zill_matematicas_3_manual_de_soluciones.pdf

CÁLCULO VECTORIAL MATEMÁTICAS 3 MANUAL DE SOLUCIONES

Chapter 11

UNIDAD 1 VECTORES Y ESPACIO TRIDIMENSIONAL

Vectors and 3-Space

PROBLEMAS 1.1

11.1

Vectors in 2-Space

1.

(a)

6i + 12j

2.

(a)

3, 3

3.

(a)

12, 0

4.

(a)

1 2i

5.

(a)

−9i + 6j

6.

(a)

3, 9

7.

(a)

−6i + 27i

8.

(a)

21, 30

9.

(a)

4, −12 − −2, 2 = 6, −14

10.

(a)

(4i + 4j) − (6i − 4j) = −2i + 8j

11.

(a)

(4i − 4j) − (−6i + 8j) = 10i − 12j

12.

(a)

8, 0 − 0, −6 = 8, 6

13.

(a)

16, 40 − −4, −12 = 20, 52

14.

(a)

8, 12 − 10, 6 = −2, 6

− 12 j

(b)

i + 8j 3, 4

(b)

(c) (c)

4, −5

(b)

2 3i

(b) (b)

+ 23 j

(b) (b)

3

(e)

√ 5

4, 5

(d)

√ 41

− 13 i − j

(c)

6, 18

−4i + 18j (c)

6, 8

(d)

√ 3 10

√ 4 10 0

(e)

(e) (e)

10/3 √

(b)

√ 2 85 10

(−3i + 3j) − (−15i + 20j) = 12i − 17j

−6, 0 − 0, −15 = −6, 15 −12, −30 − −10, −30 = −2, 0

−6, −9 − 25, 15 = −31, −24 740

34

√ 6 10

(e)

√ 4 13

√

(e)

(−3i − 3j) − (−15i − 10j) = 18i − 17j

(b)

(b) (b)

(d)

41

−3, 9 − −5, 5 = 2, 4

(b)

(b)

(d) (d)

√

(e)

√ 2 2/3

(d)

−3i − 5j

(c)

8, 12

(e)

5

(c)

(c)

65

(d)

−3i + 9j

0

√

(d)

−1, −2 (c)

−4, −12

(b)

3i

11.1. VECTORS IN 2-SPACE

741

15.

16. P2 5 -5

P1P2

5

P1 P1P2

P1 -5

5

−−−→ P1 P2 = 2, 5

P2

−−−→ P1 P2 = 6, −4

17.

18. 5

P2

5

P1

P1

P 1P 2 P2

P1P2

5

5

−−−→ P1 P2 = 2, 2

−−−→ P1 P2 = 2, −3

−−−→ −−→ −−→ 19. Since P1 P2 = OP2 − OP1 , terminal point is (1, 18) −−−→ −−→ −−→ 20. Since P1 P2 = OP2 − OP1 , point is (9, 8)

−−→ −−−→ −−→ OP2 = P1 P2 + OP1 = (4i + 8j) + (−3i + 10j) = i + 18j, and the −−→ −−→ −−−→ OP1 = OP2 + P1 P2 = 4, 7 − −5, −1 = 9, 8, and the initial

21. a(= −a), b = (− 14 a), c(= 52 a), e(= 2a), and f (= − 12 a) are parallel to a. 22. We want −3b = a, so c = −3(9) = −27 23. 6, 15 24. 5, 2 25. |a| = 26. |a| =

√ √

√ 4 + 4 = 2 2; (a) 9 + 16 = 5; (a)

27. |a| = 5; (a) 28. |a| =

√

u=

    1 1 1 1 1 ; (b) −u = − √ , − √ u = √ 2, 2 = √ , √ 2 2 2 2 2 2     −3 4 3 4 1 , ,− u = −3, 4 = ; (b) −u = 5 5 5 5 5

1 0, −5 = 0, −1; 5

1 + 3 = 2; (a)

√ 1 u = 1, − 3 = 2

(b) 

−u = 0, 1

√  3 1 ,− ; 2 2

 (b)

−u =

√  1 3 − , 2 2

CHAPTER 11. VECTORS AND 3-SPACE

742 29. |a + b| = |5, 12| = 30. |a + b| = |−5, 4| = 31. |a| = 32. |a| =

√

9 + 49 =



1 4

+

1 4

=

√

√

25 + 144 = 13;

√

25 + 16 =

58;

√1 2

√

58;

3b-a

41;

1 13 5, 12

u=

b=3

( 1/1√2



( 12 i

=



√1 −5, 4 41

 b = 2 ( √158 3i + 7j) =

3 33. − 4 a = −3 − 15/2

35.

√

u=

−

√6 i 58 1 2 j)

+

5 12 13 , 13



 = − √541 , √441 √14 j 58

√ √ 3 2 3 2 i− j = 2 2 34. 5(a + b) = 50, 1 = 0, 5 36.

b = a+(b+c)

3b

b+c

a

a

b

c

37. x = −(a + b) = −a − b

38. x = 2(a − b) = 2a − 2b

39.

40. c

-c

b a

b = (−c) − a; (b + c) + a = 0; a+b+c=0

d

e

b

a

From Problem 39, e + c + d = 0. but b = e − a and e = a + b, so (a + b) + c + d = 0.

41. From 2i + 3j = k1 b + k2 c = k1 (i + j) + k2 (i − j) = (k1 + k2 )i + (k1 − k2 )j we obtain the system of equations k1 + k2 = 2, k1 − k2 = 3. Solving, we find k1 = 52 and k2 = − 12 . Then a = 52 b − 12 c. 42. From 2i + 3j = k1 b + k2 c = k1 (−2i + 4j) + k2 (5i + 7j) = (−2k1 + 5k2 )i + (4k1 + 7k2 )j we 1 and obtain the system of equations −2k1 + 5k2 = 2, 4k1 + 7k2 = 3. Solving, we find k1 = 34 7 k2 = 17 . 43. From y  = 12 x we see that the slope of the tangent line at (2, 2) is 1. A vector with slope 1 is √ i + j. A unit vector is (i + j)/|i + j| = (i + j)/ 2 = √12 i + √12 j. Another unit vector tangent to the curve is − √12 i − √12 j. 44. From y  = −2x + 3 we see that the slope of the tangent √ line at (0, 0) is 3. A vector with slope 3 is i + 3j. A unit vector is (i + 3j)/|i + 3j| = (i + 3j)/ 10 = √110 i + √110 j. Another unit vector tangent to the curve is − √110 i − √110 j.

11.1. VECTORS IN 2-SPACE 45.

743 P2

(a) Since the shortest distance between two point is a straight line, |a + b| ≤ |a| + |b|.

a P1

b c

P3

(b) When P2 lies on the line segment between P1 and P3 , |a + b| = |a| + |b|. 46. Since y = 2a(L2 + y 2 )3/2 is an odd function on [−a, a], Fy = 0. Now using the fact that L/(L2 + y 2 )3/2 is an even function, we have

a −a

L dy L = 2 2 3/2 a 2a(L + y )

0

(L2

dy + y 2 )3/2

y = L tan θ, dy = L sec2 θ dθ

tan−1 a/L L sec2 θ dθ sec2 θ dθ 1 = La 0 sec3 θ L3 (1 + tan2 θ)3/2 0 tan−1 a/L

tan−1 a/L 1 1 cos θ dθ = = sin θ La 0 La 0 a 1 1 √ = = √ La L2 + a2 L L2 + a2 . L = a

a

tan−1 a/L

√ √ Then Fx = qQ/4π0 L L2 + a2 and F = (qQ/4π0 L L2 + a2 )i. 47. (a) Since Ff = −Fg , |Fg | = |Ff | = μ|Fn | and tan θ = |Fg |/|Fn | = μ|Fn |/|Fn | = μ (b) θ = arctan 0.6 ≈ 31◦ 48. Since w + F1 + F2 = 0, −200j + |F1 | cos 20◦ i + |F2 | sin 20◦ j − |F2 | cos 15◦ i + |F2 | sin 15◦ j = 0

or

(|F1 | cos 20◦ − |F2 | cos 15◦ )i + (|F1 | sin 20◦ − |F2 | sin 15◦ − 200)j = 0.

Thus, |F1 | cos 20◦ − |F2 | cos 15◦ = 0; |F1 | sin 20◦ + |F2 | sin 15◦ = 0. Solving this system for |F1 | and |F2 |, we obtain |F1 | =

200 cos 15◦ 200 cos 15◦ 200 cos 15◦ = = ≈ 336.8 lb sin 15◦ cos 20◦ + cos 15◦ sin20◦ sin(15◦ + 20◦ ) sin 35◦

and |F2 | =

sin 15◦

200 cos 20◦ 200 cos 20◦ = ≈ 327.7 lb. ◦ ◦ ◦ cos 20 + cos 15 sin20 sin 35◦

√ √ √ √ √ 49. Since F 2 = 200(i + j)/ 2 = 100 2i + 100 2j, F3 ) = F2 F1 = (100 2 − 200)i + 100 2j and  √ √ √ |F3 | = (100 2 − 200)2 + (100 2)2 = 200 2 − 2 ≈ 153 lb.

CHAPTER 11. VECTORS AND 3-SPACE

744 −→ 50. We have OA = 150 cos 20◦ i + 150 sin 20◦ j, 240 cos 190◦ i + 240 sin 190◦ j. Then

−−→ AB = 200 cos 113◦ i + 200 sin 113◦ j,

−−→ BC =

r = (150 cos 20◦ + 200 cos 133◦ + 240 cos 190◦ )i + (150 sin 20◦ + 200 sin 113◦ + 240 sin 190◦ j and ≈ −173.55i + 193.73j |r| ≈ 260.09 miles. 51. P2

P2 M

N P1

P1

−−→ Place one corner of the parallelogram at the origin, and let two adjacent sides be OP1 and −−→ OP2 . Let M be the midpoint of the diagonal connecting P1 and P2 and N be the midpoint −−→ −−→ −−→ −−→ −−→ of the other diagonal. By Problem 37, OM = 12 (OP1 + OP2 ). Since OP1 + OP2 is the main −−→ −−→ −−→ −−→ −−→ diagonal of the parallelogram and N is its midpoint, ON = 12 (OP1 + OP2 . Thus, OM = ON and the diagonals bisect each other. −−→ −−→ −→ −−→ −−→ −−→ −→ B 52. By Problem 39, AB + BC + CA = 0 and AD + DE + ED + CA = −−→ −−→ −→ 0. From the first equation AB + BC = −CA. Since D and E −−→ −−→ −−→ −−→ −−→ −−→ D E are midpoint, AD = 12 AB and EC = 12 BC. Then, 12 AB + DE + − − → − → 1 2 BC + CA = 0 and C A −→ 1 −→ −−→ −→ 1 −−→ −−→ 1 −→ AB + BC = −CA − −CA = − CA. DE = −CA − 2 2 2 Thus, the line segment joining the midpoints D and E is parallel to the side AC and half its length.

PROBLEMAS 1.2 and Vectors 11.2 3-Space 1.- 6.

(0,0,4)

(1,1,5)

(5,-4,3) (3,4,0)

(6,-2,0)

(6,0,0)

11.2. 3-SPACE AND VECTORS

745

7. A plane is perpendicular to the z-axis, 5 units above the xy-plane. 8. A plane perpendicular to the x-axis, 1 unit in front of the yz-plane. 9. A line perpendicular to the xy-plane at (2, 3, 0). 10. A single point located at (4, −1, 7). 11. (2, 0, 0), (2, 5, 0), (2, 0, 8), (2, 5, 8), (0, 5, 0), (0, 5, 8), (0, 0, 8), (0, 0, 0) 12. z

(-1,3,7)

(-1,6,7)

(3,3,7)

(3,6,7) (-1,3,4)

(3,3,4)

(-1,6,4)

(3,6,4)

y

x

13. (a) xy-plane: (−2, 5, 0), xz-plane: (−2, 0, 4), yz-plane: (0, 5, 4); (b) (−2, 5, −2) (c) Since the shortest distance between a point and a plane is a perpendicular line, the point in the plane x = 3 is (3, 5, 4). 14. We find planes that are parallel to coordinate planes. (a) z = −5; (b) x = 1 and y = −1; (c)

z=2

15. The union of the planes x = 0, y = 0, and z = 0. 16. The origin (0, 0, 0). 17. The point (−1, 2, −3). 18. The union of the planes x = 2 and z = 8. 19. The union of the planes z = 5 and z = −5. 20. The line through the points (1, 1, 1), (−1, −1, −1), and the origin.  √ 21. d = (3 − 6)2 + (−1 − 4)2 + (2 − 8)2 = 70

CHAPTER 11. VECTORS AND 3-SPACE

746

23. 24. 25.

26.

27.

28.

29.

30.

31.

32.

33. 34.



√ (−1 − 0)2 + (−3 − 4)2 + (5 − 3)2 = 3 6  (a) 7; (b) d = (−3)2 + (−4)2 = 5  (a) 2; (b) d = (−6)2 + 22 + (−3)2 = 7  √ d(P1 , P2 ) = (32 + 62 + (−6)2 = 9; d(P1 , P3 ) = 22 + 12 + 22 = 3 √ d(P2 , P3 ) = ((2 − 3)2 + (1 − 6)2 + (2 − (−6))2 = 90. The triangle is a right triangle.  √ √ √ d(P1 , P2 ) = (12 + 22 + 42 = 21; d(P1 , P3 ) = 32 + 22 + (2 2)2 = 21  √ √ d(P2 , P3 ) = ((3 − 1)2 + (2 − 2)2 + (2 2 − 4)2 = 28 − 16 2. The triangle is an isosceles triangle.  √ d(P1 , P2 ) = ((4 − 1)2 + (1 − 2)2 + (3 − 3)2 = 10 √ d(P1 , P3 ) = (4 − 1)2 + (6 − 2)2 + (4 − 3)2 = 26 √ d(P2 , P3 ) = ((4 − 4)2 + (6 − 1)2 + (4 − 3)2 = 26; The triangle is an isosceles triangle.  d(P1 , P2 ) = ((1 − 1)2 + (1 − 1)2 + (1 − (−1))2 = 2 d(P1 , P3 ) = (0 − 1)2 + (−1 − 1)2 + (1 − (−1))2 = 3 √ d(P2 , P3 ) = ((0 − 1)2 + (−1 − 1)2 + (1 − 1)2 = 5; The triangle is a right triangle.  √ d(P1 , P2 ) = ((−2 − 1)2 + (−2 − 2)2 + (−3 − 0)2 = 34 √ d(P1 , P3 ) = (7 − 1)2 + (10 − 2)2 + (6 − 0)2 = 2 34 √ d(P2 , P3 ) = ((7 − (−2))2 + (10 − (−2))2 + (6 − (−3))2 = 3 34 Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.  √ d(P1 , P2 ) = ((0 − 1)2 + (3 − 2)2 + (2 − (−1))2 = 11 √ d(P1 , P3 ) = (1 − 1)2 + (1 − 2)2 + ((−3) − (−1))2 = 5 √ d(P2 , P3 ) = ((1 − 0)2 + (1 − 3)2 + ((−3) − 2)2 = 30 Since adding any two of the above distances will not result in the third, the points cannot be collinear.  √ d(P1 , P2 ) = ((−4) − 1)2 + ((−3) − 0)2 + (5 − 4)2 = 35 √ d(P1 , P3 ) = ((−7) − 1)2 + ((−4) − 0)2 + (8 − 4)2 = 96 √ d(P2 , P3 ) = ((−7) − (−4))2 + ((−4) − (−3))2 + (8 − 5)2 = 19 Since adding any two of the above distances will not result in the third, the points cannot be collinear.  √ d(P1 , P2 ) = (1 − 2)2 + (4 − 3)2 + (4 − 2)2 = 6 √ d(P1 , P3 ) = (5 − 2)2 + (0 − 3)2 + (−4 − 2)2 = 3 6 √ d(P2 , P3 ) = (5 − 1)2 + (0 − 4)2 + (−4 − 4)2 = 4 6 Since d(P1 , P2 ) + d(P1 , P3 ) = d(P2 , P3 ), the points P1 , P2 , and P3 are collinear.  √ (2 − x)2 + (1 − 2)2 + (1 − 3)2 = 21 −→ x2 − 4x + 9 = 21 −→ x2 − 4x + 4 = 16 −→ (x − 2)2 = 16 −→ x = 2 + −4 or x = 6, −2  (0 − x)2 + (3 − x)2 + (5 − 1)2 = 5 −→ 2x2 − 6x + 25 = 25 −→ x2 − 3x = 0 −→ x = 0, 3

22. d =

11.2. 3-SPACE AND VECTORS

747



 1 + 7 3 + (−2) 1/2 + 5/2 , , 35. = (4, 1/2, 3/2) 2 2 2   0 + 4 5 + 1 −8 + (−6) , , = (2, 3, −7) 36. 2 2 2 37. (x1 + 2)/2 = −1, x1 = −4, (y1 + 3)/2 = −4, y1 = −11; The coordinates of P1 are (−4, −11, 10).

(z1 + 6)/2 = 8, z1 = 10

38. (−3 + (−5))/2 = x3 = −4, (4 + 8)/2 = y3 = 6; (1 + 3)/2 = z3 = 2. The coordinates of P3 are (−4, 6, 2).   −3 + (−4) 4 + 6 1 + 2 , , (a) = (−7/2, 5, 3/2) 2 2 2   −4 + (−5) 6 + 8 2 + 3 , , = (−9/2, 7, 5/2) (b) 2 2 2 −−−→ 39. P1 P2 = −3, −6, 1 −−−→ 41. P1 P2 = 2, 1, 1

−−−→ 40. P1 P2 = 8, −5/2, 8 −−−→ 42. P1 P2 = −3, −3, 7

43.

44. z

z

y y

x x

45.

46. z

z

y

y x

x

CHAPTER 11. VECTORS AND 3-SPACE

748

47. Since the k component is zero, while the i and j components are nonzero, the vector lies in the xy-plane. 48. Since the j components is the only nonzero component, the vector lies on the y-axis. 49. Since the vector is a scalar multiple of k, the vector lies on the z-axis. 50. Since the i component is zero while the j and k components are nonzero, the vector lies in the yz-plane. 51. a + (b + c) = 2, 4, 12 52. 2a − (b − c) = 2, −6, 4 − −3, −5, −8 = 5, −1, 12 53. b + 2(a − 3c) = −1, 1, 1 + 2−5, −21, −25 = −11, −41, −49 54. 4(a + 2c) − 6b = 45, 9, 20 − −6, 6, 6 = 26, 30, 74 √ √ 55. |a + c| = |3, 3, 11| = 9 + 9 + 121 = 139 √ √ √ 56. |c||2b| = ( 4 + 36 + 81)(2)( 1 + 1 + 1) = 22 3     b a 1 |b| = 1 + 5 = 6 57.   + 5   = |a| |b| |b| √ √ 58. |b|a + |a|b = 1 + 1 + 11, −3, 2 + 1 + 9 + 4−1, 1, 1 √ √ √ √ √ √ =  3, −3 3, 2 3 + − 14, 14, 14 √ √ √ √ √ √ =  3 − 14, −3 3 + 14, 2 3 + 14 59. |a| = 60. |a| =

√ √

u=−

100 + 25 + 100 = 15; 1+9+4=

√

14;

1 3 2 1 u = √ (i − 3j + 2k) = √ i − √ j + √ k 14 14 14 14

61. b = 4a = 4i − 4j + 4k 62. |a| =

√

36 + 9 + 4 = 7;

b=−

63. z

a

1 (a+b) 2 b y x

1 10, −5, 10 = −2/3, 1/3, −2/3 15

1 2

    1 3 3 1 ,− , −6, 3, −2 = 7 7 14 7

11.2. 3-SPACE AND VECTORS

749

64. Following the hint, we first complete the following: 

2  2  2 y1 + y 2 z1 + z2 x1 + x2 − x1 + − y1 + − z1 2 2 2  2  2  2 y1 − y2 z 1 − z2 x1 − x2 = + + 2 2 2  1 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 2  2  2  2 x1 + x2 y1 + y2 z1 + z 2 d(M, P2 ) = + y2 − + z2 − x2 − 2 2 2  2  2  2 y1 − y2 z1 − z2 x1 − x2 = + + 2 2 2  1 = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 2  d(P1 , P2 ) = (x2 − x1 )2 + (y2 − y1 )2 + (z2 − z1 )2 From the above, we see that d(P1 , P2 ) = d(P1 , M ) + d(M, P2 ). Therefore, M is collinear with P1 and P2 and lies between them. Since d(P1 , M ) = d(M, P2 ), we also have that M is equidistant from P1 and P2 . Therefore, M is the midpoint of the line segment joining P1 and P2 . √ 3 1 √ 65. xP = 1, yP = cos 30◦ + sin 30◦ = + ( 3 + 1), 2 √2 1 √ 3 1 ◦ ◦ = ( 3 − 1), xP = − sin 30 + cos 30 = − + 2 2√ 2 √ √ 2 1 √ 2 1 √ 1 √ 1 √ ◦ ◦ − ( 3 − 1) = (3 2 − 6), yr = ( 3 + 1), xR = cos 45 − ( 3 − 1) sin 45 = 2 2 4 2 √2 √2 √ 1 √ 2 1 √ 2 1 √ ◦ ◦ + ( 3 − 1) = ( 2 + 6), zr = sin 45 + ( 3 − 1) cos 45 = 2 2 2 2 4 √ √ √ 1 1 √ 3 1 √ 1 √ 1 √ ◦ ◦ xS = (3 2 − 6) cos 60 + ( 3 + 1) sin 60 = (3 2 − 6) + ( 3 + 1) 4 2 4 2 2 2 √ √ 1 √ = (3 2 − 6 + 6 + 2 3), 8 √ √ √ 3 1 √ 1 √ 1 √ 1 √ 1 ◦ ◦ + ( 3 + 1) yS = − (3 2 − 6) sin 60 + ( 3 + 1) cos 60 = − (3 2 − 6) 4 2 4 2 2 2 √ √ √ 1 = (−3 6 + 3 2 + 2 3 + 2), 8 √ 1 √ zs = ( 2 + 6) 4 d(P1 , M ) =

Thus, xS ≈ 1.4072, yS ≈ 0.2948,    1 0 0   66. (a) MP =  0 cos α sin α  ,  0 − sin α cos α 

zS ≈ 0.9659.   cos β  MR =  0  sin β

0 1 0

− sin β 0 cos β

     

CHAPTER 11. VECTORS AND 3-SPACE

750 ⎡

⎤ ⎡ ⎤ ⎡ ⎤ x xP xS (b) Mγ MR MP ⎣ y ⎦ = Mγ MR ⎣ yP ⎦ = ⎣ yS ⎦ zP zS z ⎤⎡ ⎤ ⎡ ⎤ ⎡ ⎤ ⎡ ⎤⎡ ⎤ ⎡ 1 √0 0 1 1 1 1 0 0 1 √ ⎢ 3 1 ⎥⎣ 1 ⎦ = ⎣ 12 (√3 + 1) ⎦ sin 30◦ ⎦ ⎣ 1 ⎦ = ⎣ 0 (c) MP ⎣ 1 ⎦ = ⎣ 0 cos 30◦ 2 √2 ⎦ ◦ ◦ 1 3 0 − sin 30 cos 30 1 1 1 0 − 12 2 ( 3 − 1) 2⎡ √ √ ⎤⎡ ⎡ ⎤ ⎡ ⎤ ⎤ ⎤ ⎡ 2 1 1 1 0 − 22 cos 45◦ 0 − sin 45◦ 2 √ √ ⎥ ⎦ ⎣ 1 ( 3 + 1) ⎦ = ⎢ 0 1 0 0 ⎦ ⎣ 12 (√3 + 1) ⎦ MR MP ⎣ 1 ⎦ = ⎣ ⎣ √0 1 2 √ √ ◦ ◦ 1 1 2 2 0 cos 45 sin 45 1 0 2 ( 3 − 1) 2 ( 3 − 1) 2 2 √ ⎤ ⎡ 1 √ 6) 4 (3 √2 − = ⎣ 12 √ ( 3 +√1) ⎦ 1 6) 4( 2 + √ ⎤ ⎤⎡ 1 √ ⎡ ⎤ ⎡ ◦ 6) sin 60◦ 0 1 cos 60 4 (3 √2 − Mγ MR MP ⎣ 1 ⎦ = ⎣ sin sin 60◦ cos sin 60◦ 0 ⎦ ⎣ 12 √ ( 3 +√1) ⎦ 1 0 0 1 1 6) 4( 2 + ⎤⎡ ⎡ √ √ √ √ √ √ ⎤ ⎡ ⎤ 1 1 3 1 6) 6 + 6 +√2 3) 0 4 (3 √2 − 8 (3 √2 − √ 2 ⎥ ⎢ 2√3 1 1 =⎣ − ( 3 +√1) ⎦ = ⎣ 18 (−3 6 +√3 2 √ + 2 3 + 2) ⎦ 0 ⎦ ⎣ 2√ 2 2 1 1 6) 6) 0 0 1 4( 2 + 4( 2 +

PROBLEMAS 1.3 11.3 Dot Product 1. a · b = 2(−1) + (−3)2 + 4(5) = 12 2. b · c = (−1)3 + 2(6) + 5(−1) = 4 3. a · c = 2(3) + (−3)6 + 4(−1) = −16 4. a · (b + c) = 2(2) + (−3)8 + 4(4) = −4 5. a · (4b) = 2(−4) + (−3)8 + 4(20) = 48 6. b · (a − c) = (−1)(−1) + 2(−9) = 5(5) = 8 7. a · a = 22 + (−3)2 + 42 = 29 8. (2b) · (3c) = (−2)9 + 4(18) + 10(−3) = 24 9. a · (a + b + c) = 2(4) + (−3)5 + 4(8) = 25 10. (2a) · (a − 2b) = 4(4) + (−6)(−7) + 8(−6) = 10     a·b 2(−1) + (−3)2 + 4(5) 12 −1, 2, 5 = −2/5, 4/5, 2 −1, 2, 5 = 11. b= b·b (−1)2 + 22 + 52 30 12. (c · b)a = [3(−1) + 6(2) + (−1)5]2, −3, 4 = 42, −3, 4 = 8, −12, 16 √ 13. a · b = 10(5) cos(π/4) = 25 2

11.3. DOT PRODUCT

751

√ 14. a · b = 6(12) cos(π/6) = 36 3 15. a · b = |a||b| cos θ = (2)(3) cos(2π/3) = 6(−1/2) = −3

√  √ 16. a · b = |a||b| cos(θ) = (4)(1) cos(5π.6) = 4 − 23 = −2 3 √ √ 5a · b = 3(2) + (−1)2 = 4; |a| = 10, |b| = 2 2 1 1 4 √ = √ −→ θ = arccos √ ≈ 1.11 rad ≈ 63.45◦ cos θ = √ ( 10)(2 2 5 5 √ √ 18. 5a · b = 2(−3) + 1(−4) = −10; |a| = 5, |b| = 5 √ −10 2 cos θ = √ = − √ −→ θ = arccos(−2/ 5) ≈ 2.68 rad ≈ 153.43◦ ( 5)5 5 √ √ √ 19. 5a · b = 2(−1) + 4(−1) + 0(4) = −6; |a| = 2 5, |b| = 3 2 √ −6 1 √ = − √ −→ θ = arccos(−1/ 10) ≈ 1.89 rad ≈ 108.43◦ cos θ = √ (2 5)(3 2 10 √ √ √ 20. 5a · b = 12 (2) + 12 (−4) + 32 (6) = 8; |a| = 11/2, |b| = 2 14 √ 8 8 √ =√ −→ θ = arccos(8/ 154) ≈ 0.87 rad ≈ 49.86◦ cos θ = √ ( 11/2)(2 14) 154 17.

