SOL Cálculo 2 de Varias Variables, 9na Edición - Ron Larson & Bruce H. Edwards (in)
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Transcript of SOL Cálculo 2 de Varias Variables, 9na Edición - Ron Larson & Bruce H. Edwards (in)
LIBROS UNIVERISTARIOS Y SOLUCIONARIOS DE MUCHOS DE ESTOS LIBROS GRATIS EN
DESCARGA DIRECTA
SIGUENOS EN:
VISITANOS PARA DESCARGARLOS GRATIS.
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C H A P T E R 1 0Vectors and the Geometry of Space
Section 10.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 474
Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 479
Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 483
Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 487
Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 491
Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 496
Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 499
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 503
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 507
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C H A P T E R 1 0Vectors and the Geometry of Space
Section 10.1 Vectors in the PlaneSolutions to Even-Numbered Exercises
474
2. (a)
(b)x
−1−2
−6
−5
−4
−3
−2
−1−3 2 31
v
(0, 6)−
y
v � �3 � 3, �2 � 4� � �0, �6� 4. (a)
(b)
x−3 −1−2
3
2
1
( 3, 2)−
v
y
v � � � 1 � 2, 3 � 1� � ��3, 2�
6.
u � v
v � �7 � 2, 7 � ��1�� � �5, 8�
u � �1 � ��4�, 8 � 0� � �5, 8� 8.
u � v
v � �25 � 0, 10 � 13� � �15, �3�
u � �11 � ��4�, �4 � ��1�� � �15, �3�
10. (b)
(a) and (c).
v
(1, 12)
(3, 6)
(2, −6)
−1
−4
2
4
6
8
10
12
−6
1 2 3 4 5 6 7
y
x
v � �3 � 2, 6 � ��6�� � �1, 12� 12. (b)
(a) and (c).
x
v
−6 −4 −2 2
4
2
−2
( 5, 3)−
(0, 4)−
( 5, 1)− −
y
v � ��5 � 0, �1 � ��4�� � ��5, 3�
14. (b)
(a) and (c).
xv
8642−2−4−6−8
3
2
1
−2
−3
( 10, 0)−
(7, 1)−( , 1)−3 −
y
v � ��3 � 7, �1 � ��1�� � ��10, 0� 16. (b)
(a) and (c).
x
v
1.000.750.25 1.250.50
1.25
1.00
0.75
0.50
0.25
(0.72, 0.65)
(0.84, 1.25)
(0.12, 0.60)y
v � �0.84 � 0.12, 1.25 � 0.60� � �0.72, 0.65�
18. (a)
—CONTINUED—
xv
4v
12
20
84−4−8−12
( 4, 20)−
( 1, 5)−
y
4v � ��4, 20� (b)
x
v
− v 43
2 2
2
1 5
1−1−2−3−4
1
−2
−3
( 1, 5)−
, −( (
y
�12 v � �1
2 , �52 �
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Section 10.1 Vectors in the Plane 475
18. —CONTINUED—
(c)
x
v
0v
3
6
21−1−2−3
( 1, 5)−
y
0v � �0, 0� (d)
xv
15105−5−10
−10
−15
−20
−25
−30
−15
( 1, 5)−
(6, 30)−
−6v
y
�6v � �6, �30�
20. Twice as long as given vector
x
u
2u
y
u. 22.
x
u
u v+ 2
2v
y
24. (a)
(b)
(c) 2u � 5v � 2��3, �8� � 5�8, 25� � �34, 109�
v � u � �8, 25� � ��3, �8� � �11, 33�
23u �
23��3, �8� � ��2, �16
3 � 26.
x
2
1
−1
vw
u321
y
� 3i � j � �3, 1�
v � �2i � j� � �i � 2j�
28.
−4 −2
2
−6
−8
−10
−12
4 6 8 10
5u−3w
v
y
x
v � 5u � 3w � 5�2, �1� � 3�1, 2� � �7, �11� 30.
Q � �7, �7�
u2 � �7
u1 � 7
u2 � 2 � �9
u1 � 3 � 4
32. �v� � �144 � 25 � 13 34. �v� � �100 � 9 � �109 36. �v� � �1 � 1 � �2
38.
unit vector v �u
�u��
�5, 15�5�10
� � 1�10
, 3
�10 �u� � �52 � 152 � �250 � 5�10 40.
unit vector v �u
�u��
��6.2, 3.4�5�2
� ��1.24�2
, 0.68�2
�u� � ���6.2�2 � �3.4�2 � �50 � 5�2
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476 Chapter 10 Vectors and the Geometry of Space
42.
(a)
(b)
(c)
(d)
(e)
(f)
� u � v�u � v� � � 1
u � v
�u � v��
1
�13 �3, �2�
� v�v� � � 1
v
�v��
1
3�2 �3, �3�
� u�u� � � 1
u
�u�� �0, 1�
�u � v� � �9 � 4 � �13
u � v � �3, �2�
�v� � �9 � 9 � 3�2
�u� � �0 � 1 � 1
u � �0, 1�, v � �3, �3� 44.
(a)
(b)
(c)
(d)
(e)
(f)
� u � v�u � v� � � 1
u � v
�u � v��
1
5�2 �7, 1�
� v�v� � � 1
v
�v��
1
5�2 �5, 5�
� u�u� � � 1
u
�u��
1
2�5 �2, �4�
�u � v� � �49 � 1 � 5�2
u � v � �7, 1�
�v� � �25 � 25 � 5�2
�u� � �4 � 16 � 2�5
u � �2, �4�, v � �5, 5�
46.
�u � v� ≤ �u� � �v�
�u � v� � 2
u � v � ��2, 0�
�v� � �5 2.236
v � �1, �2�
�u� � �13 3.606
u � ��3, 2� 48.
v � ��2�2, 2�2�
4� u�u�� � 2�2 ��1, 1�
u
�u��
1
�2 ��1, 1�
50.
v � �0, 3�
3� u�u�� � �0, 3�
u
�u��
13
�0, 3� 52.
� �52
i �5�3
2j
v � 5 �cos 120��i � �sin 120��j�
54.
0.9981i � 0.0610j � �0.9981, 0.0610�
v � �cos 3.5��i � �sin 3.5��j 56.
u � v � 5i � �3 j
v � i � �3 j
u � 4i
58.
u � v � 10 cos�0.5�� i
v � 5 cos�0.5�� i � 5 sin�0.5�� j
� 5 cos�0.5�� i � 5 sin�0.5�� j
u � 5 cos��0.5�� i � 5 sin��0.5�� j 60. See page 718:
u
ku
(ku1, ku2)
(u1, u2)
u1
ku1
u2
ku2u
v
u + v
(u1 + v1, u2 + v2)
(v1, v2)
(u1, u2)
u1
u2
v1
v2
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62. See Theorem 10.1, page 719.
For Exercises 64–68, au � bw � a�i � 2j� � b�i � j� � �a � b�i � �2a � b�j.
64. Therefore, Solvingsimultaneously, we have a � 1, b � �1.
2a � b � 3.a � b � 0,v � 3j. 66. Therefore, Solvingsimultaneously, we have a � 2, b � 1.
2a � b � 3.a � b � 3,v � 3i � 3j.
68. Therefore,Solving simultaneously, we have a � 2, b � �3.
2a � b � 7.a � b � �1,v � �i � 7j.
70. at
(a) Let then
(b) Let then
w�w�
� ±1
�145 �12, �1�.
w � �12, �1�,m � �112 .
w�w�
� ±1
�145 �1, 12�.
w � �1, 12�,m � 12.
x � �2.y � x3, y� � 3x2 � 12 72.
(a) Let then
(b) Let then
w�w�
� ±1
�5 ��2, 1�.
w � ��2, 1�,m � �12.
w�w�
� ±1
�5 �1, 2�.
w � �1, 2�,m � 2.
f��x� � sec2 x � 2 at x ��
4.
f �x� � tan x
74.
v � �u � v� � u � ��3 � 2�3� i � �3�3 � 2� j
u � v � �3i � 3�3 j
u � 2�3 i � 2j 76. magnitude
direction �8.26�
63.5
78.
�R � �F1�F2�F3 163.0�
�R� � �F1 � F2 � F3� 4.09
�F3� � 3, �F3� 200�
�F2� � 4, �F2� 140�
�F1� � 2, �F1� �10�
80.
tan � �250 � 100�2
250�3 � 100�2 ⇒ � 10.7�
�F1 � F2� � ��250�3 � 100�2�2� �250 � 100�2�2 584.6 lb
� �250�3 � 100�2�i � �250 � 100�2 �j
F1 � F2 � �500 cos 30�i � 500 sin 30�j� � �200 cos��45��i � 200 sin��45�� j�
Section 10.1 Vectors in the Plane 477
82.
�R � arctan��200 � 315�2200�3 � 35�2� 0.6908 39.6�
�R� � ��200�3 � 35�2�2� ��200 � 315�2�2 385.2483 newtons
� 200�3 � 140�2 � 175�2�i � �200 � 140�2 � 175�2�j� 350�cos�135��i � sin�135��j�� 280�cos�45��i � sin�45��j�� F1 � F2 � F3 � 400�cos��30��i � sin��30��j��
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478 Chapter 10 Vectors and the Geometry of Space
84.
(a)
(b)
00
40
�2
� �500 � 400 cos �
� �400 � 400 cos � � 100 cos2 � � 100 sin2 �
�F1 � F2� � ��20 � 10 cos �, 10 sin ���
F1 � �20, 0�, F2 � 10�cos �, sin ��
(c) The range is
The maximum is 30, which occur at and
The minimum is 10 at
(d) The minimum of the resultant is 10.
� � �.
� � 2�.� � 0
10 ≤ �F1 � F2� ≤ 30.
86.
P2 � �1, 2� � 2�2, 1� � �5, 4�
P1 � �1, 2� � �2, 1� � �3, 3�
13
u � �2, 1�
u � �7 � 1, 5 � 2� � �6, 3�
88.
Vertical components:
Horizontal components:
Solving this system, you obtain
and �v� 3611.2.�u� 2169.4
�u� cos �1 � �v� cos �2 � 0
�u� sin �1 � �v� sin �2 � 5000
v � �v��cos �2 i � sin �2 j�
u � �u��cos �1 i � sin �1 j�
�2 � arctan� 24�10� � � 1.9656 or 112.6�
�1 � arctan�2420� 0.8761 or 50.2�
A B
C
v u
y
x
θ1
θ2
90. To lift the weight vertically, the sum of the vertical components of u and v mustbe 100 and the sum of the horizontal components must be 0.
Thus, or
And or
�u��12� � �v� cos 110� � 0
�u� cos 60� � �v� cos 110� � 0
�u���32 � � �v� sin 110� � 100.
�u� sin 60� � �v� sin 110� � 100,
v � �v� �cos 110�i � sin 110�j�
u � �u� �cos 60�i � sin 60�j�
100 lb
20° 30°
uv
Multiplying the last equation by and adding to the first equation gives
Then, gives
(a) The tension in each rope:
(b) Vertical components:
�v� sin 110� 61.33 lb.
�u� sin 60� 38.67 lb.
�u� � 44.65 lb, �v� � 65.27 lb.
�u� 44.65 lb.
�u��12� � 65.27 cos 110� � 0
�u��sin 110� � �3 cos 110�� � 100 ⇒ �v� 65.27 lb.
��3�
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Section 10.2 Space Coordinates and Vectors in Space 479
92.
Direction North of East:
Speed: � 336.35 mph
� N 84.46� E
tan � �35.36
364.64 ⇒ � � 5.54�
u � v � �400 � 25�2�i � 25�2j � 364.64i � 35.36j
v � 50�cos 135�i � sin 135�j� � �25�2i � 25�2j �wind�
u � 400i�plane�
94.
�v� � �sin2 � � cos2 � � 1
�u� � �cos2 � � sin2 � � 1,
96. Let u and v be the vectors that determine the parallelogram, as indicated in the figure. The two diagonals are and Therefore, But,
Therefore, and Solving we have
u
s
r
v
x � y �12 .x � y � 0.x � y � 1
� �x � y�u � �x � y�v. � x�u � v� � y�v � u�
u � r � s
s � y�v � u�.r � x�u � v�,v � u.u � v
98. The set is a circle of radius 5, centered at the origin.
�u� � ��x, y�� � �x2 � y2 � 5 ⇒ x2 � y2 � 25
100. True 102. False
a � b � 0
104. True
Section 10.2 Space Coordinates and Vectors in Space
2.
x y
(3, 2, 5)−
23, 4, 2−( (
8
z 4.
x y
(4, 0, 5)
(0, 4, 5)−
8
z
6.
B��3, 1, 4�
A�2, �3, �1� 8.
�7, �2, �1�
x � 7, y � �2, z � �1: 10. x � 0, y � 3, z � 2: �0, 3, 2�
12. The x-coordinate is 0. 14. The point is 2 units in front of the xz-plane.
16. The point is on the plane z � �3. 18. The point is behind the yz-plane.
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480 Chapter 10 Vectors and the Geometry of Space
20. The point is in front of the plane x � 4. 22. The point is 4 units above the xy-plane, and aboveeither quadrant II or IV.
�x, y, z�
24. The point could be above the xy-plane, and thus above quadrants I or III,or below the xy-plane, and thus below quadrants II or IV.
26.
� �16 � 64 � 16 � �96 � 4�6
d � ��2 � ��2��2 � ��5 � 3�2 � ��2 � 2�2 28.
� �4 � 49 � 9 � �62
d � ��4 � 2�2 � ��5 � 2�2 � �6 � 3�2
30.
Since the triangle is isosceles.AB � AC,BC � �16 � 16 � 0 � 4�2
AC � �4 � 4 � 1 � 3
AB � �4 � 4 � 1 � 3
A�5, 3, 4�, B�7, 1, 3�, C�3, 5, 3� 32.
Neither
BC � �0 � 4 � 9 � �13
AC � �25 � 0 � 9 � �34
AB � �25 � 4 � 0 � �29
A�5, 0, 0�, B�0, 2, 0�, C�0, 0, �3�
34. The y-coordinate is changed by 3 units:
�5, 6, 4�, �7, 4, 3�, �3, 8, 3�
36. 4 � 82
, 0 � 8
2,
�6 � 202 � � �6, 4, 7�
38. Center:
Radius: 5
x2 � y2 � z2 � 8x � 2y � 2z � 7 � 0
�x � 4�2 � �y � 1�2 � �z � 1�2 � 25
�4, �1, 1� 40. Center:
(tangent to yz-plane)
�x � 3�2 � �y � 2�2 � �z � 4�2 � 9
r � 3
��3, 2, 4�
42.
Center:
Radius:�109
2
�92
, 1, �5�
x �92�
2
� �y � 1�2 � �z � 5�2 �1094
x2 � 9x �814 � � �y2 � 2y � 1� � �z2 � 10z � 25� � �19 �
814
� 1 � 25
x2 � y2 � z2 � 9x � 2y � 10z � 19 � 0
44.
Center:
Radius: 3
12
, 4, �1�
x �12�
2
� �y � 4�2 � �z � 1�2 � 9
x2 � x �14� � �y2 � 8y � 16� � �z2 � 2z � 1� � �
334
�14
� 16 � 1
x2 � y2 � z2 � x � 8y � 2z �334
� 0
4x2 � 4y2 � 4z2 � 4x � 32y � 8z � 33 � 0
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Section 10.2 Space Coordinates and Vectors in Space 481
46.
Interior of sphere of radius 4 centered at �2, �3, 4�.
�x � 2�2 � �y � 3�2 � �z � 4�2 < 16
�x2 � 4x � 4� � �y2 � 6y � 9� � �z2 � 8z � 16� < 4 � 9 � 16 � 13
x2 � y2 � z2 < 4x � 6y � 8z � 13
48. (a)
(b)
x y
⟨ − ⟩4, 5, 2
8
z
� 4i � 5j � 2k � �4, �5, 2�
v � �4 � 0�i � �0 � 5� j � �3 � 1�k 50. (a)
(b)
x y
⟨ ⟩0, 0, 44
z
� 4k � �0, 0, 4�
v � �2 � 2�i � �3 � 3� j � �4 � 0�k
52.
Unit vector:��5, 12, �5�
�194� � �5
�194,
12�194
, �5�194
���5, 12, �5�� � �25 � 144 � 25 � �194
��1 � 4, 7 � ��5�, �3 � 2� � ��5, 12, �5� 54.
Unit vector: � 1�73
, 6
�73,
�6�73
��1, 6, �6�� � �1 � 36 � 36 � �73
�2 � 1, 4 � ��2�, �2 � 4� � �1, 6, �6�
56. (b)
(a) and (c).
� �6i � 4j � 9k � ��6, 4, 9�
v � ��4 � 2�i � �3 � 1� j � �7 � 2�k
x y
12( 4, 3, 7)−
( 6, 4, 9)−
(2, 1, 2)− −
z
58.
Q � �1, �83, 3�
�q1, q2, q3� � �0, 2, 52� � �1, �23, 12�
60. (a)
(c)
x y
2
21⟨ ⟨1, 1,−
z
12 v � �1, �1, 12 �
x y
4
⟨− − ⟩2, 2, 1
z
�v � ��2, 2, �1� (b)
(d)
x y
8
25⟨ ⟨5, 5,−
z
52 v � �5, �5, 52�
x y
8
⟨ − ⟩4, 4, 2
z
2v � �4, �4, 2�
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482 Chapter 10 Vectors and the Geometry of Space
62. � �7, 0, �4�z � u � v � 2w � �1, 2, 3� � �2, 2, �1� � �8, 0, �8�
64. z � 5u � 3v �12 w � �5, 10, 15� � �6, 6, �3� � �2, 0, �2� � ��3, 4, 20�
66.
z � �0, �2, �3�
9 � 3z3 � 0 ⇒ z3 � �3
6 � 3z2 � 0 ⇒ z2 � �2
0 � 3z1 � 0 ⇒ z1 � 0
�0, 6, 9� � �3z1, 3z2, 3z3� � �0, 0, 0�
2u � v � w � 3z � 2�1, 2, 3� � �2, 2, �1� � �4, 0, �4� � 3�z1, z2, z3� � �0, 0, 0�
68. (b) and (d) are parallel since and 34 i � j �98 k �
32 �1
2 i �23 j �
34 k�.�i �
43 j �
32 k � �2�1
2 i �23 j �
34 k�
70.
(b) is parallel since ��z�z � �14, 16, �6�.
z � ��7, �8, 3� 72.
Therefore, and are parallel.
The points are collinear.
→PR
→PQ
�3, �1, 2� � �12��6, 2, �4�
PR\
� �3, �1, 2�
PQ\
� ��6, 2, �4�
P�4, �2, 7�, Q��2, 0, 3�, R�7, �3, 9�
74.
Since and are not parallel, the points are notcollinear.
PR\
PQ\
PR\
� �2, �6, 4�
PQ\
� �1, 3, �2�
P�0, 0, 0�, Q�1, 3, �2�, R�2, �6, 4� 76.
Since and the given points form thevertices of a parallelogram.
AD\
� BC\
,AB\
� DC\
BC\
� �2, 3, �7�
AD\
� �2, 3, �7�
DC\
� �8, �2, �5�
AB\
� �8, �2, �5�
A�1, 1, 3�, B�9, �1, �2�, C�11, 2, �9�, D�3, 4, �4�
78. �v� � �1 � 0 � 9 � �10 80.
�v� � �16 � 9 � 49 � �74
v � ��4, 3, 7� 82.
�v� � �1 � 9 � 4 � �14
v � �1, 3, �2�
84.
(a)
(b) �u
�u�� �
110
�6, 0, 8�
u�u�
�110
�6, 0, 8�
�u� � �36 � 0 � 64 � 10
u � �6, 0, 8� 86.
(a)
(b) �u
�u�� ��1, 0, 0�
u�u�
� �1, 0, 0�
�u� � 8
u � �8, 0, 0�
88. (a)
(b)
(c)
(d) �v� � 9.014
�u� � 5.099
�u � v� � 8.732
u � v � �4, 7.5, �2� 90.
c � ±3�14
14
14c2 � 9
�cu� � �c2 � 4c2 � 9c2 � 3
cu � �c, 2c, 3c�
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Section 10.3 The Dot Product of Two Vectors 483
92. v � 3u
�u�� 3� 1
�3,
1
�3,
1
�3� � � 3
�3,
3
�3,
3
�3� 94.
� ���707
, 3�70
14, �7014 �
v � �5 u
�u�� �5� �2
�14,
3
�14,
1
�14�
96. or
x y
8
25 2
( + )i k
z
v � 5�cos 135�i � sin 135�k� �5�2
2��i � k�
v � 5�cos 45�i � sin 45�k� �5�2
2�i � k� 98.
�1, 2, 5� � �103 , 4, �2� � �13
3 , 6, 3� 23 v � �10
3 , 4, �2 v � �5, 6, �3
100. is directed distance to yz-plane.
is directed distance to xz-plane.
is directed distance to xy-plane.z0
y0
x0 102. �x � x0�2 � �y � y0�2 � �z � z0�2 � r2
104. A sphere of radius 4 centered at
sphere�x � x1�2 � �y � y1�2 � �z � z1�2 � 16
� ��x � x1�2 � �y � y1�2 � �z � z1�2 � 4
�v� � ��x � x2, y � y1, z � z1�
�x1, y1, z1�. 106. As in Exercise 105(c), will be a verticalasymptote. Hence, lim
r0→a� T � �.
x � a
108.
306i � 204j � 409k
F 4.085�75i � 50j � 100k� c 4.085
c2 � 16.689655
302,500 � 18,125c2
550 � �c�75i � 50j � 100k�� 110. Let A lie on the y-axis and the wall on the x-axis. Then
and
Thus,
�F� 860.0 lb
��185.5, �720.1, 432.1
� ��423.1, �423.1, 253.9
F � F1 � F2 �237.6, �297.0, 178.2
F1 � 420→AB
�→AB�
, F2 � 650→AC
�→AC�
�→AB� � 10�2, �
→AC� � 2�59
→AC � ��10, �10, 6.
→AB � �8, �10, 6,
C � ��10, 0, 6�B � �8, 0, 6�,A � �0, 10, 0�,
Section 10.3 The Dot Product of Two Vectors
2.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2�22� � 44
�u � v�v � 22��2, 3 � ��44, 66
�u�2 � 116
u � u � 4�4� � 10�10� � 116
u � v � 4��2� � 10�3� � 22
u � �4, 10, v � ��2, 3 4.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2
�u � v�v � i
�u�2 � 1
u � u � 1
u � v � 1
u � i, v � i
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484 Chapter 10 Vectors and the Geometry of Space
14.
� � arccos��24
�1 � �3�� � 105�
��32 �
�22 � �
12 �2
2 � ��24
�1 � �3�
cos � �u � v
�u� �v�
v � cos 3
4 �i � sin 3
4 �j � ��22
i ��22
j
u � cos
6�i � sin
6�j ��32
i �12
j 16.
� �
2
cos � �u � v
�u� �v��
3�2� � 2��3� � 0�u� �v�
� 0
u � 3i � 2j � k, v � 2i � 3j
18.
� � arccos 3�2114 � 10.9�
�9
�14�6�
9
2�21�
3�2114
cos � �u � v
�u� �v�
u � 2i � 3j � k, v � i � 2j � k 20.
not parallel
orthogonalu � v � 0 ⇒
u cv ⇒
u � �2, 18, v � �32
, �16�
28.
cos2 � � cos2 � � cos2 �2535
�9
35�
135
� 1
cos ��1�35
cos � �3
�35
cos � �5
�35
u � �5, 3, �1 �u� � �35
6.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2��5� � �10
�u � v�v � �5�i � 3j � 2k� � �5i � 15j � 10k
�u�2 � 9
u � u � 2�2� � 1�1� � ��2���2� � 9
u � v � 2�1� � 1��3� � ��2��2� � �5
u � 2i � j � 2k, v � i � 3j � 2k 8.
Increase prices by 4%:
New total amount:
� $17,824.61
1.04�u � v� � 1.04�17,139.05�
1.04�2.22, 1.85, 3.25.
v � �2.22, 1.85, 3.25
u � �3240, 1450, 2235
10.
u � v � �40��25� cos 5
6� �500�3
u � v�u� �v�
� cos � 12.
� �
4
cos � �u � v
�u� �v��
5
�10�5�
1
�2
u � �3, 1, v � �2, �1
22.
parallelu � �16 v ⇒
u � �13 �i � 2j�, v � 2i � 4j 24.
not parallel
orthogonalu � v � 0 ⇒
u cv ⇒
u � �2i � 3j � k, v � 2i � j � k
26.
not parallel
orthogonalu � v � 0 ⇒
u cv ⇒
v � �sin �, �cos �, 0u � �cos �, sin �, �1,
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Section 10.3 The Dot Product of Two Vectors 485
32.
cos �5
5�2�
1�2
⇒
4 or 45�
cos � �3
5�2 ⇒ � 1.1326 or 64.9�
cos � ��4
5�2 ⇒ � 2.1721 or 124.4�
u � ��4, 3, 5 �u� � �50 � 5�2 34.
cos �1
�41 ⇒ � 1.4140 or 81.0�
cos � �6
�41 ⇒ � 0.3567 or 20.4�
cos � ��2�41
⇒ � 1.8885 or 108.2�
u � ��2, 6, 1 �u� � �41
36.
cos 65.4655
�F� ⇒ 74.31�
cos � �36.062
�F� ⇒ � 98.57�
cos � �230.239
�F� ⇒ � 162.02�
�F� 242.067 lb
� ��230.239, �36.062, 65.4655
13.0931��20, �10, 5 � 6.3246�5, 15, 0
F � F1 � F2
F2: C2 �100�F2�
6.3246
F1: C1 �300�F1�
13.0931 38.
� � arccos �63
35.26�
cos � �s�2
s�3�
�63
�v2� � s�2
v2 � �s, s, 0
�v1� � s�3
y
x
v1
v2
( , , 0)s s
( , , )s s s
z v1 � �s, s, s
40.
and
100�2 � 6C2 � 6C3 � 0 ⇒ C2 � C3 �25�2
3 N
�4C2 � 4C3 � 0 ⇒ C2 � C3
F � F1 � F2 � F3 � 0
F � �0, 0, w
F2 � C3�4, �6, 10
F2 � C2 ��4, �6, 10
F1 � �0, 100�2, 100�2 �F1� � 200 � C1 10�2 ⇒ C1 � 10�2F1 � C1�0, 10, 10�. 42. w2 � u � w1 � �9, 7 � �3, 9 � �6, �2
44. w2 � u � w1 � �8, 2, 0 � �6, 3, �3 � �2, �1, 3
30.
cos2 � � cos2 � � cos2 �a2
a2 � b2 � c2 �b2
a2 � b2 � c2 �c2
a2 � b2 � c2 � 1
cos �c
�a2 � b2 � c2
cos � �b
�a2 � b2 � c2
cos � �a
�a2 � b2 � c2
u � �a, b, c, �u� � �a2 � b2 � c2
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486 Chapter 10 Vectors and the Geometry of Space
50. The vectors u and v are orthogonal if
The angle between u and v is given by
cos � �u � v
�u� �v�.
�
u � v � 0. 52. (a) and (b) are defined.
54. See figure 10.29, page 739. 56. Yes,
�u� � �v�
1
�v��
1�u�
�u � v� �v��v�2 � �v � u� �u�
�u�2
� u � v�v�2 v � � � v � u
�u�2 u�
58. (a)
(b) �u� 9.165, �v� 5.745, � � 90�
�u� � 5, �v� 8.602, � 91.33� 60. (a)
(b) ��2126
, 6326
, 4213�
�6417
, 1617�
62. Because u appears to be a multiple of v, the projection of u onto v is u. Analytically,
��2652
�6, 4 � ��3, �2 � u.
projv u �u � v�v�2 v �
��3, �2 � �6, 4�6, 4 � �6, 4
�6, 4
64. Want
and are orthogonal to u.�v � �3i � 8jv � 3i � 8j
u � v � 0.u � �8i � 3j. 66. Want
and are orthogonal to u.�v � �0, �6, �3v � �0, 6, 3
u � v � 0.u � �0, �3, 6.
68.
�projvOA\
� � 20
projvOA\
�2012 �0, 0, 1 � �0, 0, 20
OA\
� �10, 5, 20, v � �0, 0, 1 70.
W � F � v � 1250 cos 20� 1174.6 ft � lb
v � 50i
F � 25�cos 20�i � sin 20�j�
72.
W � PQ\
� V\
� 74
V\
� ��2, 3, 6
PQ\
� ��4, 2, 10 74. True
and are orthogonal.u � v
� 0 � 0 � 0 ⇒ w
w � �u � v� � w � u � w � v
48.
(a)
(b)
� ��2013
, 0, 3013�
w2 � u � w1 � �1, 0, 4 � �3313
, 0, 2213�
w1 � u � v�v�2 �v �
1113
�3, 0, 2 � �3313
, 0, 2213�
u � �1, 0, 4, v � �3, 0, 246.
(a)
(b) w2 � u � w1 � �2, �3
w1 � u � v�v�2 �v � 0v � �0, 0
u � �2, �3, v � �3, 2
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Section 10.4 The Cross Product of Two Vectors in Space
76. (a)
x
y
( , 0, )k k
( , , 0)k k
(0, , )k k
kk
k
z
(b) Length of each edge:
(c)
� � arccos�12� � 60�
cos � �k2
�k�2��k�2� �12
�k2 � k2 � 02 � k�2
(d)
� � 109.5�
cos � �
�k 2
4
�k2�
2
� 3� �
13
r2
\
� �0, 0, 0� � k2
, k2
, k2 � �
k2
, �k2
, �k2
r1
\
� �k, k, 0� � k2
, k2
, k2 � k
2,
k2
, �k2
78. The curves and intersect at and at
At is tangent to and is tangent to The anglebetween these vectors is
At is tangent to and is tangent to To find the angle between these vectors,
cos � �1
�5
1
�10�3 � 2� �
1
�2 ⇒ � � 45�.
y2.�3��10��1, 1�3� � �1��10��3, 1�y1�1��5��1, 2��1, 1�:
90�.y2.�0, 1� y1�1, 0��0, 0�:
x(0, 0)
(1, 1)
2
1
1 2
y2
y1y�1, 1�.�0, 0�y2 � x1�3y1 � x2
80.
≤ �u� �v� since cos � ≤ 1.
� �u� �v� cos � u � v � �u� �v� cos �
u � v � �u� �v� cos �
82. Let as indicated in the figure. Because is a scalar multiple of v, you can write
Taking the dot product of both sides with v produces
Thus, and w1 � projv u � cv �u � v�v�2 v.u � v � c�v�2 ⇒ c �
u � v�v�2
� c�v�2, since w2 and v are orthogonol.
� cv � v � w2 � vu � v � �cv � w2� � v
u � w1 � w2 � cv � w2.
θ
w1
w2u
v
w1w1 � projv u,
2.
x y
i
j
k
11
1
−1
z
i j � i10
j01
k00 � k 4.
x y
−i
j
k
11
1
−1
z
k j � i00
j01
k10 � �i 6.
x y
i
j
k
11
1
−1
z
k i � i01
j00
k10 � j
Section 10.4 The Cross Product of Two Vectors in Space 487
http://librosysolucionarios.net
8. (a)
(b)
(c) v v � 0
v u � ��u v� � �15, �16, �9�
u v � i32
j03
k5
�2 � ��15, 16, 9� 10. (a)
(b)
(c) v v � 0
v u � ��u v� � ��8, 5, �17�
u v � i31
j�2
5
k�2
1 � �8, �5, 17�
12.
� 0 ⇒ v � u v
v � �u v� � �0���2� � �1��0� � �0���1�
� 0 ⇒ u � u v
u � �u v� � ��1���2� � �1��0� � �2���1�
u v � i�1
0
j11
k20 � �2i � k � ��2, 0, �1�
u � ��1, 1, 2�, v � �0, 1, 0� 14.
� 0 ⇒ v � u v
v � �u v� � 7�0� � �0��42� � �0��0�
� 0 ⇒ u � u v
u � �u v� � ��10��0� � �0��42� � 6�0�
u v � i�10
7
j00
k60 � 42j � �0, 42, 0�
u � ��10, 0, 6�, v � �7, 0, 0�
16.
v � �u v� � �2�6� � 1��1� � 1�13� � 0 ⇒ v � �u v�
u � �u v� � 1�6� � 6��1� � 0 ⇒ u � �u v�
u v � i1
�2
j61
k01 � 6i � j � 13k
18.
x
y
v
u4
64
12
3
1
32
456
z 20.
x
y
v
u4
64
12
3
1
32
456
z
22.
� 5
3�22,
2
3�22,
13
3�22
u v�u v�
�1
36�22�60, 24, 156�
u v � �60, 24, 156�
v � �10, �12, �2�
u � ��8, �6, 4� 24.
u v�u v�
� �0, 1, 0�
u v � 0, 13
, 0
v �12
i � 6k
u �23
k
26. (a)
(b)
�u v� � 72.498
u v � ��50, 40, �34�
�u v� � 52.650
u v � ��18, �12, 48� 28.
A � �u v� � ��j � k� � �2
u v � i10
j11
k11 � �j � k
v � j � k
u � i � j � k
488 Chapter 10 Vectors and the Geometry of Space
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Section 10.4 The Cross Product of Two Vectors in Space 489
30.
A � �u v� � ��0, 0, 3�� � 3
u v � i2
�1
j�1
2
k00 � �0, 0, 3�
v � ��1, 2, 0�
u � �2, �1, 0�
32.
Since and the figure is a parallelogram.
and are adjacent sides and
Area � �AB\
AC\
� � �920 � 2�230
AB\
AC\
� i41
j8
�3
k�2
3 � �18, �14, �20�.
AC\
AB\
AC\
� BD\
,AB\
� CD\
AB\
� �4, 8, �2�, AC\
� �1, �3, 3�, CD\
� �4, 8, �2�, BD\
� �1, �3, 3�
A�2, �3, 1�, B�6, 5, �1�, C�3, �6, 4�, D�7, 2, 2�
34.
A �12
�AB\
AC\
� �12�44 � �11
AB\
AC\
� i�2�3
j45
k�2�4 � �6i � 2j � 2k
AB\
� ��2, 4, �2�, AC\
� ��3, 5, �4�
A�2, �3, 4�, B�0, 1, 2�, C��1, 2, 0� 36.
A �12
�AB\
AC\
� �52
AB\
AC\
� i�3�1
j�1�2
k00 � 5k
AB\
� ��3, �1, 0�, AC\
� ��1, �2, 0�
A�1, 2, 0�, B��2, 1, 0�, C�0, 0, 0�
38.
�PQ\
F� � 160�3 ft � lb
PQ\
F � i00
j0
�1000�3
k0.16
�1000 � 160�3 i
PQ\
� 0.16k
x
y
60°
PQF
0.16 ft
zF � �2000�cos 30� j � sin 30�k� � �1000�3 j � 1000k
40. (a) is to the left of and one foot upwards:
F � �200�cos � j � sin �k�
AB\
��54 j � k
A,�1512 � �
54B
(b)
� 25�10 sin � � 8 cos ��
�AB\
F� � 250 sin � � 200 cos � � �250 sin � � 200 cos ��i
AB\
F � i00
j�5�4
�200 cos �
k1
�200 sin � (c) For
� 25�5 � 4�3� � 298.2.
�AB\
F� � 25�10�12� � 8��3
2 ��� � 30�,
(d) If
The vectors are orthogonal.
⇒ � � 51.34�.
dTd�
� 25�10 cos � � 8 sin �� � 0 ⇒ tan � �54
T � �AB\
F�,
(e) The zero is the angle making parallelto
0 180
−300
400
F.AB
\
� � 141.34�,
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490 Chapter 10 Vectors and the Geometry of Space
48.
V � u � �v w� � 3
u � �v w� � 110 101
021 � �3
w � �0, 1, 1�
v � �1, 0, 2�
u � �1, 1, 0� 50. See Theorem 10.8, page 746.
52. Form the vectors for two sides of the triangle, and compute their cross product:
�x2 � x1, y2 � y1, z2 � z1� �x3 � x1, y3 � y1, z3 � z1�
54. False, let
Then,
but v w.u v � u w � 0,
u � �1, 0, 0�, v � �1, 0, 0�, w � ��1, 0, 0�.
56.
u � �v � w� � u1�v2w3 � v3w2� � u2�v1w3 � v3w1� � u3�v1w2 � v2w1� � u1
v1
w1
u2
v2
w2
u3
v3
w3 v w � �v2w3 � v3w2�i � �v1w3 � v3w1�j � �v1w2 � v2w1�k
u � u1i � u2 j � u3k
u � �u1, u2, u3�, v � �v1, v2, v3�, w � �w1, w2, w3�
58. is a scalar.
� c��u2v3 � u3v2�i � �u1v3 � u3v1�j � �u1v2 � u2v1�k� � c�u v�
� �cu2v3 � cu3v2�i � �cu1v3 � cu3v1�j � �cu1v2 � cu2v1�k
�cu� v � icu1
v1
jcu2
v2
kcu3
v3 u � �u1, u2, u3�, v � �v1, v2, v3�, c
60.
� u � �v w�
� u1�v2w3 � w2v3� � u 2�v1w3 � w1v3� � u 3�v1w2 � w1v2 �
� w1�u 2v3 � v2u 3� � w2�u 1v3 � v1u 3� � w3�u1v2 � v1u 2�
�u v� � w � w � �u v� � w1
u1
v1
w2
u2
v2
w3
u3
v3 u � �v w� � u1
v1
w1
u2
v2
w2
u3
v3
w3
46.
V � u � �v w� � 72
u � �v w� � 10
�4
360
16
�4 � �72
42. u � �v w� � 120 110
101 � �1 44. u � �v w� � 210 0
12
012 � 0
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Section 10.5 Lines and Planes in Space
62. If u and v are scalar multiples of each other, for some scalar
If then Thus, and u and v are parallel. Therefore,
for some scalar c.u � cv
sin � � 0, � � 0,�Assume u � 0, v � 0.��u� �v� sin � � 0.u � v � 0,
u � v � �cv� � v � c�v � v� � c�0� � 0
c.u � cv
64.
� �u � w�v � �u � v�w
� �a1a3 � b1b3 � c1c3��a2, b2, c2� � �a1a2 � b1b2 � c1c2��a3, b3, c3�
�c2�a1a3 � b1b3 � c1c3� � c3�a1a2 � b1b2 � c1c2��k
�b2�a1a3 � b1b3 � c1c3� � b3�a1a2 � b1b2 � c1c2�� j �
� �a2�a1a3 � b1b3 � c1c3� � a3�a1a2 � b1b2 � c1c2��i �
�a1�a3c2 � a2c3� � b1�b2c3 � b3c2��k
u � �v � w� � �b1�a2b3 � a3b2� � c1�a3c2 � a2c3��i � �a1�a2b3 � a3b2� � c1�b2c3 � b3c2�� j �
u � �v � w� � ia1
�b2c3 � b3c2�
jb1
�a3c2 � a2c3�
kc1
�a2b3 � a3b2� v � w � i
a2
a3
jb2
b3
kc2
c3 � �b2c3 � b3c2�i � �a2c3 � a3c2�j � �a2b3 � a3b2�k
u � �a1, b1, c1�, v � �a2, b2, c2�, w � �a3, b3, c3�
Section 10.5 Lines and Planes in Space 491
2.
(a)
x y
z
x � 2 � 3t, y � 2, z � 1 � t
(b) When we have When we have
The components of the vector and the coefficients of areproportional since the line is parallel to
(c) when Thus, and
Point:
when Point: 0, 2, 13�t �
23
.x � 0
��1, 2, 0�y � 2.x � �1t � 1.z � 0
PQ\
.t
PQ\
� ��6, 0, �2�
Q � ��4, 2, �1�.t � 2P � �2, 2, 1�.t � 0
4. Point:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric:x
�4�
y5
�z2
x � �4t, y � 5t, z � 2t
�4, 5, 2
v � ��2, 52
, 1 �0, 0, 0� 6. Point:
Direction vector:
Direction numbers: 0, 2, 1
(a) Parametric:
(b) Symmetric:y2
� z � 2, x � �3
x � �3, y � 2t, z � 2 � t
v � �0, 6, 3�
��3, 0, 2�
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492 Chapter 10 Vectors and the Geometry of Space
8. Point:
Directions numbers:
(a) Parametric:
(b) Symmetric:x � 3
3�
y � 5�2
� z � 4
x � �3 � 3t, y � 5 � 2t, z � 4 � t
3, �2, 1
��3, 5, 4� 10. Points:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric: x � 2 �y
�4�
z � 25
x � 2 � t, y � �4t, z � 2 � 5t
1, �4, 5
�1, �4, 5�
�2, 0, 2�, �1, 4, �3�
12. Points:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric:x2
�y2
�z � 25
�5
x � 2t, y � 2t, z � 25 � 5t
2, 2, �5
�10, 10, �25�
�0, 0, 25�, �10, 10, 0� 14. Point:
Direction vector:
Direction numbers:
Parametric: x � 2 � 3t, y � 3 � 2t, z � 4 � t
3, 2, �1
v � 3i � 2j � k
�2, 3, 4�
16. Points:
Direction vector:
Direction numbers: 2, 2, 1
Parametric:
Symmetric:
(a) Not on line
(b) On line
(c) Not on line �32
��32
� �1�
1 �12
� 1�
x � 22
�y2
�z � 3
1
x � 2 � 2t, y � 2t, z � �3 � t
v � 2i � 2j � k
�2, 0, �3�, �4, 2, �2� 18. on line
on line
and are identical.L2L1
L4: v � ��2, 1, 1.5�
�8, �5, �9�L3: v � ��8, 4, �6�
L2: v � �2, 1, 5�
�8, �5, �9�L1: v � �4, �2, 3�
20. By equating like variables, we have
(i) (ii) and (iii)
From (i) we have and consequently from (ii), and from (iii), The lines do not intersect.t � �3.t �12s � �t,
2t � 4 � �s � 1.4t � 1 � 2s � 4,�3t � 1 � 3s � 1,
22. Writing the equations of the lines in parametric form we have
By equating like variables, we have Thus, and thepoint of intersection is
(First line)
(Second line)
cos � � u � v�u� �v�
�4
�46�21�
4
�966�
2�966483
v � �2, 1, 4�
u � ��3, 6, 1�
�5, �4, 2�.t � �1, s � 12 � 3t � 3 � 2s, 2 � 6t � �5 � s, 3 � t � �2 � 4s.
x � 3 � 2s y � �5 � s z � �2 � 4s.
x � 2 � 3t y � 2 � 6t z � 3 � t
24.
Point of intersection: �3, 2, 2�
z � t z � �2s � 4
y � �4t � 10 y � 3s � 11 z
x y(3, 2, 2)
3
3
3
2
2
2
−2−3
x ty tz t
= 2 1= 4 + 10=
−−
x sy sz s
= 5 12= 3 + 11= 2 4
− −
− −
x � 2t � 1 x � �5s � 12
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Section 10.5 Lines and Planes in Space 493
26.
(a)
(b)
The components of the cross product are proportional(for this choice of and they are the same) tothe coefficients of the variables in the equation. Thecross product is parallel to the normal vector.
R,P, Q,
PQ\
� PR\
� i23
j02
k�1�3 � �2, 3, 4�
PQ\
� �2, 0, �1�, PR\
� �3, 2, �3�
P � �0, 0, 1�, Q � �2, 0, 0�, R � �3, 2, �2�
2x � 3y � 4z � 4 28. Point:
z � 3 � 0
0�x � 1� � 0�y � 0� � 1�z � ��3�� � 0
n � k � �0, 0, 1��1, 0, �3�
30. Point:
Normal vector:
�3x � 2z � 0
�3�x � 0� � 0�y � 0� � 2�z � 0� � 0
n � �3i � 2k
(0, 0, 0� 32. Point:
Normal vector:
4x � y � 3z � 8
4�x � 3� � �y � 2� � 3�z � 2� � 0
v � 4i � j � 3k
�3, 2, 2�
34. Let u be vector from to
Let v be vector from to
Normal vector:
�6x � 2y � z � �8
�6�x � 2� � 2�y � 3� � 1�z � 2� � 0
� �3��6, 2, 1�
u � v � i1
�1
j1
�4
k42 � �18, �6, �3�
�1, �1, 0�: ��1, �4, 2�.�2, 3, �2�
�3, 4, 2�: �1, 1, 4�.�2, 3, �2� 36. Normal vector: v � i, 1�x � 1� � 0, x � 1�1, 2, 3�,
38. The plane passes through the three points
The vector from to
The vector from to
Normal vector:
x � �3z � 0
u � v � i0
�3
j10
k01 � i � �3k
v � �3 i � k��3, 0, 1�:�0, 0, 0�u � j�0, 1, 0�:�0, 0, 0�
��3, 0, 1�.�0, 1, 0��0, 0, 0�, 40. The direction of the line is Choose any
point on the line, for example and let v be the vector from to the given point
Normal vector:
x � 2z � 0
�x � 2� � 2�z � 1� � 0
u � v � i22
j�1�2
k11 � i � 2k
v � 2i � 2j � k
�2, 2, 1�:�0, 4, 0��,��0, 4, 0�,
u � 2i � j � k.
42. Let v be the vector from to
Let n be the normal to the given plane:
Since v and n both lie in the plane the normal vector tois:
20x � 18y � 3z � 27
20�x � 3� � 18�y � 2� � 3�z � 1� � 0
� 2�20i � 18j � 3k�
v � n � i06
j�1
7
k�6
2 � 40i � 36j � 6k
PP,
n � 6i � 7j � 2k
v � �j � 6k �3, 1, �5�:�3, 2, 1� 44. Let and let v be the vector from to
Since u and v both lie in the plane the normal vector tois:
3x � 7y � 26
3�x � 4� � 7�y � 2� � 0
u � v � i0
�7
j03
k16 � �3i � 7j � ��3i � 7j�
PP,
v � �7i � 3j � 6k��3, 5, 7�:�4, 2, 1�u � k
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494 Chapter 10 Vectors and the Geometry of Space
46. The normal vectors to the planes are Since the planes are parallel, but not equaln2 � �3n1,n2 � ��9, �3, 12�.n1 � �3, 1, �4�,
48. The normal vectors to the planes are
Therefore, � � arccos 1
�6� � 65.9.
cos � �n1 � n2�n1� �n2�
�3 � 8 � 2�14�21
�1
�6.
n1 � 3i � 2j � k, n2 � i � 4j � 2k,
50. The normal vectors to the planes are
Thus, and the planes are orthogonal.� �
2
cos � �n1 � n2�n1� �n2�
� 0.
n1 � �2, 0, �1�, n2 � �4, 1, 8�,
52.
xy2 3
23
3
z
3x � 6y � 2z � 6 54.
x
y1
−4
3
4
1
2
z
2x � y � z � 4 56.
x
y34
4
z
x � 2y � 4
58.
x y55
8
z
z � 8 60.
x y
12
2
1
12
3
Generated by Mathematica
z
x � 3z � 3 62.
xy2
1
1
2
3
Generated by Mathematica
z
2.1x � 4.7y � z � 3 � 0
64. on plane
on plane
on plane
and are parallel.P3P1, P2,
P4: n � �12, �18, 6� or ��2, 3, �1�
�0, 0, 56�P3: n � ��20, 30, 10� or ��2, 3, 1�
�0, 0, �23�P2: n � �6, �9, �3� or ��2, 3, 1�
�0, 0, 910�P1: n � ��60, 90, 30� or ��2, 3, 1�
66. If is xy-plane.
If is a plane parallel to
x-axis and passing through the points and �0, 1, �c�.
�0, 0, 0�
c � 0, cy � z � 0 ⇒ y ��1c
z
c � 0, z � 0 68. The normals to the planes are . and
The direction vector for the line is
Now find a point of intersection of the planes.
Let
x � �4 � 16t, y � �9 � 31t, z � 2 � 3t
y � �9, z � 2 ⇒ x � �4 ⇒ ��4, �9, 2�.
6x
�6x
�
�
3y
6y
3y
�
�
�
z
30z
31z
�
�
�
5
30
35
⇒⇒
6x
�x
�
�
3y
y
�
�
z
5z
�
�
5
5
n1 � n2 � i6
�1
j�3
1
k15 � ��16, �31, 3�.
n2 � ��1, 1, 5�.n1 � �6, �3, 1�
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Section 10.5 Lines and Planes in Space 495
70. Writing the equation of the line in parametric form andsubstituting into the equation of the plane we have:
Substituting into the parametric equations for theline we have the point of intersection Theline does not lie in the plane.
��1, �1, 0�.t � �
12
2�1 � 4t� � 3�2t� � �5, t ��12
x � 1 � 4t, y � 2t, z � 3 � 6t
72. Writing the equation of the line in parametric form andsubstituting into the equation of the plane we have:
Substituting into the parametric equations for theline we have the point of intersection Theline does not lie in the plane.
�4, �1, �2�.t � 0
5�4 � 2t� � 3��1 � 3t� � 17, t � 0
x � 4 � 2t, y � �1 � 3t, z � �2 � 5t
74. Point:
Plane:
Normal to plane:
Point in plane:
Vector:
D � PQ\
� n�n�
� �8�81
�8
9
PQ\
� ��1, 0, 0�
P�1, 0, 0�
n � �8, �4, 1�
8x � 4y � z � 8
Q�0, 0, 0� 76. Point:
Plane:
Normal to plane:
Point in plane:
Vector:
D � PQ\
� n�n�
� �1�6
�1�6
��66
PQ\
� ��1, 2, 1�
P�4, 0, 0�
n � �1, �1, 2�
x � y � 2z � 4
Q�3, 2, 1�
78. The normal vectors to the planes are andSince the planes are parallel.
Choose a point in each plane.
is a point in
is a point in
D � PQ\
� n1�n1�
�11
�113�
11�113
113
PQ\
� �5, 0, �1�
4x � 4y � 9z � 18.Q � �0, 0, 2�
4x � 4y � 9z � 7.P � ��5,0, 3�
n1 � n2,n2 � �4, �4, 9�.n1 � �4, �4, 9� 80. The normal vectors to the planes are and
Since the planes are parallel.Choose a point in each plane.
is a point in is a point in
PQ\
� �3, 0, 0�, D �PQ
\
� n1�n1�
�6
�20�
3�55
2x � 4z � 10.Q � �5, 0, 0�2x � 4z � 4.P � �2, 0, 0�
n1 � n2,n2 � �2, 0, �4�.n1 � �2, 0, �4�
82. is the direction vector for the line.
is a point on the line
D ��PQ
\
� u��u�
��5�9
��5
3
PQ\
� u � i12
j11
k22 � �0, 2, �1�
PQ\
� �1, 1, 2�
�let t � 0�.P � �0, �3, 2�
u � �2, 1, 2� 84. The equation of the plane containing andhaving normal vector is
You need n and P to find the equation.
a�x � x1� � b�y � y1� � c�z � z1� � 0.
n � �a, b, c�P�x1, y1, z1�
86. plane parallel to yz-plane containing
plane parallel to xz-plane containing
plane parallel to xy-plane containing �0, 0, c�z � c:
�0, b, 0�y � b:
�a, 0, 0�x � a: 88. (a) represents a line parallel to v.
(b) represents a line through the terminal point ofu parallel to v.
(c) represent the plane containing u and v.su � tv
u � tv
tv
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Section 10.6 Surfaces in Space
90. On one side we have the points and
On the adjacent side we have the points and
� � arccos165
� 89.1�
cos � ��n1 � n2��n1� �n2�
�36
2340�
165
n2 � � i0
�1
j6
�1
k08� � 48i � 6k
x
y4
66
4
6
2
(0, 6, 0)
(6, 0, 0)
(0, 0, 0)
( 1, 1, 8)− −
z��1, �1, 8�.�0, 0, 0�, �0, 6, 0�,
n1 � � i6
�1
j0
�1
k08� � �48j � 6k
��1, �1, 8�.�0, 0, 0�, �6, 0, 0�,
92. False. They may be skew lines. (See Section Project)
2. Hyperboloid of two sheets
Matches graph (e)
4. Elliptic cone
Matches graph (b)
6. Hyperbolic paraboloid
Matches graph (a)
8.
Plane parallel to theyz-coordinate plane
x y4 2
4
4
zx � 4 10.
The y-coordinate is missing so we have a cylindricalsurface with rulings parallel to the y-axis. The generatingcurve is a circle.
x
y
8
8
6
4
z
x2 � z2 � 25
496 Chapter 10 Vectors and the Geometry of Space
12.
The x-coordinate is missing so we have a cylindrical surface withrulings parallel to the x-axis. The
generating curve is a parabola.
xy8 84
12
8
4
z
z � 4 � y2 14.
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a hyperbola.
x y
5
55
z
y2
4�
z2
4� 1
y2 � z2 � 4 16.
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis.The generating curve is the exponential curve.
x y
20
15
10
5
1 2 3 43
z
z � ey
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18.
(a) From
y
z
�10, 0, 0�:
y2 � z2 � 4
(b) From
y
z
�0, 10, 0�: (c) From
yx
3
3
3
z
�10, 10, 10�:
20.
Ellipsoid
xy-trace: ellipse
xz-trace: ellipse
yz-trace: circley2 � z2 � 25
x2
16�
z2
25� 1
x2
16�
y2
25� 1
x
y
z
34
5
21
3 4 5
12
3
4
5
21
x2
16�
y2
25�
z2
25� 1 22.
Hyperboloid of two sheets
xy-trace: none
xz-trace: hyperbola
yz-trace: hyperbola
ellipsez � ±�10: x2
9�
y2
36� 1
z2 �y2
4� 1
z2 � x2 � 1x
y5
5
5
zz2 � x2 �y2
4� 1
Section 10.6 Surfaces in Space 497
30.
Hyperboloid of one sheet with center �3, 2, �3�.
�x � 3�2
4�9�
�y � 2�2
4�
�z � 3�2
4�9� 1
9�x � 3�2 � �y � 2�2 � 9�z � 3�2 � 4
9�x2 � 6x � 9� � �y2 � 4y � 4� � 9�z2 � 6z � 9� � 81 � 4 � 81
9x2 � y2 � 9z2 � 54x � 4y � 54z � 4 � 0
x y
z
2 1
4
5432
24.
Elliptic paraboloid
xy-trace: point
xz-trace: parabola
yz-trace: parabola
x
y
z
32
1
4
21
z � 4y2
z � x2
�0, 0, 0�
z � x2 � 4y2 26.
Hyperbolic paraboloid
xy-trace:
xz-trace:
yz-trace:
xy10
10
2420
28
z
z � �13 y2
z �13 x2
y � ±x
3z � �y2 � x2 28.
Elliptic Cone
xy-trace:
xz-trace:
yz-trace: point:
xy5
5
5
z
�0, 0, 0�
x � ±�2z
x � ±�2y
x2 � 2y2 � 2z2
32.
x y
4
12
2
−2
1
z
z � x2 � 0.5y2 34.
xy
84
4
8
−8
−8
−4
3
z
z � ±�4y � x2
z2 � 4y � x2 36.
x y
4
−4−4−3 −3
4
4
z
�ln�x2 � y2� � z
x2 � y2 � e�z
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38.
yx
2 2
2
4 4
4
z
z ��x
8 � x2 � y2 40.
yx
10
10
20
1020
20
z
z � ±�98
x2 �12
y2 � 9
9x2 � 4y2 � 8z2 � 72 42.
yx 4 4
5
3
z
x � 0, y � 0, z � 0
y � �4 � x2
z � �4 � x2
44.
z � 0
y � 2z
x
y
3
−3
3
3
zz � �4 � x2 � y2 46. and therefore,
x2 � z2 � 9y2.
z � r�y� � 3y;x2 � z2 � r�y�2
48. and therefore,
x2 � 4y2 � 4z2 � 4.y2 � z2 �14
�4 � x2�,
z � r�x� �12�4 � x2;y2 � z2 � r�x�2 50. and therefore,
x2 � y2 � e2z.
y � r�z� � ez;x2 � y2 � r�z�2
498 Chapter 10 Vectors and the Geometry of Space
60.
(a) When we have
Focus:
(b) When we have
Focus: �2, 0, 3�
z � 2 �y2
4, 4�z � 2� � y2.
x � 2
�0, 4, 92�
z �x2
2� 4, 4�1
2� �z � 4� � x2.y � 4
z �x2
2�
y2
4
62. If is on the surface, then
Elliptic paraboloid shifted up 2 units. Traces parallel to xy-plane are circles.
8z � x2 � y2 � 16 ⇒ z �x2
8�
y2
8� 2
z2 � x2 � y2 � z2 � 8z � 16
z2 � x2 � y2 � �z � 4�2
�x, y, z�
58.
y
2π π
0.5
1.0
z
� 2� sin y � y cos y��
0� 2� 2
V � 2���
0y sin y dy
52.
Equation of generating curve:x � cos y or z � cos y
x2 � z2 � cos2 y 54. The trace of a surface is the inter-section of the surface with a plane.You find a trace by setting onevariable equal to a constant, suchas or z � 2.x � 0
56. About x-axis:
About y-axis:
About z-axis: x2 � y2 � r�z�2
x2 � z2 � r�y�2
y2 � z2 � r�x�2
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Section 10.7 Cylindrical and Spherical Coordinates 499
64.
(a)
z � �0.775x2 � 0.007y2 � 22.15x � 0.54y � 45.4
(b)
(d) The traces parallel to the yz-plane (x constant) areconcave upward. That is, for fixed x (worker’scompensation), the rate of increase of z (Medicare)is increasing with respect to y (public assistance).
x
y
z
250
200
150
100
200100
1020
66. Equating twice the first equation with the second equation,
3x � 4y � 6, a plane
4y � 8 � �3x � 2
2x2 � 6y2 � 4z2 � 4y � 8 � 2x2 � 6y2 � 4z2 � 3x � 2
(c) For y constant, the traces parallel to the xz-plane areconcave downward. That is, for fixed y (publicassistance), the rate of increase of z (Medicare) isdecreasing with respect to x (worker’s compensation).
Year 1980 1985 1990 1995 1996 1997
z 37.5 72.2 111.5 185.2 200.1 214.6
Model 37.8 72.0 112.2 185.8 204.5 214.7
Section 10.7 Cylindrical and Spherical Coordinates
2. cylindrical
rectangular�0, 4, �2�,z � �2
y � 4 sin �
2� 4
x � 4 cos �
2� 0
�4, �
2, �2�, 4.
�3�2, �3�2, 2�z � 2
y � 6 sin���
4� � �3�2
x � 6 cos���
4� � 3�2
�6, ��
4, 2�, cylindrical 6. cylindrical
rectangular�0, �1, 1�,z � 1
y � sin 3�
2� �1
x � cos 3�
2� 0
�1, 3�
2, 1�,
8. rectangular
cylindrical�4, ��
4, 4�,
z � 4
� � arctan��1� � ��
4
r � ��2�2�2 � ��2�2�2 � 4
�2�2, �2�2, 4�, 10.
�4, ��
6, 1�, cylindrical
z � 1
� � arctan��1�3� �
5�
6
r � �12 � 4 � 4
�2�3, �2, 6�, rectangular 12. rectangular
cylindrical��13, �arctan23
, �1�,
z � �1
� � arctan��23 � � �arctan
23
r � ���3�2 � 22 � �13
��3, 2, �1�,
14. rectangular equation
cylindrical equationz � r2 � 2
z � x2 � y2 � 2 16. rectangular equation
cylindrical equation r � 8 cos �
r2 � 8r cos �
x2 � y2 � 8x
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500 Chapter 10 Vectors and the Geometry of Space
18.
Same
x
y3 32 1
3
z
z � 2 20.
x
y22
−2−2
4
z
x2 � y2 �z2
4� 0
�x2 � y2 �z2
r �z2
22.
x
y3
2
−2
2
z
�x � 1�2 � y2 � 1
x2 � y2 � 2x � 0
x2 � y2 � 2x
r 2 � 2r cos �
r � 2 cos �
30. rectangular
spherical�4, �, �
2�,
� � arccos 0 ��
2
� � �
� � ���4�2 � 02 � 02 � 4
��4, 0, 0�,
32. spherical
rectangular��2.902, 2.902, 11.276�,
z � 12 cos �
9� 11.276
y � 12 sin �
9 sin
3�
4� 2.902
x � 12 sin �
9 cos
3�
4� �2.902
�12, 3�
4,
�
9�, 34. spherical
rectangular�0, 0, �9�,z � 9 cos � � �9
y � 9 sin � sin �
4� 0
x � 9 sin � cos �
4� 0
�9, �
4, ��,
36. spherical
rectangular��6, 0, 0�,
z � 6 cos �
2� 0
y � 6 sin �
2 sin � � 0
x � 6 sin �
2 cos � � �6
�6, �, �
2�,38. (a) Programs will vary.
(b)
�x, y, z� � �1.295, 2.017, 4.388���, �, �� � �5, 1, 0.5�
24.
xy
z
123
9
654321
z � x2
z � r2 cos2 � 26. rectangular
spherical��3, �
4, arccos
1
�3�,
� � arccos 1
�3
� � arctan 1 ��
4
� � �12 � 12 � 12 � �3
�1, 1, 1�,
28. rectangular
spherical�2�10, �
4, arccos
2
�5�,
� � arccos 2
�5
� � arctan 1 ��
4
� � �22 � 22 � �4�2�2 � 2�10
�2, 2, 4�2�,
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Section 10.7 Cylindrical and Spherical Coordinates 501
40. rectangular equation
(cone) spherical equation � ��
3
cos � �12
1 � 4 cos2 �
�2 � 4 �2 cos2 �
x2 � y2 � z2 � 4z2
x2 � y2 � 3z2 � 0 42. rectangular equation
spherical equation � � 10 csc � sec �
� sin � cos � � 10
x � 10
44.
x
y3
−3
−3
3
z
x � y � 0
�1 �yx
tan � �yx
� �3�
446.
xy-plane
x
y33
−3−3
−3
−2
3
2
z
z � 0
0 �z
�x2 � y2 � z2
cos � �z
�x2 � y2 � z2
� ��
248.
x
y3 32 1
3
z
z � 2
� cos � � 2
� � 2 sec �
50.
x
y664
6
4
z
x � 4
� sin � cos � � 4
�4
sin � cos �
� � 4 csc � sec � 52. cylindrical
spherical�3, ��
4,
�
2�,
� � arccos�09� �
�
2
� � ��
4
� � �32 � 02 � 3
�3, ��
4, 0�, 54. cylindrical
spherical�2�2, 2�
3,
3�
4 �,
� � arccos��1
�2� �3�
4
� �2�
3
� � �22 � ��2�2 � 2�2
�2, 2�
3, �2�,
56. cylindrical
spherical�4�2, �
3,
�
4�,
� � arccos 1
�2�
�
4
� ��
3
� � ���4�2 � 42 � 4�2
��4, �
3, 4�, 58. cylindrical
spherical�5, �
2, arccos
35�,
� � arccos 35
� ��
2
� � �42 � 32 � 5
�4, �
2, 3�, 60. spherical
cylindrical�4, �
18, 0�,
z � 4 cos �
2� 0
� ��
18
r � 4 sin �
2� 4
�4, �
18,
�
2�,
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502 Chapter 10 Vectors and the Geometry of Space
Rectangular Cylindrical Spherical
68.
70.
72.
74.
76.
78.
Note: Use the cylindrical coordinate
80. �9.169, 1.3, 2.022��8.25, 1.3, �4��2.207, 7.949, �4�
�2, 5�
6, 3��
�3.606, 2.618, 0.588���2, 11�
6, 3���1.732, 1, 3�
�6.403, �1.571, 0.896��5, �1.571, 4��0, �5, 4�
�6.708, 0.785, 2.034��6, 0.785, �3��3�2, 3�2, �3��7.5, 0.25, 1��6.311, 0.25, 4.052��6.115, 1.561, 4.052�
�11.662, �0.750, 1.030��10, �0.75, 6��7.317, �6.816, 6�
�7.000, �0.322, 2.014��6.325, �0.322, �3��6, �2, �3�
62. spherical
cylindrical�9, �
3, 9�3�,
z � � cos � � 18 cos �
3� 9�3
� ��
3
r � � sin � � 18 sin �
3� 9
�18, �
3,
�
3�, 64. spherical
cylindrical�0, �5�
6, �5�,
z � 5 cos � � �5
� � �5�
6
r � 5 sin � � 0
�5, �5�
6, ��, 66. spherical
cylindrical�7�22
, �
4, �
7�22 �,
z � 7 cos 3�
4� �
7�22
� ��
4
r � 7 sin 3�
4�
7�22
�7, �
4,
3�
4 �,
82.
Plane
Matches graph (e)
� ��
484.
Cone
Matches graph (a)
� ��
486.
Plane
Matches graph (b)
� � 4 sec �, z � � cos � � 4
88. Cylinder with z-axis symmetry
Plane perpendicular to xy-plane
Plane parallel to xy-planez � c
� � b
r � a 90. Sphere
Vertical half-plane
Half-cone� � c
� � b
� � a
92.
(a)
(b)
tan � �12
, � � arctan 12
4 sin2 � � cos2 �, tan2 � �14
,
4��2 sin2 � cos2 � � �2 sin2 � sin2 �� � �2 cos2 �,
4r2 � z2, 2r � z
4�x2 � y2� � z2 94.
(a)
(b)
� � csc � cot �� �cos �sin2 �
,
� sin2 � � cos �,�2 sin2 � � � cos �,
r 2 � z
x2 � y2 � z
96.
(a)
(b)
� � 4 csc ��� sin � � 4��� sin � � 4� � 0,
�2 sin2 � � 16 � 0,�2 sin2 � � 16,
r2 � 16, r � 4
x2 � y2 � 16 98.
(a)
(b) � � 4 csc � csc �� sin � sin � � 4,
r � 4 csc �r sin � � 4,
y � 4
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Review Exercises for Chapter 10 503
100.
y
x4
−4
4
4
3
z
0 ≤ z ≤ r cos �
0 ≤ r ≤ 3
��
2 ≤ � ≤
�
2102.
y
x5
−5
5
4
3
z
z2 ≤ �r 2 � 6r � 8
2 ≤ r ≤ 4
0 ≤ � ≤ 2� 104.
y
x2
−2−2
2
2
z
0 ≤ � ≤ 1
�
4 ≤ � ≤
�
2
0 ≤ � ≤ 2�
106. Cylindrical: 0.75 ≤ r ≤ 1.25, z � 8
108. Cylindrical
��9 � r2 ≤ z ≤ �9 � r2
0 ≤ � ≤ 2�
12 ≤ r ≤ 3
y
x4
−4
−4
4
z 110. plane
sphere
The intersection of the plane and the sphere is a circle.
� � 4
� � 2 sec � ⇒ � cos � � 2 ⇒ z � 2
Review Exercises for Chapter 10
2.
(a)
(b)
(c) 2u � v � 14i � �4i � 5j� � 18i � 5j
�v� � �42 � 52 � �41
v � PR\
� �4, 5� � 4i � 5ju � PQ\
� �7, 0� � 7i,
R � �2, 4�Q � �5, �1�P � ��2, �1�, 4.
� ��2
4i �
�24
j
v � �v� cos � i � �v� sin � j �1
2 cos 225 i �
1
2 sin 225 j
6. (a) The length of cable POQ is L.
Tension:
Also,
Domain: L > 18 inches
⇒ T �250
��L2�4� � 81
L2
�250L
�L2 � 324cy � 250 ⇒ T �
250y�81 � y2
T � c �OQ\
� � c�81 � y2
L � 2�92 � y2 ⇒ �L2
4� 81 � y
OQ\
� 9i � yj
Q
O
P
18 in.
500 lb
x
θ
−9 9
y
(b)
(c)
18 250
1000
L 19 20 21 22 23 24 25
T 780.9 573.54 485.36 434.81 401.60 377.96 360.24
(d) The line intersects thecurve at
inches.L � 23.06
T � 400 (e)
The maximum tension is 250pounds in each side of the cablesince the total weight is 500 pounds.
limL→�
T � 250
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504 Chapter 10 Vectors and the Geometry of Space
10. Looking towards the xy-plane from the positive z-axis.The point is either in the second quadrant or in the fourth quadrant The z-coordinatecan be any number.
�x > 0, y < 0�.�x < 0, y > 0�
12. Center:
Radius:
�x � 2�2 � �y � 3�2 � �z � 2�2 � 17
��2 � 0�2 � �3 � 0�2 � �2 � 4�2 � �4 � 9 � 4 � �17
0 � 42
, 0 � 6
2,
4 � 02 � �2, 3, 2�
14.
Center:
Radius: 2
�5, �3, 2��x � 5�2 � �y � 3�2 � �z � 2�2 � 4
x
y2
86
4
2
6
4
z�x2 � 10x � 25� � �y2 � 6y � 9� � �z2 � 4z � 4� � �34 � 25 � 9 � 4
16.
x
y1
34
21
56
1
345678
2
z
(3, −3, 8)
(6, 2, 0)
v
v � �3 � 6, �3 � 2, 8 � 0� � ��3, �5, 8� 18.
Since v and w are not parallel, the points do not lie in astraight line.
w � �11 � 5, 6 � 4, 3 � 7� � �6, 10, �4�
v � �8 � 5, �5 � 4, 5 � 7� � �3, �1, �2�
20. 8 �6, �3, 2��49
�8
7�6, �3, 2� � �48
7, �
24
7,
16
7 � 22.
(a)
(b)
(c) v v � 9 � 36 � 9 � 54
u v � ��2��3� � �6��6� � ��2���3� � 36
v � PR\
� �3, 6, �3� � 3i � 6j � 3k
u � PQ\
� ��2, 6, �2� � �2i � 6j � 2k,
R � �5, 5, 0�Q � �0, 5, 1�,P � �2, �1, 3�,
24.
Since the vectors are parallel.v � �4u,
v � �16, �12, 24�u � ��4, 3, �6�, 26.
is orthogonal to v.
� ��
2
u v � 0 ⇒
v � �3, 2, �2�u � �4, �1, 5�,
28.
� 83.9
cos � � �u v��u� �v�
�1
3�10
�v� � 3
�u� � �10
u v � �1
v � �2, �2, 1�
u � �1, 0, �3� 30.
� 300�3 ft lb
W � F PQ\
� �F� �PQ\
� cos � � �75��8�cos 30
8. �0, �7, 0�y � �7:x � z � 0,
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Review Exercises for Chapter 10 505
In Exercises 32–40, w � <�1, 2, 2>.v � <2, �4, �3>,u � <3, �2, 1>,
32.
� � arccos 11
�14�29 56.9
cos � � �u v��u� �v�
�11
�14�2934. Work � �u w� � ��3 � 4 � 2� � 5
36.
Thus, u � v � ��v � u�.
� �10i � 11j � 8kv � u � � i23
j�4�2
k�3
1�u � v � � i
32
j�2�4
k1
�3� � 10i � 11j � 8k
38.
�u � v� � �u � w� � 4i � 4j � 4k � u � �v � w�
u � w � � i3
�1
j�2
2
k12� � �6i � 7j � 4k
u � v � � i32
j�2�4
k1
�3� � 10i � 11j � 8k
u � �v � w� � �3, �2, 1� � �1, �2, �1� � � i31
j�2�2
k1
�1� � 4i � 4j � 4k
40. (See Exercise 35)Area triangle �12
�v � w� �12���2�2 � ��1�2 �
�52
42. V � �u �v � w�� � �200 12
�1
012� � 2�5� � 10 44. Direction numbers: 1, 1, 1
(a)
(b) x � 1 � y � 2 � z � 3
z � 3 � ty � 2 � t,x � 1 � t,
46.
Direction numbers: 21, 11, 13
(a)
(b)x
21�
y � 111
�z � 4
13
z � 4 � 13ty � 1 � 11t,x � 21t,
u � v � � i2
�3
j�5
1
k14� � �21i � 11j � 13k 48.
27x � 4y � 32z � �33
�27�x � 3� � 4� y � 4� � 32�z � 2� � 0
n � PQ\
� PR\
� � i04
j85
k�1�4� � �27i � 4j � 32k
PR\
� �4, 5, �4�PQ\
� �0, 8, �1�,
R � �1, 1, �2�Q � ��3, 4, 1�,P � ��3, �4, 2�,
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506 Chapter 10 Vectors and the Geometry of Space
50. The normal vectors to the planes are the same,
Choose a point in the first plane, Choose apoint in the second plane,
D � �PQ\
n��n�
� ��5��35
�5
�35�
�357
PQ\
� �0, 0, �5�
Q � �0, 0, �3�.P � �0, 0, 2�.
n � �5, �3, 1�.
52.
point on line
D ��PQ
\
� u��u�
��264�6
� 2�11
PQ\
� u � � i�6
1
j�2�2
k�2�1� � ��2, �8, 14�
PQ\
� ��6, �2, �2�
P � �1, 3, 5�
u � �1, �2, �1� direction vector
Q��5, 1, 3� point
54.
Since the x-coordinate is missing,we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola in the yz-coordinate plane.
y
x
42
1
2
3
z
y � z2 56.
Since the x-coordinate is missing,we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is
yx
2 2
4
−2
z
y � cos z.
y � cos z 58.
Cone
xy-trace: point
xz-trace:
yz-trace:
xy33 2
−3 −3
4
z
x2 � y2 � 9z � 4,
z � ±4y3
z � ±4x3
�0,0, 0�
16x2 � 16y2 � 9z2 � 0
60.
Hyperboloid of one sheet
xy-trace:
xz-trace:
yz-trace:y2
4�
z2
100� 1
x2
25�
z2
100� 1
x2
25�
y2
4� 1
xy
5
−5
12
zx2
25�
y2
4�
z2
100� 1 62. Let and revolve the curve about
the x-axis.y � r�x� � 2�x
64. rectangular
(a) cylindrical
(b) spherical�302
, �
3, arccos
3
�10,� � arccos 3
�10,� �
�
3,� ���3
4 2
� 34
2
� 3�32
2
��30
2,
�32
, �
2,
3�32 ,z �
3�32
,� � arctan�3 ��
3,r ���3
4 2
� 34
2
��32
,
�34
, 34
, 3�3
2 ,
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Review Exercises for Chapter 10 503
100.
y
x4
−4
4
4
3
z
0 ≤ z ≤ r cos �
0 ≤ r ≤ 3
��
2 ≤ � ≤
�
2102.
y
x5
−5
5
4
3
z
z2 ≤ �r 2 � 6r � 8
2 ≤ r ≤ 4
0 ≤ � ≤ 2� 104.
y
x2
−2−2
2
2
z
0 ≤ � ≤ 1
�
4 ≤ � ≤
�
2
0 ≤ � ≤ 2�
106. Cylindrical: 0.75 ≤ r ≤ 1.25, z � 8
108. Cylindrical
��9 � r2 ≤ z ≤ �9 � r2
0 ≤ � ≤ 2�
12 ≤ r ≤ 3
y
x4
−4
−4
4
z 110. plane
sphere
The intersection of the plane and the sphere is a circle.
� � 4
� � 2 sec � ⇒ � cos � � 2 ⇒ z � 2
Review Exercises for Chapter 10
2.
(a)
(b)
(c) 2u � v � 14i � �4i � 5j� � 18i � 5j
�v� � �42 � 52 � �41
v � PR\
� �4, 5� � 4i � 5ju � PQ\
� �7, 0� � 7i,
R � �2, 4�Q � �5, �1�P � ��2, �1�, 4.
� ��2
4i �
�24
j
v � �v� cos � i � �v� sin � j �1
2 cos 225 i �
1
2 sin 225 j
6. (a) The length of cable POQ is L.
Tension:
Also,
Domain: L > 18 inches
⇒ T �250
��L2�4� � 81
L2
�250L
�L2 � 324cy � 250 ⇒ T �
250y�81 � y2
T � c �OQ\
� � c�81 � y2
L � 2�92 � y2 ⇒ �L2
4� 81 � y
OQ\
� 9i � yj
Q
O
P
18 in.
500 lb
x
θ
−9 9
y
(b)
(c)
18 250
1000
L 19 20 21 22 23 24 25
T 780.9 573.54 485.36 434.81 401.60 377.96 360.24
(d) The line intersects thecurve at
inches.L � 23.06
T � 400 (e)
The maximum tension is 250pounds in each side of the cablesince the total weight is 500 pounds.
limL→�
T � 250
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504 Chapter 10 Vectors and the Geometry of Space
10. Looking towards the xy-plane from the positive z-axis.The point is either in the second quadrant or in the fourth quadrant The z-coordinatecan be any number.
�x > 0, y < 0�.�x < 0, y > 0�
12. Center:
Radius:
�x � 2�2 � �y � 3�2 � �z � 2�2 � 17
��2 � 0�2 � �3 � 0�2 � �2 � 4�2 � �4 � 9 � 4 � �17
0 � 42
, 0 � 6
2,
4 � 02 � �2, 3, 2�
14.
Center:
Radius: 2
�5, �3, 2��x � 5�2 � �y � 3�2 � �z � 2�2 � 4
x
y2
86
4
2
6
4
z�x2 � 10x � 25� � �y2 � 6y � 9� � �z2 � 4z � 4� � �34 � 25 � 9 � 4
16.
x
y1
34
21
56
1
345678
2
z
(3, −3, 8)
(6, 2, 0)
v
v � �3 � 6, �3 � 2, 8 � 0� � ��3, �5, 8� 18.
Since v and w are not parallel, the points do not lie in astraight line.
w � �11 � 5, 6 � 4, 3 � 7� � �6, 10, �4�
v � �8 � 5, �5 � 4, 5 � 7� � �3, �1, �2�
20. 8 �6, �3, 2��49
�8
7�6, �3, 2� � �48
7, �
24
7,
16
7 � 22.
(a)
(b)
(c) v v � 9 � 36 � 9 � 54
u v � ��2��3� � �6��6� � ��2���3� � 36
v � PR\
� �3, 6, �3� � 3i � 6j � 3k
u � PQ\
� ��2, 6, �2� � �2i � 6j � 2k,
R � �5, 5, 0�Q � �0, 5, 1�,P � �2, �1, 3�,
24.
Since the vectors are parallel.v � �4u,
v � �16, �12, 24�u � ��4, 3, �6�, 26.
is orthogonal to v.
� ��
2
u v � 0 ⇒
v � �3, 2, �2�u � �4, �1, 5�,
28.
� 83.9
cos � � �u v��u� �v�
�1
3�10
�v� � 3
�u� � �10
u v � �1
v � �2, �2, 1�
u � �1, 0, �3� 30.
� 300�3 ft lb
W � F PQ\
� �F� �PQ\
� cos � � �75��8�cos 30
8. �0, �7, 0�y � �7:x � z � 0,
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Review Exercises for Chapter 10 505
In Exercises 32–40, w � <�1, 2, 2>.v � <2, �4, �3>,u � <3, �2, 1>,
32.
� � arccos 11
�14�29 56.9
cos � � �u v��u� �v�
�11
�14�2934. Work � �u w� � ��3 � 4 � 2� � 5
36.
Thus, u � v � ��v � u�.
� �10i � 11j � 8kv � u � � i23
j�4�2
k�3
1�u � v � � i
32
j�2�4
k1
�3� � 10i � 11j � 8k
38.
�u � v� � �u � w� � 4i � 4j � 4k � u � �v � w�
u � w � � i3
�1
j�2
2
k12� � �6i � 7j � 4k
u � v � � i32
j�2�4
k1
�3� � 10i � 11j � 8k
u � �v � w� � �3, �2, 1� � �1, �2, �1� � � i31
j�2�2
k1
�1� � 4i � 4j � 4k
40. (See Exercise 35)Area triangle �12
�v � w� �12���2�2 � ��1�2 �
�52
42. V � �u �v � w�� � �200 12
�1
012� � 2�5� � 10 44. Direction numbers: 1, 1, 1
(a)
(b) x � 1 � y � 2 � z � 3
z � 3 � ty � 2 � t,x � 1 � t,
46.
Direction numbers: 21, 11, 13
(a)
(b)x
21�
y � 111
�z � 4
13
z � 4 � 13ty � 1 � 11t,x � 21t,
u � v � � i2
�3
j�5
1
k14� � �21i � 11j � 13k 48.
27x � 4y � 32z � �33
�27�x � 3� � 4� y � 4� � 32�z � 2� � 0
n � PQ\
� PR\
� � i04
j85
k�1�4� � �27i � 4j � 32k
PR\
� �4, 5, �4�PQ\
� �0, 8, �1�,
R � �1, 1, �2�Q � ��3, 4, 1�,P � ��3, �4, 2�,
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506 Chapter 10 Vectors and the Geometry of Space
50. The normal vectors to the planes are the same,
Choose a point in the first plane, Choose apoint in the second plane,
D � �PQ\
n��n�
� ��5��35
�5
�35�
�357
PQ\
� �0, 0, �5�
Q � �0, 0, �3�.P � �0, 0, 2�.
n � �5, �3, 1�.
52.
point on line
D ��PQ
\
� u��u�
��264�6
� 2�11
PQ\
� u � � i�6
1
j�2�2
k�2�1� � ��2, �8, 14�
PQ\
� ��6, �2, �2�
P � �1, 3, 5�
u � �1, �2, �1� direction vector
Q��5, 1, 3� point
54.
Since the x-coordinate is missing,we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is a parabola in the yz-coordinate plane.
y
x
42
1
2
3
z
y � z2 56.
Since the x-coordinate is missing,we have a cylindrical surface with rulings parallel to the x-axis. The generating curve is
yx
2 2
4
−2
z
y � cos z.
y � cos z 58.
Cone
xy-trace: point
xz-trace:
yz-trace:
xy33 2
−3 −3
4
z
x2 � y2 � 9z � 4,
z � ±4y3
z � ±4x3
�0,0, 0�
16x2 � 16y2 � 9z2 � 0
60.
Hyperboloid of one sheet
xy-trace:
xz-trace:
yz-trace:y2
4�
z2
100� 1
x2
25�
z2
100� 1
x2
25�
y2
4� 1
xy
5
−5
12
zx2
25�
y2
4�
z2
100� 1 62. Let and revolve the curve about
the x-axis.y � r�x� � 2�x
64. rectangular
(a) cylindrical
(b) spherical�302
, �
3, arccos
3
�10,� � arccos 3
�10,� �
�
3,� ���3
4 2
� 34
2
� 3�32
2
��30
2,
�32
, �
2,
3�32 ,z �
3�32
,� � arctan�3 ��
3,r ���3
4 2
� 34
2
��32
,
�34
, 34
, 3�3
2 ,
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Problem Solving for Chapter 10 507
70.
(a) Cylindrical:
(b) Spherical: � � 4
r2 � z2 � 16
x2 � y2 � z2 � 16
66. cylindrical
spherical�54�3, �5�
6,
�
3�,
� � arccos�27�354�3� � arccos
12
��
3
� � �5�
6
� � �6561 � 2187 � 54�3
�81, �5�
6, 27�3�, 68. spherical
cylindrical�6�3, ��
2, �6�,
z � � cos � � 12 cos�2�
3 � � �6
� � ��
2
r2 � �12 sin�2�
3 ��2
⇒ r � 6�3
�12, ��
2,
2�
3 �,
Problem Solving for Chapter 10
2.
(a)
(c) ±��22
, ��22 �
−4 −2
−2
2
4
−4
2 4
y
x
f �x� � x
0
�t 4 � 1 dt
(b)
(d) The line is y � x: x � t, y � t.
u �1�2
�i � j� � ��2
2, �2
2 �
� ��
4
f�0� � 1 � tan �
f�x� � �x 4 � 1
4. Label the figure as indicated.
because
in a rhombus.a � b
�a � b� �b � a� � b2 � a2 � 0,
SQ\
� b � a
PR\
� a � ba
b
S
P Q
R
6.
Figure is a square.
Thus, and the points P form a circle of radiusin the plane with center at P.n
→PP0 � n
�n � PP\
0� � �n � PP\
0�
P0
n + PP0
P
n n
n − PP0
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508 Chapter 10 Vectors and the Geometry of Space
8. (a)
(b) At height
�43
�abc
�2�ab
c2 �c2d �d3
3 �c
0
V � 2c
0
�abc2 �c2 � d2� dd
Area � ���a2�c2 � d2�c2 ��b2�c2 � d2�
c2 � ��abc2 �c2 � d2�
x2
a2�c2 � d2�c2
�y2
b2�c2 � d 2�c2
� 1.
x2
a2 �y2
b2 � 1 �d2
c2 �c2 � d2
c2
x2
a2 �y2
b2 �d 2
c2 � 1
z � d > 0,
V � 2r
0� �r2 � x2� dx � 2��r2x �
x3
3 �r
0�
43
�r 3
12.
(a) direction vector for line
point on line
(b)
The minimum is at s � �1.D 2.2361
−11
−4
10
10
D �PQ
\
� uu
���7 � s�2 � ��6 � 2s�2 � 25
�21
PQ\
� u � � i1
�2
j21
ks � 1
4� � �7 � s�i � ��6 � 2s�j � 5k
PQ\
� �1, 2, s � 1�
P � �3, 1, �1�
u � ��2, 1, 4�
x � �t � 3, y �12
t � 1, z � 2t � 1; Q � �4, 3, s�
(c) Yes, there are slant asymptotes. Using we have
slant asymptotes.y � ±�105
21�s � 1�
��5�21
��x � 1�2 � 21 → ±� 5
21�x � 1�
D�s� �1
�21�5x2 � 10x � 110 �
�5�21
�x2 � 2x � 22
s � x,
10. (a)
Cylinder
r � 2 cos � (b)
Hyperbolic paraboloid
z2 � x2 � y2
z � r2 cos 2�
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(b) As in part (a),
Domain: 0 < � < 90�
⇒ T �3000cos �
6000i � 2T cos �
16. (a) Los Angeles:
Rio de Janeiro:
(b) Los Angeles:
Rio de Janeiro:
(c)
radians
(d) miles
—CONTINUED—
s � 4000�1.59� 6366
� 91.18� 1.59
cos � �u v
u v �
��1568.2��2685.2� � ��2919.7���2523.3� � �2239.7���1556.5��4000��4000�
�2685.2, �2523.3, �1556.5�
z � 4000 cos 112.90�
y � 4000 sin 112.90� sin��43.22��
x � �4000 sin 112.90� cos��43.22��
��1568.2, �2919.7, 2239.7�
z � 4000 cos 55.95�
y � 4000 sin 55.95� sin��118.24��
x � 4000 sin 55.95� cos��118.24��
�4000, �43.22�, 112.90��
�4000, �118.24�, 55.95��
T 3046.3 3192.5 3464.1 3916.2 4667.2 6000.0
60�50�40�30�20�10��
14. (a) The tension T is the same in each tow line.
⇒ T �6000
2 cos 20� 3192.5 lbs
� 2T cos 20�i
6000i � T�cos 20� � cos��20��i � T�sin 20� � sin��20���j
(c)
(d)
0 900
10,000
(e) As increases, there is less force applied in the direction of motion.
�
Problem Solving for Chapter 10 509
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510 Chapter 10 Vectors and the Geometry of Space
20. Essay.18. Assume one of a, b, c, is not zero, say a. Choose a pointin the first plane such as The distancebetween this point and the second plane is
���d1 � d2�
�a2 � b2 � c2�
�d1 � d2��a2 � b2 � c2
.
D ��a��d1�a� � b�0� � c�0� � d2�
�a2 � b2 � c2
��d1�a, 0, 0�.
16. —CONTINUED—
(e) For Boston and Honolulu:
a. Boston:
Honolulu:
b. Boston:
Honolulu:
(f)
radians
(g) miless � 4000�1.28� 5120
� 73.5� 1.28
cos � �u v
u v �
�959.4���3451.7� � ��2795.7���1404.4� � �2695.1��1453.7��4000��4000�
��3451.7, �1404.4, 1453.7�
z � 4000 cos 68.69�
y � 4000 sin 68.69� sin��157.86��
x � �4000 sin 68.69� cos��157.86��
�959.4, �2795.7, 2695.1�
z � 4000 cos 47.64�
y � 4000 sin 47.64� sin��71.06��
x � 4000 sin 47.64� cos��71.06��
�4000, �157.86�, 68.69��
�4000, �71.06�, 47.64��
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C H A P T E R 1 0Vectors and the Geometry of Space
Section 10.1 Vectors in the Plane . . . . . . . . . . . . . . . . . . . . 227
Section 10.2 Space Coordinates and Vectors in Space . . . . . . . . . . 232
Section 10.3 The Dot Product of Two Vectors . . . . . . . . . . . . . . 238
Section 10.4 The Cross Product of Two Vectors in Space . . . . . . . . 241
Section 10.5 Lines and Planes in Space . . . . . . . . . . . . . . . . . 244
Section 10.6 Surfaces in Space . . . . . . . . . . . . . . . . . . . . . . 249
Section 10.7 Cylindrical and Spherical Coordinates . . . . . . . . . . . 252
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 256
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 261
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227
C H A P T E R 1 0Vectors and the Geometry of Space
Section 10.1 Vectors in the PlaneSolutions to Odd-Numbered Exercises
1. (a)
(b)
5432
1
1
3
2
4
5
x
v
(4, 2)
y
v � �5 � 1, 3 � 1� � �4, 2� 3. (a)
(b)
4
2
−2
−2
−4
−4−6−8x
v(−7, 0)
y
v � ��4 � 3, �2 � ��2�� � ��7, 0�
5.
u � v
v � �1 � ��1�, 8 � 4� � �2, 4�
u � �5 � 3, 6 � 2� � �2, 4� 7.
u � v
v � �9 � 3, 5 � 10� � �6, �5�
u � �6 � 0, �2 � 3� � �6, �5�
9. (b)
(a) and (c).
4
4
2
2x
v
(5, 5)
(4, 3)
(1, 2)
y
v � �5 � 1, 5 � 2� � �4, 3� 11. (b)
(a) and (c).
10
2
4
6
2−4x
v
y
(−4, −3)
(6, −1)
(10, 2)
v � �6 � 10, �1 � 2� � ��4, �3�
13. (b)
(a) and (c).
64
6
4
2
2x
v
(6, 6)
(0, 4)
(6, 2)
y
v � �6 � 6, 6 � 2� � �0, 4� 15. (b)
(a) and (c).
21
3
2
−1−2x
5,,−1
v
3( (
1, 3
2( (
3,
243( (
y
v � �12 �
32 , 3 �
43 � � ��1, 53 �
17. (a)
—CONTINUED—6
6
4
2
2 4x
v
v2
(4, 6)
(2, 3)
y
2v � �4, 6� (b)
4
4
−4
−4
−8
−8
xv
−3v
(2, 3)
(−6, −9)
y
�3v � ��6, �9�
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17. —CONTINUED—
(c)
128
8
4
4
122
x
7,
(2, 3)
( (
v
v
21
27
y
72 v � �7, 21
2 � (d)
3
2
1
1 2 3x
v
v
(2, 3)
y
4, 2
3
23
( (
23 v � �4
3 , 2�
19.
x
−u
y 21.
x−v
u
u v−
y
23. (a)
(b)
(c) 2u � 5v � 2�4, 9� � 5�2, �5� � �18, �7�
v � u � �2, �5� � �4, 9� � ��2, �14�
23u �
23�4, 9� � �8
3, 6� 25.
32
−1
1
−2
−3
32
32
v = u
x
u
u
y
� �3, �32 �
v �32 �2i � j� � 3i �
32 j
27.
� 4i � 3j � �4, 3�4
2
4
−2
6
v
v = u + 2w
w2
x
u
yv � �2i � j� � 2�i � 2j� 29.
Q � �3, 5�
u2 � 5
u1 � 3
u2 � 2 � 3
u1 � 4 � �1
31. �v� � �16 � 9 � 5 33. �v� � �36 � 25 � �61 35. �v� � �0 � 16 � 4
37.
unit vector � ��1717
, 4�17
17
v �u
�u��
�3, 12��153
� � 3�153
, 12
�153 �u� � �32 � 122 � �153 39.
unit vector � �3�3434
, 5�34
34
v �u
�u��
��32�, �52���342
� � 3�34
, 5
�34
�u� ���32�
2
� �52�
2
��34
2
228 Chapter 10 Vectors and the Geometry of Space
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41.
(a)
(b)
(c)
(d)
(e)
(f)
� u � v�u � v� � � 1
u � v
�u � v�� �0, 1�
� v�v� � � 1
v
�v��
1
�5 ��1, 2�
� u�u� � � 1
u
�u��
1
�2 �1, �1�
�u � v� � �0 � 1 � 1
u � v � �0, 1�
�v� � �1 � 4 � �5
�u� � �1 � 1 � �2
�u� � �1, �1�, v � ��1, 2� 43.
(a)
(b)
(c)
(d)
(e)
(f)
� u � v�u � v� � � 1
u � v
�u � v��
2
�85 �3,
72
� v�v� � � 1
v
�v��
1
�13 �2, 3�
� u�u� � � 1
u
�u��
2
�5 �1,
12
�u � v� ��9 �494
��85
2
u � v � �3, 72
�v� � �4 � 9 � �13
�u� ��1 �14
��52
u � �1, 12, v � �2, 3�
45.
�u � v� ≤ �u� � �v�
�u � v� � �74 8.602
u � v � �7, 5�
�v� � �41 6.403
v � �5, 4�
�u� � �5 2.236
u � �2, 1� 47.
v � �2�2, 2�2�
4� u�u�� � 2�2 �1, 1�
u
�u��
1
�2 �1, 1�
49.
v � �1, �3 �
2� u�u�� �
1
�3 ��3, 3�
u
�u��
1
2�3 ��3, 3� 51. v � 3��cos 0��i � �sin 0��j� � 3i � �3, 0�
53.
� ��3i � j � ���3, 1�v � 2��cos 150��i � �sin 150��j� 55.
u � v � �2 � 3�22 � i �
3�22
j
v �3�2
2 i �
3�22
j
u � i
Section 10.1 Vectors in the Plane 229
http://librosysolucionarios.net
57.
u � v � �2 cos 4 � cos 2� i � �2 sin 4 � sin 2�j
v � �cos 2�i � �sin 2�j
u � 2�cos 4�i � 2�sin 4�j 59. A scalar is a real number. A vector is represented by a directed line segment. A vector has both length anddirection.
61. To normalize , you find a unit vector in the direction of
u �v
�v�.
v:uv
For Exercises 63–67, au � bw � a�i � 2j� � b�i � j� � �a � b�i � �2a � b�j.
63. Therefore, Solvingsimultaneously, we have b � 1.a � 1,
2a � b � 1.a � b � 2,v � 2i � j. 65. Therefore, Solvingsimultaneously, we have a � 1, b � 2.
2a � b � 0.a � b � 3,v � 3i.
67. Therefore, Solvingsimultaneously, we have a �
23 , b �
13 .
2a � b � 1.a � b � 1,v � i � j.
69.
(a) Let then
(b) Let then
w�w�
� ±1
�10 �3, �1�.
w � �3, �1�,m � �13 .
w�w�
� ±1
�10 �1, 3�.
w � �1, 3�,m � 3.
y � x3, y� � 3x2 � 3 at x � 1. 71.
(a) Let then
(b) Let then
w�w�
� ±15
�3, 4�
w � �3, 4�,m �43.
w�w�
� ±15
��4, 3�.
w � ��4, 3�,m � �34 .
f��x� ��x
�25 � x2�
�34
at x � 3.
f �x� � �25 � x2
73.
v � �u � v� � u � ��22
i ��22
j
u � v � �2 j
u ��22
i ��22
j 75. Programs will vary.
77.
�R � �F1�F2�F3 132.5�
�R� � �F1 � F2 � F3� 1.33
�F3� � 2.5, �F3� 110�
�F2� � 3, �F2� �125�
�F1� � 2, �F1� 33�
79. (a)
Direction:
Magnitude:
—CONTINUED—
�430.882 � 902 � 440.18 newtons
� � arctan� 90430.88� � 0.206� � 11.8��
180�cos 30i � sin 30j� � 275i � 430.88i � 90j (b)
� � arctan� 180 sin �275 � 180 cos ��
M � ��275 � 180 cos ��2 � �180 sin ��2
230 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
(c)
M 455 440.2 396.9 328.7 241.9 149.3 95
037.1�40.1�33.2�23.1�11.8�0��
180�150�120�90�60�30�0��
(d)
0 1800
α
50
0 1800
M
500 (e) M decreases because the forces change from acting inthe same direction to acting in the opposite directionas increases from 0� to 180�.�
81.
�R � �F1�F2�F3 71.3�
�R� � �F1 � F2 � F3� 228.5 lb
� �752�3 � 50�2 �
1252 �i � �75
2� 50�2 �
1252�3�j
F1 � F2 � F3 � �75 cos 30� i � 75 sin 30�j� � �100 cos 45�i � 100 sin 45� j� � �125 cos 120� i � 125 sin 120� j�
83. (a) The forces act along the same direction.
(c) No, the magnitude of the resultant can not be greaterthan the sum.
� � 0�. (b) The forces cancel out each other. � � 180�.
85.
x8642
8
6
4
2
−4
( 4, 1)− −
(1, 2)
(3, 1)
(8, 4)
y
��4, �1�, �6, 5�, �10, 3�
x8642
8
6
4
2
−4
−2−4 −2
(1, 2)
(3, 1)
(8, 4)(6, 5)
y
x8 1064
8
6
4
2
−4
−2−2
(1, 2)
(3, 1)
(8, 4)
(10, 3)
y
87.
Vertical components:
Horizontal components:
Solving this system, you obtain
and �v� 1758.8.�u� 1305.5
�u� cos 30� � �v� cos 130� � 0
�u� sin 30� � �v� sin 130� � 2000
v �→CA � �v��cos 130� i � sin 130� j�
A B
C30°
30°50° 130°
uv
y
x
u �→CB � �u��cos 30� i � sin 30� j�
89. Horizontal component
Vertical component � �v� sin � � 1200 sin 6� 125.43 ftsec
� �v� cos � � 1200 cos 6� 1193.43 ftsec
79. —CONTINUED—
Section 10.1 Vectors in the Plane 231
http://librosysolucionarios.net
91.
�u � v� � ���692.53�2 � �547.64�2 � 882.9 km�hr.
� � arctan� 547.64�692.53 � �38.34�. 38.34� North of West.
� �692.53 i � 547.64 j
u � v � 900 cos 148� � 100 cos 45��i � 900 sin 148� � 100 sin 45�� j
v � 100cos 45� i � sin 45� j�
u � 900cos 148� i � sin 148� j�
93.
and
Finally, T2 � 157.32
T3�0.97495� � 3600 ⇒ T3 � 3692.48T2 ��T3 cos 92�
cos 35� ⇒
T3 cos 92�
cos 35� sin 35� � T3 sin 92� � 3600
�T2 cos 35� � T3 sin 92� � 3600
T2 cos 35� � T3 cos 92� � 0
�3600j � T2�cos 35�i � sin 35� j� � T3�cos 92�i � sin 92�j� � 0
F1 � F2 � F3 � 0
95. Let the triangle have vertices at and Let u be the vector joining and as indicated in the figure. Then v, the vector joining the midpoints, is
�b2
i �c2
j �12
�bi � cj� �12
u
v � �a � b2
�a2i �
c2
j
�b, c�,�0, 0�
x
a b+ c
a
2 2
2
,( (
(( , 0
( , )b c
( , 0)a(0, 0)
u
v
y�b, c�.�a, 0�,�0, 0�,
97.
Thus, and w bisects the angle between u and v.�w � ��u � �v��2
tan �w �
sin��u � �v
2 cos��u � �v
2 cos��u � �v
2 cos��u � �v
2 � tan��u � �v
2
� 2�u� �v��cos��u � �v
2 cos��u � �v
2 i � sin��u � �v
2 cos��u � �v
2 j � �u� �v� �cos �u � cos �v�i � �sin �u � sin �v� j� � �u��v� cos �v i � �v� sin �v j� � �v��u� cos �ui � �u� sin �u j�
w � �u�v � �v�u
99. True 101. True 103. False
�a i � bj� � �2 �a�
Section 10.2 Space Coordinates and Vectors in Space
1.
x
y4324
12
3
3
456
z
(2, 1, 3) (−1, 2, 1)
3.
x
y32
−2
1
4
12
3
3
21
−2
−3
z
(5, −2, 2)
(5, −2, −2)
232 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
5.
B��1, �2, 2�
A�2, 3, 4� 7. x � �3, y � 4, z � 5: ��3, 4, 5� 9. y � z � 0, x � 10: �10, 0, 0�
11. The z-coordinate is 0. 13. The point is 6 units above the xy-plane.
15. The point is on the plane parallel to the yz-plane thatpasses through x � 4.
17. The point is to the left of the xz-plane.
19. The point is on or between the planes and y � �3.y � 3 21. The point is 3 units below the xy-plane, and beloweither quadrant I or III.
�x, y, z�
23. The point could be above the xy-plane and thus above quadrants II or IV,or below the xy-plane, and thus below quadrants I or III.
25.
� �25 � 4 � 36 � �65
d � ��5 � 0�2 � �2 � 0�2 � �6 � 0�2 27.
� �25 � 0 � 36 � �61
d � ��6 � 1�2 � ��2 � ��2��2 � ��2 � 4�2
29.
Right triangle
�BC�2 � �AB�2 � �AC�2
�BC� � �0 � 36 � 9 � 3�5
�AC� � �4 � 16 � 16 � 6
�AB� � �4 � 4 � 1 � 3
A�0, 0, 0�, B�2, 2, 1�, C�2, �4, 4� 31.
Since the triangle is isosceles.�AB� � �AC�,�BC� � �36 � 4 � 0 � 2�10
�AC� � �4 � 16 � 16 � 6
�AB� � �16 � 4 � 16 � 6
A�1, �3, �2�, B�5, �1, 2�, C��1, 1, 2�
33. The z-coordinate is changed by 5 units:
�0, 0, 5�, �2, 2, 6�, �2, �4, 9�
35. �5 � ��2�2
, �9 � 3
2,
7 � 32 � �3
2, �3, 5
37. Center:
Radius: 2
x2 � y2 � z2 � 4y � 10z � 25 � 0
�x � 0�2 � �y � 2�2 � �z � 5�2 � 4
�0, 2, 5� 39. Center:
Radius:
x2 � y2 � z2 � 2x � 6y � 0
�x � 1�2 � �y � 3�2 � �z � 0�2 � 10
�10
�2, 0, 0� � �0, 6, 0�2
� �1, 3, 0�
41.
Center:
Radius: 5
�1, �3, �4�
�x � 1�2 � �y � 3�2 � �z � 4�2 � 25
�x2 � 2x � 1� � �y2 � 6y � 9� � �z2 � 8z � 16� � �1 � 1 � 9 � 16
x2 � y2 � z2 � 2x � 6y � 8z � 1 � 0
Section 10.2 Space Coordinates and Vectors in Space 233
http://librosysolucionarios.net
43.
Center:
Radius: 1
�13
, �1, 0
�x �13
2
� �y � 1�2 � �z � 0�2 � 1
�x2 �23
x �19 � �y2 � 2y � 1� � z2 � �
19
�19
� 1
x2 � y2 � z2 �23
x � 2y �19
� 0
9x2 � 9y2 � 9z2 � 6x � 18y � 1 � 0 45.
Solid ball of radius 6 centered at origin.
x2 � y2 � z2 ≤ 36
47. (a)
(b)
x
y432
11
−3
−2
23
2
1
3
4
5
z
−2, 2, 2
� �2i � 2j � 2k � ��2, 2, 2�
v � �2 � 4�i � �4 � 2�j � �3 � 1�k 49. (a)
(b)
x
y432
11
−3
−2
23
2
1
3
4
5
z
−3, 0, 3
� �3i � 3k � ��3, 0, 3�
v � �0 � 3�i � �3 � 3� j � �3 � 0�k
51.
Unit vector:�1, �1, 6��38
� � 1�38
, �1�38
, 6
�38���1, �1, 6�� � �1 � 1 � 36 � �38
�4 � 3, 1 � 2, 6 � 0� � �1, �1, 6� 53.
Unit vector: ��1�2
, 0, �1�2�
���1, 0, �1�� � �1 � 1 � �2
��5 � ��4�, 3 � 3, 0 � 1� � ��1, 0, �1�
55. (b)
(a) and (c).
x
y42
−2
2
4
2
3
4
5
z
(−1, 2, 3)(3, 3, 4)
(0, 0, 0)
(4, 1, 1)v
� 4i � j � k � �4, 1, 1�
v � �3 � 1�i � �3 � 2�j � �4 � 3�k 57.
Q � �3, 1, 8�
�q1, q2, q3� � �0, 6, 2� � �3, �5, 6�
59. (a)
—CONTINUED—
x
y21
1−2
23
4
2
3
4
5
z
2, 4, 4
2v � �2, 4, 4� (b)
x
32
1
−3−2
−2−3
23
2
−2
−3
3
z
−1, −2, −2
�v � ��1, �2, �2�
234 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
59. —CONTINUED—
(c)
x
1
−3−2
−2−3
23
y
2
−2
−3
3
z
3, 3, 3
2
32 v � �3
2 , 3, 3� (d)
x
12
3
−3−2
−2−3
21
3y
2
1
−2
−1
−3
3
z
0, 0, 0
0v � �0, 0, 0�
61. z � u � v � �1, 2, 3� � �2, 2, �1� � ��1, 0, 4�
63. z � 2u � 4v � w � �2, 4, 6� � �8, 8, �4� � �4, 0, �4� � �6, 12, 6�
65.
z � �72 , 3, 52 �
2z3 � 9 � �4 ⇒ z3 �52
2z2 � 6 � 0 ⇒ z2 � 3
2z1 � 3 � 4 ⇒ z1 �72
2z � 3u � 2�z1, z2, z3� � 3�1, 2, 3� � �4, 0, �4� 67. (a) and (b) are parallel since
and �2, 43 , �103 � �
23 �3, 2, �5�.
��6, �4, 10� � �2�3, 2, �5�
69.
(a) is parallel since �6i � 8j � 4k � 2z.
z � �3i � 4j � 2k 71.
Therefore, and are parallel. The points arecollinear.
PR\
PQ\
�3, 6, 9� �32 �2, 4, 6�
PR\
� �2, 4, 6�
PQ\
� �3, 6, 9�
P�0, �2, �5�, Q�3, 4, 4�, R�2, 2, 1�
73.
Since and are not parallel, the points are notcollinear.
PR\
PQ\
PR\
� ��1, �1, 1�
PQ\
� �1, 3, �4�
P�1, 2, 4�, Q�2, 5, 0�, R�0, 1, 5� 75.
Since and , the given points form thevertices of a parallelogram.
AC\
� BD\
AB\
� CD\
BD\
� ��2, 1, 1�
AC\
� ��2, 1, 1�
CD\
� �1, 2, 3�
AB\
� �1, 2, 3�
A�2, 9, 1�, B�3, 11, 4�, C�0, 10, 2�, D�1, 12, 5�
77. �v� � 0 79.
�v� � �1 � 4 � 9 � �14
v � �1, �2, �3� 81.
�v� � �0 � 9 � 25 � �34
v � �0, 3, �5�
83.
(a)
(b) �u
�u�� �
13
�2, �1, 2�
u�u�
�13
�2, �1, 2�
�u� � �4 � 1 � 4 � 3
u � �2, �1, 2� 85.
(a)
(b) �u
�u�� �
1
�38�3, 2, �5�
u�u�
�1
�38�3, 2, �5�
�u� � �9 � 4 � 25 � �38
u � �3, 2, �5� 87. Programs will vary.
Section 10.2 Space Coordinates and Vectors in Space 235
http://librosysolucionarios.net
93. v �32
u
�u��
32�
23
, �23
, 13� � �1, �1,
12�
95.
x
−2
−2−1
2
1y
2
1
−2
−1
z
0, 3, 1
0, 3, −1
� �3 j ± k � �0, �3, ±1�v � 2cos�±30��j � sin�±30��k� 97.
�4, 3, 0� � ��2, �4, 2� � �2, �1, 2�
23v � ��2, �4, 2�
v � ��3, �6, 3�
99. (a)
(c)
w � u � v
a � 1, b � 1
ai � �a � b�j � bk � i � 2j � k
1
1
1
v
u
yx
z (b)
Thus, a and b are both zero.
(d)
Not possible
a � 1, a � b � 2, b � 3
a i � �a � b�j � bk � i � 2j � 3k
a � 0, a � b � 0, b � 0
w � au � bv � ai � �a � b�j � bk � 0
101. d � ��x2 � x1�2 � �y2 � y1�2 � �z2 � z1�2 103. Two nonzero vectors u and v are parallel if for some scalar c.
u � cv
105. (a) The height of the right triangle is The vector is given by
The tension vector T in each wire is
Hence, and
(b)
—CONTINUED—
�8
�L2 � 182�182 � �L2 � 182� �
8L
�L2 � 182 T � �T� �
8h�182 � h2
T �8h
�0, �18, h�
T � c�0, �18, h� where ch �243
� 8.
PQ\
� �0, �18, h�.
PQ\
Q
P
L
(0, 0, 0)
(0, 18, 0)
18
(0, 0, )hh � �L2 � 182.
L 20 25 30 35 40 45 50
T 18.4 11.5 10 9.3 9.0 8.7 8.6
89.
c � ±53
9c2 � 25
�cv� � �4c2 � 4c2 � c2 � 5
cv � �2c, 2c, �c� 91.
� �0, 10�2
, 10�2�
v � 10u
�u�� 10�0,
1
�2,
1
�2�
236 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
107. Let be the angle between v and the coordinate axes.
y
x
0.2
0.2
0.4
0.6
0.4
0.4
0.6
3 3 33 3 3
, ,( (
z
v ��33
�i � j � k� ��33
�1, 1, 1�
cos � �1
�3�
�33
�v� � �3 cos � � 1
v � �cos ��i � �cos ��j � �cos ��k
�
109.
Thus:
Solving this system yields and Thus:
�F3� � 226.521N
�F2� � 157.909N
�F1� � 202.919N
C3 � �11269 .C1 �
10469 , C2 �
2823,
115� C1 � C2 � C3� � 500
70C1 � 65C3 � 0
� 60C2 � 45C3 � 0
F � F1 � F2 � F3 � �0, 0, 500�
AD\
� �45, �65, 115�, F3 � C3�45, �65, 115�
AC\
� ��60, 0, 115�, F2 � C2��60, 0, 115�
AB\
� �0, 70, 115�, F1 � C1�0, 70, 115�
105. —CONTINUED—
(c)
is a vertical asymptote and is ahorizontal asymptote.
(d)
(e) From the table, implies inches.L � 30T � 10
limL→�
8L
�L2 � 182� lim
L →�
8
�1 � �18�L�2� 8
limL→18�
8L
�L2 � 182� �
y � 8x � 18
0 1000
30 L = 18
T = 8
111.
Sphere; center: radius:2�11
3�43
, 3, �13,
449
� �x �43
2
� �y � 3�2 � �z �13
2
�6 �169
� 9 �19
� �x2 �83
x �169 � �y2 � 6y � 9� � �z2 �
23
z �19
0 � 3x2 � 3y2 � 3z2 � 8x � 18y � 2z � 18
x2 � y2 � z2 � 2y � 2z � 2 � 4�x2 � y2 � z2 � 2x � 4y � 5�
�x2 � �y � 1�2 � �z � 1�2 � 2��x � 1�2 � �y � 2�2 � z2
d�AP� � 2d�BP�
Section 10.2 Space Coordinates and Vectors in Space 237
http://librosysolucionarios.net
Section 10.3 The Dot Product of Two Vectors
1.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2��6� � �12
�u � v�v � �6�2, �3� � ��12, 18�
�u�2 � 25
u � u � 3�3� � 4�4� � 25
u � v � 3�2� � 4��3� � �6
u � �3, 4�, v � �2, �3� 3.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2�2� � 4
�u � v�v � 2�0, 6, 5� � �0, 12, 10�
�u�2 � 29
u � u � 2�2� � ��3���3� � 4�4� � 29
u � v � 2�0� � ��3��6� � �4��5� � 2
u � �2, �3, 4�, v � �0, 6, 5�
5.
(a)
(b)
(c)
(d)
(e) u � �2v� � 2�u � v� � 2
�u � v�v � v � i � k
�u�2 � 6
u � u � 2�2� � ��1���1� � �1��1� � 6
u � v � 2�1� � ��1��0� � 1��1� � 1
u � 2i � j � k, v � i � k 7.
This gives the total amount that the person earned on hisproducts.
u � v � $17,139.05
v � �2.22, 1.85, 3.25�
u � �3240, 1450, 2235�
9.
u � v � �8��5� cos �
3� 20
u � v�u� �v�
� cos � 11.
� ��
2
cos � �u � v
�u� �v��
0
�2�8� 0
u � �1, 1�, v � �2, �2�
13.
� � arccos��1
5�2 98.1�
cos � �u � v
�u� �v��
�2
�10�20�
�1
5�2
u � 3i � j, v � �2i � 4j 15.
� � arcos �23
61.9�
cos � �u � v
�u� �v��
2
�3�6�
�23
u � �1, 1, 1�, v � �2, 1, �1�
17.
� � arccos��8�13
65 116.3�
cos � �u � v
�u� �v��
�8
5�13�
�8�1365
u � 3i � 4j, v � �2j � 3k 19.
not parallel
not orthogonal
Neither
u � v � 4 0 ⇒
u cv ⇒
u � �4, 0�, v � �1, 1�
21.
not parallel
orthogonalu � v � 0 ⇒
u cv ⇒
u � �4, 3�, v � �12
, �23� 23.
not parallel
not orthogonal
Neither
u � v � �8 0 ⇒
u cv ⇒
u � j � 6k, v � i � 2j � k
25.
not parallel
orthogonalu � v � 0 ⇒
u cv ⇒
u � �2, �3, 1�, v � ��1, �1, �1�
238 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
27.
cos2 � cos2 � � cos2 � �19
�49
�49
� 1
cos � �23
cos � �23
cos �13
u � i � 2j � 2k, �u� � 3 29.
cos2 � cos2 � � cos2 � � 0 �9
13�
413
� 1
cos � � �2
�13
cos � �3
�13
cos � 0
u � �0, 6, �4�, �u� � �52 � 2�13
31.
cos � ��2�17
⇒ � 2.0772 or 119.0�
cos � �2
�17 ⇒ � 1.0644 or 61.0�
cos �3
�17 ⇒ 0.7560 or 43.3�
u � �3, 2, �2� �u� � �17 33.
cos � �2
�30 ⇒ � 1.1970 or 68.6�
cos � �5
�30 ⇒ � 0.4205 or 24.1�
cos ��1�30
⇒ 1.7544 or 100.5�
u � ��1, 5, 2� �u� � �30
35.
cos � �14.1336
�F� ⇒ � 96.53�
cos � 59.5246
�F� ⇒ � 61.39�
cos 108.2126
�F� ⇒ 29.48�
�F� 124.310 lb
� �108.2126, 59.5246, �14.1336�
4.3193�10, 5, 3� � 5.4183�12, 7, �5�
F � F1 � F2
F2: C2 �80
�F2� 5.4183
F1: C1 �50
�F1� 4.3193 37. Let
y
x
v
s
s
s
z
� � � � � arccos� 1
�3 54.7�
cos � cos � � cos � �s
s�3�
1
�3
�v� � s�3
v � �s, s, s�
s � length of a side.
39.
�1
�2 ⇒ � � � � 45�
cos � � cos � �10
�02 � 102 � 102
cos �0
�02 � 102 � 102� 0 ⇒ � 90�
OA\
� �0, 10, 10� 41. w2 � u � w1 � �6, 7� � �2, 8� � �4, �1�
43. w2 � u � w1 � �0, 3, 3� � ��2, 2, 2� � �2, 1, 1� 45.
(a)
(b) w2 � u � w1 � ��12
, 52�
w1 � �u � v�v�2 v �
1326
�5, 1� � �52
, 12�
u � �2, 3�, v � �5, 1�
Section 10.3 The Dot Product of Two Vectors 239
http://librosysolucionarios.net
47.
(a)
(b) w2 � u � w1 � �2, �8
25,
625�
�1125
�0, 3, 4� � �0, 3325
, 4425�
w1 � �u � v�v�2 v
u � �2, 1, 2�, v � �0, 3, 4� 49. u � v � �u1, u2, u3� � �v1, v2, v3� � u1v1 � u2v2 � u3v3
53. See page 738. Direction cosines of are
are the direction angles. See Figure 10.26., �, and �
cos �v1
�v�, cos � �
v2
�v�, cos � �
v3
�v�.
v � �v1, v2, v3�
57. Programs will vary. 59. Programs will vary.
55. (a) and v are parallel.
(b) and v
are orthogonal.
�u � v�v�2 v � 0 ⇒ u � v � 0 ⇒ u
�u � v�v�2 v � u ⇒ u � cv ⇒ u
61. Because u appears to be perpendicular to v, the projection of u onto v is 0. Analytically,
projv u �u � v�v�2 v �
�2, �3� � �6, 4���6, 4��2 �6, 4� � 0�6, 4� � 0.
63. Want
and are orthogonal to u.�v � �8i � 6jv � 8i � 6j
u � v � 0.u �12
i �23
j. 65. Want
and are orthogonal to u.�v � �0, �2, �1�v � �0, 2, 1�
u � v � 0.u � �3, 1, �2�.
51. (a) Orthogonal, � ��
2(b) Acute, 0 < � <
�
2(c) Obtuse,
�
2< � < �
71.
W � PQ\
� v � 72
v � �1, 4, 8�
PQ\
� �4, 7, 5�
73. False. Let and Then and .u � w � 10 � 20 � 30u � v � 2 � 28 � 30w � �5, 5�.v � �1, 7�u � �2, 4�,
67. (a) Gravitational Force
�w1� 8335.1 lb
�8335.1�cos 10� i � sin 10� j�
w1 �F � v�v�2 v � �F � v�v � ��48,000��sin 10��v
v � cos 10� i � sin 10� j
F � �48,000 j
69.
W � F � v � 425 ft � lb
v � 10i
F � 85�12
i ��32
j
(b)
�w2� 47,270.8 lb
� 8208.5 i � 46,552.6 j
w2 � F � w1 � �48,000 j � 8335.1�cos 10� i � sin 10� j�
75. In a rhombus, The diagonals are and
Therefore, the diagonals are orthogonal.
� �u�2 � �v�2 � 0
� u � u � v � u � u � v � v � v
�u � v� � �u � v� � �u � v� � u � �u � v� � v u
v
u v+
u v−u � v.u � v�u� � �v�.
240 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
77.
The angle between u and v is Also,
cos�� � �� �u � v
�u� �v��
cos � cos � � sin � sin ��1��1� � cos � cos � � sin � sin �.
�Assuming that � > ��.� � �.
u � �cos �, sin �, 0�, v � �cos �, sin �, 0�
79.
� �u�2 � �v�2 � 2u � v
� �u�2 � u � v � u � v � �v�2
� u � u � v � u � u � v � v � v
� �u � v� � u � �u � v� � v
�u � v�2 � �u � v� � �u � v� 81.
Therefore, �u � v� ≤ �u� � �v�.
≤ ��u� � �v��2
≤ �u�2 � 2�u� �v� � �v�2 from Exercise 66
� �u�2 � 2u � v � �v�2
� u � u � v � u � u � v � v � v
� �u � v� � u � �u � v� � v
�u � v�2 � �u � v� � �u � v�
Section 10.4 The Cross Product of Two Vectors in Space
1.
x y
i
j
−k1
1
1
−1
z
j � i � � i01
j10
k00� � �k 3.
x y
i
j
k
11
1
−1
z
j � k � � i00
j10
k01� � i 5.
x y1
−1
1
1
−1
i
k−j
z
i � k � � i10
j00
k01� � �j
7. (a)
(b)
(c) v � v � � i33
j77
k22� � 0
v � u � ��u � v� � �22, �16, 23�
u � v � � i�2
3
j37
k42� � ��22, 16, �23� 9. (a)
(b)
(c) v � v � 0
v � u � ��u � v� � ��17, 33, 10�
u � v � � i71
j3
�1
k25� � �17, �33, �10�
11.
v � �u � v� � 1��1� � ��2���1� � �1���1� � 0 ⇒ v � u � v
� 0 ⇒ u � u � vu � �u � v� � 2��1� � ��3���1� � �1���1�
u � v � � i21
j�3�2
k11� � �i � j � k � ��1, �1, �1�
u � �2, �3, 1�, v � �1, �2, 1�
Section 10.4 The Cross Product of Two Vectors in Space 241
http://librosysolucionarios.net
13.
� 0 ⇒ v � u � v
v � �u � v� � �2�0� � 5�0� � 0�54�
� 0 ⇒ u � u � v
u � �u � v� � 12�0� � ��3��0� � 0�54�
u � v � � i12
�2
j�3
5
k00� � 54k � �0, 0, 54�
u � �12, �3, 0�, v � ��2, 5, 0� 15.
� 0 ⇒ v � u � v
v � �u � v� � 2��2� � 1�3� � ��1���1�
� 0 ⇒ u � u � v
u � �u � v� � 1��2� � 1�3� � 1��1�
u � v � � i12
j11
k1
�1� � �2i � 3j � k � ��2, 3, �1�
u � i � j � k, v � 2i � j � k
17.
x
y
v
u4
64
12
3
1
32
456
z 19.
x
y
v
u4
64
12
3
1
32
456
z
21.
u � v�u � v�
� � �140
24,965,
�46
24,965,
57
24,965
u � v � ��70, �23, 572
v � ��1, 8, 4�
u � �4, �3.5, 7� 23.
� ��71
7602, �
44
7602,
25
7602
u � v�u � v�
�20
7602��
7120
, �115
, 54
u � v � ��7120
, �115
, 54
v �12
i �34
j �1
10k
u � �3i � 2j � 5k
25. Programs will vary.
27.
A � �u � v� � �i� � 1
u � v � � i00
j11
k01� � i
v � j � k
u � j 29.
A � �u � v� � ��8, �10, 4�� � 180 � 65
u � v � � i31
j22
k�1
3� � �8, �10, 4�
v � �1, 2, 3�
u � �3, 2, �1�
31.
Since and the figure is a parallelo-gram. and are adjacent sides and
A � �AB\
� AC\
� � 332 � 283
AB\
� AC\
� � i15
j24
k31� � �10i � 14j � 6k.
AC\
AB\
AC\
� BD\
,AB\
� CD\
BD\
� �5, 4, 1� CD
\
� �1, 2, 3�,AB\
� �1, 2, 3�, AC\
� �5, 4, 1�,
A�1, 1, 1,�, B�2, 3, 4�, C�6, 5, 2�, D�7, 7, 5� 33.
A �12
�AB\
� AC\
� �12117 �
3213
AB\
� AC\
� � i1
�3
j20
k30� � �9j � 6k
AB\
� �1, 2, 3�, AC\
� ��3, 0, 0�
A�0, 0, 0�, B�1, 2, 3�, C��3, 0, 0�
242 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
35.
Area �12
�AB\
� AC\
� �1216,742
AB\
� AC\
� � i�3
2
j1213
k5
�4� � ��113, �2, �63�
AB\
� ��3, 12, 5�, AC\
� �2, 13, �4�
A�2, �7, 3�, B��1, 5, 8�, C�4, 6, �1� 37.
x
y
40°
12
ft
PQ
F
z
�PQ\
� F� � 10 cos 40 � 7.66 ft � lb
PQ\
� F � � i00
jcos 40�2
0
ksin 40�2
�20 � � �10 cos 40i
PQ\
�12
�cos 40 j � sin 40k�
F � �20k
39. (a)
(b) When
(c) Let
when
This is what we expected. When the pipe wrench is horizontal. � 90
� 90.dTd
� 90 cos � 0
T � 90 sin .
� 45: �OA\
� F� � 90 22 � � 452 � 63.64.
�OA\
� F� � 90 sin
OA\
� F � � i00
j0
�60 sin
k3�2
�60 cos � � 90 sin i
F � �60�sin j � cos k�
00
180
100
x
y
OA F
1.5 ftθ
z OA\
�32
k
41. u � �v � w� � �100 010
001� � 1 43. u � �v � w� � �200 0
30
101� � 6 45.
V � �u � �v � w�� � 2
u � �v � w� � �101 110
011� � 2
47.
V � �u � �v � w�� � 75
u � �v � w� � �302 050
015� � 75
w � �2, 0, 5�
v � �0, 5, 1�
u � �3, 0, 0�
49. � �u2v3 � u3v2�i � �u1v3 � u 3v1�j � �u1v2 � u 2v1�k u � v � �u1, u2, u3� � �v1, v2, v3�
51. The magnitude of the cross product will increase by a factor of 4.
53. If the vectors are ordered pairs, then the cross product does not exist. False.
55. True
Section 10.4 The Cross Product of Two Vectors in Space 243
http://librosysolucionarios.net
57.
� �u � v� � �u � w�
�u1w3 � u3w1�j � �u1w2 � u2w1�k
� �u2v3 � u3v2�i � �u1v3 � u3v1�j � �u1v2 � u2v1�k � �u2w3 � u3w2�i �
� �u2�v3 � w3� � u3�v2 � w2�� i � �u1�v3 � w3� � u3�v1 � w1�� j � �u1�v2 � w2� � u2�v1 � w1��k
u � �v � w� � � iu1
v1 � w1
ju2
v2 � w2
ku3
v3 � w3�u � �u1, u2, u3�, v � �v1, v2, v3�, w � �w1, w2, w3�
59.
u � u � � iu1
u1
ju2
u2
ku3
u3� � �u2u3 � u3u2�i � �u1u3 � u3u1�j � �u1u2 � u2u1�k � 0
u � �u1, u2, u3�
61.
Thus, and u � v � v.u � v � u
�u � v� � v � �u2v3 � u3v2�v1 � �u3v1 � u1v3�v2 � �u1v2 � u2v1�v3 � 0
�u � v� � u � �u2v3 � u3v2�u1 � �u3v1 � u1v3�u2 � �u1v2 � u2v1�u3 � 0
u � v � �u2v3 � u3v2�i � �u1v3 � u3v1�j � �u1v2 � u2v1�k
63.
If u and v are orthogonal, and Therefore, u � v � u v.sin � � 1.� � �2
u � v � u v sin �
1.
(a)
y
x
z
x � 1 � 3t, y � 2 � t, z � 2 � 5t (b) When we have When we have
The components of the vector and the coefficients of areproportional since the line is parallel to
(c) when Thus, and Point:
when Point:
when Point: ��15
, 125
, 0�t � �25
.z � 0
�0, 73
, 13�t � �
13
.x � 0
�7, 0, 12�z � 12.x � 7t � 2.y � 0
PQ\
.t
PQ\
� �9, �3, 15�
Q � �10, �1, 17�.t � 3P � �1, 2, 2�.t � 0
3. Point:
Direction vector:
Direction numbers: 1, 2, 3
(a) Parametric:
(b) Symmetric: x �y2
�z3
x � t, y � 2t, z � 3t
v � �1, 2, 3�(0, 0, 0� 5. Point:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric:x � 2
2�
y4
�z � 3�2
x � �2 � 2t, y � 4t, z � 3 � 2t
2, 4, �2
v � �2, 4, �2���2, 0, 3�
Section 10.5 Lines and Planes in Space
244 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
7. Point:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric:x � 1
3�
y�2
�z � 1
1
x � 1 � 3t, y � �2t, z � 1 � t
3, �2, 1
v � 3i � 2j � k
�1, 0, 1� 9. Points:
Direction vector:
Direction numbers:
(a) Parametric:
(b) Symmetric:x � 5
17�
y � 3�11
�z � 2�9
x � 5 � 17t, y � �3 � 11t, z � �2 � 9t
17, �11, �9
v �173
i �113
j � 3k
��23
, 23
, 1��5, �3, �2�,
11. Points:
Direction vector:
Direction numbers: 8, 5, 12
(a) Parametric:
(b) Symmetric:x � 2
8�
y � 35
�z
12
x � 2 � 8t, y � 3 � 5t, z � 12t
�8, 5, 12�
�2, 3, 0�, �10, 8, 12� 13. Point:
Direction vector:
Direction numbers: 0, 0, 1
Parametric: x � 2, y � 3, z � 4 � t
v � k
�2, 3, 4�
15. Point:
Direction vector:
Direction numbers:
Parametric:
Symmetric:
(a) On line
(b) On line
(c) Not on line
(d) Not on line �6 � 24
�2 � 1
�1 ��y 3�
x � 24
�z � 1�1
, y � 3
x � �2 � 4t, y � 3, z � 1 � t
4, 0, �1
v � 4i � k
(�2, 3, 1� 17. on line
on line
not on line
not parallel to
Hence, and are identical.
and are parallel.L3L1 � L2
L2L1
L1, L2, nor L3L4: v � �6, 4, �6�
�6, �2, 5�L3: v � ��6, 4, 8�
�6, �2, 5�L2: v � �6, �4, �8�
�6, �2, 5�Li: v � ��3, 2, 4�
19. At the point of intersection, the coordinates for one line equal the corresponding coordinates for the other line. Thus,
(i) (ii) and (iii)
From (ii), we find that and consequently, from (iii), Letting we see that equation (i) is satisfied and therefore the two lines intersect. Substituting zero for or for we obtain the point
(First line)
(Second line)
cos � � �u � v�u v
�8 � 1
17 9�
7
3 17�
7 1751
v � 2i � 2j � k
u � 4i � k
(2, 3, 1�.t,ss � t � 0,t � 0.s � 0
�t � 1 � s � 1.3 � 2s � 3,4t � 2 � 2s � 2,
21. Writing the equations of the lines in parametric form we have
For the coordinates to be equal, and Solving this system yields and When using these values for and the coordinates are not equal. The lines do not intersect.zt,s
s �117 .t �
1772 � t � �2 � s.3t � 1 � 4s
x � 1 � 4s y � �2 � s z � �3 � 3s.
x � 3t y � 2 � t z � �1 � t
23.
Point of intersection: �7, 8, �1�
z � �t � 1 z � 2s � 1
y � 5t � 2 y � s � 8
x y
68
10
4
2
4
−8
−24
68
10
(7, 8, 1)−
zx � 2t � 3 x � �2s � 7
Section 10.5 Lines and Planes in Space 245
http://librosysolucionarios.net
25.
(a)
PQ\
� �0, �2, 1�, PR\
� �3, 4, 0�
P � �0, 0, �1�, Q � �0, �2, 0�, R � �3, 4, �1�
4x � 3y � 6z � 6
(b)
The components of the cross product are proportionalto the coefficients of the variables in the equation. Thecross product is parallel to the normal vector.
PQ\
� PR\
� � i03
j�2
4
k10� � ��4, 3, 6�
27. Point:
x � 2 � 0
1�x � 2� � 0�y � 1� � 0�z � 2� � 0
n � i � �1, 0, 0��2, 1, 2� 29. Point:
Normal vector:
2x � 3y � z � 10
2�x � 3� � 3�y � 2� � 1�z � 2� � 0
n � 2i � 3j � k
�3, 2, 2�
31. Point:
Normal vector:
x � y � 2z � 12
�x � y � 2z � 12 � 0
�1�x � 0� � 1�y � 0� � 2�z � 6� � 0
n � �i � j � 2k
�0, 0, 6� 33. Let u be the vector from to
Let v be the vector from to
Normal vector:
3x � 9y � 7z � 0
�3�x � 0� � 9�y � 0� � 7�z � 0� � 0
� �3i � ��9�j � 7k
u � v � � i1
�2
j23
k33�
v � �2i � 3j � 3k��2, 3, 3�:�0, 0, 0�
u � i � 2j � 3k�1, 2, 3�:�0, 0, 0�
35. Let u be the vector from to
Let v be the vector from to
Normal vector:
4x � 3y � 4z � 10
4�x � 1� � 3�y � 2� � 4�z � 3� � 0
� 4i � 3j � 4k�12 u� � ��v� � � i
12
j04
k�1
1�v � �2i � 4j � k��1, �2, 2�:�1, 2, 3�
u � 2i � 2k�3, 2, 1�:�1, 2, 3�
37. Normal vector: v � k, 1�z � 3� � 0, z � 3�1, 2, 3�, 39. The direction vectors for the lines are
Normal vector:
Point of intersection of the lines:
x � y � z � 5
�x � 1� � �y � 5� � �z � 1� � 0
��1, 5, 1�
� �5�i � j � k�u � v � � i�2�3
j14
k1
�1�v � �3i � 4j � k.
u � �2i � j � k,
41. Let v be the vector from to
Let n be a vector normal to the plane
Since v and n both lie in the plane p, the normal vector to p is
7x � y � 11z � 5
7�x � 2� � 1�y � 2� � 11�z � 1� � 0
v � n � � i32
j1
�3
k21� � 7i � j � 11k
n � 2i � 3j � k2x � 3y � z � 3:
v � 3i � j � 2k�2, 2, 1�:��1, 1, �1�
246 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
43. Let and let v be the vector from to
Since u and v both lie in the plane the normal vector tois:
y � z � �1
�y � ��2�� � �z � ��1�� � 0
u � v � � i11
j07
k07� � �7j � 7k � �7�j � k�
PP,
v � i � 7j � 7k�2, 5, 6�:�1, �2, �1�u � i 45. The normal vectors to the planes are
Thus, and the planes are orthogonal.� � �2
n1 � �5, �3, 1�, n2 � �1, 4, 7�, cos� ��n1 � n2� n1 n2
� 0.
