1) and (1ß 1).

78. (i)

79. ™ f œ 3x# i  2yj and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê 3x# i  2yj œ -(2xi  2yj) Ê 3x# œ 2x- and 2y œ 2y- Ê - œ 1 or y œ 0. CASE 1: - œ 1 Ê 3x# œ 2x Ê x œ 0 or x œ 23 ; x œ 0 Ê y œ „ 1 yielding the points (0ß 1) and (!ß 1); x œ Ê yœ „

È5 3

yielding the points Š 23 ß

È5 3 ‹

and Š 23 ß 

2 3

È5 3 ‹.

CASE 2: y œ 0 Ê x#  1 œ 0 Ê x œ „ 1 yielding the points (1ß 0) and (1ß 0). Evaluations give f a!ß „ 1b œ 1, f Š 23 ß „

È5 3 ‹

œ

23 27

, f("ß 0) œ 1, and f("ß 0) œ 1. Therefore the absolute

maximum is 1 at a!ß „ 1b and (1ß 0), and the absolute minimum is 1 at ("ß !). 80. ™ f œ yi  xj and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê yi  xj œ -(2xi  2yj) Ê y œ 2-x and xy œ 2-y Ê x œ 2-(2-x) œ 4-# x Ê x œ 0 or 4-# œ 1. CASE 1: x œ 0 Ê y œ 0 but (0ß 0) does not lie on the circle, so no solution. CASE 2: 4-# œ 1 Ê - œ "# or - œ  "# . For - œ "# , y œ x Ê 1 œ x#  y# œ 2x# Ê x œ C œ „ È" yielding the 2

points

Š È"2

ß

" È2 ‹

and Š

" È2

,

" È2 ‹ .

" #

#

#

#

For - œ  , y œ x Ê 1 œ x  y œ 2x Ê x œ „

" È2

and

y œ x yielding the points Š È"2 ß È"2 ‹ and Š È"2 ,  È"2 ‹ . Evaluations give the absolute maximum value f Š È"2 ß È"2 ‹ œ f Š È"2 ß  È"2 ‹ œ

" #

and the absolute minimum

value f Š È"2 ß È"2 ‹ œ f Š È"2 ß  È"2 ‹ œ  #" . 81. (i) f(xß y) œ x#  3y#  2y on x#  y# œ 1 Ê ™ f œ 2xi  (6y  2)j and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê 2xi  (6y  2)j œ -(2xi  2yj) Ê 2x œ 2x- and 6y  2 œ 2y- Ê - œ 1 or x œ 0. CASE 1: - œ 1 Ê 6y  2 œ 2y Ê y œ  "# and x œ „

È3 #

yielding the points Š „

È3 #

ß  "# ‹ .

CASE 2: x œ 0 Ê y# œ 1 Ê y œ „ 1 yielding the points a!ß „ 1b . Evaluations give f Š „

È3 #

ß  "# ‹ œ

" #

, f(0ß 1) œ 5, and f(!ß 1) œ 1. Therefore

" #

and 5 are the extreme

values on the boundary of the disk. (ii) For the interior of the disk, fx (xß y) œ 2x œ 0 and fy (xß y) œ 6y  2 œ 0 Ê x œ 0 and y œ  "3 Ê ˆ!ß  13 ‰ is an interior critical point with f ˆ!ß  "3 ‰ œ  3" . Therefore the absolute maximum of f on the disk is 5 at (0ß 1) and the absolute minimum of f on the disk is  "3 at ˆ!ß  3" ‰ .

Chapter 14 Practice Exercises

933

82. (i) f(xß y) œ x#  y#  3x  xy on x#  y# œ 9 Ê ™ f œ (2x  3  y)i  (2y  x)j and ™ g œ 2xi  2yj so that ™ f œ - ™ g Ê (2x  3  y)i  (2y  x)j œ -(2xi  2yj) Ê 2x  3  y œ 2x- and 2y  x œ 2yÊ 2x("  -)  y œ 3 and x  2y(1  -) œ 0 Ê 1  - œ

x 2y

x and (2x) Š 2y ‹  y œ 3 Ê x#  y# œ 3y

Ê x# œ y#  3y. Thus, 9 œ x#  y# œ y#  3y  y# Ê 2y#  3y  9 œ 0 Ê (2y  3)(y  3) œ 0 Ê y œ 3, 3# . For y œ 3, x#  y# œ 9 Ê x œ 0 yielding the point (0ß 3). For y œ 3# , x#  y# œ 9 Ê x# 

9 4

œ 9 Ê x# œ

Ê xœ „

27 4

È

¸ 20.691, and f Š 3 # 3 , 3# ‹ œ 9 

27È3 4

3È 3 #

È

. Evaluations give f(0ß 3) œ 9, f Š 3 # 3 ß 3# ‹ œ 9 

27È3 4

¸ 2.691.

(ii) For the interior of the disk, fx (xß y) œ 2x  3  y œ 0 and fy (xß y) œ 2y  x œ 0 Ê x œ 2 and y œ 1 Ê (2ß 1) is an interior critical point of the disk with f(2ß 1) œ 3. Therefore, the absolute maximum of f on the disk is 9 

27È3 4

È

at Š 3 # 3 ß 3# ‹ and the absolute minimum of f on the disk is 3 at (2ß 1).

83. ™ f œ i  j  k and ™ g œ 2xi  2yj  2zk so that ™ f œ - ™ g Ê i  j  k œ -(2xi  2yj  2zk) Ê 1 œ 2x-, 1 œ 2y-, 1 œ 2z- Ê x œ y œ z œ -" . Thus x#  y#  z# œ 1 Ê 3x# œ 1 Ê x œ „ È"3 yielding the points Š È"3 ß  È"3 ,

" È3 ‹

and Š È"3 ,

f Š È"3 ß  È"3 ß È"3 ‹ œ

3 È3

" È3

,  È"3 ‹ . Evaluations give the absolute maximum value of

œ È3 and the absolute minimum value of f Š È"3 ß È"3 ß  È"3 ‹ œ È3.

84. Let f(xß yß z) œ x#  y#  z# be the square of the distance to the origin and g(xß yß z) œ z#  xy  4. Then ™ f œ 2xi  2yj  2zk and ™ g œ yi  xj  2zk so that ™ f œ - ™ g Ê 2x œ -y, 2y œ -x, and 2z œ 2-z Ê z œ 0 or - œ 1. CASE 1: z œ 0 Ê xy œ 4 Ê x œ  4y and y œ  4x Ê 2 Š 4y ‹ œ -y and 2 ˆ 4x ‰ œ -x Ê

8 -

œ y# and

8 -

œ x#

Ê y# œ x# Ê y œ „ x. But y œ x Ê x# œ 4 leads to no solution, so y œ x Ê x# œ 4 Ê x œ „ 2 yielding the points (2ß 2ß 0) and (2ß 2ß 0). CASE 2: - œ 1 Ê 2x œ y and 2y œ x Ê 2y œ  ˆ y# ‰ Ê 4y œ y Ê y œ 0 Ê x œ 0 Ê z#  4 œ 0 Ê z œ „ 2 yielding the points (0ß 0ß 2) and (!ß 0ß 2). Evaluations give f(2ß 2ß 0) œ f(2ß 2ß 0) œ 8 and f(!ß !ß 2) œ f(!ß !ß 2) œ 4. Thus the points (!ß !ß 2) and (!ß !ß 2) on the surface are closest to the origin.

85. The cost is f(xß yß z) œ 2axy  2bxz  2cyz subject to the constraint xyz œ V. Then ™ f œ - ™ g Ê 2ay  2bz œ -yz, 2ax  2cz œ -xz, and 2bx  2cy œ -xy Ê 2axy  2bxz œ -xyz, 2axy  2cyz œ -xyz, and 2bxz  2cyz œ -xyz Ê 2axy  2bxz œ 2axy  2cyz Ê y œ ˆ bc ‰ x. Also 2axy  2bxz œ 2bxz  2cyz Ê z œ ˆ ac ‰ x. Then x ˆ bc x‰ ˆ ac x‰ œ V Ê x$ œ #

Height œ z œ ˆ ac ‰ Š cabV ‹

"Î$

#

c# V ab

œ Š abcV ‹

#

Ê width œ x œ Š cabV ‹

"Î$

"Î$

#

, Depth œ y œ ˆ bc ‰ Š cabV ‹

"Î$

#

œ Š bacV ‹

"Î$

, and

.

86. The volume of the pyramid in the first octant formed by the plane is V(aß bß c) œ

" 3

ˆ #" ab‰ c œ

" 6

abc. The point

(2ß 1ß 2) on the plane Ê  "b  2c œ 1. We want to minimize V subject to the constraint 2bc  ac  2ab œ abc. ac ab Thus, ™ V œ bc 6 i  6 j  6 k and ™ g œ (c  2b  bc)i  (2c  2a  ac)j  (2b  a  ab)k so that ™ V œ ac ab abc Ê bc 6 œ -(c  2b  bc), 6 œ -(2c  2a  ac), and 6 œ -(2b  a  ab) Ê 6 œ -(ac  2ab  abc), abc abc 6 œ -(2bc  2ab  abc), and 6 œ -(2bc  ac  abc) Ê -ac œ 2-bc and 2-ab œ 2-bc. Now - Á 0 since 2 a

a Á 0, b Á 0, and c Á 0 Ê ac œ 2bc and ab œ bc Ê a œ 2b œ c. Substituting into the constraint equation gives y 2 2 2 x z a  a  a œ 1 Ê a œ 6 Ê b œ 3 and c œ 6. Therefore the desired plane is 6  3  6 œ 1 or x  2y  z œ 6. 87. ™ f œ (y  z)i  xj  xk , ™ g œ 2xi  2yj , and ™ h œ zi  xk so that ™ f œ - ™ g  . ™ h Ê (y  z)i  xj  xk œ -(2xi  2yj)  .(zi  xk) Ê y  z œ 2-x  .z, x œ 2-y, x œ .x Ê x œ 0

™g

934

Chapter 14 Partial Derivatives

or . œ 1. CASE 1: x œ 0 which is impossible since xz œ 1. CASE 2: . œ 1 Ê y  z œ 2-x  z Ê y œ 2-x and x œ 2-y Ê y œ (2-)(2-y) Ê y œ 0 or 4-# œ 1. If y œ 0, then x# œ 1 Ê x œ „ 1 so with xz œ 1 we obtain the points (1ß 0ß 1) and (1ß 0ß 1). If 4-# œ 1, then - œ „ "# . For - œ  "# , y œ x so x#  y# œ 1 Ê x# œ Ê xœ „

" È2

" #

with xz œ 1 Ê z œ „ È2, and we obtain the points Š È"2 ß  È"2 ß È2‹ and

Š È"2 ß È"2 ß È2‹ . For - œ

" #

, y œ x Ê x# œ

" #

Ê xœ „

" È2

with xz œ 1 Ê z œ „ È2,

and we obtain the points Š È"2 ß È"2 , È2‹ and Š È"2 ß  È"2 ß È2‹ . Evaluations give f(1ß 0ß 1) œ 1, f(1ß 0ß 1) œ 1, f Š È"2 ß  È"2 ß È2‹ œ f Š È"2 ß È"2 ß È2‹ œ

3 #

, and f Š È"2 ß  È"2 ß È2‹ œ

3 #

" #

, f Š È"2 ß È"2 , È2‹ œ

. Therefore the absolute maximum is

Š È"2 ß È"2 ß È2‹ and Š È"2 ß  È"2 ß È2‹ , and the absolute minimum is

" #

3 #

" #

,

at

at Š È"2 ß È"2 ß È2‹ and

Š È"2 ß  È"2 ß È2‹ . 88. Let f(xß yß z) œ x#  y#  z# be the square of the distance to the origin. Then ™ f œ 2xi  2yj  2zk , ™ g œ i  j  k , and ™ h œ 4xi  4yj  2zk so that ™ f œ - ™ g  . ™ h Ê 2x œ -  4x., 2y œ -  4y., and 2z œ -  2z. Ê - œ 2x(1  2.) œ 2y(1  2.) œ 2z(1  2.) Ê x œ y or . œ "# . CASE 1: x œ y Ê z# œ 4x# Ê z œ „ 2x so that x  y  z œ 1 Ê x  x  2x œ 1 or x  x  2x œ 1 (impossible) Ê x œ "4 Ê y œ 4" and z œ #" yielding the point ˆ 4" ß 4" ß #" ‰ . CASE 2: . œ

" #

Ê - œ 0 Ê 0 œ 2z(1  1) Ê z œ 0 so that 2x#  2y# œ 0 Ê x œ y œ 0. But the origin

(!ß 0ß 0) fails to satisfy the first constraint x  y  z œ 1. Therefore, the point ˆ "4 ß 4" ß #" ‰ on the curve of intersection is closest to the origin. 89. (a) y, z are independent with w œ x# eyz and z œ x#  y# Ê œ a2xeyz b

`x `y

`w `y

`w `x `w `y `w `z `x `y  `y `y  `z `y œ 2x `` xy  2y Ê `` xy œ yx

œ

 azx# eyz b (1)  ayx# eyz b (0); z œ x#  y# Ê 0

; therefore,

Š ``wy ‹ œ a2xeyz b ˆ yx ‰  zx# eyz œ a2y  zx# b eyz z

(b) z, x are independent with w œ x# eyz and z œ x#  y# Ê œ a2xeyz b (0)  azx# eyz b

`y `z

`w `z

œ

`w `x `x `z



 ayx# eyz b (1); z œ x#  y# Ê 1 œ 0  2y

1 ˆ ``wz ‰ œ azx# eyz b Š 2y ‹  yx# eyz œ x# eyz Šy  x

`w `y `w `z `y `z  `z `z `y `y " ` z Ê ` z œ  #y

; therefore,

z 2y ‹

(c) z, y are independent with w œ x# eyz and z œ x#  y# Ê

`w `z

œ

œ a2xeyz b `` xz  azx# eyz b (0)  ayx# eyz b (1); z œ x#  y# Ê 1 1 ‰ ˆ ``wz ‰ œ a2xeyz b ˆ 2x  yx# eyz œ a1  x# yb eyz

`w `x `w `y `w `z `x `z  `y `z  `z `z œ 2x `` xz  0 Ê `` xz œ #"x

; therefore,

y

90. (a) T, P are independent with U œ f(Pß Vß T) and PV œ nRT Ê ``UT œ ``UP `` TP  ‰ ˆ ``VT ‰  ˆ ``UT ‰ (1); PV œ nRT Ê P ``VT œ nR Ê ``VT œ œ ˆ ``UP ‰ (0)  ˆ `` U V U ‰ ˆ nR ‰ ˆ ``UT ‰ œ ˆ `` V P  P

`U `V `U `T `V `T  `T `T nR P ; therefore,

`U `T

`U `P `U `V (b) V, T are independent with U œ f(Pß Vß T) and PV œ nRT Ê `` U V œ `P `V  `V `V  U‰ œ ˆ ``UP ‰ ˆ ``VP ‰  ˆ `` V (1)  ˆ ``UT ‰ (0); PV œ nRT Ê V ``VP  P œ (nR) ˆ ``VT ‰ œ 0 Ê

ˆ `` U ‰ V T

œ

ˆ ``UP ‰ ˆ VP ‰



`U `V

91. Note that x œ r cos ) and y œ r sin ) Ê r œ Èx#  y# and ) œ tan" ˆ yx ‰ . Thus, `w `x

œ

`w `r `r `x



`w `) `) `x

œ ˆ ``wr ‰ Š Èx#x
y# ‹  ˆ ``w) ‰ Š x#
yy# ‹ œ (cos ))

`w `r

 ˆ sinr ) ‰

`w `)

;

`U `T `T `V `P P `V œ  V

; therefore,

Chapter 14 Practice Exercises `w `y

œ

`w `r `r `y `u `x

92. zx œ fu 93.

`u `y

`v `x

 fv

`w `x

`u `x œ a " `w " `w a `x œ b `y

œ

and

`w `z

œ

2 r
s

œ

Ê Ê

œ

2 x#
y#
2z

and

`w `s

œ

`w dw ` x œ du b ``wx œ a

" (r
s)#

`w `x `x `s

`w `y `y `s



Solving this system yields Ê ae cos vb

`u `y

`u `x

 ae sin vb `u `y

`v `y

`v `y

 fv

dw ` u du ` y

œ

2(r
s) 2 ar#
2rs
s# b

œ

" r
s

œ

`w `x `x `r

dw du

`w `z `z `s

and

œ

`w `y `y `r

 " r
s





rs (r
s)#

,

œb

`w `y

`w `z `z `r

 ˆ cosr ) ‰

dw du

œ œ

Ê

2y x#
y#
2z " r
s



`v u ` x œ 1; e sin v `v u sin v. ` x œ e

œ

œ

dw du

2(r  s) #(r
s)#

rs (r
s)#

and

" `w b `y

œ

rs (r
s)#

œ

dw du

,

 ’ (r
" s)# “ (2s) œ

2r
2s (r
s)#

2 r
s `u `x

 aeu cos vb

`v `x

œ 0.

Similarly, e cos v  x œ 0 u

œ 0 and e sin v  y œ 0 Ê aeu sin vb `v `y

`w `)

 y œ 0 Ê aeu sin vb

u

œ eu sin v and

" `w a `x

 ’ (r
" s)# “ (2r) œ

aeu sin vb

cos v and

`w `r

œ bfu  bfv

œ

`u `x  u

œe

u

second system yields



`u `y

`w `y

œa

`w `r

Ê

95. eu cos v  x œ 0 Ê aeu cos vb u

`u `x `w `y

2(r
s) (r
s)#
(r  s)#
4rs -

œ

2x x#
y#
2z

œ ˆ ``wr ‰ Š Èx#y
y# ‹  ˆ ``w) ‰ Š x#
x y# ‹ œ (sin ))

œ afu  afv , and zy œ fu

œ b and

Ê 94.

`w `) ` ) )y



935

`u `y

œ eu cos v. Therefore Š `` ux i 

 aeu cos vb `u `y

`v `y

j‹ † Š `` vx i 

œ 1. Solving this

`v `y

j‹

œ caeu cos vb i  aeu sin vb jd † caeu sin vb i  aeu cos vb jd œ 0 Ê the vectors are orthogonal Ê the angle between the vectors is the constant 1# . 96.

`g `)

œ

Ê

`f `x `x `) ` #g ` )#



`f `y `y `)

œ (r sin )) #

œ (r sin )) Š `` xf#

œ (r sin )) Š `` x) 

`y `) ‹

`x `)



`f `x

 (r cos ))

` #f ` y ` y` x ` ) ‹

`f `y `f `x

 (r cos ))

 (r cos ))  (r cos )) Š `` x) 

#

 (r cos )) Š ``x`fy

`y `) ‹

`x `)



` #f ` y ` y# ` ) ‹

 (r sin ))

`f `y

 (r sin ))

œ (r sin )  r cos ))(r sin )  r cos ))  (r cos )  r sin )) œ (2)(2)  (0  2) œ 4  2 œ 2 at (rß )) œ ˆ2ß 1# ‰ . 97. (y  z)#  (z  x)# œ 16 Ê ™ f œ 2(z  x)i  2(y  z)j  2(y  2z  x)k ; if the normal line is parallel to the yz-plane, then x is constant Ê `` xf œ 0 Ê 2(z  x) œ 0 Ê z œ x Ê (y  z)#  (z  z)# œ 16 Ê y  z œ „ 4. Let x œ t Ê z œ t Ê y œ t „ 4. Therefore the points are (tß t „ 4ß t), t a real number.

98. Let f(xß yß z) œ xy  yz  zx  x  z# œ 0. If the tangent plane is to be parallel to the xy-plane, then ™ f is perpendicular to the xy-plane Ê ™ f † i œ 0 and ™ f † j œ 0. Now ™ f œ (y  z  1)i  (x  z)j  (y  x  2z)k so that ™ f † i œ y  z  1 œ 0 Ê y  z œ 1 Ê y œ 1  z, and ™ f † j œ x  z œ 0 Ê x œ z. Then z(1  z)  ("  z)z  z(z)  (z)  z# œ 0 Ê z  2z# œ 0 Ê z œ "# or z œ 0. Now z œ "# Ê x œ  "# and y œ Ê ˆ "# ß "# ß "# ‰ is one desired point; z œ 0 Ê x œ 0 and y œ 1 Ê (0ß 1ß 0) is a second desired point. 99. ™ f œ -(xi  yj  zk) Ê

`f `x

œ -x Ê f(xß yß z) œ

" #

-x#  g(yß z) for some function g Ê -y œ

`f `y

œ

`g `y

-y#  h(z) for some function h Ê -z œ `` zf œ `` gz œ hw (z) Ê h(z) œ #" -z#  C for some arbitrary constant C Ê g(yß z) œ "# -y#  ˆ "# -z#  C‰ Ê f(xß yß z) œ "# -x#  "# -y#  "# -z#  C Ê f(0ß 0ß a) œ "# -a#  C Ê g(yß z) œ

" #

and f(0ß 0ß a) œ ‰ 100. ˆ df ds u (0 0 0) ß

ß ß

" #

-(a)#  C Ê f(0ß 0ß a) œ f(0ß 0ß a) for any constant a, as claimed.

œ lim sÄ0 œ lim sÄ0

f(0
su" ß 0
su# ß 0
su$ )f(0ß 0ß 0)

s

És# u#"
s# u##
s# u#$  0

s

,s
0

,s
0

" #

936

Chapter 14 Partial Derivatives sÉu#"
u##
u#$

œ lim œ lim kuk œ 1; s sÄ0 sÄ0 y however, ™ f œ Èx#
xy#
z# i  Èx#
y#
z# j  Èx#
zy#
z# k fails to exist at the origin (0ß 0ß 0) 101. Let f(xß yß z) œ xy  z  2 Ê ™ f œ yi  xj  k . At (1ß 1ß 1), we have ™ f œ i  j  k Ê the normal line is x œ 1  t, y œ 1  t, z œ 1  t, so at t œ 1 Ê x œ 0, y œ 0, z œ 0 and the normal line passes through the origin. 102. (b) f(xß yß z) œ x#  y#  z# œ 4 Ê ™ f œ 2xi  2yj  2zk Ê at (2ß 3ß 3) the gradient is ™ f œ 4i  6j  6k which is normal to the surface (c) Tangent plane: 4x  6y  6z œ 8 or 2x  3y  3z œ 4 Normal line: x œ 2  4t, y œ 3  6t, z œ 3  6t

CHAPTER 14 ADDITIONAL AND ADVANCED EXERCISES fx (0ß h)  fx (0ß 0) h

1. By definition, fxy (!ß 0) œ lim

hÄ0

so we need to calculate the first partial derivatives in the

numerator. For (xß y) Á (0ß 0) we calculate fx (xß y) by applying the differentiation rules to the formula for

fy (xß y) œ hÄ0

`w `x

#

x  xy x#
y#

00 h

œ lim 2.

$

x# y  y$ x#
y#

ax#
y# b (2x)  ax#  y# b (2x) ax #
y # b #

4x# y$ Ê fx (0ß h) a x #
y # b# f(0ß0) For (xß y) œ (0ß 0) we apply the definition: fx (!ß 0) œ lim f(hß 0)  œ lim 0 h 0 œ h hÄ0 hÄ0 fy (hß 0)  fy (!ß 0) h  0 fxy (0ß 0) œ lim œ 1. Similarly, fyx (0ß 0) œ lim , so for (xß y) Á h h hÄ0 hÄ0

f(xß y): fx (xß y) œ



 (xy)

$ #

4x y ax #
y # b #

Ê fy (hß 0) œ

$

x# y  y$ x#
y#

œ



$

œ  hh# œ h. 0. Then by definition (0ß 0) we have

œ h; for (xß y) œ (0ß 0) we obtain fy (0ß 0) œ lim

h h#

h0 h

œ 0. Then by definition fyx (0ß 0) œ lim

hÄ0

œ 1  ex cos y Ê w œ x  ex cos y  g(y);

`w `y

hÄ0

f(0ß h)  f(!ß 0) h

œ 1. Note that fxy (0ß 0) Á fyx (0ß 0) in this case.

œ ex sin y  gw (y) œ 2y  ex sin y Ê gw (y) œ 2y

Ê g(y) œ y#  C; w œ ln 2 when x œ ln 2 and y œ 0 Ê ln 2 œ ln 2  eln 2 cos 0  0#  C Ê 0 œ 2  C Ê C œ 2. Thus, w œ x  ex cos y  g(y) œ x  ex cos y  y#  2. 3. Substitution of u  u(x) and v œ v(x) in g(uß v) gives g(u(x)ß v(x)) which is a function of the independent variable x. Then, g(uß v) œ 'u f(t) dt Ê v

œ Š ``u

#

#

fzz œ Š ddr#f ‹ ˆ ``zr ‰  ` #r ` y# #

œ

` g du ` u dx



'uv f(t) dt‹ dudx  Š ``v 'uv f(t) dt‹ dvdx

` g dv ` v dx

œ Š ``u

#

df ` # r dr ` x#

. Similarly, fyy œ Š ddr#f ‹ Š ``yr ‹ 

Ê

` #r ` x#

y #
z# 3 ˆÈ x #
y #
z # ‰

v ` ' dv du dv dv du 'vu f(t) dt‹ du dx  Š ` v u f(t) dt‹ dx œ f(u(x)) dx  f(v(x)) dx œ f(v(x)) dx  f(u(x)) dx

4. Applying the chain rules, fx œ

Ê

dg dx

œ

df ` # r dr ` z#

x #
z# 3 ˆÈ x #
y #
z # ‰ #

df ` r dr ` x

#

Ê fxx œ Š ddr#f ‹ ˆ ``xr ‰ 

. Moreover,

; and

`r `z

œ

‰ Ê Š ddr#f ‹ Š x#
xy#
z# ‹  ˆ df dr Œ ˆÈ

`r `x

œ

x È x #
y #
z#

z È x #
y #
z# y #
z# x #
y #
z# ‰

Ê

` #r ` z# d# f

3

œ

œ

#

x#
y# 3 ˆÈ x #
y #
z# ‰ y#

;

`r `y

œ

#

y È x #
y #
z#

. Next, fxx  fyy  fzz œ 0 x #
z#


 Š dr# ‹ Š x#
y#
z# ‹  ˆ dr ‰ Œ ˆÈx#
y#
z# ‰3
df

df ` # r dr ` y#

and

Chapter 14 Practice Exercises #

Ê df dr

d dr

x#
y#

#

‰  Š ddr#f ‹ Š x#
zy#
z# ‹  ˆ df dr Œ ˆÈ

x #
y #
z# ‰

(f w ) œ ˆ 2r ‰ f w , where f w œ

œ Cr# Ê f(r) œ  Cr  b œ

Ê

df dr

3

df f

d# f dr#


œ0 Ê

 Š Èx#
2y#
z# ‹

df dr

d# f dr#

œ0 Ê



2 df r dr

937

œ0

œ  2 rdr Ê ln f w œ 2 ln r  ln C Ê f w œ Cr# , or

w

w

 b for some constants a and b (setting a œ C)

a r

5. (a) Let u œ tx, v œ ty, and w œ f(uß v) œ f(u(tß x)ß v(tß y)) œ f(txß ty) œ tn f(xß y), where t, x, and y are independent variables. Then ntnc1 f(xß y) œ ``wt œ ``wu ``ut  ``wv ``vt œ x ``wu  y ``wv . Now, `w `w `u `w `v `w `w ˆ `w ‰ ˆ `w ‰ ˆ " ‰ ˆ ``wx ‰ . Likewise, ` x œ ` u ` x  ` v ` x œ ` u (t)  ` v (0) œ t ` u Ê ` u œ t `w `y

œ

`w `u `u `y



`w `v `v `y

ntnc1 f(xß y) œ x `w `x

Ê

œ

`f `x

`w `u

`w `v

y

`w `y

and

œ ˆ ``wu ‰ (0)  ˆ ``wv ‰ (t) Ê œ

n(n  1)t

`f `x

Ê nf(xß y) œ x

Also from part (a), œ

` `y

œ t#

ˆt

`w ‰ `v

` #w ` v` u

œt

` w `u ` u# ` t

#

` w ` x#

` #w ` x#

`w `v . ` w `v ` v` u ` t

y

x

ˆ ``wx ‰ œ

t

` #w ` v ` v# ` y

œ

` #w ` u#

` `x

` #w ` y#

y

ˆt

`w ‰ `u

` #w ` v#

œ t#

, ˆ t"# ‰

`f `x

y

`f `y

, as claimed.

Differentiating with respect to t again we obtain

#

` `x

œ

` #w ` u ` u` v ` y

Ê ˆ t"# ‰

`w `u

#

f(xß y) œ x

œ ˆ "t ‰ Š ``wy ‹ . Therefore,

œ ˆ xt ‰ ˆ ``wx ‰  ˆ yt ‰ Š ``wy ‹. When t œ 1, u œ x, v œ y, and w œ f(xß y)

(b) From part (a), ntnc1 f(xß y) œ x nc2

`w `v

œ

` #w ` u ` u` v ` t

œt

, and

` #w ` v#

` #w ` v ` v# ` t

y

#

` w `u ` u# ` x

t

` #w ` y` x

` `y

œ

, and ˆ t"# ‰

#

` w `v ` v` u ` x

ˆ ``wx ‰ œ

` #w ` y` x

œ

` #w ` u#

œ x#

 2xy #

#

œ t#

` w ` u#

,

` `y

`w ‰ `u

œt

ˆt

` #w ` u` v

` w ` y#

œ

 y# ` `y

` #w ` v#

.

Š ``wy ‹

` #w ` u ` u# ` y

t

` #w ` v ` v` u ` y

` #w ` v` u

‰ Š ``y`wx ‹  Š yt# ‹ Š `` yw# ‹ for t Á 0. When t œ 1, w œ f(xß y) and Ê n(n  1)tnc2 f(xß y) œ Š xt# ‹ Š `` xw# ‹  ˆ 2xy t# #

#

#

#

#

#

#

#

we have n(n  1)f(xß y) œ x# Š `` xf# ‹  2xy Š ``x`fy ‹  y# Š `` yf# ‹ as claimed. 6. (a) lim

rÄ0

sin 6r 6r

œ lim

tÄ0

(b) fr (0ß 0) œ lim

hÄ0

œ lim

hÄ0

(c) f) (rß )) œ lim

hÄ0

sin t t

œ 1, where t œ 6r

f(0
hß 0)  f(0ß 0) h 36 sin 6h 12

œ lim

hÄ0

œ0

f(rß )
h)  f(rß )) h

ˆ sin6h6h ‰ 1 h

œ lim

hÄ0

sin 6h  6h 6h#

6 cos 6h  6 12h

i

y È x #
y #
z#

hÄ0

(applying l'Hopital's rule twice) s

œ lim

hÄ0

ˆ sin6r6r ‰  ˆ sin6r6r ‰ h

œ lim

0

hÄ0 h

7. (a) r œ xi  yj  zk Ê r œ krk œ Èx#  y#  z# and ™ r œ œ

œ lim

œ0

x È x #
y #
z#

j

z È x #
y #
z#

k

r r

(b) rn œ ˆÈx#  y#  z# ‰

n

Ê ™ arn b œ nx ax#  y#  z# b œ nrnc2 r (c) Let n œ 2 in part (b). Then

" #

ÐnÎ2Ñ
1

i  ny ax#  y#  z# b

ÐnÎ2Ñ
1

j  nz ax#  y#  z# b

™ ar# b œ r Ê ™ ˆ "# r# ‰ œ r Ê

r# #

œ

" #

ÐnÎ2Ñ
1

k

ax#  y#  z# b is the function.

(d) dr œ dxi  dyj  dzk Ê r † dr œ x dx  y dy  z dz, and dr œ rx dx  ry dy  rz dz œ

x r

dx 

y r

dy 

z r

dz

Ê r dr œ x dx  y dy  z dz œ r † dr (e) A œ ai  bj  ck Ê A † r œ ax  by  cz Ê ™ (A † r) œ ai  bj  ck œ A 8. f(g(t)ß h(t)) œ c Ê 0 œ

df dt

œ

` f dx ` x dt



` f dy ` y dt

œ Š `` xf i 

`f `y

j‹ † Š dx dt i 

dy dt

j‹ , where

dx dt

i

dy dt

j is the tangent vector

Ê ™ f is orthogonal to the tangent vector 9. f(xß yß z) œ xz#  yz  cos xy  1 Ê ™ f œ az#  y sin xyb i  (z  x sin xy)j  (2xz  y)k Ê ™ f(0ß 0ß 1) œ i  j Ê the tangent plane is x  y œ 0; r œ (ln t)i  (t ln t)j  tk Ê rw œ ˆ "t ‰ i  (ln t  1)j  k ; x œ y œ 0, z œ 1 Ê t œ 1 Ê rw (1) œ i  j  k . Since (i  j  k) † (i  j) œ rw (1) † ™ f œ 0, r is parallel to the plane, and r(1) œ 0i  0j  k Ê r is contained in the plane.

938

Chapter 14 Partial Derivatives

10. Let f(xß yß z) œ x$  y$  z$  xyz Ê ™ f œ a3x#  yzb i  a3y#  xzb j  a3z#  xyb k Ê ™ f(0ß 1ß 1) œ i  3j  3k $

Ê the tangent plane is x  3y  3z œ 0; r œ Š t4  2‹ i  ˆ 4t  3‰ j  (cos (t  2)) k #

Ê rw œ Š 3t4 ‹ i  ˆ t4# ‰ j  (sin (t  2)) k ; x œ 0, y œ 1, z œ 1 Ê t œ 2 Ê rw (2) œ 3i  j . Since rw (2) † ™ f œ 0 Ê r is parallel to the plane, and r(2) œ i  k Ê r is contained in the plane. 11.

`z `x

œ 3x#  9y œ 0 and

`z `y

œ 3y#  9x œ 0 Ê y œ

" 3

#

x# and 3 ˆ "3 x# ‰  9x œ 0 Ê

" 3

x%  9x œ 0

Ê x ax$  27b œ 0 Ê x œ 0 or x œ 3. Now x œ 0 Ê y œ 0 or (!ß 0) and x œ 3 Ê y œ 3 or (3ß 3). Next ` #z ` x#

œ 6x,

` #z ` y#

œ 6y, and

and for (3ß 3),

` #z ` #z ` x# ` y#

` #z ` x` y #

` #z ` #z ` x# ` y#

œ 9. For (!ß 0), #

 Š ``x`zy ‹ œ 243
0 and

` #z ` x#

#

#

 Š ``x`zy ‹ œ 81 Ê no extremum (a saddle point),

œ 18
0 Ê a local minimum.

12. f(xß y) œ 6xyeÐ2x
3yÑ Ê fx (xß y) œ 6y(1  2x)eÐ2x
3yÑ œ 0 and fy (xß y) œ 6x(1  3y)eÐ2x
3yÑ œ 0 Ê x œ 0 and y œ 0, or x œ "# and y œ 3" . The value f(0ß 0) œ 0 is on the boundary, and f ˆ #" ß 3" ‰ œ e"2 . On the positive y-axis, f(0ß y) œ 0, and on the positive x-axis, f(xß 0) œ 0. As x Ä _ or y Ä _ we see that f(xß y) Ä 0. Thus the absolute maximum of f in the closed first quadrant is e"2 at the point ˆ #" ß 3" ‰ .

13. Let f(xß yß z) œ P! (x! ß y! ß y! ) is

y# x# a#  b# !‰ ˆ 2x a# x

 

z# c#  1 !‰ ˆ 2y b# y

Ê ™fœ  ˆ 2zc#! ‰ z

2y 2x a# i  b# j  # 2y#! ! œ 2x a#  b# 

#

2z c# k Ê an equation of the plane tangent 2z#! ˆ x! ‰ ˆ y! ‰ ˆ z! ‰ c# œ 2 or a# x  b# y  c# z œ 1.

#

at the point

#

The intercepts of the plane are Š xa! ß 0ß 0‹ , Š0ß by! ß 0‹ and Š!ß !ß cz! ‹ . The volume of the tetrahedron formed #

#

#

by the plane and the coordinate planes is V œ ˆ "3 ‰ ˆ #" ‰ Š xa! ‹ Š by! ‹ Š cz! ‹ Ê we need to maximize V(xß yß z) œ ’

(abc)# 6

(abc)# " 6 “ Š x# yz ‹

y# x# z# a#  b#  c# œ 1. (abc)# " 2z 6 “ Š xyz# ‹ œ c# -.

(xyz)" subject to the constraint f(xß yß z) œ œ

2x a#

-, ’

(abc)# " 6 “ Š xy# z ‹

œ

2y b#

-, and ’

Thus, Multiply the first equation

by a# yz, the second by b# xz, and the third by c# xy. Then equate the first and second Ê a# y# œ b# x# Ê y œ ba x, x
0; equate the first and third Ê a# z# œ c# x# Ê z œ ca x, x
0; substitute into f(xß yß z) œ 0 Ê xœ

a È3

Ê yœ

b È3

Ê zœ

c È3

Ê Vœ

È3 #

abc.

14. 2(x  u) œ -, 2(y  v) œ -, 2(x  u) œ ., and 2(y  v) œ 2.v Ê x  u œ v  y, x  u œ  .# , and y  v œ .v Ê x  u œ .v œ  .# Ê v œ

" #

or . œ 0.

CASE 1: . œ 0 Ê x œ u, y œ v, and - œ 0; then y œ x  1 Ê v œ u  1 and v# œ u Ê v œ v#  1 1 „ È1  4 Ê no real solution. # " " " # v œ # and u œ v Ê u œ 4 ; x  4 œ #"  y and y œ x  1 Ê x  4" œ # # Ê x œ  "8 Ê y œ 78 . Then f ˆ 8" ß 87 ß 4" ß #" ‰ œ ˆ 8"  4" ‰  ˆ 87  #" ‰ is 38 È2. (Notice that f has no maximum value.)

Ê v#  v  1 œ 0 Ê v œ

CASE 2:

15. Let (x! ß y! ) be any point in R. We must show lim

Ðh ß k Ñ Ä Ð 0 ß 0 Ñ

lim

Ðx ß y Ñ Ä Ð x ! ß y ! Ñ

x 

" #

#

Ê 2x œ  4"

œ 2 ˆ 83 ‰ Ê the minimum distance

f(xß y) œ f(x! ß y! ) or, equivalently that

kf(x!  hß y!  k)  f(x! ß y! )k œ 0. Consider f(x!  hß y!  k)  f(x! ß y! )

œ [f(x!  hß y!  k)  f(x! ß y!  k)]  [f(x! ß y!  k)  f(x! ß y! )]. Let F(x) œ f(xß y!  k) and apply the Mean Value Theorem: there exists 0 with x!  0  x!  h such that Fw (0 )h œ F(x!  h)  F(x! ) Ê hfx (0ß y!  k) œ f(x!  hß y!  k)  f(x! ß y!  k). Similarly, k fy (x! ß () œ f(x! ß y!  k)  f(x! ß y! ) for some ( with y!  (  y!  k. Then kf(x!  hß y!  k)  f(x! ß y! )k Ÿ khfx (0ß y!  k)k  kkfy (x! ß ()k . If M, N are positive real numbers such that kfx k Ÿ M and kfy k Ÿ N for all (xß y) in the xy-plane, then kf(x!  hß y!  k)  f(x! ß y! )k Ÿ M khk  N kkk . As (hß k) Ä 0, kf(x!  hß y!  k)  f(x! ß y! )k Ä 0 Ê lim kf(x!  hß y!  k)  f(x! ß y! )k Ðh ß k Ñ Ä Ð 0 ß 0 Ñ

Chapter 14 Practice Exercises

939

œ 0 Ê f is continuous at (x! ß y! ). 16. At extreme values, ™ f and v œ

dr dt

df dt

are orthogonal because

œ ™f†

dr dt

œ 0 by the First Derivative Theorem for

Local Extreme Values. 17.

`f `x

œ 0 Ê f(xß y) œ h(y) is a function of y only. Also,

Moreover,

`f `y

œ

`g `x

`g `y

œ

`f `x

œ 0 Ê g(xß y) œ k(x) is a function of x only.

Ê hw (y) œ kw (x) for all x and y. This can happen only if hw (y) œ kw (x) œ c is a constant.

Integration gives h(y) œ cy  c" and k(x) œ cx  c# , where c" and c# are constants. Therefore f(xß y) œ cy  c" and g(xß y) œ cx  c# . Then f(1ß 2) œ g(1ß 2) œ 5 Ê 5 œ 2c  c" œ c  c# , and f(0ß 0) œ 4 Ê c" œ 4 Ê c œ Ê c# œ

9 #

. Thus, f(xß y) œ

" #

y  4 and g(xß y) œ

" #

" #

x  9# .

18. Let g(xß y) œ Du f(xß y) œ fx (xß y)a  fy (xß y)b. Then Du g(xß y) œ gx (xß y)a  gy (xß y)b œ fxx (xß y)a#  fyx (xß y)ab  fxy (xß y)ba  fyy (xß y)b# œ fxx (xß y)a#  2fxy (xß y)ab  fyy (xß y)b# . 19. Since the particle is heat-seeking, at each point (xß y) it moves in the direction of maximal temperature increase, that is in the direction of ™ T(xß y) œ aec2y sin xb i  a2ec2y cos xb j . Since ™ T(xß y) is parallel to 2ec2y cos x ec2y sin x œ È œ 2 ln #2

the particle's velocity vector, it is tangent to the path y œ f(x) of the particle Ê f w (x) œ

2 cot x.

Integration gives f(x) œ 2 ln ksin xk  C and f ˆ 14 ‰ œ 0 Ê 0 œ 2 ln ¸sin 14 ¸  C Ê C

œ ln Š È22 ‹

#

œ ln 2. Therefore, the path of the particle is the graph of y œ 2 ln ksin xk  ln 2. 20. The line of travel is x œ t, y œ t, z œ 30  5t, and the bullet hits the surface z œ 2x#  3y# when 30  5t œ 2t#  3t# Ê t#  t  6 œ 0 Ê (t  3)(t  2) œ 0 Ê t œ 2 (since t
0). Thus the bullet hits the surface at the point (2ß 2ß 20). Now, the vector 4xi  6yj  k is normal to the surface at any (xß yß z), so that n œ 8i  12j  k is normal to the surface at (2ß 2ß 20). If v œ i  j  5k , then the velocity of the particle †25 ‰ after the ricochet is w œ v  2 projn v œ v  Š 2knvk†#n ‹ n œ v  ˆ 2209 n œ (i  j  5k)  ˆ 400 209 i 

œ  191 209 i 

391 209

j

995 209

600 209

j

50 209

k‰

k.

21. (a) k is a vector normal to z œ 10  x#  y# at the point (!ß 0ß 10). So directions tangential to S at (!ß 0ß 10) will be unit vectors u œ ai  bj . Also, ™ T(xß yß z) œ (2xy  4) i  ax#  2yz  14b j  ay#  1b k Ê ™ T(!ß 0ß 10) œ 4i  14j  k . We seek the unit vector u œ ai  bj such that Du T(0ß 0ß 10) œ (4i  14j  k) † (ai  bj) œ (4i  14j) † (ai  bj) is a maximum. The maximum will occur when ai  bj has the same direction as 4i  14j , or u œ È"53 (2i  7j). (b) A vector normal to S at (1ß 1ß 8) is n œ 2i  2j  k . Now, ™ T(1ß 1ß 8) œ 6i  31j  2k and we seek the unit vector u such that Du T(1ß 1ß 8) œ ™ T † u has its largest value. Now write ™ T œ v  w , where v is parallel to ™ T and w is orthogonal to ™ T. Then Du T œ ™ T † u œ (v  w) † u œ v † u  w † u œ w † u. Thus Du T(1ß 1ß 8) is a maximum when u has the same direction as w . Now, w œ ™ T  Š ™knTk#†n ‹ n 62
2 ‰ œ (6i  31j  2k)  ˆ 124
(2i  2j  k) œ ˆ6 
4
1

œ  98 9 i

127 9

j

58 9

k Ê uœ

w kwk

152 ‰ i 9

 ˆ31 

152 ‰ j 9

 ˆ2 

76 ‰ 9 k

" œ  È29,097 (98i  127j  58k).

22. Suppose the surface (boundary) of the mineral deposit is the graph of z œ f(xß y) (where the z-axis points up into the air). Then  `` xf i  `` yf j  k is an outer normal to the mineral deposit at (xß y) and `` xf i  `` yf j points in the direction of steepest ascent of the mineral deposit. This is in the direction of the vector

`f `x

i

`f `y

j at (0ß 0)

(the location of the 1st borehole) that the geologists should drill their fourth borehole. To approximate this vector we use the fact that (0ß 0ß 1000), (0ß 100ß 950), and (100ß !ß 1025) lie on the graph of z œ f(xß y). The plane containing these three points is a good approximation to the tangent plane to z œ f(xß y) at the point

940

Chapter 14 Partial Derivatives

â â j k â â i â â (0ß 0ß 0). A normal to this plane is â 0 "00 50 â œ 2500i  5000j  10,000k, or i  2j  4k. So at â â â "00 0 25 â (!ß 0) the vector

`f `x

in the direction of

i " È5

`f `y

j is approximately i  2j . Thus the geologists should drill their fourth borehole

(i  2j) from the first borehole.

23. w œ ert sin 1x Ê wt œ rert sin 1x and wx œ 1ert cos 1x Ê wxx œ 1# ert sin 1x; wxx œ positive constant determined by the material of the rod Ê 1# ert sin 1x œ

" c#

" c#

wt , where c# is the

arert sin 1xb

# #

Ê ar  c# 1# b ert sin 1x œ 0 Ê r œ c# 1# Ê w œ ec 1 t sin 1x 24. w œ ert sin kx Ê wt œ rert sin kx and wx œ kert cos kx Ê wxx œ k# ert sin kx; wxx œ Ê k# ert sin kx œ

" c#

" c#

wt # #

arert sin kxb Ê ar  c# k# b ert sin kx œ 0 Ê r œ c# k# Ê w œ ec k t sin kx. # #

Now, w(Lß t) œ 0 Ê ec k t sin kL œ 0 Ê kL œ n1 for n an integer Ê k œ # # # # As t Ä _, w Ä 0 since ¸sin ˆ nL1 x‰¸ Ÿ 1 and ec n 1 tÎL Ä 0.

n1 L

# # # # Ê w œ ec n 1 tÎL sin ˆ nL1 x‰ .

CHAPTER 15 MULTIPLE INTEGRALS 15.1 DOUBLE INTEGRALS 1.

'03 '02 a4
y# b dy dx œ '03 ’4y
y3 “ # dx œ 163 '03 dx œ 16

2.

'03 'c02 a(x# y
2xyb dy dx œ '03 ’ x 2y


xy# “

œ '0 a4x
2x# b dx œ ’2x#


œ0

$

!

# #

3

3.

$ 2x$ 3 “!

'c01 'c11 (x  y  1) dx dy œ 'c01 ’ x2

#

4.

5.


#

 yx  x“

œ 'c1 (2y  2) dy œ cy#  2yd
" œ 1 0

!

dx

"
"

dy

!

'121 '01 (sin x  cos y) dx dy œ '121 c(
cos x)  (cos y)xd 1! dy 21 œ '1 (1 cos y  2) dy œ c1 sin y  2yd #1 1 œ 21

'01 '0x (x sin y) dy dx œ '01 c
x cos yd !x dx 1 1 œ '0 (x
x cos x) dx œ ’ x2
(cos x  x sin x)“ #

œ

6.

1# #

!

2

'01 '0sin x y dy dx œ '01 ’ y2 “ sin x dx œ '01 "# sin# x dx #

œ

" 4

!

'0 (1
cos 2x) dx œ "4
x
2" sin 2x‘ !1 œ 14 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

942 7.

Chapter 15 Multiple Integrals

'1ln 8 '0ln yexby dx dy œ '1ln 8 cexbyd !ln y dy œ '1ln 8 ayey
eyb dy œ c(y
1)ey
ey d 1ln 8 œ 8(ln 8
1)
8  e œ 8 ln 8
16  e

'12 'yy

#

8.

dx dy œ '1 ay#
yb dy œ ’ y3
2

$

œ ˆ 83
2‰
ˆ "3
#" ‰ œ

7 3




œ

3 #

5 6

'01 '0y 3y$ exy dx dy œ '01 c3y# exy d 0y #

9.

# y# # “"

#

dy

œ '0 Š3y# ey
3y# ‹ dy œ ’ey
y$ “ œ e
2 1

$

"

$

!

10.

Èx

'14 '0 œ

3 #

3 #

eyÎÈx dy dx œ

'14
32 Èx eyÎÈx ‘ 0Èx dx

% (e
1) '1 Èx dx œ
32 (e
1) ˆ 23 ‰ x$Î# ‘ " œ 7(e
1) 4

11.

'12 'x2x

x y

dy dx œ '1 cx ln yd x2x dx œ (ln 2) '1 x dx œ

12.

'12 '12

1 xy

13.

'01 '01cx ax#  y# b dy dx œ '01 ’x# y  y3 “ "
x dx œ '01 ’x# (1
x)  (1
3x) “ dx œ '01 ’x#
x$  (1
3x) “ dx

2

dy dx œ '1

2

2

" x

3 #

ln 2

(ln 2
ln 1) dx œ (ln 2) '1 dx œ (ln 2)# 2

$

$

$

0

$

œ ’ x3


x% 4




" (1
x)% “ 1# !

œ ˆ 3"


" 4


0‰
ˆ0
0


" ‰ 1#

œ

" 6

14.

'01 '01 y cos xy dx dy œ '01 csin xyd 1! dy œ '01 sin 1y dy œ

1" cos 1y‘ "! œ
1" (
1
1) œ 12

15.

'01 '01cu ˆv
Èu‰ dv du œ '01 ’ v2

#

œ '0 Š "#
u  1

u# #


vÈ u “

"
u 0


u"Î#  u$Î# ‹ du œ ’ 2u


du œ '0 ’ 1
2u# u
Èu(1
u)“ du 1

u# #



u$ 6

#

"


32 u$Î#  52 u&Î# “ œ !

" #




" #



" 6




2 3



Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 5

œ
"# 

2 5

" œ
10

Section 15.1 Double Integrals 16.

'12 '0ln t es ln t ds dt œ '12 ces ln td 0ln t dt œ '12 (t ln t
ln t) dt œ ’ t2

#

œ (2 ln 2
1
2 ln 2  2)
ˆ
"4  1‰ œ 17.

" 4

ln t


t# 4


t ln t  t“

# "

'c02 'vcv 2 dp dv œ 2'c02 cpd v
v dv œ 2'c02
2v dv œ
2 cv# d c2 œ 8 0

18.

È1cs

'01 '0

#

È1cs

8t dt ds œ '0 c4t# d 0 1

œ '0 4 a1
s# b ds œ 4 ’s
1

19.

#

ds

" s$ 3 “!

œ

8 3

'c11ÎÎ33 '0sec t 3 cos t du dt œ '
11ÎÎ33 c(3 cos t)ud 0sec t 1Î3

œ 'c1Î3 3 dt œ 21

20.

'03 '14
2u 4
v 2u dv du œ '03
2uv
4 ‘ 14
2u du 3 $ œ '0 (3
#u) du œ c3u
u# d ! œ ! #

21.

'24 '0Ð4
y)Î2 dx dy

22.

'
02 '0x2 dy dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

943

944

Chapter 15 Multiple Integrals

23.

'01 'xx dy dx

24.

'01 '1cy1cydx dy

25.

'1e 'ln1ydx dy

26.

'12 '0ln x dy dx

27.

'09 '0

28.

'04 '0

#

È

1 2

È9cy

È4cx

16x dx dy

y dy dx

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.1 Double Integrals È1cx

29.

'c11 '0

30.

'c22 '0

31.

'01 'x1 siny y dy dx œ '01 '0y siny y dx dy œ '01 sin y dy œ 2

32.

#

È4cy

3y dy dx

#

6x dx dy

'02 'x2 2y# sin xy dy dx œ '02 '0y2y# sin xy dx dy 2 2 œ '0 c
2y cos xyd 0y dy œ '0 a
2y cos y#  2yb dy #

œ c
sin y#  y# d ! œ 4
sin 4

33.

'01 'y1 x# exy dx dy œ '01 '0x x# exy dy dx œ '01 cxexyd 0x dx œ '0 axex
xb dx œ ’ "2 ex
1

È4cy

'02 '04cx 4xe
y dy dx œ '04 '0 #

34.

#

#

2y

œ '0 ’ #x(4e
y) “ 4

35.

'02

# 2y

Èln 3 Èln 3

'y/2

Èln 3

œ '0

È4cy

0

dy œ '0

4 2y e

Èln 3

ex dx dy œ '0 #

#

#

#

Èln 3

2xex dx œ cex d 0

" x# # “!

xe2y 4
y

œ

e
2 #

dx dy 2y

%

dy œ ’ e4 “ œ !

'02x ex

#

e)
" 4

dy dx

œ eln 3
1 œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

945

946 36.

Chapter 15 Multiple Integrals

'03 'È1xÎ3 ey

$

dy dx œ '0

1

'03y

#

$

ey dx dy

œ '0 3y# ey dy œ cey d ! œ e
1 1

37.

$

$

"

'01Î16 'y1Î2 cos a161x& b dx dy œ '01Î2 '0x

%

"Î%

cos a161x& b dy dx

161x b œ '0 x% cos a161x& b dx œ ’ sin a80 “ 1 1Î2

38.

'08 'È2x œ '0

39.

" y % 1

$

2

&

$

dy dx œ '0

2

dy œ

y y % 1

" 4

"Î# !

œ

" 801

'0y y "1 dx dy $

%

#

cln ay%  1bd ! œ

ln 17 4

' ' ay
2x# b dA R

xb1

œ 'c1 'cxc1 ay
2x# b dy dx  '0 0

1

'x1cc1x ay
2x# b dy dx

x
" 1x œ 'c1
"2 y#
2x# y‘ x1 dx  '0
2" y#
2x# y‘ x1 dx 0

1

œ 'c1
"# (x  1)#
2x# (x  1)
"# (
x
1)#  2x# (
x
1)‘dx 0

 '0
"# (1
x)#
2x# (1
x)
"# (x
1)#  2x# (x
1)‘ dx 1

œ
4 'c1 ax$  x# b dx  4 '0 ax$
x# b dx 0

1

%

œ
4 ’ x4 

40.

0

x$ 3 “ c1

" x$ 3 “!

%

 4 ’ x4


%

œ 4 ’ (
41) 

(
1)$ 3 “

3  4 ˆ 4"
3" ‰ œ 8 ˆ 12


4 ‰ 12

8 œ
12 œ
23

2
x

' ' xy dA œ ' ' xy dy dx  ' ' xy dy dx 0 x 2Î3 x 2Î3

R

2x

2Î3

1

2x 2
x œ '0
"2 xy# ‘ x dx  '2Î3
2" xy# ‘ x dx 1

œ '0 ˆ2x$
"# x$ ‰ dx  '2Î3
"# x(2
x)#
"# x$ ‘ dx 2Î3

1

œ '0

2Î3

3 #

x$ dx  '2Î3 a2x
x# b dx 1

" 2Î3 2‰ 8 ‰‘ ‰ ˆ
4 ˆ 2 ‰ ˆ 27 œ
38 x% ‘ 0 
x#
23 x$ ‘ #Î$ œ ˆ 38 ‰ ˆ 16 œ 81  1
3
9
3

41. V œ '0

1

'x2cx ax#  y# b dy dx œ '01 ’x# y  y3 “ 2cx dx œ '01 ’2x#
7x3

œ ˆ 23


$

x

7 12

2cx#

42. V œ 'c2 'x 1

œ ˆ 23





" 5

" ‰ 12


ˆ0
0


16 ‰ 12

œ

x# dy dx œ 'c2 cx# yd x


4" ‰
ˆ
16 3 

32 5




16 ‰ 4





(2
x)$ 3 “

27 81


ˆ 36 81


16 ‰ 81

$

dx œ ’ 2x3


7x% 12

4 3

2cx#

1

$

6 81

dx œ 'c2 a2x#
x%
x$ b dx œ
23 x$
15 x&
14 x% ‘
#

40 œ ˆ 60


1

12 60

"




15 ‰ 60


ˆ
320 60 

384 60




240 ‰ 60

œ

189 60

œ

63 20

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ


13 81 " (2
x)% 12 “ !

Section 15.1 Double Integrals 4cx#

43. V œ 'c4 '3x 1

4cx (x  4) dy dx œ 'c4 cxy  4yd 3x dx œ 'c4 cx a4
x# b  4 a4
x# b
3x#
12xd dx 1

1

#

œ 'c4 a
x$
7x#
8x  16b dx œ

14 x%
73 x$
4x#  16x‘
% œ ˆ
"4
1

œ

"

" 4




157 3

44. V œ '0

2

œ

‰  12‰
ˆ 64 3
64

625 12

È4cx

'0

7 3

#

È4cx

(3
y) dy dx œ '0 ’3y
2

œ ’ 32 xÈ4
x#  6 sin
" ˆ x# ‰
2x 

y# 2 “0

# x$ 6 “!

dx œ '0 ’3È4
x#
Š 4
#x ‹“ dx

#

2

œ 6 ˆ 1# ‰
4 

#

8 6

œ 31


16 6

œ

9 1
8 3

45. V œ '0

'03 a4
y# b dx dy œ '02 c4x
y# xd $! dy œ '02 a12
3y# b dy œ c12y
y$ d #! œ 24
8 œ 16

46. V œ '0

'04cx

2

2

#

2

œ
8x
43 x$  47. V œ '0

2

4cx#

a4
x#
yb dy dx œ '0 ’a4
x# b y
" 10

#

x& ‘ ! œ 16




32 3

œ

32 10

y# 2 “!

480
32096 30

œ

# (2
x)% 4 “!

xb1

0

a4
x# b dx œ '0 Š8
4x#  2

#

x% #‹

1

1Îx

2

50. V œ 4 '0

1Î3

'x1cc1x (3
3x) dy dx œ 6 'c01 a1
x# b dx  6 '01 (1
x)# dx œ 4  2 œ 6

2

2

2

$

0

1Î$

'1_ 'ec1 x"y dy dx œ '1_ ’ lnx y “ "

52.

'c11 'c1/1/È11ccxx

$

È


ˆ
1
x" ‰‘ œ 2 '1 ˆ1  x" ‰ dx

$

c7 ln ksec x  tan xk  sec x tan xd !

x

" x

'0sec x a1  y# b dy dx œ 4 '01Î3 ’y  y3 “ sec x dx œ 4 '01Î3 Šsec x  sec3 x ‹ dx

51.

$

ec x

bÄ_

1

1/ ˆ1cx# ‰

1Î#

c1/ a1c

_ _ 'c_ '
_ ax 1b"ay 1b -dx dy œ 2 '0_ Š y 21 ‹ Š #

’7 ln Š2  È3‹  2È3“

2 3

_

(2y  1) dy dx œ 'c1 cy#  ydº #

#

œ

dx œ '1
ˆ
x$x ‰ dx œ
lim

œ 4 lim c csin
" b
0d œ 21 bÄ1 #

#

x# b1Î#

lim

bÄ_

dx œ 'c1 È 2


"x ‘ b œ
lim 1

1

1
x #

bÄ_

ˆ "b
1‰ œ 1

dx œ 4 lim c csin
" xd ! bÄ1 b

tan
" b
tan
" 0‹ dy œ 21 lim

bÄ_

'0b y "1 dy #

œ 21 Š lim tan
" b
tan
" 0‹ œ (21) ˆ 1# ‰ œ 1# bÄ_

54.

'0_ '0_ xecÐx
2yÑ dx dy œ '0_ e2y _

œ '0 ec2y dy œ 55.

" # b lim Ä_

lim

bÄ_

_

c
xex
ex d b0 dy œ '0 e2y lim

a
ec2b  1b œ

bÄ_

a
beb
eb  1b dy

" #

' ' f(xß y) dA ¸ "4 f ˆ
#" ß 0‰  8" f(0ß 0)  8" f ˆ 4" ß 0‰  4" f ˆ #" ß 0‰  4" f ˆ
#" ß #" ‰  8" f ˆ!ß #" ‰  8" f ˆ 4" ß #" ‰ R

œ

" 4

dx

128 15

49. V œ '1 'c1Îx (x  1) dy dx œ '1 cxy  yd
1Î1xÎx dx œ '1
1  œ 2 cx  ln xd #" œ 2(1  ln 2)

53.

" #

œ 20

48. V œ 'c1 'cxc1 (3
3x) dy dx  '0

2 3

2

2 x '02cx a12
3y# b dy dx œ '02 c12y
y$ d #
dx œ '0 c24
12x
(2
x)$ d dx !

œ ’24x
6x# 

œ

dx œ '0

ˆ
#" 

" #

 0‰  8" ˆ0 

" 4



" #

 34 ‰ œ

3 16

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

947

948 56.

Chapter 15 Multiple Integrals

' ' f(xß y) dA ¸ "4
f ˆ 74 ß 94 ‰  f ˆ 94 ß 94 ‰  f ˆ 54 ß 114 ‰  f ˆ 74 ß 114 ‰  f ˆ 94 ß 114 ‰  f ˆ 114 ß 114 ‰  f ˆ 45 ß 134 ‰  f ˆ 47 ß 134 ‰ R

œ

" 16

‰ ˆ 11 13 ‰ ˆ 7 15 ‰ ˆ 9 15 ‰‘  f ˆ 94 ß 13 4 f 4 ß 4 f 4 ß 4 f 4 ß 4 (25  27  27  29  31  33  31  33  35  37  37  39) œ

57. The ray ) œ

1 6

È3 È4cx È 'xÎÈ3 È4
x# dy dx œ '0 3 ’a4
x# b
Èx3 È4
x# “ dx œ ”4x


R

58.

#

'2_ '02 ax
xb "(y
1) #

œ 6 ’ lim 1



0

_

dx œ 6 '2

cln (x
1)
ln xd 2b œ 6 lim

lim

bÄ_

bÄ_

dx x(x
1)

[ln (b
1)
ln b
ln 1  ln 2]

$

x

7 12




$



" ‰ 1#

(2
x)$ 3 “

dx œ


ˆ0
0


$ ’ 2x3

16 ‰ 12

œ




ˆ1
1" ‰ y 1y# dy 0 1
1 ˆ 21 ‰ ln 5 
"

'

21

ˆ2
1y ‰ 1 y # 2
"

2 tan


"

21
2 tan

21


2


7x% 12




" (2
x)% 12 “ !

4 3

'02 atan
" 1x
tan
" xb dx œ '02 'x1x 1"y

œ 2 tan

3 ‰ x #
x

'x2cx ax#  y# b dy dx œ '01 ’x# y  y3 “ 2cx dx

œ ˆ 23


œ'

2

"Î$

dy dx œ '0

2

#

'yyÎ1

" 1 y #

dx dy  '2

21

# #
" ‰
dy œ ˆ 12
" y 1 cln a1  y bd !  2 tan " 21

" 21

ln a1  41# b
2 tan
" 2  #

ln a1  41 b 

" #1

" #1

'y2Î1 1"y

#

dx dy 21

ln a1  y# b‘ 2

ln 5

ln 5 #

61. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all points where the integrand is negative. These criteria are met by the points (xß y) such that 4
x#
2y#   0 or x#  2y# Ÿ 4, which is the ellipse x#  2y# œ 4 together with its interior. 62. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all points where the integrand is positive. These criteria are met by the points (xß y) such that x#  y#
9 Ÿ 0 or x#  y# Ÿ 9, which is the closed disk of radius 3 centered at the origin. 63. No, it is not possible By Fubini's theorem, the two orders of integration must give the same result.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

x È3

.

È3

$Î#

a4
x # b 3È 3

ln ˆ1
"b ‰  ln 2“ œ 6 ln 2

1 œ '0 ’2x#
7x3

2

_


1) ' ˆ x#3
x  dy dx œ '2 ’ 3(y ax#
xb “ dx œ 2

'2b ˆ x
" 1
x" ‰ dx œ 6

bÄ_

59. V œ '0

_

#Î$

bÄ_

œ

x$ 3

20È3 9

œ 6 lim

60.

œ 24

meets the circle x#  y# œ 4 at the point ŠÈ3ß 1‹ Ê the ray is represented by the line y œ

Thus, ' ' f(xß y) dA œ '0 œ

384 16

• 0

Section 15.1 Double Integrals 64. One way would be to partition R into two triangles with the line y œ 1. The integral of f over R could then be written as a sum of integrals that could be evaluated by integrating first with respect to x and then with respect to y:

' ' f(xß y) dA R

œ '0

1

'22ccÐ2yyÎ2Ñ f(xß y) dx dy  '12 'y2c
Ð1 yÎ2Ñ f(xß y) dx dy.