√

21. a and f, b and e, c and d 22. (a) a × b = 2 · 3 + (−c)2 + 3(4) = 0 −→ c = 9 (b) a · b = c(−3) + 12 (4) + c2 = c2 − 3c + 2 = (c − 2)(c − 1) = 0 −→ c = 1, 2 23. Solving the system of equations 3x1 + y1 − 1 = 0, −3x1 + 2y1 + 2 = 0 gives x1 = 4/9 and y1 = −1/3. Thus, v = 4/9, −1/3, 1. 24. If a and b represent adjacent sides of the rhombus, then |a| = |b|, then diagonals of the rhombus are a + b and a − b, and (a + b) · (a − b) = a · a − a · b + b · a − b · b = |a|2 − |b|2 = 0. Thus, the diagonals are perpendicular. 25. Since c·a=

 b−

 a·b a·b a·b 2 a ·a=a·a− (a · a) = b · a − |a| = b · a − a · b = 0, 2 2 |a| |a| |a|2

the vectors c and a are orthogonal. √ √ 26. a · b = 1(1) + c(1) = c + 1; |a| = 1 + c2 , |b| = 2 √ c+1 1 √ =⇒ 1 + c2 = c + 1 −→ 1 + c2 = c2 + 2c + 1 =⇒ c = 0 cos 45◦ = √ = √ 2 2 1+c 2 √ √ √ √ 27. |a| = 14; cos α − 1/ 14, α ≈ 74.50◦ ; cos β = 2/ 14, β ≈ 57.69◦ ; cos γ = 3/ 14, γ = 36.70◦

752

CHAPTER 11. VECTORS AND 3-SPACE

28. |a| = 9; cos α = 2/3, α ≈ 48.19◦ ; cos β = 2/3, β ≈ 48.19◦ ; cos γ = −1/3, γ = 109.47◦ √ 29. |a| = 2; cos α = 1/2, α ≈ 60◦ ; cos β = 0, β ≈ 90◦ ; cos γ = − 3/2, γ = 150◦ √ √ √ √ 30. |a| = 78; cos α = 5/ 78, α ≈ 55.52◦ ; cos β = 7/ 78, β ≈ 37.57◦ ; cos γ = 2/ 78, γ = 76.91◦ −−→ −−→ 31. Let θ be the angle between AD and AB and a be the length of an edge of an edge of the −−→ −−→ cube. Then AD = ai + aj + ak, AB = ai and −−→ −−→ AD · AB 1 a2 cos θ −−→ −−→ = √ √ = √ 2 2 3 3a a |AD||AB| −−→ −→ so θ ≈ 0.955317 radian or 54.7356◦ . Letting φ be the angle between AD and AC and noting −→ that AC = ai + aj we have  −−→ −→ 2 AD · AC a2 + a2 cos θ −−→ −→ = √ √ = 2 2a2 3 3a |AD||AC| so φ ≈ 0.61548 radian or 35.2644◦ √ √ √ 32. a = 5, 7, 4;√ |a| = 3 10; cos α = 5/3 10, α = 58.19◦ ; cos β = 7/3 10, β = 42.45◦ ; cos γ = 4/3 10, γ ≈ 65.06◦ 33. compb a = a · b/|b| = 1, −1, 3 · 2, 6, 3/7 = 5/7 √ √ 34. compa b = b · a/|a| = 2, 6, 3 · 1, −1, 3/ 11 = 5/ 11

√ √ 35. b − a = 1, 7, 0; compa (b − a) = (b − a) · a/|a| = 1, 7, 0 · 1, −1, 3/ 11 = −6/ 11

36. a + b = 3, 5, 6; 2b = 4, 12, 6; comp2b (a + b) · 2a/|2b| = 3, 5, 6 · 4, 12, 6/14 = 54/7 √ √ −−→ −−→ −−→ −−→ − → a = a · OP /|OP | = (4i + 6j) · (3i + 10j)/ 109 = 37. OP = 3i + 10j; |OP | = 109; comp− OP √ 72/ 109 √ √ −−→ −−→ −−→ −−→ − → a = a · OP /|OP | = 2, 1, −1 · 1, −1, 1/ 3 = 0 38. OP = 1, −1, 1; |OP | = 3; comp− OP 39. (a) compb a = a · b/|b| = (−5i + 5j) · (−3i + 4j)/5 = 7 projb a = (compb a)b/|b| = 7(−3i + 4j)/5 = − 21 5 i+

28 5 j = − 45 i

28 (b) projb⊥ a = a − projb a = (−5i + 5j) − (− 21 − 35 j 5 + 5 j) √ √ 40. (a) compb a = a · b/|b| = (4i + 2j)√· (−3i + j)/ √ 10 = − 10 projb a = (compb a)b/|b| = − 10(−3i + j)/ 10 = 3i − j

(b) projb⊥ a = a − projb a = (4i + 2j) − (3i − j) = i + 3j 41. (a) compb a = a · b/|b| = (−i − 2j + 7k) · (6i − 3j − 2k)/7 = −2 6 4 projb a = (compb a)b/|b| = −2(6i − 3j − 2k)/7 = − 12 7 i + 7j + 7k

6 4 5 (b) projb⊥ a = a − projb a = (−i − 2j + 7k) − (− 12 7 i − 7 j + 7 k) = 7 i −

20 7 j

+

45 7 k

11.3. DOT PRODUCT

753

42. (a) compb a = a · b/|b| = 1, 1, 1 · −2, 2, −1/3 = −1/3 projb a = (compb a)b/|b| = − 13 −2, 2, −1/3 = 2/9, −2/9, 1/9 (b) projb⊥ a = a − projb a = 1, 1, 1 − 2/9, −2/9, 1/9 = 7/9, 11/9, 8/9 43. a + b = 3i + 4j; |a + b| = 5; compa+b a = a · (a + b)/|a + b| = (4i + 3j) · (3i + 4j)/5 = 72 96 24/5; proj(a+b) a = (comp(a+b) a)(a + b)/|a + b| = 24 5 (3i + 4j)/5 = 25 i + 25 j √ √ √ 44. a − b = 5i + 2j; |a − b| = 29; comp(a−b) b = b · (a − b)/|a − b| = (−i + j)/ 29 = −3/ 29 √ 6 3 15 proj(a−b) b = (comp(a−b) )(a − b)/|a − b| = − √ (5i + 2j)/ 29 = − i − j 29 29 29   6 14 35 15 proj(a−b)⊥ b = b − proj(a−b) b = (−i + j) − − i − j = − i + j 29 29 29 29 45. We identify F = 29, θ = 60◦ and |d| = 100. Then W = |F||d| cos θ = 20(100)( 12 ) = 1000 ft-lb. 46. W = F · d = |F||d| cos θ Since the force is acting at a 45◦ angle to the direction of motion, we have θ = 45◦ . Therefore, √ W = (3000)(400) cos 45◦ = 600, 000 2 ft-lb. 47. We identify d = −i + 3j + 8k. Then W = F · d = 4, 3, 5 · −1, 3, 8 = 45 N-m. 48. (a) Since w and d are orthogonal, W = w · d = 0.

√ (b) We identify θ = 0◦ . Then W = |F||d| cos θ = 30( 42 + 32 ) = 150 N-m.     78 49. Using d = 6i + 2j and F = 3 35 i + 45 j , W = F · d = 95 , 12 5 · 6, 2 = 5 ft-lb. 50. Let a and b be vectors from the center of the carbon atom to the center of two distinct hydrogen atoms. The distance between two hydrogen atoms is then  √ |b − a| = (b − a) · (b − a) = b · b − 2a · b + a · a  = |b|2 + |a|2 − 2|a||b| cos θ  = (1.1)2 + (1.1)2 − 2(1.1)(1.1) cos 109.5◦  = 1.21 + 1.21 − 2.42(−0.333807) ≈ 1.80 angstroms. 51. If a and b are orthogonal, then a · b = 0 and a 2 b2 a 3 b3 a 1 b1 + + a b a b a b 1 1 = (a1 b1 + a2 b2 a3 b3 ) = (a · b) = 0. |a||b| |a||b|

cos α1 cos α2 + cos β1 cos β2 + cos γ1 cos γ2 =

52. We want cos α = cos β = cos γ or a1 = a2 = a3 . Letting a1 = a2 = a3 = 1 we obtain the 1 1 1 vector i + j + k. A unit vector in the same direction is √ i + √ j + √ k. 3 3 3

CHAPTER 11. VECTORS AND 3-SPACE

754

53. For the following, let a = a1 , a2 , a3 , b = b1 , b2 , b3 , and let k be any scalar. Proof of (i): If a = 0 = 0, 0, 0, then a · b = (0)b1 + (0)b2 + (0)b3 = 0. Similarly, if b = 0, then a · b = a1 (0) + a2 (0) + a3 (0) = 0 Proof of (ii): Using the Commutative Property of real numbers, we have a · b = a 1 b1 + a 2 b2 + a 3 b 3 = b1 a 1 + b2 a 2 + b3 a 3 =b·a Proof of (iv): a · (kb) = a1 , a2 , a3  · kb2 , kb2 , kb3  = a1 kb1 + a2 kb2 + a3 kb3 = k(a1 b1 + a2 b2 + a3 b3 ) = k(a · b) = la1 , ka2 , ka3  · b1 , b2 , b3  = ka1 b1 + ka2 b2 + ka3 b3 = k(a1 b1 + a2 b2 + a3 b3 ) = k(a · b) Therefore, a · (kb) = (ka) · b = k(a · b) Proof of (v): Since x2 ≥ 0 for any real number x, we have a · a = a21 + a22 + a23 ≥ 0. 54. Using the fact that | cos θ| < 1, we have |a · b| = |a||b|| cos θ| ≤ |a||b|. 55. |a + b|2 = (a + b) · (a + b) = a · a + 2a · b + b · b = |a|2 + 2a · b + |b|2 ≤ |a|2 + 2|a · b| + |b|2

since x ≤ |x|

≤ |a|2 + 2|a||b| + |b|2 = (|a| + |b|)2 by Problem 54 Thus, since |a + b| and |a| + |b| are positive, |a + b| ≤ |a| + |b|. 56. Let P1 (x1 , y1 ) and P2 (x2 , y2 ) be distinct points on the line ax + by = −c. Then −−−→ n·P1 P2 = a, b·x2 −x1 , y2 −y1  = ax2 −ax1 = by2 −by1 (ax2 +by2 )−(ax1 +by1 ) = −c−(−c) = 0, and the vectors are perpendicular. Thus, n is perpendicular to the line. −−−→ 57. Let θ be the angle between n and P2 P1 . Then −−−→ −−−→ |a, b · x1 − x2 , y1 − y2 | |n · P2 P1 | |ax1 − ax2 + by1 = by2 | √ √ = = d = ||P1 P2 | cos θ| = 2 2 |n| a +b a 2 + b2 |ax1 + by1 − (ax2 + by2 )| |ax1 + by1 − (−c)| |ax1 + by1 + c| √ √ √ = = = . 2 2 2 2 a +b a +b a 2 + b2 58. (a) Since N = x, y is a unit normal, T = −y, x is a unit tangent at P (x, y). Now we

11.4. CROSS PRODUCT

755

compute −−→ N · P O = x, y · c − x, d − y = cx + dy − (x2 + y 2 ) = cx + dy − 1 −−→ T · OP = −y, x · c − x, d − y = dx − cy −→ N · P S = x, y · a − x, b − y = ax + by − (x2 + y 2 ) = ax + by − 1 −→ −T · P S = y, −x · a − x, b − y = ay − bx. Now, since |N| = |T| = 1, −→ −−→ −→ −−→ N · PS T · PO −T · P S N · PO = cos θ = , and = cos φ = −−→ −→ −−→ −→ , |P O| |P S| |P O| |P S| we have

−→ −−→ ax + by − 1 N · PS cx + dy − 1 N · PO = −−→ −→ or dx − cy = ay − bx . TP O −TP S

(b) With a = 2, b = 0, c = 0, and d = 3 the equation in (a) becomes (3y − 1)/3x = 2 2 (2x − 1)/2y or − 2y. Substituting y 2 = 1 − x2 we obtain 6x2 − 3x = √ 6x − 3x = 6y √ 2 2 2 6(1 − x ) − 2 1 − x or 12x − 3x − 6 = −2 1 − x2 . Squaring both sides, we obtain 144x4 − 72x3 + 36x + 32 = 0. (c) Newton’s method gives us the roots −0.6742, −0.48302, ; 0.76379, and 0.89343. Since S and O are on the√positive x- and y-axes, respectively,√we can ignore the negative roots. Computing y = 1 − 0.763792 ≈ 0.645465 and y = 1 − 0.893432 ≈ 0.449202 we see that only P (x, y) = (0.76379, 0.645465) satisfies (3y − 1)/3x = (2x − 1)/2y.

PROBLEMAS 11.4 Cross1.4 Product 1.

2.

3.

4.

5.

   a × b =      a × b =      a × b =      a × b =      a × b =  

              −1 0   i −  1 0  j +  1 −1  k = −5i − 5j + 3k =  0 3   0 5    3 5         i j k    2 1   2 0  1 0        k = −i + 2j − 4k 2 1 0 = j+ i− 4 0  4 −1  0 −1   4 0 −1       i j k    1 1   1 −3  −3 1       k = −12, −2, 6  1 −3 1  =  i− j+ 0 4  2 4  2 0   2 0 4       i j k    1 1   1 1  1 1        k = 1, −8, 7 1 1 1 = i− j+ −5 2  2 3  −5 3   −5 2 3       i j k    2  2 −1  −1 2  2     2 −1 2  =  i − j +  −1 −1   −1 3  k = −5i + 5k 3 −1  −1 3 −1  i 1 0

j −1 3

k 0 5

CHAPTER 11. VECTORS AND 3-SPACE

756    i j k     1  6. a × b =  4 1 −5  =  3  2 3 −1 

 j k   0 0 1/2  =  6 6 0 

  i  7. a × b =  1/2  4   i  8. a × b =  0  2   i  9. a × b =  2  −3   i  10. a × b =  8  1

   4 −5   i −  2  −1

j 5 −3

k 0 4

j 2 −3

     5 =   −3  k −4 6

   1/2 1/2  j +  0  4

   1/2 1/2  i −  4 0     0 0   i −   2 4

   0 0   j +   2 4

       2 −4   i −  2 =   −3 6   −3 

 j k   1 1 −6  =  −2 −2 10 

   8 −6   i −   1 10

−−−→ P1 P3 = (1,  2, 2) i j k   1 0 1 1  =  2 1 2 2 

  i j  13. a × b =  2 7  1 1 is perpendicular

 k   7 −4 −4  =  1 −1 −1  to both a and b.

  i j  14. a × b =  −1 −2  4 −1 is perpendicular to

   0 1   i −  1  2     i −  2  1 

   8 −6   j +   1 10

 2  k = 0, 0, 0 −3 

 1  k = −2, −86, −17 −2 

   −2 −4   j +   −3 1

   0 1   j +   1 2

   2 −4  j +  1 −1 

   k    −1 −2 4     4 = i −  4 −1 0  0 both a and b.

 0  k = −3, 2, 3 6 

 5  k = 20, 0, −10 −3 

   2 −4  j +  −3 6 

−−−→ −−−→ 11. P1 P2 = (−2, 2, −4); P1 P3 = (−3,  1, 1)       i j k    2 −4  −−−→ −−−→   i −  −2 P1 P2 × P1 P3 =  −2 2 −4  =    −3 1 1  −3 1 1  −−−→ 12. P1 P2 = (0, 1, 1);  −−−→ −−−→  P1 P2 × P1 P3 =  

 1  k = 14i − 6j + 10k 3 

   4 −5   j +  2  −1

 2  k = 6i + 14j + 4k 1 

 1  k=j−k 2   7  k = −3i − 2j − 5k 1 

   −1 4  j +  4 0 

   i     j k     5 1   5 −2 1        15. a × b =  5 −2 1  =   i −  2 −7  j +  2 0 −7  2 0 −7  a · (a × b) = 5, −2, 1 · 14, 37, 4 = 70 − 74 + 4 = 0 b · (a × b) = 2, 0, −7 · 14, 37, 4 = 28 + 0 − 28 = 0

 −2  k = 4, 16, 9 −1 

 −2  k = 14, 37, 4 0 

11.4. CROSS PRODUCT

757

      i j k    1/2 −1/4   −1/4 −4  k  i + 16. a × b =  1/2 −1/4 −4  =   2 −2  −2 6   2 −2 6  19 1 = − i − 11j − k 2 2  1 1 19 1 19 11 i − j − 4k) · (− i − 11j − k = − + +2=0 a · (a × b) = 2 4 2 2 4 4  19 1 b · (a × b) = (2i − 2j + 6k) · − i − 11j − k = −19 + 22 − 3 = 0 2 2         i j k         1 1   i −  2 1 j +  2 1 k = j − k 17. (a) b × c =  2 1 1  =      3 1  3 1 1 1  3 1 1       i    j k    1 2     −1 2    j +  1 −1  k = −i + j + k i − a × (b × c) =  1 −1 2  =      1 −1 0 −1 0 1   0 1 −1  (b) a · c = (i − j + 2k) · (3i + j + k) = 4; (a · c)b = 4(2i + j + k) = 8i + 4j + 4k a · b = (i − j + 2k) · (2i + j + k) = 3; (a · b)c = 3(3i + j + k) = 9i + 3j + 3k a × (b × c) = (a · c)b − (a · b)c = (8i + 4j + 4k) − (9i + 3j + 3k) = −i + j + k         i j k    1 −1     2 −1    j +  1 2  k = 21i − 7j + 7k i − 18. (a) b × c =  1 2 −1  =      5 8 −1 8 −1 5   −1 5 8        i   j k    3   3 −4  0  0 −4        3 0 −4 a × (b × c) =   =  −7 7  i −  21 7  j +  21 −7  k  21 −7 7  = −28i − 105j − 21k (b) a · c = (3i − 4k) · (−i + 5j + 8k) = −35; (a · c)b = −35(i + 2j − k) = −35i − 70j + 35k a · b = (3i − 4k) · (i + 2j − k) = 7; (a · b)c = 7(−i + 5j + 8k) = −7i + 35j + 56k a×(b×c) = (a·c)b−(a·b)c = (−35i−70j+35k)−(−7i+35j+56k) = −28i−105j−21k 19. (2i) × j = 2(i × j) = 2k 20. i × (−3k) = −3(i × k) = −3(−j) = 3j 21. k × (2i − j) = k × (2i) + k × (−j) = 2(k × i) − (k × j) = 2j − (−i) = i + 2j 22. i × (j × k) = i × i = 0 23. [(2k) × (3j)] × (4j) = [2 · 3(k × j) × (4j)] = 6(−i) × 4j = (−6)(4)(i × j) = −24k 24. (2i − j + 5k) × i = (2i × i) + (−j × i) + (5k × i) = 2(i × i) + (i × j) + 5(k × i) = 5j + k 25. (i + j) × (i + 5k) = [(i + j) × i] + [(i + j) × 5k] = (i × i) + (j × i) + (i × 5k) + (j × 5k) = −k + 5(−j) + 5i = 5i − 5j − k 26. i × k − 2(j × i) = −j − 2(−k) = −j + 2k

CHAPTER 11. VECTORS AND 3-SPACE

758 27. k · (j × k) = k · i = 0

28. i · [j × (−k)] = i · [−(j × k)] = i · (−i) = −(i · i) = −1 √ 29. |4j − 5(i × j)| = |4j − 5k| = 41 30. (i × j) · (3j × i) = k · (−3k) = −3(k · k) = −3 31. i × (i × j) = i × k = −j

32. (i × j) × i = k × i = j

33. (i × i) × j = 0 × j = 0

34. (i · i)(i × j) = 1(k) = k

35. 2j · [i × (j − 3k)] = 2j · [(i × j) + (i × (−3k)] = 2j · [k + 3(k × i)] = 2j · (k + 3j) = 2j · k + 2j · 3j = 2(j · k) + 6(j · j) = 2(0) + 6(1) = 6 36. (i × k) × (j × i) = (−j) × (−k) = (−1)(−1)(j × k) = j × k = i 37. a × (3b) = 3(a × b) = 3(4i − 3j + 6k) = 12i − 9j + 18k 38. b × a = −a × b = −(a × b) = −4i + 3j − 6k 39. (−a) × b = −(a × b) = −4i + 3j − 6k  √ 40. |a × b| = 42 + (−3)2 + 62 = 61      i j k    4  −3 6      41. (a × b) × c =  4 −3 6  =  i− 2 4 −1  2 4 −1 

   4 6  j +  2 −1 

 −3  k = −21i + 16j + 22k 4 

42. (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10 43. a · (b × c) = (a × b) · c = 4(2) + (−3)4 + 6(−1) = −10 44. (4a) · (b × c) = (4a × b) · c = 16(2) + (−12)4 + 24(−1) = −40 −−→ 45. (a) Let A = (1, 3, 0), B = (2, 0, 0), C = (0, 0, 4), and D = (1, −3, 4). Then AB = i − 3j, −→ −−→ −−→ −−→ −−→ AC = −i − 3j + 4k, CD = i − 3j, and BD = −i − 3j + 4k. Since AB = CD and −→ −−→ AC = BD, the quadrilateral is a parallelogram. (b) Computing  k  0  = −12i − 4j − 6k 4  √ we find that the area is | − 12i − 4j − 6k| = 144 + 16 + 36 = 14.   i −−→ −→  AB × AC =  1  −1

j −3 −3

46. (a) Let A = (3, 4, 1), B = (−1, 4, 2), C = (2, 0, 2), and D = (−2, 0, 3). Then −−→ −→ −−→ −−→ AB = −4i + k, AC = −i − 4j + k, CD = −4i + k, and BD = −i − 4j + k. Since −−→ −−→ −→ −−→ AB = CD and AC = BD, the quadrilateral is a parallelogram.

11.4. CROSS PRODUCT

759

(b) Computing  j k  0 1  = 4i + 3j + 16k −4 1  √ √ we find that the area is |4i + 3j + 16k| = 16 + 9 + 256 = 281 ≈ 16.76.   i −−→ −→  AB × AC =  −4  −1

−−−→ −−−→ 47. P1 P2 = j; P2 P3 = −j + k  i j k −−−→ −−−→  P 1 P2 × P 2 P 3 =  0 1 0  0 −1 1 A = 12 |i| = 12 sq. unit

     1 =   −1 

   0 0   i −   0 1

   0 0   j +   0 1

−−−→ −−−→ 48. P1 P2 = j + 2k; P2 P3 = 2i +j − 2k  i j k     1 2 −−−→ −−−→  P1 P2 × P2 P3 =  0 1 2  =  1 −2  2 1 −2  A = 12 | − 4i + 4j − 2k| = 3 sq. units

    i −  0  2 

−−−→ −−−→ 49. P1 P2 = −3j − k;  P2 P3 = −2i − k  i j k   −−−→ −−−→  −3 P1 P2 × P2 P3 =  0 −3 −1  =  0  −2 0 −1 

  −1   0 i− −1   −2

A = 12 |3i + 2j − 6k| =

−−−→ 50. P1 P2 = −i + 3k;  −−−→ −−−→  P 1 P2 × P 2 P3 =  

7 2

   0 2  j +  2 −2 

 1  k=i −1 

 1  k = −4i + 4j − 2k 1 

  −1   0 j+ −1   −2

 −3  k = 3i+2j−6k 0 

sq. units

−−−→ P2 P3 = 2i +4j − k i j k   0 −1 0 3  =  4 2 4 −1 

A = 12 | − 12i + 5j − 4k| =

√

185 2

   −1 3  i −  2 −1 

   −1 3  j +  2 −1 

0 4

   k = −12i + 5j − 4k 

sq. units

        i j k      −1 0   4 0   j +  −1 4  k = 8i + 2j − 10k  i − 51. b × c =  −1 4 0  =      2 2  2 2 2 2  2 2 2  v = |a · (b × c)| = |(i + j) · (8i + 2j − 10k)| = |8 + 2 + 0| = 10 cu. units    i j k          4 1        i −  1 1  j +  1 4  k = 19i − 4j − 3k 52. b × c =  1 4 1  =      1 1  1 5 1 5  1 1 5  v = |a · (b × c)| = |(3i + j + k) · (19i − 4j − 3k)| = |57 − 4 − 3| = 50 cu. units    i      j k     −2 −6   −2 6  6 −6         k = 21i − 14j − 21k 53. b × c =  −2 6 −6  =  i− 5 1 j +  5 3 12  3  2 2 2  5 3 1  2 2 a · (b × c) = (4i + 6j) · (21i − 14j − 21k) = 84 − 84 + 0 = 0. The vectors are coplanar.

760

CHAPTER 11. VECTORS AND 3-SPACE

     i    j k      −2 1   1 1   j +  −2 13  k = − 7 i − 4j − 3k  i − 54. b × c =  −2 1 1  =  3 2      0 −2 0 −2 3 2 2  0 −2  2 a · (b × c) = (i + 2j − 4k) · (− 27 i − 4j − 3k) = − 72 − 8 + 12 = 0. The vectors are not coplanar. −−−→ −−−→ 55. The four points will be coplanar if the three vectors P1 P2 = 3, −1, −1, P2 P3 = −3, −5, 13, −−−→ and P3 P4 = −8,   7, −6 are coplanar.       i j k    −−−→ −−−→  −5 13   −3 13   −3 −5    k = −61, −122, −61 j+ i− P2 P3 ×P3 P4 =  −3 −5 13  =  7 −6   −8 −6   −8 7   −8 7 −6  −−−→ −−−→ −−−→ P1 P2 · (P2 P3 × P3 P4 ) = 3, −1, −1 · −61, −122, −61 = −183 + 122 + 61 = 0 The four points are coplanar. −−−→ −−−→ 56. The four points will be coplanar if the three vectors P1 P2 = −3, 3, −1, P2 P3 = 1, 2, −6, −−−→ and P3 P4 = 4,−6, 5 are coplanar.        i j k   −−−→ −−−→  2 −6   1 −6   1 2    i− j+ k = −26, −29, −14 P2 P3 × P3 P4 =  1 2 −6  =  −6 5   4 5   4 −6   4 −6 5  −−−→ −−−→ −−−→ P1 P2 · (P2 P3 × P3 P4 ) = −3, 3, −1 · −26, −29, −14 = 78 − 87 + 14 = 5 The four points are not coplanar. 57. (a) Since θ = 90◦ , |a × b| = |a||b|| sin 90◦ | = 6.4(5) = 32. (b) The direction of a × b is into the fourth quadrant of the xy-plane or to the left of the plane determined by a and b as shown in Figure 11.4.9 in the text. It makes an angle of 30◦ with the positive x-axis. √ √ (c) We identify n = ( 3i − j)/2. Then a × b = 32n = 16 3i − 16j. √ √ √ 1.4, a × b = 27(8) sin 120◦ n = 24 3( 3/2)n = 36n. By the right-hand 58. Using Definition 11.4, rule, n = j or n = −j. Thus, a × b = 36j or −36j.         i j k         5 6    i −  4 6  j +  4 5  k = −30, 30, −3 59. b × c =  4 5 6  =      8 3 7 3 7 8   7 8 3     i j k           1 3       i −  1 3  j +  1 2  k = −3, 6, −3 a × b =  1 2 3  =      5 6 4 6 4 5   4 5 6          i j k    2 3   1 3   1 2     2 3 = i− j+ k = −96, −93, 96 a×(b×c) =  1 30 −3   −33 −3   −30 30   −33 30 −3         i  j k    −3 6   −3 −3   6 −3   k = 42, −12, −66       j+ i− (a × b) × c =  −3 6 −3  =  7 8  7 3  8 3   7 8 3  Therefore, a × (b × c) = (a × b) × c

11.4. CROSS PRODUCT

761

60. For the following, let a = a1 , a2 , a3 , b = b1 , b2 , b3 , and c = c1 , c2 , c3  be arbitrary vectors. Proof of (iv): (a + b) × c = −c × (ab ) by property (ii) = (−c × a) + (−c × b) by property (iii) = (a × c) + (b × c) by property (ii) Proof of (v): Let k be any scalar. Then   i j k  a2 a3 a × (kb) =  a1  kb1 kb2 kb3

     

= (a2 kb3 − a3 kb2 )i − (a1 kb3 − a3 kb1 )j + (a1 kb2 − a3 kb1 )k = k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k    i j k   (ka) × b =  ka1 ka2 ka3   b1 b2 b3  = (ka2 b3 − ka3 b2 )i − (ka1 b3 − ka3 b1 )j + (ka1 b2 − ka2 b1 )k = k(a2 b3 − a3 b2 )i − k(a − 1b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k Using Definition 11.4.1, we have k(a × b) = k[(a2 b3 − a3 b2 )i − (a1 b3 − a3 b1 )j + (a1 b2 − a2 b1 )k] = k(a2 b3 − a3 b2 )i − k(a1 b3 − a3 b1 )j + k(a1 b2 − a2 b1 )k Therefore, a × (kb) = (ka) × b = k(a × b). Proof of (vii): From Equation 11.4.7, we have   a1  a · (a × b) =  a1  b1

a2 a2 b2

a3 a3 b3

     

However, using Property (ii) of determinants in Appendix I in the text, we see that a·(a×b = 0.) Proof of (viii): From Equation 11.4.7, we have    b 1 b2 b3    b · (a × b) =  a1 a2 a3   b1 b2 b3  However, using Property (ii) of determinants in Appendix I in the text, we see that b × (a × b) = 0. 61. Using equation 9 in the text,   a1 a2  a · (b × c) =  b1 b2  c1 c2

a3 b3 c3

   c1     and (a × b) · c = c · (a × b) =  a1    b1 

c2 a2 b2

c3 a3 b3

   .  