47. The normal vectors to the planes are
Therefore, � � arccos�4 138414 � � 83.5.
cos � ��n1 � n2�n1 n2
��5 � 3 � 6� 46 27
�4 138
414.
n1 � i � 3j � 6k, n2 � 5i � j � k,
49. The normal vectors to the planes are and Since the planes areparallel, but not equal.
n2 � 5n1,n2 � �5, �25, �5�.n1 � �1, �5, �1�
51.
xy
6
6
4
6
4
z
4x � 2y � 6z � 12 53.
x
y−1
−4
3
3
2
z
2x � y � 3z � 4 55.
x y6
6
6
z
y � z � 5
57.
x y55
3
z
x � 5 59.
yx
24
6
−6
24
6
Generated by Maple
z
2x � y � z � 6 61.
Generated by Maple
y
x
1
−2
−12
z
�5x � 4y � 6z � 8 � 0
63. on plane
not on plane
on plane
and are identical.
is parallel to P2.P1 � P4
P4P1
�1, �1, 1�P4: n � �75, �50, 125�
P3: n � ��3, 2, 5�
�1, �1, 1�P2: n � ��6, 4, �10�
�1, �1, 1�P1: n � �3, �2, 5� 65. Each plane passes through the points
and �0, 0, c�.�c, 0, 0�, �0, c, 0�,
Section 10.5 Lines and Planes in Space 247
http://librosysolucionarios.net
67. The normals to the planes are andThe direction vector for the line is
Now find a point of intersection of the planes.
Substituting 2 for in the second equation, we haveor Letting a point
of intersection is
x � 2, y � 1 � t, z � 1 � 2t
�2, 1, 1�.y � 1,z � 2y � 1.�4y � 2z � �2
x
x � 2
7x � 14
x � 4y � 2z � 0
6x � 4y � 2y � 14
n2 � n1 � � i13
j�4
2
k2
�1� � 7� j � 2k�.
n2 � i � 4j � 2k.n1 � 3i � 2j � k 69. Writing the equation of the line in parametric form and
substituting into the equation of the plane we have:
Substituting into the parametric equations for theline we have the point of intersection The linedoes not lie in the plane.
�2, �3, 2�.t � 32
2�12
� t� � 2��32
� t� � ��1 � 2t� � 12, t �32
x �12
� t, y ��32
� t, z � �1 � 2t
71. Writing the equation of the line in parametric form andsubstituting into the equation of the plane we have:
contradiction
Therefore, the line does not intersect the plane.
2�1 � 3t� � 3��1 � 2t� � 10, �1 � 10,
x � 1 � 3t, y � �1 � 2t, z � 3 � t
73. Point:
Plane:
Normal to plane:
Point in plane:
Vector
D � �PQ\
� n�n
� ��12� 14
�6 14
7
PQ\
� ��6, 0 0�P�6, 0, 0�
n � �2, 3, 1�2x � 3y � z � 12 � 0
Q�0, 0, 0�
75. Point:
Plane:
Normal to plane:
Point in plane:
Vector:
D � �PQ\
� n�n
�11 6
�11 6
6
PQ\
� �2, 8, �1�
P�0, 0, 5�
n � �2, 1, 1�
2x � y � z � 5
Q�2, 8, 4� 77. The normal vectors to the planes are andSince the planes are parallel.
Choose a point in each plane.
is a point in is a point in
PQ\
� ��4, 0, 0�, D ��PQ
\
� n1n1
�4
26�
2 2613
x � 3y � 4z � 6.Q � �6, 0, 0�x � 3y � 4z � 10.P � �10, 0, 0�
n1 � n2,n2 � �1, �3, 4�.n1 � �1, �3, 4�
79. The normal vectors to the planes are andSince the planes are
parallel. Choose a point in each plane.
is a point in
is a point in
D � �PQ\
� n1�n1
� ��272� 94
�27
2 94�
27 94
188
PQ\
� �256
, 1, �1�
6x � 12y � 14z � 25.Q � �256
, 0, 0��3x � 6y � 7z � 1.P � �0, �1, 1�
n2 � �2n1,n2 � �6, �12, �14�.n1 � ��3, 6, 7� 81. is the direction vector for the line.
is the given point, and is on theline. Hence, and
D �PQ
\
� uu
� 149 17
� 2533
17
PQ\
� u � � i34
j20
k�3�1� � ��2, �9, �8�
PQ\
� �3, 2, �3�P��2, 3, 1�Q�1, 5, �2�
u � �4, 0, �1�
83. The parametric equations of a line L parallel to and passing through the point are
The symmetric equations are
x � x1
a�
y � y1
b�
z � z1
c.
x � x1 � at, y � y1 � bt, z � z1 � ct.
P�x1, y1, z1�v � �a, b, c,� 85. Solve the two linear equations representing the planes
to find two points of intersection. Then find the linedetermined by the two points.
248 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
Section 10.6 Surfaces in Space
89. (a)
(b) An increase in x or y will cause a decrease in z. In fact,any increase in two variables will cause a decrease in the third.
(c)
x y3030
30
z
(0, 0, 28.7)
(15.7, 0, 0)(0, 26.3, 0)
z � 28.7 � 1.83x � 1.09y
91. True
Year 1980 1985 1990 1994 1995 1996 1997
z (approx.) 16.16 14.23 9.81 8.60 8.42 8.27 8.23
1. Ellipsoid
Matches graph (c)
3. Hyperboloid of one sheet
Matches graph (f)
5. Elliptic paraboloid
Matches graph (d)
7.
Plane parallel to thexy-coordinate plane
xy
3 22
2
zz � 3 9.
The x-coordinate is missing so we have a cylindrical surface with rulings parallel to the x-axis. The generatingcurve is a circle.
x y47 6
4
z
y2 � z2 � 9
11.
The z-coordinate is missing so we have a cylindrical sur-face with rulings parallel to the z-axis. The generatingcurve is a parabola.
xy
44
3 32
4
z
y � x2 13.
The z-coordinate is missing so we have a cylindrical surface with rulings parallel to the z-axis. The generatingcurve is an ellipse.
xy
233
−3
2
3
z
x2
1�
y2
4� 1
4x2 � y2 � 4
87. (a) Sphere
x2 � y2 � z2 � 6x � 4y � 10z � 22 � 0
�x � 3�2 � �y � 2�2 � �z � 5�2 � 16
(b) Parallel planes
4x � 3y � z � 10 ± 4�n� � 10 ± 4�26
Section 10.6 Surfaces in Space 249
http://librosysolucionarios.net
15.
The x-coordinate is missing so we have a cylindricalsurface with rulings parallel to the x-axis. The generatingcurve is the sine curve.
z � sin y
x
y3
3
4
2
1
z
17.
(a) You are viewing the paraboloid from the x-axis:
(b) You are viewing the paraboloid from above, but not on the z-axis:
(c) You are viewing the paraboloid from the z-axis:
(d) You are viewing the paraboloid from the y-axis: �0, 20, 0��0, 0, 20�
�10, 10, 20��20, 0, 0�
x � x2 � y2
19.
Ellipsoid
xy-trace: ellipse
xz-trace: circle
yz-trace: ellipsey2
4�
z2
1� 1
x2 � z2 � 1
x2
1�
y2
4� 1
xy2
2
2
−2
zx2
1�
y 2
4�
z2
1� 1 21.
Hyperboloid on one sheet
xy-trace: hyperbola
xz-trace: circle
yz-trace: hyperbola�y2
4� 4z2 � 1
4�x2 � z2� � 1
4x2 �y2
4� 1
4x2 �y2
4� 4z2 � 1
x y3
2
−2−3
3
2
3
−3
−2
z16x2 � y2 � 16z2 � 4
29.
Ellipsoid with center �1, 2, 0�.
�x � 1�2
1�
�y � 2�2
16�9�
z2
1� 1
16�x � 1�2 � 9�y � 2�2 � 16z2 � 16
16�x2 � 2x � 1� � 9�y2 � 4y � 4� � 16z2 � �36 � 16 � 36
xy4
2
1
1
2
2
−2
z 16x2 � 9y2 � 16z2 � 32x � 36y � 36 � 0
23.
Elliptic paraboloid
xy-trace:
xz-trace:
point
yz-trace:
xy3 4
21
−3
3
2
1
3
−3
−2
z
y � 1: x2 � z2 � 1
y � z2
�0, 0, 0�x2 � z2 � 0,
y � x2
x2 � y � z2 � 0 25.
Hyperbolic paraboloid
xy-trace:
xz-trace:
yz-trace:
x y2 23 3
3
z
y � ±1: z � 1 � x2
z � y2
z � �x2
y � ±x
x2 � y 2 � z � 0 27.
Elliptic Cone
xy-trace: point
xz-trace:
yz-trace:
xy2 2
1
2
−2
−2
z
z � ±1: x2 �y2
4� 1
z �±12
y
z � ±x
�0, 0, 0�
z2 � x2 �y2
4
250 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
31.
z
x
3
3 y
π
z � 2 sin x 33.
yx
21
−1−2
5
z
z � ±�x2 � 4y2
z2 � x2 � 4y2 35.
x y4
4
4
z
y � ±�4z2 � x2
x2 � y2 � �2z�
2
37.
33 4 5
5
45
yx
z
z � 4 � �xy 39.
yx
−2
246
8
−4
−8−64
8
6
−2
−4
−6
−8
z
z � ±�y2
4� x2 � 4
4x2 � y2 � 4z2 � �16 41.
xy1
2
2
3
2
−2
−2
z
x2 � y2 � 1
2�x2 � y2 � 2
z � 2
z � 2�x2 � y2
43.
z � 0
x � z � 2
x
y32
4
2
3
3
zx2 � y2 � 1 45. and therefore,
x2 � z2 � 4y.
z � r�y� � ±2�y;x2 � z2 � r�y��2
47. and therefore,
4x2 � 4y2 � z2.x2 � y2 �z2
4,
y � r�z� �z2
;x2 � y2 � r�z��2 49. and therefore,
y2 � z2 �4x2.y2 � z2 � �2
x�2
,
y � r�x� �2x
;y2 � z2 � r�x��2
51.
Equation of generating curve: y � �2z or x � �2z
x2 � y2 � ��2z �2
x2 � y2 � 2z � 0 53. Let C be a curve in a plane and let L be a line not in aparallel plane. The set of all lines parallel to L andintersecting C is called a cylinder.
55. See pages 765 and 766. 57.
� 2��4x3
3�
x4
4 4
0�
128�
3
x1 2 3 4
4
3
2
1
p x( )
h x( )
z
V � 2��4
0x�4x � x2� dx
Section 10.6 Surfaces in Space 251
http://librosysolucionarios.net
59.
(a) When we have
Major axis:
Minor axis:
Foci: �0, ±2, 2�c2 � a2 � b2, c2 � 4, c � 2
2�4 � 4
2�8 � 4�2
or 1 �x2
4�
y2
82 �
x2
2�
y2
4,z � 2
z �x2
2�
y2
4
(b) When we have
Major axis:
Minor axis:
Foci: �0, ±4, 8�c2 � 32 � 16 � 16, c � 4
2�16 � 8
2�32 � 8�2
or 1 �x2
16�
y 2
32.8 �
x2
2�
y 2
4,z � 8
61. If is on the surface, then
Elliptic paraboloid
Traces parallel to xz-plane are circles.
x2 � z2 � 8y
y2 � 4y � 4 � x2 � y2 � 4y � 4 � z2
�y � 2�2 � x2 � �y � 2)2 � z2
�x, y, z� 63.
x
y4000
4000
4000
z
x2
39632 �y2
39632 �z2
39422 � 1
65.
Letting you obtain the two intersecting linesand
z � 2abt � a2b2.y � bt � ab2x � at,z � 0y � �bt,x � at,
x � at,
y � ±ba �x �
a2b2 � �
ab2
2
�x �
a2b2 �
2
a2 ��y �
ab2
2 �2
b2
1a2�x2 � a2 bx �
a4b2
4 � �1b2�y2 � ab2y �
a2b4
4 �
bx � ay �y2
b2 �x2
a2
z �y2
b2 �x2
a2 , z � bx � ay 67. The Klein bottle does not have both an “inside” and an“outside.” It is formed by inserting the small open endthrough the side of the bottle and making it contiguouswith the top of the bottle.
Section 10.7 Cylindrical and Spherical Coordinates
1. cylindrical
rectangular�5, 0, 2�,z � 2
y � 5 sin 0 � 0
x � 5 cos 0 � 5
�5, 0, 2�, 3. cylindrical
rectangular�1, �3, 2�,z � 2
y � 2 sin �
3� �3
x � 2 cos �
3� 1
�2, �
3, 2�, 5. cylindrical
rectangular��2�3, �2, 3�,z � 3
y � 4 sin 7�
6� �2
x � 4 cos 7�
6� �2�3
�4, 7�
6, 3�,
252 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
7. rectangular
cylindrical�5, �
2, 1�,
z � 1
� � arctan 50
��
2
r � ��0�2 � �5�2 � 5
�0, 5, 1�, 9. rectangular
cylindrical�2, �
3, 4�,
z � 4
� � arctan�3 ��
3
r � �12 � ��3�2 � 2
�1, �3, 4�, 11. rectangular
cylindrical�2�2, ��
4, �4�,
z � �4
� � arctan��1� � ��
4
r � �22 � ��2�2 � 2�2
�2, �2, �4�,
13. rectangular equation
cylindrical equation r2 � z2 � 10
x2 � y2 � z2 � 10 15. rectangular equation
cylindrical equation r � sec � � tan �
sin � � r cos2 �
r sin � � �r cos ��2
y � x2
17.
xy3
23
2
3
2
−2
−3
−3
z
x2 � y2 � 4
�x2 � y2 � 2
r � 2 19.
x � �3y � 0
x � �3y
1
�3�
yx
tan �
6�
yx
x
y
2 12
1
2
−2
−2
z� ��
621.
x
y12 2
1
2
−2
−2
−1
z
x2 � �y � 1�2 � 1
x2 � y2 � 2y � 0
x2 � y2 � 2y
r 2 � 2r sin �
r � 2 sin �
23.
x2 � y2 � z2 � 4
x y21
2
1
2
−2
−1
−2
zr2 � z2 � 4 25. rectangular
spherical�4, 0, �
2�,
� � arccos 0 ��
2
� � arctan 0 � 0
� �42 � 02 � 02 � 4
�4, 0, 0�,
27. rectangular
spherical�4�2, 2�
3,
�
4�,
� � arccos 1
�2�
�
4
� � arctan���3� �2�
3
� ���2�2 � �2�3�2 � 42 � 4�2
��2, 2�3, 4�, 29. rectangular
spherical�4, �
6,
�
6�,
� � arccos �32
��
6
� � arctan 1
�3�
�
6
� �3 � 1 � 12 � 4
��3, 1, 2�3�,
Section 10.7 Cylindrical and Spherical Coordinates 253
http://librosysolucionarios.net
31. spherical
rectangular��6, �2, 2�2 �,z � 4 cos
�
4� 2�2
y � 4 sin �
4 sin
�
6� �2
x � 4 sin �
4 cos
�
6� �6
�4, �
6,
�
4�, 33. spherical
rectangular�0, 0, 12�,z � 12 cos 0 � 12
y � 12 sin 0 sin���
4 � � 0
x � 12 sin 0 cos���
4 � � 0
�12, ��
4, 0�, 35. spherical
rectangular�52
, 52
, �5�2
2 �,
z � 5 cos 3�
4� �
5�22
y � 5 sin 3�
4 sin
�
4�
52
x � 5 sin 3�
4 cos
�
4�
52
�5, �
4,
3�
4 �,
37. (a) Programs will vary.
(b)
�, �, �� � �5.385, �0.927, 1.190��x, y, z� � �3, �4, 2�
39. rectangular equation
spherical equation 2 � 36
x2 � y2 � z2 � 36
41. rectangular equation
spherical equation � 3 csc �
sin � � 3
2 sin2 � � 9
2 sin2 � cos2 � � 2 sin2 � sin2 � � 9
x2 � y2 � 9 43.
x2 � y2 � z2 � 4
x y21
2
1
2
−2
−1
−2
z � 2
45.
3x2 � 3y2 � z2 � 0
34
�z2
x2 � y2 � z2
�32
�z
�x2 � y2 � z2
cos � �z
�x2 � y2 � z2
x
y
21 1
2
2
−2−1
−1
−1
−2
z� ��
647.
x2 � y2 � �z � 2�2 � 4
x2 � y2 � z2 � 4z � 0
�x2 � y2 � z2 �4z
�x2 � y2 � z2
xy
3 32 1
2
5
4
3
2
−2 −3
z � 4 cos �
49.
x y1
21
2
1
2
−2
−2 −2
−1
z
x2 � y2 � 1
�x2 � y2 � 1
sin � � 1
� csc � 51. cylindrical
spherical�4, �
4,
�
2�,
� � arccos 0 ��
2
� ��
4
� �42 � 02 � 4
�4, �
4, 0�, 53. cylindrical
spherical�4�2, �
2,
�
4�,
� � arccos� 4
4�2� ��
4
� ��
2
� �42 � 42 � 4�2
�4, �
2, 4�,
254 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
61. spherical
cylindrical�36, �, 0�,
z � cos � � 36 cos �
2� 0
� � �
r � sin � � 36 sin �
2� 36
�36, �, �
2�, 63. spherical
cylindrical�3�3, ��
6, 3�,
z � 6 cos �
3� 3
� � ��
6
r � 6 sin �
3� 3�3
�6, ��
6,
�
3�, 65. spherical
cylindrical�4, 7�
6, 4�3�,
z � 8 cos �
6�
8�32
� �7�
6
r � 8 sin �
6� 4
�8, 7�
6,
�
6�,
Rectangular Cylindrical Spherical
67.
69.
71.
73.
75.
77.
79.
[Note: Use the cylindrical coordinates �3.5, 5.642, 6��
�6.946, 5.642, 0.528���3.5, 2.5, 6��2.804, �2.095, 6�
�7.071, 2.356, 2.356��5, 3�
4, �5���3.536, 3.536, �5�
�3.206, 0.490, 2.058��2.833, 0.490, �1.5��52
, 43
, �32 �
�4.123, �0.588, 1.064��3.606, �0.588, 2��3, �2, 2�
�20, 2�
3,
�
4��14.142, 2.094, 14.142���7.071, 12.247, 14.142�
�9.434, 0.349, 0.559��5, �
9, 8��4.698, 1.710, 8�
�7.810, 0.983, 1.177��7.211, 0.983, 3��4, 6, 3�
81.
Cylinder
Matches graph (d)
r � 5 83.
Sphere
Matches graph (c)
� 5 85.
Paraboloid
Matches graph (f)
r2 � z, x2 � y2 � z
55. cylindrical
spherical
�2�13, ��
6, arccos
3
�13�,
� � arccos 3
�13
� ���
6
� �42 � 62 � 2�13
�4, ��
6, 6�,
57. cylindrical
spherical�13, �, arccos 513�,
� � arccos 513
� � �
� �122 � 52 � 13
�12, �, 5�,59. spherical
cylindrical�10, �
6, 0�,
z � 10 cos �
2� 0
� ��
6
r � 10 sin �
2� 10
�10, �
6,
�
2�,
87. Rectangular to cylindrical:
Cylindrical to rectangular:
z � z
y � r sin �
x � r cos �
z � z
tan � �yx
r2 � x2 � y2 89. Rectangular to spherical:
Spherical to rectangular:
z � cos �
y � sin � sin �
x � sin � cos �
� � arccos� z�x2 � y2 � z2�
tan � �yx
2 � x2 � y2 � z2
Section 10.7 Cylindrical and Spherical Coordinates 255
http://librosysolucionarios.net
91.
(a)
(b) �2 � 16, � � 4
r 2 � z2 � 16
x2 � y2 � z2 � 16 93.
(a)
(b)
� � 2 cos �
��� � 2 cos �� � 0,�2 � 2� cos � � 0,
r 2 � z2 � 2z � 0, r 2 � �z � 1�2 � 1
x2 � y2 � z2 � 2z � 0
95.
(a)
(b)
� � 4 sin � csc �� �4 sin �sin �
,
� sin ��� sin � � 4 sin �� � 0,
�2 sin2 � � 4� sin � sin �,
r � 4 sin �r 2 � 4r sin �,
x2 � y2 � 4y 97.
(a)
(b)
�2 �9 csc2 �
cos2 � � sin2 �
�2 sin2 � �9
cos2 � � sin2 �,
�2 sin2 � cos2 � � �2 sin2 � sin2 � � 9,
r 2 �9
cos2 � � sin2 �
r 2 cos2 � � r 2 sin2 � � 9,
x2 � y2 � 9
99.
x
y23
1
2 3
5
3
2
z
0 ≤ z ≤ 4
0 ≤ r ≤ 2
0 ≤ � ≤ �
2101.
x ya a
−a −a
a
z
r ≤ z ≤ a
0 ≤ r ≤ a
0 ≤ � ≤ 2� 103.
x
y
a
30°
z
0 ≤ � ≤ a sec �
0 ≤ � ≤ �
6
0 ≤ � ≤ 2�
105. Rectangular
0 ≤ z ≤ 10
0 ≤ y ≤ 10
0 ≤ x ≤ 10
y
x
1010
10
z 107. Spherical
4 ≤ � ≤ 6
y
x
8
−8
−8
8
z
109.
The curve of intersection is the ellipse formed by the intersection of the plane and the cylinder r � 1.z � y
z �yr
�y1
� y
z � sin �, r � 1
Review Exercises for Chapter 10
1.
(a)
(b)
(c) 2u � v � �6, �2� � �4, 2� � �10, 0� � 10i
�v� � �42 � 22 � 2�5
v � PR\
� �4, 2� � 4i � 2j
u � PQ\
� �3, �1� � 3i � j,
R � �5, 4�Q � �4, 1�,P � �1, 2�, 3.
� �4i � 4�3j
v � �v� cos � i � �v� sin � j � 8 cos 120 i � 8 sin 120 j
256 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
5.
�10
�11� 3.015 ft�
2
�115 y �
2
tanarccos�56��
tan � �2y ⇒ y �
2tan �
� � arccos�56
θ
y
100 lb120 lb
2 ft120 cos � � 100
7. ��5, 4, 0�x � �5:y � 4,z � 0,
11. �x � 3�2 � �y � 2�2 � �z � 6�2 � �152
2
13.
Center:
Radius: 3
x
y543
234
45
6 6
z
�2, 3, 0��x � 2�2 � �y � 3�2 � z2 � 9
�x2 � 4x � 4� � �y2 � 6y � 9� � z2 � �4 � 4 � 9 15.
x
y321
545
3
1
32
−2
z
(2, −1, 3)
(4, 4, −7)
v � �4 � 2, 4 � 1, �7 � 3� � �2, 5, �10�
17.
Since the points lie in a straight line.�2w � v,
w � �5 � 3, 3 � 4, �6 � 1� � �2, �1, �5�
v � ��1 � 3, 6 � 4, 9 � 1� � ��4, 2, 10� 19. Unit vector:u
�u��
�2, 3, 5��38
� � 2�38
, 3
�38,
5�38�
21.
(a)
(b)
(c) v v � 9 � 36 � 45
u v � ��1���3� � 4�0� � 0�6� � 3
v � PR\
� ��3, 0, 6� � �3i � 6k
u � PQ\
� ��1, 4, 0� � �i � 4j,
R � �2, 0, 6�Q � �4, 4, 0�,P � �5, 0, 0�, 23.
Since the vectors are orthogonal.u v � 0,
v � ��1, 4, 5�u � �7, �2, 3�,
9. Looking down from the positive x-axis towards the yz-plane,the point is either in the first quadrant or inthe third quadrant The x-coordinate can beany number.
�y < 0, z < 0�.�y > 0, z > 0�
25.
� � arccos �2 � �6
4� 15
cos � � �u v��u� �v�
��5�22��1 � �3 �
5�2� ��2 � �6
4
�v� � 2
�u� � 5
u v �5�2
2�1 � �3�
v � 2�cos 2�
3i � sin
2�
3j � �i � �3 j
u � 5�cos 3�
4i � sin
3�
4j �
5�22
�i � j� 27.
is parallel to v and in the oppositedirection.
� � �
u � �5v ⇒ u
v � ��2, 1, �3�u � �10, �5, 15�,
Review Exercises for Chapter 10 257
http://librosysolucionarios.net
29. There are many correct answers. For example: v � ± �6, �5, 0�.
In Exercises 31–39, w � <�1, 2, 2>.v � <2, �4, �3>,u � <3, �2, 1>,
31.
� 14 � ��14 �2� �u�2
u u � 3�3� � ��2���2� � �1��1� 33.
� ��1514
, 57
, �5
14�
� ��1514
, 1014
, �5
14� � �
514
�3, �2, 1�
projuw � �u w�u�2 u
35.
n�n�
�1
�5��2i � j�
�n� � �5
n � v � w � � i2
�1
j�4
2
k�3
2� � �2i � j 37.
� ��3, �2, 1� ��2, �1, 0�� � ��4� � 4
V � �u �v � w��
39. (See Exercises 36, 38)
� �285
Area parallelogram � �u � v� � �102 � 112 � ��8�2
41.
�F� � 100�1 � tan2 20 � 100 sec 20 � 106.4 lb
F �100
cos 20�cos 20j � sin 20k� � 100�j � tan 20k�
c �100
cos 20
200 � �PQ\
� F� � 2c cos 20
PQ\
� F � � i00
j0
c cos 20
k2
c sin 20� � �2c cos 20i
PQ\
� 2k
x
y
70°
PQF
2 ft
zF � c�cos 20j � sin 20k�
43.
(a)
(b) None
z � 3y � 2 � t,x � 1,
v � j 45.
Solving simultaneously, we have Substituting into the second equation we have Substitutingfor x in this equation we obtain two points on the line ofintersection, The direction vector ofthe line of intersection is
(a)
(b) z � 1x � y � 1,
z � 1y � �1 � t,x � t,
v � i � j.�1, 0, 1�.�0, �1, 1�,
y � x � 1.z � 1z � 1.
x � y � 2z � 33x � 3y � 7z � �4,
258 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
47. The two lines are parallel as they have the same directionnumbers, 1, 1. Therefore, a vector parallel to theplane is A point on the first line is
and a point on the second line is The vector connecting these two pointsis also parallel to the plane. Therefore, a normal to theplane is
Equation of the plane:
x � 2y � 1
�x � 1� � 2y � 0
� �2i � 4j � �2�i � 2j�.
v � u � � i�2
2
j1
�1
k1
�3�u � 2i � j � 3k
��1, 1, 2�.�1, 0, �1�v � �2i � j � k.
�2,49.
A point P on the plane is
D � �PQ\
n��n�
�87
n � �2, �3, 6�
PQ\
� ��2, 0, 2�
�3, 0, 0�.
2x � 3y � 6z � 6
Q � �1, 0, 2�
51.
normal to plane
D � �PQ\
n��n�
�10�30
��30
3
PQ\
� ��2, �2, 4�
n � �2, �5, 1�
P�5, 0, 0� point on plane
Q�3, �2, 4� point 53.
Plane
Intercepts:
x
y
6
3
3(0, 0, 2)
(6, 0, 0)
(0, 3, 0)
z
�0, 0, 2��0, 3, 0�,�6, 0, 0�,
x � 2y � 3z � 6
55.
Plane with rulings parallel to the x-axis
x
y
6
2
2
z
y �12
z 57.
Ellipsoid
xy-trace:
xz-trace:
yz-trace:y2
9� z2 � 1
x2
16� z2 � 1
x2
16�
y2
9� 1
x
y
54
2
−2
−4
zx2
16�
y2
9� z2 � 1
59.
Hyperboloid of two sheets
xy-trace:
xz-trace: None
yz-trace:y2
9� z2 � 1
y2
4�
x2
16� 1 x
y5 5
2
−2
zx2
16�
y2
9� z2 � �1
Review Exercises for Chapter 10 259
http://librosysolucionarios.net
61. (a)
(b)
(c)
� 4� �31�
64�
225�
64� 11.04 cm3
� 2��x2 �x4
8 �2
12
� 2��2
12�2x �
12
x3 dx
V � 2��2
12 x�3 � �1
2x2 � 1 � dx
� 4� � 12.6 cm3
� 2��x2 �x4
8 �2
0
� 2��2
0�2x �
12
x3 dx
V � 2��2
0 x�3 � �1
2x2 � 1 � dx
x2 � y2 � 2z � 2 � 0
� �2�z � 1��2
x2 � y2 � r�z��2
x
3
2
1
1 2 3
y
x
3
2
1
1 2 3
y
x
y1
3
−2
22
3
4
z
63. rectangular
(a) cylindrical
(b) spherical�2�5, 3�
4, arccos
�55 ,� � arccos
2
2�5� arccos
1
�5,� �
3�
4,� � ���2�2 �2
� �2�2 �2� �2�2 � 2�5,
�4, 3�
4, 2 ,z � 2,� � arctan��1� �
3�
4,r � ���2�2�2
� �2�2 �2� 4,
��2�2, 2�2, 2�,
65. cylindrical
spherical�50�5, ��
6, 63.4 ,
� � arccos� 50
50�5 � arccos� 1�5 � 63.4
� � ��
6
� � �1002 � 502 � 50�5
�100, ��
6, 50 , 67. spherical
cylindrical�25 �22
, ��
4, �
25�22 ,
z � � cos � � 25 cos 3�
4� �25
�22
� � ��
4
r2 � �25 sin�3�
4 2
⇒ r � 25�22
�25, ��
4,
3�
4 ,
69.
(a) Cylindrical:
(b) Spherical: � � 2 sec 2� cos � csc2�� sin2 � cos 2� � 2 cos � � 0,�2 sin2 � cos2 � � �2 sin2 � sin2 � � 2� cos �,
r2 cos 2� � 2zr2 cos2 � � r2 sin2 � � 2z,
x2 � y2 � 2z
260 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
91.
(a)
(b) �2 � 16, � � 4
r 2 � z2 � 16
x2 � y2 � z2 � 16 93.
(a)
(b)
� � 2 cos �
��� � 2 cos �� � 0,�2 � 2� cos � � 0,
r 2 � z2 � 2z � 0, r 2 � �z � 1�2 � 1
x2 � y2 � z2 � 2z � 0
95.
(a)
(b)
� � 4 sin � csc �� �4 sin �sin �
,
� sin ��� sin � � 4 sin �� � 0,
�2 sin2 � � 4� sin � sin �,
r � 4 sin �r 2 � 4r sin �,
x2 � y2 � 4y 97.
(a)
(b)
�2 �9 csc2 �
cos2 � � sin2 �
�2 sin2 � �9
cos2 � � sin2 �,
�2 sin2 � cos2 � � �2 sin2 � sin2 � � 9,
r 2 �9
cos2 � � sin2 �
r 2 cos2 � � r 2 sin2 � � 9,
x2 � y2 � 9
99.
x
y23
1
2 3
5
3
2
z
0 ≤ z ≤ 4
0 ≤ r ≤ 2
0 ≤ � ≤ �
2101.
x ya a
−a −a
a
z
r ≤ z ≤ a
0 ≤ r ≤ a
0 ≤ � ≤ 2� 103.
x
y
a
30°
z
0 ≤ � ≤ a sec �
0 ≤ � ≤ �
6
0 ≤ � ≤ 2�
105. Rectangular
0 ≤ z ≤ 10
0 ≤ y ≤ 10
0 ≤ x ≤ 10
y
x
1010
10
z 107. Spherical
4 ≤ � ≤ 6
y
x
8
−8
−8
8
z
109.
The curve of intersection is the ellipse formed by the intersection of the plane and the cylinder r � 1.z � y
z �yr
�y1
� y
z � sin �, r � 1
Review Exercises for Chapter 10
1.
(a)
(b)
(c) 2u � v � �6, �2� � �4, 2� � �10, 0� � 10i
�v� � �42 � 22 � 2�5
v � PR\
� �4, 2� � 4i � 2j
u � PQ\
� �3, �1� � 3i � j,
R � �5, 4�Q � �4, 1�,P � �1, 2�, 3.
� �4i � 4�3j
v � �v� cos � i � �v� sin � j � 8 cos 120 i � 8 sin 120 j
256 Chapter 10 Vectors and the Geometry of Space
http://librosysolucionarios.net
5.
�10
�11� 3.015 ft�
2
�115 y �
2
tanarccos�56��
tan � �2y ⇒ y �
2tan �
� � arccos�56
θ
y
100 lb120 lb
2 ft120 cos � � 100
7. ��5, 4, 0�x � �5:y � 4,z � 0,
11. �x � 3�2 � �y � 2�2 � �z � 6�2 � �152
2
13.
Center:
Radius: 3
x
y543
234
45
6 6
z
�2, 3, 0��x � 2�2 � �y � 3�2 � z2 � 9
�x2 � 4x � 4� � �y2 � 6y � 9� � z2 � �4 � 4 � 9 15.
x
y321
545
3
1
32
−2
z
(2, −1, 3)
(4, 4, −7)
v � �4 � 2, 4 � 1, �7 � 3� � �2, 5, �10�
17.
Since the points lie in a straight line.�2w � v,
w � �5 � 3, 3 � 4, �6 � 1� � �2, �1, �5�
v � ��1 � 3, 6 � 4, 9 � 1� � ��4, 2, 10� 19. Unit vector:u
�u��
�2, 3, 5��38
� � 2�38
, 3
�38,
5�38�
21.
(a)
(b)
(c) v v � 9 � 36 � 45
u v � ��1���3� � 4�0� � 0�6� � 3
v � PR\
� ��3, 0, 6� � �3i � 6k
u � PQ\
� ��1, 4, 0� � �i � 4j,
R � �2, 0, 6�Q � �4, 4, 0�,P � �5, 0, 0�, 23.
Since the vectors are orthogonal.u v � 0,
v � ��1, 4, 5�u � �7, �2, 3�,
9. Looking down from the positive x-axis towards the yz-plane,the point is either in the first quadrant or inthe third quadrant The x-coordinate can beany number.
�y < 0, z < 0�.�y > 0, z > 0�
25.
� � arccos �2 � �6
4� 15
cos � � �u v��u� �v�
��5�22��1 � �3 �
5�2� ��2 � �6
4
�v� � 2
�u� � 5
u v �5�2
2�1 � �3�
v � 2�cos 2�
3i � sin
2�
3j � �i � �3 j
u � 5�cos 3�
4i � sin
3�
4j �
5�22
�i � j� 27.
is parallel to v and in the oppositedirection.
� � �
u � �5v ⇒ u
v � ��2, 1, �3�u � �10, �5, 15�,
Review Exercises for Chapter 10 257
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29. There are many correct answers. For example: v � ± �6, �5, 0�.
In Exercises 31–39, w � <�1, 2, 2>.v � <2, �4, �3>,u � <3, �2, 1>,
31.
� 14 � ��14 �2� �u�2
u u � 3�3� � ��2���2� � �1��1� 33.
� ��1514
, 57
, �5
14�
� ��1514
, 1014
, �5
14� � �
514
�3, �2, 1�
projuw � �u w�u�2 u
35.
n�n�
�1
�5��2i � j�
�n� � �5
n � v � w � � i2
�1
j�4
2
k�3
2� � �2i � j 37.
� ��3, �2, 1� ��2, �1, 0�� � ��4� � 4
V � �u �v � w��
39. (See Exercises 36, 38)
� �285
Area parallelogram � �u � v� � �102 � 112 � ��8�2
41.
�F� � 100�1 � tan2 20 � 100 sec 20 � 106.4 lb
F �100
cos 20�cos 20j � sin 20k� � 100�j � tan 20k�
c �100
cos 20
200 � �PQ\
� F� � 2c cos 20
PQ\
� F � � i00
j0
c cos 20
k2
c sin 20� � �2c cos 20i
PQ\
� 2k
x
y
70°
PQF
2 ft
zF � c�cos 20j � sin 20k�
43.
(a)
(b) None
z � 3y � 2 � t,x � 1,
v � j 45.
Solving simultaneously, we have Substituting into the second equation we have Substitutingfor x in this equation we obtain two points on the line ofintersection, The direction vector ofthe line of intersection is
(a)
(b) z � 1x � y � 1,
z � 1y � �1 � t,x � t,
v � i � j.�1, 0, 1�.�0, �1, 1�,
y � x � 1.z � 1z � 1.
x � y � 2z � 33x � 3y � 7z � �4,
258 Chapter 10 Vectors and the Geometry of Space
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47. The two lines are parallel as they have the same directionnumbers, 1, 1. Therefore, a vector parallel to theplane is A point on the first line is
and a point on the second line is The vector connecting these two pointsis also parallel to the plane. Therefore, a normal to theplane is
Equation of the plane:
x � 2y � 1
�x � 1� � 2y � 0
� �2i � 4j � �2�i � 2j�.
v � u � � i�2
2
j1
�1
k1
�3�u � 2i � j � 3k
��1, 1, 2�.�1, 0, �1�v � �2i � j � k.
�2,49.
A point P on the plane is
D � �PQ\
n��n�
�87
n � �2, �3, 6�
PQ\
� ��2, 0, 2�
�3, 0, 0�.
2x � 3y � 6z � 6
Q � �1, 0, 2�
51.
normal to plane
D � �PQ\
n��n�
�10�30
��30
3
PQ\
� ��2, �2, 4�
n � �2, �5, 1�
P�5, 0, 0� point on plane
Q�3, �2, 4� point 53.
Plane
Intercepts:
x
y
6
3
3(0, 0, 2)
(6, 0, 0)
(0, 3, 0)
z
�0, 0, 2��0, 3, 0�,�6, 0, 0�,
x � 2y � 3z � 6
55.
Plane with rulings parallel to the x-axis
x
y
6
2
2
z
y �12
z 57.
Ellipsoid
xy-trace:
xz-trace:
yz-trace:y2
9� z2 � 1
x2
16� z2 � 1
x2
16�
y2
9� 1
x
y
54
2
−2
−4
zx2
16�
y2
9� z2 � 1
59.
Hyperboloid of two sheets
xy-trace:
xz-trace: None
yz-trace:y2
9� z2 � 1
y2
4�
x2
16� 1 x
y5 5
2
−2
zx2
16�
y2
9� z2 � �1
Review Exercises for Chapter 10 259
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61. (a)
(b)
(c)
� 4� �31�
64�
225�
64� 11.04 cm3
� 2��x2 �x4
8 �2
12
� 2��2
12�2x �
12
x3 dx
V � 2��2
12 x�3 � �1
2x2 � 1 � dx
� 4� � 12.6 cm3
� 2��x2 �x4
8 �2
0
� 2��2
0�2x �
12
x3 dx
V � 2��2
0 x�3 � �1
2x2 � 1 � dx
x2 � y2 � 2z � 2 � 0
� �2�z � 1��2
x2 � y2 � r�z��2
x
3
2
1
1 2 3
y
x
3
2
1
1 2 3
y
x
y1
3
−2
22
3
4
z
63. rectangular
(a) cylindrical
(b) spherical�2�5, 3�
4, arccos
�55 ,� � arccos
2
2�5� arccos
1
�5,� �
3�
4,� � ���2�2 �2
� �2�2 �2� �2�2 � 2�5,
�4, 3�
4, 2 ,z � 2,� � arctan��1� �
3�
4,r � ���2�2�2
� �2�2 �2� 4,
��2�2, 2�2, 2�,
65. cylindrical
spherical�50�5, ��
6, 63.4 ,
� � arccos� 50
50�5 � arccos� 1�5 � 63.4
� � ��
6
� � �1002 � 502 � 50�5
�100, ��
6, 50 , 67. spherical
cylindrical�25 �22
, ��
4, �
25�22 ,
z � � cos � � 25 cos 3�
4� �25
�22
� � ��
4
r2 � �25 sin�3�
4 2
⇒ r � 25�22
�25, ��
4,
3�
4 ,
69.
(a) Cylindrical:
(b) Spherical: � � 2 sec 2� cos � csc2�� sin2 � cos 2� � 2 cos � � 0,�2 sin2 � cos2 � � �2 sin2 � sin2 � � 2� cos �,
r2 cos 2� � 2zr2 cos2 � � r2 sin2 � � 2z,
x2 � y2 � 2z
260 Chapter 10 Vectors and the Geometry of Space
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Problem Solving for Chapter 10
1.
Then,
The other case, is similar.sin A�a�
�sin B�b�
�sin C�c�
.
��a � b�
�a� �b� �c�
sin A�a�
��b � c�
�a� �b� �c�
�a � b� � �a� �b� sin C
�b � c� � �b� �c� sin A
�a � b� � �b � c�
�b � a� � �b � c� � 0
b � �a � b � c� � 0
b
c a
a � b � c � 0 3. Label the figure as indicated.
From the figure, you see that
and
Since and is a parallelogram.
a
abS
P
Q
R
12
b12
+a1
2b1
2−
SR\
� PQ\
, PSRQSP\
� RQ\
SR\
�12
a �12
b � PQ\
.
SP\
�12
a �12
b � RQ\
5. (a) direction vector of line determined by and
(b) The shortest distance to the line segment is �P1Q� � ��2, 0, �1�� � �5.
���1, �2, 2��
�2�
3�2
�3�2
2
���2, 0, �1� � �0, 1, 1��
�2
D ��P1Q
\
� u��u�
P2.P1u � �0, 1, 1�
x
y4324
12
3
3
456
z
P1
P2
Q
7. (a)
Note:
(b) (slice at )
At figure is ellipse of area
(c) V �12
��abk�k �12
�base��height�
V � �k
0�abc � dc � �abc2
2 k
0�
�abk2
2
� ��ca���cb� � �abc.
z � c,
x2
��ca�2 �y2
��cb�2 � 1
z � cx2
a2 �y2
b2 � z:
12
�base��altitude� �12
��1� �12
�
V � ��1
0��2�2
dz � �z2
21
0�
12
� 9. (a)
Torus
(b)
Sphere
x
y
z
32
1
−2
−2−3
32
1
� � 2 cos
x
y
z
3
−3
2
−2
3
� � 2 sin
11. From Exercise 64, Section 10.4, �u � v� � �w � z� � ��u � v� � z�w � ��u � v� � w�z.
Problem Solving for Chapter 10 261
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13. (a)
Downward force
If and
and
(b) From part (a), and
Domain:
(c)
0 ≤ ≤ 90�
�T� � sec .�u� � tan
�u� �12 2
�3� � 0.5774 lb
⇒ �T� �2
�3� 1.1547 lb
1 � ��3�2��T� � 30�, �u� � �1�2��T�
1 � cos �T�
�u� � sin �T�
0 � u � w � T � �u� i � j � �T���sin i � cos j�
� �T���sin i � cos j�
T � �T��cos�90� � �i � sin�90� � �j�
w � �j
u � �u��cos 0 i � sin 0 j� � �u� i (d)
(e) Both are increasing functions.
(f) and lim →��2�
�u� � �.lim →��2�
T � �
0 600
T
u
2.5
T 1 1.0154 1.0642 1.1547 1.3054 1.5557 2
0 0.1763 0.3640 0.5774 0.8391 1.1918 1.7321�u�
60�50�40�30�20�10�0�
15. Let the angle between u and v. Then
For and and
Thus, sin� � �� � �v � u� � sin cos � � cos sin �.
v � u � � icos �cos
jsin �sin
k00� � �sin cos � � cos sin ��k.
�u� � �v� � 1v � �cos �, sin �, 0�,u � �cos , sin , 0�
sin� � �� ��u � v��u� �v�
��v � u��u� �v�
.
� � �,
19. and are two sets of direction numbers for the same line. The line is parallel to both and Therefore, u and v are parallel, and there exists a scalar d such that
a1 � a2d, b1 � b2d, c1 � c2d.a1i � b1 j � c1k � d�a2i � b2 j � c2k�,u � dv,v � a2i � b2 j � c2k.
u � a1i � b1j � c1ka2, b2, c2a1, b1, c1,
17. From Theorem 10.13 and Theorem 10.7 (6) we have
� �w � �u � v���u � v�
� ��u � v� � w��u � v�
� �u � �v � w���u � v�
.
D � �PQ\
� n��n�
262 Chapter 10 Vectors and the Geometry of Space
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C H A P T E R 1 1Vector-Valued Functions
Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . 268
Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . 273
Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . 278
Section 11.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . . . . 283
Section 11.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . 289
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 298
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 304
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C H A P T E R 1 1Vector-Valued Functions
Section 11.1 Vector-Valued FunctionsSolutions to Even-Numbered Exercises
268
2.
Component functions:
Domain: ��2, 2�h�t� � �6t
g�t� � t 2
f �t� � �4 � t 2
r�t� � �4 � t2 i � t 2j � 6tk 4.
Component functions:
Domain: ���, ��h�t� � t
g�t� � 4 cos t
f �t� � sin t
r�t� � sin t i � 4 cos tj � tk
6.
Domain: �0, ��
� �ln t � 1�i � t j
� �ln t � 1�i � �5t � 4t�j � ��3t 2 � 3t 2�k
r�t� � F�t� � G�t� � �ln t i � 5t j � 3t 2k� � �i � 4t j � 3t 2k�
8.
Domain: ���, �1�, ��1, ��
r�t� � F�t� � G�t� � � i
t3
3�t
j
�t 1 t � 1
k
t
t � 2� � ��t�t � 2� �t
t � 1i � �t 3�t � 2� � t 3�t�j � � t 3
t � 1� t 3�tk
10.
(a)
(b)
(c)
(d) r��
6� �t � r��
6 � cos��
6� �ti � 2 sin��
6� �tj � �cos��
6i � 2 sin �
6j
� �cos i � 2 sin jr� � �� � cos� � ��i � 2 sin� � ��j
r��
4 ��22
i � �2 j
r�0� � i
r�t� � cos t i � 2 sin tj
12.
(a)
(b)
(c)
(d)
� ��9 � �t � 3�i � ��9 � �t�32 � 27�j � �e���9��t�4� � e�94�k
r�9 � �t� � r�9� � ��9 � �t �i � �9 � �t�32j � e���9��t�4�k � �3i � 27j � e�94k�
r�c � 2� � �c � 2i � �c � 2�32j � e���c�2�4�k
r�4� � 2i � 8j � e�1k
r�0� � k
r�t� � �t i � t32j � e�t4k
14.
� �t � 9t2 � 16t2 � �t�1 � 25t�
�r�t�� � ���t �2 � �3t�2 � ��4t�2
r�t� � �t i � 3t j � 4tk
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Section 11.1 Vector-Valued Functions 269
16. a scalar.
The dot product is a scalar-valued function.
� t 3 � 2t2,r�t� u�t� � �3 cos t��4 sin t� � �2 sin t���6 cos t� � �t � 2��t 2�
18.
Thus, Matches (c)x2 � y2 � 1.
z � t2y � sin��t�,x � cos��t�,
�1 ≤ t ≤ 1r�t� � cos��t� i � sin��t�j � t 2k, 20.
Thus, and Matches (a)y � ln x.z �23 x
z �2t3
y � ln t,x � t,
0.1 ≤ t ≤ 5r�t� � t i � ln tj �2t3
k,
22.
(a)
2
3
31
1
2
y
x
�0, 0, 20�
z � 2 ⇒ x � yy � t,x � t,
r�t� � ti � tj � 2k
(b)
2 31
3
1
y
z
�10, 0, 0� (c)
y
x
2
2
2 3
3
1−1
−1−2
−2
−2
−3
−3
−3
z
�5, 5, 5�
24.
Domain:
−4 −3 −2
−2
2
3
4
5
6
−1 1 2 3 4
y
x
t ≥ 0
y � �1 � x
x � 1 � t, y � �t 26.
−1−1
1
2
3
4
5
1 2 3 4 5
y
x
x � t2 � t, y � t2 � t 28.
1
−1
−1 1x
y
x2 � y2 � 4
y � 2 sin t
x � 2 cos t
30.
−3 −2
−2
2
3
−3
2 3
y
x
x23 � y23 � 223
� 1
�x2
23
� �y2
23
� cos2 t � sin2 t
x � 2 cos3 t, y � 2 sin3 t 32.
Line passing through the points:
,
(0, 5, 0)−
yx 6
6
4 4
4
−4
−4−2
−6
−6 −6
2 2
2
52
152, 0,( )
z
�52
, 0, 152 �0, �5, 0�
y � 3t
y � 2t � 5
x � t 34.
Elliptic helix
x
y44
4
z
z �t2
x2
9�
y2
16� 1
z �t2
y � 4 sin t,x � 3 cos t,
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270 Chapter 11 Vector-Valued Functions
36.
1
1
−1
2
−2
3
2 3 4
5
−4 −3 −2 −1
z
y
x
−3
x �y2
4, z �
34
y
x � t2, y � 2t, z �32
t 38.
Helix along a hyperboloid of one sheet
xy32
423
4
z
z � t
x2 � y2 � 1 � t 2 � 1 � z2 or x2 � y2 � z2 � 1
z � t
y � sin t � t cos t
x � cos t � t sin t
t 0 1 2
x 4 1 0 1 4
y 0 2 4
z 0 332�
32�3
�2�4
�1�2
40.
Parabola
x y33
221
1
1
−3
−3
−2
−2−1
z
r�t� � t i ��32
t2j �12
t2k 42.
Ellipse
x y
2
2
1
1
1
−2
−1
−1
−1
z
r�t� � ��2 sin t i � 2 cos tj � �2 sin tk
44.
1 1
1
2
3
4
5
2 3 4 52
34
5
z
y
x
r�t� � t i � t2 j �12
t3h (a) is a translation 2 units to the left along the y-axis.
−21 1
2
3
4
5
2 323
45
z
y
x
u�t� � r �t� � 2j (b) has the roles
of x and y interchanged. The graphis a reflection in the plane
−21 1
2
3
4
5
2 323
45
z
y
x
x � y.
u�t� � t2i � t j �12
t3k
(c) is an upwardshift 4 units.
1 1
1
2
3
4
5
2 3 4 52
34
5
z
y
x
u�t� � r �t� � 4k (d) shrinks the
z-value by a factor of 4. The curverises more slowly.
1
1
2
3
4
5
52
34
5
z
y
x
u�t� � ti � t2 j �18
t3k (e) reverses the orientation.
1
1
2
3
4
5
5321
23
45
z
y
x
u�t� � r ��t�
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Section 11.1 Vector-Valued Functions 271
46.
Let then
r�t� � ti �13
�2t � 5�j
y �13
�2t � 5�.x � t,
2x � 3y � 5 � 0 48.
Let then
r�t� � ti � �4 � t2�j
y � 4 � t2.x � t,
y � 4 � x2
50.
Let
r�t� � �2 � 2 cos t�i � 2 sin t j
x � 2 � 2 cos t, y � 2 sin t.
�x � 2�2 � y2 � 4 52.
Let
r�t� � 4 cos ti � 3 sin tj
x � 4 cos t, y � 3 sin t.
x2
16�
y2
9� 1
54. One possible answer is
, 0 ≤ t ≤ 2
Note that r�2�� � 1.5i � 2k.
�r�t� � 1.5 cos t i � 1.5 sin tj �1�
tk
56.
(Other answers possible)
r3�t� � 5�2�1 � t�i � 5�2�1 � t�j, 0 ≤ t ≤ 1 �r3�0� � 5�2i � 5�2 j, r3�1� � 0
r2�t� � 10�cos t i � sin tj�, 0 ≤ t ≤ �
4 �r2�0� � 10i, r2��
4 � 5�2i � 5�2 j r1�t� � ti, 0 ≤ t ≤ 10 �r1�0� � 0, r1�10� � 10i�
58.
(Other answers possible)
r2�t� � �1 � t�i � �1 � t�j, 0 ≤ t ≤ 1 �y � x�
r1�t� � ti � �t j, 0 ≤ t ≤ 1 �y � �x �
60.
Therefore, or
r�t� � 2 cos ti � 2sin tj � 4k
z � 4.y � 2 sin t,x � 2 cos t,
x2 � y2 � 4
xy22
6
zz � 4z � x2 � y2,
62.
If then and
r�t� � t2i �12�16 � 4t4 � t2j � tk
y �12�16 � 4t4 � t2.x � t2z � t,
x
y22
4
4
2
z4x2 � 4y2 � z2 � 16, x � z2
t 0 1 1.2
x 1.69 1.44 1 0 1 1.44
y 0.85 1.25 1.66 2 1.66 1.25
z 0 1 1.2�1�1.2�1.3
�1�1.2�1.3
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272 Chapter 11 Vector-Valued Functions
64.
Let then and
r�t� � �2 � sin t�i � �2 � sin t�j � �2 cos tk
x
y44
4
z
z � �2�1 � sin2 t� � �2 cos t.y � 2 � sin tx � 2 � sin t,
x � y � 4x2 � y2 � z2 � 10,
t 0
x 1 2 3 2
y 3 2 1 2
z 0 0 ��2�62
�2�62
32
52
52
32
��
2�
6�
�
6�
�
2
68.
xy
4080
120
4080120
120
160
80
40
z
x2 � y2 � �e�t cos t�2 � �e�t sin t�2 � e�2t � z2 70.
since
(L’Hôpital’s Rule)limt→0
sin t
t� lim
t→0 cos t
1� 1
limt→0
�eti �sin t
tj � e�t k � i � j � k
72.
since
(L’Hôpital’s Rule)limt→1
ln t
t2 � 1� lim
t→1 1t2t
�12
.
limt→1
��t i �ln t
t 2 � 1j � 2t 2 k � i �
12
j � 2k 74.
since
and limt→�
t
t 2 � 1� 0.lim
t→� 1t
� 0,limt→�
e�t � 0,
limt→�
�e�t i �1tj �
tt 2 � 1
k � 0
66. (first octant)
Let then
and
r�t� � ti �4t
j �1t��t4 � 16t2 � 16k
��8 � 4�3 ≤ t ≤ �8 � 4�3
z �1t��t4 � 16t2 � 16
x2 � y2 � z2 � t2 �16t2 � z2 � 16.y �
4t
x � t,
x
y44
4
zxy � 4x2 � y2 � z2 � 16,
t 1.5 2 2.5 3.0 3.5
x 1.0 1.5 2 2.5 3.0 3.5 3.9
y 3.9 2.7 2 1.6 1.3 1.1 1.0
z 0 2.6 2.8 2.7 2.3 1.6 0
�8 � 4�3�8 � 4�3
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Section 11.2 Differentiation and Integration of Vector-Valued Functions
86. Let and Then:
� limt→c
r�t� � limt→c
u�t�
� �limt→c x1�t�i � lim
t→c y1�t�j � lim
t→c z1�t�k� � �limt→c x2�t�i � lim
t→c y2�t�j � lim
t→c z2�t�k�
� limt→c
x1�t� limt→c x2�t� � lim
t→c y1�t� limt→c
y2�t� � limt→c
z1�t� limt→c z2�t�
limt→c
�r�t� � u�t�� � limt→c
�x1�t�x2�t� � y1�t�y2�t� � z1�t�z2�t��
u�t� � x2�t�i � y2�t�j � z2�t�k.r�t� � x1�t�i � y1�t�j � z1�t�k
88. Let
and Then r is not continuous at whereas, is continuous for all t.�r� � 1
c � 0,r�t� � f �t�i.
f �t� � �1,�1,
if t ≥ 0if t < 0
90. False. The graph of represents a line.x � y � z � t3
80.
Continuous on �0, ��
r�t� � �8, t, 3t 82. No. The graph is the same because For example, if is on the graph of r, then is thesame point.
u�2�r�0�r�t� � u�t � 2�.
84. A vector-valued function r is continuous at ifthe limit of exists as and
The function is not continuous at t � 0.r�t� � � i � j�i � j
t ≥ 0t < 0
limt→a
r�t� � r�a�.
t → ar�t�t � a
76.
Continuous on �1, ��
r�t� � t i � t � 1 j 78.
Continuous on or t > 1: �1, ��.t � 1 > 0
r�t� � �2e�t, e�t, ln�t � 1�
Section 11.2 Differentiation and Integration of Vector-Valued Functions 273
2.
is tangent to the curve.
x−2−3 1 2 3
3
4
4
−2
−3
−4
−4
1
2
(1, 1)r
r'
y
r��t0�
r��1� � i � 3j
r��t� � i � 3t2j
r�1� � i � j
y � x3
y�t� � t3x�t� � t,
t0 � 1r�t� � ti � t3 j, 4.
is tangent to the curve.
x
−1
1
2
8
r
r'
y
r��t0�
r��2� � 4 i �14
j
r��t� � 2t i �1t2 j
r�2� � 4i �12
j
x �1y2
y�t� �1t
x�t� � t2,
t0 � 2r�t� � t2i �1t
j,
http://librosysolucionarios.net
(b)
r�1.25� � r�1� � 0.25i � 0.5625j
r�1.25� � 1.25i � 2.4375j
r�1� � i � 3j
(c)
This vector approximates r��1�.
r�1.25� � r�1�
1.25 � 1�
0.25i � 0.5625j0.25
� i � 2.25j
r��1� � i � 2j
r��t� � i � 2tj
6.
(a)
x
2
5
3
1
3−1−3−1
r(1)
r(1.25)
r r(1.25) (1)−
y
r�t� � t i � �4 � t 2�j
274 Chapter 11 Vector-Valued Functions
8.
x
y
2
2
−2 −2
−4
−4
4
−6
−6r
r'
z
r��2� � i � 4j
r�2� � 2 i � 4 j �32
k
r��t� � i � 2tj
z �32
y � x2,
t0 � 2r�t� � ti � t2j �32
k,
10.
r��t� � �1t2 i � 16j � tk
r�t� �1t
i � 16tj �t2
2k 12.
r��t� �2t
i � �2tt �t2
2t�j �2
tk
r�t� � 4t i � t2t j � ln t2 k
14.
r��t� � �t sin t, t cos t, 2t
r�t� � �sin t � t cos t, cos t � t sin t, t2 16.
r��t� � 1
1 � t2, �
1
1 � t2, 0�
r�t� � �arcsin t, arccos t, 0
18.
(a)
(b) r��t� � r��t� � �2t � 1��2� � �2t � 1��2� � 8t
r��t� � 2 i � 2j
r��t� � �2t � 1�i � �2t � 1�j
r�t� � �t2 � t�i � �t2 � t�j 20.
(a)
(b)
� 55 sin t cos t
r��t� � r��t� � ��8 sin t���8 cos t� � 3 cos t��3 sin t�
r��t� � �8 cos t i � 3 sin tj
r��t� � �8 sin ti � 3 cos tj
r�t� � 8 cos ti � 3 sin tj
22.
(a)
(b) r��t� � r��t� � 0
r��t� � 0
r��t� � i � 2j � 3k
r�t� � ti � �2t � 3�j � �3t � 5�k 24.
(a)
(b) r��t� � r��t� � �e�2t � 4t � 2 sec4 t tan t
r��t� � �e�t, 2, 2 sec2 t tan t
r��t� � ��e�t, 2t, sec2 t
r�t� � �e�t, t2, tan�t�
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Section 11.2 Differentiation and Integration of Vector-Valued Functions 275
26.
r��1�4�
�r��1�4�� �1
1024 � 81e3�8�32j � 9e3�16k�
� r��14� � �22 � � 9
16e3�16 �
2
�4 �81
256e3�8 �
1024 � 81e3�8
16
r��14� � 2 i �
916
e3�16k
r��t� � 2i �9
16e0.75t k
r��1�4�
�r��14�� �1
20 � 9e3�8�4i � 2j � 3e3�16k�
� r��14� � �12 � �1
2�2
� �34
e3�16�2
�54
�9
16e3�8 �
20 � 9e3�8
4
r��14� � i �
12
j � 0.75e0.1875k � i �12
j �34
e3�16k
r��t� � i � 2tj � 0.75e0.75t k
1
2 −1
−2
1
2
−1
−2
2
−1
−2
yx
r'' 14( )r' 1
4( )r'' 1
4( )r' 14( )
zt0 �
14
r�t� � t i � t2j � e0.75tk,
28.
Not continuous when
Smooth on �1, �����, 1�,t � 1
r��t� � �1
�t � 1�2 i � 3j
r�t� �1
t � 1i � 3tj 30.
any integer
Smooth on ��2n � 1�, �2n � 1��
r���2n � 1�� � 0, n
r���� � �1 � cos ��i � sin �j
r��� � �� � sin ��i � �1 � cos ��j
32.
for any value of t.
r is not continuous when
Smooth on ���, �2�, ��2, ��.
t � �2.
r��t� � 0
r��t� �16 � 4t3
�t3 � 8�2 i �32t � 2t 4
�t3 � 8�2 j
r�t� �2t
8 � t3i �
2t2
8 � t3j 34.
r is smooth for all t: ���, ��
r��t� � eti � e�tj � 3k � 0
r�t� � eti � e�tj � 3tk 36.
r is smooth for all t > 0: �0, ��
r��t� �1
2ti � 2tj �
14
k � 0
r�t� � t i � �t2 � 1�j �14
tk
38.
(a)
(c)
—CONTINUED—
Dt�r�t� � u�t�� � 0, t � 0
r�t� � u�t� � 1 � 4 sin2 t � 4 cos2 t � 5
r��t� � i � 2 cos tj � 2 sin tk
u�t� �1ti � 2 sin tj � 2 cos tk
r�t� � ti � 2 sin tj � 2 cos tk
(b)
(d)
Dt�3r�t� � u�t�� � �3 �1t2�i � 4 cos tj � 4 sin tk
3r�t� � u�t� � �3t �1t �i � 4 sin tj � 4 cos tk
r��t� � �2 sin tj � 2 cos tk
http://librosysolucionarios.net
38. —CONTINUED—
(e)
(f)
Dt��r�t��� �1
2�t 2 � 4��1�2�2t� �
tt 2 � 4
�r�t�� � t2 � 4
� �2 cos t�t �1t � � 2 sin t�1 �
1t 2��k
Dt�r�t� � u�t�� � ��2 sin t�1t
� t� � 2 cos t��1t 2 � 1��j
� 2 cos t�1t
� t�j � 2 sin t�t �1t �k
r�t� u�t� � � it
1�t
j2 sin t2 sin t
k2 cos t2 cos t�
40.
maximum at .
for any t.� �
2
�22 �t � 0.707�� 19.47��� � 0.340
� � arccos 2t3 � t
t 4 � t24t 2 � 1
cos � �2t 3 � t
t 4 � t 24t2 � 1
�r�t�� � t 4 � t 2, �r��t�� � 4t 2 � 1
r�t� � r��t� � 2t 3 � t
r��t� � 2t i � j
0
−0.5
1.0
r�t� � t 2 i � tj
42.
�1
2ti �
3t2 j � 2k
� lim�t→0
� 1
t � �t � ti �
3�t � �t�t j � 2k�
� lim�t→0
� �t�t�t � �t � t� i �
�3�t�t � �t�t��t� j � 2k�
� lim�t→0
�t � �t � t�t
i �
3t � �t
�3t
�tj � 2k�
� lim�t→0
�t � �t i �
3t � �t
j � 2�t � �t�k� � �t i �3t
j � 2tk��t
r��t� � lim�t→0
r�t � �t� � r�t�
�t
276 Chapter 11 Vector-Valued Functions
44. ��4t3i � 6tj � 4tk� dt � t 4i � 3t2j �83
t3�2 k � C 46.
(Integration by parts)
��ln t i �1t
j � k� dt � �t ln t � t�i � ln tj � tk � C
48. ��eti � sin tj � cos tk� dt � eti � cos tj � sin tk � C
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Section 11.2 Differentiation and Integration of Vector-Valued Functions 277
50. ��e�t sin t i � e�t cos t j� dt �e�t
2��sin t � cos t�i �
e�t
2��cos t � sin t� j � C
52. �1
�1�t i � t3j � 3tk� dt � �t2
2i�
1
�1� �t 4
4j�
1
�1� �3
4t 4�3k�
1
�1� 0
54.
� 2i � �e2 � 1�j � �e2 � 1�k
�2
0�ti � etj � tetk� dt � �t2
2i�
2
0� �etj�
2
0� ��t � 1�et k�
2
056.
r�t� � i � �2 � t3�j � 4t3�2k
r�0� � C � i � 2j
r�t� � ��3t2j � 6tk� dt � t3j � 4t3�2k � C
58.
r�t� � �4 cos t � 4�i � 4 j � 3 sin tk
r�0� � 4i � C2 � 4j ⇒ C2 � 4j � 4i
r�t� � 4 cos t i � 3 sin tk � C2
r��0� � 3k � 3k � C1 ⇒ C1 � 0
r��t� � �4 sin t i � 3 cos tk � C1
r��t� � �4 cos ti � 3 sin tk
60.
r�t� � �2 �
4� arctan t�i � �1 �
1t �j � ln tk
r�1� �
4i � j � C � 2i ⇒ C � �2 �
4�i � j
r�t� � �� 11 � t2
i �1t2 j �
1t
k� dt � arctan t i �1t
j � ln tk � C
62. To find the integral of a vector-valued function, youintegrate each component function separately. Theconstant of integration C is a constant vector.
64. The graph of does not change position relative to thexy-plane.
u�t�
66. Let and
� r��t� ± u��t�
� �x1��t�i � y1��t�j � z1��t�k� ± �x2��t�i � y2��t�j � z2��t�k�
Dt�r�t� ± u�t�� � �x1��t� ± x2��t��i � �y1��t� ± y2��t�� j � �z1��t� ± z2��t��k
r�t� ± u�t� � �x1�t� ± x2�t��i � �y1�t� ± y2�t�� j � �z1�t� ± z2�t��k
u�t� � x2�t�i � y2�t�j � z2�t�k.r�t� � x1�t�i � y1�t�j � z1�t�k
68. Let and
� r�t� u��t� � r��t� u�t�
��y1��t�z2�t� � z1��t�y2�t��i � �x1��t�z2�t� � z1��t�x2�t��j � �x1��t�y2�t� � y1��t�x2�t��k�
� �� y1�t�z2��t� � z1�t�y2��t��i � �x1�t�z2��t� � z1�t�x2��t�� j � �x1�t�y2��t� � y1�t�x2��t��k� �
�x1�t�y2��t� � x1��t�y2�t� � y1�t�x2��t� � y1��t�x2�t��k
Dt�r�t� u�t�� � �y1�t�z2��t� � y1��t�z2�t� � z1�t�y2��t� � z1��t�y2�t�� i � �x1�t�z2��t� � x1��t�z2�t� � z1�t�x2��t� � z1��t�x2�t�� j �
r�t� u�t� � �y1�t�z2�t� � z1�t�y2�t��i � �x1�t�z2�t� � z1�t�x2�t�� j � �x1�t�y2�t� � y1�t�x2�t��k
u�t� � x2�t�i � y2�t�j � z2�t�k.r�t� � x1�t�i � y1�t�j � z1�t�k
http://librosysolucionarios.net
Section 11.3 Velocity and Acceleration
70. Let Then
� r�t� � r��t� � �y�t�z��t� � z�t�y��t��i � �x�t�z��t� � z�t�x��t�� j � �x�t�y��t� � y�t�x��t��k
�x�t�y��t� � x��t�y��t� � y�t�x��t� � y��t�x��t��k
Dt�r�t� � r��t�� � �y�t�z��t� � y��t�z��t� � z�t�y��t� � z��t�y��t��i � �x�t�z��t� � x��t�z��t� � z�t�x��t� � z��t�x��t�� j �
r�t� � r��t� � �y�t�z��t� � z�t�y��t��i � �x�t�z��t� � z�t�x��t�� j � �x�t�y��t� � y�t�x��t��k
r��t� � x��t�i � y��t�j � z��t�k.r�t� � x�t�i � y�t�j � z�t�k.
72. Let If is constant,then:
Therefore, r�t� � r��t� � 0.
2�r�t� � r��t�� � 0
2�x�t�x��t� � y�t�y��t� � z�t�z��t�� � 0
2x�t�x��t� � 2y�t�y��t� � 2z�t�z��t� � 0
Dt�x2�t� � y2�t� � z2�t�� � Dt�C�
x2�t� � y2�t� � z2�t� � C
r�t� � r�t�r�t� � x�t�i � y�t�j � z�t�k. 74. False
(See Theorem 11.2, part 4)
Dt�r�t� � u�t�� � r�t� � u��t� � r��t� � u�t�
2.
y � 6 � xy � t,x � 6 � t,
a�t� � r��t� � 0
v�t� � r��t� � � i � j
2
2
4
4
x
(3, 3)v
yr�t� � �6 � t�i � t j 4.
At
a�1� � 2i � 6j
v�1� � 2i � 3j
�1, 1�, t � 1.
x � y2�3x � t2, y � t3
a�t� � r��t� � 2i � 6tj
v�t� � r��t� � 2ti � 3t2j
(1, 1)
a
v
2
1
2
3
4
5
6
7
8
−1 3 4 5 6 7 8
y
x
r�t� � t2i � t3j
6.
Ellipse
At
a�0� � �3i
v�0� � 2j
va
(3, 0)
y
x
1
−1−1 1 2−2
−3
3
�3, 0�, t � 0.
x � 3 cos t, y � 2 sin t, x2
9�
y2
4� 1
a�t� � �3 cos ti � 2 sin tj
v�t� � �3 sin ti � 2 cos tj
r�t� � 3 cos ti � 2 sin tj 8.
At
a�0� � �1, 1� � i � j
v�0� � ��1, 1� � �i � j
t � 0.�1, 1�,
y �1x
y � et,x � e�t �1et,
a�t� � r��t� � �e�t, et�
v�t� � r��t� � ��e�t, et�
1
1
2
2x
(1, 1)
v a
yr�t� � �e�t, et�
278 Chapter 11 Vector-Valued Functions
10.
a�t� � 0
s�t� � v�t� � 16 � 16 � 4 � 6
v�t� � 4i � 4j � 2k
r�t� � 4t i � 4tj � 2tk
http://librosysolucionarios.net
12.
a�t� �12
k
s�t� �9 � 1 �14
t2 �10 �14
t2
v�t� � 3i � j �12
tk
r�t� � 3ti � tj �14
t2k 14.
a�t� � 2i �3
2tk
s�t� � 4t2 � 1 � 9t � 4t2 � 9t � 1
v�t� � 2ti � j � 3t k
r�t� � t2 i � tj � 2t 3�2k
16.
a�t� � �2et sin ti � 2et cos tj � etk
� et3
s�t� � e2t�cos t � sin t�2 � e2t�cos t � sin t�2 � e2t
v�t� � �et cos t � et sin t�i � �et sin t � et cos t�j � etk
r�t� � �et cos t, et sin t, et� 18. (a)
(b)
� �3.100, 3.925, 3.925�
r�3 � 0.1� � �3 � 0.1, 4 �34
�0.1�, 4 �34
�0.1�
y � z � 4 �34
tx � 3 � t,
r��3� � �1, �34
, �34
r��t� � �1, �t
25 � t2,
�t
25 � t2 r�t� � � t, 25 � t2, 25 � t2 �, t0 � 3
Section 11.3 Velocity and Acceleration 279
20.
r�2� � 4i � 8j � 6k
r�0� � C � 0 ⇒ r�t� � t2i � 4tj �32
t2k
r�t� � ��2ti � 4j � 3tk� dt � t2i � 4tj �32
t2k � C
v�0� � C � 4j ⇒ v�t� � 2ti � 4j � 3tk
v�t� � ��2i � 3k� dt � 2ti � 3tk � C
a�t� � 2i � 3k 22.
r�2� � �cos 2�i � �sin 2�j � 2k
r�t� � cos t i � sin tj � tk
r�0� � i � C � i ⇒ C � 0
� cos ti � sin tj � tk � C
r�t� � ���sin t i � cos tj � k� dt
v�t� � �sin t i � cos t j � k
v�0� � j � C � j � k ⇒ C � k
v�t� � ���cos t i � sin t j� dt � �sin t i � cos t j � C
a�t� � �cos t i � sin tj, v�0� � j � k, r�0� � i
24. (a) The speed is increasing.
(b) The speed is decreasing.
26.
The maximum height occurs when which implies that The maximum height reached by the projectile is
The range is determined by setting which implies that
Range: x � 4502��4502 � 405,192�32 � � 25,315.500 feet
t ��4502 � 405,192
�32� 39.779 seconds.
y�t� � 3 � 4502t � 16t2 � 0
y � 3 � 4502�225216 � � 16�2252
16 �2
�50,649
8� 6331.125 feet.
t � �2252��16. y��t� � 4502 � 32t � 0,
� 4502t i � �3 � 4502t � 16t2�jr�t� � �900 cos 45�t i � �3 � �900 sin 45�t � 16t2�j
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28. 50 mph ft/sec
The ball is 90 feet from where it is thrown when
The height of the ball at this time is
y � 5 � �2203
sin 15�� 2722 cos 15� � 16� 27
22 cos 15�2
� 3.286 feet.
x �2203
cos 15t � 90 ⇒ t �27
22 cos 15� 1.2706 seconds.
r�t� � �2203
cos 15�t i � �5 � �2203
sin 15�t � 16t2�j
�2203
280 Chapter 11 Vector-Valued Functions
30.
From Exercise 34 we know that tan is the coefficient of x. Therefore, tan Also
negative of coefficient of
or ft/sec
Position function.
When ,
direction
Speed � v�324 � � 8225 � 4 � 858 ft�sec
v�324 � � 402 i � �402 � 242�j � 82�5i � 2j�
v�t� � 402i � �402 � 32t�j
t �60
402�
324
402t � 60
r�t� � �402t�i � �402t � 16t2�j.
v0 � 8016v0
2�2� � 0.005
x216v0
2 sec2 �
� ���4� rad � 45. � 1,
y � x � 0.005x2
32. Wind:
When and feet.
Thus, the ball clears the 10-foot fence.
y � 16.7x � 375, t � 2.98
00
450
50
r�t� � �140�cos 22�t �17615 �i � �2.5 � �140 sin 22�t � 16t2�j
8 mph �17615
ft�sec
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34. feet, 30 yards feet
(a) when
v0 � 54.088 feet per second
v02 �
129,600cos2 35�90 tan 35 � 3�
90 tan 35 � 3 �129,600
v02 cos2 35
7 � �v0 sin 35�� 90v0 cos 35� � 16� 90
v0 cos 35�2
� 4
t �90
v0 cos 35
7 � �v0 sin 35�t � 16t2 � 4v0 cos 35t � 90
r�t� � �v0 cos 35�ti � �7 � �v0 sin 35�t � 16t2� j
� 90 � 35,h � 7
(b) The maximum height occurs when
second
At this time, the height is feet.y�0.969� � 22.0
t �v0 sin 35
32� 0.969
y��t� � v0 sin 35 � 32t � 0.
(c)
secondst �90
54.088 cos 35� 2.0
x�t� � 90 ⇒ �v0 cos 35�t � 90
Section 11.3 Velocity and Acceleration 281
36. Place the origin directly below the plane. Then and
At time of impact, 30,000 seconds.
tan � �30,000
34,294.6� 0.8748 ⇒ � � 0.7187�41.18�
v�43.3� � 1596 ft sec � 1088 mph
v�43.3� � 792i � 1385.6j
r�43.3� � 34,294.6i
⇒ t � 43.3� 16t2 � 0 ⇒ t2 � 1875
v�t� � 792i � 32tj.
� 792ti � �30,000 � 16t2�j
r�t� � �v0 cos �ti � �30,000 � �v0 sin �t � 16t2�jα
α
30,000
34,295(0, 0)
v0 � 792 � 0,
38. From Exercise 37, the range is
Hence, x � 150 �v0
2
32 sin�24� ⇒ v0
2 �4800
sin 24 ⇒ v0 � 108.6 ft�sec.
x �v0
2
32 sin 2.
40. (a)
when
Range:
The range will be maximum when
or
��
4 rad.2 �
�
2,
dxdt
� �v02
32�2 cos 2 � 0
x � v0 cos �v0 sin 32 � � �v0
2
32� sin 2
t �v0 sin
16.t�v0 sin � 16t� � 0
r�t� � t�v0 cos �i � �tv0 sin � 16t2�j (b)
when
Maximum height:
Minimum height when or ��
2.sin � 1,
y�v0 sin 32 � �
v02 sin2
32� 16
v02 sin2 322 �
v02 sin2
64
t �v0 sin
32.
dydt
� v0 sin � 32t � 0
y�t� � tv0 sin � 16t2
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42.
when For this value of t,
⇒ v0 � 42.2 m�sec
50 tan 8 ��4.9��2500�v0
2 cos2 8 ⇒ v0
2 ��4.9�50
tan 8 cos2 8� 1777.698
�v0 sin 8�� 50v0 cos 8� � 4.9� 50
v0 cos 8�2
� 0
y � 0:�v0 cos 8�t � 50 ⇒ t �50
v0 cos 8.x � 50
� �v0 cos 8�t i � ��v0 sin 8�t � 4.9t2� j
r�t� � �v0 cos �t i � �h � �v0 sin �t � 4.9t2� j
44.
Speed and has a maximum value of when
55 mph since since
Therefore, the maximum speed of a point on the tire is twice the speed of the car:
2�80.67� ft�sec � 110 mph
b � 1��� 80.67 ft�sec � 80.67 rad�sec � �
3�, . . . .�t � �,2b�� v�t� � 2b�1 � cos �t
v�t� � b���1 � cos �t�i � �sin �t�j�
r�t� � b��t � sin �t�i � b�1 � cos �t�j
282 Chapter 11 Vector-Valued Functions
46. (a)
� b2�2�sin2��t� � cos2��t�� � b�
Speed � v � b2�2 sin2��t� � b2�2 cos2��t� (b)
The graphing utility draws the circle faster for greater values of �.
10−10
−10
10
48. a�t� � b�2cos��t�i � sin��t�j � b�2
50.
Let n be normal to the road.
Dividing the second equation by the first:
� arctan� 6053000� � 11.4.
tan �605
3000
n sin � 605
n cos � 3000
F � m�b�2� �300032
�300�� 44300�
2
� 605 lb
a�t� � b�2
� �v�t�
b�
44300
rad�sec
3000
θ
n
605v�t� � 30 mph � 44 ft�sec
52. feet, feet per second, . From Exercise 47,
At this time, feet.x�t� � 69.02
t �45 sin 42.5 � �45�2 sin2 42.5 � 2�32��6�
32� 2.08 seconds.
� 42.5v0 � 45h � 6
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Section 11.4 Tangent Vectors and Normal Vectors 283
54.
m and b are constants.
C is a constant.
Thus,
a�t� � x��t�i � mx��t�j � 0.
x��t� � 0
x��t� �C
�1 � m2
s�t� � ��x��t��2 � �mx��t��2 � C,
v�t� � x��t�i � mx��t�j
r�t� � x�t�i � �m�x�t�� � b� j
y�t� � m�x�t�� � b,
r�t� � x�t�i � y�t�j 56.
Velocity:
Acceleration:
In general, if then:
Velocity:
Acceleration: r3��t� � �2r1���t�r3��t� � �r1���t�
r3�t� � r1��t�,
r2��t� � 4r1��2t�r2��t� � 2r1��2t�
r2�t� � r1�2t�
r1�t� � x�t�i � y�t�j � z�t�k
Section 11.4 Tangent Vectors and Normal Vectors
2.
T�1� �1
�9 � 16�3i � 4j� �
3
5i �
4
5j
T�t� �r��t�
�r��t���
1�9t4 � 16t2
�3t2i � 4tj�
�r��t�� � �9t4 � 16t2
r��t� � 3t2i � 4tj
r�t� � t3i � 2t2j 4.
T��
3 ��3�3i � j
�36�34� � �14��
1�28
��3�3i � j�
T�t� �r��t�
�r��t���
�6 sin ti � 2 cos tj�36 sin2 t � 4 cos2 t
�r��t�� � �36 sin2 t � 4 cos2 t
r��t� � �6 sin ti � 2 cos tj
r�t� � 6 cos ti � 2 sin tj
6.
When
Direction numbers:
Parametric equations: x � 2t � 1, y � t � 1, z �43
a � 2, b � 1, c � 0
T�1� �r��1�
�r��1���
2i � j�5
��5
5�2i � j�
t � 1, r��t� � r��1� � 2i � j �t � 1 at �1, 1, 43�.
r��t� � 2ti � j
r�t� � t2i � tj �43
k 8.
When at .
Direction numbers:
Parametric equations:
z � �1
�3t � �3
y � t � 1,x � t � 1,
c � �1
�3b � 1,a � 1,
T�1� �r��1�
�r��1�� ��21
7 1, 1, �1
�3�
�1, 1, �3 ���t � 1r��1� � 1, 1, �1
�3�,t � 1,
r��t� � 1, 1, �t
�4 � t2�r�t� � � t, t, �4 � t2 �
10.
When
Direction numbers:
Parametric equations: z � 2�3 t � 1y � � t � �3,x � �3t � 1,
c � 2�3b � �1,a � �3,
T��
6 �r���6�
�r���6�� �14
��3, �1, 2�3 �
�t ��
6 at �1, �3, 1��.r���
6 � ��3, �1, 2�3 �,t ��
6,
r��t� � �2 cos t, �2 sin t, 8 sin t cos t�
r�t� � �2 sin t, 2 cos t, 4 sin2 t�
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284 Chapter 11 Vector-Valued Functions
12.
When
Direction numbers:
Parametric equations: z � t ��
4y � 4,x � �6t,
c � 1b � 0,a � �6,
T��
2 �r���2�
�r���2�� �2
�37��3i �12
k �1
�37��6i � k�
�t ��
2 at �0, 4,
�
4�.r���
2 � �3i �12
k,t ��
2,
r��t� � �3 sin t i � 4 cos tj �12
k
4
4
4
1
1
2
3
5
5
5 yx
zr�t� � 3 cos ti � 4 sin t j �12
k
14.
Parametric equations:
� �0.9, 2, 0.2�
r�t0 � 0.1� � r�0 � 0.1� � �1 � 0.1, 2, 2�0.1��z�s� � 2sy�s� � 2,x�s� � 1 � s,
T�0� �r��0�
�r��0�� ��i � 2k�5
�r��0�� � �5r��0� � �i � 2k,r�0� � i � 2j,
r��t� � �e�ti � 2 sin t j � 2 cos tk
r�t� � e�ti � 2 cos tj � 2 sin tk, t0 � 0
16.
Hence the curves intersect.
cos �r��0� u��0�
�r��0�� �u��0�� � 0 ⇒ ��
2
u��0� � ��1, 0, 1�
u��s� � �sin s cos s � cos s, �sin s cos s � cos s, 12
cos 2s �12�
r��0� � �1, 0, 1� r��t� � �1, �sin t, cos t�,
u�0� � �0, 1, 0�
r�0� � �0, 1, 0�
18.
N�2� �T��2�
�T��2���
1�13
�3i � 2 j�
T��t� �72t
�t4 � 36�32 i �12t3
�t4 � 36�32 j
�t2
�t4 � 36 �i �6
t2j
T�t� �r��t�
�r��t���
1�1 � �36t4� �i �
6
t2j
r��t� � i �6t2
j
r�t� � ti �6t
j, t � 3 20.
The unit normal vector is perpendicular to this vector andpoints toward the z-axis:
N�t� ��2 cos ti � sin tj�sin2 t � 4 cos2 t
.
T�t� �r�t�
�r��t���
�sin t i � 2 cos tj�sin2 t � 4 cos2 t
r��t� � �sin t i � 2 cos t j
r�t� � cos ti � 2 sin tj � k, t � ��
4
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Section 11.4 Tangent Vectors and Normal Vectors 285
22.
is undefined.
The path is a line and the speed is constant.
N�t� �T��t�
�T��t��
T��t� � O
T�t� �v�t�
�v�t�� �1
�5�2i � j�
a�t� � O
v�t� � 4i � 2j
r�t� � 4t i � 2tj 24.
is undefined.
The path is a line and the speed is variable.
N�t� �T��t�
�T��t��
T��t� � O
T�t� �v�t�
�v�t�� �2t j2t
� j
a�t� � 2j
v�t� � 2tj
r�t� � t2j � k
26.
aN � a N � �2
aT � a T � �2
N�1� ��22
i ��22
j
�1
�t2 � 1�i � tj�
N�t� �T��t�
�T��t�� �
1�t2 � 1�32 i �
�t�t2 � 1�32 j
1t2 � 1
T�1� �1�2
�i � j� ��2
2i �
�2
2j
T�t� �v�t�
�v�t���
1�4t2 � 4
�2ti � 2j� �1
�t2 � 1�ti � j�
a�t� � 2i, a�1� � 2i
v�t� � 2ti � 2j, v�1� � 2i � 2j
r�t� � t2i � 2tj, t � 1 28.
Motion along r is counterclockwise. Therefore,
aN � a N � a�2
aT � a T � 0
N�0� � �i.
�t�
T�0� �v�0�
�v�0�� � j
a�0� � �a�2i
a�t� � �a�2 cos��t�i � b�2 sin��t�j
v�0� � b�j
v�t� � �a� sin��t�i � b� cos��t�j
r�t� � a cos��t�i � b sin��t�j
30.
Motion along r is clockwise. Therefore,
aN � a N ��2
�2�1 � cos �t0
aT � a T ��2 sin �t0
�2�1 � cos �t0�
�2
�2�1 � cos �t0
N ��sin �t0�i � �1 � cos �t0�j
�2�1 � cos �t0.
T �v
�v��
�1 � cos �t0�i � �sin �t0� j�2�1 � cos �t0
a�t0� � �2��sin �t0�i � �cos �t0�j�
v�t0� � ���1 � cos �t0�i � �sin �t0�j�
r�t0� � ��t0 � sin �t0�i � �1 � cos �t0�j
32. points in the direction that r is moving. points in the direction that r is turning,toward the concave side of the curve.
x
a
a
N
T
yN�t�T�t�
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286 Chapter 11 Vector-Valued Functions
34. If the angular velocity is halved,
is changed by a factor of 14 .aN
aN � a��
22
�a�2
4.
�
36.
N��
4 ��22
�� i � j�
T��
4 ��22
�� i � j�
r��
4 � �2i � �2 j1
−1
−1 1x
N
T2 , 2 )(
yN�t� � �cos ti � sin tj
T�t� �12
��2 sin ti � 2 cos tj� � �sin ti � cos tj
r��t� � �2 sin ti � 2 cos t j
x � 2 cos t, y � 2 sin t ⇒ x2 � y2 � 4
r�t� � 2 cos ti � 2 sin t j, t0 ��
438.
are not defined.aT, aN
N�t� �T�
�T� � is undefined.
T�t� �v
�v��
13
�2i � 2j � k�
a�t� � 0
v�t� � 4i � 4j � 2k
r�t� � 4ti � 4t j � 2tk
40.
aN � a N � �2
aT � a T � �3
N�0� ��2
2i �
�22
j
N�t� �1�2
���sin t � cos t�i � ��cos t � sin t�j�
T�0� �1�3
�i � j � k�
T�t� �v
�v��
1�3
��cos t � sin t�i � ��sin t � cos t�j � k�
a�0� � 2i � k
a�t� � 2et cos ti � 2et sin tj � etk
v�0� � i � j � k
v�t� � �et cos t � et sin t�i � ��et sin t � et cos t�j � etk
r�t� � et sin t i � et cos tj � etk 42.
y
x 4 6 824
2
4
N
T
z
aN � a N �37
�5513�
�37
�149
aT � a T �74
�149
�1
�5513��74i � 6j � k�
N�2� �1
�37�149��74i � 6j � k�
�1
�37�1 � 37t2��37ti � 6j � k�
N�t� �T�
�T� ��
1�1 � 37t2�32��37ti � 6j � k�
�371 � 372
T�2� �1
�149�i � 12j � 2k�
T�t� �v
�v��
1
�1 � 37t2�i � 6tj � tk�
a�t� � 6 j � k
v�2� � i � 12j � 2k
v�t� � i � 6tj � tk
r�t� � ti � 3t2j �t2
2k
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Section 11.4 Tangent Vectors and Normal Vectors 287
44. The unit tangent vector points in the direction of motion. 46. If then the speed is constant.aT � 0,
48. (a)
When When
(b) Since for all values of t, the speed is increasing when and t � 2.t � 1aT � � 2 > 0
aN � 2�3.aT � �2,t � 2,aN � �3.aT � �2,t � 1,
aN � ��a�2 � aT2 � �� 4�1 � �2t 2� � � 4 � � 3t
aT � a T � cos �t�� 2 cos �t � �3t sin �t� � sin �t�� 2 sin �t � �3t cos �t� � � 2
T�t� �v�t�
�v�t�� � �cos �t, sin �t�
a�t� � �� 2 cos �t � � 3t sin �t, �2 sin �t � �3t cos �t�
v�t� � ��� sin �t � � sin �t � � 2t cos �t, � cos �t � � cos �t � �2t sin � t� � ��2t cos �t, � 2t sin �t�
r�t� � �cos �t � �t sin �t, sin �t � �t cos �t�
50.
B�1� � T�1� � N�1� �
i
�66
��2
2
j
�63
0
k
�66
�22
��33
i ��33
j ��33
k ��33
�i � j � k�
N�1� �1
�6�3��3i � 3k� �
�22
��i � k�
T�1� �1
�6�i � 2j � k�
r�1� � i � j �13
k
N�t� �1
�1 � 4t2 � t4�1 � t2 � t4���2t � t3�i � �1 � t4�j � �t � 2t3�k�
T�t� �1
�1 � 4t2 � t4�i � 2t j � t2k�
r��t� � i � 2tj � t2k
112
12
−
12 N
B
T131, 1,( )
xy
zr�t� � ti � t2j �t3
3k, t0 � 1
52. (a)
(c)
—CONTINUED—
a�t� � �32j � 4�64t2 � 200t � 625
Speed � �v�t�� � �2500�3� � �50 � 32t�2
v�t� � 50�3 i � �50 � 32t�j
� 50�3t i � �5 � 50t � 16t2� j
� �100 cos 30��t i � �5 � �100 sin 30��t � 16t2� j
r�t� � �v0 cos �t i � �h � �v0 sin �t �12
gt2� j (b)
Maximum height 44.0625
Range 279.0325��
300−20
−10
60
t 0.5 1.0 1.5 2.0 2.5 3.0
Speed 93.04 88.45 86.63 87.73 91.65 98.06
(d)
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288 Chapter 11 Vector-Valued Functions
52. —CONTINUED—
(e)
aTT � aNN � �32j
aN � a N �400�3
�64t2 � 200t � 625
aT � a T �16�16t � 25�
�64t2 � 200t � 625
N�t� ��25 � 16t�i � 25�3
2�64t2 � 200t � 625
T�t� �25�3i � �25 � 16t�j
2�64t2 � 200t � 625(f )
The speed is increasing when and have opposite signs.
aNaT
30
−50
50
54.
Motion along r is clockwise, therefore
aN � a N �1760
�4t2 � 3025
aT � a T �64t
�4t2 � 3025
N�t� ��2ti � 55j�4t2 � 3025
T�t� �880i � 32tj
16�4t2 � 3025�
55i � 2tj�4t2 � 3025
a�t� � �32j
v�t� � 880i � 32tj
r�t� � 880ti � ��16t2 � 36,000�j
600 mph � 880 ftsec 56.
(a)
(b) By Newton’s Law:
v ��GMr
v2 �GM
r,
mv2
r�
GMmr2 ,
F � m�a�t�� � m�r�2� �mr
�r2�2� �mv2
r
�a�t�� � r�2
a�t� � ��r�2 cos �t�i � �r�2 sin �t�j
�v�t�� � r��1 � r� � v
v�t� � ��r� sin �t�i � �r� cos �t�j
r�t� � �r cos �t�i � �r sin �t�j
58. v ��9.56 � 104
4200� 4.77 misec
60. Let distance from the satellite to the center of the earth Then:
v �2��26,245�24�3600� � 1.92 misec � 6871 mph
x3 ��9.56 � 104��24�2�3600�2
4�2 ⇒ x � 26,245 mi
4� 2x2
�24�2�3600�2 �9.56 � 104
x
v �2�x
t�
2�x24�3600� ��9.56 � 104
x
�x � r � 4000�.x �
62.
m and b are constants.
Hence, T��t� � 0.
T�t� �v�t�
�v�t�� �± �i � mj��1 � m2
, constant
�v�t�� � ��x��t��2 � �mx��t��2 � �x��t���1 � m2
v�t� � x��t�i � mx��t�j
r�t� � x�t�i � �m�x�t�� � b� j
y�t� � m�x�t�� � b,
r�t� � x�t�i � y�t�j 64.
Since we have aN � ��a�2 � aT2.aN > 0,
aN2 � �a�2 � aT
2
� aT2 � aN
2
� aT2�T�2 � 2aTaNT N � aN
2�N�2
� �aTT � aNN� �aTT � aNN�
�a�2 � a a
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Section 11.5 Arc Length and Curvature
Section 11.5 Arc Length and Curvature 289
2.
�14
�8�65 � ln�8 � �65 �� � 16.819
�14�2t�1 � 4t2 � ln2t � �1 � 4t
4
0
s � �4
0 �1 � 4t2 dt
dzdt
� 2tdydt
� 0,dxdt
� 1,
x
12
8
16
4
4
21 3(0, 0)
(4, 16)
yr�t� � ti � t2k
4.
� �2�
0a dt � �at
2�
0� 2�a
s � �2�
0
�a2 sin2 t � a2 cos2 t dt
dydt
� a cos tdxdt
� �a sin t,
x
a
a
yr�t� � a cos ti � a sin tj
6. (a)
when
Maximum height when or � ��
2.sin � � 1,
t �v0 sin �
g. y��t� � v0 sin � � gt � 0
y�t� � �v0 sin ��t �12
gt2
r�t� � �v0 cos ��ti � ��v0 sin ��t �1
2gt2j (b)
Range:
The range is a maximum for or � ��
4.sin 2� � 1,x�t�
x�t� � �v0 cos ���2v0 sin �g �
v02
g sin2 �
y�t� � �v0 sin ��t �12
gt2 � 0 ⇒ t �2v0 sin �
g
(c)
Since we have
Using a computer algebra system, is a maximum for � � 0.9855 � 56.5�.s���
s��� � �6 sin �
0�962 � �6144 sin ��t � 1024t2
1�2
dt.
v0 � 96 ft�sec,
s��� � �2v0 sin ��g
0�v0
2 � 2v0g sin �t � g2t21�2
dt
� v02 � 2v0g sin �t � g2t2
� v02 cos2 � � v0
2 sin2 � � 2v02g sin �t � g2t2
x��t�2 � y��t�2 � v02 cos2 � � �v0 sin � � gt�2
y��t� � v0 sin � � gt
x��t� � v0 cos �
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290 Chapter 11 Vector-Valued Functions
8.
1
2
3
4
5
2 3 4 523
5
z
y
x
(0, 2, 0), 0, 23π
2( )
� ���2
0
�13 dt � �13t��2
0�
�13�
2
s � ���2
0
�32 � ��2 sin t�2 � �2 cos t�2 dt
dxdt
� 3, dydt
� �2 sin t, dzdt
� 2 cos t
r�t� � �3t, 2 cos t, 2 sin t� 10.
1
2
3
232
3
z
y
x
(1, 0, 0)
, 1,π2
π4( )2
� ���2
0
�5t2 dt � �5 t2
2��2
0�
�5�2
8
s � ���2
0
��t cos t�2 � �t sin t�2 � �2t�2 dt
dxdt
� t cos t, dydt
� t sin t, dzdt
� 2t
r�t� � �cos t � t sin t, sin t � t cos t, t2�
12.
� �2
0
�� 2 � 9t4 dt � 11.15
s � �2
0
��� cos �t�2 � ��� sin �t�2 � �3t2�2 dt
dzdt
� 3t2dydt
� �� sin �t,dxdt
� � cos �t,
r�t� � sin �ti � cos �tj � t3k 14.
(a)
(b)
(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain
s � �2
0�r��t�� dt � 7.0105.
r�2.0� � �0, 2, 2�
r�1.5� � �2.296, 1.848, 1.5�
r�1.0� � �4.243, 1.414, 1.0�
r�0.5� � �5.543, 0.765, 0.5�
r�0� � �6, 0, 0�
distance � �62 � 22 � 22 � �44 � 2�11 � 6.633
r�2� � 2j � 2k � �0, 2, 2�
r�0� � 6i � �6, 0, 0�
r�t� � 6 cos��t4 i � 2 sin��t
4 j � tk, 0 ≤ t ≤ 2
16.
(a)
—CONTINUED—
� �t
0
�16u � 9u2 du � �t
0 5u du �
52
t2 � �t
0
��4u sin u�2 � �4u cos u�2 � �3u�2 du
s � �t
0
��x��u��2 � �y��u��2 � �z��u��2 du
r�t� � �4�sin t � t cos t�, 4�cos t � t sin t�, 32
t 2�
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16. —CONTINUED—
(b)
r�s� � 4�sin�2s5
��2s5
cos�2s5 i � 4�cos�2s
5��2s
5 sin�2s
5 j �3s5
k
z �32��2s
5 2
�3s5
y � 4�cos�2s5
��2s5
sin�2s5
x � 4�sin�2s5
��2s5
cos�2s5
t ��2s5
(c) When
(d) �r��s�� � ��45
sin�2s5
2
� �45
cos�2s5
2
� �35
2
��1625
�9
25� 1
��6.956, 14.169, 1.342�
z �3�5
5� 1.342
� 14.169y � 4�cos�2�55
��2�55
sin�2�55
� �6.956x � 4�sin�2�55
��2�55
cos�2�55
s � �5: When
�2.291, 6.029, 2.400�
z �125
� 2.4
y � 4�cos�85
��85
sin�85 � 6.029
x � 4�sin�85
��85
cos�85 � 2.291
s � 4:
Section 11.5 Arc Length and Curvature 291
18.
and
(The curve is a line.)T��s� � 0 ⇒ K � �T��s�� � 0
T�s� � r��s�
�r��s�� � 1r��s� � i
r�s� � �3 � s�i � j
20.
K � �T��s�� �4
25� 52s
�2�10s
25s
T��s� �4
25 � 5
2s cos�2s
5i �
425� 5
2s sin �2s
5j
T�s� � r��s� �45
sin�2s5
i �45
cos�2s5
j �35
k
r�s� � 4�sin�2s5
��2s5
cos�2s5 i � 4�cos�2s
5��2s
5 sin�2s
5 j �3s5
k
22.
K ��T��t���r��t�� � 0
T��t� � 0
T�t� � j
v�t� � 2tj
r�t� � t2j � k
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292 Chapter 11 Vector-Valued Functions
24.
K �a N�v�2 �
2
5�5
N�1� �1
�5��2i � j�
N�t� �1
�1 � 4t2��2ti � j�
T�t� �i � 2tj
�1 � 4t2
a�1� � 2j
a�t� � 2 j
v�1� � i � 2j
v�t� � i � 2t j
r�t� � t i � t2j 26.
�2
�4 sin2 �t � cos2 �t�3�2
K ��T��t���r��t�� �
2�
4 sin2 �t � cos2 �t
��4 sin2 �t � cos2 �t
T��t� ��2� cos �ti � 4� sin �tj
�4 sin2 �t � cos2 �t�3�2
T�t� ��2 sin �t i � cos �tj�4 sin2 �t � cos2 �t
�r��t�� � ��4 sin2 �t � cos2 �t
r��t� � �2� sin �ti � � cos �tj
r�t� � 2 cos �ti � sin �tj
28.
�ab
�a2 sin2�t� � b2 cos2�t��3�2
K ��T��t���r��t�� �
ab
a2 sin2�t� � b2cos2�t��a2 sin2�t� � b2 cos2�t�
T��t� ��ab2 cos�t�i � a2b sin�t�j
�a2 sin2�t� � b2 cos2�t��3�2
T�t� ��a sin�t� i � b cos�t�j�a2 sin2�t� � b2 cos2�t�
r��t� � �a sin�t�i � b cos�t�j
r�t� � a cos�t�i � b sin �t�j 30.
From Exercise 22, Section 11.4, we have:
��a2
�2 �1 � cos t
2a22�1 � cos t� ��2
4a�1 � cos t
K �a�t� N�t�
�v�t��2
a N �a2
�2 �1 � cos t
a�1 � cos t��r�t� � �a�t � sin t�,
32.
K ��T��t���r��t�� � 0
T��t� � 0
T�t� �13
�2i � 2j � k�
r��t� � 4i � 4j � 2k
r�t� � 4t i � 4tj � 2tk 34.
��17
�1 � 17t2�3�2
K ��T��t���r��t�� �
�289t2 � 17�1 � 17t2�3�2 ��1 � 17t2�1�2
T��t� �4i � 17tj � k�1 � 17t2�3�2
T�t� �4ti � j � tk�1 � 17t2
r��t� � 4ti � j � tk
r�t� � 2t2i � tj �12
t2k
36.
K ��T��t���r��t�� �
�1��3����cos t � sin t�2 � ��sin t � cos t�2
�3et ��23et
T��t� �1
�3���cos t � sin t�i � ��sin t � cos t�j�
T�t� �1
�3���sin t � cos t�i � �cos t � sin t�j � k�
r��t� � ��et sin t � et cos t�i � �et cos t � et sin t�j � etk
r�t� � et cos ti � et sin tj � etk
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Section 11.5 Arc Length and Curvature 293
38.
Since and the radius of curvature is undefined.K � 0,y� � 0,
y � mx � b
40.
(radius of curvature) 1K
�53�2
8
K � y��1 � �y��2�3�2 �
8�1 � 4�3�2 �
853�2
y� �8x3, y��1� � 8
y� � 2 �4x2, y��1� � �2
y � 2x �4x, x � 1 42.
At
(radius of curvature) 1K
�163
K � �3�16�1 � 02�3�2 �
316
y� � �3
16
y� � 0x � 0:
y� ���9 � �16y��2�
16y
y� ��9x16y
y �34�16 � x2
44. (a)
At
Center:
Equation:
(b) The circles have different radii since the curvature isdifferent and
r �1K
.
x2 � �y �38
2
�9
64
�0, 38
r �1K
�38
K �8�3
�1 � 02�3�2 �83
y� �7227
�83
y� � 0x � 0:
y� �72�1 � x2��x2 � 3�3
y� �24x
�x2 � 3�2
y �4x2
x2 � 346.
The slope of the tangent line at is
The slope of the normal line is
Equation of normal line:
The center of the circle is on the normal line unitsaway from the point
Since the circle is below the curve, and
Center of circle:
Equation of circle:
2
−2
−4
42 6x
(1, 0)
y
�x � 3�2 � �y � 2�2 � 8
�3, �2�y � �2.x � 3
x � 3 or x � �1
2�x � 3��x � 1� � 0
2�x2 � 2x � 3� � 0
2x2 � 4x � 2 � 8
�1 � x�2 � �x � 1�2 � 8
��1 � x�2 � �0 � y�2 � 2�2
�1, 0�.2�2
y � ��x � 1� � �x � 1
�1.
y��1� � 1.�1, 0�
r �1K
� 23�2 � 2�2K � �1�1 � �1�2�3�2 �
123�2,
y��1� � 1, y��1� � �1
y� �1x, y� � �
1x2
y � ln x, x � 1
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294 Chapter 11 Vector-Valued Functions
48.
The slope of the tangent line at is
The slope of the normal line is
Equation of normal line: or y � �x �43y �
13 � ��x � 1��1.
y��1� � 1.�1, 13�
r �1K
� �2K �2
�1 � 1�3�2 �1
�2,
y��1� � 1, y��1� � 2
y� � x2, y��1� � 2x
y �13
x3, x � 1
The center of the circle is on the normal line units away from the point
Since the circle is above the curve, and
Center of circle:
Equation of circle: x2 � �y �43�2
� 2
�0, 43�y �
43 .x � 0
x � 0 or x � 2
�x � 1�2 � 1
�1 � x�2 � �x � 1�2 � 2
��1 � x�2 � �13 � y�2
� �2
1
2
3
−1
−2 1 2x
1, 13( )
y�1, 13�.�2
50.
2−2−4
−3
3
4
−4
4
y
xA
B
52.
(a) K is maximum at
(b) limx→�
K � 0
� �14�45
, �1
4�453 . � 14�45
, 1
4�453 ,
K � 6x�1 � 9x4�3�2
y� � 6xy� � 3x2,y � x3,
54.
(a) K has a maximum when
(b) limx→�
K � 0
x �1
�2.
dKdx
��2x2 � 1�x2 � 1�5�2
K � �1�x2
�1 � �1�x�2�3�2 �x
�x2 � 1�3�2
y� � �1x2y� �
1x,y � ln x, 56.
for
Curvature is 0 at ��
2� K�, 0 .
x ��
2� K�. K � y�
�1 � �y��2�3�2 � �cos x�1 � sin2 x�3�2 � 0
y� � �cos x
y� � �sin x
y � cos x
58.
K � cosh x�1 � �sinh x�2�3�2 �
cosh x�cosh2 x�3�2 �
1cosh2 x
�1y2
y� �ex � e�x
2� cosh x
y� �ex � e�x
2� sinh x
y � cosh x �ex � e�x
260. See page 828.
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Section 11.5 Arc Length and Curvature 295
62.
At the smooth relative extremum so Yes, for example, has a curvature of 0 at its relative minimumThe curvature is positive for any other point of the curvature.�0, 0�.
y � x4K � y�.y� � 0,
K � y��1 � �y��2�3�2
64.
We observe that is a solution point to both equations. Therefore, the point P is the origin.
At
and
Since the curves have a common tangent at P, or Therefore, Since the curves have the same curvature at P,
Therefore, or In order that the curves intersect at only one point, the parabola must be concave downward. Thus,
and
and y2 �x
x � 2y1 �
14
x�2 � x�
b �1
2a� 2.a �
14
a � ±14 .2a � ±1
2
K2�0� � y2��0��1 � �y2�0��2�3�2 � �1�2
�1 � �1�2�2�3�2K1�0� � y1��0�
�1 � �y1�0��2�3�2 � �2a�1 � �1�2�2�3�2
K1�0� � K2�0�.y1��0� �
12 .ab �
12 .y1��0� � y2��0�
y2��0� �2
�0 � 2�2 �12
.y1��0� � ab
P,
y2 �x
x � 2, y2� �
2�x � 2�2, y2� �
�4�x � 2�3
y1 � ax�b � x�, y1� � a�b � 2x�, y1� � �2a
�0, 0�2
4
−4
−2−4 2 4x
y2
y2
y1
P
yy2 �x
x � 2y1 � ax�b � x�,
66.
(a)
(rotated about y-axis)
x
y2
4
4
2
24
z
0 ≤ x ≤ 5y �14
x8�5,
(c)
500
1
K �
625x2�5
�1 �4
25x6�5
3�2 �6
25x2�5�1 �4
25x6�5
3�2
y� �6
25x�2�5 �
625x2�5y� �
25
x3�5,
(b) (shells)
�5�
36 518�5 � 143.25 cm3
��
2�5
0 x13�5 dx �
�
2�x18�5
18�55
0
V � �5
0 2�x�1
4x8�5 dx
(d) No, the curvature approaches as Hence, anyspherical object will hit the sides of the goblet beforetouching the bottom �0, 0�.
x → 0�.�
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296 Chapter 11 Vector-Valued Functions
68.
When
At
s � �32 �
c�K
�30�4�2�K
� 56.27 mi hr.
K �3
[1 � �81�16��3�2 � 0.201x �32
,
30 � 4�2c ⇒ c �304�2
s �c
�1��2� 4�2c
K �1
�2x � 1:
K � 2x�1 � x4�3�2
y� � 2x
y� � x2
y �13
x3
s �c
�K
70.
K � x�y� � y�x���x��2 � �y��2�3�2 � f 2��� � f ��� f���� � 2� f�����2
� f 2��� � � f�����2�3�2 � r 2 � rr� � 2�r��2�r 2 � �r��2�3�2
y���� � �f ��� sin � � f���� cos � � f���� cos � � f���� sin � � �f��� sin � � 2 f���� cos � � f���� sin �
x���� � �f ��� cos � � f���� sin � � f���� sin � � f���� cos � � �f ��� cos � � 2 f���� sin � � f���� cos �
y���� � f ��� cos � � f���� sin �
x���� � �f ��� sin � � f���� cos �
y��� � f ��� sin �
x��� � f ��� cos �
� f ��� cos � i � f ��� sin �jr��� � r cos �i � r sin �j
72.
K � 2�r��2 � rr� � r2��r��2 � r2�3�2 �
2 � � 2
�1 � � 2�3�2
r� � 0
r� � 1
r � � 74.
K � 2�r��2 � rr� � r2��r��2 � r2�3�2 �
2e2�
�2e2��3�2 �1
�2e�
r� � e�
r� � e�
r � e�
76. At the pole,
� 2�r��2r�3 �
2
r�
K � 2�r��2 � rr� � r2��r��2 � r2�3�2
r � 0. 78.
At the pole,
and
K �2
r����6� �2
�18 �19
.
r���
6 � �18,� ��
6,
r� � �18 sin 3�
r � 6 cos 3�
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Section 11.5 Arc Length and Curvature 297
80.
as t →±�K → 0
�3t2
t3�9t2 � 1�3�2 �3
t�9t2 � 1�3�2
K � �3t2��1� � �t��6t���3t2�2 � �t�2�3�2
y��t� � 1y��t� � t,y�t� �12
t2,
x��t� � 6tx��t� � 3t2,x�t� � t3,
4−40
5
82. (a)
aN � K�dsdt
2
�2
3�1 � t2�2 9�1 � t2�2 � 6
aT �d 2sdt2 � 6t
K �2
3�1 � t2�2
d 2sdt2 � 6t
dsdt
� �v�t�� � 3�1 � t2�,
v�t� � 6ti � �3 � 3t2�j
r�t� � 3t2i � �3t � t3�j (b)
aN � K�ds
dt 2
��5
�5t2 � 1�3�2�5t2 � 1� �
�5�5t2 � 1
aT �d2s
dt2�
5t�5t2 � 1
K ��r��t� � r��t��
�r��t��3 ��5
�5t2 � 1�3�2
r��t� � r��t� � v�t� � a�t� � i10
j2t2
kt1 � �j � 2k
a�t� � 2j � k
d2s
dt2�
5t�5t2 � 1
dsdt
� �v�t�� � �5t2 � 1
v�t� � i � 2tj � tk
r�t� � ti � t2j �12
t2k
84. (a) by the Chain Rule
(b)
Since and we have:
from (a)
Therefore,
(c) K ��r��t� � r��t��
�r��t�3��
�r��t� � r��t���r��t���r��t��2 �
�v�t� � a�t���v�t��
�r��t��2 �a�t� N�t�
�r��t��2
�r��t� � r��t���r��t��3 � K.
�r��t� � r��t�� � �r��t��2�T�t� � T��t�� � �r��t��2�T�t�� �T��t�� � �r��t��2�1�K�r��t��
r��t� � r��t� � �r��t��2�T�t� � T��t��
dsdt
� �r��t��,T�t� � T�t� � 0
r��t� � r��t� � �dsdt �
d 2sdt2 �T�t� � T�t�� � �ds
dt 2
�T�t� � T��t��
r��t� � �d 2sdt2 T�t� �
dsdt
T��t�
r��t� �dsdt
T�t�
T�t� �r��t�
�r��t�� �r��t�ds�dt
� � d T�dtds�dt � �
�T��t���v�t�� �
�T��t���r��t��
K � �T��s�� � � dTds � � � dT
dt
dtds � ,
http://librosysolucionarios.net
86.
Since r is a constant multiple of a, they are parallel. Since is parallel to Also,
Thus, is a constant vector which we will denote by L.r � r�
� ddt��r � r�� � r� � r� � r � r� � 0 � 0 � 0.
r � r� � 0.r,a � r�
a � �GMr3 r
F � ma ⇒ ma ��GmM
r3 r
88.
Thus, is a constant vector which we will denote by e.� r�
GM� � L � �rr�
�1r3��r � r�� � r � �r � r�� � r � 0
� �rr3 � �r � r�� �
1r3��r � r�� � r
�1
GM0 � ��GMrr3 � � �r � r��� �
1r3��r � r�� � r
ddt
r�
GM� L �
rr� �
1GM
�r� � 0 � r� � L� �1r3��r � r�� � r
90.
Let:
Then:
� r2 d�
dtk and �L� � �r � r�� � r2
d�
dt.
r � r� � i
r cos �
�r sin � d�dt
j
r sin �
r cos � d�dt
k
0
0 �dr
dt�
drd�
d�
dt �r� � r��sin �i � cos �j�d�
dt
r � r�cos �i � sin �j�
�L� � �r � r� � 92. Let P denote the period. Then
Also, the area of an ellipse is where 2a and 2b are thelengths of the major and minor axes.
�42
GMa3 � Ka3�
4 2��L�2�GM ��L�2 a3
�4 2ed�L�2 a3�
4 2a4
�L� 2 �eda �
P2 �42a2
�L�2 �a2 � c2� �42a2
�L�2a2�1 � e2�
P �2ab�L�
ab �12
�L�P
ab
A � �P
0
dAdt
dt �12
�L�P.
Review Exercises for Chapter 11
2.
(a) Domain: and
(b) Continuous except at t � 4
�4, ���0, 4�
r�t� � �t i �1
t � 4j � k 4.
(a) Domain:
(b) Continuous for all t
���, ��
r�t� � �2t � 1�i � t2j � tk
298 Chapter 11 Vector-Valued Functions
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6. (a)
(b)
(c)
(d)
� ��3 cos �t � 3� i � sin �t � �tk
r� � �t� � r�� � �3 cos� � �t�i � �1 � sin� � �t��j � � � �t�k� � ��3i � j � k�
r�s � � � 3 cos�s � �i � �1 � sin�s � ��j � �s � �k
r�
2� � �
2k
r�0� � 3i � j
8.
y �x
x � 1
x�t� � t, y�t� �t
t � 1
x
2
2
3
4
4
1
1 3−1
−2
−2
y
r�t� � ti �t
t � 1j 10.
z � y2y �12 x,
z � t2,y � t,x � 2t,
xy3
4
5
zr�t� � 2ti � tj � t2k
t 0 1 2
x 0 2 4
y 0 1 2
z 0 1 1 4
�1
�2
�1
12.
x2 � z2 � 4
z � 2 sin ty � t,x � 2 cos t,
x
y
2
3
2π
zr�t� � 2 cos ti � tj � 2 sin tk
t 0
x 2 0 0
y 0
z 0 2 0 �2
3
2
2
�2
3
2
2
14.
yx
−1
11
1
2 2
2
33
3
4
5
6
−1−2−2
−3−3
z
r�t� �12 ti � �tj �
14 t 3k
16. One possible answer is:
r3�t� � �4 � t� j, 0 ≤ t ≤ 4
r2�t� � 4 cos ti � 4 sin tj, 0 ≤ t ≤
2
r1�t� � 4ti, 0 ≤ t ≤ 1
18. The x- and y-components are 2 cos t and 2 sin t. At
the staircase has made of a revolution and is 2 metershigh. Thus, one answer is
r�t� � 2 cos ti � 2 sin tj �4
3tk.
34
t �3
2,
20.
r�t� � ti � tj � �4 � t2k
r�t� � ti � tj � �4 � t2k
x
y
34
5
zz � ±�4 � t2y � t,x � t,
t � xx � y � 0,x2 � z2 � 4, 22.
� 2i � j � k
limt→0
�sin 2tt
i � e�tj � etk� � �limt→0
2 cos 2t
1 �i � j � k
Review Exercises for Chapter 11 299
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24.
(a)
(c)
(e)
(f)
Dt�r�t� � u�t�� � ��1t sin t �
1t2 cos t � t sin t � cos t�i � �1
t cos t �
1t2 sin t � t cos t � sin t�j
r�t� � u�t� � �1t cos t � t cos t�i � �1
t sin t � t sin t�j
Dt��r�t��� �t
�1 � t2
�r�t�� � �1 � t2
Dt�r�t� u�t�� � 0
r�t� u�t� � 2
r��t� � cos ti � sin tj � k
u�t� � sin ti � cos t j �1t
kr�t� � sin ti � cos tj � tk,
(b)
(d)
Dt�u�t� � 2r�t�� � �cos t i � sin tj � ��1t2 � 2�k
u�t� � 2r�t� � �sin ti � cos tj � �1t
� 2t�k
r��t� � �sin t i � cos t j
26. The graph of u is parallel to the yz-plane.
28. ��ln t i � t ln tj � k� dt � �t ln t � t�i �t2
4��1 � 2 ln t�j � tk � C
30. ��tj � t2k� � �i � tj � tk� dt � ���t2 � t3�i � t2j � tk� dt � �t3
3�
t4
4�i �t3
3j �
t2
2k � C
32.
r�t� � ln sec t � tan t i � ln cos t j � �t3
3� 3�k
r�0� � C � 3k
� ln sec t � tan t i � ln cos t j �t3
3k � Cr�t� � ��sec t i � tan tj � t2k� dt
34. �23
j � �sin 1 � cos 1�k�1
0��t j � t sin t k� dt � 2
3t 3�2j � �sin t � t cos t�k�
1
0
36.
� �23
k
�1
�1�t3i � arcsin tj � t2k� dt � t4
4i � �t arcsin t � �1 � t2�j �
t3
3k�
1
�1
38.
r��t� � a�t� � �0, �2 sec2 t tan t, et�
�v�t�� � �1 � sec4 t � e2t
r��t� � v�t� � �1, �sec2 t, et�
r�t� � �t, �tan t, et� 40.
direction numbers
Since the parametric equations are
r�t0 � 0.1� � r�0.1� � �3, 0.1, �0.2�
z � �2t.y � t,x � 3,r�0� � �3, 0, 0�,
r��0� � �0, 1, �2�
r��t� � �3 sinh t, cosh t, �2�
t0 � 0r�t� � �3 cosh t, sinh t, �2t�,
300 Chapter 11 Vector-Valued Functions
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42.
4163
� v02 ⇒ v0 � 11.776 ft sec
�3v0
2
104 �
v02
16
6
2�13
4
2�13 6
4
2 13
Range � 4 �v0
2
16 sin � cos �
44.
(a)
Maximum height 5.1 m; Range 35.3 m
(c)
Maximum height 15.3 m; Range 35.3 m
(Note that gives the longest range)45�
��
4500
20
r�t� � ��20 cos 60��t�i � ��20 sin 60��t � 4.9t2� j
��
4500
20
r�t� � ��20 cos 30��t�i � ��20 sin 30��t � 4.9t2� j
r�t� � ��v0 cos ��t�i � ��v0 sin ��t �12 �9.8�t2� j
46.
does not exist
does not exist a N
a T � 0
N�t�
T�t� �15
�4i � 3j�
a�t� � 0
�v� � 5
v�t� � 4 i � 3j
r�t� � �1 � 4t�i � �2 � 3t�j 48.
�4
�t � 1���t � 1�4 � 1
a N �4�t � 1�2
�t � 1�3��t � 1�4 � 1
a T ��4
�t � 1�3��t � 1�4 � 1
N�t� �i � �t � 1�2j��t � 1�4 � 1
T�t� ��t � 1�2i � j��t � 1�4 � 1
a�t� �4
�t � 1�3 j
�v�t�� �2��t � 1�4 � 1
�t � 1�2
v�t� � 2i �2
�t � 1�2 j
r�t� � 2�t � 1�i �2
t � 1j
(b)
Maximum height 10.2 m; Range 40.8 m��
4500
20
r�t� � ��20 cos 45��t�i � ��20 sin 45��t � 4.9t2� j
Review Exercises for Chapter 11 301
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50.
a�t� N�t� �t2 � 2
�t2 � 1
a�t� T�t� �t
�t2 � 1
N�t� ���t cos t � sin t�i � ��t sin t � cos t�j
�t2 � 1
T�t� �v�t�
�v�t�� ���t sin t � cos t�i � �t cos t � sin t�j
�t2 � 1
a�t� � r��t� � ��t cos t � 2 sin t�i � ��t sin t � 2 cos t�j
�v�t�� � speed � ���t sin t � cos t�2 � �t cos t � sin t�2 � �t2 � 1
v�t� � r��t� � ��t sin t � cos t�i � �t cos t � sin t�j
r�t� � t cos ti � t sin t j
52.
a N �4
t�2�2t 4 � 1
a T ��2
t3�2t 4 � 1
N�t� �i � j � 2t2k�2�2t 4 � 1
T�t� �t2i � t2j � k�2t4 � 1
a�t� �2t3 k
�v�t�� ��2t 4 � 1
t2
v�t� � i � j �1t2k
r�t� � �t � 1�i � tj �1t
k 54.
When
Direction numbers when
z � 8t �163y � 4t � 4,x � t � 2,
c � 8b � 4,a � 1,t � 2,
r��t� � i � 2tj � 2t2k
z �163 .y � 4,x � 2,t � 2,
z �23 t3y � t2,x � t,r�t� � t i � t2j �
23 t3k,
56. Factor of 4
58.
� ln��10 � 3� � 3�10 � 11.3053
� ln �t2 � 1 � t � t�t2 � 1�3
0
s � �b
a
�r��t�� dt � �3
0
�4t2 � 4 dt
r��t� � 2ti � 2k
z
y(9, 0, 6)
(0, 0, 0)
x
2
2112
34
56
7
9
−2
3
4
5
6
r�t� � t2i � 2tk, 0 ≤ t ≤ 3
302 Chapter 11 Vector-Valued Functions
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62.
4
3
2
1
13
12
2 4
z
y
x
(2, 4, 4)
� �21 �54
ln 5 �54
ln��105 � 4�5� � 6.2638
s � �b
a
�r��t�� dt � �2
0
�5 � 4t2 dt
r��t� � i � 2tj � 2k, �r��t�� � �5 � 4t2
r�t� � ti � t2j � 2tk, 0 ≤ t ≤ 2
64.
� �17 �14
ln��17 � 4� � 4.6468
s � �b
a
�r��t�� dt � ��2
0
�4t2 � 1 dt
x��t� � �2t sin t, 2t cos t, 1�, �r��t�� � �4t2 � 1
r�t� � �2�sin t � t cos t�, 2�cos t � t sin t�, t�, 0 ≤ t ≤
266.
� �2�
0 et dt � �2et�
0� �2�e � 1�
s � �
0�r��t�� dt
� �2et
�r��t�� � ��et cos t � et sin t�2 � ��et sin t � et cos t�2
r��t� � �et cos t � et sin t� i � ��et sin t � et cos t�k
0 ≤ t ≤ r�t� � et sin ti � et cos tk,
68.
K ��r��t� � r��t��
�r��t��3 �3�2t3�2
�1 � 9t�3�2�t3�2 �3
2�1 � 9t�3�2
r� � r� � i1
�t
�1 t
�3�2
2
j
3
0
k
0
0 �32
t �3�2k; �r� � r�� �3
2t3�2
r��t� � �12
t �3�2i
r��t� �1�t
i � 3j, �r��t�� ��1
t� 9 ��1 � 9t
t
r�t� � 2�ti � 3tj
60.
−8 −6 −4
−4
2468
−6−8
−2 2 4 6 8
y
x
s � �2
010 dt � 20
�r��t�� � 10
r��t� � �10 sin ti � 10 cos tj
r�t� � 10 cos ti � 10 sin tj
70.
K ��r� � r��
�r��3 ��725�29�3�2 �
�25 29
29�29�
529
�r� � r� � � �725
r� � r� � i20
j�5 sin t�5 cos t
k5 cos t
�5 sin t � 25i � 10 sin tj � 10 cos tk
r��t� � 5 cos tj � 5 sin tk
r��t� � 2i � 5 sin tj � 5 cos tk, �r��t�� � �29
r�t� � 2ti � 5 cos tj � 5 sin tk
Review Exercises for Chapter 11 303
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86.
Since r is a constant multiple of a, they are parallel. Since is parallel to Also,
Thus, is a constant vector which we will denote by L.r � r�
� ddt��r � r�� � r� � r� � r � r� � 0 � 0 � 0.
r � r� � 0.r,a � r�
a � �GMr3 r
F � ma ⇒ ma ��GmM
r3 r
88.
Thus, is a constant vector which we will denote by e.� r�
GM� � L � �rr�
�1r3��r � r�� � r � �r � r�� � r � 0
� �rr3 � �r � r�� �
1r3��r � r�� � r
�1
GM0 � ��GMrr3 � � �r � r��� �
1r3��r � r�� � r
ddt
r�
GM� L �
rr� �
1GM
�r� � 0 � r� � L� �1r3��r � r�� � r
90.
Let:
Then:
� r2 d�
dtk and �L� � �r � r�� � r2
d�
dt.
r � r� � i
r cos �
�r sin � d�dt
j
r sin �
r cos � d�dt
k
0
0 �dr
dt�
drd�
d�
dt �r� � r��sin �i � cos �j�d�
dt
r � r�cos �i � sin �j�
�L� � �r � r� � 92. Let P denote the period. Then
Also, the area of an ellipse is where 2a and 2b are thelengths of the major and minor axes.
�42
GMa3 � Ka3�
4 2��L�2�GM ��L�2 a3
�4 2ed�L�2 a3�
4 2a4
�L� 2 �eda �
P2 �42a2
�L�2 �a2 � c2� �42a2
�L�2a2�1 � e2�
P �2ab�L�
ab �12
�L�P
ab
A � �P
0
dAdt
dt �12
�L�P.
Review Exercises for Chapter 11
2.
(a) Domain: and
(b) Continuous except at t � 4
�4, ���0, 4�
r�t� � �t i �1
t � 4j � k 4.
(a) Domain:
(b) Continuous for all t
���, ��
r�t� � �2t � 1�i � t2j � tk
298 Chapter 11 Vector-Valued Functions
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6. (a)
(b)
(c)
(d)
� ��3 cos �t � 3� i � sin �t � �tk
r� � �t� � r�� � �3 cos� � �t�i � �1 � sin� � �t��j � � � �t�k� � ��3i � j � k�
r�s � � � 3 cos�s � �i � �1 � sin�s � ��j � �s � �k
r�
2� � �
2k
r�0� � 3i � j
8.
y �x
x � 1
x�t� � t, y�t� �t
t � 1
x
2
2
3
4
4
1
1 3−1
−2
−2
y
r�t� � ti �t
t � 1j 10.
z � y2y �12 x,
z � t2,y � t,x � 2t,
xy3
4
5
zr�t� � 2ti � tj � t2k
t 0 1 2
x 0 2 4
y 0 1 2
z 0 1 1 4
�1
�2
�1
12.
x2 � z2 � 4
z � 2 sin ty � t,x � 2 cos t,
x
y
2
3
2π
zr�t� � 2 cos ti � tj � 2 sin tk
t 0
x 2 0 0
y 0
z 0 2 0 �2
3
2
2
�2
3
2
2
14.
yx
−1
11
1
2 2
2
33
3
4
5
6
−1−2−2
−3−3
z
r�t� �12 ti � �tj �
14 t 3k
16. One possible answer is:
r3�t� � �4 � t� j, 0 ≤ t ≤ 4
r2�t� � 4 cos ti � 4 sin tj, 0 ≤ t ≤
2
r1�t� � 4ti, 0 ≤ t ≤ 1
18. The x- and y-components are 2 cos t and 2 sin t. At
the staircase has made of a revolution and is 2 metershigh. Thus, one answer is
r�t� � 2 cos ti � 2 sin tj �4
3tk.
34
t �3
2,
20.
r�t� � ti � tj � �4 � t2k
r�t� � ti � tj � �4 � t2k
x
y
34
5
zz � ±�4 � t2y � t,x � t,
t � xx � y � 0,x2 � z2 � 4, 22.
� 2i � j � k
limt→0
�sin 2tt
i � e�tj � etk� � �limt→0
2 cos 2t
1 �i � j � k
Review Exercises for Chapter 11 299
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24.
(a)
(c)
(e)
(f)
Dt�r�t� � u�t�� � ��1t sin t �
1t2 cos t � t sin t � cos t�i � �1
t cos t �
1t2 sin t � t cos t � sin t�j
r�t� � u�t� � �1t cos t � t cos t�i � �1
t sin t � t sin t�j
Dt��r�t��� �t
�1 � t2
�r�t�� � �1 � t2
Dt�r�t� u�t�� � 0
r�t� u�t� � 2
r��t� � cos ti � sin tj � k
u�t� � sin ti � cos t j �1t
kr�t� � sin ti � cos tj � tk,
(b)
(d)
Dt�u�t� � 2r�t�� � �cos t i � sin tj � ��1t2 � 2�k
u�t� � 2r�t� � �sin ti � cos tj � �1t
� 2t�k
r��t� � �sin t i � cos t j
26. The graph of u is parallel to the yz-plane.
28. ��ln t i � t ln tj � k� dt � �t ln t � t�i �t2
4��1 � 2 ln t�j � tk � C
30. ��tj � t2k� � �i � tj � tk� dt � ���t2 � t3�i � t2j � tk� dt � �t3
3�
t4
4�i �t3
3j �
t2
2k � C
32.
r�t� � ln sec t � tan t i � ln cos t j � �t3
3� 3�k
r�0� � C � 3k
� ln sec t � tan t i � ln cos t j �t3
3k � Cr�t� � ��sec t i � tan tj � t2k� dt
34. �23
j � �sin 1 � cos 1�k�1
0��t j � t sin t k� dt � 2
3t 3�2j � �sin t � t cos t�k�
1
0
36.
� �23
k
�1
�1�t3i � arcsin tj � t2k� dt � t4
4i � �t arcsin t � �1 � t2�j �
t3
3k�
1
�1
38.
r��t� � a�t� � �0, �2 sec2 t tan t, et�
�v�t�� � �1 � sec4 t � e2t
r��t� � v�t� � �1, �sec2 t, et�
r�t� � �t, �tan t, et� 40.
direction numbers
Since the parametric equations are
r�t0 � 0.1� � r�0.1� � �3, 0.1, �0.2�
z � �2t.y � t,x � 3,r�0� � �3, 0, 0�,
r��0� � �0, 1, �2�
r��t� � �3 sinh t, cosh t, �2�
t0 � 0r�t� � �3 cosh t, sinh t, �2t�,
300 Chapter 11 Vector-Valued Functions
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42.
4163
� v02 ⇒ v0 � 11.776 ft sec
�3v0
2
104 �
v02
16
6
2�13
4
2�13 6
4
2 13
Range � 4 �v0
2
16 sin � cos �
44.
(a)
Maximum height 5.1 m; Range 35.3 m
(c)
Maximum height 15.3 m; Range 35.3 m
(Note that gives the longest range)45�
��
4500
20
r�t� � ��20 cos 60��t�i � ��20 sin 60��t � 4.9t2� j
��
4500
20
r�t� � ��20 cos 30��t�i � ��20 sin 30��t � 4.9t2� j
r�t� � ��v0 cos ��t�i � ��v0 sin ��t �12 �9.8�t2� j
46.
does not exist
does not exist a N
a T � 0
N�t�
T�t� �15
�4i � 3j�
a�t� � 0
�v� � 5
v�t� � 4 i � 3j
r�t� � �1 � 4t�i � �2 � 3t�j 48.
