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Page 2: Solucionario larson (varias variables)

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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Page 7: Solucionario larson (varias variables)

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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Page 8: Solucionario larson (varias variables)

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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Page 9: Solucionario larson (varias variables)

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�

�1

10�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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Page 12: Solucionario larson (varias variables)

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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Page 13: Solucionario larson (varias variables)

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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Page 14: Solucionario larson (varias variables)

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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Page 15: Solucionario larson (varias variables)

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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Page 16: Solucionario larson (varias variables)

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

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Page 17: Solucionario larson (varias variables)

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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Page 18: Solucionario larson (varias variables)

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�w

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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

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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 ww

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Page 26: Solucionario larson (varias variables)

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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Page 27: Solucionario larson (varias variables)

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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Page 28: Solucionario larson (varias variables)

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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Page 29: Solucionario larson (varias variables)

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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Page 30: Solucionario larson (varias variables)

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�,w

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Page 31: Solucionario larson (varias variables)

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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Page 32: Solucionario larson (varias variables)

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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Page 33: Solucionario larson (varias variables)

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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Page 34: Solucionario larson (varias variables)

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�,w

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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 ,w

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Page 36: Solucionario larson (varias variables)

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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Page 37: Solucionario larson (varias variables)

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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Page 38: Solucionario larson (varias variables)

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�,w

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Page 39: Solucionario larson (varias variables)

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 ,w

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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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Page 41: Solucionario larson (varias variables)

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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Page 42: Solucionario larson (varias variables)

(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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Page 45: Solucionario larson (varias variables)

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

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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

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Page 49: Solucionario larson (varias variables)

(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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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Page 71: Solucionario larson (varias variables)

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

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Page 72: Solucionario larson (varias variables)

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

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Page 73: Solucionario larson (varias variables)

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

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Page 74: Solucionario larson (varias variables)

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

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Page 75: Solucionario larson (varias variables)

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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Page 76: Solucionario larson (varias variables)

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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Page 78: Solucionario larson (varias variables)

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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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

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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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Page 91: Solucionario larson (varias variables)

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,

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Page 93: Solucionario larson (varias variables)

(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�w

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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

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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�kw

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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

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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

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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 � 16t2w

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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�w

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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�w

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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�.�w

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Page 115: Solucionario larson (varias variables)

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 � ,

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Page 117: Solucionario larson (varias variables)

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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Page 118: Solucionario larson (varias variables)

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

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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Page 119: Solucionario larson (varias variables)

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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Page 124: Solucionario larson (varias variables)

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

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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Page 128: Solucionario larson (varias variables)

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

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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

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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

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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

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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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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

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

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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

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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

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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

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

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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

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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

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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

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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�.

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

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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

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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

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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�.

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

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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

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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

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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

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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 ww

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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 yw

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Page 177: Solucionario larson (varias variables)

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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Page 178: Solucionario larson (varias variables)

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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Page 179: Solucionario larson (varias variables)

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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Page 180: Solucionario larson (varias variables)

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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Page 181: Solucionario larson (varias variables)

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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Page 182: Solucionario larson (varias variables)

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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Page 183: Solucionario larson (varias variables)

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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Page 184: Solucionario larson (varias variables)

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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Page 185: Solucionario larson (varias variables)

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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Page 186: Solucionario larson (varias variables)

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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Page 187: Solucionario larson (varias variables)

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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Page 188: Solucionario larson (varias variables)

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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Page 189: Solucionario larson (varias variables)

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�rw

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Page 190: Solucionario larson (varias variables)

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

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Page 191: Solucionario larson (varias variables)

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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Page 192: Solucionario larson (varias variables)

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� � 0w

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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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Page 197: Solucionario larson (varias variables)

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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Page 199: Solucionario larson (varias variables)

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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Page 200: Solucionario larson (varias variables)

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

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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�

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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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Page 210: Solucionario larson (varias variables)

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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Page 212: Solucionario larson (varias variables)

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,w

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Page 214: Solucionario larson (varias variables)