Partitioning R with the line x œ 1 would let us write the integral of f over R as a sum of iterated integrals with order dy dx. 65.

'
bb '
bb e
x
y #

#

dx dy œ '
b '
b e
y e
x dx dy œ '
b e
y Œ'
b e
x dx dy œ Œ'
b e
x dx Œ'
b e
y dy b

b

#

#

b

#

#

b

#

b

#

#

b

#

#

# # # œ Œ'cb ecx dx œ Œ2 '0 ecx dx œ 4 Œ'0 ecx dx ; taking limits as b Ä _ gives the stated result.

b

66.

'01 '03 (y
x1)

dy dx œ '0

3

#

œ

b

#Î$

" 3 b lim Ä 1c

'0

b

dy (y
1)#Î$



'01 (y
x1)

dx dy œ '0

3

#

#Î$

" 3

b

'b

3

lim

b Ä 1b

dy (y
1)#Î$

œ

" (y
1)#Î$

lim

b Ä 1c

$

"

’ x3 “ dy œ !

" 3

'03 (y
dy1)

#Î$


(y
1)"Î$ ‘ b  lim
(y
1)"Î$ ‘ 3 0 b b Ä 1b

œ ’ lim c (b
1)"Î$
(
1)"Î$ “
’ lim b (b
1)"Î$
(2)"Î$ “ œ (0  1)
Š0
$È2‹ œ 1  $È2 bÄ1 bÄ1 67-70. Example CAS commands: Maple: f := (x,y) -> 1/x/y; q1 := Int( Int( f(x,y), y=1..x ), x=1..3 ); evalf( q1 ); value( q1 ); evalf( value(q1) ); 71-76. Example CAS commands: Maple: f := (x,y) -> exp(x^2); c,d := 0,1; g1 := y ->2*y; g2 := y -> 4; q5 := Int( Int( f(x,y), x=g1(y)..g2(y) ), y=c..d ); value( q5 ); plot3d( 0, x=g1(y)..g2(y), y=c..d, color=pink, style=patchnogrid, axes=boxed, orientation=[-90,0], scaling=constrained, title="#71 (Section 15.1)" ); r5 := Int( Int( f(x,y), y=0..x/2 ), x=0..2 ) + Int( Int( f(x,y), y=0..1 ), x=2..4 ); value( r5); value( q5-r5 ); 67-76. Example CAS commands: Mathematica: (functions and bounds will vary) You can integrate using the built-in integral signs or with the command Integrate. In the Integrate command, the integration begins with the variable on the right. (In this case, y going from 1 to x).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

949

950

Chapter 15 Multiple Integrals

Clear[x, y, f] f[x_, y_]:= 1 / (x y) Integrate[f[x, y], {x, 1, 3}, {y, 1, x}] To reverse the order of integration, it is best to first plot the region over which the integration extends. This can be done with ImplicitPlot and all bounds involving both x and y can be plotted. A graphics package must be loaded. Remember to use the double equal sign for the equations of the bounding curves. Clear[x, y, f] <
'13 '1x xy" dy dx ¸ 0.603

68.

'01 '01 ec x by

69.

'01 '01 tan
" xy dy dx ¸ 0.233

70.

'c11 '0

'0 '2ye 4

œ '0

2

œ

 "4

x#

dx dy

'0

x/2




ex dy dx
'2

4

#

"ˆ 4 4 e

#

È1cx

#

#‰

dy dx ¸ 0.558

3È1  x#  y# dy dx ¸ 3.142

The following graphs was generated using Mathematica.

71. Evaluate the integrals: 1

ˆ

'01 ex

#

dy dx

 2È1 erfia2b
2È1 erfia4b‰

¸ 1.1494 ‚ 106

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.2 Areas, Moments, and Centers of Mass 72. Evaluate the integrals:

Èy

'0 'x x cosay2 bdy dx œ '0 '0 3

9

9

2

œ

sina81b 4

x cosay2 bdx dy

¸ 0.157472

73. Evaluate the integrals:

'0 'y 2

4

È2y

3

œ

67,520 693

ax2 y  xy2 bdx dy œ '0

8

4cy#

'0 '0

Èx

'x /32 ax2 y  xy2 bdy dx 3

The following graphs was generated using Mathematica.

2

¸ 97.4315

74. Evaluate the integrals: 2

The following graphs was generated using Mathematica.

exy dx dy œ '0

4

È4cx

'0

exy dy dx

The following graphs was generated using Mathematica.

¸ 20.5648

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

951

952

Chapter 15 Multiple Integrals

75. Evaluate the integrals:

The following graphs was generated using Mathematica.

'1 '0 x1 y dy dx 1 2 4 2 œ '0 '1 x1 y dx dy
'1 'Èy x1 y dx dy 2

x#

‰ 1
lnˆ 27 4 ¸ 0.909543

76. Evaluate the integrals:

'1 'y 2

8 3

1 È x 2 y 2

dx dy œ '1

8

Èx

'1

The following graphs was generated using Mathematica.

3

1 È x 2 y 2

dy dx

¸ 0.866649

15.2 AREAS, MOMENTS, AND CENTERS OF MASS 1.

'02 '02cx dy dx œ '02 (2  x) dx œ ’2x  x2 “ # œ 2, #

2 2cy 2 or '0 '0 dx dy œ '0 (2  y) dy œ 2

2.

!

'02 '2x4 dy dx œ '02 (4  2x) dx œ c4x  x# d 02 œ 4, 4 yÎ2 4 or '0 '0 dx dy œ '0 y# dy œ 4

'c12 'yccy2 dx dy œ 'c12 ay#  y
2b dy #

3.

$

œ ’ y3  œ ˆ "3 

y# # " #


2y“

"


#


2‰  ˆ 83  2  4‰ œ

9 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.2 Areas, Moments, and Centers of Mass

'02 'cyycy

dx dy œ '0 a2y  y# b dy œ ’y# 

#

4.

œ4

2

œ

8 3

# y$ 3 “!

4 3

5.

'0ln 2 '0e dy dx œ '0ln 2 ex dx œ cex d 0ln 2 œ 2  1 œ 1

6.

'1e 'ln2 xln x dy dx œ '1e ln x dx œ cx ln x  xd 1e

x

œ (e  e)  (0  1) œ 1

'01 'y2ycy

dx dy œ '0 a2y  2y# b dy œ
y#  23 y$ ‘ !

#

7.

1

#

œ

"

" 3

'c11 '2yy cc12 dx dy œ 'c11 ay#  1  2y#
2b dy #

8.

#

œ 'c1 a1  y# b dy œ ’y  1

9.

" y$ 3 “
"

œ

4 3

'06 'y2yÎ3 dx dy œ '06 Š2y  y3 ‹ dy œ ’y#  y9 “ ' #

#

œ 36 

$

!

216 9

œ 12

'03 'c2xx cx dy dx œ '03 a3x  x# b dx œ
32 x#  "3 x$ ‘ $! #

10.

œ

27 #

9œ

9 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

953

954 11.

Chapter 15 Multiple Integrals

'01Î4 'sincosx x dy dx 1Î4 œ '0 (cos x  sin x) dx œ csin x
cos xd 01Î4 œŠ

12.

È2 #




È2 # ‹

 (0
1) œ È2  1

'c21 'yyb2 dx dy œ 'c21 ay
2  y# b dy œ ’ y2

#

œ ˆ2
4  83 ‰  ˆ "#  2
3" ‰ œ 5 

13.


2y 

#

" #

œ

2

y$ 3 “ c1

9 #

'c01 'c12x
x dy dx
'02 'c1x
Îx2 dy dx

œ '
1 (1
x) dx
'0 ˆ1  x# ‰ dx 0

2

œ ’x


14.

0

x# 2 “ c1


’x 

2

x# 4 “0

œ  ˆ1
"# ‰
(2  1) œ

3 #

È

'02 'x0c4 dy dx
'04 '0 x dy dx 2 4 œ '0 a4  x# b dx
'0 x"Î# dx #

œ ’4x 

2

x$ 3 “0

15. (a) average œ œ

" 1#

2 1#

" 1#

" # Š 1# ‹

'0 '0 1

1Î2

" 1#


 sin ˆx
1# ‰
sin x‘ 1 œ 0

'01 ax  x$ b dx œ 12 ’ x2

#

" 4

#

[( sin 21
sin 1)  ( sin 1
sin 0)] œ 0



2 1#

2
ˆ 31 1#  sin #

'01 c cos (x
y)d !1Î# dx œ 12 '01
 cos ˆx
1# ‰
cos x‘ dx #


sin 1‰  ˆ sin

1 #


sin 0‰‘ œ

4 1#

'01 xy dy dx œ '01 ’ xy2 “ " dx œ '01 #x dx œ "4 œ 0.25; #

average value over the quarter circle œ

17. average height œ

32 3

sin (x
y) dy dx œ

1

2 1

œ

#

1

16. average value over the square œ '0

œ

16 3

'01 '01 sin (x
y) dy dx œ 1" '01 c cos (x
y)d 01 dx œ 1" '01 [ cos (x
1)
cos x] dx

c sin (x
1)
sin xd 0 œ

(b) average œ œ

4

23 x$Î# ‘ 0 œ ˆ8  83 ‰


" x4 4 “!

œ

È1cx

!

'0 '0

" #1

¸ 0.159. The average value over the square is larger.

1

#

xy dy dx œ

4 1

'01 ’ xy2 “

È1cx

ˆ 14 ‰

"

#

#

dx

0

'02 '02 ax#
y# b dy dx œ "4 '02 ’x# y
y3 “ # dx œ 4" '02 ˆ2x#
38 ‰ dx œ #" ’ x3 $

$

!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley




# 4x 3 “!

œ

8 3

Section 15.2 Areas, Moments, and Centers of Mass

œ œ

'ln2 2ln 2 'ln2 2ln 2

" (ln 2)#

18. average œ

' 2 ln 2

" " (ln 2)# x ln 2 " ˆ ln # ‰ (ln 2


19. M œ '0

1

2 ln 2

ln ln 2  ln ln 2) œ 1

#

and y œ

5 14

20. M œ $ '0

3

Iy œ $ '0

3

" #

œ ˆ ln"# ‰ cln xd ln 2

dx x

'x2cx 3 dy dx œ 3'01 a2  x#  xb dx œ 7# ; My œ '01 'x2cx

Ê xœ

œ

ln 2

2 ln 2

ln 2

1

2

'ln2 2ln 2 ’ lnxy “ 2 ln 2 dx

" (ln 2)#

dy dx œ

(ln 2
ln ln 2  ln ln 2) dx œ ˆ ln"2 ‰'

œ 3'0 a2x  x$  x# b dx œ

21. M œ '0

" xy

'03

; Mx œ '0

1

5 4

'x2cx

#

3y dy dx œ

#

3x dy dx œ 3 '0 cxyd x2cx dx 1

'01 cy# d x2cx

3 #

#

dx œ

3 #

#

'01 a4  5x#
x% b dx œ 195

38 35

dy dx œ $ '0 3 dx œ 9$ ; Ix œ $ '0 3

3

'03 y# dy dx œ $ '03 ’ y3 “ 3 dx œ 27$ ; Rx œ É MI œ È3; $

x

0

'03 x# dy dx œ $ '03 cx# yd $! dx œ $ '03 3x# dx œ 27$ ; Ry œ É MI œ È3 y

'y4Î
2ydx dy œ '02 Š4  y  y# ‹ dy œ 143 ; My œ '02 'y4Î
2y #

#

#

x dx dy œ

4cy

" #

'02 cx# d y4
Îy2 dy #

'0 Š16  8y
y#  y4 ‹ dy œ 128 ' 'y Î2 y dx dy œ '0 Š4y  y#  y# ‹ dy œ 103 15 ; Mx œ 0 2

2

%

2

$

#

Ê xœ 22. M œ '0

3

and y œ

64 35

5 7

'03cx dy dx œ '03 (3  x) dx œ 9# ; My œ '03 '03cx x dy dx œ '03 cxyd 03cx dx œ '03 a3x  x# b dx œ 9#

Ê x œ 1 and y œ 1, by symmetry 23. M œ 2'0

1

È1cx

'0

#

dy dx œ 2 '0 È1x# dx œ 2 ˆ 14 ‰ œ 1

œ '0 a1  x# b dx œ ’x  1

24. M œ

125$ 6

; My œ $ '0

5

Mx œ $ '0

5

6xcx

'x

Èa cx

25. M œ '0 '0 a

#

Ê xœyœ 26. M œ '0

1

œ

" 4

#

4a 31

#

"

x$ 3 “!

'x6xcx

#

y dy dx œ

dy dx œ

1a# 4

œ

2 3

Ê yœ

4 31

; Mx œ 2 '0

1

È1cx

'0

#

'0

$ #

6xcx

c y# d x

; My œ

#

Èa cx

'0a '0

5

#

#

#

625$ 1#

'0 a35x#  12x$
x% b dx œ 6256 $ 5

$ #

dx œ

1

#

dx

and x œ 0, by symmetry

5

5

È1cx

y dy dx œ '0 cy# d 0

x dy dx œ $ '0 cxyd x6xcx dx œ $ '0 a5x#  x$ b dx œ

;

Ê xœ

5 #

and y œ 5

a a È# # x dy dx œ '0 cxyd 0 a cx dx œ '0 xÈa#  x# dx œ

a$ 3

, by symmetry

'0sin x dy dx œ '01 sin x dx œ 2; Mx œ '01 '0sin x y dy dx œ "# '01 cy# d 0sin x dx œ "# '01 sin# x dx

'01 (1  cos 2x) dx œ 14 È4cx

Ê xœ

1 #

and y œ

È4cx

27. Ix œ 'c2 'cÈ4cx# y# dy dx œ 'c2 ’ y3 “ dx œ cÈ4cx# I o œ I x
I y œ 81 2

28. Iy œ '1

21

#

2

$

0

#

ˆ

‰

#

2 3

'c22 a4  x# b$Î# dx œ 41; Iy œ 41, by symmetry;

#

ex

0

'b xe 0

lim

1 8

'0 sin x Îx x# dy dx œ '121 asin# x  0b dx œ "# '121 (1  cos 2x) dx œ 1#

29. M œ 'c_ '0 dy dx œ '
_ ex dx œ œ

1 #

b Ä
_

x

dx œ

lim

b Ä
_

lim

b Ä
_

'b0 ex dx œ 1 

cxex  ex d b0 œ 1 

eb œ 1; My œ '
_ '0 x dy dx œ '
_ xex dx b Ä
_ 0

ex

0

lim

abeb  eb b œ 1; Mx œ 'c_ '0 y dy dx b Ä
_ 0

ex

lim

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

955

956

Chapter 15 Multiple Integrals œ

30.

0 'c_ e2x dx œ "#

" #

_ e My œ '0 '0


x# Î2

lim

b Ä
_

'b0 e2x dx œ 4"

x dy dx œ lim

bÄ_

ycy#

" 4

Ê x œ 1 and y œ

'0b xe
x#Î2 dx œ 




lim

bÄ_

ycy#

" ex# Î2

b

 1‘ 0 œ 1

y 31. M œ '0 'cy (x
y) dx dy œ '0 ’ x2
xy“ dy œ '0 Š y2  2y$
2y# ‹ dy œ ’ 10  cy 2

2

#

ycy

2

%

ycy

&

Ix œ '0 'cy y# (x
y) dx dy œ '0 ’ x 2y
xy$ “ dy œ '0 Š y2  2y&
2y% ‹ dy œ cy #

2

2

#

# #

2

'

y% #

64 105




# 2y$ 3 “!

œ

8 15

;

;

Ix Rx œ É M œ É 78 œ 2É 72

È3Î2

È12
4y

32. M œ 'cÈ3Î2 '4y#

#

È3Î2

5x dx dy œ 5 '
È3Î2 ’ x2 “ #

È12
4y

#

4y#

dy œ

È

'
È33ÎÎ22 a12  4y#  16y% b dy œ 23È3

5 #

'x2cx (6x
3y
3) dy dx œ '01
6xy
3# y#
3y‘ x2cx dx œ '01 a12  12x# b dx œ 8; 1 2cx 1 1 2cx My œ '0 'x x(6x
3y
3) dy dx œ '0 a12x  12x$ b dx œ 3; Mx œ '0 'x y(6x
3y
3) dy dx 1 3 17 œ '0 a14  6x  6x#  2x$ b dx œ 17 # Ê x œ 8 and y œ 16

33. M œ '0

1

34. M œ '0

1

'y2ycy (y
1) dx dy œ '01 a2y  2y$ b dy œ "# ; Mx œ '01 'y2ycy #

#

#

My œ '0

1

#

2ycy#

'y

#

x(y
1) dx dy œ '0 a2y#  2y% b dy œ 1

œ 2 '0 ay$  y& b dy œ 1

4 15

Ê xœ

8 15

y(y
1) dx dy œ '0 a2y#  2y% b dy œ 1

and y œ

8 15

; Ix œ '0

1

2ycy#

'y

4 15

;

y# (y
1) dx dy

#

" 6

35. M œ '0

'06 (x
y
1) dx dy œ '01 (6y
24) dy œ 27; Mx œ '01 '06 y(x
y
1) dx dy œ '01 y(6y
24) dy œ 14; 1 6 1 14 ' 1' 6 # My œ '0 '0 x(x
y
1) dx dy œ '0 (18y
90) dy œ 99 Ê x œ 11 3 and y œ 27 ; Iy œ 0 0 x (x
y
1) dx dy 1 I ‰ ÉM œ 216 '0 ˆ y3
11 œ4 6 dy œ 432; Ry œ 1

y

36. M œ 'c1 'x# (y
1) dy dx œ  'c1 Š x#
x#  3# ‹ dx œ 1

œ

48 35

1

1

%

; My œ 'c1 'x# x(y
1) dy dx œ 'c1 Š 3x #  1

œ 'c1 Š 3x2  1

#

x' 2

1

1

 x% ‹ dx œ

16 35

x& #

x#

; Mx œ 'c1 'x# y(y
1) dy dx œ 'c1 Š 56  1

1

 x$ ‹ dx œ 0 Ê x œ 0 and y œ

%

31 15

x

1

œ 'c1 Š 7x#
x% ‹ dx œ 1

'

38. M œ '0

20

My œ '0

2 3

x% #‹

dx

1

1

x#

1

&

13 31

'

x% 2‹

dx œ

13 15

; Iy œ 'c1 '0 x# (7y
1) dy dx 1

x

I

21 ; Ry œ É My œ É 31

#

'c1 1

x ˆ1


x ‰ 20

;

#

'c11 ˆ1
20x ‰ dy dx œ '020 ˆ2
10x ‰ dx œ 60; Mx œ '020 'c11 y ˆ1
20x ‰ dy dx œ '020 ’ˆ1
#x0 ‰ Š y# ‹“ "

20

œ

7 5



; Iy œ 'c1 'x# x# (y
1) dy dx 1

; Mx œ 'c1 '0 y(7y
1) dy dx œ 'c1 Š 7x3


My œ 'c1 '0 x(7y
1) dy dx œ 'c1 Š 7x#
x$ ‹ dx œ 0 Ê x œ 0 and y œ 1

9 14

x' 3

I

1

#

1

3 ; Ry œ É My œ É 14

37. M œ 'c1 '0 (7y
1) dy dx œ 'c1 Š 7x#
x# ‹ dx œ 1

32 15

dy dx œ '0 Š2x
20

x# 10 ‹

'020 ˆ1
20x ‰ dx œ 20; Rx œ É MI œ É 3"

dx œ

2000 3

Ê xœ

100 9

and y œ 0; Ix œ '0

20

'c1

x

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1


"

y# ˆ1


x ‰ 20

dx œ 0;

dy dx

Section 15.2 Areas, Moments, and Centers of Mass 39. M œ '0 'cy (y
1) dx dy œ '0 a2y#
2yb dy œ 1

y

1

5 3

; Mx œ '0 'cy y(y
1) dx dy œ 2 ' ay$
y# b dy œ 0 1

1

œ

y

1

Ix Ê Rx œ É M œ

9 10

Io œ Ix
Iy œ

3È 6 10

; Iy œ '0 'cy x# (y
1) dx dy œ 1

40. M œ '0 'cy a3x#
1b dx dy œ '0 a2y$
2yb dy œ 1

y

1

È5 3

41.

1

dy dx œ 10,000 a1  e
# b 'c5

Io Ê Ro œ É M œ

I

Ê Ry œ É My œ

3È 2 10

;

32 45

1

16 15

dx kxk 1 #

1

11 30

I

11 Ê Ry œ É My œ É 45 ;

œ 10,000 a1  e
# b ’'c5 !


'!

5

dx 1
x#

dx 1  x#

“

! & œ 10,000 a1  ec2 b
2 ln ˆ1  x# ‰‘
&
10,000 a1  ec2 b
2 ln ˆ1
x# ‰‘ ! œ 10,000 a1  e
# b
2 ln ˆ1
5# ‰‘
10,000 a1  e
# b
2 ln ˆ1
5# ‰‘ œ 40,000 a1  e
# b ln ˆ 27 ‰ ¸ 43,329

'01 'y2ycy 100(y
1) dx dy œ '01 c100(y
1)xd y2ycy #

42.

#

#

#

#

œ 200 ’ y2 

" y% 4 “!

a ˆ1cx# ‰

43. M œ 'c1 '0 1

œ

2a# #

dy œ '0 100(y
1) a2y  2y# b dy œ 200 '0 ay  y$ b dy 1

1

œ (200) ˆ "4 ‰ œ 50

dy dx œ 2a '0 a1  x# b dx œ 2a ’x  1

'01 a1  2x#
x% b dx œ a# ’x  2x3

$




" x& 5 “!

œ

" x$ 3 “!

8a# 15

œ

4a 3

a ˆ1cx# ‰

; Mx œ 'c1 '0 1

y dy dx

#

Ê yœ

Mx M

œ

Š 8a 15 ‹ Š 4a 3‹

œ

. The angle ) between the

2a 5

x-axis and the line segment from the fulcrum to the center of mass on the y-axis plus 45° must be no more than 5 ‰ 1 90° if the center of mass is to lie on the left side of the line x œ 1 Ê )
14 Ÿ 1# Ê tan
" ˆ 2a 5 Ÿ 4 Ê aŸ # . Thus, if 0  a Ÿ 44. f(a) œ Ia œ '0

4

5 #

, then the appliance will have to be tipped more than 45° to fall over.

'02 (y  a)# dy dx œ '04 ’ (2
3a)

$

a$ 3“




dx œ

4 3

c(2  a)$
a$ d ; thus f w (a) œ 0 Ê 4(2  a)#
4a#

œ 0 Ê a#  (2  a)# œ 0 Ê 4
4a œ 0 Ê a œ 1. Since f w w (a) œ 8(2  a)
8a œ 16  0, a œ 1 gives a minimum value of Ia . 45. M œ '0

1

œ '0

1

È

'c11ÎÎÈ11

xx

2x È 1
x #

# #

dy dx œ '0

1

dx œ ’2 a1 

46. (a) I œ 'cLÎ2 $ x# dx œ LÎ2

(b) I œ '0 $ x# dx œ L

47. (a)

" #

œ M œ '0

1

$ L$ 12

$ L$ 3

'y2ycy

#

#

2 È 1
x #

" dx œ c2 sin
" xd ! œ 2 ˆ 1#  0‰ œ 1; My œ '0

"Î# " x# b “ !

1

œ2 Ê xœ $

L Ê R œ É $12 † $

Ê R œ É $L3 †

" $L

" $L

œ

œ

2 1

È

'
11ÎÎÈ11

xx

# #

x dy dx

and y œ 0 by symmetry

L 2È 3

L È3

$ dx dy œ 2$ '0 ay  y# b dy œ 2$ ’ y2  1

;

; Ix œ '0 'cy y# a3x#
1b dx dy œ '0 a2y&
2y$ b dy œ y

2 È5 5

y

1

1

'c55 'c02 10,000e x #

y

y

1

y

6 5

k k

1

; Iy œ '0 'cy x# a3x#
1b dx dy œ 2 '0 ˆ 35 y&
3" y$ ‰ dy œ

Io œ Ix
Iy œ

1

; Ix œ '0 'cy y# (y
1) dx dy œ '0 a2y%
2y$ b dy

'01 a2y%
2y$ b dy œ 103

1

y

Ix Ê Rx œ É M œ

;

; Mx œ '0 'cy y a3x#
1b dx dy œ '0 a2y%
2y# b dy œ

3 #

My œ '0 'cy x a3x#
1b dx dy œ 0 Ê x œ 0 and y œ 1

" 3

7 10

7 6

3È 2 5

Io Ê R! œ É M œ

6 5

y

1

y

My œ '0 'cy x(y
1) dx dy œ '0 0 dy œ 0 Ê x œ 0 and y œ

957

#

" y$ 3 “!

œ 2$ ˆ "6 ‰ œ

$ 3

Ê $œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

3 #

5 6

958

Chapter 15 Multiple Integrals #

(b) average value œ

'01 'y2y#
y

(yb1) dx dy

2y
y#

'0 'y# 1

dx dy

œ

Š "# ‹ Š "3 ‹

œ

3 #

œ $ , so the values are the same

48. Let (xi ß yi ) be the location of the weather station in county i for i œ 1ß á ß 254. The average temperature 254

! T(xi ßyi ) ?i A

in Texas at time t! is approximately

iœ1

A

, where T(xi ß yi ) is the temperature at time t! at the

weather station in county i, ?i A is the area of county i, and A is the area of Texas. 49. (a) x œ

My M

œ 0 Ê My œ ' ' x$ (xß y) dy dx œ 0 R

(b) IL œ ' ' (x
h)# $ (xß y) dA œ ' ' x# $ (xß y) dA
' ' 2hx $ (xß y) dA  ' ' h# $ (xß y) dA R

R

œ Iy
0  h#

R

' ' $ (xß y) dA œ IcÞmÞ  mh#

R

R

23 11 ‰# ˆ 5 ‰# œ 35 50. (a) IcÞmÞ œ IL
mh# Ê Ixœ5/7 œ Iy
mh# œ 39 ; Iyœ11Î14 œ Ix
mh# œ 12
14 ˆ 14 œ 5
14 7 # # ˆ 2 ‰ œ 95 ; Iyœ2 œ Iyœ11Î14  mh# œ 47 ˆ 17 ‰ œ 24 (b) Ixœ1 œ Ixœ5Î7  mh# œ 23 35  14 7 14  14 14

51. Mxp"
p# œ ' ' y dA"  ' ' y dA# œ Mx"  Mx# Ê x œ R"

R#

thus c œ x i  y j œ œ

" m"
m#

" m"
m#

Mx"
Mx# m"
m#

caMx"  Mx# b i  aMy"  My# b jd œ

cm" ax" i  y" jb  m# ax# i  y# jbd œ

m" c"
m# c# m"
m#

; likewise, y œ

" m"
m#

My"
My# m"
m#

47 14

;

cam" x"  m# x# b i  am" y"  m# y# b jd

52. From Exercise 51 we have that Pappus's formula is true for n œ 2. Assume that Pappus's formula is true for k
1

n œ k
1, i.e., that c(k
1) œ

! mi ci iœ1 k
1

! mi

. The first moment about x of k nonoverlapping plates is

iœ1

k
1

!




iœ1

MxcÐk
1Ñ
Mxk

' ' y dAi   ' ' y dAk œ MxcÐk
1Ñ  Mxk Ê x œ Ri

k
1

! mi 
mk

Rk

thus c(k) œ x i  y j œ

" k

! mi




; similarly, y œ

iœ1

MycÐk
1Ñ
Myk k
1

! mi 
mk




iœ1

ˆMxcÐk
1Ñ  Mxk ‰ i  ˆMycÐk
1Ñ  Myk ‰ j‘

iœ 1

œ

" k

! mi

k
1

k
1

iœ1

iœ1

”ŒŒ ! mi  xc  mk xk  i  ŒŒ ! mi  yc  mk yk  j•

iœ 1

k
1

œ

" k

! mi iœ1

œ

–


! mi  axc i  yc jb  mk axk i  yk jb— œ

(b) c œ (c) c œ (d) c œ 54. c œ

! mi  c(k1)
mk ck




iœ1

k
1

! mi

iœ1

m" c"
m# c#
á
mk
1 ck
1
mk ck m"
m#
á
mk
1
mk

53. (a) c œ

Ê

k
1

iœ1

, and by mathematical induction the statement follows.

8(i
3j)
2(3i
3.5 j)
31j œ 14i 10 Ê x œ 75 and y œ 31 8
2 10 8(i
3j)
6(5i
2 j) 38i
36j 19 18 œ Ê x œ and y œ 14 14 7 7 2(3i
3.5j)
6(5i
2 j) 36i
19j 9 19 œ Ê x œ and y œ 8 8 2 8 8(i
3j)
2(3i
3.5 j)
6(5i
2j) 44i
43j 11 œ 16 Ê x œ 4 and 16

15 ˆ 34 i
7j‰
48(12i
j) œ 15(3i
28j4) †
6348(48i
4j) 15
48 17 x œ 261 #8 and y œ 7

œ

2349 i
612 j 4†63

yœ

œ

43 16

261 i
68j 4†7

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

;

Section 15.3 Double Integrals in Polar Form 55. Place the midpoint of the triangle's base at the origin and above the semicircle. Then the center of 4a ‰ mass of the triangle is ˆ0ß h3 ‰ , and the center of mass of the disk is ˆ0ß
31 from Exercise 25. From #

Pappus's formula, c œ

(ah) ˆ h3 j‰
Š 12a ‹ ˆ 34a1 j‰

œ

#

Šah
1#a ‹

# $ Š ah
3 2a ‹ j

#

Šah
1#a ‹

, so the centroid is on the boundary

if ah#
2a$ œ 0 Ê h# œ 2a# Ê h œ aÈ2 . In order for the center of mass to be inside T we must have ah#
2a$  0 or h  aÈ2. 56. Place the midpoint of the triangle's base at the origin and above the square. From Pappus's formula, cœ

h #ˆ Š sh # ‹Š 3 j‹
s c

Š sh #

j‰

s 2

, so the centroid is on the boundary if


s# ‹

sh# 6




s$ #

œ 0 Ê h#
3s# œ 0 Ê h œ sÈ3.

15.3 DOUBLE INTEGRALS IN POLAR FORM

È1cx

1.

'c11 '0

2.

'c11 'cÈ11ccxx dy dx œ '021 '01 r dr d) œ "# '021 d) œ 1

3.

'01 '0

4.

'c11 'cÈ11ccyy ax#  y# b dx dy œ '021 '01 r$ dr d) œ "4 '021 d) œ 1#

5.

'caa 'cÈaa ccxx dy dx œ '021 '0a r dr d) œ a2 '021 d) œ 1a#

6.

'02 '0

7.

'06 '0yx dx dy œ '11ÎÎ42 '06 csc

8.

'02 '0x y dy dx œ '01Î4 '02 sec

9.

'c01 'c0È1 c x 1
È2x
y

È

#

dy dx œ '0

1

'01 r dr d) œ "# '01 d) œ 1#

#

#

È1cy

ax#  y# b dx dy œ '0

1Î2

#

È

'01

" 4

r$ dr d) œ

'01Î2

d) œ

1 8

#

#

È

#

#

#

#

#

È4cy

ax#  y# b dx dy œ '0

1Î2

#

#

#

)

'02 r$ dr d) œ 4 '01Î2 d) œ 21

r# cos ) dr d) œ 72 '1Î4 cot ) csc# ) d) œ
36 ccot# )d 1Î4 œ 36

)

1Î2

r# sin ) dr d) œ

dy dx œ '1

31Î2

#

8 3

1Î2

'0 Î4 tan ) sec# ) d) œ 43 1

'01 1 2r
r dr d) œ 2 '131Î2 '01 ˆ1
1
" r ‰ dr d) œ 2 '131Î2 (1
ln 2) d)

œ (1
ln 2)1 10.

'c11 'c0È1 c y

4 È x #
y # 1
x #
y #

#

œ 41
1

'0ln 2 '0

12.

'01 '0

13.

'02 '0

#

È1 c x

#

#

eÈx

#

 y#

'01

dx dy œ '0

1Î2

1Î2

e
ˆx y ‰ dy dx œ '0

È1 c (x c 1)

œ ) 

31Î2

4r# 1
r#

dr d) œ 4 '1Î2

31Î2

'01 ˆ1
1
" r ‰ dr d) œ 4 '13Î12Î2 ˆ1
14 ‰ d) #

#

È(ln 2) c y

11.

dx dy œ '1Î2

#

#

sin 2) 2

#

x
y x#
y#

dy dx œ '0

'01 rer

1Î2

1Î2

 sin# )‘ 0 œ

1
2 #

'0ln 2

'02 cos

œ

1 #

)

#

rer dr d) œ '0 (2 ln 2
1) d) œ 1Î2

dr d) œ
"#

r(cos )
sin )) r#

1 #

(2 ln 2
1)

'01Î2 ˆ "e
1‰ d) œ 1(e4e 1)

r dr d) œ '0 a2 cos# )  2 sin ) cos )b d) 1Î2

1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

959

960

Chapter 15 Multiple Integrals

14.

'02 '
0È1
(y
1)

15.

'c11 'cÈ11ccyy

16.

'c11 'cÈ11ccxx

17.

'01Î2 '02

È

xy# dx dy œ

'11Î2 '02 sin

dy dx œ 4 '0

1Î2

#

2 a1
x #
y # b #

#

È2
sin 2)

1 2

'11
cos

19. A œ 2 '0

'012 cos 3

1Î2

1Î6

)

23. Mx œ '0

1

'01ccos

)

Èa c x #

#

r dr d) œ

Èa c x

#

1 2

1

) # d) œ

#

25. M œ 2 '1Î6

1Î2

26. Io œ '1Î2

31Î2

'11
cos

Î

r dr d) œ 2 '0 ˆ 3#
2 cos )  1 2

27. M œ 2 '0

1

'01 b cos

) œ 2 '0 ˆ 4 cos  3 1

28. Io œ '0

21

'01 b cos

)

31 2

d) œ


4

1

21

'0a

r& sin# ) dr d) œ

ka' 6

2) '021 1  cos d) œ ka6 1 ; # '

'0a r& dr d) œ ka6 '021 d) œ ka3 1 '

'

dr d) œ 2 '1Î6 (6 sin )
3) d) œ 6 c
2 cos )
)d 11ÎÎ62 œ 6È3
21 1Î2

r dr d) œ )

cos 2) ‰ #

'0 (1
cos ))$ sin ) d) œ 4

#

)

8
1 4

1

21

)

d) œ 1

'0 Î2 ˆ 3#  2 sin )
cos#2) ‰ d) œ 381  1

3r# sin ) dr d) œ

'36 sin

1Î2

641$ 27

Io œ 'ca 'cÈa# c x# k ax#  y# b dy dx œ k '0 a


1
" r# ‘ " d) œ 2 ' ! 0

1Î2

" #

'13Î2Î2 acos# )
2 cos )b d) œ "#  sin42)  #)
2 sin )‘ 3Î2Î2 œ 2  14 1

1

1

r dr d) œ '0 (1  cos ))# d) œ 1

15 24

 cos 2)
sin# ) cos ) 

r$ dr d) œ

" 4

31 #

; My œ 2'0

cos 4) ‰ 4

1

d) œ

51 4

'01 b cos

Ê xœ

)

r# cos ) dr d)

5 6

and y œ 0, by symmetry

'02 (1  cos ))% d) œ 351 16 1

29. average œ

4 1 a#

'01Î2 '0a rÈa#
r# dr d) œ 314a '01Î2 a$ d) œ 2a3

30. average œ

4 1 a#

'01Î2 '0a r# dr d) œ 314a '01Î2 a$ d) œ 2a3

31. average œ

" 1 a#

'caa 'cÈaa ccxx

#

#

È

#

#

# #

Èx#  y# dy dx œ

" 1 a#

4 5

(ln 4
1) d) œ 1(ln 4
1)

(2
sin 2)) d) œ 2(1
1)

24. Ix œ 'ca 'cÈa# c x# y# ck ax#  y# bd dy dx œ k '0 a

dr d) œ 4'0

Î œ


1 Î6

'01
sin

)

2r a1
r# b#

1

1 2

r dr d) œ 144 '0 cos# 3) d) œ 121

21. A œ '0

'01
cos

1

1Î2

Î

)

1Î2

'1Î2 sin( ) cos ) d) œ 45 csin) )d

32 5

ln ar#  1b r dr d) œ 2'0

1 2

'04 Î3 r dr d) œ 89 '02 )

'01

'01

r dr d) œ '0 a2 cos )  cos# )b d) œ

)

20. A œ '0

22. A œ 4 '0

Î

r dr d) œ 2 '0

18. A œ 2 '0

1Î2

sin# ) cos ) r% dr d) œ 1Î2

#

21

)

ln ax#  y#  1b dx dy œ 4 '0

#

È

#

'021 '0a r# dr d) œ 3a1 '021 d) œ 2a3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.3 Double Integrals in Polar Form

œ

" 1

' ' c(1
x)#  y# d dy dx œ 1" ' ' c(1
r cos ))#  r# sin# )d r dr d) 0 0 21

" 1

32. average œ

'0 '0 ar 21

Èe

1

R

$


2r# cos )  rb dr d) œ

Š lnrr ‹ r dr d) œ '0

Èe

'021 ˆ 34
2 cos3 ) ‰ d) œ 1"  34 )
2 sin3 ) ‘ 021 œ #3

" 1

33.

'021 '1

34.

'021 '1e Š lnrr ‹ dr d) œ '021 '1e ˆ 2 lnr r ‰ dr d) œ '021 c(ln r)# d 1e d) œ '021 d) œ 21

21

#

2 ln r dr d) œ 2'0 cr ln r
rd1e d) œ 2'0 Èe ˆ "#
1‰  1‘ d) œ 21ˆ2
Èe‰ 21

21

"Î#

#

35. V œ 2 '0

1Î2

œ

2 3

œ

'11
cos

r# cos ) dr d) œ

21 È 2 3

È2 cos 2)

1Î4

'0




32 3

'0 Î2 a3 cos# )  3 cos$ )  cos% )b d) 1

2 3

'0

'0

e
ax

È1

" a1
x #
y # b #

1 4 b lim Ä_

#

 y# b

lim

ˆ1


dt œ

dx dy œ '0

51 8

'0 Î4 (2
2 cos 2))$Î#
2$Î# ‘ d) 1

#

3 4

:

" #

'01Î2 d) œ 14

21

Over the disk x#  y# Ÿ 1:

Î

'0_

r a1
r# b#

lim

x#  y# Ÿ 1

1 # b lim Ä_

dr d) œ

1 4

'' R

'' R

a Ä 1c

Ê Iœ

'0b rer

#

lim

bÄ_

dr• d)

È1 #

'0_ et# dt œ Š È21 ‹ Š È#1 ‹ œ 1, from part (a)

2 È1

œ

#

" 1  x#  y#

dA œ '0

21

œ '0 ˆ
"# ln 4" ‰ d) œ (ln 2) '0 d) œ 1 ln 4

21



'0_ Šer ‹ r dr d) œ '01Î2 ”

Šeb
1‹ d) œ

1 2

" ‰ 1
b#

Î

1 2

dx dy œ '0

39. Over the disk x#  y# Ÿ

œ '0

4 3

0

bÄ_ x
t# 2e

(b) x lim Ä_

'0_ '0_

rÈ2
r# dr d) œ
43

œ

$

'01Î2

œ
"#

sin 4) ‘ 1Î2 32 0

'01Î4 a1
cos# )b sin ) d) œ 213È2
323 ’ cos3 )
cos )“ 1Î4 œ 61È2
409 È2  64

_ _

37. (a) I# œ '0

œ

)

 "85)  sin 2)  3 sin )
sin$ ) 

36. V œ 4 '0

38.

'1

1

È3Î2

'0

'0b

r a1
r# b#

dr œ

1 4 b lim Ä_


1
" r# ‘ b 0

È3Î2

dr d) œ '0 
2" ln a1
r# b‘ 0 21

r 1  r#

21

" 1  x#  y#

dA œ '0

21

'01 1 r r

#

dr d) œ '0 ’ lim c aÄ1 21

'0a 1 r r

#

dr“ d)


"# ln a1
a# b‘ d) œ 21 † lim c 
"# ln a1
a# b‘ œ 21 † _, so the integral does not exist over aÄ1

40. The area in polar coordinates is given by A œ '! '0 r dr d) œ '! ’ r2 “ "

fÐ)Ñ

"

#

fÐ)Ñ 0

d) œ

" #

'!" f # ()) d) œ '!"

where r œ f())

'021 '0a c(r cos )
h)#  r# sin# )d r dr d) œ 1"a '021 '0a ar$
2r# h cos )  rh# b dr d) '021 Š a4
2a h3cos )  a #h ‹ d) œ 1" '021 Š a4
2ah 3cos )  h# ‹ d) œ 1" ’ a4)
2ah 3sin )  h#) “ 21

41. average œ œ

" 1 a#

œ

" #

d)

" 1 a#

%

#

$

# #

#

#

#

#

0

aa#  2h# b

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" # # r

d) ,

961

962

Chapter 15 Multiple Integrals

42. A œ '1Î4

31Î4

œ

" #

'csc2 sin )

)

r dr d) œ

" #

' 3Î4Î4 a4 sin# )
csc# )b d) 1

1

c2)
sin 2)  cot )d 131Î4Î4 œ

1 #

44-46. Example CAS commands: Maple: f := (x,y) -> y/(x^2+y^2); a,b := 0,1; f1 := x -> x; f2 := x -> 1; plot3d( f(x,y), y=f1(x)..f2(x), x=a..b, axes=boxed, style=patchnogrid, shading=zhue, orientation=[0,180], title="#43(a) (Section 15.3)" ); # (a) q1 := eval( x=a, [x=r*cos(theta),y=r*sin(theta)] ); # (b) q2 := eval( x=b, [x=r*cos(theta),y=r*sin(theta)] ); q3 := eval( y=f1(x), [x=r*cos(theta),y=r*sin(theta)] ); q4 := eval( y=f2(x), [x=r*cos(theta),y=r*sin(theta)] ); theta1 := solve( q3, theta ); theta2 := solve( q1, theta ); r1 := 0; r2 := solve( q4, r ); plot3d(0,r=r1..r2, theta=theta1..theta2, axes=boxed, style=patchnogrid, shading=zhue, orientation=[-90,0], title="#43(c) (Section 15.3)" ); fP := simplify(eval( f(x,y), [x=r*cos(theta),y=r*sin(theta)] )); # (d) q5 := Int( Int( fP*r, r=r1..r2 ), theta=theta1..theta2 ); value( q5 ); Mathematica: (functions and bounds will vary) For 43 and 44, begin by drawing the region of integration with the FilledPlot command. Clear[x, y, r, t] <
Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.4 Triple Integrals in Rectangular Coordinates 15.4 TRIPLE INTEGRALS IN RECTANGULAR COORDINATES 1.

'01 '01cx 'x1bz Fax, y, zb dy dz dx œ '0 ’a1
xb
xa1
xb
1

2.

3.

4.

1

'01cx 'x1bz

“dx œ '0

1

a1  x b 2 2

dy dz dx œ '0

1

a1  x b 2 dx 2

'01cx a1
x
zb dz dx "

3

œ ’
a1 6 xb “ œ 0

1 6

'01 '02 '03 dz dy dx œ '01 '02 3 dy dx œ '01 6 dx œ 6, '02 '01 '03 dz dx dy, '03 '02 '01 dx dy dz, '02 '03 '01 dx dz dy, '03 '01 '02 dy dx dz, '01 '03 '02 dy dz dx '01 '02c2x '03c3xc3yÎ2 dz dy dx 1 2
2x œ '0 '0 ˆ3
3x
32 y‰ dy dx 1 œ '0 3(1
x) † 2(1
x)
34 † 4(1
x)# ‘ dx 1 " œ 3 '0 (1
x)# dx œ c
(1
x)$ d ! œ 1, '02 '01cyÎ2 '03
3x
3yÎ2 dz dx dy, '01 '03
3x '02
2x
2zÎ3 dy dz dx, '03 '01
zÎ3 '02
2x
2zÎ3 dy dx dz, '02 '03
3yÎ2 '01
yÎ2
zÎ3 dx dz dy, '03 '02
2zÎ3 '01
yÎ2
zÎ3 dx dy dz È4
x

'02 '03 '0 2

È

dz dy dx œ '0

#

È4cx

'0 '0 '0 3

5.

œ '0

2

dz dx dy, '0

#

2

'03 È4
x# dy dx œ '02 3È4
x# dx œ 3# ’xÈ4
x#  4 sin" x# “ 2 œ 6 sin" 1 œ 31, È4cx

'0

#

2

È4cx

'c22 'cÈ44ccxx 'x8bcyx cy dz dy dx œ 4 '02 '0 #

#

#

œ 4'0

2

œ 8 '0

'0

2

œ 8'0

#

È4cx

È4cx

'0

1Î2

#

#

#

#

'x8bcyx cy #

#

#

È4cz

'0 dy dx dz, '0 '0 '0 3

2

3

0

#

dx dy dz, '0

3

È4cz

'02 '0

#

dx dz dy

#

#

dz dy dx

c8
2 ax#  y# bd dy dx #

a4
x#
y# b dy dx

'02 a4
r# b r dr d) œ 8 '01Î2 ’2r#
r4 “ # d) %

!

œ 32 '0 d) œ 32 ˆ 1# ‰ œ 161, 1Î2

È

È4cz

'0 dy dz dx, '0 '0 3

'c22 'cÈ44ccyy 'x8bcyx cy dz dx dy, #

#

#

#

#

#

È È 'c2 'y 'cÈzzccyy dx dz dy  'c22 '48cy 'cÈ88cczzccyy dx dz dy, È È È È '04 'cÈzz 'cÈzzccyy dx dy dz  '48 'cÈ88cczz 'cÈ88cczzccyy dx dy dz, È È È È '04 'cÈzz 'cÈzzccxx dy dx dz  '48 'cÈ88cczz 'cÈ88cczzccxx dy dx dz 2

4 #

#

#

#

#

#

#

#

#

#

#

#

È

'c22 'x4 'cÈzzccxx #

# #

8cx#

dy dz dx  'c2 '4 2

È

'cÈ88cczzccxx

#

#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

# #

dy dz dx,

963

964

Chapter 15 Multiple Integrals

6. The projection of D onto the xy-plane has the boundary x#  y# œ 2y Ê x#  (y
1)# œ 1, which is a circle. Therefore the two integrals are:

È

È

'02 'cÈ2y2yccyy 'x2yby

'c11 '11cbÈ11ccxx 'x2yby dz dy dx

#

#

dz dx dy and #

#

#

#

#

#

7.

'01 '01 '01 ax#  y#  z# b dz dy dx œ '01 '01

8.

'0 2 '03y'x8bc3yx cy dz dx dy œ '0 2 '03ya8
2x#
4y# b dx dy œ '0

È

#

#

È2

1

0

È

#

È2

#

œ '0 a24y
18y$
12y$ b dy œ 12y#
9.

ˆx#  y#  "3 ‰ dy dx œ ' ˆx#  23 ‰ dx œ 1

"5 2

8x
23 x$
4xy# ‘ 3y dy 0

È2

y% ‘ 0 œ 24
30 œ
6

'1e '1e '1e xyz" dx dy dz œ '1e '1e ’ lnyzx “ e dy dz œ '1e '1e yz" dy dz œ '1e ’ lnzy “ e dz œ '1e "z dz œ 1 1

10.

'01 '03c3x '03c3xcy

1

"

11.

12.

'01 '01 '01

'03c3x (3
3x
y) dy dx œ '01 (3
3x)#
"# (3
3x)# ‘ dx œ 9# '01 (1
x)# dx

dz dy dx œ '0

œ
3# c(1
x)$ d ! œ

1

3 #

y sin z dx dy dz œ '0

1

'01 1y sin z dy dz œ 1# '01 sin z dz œ 1# $

$

(1
cos 1)

'c11 'c11 'c11 (x  y  z) dy dx dz œ 'c11 'c11 xy  2" y#  zy‘ "" dx dz œ 'c11 'c11 (2x  2z) dx dz œ 'c11 cx#  2zxd "" dz 1 œ 'c1 4z dz œ 0 È9 c x

È9 c x

'03 '0

14.

'02 'cÈ44ccyy '02xby dz dx dy œ '02 'cÈ44ccyy (2x  y) dx dy œ '02 cx#  xyd È c 4cy

È

#

'0

#

3

È

È9
x# dy dx œ ' a9
x# b dx œ ’9x
0 3

È4cy

#

#

$Î# #

“ œ !

2 3

(4)$Î# œ

#

$ x$ 3 “!

œ 18

dy œ '0 a4
y# b 2

"Î#

(2y) dy

16 3

'01 '02cx '02cxcydz dy dx œ '01 '02cx (2
x
y) dy dx œ '01 (2
x)#
"# (2
x)# ‘ dx œ "# '01 (2
x)# dx "

8 6

œ

7 6

'01 '01cx '34cx cyx dz dy dx œ '01 '01cx #

#

$ "

" œ ’
12 a1
x# b “ œ !

17.

#

#

#

œ 
"6 (2
x)$ ‘ ! œ
"6  16.

'0

#

œ ’
23 a4
y# b 15.

dz dy dx œ '0

È9 c x

13.

#

x a1
x#
yb dy dx œ '0 x ’a1
x# b
"# a1
x# b“ dx œ '0 1

#

" 12

'01 '01 '01 cos (u  v  w) du dv dw œ '01 '01 [sin (w  v  1)
sin (w  v)] dv dw 1 œ '0 [(
cos (w  21)  cos (w  1))  (cos (w  1)
cos w)] dw Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

" #

#

x a1
x# b dx

Section 15.4 Triple Integrals in Rectangular Coordinates œ c
sin (w  21)  sin (w  1)
sin w  sin (w  1)d 1! œ 0 18.

'1e '1e '1e ln r ln s ln t dt dr ds œ '1e '1e (ln r ln s) ct ln t
td 1e dr ds œ '1e (ln s) cr ln r
rd 1e ds œ cs ln s
sd 1e œ 1

19.

1Î4 2t ln sec v '01Î4 '0ln sec v '
_ ex dx dt dv œ '0 '0

œ '0

1Î4

20.

# Š sec# v

È4cq

'07 '02 '0 œ

8 ln 8 3

#

" #‹




dv œ




 tan2 v

dp dq dr œ '0

7

q r
1

1Î4

b Ä
_

v ‘ 1Î% 2 !

œ

'02 qÈr4
1 q

#

" #

dq dr œ '0

7

$Î# #

“ dr œ !

8 3

'07 r
"1 dr

È

(b)

dx dz dy

(e)

'01 'cÈÈyy '01cydz dx dy

(b)

'01 '01 'cc1

Èz

'01 '01 'cc1

'01 'cÈ11cczz 'x1cz dy dx dz

'c01 '0y '01

dy dz dx

#

Èz

'c01 '01 '0y

#

(d)

’
a4
q# b

'c11 '01cx 'x1cz dy dz dx '01 '01cy'cÈÈyy

22. (a)

" 3(r
1)

œ 8 ln 2

#

(d)

'0ln sec v e2t dt dv œ '01Î4 ˆ "# e2 ln sec v
"# ‰ dv

1 8




#

21. (a)

ae2t
eb b dt dv œ '0

lim

#

dx dz dy

(e)

23. V œ '0 'c1 '0 dz dy dx œ '0 'c1 y# dy dx œ 1

y#

1

dy dx dz

1

1

2 3

(c)

'01 '01cz 'cÈÈyy

(c)

'01 'cc1 z '01

È

dx dy dz

dx dy dz

dz dx dy

'01 dx œ 23

24. V œ '0

'01cx '02c2z dy dz dx œ '01 '01cx (2
2z) dz dx œ '01 c2z
z# d 01cx dx œ '01 a1
x# b dx œ ’x
x3 “ " œ 32

25. V œ '0

'0

1

4

$

!

È4cx

È4cx

'02cy dz dy dx œ '04 '0 %

œ 
43 (4
x)$Î#  4" (4
x)# ‘ ! œ 26. V œ 2 '0 'cÈ1cx# 1

27. V œ '0

1

0

4 3

(2
y) dy dx œ '0 ’2È4
x
ˆ 4#x ‰“ dx 4

(4)$Î#
4" (16) œ

'0cy dz dy dx œ
2 '01 'c0È1cx

32 3


4œ

20 3

y dy dx œ '0 a1
x# b dx œ 1

#

2 3

'02c2x '03c3xc3yÎ2 dz dy dx œ '01 '02
2x ˆ3
3x
3# y‰ dy dx œ '01 6(1
x)#
34 † 4(1
x)# ‘ dx

œ '0 3(1
x)# dx œ c
(1
x)$ d ! œ 1 1

"

'01cx '0cos Ð1xÎ2Ñ dz dy dx œ '01 '01
x cos ˆ 1#x ‰ dy dx œ '01 ˆcos 1#x ‰ (1
x) dx 1Î2 1 1 " œ '0 cos ˆ 1#x ‰ dx
'0 x cos ˆ 1#x ‰ dx œ  12 sin 1#x ‘ !
14 '0 u cos u du œ 12
14

28. V œ '0

1

#

œ

2 1




4 1#

29. V œ 8 '0

1

30. V œ '0

2

œ

" #

ˆ 1#
1‰ œ

È1cx

'0

#

ccos u  u sin ud 01Î2

4 1#

È1cx

'0

#

#

dz dy dx œ 8 '0

1

È1cx

'0

#

È1
x# dy dx œ 8 ' a1
x# b dx œ 0 1

16 3

'04cx '04cx cydz dy dx œ '02 '04cx a4
x#
yb dy dx œ '02 ’a4
x# b#
"# a4
x# b# “ dx #

#

#

'02 a4
x# b# dx œ '02 Š8
4x#  x# ‹ dx œ 128 15 %

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

965

966

Chapter 15 Multiple Integrals

31. V œ '0

4

c Î

'0

ˆÈ16 y# ‰ 2

'04
ydx dz dy œ '04 '0

œ '0 2È16
y# dy
4

" #

'0

4

È4cx

3cx

32. V œ 'c2 'cÈ4cx# '0 #

(4
y) dz dy œ '0

4

È16y# #

%

È4cx

dz dy dx œ 'c2 'cÈ4cx# (3
x) dy dx œ 2 'c2 (3
x)È4
x# dx 2

2

% !

32 3 #

2

œ 3 'c2 2È4
x# dx
2 'c2 xÈ4
x# dx œ 3 ’xÈ4
x#  4 sin" x2 “ 2

œ 12 sin" 1
12 sin" (
1) œ 12 ˆ 1# ‰
12 ˆ
1# ‰ œ 121 33.

(4
y) dy

$Î# yÈ16
y# dy œ yÈ16
y#  16 sin" y4 ‘ !  ’ 6" a16
y# b “

œ 16 ˆ 1# ‰
"6 (16)$Î# œ 81
2


Î

ˆÈ16 y# ‰ 2

#

#

 ’ 23 a4
x# b

$Î# #

“

#

'02 '02cx 'Ð24
cx2x
cyÑÎ2y2 dz dy dx œ '02 '02cx ˆ3
3x#
3y# ‰ dy dx œ '0 3 ˆ1
x# ‰ (2
x)
34 (2
x)# ‘ dx 2

œ '0 ’6
6x  2

œ ’6x
3x# 

34. V œ '0

4

3x# #

x$ 2

3(2x)# “ 4






# (2x)$ 4 “!

dx

œ (12
12  4  0)


2$ 4

œ2

'z8 'z8cz dx dy dz œ '04 'z8 (8
2z) dy dz œ '04 (8
2z)()
z) dz œ '04 a64
24z  2z# b dz %

œ 64z
12z#  23 z$ ‘ ! œ

È4cx Î2

35. V œ 2 'c2 '0 2

#

'0x
2

320 3

È4x Î2

dz dy dx œ 2 'c2 '0 2

#

(x  2) dy dx œ 'c2 (x  2)È4
x# dx 2

$Î# œ 'c2 2È4
x# dx  'c2 xÈ4
x# dx œ ’xÈ4
x#  4 sin" x# “  ’
"3 a4
x# b “ # # œ 4 ˆ 1# ‰
4 ˆ
1# ‰ œ 41 2

#

2

36. V œ 2 '0

1

'01cy '0x by dz dx dy œ 2 '01 '01cy ax#  y# b dx dy œ 2'01 ’ x3 #

#

#

#

$

 xy# “

1cy#

1

œ

2 3

1

#

’y


" y( 7 “!

œ ˆ 23 ‰ ˆ 67 ‰ œ

dy

0

œ 2 '0 a1
y# b ’ "3 a1
y# b  y# “ dy œ 2 '0 a1
y# b ˆ 3"  3" y#  3" y% ‰ dy œ

#

2 3

'01 a1
y' b dy

4 7

37. average œ

" 8

'02 '02 '02

38. average œ

" 2

'01 '01 '02 (x  y
z) dz dy dx œ 2" '01 '01 (2x  2y
2) dy dx œ 2" '01 (2x
1) dx œ 0

39. average œ '0

1

40. average œ 41.

'04 '01 '2y2 œ (sin

" 8

ax#  9b dz dy dx œ

" 8

'02 '02

a2x#  18b dy dx œ

" 8

'02 a4x#  36b dx œ 313

'01 '01 ax#  y#  z# b dz dy dx œ '01 '01 ˆx#  y#  "3 ‰ dy dx œ '01 ˆx#  23 ‰ dx œ 1

'02 '02 '02

xyz dz dy dx œ

4 cos ax# b dx dy dz œ 2È z % 4)z"Î# ‘ ! œ 2 sin 4

" 4

'02 '02

xy dy dx œ

" 2

'02 x dx œ 1

4 2 ax b '04 '02 '0xÎ2 4 cos dy dx dz œ '0 '0 2È z #

x cos ax# b Èz

dx dz œ '0 ˆ sin# 4 ‰ z"Î# dz 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.4 Triple Integrals in Rectangular Coordinates 42.

'01 '01 'x1 12xz ezy #

#

dy dx dz œ '0

1

'01 '0Èy

12xz ezy dx dy dz œ '0

'01 6yz ezy

1

#

dy dz œ '0 ’3ezy “ dz 1

#

'01 'È1z '0ln 3 $

1e2x sin a1y# b y#

dx dy dz œ '0 'È$ z 1

1

41 sin a1y# b y#

dy dz œ '0

1

'0y

$

41 sin a1y# b y#

#

" !

1 œ 3'0 aez
zb dz œ 3 cez
1d "! œ 3e
6

43.

967

dz dy

œ '0 41y sin a1y# b dy œ c
2 cos a1y# bd ! œ
2(
1)  2(1) œ 4 1

"

2z '02 '04cx '0x sin ' 2 ' 4cx 4  z dy dz dx œ 0 0 #

44.

%

œ 
"4 cos 2z‘ ! œ 
4" 

x sin 2z 4z %

sin# z‘ ! œ

dz dx œ '0

4

È4cz

'0

2z ‰ 2z ‰ " ' ˆ sin ˆ sin 4  z x dx dz œ 4  z # (4
z) dz 4

0

sin# 4 #

'01 '04cacx 'a4cx cydz dy dx œ 154 Ê '01 '04cacx a4
x#
y
ab dy dx œ 154 1 1 1 # # # 4 4 Ê '0 ’a4
a
x# b
"# a4
a
x# b “ dx œ 15 Ê "# '0 a4
a
x# b dx œ 15 Ê '0 c(4
a)#
2x# (4
a)  x% d dx #

45.

" #

#

œ

8 15

#

#

Ê ’(4
a)# x
23 x$ (4
a) 

" x& 5 “!

œ

8 15

Ê (4
a)#
32 (4
a) 

" 5

œ

8 15

Ê 15(4
a)#
10(4
a)
5 œ 0

Ê 3(4
a)#
2(4
a)
1 œ 0 Ê [3(4
a)  1][(4
a)
1] œ 0 Ê 4
a œ
"3 or 4
a œ 1 Ê a œ 46. The volume of the ellipsoid

x# a#



y# b#



z# c#

œ 1 is

4abc1 3

so that

4(1)(2)(c)1 3

13 3

or a œ 3

œ 81 Ê c œ 3.