�4
�t � 1���t � 1�4 � 1
a N �4�t � 1�2
�t � 1�3��t � 1�4 � 1
a T ��4
�t � 1�3��t � 1�4 � 1
N�t� �i � �t � 1�2j��t � 1�4 � 1
T�t� ��t � 1�2i � j��t � 1�4 � 1
a�t� �4
�t � 1�3 j
�v�t�� �2��t � 1�4 � 1
�t � 1�2
v�t� � 2i �2
�t � 1�2 j
r�t� � 2�t � 1�i �2
t � 1j
(b)
Maximum height 10.2 m; Range 40.8 m��
4500
20
r�t� � ��20 cos 45��t�i � ��20 sin 45��t � 4.9t2� j
Review Exercises for Chapter 11 301
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50.
a�t� N�t� �t2 � 2
�t2 � 1
a�t� T�t� �t
�t2 � 1
N�t� ���t cos t � sin t�i � ��t sin t � cos t�j
�t2 � 1
T�t� �v�t�
�v�t�� ���t sin t � cos t�i � �t cos t � sin t�j
�t2 � 1
a�t� � r��t� � ��t cos t � 2 sin t�i � ��t sin t � 2 cos t�j
�v�t�� � speed � ���t sin t � cos t�2 � �t cos t � sin t�2 � �t2 � 1
v�t� � r��t� � ��t sin t � cos t�i � �t cos t � sin t�j
r�t� � t cos ti � t sin t j
52.
a N �4
t�2�2t 4 � 1
a T ��2
t3�2t 4 � 1
N�t� �i � j � 2t2k�2�2t 4 � 1
T�t� �t2i � t2j � k�2t4 � 1
a�t� �2t3 k
�v�t�� ��2t 4 � 1
t2
v�t� � i � j �1t2k
r�t� � �t � 1�i � tj �1t
k 54.
When
Direction numbers when
z � 8t �163y � 4t � 4,x � t � 2,
c � 8b � 4,a � 1,t � 2,
r��t� � i � 2tj � 2t2k
z �163 .y � 4,x � 2,t � 2,
z �23 t3y � t2,x � t,r�t� � t i � t2j �
23 t3k,
56. Factor of 4
58.
� ln��10 � 3� � 3�10 � 11.3053
� ln �t2 � 1 � t � t�t2 � 1�3
0
s � �b
a
�r��t�� dt � �3
0
�4t2 � 4 dt
r��t� � 2ti � 2k
z
y(9, 0, 6)
(0, 0, 0)
x
2
2112
34
56
7
9
−2
3
4
5
6
r�t� � t2i � 2tk, 0 ≤ t ≤ 3
302 Chapter 11 Vector-Valued Functions
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62.
4
3
2
1
13
12
2 4
z
y
x
(2, 4, 4)
� �21 �54
ln 5 �54
ln��105 � 4�5� � 6.2638
s � �b
a
�r��t�� dt � �2
0
�5 � 4t2 dt
r��t� � i � 2tj � 2k, �r��t�� � �5 � 4t2
r�t� � ti � t2j � 2tk, 0 ≤ t ≤ 2
64.
� �17 �14
ln��17 � 4� � 4.6468
s � �b
a
�r��t�� dt � ��2
0
�4t2 � 1 dt
x��t� � �2t sin t, 2t cos t, 1�, �r��t�� � �4t2 � 1
r�t� � �2�sin t � t cos t�, 2�cos t � t sin t�, t�, 0 ≤ t ≤
266.
� �2�
0 et dt � �2et�
0� �2�e � 1�
s � �
0�r��t�� dt
� �2et
�r��t�� � ��et cos t � et sin t�2 � ��et sin t � et cos t�2
r��t� � �et cos t � et sin t� i � ��et sin t � et cos t�k
0 ≤ t ≤ r�t� � et sin ti � et cos tk,
68.
K ��r��t� � r��t��
�r��t��3 �3�2t3�2
�1 � 9t�3�2�t3�2 �3
2�1 � 9t�3�2
r� � r� � i1
�t
�1 t
�3�2
2
j
3
0
k
0
0 �32
t �3�2k; �r� � r�� �3
2t3�2
r��t� � �12
t �3�2i
r��t� �1�t
i � 3j, �r��t�� ��1
t� 9 ��1 � 9t
t
r�t� � 2�ti � 3tj
60.
−8 −6 −4
−4
2468
−6−8
−2 2 4 6 8
y
x
s � �2
010 dt � 20
�r��t�� � 10
r��t� � �10 sin ti � 10 cos tj
r�t� � 10 cos ti � 10 sin tj
70.
K ��r� � r��
�r��3 ��725�29�3�2 �
�25 29
29�29�
529
�r� � r� � � �725
r� � r� � i20
j�5 sin t�5 cos t
k5 cos t
�5 sin t � 25i � 10 sin tj � 10 cos tk
r��t� � 5 cos tj � 5 sin tk
r��t� � 2i � 5 sin tj � 5 cos tk, �r��t�� � �29
r�t� � 2ti � 5 cos tj � 5 sin tk
Review Exercises for Chapter 11 303
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Problem Solving for Chapter 11
72.
At x � 0, K �1�4
�5�4�3�2�
2
53�2�
2
5�5�
2�5
25, r �
5�5
2.
K � �y���1 � �y��2�3�2 �
14
e�x�2
1 �14
e�x3�2
y� � �12
e�x�2, y� �14
e�x�2
y � e�x�2 74.
At and r �5�5
4.x �
�
4, K �
453�2
�4
5�5�
4�5
25
K � �y���1 � �y��2�3�2 � �2 sec2 x tan x�
�1 � sec4 x]3�2
y� � 2 sec2 x tan x
y� � sec2 x
y � tan x
2.
Slope at
origin
on curve.
Thus, the radius of curvature, is three times the
distance from the origin to the tangent line.
1K
,
K ��T��t���r��t�� �
1
�3 cos t sin t�
D ��PQ
\
� T��T�
� �3 cos t sin t�
� �cos3 t sin t � sin3 t cos t�k
PQ\
� T � � icos3 t
�cos t
jsin3 tsin t
k00�
P � �cos3 t, sin3 t, 0�
Q�0, 0, 0�
T��t� � sin ti � cos tj
T�t� �r��t�
�r��t�� � �cos ti � sin tj
�r��t�i � �3 cos t sin t� r��t� � �3 cos2 t sin ti � 3 sin2 t cos tj
r�t� � cos3 ti � sin3 tj
P�x, y�. y� ��y1�3
x1�3
23
x�1�3 �23
y�1�3y� � 0
x2�3 � y2�3 � a2�3 4. Bomb:
Projectile:
At 1600 feet: Bomb:
Projectile will travel 5 seconds:
Horizontal position:
At bomb is at
At projectile is at
Thus,
Combining,
v0 �1800cos
� 1843.9 ft�sec
v0 sin v0 cos
�400
1800 ⇒ tan �
29 ⇒ � 12.5.
v0 cos � 1800.
5v0 cos � 9000
�v0 cos �5.t � 5,
5000 � 400�10� � 9000.t � 10,
v0 sin � 400.
5�v0 sin � � 16�25� � 1600
3200 � 16t2 � 1600 ⇒ t � 10
r2�t� � �v0 cos �t, �v0 sin �t � 16t2�
r1�t� � 5000 � 400t, 3200 � 16t2�
304 Chapter 11 Vector-Valued Functions
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6.
s2 � 9�2 � 16 cos2
2� 16 sin2
2� 16
� �1K
�
4 sin
23
�34
sin2
2
sin3
2
�3
4 sin
2
� �3 � 3 cos �8 sin3
2
� �2 sin2 � �1 � cos ��cos � � �1 � cos �2�8 sin3
2
K � �2�r��2 � rr� � r2���r��2 � r2�3�2
� �t
�
2 sin
2 d � �4 cos
2t
�� �4 cos
t2
s�t� � �t
�
��1 � cos �2 � sin2 d � �t
�
�2 � 2 cos d
r� � sin
r � 1 � cos
8. (a)
—CONTINUED—
� d2rdt2
� r�d
dt �2
ur � 2drdt
d
dt� r
d2
dt2 u
a � �a � ur�ur � �a � u�u
a � a � u � a � ��sin i � cos j� � 2drdr
d
dt� r
d2
dt2
�d2rdt2
� r�d
dt �2
� d2rdt2
sin2 � 2drdt
sin cos d
dt� r sin2 �d
dt �2
� r cos sin d2
dt2
� d2rdt2
cos2 � 2 drdt
sin cos d
dt� r cos2 �d
dt �2
� r cos sin d2
dt2 ar � a � ur � a � �cos i � sin j�
� d2rdt2
sin �drdt
cos d
dt�
drdt
cos d
dt� r sin �d
dt �2
� r cos d2
dt2
a �d2rdt2
� d2rdt2
cos �drdt
sin d
dt�
drdt
sin d
dt� r cos �d
dt �2
� r sin d2
dt2 i
drdt
� drdt
cos � r sin d
dti � drdt
sin � r cos d
dtj
r � r cos i � r sin j
r � xi � yj position vector
Problem Solving for Chapter 11 305
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8. —CONTINUED—
(b)
Therefore,
Radial component:
Angular component: 0
�8753
�2
a � �42000� �
12�2
ur � �875
3�2ur .
d
dt�
�
12,
d2
dt2 � 0
r � 42,000, drdt
� 0, d2rdt2
� 0
r � 42,000 cos��t12�i � 42,000 sin��t
12�j
12.
At the point K �120
�89�3�2 ⇒ r �1K
��89�3�2
120� 7.�4, 1�,
K � � 15128
x1�2
�1 �25
4096x3�
3�2�y� �
15128
x1�2
y� �5
64x3�2
y �1
32x5�2
14. (a) Eliminate the parameter to see that the Ferris wheel has a radius of 15 meters and is centered at At the friend is located at which is the low point on the Ferris wheel.
(b) If a revolution takes seconds, then
and so seconds. The Ferris wheel makes three revolutions per minute.
(c) The initial velocity is The speed is The angle ofinclination is radians or
(d) Although you may start with other values, is a fine choice. The graph at the right shows two points of intersection. At sec the friend is near thevertex of the parabola, which the object reaches when
Thus, after the friend reaches the low point on the Ferris wheel, wait sec before throwing the object in order to allow it to be within reach.
(e) The approximate time is 3.15 seconds after starting to rise from the low point on the Ferris wheel. The friendhas a constant speed of The speed of the object at that time is
�r�2�3.15�� � �8.032 � �11.47 � 9.8�3.15 � 2��2 � 8.03 m�sec.
�r�1�t�� � 15 m�sec.
t0 � 2
t � t0 � �11.47
2��4.9� � 1.17 sec.
t � 3.15
00
20
30
t0 � 0
55.arctan�11.47�8�03� � 0.96�8.032 � 11.472 � 14 m�sec.r�2�t0� � �8.03i � 11.47j.
t � 20
��t � t�10
��t10
� 2�
t
r1�0� � j,t � 0,16j.
306 Chapter 11 Vector-Valued Functions
−1
−2
z
y
x
2
2
2
1
−1
−1
−2
−2
10.
B��
4� � k
N��
4� � ��22
i ��22
j
At t ��
4, T��
4� � ��22
i ��22
j
B � T � N � k
N � �cos ti � sin tj
T� � �cos ti � sin tj
T � �sin ti � cos tj
r��t� � �sin ti � cos tj, �r��t�� � 1
r�t� � cos ti � sin tj � k, t ��
4
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C H A P T E R 1 1Vector-Valued Functions
Section 11.1 Vector-Valued Functions . . . . . . . . . . . . . . . . . . . . . . 39
Section 11.2 Differentiation and Integration of Vector-Valued Functions . . . . 44
Section 11.3 Velocity and Acceleration . . . . . . . . . . . . . . . . . . . . . 48
Section 11.4 Tangent Vectors and Normal Vectors . . . . . . . . . . . . . . . 54
Section 11.5 Arc Length and Curvature . . . . . . . . . . . . . . . . . . . . .60
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 68
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
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39
C H A P T E R 1 1Vector-Valued Functions
Section 11.1 Vector-Valued FunctionsSolutions to Odd-Numbered Exercises
1.
Component functions:
Domain: ���, 0� � �0, ��
h�t� � �1t
g�t� � �4t
f �t� � 5t
r�t� � 5t i � 4t j �1t
k 3.
Component functions:
Domain: �0, ��h�t� � �t
g�t� � �et
f �t� � ln t
r�t� � ln t i � et j � tk
5.
Domain: �0, ��
r�t� � F�t� � G�t� � �cos t i � sin t j � �tk� � �cos t i � sin t j� � 2 cos t i � �tk
7.
Domain: ���, ��
r�t� � F�t� � G�t� � � isin t
0
jcos tsin t
k0
cos t � � cos2 t i � sin t cos t j � sin2 tk
9.
(a)
(b)
(c)
(d)
� �2� t �12 �� t�2�i � �� t�j
� �2 � 2�t �12 ��t�2�i � �1 � �t�j � 2i � j
r�2 � �t� � r�2� �12 �2 � �t�2i � �2 � �t � 1�j � �2i � j�
r�s � 1� �12 �s � 1�2i � �s � 1 � 1�j �
12 �s � 1�2i � sj
r�0� � j
r�1� �12 i
r�t� �12 t 2i � �t � 1�j
11.
(a)
(b) is not defined. does not exist.
(c)
(d)
� ln�1 � � t�i � � 11 � �t
� 1�j � �3�t�k
r�1 � � t� � r�1� � ln�1 � �t�i �1
1 � �tj � 3�1 � �t�k � �0i � j � 3k�
r�t � 4� � ln�t � 4�i �1
t � 4j � 3�t � 4�k
��ln��3�r��3�
r�2� � ln 2i �12
j � 6k
r�t� � ln t i �1t
j � 3tk
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13.
r�t� � ��sin 3t�2 � �cos 3t�2 � t 2 � �1 � t 2
r�t� � sin 3ti � cos 3t j � tk 15.
a scalar.
The dot product is a scalar-valued function.
� 5t 3 � t 2, � 3t 3 � t 2 � 2t 3 � 4t 3
r�t� � u�t� � �3t � 1��t 2� � �14 t 3���8� � 4�t 3�
17.
Thus, Matches (b)z � x2.
z � t2y � 2t,x � t,
�2 ≤ t ≤ 2r�t� � t i � 2tj � t 2k, 19.
Thus, Matches (d)y � x2.
z � e0.75ty � t2,x � t,
�2 ≤ t ≤ 2r�t� � t i � t 2j � e0.75t k,
21. (a) View from the negative x-axis:
(c) View from the z-axis: �0, 0, 20�
��20, 0, 0� (b) View from above the first octant:
(d) View from the positive x-axis: �20, 0, 0�
�10, 20, 10�
23.
x64
2
−2
−4
y
y �x3
� 1
y � t � 1
x � 3t 25.
x3 4 5
32
45
−1−4
−3−2
y
21−2−5 −3
67
y � x23
x � t3, y � t2 27.
Ellipse
2 3−2−3
1
2
x
y
x2 �y2
9� 1
x � cos , y � 3 sin
29.
Hyperbola
x1296−6
−6
−3
−9
−12
−9−12
12
9
6
3
y
x2
9�
y2
4� 1
x � 3 sec , y � 2 tan 31.
Line passing through the points:
x
y
(0, 6, 5)
(1, 2, 3)(2, 2, 1)−
43 5 6
4
3
5
1
3
z
�1, 2, 3��0, 6, 5�,
z � 2t � 3
y � 4t � 2
x � �t � 1 33.
Circular helix
x
y3
−33
7
z
z � t
x2
4�
y2
4� 1
z � ty � 2 sin t,x � 2 cos t,
35.
z � e�t
x
y3
−3
3
6
zx2 � y2 � 4
z � e�ty � 2 cos t,x � 2 sin t, 37. ,
z �23 x 3y � x2,
x y5
2
6
4
2
−2
−4
−6
2, 4,
− −2, 4,
(
(
)
)
16
16
3
3
zz �23 t3y � t2x � t,
t 0 1 2
x 0 1 2
y 4 1 0 1 4
z 0 163
23�
23�
163
�1�2
�1�2
40 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
39.
Parabola
yx
23
−2−3
−1−2
−3
−2
−3
−4
−5
1
z
r�t� � �12
t2i � tj ��32
t2k 41.
Helix
y
x
2
3
4
−1
2 1−2
2
z
r�t� � sin t i � ��32
cos t �12
t�j � �12
cos t ��32 �k
43.
yx
π2
π
2
−2−2
2
z (a)
The helix is translated 2 units back on the x-axis.
yx
π2
π
1
−3
12
−2−2
z (b)
The height of the helixincreases at a faster rate.
yx
π8
π4
2
−2−2
2
z
(d)
The axis of the helix isthe x-axis.
y
xπ2
π
−2
2
2
−2
z (e)
The radius of the helix is increased from 2 to 6.
z
yx
π
6
−6
6
−6
(c)
The orientation of the helix is reversed.
yx
π2
π
2
−2−2
2
z
51.
Let
r�t� � 4 sec t i � 2 tan tj
x � 4 sec t, y � 2 tan t.
x2
16�
y2
4� 1
53. The parametric equations for the line are
One possible answer is
r�t� � �2 � 2t�i � �3 � 5t�j � 8tk.
z � 8t.y � 3 � 5t,x � 2 � 2t,
xy
4 5 6 7 8
12
34
4321
5678
z
(0, 8, 8)
(2, 3, 0)
55.
(Other answers possible)
r3�t� � �6 � t�j, 0 ≤ t ≤ 6 �r3�0� � 6j, r3�6� � 0�
r2�t� � �4 � 4t�i � 6tj, 0 ≤ t ≤ 1 �r2�0� � 4i, r2�1� � 6j�
r1�t� � t i, 0 ≤ t ≤ 4 �r1�0� � 0, r1�4� � 4i�
45.
Let then
r�t� � ti � �4 � t�j
y � 4 � t.x � t,
y � 4 � x 47.
Let then
r�t� � ti � �t � 2�2 j
y � �t � 2�2.x � t,
y � �x � 2�2
49.
Let then
r�t� � 5 cos ti � 5 sin t j
y � 5 sin t.x � 5 cos t,
x2 � y2 � 25
Section 11.1 Vector-Valued Functions 41
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57.
(Other answers possible)
r3�t� � �4 � t�j, 0 ≤ t ≤ 4
r2�t� � �2 � t�i, 0 ≤ t ≤ 2
r1�t� � ti � t2j, 0 ≤ t ≤ 2 �y � x2� 59.
Let then and Therefore,
x
y
z
5
1 2 3−3
32
2, 2, 4− − 2, 2, 4) )( (
r�t� � ti � tj � 2t2k
z � 2t2.y � �t,x � t,
z � x2 � y2 � 2t2.y � �x � �tx � t,
x � y � 0z � x2 � y2,
61.
r�t� � 2 sin t i � 2 cos tj � 4 sin2 tk
z � x2 � 4 sin2 t
y � 2 cos tx � 2 sin t,
x
y
z
3
−3
3
4
z � x2x2 � y2 � 4,
t 0
x 0 1 2 0
y 2
z 0 1 2 4 2 0
�2��20�2�3
�2�2
3
4
2
4
6
63.
Let then and .
and
r�t� � �1 � sin t� i � �2 cos tj � �1 � sin t�k
r�t� � �1 � sin t� i � �2 cos tj � �1 � sin t�k
z � 1 � sin t
y � ±�2 cos tx � 1 � sin t,
y � ±�2 cos ty2 � 2 cos2 t,
�1 � sin t�2 � y2 � �1 � sin t�2 � 2 � 2 sin2 t � y2 � 4
x2 � y2 � z2 � 4z � 2 � x � 1 � sin tx � 1 � sin t,
x
y
z
3
−3
−3
3
3
x � z � 2x2 � y2 � z2 � 4,
t 0
x 0 1 2
y 0
z 2 1 012
32
0±�62
±�2±�62
32
12
2
6�
6�
2
65.
Subtracting, we have or
Therefore, in the first octant, if we let then
r�t� � ti � tj � �4 � t2k
z � �4 � t2.y � t,x � t,x � t,
y � ±x.x2 � y2 � 0
x
y
z
4(2, 2, 0)
(0, 0, 2)
23
4
3
y2 � z2 � 4x2 � z2 � 4,
42 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
67.
x y
4
4
8
12
16
812
167 6 5
z
y2 � z2 � �2t cos t�2 � �2t sin t�2 � 4t2 � 4x2 69.
since
(L’Hôpital’s Rule)limt→2
t2 � 4t 2 � 2t
� limt→2
2t
2t � 2� 2.
limt→2
�t i �t2 � 4t2 � 2t
j �1t
k� � 2i � 2j �12
k
71.
since
(L’Hôpital’s Rule)limt→0
1 � cos t
t� lim
t→0 sin t
1� 0.
limt→0
�t2 i � 3t j �1 � cos t
tk� � 0 73.
does not exist since does not exist.limt→0
1t
limt→0
�1t
i � cos t j � sin tk�
75.
Continuous on �0, �����, 0�,
r�t� � t i �1t
j 77.
Continuous on ��1, 1
r�t� � t i � arcsin t j � �t � 1�k
79.
Discontinuous at
Continuous on ��
2� n,
2� n�
t �
2� n
r�t� � �e�t, t2, tan t� 81. See the definition on page 786.
83.
(a)
(b)
(c) s�t� � r�t� � 5j � t2i � �t � 2�j � tk
s�t� � r�t� � 2i � �t2 � 2�i � �t � 3�j � tk
s�t� � r�t� � 2k � t2i � �t � 3�j � �t � 3�k
r�t� � t2i � �t � 3�j � tk
85. Let and Then:
� limt→c
r�t� � limt→c
u�t�
� �limt→c x1�t�i � lim
t→c y1�t�j � lim
t→c z1�t�k � �limt→c
x2�t�i � limt→c
y2�t�j � limt→c
z2�t�k � �limt→c
x1�t� limt→c y2�t� � lim
t→c x2�t� limt→c
y1�t� k
� �limt→c y1�t� limt→c
z2�t� � limt→c
y2�t� limt→c z1�t� i � �limt→c
x1�t� limt→c z2�t� � lim
t→c x2�t� limt→c
z1�t� j
limt→c
�r�t� � u�t� � limt→c
��y1�t�z2�t� � y2�t�z1�t� i � �x1�t�z2�t� � x2�t�z1�t� j � �x1�t�y2�t� � x2�t�y1�t� k�
u�t� � x2�t�i � y2�t�j � z2�t�k.r�t� � x1�t� � y1�t�j � z1�t�k
87. Let Since r is continuous atthen
are defined at c.
Therefore, is continuous at c.r
limt→c
r � ��x�c��2 � �y�c��2 � �z�c��2 � r�c�
r � ��x�t��2 � �y�t��2 � �z�t��2
r�c� � x�c�i � y�c�j � z�c�k ⇒ x�c�, y�c�, z�c�
limt→c
r�t� � r�c�.t � c,r�t� � x�t�i � y�t�j � z�t�k. 89. True
Section 11.1 Vector-Valued Functions 43
http://librosysolucionarios.net
Section 11.2 Differentiation and Integration of Vector-Valued Functions
1.
is tangent to the curve.r��t0�
r��2� � 4i � j
r��t� � 2t i � j
r�2� � 4i � 2j
x � y2
y�t� � tx�t� � t2,
864
4
2
2
−2
−4
r ′
xr
(4, 2)
yt0 � 2r�t� � t2i � t j, 3.
is tangent to the curve.r��t0�
r���
2� � �i
r��t� � �sin t i � cos t j
r��
2� � j
x2 � y2 � 1
y�t� � sin tx�t� � cos t,
x
r
1
(0, 1)
y
r ′
t0 ��
2r�t� � cos ti � sin t j,
5.
(a)
(b)
r�12� � r�1
4� �14
i �3
16j
r�12� �
12
i �14
j
r�14� �
14
i �1
16j
x
rr
41
41
21
21
816
616
216
416
168
166
r
r
164
216
y
r�t� � t i � t 2j
(c)
This vector approximates r��14�.
r�1�2� � r�1�4��1�2� � �1�4� �
�1�4�i � �3�16�j1�4
� i �34
j
r��14� � i �
12
j
r��t� � i � 2tj
7.
x y
)( 3π2
21
2
−2
2π
πr
r ′
0, −2,
z
r��3�
2 � � 2i � k
r�3�
2 � � �2j �3�
2k
r��t� � �2 sin ti � 2 cos tj � k
z � tx2 � y2 � 4,
t0 �3�
2r�t� � 2 cos ti � 2 sin t j � tk,
9.
r��t� � 6i � 14tj � 3t2k
r�t� � 6ti � 7t2j � t3k 11.
r��t� � �3a cos2 t sin t i � 3a sin2 t cos tj
r�t� � a cos3 ti � a sin3 tj � k
13.
r��t� � �e�t i
r�t� � e�t i � 4 j 15.
r��t� � �sin t � t cos t, cos t � t sin t, 1�
r�t� � �t sin t, t cos t, t�
17.
(a)
r��t� � 6t i � j
r��t� � 3t2i � tj
r�t� � t3i �1
2t2j
(b) r��t� � r��t� � 3t2�6t� � t � 18t3 � t
44 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
19.
(a)
(b)
� 0
r��t� � r��t� � ��4 sin t���4 cos t� � 4 cos t��4 sin t�
r��t� � �4 cos t i � 4 sin tj
r��t� � �4 sin ti � 4 cos tj
r�t� � 4 cos ti � 4 sin tj 21.
(a)
(b) r��t� � r��t� � t�1� � 1�0� �12
t2�t� � t �t3
2
r��t� � i � tk
r��t� � ti � j �1
2t2k
r�t� �1
2t2i � tj �
1
6t3k
23.
(a)
(b) r��t� � r��t� � �t cos t��cos t � t sin t� � �t sin t��sin t � t cos t� � t
r��t� � �cos t � t sin t, sin t � t cos t, 0�
� �t cos t, t sin t, 1�
r��t� � ��sin t � sin t � t cos t, cos t � cos t � t sin t, 1�
r�t� � �cos t � t sin t, sin t � t cos t, t�
25.
r���1�4�
r���1�4� �1
2� 4 � 4��2� 2i � 2� 2j � 4k�
r���14� ���
2� 2
2 �2
� �2� 2
2 �2
� �2�2 � �4 � 4
r���14� � �
2� 2
2i �
2� 2
2j � 2k
r��t� � �� 2 cos��t�i � � 2 sin��t�j � 2k
r���1�4�
r���14� �1
4� 2 � 1�2� i � 2�j � k�
r��14� ��2�
2 �2
� �2�
2 �2
� ��12�
2
��2 �14
�4� 2 � 1
2
r���14� �
2�
2i �
2�
2j �
12
k
r��t� � �� sin��t�i � � cos��t�j � 2tk
x y
r′′
r′
r′′
r′
z
t0 � �14
r�t� � cos��t�i � sin��t�j � t 2k,
27.
Smooth on �0, ���, 0�,
r��0� � 0
r��t� � 2t i � 3t 2j
r�t� � t 2i � t 3j 29.
Smooth on n any integer.�n�
2,
�n � 1��2 �,
r��n�
2 � � 0
r���� � �6 cos2 � sin � i � 9 sin2 � cos � j
r��� � 2 cos3 �i � 3 sin3 � j
31.
for any value of
Smooth on ��, �
�r���� � 0
r���� � �1 � 2 cos ��i � �1 � 2 sin ��j
r��� � �� � 2 sin ��i � �1 � 2 cos ��j 33.
r is smooth for all ��, 0�, � �0, �t � 0:
r��t� � i �1t2 j � 2tk � 0
r�t� � �t � 1�i �1t
j � t2k
Section 11.2 Differentiation and Integration of Vector-Valued Functions 45
http://librosysolucionarios.net
35.
r is smooth for all
Smooth on intervals of form ���
2� n�,
�
2� n��
t ��
2� n� �
2n � 12
�.
r��t� � i � 3j � sec2 tk � 0
r�t� � t i � 3t j � tan tk
37.
(a)
(c)
(e)
Dt�r�t� u�t�� � 8t3i � �12t 2 � 4t 3�j � �3t2 � 24t�k
r�t� u�t� � 2t 4i � �t 4 � 4t 3�j � �t3 � 12t2�k
Dt�r�t� � u�t�� � 8t � 9t2 � 5t4
r�t� � u�t� � 4t2 � 3t3 � t5
r��t� � i � 3j � 2tk
u�t� � 4ti � t2j � t3kr�t� � ti � 3tj � t2k,
(b)
(d)
(f)
Dt�r�t�� �10 � 2t 2
10 � t 2
r�t� � 10t 2 � t 4 � t10 � t 2
Dt�3r�t� � u�t�� � �i � �9 � 2t�j � �6t � 3t 2�k
3r�t� � u�t� � �t i � �9t � t 2�j � �3t2 � t3�k
r��t� � 2k
39.
maximum at and
minimum at and
for t n � 0, 1, 2, 3, . . .� n �
2,� �
�
2�1.571�
t � 5.498�7�
4 �.t � 2.356�3�
4 �� � 1.287
t � 0.785��
4�.t � 3.927�5�
4 �� � 1.855
� � arccos �7 sin t cos t
�9 sin2 t � 16 cos2 t��9 cos2 t � 16 sin2 t��
cos � �r�t� � r��t�
r�t� r��t� ��7 sin t cos t
9 sin2 t � 16 cos2 t9 cos2 t � 16 sin2 t
r�t� � r��t� � 9 sin t cos t � 16 cos t sin t � �7 sin t cos t
r��t� � 3 cos ti � 4 sin tj
−1 70
π r�t� � 3 sin t i � 4 cos tj
41.
� lim�t→0
3i � �2t � �t�j � 3i � 2tj
� lim�t→0
�3�t�i � �2t��t� � ��t�2�j
�t
� lim�t→0
�3�t � �t� � 2�i � �1 � �t � �t�2� j � �3t � 2�i � �1 � t2�j
�t
r��t� � lim�t→0
r�t � �t� � r�t�
�t
43. ��2t i � j � k� dt � t2i � tj � tk � C 45. ��1t
i � j � t3�2k� dt � ln ti � tj �25
t5�2k � C
47. ���2t � 1�i � 4t3j � 3tk� dt � �t2 � t�i � t 4j � 2t3�2k � C
49. � sec2 t i �1
1 � t2 j� dt � tan t i � arctan t j � C
46 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
51. �1
0�8t i � t j � k� dt � 4t2i�
1
0� t2
2j�
1
0� tk�
1
0� 4i �
12
j � k
53. ���2
0��a cos t�i � �a sin t�j � k� dt � a sin t i�
��2
0� a cos t j�
��2
0� tk�
��2
0� ai � aj �
�
2k
55.
r�t� � 2e2ti � 3�et � 1�j
r�0� � 2i � 3j � C � 2i ⇒ C � �3j
r�t� � ��4e2t i � 3etj� dt � 2e2t i � 3et j � C 57.
r�t� � 6003 t i � �600t � 16t 2�j
r�0� � C � 0
� 6003 ti � �600t � 16t2�j � C
r�t� � ��6003 i � �600 � 32t�j� dt
r��t� � 6003 i � �600 � 32t�j
r��0� � C1 � 6003 i � 600j
r��t� � ��32j dt � �32t j � C1
59.
r�t� � �1 �12
e�t2�i � �e�t � 2�j � �t � 1�k � �2 � e�t2
2 �i � �e�t � 2�j � �t � 1�k
r�0� � �12
i � j � C �12
i � j � k ⇒ C � i � 2j � k
r�t� � ��te�t2i � e�tj � k� dt � �12
e�t2i � e�tj � tk � C
61. See “Definition of the Derivative of a Vector-ValuedFunction” and Figure 11.8 on page 794.
63. At the graph of is increasing in the x, y, and zdirections simultaneously.
u�t�t � t0,
65. Let Then and
� c�x��t�i � y��t�j � z��t�k� � cr��t�.
Dt�cr�t�� � cx��t�i � cy��t�j � cz��t�k
cr�t� � cx�t�i � cy�t�j � cz�t�kr�t� � x�t�i � y�t�j � z�t�k.
67. Let then
� f �t�r��t� � f��t�r�t�
� f �t��x��t�i � y��t�j � z��t�k� � f��t��x�t�i � y�t�j � z�t�k�
Dt� f �t�r�t�� � � f �t�x��t� � f��t�x�t��i � � f �t�y��t� � f��t�y�t�� j � � f �t�z��t� � f��t�z�t��k
f �t�r�t� � f �t�x�t�i � f �t�y�t�j � f �t�z�t�k.r�t� � x�t�i � y�t�j � z�t�k,
69. Let Then and
(Chain Rule)
� f��t�r�� f �t��. � f��t��x�� f �t��i � y�� f �t��j � z�� f �t��k�
Dt�r� f �t��� � x�� f �t�� f��t�i � y�� f �t�� f��t�j � z�� f �t�� f��t�k
r� f �t�� � x� f �t��i � y� f �t��j � z� f �t��kr�t� � x�t�i � y�t�j � z�t�k.
Section 11.2 Differentiation and Integration of Vector-Valued Functions 47
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Section 11.3 Velocity and Acceleration
71. Let and Then:
� r��t� � �u�t� � v�t�� � r�t� � �u��t� � v�t�� � r�t� � �u�t� � v��t��
�x1�t��y2�t�z3��t� � y3��t�z2�t�� � y1�t���x2�t�z3��t� � z2�t�x3��t�� � z1�t��x2�t�y3��t� � y2�t�x3��t���
�x1�t��y2��t�z3�t� � y3�t�z2��t�� � y1�t���x2��t�z3�t� � z2��t�x3�t�� � z1�t��x2��t�y3�t� � y2��t�x3�t��� �
� �x1��t��y2�t�z3�t� � y3�t�z2�t�� � y1��t���x2�t�z3�t� � z2�t�x3�t�� � z1��t��x2�t�y3�t� � y2�t�x3�t��� �
z1�t�y2��t�x3�t� � z1��t�y2�t�x3�t� z1��t�x2�t�y3�t� � z1�t�y2�t�x3��t� �
y1�t�z2�t�x3��t� � y1�t�z2��t�x3�t� � y1��t�z2�t�x3�t� � z1�t�x2�t�y3��t� � z1�t�x2��t�y3�t� �
x1�t�y3��t�z2�t� � x1��t�y3�t�z2�t� � y1�t�x2�t�z3��t� � y1�t�x2��t�z3�t� � y1��t�x2�t�z3�t� �
Dt�r�t� � �u�t� � v�t��� � x1�t�y2�t�z3��t� � x1�t�y2��t�z3�t� � x1��t�y2�t�z3�t� � x1�t�y3�t�z2��t� �
r�t� � �u�t� � v�t�� � x1�t��y2�t�z3�t� � z2�t�y3�t�� � y1�t��x2�t�z3�t� � z2�t�x3�t�� � z1�t��x2�t�y3�t� � y2�t�x3�t��
v�t� � x3�t�i � y3�t�j � z3�t�k.u�t� � x2�t�i � y2�t�j � z2�t�k,r�t� � x1�t�i � y1�t�j � z1�t�k,
73. False. Let
�r��t�� � 1
r��t� � �sin t i � cos tj
ddt
��r�t��� � 0
�r�t�� � 2
r�t� � cos t i � sin tj � k.
1.
At
v�1� � 3i � j, a�1� � 0
�3, 0�, t � 1.
y �x3
� 1y � t � 1,x � 3t,
a�t� � r��t� � 0
v�t� � r��t� � 3i � j
64
2
−2
−4
v
x(3, 0)
yr�t� � 3t i � �t � 1�j 3.
At
a�2� � 2i
v�2� � 4i � j
t � 2.�4, 2�,
x � y2y � t,x � t2,
a�t� � r��t� � 2i
v�t� � r��t� � 2t i � j
864
4
2
2
−4
−2
v
x
a(4, 2)
yr�t� � t2 i � t j
5.
At
( 2, 2)
3
3
−3
−3x
v
a
y
a
4� � �2 i � 2j
v
4� � �2 i � 2j
t �
4.�2, 2 �,
x2 � y2 � 4y � 2 sin t,x � 2 cos t,
a�t� � r��t� � �2 cos ti � 2 sin tj
v�t� � r��t� � �2 sin t i � 2 cos tj
r�t� � 2 cos t i � 2 sin t j 7.
(cycloid)
At
v
x2
a
4
2
π
π
π
( , 2)
y
a�� � �0, �1 � �j
v�� � �2, 0 � 2i
t � .�, 2�,
y � 1 � cos tx � t � sin t,
a�t� � r��t� � �sin t, cos t
v�t� � r��t� � �1 � cos t, sin t
r�t� � �t � sin t, 1 � cos t
48 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
9.
a�t� � 0
s�t� � �v�t�� � 1 � 4 � 9 � 14
v�t� � i � 2j � 3k
r�t� � t i � �2t � 5�j � 3tk 11.
a�t� � 2j � k
s�t� � 1 � 4t2 � t2 � 1 � 5t2
v�t� � i � 2tj � tk
r�t� � t i � t2j �t2
2k
13.
a�t� � �9
�9 � t2�3�2 k
s�t� �1 � 1 �t2
9 � t2 �18 � t2
9 � t2
v�t� � i � j �t
9 � t2k
r�t� � t i � tj � 9 � t2 k 15.
a�t� � �0, �3 cos t, �3 sin t � �3 cos tj � 3 sin tk
s�t� � 16 � 9 sin2 t � 9 cos2 t � 5
v�t� � �4, �3 sin t, 3 cos t � 4i � 3 sin tj � 3 cos tk
r�t� � �4t, 3 cos t, 3 sin t
17. (a)
z �14
�34
ty � �1 � 2t,x � 1 � t,
r��1� � �1, �2, 34�
r��t� � �1, �2t, 3t2
4 �
r�t� � �t, �t2, t3
4�, t0 � 1 (b)
� �1.100, �1.200, 0.325
r�1 � 0.1� � �1 � 0.1, �1 � 2�0.1�, 14
�34
�0.1��
19.
r�2� � 2�i � j � k� � 2i � 2j � 2k
r�0� � C � 0, r�t� �t2
2�i � j � k�,
r�t� � ��ti � tj � tk� dt �t 2
2�i � j � k� � C
v�0� � C � 0, v�t� � t i � tj � tk, v�t� � t�i � j � k�
v�t� � ��i � j � k� dt � t i � tj � tk � C
a�t� � i � j � k, v�0� � 0, r�0� � 0 21.
r�2� �173
j �23
k
r�t� � t3
6�
92
t �143 �j � t3
6�
12
t �13�k
r�1� �143
j �13
k � C � 0 ⇒ C � �143
j �13
k
� t3
6�
92
t�j � t3
6�
12
t�k � C
r�t� � ��t2
2�
92�j � t2
2�
12�k� dt
v�t� � t2
2�
92�j � t2
2�
12�k
v�1� �12
j �12
k � C � 5j ⇒ C �92
j �12
k
v�t� � ��tj � tk� dt �t2
2j �
t2
2k � C
a�t� � tj � tk, v�1� � 5j, r�1� � 0
23. The velocity of an object involves both magnitude and direction of motion,whereas speed involves only magnitude.
25.
� 443 t i � �10 � 44t � 16t2�j
00
300
50r�t� � �88 cos 30�ti � �10 � �88 sin 30�t � 16t2� j
Section 11.3 Velocity and Acceleration 49
http://librosysolucionarios.net
27.
when
The maximum height is reached when the derivative of the vertical component is zero.
Maximum height: y534 � � 3 � 40353
4 � � 16534 �
2
� 78 feet
t �403
32�
534
y��t� � 403 � 32t � 0
y�t� � 3 �tv0
2� 16t2 � 3 �
406
2t � 16t 2 � 3 � 403t � 16t2
v0 � 406 � 97.98 ft�secv0 � 9600 � 406,v02 � 300�32�,
300 �3002�32�
v02 � 0
v0
23002
v0� � 163002
v0�
2
� 0,t �3002
v0,
3 �v0
2t � 16t2 � 3.
v0
2t � 300
r�t� � �v0 cos ��ti � �h � �v0 sin ��t �12
gt2�j �v0
2t i � 3 �
v0
2t � 16t2�j
29. or
y �x
v0 cos ��v0 sin �� � 16 x2
v02 cos2 �� � h � �tan ��x � 16
v02 sec2 ��x2 � h
y�t� � t�v0 sin �� � 16t2 � h
t �x
v0 cos �x�t� � t�v0 cos ��
31. or
(a)
(c) and
feet.y�45.8375� � 14.4
y� � �0.008x � 0.3667 � 0 ⇒ x � 45.8375
y � �0.004x2 � 0.3667x � 6
r�t� � ti � ��0.004t2 � 0.3667t � 6�j,
(b)
00
120
18
(d) From Exercise 29,
⇒ v0 � 67.4 ft�sec.
16 sec2 �v0
2 � 0.004 ⇒ v02 �
16 sec2 �0.004
�4000cos2 �
tan � � 0.3667 ⇒ � � 20.14
33.
(a)
(b) Graphing these curves together with shows that
—CONTINUED—
�0 � 20.y � 10
50000
100
r�t� � 4403
cos �0�t i � �3 � 4403
sin �0�t � 16t2�j
100 mph � 100 miles
hr �5280 feetmile���3600 sec�hour� �
4403
ft�sec
50 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
33. —CONTINUED—
(c) We want
and
From the minimum angle occurs when Substituting this for t in yields:
� � tan�148,400 � 1,464,332,80028,800 � � 19.38
tan � �48,400 ± 48,4002 � 4�14,400��15,247�
2�14,400�
14,400 tan2 � � 48,400 tan � � 15,247 � 0
14,400
121�1 � tan2 �� � 400 tan � � 7 � 0
400 tan � �14,400
121 sec2 � � 7
3 � 4403
sin �� 3011 cos �� � 16 30
11 cos ��2
� 10
y�t�t � 30��11 cos ��.x�t�,
y�t� � 3 � 4403
sin ��t � 16t2 ≥ 10.x�t� � 4403
cos ��t ≥ 400
35.
(a) We want to find the minimum initial speed v as a function of the angle Since the bale must be thrown to the position , we have
from the first equation. Substituting into the second equation and solving for v, we obtain:
We minimize
Substituting into the equation for v, feet per second.
(b) If
From part (a), v2 �512
2�2�2��2�2� � �2�2�2�
512
1�2� 1024 ⇒ v � 32 ft�sec.
8 � �v sin ��t � 16t2 � v22
t � 16t2
16 � �v cos ��t � v22
t
� � 45,
v � 28.78
� � 1.01722 � 58.28
tan�2�� � �2
f���� � 0 ⇒ 2 cos�2�� � sin�2�� � 0
f���� � �512 2 cos2 � � 2 sin2 � � 2 sin � cos �
�2 sin � cos � � cos2 ��2
f ��� �512
2 sin � cos � � cos2 �.
v2 �512
2 sin � cos � � cos2 �
�2 sin � cos � � cos2 �
512 1v2 � 2
sin �cos �
� 1�cos2 �512
512 1
v2 cos2 �� 2
sin �cos �
� 1
1 � 2 sin �cos �
� 512 1v2 cos2 ��
8 � �v sin �� 16v cos �� � 16 16
v cos ��2
t � 16��v cos ��
8 � �v sin ��t � 16t2.
16 � �v cos ��t
�16, 8��.
r�t� � �v cos ��t i � ��v sin ��t � 16t2�j
Section 11.3 Velocity and Acceleration 51
http://librosysolucionarios.net
37.
when and
The range is
.
Hence,
⇒ sin 2� �1
15 ⇒ � � 1.91.x �
12002
32 sin�2�� � 3000
x � �v0 cos ��t � �v0 cos ��v0 sin �16
�v0
2
32 sin 2�
t �v0 sin �
16.t � 0�v0 sin ��t � 16t2 � 0
r�t� � �v0 cos ��t i � ��v0 sin ��t � 16t2�j
39. (a)
Maximum height: 2.052 feet
Range: 46.557 feet
(c)
Maximum height: 34.031 feet
Range: 136.125 feet
(e)
Maximum height: 51.074 feet
Range: 117.888 feet
00
140
60
r�t� � �33t�i � �57.16t � 16t2�j
r�t� � �66 cos 60�ti � �0 � �66 sin 60�t � 16t2�j
v0 � 66 ft�sec� � 60,
00
200
40
r�t� � �46.67t�i � �46.67t � 16t2�j
r�t� � �66 cos 45�ti � �0 � �66 sin 45�t � 16t2�j
v0 � 66 ft�sec� � 45,
00
50
5
r�t� � �65t�i � �11.46t � 16t2�j
r�t� � �66 cos 10�ti � �0 � �66 sin 10�t � 16t2�j
v0 � 66 ft�sec� � 10, (b)
Maximum height: 10.043 feet
Range: 227.828 feet
(d)
Maximum height: 166.531 feet
Range: 666.125 feet
(f )
Maximum height: 249.797 feet
Range: 576.881 feet
00
600
300
r�t� � �73t�i � �126.44t � 16t2�j
r�t� � �146 cos 60�ti � �0 � �146 sin 60�t � 16t2�j
v0 � 146 ft�sec� � 60,
00
800
200
r�t� � �103.24t�i � �103.24t � 16t2�j
r�t� � �146 cos 45�ti � �0 � �146 sin 45�t � 16t2�j
v0 � 146 ft�sec� � 45,
00
300
15
r�t� � �143.78t�i � �25.35t � 16t2�j
r�t� � �146 cos 10�ti � �0 � �146 sin 10�t � 16t2�j
v0 � 146 ft�sec� � 10,
52 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
41.
The projectile hits the ground when
The range is therefore meters.
The maximum height occurs when
The maximum height is
meters.� 129.1y � 1.5 � �100 sin 30��5.102� � 4.9�5.102�2
100 sin 30 � 9.8t ⇒ t � 5.102 sec
dy�dt � 0.
�100 cos 30��10.234� � 886.3
�4.9t2 � 100�12�t � 1.5 � 0 ⇒ t � 10.234 seconds.
� �100 cos 30�ti � �1.5 � �100 sin 30�t � 4.9t2�j
r�t� � �v0 cos ��t i � �h � �v0 sin ��t � 4.9t2�j
43.
(a) when �t � 0, 2, 4, . . . .�v�t�� � 0
�a�t�� � b�2
�v�t�� � 2 b�1 � cos��t�
a�t� � �b�2 sin �t�i � �b�2 cos �t�j � b�2�sin��t�i � cos��t�j�
v�t� � b�� � � cos �t�i � b� sin �t j � b��1 � cos �t�i � b� sin �tj
r�t� � b��t � sin�t� i � b�1 � cos �t�j
(b) is maximum whenthen �v�t�� � 2b�.
�t � , 3, . . . ,�v�t��
45.
Therefore, and are orthogonal.v�t�r�t�
r�t� � v�t� � �b2� sin��t� cos��t� � b2� sin��t� cos��t� � 0
v�t� � �b� sin��t�i � b� cos��t�j
47.
is a negative multiple of a unit vector from to and thus is directed toward the origin.a�t��cos �t, sin �t��0, 0�a�t�
a�t� � �b�2 cos��t�i � b�2 sin��t�j � �b�2�cos��t�i � sin��t�j� � ��2r�t�
49.
�v�t�� � b� � 810 ft�sec
� � 410 rad�sec
F � m��2b� �1
32�2�2� � 10
1 � m�32�
�a�t�� � �2b
51. To find the range, set then By the Quadratic Formula, (discount the negative value)
At this time,
�v0
2 cos �g sin � �sin2 � �
2ghv0
2 �.
x�t� � v0 cos �v0 sin � � v02 sin2 � � 2gh
g � �v0 cos �
g v0 sin � �v02sin2 � �
2ghv0
2 ��
t �v0 sin � � ��v0 sin ��2 � 4��1�2�g���h�
2��1�2�g� �v0 sin � � v0
2 sin2 � � 2ghg
.
0 � �12 g�t2 � �v0 sin ��t � h.y�t� � h � �v0 sin ��t �
12 gt2 � 0
Section 11.3 Velocity and Acceleration 53
http://librosysolucionarios.net
53. Position vector
Velocity vector
Acceleration vector
C is a constant.
Orthogonal
v�t� � a�t� � 0
2�x��t�x��t� � y��t�y��t� � z��t�z��t�� � 0
2x��t�x��t� � 2y��t�y��t� � 2z��t�z��t� � 0
ddt
�x��t�2 � y��t�2 � z��t�2� � 0
� C,
Speed � �v�t�� � ��x��t�2 � y��t�2 � z��t�2
a�t� � x��t�i � y��t�j � z��t�k
v�t� � x��t�i � y��t�j � z��t�k
r�t� � x�t�i � y�t�j � z�t�k
55.
(a)
(c)
−9 9
−6
6
a�t� � v��t� � �6 cos t i � 3 sin t j
� 3�3 sin2 t � 1
� 3�4 sin2 t � cos2 t
�v�t�� � �36 sin2 t � 9 cos2 t
v�t� � r��t� � �6 sin t i � 3 cos t j
r�t� � 6 cos t i � 3 sin tj
(b)
(d) The speed is increasing when the angle between v and a is in the interval
The speed is decreasing when the angle is in the interval
�
2, �.
�0, �
2�.
t 0
Speed 3 332�136
32�10
�2�
3�
2�
4
Section 11.4 Tangent Vectors and Normal Vectors
1.
T�1� �1�2
�i � j� ��2
2i �
�2
2j
T�t� �r��t�
�r��t���
2ti � 2j2�t2 � 1
�1
�t2 � 1�ti � j�
r��t� � 2ti � 2j, �r��t�� � �4t2 � 4 � 2�t2 � 1
r�t� � t2i � 2tj 3.
T�
4� � ��22
i ��22
j
T�t� �r��t�
�r��t��� �sin ti � cos tj
�r��t�� � �16 sin2 t � 16 cos2 t � 4
r��t� � �4 sin ti � 4 cos tj
r�t� � 4 cos ti � 4 sin tj
5.
When at .
Direction numbers:
Parametric equations: z � ty � 0,x � t,
c � 1b � 0,a � 1,
T�0� �r��0�
�r��0�� ��22
�i � k�
�0, 0, 0���t � 0r��0� � i � k,t � 0,
r��t� � i � 2tj � k
r�t� � t i � t2j � tk 7.
When at .
Direction numbers:
Parametric equations: z � ty � 2t,x � 2,
c � 1b � 2,a � 0,
T�0� �r��0�
�r��0�� ��55
�2j � k�
�2, 0, 0���t � 0r��0� � 2j � k,t � 0,
r��t� � �2 sin t i � 2 cos tj � k
r�t� � 2 cos t i � 2 sin tj � tk
54 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
9.
When
Direction numbers:
Parametric equations: z � 4y � �2t � �2,x � ��2t � �2,
c � 0b � �2,a � ��2,
T�
4� �r��� 4�
�r��� 4�� �12
���2, �2, 0�
�t ��
4 at ��2, �2, 4�.r��
4� � ���2, �2, 0�,t ��
4,
r��t� � ��2 sin t, 2 cos t, 0�
r�t� � �2 cos t, 2 sin t, 4�
11.
When
Direction numbers:
Parametric equations: z � 18t � 18y � 6t � 9,x � t � 3,
c � 18b � 6,a � 1,
T�3� �r��3�
�r��3�� �1
19�1, 6, 18�
�t � 3 at �3, 9, 18��.r��3� � �1, 6, 18�,t � 3,
r��t� � �1, 2t, 2t2�
x
y
33−3
6
6
9
9
12
12
15
15
18
18
zr�t� � �t, t2, 23
t3�
13.
Tangent line:
� �1.1, 0.1, 1.05�
r�t 0 � 0.1� � r�1.1� � 1.1i � 0.1j � 1.05k
z � 1 �12
ty � t,x � 1 � t,
T�1� �r��t�
�r��t�� �i � j � �1 2�k�1 � 1 � �1 4�
�23
i �23
j �13
k
r��t� � i �1tj �
1
2�tk � r��1� � i � j �
12
k
t0 � 1r�t� � ti � ln tj � �tk, 15.
Hence the curves intersect.
cos �r��4� � u��8�
�r��4�� �u��8�� �16.2916716.29513
⇒ � 1.2
u��8� � �14
, 2, 112�u��s� � �1
4, 2,
13
s�2 3�,
r��4� � �1, 8, 12�r��t� � �1, 2t,
12�,
u�8� � �2, 16, 2�
r�4� � �2, 16, 2�
17.
N�2� �T��2�
�T��2���
1�5
��2i � j� ��2�5
5i �
�5
5j
T��2� ��253 2 i �
153 2 j
T��t� ��t
�t2 � 1�3 2 i �1
�t2 � 1�3 2 j
T�t� �r��t�
�r��t���
i � tj�1 � t2
r��t� � i � tj
r�t� � ti �12
t2j, t � 2 19.
N3�
4 � ��22
i ��22
j
T��t� � �cos t i � sin t j, �T�t�� � 1
T�t� �r��t�
�r��t��� �sin ti � cos tj
r��t� � �6 sin ti � 6 cos tj
r�t� � 6 cos ti � 6 sin tj � k, t �3�
4
Section 11.4 Tangent Vectors and Normal Vectors 55
http://librosysolucionarios.net
23.
is undefined.
The path is a line and the speed is variable.
N�t� �T��t�
�T��t��
T��t� � O
T�t� �v�t�
�v�t�� �8ti8t
� i
a�t� � 8i
v�t� � 8t i
r�t� � 4t2 i
25.
aN � a � N � �2
aT � a � T � ��2
N�1� �1
�2�i � j� �
�22
�i � j�
�1
�t4 � 1�i � t2j�
N�t� �T��t�
�T��t�� �
2t�t4 � 1�3 2 i �
2t3
�t4 � 1�3 2 j
2t�t4 � 1�
T�1� �1
�2�i � j� �
�22
�i � j�
T�t� �v�t�
�v�t�� �t2
�t4 � 1i �1t2 j� �
1
�t4 � 1�t2i � j�
a�1� � 2ja�t� �2t3 j,
v�1� � i � j,v�t� � i �1t2 j,r�t� � t i �
1t
j, 27.
At
Motion along r is counterclockwise. Therefore,
aN � a � N � �2e� 2
aT � a � T � �2e� 2
N �1
�2��i � j� � �
�22
�i � j�.
T �v
�v��
1
�2��i � j� �
�22
��i � j�.t ��
2,
a�t� � et��2 sin t�i � et�2 cos t�j
v�t� � et�cos t � sin t�i � et�cos t � sin t�j
r�t� � �et cos t�i � �et sin t�j
29.
Motion along r is counterclockwise. Therefore
aN � a � N � �2��t0� � �3t0
aT � a � T � �2
N�t0� � ��sin �t0�i � �cos �t0�j.
T�t0� �v
�v�� �cos �t0�i � �sin �t0�j
a�t0� � �2��cos �t0 � �t0 sin �t0�i � ��t0 cos �t0 � sin �t0�j�
v�t0� � ��2t0 cos �t0�i � ��2t0 sin �t0�j
r�t0� � �cos �t0 � �t0 sin �t0�i � �sin �t0 � �t0 cos �t0�j
21.
is undefined.
The path is a line and the speed is constant.
N�t� �T��t�
�T��t��
T��t� � O
T�t� �v�t�
�v�t�� �4i4
� i
a�t� � O
v�t� � 4i
r�t� � 4t i
56 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
35.
N�2� ��1717
�i � 4j�
T�2� ��1717
�4i � j�
r�2� � 2i �12
j
N�t� �i � t2j�t4 � 1
N
x32
T
3
2
1
,,2
1
21
yT�t� �t2i � j�t4 � 1
r��t� � i �1t2 j
x � t, y �1t ⇒ xy � 1
r�t� � ti �1t
j, t0 � 2 37.
are not defined.aT, aN
N�t� �T�
�T� � is undefined.
T�t� �v
�v��
1
�14�i � 2j � 3k� �
�1414
�i � 2j � 3k�
a�t� � 0
v�t� � i � 2j � 3k
r�t� � ti � 2t j � 3tk
31.
aN � a � N � a�2
aT � a � T � 0
N�t� �T��t�
�T��t�� � �cos��t�i � sin��t�j
T�t� �v�t�
�v�t�� � �sin��t�i � cos��t�j
a�t� � �a�2 cos��t�i � a�2 sin��t�j
v�t� � �a� sin��t�i � a� cos��t�j
r�t� � a cos��t�i � a sin��t�j 33. Speed:
The speed is constant since aT � 0.
�v�t�� � a�
39.
aN � a � N ��30
6
aT � a � T �5�6
6
N�1� ��3030
��5i � 2j � k�
N�t� �T�
�T� ��
�5t i � 2j � k�1 � 5t2�3 2
�51 � 5t2
��5ti � 2j � k�5�1 � 5t2
T�1� ��66
�i � 2j � k�
T�t� �v
�v��
1
�1 � 5t2�i � 2tj � tk�
a�t� � 2j � k
v�1� � i � 2j � k
v�t� � i � 2tj � tk
r�t� � ti � t2j �t2
2k
x
y2π
4π
3
3
zTN
41.
aN � a � N � 3
aT � a � T � 0
N�
2� � �k
N�t� �T�
�T� �� �cos tj � sin tk
T�
2� �15
�4i � 3j�
T�t� �v
�v��
15
�4i � 3 sin tj � 3 cos tk�
a�
2� � �3k
a�t� � �3 cos t j � 3 sin tk
v�
2� � 4i � 3j
v�t� � 4i � 3 sin tj � 3 cos tk
r�t� � 4ti � 3 cos tj � 3 sin tk
Section 11.4 Tangent Vectors and Normal Vectors 57
http://librosysolucionarios.net
45. If then the motion is in a straight line.aN � 0,
43.
If then is the tangential component of accelerationand is the normal component of acceleration.aN
aTa�t� � aTT�t� � aNN�t�,
N�t� �T��t�
�T��t��
T�t� �r��t�
�r��t��
47.
The graph is a cycloid.
(a)
When
When
When
(b)
When the speed in increasing.
When the height is maximum.
When the speed is decreasing.aT � ��2�2
2 < 0 ⇒t �
32
:
aT � 0 ⇒t � 1:
aT ��2�2
2 > 0 ⇒t �
12
:
dsdt
�� 2 sin �t
�2�1 � cos �t�� aT
Speed: s � �v�t�� � ��2�1 � cos �t�
aN ��2� 2
2aT � �
�2�2
2,t �
32
:
aN � �2aT � 0,t � 1:
aN ��2� 2
2aT �
�2
�2�
�2�2
2,t �
12
:
aN � a � N �1
�2�1 � cos �t���2 sin2 �t � �2 cos �t��1 � cos �t�� �
�2�1 � cos �t��2�1 � cos �t�
��2�2�1 � cos �t�
2
aT � a � T �1
�2�1 � cos �t��� 2 sin �t�1 � cos �t� � �2 cos �t sin �t� �
� 2 sin �t
�2�1 � cos � t�
N�t� �T��t�
�T��t�� �1
�2�1 � cos �t��sin �t, �1 � cos �t�
T�t� �v�t�
�v�t�� �1
�2�1 � cos �t��1 � cos �t, sin �t�
a�t� � �� 2 sin �t, � 2 cos �t�
v�t� � �� � � cos �t, � sin �t�
x
t = 1t =
21
t = 23
y r�t� � �� t � sin � t, 1 � cos �t�
r�t� � ��t � sin �t, 1 � cos �t�
58 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
49.
B�
2� � T�
2� � N�
2� � � i
�4�17 170
j
0
�1
k
�17 170 � �
�1717
i �4�17
17k �
�1717
�i � 4k�
N�
2� � �j
T�
2� �2�17
17 �2i �12
k� ��1717
��4i � k�
r�
2� � 2j ��
4k
N�t� � �cos ti � sin tj
T�t� �2�17
17 �2 sin ti � 2 cos t j �12
k�
r��t� � �2 sin ti � 2 cos tj �12
k
1
1
2
−1
−2
2
3
−2
yx
N
B T
0, 2, π2( )
zr�t� � 2 cos ti � 2 sin tj �t2
k, t0 ��
2
51. From Theorem 11.3 we have:
(Motion is clockwise.)
Maximum height when (vertical component of velocity)
At maximum height, and aN � 32.aT � 0
v0 sin � 32t � 0;
aN � a � N �32v0 cos
�v02 cos2 � �v0 sin � 32t�2
aT � a � T ��32�v0 sin � 32t�
�v02 cos2 � �v0 sin � 32t�2
N�t� ��v0 sin � 32t�i � v0 cos j
�v02 cos2 � �v0 sin � 32t�2
T�t� ��v0 cos �i � �v0 sin � 32t�j�v0
2 cos2 � �v0 sin � 32t�2
a�t� � �32 j
v�t� � v0 cos i � �v0 sin � 32t�j
r�t� � �v0t cos �i � �h � v0t sin � 16t2�j
53.
(a)
(b) and
because the speed is constant.aT � 0
aN � 1000�2aT � 0
� ��100��2 � 16 � 4�625�2 � 1 � 314 mi hr
�r��t�� � ��100��2 sin2�10� t� � �100��2 cos2�10�t� � 16
r��t� � ��100� sin�10�t�, 100� cos�10�t�, 4�
0 ≤ t ≤ 120r�t� � �10 cos 10�t, 10 sin 10�t, 4 � 4t�,
55.
From Exercise 31, we know and
(a) Let Then
or the centripetal acceleration is increased by a factorof 4 when the velocity is doubled.
a � N � a�02 � a�2��2 � 4a�2
�0 � 2�.
a � N � a�2.a � T � 0
r�t� � �a cos �t�i � �a sin �t�j
(b) Let Then
or the centripetal acceleration is halved when theradius is halved.
a � N � a0�2 � a
2��2 � 12�a�2
a0 � a 2.
Section 11.4 Tangent Vectors and Normal Vectors 59
http://librosysolucionarios.net
61. Let be the unit tangent vector. Then
and is rotated counterclockwise through an angle of from T.
If then the curve bends to the left and M has the same direction as . Thus, M has the same direction as
which is toward the concave side of the curve.
x
M
T
φ
y
N �T�
�T��,
T�d��dt > 0,
��2M � �sin �i � cos �j � cos�� � ���2��i � sin�� � ���2�� j
T��t� �dTdt
�dTd�
d�
dt� ��sin �i � cos �j�d�
dt� M
d�
dt.
T�t� � cos �i � sin �j
If then the curve bends to the right and M has theopposite direction as Thus,
again points to the concave side of the curve.
x
MN
Tφ
y
N �T�
�T��
T�.d��dt < 0,
63. Using and we have:
Thus, aN ��v � a�
�v�.
� �v�aN
�v � a� � �v�aN�T � N�
� �v�aN�T � N�
� �v�aT�T � T� � �v�aN�T � N�
v � a � �v�T � �aTT � aNN�
�T � N� � 1,T � T � O,a � aTT � aNN,
Section 11.5 Arc Length and Curvature
1.
8 12
8
4
12
4x
(4, 12)
(0, 0)
y
� �10t4
0� 410
� 10�4
0 dt
s � �4
0 1 � 9 dt
dzdt
� 0dydt
� 3,dxdt
� 1,
r�t� � ti � 3tj 3.
xa
a
−a
−a
y
� 3a���2
02 sin 2t dt � ��3a cos 2t
��2
0� 6a
� 12a���2
0sin t cos t dt
s � 4���2
0 �3a cos2 t��sin t�2 � �3a sin2 t cos t�2 dt
dydt
� 3a sin2 t cos tdxdt
� �3a cos2 t sin t,
r�t� � a cos3 ti � a sin3 tj
59. v �9.56 � 104
4385� 4.67 mi�sec57. v �9.56 � 104
4100� 4.83 misec
60 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
5. (a)
(b)
Maximum height:
(c)
Range:
(d) s � �4.4614
0�502�2
� �502 � 32t�2dt � 362.9 feet
502�4.4614� � 315.5 feet
3 � 502t � 16t2 � 0 ⇒ t � 4.4614
3 � 502 25216 � � 16 252
16 �2
� 81.125 ft
502 � 32t � 0 ⇒ t �252
16
v�t� � 502i � �502 � 32t�j � 502ti � �3 � 502t � 16t2� j
� �100 cos 45�ti � �3 � �100 sin 45�t �12
�32�t2j
r�t� � �v0 cos ��ti � �h � �v0 sin ��t �1
2gt2j
7.
� 214 � �2
0
14 dt � �14 t2
0
s � �2
0
22 � ��3�2 � 12 dt
dzdt
� 1dydt
� �3,dxdt
� 2
x
y
2
4
2
−2
(0, 0, 0)
(4, 6, 2)−
zr�t� � 2ti � 3tj � tk
9.
xy
( , 0, 2 )a b2 b
b
( , 0, 0)a
ππ
π
z
� �2�
0 a2 � b2 dt � �a2 � b2 t
2�
0� 2�a2 � b2
s � �2�
0
a2 sin2 t � a2 cos2 t � b2 dt
dzdt
� bdydt
� a cos t,dxdt
� �a sin t,
r�t� � a cos ti � a sin tj � btk 11.
� �3
1
4t4 � t2 � 1t
dt � 8.37
� �3
14t4 � t2 � 1
t2 dt
s � �3
1�2t�2 � �1�2 � 1
t �2
dt
dzdt
�1t
dydt
� 1,dxdt
� 2t,
r�t� � t2i � tj � ln tk
13. ,
(a)
—CONTINUED—
distance � 22 � 42 � 82 � 84 � 221 � 9.165
r�2� � �2, 0, 8�r�0� � �0, 4, 0�,
0 ≤ t ≤ 2r�t� � ti � �4 � t2�j � t3k
Section 11.5 Arc Length and Curvature 61
http://librosysolucionarios.net
13. —CONTINUED—
(b)
(c) Increase the number of line segments.
(d) Using a graphing utility, you obtain 9.57057.
� 9.529 � 0.5728 � 1.2562 � 2.7300 � 4.9702
�0.5�2 � �1.75�2 � �4.625�2
distance � �0.5�2 � �.25�2 � �.125�2 � �.5�2 � �.75�2 � �.875�2 � �0.5�2 � �1.25�2 � �2.375�2 �
r�2� � �2, 0, 8�
r�1.5� � �1.5, 1.75, 3.375�
r�1� � �1, 3, 1�
r�0.5� � �0.5, 3.75, .125�
r�0� � �0, 4, 0�
15.
(a) (b)
(c) When
When
(d) �45
�15
� 1�r��s�� � �2
5 sin s
5��2
� 2
5 cos s
5��2
� 1
5�2
��0.433, 1.953, 1.789�
z �4
5� 1.789
y � 2 sin 4
5� 1.953
x � 2 cos 4
5� �0.433s � 4:
�1.081, 1.683, 1.000�
z � 1
y � 2 sin 1 � 1.683
x � 2 cos 1 � 1.081s � 5:
r�s� � 2 cos s
5�i � 2 sin s
5�j �s
5k � �t
0
5 du � �5ut
0� 5t
z �s
5y � 2 sin s
5�,x � 2 cos s
5�, � �t
0
��2 sin u�2 � �2 cos u�2 � �1�2 du
s
5� ts � �t
0
�x��u��2 � �y��u��2 � �z��u��2 du
r�t� � �2 cos t, 2 sin t, t�
17.
and
(The curve is a line.)T��s� � 0 ⇒ K � �T��s�� � 0
T�s� �r��s�
�r��s�� � r��s�
�r��s�� �12
�12
� 1r��s� �22
i �22
j
r�s� � 1 �22
s�i � 1 �22
s�j 19.
K � �T��s�� �25
T��s� � �25
cos s
5�i �25
sin s
5�j
T�s� � r��s� � �2
5 sin s
5�i �2
5 cos s
5�j �15
k
r�s� � 2 cos s
5�i � 2 sin s
5�j �s
5k
62 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
23.
K �a � N�v�2 �
22
N�1� �1
2�i � j�
N�t� �1
�t4 � 1�1�2 �i � t2j�
T�t� �t2i � jt4 � 1
a�1� � 2j
a�t� �2t 3 j
v�1� � i � j
v�t� � i �1t2 j
r�t� � t i �1tj
25.
K ��T��t���r��t�� �
2�
8��
14
T��t� � �2� cos�2�t�i � 2� sin�2�t�j
T�t� � �sin�2�t�i � cos�2�t�j
r��t� � �8� sin�2�t�i � 8� cos�2�t�j
r�t� � 4 cos�2�t�i � 4 sin�2�t�j 27.
K ��T��t���r��t�� �
a �
1a
T��t� � � cos� t�i � sin� t�j
T�t� � �sin� t�i � cos� t�j
r��t� � �a sin� t�i � a cos� t�j
r�t� � a cos� t�i � a sin� t�j
29.
K ��T��t���r��t�� �
1
2et �22
e�t
T��t� �1
2���cos t � sin t�i � ��sin t � cos t�j�
T�t� �1
2���sin t � cos t�i � �cos t � sin t�j�
r��t� � ��et sin t � et cos t�i � �et cos t � et sin t�j
r�t� � et cos t i � et sin tj 31.
From Exercise 21, Section 11.4, we have:
K �a�t� � N�t�
�v�2 � 3t 4t2 �
1 t
a � N � 3t
r�t� � �cos t � t sin t, sin t � t cos t�
33.
�
5�1 � 5t2�1 � 5t2
�5
�1 � 5t2�3�2
K ��T��t���r��t��
T��t� ��5t i � 2j � k
�1 � 5t2�3�2
T�t� �i � 2tj � tk1 � 5t2
r��t� � i � 2t j � tk
r�t� � ti � t2j �t 2
2k 35.
K ��T��t���r��t�� �
3�55
�3
25
T��t� �15
��3 cos tj � 3 sin tk�
T�t� �15
�4i � 3 sin tj � 3 cos tk�
r��t� � 4i � 3 sin tj � 3 cos tk
r�t� � 4ti � 3 cos tj � 3 sin tk
21.
(The curve is a line.) K ��T��t���r��t�� � 0
T��t� � 0
T�t� �15
�2i � j�
v�t� � 4i � 2j
r�t� � 4t i � 2tj
Section 11.5 Arc Length and Curvature 63
http://librosysolucionarios.net
41.
At
(radius of curvature) 1K
� a
K �1�a
�1 � 02�3�2 �1a
y� �1a
y� � 0x � 0:
y� ���2x2 � a2��a2 � x2�3�2
y� ��x
a2 � x2
y � a2 � x2 43. (a) Point on circle:
Center:
Equation: x ��
2�2
� y2 � 1
�
2, 0�
�
2, 1�
(b) The circles have different radii since the curvature is different and
r �1K
.
45.
Radius of curvature Since the tangent line ishorizontal at the normal line is vertical. The centerof the circle is unit above the point at
Circle:
(1, 2)−6
−4
4
6
�x � 1�2 � y �52�
2
�14
�1, 5�2�.�1, 2�1�2�1, 2�,
� 1�2.
K �2
�1 � 02�3�2 � 2
y � x �1x, y� � 1 �
1x2, y� �
2x3
47.
The slope of the tangent line at is
The slope of the normal line is
Equation of normal line: or
The center of the circle is on the normal line unitsaway from the point
Since the circle is above the curve, and
Center of circle:
Equation of circle:
(0, 1)−6
0
6
3
�x � 2�2 � �y � 3�2 � 8
��2, 3�y � 3.x � �2
x � ±2
x2 � 4
x2 � x2 � 8
�0 � x�2 � �1 � y�2 � 22
�0, 1�.22
y � �x � 1y � 1 � �x
�1.
y��0� � 1.�0, 1�
r �1K
� 22K �1
�1 � 12�3�2 �1
23�2 �1
22,
y��0� � 1, y��0� � 1
y� � ex, y� � ex
y � ex, x � 0
39.
(radius of curvature)1K
�173�2
4� 17.523
K �4
�1 � ��4�2�3�2 �4
173�2 � 0.057
y� � 4
y� � 4x
y � 2x2 � 337.
Since and the radius of curvatureis undefined.
K � 0,y� � 0,
y � 3x � 2
64 Chapter 11 Vector-Valued Functions
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49.
−2
x
y
π
π
π
AB
51.
(a) K is maximum when or at the vertex
(b) limx→�
K � 0
�1, 3�.x � 1
K �2
�1 � �2�x � 1��2�3�2 �2
�1 � 4�x � 1�2�3�2
y� � 2y� � 2�x � 1�,y � �x � 1�2 � 3,
53.
(a) as No maximum
(b) limx→�
K � 0
x ⇒ 0.K ⇒ �
K � � ��2�9�x�4�3
�1 � �4�9�x�2�3�3�2� � � 6x1�3�9x2�3 � 4�3�2�
y� � �29
x�4�3y� �23
x�1�3,y � x2�3, 55.
at
Curvature is 0 at �1, 3�.
x � 1. K � �y���1 � �y��2�3�2 � �6�x � 1��
�1 � 9�x � 1�4�3�2 � 0
y� � 6�x � 1�
y� � 3�x � 1�2
y � �x � 1�3 � 3
63. Endpoints of the major axis:
Endpoints of the minor axis:
Therefore, since K is largest when and smallest when x � 0.x � ±2�2 ≤ x ≤ 2,
K � ��1�4y3��1 � ��x�4y�2�3�2 � ��16�
�16y2 � x2�3�2 �16
�12y2 � 4�3�2 �16
�16 � 3x2�3�2
y� ��4y���1� � ��x��4y��
16y2 ��4y � �x2�y�
16y2 ���4y2 � x2�
16y3 ��14y3
y� � �x4y
2x � 8yy� � 0
x2 � 4y2 � 4
�0, ±1��±2, 0�
57.
The curvature is zero when y� � 0.
K � �y���1 � �y��2�3�2
59. s � �b
a
�r��t�� dt 61. The curve is a line.
65.
(a)
(b) For At , the circle of curvature has radius Using the symmetry of the graph of f, you obtain
For At the circle of curvature has radius
Using the graph of f, you see that the center of curvature is Thus,
To graph these circles, use
and
—CONTINUED—
y �12
± 54
� x2.y � �12
± 14
� x2
x2 � y �12�
2
�54
.−3 3
−2
f
2�0, 12�.
52
�1K
.
�1, 0�,f �1� � 0.K � �25 ��5.x � 1,
x2 � y �12�
2
�14
.
12 .�0, 0�f �0� � 0.K � 2.x � 0,
K �2�6x2 � 1�
�16x6 � 16x4 � 4x2 � 1�3�2
f �x� � x4 � x2
Section 11.5 Arc Length and Curvature 65
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65. —CONTINUED—
(c) The curvature tends to be greatest near the extrema of f, and K decreases as However, f and K do not have the same critical numbers.
Critical numbers of f:
Critical numbers of K: ±0.4082± .7647,x � 0,
±22
� ±0.7071x � 0, 3−3
−2
5x →±�.
67. (a) Imagine dropping the circle into the parabola The circle will drop to the point where the tangents to the circle and parabola are equal.
and
Taking derivatives, and Hence,
Thus,
.
Thus,
Finally, and the center of the circle is 16.25 units from the vertex of the parabola. Since the radius ofthe circle is 4, the circle is 12.25 units from the vertex.
(b) In 2-space, the parabola has a curvature of at The radius of the largest sphere that willtouch the vertex has radius � 1�K �
12 .
�0, 0�.K � 2 �or z � x2�z � y2
k � x2 �12 � 16.25,
x2 � �x2 � k�2 � x2 � �12�
2
� 16 ⇒ x2 � 15.75.
�xy � k
� 2x ⇒ �x � 2x�y � k� ⇒ �1 � 2�x2 � k� ⇒ x2 � k � �12
�y � k�y� � �x ⇒ y� ��x
y � k.
y� � 2x.2x � 2�y � k�y� � 0
x2 � �y � k�2 � 16 ⇒ x2 � �x2 � k�2 � 16y � x2
x5 10−5−10
10
15
yy � x2.x2 � �y � k�2 � 16
69. Given
The center of the circle is on the normal line at a distance of R from
Equation of normal line:
y0 � y � z
y � y0 � �1y�
�x � �x � y�z�� � �z
x0 � x � y�z
x � x0 �y��1 � �y��2�
y�� y�z
�x � x0�2 ��y��2�1 � �y��2�2
�y��2
�x � x0�2�1 �1
�y��2 ��1 � �y��2�3
�y��2
�x � x0�2 � ��1y�
�x � x0�2
��1 � �y��2�3�2
�y��
y � y0 � �1y�
�x � x0�
�x, y�.
R �1K
K � �y���1 � �y��2�3�2y � f �x�:
Thus,
For
When
Center of curvature:
(See Exercise 47)
��2, 3�
y0 � y � z � 1 � 2 � 3
x0 � x � y�z � 0 � �1��2� � �2x � 0:
z �1 � e2x
ex � e�x � ex.y� � ex,y� � ex,y � ex,
�x0, y0� � �x � y�z, y � z�.
66 Chapter 11 Vector-Valued Functions
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71.
�3�1 � sin ��
8�1 � sin ��3�
3
22�1 � sin ��
� �2 cos2 � � �1 � sin ����sin �� � �1 � sin ��2��cos2 � � �1 � sin ��2�3
K � �2�r��2 � rr� � r2���r��2 � r2�3�2
r� � �sin �
r� � cos �
r � 1 � sin � 73.
�2a2
a3 �2a
, a > 0
� �2a2 cos2 � � a2 sin2 � � a2 sin2 ���a2 cos2 � � a2 sin2 ��3
K � �2�r �2 � rr� � r2���r��2 � r2�3�2
r� � �a sin �
r� � a cos �
r � a sin �
75.
�1
ea�a2 � 1
K � �2�r��2 � rr� � r2���r��2 � r2�3�2 � �2a2e2a� � a2e2a� � e2a��
�a2e2a� � e2a��3�2
r� � a2ea�
r� � aea�
r � ea�, a > 0
(a) As
(b) As K ⇒ 0.a ⇒ �,
K ⇒ 0.� ⇒ �,
77.
At the pole: K �2
�r��0�� �28
�14
r� � 8 cos 2�
r � 4 sin 2�
79.
� � f��t�g��t� � g��t�f ��t���� f��t��2 � �g��t��2�3�2
� �f��t�g��t� � g��t�f��t�� f��t��3 �
�� f��t��2 � �g��t��2
� f��t��2 �3
K � �y���1 � �y��2�3�2 �
�f��t�g��t� � g��t� f��t�� f��t��3 �
�1 � g��t�f��t� �
2
3�2
�f��t�g��t� � g��t�f��t�
�f��t��3
y� �
ddt�
g��t�f��t� dxdt
�
f��t�g��t� � g��t�f��t��f��t��2
f��t�
y� �dydx
�
dydtdxdt
�g��t�f��t�
y � g�t�
x � f �t� 81.
�1
2a2 � 2 cos ��
14a
csc �
2�
�1 � cos ≥ 0� �1a
1 � cos �
22�1 � cos ��3�2
�1a
�cos � � 1��2 � 2 cos ��3�2
� �a2�1 � cos �� cos � � a2 sin2 ���a2�1 � cos ��2 � a2 sin2 ��3�2
K � �x����y���� � y����x������x����2 � y����2�3�2
x���� � a sin � y���� � a cos �
x���� � a�1 � cos �� y���� � a sin �
x��� � a�� � sin �� y��� � a�1 � cos ��
Minimum:
Maximum: none �K →� as � → 0�
�� � ��14a
83. aN � mK dsdt�
2
� 5500 lb32 ft sec2 � 1
100 ft� 30�5280� ft3600 sec �
2
� 3327.5 lb
Section 11.5 Arc Length and Curvature 67
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85. Let Then and Then,
� x�t�x��t� � y�t�y��t� � z�t�z��t� � r � r�.
�2x�t�x��t� � 2y�t�y��t� � 2z�t�z��t���r�drdt� � ��x�t�2 � �y�t�2 � �z�t�21
2��x�t�2 � �y�t�2 � �z�t�2��1 2 �
r� � x��t�i � y��t�j � z��t�k.r � �r� � ��x�t�2 � �y�t�2 � �z�t�2r � x�t�i � y�t�j � z�t�k.
87. Let where and are functions of and
(using Exercise 77)
�1r3� i
yz� � y�zx
j��xz� � x�z�
y
kxy� � x�y
z � �1r3��r � r� � r�
�1r3��x�y2 � x�z2 � xyy� � xzz��i � �x2y� � z2y� � xx�y � zz�y�j � �x2z� � y2z� � xx�z � yy�z�k
��x2 � y2 � z2��x�i � y�j � z�k� � �xx� � yy� � zz���xi � yj � zk�
r3
ddt
rr� �
rr� � r�dr dt�r2 �
rr� � r��r � r�� rr2 �
r2r� � �r � r��rr3
r � �r�.t,zy,x,r � xi � yj � zk
89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have
Since and are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate thecoordinate system so that e lies along the positive x-axis and is the angle between e and r. Let
Then Also,
Thus,
and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse.
�L�2 GM1 � e cos �
� r
� GMr � e �r � r
r � � GM�re cos � � r � r � �r� � L� � r � GM�e �rr��
�L�2 � L � L � �r � r�� � L
r � e � �r� �e� cos � � re cos �.e � �e�.�
rr� � L
r� � L � GM�rr
� e�.
xθ
r
e
Sun
Planet
y
91.
Thus,
and r sweeps out area at a constant rate.
dAdt
�dAd�
d�
dt�
12
r2 d�
dt�
12
�L�
A �12�
r2 d�
Review Exercises for Chapter 11
1.
(a) Domain: n an integer
(b) Continuous except at , n an integert � n�
t � n�,
r�t� � ti � csc tk 3.
(a) Domain:
(b) Continuous for all t > 0
�0, �
r�t� � ln ti � tj � tk
68 Chapter 11 Vector-Valued Functions
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5. (a)
(b)
(c)
(d)
� 2�t i � �t��t � 2�j �13��t 3 � 3�t2 � 3�t�k
r�1 � �t� � r�1� � ��2�1 � �t� � 1i � �1 � �t2j �13 �1 � �t3k� � �3i � j �
13 k�
� �2c � 1�i � �c � 1�2j �13 �c � 1�3k
r�c � 1� � �2�c � 1� � 1�i � �c � 1�2j �13 �c � 1�3k
r��2� � �3i � 4j �83 k
r�0� � i
13.
y
x
1 21
2
3
3
2
1
z
r�t� � ti � ln tj �12 t2k 15. One possible answer is:
r3�t� � �4 � t�i, 0 ≤ t ≤ 4
r2�t� � 4i � �3 � t�j, 0 ≤ t ≤ 3
r1�t� � 4ti � 3tj, 0 ≤ t ≤ 1
17. The vector joining the points is One path is
r�t� � ��2 � 7t, �3 � 4t, 8 � 10t�.
�7, 4, �10�. 19.
x
y
5
1 2 3−3
32
z
r�t� � ti � tj � 2t2k
z � 2t2y � �t,x � t,
t � xx � y � 0,z � x2 � y2,
21. limt→2�
�t2i � �4 � t2j � k� � 4i � k
7.
1
1−1x
y
�1 ≤ x ≤ 1
y � 2�1 � x2�
x2 �y2
� 1
y�t� � 2 sin2 tx�t� � cos t,
r�t� � cos ti � 2 sin2 tj 9.
x
y2
1
2
1
z
z � t2 ⇒ z � y2
y � t
x � 1
r�t� � i � tj � t2k 11.
y
x
1 212
3
−2
2
3
1
z
z � 1y � sin t,x � 1,
r�t� � i � sin tj � k
t 0
x 1 1 1 1
y 0 1 0
z 1 1 1 1
�1
3�
2�
�
2
Review Exercises for Chapter 11 69
http://librosysolucionarios.net
25. and are increasing functions at and is adecreasing function at t � t0.
z�t�t � t0,y�t�x�t� 27. ��cos t i � t cos tj� dt � sin t i � �t sin t � cos t�j � C
29. ��cos t i � sin tj � tk� dt � ��1 � t2 dt �12
�t�1 � t2 � ln�t � �1 � t2� � C
31.
r�t� � �t2 � 1�i � �et � 2�j � �e�t � 4�k
r�0� � j � k � C � i � 3j � 5k ⇒ C � i � 2j � 4k
r�t� � ��2ti � etj � e�tk� dt � t2i � etj � e�tk � C 33. �2
�2�3t i � 2t2j � t3k� dt � 3t2
2i �
2t3
3j �
t4
4k�
2
�2�
323
j
35. � �2e � 2�i � 8j � 2k �2
0�et 2i � 3t2j � k� dt � 2et 2i � t3j � tk�
2
0
37.
� �3 cos t�2 sin2 t � cos2 t�, 3 sin t�2 cos2 t � sin2 t�, 0�
6 sin t cos2 t � 3 sin2 t��sin t�, 0�a�t� � v��t� � ��6 cos t��sin2 t� � ��3 cos2 t� cos t,
� 3�cos2 t sin2 t � 1
� 3�cos2 t sin2 t�cos2 t � sin2 t� � 1
�v�t�� � �9 cos4 t sin2 t � 9 sin4 t cos2 t � 9
v�t� � r��t� � ��3 cos2 t sin t, 3 sin2 t cos t, 3�
r�t� � �cos3 t, sin3 t, 3t�
39.
direction numbers
Since the parametric equations are
r�t0 � 0.1� � r�4.1� � �0.1, 16.8, 2.05�
z � 2 �12 t.y � 16 � 8t,x � t,
r�4� � �0, 16, 2�,
r��4� � �1, 8, 12�
r��t� � � 1t � 3
, 2t, 12�
t0 � 4r�t� � �ln�t � 3�, t2, 12
t�, 41. Range feet� x �v0
2
32 sin 2� �
�75�2
32 sin 60� � 152
23.
(a)
(c)
(e)
Dt��r�t�� �10t � 1
�10t2 � 2t � 1
�r�t�� � �10t2 � 2t � 1
Dt�r�t� � u�t� � 3t2 � 4t
r�t� � u�t� � 3t2 � t2�t � 1� � t3 � 2t2
r��t� � 3i � j
u�t� � ti � t2j �23
t3kr�t� � 3ti � �t � 1�j,
(b)
(d)
(f )
Dt�r�t� � u�t� � �83
t3 � 2t 2�i � 8t3j � �9t2 � 2t � 1�k
r�t� � u�t� �23
�t4 � t3�i � 2t4j � �3t3 � t 2 � t�k
Dt�u�t� � 2r�t� � �5i � �2t � 2�j � 2t 2k
u�t� � 2r�t� � �5ti � �t2 � 2t � 2�j �23
t3k
r��t� � 0
43. ⇒ v0 ���80��9.8�sin 40�
� 34.9 m �secRange � x �v0
2
9.8 sin 2� � 80
70 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
45.
does not exist
does not exist
(The curve is a line.)
a � N
a � T � 0
N�t�
T�t� � i
a�t� � 0
�v�t�� � 5
v�t� � 5i
r�t� � 5ti 47.
a � N �1
2t�4t � 1
a � T ��1
4t�t�4t � 1
N�t� �i � 2�t j�4t � 1
T�t� �i � �1 2�t� j
��4t � 1 � 2�t�
2�t i � j�4t � 1
a�t� � �1
4t�tj
�v�t�� ��4t � 1
2�t
v�t� � i �1
2�tj
r�t� � ti � �tj
49.
a � N �2
�e2t � e�2t
a � T �e2t � e�2t
�e2t � e�2t
N�t� �e�t i � etj�e2t � e�2t
T�t� �eti � e�tj�e2t � e�2t
a�t� � et i � e�t j
�v�t�� � �e2t � e�2t
v�t� � eti � e�tj
r�t� � eti � e�tj 51.
a � N �5
�5�1 � 5t2�
�5
�1 � 5t2
a � T �5t
�1 � 5t2
N�t� ��5t i � 2j � k�5�1 � 5t2
T�t� �i � 2tj � tk�1 � 5t2
a�t� � 2j � k
�v� � �1 � 5t2
v�t� � i � 2tj � tk
r�t� � ti � t2j �12
t2k
53.
When
Direction numbers when
z � t �3�
4y � ��2t � �2,x � ��2t � �2,
c � 1b � ��2,a � ��2,t �3�
4,
r��t� � �2 sin ti � 2 cos tj � k
z �3�
4.y � �2,x � ��2,t �
3�
4,
z � ty � 2 sin t,x � 2 cos t,r�t� � 2 cos ti � 2 sin tj � tk,
55. v ��9.56 � 104
4600� 4.56 mi �sec
57.
� �13t�5
0� 5�13
s � �b
a
�r��t�� dt � �5
0
�4 � 9 dt
r��t� � 2i � 3j2−2−4
−4
2
−6
−8
−10
−12
−14
−16
4 6 8 10 12 14x
y
(0, 0)
(10, −15)
r�t� � 2ti � 3tj, 0 ≤ t ≤ 5
Review Exercises for Chapter 11 71
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59.
s � 4�� 2
030 cos t � sin t dt � 120
sin2 t2 �
� 2
0� 60
� 30�cos t sin t� �r��t�� � 30�cos4 t sin2 t � sin4 t cos2 t
r��t� � �30 cos2 t sin ti � 30 sin2 t cos tj
−10
10
2
−10 10−2 2x
y r�t� � 10 cos3 ti � 10 sin3 tj
61.
s � �b
a
�r��t�� dt � �3
0
�9 � 4 � 16 dt � �3
0
�29 dt � 3�29
r��t� � �3i � 2j � 4k
xy
6 8
2 4
10
2
2
64
8
1012
z
(0, 0, 0)
(−9, 6, 12) r�t� � �3ti � 2tj � 4tk, 0 ≤ t ≤ 3
63.
x
y
z
π
4
468
68
(8, 0, 0)
(0, 8, )2
π2
s � �b
a
�r��t�� dt � �� 2
0
�65 dt ���65
2
r��t� � < �8 sin t, 8 cos t, 1�, �r��t�� � �65
r�t� � �8 cos t, 8 sin t, t�, 0 ≤ t ≤�
265.
��52 ��
0 dt � �5
2t�
�
0�
�52
�
� ��
0�1
4� cos2 t � sin2 t dt
s � ��
0�r��t�� dt
r��t� �12
i � cos tj � sin tk
0 ≤ t ≤ �r�t� �12
ti � sin tj � cos tk,
67.
Line
k � 0
r�t� � 3ti � 2tj 69.
K ��r� � r��
�r��3 ��20
�5t2 � 4�3 2 �2�5
�4 � 5t2�3 2
r� � r� � � i20
jt1
k2t2� � �4j � 2k, �r� � r�� � �20
r��t� � j � 2k
r��t� � 2i � tj � 2tk, �r�� � �5t2 � 4
r�t� � 2ti �12
t2j � t2k
71.
At and r � 173 2 � 17�17.x � 4, K �1
173 2
K � �y���1 � �y��23 2 �
1�1 � x2�3 2
y� � 1
y� � x
y �12
x2 � 273.
At and r � 2�2.x � 1, K �1
23 2�
1
2�2�
�2
4
K � �y���1 � �y��23 2 �
1 x2
�1 � �1 x�2]3 2
y� �1x, y� � �
1x2
y � ln x
75. The curvature changes abruptly from zero to a nonzeroconstant at the points B and C.
72 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
85. Let Then and Then,
� x�t�x��t� � y�t�y��t� � z�t�z��t� � r � r�.
�2x�t�x��t� � 2y�t�y��t� � 2z�t�z��t���r�drdt� � ��x�t�2 � �y�t�2 � �z�t�21
2��x�t�2 � �y�t�2 � �z�t�2��1 2 �
r� � x��t�i � y��t�j � z��t�k.r � �r� � ��x�t�2 � �y�t�2 � �z�t�2r � x�t�i � y�t�j � z�t�k.
87. Let where and are functions of and
(using Exercise 77)
�1r3� i
yz� � y�zx
j��xz� � x�z�
y
kxy� � x�y
z � �1r3��r � r� � r�
�1r3��x�y2 � x�z2 � xyy� � xzz��i � �x2y� � z2y� � xx�y � zz�y�j � �x2z� � y2z� � xx�z � yy�z�k
��x2 � y2 � z2��x�i � y�j � z�k� � �xx� � yy� � zz���xi � yj � zk�
r3
ddt
rr� �
rr� � r�dr dt�r2 �
rr� � r��r � r�� rr2 �
r2r� � �r � r��rr3
r � �r�.t,zy,x,r � xi � yj � zk
89. From Exercise 86, we have concluded that planetary motion is planar. Assume that the planet moves in the xy-plane with the sun at the origin. From Exercise 88, we have
Since and are both perpendicular to L, so is e. Thus, e lies in the xy-plane. Situate thecoordinate system so that e lies along the positive x-axis and is the angle between e and r. Let
Then Also,
Thus,
and the planetary motion is a conic section. Since the planet returns to its initial position periodically, the conic is an ellipse.
�L�2 GM1 � e cos �
� r
� GMr � e �r � r
r � � GM�re cos � � r � r � �r� � L� � r � GM�e �rr��
�L�2 � L � L � �r � r�� � L
r � e � �r� �e� cos � � re cos �.e � �e�.�
rr� � L
r� � L � GM�rr
� e�.
xθ
r
e
Sun
Planet
y
91.
Thus,
and r sweeps out area at a constant rate.
dAdt
�dAd�
d�
dt�
12
r2 d�
dt�
12
�L�
A �12�
r2 d�
Review Exercises for Chapter 11
1.
(a) Domain: n an integer
(b) Continuous except at , n an integert � n�
t � n�,
r�t� � ti � csc tk 3.
(a) Domain:
(b) Continuous for all t > 0
�0, �
r�t� � ln ti � tj � tk
68 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
5. (a)
(b)
(c)
(d)
� 2�t i � �t��t � 2�j �13��t 3 � 3�t2 � 3�t�k
r�1 � �t� � r�1� � ��2�1 � �t� � 1i � �1 � �t2j �13 �1 � �t3k� � �3i � j �
13 k�
� �2c � 1�i � �c � 1�2j �13 �c � 1�3k
r�c � 1� � �2�c � 1� � 1�i � �c � 1�2j �13 �c � 1�3k
r��2� � �3i � 4j �83 k
r�0� � i
13.
y
x
1 21
2
3
3
2
1
z
r�t� � ti � ln tj �12 t2k 15. One possible answer is:
r3�t� � �4 � t�i, 0 ≤ t ≤ 4
r2�t� � 4i � �3 � t�j, 0 ≤ t ≤ 3
r1�t� � 4ti � 3tj, 0 ≤ t ≤ 1
17. The vector joining the points is One path is
r�t� � ��2 � 7t, �3 � 4t, 8 � 10t�.
�7, 4, �10�. 19.
x
y
5
1 2 3−3
32
z
r�t� � ti � tj � 2t2k
z � 2t2y � �t,x � t,
t � xx � y � 0,z � x2 � y2,
21. limt→2�
�t2i � �4 � t2j � k� � 4i � k
7.
1
1−1x
y
�1 ≤ x ≤ 1
y � 2�1 � x2�
x2 �y2
� 1
y�t� � 2 sin2 tx�t� � cos t,
r�t� � cos ti � 2 sin2 tj 9.
x
y2
1
2
1
z
z � t2 ⇒ z � y2
y � t
x � 1
r�t� � i � tj � t2k 11.
y
x
1 212
3
−2
2
3
1
z
z � 1y � sin t,x � 1,
r�t� � i � sin tj � k
t 0
x 1 1 1 1
y 0 1 0
z 1 1 1 1
�1
3�
2�
�
2
Review Exercises for Chapter 11 69
http://librosysolucionarios.net
25. and are increasing functions at and is adecreasing function at t � t0.
z�t�t � t0,y�t�x�t� 27. ��cos t i � t cos tj� dt � sin t i � �t sin t � cos t�j � C
29. ��cos t i � sin tj � tk� dt � ��1 � t2 dt �12
�t�1 � t2 � ln�t � �1 � t2� � C
31.
r�t� � �t2 � 1�i � �et � 2�j � �e�t � 4�k
r�0� � j � k � C � i � 3j � 5k ⇒ C � i � 2j � 4k
r�t� � ��2ti � etj � e�tk� dt � t2i � etj � e�tk � C 33. �2
�2�3t i � 2t2j � t3k� dt � 3t2
2i �
2t3
3j �
t4
4k�
2
�2�
323
j
35. � �2e � 2�i � 8j � 2k �2
0�et 2i � 3t2j � k� dt � 2et 2i � t3j � tk�
2
0
37.
� �3 cos t�2 sin2 t � cos2 t�, 3 sin t�2 cos2 t � sin2 t�, 0�
6 sin t cos2 t � 3 sin2 t��sin t�, 0�a�t� � v��t� � ��6 cos t��sin2 t� � ��3 cos2 t� cos t,
� 3�cos2 t sin2 t � 1
� 3�cos2 t sin2 t�cos2 t � sin2 t� � 1
�v�t�� � �9 cos4 t sin2 t � 9 sin4 t cos2 t � 9
v�t� � r��t� � ��3 cos2 t sin t, 3 sin2 t cos t, 3�
r�t� � �cos3 t, sin3 t, 3t�
39.
direction numbers
Since the parametric equations are
r�t0 � 0.1� � r�4.1� � �0.1, 16.8, 2.05�
z � 2 �12 t.y � 16 � 8t,x � t,
r�4� � �0, 16, 2�,
r��4� � �1, 8, 12�
r��t� � � 1t � 3
, 2t, 12�
t0 � 4r�t� � �ln�t � 3�, t2, 12
t�, 41. Range feet� x �v0
2
32 sin 2� �
�75�2
32 sin 60� � 152
23.
(a)
(c)
(e)
Dt��r�t�� �10t � 1
�10t2 � 2t � 1
�r�t�� � �10t2 � 2t � 1
Dt�r�t� � u�t� � 3t2 � 4t
r�t� � u�t� � 3t2 � t2�t � 1� � t3 � 2t2
r��t� � 3i � j
u�t� � ti � t2j �23
t3kr�t� � 3ti � �t � 1�j,
(b)
(d)
(f )
Dt�r�t� � u�t� � �83
t3 � 2t 2�i � 8t3j � �9t2 � 2t � 1�k
r�t� � u�t� �23
�t4 � t3�i � 2t4j � �3t3 � t 2 � t�k
Dt�u�t� � 2r�t� � �5i � �2t � 2�j � 2t 2k
u�t� � 2r�t� � �5ti � �t2 � 2t � 2�j �23
t3k
r��t� � 0
43. ⇒ v0 ���80��9.8�sin 40�
� 34.9 m �secRange � x �v0
2
9.8 sin 2� � 80
70 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
45.
does not exist
does not exist
(The curve is a line.)
a � N
a � T � 0
N�t�
T�t� � i
a�t� � 0
�v�t�� � 5
v�t� � 5i
r�t� � 5ti 47.
a � N �1
2t�4t � 1
a � T ��1
4t�t�4t � 1
N�t� �i � 2�t j�4t � 1
T�t� �i � �1 2�t� j
��4t � 1 � 2�t�
2�t i � j�4t � 1
a�t� � �1
4t�tj
�v�t�� ��4t � 1
2�t
v�t� � i �1
2�tj
r�t� � ti � �tj
49.
a � N �2
�e2t � e�2t
a � T �e2t � e�2t
�e2t � e�2t
N�t� �e�t i � etj�e2t � e�2t
T�t� �eti � e�tj�e2t � e�2t
a�t� � et i � e�t j
�v�t�� � �e2t � e�2t
v�t� � eti � e�tj
r�t� � eti � e�tj 51.
a � N �5
�5�1 � 5t2�
�5
�1 � 5t2
a � T �5t
�1 � 5t2
N�t� ��5t i � 2j � k�5�1 � 5t2
T�t� �i � 2tj � tk�1 � 5t2
a�t� � 2j � k
�v� � �1 � 5t2
v�t� � i � 2tj � tk
r�t� � ti � t2j �12
t2k
53.
When
Direction numbers when
z � t �3�
4y � ��2t � �2,x � ��2t � �2,
c � 1b � ��2,a � ��2,t �3�
4,
r��t� � �2 sin ti � 2 cos tj � k
z �3�
4.y � �2,x � ��2,t �
3�
4,
z � ty � 2 sin t,x � 2 cos t,r�t� � 2 cos ti � 2 sin tj � tk,
55. v ��9.56 � 104
4600� 4.56 mi �sec
57.
� �13t�5
0� 5�13
s � �b
a
�r��t�� dt � �5
0
�4 � 9 dt
r��t� � 2i � 3j2−2−4
−4
2
−6
−8
−10
−12
−14
−16
4 6 8 10 12 14x
y
(0, 0)
(10, −15)
r�t� � 2ti � 3tj, 0 ≤ t ≤ 5
Review Exercises for Chapter 11 71
http://librosysolucionarios.net
59.
s � 4�� 2
030 cos t � sin t dt � 120
sin2 t2 �
� 2
0� 60
� 30�cos t sin t� �r��t�� � 30�cos4 t sin2 t � sin4 t cos2 t
r��t� � �30 cos2 t sin ti � 30 sin2 t cos tj
−10
10
2
−10 10−2 2x
y r�t� � 10 cos3 ti � 10 sin3 tj
61.
s � �b
a
�r��t�� dt � �3
0
�9 � 4 � 16 dt � �3
0
�29 dt � 3�29
r��t� � �3i � 2j � 4k
xy
6 8
2 4
10
2
2
64
8
1012
z
(0, 0, 0)
(−9, 6, 12) r�t� � �3ti � 2tj � 4tk, 0 ≤ t ≤ 3
63.
x
y
z
π
4
468
68
(8, 0, 0)
(0, 8, )2
π2
s � �b
a
�r��t�� dt � �� 2
0
�65 dt ���65
2
r��t� � < �8 sin t, 8 cos t, 1�, �r��t�� � �65
r�t� � �8 cos t, 8 sin t, t�, 0 ≤ t ≤�
265.
��52 ��
0 dt � �5
2t�
�
0�
�52
�
� ��
0�1
4� cos2 t � sin2 t dt
s � ��
0�r��t�� dt
r��t� �12
i � cos tj � sin tk
0 ≤ t ≤ �r�t� �12
ti � sin tj � cos tk,
67.
Line
k � 0
r�t� � 3ti � 2tj 69.
K ��r� � r��
�r��3 ��20
�5t2 � 4�3 2 �2�5
�4 � 5t2�3 2
r� � r� � � i20
jt1
k2t2� � �4j � 2k, �r� � r�� � �20
r��t� � j � 2k
r��t� � 2i � tj � 2tk, �r�� � �5t2 � 4
r�t� � 2ti �12
t2j � t2k
71.
At and r � 173 2 � 17�17.x � 4, K �1
173 2
K � �y���1 � �y��23 2 �
1�1 � x2�3 2
y� � 1
y� � x
y �12
x2 � 273.
At and r � 2�2.x � 1, K �1
23 2�
1
2�2�
�2
4
K � �y���1 � �y��23 2 �
1 x2
�1 � �1 x�2]3 2
y� �1x, y� � �
1x2
y � ln x
75. The curvature changes abruptly from zero to a nonzeroconstant at the points B and C.
72 Chapter 11 Vector-Valued Functions
http://librosysolucionarios.net
Problem Solving for Chapter 11
1.
(a)
(b)
At
(c) K � �a � � �length�
t � a, K � �a.
K ���t cos2��t2
2 � � �t sin2��t2
2 ��1
� �t
x��t� � ��t sin��t2
2 �, y��t� � �t cos��t2
2 �
s � �a
0
�x��t�2 � y��t�2 dt � �a
0dt � a
x��t� � cos��t2
2 �, y��t� � sin��t2
2 �
x�t� � �t
0
cos��u2
2 � du, y�t� � �t
0
sin��u2
2 � du 3. Bomb:
Projectile:
At 1600 feet: Bomb:
seconds.
Projectile will travel 5 seconds:
Horizontal position:
At bomb is at
At projectile is at
Thus,
Combining,
v0 �200
cos � 447.2 ftsec
v0 sin �v0 cos �
�400200
⇒ tan � � 2 ⇒ � 63.4.
v0 cos � � 200.
5v0 cos �.t � 5,
5000 � 400�10� � 1000.t � 10,
v0 sin � � 400.
5�v0 sin �� � 16�25� � 1600
3200 � 16t2 � 1600 ⇒ t � 10
r2�t� � ��v0 cos ��t, �v0 sin ��t � 16t2�
r1�t� � �5000 � 400t, 3200 � 16t2�
5.
Thus, and
s2 � 2 � 16 cos2� t2� � 16 sin2� t
2� � 16.
�1K
� 4 sin t2
�1
4 sin �
2
K �1��1 � cos ��cos � � sin � sin ��
�2 sin �
2�3 � �1 cos � � 1�
8 sin3 �
2
x� ��� � sin �, y� ��� � cos �
s�t� � �t
�
2 sin �
2 d� � �4 cos
�
2�t
�� �4 cos
t2
� �2 � 2 cos � ��4 sin2 �
2
�x����2 � y����2 � ��1 � cos ��2 � sin2 �
x���� � 1 � cos �, y���� � sin �, 0 ≤ � ≤ 2�
7.
⇒ ddt
�r�t�� �r�t� � r��t�
�r�t�� � r�t� � r��t� � r��t� � r�t�
ddt
��r�t���2 � 2�r�t�� ddt
�r�t��
�r2�t�� � r�t� � r�t�
Problem Solving for Chapter 11 73
http://librosysolucionarios.net
9.
x
y
z
π3
123
4 4
π6 T
T
B
B
N
N
B��
2� �35
i �45
k
N��
2� � �j
At t ��
2, T��
2� � �45
i �35
k
B � T � N �35
sin ti �35
cos tj �45
k
N � �cos ti � sin tj
T� � �45
cos ti �45
sin tj
T � �45
sin ti �45
cos tj �35
k
r��t� � �4 cos ti � 4 sin tj
r��t� � �4 sin ti � 4 cos tj � 3k, �r��t�� � 5
r�t� � 4 cos ti � 4 sin tj � 3tk, t ��
211. (a)
Hence, and
for some scalar
(b) Using Exercise 10.3, number 64,
Now,
Finally,
� �KT � B.
� �B � KN� � �� N � T�
N��s� �dds
�B � T� � �B � T�� � �B� � T�
KN � � dTds �
T��s��T��s�� � T��s� �
dTds
.
� N.
� ���T � N�T � �T � T�N�
B � T � �T � N� � T � �T � �T � N�
� �T
� ���N � N�T � �N � T�N�
B � N � �T � N� � N � �N � �T � N�
B � T � N.
.
dBds
� T ⇒ dBds
� � NdBds
� B
� �T � T� � N� � T � �T� �T�
�T��� � 0
T �dBds
� T � �T � N�� � T � �T� � N�
dBds
�dds
�T � N� � �T � N�� � �T� � N�
constant length ⇒ dBds
� B�B� � �T � N� � 1
13.
(a)
(c)
(e) limt→�
K � 0
K�2� 0.51
K�1� ����2 � 2���2 � 1�32 1.04
K�0� � 2�
K ���� 2t2 � 2���2t2 � 1�32
−3 3
−2
2
0 ≤ t ≤ 2r�t� � �t cos �t, t sin �t�,
(b)
(graphing utility)
(d)
(f ) As the graph spirals outward and the curvaturedecreases.
t →�,
0 50
5
� �2
0
��2t2 � 1 dt 6.766
Length � �2
0�r��t�� dt
74 Chapter 11 Vector-Valued Functions
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C H A P T E R 1 2Functions of Several Variables
Section 12.1 Introduction to Functions of Several Variables . . . . . . . 308
Section 12.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . 312
Section 12.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . 315
Section 12.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 321
Section 12.5 Chain Rules for Functions of Several Variables . . . . . .325
Section 12.6 Directional Derivatives and Gradients . . . . . . . . . . . 330
Section 12.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 334
Section 12.8 Extrema of Functions of Two Variables . . . . . . . . . . 340
Section 12.9 Applications of Extrema of Functions of Two Variables . 345
Section 12.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 350
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 355
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 361
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C H A P T E R 1 2Functions of Several Variables
Section 12.1 Introduction to Functions of Several VariablesSolutions to Even-Numbered Exercises
308
2.
No, z is not a function of x and y. For example,corresponds to both z � ±2�x, y� � �1, 0�
xz2 � 2xy � y2 � 4 4.
Yes, z is a function of x and y.
z � 8 � x ln y
z � x ln y � 8 � 0
6.
(a)
(b)
(c)
(d)
(e)
(f) f �t, 1� � 4 � t 2 � 4 � �t 2
f �x, 0� � 4 � x2 � 0 � 4 � x2
f �1, y� � 4 � 1 � 4y2 � 3 � 4y2
f �2, 3� � 4 � 4 � 36 � �36
f �0, 1� � 4 � 0 � 4 � 0
f �0, 0� � 4
f �x, y� � 4 � x2 � 4y2 8.
(a)
(b)
(c)
(d)
(e)
(f)
� ln 2 � ln e � �ln 2� � 1
g�e, e� � ln�e � e� � ln 2e
g�2, �3� � ln�2 � 3� � ln 1 � 0
g�0, 1� � ln�0 � 1� � 0
g�e, 0� � ln�e � 0� � 1
g�5, 6� � ln�5 � 6� � ln 11
g�2, 3� � ln�2 � 3� � ln 5
g�x, y� � ln�x � y�
10.
(a)
(b) f �6, 8, �3� � �6 � 8 � 3 � �11
f �0, 5, 4� � �0 � 5 � 4 � 3
f �x, y, z� � �x � y � z 12.
(a)
(b) V�5, 2� � ��5�2�2� � 50�
V�3, 10� � ��3�2�10� � 90�
V�r, h� � �r2h
14.
(a) g�4, 1� � �1
4 1t dt � �ln�t��
1
4� �ln 4
g�x, y� � �y
x
1t dt
(b) g�6, 3� � �3
6 1t dt � �ln�t��
3
6� ln 3 � ln 6 � ln1
2
16.
(a)
(b)
��y�3x � 2y � �y�
�y� 3x � 2y � �y, �y � 0
�3xy � 3x��y� � y2 � 2y��y� � ��y�2 � 3xy � y2
�y
f �x, y � �y� � f �x, y�
�y�
�3x�y � �y� � �y � �y�2� � �3xy � y2��y
�3xy � 3��x�y � y2 � 3xy � y2
�x�
3��x�y�x
� 3y, �x � 0
f �x � �x, y� � f �x, y�
�x�
�3�x � �x�y � y2� � �3xy � y2��x
f �x, y� � 3xy � y2
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Section 12.1 Introduction to Functions of Several Variables 309
18.
Domain:
Range: 0 ≤ z ≤ 2
�x, y�: x2
4�
y2
1 ≤ 1�
x2
4�
y2
1 ≤ 1
x2 � 4y2 ≤ 4
4 � x2 � 4y2 ≥ 0
f �x, y� � �4 � x2 � 4y2 20.
Domain:
Range: 0 ≤ z ≤ �
�x, y�: �1 ≤ yx ≤ 1�
f �x, y� � arccos yx
22.
Domain:
Range: all real numbers
��x, y�: xy > 6�
xy > 6
xy � 6 > 0
f �x, y� � ln�xy � 6�
30. (a) Domain: is any real number,
is any real number
Range:
(b) when which represents points on the y-axis.
(c) No. When x is positive, z is negative. When x is negative, z is positive. The surface does not pass through the first octant,the octant where y is negative and x and z are positive, the octant where y is positive and x and z are negative, and theoctant where x, y and z are all negative.
x � 0z � 0
�2 ≤ z ≤ 2
�y
��x, y�: x
32.Plane
Domain: entire xy-plane
Range:
x
y2 33
44
6
z
�� < z < �
f �x, y� � 6 � 2x � 3y 34.
Plane:
x
y
4
4
432
−2−3−4
−4
z
z �12 x
g�x, y� �12 x 36.
Cone
Domain of f : entire xy-plane
Range:
x
y1122
33
2
z
z ≥ 0
z �12�x2 � y2
38.
Domain of f : entire xy-plane
Range:
x
y
5
25
20
15
10
5
z
z ≥ 0
f �x, y� � xy,
0,
x ≥ 0, y ≥ 0
elsewhere40.
Semi-ellipsoid
Domain: set of all points lying on or inside the ellipse
Range:
4 4
4
−2
−4
yx
z
0 ≤ z ≤ 1
�x2�9� � �y2�16� � 1
f �x, y� �1
12�144 � 16x2 � 9y2
24.
Domain:
Range: all real numbers
��x, y�: x � y�
z �xy
x � y26.
Domain: is any real number,
y is any real number
Range: z ≥ 0
���x, y�: x
f �x, y� � x2 � y2 28.
Domain:
Range: all real numbers
��x, y�: y ≥ 0�
g�x, y� � x�y
42.
4
4
−4
−4 −4
yx
z
f �x, y� � x sin y
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310 Chapter 12 Functions of Several Variables
(d) The graph of g is lower than the graph of f. Ifis on the graph of f, then is on the graph
of g.
12 zz � f �x, y�
(e)
x
y
5
25
20
15
10
5
z
46.
Level curves:
Hyperbolas centered at
Matches (d)
�0, 0�
x2 � y2 � 1 � ln c
ln c � 1 � x2 � y2
c � e1�x2�y2
z � e1�x2�y2 48.
Level curves:
Ellipses
Matches (a)
x2 � 2y2 � 4 cos�1 c
cos�1 c �x2 � 2y2
4
c � cosx2 � 2y2
4
z � cosx � 2y2
4 50.
The level curves are of the formor
Thus, the level curves arestraight lines with a slope of
x−2
3
c = 0c = 2c = 4c = 6
c = 8
c = 10
y
�23 .
6 � c.2x � 3y �6 � 2x � 3y � c
f �x, y� � 6 � 2x � 3y
52.
The level curves are ellipses of the form
(except is the point ).
−3
−3
3
3
y
x
c = 0c = 2
c = 4c = 6
c = 8
�0, 0�x2 � 2y2 � 0x2 � 2y2 � c
f �x, y� � x2 � 2y2 54.
The level curves are of the form
Thus, the level curves are hyperbolas.
−2 −1
−1
1
2
−2
11
2
2
c = 3c = 2
c =
c = 4
13c =
14c =
y
x
e xy�2 � c, or ln c �xy2
.
f �x, y� � exy�2
44.
(a)
x
y
5
25
20
15
10
5
z
f �x, y� � xy, x ≥ 0, y ≥ 0
(b) g is a vertical translation of f3 units downward
(c) g is a reflection of f in theplanexy-
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Section 12.1 Introduction to Functions of Several Variables 311
56.
The level curves are of theform
Thus, the level curves are parallel lines of slope 1 passingthrough the fourth quadrant.
y � x � ec
ec � x � y
c � ln�x � y�
−6
−4
6x
c = 2−
c = 1−
c = 2
c =
c = −
c = ±
c = 1
c = 0
1
1
3
2
2
2
yf �x, y� � ln�x � y� 58.
−6
−4
6
4f �x, y� � �xy�
60.
−1
−1
1
1
h�x, y� � 3 sin��x� � �y�� 62. The graph of a function of two variables is the set of allpoints for which and is in thedomain of f. The graph can be interpreted as a surface inspace. Level curves are the scalar fields forc, a constant.
f �x, y� � c,
�x, y�z � f �x, y��x, y, z�
64.
The level curves are the lines
or
These lines all pass through the origin.
y �1c
xc �xy
f �x, y� �xy
66. The surface could be an ellipsoid centered at One possible function is
f �x, y� � x2 ��y � 1�2
4� 1.
�0, 1, 0�.
68. A�r, t� � 1000ert
Number of years
Rate 5 10 15 20
0.08 $1491.82 $2225.54 $3320.12 $4953.03
0.10 $1648.72 $2718.28 $4481.69 $7389.06
0.12 $1822.12 $3320.12 $6049.65 $11,023.18
0.14 $2013.75 $4055.20 $8166.17 $16,444.65
70.
Plane
yx
12
3
2
34
z
4 � 4x � y � 2z
c � 4
f �x, y, z� � 4x � y � 2z 72.
Elliptic paraboloid
Vertex:
xy5
3
5
z
�0, 0, �1�
1 � x2 �14 y2 � z
c � 1
f �x, y, z� � x2 �14 y2 � z 74.
or
y
x
2
8
4
z
z � sin x0 � sin x � z
c � 0
f �x, y, z� � sin x � z
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76.
(a) hr min
(c) hr min� 10W�12, 6� �1
12 � 6�
16
� 12W�15, 10� �1
15 � 10�
15
W�x, y� �1
x � y, y < x
(b) hr min
(d) hr min� 30W�4, 2� �1
4 � 2�
12
� 20W�12, 9� �1
12 � 9�
13
78.
� 100�2�0.6x0.6�2�0.4y0.4 � 100�2�0.6�2�0.4x0.6y0.4 � 2�100x0.6y0.4� � 2f �x, y�
f �2x, 2y� � 100�2x�0.6�2y�0.4
f �x, y� � 100x0.6y0.4
80.
r
l
V � �r 2l �43
�r3 ��r 2
3�3l � 4r�
88. True 90. True
84. Southwest 86. Latitude and land versus ocean location have the greatesteffect on temperature.
82. (a)
(b) x has the greater influence because its coefficient is larger than that of
(c)
This function gives the shareholder’s equity z in terms of net sales x and assumes constant assets of y � 25.
� 0.143x � 1.102
f �x, 25� � 0.143x � 0.024�25� � 0.502
y�0.024�.�0.143�
Year 1995 1996 1997 1998 1999 2000
z 12.7 14.8 17.1 18.5 21.1 25.8
Model 13.09 14.79 16.45 18.47 21.38 25.78
Section 12.2 Limits and Continuity
2. Let be given. We need to find such that
whenever Take
Then if we have
�x � 4� < �.
��x � 4�2 < �
0 < ��x � 4�2 � �y � 1�2 < � � �,
� � �.0 < ��x � a�2 � �y � b�2 � ��x � 4�2 � �y � 1�2 < �.
� f �x, y� � L� � �x � 4� < �� > 0� > 0
4. lim�x, y�→�a, b�
�4f �x, y�g�x, y� �
4� lim�x, y�→�a, b�
f �x, y�lim
�x, y�→�a, b� g�x, y� �
4�5�3
�203
312 Chapter 12 Functions of Several Variables
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6. lim�x, y�→�a, b�
�f �x, y� � g�x, y�f �x, y� �
lim�x, y�→�a, b�
f �x, y� � lim�x, y�→�a, b�
g�x, y�
lim�x, y�→�a, b�
f �x, y� �5 � 3
5�
25
8.
Continuous everywhere
lim�x, y�→�0, 0�
�5x � y � 1� � 0 � 0 � 1 � 1 10.
Continuous for x � y > 0
lim�x, y�→�1, 1�
x
�x � y�
1�1 � 1
��22
12.
Continuous everywhere
lim�x, y�→��4, 2�
y cos�xy� � 2 cos �
2� 0 14.
Continuous except at �0, 0�
lim�x, y�→�1, 1�
xy
x2 � y2 �12
16.
Continuous everywhere
lim�x, y, z�→�2, 0, 1�
xeyz � 2e0 � 2 18.
Continuous everywhere
lim�x, y�→�0, 0�
x2
�x2 � 1��y2 � 1� �0
�0 � 1��0 � 1� � 0
f �x, y� �x2
�x2 � 1��y2 � 1�
20.
The limit does not exist.
Continuous except at �0, 0�
lim�x, y�→�0, 0�
�1 �cos�x2 � y2�
x2 � y2 � ��
22.
Continuous except at
Path:
Path:
The limit does not exist because along the path the function equals 0, whereas along the path the function tends to infinity.
y � xy � 0
y � x
y � 0
�0, 0�
f �x, y� �y
x2 � y2
1 5 50 50012f �x, y�
�0.001, 0.001��0.01, 0.01��0.1, 0.1��0.5, 0.5��1, 1��x, y�
Section 12.2 Limits and Continuity 313
24.
Continuous except at
Path:
Path:
The limit does not exist because along the line the function tends to infinity, whereas along the line the function tends to 2.
y � xy � 0
y � x
y � 0
�0, 0�
f �x, y� �2x � y2
2x2 � y
1.17 1.95 1.995 2.013f �x, y�
�0.0001, 0.0001��0.001, 0.001��0.01, 0.01��0.25, 0.25��1, 1��x, y�
1 4 100 1000 1,000,000f �x, y�
�0.000001, 0��0.001, 0��0.01, 0��0.25, 0��1, 0��x, y�
0 0 0 0 0f �x, y�
�0.001, 0��0.01, 0��0.1, 0��0.5, 0��1, 0��x, y�
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26.
Hence,
f is continuous at whereas g is not continuous at �0, 0�.
�0, 0�,
lim�x, y�→�0, 0�
f �x, y� � lim�x, y�→�0, 0�
g�x, y� � 0.
lim�x, y�→�0, 0�
4x2y2
�x2 � y2 � 0 28.
Does not exist
x 64
y6
2
z
lim�x, y�→�0, 0�
�sin 1x
� cos 1x�
30.
Does not exist
x y4
4
18
z
lim�x, y�→�0, 0�
x2 � y2
x2y32.
The limit equals 0.
55
5
x
y
z
f �x, y� �2xy
x2 � y2 � 1
34. lim�x, y�→�0, 0�
xy2
x2 � y2 � limr→0
�r cos ��r2 sin2 �
r2 � limr→0
�r cos sin2 � � 0
36. lim�x, y�→�0, 0�
x2y2
x2 � y2 � limr→0
r4 cos2 sin2
r2 � limr→0
r2 cos2 sin2 � 0
38.
Continuous for x2 � y2 9
f �x, y, z� �z
x2 � y2 � 940.
Continuous everywhere
f �x, y, z� � xy sin z
314 Chapter 12 Functions of Several Variables
42.
Continuous except at �0, 0�
�1
x2 � y2
f �g�x, y�� � f �x2 � y2�
g�x, y� � x2 � y2
f �t� �1t
44.
Continuous for x2 � y2 4
f �g�x, y�� � f �x2 � y2� �1
4 � x2 � y2
g�x, y� � x2 � y2
f �t� �1
4 � t
46.
(a)
(b)
� lim�y→0
2y�y � ��y�2
�y� lim
�y→0 �2y � �y� � 2y
lim�y→0
f �x, y � �y� � f �x, y�
�y� lim
�y→0 �x2 � �y � �y�2� � �x2 � y2�
�y
� lim�x→0
2x�x � ��x�2
�x� lim
�x→0 �2x � �x� � 2x
lim�x→0
f �x � �x, y� � f �x, y�
�x� lim
�x→0 ��x � �x�2 � y2� � �x2 � y2�
�x
f �x, y� � x2 � y2
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Section 12.3 Partial Derivatives 315
48.
(a)
(b)
(L’Hôpital’s Rule)
�3y � 1
2�y
�32
y1�2 �12
y�1�2
� lim�y→0
�y � �y�3�2 � y3�2
�y� lim
�y→0 �y � �y�1�2� y1�2
�y
lim�y→0
f �x, y � �y� � f �x, y�
�y� lim
�y→0 �y � �y�3�2 � �y � �y�1�2 � �y3�2 � y1�2�
�y
lim�x→0
f �x � �x, y� � f �x, y�
�x� lim
�x→0 �y �y � 1� � �y �y � 1�
�x� 0
f �x, y� � �y �y � 1�
50. See the definition on page 854.
52.
(a) Along
If then and the limit does not exist.
(c) No, the limit does not exist. Different paths result indifferent limits.
y � 0a � 0,
� limx→0
x2�1 � a2�
ax2 �1 � a2
a, a � 0
y � ax: lim�x, ax�→�0, 0�
x2 � �ax�2
x�ax�
lim�x, y�→�0, 0�
x2 � y2
xy
54. Given that is continuous, then which means that for each there corresponds
a such that whenever
Let then for every point in the corresponding neighborhood since
⇒ 32
f �a, b� < f �x, y� < 12
f �a, b� < 0.
� f �x, y� � f �a, b�� < � f �a, b��2
⇒ �� f �a, b��2
< f �x, y� � f �a, b� < � f �a, b��2
�f �x, y� < 0� � � f �a, b���2,
0 < ��x � a�2 � �y � b�2 < �.
� f �x, y� � f �a, b�� < �� > 0
� > 0,lim�x, y�→�a, b�
f �x, y� � f �a, b� < 0,f �x, y�
56. False. Let
See Exercise 21.
f �x, y� �xy
x2 � y2. 58. True
Section 12.3 Partial Derivatives
(b) Along
limit does not exist
lim�x, x2�→�0, 0�
x2 � �x2�2
x�x2� � limx→0
1 � x2
xy � x2:
2. fy��1, �2� < 0 4. fx��1, �1� � 0 6.
fy�x, y� � �6y
fx�x, y� � 2x
f �x, y� � x2 � 3y2 � 7 8.
zy
� 4y�x
z
x�
y2
�x
z � 2y2�x
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316 Chapter 12 Functions of Several Variables
20.
gy�x, y� �12
2y
x2 � y2 �y
x2 � y2
gx�x, y� �12
2x
x2 � y2 �x
x2 � y2
g�x, y� � ln �x2 � y2 �12
ln�x2 � y2� 22.
f
y�
1
2 �2x � y3��1�2�3y2� �
3y2
2�2x � y3
f
x�
1
2�2x � y3��1�2�2� �
1�2x � y3
f �x, y� � �2x � y3
24.
zy
� �3 sin 3x sin 3y
zx
� 3 cos 3x cos 3y
z � sin 3x cos 3y 26.
zy
� �2y sin�x2 � y2�
zx
� �2x sin�x2 � y2�
z � cos�x2 � y2�
28.
fy�x, y� � 2
fx�x, y� � �2
� �y
x
2 dt � �2ty
x� 2y � 2x
� �y
x
�2t � 1� dt � �y
x
�2t � 1� dt
f �x, y� � �y
x
�2t � 1� dt � �x
y
�2t � 1� dt
16.
zy
��2y
x2 � y2
zx
�1
x2 � y2 �2x� �2x
x2 � y2
z � ln�x2 � y2�
30.
� lim�y→0
x2 � 2x�y � �y� � �y � �y�2 � x2 � 2xy � y2
�y� lim
�y→0 ��2x � 2y � �y� � 2�y � x�
fy
� lim�y→0
f �x, y � �y� � f �x, y�
�y
� lim�x→0
�x � �x�2 � 2�x � �x�y � y2 � x2 � 2xy � y2
�x� lim
�x→0 �2x � �x � 2y� � 2�x � y�
fx
� lim�x→0
f �x � �x, y� � f �x, y�
�x
f �x, y� � x2 � 2xy � y2 � �x � y�2
10.
zy
� 3y2 � 8xy
z
x� �4y2
z � y3 � 4xy2 � 1 12.
zy
� xex�y�xy2� � �
x2
y2 ex�y
zx
�xy
ex�y � ex�y � ex�y xy
� 1� z � xex�y 14.
zy
�12
xxy
�12y
zx
�12
yxy
�12x
z � ln �xy �12
ln�xy�
18.
fy�x, y� ��x2 � y2��x� � �xy��2y�
�x2 � y2�2 �x3 � xy2
�x2 � y2�2
fx�x, y� ��x2 � y2�� y� � �xy��2x�
�x2 � y2�2 �y3 � x2y
�x2 � y2�2
f �x, y� �xy
x2 � y2
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Section 12.3 Partial Derivatives 317
34.
At
At ��2, 1�: hy��2, 1� � �2
hy�x, y� � �2y
��2, 1�: hx��2, 1� � �4
hx�x, y� � 2x
h�x, y� � x2 � y2 36.
At
At
4,
3�, z
y� sin
6� �12
z
y� �sin�2x � y���1� � sin�2x � y�
4,
3�, z
x� �2 sin
6� � �1
z
x� �2 sin�2x � y�
z � cos�2x � y�
44.
Intersecting curve:
At
x
y
40
44
z
�1, 3, 0�: zy
� �2�3� � �6
zy
� �2y
z � 9 � y2
z � 9x2 � y2, x � 1, �1, 3, 0� 46.
Solving for x in the second equation, you obtain
or
or
Points: �0, 0�, 432�3,
431�3�
x �14
1632�3�⇒ x � 0
y �4
31�33y4 � 64y ⇒ y � 0
3�y 2�4�2 � 4y.x � y 2�4,
3y2 � 12x � 0 ⇒ y2 � 4x
9x2 � 12y � 0 ⇒ 3x2 � 4yfx � fy � 0:
fy�x, y� � �12x � 3y2fx�x, y� � 9x2 � 12y,
38.
At is undefined.
At is undefined.�1, 1�, fy
fy�x, y� ��x
�1 � x2y2
�1, 1�, fx
fx�x, y� ��y
�1 � x2y2
f �x, y� � arccos�xy� 40.
At
At �1, 1�, fy�1, 1� �89
fy�x, y� �24x3
�4x2 � 5y2�3�2
�1, 1�, fx�1, 1� �3027
�109
fx�x, y� �30y3
�4x2 � 5y2�3�2
f �x, y� �6xy
�4x2 � 5y242.
Intersecting curve:
At
x y
20
44
z
�2, 1, 8�: zx
� 2�2� � 4
zx
� 2x
z � x2 � 4
z � x2 � 4y2, y � 1, �2, 1, 8�
32.
fy
� lim�y→0
f �x, y � �y� � f �x, y�
�y� lim
�y→0
1x � y � �y
�1
x � y�y
� lim�y→0
�1
�x � y � �y��x � y� ��1
�x � y�2
fx
� lim�x→0
f �x � �x, y� � f �x, y�
�x� lim
�x→0
1x � �x � y
�1
x � y�x
� lim�x→0
�1
�x � �x � y��x � y� ��1
�x � y�2
f �x, y� �1
x � y
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318 Chapter 12 Functions of Several Variables
54.
Gz�x, y, z� �z
�1 � x2 � y2 � z2�3�2
Gy�x, y, z� �y
�1 � x2 � y2 � z2�3�2
Gx�x, y, z� �x
�1 � x2 � y2 � z2�3�2
G�x, y, z� �1
�1 � x2 � y2 � z2
56.
fz�x, y, z� � �5xy � 20yz
fy�x, y, z� � 3x2 � 5xz � 10z2
fx�x, y, z� � 6xy � 5yz
f �x, y, z� � 3x2y � 5xyz � 10yz2 58.
2z
xy� �12xy
2zy2 � �6x2 � 12y2
zy
� �6x2y � 4y3
2z
yx� �12xy
2zx2 � 12x2 � 6y2
zx
� 4x3 � 6xy2
z � x4 � 3x2y2 � y4 60.
Therefore,2z
yx�
2zxy
.
2z
xy�
1�x � y�2
2zy2 � �
1�x � y�2
zy
��1
x � y�
1y � x
2z
yx�
1�x � y�2
2zx2 � �
1�x � y�2
zx
�1
x � y
z � ln�x � y�
62.
2z
xy� 2ey � 3e�x
2zy2 � 2xey
zy
� 2xey � 3e�x
2z
yx� 2ey � 3ye�x
2zx2 � �3ye�x
zx
� 2ey � 3ye�x
z � 2xey � 3ye�x 64.
2z
xy� 2 sin�x � 2y�
2zy2 � �4 sin�x � 2y�
zy
� �2 cos�x � 2y�
2z
yx� 2 sin�x � 2y�
2zx2 � �sin�x � 2y�
zx
� cos�x � 2y�
z � sin�x � 2y� 66.
Therefore,
if x � y � 0zx
�zy
� 0
2zyx
�2z
xy.
2z
xy�
�xy�9 � x2 � y2�3�2
2zy2 �
x2 � 9�9 � x2 � y2�3�2
zy
��y
�9 � x2 � y2
2z
yx�
�xy�9 � x2 � y2�3�2
2zx2 �
y2 � 9�9 � x2 � y2�3�2
zx
��x
�9 � x2 � y2
z � �9 � x2 � y2
52.
wz
�3x
x � y
wy
��3xz
�x � y�2
wx
��x � y��3z� � 3xz
�x � y�2 �3yz
�x � y�2
w �3xz
x � y
48.
Points: �0, 0�
fy�x, y� �2y
x2 � y2 � 1� 0 ⇒ y � 0
fx�x, y� �2x
x2 � y2 � 1� 0 ⇒ x � 0 50. (a) The graph is that of
(b) The graph is that of fy.
fx.
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Section 12.3 Partial Derivatives 319
68.
There are no points for which zx � zy � 0.
2z
xy�
�x � y�2�2x� � x2�2��x � y��x � y�4 �
�2xy�x � y�3
2zy2 �
2x2
�x � y�3
zy
� �x�x � y� � xy
�x � y�2 �x2
�x � y�2
2z
yx�
�x � y�2��2y� � y2�2��x � y���1��x � y�4 �
�2xy�x � y�3
2zx2 �
2y2
�x � y�3
zx
�y�x � y� � xy
�x � y�2 ��y2
�x � y�2
z �xy
x � y70.