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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Page 215: Solucionario larson (varias variables)

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

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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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Page 218: Solucionario larson (varias variables)

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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Page 219: Solucionario larson (varias variables)

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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Page 222: Solucionario larson (varias variables)

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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Page 223: Solucionario larson (varias variables)

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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Page 224: Solucionario larson (varias variables)

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,w

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Page 225: Solucionario larson (varias variables)

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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Page 226: Solucionario larson (varias variables)

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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Page 227: Solucionario larson (varias variables)

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�,w

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Page 228: Solucionario larson (varias variables)

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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Page 229: Solucionario larson (varias variables)

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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Page 230: Solucionario larson (varias variables)

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,w

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Page 231: Solucionario larson (varias variables)

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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Page 232: Solucionario larson (varias variables)

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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Page 233: Solucionario larson (varias variables)

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�,w

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Page 234: Solucionario larson (varias variables)

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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Page 235: Solucionario larson (varias variables)

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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Page 236: Solucionario larson (varias variables)

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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Page 237: Solucionario larson (varias variables)

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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Page 238: Solucionario larson (varias variables)

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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Page 239: Solucionario larson (varias variables)

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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Page 240: Solucionario larson (varias variables)

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 � 2w

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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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Page 242: Solucionario larson (varias variables)

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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Page 243: Solucionario larson (varias variables)

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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Page 244: Solucionario larson (varias variables)

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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Page 248: Solucionario larson (varias variables)

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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Page 249: Solucionario larson (varias variables)

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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Page 250: Solucionario larson (varias variables)

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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Page 251: Solucionario larson (varias variables)

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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Page 252: Solucionario larson (varias variables)

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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Page 256: Solucionario larson (varias variables)

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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Page 258: Solucionario larson (varias variables)

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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Page 259: Solucionario larson (varias variables)

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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Page 261: Solucionario larson (varias variables)

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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Page 263: Solucionario larson (varias variables)

(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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Page 264: Solucionario larson (varias variables)

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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Page 265: Solucionario larson (varias variables)

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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Page 266: Solucionario larson (varias variables)

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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Page 267: Solucionario larson (varias variables)

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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Page 268: Solucionario larson (varias variables)

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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Page 269: Solucionario larson (varias variables)

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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Page 270: Solucionario larson (varias variables)

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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Page 271: Solucionario larson (varias variables)

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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Page 274: Solucionario larson (varias variables)

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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Page 275: Solucionario larson (varias variables)

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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Page 276: Solucionario larson (varias variables)

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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Page 277: Solucionario larson (varias variables)

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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Page 278: Solucionario larson (varias variables)

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 � 9.3018w

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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 � 0w

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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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Page 283: Solucionario larson (varias variables)

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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Page 284: Solucionario larson (varias variables)

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

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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Page 285: Solucionario larson (varias variables)

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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Page 286: Solucionario larson (varias variables)

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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Page 287: Solucionario larson (varias variables)

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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Page 288: Solucionario larson (varias variables)

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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Page 289: Solucionario larson (varias variables)

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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Page 290: Solucionario larson (varias variables)

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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Page 291: Solucionario larson (varias variables)

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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Page 292: Solucionario larson (varias variables)

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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Page 293: Solucionario larson (varias variables)

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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Page 294: Solucionario larson (varias variables)

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

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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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Page 299: Solucionario larson (varias variables)

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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Page 301: Solucionario larson (varias variables)

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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Page 303: Solucionario larson (varias variables)

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−

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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Page 304: Solucionario larson (varias variables)

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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Page 305: Solucionario larson (varias variables)

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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Page 306: Solucionario larson (varias variables)

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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Page 307: Solucionario larson (varias variables)

14.

x

y

22π

23π

23π

25π

π

π π

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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Page 308: Solucionario larson (varias variables)

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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Page 309: Solucionario larson (varias variables)

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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Page 310: Solucionario larson (varias variables)

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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Page 311: Solucionario larson (varias variables)

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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Page 312: Solucionario larson (varias variables)