47. To minimize the integral, we want the domain to include all points where the integrand is negative and to exclude all points where it is positive. These criteria are met by the points (xß yß z) such that 4x#  4y#  z#
4 Ÿ 0 or 4x#  4y#  z# Ÿ 4, which is a solid ellipsoid centered at the origin. 48. To maximize the integral, we want the domain to include all points where the integrand is positive and to exclude all points where it is negative. These criteria are met by the points (xß yß z) such that 1
x#
y#
z#   0 or x#  y#  z# Ÿ 1, which is a solid sphere of radius 1 centered at the origin. 49-52. Example CAS commands: Maple: F := (x,y,z) -> x^2*y^2*z; q1 := Int( Int( Int( F(x,y,z), y=-sqrt(1-x^2)..sqrt(1-x^2) ), x=-1..1 ), z=0..1 ); value( q1 ); Mathematica: (functions and bounds will vary) Due to the nature of the bounds, cylindrical coordinates are appropriate, although Mathematica can do it as is also. Clear[f, x, y, z]; f:= x2 y2 z Integrate[f, {x,
1,1}, {y,
Sqrt[1
x2 ], Sqrt[1
x2 ]}, {z, 0, 1}] N[%] topolar={x Ä r Cos[t], y Ä r Sin[t]}; fp= f/.topolar //Simplify Integrate[r fp, {t, 0, 21}, {r, 0, 1},{z, 0, 1}] N[%]

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

968

Chapter 15 Multiple Integrals

15.5 MASSES AND MOMENTS IN THREE DIMENSIONS cÎ2

bÎ2

aÎ2

cÎ 2

cÎ2

$

#


c# 12

; likewise Ry œ É a

bÎ2

cÎ2

bÎ2

1. Ix œ '
cÎ2 '
bÎ2 '
aÎ2 ay#
z# b dx dy dz œ a '
cÎ2 '
bÎ2 ay#
z# b dy dz œ a '
cÎ2 ’ y3
yz# “ dz
bÎ2 b b œ a '
cÎ2 Š 12
bz# ‹ dz œ ab ’ 12 z


Rx œ É b

#

œ '
3 '
2 ’ 8y3  3

4

#

4

2y$ 3




$

Ð4
2yÑÎ3

Iz œ 'c3 'c2 'c4Î3 3

4

3. Ix œ '0 '0 a

œ

M 3

b

and Rz œ É a

#


b# 12

c$ 12 ‹

œ

abc 1#

8(2  y)$ 81

x# (4  2y) 3







64 81 “

4x# 3

4

Ð4
2yÑÎ3

3

3

dy dx œ 'c3 ˆ12x#
3

64 81 “

ax#
y# b dz dy dx œ '
3 '
2 ax#
y# b ˆ 83  3

4

M 3

aa#
c# b and Iz œ

M 3

2y ‰ 3

dy dx œ 12 '
3 ax#
2b dx œ 360 3

$

Myz œ '0

x dz dy dx œ '0

dx œ

abc ab#
c# b 3

aa#
b# b , by symmetry

1cx

 x
"# ‹ dx œ

" 6

;

'0 x(1  x  y) dy dx œ "# '0 ax$  2x#
xb dx œ 24" 1 1cx 1cxcy Ê x œ y œ z œ "4 , by symmetry; Ix œ '0 '0 '0 ay#
z# b dz dy dx 1 1cx 1 " " œ '0 '0 ’y#  xy#  y$
(1  x3 y) “ dy dx œ "6 '0 (1  x)% dx œ 30 Ê Iy œ Ix œ 30 , by symmetry 1

'0 '0

c$ b 3 ‹




#

1cxcy

ax#
z# b dz dy dx

dx œ 280;

'01cx '01cxcy dz dy dx œ '01 '01cx (1  x  y) dy dx œ '01 Š x# 1cx

Ð4
2yÑÎ3

4

32 ‰ 3

$

1

ab#
c# b ;

ay#
z# b dz dy dx

'0c ay#
z# b dz dy dx œ '0a '0b Šcy#
c3 ‹ dy dx œ '0a Š cb3

4. (a) M œ '0

M 12

' ' ' dy dx œ '
3 104 3 dx œ 208; Iy œ
3
2
4Î3




ab#
c# b where M œ abc; Iy œ

ab#
c# b œ

, by symmetry

3

œ 'c3 'c2 ’ (4 812y)
3


c# 12

#

œ ab Š b12c


is the top of the wedge Ê Ix œ '
3 '
2 '
4Î3

4  2y 3

2. The plane z œ

#

cÎ2

z$ 3 “
cÎ2

$

1

1

$

È5 5

Ix (b) Rx œ É M œ É 5" œ

¸ 0.4472; the distance from the centroid to the x-axis is É0#


" 16




" 16

œ É 8" œ

È2 4

¸ 0.3536 5. M œ 4 '0

1

œ 2 '0

'01 '4y4 dz dy dx œ 4 '01 '01 a4  4y# b dy dx œ 16 '01 23 dx œ 323 ; Mxy œ 4 '01 '01 '4y4 #

#

z dz dy dx

12 '0 a16  16y% b dy dx œ 128 '0 dx œ 128 5 5 Ê z œ 5 , and x œ y œ 0, by symmetry; 1 1 4 1 1 64y 7904 % ' 1 1976 ‰ Ix œ 4 '0 '0 '4y ay#
z# b dz dy dx œ 4 '0 '0 ’ˆ4y#
64 3  Š4y
3 ‹“ dy dx œ 4 0 105 dx œ 105 ; 1

1

1

'

#

Iy œ 4 '0

1

œ

4832 63

'01 '4y4 ax#
z# b dz dy dx œ 4 '01 '01 ’ˆ4x#
643 ‰  Š4x# y#
64y3 ‹“ dy dx œ 4 '01 ˆ 38 x#
128 ‰ dx 7 '

#

; Iz œ 4 '0

1

œ 16 '0 Š 2x3
1

#

'01 '4y4 ax#
y# b dz dy dx œ 16 '01 '01 ax#  x# y#
y#  y% b dy dx

2 15 ‹

#

dx œ

ŠÈ4
x# ‹Î2

256 45

2
x

6. (a) M œ 'c2 'ŠcÈ4cx#‹Î2 '0 2

x dz dy dx œ '
2 'Š
È4
x#‹Î2 x(2  x) dy dx œ '
2 x(2  x) ŠÈ4  x# ‹ dx œ 21;

ŠÈ4
x# ‹Î2

2
x

y dz dy dx œ '
2 'Š
È4
x#‹Î2 y(2  x) dy dx

œ

" #



4 x # 4 “

'c22 (2  x) ’ 44x

#

2

2

ŠÈ4
x# ‹Î2

2

2
x

Mxz œ 'c2 'ŠcÈ4cx#‹Î2 '0 2

2

ŠÈ4
x# ‹Î2

Myz œ 'c2 'ŠcÈ4cx#‹Î2 '0 2

ŠÈ4
x# ‹Î2

dz dy dx œ '
2 'Š
È4
x#‹Î2 (2  x) dy dx œ '
2 (2  x) ŠÈ4  x# ‹ dx œ 41; 2

ŠÈ4
x# ‹Î2

dx œ 0 Ê x œ  "# and y œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.5 Masses and Moments in Three Dimensions ŠÈ4
x# ‹Î2

2
x

(b) Mxy œ 'c2 'ŠcÈ4cx#‹Î2 '0 2

œ 51 Ê z œ 7. (a) M œ 4 '0

2

È4cx

Mxy œ '0

#

'
22 'ŠŠ
È44

xx ‹‹ÎÎ22 È

#

#

" #

(2  x)# dy dx œ

'
22 (2  x)# ŠÈ4  x# ‹ dx

'x4
y dz dy dx œ 4 '01Î2 '02 'r 4 r dz dr d) œ 4 '01Î2 '02 a4r  r$ b dr d) œ 4 '01Î2 4 d) œ 81; #

#

#

'0 'r zr dz dr d) œ '0 '0 2

" #

5 4

'0

21

z dz dy dx œ

21

4

2

#

by symmetry (b) M œ 81 Ê 41 œ '0

21

r #

a16  r% b dr d) œ

È

Èc

'0 c 'r c r dz dr d) œ '021 '0 #

32 3

'021 d) œ 6431

acr  r$ b dr d) œ '0

21

Ê zœ

8 3

c# 4

c# 1 #

d) œ

, and x œ y œ 0, Ê c# œ 8 Ê c œ 2È2,

since c  0 8. M œ 8; Mxy œ 'c1 '3 'c1 z dz dy dx œ 'c1 '3 ’ z2 “ dy dx œ 0; Myz œ 'c1 '3 'c1 x dz dy dx " 1

5

1

1

5

"

#

1

5

1

œ 2 'c1 '3 x dy dx œ 4 'c1 x dx œ 0; Mxz œ 'c1 '3 'c1 y dz dy dx œ 2 'c1 '3 y dy dx œ 16 'c1 dx œ 32 1 5 1 1 5 1 Ê x œ 0, y œ 4, z œ 0; Ix œ ' ' ' ay#
z# b dz dy dx œ ' ' ˆ2y#
23 ‰ dy dx œ 23 ' 100 dx œ 400 3 ; 1

5

1

1

c1

3

5

1

1

c1

c1

5

1

c1

3

1 5 1 1 5 1 Iy œ 'c1 '3 'c1 ax#
z# b dz dy dx œ 'c1 '3 ˆ2x#
23 ‰ dy dx œ 43 'c1 a3x#
1b dx œ 16 3 ; 1 5 1 1 5 1 400 ‰ Iz œ 'c1 '3 'c1 ax#
y# b dz dy dx œ 2 'c1 '3 ax#
y# b dy dx œ 2 'c1 ˆ2x#
98 3 dx œ 3

Ê Rx œ Rz œ É 50 3

and Ry œ É 23 Ð2
yÑÎ2

9. The plane y
2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1 2

œ 'c2 'c2 ’ (y  6)#(4  y)
2

Mœ

" #

4

#

(2  y)$ 24

4

4

$

49 3 ‹

dt œ 1386;

IL (3)(6)(4) œ 36 Ê RL œ É M œ É 77 #

2

" #

c(y  6)#
z# d dz dy dx

#
"3 “ dy dx; let t œ 2  y Ê IL œ 4 'c2 Š 13t 24
5t
16t


Ð2
yÑÎ2

10. The plane y
2z œ 2 is the top of the wedge Ê IL œ 'c2 'c2 'c1 œ

4

c(x  4)#
y# d dz dy dx

'c22 'c42 ax#  8x
16
y# b (4  y) dy dx œ 'c22 a9x#  72x
162b dx œ 696; M œ "# (3)(6)(4) œ 36

IL Ê RL œ É M œ É 58 3

11. M œ 8; IL œ '0

4

'02 '01 cz#
(y  2)# d dz dy dx œ '04 '02 ˆy#  4y
133 ‰ dy dx œ 103 '04 dx œ 403

IL Ê RL œ É M œ É 53

12. M œ 8; IL œ '0

4

'02 '01 c(x  4)#
y# d dz dy dx œ '04 '02 c(x  4)#
y# d dy dx œ '04
2(x  4)#
83 ‘ dx œ 160 3

IL Ê RL œ É M œ É 20 3

'02cx '02cxcy 2x dz dy dx œ '02 '02cx a4x  2x#  2xyb dy dx œ '02 ax$  4x#
4xb dx œ 43 2 2cx 2cxcy 2 2cx 2 8 8 Mxy œ '0 '0 '0 2xz dz dy dx œ '0 '0 x(2  x  y)# dy dx œ '0 x(23 x) dx œ 15 ; Mxz œ 15 by 2 2cx 2cxcy 2 2cx 2 # symmetry; Myz œ '0 '0 '0 2x# dz dy dx œ '0 '0 2x# (2  x  y) dy dx œ '0 a2x  x# b dx œ 16 15

13. (a) M œ '0

2

(b)

$

Ê xœ

4 5

969

, and y œ z œ

2 5

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

970

Chapter 15 Multiple Integrals

14. (a) M œ '0

2

È

'0 x '04cx

(b) Myz œ '0

2

Ê xœ

È

256È2k 231

œ

k 4

kxy dz dy dx œ k'0

'0 x '04cx

2

#

2

2

È

'0 x '04cx

40È2 77

Ê yœ

Èx

'0

kx# y dz dy dx œ k '0

; Mxz œ '0

5 4

œ

#

#

xy a4  x# b dy dx œ

Èx

'0

x# y a4  x# b dy dx œ

kxy# dz dy dx œ k'0

2

È

'0 x '04cx

; Mxy œ '0

2

'02 a16x#  8x%
x' b dx œ 256k 105

#

Ê zœ

'02 a4x#  x% b dx œ 32k 15

k #

Èx

'0

k #

'02 a4x$  x& b dx œ 8k3

xy# a4  x# b dy dx œ

kxyz dz dy dx œ '0

2

Èx

'0

k 3

'02 ˆ4x&Î#  x*Î# ‰ dx #

xy a4  x# b dy dx

8 7

15. (a) M œ '0

'01 '01 (x
y
z
1) dz dy dx œ '01 '01 ˆx
y
3# ‰ dy dx œ '01 (x
2) dx œ 5# 1 1 1 1 1 1 4 ‰ Mxy œ '0 '0 '0 z(x
y
z
1) dz dy dx œ "# '0 '0 ˆx
y
53 ‰ dy dx œ "# '0 ˆx
13 6 dx œ 3 1

(b)

Ê Mxy œ Myz œ Mxz œ (c) Iz œ '0

1

4 3

, by symmetry Ê x œ y œ z œ

8 15

'01 '01 ax#
y# b (x
y
z
1) dz dy dx œ '01 '01 ax#
y# b ˆx
y
3# ‰ dy dx

œ '0 ˆx$
2x#
"3 x
34 ‰ dx œ 1

Ê Ix œ Iy œ Iz œ

11 6

11 6

, by symmetry

Iz (d) Rx œ Ry œ Rz œ É M œ É 11 15

16. The plane y
2z œ 2 is the top of the wedge. Ð2
yÑÎ2

(a) M œ 'c1 'c2 'c1 1

4

4

Ð2
yÑÎ2

Mxz œ 'c1 'c2 'c1 1

4

Ð2
yÑÎ2

Mxy œ 'c1 'c2 'c1 1

4

Ð2
yÑÎ2

(c) Ix œ 'c1 'c2 'c1 1

4

Ð2
yÑÎ2

Iy œ 'c1 'c2 'c1 1

4

Ð2
yÑÎ2

Iz œ 'c1 'c2 'c1 1

4

1

Ð2
yÑÎ2

(b) Myz œ 'c1 'c2 'c1 1

(x
1) dz dy dx œ 'c1 'c2 (x
1) ˆ2  y# ‰ dy dx œ 18 4

x(x
1) dz dy dx œ 'c1 'c2 x(x
1) ˆ2  y# ‰ dy dx œ 6; 1

4

y(x
1) dz dy dx œ 'c1 'c2 y(x
1) ˆ2  y# ‰ dy dx œ 0; 1

z(x
1) dz dy dx œ

" #

4

'c11 'c42 (x
1) Š y4

#

 y‹ dy dx œ 0 Ê x œ

(x
1) ay#
z# b dz dy dx œ 'c1 'c2 (x
1) ’2y#
1

4

(x
1) ax#
z# b dz dy dx œ 'c1 'c2 (x
1) ’2x#
1

4

$

1 3

, and y œ z œ 0

$
3" ˆ"  2y ‰ “ dy dx œ 45;

" 3



y #

" 3



x# y #

$
3" ˆ"  2y ‰ “ dy dx œ 15;

(x
1) ax#
y# b dz dy dx œ 'c1 'c2 (x
1) ˆ2  y# ‰ ax#
y# b dy dx œ 42 1

4

I

Ix Iz (d) Rx œ É M œ É 5# ß Ry œ É My œ É 56 ß and Rz œ É M œ É 37

17. M œ '0

1

Èz

'zc1c1z '0

(2y
5) dy dx dz œ '0

1

'zc1c1z ˆz
5Èz‰ dx dz œ '01 2 ˆz
5Èz‰ (1  z) dz

" $Î# œ 2 '0 ˆ5z"Î#
z  5z$Î#  z# ‰ dz œ 2
10
"# z#  2z&Î#  3" z$ ‘ ! œ 2 ˆ 93  #3 ‰ œ 3 3 z 1

È4cx

16c2 ˆx# by# ‰

18. M œ 'c2 'cÈ4cx# '2 ax#by#b 2

œ 4 '0

21

#

È4cx

Èx#
y# dz dy dx œ ' ' È # Èx#
y# c16  4 ax#
y# bd dy dx c2 c 4cx 2

'02 r a4  r# b r dr d) œ 4 '021 ’ 4r3

$

#

 r5 “ d) œ 4 '0 &

#

21

!

64 15

d) œ

5121 15

19. (a) Let ?Vi be the volume of the ith piece, and let (xi ß yi ß zi ) be a point in the ith piece. Then the work done by gravity in moving the ith piece to the xy-plane is approximately Wi œ mi gzi œ (xi
yi
zi
1)g ?Vi zi Ê the total work done is the triple integral W œ '0

'01 '01 (x
y
z
1)gz dz dy dx 1 1 1 1 1 " " œ g '0 '0
"# xz#
"# yz#
3" z$
#" z# ‘ ! dy dx œ g '0 '0 ˆ #" x
#" y
56 ‰ dy dx œ g '0
2" xy
4" y#
56 y‘ ! dx 1

œ g '0 ˆ "# x
1

13 ‰ 12

#

dx œ g ’ x4


13 12

"

16 ‰ x“ œ g ˆ 12 œ !

4 3

g

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.5 Masses and Moments in Three Dimensions

971

8 8 8 ‰ (b) From Exercise 15 the center of mass is ˆ 15 ß 15 ß 15 and the mass of the liquid is 5# Ê the work done by 8 ‰ gravity in moving the center of mass to the xy-plane is W œ mgd œ ˆ 52 ‰ (g) ˆ 15 œ 34 g, which is the same as

the work done in part (a). 20. (a) From Exercise 19(a) we see that the work done is W œ g '0

2

œ kg '0

2

œ

kg 4

Èx "

'0

#

#

xy a4  x# b dy dx œ

8 & " (‘ # $
16 3 x  5 x
7 x ! œ

kg 4

'0

2

È

'0 x '04cx

#

x# a4  x# b dx œ

kg 4

#

kxyz dz dy dx

'0 a16x#  8x%
x' b dx 2

256k†g 105 È

(b) From Exercise 14 the center of mass is Š 54 ß 4077 2 ß 78 ‹ and the mass of the liquid is gravity in moving the center of mass to the xy-plane is W œ mgd œ 21. (a) x œ

Myz M

ˆ 32k ‰ ˆ8‰ 15 (g) 7

32k 15

œ

Ê the work done by

256k†g 105

' ' ' x$ (xß yß z) dx dy dz œ 0 Ê Myz œ 0

œ0 Ê

R

(b) IL œ ' ' ' kv  hik# dm œ ' ' ' k(x  h) i
yjk# dm œ ' ' ' ax#  2xh
h#
y# b dm D

D

D

œ ' ' ' ax#
y# b dm  2h ' ' ' x dm
h# ' ' ' dm œ Ix  0
h# m œ Ic m
h# m Þ

D

D

22. IL œ Ic m
mh# œ Þ

Þ

2 5

ma#
ma# œ

7 5

ma# #

23. (a) (xß yß z) œ ˆ #a ß #b ß #c ‰ Ê Iz œ Ic m
abc ŠÉ a4
Þ

œ

abc aa#
b# b 3



abc aa#
b# b 4

# (b) IL œ IcÞmÞ
abc ΃ a4


œ

abc aa#
7b# b 3

3

4

Þ

b# 4‹

#

Ê IcÞmÞ œ Iz 

#
b # abc aa#
b# b ; RcÞmÞ œ É IcMÞmÞ œ É a 12 1# # # # # # ˆ b#  2b‰#  œ abc aa12
b b
abc aa 4
9b b

abc aa#
b# b 4

œ

IL ; RL œ É M œ Éa

Ð4
2yÑÎ3

24. M œ 'c3 'c2 'c4Î3

Þ

D

#

abc a4a#
28b# b 1#


7b# 3

dz dy dx œ '
3 '
2 23 (4  y) dy dx œ '
3 23 ’4y  3

œ

4

3

#

% y# 2 “ #

dx œ 12 '
3 dx œ 72; 3

x œ y œ z œ 0 from Exercise 2 Ê Ix œ IcÞmÞ
72 ŠÈ0#
0# ‹ œ IcÞmÞ Ê IL œ IcÞmÞ
72 ŠÉ16


16 9 ‹

#

‰ œ 1488 œ 208
72 ˆ 160 9 25. MyzB"
B# œ ' ' ' x dV"
' ' ' x dV# œ MÐyzÑ"
MÐyzÑ# Ê x œ MÐyzÑ"
MÐyzÑ# m"
m# ; similarly, B"

B#

y œ MÐxzÑ"
MÐxzÑ# m"
m# and z œ MÐxyÑ"
MÐxyÑ# m"
m# Ê c œ xi
yj
zk œ m
" m
ˆMÐyzÑ"
MÐyzÑ# ‰ i
ˆMÐxzÑ"
MÐxzÑ# ‰ j
ˆMÐxyÑ"
MÐxyÑ# ‰ k‘ "

œ œ

#

" m"
m# " m"
m#

cam" x"
m# x# b i
am" y"
m# y# b j
am" z"
m# z# b kd cm" ax" i
y" j
z" kb
m# ax# i
y# j
z# kbd œ

i
13j
13 13 13 2 k Ê x œ 13 7 14 , y œ 7 , z œ 14 2i
7j
12 k " ‰ k‰
12 ˆi
11 Ê x œ 1, y œ 27 , z œ 41 # j  # k 12
12 œ 2 13i
74j  5k " ‰ 11 " ‰ 37 5 ˆ Ê x œ 13 # k
12 i
# j  # k 2
12 œ 14 14 , y œ 7 , z œ  14 25i
92j
7k " ‰ k‰
2 ˆ "# i
4j
"# k‰
12 ˆi
11 Ê x œ 25 # j  # k 12
2
12 œ 26 26 ,

26. (a) c œ 12 ˆi
32 j
k‰
2 ˆ "# i
4j
"# k‰12
2 œ (b) c œ 12 ˆi
32 j
(c) c œ 2 ˆ "# i
4j
(d) c œ 12 ˆi
32 j


m" c"
m# c# m"
m#

13 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

yœ

46 13

,zœ

7 26

972

Chapter 15 Multiple Integrals #

$

#

Š 1a3 h ‹ Š h4 k‹
Š 213a ‹ Šc 3a 8 k‹

27. (a) c œ

Š a 31 ‹ Š h

œ

m"
m#

#
3a# 4

k‹

m"
m#

1 a# h 3

, where m" œ

and m# œ

21 a$ 3

; if

h#  3a# 4

œ 0, or h œ aÈ3, then the centroid is on the common base (b) See the solution to Exercise 55, Section 15.2, to see that h œ aÈ2. #

28. c œ

#

Š s 3h ‹ Š 4h k‹
s$ ˆc #s k‰

s Š 12 ‹
ah# c 6s# b k‘

œ

m"
m#

m"
m#

, where m" œ

s# h 3

and m# œ s$ ; if h#  6s# œ 0,

or h œ È6s, then the centroid is in the base of the pyramid. The corresponding result in 15.2, Exercise 56, is h œ È3s. 15.6 TRIPLE INTEGRALS IN CYLINDRICAL AND SPHERICAL COORDINATES 1.

'021 '01 'r

È2cr

#

dz r dr d) œ '0

21

œ '0 Š 2 3  23 ‹ d) œ 21

2.

È

'021 '03 'r Î318
r

4.

'03 ’r a18  r# b"Î#  r3 “ dr d) œ '021 ’ "3 a18  r# b$Î#  12r “ $ d) $

%

!

2 1

$

’ 12)1#


21

21

)& 201% “ !

È

'01 '0 Î '
3È44

rr ) 1

dz r dr d) œ '0

#

1

%

ˆ

# ‰
"Î#

)

) Î1 !

1

'0 Î

) 1

œ 4 '0 Š 21)#  1

#

21

4 3‹

1

)

1

d) œ

'02 Š 4)1 1

3 #

# #




" #

c9 a4  r# b  a4  r# bd r dr d) œ 4 '0

)% 41 % ‹

1

d)

'0 Î a4r  r$ b dr d) ) 1

371 15

d) œ

'01 ’r a2  r# b"Î#  r# “ dr d) œ 3 '021 ’ a2  r# b"Î#  r3 “ " d) $

!

d) œ 1 Š6È2  8‹

6.

'021 '01 'c11ÎÎ22 ar# sin# )
z# b dz r dr d) œ '021 '01 ˆr$ sin# )
12r ‰ dr d) œ '021 Š sin4 )
24" ‹ d) œ 13

7.

'021 '03 '0zÎ3

8.

'c11 '021 '01bcos

9.

'01 '0 z '021 ar# cos# )
z# b r d) dr dz œ '01 '0

#

r$ dr dz d) œ '0

21

)

21 z 3 '03 324 dz d) œ '0 20 d) œ 3101 %

4r dr d) dz œ 'c1 '0 2(1
cos ))# d) dz œ 'c1 61 d) œ 121 1

21

È 1

È

1

Èz

œ '0 ’ 14r
1r# z# “

10.

4) % 161% ‹

171 5

3 dz r dr d) œ 3 '0

œ 3 '0 ŠÈ2  21

'0 Î2 a3r
24r$ b dr d) œ '02
32 r#
6r% ‘ !Î2

1

#

'021 '01 'r 2cr

œ

z dz r dr d) œ '0

#

œ 4 '0 ’2r#  r4 “

5.

3

91 Š8È2  7‹

)

3 #

!

21

'021 '0 Î2 '03
24r œ

$

41 ŠÈ2  "‹

dz r dr d) œ '0

#

#

œ 3.

$Î#

'01 ’r a2  r# b"Î#  r# “ dr d) œ '021 ’ "3 a2  r# b$Î#  r3 “ " d)

%

Èz

!

1

#

$

È

#

$Î#

r# sin 2) 4

dz œ '0 Š 14z
1z$ ‹ dz œ ’ 112z


'02 'rc24cr '021 (r sin )
1) r d) dz dr œ '02 'rc24cr œ 21 ’ "3 a4  r# b

#

’ r2)




r$ 3

#


z# )“

" 1 z% 4 “!

#1 !

r dr dz œ '0

1

Èz

'0

a1r$
21rz# b dr dz

1 3

œ

21r dz dr œ 21'0 ’r a4  r# b 2

"Î#

 r#
2r“ dr

#


r# “ œ 21
 83
4
3" (4)$Î# ‘ œ 81 !

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates È4cr

'021 '01 '0

11. (a)

È3

#

dz r dr d)

È4cz

'021 '0 '01 r dr dz d)
'021 'È23 '0

(b)

È4cr

'01 '0

(c)

#

#

r dr dz d)

'021 r d) dz dr

'021 '01 'r 2cr dz r dr d) #

12. (a)

È2cz

(b)

'021 '01 '0z r dr dz d)
'021 '12 '0

(c)

'0 'r '0

2cr#

1

21

r dr dz d)

r d) dz dr

13.

'c11ÎÎ22 '0cos '03r

14.

'c11ÎÎ22 '01 '0r cos

15.

'01 '02 sin '04cr sin

17.

'c1ÎÎ22 '11cos '04

19.

'0 Î4 '0sec '02
r sin

21.

'01 '01 '02 sin 9 3# sin 9 d3 d9 d) œ 83 '01 '01 sin% 9 d9 d) œ 83 '01 Š’ sin 94cos 9 “ 1
34 '01 sin# 9 d9‹ d)

)

)

#

Î

r$ dz dr d) œ '
1Î2 '0 r% cos ) dr d) œ

)

1

f(rß )ß z) dz r dr d) 1 2

1

)

" 5

'
1ÎÎ22 cos ) d) œ 25 1

f(rß )ß z) dz r dr d)

16.

'
ÎÎ22 '03 cos '05
r cos

f(rß )ß z) dz r dr d)

18.

'
1ÎÎ22 'cos2 cos '03
r sin

20.

' ÎÎ42 '0csc '02
r sin

)

)

1

)

f(rß )ß z) dz r dr d)

1

)

1

1

)

23.

24.

)

)

1

)

1

)

f(rß )ß z) dz r dr d) f(rß )ß z) dz r dr d)

f(rß )ß z) dz r dr d)

$

!

1 1 1 1 1 œ 2 '0 '0 sin# 9 d9 d) œ '0
)  sin#2) ‘ ! d) œ '0 1 d) œ 1#

22.

)

'021 '01Î4 '02 (3 cos 9) 3# sin 9 d3 d9 d) œ '021 '01Î4 4 cos 9 sin 9 d9 d) œ '021 c2 sin# 9d !1Î% d) œ '021 d) œ 21 '021 '01 '0Ð1
cos 9ÑÎ2 3# sin 9 d3 d9 d) œ 24" '021 '01 (1  cos 9)$ sin 9 d9 d) œ 96" '021 c(1  cos 9)% d !1 d) 21 21 " ' 16 ' % œ 96 a 2  0 b d ) œ d) œ 6" (21) œ 13 96 0 0 '031Î2 '01 '01 53$ sin$ 9 d3 d9 œ

5 6

d) œ

5 4

'031Î2 '01 sin$ 9 d9 d) œ 54 '031Î2 Š’ sin 93cos 9 “ 1
23 '01 sin 9 d9‹ d)

'031Î2 c cos 9d 1! d) œ 53 '031Î2 d) œ 5#1

#

!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

973

974 25.

Chapter 15 Multiple Integrals

'021 '01Î3 'sec2 9 33# sin 9 d3 d9

d) œ '0

21

œ '0
(4  2)  ˆ8  "# ‰‘ d) œ 21

26.

27.

5 #

'01Î3 a8  sec$ 9b sin 9 d9 d) œ '021
8 cos 9  "2 sec# 9‘ !1Î$ d)

'021 d) œ 51

'021 '01Î4 '0sec 9 3$ sin 9 cos 9 d3 d9 d) œ "4 '021 '01Î4 tan 9 sec# 9 d9 d) œ "4 '021
2" tan# 9‘ !1Î% d) 21 œ "8 '0 d) œ 14 2 0 '02 'c01 '11ÎÎ42 3$ sin 29 d9 d) d3 œ '02 '
01 3$
 cos229 ‘ 11Î# d) d3 œ '0 '
1 3# Î%

$

d) d3 œ '0

2 $ 3 1

#

d3

#

%

œ ’ 138 “ œ 21 !

28.

'11ÎÎ63 'csc2 csc9 9 '021 3# sin 9 d) d3 d9 œ 21 '11ÎÎ63 'csc2 csc9 9 3# sin 9 d3 d9 œ 231 '11ÎÎ63 c3$ sin 9d csc2 csc9 9 d9 œ

29.

141 3

281 3È 3

'01 '01 '01Î4 123 sin$ 9 d9 d) d3 œ '01 '01 Œ123 ’  sin 39 cos 9 “ 1Î%
83 '01Î4 sin 9 d9 d) d3 #

!

œ '0

1

œ

30.

'11ÎÎ63 csc# 9 d9 œ

'0 Š 1

23 È2

 83 ccos

1Î% 9d ! ‹

d) d3 œ '0

1

'0 Š83  1

103 È2 ‹

d) d3 œ 1'0 Š83  1

103 È2 ‹

d3 œ 1 ’43# 

" 53# È2 “ !

Š4È2  5‹ 1 È2

'11ÎÎ62 '
11ÎÎ22 'csc2 9 53% sin$ 9 d3 d) d9 œ '11ÎÎ62 '
11ÎÎ22 a32  csc& 9b sin$ 9 d) d9 œ '11ÎÎ62 '
11ÎÎ22 a32 sin$ 9  csc# 9b d) d9 œ 1 '1Î6 a32 sin$ 9  csc# 9b d9 œ 1 ’ 32 sin 39 cos 9 “ 1Î2

œ

#

È 1 Š 3224 3 ‹



641 3

ccos

1Î# 9d 1Î'

 1 ŠÈ 3 ‹ œ

È3 3

1


1Î# 1Î'




641 3

'11ÎÎ62 sin 9 d9
1 ccot 9d 11Î# Î'

È ˆ 6431 ‰ Š #3 ‹

œ

331È3 3

œ 111È3

31. (a) x#
y# œ 1 Ê 3# sin# 9 œ 1, and 3 sin 9 œ 1 Ê 3 œ csc 9; thus

'021 '01Î6 '02 3# sin 9 d3 d9 d)
'021 '11ÎÎ62 '0csc 9 3# sin 9 d3 d9 d) '021 '12 '1sinÎ6

(b)

3# sin 9 d9 d3 d)
'0

21

'02 '01Î6 3# sin 9 d9 d3 d)

'021 '01Î4 '0sec 9 3# sin 9 d3 d9 d) '021 '01 '01Î4 3# sin 9 d9 d3 d)

32. (a) (b)


'0

21

33. V œ '0

21

œ


" Ð1Î3Ñ

" 3

È

'1 2 'cos1Î
"4 Ð"Î3Ñ 3# sin 9 d9 d3 d)

'01Î2 'cos2 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a8  cos$ 9b sin 9 d9 d)

31 ‰ '021 ’8 cos 9
cos4 9 “ 1Î# d) œ 3" '021 ˆ8  4" ‰ d) œ ˆ 12 (21) œ 3161 %

!

'01Î2 '11
cos 9 3# sin 9 d3 d9 d) œ "3 '021 '01Î2 a3 cos 9
3 cos# 9
cos$ 9b sin 9 d9 d) 21 21 1Î# 111 ' 21 ˆ 11 ‰ œ "3 '0
 3# cos# 9  cos$ 9  14 cos% 9‘ ! d) œ 3" '0 ˆ 32
1
4" ‰ d) œ 11 12 0 d) œ 12 (21) œ 6

34. V œ '0

21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 35. V œ '0

21

œ

" 12

(2)

36. V œ '0

21

œ

" 12

%

'01Î2 '01
cos 9

'0

21

21

d) œ

1Î# 9) '021 '01Î2 (1  cos 9)$ sin 9 d9 d) œ 3" '021 ’ ("  cos d) “ 4 %

!

1 6

3# sin 9 d3 d9 d) œ " 6

" 3

(21) œ

8 3

'021 '11ÎÎ42 cos$ 9 sin 9 d9 d) œ 83 '021 ’ cos4 9 “ 1Î# d) %

1Î%

1 3

21 4 ' 81 '11ÎÎ32 '02 3# sin 9 d3 d9 d) œ 83 '021 '11ÎÎ32 sin 9 d9 d) œ 83 '021 c cos 9d 11Î# Î$ d) œ 3 0 d) œ 3

1Î2

(c) 8 '0

2

(c)

'0

21

3# sin 9 d3 d9 d) œ

(21) œ

'11ÎÎ42 '02 cos 9

39. (a) 8'0

40. (a)

" 12

d) œ

" ‰ ˆ 83 ‰ ˆ 16

38. V œ '0

!

'021 d) œ 43 (21) œ 831

%

21

37. V œ '0 œ

1 9) '01 '01ccos 9 3# sin 9 d3 d9 d) œ 3" '021 '01 (1  cos 9)$ sin 9 d9 d) œ 3" '021 ’ ("  cos “ d) 4

'01Î2 '02 3# sin 9 d3 d9 d) È4cx

'0

#

È4cx cy

'0

È

'01Î2 '03Î 2 'r

#

È9
r

#

(b) 8'0

1Î2

È4cr

'02 '0

21

dz r dr d)

#

dz dy dx

dz r dr d)

(b)

'01Î2 '01Î4 '03 3# sin 9 d3 d9 d)

'01Î2 '01Î4 '03 3# sin 9 d3 d9 d) œ 9 '01Î2 '01Î4 sin 9 d9 d) œ 9 '01Î2 Š È"

41. (a) V œ '0

#

'01Î3 'sec2 9 3# sin 9 d3 d9 d)

(b) V œ '0

21

È3 È3cx È4cx cy (c) V œ 'cÈ3 'cÈ3cx '1 dz dy dx È3 21 "Î# #

#

È

91 Š2  È2‹

 1‹ d) œ

2

È4cr

'0 3 '1

#

4

dz r dr d)

#

#

(d)

V œ '0 '0 œ

#

’r a4  r b

# $Î#

'021 d) œ 531

5 6

È



r# #•

È$ !

d) œ '0 Š "3  21

3 #




4$Î# 3 ‹

d)

'01 '0 1cr r# dz r dr d) 1Î2 21 1 Iz œ '0 '0 '0 a3# sin# 9b a3# sin 9b d3 d9 d), since r# œ x#
y# œ 3# sin# 9 cos# )
3# sin# 9 sin# )

42. (a) Iz œ '0

21

(b)

21  r“ dr d) œ '0 ” a4 3r b

#

œ 3# sin# 9 (c) Iz œ '0

21

œ

2 15

1Î2

1Î2

44. V œ 4'0 œ 4 '0

!

41 15

'01 'r 4

14r

1Î2

dz r dr d) œ 4 '0

#

%

d) œ

1Î2

1Î2

#

(21) œ

43. V œ 4 '0 œ 4 '0

'01Î2 5" sin$ 9 d9 d) œ 5" '021 Œ’ sin 93cos 9 “ 1Î#
23 '01Î2 sin 9 d9 d) œ 152 '021 c cos 9d !1Î# d)

81 3

'01 '
1È
r1
r

ˆ "#



'01 a5r  4r$  r& b dr d) œ 4 '01Î2 ˆ 5#  1  "6 ‰ d)

" 3




1Î2

#

dz r dr d) œ 4 '0

"‰ 3

d) œ 2'0 d) œ 1Î2

'01 Šr  r#
rÈ1r# ‹ dr d) œ 4 '01Î2 ’ r2

2 ˆ 1# ‰

#



r$ 3

 "3 a1  r# b

œ1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

$Î# "

“ d) !

975

976

Chapter 15 Multiple Integrals

45. V œ '31Î2 '0 21

3 cos )

'0cr sin

#1

œ
94 cos% )‘ $1Î# œ c3 cos )

46. V œ 2 '1Î2 '0 1

œ 18 Œ’ cos 47. V œ '0

1Î2

#

dz r dr d) œ '31Î2 '0

)

21

0œ

9 4

1

È1
r

'0sin '0

#

)

r# dr d) œ

2 3

'1Î2 27 cos$ ) d) 1

'11Î2 cos ) d) œ 12 csin )d 11Î# œ 12 Î

dz r dr d) œ '0

1 2

'0sin

)

Î

rÈ1r# dr d) œ '0 ’ "3 a1  r# b 1 2

$Î# sin )

“

!

d)

#

!

œ  csin 2 9

48. V œ '0

1Î2

1Î# )d !

1 6




'0cos '03 )

œ

È1
r

1Î#

œ

Î

1 2

1Î2


23 ccos )d !

4
3 1 18

dz r dr d) œ '0

#

œ '0 ’ a1  cos# )b 1 #

2 3

21

'01Î2 ’a1  sin# )b$Î#  1“ d) œ  3" '01Î2 acos$ )  1b d) œ  3" Œ’ cos )3 sin ) “ 1Î#
32 '01Î2 cos ) d)

3) ‘ !1Î#

œ  3"

œ




r# sin ) dr d) œ '31Î2 a9 cos$ )b (sin )) d)

9 4

'0r dz r dr d) œ 2 '1Î2 '0c3 cos

1 ) sin ) “ 3 1Î#

)

3 cos )

$Î#

1 #

'0cos

)

Î

3rÈ1r# dr d) œ '0 ’ a1  r# b 1 2


1“ d) œ '0 a1  sin$ )b d) œ ’)




1Î2

2 3

œ

'12Î13Î3 '0a 3# sin 9 d3 d9 d) œ '021 '12Î13Î3

50. V œ '0

'01Î2 '0a 3# sin 9 d3 d9 d) œ a3 '01Î6 '01Î2

51. V œ '0

'01Î3 'sec2 9

1Î6

21



“

2 3

!

d)

'01Î2 sin ) d)

31  4 6

49. V œ '0

21

1Î# sin# ) cos ) “ 3 !

$Î# cos )

$

a$ 3

a$ 3

sin 9 d9 d) œ sin 9 d9 d) œ

a$ 3

'021 c cos 9d #11Î$Î$ d) œ a3 '021 ˆ "#
"# ‰ d) œ 213a $

'01Î6 d) œ a181 $

3# sin 9 d3 d9 d)

'021 '01Î3 a8 sin 9  tan 9 sec# 9b d9 d) 21 1Î$ œ 3" '0
8 cos 9  2" tan# 9‘ ! d) 21 21 œ "3 '0
4  #" (3)
8‘ d) œ 3" '0 5# d) œ 56 (21) œ 531 œ

" 3

52. V œ 4 '0

1Î2

'01Î4 'sec2 sec9 9 3# sin 9 d3 d9 d)

œ

4 3

'01Î2 '01Î4 a8 sec$ 9  sec$ 9b sin 9 d9 d)

'01Î2 '01Î4 sec$ 9 sin 9 d9 d) œ 283 '01Î2 '01Î4 tan 9 sec# 9 d9 d) œ 283 '01Î2
2" tan# 9‘ !1Î% d) 1Î2 ' œ 14 d) œ 731 3 0 œ

28 3

53. V œ 4 '0

'01 '0r

54. V œ 4 '0

'01 'r r
1 dz r dr d) œ 4 '01Î2 '01 r dr d) œ 2 '01Î2 d) œ 1

55. V œ 8 '0

'1 2 '0r

56. V œ 8 '0

'1 2 '0

1Î2

1Î2

1Î2

œ

8 3

'0

dz r dr d) œ 4 '0

1Î2

'01 r$ dr d) œ '01Î2 d) œ 1#

#

#

1Î2

1Î2

#

È

1Î2

È2
r

È

d) œ

dz r dr d) œ 8 '0 #

È2

'1

1Î2

dz r dr d) œ 8 '0

r# dr d) œ 8 Š 2

È2

'1

È2" ‹ 3

'01Î2 d) œ 41 Š2 3 2"‹ È

1Î2

rÈ2  r# dr d) œ 8 '0 ’ "3 a2  r# b

$Î#

“

È# 1

d)

41 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

$

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 57. V œ '0

'02 '04cr sin

58. V œ '0

'02 '04cr cos cr sin

21

21

)

dz r dr d) œ '0

21

)

)

'02 a4r  r# sin )b dr d) œ 8 '02 ˆ1  sin3 ) ‰ d) œ 161 1

dz r dr d) œ '0

21

'02 c4r  r# (cos )
sin ))d dr d) œ 83 '02

1

(3  cos )  sin )) d) œ 161

59. The paraboloids intersect when 4x#
4y# œ 5  x#  y# Ê x#
y# œ 1 and z œ 4 Ê V œ 4 '0

1Î2

'01 '4r5
r

#

#

1Î2

dz r dr d) œ 4 '0

'01 a5r  5r$ b dr d) œ 20 '01Î2 ’ r2

#

1Î2

 r4 “ d) œ 5'0 %

" !

60. The paraboloid intersects the xy-plane when 9  x#  y# œ 0 Ê x#
y# œ 9 Ê V œ 4 '0

1Î2

'13 '09
r

#

dz r dr d) œ 4 '0

1Î2

'13 a9r  r$ b dr d) œ 4 '01Î2 ’ 9r2

dz r dr d) œ 8 '0

'01 r a4  r# b"Î# dr d) œ 8 '021 ’ "3 a4  r# b$Î# “ " d)

#

%

œ 64 '0 d) œ 321 1Î2

61. V œ 8 '0

21

È4cr

'01 '0

'0

21

œ  83

#

21

ˆ3$Î#  8‰ d) œ

1Î2

 r4 “ d) œ 4 '0 ˆ 81 4  $ "

d) œ

17 ‰ 4

51 #

d)

!

41 Š8  3È3‹ 3

62. The sphere and paraboloid intersect when x#
y#
z# œ 2 and z œ x#
y# Ê z#
z  2 œ 0 Ê (z
2)(z  1) œ 0 Ê z œ 1 or z œ 2 Ê z œ 1 since z   0. Thus, x#
y# œ 1 and the volume is given by the triple integral V œ 4 '0

1Î2

œ 4 '0 ’ "3 a2  r# b 1Î2

63. average œ

"

64. average œ œ

3 21

ˆ 431 ‰

È2
r

#

#

1Î2

dz r dr d) œ 4 '0

 r4 “ d) œ 4 '0 Š 2 3 2 

$Î#

%

"

1Î2

È

!

'021 '01 'c11

" #1

'01 'r

r# dz dr d) œ

È

'021 '01 'cÈ11ccrr

# #

" #1

7 12 ‹

d) œ

'01 ’r a2  r# b"Î#  r$ “ dr d)

1 Š8È2  7‹ 6

'021 '01 2r# dr d) œ 3"1 '021 d) œ 23

r# dz dr d) œ

3 41

'021 '01 2r# È1  r# dr d)

'021 ’ 8" sin" r  8" rÈ1  r# a1  2r# b“ " d) œ 1631 '021 ˆ 1#
0‰ d) œ 323 '021 d) œ ˆ 323 ‰ (21) œ 3161 !

65. average œ

" ˆ 431 ‰

'021 '01 '01 3$ sin 9 d3 d9 d) œ 1631 '021 '01 sin 9 d9 d) œ 831 '021 d) œ 43

66. average œ

" ˆ 231 ‰

'021 '01Î2 '01 3$ cos 9 sin 9 d3 d9 d) œ 831 '021 '01Î2

œ

3 161

cos 9 sin 9 d9 d) œ

3 81

'021 ’ sin2 9 “ 1Î# d)

'021 d) œ ˆ 1631 ‰ (21) œ 38

#

!

67. M œ 4 '0

'01 '0r dz r dr d) œ 4 '01Î2 '01 r# dr d) œ 43 '01Î2 d) œ 231 ; Mxy œ '021 '01 '0r z dz r dr d) 21 1 21 M œ "# '0 '0 r$ dr d) œ 18 '0 d) œ 14 Ê z œ M œ ˆ 14 ‰ ˆ 231 ‰ œ 38 , and x œ y œ 0, by symmetry 1Î2

xy

68. M œ '0

'02 '0r dz r dr d) œ '01Î2 '02 r# dr d) œ 83 '01Î2 d) œ 431 ; Myz œ '01Î2 '02 '0r x dz r dr d) 1Î2 1Î2 1Î2 1Î2 2 2 r 2 œ '0 '0 r$ cos ) dr d) œ 4 '0 cos ) d) œ 4; Mxz œ '0 '0 '0 y dz r dr d) œ '0 '0 r$ sin ) dr d) 1Î2 1Î2 1Î2 1Î2 2 r 2 M œ 4 '0 sin ) d) œ 4; Mxy œ '0 '0 '0 z dz r dr d) œ "# '0 '0 r$ dr d) œ 2 '0 d) œ 1 Ê x œ M 1Î2

yz

yœ

Mxz M

œ

3 1

, and z œ

Mxy M

œ

3 4

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

3 1

,

977

978

Chapter 15 Multiple Integrals ; Mxy œ '0

21

81 3

69. M œ

œ 4 '0 ’ sin2 9 “ 21

#

'11ÎÎ32 '02 z3# sin 9 d3 d9 d) œ '021 '11ÎÎ32 '02 3$ cos 9 sin 9 d3 d9 d) œ 4 '021 '11ÎÎ32

d) œ 4 '0 ˆ "#  38 ‰ d) œ

1Î#

21

1Î$

" #

'021 d) œ 1

Ê zœ

Mxy M

œ (1) ˆ 831 ‰ œ

3 8

cos 9 sin 9 d9 d)

, and x œ y œ 0,

by symmetry 70. M œ '0

'01Î4 '0a 3# sin 9 d3 d9 d) œ a3 '021 '01Î4 sin 9 d9 d) œ a3 '021 2 #È2 d) œ 1a Š23 2‹ ; 1Î4 1Î4 21 a 21 21 a ' Mxy œ '0 '0 '0 3$ sin 9 cos 9 d3 d9 d) œ a4 '0 '0 sin 9 cos 9 d9 d) œ 16 d) œ 18a 0 21

$

$

$

%

Mxy M

Ê zœ

%

È2

‰ 2
œ Š 18a ‹ – $ 3 È — œ ˆ 3a 8 Š # 1a Š2 2‹ %

‹œ

3 Š2
È2‹ a 16

%

, and x œ y œ 0, by symmetry

È

71. M œ '0

È

È

'04 '0 r dz r dr d) œ '021 '04 r$Î# dr d) œ 645 '021 d) œ 1285 1 ; Mxy œ '021 '04 '0 r z dz r dr d) 21 4 '021 d) œ 6431 Ê z œ MM œ 65 , and x œ y œ 0, by symmetry œ "# '0 '0 r# dr d) œ 32 3 21

xy

1Î3

È1
r

1Î3

1Î3

72. M œ 'c1Î3 '0 '
È1
r# dz r dr d) œ '
1Î3 '0 2rÈ1  r# dr d) œ '
1Î3 ’ 23 a1  r# b 1

#

1

$Î# "

“ d) !

È1
r 1Î3 1 œ 23 'c1Î3 d) œ ˆ 23 ‰ ˆ 231 ‰ œ 491 ; Myz œ '
1Î3 '0 '
È1
r r# cos ) dz dr d) œ 2 '
1Î3 '0 r# È1  r# cos ) dr d) 1Î3

1Î3

#

1

#

1Î3

œ 2 'c1Î3 ’ 18 sin" r  "8 rÈ1  r# a1  2r# b“ cos ) d) œ " !

Myz M

Ê xœ

œ

9È 3 32

1 8

'
11ÎÎ33 cos ) d) œ 18 csin )d
1Î13Î3 œ ˆ 18 ‰ Š2 † È#3 ‹ œ 1È8 3

, and y œ z œ 0, by symmetry

73. Iz œ '0

'12 '04 ax#
y# b dz r dr d) œ 4 '021 '12 r$ dr d) œ '021 15 d) œ 301; M œ '021 '12 '04 dz r dr d) 21 2 21 I œ '0 '1 4r dr d) œ '0 6 d) œ 121 Ê Rz œ É M œ É #5 21

z

74. (a) Iz œ '0

'01 'c11 r$ dz dr d) œ 2 '021 '01 r$ dr d) œ "# '021 d) œ 1 21 1 1 21 1 21 Ix œ '0 '0 'c1 ar# sin# )
z# b dz r dr d) œ 2 '0 '0 ˆ2r$ sin# )
2r3 ‰ dr d) œ '0 Š sin# )
"3 ‹ d) 21

(b)

#

œ
4) 

sin 2) 8

#1
3) ‘ ! œ

1 #




21 3

œ

71 6

75. We orient the cone with its vertex at the origin and axis along the z-axis Ê 9 œ which is through the vertex and parallel to the base of the cone Ê Ix œ '0

21

œ '0

%

76. Iz œ '0

21

œ 2 '0

'0a

21

77. Iz œ '0

21

21

. We use the the x-axis

1

'01 Šr$ sin# )  r% sin# )
3r  r3 ‹ dr d) œ '021 Š sin20 )
10" ‹ d) œ
40)  sin802)
10) ‘ #!1 œ #10
15 œ 14

21

œ '0

1 4

'0 'r ar# sin# )
z# b dz r dr d) 1

2 15

Èa cr #

'

#

cÈa# cr#

r$ dz dr d) œ '0

21

a& d) œ

81 a 15

#

'0a 2r$ Èa#  r# dr d) œ 2 '021 ’Š r5

#

ˆh‰ a

%

a

r& 5a “ !

2a# # 15 ‹ aa

 r# b

$Î# a

“ d)

&

'0a ' h r ax#
y# b dz r dr d) œ '021 '0a

h ’ r4 



d) œ '0 h Š a4  21

%

a& 5a ‹

h

' hr a

d) œ

' h r$ dz dr d) œ '021 '0a hr a

ha% 20

'021 d) œ 110ha

Šhr$ 

hr% a ‹

dr d)

%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

!

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates 78. (a) M œ '0

'01 '0r z dz r dr d) œ '021 '01 "# r& dr d) œ 12" '021 d) œ 16 ; Mxy œ '021 '01 '0r 21 1 21 " ' 1 œ "3 '0 '0 r( dr d) œ 24 d) œ 12 Ê z œ #" , and x œ y œ 0, by symmetry; 0 21

Iz œ '0

21

#

'01 '0r

(b) M œ '0

21

#

'01 '0r

zr$ dz dr d) œ

'021 '01

" #

r# dz dr d) œ '0

21

#

r( dr d) œ

'021 d) œ 18

" 16

#

z# dz r dr d)

È3 #

Iz Ê Rz œ É M œ

'01 r% dr d) œ "5 '021 d) œ 215 ; Mxy œ '021 '01 '0r

#

zr# dz dr d)

'021 '01 r' dr d) œ 14" '021 d) œ 17 Ê z œ 145 , and x œ y œ 0, by symmetry; Iz œ '021 '01 '0r 21 1 21 I œ '0 '0 r' dr d) œ "7 '0 d) œ 271 Ê Rz œ É M œ É 57 " 2

œ

979

#

r% dz dr d)

z

79. (a) M œ '0

'01 'r 1 z dz r dr d) œ "# '021 '01 ar  r$ b dr d) œ 8" '021 d) œ 14 ; Mxy œ '021 '01 'r 1 z# dz r dr d) 21 1 21 21 1 1 " ' œ 3" '0 '0 ar  r% b dr d) œ 10 d) œ 15 Ê z œ 45 , and x œ y œ 0, by symmetry; Iz œ '0 '0 'r zr$ dz dr d) 0 21 1 21 I 1 " ' œ "# '0 '0 ar$  r& b dr d) œ 24 d) œ 12 Ê Rz œ É M œ É 3" 0 21

z

(b) M œ '0

'01 'r 1 z# dz r dr d) œ 15 from part (a); Mxy œ '021 '01 'r 1 z$ dz r dr d) œ 4" '021 '01 ar  r& b dr d) 21 21 1 1 21 1 " ' 1 5 " ' ' # $ ' ' ' œ 12 d ) œ Ê z œ , and x œ y œ 0, by symmetry; I z r dz dr d ) œ ar$  r' b dr d) z œ 6 6 3 0 0 0 r 0 0 21 I 5 " ' 1 œ 28 d) œ 14 Ê Rz œ É M œ É 14 0 21

z

80. (a) M œ '0

'01 '0a 3% sin 9 d3 d9 d) œ a5 '021 '01 sin 9 d9 d) œ 2a5 '021 d) œ 415a ; 1 1 1 21 a 21 21 Iz œ '0 '0 '0 3' sin$ 9 d3 d9 d) œ a7 '0 '0 a1  cos# 9b sin 9 d9 d) œ a7 '0 ’ cos 9
cos3 9 “ d) 21

&

&

&

(

'0

21

4a( #1

œ

$

(

!

8a( 1 21

d) œ

Ê Rz œ

œ

Iz ÉM

(b) M œ '0

É 10 21

a

'01 '0a 3$ sin# 9 d3 d9 d) œ a4 '021 '01 (1cos# 29) d9 d) œ 18a '021 d) œ 1 4a ; 1 1 21 a 21 Iz œ '0 '0 '0 3& sin% 9 d3 d9 d) œ a6 '0 '0 sin% 9 d9 d) 1 1 21 21 21 1 œ a6 '0 Š’  sin 49 cos 9 “
34 '0 sin# 9 d9‹ d) œ a8 '0
9#  sin429 ‘ ! d) œ 116a '0 d) 21

%

%

# %

'

$

'

'

'

!

a' 1 # 8

œ 81. M œ '0

21

Ê Rz œ

'0a '0

h a

Èa cr #

#

Iz ÉM

œ

a È2

dz r dr d) œ '0

21

'0a

rÈa#  r# dr d) œ

h a

Èa cr

'021 ’ 3" aa#  r# b$Î# “ a d)

h a

!

'0 3 d) œ '0 '0 aa# r  r$ b dr d) ; Mxy œ '0 '0 '0 z dz r dr d) œ 21 h ' œ 2a Š a#  a4 ‹ d) œ a h4 1 Ê z œ Š 1a4h ‹ ˆ 2ha3 1 ‰ œ 38 h, and x œ y œ 0, by symmetry 0 œ

21 $ a

h a

21

2ha# 1 3

#

%

%

h a

a

#

# #

#

21

h# 2a#

a

# #

#

#

82. Let the base radius of the cone be a and the height h, and place the cone's axis of symmetry along the z-axis with the vertex at the origin. Then M œ œ

h# #

'021 ’ r2

#



a r% 4a# “ !

d) œ

h# #

'021 Š a#

#



1a# h 3

and Mxy œ '0

a# 4‹

d) œ

21

h # a# 8

'0a ' h r z dz r dr d) œ "# '021 '0a Šh# r  ha

21

Mxy œ '0

'0a '0h (z
1) dz r dr d) œ '021 '0a

'021 d) œ h a4 1 # #

21

'0a '0h

Ê z œ ’ 1a

#

az#
zb dz r dr d) œ '0

21

a2h$
3h# b “ ’ 1a# ah2#
2hb “ 6

œ

#

Š h#
h‹ r dr d) œ

'0a

2h#
3h 3h
6

$

Š h3


h# # ‹r

r$ ‹ dr d)

a

Ê zœ

Mxy M

# #

œ Š h a4 1 ‹ ˆ 1a3# h ‰ œ

x œ y œ 0, by symmetry Ê the centroid is one fourth of the way from the base to the vertex 83. M œ '0

#

#

ˆh‰

a# ah#
2hb 4

dr d) œ

'021 d) œ 1a ah#
2hb ;

a# a2h$
3h# b 12

#

#

'021 d) œ 1a a2h6
3h b #

$

, and x œ y œ 0, by symmetry;

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

#

3 4

h, and

980

Chapter 15 Multiple Integrals

Iz œ '0

21

'0a '0h

(z
1)r$ dz dr d) œ Š h

Iz Rz œ É M œ É 1a

%

ah#
2hb 4

†

2 1a# ah#
2hb

#


2h # ‹

œ

'021 '0a r$ dr d) œ Š h
# 2h ‹ Š a4 ‹'021 d) œ 1a ah4
2hb ; #

%

%

#

a È2

84. The mass of the plant's atmosphere to an altitude h above the surface of the planet is the triple integral M(h) œ '0

'01 'Rh .! ecÐ3RÑ 3# sin 9 d3 d9 d) œ 'Rh '021 '01 .! ecÐ3RÑ 3# sin 9 d9 d) d3 h 21 h 21 h 1 œ 'R '0
.! ecÐ3RÑ 3# ( cos 9)‘ ! d) d3 œ 2 'R '0 .! ecR ec3 3# d) d3 œ 41.! ecR 'R ec3 3# d3 21


c3

#

œ 41.! ecR ’ 3 ec œ 41.! ecR Š



h e
ch #



c

23e
c3 c#



2he
ch c#



h 2e
c3 c$ “ R 2e
ch c$




(by parts) R# e
cR c




2Re
cR c#




2e
cR c$ ‹ .

The mass of the planet's atmosphere is therefore M œ lim

hÄ_

#

M(h) œ 41.! Š Rc


2R c#




2 c$ ‹ .

85. The density distribution function is linear so it has the form $ (3) œ k3
C, where 3 is the distance from the center of the planet. Now, $ (R) œ 0 Ê kR
C œ 0, and $ (3) œ k3  kR. It remains to determine the constant k: M œ '0

21

'01 '0R (k3  kR) 3# sin 9 d3 d9 d) œ '021 '01 ’k 34

%

1 21 œ '0 '0 k Š R4

%



R% 3 ‹

Ê $ (3) œ  13M 3
R%

$

R

 kR 33 “ sin 9 d9 d) !

21 21 sin 9 d9 d) œ '0  1k# R% c cos 9d 1! d) œ '0  6k R% d) œ  k13R

3M 1R%

‰R œ R . At the center of the planet 3 œ 0 Ê $ (0) œ ˆ 13M R%

3M 1R$

%

Ê k œ  13M R%

.

86. x2
y2 œ a2 Ê a3 sin 9 cos )b2
a3 sin 9 sin )b2 œ a2 Ê a32 sin2 9bacos2 )
sin2 )b œ a2 Ê 32 sin2 9 œ a2 Ê 3 sin 9 œ a or 3 sin 9 œ a Ê 3 sin 9 œ a or 3 œ a csc 9, since 0 Ÿ 9 Ÿ 1 and 3   0. 87. (a) A plane perpendicular to the x-axis has the form x œ a in rectangular coordinates Ê r cos ) œ a Ê r œ Ê r œ a sec ), in cylindrical coordinates. (b) A plane perpendicular to the y-axis has the form y œ b in rectangular coordinates Ê r sin ) œ b Ê r œ Ê r œ b csc ), in cylindrical coordinates. 88. ax
by œ c Ê aar cos )b
bar sin )b œ c Ê raa cos )
b sin )b œ c Ê r œ

a cos ) b sin )

c a cos )
b sin ) .

89. The equation r œ fazb implies that the point ar, ), zb œ afazb, ), zb will lie on the surface for all ). In particular afazb, )
1, zb lies on the surface whenever afazb, ), zb does Ê the surface is symmetric with respect to the z-axis.

90. The equation 3 œ fa9b implies that the point a3, 9, )b œ afa9b, 9, )b lies on the surface for all ). In particular, if afa9b, 9, )b lies on the surface, then afa9b, 9, )
1b lies on the surface, so the surface is symmetric wiith respect to the z-axis. 15.7 SUBSTITUTIONS IN MULTIPLE INTEGRALS 1. (a) x  y œ u and 2x
y œ v Ê 3x œ u
v and y œ x  u Ê x œ ` (xßy) ` (ußv)

œ»

" 3  23

" 3 " 3

»œ

" 9




2 9

œ

" 3

(u
v) and y œ

" 3

(2u
v);

" 3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates (b) The line segment y œ x from (!ß 0) to (1ß 1) is x  y œ 0 Ê u œ 0; the line segment y œ 2x from (0ß 0) to (1ß 2) is 2x
y œ 0 Ê v œ 0; the line segment x œ 1 from (1ß 1) to ("ß 2) is (x  y)
(2x
y) œ 3 Ê u
v œ 3. The transformed region is sketched at the right. 2. (a) x
2y œ u and x  y œ v Ê 3y œ u  v and x œ v
y Ê y œ ` (xßy) ` (ußv)

œ»

" 3 1 3

2 3  "3

" »œ9

2 9

" 3

(u  v) and x œ

" 3

(u
2v);

œ  3"

(b) The triangular region in the xy-plane has vertices (0ß 0), (2ß 0), and ˆ 23 ß 23 ‰ . The line segment y œ x from (0ß 0) to ˆ 23 ß 23 ‰ is x  y œ 0 Ê v œ 0; the line segment y œ 0 from (0ß 0) to (#ß 0) Ê u œ v; the line segment x
2y œ 2 from ˆ 23 ß 23 ‰ to (2ß 0) Ê u œ 2. The transformed region is sketched at the right. 3. (a) 3x
2y œ u and x
4y œ v Ê 5x œ 2u
v and y œ ` (xßy) ` (ußv)

œ»



2 5 1 10

 15 3 10

»œ

6 50



1 50

œ

" #

(u  3x) Ê x œ

" 5

(2u  v) and y œ

" 10

" 10

(b) The x-axis y œ 0 Ê u œ 3v; the y-axis x œ 0 Ê v œ 2u; the line x
y œ 1 " Ê "5 (2u  v)
10 (3v  u) œ 1 Ê 2(2u  v)
(3v  u) œ 10 Ê 3u
v œ 10. The transformed region is sketched at the right.

4. (a) 2x  3y œ u and x
y œ v Ê x œ u
3v and y œ v
x Ê x œ u  3v and y œ u  2v; " 3 ` (xßy) ` (ußv) œ º 1 2 º œ 2  3 œ 1 (b) The line x œ 3 Ê u  3v œ 3 or u
3v œ 3; x œ 0 Ê u
3v œ 0; y œ x Ê v œ 0; y œ x
1 Ê v œ 1. The transformed region is the parallelogram sketched at the right.

5.

'04 'yÐÎy2Î2Ñ
1 ˆx  y# ‰ dx dy œ '04 ’ x2

#

œ

" #

y



'0 (y
1  y) dy œ '0 dy œ 4

" #

4

xy 2 # “y


"

dy œ

2

" #

" #

'04 ’ˆ y#
1‰#  ˆ y# ‰#  ˆ y#
1‰ y
ˆ y# ‰ y“ dy

(4) œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

(3v  u);

981

982 6.

Chapter 15 Multiple Integrals

' ' a2x#  xy  y# b dx dy œ ' ' (x  y)(2x
y) dx dy R

ßy) œ ' ' uv ¹ `` (x (ußv) ¹ du dv œ

G

R

' ' uv du dv;

" 3

G

We find the boundaries of G from the boundaries of R, shown in the accompanying figure: xy-equations for

Corresponding uv-equations

Simplified

for the boundary of G

uv-equations

the boundary of R y œ 2x
4

" 3

(2u
v) œ  (u
v)
4

vœ4

y œ 2x
7

" 3

(2u
v) œ  23 (u
v)
7

vœ7

yœx2

" 3

uœ2

yœx
1

" 3

Ê

7.

2 3

(2u
v) œ

1 3

(u
v)  2

(2u
v) œ

1 3

(u
v)
1

u œ 1

' ' uv du dv œ 3" ' ' uv dv du œ 3" ' u ’ v2# “ du œ c1 4 c1 2

" 3

7

(

2

11 #

%

G

'c21 u du œ ˆ 11# ‰ ’ u2 “ # #

"

‰ œ ˆ 11 4 (4  1) œ

33 4

' ' a3x#
14xy
8y# b dx dy R

œ ' ' (3x
2y)(x
4y) dx dy R

ßy) œ ' ' uv ¹ `` (x (ußv) ¹ du dv œ

G

" 10

' ' uv du dv; G

We find the boundaries of G from the boundaries of R, shown in the accompanying figure: xy-equations for the boundary of R

Simplified

for the boundary of G

uv-equations

3 #

yœ x
1

" 10

(3v  u) œ 

(2u  v)
1

uœ2

y œ  #3 x
3

" 10

3 (3v  u) œ  10 (2u  v)
3

uœ6

" 4

yœ x

" 10

(3v  u) œ 

vœ0

y œ  "4 x
1

" 10

1 (3v  u) œ  20 (2u  v)
1

Ê

8.

Corresponding uv-equations

" 10

' ' uv du dv œ

" 10

G

3 10

1 20

(2u  v)

vœ4

'26 '04 uv dv du œ 10" '26 u ’ v2 “ % du œ 45 '26 u du œ ˆ 45 ‰ ’ u2 “ ' œ ˆ 54 ‰ (18  2) œ 645 #

#

!

#

' ' 2(x  y) dx dy œ ' ' 2v ¹ `` (x(ußßy)v) ¹ du dv œ ' ' 2v du dv; the region G is sketched in Exercise 4 R

G

G

3c3v

" Ê ' ' 2v du dv œ '0 'c3v 2v du dv œ '0 2v(3  3v
3v) dv œ '0 6v dv œ c3v# d ! œ 3 1

1

1

G

9. x œ

u v

v" uv# œ v" u
v" u œ 2u v ; v u º Ê v œ 1, and y œ 4x Ê v œ 2; xy œ 1 Ê u œ 1, and xy œ 9 Ê u œ 3; thus

and y œ uv Ê

y œ x Ê uv œ

u v

y x

œ v# and xy œ u# ;

` (xßy) ` (ußv)

œ J(uß v) œ º

' ' ŠÉ yx
Èxy‹ dx dy œ ' ' (v
u) ˆ 2uv ‰ dv du œ ' ' Š2u
2uv # ‹ dv du œ ' c2uv
2u# ln vd #" du 1 1 1 1 1 3

2

3

2

3

R

œ '1 a2u
2u# ln 2b du œ
u#
23 u# ln 2‘ " œ 8
23 (26)(ln 2) œ 8
3

$

52 3

(ln 2)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 15.6 Triple Integrals in Cylindrical and Spherical Coordinates ` (xßy) ` (ußv)

10. (a)

œ J(uß v) œ º

" 0 œ u, and v uº

the region G is sketched at the right

(b) x œ 1 Ê u œ 1, and x œ 2 Ê u œ 2; y œ 1 Ê uv œ 1 Ê v œ "u , and y œ 2 Ê uv œ 2 Ê v œ

'1 '1 2

œ

2

3 #

y x

dy dx œ '1

2

'1Îu ˆ uvu ‰ u dv du œ '1 '1Îu uv dv du œ '1 2Îu

1Îu

du œ '1 u ˆ u2# 

" ‰ 2u#

; thus,

du

1

R

` (xßy) ` (ußv)

u ’ v2 “

2

2

21

12.

2Îu

#

#

I! œ ' ' ax#
y# b dA œ '0 œ

2

'12 u ˆ u" ‰ du œ 3# cln ud #" œ 3# ln 2; '12 '12 yx dy dx œ '12 ’ x1 † y2 “ 2 dx œ 3# '12 dxx œ 3# cln xd #" œ 3# ln 2

11. x œ ar cos ) and y œ ar sin ) Ê

ab 4

2Îu

2

2 u

` (xßy) ` (rß))

œ J(rß )) œ º

ar sin ) œ abr cos# )
abr sin# ) œ abr; br cos ) º

'01 r# aa# cos# )
b# sin# )b kJ(rß ))k dr d) œ '021 '01 abr$ aa# cos# )
b# sin# )b dr d)

'021 aa# cos# )
b# sin# )b d) œ ab4 ’ a2)
a #

œ J(uß v) œ º

a cos ) b sin )

#

sin 2) 4




b# ) 2



21

b# sin 2) “ 4 !