CHAPTER 11. VECTORS AND 3-SPACE

762

The second determinant can be obtained from the first by an interchange of the second and third rows followed by an interchange of the new first and second rows. Using Property (iii) of determinates in Appendix I in the text, we see that a · (b × c) = (a × b) · c. 62. b × c = (b2 c3 − b3 c2 )i − (b1 c3 − b3 c1 )j + (b1 c2 − b2 c1 )k a × (b × c) = [a2 (b1 c2 − b2 c1 ) + a3 (b1 c3 − b3 c1 )]i − [a1 (b1 c2 − b2 c1 ) − a3 (b2 c3 − b3 c2 )]j + [−a1 (b1 c3 − b3 c1 ) − a2 (b2 c3 − b3 c2 )]k = (a2 b1 c2 − a2 b2 c1 + a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j − (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k (a × c)b − (a · b)c = (a1 c1 + a2 c2 + a3 c3 )(b1 i + b2 j + b3 k) − (a1 b1 + a2 b2 + a3 b3 )(c1 i + c2 j + c3 k) = (a2 b1 c2 − a2 b2 c1 + a3 b1 c3 − a3 b3 c1 )i − (a1 b1 c2 − a2 b2 c1 − a3 b2 c3 + a3 b3 c2 )j − (a1 b1 c3 − a1 b3 c1 + a2 b2 c3 − a2 b3 c2 )k 63. a × (b × c) + b × (c × a) + c × (a × b) = (a · c)b − (a · b)c − (b · c)a + (c · b)a − (c · a)b = [(a · c)b − (c · a)b] + [(b · a)c − (a · b)c] + [(c · b)a − (b · c)a] = 0 64. If either a, b or c is the zero vector, the result is trivial. Therefore, assume all three are nonzero. If b is a scalar multiple of c, then b × c = 0 and a · (b × c) = a · 0 = 0 If b is not a scalar multiple of c, then b × c is orthogonal to the plane containing b and c. This implies b × c is orthogonal to a since a lies in the same plane as b and c. Hence, by Theorem 11.3.3, we have a · (b × c) = 0. 65. (a) We first note that a × b = k, b × c = 12 (i − k), c × a = 1 1 1 2 , b · (c × a) = 2 , and c · (a × b) = 2 . Then A=

1 2 (i

− k) 1 2

= i − k, B =

1 2 (j

− k) 1 2

= j − k,

1 2 (j

− k), a · (b × c) =

and C =

k 1 2

= 2k.

(b) We need to compute A · (B × C). Using the formula from Problem 62 we have ([c × a) · b]a − [(c × a) · a]b (c × a) × (a × b) = [b · (c × a)][c × (a × b)] [b · (c × a)][c × (a × b)] a = since (c × a) · a = 0. c · (a × b

B×C=

Then A · (B × C) =

a 1 b×c · = a · (b × c) c · (a × b) c · (a × b)

and the volume of the unit cell of the reciprocal lattice is the reciprocal of the volume of the unit cell of the original lattice.

PROBLEMAS 11.5 Lines1.5in 3-Space 1. x, y, z = 4, 6, −7 + t3, 12 , − 32 

11.5. LINES IN 3-SPACE

763

2. x, y, z = 1, 8, −2 + t−7, −7, 0 3. x, y, z = 0, 0, 0 + t5, 9, 4 4. x, y, z = 0, −3, 10 + t12, −5, −6 −−→ −−→ The equation of a line through P1 and P2 is 3-space with r1 = OP1 and r2 = OP2 can be expressed as r = r1 + t(ka) or r = r2 + t(ka) where a = r2 − r1 and k is any non-zero scalar. Thus, the form of the equation of a line is not unique. (See the alternative solution to Problem 5.) 5. a = 1 − 3, 2 − 5, 1 − (−2) = −2, −3, 3;

x, y, z = 1, 2, 1 + t−2, −3, 3

Alternate Solution: a = −31, 5 − 2, −2 − 1 = 2, 3, −3; 6. a = 0 − (−2), 4 − 6, 5 − 3 = 2, −2, 2;

x, y, z = 3, 5, −2 + t2, 3, −3

x, y, z = 0, 4, 5 + t−2, −2, 2

7. a = 1/2 − (−3/2), −1/2 − 5/2, 1 − (−1/2) = 2, −3, 3/2; x, y, z = 1/2, −1/2, 1 + t2, −3, 3/2 8. a = 10 − 5, 2 − (−3), 10 − 5 = 5, 5, −15;

x, y, z = 10, 2, −10 + t5, 5, −15

9. a = 1 − (−4), 1 − 1, −1 − (−1) = 5, 0, 0;

x, y, z = 1, 1, −1 + t5, 0, 0

10. a = 3 − 5/2, 2 − 1, 1 − (−2) = 1/2, 1, 3;

x, y, z = 3, 2, 1 + t1/2, 1, 3

11. a = 2 − 6, 3 − (−1), 5 − 8 = −4, 4, −3; x = 2 − 4t, y = 3 + 4t, z = 5 − 3t 12. a = 2 − 0, 0 − 4, 0 − 9 = 2, −4, −9; x = 2 + 2t, y = −4t, z = −9t 13. a = 1 − 3, 0 − (−2), 0 − (−7) = −2, 2, 7; x = 1 − 2t, y = 2t, z = 7t 14. a = 0 − (−2), 0 − 4, 5 − 0 = 2, −4, 5; x = 2t, y = −4t, z = 5 + 5t 15. a = 4−(−)6, 1/2−(−1/4), 1/3−1/6 = 10, 3/4, 1/6; x = 4+10t, y =

1 3 1 1 + t, z = + t 2 4 3 6

16. a = −3 − 4, 7 − (−8), 9 − 1(−1) = −7, 15, 10; x = −3 − 7t, y = 7 + 15t, z = 9 + 10t 17. a1 = 10 − 1 = 9, a2 = 14 − 4 = 10, a3 = −2 − (−9) = 7;

y − 14 z+2 x − 10 = = 9 10 7

18. a1 = 1 − 2/3 = 1/3, a2 = 3 − 0 = 3, a3 = 1/4 − (1/4) = 1/2; 19. a1 = −7 − 4 = −11, a2 = 2 − 2 = 0, a3 = 5 − 1 = 4; 20. a1 = 1 − (−5) = 6, a2 = 1 − (−2), a3 = 2 − (−4);

y−3 z − 1/4 x−1 = = 1/3 3 1/2

z−5 x−7 = , y=2 −11 4

y−1 z−2 x−1 = = 6 3 6

21. a1 = 5 − 5 = 0, a2 = 10 − 1 = 9, a3 = −2 − (−14) = 12; x = 5,

z+2 y − 10 = 9 12

764

CHAPTER 11. VECTORS AND 3-SPACE

22. a1 = 5/6 − 1/3 = 1/2, a2 = −1/4 − 3/8 = 5/8, a3 = 1/5 − 1/10 = 1/10; x − 5/6 y + 1/4 z − 1/5 = = 1/2 −5/8 1/10 23. Writing the given line in the form x/2 = (y − 1)/(−3) = (z − 5)/6, we see that a direction vector is 2, −3, 6. Parametric equations for the lines are x = 6+2t, y = 4−3t, z = −2+6t. 24. A direction vector is 5, 1/3, −2. Symmetric equations for the line are (x − 4)/5 = (y + 11)(1/3) = (z + 7)/(−2). 25. A direction vector parallel to both the xy- and xy-planes is i = 1, 0, 0. Parametric equations for the line are x = 2 + t, y = −2, z = 15. 26. (a) Since the unit vector j = 0, 1, 0 lies along the y-axis, we have x = 1, y = 2 + t, z = 8. (b) Since the unit vector k = 0, 0, 1 is perpendicular to the xy-plane, we have z = 1, y = 2, z = 8 + t. 27. Both lines go through the points (0, 0, 0) and (6, 6, 6). Since two points determine a line, the lines are the same. 28. The direction vector of line L1 is v1 = 3, 6, −9. The direction vector of line L2 is v2 = −1, −2, 3. Since v1 = −3v2 , lines L1 and L2 are parallel. Hence, if we can find a point that lies on both lines, then they must be parallel. Letting t = 0 for L1 and t = 3 for L2 , we see that the point (2, −5, 4) lies on both lines. Therefore L1 and L2 are the same. −7 − 3 = −5. 29. (a) Equating the x components, we have x = 3 + 2t, = −7, which gives t = 2 We can check our work by plugging this value of t into the y and z components to get y = 4 − (−5) = 9 and z = −1 + 6(−5) = −31 (b) Equating the x components, we have x = 5 − x = −7 which gives s = 5 + 7 = 12. We can check our work by plugging this value of s into the y and z components to get 1 y = 3 + (12) = 9 and z = 5 − 3(12) = −31 2 30. a and f are parallel since 9, −12, 6 = −3−3, 4, 2. c and d are orthogonal since 2, −3, 4 · 1, 4, 5/2 = 0. 31. In the xy-plane, z = 9+3t = 0 and t = −3. Then x = 4−2(−3) = 10 and y = 1+2(−3) = −5. The point is (10, −5, 0). In the xz-plane, y = 1+2t = 0 and t = −1/2. Then x = 4−2(−1/2) = 5 and z = 9 + 3(−1/2) = 15/2. The point is (5, 0, 15/2). In the yz-plane, x = 4 − 2t = 0 and t = 2. Then y = 1 + 2(2) = 5 and z = 9 + 3(2) = 15. The point is (0, 5, 15). 32. The parametric equations for the line are x = 1+2t, y = −2+3t, z = 4+2t. In the xy-plane, z = 4 + 2t = 0 and t = −2. Then x = 1 + 2(−2) = −3 and y = −2 + 3(−2) = −8. The point is (−3, −8, 0). In the xz-plane, y = −2 + 3t = 0 and t = 2/3. Then x = 1 + 2(2/3) = 7/3 and z = 4 + 2(2/3) = 16/3. The point is (7/3, 0, 16/3). In the yz-plane, x = 1 + 2t = 0 and t = −1/2. Then y = −2 + 3(−1/2) = −7/2 and z = 4 + 2(−1/2) = 3. The point is (0, −7/2, 3).

11.5. LINES IN 3-SPACE

765

33. Solving the system 4 + t = 6 + 2s, 5 + t = 11 + 4s, −1 + 2t = −3 + s, or t − 2s = 2, t − 4s = 6, 2t − s = −2 yields s = −2 and t = −2 in all three equations. Thus, the lines intersect at the point x = 4 + (−2) = 2, y = 5 + (−2) = 3, z = −1 + 2(−2) = −5, or (2, 3, −5). 34. Solving the system 1 + t = 2 − s, 2 − t = 1 + s, 3t = 6s, or t + s = 1, t + s = 1, t − 2s = 0 yields s = 1/3 and t = 2/3 in all three equations. Thus, the lines intersect at the point x = 1 + 2/3 = 5/3, y = 2 − 2/3 = 4/3, z = 3(2/3) = 2, or (5/3, 4/3, 2). 35. The system of equations 2 − t = 4 + s, 3 + t = 1 + s, 1 + t = 1 − s, or t + s = −2, t − s = −2, t + s = 0 has no solution since −2 = 0. Thus, the lines do not intersect. 36. Solving the system 3 + t = 2 + 2s, 2 + t = −2 + 3s, 8 + 2t = −2 + 8s, or t + 2s = 1, t − 3s = −4, 2t − 8s = −10 yields s = 1 and t = −1 in all three equations. Thus, the lines intersect at the point x = 3 − (−1) = 4, y = 2 + (−1) = 1, z = 8 + 2(−1) = 6, or (4, 1, 6). 37. Using the first two points, we determine the line x = 4 + 6t, y = 3 + 12t, z = −5 − 6t. Letting t = −5/6 we see that (−1, −7, 0) is on the line. Thus, the points lie on the same line. 38. Using the first two points, we determine the line x = −1 − 12t, y = 6 + z = 6 − 8t.   4t,

5 when Setting x = −2 in the first equation, we obtain t = 1/4. Since z = 6 − 8 14 = 4 = t = 1/4, the points do not lie on the same line. 39. A direction vector for the line is 6−2, −1−5, 3−9 = 4, −6, −6. Thus, parametric equations for the line segment are x = 2 + 4t, y = 5 − 6t, z = 9 − 6t, where 0 ≤ t ≤ 1. 40. The midpoint of the first line segment, obtained by letting t = 3/2, is (4, 1/2, −1/2). The midpoint of the second line segment, obtained by letting t = 0, is (−2, 6, 5). A direction vector for the line segment connecting the midpoints is −2−4, 6−1/2, 5−(−1/2) = −6, 11/2, 11/2. Thus, parametric equations for the line segment are x = 4 − 6t, y = 1/2 + (11/2)t, z = −1/2 + (11/2)t, where 0 ≤ t ≤ 1. 41. a = −1, 2, −2, θ = arccos

b = 2, 3, −6,

16 ≈ 40.37◦ 21

a · b = 16,

|a| = 3,

|b| = 7;

cos θ =

16 a·b = ; |a||b| 3·7

√ √ 42. a = 2, 7, −1, b = −2, 1, 4, a · b = −1, |a| = 3  6, |b| = 21;  1 a·b 1 1 =− √ √ cos θ = = − √ ; θ = arccos − √ √ ≈ 91.70◦ |a||b| (3 6)( 21) 9 14 (3 6)( 21) 43. A direction vector perpendicular to the given lines will be 1, 1, 1 × −1, 1, −5 = −6, 3, 3. Equations of the lines are x = 4 − 6t, y = 1 + 3t, z = 6 + 3t. 44. The direction vectors of the given lines are 3, 2, 4 and 6, 4, 8 = 23, 2, 4. These are parallel, so we need a third vector parallel to the plane containing the lines which is not parallel to them. The point (1, −1, 0) is on the first line and (−4, 6, 10) is on the second line. A third vector is then 1, −1, 0 − −4, 6, 10 = 5, −7, −10. Now a direction vector perpendicular to the plane is 3, 2, 4 × 5, −, 7−, 10 = 8, 50, −31. Equations of the line through (1, −1, 0) and perpendicular to the plane are x = 1 + 8t, y = −1 + 50t, z = −31t.

CHAPTER 11. VECTORS AND 3-SPACE

766

45. In the system −3 + t = 4 + s, 7 + 3t = 8 − 2s, 5 + 2t = 10 − 4s, or t − s = 7, 3t + 2s = 1, 2t + 4s = 5, the first and second equations have solution t = 3 and s = −4. Substituting into the third equation, we find 2(3) = 4(−4) = 6 − 16 = −10 = 5. The direction vectors of the lines are 1, 3, 2 and 1, −2, −4, so the lines are not parallel. Thus, the lines are skew. 46. In the system 6 + 2t = 7 + 8s, 6t = 4 − 4s, −8 + 10t = 3 − 24s, or 2t − 8s = 1, 6t + 4s = 4, 10t + 24s = 11, the second and third equations have solution t = 1/2 and s = 1/4. Substituting into the first equation, we find 2(1/2) − 8(1/4) = −1 = 1. The direction vectors of the lines are 2, 6, 10 and 8, −4, , −24, so the lines are not parallel. Thus, the lines are skew. −−−→ −−−→ −−−→ −−−→ 47. The vector (P1 P2 × P3 P4 )/|P1 P2 × P3 P4 | is a unit vector perpendicular to the two planes. To find the shortest distance between the planes we compute the absolute value of the component −−−→ of P1 P3 on this unit vector. Then  −−−→  d =  P 1 P3 · 

−−−→ −−−→ −−−→ −−−→ −−−→  P1 P2 × P3 P4  |P1 P3 · P1 P2 × P3 P4 | −−−→ −−−→  = −−−→ −−−→ | P 1 P2 × P 3 P4 |  | P 1 P 2 × P 3 P4 |

−−−→ 48. We take P1 = (−3, 7, 5), P2 = (−2, 10, 7), P3 + (4, 8, 10), and P4 = (5, 6, 6). Then P1 P3 = −−−→ −−−→ −−−→ −−−→ 7, 1, 5, P1 P2 = 1, 3, 2, P3 P4 = 1, −2, −4, and P1 P2 × P3 P4 = −8, 6, −5. The distance between the lines is then d=

√ 75 |7, 1, 5 · −8, 6, −5| = √ = 3 5. |−8, 6, −5| 5 5

PROBLEMAS 11.6 Planes1.6 1. 2(x − 5) − 3(y − 1) + 4(z − 3) = 0;

2x − 3y + 4z = 19

2. 4(x − 1) − 2(y − 2) + 0(z − 5) = 0;

4x − 2y = 0

3. −5(x − 6) + 0(y − 10) + 3(z + 7) = 0;

−5x + 3z = −51

4. 6x − y + 3z = 0 5. 6(x − 1/2) + 8(y − 3/4) − 4(z − 1/2) = 0; 6. −(x + 1) + (y − 1) − (z − 0) = 0;

6x + 8y − 4z = 11

−x + y − z = 2

7. From the points (3, 5, 2) and (2, 3, 1) we obtain the vector u = i + 2j + k. From the points (2, 3, 1) and (−1, −1, 4) we obtain the vector v = 3i + 4j − 3k. From the points (−1, −1, 4) and (x, y, z) we obtain the vector w = (x + 1)i + (y + 1)j + (z − 4)k. Then, a normal vector is   i  u × v =  1  3

j 2 4

k 1 −3

    = −10i + 6j − 2k  

A vector equation of the plane is −10(x + 1) + 6(y + 1) − 2(z − 4) = 0 or 5x − 3y + z = 2.

11.6. PLANES

767

8. From the points (0, 1, 0) and (0, 1, 1) we obtain the vector u = k. From the points (0, 1, 1) and (1, 3, −1) we obtain the vector v = i + 2j − 2k. From the points (1, 3, −1) and (x, y, z) we obtain the vector w = (x − 1)i + (y − 3)j + (z + 1)k. Then, a normal vector is   i  u × v =  0  1

j 0 2

k 1 −2

    = −2i + j  

A vector equation of the plane is −2(x − 1) + (y − 3) + 0(z + 1) = 0 or −2x + y = 1. 9. From the points (0, 0, 0) and (1, 1, 1) we obtain the vector u = i + j + k. From the points (1, 1, 1) and (3, 2, −1) we obtain the vector v = 2i + j − 2k. From the points (3, 2, −1) and (x, y, z) we obtain the vector w = (x − 3)i + (y − 2)j + (z + 1)k. Then, a normal vector is   i  u × v =  1  2

j 1 1

k 1 −2

    = −3i + 4j − k  

A vector equation of the plane is −3(x − 3) + 4(y − 2) − (z + 1) = 0 or −3x + 4y − z = 0. 10. The three points are not collinear and all satisfy x = 0, which is the equation of the plane. 11. From the points (1, 2, −1) and (4, 3, 1) we obtain the vector u = 3i + j + 2k. From the points (4, 3, 1) and (7, 4, 3) we obtain the vector v = 3i + j + 2k. From the points (7, 4, 3) and (x, y, z) we obtain the vector w = (x − 7)i + (y − 4)j + (z − 3)k. Since u × v = 0, the points are collinear. 12. From the points (2, 1, 2) and (4, 1, 0) we obtain the vector u = 2i−2k. From the points (4, 1, 0) and (5, 0, −5) we obtain the vector v = i − j − 5k. From the points (5, 0, −5) and (x, y, z) we obtain the vector w = (x − 5)i + yj + (z + 5)k. Then, a normal vector is   i  u × v =  2  1

j 0 −1

k −2 −5

    = −2i + 8j − 2k  

A vector equation of the plane is −2(x − 5) + 8y − 2(z + 5) = 0 or x − 4y = z = 0. 13. A normal vector to x + y − 4z = 1 is 1, 1, −4. The equation of the parallel plane is (x − 2) + (y − 3) − 4(z + 5) = 0 or x + y − 4z = 25. 14. A normal vector to 5x − y + z = 6 is 5, −1, 1. The equation of the parallel plane is 5(x − 0) − (y − 0) + (z − 0) = 0 or 5x − y + z = 0. 15. A normal vector to the xy-plane is 0, 0, 1. The equation of the parallel plane is z − 12 = 0 or z = 12. 16. A normal vector is 0, 1, 0. The equation of the plane is y + 5 = 0 or y = −5.

768

CHAPTER 11. VECTORS AND 3-SPACE

17. Direction vectors of the lines are 3, −1, 1 and 4, 2, 1. A normal vector to the plane is 3, −1, 1 × 4, 2, 1 = −3, 1, 10. A point on the first line, and thus in the plane, is (1, 1, 2). The equation of the plane is −3(x − 1) + (y − 1) + 10(z − 2) = 0 or −3x + y + 10z = 18. 18. Direction vectors of the lines are 2, −1, 6 and 1, 1, −3. A normal vector to the plane is 2, −1, 6×1, 1, −3 = −3, 12, 3. A point on the first line, and thus in the plane, is (1, −1, 5). The equation of the plane is −3(x − 1) + 12(y + 1) + 3(z − 5) = 0 or −x + 4y + z = 0. 19. A direction vector for the two lines is 1, 2, 1. Points on the lines are (1, 1, 3) and (3, 0, −2). Thus, another vector parallel to the plane is 1 − 3, 1 − 0, 3 + 2 = −2, 1, 5. A normal vector to the plane is 1, 2, 1 × −2, 1, 5 = 9, −7, 5. Using the point (3, 0, −2) in the plane, the equation of the plane is 9(x − 3) − 7(y − 0) + 5(z + 2) = 0 or 9x − 7y + 5z = 17. 20. A direction vector for the line is 3, 2, −1. Letting t = 0, we see that the origin is on the line and hence in the plane. Thus, another vector parallel to the plane is 4 − 0, 0 − 0, −6 − 0 = 4, 0, −6. A normal vector to the plane is 3, 2, −2 × 4, 0, −6 = −12, 10, −8. The equation of the plane is −12(x − 0) + 10(y − 0) − 8(z − 0) = 0 or 6x − 5y + 4z = 0. 21. A direction vector for the line, and hence a normal vector for the plane, is −3, 1, −1/2. The equation of the plane is −3(x − 2) + (y − 4) − 12 (z − 8) = 0 or −3x + y − 12 z = −6. 22. A normal vector to the plane is 2 − 1, 6 − 0, −3 + 2 = 1, 6, −1. The equation of the plane is (x − 1) + 6(y − 1) − (z − 1) = 0 or x + 6y − z = 6. 23. Normal vectors to the plane are (a) 2, −1, 3, (b) 1, 2, 2, (c) 1, 1, −3/2, (d) −5, 2, 4, (e) −8, −8, 12, (f ) −2, 1, −2. Parallel planes are (c) and (e), and (a) and (f ). Perpendicular planes are (a) and (d), (b) and (c), (b) and (e), and (d) and (f ). 24. A normal vector to the plane is −7, 2, 3. This is the direction vector for the line and the equations of the line are x − 4 − 7t, y = 1 + 2t, z = 7 + 3t. 25. A direction vector of the line is −6, 9, 3, and the normal vectors of the plane are (a) 4, 1, 2, (b) 2, −3, 1, (c) 10, −15, −5, (d) −4, 6, 2. Vectors (c) and (d) are multiples of the direction vector and hence the corresponding planes are perpendicular to the line. 26. A direction vector of the line is −2, 4, 1, and the normal vectors of the plane are (a) 1, −1, 3, (b) 6, −3, 0, (c) 1, −2, 5, (d) −2, 1, −2. Since the dot product of each normal vector with the direction vector is non-zero, none of the planes are parallel to the line. 27. Letting z = t in both equations and solving 5x − 4y = 8 + 9t, x = 2 + t, y = 12 − t, z = t.

x + 4y = 4 − 3t, we obtain

28. Letting y = t in both equations and solving x − z = 2 − 2t, 3x + 2z = 1 + t, we obtain x = 1 − 35 t, y = t, z = −1 + 75 t or, letting t = 5s, x = 1 − 3s, y = 5s, z = −1 + 7s. 29. Letting z = t in both equations and solving 4x − 2y = 1 + t, x = 12 − 12 t, y = 12 − 32 t, z = t.

x + y = 1 − 2t, we obtain

30. Letting z = t and using y = 0 in the first equation, we obtain x = − 12 t, y = 0, z = t.

11.6. PLANES

769

31. Substituting the parametric equations into the equation of the plane, we obtain 2(1+2t)23(2− t) + 2(−3t) = −7 or t = −3. Letting t = −3 in the equation of the line, we obtain the point of intersection (−5, 5, 9). 32. Substituting the parametric equations into the equation of the plane, we obtain (3 − 2t) + (1 + 6t) − 4(2 − 12 ) = 12 or 2t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (3, 1, 2). 33. Substituting the parametric equations into the equation of the plane, we obtain 1+2−(1+t) = 8 or t = −6. Letting t = −6 in the equation of the line, we obtain the point of intersection (1, 2, −5). 34. Substituting the parametric equations into the equation of the plane, we obtain 4 + t − 3(2 + t) + 2(1 + 5t) = 0 or t = 0. Letting t = 0 in the equation of the line, we obtain the point of intersection (4, 2, 1). In Problems 35 and 26, the cross product of the normal vectors to the two planes will be a vector parallel to both planes, and hence a direction vector for a line parallel to the two planes. 35. Normal vectors are 1, 1, −4 and 2, −1, 1. A direction vector is 1, 1, −4 × 2, −1, 1 = −3, −9, −3 = −31, 3, 1. Equations of the line are x = 5 + t,

y = 6 + 3t,

z = −12 + t.

36. Normal vectors are 2, 0, 1 and −1, 3, 1. A direction vector is 2, 0, 1 × −1, 3, 1 = −3, −3, 6 = −31, 1, −2. Equations of the line are x = −3 + t, y = 5 + t, z = −1 − 2t. In Problems 37 and 38, the cross product of the direction vector of the line with the normal vector of the given plane will be a normal vector to the desired plane. 37. A direction vector of the line is 3, −1, 5 and a normal vector to the given plane is 1, 1, 1. A normal vector to the desired plane is 3, −1, 5 × 1, 1, 1 = −6, 2, 4. A point on the line, and hence in the plane is 4, 0, 1. The equation of the plane is −6(x − 4) + 2(y − 0) + 4(z − 1) = 0 or 3x − y − 2x = 10. 38. A direction vector of the line is 3, 5, 2 and a normal vector to the given plane is 2, −4, −1. A normal vector to the desired plane is −3, 5, 2 × 2, −4, −1 = 3, 1, 2. A point on the line, and hence in the plane is 2, −2, 8. The equation of the plane is 3(x−2)+(y +2)+2(z −8) = 0 or 3x + y + 2x = 20.

CHAPTER 11. VECTORS AND 3-SPACE

770

40.

39.

z

z 10

6

y x

6

2

5

y

x

41.

42. z

z

2

4

y 6

4

y

x x

-6

43.

44. z

z

6

4 2 y

1 x

2

y x

45. (a) A direction vector for the line is a = −2i + j − k and a normal vector for the plane is n = i + j − k. Since a · n = −2 + 1 + 1 = 0, the line is perpendicular to n and thus parallel

11.6. PLANES

771

to the plane. Since (0, 0, 0) is on the line and (0, 0, −1) is in the plane, the line is above the plane. (b) A normal vector for the plane is n − 3i − 4j + 2k. Since a · n = 6 − 4 − 2 = 0, the line is parallel to the plane. Since (0, 0, 0) is one the line and (0, 0, 4) is in the plane, the line is below the plane. −−−→ 46. The distance D will be the absolute value of compn P0 P1 . Thus, using ax1 + by1 + cz1 = −d,     −−−→ n   x2 − x1 , y2 − y1 , z2 − z1  · a, b, c  =  √ D = P0 P1 ·  |n|   a 2 + b2 + c 2 |ax2 + by2 + cz2 − (zx1 + by1 + cz1 )| |ax2 + by2 + cz2 + d| √ √ = = . a 2 + b2 + c 2 a 2 + b2 + c 2 47. Using, Problem 46, D =

|1(2) − 3(1) + 1(4) − 6| 3 √ =√ . 1+9+1 11

48. (a) The normal vectors are n1 = i − 2j + 3k and n2 = −4i + 8j − 12k = −4n. Since n1 and n2 are parallel, the planes are parallel. (b) To find the distance between the planes we choose (0, 0, 1) on the first plane. Then, using Problem 46, the distance between the planes is D=

19 | − 4(0) + +8(0) − 12(1) − 7|  ≈ 1.27. =√ 2 2 2 224 (−4) + 8 + (−12)

49. Normal vectors are 1, −3, 2 and −1, 1, 1. Then cos θ =

−2 2 1, −3, 2 · −1, 1, 1 = √ √ = −√ |1, −3, 2||−1, 1, 1| 14 3 42

√ and θ = arccos(−2/ 42) ≈ 107.98◦ 50. Normal vectors are , 2, 6, 3 and 4, −2, 4. Then cos θ =

8 4 2, 6, 3 · 4, −2, 4 = = |2, 6, 3||4, −2, 4| 7(6) 21

and θ = arccos(−4/21) ≈ 79.02◦ 51. Let the bottoms of the table legs be represented by points in 3-space. The rocking of a fourlegged table occurs when these four points are not coplanar. Hence, not all four legs can rest on the plane of the floor simultaneously. However, a three-legged table cannot have this problem. Given any three points in space, a plane can be found passing through them. Therefore, the bottoms of the legs in a three-legged table are coplanar. This implies that they will all rest on the plane of the floor, even if the legs are of uneven lengths.

CHAPTER 11. VECTORS AND 3-SPACE

772

52. Let n1 = 1, −1, 2 which is normal to the plane x − y + 2z = 1. Let n2 = 1, 1, 1 which is normal to the plane x + y + z = 3. Since L is perpendicular to both n1 and n2 , L must be parallel to n1 × n2 .   i  n1 × n2 =  1  1

j −1 1

k 2 1

    − 3i + j + 2k  

Therefore, L is parallel to v = −3, 1, 2. To completely determine L, we need a point which L passes through. Hence we need a point (x, y, z) which satisfies the equations of both planes. Since it satisfies both equations, it must satisfy their sum: x − y + 2z = 1 x+y+z =3 2x + 3z = 4 So the coordinates of the point satisfy 2x + 3z = 4. Since v as a nonzero k-component, the line L passes through every possible z-value. This implies the existence of a point on the intersection of the two planes with a z-value of zero. Letting z = 0, we must have x = 2 since 2x + 3z = 4 for every point on L. Plugging x = 2 and z = 0 into the equation of the first place, we get y = 1. Therefore (2, 1, 0) lies on the line L. Using this point and the parallel vector v, the parametric equations of L are x = 2 − 3t; y = 1 + t; z = 2t To show that this answer is equivalent to that found in Example 8, first note that both lines 1 pass through (2, 1, 0). Also, the parallel vector used in Example 8 is −3/2, 1/2, 1 = v. 2 Therefore, the two solutions are the same.