Therefore, fxyy � fyxy � fyyx � 0.
fyxy�x, y, z� � 0
fxyy�x, y, z� � 0
fyyx�x, y, z� � 0
fyx�x, y, z� � �3
fxy�x, y, z� � �3
fyy�x, y, z� � 0
fy�x, y, z� � �3x � 4z
fx�x, y, z� � 2x � 3y
f �x, y, z� � x2 � 3xy � 4yz � z3
72.
fyxy�x, y, z� ��12z
�x � y�4
fxyy�x, y, z� ��12z
�x � y�4
fyyx�x, y, z� ��12z
�x � y�4
fyx�x, y, z� �4z
�x � y�3
fxy�x, y, z� �4z
�x � y�3
fyy�x, y, z� �4z
�x � y�3
fy�x, y, z� ��2z
�x � y�2
fx�x, y, z� ��2z
�x � y�2
f �x, y, z� �2z
x � y74.
Therefore,
� 0.
2zx2 �
2zy2 � �sin xey � e�y
2 � � sin xey � e�y
2 �
2zy2 � sin xey � e�y
2 �
zy
� sin xey � e�y
2 �
2zx2 � �sin xey � e�y
2 �
zx
� cos xey � e�y
2 �
z � sin xey � e�y
2 �
76.
From Exercise 53, we have
2zx2 �
2zy2 �
2xy�x2 � y2�2 �
�2xy�x2 � y2�2 � 0.
z � arctan yx
78.
Therefore,2zt2 � c2 2z
x2.
2zx2 � �w2 sin�wct� sin�wx�
zx
� w sin�wct� cos�wx�
2zt2 � �w2c2 sin�wct� sin�wx�
zt
� wc cos�wct� sin�wx�
z � sin�wct� sin�wx�
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320 Chapter 12 Functions of Several Variables
80.
Therefore,zt
� c2 2zx2.
2zx2 � �
1c2 e�t sin
xc
zx
�1c
e�t cos xc
zt
� �e�t sin xc
z � e�t sin xc
82. If then to find you consider y constant and differentiatewith respect to x. Similarly, to find
you consider x constant and differentiate with respect to y.fy,
fxz � f �x, y�, 84. The plane satisfies
and
y
x
2
4
42
4
z
fy
> 0.fx
< 0
z � �x � y � f �x, y�
86. In this case, the mixed partials are equal,
See Theorem 12.3.
fxy � fyx.
88.
(a)
At
(b)
At � 60�2�0.7 � 97.47�x, y� � �1000, 500�, f
x� 601000
500 �0.7
fx
� 60x0.7y�0.7 � 60xy�
0.7
�x, y� � �1000, 500�, f
x� 140 500
1000�0.3
� 14012�
0.3
� 113.72
fx
� 140x�0.3y0.3 � 140yx�
0.3
f �x, y� � 200x0.7y0.3
90.
The rate of inflation has the greater negative influence on the growth of the investment. (See Exercise 61 in Section 12.1.)
VR�0.03, 0.28� � �1391.17
VR�I, R� � 10,000�1 � 0.10�1 � R�1 � I
9
��0.101 � I � �1000
1 � 0.10�1 � R��9
�1 � I �10
VI�0.03, 0.28� � �14,478.99
VI�I, R� � 10,000�1 � 0.10�1 � R�1 � I
9
��1 � 0.10�1 � R�
�1 � I �2 � �10,000 1 � 0.10�1 � R��10
�1 � I �11
V�I, R� � 1000�1 � 0.10�1 � R�1 � I
10
94.
(a)
(b)
(c) and The personshould consume one more unit of y because the rate ofdecrease of satisfaction is less for y.
(d)
xy
z
−2
22 1
1
Uy�2, 3� � �16.Ux�2, 3� � �17
Uy � x � 6y
Ux � �10x � y
U � �5x2 � xy � 3y292.
(a)
(b) The humidity has a greater effect on A since its coefficient is larger than that of t.�22.4
Ah
�30�, 0.80� � �22.4 � 1.20�30�� � 13.6
Ah
� �22.4 � 1.20t
At
�30�, 0.80� � 0.885 � 1.20�0.80� � 1.845
At
� 0.885 � 1.20h
A � 0.885t � 22.4h � 1.20th � 0.544
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Section 12.4 Differentials 321
98. False
Let z � x � y � 1.
100. True
102.
By the Second Fundamental Theorem of Calculus,
�f�y
�ddy�
y
x
�1 � t3 dt � �1 � y3.
� �ddx�
x
y
�1 � t3 dt � ��1 � x3 �f�x
�ddx�
y
x
�1 � t3 dt
f �x, y� � �y
x
�1 � t3 dt
Section 12.4 Differentials
2.
dz �2xy
dx �x2
y2 dy
z �x2
y4.
dw �1
z � 2y dx �
z � 2x�z � 2y�2 dy �
x � y�z � 2y�2 dz
w �x � yz � 2y
6.
dz � 2x�ex2�y2� e�x2�y2
2 � dx � 2y�ex2�y2� e�x2�y2
2 � dy � �ex2�y2� e�x2�y2��x dx � y dy�
z � �12��ex2�y2
� e�x2�y2�
8.
dw � �ey sin x dx � ey cos x dy � 2z dz
w � ey cos x � z2 10.
�2x2yz � y cos yz� dz
dw � 2xyz2 dx � �x2z2 � z cos yz� dy �
w � x2yz2 � sin yz
12. (a)
(b)
�x dx � y dy�x2 � y2
�0.05 � 2�0.1�
�5� 0.11180
dz �x
�x2 � y2 dx �
y�x2 � y2
dy
�z � 0.11180
f �1.05, 2.1� � �5.5125 � 2.3479
f �1, 2� � �5 � 2.2361 14. (a)
(b)
� e2�0.05� � e2�0.1� � 1.1084
dz � ey dx � xey dy
�z � 1.1854
f �1.05, 2.1� � 1.05e2.1 � 8.5745
f �1, 2� � e2 � 7.3891
96. (a)
�2z�y2 � 0.014
�z�y
� 0.014y � 0.54
�2z�x2 � �1.55
�z�x
� �1.55x � 22.15 (b) Concave downward
The rate of increase of Medicareexpenses is declining withrespect to worker’s compensationexpenses �x�.
�z�
��2z�x2 < 0� (c) Concave upward
The rate of increase of Medicareexpenses is increasing withrespect to public assistanceexpenses �y�.
�z�
��2z�y2 > 0�
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322 Chapter 12 Functions of Several Variables
16. (a)
(b)
�12
�0.05� �14
�0.1� � 0
dz �1y dx �
xy2 dy
�z � 0
f �1.05, 2.1� �1.052.1
� 0.5
f �1, 2� �12
� 0.5
18. Let Then:
�2.03�2�1 � 8.9�3 � 22�1 � 9�3 � 2�2��1 � 9�3�0.03� � 3�2�2�1 � 9�2��0.1� � 0
dz � 2x�1 � y�3 dx � 3x2�1 � y�2 dydy � �0.1.dx � 0.03,y � 9,x � 2,z � x2�1 � y�3,
20. Let Then:
sin�1.05�2 � �0.95�2 � sin 2 � 2�1� cos�12 � 12��0.05� � 2�1� cos�12 � 12���0.05� � 0
dz � 2x cos�x2 � y2� dx � 2y cos�x2 � y2� dydy � �0.05.dx � 0.05,x � y � 1,z � sin�x2 � y2�,
22. In general, the accuracy worsens as and increase.�y�x 24. If then is the propagated error,
and is the relative error.�zz
�dzz
�z � dzz � f �x, y�,
26.
∆r
r
∆h
2 rhdrπ
∆ −V dVr dh2π
dV � 2�rh dr � �r2 dh
V � �r2h
28.
S�8, 20� � 541.3758
��
�r2 � h2�2r 2 � h2� dr � �rh� dh
dS � �2r 2 � h2
�r 2 � h2 dr � �
rh�r 2 � h2
dh
dSdh
� �r�r2 � h2��1�2h � �rh
�r 2 � h2
�� �r 2 � h2� � �r 2
�r 2 � h2�1�2 � �2r2 � h2
�r2 � h2
dSdr
� ��r2 � h2�1�2 � �r2�r2 � h2��1�2
r � 8, h � 20
20
8
S � �r�r2 � h2
0.1 0.1 10.0341 10.0768 0.0427
0.1 5.3671 5.3596
0.001 0.002 0.12368 0.12368
0.0002 �0.286 � 10�7�0.00303�0.00303�0.0001
0.683 � 10�5
�0.0075�0.1
�S � dS�SdS�h�r
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Section 12.4 Differentials 323
30.
Maximum propagated error
dCC
�±4.25
�30.24� ±0.14
� ±2.79 ± 1.46 � ±4.25
� �0.1516231�2 � 0.0204��8 � 91.4��±3� � 0.0817�3.71�23 � 5.81 � 0.25�23���±1�
dC � Cv dv � CT dT
�C�T
� 0.0817�3.71�v � 5.81 � 0.25v�
� �0.1516v1�2 � 0.0204 �T � 91.4�
�C�v
� 0.0817��3.71�12
v�1�2 � 0.25 �T � 91.4�
32.
Using the worst case scenario, and you see that
�d� ≤ 0.00194 � 0.00515 � 0.0071.
dy � 0.05,dx � �0.05
��3.2
8.52 � 3.22 dx �8.5
8.52 � 3.22 dy ��y
x2 � y2 dx �x
x2 � y2 dy
� arctan�yx� ⇒ d �
�yx2
1 � �yx�
2 dx �
1x
1 � �yx�
2 dy
�dr� ≤ �1.288��0.05� � 0.064
� 0.9359 dx � 0.3523 dy �8.5
�8.52 � 3.22 dx �
3.2�8.52 � 3.22
dy
r � �x2 � y2 ⇒ dr �x
�x2 � y2 dx �
y�x2 � y2
dy
�dy� ≤ 0.05�dx� ≤ 0.05,�x, y� � �8.5, 3.2�,
34.
Note: The maximum error will occur when dv and dr differ in signs.
daa
� 2dvv
�drr
� 2�0.03� � ��0.02� � 0.08 � 8%
da �2vr
dv �v 2
r 2 dr
a �v 2
r
36. (a) Using the Law of Cosines:
(b)
�12
11512.79�1�2±1774.79 � ±8.27 ft
�12�3302 � 4202 � 840�330��cos
�
20� �1�2
��2�330� � 840 cos �
20��6� � 840�330��sin �
20���
180�
da �12�b2 � 4202 � 840b cos
�1�2
�2b � 840 cos � db � 840b sin d
a � �b2 � 4202 � 2b�420�cos
a � 107.3 ft.
� 3302 � 4202 � 2�330��420�cos 9
a2 � b2 � c2 � 2bc cos A330 ft
420 ft
9˚
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324 Chapter 12 Functions of Several Variables
38.
When and we have �R �152
�10 � 15�2 �0.5� �102
�10 � 15�2 ��2� � �0.14 ohm.R2 � 15,R1 � 10
�R � dR ��R�R1
dR ��R�R2
dR2 �R2
2
�R1 � R2�2 �R1 �R1
2
�R1 � R2�2 �R2
dR2 � �R2 � �2
dR1 � �R1 � 0.5
R �R1R2
R1 � R2
1R
�1R1
�1R2
40.
When and we have �T � ��
32.09� 2.532.09
�0.15� ��
��2.5��32.09���0.02� � �0.0111 sec.L � 2.5,g � 32.09
�T � dT ��T�g
dg ��T�L
dL � ��
g�Lg
�g ��
�Lg �L
dL � �L � 2.48 � 2.5 � �0.02
dg � �g � 32.24 � 32.09 � 0.15
T � 2��Lg
42.
As and �2 → 0.��x, �y� → �0, 0�, �1 → 0
� fx�x, y� �x � fy�x, y� �y � �1�x � �2�y where �1 � �x and �2 � �y.
� 2x��x� � 2y��y� � �x��x� � �y��y�
� x2 � 2x��x� � ��x�2 � y2 � 2y��y� � ��y�2 � �x2 � y2�
�z � f �x � �x, y � �y� � f �x, y�
z � f �x, y� � x2 � y2
44.
As and �2 → 0.��x, �y� → �0, 0�, �1 → 0
� fx�x, y� �x � fy�x, y� �y � �1�x � �2�y where �1 � 0 and �2 � 3y��y� � ��y�2.
� 5��x� � �3y2 � 10���y� � 0��x� � �3y��y� � ��y�2� �y
� 5x � 5�x � 10y � 10�y � y3 � 3y2��y� � 3y��y �2 � ��y�3 � �5x � 10y � y3�
�z � f �x � �x, y � �y� � f �x, y�
z � f �x, y� � 5x � 10y � y3
46.
(a)
Thus, the partial derivatives exist at �0, 0�.
fy�0, 0� � lim�y→0
f �0, �y� � f �0, 0�
�y� lim
�y→0 0 � 0
�y� 0
fx�0, 0� � lim�x→0
f ��x, 0� � f �0, 0�
�x� lim
�x→0 0 � 0
�x� 0
f �x, y� � � 5x2yx3 � y3
,
0 ,
�x, y� � �0, 0�
�x, y� � �0, 0�
(b) Along the line
Along the line
Thus, f is not continuous at Therefore f is not differentiable at
(See Theorem 12.5)
�0, 0�.�0, 0�.
x � 0, lim�x, y�→�0, 0�
f �x, y� � 0.
y � x: lim�x, y�→�0, 0�
f �x, y� � limx→0
5x3
2x3 �52
.
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Section 12.5 Chain Rules for Functions of Several Variables
Section 12.5 Chain Rules for Functions of Several Variables 325
2.
��x sin t � yet
�x2 � y2�
�cos t sin t � e2t
�cos2 t � e2t
dw
dt�
x�x2 � y2
��sin t� �y
�x2 � y2 et
x � cos t, y � et
w � �x2 � y2 4.
� tan t � cot t �1
sin t cos t
dwdt
� ��1x ���sin t� � �1
y��cos t�
y � sin t
x � cos t
w � ln yx
6.
(a)
(b)dwdt
� �2t sin�t2 � 1�w � cos�t2 � 1�,
� �2t sin�x � y� � �2t sin�t2 � 1�
dwdt
� �sin�x � y��2t� � sin�x � y��0�
y � 1x � t2,w � cos�x � y�,
8.
z � arccos t
y � t2
x � t
w � xy cos z
(a)
(b)dwdt
� 4t3w � t 4,
� t2�t � � t�t��2t� � t�t2��1 � t2� �1
�1 � t2� � t3 � 2t3 � t3 � 4t3
dwdt
� � y cos z��1� � �x cos z��2t� � ��xy sin z���1
�1 � t2�
10.
(a)
(b)
dwdt
� �2t3���e�t� � �e�t��6t2� � 2t2e�t��t � 3�
w � �t2��2t��e�t � � 2t3e�t
� 2t2e�t�2 � 1 � t� � 2t2e�t�3 � t�
� �2t��e�t ��2t� � �t2��e�t ��2� � �t2��2t���e�t �
dwdt
� yz�2t� � xz�2� � �xy���e�t �
z � e�ty � 2t,x � t2,w � xyz,
12.
f��t� � 48�8 � 2�2 � 2�6 � f��1�
� 48t�8 � 2�2 � 2�6
Distance � f �t� � ��x2 � x1�2 � �y2 � y1�2 � ��48 � ��3 � �2��2� �48t�1 � �2��2
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326 Chapter 12 Functions of Several Variables
14.
At d 2wdt 2 �
4�24� � �7��4�16
�6816
� 4.25t � 1:
d 2wdt 2 �
�t � 1�2�12t3 � 12t2� � �3t4 � 4t3�2�t � 1��t � 1�4
�3t4 � 4t3
�t � 1�2
��t � 1��4t3� � t4
�t � 1�2
�2t 2�2t�t � 1
�t 4
�t � 1�2
�2xy
�2t� ��x2
y2 �1�
dwdt
��w�x
dxdt
��w�y
dydt
t � 1
y � t � 1,
x � t2,
w �x2
y, 16.
When and and �w�t
� 3e�e2 � 1�.�w�s
� �6et � 1,s � 0
� 3et�e2t � e2s�
�w�t
� �6xy�0� � �3y2 � 3x2��et �
�w�s
� �6xy�es� � �3y2 � 3x2��0� � �6e2s� t
y � et
x � es
w � y3 � 3x2y
18.
When and and �w�t
� 0.�w�s
� 0t ��
2,s � 0
� �cos�2x � 3y� � �cos�5s � t�
�w�t
� 2 cos�2x � 3y� � 3 cos�2x � 3y�
� 5 cos�2x � 3y� � 5 cos�5s � t�
�w�s
� 2 cos�2x � 3y� � 3 cos�2x � 3y�
y � s � t
x � s � t
w � sin�2x � 3y�
20.
(a)
(b)
�w��
� 0 �w�r
��5r
�25 � 5r2;
w � �25 � 5r2
��5r2 sin2 � cos � � 5r2 sin � cos �
�25 � 5x2 � 5y2� 0
�w��
��5x
�25 � 5x2 � 5y2 ��r sin �� �
�5y�25 � 5x2 � 5y2
�r cos ��
��5r cos2 � � 5r sin2 ��25 � 5x2 � 5y2
��5r
�25 � 5r2
�w�r
��5x
�25 � 5x2 � 5y2 cos � �
�5y�25 � 5x2 � 5y2
sin �
w � �25 � 5x2 � 5y2, x � r cos �, y � r sin �
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Section 12.5 Chain Rules for Functions of Several Variables 327
22.
(a)
(b)
�w��
��2r2
�3
�w�r
�2r�2
w �yzx
��r � ���r � ��
�2 �r2
�2 � 1
�2��2 � r2�
�3 �2�
��2r2
�3
���r � ���r � ��
�4 �2�� ��r � �� � �r � ��
�2
�w��
��yzx2 �2�� �
zx
�1� �yx
��1�
�w�r
��yzx2 �0� �
zx
�1� �yx
�1� �z � y
x�
2r�2
w �yzx
, x � �2, y � r � �, z � r � � 24.
� �6s2t2 � 2s3t� sin�st2 � 2t3�
� �2s2t�s � 2t� sin�st2 � 2t3� � 2s2t2 sin�st2 � 2t3�
�w�t
� cos�yz��0� � xz sin�yz��2t� � xy sin�yz���2�
� cos�st2 � 2t3�2s � s2t2 sin�st2 � 2t3�
�w�s
� cos�yz��2s� � xz sin�yz��0� � xy sin�yz��1�
w � x cos yz, x � s2, y � t2, z � s � 2t
26.
� 2t sin2 s � 2t cos2 s � 4s2t3 � 2t � 4s2t3
�w�t
� 2x sin s � 2y cos s � 2z�2st�
� 2t2 sin s cos s � 2t2 sin s cos s � 2st4 � 2st4
�w�s
� 2x � cos s � 2y��t sin s� � 2z�t2�
w � x2 � y2 � z2, x � t sin s, y � t cos s, z � st2 28.
dydx
� �Fx�x, y�Fy�x, y� � �
�sin x � y sec2 xyx sec2 xy
cos x � tan xy � 5 � 0
30.
�y 2 � x 2
2xy � 2yx4 � 4x2y3 � 2y5
�y 2 � x 2
2xy � 2y�x 2 � y 2�2
� ��y 2 � x 2��x 2 � y 2�2
��2xy��x 2 � y 2�2 � 2y
dydx
� �Fx�x, y�Fy�x, y�
xx2 � y2 � y2 � 6 � 0 32.
�z�y
� �Fy
Fz
� �x � zx � y
�z�x
� �Fx
Fz
� �y � zx � y
Fz � x � y
Fy � z � x
Fx � z � y
F�x, y, z� � xz � yz � xy
34.
�z�y
� �Fy
Fz
�ex cos� y � z�
1 � ex cos� y � z�
�z�x
� �Fx
Fz
�ex sin�y � z�
1 � ex cos�y � x�
Fz � ex cos�y � z� � 1
Fy � ex cos�y � z�
Fx � ex sin�y � z�
F�x, y, z� � ex sin�y � z� � z 36.
(i) implies
(ii) implies�z�y
� �1.�1 ��z�y� cos�y � z� � 0
�z�x
� �1
cos�y � z� � �sec�y � z�.
1 ��z�x
cos�y � z� � 0
x � sin�y � z� � 0
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328 Chapter 12 Functions of Several Variables
38.
(i)
(ii)�z�y
��Fy�x, y, z�Fz�x, y, z� � �
xy
� 2yz
y2 � 2z� �
x � 2y2zy3 � 2yz
�z�x
��Fx�x, y, z�Fz�x, y, z� �
�ln yy2 � 2z
x ln y � y2z � z2 � 8 � 0 40.
�w
�z� �
Fz
Fw
�2z
5y � 20w
�w
�y� �
Fy
Fw
�5w � 2y
20w � 5y
�w�x
� �Fx
Fw
��2x
�5y � 20w�
2x5y � 20w
Fx � 2x, Fy � 2y � 5w, Fz � 2z, Fw � �5y � 20w
x2 � y2 � z2 � 5yw � 10w2 � 2 � F�x, y, z, w�
42.
�w�z
��Fz
Fw
��1
2�y � z
��1
2�x � y�
1
2�y � z
�w�y
��Fy
Fw
��12
�x � y��12 �12
�y � z��12
�w�x
��Fx
Fw
�12
�x � y��12
1�
1
2�x � y
F�x, y, z, w� � w � �x � y � �y � z � 0 44.
Degree: 3
� 3x3 � 9xy2 � 3y3 � 3f �x, y�
x fx�x, y� � y fy�x, y� � x�3x2 � 3y2� � y��6xy � 3y2�
� t3�x3 � 3xy2 � y3� � t3f �x, y�
f �tx, ty� � �tx�3 � 3�tx��ty�2 � �ty�3
f �x, y� � x3 � 3xy2 � y3
46.
Degree: 1
�x2
�x2 � y2� f �x, y�
�x4 � x2y2
�x2 � y2�32 �x2�x2 � y2��x2 � y2�32
x fx�x, y� � y fy�x, y� � x x3 � 2xy2
�x2 � y2�32� � y �x2y�x2 � y2�32�
f �tx, ty� ��tx�2
��tx�2 � �ty�2� t� x2
�x2 � y2� � tf �x, y�
f �x, y� �x2
�x2 � y2
48.
(Page 878) �w�t
��w�x
�x�t
��w�y
�y�t
�w�s
��w�x
�x�s
��w�y
�y�s 50.
�z�y
� � fy�x, y, z�fz�x, y, z�
�z�x
� � fx�x, y, z�fz�x, y, z�
dydx
� � fx�x, y�fy�x, y�
52. (a)
(b)
dSdt
� 2��2r � h� drdt
� rdhdt� � 2� ��24 � 36��6� � 12��4�� � 624� in.2min
S � 2�r�r � h�
dVdt
� ��2rhdrdt
� r 2dhdt � � �r�2h
drdt
� rdhdt � � ��12��2�36��6� � 12��4�� � 4608� in.3min
V � �r2h
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Section 12.5 Chain Rules for Functions of Several Variables 329
54. (a)
(b)
� 320�2� cm2min
��25 � 15�2 � 102 � �25 � 15� 25 � 15
��25 � 15�2 � 102��4� � �25 � 15� 10
��25 � 15�2 � 102�12��
� � ��25 � 15�2 � 102 � �25 � 15� 25 � 15
��25 � 15�2 � 102��4� �
�R � r� h
��R � r�2 � h2 dhdt�
dSdt
� � ��R � r�2 � h2 � �R � r� �R � r���R � r�2 � h2�dr
dt� ��R � r�2 � h2 � �R � r� �R � r�
��R � r�2 � h2�dRdt
�
S � � �R � r���R � r�2 � h2
��
3�19,500� � 6,500� cm3min
��
3 �2�15� � 25��10��4� � �15 � 2�25���10��4� � ��15�2 � �15��25� � �25�2��12��
dVdt
��
3 �2r � R�h drdt
� �r � 2R�h dRdt
� �r2 � rR � R2�dhdt�
V ��
3�r2 � rR � R2�h
56.
dTdt
�1
mRVdpdt
� pdVdt �
T �1
mR�pV �
pV � mRT 58.
Let then
and
Now, let and we have Thus,
�f�x
x ��f�y
y � nf �x, y�.
v � y.u � x,t � 1
g��t� � ntn�1f �x, y�.
g��t� ��f�u
dudt
��f�v
dvdt
��f�u
x ��f�v
y
v � yt,u � xt,
g�t� � f �xt, yt� � t n f �x, y�
60.
�w�x
��w�y
� 0
�w�y
� �x � y� cos�y � x� � sin�y � x�
�w�x
� ��x � y� cos�y � x� � sin�y � x�
w � �x � y� sin�y � x�
62.
Therefore, ��w�x �
2
� ��w�y �
2
� ��w�r �
2
�1r2 ��w
���2
.
��w�r �
2
� � 1r2���w
���2
� 0 �1r2 �1� �
1r2
��w�x �
2
� ��w�y �
2
�y2
�x2 � y2�2 �x2
�x2 � y2�2 �1
x2 � y2 �1r2
�w�x
��y
x2 � y2, �w�y
�x
x2 � y2, �w�r
� 0, �w��
� 1
� arctan� r sin �r cos �� � arctan�tan �� � � for �
�
2 < � <
�
2
w � arctan yx, x � r cos �, y � r sin �
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Section 12.6 Directional Derivatives and Gradients
64. Note first that
Thus,
Thus,�v�r
� �1r �u��
.
�u��
�x
x2 � y2��r sin �� �y
x2 � y2 �r cos �� ��r 2 sin � cos � � r 2 sin � cos �
r 2 � 0
�v�r
��y
x2 � y2 cos � �x
x2 � y2 sin � ��r sin � cos � � r sin � cos �
r 2 � 0
�u�r
�1r �v��
.
�v��
��y
x2 � y2 ��r sin �� �x
x2 � y2 �r cos �� �r 2 sin2 � � r 2 cos2 �
r 2 � 1
�u�r
�x
x2 � y2 cos � �y
x2 � y2 sin � �r cos2 � � r sin2 �
r 2 �1r
�u�y
� ��v�x
�y
x 2 � y 2.
�u�x
��v�y
�x
x 2 � y 2
2.
Du f �4, 3� � �f �4, 3� � u � 24�2 �272�2 �
212�2
u �v
�v��
�22
i ��22
j
�f �4, 3� � 48i � 27j
�f �x, y� � 3x2i � 3y2j
f �x, y� � x3 � y3, v ��2
2�i � j� 4.
Du f �1, 1� � �f �1, 1� � u � 1
u �v
�v�� �j
�f �1, 1� � i � j
�f �x, y� �1y
i �xy2 j
v � �j
f �x, y� �xy
330 Chapter 12 Functions of Several Variables
6.
Dug�1, 0� � �g�1, 0� � u ��5�26
��5�26
26
u �v
�v��
1�26
i �5
�26j
�g�1, 0� � �j
�g�x, y� ��y
�1 � �xy�2i �
�x�1 � �xy�2
j
g�x, y� � arccos xy, v � i � 5j 8.
Du h�0, 0� � �h�0, 0� � u � 0
�h�0, 0� � 0
�h � �2xe��x2�y2�i � 2ye��x2�y2�j
v � i � j
h�x, y� � e��x2�y2�
10.
Du f �1, 2, �1� � �f �1, 2, �1� � u � �67�14
u �v
�v��
1
�14i �
2
�14j �
3
�14k
�f �1, 2, �1� � 2i � 4j � 2k
�f � 2x i � 2yj � 2zk
v � i � 2j � 3k
f �x, y, z� � x2 � y2 � z2 12.
Du h�2, 1, 1� � �h�2, 1, 1� � u �83
u �v
�v��
23
i �13
j �23
k
�h�2, 1, 1� � i � 2j � 2k
�h � yz i � xz j � xyk
v � �2, 1, 2�
h�x, y, z� � xyz
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14.
� �1
2�x � y�2 ��3y � x�
Du f � �f � u � ��3y
2�x � y�2 �x
2�x � y�2
�f � �y
�x � y�2 i �x
�x � y�2 j
u ��32
i �12
j
f �x, y� �y
x � y16.
Dug � �12
ey ��32
xey �ey
2��3x � 1�
�g � ey i � xey j
u � �12
i ��32
j
g�x, y� � xey
18.
At Du f � 0.�0, �,
�1
�5 sin�x � y� �
�55
sin�x � y�
Du f � �1
�5 sin�x � y� �
2
�5 sin�x � y�
u �v
�v��
1
�5i �
2
�5j
�f � �sin�x � y�i � sin�x � y�j
v �
2i � j
f �x, y� � cos�x � y� 20.
At
Du g � �g � u � �4
�5�
4
�5� �
8
�5
u �v
�v�� �
1
�5i �
2
�5j
�g � 4i � 2j � 8k.�2, 4, 0�,
�g � yez i � xez j � xyez k
v � �2i � 4j
g�x, y, z� � xyez
Section 12.6 Directional Derivatives and Gradients 331
22.
�g�2, 0� � 2i � 2j
�g�x, y� � ��2yx
eyx � 2eyxi � 2eyxj
g�x, y� � 2xeyx 24.
�z�2, 3� � 4i � j
�z�x, y� �2x
x2 � yi �
1x2 � y
j
z � ln�x2 � y�
26.
�w�4, 3, �1� � tan 2i � 4 sec2 2j � 4 sec2 2k
�w�x, y, z� � tan�y � z�i � x sec2�y � z�j � x sec2�y � z�k
w � x tan�y � z�
28.
Du f � �f � u � �36�53
�14�53
� �50�53
� �50�53
53
�f �x, y� � 6xi � 2yj, �f �3, 1� � 18i � 2j
PQ\
� �2i � 7j, u � �2
�53i �
7�53
j
30.
Du f � �f � u �2�5
�2�5
5
�f �0, 0� � 2i
�f �x, y� � 2 cos 2x cos yi � sin 2x sin yj
PQ\
�
2i � j, u �
1�5
i �2�5
j
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32.
��3�2 2 � 2�3 � 3�
6� �h�0,
3 � ��3 2
36�
9 � 6�3 � 3 2
36
�h�0,
3 ��3
6i � �3 � �3
6 j
�h�x, y� � �y sin�x � y�i � �cos�x � y� � y sin�x � y�� j
h�x, y� � y cos�x � y�
34.
��g�0, 5�� � 1
�g�0, 5� � j
�g�x, y� � �2xye�x2i � e�x2 j
g�x, y� � ye�x2 36.
��w�0, 0, 0�� � 0
�w�0, 0, 0� � 0
�w �1
��1 � x2 � y2 � z2�3�xi � yj � zk�
w �1
�1 � x2 � y2 � z2
38.
��w�2, 1, 1�� � �33
�w�2, 1, 1� � i � 4j � 4k
�w � y2z2i � 2xyz2j � 2xy2zk
w � xy2z2
332 Chapter 12 Functions of Several Variables
For Exercises 40 – 46, and D� f x, y� � ��13� cos � � �1
2� sin �.f x, y� � 3 �x3
�y2
40. (a)
(b) D23 f �3, 2� � ��13��
12 � �1
2�32
�2 � 3�3
12
D4 f �3, 2� � ��13�22
� �12�22
� �5�212
42. (a)
(b)
Du f � �f � u �15
�25
�35
u � �35
i �45
j
�v� � �9 � 16 � 5
v � �3i � 4j
� ��13
1
�2� �1
21
�2� �
5�212
Du f � �f � u
u � � 1
�2�i � j� 44. �f � ��13i � �1
2j
46.
Therefore, and is the direction of greatest rate of change of Hence, in a direction orthogonal to the rate of change of is 0.f�f,
f.�f�f � u � 0.Du f �3, 2� �u � �1�13 ��3i � 2j�
�f��f �
�1
�13��2i � 3j�
�f � �13
i �12
j
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For Exercises 48 and 50, and D� f x, y� � �2x cos � � 2y sin � � �2 x cos � � y sin ��.f x, y� � 9 � x2 � y2
48. (a)
(b) D3 f �1, 2� � �2�12
� �3 � ��1 � 2�3 �
D�4 f �1, 2� � �2��22
� �2 � �2 50.
Therefore,
and
Du f �1, 2� � �f �1, 2� � u � 0.
u � �1�5���2i � j�
�f �1, 2���f �1, 2�� �
1
�5��i � 2j�
�f �1, 2� � �2i � 4j
52. (a) In the direction of the vector
(b)
(Same direction as in part (a).)
(c) the direction opposite that of the gradient.��f � �
12 i �
12 j,
�f �1, 2� �12
i �12
j
�f �12
y 1
2�xi �
12�x j �
y
4�xi �
12�x j
i � j.
(b)
x
2
1
3
4
1 2−1−2
y
�f ��3, 2� ���3
2i
�f ��16xy
�1 � x2 � y2�2 i �8 � 8x2 � 8y2
�1 � x2 � y2�2 j
Section 12.6 Directional Derivatives and Gradients 333
56.
�f �0, 0� � �2i � 3j
0 � 2x � 3y
6 � 2x � 3y � 6
�f �x, y� � �2i � 3j
P � �0, 0�c � 6,
f �x, y� � 6 � 2x � 3y
54. (a)
Circle: center: radius:
(c) The directional derivative of is 0 in the directions (d)
−6
−6
6
6yx
z± j.f
�3�0, 2�,
� y � 2�2 � x2 � 3
4 � y2 � 4y � 4 � x2 � 1
⇒ 4y � 1 � x2 � y2
f �x, y� �8y
1 � x2 � y2 � 2
58.
�f ��1, 3� � 3i � j
xy � �3
�f �x, y� � y i � xj
P � ��1, 3�c � �3,
f �x, y� � xy
http://librosysolucionarios.net
334 Chapter 12 Functions of Several Variables
74.
3x 2
16� e�4t � y ⇒ u �
316
x 2
y�t� � 3e�4tx�t� � 4e�2t
3 � y�0� � C24 � x�0� � C1
y�t� � C2e�4tx�t� � C1e
�2t
dydt
� �4ydxdt
� �2x
P � �4, 3�T�x, y� � 100 � x2 � 2y2,
64.
or
5�h � ��5i � 12j�
�h�500, 300� � �i � 2.4j
�h � �0.002x i � 0.008y j
h�x, y� � 5000 � 0.001x2 � 0.004y2 66. The directional derivative gives the slope of a surface at apoint in an arbitrary direction u � cos � i � sin � j.
76. (a)
(b)
There will be no change in directions perpendicular tothe gradient:
(c) The greatest increase is in the direction of the gradi-ent: �3i �
12 j
± �i � 6j�
�T �3, 5� � 400e�7��3i �12 j�
�T �x, y� � 400e��x2�y��2���x�i �12 j�
6
500
yx 6
z
78. False
when
u � �cos �
4� i � �sin �
4� j.
Du f �x, y� � 2 > 1
80. True
68. See the definition, page 887. 70. The gradient vector is normal to the level curves.
See Theorem 12.12.
72. The wind speed is greatest at A.
60.
�1313
�3i � 2j�
�f �1, 1��f �1, 1� �
1
13�3i � 2j�
�f �1, 1� � 6i � 4j
�f �x, y� � 6xi � 4y j
f �x, y� � 3x2 � 2y21
1−1
−1
x
y3x2 � 2y2 � 1 62.
�1717
�i � 4j�
�f �5, 0��f �5, 0� �
1
17�i � 4j�
�f �5, 0� � i � 4j
�f �x, y� � ey i � �xey � 1�j
f �x, y� � xey � y
x2 4 6
2
4
6
yxey � y � 5
Section 12.7 Tangent Planes and Normal Lines
2.
Sphere, radius 5, centered at origin.
x2 � y2 � z2 � 25
F�x, y, z� � x2 � y2 � z2 � 25 � 0 4.
Hyperbolic paraboloid 16x2 � 9y2 � 144z � 0
F�x, y, z� � 16x2 � 9y2 � 144z � 0
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Section 12.7 Tangent Planes and Normal Lines 335
6.
�1
11�3i � j � k� �
1111
�3i � j � k�
n ��F
�F �
1
44�6i � 2j � 2k�
�F�3, 1, 1� � 6i � 2j � 2k
�F�x, y, z� � 2xi � 2yj � 2zk
F�x, y, z� � x2 � y2 � z2 � 11 8.
�145145
�12i � k�
n ��F
�F �
1
145�12i � k�
�F�2, 1, 8� � 12i � k
�F�x, y, z� � 3x2i � k
F�x, y, z� � x3 � z
10.
n ��F
�F �
113
�4i � 3j � 12k�
�F�2, �1, 2� � 4i � 3j � 12k
�F�x, y, z� � 2xi � 3j � 3z2k
F�x, y, z� � x2 � 3y � z3 � 9 12.
n ��F
�F �
117
�12i � 12j � k�
�F�2, 2, 3� � 12i � 12j � k
�F�x, y, z� � 2xzex2�y2i � 2yzex2�y2j � ex2�y2k
F�x, y, z� � zex2�y2� 3
14.
�1010
�3 i � 3 j � 2k�
�1
10�3 i � 3 j � 2k�
n ��F
�F �
2
10 �32
i �32
j � k��F��
3,
�
6, �
32� �
32
i �32
j � k
�F�x, y, z� � cos�x � y�i � cos�x � y�j � k
F�x, y, z� � sin�x � y� � z � 2
18.
y � z � 0
Gz�1, 0, 0� � �1 Gy�1, 0, 0� � 1 Gx�1, 0, 0� � 0
Gz�x, y, z� � �1Gy�x, y, z� �1�x
1 � �y2�x2� �x
x2 � y2Gx�x, y, z� ���y�x2�
1 � �y2�x2� ��y
x2 � y2
G�x, y, z� � arctan yx
� z
g�x, y� � arctan yx, �1, 0, 0�
16.
2x � y � z � 2
�2x � y � z � 2 � 0
�2�x � 1� � �y � 2� � �z � 2� � 0
Fz�1, 2, 2� � �1 Fy�1, 2, 2� � 1 Fx�1, 2, 2� � �2
Fz�x, y, z� � �1Fy�x, y, z� �1x
Fx�x, y, z� � �yx2
F�x, y, z� �yx
� z
f �x, y� �yx, �1, 2, 2�
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336 Chapter 12 Functions of Several Variables
20.
2x � 3y � 3z � 6
�23 x � y � z � 2 � 0
�23 �x � 3� � �y � 1� � �z � 1� � 0
Fz�x, y, z� � �1Fy�x, y, z� � �1,Fx�x, y, z� � �23 ,
F�x, y, z� � 2 �23 x � y � z
f �x, y� � 2 �23 x � y, �3, �1, 1�
22.
2x � 2y � z � �1
�2x � 2y � z � 1 � 0
�2�x � 1� � 2�y � 2� � �z � 1� � 0
Fz�1, 2, 1� � �1Fy�1, 2, 1� � 2 Fx�1, 2, 1� � �2
Fz�x, y, z� � �1Fy�x, y, z� � �2x � 2y Fx�x, y, z� � 2x � 2y
F�x, y, z� � x2 � 2xy � y2 � z
z � x2 � 2xy � y2, �1, 2, 1�
24.
42y � 8z � 2�� � 4�
�22
y � z �2�
8�
22
� 0
�22 �y �
�
4� � �z �22 � � 0
Hz�5, �
4, 22 � � �1Hy�5,
�
4, 22 � � �
22
Hx�5, �
4, 22 � � 0
Hz�x, y, z� � �1 Hy�x, y, z� � �sin y Hx�x, y, z� � 0
H�x, y, z� � cos y � z
h�x, y� � cos y, �5, �
4, 22 �
26.
x � 3y � 4z � 0
�x � 1� � 3�y � 3� � 4�z � 2� � 0
2�x � 1� � 6�y � 3� � 8�z � 2� � 0
Fz�1, 3, �2� � �8Fy�1, 3,�2� � �6Fx�1, 3, �2� � 2
Fz�x, y, z� � 4zFy�x, y, z� � �2y Fx�x, y, z� � 2x
F�x, y, z� � x2 � y2 � 2z2
x2 � 2z2 � y2, �1, 3, �2�
28.
�x � y � 8z � 16
x � y � 8z � �16
�x � 4� � 1�y � 4� � 8�z � 2� � 0
Fz�4, 4, 2� � �8Fy�4, 4, 2� � �1 Fx�4, 4, 2� � 1
Fz�x, y, z� � �2yFy�x, y, z� � �2z � 3 Fx�x, y, z� � 1
F�x, y, z� � x � 2yz � 3y
x � y�2z � 3�, �4, 4, 2�
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Section 12.7 Tangent Planes and Normal Lines 337
30.
Direction numbers: 1, 2, 2
Plane:
Line:x � 1
1�
y � 22
�z � 2
2
�x � 1� � 2�y � 2� � 2�z � 2� � 0, x � 2y � 2z � 9
Fz�1, 2, 2� � 4 Fy�1, 2, 2� � 4 Fx�1, 2, 2� � 2
Fz�x, y, z� � 2zFy�x, y, z� � 2yFx�x, y, z� � 2x
F�x, y, z� � x2 � y2 � z2 � 9
x2 � y2 � z2 � 9, �1, 2, 2�
32.
Direction numbers:
Plane
5x � 13y � 12z � 0
5�x � 5� � 13�y � 13� � 12�z � 12� � 0
5, �13, �12
Fz�x, y, z� � �24Fy�5, 13, �12� � �26Fx�5, 13, �12� � 10
Fz�x, y, z� � 2zFy�x, y, z� � �2yFx�x, y, z� � 2x
F�x, y, z� � x2 � y2 � z2
x2 � y2 � z2 � 0, �5, 13, �12�
34.
Direction numbers: 10, 5, 2
Plane:
Line:x � 1
10�
y � 25
�z � 5
2
10�x � 1� � 5�y � 2� � 2�z � 5� � 0, 10x � 5y � 2z � 30
Fz�1, 2, 5� � 2 Fy�1, 2, 5� � 5 Fx�1, 2, 5� � 10
Fz�x, y, z� � xyFy�x, y, z� � xzFx�x, y, z� � yz
F�x, y, z� � xyz � 10
xyz � 10, �1, 2, 5�
Line:x � 5
5�
y � 13�13
�z � 12�12
36. See the definition on page 897. 38. For a sphere, the common object is the center of thesphere. For a right circular cylinder, the common object isthe axis of the cylinder.
40.
(a)
Direction numbers: 1, 4,
(b) cos � � ��F � �G��F �G
�3
212�
3
42�
4214
; not orthogonal
�4, x � 2
1�
y � 14
�z � 5�4
�F �G � � i40
j�2�1
k�1�1� � i � 4j � 4k
�G�2, �1, 5� � �j � k�F�2, �1, 5� � 4i � 2j � k
�G�x, y, z� � �j � k �F�x, y, z� � 2xi � 2yj � k
G�x, y, z� � 4 � y � z F�x, y, z� � x2 � y2 � z
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338 Chapter 12 Functions of Several Variables
42.
Direction numbers:
Tangent line
Not orthogonalcos � � ��F � �G��F �G
���8�5�238
��8
576
x � 31
�y � 4�17
�z � 5�13
1, �17, �13
�F �G � � i3�55
j4�5�2
k�13 � �
25
i �345
j �265
k
�G�3, 4, 5� � 5i � 2j � 3k �F�3, 4, 5� �35
i �45
j � k
�G�x, y, z� � 5i � 2j � 3k �F�x, y, z� �x
x2 � y2i �
yx2 � y2
j � k
G�x, y, z� � 5x � 2y � 3z � 22 F�x, y, z� � x2 � y2 � z
44.
(a)
Direction numbers: 25,
(b) cos � � ��F � �G��F �G
� 0; orthogonal
�13, �2, x � 1
25�
y � 2�13
�z � 5�2
�F �G � � i21
j41
k�1
6� � 25i � 13j � 2k
�G�1, 2, 5� � i � j � 6k�F�1, 2, 5� � 2i � 4j � k
�G�x, y, z� � i � j � 6k �F�x, y, z� � 2xi � 2y j � k
G�x, y, z� � x � y � 6z � 33 F�x, y, z� � x2 � y2 � z
46. (a)
(b)
To find points of intersection, let Then
The normals to and at this point are and, which are orthogonal.
Similarly, and and the normals are andwhich are also orthogonal.
(c) No, showing that the surfaces are orthogonal at 2 points does not imply that they are orthogonal at every point ofintersection.
��1�2 �j � k,2 j � k�g�1, �2 � 14 � � ��1�2 �j�f �1, �2 � 14 � � 2 j
�1�2�j � k�2j � kgf�g�1, �2 � 14� � �1�2�j.�f �1, �2 � 14� � �2 j,
y � �2 ± 14
�y � 2�2 � 14
3�y � 2�2 � 42
x � 1.
�x � 1�2 � 42 � 3�y � 2�2
�x2 � 2x � 1� � 42 � 3�y2 � 4y � 4�
x2 � 2x � 31 � 3y2 � 12y
32 � 2x2 � 2y2 � 4x � 8y � 1 � 3x2 � y2 � 6x � 4y
16 � x2 � y2 � 2x � 4y �12
�1 � 3x2 � y2 � 6x � 4y�
f �x, y� � g�x, y�
g�x, y� �221 � 3x2 � y2 � 6x � 4y
x
y5
5
5
fg
zf �x, y� � 16 � x2 � y2 � 2x � 4y
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Section 12.7 Tangent Planes and Normal Lines 339
48.
� � arccos�31111 � � 25.24
cos � � ��F�2, 2, 2� � k��F�2, 2, 2�
� ��12�176
�311
11
�F�2, 2, 2� � 4i � 4j � 12k
�F � 2yi � 2xj � 3z2k
F�x, y, z� � 2xy � z3, �2, 2, 2� 50.
� � arccos 0 � 90
cos � � ��F�2, 1, 3� � k��F�2, 1, 3� � 0
�F�2, 1, 3� � 4i � 2j
�F�x, y, z� � 2xi � 2yj
F�x, y, z� � x2 � y2 � 5, �2, 1, 3�
52.
�12 , �1, �31
4 �z � 3�1
2�2� 2��1�2 � 3�1
2� � 4��1� � 5 � �314
4y � 4 � 0, y � �1
6x � 3 � 0, x �12
�F�x, y, z� � �6x � 3�i � �4y � 4�j � k
3 3
−2−3−3
30
25
yx
z F�x, y, z� � 3x2 � 2y2 � 3x � 4y � z � 5
54.
z � 5e�2ty � �t � 2x � �3t � 2
z�0� � C3 � 5y�0� � C2 � 2x�0� � C1 � 2
z�t� � C3e�2ty�t� � �t � C2x�t� � �3t � C1
dzdt
� �2zdydt
� �1dxdt
� �3
T �x, y, z� � 100 � 3x � y � z2, �2, 2, 5� 56.
Plane:
x0x
a2 �y0 y
b2 �z0z
c2 �x0
2
a2 �y0
2
b2 �z0
2
c2 � 1
2x0
a2 �x � x0� �2y0
b2 �y � y0� �2z0
c2 �z � z0� � 0
Fz�x, y, z� ��2zc2
Fy�x, y, z� �2yb2
Fx�x, y, z� �2xa2
F�x, y, z� �x2
a2 �y2
b2 �z2
c2 � 1
58.
Tangent plane at
Therefore, the plane passes through the origin �x, y, z� � �0, 0, 0�.
f �y0
x0� �
y0
x0f��y0
x0��x � f��y0
x0�y � z � 0
f �y0
x0� �
y0
x0f��y0
x0��x � x0 f �y0
x0� � y0 f��y0
x0� � yf��y0
x0� � y0 f��y0
x0� � z � x0 f �y0
x0� � 0
f �y0
x0� �
y0
x0f��y0
x0���x � x0� � f��y0
x0��y � y0� � �z � z0� � 0
�x0, y0, z0�:
Fx�x, y, z� � �1
Fy�x, y, z� � x f��yx��
1x� � f��y
x�
Fx�x, y, z� � f �yx� � x f��y
x���yx2� � f �y
x� �yx
f��yx�
F�x, y, z� � x f �yx� � z
z � x f �yx�
http://librosysolucionarios.net
340 Chapter 12 Functions of Several Variables
60.
(a)
(b)
(c) If This is the second–degree Taylor polynomial for
If This is the second–degree Taylor polynomial for cos x.P2�x, 0� � 1 �12 x2.y � 0,
cos y.P2�0, y� � 1 �12 y2.x � 0,
� 1 �12 x2 � xy �
12 y2
P2�x, y� � f �0, 0� � fx�0, 0�x � fy�0,0�y �12 fxx�0, 0�x2 � fxy�0, 0�xy �
12 fyy�0, 0�y2
P1�x, y� � f �0, 0� � fx�0, 0�x � fy�0, 0�y � 1
fxy�x, y� � �cos�x � y�fyy�x, y� � �cos�x � y�,fxx�x, y� � �cos�x � y�,
fy�x, y� � �sin�x � y� fx�x, y� � �sin�x � y�
f �x, y� � cos�x � y�
(d) (e)
x y5
5
5
z
x y
0 0 1 1 1
0 0.1 0.9950 1 0.9950
0.2 0.1 0.9553 1 0.9950
0.2 0.5 0.7648 1 0.7550
1 0.5 0.0707 1 0.1250�
P2�x, y�P1�x, y�f �x, y�
Section 12.8 Extrema of Functions of Two Variables
62. Given then:
�1
�� fx�x0, y0��2 � � fy�x0, y0��2 � 1
� ��1�� � fx�x0, y0��2 � � fy�x0, y0��2 � ��1�2
cos � ���F�x0, y0, z0� � k��F�x0, y0, z0� k
�F�x0, y0, z0� � fx�x0, y0�i � fy�x0, y0�j � k
F�x, y, z� � f �x, y� � z � 0
z � f �x, y�,
2.
Relative maximum:
gy � �2�y � 2� � 0 ⇒ y � �2
gx � �2�x � 3� � 0 ⇒ x � 3
�3, �2, 9�
x
y1
4
2
6
6
8
(3, 2, 9)−zg�x, y� � 9 � �x � 3�2 � �y � 2�2 ≤ 9
4.
Relative maximum:
Check:
At the critical point and Therefore, is a relative maximum.�2, 0, 5�fxx fyy � � fxy�2 > 0.fxx < 0�2, 0�,
fxx � �25 � y2
�25 � �x � 2�2 � y2�32 , fyy � �25 � �x � 2�2
�25 � �x � 2�2 � y2�32 , fxy � �y�x � 2�
�25 � �x � 2�2 � y2�32
fy � �y
�25 � �x � 2�2 � y2� 0 ⇒ y � 0
fx � �x � 2
�25 � �x � 2�2 � y2� 0 ⇒ x � 2
�2, 0, 5�
xy
(2, 0 , 5)
5
5
5
zf �x, y� � �25 � �x � 2�2 � y2 ≤ 5
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Section 12.8 Extrema of Functions of Two Variables 341
6.
Relative maximum:
Check:
At the critical point and Therefore, is a relative maximum.�2, 4, 9�fxx fyy � � fxy�2 > 0.fxx < 0�2, 4�,
fxx � �2, fyy � �2, fxy � 0
fy � �2y � 8 � 0 ⇒ y � 4
fx � �2x � 4 � 0 ⇒ x � 2
�2, 4 , 9�
yx
(2, 4 , 9)
42
8
6
8
6
zf �x, y� � �x2 � y2 � 4x � 8y � 11 � ��x � 2�2 � �y � 4�2 � 9 ≤ 9
8.
At the critical point and
Therefore, is a relative maximum.�5, �3, 8�
fxx fyy � f 2xy > 0.�5, �3�, fxx < 0
fxx � �2, fyy � �10, fxy � 0
fxfy
�
�
�2x
�10y
�
�
10
30
�
�
0
0� x � 5, y � �3
f �x, y� � �x2 � 5y2 � 10x � 30y � 62
10.
At the critical point and Therefore, is a relative minimum.��6, 2, 0�fxx fyy � � fxy�2 > 0.fxx > 0��6, 2�,
fxx � 2, fyy � 20, fxy � 6
fy � 6x � 20y � 4 � 0�fx � 2x � 6y � 0
f �x, y� � x2 � 6xy � 10y2 � 4y � 4
Solving simultaneously yields and y � 2.x � �6
12.
when
when
,
At the critical point and
Therefore, is a relative
maximum.
�12 , �1, 31
4 �fxx fyy � � fxy�2 > 0.
fxx < 0�12 , �1�,fxy � 0fyy � �4fxx � �6,
y � �1.fy � �4y � 4 � 0
x �12 .fx � �6x � 3 � 0
f �x, y� � �3x2 � 2y2 � 3x � 4y � 5 14.
Since for all is a relativeminimum.
�x, y�, �0, 0, 2�h�x, y� ≥ 2
hx
hy
�
�
2x3�x2 � y2�23
2y
3�x2 � y2�23
�
�
0
0�
x � 0, y � 0
h�x, y� � �x2 � y2�13 � 2
16.
Since for all the relative minima of f consist ofall points satisfyingx � y � 0.
�x, y�
�x, y�,f �x, y� ≥ �2
f �x, y� � �x � y� � 2 18.
Relative maximum:
Saddle points:
z
x
y3
3
20
40
�0, 2, �3�, �±�3, �1, �3��0, 0, 1�
f �x, y� � y3 � 3yx2 � 3y2 � 3x2 � 1
20.
Saddle point: �0, 0, 1�
yx 33
100
zz � exy
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342 Chapter 12 Functions of Several Variables
22.
At the critical point and Therefore, is a relative maximum.�40, 40, 4800�gxx gyy � �gxy�2 > 0.gxx < 0�40, 40�,
gxy � �1gyy � �2,gxx � �2,
gy � 120 � x � 2y � 0
gx � 120 � y � 2x � 0�g�x, y� � 120x � 120y � xy � x2 � y2
Solving simultaneously yields and y � 40.x � 40
24.
At the critical point Therefore, is a saddle point.�0, 0, 0�gxx gyy � �gxy�2 < 0.�0, 0�,
gxx � 0, gyy � 0, gxy � 1
gy � x
gx � y�g�x, y� � xy
and y � 0x � 0
26.
At saddle point.
At and relative maximum.
At and relative maximum.fxx < 0 ⇒ ��1, �1, 2���1, �1�, fxx fyy � � fxy�2 > 0
fxx < 0 ⇒ �1, 1, 2��1, 1�, fxx fyy � � fxy�2 > 0
�0, 0�, fxx fyy � � fxy�2 < 0 ⇒ �0, 0, 1�
fxx � �6x2, fyy � �6y2, fxy � 2
fxfy
�
�
2y
2x
�
�
2x3
2y3� Solving by substitution yields 3 critical points:
�0, 0�, �1, 1�, ��1, �1�
f �x, y� � 2xy �12
�x4 � y2� � 1
Solving yields the critical points �±�62
, 0 .�0, ±�22 ,�0, 0�,
30.
Relative minima at all points and �x, �x�, x � 0.�x, x�x y5
5
2
z
z ��x2 � y2�2
x2 � y2 ≥ 0. z � 0 if x2 � y2 � 0.
32. and
has a relative maximum at �x0, y0�.f
fxx fyy � � fxy�2 � ��3���8� � 22 > 0fxx < 0 34. and
has a relative minimum at �x0, y0�.f
fxx fyy � � fxy�2 � �25��8� � 102 > 0fxx > 0
36. See Theorem 12.17.
28.
At the critical point Therefore, is a saddle point. At the critical points
and Therefore, are relative maxima. At the critical points
and Therefore, are relative minima.�±�62, 0, ��ee�fxx fyy � � fxy�2 > 0.
fxx > 0�±�62, 0�,�0, ±�22, �e �fxx fyy � � fxy�2 > 0.fxx < 0
�0, ±�22�,�0, 0, e2�fxx fyy � �fxy�2 < 0.�0, 0�,
fxy � ��4x3y � 4xy3 � 2xy�e1�x2�y2
fyy � �4y4 � 4x2y2 � 2x2 � 8y2 � 1�e1�x2�y2
fxx � ��4x4 � 4x2y2 � 12x2 � 2y2 � 3�e1�x2�y2
fy � �2x2y � 2y3 � y�e1�x2�y2� 0
fx � �2x3 � 2xy2 � 3x�e1�x2�y2� 0�
f �x, y� � �12
� x2 � y2 e1�x2�y2
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Section 12.8 Extrema of Functions of Two Variables 343
42. and are relative extrema. and are saddle points.DCBA 44. if and have opposite signs.Hence, is a saddle point. For example,consider and �a, b� � �0, 0�.f �x, y� � x2 � y2
�a, b, f �a, b��fyyfxxd � fxx fyy � fxy
2 < 0
46.
At and the test fails. is a saddle point.�1, �2, 0�fxx fyy � � fxy�2 � 0�2, �3�,
fxx � 6x � 12, fyy � 6y � 18, fxy � 0
fy � 3y2 � 18y � 27 � 0�fx � 3x2 � 12x � 12 � 0
f �x, y� � x3 � y3 � 6x2 � 9y2 � 12x � 27y � 19
Solving yields and y � �3.x � 2
48.
At is undefined and the test fails.
Absolute minimum: �1, �2, 0�
fxx fyy � � fxy�2�1, �2�,
fxy ��x � 1��y � 2�
��x � 1�2 � �y � 2�2�32fyy ��x � 1�2
��x � 1�2 � �y � 2�2�32 ,fxx ��y � 2�2
��x � 1�2 � �y � 2�2�32 ,
fy �y � 2
��x � 1�2 � �y � 2�2� 0
fx �x � 1
��x � 1�2 � �y � 2�2� 0�
f �x, y� � ��x � 1�2 � �y � 2�2 ≥ 0
Solving yields and y � �2.x � 1
38. Extrema at all �x, y�
x
y2
3
3
4
4
4
z 40. Relative maximum
x
y3 4
4
4
(2, 1, 4)
z
50.
At is undefined and the test fails.
Absolute minimum: �0, 0, 0�
fxx fyy � � fxy�2�0, 0�,
fxy ��8xy
9�x2 � y2�43fyy �4�3x2 � y2�
9�x2 � y2�43 ,fxx �4�x2 � 3y2�
9�x2 � y2�43 ,
fy �4y
3�x2 � y2�13
fx �4x
3�x2 � y2�13�f �x, y� � �x2 � y2�23 ≥ 0
and are undefined at , The critical point is �0, 0�.y � 0.x � 0fyfx
52.
fz � �2x�y � 1�2�z � 2� � 0
fy � �2x2�y � 1��z � 2�2 � 0 �fx � �2x� y � 1�2�z � 2�2 � 0
f �x, y, z� � 4 � �x�y � 1��z � 2��2 ≤ 4
Solving yields the critical points These points are all absolute maxima.
�0, a, b�, �c, 1, d�, �e, f, �2�.
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344 Chapter 12 Functions of Several Variables
56.
On the line
On the curve
and the maximum is 1, the minimum is
Absolute maximum: 1 at and on
Absolute minimum: at ��12 , 14��
1116 � �0.6875
y � 1�1, 1��
1116 .
f �x, y� � f �x� � 2x � 2x�x2� � �x2�2 � x4 � 2x3 � 2x
�1 ≤ x ≤ 1y � x2,
f �x, y� � f �x� � 2x � 2x � 1 � 1.
�1 ≤ x ≤ 1,y � 1,
fy � 2y � 2x � 0 ⇒ y � x ⇒ x � 1� f �1, 1� � 1 fx � 2 � 2y � 0 ⇒ y � 1
x
2
1−1
( 1, 1)− (1, 1)
yf �x, y� � 2x � 2xy � y2
54.
On the line
and the maximum is 1, the minimum is 0. On the line
and the maximum is 16, the minimum is 0. On the line
and the maximum is 16, the minimum is 0.
Absolute maximum: 16 at
Absolute minimum: 0 at and along the line y � 2x.�1, 2��2, 0�
f �x, y� � f �x� � �2x � ��2x � 4��2 � �4x � 4�2
1 ≤ x ≤ 2,y � �2x � 4,
f �x, y� � f �x� � �2x � ��12 x � 1��2
� �52 x � 1�2
0 ≤ x ≤ 2,y � �12 x � 1,
f �x, y� � f �x� � �2x � �x � 1��2 � �x � 1�2
0 ≤ x ≤ 1,y � x � 1,
fy � �2�2x � y� � 0 ⇒ 2x � y
fx � 4�2x � y� � 0 ⇒ 2x � y
x
1
2
31 2
3
(0, 1)
(1, 2)
(2, 0)
y x= 2
yf �x, y� � �2x � y�2
60.
Along and
Along and
Along on
Thus, the maximum is and the minimum is f �4, 2� � �11.f �4, 0� � 21
�0, 4�.y � �x, 0 ≤ x ≤ 4, f � x2 � 4x32 � 5, f � 2x � 6x12 � 0
f �4, 2� � �11.x � 4, 0 ≤ y ≤ 2, f � 16 � 16y � 5, f � �16 � 0
f �4, 0� � 21.f � x2 � 5y � 0, 0 ≤ x ≤ 4,
f �0, 0� � 5
fy � �4x � 0
fx � 2x � 4y � 0� x � y � 0
x
1
4
2
31 42
3
yf �x, y� � x2 � 4xy � 5, R � ��x, y�: 0 ≤ x ≤ 4, 0 ≤ y ≤ �x�
58.
Along
Along
Along
Along
Thus, the maxima are and and the minima are f �x, �x� � 0, �1 ≤ x ≤ 1.f �2, 1� � 9,f ��2, �1� � 9
x � �2, �1 ≤ y ≤ 1, f � 4 � 4y � y2, f � 2y � 4 � 0.
x � 2, �1 ≤ y ≤ 1, f � 4 � 4y � y2, f � 2y � 4 � 0.
f � x2 � 2x � 1, f � 2x � 2 � 0 ⇒ x � 1, f ��2, �1� � 9, f �1, �1� � 0, f �2, �1� � 1.
y � �1, �2 ≤ x ≤ 2,
f � x2 � 2x � 1, f � 2x � 2 � 0 ⇒ x � �1, f ��2, 1� � 1, f ��1, 1� � 0, f �2, 1� � 9.
y � 1, �2 ≤ x ≤ 2,
f �x, �x� � x2 � 2x2 � x2 � 0
fy � 2x � 2y � 0
fx � 2x � 2y � 0� y � �x2
−1
−2
1x
yf �x, y� � x2 � 2xy � y2, R � ��x, y�: �x� ≤ 2, �y� ≤ 1�
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Section 12.9 Applications of Extrema of Functions of Two Variables 345
62.
For also, and
For and the point is outside
For and the maximum occurs at
Absolute maximum is
The absolute minimum is In fact,
x1
1
R
y
f �0, y� � f �x, 0� � 0�0 � f �0, 0�. �
89
� f��22
, �22 �.
y ��22
.x ��22
,f �x, y� � f �x, �1 � x2 � �4x�1 � x2
2 � x2 � x4 ,x2 � y2 � 1,
R.�1, 1�y � 1,x � 1
f �0, 0� � 0.y � 0,x � 0,
fy �4�1 � y2�x
�x2 � 1��y2 � 1�2 � 0 ⇒ y � 1 or x � 0
fx �4�1 � x2�y
�y2 � 1��x2 � 1�2 � 0 ⇒ x � 1 or y � 0
f �x, y� �4xy
�x2 � 1��y2 � 1�, R � ��x, y�: x ≥ 0, y ≥ 0, x2 � y2 ≤ 1�
64. False
Let
Relative minima:
Saddle point: �0, 0, 0��±1, 0, �1�
f �x, y� � x4 � 2x2 � y2.
Section 12.9 Applications of Extrema of Functions of Two Variables
2. A point on the plane is given by Thesquare of the distance from to a point on theplane is given by
From the equations and we obtain the system
Solving simultaneously, we have and the distance is
��1614
� 1�2
� �3114
� 2�2
� �4314
� 3�2
�1
�14.
z �4314y �
3114 ,x �
1614 ,
6x � 10y � 29.
5x � 6y � 19
Sy � 0,Sx � 0
Sy � 2�y � 2� � 2�9 � 2x � 3y���3�.
Sx � 2�x � 1� � 2�9 � 2x � 3y���2�
S � �x � 1�2 � � y � 2�2 � �9 � 2x � 3y�2
�1, 2, 3��x, y, 12 � 2x � 3y�. 4. A point on the paraboloid is given by The
square of the distance from to a point on theparaboloid is given by
From the equations and we obtain the system
Solving as in Exercise 3, we have and the distance is
��1.235 � 5�2 � �1.525�2 4.06.
z 1.525y � 0,x 1.235,
2y3 � 2x2y � y � 0.
2x3 � 2xy2 � x � 5 � 0
Sy � 0,Sx � 0
Sy � 2y � 4y�x2 � y2� � 0.
Sx � 2�x � 5� � 4x�x2 � y2� � 0
S � �x � 5�2 � y2 � �x2 � y2�2
�5, 0, 0��x, y, x2 � y2�.
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346 Chapter 12 Functions of Several Variables
10. Let and be the length, width, and height, respectively. Then and The volume is given by
In solving the system and we note by the symmetry of the equations that Substituting into yields
and z �14�2C0.y �
13�2C0 ,x �
13�2C0 ,2C0 � 9x2,
x2�2C0 � 9x2�16x2 � 0,
Vx � 0y � xy � x.Vy � 0,Vx � 0
Vy �x2�2C0 � 3y2 � 6xy�
4�x � y�2 .
Vx �y2�2C0 � 3x2 � 6xy�
4�x � y�2
V � xyz �C0 xy � 1.5x2y2
2�x � y�
z �C0 � 1.5xy
2�x � y� .C0 � 1.5xy � 2yz � 2xzzx, y,
12. Consider the sphere given by and let a vertex of the rectangular box be Then the volume is given by
Solving the system
yields the solution x � y � z � r�3.
x2 � 2y2 � r 2
2x2 � y2 � r 2
Vy � 8�xy�y
�r 2 � x2 � y2� x�r 2 � x2 � y2� �
8x
�r 2 � x2 � y2�r 2 � x2 � 2y2� � 0.
Vx � 8�xy�x
�r 2 � x2 � y2� y�r 2 � x2 � y2� �
8y
�r 2 � x2 � y2�r 2 � 2x2 � y2� � 0
V � �2x��2y��2�r 2 � x2 � y2� � 8xy�r 2 � x2 � y2
�x, y, �r 2 � x2 � y2 �.x2 � y2 � z2 � r 2
6. Since Therefore,
Ignoring the solution and substituting into we have
Therefore, and z � 8.y � 16,x � 8,
4x�x � 8� � 0.
64x � 2x2 � 3x�32 � 2x� � 0
Py � 0,y � 32 � 2xy � 0
Py � 64xy � 2x2y � 3xy2 � y�64x � 2x2 � 3xy� � 0.
Px � 32y2 � 2xy2 � y3 � y2�32 � 2x � y� � 0
P � xy2z � 32xy2 � x2y2 � xy3
z � 32 � x � y.x � y � z � 32, 8. Let and be the numbers and let Since we have
Solving simultaneously yields and z �13 .y �
13 ,x �
13 ,
Sy � 2y � 2�1 � x � y� � 0 x � 2y � 1.
Sx � 2x � 2�1 � x � y� � 0� 2x � y � 1
S � x2 � y2 � �1 � x � y�2
x � y � z � 1,S � x2 � y2 � z2.zx, y,
14. Let and be the length, width, and height, respectively.Then the sum of the two perimeters of the two cross sections is given by
or
The volume is given by
Solving the system and we obtain the solution
inches, inches, and inches.z � 9y � 18x � 18
y � 4z � 54,2y � 2z � 54
Vz � 54y � y2 � 4yz � y�54 � y � 4z� � 0.
Vy � 54z � 2yz � 2z2 � z�54 � 2y � 2z� � 0
V � xyz � 54yz � y2z � 2yz2
x � 54 � y � 2z.�2x � 2z� � �2y � 2z� � 108
zx, y,
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Section 12.9 Applications of Extrema of Functions of Two Variables 347
16.
From we have
From we obtain
Then cos � �12 ⇒ � � 60�.
x � 10
3x2 � 30x � 0
30�2x � 15� � 2x�2x � 15� � 2�2x � 15�2 � x2 � 0
30x�2x � 15x � � 2x2�2x � 15
x � � x2�2�2x � 15x �
2
� 1� � 0
�A��
� 0
15 � 2x � x cos � � 0 ⇒ cos � �2x � 15
x.
�A�x
� 0
�A��
� 30 cos � � 2x2 cos � � x2�2 cos2 � � 1� � 0
�A�x
� 30 sin � � 4x sin � � 2x sin � cos � � 0
� 30x sin � � 2x2 sin � � x2 sin � cos �
A �1
2��30 � 2x� � �30 � 2x� � 2x cos � x sin �
18.
implies that .
Solving gives
and hence and
�69
�23
.
P�13
, 13� � �2�1
9� � 2�13� � 2�1
3� � 2�19� � 2�1
9�
p � q �13
p � 2q � 1
q � 2p � 1
�P�p
��P�q
� 0
�P�p
� �2q � 2 � 4p; �P�q
� �2p � 2 � 4q
� �2pq � 2p � 2q � 2p2 � 2q2
� 2pq � 2p � 2p2 � 2pq � 2q � 2pq � 2q2
P� p, q� � 2pq � 2p�1 � p � q� � 2q�1 � p � q�
r � 1 � p � qp � q � r � 1
P� p, q, r� � 2pq � 2pr � 2qr. 20.
Solving this system yields p1 � $2296.67, p2 � $4250.
�1.5p1 � p2 � 805
3p1 � 1.5p2 � 515
Rp2� 805 � 1.5p1 � p2 � 0
Rp1� 515 � 1.5p2 � 3p1 � 0
R � 515p1 � 805p2 � 1.5p1p2 � 1.5p12 � p2
2
22.
when
The sum of the distance is minimized when y �2�3 � �3�
3 0.845.
y � 2 �2�3
3�
6 � 2�33
.dSdy
� 1 �2�y � 2�
�4 � �y � 2�2� 0
� y � 2�4 � �y � 2�2
S � d1 � d2 � d3 � ��0 � 0�2 � �y � 0�2 � ��0 � 2�2 � �y � 2�2 � ��0 � 2�2 � �y � 2�2
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348 Chapter 12 Functions of Several Variables
24. (a)
The surface appears to have a minimum near
(b)
Sy �y
��x � 4�2 � y2�
y � 6
��x � 1�2 � �y � 6�2�
y � 2
��x � 12�2 � �y � 2�2
Sx �x � 4
��x � 4�2 � y2�
x � 1
��x � 1�2 � �y � 6�2�
x � 12
��x � 12�2 � �y � 2�2
�x, y� � �1, 5�.
22
30
−2 −4
44
68
x
y
zS � ��x � 4�2 � y2 � ��x � 1�2 � �y � 6�2 � ��x � 12�2 � �y � 2�2
(c) Let Then
Direction
(d)
(e)
Note: Minimum occurs at
(f ) points in the direction that decreases most rapidly.S��S�x, y�
�x, y� � �1.2335, 5.0694�
y4 5.06x4 1.23,t 1.04,
y3 5.06,x3 1.24,t 3.56,
y2 5.03x2 1.24t 0.94
6.6�
��S�1, 5� � 0.258i � 0.03j
�x1, y1� � �1, 5�.
26. See the last paragraph on page 915 and Theorem 12.18.
28. (a)
y �3
10x � 1
a �4�6� � 0�4�4�20� � �0�2 �
310
, b �14 �4 �
310
�0�� � 1,
x y xy
0 0 9
1 1
1 1 1 1
3 2 6 9
� xi2 � 20�xiyi � 6� yi � 4� xi � 0
�1�1
�3
x2 (b)
�15
S � � 110
� 0�2
� � 710
� 1�2
� �1310
� 1�2
� �1910
� 2�2
30. (a)
(b) S � �34
� 0�2
� ��14
� 0�2
� �14
� 0�2
� �34
� 1�2
� �54
� 1�2
� �54
� 2�2
� �74
� 2�2
� �94
� 2�2
�32
a �8�37� � �28��8�8�116� � �28�2 �
72144
�12
, b �18�8 �
12
�28�� � �34
, y �12
x �34
x y xy
3 0 0 9
1 0 0 1
2 0 0 4
3 1 3 9
4 1 4 16
4 2 8 16
5 2 10 25
6 2 12 36
�xi2 � 116� xiyi � 37� yi � 8� xi � 28
x2
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Section 12.9 Applications of Extrema of Functions of Two Variables 349
32.
6
7
−1
−1
y �32
x �32
b �13 �9 �
32
�9�� � �96
� �32
a �3�39� � 9�9�3�35� � �9�2 �
3624
�32
� xi2 � 35�xiyi � 39,
� yi � 9,� xi � 9,
�1, 0�, �3, 3�, �5, 6� 34.
−1
−1
14
9
y �2953
x �425318
b �16 �31 �
2953
42� �425318
1.3365
a �6�275� � �42��31�
6�400� � �42�2 �2953
0.5472
� xi2 � 400� xi yi � 275
� yi � 31� xi � 42
�6, 4�, �1, 2�, �3, 3�, �8, 6�, �11, 8�, �13, 8�; n � 6
36. (a)
(b) When y � �240�1.40� � 685 � 349.x � 1.40,
y � �240x � 685
b �13
�1,155 � ��240��3.75� � 685
a �3�1,413.75� � �3.75��1,155�
3�4.8125� � �3.75�2 � �240
� xiyi � 1,413.75
� xi2 � 4.8125,� xi � 3.75, � yi � 1,155,
�1.00, 450�, �1.25, 375�, �1.50, 330� 38. (a)
(b) For each 1 point increase in the percent yincreases by about 1.83 (slope of line).
�x�,
y � 1.8311x � 47.1067
40.
as long as for all (Note: If for all then is the least squares regression line.)
since
As long as the given values for and yield a minimum.bad 0,
n�n
i�1xi
2 ≥ ��n
i�1xi�2
.d � SaaSbb � Sab2 � 4n�
n
i�1xi
2 � 4��n
i�1xi�2
� 4�n�n
i�1xi
2 � ��n
i�1xi�2� ≥ 0
x � 0i,xi � 0i.xi 0Saa�a, b� > 0
Sab�a, b� � 2�n
i�1xi
Sbb�a, b� � 2n
Saa�a, b� � 2�n
i�1xi
2
Sb�a, b� � 2a�n
i�1xi � 2nb � 2�
n
i�1yi
Sa�a, b� � 2a�n
i�1xi
2 � 2b�n
i�1xi � 2�
n
i�1xiyi
S�a, b� � �n
i�1�axi � b � yi �2
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350 Chapter 12 Functions of Several Variables
42.
y � �524 x2 �
310 x �
416c �
416 ,b � �
310 ,a � �
524 ,
40a � 4c � 1940b � �12,544a � 40c � 160,
� xi2yi � 160
� xiyi � �12
� xi4 � 544
� xi3 � 0
� xi2 � 40
� yi � 19
−4
9−9
(2, 6)
(4, 2)
( 2, 6)−( 4, 5)−
8 � xi � 0
�4, 2��2, 6�,��2, 6�,��4, 5�, 44.
y � �54 x2 �
920 x �
19920c �
19920 ,b �
920 ,a � �
54 ,
14a � 6b � 4c � 25
36a � 14b � 6c � 21
98a � 36b � 14c � 33
� xi2yi � 33
� xiyi � 21
� xi4 � 98
� xi3 � 36
� xi2 � 14
� yi � 25
−1
9−9
(0, 10)
(3, 0)
(2, 6)
(1, 9)
11 � xi � 6
�0, 10�, �1, 9�, �2, 6�, �3, 0�
Section 12.10 Lagrange Multipliers
46. (a)
(b)
(c)
(d) For the linear model, gives billion.
For the quadratic model, gives billion.
As you extrapolate into the future, the quadratic modelincreases more rapidly.
y � 6.96x � 50
y � 6.86x � 50
−50
45
7
y � 0.0001429x2 � 0.07229x � 2.9886
y � 0.078x � 2.96 48. (a)
(b)
(c) No. For Note that there is avertical asymptote at x � 56.6.
y � �100.x � 60,
6000
50
y �1
�0.0029x � 0.1640
1y
� ax � b � �0.0029x � 0.1640
2. Maximize
Constraint:
f �1, 2� � 2
� � 1, x � 1, y � 2
2x � y � 4 ⇒ 4� � 4
x � �
y � 2�
y i � xj � 2� i � � j
�f � ��g
2x � y � 4
f �x, y� � xy. 4. Minimize
Constraint:
f �12 , 1� �
54
� �12 , x �
12 , y � 1
2x � 4y � 5 ⇒ 10� � 5
2y � 4� ⇒ y � 2�
2x � 2� ⇒ x � �
2x i � 2yj � 2� i � 4� j
�f � ��g
2x � 4y � 5
f �x, y� � x2 � y2.
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Section 12.10 Lagrange Multipliers 351
6. Maximize
Constraint:
or
If then and
If
Maximum.f ��2, 1� � 2 � 1 � 1
�2y � 2� � �2 ⇒ y � 1 ⇒ x2 � 2 ⇒ x � �2.
� � �1,
f �0, 0� � 0.y � 0x � 0,
� � �12x � �2x� ⇒ x � 0
2x i � 2yj � �2x� i � 2� j
�f � ��g
2y � x2 � 0
f �x, y� � x2 � y2. 8. Minimize
Constraint:
f � 3�4, 3 3�4
2 � �9 3�4 � 20
2
x � 3�4, y �3 3�4
2
x3 � 4
x2y � 6 ⇒ x2�3x2 � � 6
3 � 2xy� ⇒ � � 3 2xy
1 � x2� ⇒ � � 1
x23x2 � 2xy ⇒ y �
3x2
�x � 0�
3i � j � 2xy� i � x2� j
�f � ��g
x2y � 6
f �x, y� � 3x � y � 10.
10. Note: is minimum when isminimum.
Minimize
Constraint:
f �32
, 3� ��g�32
, 3� �3�5
2
x �32
, y � 3
2x � 4y � 15 ⇒ 10 x � 15
2x � 2�
2y � 4� y � 2x
2x � 4y � 15
g�x, y� � x2 � y2.
g�x, y�f �x, y� � �x2 � y2 12. Minimize .
Constraint:
f �4, 8� � 16
x � 4, y � 8
xy � 32 ⇒ 2x2 � 32
2 � y�
1 � x� y � 2x
xy � 32
f �x, y� � 2x � y
14. Maximize or minimize
Constraint:
Case 1: On the circle
Maxima:
Minima:
Case 2: Inside the circle
At
Saddle point:
Combining the two cases, we have a maximum of at and a minimum of at �±�22
, ±�22 �.e�18�±�2
2, �
�22 �e18
f �0, 0� � 1
fxx fyy � � fxy�2 < 0.�0, 0�,
fxx �y2
16e�xy4, fyy �
x2
16e�xy4, fxy � e�xy� 1
16xy �
14�
fx � ��y4�e�xy4 � 0
fy � ��x4�e�xy4 � 0 ⇒ x � y � 0
f �±�22
, ±�22 � � e�18 � 0.8825
f �±�22
, ��22 � � e18 � 1.1331
x2 � y2 � 1 ⇒ x � ±�22
��y4�e�xy4 � 2x�
��x4�e�xy4 � 2y� ⇒ x2 � y2
x2 � y2 � 1
x2 � y2 ≤ 1
f �x, y� � e�xy4.
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352 Chapter 12 Functions of Several Variables
16. Maximize .
Constraint:
f �2, 2, 2� � 8
x � y � z � 6 ⇒ x � y � z � 2
yz � �
xz � �
xy � � x � y � z
x � y � z � 6
f �x, y, z� � xyz 18. Minimize
Constraint:
Then
f �3, 5� � 9 � 30 � 25 � 70 � 70 � 4
x � 3, y � 5.
� � � 12 � 8 ⇒ � � �4
x � y �12
�� � 10� �12
�� � 14�
2x � 10 � �
2y � 14 � �
x � y � 8
x � �12��� � 10�y � �12��� � 14�
x � y � 10
x2 � 10x � y2 � 14y � 70
20. Minimize
Constraints:
f �6, 6, 0� � 72
x � 6, z � 0
⇒ 92
x � 27 ⇒ x � 6 2x � 2�12 � x� � �3 �x2�
x � y � 12 ⇒ y � 12 � x
x � 2z � 6 ⇒ z �6 � x
2� 3 �
x2
2x � � �
2y �
2z � 2�2x � 2y � z
2x i � 2yj � 2zk � ��i � 2k� � �i � j�
�f � ��g � �h
x � y � 12
x � 2z � 6
f �x, y, z� � x2 � y2 � z2. 22. Maximize .
Constraints:
x � 2y � 0 ⇒ y �x2
x2 � z2 � 5 ⇒ z � �5 � x2
xy � 2z� ⇒ � �xy2z
xz � �2 ⇒ � �xy2
yz � 2x� �
yz i � xz j � xyk � ��2x i � 2zk� � �i � 2j�
�f � ��g � �h
x � 2y � 0
x2 � z2 � 5
f �x, y, z� � xyz
Note: does not yield a maximum.f �0, 0, �5 � � 0
f��103
, 12�10
3, �5
3 � �5�15
9
x � 0 or x ��103
, y �12�10
3, z ��5
3
0 � 3x3 � 10x � x�3x2 � 10�
2x�5 � x2� � x3
x�5 � x2 �x3
2�5 � x2
x�5 � x2
2�
x3
2�5 � x2�
x�5 � x2
2
yz � 2x�xy2z� �
xz2
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Section 12.10 Lagrange Multipliers 353
24. Minimize the square of the distance subject to the constraint
Using a graphing utility, we obtain and or, by the Quadratic Formula,
Using the smaller value, we have and
The point on the circle is
and the desired distance is
The larger x-value does not yield a minimum.
d ��16�1 ��2929 �
2
� �10�2929
� 10�2
� 8.77.
�4�1 ��2929 �,
10�2929 �
y �10�29
29� 1.8570.x � 4�1 �
�2929 �
x �58 ± �582 � 4�294��112�
2�294� �58 ± 2�29
292� 4 ±
4�2929
.
x � 4.7428x � 3.2572
294
x2 � 58x � 112 � 0
�x � 4�2 � y2 � 4 ⇒ �x2 � 8x � 16� � �254
x2 � 50x � 100� � 4
2x � 2�x � 4��
2�y � 10� � 2y� x
x � 4�
y � 10
y ⇒ y � �
5
2x � 10
�x � 4�2 � y2 � 4.f �x, y� � x2 � �y � 10�2
26. Minimize the square of the distance
subject to the constraint
The point on the plane is and the desired distance is
d � ��2 � 4�2 � 02 � 22 � 2�2.
�2, 0, 2�
z � 2y � 0,x � 2,�x2 � y2 � z � 0,
2z � ��
2y �y
�x2 � y2� �
yz�
2�x � 4� �x
�x2 � y2� �
xz�
�x2 � y2 � z � 0.
f �x, y, z� � �x � 4�2 � y2 � z2
28. Maximize subject to the constraintsand
The maximum value of f occurs when at the pointof ��4, 0, 4�.
z � 4
z �43 or z � 4
�3z � 4��z � 4� � 0
3z2 � 16z � 16 � 0
�4 � 2z�2 � 02 � z2 � 0
x � 2z � 4 ⇒ x � 4 � 2z
x2 � y2 � z2 � 0
1 � �2z� � 2
0 � 2y� ⇒ y � 0
0 � 2x� �
x � 2z � 4.x2 � y2 � z2 � 0f �x, y, z� � z
30. See explanation at the bottom of page 922.
32. Maximize subject to the constraint
Volume is maximum when
and z ��2C
4.x � y �
�2C3
x ��2C
3
1.5xy � 2xz � 2yz � C ⇒ 1.5x2 �32
x2 �32
x2 � C
yz � �1.5y � 2z��xz � �1.5x � 2z��xy � �2x � 2y��
x � y and z �3
4x
1.5xy � 2xz � 2yz � C.V�x, y, z� � xyz 34. Minimize subject to the
constraint
Dimensions: and h � 2 3�V0
2r � 3�V0
2
r2h � V0 ⇒ 2r3 � V0
2h � 4r � 2rh�
2r � r2� h � 2r
r2h � V0.A�, r� � 2rh � 2r2
2y � �2y
2�x � 4� � �2x
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354 Chapter 12 Functions of Several Variables
36. (a) Maximize subject to the constraint
Therefore,
3�xyz ≤ x � y � z
3.
3�xyz ≤ S3
xyz ≤ S3
27
xyz ≤ �S3��
S3��
S3�, x, y, z > 0
x � y � z � S ⇒ x � y � z �S3
yz � �
xz � �
xy � � x � y � z
x � y � z � S.
P�x, y, z� � xyz (b) Maximize subject to the constraint
Therefore,
n�x1x2x3 . . . xn ≤ x1 � x2 � x3 � . . . � xn
n.
n�x1x2x3 . . . xn ≤ Sn
x1x2x3 . . . xn ≤ �Sn�
n
x1x2x3 . . . xn ≤ �Sn��
Sn��
Sn� . . . �S
n�, xi ≥ 0
�n
i�1 xi � S ⇒ x1 � x2 � x3 � . . . � xn �
Sn
x2x3 . . . xn � �
x1x3 . . . xn � �
x1x2 . . . xn � �
�x1x2x3 . . . xn�1 � �
x1 � x2 � x3 � . . . � xn
�n
i�1xi � S.
P � x1x2x3 . . . xn
38. Case 1: Minimize subject to the constraint
Case 2: Minimize subject to the constraint
h �l2
or l � 2h
l � 2� ⇒ � �l2
, h �l4
�l2
�l4
h �l4
� �� �
2��
2h � l � �l2 � � P.A�l, h� � lh � �l 2
8 � l � 2h
2 � l� ⇒ � �2l, 1 �
2�
2hl
�
2
l
h
1 �
2� �h �
l4 ��
lh � �l 2
8 � � A.P�l, h� � 2h � l � �l2 �
40. Maximize subject to the constraints and
If , then and
Thus, and
If then and
Therefore, the maximum temperature is 150.
T��502
, 0, �50
2 � � 100 �504
� 112.5
x � z � �502.x 2 � z2 � 2x 2 � 50y � 0,
T �0, �50, 0� � 100 � 50 � 150
y � �50.x � z � 0
z � 0. � 0,� � 1y � 0
2x � 2x� �
2y � 2y�
0 � 2z� �
x � z � 0.x2 � y2 � z2 � 50T�x, y, z� � 100 � x2 � y2 42. Maximize
Constraint:
P�25003
, 5000
3 � � $126,309.71.
48x � 36y�2x� � 100,000 ⇒ x �2500
3, y �
50003
yx
� 2 ⇒ y � 2x
�yx�
0.6
�yx�
0.4
� �48�
40 �� 6036��
60x0.4y�0.4 � 36� ⇒ �xy�
0.4
�36�
60
40x�0.6y0.6 � 48� ⇒ �yx�
0.6
�48�
40
48x � 36y � 100,000.
P�x, y� � 100x0.4y0.6
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Review Exercises for Chapter 12 355
44. Minimize subject to the constraint
Therefore, C�209.65, 186.35� � $16,771.94.
y �89�
200�8�9�0.4� � 186.35
x �200
�8�9�0.4 � 209.65
100x0.6y0.4 � 20,000 ⇒ x0.6�89
x0.4
� 200
yx
�89
⇒ y �89
x
�yx
0.4
�yx
0.6
� � 4860��
40�
36
36 � 40x0.6y�0.6� ⇒ �xy
0.6
�36
40�
48 � 60x�0.4y0.4� ⇒ �yx
0.4
�48
60�
100x0.6y0.4 � 20,000.C�x, y� � 48x � 36y
Review Exercises for Chapter 12
46.
Constraint:
(a) Level curves of are lines of form
Using you obtain
and
Constraint is an ellipse.
−10 10
−8
8
f �7, 3� � 28 � 9 � 37.y � 3,x � 7,
y � �43
x � 12.3,
y � �43
x � C.
f �x, y� � 4x � 3y
x2
64�
y2
36� 1
x, y > 0f �x, y� � ax � by,
(b) Level curves of are lines of form
Using you obtain
and f �4, 5.2� � 62.8.y � 5.2,x � 4,
y � �49
x � 7,
y � �49
x � C.
f �x, y� � 4x � 9y
2. Yes, it is the graph of a function.
4.
The level curves are of the form
The level curves arehyperbolas.
ec � xy.
c � ln xy c = 0
c = 1
c = 2 c = 2
−3 3
−2
21 3
f �x, y� � ln xy 6.
The level curves are of the form
The level curves are passing through the origin with slope
1 � cc
.
y � �1 � cc x.
c �x
x � y
c = 2
c = 2−
c = 1−
c = 1
c = 12
c = − 32
c = − 2
c = 32
−3 3
−2
2
1
f �x, y� �x
x � y
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356 Chapter 12 Functions of Several Variables
8.
yx
5 5
60
z
g�x, y� � y1�x 10.
Elliptic cone
x y5
2
2
z
f �x, y, z� � 9x2 � y2 � 9z2 � 0 12.
Does not exist
Continuous except when y � ± x.
lim�x, y�→�1, 1�
xy
x2 � y2
14.
Continuous everywhere
lim�x, y�→�0, 0�
y � xe�y2
1 � x2 �0 � 01 � 0
� 0 16.
fy �x2
�x � y�2
�y2
�x � y�2 fx �y�x � y� � xy
�x � y�2
f �x, y� �xy
x � y18.
�z�y
�2y
x2 � y2 � 1
�z�x
�2x
x2 � y2 � 1
z � ln�x2 � y 2 � 1�
20.
�w�z
�z
�x2 � y2 � z2
�w�y
�y
�x2 � y2 � z2
�w�x
�12
�x2 � y2 � z2��1�2�2x� �x
�x2 � y2 � z2
w � �x2 � y2 � z2 22.
fz �z
�1 � x2 � y2 � z2�3�2
fy �y
�1 � x2 � y2 � z2�3�2
�x
�1 � x2 � y2 � z2�3�2
fx � �12
�1 � x2 � y2 � z2��3�2��2x�
f �x, y, z� �1
�1 � x2 � y2 � z2
24.
�u�t
� �kc�sin akx� sin kt
�u�x
� akc�cos akx� cos kt
u�x, t� � c�sin akx� cos kt 26.
At
Slope in x-direction.
At
Slope in y-direction.
�z�y
� 4.�2, 0, 0�,�z�y
�x2
1 � y.
�z�x
� 0.�2, 0, 0�, �z�x
� 2x ln�y � 1�.
z � x2 ln�y � 1�
28.
hyx ���x � y�2 � 2y�x � y�
�x � y�4 �x � y
�x � y�3
hxy ��x � y�2 � 2y�x � y�
�x � y�4 �x � y
�x � y�3
hyy �2x
�x � y�3
hxx ��2y
�x � y�3
hy ��x
�x � y�2
hx �y
�x � y�2
h�x, y� �x
x � y30.
gyx � 2 cos�x � 2y�
gxy � 2 cos�x � 2y�
gyy � �4 cos�x � 2y�
gxx � �cos�x � 2y�
gy � 2 sin�x � 2y�
gx � �sin�x � 2y�
g�x, y� � cos�x � 2y�
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Review Exercises for Chapter 12 357
32.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �6x
�z�y
� �6xy
�2z�x2 � 6x
�z�x
� 3x2 � 3y2
z � x3 � 3xy2 34.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �ex sin y
�z�y
� ex cos y
�2z�x2 � ex sin y
�z�x
� ex sin y
z � ex sin y
36.
�y3
�x2 � y2�3�2 dx �x3
�x2 � y2�3�2 dy � ��x2 � y2 y � xy�x��x2 � y2 �x2 � y2 � dx � ��x2 � y2x � xy�y��x2 � y2 �
x2 � y2 � dy
dz ��z�x
dx ��z�y
dy
z �xy
�x2 � y2
38. From the accompanying figure we observe
Letting and
(Note that we express the measurement of the angle in radians.) The maximum error is approximately
dh � tan�11�
60 �±12 � 100 sec2�11�
60 �± �
180 � ±0.3247 ± 2.4814 � ±2.81 feet.
d� � ±�
180.� �
11�
60,dx � ±
12
,x � 100,
dh ��h�x
dx ��h��
d� � tan � dx � x sec2 � d�.
tan � �hx or h � x tan �
x
h
θ
40.
���2r2 � h2��r2 � h2
dr ��rh
�r2 � h2 dh �
� �8 � 25��29 �±1
8 �10�
�29�±18 � ±
43�
8�29
dA � ���r2 � h2 ��r2
�r2 � h2 dr ��rh
�r2 � h2 dh
A � �r�r2 � h2
42.
Chain Rule:
Substitution:
� sin t�1 � 2 cos t�dudt
� 2 sin t cos t � sin t
u � sin2 t � cos t
� sin t�1 � 2 cos t�
� sin t � 2�sin t� cos t
� �1��sin t� � 2y�cos t�
dudt
��u�x
�x�t
��u�y
�y�t
y � sin tx � cos t,u � y2 � x,
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358 Chapter 12 Functions of Several Variables
44.
Chain Rule:
�4r2t � rt2 � 4r3
�2r � t�2
�yz
�1� �xz
�r� �xyz2 ��1�
�w�t
��w�x
�x�t
��w�y
�y�t
��w�z
�z�t
�4r2t � 4rt2 � t3
�2r � t�2
�2rt
2r � t�
�2r � t�t2r � t
�2�2r � t��rt�
�2r � t�2
�yz
�2� �xz
�t� �xyz2 �2�
�w�r
��w�x
�x�r
��w�y
�y�r
��w�z
�z�r
w �xyz
, x � 2r � t, y � rt, z � 2r � t
Substitution:
�w�t
�4r2t � rt2 � 4r3
�2r � t�2
�w�r
�4r2t � 4rt2 � t3
�2r � t�2
w �xyz
��2r � t��rt�
2r � t�
2r2t � rt2
2r � t
46.
�z�y
�sin z
2xz � y cos z
2xz �z�y
� y cos z �z�y
� sin z � 0
�z�x
�z2
y cos z � 2xz
2xz �z�x
� z2 � y cos z �z�x
� 0
xz2 � y sin z � 0
48.
� �4�5
5�
2�55
� �2�5
5Du f �1, 4� � f �1, 4� u
u �1
�5v �
2�55
i ��55
j
f �1, 4� � �2i � 2j
f � �2xi �12
yj
f �x, y� �14
y2 � x2 50.
� 4�3 � �3 � 0 � 5�3
Duw�1, 0, 1� � w�1, 0, 1� u
u �1
�3v �
�33
i ��33
j ��33
k
w�1, 0, 1� � 12i � 3j
w � �12x � 3y�i � �3x � 8yz�j � ��4y2�k
w � 6x2 � 3xy � 4y2z
52.
�z�2, 1�� � 4
z�2, 1� � 4 j
z �x2 � 2xy�x � y�2 i �
x2
�x � y�2 j
z �x2
x � y54.
�z�2, 1�� � 4�2
z�2, 1� � 4i � 4 j
z � 2xy i � x2j
z � x2y
56.
Normal vector: j
f ��
2, 1 � 2j
f �x, y� � 4y cos xi � �4 sin x � 2y�j
f �x, y� � 4y sin x � y2
4y sin x � y2 � 3 58.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
y � 33
�z � 4
4.x � 2,
3y � 4z � 25,3� y � 3� � 4�z � 4� � 0
F�2, 3, 4� � 6j � 8k � 2�3j � 4k�
F � 2yj � 2zk
F�x, y, z� � y2 � z2 � 25 � 0
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Review Exercises for Chapter 12 359
60.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 11
�y � 2
2�
z � 22
.
x � 2y � 2z � 9,
�x � 1� � 2� y � 2� � 2�z � 2� � 0
F�1, 2, 2� � 2i � 4j � 4k � 2�i � 2j � 2k�
F � 2xi � 2y j � 2zk
F�x, y, z� � x2 � y2 � z2 � 9 � 0 62.
Therefore, the equation of the tangent line is
x � 41
�y � 4
1�
z � 9�8
.
F � G � i81
j0
�1
k10 � i � j � 8k
F�4, 4, 9� � 8i � k
G � i � j
F � 2y i � k
G�x, y, z� � x � y � 0
F�x, y, z� � y2 � z � 25 � 0
64. (a)
(c) If you obtain the 2nd degree Taylor polynomial (d)for cos x.
(e)
The accuracy lessens as the distance from increases.�0, 0�
2
2
3
211
−1
−1−2
−2
x
y
z
1
2
1
−1
y
x
z
1
2
1
−1
y
x
z
y � 0,
P1�x, y� � 1 � y
fy � cos y, fy�0, 0� � 1
fx � �sin x, fx�0, 0� � 0
f �x, y� � cos x � sin y, f �0, 0� � 1
66.
Therefore, is a relative minimum.��4, 43 , �2�fxx fyy � � f xy�2 � 4�18� � �6�2 � 36 > 0
fxy � 6
fyy � 18
fxx � 4
4��3y� � 6y � �8 ⇒ y �43 , x � �4
x � �3y fy � 6x � 18y � 0,
fx � 4x � 6y � 8 � 0
f �x, y� � 2x2 � 6xy � 9y2 � 8x � 14
(b)
P2�x, y� � 1 � y �12 x2
fxy � 0, fxy�0, 0� � 0
fyy � �sin y, fyy�0, 0� � 0
fxx � �cos x, fxx�0, 0� � �1
x y
0 0 1.0 1.0 1.0
0 0.1 1.0998 1.1 1.1
0.2 0.1 1.0799 1.1 1.095
0.5 0.3 1.1731 1.3 1.175
1 0.5 1.0197 1.5 1.0
P2�x, y�P1�x, y�f �x, y�
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360 Chapter 12 Functions of Several Variables
68.
Critical Points:
At ,
is a relative maximum.
At ,
is a saddle point.
At ,
is a saddle point.
At ,
is a relative minimum.��10, �14, �749.4�zxx < 0.zxx zyy � �xxy�2 � �6��4.2� � 02 > 0,��10, �14�
��10, 14, �200.6�zxx zyy � �zxy�2 � �6���4.2� � 02 < 0.��10, 14�
�10, �14, �349.4�zxx zyy � �zxy�2 � ��6��4.2� � 02 < 0.�10, �14�
�10, 14, 199.4�zxx < 0.zxx zyy � �zxy�2 � ��6���4.2� � 02 > 0,�10, 14�
zxy � 0zyy � �0.3y,zxx � �0.6x,
��10, �14���10, 14�,�10, �14�,�10, 14�,
zy � 50 � 0.15y2 � 20.6 � 0, y � ±14
zx � 50 � 0.3x2 � 20 � 0, x � ±10
z � 50�x � y� � �0.1x3 � 20x � 150� � �0.05y3 � 20.6y � 125�
70. The level curves indicate that there is a relative extremum at A, the center of the ellipse in the second quadrant,and that there is a saddle point at B, the origin.
72. Minimize subject to the constraint
C�377.5, 622.5� � 104,997.50
x2 � 622.5
x1 � 377.5
8x1 � 3020
5x1 � 3x2 � 20
x1 � x2 � 1000 ⇒ 3x1 � 3x2 � 3000
0.50x1 � 10 � �
0.30x2 � 12 � � 5x1 � 3x2 � 20
x1 � x2 � 1000.C�x1, x2� � 0.25x12 � 10x1 � 0.15x2
2 � 12x2
74. Minimize the square of the distance:
Clearly and hence: Using a computer algebra system,Thus,distance � 2.08
�distance�2 � �0.6894 � 2�2 � �0.6894 � 2�2 � �2�0.6894�2�2 � 4.3389.x � 0.6894.4x3 � x � 2 � 0.x � y
fx � 2�x � 2� � 2�x2 � y2�2x � 0
fy � 2�y � 2� � 2�x2 � y2�2y � 0 x � 2 � 2x3 � 2xy2 � 0
y � 2 � 2y3 � 2x2y � 0
f �x, y, z� � �x � 2�2 � �y � 2�2 � �x2 � y2 � 0�2.
76. (a)
(b) When km/hr, km.y � 57.8x � 80
y � 0.0045x2 � 0.0717x � 23.2914c � 23.2914,b � 0.0717,a � 0.0045,
34,375a � 375b � 5c � 297
3,515,625a � 34,375b � 375c � 26,900
382,421,875a � 3,515,625b � 34,375c � 2,760,000
� xi3 � 3,515,625� xi
2 � 34,375,
� xi2yi � 2,760,000,
� yi � 297,
� xi yi � 26,900,
� xi � 375,
� xi4 � 382,421,875,
�125, 102��100, 75�,�75, 54�,�50, 38�,�25, 28�,
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Review Exercises for Chapter 12 355
44. Minimize subject to the constraint
Therefore, C�209.65, 186.35� � $16,771.94.
y �89�
200�8�9�0.4� � 186.35
x �200
�8�9�0.4 � 209.65
100x0.6y0.4 � 20,000 ⇒ x0.6�89
x0.4
� 200
yx
�89
⇒ y �89
x
�yx
0.4
�yx
0.6
� � 4860��
40�
36
36 � 40x0.6y�0.6� ⇒ �xy
0.6
�36
40�
48 � 60x�0.4y0.4� ⇒ �yx
0.4
�48
60�
100x0.6y0.4 � 20,000.C�x, y� � 48x � 36y
Review Exercises for Chapter 12
46.
Constraint:
(a) Level curves of are lines of form
Using you obtain
and
Constraint is an ellipse.
−10 10
−8
8
f �7, 3� � 28 � 9 � 37.y � 3,x � 7,
y � �43
x � 12.3,
y � �43
x � C.
f �x, y� � 4x � 3y
x2
64�
y2
36� 1
x, y > 0f �x, y� � ax � by,
(b) Level curves of are lines of form
Using you obtain
and f �4, 5.2� � 62.8.y � 5.2,x � 4,
y � �49
x � 7,
y � �49
x � C.
f �x, y� � 4x � 9y
2. Yes, it is the graph of a function.
4.
The level curves are of the form
The level curves arehyperbolas.
ec � xy.
c � ln xy c = 0
c = 1
c = 2 c = 2
−3 3
−2
21 3
f �x, y� � ln xy 6.
The level curves are of the form
The level curves are passing through the origin with slope
1 � cc
.
y � �1 � cc x.
c �x
x � y
c = 2
c = 2−
c = 1−
c = 1
c = 12
c = − 32
c = − 2
c = 32
−3 3
−2
2
1
f �x, y� �x
x � y
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356 Chapter 12 Functions of Several Variables
8.
yx
5 5
60
z
g�x, y� � y1�x 10.
Elliptic cone
x y5
2
2
z
f �x, y, z� � 9x2 � y2 � 9z2 � 0 12.
Does not exist
Continuous except when y � ± x.
lim�x, y�→�1, 1�
xy
x2 � y2
14.
Continuous everywhere
lim�x, y�→�0, 0�
y � xe�y2
1 � x2 �0 � 01 � 0
� 0 16.
fy �x2
�x � y�2
�y2
�x � y�2 fx �y�x � y� � xy
�x � y�2
f �x, y� �xy
x � y18.
�z�y
�2y
x2 � y2 � 1
�z�x
�2x
x2 � y2 � 1
z � ln�x2 � y 2 � 1�
20.
�w�z
�z
�x2 � y2 � z2
�w�y
�y
�x2 � y2 � z2
�w�x
�12
�x2 � y2 � z2��1�2�2x� �x
�x2 � y2 � z2
w � �x2 � y2 � z2 22.
fz �z
�1 � x2 � y2 � z2�3�2
fy �y
�1 � x2 � y2 � z2�3�2
�x
�1 � x2 � y2 � z2�3�2
fx � �12
�1 � x2 � y2 � z2��3�2��2x�
f �x, y, z� �1
�1 � x2 � y2 � z2
24.
�u�t
� �kc�sin akx� sin kt
�u�x
� akc�cos akx� cos kt
u�x, t� � c�sin akx� cos kt 26.
At
Slope in x-direction.
At
Slope in y-direction.
�z�y
� 4.�2, 0, 0�,�z�y
�x2
1 � y.
�z�x
� 0.�2, 0, 0�, �z�x
� 2x ln�y � 1�.
z � x2 ln�y � 1�
28.
hyx ���x � y�2 � 2y�x � y�
�x � y�4 �x � y
�x � y�3
hxy ��x � y�2 � 2y�x � y�
�x � y�4 �x � y
�x � y�3
hyy �2x
�x � y�3
hxx ��2y
�x � y�3
hy ��x
�x � y�2
hx �y
�x � y�2
h�x, y� �x
x � y30.
gyx � 2 cos�x � 2y�
gxy � 2 cos�x � 2y�
gyy � �4 cos�x � 2y�
gxx � �cos�x � 2y�
gy � 2 sin�x � 2y�
gx � �sin�x � 2y�
g�x, y� � cos�x � 2y�
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Review Exercises for Chapter 12 357
32.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �6x
�z�y
� �6xy
�2z�x2 � 6x
�z�x
� 3x2 � 3y2
z � x3 � 3xy2 34.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �ex sin y
�z�y
� ex cos y
�2z�x2 � ex sin y
�z�x
� ex sin y
z � ex sin y
36.
�y3
�x2 � y2�3�2 dx �x3
�x2 � y2�3�2 dy � ��x2 � y2 y � xy�x��x2 � y2 �x2 � y2 � dx � ��x2 � y2x � xy�y��x2 � y2 �
x2 � y2 � dy
dz ��z�x
dx ��z�y
dy
z �xy
�x2 � y2
38. From the accompanying figure we observe
Letting and
(Note that we express the measurement of the angle in radians.) The maximum error is approximately
dh � tan�11�
60 �±12 � 100 sec2�11�
60 �± �
180 � ±0.3247 ± 2.4814 � ±2.81 feet.
d� � ±�
180.� �
11�
60,dx � ±
12
,x � 100,
dh ��h�x
dx ��h��
d� � tan � dx � x sec2 � d�.
tan � �hx or h � x tan �
x
h
θ
40.
���2r2 � h2��r2 � h2
dr ��rh
�r2 � h2 dh �
� �8 � 25��29 �±1
8 �10�
�29�±18 � ±
43�
8�29
dA � ���r2 � h2 ��r2
�r2 � h2 dr ��rh
�r2 � h2 dh
A � �r�r2 � h2
42.
Chain Rule:
Substitution:
� sin t�1 � 2 cos t�dudt
� 2 sin t cos t � sin t
u � sin2 t � cos t
� sin t�1 � 2 cos t�
� sin t � 2�sin t� cos t
� �1��sin t� � 2y�cos t�
dudt
��u�x
�x�t
��u�y
�y�t
y � sin tx � cos t,u � y2 � x,
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358 Chapter 12 Functions of Several Variables
44.
Chain Rule:
�4r2t � rt2 � 4r3
�2r � t�2
�yz
�1� �xz
�r� �xyz2 ��1�
�w�t
��w�x
�x�t
��w�y
�y�t
��w�z
�z�t
�4r2t � 4rt2 � t3
�2r � t�2
�2rt
2r � t�
�2r � t�t2r � t
�2�2r � t��rt�
�2r � t�2
�yz
�2� �xz
�t� �xyz2 �2�
�w�r
��w�x
�x�r
��w�y
�y�r
��w�z
�z�r
w �xyz
, x � 2r � t, y � rt, z � 2r � t
Substitution:
�w�t
�4r2t � rt2 � 4r3
�2r � t�2
�w�r
�4r2t � 4rt2 � t3
�2r � t�2
w �xyz
��2r � t��rt�
2r � t�
2r2t � rt2
2r � t
46.
�z�y
�sin z
2xz � y cos z
2xz �z�y
� y cos z �z�y
� sin z � 0
�z�x
�z2
y cos z � 2xz
2xz �z�x
� z2 � y cos z �z�x
� 0
xz2 � y sin z � 0
48.
� �4�5
5�
2�55
� �2�5
5Du f �1, 4� � f �1, 4� u
u �1
�5v �
2�55
i ��55
j
f �1, 4� � �2i � 2j
f � �2xi �12
yj
f �x, y� �14
y2 � x2 50.
� 4�3 � �3 � 0 � 5�3
Duw�1, 0, 1� � w�1, 0, 1� u
u �1
�3v �
�33
i ��33
j ��33
k
w�1, 0, 1� � 12i � 3j
w � �12x � 3y�i � �3x � 8yz�j � ��4y2�k
w � 6x2 � 3xy � 4y2z
52.
�z�2, 1�� � 4
z�2, 1� � 4 j
z �x2 � 2xy�x � y�2 i �
x2
�x � y�2 j
z �x2
x � y54.
�z�2, 1�� � 4�2
z�2, 1� � 4i � 4 j
z � 2xy i � x2j
z � x2y
56.
Normal vector: j
f ��
2, 1 � 2j
f �x, y� � 4y cos xi � �4 sin x � 2y�j
f �x, y� � 4y sin x � y2
4y sin x � y2 � 3 58.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
y � 33
�z � 4
4.x � 2,
3y � 4z � 25,3� y � 3� � 4�z � 4� � 0
F�2, 3, 4� � 6j � 8k � 2�3j � 4k�
F � 2yj � 2zk
F�x, y, z� � y2 � z2 � 25 � 0
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Review Exercises for Chapter 12 359
60.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 11
�y � 2
2�
z � 22
.
x � 2y � 2z � 9,
�x � 1� � 2� y � 2� � 2�z � 2� � 0
F�1, 2, 2� � 2i � 4j � 4k � 2�i � 2j � 2k�
F � 2xi � 2y j � 2zk
F�x, y, z� � x2 � y2 � z2 � 9 � 0 62.
Therefore, the equation of the tangent line is
x � 41
�y � 4
1�
z � 9�8
.
F � G � i81
j0
�1
k10 � i � j � 8k
F�4, 4, 9� � 8i � k
G � i � j
F � 2y i � k
G�x, y, z� � x � y � 0
F�x, y, z� � y2 � z � 25 � 0
64. (a)
(c) If you obtain the 2nd degree Taylor polynomial (d)for cos x.
(e)
The accuracy lessens as the distance from increases.�0, 0�
2
2
3
211
−1
−1−2
−2
x
y
z
1
2
1
−1
y
x
z
1
2
1
−1
y
x
z
y � 0,
P1�x, y� � 1 � y
fy � cos y, fy�0, 0� � 1
fx � �sin x, fx�0, 0� � 0
f �x, y� � cos x � sin y, f �0, 0� � 1
66.
Therefore, is a relative minimum.��4, 43 , �2�fxx fyy � � f xy�2 � 4�18� � �6�2 � 36 > 0
fxy � 6
fyy � 18
fxx � 4
4��3y� � 6y � �8 ⇒ y �43 , x � �4
x � �3y fy � 6x � 18y � 0,
fx � 4x � 6y � 8 � 0
f �x, y� � 2x2 � 6xy � 9y2 � 8x � 14
(b)
P2�x, y� � 1 � y �12 x2
fxy � 0, fxy�0, 0� � 0
fyy � �sin y, fyy�0, 0� � 0
fxx � �cos x, fxx�0, 0� � �1
x y
0 0 1.0 1.0 1.0
0 0.1 1.0998 1.1 1.1
0.2 0.1 1.0799 1.1 1.095
0.5 0.3 1.1731 1.3 1.175
1 0.5 1.0197 1.5 1.0
P2�x, y�P1�x, y�f �x, y�
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360 Chapter 12 Functions of Several Variables
68.
Critical Points:
At ,
is a relative maximum.
At ,
is a saddle point.
At ,
is a saddle point.
At ,
is a relative minimum.��10, �14, �749.4�zxx < 0.zxx zyy � �xxy�2 � �6��4.2� � 02 > 0,��10, �14�
��10, 14, �200.6�zxx zyy � �zxy�2 � �6���4.2� � 02 < 0.��10, 14�
�10, �14, �349.4�zxx zyy � �zxy�2 � ��6��4.2� � 02 < 0.�10, �14�
�10, 14, 199.4�zxx < 0.zxx zyy � �zxy�2 � ��6���4.2� � 02 > 0,�10, 14�
zxy � 0zyy � �0.3y,zxx � �0.6x,
��10, �14���10, 14�,�10, �14�,�10, 14�,
zy � 50 � 0.15y2 � 20.6 � 0, y � ±14
zx � 50 � 0.3x2 � 20 � 0, x � ±10
z � 50�x � y� � �0.1x3 � 20x � 150� � �0.05y3 � 20.6y � 125�
70. The level curves indicate that there is a relative extremum at A, the center of the ellipse in the second quadrant,and that there is a saddle point at B, the origin.
72. Minimize subject to the constraint
C�377.5, 622.5� � 104,997.50
x2 � 622.5
x1 � 377.5
8x1 � 3020
5x1 � 3x2 � 20
x1 � x2 � 1000 ⇒ 3x1 � 3x2 � 3000
0.50x1 � 10 � �
0.30x2 � 12 � � 5x1 � 3x2 � 20
x1 � x2 � 1000.C�x1, x2� � 0.25x12 � 10x1 � 0.15x2
2 � 12x2
74. Minimize the square of the distance:
Clearly and hence: Using a computer algebra system,Thus,distance � 2.08
�distance�2 � �0.6894 � 2�2 � �0.6894 � 2�2 � �2�0.6894�2�2 � 4.3389.x � 0.6894.4x3 � x � 2 � 0.x � y
fx � 2�x � 2� � 2�x2 � y2�2x � 0
fy � 2�y � 2� � 2�x2 � y2�2y � 0 x � 2 � 2x3 � 2xy2 � 0
y � 2 � 2y3 � 2x2y � 0
f �x, y, z� � �x � 2�2 � �y � 2�2 � �x2 � y2 � 0�2.
76. (a)
(b) When km/hr, km.y � 57.8x � 80
y � 0.0045x2 � 0.0717x � 23.2914c � 23.2914,b � 0.0717,a � 0.0045,
34,375a � 375b � 5c � 297
3,515,625a � 34,375b � 375c � 26,900
382,421,875a � 3,515,625b � 34,375c � 2,760,000
� xi3 � 3,515,625� xi
2 � 34,375,
� xi2yi � 2,760,000,
� yi � 297,
� xi yi � 26,900,
� xi � 375,
� xi4 � 382,421,875,
�125, 102��100, 75�,�75, 54�,�50, 38�,�25, 28�,
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Problem Solving for Chapter 12 361
78. Optimize subject to the constraint
If If then
Maximum:
Minimum: f �0, 1� � 0
f �43 , 13� �
1627
x �43 .y �
13 ,x � 4y,y � 1.x � 0,
x � 2y � 2
2xy � �
x2 � 2�� x2 � 4xy ⇒ x � 0 or x � 4y
x � 2y � 2.f �x, y� � x2y
Problem Solving for Chapter 12
2.
Material
Hence,
Then,
The tank is a sphere of radius r � 5�6��
1�3
.
h �1000 � �4�3���750���
�r2 � 0.
r3 �750�
⇒ r � 5�6��
1�3
.
r3�83
�� � 2000
8�r �163
�r �2000
r 2
dMdr
� 8�r �2000
r2 �163
�r � 0
� 4�r2 �2000
r�
83
�r2
M � 4�r2 � 2�r�1000 � �4�3��r3
�r2 �
V � 1000 ⇒ h �1000 � �4�3��r3
�r2
� M � 4�r2 � 2�rh
V �4
3�r3 � �r2h
4. (a) As and hence
(b) Let be a point on the graph of f.
The line through this point perpendicular to g is
This line intersects g at the point
The square of the distance between these two points is
h is a maximum for Hence, the point on f farthest from g is � 13�2
, �1
3�2�.x0 �13�2
.
h�x0� �12
�x0 � 3�x03 � 1�2
.
�12
x0 � 3�x03 � 1, 1
2x0 � 3�x0
3 � 1�.
y � �x � x0 � 3�x03 � 1.
�x0, �x03 � 1�1�3�
limx→�
f �x� � g�x� � limx→��
f �x� � g�x� � 0.−6
−4
6
4x →±�, f �x� � �x3 � 1�1�3 → x
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362 Chapter 12 Functions of Several Variables
6.
Then
Setting you obtain x � y � z � 10.Hx � Hy � 0,
H � 6k�xy �1000
y�
1000x �.
V � xyz � 1000 ⇒ z �1000
xy.
� k�6xy � 6xz � 6yz�
Heat Loss � H � k�5xy � xy � 3xz � 3xz � 3yz � 3yz�
8. (a)
ellipse
(b) On
Inside:
T�±�32
, �12� �
494
maximum
T�0, 12� �
394
minimum
Tx � 4x � 0, Ty � 2y � 1 � 0 ⇒ �0, 12�
T��y� � �2y � 1 � 0 ⇒ y � �12
, x � ±�32
.
x2 � y2 � 1, T�x, y� � T�y� � 2�1 � y2� � y2 � y � 10 � 12 � y2 � y
x2
1�8�
�y � �1�2��2
1�4� 1
2x2 � �y �12�
2
�14
2x2 � y2 � y �14
�14
− 12
12
− 12
12
y
x
T�x, y� � 2x2 � y2 � y � 10 � 10
10.
Similarly,
Similarly,
Now observe that
Thus, Laplaces equation in cylindrical coordinates, is 2ur2 �
1r ur
�1r2
2u 2 �
2uz2 � 0.
�2ux2 �
2uy2 �
2uz2.
� �2ux2 sin2 �
2uy2 cos2 � 2
2uxy
sin cos �1r ux
cos �1r uy
sin � �2uz2
2ur2 �
1r ur
�1r2
2u 2 �
2uz2 � �2u
x2 cos2 �2uy2 sin2 � 2
2uxy
cos sin � �1r�
ux
cos �uy
sin �
2ur2 �
2ux2 cos2 �
2uy2 sin2 � 2
2 uxy
cos sin .
�2ux2 r2 sin2 �
2uy2 r2 cos2 � 2
2uxy
r2 sin cos �ux
r cos �uy
r sin
� �r cos �� 2uyx
x
�2uy2
y
�2uyz
z� � r
uy
sin
2u 2 � ��r sin ��2u
x2 x
�2u
xy y
�2uxz
z� � r
ux
cos
ur
�ux
cos �uy
sin .
�ux
��r sin � �uy
r cos
u
�ux
x
�uy
y
�uz
z
x � r cos , y � r sin , z � z
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Problem Solving for Chapter 12 363
12. (a)
� 16t�t2 � 4�2t � 16
� �4096t2 � 1024�2t3 � 256t4
d � �x2 � y2 � ��32�2t�2 � �32�2t � 16t2�2 (b)dddt
�32�t2 � 3�2t � 8��t2 � 4�2t � 16
(c) When
dddt
�32�12 � 6�2��20 � 8�2
38.16 ft�sec
t � 2: (d)
when seconds. No. The projectile is at itsmaximum height when t � �2.
t 1.943
d 2ddt2
�32�t3 � 6�2t2 � 36t � 32�12�
�t2 � 4�2t � 16�3�2� 0
14. Given that is a differentiable function such that then and Therefore, the tangent plane is or which is horizontal.z � z0 � f �x0, y0���z � z0� � 0
fy�x0, y0� � 0.fx�x0, y0� � 0�f �x0, y0� � 0,f
16.
y � r sin � 5 sin �
18 0.868
x � r cos � 5 cos �
18 4.924
dr � ±0.05, d � ±0.05
x
1
4
5
2
31 4 52
3
518π
=θ 18π θ(r, ) = 5,( )
y�r, � � �5, �
18�
(a) should be more effected by changes in
is more effected by changes in because0.985 > 0.868.
rdx
�0.985�dr � 0.868 d
dx � �cos �dr � ��r sin �d
r.dx (b) should be more effected by changes in .
is more effected by because 4.924 > 0.174.dy
0.174 dr � 4.924 d
dy � sin dr � r cos d
dy
18.
Then,2ut2 �
2ux2.
2ux2 �
12
�sin�x � t� � sin�x � t�
ux
�12
cos�x � t� � cos�x � t�
2ut2 �
12
�sin�x � t� � sin�x � t�
ut
�12
�cos�x � t� � cos�x � t�
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C H A P T E R 1 2Functions of Several Variables
Section 12.1 Introduction to Functions of Several Variables . . . . . . . 76
Section 12.2 Limits and Continuity . . . . . . . . . . . . . . . . . . . . 80
Section 12.3 Partial Derivatives . . . . . . . . . . . . . . . . . . . . . . 83
Section 12.4 Differentials . . . . . . . . . . . . . . . . . . . . . . . . . 88
Section 12.5 Chain Rules for Functions of Several Variables . . . . . . . 92
Section 12.6 Directional Derivatives and Gradients . . . . . . . . . . . . 98
Section 12.7 Tangent Planes and Normal Lines . . . . . . . . . . . . . 103
Section 12.8 Extrema of Functions of Two Variables . . . . . . . . . . 109
Section 12.9 Applications of Extrema of Functions of Two Variables . 113
Section 12.10 Lagrange Multipliers . . . . . . . . . . . . . . . . . . . . 119
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 123
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 129
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C H A P T E R 1 2Functions of Several Variables
Section 12.1 Introduction to Functions of Several VariablesSolutions to Odd-Numbered Exercises
76
1.
Yes, z is a function of x and y.
z �10 � xyx2 � y
z�x2 � y� � 10 � xy
x2z � yz � xy � 10 3.
No, z is not a function of x and y. For example,corresponds to both z � ±1.�x, y� � �0, 0�
x2
4�
y2
9� z2 � 1
5.
(a) f �3, 2� �32
f �x, y� �xy
(b) f ��1, 4� � �14
(c) f �30, 5� �305
� 6
(d) f �5, y� �5y
(e) f �x, 2� �x2
(f) f �5, t� �5t
7.
(a)
(b)
(c)
(d)
(e)
(f) f �t, t� � tet
f �x, 2� � xe2
f �5, y� � 5ey
f �2, �1� � 2e�1 �2e
f �3, 2� � 3e2
f �5, 0� � 5e0 � 5
f �x, y� � xey 9.
(a)
(b) h�1, 0, 1� ��1��0�
1� 0
h�2, 3, 9� ��2��3�
9�
23
h�x, y, z� �xyz
11.
(a)
(b) f �3, 1� � 3 sin 1
f �2, �
4� � 2 sin �
4� �2
f �x, y� � x sin y 13.
(a)
(b) g�1, 4� � �4
1�2t � 3� dt � �t 2 � 3t
4
1� 6
g�0, 4� � �4
0�2t � 3� dt � �t 2 � 3t
4
0� 4
g�x, y� � �y
x
�2t � 3� dt
15.
(a)
(b)x2 � 2y � 2�y � x2 � 2y
�y�
�2�y�y
� �2, �y � 0 f �x, y � �y� � f �x, y�
�y�
x2 � 2�y � �y�� � �x2 � 2y��y
�
�x2 � 2x��x� � ��x�2 � 2y � x2 � 2y
�x�
�x�2x � �x��x
� 2x � �x, �x � 0
f �x � �x, y� � f �x, y�
�x�
�x � �x�2 � 2y� � �x2 � 2y��x
f �x, y� � x2 � 2y
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Section 12.1 Introduction to Functions of Several Variables 77
17.
Domain:
Range: 0 ≤ z ≤ 2
��x, y�: x2 � y2 ≤ 4
x2 � y2 ≤ 4
4 � x2 � y2 ≥ 0
f �x, y� � �4 � x2 � y2 19.
Domain:
Range: ��
2 ≤ z ≤
�
2
��x, y�: �1 ≤ x � y ≤ 1
f �x, y� � arcsin�x � y� 21.
Domain:
Range: all real numbers
��x, y�: y < �x � 4
x � y < 4
4 � x � y > 0
f �x, y� � ln�4 � x � y�
23.
Domain: and
Range: all real numbers
y � 0 ��x, y�: x � 0
z �x � y
xy25.
Domain:
Range: z > 0
��x, y�: y � 0
f �x, y� � ex�y 27.
Domain: and
Range: all real numbers except zero
y � 0 ��x, y�: x � 0
g�x, y� �1xy
29.
(a) View from the positive x-axis:
(c) View from the first octant: �20, 15, 25�
�20, 0, 0�
f �x, y� ��4x
x2 � y2 � 1
(b) View where x is negative, y and z are positive:
(d) View from the line in the xy-plane: �20, 20, 0�y � x
��15, 10, 20�
31.
Plane: z � 5
x
y422
4
4
zf �x, y� � 5 33.
Since the variable x is missing, the surface is a cylinderwith rulings parallel to the x-axis. The generating curve is
The domain is the entire xy-plane and the range is
x
y2 31
4
4
5
z
z ≥ 0.z � y2.
f �x, y� � y2
35.
Paraboloid
Domain: entire xy-plane
Range: z ≤ 4
x
y2 3
−3
3
4
zz � 4 � x2 � y2 37.
Since the variable y is missing, thesurface is a cylinder with rulings parallel to the y-axis. The generatingcurve is The domain is theentire xy-plane and the range is z > 0.
z � e�x.
x
y44
2
4
6
8
z
f �x, y� � e�x
41.
x
y
z
f �x, y� � x2e��xy�2�39.
Hyperbolic paraboloid
Domain: entire xy-plane
Range: �� < z < �
x
y
zz � y2 � x2 � 1
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78 Chapter 12 Functions of Several Variables
45.
Level curves:
Circles centered at
Matches (c)
�0, 0�
x2 � y2 � 1 � ln c
ln c � 1 � x2 � y2
c � e1�x2�y2
z � e1�x2�y247.
Level curves:
Parabolas
Matches (b)
y � x2 ± ec
±ec � y � x2
c � ln�y � x2�
z � ln�y � x2� 49.
Level curves are parallel lines ofthe form
4
4
2
2
−2
−2x
c = −1 c = 0
c = 2
c = 4
y
x � y � c.
z � x � y
51.
The level curves are of the form
Thus, the level curves are circles of radius 5 or less,centered at the origin.
6
6
2−2
−2
2
−6
−6x
y
c = 5c = 4
c = 3c = 2
c = 1
c = 0
x2 � y2 � 25 � c2.
c � �25 � x2 � y2,
f �x, y� � �25 � x2 � y2 53.
The level curves are hyperbolas of the form
1
1
−1
−1x
c = 6c = 5c = 4c = 3c = 2c = 1
c = −1c = −2c = −3c = −4c = −5c = −6
y
xy � c.
f �x, y� � xy
43.
(a)
(b) g is a vertical translation of f two units upward
x
y22 1
−2
4
5
z
f �x, y� � x2 � y2
(c) g is a horizontal translation of f two units to the right.The vertex moves from to
(d) g is a reflection of f in the xy-plane followed by a ver-tical translation 4 units upward.
(e)
x
y2
2
4
5
z
z = f (x, 1)
x
y22
4
5
z
z = f (1, y)
�0, 2, 0�.�0, 0, 0�
55.
The level curves are of the form
Thus, the level curves are circles passing through the origin and centered at �1�2c, 0�.
�x �12c�
2
� y2 � � 12c�
2
x2 �xc
� y2 � 0x
2
12
2c = 1
c = −1
c = −2
c = −
32
c = −
32
c = 12
c =
c = 2
y
c �x
x2 � y2
f �x, y� �x
x2 � y2 57.
−9
−6
9
6f �x, y� � x2 � y2 � 2
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Section 12.1 Introduction to Functions of Several Variables 79
65. The surface is sloped like a saddle. The graph is notunique. Any vertical translation would have the same levelcurves. One possible function is
f �x, y� � x2 � y2.
67. V�I, R� � 1000�1 � 0.10�1 � R�1 � I
10
Inflation Rate
Tax Rate 0 0.03 0.05
0 2593.74 1929.99 1592.33
0.28 2004.23 1491.34 1230.42
0.35 1877.14 1396.77 1152.40
75.
(a) board-feetN�22, 12� � �22 � 44 �
2
�12� � 243
N�d, L� � �d � 44 �
2
L
(b) board-feetN�30, 12� � �30 � 44 �
2
�12� � 507
59.
−6
−4
6
4
g�x, y� �8
1 � x2 � y2 61. See Definition, page 838. 63. No, The following graphs are nothemispheres.
z � x2 � y2
z � e��x2�y2�
69.
Plane
x
y
−3
6
3
z
6 � x � 2y � 3z
c � 6
f �x, y, z� � x � 2y � 3z 71.
Sphere
xy
−4
−4
4
44
z
9 � x2 � y2 � z2
c � 9
f �x, y, z� � x2 � y2 � z2 73.
Elliptic cone
xy
−2
−2
2
212
z
0 � 4x2 � 4y2 � z2
c � 0
f �x, y, z� � 4x2 � 4y2 � z2
77.
The level curves are of the form
The level curves are circles centered at the origin.
30
30
y
x
−30
c = 600c = 500c = 400
c = 300c = 200c = 100c = 0
−30
x2 � y 2 �600 � c
0.75.
c � 600 � 0.75x2 � 0.75y2
T � 600 � 0.75x2 � 0.75y2 79.
y
zx
� 0.75xy � 0.80�xz � yz�
base � front & back � two ends
C � 0.75xy � 2�0.40�xz � 2�0.40�yz
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81.
(a)
(b)
The level curves are of the form:
Thus, the level curves are lines through the origin with slope 5203c
.
V �5203c
T
c � �5203 ��T
V �
P �kTV
�5203 �T
V �
k �20�2600�
300�
5203
PV � kT, 20�2600� � k�300�
83. (a) Highest pressure at C
(b) Lowest pressure at A
(c) Highest wind velocity at B
85. (a) The boundaries between colors represent level curves
(b) No, the colors represent intervals of different lengths,as indicated in the box
(c) You could use more colors, which means usingsmaller intervals
87. False. Let
but 1 � 2f �1, 2� � f �2, 1�,
f �x, y� � 2xy
89. False. Let
Then, f �2x, 2y� � 5 � 22 f �x, y�.
f �x, y� � 5.
Section 12.2 Limits and Continuity
1. Let be given. We need to find such that
whenever Take
Then if we have
�y � b� < �.
��y � b�2 < �
0 < ��x � a�2 � �y � b�2 < � � �,
� � �.0 < ��x � a�2 � �y � b�2 < �.
� f �x, y� � L� � �y � b� < �� > 0� > 0
3. lim�x, y�→�a, b�
� f �x, y� � g�x, y� � lim�x, y�→�a, b�
f �x, y� � lim�x, y�→�a, b�
g�x, y� � 5 � 3 � 2
5. lim�x, y�→�a, b�
� f �x, y�g�x, y� � lim�x, y�→�a, b�
f �x, y�� lim�x, y�→�a, b�
g�x, y�� � 5�3� � 15
7.
Continuous everywhere
lim�x, y�→�2, 1�
�x � 3y2� � 2 � 3�1�2 � 5 9.
Continuous for x � y
lim�x, y�→�2, 4�
x � yx � y
�2 � 42 � 4
� �3
11.
Continuous for xy � �1, y � 0, �x�y� ≤ 1
lim�x, y�→�0, 1�
arcsin�x�y�
1 � xy� arcsin 0 � 0 13.
Continuous everywhere
lim�x, y�→��1, 2�
e xy � e�2�1e2
15.
Continuous for x � y � z ≥ 0
lim�x, y, z�→�1, 2, 5�
�x � y � z � �8 � 2�2 17.
Continuous everywhere
lim�x, y�→�0, 0�
exy � 1
80 Chapter 12 Functions of Several Variables
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19.
The limit does not exist.
Continuous except at �0, 0�
lim�x, y�→�0, 0�
ln�x2 � y2� � ln�0� � ��
21.
Continuous except at
Path:
Path:
The limit does not exist because along the path the function equals 0, whereas along the path the function equals 12 .
y � xy � 0
y � x
y � 0
�0, 0�
f �x, y� �xy
x2 � y2
0 0 0 0 0f �x, y�
�0.001, 0��0.01, 0��0.1, 0��0.5, 0��1, 0��x, y�
12
12
12
12
12f �x, y�
�0.001, 0.001��0.01, 0.01��0.1, 0.1��0.5, 0.5��1, 1��x, y�
23.
Continuous except at
Path:
Path:
The limit does not exist because along the path the function equals whereas along the path the function equals 12 .
x � �y2�12 ,x � y2
x � �y2
x � y2
�0, 0�
f �x, y� � �xy2
x2 � y4
12
12
12
12
12f �x, y�
��0.000001, 0.001���0.0001, 0.01���0.01, 0.1���0.25, 0.5���1, 1��x, y�
25.