œ

ab1 aa#
b# b 4

È1cu# 1 a 0 œ ab; A œ ' ' dy dx œ ' ' ab du dv œ 'c1 'cÈ1cu# ab dv du º 0 b R G

œ 2ab 'c1 È1u# du œ 2ab ’ u2 È1  u#
1

" #

sin" u“

"

"

œ ab csin" 1  sin" (1)d œ ab
1#  ˆ 1# ‰‘ œ ab1

13. The region of integration R in the xy-plane is sketched in the figure at the right. The boundaries of the image G are obtained as follows, with G sketched at the right:

xy-equations for

Corresponding uv-equations

Simplified

for the boundary of G

uv-equations

the boundary of R xœy

" 3

(u
2v) œ

x œ 2  2y

" 3

(u
2v) œ 2  23 (u  v)

yœ0

0œ

Also, from Exercise 2,

` (xßy) ` (ußv)

1 3

1 3

(u  v)

(u  v)

œ J(uß v) œ  "3 Ê

vœ0 uœ2 vœu

'02Î3 'y2
2y (x
2y) eÐy
xÑ dx dy œ '02 '0u ue
v ¸ 3" ¸ dv du

œ

" 3

'02 u cecv d !u du œ 3" '02 u a1  ecu b du œ 3" ’u au
ecu b  u#

œ

" 3

a3ec2
1b ¸ 0.4687

#

#


ecu “ œ !

" 3

c2 a2
ec2 b  2
ec2  1d

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

983

984

Chapter 15 Multiple Integrals

14. x œ u
` (xßy) ` (ußv)

v #

and y œ v Ê 2x  y œ (2u
v)  v œ 2u and

" "# v º œ 1; next, u œ x  # 0 " and v œ y, so the boundaries of the region of

œ J(uß v) œ º

œx

y #

integration R in the xy-plane are transformed to the boundaries of G: xy-equations for

Corresponding uv-equations

Simplified

for the boundary of G

uv-equations

œ

uœ0

the boundary of R xœ xœ

u


y # y #


2

u


yœ0

œ


2

uœ2 vœ0

vœ2

2

" 4

œ

v # v #

vœ0

yœ2 Ê '0

v # v #

ÐyÎ2Ñ
2

'yÎ2

ae  16

vœ2

y$ (2x  y) eÐ2xyÑ dx dy œ '0

2

#

% # 1b ’ v4 “ !

'0

2

v$ (2u) e4u du dv œ '0 v$ ’ 4" e4u “ dv œ 2

#

#

# !

" 4

'02 v$ ae16  1b dv

œe 1 16

15. (a) x œ u cos v and y œ u sin v Ê

` (xßy) ` (ußv)

ϼ

cos v u sin v œ u cos# v
u sin# v œ u sin v u cos v º

(b) x œ u sin v and y œ u cos v Ê

` (xßy) ` (ußv)

ϼ

sin v u cos v œ u sin# v  u cos# v œ u cos v u sin v º

â â â cos v u sin v 0 â â â z) u cos v 0 â œ u cos# v
u sin# v œ u œ â sin v 16. (a) x œ u cos v, y œ u sin v, z œ w Ê ``(u(xßßvyßßw) â â 0 "â â 0 â â â2 0 0â â â z) (b) x œ 2u  1, y œ 3v  4, z œ "# (w  4) Ê ``(u(xßßvyßßw) œ â 0 3 0 â œ (2)(3) ˆ #" ‰ œ 3 â â â 0 0 "# â â â sin 9 cos ) â 17. â sin 9 sin ) â â cos 9 œ (cos 9) º

3 cos 9 cos ) 3 cos 9 sin ) 3 sin 9 3 cos 9 cos ) 3 cos 9 sin )

â 3 sin 9 sin ) â â 3 sin 9 cos ) â â 0 â 3 sin 9 sin ) sin 9 cos )
(3 sin 9) º 3 sin 9 cos ) º sin 9 sin )

3 sin 9 sin ) 3 sin 9 cos ) º

œ a3# cos 9b asin 9 cos 9 cos# )
sin 9 cos 9 sin# )b
a3# sin 9b asin# 9 cos# )
sin# 9 sin# )b œ 3# sin 9 cos# 9
3# sin$ 9 œ a3# sin 9b acos# 9
sin# 9b œ 3# sin 9 18. Let u œ g(x) Ê J(x) œ

du dx

œ gw (x) Ê 'a f(u) du œ 'g(a) f(g(x))gw (x) dx in accordance with Theorem 6 in b

g(b)

Section 5.6. Note that gw (x) represents the Jacobian of the transformation u œ g(x) or x œ g" (u). 19.

'03 '04 'y1Î
Ð2 yÎ2Ñ ˆ 2x # y
3z ‰ dx dy dz œ '03 '04 ’ x2

#

œ '0 ’ (y
4 1)  3

#

y# 4




% yz 3 “!

dz œ '0 ˆ 94
3

4z 3



xy #




"
ÐyÎ2Ñ xz 3 “ yÎ2

 "4 ‰ dz œ '0 ˆ2
3

dy dz œ '0

4z ‰ 3

3

'04
"# (y
1)  y#
3z ‘ dy dz

dz œ ’2z


$ 2z# 3 “!

œ 12

â â âa 0 0â # # # â â 20. J(uß vß w) œ â 0 b 0 â œ abc; the transformation takes the ellipsoid region xa#
yb#
zc# Ÿ 1 in xyz-space â â â0 0 câ into the spherical region u#
v#
w# Ÿ 1 in uvw-space ˆwhich has volume V œ 43 1‰ Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 15 Practice Exercises Ê V œ ' ' ' dx dy dz œ ' ' ' abc du dv dw œ R

G

985

41abc 3

â â âa 0 0â â â 21. J(uß vß w) œ â 0 b 0 â œ abc; for R and G as in Exercise 19, ' ' ' kxyzk dx dy dz â â R â0 0 câ œ ' ' ' a# b# c# uvw dw dv du œ 8a# b# c# G

œ

4a# b# c# 3

'01Î2 '01Î2

'01Î2 '01Î2 '01 (3 sin 9 cos ))(3 sin 9 sin ))(3 cos 9) a3# sin 9b d3 d9 d) a # b # c# 3

sin ) cos ) sin$ 9 cos 9 d9 d) œ

'01Î2 sin ) cos ) d) œ a b6 c

# # #

â 1 â â 22. u œ x, v œ xy, and w œ 3z Ê x œ u, y œ vu , and z œ "3 w Ê J(uß vß w) œ â  uv# â â 0

0 " u

0

0 ââ 0 ââ œ " â â 3

" 3u

;

' ' ' ax# y
3xyzb dx dy dz œ ' ' '
u# ˆ vu ‰
3u ˆ vu ‰ ˆ w3 ‰‘ kJ(uß vß w)k du dv dw œ "3 ' ' ' ˆv
vw ‰ du dv dw u 0 0 1 3

R

2

2

G

œ

" 3

'0 '0 (v
vw ln 2) dv dw œ 3" '03 (1
w ln 2) ’ v2 “ # dw œ 32 '03 (1
w ln 2) dw œ 32 ’w
w2

œ

2 3

ˆ3


3

2

#

#

!

9 #

ln 2“

$ !

ln 2‰ œ 2
3 ln 2 œ 2
ln 8

23. The first moment about the xy-coordinate plane for the semi-ellipsoid,

x# a#




y# b#




z# c#

œ 1 using the

transformation in Exercise 21 is, Mxy œ ' ' ' z dz dy dx œ ' ' ' cw kJ(uß vß w)k du dv dw D

œ abc#

G

' ' ' w du dv dw œ aabc# b † aMxy of the hemisphere x#
y#
z# œ 1, z   0b œ G

the mass of the semi-ellipsoid is

2abc1 3

#

3 ‰ Ê z œ Š abc4 1 ‹ ˆ 2abc 1 œ

3 8

abc# 1 4

;

c

24. A solid of revolutions is symmetric about the axis of revolution, therefore, the height of the solid is solely a function of r. That is, y œ faxb œ farb. Using cylindrical coordinates with x œ r cos ), y œ y and z œ r sin ), we have V œ ' ' ' r dy d) dr œ 'a G

b

'021 '0farb

r dy d) dr œ 'a

b

'021 c r y df0arb d) dr œ 'ab '021 r farb d) dr œ 'ab c r)farb d021 dr

'a 21rfarbdr. In the last integral, r is a dummy or stand-in variable and as such it can be replaced by any variable name. b Choosing x instead of r we have V œ 'a 21xfaxbdx, which is the same result obtained using the shell method. b

CHAPTER 15 PRACTICE EXERCISES 1.

'110 '01Îyyexy dx dy œ '110 cexy d !"Îy dy 10 œ '1 (e  1) dy œ 9e  9

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

986

Chapter 15 Multiple Integrals

'01 '0x eyÎx dy dx œ '01 x
eyÎx ‘ !x $

2.

$

dx

œ '0 Šxex  x‹ dx œ ’ "2 ex  1

3.

#

È

" x# # “!

#

'03Î2 '
È99

4t4t

3Î2

t ds dt œ '0

# #

È9
4t

œ

#

ctsd È # dt
9
4t

œ '0 2tÈ9  4t# dt œ ’ "6 a9  4t# b 3Î2

" 6

œ  ˆ0$Î#  9$Î# ‰ œ

4.

27 6

œ

#

Èy

" #

“

!

dy

'01 y ˆ4  4Èy
y  y‰ dy

œ '0 ˆ2y  2y$Î# ‰ dy œ ’y#  1

'c02 '2x4 cb x4

#

5.

$Î# $Î#

9 #

'01 'È2cy Èy xy dx dy œ '01 y ’ x2 “ 2cÈy œ

e2 #

" 4y&Î# 5 “!

" 5

œ

dy dx œ 'c2 ax#  2xb dx 0

$

œ ’ x3  x# “

!

#

œ  ˆ 83  4‰ œ

4 3

'04 'c(Èy c4 c4)/2y dx dy œ '04 ˆ y c2 4
È4  y‰ dy 4

2

œ ’ y2  2y  23 a4  yb3/2 “ œ 4  8
0

œ 4
6.

œ

16 3

2 3

† 43/2

4 3

'01 'yÈy Èx dx dy œ '01
23 x$Î# ‘ yÈy dy œ œ

2 3 2 3

'01 ˆy$Î%  y$Î# ‰ dy œ 23
47 y(Î%  25 y&Î# ‘ "! ˆ 47  25 ‰ œ

4 35

'01 'xx Èx dy dx œ '01 x1/2 ax  x2 b dx œ '01 ˆx3/2  x5/2 ‰ dx 2

1

œ
25 x5/2  27 x7/2 ‘0 œ 7.

È9
x

'c33 '0Ð1Î2Ñ

#

2 5



œ

2 7

y dy dx œ 'c3 ’ y2 “ 3

#

œ 'c3 "8 a9  x# b dx œ ’ 9x 8  3

œ ˆ 27 8 

È

27 ‰ 24

'03Î2 'È99

4y4y #

#

4 35

 ˆ 27 8


27 ‰ 24

3Î2

È

Ð1Î2Ñ 9
x#

!

dx

$ x$ 24 “ $

œ

27 6

œ

9 #

y dx dy œ '0 2yÈ94y# dy

œ  "4 † 23 a94y# b3/2 º

3/2

œ 0

" 6

† 93/2 œ

27 6

œ

9 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 15 Practice Exercises

'02 '04
x 2x dy dx œ '02 c2xyd 04
x dx 2 2 œ '0 a2xa4  x2 bb dx œ '0 a8x  2x3 b dx 2

8.

2

2

x4 2 “!

œ ’4x2 

È4 c y

'04 '0

œ 16 

œ8

16 2

È4 c y

2x dx dy œ '0 cx2 d 0 4

œ '0 a4  yb dy œ ’ 4y  4

dy

4

y2 2

“ œ 16 

16 2

0

œ8

9.

'01 '2y2 4 cos ax# b dx dy œ '02 '0x/2 4 cos ax# b dy dx œ '02 2x cos ax# b dx œ csin ax# bd #! œ sin 4

10.

'02 'y1Î2 ex

11.

'08 'È2x

y%

12.

'01 'È1y

21 sin a1x# b x#

$

$

#

dx dy œ '0

1

"
1

'02x ex

dy dx œ '0

2

'0y

$

1

" y%
1

dx dy œ '0

1

4 c x#

dy dx œ '0 2xex dx œ cex d ! œ e  1

#

#

dx dy œ

'02 y 4y
1 dy œ ln417 $

" 4

%

'0x 21 sinx a1x b dy dx œ '01 21x sin a1x# b dx œ c cos a1x# bd "! œ (1)  (1) œ 2 $

#

#

13. A œ 'c2 '2x b 4 dy dx œ 'c2 ax#  2xb dx œ 0

15. V œ '0

1

"

#

0

14. A œ '1

4

4 3

'x2 c x ax#
y# b dy dx œ '01 ’x# y
y3 “ 2cx dx œ '01 ’2x#
(23x) $

$

x

" 12

œ ˆ 23 



7 ‰ 12

6 c x#

16. V œ 'c3 'x 2

"

6 c x#

dx œ 'c3 a6x#  x%  x$ b dx œ 2

xy dy dx œ '0 ’ xy2 “ dx œ '0 1

#

"

1

!

È1 c x

ˆ1‰ 4

'01



4

7x$ 3 “

$

dx œ ’ 2x3 

(2x)% 12



37 6

" 7x% 12 “ !

'01 '0

#

xy dy dx œ

4 1

'01 ’ xy2 “ #

x 2

dx œ

È1 c x

#

!

125 4

" 4

dx œ

2 1

'01 ax  x$ b dx œ #"1

'22Îx dy dx œ '12 ˆ2  2x ‰ dx œ 2  ln 4; My œ '12 '22Îx x dy dx œ '12 x ˆ2  2x ‰ dx œ 1;

Mx œ '1

2

'22Îx

y dy dx œ '1 ˆ2  2

2y c y#

2‰ x#

dx œ 1 Ê x œ y œ

20. M œ '0 'c2y dx dy œ '0 a4y  y# b dy œ 4

4

2y c y

32 3

#

4

4

# #

4

" #  ln 4 2y c y#

; Mx œ '0 'c2y y dx dy œ '0 a4y#  y$ b dy œ ’ 4y3 

y My œ '0 'c2y x dx dy œ '0 ’ a2y#y b  2y# “ dy œ ’ 10  2

dx dy œ '1 ˆÈy  2
y‰ dy œ

4 3 2

18. average value œ 2

œ

x# dy dx œ 'c3 cx# yd x 1

19. M œ '1

2% 12




17. average value œ '0

21. Io œ '0

'2Ècyy

&

% y% 2 “!

4

$

œ  128 5 Ê xœ

My M

œ  12 5 and y œ

% y% 4 “! Mx M

œ

œ2

'2x4 ax#
y# b (3) dy dx œ 3 '02 Š4x#
643  14x3 ‹ dx œ 104 $

'c22 'c11 ax#
y# b dy dx œ 'c22 ˆ2x#
23 ‰ dx œ 403 a b a 'b'a Ix œ 'ca 'cb y# dy dx œ 'ca 2b3 dx œ 4ab 3 ; Iy œ cb ca

22. (a) Io œ (b)

$

œ

4ab 3

$




$

4a b 3

œ

#

#

4ab ab
a b 3

$

x# dx dy œ 'cb b

2a$ 3

dy œ

4a$ b 3

Ê Io œ Ix
Iy

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

64 3

;

987

988

Chapter 15 Multiple Integrals

23. M œ $ '0

3

'02xÎ3 dy dx œ $ '03 2x3 dx œ 3$ ; Ix œ $ '03 '02xÎ3 y# dy dx œ 818$ '03 x$ dx œ ˆ 818$ ‰ Š 34 ‹ œ 2$ %

24. M œ '0

Ê Rx œ É 32

"3 'xx (x
1) dy dx œ '01 ax  x$ b dx œ "4 ; Mx œ '01 'xx y(x
1) dy dx œ #" '01 ax$  x&
x#  x% b dx œ 120 ; 1 x 1 1 x 2 8 13 My œ '0 'x x(x
1) dy dx œ '0 ax#  x% b dx œ 15 Ê x œ 15 and y œ 30 ; Ix œ '0 'x y# (x
1) dy dx 1 I 17 ' 1' x # '1 $ & œ "3 '0 ax%  x(
x$  x' b dx œ 280 Ê Rx œ É M œ É 17 70 ; Iy œ 0 x x (x
1) dy dx œ 0 ax  x b dx 1

#

#

#

#

x

#

œ

Iy ÉM

Ê Ry œ

1 12

œ É 13

25. M œ 'c1 'c1 ˆx#
y#
"3 ‰ dy dx œ 'c1 ˆ2x#
43 ‰ dx œ 4; Mx œ 'c1 'c1 y ˆx#
y#
"3 ‰ dy dx œ 'c1 0 dx œ 0; 1 1 1 My œ ' ' x ˆx#
y#
3" ‰ dy dx œ ' ˆ2x$
43 x‰ dx œ 0 1

1

1

c1 c1

1

1

1

c1

26. Place the ?ABC with its vertices at A(0ß 0), B(bß 0) and C(aß h). The line through the points A and C is yœ

h a

x; the line through the points C and B is y œ

œ b$ '0 ˆ1  yh ‰ dy œ h

È

27.

'
11 '
È11

xx

28.

'c11 'cÈ11ccyy

È

" #

h

21

2 a1
x #
y # b

(x  b). Thus, M œ '0

h

'ayÐaÎh
bÑyÎh  b $ dx dy

'ayÐaÎ
h bÑyÎh  b y# $ dx dy œ b$ '0h Šy#  yh ‹ dy œ $1bh# $

'01

2r a1
r # b#

ln ax#
y#
1b dx dy œ '0

21

# #

œ

; Ix œ '0

dy dx œ '0

# #

$ bh #

h ab

" dr d) œ '0
 1
" r# ‘ ! d) œ 21

'01 r ln ar#
1b dr d) œ '021 '12

" #

" #

$

Ix ; Rx œ É M œ

h È6

'021 d) œ 1

ln u du d) œ

" #

'021 cu ln u  ud #" d)

'021 (2 ln 2  1) d) œ [ln (4)  1] 1 1Î3

29. M œ 'c1Î3 '0 r dr d) œ 3

9 #

'
11ÎÎ33 d) œ 31; My œ '
11ÎÎ33 '03 r# cos ) dr d) œ 9 '
11ÎÎ33 cos ) d) œ 9È3

Ê xœ

3È 3 1

and y œ 0 by symmetry 30. M œ '0

1Î2

yœ

13 31

'13 r dr d) œ 4 '01Î2 d) œ 21; My œ '01Î2 '13

26 3

'01Î2 cos ) d) œ 263

Ê xœ

by symmetry

31. (a) M œ 2 '0

1Î2

'11bcos

)

1Î2

1Î2

1
cos )

My œ 'c1Î2 '1

1
cos 2) ‰ #

1Î2

Ê xœ

d) œ

8
1 4

;

(r cos )) r dr d)

œ 'c1Î2 Šcos# )
cos$ )
32
151 24

(b)

r dr d)

œ '0 ˆ2 cos )


œ

r# cos ) dr d) œ

151
32 61
48

cos% ) 3 ‹

d)

, and

y œ 0 by symmetry 32. (a) M œ 'c! '0 r dr d) œ 'c!

d) œ a# !; My œ 'c! '0 (r cos )) r dr d) œ 'c! ! 2a sin ! Ê x œ 2a 3sin œ0 ! , and y œ 0 by symmetry; lim c x œ lim c 3! !

a

!

!

a# #

!Ä1

!

a

a$ cos ) 3

d) œ

2a$ sin ! 3

!Ä1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

13 31

, and

,

Chapter 15 Practice Exercises (b) x œ

and y œ 0

2a 51

Ècos 2)

33. ax#
y# b  ax#  y# b œ 0 Ê r%  r# cos 2) œ 0 Ê r# œ cos 2) so the integral is 'c1Î4 '0 1Î4

#

1Î4

œ 'c1Î4 ’ 2 a1 "
r# b “ œ

" #

1Î4

"
y # b#

œ '0 ’ "#  1Î3

œ

" #

dx dy œ '0

1Î3

" # a1
sec# )b “

’ È"2 tan"

''

"
y # b#

a1
x #

R

œ '0 35.

1Î4 " " ' " '
11ÎÎ44 ˆ1  1
cos ‰ ˆ ‰ 2) d) œ #
1Î4 1  # cos ) d) #

#

a1
x #

R

(b)

!

d) œ "#

dr d)

'c1Î4 Š1  sec# ) ‹ d) œ "#
)  tan2 ) ‘
1Î14Î4 œ 14 2

''

34. (a)

Ècos 2)

r a1
r # b #

1Î2

lim

bÄ_

u È2 “

È$ !

d) œ È2 4

œ

dx dy œ '0

1Î2

’ "# 

'0sec

)

dr d) œ '0

1Î3

r a1
r# b#

'01Î3 1
secsec) ) d); ” #

" #

#

" # a1
b# b “d)

dr d) œ '0

œ

'0

1Î2

" #

Î

1 2

r a1
r# b#

d) œ

sec )

!

d)

u œ tan ) Ä du œ sec# ) d) •

tan" É 3#

'0_

’ 2 a1 "
r# b “

" #

È3

'0

du 2
u #

b

lim

bÄ_

’ 2 a1 "
r# b “ d) 0

1 4

'01 '01 '01 cos (x
y
z) dx dy dz œ '01 '01 [sin (z
y
1)  sin (z
y)] dy dz 1 œ '0 [ cos (z
21)
cos (z
1)  cos z
cos (z
1)] dz œ 0

36.

'lnln67 '0ln 2 'lnln45 eÐx
y
zÑ dz dy dx œ 'lnln67 '0ln 2 eÐx
yÑ dy dx œ 'lnln67 ex dx œ 1

37.

'01 '0x '0xby (2x  y  z) dz dy dx œ '01 '0x Š 3x#

38.

'1e '1x '0z 2yz dy dz dx œ '1e '1x "z dz dx œ '1e ln x dx œ cx ln x  xd 1e œ 1

#

#

#



3y# # ‹

dy dx œ '0 Š 3x#  1

%

x' #‹

dx œ

8 35

$

39. V œ 2 '0

1Î2

40. V œ 4 '0

2

'
0cos y '0
2x dz dx dy œ 2 '01Î2 '
0cos y È4cx

'0

œ ’x a4  x# b 41. average œ

" 3

#

'04cx

$Î#

#

dz dy dx œ 4 '0

2

È4cx

'0

#

1Î2

2x dx dy œ 2 '0 cos# y dy œ 2 ’ y2


a4  x# b dy dx œ 4 '0 a4  x# b 2

$Î#

1Î2

sin 2y 4 “!

œ

1 #

dx

#


6xÈ4  x#
24 sin" x2 “ œ 24 sin" 1 œ 121 !

'01 '03 '01

30xzÈx#
y dz dy dx œ

" 3

'01 '03 15xÈx#
y dy dx œ 3" '03 '01 15xÈx#
y dx dy

œ

" 3

'03 ’5 ax#
yb$Î# “ " dy œ "3 '03
5(1
y)$Î#  5y$Î# ‘ dy œ 3"
2(1
y)&Î#  2y&Î# ‘ $! œ 3"
2(4)&Î#  2(3)&Î#  2‘

œ

" 3


2 ˆ31  3&Î# ‰‘

!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

989

990

Chapter 15 Multiple Integrals

42. average œ

43. (a)

'021 '01 '0a

3 4 1 a$

È

È

È

3$ sin 9 d3 d9 d) œ

'021 '01 sin 9 d9 d) œ 83a1 '021 d) œ 3a4

3a 161

'cÈ22 'cÈ22ccyy 'Èx4bcyx cy 3 dz dx dy #

#

(b)

'0 '0 '0

(c)

'021 '0 2 'r

21

#

#

1Î4

2

È

È4cr

#

#

33# sin 9 d3 d9 d) 3 dz r dr d) œ 3 '0

21

#

È2

'0

’r a4  r# b

"Î#

 r# “ dr d) œ 3 '0 ’ "3 a4  r# b 21

$Î#

$

 r3 “

œ '0 ˆ2$Î#  2$Î#
4$Î# ‰ d) œ Š8  4È2‹'0 d) œ 21 Š8  4È2‹ 21

21

'c11ÎÎ22 '01 '
rr

1Î2

#

44. (a)

'c11ÎÎ22 '01 '
rr

#

45. (a) (b)

46. (a) (c) (d)

47.

#

!

d)

21(r cos ))(r sin ))# dz r dr d) œ '
1Î2 '0 '
r# 21r$ cos ) sin# ) dz r dr d) 1Î2

#

(b)

È#

21r$ cos ) sin# ) dz r dr d) œ 84 '0

1

r#

'01 r' sin# ) cos ) dr d) œ 12'01Î2 sin# ) cos ) d) œ 4

'021 '01Î4 '0sec 9 3# sin 9 d3 d9 d) '021 '01Î4 '0sec 9 3# sin 9 d3 d9 d) œ 3" '021 '01Î4 (sec 9)(sec 9 tan 9) d9 d) œ 3" '021
2" tan# 9‘ !1Î4 d) œ 6" '021 d) œ 13 È

È

1Î2 1 r '01 '0 1cx '0 x
y (6
4y) dz dy dx (b) '0 '0 '0 (6
4r sin )) dz r dr d) '01Î2 '11ÎÎ42 '0csc 9 (6
43 sin 9 sin )) a3# sin 9b d3 d9 d) #

#

#

'01Î2 '01 '0r (6
4r sin )) dz r dr d) œ '01Î2 '01 a6r#
4r$ sin )b dr d) œ '01Î2 c2r$
r% sin )d "! d) 1Î2 1Î# œ '0 (2
sin )) d) œ c2)  cos )d ! œ 1
1 È

È3 È3cx

È4cx cy

'01 'È13ccxx '1 #

#

z# yx dz dy dx
'1

#

#

'0

#

È4cx cy

'1

#

#

z# yx dz dy dx

48. (a) Bounded on the top and bottom by the sphere x#
y#
z# œ 4, on the right by the right circular cylinder (x  1)#
y# œ 1, on the left by the plane y œ 0 (b)

È

'01Î2 '02 cos 'cÈ44ccrr dz r dr d) )

#

#

49. (a) V œ '0

21

È8cr

'02 '2

#

dz r dr d) œ '0

21

'02 ŠrÈ8  r#  2r‹ dr d) œ '021 ’ "3 a8  r# b$Î#  r# “ # d) !

œ '0
 "3 (4)$Î#  4
3" (8)$Î# ‘ d) œ '0 21

21

(b) V œ '0

21

œ œ 50. Iz œ '0 32 5

Š2  3
2È8‹ d) œ

'0 '2 sec 9 3# sin 9 d3 d9 d) œ 83 '0 '0 1Î4

21

1Î4

4 3

Š4È2  5‹ '0 d) œ

Š2È2 sin 9  sec$ 9 sin 9‹ d9 d)

8 3

'021 '01Î4 Š2È2 sin 9  tan 9 sec# 9‹ d9 d) œ 83 '021 ’2È2 cos 9  "# tan# 9“ 1Î% d)

8 3

'0

21

œ

È8

4 3

21

!

21

Š 2 

" #


2È2‹ d) œ

8 3

'0

21

Š 5
#4

È2

‹ d) œ

81 Š4È2  5‹ 3

'01Î3 '02 (3 sin 9)# a3# sin 9b d3 d9 d) œ '021 '01Î3 '02 3% sin$ 9 d3 d9 d)

'021 '01Î3 asin 9  cos# 9 sin 9b d9 d) œ 325 '021 ’ cos 9
cos3 9 “ 1Î$ d) œ 831 $

!

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

81 Š4È2  5‹ 3

Chapter 15 Additional and Advanced Exercises 51. With the centers of the spheres at the origin, Iz œ '0

21

'01 'ab $ (3 sin 9)# a3# sin 9b d3 d9 d)

'021 '01 sin$ 9 d9 d) œ $ ab 5 a b '021 '01 asin 9  cos# 9 sin 9b d9 d) 1 21 21 œ $ ab 5 a b '0 ’ cos 9
cos3 9 “ d) œ 4$ ab15 a b '0 d) œ 81$ ab15 a b $ ab &  a & b 5

œ

&

&

&

&

&

$

&

&

&

!

'01 '01ccos (3 sin 9)# a3# sin 9b d3 d9 d) œ '02 '0 '01ccos 3% sin$ 9 d3 d9 d) 1 1 21 21 œ "5 '0 '0 (1  cos 9)& sin$ 9 d9 d) œ '0 '0 (1  cos 9)' (1
cos 9) sin 9 d9 d);

52. Iz œ '0

21

)

1

u œ 1  cos 9 ” du œ sin 9 d9 • Ä " 5

œ

'021 2 56†2 $

&

d) œ

32 35

" 5

1

)

'021 '02 u' (2  u) du d) œ 5" '021 ’ 2u7

(

# u) 8 “!



d) œ

" 5

'021 ˆ 7"  8" ‰ 2) d)

'021 d) œ 64351

53. x œ u
y and y œ v Ê x œ u
v and y œ v " " Ê J(uß v) œ º œ 1; the boundary of the 0 "º image G is obtained from the boundary of R as follows:

xy-equations for

Corresponding uv-equations

Simplified

the boundary of R

for the boundary of G

uv-equations

yœx yœ0 Ê

vœu
v

_

uœ0

vœ0

_ _

'0 '0 esx f(x  yß y) dy dx œ '0 '0 x

vœ0 esÐu
vÑ f(uß v) du dv $s  "t !$  "#

54. If s œ !x
" y and t œ # x
$ y where (!$  "# )# œ ac  b# , then x œ and J(sß t) œ œ

" Èac  b#

" (!$  "# )#

$ º #

'021 '0_ rer

#

" œ ! º

dr d) œ

" !$  "#

" #Èac  b#

Ê

_ _ 'c_ '
_ e as
t b È

'021 d) œ È

#

1 ac  b#

#

" ac  b#

,yœ

,

ds dt

1 Èac  b#

. Therefore,

# s
!t !$  "#

œ 1 Ê ac  b# œ 1# .

CHAPTER 15 ADDITIONAL AND ADVANCED EXERCISES 6cx#

1. (a) V œ 'c3 'x 2

6cx#

(c) V œ 'c3 'x 2

6cx#

(b) V œ 'c3 'x 2

x# dy dx 6cx#

x# dy dx œ 'c3 'x 2

a6x#  x%  x$ b dx œ ’2x$ 

x& 5



'0x

#

dz dy dx #

x% 4 “ $

œ

125 4

2. Place the sphere's center at the origin with the surface of the water at z œ 3. Then 9 œ 25  x#  y# Ê x#
y# œ 16 is the projection of the volume of water onto the xy-plane

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

991

992

Chapter 15 Multiple Integrals Ê V œ '0

21

'04 'ccÈ325cr

dz r dr d) œ '0

21

#

'04 ŠrÈ25  r#  3r‹ dr d) œ '021 ’ "3 a25  r# b$Î#  3# r# “ % d)

21 21 œ '0
 "3 (9)$Î#  24
3" (25)$Î# ‘ d) œ '0

3. Using cylindrical coordinates, V œ '0

21

œ '0 ˆ1  21

4. V œ 4 '0

1Î2

" 3

cos ) 

'01 'r

È2
r

#

#

œ 4 '0 Š 3"  1Î2

" 4

" 3

'01 '02crÐcos

sin )‰ d) œ
)  1Î2

dz r dr d) œ 4 '0 2È 2 3 ‹




d) œ Š 8

!

26 3

" 3

)

521 3

d) œ
sin )Ñ

sin )


dz r dr d) œ '0

21

'01 a2r  r# cos )  r# sin )b dr d)

#1

" 3

cos )‘ ! œ 21

'01 ŠrÈ2  r#  r$ ‹ dr d) œ 4'01Î2 ’ "3 a2  r# b$Î#  r4 “ " d) %

!

'

1Î2

È 2 7 ‹ 0 3

1 Š8È2  7‹

d) œ

6

5. The surfaces intersect when 3  x#  y# œ 2x#
2y# Ê x#
y# œ 1. Thus the volume is V œ 4 '0

1

È1 c x

'0

6. V œ 8 '0

1Î2

œ

64 3

#

'2x3cx2ycy #

#

1Î2

dz dy dx œ 4 '0

#

#

'01 '2r3
r #

#

1Î2

dz r dr d) œ 4 '0

'01 a3r  3r$ b dr d) œ 3'01Î2 d) œ 31#

'01Î2 '02 sin 9 3# sin 9 d3 d9 d) œ 643 '01Î2 '01Î2 sin% 9 d9 d)

'01Î2 ” sin 94cos 9 ¹1Î#
43 '01Î2 sin# 9 d9• d) œ 16 '01Î2
92  sin429 ‘ 1! Î# d) œ 41 '01Î2 d) œ 21# $

!

7. (a) The radius of the hole is 1, and the radius of the sphere is 2.

(b) V œ 2 '0

21

8. V œ '0

1

È4cz

È

'0 3 '1 È9cr

'03 sin '0 )

#

r dr dz d) œ '0

21

#

dz r dr d) œ '0

'03 sin

1

œ '0 ’ "3 a9  9 sin# )b 1

$Î#

)

È3

'0

a3  z# b dz d) œ 2È3 '0 d) œ 4È31 21

rÈ9  r# dr d) œ '0 ’ "3 a9  r# b 1


3" (9)$Î# “ d) œ 9'0 ’1  a1  sin# )b 1

œ '0 a1  cos )
sin# ) cos )b d) œ 9 ’)  sin )
1

9. The surfaces intersect when x#
y# œ coordinates is V œ 4 '0

1Î2

œ

" #

'0

1Î2

10. V œ '0

1Î2

œ

15 4

'0

d) œ


Î

ˆr# 1‰ 2 #

“

!

d)

“ d) œ 9'0 a1  cos$ )b d) 1

œ 91

Ê x#
y# œ 1. Thus the volume in cylindrical 1Î2

dz r dr d) œ 4 '0

'01 Š #r  r# ‹ dr d) œ 4'01Î2 ’ r4 $

#

%

"

 r8 “ d) !

1 4

'12 '0r sin

1Î2

'0 'r 1

x#
y#
1 #

1 sin$ ) 3 “!

$Î#

$Î# 3 sin )

#

) cos )

dz r dr d) œ '0

sin ) cos ) d) œ

1Î2

15 4

'12 r$ sin ) cos ) dr d) œ '0 Î2 ’ r4 “ # sin ) cos ) d)

1Î# # ’ sin2 ) “ !

1

%

"

œ

15 8

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 15 Additional and Advanced Exercises 11.

'0_ ec

ax

ecbx x

_

dx œ '0

'ab exy dy dx œ 'ab '0_ exy dx dy œ 'ab Š

œ 'a lim ’ e y “ dy œ 'a lim Š "y  cxy

b

t

tÄ_

b

tÄ_

!

ecyt y ‹

dy œ 'a

b

" y

lim

tÄ_

'0t exy dx‹ dy

dy œ cln yd ab œ ln ˆ ba ‰

12. (a) The region of integration is sketched at the right Ê '0

a sin "

œ '0

"

È

'y cota "c y #

#

ln ax#
y# b dx dy

'0a r ln ar# b dr d);

u œ r# ” du œ 2r dr • Ä

" #

'0" '0a ln u du d) #

'0" [u ln u  u] !a d) " œ "# '0 ’2a# ln a  a#  lim " #

œ

(b)

13.

#

'0

a cos "

'0

t ln t“ d) œ

a# #

ln ax
y b dy dx
'a cos " '0

#

tÄ0

(tan ")x

#

Èa cx

a

#

'0" (2 ln a  1) d) œ a# " ˆln a  "# ‰ #

ln ax#
y# b dy dx

'0x '0u emÐxtÑ f(t) dt du œ '0x 't x emÐxtÑ f(t) du dt œ '0x (x  t)emÐxtÑ f(t) dt; also '0x '0v '0u emÐxtÑ f(t) dt du dv œ '0x 't x 't v emÐxtÑ f(t) du dv dt œ '0x 't x (v  t)emÐxtÑ f(t) dv dt x x x œ '0
"2 (v  t)# emÐxtÑ f(t)‘ t dt œ '0 (x # t) emÐxtÑ f(t) dt #

14.

'01 f(x) Š'0x g(xy)f(y) dy‹ dx œ '01 '0x œ '0

1

g(xy)f(x)f(y) dy dx

'y1 g(xy)f(x)f(y) dx dy œ '01 f(y) Œ'y1 g(xy)f(x) dx dy;

'01 '01 g akxykb f(x)f(y) dx dy œ '01 '0x g(xy)f(x)f(y) dy dx
'01 'x1 g(yx)f(x)f(y) dy dx 1 1 1 1 œ '0 'y g(xy)f(x)f(y) dx dy
'0 'x g(yx)f(x)f(y) dy dx œ '0

1

'y1 g(xy)f(x)f(y) dx dy
'01 'y1 g(xy)f(y)f(x) dx dy

ðóóóóóóóóóóóóñóóóóóóóóóóóóò simply interchange x and y variable names

œ 2'0

1

'y1 g(xy)f(x)f(y) dx dy, and the statement now follows.

15. Io (a) œ '0 '0 a

œ

a# 4




" 1#

xÎa#

ax#
y# b dy dx œ '0 ’x# y
a

a# ; Iwo (a) œ

" #

xÎa#

y$ 3 “!

a  6" a$ œ 0 Ê a% œ

dx œ '0 Š xa#


" 3

a

$

Ê a œ %É 3" œ

x$ 3a' ‹ " % È 3

%

x dx œ ’ 4a #


a

x% 12a' “ !

. Since Iwwo (a) œ

" #


#" a%  0, the

value of a does provide a minimum for the polar moment of inertia Io (a). 16. Io œ '0

2

'2x4 ax#
y# b (3) dy dx œ 3 '02 Š4x#  14x3

$




64 3 ‹

dx œ 104

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

993

994

Chapter 15 Multiple Integrals

17. M œ 'c) 'b sec ) r dr d) œ )

'c Š a# )

a

#

)



b# #

sec# )‹ d)

œ a# )  b# tan ) œ a# cos" ˆ ba ‰  b# Š

È a#  b# ‹ b

œ a# cos" ˆ ba ‰  bÈa#  b# ; Io œ 'c) 'b sec ) r$ dr d) )

a

'c aa%
b% sec% )b d) œ "4 'c ca%
b% a1
tan# )b asec# )bd d) )

" 4

œ

)

)

)

) b% tan$ ) “ 3 )

œ

" 4

œ

% $ a% ) b% tan ) )  b tan #  # 6 " % " $È # " ˆ b ‰ a # a cos a  # b

œ

’a% )  b% tan ) 

2
ˆy# Î2‰

 b#  6" b$ aa#  b# b

18. M œ 'c2 '1cay#Î4b dx dy œ '
2 Š1  2

œ 'c2 ’ x2 “ 2

œ 19.

3 16

#

2c ˆy# Î2‰

1
ay# Î4b

ˆ32 

64 3




2

y# 4‹

2

3 ‰ ˆ 32†8 ‰ œ ˆ 16 œ 15

32 ‰ 5

# y$ 12 “ #

dy œ ’y 

3 dy œ 'c2 32 a4  y# b dy œ 48 15

$Î#

œ

8 3

2

'c22 a16  8y#
y% b dy œ 163 ’16y  8y3

3 32

Ê xœ

My M

‰ ˆ 38 ‰ œ œ ˆ 48 15

6 5

# #

# #

# #

" ab

# #

# y& 5 “!

# #

# #

# #

!

 1‹


" #ab

# #

Šea b  1‹

# #

Šea b  1‹

ßy) 'yy 'xx ``F(x ' y ` F(xßy) x x ` y dx dy œ y ’ ` y “ "

!




# #

!

œ

$

, and y œ 0 by symmetry

'0a '0b emax ab x ßa y b dy dx œ '0a '0bxÎa eb x dy dx
'0b '0ayÎb ea y dx dy a b a b " " œ '0 ˆ ba x‰ eb x dx
'0 ˆ ba y‰ ea y dy œ ’ 2ab eb x “
’ 2ba ea y “ œ #"ab Šeb a # #

20.

2
ˆy# Î2‰

; My œ '
2 '1
ay#Î4b x dx dy

"

#

"

!

!

" ßy) dy œ 'y ’ ` F(x  `y

y"

"

!

x!

` F(x! ßy) `y “

dx œ cF(x" ß y)  F(x! ß y)d yy!"

œ F(x" ß y" )  F(x! ß y" )  F(x" ß y! )
F(x! ß y! ) 21. (a) (i) (ii) (iii) (iv)

Fubini's Theorem Treating G(y) as a constant Algebraic rearrangement The definite integral is a constant number

(b)

'0ln 2 '01Î2 ex cos y dy dx œ Œ'0ln 2 ex dx Œ'01Î2 cos y dy œ aeln 2  e0 b ˆsin 1#  sin 0‰ œ (1)(1) œ 1

(c)

'12 'c11 yx

#

dx dy œ Œ'1

2

" y#

dy Œ'c1 x dx œ ’ y" “ ’ x2 “ #

1

#

"

" "

œ ˆ #"
1‰ ˆ #"  #" ‰ œ 0

22. (a) ™ f œ xi
yj Ê Du f œ u" x
u# y; the area of the region of integration is Ê average œ 2'0

1

#

œ 2 ’u" Š x2  (b) average œ

" area

_ _

23. (a) I# œ '0 œ  "#

'0

'01Î2

_

x$ 3‹

1cx

'0

(u" x
u# y) dy dx œ 2 '0
u" x(1  x)
"# u# (1  x)# ‘ dx

 ˆ "# u# ‰

" (1x)$ 3 “!

' ' (u" x
u# y) dA œ R 1Î2

e
ˆx y ‰ dx dy œ '0 #

lim

bÄ_

" #

1

#

_

(b) > ˆ "# ‰ œ '0 t"Î# et dt œ '0 ay# b

u" area

" 3

(u"
u# )

M u# ' ' ' ' x dA
area y dA œ u" Š My ‹
u# ˆ MMx ‰ œ u" x
u# y

R

R

'0_ ae
r b r dr d) œ '01Î2 ”

aecb  1b d) œ #

œ 2 ˆ 6" u"
6" u# ‰ œ

#

" #

'01Î2 d) œ 14

"Î# y#

e

Ê Iœ

_

lim

bÄ_

'0b re
r

#

dr• d)

È1 #

(2y) dy œ 2 '0 ey dy œ 2 Š #

È1 # ‹

œ È1, where y œ Èt

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 15 Additional and Advanced Exercises 24. Q œ '0

21

'0R kr# (1  sin )) dr d) œ kR3 '021 (1  sin )) d) œ kR3 $

25. For a height h in the bowl the volume of water is V œ œ

È

È

Èh

'cÈhh 'cÈhhccxx ah  x#  y# b dy dx œ '021 '0 #

$

c)
cos )d #!1 œ

È

È

'cÈhh 'cÈhhccxx 'xhby #

#

#

#

dz dy dx

ah  r# b r dr d) œ '0 ’ hr2  r4 “ 21

#

21kR$ 3

#

%

d) œ '0

Èh

21

!

h# 4

d) œ

h# 1 #

.

Since the top of the bowl has area 101, then we calibrate the bowl by comparing it to a right circular cylinder h# 1 #

whose cross sectional area is 101 from z œ 0 to z œ 10. If such a cylinder contains to a depth w then we have 101w œ rain, w œ 3 and h œ È60.

h# 1 #

h# 20

Ê wœ

cubic inches of water . So for 1 inch of rain, w œ 1 and h œ È20; for 3 inches of

26. (a) An equation for the satellite dish in standard position is z œ "# x#
"# y# . Since the axis is tilted 30°, a unit vector v œ 0i
aj
bk normal to the plane of the È3 # È
#3

water level satisfies b œ v † k œ cos ˆ 16 ‰ œ Ê a œ È1  b# œ  "# Ê v œ  "# j Ê  "# (y  1)
Ê zœ

" È3

È3 #

y
Š #" 

k

ˆz  "# ‰ œ 0 " È3 ‹

is an equation of the plane of the water level. Therefore the volume of water is V œ ' ' R

x#
y#  23 y  1
and " œ

2 3

2 È3

1 2

3

#

1

3

#

1 2

#

Ê Vœ ' ' c

Š 23 yb1c È23 cy# ‹

œ y; y œ 1 Ê

dz dy

!

y# and

dz dy

"Î#

Š 23 yb1c È23 cy# ‹

"

3

" #

1

1 2

œ 0. When x œ 0, then y œ ! or y œ " , where ! œ

 Ê 49  4 Š È2  1‹

(b) x œ 0 Ê z œ

' Èx byby c È dz dy dx, where R is the interior of the ellipse

' È bb cÈ 1

3

"Î#

1 2

y

x#

1 2

1 2

2 3


Ê 49  4 Š È2  1‹ 3

#

1 3

1 dz dx dy

y#

œ 1 Ê the tangent line has slope 1 or a 45° slant

Ê at 45° and thereafter, the dish will not hold water. 27. The cylinder is given by x#
y# œ 1 from z œ 1 to _ Ê œ '0

21

_

'0 '1 1

z ar#
z# b&Î#

'0 '0 ' 21

dz r dr d) œ a lim Ä_

1

' ' ' z ar#
z# b&Î# dV D

a

rz 1 ar#
z# b&Î#

dz dr d)

œ a lim Ä_

'021 '01 ’ˆ "3 ‰

œ a lim Ä_

'021 ’ "3 ar#
a# b"Î#  3" ar#
1b"Î# “ " d) œ a lim ' 21 ’ 3" a1
a# b"Î#  3" ˆ2"Î# ‰  3" aa# b"Î#
3" “ d) Ä_ 0

a

r “ ar#
z# b$Î# 1

dr d) œ a lim Ä_

'021 '01 ’ˆ 3" ‰

r ar#
a# b$Î#


ˆ 3" ‰

r “ ar#
1b$Î#

dr d)

!

œ a lim 21 ’ "3 a1
a# b Ä_

"Î#

 3" Š

È2 # ‹

 3" ˆ "a ‰
3" “ œ 21 ’ 3"  ˆ 3" ‰

È2 # “.

28. Let's see? The length of the "unit" line segment is: L œ 2'0 dx œ 2. 1

The area of the unit circle is: A œ 4'0

1

È1 c x

'0

The volume of the unit sphere is: V œ 8'0

1

2

dy dx œ 1.

È1 c x

'0

2

È1 c x c y

'0

2

2

dz dy dx œ 43 1.

Therefore, the hypervolume of the unit 4-sphere should be: Vhyper œ 16'0

1

È1cx

'0

2

È1cx cy

'0

2

2

È1cx cy cz

'0

2

2

2

dw dz dy dx.

Mathematica is able to handle this integral, but we'll use the brute force approach. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

995

996

Chapter 15 Multiple Integrals

Vhyper œ 16'0

1

È1cx

œ 16'0

'0

œ 16'0

'0

œ 16'0

'0

1

1

1

È1cx È1cx

È1cx

'0

2

2

2

2

È1cx cy

'0

È1cx cy

'0

2

2

È1cx cy cz

'0

2

2

2

dw dz dy dx œ 16'0

1

2

z2 È 1  x2  y2 É 1  1 c x2 c y2

dz dy dx œ –

È1cx

'0

È1cx cy

'0

2

2

È 1  x 2  y2  z2

œ cos )

1

È1cx

'0

2

a1  x2  y2 b'1/2 sin2 ) d) dy dx 0

3/2  x2  y2 b dy dx œ 41'0 ŠÈ1  x2  x2 È1  x2  3" a1  x2 b ‹ dx 1

1  x3 $

‘ dx œ 83 1' a1  x2 b3/2 dx œ ” 0 1

0 x œ cos ) œ  83 1'1/2 sin4 ) d) • dx œ sin ) d)

2 ) ‰2 œ  83 1'1/2 ˆ 1  cos d) œ  23 1'1/2 a1  2 cos 2)
cos2 2)bd) œ  23 1'1/2 ˆ 3#  2 cos 2)
2 0

dz dy dx

— dz œ È1  x2  y2 sin ) d)

0

1 4 a1

2

z È 1  x2  y2

a1  x2  y2 b'1/2 È1  cos2 ) sin ) d) dy dx œ 16'0

œ 41'0 È1  x2
a1  x2 b  1

2

0

0

cos 4) ‰ d) 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ

12 2

CHAPTER 16 INTEGRATION IN VECTOR FIELDS 16.1 LINE INTEGRALS 1. r œ ti
("  t)j Ê x œ t and y œ 1  t Ê y œ 1  x Ê (c) 2. r œ i
j
tk Ê x œ 1, y œ 1, and z œ t Ê (e) 3. r œ (2 cos t)i
(2 sin t)j Ê x œ 2 cos t and y œ 2 sin t Ê x#
y# œ 4 Ê (g) 4. r œ ti Ê x œ t, y œ 0, and z œ 0 Ê (a) 5. r œ ti
tj
tk Ê x œ t, y œ t, and z œ t Ê (d) 6. r œ tj
(2  2t)k Ê y œ t and z œ 2  2t Ê z œ 2  2y Ê (b) 7. r œ at#  1b j
2tk Ê y œ t#  1 and z œ 2t Ê y œ

z# 4

 1 Ê (f)

8. r œ (2 cos t)i
(2 sin t)k Ê x œ 2 cos t and z œ 2 sin t Ê x#
z# œ 4 Ê (h) 9. r(t) œ ti
(1  t)j , 0 Ÿ t Ÿ 1 Ê

dr dt

œ i  j Ê ¸ ddtr ¸ œ È2 j ; x œ t and y œ 1  t Ê x
y œ t
("  t) œ 1

Ê 'C f(xß yß z) ds œ '0 f(tß 1  tß 0) ¸ ddtr ¸ dt œ '0 (1) ŠÈ2‹ dt œ ’È2 t“ œ È2 1

"

1

!

10. r(t) œ ti
(1  t)j
k , 0 Ÿ t Ÿ 1 Ê œ t  (1  t)
1  2 œ 2t  2 Ê

dr dt

œ i  j Ê ¸ ddtr ¸ œ È2; x œ t, y œ 1  t, and z œ 1 Ê x  y
z  2

'C f(xß yß z) ds œ '01 (2t  2) È2 dt œ È2 ct#  2td "! œ È2

11. r(t) œ 2ti
tj
(2  2t)k , 0 Ÿ t Ÿ 1 Ê

dr dt

œ 2i
j  2k Ê ¸ ddtr ¸ œ È4
1
4 œ 3; xy
y
z

œ (2t)t
t
(2  2t) Ê 'C f(xß yß z) ds œ '0 a2t#  t
2b 3 dt œ 3
23 t$  "# t#
2t‘ ! œ 3 ˆ 23  1

12. r(t) œ (4 cos t)i
(4 sin t)j
3tk , 21 Ÿ t Ÿ 21 Ê

"

dr dt

1 œ c20td #
# 1 œ 801

dr dt

œ È1
9
4 œ È14 ; x
y
z œ (1  t)
(2  3t)
(3  2t) œ 6  6t Ê

œ  i  3 j  2k

'C f(xß yß z) ds

œ '0 (6  6t) È14 dt œ 6È14 ’t  t2 “ œ Š6È14‹ ˆ "# ‰ œ 3È14 1

#

" !

14. r(t) œ ti
tj
tk , 1 Ÿ t Ÿ _ Ê

_

dr dt

13 #

'C f(xß yß z) ds œ 'c2211 (4)(5) dt

13. r(t) œ (i
2j
3k)
t(i  3j  2k) œ (1  t)i
(2  3t)j
(3  2t)k , 0 Ÿ t Ÿ 1 Ê Ê


2‰ œ

œ (4 sin t)i
(4 cos t)j
3k

Ê ¸ ddtr ¸ œ È16 sin# t
16 cos# t
9 œ 5; Èx#
y# œ È16 cos# t
16 sin# t œ 4 Ê

¸ ddtr ¸

" #

œ i
j
k Ê ¸ ddtr ¸ œ È3 ;

È3 x #  y #  z#

_ Ê 'C f(xß yß z) ds œ '1 Š 3t#3 ‹ È3 dt œ
 1t ‘ " œ lim ˆ b"
1‰ œ 1

œ

È3 t#  t#  t#

œ

È3 3t#

È

bÄ_

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

998

Chapter 16 Integration in Vector Fields

15. C" : r(t) œ ti
t# j , 0 Ÿ t Ÿ 1 Ê

œ i
2tj Ê ¸ ddtr ¸ œ È1
4t# ; x
Èy  z# œ t
Èt#  0 œ t
ktk œ 2t

dr dt

$Î# since t   0 Ê 'C f(xß yß z) ds œ '0 2tÈ1
4t# dt œ ’ "6 a"
4t# b “ œ "

1

!

"

C# : r(t) œ i
j
tk, 0 Ÿ t Ÿ 1 Ê

dr dt

1

"

#

"

5 6

#

16. C" : r(t) œ tk , 0 Ÿ t Ÿ 1 Ê

(5)$Î# 

" 6

œ

" 6

Š5È5  1‹ ;

œ k Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 1
È1  t# œ 2  t#

Ê 'C f(xß yß z) ds œ '0 a2  t# b (1) dt œ
2t  "3 t$ ‘ ! œ 2 

œ 'C f(xß yß z) ds
'C f(xß yß z) ds œ

" 6

È5


" 3

œ

5 3

; therefore 'C f(xß yß z) ds

3 #

œ k Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 0
È0  t# œ t#

dr dt

Ê 'C f(xß yß z) ds œ '0 at# b (1) dt œ ’ t3 “ œ  "3 ; 1

"

$

!

"

C# : r(t) œ tj
k, 0 Ÿ t Ÿ 1 Ê

œ j Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ 0
Èt  1 œ Èt  1

dr dt

" Ê 'C f(xß yß z) ds œ '0 ˆÈt  1‰ (1) dt œ
23 t$Î#  t‘ ! œ 1

#

C$ : r(t) œ ti
j
k , 0 Ÿ t Ÿ 1 Ê

Ê 'C f(xß yß z) ds œ '0 (t)(1) dt œ ’ t2 “ œ "

#

œ  "6

Ê

" #

!

$

17. r(t) œ ti
tj
tk , 0  a Ÿ t Ÿ b Ê

 1 œ  "3 ;

œ i Ê ¸ ddtr ¸ œ 1; x
Èy  z# œ t
È1  1 œ t

dr dt

1

2 3

dr dt

Ê

'C f(xß yß z) ds œ 'C

œ i
j
k Ê ¸ ddtr ¸ œ È3 ;

"

f ds
'C f ds
'C f ds œ  "3
ˆ 3" ‰


xyz x #  y #  z#

#

œ

'C f(xß yß z) ds œ 'ab ˆ 1t ‰ È3 dt œ ’È3 ln ktk “ b œ È3 ln ˆ ba ‰ , since 0  a Ÿ b

$

ttt t#  t#  t#

œ

" #

1 t

a

18. r(t) œ (a cos t) j
(a sin t) k , 0 Ÿ t Ÿ 21 Ê Èx#
z# œ È0
a# sin# t œ œ 1

dr dt

œ (a sin t) j
(a cos t) k Ê ¸ ddtr ¸ œ Èa# sin# t
a# cos# t œ kak ;

1 21  kak sin t, 0 Ÿ t Ÿ 1 Ê 'C f(xß yß z) ds œ '0  kak# sin t dt
'1 kak# sin t dt kak sin t, 1 Ÿ t Ÿ 21

#1

œ ca# cos td !  ca# cos td 1 œ ca# (1)  a# d  ca#  a# (1)d œ 4a# 19. r(x) œ xi
yj œ xi


x# #

j, 0 Ÿ x Ÿ 2 Ê

œ '0 (2x)È1
x# dx œ ’ 23 a1
x# b 2

#

dr ¸ œ i
xj Ê ¸ dx œ È1
x# ; f(xß y) œ f Šxß x# ‹ œ

dr dx

$Î# #

“ œ !

2 3

ˆ5$Î#  1‰ œ

10È5
2 3

20. r(t) œ a1  tbi
1# a1  tb2 j, 0 Ÿ t Ÿ 1 Ê ¸ ddtr ¸ œ É1
a1  tb# ; f(xß y) œ f Ša1  tbß 1# a1  tb2 ‹ œ Ê

'C f ds œ '01 a1
tb 

œ0

1 4 a1
tb #

É1  a1
tb

ˆ "#



" ‰ #0

œ

4

œ 2x Ê 'C f ds

x$ # Š x# ‹

É1
a1  tb# dt œ ' Ša1  tb
14 a1  tb4 ‹ dt œ ’ "# a1  tb2  0 1

a1
tb  14 a1
tb4 É1  a1
tb#

1 20 a1

 tb5 “

" !

11 #0

21. r(t) œ (2 cos t) i
(2 sin t) j , 0 Ÿ t Ÿ

1 #

Ê

dr dt

œ (2 sin t) i
(2 cos t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 cos tß 2 sin t)

œ 2 cos t
2 sin t Ê 'C f ds œ '0 (2 cos t
2 sin t)(2) dt œ c4 sin t  4 cos td ! 1Î2

22. r(t) œ (2 sin t) i
(2 cos t) j , 0 Ÿ t Ÿ œ 4 sin# t  2 cos t Ê

1Î#

1 4

Ê

dr dt

œ 4  (4) œ 8

œ (2 cos t) i
(2 sin t) j Ê ¸ ddtr ¸ œ 2; f(xß y) œ f(2 sin tß 2 cos t)

'C f ds œ '01Î4 a4 sin# t  2 cos t b (2) dt œ c4t  2 sin 2t  4 sin td 01Î%

œ 1  2Š1
È2‹

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.1 Line Integrals 23. r(t) œ at#  1b j
2tk , 0 Ÿ t Ÿ 1 Ê

dr dt

999

œ 2tj
2k Ê ¸ ddtr ¸ œ 2Èt#
1; M œ 'C $ (xß yß z) ds œ '0 $ (t) Š2Èt#
1‹ dt 1

3/2 œ '0 ˆ 3# t‰ Š2Èt#
1‹ dt œ ’at#
1b “ œ 2$Î#  1 œ 2È2  1 "

1

!

24. r(t) œ at#  1b j
2tk , 1 Ÿ t Ÿ 1 Ê ddtr œ 2tj
2k Ê ¸ dr ¸ œ 2Èt#
1; M œ ' $ (xß yß z) ds dt

C

œ 'c1 ˆ15Èat#  1b
2‰ Š2Èt#
1‹ dt 1

œ 'c1 30 at#
1b dt œ ’30 Š t3
t‹“ 1

$

"
"

œ 60 ˆ "3
1‰ œ 80;

Mxz œ 'C y$ (xß yß z) ds œ 'c1 at#  1b c30 at#
1bd dt 1

œ 'c1 30 at%  1b dt œ ’30 Š t5  t‹“ 1

&

œ 48 Ê y œ

Mxz M

"


"

œ 60 ˆ "5  1‰

3 ' ' œ  48 80 œ  5 ; Myz œ C x$ (xß yß z) ds œ C 0 $ ds œ 0 Ê x œ 0; z œ 0 by symmetry (since $ is

independent of z) Ê (xß yß z) œ ˆ!ß  35 ß 0‰ 25. r(t) œ È2t i
È2t j
a4  t# b k , 0 Ÿ t Ÿ 1 Ê

dr dt

œ È2i
È2j  2tk Ê ¸ ddtr ¸ œ È2
2
4t# œ 2È1
t# ;

(a) M œ 'C $ ds œ '0 (3t) Š2È1
t# ‹ dt œ ’2 a1
t# b 1

(b)

$Î# "

“ œ 2 ˆ2$Î#  1‰ œ 4È2  2 !

" 1 M œ 'C $ ds œ '0 (1) Š2È1
t# ‹ dt œ ’tÈ1
t#
ln Št
È1
t# ‹“ œ ’È2
ln Š1
È2‹“  (0
ln 1) !

œ È2
ln Š1
È2‹ 26. r(t) œ ti
2tj
23 t$Î# k , 0 Ÿ t Ÿ 2 Ê

dr dt

œ i
2j
t"Î# k Ê ¸ ddtr ¸ œ È1
4
t œ È5
t;

# M œ 'C $ ds œ '0 ˆ3È5
t‰ ˆÈ5
t‰ dt œ '0 3(5
t) dt œ
32 (5
t)# ‘ ! œ 2

2

Myz œ 'C x$ ds œ '0 t[3(5
t)] dt œ '0 a15t
3t# b dt œ
"25 t#
2

2

# t$ ‘ !

3 #

a7#  5# b œ

2

œ '0 ˆ10t$Î#
2t&Î# ‰ dt œ
4t&Î#
2

œ

38 36

œ

19 18

,yœ

Mxz M

œ

76 36

œ

19 9

4 (Î# ‘ # 7 t !

, and z œ

2

œ 4(2)&Î#
47 (2)(Î# œ 16È2


Mxy M

œ

144È2 7†36

27. Let x œ a cos t and y œ a sin t, 0 Ÿ t Ÿ 21. Then

dx dt

œ

4 7

(24) œ 36;

œ 30
8 œ 38;

Mxz œ 'C y$ ds œ '0 2t[3(5
t)] dt œ 2 '0 a15t
3t# b dt œ 76; Mxy œ 'C z$ ds œ '0 2

3 #

32 7

È2 œ

dz dt

œ0

2 $Î# [3(5 3 t

144 7


t)] dt

È2 Ê x œ

Myz M

È2

œ a sin t,

dy dt

œ a cos t,

‰
Š dy ˆ dz ‰ dt œ a dt; Iz œ ' ax#
y# b $ ds œ ' aa# sin# t
a# cos# tb a$ dt Ê Êˆ dx dt dt ‹
dt C 0 #

#

21

#

Iz œ '0 a$ $ dt œ 21$ a$ ; M œ 'C $ (xß yß z) ds œ '0 $ a dt œ 21$ a Ê Rz œ É M œ É 2211aa$$ œ a. 21

21

28. r(t) œ tj
(2  2t)k , 0 Ÿ t Ÿ 1 Ê

dr dt

$

œ j  2k Ê ¸ ddtr ¸ œ È5; M œ 'C $ ds œ '0 $ È5 dt œ $ È5; 1

" Ix œ 'C ay#
z# b $ ds œ '0 ct#
(2  2t)# d $ È5 dt œ '0 a5t#  8t
4b $ È5 dt œ $ È5
53 t$  4t#
4t‘ ! œ 1

1

" Iy œ 'C ax#
z# b $ ds œ '0 c0#
(2  2t)# d $ È5 dt œ '0 a4t#  8t
4b $ È5 dt œ $ È5
43 t$  4t#
4t‘ ! œ 1

1

Iz œ 'C ax#
y# b $ ds œ '0 a0#
t# b $ È5 dt œ $ È5 ’ t3 “ œ 1

$

" !

Iz and Rz œ É M œ

" 3

I

5 3 4 3

$ È5 ; $ È5 ;

Ix $ È5 Ê Rx œ É M œ É 35 , Ry œ É My œ É 34 œ

" È3

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

2 È3

,

1000 Chapter 16 Integration in Vector Fields 29. r(t) œ (cos t)i
(sin t)j
tk , 0 Ÿ t Ÿ 21 Ê

œ ( sin t)i
(cos t)j
k Ê ¸ ddtr ¸ œ Èsin# t
cos# t
1 œ È2;

dr dt

(a) M œ 'C $ ds œ '0 $ È2 dt œ 21$ È2; Iz œ 'C ax#
y# b $ ds œ '0 acos# t
sin# tb $ È2 dt œ 21$ È2 21

21

Iz Ê Rz œ É M œ1

(b) M œ 'C $ (xß yß z) ds œ '0 $ È2 dt œ 41$ È2 and Iz œ 'C ax#
y# b $ ds œ '0 $ È2 dt œ 41$ È2 41

41

Iz Ê Rz œ É M œ1

30. r(t) œ (t cos t)i
(t sin t)j


2È2 $Î# k, 3 t

0ŸtŸ1 Ê

dr dt

œ (cos t  t sin t)i
(sin t
t cos t)j
È2t k

" Ê ¸ ddtr ¸ œ È(t
1)# œ t
1 for 0 Ÿ t Ÿ 1; M œ 'C $ ds œ '0 (t
1) dt œ
"2 (t
1)# ‘ ! œ 1

Mxy œ 'C z$ ds œ '0 Š 2 3 2 t$Î# ‹ (t
1) dt œ È

1

œ

2È 2 3

ˆ 72
52 ‰ œ

2È 2 3

ˆ 24 ‰ 35 œ

16È2 35

Ê zœ

2È 2 3

'0 ˆt&Î#
t$Î# ‰ dt œ 2È3 2

Mxy M

œ Š 1635 2 ‹ ˆ 23 ‰ œ

32È2 105

œ '0 at# cos# t
t# sin# tb (t
1) dt œ '0 at$
t# b dt œ ’ t4
t3 “ œ 1

1

%

"

$

!