53. (a) The plane should pass through the midpoint of the −2, 3) and  line segment joining (1,   1 + 2 −2 + 5 3 − 1 3 3 (2, 5, −2). This is given in Problem 11.2.64 as M = , , , ,1 . = 2 2 2 2 2 The vector joining (1, −2, 3) and (2, 5, −1) should be perpendicular to the plane. This vector is n = 1, 7, −4. Using the point (3/2, 3/2, 1) and the normal vector n, the equation of the plane is given by z + 7y − 4z = 8. (b) The distance from the plane to either of the two points is equal to half the length of the 1 2 1√ line segment joining the two points. This is given by 1 + 72 + (−4)2 = 66 2 2

11.7. CYLINDERS AND SPHERES

773

PROBLEMAS 1.7 11.7 Cylinders and Spheres 1.

2.

z

3.

z

z

y

y y

2

x

x

x

4.

5.

6. z

z

z

5

3

y

y y

x

x

x

7.

8.

9. z

z

z

y

y 1 y x

x x

CHAPTER 11. VECTORS AND 3-SPACE

774

11.

10.

12.

z

z

z

6 y

3 y

x

y

1 x x

13.

14.

15.

z

z

z

y

y y

x

x x

16.

17.

z

18. z

z

3 y

y x

y x

x

center: (0, 0, 3) radius: 4

11.7. CYLINDERS AND SPHERES

775

19.

20. z

z

y

x

y

x

center: (−3, −4, 5) radius: 2

center: (1, 1, 1) radius: 1

21. (x2 + 8x + 16) + (y 2 − 6x + 9) + (z 2 − 4z + 4) = 7 + 16 + 9 + 4 (x + 4)2 + (y − 3)2 + (z − 2)2 = 36; center: (−4, 3, 2); radius: 6 22. 4(x2 + x + 1/4) + 4y 2 + 4(z 2 − 3z + 9/4) = −9 + 1 + 9 (x + 1/2)2 + y 2 + (z − 3/2)2 = 1/4; center: (−1/2, 0, 3/2); 23. x2 + y 2 + (z 2 − 16z = 64) = 64; 2

center: (0, 0, 8);

2

2

24. (x − x + 1/4) + (y + y + 1/4)√+ z = 1/4 + 1/4; center: (1/2, −1/2, 0); radius: 2/2

radius: 1/2

radius: 8 (x − 1/2)2 + (y + 1/2)2 + z 2 = 1/2

25. (x + 1)2 + (y − 4)2 + (z − 6)2 = 3 26. x2 + (y − 3)2 + z 2 = 25/16 27. (x − 1)2 + (y − 1)2 + (z − 4)2 = 16 28. (x − 5)2 + (y − 2)2 + (z − 2)2 = 52 29. There are two solutions: one sphere is inside the given sphere and the other is outside. x2 + (y − 8)2 + z 2 = 4 or x2 + (y − 4)2 + z 2 = 4.  30. (2t)2 + (3t)2 + (6t)2 = 21; t = 3; a = 2t = 6; b = 3t = 9; c = 6t = 18 (x − 6)2 + (y − 9)2 + (z − 18)2 = 25  √ 31. The center is at (1, 4, 2) and the radius is (1 − 0)2 + (4 + 4)2 (2 − 7)2 = 3 10. The equation is (x − 1)2 + (y − 4)2 + (z − 2)2 = 90.  √ 32. The radius is (−3 − 0)2 + (1 − 0)2 + (2 − 0)2 = 14. The equation is (x + 3)2 + (y − 1)2 + (z − 2)2 = 14. 33. The upper half of the sphere x2 + y 2 + (z − 1)2 = 4; a hemisphere 34. A circle√on the sphere x2 + y 2 + (z − 1)2 = 4; the circle is parallel to the xy-plane and has radius 3. 35. All points on and outside the unit sphere centered at the origin

CHAPTER 11. VECTORS AND 3-SPACE

776

36. All points inside the sphere of radius 1 centered at (1, 2, 3), except the center 37. x2 + y 2 + z 2 = 1 represents a sphere of radius 1 and x2 + y 2 + z 2 = 9 represents a sphere of radius 3. Therefore 1 ≤ x2 + y 2 + z 2 ≤ 9 represents the set of points lying between these two spheres. Thus, the geometric object is a hollowed out ball with outer radius 3 and inner radius 1. 38. This set of points is identical to the found in Problem 11.7.37, with the added restriction z ≥ 0. This restriction will remove points with negative z-coordinates, leaving only the upper half of the hollowed out ball.

PROBLEMAS 1.8 Surfaces 11.8 Quadric 1. paraboloid

2. elliptical cone z

z

y x

y x

3. x2 /4 + y 2 + z 2 /9 = 1; ellipsoid

4. −x2 /4 − y 2 /4 + z 2 /4 = 1 hyperboloid of two sheets

z z

y x

x

y

11.8. QUADRIC SURFACES

777

5. x2 /4 − y 2 /144 + z 2 /16 = 1 hyperboloid of one sheet

6. x2 /25 + y 2 /25 + z 2 /100 = 1 ellipsoid

z

z 10

4

y

5

x

y

5

x

8. y 2 /9 − x2 /16 = z hyperbolic paraboloid

7. elliptical cone

z

z

y

y

x

x

10. x2 + y 2 = −9z paraboloid

9. hyperbolic paraboloid

z

z

y y

x x

CHAPTER 11. VECTORS AND 3-SPACE

778 11. x2 /4 − y 2 /4 − z 2 /4 = 1 hyperboloid of two sheets

12. −z 2 /9 + y 2 + z 2 /9 = 1 hyperboloid of one sheet

z

z

3

y

y

1

x

x

z2 =x 1/4 paraboloid

13. y 2 +

14. hyperboloid of one sheet z

z

y

y x

x

15.

16. z

z

4 x

3 y x

y

11.8. QUADRIC SURFACES

17.

779

18.

z

z

10 y y

x

x

 19. The equation can be written as x2 + (± y 2 + z 2 )2 = 1. The surface is generated by revolving the circles x2 + y 2 = 1 or x2 + z 2 = 1 about the x-axis. [Alternatively, the surface is generated by revolving the circles x2 + y 2 = 1 or y 2 + z 2 = 1 about the y-axis, or the circles x2 + z 2 = 1 or y 2 + z 2 = 1 about the z-axis.]  20. The equation can be written as −9x2 + (±2 y 2 + z 2 02 = 36. The surface is generated by revolving the hyperbolas = 9x2 + 4y 2 = 36 or −9x2 + 4z 2 = 36 about the x-axis. √

2

2 2

21. The equation can be written as y = e± x +z ) . The surface is generated by revolving the 2 2 curves y = ex or y = ez about the y-axis.  22. The equation can be written as (± x2 + y 2 )2 = sin2 z. The surface is generated by revolving the curves x2 = sin2 z or y 2 = sin2 z about the z-axis.  √ 23. Replacing x by ± x2 + y 2 we have y = ±2 x2 + z 2 or y 2 = 4x2 + 4z 2 . √ √ 24. Replacing z by x2 + z 2 we have y = ( x2 + z 2 )1/2 or y 4 = x2 + z 2 ; y ≥ 0. 25. Replacing z by ±

  y 2 + z 2 we have ± y 2 + z 2 = 9 − x2 or y 2 + z 2 = (9 − x2 )2 , x ≥ 0.

26. Replacing y by

  x2 + y 2 we have z = 1 + ( x2 + y 2 )2 or z = 1 + x2 + y 2 .

27. Replacing z by

  y 2 + z 2 we have x2 − (± y 2 + z 2 )2 = 4 or x2 − y 2 − z 2 = 4.

28. Replacing x by ± 29. Replacing y by





30. Replacing y by ±

x2 + y 2 we have 3(±



x2 + y 2 we have z = ln



y 2 + z 2 we have x(±

x2 + y 2 )2 + 4z 2 = 12 or 3x2 + 3y 2 + 4z 2 = 12.





x2 + y 2 .

y 2 + z 2 ) = 1 or x2 (y 2 + z 2 ) = 1.

31. The surface is Problem 11 is a surface of revolution about the x-axis. The surface in Problem 2 is a surface of revolution about the y-axis. The surface is Problems 1, 4, 6, 10, and 14 are surfaces of revolution about the z

CHAPTER 11. VECTORS AND 3-SPACE

780 32. z

π

1

y

x

33. The first equation is the lower nappe of the cone (z + 2)2 = x2 + y 2 whose axis of revolution is the z-axis and whose vertex is at (0, 0, −2). 34. The first equation is the right-hand of the cone (y − 1)2 = x2 + z 2 whose axis of revolution is the y-axis and whose vertex is at (0, 1, 0). 2 2 2 2 35. (a) Writing the equation of the ellipse in√the form √ x /(c − z)a + y /(c − z)b = 1 we see that the area of a cross-section is πa c − zb c − z = πab(c − z). c   c 1 (b) V = 0 πab(c − x)dz = πab − 12 (c − z)2 0 = πabc2 2

36. (a) Using the formula for the area of an ellipse given in Problem 35(a) we see that a horizontal cross-sectional area of the ellipsoid is πab(1 − z 2 /c2 ). Then

V =2

c 0



z πab 1 − 2 dz = 2πab c

c 4 z 3  z − 2  = πabc. 3c 3 0



(b) When a = b = c the volume is 43 πa3 , which is the formula for the volume of a sphere. 37. Expressing the line in the form (x − 2)/4 = (y + 2)/(−6) = (z − 6)/3 we see that parametric equations for the line are x = 2 + 4t, y = −2 = 6t, z = 6 + 3t. Writing the equation of the ellipse as 36x2 + 9y 2 + 4z 2 = 324 and substituting, we obtain 36(2 + 4t)2 + 9(−2 − 6t)2 + 4(6 + 3t)2 = 936t2 + 936t + 324 or 936t(t + 1) = 0. When t = 0 we obtain the point (2, −2, 6), and when t = −1 we obtain the point (−2, 4, 3).

PROBLEMAS DEin REPASO DE LA UNIDAD 1 Chapter 11 Review A. True/False 1. True 2. False; the points must be non-collinear. 3. False; since a normal to the plane is 2, 3, −4 which is not a multiple of the direction vector 5, −2, 2 of the line.

CHAPTER 11 IN REVIEW 4. True 7. True

781 5. True 8. True

6. True 9. True

10. True; since a × b and c × d are both normal to the plane and hence parallel (unless a × b = 0 or c × d.) 11. True. The normal vector of the first plane is 1, 2, −1 while the normal vector of the second plane is −2, −4, 2. Since the second vector is a scalar multiple of the first, the planes are parallel. 12. False. Look at Figure 11.5.3 in the text. 13. True. This is a parabolic cylinder similar to that shown in Figure 11.7.6. 14. True. In the yz-plane, we have x = 0. Therefore, the equation of the surface becomes 1 or y 2 + z 2 = 2.

y2 2

2

+ z2 =

15. False. Find the equation of the plane containing the first three points, P1 (0, 1, 2), P2 (1, −1, 1), −−−→ −−−→ and P3 (3, 2, 6). This plane must contain the   vectors P1 P2 = 1, −2, 1 and P1 P3 = 3, 1, 4.  i j k  −−−→ −−−→  Define n = P1 P2 × P1 P3 =  1 −2 −1  = −7, −7, −7. Then n must be normal to the  3 1 4  plane. Using n and the point P1 , the equation of the plane becomes −7x − 7y + 7z = 7 or z +y −z = −1. The fourth point P4 (2, 1, 2) does not lie on the plane since (2)+(1)−(2) = −1. 16. True 17. False. The trace in the yz-plane is described by the equation 9y 2 + z 2 = 1 which represents an ellipse. 18. True. This ellipsoid results from revolving the graph of the ellipse x2 + 9y 2 = 1 about the y-axis. 19. True. |a × b| = |a||b|| sin θ| = |a||b| since θ = 90◦ 20. False. Let a = i,

b = j, and c = k. Then a · b = a · c = 0 but b = c.

B. Fill in the Blanks 1. 9i + 2j + 2k 2. orthogonal 3. −5(k × j) = −5(−i) = 5i 4. i · (i × j) = i × k = 0  5. (−12)2 + 42 + 62 = 14 6. k × (i + 2j − 5k) = k × i + 2(k × j) − 5(k × k) = j − 2i − 5(0) = −2, 1, 0

CHAPTER 11. VECTORS AND 3-SPACE

782   2 7.  4

 −5  = 2(3) − (−5)(4) = 6 + 20 = 26 3 

8. (−1 − 20)i − (−2 − 0)j + (8 − 0)k = −21i + 2j + 8k 9. −6i + j − 7k 10. The smallest component is the j-component with magnitude 3. Therefore, the sphere cannot have a radius larger than 3 or its interior will intersect the xz-plane. Thus, we need a sphere with radius 3 and center (4, 3, 7). The equation is (x − 4)2 + (y − 3)2 + (z − 7)2 = 9 11. Writing the line in parametric form, we have x = 1 + t, y = −2 + 3t z = −1 + 2t. Substituting into the equation of the plane yields (1 + t) + 2(−2 + 3t) − (−1 + 2t) = 13 or t = 3. Thus, the point of intersection is x = 1 + 3, y = −2 + 3(3) = 7, z = −1 + 2(3) = 5, or (4, 7, 5). 12. |a| =



√ 42 + 32 + (−5)2 = 5 2;

13. x2 − 2 = 3,

x2 = 5;

4 3 1 1 u = − √ (4i + 3j − 5k) = − √ i − √ j + √ k 5 2 5 2 5 2 2

y2 − 1 = 5,

y2 = 6;

z2 − 7 = −4,

z2 = 3;

P2 = (5, 6, 3)

14. (5, 1/2, 5/2) √ 15. (7.2)(10) cos 135◦ = −36 2 16. 2b = −2, 4, 2;

4c = 0, −8, 8;

a · (2b + 4c) = 3, 1, 0 · −2, −4, 10 = −10

17. 12, −8, 6 18. cos θ = 19. A =

a·b 1 = √ √ = 1/2; |a||b| 2 2

θ = 60◦

√ 1 |5i − 4j − 7k| = 3 10/2 2

20. (x + 5)2 + (y − 7)2 + (z + 9)2 = 6 21. | − 5 − (−3)| = 2 22. parallel: −2c = 5,

c = −5/2; orthogonal: 1(−2) + 3(−6) + c(5) = 0,

c=4

23. The equation can be transformed into something more recognizable by completing the square: x2 + 2y 2 + 2z 2 − 4y − 12z = 0 =⇒ x2 + 2(y 2 − 2y) + 2(z 2 − 6z) = 0 =⇒ x2 + 2(y 2 − 2y + 1) + 2(z 2 − 6z = 9) = 20 =⇒ x2 + 2(y − 1)2 + 2(z − 3)2 = 20 This is the equation of an ellipsoid centered at (0, 1, 3). 24. Letting z = 1, the trace is described by the equation y = x2 − 1, which is a parabola.

CHAPTER 11 IN REVIEW

783

C. Exercises

       i j k    1 0   1  1 0  +     i− 1. a × b =  1 1 0  =  1 1   1 −2 1   1 −2 1  A unit vector perpendicular to both a and b is

 1  k = i − j + 3k −2 

a×b 1 1 3 1 =√ (i − j − 3k = √ i − √ j − √ k. |a × b| 1+1+9 11 11 11    √  2  2 2 1 1 3 3 1 2. The magnitude of a is given by |a| = = . Letting + + − = 2 2 2 4 2 α, β, and γ represent the  1 angles between a and i, j,1 and k respectively, we have cos α = 1 − 1 1 1

√2  = √ , cos β = √2  = √ , and cos γ = √ 2  = − √ . From this we are able to 3 3 3 3 3 3 2 2 2  1 compute: α = cos−1 √ ≈ 0.95532 3   1 β = cos−1 √ ≈ 0.95532  3  1 ≈ 2.18628 γ = cos−1 − √ 3 3. compb a = a · b/|b| = 1, 2, −2 · 4, 3, 0/5 = 2 4. compa b = b · a/|a| = 4, 3, 0 · 1, 2, −2/3 = 10/3 proja b = (compa ba/|a| = (10/3)1, 2, −2/3 = 10/9, 20/9, −20/9 √ 5. First we compute 2a = 2, 4, −4, |b| = 16 + 9 = 5, and 2a · b = 20. So projb 2a = 2a · b 16 12 20 4, 3, 0 =  , , 0. b= 2 |b| 25 5 5 6. compb (a − b) = (a − b) · b/|b| = −3, −1, −2 · 4, 3, 0/5 = −3 projb (a − b) = (compb (a − b))b/|b| = −34, 3, 0/5 = −12/5, −9/5, 0 projb⊥ (a−b) = (a−b)−projb (a−b) = −3, −1, −2−−12/5, −9/5, 0 = −3/5, 4/5, −10/5 7.

x2 y2 + = 1; 16 4

8.

x2 1 + z 2 = − y; 2 4

9. − 10.

elliptical cylinder paraboloid

y2 z2 x2 − + = 1; 9 9/4 9

x2 y2 (z − 5)2 + + = 1; 25 25 25

11. x2 − y 2 = 9z; 12. plane

hyperboloid of two sheets sphere

hyperbolic paraboloid

CHAPTER 11. VECTORS AND 3-SPACE

784

√ √ 13. Replacing x by ± x2 + z 2 we have (± x2 + z 2 )2 − y 2 = 1 or x2 + z 2 − y 2 = 1, which is a 2 2 2 hyperboloid of one sheet. Replacing y by (± y + z ) we have x − (± y 2 + z 2 )2 = 1 or x2 − y 2 − z 2 = 1, which is a hyperboloid of two sheets. 14. The surface is generated by revolving y = 1+x, x ≥ 0, about the y-axis or √ y = 1+z, z ≥ 0 about the z-axis. The restrictions on x and z are required since y = 1 + x2 + z 2 ≥ 1. 15. Let a = a, b, c and r = x, y, z. Then (a) (r − a) · r = x − a, y − b, z − c · x, y, z = x2 − ax + y 2 − by + z 2 − ac = 0 implies  2

b c 2 a 2 + b2 + c 2 a 2 . The surface is a sphere. + y− + z− = x− 2 x 2 4 (b) (r − a) · a = x − a, y − b, z − c · a, b, c = a(x − a) + b(y − b) + c(z − c) = 0 The surface is a plane.

16. 4, 2, −2 − 2, 4, −3 = 2, −2, 1; 2, 4, −3 − 6, 7, −5 = −4, −3, 2; 2, −2, 1 · −4, −3, 2 = 0. The points are the vertices of a right triangle. 17. A direction vector of the given line is 4, −2, 6. A parallel line containing (7, 3, −5) is (x − 7)/4 = (y − 3)/(−2) = (z + 5)/6. 18. A normal to the plane is 8, 3, −4. The line with this direction vector and through (5, −9, 3) is x = 5 + 8t, y = −9 + 3t, z = 3 − 4t. 19. The direction vectors are −2, 3, 1 and 2, 1, 1. Since −2, 3, 1 · 2, 1, 1 = 0, the lines are orthogonal. Solving 1 − 2t = x = 1 + 2s, 3t = y = −4 + s, we obtain t = −1 and s = 1. The point (3, −3, 0) obtained by letting t = −1 and s = 1 is common to the two lines, so they do intersect. 20. Vectors in the plane are 2, 3, 1 and 1, 0, 2. A normal vector is 2, 3, 1×1, 0, 2 = 6, −3, −3 = 32, −1, −1. An equation of the plane is 2x − y − z = 0. 21. The lines are parallel with direction vector 1, 4, −2. Since (0, 0, 0) is on the first line and (1, 1, 3) is on the second line, the vector 1, 1, 3 is in the plane. A normal vector to the plane is thus 1, 4, −2 × 1, 1, 3 = 14, −5, −3. An equation of the plane is 14x − 5y − 3z = 0. 22. Letting z = t in the equations of the plane and solving −x + y = 4 + 8t, 3x − y = −2t, we obtain x = 2 + 3t, y = 6 + 11t, z = t. Thus, a normal to the plane is 3, 11, 1 and an equation of the plane is 3(x − 1) + 11(y − 7) + (z + 1) = 0 or 3x + 11y + z = 79. 23. A normal vector is (i − 2j) × (2i + 3k) = −6i − 3j + 4k. Thus, an equation of the plane is −6(z − 1) − 3(y + 1) + 4(z − 2) = 0 or 6x + 3y − 4z = −5. 24. The points at the ends of the diameter, obtained from t = −1 and t = 0 are (2, 4, 2) and (4, 7, 8). The center of the √ sphere is the midpoint of the line segment or (3, 11/2, 5). The diameter of the sphere is 22 + 32 + 62 = 7. The equation is (x−3)2 +(y −11/2)2 +(z −5)2 = 49/4.

CHAPTER 11 IN REVIEW

785

25. We compute (a × b) · c. First a × b = −3i + 3j − 3k. Then (a × b) · c = −3(4) + 3(5) − 3(1) = 0, and the vectors are coplanar. 1 |c|. Since 2 2 2 2 a · b = 0 we have |b − a| = (b − a) · (b − a) = a · a + 2a · b + b · b = |a| + |b| , we have 1 1 2 1 |d| = 12 |a + b| = |a| + |b|2 = |b − a| = |c|. 2 2 2

26. Let d be the vector from the right angle to M . We want to show that |d| =

27. (a) We have v = vj and B = Bi. Then F = q(v × B = q(vj × Bi) = q(−vBk) = −qvBk. (b) We first note that L = mr × v and r × v = 0. Then r × L = r × (mr × v) = m[r × (r × v)] = m[(r · v)r − (r · r)v] = −m|r|2 v, and so v = −

1 1 (r × L) = (L × r). 2 m|r| m|r|2

√ √ 10 a = √ (i + j) = 5 2i + 5 2j; d = 7, 4, 0 − 4, 1, 0 = 3i + 3j |a| 2√ √ √ W = F · d = 15 2 + 15 2 = 30 2 N-m √ √ √ √ 29. F = 5 2i + 5 2j + 50i = (5 2 + 50)i + 5 2j; d = 3i + 3j 28. F = 10

√ √ √ W = 15 2 + 150 + 15 2 = 30 2 + 150N-m ≈ 192.4 N-m 30. Let |F1 | = F1 and |F2 | = F2 . Then F1 = F1 [(cos 45◦)i + (sin 45◦ )j] and F2 = F2 [(cos 120◦ )i + √   1 3 1 1 ◦ j . Since w + F1 + F2 = 0, and F2 = F2 − i + (sin 120 )j] or F1 = F1 √ i + √ 2 2 2 2  F1

 √  1 3 1 1 √ i + √ + F2 − i + j 2 2 2 2

 = 50j,

 √  1 3 1 1 √ F 1 − F 2 i + √ F1 + F2 2 2 2 2

√ 3 1 1 1 √ F1 − F2 = 0, √ F1 + F2 = 50. 2 2 2 2 √ √ √ Solving, we obtain F1 = 25( 6 − 2) ≈ 25.9lb and F2 = 50( 3 − 1) ≈ 36.6lb. and

j = 50j

CÁLCULO VECTORIAL √ MATEMÁTICAS 3

10.2. PARAMETRIC EQUATIONS

685

87. Since a = 4 and b = 20, we have c2 = a2 + b2 = 16 + 20 = 36 and hence c = 6. Thus the foci occur at F1 = (−6, 0) and F2 = (6, 0). The line joining (−6, −5) and F2 is given by 5 5 y= x − . The ray of light travels southwest along this line. 12 2 √ √ 88. (a) The distance from (0, b) to (a, 0) is a2 + b2 . Thus R = a2 + b2 = r. √ √ 2 2 2 2 (b) From A = a + √ r and B = R − b = a + b + r − b = a + b + (A − a) − b = 2 2 A − (a + b) + a + b we see that � � � � � A − B = a + b − a2 + b2 = (a + b)2 − a2 + b2 = a2 + 2ab + b2 − a2 + b2 > 0.

MANUAL DE SOLUCIONES

UNIDAD 2 ECUACIONES PARAMÉTRICAS Y COORDENADAS POLARES Thus, A > B. PROBLEMAS 2.1

10.2 1.

2.

Parametric Equations

t x y

-3 -5 6

-2 -3 2

t x y

0 1 0

π √6 3 2

-1 -1 0 π √4 2 2

1/4

1/2

0 1 0

1 3 2 π 3 1 2

3/4

3.

2 5 6

3 7 12

π 2

0 1

5π 6 √ - 23

1/4

7π √4 2 2

1/2

4.

5. y

y

y

x x

x

6.

7. y

8. y

y

x

x

x

686

CHAPTER 10. CONICS AND POLAR COORDINATES

9.

10. y

y

x

x

11. y = (t2 )2 + 3t2 − 1 = x2 + 3x − 1; 12. − 12 y = t3 + t;

x = − 12 y + 4;

y = x2 + 3x − 1,

2x + y = 4

13. x = cos 2t = cos2 t − sin2 t = 1 − 2 sin2 t = 1 − 2y 2 ; 14. ln x = t;

x≥0

y = 1 − 2y 2 ,

x > 1. Alternatively, ey = t;

y = ln(ln x),

−1 ≤ y ≤ 1

y

x = ee , x > 1

15. t = x1/3 ; y = 3 ln x1/3 ; y = ln x, x > 0

16. x2 tan2 x, y 2 = sec2 t; x2 + 1 = tan2 t + 1 = sec2 t = y 2 ; y 2 − x2 = 1. y ≥ 1 17. y

y

x

y=x

x

x = sin t y = sin t

10.2. PARAMETRIC EQUATIONS

687

18. y

y

x

x

√ x=− t y=t t≥0

y = x2

19. y

y

x

x

y=

x2 −1 4

x = 2t y = t2 − 1 −1 ≤ t ≤ 2

20. y

y

x

x

y = −x2

x = at y = −e2t t≥0

688

CHAPTER 10. CONICS AND POLAR COORDINATES

21. y

y

x

x

x2 − y 2 = 1

x = cosh t y = sinh t

22. y

y

x

x

x = x = t2 − 1 y = 2t2 − 4

y = 2x − 2

23. y

y

a

a

a

x

a

x = a cos t y = a sin t a>0 0≤t≤π

x = a sin t y = a cos t a>0 0≤t≤π

x

10.2. PARAMETRIC EQUATIONS

689

24. y

y a x

b

-b

-a x

x = a cos t y = b sin t a>b>0 π ≤ t ≤ 2π

x = a sin t y = b cos t a>b>0 π ≤ t ≤ 2π

25. y

y

a

a

a

a

x

x

x = a cos t y = a sin t a>0 −π/2 ≤ t ≤ π/2

x = a cos 2t y = a sin 2t a>0 −π/2 ≤ t ≤ π/2

26. y

y

a

a

a

x = a cos t/2 y = a sin t/2 a>0 0 ≤ t ≤ π/2

x

a

x = a cos(−t/2) y = a sin(−t/2) a>0 −π ≤ t ≤ 0

x

690

CHAPTER 10. CONICS AND POLAR COORDINATES

27.

28. y

y

x x

29. This is the same as x = 1/y or xy = 1. The graphs are the same. 30. Since x = t1/2 ≥ 0 for all t, x can never be -1. But (−1, −1) is on the xy = 1, so the graphs are not the same. 31. Since | cos t| ≥ 1, x can never be 2. But (2, 1/2) is on xy = 1, so the graphs are not the same. 32. Since t2 + 1 ≥ 1 for all t, x can never be -1. But (−1, −1) is on xy = 1, so the graphs are not the same. 33. Since e−2t > 0 for all t, x can never be -1. But (−1, −1) is on xy = 1, so the graphs are not the same. 34. This is the same as x = 1/y or xy = 1. The graphs are the same. 35. From sin φ = Ly we have t = L sin φ. Since (x, y) is on the circle x2 + y 2 = r2 , � � ± r2 − y 2 = ± r2 − l2 sin2 φ.

x =

36. From the figure in the text, we see that x = r cos 3θ + R cos θ and y = r sin 3θ + r sin θ. (The actual curve generated by these equations will have the general appearance of the curve in 2.1.12 in the text only when R > 3r.) Figure 10.2.12 37. From the figure we see that β = θ − π/2 and α = β = θ − π/2. The length of the line segment from R to P is equal to the arc of the circle subtended by θ; that is aθ. Now, x = aθ sin α = aθ sin(θ − π/2) = −aθ cos θ, t = a cos β = a cos(θ − π/2) = a sin θ, b = a sin β = a sin(θ − π/2) = −a cos θ, and c = aθ cos α = aθ cos(θ − π/2) = aθ sin θ. Thus

P aθ

y = s + t = −aθ cos θ = a(sin θ − θ cos θ).

y

x

R b a β

t

θ x c

x = c − b = aθ sin θ − (−a cos θ) = a(cos θ + θ sin θ)

s

α

10.2. PARAMETRIC EQUATIONS

38. The hypotenuse of the triangle OAB is b − a and OB = (b − a) cos θ. The actue angle at A in the right triangle with hypotenuse AP is φ − θ. Thus, BC = a cos(φ − θ) and x = OB + BC = (b − a) cos θ + a cos(φ − θ). Similarly, y = AB − AD = (b − a) sin θ − a sin(φ − θ). Now, the arc on the smaller circle subtended by the angle φ has length aφ and the arc on the larger circle subended by the angle θ has length bθ. From the definition of the hypocycloid, aφ = bθ. Then φ = bθ a and φ − θ = � bθ � − θ. Thus, the parametric equations of the a hypocycloid are x = (b−a) cos θ+a cos[(b−a)/a]θ, y = (b − a) sin θ − a sin[(b − a)/a]θ.