(same limit for g)
Thus, f is not continuous at whereas g is continuousat �0, 0�.
�0, 0�,
� lim�x, y�→�0, 0�
�1 �2xy2
x2 � y2� � 1
lim�x, y�→�0, 0�
f �x, y� � lim�x, y�→�0, 0�
�x2 � 2xy2 � y2
x2 � y2 � 27.
x
y
z
lim�x, y�→�0, 0�
�sin x � sin y� � 0
29.
Does not exist
x
y
z
lim�x, y�→�0, 0�
x2y
x4 � 4y2 31.
The limit does not exist. Use the paths and
z
x
y
x � y.x � 0
f �x, y� �10xy
2x2 � 3y2
�12�
12�
12�
12�
12f �x, y�
�0.000001, 0.001��0.0001, 0.01��0.01, 0.1��0.25, 0.5��1, 1��x, y�
Section 12.2 Limits and Continuity 81
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33. lim�x, y�→�0, 0�
sin�x2 � y2�
x2 � y2 � limr→0
sin r2
r2 � limr→0
2r cos r2
2r� lim
r→0 cos r2 � 1
35. lim�x, y�→�0, 0�
x3 � y3
x2 � y2 � limr→0
r3 �cos3 � sin3 �
r2 � limr→0
r�cos3 � sin3 � � 0
37.
Continuous except at �0, 0, 0�
f �x, y, z� �1
�x2 � y2 � z239.
Continuous everywhere
f �x, y, z� �sin z
ex � ey
41.
Continuous everywhere
� 9x2 � 12xy � 4y2
� �3x � 2y�2
f �g�x, y�� � f �3x � 2y�
g�x, y� � 3x � 2y
f �t� � t 2 43.
Continuous for y �3x2
f �g�x, y�� � f �3x � 2y� �1
3x � 2y
g�x, y� � 3x � 2y
f �t� �1t
45.
(a)
(b)
� limy→0
�4y
y� lim
y→0 ��4� � �4
limy→0
f �x, y � y� � f �x, y�
y� lim
y→0 �x2 � 4�y � y� � �x2 � 4y�
y
� limx→0
2xx � �x�2
x� lim
x→0 �2x � x� � 2x
limx→0
f �x � x, y� � f �x, y�
x� lim
x→0 ��x � x�2 � 4y � �x2 � 4y�
x
f �x, y� � x2 � 4y
47.
(a)
(b)
� limy→0
xy � 3y
y� lim
y→0 �x � 3� � x � 3
limy→0
f �x, y � y� � f �x, y�
y� lim
y→0 �2x � x�y � y� � 3�y � y� � �2x � xy � 3y�
y
� limx→0
2x � xy
x� lim
x→0 �2 � y� � 2 � y
limx→0
f �x � x, y� � f �x, y�
x� lim
x→0 �2�x � x� � �x � x�y � 3y � �2x � xy � 3y�
x
f �x, y� � 2x � xy � 3y
49. See the definition on page 851.
Show that the value of is not the same
for two different paths to �x0, y0�.
lim�x, y�→�x0, y0�
f �x, y�
51. No.
The existence of has no bearing on the existence ofthe limit as �x, y� → �2, 3�.
f �2, 3�
82 Chapter 12 Functions of Several Variables
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Section 12.3 Partial Derivatives
Section 12.3 Partial Derivatives 83
1. fx�4, 1� < 0 3. fy�4, 1� > 0 5.
fy�x, y� � �3
fx�x, y� � 2
f �x, y� � 2x � 3y � 5
7.
�z�y
�x
2�y
�z�x
� �y
z � x�y 9.
�z�y
� �5x � 6y
�z�x
� 2x � 5y
z � x2 � 5xy � 3y2 11.
�z�y
� 2x2e2y
�z�x
� 2xe2y
z � x2e2y
13.
�z�y
�2y
x2 � y2
�z�x
�2x
x2 � y2
z � ln�x2 � y2� 15.
�z�y
�1
x � y�
1x � y
�2x
x2 � y2
�z�x
�1
x � y�
1x � y
� �2y
x2 � y2
z � ln�x � yx � y� � ln�x � y� � ln�x � y�
17.
�z�y
� �x2
2y2 �8yx
��x3 � 16y3
2xy2
�z�x
�2x2y
�4y2
x2 �x3 � 4y3
x2y
z �x2
2y�
4y2
x19.
hy�x, y� � �2ye��x2�y2�
hx�x, y� � �2xe��x2�y2�
h�x, y� � e��x2�y2�
21.
fy�x, y� �12
�x2 � y2��1�2 �2y� �y
�x2 � y2
fx�x, y� �12
�x2 � y2��1�2 �2x� �x
�x2 � y2
f �x, y� � �x2 � y2 23.
�z�y
� �sec2�2x � y�
�z�x
� 2 sec2�2x � y�
z � tan�2x � y�
53. Since then for there corresponds such that whenever
Since then for there corresponds such that whenever
Let be the smaller of and By the triangle inequality, whenever we have
Therefore, lim�x, y�→�a, b�
� f �x, y� � g�x, y� � L1 � L 2.
≤ f �x, y� � L1 � g�x, y� � L 2 <�
2�
�
2� �. f �x, y� � g�x, y� � �L1 � L 2� � � f �x, y� � L1� � �g�x, y� � L 2�
��x � a�2 � �y � b�2 < �,�2.�1�
0 < ��x � a�2 � �y � b�2 < �2.
g�x, y� � L2 < ��2�2 > 0��2 > 0,lim�x, y�→�a, b�
g�x, y� � L2,
0 < ��x � a�2 � �y � b�2 < �1.
f �x, y� � L1 < ��2�1 > 0��2 > 0,lim�x, y�→�a, b�
f �x, y� � L1,
55. True 57. False. Let
See Exercise 19.
f �x, y� � �ln�x2 � y2�,0,
�x, y� � �0, 0�x � 0, y � 0
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84 Chapter 12 Functions of Several Variables
29.
�f�y
� limy→0
f �x, y � y� � f �x, y�
y� lim
y→0 2x � 3�y � y� � 2x � 3y
y� lim
y→0 3yy
� 3
�f�x
� limx→0
f �x � x, y� � f �x, y�
x� lim
x→0 2�x � x� � 3y � 2x � 3y
x� lim
x→0 2xx
� 2
f �x, y� � 2x � 3y
31.
� limy→0
1
�x � y � y � �x � y�
1
2�x � y
� limy→0
��x � y � y � �x � y ���x � y � y � �x � y �
y��x � y � y � �x � y
�f�y
� limy→0
f �x, y � y� � f �x, y�
y� lim
y→0 �x � y � y � �x � y
y
� limx→0
1
�x � x � y � �x � y�
1
2�x � y
� limx→0
��x � x � y � �x � y ���x � x � y � �x � y �
x��x � x � y � �x � y �
�f�x
� limx→0
f �x � x, y� � f �x, y�
x� lim
x→0 �x � x � y � �x � y
x
f �x, y� � �x � y
33.
At
At �1, 1�: gy�1, 1� � �2
gy�x, y� � �2y
�1, 1�: gx�1, 1� � �2
gx�x, y� � �2x
g�x, y� � 4 � x2 � y2 35.
At
At �0, 0�: �z�y
� 0
�z�y
� �e�x sin y
�0, 0�: �z�x
� �1
�z�x
� �e�x cos y
z � e�x cos y
25.
� ey�x cos xy � sin xy�
�z�y
� ey sin xy � xey cos xy
�z�x
� yey cos xy
z � ey sin xy 27.
[You could also use the Second Fundamental Theorem ofCalculus.]
fy�x, y� � y2 � 1
fx�x, y� � �x2 � 1 � 1 � x2
� �t3
3� t
y
x� �y3
3� y� � �x3
3� x�
f �x, y� � �y
x
�t 2 � 1� dt
37.
At
At �2, �2�: fy�2, �2� �14
fy�x, y� �1
1 � � y2�x2� �1x� �
xx2 � y2
�2, �2�: fx�2, �2� �14
fx�x, y� �1
1 � � y2�x2� ��yx2� �
�yx2 � y2
f �x, y� � arctan yx
39.
At
At �2, �2�: fy�2, �2� �14
fy�x, y� �x�x � y� � xy
�x � y�2 �x2
�x � y�2
�2, �2�: fx�2, �2� � �14
fx�x, y� �y�x � y� � xy
�x � y�2 ��y2
�x � y�2
f �x, y� �xy
x � y
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Section 12.3 Partial Derivatives 85
41.
Intersecting curve:
At
yx
x = 210
88
z
�2, 3, 6�: �z�y
��3
�45 � 9� �
12
�z�y
��y
�45 � y2
z � �45 � y2
�2, 3, 6�z � �49 � x2 � y2, x � 2, 43.
Intersecting curve:
At
z
x
y
160
2
43 4
y = 3
�1, 3, 0�: �z�x
� 18�1� � 18
�z�x
� 18x
z � 9x2 � 9
z � 9x2 � y2, y � 3, �1, 3, 0�
45.
Solving for x and y,
and y � 4.x � �6
4x � 2y � �16
2x � 4y � 4fx � fy � 0:
fy�x, y� � 4x � 2y � 16fx�x, y� � 2x � 4y � 4, 47.
and
and
Points: �1, 1�
y � y4 ⇒ y � 1 � x
x �1y2y �
1x2
�1y2 � x � 0�
1x2 � y � 0fx � fy � 0:
fy�x, y� � �1y2 � xfx�x, y� � �
1x2 � y,
49. (a) The graph is that of
(b) The graph is that of fx.
fy. 51.
�w�z
�z
�x2 � y2 � z2
�w�y
�y
�x2 � y2 � z2
�w�x
�x
�x2 � y2 � z2
w � �x2 � y2 � z2 53.
Fz�x, y, z� �z
x2 � y2 � z2
Fy�x, y, z� �y
x2 � y2 � z2
Fx�x, y, z� �x
x2 � y2 � z2
�12
ln�x2 � y2 � z2�
F�x, y, z� � ln �x2 � y2 � z2
55.
Hz�x, y, z� � 3 cos�x � 2y � 3z�
Hy�x, y, z� � 2 cos�x � 2y � 3z�
Hx�x, y, z� � cos�x � 2y � 3z�
H�x, y, z� � sin�x � 2y � 3z� 57.
�2z
�x�y� �2
�2z�y2 � 6
�z�y
� �2x � 6y
�2z
�y�x� �2
�2z�x2 � 2
�z�x
� 2x � 2y
z � x2 � 2xy � 3y2 59.
�2z
�x�y�
�xy�x2 � y2�3�2
�2z�y2 �
x2
�x2 � y2�3�2
�z�y
�y
�x2 � y2
�2z
�y�x�
�xy�x2 � y2�3�2
�2z�x2 �
y2
�x2 � y2�3�2
�z�x
�x
�x2 � y2
z � �x2 � y2
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86 Chapter 12 Functions of Several Variables
61.
�2z
�x�y� ex sec2 y
�2z�y2 � 2ex sec2 y tan y
�z�y
� ex sec2 y
�2z
�y�x� ex sec2 y
�2z�x2 � ex tan y
�z�x
� ex tan y
z � ex tan y 63.
�2z
�x�y�
�x2 � y2� � x�2x��x2 � y2�2 �
y 2 � x2
�x2 � y2�2
�2z�y2 �
�2xy�x2 � y2�2
�z�y
�1
1 � � y2�x2� �1x� �
xx2 � y2
�2z
�y�x�
��x2 � y2� � y�2y��x2 � y2�2 �
y2 � x2
�x2 � y2�2
�2z�x2 �
2xy�x2 � y2�2
�z�x
�1
1 � � y2�x2� ��yx2� �
�yx2 � y2
z � arctan yx
65.
Therefore,
There are no points for which because
�z�x
� sec y � 0.
zx � 0 � zy,
�2z�y�x
��2z
�x�y.
�2z
�x�y� sec y tan y
�2z�y2 � x sec y�sec2 y � tan2 y�
�z�y
� x sec y tan y
�2z
�y�x� sec y tan y
�2z�x2 � 0
�z�x
� sec y
z � x sec y 67.
There are no points for which zx � zy � 0.
�2z
�x�y�
4xy�x2 � y2�2
�2z�y2 �
2�y2 � x2��x2 � y2�2
�z�y
� �2y
x2 � y2
�2z
�y�x�
4xy�x2 � y2�2
�2z�x2 �
x4 � 4x2y2 � y4
x2�x2 � y2�2
�z�x
�1x
�2x
x2 � y2 �y2 � x2
x�x2 � y2�
z � ln� xx2 � y2� � ln x � ln�x2 � y2�
69.
Therefore, fxyy � fyxy � fyyx � 0.
fyxy�x, y, z� � 0
fxyy�x, y, z� � 0
fyyx�x, y, z� � 0
fyx�x, y, z� � z
fxy�x, y, z� � z
fyy�x, y, z� � 0
fy�x, y, z� � xz
fx�x, y, z� � yz
f �x, y, z� � xyz 71.
Therefore, fxyy � fyxy � fyyx.
fyxy�x, y, z� � z2e�x sin yz
fxyy�x, y, z� � z2e�x sin yz
fyyx�x, y, z� � z2e�x sin yz
fyx�x, y, z� � �ze�x cos yz
fxy�x, y, z� � �ze�x cos yz
fyy�x, y, z� � �z2e�x sin yz
fy�x, y, z� � ze�x cos yz
fx�x, y, z� � �e�x sin yz
f �x, y, z� � e�x sin yz 73.
Therefore,�2z�x2 �
�2z�y2 � 0 � 0 � 0.
�2z�y2 � 0
�z�y
� 5x
�2z�x2 � 0
�z�x
� 5y
z � 5xy
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Section 12.3 Partial Derivatives 87
75.
Therefore,�2z�x2 �
�2z�y2 � ex sin y � ex sin y � 0.
�2z�y2 � �ex sin y
�z�y
� ex cos y
�2z�x2 � ex sin y
�z�x
� ex sin y
z � ex sin y 77.
Therefore,�2z�t2 � c2 �2z
�x2.
�2z�x2 � �sin�x � ct�
�z�x
� cos�x � ct�
�2z�t2 � �c2 sin�x � ct�
�z�t
� �c cos�x � ct�
z � sin�x � ct�
79.
Therefore,�z�t
� c2 �2z�x2.
�2z�x2 � �
1c2 e�t cos
xc
�z�x
� �1c
e�t sin xc
�z�t
� �e�t cos xc
z � e�t cos xc
81. See the definition on page 859.
83.
denotes the slope of the surface in the x-direction.
denotes the slope of the surface in the y-direction. �f�y
�f�x
x y
Plane: x = x0
(x0, y0, z0)z
x
Plane: y = y0
y
(x0, y0, z0)z 85. The plane satisfies
and
x
y
−6
6
8
z
�f�y
> 0.�f�x
> 0
z � x � y � f �x, y�
87. (a)
�C�y �80, 20�
� 16�4 � 205 � 237
�C�y
� 16�xy
� 205
�C�x �80, 20�
� 16�14
� 175 � 183
�C�x
� 16�yx
� 175
C � 32�xy � 175x � 205y � 1050 (b) The fireplace-insert stove results in the costincreasing at a faster rate because
�C�y
> �C�x
.
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89. An increase in either price will cause a decrease indemand.
91.
�T�y
� �3y ��T�y
�2, 3� � �9��m
�T�x
� �1.2x, �T�x
�2, 3� � �2.4��m
T � 500 � 0.6x2 � 1.5y2
93.
� �mRTVP
� �mRTmRT
� �1
�T�P
��P�V
��V�T
� � VmR� ��
mRTV 2 � �mR
P �
V �mRT
P ⇒ �V
�T�
mRP
P �mRT
V ⇒ �P
�V� �
mRTV 2
T �PVmR
⇒ �T�P
�V
mR
PV � mRT 95. (a)
(b) As the consumption of skim milk increases,the consumption of whole milk decreases.
Similarly, as the consumption of reduced-fat milkincreases, the consumption of whole milk
decreases.�z��y�
�z��x�
�z�x
� �1.09
�z�x
� �1.83
97.
(a)
(b)
(c)
(d) or or both are not continuous at �0, 0�.fxyfyx
fyx�0, 0� ��
�x ��f�y���0, 0�
� lim�x→0
fy��x, 0� � fy�0, 0�
�x� lim
�x→0
�x ���x�4����x�2�2��x� � lim
�x→0 1 � 1
fxy�0, 0� ��
�y ��f�x���0, 0�
� lim�y→0
fx�0, �y� � fx�0, 0�
�y� lim
�y→0 �y ����y�4����y�2�2��y� � lim
�y→0 ��1� � �1
fy�0, 0� � lim�y→0
f �0, �y� � f �0, 0��y
� lim�y→0
0����y�2 � 0
�y� 0
fx�0, 0� � lim�x→0
f ��x, 0� � f �0, 0�
�x� lim
�x→0 0����x�2 � 0
�x� 0
fy�x, y� ��x2 � y2��x3 � 3xy2� � �x3y � xy3��2y�
�x2 � y2�2 �x�x4 � 4x2y2 � y4�
�x2 � y2�2
fx�x, y� ��x 2 � y 2��3x2y � y3� � �x3y � xy3��2x�
�x2 � y2�2 �y�x4 � 4x2y2 � y4�
�x2 � y2�2
f �x, y� � xy�x2 � y2�
x2 � y2 , �x, y� �0, 0�
0, �x, y� � �0, 0�
99. True 101. True
Section 12.4 Differentials
1.
dz � 6xy3 dx � 9x2y2 dy
z � 3x2y3 3.
�2
�x2 � y2�2 �x dx � y dy�
dz �2x
�x2 � y2�2 dx �2y
�x2 � y2�2 dy
z ��1
x2 � y2
88 Chapter 12 Functions of Several Variables
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5.
dz � �cos y � y sin x� dx � ��x sin y � cos x� dy � �cos y � y sin x� dx � �x sin y � cos x� dy
z � x cos y � y cos x
7.
dz � �ex sin y� dx � �ex cos y� dy
z � ex sin y 9.
dw � 2z3y cos x dx � 2z3 sin x dy � 6z2y sin x dz
w � 2z3y sin x
11. (a)
(b)
� �2�0.05� � 4�0.1� � �0.5
dz � �2x dx � 2y dy
�z � f �1.05, 2.1� � f �1, 2� � �0.5125
f �1.05, 2.1� � 3.4875
f �1, 2� � 4 13. (a)
(b)
� �sin 2��0.05� � �cos 2��0.1� � 0.00385
dz � sin y dx � x cos y dy
�z � f �1.05, 2.1� � f �1, 2� � �0.00293
f �1.05, 2.1� � 1.05 sin 2.1
f �1, 2� � sin 2
15. (a)
(b)
� 3�0.05� � 4�0.1� � �0.25
dz � 3 dx � 4 dy
�z � �0.25
f �1.05, 2.1� � �5.25
f �1, 2� � �5
17. Let Then:
��5.05�2 � �3.1�2 � �52 � 32 �5
�52 � 32�0.05� �
3�52 � 32
�0.1� �0.55�34
� 0.094
dz �x
�x2 � y2 dx �
y�x2 � y2
dydy � 0.1.dx � 0.05,y � 3,x � 5,z � �x2 � y2,
19. Let Then:
1 � �3.05�2
�5.95�2 �1 � 32
62 � �2�3�62 �0.05� �
2�1 � 32�63 ��0.05� � �0.012
dz � �2xy2 dx �
�2�1 � x2�y3 dydy � �0.05.dx � 0.05,y � 6,x � 3,z � �1 � x2��y2,
21. See the definition on page 869. 23. The tangent plane to the surface at the point Pis a linear approximation of z.
z � f �x, y�
25.
dA∆AAd
ll ∆
Ad
h∆
h
dA � l dh � h dl
A � lh
27.
dV �2rh
3 dr �
r2
3 dh �
r3
�2h dr � r dh�
h � 6
r � 3
V �r2h
30.1 0.1 4.7124 4.8391 0.1267
0.1 2.8274 2.8264
0.001 0.002 0.0565 0.0566 0.0001
0.0002 0.0000�0.0019�0.0019�0.0001
�0.0010�0.1
�V � dV�VdV�h�r
Section 12.4 Differentials 89
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29. (a)
(b)
Maximum propagated error:
Relative error:dzz
�±0.73
��1.83��7.2� � 1.09�8.5� � 28.7�
±0.736.259
� ±0.1166 � 11.67%
±0.73
� ±0.73
� �1.83�±0.25� � ��1.09��±0.25�
dz ��z�x
dx ��z�y
dy
dz � �1.83 dx � 1.09 dy
31.
� 0.10 � 10% � 2�0.04� � �0.02�
dVV
� 2drr
�dhh
V � r2h � dV � �2rh� dr � �r2� dh
33.
� 12 �4�sin 45���± 1
16� � 3�sin 45���± 116� � 12�cos 45���±0.02� � ±0.24 in.2
dA �12 ��b sin C� da � �a sin C� db � �ab cos C� dC
A �12 ab sin C
35. (a)
is maximum when or
(b)
� 1809 in3 � 1.047 ft3
� 18�sin
2��16��12��12� �
182
2�16��12��cos
2��
90� �182
2 �sin
2��12�
dV � s�sin ��l ds �s2
2l�cos �� d� �
s2
2�sin �� dl
V �s2
2�sin ��l
� � �2.sin � � 1V
� 18 sin � ft3
� 31,104 sin � in.3
� �18 sin �
2��18 cos �
2��16��12�18 18
h
θ2
b2
V �12
bhl
37.
dPP
� 2dEE
�dRR
� 2�0.02� � ��0.03� � 0.07 � 7%
dP �2ER
dE �E 2
R 2 dR
P �E 2
R
90 Chapter 12 Functions of Several Variables
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41.
As ��x, �y� → �0, 0�, �1 → 0 and �2 → 0.
� fx�x, y� �x � fy�x, y� �y � �1�x � �2�y where �1 � �x and �2 � 0.
� �2x � 2� �x � �y � �x��x� � 0��y�
� 2x��x� � ��x�2 � 2��x� � ��y�
� �x2 � 2x��x� � ��x�2 � 2x � 2��x� � y � ��y�� � �x2 � 2x � y�
�z � f �x � �x, y � �y� � f �x, y�
z � f �x, y� � x2 � 2x � y
43.
As and �2 → 0.��x, �y� → �0, 0�, �1 → 0
� fx�x, y� �x � fy�x, y� �y � �1�x � �2�y where �1 � y��x� and �2 � 2x�x � ��x�2.
� 2xy��x� � x2�y � �y�x� �x � �2x�x � ��x�2 �y
� 2xy��x� � y��x�2 � x2�y � 2x��x���y� � ��x�2 �y
� �x2 � 2x��x� � ��x�2��y � �y� � x2y
�z � f �x � �x, y � �y� � f �x, y�
z � f �x, y� � x2y
45.
(a)
Thus, the partial derivatives exist at �0, 0�.
fy�0, 0� � lim�y→0
f �0, �y� � f �0, 0�
�y� lim
�y→0
0��y�2 � 0
�y� 0
fx�0, 0� � lim�x→0
f ��x, 0� � f �0, 0�
�x� lim
�x→0
0��x�4 � 0
�x� 0
f �x, y� � 3x2y ,x4 � y2
0,
�x, y� �0, 0�
�x, y� � �0, 0�
(b) Along the line
Along the curve
is not continuous at Therefore, is not differentiable at (See Theroem 12.5)�0, 0�.f�0, 0�.f
lim�x, y� →�0, 0�
f �x, y� �3x4
2x4 �32
y � x2:
lim�x, y� →�0, 0�
f �x, y� � limx →0
3x3
x4 � x2 � limx →0
3x
x2 � 1� 0 y � x:
47. Essay. For example, we can use the equation
dF ��F�m
dm ��F�a
da � a dm � m da.
F � ma:
39.
micro–henrys L � 0.00021�ln 100 � 0.75� � 8.096 10�4 ± dL � 8.096 10�4 ± 6.6 10�6
dL � 0.00021 dhh
�drr � � 0.00021 �±1�100�
100�
�±1�16�2 � � �±6.6� 10�6
L � 0.00021�ln 2hr
� 0.75�
Section 12.4 Differentials 91
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Section 12.5 Chain Rules for Functions of Several Variables
1.
dwdt
� 2xet � 2y��e�t � � 2�e2t � e�2t �
y � e�t
x � et
w � x2 � y2 3.
� �et �sec t � sec t tan t�
� et sec�� � t��1 � tan�� � t��
dwdt
� �sec y��et � � �x sec y tan y���1�
y � � � t
x � et
w � x sec y
5.
(a)
(b)dwdt
� 2 cos 2tw � 2 sin t cos t � sin 2t,
� 2�cos2 t � sin2 t� � 2 cos 2t
dwdt
� 2y cos t � x��sin t� � 2y cos t � x sin t
y � cos tx � 2 sin t,w � xy,
7.
z � et
y � et sin t
x � et cos t
w � x2 � y2 � z2
(a)
(b)dwdt
� 4e2tw � 2e2t,
dwdt
� 2x��et sin t � et cos t� � 2y�et cos t � et sin t� � 2zet � 4e2t
9.
(a)
(b)
dwdt
� 2t�t � 1� � �t 2 � 1� � 2t � 1 � 3t 2 � 1 � 3�2t 2 � 1�
w � �t � 1��t 2 � 1� � �t � 1�t � �t 2 � 1�t
� �t 2 � 1 � t� � �t � 1 � 1��2t� � �t � 1 � t 2 � 1� � 3�2t 2 � 1�
dwdt
� �y � z� � �x � z��2t� � �x � y�
z � ty � t 2 � 1,x � t � 1,w � xy � xz � yz,
11.
�1
2�116��1�2��44� �
22
2�29�
�11�29
20� �2.04
f��
2 �12
���10�2 � 42��1�2��2��10��7�� � �2��4���12��
��2�10 cos 2t � 7 cos t���20 sin 2t � 7 sin t�� � �2�6 sin 2t � 4 sin t��12 cos 2t � 4 cos t���
f��t� �12
��10 cos 2t � 7 cos t�2 � �6 sin 2t � 4 sin t�2��1�2
Distance � f �t� � ��x1 � x2�2 � �y1 � y2�2 � ��10 cos 2t � 7 cos t�2 � �6 sin 2t � 4 sin t�2
92 Chapter 12 Functions of Several Variables
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13.
At d 2wdt 2 � 0.t � 0,
��8 cos t sin t�1 � 2 sin4 t � 2 cos4 t�
�1 � 4 cos2 t sin2 t�2
d 2wdt 2 �
�1 � 4 cos2 t sin2 t���8 cos t sin t� � �2 cos2 t � 2 sin2 t��8 cos3 t sin t � 8 sin3 t cos t��1 � 4 cos2 t sin2 t�2
�2 cos2 t � 2 sin2 t1 � 4 cos2 t sin2 t
�2 sin t
1 � 4 cos2 t sin2 t��sin t� �
2 cos t1 � 4 cos2 t sin2 t
�cos t�
�2y
1 � �4x2y2� ��sin t� �2x
1 � �4x2y2� �cos t�
dwdt
��w�x
dxdt
��w�y
dydt
t � 0y � sin t,x � cos t,w � arctan�2xy�,
15.
When and
and �w�t
� �4.�w�s
� 8
t � �1,s � 2
�w�t
� 2x � 2y��1� � 2�x � y� � 4t
�w�s
� 2x � 2y � 2�x � y� � 4s
y � s � t
x � s � t
w � x2 � y2 17.
When and and �w�t
� �18.�w�s
� 0t ��
4,s � 3
�w�t
� 2x��s sin t� � 2y�s cos t� � �2s2 sin 2t
� 2s cos2 t � 2s sin2 t � 2s cos 2t
�w�s
� 2x cos t � 2y sin t
y � s sin t
x � s cos t
w � x2 � y2
19.
(a)
� 4��r � �� � �r � ��� � 8�
� 4x � 4y � 4�x � y�
�w��
� �2x � 2y��1� � ��2x � 2y���1�
�w�r
� �2x � 2y��1� � ��2x � 2y��1� � 0
y � r � �x � r � �,w � x2 � 2xy � y2,
(b)
�w��
� 8�
�w�r
� 0
� 4�2
� �r2 � 2r� � �2� � 2�r2 � �2� � �r2 � 2r� � �2�
w � �r � ��2 � 2�r � ���r � �� � �r � ��2
Section 12.5 Chain Rules for Functions of Several Variables 93
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23.
� �2st3 � 2s3t � 2st3 � 2s3t � 4st3 � 2st�s2 � 2t2�
� �s � t�st2 � �s � t�st2 � �s � t��s � t��2st�
�w�t
� yz�1� � xz��1� � xy�2st�
� 2s2t2 � s2t2 � t4 � 3s2t2 � t4 � t2�3s2 � t2�
� �s � t�st2 � �s � t�st2 � �s � t��s � t�t2
�w�s
� yz�1� � xz�1� � xy�t2�
w � xyz, x � s � t, y � s � t, z � st2 25.
� e�s� t�s� t�s�s2 � t2��s � t�2
� e�s� t�s� t���st�s � t� � st�s � t� � s�s � t�2
�s � t�2 �
� e�s� t�s� t���st
s � t�
st�s � t��s � t�2 � s�
�w�t
�zye x�y��1� � �
zxy2e x�y�1� � e x�y�s�
� e�s� t�s� t�t�s2 � 4st � t2��s � t�2
� e�s� t�s� t��st�s � t� � s2t � st2 � t�s � t�2
�s � t�2 �
� e�s� t�s� t�� sts � t
��s � t�st�s � t�2 � t�
�w�s
�zye x�y�1� � �
zxy2e x�y�1� � e x�y�t�
w � ze x�y, x � s � t, y � s � t, z � st
27.
�3y � 2x � 22y � 3x � 1
dydx
� �Fx�x, y�Fy�x, y� � �
2x � 3y � 2�3x � 2y � 1
x2 � 3xy � y2 � 2x � y � 5 � 0 29.
� �x � x2y � y3
y � xy2 � x3
dydx
� �Fx�x, y�Fy�x, y� � �
xx2 � y2 � y
yx2 � y2 � x
12
ln�x2 � y2� � xy � 4 � 0
ln �x2 � y2 � xy � 4
31.
�z�y
� �Fy
Fz
� �yz
�z�x
� �Fx
Fz
� �xz
Fz � 2z
Fy � 2y
Fx � 2x
F�x, y, z� � x2 � y2 � z2 � 25 33.
� �sec2�x � y�sec2�y � z� � 1
�z�y
� �Fy
Fz
� �sec2�x � y� � sec2�y � z�
sec2�y � z�
�z�x
� �Fx
Fz
� �sec2�x � y�sec2�y � z�
Fz � sec2�y � z�
Fy � sec2�x � y� � sec2�y � z�
Fx � sec2�x � y�
F�x, y, z� � tan�x � y� � tan�y � z� � 1
94 Chapter 12 Functions of Several Variables
21.
(a)
(b)
�w��
� 1
�w�r
� 0
w � arctan r sin �r cos �
� arctan�tan �� � �
�w��
��y
x2 � y2 ��r sin �� �x
x2 � y2 �r cos �� ���r sin ����r sin ��
r2 ��r cos ���r cos ��
r2 � 1
�w�r
��y
x2 � y2 cos � �x
x2 � y2 sin � ��r sin � cos �
r2 �r cos � sin �
r2 � 0
y � r sin �x � r cos �,w � arctan yx,
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39.
�w�z
� �Fz
Fw
� �xy � xw � ywxz � yz � 2w
�w�y
� �Fy
Fw
� �z�x � w�
xz � yz � 2w
�w�x
� �Fx
Fw
� �z�y � w�
xz � yz � 2w
Fw � xz � yz � 2w
Fz � xy � xw � yw
Fy � xz � zw
Fx � yz � zw
F�x, y, z, w� � xyz � xzw � yzw � w2 � 5 41.
�w�z
��Fz
Fw
� �y cos zy � w
z
�w�y
��Fy
Fw
�x sin xy � z cos yz
z
�w�x
��Fx
Fw
�y sin xy
z
cos xy � sin yz � wz � 20F�x, y, z, w� �
43.
Degree: 1
�xy
�x2 � y2� 1 f �x, y�
x fx�x, y� � y fy�x, y� � x y3
�x2 � y2�3�2 � y x3
�x2 � y2�3�2
f �tx, ty� ��tx��ty�
��tx�2 � �ty�2� t xy
�x2 � y2 � tf �x, y�
f �x, y� �xy
�x2 � y2
45.
Degree: 0
x fx�x, y� � y fy�x, y� � x1y
ex�y � y�xy2 ex�y � 0
f �tx, ty� � etx�ty � ex�y � f �x, y�
f �x, y� � ex�y 47. (Page 876) dwdt
��w�x
dxdt
��w�y
dydt
49. is the explicit form of a function of two variables, as in The implicit form is of the form as in z � x2 � y2 � 0.F �x, y, z� � 0,
z � x2 � y2.w � f �x, y�
51.
� 6sin �
412 �
62
2 cos �
4�
90 �3�2
2�
��210
m2�hr
dAdt
� x sin � dxdt
�x2
2 cos �
d�
dtxx
h
θ2
b2
A �12
bh � x sin �
2x cos �
2 �x2
2 sin �
Section 12.5 Chain Rules for Functions of Several Variables 95
35.
(i) implies
(ii) implies�z�y
� �z
y � z.2y
�z�y
� 2z � 2z �z�y
� 0
�z�x
� �x
y � z.2x � 2y
�z�x
� 2z �z�x
� 0
x 2 � 2yz � z2 � 1 � 0 37.
�z�y
� �Fy�x, y, z�Fz�x, y, z� �
�xxexz �
�1exz � �e�xz
�z�x
� �Fx�x, y, z�Fz�x, y, z� � �
zexz � yxexz
exz � xy � 0
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55.
� m��6��2� � �8��2�� � 28m cm2�secdIdt
�12
m�2r1
dr1
dt� 2r2
dr2
dt �
I �12
m�r12 � r2
2�
57. (a)
(b)
(c)
Thus, x21x � 2x � 81
x � 0 ⇒ 8x
� x ⇒ x � 2�2 ft.
d�
dx� 0 ⇒ 2 cos2 � � 2x sin � cos � ⇒ cos � � x sin � ⇒ tan � �
1x
d�
dx� �
Fx
F�
� �2x tan � � 2
sec2 ��x2 � 8� �2 cos2 � � 2x sin � cos �
x2 � 8
F�x, �� � �x2 � 8�tan � � 2x � 0
x2 tan � � 2x � 8 tan � � 0
x tan � � 2 � 4 �8x tan �
tan � � �2�x�1 � �2�x�tan �
�4x
tan � � tan
1 � tan � tan �
4x
tan�� � � �4x
θφ
4
6
8
x
tan �2x
59.
�w�u
��w�v
� 0
�w�v
��w�x
dxdv
��w�y
dydv
� ��w�x
��w�y
�w�u
��w�x
dxdu
��w�y
dydu
��w�x
��w�y
y � v � u
x � u � v
w � f �x, y�
96 Chapter 12 Functions of Several Variables
53. (a)
(b) (Surface area includes base.)
�648�
�10� 144� in.2�min �
36�
5�20 � 9�10� in.2�min
� ��12�10 �12
�10�6� � 144 �36
�10��4��
� ���122 � 362 �144
�122 � 362� 2�12��6� �
36�12��122 � 362
��4��
dSdt
� ���r2 � h2 �r2
�r2 � h2� 2rdr
dt�
rh
�r2 � h2 dhdt �
S � �r�r2 � h2 � �r2
dVdt
�13
�2rhdrdt
� r2 dhdt �
13
� �2�12��36��6� � �12�2��4�� � 1536� in.3�min
V �13
�r2h
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61.
(a)
(b)
2 �w�x
�w�y
sin � cos � � �w�y
2 cos2 � � �w
�x 2
� �w�y
2
�w�r
2
�1r 2�w
��2
� �w�x
2
cos2 � � 2 �w�x
�w�y
sin � cos � � �w�y
2
sin2 � � �w�x
2
sin2 � �
�w�y
��w�r
sin � ��w��
cos �
r
r�w�y
��w�r
r sin � ��w��
cos �
r sin � �w�r
� cos � �w��
��w�y
�r sin2 � � r cos2 ��
cos � �w��
��w�x
��r sin � cos �� ��w�y
�r cos2 ��
r sin � �w�r
��w�x
r sin � cos � ��w�y
r sin2 �
�w�x
��w�r
cos � ��w��
sin �
r
r �w�x
��w�r
�r cos �� ��w��
sin �
r cos � �w�r
� sin � �w��
��w�x
�r cos2 � � r sin2 ��
�sin � �w��
��w�x
�r sin2 �� ��w�x
r sin � cos �
r cos � �w�r
��w�x
r cos2 � ��w�y
r sin � cos �
�w��
��w�x
��r sin �� ��w�y
�r cos ��
�w�r
��w�x
cos � ��w�y
sin �
w � f �x, y�, x � r cos �, y � r sin �
63. Given and and
Therefore,
Therefore,�v�r
� �1r �u��
.
�u��
��u�x
��r sin �� ��u�y
�r cos �� � �r���u�y
cos � ��u�x
sin ��
�v�r
��v�x
cos � ��v�y
sin � � ��u�y
cos � ��u�x
sin �
�u�r
�1r �v��
.
�v��
��v�x
��r sin �� ��v�y
�r cos �� � r��v�y
cos � ��v�x
sin ��
�u�r
��u�x
cos � ��u�y
sin � ��v�y
cos � ��v�x
sin �
y � r sin �.x � r cos ��u�y
� ��v�x
,�u�x
��v�y
Section 12.5 Chain Rules for Functions of Several Variables 97
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Section 12.6 Directional Derivatives and Gradients
1.
�12
��5 � �3 �Du f �1, 2� � �f �1, 2� � u
u �v
�v��
12
i ��32
j
�f �1, 2� � �5i � j
�f �x, y� � �3 � 4y�i � ��4x � 5�j
v �12
�i � �3 j�
f �x, y� � 3x � 4xy � 5y 3.
Du f �2, 3� � �f �2, 3� � u �5�2
2
u �v
�v��
�22
i ��22
j
�f �2, 3� � 3i � 2j
�f �x, y� � yi � xj
v � i � j
f �x, y� � xy
5.
Du g�3, 4� � �g�3, 4� � u � �7
25
u �v
�v��
35
i �45
j
�g�3, 4� �35
i �45
j
�g �x
�x2 � y2i �
y
�x2 � y2j
v � 3i � 4j
g�x, y� � �x2 � y2 7.
Duh�1, �
2� � �h�1, �
2� � u � �e
u �v
�v�� �i
h�1, �
2� � ei
�h � ex sin yi � ex cos yj
v � �i
h�x, y� � ex sin y
9.
Du f �1, 1, 1� � �f �1, 1, 1� � u �2�6
3
u �v
�v��
�63
i ��66
j ��66
k
�f �1, 1, 1� � 2i � 2j � 2k
�f �x, y, z� � �y � z�i � �x � z�j � �x � y�k
v � 2i � j � k
f �x, y, z� � xy � yz � xz 11.
Du h�4, 1, 1� � �h�4, 1, 1� � u �� � 8
4�6�
�� � 8��624
u �v
�v�� � 1
�6,
2
�6, �
1
�6
�h�4, 1, 1� ��
4i � 2j � 2k
�h�x, y, z� � arctan yz i �xz
1 � �yz�2 j �xy
1 � �yz�2 k
v � 1, 2, �1�
h�x, y, z� � x arctan yz
13.
Du f � �f � u �2
�2x �
2
�2y � �2 �x � y�
�f � 2x i � 2y j
u �1
�2i �
1
�2j
f �x, y� � x2 � y2 15.
� �2 � �32 � cos�2x � y�
Du f � �f � u � cos�2x � y� ��32
cos�2x � y�
�f � 2 cos�2x � y� i � cos�2x � y� j
u �12
i ��32
j
f �x, y� � sin�2x � y�
98 Chapter 12 Functions of Several Variables
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17.
At Du f � �7�2.P � �3, 1�,
Du f � �2
�2x �
8
�2y � ��2�x � 4y�
u �v
�v�� �
1
�2i �
1
�2j
�f � 2x i � 8yj
v � �2i � 2j
f �x, y� � x2 � 4y2 19.
At .
Du h � �h � u �7
�19�
7�1919
u �v
�v��
1
�19�3i � 3j � k�
�h � i � j � k�1, 0, 0�,
�h �1
x � y � z�i � j � k�
v � 3i � 3j � k
h�x, y, z� � ln�x � y � z�
21.
�f �2, 1� � 3i � 10j
�f �x, y� � 3i � 10yj
f �x, y� � 3x � 5y2 � 10 23.
�z�3, �4� � �6 sin 25i � 8 sin 25j � 0.7941i � 1.0588j
�z�x, y� � �2x sin�x2 � y2�i � 2y sin�x2 � y2�j
z � cos�x2 � y2�
25.
�w�1, 1, �2� � 6i � 13j � 9k
�w�x, y, z� � 6xyi � �3x2 � 5z�j � �2z � 5y�k
w � 3x2y � 5yz � z2 27.
Dug � �g � u �2�5
�8�5
�10�5
� 2�5
�g�x, y� � 2xi � 2yj, �g�1, 2� � 2i � 4j
PQ\
� 2i � 4j, u �1�5
i �2�5
j
29.
Du f � �f � u � �2�5
� �2�5
5
�f �0, 0� � �i
�f �x, y� � �e�x cos yi � e�x sin yj
PQ\
� 2i � j, u �2�5
i �1�5
j 31.
� �h�2, �
4� � � �17
�h�2, �
4� � i � 4j
�h�x, y� � tan yi � x sec2 yj
h�x, y� � x tan y
33.
��g�1, 2�� �2�515
�g�1, 2� �13�
25
i �45
j� �2
15�i � 2j�
�g�x, y� �13
2xx2 � y2 i �
2yx2 � y2 j�
g�x, y� � ln 3�x2 � y2 �13
ln�x2 � y2� 35.
��f �1, 4, 2�� � 1
�f �1, 4, 2� �1
�21�i � 4j � 2k�
�f �x, y, z� �1
�x2 � y2 � z2�xi � yj � zk�
f �x, y, z� � �x2 � y2 � z2
37.
��f �2, 0, �4�� � �65
�f �2, 0, �4� � i � 8j
�f �x, y, z� � eyz i � xzeyz j � xyeyz k
f �x, y, z� � xeyz
Section 12.6 Directional Derivatives and Gradients 99
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For Exercises 39–45, and D� f �x, y� � ��13� cos � � �1
2� sin �.f �x, y� � 3 �x3
�y2
39.
x
y
3
6
9
(3, 2, 1)
z
f �x, y� � 3 �x3
�y2
41. (a)
(b)
�3 � 2�3
12
D���6 f �3, 2� � ��13���3
2 � � �12���
12�
�2 � 3�3
12
D4��3 f �3, 2� � ��13���
12� � �1
2����32 �
43. (a)
�15
�25
� �15
Du f � �f � u
u � �35
i �45
j
�v� � �9 � 16 � 5
v � �3i � 4j (b)
Du f � �f � u ��11
6�10� �
11�1060
u �1
�10i �
3
�10j
�v� � �10
v � i � 3j
45. ��f � � �19 �
14 �
16�13
For Exercises 47 and 49, and D� f �x, y� � �2x cos � � 2y sin � � �2�x cos � � y sin ��.f �x, y� � 9 � x2 � y2
47.
x
y
9
33
(1, 2, 4)
z
f �x, y� � 9 � x2 � y2 49.
��f �1, 2�� � �4 � 16 � �20 � 2�5
�f �1, 2� � �2i � 4j
51. (a) In the direction of the vector
(b)
(Same direction as in part (a).)
(c) the direction opposite that of the gradient.��f �25 i �
110 j,
�f �1, 2� �1
10 ��4�i �1
10 �1�j � �25 i �
110 j
�f �110 �2x � 3y�i �
110 ��3x � 2y�j
�4i � j.
100 Chapter 12 Functions of Several Variables
53.
(a)
—CONTINUED–
x
y
z
�4, �3, 7�f �x, y� � x2 � y2,
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(d)
Critical numbers:
These are the angles for which is a maximum and minimum
(e) the maximum value of at
(f )
is perpendicular to the level curve at �4, �3�.�f �4, �3� � 8i � 6j
x
y
2
−2−4 2 4 6−6
−4
−6
4
6
Generated by Mathematica
f �x, y� � x2 � y2 � 7
� 0.64.Du f �4, �3�,��f �4, �3�� � �2�4�i � 2�3�j� � �64 � 36 � 10,
�3.79�.�0.64�Du f �4, �3� � 0.64, 3.79
g�� � �8 sin � 6 cos
g�� � Du f �4, �3� � 8 cos � 6 sin
55.
�f �3, 4� � 6i � 8j
x2 � y2 � 25
�f �x, y� � 2xi � 2yj
P � �3, 4�c � 25,
f �x, y� � x2 � y2 57.
�f �1, 1� � �12
j
x2 � y2 � 2x � 0
xx2 � y2 �
12
�f �x, y� �y2 � x2
�x2 � y2�2 i �2xy
�x2 � y2�2 j
P � �1, 1�c �12
,
f �x, y� �x
x2 � y2
59.
��257257
�16i � j�
�f �2, 10���f �2, 10�� �
1
�257�16i � j�
�f �2, 10� � 16i � j
�f �x, y� � 8xi � j
f �x, y� � 4x2 � y 12
4
8
4x
y4x2 � y � 6 61.
��8585
�9i � 2j�
�f �2, �1���f �2, �1�� �
1
�85�9i � 2j�
�f �2, �1� � 36i � 8j
�f �x, y� � 18xi � 8yj
f �x, y� � 9x2 � 4y2
4
4
2
−2
−4
−4x
y9x2 � 4y2 � 40
Section 12.6 Directional Derivatives and Gradients 101
(c) Zeros:
These are the angles for which equals zero.Du f �4, 3�
� 2.21, 5.36
53. —CONTINUED—
(b)
x
y
4
2−4
−8
−12
8
12
ππ
Generated by Mathematica
Du f �4, �3� � 8 cos � 6 sin
Du f �x, y� � �f �x, y� � u � 2x cos � 2y sin
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67. Let be a function of two variables anda unit vector.
(a) If then
(b) If then Du f ��f�y
. � 90�,
Du f ��f�x
. � 0�,
u � cos i � sin jf �x, y� 69.
x y5
3
3
z
P
63.
�T�3, 4� �7
625i �
24625
j �1
625�7i � 24j�
�T �y2 � x2
�x2 � y2�2 i �2xy
�x2 � y2�2 j
T �x
x2 � y2 65. See the definition, page 885.
71.
1800
1800
A
B
1994
1671
73.
y2 � 10x
y2�t� � 100e�4tx �y2
10
y�t� � 10e�2tx�t� � 10e�4t
10 � y�0� � C210 � x�0� � C1
y�t� � C2e�2tx�t� � C1e
�4t
dydt
� �2ydxdt
� �4x
P � �10, 10�T �x, y� � 400 � 2x2 � y2,
75. (a)
(c)
(e) and �D�y
�1, 0.5� � 25� cos �
4� 55.5
�D�y
� 25� cos �y2
D�1, 0.5� � 250 � 30�1� � 50 sin �
4� 315.4 ft
x
y
300
400
12
1
2
D (b) The graph of would model the ocean floor.
(d) and
(f )
�D�1, 0.5� � 60i � 55.5j
�D � 60x i � 25� cos��y2 �j
�D�x
�1, 0.5� � 60�D�x
� 60x
�D � �250 � 30x2 � 50 sin��y�2�
77. True 79. True
81. Let Then �f �x, y, z� � ex cos yi � ex sin yj � zk.f �x, y, z� � ex cos y �z2
2� C.
102 Chapter 12 Functions of Several Variables
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Section 12.7 Tangent Planes and Normal Lines
Section 12.7 Tangent Planes and Normal Lines 103
1.
Plane 3x � 5y � 3z � 15
F�x, y, z� � 3x � 5y � 3z � 15 � 0 3.
Elliptic cone 4x2 � 9y2 � 4z2
F�x, y, z� � 4x2 � 9y2 � 4z2 � 0
5.
��33
�i � j � k�
n ��F
��F ��
1
�3�i � j � k�
�F � i � j � k
F�x, y, z� � x � y � z � 4 7.
��210
�3i � 4j � 5k�
�1
5�2�3i � 4j � 5k�
n ��F
��F ��
5
5�2 �35
i �45
j � k�
�F�3, 4, 5� �35
i �45
j � k
�F�x, y, z� �x
�x2 � y2i �
y
�x2 � y2j � k
F�x, y, z� � �x2 � y2 � z
9.
��20492049
�32i � 32j � k�
n ��F
��F ��
1
�2049�32i � 32j � k�
�F�1, 2, 16� � 32i � 32j � k
�F�x, y, z� � 2xy4 i � 4x2y3j � k
F�x, y, z� � x2y4 � z 11.
��33
�i � j � k�
n ��F
��F ��
1
�3�i � j � k�
�F�1, 4, 3� � i � j � k
�F�x, y, z� �1x
i �1
y � zj �
1y � z
k
F�x, y, z� � ln� xy � z� � ln x � ln�y � z�
13.
��113113
��i � 6�3 j � 2k�
�1
�113��i � 6�3 j � 2k�
n ��F
��F ��
2
�113 ��12
i � 3�3 j � k�
�F�6, �
6, 7� � �
12
i � 3�3 j � k
�F�x, y, z� � �sin yi � x cos yj � k
F�x, y, z� � �x sin y � z � 4
15.
6x � 2y � z � 35
0 � 6x � 2y � z � 35
�6�x � 3� � 2�y � 1� � �z � 15� � 0
Fz�3, 1, 15� � �1Fy�3, 1, 15� � �2Fx�3, 1, 15� � �6
Fz�x, y, z� � �1Fy�x, y, z� � �2y Fx�x, y, z� � �2x
F�x, y, z� � 25 � x2 � y2 � z
f �x, y� � 25 � x2 � y2, �3, 1, 15�
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104 Chapter 12 Functions of Several Variables
19.
10x � 8y � z � 9
10�x � 5� � 8�y � 4� � �z � 9� � 0
Gz�5, 4, 9� � �1 Gy�5, 4, 9� � �8 Gx�5, 4, 9� � 10
Gz�x, y, z� � �1Gy�x, y, z� � �2yGx�x, y, z� � 2x
G�x, y, z� � x2 � y2 � z
g�x, y� � x2 � y2, �5, 4, 9�
21.
2x � z � �2
Fz�0, �
2, 2� � �1Fy�0,
�
2, 2� � 0Fx�0,
�
2, 2� � 2
Fz�x, y, z� � �1 Fy�x, y, z� � ex cos y Fx�x, y, z� � ex�sin y � 1�
F�x, y, z� � ex�sin y � 1� � z
z � ex�sin y � 1�, �0, �
2, 2�
23.
3x � 4y � 25z � 25�1 � ln 5�
3�x � 3� � 4�y � 4� � 25�z � ln 5� � 0
325
�x � 3� �4
25�y � 4� � �z � ln 5� � 0
Hz�3, 4, ln 5� � �1Hy�3, 4, ln 5� �4
25Hx�3, 4, ln 5� �
325
Hz�x, y, z� � �1 Hy�x, y, z� �y
x2 � y2 Hx�x, y, z� �x
x2 � y2
H�x, y, z� � ln �x2 � y2 � z �12
ln�x2 � y2� � z
h�x, y� � ln �x2 � y2, �3, 4, ln 5�
25.
x � 4y � 2z � 18
�x � 2� � 4�y � 2� � 2�z � 4� � 0
4�x � 2� � 16�y � 2� � 8�z � 4� � 0
Fz�2, �2, 4� � 8Fy�2, �2, 4� � �16Fx�2, �2, 4� � 4
Fz�x, y, z� � 2z Fy�x, y, z� � 8y Fx�x, y, z� � 2x
F�x, y, z� � x2 � 4y2 � z2 � 36
x2 � 4y2 � z2 � 36, �2, �2, 4�
17.
3x � 4y � 5z � 0
3�x � 3� � 4�y � 4� � 5�z � 5� � 0
35
�x � 3� �45
�y � 4� � �z � 5� � 0
Fz�3, 4, 5� � �1 Fy�3, 4, 5� �45
Fx�3, 4, 5� �35
Fz�x, y, z� � �1Fy�x, y, z� �y
�x2 � y2Fx�x, y, z� �
x
�x2 � y2
F�x, y, z� � �x2 � y2 � z
f �x, y� � �x2 � y2, �3, 4, 5�
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Section 12.7 Tangent Planes and Normal Lines 105
27.
x � y � z � 1
4�x � 2� � 4�y � 1� � 4�z � 2� � 0
FZ�2, 1, �2� � 4Fy�2, 1, �2� � 4Fx�2, 1, �2� � 4
Fz�x, y, z� � �2z Fy�x, y, z� � 2xy Fx�x, y, z� � y2 � 3
F�x, y, z� � xy2 � 3x � z2 � 4
xy2 � 3x � z2 � 4, �2, 1, �2�
29.
Direction numbers:
Plane:
Line:x � 1
2�
y � 24
�z � 4
1
2�x � 1� � 4�y � 2� � �z � 4� � 0, 2x � 4y � z � 14
2, 4, 1
Fz�1, 2, 4� � 1Fy�1, 2, 4� � 4 Fx�1, 2, 4� � 2
Fz�x, y, z� � 1Fy�x, y, z� � 2yFx�x, y, z� � 2x
F�x, y, z� � x2 � y2 � z � 9
x2 � y2 � z � 9, �1, 2, 4�
31.
Direction numbers: 3, 2, 1
Plane:
Line:x � 2
3�
y � 32
�z � 6
1
3�x � 2� � 2�y � 3� � �z � 6� � 0, 3x � 2y � z � �6
Fz��2, �3, 6� � �1Fy��2, �3, 6� � �2Fx��2, �3, 6� � �3
Fz�x, y, z� � �1 Fy�x, y, z� � x Fx�x, y, z� � y
F�x, y, z� � xy � z
xy � z � 0, ��2, �3, 6�
33.
Direction numbers: 1,
Plane:
Line:x � 1
1�
y � 1�1
�z � ���4�
2
�x � 1� � �y � 1� � 2�z ��
4� � 0, x � y � 2z ��
2
�1, 2
Fz�1, 1, �
4� � �1Fy�1, 1, �
4� �12
Fx�1, 1, �
4� � �12
Fz�x, y, z� � �1 Fy�x, y, z� �x
x2 � y2 Fx�x, y, z� ��y
x2 � y2
F�x, y, z� � arctan yx
� z
z � arctan yx, �1, 1,
�
4�
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106 Chapter 12 Functions of Several Variables
35.
(a) Let
Direction numbers: 0, 0,
Line:
Tangent plane: 0�x � 1� � 0�y � 1� � 1�z � 1� � 0 ⇒ z � 1
z � 1 � ty � 1,x � 1,
�1.
�F�1, 1, 1� � �k.
�4y�1 � x2�
�y2 � 1��x2 � 1�2 i �4x�1 � y2�
�x2 � 1��y2 � 1�2 j � k
�F�x, y, z� �4y
y2 � 1�x2 � 1 � 2x2
�x2 � 1�2 �i �4x
x2 � 1�y2 � 1 � 2y2
�y2 � 1�2 �j � k
F�x, y, z� �4xy
�x2 � 1��y2 � 1� � z
z � f �x, y� �4xy
�x2 � 1��y2 � 1�, �2 ≤ x ≤ z, 0 ≤ y ≤ 3
(c)
x
y
1
2 23
−2
−1
z
x y−1
32
1
z (d) At the tangent plane is parallel to the xy-plane,
implying that the surface is level there. At
the function does not change in the x-direction.
��1, 2, �45�,
�1, 1, 1�,
(b)
Line:
Plane:
6y � 25z � 32 � 0
6y � 12 � 25z � 20 � 0
0�x � 1� �6
25�y � 2� � 1�z �
45� � 0
z � �45
� ty � 2 �6
25t,x � �1,
�F��1, 2, �45� � 0i �
�4��3��2��5�2 j � k �
625
j � k
37.
(Theorem 12.13)
Fx�x0, y0, z0��x � x0� � Fy�x0, y0, z0��y � y0� � F2�x0, y0, z0��z � z0� � 0
39.
(a)
Direction numbers: 1,
(b) cos � � �F � �G��F � ��G�
�4
�20�2�
2
�10�
�105
; not orthogonal
�2, 1, x � 2
1�
y � 1�2
�z � 2
1
�F �G � i41
j20
k0
�1 � �2i � 4j � 2k � �2�i � 2j � k�
�G�2, 1, 2� � i � k�F�2, 1, 2� � 4i � 2j
�G�x, y, z� � i � k �F�x, y, z� � 2x i � 2y j
G�x, y, z� � x � z F�x, y, z� � x2 � y2 � 5
41.
—CONTINUED—
�G�3, 3, 4� � 6j � 8k�F�3, 3, 4� � 6i � 8k
�G � 2yj � 2zk �F � 2x i � 2zk
G�x, y, z� � y2 � z2 � 25 F�x, y, z� � x2 � z2 � 25
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Section 12.7 Tangent Planes and Normal Lines 107
41. —CONTINUED—
(a)
Direction numbers: 4, 4,
(b) cos � � �F � �G��F � ��G�
�64
�10��10� �1625
; not orthogonal
�3, x � 3
4�
y � 34
�z � 4�3
�F �G � i60
j06
k88 � �48i � 48j � 36k � �12�4i � 4j � 3k�
43.
(a)
Direction numbers: 0, 1, �1, x � 2, y � 1
1�
z � 1�1
�F �G � i41
j2
�1
k2
�1 � 6j � 6k � 6� j � k�
�G�2, 1, 1� � i � j � k�F�2, 1, 1� � 4i � 2j � 2k
�G�x, y, z� � i � j � k �F�x, y, z� � 2x i � 2yj � 2zk
G�x, y, z� � x � y � z F�x, y, z� � x2 � y2 � z2 � 6
(b) cos � � �F � �G��F � ��G�
� 0; orthogonal
45.
(a)
The cross product of these gradients is parallel to the curve of intersection.
Using direction numbers you get
(b)
xy
68
8
(1, 2, 4)
z
cos � ��F � �G
��F� ��G��
�4 � 1 � 1
�6 �6�
�46
⇒ � 48.2
z � 4.y � 2 � 2t,x � 1 � t,1, �2, 0,
�F�1, 2, 4� �G�1, 2, 4� � i2
�2
j1
�1
k11 � 2i � 4j
�F�1, 2, 4� � 2i � j � k
�F�x, y, z� � 2x i �12
yj � k
F�x, y, z� � z � x2 �y2
4� 6
f �x, y� � 6 � x2 �y2
4, g�x, y� � 2x � y
�G�1, 2, 4� � �2i � j � k
�G�x, y, z� � �2i � j � k
G�x, y, z� � z � 2x � y
47.
� � arccos� 1
�209� 86.03
cos � � �F�2, 2, 5� � k��F�2, 2, 5�� �
1
�209
�F�2, 2, 5� � 12i � 8j � k
�F�x, y, z� � 6xi � 4yj � k
F�x, y, z� � 3x2 � 2y2 � z � 15, �2, 2, 5� 49.
� � arccos 1
�21 77.40
cos � � �F�1, 2, 3� � k��F�1, 2, 3�� �
1
�21
�F�1, 2, 3� � 2i � 4j � k
�F�x, y, z� � 2xi � 2yj � k
F�x, y, z� � x2 � y2 � z, �1, 2, 3�
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108 Chapter 12 Functions of Several Variables
51.
(vertex of paraboloid)�0, 3, 12�
z � 3 � 02 � 32 � 6�3� � 12
�2y � 6 � 0, y � 3
y
x
8
68
8
z�2x � 0, x � 0
�F�x, y, z� � �2xi � ��2y � 6�j � k
F�x, y, z� � 3 � x2 � y2 � 6y � z 53.
z � 10e�8kty � 3e�2ktx � 4e�4kt
z�0� � C3 � 10y�0� � C2 � 3x�0� � C1 � 4
z�t� � C3e�8kty�t� � C2e
�2ktx�t� � C1e�4kt
dzdt
� �8kzdydt
� �2kydxdt
� �4kx
T�x, y, z� � 400 � 2x2 � y2 � 4z2, �4, 3, 10�
55.
Plane:
x0x
a2 �y0 y
b2 �z0z
c2 �x0
2
a2 �y0
2
b2 �z0
2
c2 � 1
2x0
a2 �x � x0� �2y0
b2 �y � y0� �2z0
c2 �z � z0� � 0
Fz�x, y, z� �2zc2
Fy�x, y, z� �2yb2
Fx�x, y, z� �2xa2
F�x, y, z� �x2
a2 �y2
b2 �z2
c2 � 1 57.
Plane:
Hence, the plane passes through the origin.
a2x0x � b2y0y � z0z � a2x02 � b2y0
2 � z02 � 0
2a2x0�x � x0� � 2b2y0�y � y0� � 2z0�z � z0� � 0
Fz�x, y, z� � �2z
Fy�x, y, z� � 2b2y
Fx�x, y, z� � 2a2x
F�x, y, z� � a2x2 � b2y2 � z2
59.
(a)
(b)
(c) If This is the second–degree Taylor polynomial for
If This is the second–degree Taylor polynomial for
(d)
ex.P2�x, 0� � 1 � x �12 x2.y � 0,
e�y.P2�0, y� � 1 � y �12 y2.x � 0,
� 1 � x � y �12 x2 � xy �
12 y2
P2�x, y� f �0, 0� � fx�0, 0�x � fy�0,0�y �12 fxx�0, 0�x2 � fxy�0, 0�xy �
12 fyy�0, 0�y2
P1�x, y� f �0, 0� � fx�0, 0�x � fy�0, 0�y � 1 � x � y
fxy�x, y� � �ex�yfyy�x, y� � ex�y,fxx�x, y� � ex�y,
fy�x, y� � �ex�y fx�x, y� � ex�y,
f �x, y� � ex�y
x y
0 0 1 1 1
0 0 0.9048 0.9000 0.9050
0.2 0.1 1.1052 1.1000 1.1050
0.2 0.5 0.7408 0.7000 0.7450
1 0.5 1.6487 1.5000 1.6250
P2�x, y�P1�x, y�f �x, y� (e)z
f
P1
P2
y
x
4
2
2
1
−2
−2
−4
−2
61. Given where is differentiable at
and
the level surface of at is of the form for some constant Let
Then where is normal to
Therefore, is normal to F�x0, y0, z0� � C.�F�x0, y0z0�
F�x0, y0, z0� � C � 0.�G�x0, y0, z0��G�x0, y0, z0� � �F�x0, y0, z0�
G�x, y, z� � F�x, y, z� � C � 0.
C.F�x, y, z� � C�x0, y0, z0�F
�F�x0, y0, z0� � 0,�x0, y0, z0�
Fw � F�x, y, z�
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Section 12.8 Extrema of Functions of Two Variables
Section 12.8 Extrema of Functions of Two Variables 109
1.
Relative minimum:
gy � 2�y � 3� � 0 ⇒ y � 3
gx � 2�x � 1� � 0 ⇒ x � 1
�1, 3, 0�
yx3 2
4
5
1
1
(1, 3, 0)
zg�x, y� � �x � 1�2 � �y � 3�2 ≥ 0
3.
Relative minimum:
Check:
At the critical point and Therefore, is a relative minimum.�0, 0, 1�fxx fyy � � fxy�2 > 0.fxx > 0�0, 0�,
fxx �y2 � 1
�x2 � y2 � 1�3�2 , fyy �x2 � 1
�x2 � y2 � 1�3�2 , fxy ��xy
�x2 � y2 � 1�3�2
fy �y
�x2 � y2 � 1� 0 ⇒ y � 0
fx �x
�x2 � y2 � 1� 0 ⇒ x � 0
�0, 0, 1�
(0, 0, 1)
yx 2 3
5
−3
3 2
zf �x, y� � �x2 � y2 � 1 ≥ 1
5.
Relative minimum:
Check:
At the critical point and Therefore, is a relative minimum.��1, 3, �4�fxx fyy � � fxy�2 > 0.fxx > 0��1, 3�,
fxx � 2, fyy � 2, fxy � 0
fy � 2y � 6 � 0 ⇒ y � 3
fx � 2x � 2 � 0 ⇒ x � �1
��1, 3, �4�
( 1, 3, 4)− −
y
x2 1
1
7
2
−1
−2
−3
−4
1
zf �x, y� � x2 � y2 � 2x � 6y � 6 � �x � 1�2 � �y � 3�2 � 4 ≥ �4
7.
At the critical point and Therefore, is a relative minimum.��1, 1, �4�fxx fyy � � fxy�2 > 0.fxx > 0��1, 1�,
fxx � 4, fyy � 2, fxy � 2
fy � 2x � 2y � 0
fx � 4x � 2y � 2 � 0�f �x, y� � 2x2 � 2xy � y2 � 2x � 3
Solving simultaneously yields and y � 1.x � �1
9.
At the critical point and Therefore, is a relative maximum.�8, 16, 74�fxx fyy � � fxy�2 > 0.fxx < 0�8, 16�,
fxx � �10, fyy � �2, fxy � 4
fy � 4x � 2y � 0
fx � �10x � 4y � 16 � 0�f �x, y� � �5x2 � 4xy � y2 � 16x � 10
Solving simultaneously yields and y � 16.x � 8
11.
when
when
At the critical point andTherefore, is a relative
minimum.�1, 2, �1�fxx fyy � � fxy�2 > 0.
fxx > 0�1, 2�,
fxx � 4, fyy � 6, fxy � 0
y � 2.fy � 6y � 12 � 6�y � 2� � 0
x � 1.fx � 4x � 4 � 4�x � 1� � 0
f �x, y� � 2x2 � 3y2 � 4x � 12y � 13 13.
Since for all is relative minimum.�0, 0, 3��x, y�,f �x, y� ≥ 3
fx
fy
�
�
2x�x2 � y2
2y�x2 � y2
�
�
0
0� x � 0, y � 0
f �x, y� � 2�x2 � y2 � 3
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110 Chapter 12 Functions of Several Variables
15.
is the only critical point. Since for all is relative maximum.�0, 0, 4��x, y�,g�x, y� ≤ 4�0, 0�
g�x, y� � 4 � �x� � �y�
17.
Relative minimum:
Relative maximum:
x
y
5
−4
−4
4
4
z
��1, 0, 2��1, 0, �2�
z ��4x
x2 � y2 � 119.
Relative minimum:
Relative maxima:
Saddle points:
−4
44
−4
5
6
yx
z
�±1, 0, 1��0, ±1, 4�
�0, 0, 0�z � �x2 � 4y2�e1�x2�y2
21.
when
when
At the critical point Therefore, is a saddle point.�1, �2, �1�hxx hyy � �hxy�2 < 0.�1, �2�,
hxx � 2, hyy � �2, hxy � 0
y � �2.hy � �2y � 4 � �2� y � 2� � 0
x � 1.hx � 2x � 2 � 2�x � 1� � 0
h�x, y� � x2 � y2 � 2x � 4y � 4
23.
At the critical point Therefore, is a saddle point.�0, 0, 0�hxx hyy � �hxy�2 < 0.�0, 0�,
hxx � 2, hyy � �2, hxy � �3
hy � �3x � 2y � 0�hx � 2x � 3y � 0
h�x, y� � x2 � 3xy � y2
Solving simultaneously yields and y � 0.x � 0
Solving by substitution yields two critical points and �1, 1�.�0, 0�
25.
At the critical point Therefore, is a saddle point. At the critical point andTherefore, is a relative minimum.�1, 1, �1�fxx fyy � � fxy�2 > 0.
fxx � 6 > 0�1, 1�,�0, 0, 0�fxx fyy � � fxy�2 < 0.�0, 0�,
fxx � 6x, fyy � 6y, fxy � �3
fy � 3��x � y2� � 0�fx � 3�x2 � y� � 0
f �x, y� � x3 � 3xy � y3
27.
fy � e�x cos y � 0
fx � �e�x sin y � 0�f �x, y� � e�x sin y
Since for all and and are never both zero for agiven value of there are no critical points.y,
cos ysin yxe�x > 0
29.
Relative minimum at all points �x, x�, x � 0.
z
yx
33
40
60
z ��x � y�4
x2 � y2 ≥ 0. z � 0 if x � y � 0.
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Section 12.8 Extrema of Functions of Two Variables 111
31.
Insufficient information.
fxx fyy � � fxy�2 � �9��4� � 62 � 0 33.
has a saddle point at �x0, y0�.f
fxx fyy � � fxy�2 � ��9��6� � 102 < 0
35. (a) The function f defined on a region R containing has a relative minimum at if for all in R.
(b) The function f defined on a region R containing has a relative maximum at if for all in R.
(c) A saddle point is a critical point which is not a relative extremum.
(d) See definition page 906.
�x, y�f �x, y� ≤ f �x0, y0��x0, y0��x0, y0�
�x, y�f �x, y� ≥ f �x0, y0��x0, y0��x0, y0�
43.
⇒ fxy2 < 16 ⇒ �4 < fxy < 4
d � fxx fyy � fxy2 � �2��8� � fxy
2 � 16 � fxy2 > 0 45.
At and the test fails. is a saddle point.
�0, 0, 0�fxx fyy � � fxy�2 � 0�0, 0�,
fxx � 6x, fyy � 6y, fxy � 0
fy � 3y2 � 0
fx � 3x2 � 0�f �x, y� � x3 � y3
47.
At both and and the test fails.
Absolute minima: and �b, �4, 0��1, a, 0�
fxx fyy � � fxy�2 � 0�b, �4�,�1, a�
fxy � 4�x � 1��y � 4�fyy � 2�x � 1�2,fxx � 2�y � 4�2,
fy � 2�x � 1�2�y � 4� � 0
fx � 2�x � 1��y � 4�2 � 0�f �x, y� � �x � 1�2�y � 4�2 ≥ 0
Solving yields the critical points and �b, �4�.�1, a�
49.
At is undefined and the test fails.
Absolute minimum: 0 at �0, 0�
�0, 0�, fxx fyy � � fxy�2
fxx � �2
9x 3�3, fyy � �
2
9y 3�y, fxy � 0
fy �2
3 3�y
fx �2
3 3�x �
f �x, y� � x2�3 � y2�3 ≥ 0
and are undefined at The critical point is �0, 0�.y � 0.x � 0,fyfx
37. No extrema
x
y
2
30
45
60
75
2
z 39. Saddle point
xy
6
7
−3
36
z 41. The point will be a saddle point. The function could be
f �x, y� � x2 � y2.
A
51.
Absolute minimum: 0 at �0, 3, �1�
fz � 2�z � 1� � 0
fy � 2�y � 3� � 0�fx � 2x � 0
f �x, y, z� � x2 � �y � 3�2 � �z � 1�2 ≥ 0
Solving yields the critical point �0, 3, �1�.
Solving yields x � y � 0
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112 Chapter 12 Functions of Several Variables
53. has no critical points. On the line
and the maximum is 10, the minimum is 5. On the line
and the maximum is 6, the minimum is 5. On the line
and the maximum is 10, the minimum is 6.
Absolute maximum: 10 at
Absolute minimum: 5 at �1, 2��0, 1�
f �x, y� � f �x� � 12 � 3x � 2��12 x � 1� � �2x � 10
0 ≤ x ≤ 2,y � �12 x � 1,
f �x, y� � f �x� � 12 � 3x � 2��2x � 4� � x � 4
y � �2x � 4, 1 ≤ x ≤ 2,
f �x, y� � f �x� � 12 � 3x � 2�x � 1� � �5x � 10
2
2
31
3
1
x
y x= + 1
y x= 2 + 4−(0, 1)
(1, 2)
(2, 0)
12
y x= + 1−
y0 ≤ x ≤ 1,y � x � 1,f �x, y� � 12 � 3x � 2y
55.
On the line
and the maximum is 28, the minimum is 16. On the curve
and the maximum is 28, the minimum is
Absolute maximum: 28 at
Absolute minimum: at �0, 1��2
�±2, 4��
18 .
f �x, y� � f �x� � 3x2 � 2�x2�2 � 4x2 � 2x4 � x2 � x2�2x2 � 1��2 ≤ x ≤ 2,y � x2,
f �x, y� � f �x� � 3x2 � 32 � 16 � 3x2 � 16
�2 ≤ x ≤ 2,y � 4,
fy � 4y � 4 � 0 ⇒ y � 1
fx � 6x � 0 ⇒ x � 0�
x
2
1
3
1 2−1−2
( 2, 4)− (2, 4)
yf �x, y� � 3x2 � 2y2 � 4y
f �0, 1� � �2
57.
Along
Thus, and
Along
Thus,
Along
Along
Thus, the maxima are and and the minima are and f �12 , �1� � �
14 .f ��
12 , 1� � �
14f ��2, �1� � 6f �2, 1� � 6
x � �2, �1 ≤ y ≤ 1, f � 4 � 2y ⇒ f� � �2 � 0.
x � 2, �1 ≤ y ≤ 1, f � 4 � 2y ⇒ f� � 2 � 0.
f �2, �1� � 2.f �12 , �1� � �
14 ,f ��2, �1� � 6,
f� � 2x � 1 � 0 ⇒ x �12 .f � x2 � x,�2 ≤ x ≤ 2,y � �1,
f �2, 1� � 6.f ��12 , 1� � �
14f ��2, 1� � 2,
f� � 2x � 1 � 0 ⇒ x � �12 .f � x2 � x,�2 ≤ x ≤ 2,y � 1,
f �0, 0� � 0
fy � x � 0
fx � 2x � y � 0� x � y � 02
−1
−2
1x
yR � ��x, y�: �x� ≤ 2, �y� ≤ 1f �x, y� � x2 � xy,
59.
On the boundary we have and Thus,
Then, implies or
and
Thus, the maxima are and and the minima are f �x, �x� � 0, �x� ≤ 2.f ��2, �2� � 16,f �2, 2� � 16
f �2, �2� � f ��2, 2� � 0f �2, 2� � f ��2, �2� � 16
x � ±2.16 � 4x2f� � 0
f� � ± �8 � x2��1�2��2x� � 2�8 � x2�1�2� � ±16 � 4x2
�8 � x2.
f � x2 ± 2x�8 � x2 � �8 � x2� � 8 ± 2x�8 � x2
y � ±�8 � x2.y2 � 8 � x2x2 � y2 � 8,
f �x, �x� � x2 � 2x2 � x2 � 0
fy � 2x � 2y � 0
fx � 2x � 2y � 0� y � �x2
4
−2
−4
−2−4 2 4x
yf �x, y� � x2 � 2xy � y2, R � ��x, y�: x2 � y2 ≤ 8
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61.
For also, and
For
The absolute maximum is
The absolute minimum is In fact, f �0, y� � f �x, 0� � 0��0 � f �0, 0�.
1 � f �1, 1�.
f �1, 1� � 1.y � 1,x � 1,
f �0, 0� � 0.y � 0,x � 0,
fy �4�1 � y2�x
�x2 � 1��y2 � 1�2 ⇒ x � 0 or y � 1
fx �4�1 � x2�y
�y2 � 1��x2 � 1� � 0 ⇒ x � 1 or y � 0
x1
1
R
yf �x, y� �4xy
�x2 � 1��y2 � 1�, R � ��x, y�: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1�
63. False
Let
is a relative maximum, but and do not exist.fy�0, 0�fx�0, 0��0, 0, 1�f �x, y� � �1 � x � y�.
Section 12.9 Applications of Extrema of Functions of Two Variables 113
Section 12.9 Applications of Extrema of Functions of Two Variables
1. A point on the plane is given by Thesquare of the distance from the origin to this point is
From the equations and we obtain the system
Solving simultaneously, we have
Therefore, the distance from
the origin to is
��127
2
� �187
2
� �67
2
�6�14
7.
�127 , 18
7 , 67�z � 12 �
247 �
547 �
67.
y �187x �
127 ,
3x � 5y � 18.
5x � 6y � 24
Sy � 0,Sx � 0
Sy � 2y � 2�12 � 2x � 3y���3�
Sx � 2x � 2�12 � 2x � 3y���2�
S � x2 � y2 � �12 � 2x � 3y�2
�x, y, 12 � 2x � 3y�. 3. A point on the paraboloid is given by Thesquare of the distance from to a point on theparaboloid is given by
From the equations and we obtain the system
Multiply the first equation by y and the second equationby x, and subtract to obtain Then, we have
and the distance is
��1 � 5�2 � �1 � 5�2 � �2 � 0�2 � 6.
z � 2y � 1,x � 1,x � y.
2y3 � 2x2y � y � 5 � 0
2x3 � 2xy2 � x � 5 � 0
Sy � 0,Sx � 0
Sy � 2�y � 5� � 4y�x2 � y2� � 0.
Sx � 2�x � 5� � 4x�x2 � y2� � 0
S � �x � 5�2 � � y � 5�2 � �x2 � y2�2
�5, 5, 0��x, y, x2 � y2�.
5. Let and be the numbers. Since
Solving simultaneously yields and z � 10.y � 10,x � 10,
Py � 30x � x2 � 2xy � x�30 � x � 2y� � 0 x � 2y � 30
Px � 30y � 2xy � y2 � y�30 � 2x � y� � 0 2x � y � 30
P � xyz � 30xy � x2y � xy2
z � 30 � x � y.x � y � z � 30,zx, y
7. Let and be the numbers and let Since we have
Solving simultaneously yields and z � 10.y � 10,x � 10,
Sy � 2y � 2�30 � x � y���1� � 0 x � 2y � 30.
Sx � 2x � 2�30 � x � y���1� � 0 2x � y � 30
S � x2 � y2 � �30 � x � y�2
x � y � z � 30,S � x2 � y2 � z2.zx, y,
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114 Chapter 12 Functions of Several Variables
11. Let Then
Solving this system simultaneously yields and substitution yields Therefore, the solution isa � b � c � k�3.
b � k�3.a � b
Vb �4�
3�ka � a2 � 2ab� � 0 ka � a2 � 2ab � 0.
Va �4�
3�kb � 2ab � b2� � 0 kb � 2ab � b2 � 0
�43
� �kab � a2b � ab2�
V �4� abc
3�
43
� ab�k � a � b�
a � b � c � k. 13. Let and be the length, width, and height,respectively and let be the given volume.
Then and The surface area is
Solving simultaneously yields and z � 3�V0.
x � 3�V0, y � 3�V0,
Sy � 2�x �V0
y2 � 0 xy2 � V0 � 0.
Sx � 2�y �V0
x2 � 0 x2y � V0 � 0
S � 2xy � 2yz � 2xz � 2�xy �V0
x�
V0
y z � V0�xy.V0 � xyz
V0
zx, y,
15. The distance from to is The distance from to is The distance from to is
Cy � 2k� y � x
��y � x�2 � 1 � k � 0 ⇒ y � x
��y � x�2 � 1�
12
Cx � 3k� x
�x2 � 4 � 2k� ��y � x���y � x�2 � 1 � 0
C � 3k�x2 � 4 � 2k��y � x�2 � 1 � k�10 � y�
10 � y.SR��y � x�2 � 1.RQ�x2 � 4.QP
9. Let and be the length, width, and height, respectively. Then the sum of the length and girth is given byor The volume is given by
Solving the system and we obtain the solution inches, inches, and inches.z � 18y � 18x � 362y � 4z � 108,4y � 2z � 108
Vz � 108y � 2y2 � 4yz � y�108 � 2y � 4z� � 0.
Vy � 108z � 4yz � 2z2 � z�108 � 4y � 2z� � 0
V � xyz � 108zy � 2zy2 � 2yz2
x � 108 � 2y � 2z.x � �2y � 2z� � 108zx, y,
Therefore, km and kms.y �2�3 � 3�2
6� 1.284x �
�22
� 0.707
y �1
�3�
1
�2�
2�3 � 3�26
�y � x�2 �13
4�y � x�2 � �y � x�2 � 1
2�y � x� � ��y � x�2 � 1
x ��22
x2 �12
9x2 � x2 � 4
3x � �x2 � 4
x
�x2 � 4�
13
3k� x
�x2 � 4 � 2k��12 � 0
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Section 12.9 Applications of Extrema of Functions of Two Variables 115
17. Let h be the height of the trough and r the length of the slanted sides. We observe that the area of a trapezoidal cross section isgiven by
where and Substituting these expressions for and we have
Now
Substituting the expression for from into the equation we have
Therefore, the first partial derivatives are zero when and (Ignore the solution ) Thus, thetrapezoid of maximum area occurs when each edge of width is turned up from the horizontal.60�w�3
r � � � 0.r � w�3.� � ��3
r 2�2 cos � � 1� � 0 or cos � �12
.
r 2�4 � 2 cos ��cos � � 2r 2 cos � � r 2�2 cos2 � � 1� � 0
A��r, �� � 0,Ar�r, �� � 0w
A��r, �� � wr cos � � 2r 2 cos � � r 2 cos 2� � 0.
Ar�r, �� � w sin � � 4r sin � � 2r sin � cos � � sin ��w � 4r � 2r cos �� � 0 ⇒ w � r �4 � 2 cos ��
A�r, �� � �w � 2r � r cos ���r sin �� � wr sin � � 2r 2 sin � � r 2 sin � cos �
h,xh � r sin �.x � r cos �
A � h �w � 2r� � ��w � 2r� � 2x�2 � � �w � 2r � x�h
19.
Solving this system yields and
and
Thus, revenue is maximized when and x2 � 6.x1 � 3
Rx1x1Rx2x2
� �Rx1x2�2 > 0Rx1x1
< 0
Rx2x2� �16
Rx1x2� �2
Rx1x1� �10
x2 � 6.x1 � 3
Rx2� �16x2 � 2x1 � 102 � 0, x1 � 8x2 � 51
Rx1� �10x1 � 2x2 � 42 � 0, 5x1 � x2 � 21
R�x1, x2� � �5x12 � 8x2
2 � 2x1x2 � 42x1 � 102x2
21.
and
Therefore, profit is maximized when and x2 � 110.x1 � 275
Px1x1Px2x2
� �Px1x2�2 > 0Px1x1
< 0
Px2x2� �0.10
Px1x2� 0
Px1x1� �0.04
Px2� �0.10x2 � 11 � 0, x2 � 110
Px1� �0.04x1 � 11 � 0, x1 � 275
� �0.02x12 � 0.05x2
2 � 11x1 � 11x2 � 775
� 15x1 � 15x2 � �0.02x12 � 4x1 � 500� � �0.05x2
2 � 4x2 � 275�
P�x1, x2� � 15�x1 � x2� � C1 � C2
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116 Chapter 12 Functions of Several Variables
23. (a)
From the graph we see that the surface has a minimum.
(b)
(c)
(d)
Using a computer algebra system, we find that the minimum occurs when Thus,
(e)
Using a computer algebra system, we find that the minimum occurs when Thus
Using a computer algebra system, we find that the minimum occurs when Thus,
Note: The minimum occurs at
(f ) points in the direction that decreases most rapidly. You would use for maximization problems.�S�x, y�S��S�x, y�
�x, y� � �0.0555, 0.3992�
�x4, y4� � �0.06, 0.44�.t � 0.44.
� ���3.90 � 0.09t�2 � ��1.55 � 0.01t�2
S�0.10 � 0.09t, 0.45 � 0.01t� � ��0.10 � 0.09t�2 � �0.45 � 0.01t�2 � ��2.10 � 0.09t�2 � ��1.55 � 0.01t�2
�x4, y4� � �x3 � Sx�x3, y3�t, y3 � Sy�x3, y3�t� � �0.10 � 0.09t, 0.44 � 0.01t�
�x3, y3� � �0.10, 0.44�.t � 1.78.
� ���3.95 � 0.03t�2 � ��1.10 � 0.26t�2
S�0.05 � 0.03t, 0.90 � 0.26t� � ��0.05 � 0.03t�2 � �0.90 � 0.26t�2 � ��2.05 � 0.03t�2 � ��1.10 � 0.26t�2
�x3, y3� � �x2 � Sx�x2, y2�t, y2 � Sy�x2, y2�t� � �0.05 � 0.03t, 0.90 � 0.26t�
�x2, y2� � �0.05, 0.90�.t � 1.344.
��10 � �2�105
� 4�2t � �1 �2�5
5�
25t2
��10 � �2�105
� 2�2t � �1 �2�5
5�
25t2
S�1 �1
�2t, 1 � � 2
�10�
1
�2 t ��2 � �2�105
� 2�2t � �1 �2�5
5�
25t2
�x2, y2� � �x1 � Sx�x1, y1�t, y1 � Sy�x1, y1�t� � �1 �1
�2t, 1 � � 2
�10�
1�2t
tan � ��2��10� � �1��2�
�1��2� 1 �
2
�5 ⇒ � � 186.027�
��S�1, 1� � �Sx�1, 1�i � Sy�1, 1�j � �1
�2i � � 1
�2�
2
�10j
Sy�x, y� �y
�x2 � y2�
y � 2
��x � 2�2 � �y � 2�2�
y � 2
��x � 4�2 � �y � 2�2
Sx�x, y� �x
�x2 � y2�
x � 2
��x � 2�2 � �y � 2�2�
x � 4
��x � 4�2 � �y � 2�2
xy
468
24
20
4
22 4 6 8
S
� �x2 � y2 � ��x � 2�2 � �y � 2�2 � ��x � 4�2 � �y � 2�2
� ��x � 0�2 � �y � 0�2 � ��x � 2�2 � �y � 2�2 � ��x � 4�2 � �y � 2�2
S�x, y� � d1 � d2 � d3
25. Write the equation to be maximized or minimized as a function of two variables. Set the partial derivativesequal to zero (or undefined) to obtain the critical points. Use the Second Partials Test to test for relative extremausing the critical points. Check the boundary points, too.
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Section 12.9 Applications of Extrema of Functions of Two Variables 117
27. (a)
(b)
�16
S � ��32
�43
� 02
� �43
� 12
� �32
�43
� 32
y �34
x �43
a �3�6� � 0�4�3�8� � 02 �
34
, b �13 4 �
34
�0�� �43
,
x y xy
0 0 4
0 1 0 0
2 3 6 4
� xi2 � 8� xiyi � 6� yi � 4� xi � 0
�2
x229. (a)
(b) S � �4 � 4�2 � �2 � 3�2 � �2 � 1�2 � �0 � 0�2 � 2
y � �2x � 4
a �4�4� � 4�8�4�6� � 42 � �2, b �
14
�8 � 2�4�� � 4,
x y xy
0 4 0 0
1 3 3 1
1 1 1 1
2 0 0 4
� xi2 � 6� xiyi � 4� yi � 8� xi � 4
x2
31.
−2 10
−1
7
(0, 0)(1, 1)
(4, 2)
(3, 4)
(5, 5)
y x= +37 743 43
y �3743
x �7
43
b �15 12 �
3743
�13�� �7
43
a �5�46� � 13�12�5�51� � �13�2 �
7486
�3743
�xi2 � 51�xiyi � 46,
� yi � 12,� xi � 13,
�0, 0�, �1, 1�, �3, 4�, �4, 2�, �5, 5� 33.
−4 18
−6
(0, 6)
(4, 3)
(5, 0)
(8, 4)− (10, 5)−
y = − x +175148
945148
8
y � �175148
x �945148
b �15 0 � ��
175148�27�� �
945148
a �5��70� � �27��0�
5�205� � �27�2 ��350296
� �175148
� xi2 � 205� xiyi � �70,
� yi � 0,� xi � 27,
�0, 6�, �4, 3�, �5, 0�, �8, �4�, �10, �5�
35. (a)
(b)
(c) For each one-year increase in age, the pressurechanges by 1.7236 (slope of line).
0100
100
240
y � 1.7236x � 79.7334 37.
When bushels per acre.y � 41.4x � 1.6,
y � 14x � 19b � 19,a � 14,
� xi2 � 13.5� xi � 7, � yi � 174, � xiyi � 322,
�1.0, 32�, �1.5, 41�, �2.0, 48�, �2.5, 53�
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118 Chapter 12 Functions of Several Variables
39.
a�n
i�1xi
2 � b�n
i�1xi � cn � �
n
i�1yi
a�n
i�1xi
3 � b�n
i�1xi
2 � c�n
i�1xi � �
n
i�1xiyi
a�n
i�1xi
4 � b�n
i�1xi
3 � c�n
i�1xi
2 � �n
i�1xi
2yi
Sc
� �2�n
i�1� yi � axi
2 � bxi � c� � 0
Sb
� �n
i�1�2xi� yi � axi
2 � bxi � c� � 0
Sa
� �n
i�1�2xi
2� yi � axi2 � bxi � c� � 0
S�a, b, c� � �n
i�1 � yi � axi
2 � bxi � c�2 41.
y �37 x2 �
65 x �
2635c �
2635 ,b �
65 ,a �
37 ,
10a � 5c � 810b � 12,34a � 10c � 22,
� xi2yi � 22
� xiyi � 12
� xi4 � 34
�xi3 � 0
� xi2 � 10
� yi � 8
−2
6−9( 2, 0)−
( 1, 0)−
(0, 1)
(1, 2)
(2, 5)
8 � xi � 0
��2, 0�, ��1, 0�, �0, 1�, �1, 2�, �2, 5�
43.
−5
−2
7(0, 0)
(2, 2)
(3, 6)
(4, 12)
14
y � x2 � xc � 0,b � �1,a � 1,
29a � 9b � 4c � 20
99a � 29b � 9c � 70
353a � 99b � 29c � 254
� xi2yi � 254
� xiyi � 70
�xi4 � 353
� xi3 � 99
� xi2 � 29
� yi � 20
� xi � 9
�0, 0�, �2, 2�, �3, 6�, �4, 12� 45.
−1 14
−20
120
y � �25112 x2 �
54156 x �
2514 � �0.22x2 � 9.66x � 1.79
220a � 30b � 6c � 230
1,800a � 220b � 30c � 1,670
15,664a � 1,800b � 220c � 13,500
� xi2yi � 13,500
� xiyi � 1,670,
� xi4 � 15,664,
� xi3 � 1,800,
� xi2 � 220,
� yi � 230,
� xi � 30,
�0, 0�, �2, 15�, �4, 30�, �6, 50�, �8, 65�, �10, 70�
47. (a)
(b)
(c)
(d) Same answers.
−2 24
−2,000
14,000
P � e�0.1499h�9.3018 � 10,957.7e�0.1499h
ln P � �0.1499h � 9.3018
ln P � �0.1499h � 9.3018
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Section 12.10 Lagrange Multipliers
Section 12.10 Lagrange Multipliers 119
1. Maximize
Constraint:
f �5, 5� � 25
x � y � 10 � ⇒ x � y � 5
y � �
x � ��x � y
y i � xj � ��i � j�
�f � ��g
x � y � 10
2
2
4
4
6
6
8
8
10
10
12
12
xconstraint
level curves
yf �x, y� � xy. 3. Minimize
Constraint:
f �2, 2� � 8
x � y � 4 ⇒ x � y � 2
2x � �
2y � ��x � y
2x i � 2yj � � i � � j
�f � ��g
x � y � 4
x
4
4−4
−4
constraint
level curves
yf �x, y� � x2 � y2.
5. Minimize
Constraint:
f �2, 4� � �12
� � 4, x � 2, y � 4
x � 2y � �6 ⇒ �32
� � �6
�2y � �2� ⇒ y � �
2x � � ⇒ x ��
2
2x i � 2yj � � i � 2� j
�f � ��g
x � 2y � �6
f �x, y� � x2 � y2. 7. Maximize
Constraint:
f �25, 50� � 2600
x � 25, y � 50
2x � y � 100 ⇒ 4x � 100
2 � 2y � 2� ⇒ y � � � 1
� � 12x � 1 � � ⇒ x �
2�y � 2x
�2 � 2y�i � �2x � 1�j � 2� i � � j
�f � ��g
2x � y � 100
f �x, y� � 2x � 2xy � y.
9. Note: is maximum when is maximum.
Maximize
Constraint:
f �1, 1� � �g�1, 1� � 2
x � y � 2 ⇒ x � y � 1
�2x � �
�2y � �� x � y
x � y � 2
g�x, y� � 6 � x2 � y2.
g�x, y�f �x, y� � �6 � x2 � y2 11. Maximize .
Constraint:
f �2, 2� � e4
x � y � 2
x2 � y2 � 8 ⇒ 2x2 � 8
yexy � 2x�
xexy � 2y�� x � y
x2 � y2 � 8
f �x, y� � exy
13. Maximize or minimize
Constraint:
Case 1: On the circle
Maxima:
Minima: f �±�22
, ��22 � � �
12
f �±�22
, ±�22 � �
52
x2 � y2 � 1 ⇒ x � ±�22
, y � ±�22
2x � 3y � 2x�
3x � 2y � 2y�� x2 � y2
x2 � y2 � 1
x2 � y2 ≤ 1
f �x, y� � x2 � 3xy � y2. Case 2: Inside the circle
Saddle point:
By combining these two cases, we have a maximum of at
and a minimum of at
�±�22
, ��22 �.
�12
�±�22
, ±�22 �
52
f �0, 0� � 0
fxx � 2, fyy � 2, fxy � 3, fxx fyy � � fxy�2 ≤ 0
fx � 2x � 3y � 0
fy � 3x � 2y � 0� x � y � 0
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120 Chapter 12 Functions of Several Variables
15. Minimize .
Constraint:
f �2, 2, 2� � 12
x � y � z � 6 ⇒ x � y � z � 2
2x � �
2y � �
2z � �� x � y � z
x � y � z � 6
f �x, y, z� � x2 � y2 � z2 17. Minimize .
Constraint:
f �13 , 13 , 13� �
13
x � y � z � 1 ⇒ x � y � z �13
2x � �
2y � �
2z � �� x � y � z
x � y � z � 1
f �x, y, z� � x2 � y2 � z2
19. Maximize .
Constraints:
f �8, 16, 8� � 1024
y � 16
x � y � z � 32
x � y � z � 0 � 2x � 2z � 32 ⇒ x � z � 8
yz � � � �
xz � � � �
xy � � � �� yz � xy ⇒ x � z
yz i � xz j � xyk � ��i � j � k� � ��i � j � k�
�f � ��g � ��h
x � y � z � 0
x � y � z � 32
f �x, y, z� � xyz 21. Maximize .
Constraints:
f �3, 32
, 1� � 6
x � 3, y �32
, z � 1
x �x3
�83�3 �
x2�
x � 3z � 0 ⇒ z �x3
x � 2y � 6 ⇒ y � 3 �x2
y � � � �
x � z � 2�
y � �3�� y �
34
� ⇒ x � z �8
3y
y i � �x � z� j � yk � ��i � 2j� � ��i � 3k�
�f � ��g � ��h
x � 3z � 0
x � 2y � 6
f �x, y, z� � xy � yz
23. Minimize the square of the distance subject to the constraint
The point on the line is and the desired distance is
d ����2
13�2
� ��3
13�2
��1313
.
��213 , � 3
13�
2x � 3y � �1 ⇒ x � �2
13, y � �
313
2x � 2�
2y � 3�� y �3x2
2x � 3y � �1.f �x, y� � x2 � y2 25. Minimize the square of the distance
subject to the constraint
The point on the plane is and the desired distance is
d � ��1 � 2�2 � �0 � 1�2 � �0 � 1�2 � �3.
�1, 0, 0�
x � 1, y � z � 0
x � y � z � 1 ⇒ x � 2�x � 1� � 1
2�x � 2� � �
2�y � 1� � �
2�z � 1� � �� y � z and y � x � 1
x � y � z � 1.
f �x, y, z� � �x � 2�2 � �y � 1�2 � �z � 1�2
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Section 12.10 Lagrange Multipliers 121
27. Maximize subject to the constraintsand
Choosing the positive value for y we have the point
�10 � 2�26515
, 5 � �265
15,
�1 � �2653 �.
y �5 ± �265
15
15y2 � 10y � 16 � 0
30y2 � 20y � 32 � 0
�2y�2 � y2 � �5y � 2�2 � 36
2x � y � z � 2 ⇒ z � 2x � y � 2 � 5y � 2
x2 � y2 � z2 � 36
0 � 2x� � 2�
0 � 2y� � �
1 � 2z� � �� x � 2y
2x � y � z � 2.x2 � y2 � z2 � 36f �x, y, z� � z 29. Optimization problems that have restrictions or contstraints
on the values that can be used to produce the optimalsolution are called contrained optimization problems.
31. Maximize subject to the constraint
Volume is maximum when the dimensions areinches
xy
z
36 18 18
x � 36, y � z � 18
x � 2y � 2z � 108 ⇒ 6y � 108, y � 18
yz � �
xz � 2�
xy � 2�� y � z and x � 2y
x � 2y � 2z � 108.V�x, y, z� � xyz 33. Minimize subject
to the constraint
Dimensions: feet
x
y
z
3�360 3�360 43 3�360
x � y � 3�360, z �43 3�360
xyz � 480 ⇒ 43 y3 � 480
8y � 6z � yz�
8x � 6z � xz�
6x � 6y � xy�� x � y, 4y � 3z
xyz � 480.C�x, y, z� � 5xy � 3�2xz � 2yz � xy�
35. Maximize subject to the constraint
Therefore, the dimensions of the box are 2�3a
3
2�3b3
2�3c
3.
x �a
�3, y �
b
�3, z �
c
�3
x2
a2 �y2
b2 �z2
c2 � 1 ⇒ 3x2
a2 � 1, 3y2
b2 � 1, 3z2
c2 � 1
8xy �2zc2�
x2
a2 �y2
b2 �z2
c28xz �2yb2��
8yz �2xa2�
x2
a2 �y2
b2 �z2
c2 � 1.V�x, y, z� � �2x��2y��2z� � 8xyz
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122 Chapter 12 Functions of Several Variables
37. Using the formula Time , minimize subject to the constraint
Since and , we have
orsin 1
v1�
sin 2
v2.
x��d12 � x2
v1�
y��d22 � y2
v2
sin 2 �y
�d22 � y2
sin 1 �x
�d12 � x2
x � y � a
Medium 2Q
d2
d1
xy
a
1θ
2θ
Medium 1P x
v1�d12 � x2
�y
v2�d22 � y2
x
v1�d22 � x 2
� �
y
v2�d22 � y 2
� ��x � y � a.T�x, y� �
�d12 � x2
v1�
�d22 � y2
v2
DistanceRate
�
39. Maximize
Constraint:
P�13 , 13 , 13� � 2�1
3��13� � 2�1
3��13� � 2�1
3��13� �
23 .
q � r �23
p � q � r � 1� ⇒ p �13 , q �
13 , r �
13
p � q � r � 1
⇒ � �43
2q � 2r � �
2p � 2r � �
2p � 2q � �� ⇒ 3� � 4�p � q � r� � 4�1�
�P � ��g
p � q � r � 1
P� p, q, r� � 2pq � 2pr � 2qr. 41. Maximize
subject to the constraint
Therefore, P�31256 , 6250
3 � 147,314.
x �3125
6, y �
62503
48x � 36y � 100,000 ⇒ 192x � 100,000
y � 4x
yx
� 4
�yx�
0.75
�yx�
0.25
� �48�
25 �� 7536��
75x0.25y�0.25 � 36� ⇒ �xy�
0.25
�36�
75
25x�0.75y0.75 � 48� ⇒ �yx�
0.75
�48�
25
48x � 36y � 100,000.
P�x, y� � 100x0.25y0.75
43. Minimize subject to the constraint
Therefore, C�50�2, 200�2 � $13,576.45.
y � 4x � 200�2
x �20040.75 �
200
2�2� 50�2
100x0.25y0.75 � 20,000 ⇒ x0.25�4x�0.75 � 200
yx
� 4 ⇒ y � 4x
�yx�
0.75
�yx�
0.25
� � 4825���
75�
36 �
36 � 75x0.25y�0.25� ⇒ �xy�
0.25
�36
75�
48 � 25x�0.75y0.75� ⇒ �yx�
0.75
�48
25�
100x0.25y0.75 � 20,000.C�x, y� � 48x � 36y
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Review Exercises for Chapter 12
45. (a) Maximize subject to the constraint
(b)
� �cos � cos � cos�� � ��
� cos � cos ��cos � cos�� � �� � sin � sin�� � ���
g�� � �� � cos � cos � cos �� � �� � ���
� � � � � � � ⇒ � � � � �� � ��
α β3
3
2
3
γg��
3,
�
3.
�
3� �18
� � � � � � � ⇒ � � � � � ��
3
�sin � cos � cos � �
�cos � sin � cos � �
�cos � cos � sin � � � tan � � tan � � tan � ⇒ � � � � �
� � � � � � �.� cos � cos � cos �g��, �, ��
Review Exercises for Chapter 12 123
1. No, it is not the graph of a function.
3.
The level curves are of the form
The level curves are circles centered at the origin.
x
y
Generated by Mathematica
2−2
c = 10
c = 1
−2
2
ln c � x2 � y2.
c � ex2�y2
f �x, y� � ex2�y2 5.
The level curves are of the form
The level curves are hyperbolas.
x
y
Generated by Mathematica
1
4
4
c = 12
c = −12 c = −2c = 2
−4 1−1
−4
1 �x2
c�
y2
c.
c � x2 � y2
f �x, y� � x2 � y2
7.
x
y3 3
3
−3
−3
−3
z
f �x, y� � e��x2�y2� 9.
Elliptic paraboloid
y � x2 � z2 � 1
xy3
2
3
z f �x, y, z� � x2 � y � z2 � 1
11.
Continuous except at �0, 0�.
lim�x, y�→�1, 1�
xy
x2 � y2 �12
13.
For for
For for
Thus, the limit does not exist. Continuous except at �0, 0�.
x 0�4x 2yx4 � y 2 � 0,y � 0,
x 0�4x 2yx4 � y 2 �
�4x4
x4 � x4 � �2,y � x 2,
lim�x, y�→�0, 0�
�4x2y
x4 � y2
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124 Chapter 12 Functions of Several Variables
17.
�z�y
� xey � ex
�z�x
� ey � yex
z � xey � yex
19.
gy �x�x2 � y2��x2 � y2�2
�y� y2 � x2��x2 � y2�2 gx �
y�x2 � y2� � xy�2x��x2 � y2�2
g�x, y� �xy
x2 � y221.
fz � arctan yx
fy �z
1 � �y2x2��1x� �
xzx2 � y2
fx �z
1 � �y2x2���yx2� �
�yzx2 � y2
f �x, y, z� � z arctan yx
23.
�u�t
� �cn2e�n2t sin�nx�
�u�x
� cne�n2t cos�nx�
u�x, t� � ce�n2t sin�nx�
31.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �2
�z�y
� �2y
�2z�x2 � 2
�z�x
� 2x
z � x2 � y2
33.
Therefore,�2z�x2 �
�2z�y2 � 0.
� �2y 3x2 � y2
�x2 � y2�3
�2z�y2 �
�x2 � y2�2��2y� � 2�x2 � y2��x2 � y2��2y��x2 � y2�4
�z�y
��x2 � y2� � 2y
�x2 � y2�2 �x2 � y2
�x2 � y2�2
�2z�x2 � �2y �4x2
�x2 � y2�3 �1
�x2 � y2�2� � 2y 3x2 � y2
�x2 � y2�3
�z�x
��2xy
�x2 � y2�2
z �y
x2 � y2
15.
fy � �ex sin y
fx � ex cos y
f �x, y� � ex cos y
25.
x
y
3
3
−1
3
z
27.
fyx � �1
fxy � �1
fyy � 12y
fxx � 6
fy � �x � 6y2
fx � 6x � y
f �x, y� � 3x2 � xy � 2y3 29.
hyx � cos y � sin x
hxy � cos y � sin x
hyy � �x sin y
hxx � �y cos x
hy � x cos y � cos x
hx � sin y � y sin x
h�x, y� � x sin y � y cos x
35.
dz ��z�x
dx ��z�y
dy � �sin yx
�yx cos
yx� dx � �cos
yx� dy
z � x sin yx
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Review Exercises for Chapter 12 125
37.
Percentage error:dzz
�1726
13� 0.0503 � 5%
dz �xz dx �
yz dy �
513�
12� �
1213�
12� �
1726
� 0.654 cm
2z dx � 2x dx � 2y dy
z2 � x2 � y2 39.
� ±56� ± 1
6� � ±� in.3
dV �23�rh dr �
13�r2 dh �
23� �2��5��±1
8� �13� �2�2�±1
8� V �
13�r2h
41.
Chain Rule:
Substitution:
dwdt
�2�2t � 3��2� � 2�4 � t�
�2t � 3�2 � �4 � t�2 �10t � 4
5t2 � 4t � 25
w � ln�x2 � y2� � ln��2t � 3�2 � �4 � t�2�
�10t � 4
5t2 � 4t � 25
�2�2t � 3�2
�2t � 3�2 � �4 � t�2 �2�4 � t�
�2t � 3�2 � �4 � t�2
�2x
x2 � y2 �2� �2y
x2 � y2 ��1�
dwdt
��w�x
dxdt
��w�y
dydt
w � ln�x2 � y2�, x � 2t � 3, y � 4 � t
43.
Chain Rule:
Substitution:
�u�t
� 2t
�u�r
� 2r
u�r, t� � r 2 cos2 t � r 2 sin2 t � t2 � r 2 � t2
� 2t
� 2��r2 sin t cos t � r2 sin t cos t� � 2t
� 2x��r sin t� � 2y�r cos t� � 2z
�u�t
��u�x
�x�t
��u�y
�y�t
��u�z
�z�t
� 2�r cos2 t � r sin2 t� � 2r
� 2x cos t � 2y sin t � 2z�0�
�u�r
��u�x
�x�r
��u�y
�y�r
��u�z
�z�r
z � ty � r sin t,x � r cos t,u � x2 � y2 � z2,
45.
�z�y
��x2 � 2z
�2y � x � 2z�
x2 � 2zx � 2y � 2z
x2 � 2y �z�y
� 2z � x �z�y
� 2z �z�y
� 0
�z�x
��2xy � z
�2y � x � 2z�
2xy � zx � 2y � 2z
2xy � 2y �z�x
� x �z�x
� z � 2z �z�x
� 0
x2y � 2yz � xz � z2 � 0
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126 Chapter 12 Functions of Several Variables
47.
Du f �2, 1� � �f �2, 1� u � 2 2 � 2 2 � 0
u �1
2v �
22
i � 22
j
�f �2, 1� � 4 i � 4j
�f � 2xyi � x2j
f �x, y� � x2y 49.
�43
�43
�23
�23
Duw�1, 2, 2� � �w�1, 2, 2� u
u �13
v �23
i �13
j �23
k
�w�1, 2, 2� � 2i � 4 j � k
�w � z i � 2yj � xk
w � y2 � xz
51.
��z�1, 1�� �12
�z�1, 1� � �12
i � ��12
, 0�
�z � �2xy
�x2 � y2�2 i �x2 � y2
�x2 � y2�2 j
z �y
x2 � y2 53.
� �z�0, �
4� � � 1
�z�0, �
4� � � 22
i � 22
j � �� 22
, � 22 �
�z � �e�x cos yi � e�x sin y j
z � e�x cos y
55.
Unit normal:54i � 16j
�54i � 16j��
1 793
�27i � 8j�
�f �3, 2� � 54i � 16j
�f �x, y� � 18xi � 8yj
f �x, y� � 9x2 � 4y2
9x2 � 4y2 � 65 57.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 2
4�
y � 14
�z � 4�1
.
4x � 4y � z � 8,
4�x � 2� � 4� y � 1� � �z � 4� � 0
�F�2, 1, 4� � 4i � 4j � k
�F � 2xy i � x2j � k
F�x, y, z� � x2y � z � 0
59.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 2, y � �3, z � 4 � t.
z � 4,z � 4 � 0
�F�2, �3, 4� � k
�F � �2x � 4�i � �2y � 6�j � k
F�x, y, z� � x2 � y2 � 4x � 6y � z � 9 � 0 61.
Therefore, the equation of the tangent line is
z � 3.x � 2
1�
y � 12
,
�F � �G � � i40
j�2
0
k�1�1� � 2�i � 2j�
�F�2, 1, 3� � 4i � 2j � k
�G � �k
�F � 2x i � 2yj � k
G�x, y, z� � 3 � z � 0
F�x, y, z� � x2 � y2 � z � 0
63.
Normal vector to plane.
� � 36.7�
cos � � �n k��n�
�6
56�
3 14
14
�f �2, 1, 3� � 4i � 2j � 6k
�f �x, y, z� � 2xi � 2yj � 2zk
f �x, y, z� � x2 � y2 � z2 � 14
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Review Exercises for Chapter 12 127
65.
From we have Substituting this into we have Thus, or
At the critical point , Therefore, is a saddle point.
At the critical point , and Therefore, is a relative minimum.�32 , 94 , �27
16�fxx > 0.fxx fyy � � fxy�2 > 0�32 , 94�
�0, 0, 0�fxx fyy � � fxy�2 < 0.�0, 0�
32 .x � 0�3x � 2x2 � x�2x � 3� � 0.fy � 0,y � x2.fx � 0,
fxy � �3
fyy � 2
fxx � 6x
fy � �3x � 2y � 0
fx � 3x2 � 3y � 3�x2 � y� � 0
yx
30
2
−30
zf �x, y� � x3 � 3xy � y2
67.
Thus, or and substitution yields the critical point
At the critical point and Thus, is a relative minimum.�1, 1, 3�� 3 > 0.fxx fyy � � fxy�2fxx � 2 > 0�1, 1�,
fyy �2y3
fxy � 1
fxx �2x3
�1, 1�.x � yx2y � xy2
fy � x �1y2 � 0, xy2 � 1
fx � y �1x2 � 0, x2y � 1
x
y
−24
−20
3 44
20
(1, 1, 3)
zf �x, y� � xy �1x
�1y
69. The level curves are hyperbolas. There is a critical point at , but there are no relative extrema. The gradient is normal to thelevel curve at any given point at �x0, y0�.
�0, 0�
71.
Solving this system yields and
Therefore, profit is maximum when and x2 � 157.x1 � 94
Px1x1 Px2x2
� �Px1x2�2 > 0
Px1x1 < 0
Px2x2� �0.86
Px1x2� �0.8
Px1x1� �0.9
x2 � 157.x1 � 94
0.8x1 � 0.86x2 � 210
Px2� �0.86x2 � 0.8x1 � 210 � 0
0.9x1 � 0.8x2 � 210
Px1� �0.9x1 � 0.8x2 � 210 � 0
� �0.45x12 � 0.43x2
2 � 0.8x1x2 � 210x1 � 210x2 � 11,500
� �225 � 0.4�x1 � x2���x1 � x2� � �0.05x12 � 15x1 � 5400� � �0.03x2
2 � 15x2 � 6100�
P�x1, x2� � R � C1 � C2
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128 Chapter 12 Functions of Several Variables
73. Maximize subject to the constraint
f �49.4, 253� � 13,201.8
y � 253
x � 49.4
10x � 494
5x � y � �6
20x � 4y � 2000 ⇒ 5x � y � 500
4 � y � 20
x � 2 � 4� 5x � y � �6
20x � 4y � 2000.f �x, y� � 4x � xy � 2y
77. Optimize subject to the constraint
Maximum: f �13 , 13 , 13� �
13
x � y � z � 1 ⇒ x � y � z �13
y � z �
x � z �
x � y � � x � y � z
x � y � z � 1.f �x, y, z� � xy � yz � xz
75. (a)
(b)
Yes, the data appears more linear.
(c)
(d)
The logarithmic model is a better fit.
−1 10
−5
25
y � 8.37 ln t � 1.54
−1 3
−5
20
−2 11
−5
30
y � 2.29t � 2.34
79.
Constraint:
Hence, x � 22
, y � 33
, z � 10 � 22
� 33
� 8.716 m.
4y2 � y2 � 1 ⇒ y2 �13
9x2 � x2 � 4 ⇒ x2 �12
1 �
2y � y2 � 1
3x � x2 � 4
3x x2 � 4
i �2y
y2 � 1j � k � �i � j � k�
�C � �g
x � y � z � 10
C � 3 x2 � 4 � 2 y2 � 1 � 2
PQ � x2 � 4, QR � y2 � 1, RS � z; x � y � z � 10
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Review Exercises for Chapter 12
45. (a) Maximize subject to the constraint
(b)
� �cos � cos � cos�� � ��
� cos � cos ��cos � cos�� � �� � sin � sin�� � ���
g�� � �� � cos � cos � cos �� � �� � ���
� � � � � � � ⇒ � � � � �� � ��
α β3
3
2
3
γg��
3,
�
3.
�
3� �18
� � � � � � � ⇒ � � � � � ��
3
�sin � cos � cos � �
�cos � sin � cos � �
�cos � cos � sin � � � tan � � tan � � tan � ⇒ � � � � �
� � � � � � �.� cos � cos � cos �g��, �, ��
Review Exercises for Chapter 12 123
1. No, it is not the graph of a function.
3.
The level curves are of the form
The level curves are circles centered at the origin.
x
y
Generated by Mathematica
2−2
c = 10
c = 1
−2
2
ln c � x2 � y2.
c � ex2�y2
f �x, y� � ex2�y2 5.
The level curves are of the form
The level curves are hyperbolas.
x
y
Generated by Mathematica
1
4
4
c = 12
c = −12 c = −2c = 2
−4 1−1
−4
1 �x2
c�
y2
c.
c � x2 � y2
f �x, y� � x2 � y2
7.
x
y3 3
3
−3
−3
−3
z
f �x, y� � e��x2�y2� 9.
Elliptic paraboloid
y � x2 � z2 � 1
xy3
2
3
z f �x, y, z� � x2 � y � z2 � 1
11.
Continuous except at �0, 0�.
lim�x, y�→�1, 1�
xy
x2 � y2 �12
13.
For for
For for
Thus, the limit does not exist. Continuous except at �0, 0�.
x 0�4x 2yx4 � y 2 � 0,y � 0,
x 0�4x 2yx4 � y 2 �
�4x4
x4 � x4 � �2,y � x 2,
lim�x, y�→�0, 0�
�4x2y
x4 � y2
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124 Chapter 12 Functions of Several Variables
17.
�z�y
� xey � ex
�z�x
� ey � yex
z � xey � yex
19.
gy �x�x2 � y2��x2 � y2�2
�y� y2 � x2��x2 � y2�2 gx �
y�x2 � y2� � xy�2x��x2 � y2�2
g�x, y� �xy
x2 � y221.
fz � arctan yx
fy �z
1 � �y2x2��1x� �
xzx2 � y2
fx �z
1 � �y2x2���yx2� �
�yzx2 � y2
f �x, y, z� � z arctan yx
23.
�u�t
� �cn2e�n2t sin�nx�
�u�x
� cne�n2t cos�nx�
u�x, t� � ce�n2t sin�nx�
31.
Therefore,�2z�x2 �
�2z�y2 � 0.
�2z�y2 � �2
�z�y
� �2y
�2z�x2 � 2
�z�x
� 2x
z � x2 � y2
33.
Therefore,�2z�x2 �
�2z�y2 � 0.
� �2y 3x2 � y2
�x2 � y2�3
�2z�y2 �
�x2 � y2�2��2y� � 2�x2 � y2��x2 � y2��2y��x2 � y2�4
�z�y
��x2 � y2� � 2y
�x2 � y2�2 �x2 � y2
�x2 � y2�2
�2z�x2 � �2y �4x2
�x2 � y2�3 �1
�x2 � y2�2� � 2y 3x2 � y2
�x2 � y2�3
�z�x
��2xy
�x2 � y2�2
z �y
x2 � y2
15.
fy � �ex sin y
fx � ex cos y
f �x, y� � ex cos y
25.
x
y
3
3
−1
3
z
27.
fyx � �1
fxy � �1
fyy � 12y
fxx � 6
fy � �x � 6y2
fx � 6x � y
f �x, y� � 3x2 � xy � 2y3 29.
hyx � cos y � sin x
hxy � cos y � sin x
hyy � �x sin y
hxx � �y cos x
hy � x cos y � cos x
hx � sin y � y sin x
h�x, y� � x sin y � y cos x
35.
dz ��z�x
dx ��z�y
dy � �sin yx
�yx cos
yx� dx � �cos
yx� dy
z � x sin yx
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Review Exercises for Chapter 12 125
37.
Percentage error:dzz
�1726
13� 0.0503 � 5%
dz �xz dx �
yz dy �
513�
12� �
1213�
12� �
1726
� 0.654 cm
2z dx � 2x dx � 2y dy
z2 � x2 � y2 39.
� ±56� ± 1
6� � ±� in.3
dV �23�rh dr �
13�r2 dh �
23� �2��5��±1
8� �13� �2�2�±1
8� V �
13�r2h
41.
Chain Rule:
Substitution:
dwdt
�2�2t � 3��2� � 2�4 � t�
�2t � 3�2 � �4 � t�2 �10t � 4
5t2 � 4t � 25
w � ln�x2 � y2� � ln��2t � 3�2 � �4 � t�2�
�10t � 4
5t2 � 4t � 25
�2�2t � 3�2
�2t � 3�2 � �4 � t�2 �2�4 � t�
�2t � 3�2 � �4 � t�2
�2x
x2 � y2 �2� �2y
x2 � y2 ��1�
dwdt
��w�x
dxdt
��w�y
dydt
w � ln�x2 � y2�, x � 2t � 3, y � 4 � t
43.
Chain Rule:
Substitution:
�u�t
� 2t
�u�r
� 2r
u�r, t� � r 2 cos2 t � r 2 sin2 t � t2 � r 2 � t2
� 2t
� 2��r2 sin t cos t � r2 sin t cos t� � 2t
� 2x��r sin t� � 2y�r cos t� � 2z
�u�t
��u�x
�x�t
��u�y
�y�t
��u�z
�z�t
� 2�r cos2 t � r sin2 t� � 2r
� 2x cos t � 2y sin t � 2z�0�
�u�r
��u�x
�x�r
��u�y
�y�r
��u�z
�z�r
z � ty � r sin t,x � r cos t,u � x2 � y2 � z2,
45.
�z�y
��x2 � 2z
�2y � x � 2z�
x2 � 2zx � 2y � 2z
x2 � 2y �z�y
� 2z � x �z�y
� 2z �z�y
� 0
�z�x
��2xy � z
�2y � x � 2z�
2xy � zx � 2y � 2z
2xy � 2y �z�x
� x �z�x
� z � 2z �z�x
� 0
x2y � 2yz � xz � z2 � 0
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126 Chapter 12 Functions of Several Variables
47.
Du f �2, 1� � �f �2, 1� u � 2 2 � 2 2 � 0
u �1
2v �
22
i � 22
j
�f �2, 1� � 4 i � 4j
�f � 2xyi � x2j
f �x, y� � x2y 49.
�43
�43
�23
�23
Duw�1, 2, 2� � �w�1, 2, 2� u
u �13
v �23
i �13
j �23
k
�w�1, 2, 2� � 2i � 4 j � k
�w � z i � 2yj � xk
w � y2 � xz
51.
��z�1, 1�� �12
�z�1, 1� � �12
i � ��12
, 0�
�z � �2xy
�x2 � y2�2 i �x2 � y2
�x2 � y2�2 j
z �y
x2 � y2 53.
� �z�0, �
4� � � 1
�z�0, �
4� � � 22
i � 22
j � �� 22
, � 22 �
�z � �e�x cos yi � e�x sin y j
z � e�x cos y
55.
Unit normal:54i � 16j
�54i � 16j��
1 793
�27i � 8j�
�f �3, 2� � 54i � 16j
�f �x, y� � 18xi � 8yj
f �x, y� � 9x2 � 4y2
9x2 � 4y2 � 65 57.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 2
4�
y � 14
�z � 4�1
.
4x � 4y � z � 8,
4�x � 2� � 4� y � 1� � �z � 4� � 0
�F�2, 1, 4� � 4i � 4j � k
�F � 2xy i � x2j � k
F�x, y, z� � x2y � z � 0
59.
Therefore, the equation of the tangent plane is
or
and the equation of the normal line is
x � 2, y � �3, z � 4 � t.
z � 4,z � 4 � 0
�F�2, �3, 4� � k
�F � �2x � 4�i � �2y � 6�j � k
F�x, y, z� � x2 � y2 � 4x � 6y � z � 9 � 0 61.
Therefore, the equation of the tangent line is
z � 3.x � 2
1�
y � 12
,
�F � �G � � i40
j�2
0
k�1�1� � 2�i � 2j�
�F�2, 1, 3� � 4i � 2j � k
�G � �k
�F � 2x i � 2yj � k
G�x, y, z� � 3 � z � 0
F�x, y, z� � x2 � y2 � z � 0
63.
Normal vector to plane.
� � 36.7�
cos � � �n k��n�
�6
56�
3 14
14
�f �2, 1, 3� � 4i � 2j � 6k
�f �x, y, z� � 2xi � 2yj � 2zk
f �x, y, z� � x2 � y2 � z2 � 14
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Review Exercises for Chapter 12 127
65.
From we have Substituting this into we have Thus, or
At the critical point , Therefore, is a saddle point.
At the critical point , and Therefore, is a relative minimum.�32 , 94 , �27
16�fxx > 0.fxx fyy � � fxy�2 > 0�32 , 94�
�0, 0, 0�fxx fyy � � fxy�2 < 0.�0, 0�
32 .x � 0�3x � 2x2 � x�2x � 3� � 0.fy � 0,y � x2.fx � 0,
fxy � �3
fyy � 2
fxx � 6x
fy � �3x � 2y � 0
fx � 3x2 � 3y � 3�x2 � y� � 0
yx
30
2
−30
zf �x, y� � x3 � 3xy � y2
67.
Thus, or and substitution yields the critical point
At the critical point and Thus, is a relative minimum.�1, 1, 3�� 3 > 0.fxx fyy � � fxy�2fxx � 2 > 0�1, 1�,
fyy �2y3
fxy � 1
fxx �2x3
�1, 1�.x � yx2y � xy2
fy � x �1y2 � 0, xy2 � 1
fx � y �1x2 � 0, x2y � 1
x
y
−24
−20
3 44
20
(1, 1, 3)
zf �x, y� � xy �1x
�1y
69. The level curves are hyperbolas. There is a critical point at , but there are no relative extrema. The gradient is normal to thelevel curve at any given point at �x0, y0�.
�0, 0�
71.
Solving this system yields and
Therefore, profit is maximum when and x2 � 157.x1 � 94
Px1x1 Px2x2
� �Px1x2�2 > 0
Px1x1 < 0
Px2x2� �0.86
Px1x2� �0.8
Px1x1� �0.9
x2 � 157.x1 � 94
0.8x1 � 0.86x2 � 210
Px2� �0.86x2 � 0.8x1 � 210 � 0
0.9x1 � 0.8x2 � 210
Px1� �0.9x1 � 0.8x2 � 210 � 0
� �0.45x12 � 0.43x2
2 � 0.8x1x2 � 210x1 � 210x2 � 11,500
� �225 � 0.4�x1 � x2���x1 � x2� � �0.05x12 � 15x1 � 5400� � �0.03x2
2 � 15x2 � 6100�
P�x1, x2� � R � C1 � C2
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128 Chapter 12 Functions of Several Variables
73. Maximize subject to the constraint
f �49.4, 253� � 13,201.8
y � 253
x � 49.4
10x � 494
5x � y � �6
20x � 4y � 2000 ⇒ 5x � y � 500
4 � y � 20
x � 2 � 4� 5x � y � �6
20x � 4y � 2000.f �x, y� � 4x � xy � 2y
77. Optimize subject to the constraint
Maximum: f �13 , 13 , 13� �
13
x � y � z � 1 ⇒ x � y � z �13
y � z �
x � z �
x � y � � x � y � z
x � y � z � 1.f �x, y, z� � xy � yz � xz
75. (a)
(b)
Yes, the data appears more linear.
(c)
(d)
The logarithmic model is a better fit.
−1 10
−5
25
y � 8.37 ln t � 1.54
−1 3
−5
20
−2 11
−5
30
y � 2.29t � 2.34
79.
Constraint:
Hence, x � 22
, y � 33
, z � 10 � 22
� 33
� 8.716 m.
4y2 � y2 � 1 ⇒ y2 �13
9x2 � x2 � 4 ⇒ x2 �12
1 �
2y � y2 � 1
3x � x2 � 4
3x x2 � 4
i �2y
y2 � 1j � k � �i � j � k�
�C � �g
x � y � z � 10
C � 3 x2 � 4 � 2 y2 � 1 � 2
PQ � x2 � 4, QR � y2 � 1, RS � z; x � y � z � 10
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Problem Solving for Chapter 12
Problem Solving for Chapter 12 129
1. (a) The three sides have lengths 5, 6, and 5.
Thus, and
(b) Let subject to the constraint (perimeter).
Using Lagrange multipliers,
From the first 2 equations
Similarly, and hence which is an equilateral triangle.
(c) Let subject to constant.
Using Lagrange multipliers,
Hence, and a � b � c.s � a � s � b ⇒ a � b
1 � ��s�s � a��s � b�
1 � ��s�s � a��s � c�
1 � ��s�s � b��s � c�
�Area�2 � s�s � a��s � b��s � c�f �a, b, c� � a � b � c,
a � b � cb � c
s � b � s � a ⇒ a � b.
�s�s � b��s � b� � �
�s�s � a��s � c� � �
�s�s � b��s � c� � �
a � b � c � constantf �a, b, c� � �area�2 � s�s � a��s � b��s � c�,
A � �8�3��2��3� � 12 s �162 � 8
3. (a)
Tangent plane:
y0z0x � x0z0 y � x0 y0z � 3x0 y0z0 � 3
y0z0�x � x0� � x0z0�y � y0� � x0 y0�z � z0� � 0
Fx � yz, Fy � xz, Fz � xy
F�x, y, z� � xyz � 1 � 0 (b)
�13�
12
3
y0z0
3x0z0
�� 3x0 y0
� �92
x
y
3
3
3
z
3x
0y
0
3x
0y
0
3y
0z
0
3x
0z
0
Tangent plane
Base
V �13
�base��height�
5. We cannot use Theorem 12.9 since is not a differentiable function of and . Hence, we use the definition ofdirectional derivatives.
which does not exist.
If then
which implies that the directional derivative exists.
Du f �0, 0� � limt →0
f�0 �t
�2, 0 �
t�2� � 2
t� lim
t →0
1t �
2t2
t2� 2� � 0
f �0, 0� � 2,
� limt→0
1t �4� t
�2�� t�2�
�t2
2� � �t2
2� � � limt →0
1t�
2t2
t2 � � limt →0
2t
Du f �0, 0� � limt →0
f �0 � � t�2�, 0 � � t
�2�� � f �0, 0�
t
Du f �x, y� � limt →0
f �x � t cos �, y � t sin �� � f �x, y�
t
yxf
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130 Chapter 12 Functions of Several Variables
7.
By symmetry,
Thus, and z �53
3�150.x � y � 2 3�150
x � y ⇒ x3 � y3 � 1200.
Hx � 5y �6000
x2 � 0 ⇒ 5yx2 � 6000
z �1000
xy ⇒ H � k�5xy �
6000y
�6000
x �.
H � k�5xy � 6xz � 6yz� 9. (a)
(b)
� Cxay1�a�t� � t f �x, y�
f �tx, ty� � C�tx�a�ty�1�a � Ctaxat1�ay1�a
� Cxay1�a � f
� Ca � C�1 � a�xay1�a
x �f�x
� y �f�y
� Caxay1�a � C�1 � a�xay1�a
�f�x
� Caxa�1y1�a, �f�y
� C�1 � a�xay�a
11. (a)
(b)
� � arctan� yx � 50� � arctan�32�2t � 16t2
32�2t � 50 �tan � �
yx � 50
y � 64�sin 45�t � 16t2 � 32�2t � 16t2
x � 64�cos 45�t � 32�2t
(c)
(d)
No. The rate of change of is greatest when the projectile is closest to the camera.
(e) when
No, the projectile is at its maximum height when or seconds.t � �2 � 1.41dy�dt � 32�2 � 32t � 0
t ��25 � �252 � 4�8�2���25�2�
2�8�2� � 0.98 second.
8�2t2 � 25t � 25�2 � 0
d�
dt� 0
�
0 4
−5
30
��16�8�2t2 � 25t � 25�2�
64t4 � 256�2t3 � 1024t2 � 800�2t � 625
d�
dt�
1
1 � �32�2t � 16t2
32�2t � 50 �2
�64�8�2t2 � 25t � 25�2�
�32�2t � 50�2
13. (a) There is a minimum at maxima at and saddle point at
Solving the two equations and you obtain the following critical points:Using the second derivative test, you obtain the results above.
—CONTINUED—
�0, 0�.�±1, 0�,�0, ±1�,2y3 � x2y � 2y � 0,x3 � 2xy2 � x � 0
� e��x2�y2��4y3 � 2x2y � 4y � 0 ⇒ 2y3 � x2y � 2y � 0
� e��x2�y2��x2 � 2y2���2y� � 4y
fy � �x2 � 2y2�e��x2�y2���2y� � �4y�e��x2�y2�
� e��x2�y2��2x3 � 4xy2 � 2x � 0 ⇒ x3 � 2xy2 � x � 0
� e��x2�y2��x2 � 2y2���2x� � 2x
fx � �x2 � 2y2�e��x2�y2���2x� � �2x�e��x2�y2�
x
y
1
22
z�±1, 0, 1�e�:�0, ±1, 2�e��0, 0, 0�,
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13. —CONTINUED—
(b) As in part (a), you obtain
The critical numbers are These yield
minima
maxima
saddle
x
y12
−1
1
z
�0, 0, 0�
�0, ±1, 2�e��±1, 0, �1�e�
�±1, 0�.�0, ±1�,�0, 0�,
fy � e��x2�y2�2y�2 � x2 � 2y2�fx � e��x2�y2�2x�x2 � 1 � 2y2�
(c) In general, for you obtain
minimum
maxima
saddle
For you obtain
minima
maxima
saddle�0, 0, 0�
�0, ±1, ��e�
�±1, 0, ��e�
� < 0,
�±1, 0, ��e�
�0, ±1, ��e�
�0, 0, 0�
� > 0
Problem Solving for Chapter 12 131
15. (a)
1 cm
6 cm (b)
1 cm
6 cm
(c) The height has more effect since the shaded region in(b) is larger than the shaded region in (a).
(d)
If and then
If and then dA � 6�0.01� � 0.06.dl � 0,dh � 0.01
dA � 1�0.01� � 0.01.dh � 0,dl � 0.01
A � hl ⇒ dA � l dh � h dl
17. Essay
19.
Let and Then
Thus,�2u�t2
� c2�2u�x2.
�2u�x2 �
12
d 2fdr2 �1�2 �
12
d 2fds2 �1�2 �
12�
d 2fdr2 �
d 2fds2�
�u�x
��u�r
�r�x
��u�s
�s�x
�12
dfdr
�1� �12
dfds
�1�
�2u�t2
�12
d 2fdr2 ��c�2 �
12
d 2fds2 �c�2 �
c2
2 �d2fdr2 �
d 2fds2�
�u�t
��u�r
�r�t
��u�s
�s�t
�12
dfdr
��c� �12
dfds
�c�
u�r, s� �12
f �r� � f �s�.s � x � ct.r � x � ct
u�x, t� �12
f �x � ct� � f �x � ct�
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C H A P T E R 1 3Multiple Integration
Section 13.1 Iterated Integrals and Area in the Plane . . . . . . . . . . . . . 365
Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 369
Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 375
Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 379
Section 13.5 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 385
Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 388
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 393
Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 397
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 400
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 405
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365
C H A P T E R 1 3Multiple Integration
Section 13.1 Iterated Integrals and Area in the PlaneSolutions to Even-Numbered Exercises
2. �x2
x
yx dy � �1
2 y2
x �x2
x�
12 �
x 4
x�
x2
x � �x2
�x 2 � 1� 4. �cos y
0 y dx � �yx�
cos y
0� y cos y
6. �x
x3
�x2 � 3y2� dy � �x2y � y3�x
x3
� �x2x � �x�3� � �x2x3 � �x3�3� � x52 � x32 � x5 � x9
8. �1�y2
�1�y2
�x2 � y 2� dx � �13
x3 � y2x�1�y2
�1�y2� 2�1
3�1 � y2�32 � y2�1 � y2�12� �
21 � y 2
3 �1 � 2y 2�
10.