" 4

a2#  1# b œ

3 #

;


27 t(Î#
25 t&Î# ‘ " !

1

È

" #

; Iz œ 'C ax#
y# b $ ds




" 3

œ

7 12

Iz 7 Ê Rz œ É M œ É 18

31. $ (xß yß z) œ 2  z and r(t) œ (cos t)j
(sin t)k , 0 Ÿ t Ÿ 1 Ê M œ 21  2 as found in Example 4 of the text;

Ix also ¸ ddtr ¸ œ 1; Ix œ 'C ay#
z# b $ ds œ '0 acos# t
sin# tb (2  sin t) dt œ '0 (2  sin t) dt œ 21  2 Ê Rx œ É M 1

1

œ1 32. r(t) œ ti


2È2 $Î# j 3 t




t# #

k, 0 Ÿ t Ÿ 2 Ê

dr dt

œ i
È2 t"Î# j
tk Ê ¸ ddtr ¸ œ È1
2t
t# œ È(1
t)# œ 1
t for

0 Ÿ t Ÿ 2; M œ 'C $ ds œ '0 ˆ t"1 ‰ (1
t) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t"1 ‰ (1
t) dt œ ’ t2 “ œ 2; 2

Mxz œ 'C y$ ds œ '

2È2 $Î# 3 t 0

yœ

Mxz M

œ

16 15

2

, and z œ

Mxy M

œ

2

dt œ # 3

# È ’ 4152 t&Î# “ !

œ

$

œ '0 ˆt#
89 t$ ‰ dt œ ’ t3
92 t% “ œ # !

Iz Rz œ É M œ

2 3

; Mxy œ 'C z$ ds œ '0

2 # t

2

2

$

32 15

#

dt œ

#

$ # ’ t6 “ !

; Ix œ 'C ay#
z# b $ ds œ '0 ˆ 89 t$
4" t% ‰ dt œ ’ 29 t%


Iy œ 'C ax#
z# b $ ds œ '0 ˆt#
"4 t% ‰ dt œ ’ t3
2

2

8 3




32 9

œ

56 9

# t& 20 “ !

œ

8 3




32 20

œ

64 15

Ix Ê Rx œ É M œ

2 3

!

œ

# t& 20 “ !

% 3

œ

Ê xœ

Myz M




œ

32 9

; Iz œ 'C ax#
y# b $ ds Iy 32 ÉM É 29 œ É 15 , and 5 , Ry œ

È7

33-36. Example CAS commands: Maple: f := (x,y,z) -> sqrt(1+30*x^2+10*y); g := t -> t; h := t -> t^2; k := t -> 3*t^2; a,b := 0,2; ds := ( D(g)^2 + D(h)^2 + D(k)^2 )^(1/2): 'ds' = ds(t)*'dt'; F := f(g,h,k): 'F(t)' = F(t); Int( f, s=C..NULL ) = Int( simplify(F(t)*ds(t)), t=a..b ); `` = value(rhs(%)); Mathematica: (functions and domains may vary) Clear[x, y, z, r, t, f] f[x_,y_,z_]:= Sqrt[1
30x2
10y]

#

# (a) # (b) # (c)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

32 20

œ 1, 232 45

;

Section 16.2 Vector Fields, Work, Circulation, and Flux 1001 {a,b}= {0, 2}; x[t_]:= t y[t_]:= t2 z[t_]:= 3t2 r[t_]:= {x[t], y[t], z[t]} v[t_]:= D[r[t], t] mag[vector_]:=Sqrt[vector.vector] Integrate[f[x[t],y[t],z[t]] mag[v[t]], {t, a, b}] N[%] 16.2 VECTOR FIELDS, WORK, CIRCULATION, AND FLUX 1. f(xß yß z) œ ax#
y#
z# b `f `y


"Î#

#
$Î#

œ  y a x#
y #
z b

and

similarly,

œ

y x #  y #  z#

and

`f `z

" #

2. f(xß yß z) œ ln Èx#
y#
z# œ `f `y

`f `x

Ê

`f `z


$Î#

#
$Î#

œ  z a x#
y #
z b

`f `x

ln ax#
y#
z# b Ê

œ

3. g(xß yß z) œ ez  ln ax#
y# b Ê

œ  "# ax#
y#
z# b

Ê ™fœ

z x #  y #  z#

`g `x

`g `y

œ  x# 2x  y# ,

(2x) œ x ax#
y#
z# b

Ê ™fœ

œ

" #


$Î#

; similarly,


x i
y j
z k ax#  y#  z# b$Î#

Š x# y"# z# ‹ (2x) œ

x x #  y #  z#

;

x i  y j  zk x#  y#  z#

œ  x# 2y  y# and

`g `z

œ ez

z Ê ™ g œ Š x#
2xy# ‹ i  Š x# 2y  y# ‹ j
e k

`g `x

4. g(xß yß z) œ xy
yz
xz Ê

œ y
z,

`g `y

`g `z

œ x
z, and

œ y
x Ê ™ g œ (y
z)i
(B
z)j
(x
y)k

5. kFk inversely proportional to the square of the distance from (xß y) to the origin Ê È(M(xß y))#
(N(xß y))# œ

k x#  y#

, k  0; F points toward the origin Ê F is in the direction of n œ

Ê F œ an , for some constant a  0. Then M(xß y) œ Ê È(M(xß y))#
(N(xß y))# œ a Ê a œ

k x#  y#


ax È x#  y#

Ê Fœ


x È x#  y#

and N(xß y) œ


kx ax#  y# b$Î#

i

i

y È x#  y#

j


ay È x#  y#

ky ax#  y# b$Î#

j , for any constant k  0

6. Given x#
y# œ a#
b# , let x œ Èa#
b# cos t and y œ Èa#
b# sin t. Then r œ ŠÈa#
b# cos t‹ i  ŠÈa#
b# sin t‹ j traces the circle in a clockwise direction as t goes from 0 to 21 Ê v œ ŠÈa#
b# sin t‹ i  ŠÈa#
b# cos t‹ j is tangent to the circle in a clockwise direction. Thus, let F œ v Ê F œ yi  xj and F(0ß 0) œ 0 . 7. Substitute the parametric representations for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F , and calculate the work W œ 'C F † (a) F œ 3ti
2tj
4tk and

dr dt

(b) F œ 3t# i
2tj
4t% k and œ

7 3


2œ

dr dt

.

œi
j
k Ê F† dr dt

dr dt

œ 9t Ê W œ '0 9t dt œ 1

œ i
2tj
4t$ k Ê F †

dr dt

œ 7t#
16t( Ê W œ '0 a7t#
16t( b dt œ
73 t$
2t) ‘ ! 1

13 3

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ 3ti
2tj and F# œ 3i
2j
4tk and

d r# dt

œ k Ê F# †

d r# dt

d r" dt

9 #

œ i
j Ê F" †

d r" dt

"

œ 5t Ê W" œ '0 5t dt œ 1

œ 4t Ê W# œ '0 4t dt œ 2 Ê W œ W"
W# œ 1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

9 #

5 #

;

1002 Chapter 16 Integration in Vector Fields 8. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †

dr dt

.

" ‰ (a) F œ ˆ t#  1 j and

dr dt

œi
j
k Ê F†

" ‰ (b) F œ ˆ t#  1 j and

dr dt

œ i
2tj
4t$ k Ê F †

dr dt

Ê F# †

d r# dt

œ 0 Ê W œ '0

" t#  1

dt œ

1

œ

2t t#  1 Ê W and ddtr" œ i
j

dr dt

" ‰ (c) r" œ ti
tj and r# œ i
j
tk ; F" œ ˆ t#  1 j 1

Ê W œ '0

" t#  1

œ

" t#  1

œ '0

"

dt œ ctan
" td ! œ

1

Ê

2t t#  1 F" † ddtr"

1 4 "

dt œ cln at#
1bd ! œ ln 2 œ

" t#  1

; F# œ

" #

j and

œk

d r# dt

1 4

9. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate the work W œ 'C F † (a) F œ Èti  2tj
Ètk and (b) F œ t# i  2tj
tk and

dr dt

dr dt

dr dt

.

œi
j
k Ê F†

" œ 2Èt  2t Ê W œ '0 ˆ2Èt  2t‰ dt œ
43 t$Î#  t# ‘ ! œ 1

dr dt

œ i
2tj
4t$ k Ê F †

dr dt

œ 4t%  3t# Ê W œ '0 a4t%  3t# b dt œ
45 t&  t$ ‘ ! œ  "5 1

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ 2tj
Èt k and œ 1; F# œ Èti  2j
k and

d r# dt

œ k Ê F# †

" 3

œ i
j Ê F" †

d r" dt

d r" dt

"

œ 2t Ê W" œ '0 2t dt 1

œ 1 Ê W# œ '0 dt œ 1 Ê W œ W"
W# œ 0 1

d r# dt

10. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †

dr dt

. œ 3t# Ê W œ '0 3t# dt œ 1 1

(a) F œ t# i
t# j
t# k and

dr dt

œi
j
k Ê F†

(b) F œ t$ i  t' j
t& k and

dr dt

œ i
2tj
4t$ k Ê F †

%

œ ’ t4


t) 4

"


94 t* “ œ !

dr dt

œ t$
2t(
4t) Ê W œ '0 at$
2t(
4t) b dt 1

17 18

(c) r" œ ti
tj and r# œ i
j
tk ; F" œ t# i and F# œ i
tj
tk and

dr dt

d r# dt

œ k Ê F# †

d r# dt

œ i
j Ê F" †

d r" dt

œ t Ê W# œ '0 t dt œ 1

d r" dt " #

œ t# Ê W" œ '0 t# dt œ 1

Ê W œ W"
W# œ

" 3

;

5 6

11. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate the work W œ 'C F †

dr dt

. œ 3t#
1 Ê W œ '0 a3t#
1b dt œ ct$
td ! œ 2

(a) F œ a3t#  3tb i
3tj
k and

†

(b) F œ a3t#  3tb i
3t% j
k

Ê F†

dr dt

1

"

dr & $ # dt œ 6t
4t
3t  3t 1 " Ê W œ 0 a6t&
4t$
3t#  3tb dt œ
t'
t%
t$  3# t# ‘ ! œ 3# r" œ ti
tj and r# œ i
j
tk ; F" œ a3t#  3tb i
k and ddtr" œ i
j Ê F" † ddtr" œ 3t# 1 " Ê W" œ 0 a3t#  3tb dt œ
t$  32 t# ‘ ! œ  "# ; F# œ 3tj
k and ddtr# œ k Ê F# † ddtr# Ê W œ W"
W# œ 12

'

(c)

dr dt œ i
j
k Ê F and ddtr œ i
2tj
4t$ k

'

 3t œ 1 Ê W# œ '0 dt œ 1

12. Substitute the parametric representation for r(t) œ x(t)i
y(t)j
z(t)k representing each path into the vector field F, and calculate the work W œ 'C F † (a) F œ 2ti
2tj
2tk and

dr dt

dr dt

.

œi
j
k Ê F†

(b) F œ at#
t% b i
at%
tb j
at
t# b k and

dr dt

dr dt

œ 6t Ê W œ '0 6t dt œ c3t# d ! œ 3

œ i
2tj
4t$ k Ê F †

Ê W œ '0 a6t&
5t%
3t# b dt œ ct'
t&
t$ d ! œ 3 1

1

"

dr dt

œ 6t&
5t%
3t#

"

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1

Section 16.2 Vector Fields, Work, Circulation, and Flux 1003 (c) r" œ ti
tj and r# œ i
j
tk ; F" œ ti
tj
2tk and F# œ (1
t)i
(t
1)j
2k and

d r# dt

œ k Ê F# †

dr" dt

œ i
j Ê F" †

d r" dt

Ê F†

dr dt

œ 2t$ Ê work œ '0 2t$ dt œ

1

œ 2 Ê W# œ '0 2 dt œ 2 Ê W œ W"
W# œ 3 1

d r# dt

13. r œ ti
t# j
tk , 0 Ÿ t Ÿ 1, and F œ xyi
yj  yzk Ê F œ t$ i
t# j  t$ k and 1

œ 2t Ê W" œ '0 2t dt œ ";

œ i
2tj
k

dr dt

" #

14. r œ (cos t)i
(sin t)j
6t k , 0 Ÿ t Ÿ 21, and F œ 2yi
3xj
(x
y)k Ê F œ (2 sin t)i
(3 cos t)j
(cos t
sin t)k and œ 3 cos# t  2sin2 t
œ
32 t


3 4

" 6

sin 2t  t


cos t
" 6




sin 2t 2

" 6

dr dt

œ ( sin t)i
(cos t)j
6" k Ê F †

sin t Ê work œ '0 ˆ3 cos# t  2 sin2 t


sin t 

" 6

#1 t‘ !

cos

21

" 6

cos t


" 6

dr dt

sin t‰ dt

œ1

15. r œ (sin t)i
(cos t)j
tk , 0 Ÿ t Ÿ 21, and F œ zi
xj
yk Ê F œ ti
(sin t)j
(cos t)k and dr dt

œ (cos t)i  (sin t)j
k Ê F †

œ
cos t
t sin t 




t 2

sin 2t 4

dr dt

œ t cos t  sin# t
cos t Ê work œ '0 at cos t  sin# t
cos tb dt 21

#1


sin t‘ ! œ 1

16. r œ (sin t)i
(cos t)j
6t k , 0 Ÿ t Ÿ 21, and F œ 6zi
y# j
12xk Ê F œ ti
acos# tbj
(12 sin t)k and dr dt

œ (cos t)i  (sin t)j
"6 k Ê F †

dr dt

œ t cos t  sin t cos# t
2 sin t

Ê work œ '0 at cos t  sin t cos# t
2 sin tb dt œ
cos t
t sin t
21

1 3

#1

cos$ t  2 cos t‘ ! œ 0

17. x œ t and y œ x# œ t# Ê r œ ti
t# j , 1 Ÿ t Ÿ 2, and F œ xyi
(x
y)j Ê F œ t$ i
at
t# b j and dr dt

œ i
2tj Ê F †

dr dt

œ t$
a2t#
2t$ b œ 3t$
2t# Ê 'C xy dx
(x
y) dy œ 'C F †

#

œ
34 t%
23 t$ ‘
" œ ˆ12


16 ‰ 3

 ˆ 34  23 ‰ œ

45 4




18 3

œ

dr dt

dt œ 'c" a3t$
2t# b dt #

69 4

18. Along (0ß 0) to (1ß 0): r œ ti , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ ti
tj and

dr dt

œi Ê F†

dr dt

œ t;

Along (1ß 0) to (0ß 1): r œ (1  t)i
tj , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ (1  2t)i
j and dr dr dt œ i
j Ê F † dt œ 2t; Along (0ß 1) to (0ß 0): r œ (1  t)j , 0 Ÿ t Ÿ 1, and F œ (x  y)i
(x
y)j Ê F œ (t  1)i
(1  t)j and dr dt

œ j Ê F †

dr dt

œ t  1 Ê 'C (x  y) dx
(x
y) dy œ '0 t dt
'0 2t dt
'0 (t  1) dt œ '0 (4t  1) dt 1

1

1

1

"

œ c2t#  td ! œ 2  1 œ 1 19. r œ xi
yj œ y# i
yj , 2   y   1, and F œ x# i  yj œ y% i  yj Ê Ê

c1

'C F † T ds œ '2

F†

dr dy

dr dt

œ 2yi
j and F †

dr dy

œ 2y&  y


" 4‰ dy œ '2 a2y&  yb dy œ
3" y'  #" y# ‘ # œ ˆ 3"  #" ‰  ˆ 64 3  # œ

20. r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ Ê F†

c1

dr dy

1 #

, and F œ yi  xj Ê F œ (sin t)i  (cos t)j and

œ  sin# t  cos# t œ 1 Ê

'C F † dr œ '0

1Î2

dr dt

3 #



63 3

œ  39 #

œ ( sin t)i
(cos t)j

(1) dt œ  1#

21. r œ (i
j)
t(i
2j) œ (1
t)i
(1
2t)j , 0 Ÿ t Ÿ 1, and F œ xyi
(y  x)j Ê F œ a1
3t
2t# b i
tj and dr dt

œ i
2j Ê F †

dr dt

œ 1
5t
2t# Ê work œ 'C F †

dr dt

dt œ '0 a1
5t
2t# b dt œ
t
52 t#
23 t$ ‘ ! œ 1

22. r œ (2 cos t)i
(2 sin t)j , 0 Ÿ t Ÿ 21, and F œ ™ f œ 2(x
y)i
2(x
y)j Ê F œ 4(cos t
sin t)i
4(cos t
sin t)j and ddtr œ (2 sin t)i
(2 cos t)j Ê F †

"

dr dt

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

25 6

1004 Chapter 16 Integration in Vector Fields œ 8 asin t cos t
sin# tb
8 acos# t
cos t sin tb œ 8 acos# t  sin# tb œ 8 cos 2t Ê work œ 'C ™ f † dr œ 'C F †

dt œ '0 8 cos 2t dt œ c4 sin 2td #!1 œ 0 21

dr dt

23. (a) r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ 21, F" œ xi
yj , and F# œ yi
xj Ê F" œ (cos t)i
(sin t)j , and F# œ ( sin t)i
(cos t)j Ê F" †

dr dt

dr dt

œ ( sin t)i
(cos t)j ,

œ 0 and F# †

dr dt

œ sin# t
cos# t œ 1

Ê Circ" œ '0 0 dt œ 0 and Circ# œ '0 dt œ 21; n œ (cos t)i
(sin t)j Ê F" † n œ cos# t
sin# t œ 1 and 21

21

F# † n œ 0 Ê Flux" œ '0 dt œ 21 and Flux# œ '0 0 dt œ 0 21

21

(b) r œ (cos t)i
(4 sin t)j , 0 Ÿ t Ÿ 21 Ê F# œ (4 sin t)i
(cos t)j Ê F" †

dr dt

dr dt

œ ( sin t)i
(4 cos t)j , F" œ (cos t)i
(4 sin t)j , and

œ 15 sin t cos t and F# †

dr dt

œ 4 Ê Circ" œ '0 15 sin t cos t dt 21

#1 œ
"25 sin# t‘ ! œ 0 and Circ# œ '0 4 dt œ 81; n œ Š È417 cos t‹ i
Š È"17 sin t‹ j Ê F" † n 21

œ

4 È17

cos# t


4 È17

sin# t and F# † n œ  È1517 sin t cos t Ê Flux" œ '0 (F" † n) kvk dt œ '0 Š È417 ‹ È17 dt 21

21

# ‘ #1 œ 81 and Flux# œ '0 (F# † n) kvk dt œ '0 Š È1517 sin t cos t‹ È17 dt œ
 15 2 sin t ! œ 0 21

21

24. r œ (a cos t)i
(a sin t)j , 0 Ÿ t Ÿ 21, F" œ 2xi  3yj , and F# œ 2xi
(x  y)j Ê

œ (a sin t)i
(a cos t)j ,

dr dt

F" œ (2a cos t)i  (3a sin t)j , and F# œ (2a cos t)i
(a cos t  a sin t)j Ê n kvk œ (a cos t)i
(a sin t)j , F" † n kvk œ 2a# cos# t  3a# sin# t, and F# † n kvk œ 2a# cos# t
a# sin t cos t  a# sin# t Ê Flux" œ '0 a2a# cos# t  3a# sin# tb dt œ 2a#
2t
21

sin 2t ‘ #1 4 !

 3a#
2t 

Flux# œ '0 a2a# cos# t  a# sin t cos t  a# sin# tb dt œ 2a#
2t
21

25. F" œ (a cos t)i
(a sin t)j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

sin 2t ‘ #1 4 ! d r" dt




sin 2t ‘ #1 4 !

œ 1a# , and

a# #

#1

csin# td !  a#
2t 

sin 2t ‘ #1 4 !

œ 1a#

œ 0 Ê Circ" œ 0; M" œ a cos t,

N" œ a sin t, dx œ a sin t dt, dy œ a cos t dt Ê Flux" œ 'C M" dy  N" dx œ '0 aa# cos# t
a# sin# tb dt œ '0 a# dt œ a# 1;

1

1

F # œ ti ,

d r# dt

œ i Ê F# †

d r# dt

œ t Ê Circ# œ 'ca t dt œ 0; M# œ t, N# œ 0, dx œ dt, dy œ 0 Ê Flux# a

œ 'C M# dy  N# dx œ 'ca 0 dt œ 0; therefore, Circ œ Circ"
Circ# œ 0 and Flux œ Flux"
Flux# œ a# 1 a

26. F" œ aa# cos# tb i
aa# sin# tb j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

d r" dt

œ a$ sin t cos# t
a$ cos t sin# t

Ê Circ" œ '0 aa$ sin t cos# t
a$ cos t sin# tb dt œ  2a3 ; M" œ a# cos# t, N" œ a# sin# t, dy œ a cos t dt, 1

$

dx œ a sin t dt Ê Flux" œ 'C M" dy  N" dx œ '0 aa$ cos$ t
a$ sin$ tb dt œ 1

F # œ t# i ,

d r# dt

œ i Ê F# †

d r# dt

œ t# Ê Circ# œ 'ca t# dt œ a

2a$ 3

4 3

a$ ;

; M# œ t# , N# œ 0, dy œ 0, dx œ dt

Ê Flux# œ 'C M# dy  N# dx œ 0; therefore, Circ œ Circ"
Circ# œ 0 and Flux œ Flux"
Flux# œ 27. F" œ (a sin t)i
(a cos t)j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

d r" dt

4 3

a$

œ a# sin# t
a# cos# t œ a#

Ê Circ" œ '0 a# dt œ a# 1 ; M" œ a sin t, N" œ a cos t, dx œ a sin t dt, dy œ a cos t dt 1

Ê Flux" œ 'C M" dy  N" dx œ '0 aa# sin t cos t
a# sin t cos tb dt œ 0; F# œ tj , 1

dr# dt

œ i Ê F# †

d r# dt

œ0

Ê Circ# œ 0; M# œ 0, N# œ t, dx œ dt, dy œ 0 Ê Flux# œ 'C M# dy  N# dx œ 'ca t dt œ 0; therefore, a

Circ œ Circ"
Circ# œ a# 1 and Flux œ Flux"
Flux# œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.2 Vector Fields, Work, Circulation, and Flux 1005 28. F" œ aa# sin# tb i
aa# cos# tb j ,

d r" dt

œ (a sin t)i
(a cos t)j Ê F" †

Ê Circ" œ '0 aa$ sin$ t
a$ cos$ tb dt œ 1

4 3

d r" dt

œ a$ sin$ t
a$ cos$ t

a$ ; M" œ a# sin# t, N" œ a# cos# t, dy œ a cos t dt, dx œ a sin t dt

Ê Flux" œ 'C M" dy  N" dx œ '0 aa$ cos t sin# t
a$ sin t cos# tb dt œ 1

2 3

a$ ; F# œ t# j ,

d r# dt

œ i Ê F# †

d r# dt

œ0

Ê Circ# œ 0; M# œ 0, N# œ t# , dy œ 0, dx œ dt Ê Flux# œ 'C M# dy  N# dx œ 'ca t# dt œ  23 a$ ; therefore, a

Circ œ Circ"
Circ# œ

4 3

a$ and Flux œ Flux"
Flux# œ 0

29. (a) r œ (cos t)i
(sin t)j , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê F œ (cos t
sin t)i  acos# t
sin# tb j Ê F †

dr dt

(b)

sin 2t 1 ‘1 4  sin t ! œ  # r œ (1  2t)i , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê ddtr œ 2i and F œ (1  2t)i  (1  2t)# j Ê 1 " F † ddtr œ 4t  2 Ê C F † T ds œ 0 (4t  2) dt œ c2t#  2td ! œ 0 r" œ (1  t)i  tj , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê ddtr" œ i  j and F œ (1  2t)i  a1  2t

'

(c)

œ (sin t)i
(cos t)j and

œ  sin t cos t  sin# t  cos t Ê 'C F † T ds

œ '0 a sin t cos t  sin# t  cos tb dt œ
 "2 sin# t  1

dr dt

Ê F†




t #

'

œ (2t  1)
a1  2t
2t# b œ 2t# Ê Flow" œ 'C F †

d r" dt

"

0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê œ i  a2t#  2t
1b j Ê F † "

œ
t#  23 t$ ‘ ! œ

" 3

dr# dt

dr# dt

dr" dt

œ '0 2t# dt œ 1

2 3


2t# b j

; r# œ ti
(t  1)j ,

œ i
j and F œ i  at#
t#  2t
1b j

œ 1  a2t#  2t
1b œ 2t  2t# Ê Flow# œ 'C F †

dr# dt

#

Ê Flow œ Flow"
Flow# œ

2 3




" 3

œ '0 a2t  2t# b dt 1

œ1

30. From (1ß 0) to (0ß 1): r" œ (1  t)i
tj , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê

d r" dt

œ i
j ,

F œ i  a1  2t
2t# b j , and n" kv" k œ i
j Ê F † n" kv" k œ 2t  2t# Ê Flux" œ '0 a2t  2t# b dt 1

"

œ
t#  23 t$ ‘ ! œ

" 3

;

From (0ß 1) to (1ß 0): r# œ ti
(1  t)j , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê

d r# dt

œ i  j ,

#

F œ (1  2t)i  a1  2t
2t b j , and n# kv# k œ i
j Ê F † n# kv# k œ (2t  1)
a1
2t  2t# b œ 2
4t  2t# Ê Flux# œ '0 a2
4t  2t# b dt œ
2t
2t#  23 t$ ‘ ! œ  23 ; 1

"

From (1ß 0) to (1ß 0): r$ œ (1
2t)i , 0 Ÿ t Ÿ 1, and F œ (x
y)i  ax#
y# b j Ê #

d r$ dt

œ 2i ,

#

F œ (1
2t)i  a1  4t
4t b j , and n$ kv$ k œ 2j Ê F † n$ kv$ k œ 2 a1  4t
4t b Ê Flux$ œ 2 '0 a1  4t
4t# b dt œ 2
t  2t#
43 t$ ‘ ! œ 1

31. F œ  Èx#y y# i


"

x È x#  y#

2 3

Ê Flux œ Flux"
Flux#
Flux$ œ

j on x#
y# œ 4;

at (2ß 0), F œ j ; at (0ß 2), F œ i ; at (2ß 0), È F œ j ; at (!ß 2), F œ i ; at ŠÈ2ß È2‹ , F œ  #3 i
"# j ; at ŠÈ2ß È2‹ , F œ Fœ

È3 #

È3 #

i
"# j ; at ŠÈ2ß È2‹ ,

i  "# j ; at ŠÈ2ß È2‹ , F œ

È3 #

i  "# j

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 3



2 3




2 3

œ

" 3

1006 Chapter 16 Integration in Vector Fields 32. F œ xi
yj on x#
y# œ 1; at (1ß 0), F œ i ; at (1ß 0), F œ i ; at (0ß 1), F œ j ; at (0ß 1), F œ j ; at Š "# ß

È3 # ‹,

Fœ

" #

at Š "# ß

È3 # ‹,

F œ  "# i


at Š "# ß 

È3 # ‹,

Fœ

at Š "# ß 

È3 # ‹,

" #

i

i
È3 #

È3 #

È3 #

j;

j;

j;

F œ  "# i 

È3 #

j.

33. (a) G œ P(xß y)i
Q(xß y)j is to have a magnitude Èa#
b# and to be tangent to x#
y# œ a#
b# in a counterclockwise direction. Thus x#
y# œ a#
b# Ê 2x
2yyw œ 0 Ê yw œ  xy is the slope of the tangent line at any point on the circle Ê yw œ  ba at (aß b). Let v œ bi
aj Ê kvk œ Èa#
b# , with v in a counterclockwise direction and tangent to the circle. Then let P(xß y) œ y and Q(xß y) œ x Ê G œ yi
xj Ê for (aß b) on x#
y# œ a#
b# we have G œ bi
aj and kGk œ Èa#
b# . (b) G œ ˆÈx#
y# ‰ F œ ŠÈa#
b# ‹ F . 34. (a) From Exercise 33, part a, yi
xj is a vector tangent to the circle and pointing in a counterclockwise direction Ê yi  xj is a vector tangent to the circle pointing in a clockwise direction Ê G œ Èyxi #
xjy# is a unit vector tangent to the circle and pointing in a clockwise direction. (b) G œ F 35. The slope of the line through (xß y) and the origin is pointing away from the origin Ê F œ 

xi  yj È x#  y#

y x

Ê v œ xi
yj is a vector parallel to that line and

is the unit vector pointing toward the origin.

36. (a) From Exercise 35,  Èxxi #yjy# is a unit vector through (xß y) pointing toward the origin and we want kFk to have magnitude Èx#
y# Ê F œ Èx#
y# Š Èxxi #yjy# ‹ œ xi  yj . (b) We want kFk œ

C È x#  y#

37. F œ 4t$ i
8t# j
2k and 38. F œ 12t# j
9t# k and

dr dt

where C Á 0 is a constant Ê F œ dr dt

œ i
2tj Ê F †

œ 3j
4k Ê F †

39. F œ (cos t  sin t)i
(cos t)k and

dr dt

dr dt

dr dt

40. F œ (2 sin t)i  (2 cos t)j
2k and

œ 12t$ Ê Flow œ '0 12t$ dt œ c3t% d ! œ 48 2

#

1

œ ( sin t)i
(cos t)k Ê F †

dr dt

yj Š Èxxi #yjy# ‹ œ C Š xx#i   y# ‹.

œ 72t# Ê Flow œ '0 72t# dt œ c24t$ d ! œ 24

Ê Flow œ '0 ( sin t cos t
1) dt œ
"2 cos# t
1

C È x#  y#

1 t‘ !

dr dt

"

œ  sin t cos t
1

œ ˆ "#
1‰  ˆ "#
0‰ œ 1

œ (2 sin t)i
(2 cos t)j
2k Ê F †

dr dt

œ 4 sin# t  4 cos# t
4 œ 0

Ê Flow œ 0 41. C" : r œ (cos t)i
(sin t)j
tk , 0 Ÿ t Ÿ Ê F†

dr dt

1 #

Ê F œ (2 cos t)i
2tj
(2 sin t)k and

dr dt

œ ( sin t)i
(cos t)j
k

œ 2 cos t sin t
2t cos t
2 sin t œ  sin 2t
2t cos t
2 sin t Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.2 Vector Fields, Work, Circulation, and Flux 1007 Ê Flow" œ '0

1Î2

C# : r œ j


1 #

1Î#

( sin 2t
2t cos t
2 sin t) dt œ
"2 cos 2t
2t sin t
2 cos t  2 cos t‘ !

(1  t)k , 0 Ÿ t Ÿ 1 Ê F œ 1(1  t)j
2k and

dr dt

Ê Flow# œ '0 1 dt œ c1td "! œ 1;

1 #

œ k Ê F†

dr dt

œ 1
1 ;

œ 1

1

C$ : r œ ti
(1  t)j , 0 Ÿ t Ÿ 1 Ê F œ 2ti
2(1  t)k and

dr dt

œij Ê F†

Ê Flow$ œ '0 2t dt œ ct# d ! œ 1 Ê Circulation œ (1
1)  1
1 œ 0 1

42. F †

œx

dr dt

dx dt


y

dr dt

œ 2t

"

dy dt


z

` f dx ` x dt

œ

dz dt




` f dy ` y dt

by the chain rule Ê Circulation œ 'C F †

dr dt




` f dz ` z dt

dt œ 'a

, where f(xß yß z) œ b

d dt afaratbbb

" #

ax#
y#
x# b Ê F †

dr dt

œ

d dt afaratbbb

dt œ farabbb  faraabb. Since C is an entire ellipse,

rabb œ raab, thus the Circulation œ 0. 43. Let x œ t be the parameter Ê y œ x# œ t# and z œ x œ t Ê r œ ti
t# j
tk , 0 Ÿ t Ÿ 1 from (0ß 0ß 0) to (1ß 1ß 1) Ê œ

dr dt

œ i
2tj
k and F œ xyi
yj  yzk œ t$ i
t# j  t$ k Ê F †

œ t$
2t$  t$ œ 2t$ Ê Flow œ '0 2t$ dt 1

" #

44. (a) F œ ™ axy# z$ b Ê F † œ 'a

(b)

dr dt

b

d dt afaratbbb

dr dt

œ

` f dx ` x dt




` f dy ` y dt

` z dz ` z dt




œ

df dt

, where f(xß yß z) œ xy# z$ Ê )C F †

dr dt

dt

dt œ farabbb  faraabb œ 0 since C is an entire ellipse.

Ð2ß1ß
1Ñ

'C F † ddtr œ 'Ð1ß1ß1Ñ

Ð#ß"ß
"Ñ

axy# z$ b dt œ cxy# z$ d Ð"ß"ß"Ñ œ (2)(1)# (1)$  (1)(1)# (1)$ œ 2  1 œ 3

d dt

45. Yes. The work and area have the same numerical value because work œ 'C F † dr œ 'C yi † dr œ 'b [f(t)i] †
i
a

df dt

j‘ dt

[On the path, y equals f(t)]

œ 'a f(t) dt œ Area under the curve b

46. r œ xi
yj œ xi
f(x)j Ê from the origin Ê F † Ê

'C

dr dx

œ

F † T ds œ 'C F †

dr dx

dr dx

[because f(t)  0]

œ i
f w (x)j ; F œ

kx È x#  y#




k†y†f (x) È x#  y#

dx œ 'a k b

w

d dx

k È x#  y#

œ

(xi
yj) has constant magnitude k and points away

kx  k†f(x)†f (x) Èx#  [f(x)]# w

œk

d dx

Èx#
[f(x)]# , by the chain rule

Èx#
[f(x)]# dx œ k
Èx#
[f(x)]# ‘ b a

œ k ˆÈb#
[f(b)]#  Èa#
[f(a)]# ‰ , as claimed. 47-52. Example CAS commands: Maple: with( LinearAlgebra );#47 F := r -> < r[1]*r[2]^6 | 3*r[1]*(r[1]*r[2]^5+2) >; r := t -> < 2*cos(t) | sin(t) >; a,b := 0,2*Pi; dr := map(diff,r(t),t); F(r(t)); q1 := simplify( F(r(t)) . dr ) assuming t::real; q2 := Int( q1, t=a..b ); value( q2 ); Mathematica: (functions and bounds will vary): Exercises 47 and 48 use vectors in 2 dimensions Clear[x, y, t, f, r, v] f[x_, y_]:= {x y6 , 3x (x y5
2)}

# (a) # (b) # (c)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1008 Chapter 16 Integration in Vector Fields {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= Sin[t] r[t_]:={x[t], y[t]} v[t_]:= r'[t] integrand= f[x[t], y[t]] . v[t] //Simplify Integrate[integrand,{t, a, b}] N[%] If the integration takes too long or cannot be done, use NIntegrate to integrate numerically. This is suggested for exercises 49 - 52 that use vectors in 3 dimensions. Be certain to leave spaces between variables to be multiplied. Clear[x, y, z, t, f, r, v] f[x_, y_, z_]:= {y
y z Cos[x y z], x2
x z Cos[x y z], z
x y Cos[x y z]} {a, b}={0, 21}; x[t_]:= 2 Cos[t] y[t_]:= 3 Sin[t] z[t_]:= 1 r[t_]:={x[t], y[t], z[t]} v[t_]:= r'[t] integrand= f[x[t], y[t],z[t]] . v[t] //Simplify NIntegrate[integrand,{t, a, b}] 16.3 PATH INDEPENDENCE, POTENTIAL FUNCTIONS, AND CONSERVATIVE FIELDS 1.

`P `y

œxœ

2.

`P `y

œ x cos z œ

3.

`P `y

œ 1 Á 1 œ

5.

`N `x

œ0Á1œ

6.

`P `y

œ0œ

7.

`f `x

œ 2x Ê f(xß yß z) œ x#
g(yß z) Ê

Ê 8.

`f `x

`f `z

`N `z

`N `z

,

,

`M `z `N `z

œyœ ,

`N `z

`M `y `M `z

`M `z

`P `x

,

`N `x

`M `y

œzœ

œ y cos z œ

`P `x

,

`f `x

`N `x

œ0œ

`P `x

,

`N `x

`M `y

œ ex sin y œ `f `y

œ

`f `z

`M `y

Ê Not Conservative

`g `y

œ 3y Ê g(yß z) œ


h(z) Ê

`f `z

œ 2xe

y2z

w

œx


`f `y œ y2z

3y# #


h(z)


2z#
C `g `y

œx
z Ê

œ z Ê g(yß z) œ zy
h(z)

w

xey2z


`f `y

`g `y

œ xey2z Ê

`g `y

œ 0 Ê f(xß yß z)

w

Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xey2z
C


h (z) œ 2xe

`f `z

`g `y


h(z) Ê f(xß yß z) œ x#


w

œ y sin z Ê f(xß yß z) œ xy sin z
g(yß z) Ê

Ê f(xß yß z) œ xy sin z
h(z) Ê

`f `y

3y #

#

3y# #

œ x
y
h (z) œ x
y Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

œ ey2z Ê f(xß yß z) œ xey2z
g(yß z) Ê

œ xy sin z
C

œ 1 Á 1 œ

Ê Conservative

œ hw (z) œ 4z Ê h(z) œ 2z#
C Ê f(xß yß z) œ x#


y2z

10.

Ê Conservative

4.

œ y
z Ê f(xß yß z) œ (y
z)x
g(yß z) Ê

œ xe

`M `y

Ê Not Conservative

œ (y
z)x
zy
C `f `x

œ sin z œ

Ê Not Conservative

Ê f(xß yß z) œ (y
z)x
zy
h(z) Ê

9.

`N `x

Ê Conservative

œ x sin z
w

`g `y

œ x sin z Ê

`g `y

œ 0 Ê g(yß z) œ h(z)

w

œ xy cos z
h (z) œ xy cos z Ê h (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1009 11.

`f `z

œ

Ê f(xß yß z) œ

z y #  z#

" #

œ (x ln x  x)
tan (x
y)
h(y) Ê f(xß yß z) œ `f `y

Ê

12.

œ

" #

`f `x

œ

œ

y y #  z# # #

" #

`g `x œ #

œ

ln x
sec# (x
y) Ê g(xß y)

ln ay#
z b
(x ln x  x)
tan (x
y)
h(y)


sec# (x
y)
hw (y) œ sec# (x
y)


y y #  z#

Ê hw (y) œ 0 Ê h(y) œ C Ê f(xß yß z)

ln ay
z b
(x ln x  x)
tan (x
y)
C y 1  x# y# `g `y

œ

z È 1
y # z#

Ê

`f `z

œ

y È 1
y # z#


hw (z) œ

Ê f(xß yß z) œ tan


"

(xy)
sin

`f `z

" z




y È 1
y # z#

`g `y




x 1  x# y#

œ

" z

Ê hw (z) œ

`P `y

`N `z

`M `P `N `M `z œ 0 œ `x , `x œ 0 œ `y `g `f # ` y œ ` y œ 2y Ê g(yß z) œ y


œ0œ

œ 2x Ê f(xß yß z) œ x#
g(yß z) Ê

`P `y

14. Let F(xß yß z) œ yzi
xzj
xyk Ê

œ yz Ê f(xß yß z) œ xyz
g(yß z) Ê `f `z

œ xyz
h(z) Ê Ð3ß5ß0Ñ

'Ð1ß1ß2Ñ

Ê

`N `z

œxœ

Ê M dx
N dy
P dz is

,

,

`f `y

`M `z

œyœ `g `y

œ xz


`P `x

,

`N `x

`M `y

œzœ `g `y

œ xz Ê

h(z) Ê f(xß yß z) œ x#
y# œ h(z)

'Ð0Ð2ß0ß3ß0ß
Ñ 6Ñ 2x dx
2y dy
2z dz

œ f(2ß 3ß 6)  f(!ß !ß !) œ 2#
3#
(6)# œ 49

exact;

z È 1
y # z#

Ê h(z) œ ln kzk
C

œ hw (z) œ 2z Ê h(z) œ z#
C Ê f(xß yß z) œ x#
y#
z#
C Ê

`f `x




x 1  x# y#

(yz)
ln kzk
C

13. Let F(xß yß z) œ 2xi
2yj
2zk Ê `f `x

œ

Ê g(yß z) œ sin
" (yz)
h(z) Ê f(xß yß z) œ tan
" (xy)
sin
" (yz)
h(z)
"

exact;

`f `y

Ê f(xß yß z) œ tan
" (xy)
g(yß z) Ê

Ê

Ê

`f `x

ln ay#
z# b
g(xß y) Ê

Ê M dx
N dy
P dz is

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z)

œ xy
hw (z) œ xy Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xyz
C

yz dx
xz dy
xy dz œ f(3ß 5ß 0)  f(1ß 1ß 2) œ 0  2 œ 2

15. Let F(xß yß z) œ 2xyi
ax#  z# b j  2yzk Ê Ê M dx
N dy
P dz is exact;

`f `x

`P `y

`N `z

œ 2z œ

,

`M `z

œ0œ

`P `x

œ 2xy Ê f(xß yß z) œ x# y
g(yß z) Ê

Ê g(yß z) œ yz#
h(z) Ê f(xß yß z) œ x# y  yz#
h(z) Ê

`f `z

,

`N `x

`f `y w

œ 2x œ

œ x#


`g `y

`M `y `g `y

œ x#  z# Ê

œ z#

œ 2yz
h (z) œ 2yz Ê hw (z) œ 0 Ê h(z) œ C

Ê f(xß yß z) œ x# y  yz#
C Ê 'Ð0ß0ß0Ñ 2xy dx
ax#  z# b dy  2yz dz œ f("ß #ß $)  f(!ß !ß !) œ 2  2(3)# œ 16 Ð1ß2ß3Ñ

16. Let F(xß yß z) œ 2xi  y# j  ˆ 1 4 z# ‰ k Ê Ê M dx
N dy
P dz is exact; Ê f(xß yß z) œ x#  œ x# 

y$ 3

œ ˆ9 

27 3

y$ 3


h(z) Ê

`f `x

`P `y

œ0œ

`N `z

`f `z

,

`N `x

œ0œ `f `y

œ

`M `y

`g `y

$

œ y# Ê g(yß z) œ  y3
h(z)

4 1
z#

dz œ f(3ß 3ß 1)  f(!ß !ß !)

 (!  !  0) œ 1

17. Let F(xß yß z) œ (sin y cos x)i
(cos y sin x)j
k Ê Ê M dx
N dy
P dz is exact; œ cos y sin x Ê

`P `x

œ0œ

œ hw (z) œ  1 4 z# Ê h(z) œ 4 tan
" z
C Ê f(xß yß z)

Ð3ß3ß1Ñ

4†

`M `z

œ 2x Ê f(xß yß z) œ x#
g(yß z) Ê

 4 tan
" z
C Ê 'Ð0ß0ß0Ñ 2x dx  y# dy  1‰ 4

,

`g `y

`f `x

`P `y

œ0œ

`N `z

,

`M `z

œ0œ

`P `x

,

`N `x

œ cos y cos x œ

œ sin y cos x Ê f(xß yß z) œ sin y sin x
g(yß z) Ê

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ sin y sin x
h(z) Ê

Ê f(xß yß z) œ sin y sin x
z
C Ê

' 100101

Ð ß ß Ñ

Ð ß ß Ñ

`f `z

`f `y

`M `y

œ cos y sin x


`g `y

œ hw (z) œ 1 Ê h(z) œ z
C

sin y cos x dx
cos y sin x dy
dz œ f(0ß 1ß 1)  f(1ß !ß !)

œ (0
1)  (0
0) œ 1 18. Let F(xß yß z) œ (2 cos y)i
Š "y  2x sin y‹ j
ˆ "z ‰ k Ê Ê M dx
N dy
P dz is exact; œ

" y

 2x sin y Ê

`g `y

œ

" y

`f `x

`P `y

œ0œ

`N `z

,

`M `z

œ0œ

`P `x

œ 2 cos y Ê f(xß yß z) œ 2x cos y
g(yß z) Ê

, `f `y

`N `x

œ 2 sin y œ

œ 2x sin y


Ê g(yß z) œ ln kyk
h(z) Ê f(xß yß z) œ 2x cos y
ln kyk
h(z) Ê

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

`f `z

`M `y

`g `y

œ hw (z) œ

" z

1010 Chapter 16 Integration in Vector Fields Ê h(z) œ ln kzk
C Ê f(xß yß z) œ 2x cos y
ln kyk
ln kzk
C

Ê 'Ð0ß2ß1Ñ

Ð1ß1Î2ß2Ñ

2 cos y dx
Š "y  2x sin y‹ dy
1 #

œ ˆ2 † 0
ln

" z

dz œ f ˆ1ß 1# ß 2‰  f(!ß #ß ")


ln 2‰  (0 † cos 2
ln 2
ln 1) œ ln #

19. Let F(xß yß z) œ 3x# i
Š zy ‹ j
(2z ln y)k Ê Ê M dx
N dy
P dz is exact;

`f `x

`P `y

œ

2z y

1 # `N `z

œ

`M `z

,

œ0œ

`P `x

`f `y

œ 3x# Ê f(xß yß z) œ x$
g(yß z) Ê

Ê f(xß yß z) œ x$
z# ln y
h(z) Ê œ x$
z# ln y
C Ê 'Ð1ß1ß1Ñ 3x# dx
Ð1ß2ß3Ñ

`N `x

,

œ0œ

œ

`g `y

œ

`M `y z# y

Ê g(yß z) œ z# ln y
h(z)

`f `z

œ 2z ln y
hw (z) œ 2z ln y Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z)

z# y

dy
2z ln y dz œ f(1ß 2ß 3)  f("ß "ß ")

œ (1
9 ln 2
C)  (1
0
C) œ 9 ln 2 #

`P `y

20. Let F(xß yß z) œ (2x ln y  yz)i
Š xy  xz‹ j  (xy)k Ê Ê M dx
N dy
P dz is exact; x# y

œ

`g `y

 xz Ê

`f `x

`N `z

œ x œ

,

`M `z

œ y œ

`P `x

,

`N `x

œ 2x ln y  yz Ê f(xß yß z) œ x# ln y  xyz
g(yß z) Ê `f `z

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ x# ln y  xyz
h(z) Ê

œ

2x y

`f `y

œ

zœ x# y

`M `y

 xz


`g `y

œ xy
hw (z) œ xy Ê hw (z) œ 0

Ê h(z) œ C Ê f(xß yß z) œ x# ln y  xyz
C Ê 'Ð1ß2ß1Ñ (2x ln y  yz) dx
Š xy  xz‹ dy  xy dz Ð2ß1ß1Ñ

#

œ f(2ß 1ß 1)  f("ß 2ß 1) œ (4 ln 1  2
C)  (ln 2  2
C) œ  ln 2 21. Let F(xß yß z) œ Š "y ‹ i
Š 1z 

x y# ‹ j

Ê M dx
N dy
P dz is exact; Ê

`g `y

œ

" z

Ê g(yß z) œ

Ê f(xß yß z) œ

x y




y z

 ˆ zy# ‰ k Ê

`f `x

" y

œ


C Ê 'Ð1ß1ß1Ñ

Ð2ß2ß2Ñ

" y

œ  z"# œ

Ê f(xß yß z) œ


h(z) Ê f(xß yß z) œ

y z

`P `y

x y

dx
Š 1z 




y z

x y# ‹

x y

`N `z

`M `z

,

`f `y  zy#


g(yß z) Ê `f `z


h(z) Ê dy 

y z#

`P `x

œ0œ

œ

,

`N `x

œ  y1# œ

œ  yx#


`g `y

œ

" z

`M `y



x y#


hw (z) œ  zy# Ê hw (z) œ 0 Ê h(z) œ C

dz œ f(2ß 2ß 2)  f("ß 1ß 1) œ ˆ 2#


2 #


C‰  ˆ 1"


œ0 22. Let F(xß yß z) œ Ê `f `x

`P `y

2xi  2yj  2zk x #  y #  z#

œ  4yz œ 3%

`N `z

,

`M `z

Šand let 3# œ x#
y#
z# Ê

œ

2x x#  y#  z# Ê f(xß yß z) œ Ê `` gy œ 0 Ê g(yß z) œ h(z) w œ x#  2z y#  z# Ê h (z) œ 0 Ê

Ð2ß2ß2Ñ

Ê 'Ð
1ß
1ß
1Ñ

`P `x

œ  4xz œ 3%

2x dx  2y dy  2z dz x #  y #  z#

,

`N `x

œ  4xy œ 3%

`3 `x

`M `y

œ

x 3

`3 `y

,

œ

y 3

`3 `z

,

œ 3z ‹

Ê M dx
N dy
P dz is exact; `f `y

ln ax#
y#
z# b
g(yß z) Ê

œ

2y x #  y #  z#

Ê f(xß yß z) œ ln ax#
y#
z# b
h(z) Ê

`g `y




`f `z

œ

œ

2y x #  y #  z#


hw (z)

2z x #  y #  z#

h(z) œ C Ê f(xß yß z) œ ln ax#
y#
z# b
C

œ f(2ß 2ß 2)  f("ß 1ß 1) œ ln 12  ln 3 œ ln 4

23. r œ (i
j
k)
t(i
2j  2k) œ (1
t)i
(1
2t)j
(1  2t)k, 0 Ÿ t Ÿ 1 Ê dx œ dt, dy œ 2 dt, dz œ 2 dt Ð2ß3ß
1Ñ

Ê 'Ð1ß1ß1Ñ y dx
x dy
4 dz œ '0 (2t
1) dt
(t
1)(2 dt)
4(2) dt œ '0 (4t  5) dt œ c2t#  5td ! œ 3 1

1

24. r œ t(3j
4k), 0 Ÿ t Ÿ 1 Ê dx œ 0, dy œ 3 dt, dz œ 4 dt Ê

' 000304

Ð ß ß Ñ

Ð ß ß Ñ

"

#

x# dx
yz dy
Š y# ‹ dz

œ '0 a12t# b (3 dt)
Š 9t# ‹ (4 dt) œ '0 54t# dt œ c18t# d ! œ 18 1

25.

`P `y

œ0œ

1

#

`N `z

,

`M `z

œ 2z œ

`P `x

,

`N `x

,

`M `z

"

œ0œ

`M `y

Ê M dx
N dy
P dz is exact Ê F is conservative

Ê path independence 26.

`P `y

œ  ˆÈ

yz x #  y #  z# ‰

$

œ

`N `z

œ  ˆÈ

xz $ x #  y #  z# ‰

œ

`P `x

,

`N `x

œ  ˆÈ

xy x #  y #  z# ‰

$

œ

`M `y

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" 1


C‰

Section 16.3 Path Independence, Potential Functions, and Conservative Fields 1011 Ê M dx
N dy
P dz is exact Ê F is conservative Ê path independence 27.

`P `y `f `x

œ0œ œ

2x y

`N `z

,

œ0œ

Ê f(xß y) œ

Ê f(xß y) œ 28.

`M `z

x# y



" y

`N `z

,

`M `z

#

x y

`P `x

`N `x

,

œ  2x y# œ

`P `x

`f `x

œ e ln y Ê f(xß yß z) œ e ln y
g(yß z) Ê

œ

ex y

" y#

Ê gw (y) œ

Ê g(y) œ  "y
C


1 y ‹

œ cos z œ x

`N `x

#

`P `y

,

1
x# y#

œ  yx#
gw (y) œ


C Ê F œ ™ Šx œ0œ

Ê F is conservative Ê there exists an f so that F œ ™ f;

#

`f `y


g(y) Ê

`M `y

œ

`M `y

x

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

ex y

œ




œ y sin z
h(z) Ê f(xß yß z) œ e ln y
y sin z
h(z) Ê x

`g ex `y œ y `f `z œ y x

`g `y


sin z Ê

œ sin z Ê g(yß z)

w

cos z
h (z) œ y cos z Ê hw (z) œ 0

Ê h(z) œ C Ê f(xß yß z) œ ex ln y
y sin z
C Ê F œ ™ ae ln y
y sin zb 29.

`P `y

œ0œ

`N `z

`f `x

œ x#
y Ê f(xß yß z) œ

,

`M `z

Ê f(xß yß z) œ œ

" 3

x$
xy


œ0œ

" $ 3 x
xy " $ z 3 y
ze

(a) work œ 'A F † B

dr dt

`P `x

, " 3 " 3

`N `x

`M `y

œ1œ

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

x$
xy
g(yß z) Ê


y$
h(z) Ê

œx


`g `y

œ y#
x Ê

`f `z

`g `y z

œ y# Ê g(yß z) œ

" 3

y$
h(z)

œ hw (z) œ zez Ê h(z) œ zez  e
C Ê f(xß yß z)  ez
C Ê F œ ™ ˆ 3" x$
xy
3" y$
zez  ez ‰

dt œ 'A F † dr œ
"3 x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ ˆ 3"
0
0
e  e‰  ˆ 3"
0
0  1‰ B

Ð"ß!ß"Ñ

œ1

(b) work œ 'A F † dr œ
"3 x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ 1 B

Ð"ß!ß"Ñ

(c) work œ 'A F † dr œ
"3 x$
xy
3" y$
zez  ez ‘ Ð"ß!ß!Ñ œ 1 B

Ð"ß!ß"Ñ

Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 0) to (1ß 0ß 1). B

30.

`P `y

œ xeyz
xyzeyz
cos y œ

that F œ ™ f;

`f `x

`N `z

,

`M `z

œ yeyz œ

`P `x

`N `x

,

œ zeyz œ

œ eyz Ê f(xß yß z) œ xeyz
g(yß z) Ê

`f `y

`M `y

Ê F is conservative Ê there exists an f so

œ xzeyz


Ê g(yß z) œ z sin y
h(z) Ê f(xß yß z) œ xe
z sin y
h(z) Ê yz

`g `y

`f `z

œ xzeyz
z cos y Ê

`g `y

œ z cos y

w

œ xye
sin y
h (z) œ xyeyz
sin y yz

Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ xeyz
z sin y
C Ê F œ ™ axeyz
z sin yb

(a) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

(b) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

(c) work œ 'A F † dr œ cxeyz
z sin yd Ð"ß!ß"Ñ B

Ð"ß1Î#ß!Ñ

œ (1
0)  (1
0) œ 0 œ0 œ0

Note: Since F is conservative, 'A F † dr is independent of the path from (1ß 0ß 1) to ˆ1ß 1# ß 0‰ . B

31. (a) F œ ™ ax$ y# b Ê F œ 3x# y# i
2x$ yj ; let C" be the path from (1ß 1) to (0ß 0) Ê x œ t  1 and y œ t
1, 0 Ÿ t Ÿ 1 Ê F œ 3(t  1)# (t
1)# i
2(t  1)$ (t
1)j œ 3(t  1)% i  2(t  1)% j and r" œ (t  1)i
(t
1)j Ê dr" œ dt i  dt j Ê

'C

"

F † dr" œ '0 c3(t  1)%
2(t  1)% d dt 1

œ '0 5(t  1)% dt œ c(t  1)& d ! œ 1; let C# be the path from (0ß 0) to (1ß 1) Ê x œ t and y œ t, 1

"

0 Ÿ t Ÿ 1 Ê F œ 3t% i
2t% j and r# œ ti
tj Ê dr# œ dt i
dt j Ê 'C F † dr# œ '0 a3t%
2t% b dt 1

1 œ '0 5t% dt œ 1

Ê 'C F † dr œ 'C F † dr"
'C "

#

#

F † dr# œ 2 Ð1ß1Ñ

(b) Since f(xß y) œ x$ y# is a potential function for F, 'Ð
1ß1Ñ F † dr œ f(1ß 1)  f(1ß 1) œ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1012 Chapter 16 Integration in Vector Fields 32.

`P `y `f `x

œ0œ

`N `z

`M `z

,

œ0œ

`P `x

,

`N `x

œ 2x sin y œ

œ 2x cos y Ê f(xß yß z) œ x# cos y
g(yß z) Ê

Ê f(xß yß z) œ x# cos y
h(z) Ê (a) (b) (c) (d)

`M `y

`f `z

Ê F is conservative Ê there exists an f so that F œ ™ f; `f `y

œ x# sin y


`g `y

œ x# sin y Ê

`g `y

œ 0 Ê g(yß z) œ h(z)

œ hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x# cos y
C Ê F œ ™ ax# cos yb

'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð!ß"Ñ Ð"ß!Ñ œ 0  1 œ 1 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð
"ß1Ñ œ 1  (1) œ 2 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð
"ß!Ñ œ 1  1 œ 0 'C 2x cos y dx  x# sin y dy œ cx# cos yd Ð"ß!Ñ Ð"ß!Ñ œ 1  1 œ 0

33. (a) If the differential form is exact, then all x, and

`N `x

œ

`M `y

`P `y

`N `z

œ

Ê 2ay œ cy for all y Ê 2a œ c,

`M `z

œ

`P `x

Ê 2cx œ 2cx for

Ê by œ 2ay for all y Ê b œ 2a and c œ 2a

(b) F œ ™ f Ê the differential form with a œ 1 in part (a) is exact Ê b œ 2 and c œ 2 34. F œ ™ f Ê g(xß yß z) œ 'Ð0ß0ß0Ñ F † dr œ 'Ð0ß0ß0Ñ ™ f † dr œ f(xß yß z)  f(0ß 0ß 0) Ê ÐxßyßzÑ

`g `z

œ

`f `z

ÐxßyßzÑ

`g `x

œ

`f `x

 0,

`g `y

œ

`f `y

 0, and

 0 Ê ™ g œ ™ f œ F, as claimed

35. The path will not matter; the work along any path will be the same because the field is conservative. 36. The field is not conservative, for otherwise the work would be the same along C" and C# . 37. Let the coordinates of points A and B be axA , yA , zA b and axB , yB , zB b, respectively. The force F œ ai
bj
ck is conservative because all the partial derivatives of M, N, and P are zero. Therefore, the potential function is fax, y, zb œ ax
by
cz
C, and the work done by the force in moving a particle along any path from A to B is faBb  faAb œ f axB , yB , zB b  faxA , yA , zA b œ aaxB
byB
czB
Cb  aaxA
byA
czA
Cb Ä œ aaxB  xA b
bayB  yA b
cazB  zA b œ F † BA 38. (a) Let GmM œ C Ê F œ C ’ `P `y

Ê


3yzC ax#  y#  z# b&Î#

`f `x

œ

œ

xC ax#  y#  z# b$Î# yC Ê `` gy œ ax#  y#  z# b$Î#

some f; œ

œ

`N `z

,

x ax#  y#  z# b$Î# `M `z

œ

i


y ax#  y#  z# b$Î#


3xzC ax#  y#  z# b&Î#

Ê f(xß yß z) œ 

œ

,

C ax#  y#  z# b"Î#

0 Ê g(yß z) œ h(z) Ê

Ê h(z) œ C" Ê f(xß yß z) œ 

`P `x

C ax#  y#  z# b"Î#

`f `z

œ

j


`N `x

œ

z ax#  y#  z# b$Î#
3xyC ax#  y#  z# b&Î#


g(yß z) Ê zC ax#  y#  z# b$Î#

`f `y

k“ `M `y

œ

œ

Ê F œ ™ f for

yC ax#  y#  z# b$Î#


hw (z) œ


C" . Let C" œ 0 Ê f(xß yß z) œ




zC ax#  y#  z# b$Î# GmM ax#  y#  z# b"Î#

`g `y

is a potential

function for F. (b) If s is the distance of (xß yß z) from the origin, then s œ Èx#
y#
z# . The work done by the gravitational field F is work œ 'P F † dr œ ’ Èx#GmM “  y #  z# P#

T#

"

T"

œ

GmM

s#



GmM

s"

œ GmM Š s"# 

"

s" ‹ ,

as claimed.

16.4 GREEN'S THEOREM IN THE PLANE 1. M œ y œ a sin t, N œ x œ a cos t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y

`M `x

œ 0,

`M `y

œ 1,

œ 0;

Equation (11):

)C M dy  N dx œ '021 [(a sin t)(a cos t)  (a cos t)(a sin t)] dt œ '021 0 dt œ 0;

' ' Š ``Mx
``Ny ‹ dx dy œ ' ' 0 dx dy œ 0, Flux R

`N `x

R

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 1, and

Section 16.4 Green's Theorem in the Plane 1013 Equation (12):

)C M dx
N dy œ '021 [(a sin t)(a sin t)  (a cos t)(a cos t)] dt œ '021 a# dt œ 21a# ; Èa c x

' ' Š ``Nx  ``My ‹ dx dy œ ' ' ca cc a

R

#

œ 2a

ˆ 1#




1‰ #

#

#

2 dy dx œ 'ca 4Èa#  x# dx œ 4 ’ x2 Èa#  x#
a

sin
" xa “

a

ca

#

œ 2a 1, Circulation

2. M œ y œ a sin t, N œ 0, dx œ a sin t dt, dy œ a cos t dt Ê Equation (11):

a# #

)C M dy  N dx œ '0

21

`M `x

`M `y

œ 0,

`N `x

œ 1,

œ 0, and

`N `y

œ 0;

a# sin t cos t dt œ a#
"2 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Flux #1

R

21 #1 Equation (12): )C M dx
N dy œ '0 aa# sin# tb dt œ a#
2t  sin4 2t ‘ ! œ 1a# ; ' ' Š ``Nx  ``My ‹ dx dy

œ ' ' 1 dx dy œ '0

21

R

'0

a

r dr d) œ '0  21

R

a# #

#

d) œ 1a , Circulation

3. M œ 2x œ 2a cos t, N œ 3y œ 3a sin t, dx œ a sin t dt, dy œ a cos t dt Ê `N `y

`M `x

œ 2,

`M `y

œ 0,

`N `x

œ 0, and

œ 3;

Equation (11):

)C M dy  N dx œ '021 [(2a cos t)(a cos t)
(3a sin t)(a sin t)] dt

œ '0 a2a# cos# t  3a# sin# tb dt œ 2a#
2t
21

sin 2t ‘ #1 4 !

 3a#
2t 

sin 2t ‘ #1 4 !

œ 21a#  31a# œ 1a# ;

' ' Š ``Mx
``Ny ‹ œ ' ' 1 dx dy œ ' ' r dr d) œ '  a## d) œ 1a# , Flux 0 0 0 21

R

a

21

R

Equation (12):

)C M dx
N dy œ '021 [(2a cos t)(a sin t)
(3a sin t)(a cos t)] dt

#1 œ '0 a2a# sin t cos t  3a# sin t cos tb dt œ 5a#
12 sin# t‘ ! œ 0; ' ' 0 dx dy œ 0, Circulation 21

R

4. M œ x# y œ a$ cos# t, N œ xy# œ a$ cos t sin# t, dx œ a sin t dt, dy œ a cos t dt Ê ``Mx œ 2xy, ``My œ x2 , ``Nx œ y# , and ``Ny œ 2xy; Equation (11):

)C M dy  N dx œ '021 aa% cos$ t sin t
a% cos t sin$ tb œ ’ a4

%

cos% t


a% 4

sin% t“

' ' Š ``Mx
``Ny ‹ dx dy œ ' ' (2xy
2xy) dx dy œ 0, Flux R

#1 !