691

a

0

θ

A

θ

a

φ D B C

P

b

39. (a) When b = 4a, the equations become x = 3a cos θ + a cos 3θ, y = 3a sin θ − a sin 3θ. Using the identities cos 3θ = 4 cos3 θ − 3 cos θ and sin 3θ = 3 sin θ − 4 sin3 θ, the parametric equations of the hypocycloid of four cusps become x = 4a cos3 θ = b cos3 θ, y = 4a sin3 θ = b sin3 θ.

(b) y

x

(c) Writing x2/3 = b2/3 cos2 θ, y 2/3 = b2/3 sin2 θ we obtain x2/3 + y 2/3 = b2/3 .

692

CHAPTER 10. CONICS AND POLAR COORDINATES

40. The hypotenuse of triangle OAB is a + b and OB = (a + b) cos θ.� The acute angle at A in tri� angle ADP is φ − 12 − θ = (φ + θ) − π2 . Thus, BC = DP = a sin(φ + θ − π/2) = −a cos(φ + θ) and x = OB + BC = (a + b) cos θ − a cos(φ + θ). Similarly, � π� y = AB − AD = (a + b) sin θ − a cos φ + θ − 2 = (a + b) sin θ − a sin(φ + θ). Now, the arc on the smaller circle subtended by the angle φ has length aφ and the arc on the large circle subtended by the angle θ has length bθ. From the definition of the epicycloid, �aφ = �bθ. a+b bθ Then φ = bθ θ. a and φ + θ = a + θ = a Thus, the parametric equations � of the � epicycloid a+b are x = (a + b) cos θ − a cos θ, y = (a + a � � a+b b) sin θ − a sin θ. a

a b

φ

A D

a P

θ O

b B

C

41. (a) When b = 3a, the equations become x = 4a cos θ − a cos 4θ, y = 4a sin θ − a sin 4θ. (b) y

x

42. (a) The Q be the point (0, 2a) at the top of the circle and let (x, y) be the coordinates of point P. Then the measure of angle ∠OBQ is equal to θ. The gives tan θ =

2a x

or

x=

2a . tan θ

Also, the measure of angle ∠AQO is equal to θ. Letting r represent the length of the segment AO, we have r sin θ = or r = 2a sin θ. 2a Since y = r sin θ, we have y = (2a sin θ) sin θ = 2a sin2 θ.

10.2. PARAMETRIC EQUATIONS (b) Rewrite x as x =

693

2a cos θ . Then sin θ x2 y =

4a2 cos2 θ (2a sin2 θ) = 8a3 cos2 θ sin2 θ

and 4a2 y = 8a3 sin2 θ. This gives x2 y + 4a2 y = 8a3 cos2 θ + 8a3 sin2 θ y(x2 + 4a2 ) = 8a3 y=

43.

8a3 x2 + 4a2

44.

45. y

y

x

x

46.

y

47. y

x

48. y

y

x x

x

x − x1 . Plugging this into the equation x2 − x1 � � x − x1 y = y1 + (y2 − y1 ) x2 − x1 � � y 2 − y1 = y1 + (x − x1 ) x2 − x1

49. Using the equation from x to solve for t, we have t = for y yields

which is the equation of a line joining (x1 , y1 ) and (x2 , y2 ). When 0 ≤ t ≤ 1, we get the line segment with endpoints (x1 , y1 ) and (x2 , y2 ).

694

CHAPTER 10. CONICS AND POLAR COORDINATES

50. (a) x = −2 + (4 − (−2))t = −2 + 6t y = 5 + (8 − 5) = 5 + 3t (b) y = 21 x + 6

(c) x = −2 + 6t, y = 5 + 3t; 0 ≤ t ≤ 1 51. If the launch point is designated as the origin, then the equations describing the skier’s motion from launch until landing are given by x = 75t

and

y = −16t2

where t = 0 at the moment of launch. At the moment of impact, we have tan 33◦ = Thus, t =

|y| 16 = t. |x| 15

75 tan 33◦ ≈ 3.044, and therefore x ≈ 228.3 ft, y ≈ −148.25 ft. 16

PROBLEMAS 2.2 and Parametric Equations 10.3 Calculus dx 1. = 3t2 − 2t; dt 2.

dx 4 = − 2; dt t

3.

dx t =√ ; 2 dt t +1

4.

5.

6.

7.

dy = 2t + 5; dt

dy = 6t2 − 1; dt dy = 4t3 ; dt

� dy �� 3 = � dx t=−1 5 � dy 6t2 − 1 6t4 − t2 dy �� 92 = = − ; =− = −23 dx −4/t2 4 dx �t=2 4 � � √ dy 4t3 dy �� 2 2 = √ = 4t t + 1; = 4(3) 4 = 24 � √ 2 dx dx t/ t + 1 t= 3 dy 2t + 5 = 2 ; dx 3t − 2t

dx dy dy 4e−4t = 2e2t ; = −4e−4t ; = − 2t = −2e−6t dt � dt dx 2e −6 dy �� 1 = −2e−6 ln 2 = −2eln 2 = −2(2−6 ) = − dx �t=ln 2 32 dx dy = 2 cos θ(− sin θ); cos θ; dθ � dθ dy �� 1 =− = −1 � dx θ=π/6 2(1/2)

dy cos θ 1 = =− dx −2 sin θ cos θ 2 sin θ

dx dy dy 2 sin θ sin θ = 2 − 2 cos θ; = 2 sin θ; = = dθ dθ dx 2 − 2 cos θ 1 − cos θ √ √ � √ dy �� 2/2 2 √ √ = 2+1 = = � dx θ=π/4 1 − 2/2 2− 2

dy 12t 4t = 2 = 2 . dx 3t + 3 t +1 At t = −1 we observe x = −4, y = 7, and m = dy/dx = −2. The tangent line is y = −2x − 1.

10.3. CALCULUS AND PARAMETRIC EQUATIONS 8.

9.

10.

11.

12.

13.

695

dy 2t + 1/t 1 = =t+ . dx 2 2t 3 At t = 1 we observe x = 6, y = 1, and m = dy/dx = 3/2. The tangent line is y = x − 8. 2 dy 2t 4 4 = . At (2, 4), t = −2 and m = dy/dx = 4/3. The tangent line is y = x + . dt (2t + 1) 3 3 √ √ dy (4t3 − 2t) 1 = = 1 − 2 . At (0, 6), t = 3 or − 3 and m = 5/6. The tangent line is 3 dx 4t 2t 5 y = x + 6. 6 dy −2 sin t = . When y = 1, cos t = 1/2 and t = π/3 or 5π/3. For t = π/3, x = 4 sin(2π/3) = dx 8 cos 2t √ √ √ dy −2 sin(π/3) 3 4( 3/2) = 2 3, and m = = = . dx 8 cos(2π/3) 4 dy 3t2 3t 3t = = . The slope of the tangent line is -3. Solving = −3, we obtain t = −2. At dx 2t 2 2 t = −2 we observe x = 4 and y = −7. The point on the graph is (4, 7). dx dy dy 2t − 4 = 2; = 2t − 4; = =t−2 dt dt dx 2 We want t − 2 = 3. Then t = 5 and the point of tangency is (5, 8). The equation of the tangent line is y − 8 = 3(x − 5) or y = 3x − 7.

14. For θ = π/2 we observe x = −2/π and y = 1 − 1 = 0. For θ = −π/2 we observe x = −2/π and y = −1 + 1 = 0. Thus, the curve intersects itself when θ = π/2 and θ = −π/2. dx dy 2 dy cos θ − 2/π 2 = − sin θ; = cos θ − ; = = csc θ − cot θ dθ dθ π dx − sin θ π� dy �� 2 When θ = π/2, the slope of the tangent line is = and its equation is y − 0 = dx �θ=π/2 π � � � dy �� 2 2 2 2 4 x+ or y = x+ 2 . When θ = −π/2, the slope of the tangent line is =− π π π π dx �θ=−π/2 π � � 2 2 2 4 and its equation is y − 0 = − x+ or y = − x − 2 . π π π π 15.

dx dy dy 2t = 3t2 − 1; = 2t; = 2 . The tangent line is dt dt dt 3t − 1 2 horizontal when √ 2t = 0 or t = 0, and vertical when 3t −1 = 0 or t = ±1/ 3. Thus,√there is a horizontal √ tangent at (0, 0) and vertical at (−2/3 3, 1/3) and (2/3 3, 1/3).

y

2

1

x

696

16.

17.

18.

CHAPTER 10. CONICS AND POLAR COORDINATES dx 3 dy dy 2t − 2 16t − 16 = t2 ; = 2y − 2; = = . The 2 dt 8 dt dx 3t /8 3t2 tangent line is horizontal when 16t − 16 = 0 or t = 1, and vertical when 3t2 = 0 or t = 0. Thus, there is a horizontal tangent at (9/8, −1) and a vertical tangent at (1, 0).

dx dy dy −3 sin 3t = cos t; = −3 sin 3t; = . The dt dt dx cos t tangent line is horizontal when sin 3t = 0 or t = 0, π/3, 2π/3, π, 4π/3, 5π/3, 2π and vertical when cos t = 0 or t = π/2, there are horizontal √ 3π/2. Thus, √ √ tangents √ at (0, 1), ( 3/2, −1), ( 3/2, 1), (0, −1), (− 3/2, 1), ( 3/2, 1) and vertical tangents at (1, 0) and (−1, 0). dy 18t2 = = 3t; dx 6t

20.

dy cos t = = − cot t; dx − sin t

22.

1

dx dy dy = 1; = 3t2 − 6t; = 3t2 − 6t. The tangent line is dt dt dx horizontal when 3t2 − 6t = 3t(t − 2) = 0 or t = 0, 2. Thus, there are horizontal tangents at (−1, 0) and (1, −4). There are no vertical tangents.

19.

21.

y

d2 y 3 1 = = ; 2 dx 6t 2t

1

x

y 1 x

1

y 1

1

x

d3 y −1/2t2 1 = =− 3 3 dx 6t 12t

d2 y csc2 t = = − csc3 t; dx2 − sin t

d3 y 3 csc3 t cot t = = −3 csc4 t cot t dx3 − sin t

dy 2e2t + 3e3t d2 y −6e3t − 12e4t 3t 4t = = −2e − 3e ; = = 6e4t + 12e5t dx −e−t dx2 −e−t d3 y 24e4t + 60e5t = = −24e5t − 60e6t 3 dx −e−t dy t−1 = ; dx t+1

d2 y 2/(t + 1)2 2 = = ; 2 dx t+1 (t + 1)3

d3 y −6/(t + 1)4 6 = =− 3 dx t+1 (t + 1)5

� � d2 y dy � /dt (32 − 16t)/3t3 256 − 128t 128 2 − t = = = = . Then dx2 dx/dt 3t2 /8 9t5 9 t5 d2 y/dx2 is 0 when t = 2 and undefined when t = 0. The graph is concave downward on (−∞, 0) and (2, ∞) and concave upward on (0, 2).

23. Using Problem 16,

t y ��

-

0 und

+

2 0

-

10.3. CALCULUS AND PARAMETRIC EQUATIONS 24.

697

dx dy dy d2 y dy � /dt 6t + 6 = 2; = 6t2 + 12t + 4; = 3t2 + 6t + 2; = = = 3t + 3 2 dt dt dx dx dx/dt 2 Solving 3t + 3 = 0 we obtain t = −1. Since d2 y/dx2 < 0 for t < −1 and d2 y/dx2 > 0 for t > −1, the graph has a point of inflection when t = −1 or at (3, 0). �

�

2

25. x (t) = 5t ,

2

y (t) = 12t ;

26. x� (t) =√t2 , y � (t) = t � 3� � s= t4 + t2 dt = �

0

4

1

π

s= =

√

3

t 0

�

2 0

�

25t4

t2 + 1dt

+

144t4 dt

= 13

�

2

�2 13 3 �� 104 t dt = t � = 3 3 0 2

0

u = t2 + 1, du = 2tdt

�4 1 1/2 1 3/2 �� 1 7 u du = u � = (8 − 1) = 2 3 3 3 1

27. x� (t)�= et cos t + et sin t; �

s=

�

0

y � (t) = −et sin t + et cos t

[e2t (cos2 t + 2 sin t cos t + sin2 t) + e2t (sin2 t − 2 sin t cos t + cos2 t)]1/2 dt

π

et (2)1/2 dt = 0

28. x� (θ) = a(1 − cos θ), �

�

√

�π √ � 2et � = 2(eπ − 1) 0

y � (θ) = a sin θ. Using symmetry, �

� 1 − 2 cos θ + cos2 θ + sin2 θdθ 0 0 �√ � √ � π√ √ � π√ 1 + cos θ √ = 2 2a 1 − cos θdθ = 2 2a 1 − cos θ dθ 1 + cos θ 0 0 √ � � π π √ √ 1 − cos2 θ sin θ √ √ = 2 2a dθ = 2 2a dθ u = 1 + cos θ, du = − sin θdθ 2 1 + cos θ 1 + cos θ 0 0 � 2 √ � 0 −du √ � 0 −1/2 √ √ = 2 2a = 2 2a u du = 2 2a lim+ u1/2 du u b→0 2 2 b � √ � � √ √ √ √ 2 = 2 2a lim 2 u�b = 2 2a lim (2 2 − 2 b) = 8a.

s=2

π

a2 (1 − cos θ)2 + a2 sin2 θdθ = 2a

b→0+

π

b→0+

29. x� (θ) = −3b cos2 θ sin θ; y � (θ) = 3b sin2 θ cos θ � π/2 � s= (9b2 cos4 θ sin2 θ + 9b2 sin4 θ cos2 θ)1/2 dθ = 3|b| 0

= 3|b|

�

π/2

0

π/2

sin θ cos θ 0

�π/2 � 1 3 3 3 sin 2θdθ = − |b| cos 2θ�� = − |b|(−1 − 1) = |b| 2 4 4 2 0

30. x� (θ) = −4a sin θ + 4a sin 4θ;

y � (θ) = 4a cos θ − 4a cos 4θ

�

cos2 θ + sin2 θdθ

698

CHAPTER 10. CONICS AND POLAR COORDINATES s=

�

2π/3 0

= 4|a| = 4|a|

� �

[(−4a sin θ + 4a sin 4θ)2 + (4a cos θ − 4a cos 4θ)2 ]1/2 dθ 2π/3 0 2π/3

(2 − 2 sin θ sin 4θ − 2 cos θ cos 4θ)1/2 dθ

0

� √ = 4 2|a| � √ = 4 2|a| � √ = 4 2|a| = 8|a|

(sin2 θ − 2 sin θ sin 4θ + sin2 4θ + cos2 θ − 2 cos θ cos 4θ + cos2 4θ)1/2 dθ

�

2π/3

[1 − (cos 4θ cos θ + sin 4θ sin θ)]1/2 dθ

0 2π/3 0 2π/3 0

� �

� √ 1 − cos(4θ − θ)dθ = 4 2|a| 2 sin2 3θ/2dθ = 8|a|

�

2π/3 0

2π/3

sin 0

√

1 − cos 3θdθ

3θ dθ 2

��2π/3 2 3θ �� 16 32 − cos = − |a|(cos π − cos 0) = |a| 3 2 �0 3 3

31. (a) Setting x = 0 we have t2 −√ 4t − 2 = 0 which implies t = 2 ± y ≈ −0.6551. When t = 2 + 6, y ≈ 1390.66.

√

6. When t = 2 −

√

6,

(b) Using Newton’s Method to solve t5 −4t3 −1 = 0 we obtain t ≈ −1.96687, −0.654175, 2.02968 with corresponding x values 9.73606, 1.04465, −5.99912.

32. If y = F (x) and x = f (t), then g(y) = y = F (f (t)). Thus, A= =

� �

x2

F (x)dx x = f (t), dx = f � (t)dt

x1 b

�

F (f (t))f (t)dt = a

�

b

g(t)f � (t)dt.

a

2.1, f (θ) = a(θ − sin θ) and g(θ) = a(1 − cos θ) for 0 ≤ θ ≤ 2π. 33. From Example 7 in Section 10.2, � Then f (θ) = a(1 − cos θ), and using symmetry, �

π

a2 (1 − cos θ)2 dθ = 2a2

�

π

(1 − 2 cos θ + cos2 θ)dθ 0 0 � ��π � π � 1 1 3 1 2 = 2a (1 − 2θ + + cos 2θ)dθ = 2a2 θ − 2 sin θ + sin 2θ �� 2 2 2 4 0 0 � � 3 = 2a2 π = 3(πa2 ). 2

A=2

Thus, the area under an arch of the cycloid is three times the area of the circle.

10.4. POLAR COORDINATE SYSTEM

699

PROBLEMAS 10.4 Polar 2.3 Coordinate System 1.

2. π (3,π)

Ο polar axis

Ο

polar axis

−π/2

(2,−π/2)

3.

4.

Ο

π/6 polar axis

(−1,π/6)

π/2 Ο

polar axis

(−1/2,π/2)

5.

6. (−4,−π/6) 7π/4

Ο polar axis

Ο

7.

(a)

8.

(a)

9. 10.

(a) (a)

� � � �

−π/6

2, − 5π 4 5, − 3π 2 4, − 3, −

� �

� 5π 3

� 7π 4

polar axis

(b) (b) (b) (b)

�

� 2, 11π 4 � 5π � 5, 2 � 7π � 4, 3 � 9π � 3, 4

(2/3,7π/4)

(c) (c) (c) (c)

�

� −2, 7π 4 � � −5, 3π 2 � � −4, 4π 3 � � −3, 5π 4

(d) (d) (d) (d)

�

� −2, − π4 � � −5, − π2 � � −4, − 2π 3 � � −3, − 3π 4

700

CHAPTER 10. CONICS AND POLAR COORDINATES

11.

(a)

12.

(a)

� �

1, − 11π 6 3, − 5π 6

�

�

(b) (b)

�

�

1, 13π 6

3, 19π 6

�

�

(c) (c)

�

�

−1, 7π 6

−3, π6

�

�

(d) (d)

�

�

−1, − 5π 6

−3, − 11π 6

�

�

13. With r = 1/2 and θ = 2π/3 we have cos 2π/3 = 1/2(−1/2) = −1/4, y = 1/2 sin 2π/3 = � x = 1/2 √ � √ √ 1 3 1/2( 3/2) = 3/4. The point is − , in rectangular coordinates. 4 4 √ √ 14. With r = −1 and θ = 7π/4 we have x = −1 7π/4 � = −1( 2/2) = − 2/2, y = � cos √ √ √ √ 2 2 −1 sin 7π/4 = −1(− 2/2) = 2/2. The point is − , in rectangular coordinates. 2 2 15. With √ r = −6 and√θ = −π/3 we have � x =√−6�cos(−π/3) = −6(1/2) = −3, y = −6 sin(−π/3) = −6(− 3/2) = 3 3. The point is −3, 3 3 in rectangular coordinates. √ √ √ √ √ 16. With r = 2 and θ = 11π/6 we have x = � 2 cos 11π/6 �= 2( 3/2) = 6/2, y = √ √ √ √ √ 6 2 2 sin 11π/6 = 2(−1/2) = − 2/2. The point is ,− in rectangular coordinates. 2 2

√ √ 17. With √ r = 4 and√θ = 5π/4 we have √5π/4 � x√= 4 cos � = 4(− 2/2) = −2 2, y = 4 sin 5π/4 = 4(− 2/2) = −2 2. The point is −2 2, −2 2 in rectangular coordinates.

18. With r = −5 and θ = π/2 we have x = −5 cos π/2 = 0, y = −5 sin π/2 = −5. The point is (0, −5) in rectangular coordinates. 19. With� x√= −2 and have r�2 = 8 and tan θ = 1. � y = −2�we √ 3π (a) 2 2, − 4 (b) −2 2, π4

20. With� x = 0� and y = −4 r2 = 16 and tan θ undefined. � we have � (a) 4, − π2 (b) −4, π2 √ √ 21. With� x = 1� and y = − 3 we�have r2 = 4 and tan θ = − 3. � (a) 2, − π3 (b) −2, 3π 3

√ √ √ 22. With� x√= �6 and y = � 2 we have �r2 = 8 and tan θ = 1/ 3. √ (a) 2 2, π6 (b) −2 2, − 5π 6 23. With x = 7 and y = 0 we have r2 = 49 and tan θ = 0. (a) (7, 0) (b) (−7, π) or (−7, −π)

24. With�√x = 1 and �y = 2�√ we have r2� = 5 and tan θ = 2. � √ � � √ � −1 (a) 5, tan 2 or 5, 1.1071 (b) − 5, −π + tan−1 2 or − 5, −2.0344

10.4. POLAR COORDINATE SYSTEM 25.

701 26.

y

y

x

x 2

2

polar axis

4

27.

4

polar axis

28. y

y

x

x

polar axis

polar axis

29.

30. y

y

x 1

polar axis

In Problems 31-40, we use x = r cos θ and y = r sin θ. 31. r sin θ = 5;

r = 5 csc θ

32. r cos θ + 1 = 0;

r = − sec θ

2 4

x

polar axis

702

CHAPTER 10. CONICS AND POLAR COORDINATES

33. r sin θ = 7r cos θ;

tan θ = 7;

34. 3r cos θ + 8r sin θ + 6 = 0;

θ = tan−1 7

r(3 cos θ + 8 sin θ) = −6;

r=−

6 (3 cos θ + 8 sin θ)

35. r2 sin2 θ = −4r cos θ; r2 (1 − cos2 θ) + 4r cos θ = 0; r2 − (r2 cos2 θ − 4r cos θ + 4) = 0; r2 − (r cos θ − 2)2 = 0; [r − (r cos θ − 2)][r + (r cos θ) − 2] = 0 −2 2 Solving for r, we obtain r = or r = . Since replacement of (r, θ) by (1 − cos θ) (1 + cos θ) (−r, θ + π) in the first equation gives the second equation, we take the polar equation to be 2 r= . 1 + cos θ 36. r2 cos2 θ−12r sin θ−36 = 0; r2 (1−sin2 θ)−12r sin θ−36 = 0; r2 −(r2 sin2 θ+12r sin θ+36) = 0; r2 − (r sin θ + 6)2 = 0; [r − (r sin θ + 6)][r + (r sin θ + 6)] = 0 6 −6 Solving for r, we obtain r = or r = . Since replacement of (r, θ) by (1 − sin θ) (1 + sin θ) (−r, θ + π) in the second equation gives the first equation, we take the polar equation to be 6 r= . (1 − sin θ) 37. r2 = 36. Since r = −6 has the same graph as r = 6, we take the equation to be r = 6. 38. (r cos θ)2 − (r sin θ)2 = 1 r2 cos2 θ − r2 sin2 θ = 1 � � r2 cos2 θ − sin2 θ = 1 � � r2 1 − 2 sin2 θ = 1

√ 39. r2 + r cos θ = r2 = ±r; r(r + cos θ ∓ 1) = 0. Solving for r, we obtain r = 0 or r = ±1 − cos θ. Since replacement of (r, θ) in r = −1 − cos θ by (−r, θ + π) gives r = 1 − cos θ, and since θ = 0 gives r = 0, we take the polar equation to be r = 1 − cos θ. 40. r3 cos3 θ + r3 sin3 θ − r2 sin θ cos θ = 0; r2 [r(cos3 θ + sin3 θ) − 12 sin 2θ] = 0 sin 2θ Solving for r gives r = 0 or r = . Since π = 0 gives r = 0 in the second 2(cos3 θ + sin3 θ) sin 2θ equation, we take the polar equation to be r = . 2(cos3 θ + sin3 θ) In Problems 41-52, we use r2 = x2 + y 2 , r cos θ = x, r sin θ = y, and tan θ = y/x. 41. r cos θ = 2;

x=2

42. x = −4 43. r = 12 sin θ cos θ;

r3 = 12r sin θr cos θ;

44. 2(x2 + y 2 )1/2 = y/x; 45. r2 = 8 sin θ cos θ;

(x2 + y 2 )3/2 = 12xy;

4x2 (x2 + y 2 ) = y 2

r4 = 8r sin θr cos θ;

(x2 + y 2 )2 = 8xy

(x2 + y 2 )3 = 144x2 y 2

10.4. POLAR COORDINATE SYSTEM 46. r2 (cos2 θ − r2 sin2 θ) = 16; 47. r2 + 5r sin θ = 0;

703

r2 cos2 θ − r2 sin2 θ = 16;

x2 − y 2 = 16

x2 + y 2 + 5y = 0

48. r2 = 2r+r cos θ; x2 +y 2 = 2(x2 +y 2 )1/2 +x; x2 +y 2 −x = 2(x2 +y 2 )1/2 ; (x2 +y 2 −x)2 = 4(x2 +y 2 ) 49. r + 3r cos θ = 2; (x2 + y 2 )1/2 + 3x = 2; (x2 + y 2 )1/2 = 2 − 3x; x2 + y 2 = 4 − 12x + 9x2 ; 8x2 − y 2 − 12x + 4 = 0 � � 50. 4r − r sin θ = 10; 4 x2 + y 2 − y = 10; 4 x2 + y 2 = y + 10; 16(x2 + y 2 ) = y 2 + 20y + 100; 16x2 + 15y 2 − 20y − 100 = 0 51. 3r cos θ + 8r sin θ = 5; 3x + 8y = 5 52. r cos θ = 3 cos θ + 3; r2 cos θ = 3r cos θ + 3r; r(r cos θ − 3) = 3r cos θ; (x2 + y 2 )1/2 (x − 3) = 3x; (x2 + y 2 )(x − 3)2 = 9x2 � � 53. (x2 − x1 )2 + (y2 − y1 )2 = (r2 cos θ2 − r1 cos θ1 )2 + (r2 sin θ2 − r1 sin θ1 )2 � = r22 cos2 θ2 − 2r2 r1 cos θ2 cos θ1 + r12 cos2 θ1 + r22 sin2 θ2 − 2r2 r2 sin θ2 sin θ1 + r12 sin2 θ1 � = r22 + r12 − 2r1 r2 (cos θ2 cos θ1 + sin θ2 sin θ1 ) � = r22 + r12 − 2r1 r2 cos(θ2 − θ1 ) 54. Consider the general linear function y = ax + b. Transforming this function into polar coordinates, we have r sin θ = ar cos θ + b. To find a line passing through (r1 , θ1 ) and (r2 , θ2 ), we would need to solve the following system for a and b: r1 sin θ1 = ar1 cos θ1 + br2 sin θ2 = ar2 cos θ2 + b � � � � To find the line passing through 3, 3π and 1, π4 , we solve the system 4 √ √ 3 2 2 = −3a +b 2 2 √ √ 2 2 =a +b 2 2 for a and b. This yields a = − 21 , b =

√ 3 2 4 .

Thus, the equation of the line is √ 1 3 2 r sin θ = − r cos θ + 2 4

or

√

3 42 r= sin θ + 21 cos θ

√ 3 2 The y-intercept occurs when r cos θ = 0 and hence r sin θ = . Together, these yield θ = π2 4 � √ � √ 3 2 3 2 π and r = . The y-intercept is thus , in polar coordinates. The x-intercept 4 4 2

704

CHAPTER 10. CONICS AND POLAR COORDINATES √ √ 3 2 3 2 occurs when r sin θ = 0 and hence r cos θ = . Together, these yield θ = 0 and r = . 2 2 � √ � 3 2 The x-intercept is thus ,0 . 2

55. Solutions of f (θ) = 0 are θ values at which the graph of r = f (θ) passes through the origin.

PROBLEMAS 2.4 of Polar Equations 10.5 Graphs 1.

2.

polar axis

3.

1

circle

polar axis

circle

4.

line through origin

5.

6.

polar axis

polar axis

7.

polar axis

spiral

line through origin

spiral

8.

9.

polar axis

cardioid

polar axis

polar axis

polar axis

cardioid

cardioid

10.5. GRAPHS OF POLAR EQUATIONS 10.

705

11.

12.

polar axis

cardioid

limacon with an interior loop

limacon with an interior loop

13.

14.

15.

polar axis

polar axis

dimpled limacon 16.

polar axis

dimpled limacon 17.

convex limacon 18.

polar axis

polar axis

polar axis

rose curve

convex limacon 19.

rose curve

20.

21.

polar axis

polar axis

rose curve

polar axis

polar axis

rose curve

polar axis

rose curve

706

CHAPTER 10. CONICS AND POLAR COORDINATES

22.

23.

24.

polar axis

polar axis

polar axis

rose curve

circle

circle with center on x-axis

25.

26.

27.

circle with center on y-axis 28.

polar axis

polar axis

polar axis

circle with center on y-axis

lemniscate

29.