� ���cos x �13
cos3 x� cos y��2
y� �cos y �
13
cos3 y� cos y
��2
y
sin3 x cos y dx � ��2
y
�1 � cos2 x� sin x cos y dx
12.
� �1
�1�4x2 �
163 � dx � �4x3
3�
163
x�1
�1� �4
3�
163 � � ��
43
�163 � � �8
�1
�1�2
�2�x2 � y2� dy dx � �1
�1�x2y �
y3
3 �2
�2dx � �1
�1�2x2 �
83
� 2x2 �83� dx
14.
� �4
�4
64 � x3 x2 dx � ��29
�64 � x3�32�4
�4� 0 �
29
�128�32 �2048
92
�4
�4�x2
0
64 � x3 dy dx � �4
�4�y64 � x3�
x2
0dx
16.
� �2
0�10y �
143
y3 � 2y3� dy � �5y2 �7y4
6�
y4
2 �2
0� 20 �
563
� 8 �1403
�2
0�2y
y
�10 � 2x2 � 2y2� dx dy � �2
0�10x �
2x3
3� 2y2x�
2y
ydy � �2
0��20y �
163
y3 � 4y3� � �10y �23
y3 � 2y3�� dy
18. �2
0�2y�y2
3y2�6y
3y dx dy � �2
0 �3xy�
2y�y2
3y2�6y dy � 3�2
0�8y2 � 4y3� dy � �3�8
3y3 � y4��
2
0� 16
20. ��2
0�2 cos�
0 r dr d� � ��2
0 �r 2
2 �2 cos �
0 d� � ��2
0 2 cos2 � d� � �� �
12
sin 2���2
0�
�
2
22.
� ��4
0 cos3 sin � d� � ��
cos4 �4 �
�4
0� �
14��
12�
4� 1� �
316
��4
0�cos �
0 3r 2 sin � dr d� � ��4
0 �r3 sin ��
cos �
0 d�
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24. �3
0��
0
x2
1 � y2 dy dx � �3
0 �x2 arctan y�
�
0 dx � �3
0 x2��
2� dx � ��
2�
x3
3 �3
0�
9�
2
26. ��
0��
0 xye��x2�y2� dx dy � ��
0 ��
12
ye��x2�y2���
0 dy � ��
0 12
ye�y2 dy � ��14
e�y2��
0�
14
28.
A � �3
1�3
1 dx dy � �3
1 �x�
3
1 dy � �3
1 2 dy � �2y�
3
1� 4
2
2
31
3
1
x
y
A � �3
1�3
1 dy dx � �3
1 �y�
3
1 dx � �3
1 2 dx � �2x�
3
1� 4
30.
� �3y�12
0� ��
1y
� y�1
12� 2
� �12
0 3 dy � �1
12 � 1
y 2 � 1� dy
� �12
0 �x�
5
2 dy � �1
12 �x�
1��1y2�
2 dy
x2 5
3
3
5
4
2
41
1
1
x − 1y =
y
A � �12
0�5
2 dx dy � �1
12�1��1y2�
2 dx dy
A � �5
2�1x�1
0 dy dx � �5
2 �y�
1x�1
0 dx � �5
2
1x � 1
dx � �2x � 1�5
2� 2
32.
�y � 2 sin �, dy � 2 cos � d�, 4 � y 2 � 2 cos ��
� �2�� �12
sin 2����2
0� �
� 4��2
0 cos2 � d� � 2��2
0 �1 � cos 2�� d�
A � �2
0�4�y 2
0 dx dy � �2
0 4 � y2 dy
�x � 2 sin �, dx � 2 cos � d�, 4 � x2 � 2 cos ��
� �2�� �12
sin 2����2
0� �
� 2��2
0 �1 � cos 2�� d�
� 4��2
0 cos2 � d�
� �2
0 4 � x2 dx
x
2
1 2
1
y x= 4 − 2
y
A � �2
0�4�x2
0 dy dx 34.
�35
�32� � 16 �165
� �35
y53 �y2
4 �8
0
� �8
0 �y23 �
y2� dy
A � �8
0�y23
y2 dx dy
� 16 �25
�32� �165
� � x2 �25
x52�4
0
� �4
0�2x � x32� dx
� �4
0 �y�
2x
x32
dx
−1 1
1
2
3
4
5
6
7
8
2 3 4 5 6 7 8
y
x
(4, 8)
y = x3/2
y = 2x A � �4
0�2x
x32
dy dx
366 Chapter 13 Multiple Integration
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36.
x
4
6
2 4 6 8
2
−2
(3, 3)
(9, 1)
y x=
y = 9x
y
� �9y �12
y 2�1
0� �9 ln y �
12
y 2�3
1�
92
�1 � ln 9�
� �1
0 �9 � y� dy � �3
1 �9
y� y� dy
� �1
0 �x�
9
y dy � �3
1 �x�
9y
y dy
A � �1
0�9
y
dx dy � �3
1�9y
y
dx dy
�92
�1 � ln 9�
� �12
x2�3
0� �9 ln x�
9
3�
92
� 9�ln 9 � ln 3�
� �3
0 �y�
x
0 dx � �9
3 �y�
9x
0 dx � �3
0 x dx � �9
3 9x dx
A � �3
0�x
0 dy dx � �9
3�9x
0 dy dx 38.
x
2
3
4
1 2 3 4
1
y x= 2
y x=
y
A � �2
0�2x
x
dy dx � �2
0 �2x � x� dx � �x2
2 �2
0� 2
� 1 � �4 � 3� � 2
� �y2
4 �2
0� �2y �
y 2
4 �4
2
� �2
0 y2
dy � �4
2 �2 �
y2� dy
A � �2
0�y
y2 dx dy � �4
2�2
y2 dx dy
40.
x
2
3
4
1 2 3 4
1
y x= 2
y
� �2
0�x2
0 f �x, y� dy dx
�4
0�2
y f �x, y� dx dy, y ≤ x ≤ 2, 0 ≤ y ≤ 4 42.
−1
−1
1
2
3
4
1 2 3 4
y
x
� �4
0�4�y
0f �x, y� dx dy
�2
0�4�x2
0f �x, y� dy dx, 0 ≤ y ≤ 4 � x2, 0 ≤ x ≤ 2
44.
−1 1 2
2
3
y
x
� �e�2
0�2
�1f �x, y� dx dy � �e
e�2��ln y
�1f �x, y� dx dy
�2
�1�e�x
0f �x, y� dy dx, 0 ≤ y ≤ e�x, �1 ≤ x ≤ 2 46.
2
x
21
23
4π
4π−
y
� �1
0 �arccos y
�arccos y f �x, y� dx dy
��2
��2�cos x
0 f �x, y� dy dx, 0 ≤ y ≤ cos x, �
�
2 ≤ x ≤
�
2
Section 13.1 Iterated Integrals and Area in the Plane 367
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48.
x
2
3
4
1 2 3 4
1
y
�2
1�4
2 dx dy � �4
2�2
1 dy dx � 2 50.
−1
−1
1
1
y
x
�2
�2�4�y2
�4�y2
dx dy � 4�
�2
�2�4�x2
�4�x2
dy dx � �2
�2�4 � x2 � 4 � x2� dx � 4�
52.
�2
0�6�y
2y
dx dy � �2
0�6 � 3y� dy � �6y �
3y2
2 �2
0� 6
y = x2 y = 6 − x
y
x−1
−11
1
2
3
4
5
6
2 3 4 5 6
(4, 2)
�4
0�x2
0dy dx � �6
4�6�x
0dy dx � �4
0
x2
dx � �6
4�6 � x� dx � 4 � 2 � 6
54.
�3
0�y2
0dx dy � �3
0y2 dy � �y3
3 �3
0� 9
−1
−2
12345
−3−4−5
1 2 3 4 5 6 7 8 9
y
x
y = x
�9
0�3
x dy dx � �9
0�3 � x� dx � �3x �
23
x32�9
0� 27 � 18 � 9
56. �2
�2�4�y2
0 dx dy � �4
0�4�x
�4�x
dy dx �323
x
2
1 2 3
1
−1
−2x y= 4 − 2
y
58. The first integral arises using vertical representative rectangles. The second integral arises using horizontal representative rectangles.
� �14
cos�4� �12
sin�4� �14
�4
0�y
y2 x sin y dx dy � �4
0 �1
2 y sin�y� �
18
y2 sin�y�� dy
� �14
cos�4� �12
sin�4� �14
x1
1
2
2
3
4(2, 4)
y
�2
0�2x
x2
x sin y dy dx � �2
0 ��x cos�2x� � x cos�x2�� dx
368 Chapter 13 Multiple Integration
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Section 13.2 Double Integrals and Volume
60.
� �2
0 �xe�y2�
y
0 dy � �2
0 ye�y2 dy � ��
12
e�y2�2
0� �
12
�e�4� �12
e0 �12 �1 �
1e4� 0.4908
�2
0�2
x
e�y 2 dy dx � �2
0�y
0 e�y 2 dx dy
62.
� �4
0 �yx sin x�x
0 dx � �4
0 x sin x dx � �sin x � x cos x�
4
0� sin 4 � 4 cos 4 1.858
�2
0�4
y 2
x sin x dx dy � �4
0�x
0 x sin x dy dx
64. �1
0�2y
y
sin �x � y� dx dy �sin 2
2�
sin 33
0.408 66. �a
0�a�x
0 �x2 � y 2� dy dx �
a4
6
68. (a)
(b)
(c) Both orders of integration yield 1.11899.
�2
0�2
4�y 2
xy
x2 � y 2 � 1 dx dy � �3
2�2
0
xyx2 � y 2 � 1
dx dy � �4
3�16�4y
0
xyx2 � y 2 � 1
dx dy
y � 4 �x2
4 ⇔ x � 16 � 4y
x1
1
2
2
3
4
yy � 4 � x2 ⇔ x � 4 � y 2
70. �2
0�2
x
16 � x3 � y3 dy dx 6.8520 72. ���2
0�1�sin �
015�r dr d� �
45�2
32�
1358
30.7541
78. False, let f �x, y� � x.
74. A region is vertically simple if it is bounded on the left and right by vertical lines, and bounded on the top and bottom byfunctions of x. A region is horizontally simple if it is bounded on the top and bottom by horizontal lines, and bounded on theleft and right by functions of y.
76. The integrations might be easier. See Exercise 59-62.
For Exercises 2 and 4, and the midpoints of the squares are
�12
, 12 , �3
2,
12 , �5
2,
12 , �7
2,
12 , �1
2,
32 , �3
2,
32 , �5
2,
32 , �7
2,
32 .
x
2
3
4
1 2 3 4
1
y�xi � �yi � 1
2.
�4
0�2
0 12
x2y dy dx � �4
0 �x2y2
4 �2
0 dx � �4
0 x2 dx �
x3
3 �4
0�
643
21.3
��
i�1 f �xi, yi��xi �yi �
116
�9
16�
2516
�4916
�3
16�
2716
�7516
�14716
� 21
f �x, y� �12
x2y
Section 13.2 Double Integrals and Volume 369
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4.
� �4
0
ln 3x � 1
dx � �ln 3 ln�x � 1��4
0� �ln 3��ln 5� 1.768
�4
0�2
0
1�x � 1��y � 1� dy dx � �4
0 � 1
x � 1ln�y � 1��
2
0 dx
�8
i�1 f �xi, yi� �xi �yi �
49
�4
15�
421
�4
27�
415
�4
25�
435
�4
45�
79364725
1.680
f �x, y� �1
�x � 1��y � 1�
6. �2
0�2
0f �x, y� dy dx 4 � 2 � 8 � 6 � 20
8.
�� 2
8
� ��
8 �x �
12
sin 2x���
0
��
8��
0 �1 � cos 2x� dx
� ��
0 12
sin2 x��
2� dx
2
2
31
3
1
x
y
��
0���2
0 sin2 x cos2 y dy dx � ��
0 �1
2 sin2 x�y �
12
sin 2y����2
0 dx
10.
�102427
�2569
�25627
� �2y9�2
27�
y6
144�4
0
� �4
0�y7�2
3�
y5
24� dy
1
1
2
3
4
2 3 4
y
x
(2, 4) �4
0�y
�1�2�yx2y2 dx dy � �4
0�x3y2
3 �y
�1�2�y dy
12.
�12
�e � e�1�
� �ey �12
e2y�1�1
0
� �1
0 �e � e2y�1� dy
x−1 1
2
y x= + 1−y x= + 1
y
�1
0�0
y�1 ex�y dx dy � �1
0�1�y
0 ex�y dx dy � �1
0 �ex�y�
0
y�1 dy � �1
0 �ex�y�
1�y
0 dy
370 Chapter 13 Multiple Integration
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14.
x
y
22π
23π
23π
25π
π
π π
2π
2π−
23π− −
� 0
� ��
��
sin x dx
� ��
��
��sin x cos y���2
0 dx
���2
0��
��
sin x sin y dx dy � ��
�����2
0 sin x sin y dy dx 16.
For the first integral, we obtain:
1
1
2
3
4
2 3 4
y
x
� ��5 � 8� � e4 � e4 � 13.
� ��e4�x�1 � x� �x2
2 �4
0
�4
0�xey�
4�x
�x 0
dx � �4�x
0�xe4 � x�dx
�4
0�4�x
0xey dy dx � �4
0�4�y
0xey dx dy
18.
� �14
ln�1 � x2��4
0�
14
ln�17�
�12
�4
0
x1 � x2 dx
�12�
4
0 � y 2
1 � x2�x
0 dx
x
2
3
4
1 2 3 4
1
y x=
y
�2
0�4
y 2
y
1 � x2 dx dy � �4
0�x
0
y1 � x2 dy dx
22.
x
2
3
4
1 2 3 4
1
y
� �4
0 8 dx � 32
�4
0�2
0 �6 � 2y� dy dx � �4
0 �6y � y 2�
2
0 dx 24.
1
1
2
2
y = x
y
x
�2
0�x
04 dy dx � �2
04x dx � 2x2�
2
0� 8
20.
� ��x4
�4 � x2�3�2 �12 �x4 � x2 � 4 arcsin
x2� �
112�x�4 � x2�3�2 � 6x4 � x2 � 24 arctan
x2��
2
�2� 4�
� �2
�2�x24 � x2 �
13
�4 � x2�3�2� dx
� �2
�2�x2y �
13
y3�4�x2
0 dx
� �2
�2�4�x2
0 �x2 � y 2� dy dx
x−2 −1 1
1
2
3
4
x y= 4 − 2x y= 4− − 2
y�2
0�4�y 2
4�y 2
�x2 � y 2� dx dy
Section 13.2 Double Integrals and Volume 371
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26.
1
1
2
2
y = 2 − x
y
x
� �16
�x � 2�3�2
0�
43
� �2
0
12
�2 � x�2 dx
�2
0�2�x
0�2 � x � y� dy dx � �2
0�2y � xy �
y2
2 �2�x
0 dx 28.
x
2
1 2
1 y x=
y
� 4
� �2y2 �y 4
4 �2
0
�2
0�y
0 �4 � y 2� dx dy � �2
0 �4y � y3� dy
30. � ��
0 2e�x�2 dx � ��4e�x�2��
0� 4��
0��
0 e��x�y��2 dy dx � ��
0 ��2e��x�y��2��
0 dx
32. �1
0�x
0 1 � x2 dy dx �
13
34.
� �13
x3�5
0�
1253
� �5
0 �xy�
x
0 dx � �5
0 x2 dx
V � �5
0�x
0 x dy dx
x2 5
3
3
5
4
2
41
1
y x=
y
36.
�4�r3
3
� �2��r2x �13
x3��r
0
� 4��
2��r
0 �r2 � x2� dx
� 4�r
0 ��yr2 � x2 � y 2 � �r2 � x2� arcsin
yr2 � x2��r2�x2
0 dx
x
r
r
y r x= 2 2−
y V � 8�r
0�r2�x2
0 r2 � x2 � y 2 dy dx
38.
� 32 �643
�325
�25615
� �16x � 8x3
3�
x5
5 �2
0
� �2
0�16 � 8x2 � x4� dx
� �2
0�4 � x2��4 � x2� dx
1 2
1
2
3
4
3 4
y
x
y = 4 − x2
V � �2
0�4�x2
0�4 � x2� dy dx 40.
� ��x2 �
2
0� �
� �2
0 �
2 dx
� �2
0 �arctan y��
0 dx
x
2
1 2
1
y V � �2
0��
0
11 � y 2 dy dx
372 Chapter 13 Multiple Integration
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42.
�5�
2
� ��
2y�
5
0
� �5
0 �
2 dy
x2 5
3
3
5
4
2
41
1
y
V � �5
0��
0 sin2 x dx dy 44. V � �9
0�9�y
0 9 � y dx dy �
812
46. V � �16
0�4�y
0ln�1 � x � y� dx dy 38.25
48.
� c���ab2
�ab3 � � ��
ab2
�ab6 �� �
abc6
� c��ab2 �1 �
xa�
2�
x2b2a
�x3b3a2 �
ab6 �1 �
xa�
3�a
0
� c�a
0 �b�1 �
xa� �
xba �1 �
xa� �
b2
2b �1 �xa�
2� dx
� c�a
0 �y �
xya
�y 2
2b�b�1��x�a��
0 dx
V � �R� f �x, y� dA � �a
0�b�1��x�a��
0 c�1 �
xa
�yb� dy dx
z � c�1 �xa
�yb�
xya
a
a
R
z
xa
�yb
�zc
� 1
50.
x2 5
6
3
10
8
4
41
2
y e= x
y
� �10
1 dy � �y�
10
1� 9
� �10
1 � x
ln y�ln y
0 dy
�ln 10
0�10
ex
1ln y
dy dx � �10
1�ln y
0
1ln y
dx dy 52.
� 2�cos 2 � 2 sin 2 � 1�
� 2�cos y � y sin y�2
0
� 2�2
0y cos y dy
(2, 2)
y = x212
1
1
2
2
y
x
� �2
0
y cos y2y dy
�2
0�2
�1�2�x2
y cos y dy dx � �2
0�2y
0
y cos y dx dy
54. Average �18
�4
0�2
0 xy dy dx �
18
�4
0 2x dx � �x2
8 �4
0� 2 56.
� �e � 1�2
� e2 � 2e � 1
� 2� ex�1 �12
e2x�1
0� 2�e2 �
12
e2 � e �12�
Average �1
1�2�1
0�1
x
e x�y dy dx � 2�1
0 e x�1 � e2x dx
Section 13.2 Double Integrals and Volume 373
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58. The second is integrable. The first contains
which does not have an elementary antiderivation.
�sin y2 dy 60. (a) The total snowfall in the county R.
(b) The average snowfall in R.
62. Average �1
150 �60
45�50
40 �192x � 576y � x2 � 5y 2 � 2xy � 5000� dx dy 13,246.67
64. for all and
P�0 ≤ x ≤ 1, 1 ≤ y ≤ 2� � �1
0�2
1
14
xy dy dx � �1
0 3x8
dx �3
16.
��
����
��
f �x, y� dA � �2
0�2
0 14
xy dy dx � �2
0 x2
dx � 1
�x, y�f �x, y� ≥ 0
66. for all and
� ��12
e�2x � e�x�1�1
0�
12
e�2 � e�1 �12
�12
�e�1 � 1�2 0.1998.
P�0 ≤ x ≤ 1, x ≤ y ≤ 1� � �1
0�1
x
e�x�y dy dx � �1
0 �e�x�y�
1
x dx � �1
0 �e�2x � e�x�1� dx
� ��
0 limb→�
��e�x�y�b
0 dx � ��
0 e�x dx � lim
b→� ��e�x�
b
0� 1
��
��
��
��
f �x, y� dA � ��
0 ��
0 e�x�y dy dx
�x, y�f �x, y� ≥ 0
68. Sample Program for TI-82:
Program: DOUBLE
: Input A
: Input B
: Input M
: Input C
: Input D
: Input N
:
:
:
: For
: For
:
:
:
: End
: End
: Disp V
V � sin �x � y � G H → V
C � 0.5H�2J � 1� → y
A � 0.5G�2I � 1� → x
�J, 1, N, 1��I, 1, M, 1�
�D � C��N → H
�B � A��M → G
0 → V
70.
(a) 129.2018
(b) 129.2756
m � 10, n � 20�2
0�4
0 20e�x3�8 dy dx
374 Chapter 13 Multiple Integration
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72.
(a) 13.956
(b) 13.9022
m � 6, n � 4�4
1�2
1 �x3 � y3 dx dy 74.
Matches a.
xy
3
4
3 3
z
V � 50
76. True
78.
Thus,
� �2
1 1y dy � �ln y�
2
1� ln 2.
� �2
1 ��
e�xy
y ��
0 dy
� �2
1��
0 e�xy dx dy
��
0 e�x � e�2x
x dx � ��
0�2
1 e�xy dx dy
�2
1 e�xy dy � ��
1x
e�xy�2
1�
e�x � e�2x
x
Section 13.3 Change of Variables: Polar Coordinates
2. Polar coordinates 4. Rectangular coordinates
6. R � ��r, �: 0 ≤ r ≤ 4 sin �, 0 ≤ � ≤ � 8. R � ��r, �: 0 ≤ r ≤ r cos 3�, 0 ≤ � ≤ �
10.
02 3 41
π2
�163
� ��643 �
sin2 �2 �
� 4
0
�� 4
0�4
0 r 2 sin � cos dr d� � �� 4
0 �r3
3 sin � cos ��
4
0 d� 12.
02 31
π2
��
4 �1 �1e9�
� ��12
�e�9 � 1��� 2
0
�� 2
0�3
0 re�r2 dr d� � �� 2
0 ��
12
e�r2�3
0 d�
Section 13.3 Change of Variables: Polar Coordinates 375
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14.
� �16
�1 � cos ��3�� 2
0�
16
� �� 2
0 sin �
2�1 � cos �2 d�
01
( , ) = (0, 1)x y
π2
�� 2
0�1�cos �
0 �sin �r dr d� � �� 2
0 ��sin �r
2
2 �1�cos �
0 d�
16. �a
0��a2�x2
0 x dy dx � �� 2
0�a
0 r2 cos � dr d� �
a3
3 �� 2
0 cos � d� � �a3
3 sin ��
� 2
0�
a3
3
18.
� �� 4
0 �2�2 3
3 d� � ��2�2 3
3 ��
� 4
0
��2�2 3
3�
�
4�
4�2�
3
�2
0��8�y2
y
�x2 � y 2 dx dy � �� 4
0�2�2
0 r2 dr d�
20.
� 64�� 2
0 �sin4 � � sin6 � d� �
646
�sin5 � cos � �sin3 � cos �
4�
38
�� � sin � cos ��� 2
0� 2�
�4
0��4y�y2
0 x2 dx dy � �� 2
0�4 sin �
0 r3 cos2 � dr d� � �� 2
0 64 sin4 � cos2 d�
22.
�62516
� �6258
sin2 ��� 4
0
� �� 4
0 6254
sin � cos d�
��5�2 2
0�x
0 xy dy dx � �5
�5�2 2 ��25�x2
0 xy dy dx � �� 4
0�5
0 r3 sin � cos � dr d�
24.
� ��1 � e�25 2��� 2
�� 2� � �1 � e�25 2
� �� 2
�� 2�1 � e�25 2 d�
−5 −4 −3 −2
−2
12345
−3−4−5
−1 1 2 3 4
y
x
�� 2
�� 2 �5
0e�r2 2r dr d� � �� 2
�� 2��e�r2 2�
5
0 d�
26.
� �� 2
0�3
0 �9r � r3 dr d� � �� 2
0 �9
2r2 �
14
r 4�3
0 d� �
814 �
� 2
0 d� �
81�
8
�3
0��9�x2
0
�9 � x2 � y2 dy dx � �� 2
0�3
0 �9 � r2r dr d�
376 Chapter 13 Multiple Integration
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28.
� 7��
4� �7�
4
� 4�� 4
0 74
d�
V � 4�� 2
0�1
0�r2 � 3r dr d� � 4�� 2
0�r4
4�
3r2
2 �1
0 d� 30.
� 4��ln 4 �34�
� 2�2�
0 �ln 4 �
34�d�
� 2�2�
0 �r2
4��1 � 2 ln r�
2
1 d�
� 2�2�
0�2
1 r ln r dr d�
�R� ln�x2 � y 2 dA � �2�
0�2
1 �ln r2r dr d�
32. � �2�
0 5�15 d� � 10�15�V � �2�
0�4
1 �16 � r2 r dr d� � �2�
0 ��
13
��16 � r23�4
1 d�
34.
(8 times the volume in the first octant)
� 8�� 2
0 a3
3 d� � �8a3
3��
� 2
0�
4�a3
3
� 8�� 2
0 ��
12
�23
�a2 � r23 2�a
0 d�
V � 8�� 2
0�a
0 �a2 � r2 r dr d�
x2 � y 2 � z2 � a2 ⇒ z � �a2 � �x2 � y 2 � �a2 � r2
36.
(a)
(b)
(c) V � 2�2�
0�1 2�1�cos2 �
1 4
94r2 � 36
r dr d� � 0.8000
Perimeter � 2 ��
0 �1
4�1 � cos2 �2 � cos2 � sin2 � d� � 5.21
drd�
� �cos � sin �
xy
1
11
z r �12
�1 � cos2 � �12
�12
cos2 �
Perimeter � �
�r2 � �drd��
2 d�.
�94r2 � 36
≤ z ≤ 9
4r2 � 36
−0.7
−1 1
0.714
≤ r ≤ 12
�1 � cos2��94�x2 � y2 � 9 ≤ z ≤
94�x2 � y2 � 9;
38. A � �2�
0�4
2 r dr d� � �2�
0 6 d� � 12�
40.
�12�4� � 4 cos � �
12
� �14
sin 2��2�
0�
12
�8� � 4 � � � 4� �9�
2
�2�
0�2�sin �
0r dr d� �
12�
2�
0�2 � sin �2 d� �
12�
2�
0�4 � 4 sin � � sin2 � d� �
12�2�
0�4 � 4 sin � �
1 � cos 2�
2 � d�
Section 13.3 Change of Variables: Polar Coordinates 377
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42. 8�� 4
0�3 cos 2�
0 r dr d� � 4 �� 4
0 9 cos2 2� d� � 18 �� 4
0 �1 � cos 4� d� � 18�� �
14
sin 4��� 4
0�
9�
2
44. See Theorem 13.3. 46. (a) Horizontal or polar representative elements
(b) Polar representative element
(c) Vertical or polar
48. (a) The volume of the subregion determined by the point is base height Adding up the 20volumes, ending with you obtain
(b)
(c) �7.48�24103.5 � 179,621 gallons
�56�24013.5 � 1,344,759 pounds
�5�
4�150 � 555 � 1250 � 2135 � 2025� �
5�
4�6115� � 24013.5 ft3
� 35�12 � 15 � 18 � 16 � 45�9 � 10 � 14 � 12�
V � 10 ��
8�5�7 � 9 � 9 � 5 � 15�8 � 10 � 11 � 8 � 25�10 � 14 � 15 � 11
�45 � 10 � � 8�12,� �5 � 10 � � 8�7.��5, � 16, 7
50. �� 4
0�4
0 5e�r� r dr d� � 87.130 52.
Answer (a)
� 94 � � 3 � 21
x
y4
4
2
2
4
6
zVolume � base � height
54. True
56. (a) Let then
(b) Let then ��
��
e�4x2 dx � ��
��
e�u2 12
du �12
��.u � 2x,
��
��
e�x2 dx � ��
��
e�u2 2 1�2
du �1�2
��2� � ��.u � �2x,
58.
For to be a probability density function,
k �4�
.
k�
4� 1
f �x, y
��
0��
0 ke��x2�y2 dy dx � �� 2
0��
0 ke�r2 r dr d� � �� 2
0 ��
k2
e�r2��
0 d� � �� 2
0 k2
d� �k�
4
60. (a)
(b)
(c) 2�� 2
0�4 cos �
0 f r dr d�
4�2
0��4��x�22
0 f dy dx x
2
1
−2
−1
3
1
( 2) + 2 = 4x y− 2y
4�2
0�2��4�y2
2 f dx dy
378 Chapter 13 Multiple Integration
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Section 13.4 Center of Mass and Moments of Inertia
2.
� �112
�0 � 93� �2434
� ��14
�9 � x2�3
3 �3
0
� �3
0
x�9 � x2�2
2 dx
m � �3
0�9�x2
0
xy dy dx � �3
0�xy2
2 �9�x2
0 dx
4.
�12�
812
�814
� 54� �2978
�12��2�9 � x2�3�2 �
9x2
2�
x4
4 �3
0
�12�
3
0�6x9 � x2 � 9x � x3 dx
� �3
0
x2
��3 � 9 � x2� � 9� dx
m � �3
0�3�9�x2
3xy dy dx � �3
0�x
y2
2 �3�9�x2
3 dx
6. (a)
y �Mx
m�
ka2b3�6ka2b2�4
�23
b
x �My
m�
ka3b2�6ka2b2�4
�23
a
My � �a
0�b
0 kx2y dy dx �
ka3b2
6
Mx � �a
0�b
0 kxy2 dy dx �
ka2b3
6
m � �a
0�b
0 kxy dy dx �
ka2b2
4(b)
y �Mx
m�
�kab2�12��2a2 � 3b2��kab�3��a2 � b2� �
b4 �
2a2 � 3b2
a2 � b2 �
x �My
m�
�ka2b�12��3a2 � 2b2��kab�3��a2 � b2� �
a4 �
3a2 � 2b2
a2 � b2 �
My � �a
0�b
0 k�x3 � xy 2� dy dx �
ka2b12
�3a2 � 2b2�
Mx � �a
0�b
0 k�x2y � y3� dy dx �
kab2
12�2a2 � 3b2�
m � �a
0�b
0 k�x2 � y 2� dy dx �
kab3
�a2 � b2�
8. (a)
—CONTINUED—
x � y �Mx
m�
ka3�6ka2�2
�a3
My � Mx by symmetry
Mx � �a
0�a�x
0 ky dy dx �
ka3
6
xa
ay a x= −
y m �a2k2
Section 13.4 Center of Mass and Moments of Inertia 379
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10. The x-coordinate changes by h units horizontally and k units vertically. This is not true for variable densities.
12. (a)
y �4a3�
by symmetry
x �My
m�
ka3�3k�a2�4
�4a3�
� ��k3
�a2 � x2�3�2�a
0�
ka3
3
� k�a
0 xa2 � x2 dx
My � �a
0�a2�x2
0
kx dy dx
m � �a
0 �a2�x2
0 k dy dx �
k�a2
4(b)
x � y �My
m�
ka5
5 �
8ka4�
�8a5�
My � Mx by symmetry
� ���2
0�a
0 kr4 sin � dr d� �
ka5
5
Mx � �a
0�a2�x2
0 k�x2 � y 2� y dy dx
� ���2
0�a
0 kr3 dr d� �
ka4�
8
m � �a
0�a2�x2
0 k�x2 � y 2� dy dx
8. —CONTINUED—
(b)
y �2a5
by symmetry
x �My
m�
a5�15a4�6
�2a5
� �a
0 �ax3 � x 4 �
13
a3x � a2x2 � ax3 �13
x 4� dx �13
�a
0 �a3x � 3a2x2 � 6ax3 � 4x4� dx �
a5
15
My � �a
0�a�x
0 �x3 � xy 2� dy dx
� �a
0 �x2y �
y3
3 �a�x
0 dx � �a
0 �ax2 � x3 �
13
�a � x�3� dx �a4
6
m � �a
0�a�x
0 �x2 � y 2� dy dx
14.
y �Mx
m�
16k32k
�5� �52
x �My
m�
32k
3�
5
32k�
5
3
My � �2
0�x3
0kx2 dy dx �
32k3
Mx � �2
0�x3
0kxy dy dx � 16k
m � �2
0�x3
0kx dy dx � �2
0kx4 dx �
32k5
16.
x
2
3
4
1 2 3 4
1
y = 4x
y
y �My
m�
24k30k
�45
x �My
m�
84k30k
�145
My � �4
1�4�x
0 kx3 dy dx � 84k
Mx � �4
1�4�x
0 kx2 y dy dx � 24k
m � �4
1�4�x
0 kx2 dy dx � 30k
380 Chapter 13 Multiple Integration
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18.
x−6 −3 3 6
6
3
12
y x= 9 − 2
y
y �Mx
m�
139,968k35
�35
23,328k� 6
Mx � �3
�3�9�x2
0 ky3 dy dx �
139,968k35
m � �3
�3�9�x2
0 ky 2 dy dx �
23,328k35
x � 0 by symmetry 20.
x
1
L2
y = cos πxL
t
y �Mx
m�
kL8
��
kL�
�
8
x �My
m�
L 2�� � 2�k2� 2 �
�
kL�
L�� � 2�2�
My � �L�2
0 �cos �x�L
0 kx dy dx �
L 2�� � 2�k2� 2
Mx � �L�2
0�cos �x�L
0 ky dy dx �
kL8
m � �L�2
0�cos �x�L
0 k dy dx �
kL�
22.
y �Mx
m�
ka4�2 � 2 �8
�12
ka3��
3�2 � 2�a2�
x �My
m�
ka428
�12
ka3��
32a2�
My � �R� kx2 � y 2 dA � ���4
0�a
0 kr3 cos � d� �
ka428
Mx � �R� kx2 � y 2 y dA � ���4
0�a
0 kr3 sin � d� �
ka4�2 � 2�8
0a
r a=y x=
π2
m � �R� kx2 � y 2 dA � ���4
0�a
0 kr2 dr d� �
ka3�
12
24.
y �Mx
m�
k6
�2k
�13
x �My
m�
k1
�2k
� 2
My � �e
1�ln x
0 kx
x dy dx � k
Mx � �e
1�ln x
0 kx
y dy dx �k6
m � �e
1�ln x
0 kx dy dx �
k2
2
2
3e1
3
1
x
y x= ln
y
Section 13.4 Center of Mass and Moments of Inertia 381
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26.
x �My
m�
5k�
4�
23k�
�56
�5k�
4
�k3�
2�
0 �cos � �
32
�1 � cos2 �� � 3 cos ��1 � sin2 �� �14
�1 � cos 2��2 d�
�k3�
2�
0 cos ��1 � 3 cos � � 3 cos2 � � cos3 �� d�
My � �R� kx dA � �2�
0�1�cos �
0 kr2 cos � dr d�
m � �R� k dA � �2�
0�1�cos �
0 kr dr d� �
3�k2
01
r = 1 + cosθπ2
y � 0 by symmetry
28.
y �Ix
m�bh3�12
bh�2�
h6
�66
h
x �Iy
m�b3h�12
bh�2�
b6
�66
b
Iy � �b
0�h��hx�b�
0 x2 dy dx �
b3h12
Ix � �b
0�h��hx�b�
0 y2 dy dx �
bh3
12
m � �b
0�h��hx�b�
0 dy dx �
bh2
30.
x � y �Ix
m�a4�
8�
2�a2 �
a2
I0 � Ix � Ix �a4�
8�
a4�
8�
a4�
4
Iy � �R� x2 dA � ��
0�a
0 r3 cos2 � dr d� �
a4�
8
Ix � �R� y 2 dA � ��
0�a
0 r3 sin2 � dr d� �
a4�
8
m ��a2
2
32.
y �Ix
m�ab3�
4�
1�ab
�b2
x �Iy
m�a3b�
4�
1�ab
�a2
I0 � Iy � Ix �a3b�
4�
ab3�
4�
ab�
4�a2 � b2�
Iy � 4 �b
0��a�b�b2�y 2
0 x2 dx dy �
a3b�
4
�4b3
3a3 �a2
2 �xa2 � x2 � a2 arcsin xa� �
18
�x�2x2 � a2�a2 � x2 � a4 arcsin xa��
a
0�
ab3�
4
� 4�a
0
b3
3a3 �a2 � x2�3�2 dx �4b3
3a3 �a
0 �a2a2 � x2 � x2a2 � x2 dx
Ix � 4�a
0��b�a�a2�x2
0 y 2 dy dx
m � �ab
382 Chapter 13 Multiple Integration
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34.
y �Ix
m�4ka5�15
2ka3�3�2a2
5�
2a10
x �Iy
m�2ka5�15
2ka3�3�a2
5�
a5
I0 � Ix � Iy �2ka5
5
Iy � k�a
�a�a2�x2
0 x2y dy dx �
2ka5
15
Ix � k�a
�a�a2�x2
0
y3 dy dx �4ka5
15
� k�a
0 �a2 � x2� dx �
2ka3
3
m � 2k�a
0�a2�x2
0 y dy dx
� � ky 36.
y �Ix
mk�60k�24
�25
x �Iy
m�k�48
k�24�
12
I0 � Ix � Iy �9k
240�
3k80
Iy � k�1
0�x
x2
x3y dy dx �k2�
1
0 �x5 � x7� dx �
k48
Ix � k�1
0�x
x2
xy3 dy dx �k4�
1
0 �x5 � x9� dx �
k60
m � k�1
0�x
x2
xy dy dx �k2�
1
0 �x3 � x5� dx �
k24
� � kxy
38.
y �Ix
m� x �395
891
x �Iy
m� 158
2079�
356
�395891
I0 � Ix � Iy �316
2079
Iy � �1
0�x
x2
�x2 � y 2�x2 dy dx �158
2079
Ix � �1
0�x
x 2
�x2 � y 2� y 2 dy dx �158
2079
m � �1
0�x
x2
�x2 � y 2� dy dx �6
35
� � x2 � y 2 40.
y �Ix
m�32,768k
65�
21512k
�81365
65
x �Iy
m�2048k
45�
21512k
�2815
�2105
15
I0 � Ix � Iy �321,536k
585
Iy � 2�2
0�4x
x3
kx2 y dy dx �2048k
45
Ix � 2�2
0�4x
x3
ky3 dy dx �32,768k
65
m � 2�2
0�4x
x3
ky dy dx �512k21
� � ky
42. I � �4
0�2
0 k�x � 6�2 dy dx � �4
0 2k�x � 6�2 dx � �2k
3�x � 6�3�
4
0�
416k3
44.
� 2k�7a5
15�
a5�
8 � � ka5�56 � 15�
60 � � 2k�14�a5 �
23
a5 �15
a5� �2a3 �a4�
4�
a4�
16 � �a2
2 �a3 �a3
3 ��
�18 �x�2x2 � a2�a2 � x2 � a4 arcsin
xa�� �
a2
2 �a2x �x3
3 ��a
�a
� k�14�a4x �
2a2x3
3�
x5
5 � �2a3 �a2
2 �xa2 � x2 � a2 arcsin xa�
� �a
�a
k�14
�a4 � 2a2x2 � x 4� �2a3
�a2a2 � x2 � x2a2 � x2� �a2
2�a2 � x2�� dx
� �a
�a
k�y4
4�
2ay3
3�
a2y 2
2 �a2�x2
0 dx
I � �a
�a�a2�x2
0 ky�y � a�2 dy dx
Section 13.4 Center of Mass and Moments of Inertia 383
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46.
�2k3 �32 � 32 �
1925
�1287 � �
1408k105
�k3�
2
�2 �16 � 12x2 � 6x4 � x6� dx � �k
3�16x � 4x3 �65
x5�
17
x7��2
�2
I � �2
�2�4�x2
0 k�y � 2�2 dy dx � �2
�2 �k
3�y � 1�3�
4�x2
0 dx � �2
�2 k3
��2 � x2� � 8 dx
48.
will be the same.�x, y�
��x, y� � k 2 � x . 50. Both and will decreaseyx��x, y� � k�4 � x��4 � y�.
52. Moment of inertia about x-axis.
Moment of inertia about y-axis.Iy � �R�x2 ��x, y� dA
Ix � �R�y2 ��x, y� dA
54. Orient the xy-coordinate system so that L is along the y-axis and R is in the first quadrant. Then the volume of the solid is
By our positioning, Therefore, V � 2� rA.x � r.
� 2� xA.
� 2� ��R� x dA
�R� dA ��R
� dA
� 2� �R� x dA
x
LR
( , )x y
y
V � �R� 2� x dA
56.
ya �a2
�a3b�12
�L � �a�2�ab�
a�3L � 2a�3�2L � a�
Iy � �b
0�a
0 �y �
a2�
2 dy dx �
a3b12
y �a2
, A � ab, h � L �a2
58.
ya � ��a4��4�
L�a2 � �a2
4L
�a4�
4
� �2�
0 a4
4 sin2 � d�
� �2�
0�a
0 r3 sin2 � dr d�
Iy � �a
�a�a2�x2
�a2�x2 y
2 dy dx
y � 0, A � �a2, h � L
384 Chapter 13 Multiple Integration
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Section 13.5 Surface Area
2.
2
2
31
3
1
x
R
y
S � �3
0�3
0 �14 dy dx � �3
0 3�14 dx � 9�14
�1 � � fx �2 � � fy �2 � �14
fx � 2, fy � �3
f �x, y� � 15 � 2x � 3y 4.
1
2
−1
−2
−2 −1 1x
y x= 9 − 2
y x= 9− − 2
R
y
� �2�
0�3
0 �14 r dr d� � 9�14�
S � �3
�3��9�x2
��9�x2
�14 dy dx
�1 � � fx �2 � � fy �2 � �14
fx � 2, fy � �3
R � ��x, y�: x2 � y 2 ≤ 9�
f �x, y� � 10 � 2x � 3y
6.
square with vertices
� �34
�2y�1 � 4y2 � ln2y � �1 � 4y2�3
0�
34
�6�37 � ln6 � �37�
S � �3
0�3
0
�1 � 4y 2 dx dy � �3
0 3�1 � 4y 2 dy
�1 � � fx �2 � � fy �2 � �1 � 4y2
fx � 0, fy � 2y
�0, 0�, �3, 0�, �0, 3�, �3, 3�R �
2
2
31
3
1
x
R
yf �x, y� � y 2
8.
x
2
1 2
1 R
y = 2 − x
y
�125�3 �
85
� 2 � 33�2 �25
� 35�2 � 2 �25
� �2�1 � y�3�2 �25
�1 � y�5�22
0
S � �2
0�2�y
0 �1 � y dx dy � �2
0
�1 � y �2 � y� dy
�1 � fx2 � fy
2 � �1 � y
fx � 0, fy � y1�2
f �x, y� � 2 �23
y3�2 10.
1
−1
−1 1x
R
y
� �2�
0
112
�173�2 � 1� d� ��
6�17�17 � 1�
� �2�
0� 1
12�1 � 4r2 �3�2
2
0 d�
S � �2�
0�2
0
�1 � 4r2 r dr d�
�1 � � fx �2 � � fy �2 � �1 � 4x2 � 4y 2
fx � 2x, fy � �2y
f �x, y� � 9 � x2 � y 2
Section 13.5 Surface Area 385
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12.
� �2�
0�4
0 �1 � r2 r dr d� �
2�
3�17�17 � 1�
S � �4
�4��16�x2
��16�x2
�1 � y 2 � x2 dy dx
�1 � � fx �2 � � fy �2 � �1 � y 2 � x2
fx � y, fy � x
R � ��x, y�: x2 � y 2 ≤ 16�2
−2
−2 2x
x y2 2+ = 16
yf �x, y� � xy
14. See Exercise 13.
� �2�
0�a
0
a�a2 � r2
r dr d� � 2�a2 S � �a
�a��a2�x2
��a2�x2
a
�a2 � x2 � y 2 dy dx
16.
42
4
2
6
6
x
y x= 16 − 2
y
� ���2
0�4
0 �1 � 4r2 r dr d� �
�
24�65�65 � 1�
S � �4
0��16�x2
0 �1 � 4�x2 � y 2� dy dx
�1 � fy2 � fy
2 � �1 � 4x2 � 4y 2
z � 16 � x2 � y 2 18.
1
−1
−1 1x
y
x2 + y2 = 4
S � �2�
0�2
0
�5r dr d� � 4��5
�1 � fx2 � fy
2 ��1 �4x2
x2 � y2 �4y2
x2 � y2 � �5
z � 2�x2 � y2
20.
triangle with vertices
2
2
31
3
1
x
R
y x=
y
S � �2
0�x
0 �5 � 4y2 dy dx �
112
�21�21 � 5�5 �
�1 � � fx �2 � � fy �2 � �5 � 4y2
�0, 0�, �2, 0�, �2, 2�R �
f �x, y� � 2x � y 2 22.
2
−2
−2 2x
x y2 2+ = 16
y
� �2�
0�4
0 �1 � 4r2 dr d� �
�65�65 � 1��6
S � �4
�4��16�x2
��16�x2
�1 � 4x2 � 4y 2 dy dx
�1 � � fx �2 � � fy �2 � �1 � 4x2 � 4y 2
fx � 2x, fy � 2y
0 ≤ x2 � y 2 ≤ 16
R � ��x, y�: 0 ≤ f �x, y� ≤ 16�
f �x, y� � x2 � y 2
386 Chapter 13 Multiple Integration
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24.
S � �1
0�1
0 �1 � ��x � sin x�2 dy dx � 1.02185
�1 � � fx �2 � � fy �2 ��1 � ��x � sin x�2
fx � x1�2 � sin x, fy � 0
R � ��x, y�: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1�
f �x, y� �23 x3�2 � cos x 26. Surface
Matches (c)
xy
33
3
2
z
area � �9��
28.
� �1
0 �1 � y3 dy � 1.1114
S � �1
0�1
0 �1 � y3 dx dy
�1 � � fx �2 � � fy �2 � �1 � y3
fx � 0, fy � y3�2
R � ��x, y�: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1�
f �x, y� �25 y5�2 30.
S � �4
0�x
0 �1 � 13�x2 � y 2� dy dx
� �1 � 13�x2 � y 2�
�1 � � fx � � � fy �2 � �1 � �2x � 3y�2 � �3x � 2y�2
fx � 2x � 3y, fy � �3x � 2y � ��3x � 2y�
R � ��x, y�: 0 ≤ x ≤ 4, 0 ≤ y ≤ x�
f �x, y� � x2 � 3xy � y 2
32.
S � ����2
����2 ��(��2)�x2
��(��2)�x2
�1 � 4�x2 � y 2� sin2�x2 � y 2� dy dx
�1 � � fx �2 � � fy �2 � �1 � 4x2 sin2�x2 � y 2� � 4y 2sin2�x2 � y 2� � �1 � 4 sin2�x2 � y 2���x2 � y 2�
fx � �2x sin�x2 � y 2�, fy � �2y sin�x2 � y 2�
R � ��x, y�: x2 � y 2 ≤ �
2�f �x, y� � cos�x2 � y 2�
34.
S � �4
0�x
0 �1 � e�2x dy dx
� �1 � e�2x
�1 � � fx �2 � � fy �2 � �1 � e�2x sin2 y � e�2x cos2 y
fx � �e�x sin y, fy � e�x cos y
R � ��x, y�: 0 ≤ x ≤ 4, 0 ≤ y ≤ x�
f �x, y� � e�x sin y 36. (a) Yes. For example, let R be the square given by
and S the square parallel to R given by
(b) Yes. Let R be the region in part (a) and S the surfacegiven by
(c) No.
f �x, y� � xy.
0 ≤ x ≤ 1, 0 ≤ y ≤ 1, z � 1.
0 ≤ x ≤ 1, 0 ≤ y ≤ 1,
38.
S � �R� �1 � � fx �2 � � fy �2 dA � �
R� �k2 � 1 dA � �k2 � 1 �
R� dA � A�k2 � 1 � �r2�k2 � 1
�1 � � fx �2 � � fy �2 ��1 �k2x2
x2 � y 2 �k2y 2
x2 � y 2 � �k2 � 1
f �x, y� � k�x2 � y 2
Section 13.5 Surface Area 387
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42. False. The surface area will remain the same for any vertical translation.
40. (a)
(c)
S � 2�50
0�15
0
�1 � fy2 � fx
2 dy dx � 3087.58 sq ft
fx � 0, fy � �1
25y2 �
825
y �1615
f �x, y� � �1
75y3 �
425
y2 �1615
y � 25
z ��175
y3 �4
25y2 �
1615
y � 25 (b)
cubic feet
(d) Arc length
Surface area of roof � 2�50��30.8758� � 3087.58 sq ft
� 30.8758
� 100�266.25� � 26,625
V � 2�50��15
0��
175
y3 �4
25y2 �
1615
y � 25� dy
Section 13.6 Triple Integrals and Applications
2.
�23
�1
�1�1
�1 y 2z2 dy dz �
29
�1
�1 y3z2
1
�1 dz �
49
�1
�1 z2 dz � 4
27z3
1
�1�
827
�1
�1�1
�1�1
�1 x2y 2z2 dx dy dz �
13�
1
�1�1
�1 x3y 2z2
1
�1 dy dz
4.
�12
�9
0 xy 2 � 3x3
y�3
0 dy �
218
�9
0 y3 dy � 1
36y 4
9
0�
7294
�9
0�y�3
0��y2�9x2
0 z dz dx dy �
12
�9
0�y�3
0 � y 2 � 9x2� dx dy
6.
� �4
1 1
x �ln z�2
2 e2
1 dx � �4
1 2x dx � 2 ln �x�
4
1� 2 ln 4
�4
1�e2
1�1�xz
0 ln z dy dz dx � �4
1�e2
1 �ln z�y
1�xz
0 dz dx � �4
1�e2
1 ln zxz
dz dx
8. ���2
0�y�2
0�1�y
0 sin y dz dx dy � ���2
0�y�2
0 sin y
y dx dy �
12
���2
0 sin y dy � �
12
cos y��2
0�
12
10. ��2
0��2�x2
0�4�y2
2x2�y2
y dz dy dx � ��2
0��2�x2
0 �4y � 2x2y � 2y3� dy dx �
16�215
12.
� �6
0�3�(x�2)
0 12 �
6 � x � 2y3 �
2
e�x2y 2 dy dx � 2.118
�3
0�2�(2y�3)
0�6�2y�3z
0 ze�x2y 2 dx dz dy � �6
0�(6�x)�2
0�(6�x�2y)�3
0 ze�x2y 2 dz dy dx
14. �3
0�2x
0�9�x2
0
dz dy dx
388 Chapter 13 Multiple Integration
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16.
�4
�4 ��16�x2
��16�x2 ��80�x2�y2
1�2�x2�y2�
dz dy dx
⇒ z � 8 ⇒ x2 � y2 � 2z � 16x2 � y2 � z2 � 2z � z2 � 80 ⇒ z2 � 2z � 80 � 0 ⇒ �z � 8��z � 10� � 0
z �12
�x2 � y2� ⇒ 2z � x2 � y2
18. �1
0�1
0�xy
0 dz dy dx � �1
0�1
0 xy dy dx � �1
0 x2
dx � x2
4 1
0�
14
20.
� 49x�36 � x2 � 324 arcsin�x6� �
16
x�36 � x2�3�26
0� 4�162�� � 648�
� 4�6
036�36 � x2 � x2�36 � x2 �
13
�36 � x2�3�2 dx
4�6
0��36�x2
0�36�x2
�y2
0dz dy dx � 4�6
0��36�x2
0�36 � x2 � y2�dy dx � 4�6
036y � x2y �
y3
3 �36�x2
0 dx
22.
� �2
0 �18 � 9x � 2x2 � x3� dx � 18x �
92
x2 �23
x3 �14
x 42
0�
503
�2
0�2�x
0�9�x2
0 dz dy dx � �2
0�2�x
0 �9 � x2� dy dx � �2
0 �9 � x2��2 � x� dx
24. Top plane:
Side cylinder:
�3
0��9�y2
0�6�x�y
0dz dx dy
x2 � y2 � 9
x
y
6
3
6
63
zx � y � z � 6 26. Elliptic cone:
�4
0�4
z��y2�z2�2
0 dx dy dz
x
y
5
5
4
3
3
2
2
1
1
z
4x2 � z2 � y 2
28.
� �4
0�2��y
0��y
0 xyz dx dz dy � �4
0�2
2��y
�2�z
0 dx dz dy� �
10421 �
� �2
0�(2�z)2
0��y
0 xyz dx dy dz � �2
0�4
(2�z)2�2�z
0 xyz dx dy dz
� �2
0�2�z
0�4
x2
xyz dy dx dz
� �2
0�2�x
0�4
x2
xyz dy dz dx
� �4
0��y
0�2�x
0 xyz dz dx dy
��Q
� xyz dV � �2
0�4
x2�2�x
0 xyz dz dy dx
x
y442
4
2
(2, 4)
zQ � �x, y, z�: 0 ≤ x ≤ 2, x2 ≤ y ≤ 4, 0 ≤ z ≤ 2 � x�
Section 13.6 Triple Integrals and Applications 389
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30.
� �6
0�1
0��1�x2
0xyz dy dx dz
� �1
0�6
0��1�x2
0xyz dy dz dx
� �6
0�1
0��1�y2
0xyz dx dy dz
� �1
0�6
0��1�y2
0xyz dx dz dy
� �1
0��1�y2
0�6
0xyz dz dx dy
��Q
�xyz dV � �1
0��1�x2
0�6
0xyz dz dy dx
x
y2
1
6
21
zQ � �x, y, z�: 0 ≤ x ≤ 1, y ≤ 1 � x2, 0 ≤ z ≤ 6�
32.
y �Mxz
m� 2
Mxz � k�5
0�5�x
0�1�5�15�3x�3y�
0y2 dz dy dx �
1254
k
m � k�5
0�5�x
0�1�5�15�3x�3y�
0y dz dy dz �
1258
k 34.
y �Mxz
m�
kab2c�24kabc�6
�b4
Mxz � k�b
0�a[1�(y�b)]
0�c[1�(y�b)�(x�a)]
0 y dz dx dy �
kab2c24
m � k�b
0�a�1�(y�b)�
0�c�1�(y�b)�(x�a)�
0 dz dx dy �
kabc6
36.
z �Mxy
m�
kabc3�3kabc2�2
�2c3
y �Mxz
m�
kab2c2�4kabc2�2
�b2
x �Myz
m�
ka2bc2�4kabc2�2
�a2
Mxz � k�a
0�b
0�c
0 yz dz dy dx �
kab2c2
4
Myz � k�a
0�b
0�c
0 xz dz dy dx �
ka2bc2
4
Mxy � k�a
0�b
0�c
0 z2 dz dy dx �
kabc3
3
m � k�a
0�b
0�c
0 z dz dy dx �
kabc2
2
38. will be greater than whereas and will be unchanged.yx8�5,z
40. and will all be greater than their original values.zx, y
390 Chapter 13 Multiple Integration
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42.
z �Mxy
m�
k�
16k�3�
3�
16
y �Mxz
m�
2k�
16k�3�
3�
8
x �Myz
m�
016k�3
� 0
Mxy � 2k�2
0��4�x2
0�y
0 z dz dy dx � k�
Mxz � 2k�2
0��4�x2
0�y
0 y dz dy dx � 2k�
Myz � k�2
�2 ��4�x2
0�y
0 x dz dy dx � 0
� k�2
0 �4 � x2� dx �
16k3
m � 2k�2
0��4�x2
0�y
0 dz dy dx
44.
z �Mxy
m� k�1
2�
�
4��k� �2 � �
4�
y �Mxz
m�
k ln 4k�
�ln 4�
� k �2
0�1
0
1�y 2 � 1�2 dy dx � k �2
0 y
2�y 2 � 1� �12
arctan y1
0 dx � k�1
4�
�
8� �2
0 dx � k�1
2�
�
4�
Mxy � 2k �2
0�1
0�1�(y2�1)
0 z dz dy dx
Mxz � 2k�2
0�1
0�1�(y2�1)
0 y dz dy dx � 2k �2
0�1
0
yy 2 � 1
dy dx � k �2
0 �ln 2� dx � k ln 4
m � 2k�2
0�1
0�1�(y2�1)
0 dz dy dx � 2k �2
0�1
0
1y 2 � 1
dy dx � 2k��
4� �2
0dx � k�
x � 0
46.
z �Mxy
m�
10k10k
� 1
y �Mxz
m�
15k�210k
�34
x �Myz
m�
25k�210k
�54
Mxy � k�5
0��(3�5)x�3
0�(1�15)(60�12x�20y)
0 z dz dy dx � 10k
Mxz � k�5
0��(3�5)x�3
0��(1�15)(60�12x�20y)
0 y dz dy dx �
15k2
Myz � k�5
0��(3�5)x�3
0�(1�15)(60�12x�20y)
0 x dz dy dx �
25k2
m � k�5
0��(3�5)x�3
0�(1�15)(60�12x�20y)
0 dz dy dx � 10k
x2 5
3
3
5
4
2
41
1
y x= (5 - )35
yf �x, y� �1
15�60 � 12x � 20y�
Section 13.6 Triple Integrals and Applications 391
http://librosysolucionarios.net
48. (a)
(b)
Iz � Iyz � Ixz �7ka7
180
Iy � Ixy � Iyz �a7k30
Ix � Ixy � Ixz �a7k30
Iyz � Ixz by symmetry
Ixz � k�a�2
�a�2�a�2
�a�2�a�2
�a�2 y 2�x2 � y 2� dz dy dx � ka�a�2
�a�2�a�2
�a�2 �x2y 2 � y 4� dy dx �
7ka7
360
Ixy � k�a�2
�a�2�a�2
�a�2�a�2
�a�2 z2�x2 � y 2� dz dy dx �
a3k12�
a�2
�a�2�a�2
�a�2 �x2 � y 2� dy dx �
a7k72
Ix � Iy � Iz �ka5
12�
ka5
12�
ka5
6
Ixz � Iyz �ka5
12 by symmetry
Ixy � k�a�2
�a�2�a�2
�a�2�a�2
�a�2 z2 dz dy dx �
ka5
12
50. (a)
—CONTINUED—
Ix � Ixz � Ixy �2048k
9, Iy � Iyz � Ixy �
8192k21
, Iz � Iyz � Ixz �63,488k
315
� k�4
0�2
0 12
x2�16 � 8y 2 � y4� dy dx �k2�
4
0 x2�16y �
8y3
3�
y5
5 �2
0 dx �
k2�
4
0 25615
x2 dx �8192k
45
Iyz � k�4
0�2
0�4�y2
0 x2z dz dy dx � k�4
0�2
0 12
x2�4 � y 2�2 dy dx
� k�4
0�2
0 12
�16y 2 � 8y4 � y6� dy dx �k2
�4
016y3
3�
8y5
5�
y7
7 2
0 dx �
k2�
4
0 1024105
dx �2048k105
Ixz � k�4
0�2
0�4�y2
0 y 2z dz dy dx � k�4
0�2
0 12
y 2�4 � y 2�2 dy dx
�k4�
4
0 256y �
256y3
3�
96y5
5�
16y7
7�
y9
9 2
0 dx � k�4
0 16,384
945 dx �
65,536k315
�k4�
4
0�2
0 �256 � 256y 2 � 96y4 � 16y6 � y8� dy dx
Ixy � k�4
0�2
0�4�y2
0 z3 dz dy dx � k�4
0�2
0 14
�4 � y 2�4 dy dx
392 Chapter 13 Multiple Integration
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Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates
50. —CONTINUED—
(b)
Ix � Ixz � Ixy �48,128k
315, Iy � Iyz � Ixy �
118,784k315
, Iz � Ixz � Iyz �11,264k
35
� k�4
0�2
0�4�y2
0 4x2 dz dy dx � k�4
0�2
0�4�y2
0 x2z dz dy dx �
4096k9
�8192k
45�
4096k15
Iyz � k�4
0�2
0�4�y2
0 x2�4 � z� dz dy dx
� k�4
0�2
0�4�y2
0 4y 2 dz dy dx � k�4
0�2
0�4�y2
0 y 2z dz dy dx �
1024k15
�2048k105
�1024k
21
Ixz � �4
0�2
0�4�y2
0 y 2�4 � z� dz dy dx
� k�4
0�2
0�4�y2
0 4z2 dz dy dx � k�4
0�2
0�4�y2
0 z3 dz dy dx �
32,768k105
�65,536k
315�
32,768k315
Ixy � �4
0�2
0�4�y2
0 z2�4 � z� dz dy dx
52.
Iz � Ixz � Iyz �1
12m�a2 � c2�
Iy � Ixy � Iyz �1
12m�b2 � c2�
Ix � Ixy � Ixz �1
12m�a2 � b2�
Iyz � �c�2
�c�2�a�2
�a�2�b�2
�b�2 x2 dz dy dx � ab�c�2
�c�2 x2 dx �
abc3
12�
112
c2�abc� �1
12mc2
Ixz � �c�2
�c�2�a�2
�a�2�b�2
�b�2 y 2 dz dy dx � b�c�2
�c�2�a�2
�a�2 y 2 dy dx �
ba3
12�c�2
�c�2 dx �
ba3c12
�1
12a2�abc� �
112
ma2
Ixy � �c�2
�c�2�a�2
�a�2�b�2
�b�2 z2 dz dy dx �
b3
12�c�2
�c�2�a�2
�a�2 dy dx �
112
b2�abc� �1
12mb2
54. �1
�1��1�x2
��1�x2
�4�x2�y2
0 kx2�x2 � y 2� dz dy dx 56. 6
58. Because the density increases as you move away from the axis of symmetry, the moment of intertia will increase.
2.
�12
���4
0�2
0�4r � 4r 2 � r 3�dr d� �
12
���4
0 �2r 2 �
4r 3
3�
r4
4 �2
0
d� �23
���4
0d� �
�
6
���4
0�2
0�2�r
0 rz dz dr d� � ���4
0�2
0�rz2
2 �2�r
0
dr d�
4. ���2
0��
0�2
0e��3
�2 d� d� d � ���2
0��
0 ��
13
e��3�2
0
d� d � ���2
0��
0 13
�1 � e�8� d� d �� 2
6�1 � e�8�
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 393
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6.
�5�236
���4
0sin cos d � �5�2
36 sin2
2 ���4
0
�5�2144
�13
���4
0sin cos �sin � �
sin3 �3 �
��4
0
d
�13
���4
0���4
0sin cos cos ��1 � sin2 ��d� d
���4
0���4
0�cos �
0�2 sin cos d� d� d �
13
���4
0���4
0cos3 � sin cos d� d
8. ���2
0��
0�sin �
0�2 cos ��2 d� d� d �
89
10.
� �2�
0 94
d� �9�
2
� �2�
0 �3r2
2�
r4
4 ���3
0
d�
y
x
2233
4
z�2�
0��3
0�3�r2
0
r dz dr d� � �2�
0��3
0
r �3 � r 2�dr d�
12.
�468�
3
�1173
�2�
0��cos �
�
0 d�
yx 77
7 r = 5
r = 2
z�2�
0��
0�5
2�2 sin d� d d� �
1173
�2�
0��
0sin d d�
14. (a)
(b) ���2
0���6
0�4
0�3 sin2 d� d d� � ���2
0���2
��6�2 csc
4�3 sin2 d� d d� �
8� 2
3� 2��3
���2
0�2
0��16�r2
0
r 2 dz dr d� �8� 2
3� 2��3
16. (a) ���2
0�1
0��1�r2
0
r�r 2 � z2 dz dr d� ��
8(b) ���2
0���2
0�1
0�3 sin d� d d� �
�
8
18.
(Volume of lower hemisphere) (Volume in the first octant)
�128�
3�
64�2�
3�
64�
3�2 � �2�
�128�
3� 4�8�2�
3�
8�2�
3 �
�128�
3� 4�8�2�
3� ���2
0��
13
�16 � r2�3�2�4
2�2 d��
V �128�
3� 4����2
0�2�2
0 r2 dr d� � ���2
0�4
2�2r�16 � r2 dr d��
� 4
yx7
7
7
z V �23
� �4�3 � 4����2
0�2�2
0�r
0 r dz dr d� � ���2
0�4
2�2��16�r2
0 r dz dr d��
394 Chapter 13 Multiple Integration
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20.
�8�
3�2 � �2�
� �2�
0��
13
�4 � r2�3�2 �r3
3 ��2
0 d�
� �2�
0��2
0 �r�4 � r2 � r2� dr d�
V � �2�
0��2
0��4�r2
r
r dz dr d� 22.
� 3k� �1 � e�4�
� ���2
0��6ke�4 � 6k� d�
� ���2
0 ��6ke�r2�
2
0
���2
0�2
0�12e�r 2
0
k r dz dr d� � ���2
0�2
0
12ke�r2
r dr d�
24.
z �Mxy
m�
kr02h2�
12 � 3�r0
2hk� �h4
�2kh2
r02 �r0
4
12���
2� �kr0
2 h2�
12
�2kh2
r02 ���2
0�r0
0�r0
2r � 2r0r2 � r3�dr d�
Mxy � 4k���2
0�r0
0�h�r0�r��r0
0zr dz dr d�
m �13
�r02hk from Exercise 23
x � y � 0 by symmetry 26.
z �Mxy
m�
k�r02h3�30
k�r02h2�12
�2h5
�1
30k�r0
2h3
Mxy � 4k���2
0�r0
0�h(r0�r)�r0
0 z2r dz dr d�
�1
12k�r0
2h2
m � 4k���2
0�r0
0�h(r0�r)�r0
0 zr dz dr d�
x � y � 0 by symmetry
� � kz
28.
�1
15r0
5�kh
� 4kh1
30r0
5 �
2
� 4kh���2
0
130
r05 d�
� 4kh���2
0�r0
5
5�
r05
6 � d�
� 4kh���2
0�r 5
5�
r 6
6r0�r0
0 d�
� 4kh���2
0�r0
0 r0 � r
r0 r 4 dr d�
� 4k���2
0�r0
0�h�r0�r��r0
0 r 4 dz dr d�
Iz � ��Q
��x2 � y2���x, y, z� dV 30.
�32
ma2
�32
k�a4h
Iz � 2k���2
0�2a sin �
0�h
0 r3 dz dr d�
m � k�a2h
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 395
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34.
� k�a4 �
4�
14
k� 2a4
� �k�a4�12
�14
sin 2����2
0
� k�a4���2
0 sin2 d
� 2ka4���2
0���2
0 sin2 d� d
m � 8k���2
0���2
0�a
0 �3 sin2 d� d� d 36.
z �Mxy
m�
k��R4 � r4��42k��R3 � r3��3
�3�R4 � r4�8�R3 � r3�
�14
k��R4 � r4� � ��18
k��R4 � r4� cos 2���2
0
�14
k��R4 � r4����2
0 sin 2 d
�12
k�R4 � r4����2
0���2
0 sin 2 d� d
Mxy � 4k���2
0���2
0�R
r
�3 cos sin d� d� d
m � k�23
�R3 �23
�r3� �23
k��R3 � r3�
x � y � 0 by symmetry
38.
�4k�
15�R5 � r5�
� �2k�
5�R5 � r5���cos �
cos3 3 ��
��2
0
�2k�
5�R5 � r5����2
0 sin �1 � cos2 � d
�4k5
�R5 � r5����2
0���2
0 sin3 d� d
Iz � 4k���2
0���2
0�R
r
�4 sin3 d� d� d 40.
z � � cos cos �z
�x2 � y2 � z2
y � � sin sin � tan � �yx
x � � sin cos � �2 � x2 � y2 � z2
42. ��2
�1
�2
1
��2
�1
f �� sin cos �, � sin sin �, � cos ��2 sin d� d d�
44. (a) You are integrating over a cylindrical wedge.
46. The volume of this spherical block can be determined as follows. One side is length Another side is Finally, the third side is given by the length of an arc of angle in a circle of radius Thus:
� �2 sin ��
V ��������� sin �
� sin .��.
x
y
θφρi i isin ∆
ρ∆ i φρi i∆
z�.
32. (includes upper and lower cones)
� �1 ��22 �4�
3�b3 � a3� �
2�
3�2 � �2 ��b3 � a3�
� �4�
3�b3 � a3���cos ��
��4
0
�4�
3�b3 � a3����4
0 sin d
yxa
ab
b
b
z
�83
�b3 � a3����4
0���2
0 sin d� d
V � 8���4
0���2
0�b
a
�2 sin d� d� d
(b) You are integrating over a spherical block.
396 Chapter 13 Multiple Integration
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Section 13.8 Change of Variables: Jacobians
2.
�x�u
�y�v
��y�u
�x�v
� ad � cb
y � cu � dv
x � au � bv 4.
�x�u
�y�v
��y�u
�x�v
� �v � 2�u � vu � �2u
y � uv
x � uv � 2u
6.
�x�u
�y�v
��y�u
�x�v
� �1��1� � �0��0� � 1
y � v � a
x � u � a 8.
�x�u
�y�v
��y�u
�x�v
� �1v��1� � �1���
uv2� �
1v
�uv2 �
u � vv2
y � u � v
x �uv
�3, �6��6, 3�
�0, �6��2, 2�
�3, 0��4, 1�
�0, 0��0, 0�
�u, v��x, y�10.
v � x � 4y
u � x � y
y �13
�u � v�
x �13
�4u � v�(0, 0) (3, 0)
(3, −6)(0, −6)
−1−1
1
−2
−3
−4
−5
−6
1 2 4 5 6
v
u
12.
� 15�23
�263 � � �120
� �152 �2
3u3 �
263
u�1
�1
� �1
�1
152 �2u2 �
263 � du
� �1
�1��
152 �v3
3� u2v��
3
1 du
� �1
�1�3
1�
152
�v2 � u2� dv du
� �1
�1�3
160�1
2�u � v����
12
�u � v���12� dv du
�R�60xy dA
�x�u
�y�v
��y�u
�x�v
�12 �
12� � ��
12��
12� �
12
y � �12
�u � v�, v � x � y
x �12
�u � v�, u � x � y
(1, 3)(−1, 3)
(−1, 1) (1, 1)
−2 −1
−1
2
1 2
v
u
�1, 1��1, 0�
��1, 3��1, 2�
�1, 3��2, 1�
��1, 1��0, 1�
�u, v��x, y�
Section 13.8 Change of Variables: Jacobians 397
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18.
� ��
0 �1
2 �u3
3 � 1 � cos 2v
2 �2�
� dv � �7�3
12 �v �12
sin 2v���
0�
7�4
12
�R� �x � y�2 sin2 �x � y� dA � ��
0�2�
�
u2 sin2 v�12� du dv
��x, y���u, v� � �
12
x �12
�u � v�,
u � x � y � 2�,
2
x
x y− = 0
x y− =x y+ =
π
π
π
π
x y+ = 2π
π2π
2π3
2π3
y u � x � y � �,
y �12
�u � v�
v � x � y � �
v � x � y � 0
20.
� �8
016v 32 dv � �2
5�16v52�8
0�
40965
2
�R� �3x � 2y��2y � x�32 dA � �8
0�16
0 uv32�1
8� du dv
�x�u
�y�v
��y�u
�x�v
�14�
38� �
18��
14� �
18
x �14
�u � v�,
u � 3x � 2y � 16,
x2
3
3
5
2
41−1
−1−2
2 = 0y x−
2 = 8y x−
3 + 2 = 16x y
3 + 2 = 0x y
y
(0, 0)
(−2, 3)
(2, 5)
(4, 2)
u � 3x � 2y � 0,
y �18
�u � 3v�
v � 2y � x � 8
v � 2y � x � 0
16.
�R� y sin xy dA � �4
1�4
1 v�sin u� 1
v dv du � �4
1 3 sin u du � ��3 cos u�
4
1� 3�cos 1 � cos 4� � 3.5818
�x�u
�y�v
��y�u
�x�v
�1v
y � v
x �uv
14.
� �2
0 2u�1 � eu�2� du � 2�u2
2� ueu�2 � eu�2�
2
0� 2�1 � e�2�
�R� 4�x � y�ex�y dA � �2
0�0
u�2 4uev �1
2� dv du
�x�u
�y�v
��y�u
�x�v
� �12
y �12
�u � v�
x �12
�u � v�u
1 2
−1
−2
v u= 2−
v
398 Chapter 13 Multiple Integration
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22.
� �4
1�1
2ln�1 � v2��
4
1 1u
du � �12
�ln 17 � ln 2 ln u�4
1�
12 �ln
172 ��ln 4�
�R�
xy1 � x2y 2 dA � �4
1�4
1
v1 � v2 �1
u� dv du
�x�u
�y�v
��y�u
�x�v
�1u
x � u,
u � x � 4,
u � x � 1,
y �vu
v � xy � 4
v � xy � 1
x
2
3
4
1 2 3 4
1
x = 4
x = 1
xy = 4
xy = 1
y
24. (a)
Let and
Let
(b)
Let and
Let
� 2� Aab��2�
� 0� � �0 �4
�2�� �4�� � 2�Aab
�
Aab�2�
0�1
0 cos��
2r�r dr d� � Aab�2r
� sin��r
2 � �4
�2 cos��r2 ��
1
0 �2��
u � r cos �, v � r sin �.
�R� f �x, y� dA � �1
�1�1�u2
�1�u2
A cos��
2u2 � v2� ab dv du
y � bv.x � au
R: x2
a2 �y 2
b2 ≤ 1
f �x, y� � A cos ��
2 x2
a2 �y 2
b2�
� 12�398
� �7
16 sin 2��
2�
0� 12�39�
4 � � 117�
� 12�2�
0 �8 � 4�1 � cos 2�
2 � �94 �
1 � cos 2�
2 �� d� � 12�2�
0 �39
8�
78
cos 2�� d�
� 12�2�
0 �8r2 � 4r4 cos2 � �
94
r4 sin2 ��1
0 d� � 12�2�
0 �8 � 4 cos2 � �
94
sin2 �� d�
� �2�
0�1
0 �16 � 16r2 cos2 � � 9r2 sin2 �� 12r dr d�
u � r cos �, v � r sin �.�� �R� �16 � x2 � y 2� dA � �1
�1�1�u2
�1�u2
�16 � 16u2 � 9v2� 12dv du
y � 3v.x � 4u
V � �R� f �x, y� dA
R: x2
16�
y 2
9 ≤ 1
f �x, y� � 16 � x2 � y 2
Section 13.8 Change of Variables: Jacobians 399
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Review Exercises for Chapter 13
2. �2y
y
�x 2 � y 2� dx � �x3
3� xy 2�
2y
y�
10y3
3
4. � �2
0�4x 2 � 2x3 � 2x 4� dx � �4
3x3 �
12
x4 �25
x5�2
0�
8815�2
0�2x
x2
�x 2 � 2y� dy dx � �2
0�x 2y � y 2�
2x
x2 dx
6. � �y�4 � y 2 � 4 arcsin y2�
�3
0� �3 �
4�
3��3
0�2��4�y2
2��4�y2
dx dy � 2��3
0 �4 � y 2 dy
8.
�12
�2
0�6 � 3y� dy � �1
2�6y �32
y2�2
0� 3
A � �2
0��6�y�2
y
dx dy
�2
0�x
0 dy dx � �3
2�6�2x
0 dy dx � �2
0��6�y�2
y
dx dy
10.
A � �4
0�6x�x2
x2�2x
dy dx � �4
0�8x � 2x2� dx � �4x2 �
23
x3�4
0�
643
�4
0�6x�x2
x2�2x
dy dx � �0
�1�1��1�y
1��1�y
dy dx � �8
0�1��1�y
3��9�y
dx dy � �9
8�3��9�y
3��9�y
dx dy
12. A � �2
0�y2�1
0 dx dy � �1
0�2
0 dy dx � �5
1�2
�x�1 dy dx �
143
14. A � �3
0�2y�y2
�y
dx dy � �0
�3�1��1�x
�x
dy dx � �1
0�1��1�x
1��1�x
dy dx �92
16. Both integrations are over the common region R shown in the figure. Analytically,
�3
0�2x3
0 ex�y dy dx � �5
3�5�x
0 ex�y dy dx � �3
5e5 � e3 �
25 � �e5 � e3� �
85
e5 �25
�2
0�5�y
3y2 ex�y dx dy �
25
�85
e5
x1
1
2
2
3
3
4
4
5
5
(3, 2)
y
26. See Theorem 13.5. 28.
��x, y, z���u, v, w� � �401 �1
40
0�1
1� � 17
x � 4u � v, y � 4v � w, z � u � w
30.
� 1�r cos2 � � r sin2 � � r��x, y, z���r, �, z� � �cos �
sin �0
�r sin �r cos �
0
001�
x � r cos �, y � r sin �, z � z
400 Chapter 13 Multiple Integration
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22.
Since we have
P � �0.5
0�0.25
0 8xy dy dx � 0.03125
k � 8.k8 � 1,
� �kx4
8 �1
0�
k8
� �1
0 kx3
2 dx
�1
0�x
0 kxy dy dx � �1
0�kxy2
2 �x
0 dx 24. False, �1
0�1
0x dy dx � �2
1�2
1x dy dx
26. True, �1
0�1
0
11 � x2 � y2 dx dy < �1
0�1
0
11 � x2 dx dy �
�
4
28. � ��2
0�r 4
4 �4
0 d� � ��2
0 64 d� � 32��4
0��16�y2
0�x 2 � y 2� dx dy � ��2
0�4
0r3 dr d�
30.
�43
��R2 � b2�32
�83
�R2 � b2�32��2
0 d�
� �83�
�2
0��R2 � r2�32�
R
b d�
V � 8��2
0�R
b
�R2 � r2r dr d� 32.
The polar region is given by andHence,
x1
1
2
3
4
4, )( 12/ 138/ 13
8/ 13
x y2 2+ =1623
xy =
θ
y
�arctan�32�
0�4
0�r cos ���r sin ��r dr d� �
28813
0 ≤ � ≤ 0.9828.0 ≤ r ≤ 4
tan � �12�13
8�13�
32
⇒ � � 0.9828
18.
� �12
x3�3
0�
272
�32�
3
0 x2 dx
� �3
0�xy �
12
y2�x
0 dx
V � �3
0�x
0�x � y� dy dx 20. Matches (c)
x
y
2
2
1
2
3
z
Review Exercises for Chapter 13 401
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36.
y ��Ix
m��16,384k315
128k15��128
21
x ��Iy
m��512k105
128k15��4
7
m � �R��x, y� dA � �2
0�4�x2
0ky dy dx �
12815
k
I0 � Ix � Iy �16,384k
315�
512k105
�17,920
315k �
5129
k
Iy � �R�x2 �x, y� dA � �2
0�4�x2
0kx2y dy dx �
512105
k
Ix � �R�y2 �x, y� dA � �2
0�4�x2
0ky3 dy dx �
16,384315
k
38.
� 6�2 � ln�4 � 3�2� �9�2
2� ln�2 �
�26
�5�2
3� ln�2�2 � 3�
� �12
�4�18 � 2 ln�4 � �18 �� �1
12�18�18 �� � �ln�2 �
2�212 �
� �12
�2y�2 � 4y2 � 2 ln�2y � �2 � 4y2�� �1
12�2 � 4y2�32�2
0
S � �2
0�2
y
�2 � 4y2 dx dy � �2
0�2�2 � 4y2 � y�2 � 4y2 dy
�1 � � fx�2 � � fy�2 � �2 � 4y2
fx � �1, fy � �2y
R � ��x, y�: 0 ≤ x ≤ 2, 0 ≤ y ≤ x�
f �x, y� � 16 � x � y2
34.
y �Mx
m�
17kh2L80
12
7khL�
51h140
x �My
m�
5khL2
24
127khL
�5L14
�kh2
�L
0�2x �
x2
L�
x3
L2dx �kh2 �x2 �
x3
3L�
x4
4L2�L
0
�kh2
5L2
12�
5khL2
24
My � k�L
0��h2��2��xL���x2L2�
0x dy dx
�kh2
8
17L10
�17kh2L
80 �
kh2
8 �4x �2x2
L�
x3
L2 �x 4
2L3 �x5
5L4�L
0
�kh2
8 �L
0 �4 �
4xL
�3x2
L2 �2x3
L3 �x4
L4�dx
�kh2
8 �L
0�2 �
xL
�x2
L22
dx
Mx � k�L
0��h2��2��xL���x2L2�
0y dy dx
x
h
L
y = 2 − −h2
xL
x2
L2( (
y m � k�L
0��h2��2��xL���x2L2�
0dy dx �
kh2
�L
0�2 �
xL
�x2
L2dx �7khL12
402 Chapter 13 Multiple Integration
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44.
� ��2
0��5 � arctan 5���cos ���
�2
0 d� �
�
2�5 � arctan 5�
� ��2
0��2
0� � arctan �
5
0 sin � d� d�
�5
0��25�x2
0��25�x2�y2
0
11 � x2 � y2 � z2 dz dy dx � ��2
0��2
0�5
0
2
1 � 2 sin � d d� d�
46. �2
0��4�x2
0��4�x2�y2
0 xyz dz dy dx �
43
48.
� 8�4� � 2 sin 2� �14
sin3 � cos � �34�
12
� �14
sin 2���2
0�
29�
2
� 2��2
0�32 sin2 � � 4 sin4 �� d� � 8��2
0�8 sin2 � � sin4 �� d�
� 2��2
0�2 sin �
0 r�16 � r2� dr d�V � 2��2
0�2 sin �
0�16�r2
0 r dz dr d�
50.
z �Mxy
m�
�kc2a4162kca33
�3�ca
32
y �Mxz
m�
�kca482kca33
�3�a16
x � 0
Mxy � 2k��2
0�a
0�cr sin �
0rz dz dr d� � kc2��2
0�a
0r3 sin2 � dr d� �
14
kc2 a4��2
0sin2 � d� �
116
�kc2a4
Mxz � 2k��2
0�a
0�cr sin �
0r2 sin � dz dr d� � 2kc��2
0�a
0r3 sin2 � dr d� �
12
kca4��2
0sin2 � d� �
18
�kca4
m � 2k��2
0�a
0�cr sin �
0r dz dr d� � 2kc��2
0�a
0r 2 sin � dr d� �
23
kca3��2
0sin � d� �
23
kca3
40. (a) Graph of
over region R
x
y50
50
50
R
z
� 25 �1 � e��x2�y2�1000 cos2 �x2 � y 2
1000 �f �x, y� � z
(b) Surface area
Using a symbolic computer program, you obtain surfacearea sq. ft.� 4,540
� �R��1 � fx�x, y�2 � fy�x, y�2 dA
42. �12�
2�
0�2
0r5 dr d� �
163 �
2�
0d� �
32�
3 �2
�2��4�x2
��4�x2 �(x2�y2)�2
0�x 2 � y 2� dz dy dx � �2�
0�2
0�r 22
0r3 dz dr d�
Review Exercises for Chapter 13 403
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52.
z �Mxy
m� �
81�
4
1162�
� �18
� �2�
0�3
0 �1
2r 3 �
92
r� dr d� � �2�
0 �1
8r 4 �
94
r 2�3
0
d� � ��818
��2�
0
� �814
�
Mxy � �2�
0�3
0�4
��25�r2
zr dz dr d� � �2�
0�5
3��25�r2
��25�r2
zr dz dr d� � �2�
0�3
0�8 �
12
�25 � r2��r dr d� � 0
x � y � 0 by symmetry
�500�
3� 2���
13
�25 � r2�32 � 2r 2�3
0
�500�
3� 2���
643
� 18 �1253 � �
500�
3�
14�
3� 162�
m �500�
3� �3
0�2�
0��25�r2
4r dz d� dr �
500�
3� �3
0�2�
0�r�25 � r2 � 4r�d� dr
54.
�4k�a6
9
Iz � k��
0�2�
0�a
02 sin2 ���2 sin � d d� d�
56.
�815
�a
� �a
�a
��1�z2�a2
��1�z2�a2
��1�y2�z2�a2
��1�y2�z2�a2
�x2 � y 2� dx dy dz
Iz � ��Q��x2 � y2�dV
x2 � y2 �z2
a2 � 1 58.
Since represents a paraboloid with vertexthis integral represents the volume of the solid
below the paraboloid and above the semi-circlein the xy-plane.y � �4 � x2
�0, 0, 1�,z � 1 � r 2
��
0�2
0�1�r2
0r dz dr d�
60.
� �2u���2v� � �2u��2v� � �8uv
��x, y���u, v� �
�x�u
�y�v
��y�u
�x�v
62.
Boundary in xy-plane Boundary in uv-plane
� 4 arctan 5 � � � 4 arctan v�5
1
� �5
1
41 � v2 dv � �5
1�5
1
11 � v2 du dv �
R� x
1 � x2y2 dA � �5
1�5
1
u1 � u2�vu�2 �1
u du dv
v � 5xy � 5
v � 1xy � 1
u � 5x � 5
u � 1x � 1
x � u, y �vu ⇒ u � x, v � xy
y = 1x
y = 1x
x = 5
x = 1
1
1
2
3
4
5
4 5
y
x
��x, y���u, v� �
�x�u
�y�v
��x�v
�y�u
� 1�1u � 0 �
1u
404 Chapter 13 Multiple Integration
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Review Exercises for Chapter 13
2. �2y
y
�x 2 � y 2� dx � �x3
3� xy 2�
2y
y�
10y3
3
4. � �2
0�4x 2 � 2x3 � 2x 4� dx � �4
3x3 �
12
x4 �25
x5�2
0�
8815�2
0�2x
x2
�x 2 � 2y� dy dx � �2
0�x 2y � y 2�
2x
x2 dx
6. � �y�4 � y 2 � 4 arcsin y2�
�3
0� �3 �
4�
3��3
0�2��4�y2
2��4�y2
dx dy � 2��3
0 �4 � y 2 dy
8.
�12
�2
0�6 � 3y� dy � �1
2�6y �32
y2�2
0� 3
A � �2
0��6�y�2
y
dx dy
�2
0�x
0 dy dx � �3
2�6�2x
0 dy dx � �2
0��6�y�2
y
dx dy
10.
A � �4
0�6x�x2
x2�2x
dy dx � �4
0�8x � 2x2� dx � �4x2 �
23
x3�4
0�
643
�4
0�6x�x2
x2�2x
dy dx � �0
�1�1��1�y
1��1�y
dy dx � �8
0�1��1�y
3��9�y
dx dy � �9
8�3��9�y
3��9�y
dx dy
12. A � �2
0�y2�1
0 dx dy � �1
0�2
0 dy dx � �5
1�2
�x�1 dy dx �
143
14. A � �3
0�2y�y2
�y
dx dy � �0
�3�1��1�x
�x
dy dx � �1
0�1��1�x
1��1�x
dy dx �92
16. Both integrations are over the common region R shown in the figure. Analytically,
�3
0�2x3
0 ex�y dy dx � �5
3�5�x
0 ex�y dy dx � �3
5e5 � e3 �
25 � �e5 � e3� �
85
e5 �25
�2
0�5�y
3y2 ex�y dx dy �
25
�85
e5
x1
1
2
2
3
3
4
4
5
5
(3, 2)
y
26. See Theorem 13.5. 28.
��x, y, z���u, v, w� � �401 �1
40
0�1
1� � 17
x � 4u � v, y � 4v � w, z � u � w
30.
� 1�r cos2 � � r sin2 � � r��x, y, z���r, �, z� � �cos �
sin �0
�r sin �r cos �
0
001�
x � r cos �, y � r sin �, z � z
400 Chapter 13 Multiple Integration
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22.
Since we have
P � �0.5
0�0.25
0 8xy dy dx � 0.03125
k � 8.k8 � 1,
� �kx4
8 �1
0�
k8
� �1
0 kx3
2 dx
�1
0�x
0 kxy dy dx � �1
0�kxy2
2 �x
0 dx 24. False, �1
0�1
0x dy dx � �2
1�2
1x dy dx
26. True, �1
0�1
0
11 � x2 � y2 dx dy < �1
0�1
0
11 � x2 dx dy �
�
4
28. � ��2
0�r 4
4 �4
0 d� � ��2
0 64 d� � 32��4
0��16�y2
0�x 2 � y 2� dx dy � ��2
0�4
0r3 dr d�
30.
�43
��R2 � b2�32
�83
�R2 � b2�32��2
0 d�
� �83�
�2
0��R2 � r2�32�
R
b d�
V � 8��2
0�R
b
�R2 � r2r dr d� 32.
The polar region is given by andHence,
x1
1
2
3
4
4, )( 12/ 138/ 13
8/ 13
x y2 2+ =1623
xy =
θ
y
�arctan�32�
0�4
0�r cos ���r sin ��r dr d� �
28813
0 ≤ � ≤ 0.9828.0 ≤ r ≤ 4
tan � �12�13
8�13�
32
⇒ � � 0.9828
18.
� �12
x3�3
0�
272
�32�
3
0 x2 dx
� �3
0�xy �
12
y2�x
0 dx
V � �3
0�x
0�x � y� dy dx 20. Matches (c)
x
y
2
2
1
2
3
z
Review Exercises for Chapter 13 401
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36.
y ��Ix
m��16,384k315
128k15��128
21
x ��Iy
m��512k105
128k15��4
7
m � �R��x, y� dA � �2
0�4�x2
0ky dy dx �
12815
k
I0 � Ix � Iy �16,384k
315�
512k105
�17,920
315k �
5129
k
Iy � �R�x2 �x, y� dA � �2
0�4�x2
0kx2y dy dx �
512105
k
Ix � �R�y2 �x, y� dA � �2
0�4�x2
0ky3 dy dx �
16,384315
k
38.
� 6�2 � ln�4 � 3�2� �9�2
2� ln�2 �
�26
�5�2
3� ln�2�2 � 3�
� �12
�4�18 � 2 ln�4 � �18 �� �1
12�18�18 �� � �ln�2 �
2�212 �
� �12
�2y�2 � 4y2 � 2 ln�2y � �2 � 4y2�� �1
12�2 � 4y2�32�2
0
S � �2
0�2
y
�2 � 4y2 dx dy � �2
0�2�2 � 4y2 � y�2 � 4y2 dy
�1 � � fx�2 � � fy�2 � �2 � 4y2
fx � �1, fy � �2y
R � ��x, y�: 0 ≤ x ≤ 2, 0 ≤ y ≤ x�
f �x, y� � 16 � x � y2
34.
y �Mx
m�
17kh2L80
12
7khL�
51h140
x �My
m�
5khL2
24
127khL
�5L14
�kh2
�L
0�2x �
x2
L�
x3
L2dx �kh2 �x2 �
x3
3L�
x4
4L2�L
0
�kh2
5L2
12�
5khL2
24
My � k�L
0��h2��2��xL���x2L2�
0x dy dx
�kh2
8
17L10
�17kh2L
80 �
kh2
8 �4x �2x2
L�
x3
L2 �x 4
2L3 �x5
5L4�L
0
�kh2
8 �L
0 �4 �
4xL
�3x2
L2 �2x3
L3 �x4
L4�dx
�kh2
8 �L
0�2 �
xL
�x2
L22
dx
Mx � k�L
0��h2��2��xL���x2L2�
0y dy dx
x
h
L
y = 2 − −h2
xL
x2
L2( (
y m � k�L
0��h2��2��xL���x2L2�
0dy dx �
kh2
�L
0�2 �
xL
�x2
L2dx �7khL12
402 Chapter 13 Multiple Integration
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44.
� ��2
0��5 � arctan 5���cos ���
�2
0 d� �
�
2�5 � arctan 5�
� ��2
0��2
0� � arctan �
5
0 sin � d� d�
�5
0��25�x2
0��25�x2�y2
0
11 � x2 � y2 � z2 dz dy dx � ��2
0��2
0�5
0
2
1 � 2 sin � d d� d�
46. �2
0��4�x2
0��4�x2�y2
0 xyz dz dy dx �
43
48.
� 8�4� � 2 sin 2� �14
sin3 � cos � �34�
12
� �14
sin 2���2
0�
29�
2
� 2��2
0�32 sin2 � � 4 sin4 �� d� � 8��2
0�8 sin2 � � sin4 �� d�
� 2��2
0�2 sin �
0 r�16 � r2� dr d�V � 2��2
0�2 sin �
0�16�r2
0 r dz dr d�
50.
z �Mxy
m�
�kc2a4162kca33
�3�ca
32
y �Mxz
m�
�kca482kca33
�3�a16
x � 0
Mxy � 2k��2
0�a
0�cr sin �
0rz dz dr d� � kc2��2
0�a
0r3 sin2 � dr d� �
14
kc2 a4��2
0sin2 � d� �
116
�kc2a4
Mxz � 2k��2
0�a
0�cr sin �
0r2 sin � dz dr d� � 2kc��2
0�a
0r3 sin2 � dr d� �
12
kca4��2
0sin2 � d� �
18
�kca4
m � 2k��2
0�a
0�cr sin �
0r dz dr d� � 2kc��2
0�a
0r 2 sin � dr d� �
23
kca3��2
0sin � d� �
23
kca3
40. (a) Graph of
over region R
x
y50
50
50
R
z
� 25 �1 � e��x2�y2�1000 cos2 �x2 � y 2
1000 �f �x, y� � z
(b) Surface area
Using a symbolic computer program, you obtain surfacearea sq. ft.� 4,540
� �R��1 � fx�x, y�2 � fy�x, y�2 dA
42. �12�
2�
0�2
0r5 dr d� �
163 �
2�
0d� �
32�
3 �2
�2��4�x2
��4�x2 �(x2�y2)�2
0�x 2 � y 2� dz dy dx � �2�
0�2
0�r 22
0r3 dz dr d�
Review Exercises for Chapter 13 403
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52.
z �Mxy
m� �
81�
4
1162�
� �18
� �2�
0�3
0 �1
2r 3 �
92
r� dr d� � �2�
0 �1
8r 4 �
94
r 2�3
0
d� � ��818
��2�
0
� �814
�
Mxy � �2�
0�3
0�4
��25�r2
zr dz dr d� � �2�
0�5
3��25�r2
��25�r2
zr dz dr d� � �2�
0�3
0�8 �
12
�25 � r2��r dr d� � 0
x � y � 0 by symmetry
�500�
3� 2���
13
�25 � r2�32 � 2r 2�3
0
�500�
3� 2���
643
� 18 �1253 � �
500�
3�
14�
3� 162�
m �500�
3� �3
0�2�
0��25�r2
4r dz d� dr �
500�
3� �3
0�2�
0�r�25 � r2 � 4r�d� dr
54.
�4k�a6
9
Iz � k��
0�2�
0�a
02 sin2 ���2 sin � d d� d�
56.
�815
�a
� �a
�a
��1�z2�a2
��1�z2�a2
��1�y2�z2�a2
��1�y2�z2�a2
�x2 � y 2� dx dy dz
Iz � ��Q��x2 � y2�dV
x2 � y2 �z2
a2 � 1 58.
Since represents a paraboloid with vertexthis integral represents the volume of the solid
below the paraboloid and above the semi-circlein the xy-plane.y � �4 � x2
�0, 0, 1�,z � 1 � r 2
��
0�2
0�1�r2
0r dz dr d�
60.
� �2u���2v� � �2u��2v� � �8uv
��x, y���u, v� �
�x�u
�y�v
��y�u
�x�v
62.
Boundary in xy-plane Boundary in uv-plane
� 4 arctan 5 � � � 4 arctan v�5
1
� �5
1
41 � v2 dv � �5
1�5
1
11 � v2 du dv �
R� x
1 � x2y2 dA � �5
1�5
1
u1 � u2�vu�2 �1
u du dv
v � 5xy � 5
v � 1xy � 1
u � 5x � 5
u � 1x � 1
x � u, y �vu ⇒ u � x, v � xy
y = 1x
y = 1x
x = 5
x = 1
1
1
2
3
4
5
4 5
y
x
��x, y���u, v� �
�x�u
�y�v
��x�v
�y�u
� 1�1u � 0 �
1u
404 Chapter 13 Multiple Integration
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Problem Solving for Chapter 13
2.
��a2 � b2 � c2
c A�R�
��a2 � b2 � c2
c �
R� dA
S � �R��1 �
a2
c2 �b2
c2 dA
�1 � fx2 � fy
2 ��1 �a2
c2 �b2
c2
fx � �ac, fy � �
bc
z �1
c�d � ax � by� Plane 4.
The distribution is not uniform. Less water in region ofgreater area.
In one hour, the entire lawn receives
�2�
0�10
0� r
16�
r2
160�r dr d� �125�
12� 32.72 ft3.
B � �2�
0�10
9� r
16�
r2
160�r dr d� �523�
960� 1.71 ft3
A: �2�
0�5
4� r
16�
r2
160�r dr d� �1333�
960� 4.36 ft3
10. Let
Let
(PS #9) � 2���
4 � ���
2 � 2��
0u2e�u2 du �1
0
�ln�1x� dx � ��
0u e�u2�2u du�
u � �v, u2 � v, 2u du � dv.
�1
0
�ln�1x� dx � �0
�
�v ��e�v� dv � ��
0
�ve�v dv
ev �1x, x � e�v, dx � �e�v dv
v � ln�1x�, dv � �
dxx
.
12. Essay 14. The greater the angle between the given plane and the xy-plane, the greater the surface area. Hence:
z2 < z1 < z4 < z3
6. (a)
(b) V � �2�
0��4
0�2�2
2 sec �2 sin � d d� d� �
8�
3�4�2 � 5�
V � �2�
0�2
0��8�r2
2r dz dr d� �
8�
3�4�2 � 5�
8. Volume � 5 � 6 � 5 � 5�4 � 84 m3
Problem Solving for Chapter 13 405
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C H A P T E R 1 3Multiple Integration
Section 13.1 Iterated Integrals and Area in the Plane . . . . . . . . . . . . . 133
Section 13.2 Double Integrals and Volume . . . . . . . . . . . . . . . . . . . 137
Section 13.3 Change of Variables: Polar Coordinates . . . . . . . . . . . . . 143
Section 13.4 Center of Mass and Moments of Inertia . . . . . . . . . . . . . 146
Section 13.5 Surface Area . . . . . . . . . . . . . . . . . . . . . . . . . . . 153
Section 13.6 Triple Integrals and Applications . . . . . . . . . . . . . . . . . 157
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates . . . . 162
Section 13.8 Change of Variables: Jacobians . . . . . . . . . . . . . . . . . . 166
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 169
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174
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133
C H A P T E R 1 3Multiple Integration
Section 13.1 Iterated Integrals and Area in the PlaneSolutions to Odd-Numbered Exercises
1. �x
0 �2x � y� dy � �2xy �
12
y 2�x
0�
32
x2 3. �2y
1 yx dx � �y ln x�
2y
1� y ln 2y � 0 � y ln 2y
5. ��4�x2
0 x2y dy � �1
2x2y 2��4�x2
0�
4x2 � x 4
2
7. �12
y �ln2y � ln2ey �y2
��ln y�2 � y 2�y
ey y ln x
x dx � �1
2y ln2 x�
y
ey
9.
u � y, du � dy, dv � e�y x dy, v � �xe�yx
�x3
0 ye�yx dy � ��xye�yx�
x3
0� x �x3
0 e�yx dy � �x 4 e�x2
� �x2e�yx�x3
0� x2�1 � e�x2
� x2e�x2�
11. �1
0�2
0 �x � y� dy dx � �1
0 �xy �
12
y 2�2
0 dx � �1
0 �2x � 2� dx � �x2 � 2x�
1
0� 3
13. �1
0�x
0 �1 � x2 dy dx � �1
0 �y�1 � x2�
x
0 dx � �1
0 x�1 � x2 dx � ��
12 �
23��1 � x2 �32�
1
0�
13
15.
� �2
1 �64
3� 8y 2 � 4� dy �
43
�2
1 �19 � 6y 2� dy � �4
3�19y � 2y3��
2
1�
203
�2
1�4
0 �x2 � 2y 2 � 1� dx dy � �2
1 �1
3x3 � 2xy 2 � x�
4
0 dy
17.
� �1
0 �1
2�1 � y 2� � y�1 � y 2� dy � �1
2y �
16
y3 �12 �
23��1 � y 2�32�
1
0�
23
�1
0��1�y 2
0 �x � y� dx dy � �1
0 �1
2x2 � xy��1�y 2
0 dy
19. �2
0��4�y2
0
2�4 � y2
dx dy � �2
0 � 2x�4 � y2�
�4�y2
0 dy � �2
0 2 dy � �2y�
2
0� 4
21.
�14
��2
0 �� � � cos 2�� d� �
14
�� 2
2� �1
4 cos 2� �
�
2 sin 2���
�2
0�
�2
32�
18
��2
0�sin �
0 �r dr d� � ��2
0 ��
r2
2 �sin �
0 d� � ��2
0 12
� sin2 � d�
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23. ��
1�1x
0 y dy dx � ��
1 �y2
2 �1x
0 dx �
12
��
1 1x2 dx � ��
12x�
�
1� 0 �
12
�12
25.
Diverges
��
1��
1 1xy
dx dy � ��
1 �1
y ln x�
�
1 dy � ��
1 �1
y��� �
1y
�0�� dy
27.
A � �3
0�8
0 dx dy � �3
0�x�
8
0 dy � �3
0 8 dy � �8y�
3
0� 24
A � �8
0�3
0 dy dx � �8
0�y�
3
0 dx � �8
0 3 dx � �3x�
8
0� 24
x
4
6
8
2 4 6 8
2
y
29.
� �4
0 �x��4�y
0 dy � �4
0 �4 � y dy � ��4
0 �4 � y�12��1� dy � ��
23
�4 � y�32�4
0�
23
�8� �163
A � �4
0��4�y
0 dx dy
� �4x �x3
3 �2
0�
163
� �2
0 �4 � x2� dx
x−1 1
1
2
2
3
4
3
y x= 4 − 2
y
A � �2
0�4�x2
0 dy dx � �2
0 �y�
4�x2
0 dx
31.
� �12
y2 � 2y �23
�4 � y�32�3
0� �4
3�4 � y�32�
4
3�
92
� �3
0 �y � 2 � �4 � y� dy � 2�4
3 �4 � y dy
� �3
0 �x�
y�2
��4�y dy � 2�4
3 �x��4�y
0 dy
A � �3
0�y�2
��4�y dx dy � 2�4
3��4�y
0 dx dy
� �2x �12
x2 �13
x3�1
�2�
92
� �1
�2 �2 � x � x2� dx
� �1
�2 �4 � x2 � x � 2� dx
� �1
�2�y�4�x2
x�2 dx
x−2 −1 1
1
2
2
3
y x= + 2
y x= 4 − 2
(1, 3)
y
A � �1
�2�4�x2
x�2 dy dx 33.
Integration steps are similar to those above.
x
2
3
4
1 2 3 4
1
y x= (2 )− 2
y
�4
0 ��2��y �2
0 dx dy �
83
� �4x �83
x�x �x2
2 �4
0�
83
� �4
0 �4 � 4�x � x� dx
�4
0��2��x �2
0 dy dx � �4
0 �y��2��x �2
0 dx
134 Chapter 13 Multiple Integration
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35.
x
2
3
4
1 2 3 4 5
1
−1
y x= 5 −y x= 23
y
� �2
2 �5 �
5y2 � dy � �5y �
54
y 2�2
0� 5
� �2
0 �5 � y �
3y2 � dy
� �2
0�x�
5�y
3y2 dy
A � �2
0�5�y
3y2 dx dy
� �13
x2�3
0� �5x �
12
x2�5
3� 5
� �3
0 2x3
dx � �5
3 �5 � x� dx
� �3
0 �y�
2x3
0 dx � �5
3�y�
5�x
0 dx
A � �3
0�2x3
0 dy dx � �5
3�5�x
0 dy dx 37.
Therefore,
Therefore, Integration steps are similar to thoseabove.
a
b
x
bay a x= 2 2−
y
A � �ab.
A4
� �b
0��ab��b2�y2
0 dx dy �
�ab4
A � �ab.
��ab
4
�ab2
��2
0 �1 � cos 2�� d� � �ab
2 �� �
12
sin 2����2
0
�x � a sin �, dx � a cos � d��
�ba�
a
0 �a2 � x2 dx � ab��2
0 cos2 � d�
A4
� �a
0��ba��a2�x2
0 dy dx � �a
0 �y�
�ba��a2�x2
0 dx
39.
1 2 3 4
1
2
3
x
y
� �4
0�4
x
f �x, y� dy dx
�4
0�y
0 f �x, y� dx dy, 0 ≤ x ≤ y, 0 ≤ y ≤ 4 41.
−2 −1 1 2
−1
3
1
y
x
� �2
0��4�y2
��4�y2 dx dy
�2
�2��4�x2
0f �x, y�dy dx, 0 ≤ y ≤ �4 � x2, �2 ≤ x ≤ 2
43.
1
2
4
6
8
2 3
y
x
� �ln 10
0�10
exf �x, y�dy dx
�10
1�ln y
0f �x, y�dx dy, 0 ≤ x ≤ ln y, 1 ≤ y ≤ 10 45.
x−2 −1 1 2
2
3
4
y
� �1
0��y
��y f �x, y� dx dy
�1
�1�1
x2
f �x, y� dy dx, x2 ≤ y ≤ 1, 1 ≤ x ≤ 1
Section 13.1 Iterated Integrals and Area in the Plane 135
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47.
31 2
3
2
1
x
y
�1
0�2
0 dy dx � �2
0�1
0 dx dy � 2 49.
x
1
1− 1
y
�1
0��1�y2
��1�y2 dx dy � �1
�1 ��1�x2
0 dy dx �
�
2
51.
x
2
3
1 2 3 4
1
−1
y
�2
0�x
0 dy dx � �4
2�4�x
0 dy dx � �2
0�4�y
y
dx dy � 4 53.
21
2
1
x
y
�2
0�1
x2 dy dx � �1
0�2y
0 dx dy � 1
55.
2
2
1
1
3
x
(1, 1)
yx = y
x = y2
�1
0� 3�y
y2
dx dy � �1
0��x
x3
dy dx �5
12
57. The first integral arises using vertical representative rectangles. The second two integrals arise using horizontal representative rectangles.
x5
5 (5, 5)
(0, )5 2y x= 50 − 2
y x=
y
�15625
24�
�5
0�y
0 x2y 2 dx dy � �5�2
5��50�y 2
0 x2y 2 dx dy � �5
0 13
y5 dy � �5�2
5 13
�50 � y2�32 y2 dy �15625
18� �15625
18� �
1562518 �
�15625
24�
�5
0��50�x2
x
x2y 2 dy dx � �5
0 �1
3x2�50 � x2�32 �
13
x5� dx
136 Chapter 13 Multiple Integration
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Section 13.2 Double Integrals and Volume
59.
�12�
2
0 �1 � y3 y2 dy � �1
2�
13
�23
�1 � y3�3�2�2
0�
19
�27� �19
�1� �269
�2
0�2
x
x�1 � y3 dy dx � �2
0�y
0 x�1 � y3 dx dy � �2
0 ��1 � y3 �
x2
2 �y
0 dy
61.
� �1
0 x sin �x2� dx � ��
12
cos �x2��1
0� �
12
cos 1 �12
�1� �12
�1 � cos 1� 0.2298
�1
0�1
y
sin �x2� dx dy � �1
0�x
0 sin�x2� dy dx � �1
0 �y sin �x2��
x
0 dx
63. �2
0�2x
x2 �x3 � 3y 2� dy dx �
1664105
15.848 65. �4
0�y
0
2�x � 1��y � 1� dx dy � �ln 5�2 2.590
67. (a)
(b)
(c) Both integrals equal 67520�693 97.43
�8
0�x1�3
x2�32 �x2y � xy 2� dy dx
x � 4�2y ⇔ x2 � 32y ⇔ y �x2
32
x2 4 6 8
−2
2
4
(8, 2)x y= 3
x y= 4 2
yx � y3 ⇔ y � x1�3
69. �2
0�4�x2
0 exy dy dx 20.5648 71. �2�
0�1�cos �
06r2 cos � dr d� �
15�
2
75. The region is a rectangle. 77. True
73. An iterated integral is a double integral of a function of two variables. First integrate with respect to one variablewhile holding the other variable constant. Then integrate with respect to the second variable.
For Exercise 1–3, and the midpoints of the squares are
12
, 12�, 3
2,
12�, 5
2,
12�, 7
2,
12�, 1
2,
32�, 3
2,
32�, 5
2,
32�, 7
2,
32�.
x
2
3
4
1 2 3 4
1
y�xi � �yi � 1
1.
�4
0�2
0 �x � y� dy dx � �4
0 �xy �
y 2
2 �2
0 dx � �4
0 �2x � 2� dx � �x2 � 2x�
4
0� 24
�8
i�1 f �xi, yi� �xi �yi � 1 � 2 � 3 � 4 � 2 � 3 � 4 � 5 � 24
f �x, y� � x � y
Section 13.2 Double Integrals and Volume 137
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3.
�4
0 �2
0 �x2 � y 2� dy dx � �4
0 �x2y �
y3
3 �2
0 dx � �4
0 2x2 �
83� dx � �2x3
3�
8x3 �
4
0�
1603
�8
i�1 f �xi, yi� �xi �yi �
24
�104
�264
�504
�104
�184
�344
�584
� 52
f �x, y� � x2 � y 2
5.
Using the corner of the ith square furthest from the origin, you obtain 272.
� 400
�4
0�4
0
f �x, y�dy dx �32 � 31 � 28 � 23� � �31 � 30 � 27 � 22� � �28 � 27 � 24 � 19� � �23 � 22 � 19 � 14�
7.
� 8
� �2x � x2�2
0
� �2
0 �2 � 2x� dx
31 2
3
1
2
x
y
�2
0�1
0 �1 � 2x � 2y� dy dx � �2
0 �y � 2xy � y 2�
1
0 dx
9.
� 36
� �92
y �32
y 2 �5
24y3�
6
0
� �6
0 9
2� 3y �
58
y2� dy
642
6
4
2
x
(3, 6)
y �6
0�3
y�2 �x � y� dx dy � �6
0 �1
2x2 � xy�
3
y�2 dy
11.
� ��23
�a2 � x2�3�2�a
�a� 0
� �a
�a
2x�a2 � x2 dx
a
xa−a
−a
y
�a
�a��a2�x2
��a2�x2
�x � y� dy dx � �a
�a
�xy �12
y2��a2�x2
��a2�x2 dx
13.
� �254
x2�3
0�
2254
�252 �
3
0 x dx
� �3
0 �1
2x y 2�
5
0 dx
x2 5
3
3
5
4
2
41
1
y �5
0�3
0 xy dx dy � �3
0�5
0 x y dy dx
138 Chapter 13 Multiple Integration
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15.
� �12
ln 52�x�
2
0� ln
52
�12
ln 52
�2
0 dx
�12
�2
0 �ln 5x2 � ln 2x2� dx
�12
�2
0 �ln�x2 � y 2��
2x
x dx
x
2
3
4
1 2 3 4
1
x = 2
y x=
y x= 2
y
�2
0�y
y�2
yx2 � y 2 dx dy � �4
2�2
y�2
yx2 � y 2 dx dy � �2
0�2x
x
y
x2 � y 2 dy dx
17.
�2625
� ��1
0�ln x��4 � x2�2 � �4 � x�2�� dx
� ��1
0�ln x � y2�
4�x2
4�x dx
(1, 3)
1
1
2
3
4
3 4
y
x
�4
3��4�y
4�y
�2y ln x dx dy � �1
0�4�x2
4�x
�2y ln x dy dx
19.
� �2518 9y �
13
y3��3
0� 25
�2518
�3
0 �9 � y 2� dy
� �3
0 �1
2x2��25�y2
4y�3 dy
x2 5
3
3
5
4
2
41
1
x y= 25 − 2
(4, 3)x y= 4
3
y
�4
0�3x�4
0 x dy dx � �5
4��25�x2
0 x dy dx � �3
0��25�y 2
4y�3 x dx dy
21.
� �4
0 dx � 4
x
2
3
4
1 2 3 4
1
y
�4
0�2
0 y2
dy dx � �4
0 �y2
4 �2
0 dx
23.
� 8 �86
�83
� 4
� �2y2 �y3
6�
y3
3 �2
0
� �2
0 4y �
y2
2� y2�dy
1
1
2
2
y = x
y
x
�2
0�y
0�4 � x � y�dx dy � �2
0�4x �
x2
2� xy�
y
0 dy
Section 13.2 Double Integrals and Volume 139
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25.
� 12
� � 118
x3 � x2 � 6x�6
0
� �6
0 1
6x2 � 2x � 6� dx
x1
1
−12
2
3
3
4
4
5
5
6
23x + 4y = −
y
�6
0���2�3�x�4
0 12 � 2x � 3y
4 � dy dx � �6
0 �1
4 12y � 2xy �32
y 2����2�3�x�4
0 dx
27.
�38
� �y 2
2�
y 4
8 �1
0
� �1
0 y �
y3
2 � dy
�1
0�y
0 �1 � xy� dx dy � �1
0 �x �
x2y2 �
y
0 dy
x1
1
y x=
y
29. � �
0
1�x � 1�2 dx � ��
1�x � 1��
0� 1� �
0 ��
1�x � 1�2�y � 1��
0 dx�
0�
0
1�x � 1�2�y � 1�2 dy dx
31. 4�2
0��4�x2
0 �4 � x2 � y 2� dy dx � 8�
33.
x1
1y x=
y
� �18
x 4�1
0�
18
� �1
0 �1
2xy 2�
x
0 dx �
12�
1
0 x3 dx
V � �1
0�x
0 xy dy dx 35.
x−1 1
1
2
2
3
4
3
y
� �4x3
3 �2
0�
323
� �2
0 �x2y�
4
0 dx � �2
0 4x2 dx
V � �2
0�4
0 x2 dy dx
37. Divide the solid into two equal parts.
� ��23
�1 � x2�3�2�1
0�
23
� 2�1
0 x�1 � x2 dx
� 2�1
0 �y�1 � x2�
x
0 dx
V � 2�1
0�x
0 �1 � x2 dy dx
x1
1
y x=
y
140 Chapter 13 Multiple Integration
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39.
x
2
1 2
1
y x= 4 − 2
y
� ��13
�4 � x2�3�2 � 2x �16
x3�2
0�
163
� �2
0 x�4 � x2 � 2 �
12
x2� dx
� �2
0 �xy �
12
y2��4�x2
0 dx
V � �2
0��4�x2
0 �x � y� dy dx 41.
1
−1
−1 1x
x y2 2+ = 4
y
� 8�
� 4�16 �
4� �323 3�
16��
� 4���2
0 16 cos2 � �
323
cos4 �� d�
x � 2 sin � � 4�2
0 �x2�4 � x2 �
13
�4 � x2�3�2� dx,
V � 4�2
0��4�x2
0 �x2 � y 2� dy dx
43. V � 4�2
0��4�x2
0
�4 � x2 � y 2� dy dx � 8� 45. V � �2
0��0.5x�1
0
21 � x2 � y 2 dy dx 1.2315
47. f is a continuous function such that over a region R of area 1. Let the minimum value of f over Rand the maximum value of f over R. Then
Since and we have
Therefore, 0 ≤ �R� f �x, y� dA ≤ 1.
0 ≤ f �m, n��1� ≤ �R� f �x, y� dA ≤ f �M, N��1� ≤ 1.0 ≤ f �m, n� ≤ f �M, N� ≤ 1,�
R� dA � 1
f �m, n��R� dA ≤ �
R� f �x, y� dA ≤ f �M, N��
R� dA.
f �M, N� �f �m, n� �0 ≤ f �x, y� ≤ 1
49.
x1
1
12
12
y x= 2
y
� 1 � e�1�4 0.221
� �e�1�4 � 1
� ��e�x2�1�2
0
� �1�2
0 2xe�x2 dx
�1
0�1�2
y�2 e�x2 dx dy � �1�2
0�2x
0 e�x2 dy dx 51.
x
2
1
2
y x= cos
π π
y
� �12
�23
�1 � sin2 x�3�2���2
0�
13
�2�2 � 1�
� ���2
0 �1 � sin2 x�1�2 sin x cos x dx
� ���2
0�cos x
0 sin x�1 � sin2 x dy dx
�1
0�arccos y
0 sin x�1 � sin2 x dx dy
Section 13.2 Double Integrals and Volume 141
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53. Average �18�
4
0�2
0 x dy dx �
18�
4
0 2x dx � �x2
8 �4
0� 2 55.
� �14
83
y �23
y3��2
0�
83
�14
�2
0 �x3
3� xy 2�
2
0 dy �
14
�2
0 8
3� 2y 2� dy
Average �14
�2
0�2
0 �x2 � y 2� dx dy
57. See the definition on page 946. 59. The value of would be kB.�R
� f �x, y� dA
61.
�1
1250 �325
300 ��100y 0.4� x1.6
1.6�250
200 dy �
128,844.11250
�325
300 y 0.4 dy � 103.0753�y1.4
1.4�325
300 25,645.24
Average �1
1250 �325
300�250
200 100x 0.6y 0.4 dx dy
63. for all and
P�0 ≤ x ≤ 2, 1 ≤ y ≤ 2� � �2
0�2
1
110
dy dx � �2
0
110
dx �15
.
�
��
�
f �x, y� dA � �5
0�2
0
110
dy dx � �5
0 15
dx � 1
�x, y�f �x, y� ≥ 0
65. for all and
P�0 ≤ x ≤ 1, 4 ≤ y ≤ 6� � �1
0�6
4
127
�9 � x � y� dy dx � �1
0
227
�4 � x� dx �7
27.
� �3
0
127�9y � xy �
y2
2 �6
3 dx � �3
0 1
2�
19
x� dx � �x2
�x2
18�3
0� 1
�
��
�
f �x, y� dA � �3
0�6
3
127
�9 � x � y� dy dx
�x, y�f �x, y� ≥ 0
67. Divide the base into six squares, and assume the height at the center of each square is the height of the entire square.
Thus,
x
y
(15, 5, 6)(15, 15, 7)
(5, 5, 3)
(5, 15, 2)(25, 5, 4)
(25, 15, 3)
7
20
30
z
V �4 � 3 � 6 � 7 � 3 � 2��100� � 2500m3.
69.
(a) 1.78435
(b) 1.7879
m � 4, n � 8�1
0�2
0 sin �x � y dy dx 71.
(a) 11.0571
(b) 11.0414
m � 4, n � 8�6
4�2
0 y cos �x dx dy
142 Chapter 13 Multiple Integration
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Section 13.3 Change of Variables: Polar Coordinates
73.
Matches d.
xy
55
16(4, 0, 16)(4, 4, 16)
(0, 4, 0)(4, 0, 0)
(4, 4, 0)
zV � 125 75. False
V � 8�1
0��1�y2
0 �1 � x2 � y 2 dx dy
77.
� ��12
e t2�1
0� �
12
�e � 1� �12
�1 � e�
� ��1
0�t
0 e t2 dx dt � ��1
0 tet2 dt
x1
1
t
Average � �1
0 f �x� dx � �1
0�x
1 e t2 dt dx � ��1
0�1
x
et2 dt dx
1. Rectangular coordinates 3. Polar coordinates
5. R � �r, ��: 0 ≤ r ≤ 8, 0 ≤ � ≤ � 7. R � �r, ��: 0 ≤ r ≤ 3 � 3 sin �, 0 ≤ � ≤ 2� Cardioid
9.
� ��216 cos ��2�
0� 0
� �2�
0 216 sin � d�
4
0
2π
�2�
0�6
0 3r2 sin � dr d� � �2�
0 �r3 sin ��
6
0 d�
11.
�5�5�
6
� �5�53
����2
0
3210
2π
���2
0�3
2 �9 � r2 r dr d� � ���2
0 ��
13
�9 � r2�3�2�3
2 d�
13.
�332
� 2 �98
� �18
�2 � sin � � � cos � �12
���12
cos � � sin � �12
� �18
sin2 ����2
0
� ���2
0 12
��1 � sin ��2 d�
210
2π ���2
0�1�sin �
0 �r dr d� � ���2
0 ��r2
2 �1�sin �
0 d�
Section 13.3 Change of Variables: Polar Coordinates 143
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15. � �a3
3��cos ���
��2
0�
a3
3�a
0��a2�y2
0 y dx dy � ���2
0�a
0 r2 sin � dr d� �
a3
3 ���2
0 sin � d�
19. � ��4 cos6 �
6 ���2
0�
23�2
0��2x�x2
0 xy dy dx � ���2
0�2 cos �
0 r3 cos � sin � dr d� � 4���2
0 cos5 � sin d�
17. �3
0��9�x2
0 �x2 � y 2�3�2 dy dx � ���2
0�3
0 r 4 dr d� �
2435
���2
0 d� �
243�
10
21.
�4�2�
3
� ���4
0 16�2
3 d�
02 31
π2
�2
0�x
0 �x2 � y 2 dy dx � �2�2
2 ��8�x2
0 �x2 � y 2 dy dx � ���4
0�2�2
0 r2 dr d�
23.
�83
���2
0 �cos � � sin �� d� � �8
3�sin � � cos ���
��2
0�
163
�2
0��4�x2
0 �x � y� dy dx � ���2
0�2
0 �r cos � � r sin ��r dr d� � ���2
0�2
0 �cos � � sin ��r2 dr d�
25.
� ���4
0 32
� d� � �3�2
4 ���4
0�
3�2
64
� ���4
0�2
1 �r dr d�
021
2 2( ),
2
1
2
1,( (
π2
�1��2
0��4�y2
�1�y2
arctan yx dx dy � ��2
1��2 ��4�y2
y
arctan yx dx dy
27.
� ��1
16 cos 2��
��2
0�
18
�18�
��2
0 sin 2� d� �
12�
��2
0�1
0 r3 sin 2� dr d�
V � ���2
0�1
0 �r cos ���r sin ��r dr d�
29. V � �2�
0�5
0 r2 dr d� �
250�
3
31.
�1283 ���2
0 �1 � sin ��1 � cos2 ��� d� �
1283 �� � cos � �
cos3 �3 �
��2
0�
649
�3� � 4�
V � 2���2
0�4 cos �
0 �16 � r2 r dr d� � 2���2
0 ��
13
��16 � r2�3�4 cos �
0 d� � �
23
���2
0 �64 sin3 � � 64� d�
144 Chapter 13 Multiple Integration
http://librosysolucionarios.net
33.
One-half the volume of the hemisphere is
a � �4�4 � 2 3�2 � � 2�4 � 2 3�2 � 2.4332
a2 � 16 � 322�3 � 16 � 8 3�2
16 � a2 � 322�3
�16 � a2�3�2 � 32
2�
3�16 � a2�3�2 �
64�
3
�64���3.
V � �2�
0�4
a
�16 � r2 r dr d� � �2�
0 ��
13
��16 � r2�3�4
a d� �
13
��16 � a2�3�2��
35. Total Volume
Let c be the radius of the hole that is removed.
⇒ diameter � 2c � 1.2858
c � 0.6429
c2 � 0.41331
�c2
4� �0.10333
⇒ e�c2�4 � 0.90183
� �2�
0 �50�e�c2�4 � 1� d� ⇒ 30.84052 � 100��1 � e�c2�4�
� �2�
0 ��50e�r 2�4�
c
0 d�
110
V � �2�
0�c
0 25e�r2�4 r dr d�
� �1 � e�4� 100� � 308.40524
� �2�
0 �50�e�4 � 1� d�
� �2�
0 ��50e�r2�4�
4
0 d�
� V � �2�
0�4
0 25e�r2�4 r dr d�
37. � �9�� �12
sin 2� ��
0� 9�� ��
018 cos2 � d� � 9��
0 �1 � cos 2�� d�A � ��
0�6 cos �
0 r dr d�
39.
�12
�2�
0 �1 � 2 cos � �
1 � cos 2�
2 d� �12
�� � 2 sin � �12 �� �
12
sin 2� �2�
0�
3�
2
�2�
0�1�cos �
0 r dr d� �
12
�2�
0 �1 � 2 cos � � cos2 �� d�
41. 3 ���3
0�2 sin 3�
0 r dr d� �
32
���3
0 4 sin2 3� d� � 3���3
0 �1 � cos 6�� d� � 3�� �
16
sin 6����3
0� �
43. Let R be a region bounded by the graphs of andand the lines and
When using polar coordinates to evaluate a double integralover R, R can be partitioned into small polar sectors.
� � b.� � ar � g2���,r � g1��� 45. r-simple regions have fixed bounds for
-simple regions have fixed bounds for r.�
�.
Section 13.3 Change of Variables: Polar Coordinates 145
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47. You would need to insert a factor of r because of the nature of polar coordinate integrals. The plane regions would besectors of circles.
r dr d�
49.
�Note: This integral equals ����2
��4 sin �� d����5
0 r�1 � r3 dr��
���2
��4 �5
0 r�1 � r3 sin�� dr d� 56.051
51.
Answer (c)
8� � 12 300
xy4 6
46
16
z Volume � base � height 53. False
Let where R is the circular sectorand Then,
but for all r.r � 1 � 0�R
�r � 1� dA > 0
0 ≤ � ≤ �.0 ≤ r ≤ 6f r, �� � r � 1
55. (a)
(b) Therefore, I � �2�.
I2 � �
��
�
e�x2�y2��2 dA � 4���2
0�
0 e�r2�2 r dr d� � 4���2
0 ��e�r2�2�
0 d� � 4���2
0 d� � 2�
57.
� 2��200,000�e�0.49 � 1� � 400,000�1 � e�0.49� 486,788
� �2�
0�7
0 4000e�0.01r2 r dr d� � �2�
0 ��200,000e�0.01r2�
7
0 d��7
�7��49�x2
��49�x2
4000e�0.01x2�y2� dy dx
59. (a)
(b)
(c) ���3
��4�4 csc �
2 csc � f r dr d�
�2
2��3 ��3x
2 f dy dx � �4��3
2��3x
x
f dy dx � �4
4��3�4
x
f dy dx
x2 5
3
3
5
41
1
(4, 4)
(2, 2)
y x= 3y x=
, 223( (
, 443( (
y�4
2�y
y��3 f dx dy
61. A ��r2
2
2�
�r12
2� ��r1 � r2
2 �r2 � r1� � rr�
Section 13.4 Center of Mass and Moments of Inertia
3.
� �4sin2 �
2 ���2
0� 2
� ���2
04 cos � sin � d�
m � ���2
0�2
0r cos ��r sin ��r dr d� � ���2
0�2
0cos � sin � � r3 dr d�
1. � �4
0
92
x dx � �9x2
4 �4
0� 36 m � �4
0�3
0
xy dy dx � �4
0�xy2
2 �3
0 dx
146 Chapter 13 Multiple Integration
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5. (a)
(c)
x, y� � �23
a, b2�
y �Mx
m�
ka2b2�4ka2b�2
�b2
x �My
m�
ka3b�3ka2b�2
�23
a
My � �a
0�b
0 kx2 dy dx �
ka3b3
Mx � �a
0�b
0 kxy dy dx �
ka2b2
4
m � �a
0�b
0 kx dy dx � k�a
0 xb dx �
12
ka2b
x, y� � �a2
, b2�
y �Mx
m�
kab2�2kab
�b2
x �My
m�
ka2b�2kab
�a2
My � �a
0�b
0 kx dy dx �
ka2b2
Mx � �a
0�b
0 ky dy dx �
kab2
2
m � �a
0�b
0 k dy dx � kab (b)
x, y� � �a2
, 23
b�
y �Mx
m�
kab3�3kab2�2
�23
b
x �My
m�
ka2b2�4kab2�2
�a2
My � �a
0�b
0 kxy dy dx �
ka2b2
4
Mx � �a
0�b
0 ky 2 dy dx �
kab3
3
m � �a
0�b
0 ky dy dx �
kab2
2
7. (a)
—CONTINUED—
x, y� � �b2
, h3�
y �Mx
m�
kbh2�6kbh�2
�h3
�kbh2
12�
kbh2
12�
kbh2
6
Mx � �b�2
0�2hx�b
0 ky dy dx � �b
b�2��2hx�b��b
0 ky dy dx
x �b2
by symmetry
xb
h
y = 2hxb
y = − 2 ( )h x b−b
y m �k2
bh
Section 13.4 Center of Mass and Moments of Inertia 147
http://librosysolucionarios.net
(b)
(c)
y �Mx
m�
kh2b2�12kb2h�4
�h3
x �My
m�
7kb3h�48kb2h�4
�7
12b
�132
kb3h �1196
kb3h �7
48kb3h
My � �b�2
0�2hx�b
0 kx2 dy dx � �b
b�2��2hx�b��b
0 kx2 dy dx
�1
32kh2b2 �
596
kh2b2 �1
12kh2b2
Mx � �b�2
0�2hx�b
0 kxy dy dx � �b
b�2��2hx�b��b
0 kxy dy dx
�112
kb2h �16
kb2h �14
kb2h
m � �b�2
0�2hx�b
0 kx dy dx � �b
b�2��2hx�b��b
0 kx dy dx
y �Mx
m�
kbh3�12kbh2�6
�h2
x �My
m�
kb2h2�12kbh2�6
�b2
My � �b�2
0�2hx�b
0 kxy dy dx � �b
b�2��2hx�b��b
0 kxy dy dx �
kb2h2
12
Mx � �b�2
0�2hx�b
0 ky 2 dy dx � �b
b�2��2hx�b��b
0 ky 2 dy dx �
kbh3
12
m � �b�2
0�2hx�b
0 ky dy dx � �b
b�2��2hx�b��b
0 ky dy dx �
kbh2
6
7. —CONTINUED—
9. (a) The x-coordinate changes by
(b) The x-coordinate changes by x, y� � �a2
� 5, 2b3 �5:
x, y� � �a2
� 5, b2�5:
(c)
y �Mx
m�
b2
x �My
m�
2a2 � 15a � 75�3a � 10�
My � �a�5
5�b
0 kx2 dy dx �
13
ka � 5�3 b �1253
kb
Mx � �a�5
5�b
0kxy dy dx �
14
ka � 5�2b2�
254
kb2
m � �a�5
5�b
0 kx dy dx �
12
ka � 5�2b �252
kb
11. (a)
y �Mx
m�
2a3k3
�2
�a2k�
4a3�
Mx � �a
�a��a2�x2
0 yk dy dx �
2a3k3
m ��a2k
2
x � 0 by symmetry
(b)
y �Mx
m�
a5�
15� � 3216 � 3� �
x �My
m� 0
My � �a
�a��a2�x2
0 kxa � y� y dy dx � 0
Mx � �a
�a��a2�x2
0 ka � y� y 2 dy dx �
a5k120
15� � 32�
m � �a
�a��a2�x2
0 ka � y� y dy dx �
a4k24
16 � 3��
148 Chapter 13 Multiple Integration
http://librosysolucionarios.net
13.
x
2
3
1 2 3 4
1
−1
y x=
y
y �Mx
m�
256k21
�3
32k�
87
x �My
m�
32k1
�3
32k� 3
My � �4
0��x
0 kx2y dy dx � 32k
Mx � �4
0��x
0 kxy 2 dy dx �
256k21
m � �4
0��x
0 kxy dy dx �
32k3
15.
x−1 1
2
y = 11 + x2
y
y �Mx
m�
k8
2 � �� �2
k��
2 � �
4�
Mx � �1
�1�1�1�x2�
0 ky dy dx �
k8
2 � ��
m � �1
�1�1�1�x2�
0 k dy dx �
k�
2
x � 0 by symmetry
17.
x
8
4 8
−8
−4
4
x y= 16 − 2
y
x �My
m�
524,288k105
�15
8192k�
647
My � �4
�4�16�y 2
0 kx2 dx dy �
524,288k105
m � �4
�4�16�y 2
0 kx dx dy �
8192k15
y � 0 by symmetry 19.
xL
1
2
L2
y = sin πxL
y
y �Mx
m�
4kL9�
�4kL
�169�
Mx � �L
0�sin �x�L
0 ky 2 dy dx �
4kL9�
m � �L
0�sin �x�L
0 ky dy dx �
kL4
x �L2
by symmetry
21.
y �Mx
m�
ka32 � �2�6
�8
�a2k�
4a2 � �2�3�
x �My
m�
ka3�26
�8
�a2k�
4a�23�
My � �R� kx dA � ���4
0�a
0 kr2 cos � dr d� �
ka3�26
Mx � �R� ky dA � ���4
0�a
0 kr2 sin � dr d� �
ka32 � �2�6
0a
r a=y x=
π2
m ��a2k
8
Section 13.4 Center of Mass and Moments of Inertia 149
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23.
x
2
1 2
1 y e x= −
y
y �Mx
m�
ke6 � 1�9e6 �
4e4
ke4 � 1� �49�
e6 � 1e6 � e2� 0.45
x �My
m�
ke4 � 5�8e4 �
4e4
ke4 � 1� �e4 � 5
2e4 � 1� 0.46
My � �2
0�e�x
0 kxy dy dx �
k1 � 5e�4�8
Mx � �2
0�e�x
0 ky 2 dy dx �
k9
1 � e�6�
m � �2
0�e�x
0 ky dy dx �
k4
1 � e�4� 25.