œ 0;

R

)C M dx
N dy œ '021 aa% cos# t sin# t
a% cos# t sin# tb dt œ '021 a2a% cos# t sin# tb dt 21 41 %1 œ '0 "# a% sin# 2t dt œ a4 '0 sin# u du œ a4
u2  sin42u ‘ ! œ 1#a ; ' ' Š ``Nx  ``My ‹ dx dy œ ' ' ay#
x# b dx dy

Equation (12):

%

%

21 a 21 œ '0 '0 r# † r dr d) œ '0 a4

%

5. M œ x  y, N œ y  x Ê

`M `x

%

R

d) œ

1 a% #

, Circulation

œ 1,

`M `y

œ 1,

`N `x

œ 1,

`N `y

Circ œ ' ' [1  (1)] dx dy œ 0

R

œ 1 Ê Flux œ ' ' 2 dx dy œ '0

1

R

'01 2 dx dy œ 2;

R

6. M œ x#
4y, N œ x
y# Ê

`M `x

œ 2x,

`M `y

œ 4,

`N `x

œ 1,

`N `y

œ 2y Ê Flux œ ' ' (2x
2y) dx dy R

1 1 1 1 " " œ '0 '0 (2x
2y) dx dy œ '0 cx#
2xyd ! dy œ '0 (1
2y) dy œ cy
y# d ! œ 2; Circ œ ' ' (1  4) dx dy

œ '0

1

'0 3 dx dy œ 3

R

1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1014 Chapter 16 Integration in Vector Fields `M `x

7. M œ y#  x# , N œ x#
y# Ê

`M `y

œ 2x,

`N `x

œ 2y,

œ 2x,

œ 2y Ê Flux œ ' ' (2x
2y) dx dy

`N `y

R

3 x 3 $ œ '0 '0 (2x
2y) dy dx œ '0 a2x#
x# b dx œ
 "3 x$ ‘ ! œ 9; Circ œ ' ' (2x  2y) dx dy

R

3 x 3 œ '0 '0 (2x  2y) dy dx œ '0 x# dx œ 9

`M `x

8. M œ x
y, N œ  ax#
y# b Ê œ '0

1

`M `y

œ 1,

'0 (1  2y) dy dx œ '0 ax  x b dx œ x

1

œ 2x,

œ 2y Ê Flux œ ' ' (1  2y) dx dy

`N `y

; Circ œ ' ' (2x  1) dx dy œ '0

1

" 6

#

`N `x

œ 1,

R

œ '0 a2x#  xb dx œ  76

R

'0 (2x  1) dy dx x

1

`M `x

9. M œ x
ex sin y, N œ x
ex cos y Ê

Ècos 2)

1Î4

Ê Flux œ ' ' dx dy œ 'c1Î4 '0 R

œ 1
ex sin y, Î

`M `y

œ ex cos y,

1 4

1Î%

Ècos 2)

1Î4

R

R

y x

, N œ ln ax#
y# b Ê

Ê Flux œ ' ' Š x#
yy#
R

Circ œ ' ' Š x# 2x  y# 

x x#  y# ‹

R

`M `x

11. M œ xy, N œ y# Ê œ '0 Š 3x#  1

#

3x% # ‹

2y x#  y# ‹

dx œ

`M `x

dx dy œ '0

1

dx dy œ '0

1

`M `y

œ y,


y x#  y#

œ

'1

`M `y

œ

œ 0,

`M `y

œ 0,

Ê Flux œ ' ' (x sin y) dx dy œ '0

1Î2

R

œ

2x x#  y#

,

`N `y

œ

" #

2y x#  y#

1

`N `y

œ 2y Ê Flux œ ' ' (y
2y) dy dx œ '0

1

R

`M `x

`N `x

Î

1 4

#

1

12. M œ  sin y, N œ x cos y Ê

,

;

r dr d) œ '
1Î4 ˆ "# cos 2)‰ d) œ

'12 ˆ r sinr ) ‰ r dr d) œ '01 sin ) d) œ 2;

; Circ œ ' ' x dy dx œ '0

" 5

x x#  y#

" #

œ ex sin y

)‰ ˆ r cos r dr d) œ '0 cos ) d) œ 0 r#

2

`N `x

œ x,

,

`N `y

œ 1
ex cos y,

r dr d) œ '
1Î4 ˆ "# cos 2)‰ d) œ
4" sin 2)‘
1Î% œ

Circ œ ' ' a1
ex cos y  ex cos yb dx dy œ ' ' dx dy œ 'c1Î4 '0 10. M œ tan
"

`N `x

œ  cos y,

'0

1Î2

R

'xx 3y dy dx #

'xx x dy dx œ '01 ax#
x$ b dx œ  1"# #

`N `x

`N `y

œ cos y,

œ x sin y

(x sin y) dx dy œ '0 Š 18 sin y‹ dy œ  18 ; 1Î2

#

#

1Î# Circ œ ' ' [cos y  ( cos y)] dx dy œ '0 '0 2 cos y dx dy œ '0 1 cos y dy œ c1 sin yd ! œ 1 1Î2

1Î2

1Î2

R

13. M œ 3xy 

x 1  y#

`M `x

, N œ ex
tan
" y Ê

Ê Flux œ ' ' Š3y  R

" 1  y#




œ 3y  R

$

ex y

Ê

3 c x#

`M `y

`N `y

œ

" 1  y# 21

œ '0 a$ (1
cos ))$ (sin )) d) œ ’ a4 (1
cos ))% “ 14. M œ y
ex ln y, N œ

,

dx dy œ ' ' 3y dx dy œ '0

" 1  y# ‹

21

" 1  y#

œ1


ex y

,

`N `x

œ

ex y

#1 !

'0aÐ1
cos Ñ )

(3r sin )) r dr d)

œ 4a$  a4a$ b œ 0

Ê Circ œ ' ' ’ ey  Š1
x

R

ex y ‹“

œ 'c1 'x% b 1  dy dx œ  'c1 ca3  x b  ax
1bd dx œ 'c1 ax
x  2b dx œ  1

15. M œ 2xy$ , N œ 4x# y# Ê œ '0

1

'0

x$

2xy dy dx œ ' #

1

#

`M `y

œ 6xy# ,

`N `x

"!

2 33

1

2 0 3

x

dx œ

%

1

%

#

dx dy œ ' ' (1) dx dy R

44 15

œ 8xy# Ê work œ )C 2xy$ dx
4x# y# dy œ ' ' a8xy#  6xy# b dx dy R

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.4 Green's Theorem in the Plane 1015 `M `y

16. M œ 4x  2y, N œ 2x  4y Ê

œ 2,

`N `x

œ 2 Ê work œ )C (4x  2y) dx
(2x  4y) dy

œ ' ' [2  (2)] dx dy œ 4 ' ' dx dy œ 4(Area of the circle) œ 4(1 † 4) œ 161 R

R `M `y

17. M œ y# , N œ x# Ê œ '0

1

1cx

'0

œ 2y,

œ 2x Ê )C y# dx
x# dy œ ' ' (2x  2y) dy dx

`N `x

R

(2x  2y) dy dx œ '0 a3x
4x  1b dx œ cx
2x#  xd ! œ 1
2  1 œ 0 1

`M `y

18. M œ 3y, N œ 2x Ê

œ 3,

#

`N `x

œ 2 Ê )C 3y dx
2x dy œ ' ' (2  3) dx dy œ '0

`M `y

œ 6,

1

R

œ '0 sin x dx œ 2 1

19. M œ 6y
x, N œ y
2x Ê

"

$

`N `x

'0sin x 1 dy dx

œ 2 Ê )C (6y
x) dx
(y
2x) dy œ ' ' (2  6) dy dx R

œ 4(Area of the circle) œ 161 20. M œ 2x
y# , N œ 2xy
3y Ê

`M `y

œ 2y,

`N `x

œ 2y Ê

)C a2x
y# b dx
(2xy
3y) dy œ ' ' (2y  2y) dx dy œ 0 R

21. M œ x œ a cos t, N œ y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt Ê Area œ œ

'0

21

" #

aa# cos# t
a# sin# tb dt œ

" #

'0

21

'021 aab cos# t
ab sin# tb dt œ "# '021 ab dt œ 1ab

" #

)C

" #

)C x dy  y dx

x dy  y dx

a# dt œ 1a#

22. M œ x œ a cos t, N œ y œ b sin t Ê dx œ a sin t dt, dy œ b cos t dt Ê Area œ œ

" #

" #

23. M œ x œ a cos$ t, N œ y œ sin$ t Ê dx œ 3 cos# t sin t dt, dy œ 3 sin# t cos t dt Ê Area œ

'0

21

œ

" #

œ

3 16

a3 sin# t cos# tb acos# t
sin# tb dt œ


u2 

sin 2u ‘ %1 4 !

" #

'0

21

a3 sin# t cos# tb dt œ

3 8

œ

3 8

1

24. M œ x œ t# , N œ y œ

t$ 3

 t Ê dx œ 2t dt, dy œ at#  1b dt Ê Area œ

È

È

'0

21

" #

" #

'cÈ33 ’t# at#  1b  Š t3  t‹ (2t)“ dt œ "# 'cÈ33 ˆ 3" t%
t# ‰ dt œ 12
151 t&


œ

8 5

È3

25. (a) M œ f(x), N œ g(y) Ê

`M `y

œ 0,

`N `x

R

(b) M œ ky, N œ hx Ê

`M `y

œ k,

`N `x

È$

 31 t$ ‘
È$ œ

œ 0 Ê )C f(x) dx
g(y) dy œ ' ' Š ``Nx  R

œ ' ' 0 dx dy œ 0

œ h Ê )C ky dx
hx dy œ ' ' Š ``Nx 

œ ' ' (h  k) dx dy œ (h  k)(Area of the region)

3 16

'0

sin# u du

)C x dy  y dx

œ

$

sin# 2t dt œ

)C x dy  y dx

41

R

`M `y ‹

`M `y ‹

" 15

Š9È3
15È3‹

dx dy

dx dy

R

26. M œ xy# , N œ x# y
2x Ê

`M `y

œ 2xy,

`N `x

œ 2xy
2 Ê )C xy# dx
ax# y
2xb dy œ ' ' Š ``Nx 

œ ' ' (2xy
2  2xy) dx dy œ 2 ' ' dx dy œ 2 times the area of the square R

R

R

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

`M `y ‹

dx dy

1016 Chapter 16 Integration in Vector Fields 27. The integral is 0 for any simple closed plane curve C. The reasoning: By the tangential form of Green's Theorem, with M œ 4x$ y and N œ x% , )C 4x$ y dx
x% dy œ ' ' ’ ``x ax% b 

` `y

R

œ ' ' ðóóñóóò a4x$  4x$ b dx dy œ 0.

a4x$ yb“ dx dy

R

0 28. The integral is 0 for any simple closed curve C. The reasoning: By the normal form of Green's theorem, with ` ` $ $ M œ x$ and N œ y$ , )C y$ dy
x$ dx œ ' ' ”ðñò ` x ay b  ï ` y ax b • dx dy œ 0.

R

0 `M `x

29. Let M œ x and N œ 0 Ê

œ 1 and

`N `y

œ0 Ê

0

)C M dy  N dx œ ' ' Š ``Mx
``Ny ‹ dx dy

œ ' ' (1
0) dx dy Ê Area of R œ ' ' dx dy œ )C x dy; similarly, M œ y and N œ 0 Ê R

`N `x

R

œ 0 Ê )C M dx
N dy œ ' ' Š ``Nx
R

œ ' ' dx dy œ Area of R

Ê )C x dy

R

`M `y ‹

`M `y

œ 1 and

dy dx Ê )C y dx œ ' ' (0  1) dy dx Ê  )C y dx R

R

30.

'ab f(x) dx œ Area of R œ )C y dx, from Exercise 29

31. Let $ (xß y) œ 1 Ê x œ

My M

' ' x $ (xßy) dA

œ 'R '

$ (xßy) dA

' ' x dA

œ 'R '

R

dA

' ' x dA

œ

Ê Ax œ ' ' x dA œ ' ' (x
0) dx dy

R

A

R

R

R

œ )C x#

dy, Ax œ ' ' x dA œ ' ' (0
x) dx dy œ  ) xy dx, and Ax œ ' ' x dA œ ' ' ˆ 23 x
3" x‰ dx dy

œ)

#

#

" C 3

R

R

" 3

" #

x dy  xy dx Ê

C

)C x

dy œ )C xy dx œ

#

" 3

)C x

dy  xy dx œ Ax

32. If $ (xß y) œ 1, then Iy œ ' ' x# $ (xß y) dA œ ' ' x# dA œ ' ' ax#
0b dy dx œ R

R

R

R

#

R

" 3

)C

x$ dy,

' ' x# dA œ ' ' a0
x# b dy dx œ  ) x# y dx, and ' ' x# dA œ ' ' ˆ 34 x#
"4 x# ‰ dy dx C R

R

œ)

" C 4

33. M œ

`f `y

" 4

$

#

x dy  x y dx œ , N œ  `` xf Ê

`M `y

" 4

œ

)C x ` #f ` y#

,

R

$

#

dy  x y dx Ê

`N `x

" 3

œ  `` xf# Ê )C #

)C x

`f `y

R

$

dy œ  )C x# y dx œ

dx 

`f `x

" 4

)C

dy œ ' ' Š `` xf#  #

R

x$ dy  x# y dx œ Iy

` #f ` y# ‹

dx dy œ 0 for such

curves C 34. M œ

" 4

x# y
3" y$ , N œ x Ê

the ellipse

" 4

`M `y

œ

1 4

x#
y# ,

`N `x

œ 1 Ê Curl œ

`N `x



`M `y

œ 1  ˆ "4 x#
y# ‰  0 in the interior of

x#
y# œ 1 Ê work œ 'C F † dr œ ' ' ˆ1  4" x#  y# ‰ dx dy will be maximized on the region R

R œ {(xß y) | curl F}   0 or over the region enclosed by 1 œ 2y 35. (a) ™ f œ Š x# 2x  y# ‹ i
Š x#  y# ‹ j Ê M œ

2x x#  y#

,Nœ

" 4

x#
y#

2y x#  y#

; since M, N are discontinuous at (0ß 0), we

compute 'C ™ f † n ds directly since Green's Theorem does not apply. Let x œ a cos t, y œ a sin t Ê dx œ a sin t dt, dy œ a cos t dt, M œ

2 a

cos t, N œ

2 a

sin t, 0 Ÿ t Ÿ 21, so 'C ™ f † n ds œ 'C M dy  N dx

œ '0
ˆ 2a cos t‰aa cos tb  ˆ 2a sin t‰aa sin tb ‘dt œ '0 2acos2 t
sin2 tbdt œ 41. Note that this holds for any 21

21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.4 Green's Theorem in the Plane 1017 a  0, so 'C ™ f † n ds œ 41 for any circle C centered at a0, 0b traversed counterclockwise and 'C ™ f † n ds œ 41 if C is traversed clockwise.

(b) If K does not enclose the point (0ß 0) we may apply Green's Theorem: 'C ™ f † n ds œ 'C M dy  N dx œ ' ' Š ``Mx
R

`N `y ‹

dx dy œ ' ' Š ax2  y2 b2
2 ˆy 2
x 2 ‰

R

2 ˆx 2
y 2 ‰ ‹ ax 2  y 2 b 2

dx dy œ ' ' 0 dx dy œ 0. If K does enclose the point R

(0ß 0) we proceed as in Example 6: Choose a small enough so that the circle C centered at (0ß 0) of radius a lies entirely within K. Green's Theorem applies to the region R that lies between K and C. Thus, as before, 0 œ ' ' Š ``Mx
R

`N `y ‹

dx dy

œ 'K M dy  N dx
'C M dy  N dx where K is traversed counterclockwise and C is traversed clockwise.

Hence by part (a) 0 œ ’ ' M dy  N dx “  41 Ê 41 œ K

'K ™ f † n ds œ œ 0

'K M dy  N dx

œ 'K ™ f † n ds. We have shown:

if (0ß 0) lies inside K if (0ß 0) lies outside K

41

36. Assume a particle has a closed trajectory in R and let C" be the path Ê C" encloses a simply connected region R" Ê C" is a simple closed curve. Then the flux over R" is )C F † n ds œ 0, since the velocity vectors F are "

tangent to C" . But 0 œ )C F † n ds œ )C M dy  N dx œ ' ' Š ``Mx
"

"

R"

`N `y ‹

dx dy Ê Mx
Ny œ 0, which is a

contradiction. Therefore, C" cannot be a closed trajectory. 37.

'gg yy

#Ð Ñ

"Ð Ñ

`N `x

'cd 'gg yy ˆ ``Nx dx‰ dy œ 'cd [N(g# (y)ß y)  N(g" (y)ß y)] dy #Ð Ñ

dx dy œ N(g# (y)ß y)  N(g" (y)ß y) Ê

"Ð Ñ

œ 'c N(g# (y)ß y) dy  'c N(g" (y)ß y) dy œ 'c N(g# (y)ß y) dy
'd N(g" (y)ß y) dy œ 'C N dy
'C N dy d

38.

d

c

#

"

œ )C dy

Ê

'ab 'cd

dy dx œ 'a [M(xß d)  M(xß c)] dx œ 'a M(xß d) dx
'a M(xß c) dx œ 'C M dx  'C M dx.

`M `y

)C N dy œ ' '

d

R

`N `x

dx dy

b

b

b

3

Because x is constant along C# and C% , 'C M dx œ 'C M dx œ 0 Ê

 Š'C

"

M dx
'C

#

M dx
'C

$

#

M dx
'C

%

%

M dx‹ œ  )C M dx Ê 'a

b

'cd

`M `y

"

dy dx œ )C M dx.

39. The curl of a conservative two-dimensional field is zero. The reasoning: A two-dimensional field F œ Mi
Nj can be considered to be the restriction to the xy-plane of a three-dimensional field whose k component is zero, and whose i and j components are independent of z. For such a field to be conservative, we must have `N `M `N `M ` x œ ` y by the component test in Section 16.3 Ê curl F œ ` x  ` y œ 0. 40. Green's theorem tells us that the circulation of a conservative two-dimensional field around any simple closed curve in the xy-plane is zero. The reasoning: For a conservative field F œ Mi
Nj , we have ``Nx œ ``My (component test for conservative fields, Section 16.3, Eq. (2)), so curl F œ

`N `x



`M `y

œ 0. By Green's theorem,

the counterclockwise circulation around a simple closed plane curve C must equal the integral of curl F over the region R enclosed by C. Since curl F œ 0, the latter integral is zero and, therefore, so is the circulation.

The circulation )C F † T ds is the same as the work )C F † dr done by F around C, so our observation that circulation of a conservative two-dimensional field is zero agrees with the fact that the work done by a conservative field around a closed curve is always 0.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1018 Chapter 16 Integration in Vector Fields 41-44. Example CAS commands: Maple: with( plots );#41 M := (x,y) -> 2*x-y; N := (x,y) -> x+3*y; C := x^2 + 4*y^2 = 4; implicitplot( C, x=-2..2, y=-2..2, scaling=constrained, title="#41(a) (Section 16.4)" ); curlF_k := D[1](N) - D[2](M): # (b) 'curlF_k' = curlF_k(x,y); top,bot := solve( C, y ); # (c) left,right := -2, 2; q1 := Int( Int( curlF_k(x,y), y=bot..top ), x=left..right ); value( q1 ); Mathematica: (functions and bounds will vary) The ImplicitPlot command will be useful for 41 and 42, but is not needed for 43 and 44. In 44, the equation of the line from (0, 4) to (2, 0) must be determined first. Clear[x, y, f] <
k™f†pk

R

œ ' ' È4r# cos# )
4r# sin# )
1 r dr d) œ '0

21

R

œ '0

21

13 6

d) œ

13 3

È2

'0

È4r#
1 r dr d) œ ' ’ " a4r#
1b$Î# “ 12 0 21

È# !

d)

1

2. p œ k , ™ f œ 2xi
2yj  k Ê k ™ f k œ È4x#
4y#
1 and k ™ f † pk œ 1; 2 Ÿ x#
y# Ÿ 6 Ê Sœ'' R

œ'

21

" ’ 12 0

k™f k k™f†pk

dA œ ' ' È4x#
4y#
1 dx dy œ ' ' È4r#
1 r dr d) œ '0

a4r#
1b

21

R

$Î#

È'

R

“ È d) œ '0 #

21

49 6

d) œ

49 3

È

'È26 È4r#
1 r dr d)

1

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.5 Surface Area and Surface Integrals 1019 3. p œ k , ™ f œ i
2j
2k Ê k ™ f k œ 3 and k ™ f † pk œ 2; x œ y# and x œ 2  y# intersect at (1ß 1) and (1ß 1) Ê Sœ'' R

k™f k k™f†pk

dA œ ' ' R

2 c y#

dx dy œ 'c1 'y# 1

3 #

dx dy œ 'c1 a3  3y# b dy œ 4 1

3 #

4. p œ k , ™ f œ 2xi  2k Ê k ™ f k œ È4x#
4 œ 2Èx#
1 and k ™ f † pk œ 2 Ê S œ ' ' œ'' R

2È x#  1 #

È3

dx dy œ '0

È3

'0x Èx#
1 dy dx œ '0

k™f k k™f†pk

R $Î# xÈx#
1 dx œ ’ "3 ax#
1b “

È$ !

œ

" 3

dA

(4)$Î# 

" 3

œ

7 3

5. p œ k , ™ f œ 2xi  2j  2k Ê k ™ f k œ È(2x)#
(2)#
(2)# œ È4x#
8 œ 2Èx#
2 and k ™ f † pk œ 2 Ê Sœ''

dA œ ' '

k™f k k™f†pk

R

R

2 È x # 2 #

dx dy œ '0

2

'03x Èx#
2 dy dx œ '02 3xÈx#
2 dx œ ’ax#
2b$Î# “ # !

œ 6È6  2È2 6. p œ k , ™ f œ 2xi
2yj
2zk Ê k ™ f k œ È4x#
4y#
4z# œ È8 œ 2È2 and k ™ f † pk œ 2z; x#
y#
z# œ 2 and z œ Èx#
y# Ê x#
y# œ 1; thus, S œ ' ' œ È2 ' '

" È 2
ax #  y # b

R

dA œ È2 '0

21

R

'01 Èr dr d)

dA œ ' '

k™f k k™f†pk

2
r#

2È 2 #z

R

dA œ È2 ' '

Èc#
1 r dr d) œ '

21

21

0

R È c#  1 #

dA

œ È2 '0 Š1
È2‹ d) œ 21 Š2  È2‹

7. p œ k , ™ f œ ci  k Ê k ™ f k œ Èc#
1 and k ™ f † pk œ 1 Ê S œ ' ' 21 1 œ '0 '0

" z

R

k™f k k™f†pk

dA œ ' ' Èc#
1 dx dy R

d) œ 1Èc#
1

8. p œ k , ™ f œ 2xi
2zj Ê k ™ f k œ È(2x)#
(2z)# œ 2 and k ™ f † pk œ 2z for the upper surface, z   0 Ê Sœ''

dA œ ' '

k™f k k™f†pk

R "Î#

œ csin
" xd
"Î# œ

R

1 6

2 #z

 ˆ 16 ‰ œ

dA œ ' ' R

" È 1
x#

1Î2

1Î2

dy dx œ 2'c1Î2 '0

1 È 1
x#

1Î2

dy dx œ '
1Î2

" È 1
x#

dx

1 3

9. p œ i , ™ f œ i
2yj
2zk Ê k ™ f k œ È1#
(2y)#
(2z)# œ È1
4y#
4z# and k ™ f † pk œ 1; 1 Ÿ y#
z# Ÿ 4 Ê Sœ'' œ '0

21

R

k™f k k™f†pk

dA œ ' ' È1
4y#
4z# dy dz œ '0

21

R

'12 È1
4r# cos# )
4r# sin# ) r dr d)

'1 È1
4r# r dr d) œ '021 ’ 12" a1
4r# b$Î# “ # d) œ '021 1"# Š17È17  5È5‹ d) œ 16 Š17È17  5È5‹ 2

"

10. p œ j , ™ f œ 2xi
j
2zk Ê k ™ f k œ È4x#
4z#
1 and k ™ f † pk œ 1; y œ 0 and x#
y
z# œ 2 Ê x#
z# œ 2; thus, S œ ' ' R

k™f k k™f†pk

dA œ ' ' È4x#
4z#
1 dx dz œ '0

21

R

È2

'0

È4r#
1 r dr d) œ '

21

0

13 6

d) œ

13 3

1

#

# 11. p œ k , ™ f œ ˆ2x  2x ‰ i
È15 j  k Ê k ™ f k œ ʈ2x  2x ‰
ŠÈ15‹
(1)# œ É4x#
8


œ 2x
2x , on 1 Ÿ x Ÿ 2 and k ™ f † pk œ 1 Ê S œ ' ' R

k ™f k k™f†pk

4 x#

# œ Ɉ2x
2x ‰

dA œ ' ' a2x
2x
" b dx dy R

1 2 1 1 # œ '0 '1 a2x
2x
" b dx dy œ '0 cx#
2 ln xd " dy œ '0 (3
2 ln 2) dy œ 3
2 ln 2

12. p œ k , ™ f œ 3Èx i
3Èy j  3k Ê k ™ f k œ È9x
9y
9 œ 3Èx
y
1 and k ™ f † pk œ 3 Ê Sœ'' R

k™f k k™f†pk

dA œ ' ' Èx
y
1 dx dy œ '0

1

R

'01 Èx
y
1 dx dy œ '01
23 (x
y
1)$Î# ‘ "! dy

1 " 4 4 4
œ '0
23 (y
2)$Î#  32 (y
1)$Î# ‘ dy œ
15 (y
2)&Î#  15 (y
1)&Î# ‘ ! œ 15 (3)&Î#  (2)&Î#  (2)&Î#
1‘

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1020 Chapter 16 Integration in Vector Fields œ

Š9È3  8È2
1‹

4 15

13. The bottom face S of the cube is in the xy-plane Ê z œ 0 Ê g(xß yß 0) œ x
y and f(xß yß z) œ z œ 0 Ê p œ k

' ' g d5 œ ' ' (x
y) dx dy

and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê a a a œ '0 '0 (x
y) dx dy œ '0 Š a#

#

R

S $


ay‹ dy œ a . Because of symmetry, we also get a$ over the face of the cube

in the xz-plane and a$ over the face of the cube in the yz-plane. Next, on the top of the cube, g(xß yß z) œ g(xß yß a) œ x
y
a and f(xß yß z) œ z œ a Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy

' ' g d5 œ ' ' (x
y
a) dx dy œ ' ' (x
y
a) dx dy œ ' ' (x
y) dx dy
' ' a dx dy œ 2a$ . 0 0 0 0 0 0 a

a

a

a

a

a

R

S

Because of symmetry, the integral is also 2a$ over each of the other two faces. Therefore,

' ' (x
y
z) d5 œ 3 aa$
2a$ b œ 9a$ . cube

14. On the face S in the xz-plane, we have y œ 0 Ê f(xß yß z) œ y œ 0 and g(xß yß z) œ g(xß 0ß z) œ z Ê p œ j and ™ f œ j Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dz Ê

' ' g d5 œ ' ' (y
z) d5 œ ' ' z dx dz œ ' 2z dz 0 0 0 1

S

2

1

S

œ 1. On the face in the xy-plane, we have z œ 0 Ê f(xß yß z) œ z œ 0 and g(xß yß z) œ g(xß yß 0) œ y Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dy Ê ' ' g d5 œ ' ' y d5 œ '0

1

S

S

'02 y dx dy œ 1.

On the triangular face in the plane x œ 2 we have f(xß yß z) œ x œ 2 and g(xß yß z) œ g(2ß yß z) œ y
z Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê

1

S

1 œ '0 "# a1  y# b dy œ 3" .

1cy

' ' g d5 œ ' ' (y
z) d5 œ ' ' (y
z) dz dy 0 0 S

On the triangular face in the yz-plane, we have x œ 0 Ê f(xß yß z) œ x œ 0 and g(xß yß z) œ g(0ß yß z) œ y
z Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê œ '0

1

1cy

'0

' ' g d5 œ ' ' (y
z) d5 S

(y
z) dz dy œ

" 3

S

.

Finally, on the sloped face, we have y
z œ 1 Ê f(xß yß z) œ y
z œ 1 and g(xß yß z) œ y
z œ 1 Ê p œ k and ™ f œ j
k Ê k ™ f k œ È2 and k ™ f † pk œ 1 Ê d5 œ È2 dx dy Ê ' ' g d5 œ ' ' (y
z) d5 1 2 œ '0 '0 È2 dx dy œ 2È2.

S

S

Therefore, ' ' g(xß yß z) d5 œ 1
1
"3
3"
2È2 œ 38
2È2 wedge

15. On the faces in the coordinate planes, g(xß yß z) œ 0 Ê the integral over these faces is 0. On the face x œ a, we have f(xß yß z) œ x œ a and g(xß yß z) œ g(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dy dz Ê

' ' g d5 œ ' ' ayz d5 œ ' ' ayz dy dz œ 0 0 c

S

b

S

ab# c# 4

.

On the face y œ b, we have f(xß yß z) œ y œ b and g(xß yß z) œ g(xß bß z) œ bxz Ê p œ j and ™ f œ j Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dx dz Ê

' ' g d5 œ ' ' bxz d5 œ ' ' bxz dx dz œ 0 0 c

S

a

S

a# bc# 4

.

On the face z œ c, we have f(xß yß z) œ z œ c and g(xß yß z) œ g(xß yß c) œ cxy Ê p œ k and ™ f œ k Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dy dx Ê

' ' g(xß yß z) d5 œ S

' ' g d5 œ ' ' cxy d5 œ ' ' cxy dx dy œ 0 0 b

S

abc(ab  ac  bc) 4

S

a

a# b# c 4

. Therefore,

.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.5 Surface Area and Surface Integrals 1021 16. On the face x œ a, we have f(xß yß z) œ x œ a and g(xß yß z) œ g(aß yß z) œ ayz Ê p œ i and ™ f œ i Ê k ™ f k œ 1 and k ™ f † pk œ 1 Ê d5 œ dz dy Ê

' ' g d5 œ ' ' ayz d5 œ ' ' ayz dz dy œ 0. Because of the symmetry cb cc b

S

c

S

of g on all the other faces, all the integrals are 0, and ' ' g(xß yß z) d5 œ 0. S

17. f(xß yß z) œ 2x
2y
z œ 2 Ê ™ f œ 2i
2j
k and g(xß yß z) œ x
y
(2  2x  2y) œ 2  x  y Ê p œ k ,

k ™ f k œ 3 and k ™ f † pk œ 1 Ê d5 œ 3 dy dx; z œ 0 Ê 2x
2y œ 2 Ê y œ 1  x Ê ' ' g d5 œ ' ' (2  x  y) d5 S

1 1cx 1 1 œ 3 '0 '0 (2  x  y) dy dx œ 3 '0
(2  x)(1  x)  "# (1  x)# ‘ dx œ 3 '0 Š 3#  2x
x# ‹ dx œ 2

S

#

18. f(xß yß z) œ y#
4z œ 16 Ê ™ f œ 2yj
4k Ê k ™ f k œ È4y#
16 œ 2Èy#
4 and p œ k Ê k ™ f † pk œ 4 2È y#  4 4

Ê d5 œ œ 'c4 4

" 4

dx dy Ê ' ' g d5 œ 'c4 '0 ˆxÈy#
4‰ Š 4

S

ay#
4b dy œ

%

$

" #

’ y3
4y“ œ !

" #

1

ˆ 64 ‰ 3
16 œ

È y#  4 ‹ #

dx dy œ 'c4 '0 4

1

x ay #  4 b #

dx dy

56 3

19. g(xß yß z) œ z, p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † k) dA œ '0

2

R

S

'0

3

3 dy dx œ 18

20. g(xß yß z) œ y, p œ j Ê ™ g œ j Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê Flux œ ' ' F † n d5 œ ' ' (F † j) dA 7

2

21. ™ g œ 2xi
2yj
2zk Ê k ™ gk œ È4x#
4y#
4z# œ 2a; n œ k ™ g † kk œ 2z Ê d5 œ œ '0

1Î2

2a 2z

'0 Èa#  r# r dr d) œ a

œ ' ' 1 dA œ R

R

R

2a 2z

Ê F†nœ

R

S

and d5 œ

x i  y j  zk a

and d5 œ

2xi  2yj  2zk #Èx#  y#  z#

œ

x i  y j  zk a

Ê F†nœ

S

a z

dA Ê F † n œ

a z

dA Ê F † n œ

xy a



œ ' ' a ˆ az ‰ dA œ ' ' R

xy a




a z

dA œ ' '

œ '0 a# ’Èa#  r# “ d) œ !

and d5 œ

#

R

a z #

dA Ê F † n œ

a È a #
ax #  y # b

dA œ '0

1Î2

x# a

zy# a




R

x i  y j  zk a

a

zx# a

dx dy œ ' ' a dx dy œ a (Area of R) œ #

#

z# a

;

R

dA Ê Flux œ ' ' F † n d5 œ ' ' 0 d5 œ 0

x i  y j  zk a

(za) ˆ az ‰

25. From Exercise 21, n œ

1Î2

x i  y j  zk a

z a

œ

z a

Ê Flux œ ' ' ˆ za ‰ ˆ az ‰ dA R

1a# 4

24. From Exercise 21, n œ Ê Flux œ ' '

œ

1a$ 6

œ 0; k ™ g † kk œ 2z Ê d5 œ 23. From Exercise 21, n œ

2xi  2yj  2zk #Èx#  y#  z#

# dA Ê Flux œ ' ' Š za ‹ ˆ az ‰ dA œ ' ' z dA œ ' ' Èa#  ax#
y# b dx dy

22. ™ g œ 2xi
2yj
2zk Ê k ™ gk œ È4x#
4y#
4z# œ 2a; n œ

R

R

S

œ 'c1 '2 2 dz dx œ 'c1 2(7  2) dx œ 10(2
1) œ 30 2




" 4

1a

y# a




'0 È a a




#

a#
r#

z$ a

œ z Šx

#

 y #  z# ‹ a

œ az

%

z# a

œ a Ê Flux

r dr d)

1a$ #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley


xy a




xy a

1022 Chapter 16 Integration in Vector Fields #

26. From Exercise 21, n œ Ê Flux œ ' ' R

a z

x i  y j  zk a

dx dy œ ' '

and d5 œ

a z

a È a #
ax #  y # b

R

dA Ê F † n œ dx dy œ '0

1Î2

'0

a

Š xa ‹  Œ

y# z# a
Š a ‹

È x #  y #  z# a Èa#
r#

2xy
3z È4y#  1

Š aa ‹ a

œ1

1 a# #

r dr d) œ

27. g(xß yß z) œ y#
z œ 4 Ê ™ g œ 2yj
k Ê k ™ gk œ È4y#
1 Ê n œ Ê F†nœ

#

œ

2yj  k È4y#  1

; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4y#
1 dA Ê Flux

2xy
3z œ ' ' ŠÈ ‹ È4y#
1 dA œ ' ' (2xy  3z) dA; z œ 0 and z œ 4  y# Ê y# œ 4 4y#  1

R

R

Ê Flux œ ' ' c2xy  3 a4  y bd dA œ '0 'c2 a2xy  12
3y# b dy dx œ '0 cxy#  12y
y$ d
# dx 1

#

2

1

#

R

œ '0 32 dx œ 32 1

28. g(xß yß z) œ x#
y#  z œ 0 Ê ™ g œ 2xi
2yj  k Ê k ™ gk œ È4x#
4y#
1 œ È4 ax#
y# b
1 Ê nœ

2xi  2yj
k È 4 ax #  y # b  1

Ê F†nœ

8x#  8y#
2 È 4 ax #  y # b  1

; p œ k Ê k ™ g † pk œ 1 Ê d5 œ È4 ax#
y# b
1 dA

 8y
2 Ê Flux œ ' ' Š È8x ‹ È4 ax#
y# b
1 dA œ ' ' a8x#
8y#  2b dA; z œ 1 and x#
y# œ z 4 ax #  y # b  1 #

#

R

Ê x
y œ 1 Ê Flux œ '0 #

21

#

'0 a8r 1

R

#

 2b r dr d) œ 21

29. g(xß yß z) œ y  ex œ 0 Ê ™ g œ ex i
j Ê k ™ gk œ Èe2x
1 Ê n œ Ê k ™ g † pk œ ex Ê d5 œ œ ' ' 4 dA œ '0

1

R

Èe2x 1 ex

dA Ê Flux œ ' ' Š
È2e2x
2y ‹ Š x

R

'12 4 dy dz œ 4

e 1

Èe2x  1 ‹ ex

30. g(xß yß z) œ y  ln x œ 0 Ê ™ g œ  "x i
j Ê k ™ gk œ É x"#
1 œ Ê nœ

Š
" i  j‹ Œ

x È 1
x# œ x





i  x j È 1  x#

Ê Flux œ ' ' Š È2xy # ‹ Š 1x R

Ê F†nœ È 1  x# ‹ x

2xy È 1  x#

dA œ '0

1

ex i
j Èe2x  1

È 1  x# x

Ê F†nœ

dA œ ' ' R


2ex
2ex ex


2ex
2y Èe2x  1

;pœi

dA

since 1 Ÿ x Ÿ e

; p œ j Ê k ™ g † p k œ 1 Ê d5 œ

È 1  x# x

dA

'1e 2y dx dz œ '1e '01 2 ln x dz dx œ '1e 2 ln x dx

œ 2 cx ln x  xd e" œ 2(e  e)  2(0  1) œ 2 31. On the face z œ a: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 2ax since z œ a; d5 œ dx dy Ê Flux œ ' ' 2ax dx dy œ '0 '0 2ax dx dy œ a% . a

a

R

On the face z œ 0: g(xß yß z) œ z Ê ™ g œ k Ê k ™ gk œ 1; n œ k Ê F † n œ 2xz œ 0 since z œ 0;

d5 œ dx dy Ê Flux œ ' ' 0 dx dy œ 0. R

On the face x œ a: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 2ay since x œ a;

d5 œ dy dz Ê Flux œ '0 '0 2ay dy dz œ a% . a

a

On the face x œ 0: g(xß yß z) œ x Ê ™ g œ i Ê k ™ gk œ 1; n œ i Ê F † n œ 2xy œ 0 since x œ 0 Ê Flux œ 0. On the face y œ a: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 2az since y œ a;

d5 œ dz dx Ê Flux œ '0 '0 2az dz dx œ a% . a

a

On the face y œ 0: g(xß yß z) œ y Ê ™ g œ j Ê k ™ gk œ 1; n œ j Ê F † n œ 2yz œ 0 since y œ 0 Ê Flux œ 0. Therefore, Total Flux œ 3a% .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.5 Surface Area and Surface Integrals 1023 32. Across the cap: g(xß yß z) œ x#
y#
z# œ 25 Ê ™ g œ 2x i
2yj
2zk Ê k ™ gk œ È4x#
4y#
4z# œ 10 Ê nœ

™g k™g k

œ

x i  y j  zk 5

Ê Fluxcap œ ' ' F † n d5 œ ' ' Š x5z
#

cap

R

œ '0 72 d) œ 1441. 21




y# z 5

y# z 5


5z ‹ ˆ 5z ‰ dA œ ' ' ax#
y#
1b dx dy œ '0

x# z 5

Ê F†nœ


5z ; p œ k Ê k ™ g † pk œ 2z since z   0 Ê d5 œ 21

R

10 2z

dA

'0 ar#
1b r dr d) 4

Across the bottom: g(xß yß z) œ z œ 3 Ê ™ g œ k Ê k ™ gk œ 1 Ê n œ k Ê F † n œ 1; p œ k Ê k ™ g † pk œ 1 Ê d5 œ dA Ê Fluxbottom œ ' ' F † n d5 œ ' ' 1 dA œ 1(Area of the circular region) œ 161. Therefore, R

bottom

Flux œ Fluxcap
Fluxbottom œ 1281 33. ™ f œ 2x i
2yj
2zk Ê k ™ f k œ È4x#
4y#
4z# œ 2a; p œ k Ê k ™ f † pk œ 2z since z   0 Ê d5 œ œ dA; M œ ' ' $ d5 œ 8$ (surface area of sphere) œ a z

S

œ a$ ' ' dA œ a$ '0

1Î2

R

œ

a 2

'0

a

$1a$ 4

r dr d) œ

Ê zœ

Mxy M

$1a #

œ

; Mxy œ ' ' z$ d5 œ $ ' ' z ˆ za ‰ dA

#

œ

a #

. Because of symmetry, x œ y

Ê the centroid is ˆ #a ß #a ß #a ‰ .

34. ™ f œ 2y j
2zk Ê k ™ f k œ È4y#
4z# œ È4 ay#
z# b œ 6; p œ k Ê k ™ f † kk œ 2z since z   0 Ê d5 œ œ œ

3 z

dA

R

S

$ Š $14a ‹ ˆ $12a# ‰

2a 2z

dA; M œ ' ' 1 d5 œ S

'c3 '0 3

3

3 z

dx dy œ

'c3 '0 3

3

3 È 9
y#

S 3y È 9
y#

S

3

3

S

3x È 9
y#

dx dy œ

27 #

dA

dx dy œ 91; Mxy œ ' ' z d5

'c33 '03 z ˆ 3z ‰ dx dy œ 54; Mxz œ ' ' y d5 œ 'c33 '03 y ˆ 3z ‰ dx dy œ 'c33 '03

Myz œ ' ' x d5 œ 'c3 '0

6 2z

1. Therefore, x œ

Š 27 # 1‹ 91

œ

3 #

dx dy œ 0;

, y œ 0, and z œ

54 91

œ

6 1

35. Because of symmetry, x œ y œ 0; M œ ' ' $ d5 œ $ ' ' d5 œ (Area of S)$ œ 31È2 $ ; ™ f œ 2x i
2yj  2zk S

S

Ê k ™ f k œ È4x#
4y#
4z# œ 2Èx#
y#
z# ; p œ k Ê k ™ f † pk œ 2z Ê d5 œ œ

È x #  y #  ax #  y # b z

dA œ

È 2 È x#  y# z

È 2 È x#  y# ‹ z

dA Ê Mxy œ $ ' ' z Š R

œ $ ' ' È2 Èx#
y# dA œ $ '0 '1 È2 r# dr d) œ 21

2

R

141È2 3

$ Ê zœ

Œ

œ $ È2 ' ' ax#
y# b dA œ $ È2 '0

21

R

'12 r$ dr d) œ 151#È2 $

dA

dA

È

141 2 3

$


31 È 2 $

' ' ax#
y# b $ d5 œ ' ' ax#
y# b Š ‰ Ê axß yß zb œ ˆ0ß 0ß 14 9 . Next, Iz œ S

2 È x #  y #  z# #z

R

œ

14 9

È 2 È x#  y# ‹ z

Iz Ê Rz œ É M œ

$ dA

È10 #

36. f(xß yß z) œ 4x#
4y#  z# œ 0 Ê ™ f œ 8xi
8yj  2zk Ê k ™ f k œ È64x#
64y#
4z# œ 2È16x#
16y#
z# œ 2È4z#
z# œ 2È5 z since z   0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ Ê Iz œ ' ' ax#
y# b $ d5 œ $ È5 ' ' ax#
y# b dx dy œ $ È5 'c1Î2 '0 1Î2

2 cos )

dA œ È5 dA

3È51$ #

r$ dr d) œ

R

S

2È 5 z 2z

37. (a) Let the diameter lie on the z-axis and let f(xß yß z) œ x#
y#
z# œ a# , z   0 be the upper hemisphere Ê ™ f œ 2xi
2yj
2zk Ê k ™ f k œ È4x#
4y#
4z# œ 2a, a  0; p œ k Ê k ™ f † pk œ 2z since z   0 Ê d5 œ

a z

dA Ê Iz œ ' ' $ ax#
y# b ˆ az ‰ d5 œ a$ ' ' S

œ a$ '0 ’r# Èa#  r#  23 aa#  r# b 21

“ d) œ a$ '0

$Î# a

!

R

21

2 3

x#  y# È a #
ax #  y # b

a$ d) œ

41 3

dA œ a$ '0

21

'0a È r

#

a #
r#

r dr d)

a% $ Ê the moment of inertia is

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

81 3

a% $ for

1024 Chapter 16 Integration in Vector Fields the whole sphere (b) IL œ Ic.m.
mh# , where m is the mass of the body and h is the distance between the parallel lines; now, Ic.m. œ 831 a% $ (from part a) and m# œ ' ' $ d5 œ $ ' ' ˆ az ‰ dA œ a$ ' ' È # " # # dy dx œ a$ '0

'0

Ê IL œ

81 3

21

38. (a) Let z œ zœ

a

" Èa#
r# %

a $
41a# $ a# œ

4y# h%

# %

œ 2Éz# ˆ h È h #  a# a

!

a% $

h# a#

#

 a# ‰ a#

%

R

S R

S

h# a#

2xh# a#

i


2yh# a#

j  2zk

#

#

ax#
y# b œ 2Ɉ ha# ‰ ax#
y# b ˆ ha#
1‰

œ ˆ 2za ‰ Èh#
a# since z   0; p œ k Ê k ™ f † pk œ 2z Ê d5 œ

dA; M œ ' ' d5 œ ' '

21ahÈh#  a# 3

ax#
y# b  z# œ 0 Ê ™ f œ


4z# œ 2É ha% ax#
y# b


a%

Mxy œ ' ' z d5 œ ' ' z Š œ

a
ax  y b

R

21

a

Èx#
y# be the cone from z œ 0 to z œ h, h  0. Because of symmetry, x œ 0 and y œ 0;

h a

Ê k ™ f k œ É 4xa%h


œ

201 3

Èx#
y# Ê f(xß yß z) œ

h a

R

S

r dr d) œ a$ '0 ’Èa#  r# “ d) œ a$ '0 a d) œ 21a# $ and h œ a 21

Ê zœ

È h #  a# ‹ a

Mxy M

œ

2h 3

È h #  a# a

dA œ

È h #  a# a

dA œ

È h #  a# a

' ' ha Èx#
y# dx dy œ

ˆ 2z ‰ È h #  a# a 2z

dA

a1a# b œ 1aÈh#
a# ;

R

'021 '0a r# dr d)

h È h #  a# a#

‰ Ê the centroid is ˆ0ß 0ß 2h 3

(b) The base is a circle of radius a and center at (0ß 0ß h) Ê (0ß 0ß h) is the centroid of the base and the mass is M œ ' ' d5 œ 1a# . In Pappus' formula, let c" œ 2h k , c# œ hk , m" œ 1aÈh#
a# , and m# œ 1a# 3

S

Ê cœ

# 1aÈh#  a# Š 2h 3 ‹ k  1 a hk

1 a È h #  a#  1 a#

œ

2hÈh#  a#  3ah 3 ŠÈh#  a#  a‹

È

#

#

k Ê the centroid is !ß !ß 2h Èh # a # 3ah  3Š

h  a  a‹

(c) If the hemisphere is sitting so its base is in the plane z œ h, then its centroid is ˆ0ß 0ß h
#a ‰ and its mass is 21a# . In Pappus' formula, let c" œ 2h k , c# œ ˆh
a ‰ k , m" œ 1aÈh#
a# , and m# œ 21a# #

3

Ê cœ !ß !ß

a‰ #ˆ ‰ 1aÈh#  a# ˆ 2h 3 k  21 a h  # k 1 a È h #  a#  2 1 a#

2hÈh#  a#  6ah  3a# . 3 ŠÈh#  a#  2a‹ 

2hÈh#  a#  6ah  3a# 3 ŠÈh#  a#  2a‹

Ê

œ

2hÈh#  a#  6ah  3a# 3 ŠÈh#  a#  2a‹

k Ê the centroid is

Thus, for the centroid to be in the plane of the bases we must have z œ h

œ h Ê 2hÈh#
a#
6ah
3a# œ 3hÈh#
a#
6ah Ê 3a# œ hÈh#
a#

Ê 9a% œ h# ah#
a# b Ê h%
a# h#  9a% œ 0 Ê h# œ

ŠÈ37
"‹ a# #

(the positive root) Ê h œ

É2È37
2 #

39. fx (xß y) œ 2x, fy (xß y) œ 2y Ê Éfx#
fy#
1 œ È4x#
4y#
1 Ê Area œ ' ' È4x#
4y#
1 dx dy œ '0

21

È3

'0

R

È4r#
1 r dr d) œ

1 6

Š13È13  1‹

40. fy (yß z) œ 2y, fz (yß z) œ 2z Ê Éfy#
fz#
1 œ È4y#
4z#
1 Ê Area œ ' ' È4y#
4z#
1 dy dz R

21 1 œ '0 '0 È4r#
1 r dr d) œ 16 Š5È5  1‹

41. fx (xß y) œ

x È x#  y#

Ê Area œ

, fy (xß y) œ

y È x#  y#

#

Ê Éfx#
fy#
1 œ É x# x y#


y# x#  y#


1 œ È2

' ' È2 dx dy œ È2(Area between the ellipse and the circle) œ È2(61  1) œ 51È2 Rxy

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

a

Section 16.6 Parameterized Surfaces 1025 42. Over Rxy : z œ 2  23 x  2y Ê fx (xß y) œ  23 , fy (xß y) œ 2 Ê Éfx#
fy#
1 œ É 49
4
1 œ Ê Area œ

''

7 3

Rxy

dA œ

7 3

(Area of the shadow triangle in the xy-plane) œ ˆ 73 ‰ ˆ 3# ‰ œ

7 #

.

Over Rxz : y œ 1  13 x  "# z Ê fx (xß z) œ  13 , fz (xß z) œ  "# Ê Èfx#
fz#
1 œ É 19
Ê Area œ

'' Rxz

dA œ

7 6

7 6

(Area of the shadow triangle in the xz-plane) œ ˆ 76 ‰ (3) œ

7 #

Ryz

43. y œ

2 3

7 2

dA œ

7 2

(Area of the shadow triangle in the yz-plane) œ ˆ 72 ‰ (1) œ

z$Î# Ê fx (xß z) œ 0, fz (xß z) œ z"Î# Ê Èfx#
fz#
1 œ Èz
1 ; y œ

Ê Area œ '0

4

'0 Èz
1 dx dz œ '0 Èz
1 dz œ 23 Š5È5  1‹ 1

16 3

Ê

7 #

" 4


1œ

7 6

.

Over Ryz : x œ 3  3y  3# z Ê fy (yß z) œ 3, fz (yß z) œ  3# Ê Éfy#
fz#
1 œ É9
Ê Area œ ' '

7 3

9 4


1œ

7 #

.

16 3

œ

2 3

z$Î# Ê z œ 4

4

44. y œ 4  z Ê fx (xß z) œ 0, fz (xß z) œ 1 Ê Èfx#
fz#
1 œ È2 Ê Area œ œ È2 '0 a4  z# b dz œ 2

4 c z#

' ' È2 dA œ ' ' 0 0 2

È2 dx dz

Rxz

16È2 3

16.6 PARAMETRIZED SURFACES #

1. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ ˆÈx#
y# ‰ œ r# . Then r(rß )) œ (r cos ))i
(r sin ))j
r# k , 0 Ÿ r Ÿ 2, 0 Ÿ ) Ÿ 21. 2. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 9  x#  y# œ 9  r# . Then r(rß )) œ (r cos ))i
(r sin ))j
a9  r# b k ; z   0 Ê 9  r#   0 Ê r# Ÿ 9 Ê 3 Ÿ r Ÿ 3, 0 Ÿ ) Ÿ 21. But 3 Ÿ r Ÿ 0 gives the same points as 0 Ÿ r Ÿ 3, so let 0 Ÿ r Ÿ 3. È x#  y# Ê z œ #r . Then # r # Ÿ 3 Ê 0 Ÿ r Ÿ 6; to

3. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ r(rß )) œ (r cos ))i
(r sin ))j
ˆ #r ‰ k . For 0 Ÿ z Ÿ 3, 0 Ÿ 0Ÿ)Ÿ

1 #

get only the first octant, let

.

4. In cylindrical coordinates, let x œ r cos ), y œ r sin ), z œ 2Èx#
y# Ê z œ 2r. Then r(rß )) œ (r cos ))i
(r sin ))j
2rk . For 2 Ÿ z Ÿ 4, 2 Ÿ 2r Ÿ 4 Ê 1 Ÿ r Ÿ 2, and let 0 Ÿ ) Ÿ 21. 5. In cylindrical coordinates, let x œ r cos ), y œ r sin ) since x#
y# œ r2 Ê z# œ 9  ax#
y# b œ 9  r# Ê z œ È9  r# , z   0. Then r(rß )) œ (r cos ))i
(r sin ))j
È9  r# k . Let 0 Ÿ ) Ÿ 21. For the domain #

of r: z œ Èx#
y# and x#
y#
z# œ 9 Ê x#
y#
ˆÈx#
y# ‰ œ 9 Ê 2 ax#
y# b œ 9 Ê 2r# œ 9 Ê rœ

3 È2

Ê 0ŸrŸ

3 È2

.

6. In cylindrical coordinates, r(rß )) œ (r cos ))i
(r sin ))j
È4  r# k (see Exercise 5 above with x#
y#
z# œ 4, instead of x#
y#
z# œ 9). For the first octant, let 0 Ÿ ) Ÿ 1# . For the domain of r: z œ Èx#
y# and #

x#
y#
z# œ 4 Ê x#
y#
ˆÈx#
y# ‰ œ 4 Ê 2 ax#
y# b œ 4 Ê 2r# œ 4 Ê r œ È2. Thus, let È2 Ÿ r Ÿ 2 (to get the portion of the sphere between the cone and the xy-plane).

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1026 Chapter 16 Integration in Vector Fields 7. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx#
y#
z# Ê 3# œ 3 Ê 3 œ È3 Ê z œ È3 cos 9 for the sphere; z œ Ê cos 9 œ  "# Ê 9 œ 1 3

Ÿ9Ÿ

21 3

21 3

È3 #

œ È3 cos 9 Ê cos 9 œ

" #

Ê 9œ

1 3

;zœ

È3 #

Ê 

È3 #

œ È3 cos 9

. Then r(9ß )) œ ŠÈ3 sin 9 cos )‹ i
ŠÈ3 sin 9 sin )‹ j
ŠÈ3 cos 9‹ k ,

and 0 Ÿ ) Ÿ 21.

8. In spherical coordinates, x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), 3 œ Èx#
y#
z# Ê 3# œ 8 Ê 3 œ È8 œ 2È2 Ê x œ 2È2 sin 9 cos ), y œ 2È2 sin 9 sin ), and z œ 2È2 cos 9. Thus let r(9ß )) œ Š2È2 sin 9 cos )‹ i
Š2È2 sin 9 sin )‹ j
Š2È2 cos 9‹ k ; z œ 2 Ê 2 œ 2È2 cos 9 Ê cos 9 œ  È"2 Ê 9 œ

31 4

; z œ 2È2 Ê 2È2 œ 2È2 cos 9 Ê cos 9 œ 1 Ê 9 œ 0. Thus 0 Ÿ 9 Ÿ

31 4

and

0 Ÿ ) Ÿ 21 . 9. Since z œ 4  y# , we can let r be a function of x and y Ê r(xß y) œ xi
yj
a4  y# b k . Then z œ 0 Ê 0 œ 4  y# Ê y œ „ 2. Thus, let 2 Ÿ y Ÿ 2 and 0 Ÿ x Ÿ 2. 10. Since y œ x# , we can let r be a function of x and z Ê r(xß z) œ xi
x# j
zk . Then y œ 2 Ê x# œ 2 Ê x œ „ È2. Thus, let È2 Ÿ x Ÿ È2 and 0 Ÿ z Ÿ 3. 11. When x œ 0, let y#
z# œ 9 be the circular section in the yz-plane. Use polar coordinates in the yz-plane Ê y œ 3 cos ) and z œ 3 sin ). Thus let x œ u and ) œ v Ê r(u,v) œ ui
(3 cos v)j
(3 sin v)k where 0 Ÿ u Ÿ 3, and 0 Ÿ v Ÿ 21. 12. When y œ 0, let x#
z# œ 4 be the circular section in the xz-plane. Use polar coordinates in the xz-plane Ê x œ 2 cos ) and z œ 2 sin ). Thus let y œ u and ) œ v Ê r(u,v) œ (2 cos v)i
uj
(3 sin v)k where 2 Ÿ u Ÿ 2, and 0 Ÿ v Ÿ 1 (since we want the portion above the xy-plane). 13. (a) x
y
z œ 1 Ê z œ 1  x  y. In cylindrical coordinates, let x œ r cos ) and y œ r sin ) Ê z œ 1  r cos )  r sin ) Ê r(rß )) œ (r cos ))i
(r sin ))j
(1  r cos )  r sin ))k , 0 Ÿ ) Ÿ 21 and 0 Ÿ r Ÿ 3. (b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let y œ u cos v, z œ u sin v where u œ Èy#
z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0) with (xß 0ß 0) as vertex. Since x
y
z œ 1 Ê x œ 1  y  z Ê x œ 1  u cos v  u sin v, then r is a function of u and v Ê r(uß v) œ (1  u cos v  u sin v)i
(u cos v)j
(u sin v)k , 0 Ÿ u Ÿ 3 and 0 Ÿ v Ÿ 21. 14. (a) In a fashion similar to cylindrical coordinates, but working in the xz-plane instead of the xy-plane, let x œ u cos v, z œ u sin v where u œ Èx#
z# and v is the angle formed by (xß yß z), (yß 0ß 0), and (xß yß 0) with vertex (yß 0ß 0). Since x  y
2z œ 2 Ê y œ x
2z  2, then r(uß v) œ (u cos v)i
(u cos v
2u sin v  2)j
(u sin v)k , 0 Ÿ u Ÿ È3 and 0 Ÿ v Ÿ 21. (b) In a fashion similar to cylindrical coordinates, but working in the yz-plane instead of the xy-plane, let y œ u cos v, z œ u sin v where u œ Èy#
z# and v is the angle formed by (xß yß z), (xß 0ß 0), and (xß yß 0) with vertex (xß 0ß 0). Since x  y
2z œ 2 Ê x œ y  2z
2, then r(uß v) œ (u cos v  2u sin v
2)i
(u cos v)j
(u sin v)k , 0 Ÿ u Ÿ È2 and 0 Ÿ v Ÿ 21. 15. Let x œ w cos v and z œ w sin v. Then (x  2)#
z# œ 4 Ê x#  4x
z# œ 0 Ê w# cos# v  4w cos v
w# sin# v œ 0 Ê w#  4w cos v œ 0 Ê w œ 0 or w  4 cos v œ 0 Ê w œ 0 or w œ 4 cos v. Now w œ 0 Ê x œ 0 and y œ 0, which is a line not a cylinder. Therefore, let w œ 4 cos v Ê x œ (4 cos v)(cos v) œ 4 cos# v and z œ 4 cos v sin v. Finally, let y œ u. Then r(uß v) œ a4 cos# vb i
uj
(4 cos v sin v)k ,  1# Ÿ v Ÿ 1# and 0 Ÿ u Ÿ 3. Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.6 Parameterized Surfaces 1027 16. Let y œ w cos v and z œ w sin v. Then y#
(z  5)# œ 25 Ê y#
z#  10z œ 0 Ê w# cos# v
w# sin# v  10w sin v œ 0 Ê w#  10w sin v œ 0 Ê w(w  10 sin v) œ 0 Ê w œ 0 or w œ 10 sin v. Now w œ 0 Ê y œ 0 and z œ 0, which is a line not a cylinder. Therefore, let w œ 10 sin v Ê y œ 10 sin v cos v and z œ 10 sin# v. Finally, let x œ u. Then r(uß v) œ ui
(10 sin v cos v)j
a10 sin# vb k , 0 Ÿ u Ÿ 10 and 0 Ÿ v Ÿ 1. 17. Let x œ r cos ) and y œ r sin ). Then r(rß )) œ (r cos ))i
(r sin ))j
ˆ 2
r#sin ) ‰ k , 0 Ÿ r Ÿ 1 and 0 Ÿ ) Ÿ 21 )‰ Ê rr œ (cos ))i
(sin ))j  ˆ sin# ) ‰ k and r) œ (r sin ))i
(r cos ))j  ˆ r cos k # â â i j k â â â sin )  sin# ) ââ Ê rr ‚ r) œ â cos ) â ) ââ â r sin ) r cos )  r cos # œ Š
r sin#) cos )


(sin ))(r cos )) ‹i # #

Ê krr ‚ r) k œ É r4
r# œ


Š r sin#

È5 r #

#

)




r cos# ) # ‹j

Ê A œ '0

21

'01


ar cos# )
r sin# )b k œ

È5 r #

dr d) œ '0 ’ 21

È 5 r# " 4 “!

r #

j
rk

d) œ '0 d) œ 21

1È5 #

18. Let x œ r cos ) and y œ r sin ) Ê z œ x œ r cos ), 0 Ÿ r Ÿ 2 and 0 Ÿ ) Ÿ 21. Then r(rß )) œ (r cos ))i
(r sin ))j  (r cos ))k Ê rr œ (cos ))i
(sin ))j  (cos ))k and r) œ (r sin ))i
(r cos ))j
(r sin ))k â â i j k â â â â sin )  cos ) â Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) r sin ) â œ ar sin# )
r cos# )b i
(r sin ) cos )  r sin ) cos ))j
ar cos# )
r sin# )b k œ ri
rk Ê krr ‚ r) k œ Èr#
r# œ rÈ2 Ê A œ '0

21

'02 rÈ2 dr d) œ '021 ’ r È2 2 “ # d) œ '021 2È2 d) œ 41È2 #

!

19. Let x œ r cos ) and y œ r sin ) Ê z œ 2Èx#
y# œ 2r, 1 Ÿ r Ÿ 3 and 0 Ÿ ) Ÿ 21. Then r(rß )) œ (r cos ))i
(r sin ))j
2rk Ê rr œ (cos ))i
(sin ))j
2k and r) œ (r sin ))i
(r cos ))j â â i j kâ â â â sin ) 2 â œ (2r cos ))i  (2r sin ))j
ar cos# )
r sin# )b k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â œ (2r cos ))i  (2r sin ))j
rk Ê krr ‚ r) k œ È4r# cos# )
4r# sin# )
r# œ È5r# œ rÈ5 Ê A œ '0

21

'13 rÈ5 dr d) œ '021 ’ r È2 5 “ $ d) œ '021 4È5 d) œ 81È5 #

"

È x#  y#

20. Let x œ r cos ) and y œ r sin ) Ê z œ œ 3r , 3 Ÿ r Ÿ 4 and 0 Ÿ ) Ÿ 21. Then 3 r(rß )) œ (r cos ))i
(r sin ))j
ˆ 3r ‰ k Ê rr œ (cos ))i
(sin ))j
ˆ 3" ‰ k and r) œ (r sin ))i
(r cos ))j â â i j kâ â â " â # # ˆ " ‰ ˆ" ‰ Ê rr ‚ r) œ â cos ) sin ) 3 ââ œ  3 r cos ) i  3 r sin ) j
ar cos )
r sin )b k â â r sin ) r cos ) 0 â #

œ ˆ "3 r cos )‰ i  ˆ 3" r sin )‰ j
rk Ê krr ‚ r) k œ É 9" r# cos# )
9" r# sin# )
r# œ É 10r 9 œ Ê A œ '0

21

'34 rÈ310 dr d) œ '021 ’ r È6 10 “ % d) œ '021 7È610 d) œ 71È3 10 #

$

21. Let x œ r cos ) and y œ r sin ) Ê r# œ x#
y# œ 1, 1 Ÿ z Ÿ 4 and 0 Ÿ ) Ÿ 21. Then r(zß )) œ (cos ))i
(sin ))j
zk Ê rz œ k and r) œ ( sin ))i
(cos ))j â â j kâ â i â â Ê r) ‚ rz œ â  sin ) cos ) 0 â œ (cos ))i
(sin )) j Ê kr) ‚ rz k œ Ècos# )
sin# ) œ 1 â â 0 1â â 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

rÈ10 3

1028 Chapter 16 Integration in Vector Fields Ê A œ '0

21

'14 1 dr d) œ '021 3 d) œ 61

22. Let x œ u cos v and z œ u sin v Ê u# œ x#
z# œ 10, 1 Ÿ y Ÿ 1, 0 Ÿ v Ÿ 21. Then r(yß v) œ (u cos v)i
yj
(u sin v)k œ ŠÈ10 cos v‹ i
yj
ŠÈ10 sin v‹ k â i â â Ê rv œ ŠÈ10 sin v‹ i
ŠÈ10 cos v‹ k and ry œ j Ê rv ‚ ry œ â È10 sin v â â 0 œ ŠÈ10 cos v‹ i  ŠÈ10 sin v‹ k Ê krv ‚ ry k œ È10 Ê A œ '0

21

œ '0 2È10 dv œ 41È10

j 0 1

â k â È10 cos v ââ â â 0

'c11 È10 du dv œ '021 ’È10u“ "


"

21

dv

23. z œ 2  x#  y# and z œ Èx#
y# Ê z œ 2  z# Ê z#
z  2 œ 0 Ê z œ 2 or z œ 1. Since z œ Èx#
y#   0, we get z œ 1 where the cone intersects the paraboloid. When x œ 0 and y œ 0, z œ 2 Ê the vertex of the paraboloid is (0ß 0ß 2). Therefore, z ranges from 1 to 2 on the “cap" Ê r ranges from 1 (when x#
y# œ 1) to 0 (when x œ 0 and y œ 0 at the vertex). Let x œ r cos ), y œ r sin ), and z œ 2  r# . Then r(rß )) œ (r cos ))i
(r sin ))j
a2  r# b k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i
(sin ))j  2rk and â â i j k â â â â sin ) 2r â r) œ (r sin ))i
(r cos ))j Ê rr ‚ r) œ â cos ) â â 0 â â r sin ) r cos ) œ a2r# cos )b i
a2r# sin )b j
rk Ê krr ‚ r) k œ È4r% cos# )
4r% sin# )
r# œ rÈ4r#
1 Ê A œ '0

21

'01 rÈ4r#
1 dr d) œ '021 ’ 12" a4r#
1b$Î# “ " d) œ '021 Š 5È15#
" ‹ d) œ 16 Š5È5  1‹ !