30.

polar axis

polar axis

polar axis

lemniscate

lemniscate 31.

lemniscate

32. π

10

polar axis 1

polar axis

10.5. GRAPHS OF POLAR EQUATIONS

707

33. r = 2.5 34. r = −4 cos θ 35. r = 4 − 3 cos θ 36. r = 2 + 3 sin θ 37. r = 2 cos 4θ 38. r = 5 cos 2θ 39. Solving 4 sin θ = 2 we have sin θ = 1/2 and θ = π/6 and 5π/6. The points of intersection are (2, π/6) and (2, 5π/6).

polar axis

1

40. Writing sin 2θ = 2 sin θ cos θ and setting sin θ = 2 sin θ cos θ, we obtain sin θ(2 cos θ − 1) = 0. This gives θ = 0, π, π/3, and 5π/3. For θ = 0 and θ = π we obtain √the pole (0, 0). The √ other two points of intersection are ( 3/2, π/3) and (− 3/2, 5π/3).

41. Setting 1 − cos θ = 1 + cos θ, we obtain 2 cos θ = 0. This gives θ = ±π/2. Two points of intersection are (1, π/2) and (1, −π/2). From the figure we see that the pole ((0, 0) on r = 1 − cos θ and (0, π) on r = 1 + cos θ) is also a point of intersection.

42. Setting 3−3 cos θ−3 cos θ, we obtain 3 = 6 cos θ or 12 =�cos θ.� This �yields �θ = ± π3 . Two points of intersection are 32 , π3 and 32 , − π3 . From the figure we see that the pole ((0, 0) on r = 3 − 3 cos θ and (0, π/2) on r = 3 cos θ) is also a point of intersection.

1

1

polar axis

polar axis

polar axis

708

CHAPTER 10. CONICS AND POLAR COORDINATES

43. Setting 6 sin 2θ = 3, we obtain sin 2θ = 1/2. Then 2θ = π/6, 5π/6, 13π/6, and 17π/6. This gives the points of intersection (3, π/12), (3, 5π/12), (3, 13π/12), and (3, 17π/12). Writing the second equation in the form −r = 6 sin 2(θ + π), we obtain r = −6 sin 2θ. Setting −6 sin 2θ = 3, we obtain sin 2θ = −1/2. Then 2θ = −π/6, −5π/6, −13π/6, and −17π/6. This gives the points of intersection (3, −π/12), (3, −5π/12), (3, −13π/12), and (3, −17π/12).

44. Using cos 2θ = 2 cos2 θ − 1, we have 2 cos2 θ − 1 = 1 + cos θ or 2 cos2 θ −√cos θ − 2 = 0.√From the quadratic formula, 1 ± 17 1 + 17 cos θ = . Since > 1, we solve only cos θ = 4 4 √ 1 − 17 . This gives θ ≈ 2.4667 and θ ≈ 3.8165. For both of 4 these values r ≈ 0.219. Writing −r = cos 2(θ + π), we have r = − cos 2θ = −2 cos2 θ + 1. Solving this equation with r = 1 + cos θ, we have −2 cos2 θ + 1 = 1 + cos θ or cos θ(2 cos θ + 1) = 0. For cos θ = 0 we obtain θ = π/2 and θ = 3π/2. For both of these values r = 1. From cos θ = −1/2 we obtain θ = 2π/3 and θ = 4π/3. For both of these values r = 1/2. Thus, (0.219, 2.47), (0.219, 3.82), (1, π/2), (1, 3π/2), (1/2, 2π/3), and (1/2, 4π/3) are points of intersection. From the graph we see that the pole ((0, π/4) on r = cos 2θ and (0, π) on r = 1 + cos θ) is also a point of intersection.

polar axis

polar axis

1

45. Setting 4 sin θ cos2 θ = sin θ we obtain sin θ(4 cos2 θ −1) = 0. This gives θ = 0, θ = √ π/3, and θ = 2π/3. √ The points of intersection are (0, 0), ( 3/2, π/3), and ( 3/2, 2π/3).

1

polar axis

10.5. GRAPHS OF POLAR EQUATIONS

709

46. From the figure we see that the graphs intersect at the pole (which occurs for θ = π on the cardioid and θ = π/2 on the lemniscate) and at (2, 0). Setting (1 + cos θ)2 = 4 cos θ we have cos2 θ − 2 cos θ + 1 = 0 =⇒ (cos θ − 1) = 0 =⇒ cos θ = 1

polar axis

1

which yields only the point (2,0). The points of intersection in the second and third quadrants occur when π/2 < θ < 3π/2 on the cardioid and when −π/2 < θ < π/2 and r < 0 on the lemniscate. If (r2 , θ2 ) represents a point in the second or third quadrants on the cardioid, we have r2 = 1 + cos θ2 . If (r1 , θ1 ) is a point in the second or third quadrants on the lemniscate, we have r12 = 4 cos θ. At points of intersection then, we have r2 = 1 + cos θ, r12 = 4 cos θ, r2 = −r1 , and θ2 = θ1 + π. Substituting the last two equations into the first equation, we obtain −r1 = 1 + cos(θ1 + π) = 1 − cos θ1 or r12 = 1 − 2 cos θ1 + cos2 θ1 . Combining with r12 = 4 cos θ1 we have √ √ 6 ± 36 − 4 2 2 1 − 2 cos θ1 + cos θ1 = 4 cos θ1 =⇒ cos θ1 − 6 cos θ1 + 1 = 0 =⇒ cos θ1 = = 3 ± 2 2. 2 √ Since cos θ1 ≤ 1, cos θ1 = 3−2 2 and θ1 ≈ 1.40 or θ1 ≈ −1.40. In either case r1 ≈ −0.83. The point of intersection in the second quadrant occurs when θ ≈ −1.40 and r ≈ −0.83 on the lemniscate and when θ ≈ −1.40 + π ≈ 1.74 and r ≈ 0.83 on the cardioid. In the third quadrant the point of intersection occurs when θ ≈ 1.40 and r ≈ −0.83 on the lemniscate and when θ ≈ 1.40 + π ≈ 4.54 and r ≈ 0.83 on the cardioid. 47.

(a)

(b)

polar axis

polar axis

48.

(a)

(b)

polar axis

49. (d)

(c)

50. (c)

(d)

polar axis

(c)

polar axis

(d)

polar axis

polar axis

51. (b)

polar axis

52. (a)

710

CHAPTER 10. CONICS AND POLAR COORDINATES

53.

polar axis

54. For a = 0, the graph is a circle. For a = 14 , 12 , and 34 , the graph is a limacon with an interior loop. For a = 1, the graph is a cardiod. For a = 54 , 32 , and 74 , the graph is a dimpled limacon For a = 2, 94 , 52 , 11 4 , and 3, the graph is a convex limacon. As a → ∞, the graphs more closely approximate a circle. 55.

56. (r,θ)

(r,θ) θ

θ

(r, θ+π)

(−r, π−θ)

Symmetric with respect to the origin.

Symmetric with respect to the x-axis. 57.

58. (r,θ)

θ

(−r, −θ)

(r,θ) θ

(−r,θ+2π)

Symmetric with respect to the origin.

Symmetric with respect to the y-axis .

10.6. CALCULUS IN POLAR COORDINATES

711

59. Symmetric with respect to the x-axis. 60. The graph is symmetric with respect to the y-axis. 61. (a) The graphs are identical. (b) The graphs are identical.

62. From the statement in the text preceding Problem 33 in Section 4.1 we have that the component of acceleration in the direction of the ramp is −g sin θ, where g is the acceleration due to gravity and −π ≤ θ ≤ 0. Thus the distance traveled in time t along the ramp at angle θ is r = − 12 gt2 sin θ. But this is the equation of a circle of radius gt2 /4 centered at (0, −gt2 /4), whose topmost point is (0, 0) which is taken at the point of release.

PROBLEMAS 2.5 in Polar Coordinates 10.6 Calculus 1.

dy = θ cos θ + sin θ dθ dx = −θ sin θ + cos θ dθ dy θ cos θ + sin θ 2 = = − at θ = dx −θ sin θ + cos θ π

π 2.

2. At θ = 3, dy cos θ sin θ cos 3 sin 3 = − 2 = − dθ θ θ 3 9 dx − sin θ cos θ sin 3 cos 3 = − 2 =− − dθ θ θ 3 9 dy 3 cos 3 − sin 3 = dx −3 sin 3 − cos 3 3. At θ = π3 , dy = (4 − 2 sin θ) cos θ + (−2 cos θ) sin θ dθ �√ � � � √ 3 1 = 4 − 4 sin θ cos θ = 4 − 4 =4− 3 2 2 dx = −(4 − 2 sin θ) sin θ + (−2 cos θ) cos θ dθ � � � � 3 1 2 2 = −4 + 2 sin θ − 2 cos θ = −4 + 2 −2 4 4 3 1 = −4 + − = −3 √2 2 dy 4− 3 = dx −3

712

CHAPTER 10. CONICS AND POLAR COORDINATES

3π 4. At θ = , 4 dy = (2 − cos θ) cos θ + (sin θ) sin θ dθ � � � � 1 1 = 1 − cos2 θ + sin2 θ = 1 − + =1 2 2 dx = −(1 − cos θ) sin θ + (sin θ) cos θ θ = − sin θ + 2 cos θ sin θ �√ � �√ � √ 2 2 2 =− +2 2 2 2 √ 2 =1− 2 dy 1 √ = dx 1− 2 2

5. At θ = π/6, dy = sin θ cos θ + cos θ sin θ dθ �√ � � � √ 3 1 3 = 2 cos θ sin θ = 2 = 2 2 2 � � � � dx 1 3 1 = − sin2 θ + cos2 θ = − + = dθ 4 4 2 √ dy = 3 dx

6. At θ = π/4, dy = (10 cos θ) cos θ + (−10 sin θ) sin θ dθ � � � � 1 1 2 2 = 10 cos θ = 1 − sin θ = 10 − 10 2 2 =0

dx = −(10 cos θ) sin θ + (−10 sin θ) cos θ dθ �√ � �√ � 2 2 = −10 cos θ sin θ = −20 2 2 = −10 dy 0 = =0 dx −10

10.6. CALCULUS IN POLAR COORDINATES 7.

713

dy = (2 + 2 cos θ) cos θ + (−2 sin θ) sin θ dθ = 2(1 + cos2 θ − sin2 θ) dx = −(2 + 2 cos θ) sin θ + (−2 sin θ) cos θ dθ = −2 − 4 cos θ sin θ

dy 2(1 + cos2 θ − sin2 θ) 1 + cos2 θ − sin2 θ = = dx 2(−2 − 2 cos θ sin θ) −1 − 2 cos θ sin θ If the tangent line is horizontal, we must have 1 + cos2 θ − sin2 θ = 0

which requires sin θ = ±1 and thus θ = π2 or θ = 3π 2 . Hence, the polar coordinates of points on the graph with horizontal tangents are (2, π/2) and (2, 3π/2). If the tangent line is vertical, we must have 1 −1 − 2 cos θ sin θ = 0 or cos θ sin θ = − 2 3π 7π which occurs at θ = 4 or θ = 4 . Hence the polar coordinates of points on the graph with √ √ vertical tangents are (2 − 3, 3π/4) and (2 + 3, 7π/4). 8.

dy = (1 − sin θ) cos θ + (− cos θ) sin θ dθ = 1 − 2 sin θ cos θ dx = −(1 − sin θ) sin θ + (− cos θ) cos θ dθ = − sin θ + sin2 θ − cos2 θ dy 1 − 2 sin θ cos θ = dx − sin θ + sin2 θ − cos2 θ If the tangent line is horizontal, we must have 1 − 2 sin θ cos θ = 0

or

sin θ cos θ =

1 2

and = 5π 4 .� Hence, � θ√ � the√polar �coordinates of points on the graph with 2 π horizontal tangents are 1 − 2 , 4 and 1 + 22 , 5π 4 . If the tangent line is vertical, we must have − sin θ + sin2 θ − cos2 θ = 0 which occurs at θ =

π 4

− sin θ + sin2 θ − (1 − sin2 θ) = 0

2 sin2 θ − sin θ − 1 = 0

9.

11π which gives sin θ = − 12 or sin θ = 1. This occurs at θ = π2 , θ = 7π 6 , and � θπ = � � 36 .7πHence, � the polar � 3 11π � coordinates of points on the graph with vertical tangents are 0, 2 , 2 , 6 , and 2, 6 .

dy = (4 cos 3θ)(cos θ) + (−12 sin 3θ)(sin θ) dθ dx = −(4 cos 3θ)(sin θ) + (−12 sin 3θ)(cos θ) dθ

714

CHAPTER 10. CONICS AND POLAR COORDINATES π and θ = 2π 3 . At θ = 3 , we have �√ � �1� 3 (−4) + (0) 2 2 dy �√ � = � � dx −(−4) 23 + (0) 12 √ −2 3 = √ =− 3 2 3 √ and the rectangular coordinates of the point are (−2, −2 3). Hence, the equation of the √ √ 3 tangent line is y = −2 3 − (x + 2). At θ = 2π 3 , we have 3 �√ � � � 4 − 12 + (0) 23 dy �√ � = � � dx −(4) 23 + (0) − 12 √ 2 3 = √ = 3 2 3 √ and the rectangular √coordinates of the point are (−2, 2 3). Hence, the equation of the tangent √ 3 line is y = 2 3 + (x + 2). 3

The points on the graph correspond to θ =

10.

π 3

dy = (1 + 2 cos θ) cos θ + (−2 sin θ) sin θ dθ = 1 + 2 cos2 θ − 2 sin2 θ dx = −(1 + 2 cos θ) sin θ + (−2 sin θ) cos θ dθ = − sin θ − 4 cos θ sin θ At θ = π3 , we have � √ �2 � 1 �2 3 1 + 2 − 2 2 2 dy � � = 0. = √ √ � � 3 dx − 3 −4 1 2

2

2

Also, r = 2 so the rectangular coordinates of the point are (1, √ the tangent line is y = 3. 5π At θ = , we have 3 � √ �2 � �2 1 + 2 12 − 2 − 23 dy = √ � 1 � � √3 � = 0. 3 dx − 4 − 2 2 2

√

3). Hence, the equation of

√ Also, r = 2 so the rectangular coordinates of the point are (1, − 3). Hence, the equation of √ the tangent line is y = − 3. dr 11. r = 0 when sin θ = 0 which occurs at θ = 0 and θ = π. = −2 cos θ �= 0 at either θ = 0 or dθ θ = π. Therefore, θ = 0 and θ = π define tangent lines to the graph at the origin.

10.6. CALCULUS IN POLAR COORDINATES

dr = −3 sin θ �= 0 at either θ = dθ define tangent lines to the graph at the origin.

12. r = 0 when cos θ = 0 which occurs at θ = Therefore, θ =

π 2

π 2

and θ =

3π 2 .

π 2

and θ = 3π 2 √ 1 2 5π 7π dr √ 13. r = 0 when sin θ = − √ = − which occurs at θ = and θ = . = 2 cos θ �= 0 at 2 4 4 dθ 2 5π 7π 5π 7π θ= or θ = . Therefore, θ = and θ = define tangent lines to the graph at the 4 4 4 4 origin. or θ =

3π 2 .

715

1 dr which occurs at θ = π6 and θ = 5π . = −2 cos θ �= 0 at θ = 6 2 dθ 5π π 5π θ = 6 . Therefore, θ = 6 and θ = 6 define tangent lines to the graph at the origin.

14. r = 0 when sin θ =

π 6

or

π 5π 7π 9π 11π 13π 15π 17π 19π 15. r = 0 when cos 5θ = 0 which occurs at θ = 10 , 3π 10 , 10 , 10 , 10 , 10 , 10 , 10 , 10 , and 10 . dr nπ = −5 sin 5θ �= 0 at any of these θ values. Therefore, θ = defines a tangent line to the dθ 10 graph at the origin for n = 1, 3, 5, . . . , 19.

16. r = 0 when sin 2θ = 0 which occurs at θ = 0,

π 2,

dr = 4 cos 2θ �= 0 at any of these dθ define tangent lines to the graph at the

π, and

θ values. Therefore, θ = 0, θ = π2 , θ = π and θ = origin. � ��π � � 1 π 1 17. A = 4 sin2 θdθ = θ − sin 2θ �� = π 2 0 2 0

3π 2

3π 2 .

1

18. A =

1 2

1 19. A = 2

�

�

π

100 cos2 θdθ = 0

�

25θ +

2π 2

(4 + 4 cos θ) dθ = 8 0

�

��π � 25 sin 2θ �� = 25π 2 0

polar axis

polar axis

2π

(1 + 2 cos θ + cos2 θ)dθ 0 2π

= (8θ + 16 sin θ + 4θ + 2 sin 2θ)|0 = 24π polar axis

716

CHAPTER 10. CONICS AND POLAR COORDINATES

20. A = =

21. A = =

22. A = =

23. A =

1 2 �

1 2 �

1 2 �

1 2

9 = 4

24. A = =

1 2 1 6

1 = 6

�

2π

(1 − sin θ)2 dθ =

0

1 2

�

2π 0

(1 − 2 sin θ + sin2 θ)dθ

��2π � 1 1 1 3 θ + cos θ + θ − sin 2θ �� = π 2 4 8 2 0

�

2π

(3 + 2 sin θ)2 dθ = 0

1 2

�

2π

(9 + 12 sin θ + 4 sin2 θ)dθ

0

��2π � 9 1 θ − 6 cos θ + θ − sin 2θ �� = 11π 2 2 0

�

2π

(2 + cos θ)2 dθ = 0

1 2

�

�

� �

2π

9 sin2 2θdθ

0 4π

2

sin udu = 0

polar axis

2π

(4 + 4 cos θ + cos2 θ)dθ 0

��2π � 1 1 9 2θ + 2 sin θ + θ + sin 2θ �� = π 4 8 2 0 �

polar axis

polar axis

u = 2θ du = 2dθ �

��4π � 9 9 9 u− sin 2u �� = π 8 16 2 0

polar axis

π

cos2 3θdθ

u = 3θ, du = 3dθ

0 3π

cos2 udu 0

�

��3π � 1 1 π u + sin 2u �� = 2 4 4 0

polar axis

10.6. CALCULUS IN POLAR COORDINATES 1 25. A = 2

�

3π/2

717

�3π/2 29 3 2 3 �� 4θ dθ = θ � = π 3 0 4 2

0

polar axis

26. A =

1 2

�

π π/2

�π� θ

dθ = −

�π �π � π π 2 �� =− −π = � 2θ π/2 2 2 polar axis

27. A =

1 2

1 28. A = 2

�

�

π

e2θ dθ = 0

2

�π 1 2θ �� 1 1 e � = e2θ − 4 4 4 0

100e−2θ dθ

1

�2 = −25e−2θ �1 = −25(e−4 − e−2 ) = 25e−2 − 25e−4

29. A = =

1 2 �

�

π/4

tan2 θdθ = 0

1 2

�

1 2

0

(sec2 θ − 1)dθ

��π/4 � � 1 π 1 1 �� 4−π tan θ − θ � = − −0= 2 2 2 8 8 0

�

polar axis

π/4

�π/3 � 25 cot θ�� 2 π/6 π/6 �√ � √ 25 3 √ 25 3 =− − 3 = 2 3 3

30. A =

polar axis

polar axis

π/3

25 csc2 θdθ = −

polar axis

718

CHAPTER 10. CONICS AND POLAR COORDINATES

� 1 31. A = 4π/3(1 + 2 cos θ)2 dθ 2 2π/3 � 1 = 4π/3(1 + 4 cos θ + 4 cos2 θ)dθ 2 2π/3 � � ���4π/3 � 1 1 1 = θ + 4 sin θ + 4 θ + sin 2θ �� 2 2 4 2π/3 √ 2π − 3 3 = 2 � � 1 2π/3 1 4π/3 32. A = (1 + 2 cos θ)2 dθ − (1 + 2 cos θ)2 dθ 2 −2π/3 2 2π/3 � � 1 2π/3 1 4π/3 = (1 + 4 cos θ + 4 cos2 θ)dθ − (1 + 4 cos θ + 4 cos2 θ)dθ 2 −2π/3 2 2π/3 � � ���2π/3 � � ���4π/3 � � 1 1 1 1 1 1 � � = θ + 4 sin θ + 4 θ + sin 2θ − θ + 4 sin θ + 4 θ + sin 2θ � � 2 2 4 2 2 4 −2π/3 2π/3 √ √ 1 1 = (3 3 + 4π) − (2π − 3 3) 2√ 2 =3 3+π 33. Solving 2 cos 3θ = 1 in the first quadrant, we obtain cos 3θ = 1/2, 3θ = π/3, and θ = π/9. Using symmetry, �� � π/9

A=6

0

=4

�

π/3 0

(4 cos2 3θ − 1)dθ

u = 3θ, du = 3dθ π/3

(cos2 u − 1)du = (2u + sin 2u − u)|0

=

√

polar axis

π 3 + . 3 2

34. The circles intersect at θ = π/4. Using symmetry, � � � � ��π/4 � 1 π/4 2 1 1 π 1 π−2 sin θdθ = A=2 θ − sin 2θ �� = − = . 2 0 2 4 8 4 8 0

polar axis

10.6. CALCULUS IN POLAR COORDINATES 35. Solving 5 sin θ = 3 − sin θ in the first quadrant, we obtain sin θ = 1/2 and θ = π/6. Using symmetry, � � � 1 π/2 2 2 A=2 (25 sin θ − (3 − sin θ) )dθ 2 π/6 � π/2 = (24 sin2 θ + 6 sin θ − 9)dθ

719

polar axis

π/6

π/2

= (12θ − 6 sin 2θ − 6 cos θ − 9θ)|π/6 √ √ � √ 3π � π = − − 3 3 − 3 3 = π + 6 3. 2 2

36. From Problem 35, the point of intersection in the first quadrant is at θ = π/6. Using symmetry, � � � � π/3 1 π/6 1 A=2 25 sin2 θdθ + (3 − sin θ)2 dθ 2 0 2 π/6 � π/6 � π/3 2 = 25 sin θdθ + (9 − 6 sin θ + sin2 θ)dθ �

0

polar axis

π/6

��π/6 � ��π/2 � � 25 25 1 1 � = θ− sin 2θ � + 9θ + 6 sin θ + θ − sin 2θ �� 2 4 2 4 0 π/6 � � √ � √ � √ √ 25π 25 3 19π 19π 3 21π = − + − +3 3− = − 6 3. 12 8 4 12 8 4

37. Solving 4 − 4 cos θ = 6 in the second quadrant, we obtain cos θ = −1/2 and θ = 2π/3. Using symmetry, � � � � 1 π � 2 A=2 (4 − 4 cos θ) − 36 dθ 2 2π/3 � π = (16 cos2 θ − 32 cos θ − 20)dθ 2π/3

π

= (8θ + 4 sin 2θ − 32 sin θ − 20θ)|2π/3 √ √ √ = −12π − (8π − 2 3 − 16 3) = 18 3 − 4π.

polar axis

720

CHAPTER 10. CONICS AND POLAR COORDINATES

38. From Problem 37, the point of intersection in the second quadrant is at θ = 2π/3. Using symmetry, � � � � 1 2π/3 1 π 2 A=2 (4 − 4 cos θ) dθ + 36dθ 2 0 2 2π/3 � � � 2π/3 2π 2 = 16 (1 − 2 cos θ + cos θ)dθ + 36 π − 3 0

polar axis

2π/3

= (16θ − 32 sin θ + 8θ + 4 sin 2θ)|0 + 12π √ √ √ = 16π − 16 3 − 2 3 + 12π = 28π − 18 3.

39.

40.

� 2π √ � 2π dr = 0 so we have L = 0 32 dθ = 0 3dθ = 6π dθ dr = −6 sin θ so we have dθ � π � L= 36 cos2 θ + 36 sin2 θdθ 0 � π = 6dθ = 6π 0

dr 1 41. = eθ/2 ; dθ 2

�

dr dθ

�2

1 5 + r = eθ + eθ = eθ ; 4 4 2

s=

√

�4 √ √ 5 � 4 θ/2 e dθ = 5eθ/2 �0 = 4(e2 − 1) 0 2

� �2 dr dr −θ 42. = −2e ; + r2 = 4e−2θ + 4e−2θ = 8e−2θ ; dθ √ dθ �π √ √ √ �π s = 2 2 0 e−θ dθ = −2 2e−θ �0 = −2 2(e−π−1 ) = 2 2(1 − e−π ) 43.

dr = 3 sin θ so we have dθ

10.6. CALCULUS IN POLAR COORDINATES L= = = =

� � �

2π 0 2π 0 2π 0

√

18

721

� (3 − 3 cos θ)2 + (3 sin θ)2 dθ

� 9 − 18 cos θ + 9 cos2 θ + 9 sin2 θdθ √ �

18 − 18 cos θdθ 2π

0

√

1 − cos θdθ

� � � � θ 1 θ = (1 + cos θ) −→ 1 − cos θ = 2 sin2 2 2 2 � � � √ � 2π θ = 18 2 sin2 dθ 2 0 � � � 2π θ θ =6 sin dθ u = , du = 2dθ 2 2 0 � π = 12 sin udu cos2

0

π

= 12 (− cos u)|0 = 24 44.

45.

� � � � dr = sin2 θ3 cos θ3 so we have dθ � � � � � � � π� θ θ θ 6 4 2 L= sin + sin cos dθ 3 3 3 0 � � � π� θ 4 = sin dθ 3 0 � ��π/3 � π3 � 1 1 2 =3 sin udu = 3 u − sin 2u �� 2 4 0 0 √ 4π − 3 3 = 8

=

�

π 0

� � θ sin dθ 3 2

u=

θ 1 , du = dθ 3 2

(a) The lemniscate r2 = 9 cos 2θ is only defined for − π4 ≤ 5π θ ≤ π4 and 3π 4 ≤θ ≤ 4 . 1 (b) A = 2

�

π/4

1 9 cos 2θdθ + 2 −π/4

�

5π/4

9 cos 2θdθ 3π/4

u = 2θ, du = 2dθ 9 9 π/2 5π/2 (sin u)|−π/2 + (sin u)|3π/2 4 4 9 9 = + =9 2 2 =

polar axis

722

CHAPTER 10. CONICS AND POLAR COORDINATES �

cos2

�θ�

�θ�

on the interval 0 ≤ θ ≤ π. This same � � equality does not hold on the interval π < θ < 2π since cos θ2 is negative on this interval. √ � � � � In Problem 43, the equality sin2 θ2 = sin θ2 is used which holds for all θ in the interval 0 ≤ θ ≤ 2π.

46. Example 8 uses the fact that

2

= cos

2

1

47. To obtain the area, we can compute the area of half of one of the petals and then use symmetry, multiplying by 16. area of a half-petal =

1 2

1 = 4

� �

1

π/8

sin2 2θdθ

polar axis

u = 2θ, du = 2dθ

0 π/4

sin2 udu

0

� ��π/4 � 1 1 1 u − sin 2u �� 4 2 4 0 π−2 = 32 � � π−2 π−2 Total area = 16 · = 32 2 =

48. Each of the areas will equal 1 + cos θ.

polar axis

3π 2

since the graphs are simply rotations of the graph of r =

polar axis

polar axis

49. No; Let A1 denote the area of the graph of r = 2(1 + cos θ) and let A2 denote the area of the graph of� r = 1 + cos θ. Then � � � 1 2π 1 2π 2 2 A= 4(1 + cos θ) dθ = 4 (1 + cos θ) dθ 2 0 2 0 = 4A1

polar axis

10.7. CONIC SECTIONS IN POLAR COORDINATES

polar axis

723

polar axis

50. The equations of the circles are r = 1, r = 2 sin θ, and r = 2 cos θ. We split the area into three regions to get � � � 1 π/6 1 π/3 1 π A= 4 sin2 θdθ + 1dθ + 4 cos2 θdθ 2 0 2 π/6 2 π/3 � ��π/6 � ��π � � 1 1 1 1 1 π/3 =2 θ − sin 2θ �� + (θ)|π/6 + 2 θ − sin 2θ �� 2 4 2 2 4 0 π/3 √ √ π 3 π 2π 3 = − + + − 6 4 12 3 4

polar axis

51. From L = mr3 dθ = dt we obtain r2 dθ = Ldt/m. Then 1 A= 2

�

θ2 θ1

1 r dθ = 2 2

�

b a

L L dt = (b − a). m 2m

PROBLEMAS 10.7 Conic2.6 Sections in Polar Coordinates 1. Identifying e = 1, the graph is a parabola.

polar axis

2. Writing r = is an ellipse.

1 , we identify e = 1/2. The graph 1 − (1/2) sin θ

polar axis

724

CHAPTER 10. CONICS AND POLAR COORDINATES

3. Writing r = is an ellipse.

15/4 , we identify e = 1/4. The graph 1 − (1/4) cos θ polar axis

4. Writing r = parabola.

5/2 , we identify e = 1. The graph is a 1 − sin θ

polar axis

5. Identifying e = 2, the graph is a hyperbola.

polar axis

6. Writing r =

2 , we identify e = 1/3. The graph 1 + (1/3) sin θ

is an ellipse. polar axis

7. Writing r = hyperbola.

6 , we identify e = 2. The graph is a 1 + 2 cos θ

polar axis

10.7. CONIC SECTIONS IN POLAR COORDINATES

8. Writing r = parabola.

725

6 , we identify e = 1. The graph is a 1 − cos θ

polar axis

9. Writing r =

2 , we identify e = 4/5. The graph 1 + (4/5) sin θ polar axis

is an ellipse.