01
r = 2 cos 3θ
θ
θ π6
π6
=
= −
π2
x �My
m 1.17k� 3
�k� 1.12
� ���6
���6�2 cos 3�
0 kr2 cos � dr d� 1.17k
My � �R� kx dA
m � �R� k dA � ���6
���6�2 cos 3�
0 kr dr d� �
k�
3
y � 0 by symmetry
27.
y ��Ix
m��bh3
3�
1bh
��h2
3�
h�3
��33
h
x ��Iy
m��b3h
3�
1bh
��b2
3�
b�3
��33
b
Iy � �b
0�h
0 x2dy dx �
b3h3
Ix � �b
0�h
0 y 2 dy dx �
bh3
3
m � bh 29.
x � y ��Ix
m��a4�
4�
1�a2 �
a2
I0 � Ix � Iy �a4�
4�
a4�
4�
a4�
2
Iy � �R� x2 dA � �2�
0�a
0 r3 cos2 � dr d� �
a4�
4
Ix � �R� y 2 dA � �2�
0�a
0 r3 sin2 � dr d� �
a4�
4
m � �a2
31.
x � y ��Ix
m���a4
16�
4�a2 �
a2
I0 � Ix � Iy ��a4
16�
�a4
16�
�a4
8
Iy � �R� x2 dA � ���2
0�a
0 r3 cos2 � dr d� �
�a4
16
Ix � �R� y 2 dA � ���2
0�a
0 r3 sin2 � dr d� �
�a4
16
m ��a2
433.
y ��Ix
m��kab4�4
kab2�2��b2
2�
b�2
��22
b
x ��Iy
m��ka3b2�6
kab2�2��a2
3�
a�3
��33
a
I0 � Ix � Iy �3kab4 � 2kb2a3
12
Iy � k�a
0�b
0 x2y ydy dx �
ka3b2
6
Ix � k�a
0�b
0 y3 dy dx �
kab4
4
m � k�a
0�b
0 y dy dx �
kab2
2
� � ky
150 Chapter 13 Multiple Integration
http://librosysolucionarios.net
35.
y ��Ix
m��32k�3
4k��8
3�
4�6
�2�6
3
x ��Iy
m��16k�3
4k��4
3�
2�3
�2�3
3
I0 � Ix � Iy � 16k
Iy � k�2
0�4�x2
0 x3 dy dx �
16k3
Ix � k�2
0�4�x2
0 xy 2 dy dx �
32k3
m � k�2
0�4�x2
0 x dy dx � 4k
� � kx 37.
y ��Ix
m��16k
1�
332k
��32
��62
x ��Iy
m��512k
5�
332k
��485
�4�15
5
I0 � Ix � Iy �592k
5
Iy � �4
0��x
0 kx3 y dy dx �
512k5
Ix � �4
0��x
0 kxy3 dy dx � 16k
m � �4
0��x
0 kxy dy dx �
32k3
� � kxy
39.
y ��Ix
m��3k
56�
203k
��7014
x ��Iy
m�� k
18�
203k
��30
9
I0 � Ix � Iy �55k504
Iy � �1
0��x
x2
kx3 dy dx �k
18
Ix � �1
0��x
x2 kxy 2 dy dx �
3k56
m � �1
0��x
x2 kx dy dx �
3k20
� � kx
41.
� 2k��b4
8� 0 �
�a2b2
2 � �k�b2
4b2 � 4a2�
� 2k��b
�b
x2�b2 � x2 dx � 2a�b
�b
x�b2 � x2 dx � a2�b
�b
�b2 � x2 dx�
I � 2k�b
�b
��b2�x2
0
x � a�2 dy dx � 2k�b
�b
x � a�2�b2 � x2 dx
43. I � �4
0��x
0 kxx � 6�2 dy dx � �4
0 kx�x x2 � 12x � 36� dx � k�2
9x9�2 �
247
x7�2 �725
x5�2�4
0�
42,752k315
Section 13.4 Center of Mass and Moments of Inertia 151
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45.
� �k4�7a5 �
83
a5 �15
a5 � 2a5� �14
a5�� � a5k�7�
16�
1715�
� �k4
�7a4x �8a2
3x3 �
x5
5� 4a3�x�a2 � x2 � a2 arcsin
xa� �
a2�x2x2 � a2��a2 � x2 � a4 arcsin
xa��
a
0
� �k4
�a
0�7a4 � 8a2x2 � x4 � 8a3�a2 � x2 � 4ax2�a2 � x2 dx
� �k4
�a
0 �a4 � 4a3�a2 � x2 � 6a2a2 � x2� � 4aa2 � x2��a2 � x2 � a4 � 2a2x2 � x4� � a4 dx
� �k4
�a
0 �a4 � 4a3y � 6a2y2 � 4ay3 � y4��a2�x2
0 dx
� �a
0 ��
k4
a � y�4��a2�x2
0 dx I � �a
0 ��a2�x2
0 ka � y�y � a�2 dy dx � �a
0��a2�x2
0 ka � y�3 dy dx
47. will increasey�x, y� � ky. 49.
Both and will increaseyx
�x, y� � kxy.
51. Let be a continuous density function on the planar lamina R.
The movements of mass with respect to the x- and y-axes are
and
If m is the mass of the lamina, then the center of mass is
x, y� � �My
m,
Mx
m �.
My � �R�x �x, y� dA.Mx � �
R�y �x, y�dA
�x, y�
55.
ya � y �Iy
hA�
L2
�L3b�12
L�2�bL� �L3
� �b
0 ��y � L�2� 3
3 �L
0 dx �
L3b12
Iy � �b
0�L
0 �y �
L2�
2
dy dx
y �L2
, A � bL, h �L2
57.
ya �2L3
�L3b�36L2b�6
�L2
�23�
L3x27
�b
8L�2Lxb
�2L3 �
4
�b�2
0�
L3b36
�23
�b�2
0 � L
27� �2Lx
b�
2L3 �
3
� dx
�23�
b�2
0 ��y �
2L3 �
3
�L
2L x�b dx
Iy � 2�b�2
0 �L
2Lx�b
�y �2L3 �2 dy dx
y �2L3
. A �bL2
, h �L3
53. See the definition on page 968
152 Chapter 13 Multiple Integration
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Section 13.5 Surface Area
1.
triangle with vertices
x
2
1 2
1 R
y x= + 2−
y
� �3�2x �x2
2 ��2
0� 6
S � �2
0�2�x
0 3 dy dx � 3 �2
0 �2 � x� dx
1 � � fx �2 � � fy �2 � 3
fy � 2fx � 2,
�0, 0�, �2, 0�, �0, 2�R �
f �x, y� � 2x � 2y 3.
1
−1
−1 1x
y x= 4 − 2
y x= 4− − 2
R
y
S � �2
�2�4�x2
�4�x2
3 dy dx � �2�
0�2
0 3r dr d� � 12�
1 � � fx �2 � � fy �2 � 3
fx � 2, fy � 2
R � �x, y�: x2 � y 2 ≤ 4�
f �x, y� � 8 � 2x � 2y
5.
square with vertices,
� �34
�2x1 � 4x2 � ln �2x � 1 � 4x2���3
0�
34
�637 � ln�6 � 37��
S � �3
0�3
0 1 � 4x2 dy dx � �3
0 31 � 4x2 dx
1 � � fx �2 � � fy �2 � 1 � 4x2
fx � �2x, fy � 0
�0, 0�, �3, 0�, �0, 3�, �3, 3�R �
2
2
31
3
1
x
R
yf �x, y� � 9 � x2
7.
rectangle with vertices
� � 427
�4 � 9x�3 2�3
0�
427
�3131 � 8�
S � �3
0�4
0 4 � 9x
2 dy dx � �3
04�4 � 9x
2 � dx
1 � � fx �2 � � fy �2 �1 � �94�x �
4 � 9x2
fx �32
x1 2, fy � 0
�0, 0�, �0, 4�, �3, 4�, �3, 0�R �
x
2
3
4
1 2 3 4
1
R
yf �x, y� � 2 � x3 2
9.
S � �� 4
0�tan x
0 sec x dy dx � �� 4
0 sec x tan x dx � �sec x�
� 4
0� 2 � 1
1 � � fx �2 � � fy �2 � 1 � tan2 x � sec x
fx � tan x, fy � 0
R � ��x, y�: 0 ≤ x ≤ �
4, 0 ≤ y ≤ tan x�
x
2
1
4
y x= tan
π2π
R
yf �x, y� � ln�sec x�
Section 13.5 Surface Area 153
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11.
S � �1
�1�1�x2
�1�x2
2 dy dx � �2�
0�1
0 2 r dr d� � 2�
1 � � fx �2 � � fy �2 �1 �x2
x2 � y 2 �y 2
x2 � y 2 � 2
fx �x
x2 � y 2, fy �
yx2 � y 2
0 ≤ x2 � y 2 ≤ 1, x2 � y 2 ≤ 1
R � �x, y�: 0 ≤ f �x, y� ≤ 1� 1
1x
x y2 2+ = 1
yf �x, y� � x2 � y 2
13.
� �2�
0�b
0
aa2 � r2
r dr d� � 2�a�a � a2 � b2� S � �b
�b�b2�x2
�b2�x2
a
a2 � x2 � y 2 dy dx
1 � � fx �2 � � fy �2 �1 �x2
a2 � x2 � y 2 �y2
a2 � x2 � y 2 �a
a2 � x2 � y 2
fx ��x
a2 � x2 � y 2, fy �
�ya2 � x2 � y 2
R � �x, y�: x2 � y 2 ≤ b2, b < a�
ab
b
a
−b
−bx
x y b2 2 2+ ≤
yf �x, y� � a2 � x2 � y 2
15.
S � �8
0���3 2�x�12
0 14 dy dx � 4814
1 � � fx �2 � � fy �2 � 14
z � 24 � 3x � 2y
x
8
12
16
124 168
4
y
17.
� 2�2�
0�3
0
525 � r2
r dr d� � 20�
S � 2�3
�3 �9�x2
�9�x2
5
25 � �x2 � y 2� dy dx
1 � � fx �2 � � fy �2 �1 �x2
25 � x2 � y 2 �y 2
25 � x2 � y 2 �5
25 � x2 � y 21
2
2−1
−2
−2 −1 1x
x y2 2+ = 9yz � 25 � x2 � y 2
19.
triangle with vertices
S � �1
0�x
0 5 � 4x2 dy dx �
112
�27 � 55�
1 � � fx �2 � � fy �2 � 5 � 4x2
�0, 0�, �1, 0�, �1, 1�R �
x1
1 y x=
R
yf �x, y� � 2y � x2
154 Chapter 13 Multiple Integration
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21.
1
−1
−1 1x
x y2 2+ = 4
y
� �2�
0�2
0 1 � 4r2 r dr d� �
�1717 � 1��6
S � �2
�2�4�x2
�4�x2
1 � 4x2 � 4y 2 dy dx
1 � � fx �2 � � fy �2 � 1 � 4x2 � 4y 2
fx � �2x, fy � �2y
0 ≤ 4 � x2 � y 2, x2 � y 2 ≤ 4
R � �x, y�: 0 ≤ f �x, y��
f �x, y� � 4 � x2 � y 2 23.
S � �1
0�1
0 �1 � 4x2� � 4y2 dy dx � 1.8616
1 � � fx �2 � � fy �2 � 1 � 4x2 � 4y 2
fx � �2x, fy � �2y
R � �x, y�: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1�
f �x, y� � 4 � x2 � y 2
25. Surface
Matches (e)
x
y5
5
10
z
area > �4� � �6� � 24. 27.
� �1
0 1 � e2x � 2.0035
S � �1
0�1
0 1 � e2x dy dx
1 � � fx �2 � � fy �2 � 1 � e2x
fx � ex, fy � 0
R � �x, y�: 0 ≤ x ≤ 1, 0 ≤ y ≤ 1�
f �x, y� � ex
29.
square with vertices
S � �1
�1 �1
�1 1 � 9�x2 � y�2 � 9�y 2 � x�2 dy dx
fy � �3x � 3y 2 � 3�y 2 � x�fx � 3x2 � 3y � 3�x2 � y�,
�1, �1���1, �1�,��1, 1�,�1, 1�,R �
f �x, y� � x3 � 3xy � y3 31.
S � �2
�2 �4�x2
�4�x2
1 � e�2x dy dx
� 1 � e�2x
1 � fx2 � fy
2 � 1 � e�2x sin2 y � e�2x cos2 y
fx � �e�x sin y, fy � e�x cos y
f �x, y� � e�x sin y
33.
S � �4
0�10
0 1 � e2xy�x2 � y 2� dy dx
1 � � fx �2 � � fy �2 � 1 � y 2e2xy � x2e2xy � 1 � e2xy �x2 � y 2�
fx � yexy, fy � xexy
R � �x, y�: 0 ≤ x ≤ 4, 0 ≤ y ≤ 10�
f �x, y� � exy
Section 13.5 Surface Area 155
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35. See the definition on page 972. 37.
� 16�1
0
x1 � x2
dx � ��16�1 � x2�1 2�1
0� 16
� 16�1
0�x
0
11 � x2
dy dx
S � ��R
1 � fx2 � fy
2 dA
f �x, y� � 1 � x2; fx ��x
12 � x2, fy � 0
39. (a)
(b)
�1
100 �� 2
0�50
0 1002 � r2 r dr d� � 2081.53 ft2
S �1
100 �50
0 �502�x2
0 1002 � x2 � y 2 dy dx
1 � � fx �2 � � fy �2 �1 �y 2
1002 �x2
1002 �1002 � x2 � y 2
100
z � 20 �xy
100
� 30,415.74 ft3
� �10�x50 � x2 � 502 arcsin x
50� �254
x2 �x 4
800�
115
�502 � x2�3 2 � 250x �x3
30�50
0
� �50
0 �20502 � x2 �
x200
�502 � x2� �x5502 � x2 �
502 � x2
10 � dy
V � �50
0�502�x2
0 �20 �
xy100
�x � y
5 � dy dx
41. (a)
where R is the region in the first quadrant
(b)
� limb→25� ��200625 � r2�
b
4�
�
2� 100�609 cm2
� 8�R�
25625 � x2 � y 2
dA � 8 �� 2
0 �25
4
25625 � r2 r dr d�
A � �R� 1 � � fx �2 � � fy �2 dA � 8�
R� 1 �
x2
625 � x2 � y 2 �y 2
625 � x2 � y 2 dA
� 812�609 cm3
� �83
�0 � 609609� ��
2
� �4�� 2
0 �2
3�625 � r2�3 2�
25
4 d�
� 8�� 2
0�25
4 625 � r2 r dr d�
� 8�R� 625 � x2 � y 2 dA
x8 20 24
12
12
20
24
16
8
164
4
R
y
V � �R� f �x, y�
156 Chapter 13 Multiple Integration
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Section 13.6 Triple Integrals and Applications
1.
� �3
0�2
0 �1
2� y � z� dy dz � �3
0 �1
2y �
12
y 2 � yz�2
0 dz � �3z � z2�
3
0� 18
�3
0�2
0�1
0 �x � y � z� dx dy dx � �3
0�2
0 �1
2x2 � xy � xz�
1
0 dy dx
3.
� �1
0�x
0 x2y dy dx � �1
0�x2y 2
2 �x
0 dx � �1
0 x4
2 dx � � x5
10�1
0�
110
�1
0�x
0�xy
0 x dz dy dx � �1
0�x
0 �xz�
xy
0 dy dx
5.
� �4
1 ��ze�x2�
1
0 dz � �4
1 z�1 � e�1� dz � ��1 � e�1� z
2
2�4
1�
152 �1 �
1e�
�4
1�1
0�x
0 2ze�x2 dy dx dz � �4
1�1
0 ��2ze�x2�y�
x
0 dx dz � �4
1�1
0 2zxe�x2 dx dz
7.
� �4
0�x�1 � x�sin y�
�2
0 dx � �4
0x�1 � x�dx � �x2
2�
x3
3 �4
0� 8 �
643
��40
3
�4
0��2
0�1�x
0
x cos y dz dy dx � �4
0��2
0��x cos y�z�
1�x
0 dy dx � �4
0��2
0
x�1 � x�cos y dy dx
9. �2
0�4�x2
�4�x2
�x2
0 x dz dy dx � �2
0�4�x2
�4�x2
x3 dy dx �12815
11.
� �2
0 �x2
ln4��cos y��4�x2
0 dx � �2
0 x2 ln 4�1 � cos4 � x2� dx 2.44167
�2
0�4�x2
0�4
1 x2 sin y
z dz dy dx � �2
0�4�x2
0 �x2
sin y ln �z��4
1 dy dx
13. �4
0�4�x
0�4�x�y
0dz dy dx 15. �3
�3 �9�x2
�9�x2
�9�x2�y2
0dz dy dx
17.
�12�
2
�2 �4 � y 2�2 dy � �2
0 �16 � 8y 2 � y 4� dy � �16y �
83
y3 �15
y5�2
0�
25615
�2
�2�4�y2
0�x
0 dz dx dy � �2
�2�4�y2
0 x dx dy
19.
� 4��
2��a
0 �a2 � x2� dx � �2��a2x �
13
x3��a
0�
43
�a3
� 4�a
0 �ya2 � x2 � y 2 � �a2 � x2� arcsin � y
a2 � x2��a2�x2
0 dx
8�a
0�a2�x2
0�a2�x2�y2
0 dz dy dx � 8�a
0�a2�x2
0 a2 � x2 � y 2 dy dx
Section 13.6 Triple Integrals and Applications 157
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21. �2
0�4�x2
0�4�x2
0 dz dy dx � �2
0 �4 � x2�2 dx � �2
0 �16 � 8x2 � x 4� dx � �16x �
83
x3 �15
x5�2
0�
25615
23. Plane:
�3
0�(12�4z)3
0�(12�4z�3x)6
0 dy dx dz
x
y2 3
3
4
z
3x � 6y � 4z � 12 25. Top cylinder:
Side plane:
�1
0�x
0�1�y2
0 dz dy dx
x
y
1
1
1
z
x � y
y 2 � z2 � 1
27.
� �1
0�x
0�3
0 xyz dz dy dx� �
916�
� �1
0�1
y�3
0 xyz dz dx dy
� �1
0�3
0�x
0 xyz dy dz dx
� �1
0�3
0�1
y
xyz dx dz dy
��Q
� xyz dV � �3
0�1
0�1
y
xyz dx dy dz � �3
0�1
0�x
0 xyz dy dx dz
x1
1 y x=
R
yQ � ��x, y, z�: 0 ≤ x ≤ 1, 0 ≤ y ≤ x, 0 ≤ z ≤ 3�
29.
� �3
�3�9�x2
�9�x2�4
0xyz dz dy dx � � 0�
� �3
�3�4
0�9�x2
�9�x2
xyz dy dz dx
� �3
�3�9�y2
�9�y2�4
0xyz dz dx dy
� �3
�3�4
0�9�y2
�9�y2
xyz dx dz dy
� �4
0�3
�3�9�y2
�9�y2
xyz dx dy dz
�Q
��xyz dV � �4
0�3
�3�9�x2
�9�x2
xyz dy dx dz
x
y4 3
5
43
zQ � ��x, y, z�: x2 � y2 ≤ 9, 0 ≤ z ≤ 4�
158 Chapter 13 Multiple Integration
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31.
x �Myz
m�
12k8k
�32
� 12k
Myz � k�6
0�4�(2x3)
0�2�(y2)�(x3)
0 x dz dy dx
� 8k
m � k�6
0�4�(2x3)
0�2�(y2)�(x3)
0 dz dy dx
33.
z �Mxy
m� 1
� 2k�4
0 �16x � 8x2 � x3� dx �
128k3
Mxy � k�4
0�4
0�4�x
0 xz dz dy dx � k�4
0�4
0 x
�4 � x�2
2 dy dx
� 4k�4
0 �4x � x2� dx �
128k3
m � k�4
0�4
0�4�x
0 x dz dy dx � k�4
0�4
0 x�4 � x� dy dx 35.
z �Mxy
m�
kb68kb54
�b2
y �Mxz
m�
kb66kb54
�2b3
x �Myz
m�
kb66kb54
�2b3
Mxy � k�b
0�b
0�b
0 xyz dz dy dx �
kb6
8
Mxz � k�b
0�b
0�b
0 xy 2 dz dy dx �
kb6
6
Myz � k�b
0�b
0�b
0 x2y dz dy dx �
kb6
6
m � k�b
0�b
0�b
0 xy dz dy dx �
kb5
4
37. will be greater than 2, whereas and will be unchanged.zyx
39. will be greater than 0, whereas and will be unchanged.zxy
41.
z �Mxy
m�
k�r2h24k�r2h3
�3h4
�k�r2h2
4
�4kh2
3r2 �r
0 �r2 � x2�32 dx
�3kh2
r2 �r
0�r2�x2
0 �r2 � x2 � y 2� dy dx
Mxy � 4k�r
0�r2�x2
0�h
hx2�y2r
z dz dy dx
x � y � 0 by symmetry
m �13
k�r 2h
Section 13.6 Triple Integrals and Applications 159
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43.
by Wallis’s Formula
z �Mxy
m�
64k�
1�
3128k�
�32
� 64�k
�let x � 4 sin �� �1024k
3 ��2
0 cos4 � d�
� 2k�4
0�42 �x2
0 �42 � x2 � y 2� dy dx � 2k�4
0 �16y � x2y �
13
y3�42�x2
0 dx �
4k3 �
4
0 �42
� x2�32 dx
Mxy � 4k�4
0�42�x2
0�42�x2�y2
0 z dz dy dx
z � 42 � x2 � y 2
x � y � 0 by symmetry
m �128k�
3
45.
z �Mxy
m�
250k200k
�54
y �Mxz
m�
1200k200k
� 6
x �Myz
m�
1000k200k
� 5
Mxy � k�20
0��(35)x�12
0�(512)y
0 z dz dy dx � 250k
Mxz � k�20
0��(35)�x�12
0�(512)y
0 y dz dy dx � 1200k
Myz � k�20
0��(35)x�12
0�(512)y
0 x dz dy dx � 1000k
m � k�20
0��(35)x�12
0�(512)y
0 dz dy dx � 200k
x8 20
12
12
20
16
8
164
4
y x= + 12− 35
yf �x, y� �5
12y
47. (a)
(b)
Ix � Iy � Iz �ka8
8 by symmetry
�ka2
2 �a
0 �y 4z
4�
y2z3
2 �a
0 dz �
ka4
8 �a
0 �a2z � 2z3� dz � �ka4
8 �a2z2
2�
2z4
4 ��a
0�
ka8
8
Ix � k�a
0�a
0�a
0 �y 2 � z2�xyz dx dy dz �
ka2
2 �a
0�a
0 �y3z � yz3� dy dz
Ix � Iy � Iz �2ka5
3 by symmetry
� ka�a
0 �1
3y3 � z2y�
a
0 dz � ka�a
0 �1
3a3
� az2� dz � �ka�13
a3z �13
az3��a
0�
2ka5
3
Ix � k�a
0�a
0�a
0 �y 2 � z2� dx dy dz � ka�a
0�a
0 �y 2 � z2� dy dz
160 Chapter 13 Multiple Integration
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49. (a)
(b)
� 8k�4
0�32 � 8x � 4x2 � x3� dx � �8k�32x � 4x2 �
43
x3 �14
x4��4
0�
2048k3
� k�4
0 ��x2y 2
2�
y 4
4 ��4 � x��4
0 dx � k�4
0 �8x2 � 64��4 � x� dx
Iz � k�4
0�4
0�4�x
0 y�x2 � y 2� dz dy dx � k�4
0�4
0 �x2y � y3��4 � x� dx
� 8k�4
0�4x2 � x3 �
13
�4 � x�3� dx � 8k�43
x3 �14
x4 �1
12�4 � x�4�
4
0�
1024k3
Iy � k�4
0�4
0�4�x
0 y�x2 � z2� dz dy dx � k�4
0�4
0 �x2y�4 � x� �
13
y�4 � x�3� dy dx
� k��32�4 � x�2 �23
�4 � x�4�4
0�
2048k3
� k�4
0 �y 4
4�4 � x� �
y 2
6�4 � x�3�
4
0 dx � k�4
0 �64�4 � x� �
83
�4 � x�3� dx
Ix � k�4
0�4
0�4�x
0 y�y 2 � z2� dz dy dx � k�4
0�4
0 �y3�4 � x� �
13
y�4 � x�3� dy dx
� k�4
0 ��x2y �
y3
3 ��4 � x��4
0 dx � k�4
0�4x2 �
643 ��4 � x� dx � 256k
Iz � k�4
0�4
0�4�x
0 �x2 � y 2� dz dy dx � k�4
0�4
0 �x2 � y 2��4 � x� dy dx
� 4k�4
0 �4x2 � x3 �
13
�4 � x�3� dx � 4k�43
x3 �14
x 4 �1
12�4 � x�4�
4
0�
512k3
Iy � k�4
0�4
0�4�x
0 �x2 � z2� dz dy dx � k�4
0�4
0 �x2�4 � x� �
13
�4 � x�3� dy dx
� k��323
�4 � x�2 �13
�4 � x�4�4
0� 256k
� k�4
0 �y3
3�4 � x� �
y3
�4 � x�3�4
0 dx � k�4
0 �64
3�4 � x� �
43
�4 � x�3� dx
Ix � k�4
0�4
0�4�x
0 �y 2 � z2� dz dy dx � k�4
0�4
0 �y 2�4 � x� �
13
�4 � x�3� dy dx
51.
Since
—CONTINUED—
m � �a2Lk, Ixy � ma24.
�2k3 �
L2
�L2 2�a4�
4�
a4�
16 � dy �a4�Lk
4
�23�
L2
�L2 k�a2
2 �xa2 � x2 � a2 arcsin xa� �
18 �x�2x2 � a2�x2 � a2 � a4 arcsin
xa��
a
�a dy
Ixy � k�L2
�L2�a
�a�a2�x2
�a2�x2
z2 dz dx dy � k�L2
�L2�a
�a
23
�a2 � x2�a2 � x2 dx dy
Section 13.6 Triple Integrals and Applications 161
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Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates
51. —CONTINUED—
Iz � Ixz � Iyz �mL2
12�
ma2
4�
m12
�3a2 � L2�
Iy � Ixy � Iyz �ma2
4�
ma2
4�
ma2
2
Ix � Ixy � Ixz �ma2
4�
mL2
12�
m12
�3a2 � L2�
� 2k�L�2
�L�2 18�x�2x2 � a2��a2 � x2 � a4 arcsin
xa�
a
�a dy �
ka4�
4 �L�2
�L�2 dy �
ka4�L4
�ma2
4
Iyz � k�L�2
�L�2�a
�a��a2�x2
��a2�x2
x2 dz dx dy � 2k�L�2
�L�2�a
�a
x2�a2 � x2 dx dy
� 2k�L�2
�L�2 �y 2
2 x�a2 � x2 � a2 arcsin xa�
a
�a dy � k�a2�L�2
�L�2 y 2 dy �
2k�a2
3 L3
8 �1
12mL2
Ixz � k�L�2
�L�2�a
�a��a2�x2
��a2�x2
y 2 dz dx dy � 2k�L�2
�L�2�a
�a
y 2�a2 � x2 dx dy
53. �1
�1�1
�1�1�x
0 �x2 � y 2��x2 � y 2 � z2 dz dy dx 55. See the definition, page 978.
See Theorem 13.4, page 979.
57. (a) The annular solid on the right has the greater density.
(b) The annular solid on the right has the greater movement of inertia.
(c) The solid on the left will reach the bottom first. The solid on the righthas a greater resistance to rotational motion.
1.
� �4
0���2
0 2 cos � d� dz � �4
0 �2 sin ��
��2
0
dz � �4
0 2dz � 8
�4
0���2
0�2
0r cos � dr d� dz � �4
0���2
0 �r 2
2 cos ��2
0
d� dz
3.
� ���2
0�8 cos4 � � 4 cos8 ��sin � d� � ��
8 cos5 �5
�4 cos9 �
9 ���2
0
�5245
���2
0�2 cos2 �
0�4�r2
0r sin � dz dr d� � ���2
0�2 cos2 �
0r�4 � r 2�sin � dr d� � ���2
0 �2r 2 �
r4
4 sin ��2 cos2 �
0
d�
5. �2�
0���4
0�cos �
02 sin � d d� d� �
13
�2�
0���4
0 cos3 � sin � d� d� � �
112
�2�
0�cos4 ��
��4
0
d� ��
8
7. �4
0�z
0���2
0rer d� dr dz � ��e4 � 3�
162 Chapter 13 Multiple Integration
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9.
��
4�1 � e�9�
� ���2
0 12
�1 � e�9�d�
� ���2
0��
12
e�r2�3
0 d�
x
y
z
12
2
3
3
3
1
���2
0�3
0�e�r2
0r dz dr d� � ���2
0�3
0re�r2 dr d�
11.
�64�3�
3
�32�3
3 �2�
0d�
�643
�2�
0��cos ��
��2
��6
d�
x
y4
4
4
z�2�
0���2
��6�4
02 sin � d d� d� �
643
�2�
0���2
��6 sin � d� d�
13. (a)
(b) �2�
0�arctan�1�2�
0�4 sec �
03 sin2 � cos � d d� d� � �2�
0���2
arctan�1�2��cot � csc �
03 sin2 � cos � d d� d� � 0
�2�
0�2
0�4
r2
r 2 cos � dz dr d� � 0
15. (a)
(b) ���4
0�2�
0�2a cos �
a sec �3 sin2 � cos � d d� d� � 0
�2�
0�a
0�a��a2�r2
a
r 2 cos � dz dr d� � 0
17.
�43
a3���2
0 �1 � sin3 �� d� �
43
a3�� �13
cos ��sin2 � � 2����2
0�
43
a3�
2�
23 �
2a3
9�3� � 4�
V � 4���2
0�a cos �
0��a2�r2
0 r dz dr d� � 4���2
0�a cos �
0 r�a2 � r2 dr d�
19.
�2a3
9�3� � 4�
�2a3
3 �� � cos � �cos3 �
3 ��
0
�2a3
3 ��
0�1 � sin3 �� d�
� 2��
0��
13
�a2 � r2�3�2�a cos �
0 d�
� 2��
0�a cos �
0 r�a2 � r2 dr d�
V � 2��
0�a cos �
0��a2�r2
0 r dz dr d� 21.
� k�48� � 8 � 8� � 48k�
� k�24� � 4 sin � � 8 cos ��2�
0
� �2�
0k�24 � 4 cos � � 8 sin ��d�
� �2�
0k�3r3 �
r4
4 cos � �
r4
2 sin ��
2
0 d�
� �2�
0�2
0kr2�9 � r cos � � 2r sin ��dr d�
m � �2�
0�2
0�9�r cos ��2r sin �
0
�kr�r dz dr d�
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 163
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23.
�4hr0
r03
6 �
2 �13
�r02h
�4hr0
���2
0 r0
3
6 d�
�4hr0
���2
0�r0
0�r0r � r 2� dr d�
V � 4���2
0�r0
0�h�r0�r��r0
0r dz dr d�
z � h �hr0
�x2 � y2 �hr0
�r0 � r� 25.
z �Mxy
m�
k�r03h2�30
k�r03h�6
�h5
�1
30k�r0
3 h2
Mxy � 4k���2
0�r0
0�h(r0�r)�r0
0 r2 z dz dr d�
�16
k�r03h
m � 4k���2
0�r0
0�h(r0�r)�r0
0 r2 dz dr d�
x � y � 0 by symmetry
� k�x2 � y 2 � kr
27.
Since the mass of the core is fromExercise 23, we have Thus,
�3
10mr0
2
�1
103m
�r02h�r0
4h
Iz �1
10k�r0
4h
k � 3m��r02h.
m � kV � k�13�r0
2h�
�1
10k�r0
4h
�4khr0
r05
20�
2
�4khr0���2
0�r0
0 �r0r
3 � r4� dr d�
Iz � 4k���2
0�r0
0�h(r0�r)�r0
0 r3 dz dr d� 29.
�12
m�a2 � b2�
�k��b2 � a2��b2 � a2�h
2
�k��b4 � a4�h
2
� kh���2
0 �b4 � a4� d�
� 4kh���2
0�b
a
r3 dr d�
Iz � 4k���2
0�b
a�h
0 r3 dz dr d�
m � k��b2h � �a2h� � k�h�b2 � a2�
31. V � �2�
0��
0�4 sin �
0 2 sin� d d� d� � 16� 2 33.
� k�a4
� �k�a4��cos �����2
0
� k�a4���2
0 sin � d�
� 2ka4���2
0���2
0 sin � d� d�
m � 8k���2
0���2
0�a
0 3 sin � d d� d�
164 Chapter 13 Multiple Integration
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37.
�k�
192
� �25
k��16
cos6 � �18
cos8 ����2
��4
�25
k����2
��4 cos5 ��1 � cos2 �� sin � d�
�45
k���2
��4���2
0 cos5 � sin3 � d� d�
Iz � 4k���2
��4���2
0�cos �
0 4 sin3 � d d� d�
39.
z � z z � z
y � r sin � tan � �yx
x � r cos � x2 � y2 � r2 41. ��2
�1
�g2���
g1����h2�r cos �, r sin ��
h1�r cos �, r sin �� f �r cos �, r sin �, z�r dz dr d�
43. (a) right circular cylinder about z-axis
plane parallel to z-axis
plane parallel to xy-planez � z0:
� � �0:
r � r0: (b) sphere of radius
plane parallel to z-axis
cone� � �0:
� � �0:
0 � 0:
45.
� a4����2
0 d� �
a4� 2
2
� 4����2
0 �a2r2
2�
r4
4 �a
0 d�
� 8���2
0�a
0 �
2�a2 � r2�r dr d�
� 16���2
0�a
0 12�z��a2 � r2� � z2 � �a2 � r2� arcsin
z�a2 � r2�
�a2�r2
0 r dr d�
� 16���2
0�a
0��a2�r2
0 ��a2 � r2� � z2 dz�r dr d��
� 16�a
0��a2�x2
0��a2�x2�y2
0
�a2 � x2 � y 2 � z2 dz dy dx
16�a
0��a2�x2
0��a2�x2�y2
0��a2�x2�y2�z2
0 dw dz dy dx
35.
z �Mxy
m�
k�r4�42k�r3�3
�3r8
�14
k�r4 � ��18
k�r4 cos 2����2
0
�kr4�
4 ���2
0 sin 2� d�
�12
kr4���2
0���2
0 sin 2� d� d�
Mxy � 4k���2
0���2
0�r
0 3 cos � sin � d d� d�
x � y � 0 by symmetry
m �23
k�r3
Section 13.7 Triple Integrals in Cylindrical and Spherical Coordinates 165
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Section 13.8 Change of Variables: Jacobians
1.
� �12
�x�u
�y�v
��y�u
�x�v
� ��12��
12� � �1
2��12�
y �12
�u � v�
x � �12
�u � v� 3.
�x�u
�y�v
��y�u
�x�v
� �1��1� � �1���2v� � 1 � 2v
y � u � v
x � u � v2
5.
�x�u
�y�v
��y�u
�x�v
� cos2 � � sin2 � � 1
y � u sin � � v cos �
x � u cos � � v sin �
7.
�x�u
�y�v
��y�u
�x�v
� �eu sin v���eu sin v� � �eu cos v��eu cos v� � �e2u
y � eu cos v
x � eu sin v
9.
�x3
�2y9
u �x � 2v
3�
x � 2�y�3�3
v �y3
y � 3v
x � 3u � 2v
u1
1(0, 1)
(1, 0)
v
�0, 1��2, 3�
�1, 0��3, 0�
�0, 0��0, 0�
�u, v��x, y�
11.
� �1
�1�1
�1 �u2 � v2� dv du � �1
�1 2�u2 �
13� du � �2�u3
3�
u3�
1
�1�
83
�R� 4�x2 � y 2� dA � �1
�1�1
�1 4�1
4�u � v�2 �
14
�u � v�2�12� dv du
�x�u
�y�v
��y�u
�x�v
� �12���
12� � �1
2��12� � �
12
y �12
�u � v�
x �12
�u � v�
13.
�R� y�x � y� dA � �3
0�4
0 uv�1� dv du � �3
0 8u du � 36
�x�u
�y�v
��y�u
�x�v
� �1��0� � �1��1� � �1
y � u
u
2
3
4
1 2 3 4
1
vx � u � v
166 Chapter 13 Multiple Integration
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15.
Transformed Region:
� ���e�2 � e�1�2�ln u2
1�4� ��e�2 � e�1�2� �ln 2 � ln
14� � �e�1�2 � e�2�ln 8 0.9798
�R� e�xy�2 dA � �2
1�4�4
1 e�v�2� 1
2u� dv du � ��2
1�4�e�v�2
u 4
1 du � ��2
1�4�e�2 � e�1�2�1
u du
y �x4
⇒ yx
�14
⇒ u �14
y � 2x ⇒ yx
� 2 ⇒ u � 2
y �4x ⇒ yx � 4 ⇒ v � 4
y �1x ⇒ yx � 1 ⇒ v � 1
u
2
3
1 3 4
S
v
� �14�
1u
�1u� � �
12u� 1
u1�2v1�2
u1�2
v1�2
1212
v1�2
u3�2
v1�2
u1�2
��x, y���u, v� � ��x
�u�y�u
�x�v�y�v� � ��1
212
x � �v�u, y � �uv ⇒ u �yx, v � xy
R: y �x4
, y � 2x, y �1x, y �
4x
y
x
2
3
4
1 2 3 4
1
y x= 2
y x= 14
y = 1x
y = 4x
R
�R� e�xy�2 dA
17.
�12
�8
4 u�e4 � 1� du � �1
4u2�e4 � 1�
8
4� 12�e4 � 1�
�R� �x � y�ex�y dA � �8
4�4
0 uev �1
2� dv du
��x, y���u, v� � �
12
x �12
�u � v�
u � x � y � 8,
42
4
2
6
6
x
x y− = 0
x y− = 4
x y+ = 8
x y+ = 4
y u � x � y � 4,
y �12
�u � v�
v � x � y � 4
v � x � y � 0
19.
� �5
0 �1
5 �23�u3�2�v
5
0 dv � �2�5
3 �23�v3�2
5
0�
1009
�R���x � y��x � 4y� dA � �5
0�5
0 �uv �1
5� du dv
�x�u
�y�v
��y�u
�x�v
� �15���
15� � �1
5��45� � �
15
x �15
�u � 4v�,
u � x � 4y � 5,
1
2
−1
−2
−1 3 4x
x y+ 4 = 0
x y+ 4 = 5
x y− = 5
x y− = 0y u � x � 4y � 0,
y �15
�u � v�
v � x � y � 5
v � x � y � 0
Section 13.8 Change of Variables: Jacobians 167
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21.
x
a
−a
2a
v u= −
v u=
y
x
a
a
x y a+ =
y
�R� �x � y dA � �a
0�u
�u
�u�12� dv du � �a
0 u�u du � �2
5u5�2
a
0�
25
a5�2
�x�u
�y�v
��y�u
�x�v
� �12
u � x � y, v � x � y, x �12
�u � v�, y �12
�u � v�
(b)
(c)
� ab���1�2� � �ab
A � �S� ab dS
� �a��b� � �0��0� � ab
��x, y���u, v� �
�x�u
�y�v
��y�u
�x�v
25. Jacobian ���x, y���u, v� �
�x�u
�y�v
��y�u
�x�v
27.
� u2v
� �1 � v��u2v� � u�uv2�
� �1 � v� u2v�1 � w� � u2vw� � u uv2�1 � w� � uv2w���x, y, z���u, v, w� � � 1 � v
v�1 � w�vw
�u u�1 � w�
uw
0�uvuv �
x � u�1 � v�, y � uv�1 � w�, z � uvw
29.
� �2 sin
� �2 sin �cos2 � sin2 �
� �2 sin cos2 � 2 sin3
� cos �2 sin cos �sin2 � � cos2 ��� � sin sin2 �cos2 � � sin2 ���
� cos �2 sin cos sin2 � � 2 sin cos cos2 �� � sin sin2 cos2� � sin2 sin2 ��
��x, y, z���, �, � � �sin cos �
sin sin �cos
� sin sin � sin cos �
0
cos cos � cos sin �
� sin �x � sin cos �, y � sin sin �, z � cos
23.
u2 � v2 � 1
�au�2
a2 ��bv�2
b2 � 1
x2
a2 �y 2
b2 � 1, x � au, y � bv (a)
x
a
R
b
y
x2
a2 �y 2
b2 � 1
u1
S
1
v
u2 � v2 � 1
168 Chapter 13 Multiple Integration
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Review Exercises for Chapter 13
1. � x � x3 � x3 ln x 2 � x3��1 � ln x 2� � x �x2
1x ln y dy � �xy��1 � ln y��
x2
1
3. � �1
0�4x 2 � 5x � 1� dx � �4
3x3 �
52
x 2 � x�1
0�
296�1
0�1�x
0 �3x � 2y� dy dx � �1
0�3xy � y 2�
1�x
0 dx
5. �3
0��9�x2
0 4x dy dx � �3
04x�9 � x 2 dx � ��
43
�9 � x 2�3�2�3
0� 36
7.
� �1
0�3 � 3y� dy � �3y �
32
y2�1
0�
32
A � �1
0�3�3y
0 dx dy
�3
0��3�x��3
0 dy dx � �1
0�3�3y
0dx dy
9.
� 2�3
�5
�25 � x2 dx � �x�25 � x2 � 25 arcsin x5�
3
�5�
25�
2� 12 � 25 arcsin
35
67.36A � 2�3
�5��25�x2
0 dy dx
�3
�5��25�x2
��25�x2
dy dx � ��4
�5��25�y2
��25�y2
dx dy � �4
�4�3
��25�y2
dx dy � �5
4��25�y2
��25�y2
dx dy
11.
A � 4�1�2
0���1��1�4y2 ��2
��1��1�4y2 ��2 dx dy
A � 4�1
0�x�1�x2
0 dy dx � 4�1
0 x�1 � x2 dx � ��
43
�1 � x2�3�2�1
0�
43
13. A � �5
2��x�1
x�3 dy dx � 2�2
1��x�1
0 dy dx � �2
�1�y�3
y2�1 dx dy �
92
15. Both integrations are over the common region R shown in the figure. Analytically,
�43
�43�2�2
0�x�2
0�x � y� dy dx � �2�2
2��8�x2�2
0�x � y� dy dx �
53
� 43�2 �
13�
�1
0�2�2�y2
2y
�x � y� dx dy �43
�43�2 (2, 1)
y x= 12
y x= 8 − 2
1 2 3
−1
1
2
x
12
y
17.
� � 110
x5 �43
x3 � 8x�4
0�
329615
� �4
01
2x4 � 4x2 � 8� dx
� �4
0�x2y �
12
y2 � 4y�x2�4
0 dx
V � �4
0�x2�4
0�x2 � y � 4� dy dx 19.
Matches (c)
92
�3� �272
x
y
(3, 3)(3, 0)
2
2
4
4
4
6
zVolume �base��height�
Review Exercises for Chapter 13 169
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21.
Therefore,
P � �1
0�1
0 xye��x�y� dy dx 0.070
k � 1.
� ��k�x � 1�e�x��
0� k � ��
0 kxe�x dx��
0��
0 kxye��x�y� dy dx � ��
0��kxe��x�y�� y � 1��
�
0 dx
27.
�h3
3���4
0sec3 � d� �
h3
6 �sec � tan � � ln�sec � � tan �����4
0 �
h3
6 �2 � ln��2 � 1��
�h
0�x
0
�x 2 � y 2 dy dx � ���4
0�h sec �
0 r
2 dr d�
29.
� ��13
z3��h
0�
�h3
3
� � �h
0 z2 dz
� 2�h
0���2
0�1 � z2 � 1� d� dz
V � 4�h
0���2
0��1�z2
1 r dr d� dz 31. (a)
(b)
(c) V � 4���4
0�3�cos 2�
0
�9 � r 2 r dr d� 20.392
A � 4���4
0�3�cos 2�
0r dr d� � 9
−4
−6 6
4
r � 3�cos 2�
r 2 � 9�cos2 � � sin2 �� � 9 cos 2�
�r 2�2 � 9�r 2 cos2 � � r 2 sin2 ��
�x2 � y2�2 � 9�x2 � y2�
23. True 25. True
33. (a)
x
2
1 2
1
y x= 2 3
y x= 2
y
y �Mx
m�
6455
x �My
m�
3245
My � k�1
0�2x
2x3x2y dy dx �
8k45
Mx � k�1
0�2x
2x3xy2 dy dx �
16k55
m � k�1
0�2x
2x3xy dy dx �
k4
(b)
y �Mx
m�
784663
x �My
m�
9361309
My � k�1
0�2x
2x3x�x2 � y2�dy dx �
156k385
Mx � k�1
0�2x
2x3y�x2 � y2�dy dx �
392k585
m � k�1
0�2x
2x3�x2 � y2�dy dx �
17k30
170 Chapter 13 Multiple Integration
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35.
y ��Ix
m���1�6�kb3a2
�1�2�kba2 ��b2
3�
b�33
x ��Iy
m���1�4�kba4
�1�2�kba2 ��a2
2�
a�22
m � �R���x, y�dA � �a
0�b
0kx dy dx �
12
kba2
I0 � Ix � Iy �16
kb3a2 �14
kba4 �ka2b12
�2b2 � 3a2�
Iy � �R�x2 ��x, y�dA � �a
0�b
0kx3 dy dx �
14
kba4
Ix � �R�y2 ��x, y�dA � �a
0�b
0kxy2 dy dx �
16
kb3a2 37.
� �13
�653�2� 1���
��2
0�
�
6�65�65 � 1�
� 4 ���2
0�4
0 �1 � 4r 2 r dr d�
� 4�4
0��16�x2
0
�1 � 4x2 � 4y2 dy dx
S � �R��1 � � fx�2 � � fy�2 dA
39.
� �3
02�1 � 4y2 dy �
14
23
�1 � 4y2�3�2�3
0
�16
�37�3�2 � 1�
� �3
0��1 � 4y2 x�
y
�y dy
� �3
0�y
�y
�1 � 4y2 dx dy
S � �R��1 � fx
2 � fy2 dA
fx � 0, fy � �2y
f �x, y� � 9 � y2
45. �1
�1��1�x2
��1�x2
��1�x2�y2
��1�x2�y2 �x2 � y2� dz dy dx � �2�
0�1
0��1�r2
��1�r2
r3 dz dr d� �8�
15
41.
� �2�
0�3
0�9r 2 � r 4� dr d� � �2�
0�3r 3 �
r 5
5 �3
0 d� �
1625 �2�
0 d� �
324�
5
�3
�3��9�x2
��9�x2�9
x2�y2
�x 2 � y 2 dz dy dx � �2�
0�3
0�9
r2r2 dz dr d�
43.
� �a
01
3bc3 �
13
b3c � bcz2� dz �13
abc3 �13
ab3c �13
a3bc �13
abc�a2 � b2 � c 2�
�a
0�b
0�c
0�x 2 � y 2 � z2� dx dy dz � �a
0�b
01
3c3 � cy2 � cz2� dy dz
Review Exercises for Chapter 13 171
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47.
�323 �� � cos � �
13
cos3 ����2
0�
323 �
2�
23�
�323 �
��2
0�1 � sin3 �� d�
� ����2
0�4
3�4 � r2�3�2�
2 cos �
0 d�
� 4���2
0�2 cos �
0 r�4 � r2 dr d�
V � 4���2
0�2 cos �
0��4�r2
0 r dz dr d�
49.
x � y � 0 by symmetry
z �Mxy
m�
k��96k��24
�14
� k���2
��4���2
0cos5 sin d� d �
12
k����2
��4cos5 sin d � ��
112
k� cos6 ���2
��4
�k�
96
Mxy � 4k���2
��4���2
0�cos
0�3 cos sin d� d� d
�43
k���2
��4���2
0cos3 sin d� d �
23
k����2
��4cos3 sin d � ��
23
k�14
cos4 ����2
��4
�k�
24
m � 4k���2
��4���2
0�cos
0�2 sin d� d� d
51.
x � y � z �Mxy
m�
k�a4
16 6
k�a3� �3a8
Mxy � k���2
0���2
0�a
0�� cos ��2 sin d� d� d �
k�a4
16
m � k���2
0���2
0�a
0�2 sin d� d� d �
k�a3
653.
� 4k���2
0�4
3�16r3 � r5� dr d� �
833�k3
Iz � 4k���2
0�4
3�16�r2
0r3 dz dr d�
55.
(a) Disc Method
Equivalently, use spherical coordinates
—CONTINUED—
V � �2�
0�cos�1�a�h�a�
0�a
�a�h�sec �2 sin d� d d�
� � �ah2 �h3
3 � �13
�h2 3a � h� � � �a3 �a3
3� a3 � a2h �
a3
3� a2h � ah2 �
h3
3 �
� � �a2y �y3
3 �a
a�h
� � �a3 �a3
3 � � a2�a � h� ��a � h�3
3 ��
x
y a x= 2 2−
2a
a
a−a
a h−
yV � ��a
a�h
�a2 � y2�dy
0 ≤ r ≤ �2ah � h2
� �a2 � r 2
x
yaa
aa h−
h
zz � f �x, y� � �a2 � x2 � y2
172 Chapter 13 Multiple Integration
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55. —CONTINUED—
(b)
centroid:
(c) If
centroid of hemisphere:
(d)
(e)
(f ) If Iz �a3�
30�20a2 � 15a2 � 3a2� �
415
a5�h � a,
�h3
30�20a2 � 15ah � 3h2��
Iz � �2�
0�cos�1�a�h�a�
0�a
�a�h� sec ��2 sin2 ��2 sin d� d d�
x2 � y2 � �2 sin2
limh →0
z � limh →0
3�2a � h�2
4�3a � h� �3�4a2�
12a� a
0, 0, 38
a�
z �3�a�2
4�2a� �38
ah � a,
0, 0, 3�2a � h�2
4�3a � h� �
z �Mxy
V�
14
h2��2a � h�2
13
h2��3a � h��
34
�2a � h�2
3a � h
�14
h2� �2a � h�2
Mxy � �2�
0�cos�1�a�h�a�
0�a
�a�h�sec �� cos ��2 sin d� d d�
57.
Since represents (in the yz-plane) a circle ofradius 3 centered at the integral represents thevolume of the torus formed by revolving this circle about the z-axis.
�0 < � < 2���0, 3, 0�,
� � 6 sin
�2�
0��
0�6 sin
0�2 sin d� d d� 59.
� 1��3� � 2�3� � �9
�x, y��u, v� �
xu
yv
�yu
xv
61.
Boundaries in xy-plane Boundaries in uv-plane
� 5 ln 5 � 3 ln 3 � 2 2.751 � �5 ln 5 � 5� � �3 ln u � 3�
� �u ln u � u�5
3 � �5
3ln u du � �5
3�1
�1
12
ln u dv du �R�ln �x � y�dA � �5
3�1
�1ln1
2�u � v� �
12
�u � v�12�� dv du
v � 1x � y � 1
v � �1x � y � �1
u � 5x � y � 5
u � 3x � y � 3
x �12
�u � v�, y �12
�u � v� ⇒ u � x � y, v � x � y
y = −x + 5y = x + 1
y = −x + 3 y = x − 1
1
1
2
3
2 3
y
x
�x, y��u, v� �
xu
yv
�xv
yu
�12 �
12� �
12
12� � �
12
Review Exercises for Chapter 13 173
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Problem Solving for Chapter 13
1. (a)
(b) Programs will vary.
� 8�2 � �2� � 4.6863
� �163 �sec � � cos � � tan ��
��4
0
� �163 ��4
0
1cos2 �
�1 � cos2 ��3�2 � 1� d�
� 16��4
01
0
�1 � r2 cos2 � r dr d�
V � 16R�1 � x2 dA
x
y1
1
1
z
y = xR
3. (a) Let
Then
(b)
Let
� 4��6
0arctan�tan ��d� �
4�2
2 ���6
0� 2��
6 2
��2
18
I1 � 4��6
0
1�2 cos �
arctan��2 sin ��2 cos � � �2 cos � d�
u � �2 sin �, du � �2 cos � d�, 2 � u2 � 2 � 2 sin2 � � 2 cos2 �.
� �2�2
0
4�2 � u2
arctan u
�2 � u2 du
� �2�2
0
2�2 � u2�arctan
u�2 � u2
� arctan �u
�2 � u2 du
I1 � �2�2
0� 2�2 � u2
arctan v
�2 � u2�u
�u
du
1
�2 � u2� � v2 dv �
1�2 � u2
arctan v
�2 � u2� C.
a2 � 2 � u2, u � v. dua2 � u2 �
1a
arctan ua
� c.
(c)
Let
—CONTINUED—
� 4��2
��6 arctan�1 � sin �
cos � d�
I2 � 4��2
��6
1�2 cos �
arctan��2 � �2 sin ��2 cos � � �2 cos � d�
u � �2 sin �.
� �2
�2�2
4�2 � u2
arctan��2 � u�2 � u2 du
� �2
�2�2
2�2 � u2
�arctan��u � �2�2 � u2 � arctan� u � �2
�2 � u2 � du
I2 � �2
�2�2� 2�2 � u2
arctan v
�2 � u2��u��2
u��2
du
174 Chapter 13 Multiple Integration
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3. —CONTINUED—
(d)
(e)
(f )
� ��
K�0
1�K � 1�2 � �
�
n�1 1n2
� ��
K�0 1
0
yK
K � 1 dy � �
�
K�0
yK�1
�K � 1�2�1
0
� 1
01
0��
K�0�xy�K dx dy � �
�
K�0 1
0 xK�1y K
K � 1 �1
0 dy
1
01
0
11 � xy
dx dy � 1
01
01 � �xy� � �xy�2 � . . .� dx dy
11 � xy
� 1 � �xy� � �xy�2 � . . . �xy� < 1
� 2�18 � 9 � 6 � 172
�2� �4
36 �2 �
�2
9
� 2��
2 � �
�2
2 ���2
��6� 2���2
4�
�2
8 � ��2
12�
�2
72
� 4��2
��6 12 �
�
2� � d� � 2��2
��6��
2� � d�
I2 � 4��2
��6arctan�1 � sin �
cos � d� � 4��2
��6arctan�tan�1
2��
2� � d�
�� �1 � sin ��2
�1 � sin ���1 � sin �� ���1 � sin ��2
cos2 ��
1 � sin �cos �
tan�12 �
�
2� � ��1 � cos����2� � ��
1 � cos����2� � �� ��1 � sin �1 � sin �
(g)
R S
��2, 0�↔�1, 1�
� 1�2
, 1�2 ↔�0, 1�
� 1�2
, �1�2 ↔�1, 0�
�0, 0�↔�0, 0�
�x, y��u, v� � �1��2
1��2�1��2
1��2� � 1
u � v �2y�2
⇒ y �u � v�2
u � v �2x�2
⇒ x �u � v�2
u �x � y�2
, v �y � x�2
−1
1
1 2 3 4
2
−2
21
2
2, 0
1,( )
21
21
,( )−
( )S
v
u
R
y
x
� I1 � I2 ��2
18�
�2
9�
�2
6
1
01
0
11 � xy
dx dy � �2�2
0u
�u
1
1 �u2
2�
v2
2
dv du � �2
�2�2�u��2
u��2
1
1 �u2
2�
v2
2 dv du
Problem Solving for Chapter 13 175
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5. Boundary in xy-plane Boundary in uv-plane
A � R1 dA �
S1 ��x, y�
�u, v��dA �13
23�
vu
1�3
13�
uv
2�3
� �13� 13�v
u 2�3
23 �
uv
1�3��x, y��u, v� �
v � 4y �14
x2
v � 3y �13
x2
u � 2y � �2x
u � 1y � �x
7.
V � 3
02x
06�x
x
dy dz dx � 18
x
y
(0, 0, 0)
(3, 3, 6)
(3, 3, 0)(0, 6, 0)
2
4
5
6
6
3
z
9. From Exercise 55, Section 13.3,
Thus, and
�
0x2e�x2 dx � ��
12
xe�x2��
0�
12�
0e�x2 dx �
12
��
2�
��
4
�
0e�x2 dx �
��
2�
0e�x2�2 dx �
�2�
2
�
��
e�x2�2 dx � �2�
11.
These two integrals are equal to
Hence, assuming you obtain
1 � ka2 or a �1�k
.
a, k > 0,
�
0 e�x�a dx � lim
b→� ���a�e�x�a�
b
0� a.
� k�
0 e�x�a dx � �
0 e�y�a dy
�
���
��
f �x, y� dA � �
0�
0 ke��x�y��a dx dy
f �x, y� � �ke��x�y��a
0 x ≥ 0, y ≥ 0 elsewhere
13.
Area in xy-plane: x y
θ
P
∆x
∆xθcos
∆y∆y
A � l � w � � xcos � y � sec � x y
176 Chapter 13 Multiple Integration
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C H A P T E R 1 4Vector Analysis
Section 14.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 407
Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 412
Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 419
Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 423
Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 427
Section 14.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 431
Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 436
Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 439
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 442
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 448
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407
C H A P T E R 1 4Vector Analysis
Section 14.1 Vector FieldsSolutions to Even-Numbered Exercises
2. All vectors are parallel to x-axis.
Matches (d)
4. Vectors are in rotational pattern.
Matches (e)
6. Vectors along x-axis have no x-component.
Matches (f)
8.
x−2
−2
−1
1
21
y
�F� � 2
F�x, y� � 2i 10.
−5−4−3−2
12345
−2−4−5 −1 1 2 4 5
y
x
�F� � �x2 � y2
F�x, y� � x i � yj
12.
x−2−3 −1
1
2
2
3
3
1
y
�F� � �x� � c
F�x, y� � x i 14.
−2 2
4
6
8
10
y
x
�F� � �1 � �x2 � y2�2
F�x, y� � �x2 � y2� i � j
16.
x y
2
22
−2
−2−2
z
x2 � y2 � z2 � c2
�F� � �x2 � y2 � z2 � c
F�x, y, z� � x i � y j � zk 18.
−6
−6
2
4
6
−4 −2 2 4 6
y
x
F�x, y� � �2y � 3x�i � �2y � 3x�j
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24.
F�x, y, z� � ��zx2 �
zy�i � �1
z�
xzy2�j � ��
yz2 �
1x
�xy�k
fz�x, y, z� � �yz2 �
1x
�xy
fy�x, y, z� �1z
�xzy2
fx�x, y, z� � �zx2 �
zy
f �x, y, z� �yz
�zx
�xzy
26.
G�x, y, z� � �arcsin yz�i �xz
�1 � y2z2j �
xy�1 � y2z2
k
gz�x, y, z� �xy
�1 � y2z2
gy�x, y, z� �xz
�1 � y2z2
gx�x, y, z� � arcsin yz
g�x, y, z� � x arcsin yz
28.
and have continuous first partialderivatives for all
is conservative.�N�x
�1x2 �
�M�y
⇒ F
x � 0.N � ��1x�M � yx2
F�x, y� �1
x2�yi � xj� �
y
x2i �
1
xj 30.
and have continuous first partialderivatives for all
is conservative.�N�x
� 0 ��M�y
⇒ F
x, y � 0.N � �1yM � 1x
F�x, y� �1
xy�y i � x j� �
1
xi �
1
yj
32.
Conservative�N�x
� ��xy
�x2 � y2�32 ��M�y
⇒
M �x
�x2 � y2, N �
y�x2 � y2
34.
Not conservative⇒
�N�x
� ��1
�1 � x2y2�32 ��M�y
�1
�1 � x2y2�32
M �y
�1 � x2y2, N �
�x�1 � x2y2
36.
Not conservative
�
�x�2xy2� � �
2y2
�
�y1y� � �
1y2
�1y
i �2xy 2 j
F�x, y� �1y2 �y i � 2x j� 38.
Conservative
f �x, y� � x3y2 � K
fy�x, y� � 2x3y
fx�x, y� � 3x2y2
�
�x�2x3y � 6x2y
�
�y�3x2y2 � 6x2y
F�x, y� � 3x2y2 i � 2x3yj 40.
Not conservative
�
�x�x2
y2� � �2xy2
�
�y2yx � �
2x
F�x, y� �2yx
i �x2
y2 j
20.
y
x
2
2
2
1
11
z
F�x, y, z� � xi � yj � zk 22.
F�x, y� � 3 cos 3x cos 4yi � 4 sin 3x sin 4yj
fy�x, y� � �4 sin 3x sin 4y
fx�x, y� � 3 cos 3x cos 4y
f �x, y� � sin 3x cos 4y
408 Chapter 14 Vector Analysis
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46.
curl F �3, 2, 0� � 6i � 6j
curl F � � i��x
e�xyz
j�
�ye�xyz
k��z
e�xyz� � ��xz � xy�e�xyzi � ��yz � xy�e�xyzj � ��yz � xz�e�xyzk
�3, 2, 0�F�x, y, z� � e�xyz�i � j � k�,
48.
� x2 1�x � y�2 �
1�x � z�2�i � y2 1
�x � y�2 �1
�y � z�2� j � z2 1�y � z�2 �
1�x � z�2�k
� x2
�x � y�2 �x2
�x � z�2� i � �y2
�x � y�2 �y2
�y � z�2� j � �z2
�x � z�2 ��z2
�y � z�2�k
curl F � � i
�
�x
yz y � z
j
�
�y
xz x � z
k
�
�z
xy x � y �
F�x, y, z� �yz
y � zi �
xzx � z
j �xy
x � yk
50.
curl F � � i��x
�x2 � y2 � z2
j��y
�x2 � y2 � z2
k��z
�x2 � y2 � z2 � ��y � z�i � �z � x�j � �x � y�k
�x2 � y2 � z2
F�x, y, z� � �x2 � y2 � z2�i � j � k�
52.
Not conservative
curl F � � i��x
yez
j��y
xez
k��z
ez � � �xez i � yezj � 0
F�x, y, z� � ez �y i � x j � k� 54.
Conservative
f �x, y, z� � xy2z3 � K
curl F � � i��x
y2z3
j��y
2xyz3
k��z
3xy2z2 � � 0
F�x, y, z� � y2z3 i � 2xyz3 j � 3xy2z2 k
42.
Conservative
f �x, y� � �1
x2 � y2 � K
fy�x, y� �2y
�x2 � y2�2
fx�x, y� �2x
�x2 � y2�2
�
�x2y
�x2 � y2�2� � �8xy
�x2 � y2�3
�
�y2x
�x2 � y2�2� � �8xy
�x2 � y2�3
F�x, y� �2x
�x2 � y2�2 i �2y
�x2 � y2�2 j 44.
curl F �2, �1, 3� � 7i � 4j � 6k
� �z � 2x�i � x2 j � 2zk
� �z � 2x�i � �0 � x2�j � ��2z � 0�k
curl F � � i��x
x2z
j��y
�2xz
k��zyz �
�2, �1, 3�F�x, y, z� � x2z i � 2xz j � yzk,
Section 14.1 Vector Fields 409
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56.
Conservative
f �x, y, z� �12
ln�x2 � y2� � z � K
f �x, y, z� � �dz � z � p�x, y� � K3
�12
ln�x2 � y2� � h�x, z� � K2
f �x, y, z� � � yx2 � y2 dy
�12
ln�x2 � y2� � g�y, z� � K1
f �x, y, z� � � xx2 � y2 dx
fz�x, y, z� � 1
fy �x, y, z� �y
x2 � y2
fx�x, y, z� �x
x2 � y2
curl F � � i��x
x x2 � y2
j��y
y x2 � y2
k��z
1 � � 0
F�x, y, z� �x
x2 � y2 i �y
x2 � y2 j � k 58.
� e x�x � 1� � ey�y � 1�
� xe x � ex � yey � ey
div F�x, y� ��
�x�xex �
�
�y� yey
F�x, y� � xex i � yey j
60.
div F�x, y, z� ��
�x�ln�x2 � y2� �
�
�y�xy �
�
�z�ln�y2 � z2� �
2xx2 � y2 � x �
2zy2 � z2
F�x, y, z� � ln�x2 � y2� i � xy j � ln�y2 � z2�k
62.
div F�2, �1, 3� � 11
div F�x, y, z� � 2xz � y
F�x, y, z� � x2z i � 2xz j � yzk 64.
div F�3, 2, 1� �13
�12
� 1 �116
div F�x, y, z� �1x
�1y
�1z
F�x, y, z� � ln�xyz��i � j � k�
66. See the definition of Conservative Vector Field on page1011. To test for a conservative vector field, see Theorem 14.1 and 14.2.
68. See the definition on page 1016.
70.
� x�x � 2z � 1�i � z�z � 2x � 1�k
� �x � 2xz � x2�i � �y � y�j � ��z2 � 2xz � z�k
curl�F � G� � � i��x
yz
j��y
�xz2 � x2z
k��z
xy � F � G � � i
xx2
j0y
k�zz2 � � yz i � �xz2 � x2z� j � xyk
G�x, y, z� � x2 i � yj � z2 k
F�x, y, z� � x i � zk
410 Chapter 14 Vector Analysis
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72.
curl�curl F� � � i
��x
z � 2x
j
��y
x2
k
��z
�2z� � j � 2xk
curl F � � i
��xx2z
j
��y
�2xz
k
��zyz � � �z � 2x�i � x2 j � 2zk
F�x, y, z� � x2z i � 2xz j � yzk 74.
div�F � G� � 0
F � G � � ixx2
j0y
k�zz2 � � yzi � �xz2 � x2z�j � xyk
G�x, y, z� � x2 i � y j � z2k
F�x, y, z� � x i � zk
76.
div�curl F� � 2 � 2 � 0
curl F � � i��x
x2z
j��y
�2xz
k��z
yz � � �z � 2x�i � x2 j � 2zk
F�x, y, z� � x2z i � 2xz j � yzk
78. Let be a scalar function whose second partial derivatives are continuous.
curl��f � � � i
��x
�f�x
j
��y
�f�y
k
��z
�f�z� � � �2f
�y�z�
�2f�z�y�i � � �2f
�x�z�
�2f�z�x�j � � �2f
�x�y�
�2f�y�x�k � 0
�f ��f�x
i ��f�y
j ��f�z
k
f �x, y, z�
80. and
� �curl F� G � F �curl G�
� M��T�y
��S�z� � N��R
�z�
�T�x� � P��S
�x�
�R�y�� � ��P
�y�
�N�z �R � ��M
�z�
�P�x�S � ��N
�x�
�M�y �T�
� T�M�y
� M�S�z
� S�M�z
� N�R�z
� R�N�z
� N�T�x
� T�N�x
� P�S�x
� S�P�x
� P�R�y
� R�P�y
� M�T�y
div�F � G� ��
�x�NT � PS� �
�
�y�PR � MT � �
�
�z�MS � NR�
F � G � � iMR
jNS
kPT � � �NT � PS�i � �MT � PR�j � �MS � NR�k
G � R i � Sj � T k.Let F � M i � N j � Pk
82.
� f ��P�y
��N�z �i � ��P
�x�
�M�z � j � ��N
�x�
�M�y �k� � � i
�f�x
M
j�f�y
N
k�f�z
P � � f �� � F � ��f � � F
� ��f�x
N � f�N�x
��f�y
M � f�M�y �k � ��f
�yP � f
�P�y
��f�z
N � f�N�z �i � ��f
�xP � f
�P�x
��f�z
M � f�M�z �j
� � � f F� � � i��x
fM
j��y
fN
k��z
fP�Let F � M i � Nj � Pk.
Section 14.1 Vector Fields 411
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Section 14.2 Line Integrals
84.
(since the mixed partials are equal) ��2P�x�y
��2N�x�z
��2P�y�x
��2M�y�z
��2N�z�x
��2M�z�y
� 0
div�curl F� ��
�x��P�y
��N�z � �
�
�y��P�x
��M�z � �
�
�z��N�x
��M�y �
curl F � ��P�y
��N�z �i � ��P
�x�
�M�z �j � ��N
�x�
�M�y �k
Let F � M i � Nj � Pk.
In Exercises 86 and 88, and f �x, y, z � F�x, y, z � �x2 � y2 � z2.F�x, y, z� � x i � yj � zk
86.
�� 1f � �
�x�x2 � y2 � z2�3�2 i �
�y�x2 � y2 � z2�3�2 j �
�z�x2 � y2 � z2�3�2 k �
��x i � y j � zk���x2 � y2 � z2 �3
� �Ff 3
1f
�1
�x2 � y2 � z2
88.
�w�z
� �z
�x2 � y2 � z2�3�2
�w�y
� �y
�x2 � y2 � z2�3�2
�w�x
� �x
�x2 � y2 � z2�3�2
w �1f
�1
�x2 � y2 � z2
Therefore is harmonic.w �1f
�2w ��2w�x2 �
�2w�y2 �
�2w�z2 � 0
�2w�z2 �
2z2 � x2 � y2
�x2 � y2 � z2�5�2
�2w�y2 �
2y2 � x2 � z2
�x2 � y2 � z2�5�2
�2w�x2 �
2x2 � y2 � z2
�x2 � y2 � z2�5�2
2.
0 ≤ t ≤ 2�
r�t� � 4 cos t i � 3 sin t j
y � 3 sin t
x � 4 cos t
sin2 t �y2
9
cos2 t �x2
16
cos2 t � sin2 t � 1
x2
16�
y2
9� 1 4. r�t� � t i �
45t j,
5 i � �9 � t�j,�14 � t�i,
0 ≤ t ≤ 55 ≤ t ≤ 99 ≤ t ≤ 14
6. r�t� � t i � t 2 j,�4 � t� i � 4 j,�8 � t� j,
0 ≤ t ≤ 22 ≤ t ≤ 44 ≤ t ≤ 8
8.
�C
4xy ds � �2
0 4t�2 � t��1 � 1 dt � 4�2 �2
0 �2t � t 2� dt � 4�2�t 2 �
t 3
3�2
0� 4�2�4 �
83� �
16�23
r��t� � i � j0 ≤ t ≤ 2;r�t� � t i � �2 � t�j,
412 Chapter 14 Vector Analysis
http://librosysolucionarios.net
10.
�C
8xyz ds � �2
0 8�12t��5t��3��122 � 52 � 02 dt � �2
0 18,720t 2 dt � 18,720�t3
3�2
0� 49,920
r��t� � 12i � 5j0 ≤ t ≤ 2;r�t� � 12t i � 5t j � 3k,
12.
� �13
t 3�10
1� 333
� �10
1 t
2 dt
�C
�x2 � y2� ds � �10
1 �0 � t 2��0 � 1 dt
x2 4
2
4
6
8
10
−2−4
y1 ≤ t ≤ 10r�t� � t j,
14.
� 4� � ���2
0 8 dt
�C
�x2 � y2� ds � ���2
0 �4 cos2 t � 4 sin2 t����2 sin t�2 � �2 cos t�2 dt
x1 2
1
2
y
0 ≤ t ≤ �
2r�t� � 2 cos t i � 2 sin t j,
16.
��10
6�27 � 144� �
57�102
� ��10 �t2
2�
8�33
t3�2��3
0
�C
�x � 4�y � ds � �3
0 �t � 4�3t ��1 � 9 dt
x6−3 3
6
3
9 (3, 9)
y0 ≤ t ≤ 3r�t� � t i � 3t j,
18.
�C
�x � 4�y� ds � 2 � 4 �16�2
3� 2 � 8�2 �
16�23
� 8 �563�2
�C4
�x � 4�y � ds � �8
6 4�8 � t ds �
16�23
�C3
�x � 4�y � ds � �6
4��6 � t� � 4�2�� ds � 2 � 8�2
�C2
�x � 4�y � ds � �4
2�2 � 4�t � 2 � ds � 4 �
16�23
�C1
�x � 4�y � ds � �2
0 t dt � 2
C1
C2
C3
C4
1
1
2
2
y
x
(2, 2)r�t� � t i,2i � �t � 2� j,�6 � t�i � 2 j,�8 � t� j,
0 ≤ t ≤ 22 ≤ t ≤ 44 ≤ t ≤ 66 ≤ t ≤ 8
Section 14.2 Line Integrals 413
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20.
Mass � �C �x, y, z� ds � �4�
02t�13 dt � 16� 2�13
r��t� � ���3 sin t�2 � �3 cos t�2 � �2�2 � �13
r��t� � �3 sin t i � 3 cos t j � 2k
0 ≤ t ≤ 4� r�t� � 3 cos t i � 3 sin t j � 2 tk,
�x, y, z� � z 22.
� ��643
sin3 t � 8 sin2 t���2
0� �
403
�C
F dr � ���2
0 ��64 sin2 t cos t � 16 sin t cos t�dt
r��t� � �4 sin t i � 4 cos t j
F�t� � 16 sin t cos t i � 4 sin t j
0 ≤ t ≤ �
2C: r�t� � 4 cos t i � 4 sin t j,
F�x, y� � xyi � yj
24.
�C
F dr � �2
�2 �3t � 4t� dt � ��
t2
2�2
�2� 0
r��t� � i �t
�4 � t 2j
F�t� � 3t i � 4�4 � t 2 j
�2 ≤ t ≤ 2C: r�t� � t i � �4 � t 2 j,
F�x, y� � 3x i � 4y j 26.
� �83
��6
24�
83
��6
24�
163
� �83
sin3 t �83
cos3 t �t 6
24��
0
�C
F dr ���
0 �8 sin2 t cos t � 8 cos2 t sin t �
14
t5� dt
r��t� � 2 cos t i � 2 sin t j � tk
F�t� � 4 sin2 t i � 4 cos2 tj �14
t 4k
0 ≤ t ≤ �C: r�t� � 2 sin t i � 2 cos t j �12
t 2 k,
F�x, y, z� � x2 i � y2 j � z2 k
28.
�C
F dr � �2
0
1�2t 2 � e2t
�2t � e2t� dt � 6.91
dr � �i � j � et k� dt
F�t� �t i � t j � et k�2t 2 � e2t
0 ≤ t ≤ 2r�t� � t i � t j � et k,
F�x, y, z� �x i � y j � zk�x2 � y2 � z2
30.
from
� �2 cos9 t3
�6 cos7 t
7�
3 cos5 t5 �
��2
0� �
43105
Work � �C
F dr � ���2
0 ��6 cos8 t sin t � 6 cos6 t sin t � 3 cos4 t sin t� dt
� �6 cos8 t sin t � 6 cos6 t sin t � 3 cos4 t sin t
� �3 cos4 t sin t �2 cos4 t � 2 cos2 t � 1�
� �3 cos4 t sin t �cos4 t � �1 � cos2 t�2�
� �3 cos4 t sin t�cos4 t � sin4 t�
F r� � �3 cos8 t sin t � 3 cos4 t sin5 t
F�t� � cos6 t i � cos3 t sin3 t j
r��t� � �3 cos2 t sin t i � 3 sin2 t cos t j
0 ≤ t ≤ �
2 r�t� � cos3 t i � sin3 t j,
�1, 0� to �0, 1�y � sin3 tC: x � cos3 t,
F�x, y� � x2 i � xy j
414 Chapter 14 Vector Analysis
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32.
counterclockwise along the semicircle from to
Work � �C
F dr � �4��
0 cos 2t dt � ��2 sin 2t�
�
0� 0
F r� � 4 sin2 t � 4 cos2 t � �4 cos 2t
F�t� � �2 sin t i � 2 cos tj
r��t� � �2 sin t i � 2 cos t j
r�t� � 2 cos t i � 2 sin t j, 0 ≤ t ≤ �
��2, 0��2, 0�y � �4 � x2C:
F�x, y� � �y i � x j 34.
Work � �C
F dr � �1
0 90t 2 dt � 30
F r� � 90t 2
F�t� � 6t 2 i � 10t 2 j � 15t 2 k
r��t� � 5i � 3j � 2k
r�t� � 5t i � 3t j � 2tk, 0 ≤ t ≤ 1
C: line from �0, 0, 0� to �5, 3, 2�
F�x, y, z� � yz i � xz j � xyk
36.
�163
�C
F dr �1 � 03�4� �5 � 4�4� � 2�4� � 4�6� � 11�
r��t� � i � 2 t j
0 ≤ t ≤ 1r�t� � t i � t 2 j,
5 4 4 6 11F r�
i � 2ji � 1.5ji � ji � 0.5jir��t�
i � 5j1.5i � 3j2i � 2j3.5i � j5iF�x, y�
�1, 1��34 , 9
16��12 , 14��1
4 , 116��0, 0��x, y�
38.
(a)
(b)
Both paths join and The integrals are negatives of each other because the orientations are different.�3, 4�.�1, 0�
�C2
F dr � ���2
0��1 � 2 cos t�2�4 cos2 t���2 sin t� � 8 cos t sin t�1 � 2 cos t��8 cos3 t� dt � �
2565
F�t� � �1 � 2 cos t�2�4 cos2 t�i � �1 � 2 cos t��8 cos3 t� j
r2��t� � �2 sin t i � 8 cos t sin t j
r2�t� � �1 � 2 cos t�i � 4 cos2 t j, 0 ≤ t ≤�
2
�C1
F dr � �2
0��t � 1�2t2 � 2t 4�t � 1�� dt �
2563
F�t� � �t � 1�2t2 i � �t � 1�t 3 j
r1��t� � i � 2 t j
r1�t� � �t � 1�i � t2j, 0 ≤ t ≤ 2
F�x, y� � x2 y i � xy3�2 j
40.
Thus, �C
F dr � 0.
F r� � 3t 3 � 3t 3 � 0
F�t� � 3t 3 i � t j
r��t� � i � 3t 2 j
C: r�t� � t i � t 3 j
F�x, y� � �3y i � x j 42.
Thus, �C
F dr � 0.
F r� � 9 sin t cos t � 9 sin t cos t � 0
F�t� � 3 sin t i � 3 cos t j
r��t� � 3 cos t i � 3 sin t j
C: r�t� � 3 sin t i � 3 cos t j
F�x, y� � x i � yj
44.
�C
�x � 3y 2� dx � �2
0 �x � 75x 2� dx � �x 2
2� 25x3�
2
0� 202
0 ≤ x ≤ 20 ≤ t ≤ 1 ⇒ y � 5x,y � 10t,x � 2t,
Section 14.2 Line Integrals 415
http://librosysolucionarios.net
46.
� �7x2 �1253
x3�2
0� 28 �
1253
�8� �1084
3
�C
�3y � x� dx � y2 dy � �2
0 �3�5x� � x� dx � �5x�25 dx � �2
0�14x � 125x2� dx
0 ≤ x ≤ 2dy � 5 dx,0 ≤ t ≤ 1 ⇒ y � 5x,y � 10t,x � 2t,
48.
�C
�2x � y� dx � �x � 3y� dy � �2
0 3t dt � �3
2t 2�
2
0� 6
dy � dtdx � 0,
y�t� � tx�t� � 0,
x1
2
1
−1
y0 ≤ t ≤ 2r�t� � t j,
50.
�C
�2x � y� dx � �x � 3y� dy �272
� 10 �472
�C2
�2x � y� dx � �x � 3y� dy � �5
3 �2�t � 3� � 3� dt � ��t � 3�2 � 3t�
5
3� 10
dy � 0dx � dt,
y�t� � �3x�t� � t � 3,C2:
�C1
�2x � y� dx � �x � 3y� dy � �3
0 3t dt �
272
dy � �dtdx � 0,
y�t� � �tx�t� � 0,C1:
x321
C1
C2
−1
−2
−3 (2, 3)−
y
r�t� � �t j,
�t � 3�i � 3j,
0 ≤ t ≤ 3
3 ≤ t ≤ 5
52.
� �4
0�9
2t 2 �
12
t 3�2 � 2t� dt � �32
t 3 �15
t5�2 � t2�4
0� 96 �
15
�32� � 16 �5925
�C
�2x � y� dx � �x � 3y� dy � �4
0 ��2t � t3�2� � �t � 3t 3�2��3
2t1�2�� dt
dy �32
t1�2 dtdx � dt,0 ≤ t ≤ 4,y�t� � t 3�2,x�t� � t,
54.
� �52
sin2 t � 12t���2
0�
52
� 6�
� ���2
0 �5 sin t cos t � 12 cos2 t � 12 sin2 t� dt
� C
�2x � y� dx � �x � 3y� dy � ���2
0 �8 sin t � 3 cos t��4 cos t� dt � �4 sin t � 9 cos t���3 sin t� dt
dy � �3 sin t dtdx � 4 cos t dt,
0 ≤ t ≤ �
2y�t� � 3 cos t,x�t� � 4 sin t,
416 Chapter 14 Vector Analysis
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56.
line from
Lateral surface area:
�C
f �x, y� ds � �4
0 t ��2 � dt � 8�2
r��t� � �2
r��t� � i � j
0 ≤ t ≤ 4 r�t� � t i � t j,
�0, 0� to �4, 4�C:
f �x, y� � y 58.
from to
Lateral surface area:
� �sin t � cos t���2
0� 2
�C
f �x, y� ds � ���2
0 �cos t � sin t� dt
r��t� � 1
r��t� � �sin t i � cos t j
0 ≤ t ≤ �
2 r�t� � cos t i � sin t j,
�0, 1��1, 0�C: x2 � y2 � 1
f �x, y� � x � y
60.
from
Lateral surface area:
�2332
�5 �3364
ln�2 � �5 � �1
64�46�5 � 33 ln�2 � �5 �� � 2.3515
�12
�2�5 � ln�2 � �5 �� �1
64�18�5 � ln�2 � �5 ��
�1
64�2�1 � t��2�4��1 � t�2 � 1��1 � 4�1 � t�2 � ln�2�1 � t� � �1 � 4�1 � t�2��1
0
� �12�2�1 � t��1 � 4�1 � t�2 � ln�2�1 � t� � �1 � 4�1 � t�2��
1
0
� 2 �1
0 �1 � 4�1 � t�2 dt � �1
0 �1 � t�2�1 � 4�1 � t�2 dt
�C
f �x, y� ds � �1
0 �2 � �1 � t�2��1 � 4�1 � t�2 dt
r��t� � �1 � 4�1 � t�2
r��t� � �i � 2�1 � t�j
0 ≤ t ≤ 1 r�t� � �1 � t�i � �1 � �1 � t�2� j,
�1, 0� to �0, 1�C: y � 1 � x2
f �x, y� � y � 1
62.
Lateral surface area:
�C
f �x, y� ds � �2�
0 �4 cos2 t � 4 sin2 t � 4��2� dt � 8�2π
0 �1 � cos 2t� dt � �8�t �
12
sin 2t��2�
0� 16�
r��t� � 2
r��t� � �2 sin t i � 2 cos t j
0 ≤ t ≤ 2� r�t� � 2 cos t i � 2 sin t j,
C: x2 � y2 � 4
f �x, y� � x2 � y2 � 4
Section 14.2 Line Integrals 417
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64.
Lateral surface area:
Let then and
�881�
u5
5�
179u3
3 ��91
1�
850,304�91 � 71841215
� 6670.12
�40
0�20 �
14
t��1 � �94�t dt � ��91
1 �20 �
19
�u2 � 1���u��89
u� du �8
81��91
1 �u4 � 179u2� du
dt �89 u du.t �
49 �u2 � 1�u � �1 � �9
4�t,
�C
f �x, y� ds � �40
0�20 �
14
t��1 � �94�t dt
r��t� ��1 � �94�t
r��t� � i �32
t 1�2 j
0 ≤ t ≤ 40 r�t� � t i � t3�2 j,
0 ≤ x ≤ 40C: y � x3�2,
f �x, y� � 20 �14
x
66.
from to
Matches c.
S � 8
�2, 4��0, 0�y � x2C:
x
y
1
2
2
3
3
4
44
(2, 4, 0)
zf �x, y� � y
68.
(parabola)
yields the minimum work, 119.5. Along the straight line path, the work is 120.y � 0,w� � 16c � 4 � 0 ⇒ c �14
� 120 � 4c � 8c2
W � �1
�1 �60 � 15x2�c � cx2� � ��15x�c � cx2����2 cx�� dx
dy � �2cx dxdx � dx,
N � �15xy � �15x�c � cx2�
M � 15�4 � x2y� � 60 � 15x2�c � cx2�
x−1 1
cy c x= 1 − 2))
yW � �C
F dr � �C
M dx � N dy
70. See the definition, page 1024. 72. (a) Work
(b) Work is negative, since against force field.
(c) Work is positive, since with force field.
� 0
74. False, the orientation of C does not affect the form
�C
f �x, y� ds.
76. False. For example, see Exercise 32.
418 Chapter 14 Vector Analysis
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Section 14.3 Conservative Vector Fields and Independence of Path
2.
(a) 0 ≤ t ≤ 4
� �t 3
3�
t 2
2�
t 3�2
3 �4
0�
803
�C
F � dr � �4
0 �t 2 � t �
12�t� dt
Ft � t 2 � t i � t j
r1�t � i �1
2�tj
r1t � t i � �t j,
Fx, y � x2 � y2i � x j
(b)
� �w6
3�
w4
2�
w3
3 �2
0�
803
�C
F � dr � �2
0 �2ww4 � w 2 � w 2� dw
Fw � w 4 � w 2i � w 2 j
r2�w � 2w i � j
0 ≤ w ≤ 2r2w � w2 i � w j,
4.
(a)
(b)
� ��3 � ln w2
2�
2 � ln w3
3 �e3
1� �
692
� �e3
1 �3 � ln w�1
w� � 2 � ln w2�1w��dw�
C
F � dr
Fw � 3 � ln wi � 2 � ln w2 j
r2�w �1w
i �1w
j
1 ≤ w ≤ e3r2w � 2 � ln wi � 3 � ln wj,
�C
F � dr � �3
0 �3 � t � 2 � t2� dt � ��
3 � t2
2�
2 � t3
3 �3
0� �
692
Ft � 3 � ti � 2 � t2 j
r1�t � i � j
0 ≤ t ≤ 3r1t � 2 � ti � 3 � tj,
Fx, y � yi � x2 j
6.
Since is conservative.F�N�x
��M�y
,
�M�y
� 30x2y�N�x
� 30x2y
Fx, y � 15x2 y2 i � 10x3y j 8.
so is not conservative.
��P�y
�xz
� �xz
��N�z �
Fcurl F � 0
Fx, y, z � y ln z i � x ln z j �xyz
k
10.
is not conservative.curl F � 0, so F
Fx, y, z � sinyzi � xz cosyzj � xy sinyzk
12.
(a)
� �e3t� t2�3
0� e0 � e0 � 0
� �3
0 e3t� t 2
3 � 2t dt
�C
F � dr � �3
0 �� t � 3e3t� t 2
� te3t� t 2� dt
Ft � � t � 3e3t� t 2 i � te3t� t 2 j
r1�t � i � j
0 ≤ t ≤ 3r1t � t i � t � 3j,
Fx, y � yexy i � xe xy j
(b) is conservative since
The potential function is f x, y � exy � k.
�M�y
��N�x
� xyexy � exy.
Fx, y
Section 14.3 Conservative Vector Fields and Independence of Path 419
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14.
(a)
� ��ln t �3
1� �ln 3
�C
F � dr � �3
1 �
1t dt
Ft �1t
i � 2 t j
r1�t � i �1t 2 j
1 ≤ t ≤ 3r1t � t i �1t
j,
Fx, y � xy2 i � 2x2y j
(b)
�19�
3t 4
4�
7t 3
3�
7t 2
2� 3t�
2
0� �
4427
�19
�2
0 3t 3 � 7t 2 � 7t � 3 dt
�C
F � dr � �2
0 �1
9t � 1t � 32 �
29
t � 12t � 3� dt
Ft �19
t � 1t � 32 i �23
t � 12t � 3 j
r2�t � i �13
j
0 ≤ t ≤ 2r2t � t � 1i �13
t � 3j,
16.
Since is conservative. The potential function is
(a) and (d) Since C is a closed curve,
(b)
(c) �C
2x � 3y � 1 dx � 3x � y � 5 dy � �x2 � 3xy �y2
2� x � 5y�
2, e2
0, 1�
12
3 � 2e2 � e4
�C
2x � 3y � 1 dx � 3x � y � 5 dy � �x2 � 3xy �y2
2� x � 5y�
0, 1
0, �1� 10
�C
2x � 3y � 1 dx � 3x � y � 5 dy � 0.
f x, y � x2 � 3xy � y2�2 � x � 5y � k.Fx, y � 2x � 3y � 1 i � 3x � y � 5 j�M��y � �N��x � �3,
�C
2x � 3y � 1 dx � 3x � y � 5 dy
18.
Since
is conservative. The potential function is
(a)
(b) � �83�
C
x 2 � y 2 dx � 2xy dy � �x3
3� xy 2�
0, 2
2, 0
�8963
�C
x2 � y2 dx � 2xy dy � �x3
3� xy2�
8, 4
0, 0
f x, y � x3�3 � xy2 � k.
Fx, y � x2 � y2 i � 2xy j
�M��y � �N��x � 2y,
�C
x2 � y2 dx � 2xy dy 20.
Since is conservative. The potentialfunction is
(a)
(b)
�C
F � dr � �x � yz��1, 0, 2
1, 0, 0� �2
0 ≤ t ≤ 1r2t � 1 � 2ti � 2tk,
�C
F � dr � �x � yz��1, 0, 2
1, 0, 0� �2
0 ≤ t ≤ r1t � cos t i � sin t j � t 2 k,
f x, y, z � x � yz � k.F x, y, zcurl F � 0,
Fx, y, z � i � z j � yk
22.
is not conservative.
(a)
—CONTINUED—
� �t�
0 � 3�t 2 sin t�
0 � 6 �
0 t sin t dt � �t � 3t 2 sin t � 6sin t � t cos t�
0� �5
�C
F � dr � �
0 �sin2 t � cos2 t � 3t 2 cos t� dt � �
0 �1 � 3t 2 cos t� dt
Ft � �sin t i � cos t j � 3t 2 cos tk
r1�t � �sin t i � cos t j � k
0 ≤ t ≤ r1t � cos t i � sin t j � tk,
Fx, y, z
Fx, y, z � �y i � x j � 3xz2 k
420 Chapter 14 Vector Analysis
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22. —CONTINUED—
(b)
�C
F � dr � �1
0 3 3t 21 � 2t dt � 3 3 �1
0 t 2 � 2t 3 dt � 3 3�t 3
3�
t 4
2�1
0� �
3
2
Ft � 1 � 2tj � 3 2t 2 1 � 2tk
r2�t � �2 i � k
0 ≤ t ≤ 1r2t � 1 � 2t i � tk,
24.
(a)
(b)
�C
F � dr � �1
0 0 dt � 0
Ft � 16t 2 cos4tk
r2�t � 4 i � 4 j
0 ≤ t ≤ 1r2t � 4t i � 4t j,
�C
F � dr � �2
0 0 dt � 0
Ft � t 4 cos t2 k
r1�t � 2 t i � 2 t j
0 ≤ t ≤ 2r1t � t 2 i � t 2 j,
Fx, y, z � y sin z i � x sin z j � xy cos xk 26.
� 49
�C
�2x � yi � 2x � yj� � dr � �x � y2�4, 3
�3, 2
28. �C
y dx � x dy
x2 � y2 � �arctan�xy��
2�3, 2
1, 1�
3�
4�
12
30. �C
2x
x2 � y22 dx �2y
x2 � y22 dy � ��1
x2 � y2�1, 5
7, 5� �
126
�1
74�
�12481
32.
Note: Since is conservative and the potential function is the integral isindependent of path as illustrated below.
(a)
(b)
(c) �xyz�1, 0, 0
0, 0, 0� �xyz�
1, 1, 0
1, 0, 0� �xyz�
1, 1, 1
1, 1, 0� 0 � 0 � 1 � 1
�xyz�0, 0, 1
0, 0, 0� �xyz�
1, 1, 1
0, 0, 1� 0 � 1 � 1
�xyz�1, 1, 1
0, 0, 0� 1
f x, y, z � xyz � k,Fx, y, z � yz i � xz j � xyk
�C
zy dx � xz dy � xy dz
34. �C
6x dx � 4z dy � 4y � 20z dz � �3x2 � 4yz � 10z2�4, 3, 1
0, 0, 0� 46
36. is conservative.
Work � �x 2
y �1, 4
�3, 2�
14
�92
� �174
Fx, y �2xy
i �x2
y2 j
Section 14.3 Conservative Vector Fields and Independence of Path 421
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38.
Since is conservative, the work done in moving a particle along any path from P to Q is
� a1q1 � p1 � a2q2 � p2 � a3q3 � p3 � F � PQ\
.
f x, y, z � �a1x � a2 y � a3z�Q�q1, q2, q3
P� p1, p2, p3
Fx, y, z
Fx, y, z � a1i � a2 j � a3k
40.
(a)
�C
F � dr � �50
0 150 dt � 7500 ft � lbs
dr � i � j dt
0 ≤ t ≤ 50rt � t i � 50 � tj,
F � �150j
(b)
�C
F � dr � 6�50
0 50 � t dt � 7500 ft � lbs
dr � i �1
25 50 � tj dt
rt � t i �15050 � t2 j
42.
(a)
Thus,
(c)
(e)
�y
x2 � y2 i �x
x2 � y2 j � F
�arctan xy� �
1�y1 � x�y2 i �
�x�y 2
1 � x�y2 j
� �t�
0�
�C
F � dr � �
0 sin2 t � cos2 t dt
dr � �sin t i � cos t j dt
F � �sin t i � cos t j
0 ≤ t ≤ rt � cos t i � sin t j,
�N�x
��M�y
.
�N�x
�x 2 � y 2�1 � x2x
x 2 � y 22 �x 2 � y 2
x 2 � y 22
N � �x
x 2 � y 2
�M�y
�x 2 � y 21 � y 2y
x 2 � y 22 �x2 � y 2
x2 � y 22
M �y
x2 � y2
Fx, y �y
x2 � y2 i �x
x2 � y2 j
(b)
(d)
This does not contradict Theorem 14.7 since F is notcontinuous at in R enclosed by curve C.0, 0
� ��t�2
0� �2
�C
F � dr � �2
0 �sin2 t � cos2 t dt
dr � �sin t i � cos t j dt
F � sin t i � cos t j
0 ≤ t ≤ 2rt � cos t i � sin t j,
� ��t�
0� � �
C
F � dr � �
0 �sin2 t � cos2 t dt
dr � �sin t i � cos t j dt
F � sin t i � cos t j
0 ≤ t ≤ rt � cos t i � sin t j,
44. A line integral is independent of path if does not depend on the curve joining P and Q. See Theorem 14.6�C
F � dr
46. No, the amount of fuel required depends on the flight path. Fuel consumption is dependent on wind speed and direction. The vector field is not conservative.
48. True 50. False, the requirement is �M��y � �N��x.
422 Chapter 14 Vector Analysis
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Section 14.4 Green’s Theorem
2.
By Green’s Theorem, �R� ��N
�x�
�M�y � dA � �4
0�x
0 �2x � 2y� dy dx � �4
0 x2 dx �
643
.
� 0 � 64 �1283
�643
� �12
8 ��12 � t�2��dt� � �12 � t�2��dt��
�C
y2 dx � x2 dy � �4
0 �0 dt � t 2�0�� � �8
4 ��t � 4�2�0� � 16 dt�
x1
1
2
2
3
3
4
4(4, 4)
y x=
y
r�t� � t i,4 i � �t � 4�j,�12 � t�i � �12 � t�j,
0 ≤ t ≤ 44 ≤ t ≤ 88 ≤ t ≤ 12
4.
By Green’s Theorem,
� �2�
0�1
0 �2r cos � � 2r sin ��r dr d� �
23
�2�
0 �cos � � sin �� d� �
23
�0� � 0.
�R���N
�x�
�M�y � dA � �1
�1�1�x2
�1�x2
�2x � 2y� dy dx
� �sin t �sin3 t
3� cos t �
cos3 t3 �
2�
0� 0
� �2�
0 �cos t�1 � sin2 t� � sin t�1 � cos2 t�� dt
� �2�
0 �cos3 t � sin3 t� dt
�C
y2 dx � x2 dy � �2�
0 �sin2 t ��sin t dt� � cos2 t�cos t dt��
x1
1
−1
−1
x y2 2+ =1
y0 ≤ t ≤ 2�r�t� � cos t i � sin t j,
6. C: boundary of the region lying between the graphs of and
�R� ��N
�x�
�M�y � dA � �1
0�x
x3
�ex � xey� dy dx � �1
0 �xex3
� x3 ex� dx 0.22
�C
xey dx � ex dy � �1
0 �xex3
� 3x2ex� dx � �0
1 �xex � ex� dx 2.936 � 2.718 0.22
y � x3y � x
8. Since C is an ellipse with and then R is an ellipse of area Thus, Green’s Theorem yields
�C
� y � x� dx � �2x � y� dy � �R� 1 dA � Area of ellipse � 2�.
�ab � 2�.b � 1,a � 2
10. R is the shaded region of the accompanying figure.
�12
� �25 � 9� � 8�
� Area of shaded region
�C
�y � x� dx � �2x � y� dy � �R�
1 dA
−5 −4 −3 −2
−2
12
4
−3−4−5
−1 1 2 3 4 5
y
x
In Exercises 8 and 10,�N�x
��M�y
� 1.
Section 14.4 Green’s Theorem 423
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12. The given curves intersect at and Thus, Green’s Theorem yields
� �9
0�x
0 �y dy dx � �9
0 ��y2
2 �x
0 dx � �9
0 �x2
dx � ��x2
4 �9
0� �
814
�C
y2 dx � xy dy � �R� �y � 2y� dA
�9, 3�.�0, 0�
14. In this case, let Then and Green’s Theorem yields
� ���1 � cos ��4
3 �2�
0� 0.
�43
�2�
0 sin ��1 � cos ��3 d�
� 4 �2�
0�1�cos�
0 r2 sin� dr d�
�C
�x2 � y2� dx � 2xy dy � �R� 4y dA � 4 �2�
0�1�cos�
0 r sin �r dr d�
dA � r dr d�x � r cos �.y � r sin �,
16. Since we have
�R� ��N
�x�
�M�y � dA � 0.
�M�y
� �2ex sin 2y ��N�x
18. By Green’s Theorem,
�C
�e�x2�2 � y� dx � �e�y2�2 � x� dy � �R� 2 dA � 2�Area of R� � 2���6�2 � ��2��3�� � 60�.
20. By Green’s Theorem,
� �16e2 � 16e�2 � 2e � 2e�1.
� �7�e2 � e�2� � 2�e2 � e� � 7�e2 � e�2� � 2�e�1 � e�2�
� ��1
�2�2
�2�3x2ey dy dx � �1
�1��1
�2 �3x2ey dy dx
� �2
1�2
�2 �3x2ey dy dx � �1
�1�2
1 �3x2ey dy dx
�C
3x2 ey dx � ey dy � �R� �3x2ey dA
x
(1, 1)
(2, 2)
( 2, 2)− − (2, 2)−
(1, 1)−( , )−1 −1
( , )−1 1
( , )−2 2
y
22.
since is a circle with a radius of one.r � 2 cos �Work � �C
�ex � 3y� dx � �ey � 6x� dy � �R� 9 dA � 9�
C: r � 2 cos �
F�x, y� � �ex � 3y�i � �ey � 6x�j
24.
C: boundary of the region bounded by the graphs of
Work � �C
�3x2 � y� dx � 4xy2 dy � �4
0�x
0 �4y2 � 1� dy dx � �4
0 �4
3 x3�2 � x1�2� dx �17615
x � 4y � 0,y � x,
F�x, y� � �3x2 � y�i � 4xy2 j
424 Chapter 14 Vector Analysis
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26. From the figure we see that
�12
�0
2 ��4� dx � 2�2
0dx � 4
A �12
�2
0 �3
2x �
32
x� dx �12�
0
2 ��
12
x �x2
� 4� dx �12
�0�
dx � 0.C3: x � 0,
dy � �12
dxC2: y � �x2
� 4,
0 ≤ x ≤ 2dy �32
dx,C1: y �32
x, (2, 3)
3x − 2y = 0
x + 2y = 8
1
1
2
3
4
2 3 4
y
x
C1
C2
C3
28. Since the loop of the folium is formed on the interval
and
we have
�92
��
0
t5 � t 2
�t 3 � 1�3 dt �92
��
0 t2�t 3 � 1��t 3 � 1�3 dt �
32
��
0 3t2�t3 � 1��2 dt � � �3
2�t 3 � 1���
0�
32
.
A �12
��
0 �� 3t
t 3 � 1� 3�2t � t 4��t 3 � 1�2 � � 3t 2
t 3 � 1�3�1 � 2t 3��t 3 � 1�2 � dt
dy �3�2t � t4��t3 � 1�2 dt,dx �
3�1 � 2t3��t3 � 1�2 dt
0 ≤ t ≤ �,
30. See Theorem 14.9: A �12�C
x dy � y dx.
32. (a) For the moment about the x-axis, Let and By Green’s Theorem,
and
For the moment about the y-axis, Let and By Green’s Theorem,
and x �My
2A�
12A
�C
x2 dy.My � �C
x2
2 dy �
12
�C
x2 dy
M � 0.N � x2�2My � �R� x dA.
y �Mx
2A� �
12A
�C
y2 dx.Mx � �C
�y2
2 dx � �
12
�C
y2 dx
M � �y2�2.N � 0Mx � �R� y dA.
(b) By Theorem 14.9 and the fact that we have
A �12
� x dy � y dx �12
� �r cos ���r cos �� d� � �r sin ����r sin �� d� �12
�C
r2 d�.
y � r sin �,x � r cos �,
34. Since area of semicircle we have Note that and along the boundary
Let then
�x, y � � �0, 4a3��
y ��1�a2 ��
0 a2 sin2 t��a sin t dt� �
a�
��
0 sin3 t dt �
a���cos t �
cos3 t3 �
�
0�
4a3�
.
x �1
�a2 ��
0 a2 cos2 t�a cos t� dt �
a�
��
0 cos3 t dt �
a�
��
0 �1 � sin2 t� cos t dt �
a��sin t �
sin3 t3 �
�
0� 0
0 ≤ t ≤ �,y � a sin t,x � a cos t,
y � 0.dy � 0y � 01
2A�
1�a2.�
�a2
2,A �
Section 14.4 Green’s Theorem 425
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36. Since we have
Thus,
�x, y � � �b3
, c3�
��12ac�
c2�b � a�3
�c2�b � a�
3 � �c3
.
y ��12ac�C
y2 dx ��12ac�0 � �b
a
� cb � a�
2
�x � a�2 dx � ��a
b
� cb � a�
2
�x � a�2 dx�
x �1
2ac �
C
x2 dy �1
2ac��a
�a 0 � �b
a
x2 cb � a
dx � ��a
b
x2 cb � a
dx� �1
2ac�0 �2abc
3 � �b3
dy �c
b � a dx.C3: y �
cb � a
�x � a�,
dy �c
b � a dxC2: y �
cb � a
�x � a�,
dy � 0C1: y � 0,
xa
2a
−a C1
C3 C2
(b, c)
y12A
�1
2ac,A �
12
�2a��c� � ac,
38.
Note: In this case R is enclosed by where 0 ≤ � ≤ �.r � a cos 3�
A �12
��
0 a2 cos2 3� d� �
a2
2 ��
0 1 � cos 6�
2 d� �
a2
4 �� �sin 6�
6 ��
0�
�a2
4
40. In this case, and we let
Now as and we have
�63
��
2� �63�
u1 � 3u2�
�
0� � 6
3 arctan 3u��
0�
3�
3� 0 �
3�
3� 23�.
� 18 ��
0
1�31 � 3u2 du � 18 ��
0
2�3�1 � 3u2�2 du � � 6
3 arctan 3 u��
0�
123 �
12� � u
1 � 3u2 � � 3
1 � 3u2 du��
0
A � 2�12� ��
0
9�2 � cos ��2 d� � 9 ��
0
2du1 � u2
4 � 4�1 � u2
1 � u2� ��1 � u2�2
�1 � u2�2
� 18 ��
0
1 � u2
�1 � 3u2�2 du
� ⇒ �u ⇒ �
d� �2 du
1 � u2.cos � �1 � u2
1 � u2,u �sin �
1 � cos �,
0 ≤ � ≤ 2�
42. (a) Let C be the line segment joining and
—CONTINUED—
� x1�y2 � y1� � y1�x2 � x1� � x1y2 � x2y1
� ��x1�y2 � y1
x2 � x1� � y1�x�
x2
x1
� �x1�y2 � y1
x2 � x1� � y1� �x2 � x1�
�C
�y dx � x dy � �x2
x1
��y2 � y1
x2 � x1�x � x1� � y1 � x�y2 � y1
x2 � x1�� dx � �x2
x1
�x1�y2 � y1
x2 � x1� � y1� dx
dy �y2 � y1
x2 � x1 dx
y �y2 � y1
x2 � x1�x � x1� � y`
1
�x2, y2�.�x1, y1�
426 Chapter 14 Vector Analysis
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Section 14.5 Parametric Surfaces
42. —CONTINUED—
(b) Let C be the boundary of the region
Therefore,
where is the line segment joining and is the line segment joining and and isthe line segment joining and Thus,
�R� dA �
12
��x1y2 � x2y1� � �x2y3 � x3y2� � . . . � �xn�1 yn � xn yn�1� � �xn y1 � x1yn ��.
�x1, y1�.�xn, yn�Cn�x3, y3�, . . . ,�x2, y2�C2�x2, y2�,�x1, y1�C1
�R� dA �
12��C1
�y dx � x dy � �C2
�y dx � x dy � . . . � �Cn
�y dx � x dy�
A �12
�C
�y dx � x dy �12�R
� �1 � ��1�� dA � �R� dA.
44. Hexagon:
A �12 ��0 � 0� � �4 � 0� � �12 � 4� � �6 � 0� � �0 � 3� � �0 � 0�� �
212
��1, 1��0, 3�,�2, 4�,�3, 2�,�2, 0�,�0, 0�,
46. Since then
� �R� div� f �g� dA � �
R� � f div ��g� � �f � �g� dA � �
R� � f �2g � �f � �g� dA.
�C
fDN g ds � �C
f �g � N ds
�C
F � N ds � �R� div F dA,
48. �C
f �x� dx � g�y� dy � �R� � �
�xg�y� �
�
�y f �x�� dA � �
R� �0 � 0� dA � 0
2.
Matches d.
x2 � y2 � z2
r�u, v� � u cos v i � u sin v j � uk 4.
Matches a.
x2 � y2 � 16
r�u, v� � 4 cos ui � 4 sin uj � vk
6.
Paraboloid
y
4
44x
z
x2 � y2 � 4u2 ⇒ z �18
�x2 � y2�z �12
u2,
r�u, v� � 2u cos v i � 2u sin v j �12
u2 k 8.
Ellipsoid
y
5
434
3x
z
x2
9�
y2
9�
z2
25� 1
x2 � y2
9�
z2
25� cos2 v � sin2 v � 1
x2 � y2 � 9 cos2 v cos2 u � 9 cos2 v sin2 u � 9 cos2 v
r�u, v� � 3 cos v cos ui � 3 cos v sin u j � 5 sin vk
Section 14.5 Parametric Surfaces 427
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For Exercises 10 and 12,
Eliminating the parameter yields
0 ≤ z ≤ 4.z � x2 � y2,
0 ≤ v ≤ 2�.0 ≤ u ≤ 2,ru, v � u cos vi � u sin vj � u2k,
x
y2
5
2
z
10.
The paraboloid opens along the y-axis instead of the z-axis.
y � x2 � z2
0 ≤ v ≤ 2�0 ≤ u ≤ 2,s�u, v� � u cos v i � u2 j � u sin vk,
12.
The paraboloid is “wider.” The top is now the circle It was x2 � y2 � 4.x2 � y2 � 64.
z �x2 � y2
16
0 ≤ v ≤ 2�0 ≤ u ≤ 2,s�u, v� � 4u cos v i � 4u sin v j � u2 k,
14.
x2
4�
y2
16�
z2
1� 1
x y
3
−4−5
−5−5
−4
−3
5
55
43
4
z0 ≤ v ≤ 2�0 ≤ u ≤ 2�,
r�u, v� � 2 cos v cos ui � 4 cos v sin u j � sin vk, 16.
z � arctan�yx�
yx
2
−2−4 −4
42
4
4
8
z0 ≤ v ≤ 3�0 ≤ u ≤ 1,
r�u, v� � 2u cos v i � 2u sin v j � vk,
18.
xy
−1−1
11
2
z
0 ≤ v ≤ 2�0 ≤ u ≤ �
2,
r�u, v� � cos3 u cos v i � sin3 u sin v j � uk, 20.
r�u, v� � u i � v j � �6 � u � v�k
z � 6 � x � y
22.
r�u, v� � 2 cos ui � 4 sin uj � vk
4x2 � y2 � 16 24.
r�u, v� � 3 cos v cos ui � 2 cos v sin uj � sin vk
x2
9�
y2
4�
z2
1� 1
26. inside
0 ≤ v ≤ 3r�u, v� � v cos u i � v sin uj � v2 k,
x2 � y2 � 9.z � x2 � y2 28. Function:
Axis of revolution: x-axis
0 ≤ v ≤ 2�0 ≤ u ≤ 4,
z � u3 2 sin vy � u3 2 cos v,x � u,
0 ≤ x ≤ 4y � x3 2,
30. Function:
Axis of revolution: y-axis
0 ≤ v ≤ 2�0 ≤ u ≤ 2,
z � �4 � u2� sin vy � u,x � �4 � u2� cos v,
0 ≤ y ≤ 2z � 4 � y2,
428 Chapter 14 Vector Analysis
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32.
At and
Direction numbers:
Tangent plane:
x � y � 2z � 0
�x � 1� � �y � 1� � 2�z � 1� � 0
1, 1, �2
N � ru�1, 1� rv�1, 1� � � i
1
0
j
0
1
k1212 � � �
12 i �
12 j � k
rv�1, 1� � j �12
kru�1, 1� � i �12
k,
v � 1.u � 1�1, 1, 1�,
rv�u, v� � j �u
2�uvkru�u, v� � i �
v
2�uvk,
�1, 1, 1�r�u, v� � u i � v j � �uv k, 34.
At and
Direction numbers:
Tangent plane:
x � z � �2
�x � 4� � �z � 2� � 0
1, 0, 1
N � ru rv � �8i � 8k
rv��2, 0� � �4jru��2, 0� � 2i � 2k,
v � 0.u � �2��4, 0, 2�,
rv�u, v� � 2u sinh v i � 2u cosh v j
ru�u, v� � 2 cosh v i � 2 sinh v j � uk
r�u, v� � 2u cosh v i � 2u sinh v j �12
u2 k,
36.
A � �2�
0�2
0 8u�u2 � 4 du dv � �2�
0�128�2
3�
643 � dv �
128�
3�2�2 � 1�
�ru rv� � �64u4 � 256u2 � 8u�u2 � 4
ru rv � � i4 cos v
�4u sin v
j4 sin v
4u cos v
k2u0 � � �8u2 cos v i � 8u2 sin v j � 16uk
rv�u, v� � �4u sin v i � 4u cos v j
ru�u, v� � 4 cos v i � 4 sin v j � 2uk
0 ≤ v ≤ 2�0 ≤ u ≤ 2,r�u, v� � 4u cos v i � 4u sin v j � u2 k,
38.
A � �2�
0��
0 a2 sin u du dv � 4�a2
�ru rv� � a2 sin u
a2 sin2 u cos vi � a2 sin2 u sin v j � a2 sin u cos ukru rv � � ia cos u cos v
�a sin u sin v
ja cos u sin va sin u cos v
k�a sin u
0 � �
rv�u, v� � �a sin u sin v i � a sin u cos v j
ru�u, v� � a cos u cos v i � a cos u sin v j � a sin uk
0 ≤ v ≤ 2�0 ≤ u ≤ π,r�u, v� � a sin u cos v i � a sin u sin v j � a cos uk,
40.
A � �2�
0�2�
0 b�a � b cos v� du dv � 4� 2ab
�ru rv� � b�a � b cos v�
� b cos u cos v�a � b cos v�i � b sin u cos v�a � b cos v� j � b sin v�a � b cos v�k
ru rv � � i��a � b cos v� sin u
�b sin v cos u
j�a � b cos v� cos u
�b sin v sin u
k0
b cos v �rv�u, v� � �b sin v cos u i � b sin v sin u j � b cos vk
ru�u, v� � ��a � b cos v� sin u i � �a � b cos v� cos u j
0 ≤ v ≤ 2�0 ≤ u ≤ 2�,a > b,r�u, v� � �a � b cos v� cos u i � �a � b cos v� sin u j � b sin vk,
Section 14.5 Parametric Surfaces 429
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42.
A � �2�
0��
0 sin u �1 � cos2 u du dv � � �2�2 � ln��2 � 1
�2 � 1���ru rv� � sin u �1 � cos2 u
ru rv � sin u cos v i � cos u sin u j � sin u sin vk
rv�u, v� � �sin u sin v i � sin u cos vk
ru�u, v� � cos u cos v i � j � cos u sin vk
0 ≤ v ≤ 2�0 ≤ u ≤ �,r�u, v� � sin u cos v i � u j � sin u sin vk,
44. See the definition, page 1055.
(b)
x
y
3
3
−3
−3
�0, 0, 10� (c)
x y3
3
3
z
�10, 10, 10�
46. Graph of
0 ≤ u ≤ from
(a)
y
3
−3
z
�10, 0, 0�
0 ≤ v ≤ ��,
r�u, v� � u cos v i � u sin v j � vk
48. 0 ≤ v ≤ 3�0 ≤ u ≤ 1,r�u, v� � 2u cos v i � 2u sin v j � vk,
(a) If
Helix
0 ≤ z ≤ 3�
yx 22
2
4
8
10
−2−2
zx2 � y2 � 4
r�1, v� � 2 cos v i � 2 sin v j � vk
u � 1: (b) If
Line
z �2�
3
y � ��3x
r�u, 2�
3 � � �u i � �3u j �2�
3k
v �2�
3:
(c) If one parameter is held constant, the result is a curve in 3-space.
1
−1−1
−2−2
1
1
2
22yx
z
50.
Let and Then,
At and is undefined and The tangent plane at is x � 1.�1, 0, 0�rv�1, 0� � j.ru�1, 0�v � 0.u � 1�1, 0, 0�,
rv�u, v� � �u sin v i � u cos v j.
ru�u, v� � cos v i � sin v j �u
�u2 � 1k
z � �u2 � 1.y � u sin v,x � u cos v,
x2 � y2 � z2 � 1
430 Chapter 14 Vector Analysis
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Section 14.6 Surface Integrals
52.
�since u � x� � 2� �b
a
f �x��1 � � f��x��2 dx
A � �2�
0�b
a
f �u��1 � � f��u��2 du dv
�ru � rv� � f �u��1 � � f��u��2
f �u�f��u�i � f �u� cos v j � f �u� sin vkru � rv � i10
jf��u� cos v
�f �u� sin v
kf��u� sin vf �u� cos v �
rv�u, v� � �f �u� sin v j � f �u� cos vk
ru�u, v� � i � f��u� cos v j � f��u� sin vk
0 ≤ v ≤ 2�a ≤ u ≤ b,r�u, v� � ui � f �u� cos v j � f �u� sin vk,
2.
� �14�2
0�4
0�15 � x � y� dy dx � 128�14
�S��x � 2y � z� dS � �2
0�4
0�x � 2y � 15 � 2x � 3y��14 dy dx
dS � �1 � 4 � 9 dy dx � �14 dy dx�z�y
� 3,�z�x
� �2,0 ≤ y ≤ 4,0 ≤ x ≤ 2,S: z � 15 � 2x � 3y,
4. ,
� 16
x5�2�1 � x�3�2�1
0�
512
13�x3�2�1 � x�3�2�
1
0�
524�
1
0x1�2�1 � x dx
�23
14
x5�2�1 � x�3�2�1
0�
512�
1
0x3�2�1 � x dx
�23�
1
0x5�2�x � 1 dx
� �1
0�x
0 x � 2y �
23
x3�2��1 � x dy dx
�S��x � 2y � z� dS � �1
0�x
0 x � 2y �
23
x3�2��1 � �x1�2�2 � �0�2 dy dx
�z�y
� 0�z�x
� x1�20 ≤ y ≤ x,0 ≤ x ≤ 1,S: z �23
x3�2,
��218
�15�2
96�
5192
ln 1
3 � 2�2 �61�2288
�5
192 ln3 � 2�2 � 0.2536
��218
�548
32�2 �
14
ln32 � �2 �14
ln12� �
�218
�524
12� x �
12��x2 � x �
14
ln x �12� � �x2 � x�1
0
��218
�524�
1
0� x �
12�
2
�14
dx
��23
�5�218
�524�
1
0
�x � x2 dx
Section 14.6 Surface Integrals 431
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8.
�S�xy dS � �4
0�4
0xy�1 �
y2
4�
x2
4 dy dx �
390415
�160�5
3
�z�y
�12
x�z�x
�12
y,0 ≤ y ≤ 4,0 ≤ x ≤ 4,S: z �12
xy,
6.
�122x 2 �
x4
4 �2
0� 2 �
12�
2
0x�4 � x 2� dx�
S�dx dS � �2
0��4�x2
0xy dy dx
�z�x
��z�y
� 00 ≤ y ≤ �4 � x 2,0 ≤ x ≤ 2,S: z � h,
10.
�S��x 2 � 2xy� dS � ���2
0�x�2
0�x 2 � 2xy��1 � sin2 x dy dx � ���2
0 x3
4�1 � sin2 x dx � 0.52
0 ≤ y ≤ x2
0 ≤ x ≤ �
2,S: z � cos x,
12.
� �R�ka dA � ka�
R� dA � ka�2�a2� � 2ka3�
� �R� k�a2 � x 2 � y 2 a
�a2 � x 2 � y 2� dA
m � �S�kz dS � �
R� k�a2 � x 2 � y 2�1 � �x
�a2 � x 2 � y 2�2
� �y
�a2 � x 2 � y 2�2
dA
�x, y, z� � kz
y
x
a
a a
zS: z � �a2 � x 2 � y 2
14. ,
�S��x � y� dS ��2
0���2
0�2 cos u � 2 sin u�2 du dv � 16
�ru � rv� � �2 cos ui � 2 sin u j� � 2
0 ≤ v ≤ 2
0 ≤ u ≤ �
2S: r�u, v� � 2 cos u i � 2 sin u j � vk,
16.
�S��x � y� dS � ��
0�4
0�4u cos v � 4u sin v�20u du dv �
10,2403
�ru � rv� � ��12u cos v i � 12u sin v j � 16uk� � 20u
0 ≤ v ≤ �0 ≤ u ≤ 4,S: r�u, v� � 4u cos v i � 4u sin v j � 3uk,
18.
� 65�65 � 17�1712 sin2
2 ��2�
0� 0
� �2�
0�4
2 r�1 � 4r2 sin cos dr d � �2�
0 1
12�1 � 4r2�3�2�
4
2 sin cos d
�S�f �x, y, z� dS � �
S�
xyx 2 � y 2
�1 � 4x 2 � 4y 2 dy dx � �2�
0�4
2 r2 sin cos
r2�1 � 4r2 r dr d
4 ≤ x 2 � y 2 ≤ 16S: z � x 2 � y 2,
f �x, y, z� �xyz
432 Chapter 14 Vector Analysis
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20.
� 163 sin �
sin3 3 ��
�
0� 0
�163 �
�
0 cos3 d �
163 �
�
0�1 � sin2 � cos d
� 2�S��x 2 � y 2 dy dx � 2��
0�2 cos
0 r2 dr d
� �S��2�x 2 � y 2��2�x 2 � y 2�
x 2 � y 2 dy dx
�S�f �x, y, z� dS � �
S� �x 2 � y 2 � ��x 2 � y 2 �2�1 � x
�x 2 � y 2�2
� y
�x 2 � y 2�2
dy dx
�x � 1�2 � y 2 ≤ 1S: z � �x 2 � y 2,
f �x, y, z� � �x 2 � y 2 � z2
22.
0 ≤ x ≤ 3, 0 ≤ z ≤ x
Project the solid onto the xz-plane;
Let then
� 81 �23
�9 � x 2�3�2�3
0� 81 � 18 � 99
� �27�9 � x 2�3
0� �x 2�9 � x 2�
3
0� �3
0 2x�9 � x 2 dx�
v � ��9 � x 2.du � 2x dx,dv � x�9 � x 2��1�2 dx,u � x 2,
� �3
0
3
�9 � x 2 9x �x3
3 � dx � �3
0 27x�9 � x 2��1�2 dx � �3
0 x3�9 � x 2��1�2 dx
� �3
0�x
0�9 � z 2� 3
�9 � x 2 dz dx � �3
0 3
�9 � x 2 9z �z3
3 ��x
0 dx
�S�f �x, y, z� dS � �3
0�x
0 �x 2 � �9 � x 2� � z2��1 � �x
�9 � x 2�2
� �0�2 dz dx
y � �9 � x 2.
S: x 2 � y 2 � 9,
f �x, y, z� � x 2 � y 2 � z2
24.
(first octant)
� �49
x3 � 2x 2 �34 �
23
x � 2�3
�3
0� 12
� �3
0�
43
x 2 � 4x �32 �
23
x � 2�2
� dx
�S�F � N dS � �
R�F � �G dA � �3
0���2x�3��2
0�2x � 3y� dy dx
�G�x, y, z� � 2 i � 3 j � k
G�x, y, z� � 2x � 3y � z � 6
S: 2x � 3y � z � 6
2
2
31
3
1
x
R
y x= + 2− 23
yF�x, y, z� � x i � y j
Section 14.6 Surface Integrals 433
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26.
(first octant)
(improper)
� 108�
� ���2
0�6
0
36
�36 � r2r dr d
�S�F � N dS � �
R�F � �G dA � �
R�
36
�36 � x 2 � y 2 dA
F � �G �x 2
�36 � x 2 � y 2�
y 2
�36 � x 2 � y 2� z �
36
�36 � x 2 � y 2
�G�x, y, z� �x
�36 � x 2 � y 2i �
y
�36 � x 2 � y 2j � k
G�x, y, z� � z � �36 � x 2 � y 2
z � �36 � x 2 � y 2
S: x 2 � y 2 � z2 � 36
1
1
2
3
4
5
6
2 3 4 5 6
y
x
R
x2 + y2 = 62
F�x, y, z� � x i � y j � zk
28.
� 3�2�
0 23
a3 d � 2a2�2�
0 a d � 0
� 3�2�
0�r2�a2 � r2 �
23
�a2 � r2�3�2�a
0 d� � 2a2�2�
0 ��a2 � r2�
a
0 d
� 3�2�
0�a
0
r3
�a2 � r2 dr d � 2a2�2�
0�a
0
r
�a2 � r2 dr d
� �2�
0�a
0 3r2 � 2a2
�a2 � r2r dr d
�S�F � N dS � �
R�F � �G dA � �
R�
3x 2 � 3y 2 � 2a2
�a2 � x 2 � y 2 dA
F � �G �x 2
�a2 � x 2 � y 2�
y 2
�a2 � x 2 � y 2� 2�a2 � x 2 � y 2 �
3x 2 � 3y 2 � 2a2
�a2 � x 2 � y 2
�G�x, y, z� �x
�a2 � x 2 � y 2i �
y
�a2 � x 2 � y 2j � k
G�x, y, z� � z � �a2 � x 2 � y 2
S: z � �a2 � x 2 � y 2 a
a
−a
−ax
x y a2 2 2+ ≤
yF�x, y, z� � x i � y j � 2zk
30.
The flux across the bottom is zero.z � 0
� �2�
0 3
4�
12
sin cos � d � 34
�sin2
4 �2�
0�
3�
2
� �2�
0�1
0�r2 � 2r2 cos sin � 1�r dr d
�S�F � N dS � �
R�F � �G dA � �
R� �x 2 � 2xy � y 2 � 1� dA
F � �G � 2x�x � y� � 2y� y� � �1 � x 2 � y 2� � xx 2 � 2xy � y 2 � 1
�G�x, y, z� � 2x i � 2y j � k
G�x, y, z� � z � x 2 � y 2 � 1
z � 0S: z � 1 � x 2 � y 2,
F�x, y, z� � �x � y� i � y j � zk
434 Chapter 14 Vector Analysis
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32. A surface is orientable if a unit normal vector N can bedefined at every nonboundary point of S in such a way thatthe normal vectors vary continuously over the surface S.
34. Orientable
36.
� �R� 2xyz
�1 � x 2 � y 2� xy� dA � �
R�3xy dA � �1
�1��1�x2
��1�x2
3xy dy dx � 0
� �R��yz i � xz j � xyk� � x
�1 � x 2 � y 2i �
y
�1 � x 2 � y 2j � k� dA
�S�E � N dS � �
R�E � ��gx�x, y� i � gy�x, y� j � k� dA
S: z � �1 � x 2 � y 2
E � yz i � xz j � xyk
38.
(use integration by parts)
Let dv .v � ��a2 � r2du � 2r dr,� r�a2 � r2��1�2 dr,u � r2,
� 2ka 23
a3��2�� �23
a2�4�ka2� �23
a2m
� 2ka�r2�a2 � r2 �23
�a2 � r2�3�2�a
0�2��
� 2k�R��x 2 � y 2� a
�a2 � x 2 � y 2 dA � 2ka�2�
0�a
0
r3
�a2 � r2 dr d
Iz � 2�S�k�x 2 � y 2� dS
� 2ka��a2 � r2�a
0�2�� � 4�ka2
� 2k�R� a
�a2 � x 2 � y 2 dA � 2ka�2�
0�a
0
r
�a2 � r2 dr d
m � 2�S�k dS � 2k�
R��1 � �x
�a2 � x 2 � y 2�2
� �y
�a2 � x 2 � y 2�2
dA
z � ±�a2 � x 2 � y 2
x 2 � y 2 � z2 � a2
40. 0 ≤ z ≤ h
Project the solid onto the xy-plane.
��
60��1 � 4h�3�2�6h � 1� � 1� �
�1 � 4h�3�2�
60�10h � �1 � 4h�� �
�
60
� 2� h12
�1 � 4h�3�2 �1
120�1 � 4h�5�2� �
2�
120
� �2�
0��h
0r2�1 � 4r2 r dr d
� ��h
��h��h�x2
��h�x2
�x 2 � y 2��1 � 4x 2 � 4y 2 dy dx
Iz � �S��x 2 � y 2��1� dS
z
y
x
h
z � x 2 � y 2,
Section 14.6 Surface Integrals 435
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Section 14.7 Divergence Theorem
42.
� 0.5��2�
0�4
0
�16 � r2r dr d� � 0.5��2�
0 643
d� �64��
3
� �R�0.5 �z dA � �
R�0.5��16 � x 2 � y 2 d�
� �R�0.5�zk � � x
�16 � x 2 � y 2i
y
�16 � x 2 � y 2j k� dA
�S��F � N dS ��
R��F � ��gx�x, y�i � gy�x, y�j k� dA
F�x, y, z� � 0.5zk
S: z � �16 � x 2 � y 2
2. Surface Integral: There are three surfaces to the cylinder.
Bottom:
Top:
Side:
Therefore,
Divergence Theorem:
yx
h
22
z
��Q
�2z dV � �2�
0�2
0�h
0 2zr dz dr d� � 4�h2.
div F � 2 � 2 2z � 2z
�S�F � N dS � 0 4�h2 0 � 4�h2.
�S3
�F � N dS � �h
0�2�
0�8 cos2 u � 8 sin2 u� du dv � 0
F � �ru rv� � 8 cos2 u � 8 sin2 u
ru rv � 2 cos ui 2 sin uj
ru � �2 sin ui 2 cos u j, rv � k
0 ≤ v ≤ h0 ≤ u ≤ 2�,r�u, v� � 2 cos ui 2 sin u j vk,
�S2
�h2 dS � h2 �Area of circle) � 4�h2
F � N � z2N � k,z � h,
�S1
�0 dS � 0
F � N � �z2N � �k,z � 0,
436 Chapter 14 Vector Analysis
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4.
S: surface bounded by the planes and the coordinate planes
Surface Integral: There are five surfaces to this solid.
Therefore,
Divergence Theorem: Since div we have
��Q
� div F dV � �4
0�4
0�4�x
0y dz dy dx � 64.
F � y,
�S�F � N dS � �64 �
323
323 0 128 � 64.
�S5
� 1
�2�xy x y�2 dA � �4
0�4
0�xy x y� dy dx � 128
dS � �2 dAF � N �1
�2�xy x y,N �
i k�2
,x z � 4,
�S4
��xy dS � �4
0�4
00 dS � 0
F � N � �xyN � �i,x � 0,
�S3
�z dS � �4
0�4�x
0z dz dx � �4
0 �4 � x�2
2 dx �
323
F � N � zN � j,y � 4,
�S2
��z dS � �4
0�4�x
0�z dz dx � ��4
0 �4 � x�2
2 dx � �
323
F � N � �zN � �j,y � 0,
�S1
���x y� dS � �4
0�4
0��x y� dy dx � ��4
0 �4x 8� dx � �64
F � N � ��x y�N � �k,z � 0,
z � 4 � xy � 4,
F�x, y, z� � xyi z j �x y�k
6. Since div we have
�13
a6 � 2a3 34
a5.
� �a
02
3xa4 � 2a2
32
xa3� dx
��Q
� div F dV � �a
0�a
0�a
0�2xz2 � 2 3xy� dz dy dx � �a
0�a
02
3xa3
� 2a 3xya� dy dx
F � 2xz2 � 2 3xy
8. Since div we have
� �2�
0�a
0�a2r
2�
r3
2 � dr d� � �2�
0�a2r 2
4�
r4
8 �a
0 d� � �2�
0
a4
8 d� �
�a4
4.
��Q
� div F dV � �a
�a��a2�x2
��a2�x2��a2�x2�y2
0 z dz dy dx � �2�
0�a
0��a2�r2
0 zr dz dr d�
F � y z � y � z,
10. Since div we have
� �4
0�3
�3 z2
�0� dy dz � 0.��Q
�xz dV � �4
0�3
�3��9�y2
��9�y2
xz dx dy dz
F � xz,
Section 14.7 Divergence Theorem 437
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12. Since div we have
� �2�
0131,052
5 100e8� d� �
262,1045
� 200e8�
� �2�
0�16
0�8
r�2 �r2 ez�r dz dr d� � �2�
0�16
08r3 re8 �
12
r4 � rer�2� dr d�
��Q
��x 2 y 2 ez� dV � �16
0��256�x2
��256�x2
�8
�1�2��x2y2
�x 2 y 2 ez� dz dy dx
F � y 2 x 2 ez,
14. Since div we have
��Q
�3ez dV � �6
0�4
0�4�y
03ez dz dy dx � �6
0�4
03�e4�y � 1 dy dx � �6
03�e4 � 5� dx � 18�e 4 � 5�.
F � ez ez ez � 3ez,
16.
The surface S is the upper half of a hemisphere of radius 2. Since the volume is you have
�S�F � N dS � 2�Volume� �
32�
3.
12 �4
3 ��23�� � 16��3,
�S�F � N dS � ��
Q
� div F dV � ��Q
�2 dV.
div F � 2
18. Using the Divergence Theorem, we have
Now, div curl F Therefore,
�S�curl F � N dS � ��
Q
� div �curl F� dV � 0.
�x, y, z� � �z � y cos x� � �z x sin z� �y cos x x sin z� � 0.
� �xz � y sin x� i � �yz xy sin z�j �yz cos x � x cos z�k. curl F�x, y, z� � i � �x
xy cos z
j � �y
yz sin x
k � �z
xyz �
S�curl F � N dS � ��
Q
�div �curlF� dV
20. If div then source.
If div then sink.
If div then incompressible.F�x, y, z� � 0,
F�x, y, z� < 0,
F�x, y, z� > 0, 22.
Similarly, .�a
0�a
0y dz dx � �a
0�a
0z dx dy � a3
v � �a
0�a
0x dy dz � �a
0�a
0a dy dz � �a
0a2 dz � a3
24. If then div Therefore,
�S�F � N dS � ��
Q
�div F dV � ��Q
�0 dV � 0.
F � 0.F�x, y, z� � a1i a2 j a3 k,
26. If then div
1�F�
�S�F � N dS �
1�F�
��Q
�div F dV �1
�F� ��
Q
�3 dV �3
�F� ��
Q
�dV
F � 3.F�x, y, z� � x i yj zk,
28.