24. Let x œ r cos ), y œ r sin ) and z œ x#
y# œ r# . Then r(rß )) œ (r cos ))i
(r sin ))j
r# k , 1 Ÿ r Ÿ 2, 0 Ÿ ) Ÿ 21 Ê rr œ (cos ))i
(sin ))j
2rk and r) œ (r sin ))i
(r cos ))j â â i j kâ â â â sin ) 2r â œ a2r# cos )b i  a2r# sin )b j
rk Ê krr ‚ r) k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â œ È4r% cos# )
4r% sin# )
r# œ rÈ4r#
1 Ê A œ '0

21

œ'

21

È È Š 17 171#
5 5 ‹ 0

d) œ

1 6

'12 rÈ4r#
1 dr d) œ '021 ’ 12" a4r#
1b$Î# “ # d) "

Š17È17  5È5‹

25. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx#
y#
z# œ È2 on the sphere. Next, x#
y#
z# œ 2 and z œ Èx#
y# Ê z#
z# œ 2 Ê z# œ 1 Ê z œ 1 since z   0 Ê 9 œ 14 . For the lower portion of the sphere cut by the cone, we get 9 œ 1. Then r(9ß )) œ ŠÈ2 sin 9 cos )‹ i
ŠÈ2 sin 9 sin )‹ j
ŠÈ2 cos 9‹ k ,

1 4

Ÿ 9 Ÿ 1 , 0 Ÿ ) Ÿ 21

Ê r9 œ ŠÈ2 cos 9 cos )‹ i
ŠÈ2 cos 9 sin )‹ j  ŠÈ2 sin 9‹ k and r) œ ŠÈ2 sin 9 sin )‹ i
ŠÈ2 sin 9 cos )‹ j â â i j k â â â â Ê r9 ‚ r) œ â È2 cos 9 cos ) È2 cos 9 sin ) È2 sin 9 â â â â È2 sin 9 sin ) È2 sin 9 cos ) â 0 # # œ a2 sin 9 cos )b i
a2 sin 9 sin )b j
(2 sin 9 cos 9)k Ê kr9 ‚ r) k œ È4 sin% 9 cos# )
4 sin% 9 sin# )
4 sin# 9 cos# 9 œ È4 sin# 9 œ 2 ksin 9k œ 2 sin 9 Ê A œ '0

21

'11Î4 2 sin 9 d9 d) œ '021 Š2
È2‹ d) œ Š4
2È2‹ 1

26. Let x œ 3 sin 9 cos ), y œ 3 sin 9 sin ), and z œ 3 cos 9 Ê 3 œ Èx#
y#
z# œ 2 on the sphere. Next, z œ 1 Ê 1 œ 2 cos 9 Ê cos 9 œ  "# Ê 9 œ

21 3

; z œ È3 Ê È3 œ 2 cos 9 Ê cos 9 œ

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

È3 #

Ê 9œ

1 6

. Then

Section 16.6 Parameterized Surfaces 1029 1 6

r(9ß )) œ (2 sin 9 cos ))i
(2 sin 9 sin ))j
(2 cos 9)k ,

Ÿ9Ÿ

21 3 ,

0 Ÿ ) Ÿ 21

Ê r9 œ (2 cos 9 cos ))i
(2 cos 9 sin ))j  (2 sin 9)k and r) œ (2 sin 9 sin ))i
(2 sin 9 cos )) j â â i j k â â â â Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â â 0 â 2 sin 9 sin ) 2 sin 9 cos ) â œ a4 sin# 9 cos )b i
a4 sin# 9 sin )b j
(4 sin 9 cos 9)k Ê kr9 ‚ r) k œ È16 sin% 9 cos# )
16 sin% 9 sin# )
16 sin# 9 cos# 9 œ È16 sin# 9 œ 4 ksin 9k œ 4 sin 9 Ê A œ '0

21

'12Î16Î3 4 sin 9 d9 d) œ '021 Š2
2È3‹ d) œ Š4
4È3‹ 1

â âi â 27. Let the parametrization be r(xß z) œ xi
x# j
zk Ê rx œ i
2xj and rz œ k Ê rx ‚ rz œ â 1 â â0 œ 2xi
j Ê krx ‚ rz k œ È4x#
1 Ê œ '0

3

â kâ â 0â â "â

' ' G(xß yß z) d5 œ ' ' xÈ4x#
1 dx dz œ ' ’ 12" a4x#
1b$Î# “ dz 0 0 0 3

2

#

3

!

S

" 1#

17È17
" 4

Š17È17  1‹ dz œ

28. Let the parametrization be r(xß y) œ xi
yj
È4  y# k , 2 Ÿ y Ÿ 2 Ê rx œ i and ry œ j  âi j â k â â # â1 0 â 0 Ê rx ‚ ry œ â â œ y j
k Ê krx ‚ ry k œ É 4
y y#
1 œ È42
y# â 0 1  y â È 4
y# â È 4
y# â Ê

j 2x 0

y È 4
y#

k

' ' G(xß yß z) d5 œ ' ' È4  y# Š È 2 # ‹ dy dx œ 24 4
y 1 c2 4

2

S

29. Let the parametrization be r(9ß )) œ (sin 9 cos ))i
(sin 9 sin ))j
(cos 9)k (spherical coordinates with 3 œ 1 on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (cos 9 cos ))i
(cos 9 sin ))j  (sin 9)k and â â i j k â â â â r) œ ( sin 9 sin ))i
(sin 9 cos ))j Ê r9 ‚ r) œ â cos 9 cos ) cos 9 sin )  sin 9 â â â 0 â â  sin 9 sin ) sin 9 cos ) œ asin# 9 cos )b i
asin# 9 sin )b j
(sin 9 cos 9)k Ê kr9 ‚ r) k œ Èsin% 9 cos# )
sin% 9 sin# )
sin# 9 cos# 9 œ sin 9; x œ sin 9 cos ) Ê G(xß yß z) œ cos# ) sin# 9 Ê ' ' G(xß yß z) d5 œ '0

21

S

œ '0

21

œ

'0

1

u œ cos 9 Ä acos# )b a1  cos# 9b (sin 9) d9 d); ” du œ  sin 9 d9 •

'021 acos# )b ’ u3

$

 u“


" "

d) œ

4 3

'01 acos# ) sin# 9b (sin 9) d9 d)

'021 '1c1 acos# )b au#  1b du d)

'021 cos# ) d) œ 43
2)
sin42) ‘ #!1 œ 431

30. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k (spherical coordinates with 3 œ a, a   0, on the sphere), 0 Ÿ 9 Ÿ 1# (since z   0), 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i
(a cos 9 sin ))j  (a sin 9)k and â i â â r) œ (a sin 9 sin ))i
(a sin 9 cos ))j Ê r9 ‚ r) œ â a cos 9 cos ) â â a sin 9 sin )

j a cos 9 sin ) a sin 9 cos )

â k â â a sin 9 â â 0 â

œ aa# sin# 9 cos )b i
aa# sin# 9 sin )b j
(a# sin 9 cos 9)k Ê kr9 ‚ r) k œ Èa% sin% 9 cos# )
a% sin% 9 sin# )
a% sin# 9 cos# 9 œ a# sin 9; z œ a cos 9 Ê G(xß yß z) œ a# cos# 9 Ê

' ' G(xß yß z) d5 œ ' ' 21

S

0

1Î2

0

aa# cos# 9b aa# sin 9b d9 d) œ

2 3

1a%

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1030 Chapter 16 Integration in Vector Fields 31. Let the parametrization be r(xß y) œ xi
yj
(4  x  y)k Ê rx œ i  k and ry œ j  k â â âi j k â 1 1 â â Ê rx ‚ ry œ â 1 0 1 â œ i
j
k Ê krx ‚ ry k œ È3 Ê ' ' F(xß yß z) d5 œ '0 '0 (4  x  y) È3 dy dx â â S â 0 1 " â œ '0 È3 ’4y  xy  1

" y# 2 “!

dx œ '0 È3 ˆ #7  x‰ dx œ È3 ’ 27 x  1

" x# # “!

œ 3È3

32. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â sin ) " â Ê rr œ (cos ))i
(sin ))j
k and r) œ (r sin ))i
(r cos ))j Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) 0 â œ (r cos ))i  (r sin ))j
rk Ê krr ‚ r) k œ È(r cos ))#
(r sin ))#
r# œ rÈ2; z œ r and x œ r cos )

' ' F(xß yß z) d5 œ ' ' (r  r cos )) ŠrÈ2‹ dr d) œ È2 ' ' (1  cos )) r# dr d) 0 0 0 0 21

Ê F(xß yß z) œ r  r cos ) Ê

21

1

1

S

œ

21 È 2 3

33. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
a1  r# b k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j k â â â â sin ) 2r â Ê rr œ (cos ))i
(sin ))j  2rk and r) œ (r sin ))i
(r cos ))j Ê rr ‚ r) œ â cos ) â â 0 â â r sin ) r cos ) œ a2r# cos )b i
a2r# sin )b j
rk Ê krr ‚ r) k œ Éa2r# cos )b#
a2r# sin )b
r# œ rÈ1
4r# ; z œ 1  r# and

' ' H(xß yß z) d5

x œ r cos ) Ê H(xß yß z) œ ar# cos# )b È1
4r# Ê œ '0

21

'0

1

S

#

#

ar cos )b ŠÈ1


4r# ‹ ŠrÈ1




4r# ‹

dr d) œ '0

21

'01 r$ a1
4r# b cos# ) dr d) œ 11121

34. Let the parametrization be r(9ß )) œ (2 sin 9 cos ))i
(2 sin 9 sin ))j
(2 cos 9)k (spherical coordinates with 3 œ 2 on the sphere), 0 Ÿ 9 Ÿ 1 ; x#
y#
z# œ 4 and z œ Èx#
y# Ê z#
z# œ 4 Ê z# œ 2 Ê z œ È2 (since 4

z   0) Ê 2 cos 9 œ È2 Ê cos 9 œ

È2 #

Ê 9œ

1 4

, 0 Ÿ ) Ÿ 21; r9 œ (2 cos 9 cos ))i
(2 cos 9 sin ))j  (2 sin 9)k â â i j k â â â â and r) œ (2 sin 9 sin ))i
(2 sin 9 cos ))j Ê r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â â 0 â 2 sin 9 sin ) 2 sin 9 cos ) â œ a4 sin# 9 cos )b i
a4 sin# 9 sin )b j
(4 sin 9 cos 9)k Ê kr9 ‚ r) k œ È16 sin% 9 cos# )
16 sin% 9 sin# )
16 sin# 9 cos# 9 œ 4 sin 9; y œ 2 sin 9 sin ) and z œ 2 cos 9 Ê H(xß yß z) œ 4 cos 9 sin 9 sin ) Ê 1Î4 21 œ '0 '0 16 sin# 9 cos 9 sin ) d9 d) œ 0

' ' H(xß yß z) d5 œ ' ' (4 cos 9 sin 9 sin ))(4 sin 9) d9 d) 0 0 21

1Î4

S

35. Let the parametrization be r(xß y) œ xi
yj
a4  y# b k , 0 Ÿ x Ÿ 1, 2 Ÿ y Ÿ 2; z œ 0 Ê 0 œ 4  y# â â k â âi j â â 0 â œ 2yj
k Ê F † n d5 Ê y œ „ 2; rx œ i and ry œ j  2yk Ê rx ‚ ry œ â 1 0 â â â 0 1 2y â œF†

r x ‚r y kr x ‚r y k

krx ‚ ry k dy dx œ (2xy  3z) dy dx œ c2xy  3a4  y# bd dy dx Ê ' ' F † n d5 S

1 2 1 1 # œ '0 'c2 a2xy
3y#  12b dy dx œ '0 cxy#
y$  12yd
# dx œ '0 32 dx œ 32

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Section 16.6 Parameterized Surfaces 1031 36. Let the parametrization be r(xß y) œ xi
x# j
zk , 1 Ÿ x Ÿ 1, 0 Ÿ z Ÿ 2 Ê rx œ i
2xj and rz œ k â â â i j kâ â â rz # Ê rx ‚ rz œ â 1 2x 0 â œ 2xi  j Ê F † n d5 œ F † krrxx ‚ ‚rz k krx ‚ rz k dz dx œ x dz dx â â â0 0 1â Ê

' ' F † n d5 œ ' ' x# dz dx œ  43 c1 0 1

2

S

37. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k (spherical coordinates with 3 œ a, a   0, on the sphere), 0 Ÿ 9 Ÿ 1# (for the first octant) ß 0 Ÿ ) Ÿ 1# (for the first octant) Ê r9 œ (a cos 9 cos ))i
(a cos 9 sin ))j  (a sin 9)k and r) œ (a sin 9 sin ))i
(a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â œ aa# sin# 9 cos )b i
aa# sin# 9 sin )b j
aa# sin 9 cos 9b k Ê F † n d5 œ F † œ a$ cos# 9 sin 9 d) d9 since F œ zk œ (a cos 9)k Ê

' ' F † n d5 œ '

1Î2

0

S

'0

r 9 ‚r ) kr 9 ‚ r ) k

1Î2

kr9 ‚ r) k d) d9

a$ cos# 9 sin 9 d9 d) œ

1a$ 6

38. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k (spherical coordinates with 3 œ a, a   0, on the sphere), 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i
(a cos 9 sin ))j  (a sin 9)k and r) œ (a sin 9 sin ))i
(a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â œ aa# sin# 9 cos )b i
aa# sin# 9 sin )b j
aa# sin 9 cos 9b k Ê F † n d5 œ F † $

$

#

$

$

#

$

#

r 9 ‚r ) kr 9 ‚ r ) k

kr9 ‚ r) k d) d9

$

œ aa sin 9 cos 9
a sin 9 sin )
a sin 9 cos 9b d) d9 œ a sin 9 d) d9 since F œ xi
yj
zk

œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k Ê ' ' F † n d5 œ '0

21

S

'01 a$ sin 9 d9 d) œ 41a$

39. Let the parametrization be r(xß y) œ xi
yj
(2a  x  y)k , 0 Ÿ x Ÿ a, 0 Ÿ y Ÿ a Ê rx œ i  k and ry œ j  k â â âi j k â â â r ‚r Ê rx ‚ ry œ â 1 0 1 â œ i
j
k Ê F † n d5 œ F † krxx ‚ryyk krx ‚ ry k dy dx â â â 0 1 1 â œ [2xy
2y(2a  x  y)
2x(2a  x  y)] dy dx since F œ 2xyi
2yzj
2xzk œ 2xyi
2y(2a  x  y)j
2x(2a  x  y)k Ê

' ' F † n d5 S

a a a a œ '0 '0 [2xy
2y(2a  x  y)
2x(2a  x  y)] dy dx œ '0 '0 a4ay  2y#
4ax  2x#  2xyb dy dx a œ '0 ˆ 43 a$
3a# x  2ax# ‰ dx œ ˆ 43
#3  23 ‰ a% œ 13a 6 %

40. Let the parametrization be r()ß z) œ (cos ))i
(sin ))j
zk , 0 Ÿ z Ÿ a, 0 Ÿ ) Ÿ 21 (where r œ Èx#
y# œ 1 on â â j kâ â i â â the cylinder) Ê r) œ ( sin ))i
(cos ))j and rz œ k Ê r) ‚ rz œ â  sin ) cos ) 0 â œ (cos ))i
(sin ))j â â 0 1â â 0 Ê F † n d5 œ F † Ê

r ) ‚r z kr ) ‚r z k

kr) ‚ rz k dz d) œ acos# )
sin# )b dz d) œ dz d), since F œ (cos ))i
(sin ))j
zk

' ' F † n d5 œ ' ' 1 dz d) œ 21a 0 0 21

a

S

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1032 Chapter 16 Integration in Vector Fields 41. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
k and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) œ (r cos ))i
(r sin ))j  rk Ê F † n d5 œ F † F œ ar# sin ) cos )b i  rk Ê ' ' F † n d5 œ '0

r ) ‚r r kr ) ‚r r k

21

#1 " œ
 12 cos$ )
3) ‘ ! œ

S

kr) ‚ rr k d) dr œ ar$ sin ) cos# )
r# b d) dr since

'01 ar$ sin ) cos# )
r# b dr d) œ '021 ˆ "4 sin ) cos# )
"3 ‰ d)

21 3

42. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
2rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
2k and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 2 â â cos ) œ (2r cos ))i
(2r sin ))j  rk Ê F † n d5 œ F † $

#

r ) ‚r r kr ) ‚r r k

kr) ‚ rr k d) dr

$

œ a2r sin ) cos )
4r cos ) sin )
rb d) dr since

F œ ar# sin# )b i
a2r# cos )b j  k Ê ' ' F † n d5 œ '0

21

œ '0

21

S

ˆ "2

#

sin ) cos )
cos ) sin )


"‰ 2

d) œ
6" sin$ )


'01 a2r$ sin# ) cos )
4r$ cos ) sin )
rb dr d) 1 2

#1

sin# )
#" )‘ ! œ 1

43. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
rk , 1 Ÿ r Ÿ 2 (since 1 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
k and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) œ (r cos ))i
(r sin ))j  rk Ê F † n d5 œ F †

r ) ‚r r kr ) ‚r r k

kr) ‚ rr k d) dr œ ar# cos# )  r# sin# )  r$ b d) dr

œ ar#  r$ b d) dr since F œ (r cos ))i  (r sin ))j
r# k Ê ' ' F † n d5 œ '0

21

S

'12 ar#  r$ b dr d) œ  7361

44. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
r# k , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
2rk and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 2r â â cos ) œ a2r# cos )b i
a2r# sin )b j  rk Ê F † n d5 œ F †

r ) ‚r r kr ) ‚ r r k

kr) ‚ rr k d) dr œ a8r$ cos# )
8r$ sin# )  2rb d) dr

œ a8r$  2rb d) dr since F œ (4r cos ))i
(4r sin ))j
2k Ê ' ' F † n d5 œ '0

21

S

'01 a8r$  2rb dr d) œ 21

45. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k , 0 Ÿ 9 Ÿ

1 #

,0Ÿ)Ÿ

1 #

Ê r9 œ (a cos 9 cos ))i
(a cos 9 sin ))j  (a sin 9)k and r) œ (a sin 9 sin ))i
(a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â œ aa# sin# 9 cos )b i
aa# sin# 9 sin )b j
aa# sin 9 cos 9b k Ê kr9 ‚ r) k œ Èa% sin% 9 cos# )
a% sin% 9 sin# )
a% sin# 9 cos# 9 œ Èa% sin# 9 œ a# sin 9. The mass is

M œ ' ' d5 œ '0

1Î2

S

œ '0

1Î2

'0

1Î2

'01Î2 aa# sin 9b d9 d) œ a#1 ; the first moment is Myz œ ' ' x d5 #

S

(a sin 9 cos )) aa# sin 9b d9 d) œ

$

a 1 4

$

Ê xœ

Š a 41 ‹ #

Š a #1 ‹

œ

a #

Ê the centroid is located at ˆ #a ,

symmetry

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

a #

, #a ‰ by

Section 16.6 Parameterized Surfaces 1033 46. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
rk , 1 Ÿ r Ÿ 2 (since 1 Ÿ z Ÿ 2) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
k and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) œ (r cos ))i
(r sin ))j  rk Ê kr) ‚ rr k œ Èr# cos# )
r# sin# )
r# œ rÈ2. The mass is M œ ' ' $ d5 œ '0

21

S

œ

Š14È2‹1$ 3

Ê zœ

'12 $ rÈ2 dr d) œ Š3È2‹ 1$ ; the first moment is Mxy œ ' ' S



Š14È2‹ 1$ 3



œ

Š3È2‹ 1$

$ z d5 œ '0

21

'12 $ rŠrÈ2‹ dr d)

‰ Ê the center of mass is located at ˆ0ß 0ß 14 9 by symmetry. The

14 9

moment of inertia is Iz œ ' ' $ ax#
y# b d5 œ '0

21

S

'12 $ r# ŠrÈ2‹ dr d) œ Š15

È2‹ 1$ #

Ê the radius of gyration is

Iz Rz œ É M œ É #5

47. Let the parametrization be r(9ß )) œ (a sin 9 cos ))i
(a sin 9 sin ))j
(a cos 9)k , 0 Ÿ 9 Ÿ 1, 0 Ÿ ) Ÿ 21 Ê r9 œ (a cos 9 cos ))i
(a cos 9 sin ))j  (a sin 9)k and r) œ (a sin 9 sin ))i
(a sin 9 cos ))j â â i j k â â â â Ê r9 ‚ r) œ â a cos 9 cos ) a cos 9 sin ) a sin 9 â â â 0 â a sin 9 sin ) a sin 9 cos ) â œ aa# sin# 9 cos )b i
aa# sin# 9 sin )b j
aa# sin 9 cos 9b k Ê kr9 ‚ r) k œ Èa% sin% 9 cos# )
a% sin% 9 sin# )
a% sin# 9 cos# 9 œ Èa% sin# 9 œ a# sin 9. The moment of

inertia is Iz œ ' ' $ ax#
y# b d5 œ '0

21

S

'01 $ c(a sin 9 cos ))#
(a sin 9 sin ))# d aa# sin 9b d9 d)

1 1 21 21 21 1 œ '0 '0 $ aa# sin# 9b aa# sin 9b d9 d) œ '0 '0 $ a% sin$ 9 d9 d) œ '0 $ a%
ˆ "3 cos 9‰ asin# 9
2b‘ ! d) œ 8$13 a

%

48. Let the parametrization be r(rß )) œ (r cos ))i
(r sin ))j
rk , 0 Ÿ r Ÿ 1 (since 0 Ÿ z Ÿ 1) and 0 Ÿ ) Ÿ 21 â â i j kâ â â â Ê rr œ (cos ))i
(sin ))j
k and r) œ (r sin ))i
(r cos ))j Ê r) ‚ rr œ â r sin ) r cos ) 0 â â â sin ) 1 â â cos ) œ (r cos ))i
(r sin ))j  rk Ê kr) ‚ rr k œ Èr# cos# )
r# sin# )
r# œ rÈ2. The moment of inertia is Iz œ ' ' $ ax#
y# b d5 œ '0

21

S

'01 $ r# ŠrÈ2‹ dr d) œ 1$#È2

49. The parametrization r(rß )) œ (r cos ))i
(r sin ))j
rk at P! œ ŠÈ2ß È2ß 2‹ Ê ) œ 1 , r œ 2, 4

rr œ (cos ))i
(sin ))j
k œ

È2 #

È

i
#2 j
k and r) œ (r sin ))i
(r cos ))j œ È2i
È2j â i j k ââ â âÈ â Ê rr ‚ r) œ â 2/2 È2/2 1 â â â â È 2 È 2 0 â œ È2i  È2j
2k Ê the tangent plane is 0 œ ŠÈ2i  È2j
2k‹ † ’Šx  È2‹ i
Šy  È2‹ j
(z  2)k“ Ê È2x
È2y  2z œ 0, or x
y  È2z œ 0. The parametrization r(rß )) Ê x œ r cos ), y œ r sin ) and z œ r Ê x#
y# œ r# œ z# Ê the surface is z œ Èx#
y# .

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1034 Chapter 16 Integration in Vector Fields 50. The parametrization r(9ß )) œ (4 sin 9 cos ))i
(4 sin 9 sin ))j
(4 cos 9)k at P! œ ŠÈ2ß È2ß 2È3‹ Ê 3 œ 4 and z œ 2È3 œ 4 cos 9 Ê 9 œ Ê )œ

1 4

1 6

; also x œ È2 and y œ È2

. Then r9

œ (4 cos 9 cos ))i
(4 cos 9 sin ))j  (4 sin 9)k œ È6i
È6j  2k and r) œ (4 sin 9 sin ))i
(4 sin 9 cos ))j â i j k ââ â â â œ È2i
È2j at P! Ê r9 ‚ r) œ â È6 È6 2 â â â â È 2 È 2 0 â œ 2È2i
2È2j
4È3k Ê the tangent plane is Š2È2i
2È2j
4È3k‹ † ’Šx  È2‹ i
Šy  È2‹ j
Šz  2È3‹ k“ œ 0 Ê È2x
È2y
2È3z œ 16, or x
y
È6z œ 8È2. The parametrization Ê x œ 4 sin 9 cos ), y œ 4 sin 9 sin ), z œ 4 cos 9 Ê the surface is x#
y#
z# œ 16, z   0. 51. The parametrization r()ß z) œ (3 sin 2))i
a6 sin# )b j
zk È

at P! œ Š 3 # 3 ß 9# ß 0‹ Ê ) œ

1 3

and z œ 0. Then

r) œ (6 cos 2))i
(12 sin ) cos ))j œ 3i
3È3j and rz œ k at P! â i j k ââ â â â Ê r) ‚ rz œ â 3 3È3 0 â œ 3È3i
3j â â â 0 0 1â Ê the tangent plane is È Š3È3i
3j‹ † ’Šx  3 3 ‹ i
ˆy  9 ‰ j
(z  0)k“ œ 0 #

#

Ê È3x
y œ 9. The parametrization Ê x œ 3 sin 2) #

and y œ 6 sin# ) Ê x#
y# œ 9 sin# 2)
a6 sin# )b œ 9 a4 sin# ) cos# )b
36 sin% ) œ 6 a6 sin# )b œ 6y Ê x#
y#  6y
9 œ 9 Ê x#
(y  3)# œ 9 52. The parametrization r(xß y) œ xi
yj  x# k at P! œ (1ß 2ß 1) Ê rx œ i  2xk œ i  2k and ry œ j at P! â â âi j k â â â Ê rx ‚ ry œ â 1 0 2 â œ 2i
k Ê the tangent plane â â â0 " 0 â is (2i
k) † [(x  1)i
(y  2)j
(z
1)k] œ 0 Ê 2x
z œ 1. The parametrization Ê x œ x, y œ y and z œ x# Ê the surface is z œ x#

53. (a) An arbitrary point on the circle C is (xß z) œ (R
r cos u, r sin u) Ê (xß yß z) is on the torus with x œ (R
r cos u) cos v, y œ (R
r cos u) sin v, and z œ r sin u, 0 Ÿ u Ÿ 21, 0 Ÿ v Ÿ 21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.7 Stoke's Theorem 1035 (b) ru œ (r sin u cos v)i  (r sin u sin v)j
(r cos u)k and rv œ ((R
r cos u) sin v)i
((R
r cos u) cos v)j â â i j k â â â â r sin u sin v r cos u â Ê ru ‚ rv œ â r sin u cos v â â 0 â â (R
r cos u) sin v (R
r cos u) cos v œ (R
r cos u)(r cos v cos u)i  (R
r cos u)(r sin v cos u)j
(r sin u)(R
r cos u)k Ê kru ‚ rv k# œ (R
r cos u)# ar# cos# v cos# u
r# sin# v cos# u
r# sin# ub Ê kru ‚ rv k œ r(R
r cos u) Ê A œ '0

21

'021

arR
r# cos ub du dv œ '0 21rR dv œ 41# rR 21

54. (a) The point (xß yß z) is on the surface for fixed x œ f(u) when y œ g(u) sin ˆ 1#  v‰ and z œ g(u) cos ˆ 1#  v‰ Ê x œ f(u), y œ g(u) cos v, and z œ g(u) sin v Ê r(uß v) œ f(u)i
(g(u) cos v)j
(g(u) sin v)k , 0 Ÿ v Ÿ 21, aŸuŸb (b) Let u œ y and x œ u# Ê f(u) œ u# and g(u) œ u Ê r(uß v) œ u# i
(u cos v)j
(u sin v)k , 0 Ÿ v Ÿ 21, 0 Ÿ u 55. (a) Let w#


z# c#

œ 1 where w œ cos 9 and

z c

œ sin 9 Ê

x# a#




y# b#

œ cos# 9 Ê

x a

œ cos 9 cos ) and

y b

œ cos 9 sin )

Ê x œ a cos ) cos 9, y œ b sin ) cos 9, and z œ c sin 9 Ê r()ß 9) œ (a cos ) cos 9)i
(b sin ) cos 9)j
(c sin 9)k (b) r) œ (a sin ) cos 9)i
(b cos ) cos 9)j and r9 œ (a cos ) sin 9)i  (b sin ) sin 9)j
(c cos 9)k â â i j k â â â â 0 â Ê r) ‚ r9 œ â a sin ) cos 9 b cos ) cos 9 â â â a cos ) sin 9 b sin ) sin 9 c cos 9 â œ abc cos ) cos# 9b i
aac sin ) cos# 9b j
(ab sin 9 cos 9)k Ê kr) ‚ r9 k# œ b# c# cos# ) cos% 9
a# c# sin# ) cos% 9
a# b# sin# 9 cos# 9, and the result follows.

A Ê '0

21

'01 kr) ‚ r9 k d9 d) œ '021 '01 c a# b# sin# 9 cos# 9
b# c# cos# ) cos% 9
a# c# sin# ) cos% 9 d1/2 d9 d)

56. (a) r()ß u) œ (cosh u cos ))i
(cosh u sin ))j
(sinh u)k (b) r()ß u) œ (a cosh u cos ))i
(b cosh u sin ))j
(c sinh u)k 57. r()ß u) œ (5 cosh u cos ))i
(5 cosh u sin ))j
(5 sinh u)k Ê r) œ (5 cosh u sin ))i
(5 cosh u cos ))j and ru œ (5 sinh u cos ))i
(5 sinh u sin ))j
(5 cosh u)k â â i j k â â â â 0 Ê r) ‚ ru œ â 5 cosh u sin ) 5 cosh u cos ) â â â ) ) 5 sinh u cos 5 sinh u sin 5 cosh u â â œ a25 cosh# u cos )b i
a25 cosh# u sin )b j  (25 cosh u sinh u)k. At the point (x! ß y! ß 0), where x#0
y#0 œ 25 we have 5 sinh u œ 0 Ê u œ 0 and x! œ 25 cos ), y! œ 25 sin ) Ê the tangent plane is 5(x! i
y! j) † [(x  x! )i
(y  y! )j
zk] œ 0 Ê x! x  x#0
y! y  y#0 œ 0 Ê x! x
y! y œ 25 58. Let

z# c#

 w# œ 1 where

z c

œ cosh u and w œ sinh u Ê w# œ

x# a#




y# b#

Ê

x a

œ w cos ) and

y b

œ w sin )

Ê x œ a sinh u cos ), y œ b sinh u sin ), and z œ c cosh u Ê r()ß u) œ (a sinh u cos ))i
(b sinh u sin ))j
(c cosh u)k , 0 Ÿ ) Ÿ 21, _  u  _ 16.7 STOKES' THEOREM â i â â 1. curl F œ ™ ‚ F œ â ``x â # âx

j ` `y

2x

k ââ ` â ` z â œ 0i
0j
(2  0)k œ 2k and n œ k Ê curl F † n œ 2 Ê d5 œ dx dy â z# â

Ê )C F † dr œ ' ' 2 dA œ 2(Area of the ellipse) œ 41 R

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1036 Chapter 16 Integration in Vector Fields â i â â 2. curl F œ ™ ‚ F œ â ``x â â 2y Ê

)C F † dr œ ' '

k ââ ` â ` z â œ 0i
0j
(3  2)k œ k and n œ k Ê curl F † n œ 1 Ê d5 œ dx dy â z# â

j ` `y

3x

dx dy œ Area of circle œ 91

R

â i â â 3. curl F œ ™ ‚ F œ â ``x â â y œ

" È3

k ââ ` â ` z â œ xi  2xj
(z  1)k and n œ â x# â

j ` `y

xz

È3 1

(x  2x
z  1) Ê d5 œ

dA Ê )C F † dr œ ' '

" È3

R

ijk È3

Ê curl F † n

(3x
z  1) È3 dA

'01cx [3x
(1  x  y)  1] dy dx œ '01 '01cx (4x  y) dy dx œ '01 
4x(1  x)
"# (1  x)# ‘ dx 1 œ '0 ˆ "#
3x  7# x# ‰ dx œ  56 œ '0

1

â i â â ` 4. curl F œ ™ ‚ F œ â ` x â # â y
z# Ê curl F † n œ

" È3

j

â â â ` â œ (2y  2z)i
(2z  2x)j
(2x  2y)k and n œ `z â x#
y# â k

` `y

x#
z#

ijk È3

(2y  2z
2z  2x
2x  2y) œ 0 Ê )C F † dr œ ' ' 0 d5 œ 0 S

â i â â 5. curl F œ ™ ‚ F œ â ``x â # â y
z#

j

â â â â œ 2yi
(2z  2x)j
(2x  2y)k and n œ k â # #â x
y k

` `y

x#
y#

` `z

Ê curl F † n œ 2x  2y Ê d5 œ dx dy Ê )C F † dr œ 'c1 'c1 (2x  2y) dx dy œ 'c1 cx#  2xyd
" dy 1

1

1

"

œ 'c1 4y dy œ 0 1

â i â â 6. curl F œ ™ ‚ F œ â ``x â # $ âx y

k ââ ` â # # ` z â œ 0i
0j  3x y k and n œ â z â

j ` `y

1

# #

Ê curl F † n œ  x y z; d5 œ 3 4

4 z

21

œ 32 '0

21

" 4

x i  y j  zk 4

œ

dA (Section 16.5, Example 5, with a œ 4) Ê )C F † dr

œ ' ' ˆ 34 x# y# z‰ ˆ 4z ‰ dA œ 3 '0 R

2xi  2yj  2zk 2 È x #  y #  z#

'02 ar# cos# )b ar# sin# )b r dr d) œ 3'021 ’ r6 “ # (cos ) sin ))# d) '

!

sin# 2) d) œ 4 '0 sin# u du œ 4
u2  41

sin 2u ‘ %1 4 !

œ 81

7. x œ 3 cos t and y œ 2 sin t Ê F œ (2 sin t)i
a9 cos# tb j
a9 cos# t
16 sin% tb sin eÈÐ6 sin t cos tÑÐ0Ñ k at the base of the shell; r œ (3 cos t)i
(2 sin t)j Ê dr œ (3 sin t)i
(2 cos t)j Ê F † ddtr œ 6 sin# t
18 cos$ t Ê

' ' ™ ‚ F † n d5 œ ' a6 sin# t
18 cos$ tb dt œ
3t
32 sin 2t
6(sin t) acos# t
2b‘ #!1 œ 61 0 21

S

â i â â ` â 8. curl F œ ™ ‚ F œ â `x â z
" â #x Ê nœ

™f k™f k

j ` `y
"

tan

â â â ` â œ 2j ; f(xß yß z) œ 4x#
y
z# Ê ™ f œ 8xi
j
2zk `z â x
4 " z ââ k

y

and p œ j Ê k ™ f † pk œ 1 Ê d5 œ

Ê ™ ‚ F † n d5 œ 2 dA Ê

k™f k k™f†pk

dA œ k ™ f k dA; ™ ‚ F † n œ

" k™f k

(2j † ™ f) œ


2 k™f k

' ' ™ ‚ F † n d5 œ ' ' 2 dA œ 2(Area of R) œ 2(1 † 1 † 2) œ 41, where R S

R

is the elliptic region in the xz-plane enclosed by 4x#
z# œ 4.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.7 Stoke's Theorem 1037 9. Flux of ™ ‚ F œ ' ' ™ ‚ F † n d5 œ )C F † dr, so let C be parametrized by r œ (a cos t)i
(a sin t)j , S

0 Ÿ t Ÿ 21 Ê

dr dt

œ (a sin t)i
(a cos t)j Ê F †

dr dt

œ ay sin t
ax cos t œ a# sin# t
a# cos# t œ a#

Ê Flux of ™ ‚ F œ )C F † dr œ '0 a# dt œ 21a# â i â â 10. ™ ‚ (yi) œ â ``x â â y

21

j ` `y

0

k ââ ` â ` z â œ k ; n œ â 0 â

Ê ™ ‚ (yi) † n œ z; d5 œ

" z

™f k™f k

œ

2xi  2yj  2zk 2 È x #  y #  z#

œ xi
yj
zk

dA (Section 16.5, Example 5, with a œ 1) Ê ' ' ™ ‚ (yi) † n d5 S

œ ' ' (z) ˆ "z dA‰ œ  ' ' dA œ 1, where R is the disk x#
y# Ÿ 1 in the xy-plane. R

R

11. Let S" and S# be oriented surfaces that span C and that induce the same positive direction on C. Then

' ' ™ ‚ F † n" d5" œ ) F † dr œ ' ' ™ ‚ F † n# d5# C S"

12.

S#

' ' ™ ‚ F † n d5 œ ' ' ™ ‚ F † n d5
' ' ™ ‚ F † n d5, and since S" and S# are joined by the simple S

S"

S#

closed curve C, each of the above integrals will be equal to a circulation integral on C. But for one surface the circulation will be counterclockwise, and for the other surface the circulation will be clockwise. Since the integrands are the same, the sum will be 0 Ê

' ' ™ ‚ F † n d5 œ 0. S

â i j k ââ â â â 13. ™ ‚ F œ â ``x ``y ``z â œ 5i
2j
3k ; rr œ (cos ))i
(sin ))j  2rk and r) œ (r sin ))i
(r cos ))j â â â 2z 3x 5y â â â i j k â â â â r) sin ) 2r â œ a2r# cos )b i
a2r# sin )b j
rk ; n œ krrrr ‚ Ê rr ‚ r) œ â cos ) ‚r) k and d5 œ krr ‚ r) k dr d) â â 0 â â r sin ) r cos ) Ê ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) œ a10r# cos )
4r# sin )
3rb dr d) Ê

' ' ™ ‚ F † n d5

21 2 21 # œ '0 '0 a10r# cos )
4r# sin )
3rb dr d) œ '0
"30 r$ cos )
43 r$ sin )
#3 r# ‘ ! d) 21 32 ‰ œ '0 ˆ 80 3 cos )
3 sin )
6 d) œ 6(21) œ 121

S

â i j k ââ â â ` â ` ` # # 14. ™ ‚ F œ â ` x `y ` z â œ i  2j  2k ; rr ‚ r) œ a2r cos )b i
a2r sin )b j
rk and â â ây  z z  x x
zâ ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê

' ' ™ ‚ F † n d5 S

21 3 21 $ œ '0 '0 a2r# cos )  4r# sin )  2rb dr d) œ '0
 23 r$ cos )  43 r$ sin )  r# ‘ ! d) 21 œ '0 a18 cos )  36 sin )  9b d) œ 9(21) œ 181

â i â â 15. ™ ‚ F œ â ``x â # âx y

j ` `y $

â k ââ i â â â ` $ # ` z â œ 2y i
0j  x k ; rr ‚ r) œ â cos ) â â â r sin ) 3z â

j sin ) r cos )

â kâ â 1â â 0â

2y z œ (r cos ))i  (r sin ))j
rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê

' ' ™ ‚ F † n d5 œ ' ' a2ry$ cos )  rx# b dr d) œ ' ' a2r% sin3 ) cos )  r$ cos# )b dr d) 0 0 21

S

1

R

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1038 Chapter 16 Integration in Vector Fields œ '0 ˆ 25 sin3 ) cos )  21

" 4

" cos# )‰ d) œ
10 sin4 )  4" ˆ #)


sin 2) ‰‘ #1 4 !

œ  14

â i â â j k ââ i j k â â â â ` â â â ` ` sin ) 1 â 16. ™ ‚ F œ â ` x `y ` z â œ i
j
k ; rr ‚ r) œ â cos ) â â â â âx  y y  z z  xâ â r sin ) r cos ) 0 â œ (r cos ))i
(r sin ))j
rk and ™ ‚ F † n d5 œ ( ™ ‚ F) † (rr ‚ r) ) dr d) (see Exercise 13 above) Ê

' ' ™ ‚ F † n d5 œ ' ' (r cos )
r sin )
r) dr d) œ ' ’(cos )
sin )
1) r## “ d) œ ˆ 25# ‰ (21) œ 251 0 0 0 21

&

21

5

!

S

â i j k ââ â â ` â ` ` 17. ™ ‚ F œ â ` x â œ 0i
0j  5k ; `y `z â â â 3y 5  2x z#  2 â â â i j k â â âÈ â r9 ‚ r) œ â 3 cos 9 cos ) È3 cos 9 sin ) È3 sin 9 â â â â È3 sin 9 sin ) È3 sin 9 cos ) â 0 # # œ a3 sin 9 cos )b i
a3 sin 9 sin )b j
(3 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise 13 above) Ê

' ' ™ ‚ F † n d5 œ ' ' 15 cos 9 sin 9 d9 d) œ '
152 cos# 9‘ 1! Î# d) œ '  15# d) œ 151 0 0 0 0 21

1/2

21

21

S

â i j k ââ â â ` â ` 18. ™ ‚ F œ â ` x ` y ``z â œ 2zi  j  2yk ; â # â ây z# x â â â i j k â â â â r9 ‚ r) œ â 2 cos 9 cos ) 2 cos 9 sin ) 2 sin 9 â â â 0 â 2 sin 9 sin ) 2 sin 9 cos ) â œ a4 sin# 9 cos )b i
a4 sin# 9 sin )b j
(4 sin 9 cos 9)k ; ™ ‚ F † n d5 œ ( ™ ‚ F) † (r9 ‚ r) ) d9 d) (see Exercise 13 above) Ê œ '0

21

'0

1/2

' ' ™ ‚ F † n d5 œ ' ' a8z sin# 9 cos )  4 sin# 9 sin )  8y sin 9 cos )b d9 d) R

S #

a16 sin 9 cos 9 cos )  4 sin# 9 sin )  16 sin# 9 sin ) cos )b d9 d)

œ '0
 "36 sin$ 9 cos )  4 ˆ 9#  21

sin 29 ‰ (sin 4

))  16 ˆ 9# 

sin 29 ‰ (sin 4

1Î#

) cos ))‘ !

d)

# ‘ #1 ‰
"6 œ '0 ˆ 16 3 cos )  1 sin )  41 sin ) cos ) d) œ  3 sin )
1 cos )  21 sin ) ! œ 0 21

19. (a) F œ 2xi
2yj
2zk Ê curl F œ 0 Ê

)C F † dr œ ' '

™ ‚ F † n d5 œ ' ' 0 d 5 œ 0

S

S

# # $

(b) Let f(xß yß z) œ x y z Ê ™ ‚ F œ ™ ‚ ™ f œ 0 Ê curl F œ 0 Ê

)C F † dr œ ' '

™ ‚ F † n d5

S

œ ' ' 0 d5 œ 0 S

(c) F œ ™ ‚ (xi
yj
zk) œ 0 Ê ™ ‚ F œ 0 Ê )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 (d) F œ ™ f Ê ™ ‚ F œ ™ ‚ ™ f œ 0

Ê )C F † dr œ ' '

S

S

™ ‚ F † n d5 œ ' ' 0 d5 œ 0

S

20. F œ ™ f œ  "# ax#
y#
z# b #
$Î#


$Î#

(2x)i  "# ax#
y#
z# b #
$Î#

œ x ax#
y#
z b i  y a x#
y#
z b (a) r œ (a cos t)i
(a sin t)j , 0 Ÿ t Ÿ 21 Ê Ê F† œ ˆ

dr dt

œ x ax#
y#
z# b

a cos t ‰ ( a a$

sin t) 


$Î#

t‰ ˆ a sin (a a$

dr dt


$Î#

S

(2y)j  "# ax#
y#
z# b


$Î#

(2z)k

#
$Î#

j  z a x#
y #
z b k œ (a sin t)i
(a cos t)j

(a sin t)  y ax#
y#
z# b


$Î#

(a cos t)

cos t) œ 0 Ê )C F † dr œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.8 The Divergence Theorem and a Unified Theory 1039

)C F † dr œ ' '

(b)

™ ‚ F † n d5 œ ' ' ™ ‚ ™ f † n d5 œ ' ' 0 † n d5 œ ' ' 0 d 5 œ 0

S

S

S

â i â â 21. Let F œ 2yi
3zj  xk Ê ™ ‚ F œ â ``x â â 2y

S

k ââ ` â ` z â œ  3i
j  2k ; n œ â x â

j ` `y

3z

)C 2y dx
3z dy  x dz œ )C F † dr œ ' '

Ê ™ ‚ F † n œ 2 Ê

2i  2j  k 3

™ ‚ F † n d5 œ ' ' 2 d5

S

S

œ 2 ' ' d5, where ' ' d5 is the area of the region enclosed by C on the plane S: 2x
2y
z œ 2 S

S

â i â â 22. ™ ‚ F œ â ``x â â x

k ââ ` â `z â œ 0 â z â

j ` `y

y

23. Suppose F œ Mi
Nj
Pk exists such that ™ ‚ F œ Š `` Py  œ xi
yj
zk . Then ` #M ` y` z

Ê



#

` #P ` y` x

Ê Š ``x`Py 

` `x

œ 1 and

#

` P ` y` x ‹

Š `` Py  ` `z

#

`N `z ‹

Š ``Nx 


Š ``z`Nx 

œ

` `x

(x) Ê

`M `y ‹

œ

` `z

#

` N ` x` z ‹

` #P ` x` y

` #N ` z` x

(z) Ê

#


Š ``y`Mz 



#

` M ` z` y ‹

`N `z ‹ i

` #N ` x` z




ˆ ``Mz 

`P ‰ `x j ` `y

œ 1. Likewise,

` #M ` z` y


Š ``Nx 

ˆ ``Mz 

`M `y ‹ k

`P ‰ `x

œ

` `y

(y)

œ 1. Summing the calculated equations

œ 3 or 0 œ 3 (assuming the second mixed partials are

equal). This result is a contradiction, so there is no field F such that curl F œ xi
yj
zk . 24. Yes: If ™ ‚ F œ 0 , then the circulation of F around the boundary C of any oriented surface S in the domain of

F is zero. The reason is this: By Stokes's theorem, circulation œ )C F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 † n d5 S

S

œ 0. #

25. r œ Èx#
y# Ê r% œ ax#
y# b Ê F œ ™ ar% b œ 4x ax#
y# b i
4y ax#
y# b j œ Mi
Nj Ê

)C

™ ar% b † n ds œ )C F † n ds œ )C M dy  N dx œ ' ' Š ``Mx
R

`N `y ‹

dx dy

œ ' ' c4 ax#
y# b
8x#
4 ax#
y# b
8y# d dA œ ' ' 16 ax#
y# b dA œ 16 ' ' x# dA
16 ' ' y# dA R

R

R

R

œ 16Iy
16Ix . 26.

`P `y

œ 0,

`N `z

œ 0,

`M `z

œ 0,

`P `x

œ 0,

`N `x

œ

y#
x# ax #  y # b #

,

`M `y

However, x#
y# œ 1 Ê r œ (cos t)i
(sin t)j Ê Ê F œ a sin tb i
acos tb j Ê F †

œ

y#
x# a x #  y # b#

#

dr dt

œ (sin t)i
(cos t)j

#

Ê curl F œ ’ axy# 
yx# b# 

y#
x# “k a x #  y # b#

#1

dr dt

œ sin# t
cos# t œ 1 Ê )C F † dr œ ) 1 dt œ 21 which is !

not zero. 16.8 THE DIVERGENCE THEOREM AND A UNIFIED THEORY 1. F œ


y i  x j È x#  y#

Ê div F œ

xy
xy ax#  y# b$Î#

œ0

i  yj  zk) 3. F œ  GM(x Ê div F œ GM ” ax # # # $Î# ax  y  z b

 GM ” ax

#

$Î#

"Î#

 y#  z# b
3y# ax#  y#  z# b ax #  y #  z # b $

2. F œ xi
yj Ê div F œ 1
1 œ 2 #

$Î#

"Î#

 y#  z# b
3x# ax#  y#  z# b ax #  y #  z # b $

•  GM ”

$Î#

• "Î#

ax#  y#  z# b
3z# ax#  y#  z# b a x #  y #  z # b$

•

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 0.

1040 Chapter 16 Integration in Vector Fields #

# #

#

#

#

#

#

#

#

z b ax  y  z b œ GM ’ 3 ax y z b
# 3 ax# y #(Î# “œ0 ax  y  z b

4. z œ a#  r# in cylindrical coordinates Ê z œ a#  ax#
y# b Ê v œ aa#  x#  y# b k Ê div v œ 0 5.

` `x

` `y

(y  x) œ 1,

` `z

(z  y) œ 1,

œ 16 6.

` `x

ax# b œ 2x,

` `y

ay# b œ 2y,

(a) Flux œ '0

` `x

(y  x) œ 0 Ê ™ † F œ 2 Ê Flux œ 'c1 'c1 'c1 2 dx dy dz œ 2 a2$ b 1

1

1

az# b œ 2z Ê ™ † F œ 2x
2y
2z

'01 '01 (2x
2y
2z) dx dy dz œ '01 '01 cx#
2x(y
z)d "! dy dz œ '01 '01 (1
2y
2z) dy dz 1 1 " " œ '0 cy(1
2z)
y# d ! dz œ '0 (2
2z) dz œ c2z
z# d ! œ 3 1 1 1 1 1 1 1 " Flux œ 'c1 'c1 'c1 (2x
2y
2z) dx dy dz œ 'c1 'c1 cx#
2x(y
z)d
" dy dz œ 'c1 'c1 (4y
4z) dy dz 1 1 " " œ 'c1 c2y#
4yzd
" dz œ 'c1 8z dz œ c4z# d
" œ 0 1

(b)

(c) In cylindrical coordinates, Flux œ ' ' ' (2x
2y
2z) dx dy dz D

1 21 2 1 21 # œ '0 '0 '0 (2r cos )
2r sin )
2z) r dr d) dz œ '0 '0
23 r$ cos )
23 r$ sin )
zr# ‘ ! d) dz 1 21 16 ' 1
"36 sin )  163 cos )
4z)‘ #!1 dz œ '01 81z dz œ c41z# d "! œ 41 ‰ œ '0 '0 ˆ 16 3 cos )
3 sin )
4z d) dz œ 0

7.

` `x

(y) œ 0,

` `y

(xy) œ x,

` `z

(z) œ 1 Ê ™ † F œ x  1; z œ x#
y# Ê z œ r# in cylindrical coordinates

Ê Flux œ ' ' ' (x  1) dz dy dx œ '0

21

D

'02 '0r

#

(r cos )  1) dz r dr d) œ '0

21

'02 ar$ cos )  r# b r dr d)

‰
32 ‘ #1 œ '0 ’ r5 cos )  r4 “ d) œ '0 ˆ 32 5 cos )  4 d) œ 5 sin )  4) ! œ 81 21

&

%

#

21

!

8.

` `x

ax# b œ 2x,

` `y

(xz) œ 0,

` `z

(3z) œ 3 Ê ™ † F œ 2x
3 Ê Flux œ ' ' ' (2x
3) dV

œ '0

'0 '0 (23 sin 9 cos )
3) a3# sin 9b d3 d9 d) œ '0 '0 ’ 32

œ '0

'0

21

21

1 1

21

2

(8 sin 9 cos )
8) sin 9 d9 d) œ '

21


8 ˆ 92 0

1

sin 29 ‰ 4



%

D

#

sin 9 cos )
3$ “ sin 9 d9 d) !

cos )  8 cos 9‘ ! d) œ '0 (41 cos )
16) d) 1

21

œ 321 9.

` `x

ax# b œ 2x,

œ '0

1Î2

10.

` `x

` `y

'0 '0 1Î2

2

(2xy) œ 2x,

` `z

(3xz) œ 3x Ê Flux œ ' ' ' 3x dx dy dz

(33 sin 9 cos )) a3 sin 9b d3 d9 d) œ '0

1Î2

#

a6x#
2xyb œ 12x
2y,

` `y

a2y
x# zb œ 2,

Ê Flux œ ' ' ' (12x
2y
2) dV œ '0

3

D

` `z

'0

D

1Î2

12 sin# 9 cos ) d9 d) œ '0 31 cos ) d) œ 31 1Î2

a4x# y$ b œ 0 Ê ™ † F œ 12x
2y
2

'01Î2 '02 (12r cos )
2r sin )
2) r dr d) dz

1Î2 3 16 ‰ ' 3ˆ ‰ œ '0 '0 ˆ32 cos )
16 3 sin )
4 d) dz œ 0 32
21
3 dz œ 112
61

11.

` `x

` `y

(2xz) œ 2z,

œ '0

2

È16 c 4x

'0

#

(xy) œ x, 4cy

'0

` `z

az# b œ 2z Ê ™ † F œ x Ê Flux œ ' ' ' x dV

x dz dy dx œ '0

2

œ ’4x#  "2 x%
"3 a16  4x# b

È16 c 4x

'0

D #

(xy  4x) dy dx œ '0 ’ 12 x a16  4x# b  4xÈ16  4x# “ dx 2

$Î# #

“ œ  40 3 !

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.8 The Divergence Theorem and a Unified Theory 1041 12.

` `x

` `y

ax$ b œ 3x# ,

œ 3 '0

21

'0 '0 1

a

` `z

ay$ b œ 3y# ,

3 a3 sin 9b d3 d9 d) œ 3 '0 #

21

#

`3 `x

13. Let 3 œ Èx#
y#
z# . Then œ

y# 3


3,

` `z

az$ b œ 3z# Ê ™ † F œ 3x#
3y#
3z# Ê Flux œ ' ' ' 3 ax#
y#
z# b dV

(3z) œ Š `` 3z ‹ z
3 œ

Ê Flux œ ' ' ' 43 dV œ '0

21

D

" 3

Š 3y ‹ œ



#

y 3$

Ê Flux œ ' ' '

2 3

D

15.

` `x

` `z

and

Š 3z ‹ œ

dV œ '0

21

x 3

z# 3

,

`3 `y

œ

" 3



œ

y 3

a& 0 5 `3 `z

,

sin 9 d9 d) œ 3 '0

21

œ

Ê

z 3

È2

x 3

, #

z 3$

` `x

21

œ

y 3

`3 `z

,

œ

Ê ™ †Fœ

z 3 3 3

Ê

` `x

#

#



d) œ

#

œ

 Š 3x# ‹

` `y

ay$
ey sin zb œ 3y#
ey sin z,

` `z

` `x

D

Ê ™ †Fœ 21

`3 `x

œ

1 3



x# 3$

. Similarly,

È2

'01 '1

a153# b a3# sin 9b d3 d9 d)

'01 Š12È2  3‹ sin 9 d9 d) œ '021 Š24È2  6‹ d) œ Š48È2  12‹ 1

cln ax#
y# bd œ

œ '0

(3y) œ Š `` 3y ‹ y
3

a5z$
ey cos zb œ 15z#  ey sin z 21

16.

` `y

2 3

Ê ™ † F œ 15x#
15y#
15z# œ 153# Ê Flux œ ' ' ' 153# dV œ '0 21


3,

'01 3 sin 9 d9 d) œ '021 6 d) œ 121 " 3

Š 3x ‹ œ

x y z 3$

x# 3

'01 '12 Š 32 ‹ a3# sin 9b d3 d9 d) œ '021 '01 3 sin 9 d9 d) œ '021 6 d) œ 121

a5x$
12xy# b œ 15x#
12y# ,

œ '0

121a& 5


33 œ 43, since 3 œ Èx#
y#
z#

(43) a3# sin 9b d3 d9 d) œ '0

`3 `y

D 2a& 5

(3x) œ Š `` 3x ‹ x
3 œ

x #  y #  z# 3


3 Ê ™ †Fœ

'01 '1

`3 `x

14. Let 3 œ Èx#
y#
z# . Then ` `y

œ

'

1

2x x#  y#

È

'1 2 'c21

x#

2x `  y# , ` y



2z x#  y#

) ˆ 2r cos  r#


" y ‰ ˆ 2z ˆ 2z ‰ x tan x œ  x –

Š "x ‹ y #

1  ˆx‰

2z — œ  x#  y# ,


Èx#
y# Ê Flux œ ' ' ' Š x# 2x  y# 

2z r#


r‰ dz r dr d) œ '0

21

È2

'1

D

ˆ6 cos ) 

3 r

2z x#  y#

` `z

ˆzÈx#
y# ‰ œ Èx#
y#


Èx#
y# ‹ dz dy dx


3r# ‰ dr d)

œ '0 ’6 ŠÈ2  1‹ cos )  3 ln È2
2È2  1“ d) œ 21 Š 3# ln 2
2È2  1‹ 21

17. (a) G œ Mi
Nj
Pk Ê ™ ‚ G œ curl G œ Š `` Py  ` `x

œ div(curl G) œ œ

` #P ` x` y



` #N ` x` z




Š `` Py 

` #M ` y` z



`N `z ‹

` #P ` y` x







` `y

` #N ` z` x

ˆ ``Mz  

` #M ` z` y

`P ‰ `x




`N `z ‹ i

` `z


ˆ ``Mz 

Š ``Nx 

`P ‰ `x k


Š ``Nx 

`M `y ‹ k

Ê ™ † ™ ‚G

`M `y ‹

œ 0 if all first and second partial derivatives are continuous

(b) By the Divergence Theorem, the outward flux of ™ ‚ G across a closed surface is zero because outward flux of ™ ‚ G œ ' ' ( ™ ‚ G) † n d5 œ ' ' ' ™ † ™ ‚ G dV

S

[Divergence Theorem with F œ ™ ‚ G]

D

œ ' ' ' (0) dV œ 0

[by part (a)]

D

18. (a) Let F" œ M" i
N" j
P" k and F# œ M# i
N# j
P# k Ê aF"
bF# œ (aM"
bM# )i
(aN"
bN# )j
(aP"
bP# )k Ê ™ † (aF"
bF# ) œ ˆa

` M" `x


b ``Mx# ‰
Ša

œ a Š ``Mx"


` N" `y




` P" `z ‹

` N" `y


b ``Ny# ‹
ˆa


b Š ``Mx#


` N# `y




` P" `z

` P# `z ‹


b ``Pz# ‰ œ a( ™ † F" )
b( ™ † F# )

(b) Define F" and F# as in part a Ê ™ ‚ (aF"
bF# ) œ ’Ša ``Py"
b ``Py# ‹  ˆa ``Nz"
b ``Nz# ‰“ i

ˆa ``Mz"
b ``Mz# ‰  ˆa ``Px"
b ``Px# ‰‘ j

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1042 Chapter 16 Integration in Vector Fields
’ˆa ``Nx"
b ``Nx# ‰  Ša ``My"
b ``My# ‹“ k œ a ’Š ``Py" 

` N" `z ‹ i


ˆ ``Mz" 

` P" ‰ `x j


Š ``Nx" 

` M" ` y ‹ k“


b ’Š ``Py#  ``Nz# ‹ i
ˆ ``Mz#  ``Px# ‰ j
Š ``Nx#  ``My# ‹ k“ œ a ™ ‚ F"
b ™ ‚ F# â â j kâ â i â â (c) F" ‚ F# œ â M" N" P" â œ (N" P#  P" N# )i  (M" P#  P" M# )j
(M" N#  N" M# )k Ê ™ † (F" ‚ F# ) â â â M# N# P# â œ ™ † [(N" P#  P" N# )i  (M" P#  P" M# )j
(M" N#  N" M# )k] œ ``x (N" P#  P" N# )  ``y (M" P#  P" M# )
``z (M" N#  N" M# ) œ ˆP#  ŠM"

` P# `y

` M" `y


P#

œ M# Š ``Py" 
P" Š ``My# 

` N" `z ‹ ` N# `x ‹

 P"

` M# `y

 M#


N# ˆ ``Mz" 

` P" `y ‹

` P" ‰ `x

` N# `z


ˆM"


P# Š ``Nx" 


N#

` M" `y ‹

` M" `z

 N"

` N" `x

` M# `z


M" Š ``Nz# 


N"

 M#

` P# `y ‹

` P# `x

 N#

` P" `x

 P"

` N# ‰ `x

` N" ‰ `z


N" ˆ ``Px# 

` M# ‰ `z

œ F# † ™ ‚ F"  F" † ™ ‚ F#

19. (a) div(gF) œ ™ † gF œ

` `x

(gM)


` `y

(gN)


œ ŠM `` gx
N `` gy
P `` gz ‹
g Š ``Mx
(b) ™ ‚ (gF) œ ’ ``y (gP) 

` `z

`N `y

` `z




(gP) œ Šg ``Mx
M `` xg ‹
Šg ``Ny
N `` yg ‹
Šg ``Pz
P `` gz ‹

`P `z ‹

œg™ †F
™g†F

(gN)“ i

``z (gM) 

` `x

(gP)‘ j
’ ``x (gN) 

` `y

(gM)“ k

œ ŠP `` gy
g `` Py  N `` gz  g ``Nz ‹ i
ŠM `` gz
g ``Mz  P `` gx  g `` Px ‹ j
ŠN `` gx
g ``Nx  M `` gy  g ``My ‹ k œ ŠP `` gy  N `` gz ‹ i
Šg `` Py  g ``Nz ‹ i
ŠM `` gz  P `` xg ‹ j
ˆg ``Mz  g `` Px ‰ j
ŠN `` gx  M `` gy ‹ k
Šg ``Nx  g ``My ‹ k œ g ™ ‚ F
™ g ‚ F 20. Let F" œ M" i
N" j
P" k and F# œ M# i
N# j
P# k . (a) F" ‚ F# œ (N" P#  P" N# )i
(P" M#  M" P# )j
(M" N#  N" M# )k Ê ™ ‚ (F" ‚ F# ) œ ’ ``y (M" N#  N" M# ) 
’ ``x (P" M#  M" P# ) 

` `z

` `y

(P" M#  M" P# )“ i

``z (N" P#  P" N# ) 

` `x

(M" N#  N" M# )‘ j

(N" P#  P" N# )“ k ` ` ` y (M" N#  N" M# )  ` z (P" M#  M" P# ) N" ``My#  M# ``Pz"  P" ``Mz#
P# ``Mz"
M" ``Pz#

and consider the i-component only: œ N#

` M" `y

` N# `y


M"

œ ŠN#

` M" `y


P#

œ ŠM#

` M" `x


N#

 Š ``Mx"
œ ŠM#

` N" `y

` M" `x





N#

 M#

` M" `z ‹ ` M" `y

` N" `y

 ŠN"


P#



` M# `y

` M" `z ‹

` M# `z ‹


P"

 ŠM"

` M# `x


Š ``Ny#



N"

` M# `y

` P# ` z ‹ M"


P"

` M# `z ‹

` P" ` z ‹ M# .

Now, i-comp of (F# † ™ )F" œ ŠM#

` M" `y

` M" `z ‹ ;


P#

i-comp of ( ™ † F# )F" œ Š ``Mx#


 Š ``Ny"



Š ``Mx#


` `x


N#

` `y

likewise, i-comp of (F" † ™ )F# œ ŠM"

` N# `y




` P# ` z ‹ M"

` P" ` z ‹ M# ` N# `y


P#

` M# `x




` P# ` z ‹ M"

` ` z ‹ M"


N"

` M# `y

and i-comp of ( ™ † F" )F# œ Š ``Mx"



P"

` N" `y




` M# `z ‹ ; ` P" ` z ‹ M# .

Similar results hold for the j and k components of ™ ‚ (F" ‚ F# ). In summary, since the corresponding components are equal, we have the result ™ ‚ (F " ‚ F # ) œ (F # † ™ )F "  (F " † ™ )F #
( ™ † F # )F "  ( ™ † F " )F # (b) Here again we consider only the i-component of each expression. Thus, the i-comp of ™ (F" † F# ) œ ``x (M" M#
N" N#
P" P# ) œ ˆM" ``Mx#
M# ``Mx"
N" ``Nx#
N# ``Nx"
P" ``Px#
P# ``Px" ‰ i-comp of (F" † ™ )F# œ ŠM"

` M# `x


N"

` M# `y


P"

` M# `z ‹ ,

i-comp of (F# † ™ )F" œ ŠM#

` M" `x


N#

` M" `y


P#

` M" `z ‹ ,

i-comp of F" ‚ ( ™ ‚ F# ) œ N" Š ``Nx# 

` M# `y ‹

 P" ˆ ``Mz# 

` P# ‰ `x ,

and

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Section 16.8 The Divergence Theorem and a Unified Theory 1043 i-comp of F# ‚ ( ™ ‚ F" ) œ N# Š ``Nx" 

` M" `y ‹

 P# ˆ ``Mz" 

` P" ‰ `x .