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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Page 313: Solucionario larson (varias variables)

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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Page 314: Solucionario larson (varias variables)

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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Page 315: Solucionario larson (varias variables)

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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Page 316: Solucionario larson (varias variables)

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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Page 317: Solucionario larson (varias variables)

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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Page 318: Solucionario larson (varias variables)

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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Page 319: Solucionario larson (varias variables)

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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Page 320: Solucionario larson (varias variables)

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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Page 322: Solucionario larson (varias variables)

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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Page 323: Solucionario larson (varias variables)

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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Page 324: Solucionario larson (varias variables)

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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Page 325: Solucionario larson (varias variables)

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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Page 326: Solucionario larson (varias variables)

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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Page 327: Solucionario larson (varias variables)

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

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Page 328: Solucionario larson (varias variables)

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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Page 339: Solucionario larson (varias variables)

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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Page 340: Solucionario larson (varias variables)

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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Page 345: Solucionario larson (varias variables)

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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Page 346: Solucionario larson (varias variables)

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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Page 348: Solucionario larson (varias variables)

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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Page 349: Solucionario larson (varias variables)

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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Page 350: Solucionario larson (varias variables)

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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Page 354: Solucionario larson (varias variables)

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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Page 355: Solucionario larson (varias variables)

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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Page 356: Solucionario larson (varias variables)

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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Page 357: Solucionario larson (varias variables)

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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Page 358: Solucionario larson (varias variables)

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

�2�

0�6

0 3r2 sin � dr d� � �2�

0 �r3 sin ��

6

0 d�

11.

�5�5�

6

� �5�53

����2

0

3210

���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

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Page 360: Solucionario larson (varias variables)

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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Page 361: Solucionario larson (varias variables)

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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Page 362: Solucionario larson (varias variables)

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

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Page 363: Solucionario larson (varias variables)

(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

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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

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Page 366: Solucionario larson (varias variables)

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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Page 367: Solucionario larson (varias variables)

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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Page 368: Solucionario larson (varias variables)

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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Page 370: Solucionario larson (varias variables)

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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Page 372: Solucionario larson (varias variables)

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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Page 373: Solucionario larson (varias variables)

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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Page 374: Solucionario larson (varias variables)

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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Page 375: Solucionario larson (varias variables)

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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Page 376: Solucionario larson (varias variables)

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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Page 377: Solucionario larson (varias variables)

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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Page 378: Solucionario larson (varias variables)

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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Page 379: Solucionario larson (varias variables)

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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Page 380: Solucionario larson (varias variables)

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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Page 381: Solucionario larson (varias variables)

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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Page 382: Solucionario larson (varias variables)

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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Page 383: Solucionario larson (varias variables)

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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Page 384: Solucionario larson (varias variables)

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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Page 385: Solucionario larson (varias variables)

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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Page 386: Solucionario larson (varias variables)

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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Page 387: Solucionario larson (varias variables)

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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Page 393: Solucionario larson (varias variables)

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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Page 394: Solucionario larson (varias variables)

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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Page 396: Solucionario larson (varias variables)

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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Page 397: Solucionario larson (varias variables)

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

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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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Page 401: Solucionario larson (varias variables)

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

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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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Page 413: Solucionario larson (varias variables)

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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Page 414: Solucionario larson (varias variables)

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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Page 415: Solucionario larson (varias variables)

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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Page 417: Solucionario larson (varias variables)

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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Page 429: Solucionario larson (varias variables)

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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Page 431: Solucionario larson (varias variables)

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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Page 433: Solucionario larson (varias variables)

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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Page 435: Solucionario larson (varias variables)

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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Page 441: Solucionario larson (varias variables)

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 � Pkw

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Page 447: Solucionario larson (varias variables)

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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Page 448: Solucionario larson (varias variables)

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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Page 451: Solucionario larson (varias variables)

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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Page 453: Solucionario larson (varias variables)

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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Page 455: Solucionario larson (varias variables)

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,w

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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�w

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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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Page 485: Solucionario larson (varias variables)

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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