1 , we identify e = 5/2. The graph 1 + (5/2) cos θ is an hyperbola.

10. Writing r =

polar axis

11. From r =

6 , we have e = 2. Converting to a rectangular equation, we get 1 + 2 sin θ 6 1 + 2 sin θ r + 2r sin θ = 6 r=

�

x2

r = 6 − 2r sin θ

+ y 2 = 6 − 2y

x2 + y 2 = 36 − 24y + 4y 2

(y − 4)2 x2 − =1 4 12 with a = 2 and b =

√

12 so c2 = a2 + b2 = 4 + 12 = 16 so c = 4. Thus e =

c 4 = = 2. a 2

726

CHAPTER 10. CONICS AND POLAR COORDINATES

12. From r = get

10 5 = , we have e = 32 . Converting to a rectangular equation, we 2 − 3 cos θ 1 − 32 cos θ 10 2 − 3 cos θ 2r = 10 + 3r cos θ r=

2

� x2 + y 2 = 10 + 3x

4(x2 + y 2 ) = 100 + 60x + 9x2

with a = 4 and b = 13. From r = get

√

(x + 6)2 y2 − =1 16 20 20 so c2 = a2 + b2 = 16 + 20. Thus c = 6 and e =

c a

=

6 4

= 32 .

12 4 = , we have e = 23 . Converting to a rectangular equation, we 2 3 − 2 cos θ 1 − 3 cos θ 12 3 − 2 cos θ 3r = 12 + 2r cos θ r=

� 3 x2 + y 2 = 12 + 2x � �2 x − 24 y2 5 + 144 = 1 1296 25

with a =

36 5

and b =

12 √ 5

2

2

5

2

so c = a − b =

1296 25

− 144 5 =

576 25 .

Thus c =

24 5

so e =

c a

=

25/4 36/5

= 23 .

√ √ 2 3 2 1 3 14. From r = √ = , we have e = √ = . Converting to a rectangular 1 3 1 + √3 sin θ 3 + sin θ 3 equation, we get √ 2 3 r= √ 3 + sin θ √ √ 3r = 2 3 − r sin θ √ � √ 2 3 x + y2 = 2 3 − y √ x2 (y + 3) + =1 6 9 √ √ √ c 3 2 2 2 with a = 3 and b = 6 so c = a − b = 9 − 6 = 3. Hence c = 3 and e = = . a 3 15. Since e = 1, the conic is a parabola. The directrix is 3 units to the right of the focus and 3 perpendicular to the x-axis. Therefore, r = . 1 + cos θ 16. Since e = 32 , the conic is a hyperbola. The directix is 2 units above the focus and parallel to 3 the x-axis. Therefore, r = . 1 + 32 sin θ

10.7. CONIC SECTIONS IN POLAR COORDINATES

727

17. Since e = 23 , the conic is an ellipse. The directrix is 2 units below the focus and parallel to the x-axis. Therefore, r = 18. Since e =

1−

4 3 . 2 3 sin θ

1 2,

the conic is an ellipse. The directrix is 4 units to the right of the focus and 2 perpendicular to the x-axis. Therefore, r = . 1 1 + 2 cos θ

19. Since e = 2, the conic is a hyperbola. The directrix is 6 units to the right of the focus and 12 perpendicular to the x-axis. Therefore, r = . 1 + 2 cos θ 20. Since e = 1, the conic is a parabola. The driectrix is 2 units below the focus and parallel to 2 the x-axis. Therefore, r = . 1 − sin θ 21. r = 22. r =

3 � 1 + cos θ + 1+

3 2

2π 3

�

3 � � sin θ − π6

23. Since the vertex is

3 2

units below the focus, the directrix must be 3 units below the focus and 3 parallel to the x-axis. Therefore, r = . 1 − sin θ

24. Since the vertex is 2 units to the left of the focus, the directrix must be 4 units tot he left of 4 the focus and perpendicular to the x-axis. Therefore, r = . 1 − cos θ 25. Since the vertex is

1 2

units to the left of the focus, the directrix must be 1 units to the left of 1 the focus and perpendicular to the x-axis. Therefore, r = . 1 − cos θ

26. Since the vertex is 2 units to the right of the focus, the directrix must be 4 units to the right 4 of the focus and perpendicular to the x-axis. Therefore, r = . 1 + cos θ 27. Since the vertex is

1 4

28. Since the vertex is

3 2

units below the focus, the directrix must be 1/2 parallel to the x−axis. Therefore, r = . 1 − sin θ

1 2

units below the focus and

units above the focus, the directrix must be 3 units above the focus and 3 parallel to the x-axis. Therefore, r = . 1 + sin θ 4 rotated counterclockwise by π/4. The original parabola 1 + cos θ had its focus at the origin and its directrix at x = 4. The original vertex therefore had polar coordinates (2, 0). After rotation, the vertex is located at (2, π/4).

29. This is the parabola r =

728

CHAPTER 10. CONICS AND POLAR COORDINATES 5 5/3 = rotated counterclockwise by π/3. The 3 + 2 cos θ 1 + 2/3 cos θ original ellipse had vertices at θ = 0 and θ = π. The polar coordinates of the vertices were (1, 0) and (5, π). After rotation, the vertices are located at (1, π/3) and (5, 4π/3).

30. This is the ellipse r =

10 5 = rotated clockwise by π/6. The original ellipse 1 2 − sin θ 2 − 2 sin θ had vertices at θ = π/2 and θ = 3π/2. The polar coorinates of the vertices were (10, π/2) and (10.3, 3π/2). After rotation, the vertices are located at (10.π/3) and (10/3, 4π/3).

31. This is the ellipse r =

6 rotated clockwise by π/3. The original hyperbola had 1 + 2 sin θ vertices at θ = π/2 and θ = 3π/2. The polar coordinates of the vertices were (2, π/2) and (−6, 3π/2). After rotation, the vertices are located at (2, π/6) and (−6, 7π/6).

32. This is the hyperbola r =

33. Identifying ra = 12, 000 and e = 0.2, we have from (7) in the text 0.2 =

12, 000 − rp . Solving 12, 000 + rp

for rp , we obtain rp = 8, 000 km. 0.2p . When θ = 0, r = 12, 000 so 12, 000 = (1 − 0.2 cos θ) 0.2p 9, 600 = p4 . Thus, p = 48, 000 and the equation of the orbit is r = . (1 − 0.2) (1 − 0.2 cos θ)

34. The equation of the orbit is r =

35. The equation of the orbit is r =

ep . From (7) in the text, (1 − e cos θ)

1.5 × 108 − 1.47 × 108 5 = ≈ 1.67 × 10−2 . 8 8 1.52 × 10 + 1.47 × 10 299 ep When θ = 0, r = ra = 1.52 × 108 = . Thus ep ≈ 1.52 × 108 − 2.52 × 106 ≈ (1 − 1.67 × 10−2 ) (1.49 × 108 ) 1.49 × 108 and the equation of the orbit is r = . (1 − 1.67 × 10−2 cos θ) e=

0.97p . The length of the major axis is the sum (1 − 0.97 cos θ) 0.97p 0.97p 0.97p of ra = r(0) = and rp = r(π) = = . That is 0.03 1 + 0.97 1.97 � � 1 1 ra + rp = 0.97p + = 3.34 × 109 . 0.03 1.97

36. (a) The equation of the orbit is r =

Solving for p we obtain p = 1.02 × 108 . The equation of the orbit is r=

0.97(1.02 × 108 ) . (1 − 0.97 cos θ)

(b) From part (a) ra = r(0) =

9.87 × 107 9.87 × 107 = ≈ 3.29 × 109 miles 1 − 0.97 0.03

10.7. CONIC SECTIONS IN POLAR COORDINATES

729

and

rp = r(π) =

9.87 × 107 9.87 × 107 = ≈ 5.01 × 107 miles . 1 + 0.97 1.97

37.

38.

polar axis

polar axis

39.

40.

polar axis

polar axis

730

CHAPTER 10. CONICS AND POLAR COORDINATES

41.

42.

polar axis

polar axis

43. Continuing the development in the paragraph following (3) in the text, we see that r = e(d + r cos θ) yields (1 − e2 )x2 − 2e2 dx + y 2 = e2 d2 which in turn gives 2e2 d y2 x2 − x+ 2 1−e 1 − e2 � � 2 2e2 d e2 d y2 x2 − x+ + 2 2 1−e 1−e 1 − e2 � � 2 e2 d y2 x− + 1 − e2 1 − e2 � � 2 e2 d x − 1−e 2 y2 � � +� � 1 (1−e2 )2

1 1−e2

= = =

e2 d2 1 − e2

e2 d2 + 1 − e2

�

e2 d 1 − e2

�2

e2 d2 (1 − e2 ) e4 d2 1 + = 2 2 2 2 (1 − e ) (1 − e ) (1 − e2 )2

= 1,

When 0 < e < 1, both the denominators are positive and the denominator of the fraction involving the x term is smaller. Therefore, this is in the standard form for an ellipse with center and foci on the x-axis. When e > 1, the first denominator is positive while the second is negative. Therefore, this is in the standard form for a hyperbola with center and foci on the x-axis.

ed = ra . 1−e ed At θ = π, r = rp so = rp . 1−e

44. At θ = 0, r = ra so

Solving the second equation for d, we have d =

(1 + e)rp . Plugging this value for d into the e

CHAPTER 10 IN REVIEW

731

first equation, we have e

�

(1+e)rp e

�

= ra ra (1 + e)rp = ra 1−e rp + erp = ra − era

r(ra + rp ) = ra − rp ra − rp e= ra + rp

PROBLEMAS DE REPASO DE LA UNIDAD 2

Chapter 10 in Review A. True/False 1. True 2. True 3. True

4. False; there are no y-intercepts since −y 2 /b2 = 1 has no real solution. 5. True 6. True 7. False; (−r, θ) and (r, θ + π) are the same point. 8. False; since x = t2 , x ≥ 0 in the parametric form, but (−1, 2) is on the graph of y = x2 + 1. 9. True; solving x = t2 + t − 12 = (t − 3)(t + 4) = 0 we obtain t = 3 and t = 4. Since 33 − 7(3) = 27 − 21 = 6, the graph intersects the y-axis at (0, 6). 10. True 11. True 12. False; since 6 is even the graph has 12 petals. 13. False; the same point can be expressed as (−4, π/2), which does satisfy the equation. 14. True 15. True 16. True; since e = 1/15 is close to 0. 17. True 18. True; since r = −5 sec θ is equivalent to r cos θ or x = −5.

732

CHAPTER 10. CONICS AND POLAR COORDINATES

19. False; if r < 0, the point (r, θ) is in the same quadrant as the terminal side of θ + π. 20. True 21. True 22. True 23. True 24. False; this integral will compute the area inside the inner loop of the limacon twice. 25. False 26. False; r = cos θ and r = sin θ intersect at the pole which is (0, π/2) for r = cos θ and (0, 0) for r = sin θ.

B. Fill in the Blanks 1. 4p = 1/2, p = 1/8. The focus is (0, 1/8). 2. c2 = 4 + 12 = 16. The foci are (±4, 0). 3. The center is (0, 2). 4. The asymptotes are 25y 2 − 4x2 = 0 or y = ±2x/5. 5. 4p = 8, p = 2. The directrix is y = −3 − 2 = −5. 6. a2 = 36, b2 = 16. The vertices are (−1 ± 6, −7) and (−1, −7 ± 4) or (−7, −7), (5, −7), (−1, −11), and (−1, −3). 7. Completing the square, y + 10 = (x + 2)2 . The vertex is (−2, −10). 8. b2 = 9. The length of the conjugate axis is 2 · 3 = 6. 9. a2 = 4. The endpoints of the transverse axis are the vertices (4 ± 2, −1) or (2, −1) and (6, −1). 10. The major axis is on the line x = 3. 11. Completing the square, we have 25(x2 − 8x + 16) + (y 2 + 6y + 9) = −384 + 409 = 25 or (x − 4)2 + (y + 3)2 /25 = 1. The center of the ellipse is at (4, −3). 12. Setting y = 0 and solving, we have (x + 1)2 + 64 = 100, (x + 1)2 = 36, or x = ±6 − 1. The x-intercepts are -7 and 5. √ 13. Setting x = 0 and solving, we have y 2 − 4 = 1, y 2 = 5, or y = ± 5. The y-intercepts are √ ± 5. 14. Using implicit differentiation, 2yy � − y � + 3 = 0 or y � = tangent line is 15. line

3 (−1)

= −3.

3 . At (1, 1) the slope of the 1 − 2y

CHAPTER 10 IN REVIEW

733

16. x = 0 at t = ±1, so the y-intercepts occurs at (0, 3) and (0, −1). 17. circle 18. convex limacon dr 19. r = 0 at θ = 0, π/3, 2π/3, π, 4π/3, and 5π/3. = 3 cos 3θ �= 0 at any of the θ values dθ nπ mentioned. Thus, the polar equation θ = defines a tangnet to the graph at the origin for 3 n = 0, . . . , 5. 20. From r =

1 = 2 + 5 sin θ 1+

1 2 , 5 2 sin θ

we have e = 52 .

21. The focus is the origin. The directrix is 10 units below the origin and parallel to the x-axis. Therefore, the vertex is 5 units below the origin at (0, −5). 12 6 = , we see that the conic is an ellipse with a directrix 12 units 2 + cos θ 1 + 12 cos θ to the right of the focus at the origin and perpendicular to the x-axis. The two vertices must therefore occur at θ = 0 and θ = π. Thus polar coordinates of the vertices are (4, 0) and (12, π). The foci must therefore be at the origin and at (8, pi). The center is at (4, π).

22. From r =

C. Exercises

√ √ dy sin t dy ( 3/2) −1 1. = . At t = π/2, = = 3. The slope of the normal line is √ and its 3 dx 1 − cos t dx (1/2) equation is � � √ �� √ √ 1 1 π 3 1 π 3 3π y − = −√ x − − or y = − √ x + √ = − x+ . 2 3 2 3 9 3 3 3 3 2. x� (t) = 1 − cos t, y � (t) = sin t � 2π � � s= (1 − cos t)2 + sin2 tdt = 0

√ � = 2

2π

0

√

1 − cos tdt

2π 0

�

1 − 2 cos t + cos2 t + sin2 tdt

by symmetry

√ √ √ � π√ √ � π√ √ � π 1 − cos2 t 1 + cos t √ =2 2 1 − cos tdt = 2 2 1 − cos t √ dt = 2 2 dt 1 + cos t 1 + cos t 0 0 0 � π √ sin t √ =2 2 dt u = 1 + cos t, du = − sin tdt 1 + cos t 0 � � 2 0 � √ � � √ √ √ √ √ √ −du 2 √ = 2 2 lim =2 2 u−1/2 du = 2 2 lim 2 u�b = 2 2 lim (2 2 − 2 b) = 8 u b→0+ b b→0+ b→0+ 2

(3t2 − 18t) 3t = 3t 2 −9, t �= 0. Solving 2 −9 = −6 2t we obtain t = 2. Since x(2) = 8 and y(2) = −26, the point on the graph is (8, −26).

3. The slope of the line 6x+y = 8 is -6 and

dy dx

=

734 4.

CHAPTER 10. CONICS AND POLAR COORDINATES dy dt

2 = 2t = 1t . The slope of the tangent line is 1t and the point on the curve is (t2 + 1, 2t). The equation of a tangent line is then y − 2t = 1t [x − (t2 + 1)]. Since we want the tangent line to pass through (1, 5), we have 5 − 2t = 1t (1 − t2 − 1) = −t. Solving for t we obtain t = 5. The point on the curve is (x(5), y(5)) = (26, 10). Also, the tangent is vertical if t = 0. At t = 0, the point on the graph is (1, 0). The tangent line at this point will also pass through (1, 5).

5. (a) Since 4x2 (1 − x2 ) = y 2 ≥ 0, 1 − x2 ≥ 0, x2 ≤ 1, and |x| ≤ 1. (b) Letting x = sin t, we have y 2 = 4 sin2 t(1 − sin2 t) = 4 sin2 t cos2 t = sin2 2t. Parametric equations for the curve are x = sin t, y = sin 2t, for 0 ≤ t ≤ 2π. (c)

cos 2t) = (2 cos is horizontal when cos 2t = 0 or t = π/4, 3π/4, 5π/4, 7π/4. t . The tangent line√ √ √ √ The points on the graph are ( 2/2, 1), ( 2/2, −1), (− 2/2, 1) and (− 2/2, −1). dy dx

(d) y

1

x

6. The area inside the limacon is � � 1 2π 1 2π (3 + cos θ)2 dθ = (9 + 6 cos θ + cos2 θ)dθ 2 0 2 0 � ��2π � 9 1 1 19π = θ + 3 sin θ + θ + sin 2θ �� = . 2 4 8 2 0

polar axis

The circle r = 4 cos θ has radius 2, so its area is 4π. Thus, 19π the area outside the circle and inside the limacon is − 2 11π 4π = . 2

7. Solving 3 sin θ = 1 + sin θ in the first quadrant, we have sin θ = 1/2 and θ = π/6. Using symmetry, polar axis

CHAPTER 10 IN REVIEW

735

� � � � 1 π/6 1 π/2 2 2 A=2 9 sin θdθ + (1 + sin θ) dθ 2 0 2 π/6 � ��π/6 � ��π/2 � � 9 9 1 1 = θ − sin 2θ �� + θ − 2 cos θ + θ − sin 2θ �� 2 4 2 4 0 π/6 � � √ � � √ �� √ 3π 9 3 3π π 3 5π = − + − 3− = . 4 8 4 4 8 4 1 � π/2 50 cos2 θdθ, we 2 0 √ 5 2 see that the area corresponds to a semicircle of radius . 2

8. Writing A =

� π/2 0

25(1 − sin2 θ)dθ =

9. From x = 2 sin 2θ cos θ, y = 2 sin 2θ sin θ, we find

dy 2 sin 2θ cos θ + 4 cos 2θ sin θ = dx −2 sin 2θ sin θ + 4 cos 2θ cos θ

� dy �� dx �θ=π/4

and

polar axis

�√ � �√ � 2 2 2(1) + 4(0) 2 2 �√ � � √ � = −1. = 2 2 −2(1) + 4(0) 2 2

√ √ √ (a) At θ = π/4, √ x = 2 and y = √ 2. The Cartesian equation of the tangent line is y − 2 = −1(x − 2) or y = −x + 2 2. √ (b) Using x √ = r cos θ and y = r sin θ in (a), we obtain r sin θ = −r cos θ = 2 2 or 2 2 r= . sin θ + cos θ

10. By writing r =

1 (1 −

1 2

sin θ)

we see that e = 1/2 and the graph is an ellipse with major axis

along the y-axis. The vertices on this axis have polar coordinates (2, π/2) and (2/3, 3π/2). The corresponding rectangular coordinates are (0, 2) and (0, −2/3). Thus, the center of the ellipse is at (0, 2/3). Since once focus is at the origin, c = 2/3. From a√= 4/3 and c = 2/3 we find 2 3 4 12 b2 = 16 the 9 − 9 = 9 �and b = � 3 . Thus, � � vertices on the √

minor axis are at − 2 3 3 , 23

and

√

2 3 2 3 , 3

y v

1 c

.

x v

736

CHAPTER 10. CONICS AND POLAR COORDINATES

11. Multiplying both sides of the equation by r, we have r2 = r cos θ + r sin θ. The corresponding Cartesian equation is x2 + y 2 = x + y. 12. Multipying both sides by cos θ, we have r cos θ = 1 − 5 cos2 θ

5x2 x+ y 2 5x2 x−1=− + 2 x y 5x2 x2 + y 2 = − x−1 5x2 y 2 = −x2 − x−1 3 −x − 4x2 y2 = x−1 x=1−

13. 2(r cos θ)(r sin θ) = 5 5 r2 = 2 cos θ sin θ 14. Using x2 + y 2 = r2 and x = r cos θ, we have (r2 − 2r cos θ)2 = 9r2 =⇒ [r(r − 2 cos θ)]2 = 9r2 =⇒ r2 (r − 2 cos θ)2 = 9r2 =⇒ r − 2 cos θ = ±3 =⇒ r = ±3 + 2 cos θ.

Since replacement of (r, θ) by (−r, θ + π) in r = −3 + 2 cos θ gives r = 3 + 2 cos θ, we take the polar equation to be r = 3 + 2 cos θ. 15. By writing the equation as r cos θ = −1 we see that the line is x = −2. The curve is then a parabola with axis along the x-axis, directrix x = −1, and focus at the origin. Since the 1 . vertex is at (−1/2, 0), p = 1 and the equation is r = (1 − cos θ) 16. Since the transverse axis lie along the y − axis and e = 2, the form of the equation is 2p 2p r = . From 4/3 = r(3π/2) = (1+2) we see that 2p = 4 and the equation is (1 − 2 sin θ) 4 r= . (1 − 2 sin θ) 17. r = 3 sin(10θ) 18. r = 2.8 cos(7θ) y 2 x2 y 2 x2 − 2 = 1. The asymptotes for the hyperbola are − =0 100 b 100 b2 or by = ±10x. Since the given asymptotes are 3y = ±5x, we have the proportion 3b = 10 5 . y2 x2 Thus, b = 6 and the equation of the hyperbola is − = 1. 100 36

19. The form of the equation is

CHAPTER 10 IN REVIEW

737

sin θ =1 1 + cos θ cos θ x = r cos θ = =0 1 + cos θ dy cos θ(1 + cos θ) − sin θ(sin θ) cos θ + cos2 θ + sin2 θ = = dθ (1 + cos θ)2 (1 + cos θ)2 1 + cos θ =1 = (1 + cos θ)2 dx − sin θ(1 + cos θ) − cos θ(− sin θ) = dθ (1 + cos θ)2 − sin θ − sin θ cos θ + sin θ cos θ = (1 + cos θ)2 − sin θ = = −1 (1 + cos θ)2 dy 1 = = −1 dx −1 The tangent line therefore has slope -1 and passes through the point (0, 1). Its equations is y = 1 − (x − 1) or y = 1 − x.

20. At θ = π/2,

y = r sin θ =

21. Substituting y = tx into x3 + y 3 = 3axy we obtain x3 + y 3 x = 3atx2 =⇒ (1 + t3 )x = 3at =⇒ x = and y = tx = 22.

3at , 1 + t3

3at2 . (1 + t3 )

dx 3a(1 − 2t3 ) dy 3at(2 − t3 ) = ; = dt (1 + t3 )2 dt (1 + t3 )2 dy Solving dt = 0 we obtain t = 0 and t = 21/3 . Since dx is not zero at these values, the graph √ √ dt has horizontal tangent lines at (0, 0) and ( 3 2a, 3 4a).

23. (a) From x = r cos θ and y = r sin θ we obtain r3 (cos3 θ + sin3 θ) = 3ar2 cos θ sin θ or r = (3a cos θ sin θ) . (cos3 θ + sin3 θ) (b) The loop is formed from θ = 0 to θ = π/2, so the area is � � 1 π/2 9a2 cos2 θ sin2 θ 9a2 π/2 cos2 θ sin2 θ A= dθ = dθ θ 2 0 2 0 (1 + tan3 θ)2 cos6 θ (cos3 θ + sin ) � � 9a2 π/2 sin2 θ 9a2 π/2 tan2 θ = dθ = dθ 2 0 2 0 (1 + tan3 θ)2 cos4 θ (1 + tan3 θ) cos2 θ � 9a2 π/2 tan2 θ sec3 θ = dθ u = tan θ, du = sec2 θdθ 2 0 (1 + tan3 θ)θ � � ���∞ � � � � 9a2 ∞ u2 9a2 1 1 3a2 1 3a2 � = du = − = − lim − 1 = . � 2 0 (1 + u3 )2 2 3 1 + u3 2 t→∞ 1 + t3 2 0

738

CHAPTER 10. CONICS AND POLAR COORDINATES 3at 3at2 and y(t) = . Then, assuming a > 0, we have (1 + t3 ) (1 + t3 ) lim − y(t) = −∞. lim + x(t) = −∞, and lim + y(t) = ∞. Finally, using

24. From Problem 21, x(t) = lim x(t) = ∞,

t→−1−

t→−1

L’Hˆ opital’s Rule,

t→−1

t→−1

3at + 3at2 h 3a + 6at = lim = −a t→−1 t→−1 1 + t3 3t2

lim [x(t) + y(t)] = lim

t→−1

and x + y = −a or x + y + a = 0 is an asymptote. 25. Using symmetry � � � π/2 � �2 � � π/2 � 1 θ 1 2θ A=2 2 sin dθ = 4 1 − cos dθ 2 3 2 3 0 0 � � √ √ � ��π/2 3 2θ �� π 3 3 3 3 = 2 θ − sin =2 − =π− . 2 3 �0 2 2 2 2

26. The circle centered at (1, 0) has polar equation r = 2 cos θ. Solving 1 = 2 cos θ, we obtain θ = π/3. Using symmetry, � � � � � √ �� √ π 1 π/2 π π 3 π/2 A=4 (1 − 4 cos2 θ)dθ = 2(θ − 2θ − sin 2θ)|π/3 = 2 − − − − = 3− . 2 π/3 2 3 2 3 � � 27. (a) r = 2 cos θ − π4

(b) Note that r = 2 cos θ defines a circle of radius 1 centered at (1, 0). A rotation of π/4 �√ √ � � √ �2 puts the center at 22 , 22 . The new rectangular equation is therefore x − 22 + � √ �2 y − 22 = 1

28. (a) r =

1 1 + cos(θ + π/6)

(b) Using the sum of angles identity for cosine, we have r = 1 √ r= 3 1 � 1 +√ 2 cos θ − 2 sin�θ r 1 + 23 cos θ − 12 sin θ = 1 √

r + 23 r cos θ − 12 r sin θ = 1 √ � x2 + y 2 + 23 x − 12 y = 1 √ � x2 + y 2 = 1 − 23 x + 12 y √ √ 2 x2 + y 2 = 34 x2 − 3xy − 3x + y4 + y + 1 2 √ √ 3xy 1 2 + 3x + 34 y 2 − y = 1 4x + 2

1 1 + cos θ cos π/6 − sin θ sin π/6

CHAPTER 10 IN REVIEW

739

29. Taking the center of the ellipse to be at the origin, we have a = 5 × 108 and b = 3 × 108 , since c2 = a2 − b2 , c2 = 1.6 × 1017 and c = 4 × 108 . The minimum distance is a − c = 108 m and the maximum distance is a + c = 9 × 108 m. 30. To find the width, we need to first find the maximum y-value for points on the portion of the petal satisfying 0 ≤ θ ≤ π/4. We know y = r sin θ = cos 2θ sin 2θ and therefore = (cos2 θ − sin2 θ) sin θ dy = (2 cos θ(− sin θ) − 2 sin θ(cos θ)) sin θ + (cos2 θ − sin2 θ) cos θ Thus, dθ = −5 sin2 θ cos θ + cos3 θ

dy dθ

= 0 when cos θ =

= cos θ(cos2 θ − 5 sin2 θ) 0 or when cos2 θ − 5 sin2 θ = 0. Since cos θ �= 0 for 0 ≤ θ ≤ π/4, we need to find where cos2 θ − 5 sin2 θ = 0. This occurs when cos2 θ = 5 sin2 θ

θ = tan

−1

�

1 √ 5

�

1 sin2 θ = 5 cos2 θ 1 = tan2 θ 5 1 √ = tan θ 5 ≈ 0.4205

� � 1 A cursory examination of the graph tells us that the critical point θ = tan−1 √ yields a 5 √ 6 maximum for y on the interval 0 ≤ θ ≤ pi/4, and this maximum is ymax = . The width is 9 √ 2 6 therefore w = 2ymax = . 9

CÁLCULO VECTORIAL MATEMÁTICAS 3

Chapter 12 MANUAL DE SOLUCIONES UNIDAD 3 Vector-Valued Functions FUNCIONES VECTORIALES DE UNA VARIABLE REAL PROBLEMAS 3.1 Functions 12.1 Vector 2 1. Since the square root function  is only defined for nonnegative values, we must have t − 9 ≥ 0. So the domain is (−∞, −3) [3, ∞).

2. Since the natural logarithm is only defined for positive values, we must have 1 − t2 > 0. So the domain is (−1, 1). 3. Since the inverse sine function is only defined for values between -1 and 1, the domain is [−1, 1]. 4. The vector function is defined for all real numbers. 5. r(t) = sin πti + cos πtj − cos2 πtk 6. r(t) = cos2 πti + 2 sin2 πtj + t2 k 7. r(t) = e−t i + e2t j + e3t k 8. r(t) = −16t2 i + 50tj + 10k 9. x = t2 ,

y = sin t,

z = cos t

10. r(t) = t sin t(i + k) = t sin ti + 0j + t sin tk so x = t sin t, 11. x = ln t,

y = 1 + t,

12. x = 5 sin t sin 3t,

z = t3

y = 5 cos 3t,

z = 5 cos t sin 3t 786

y = 0,

z = t sin t

12.1. VECTOR FUNCTIONS

13.

787

14.

z

15.

z

z

y

y

4 y

x

x

16.

x

17.

z

18.

y

y

2

y

x

2

x

x

19.

20.

z

z

y

y x

x

21. z

y x

Note: the scale is distorted in this graph. For t = 0, the graph starts at (1, 0, 1). The upper loop shown intersects the xz-plane at about (286751, 0, 286751).