Since corresponding components are equal, we see that ™ (F" † F# ) œ (F" † ™ )F#
(F# † ™ )F"
F" ‚ ( ™ ‚ F# )
F# ‚ ( ™ ‚ F" ), as claimed. 21. The integral's value never exceeds the surface area of S. Since kFk Ÿ 1, we have kF † nk œ kFk knk Ÿ (1)(1) œ 1 and

' ' ' ™ † F d5 œ ' ' F † n d 5 D

[Divergence Theorem]

S

Ÿ ' ' kF † nk d5

[A property of integrals]

S

Ÿ ' ' (1) d5

ckF † nk Ÿ 1d

S

œ Area of S. 22. Yes, the outward flux through the top is 5. The reason is this: Since ™ † F œ ™ † (xi  2yj
(z
3)k œ 1  2
1 œ 0, the outward flux across the closed cubelike surface is 0 by the Divergence Theorem. The flux across the top is therefore the negative of the flux across the sides and base. Routine calculations show that the sum of these latter fluxes is 5. (The flux across the sides that lie in the xz-plane and the yz-plane are 0, while the flux across the xy-plane is 3.) Therefore the flux across the top is 5. 23. (a)

` `x

(x) œ 1,

` `y

(y) œ 1,

` `z

(z) œ 1 Ê ™ † F œ 3 Ê Flux œ ' ' ' 3 dV œ 3 ' ' ' dV D

D

œ 3(Volume of the solid)

(b) If F is orthogonal to n at every point of S, then F † n œ 0 everywhere Ê Flux œ ' ' F † n d5 œ 0. S

But the flux is 3(Volume of the solid) Á 0, so F is not orthogonal to n at every point. 24. ™ † F œ 2x  4y  6z
12 Ê Flux œ '0 '0 a

œ '0

a

`f `a

b

'01 (2x  4y  6z
12) dz dy dx

'0b (2x  4y
9) dy dx œ '0a a2xb  2b#
9bb dx œ a# b  2ab#
9ab œ ab(a  2b
9) œ f(aß b);

œ 2ab  2b#
9b and

`f `b

œ a#  4ab
9a so that

`f `a

œ 0 and

`f `b

œ 0 Ê b(2a  2b
9) œ 0 and

a(a  4b
9) œ 0 Ê b œ 0 or 2a  2b
9 œ 0, and a œ 0 or a  4b
9 œ 0. Now b œ 0 or a œ 0 Ê Flux œ 0; 2a  2b
9 œ 0 and a  4b
9 œ 0 Ê 3a  9 œ 0 Ê a œ 3 Ê b œ 3# so that f ˆ3ß 3# ‰ œ

27 #

maximum flux. 25.

' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV Ê D

S

D

' ' F † n d5 œ ' ' ' dV œ Volume of D

" 3

D

S

26. F œ C Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 0 dV œ 0 D

S

D

27. (a) From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' ™ # f dV œ ' ' ' 0 dV œ 0 D

S

D

D

(b) From the Divergence Theorem, ' ' f ™ f † n d5 œ ' ' ' ™ † f ™ f dV. Now, D

S

f ™ f œ ˆf

`f ‰ `x i


Šf

`f `y ‹ j


ˆf

`f ‰ `z k

#

#

#

#

œ f ™ # f
k ™ f k# œ 0
k ™ f k# since f is harmonic Ê

D

28. From the Divergence Theorem, ' ' ™ f † n d5 œ ' ' ' ™ † ™ f dV œ ' ' ' Š `` xf#
#

f(xß yß z) œ ln Èx#
y#
z# œ

" #

#

' ' f ™ f † n d5 œ ' ' ' k ™ f k# dV, as claimed. S

S

#

Ê ™ † f ™ f œ ’f `` xf#
ˆ `` xf ‰ “
”f `` yf#
Š `` yf ‹ •
’f `` zf#
ˆ `` zf ‰ “

D

ln ax#
y#
z# b Ê

D

`f `x

œ

x x #  y #  z#

,

`f `y

œ

y x #  y #  z#

,

` #f ` y#




` #f ` z# ‹

`f `z

œ

z x #  y #  z#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

dV. Now,

is the

1044 Chapter 16 Integration in Vector Fields Ê

` #f ` x#


x #  y #  z # ax #  y #  z # b #

x # y #  z # ax #  y #  z # b #

œ

œ '0

1Î2

29.

œ

œ

` #f ` y#

,

" x #  y #  z#

x #
y #  z# a x #  y #  z # b#

œ Ê

,

` #f ` z#

œ

x# y#
z# a x #  y #  z# b #

' ' ™ f † n d5 œ ' ' ' D

S

, Ê

dV x #  y #  z#

` #f ` x#




` #f ` y#

œ '0

1Î2




` #f ` z#

'01Î2 '0a 3

#

sin 9 3#

d 3 d9 d )

'01Î2 a sin 9 d9 d) œ '01Î2 ca cos 9d 1! Î# d) œ '01Î2 a d) œ 1#a

' ' f ™ g † n d5 œ ' ' ' ™ † f ™ g dV œ ' ' ' ™ † Šf D

S

œ ' ' ' Šf `` xg#
#

D

D #

`f `g `x `x

œ ' ' ' ’f Š `` xg#
#

D

` #g ` y#


f `` yg#





` #g ` z# ‹

`f `g `y `y


Š `` xf

`g `x

#


f `` zg#



`f `g `y `y

`f `g `z `z ‹




`g `x

i
f `` gy j
f `` gz k‹ dV

dV

`f `g ` z ` z ‹“

dV œ ' ' ' af ™ # g
™ f † ™ gb dV D

30. By Exercise 29, ' ' f ™ g † n d5 œ ' ' ' af ™ # g
™ f † ™ gb dV and by interchanging the roles of f and g, D

S

' ' g ™ f † n d5 œ ' ' ' ag ™ # f
™ g † ™ f b dV. Subtracting the second equation from the first yields: D

S

' ' a f ™ g  g ™ f b † n d5 œ ' ' ' af ™ # g  g ™ # f b dV since ™ f † ™ g œ ™ g † ™ fÞ D

S

31. (a) The integral ' ' ' p(tß xß yß z) dV represents the mass of the fluid at any time t. The equation says that D

the instantaneous rate of change of mass is flux of the fluid through the surface S enclosing the region D: the mass decreases if the flux is outward (so the fluid flows out of D), and increases if the flow is inward (interpreting n as the outward pointing unit normal to the surface).

'''

(b)

D

`p `t

dV œ

d dt

' ' ' p dV œ  ' ' pv † n d5 œ  ' ' ' ™ † pv dV Ê D

D

S

Since the law is to hold for all regions D, ™ † pv


`p `t

`3 `t

œ  ™ † pv

œ 0, as claimed

32. (a) ™ T points in the direction of maximum change of the temperature, so if the solid is heating up at the point the temperature is greater in a region surrounding the point Ê ™ T points away from the point Ê  ™ T points toward the point Ê  ™ T points in the direction the heat flows. (b) Assuming the Law of Conservation of Mass (Exercise 31) with k ™ T œ pv and c3T œ p, we have d dt

' ' ' c3T dV œ  ' ' k ™ T † n d5 Ê the continuity equation, ™ † (k ™ T)
``t (c3T) œ 0 D

S

Ê c3 ``Tt œ  ™ † (k ™ T) œ k ™ # T Ê

`T `t

œ

k c3

™ # T œ K ™ # T, as claimed

CHAPTER 16 PRACTICE EXERCISES 1. Path 1: r œ ti
tj
tk Ê x œ t, y œ t, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ 3  3t# and dz dt

dx dt

dy dt

œ 1,

‰#
Š dy ˆ dz ‰# dt œ È3 dt Ê ' f(xß yß z) ds œ ' È3 a3  3t# b dt œ 2È3 œ 1 Ê Êˆ dx dt dt ‹
dt C 0 #

1

Path 2: r" œ ti
tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 2t  3t#
3 and dz dt

œ 1,

dx dt

œ 1,

dy dt

‰#
Š dy ˆ dz ‰# dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 a2t  3t#
3b dt œ 3È2 ; œ 0 Ê Êˆ dx dt dt ‹
dt 0 #

1

C"

r# œ i
j
tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2  2t and #

#

'C

#

‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt Ê

'C f(xß yß z) ds œ 'C

"

#

dx dt

œ 0,

dy dt

œ 0,

dz dt

œ1

f(xß yß z) ds œ '0 (2  2t) dt œ 1 1

f(xß yß z) ds
'C f(xß yß z) œ 3È2
1 #

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

œ 1,

Chapter 16 Practice Exercises 1045 2. Path 1: r" œ ti Ê x œ t, y œ 0, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t# and

dx dt

‰
Š dy ˆ dz ‰ dt œ dt Ê ' f(xß yß z) ds œ ' t# dt œ Ê Êˆ dx dt dt ‹
dt C 0 #

#

#

1

"

r# œ i
tj Ê x œ 1, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ 1
t and #

#

#

‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt

'C

#

dx dt

#

#

#

'C

$

" 3

f(xß yß z) ds œ '0 (1
t) dt œ 1

3 2

1

Path 2: r% œ ti
tj Ê x œ t, y œ t, z œ 0 Ê f(g(t)ß h(t)ß k(t)) œ t#
t and

œ 0,

dy dt

dx dt

œ 1,

1

%

Ê

'Path 2

'C

$

f(xß yß z) ds œ %

5 6

È2


3 #

#

#

#

'C

&

#

#

#

#

#

#

'C

(

œ 0,

dx dt

dx dt

œ 0,

dz dt

œ1

œ 0,

dy dt

œ 1,

dz dt

œ0

œ 1,

dy dt

œ 0,

dz dt

œ0

1

f(xß yß z) ds œ '0 t# dt œ 1

dx dt

1 3

Ê 'Path 3 f(xß yß z) ds œ 'C f(xß yß z) ds
'C f(xß yß z) ds
'C f(xß yß z) ds œ  "#  &

dy dt

f(xß yß z) ds œ '0 (t  1) dt œ  12 ;

'

'C

È2;

œ0

1

r( œ ti
j
k Ê x œ t, y œ 1, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t# and ‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt

5 6

dz dt

f(xß yß z) ds œ '0 t dt œ  2" ;

r' œ tj
k Ê x œ 0, y œ t, z œ 1, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t  1 and ‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt

œ 1,

5È 2  9 6

œ

Path 3: r& œ tk Ê x œ 0, y œ 0, z œ t, 0 Ÿ t Ÿ 1 Ê f(g(t)ß h(t)ß k(t)) œ t and ‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt

dy dt

3 2

f(xß yß z) ds œ 'C f(xß yß z) ds
'C f(xß yß z) ds œ $

œ1

dz dt

10 3

‰#
Š dy ˆ dz ‰# dt œ È2 dt Ê ' f(xß yß z) ds œ ' È2 at#
tb dt œ Ê Êˆ dx dt dt ‹
dt C 0 r$ œ i
j
tk (see above) Ê

œ0

dz dt

3 2

$

#

œ0

dz dt

;

œ 0,

dx dt

f(xß yß z) ds œ '0 (2  t) dt œ

#

œ 1,

dy dt

Ê 'Path 1 f(xß yß z) ds œ 'C f(xß yß z) ds
'C f(xß yß z) ds
'C f(xß yß z) ds œ "

œ 0,

dy dt

;

œ 0,

r$ œ i
j
tk Ê x œ 1, y œ 1, z œ t Ê f(g(t)ß h(t)ß k(t)) œ 2  t and ‰
Š dy ˆ dz ‰ dt œ dt Ê Ê Êˆ dx dt dt ‹
dt

œ 1,

'

(

" #




" 3

œ  23

3. r œ (a cos t)j
(a sin t)k Ê x œ 0, y œ a cos t, z œ a sin t Ê fag(t)ß h(t)ß k(t)b œ Èa# sin# t œ a ksin tk and dx dt

œ 0,

Ê

dy dt

œ a sin t,

dz dt

#

#

#

‰
Š dy ˆ dz ‰ dt œ a dt œ a cos t Ê Êˆ dx dt dt ‹
dt

'C f(xß yß z) ds œ '021 a# ksin tk dt œ '01 a# sin t dt
'121 a# sin t dt œ 4a#

4. r œ (cos t
t sin t)i
(sin t  t cos t)j Ê x œ cos t
t sin t, y œ sin t  t cos t, z œ 0 Ê fag(t)ß h(t)ß k(t)b œ È(cos t
t sin t)#
(sin t  t cos t)# œ È1
t# and dx œ  sin t
sin t
t cos t dt

œ t cos t,

dy dt

œ cos t  cos t
t sin t œ t sin t,

dz dt

œ0 Ê

‰# ʈ dx dt

#

#

ˆ dz ‰ dt
Š dy dt ‹
dt

È3

œ Èt# cos# t
t# sin# t dt œ ktk dt œ t dt since 0 Ÿ t Ÿ È3 Ê 'C f(xß yß z) ds œ '0 5.

`P `y

œ  "# (x
y
z)
$Î# œ

`N `z

,

`M `z

Ê M dx
N dy
P dz is exact; œ

" Èx  y  z

Ê

`g `y

`f `x

œ  "# (x
y
z)
$Î# œ œ

" Èx  y  z

`P `x

,

`N `x

œ  "# (x
y
z)
$Î# œ

Ê f(xß yß z) œ 2Èx
y
z
g(yß z) Ê

œ 0 Ê g(yß z) œ h(z) Ê f(xß yß z) œ 2Èx
y
z
h(z) Ê

`f `z

Ð4ß
3ß0Ñ

Ê hw (x) œ 0 Ê h(z) œ C Ê f(xß yß z) œ 2Èx
y
z
C Ê 'Ð
1ß1ß1Ñ œ f(4ß 3ß 0)  f(1ß 1ß 1) œ 2È1  2È1 œ 0 œ

" Èx  y  z

tÈ1
t# dt œ

œ

`f `y

`M `y

œ

" Èx  y  z

dx  dy  dz Èx  y  z

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

7 3

" Èx  y  z


hw (z)




`g `y

1046 Chapter 16 Integration in Vector Fields 6.

`P `y

œ  #È"yz œ

`N `z

`f `y

œ x
g(yß z) Ê

7.

Ê

`f `z

Ê

' 1101 13 3

`M `z

,

`M `z

œ

œ0œ `g `y

`P `x

,

`N `x

œ0œ

`M `y

Ê M dx
N dy
T dz is exact;

`f `x

œ 1 Ê f(xß yß z)

œ  É yz Ê g(yß z) œ 2Èyz
h(z) Ê f(xß yß z) œ x  2Èyz
h(z)

œ  É yz
hw (z) œ  É yz Ê hw (z) œ 0 Ê h(z) œ C Ê f(xß yß z) œ x  2Èyz
C Ð

ß ß Ñ

Ð ß ß Ñ

dx  É yz dy  É yz dz œ f(10ß 3ß 3)  f(1ß 1ß 1) œ (10  2 † 3)  (1  2 † 1) œ 4
1 œ 5 `P `x

œ y cos z Á y cos z œ

Ê F is not conservative; r œ a2 cos tbi
a2 sin tbj  k, 0 Ÿ t Ÿ 21

Ê dr œ a2 sin tbi  a2 cos tbj Ê 'C F † dr œ '0 c a2 sin tbasina1bba2 sin tb
a2 cos tbasina1bba2 cos tbd dt 21

œ 4 sina1b'0 asin2 t
cos2 tbdt œ 81 sina1b 21

8.

`P `y

œ0œ

`N `z

,

`M `z

œ0œ

`P `x

,

`N `x

œ 3x# œ

9. Let M œ 8x sin y and N œ 8y cos x Ê œ ' ' (8y sin x  8x cos y) dy dx œ '0

1/2

R

`M `y

`M `y

'C F † dr œ 0

Ê F is conservative Ê œ 8x cos y and

'01/2

`N `x

œ 8y sin x Ê

'C 8x sin y dx  8y cos x dy

(8y sin x  8x cos y) dy dx œ '0 a1# sin x  8xb dx 1/2

œ 1 #
1 # œ 0 10. Let M œ y# and N œ x# Ê œ '0

21

`M `y

œ 2y and

`N `x

'02 (2r cos )  2r sin )) r dr d) œ '021

œ 2x Ê 'C y# dx
x# dy œ ' ' (2x  2y) dx dy R

16 3

(cos )  sin )) d) œ 0

11. Let z œ 1  x  y Ê fx (xß y) œ 1 and fy (xß y) œ 1 Ê Éfx#
fy#
1 œ È3 Ê Surface Area œ ' ' È3 dx dy R

œ È3(Area of the circular region in the xy-plane) œ 1È3 12. ™ f œ 3i
2yj
2zk , p œ i Ê k ™ f k œ È9
4y#
4z# and k ™ f † pk œ 3 Ê Surface Area œ ' ' R

È9  4y#  4z# 3

dy dz œ '0

21

È3 È9  4r#

'0

r dr d) œ

3

" 3

'021 Š 47 È21  49 ‹ d) œ 16 Š7È21  9‹

13. ™ f œ 2xi
2yj
2zk , p œ k Ê k ™ f k œ È4x#
4y#
4z# œ 2Èx#
y#
z# œ 2 and k ™ f † pk œ k2zk œ 2z since z   0 Ê Surface Area œ ' ' R È#

'021 ’È1  r# “ "Î !

2 2z

dA œ ' ' R

d) œ '0 Š1  21

" È2 ‹

" z

dA œ ' ' R

d) œ 21 Š1 

" È 1
x#
y#

dx dy œ '0

21

È2

'01Î

" È 1
r#

r dr d)

" È2 ‹

14. (a) ™ f œ 2xi
2yj
2zk , p œ k Ê k ™ f k œ È4x#
4y#
4z# œ 2Èx#
y#
z# œ 4 and k ™ f † pk œ 2z since z   0 Ê Surface Area œ ' ' R

4 2z

dA œ ' ' R

2 z

dA œ 2 '0

1/2

'02 cos

)

2 È 4
r#

r dr d) œ 41  8

(b) r œ 2 cos ) Ê dr œ 2 sin ) d); ds# œ r# d)#
dr# (Arc length in polar coordinates) Ê ds# œ (2 cos ))# d)#
dr# œ 4 cos# ) d)#
4 sin# ) d)# œ 4 d)# Ê ds œ 2 d); the height of the cylinder is z œ È4  r# œ È4  4 cos# ) œ 2 ksin )k œ 2 sin ) if 0 Ÿ ) Ÿ 1 Ê Surface Area œ

'c11/2Î2 h ds œ 2 '01/2 (2 sin ))(2 d)) œ 8

#

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 16 Practice Exercises 1047 15. f(xß yß z) œ

x a




y b




z c

œ 1 Ê ™ f œ ˆ "a ‰ i
ˆ b" ‰ j
ˆ "c ‰ k Ê k ™ f k œ É a"#


since c  0 Ê Surface Area œ ' '

É

" a#

R

" b# Š "c ‹



since the area of the triangular region R is " # kv

found by computing

" #



" c#

dA œ cÉ a"#


" b#

" c#




" b#

" c#




and p œ k Ê k ™ f † pk œ

' ' dA œ #" abcÉ a"#
b"#
c"# , R

ab. To check this result, let v œ ai
ck and w œ ai
bj; the area can be

‚ w k.

16. (a) ™ f œ 2yj  k , p œ k Ê k ™ f k œ È4y#
1 and k ™ f † pk œ 1 Ê d5 œ È4y#
1 dx dy Ê

' ' g(xß yß z) d5 œ ' ' È yz# È4y#
1 dx dy œ ' ' y ay#  1b dx dy œ ' ' ay$  yb dx dy 4y  1 c1 0 1

R

S

œ 'c1 3 ay$  yb dy œ 1

(b)

% 3 ’ y4

' ' g(xß yß z) d5 œ ' ' S

œ

$ 3 ’ y3

 y“

R

"
"

" y# # “
"



z È4y#  1

3

R

œ0

È4y#
1 dx dy œ '

1

c1

'03 ay#  1b dx dy œ 'c11 3 ay#  1b dy

œ 4

17. ™ f œ 2yj
2zk , p œ k Ê k ™ f k œ È4y#
4z# œ 2Èy#
z# œ 10 and k ™ f † pk œ 2z since z   0 5 ' ' g(xß yß z) d5 œ ' ' ax% yb ay#
z# b ˆ 5z ‰ dx dy Ê d5 œ 10 2z dx dy œ z dx dy œ R

S

œ ' ' ax% yb (25) Š È255
y# ‹ dx dy œ '0 '0 4

1

R

125y È25
y#

x dx dy œ '0

4

%

25y È25
y#

dy œ 50

18. Define the coordinate system so that the origin is at the center of the earth, the z-axis is the earth's axis (north is the positive z direction), and the xz-plane contains the earth's prime meridian. Let S denote the surface which is Wyoming so then S is part of the surface z œ aR#  x#  y# b

"Î#

. Let Rxy be the projection of S onto #

the xy-plane. The surface area of Wyoming is ' ' 1 d5 œ ' ' Ê1
ˆ `` xz ‰
Š `` yz ‹ dA #

Rxy

S

'' É Rxy

x# R #
x#


y#





1 dA œ ' '

y# R #
x#
y#

Rxy

aR #

R
y# b"Î#


x#

dA œ ')

)# "

'RRsinsin45°49° R aR#  r# b
"Î# r dr d)

(where )" and )# are the radian equivalent to 104°3w and 111°3w , respectively) œ ') R aR#  r# b )# "

"Î# R sin 49°

¹

R sin 45°

œ ') R aR#  R# sin# 45°b )#

"Î#

"

œ ()#  )" )R# (cos 45°  cos 49°) œ

71 180

 R aR#  R# sin# 49°b

R# (cos 45°  cos 49°) œ

71 180

"Î#

d)

(3959)# (cos 45°  cos 49°)

¸ 97,751 sq. mi. 19. A possible parametrization is r(9ß )) œ (6 sin 9 cos ))i
(6 sin 9 sin ))j
(6 cos 9)k (spherical coordinates); now 3 œ 6 and z œ 3 Ê 3 œ 6 cos 9 Ê cos 9 œ  " Ê 9 œ 21 and z œ 3È3 Ê 3È3 œ 6 cos 9 Ê cos 9 œ

È3 #

#

Ê 9œ

1 6

Ê

1 6

Ÿ9Ÿ

21 3 ;

3

also 0 Ÿ ) Ÿ 21 #

20. A possible parametrization is r(rß )) œ (r cos ))i
(r sin ))j  Š r# ‹ k (cylindrical coordinates); #

#

now r œ Èx#
y# Ê z œ  r# and 2 Ÿ z Ÿ 0 Ê 2 Ÿ  r# Ÿ 0 Ê 4   r#   0 Ê 0 Ÿ r Ÿ 2 since r   0; also 0 Ÿ ) Ÿ 21

21. A possible parametrization is r(rß )) œ (r cos ))i
(r sin ))j
(1
r)k (cylindrical coordinates); now r œ Èx#
y# Ê z œ 1
r and 1 Ÿ z Ÿ 3 Ê 1 Ÿ 1
r Ÿ 3 Ê 0 Ÿ r Ÿ 2; also 0 Ÿ ) Ÿ 21 22. A possible parametrization is r(xß y) œ xi
yj
ˆ3  x  y# ‰ k for 0 Ÿ x Ÿ 2 and 0 Ÿ y Ÿ 2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

" c

1048 Chapter 16 Integration in Vector Fields 23. Let x œ u cos v and z œ u sin v, where u œ Èx#
z# and v is the angle in the xz-plane with the x-axis Ê r(uß v) œ (u cos v)i
2u# j
(u sin v)k is a possible parametrization; 0 Ÿ y Ÿ 2 Ê 2u# Ÿ 2 Ê u# Ÿ 1 Ê 0 Ÿ u Ÿ 1 since u   0; also, for just the upper half of the paraboloid, 0 Ÿ v Ÿ 1 24. A possible parametrization is ŠÈ10 sin 9 cos )‹ i
ŠÈ10 sin 9 sin )‹ j
ŠÈ10 cos 9) k , 0 Ÿ 9 Ÿ 0Ÿ)Ÿ

1 #

â âi â 25. ru œ i
j , rv œ i  j
k Ê ru ‚ rv œ â " â â"

j " "

Ê Surface Area œ ' ' kru ‚ rv k du dv œ '0

1

Ruv

and

â kâ â 0 â œ i  j  2k Ê kru ‚ rv k œ È6 â "â È6 du dv œ È6

' ' axy  z# b d5 œ ' ' c(u
v)(u  v)  v# d È6 du dv œ È6' ' au#  2v# b du dv 0 0 0 0 1

26.

'01

1 2

S

1

1

1

" œ È6 '0 ’ u3  2uv# “ dv œ È6 '0 ˆ 3"  2v# ‰ dv œ È6
3" v  32 v$ ‘ ! œ  1

"

$

1

!

È6 3

œ  É 32

â â i j kâ â â â sin ) 0 â 27. rr œ (cos ))i
(sin ))j , r) œ (r sin ))i
(r cos ))j
k Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) " â œ (sin ))i  (cos ))j
rk Ê krr ‚ r) k œ Èsin# )
cos# )
r# œ È1
r# Ê Surface Area œ ' ' krr ‚ r) k dr d) œ '0

21

'0 È1
r# dr d) œ '0

21

1

’ 2r È1
r#


" #

ln Šr
È1
r# ‹“ d) œ '0 ’ "# È2
"

21

!

Rr) " #

ln Š1
È2‹“ d)

œ 1 ’È2
ln Š1
È2‹“

28.

' ' Èx#
y#
1 d5 œ ' ' Èr# cos# )
r# sin# )
1 È1
r# dr d) œ ' ' a1
r# b dr d) 0 0 0 0 21

S

œ '0 ’r
r3 “ d) œ '0 21

"

$

21

4 3

d) œ

,

`N `x

œ0œ

`M `z

œ


3xz ax#  y#  z# b
&Î#

!

`N `z

`P `y

œ0œ

30.

`P `y

œ

31.

`P `y

œ 0 Á yez œ

32.

`P `y

œ

33.

`f `x

œ 2 Ê f(xß yß z) œ 2x
g(yß z) Ê

œ0œ


3zy ax#  y#  z# b
&Î#

x (x  yz)#

œ

œ

`N `z `N `z

,

`N `z

`P `x

,

`M `z

œ


y (x  yz)#

œ

`P `x

Ê f(xß yß z) œ 2x
y#
zy
z
C `f `x

`M `y

Ê Conservative ,

`N `x


z (x  yz)#

œ

œ

`P `x


3xy ax#  y#  z# b
&Î#

œ

œ

`M `y

Ê Conservative

Ê Not Conservative

Ê f(xß yß z) œ 2x
y#
zy
h(z) Ê

34.

1

1

8 3

29.

,

`M `z

21

1

,

`N `x

œ

`g `f ` y œ ` y œ 2y
z `f w ` z œ y
h (z) œ

œ z cos xz Ê f(xß yß z) œ sin xz
g(yß z) Ê

Ê f(xß yß z) œ sin xz
ey
h(z) Ê Ê f(xß yß z) œ sin xz
ey
C

`M `y

`f `z

`f `y

œ

`g `y w

Ê Conservative

Ê g(yß z) œ y#
zy
h(z) y
1 Ê hw (z) œ 1 Ê h(z) œ z
C

œ ey Ê g(yß z) œ ey
h(z)

œ x cos xz
h (z) œ x cos xz Ê hw (z) œ 0 Ê h(z) œ C

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 16 Practice Exercises 1049 35. Over Path 1: r œ ti
tj
tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i
j
k) dt Ê F œ 2t# i
j
t# k Ê F † dr œ a3t#
1b dt Ê Work œ '0 a3t#
1b dt œ 2; 1

Over Path 2: r" œ ti
tj , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ 0 and dr" œ (i
j) dt Ê F" œ 2t# i
j
t# k Ê F" † dr" œ a2t#
1b dt Ê Work" œ '0 a2t#
1b dt œ 1

5 3

; r# œ i
j
tk , 0 Ÿ t Ÿ 1 Ê x œ 1, y œ 1, z œ t and

dr# œ k dt Ê F# œ 2i
j
k Ê F# † dr# œ dt Ê Work# œ '0 dt œ 1 Ê Work œ Work"
Work# œ 1

5 3


1œ

8 3

36. Over Path 1: r œ ti
tj
tk , 0 Ÿ t Ÿ 1 Ê x œ t, y œ t, z œ t and dr œ (i
j
k) dt Ê F œ 2t# i
t# j
k Ê F † dr œ a3t#
1b dt Ê Work œ '0 a3t#
1b dt œ 2; 1

Over Path 2: Since f is conservative, )C F † dr œ 0 around any simple closed curve C. Thus consider

'curve

F † dr œ 'C F † dr
'C F † dr , where C" is the path from (0ß 0ß 0) to (1ß 1ß 0) to ("ß "ß ") and C# is the path "

#

from (1ß 1ß 1) to (!ß !ß !). Now, from Path 1 above, 'C F † dr œ 2 Ê 0 œ 'curve F † dr œ 'C F † dr
(2) Ê 'C

"

#

"

F † dr œ 2

37. (a) r œ aet cos tb i
aet sin tb j Ê x œ et cos t, y œ et sin t from (1ß 0) to ae21 ß 0b Ê 0 Ÿ t Ÿ 21 Ê

dr dt

œ aet cos t  et sin tb i
aet sin t
et cos tb j and F œ

t‰ t‰ œ ˆ cos i
ˆ sin e2t e2t j Ê F †

dr dt

#

œ Š coset t 

sin t cos t et




sin# t et




xi  yj ax#  y# b$Î#

sin t cos t ‹ et

œ

aet cos tb i  aet sin tb j ae2t cos# t  e2t sin# tb$Î#

œ e
t

Ê Work œ '0 e
t dt œ 1  e
21 21

(b) F œ œ

xi  yj ax#  y# b$Î#

`f `x

Ê

œ

x ax#  y# b$Î#

Ê f(xß yß z) œ  ax#
y# b

Ê g(yß z) œ C Ê f(xß yß z) œ  ax#
y# b

y ax#  y# b$Î# 21


"Î#


"Î#


g(yß z) Ê

`f `y

œ

y ax#  y# b$Î#

is a potential function for F Ê




`g `y

'C F † dr

œ f ae ß 0b  f(1ß 0) œ 1  e
21 38. (a) F œ ™ ax# zey b Ê F is conservative Ê (b)

'C F † dr œ ' 1 0 0

Ð1ß0ß21Ñ

Ð ß ß Ñ

â i â â 39. ™ ‚ F œ â ``x â # ây

y

Ê

)C

6 7

F † dr œ ' ' R

â i â â 40. ™ ‚ F œ â ``x â # âx
y nœ

" È2

j


" È2

™ ax# zey b † dr œ ax# zey bk Ð"ß!ß#1Ñ  ax# zey bk Ð"ß!ß!Ñ œ 21  0 œ 21

k ââ ` â ` z â œ 2yk; unit normal to the plane is n œ â 3z# â

j ` `y

Ê ™ ‚F†nœ

)C F † dr œ 0 for any closed path C

2i  6j
3k È4  36  9

œ

2 7

i
67 j  37 k

y; p œ k and f(xß yß z) œ 2x
6y  3z Ê k ™ f † pk œ 3 Ê d5 œ 6 7

y d5 œ ' ' ˆ 67 y‰ ˆ 73 dA‰ œ ' ' 2y dA œ '0

21

R

j

R

k ™f k k™f†pk

'0 2r sin ) r dr d) œ '0

21

1

dA œ 2 3

7 3

dA

sin ) d) œ 0

â â â ` â œ 8yi ; the circle lies in the plane f(xß yß z) œ y
z œ 0 with unit normal `z â 4y#  z â k

` `y

x
y

k Ê ™ ‚F†nœ0 Ê

)C

F † dr œ ' ' ™ ‚ F † n d5 œ ' ' 0 d5 œ 0 R

R

41. (a) r œ È2ti
È2tj
a4  t# b k , 0 Ÿ t Ÿ 1 Ê x œ È2t, y œ È2t, z œ 4  t# Ê

dx dt

œ È2,

dy dt

œ È2,

dz dt

œ 2t

‰
Š dy ˆ dz ‰ dt œ È4
4t# dt Ê M œ ' $ (xß yß z) ds œ ' 3tÈ4
4t# dt œ
"4 (4
4t)$Î# ‘ Ê Êˆ dx dt dt ‹
dt ! C 0 #

#

#

1

œ 4È2  2

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

"

1050 Chapter 16 Integration in Vector Fields (b) M œ 'C $ (xß yß z) ds œ '0 È4
4t# dt œ ’tÈ1
t#
ln Št
È1
t# ‹“ œ È2
ln Š1
È2‹ "

1

!

42. r œ ti
2tj
23 t$Î# k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ 2t, z œ

2 $Î# 3 t

Ê

œ 1,

dx dt

dy dt

œ 2,

œ t"Î#

dz dt

‰
Š dy ˆ dz ‰ dt œ Èt
5 dt Ê M œ ' $ (xß yß z) ds œ ' 3È5
t Èt
5 dt Ê Êˆ dx dt dt ‹
dt C 0 #

#

#

2

œ '0 3(t
5) dt œ 36; Myz œ 'C x$ ds œ '0 3t(t
5) dt œ 38; Mxz œ 'C y$ ds œ '0 6t(t
5) dt œ 76; 2

2

Mxy œ 'C z$ ds œ '0 2t$Î# (t
5) dt œ 2

œ

4 7

144 7

2

Myz M

È2 Ê x œ

œ

38 36

œ

,yœ

Mxz M

œ

t# #

dx dt

œ 1,

19 18

76 36

œ

19 9

Mxy M

,zœ

œ

È 2‹ Š 144 7 36

È2 È

#

43. r œ ti
Š 2 3 2 t$Î# ‹ j
Š t# ‹ k , 0 Ÿ t Ÿ 2 Ê x œ t, y œ #

#

2È2 $Î# , 3 t

zœ

Ê

œ È2 t"Î# ,

dy dt

œt

dz dt

#

‰
Š dy ˆ dz ‰ dt œ È1
2t
t# dt œ È(t
1)# dt œ kt
1k dt œ (t
1) dt on the domain given. Ê Êˆ dx dt dt ‹
dt Then M œ 'C $ ds œ '0 ˆ t " 1 ‰ (t
1) dt œ '0 dt œ 2; Myz œ 'C x$ ds œ '0 t ˆ t " 1 ‰ (t
1) dt œ '0 t dt œ 2; 2

2

2

Mxz œ 'C y$ ds œ '0 Š 2 3 2 t$Î# ‹ ˆ t " 1 ‰ (t
1) dt œ '0 È

2

œ '0 Š t# ‹ ˆ t"1 ‰ (t
1) dt œ '0 2

œ

ˆ 43 ‰ #

2 # t

#

œ

2 3

2È2 $Î# 3 t

2

#

dt œ

4 3

Ê xœ

Myz M

œ

; Ix œ 'C ay#
z# b $ ds œ '0 Š 89 t$
t4 ‹ dt œ 2

%

Iz œ 'C ay#
x# b $ ds œ '0 ˆt#
89 t$ ‰ dt œ 2

Iz Rz œ É M œÊ

Š 56 9 ‹ #

œ

56 9 ;

2 #

dt œ

32 15

œ 1; y œ

232 45

2

; Mxy œ 'C z$ ds œ

Mxz M

ˆ 32 ‰ 15 #

œ

16 15

;zœ

Mxy M

; Iy œ 'C ax#
z# b $ ds œ '0 Št#
t4 ‹ dt œ 2

Ix Rx œ É M œÊ

Š 232 45 ‹ #

œ

2È29 3È 5

%

I

; Ry œ É My œ Ê

Š 64 15 ‹ #

œ

4È 2 È15

64 15

;

;

2È 7 3

44. z œ 0 because the arch is in the xy-plane, and x œ 0 because the mass is distributed symmetrically with respect #

#

#

‰
Š dy ˆ dz ‰ dt to the y-axis; r(t) œ (a cos t)i
(a sin t)j , 0 Ÿ t Ÿ 1 Ê ds œ ʈ dx dt dt ‹
dt œ È(a sin t)#
(a cos t)# dt œ a dt, since a   0; M œ 'C $ ds œ 'C (2a  y) ds œ '0 (2a  a sin t) a dt 1

œ 2a2 1  2a# ; Mxz œ 'C y$ dt œ 'C y(2a  y) ds œ '0 (a sin t)(2a  a sin t) dt œ '0 a2a# sin t  a# sin# tb dt 1

œ
2a# cos t  a# ˆ 2t 

sin 2t ‰ 4

‘ 1 œ 4a#  !

a# 1 #

Ê yœ

1

Š4a#


a# 1 # ‹

2a# 1
2a#

œ

8
1 41
4

Ê axß yß zb œ ˆ0ß 481

14 ß 0‰

45. r(t) œ aet cos tb i
aet sin tb j
et k , 0 Ÿ t Ÿ ln 2 Ê x œ et cos t, y œ et sin t, z œ et Ê dy dt

œ aet sin t
et cos tb,

dz dt

#

#

dx dt

œ aet cos t  et sin tb ,

#

‰
Š dy ˆ dz ‰ dt œ et Ê Êˆ dx dt dt ‹
dt

œ Éaet cos t  et sin tb#
aet sin t
et cos tb#
aet b# dt œ È3e2t dt œ È3 et dt; M œ 'C $ ds œ œ È3; Mxy œ 'C z$ ds œ '0

ln 2

ŠÈ3 et ‹ aet b dt œ '

ln 2

0

È3 e2t dt œ

3È 3 #

Ê zœ

Mxy M

Iz œ 'C ax#
y# b $ ds œ '0 ae2t cos# t
e2t sin# tb ŠÈ3 et ‹ dt œ '0 È3 e3t dt œ ln 2

ln 2

œ

7È 3 3

Œ

3

È3 #




È3

œ

3 #

'0ln 2 È3 et dt ;

Iz Ê Rz œ É M

È

œ Ê 73È33 œ É 73 46. r(t) œ (2 sin t)i
(2 cos t)j
3tk , 0 Ÿ t Ÿ 21 Ê x œ 2 sin t, y œ 2 cos t, z œ 3t Ê dz dt

dx dt

œ 2 cos t,

dy dt

œ 2 sin t,

‰
Š dy ˆ dz ‰ dt œ È4
9 dt œ È13 dt; M œ ' $ ds œ ' $ È13 dt œ 21$ È13; œ 3 Ê Êˆ dx dt dt ‹
dt C 0 #

#

#

21

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 16 Practice Exercises 1051 Mxy œ 'C z$ ds œ '0 (3t) Š$ È13‹ dt œ 6$1# È13; Myz œ 'C x$ ds œ '0 (2 sin t) Š$ È13‹ dt œ 0; 21

21

Mxz œ 'C y$ ds œ '0 (2 cos t) Š$ È13‹ dt œ 0 Ê x œ y œ 0 and z œ 21

Mxy M

6$1# È13 2$1È13

œ

œ 31 Ê (!ß !ß 31) is the

center of mass 47. Because of symmetry x œ y œ 0. Let f(xß yß z) œ x#
y#
z# œ 25 Ê ™ f œ 2xi
2yj
2zk Ê k ™ f k œ È4x#
4y#
4z# œ 10 and p œ k Ê k ™ f † pk œ 2z, since z   0 Ê M œ ' ' $ (xß yß z) d5 R

' ' 5 dA œ 5(Area of the circular region) œ 801; Mxy œ ' ' z$ d5 œ ' ' 5z dA ‰ œ ' ' z ˆ 10 #z dA œ R

R

R

œ ' ' 5È25  x#  y# dx dy œ '0

21

R

'0 Š5È25  r# ‹ r dr d) œ '0

21

4

490 3

d) œ

1 Ê zœ

980 3

' ' ax#
y# b $ d5 œ ' ' 5 ax#
y# b dx dy œ ' ‰ Ê axß yß zb œ ˆ0ß !ß 49 12 ; Iz œ 0

21

R

R

R

Š 980 3 1‹ 801

œ

49 12

'04 5r$ dr d) œ '021 320 d) œ 6401;

Iz 1 È Rz œ É M œ É 640 801 œ 2 2

48. On the face z œ 1: g(xß yß z) œ z œ 1 and p œ k Ê ™ g œ k Ê k ™ gk œ 1 and k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax#
y# b dA œ 2 '0

'0sec

1/4

R

)

r$ dr d) œ

2 3

; On the face z œ 0: g(xß yß z) œ z œ 0 Ê ™ g œ k and p œ k

Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax#
y# b dA œ R

; On the face y œ 0: g(xß yß z) œ y œ 0

2 3

Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax#
0b dA œ '0

1

R

'01 x# dx dz œ "3 ;

On the face y œ 1: g(xß yß z) œ y œ 1 Ê ™ g œ j and p œ j Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' ax#
1# b dA œ '0

1

R

'01 ax#
1b dx dz œ 43 ; On the face x œ 1:

g(xß yß z) œ x œ 1 Ê ™ g œ i and p œ i

Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' a1#
y# b dA œ '0

1

R

'01 a1
y# b dy dz œ 43 ; On the face

x œ 0: g(xß yß z) œ x œ 0 Ê ™ g œ i and p œ i Ê k ™ gk œ 1 Ê k ™ g † pk œ 1 Ê d5 œ dA Ê I œ ' ' a0#
y# b dA œ '0

1

R

'01 y# dy dz œ 13

49. M œ 2xy
x and N œ xy  y Ê

`M `x

œ ' ' (2y
1
x  1) dy dx œ '0

1

R

œ ' ' (y  2x) dy dx œ '0

1

R

œ ' ' (12x
2y) dx dy œ R

R

`M `y ‹

R

`M `y ‹

œ 2x,




`N `x

2 3




œ y,

" 3




`N `y

4 3

œ 12x,

`M `y

R

œ 1,

`N `x

œ 1,

`N `y

4 3




" 3

œ

14 3

œ x  1 Ê Flux œ ' ' Š ``Mx


'01 (2y
x) dy dx œ 3# ; Circ œ ' '

`M `x




R

Š ``Nx 

`M `y ‹

œ 2y Ê Flux œ ' ' Š ``Mx
R

dx dy œ ' ' (1  1) dx dy œ 0 R `M `y

œ

dx dy œ ' ' Š sinx y  R

sin y x

and

sin y x ‹

`N `x

œ

sin y x

Ê )C ln x sin y dy 

`N `y ‹

dx dy

'01 'y1 (12x
2y) dx dy œ '01 a4y#
2y  6b dy œ  113 ;

51. M œ  cosx y and N œ ln x sin y Ê œ ' ' Š ``Nx 

`M `y

2 3

'01 (y  2x) dy dx œ  "#

50. M œ y  6x# and N œ x
y# Ê

Circ œ ' ' Š ``Nx 

œ 2y
1,

Ê Iz œ

cos y x

dx

dx dy œ 0

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

`N `y ‹

dx dy

dx dy

1052 Chapter 16 Integration in Vector Fields `M `x

52. (a) Let M œ x and N œ y Ê

œ 1,

`M `y

œ 0,

`N `x

œ 0,

`N `y

œ 1 Ê Flux œ ' ' Š ``Mx
R

œ ' ' (1
1) dx dy œ 2 ' ' dx dy œ 2(Area of the region) R

`N `y ‹

dx dy

R

(b) Let C be a closed curve to which Green's Theorem applies and let n be the unit normal vector to C. Let F œ xi
yj and assume F is orthogonal to n at every point of C. Then the flux density of F at every point of C is 0 since F † n œ 0 at every point of C Ê ``Mx
``Ny œ 0 at every point of C Ê Flux œ ' ' Š ``Mx
R

`N `y ‹

dx dy œ ' ' 0 dx dy œ 0. But part (a) above states that the flux is R

2(Area of the region) Ê the area of the region would be 0 Ê contradiction. Therefore, F cannot be orthogonal to n at every point of C. 53.

` `x

` `y

(2xy) œ 2y,

(2yz) œ 2z,

(2xz) œ 2x Ê ™ † F œ 2y
2z
2x Ê Flux œ ' ' ' (2x
2y
2z) dV

` `z

D

1 1 1 1 1 1 œ '0 '0 '0 (2x
2y
2z) dx dy dz œ '0 '0 (1
2y
2z) dy dz œ '0 (2
2z) dz œ 3

54.

` `x

œ '0

21

55.

` `x

` `y

(xz) œ z,

(yz) œ z,

È25 c r

'0 '3 4

(2x) œ 2,

#

` `z

(1) œ 0 Ê ™ † F œ 2z Ê Flux œ ' ' ' 2z r dr d) dz D

2z dz r dr d) œ '0

` `y

21

` `z

(3y) œ 3,

'0 ra16  r b dr d) œ '0 4

21

#

(z) œ 1 Ê ™ † F œ 4; x#
y#
z# œ 2 and x#
y# œ z Ê z œ 1

Ê x#
y# œ 1 Ê Flux œ ' ' ' 4 dV œ 4 '0

21

œ 4'0 Š 21

56.

` `x

D

7 12

(6x
y) œ 6,


È2‹ d) œ 2 3

` `y

'01 'r

È2cr

#

#

dz r dr d) œ 4 '0

21

'01 ŠrÈ2  r#  r$ ‹ dr d)

1 Š7  8È2‹

2 3

(x  z) œ 0,

64 d) œ 1281

` `z

(4yz) œ 4y Ê ™ † F œ 6
4y; z œ Èx#
y# œ r

Ê Flux œ ' ' ' (6
4y) dV œ '0

1/2

D

'01 '0r (6
4r sin )) dz r dr d) œ '01/2 '01 a6r#
4r$ sin )b dr d)

1/2 œ '0 (2
sin )) d) œ 1
1

57. F œ yi
zj
xk Ê ™ † F œ 0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ 0 D

S

58. F œ 3xz# i
yj  z$ k Ê ™ † F œ 3z#
1  3z# œ 1 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ '0

4

È16cx Î2

'0

#

D

S

'0

yÎ2

1 dz dy dx œ '

4


x # Š 1616 ‹ 0

dx œ ’x 

% x$ 48 “ !

œ

8 3

59. F œ xy# i
x# yj
yk Ê ™ † F œ y#
x#
0 Ê Flux œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' ax
y b dV œ '0 #

D

#

21

'0 'c1 r 1

1

#

dz r dr d) œ '0

21

'0 2r 1

S $

dr d) œ '0

21

D " #

d) œ 1

60. (a) F œ (3z
1)k Ê ™ † F œ 3 Ê Flux across the hemisphere œ ' ' F † n d5 œ ' ' ' ™ † F dV œ ' ' ' 3 dV œ 3 ˆ "# ‰ ˆ 43 1a$ ‰ œ 21a$

S

D

D

(b) f(xß yß z) œ x#
y#
z#  a# œ 0 Ê ™ f œ 2xi
2yj
2zk Ê k ™ f k œ È4x#
4y#
4z# œ È4a# œ 2a since a   0 Ê n œ 2xi  2y#aj  2zk œ xi  yaj  zk Ê F † n œ (3z
1) ˆ za ‰ ; p œ k Ê ™ f † p œ ™ f † k œ 2z

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 16 Additional and Advanced Exercises 1053 k™f k k™f†pk

Ê k ™ f † pk œ 2z since z   0 Ê d5 œ

œ

2a 2z

dA œ

a z

dA Ê

Rxy

S

œ ' ' (3z
1) dx dy œ ' ' ˆ3Èa#  x#  y#
1‰ dx dy œ '0

21

Rxy

' ' F † n d5 œ ' ' (3z
1) ˆ za ‰ ˆ az ‰ dA

Rxy

'0 Š3Èa#  r#
1‹ r dr d) a

œ '0 Š a#
a$ ‹ d) œ 1a#
21a$ , which is the flux across the hemisphere. Across the base we find 21

#

F œ [3(0)
1]k œ k since z œ 0 in the xy-plane Ê n œ k (outward normal) Ê F † n œ 1 Ê Flux across the base œ ' ' F † n d5 œ ' ' 1 dx dy œ 1a# . Therefore, the total flux across the closed surface is Rxy

S #

$

#

$

a1a
21a b  1a œ 21a . CHAPTER 16 ADDITIONAL AND ADVANCED EXERCISES 1. dx œ (2 sin t
2 sin 2t) dt and dy œ (2 cos t  2 cos 2t) dt; Area œ

" #

)C x dy  y dx

'021 [(2 cos t  cos 2t)(2 cos t  2 cos 2t)  (2 sin t  sin 2t)(2 sin t
2 sin 2t)] dt 21 21 œ "# '0 [6  (6 cos t cos 2t
6 sin t sin 2t)] dt œ "# '0 (6  6 cos t) dt œ 61 œ

" #

2. dx œ (2 sin t  2 sin 2t) dt and dy œ (2 cos t  2 cos 2t) dt; Area œ

" #

)C

x dy  y dx

'0 [(2 cos t
cos 2t)(2 cos t  2 cos 2t)  (2 sin t  sin 2t)(2 sin t  2 sin 2t)] dt 21 21 #1 œ "# '0 [2  2(cos t cos 2t  sin t sin 2t)] dt œ "# '0 (2  2 cos 3t) dt œ "#
2t  23 sin 3t‘ ! œ 21 œ

" #

21

3. dx œ cos 2t dt and dy œ cos t dt; Area œ œ

" #

" #

)C

x dy  y dx œ

" #

'01 ˆ "# sin 2t cos t  sin t cos 2t‰ dt

" #

)C

'01 csin t cos# t  (sin t) a2 cos# t  1bd dt œ "# '01 a sin t cos# t
sin tb dt œ "#
3" cos$ t  cos t‘ !1 œ  3"
1 œ 23

4. dx œ (2a sin t  2a cos 2t) dt and dy œ (b cos t) dt; Area œ

x dy  y dx

'0 ca2ab cos# t  ab cos t sin 2tb  a2ab sin# t  2ab sin t cos 2tbd dt 21 21 œ "# '0 c2ab  2ab cos# t sin t
2ab(sin t) a2 cos# t  1bd dt œ "# '0 a2ab
2ab cos# t sin t  2ab sin tb dt œ

" #

œ

" #

21


2abt  23 ab cos$ t
2ab cos t‘ #1 œ 21ab !

5. (a) F(xß yß z) œ zi
xj
yk is 0 only at the point (0ß 0ß 0), and curl F(xß yß z) œ i
j
k is never 0 . (b) F(xß yß z) œ zi
yk is 0 only on the line x œ t, y œ 0, z œ 0 and curl F(xß yß z) œ i
j is never 0 . (c) F(xß yß z) œ zi is 0 only when z œ 0 (the xy-plane) and curl F(xß yß z) œ j is never 0 . 6. F œ yz# i
xz# j
2xyzk and n œ and 2xyz œ

cz R

Ê

yz x

#

œ

xz y

#

x i  y j  zk È x #  y #  z#

œ

x i  y j  zk R

, so F is parallel to n when yz# œ

œ 2xy Ê y# œ x# Ê y œ „ x and z# œ „

x#
y#
z# œ R# Ê x#
x#
2x# œ R# Ê 4x# œ R# Ê x œ „ Š R# ß R# ß 

È2R # ‹,

Š R# ß  R# ß

È2R # ‹,

Š R# ß R# ß

È2R # ‹,

Š R# ß R# ß 

È2R # ‹

Š R# ß  R# ß 

È2R # ‹,

Š R# ß  R# ß

R #

c R

cx R

, xz# œ

cy R

œ 2x# Ê z œ „ È2x. Also,

. Thus the points are: Š R# ß R# ß

È2R # ‹,

Š R# ß  R# ß 

È2R # ‹,

7. Set up the coordinate system so that (aß bß c) œ (0ß Rß 0) Ê $ (xß yß z) œ Èx#
(y  R)#
z# œ Èx#
y#
z#  2Ry
R# œ È2R#  2Ry ; let f(xß yß z) œ x#
y#
z#  R# and p œ i Ê ™ f œ 2xi
2yj
2zk Ê k ™ f k œ 2Èx#
y#
z# œ 2R Ê d5 œ

,

k™f k k™f†ik

dz dy œ

2R 2x

dz dy

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

È2R # ‹,

1054 Chapter 16 Integration in Vector Fields Ê Mass œ ' ' $ (xß yß z) d5 œ ' ' È2R#  2Ry ˆ Rx ‰ dz dy œ R ' ' Ryz

S

ÈR cy

œ 4R 'cR '0 R

#

#

È2R#
2Ry È R #
y #
z#

Ryz

È2R#
2Ry È R #
y #
z#

dz dy œ 4R 'cR É2R2  2Ry sin
1 Š ÈR2z
y2 ‹» R

dz dy

ÈR cy #

#

dy 0

1‰ 2 œ 21R'cR É2R2  2Ry dy œ 21Rˆ
3R a2R  2Ryb » R

R

3/2

cR

œ

161R$ 3

â â i j kâ â â â sin ) 0 â 8. r(rß )) œ (r cos ))i
(r sin ))j
)k , 0 Ÿ r Ÿ 1, 0 Ÿ ) Ÿ 21 Ê rr ‚ r) œ â cos ) â â â r sin ) r cos ) " â œ (sin ))i  (cos ))j
rk Ê krr ‚ r) k œ È1
r# ; $ œ 2Èx#
y# œ 2Èr# cos# )
r# sin# ) œ 2r Ê Mass œ ' ' $ (xß yß z) d5 œ '0

21

!

S

œ

41 3

'01 2rÈ1
r# dr d) œ '021 ’ 23 a1
r# b$Î# “ " d) œ '021 23 Š2È2  1‹ d)

Š 2È 2  1‹

9. M œ x#
4xy and N œ 6y Ê

`M `x

œ 2x
4y and

`N `x

œ 6 Ê Flux œ '0

b

'0a (2x
4y  6) dx dy

œ '0 aa#
4ay  6ab dy œ a# b
2ab#  6ab. We want to minimize f(aß b) œ a# b
2ab#  6ab œ ab(a
2b  6). b

Thus, fa (aß b) œ 2ab
2b#  6b œ 0 and fb (aß b) œ a#
4ab  6a œ 0 Ê b(2a
2b  6) œ 0 Ê b œ 0 or b œ a
3. Now b œ 0 Ê a#  6a œ 0 Ê a œ 0 or a œ 6 Ê (0ß 0) and (6ß 0) are critical points. On the other hand, b œ a
3 Ê a#
4a(a
3)  6a œ 0 Ê 3a#
6a œ 0 Ê a œ 0 or a œ 2 Ê (0ß 3) and (#ß ") are also critical points. The flux at (0ß 0) œ 0, the flux at (6ß 0) œ 0, the flux at (0ß 3) œ 0 and the flux at (2ß 1) œ 4. Therefore, the flux is minimized at (2ß 1) with value 4. 10. A plane through the origin has equation ax
by
cz œ 0. Consider first the case when c Á 0. Assume the plane is given by z œ ax
by and let f(xß yß z) œ x#
y#
z# œ 4. Let C denote the circle of intersection of the plane with the sphere. By Stokes's Theorem, )C F † dr œ ' ' ™ ‚ F † n d5 , where n is a unit normal to the plane. Let S â â â i j kâ â â r(xß y) œ xi
yj
(ax
by)k be a parametrization of the surface. Then rx ‚ ry œ â " 0 a â œ ai  bj
k â â â0 " bâ â i j k ââ â â â ` ` # # Ê d5 œ krx ‚ ry k dx dy œ Èa
b
1 dx dy. Also, ™ ‚ F œ â ` x ` y ``z â œ i
j
k and n œ Èai # bj
# k a b 1 â â â z x y â Ê ' ' ™ ‚ F † n d5 œ ' ' a  b
1 Èa#
b#
1 dx dy œ ' ' (a
b  1) dx dy œ (a
b  1) ' ' dx dy. Now Rxy

S #

#

#

x
y
(ax
by) œ 4 Ê

È a#  b#  1

# Š a 4 1 ‹ x#

Rxy




# Š b 4 1 ‹ y#




ellipse Ax#
Bxy
Cy# œ 1 in the xy-plane, where A œ Section 10.3, the area of the ellipse is Thus we optimize H(aß b) œ `H `b

œ

#

2(a  b
1) aa  1  b
abb aa #  b #  1 b 2 #

#

(a  b
1) a#  b#  1

21 È4AC
B#

:

`H `a

œ

œ

ˆ ab ‰ # xy a#  1 4

41 È a#  b#  1 #

Rxy

œ 1 Ê the region Rxy is the interior of the

,Bœ

ab #

, and C œ

b#  1 4

Ê )C F † dr œ h(aß b) œ

2(a  b
1) ab  1  a
abb aa #  b #  1 b 2

. By Exercise 47 in

41(a  b
1) È a#  b#  1

.

œ 0 and

œ 0 Ê a
b  1 œ 0, or b#
1
a  ab œ 0 and a#
1
b  ab œ 0

Ê a
b  1 œ 0, or a  b#
(b  a) œ 0 Ê a
b  1 œ 0, or (a  b)(a
b  1) œ 0 Ê a
b  1 œ 0 or a œ b. The critical values a
b  1 œ 0 give a saddle. If a œ b, then 0 œ b#
1
a  ab Ê a#
1
a  a# œ 0 Ê a œ 1 Ê b œ 1. Thus, the point (aß b) œ (1ß 1) gives a local extremum for )C F † dr Ê z œ x  y

Ê x
y
z œ 0 is the desired plane, if c Á 0. Note: Since h(1ß 1) is negative, the circulation about n is clockwise, so n is the correct pointing normal for Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

Chapter 16 Additional and Advanced Exercises 1055 the counterclockwise circulation. Thus ' ' ™ ‚ F † (n) d5 actually gives the maximum circulation. S

If c œ 0, one can see that the corresponding problem is equivalent to the calculation above when b œ 0, which does not lead to a local extreme. 11. (a) Partition the string into small pieces. Let ?i s be the length of the ith piece. Let (xi ß yi ) be a point in the ith piece. The work done by gravity in moving the ith piece to the x-axis is approximately Wi œ (gxi yi ?i s)yi where xi yi ?i s is approximately the mass of the ith piece. The total work done by gravity in moving the string to the x-axis is D Wi œ D gxi yi# ?i s Ê Work œ 'C gxy# ds i

(b)

i

1/2 1/2 Work œ 'C gxy# ds œ '0 g(2 cos t) a4 sin# tbÈ4 sin# t
4 cos# t dt œ 16g '0 cos t sin# t dt

1Î#

$

œ ’16g Š sin3 t ‹“

!

œ

16 3

g

' x(xy) ds ' y(xy) ds (c) x œ 'C and y œ 'C ; the mass of the string is 'C xy ds and the weight of the string is xy ds xy ds C C

g 'C xy ds. Therefore, the work done in moving the point mass at axß yb to the x-axis is W œ Šg 'C xy ds‹ y œ g 'C xy# ds œ

16 3

g.

12. (a) Partition the sheet into small pieces. Let ?i 5 be the area of the ith piece and select a point (xi ß yi ß zi ) in the ith piece. The mass of the ith piece is approximately xi yi ?i 5 . The work done by gravity in moving the ith piece to the xy-plane is approximately (gxi yi ?i 5 )zi œ gxi yi zi ?i 5 Ê Work œ ' ' gxyz d5.

(b)

1cx

' ' gxyz d5 œ g ' ' xy(1  x  y)È1
(1)#
(1)# dA œ È3g ' ' 0 0 1

S

axy  x# y  xy# b dy dx

Rxy

S

"
x œ È3g '0
"2 xy#  #" x# y#  3" xy$ ‘ ! dx œ È3g '0
6" x  #" x#
#" x$  6" x% ‘ dx 1

1

" # œ È3g
12 x  6" x$
6" x% 

" 30

"

x& ‘ ! œ È3g ˆ 1"# 

" ‰ 30

È3g 20 M œ Mxy

œ

(c) The center of mass of the sheet is the point axß yß zb where z

with Mxy œ ' ' xyz d5 and S

M œ ' ' xy d5. The work done by gravity in moving the point mass at axß yß zb to the xy-plane is S

gMz œ gM Š MMxy ‹ œ gMxy œ ' ' gxyz d5 œ S

È3g 20

.

13. (a) Partition the sphere x#
y#
(z  2)# œ 1 into small pieces. Let ?i 5 be the surface area of the ith piece and let (xi ß yi ß zi ) be a point on the ith piece. The force due to pressure on the ith piece is approximately w(4  zi )?i 5. The total force on S is approximately D w(4  zi )?i 5. This gives the actual force to be i

' ' w(4  z) d5. S

(b) The upward buoyant force is a result of the k-component of the force on the ball due to liquid pressure. The force on the ball at (xß yß z) is w(4  z)(n) œ w(z  4)n , where n is the outer unit normal at (xß yß z). Hence the k-component of this force is w(z  4)n † k œ w(z  4)k † n . The (magnitude of the) buoyant force on the ball is obtained by adding up all these k-components to obtain ' ' w(z  4)k † n d5. S

(c) The Divergence Theorem says ' ' w(z  4)k † n d5 œ ' ' ' div(w(z  4)k) dV œ ' ' ' w dV, where D S

is x#
y#
(z  2)# Ÿ 1 Ê

D

D

' ' w(z  4)k † n d5 œ w ' ' ' 1 dV œ 43 1w, the weight of the fluid if it S

D

were to occupy the region D.

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

1056 Chapter 16 Integration in Vector Fields 14. The surface S is z œ Èx#
y# from z œ 1 to z œ 2. Partition S into small pieces and let ?i 5 be the area of the ith piece. Let (xi ß yi ß zi ) be a point on the ith piece. Then the magnitude of the force on the ith piece due to liquid pressure is approximately Fi œ w(2  zi )?i 5 Ê the total force on S is approximately D Fi œ D w(2  zi )?i 5 Ê the actual force is ' ' w(2  z) d5 œ ' ' w ˆ2  Èx#
y# ‰ É1
i

Rxy

S

œ ' ' È2 w ˆ2  Èx#
y# ‰ dA œ '0

21

Rxy

œ

x# x#  y#




y# x#  y#

dA

'12 È2w(2  r) r dr d) œ '021 È2w
r#  "3 r$ ‘ #" d) œ '021 2È32w d)

4È 21 w 3

15. Assume that S is a surface to which Stokes's Theorem applies. Then )C E † dr œ ' ' ( ™ ‚ E) † n d5 œ ' '  ``Bt † n d5 œ  ``t

S

' ' B † n d5. Thus the voltage around a loop equals the negative of the rate

S

S

of change of magnetic flux through the loop. 16. According to Gauss's Law, ' ' F † n d5 œ 41GmM for any surface enclosing the origin. But if F œ ™ ‚ H S

then the integral over such a closed surface would have to be 0 by the Divergence Theorem since div F œ 0. 17.

)C

f ™ g † dr œ ' ' ™ ‚ (f ™ g) † n d5

(Stokes's Theorem)

S

œ ' ' (f ™ ‚ ™ g
™ f ‚ ™ g) † n d5

(Section 16.8, Exercise 19b)

S

œ ' ' [(f)(0)
™ f ‚ ™ g] † n d5

(Section 16.7, Equation 8)

S

œ ' ' ( ™ f ‚ ™ g) † n d5 S

18. ™ ‚ F" œ ™ ‚ F# Ê ™ ‚ (F#  F" ) œ 0 Ê F#  F" is conservative Ê F#  F" œ ™ f; also, ™ † F" œ ™ † F# Ê ™ † (F#  F" ) œ 0 Ê ™ # f œ 0 (so f is harmonic). Finally, on the surface S, ™ f † n œ (F#  F" ) † n œ F# † n  F" † n œ 0. Now, ™ † (f ™ f) œ ™ f † ™ f
f ™ # f so the Divergence Theorem gives

' ' ' k ™ f k# dV
' ' ' f ™ # f dV œ ' ' ' ™ † (f ™ f) dV œ ' ' f ™ f † n d5 œ 0, and since ™ # f œ 0 we have D

D

D

S

' ' ' k ™ f k# dV
0 œ 0 Ê ' ' ' kF#  F" k # dV œ 0 Ê F#  F" œ 0 Ê F# œ F" , as claimed. D

D

â i â â 19. False; let F œ yi
xj Á 0 Ê ™ † F œ ``x (y)
``y (x) œ 0 and ™ ‚ F œ â ``x â â x

j ` `y

y

k ââ ` â ` z â œ 0i
0j
0k œ 0 â 0 â

20. kru ‚ rv k# œ kru k# krv k# sin# ) œ kru k# krv k# a1  cos# )b œ kru k# krv k#  kru k# krv k# cos# ) œ kru k# krv k#  (ru † rv )# Ê kru ‚ rv k# œ ÈEG  F# Ê d5 œ kru ‚ rv k du dv œ ÈEG  F# du dv 21. r œ xi
yj
zk Ê ™ † r œ 1
1
1 œ 3 Ê œ ' ' r † n d5, by the Divergence Theorem " 3

' ' ' ™ † r dV œ 3 ' ' ' dV œ 3V Ê V œ "3 ' ' ' ™ † r dV D

D

S

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley

D

Chapter 16 Additional and Advanced Exercises 1057 NOTES:

Copyright (c) 2006 Pearson Education, Inc., publishing as Pearson Addison-Wesley