� ��Q
�� f �2g �f � �g� dV � ��Q
��g�2f �g � �f � dV � ��Q
�� f �2g � g�2f � dV
�S�� f DN g � gDN f � dS � �
S� f DN g dS � �
S� gDN f dS
438 Chapter 14 Vector Analysis
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Section 14.8 Stokes’s Theorem
2.
curl F � � i��x
x2
j��y
y2
k��z
x2 � � �2x j
F�x, y, z� � x2 i � y2 j � x2 k 4.
� z2 i � �y sin x � x cos y�k
curl F � � i��x
x sin y
j��y
�y cos x
k��z
yz2 �F�x, y, z� � x sin y i � y cos x j � yz2 k
6.
� 2y i � � x�1 � x2
�1
�1 � y2�k
� 2y i � � �x�1 � x2
�1
�1 � y2�k
curl F � � i � �x
arcsin y
j � �y
�1 � x2
k��z
y2 �F�x, y, z� � arcsin y i � �1 � x2 j � y2 k
8. In this case C is the circle
Let then and
Double Integral:
therefore
� 4�2
�2 �4 � x2 dx � 2�x�4 � x2 � 4 arcsin
x2�
2
�2� 8�.
�� �curl F� � NdS � �R
� 2 dA � �2
�2 ��4�x2
��4�x2
2 dy dx � 2�2
�2 2�4 � x2 dx
curl F � 2k,
dS � �1 � 4x2 � 4y2 dAN ��F
�F �
2xi � 2yj � k�1 � 4x2 � 4y2
,F�x, y, z� � z � x2 � y2 � 4,
�C
�y dx � x dy � �2�
0 4 dt � 8�.dy � 2 cos t dt,dx � �2 sin t dt,y � 2 sin t,x � 2 cos t,
Line Integral: �C
F � dr � �C
�y dx � x dy
dz � 0.z � 0,x2 � y 2 � 4,
10. Line Integral: From the accompanying figure we see that for
Hence,
—CONTINUED—
� �a
0 2y3 dy � �a
0 a4 dx � �0
a
a2 dy � �0
a
2y3 dy � �a4x�a
0� �a2y�
0
a� a5 � a3 � a3�a2 � 1�.
� �C1
0 � �C2
2y3 dy � �C3
a4 dx � �C4
a2 dy � y2�2y� dy
�C
F � dr � �C
z2 dx � x2 dy � y2 dz
dz � 2y dy.dx � 0,x � a,C4: z � y2,
dy � dz � 0z � a2,C3: y � a,
dz � 2y dydx � 0,x � 0,C2: z � y2,
dy � dz � 0z � 0,C1: y � 0,
yx
1
11
C1C2
C3
C4
z
Section 14.8 Stokes’s Theorem 439
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10. —CONTINUED—
Double Integral: Since we have
and
Furthermore, Therefore,
�S� �curl F� � N dS � �
R� �4yz � 2x� dA � �a
0 �a
0 �4y2 � 2x� dy dx � �a
0 �a 4 � 2ax� dx � �a4x � ax2�
a
0� a3�a2 � 1�.
curl F � 2y i � 2z j � 2xk.
dS � �1 � 4y2 dA.N �2yj � k�1 � 4y2
F�x, y, z� � y2 � z,
12. Let and Then and and
Hence, and Since we have �S
� �curl F� � N dS � �R
� 0 dS � 0.curl F �2x
x2 � y2 k,dS � �2 dA.F�x, y, z� � x � y
N �U V
U V �
2i � 2 j2�2
�i � j�2
.
V � AC\
� 2k,U � AB\
� i � j � k,C � �0, 0, 2�.B � �1, 1, 1�,A � �0, 0, 0�,
14.
� �3
�316x2�9 � x2 �
163
�9 � x2�3�2� dx � 0
� �3
�3 ��9�x2
��9�x2
8x2 � 8xy � 8y2� dy dx
�S
� �curl F� � N dS � �R
� 8x2 � 2y�4x � 4y�� dA
�G�x, y, z� � 2x i � 2y j � k
G�x, y, z� � x2 � y2 � z � 9
curl F � 4x i � �4x � 4y�j
z ≤ 0S: 9 � x2 � y2,F�x, y, z� � 4xz i � y j � 4xyk,
16.
�G�x, y, z� �x
�4 � x2 � y2i �
y�4 � x2 � y2
j � k
G�x, y, z� � z � �4 � x2 � y2
curl F � � i��x
x2
j��y
z2
k��z
�xyz � � ��xz � 2z�i � yzj
S: z � �4 � x2 � y2F�x, y, z� � x2 i � z2 j � xyzk,
� ��13��8�� �
43
�2�� �13
��8�� �43
��2��� � 0
�43
�4 � x2�3�2 �83
12��x�4 � x2 � 4 arcsin
x2��
2
�2 � ��8
3 18��x�2x2 � 4��4 � x2 � 16 arcsin
x2�
� �2
�2 ��8
3x2�4 � x2 � 4x�4 � x2 �
83�4 � x2� dx
� �2
�2 ��2x2�4 � x2 � 4x�4 � x2 �
23
�4 � x2��4 � x2� dx
� �2
�2 ��x2y � 2xy �
y3
3 ��4�x2
��4�x2 dx
� �R
� �x�x � 2� � y2� dA � �2
�2 ��4�x2
��4�x2
��x2 � 2x � y2� dy dx
�S
� �curl F� � N dS � �R
� � �z�x � 2�x�4 � x2 � y2
�y2z
�4 � x2 � y2� dA
440 Chapter 14 Vector Analysis
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18.
S: the first octant portion of over
�G�x, y, z� �x
�16 � x2i � k
G�x, y, z� � z � �16 � x2
x2 � y2 � 16x2 � z2 � 16
curl F � � i��x
yz
j��y
2 � 3y
k��z
x 2 � y 2 � � 2y i � �y � 2x� j � zk
F�x, y, z� � yz i � �2 � 3y� j � �x2 � y2�k
� �64 �643 � � �64
3 � � �643
� ��13
�16 � x2�3�2 � 16x �x3
3 �4
0
� �4
0 x�16 � x2 � �16 � x2�� dx
� �4
0 � x�16 � x2
y2 � �16 � x2 y��16�x2
0 dx
� �4
0 ��16�x2
0 � 2xy�16 � x2
� �16 � x2� dy dx
� �R
� � 2xy�16 � x2
� �16 � x2� dA
�S
� �curl F� � N dS � �R
� � 2xy�16 � x2
� z� dA
20.
S: the first octant portion of over We have
and
�2
15a5
� ��13
x2�a2 � x2�3�2 �2
15�a2 � x2�5�2�
a
0
� �a
0 x3�a2 � x2 dx
� �a
0 ��a2�x2
0 x3 dy dx
�S
� �curl F� � N dS � �R
� xz dA � �R
� x3 dA
dS � �1 � 4x2 dA.N �2x i � k�1 � 4x2
x2 � y2 � a2.z � x2
curl F � � i��x
xyz
j��y
y
k��z
z � � xyj � xzk
F�x, y, z� � xyz i � y j � zk
Section 14.8 Stokes’s Theorem 441
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22.
Letting and �S
� �curl F� � N dS � 0.curl F � N � 0N � k,
curl F � � i�
�x�z
j�
�y0
k�
�zy � � i � j
S: x2 � y2 � 1
F�x, y, z� � �z i � yk
24. curl F measures the rotational tendency.
See page 1084.
26.
(a)
(b)
� �2
0 � 2r3
�4 � r2 �12
sin 2�2�
0 dr � 0
� �2
0 �2�
0 2r2�cos2 � � sin2 ��
�4 � r2r d� dr
� �S
� 2�x2 � y2�
�4 � x2 � y2 dA
2�4 � x2 � y2
dA �S
� �f �x, y, z� g�x, y, z�� � N dS � �S
� � x2z�4 � x2 � y2
�y2z
�4 � x2 � y2
dS ��1 � � �x�4 � x2 � y2
2
� � �y�4 � x2 � y2
2
dA �2
�4 � x2 � y2 dA
N �x
�4 � x2 � y2i �
y�4 � x2 � y2
j � k
f g � � iyz0
jxz0
kxy1 � � xz i � yz j
g�x, y, z� � k
f �x, y, z� � yz i � xz j � xyk
�C
� f �x, y, z�g�x, y, z�� � dr � 0
0 ≤ t ≤ 2�r�t� � 2 cos t i � 2 sin t j � 0k,
f �x, y, z�g�x, y, z� � xyzk
g�x, y, z� � k
S: z � �4 � x2 � y2g�x, y, z� � z,f �x, y, z� � xyz,
Review Exercises for Chapter 14
2.
x
543
2 4
2
−2−1
−3−4−5
y
F�x, y� � i � 2yj 4.
� xeyz�2i � xz j � xyk�
F�x, y, z� � 2xeyz i � x2zeyz j � x2yey z k
f �x, y, z� � x2eyz
442 Chapter 14 Vector Analysis
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6. Since is conservative. From and partial integrationyields and which suggests that U�x, y� � �y x� � C.U � �y x� � g�x�U � �y x� � h�y�
N � �U �y � 1 x,M � �U �x � �y x2F�M �y � �1 x2 � �N �x,
8. Since is conservative. From and we obtain and which suggests that andU�x, y� � y3�1 � cos 2x� � C.
g�x� � C,h�y� � y3,U � y3�1 � cos 2x� � g�x�U � y3 cos 2x � h�y�N � �U �y � 3y 2�1 � cos 2x�,M � �U �x � �2y3 sin 2xF�M �y � �6y 2 sin 2x � �N �x,
10. Since
F is not conservative.
�N�z
� 6y ��P�y
�M�z
� 2z ��P�x
,
�M�y
� 4x ��N�x
,
12. Since
F is not conservative.
�M�z
� y cos z ��P�x
,�M�y
� sin z ��N�x
,
18. Since
(a)
(b) curl F � 0
div F � 2x � 2 sin y cos y
F � �x2 � y�i � �x � sin2 y�j: 20. Since
(a)
(b) curl F � �1y
i �1x
j
div F � �zx2 �
zy 2 � 2z � z�2 �
1x2 �
1y 2
F �zx
i �zy
j � z2 k:
14. Since
(a)
(b) curl F � 2xz j � y 2k
div F � 2xy � x2
F � xy 2j � zx2k; 16. Since
(a)
(b) curl F � 2i � 3j � k
div F � 3 � 1 � 1 � 5
F � �3x � y�i � �y � 2z�j � �z � 3x�k:
22. (a) Let then
(b)
Therefore,
� 16�5�t2
2�
t3
31
0�
8�53
.
�C
xy ds � �4
0 0 dt � �1
0 �8t � 8t 2�2�5 dt � �2
0 0 dt
ds � dt0 ≤ t ≤ 2,y � 2 � t,C3: x � 0,
ds � 2�5 dt0 ≤ t ≤ 1,y � 2t,C2: x � 4 � 4t,
ds � dt0 ≤ t ≤ 4,y � 0,C1: x � t,
�C
xy ds � �1
0 20t 2 �41 dt �
20�413
x
2
3
4
3 4
(0, 2)
(4, 0)
C2
C1
C3
y = x + 212
y
−
ds � �41 dt.0 ≤ t ≤ 1,y � 4t,x � 5t,
24.
� 8�
� �2�2�
0 t�1 � cos t dt
� �2�2�
0 �t�1 � cos t � sin t�1 � cos t � dt � �2��
23
�1 � cos t�3 22�
0� �2�2�
0t�1 � cos t dt
�C
x ds � �2�
0 �t � sin t���1 � cos t�2 � �sin t�2 dt � �2�
0 �t � sin t��2 � 2 cos t dt
dydt
� sin tdxdt
� 1 � cos t,0 ≤ t ≤ 2�,y � 1 � cos t,x � t � sin t,
Review Exercises for Chapter 14 443
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26.
�C
�2x � y� dx � �x � 3y� dy � �� 2
0 �sin t cos t�5t 2 � 6t � 2� � cos2 t�t � 1� � sin2 t�2t � 3�� dt � 1.01
dy � �cos t � t cos t � sin t� dtdx � t cos t dt,0 ≤ t ≤ �
2,y � sin t � t sin t,x � cos t � t sin t,
28.
�C
�x2 � y 2 � z2� ds � �4
0 �t 2 � t 4 � t3��1 � 4t 2 �
94
t dt � 2080.59
z��t� �32
t1 2y��t� � 2t,x��t� � 1,
0 ≤ t ≤ 4r�t� � t i � t2j � t3 2 k,
30.
from to
Lateral surface area:
�C
f �x, y� ds � �2
0 �12 � t � t2��1 � 4t2 dt � 41.532
�r��t�� � �1 � 4t 2
r��t� � i � 2t j
0 ≤ t ≤ 2r�t� � t i � t 2 j,
�2, 4��0, 0�C: y � x2
f �x, y� � 12 � x � y
32.
�C
F � dr � �2�
0 �12 � 7 sin t cos t� dt � �12t �
7 sin2 t2
2�
0� 24�
0 ≤ t ≤ 2�F � �4 cos t � 3 sin t�i � �4 cos t � 3 sin t�j,
dr � ���4 sin t�i � 3 cos t j� dt
34.
�C
F � dr � �2
0 �t � 2� dt � �t 2
2� 2t
2
0� �2
F � �4 � 2t � �4t � t 2 � i � ��4t � t 2 � 2 � t�j � 0k
dr � ��i � j �2 � t
�4t � t 2k dt
0 ≤ t ≤ 2z � �4t � t2,y � 2 � t,x � 2 � t,
36. Let
�C
F � dr � ��
0 �8 sin t � 16 sin2 t cos t� dt � ��8 cos t �
163
sin3 t�
0� 16
F � 0 i � 4 j � �2 sin t�k
dr � ��2 cos t�i � �2 sin t�j � �8 sin t cos t�k� dt
0 ≤ t ≤ �.z � 4 sin2 t,y � �2 cos t,x � 2 sin t,
38.
�C
F � dr � 4�2 � 4�
0 ≤ t ≤ �r�t� � �2 cos t � 2t sin t�i � �2 sin t � 2t cos t�j,
�C
F � dr � �C �2x � y� dx � �2y � x� dy
444 Chapter 14 Vector Analysis
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40.
�C
F � dr � �� 2
0 50033�
dt �25033
mi � ton
dr � �10 cos t i � 10 sin t j �25
33�k
F � 20k
� 10 sin t i � 10 cos t j �25
33�tk
0 ≤ t ≤ �
2 r�t� � 10 sin t i � 10 cos t j �
2000 5280� 2
tk,
42. �C
y dx � x dy �1z dz � �xy � ln�z�
�4, 4, 4)
�0, 0, 1�� 16 � ln 4
44.
(a)
Since these equations orient the curve backwards, we will use
(b) By symmetry, From Section 14.4,
�1
2�3�a2� a3�5�� �56
a y � �1
2A �
C
y 2 dx �1
2A �2�
0a3�1 � cos ��2�1 � cos �� d�
x � �a.
�a2
2 �2�
0 �2 � 2 cos � � � sin �� d� �
a2
2�6�� � 3�a2.
�a2
2 �2�
0 �1 � 2 cos � � cos2 � � � sin � � sin2 �� d�
�12
�2�
0 �a2�1 � cos ���1 � cos �� � a2�� � sin ���sin ��� d� �
12
�2�a
0 �0 � 0� d�
A �12
� � y dx � x dy�
A �12�C
x dy � y dx.
x
C1
C22 aπ
y0 ≤ � ≤ 2�y � a�1 � cos ��,x � a�� � sin ��,
46.
� �2
0 2x dx � 4
�C
xy dx � �x2 � y 2� dy � �2
0�2
0 �2x � x� dy dx 48.
� �a
�a
0 dx � 0
�C
�x2 � y 2� dx � 2xy dy � �a
�a��a2�x2
��a2�x2
4y dy dx
50.
� 0
� ��87
x2 3 �1 � x2 3�5 2 �1635
�1 � x2 3�5 21
�1
� �1
�1 83
x1 3�1 � x2 3�3 2 dx
� �1
�1 �4
3x1 3y�y 2
�1�x2 3�3 2
��1�x2 3�3 2
dx
�C
y 2 dx � x4 3 dy � �1
�1��1�x2 3�3 2
�(1�x2 3�3 2
�43
x1 3 � 2y dy dx
Review Exercises for Chapter 14 445
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52.
2 2
2
yx
z
0 ≤ v ≤ 2�0 ≤ u ≤ 4,
r �u, v� � e�u 4 cos v i � e�u 4 sin v j �u6
k
54.
�S� z dS � ��
0�2
0 sin v�2 cos2v � 4 du dv � 2��6 � �2 ln ��6 � �2
�6 � �2�ru rv � � �2 cos2v � 4
� cos v i � cos v j � 2kru rv � �i11 j 1 �1
k 0cos v �
ru �u, v� � i � j � cos vk
ru �u, v� � i � j
0 ≤ v ≤ �0 ≤ u ≤ 2,S: r�u, v� � �u � v�i � �u � v� j � sin vk,
56. (a)
(b)
�23
k�a2 � 1 a3�
� k�a2 � 1�2�
0 a3
3 d�
� k�a2 � 1�2�
0�a
0 r2 dr d�
� k�R� �a2 � 1��x2 � y2� dA
� k�R� �x2 � y2 �1 �
a2x2
x2 � y2 �a2y2
x2 � y2 dA
� �R� k�x2 � y2 �1 � gx
2 � gy2 dA
m � �S� e�x, y, z� dS
��x, y� � k�x2 � y2
S: g�x, y� � z � a2 � a�x2 � y2
z � 0 ⇒ x2 � y2 � a2
yaa
a2
x
zz � a�a � �x2 � y2�, 0 ≤ z ≤ a2
446 Chapter 14 Vector Analysis
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58.
Q: solid region bounded by the coordinate planes and the plane
Surface Integral: There are four surfaces for this solid.
Triple Integral: Since the Divergence Theorem yields.
��Q
� div F dV � ��Q
� 3 dV � 3�Volume of solid� � 3�13
�Area of base��Height� �12
�6��4��3� � 36.
div F � 3,
�14�
6
0�(12�2x) 3
0 12 dy dx � 3�6
0 �4 �
2x3 dx � 3�4x �
x2
3 6
0� 36
�S4
� N � F dS �14�R
� �2x � 3y � 4z� dy dx
dS ��1 � �14 � � 9
16dA ��29
4 dAN �
2i � 3j � 4k�29
,2x � 3y � 4z � 12,
�S3
� 0 dS � 0F � N � �x,N � �i,x � 0,
�S2
� 0 dS � 0F � N � �y,N � �j,y � 0,
�S1
� 0 dS � 0F � N � �z,N � �k,z � 0y
x
(0, 4, 0)
(6, 0, 0)
(0, 0, 3)
z2x � 3y � 4z � 12
F�x, y, z� � x i � y j � zk
60.
S: first octant portion of the plane
Line Integral:
Double Integral:
�S� �curl F� � N dS � �4
0�12�3x
0 �x � 1� dy dx � �4
0 ��3x2 � 15x � 12� dx � 8
curl F � i � �2x � 1� j
G�x, y, z� � �32
i �12
j � k
G�x, y, z� �12 � 3x � y
2� z
� �0
4 ��
32
x2 �52
x � 6 dx � �12
0 �3
2y � 6 dy � �4
0 �10x � 36� dx � 8
� �C1
�x �12 � 3x
2� x2��
32 dx � �
C2 �y �
12 � y2 dy � �
C3
�x � �12 � 3x���3�� dx
�C
F � dr � �C
�x � z� dx � �y � z� dy � x2 dz
dy � �3 dxy � 12 � 3x,dz � 0,C3: z � 0,
dz � �12
dyz �12 � y
2,dx � 0,C2: x � 0,
dz � �32
dxz �12 � 3x
2,dy � 0,C1: y � 0,
3x � y � 2z � 12
y
x
(0, 12, 0)
(0, 0, 6)
(4, 0, 0)
zF�x, y, z� � �x � z�i � �y � z�j � x2 k
Review Exercises for Chapter 14 447
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22.
Letting and �S
� �curl F� � N dS � 0.curl F � N � 0N � k,
curl F � � i�
�x�z
j�
�y0
k�
�zy � � i � j
S: x2 � y2 � 1
F�x, y, z� � �z i � yk
24. curl F measures the rotational tendency.
See page 1084.
26.
(a)
(b)
� �2
0 � 2r3
�4 � r2 �12
sin 2�2�
0 dr � 0
� �2
0 �2�
0 2r2�cos2 � � sin2 ��
�4 � r2r d� dr
� �S
� 2�x2 � y2�
�4 � x2 � y2 dA
2�4 � x2 � y2
dA �S
� �f �x, y, z� g�x, y, z�� � N dS � �S
� � x2z�4 � x2 � y2
�y2z
�4 � x2 � y2
dS ��1 � � �x�4 � x2 � y2
2
� � �y�4 � x2 � y2
2
dA �2
�4 � x2 � y2 dA
N �x
�4 � x2 � y2i �
y�4 � x2 � y2
j � k
f g � � iyz0
jxz0
kxy1 � � xz i � yz j
g�x, y, z� � k
f �x, y, z� � yz i � xz j � xyk
�C
� f �x, y, z�g�x, y, z�� � dr � 0
0 ≤ t ≤ 2�r�t� � 2 cos t i � 2 sin t j � 0k,
f �x, y, z�g�x, y, z� � xyzk
g�x, y, z� � k
S: z � �4 � x2 � y2g�x, y, z� � z,f �x, y, z� � xyz,
Review Exercises for Chapter 14
2.
x
543
2 4
2
−2−1
−3−4−5
y
F�x, y� � i � 2yj 4.
� xeyz�2i � xz j � xyk�
F�x, y, z� � 2xeyz i � x2zeyz j � x2yey z k
f �x, y, z� � x2eyz
442 Chapter 14 Vector Analysis
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6. Since is conservative. From and partial integrationyields and which suggests that U�x, y� � �y x� � C.U � �y x� � g�x�U � �y x� � h�y�
N � �U �y � 1 x,M � �U �x � �y x2F�M �y � �1 x2 � �N �x,
8. Since is conservative. From and we obtain and which suggests that andU�x, y� � y3�1 � cos 2x� � C.
g�x� � C,h�y� � y3,U � y3�1 � cos 2x� � g�x�U � y3 cos 2x � h�y�N � �U �y � 3y 2�1 � cos 2x�,M � �U �x � �2y3 sin 2xF�M �y � �6y 2 sin 2x � �N �x,
10. Since
F is not conservative.
�N�z
� 6y ��P�y
�M�z
� 2z ��P�x
,
�M�y
� 4x ��N�x
,
12. Since
F is not conservative.
�M�z
� y cos z ��P�x
,�M�y
� sin z ��N�x
,
18. Since
(a)
(b) curl F � 0
div F � 2x � 2 sin y cos y
F � �x2 � y�i � �x � sin2 y�j: 20. Since
(a)
(b) curl F � �1y
i �1x
j
div F � �zx2 �
zy 2 � 2z � z�2 �
1x2 �
1y 2
F �zx
i �zy
j � z2 k:
14. Since
(a)
(b) curl F � 2xz j � y 2k
div F � 2xy � x2
F � xy 2j � zx2k; 16. Since
(a)
(b) curl F � 2i � 3j � k
div F � 3 � 1 � 1 � 5
F � �3x � y�i � �y � 2z�j � �z � 3x�k:
22. (a) Let then
(b)
Therefore,
� 16�5�t2
2�
t3
31
0�
8�53
.
�C
xy ds � �4
0 0 dt � �1
0 �8t � 8t 2�2�5 dt � �2
0 0 dt
ds � dt0 ≤ t ≤ 2,y � 2 � t,C3: x � 0,
ds � 2�5 dt0 ≤ t ≤ 1,y � 2t,C2: x � 4 � 4t,
ds � dt0 ≤ t ≤ 4,y � 0,C1: x � t,
�C
xy ds � �1
0 20t 2 �41 dt �
20�413
x
2
3
4
3 4
(0, 2)
(4, 0)
C2
C1
C3
y = x + 212
y
−
ds � �41 dt.0 ≤ t ≤ 1,y � 4t,x � 5t,
24.
� 8�
� �2�2�
0 t�1 � cos t dt
� �2�2�
0 �t�1 � cos t � sin t�1 � cos t � dt � �2��
23
�1 � cos t�3 22�
0� �2�2�
0t�1 � cos t dt
�C
x ds � �2�
0 �t � sin t���1 � cos t�2 � �sin t�2 dt � �2�
0 �t � sin t��2 � 2 cos t dt
dydt
� sin tdxdt
� 1 � cos t,0 ≤ t ≤ 2�,y � 1 � cos t,x � t � sin t,
Review Exercises for Chapter 14 443
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26.
�C
�2x � y� dx � �x � 3y� dy � �� 2
0 �sin t cos t�5t 2 � 6t � 2� � cos2 t�t � 1� � sin2 t�2t � 3�� dt � 1.01
dy � �cos t � t cos t � sin t� dtdx � t cos t dt,0 ≤ t ≤ �
2,y � sin t � t sin t,x � cos t � t sin t,
28.
�C
�x2 � y 2 � z2� ds � �4
0 �t 2 � t 4 � t3��1 � 4t 2 �
94
t dt � 2080.59
z��t� �32
t1 2y��t� � 2t,x��t� � 1,
0 ≤ t ≤ 4r�t� � t i � t2j � t3 2 k,
30.
from to
Lateral surface area:
�C
f �x, y� ds � �2
0 �12 � t � t2��1 � 4t2 dt � 41.532
�r��t�� � �1 � 4t 2
r��t� � i � 2t j
0 ≤ t ≤ 2r�t� � t i � t 2 j,
�2, 4��0, 0�C: y � x2
f �x, y� � 12 � x � y
32.
�C
F � dr � �2�
0 �12 � 7 sin t cos t� dt � �12t �
7 sin2 t2
2�
0� 24�
0 ≤ t ≤ 2�F � �4 cos t � 3 sin t�i � �4 cos t � 3 sin t�j,
dr � ���4 sin t�i � 3 cos t j� dt
34.
�C
F � dr � �2
0 �t � 2� dt � �t 2
2� 2t
2
0� �2
F � �4 � 2t � �4t � t 2 � i � ��4t � t 2 � 2 � t�j � 0k
dr � ��i � j �2 � t
�4t � t 2k dt
0 ≤ t ≤ 2z � �4t � t2,y � 2 � t,x � 2 � t,
36. Let
�C
F � dr � ��
0 �8 sin t � 16 sin2 t cos t� dt � ��8 cos t �
163
sin3 t�
0� 16
F � 0 i � 4 j � �2 sin t�k
dr � ��2 cos t�i � �2 sin t�j � �8 sin t cos t�k� dt
0 ≤ t ≤ �.z � 4 sin2 t,y � �2 cos t,x � 2 sin t,
38.
�C
F � dr � 4�2 � 4�
0 ≤ t ≤ �r�t� � �2 cos t � 2t sin t�i � �2 sin t � 2t cos t�j,
�C
F � dr � �C �2x � y� dx � �2y � x� dy
444 Chapter 14 Vector Analysis
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40.
�C
F � dr � �� 2
0 50033�
dt �25033
mi � ton
dr � �10 cos t i � 10 sin t j �25
33�k
F � 20k
� 10 sin t i � 10 cos t j �25
33�tk
0 ≤ t ≤ �
2 r�t� � 10 sin t i � 10 cos t j �
2000 5280� 2
tk,
42. �C
y dx � x dy �1z dz � �xy � ln�z�
�4, 4, 4)
�0, 0, 1�� 16 � ln 4
44.
(a)
Since these equations orient the curve backwards, we will use
(b) By symmetry, From Section 14.4,
�1
2�3�a2� a3�5�� �56
a y � �1
2A �
C
y 2 dx �1
2A �2�
0a3�1 � cos ��2�1 � cos �� d�
x � �a.
�a2
2 �2�
0 �2 � 2 cos � � � sin �� d� �
a2
2�6�� � 3�a2.
�a2
2 �2�
0 �1 � 2 cos � � cos2 � � � sin � � sin2 �� d�
�12
�2�
0 �a2�1 � cos ���1 � cos �� � a2�� � sin ���sin ��� d� �
12
�2�a
0 �0 � 0� d�
A �12
� � y dx � x dy�
A �12�C
x dy � y dx.
x
C1
C22 aπ
y0 ≤ � ≤ 2�y � a�1 � cos ��,x � a�� � sin ��,
46.
� �2
0 2x dx � 4
�C
xy dx � �x2 � y 2� dy � �2
0�2
0 �2x � x� dy dx 48.
� �a
�a
0 dx � 0
�C
�x2 � y 2� dx � 2xy dy � �a
�a��a2�x2
��a2�x2
4y dy dx
50.
� 0
� ��87
x2 3 �1 � x2 3�5 2 �1635
�1 � x2 3�5 21
�1
� �1
�1 83
x1 3�1 � x2 3�3 2 dx
� �1
�1 �4
3x1 3y�y 2
�1�x2 3�3 2
��1�x2 3�3 2
dx
�C
y 2 dx � x4 3 dy � �1
�1��1�x2 3�3 2
�(1�x2 3�3 2
�43
x1 3 � 2y dy dx
Review Exercises for Chapter 14 445
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52.
2 2
2
yx
z
0 ≤ v ≤ 2�0 ≤ u ≤ 4,
r �u, v� � e�u 4 cos v i � e�u 4 sin v j �u6
k
54.
�S� z dS � ��
0�2
0 sin v�2 cos2v � 4 du dv � 2��6 � �2 ln ��6 � �2
�6 � �2�ru rv � � �2 cos2v � 4
� cos v i � cos v j � 2kru rv � �i11 j 1 �1
k 0cos v �
ru �u, v� � i � j � cos vk
ru �u, v� � i � j
0 ≤ v ≤ �0 ≤ u ≤ 2,S: r�u, v� � �u � v�i � �u � v� j � sin vk,
56. (a)
(b)
�23
k�a2 � 1 a3�
� k�a2 � 1�2�
0 a3
3 d�
� k�a2 � 1�2�
0�a
0 r2 dr d�
� k�R� �a2 � 1��x2 � y2� dA
� k�R� �x2 � y2 �1 �
a2x2
x2 � y2 �a2y2
x2 � y2 dA
� �R� k�x2 � y2 �1 � gx
2 � gy2 dA
m � �S� e�x, y, z� dS
��x, y� � k�x2 � y2
S: g�x, y� � z � a2 � a�x2 � y2
z � 0 ⇒ x2 � y2 � a2
yaa
a2
x
zz � a�a � �x2 � y2�, 0 ≤ z ≤ a2
446 Chapter 14 Vector Analysis
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58.
Q: solid region bounded by the coordinate planes and the plane
Surface Integral: There are four surfaces for this solid.
Triple Integral: Since the Divergence Theorem yields.
��Q
� div F dV � ��Q
� 3 dV � 3�Volume of solid� � 3�13
�Area of base��Height� �12
�6��4��3� � 36.
div F � 3,
�14�
6
0�(12�2x) 3
0 12 dy dx � 3�6
0 �4 �
2x3 dx � 3�4x �
x2
3 6
0� 36
�S4
� N � F dS �14�R
� �2x � 3y � 4z� dy dx
dS ��1 � �14 � � 9
16dA ��29
4 dAN �
2i � 3j � 4k�29
,2x � 3y � 4z � 12,
�S3
� 0 dS � 0F � N � �x,N � �i,x � 0,
�S2
� 0 dS � 0F � N � �y,N � �j,y � 0,
�S1
� 0 dS � 0F � N � �z,N � �k,z � 0y
x
(0, 4, 0)
(6, 0, 0)
(0, 0, 3)
z2x � 3y � 4z � 12
F�x, y, z� � x i � y j � zk
60.
S: first octant portion of the plane
Line Integral:
Double Integral:
�S� �curl F� � N dS � �4
0�12�3x
0 �x � 1� dy dx � �4
0 ��3x2 � 15x � 12� dx � 8
curl F � i � �2x � 1� j
G�x, y, z� � �32
i �12
j � k
G�x, y, z� �12 � 3x � y
2� z
� �0
4 ��
32
x2 �52
x � 6 dx � �12
0 �3
2y � 6 dy � �4
0 �10x � 36� dx � 8
� �C1
�x �12 � 3x
2� x2��
32 dx � �
C2 �y �
12 � y2 dy � �
C3
�x � �12 � 3x���3�� dx
�C
F � dr � �C
�x � z� dx � �y � z� dy � x2 dz
dy � �3 dxy � 12 � 3x,dz � 0,C3: z � 0,
dz � �12
dyz �12 � y
2,dx � 0,C2: x � 0,
dz � �32
dxz �12 � 3x
2,dy � 0,C1: y � 0,
3x � y � 2z � 12
y
x
(0, 12, 0)
(0, 0, 6)
(4, 0, 0)
zF�x, y, z� � �x � z�i � �y � z�j � x2 k
Review Exercises for Chapter 14 447
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Problem Solving for Chapter 14
2. (a)
(b)
Flux � 25k�2�
0���2
0sin u du dv � 50�k
�ru � rv� � sin u
ru � rv � �sin2 u cos v, sin2 u sin v, sin u cos u sin2 v � sin u cos u cos2 v�
rv � ��sin u sin v, sin u cos v, 0�
ru � �cos u cos v, cos u sin v, �sin u�
r�u, v� � �sin u cos v, sin u sin v, cos u�
� 25k�2�
0�1
0
11 � r2
r dr d� � 50�k
� k�R� 251 � x2 � y2
dA
� k�R�25�x i � y j � zk� � �x i � y j � zk� 1
1 � x2 � y2 dA
Flux � �S��kT � N dS
� x i � y j � 1 � x2 � y2 k � x i � y j � zk
� x1 � x2 � y2
i �y
1 � x2 � y2 j � k�1 � x2 � y2
N �
�zx
i �zy
j � k
zx�
2
� zy�
2
� 1
T ��25
�x2 � y2 � z2�3�2�x i � y i � zk� � �25�x i � yj � zk�
dS �z
x�2
� z
y�2
� 1 dA �1
1 � x2 � y2
z � 1 � x2 � y2, z
x�
�x1 � x2 � y2
, z
y�
�y1 � x2 � y2
4.
Iz � �C
�x2 � y2�� ds��1
0t4
4� t2� dt �
2360
Ix � �C
�y2 � z2�� ds � �1
0t2 �
89
t3� dt �59
Iy � �C
�x2 � z2�� ds � �1
0t4
4�
89
t3� dt �49
180
� ds �1
1 � t�t � 1� dt � 1
r��t� � � t, 1, 2t1�2�, �r��t�� � t � 1
r�t� � �t2
2, t,
22t 3�2
3
448 Chapter 14 Vector Analysis
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6.
Hence, the area is 4�3.
12�C
x dy � y dx � 2���2
0�1
2 sin 2t cos t � sin t cos 2t� dt � 22
3�
8. is conservative.
potential function.
Work � f �2, 4� � f �1, 1� � 8�16� � 1 � 127
f �x, y� � x3y2
F �x, y� � 3x2y2 i � 2x3y j
10.
Same as area.
W � �2�
0F � dr �
12
ab�2�� � �ab
F � dr � �12
ab sin2 t �12
ab cos2 t� dt �12
ab
F � �12
b sin t i �12
a cos t j
r��t� � �a sin t i � b cos t j
r�t� � a cos t i � b sin t j, 0 ≤ t ≤ 2�
Area � �ab
Problem Solving for Chapter 14 449
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C H A P T E R 1 4Vector Analysis
Section 14.1 Vector Fields . . . . . . . . . . . . . . . . . . . . . . . . . . . 178
Section 14.2 Line Integrals . . . . . . . . . . . . . . . . . . . . . . . . . . . 183
Section 14.3 Conservative Vector Fields and Independence of Path . . . . . . 190
Section 14.4 Green’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 193
Section 14.5 Parametric Surfaces . . . . . . . . . . . . . . . . . . . . . . . . 198
Section 14.6 Surface Integrals . . . . . . . . . . . . . . . . . . . . . . . . . 202
Section 14.7 Divergence Theorem . . . . . . . . . . . . . . . . . . . . . . . 208
Section 14.8 Stokes’s Theorem . . . . . . . . . . . . . . . . . . . . . . . . . 211
Review Exercises . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215
Problem Solving . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 220
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C H A P T E R 1 4Vector Analysis
Section 14.1 Vector FieldsSolutions to Odd-Numbered Exercises
178
1. All vectors are parallel to y-axis.
Matches (c)
3. All vectors point outward.
Matches (b)
5. Vectors are parallel to x-axis for
Matches (a)
y � n�.
7.
1
−4
−4x
y
�F� � �2
F�x, y� � i � j 9.
53
−5
5
−5x
y
x2 � y2 � c2
�F� � �x2 � y2 � c
F�x, y� � x i � yj 11.
yx 4
4
2
z
�F� � 3�y� � c
F�x, y, z� � 3yj
13.
x−2
−2
−1 1 2
2
y
x2
c2�16�
y2
c2 � 1
�F� � �16x2 � y2 � c
F�x, y� � 4x i � yj 15.
x
y
4
4
4
−4
−4
z
�F� � �3
F�x, y, z� � i � j � k 17.
x−2 −1 1 2
2
1
−1
−2
y
19.
y
x
2
2
2
1
11
z 21.
F�x, y� � �10x � 3y�i � �3x � 20y�j
fy�x, y� � 3x � 20y
fx�x, y� � 10x � 3y
f �x, y� � 5x2 � 3xy � 10y2 23.
F�x, y, z� � �2xyex2i � ex2j � k
fz � 1
fy�x, y, z� � �ex 2
fx�x, y, z� � �2xyex 2
f �x, y, z� � z � yex 2
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Section 14.1 Vector Fields 179
25.
G�x, y, z� � � xyx � y
� y ln�x � y� i � � xyx � y
� x ln�x � y� j
gz�x, y, z� � 0
gy�x, y, z� � x ln�x � y� �xy
x � y
gx�x, y, z� � y ln�x � y� �xy
x � y
g�x, y, z� � xy ln�x � y�
27.
and have continuous firstpartial derivatives.
is conservative.�N�x
� 12x ��M�y
⇒ F
N � 6�x2 � y�M � 12xy
F�x, y� � 12xyi � 6�x2 � y�j 29.
and have continuous first partialderivatives.
is conservative.�N�x
� cos y ��M�y
⇒ F
N � x cos yM � sin y
F�x, y� � sin yi � x cos yj
31.
Not conservative�N�x
� �5y2 ��M�y
� 45y2 ⇒
M � 15y3, N � �5xy2 33.
Conservative�N�x
��2�y � 2x�
y3 e2x�y ��M�y
⇒
M �2y
e2x�y, N ��2x
y2 e2x�y
35.
Conservative
f �x, y� � x2y � K
fy�x, y� � x 2
fx�x, y� � 2xy
�
�xx2� � 2x
�
�y2xy� � 2x
F�x, y� � 2xy i � x2j 37.
Conservative
f �x, y� � ex2y � K
fy�x, y� � x2ex2y
fx�x, y� � 2xye x2y
�
�xx2e x2y� � 2xe x2y � 2x3ye x2y
�
�y2xye x2y� � 2xe x2y � 2x3ye x2y
F�x, y� � xex2y�2yi � x j� 39.
Conservative
f �x, y� �12
ln�x2 � y2� � K
fy�x, y� �y
x2 � y2
fx�x, y� �x
x2 � y2
�
�x�y
x2 � y2 � �2xy
�x2 � y2�2
�
�y�x
x2 � y2 � �2xy
�x2 � y2�2
F�x, y� �x
x2 � y2 i �y
x2 � y2 j
41.
Not conservative
�
�xex sin y� � ex sin y
�
�yex cos y� � �ex sin y
F�x, y� � ex�cos y i � sin y j� 43.
curl F �1, 2, 1� � 2j � k
curl F � � i
��xxyz
j
��yy
k
��zz � � xyj � xzk
�1, 2, 1�F�x, y, z� � xyz i � y j � zk,
45.
curl F �0, 0, 3� � �2k
curl F � � i��x
ex sin y
j��y
�ex cos y
k��z
0 � � �2ex cos yk
�0, 0, 3�F�x, y, z� � ex sin y i � ex cos y j,
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180 Chapter 14 Vector Analysis
47.
� � xx2 � y2 �
��x�y2�1 � �x�y�2k �
2xx2 � y2 k� i
��x
arctan�x y
j
��y
1 ln�x2 � y2�2
k
��z
1 �curl F �
F�x, y, z� � arctan�xy i � ln�x2 � y2 j � k
49.
curl F � � i��x
sin�x � y�
j��y
sin�y � z�
k��z
sin�z � x�� � cos�y � z�i � cos�z � x�j � cos�x � y�k
F�x, y, z� � sin�x � y�i � sin�y � z�j � sin�z � x�k
51.
Not conservative
curl F � � i��x
sin y
j��y
�x cos y
k��z
1 � � �2 cos yk � 0
F�x, y, z� � sin y i � x cos y j � k 53.
Conservative
f �x, y, z� � xyez � K
fz�x, y, z� � xyez
fy�x, y, z� � xez
fx�x, y, z� � yez
curl F � � i��x
yez
j��y
xez
k��z
xyez� � 0
F�x, y, z� � ez�y i � xj � xyk�
55.
Conservative
f �x, y, z� �xy
� z2 � z � K
� z2 � z � p�x, y� � K3
f �x, y, z� � ��2z � 1�dz
f �x, y, z� � ��xy2 dy �
xy
� h�x, z� � K2
f �x, y, z� � �1y dx �
xy
� g�y, z� � K1
fz�x, y, z� � 2z � 1
fy�x, y, z� � �xy2
fx�x, y, z� �1y
curl F � � i � �x
1y
j � �y
� x y2
k � �z
2z � 1� � 0
F�x, y, z� �1y
i �xy2 j � �2z � 1�k 57.
� 12x � 2xy
div F�x, y� ��
�x6x2� �
�
�y�xy2�
F�x, y� � 6x2 i � xy2j
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Section 14.1 Vector Fields 181
59.
div F�x, y, z� ��
�xsin x� �
�
�ycos y� �
�
�zz2� � cos x � sin y � 2z
F�x, y, z� � sin x i � cos y j � z2 k
61.
div F�1, 2, 1� � 4
div F�x, y, z� � yz � 1 � 1 � yz � 2
F�x, y, z� � xyz i � y j � zk 63.
div F�0, 0, 3� � 0
div F�x, y, z� � ex sin y � ex sin y
F�x, y, z� � e x sin y i � ex cos y j
65. See the definition, page 1008. Examples include velocityfields, gravitational fields and magnetic fields.
67. See the definition on page 1014.
69.
curl �F � G� � � i��x
2xz � 3y2
j��y
3xy � z
k��z
�y � 2x2� � ��1 � 1�i � ��4x � 2x�j � �3y � 6y�k � 6xj � 3yk
F � G� i1x
j2x�y
k3yz � � �2xz � 3y2�i � �z � 3xy�j � ��y � 2x2�k
G�x, y, z� � x i � y j � zk
F�x, y, z� � i � 2x j � 3yk
71.
curl�curl F� � � i
��x
0
j
��y
xy
k
��z
�xz� � z j � yk
curl F � � i
��xxyz
j
��yy
k
��zz � � xy j � xzk
F�x, y, z� � xyz i � y j � zk 73.
div�F � G� � 2z � 3x
� �2xz � 3y2�i � �z � 3xy�j � ��y � 2x2�k
F � G � � i1x
j2x�y
k3yz �
G�x, y, z� � xi � y j � zk
F�x, y, z� � i � 2x j � 3yk
75.
div�curl F� � x � x � 0
curl F � � i��x
xyz
j��y
y
k��z
z � � xyj � xzk
F�x, y, z� � xyz i � y j � zk
77. and where M, N, P, Q, R, and S have continuous partial derivatives.
� curl F � curl G
� ��S�y
��R�z i � ��S
�x�
�Q�z j � ��R
�x�
�Q�y k � ��P
�y�
�N�z i � ��P
�x�
�M�z j � ��N
�x�
�M�y k
� � �
�y�P � S� �
�
�z�N � R�i � � �
�x�P � S� �
�
�z�M � Q�j � � �
�x�N � R� �
�
�y�M � Q�k
curl�F � G� � � i��x
M � Q
j��y
N � R
k��z
P � S �F � G � �M � Q�i � �N � R�j � �P � S�k
G � Q i � R j � SkLet F � M i � N j � Pk
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182 Chapter 14 Vector Analysis
79. Let and
� div F � div G
� ��M�x
��N�y
��P�z � ��R
�x�
�S�y
��T�z
div�F � G� ��
�x�M � R� �
�
�y�N � S� �
�
�z�P � T � �
�M�x
��R�x
��N�y
��S�y
��P�z
��T�z
G � R i � Sj � T k.F � M i � Nj � Pk
81.
(Exercise 77)
(Exercise 78)
� � � � F�
� curl� � F�
� curl�f � � curl� � F�
� f � � � F�� � curl�f � � � F��
F � M i � N j � Pk
83. then
� f div F � f F
� f ��M�x
��N�y
��N�z � ��f
�xM �
�f�y
N ��f�z
P
div� f F� ��
�x� f M � �
�
�y� fN � �
�
�z� fP� � f
�M�x
� M�f�x
� f�N�y
� N�f�y
� f�P�z
� P�f�z
f F � fM i � f Nj � fPk.Let F � M i � Nj � Pk,
In Exercises 85 and 87, and f �x, y, z� � �F�x, y, z�� � �x2 � y2 � z2.F�x, y, z� � xi � yj � zk
85.
�ln f � �x
x2 � y2 � z2 i �y
x2 � y2 � z2 j �z
x2 � y2 � z2 k �xi � yj � zkx2 � y2 � z2 �
Ff 2
ln f �12
ln�x2 � y2 � z2�
87.
� n��x2 � y2 � z2 �n�2�x i � yj � zk� � n f n�2F
� n��x2 � y2 � z2 �n�1 z�x2 � y2 � z2
k
f n � n��x2 � y2 � z2 �n�1 x�x2 � y2 � z2
i � n��x2 � y2 � z2 �n�1 y�x2 � y2 � z2
j
f n � ��x2 � y2 � z2 �n
89. The winds are stronger over Phoenix. Although the winds over both cities are northeasterly,they are more towards the east over Atlanta.
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Section 14.2 Line Integrals
Section 14.2 Line Integrals 183
1.
0 ≤ t ≤ 2�
r�t� � 3 cos t i � 3 sin t j
y � 3 sin t
x � 3 cos t
sin2 t �y2
9
cos2 t �x2
9
cos2 t � sin2 t � 1
x2
9�
y2
9� 1
x2 � y2 � 9 3. r�t� � �t i,3i � �t � 3�j,�9 � t�i � 3j,�12 � t�j,
0 ≤ t ≤ 33 ≤ t ≤ 66 ≤ t ≤ 99 ≤ t ≤ 12
5. r�t� � �t i � �t j,�2 � t�i � �2 � t�j,
0 ≤ t ≤ 11 ≤ t ≤ 2
7.
�C
�x � y� ds � �2
0 �4t � 3t���4�2 � �3�2 dt � �2
0 5t dt � �5t 2
2 �2
0� 10
r��t� � 4 i � 3j0 ≤ t ≤ 2;r�t� � 4 t i � 3tj,
9.
� ��2
0
�65�1 � 64t 2� dt � ��65t �64t 3
3 ���2
0� �65�
2�
8� 3
3 � ��65�
6�3 � 16� 2�
�C
�x2 � y2 � z2� ds � ��2
0 �sin2 t � cos2 t � 64t 2���cos t�2 � ��sin t�2 � 64 dt
r��t� � cos t i � sin t j � 8k0 ≤ t ≤ �
2;r�t� � sin t i � cos t j � 8tk,
11.
� �13
t 3�3
0� 9
� �3
0 t
2 dt
�C
�x 2 � y 2� ds � �3
0 �t2 � 02 �1 � 0 dt
x1 2 3
−1
1
2
y0 ≤ t ≤ 3r�t� � t i,
13.
��
2 � ��2
0 dt
�C
�x2 � y2� ds � ��2
0 �cos2 t � sin2 t ���sin t�2 � �cos t�2 dt
x1
1
y
0 ≤ t ≤ �
2r�t� � cos t i � sin t j,
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184 Chapter 14 Vector Analysis
15.
�19�2
6 � ��2t 2
2�
83
t 32��1
0
�C
�x � 4�y � ds � �1
0 �t � 4�t ��1 � 1 dt
x1
1 (1, 1)
y0 ≤ t ≤ 1r�t� � t i � t j,
17.
�C
�x � 4�y � ds �12
�19�2
6�
83
�19 � 19�2
6�
19�1 � �2 �6
�C3
�x � 4�y � ds � �3
24�3 � t dt � ��
83
�3 � t�32�3
2�
83
� �2�2t �t 2
2�
83
�t � 1�32�2
1�
19�26
�C2
�x � 4�y � ds � �2
1 ��2 � t� � 4�t � 1 �1 � 1 dt
�C1
�x � 4�y� ds � �1
0 t dt �
12
xC1
C2C3
(1, 0)
(0, 1)
y
r�t� � �t i,�2 � t�i � �t � 1�j,�3 � t�j,
0 ≤ t ≤ 11 ≤ t ≤ 22 ≤ t ≤ 3
19.
�2�13�
3�27 � 64�2� � 4973.8
� ��132 9t �
4t 3
3 ��4�
0 �
�132 �4�
0 �9 � 4t 2� dt
� �4�
0 12
��3 cos t�2 � �3 sin t�2 � �2t�2 �13 dtMass � �C
� �x, y, z� ds
�r��t�� � ���3 sin t�2 � �3 cos t�2 � �2�2 � �13
r��t� � �3 sin t i � 3 cos t j � 2k
0 ≤ t ≤ 4� r�t� � 3 cos t i � 3 sin tj � 2tk,
��x, y, z� �12
�x2 � y2 � z2�
21.
� �163
t3 �12
t2�1
0�
356
�C
F � dr � �1
0 �16t 2 � t� dt
r��t� � 4i � j
F�t� � 4t 2 i � t j
0 ≤ t ≤ 1C: r�t� � 4t i � tj,
F�x, y� � xyi � yj 23.
� �2 sin2 t��2
0� 2
�C
F � dr � ��2
0 ��12 sin t cos t � 16 sin t cos t� dt
r��t� � �2 sin t i � 2 cos t j
F�t� � 6 cos t i � 8 sin tj
0 ≤ t ≤ �
2C: r�t� � 2 cos t i � 2 sin t j,
F�x, y� � 3x i � 4yi
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Section 14.2 Line Integrals 185
25.
� �t5
5�
2t3
3� 2t 2�
1
0� �
1715
�C
F � dr � �1
0 �t 4 � 2t�t � 2� dt
r��t� � i � 2 t j
F�t� � t 4 i � �t � 2�j � 2 t 3k
0 ≤ t ≤ 1C: r�t� � t i � t2 j � 2k,
F�x, y, z� � x2yi � �x � z�j � xyzk 27.
� 249.49
�C
F � dr � �3
1 �t2 ln t � 12t 3 � t �ln t�2 dt
dr � i � 2 t j �1t
k� dt
F�t� � t 2 ln t i � 6t 2 j � t 2 ln2 tk
1 ≤ t ≤ 3r�t� � t i � t2j � ln tk,
F�x, y, z� � x2z i � 6yj � yz2k
29.
Work � �C
F � dr � �2
0 ��t � 6t5� dt � ��
12
t2 � t 6�2
0� �66
F � r� � �t � 6t 5
F�t� � �t i � 2t 3 j
r��t� � i � 3t 2 j
0 ≤ t ≤ 2 r�t� � t i � t 3 j,
C: y � x3 from �0, 0� to �2, 8�
F�x, y� � �x i � 2y j
31.
C: counterclockwise around the triangle whose vertices are
On
On
Total work � �C
F � dr � 1 �12
�32
� 0
Work � �C3
F � dr � �3
2 ��2�3 � t� � �3 � t� dt � �
32
F�t� � 2�3 � t�i � �3 � t�j, r��t� � �i � j On C3:
Work � �C2
F � dr � �2
1 �t � 1� dt �
12
F�t� � 2 i � �t � 1�j, r��t� � jC2:
Work � �C1
F � dr � �1
0 2t dt � 1
F�t� � 2t i, r��t� � iC1:
r�t� � �t i,i � �t � 1�j,�3 � t�i � �3 � t�j,
0 ≤ t ≤ 11 ≤ t ≤ 22 ≤ t ≤ 3
�0, 0�, �1, 0�, �1, 1�
F�x, y� � 2xi � yj
33.
Work � �C
F � dr � �2�
0 �5t dt � �10� 2
F � r� � �5t
F�t� � 2 cos t i � 2 sin t j � 5tk
r��t� � �2 sin t i � 2 cos t j � k
0 ≤ t ≤ 2�C: r�t� � 2 cos t i � 2 sin t j � tk,
F�x, y, z� � x i � yj � 5zk 35.
�C
F � dr � �2�
0 15002�
dt � �15002�
t�2�
0� 1500 ft � lb
dr � 3 cos t i � 3 sin t j �102�
k� dt
F � 150k
r�t� � 3 sin t i � 3 cos t j �102�
tk, 0 ≤ t ≤ 2�
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186 Chapter 14 Vector Analysis
37.
(a)
Both paths join and The integrals are negativesof each other because the orientations are different.
�6, 2�.�2, 0�
�C1
F � dr � �3
1�8t2 � 2t�t � 1�� dt �
2363
F�t� � 4t2 i � 2t�t � 1�j
r1��t� � 2 i � j
r1�t� � 2t i � �t � 1�j, 1 ≤ t ≤ 3
F�x, y� � x2 i � xyj
(b)
� �2363
�C2
F � dr � �2
0��8�3 � t�2 � 2�3 � t��2 � t� dt
F�t� � 4�3 � t�2 i � 2�3 � t��2 � t�j
r2��t� � �2i � j
r2�t� � 2�3 � t�i � �2 � t�j, 0 ≤ t ≤ 2
39.
Thus, �C
F � dr � 0.
F � r� � �2t � 2t � 0
F�t� � �2 t i � t j
r��t� � i � 2 j
C: r�t� � t i � 2 t j
F�x, y� � yi � x j 41.
Thus, �C
F � dr � 0.
F � r� � �t 3 � 2t 2� � 2tt �t 2
2 � � 0
F�t� � �t3 � 2t 2�i � t �t 2
2 � j
r��t� � i � 2 t j
C: r�t� � t i � t 2j
F�x, y� � �x 3 � 2x 2�i � x �y2� j
43.
�C
�x � 3y 2� dy � �10
0 y
5� 3y 2� dy � �y 2
10� y 3�
10
0� 1010
0 ≤ y ≤ 100 ≤ t ≤ 1 ⇒ y � 5x or x �y5
,y � 10t,x � 2t,
45.
�C
xy dx � y dy � �2
0�5x 2 � 25x� dx � �5x 3
3�
25x 2
2 �2
0�
1903
0 ≤ x ≤ 2dy � 5 dx, y � 5x,
�C
xy dx � y dy � �10
0y 2
25� y� dy � �y 3
75�
y 2
2 �10
0�
1903
OR
dx �15
dy0 ≤ y ≤ 10,0 ≤ t ≤ 1 ⇒ x �y5
,y � 10t,x � 2t,
47.
�C
�2x � y� dx � �x � 3y� dy � �5
0 2t dt � 25
dy � 0dx � dt,
y�t� � 0x�t� � t,
x1 2 3 4 5
1
2
3
−1
−2
y0 ≤ t ≤ 5r�t� � t i,
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Section 14.2 Line Integrals 187
49.
x32
2
1
1
3(3, 3)
C1
C2
y
�C
�2x � y� dx � �x � 3y� dy � 9 �452
�632
�C2
�2x � y� dx � �x � 3y� dy � �6
3 �3 � 3�t � 3� dt � �3t 2
2� 6t�
6
3�
452
dy � dtdx � 0,
y�t� � t � 3x�t� � 3,C2:
�C1
�2x � y� dx � �x � 3y� dy � �3
0 2t dt � 9
dy � 0dx � dt,
y�t� � 0,x�t� � t,C1:
r�t� � �t i,3i � �t � 3�j,
0 ≤ t ≤ 3 3 ≤ t ≤ 6
51.
� �1
0�6t3 � t2 � 4t � 1� dt � �3t 4
2�
t 3
3� 2t2 � t�
1
0� �
116
�C
�2x � y� dx � �x � 3y� dy � �1
0 ��2 t � 1 � t 2� � �t � 3 � 3t2���2t� dt
dy � �2t dtdx � dt,0 ≤ t ≤ 1,y�t� � 1 � t2,x�t� � t,
53.
� �2
0 �24t3 � 2t2 � 2t� dt � �6t 4 �
23
t 3 � t2�2
0�
3163
�C
�2x � y� dx � �x � 3y� dy � �2
0 �2t � 2t2� dt � �t � 6t2�4t dt
dy � 4t dtdx � dt,
0 ≤ t ≤ 2y�t� � 2t 2,x�t� � t,
55.
line from
Lateral surface area:
�C
f �x, y� ds � �1
0 5h dt � 5h
�r��t�� � 5
r��t� � 3i � 4j
0 ≤ t ≤ 1 r � 3t i � 4t j,
�0, 0� to �3, 4�C:
f �x, y� � h 57.
from
Lateral surface area:
� �sin2 t2 �
�2
0�
12
�C
f �x, y� ds � ��2
0 cos t sin t dt
�r��t�� � 1
r��t� � �sin t i � cos t j
0 ≤ t ≤ �
2 r�t� � cos t i � sin t j,
�1, 0� to �0, 1�x2 � y2 � 1C:
f �x, y� � xy
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188 Chapter 14 Vector Analysis
59.
from
Lateral surface area:
�h4
�2�5 � ln�2 � �5 � � 1.4789h
� �h4�2�1 � t��1 � 4�1 � t�2 � ln�2�1 � t� � �1 � 4�1 � t�2 ��
1
0
�C
f �x, y� ds � �1
0 h�1 � 4�1 � t�2 dt
�r��t�� � �1 � 4�1 � t�2
r��t� � �i � 2�1 � t�j0 ≤ t ≤ 1 r�t� � �1 � t� i � �1 � �1 � t�2 j,
�1, 0� to �0, 1�C: y � 1 � x2
f �x, y� � h
61.
from
You could parameterize the curve C as in Exercises 59 and 60. Alternatively, let then:
Lateral surface area:
Let and then and
� �1
12�
1120� �
1120
�5�52 �1
120�25�5 � 11� � 0.3742
� ��1
12 sin2 t�1 � 4 cos2 t�32 �
1120
�1 � 4 cos2 t�52��2
0
�C
f �x, y� ds � ��1
12 sin2 t�1 � 4 cos2 t�32�
�2
0�
16
��2
0 �1 � 4 cos2 t�32 sin t cos t dt
v � �1
12 �1 � 4 cos2 t�32.du � 2 sin t cos t dtdv � �1 � 4 cos2 t�12 sin t cos t,u � sin2 t
�C
f �x, y� ds � ��2
0cos t sin2 t�sin t�1 � 4 cos2 t � dt � ��2
0sin2 t��1 � 4 cos2 t�12 sin t cos t dt
�r��t�� � �sin2 t � 4 sin2 t cos2 t � sin t�1 � 4 cos2 t
r��t� � �sin t i � 2 sin t cos t j
0 ≤ t ≤ �
2 r�t� � cos t i � sin2 t j,
y � 1 � cos2 t � sin2 t
x � cos t,
�1, 0� to �0, 1�C: y � 1 � x2
f �x, y� � xy
63. (a)
(b) 0.2�12�� �12�
5� 7.54 cm3
� �2t � 4�t � sin t cos t��2�
0� 12� � 37.70 cm2
S � �C
f �x, y� ds � �2�
0 �1 � 4 sin2 t��2� dt
�r��t� � � 2
r��t� � �2 sin t i � 2 cos t j
0 ≤ t ≤ 2� r�t� � 2 cos t i � 2 sin t j,
f �x, y� � 1 � y2 (c)
x
y33
−3
4
5
z
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Section 14.2 Line Integrals 189
65.
Matches b
x
y
30
40
3
50
3
60
20
10
zS � 25
67. (a) Graph of:
(b) Consider the portion of the surface in the first quadrant. The curve is over the curve Hence, the total lateral surface area is
(c) The cross sections parallel to the xz-plane are rectangles of height and base Hence,
Volume � 2 �3
0 2�9 � y 2 1 � 4
y 2
9 1 �y 2
9 �� dy � 42.412 cm3
2�9 � y 2.1 � 4�y3�2�1 � y 29�
� 123�
4 � � 9� sq. cm4 �C
f �x, y� ds � 4 ��2
0 �1 � sin2 2t�3 dt
0 ≤ t ≤ �2.3 sin t j,r1�t� � 3 cos t i �z � 1 � sin2 2t
x
y34
3
3
2
1
4
z
0 ≤ t ≤ 2�r�t� � 3 cos t i � 3 sin t j � �1 � sin2 2t�k
73. False
�C
xy ds � �2 �1
0 t 2 dt
75. False, the orientations are different.
69. See the definition of Line Integral, page 1020.
See Theorem 14.4.
71. The greater the height of the surface over the curve, the greater the lateral surface area. Hence,
z3 < z1 < z2 < z4 .
x
2
3
4
1 2 3 4
1
y
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Section 14.3 Conservative Vector Fields and Independence of Path
1.
(a)
�C
F � dr � �1
0 �t 2 � 2t 4� dt �
1115
F�t� � t 2 i � t 3 j
r1��t� � i � 2t j
0 ≤ t ≤ 1r1�t� � t i � t 2 j,
F�x, y� � x2i � xy j
(b)
� �sin3 �3
�2 sin5 �
5 ���2
0�
1115
�C
F � dr � ���2
0 �sin2 � cos � � 2 sin4 � cos �� d�
F�t� � sin2 � i � sin3 � j
r2� ��� � cos � i � 2 sin � cos � j
0 ≤ � ≤ �
2r2��� � sin � i � sin2 � j,
3.
(a)
(b)
� �12�ln�7
2� 23 � ln�1
2� � �12
ln�7 � 43 � � �1.317
� �12�
3
0
1�t � �1�2� 2 � �1�4�
dt � ��12
ln ��t �12 � t 2 � t��3
0
�C
F � dr � �3
0 � t
2t � 1�
t � 1
2t � dt � �12�
3
0
1tt � 1
dt � �12
�3
0
1t 2 � t � �1�4� � �1�4�
dt
F�t� � t i � t � 1 j
r2��t� �1
2t � 1i �
1
2tj
0 ≤ t ≤ 3r2�t� � t � 1 i � t j,
� ����3
0 sec � d� � ��ln�sec � � tan ���
��3
0� �ln�2 � 3 � � �1.317
�C
F � dr � ���3
0 �sec � tan2 � � sec3 �� d� � ���3
0 �sec ��sec2 � � 1� � sec3 � d�
F��� � tan � i � sec �j
r1���� � sec � tan � i � sec2 � j
0 ≤ � ≤ �
3r1��� � sec � i � tan � j,
F�x, y� � y i � x j
5.
Since is conservative.FNx
�My
,
My
� ex cos yNx
� ex cos y
F�x, y� � ex sin yi � ex cos yj 7.
Since is not conservative.FNx
My
,
My
� �1y2
Nx
�1y2
F�x, y� �1y
i �xy2 j
9.
is conservative.curl F � 0 ⇒ F
F�x, y, z� � y2z i � 2xyz j � xy 2k
190 Chapter 14 Vector Analysis
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11.
(a)
�C
F � dr � �1
0 4t 3 dt � 1
F�t� � 2t 3 i � t 2 j
r1��t� � i � 2t j
0 ≤ t ≤ 1 r1�t� � t i � t 2j,
F�x, y� � 2xyi � x2j
(b)
�C
F � dr � �1
0 5t 4 dt � 1
F�t� � 2t 4 i � t 2 j
r2��t� � i � 3t 2 j
0 ≤ t ≤ 1r2�t� � t i � t 3 j,
13.
(a) (b) (c)
�C
F � dr � �1
0 �2t 3 dt � �
12�
C
F � dr � �1
0 �t 2 dt � �
13�
C
F � dr � 0
F�t� � t 3 i � t jF�t� � t 2 i � t jF�t� � t i � t j
r3��t� � i � 3t 2 jr2��t� � i � 2t jr1��t� � i � j
0 ≤ t ≤ 1r3�t� � t i � t 3 j,0 ≤ t ≤ 1r2�t� � t i � t 2 j,0 ≤ t ≤ 1r1�t� � t i � t j,
F�x, y� � yi � xj
15.
Since is conservative. The potential function is Therefore, wecan use the Fundamental Theorem of Line Integrals.
(a) (b)
(c) and (d) Since C is a closed curve, �C
y2 dx � 2xy dy � 0.
�C
y2 dx � 2xy dy � �x2 y��1, 0�
��1, 0�� 0�
C
y2 dx � 2xy dy � �x2 y��4, 4�
�0, 0�� 64
f �x, y� � xy2 � k.F�x, y� � y2 i � 2xy jM�y � N�x � 2y,
�C
y2 dx � 2xy dy
17.
Since
is conservative.
The potential function is
(a)
(b) �643
�C
2xy dx � �x2 � y2� dy � �x2 y �y3
3 ��0, 4�
�2, 0�
�643
�C
2xy dx � �x2 � y2� dy � �x2 y �y 3
3 ��0, 4�
�5, 0�
f �x, y� � x2y �y3
3� k.
F�x, y� � 2xyi � �x2 � y2�j
M�y � N�x � 2x,
�C
2xy dx � �x2 � y2� dy19.
Since is conservative. The potentialfunction is
(a)
(b)
�C
F � dr � �xyz��4, 2, 4�
(0, 0, 0�� 32
0 ≤ t ≤ 2r2�t� � t 2 i � t j � t 2k,
�C
F � dr � �xyz��4, 2, 4�
�0, 2, 0�� 32
0 ≤ t ≤ 4r1�t� � t i � 2j � tk,
f �x, y, z� � xyz � k.F�x, y, z�curl F � 0,
F�x, y, z� � yz i � xz j � xyk
21.
is not conservative.
(a)
—CONTINUED—
�C
F � dr � �1
0 �2t3 � 2t2 � t� dt �
23
F�t� � �2t 2 � t�i � �t 2 � 1�j � �2t 2 � 4�k
r1��t� � i � 2t j
0 ≤ t ≤ 1r1�t� � t i � t 2j � k,
F�x, y, z�
F�x, y, z� � �2y � x�i � �x2 � z�j � �2y � 4z�k
Section 14.3 Conservative Vector Fields and Independence of Path 191
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21. —CONTINUED—
(b)
� �1
0 �17t2 � 5t � �2t � 1�2 � 16�2t � 1�3 dt � �17t 3
3�
5t 2
2�
�2t � 1�3
6� 2�2t � 1�4�
1
0�
176
�C
F � dr � �1
0 �3t � t2 � �2t � 1�2 � 8t�2t � 1� � 16�2t � 1�3 dt
F�t� � 3t i � �t 2 � �2t � 1�2 j � �2t � 4�2t � 1�2 k
r2��t� � i � j � 4�2t � 1�k
0 ≤ t ≤ 1r2�t� � t i � t j � �2t � 1�2 k,
23.
is conservative. The potential function is
(a) 0 ≤ t ≤
(b)
�C
F � dr � �xyez���4, 0, 3�
�4, 0, 3�� 0
0 ≤ t ≤ 1r2�t� � �4 � 8t�i � 3k,
�C
F � dr � �xyez���4, 0, 3�
�4, 0, 3�� 0
�r1�t� � 4 cos t i � 4 sin tj � 3k,
f �x, y, z� � xyez � k.F�x, y, z�
F�x, y, z� � ez�y i � x j � xyk� 25. �C
�yi � x j� � dr � �xy��3, 8�
�0, 0�� 24
27. �C
cos x sin y dx � sin x cos y dy � �sin x sin y��3��2, ��2�
�0, ���� �1
29. �C
ex sin y dx � ex cos y dy � �ex sin y��2�, 0�
�0, 0�� 0
31.
is conservative and the potential function is
(a)
(b)
(c) �xy � 3yz � 2xz��1, 0, 0�
�0, 0, 0�� �xy � 3yz � 2xz�
�1, 1, 0�
�1, 0, 0�� �xy � 3yz � 2xz�
�1, 1, 1�
�1, 1, 0�� 0 � 1 � ��1� � 0
�xy � 3yz � 2xz��0, 0, 1�
�0, 0, 0�� �xy � 3yz � 2xz�
�1, 1, 1�
�0, 0, 1�� 0 � 0 � 0
�xy � 3yz � 2xz��1, 1, 1�
�0, 0, 0�� 0 � 0 � 0
f �x, y, z� � xy � 3yz � 2xz.F�x, y, z�
�C
�y � 2z� dx � �x � 3z� dy � �2x � 3y� dz
33. �C
�sin x dx � z dy � y dz � �cos x � yz����2, 3, 4�
�0, 0, 0�� 12 � 1 � 11
35. is conservative.
Work � �3x3y2 � y��5, 9�
�0, 0�� 30,366
F�x, y� � 9x2y2 i � �6x3y � 1�j
192 Chapter 14 Vector Analysis
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Section 14.4 Green’s Theorem 193
37.
W � �C
F � dr � �C
�� 2
4 �cos 2� t i � sin 2� t j� � 4���sin 2� t i � cos 2� t j� dt � �� 3�
C
0 dt � 0
F�t� � m � a�t� �1
32a�t� � �
� 2
4 �cos 2�t i � sin 2� t j�
a�t� � �8� 2 cos 2�t i � 8� 2 sin 2�t j
r��t� � �4� sin 2�t i � 4� cos 2�t j
r�t� � 2 cos 2�t i � 2 sin 2�t j
39. Since the sum of the potential and kinetic energies remains constant from point to point, if the kinetic energy is decreasing at arate of 10 units per minute, then the potential energy is increasing at a rate of 10 units per minute.
41. No. The force field is conservative. 43. See Theorem 14.5, page 1033.
47. False, it would be true if F were conservative. 49. True
51. Let
Then and Since
we have
Thus, F is conservative. Therefore, by Theorem 14.7, we have
for every closed curve in the plane.
�C
��f�y
dx ��f�x
dy� � �C
�M dx � N dy� � �C
F � dr � 0
�M�y
��N�x
.�2f�x2 �
�2f�y 2 � 0
�N�x
��
�x ���f�x � � �
�2f�x2.
�M�y
��
�y�� f�y � �
�2f�y2
F � M i � N j ��f�y
i ��f�x
j.
45. (a) The direct path along the line segment joining to requires less work than the path going from to and then to
(b) The closed curve given by the line segments joining and satisfies �C
F � dr 0.��4, 0���4, 0�, ��4, 4�, �3, 4�,
�3, 4�.��4, 4���4, 0��3, 4���4, 0�
Section 14.4 Green’s Theorem
1.
By Green’s Theorem, �R� ��N
�x�
�M�y � dA � �4
0�4
0�2x � 2y� dy dx � �4
0 �8x � 16� dx � 0.
� 0 � 64 � 64 � 0 � 0
� �12
8 �16��dt� � �12 � t�2�0�� � �16
12 ��16 � t�2�0� � 0��dt��
�C
y2 dx � x2 dy � �4
0 �0 dt � t2�0�� � �8
4 ��t � 4�2�0� � 16 dt�
x1
1
2
2
3
3
4
4(4, 4)
y
r�t� � t i,4 i � �t � 4�j,�12 � t�i � 4j,�16 � t�j,
0 ≤ t ≤ 44 ≤ t ≤ 88 ≤ t ≤ 1212 ≤ t ≤ 16
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194 Chapter 14 Vector Analysis
3.
By Green’s Theorem,
�R� ��N
�x�
�M�y � dA � �4
0�x
x24�2x � 2y� dy dx � �4
0 �x2 �
x3
2�
x4
16� dx �3215
.
� �4
0 � t 4
16�
t 3
2� dt � �8
4 �2�8 � t�2 dt �
2245
�1283
�3215
�C
y2 dx � x2 dy � �4
0 � t 4
16�dt� � t 2� t
2 dt�� � �8
4 ��8 � t�2��dt� � �8 � t�2��dt��
x
1
1
2
2
3
3
4
4
(4, 4)
C1
C2
y
r�t� � t i � t24 j,�8 � t�i � �8 � t�j,
0 ≤ t ≤ 44 ≤ t ≤ 8
5.
Let and
� �2
�2 �2 4 � x2 ex � xe 4�x2
� xe� 4�x2� dx � 19.99 �R� ��N
�x�
�M�y � dA � �2
�2� 4�x2
� 4�x2
�ex � xey� dy dx
�C
xey dx � ex dy � �2�
0 �2 cos te2 sin t ��2 sin t� � e2 cos t �2 cos t�� dt � 19.99
0 ≤ t ≤ 2�.y � 2 sin t,x � 2 cos t
x2 � y2 � 4C:
7.
�43
� �2
0 �2x � x2� dx
x
y x=
1
1
2
2(2, 2)
y x x= 2 −
y
�C
�y � x� dx � �2x � y� dy � �2
0�x
x2�x
dy dx
9. From the accompanying figure, we see that R is the shaded region. Thus, Green’s Theorem yields
� 56.
� 6�10� � 2�2�
� Area of R
�C
�y � x� dx � �2x � y� dy � �R� 1 dA
x
2
2
4
4
−2
−4
(1, 1)
(5, 3)( 5, 3)−
( 1, 1)−
(5, 3)−
(1, 1)−
( 5, 3)− −
( 1, 1)− −
y
11. Since the curves and intersect at and Green’s Theorem yields
� �83
�83
� 16 �323
.
� �4x � 4x2 �x3
3�
x4
2 �2
�2
� �2
�2 �4 � 8x � x2 � 2x3� dx
� �2
�2 �y � 2xy�
4�x2
0 dx
�C
2xy dx � �x � y� dy � �R� �1 � 2x� dA � �2
�2�4�x2
0 �1 � 2x� dy dx
�2, 0�,��2, 0�y � 4 � x2y � 0
In Exercises 7 and 9,�N�x
��M�y
� 1.
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Section 14.4 Green’s Theorem 195
13. Since R is the interior of the circle Green’s Theorem yields
� �a
�a� a2�x2
� a2�x2
4y dy dx � 4 �a
�a
0 dx � 0.
�C
�x2 � y2� dx � 2xy dy � �R� �2y � 2y� dA
x2 � y2 � a2,
15. Since
we have path independence and
�R���N
�x�
�M�y � dA � 0.
�M�y
�2x
x2 � y2 ��N�x
,
17. By Green’s Theorem,
� �1
0� x
x
y dy dx �12
�1
0 �x � x2� dx �
12�
x 2
2�
x3
3 �1
0�
112
.
�C
sin x cos y dx � �xy � cos x sin y� dy � �R� ��y � sin x sin y� � ��sin x sin y�� dA
19. By Green’s Theorem,
� �2�
0�3
1 �1 � r cos �r dr d � �2�
0 �4 �
263
cos � d � 8�.
�C
xy dx � �x � y� dy � �R� �1 � x� dA
21.
� �2�
0 �2 �
83
cos � d � 4�Work � �C
xy dx � �x � y� dy � �R��1 � x� dA � �2�
0�2
0 �1 � r cos �r dr d
C: x2 � y2 � 4
F�x, y� � xy i � �x � y�j
23.
C: boundary of the triangle with vertices
Work � �C
�x32 � 3y� dx � �6x � 5 y � dy � �R� 9 dA � 9�1
2��5��5� �2252
�0, 0�, �5, 0�, �0, 5�
F�x, y� � �x32 � 3y� i � �6x � 5 y �j
25. C: let By Theorem 14.9, we have
A �12�C
x dy � y dx �12
�2�
0 �a cos t�a cos t� � a sin t��a sin t�� dt �
12�
2�
0 a2 dt � �a2
2 t�
2�
0� �a2.
0 ≤ t ≤ 2�.y � a sin t,x � a cos t,
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196 Chapter 14 Vector Analysis
27. From the accompanying figure we see that
Thus, by Theorem 14.9, we have
�12
�1
�3 ��1� dx �
12
�1
�3 �x2 � 4� dx �
12
�1
�3 �3 � x2� dx �
12
�3x �x3
3 �1
�3�
323
.
�12
�1
�3 ��1� dx �
12
��3
1 ��x2 � 4� dx
A �12�
1
�3�x�2� � �2x � 1�� dx �
12�
�3
1 �x��2x� � �4 � x2�� dx
dy � �2x dx.C2: y � 4 � x2,
dy � 2 dxC1: y � 2x � 1,
x4
−4
−6
−4−6
2(1, 3)
( 3, 5)− −
y
29. See Theorem 14.8, page 1042. 31. Answers will vary.
F3�x, y� � 2xyi � x2j
F2�x, y� � x2i � y2j
F1�x, y� � yi � xj
33.
For and for Thus,
To calculate note that along Thus,
�x, y � � �0, 85�
y ��1
2�323� ��2
2 �4 � x2�2 dx �
364
�2
�2 �16 � 8x2 � x4� dx �
364 �16x �
8x3
3�
x5
5 �2
�2�
85
.
C2.y � 0y,
x �1
2�323� ��2
2 x2��2x dx� � � 3
64 ��
x4
2 ���2
2� 0.
dy � 0.C2,dy � �2x dxC1,
x �1
2A �
C1
x2 dy �1
2A �
C2
x2 dy
x1 2
1
2
3
−1−2
C1
C2
y x= 4 − 2
y
A � �2
�2 �4 � x2� dx � �4x �
x3
3 �2
�2�
323
35. Since we have On we have and on we have
Thus,
�x, y� � � 815
, 821�
� �2 �1
0 x6 dx � 2 �0
1 x2 dx � �
27
�23
�8
21.
y � �2 �C
y2 dx
� 6 �1
0 x4 dx � 2 �0
1 x2 dx �
65
�23
�8
15
x1
1
C1
C2
(1, 1)
y x � 2 �C
x2 dy � 2 �C1
x2�3x2 dx� � 2 �C2
x2 dx
dy � dx.y � x,
C2dy � 3x2 dxy � x3,C11
2A� 2.A � �1
0 �x � x3� dx � �x2
2�
x4
4 �1
0�
14
,
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39. In this case the inner loop has domain Thus,
�12
�4�3
2�3 �3 � 4 cos � 2 cos 2� d �
12�3 � 4 sin � sin 2�
4�3
2�3� � �
3 32
.
A �12
�4�3
2�3 �1 � 4 cos � 4 cos2 � d
2�
3 ≤ ≤
4�
3.
41.
(a) Let
F is conservative since
F is defined and has continuous first partials everywhere except at the origin. If C is a circle (a closed path) that does not contain the origin, then
(b) Let be a circle oriented clockwise inside C (see figure). Introduce line segmentsand as illustrated in Example 6 of this section in the text. For the region inside C and outside Green’s Theorem
applies. Note that since and have opposite orientations, the line integrals over them cancel. Thus,and
But,
Finally,
Note: If C were orientated clockwise, then the answer would have been
x
−3
−2
2
3
4
C1C2
C3
C
y
2�.
�C
F � dr � ��C1
F � dr � �2�.
� �2�
0 �sin2 t � cos2 t� dt � �t�
2�
0� 2�.
�C1
F � dr � �2�
0 ���a sin t���a sin t�
a2 cos2 t � a2 sin2 t�
��a cos t���a cos t�a2 cos2 t � a2 sin2 t � dt
�C4
F � dr � �C1
F � dr � �C
F � dr � 0.
C4 � C1 � C2 � C � C3
C3C2
C1,C3C2
C10 ≤ t ≤ 2�r � a cos t i � a sin t j,
�C
F � dr � �C
M dx � N dy � �R� ��N
�x�
�M�y � dA � 0.
�N�x
��M�y
�x2 � y2
�x2 � y2�2.
F �y
x2 � y2 i �x
x2 � y2 j.
I � �C
y dx � x dy
x2 � y2
37.
�a2
2 �2�
0 �1 � 2 cos �
12
�cos 2
2 � d �a2
2 �3
2� 2 sin �
14
sin 2�2�
0�
a2
2�3�� �
3�a2
2
A �12
�2�
0 a2�1 � cos �2 d
Section 14.4 Green’s Theorem 197
43. Pentagon:
A �12 ��0 � 0� � �4 � 0� � �12 � 2� � �1 � 4� � �0 � 0�� �
192
��1, 1��1, 4�,�3, 2�,�2, 0�,�0, 0�,
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Section 14.5 Parametric Surfaces
45.
For the line integral, use the two paths
,
, to
(a) For both integrals give 0.
(b) For n even, you obtain
(c) If n is odd and then the integral equals 0.0 < a < 1,
n � 8 : �256315 a9n � 6 : �32
35 a7n � 4 : �1615 a5n � 2 : �4
3 a3
n � 1, 3, 5, 7,
�R���N
�x�
�M�y � dA � �a
�a��a2�x2
0 �nxn�1 � nyn�1� dy dx
�C2
yn dx � xn dy � ��a
a
�a2 � x2n�2 � xn �x�a2 � x2� dx
�C1
yn dx � xn dy � 0
x � �ax � aC2: r2x � x i � �a2 � x2 j
�a ≤ x ≤ aC1: r1x � x i
x−a a
2a
y a x= 2 2−
C1
C2
y�C
yn dx � xn dy � �R� ��N
�x�
�M�y � dA
47.
� �R� f �2g � �f � �g dA � �
R�g�2f � �g � �f dA � �
R� f �2g � g�2f dA
�C
f DNg � gDN f ds � �C
fDNg ds � �C
gDN f ds
49.
�C
F � dr � �C
M dx � N dy � �R� ��N
�x�
�M�y � dA � �
R� 0 dA � 0
�N�x
��M�y
� 0 ⇒ �N�x
��M�y
� 0.
F � M i � N j
1.
Matches c.
z � xy
ru, v � u i � vj � uvk 3.
Matches b.
x2 � y2 � z2 � 4
� 2 sin vkru, v � 2 cos v cos ui � 2 cos v sin uj
5.
Plane
x
y43 55
−4
32
z
y � 2z � 0
ru, v � ui � vj �v2
k 7.
Cylinder
x
y5
5
−3
3
z
x2 � z2 � 4
ru, v � 2 cos ui � vj � 2 sin uk
198 Chapter 14 Vector Analysis
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For Exercises 9 and 11,
Eliminating the parameter yields
0 ≤ z ≤ 4.z � x2 � y2,
0 ≤ v ≤ 2�.0 ≤ u ≤ 2,r u, v� � u cos vi � u sin vj � u2k,
x
y2
5
2
z
9.
The paraboloid is reflected (inverted) through the xy-plane.
z � �x2 � y20 ≤ v ≤ 2�0 ≤ u ≤ 2,su, v � u cos v i � u sin v j � u2k,
11.
The height of the paraboloid is increased from 4 to 9.
0 ≤ v ≤ 2�0 ≤ u ≤ 3,su, v � u cos v i � u sin vj � u2k,
15.
z2
1�
x2
4�
y2
1� 1
x
y3 6 9
9
69
6
z0 ≤ v ≤ 2�0 ≤ u ≤ 2,
ru, v � 2 sinh u cos vi � sinh u sin vj � cosh uk,
17.
x
y3 3
5
221
−2
−2
−3
−3
−1
4
3
z0 ≤ v ≤ 2�0 ≤ u ≤ �,
ru, v � u � sin u cos v i � 1 � cos u sin v j � uk,
19.
ru, v � ui � vj � vk
z � y 21.
ru, v � 4 cos ui � 4 sin uj � vk
x2 � y2 � 16
23.
ru, v � ui � vj � u2k
z � x2 25. inside
0 ≤ v ≤ 3ru, v � v cos u i � v sin u j � 4k,
x2 � y2 � 9.z � 4
27. Function:
Axis of revolution: x-axis
0 ≤ v ≤ 2�0 ≤ u ≤ 6,
z �u2
sin vy �u2
cos v,x � u,
0 ≤ x ≤ 6y �x2
,
Section 14.5 Parametric Surfaces 199
13.
z �x2 � y22
16
0 ≤ v ≤ 2�0 ≤ u ≤ 1,
ru, v � 2u cos v i � 2u sin v j � u4k,
yx2
2
3
2
1
z
29. Function:
Axis of revolution: z-axis
0 ≤ v ≤ 2�0 ≤ u ≤ �,
z � uy � sin u sin v,x � sin u cos v,
0 ≤ z ≤ �x � sin z,
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31.
At and
Tangent plane:
(The original plane!)
x � y � 2z � 0
x � 1 � y � 1 � 2z � 1 � 0
N � ru0, 1 rv0, 1 � � i11
j1
�1
k01� � i � j � 2k
rv0, 1 � i � j � kru0, 1 � i � j,
v � 1.u � 01, �1, 1,
rvu, v � i � j � kruu, v � i � j,
1, �1, 1ru, v � u � vi � u � vj � vk, 33.
At and
Direction numbers:
Tangent plane:
4y � 3z � 12
4y � 6 � 3z � 4 � 0
0, 4, �3
� � i0
�4
j30
k40� � �16j � 12k
N � ru�2, �
2� rv�2, �
2�
rv�2, �
2� � �4iru�2, �
2� � 3j � 4k,
v � ��2.u � 20, 6, 4,
rvu, v � �2u sin v i � 3u cos v j
ruu, v � 2 cos v i � 3 sin v j � 2uk
0, 6, 4ru, v � 2u cos v i � 3u sin v j � u2 k,
35.
A � �1
0�2
0 �2 du dv � 2�2
�ru rv� � �2
ru rv � � i2
0
j0
�12
k012� � �j � k
rvu, v � �12
j �12
kruu, v � 2i,
0 ≤ v ≤ 10 ≤ u ≤ 2,ru, v � 2ui �v2
j �v2
k,
37.
A � �b
0�2�
0 a du dv � 2�ab
�ru rv� � a
ru rv � � i�a sin u
0
ja cos u
0
k01 � � a cos ui � a sin uj
rvu, v � k
ruu, v � �a sin ui � a cos uj
0 ≤ v ≤ b0 ≤ u ≤ 2�,ru, v � a cos ui � a sin uj � vk,
39.
A � �2�
0�b
0 a�1 � a2 u du dv � �ab2�1 � a2
�ru rv� � au�1 � a2
ru rv � � ia cos v
�au sin v
ja sin v
au cos v
k10 � � �au cos v i � au sin v j � a2uk
rvu, v � �au sin v i � au cos v j
ruu, v � a cos vi � a sin vj � k
0 ≤ v ≤ 2�0 ≤ u ≤ b,ru, v � au cos v i � au sin v j � uk,
200 Chapter 14 Vector Analysis
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41.
A � �2�
0�4
0 �u �
14
du dv ��
617�17 � 1 � 36.177
�ru rv� ��u �14
� ��u cos vi � �u sin vj �12
kru rv � � icos v
2�u
��u sin v
jsin v
2�u
�u cos v
k
1
0 �rvu, v � ��u sin v i � �u cos v j
ruu, v �cos v
2�ui �
sin v
2�uj � k
0 ≤ v ≤ 2�0 ≤ u ≤ 4,ru, v � �u cos v i � �u sin v j � uk,
43. See the definition, page 1051.
The radius of the generating circle that is revolved about the z-axis is b, and its center is a units from the axis of revolution.
47. (a)
x y
66
4−6
−4
−6
z
0 ≤ v ≤ 2�0 ≤ u ≤ 2�,
4 � cos v sin u j � sin vk,
ru, v � 4 � cos v cos ui � (b)
x y66
4
z
0 ≤ v ≤ 2�0 ≤ u ≤ 2�,
4 � 2 cos v sin uj � 2 sin vk,
ru, v � 4 � 2 cos v cos ui �
(c)
yx
3
9
−9
3
z
0 ≤ v ≤ 2� 0 ≤ u ≤ 2�,
8 � cos v sin uj � sin vk,
ru, v � 8 � cos v cos u i � (d)
yx 1212
12
−12
z
0 ≤ v ≤ 2�0 ≤ u ≤ 2�,
8 � 3 cos v sin uj � 3 sin vk,
ru, v � 8 � 3 cos v cos ui �
Section 14.5 Parametric Surfaces 201
45. (a) From
(b) From
(c) From
(d) From 10, 0, 0
0, 10, 0
10, 10, 10
�10, 10, 0
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Section 14.6 Surface Integrals
49.
� �2�
0 200 dv � 400� m2
S � �S� dS � �2�
0���3
0 400 sin u du dv � �2�
0 ��400 cos u�
��3
0 dv
� 400�sin2 u � 400 sin u
� 400�sin4 u � cos2 u sin2 u
�ru � rv� � 400�sin4 u cos2 v � sin4 u sin2 v � cos2 u sin2 u
� 400� sin2 u cos v i � sin2 u sin vj � cos u sin uk
400cos u sin u cos2 v � cos u sin u sin2 v�k � 400 sin2 u cos v i � 400 sin2 u sin vj �
ru � rv � � i20 cos u cos v
�20 sin u sin v
j20 cos u sin v20 sin u cos v
k�20 sin u
0 �rv � �20 sin u sin v i � 20 sin u cos v j
ru � 20 cos u cos v i � 20 cos u sin v j � 20 sin uk
0 ≤ v ≤ 2�0 ≤ u ≤ ��3,ru, v� � 20 sin u cos v i � 20 sin u sin v j � 20 cos uk
51.
x
y42
4
−4 −2
π2
π4
z
A � �2�
0�3
0 �4 � u2 du dv � ��3�13 � 4 ln 3 � �13
2 ��
�ru � rv� � �4 � u2
ru � rv � � icos v
�u sin v
jsin v
u cos v
k02 � � 2 sin vi � 2 cos vj � uk
rvu, v� � �u sin v i � u cos vj � 2k
ruu, v� � cos v i � sin vj
0 ≤ v ≤ 2�0 ≤ u ≤ 3,ru, v� � u cos v i � u sin vj � 2vk,
53. Essay
1.
� �2�4
0�4
04 � 2y� dy dx � 0
�S�x � 2y � z� dS � �4
0�4
0x � 2y � 4 � x��1 � �1�2 � 0�2 dy dx
�z�y
� 0�z�x
� �1,0 ≤ y ≤ 4,0 ≤ x ≤ 4,S: z � 4 � x,
202 Chapter 14 Vector Analysis
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3.
� �13
sin � �23
cos � � 5��2�
0� 10�
� �2�
0 1
3 cos � �
23
sin � � 5� d�
� �2�
0�1
0r cos � � 2r sin � � 10�r dr d�
�S�x � 2y � z� dS � �1
�1��1�x2
��1�x2
x � 2y � 10��1 � 0�2 � 0�2 dy dx
�z�x
��z�y
� 0x 2 � y 2 ≤ 1,S: z � 10,
5. (first octant)
��62 �9x 2
2� x3 �
x4
16�6
0�
27�62
��62 �6
0x 9 � 3x �
14
x2� dx
� �6�6
0�xy 2
2 �3�x�2�
0 dx
�S�xy dS � �6
0�3�x�2�
0xy�1 � �1�2 � �2�2 dy dx
x
−1
1
2
2
3 4
4
5
5
6
3
1
12
y x= 3 −
y�z�y
� �2�z�x
� �1,S: z � 6 � x � 2y,
7.
�391�17 � 1
240�S�xy dS � �2
0�2
y
xy�1 � 4x 2 dx dy
�z�y
� 0 �z�x
� �2x,
0 ≤ y ≤ x,0 ≤ x ≤ 2,S: z � 9 � x2,
9. 0 ≤ x ≤ 2, 0 ≤ y ≤ 2
�S�x 2 � 2xy� dS � �2
0�2
0x 2 � 2xy��1 � 4x 2 � 4y 2 dy dx � �11.47
S: z � 10 � x 2 � y 2,
11. (first octant)
�76�
43
x3 �16
x4 �18 4 �
23
x�4
�6
0�
3643
�76�
6
0�x 2 4 �
23
x� �13 4 �
23
x�3
� dx
�76�
6
0�4�2x�3�
0x 2 � y 2� dy dx
m � �R�x 2 � y 2��1 � �
13�
2
� �12�
2
dA
x, y, z� � x 2 � y 2
x
−1
1
2
2
3 4
4
5
5
6
3
1
23
y x= 4 −
R
y⇒ z � 2 �13
x �12
yS: 2x � 3y � 6z � 12
Section 14.6 Surface Integrals 203
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13.
�S� y � 5� dS � �2
0�1
0v � 5�
�52
du dv � 6�5
�ru � rv� � � �12
j � k � ��52
0 ≤ v ≤ 20 ≤ u ≤ 1,S: ru, v� � ui � vj �v2
k,
15.
�S�xy dS � �2
0���2
08 cos u sin u du dv � 8
�ru � rv� � �2 cos ui � 2 sin uj� � 2
0 ≤ v ≤ 20 ≤ u ≤ �
2,S: ru, v� � 2 cos ui � 2 sin uj � vk,
17.
� �2�94
� �18 � �
12
sin 2�� �43
sin ��2�
0� �2�18�
4�
�
4� �19�2�
4
� �2�2�
0 �9
4� 1
4�1 � cos 2�
2�
43
cos �� d�
� �2�2�
0 �r 4
4�
r 4
4 cos2 � �
4r 3
3 cos � � 2r2�
1
0 d�
� �2�2�
0�1
0 �r2 � r2 cos2 � � 4r cos � � 4r dr d�
� �2�2�
0�1
0 �r2 � r cos � � 2�2r dr d�
�S�f x, y, z� dS � �1
�1��1�x2
��1�x2 �x 2 � y 2 � x � 2�2�1 � 1�2 � 0�2 dy dx
x 2 � y 2 ≤ 1S: z � x � 2,
f x, y, z� � x 2 � y 2 � z2
19.
� 2�2�
0�2
0 r2 dr d� � 2�2�
0�r3
3 �2
0 d� � �16
3��
2�
0�
32�
3
� 2�2
�2��4�x2
��4�x2
�x 2 � y 2 dy dx
� �2�2
�2��4�x2
��4�x2
�x 2 � y 2�x 2 � y 2 � x 2 � y 2
x 2 � y 2 dy dx
�S�f x, y, z� dS � �2
�2��4�x2
��4�x2
�x 2 � y 2 � �x 2 � y 2 �2�1 � x
�x 2 � y 2�2
� y
�x 2 � y 2�2
dy dx
x 2 � y 2 ≤ 4S: z � �x 2 � y 2,
f x, y, z� � �x 2 � y 2 � z2
204 Chapter 14 Vector Analysis
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21.
0 ≤ x ≤ 3, 0 ≤ y ≤ 3, 0 ≤ z ≤ 9
Project the solid onto the yz-plane; 0 ≤ y ≤ 3, 0 ≤ z ≤ 9.
� 324�3
0
3
�9 � y 2 dy � �972 arcsin y
3��3
0� 972 �
2� 0� � 486�
� �3
0� 3
�9 � y 2 9z �z3
3 ��9
0 dy � �3
0�9
09 � z2� 3
�9 � y 2 dz dy
�S�f x, y, z� dS � �3
0�9
0 �9 � y2� � y 2 � z2�1 � y
�9 � y 2�2
� 0�2 dz dy
x � �9 � y 2,
S: x 2 � y 2 � 9,
f x, y, z� � x 2 � y 2 � z2
23.
(first octant)
� ��1
02 � 2x2� dx � �
43
� ��1
0�1 � x� � 3x1 � x� � 1 � x�2 dx
� �1
0��y � 3xy � y 2�
1�x
0 dx
� �1
0�1�x
0�1 � 3x � 2y� dy dx
� �1
0�1�x
0�31 � x � y� � 4 � y dy dx
�S�F N dS � �
R�F �G dA � �1
0�1�x
03z � 4 � y� dy dx
�Gx, y, z� � i � j � k
Gx, y, z� � x � y � z � 1
S: x � y � z � 1
x
1
1
y x= + 1−
R
yFx, y, z� � 3z i � 4 j � yk
25.
0 ≤ z
� �2�
0�r4
4�
9r2
2 �3
0 d� �
243�
2
� �2�
0�3
0 r2 � 9�r dr d�
� �R�x 2 � y 2 � 9� dA
� �R��2x 2 � 2y 2 � 9 � x 2 � y 2� dA
�S�F N dS � �
R�F �G dA � �
R�2x 2 � 2y 2 � z� dA
�Gx, y, z� � 2x i � 2y j � k
Gx, y, z� � x 2 � y 2 � z � 9
S: z � 9 � x 2 � y 2,
x2
2
−2
−2
−4
−4 4
4
R
x y2 2+ 9≤
yFx, y, z� � x i � yj � zk
Section 14.6 Surface Integrals 205
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27.
� ��643
sin � � 16 cos � � 10��2�
0� 20�
� �2�
0 ��
643
cos � � 16 sin � � 10� d�
� �2�
0��
83
r3 cos � � 2r3 sin � �52
r2�2
0 d�
� �2�
0�2
0��8r cos � � 6r sin � � 5r dr d�
�S�F N dS � �
R�F �G dA � �
R��8x � 6y � 5� dA
�Gx, y, z� � �2x i � 2y j � k
Gx, y, z� � �x 2 � y2 � z
x 2 � y 2 ≤ 4S: z � x 2 � y 2,1
−1
−1 1x
x y2 2+ 4≤
R
yFx, y, z� � 4i � 3j � 5k
29.
S: unit cube bounded by
The top of the cube
The bottom of the cube
The back of the cube
The left side of the cube
Therefore,
�S�F N dS �
12
� 0 � 2 � 0 �13
�13
�52
.
�S6
�F N dS � �1
0�1
0 �z2 dz dx � �
13
y � 0N � �j,
S6:
�S4
�F N dS � �1
0�1
0 �40�y dy dx � 0
x � 0N � �i,
S4:
�S2
�F N dS � �1
0�1
0 �y0� dy dx � 0
z � 0N � �k,
S2:
�S1
�F N dS � �1
0�1
0 y1� dy dx �
12
z � 1N � k,
S1:
z � 1z � 0,y � 1,y � 0,x � 1,x � 0,
y
1
11
x
zFx, y, z� � 4xyi � z2j � yzk
The front of the cube
The right side of the cube
�S5
�F N dS � �1
0�1
0 z2 dz dx �
13
y � 1N � j,
S5:
�S3
�F N dS � �1
0�1
0 41�y dy dz � 2
x � 1N � i ,
S3:
31. The surface integral of f over a surface S, where S is given by is defined as
(page 1061)
See Theorem 14.10, page 1061.
�S� f x, y, z� dS � lim
���→0 �
n
i�1 f xi, yi, zi ��Si.
z � gx, y�,
33. See the definition, page 1067.
See Theorem 14.11, page 1067.
206 Chapter 14 Vector Analysis
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37.
��2k�a4
2�
a2
2�2k�a2� �
a2m2
� �2k�2�
0�a
0 r 3 dr d� �
�2ka4
42��
Iz � �S�kx 2 � y 2� dS � �
R�kx 2 � y 2��2 dA
m � �S�k dS � k�
R��1 � x
�x 2 � y 2�2
� y
�x 2 � y 2�2
dA � k�R��2 dA � �2 k�a2
0 ≤ z ≤ az � �x 2 � y 2,
39. 0 ≤ z ≤ h
Project the solid onto the xz-plane.
� 4a3 �
2�h� � 2�a3h � 4a3�h
0�arcsin
xa�
a
0 dz
� 4a3�h
0�a
0
1
�a2 � x 2 dx dz
� 4�h
0�a
0�x 2 � a2 � x 2��1 � �x
�a2 � x 2�2
� 0�2 dx dz
Iz � 4�S�x 2 � y 2�1� dS
y � ±�a2 � x 2
x, y, z� � 1
x
ya a
h
zx 2 � y 2 � a2,
41. z ≥ 0
� 0.5�2�
064 d� � 64� � 0.5�2�
0�4
016 � r2�r dr d�
� �R�0.516 � x 2 � y 2� dA � �
R�0.5z dA
� �R�0.5zk 2x i � 2y j � k� dA�
S�F N dS ��
R�F �gxx, y�i � gyx, y�j � k� dA
Fx, y, z� � 0.5zk
S: z � 16 � x 2 � y 2,
Section 14.6 Surface Integrals 207
35. (a)
xy6
4
6
−6
−4
−6
z (b) If a normal vector at a point P onthe surface is moved around theMöbius strip once, it will point inthe opposite direction.
(c)
This is a circle.
yx
4
−4
2
2
−2
z
ru, 0� � 4 cos2u�i � 4 sin2u�j
(d) (construction) (e) You obtain a strip with a double twist and twice as long asthe original Möbius strip.
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Section 14.7 Divergence Theorem
1. Surface Integral: There are six surfaces to the cube, each with
Therefore,
Divergence Theorem: Since div the Divergence Theorem yields
��Q
�div F dV � �a
0�a
0�a
02z dz dy dx � �a
0�a
0a2 dy dx � a4.
F � 2z,
�s�F � N dS � a4 � 2a3 � 2a3 � a4.
y � a, N � j, F � N � �2y, �S6
��2a dA � �a
0�a
0�2a dz dx � �2a3
y � 0, N � �j, F � N � 2y, �S5
�0 dA � 0
x � a, N � i, F � N � 2x, �S4
�2a dy dz � �a
0�a
0 2a dy dz � 2a3
x � 0, N � �i, F � N � �2x, �S3
�0 dA � 0
z � a, N � k, F � N � z2, �S2
�a2 dA � �a
0�a
0a2 dx dy � a4
z � 0, N � �k, F � N � �z2, �S1
�0 dA � 0
dS � �1 dA.
3. Surface Integral: There are four surfaces to this solid.
Therefore,
Divergence Theorem: Since div we have
��Q
� dV � �Volume of solid� �13
�Area of base� � �Height� �13
�9��6� � 18.
F � 1,
�s�F � N dS � 0 � 36 � 9 � 45 � 18.
�S4
��2x � 5y � 3z� dz dy � �3
0�6�2y
0�18 � x � 11y� dx dy � �3
0�90 � 90y � 20y 2� dy � 45
dS � �6 dAF � N �2x � 5y � 3z
�6,N �
i � 2j � k�6
,x � 2y � z � 6,
�S3
�y dS � �3
0�6�2y
0y dz dy � �3
0�6y � 2y 2� dy � 9
dS � dA � dz dyF � N � y � 2x,N � �i,x � 0,
�S2
��z dS � �6
0�6�z
0�z dx dz � �6
0�z2 � 6z� dz � �36
dS � dA � dx dzF � N � 2y � z,N � �j,y � 0,
�S1
�0 dS � 0
F � N � �zN � �k,z � 0,
x
y
6
36
z
208 Chapter 14 Vector Analysis
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5. Since div we have
� �a
0�a
0�2ax � 2ay � a2� dy dx � �a
0�2a2x � 2a3� dx � �a2x 2 � 2a3x�
a
0� 3a4.
��Q
� div F dV � �a
0�a
0�a
0�2x � 2y � 2z� dz dy dx
F � 2x � 2y � 2z,
7. Since div
� �a
0�2�
0 12
�5 sin cos d d� � �a
0���5
2 sin2 2 �
2�
0 d� � 0.
� �a
0�2�
0��2
02�5�sin cos ��sin3 cos � d d d�
��Q
� div F dV � ��Q
�2xyz dV � �a
0�2�
0��2
02�� sin cos ��� sin sin ��� cos ��2 sin d d d�
F � 2x � 2x � 2xyz � 2xyz
9. Since div we have
� 3�43
��2�3� � 32�.��Q
�3 dV � 3�Volume of sphere�
F � 3,
11. Since div we have
��Q
�2y dV � �4
0�3
�3��9�y2
��9�y2
2y dx dy dz � �4
0�3
�34y�9 � y 2 dy dz � �4
0��
43
�9 � y 2�32�3
�3 dz � 0.
F � 1 � 2y � 1 � 2y,
13. Since div we have
��Q
�4x2 dV � �6
0�4
0�4�y
04x 2 dz dy dx � �6
0�4
04x2�4 � y� dy dx � �6
032x 2 dx � 2304.
F � 3x 2 � x 2 � 0 � 4x2,
15.
� �3
016��2 d� � �16��3
3 �3
0� 144�.
� �3
0��8��2 cos �
�
0 d�
� �3
0��
08��2 sin d d�
� �3
0��
0���3 sin2 cos � �3 sin2 sin � 4�2 sin � �
2�
0 d d�
� �3
0��
0�2�
0��3 sin2 sin � �3 sin2 cos � 4�2 sin � d d d�
� �3
0��
0�2�
0�� sin sin � � sin cos � 4��2 sin d d d�
�S�F � N dS � ��
Q
� div F dV � ��Q
��y � x � 4� dV
div F � y � 4 � x
F�x, y, z� � xyi � 4yj � xzk
Section 14.7 Divergence Theorem 209
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17. Using the Divergence Theorem, we have
Therefore, ��Q
�div �curl F� dV � 0.
div �curl F� � 0.
� �6y i � �2z � 2z�j � �4x � 4x�k � �6yi curl F�x, y, z� � i � �x
4xy � z2
j � �y
2x 2 � 6yz
k � �z
2xz �
S�curl F � N dS � ��
Q
�div �curlF� dV
19. See Theorem 14.12, page 1073.
21. Using the triple integral to find volume, we need F so that
Hence, we could have or
For dA dy dz consider then and
For dA dz dx consider then and
For dA dx dy consider then and
Correspondingly, we then have V � �S�F � N dS � �
S�x dy dz � �
S�y dz dx � �
S�z dx dy.
dS � �1 � fx2 � fy
2 dx dy.N �fx i � fy j � k
�1 � fx2 � fy
2z � f �x, y�,F � zk,�
dS � �1 � fx2 � fz
2 dz dx.N �fx i � j � fzk
�1 � fx2 � fz
2y � f �x, z�,F � yj,�
dS � �1 � fy2 � fz
2 dy dz.N �i � fy j � fzk
�1 � fy2 � fz
2x � f �y, z�,F � xi,�
F � zk.F � yj,F � x i,
div F ��M�x
��N�y
��P�z
� 1.
23. Using the Divergence Theorem, we have Let
Therefore, �S�curl F � N dS � ��
Q
�0 dV � 0.
div �curl F� ��2P�x�y
��2N�x�z
��2P�y�x
��2M�y�z
��2N�z�x
��2M�z�y
� 0.
curl F � ��P�y
��N�z i � ��P
�x�
�M�z j � ��N
�x�
�M�y k
F�x, y, z� � M i � Nj � Pk
�S�curl F � N dS � ��
Q
�div �curl F� dV.
25. If then div
�S�F � N dS � ��
Q
�div F dV � ��Q
�3 dV � 3V.
F � 3.F�x, y, z� � x i � yj � zk,
27.
� ��Q
�� f �2g � �f � �g� dV � ��Q
�div � f �g� dV � ��Q
�� f div �g � � f � �g� dV
�S�f DNg dS � �
S�f �g � N dS
210 Chapter 14 Vector Analysis
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Section 14.8 Stokes’s Theorem
Section 14.8 Stokes’s Theorem 211
1.
� �xyi � j � �yz � 2�kcurl F � � i��x
2y � z
j��y
xyz
k��z
ez �F�x, y, z� � �2y � z�i � xyz j � ezk
3.
� �2 �1
1 � x2�j � 8xkcurl F � � i��x
2z
j��y
�4x2
k��z
arctan x �F�x, y, z� � 2z i � 4x2 j � arctan xk
5.
� z�x � 2ey2�z2�i � yz j � 2yex2�y2 k
� �xz � 2zey2�z2�i � yz j � 2yex2�y2 k
curl F � � i��x
ex2�y2
j � �y
ey2�z2
k��z
xyz �F�x, y, z� � ex2�y2 i � ey2�z2j � xyzk
7. In this case, and C is the circle
Letting we have and
Double Integral: Consider
Then
N ��F
��F ��
2x i � 2yj � 2zk2�x2 � y2 � z2
� x i � yj � zk.
F�x, y, z� � x2 � y2 � z2 � 1.
C
�y dx � x dy � 2�
0 �sin2 t � cos2 t�dt � 2�.
dy � cos t dtdx � �sin t dt,y � sin t,x � cos t,
Line Integral: C
F � dr � C
�y dx � x dy
dz � 0.z � 0,x2 � y2 � 1,P � x � yN � x � z,M � �y � z,
Since
and
Now, since we have
S
�curl F� � N dS � R
2z�1z� dA �
R
2 dA � 2�Area of circle of radius 1� � 2�.
curl F � 2k,
dS ��1 �x2
z2 �y2
z2 dA �1z
dA.zy ��yz
,zx ��2x2z
��xz
,z2 � 1 � x2 � y2,
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212 Chapter 14 Vector Analysis
Double Integral:
Considering then
and
Thus,
� 4
0 0 dx � 0.
� 4
0 (12�3x)4
0�8xy � 3x2 � 12x� dy dx
� 4
0 (�3x�12)4
0 �4xy � 2x�6 � 2y �
32
x�� dy dx
S
�curl F� � N dS � R
�4xy � 2xz� dy dx
dS � �29 dA.N ��F
��F ��
3i � 4j � 2k�29
F�x, y, z� � 3x � 4y � 2z � 12,
curl F � xyj � xzk
11. Let and Then and Thus,
Surface S has direction numbers 0, 1, with equation and Since we have
� R
dA � �Area of triangle with a � 1, b � 2� � 1. S
�curl F� � N dS � R
1�2
��2� dA
curl F � �3i � j � 2k,dS � �2 dA.z � x � 0�1,
N �U V
�U V��
�2i � 2k2�2
��i � k�2
.
V � AC\
� 2j.U � AB\
� i � j � kC � �0, 2, 0�.B � �1, 1, 1�A � �0, 0, 0�,
13.
� 2
�2 4x�4 � x2 dx � 0
� 2
�2 �4�x2
��4�x2
4xy � 16y � 4x2y � 4y3 � 2x� dy dx
2
�2 �4�x2
��4�x2
4xy � 4y�4 � x2 � y2� � 2x� dy dxS
�curl F� � N dS � R
�4xy � 4yz � 2x� dA �
�G�x, y, z� � 2x i � 2yj � k
G�x, y, z� � x2 � y2 � z � 4
curl F � � i��x
z2
j��y
x2
k��z
y2 � � 2yi � 2z j � 2xk
0 ≤ zS: z � 4 � x2 � y2,F�x, y, z� � z2 i � x2 j � y2k,
9. Line Integral: From the accompanying figure we see that for
Hence,
� 3
0 y dy � 0
3 y dy � 6
0 z dz � 0
6 z dz � 0.
� C1
y dy � C2
y dy � z dz � C3
z dz
C
F � dr � C
xyz dx � y dy � z dz
dy � 0.C3: y � 0,
dx � 0C2: x � 0,
dz � 0C1: z � 0,
x
yC1
C2C3
(0, 0, 6)
(4, 0, 0)
(0, 3, 0)
2
6
2
4
4
z
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Section 14.8 Stokes’s Theorem 213
17.
over one petal of in the first octant.
� �2
0 8 sin2 cos d � �8 sin3
3 ��2
0�
83
� �2
0 4 sin cos
0 2 sin dr d
� �2
0 2 sin 2
0 2r sin
r2 r dr d
S
�curl F� � N dS � R
2y
x2 � y2 dA
�G�x, y, z� � 2i � 3j � k
G�x, y, z� � 2x � 3y � z � 9
r � 2 sin 2S: z � 9 � 2x � 3y
� � �1y�1 � �x2y2� �
yx2 � y2�k � � 2y
x2 � y2�kcurl F � � i��x
�12 ln�x2 � y2�
j��y
arctan xy
k��z
1 �F�x, y, z� � �ln�x2 � y2 i � arctan
xy
j � k
19. From Exercise 10, we have and Since we have
S
�curl F� � N dS � R
xz dA � a
0 a
0 x3 dy dx � a
0 ax3 dx � �ax4
4 �a
0�
a5
4.
curl F � xyj � xzk,dS � �1 � 4x2 dA.N �2x i � k�1 � 4x2
21.
Letting we have S
�curl F� � N dS � 0.N � k,
curl F � � i��x
1
j��y
1
k��z
�2 � � 0
F�x, y, z� � i � j � 2k 23. See Theorem 14.13, page 1081.
15.
� 2
�2 �4�x2
��4�x2
y dy dx � 0S
�curl F� � F dS � R
yz
�4 � x2 � y2 dA �
R
y�4 � x2 � y2
�4 � x2 � y2 dA
�G�x, y, z� �x
�4 � x2 � y2i �
y�4 � x2 � y2
j � k
G�x, y, z� � z � �4 � x2 � y2
curl F � � i��x
z2
j��y
y
k��z
xz � � z j
S: z � �4 � x2 � y2F�x, y, z� � z2 i � yj � xzk,
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214 Chapter 14 Vector Analysis
27. Let then
since
and
curl�C r� � � i��x
bz � cy
j��y
cx � az
k��z
ay � bx � � 2�ai � b j � ck� � 2C.
C r � � iax jby
kcz � � �bz � cy� i � �az � cx� j � �ay � bx�k
12
C
�C r� � dr �12
S
curl �C r� � N dS �12
S
2C � N dS � S
C � N dS
C � ai � bj � ck,
25. (a) (Stoke’s Theorem)
Therefore,
(b) (using part a.)
(c)
(using part a.)
� S
��f �g� � N dS � S
���f �g� � N dS � 0
� S
��f �g� � N dS � S
��g �f � � N dS
C
� f �g � g�f � � dr � C
� f �g� � dr � C
�g�f � � dr
� 0 since �f �f � 0.
C
� f �f � � dr � S
��f �f � � N dS
C
f �g � dr � S
curl f �g� � N dS � S
�f �g� � N dS.
� �
i
�f�x
�g�x
j
�f�y
�g�y
k
�f�z
�g�z � � �f �g
� ���f�x��
�g�y� � ��f
�y���g�x��k � ���f
�y���g�z� � ��f
�z���g�y��i � ���f
�x���g�z� � ��f
�z���g�x��j
� ��f � �2g�x�y� � ��f
�x���g�y�� � �f � �2g
�y�x� � ��f�y��
�g�x���k
� ��f � �2g�x�z� � ��f
�x���g�z�� � �f � �2g
�z�x� � ��f�z��
�g�x��� j
� ��f � �2g�y�z� � ��f
�y���g�z�� � �f � �2g
�z�y� � ��f�z��
�g�y���i
curl � f �g� � � i��x
f ��g�x�
j��y
f ��g�y�
k��z
f ��g�z� � f �g � f
�g�x
i � f�g�y
j � f�g�z
k
C
f �g � dr � S
curl f �g� � N dS
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Review Exercises for Chapter 14
Review Exercises for Chapter 14 215
1.
xy
3
2
4
4
3
z
F�x, y, z� � x i � j � 2k 3.
F�x, y, z� � �16x � y�i � x j � 2zk
f �x, y, z� � 8x2 � xy � z2
5. Since is not conservative.F�M��y � �1�y 2 � �N��x,
7. Since is conservative. From and partial integration yields and which suggests
and U�x, y� � 3x2y 2 � x3 � y3 � 7y � C.g�x� � �x3,h�y� � y3 � 7y,U � 3x2y 2 � y3 � 7y � g�x�U � 3x2y 2 � x3 � h�y�
N � �U��y � 6x2y � 3y 2 � 7,M � �U��x � 6xy 2 � 3x2F�M��y � 12xy � �N��x,
9. Since
F is not conservative.
�M�z
� 1 ��P�x
.
�M�y
� 4x ��N�x
,
11. Since
F is conservative. From
we obtain
U �xyz
� h�x, y� ⇒ f �x, y, z� �xyz
� KU �xyz
� g�x, z�,U �xyz
� f �y, z�,
P ��U�z
��xyz2N �
�U�y
��xy 2z
,M ��U�x
�1yz
,
�N�z
�x
y 2z2 ��P�y
,�M�z
��1yz2 �
�P�x
,�M�y
��1y 2z
��N�x
,
13. Since
(a)
(b) curl F � ��P�y
��N�z �i � ��P
�x�
�M�z � j � ��N
�x�
�M�y �k � 0i � 0j � 0k � 0
div F � 2x � 2y � 2z
F � x2i � y 2j � z2k:
15. Since
(a)
(b) curl F � xz i � yz j � �cos x � sin y � sin y � cos x�k � xz i � yz j
div F � �y sin x � x cos y � xy
F � �cos y � y cos x�i � �sin x � x sin y�j � xyzk:
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216 Chapter 14 Vector Analysis
17. Since
(a)
(b) curl F � z2 i � y 2 k
div F �1
�1 � x2� 2xy � 2yz
F � arcsin x i � xy 2j � yz2 k: 19. Since
(a)
(b) curl F �2x � 2yx2 � y 2 k
�2x � 2yx2 � y 2 � 1
div F �2x
x2 � y 2 �2y
x2 � y 2 � 1
F � ln�x2 � y 2�i � ln�x2 � y 2�j � zk:
21. (a) Let then
(b) Let then
�C
�x2 � y 2� ds � �2�
0 16�4 dt� � 128�
ds � 4 dt.0 ≤ t ≤ 2�,y � 4 sin t,x � 4 cos t,
�C
�x2 � y 2� ds � �2
�1 2t 2�2 dt � 2�2�t3
3�2
�1� 6�2
ds � �2 dt.�1 ≤ t ≤ 2,y � t,x � t,
23.
� 2� 2�1 � 2� 2�
�C
�x2 � y 2� ds � �2�
0 ��cos t � t sin t�2 � �sin t � t cos t�2��t2 cos2 t � t 2 sin2 t dt � �2�
0�t3 � t� dt
dydt
� t sin tdxdt
� t cos t,0 ≤ t ≤ 2�,y � sin t � t cos t,x � cos t � t sin t,
25. (a) Let
(b)
�C
�2x � y� dx � �x � 3y� dy � �2�
0 �9 � 9 sin t cos t� dt � 18�
0 ≤ t ≤ 2�dy � 3 cos t dt,dx � �3 sin t dt,y � 3 sin t,x � 3 cos t,
�C
�2x � y� dx � �x � 3y� dy � �1
0 �7t�2� � ��7t���3�� dt � �1
0 35t dt �
352
0 ≤ t ≤ 1y � �3t,x � 2t,
27.
�C
�2x � y� ds � ���2
0 �2�a � cos3 t� � a � sin3 t��x�t�2 � y�t�2 dt �
9a2
5
y�t� � 3a � sin2 t cos t
x�t� � �3a � cos2 t sin t
0 ≤ t ≤ �
2r �t� � a cos3 t i � a sin3 t j,�
C
�2x � y� ds,
29.
from to
Lateral surface area:
�C2
f �x, y� ds � �2
0 �5 � sin�t � 3t���10 dt � �10 �2
0 �5 � sin 4t� dt �
�104
�41 � cos 8� 32.528
�r�t�� � �10
r�t� � i � 3j
0 ≤ t ≤ 2r�t� � t i � 3t j,
�2, 6��0, 0�C: y � 3x
f �x, y� � 5 � sin�x � y�
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Review Exercises for Chapter 14 217
31.
�C
F � dr � �1
0 5t6 dt �
57
0 ≤ t ≤ 1F � t5 i � t 4j,
dr � �2t i � 3t2j� dt 33.
�C
F � dr � �2�
0 t dt � 2�2
0 ≤ t ≤ 2�F � �2 cos t�i � �2 sin t�j � tk,
dr � ���2 sin t�i � �2 cos t�j � k� dt
35. Let
�C
F � dr � �2
�2 4t2 dt � 4t3
3 2
�2�
643
F � ��t � 2t2�i � �2t2 � t�j � �2t�k
dr � �i � j � 4tk� dt.�2 ≤ t ≤ 2,z � 2t 2,y � �t,x � t,
37. For
For
�1003
� ��32� �43
�C
xy dx � �x2 � y 2� dy � �C1
xy dx � �x2 � y 2� dy � �C2
xy dx � �x2 � y 2� dy
0 ≤ t ≤ 2r2�t� � �2 � t�i � �4 � 2t�j,y � 2x,
x
2
3
4
1 2 3 4
1
(2, 4)
C1
C2
y x= 2
y x= 2y0 ≤ t ≤ 2r1�t� � t i � t 2j,y � x2,
39. is conservative.
Work � 12
x2 �23
y3�2�4, 8�
�0, 0��
12
�16� � �23�83�2 �
83
�3 � 4�2 �
F � x i � �y j
41. �C
2xyz dx � x 2z dy � x 2y dz � x2yz�1, 4, 3�
�0, 0, 0�� 12
43. (a)
(b)
(c) where
Hence,
�C
F � dr � 4�2�2� 1�1�2
� 15
f �x, y� � xy 2.F�x, y� � y 2i � 2xy j � f
� t 24
1� 15
� �4
1 �t � t� dt
�C
y 2 dx � 2xy dy � �4
1 t�1� � 2�t���t � 1
2�t dt
� 3t3 � 7t 2 � 5t1
0� 15
� �1
0 �9t 2 � 14t � 5� dt
� �1
0 3�t 2 � 2t � 1� � 2�3t 2 � 4t � 1�� dt
�C
y 2 dx � 2xy dy � �1
0 ��1 � t�2�3� � 2�1 � 3t��1 � t�� dt
45. �C
y dx � 2x dy � �2
0�2
0 �2 � 1� dy dx � �2
0 2 dx � 4 47. �
C
xy 2 dx � x2y dy � �R� �2xy � 2xy� dA � 0
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218 Chapter 14 Vector Analysis
49. �C
xy dx � x2 dy � �1
0�x
x2
x dy dx � �1
0 �x2 � x3� dx �
112
51.
0 ≤ v ≤ 2�0 ≤ u ≤ �
3,
22
4
6
4
−4
−2y
x
zr�u, v� � sec u cos v i � �1 � 2 tan u� sin v j � 2uk
53. (a)
x
y4
3−4
−3
−2
4
−4
z (b)
y
x
22 3
3
4
−4
−4 −3
−2
−1
−3
34
z
(c)
yx
24
3
−3
−2
−3
−2−3
−4
−4
3
2
3
4
z (d)
The space curve is a circle:
r�u, �
4� �3�2
2 cos u i �
3�22
sin u j ��22
k
yx
24
3
−3
−2
−3
−2−4
−4
3
1
3
4
z
(e)
Using a Symbolic integration utility,
(f) Similarly,
���4
0���2
0 �ru � rv � dv du 4.27
���2
��4�2�
0 �ru � rv � du dv 14.44
� �9 cos4 v � 81 cos2 v sin2 v
�ru � rv � � �9 cos4 v cos2 u � 9 cos4 v sin2 u � 81 cos2 v sin2 v
� �3 cos2 v cos u�i � �3 cos2 v sin u�j � �9 cos v sin v�k
� �3 cos2 v cos u�i � �3 cos2 v sin u�j � �9 cos v sin v sin2 u � 9 cos v sin v cos2 u�k
ru � rv � � i�3 cos v sin u�3 sin v cos u
j 3 cos v cos u
�3 sin v sin u
k0
cos v �rv � �3 sin v cos u i � 3 sin v sin u j � cos v k
ru � �3 cos v sin u i � 3 cos v cos u j
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Review Exercises for Chapter 14 219
55.
� �2
0�2�
0 �cos v � sin v�u2��2u � 3�2 � 1 dv du � 0
�S� �x � y� dS � �2�
0�2
0 �u cos v � u sin v� u��2u � 3�2 � 1 du dv
�ru � rv � � u��2u � 3�2 � 1
� �2u � 3�u cos v i � �2u � 3�u sin v j � ukru � ru � � icos v
�u sin v
j sin v
u cos v
k3 � 2u
0 �rv �u, v� � �u sin v i � u cos vj
ru �u, v� � cos v i � sin vj � �3 � 2u�k
x
y3
2
3
−3
−3
−2
z0 ≤ v ≤ 2�0 ≤ u ≤ 2,S: r�u, v� � u cos v i � u sin vj � �u � 1��2 � u�k,
57.
Q: solid region bounded by the coordinates planes and the plane
Surface Integral: There are four surfaces for this solid.
�16�
x 4
4�
x3
3� 12x2 � 36x
6
0� 66 �
16�
6
0 ��x3 � x2 � 24x � 36� dx
�14�
6
02x2�12 � 2x
3 � �3x2 �12 � 2x
3 �2
� 12�12 � 2x3 � � 2x�12 � 2x
3 � �32 �
12 � 2x3 �
2
dx
�14�
6
0�4�(2x�3)
0 �2x2 � 3xy � 12 � 2x � 3y� dy dx
�S4
� F � N dS �14�R
� �2x2 � 3xy � 4z� dA
dS ��1 � �14� � � 9
16�dA ��29
4 dAN �
2i � 3j � 4k�29
,2x � 3y � 4z � 12,
�S3
� 0 dS � 0F � N � �x2,N � �i,x � 0,
�S2
� 0 dS � 0F � N � �xy,N � �j,y � 0,
�S1
� 0 dS � 0F � N � �z,N � �k,z � 0
2x � 3y � 4z � 12
F�x, y, z� � x2 i � xyj � zk
Divergence Theorem: Since Divergence Theorem yields
�16
3x 4
4�
35x3
3� 48x2 � 36x
6
0� 66. �
14�
6
0 23
�3x3 � 35x2 � 96x � 36� dx
�14�
6
0 �3x � 1�4�12 � 2x� � 2x�12 � 2x
3 � �32�
12 � 2x3 �
2
dx
�14�
6
0 �3x � 1�12y � 2xy �
32
y 2(12�2x��3
0 dx
� �6
0�(12�2x)�3
0 �3x � 1��12 � 2x � 3y
4 � dy dx
��Q
� div F dV � �6
0�(12�2x)�3
0�(12�2x�3y)�4
0 �3x � 1� dz dy dx
div F � 2x � x � 1 � 3x � 1,
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220 Chapter 14 Vector Analysis
59.
S: portion of over the square in the xy-plane with vertices
Line Integral: Using the line integral we have:
Double Integral: Considering we have:
and
Hence,
�S� �curl F� � N dS � �a
0�a
0 2y 2z dy dx � �a
0�a
0 2y 4 dy dx � �a
0 2a5
5 dx �
2a6
5.
curl F � xz i � yz j.N ��f
��f ��
�2yj � k�1 � 4y 2
, dS � �1 � 4y 2 dA,
f �x, y, z� � z � y 2,
� a � a cos a � a sin a � a sin a � a cos a � a �2a6
5�
2a6
5
� a � �x cos a � a sin x�0
a� �y sin a � a cos y�
a
0� �2a
y5
5 �a
0
� �a
0 dx � �0
a
�cos a � a cos x� dx � �a
0 �sin a � a sin y� dy � �a
0 2ay 4 dy
� �C1
dx � �C2
0 � �C3
�cos a � a cos x� dx � �C4
�sin a � a sin y� dy � ay3�2y dy�
�C
F � dr � �C
�cos y � y cos x� dx � �sin x � x sin y� dy � xyz dz
dz � 2y dyz � y 2,dx � 0,C4: x � a,
dz � 0z � a2,dy � 0,C3: y � a,
dz � 2y dyz � y 2,dx � 0,C2: x � 0,
dy � 0C1: y � 0,
yx
1
aa
C1C2
C3
C4
z
�0, 0�, �a, 0�, �a, a�, �0, a�z � y 2
F�x, y, z� � �cos y � y cos x�i � �sin x � x sin y�j � xyzk
1. (a)
—CONTINUED—
� 25k�22 �
3 � 25k�2�
6
� 25k�1
0
1�1 � y2�3�2 dy �1�2
�1�2
1�1 � x2�1�2 dx
� 25k�1�2
�1�2�1
0
1�1 � y2�3�2�1 � x2�1�2 dy dx
� 25k�1�2
�1�2�1
0� x2
�x2 � y2 � z2�3�2�1 � x2�1�2 �1 � x2
�x2 � y2 � z2�3�2�1 � x2�1�2� dy dx
� 25k�R�� x2
�x2 � y2 � z2�3�2�1 � x2�1�2 �z
�x2 � y2 � z2�3�2� dA
Flux � �S��k�T � N dS
dS �1
�1 � x2 dy dx
N � xi � �1 � x2 k
�T ��25
�x2 � y2 � z2�3�2�xi � yi � zk
Problem Solving for Chapter 14
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Problem Solving for Chapter 14 221
1. —CONTINUED—
(b)
Flux � �1
0�2��3
��3
25k�v2 � 1�3�2 du dv � 25k
�2�
6
�T � �ru � rv� ��25
�v2 � 1�3�2 ��cos2 u � sin2 u� �25
�v2 � 1�3�2
��25
�v2 � 1�3�2 �cos ui � vj � sin uk
�T ��25
�x2 � y2 � z2�3�2 �xi � yj � zk
ru � rv � ��cos u, 0, �sin u�
ru � ��sin u, 0, cos u�, rv � �0, 1, 0�
r�u, v� � �cos u, v, sin u�
3.
Iz � �C
�x2 � y2� ds � �2�
0�9 cos2 t � 9 sin2 t��13 dt � 18��13
Iy � �C
�x2 � z2� ds � �2�
0�9 cos2 t � 4t2��13 dt �
13�13� �32�2 � 27�
Ix � �C
�y2 � z2� ds � �2�
0�9 sin2 t � 4t2��13 dt �
13�13� �32�2 � 27�
r�t� � ��3 sin t, 3 cos t, 2�, �r�t�� � �13
r�t� � �3 cos t, 3 sin t, 2t�
5.
Hence, the area is 3�a2.
� �3�a2
�12
a2�2�
0�� sin � � 2 cos � � 2� d�
�12
a2�2�
0�� sin � � sin2 � � 1 � 2 cos � � cos2 � d�
12�C
x dy � y dx �12�
2�
0�a�� � sin ���a sin �� d� � a�1 � cos ���a�1 � cos ��� d�
7. (a)
(b)
—CONTINUED—
� �1
0�t4 � 4t2 � 2t � 1� dt �
1315
� �1
0��1 � 2t��2t � t2� � �t4 � 2t3 � t2 � 1� dt
W � F � dr � �1
0��2t � t2�i � ��t � t2�2 � 1 j� � ��1 � 2t�i � j� dt
r�t� � �1 � 2t�i � j
r�t� � �t � t2�i � tj, 0 ≤ t ≤ 1
W � �C
F � dr � �1
0�ti � j� � j dt � �1
0dt � 1
r�t� � j
r�t� � tj, 0 ≤ t ≤ 1
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222 Chapter 14 Vector Analysis
9.
By Stoke’s Theorem,
� �S�2v � N dS.
�C
�v � r� dr � �S�curl�v � r� � N dS
curl�v � r� � �2a1, 2a2, 2a3� � 2v
� �a2z � a3 y, �a1z � a3 x, a1 y � a2 x�
v � r � �a1, a2, a3� � �x, y, z�
7. —CONTINUED—
(c)
minimum.c �52
d2Wdc2 �
115
> 0
dWdc
�115
c �16
� 0 ⇒ c �52
W � �C
F � dr �1
30c2 �
16
c � 1
� c2t 4 � 2c2t 2 � c2t � 2ct 2 � ct � 1
F � dr � �c�t � t2� � t��c�1 � 2t�� � �c2�t � t2�2 � 1��1�
r�t� � c�1 � 2t�i � j
r�t� � c�t � t2�i � t j, 0 ≤ t ≤ 1
11.
Therefore, and is conservative.F�N�x
��M�y
� mx�x2 � y2��7�2 �3x2 � 12y2� �3mx�x2 � 4y2�
�x2 � y2�7�2
� mx�x2 � y2��7�2 ��2y2 � x2���5� � �x2 � y2���2�
�N�x
� m�2y2 � x2���52
�x2 � y2��7�2�2x�� � �x2 � y2��5�2��2mx�
N �m�2y2 � x2��x2 � y2�5�2 � m�2y2 � x2��x2 � y2��5�2
� 3mx�x2 � y2��7�2��5y2 � �x2 � y2� �3mx�x2 � 4y2�
�x2 � y2�7�2
�M�y
� 3mxy��52
�x2 � y2��7�2�2y�� � �x2 � y2��5�2�3mx�
M �3mxy
�x2 � y2�5�2 � 3mxy�x2 � y2��5�2
F�x, y� � M�x, y�i � N�x, y� j �m
�x2 � y2�5�2 �3xy i � �2y2 � x2�j
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