CHAPTER 12. VECTOR-VALUED FUNCTIONS

788

23.

22.

z

z

y

10 10

x

10

y

x

24. z

y x

25. r(t) = 4, 0 + 0 − 4, 3 − 0t = (4 − 4t)i + 3tj, 0 ≤ t ≤ 1 y

x

26. r(t) = 0, 0, 0 + 1 − 0, 1 − 0, 1 − 0t = ti + tj + tk, 0 ≤ t ≤ 1

12.1. VECTOR FUNCTIONS 27. x = t, y = t, z = t2 + t2 = 2t2 ; r(t) = ti + tj + 2t2 k z

y

x

√ √ √ 28. x = t, y = 2t, z = ± t2 + 4t2 + 1 = ± 5t2 − 1; r(t) = ti + 2tj ± 5t2 − 1k z

y x

29. x = 3 cos t, z = 9 − 9 cos2 t = 0 sin2 t; y = 3 sin t; r(t) = 3 cos ti + 3 sin tj + 9 sin2 tk

z

y x

30. x = sin t, z = 1, y = cos t; r(t) = sin ti + cos tj + k

789

CHAPTER 12. VECTOR-VALUED FUNCTIONS

790

z

y x

31. x = t, y = t, z = 1 − 2t; r(t) = ti + tj + (1 − 2t)k

z

y x

32. x = 11,

y = t,

z = 3 + 2t;

r(t) = i + tj + (3 + 2t)k

z

y x

33. (b); Notice that the y and z values consistently increase while the x values oscillate rapidly between -1 and 1. The only vector fucntion that describes this behavior is (b). 34. (c); The trace of the graph on the xy−plane would look like a circle, while the z value oscillates between 0 and 1. The only vector function that describes this behavior is (c). 35. (d); Notice that the z value is contant. The only vector function that satisfies this constraint is (d). 36. (a); Notice that the x values consistently increase while the trace of the graph on the yz-plane would look like a circle. The only vector function that describes this behavior is (a).

12.1. VECTOR FUNCTIONS

791

37. Letting x = at cos t, y = bt sin t, and z = ct, we have c 2 t2 z2 = 2 = t2 = t2 cos2 t + t2 sin2 t 2 c c a2 t2 cos2 t b2 t2 sin2 = + a2 b2 x2 y2 = 2 + 2 a b

38. z

y

x

39. Letting x = aekt cos t, y = bekt sin t, and z = cekt , we have z2 c2 ekt = = e2kt = e2kt cos2 t + e2kt sin2 t c2 c2 a2 e2kt cos2 t b2 e2kt sin2 t + = a2 b2 2 2 x y + 2 a2 b

40. z

y x

41. x2 + y 2 + z 2 = a2 sin2 kt cos2 t + a2 sin2 kt sin2 t + a2 cos2 kt = a2 sin2 kt + a2 cos2 kt = a2

CHAPTER 12. VECTOR-VALUED FUNCTIONS

792 42.

k=1

k=2

k=3 z z

z

y

y

y

x

x

x

k=4

k = 10

k = 20

z

z

z

y

y

y

x

x

43. (a) z

y

x

(b) r1 (t) = ti + tj + (4 − t2 )k r2 (t) = ti − tj + (4 − t2 )k (c) z

y

x

44. C lies on the surface of the sphere of radius a.

x

12.2. CALCULUS OF VECTOR FUNCTIONS

793

45.

46.

k = 0.1

k = 0.2

k = 0.3 z

z

z

y y

y x

x

47.

x

k=2

k=4 z

z

y

y

x

48.

k=

x

1 10

k=1 z

z

y x

y x

PROBLEMAS 3.2

12.2

Calculus of Vector Functions

1. lim [t3 i + t4 j + t5 k] = 23 i + 24 j + 25 k = 8i + 16j + 32k t→2

CHAPTER 12. VECTOR-VALUED FUNCTIONS

794 2. r(t) =

ln t sin 2t i + (t − 2)5 k + k. Using L’Hˆ opital’s Rule, t 1/t  lim r(t) = +

t→0

 2 cos 2t 1/t i + (t − 2)5 j + k = 2i − 32j 1 −1/t2

3. Using opital’s Rule, we have    L’Hˆ  2t 5t − 1 2et−1 t2 − 1 5t − 1 2et−1 − 2 , , , ,  = 2, 2, 2 lim = lim = t→1 t→1 t−1 t+1 t−1 1 t+1 1 π 4. Since lim tan−1 t = , we have t→∞ 2     e−1 1 e2t 1 −1 −1 , , tan lim t = lim , , tan t t→∞ 2e2t + t 2e−t + 5 t→∞ 2 + te−2t 2 + 5et   1 π , 0, = 2 2 The last equality follows from using L’Hˆ opital’s Rule to get lim te−2t = lim

t→∞

t→∞

t 1 = lim =0 t→∞ 2e2t e2t

5. lim [−4r1 (t) + 3r2 (t)] = −4(i − 2j + k) + 3(2i + 5j + 7k) = 2i + 23j + 17k t→α

6. lim r1 (t) · r2 (t) = (i − 2j + k) = (i − 2j + k) · (2i + 5j + 7k) = −1 t→α

7. Notice that the k component ln(t − 1) is not defined at t = 1. Therefore, r(t) is not continuous at t = 1. 8. Notice that sin πt, tan πt, and cos πt are each continuous at t = 1 since the sine, cosine, and tangent function are continuous on their domains. Therefore, since each of the component functions are continuous at t = 1, we know that r(t) is continuous at t = 1. 9. r (t) = 3i + 8tj + (10t − 1)k so r (1) = 3i + 8j + 9k = 3, 8, 9 3(1.1) − 1, 4(1.1)2 , 5(1.1)2 − (1.1) − 3(1) − 1, 4(1)2 , 5(1)2 − (1) r(1.1) − r(1) = while 0.1 0.1 2.3, 4.84, 4.95 − 2, 4, 4 = 0.1 0.3, 0.84, 0.95 = 3, 8.4, 9.5 = 0.1 −5 i + (6t + 1)j − 3(1 − t)2 k (1 + 5t)2 −5 i + j + 3k = −5, 1, −3 so r (0) = 1

10. r (t) =

12.2. CALCULUS OF VECTOR FUNCTIONS

795



1 , 3(0.05)2 + (0.05), (1 + 0.05)3 r(0.05) − r(0) 1 + 5(0.05) = while 0.05 0.05 0.8, 0.0575, 0.857375 − 1, 0, 1 = 0.05 −0.2, 0.0575, −0.142625 = 0.05 = −4, 1.15, −2.8525 11. r (t) =

1 1 i − 2 j; t t

r (t) = −

12. r (t) = −t sin t, 1 − sin t;

1 k; 1 + t2

 −

1 , 3(0)2 + (0), (1 − 0)3 1 + 5(0)

1 2 i + 3j t2 t

r (t) = −t cos t − sin t, − cos t

13. r (t) = 2te2t + e2t , 3t2 , 8t − 1; 14. r (t) = 2ti + 3t2 j +



r (t) = 4te2t + 4e2t , 6t, 8

r (t) = 2i + 6tj −

15. r (t) = −2 sin ti +√6 cos tj r (π/6) = −i + 3 3j

2t k (1 + t2 )2 16. r (t) = 3t2 i + 2tj r (−1) = 3i − 2j y

y

x

x

8t k (1 + t2 )2  r (−1) = j − 2k

17. r (t) = j −

18. r (t) = −3 sin√ ti + 3 cos√tj + 2k 2 3 2 −3 r (π/4) = i+ j + 2k 2 2

z

z

y x

y x



CHAPTER 12. VECTOR-VALUED FUNCTIONS

796

1 1 8 19. r(t) = ti + j + t3 k; r(2) = 2i + 2j + k; r (t) = i + tj + t2 k; r (2) = i + 2j + 4k 2 3 3 Using the point (2, 2, 8/3) and the direction vector r (2), we have x = 2 + t, y = 2 + 2t, z = 8/3 + 4t. 20. r(t) = (t3 −t)i+

6t j+(2t+1)2 k; t+1

r(1) = 3j+9k;

r (t) = (3t2 −1)i+

6 j+(8t+4)k; (t + 1)2

3 r (1) = 2i + j + 12k. Using the point (0, 3, 9) and the direction vector r (1), we have x = 2 2t, y = 3 + 32 , z = 9 + 12t.

 √ 2 = 6 21. r (t) = et + tet , 2t + 2, 3t2 − 1 so r (0) = 1, 2, −1 and |r (0)| = 12 + 22 + (−1)  1 r (0) 1, 2, −1 2 −1 √ The unit tangent vector at t = 0 is given by  = = √ ,√ ,√ |r (0)| 6 6 6 6 To find the parametric equations of the tangent line at t = 0, we first compute r(0) = 1 2 −1 = 0, 0, 0. The tangent line is then given in vector form as p(t) = 0, 0, 0 + t √ , √ , √ 6 6 6   1 2 −1 1 2 −1 √ t, √ t, √ t or in parametric form as x = √ t, y = √ t, z = √ t. 6 6 6 6 6 6  √ 2 + (2)2 + (1)2 = 22. r (t) = 3 cos 3t, 2 sec2 2t, 1 so r (π) = −3, 2, 1 and |r (π)| = (−3) 14.   −3 r (π) −3, 2, 1 2 1 The unit tangent vector at t = π is given by  = √ = √ ,√ ,√ |r (π)| 14 14 14 14 To find the parametric equations of the tangent line at t = π, we first compute r (π) = 1, 0, π. The tangent line is then given in vector  form as −3 2 1 p(t) = 1, 0, π + t √ , √ , √ 14 14 14   −3 2 1 = 1 − √ t, √ t, π + √ t 14 14 14 −3 2 1 or in parametric form as x = 1 − √ t, y = √ t, z = π + √ t 14 14 14  √ 3 π 1 , , 23. r(π/3) = 2 2 3 r (t) = −  sin t, cos t, 1 √ 3 1 , ,1 r (π/3) = − 2 2 so the tangent  √ line is given  by √ 1 3 π 3 1 p(t) = , , +t − , ,1 2 2 3 2 2  √ √ 1 3 3 1 π = − t, + t, + t 2 2 2 2 3 24. r(0) = 6, 1, 1 r (t) = −3e−t/2 , 2e2t , 3e3t 

12.2. CALCULUS OF VECTOR FUNCTIONS

797

r (0) = −3, 2, 3 So the tangent line is given by r(t) = 6, 1, 1 + t−3, 2, 2 = 6 − 3t, 1 + 2t, 1 + 3t d [r(t) × r (t)] = r(t) × r (t) + r (t) × r (t) = r(t) × r (t) dt d d [r(t) · (tr(t))] = r(t) · (tr(t))+ = r(t) · (tr (t) + r(t)) + r (t) · (tr(t)) 26. dt dt = r(t) · (tr (t)) + r(t) · r(t) + r (t) · (tr(t)) = 2t(r(t) · r (t)) + r(t) · r(t) 25.

27.

d d [r(t) · (r (t) × r (t))] = r(t) · (r (t) × r (t)) + r (t) · (r (t) × r (t)) dt dt = r(t) · (r (t) × r (t) + r (t) × r (t)) + r (t) · (r (t) × r (t)) = r(t) · (r (t) × r (t))

28.

d d [r1 (t) × (r2 (t) × r3 (t))] = r1 (t) × (r2 (t) × r3 (t)) + r (t) × (r2 (t) × r3 (t)) dt dt = r1 (t) × (r2 (t) × r3 (t) + r2 (t) × r3 (t) + r1 (t) × (r2 (t) × r3 (t)) = r1 (t) × (r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t)) + r1 (t) × (r2 (t) × r3 (t))

d 1 [r1 (2t) + r2 ( 1t )] = 2r (2t) − 2 r2 ( 1t ) dt t d 3 2 [t r(t )] = t3 (2t)r (t2 ) + 3t2 r(t2 ) = 2t4 r (t2 ) + 3t2 r(t2 ) 30. dt 2  2   2   2 

2 2 2 1 2  3 2 3 r(t)dt = tdt i + 3t dt j + 4t dt k = t  i + t3 −1 j + t4 −1 k = i + 9j + 15k 31. 2 −1 2 −1 −1 −1 −1  4   4   4 

4 √ √ 32. r(t)dt = 2t + 1dt i + − tdt j + sin πtdt k 29.

0

0

0

0

4 4 4   1 16 2 3/2  1 26 3/2  = (2t + 1)  i − t  j − cos πt k = i− j 3 3 π 3 3 0 0 0 

 

 



2 33. r(t)dt = tet dt i + −e−2t dt j + tet dt k     1 −2t 1 t2 1 1 2 t t + c2 j + e + d3 k = et (t − 1)i + e−2t j + et k + c, = [te − e + c1 ]i + e 2 2 2 2 where c = c1 i + c2 j + c3 k. 

 

 



1 t t2 34. r(t)dt = dt i + dt j + dt k 1 + t2 1 + t2 1 + t2   

 1 1 −1 2 k ln(1 + t ) + c2 j + 1− = [tan t + c1 ]i + 2 1 + t2   1 ln(1 + t2 ) + c2 j + [t − tan−1 t + c3 ]k = [tan−1 t + c1 ]i + 2 1 = tan−1 ti + ln(1 + t2 )j + (t − tan−1 t)k + c, 2

CHAPTER 12. VECTOR-VALUED FUNCTIONS

798

where c = c1 i + c2 j + c3 k.        35. r(t) = r (t)dt = 6dt i + 6tdt j + 3t2 dt k = [6t + c1 ]i + [3t2 + c2 ]j + [t3 + c3 ]k Since r(0) = i + 2j + k = c1 i + c2 j + c3 k, c1 − 1, c2 = −2, and c3 = 1. Thus, r(t) = (6t + 1)i + (3t2 − 2)j + (t3 + 1)k          t sin t2 dt i + − cos 2tdt j = − 12 cos t2 + c1 i + − 12 sin 2t + c2 j 36. r(t) = r (t)dt = Since r(0) = 32 = (− 12 + c1 )i + c2 j, c1 = 2, and c2 = 0. Thus,

r(t) =

1 1 2 − cos t + 2 i − sin 2tj. 2 2

       37. r (t) = r (t)dt = 12tdt i + −3t−1/2 dt j + 2dt k = [6t2 + c1 ]i + [−6t1/2 + c2 ]j + [2t + c3 ]k Since r (1) = j = (6 + c1 )i + (−6 + c2 )j + (2 + c3 )k, c1 = −6, c2 = 7, and c3 = −2. Thus, r (t) = (6t2 − 6)i + (−6t1/2 + 7)j + (2t − 2)k.

r(t) =





2



(6t − 6)dt i +

r (t)dt =



(−6t

1/2

 + 7)dt j +



 (2t − 2)dt k

= [2t3 − 6t + c4 ]i + [−4t3/2 + 7t + c5 ]j + [t2 − 2t + c6 ]k. Since r(1) = 2i − k = (−4 + c4 )i + (3 + c5 )j + (−1 + c6 )k, c4 = 6, c5 = −3, and c6 = 0. Thus, r(t) = (2t3 − 6t + 6)i + (−4t3/2 + 7t − 3)j + (t2 − 2t)k. 38. r (t) =



r (t)dt =



 

 

 sec2 tdt i + cos tdt j + − sin tdt k

= [tan t + c1 ]i + [sin t + c2 ]j + [cos t + c3 ]k Since r (0) = i + j + k = c1 i + c2 j + c3 k, c1 = 1, c2 = 1, and c3 = 0. Thus, r (t) = (tan t + 1)i + (sin t + 1)j + cos tk.

r(t) =

r (t)dt =



 

 

 (tan t + 1)dt i + (sin t + 1)dt j + cos tdt k.

= [ln | sec t| + c4 ]i + [− cos t + t + c5 ]j + [sin t + c6 ]k Since r(0) = −j + 5k = (−1 + c5 )j + (c6 )k, c4 = 0, c5 = 0, and c6 = 5. Thus, r(t) = (ln | sec t| + t)i + (− cos t + t)j + (sin t + 5)k.  √ 39. r (t) = −a sin ti + a cos tj + ck; |r (t)| = (−a sin t)2 + (a cos t)2 + c2 = a2 + c2 2π √ √  2π √ s= 0 a2 + c2 dt = a2 + c2 t0 = 2π a2 + c2

12.2. CALCULUS OF VECTOR FUNCTIONS

799

40. r (t) = i + (cos t − t sin t)j + (sin t + t cos t)k √ |r (t)| = 12 + (cos t − t sin t)2 + (sin t + t cos t)2 = 2 + t2 √ √ √ √ t√ π π√ s = 0 2 + t2 dt = 2 2 + t2 + ln |t + 2 + t2 | 0 = π2 2 + π 2 + ln(π + 2 + π 2 ) − ln 2 t t t 41. r (t) = (−2e + et sin 2t)j + et k  sin 2t + e cos 2t)i + (2e cos 2t √ √ 2  2t 2 2t 2t |r (t)| = 5e cos 2t + 5e sin 2t + e = 6e2t = 6et  √ 3π √  3π √ t s= 0 6e dt = 6et  = 6(e3π − 1) 0

 √ √ √ 42. r (t) = 3i + 2 3tj + 2t2 k; |r (t)| = 32 + (2 3t)2 + (2t2 )2 = 9 + 12t2 + 4t4 = 3 + 2t2 1  1 s = 0 (3 + 2t2 )dt = 3t + 23 t3  = 3 + 23 = 11 3 0

t s 43. From r (t) = 9 cos t, −9 sin t, we find |r (t)| = 9. Therefore, s = 0 9du = 9t so that t = . By 9    s s s s substituting for t in r(t), we obtain r(s) = 9 sin , 9 cos . Note that r (s) = sin , cos 9 9 9 9     s s so that r (s) = sin2 + cos  = 1. 9 9 √ t 44. From r(t) = −5 sin t, 12, 5 cos t, we find |r (t)| = 169 = 13. Therefore, s = 0 13du = 13t s s 12 5 s so that t = . By substituting for t in r(t), we obtain r(s) = 5 cos 13 , 13 s, 13 cos 13 . 13   5 s 12 5 s  Note that r (t) = − 13 sin 13 , 13 , 13 cos 13 so that  25 144 25 25 s  sin + 169 + cos2 13 |r (s)| = =1 169 13 169 √ √ t√ 45. From r (t) = 2, −3, 4, we find |r (t)| = 29. Therefore,  s = 0 29du = 20t so that  s 2 3 4 t = √ . By substituting for t in r(t), we obtain r(s) = 1 + √ s, 5 − √ s, 2 + √ s . 29 29 29 29    9 16 4 2 3 4 Note that r (s) = √29 , − √29 , √29 so that r (s) = + + = 1. 29 29 29 t t t t 46. From r (t) = e cos t − e sin t, e sin t + e cos t, 0 we find √ √ |r (t)| = e2t cost −2e2t cos t sin t + e2t sin2 t + e2t sin2 t + 2e2t sin t cos t + e2t cos2 t = 2e2t = et 2.   √ t √ Therefore, s = 0 eu 2du = 2(et − 1) so that t = ln √s2 + 1 . By substituting for t in r(t), we obtain          √s + 1 cos(ln √s + 1 √s + 1 sin ln √s + 1 , , 1 Note that r(s) = 2  2   2    2        r (s) = √12 cos ln √s2 + 1 − √12 sin ln √s2 + 1 , √12 sin ln √s2 + 1 + √12 cos ln √s2 + 1 , 0 so that                1 cos2 ln √s + 1 − cos ln √s + 1 sin ln √s + 1 + 1 sin2 ln √s + 1  2  2  2   2   2  2  |r (s)| =  + 12 sin2 ln √s2 + 1 + sin ln √s2 + 1 cos ln √s2 + 1 + 12 cos2 ln √s2 + 1 



s s + sin2 ln √ + 1 =1 = cos2 ln √ + 1 2 2

CHAPTER 12. VECTOR-VALUED FUNCTIONS

800 d d d (r · r) = |r|2 = c2 = 0 and dt dt dt Thus, r is perpendicular to r.

47. Since

d dt (r

· r) = r · r + r · r = 2r · r , we have r · r = 0.

48. Let v = ai + bj and r(t) = x(t)i + y(t)j. Then



b a

v · r(t)dt =



b a

[ax(t) + by(t)]dt = a



b a

x(t)dt + b



b a

y(t)dt = v ·

b a

r(t)dt.

t 49. From r(t) = r0 + tv, we get r (t) = v so that |r (t)| = |v|. Therfore s = 0 |r (t)|du = t s . Substituting for t in r(t), we have r (s) = r0 + |v|du = |v|t which gives t = 0 |v| v |v| s   v . Note that r (s) = |v| so that |r (s)| = |v| = 1. |v|v = r0 + s |v|    3 −4 s , 50. (a) |3, −4| = 32 + (−4)2 = 5 so r(s) = 1, 2 + 3, −4 = 1, 2 + s 5 5 5 √ √ (b) r(t) + t1, 2, −1 and |1, 2, −1| = 1 + 4 + 1 = 6 so r(s) = 1, 1, 10 +  = 1, 1, 10  1 2 −1 s √ ,√ ,√ 6 6 6

PROBLEMAS 3.3 on a Curve 12.3 Motion 1. v(t) = 2ti + t3 j; v(1) = 2i + j; |v(1)| = a(t) = 2i + 3t2 j; a(1) = 2i + 3j

√

4+1=

√

5;

y a

v x

√ √ 2 j; v(1) = 2i − 2j; |v(1)| = 4 + 4 = 2 2; 3 t 6 a(t) = 2i + 4 j; a(1) = 2i + 6j t

2. v(t) = 2ti −

y a

v

x

12.3. MOTION ON A CURVE

801

3. v(t) = −2 sinh 2ti + 2 cosh 2tj; v(1) = 2j; |v(0)| = 2; a(t) = −4 cosh 2ti + +4 sinh 2tj; a(0) = −4i

y

v

a

x

√ 1 4. v(t) = −2 sin ti + cos tj; v(π/3) = − 3i + j; 2  √ |v(π/3)| = 3 + √ 1/4 = 13/2; a(t) = −2 cos ti − sin tj; 3 a(π/3) = −i − j 2

y v a x

5. v(t) = (2t − 2)i + k; v(2) = 2j + k; |v(2)| = a(t) = 2j; a(2) = 2j

√

4+1=

√

5;

z

v a y x

6. v(t) = i + j; v(2) = i + j + 12k; |v(2)| = a(t) = 6tk; a(2) = 12k

√

1 + 1 + 144 =

√

146;

z a v

y x

CHAPTER 12. VECTOR-VALUED FUNCTIONS

802

7. v(t) = i + 2tj + 3t2 k; √ √ mathbf v(1) = i + 2j + 3k; |v(1)| = 1 + 1 + 9 = 14; a(t) = 2j + 6tk; a(1) = 2j + 6k

z

a

v y

8. v(t) = i + 3t2 j + k; √ √ v(1) = i + 3j + k; |v(1)| = 1 + 9 + 1 = 11; a(t) = 6tj; a(1) = 6j

x

z

v a y x

9. The particle passes through the xy-plane when z(t) = t2 −5t = 0 or t = 0, 5 which gives us the points (0, 0, 0) and (25, 115, 0). v(t) = 2ti + (3t2 − 2)j + (2t − 5)k; v(0) = −2j − 5k, v(5) = 10i + 73j + 5k; a(t) = 2i + 6tj + 2k; a(0) = 2i + 30j + 2k 10. If a(t) = 0, then v(t) = c1 and r(t) = c1 t + c2 . The graph of this equation is a straight line. √ 11. Initially we are given s0 = 0 and v0 = (480 cos 30◦ )i + (480 cos 30◦ )j = 240 3i + 240j. Using a(t) = −32j we find

v(t) =

a(t)dt = −32tj + c

√ 240 3i + 240j = v(0) = c √ √ v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j

√ r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b 0 = r(0) = b. √ √ (a) The shell’s trajectory is given by r(t) = 240 3ti + (240t − 16t2 )j or x = 240 3t, y = 240 − 16t2 . (b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 900 ft. (c) Solving y(t) = 240t − 16t2 = 16t(15 − t) = 0, we see that the √ shell is at ground level when t = 0 and t = 15. The range of the shell is s(15) = 3600 3 ≈ 6235 ft.

12.3. MOTION ON A CURVE

803

(d) From (c), impact is when t = 15. The speed at impact is  √ |v(15)| = |240 3i + (240 − 32 · 15)j| = 2402 · 3 + (−240)2 = 480 ft/s. √ 12. Initially we are given s0 = 1600j and v0 = (480 cos 30◦ )i + (480 sin 30◦ )j = 240 3i + 240j. Using a(t) = −32j we find

v(t) = a(t)dt = −32tj + c √ 240 3i + 240j = v(0) = c √ √ v(t) = −32tj + 240 3i + 240j = 240 3i + (240 − 32t)j

√ r(t) = v(t)dt = 240 3ti + (240t − 16t2 )j + b 1600j = r(0) = b. √ √ (a) The shell’s trajectory is given by r(t) = 240 3ti+(240t−16t2 +1600)j or s = 240 3t, y = 240t − 16t2 + 1600. (b) Solving dy/dt = 240 − 32t = 0, we see that y is maximum when t = 15/2. The maximum altitude is y(15/2) = 2400 ft. (c) Solving y(t) = −16t2 + 240t + 1600 = −16(t − 20)(t + 5) = 0, we √ see that the shell hits the ground when t = 20. The range of the shell is x(20) = 4800 3 ≈ 8314 ft. (d) From (c), impact is when t = 20. The speed at impact is  √ √ |v(20)| = |240 3i + (240 − 32 · 20)j| = 2402 · 3 + (−400)2 = 160 13 ≈ 577 ft/s.

13. We are given s0 = 81j and v0 = 4i. Using a(t) = −32j, we have

v(t) =

a(t)dt = −32tj + c 4i = v(0) = c

r(t) =

v(t) = 4i − 32tj v(t)dt = 4ti − 16t2 j + b 81j = r(0) = b

r(t) = 4ti + (81 − 16t2 )j. Solving y(t) = 81 − 16t2 = 0, we see that the car hits the water when t = 9/4. Then  √ |v(9/4)| = |4i − 32(9/4)j| = 42 + 722 = 20 13 ≈ 72.11ft/s.

CHAPTER 12. VECTOR-VALUED FUNCTIONS

804

14. Let θ be the angle of elevation. Then v(0) = 98 cos θi + 98 sin θj. Using a(t) = −9.8j, we have

v(t) = a(t)dt = −9.8tj + c 98 cos θi + 98 sin θj = v(0) = c v(t) = 98 cos θi + (98 sin θ − 9.8t)j r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j + b. Since r(0) = 0, b = 0 and r(t) = 98t cos θi + (98t sin θ − 4.9t2 )j. Setting y(t) = 98t sin θ − 4.9t2 = t(98 sin θ − 4.9t) = 0, we see that the projectile hits the ground when t = 20 sin θ. Thus, using x(t) = 98t cos θ, 490 = s(t) = 98(20 sin θ) cos θ or sin 2θ = 0.5. Then 2θ = 30◦ or 150◦ . The angles of elevation are 15◦ and 75◦ . √ √ s 2 s 2 i+ j. Using a(t) = 15. Let s be the initial speed. Then v(0) = s cos 45◦ i + s sin 45◦ j = 2 2 −32j, we have

v(t) = a(t)dt = −32j + c √ √ s 2 s 2 i+ j = v(0) = c 2 2  √ √ s 2 s 2 i+ − 32t j v(t) = 2 2  √ √ s 2 s 2 ti + t − 16t2 j + b. r(t) = 2 2 Since r(0) = 0, b = 0 and  √ √ s 2 s 2 ti + t − 16t2 r(t) = 2 2

j.

√ √ 2 Setting √ y(t) = s 2t/2 − 16t = t(2 √2/2 − 16t) = 0 we see that the ball hits the ground when t√ = 2s/32. Thus, using x(t) = s 2t/2 and the fact that 100 yd = 300 ft, 300 = x(t) = √ s 2 √ s2 ( 2s/32) = and s = 9600 ≈ 97.98 ft/s. 2 32 16. Let s be the initial speed and θ the initial angle. Then v() = s cos θi + s sin θj. Using a(t) = −32j, we have

v(t) = a(t)dt = −32tj + c s cos θi + s sin θj = v(0) = c v(t) = s cos θi + (s sin θ − 32t)j r(t) = st cos θi + (st sin θ − 16t2 )j + b.

12.3. MOTION ON A CURVE

805

Since r(0) = 0, b = 0 and r(t) = st cos θi + (st sin θ − 16t2 )j. Setting y(t) = st sin θ − 16t2 = t(s sin θ − 16t) =, we see that the ball hits the ground when t = (s sin θ)/16. Using x(t) = st cos θi, we see that the range of the ball is

s2 sin 2θ s sin θ s2 sin θ cos θ = . x = 16 16 32 √ 2 ◦ 2 ◦ ◦ 2 ◦ For √ θ2 = 30 , the range is s sin 60 /32 = ◦3s /64 and for θ = 60 the range is s sin 120 /32 = 3s /64. In general, when the angle is 90 − θ then range is [s2 sin 2(90◦ − θ)]/32 = s2 [sin(180◦ − 2θ)]/32 = s2 (sin 2θ)/32. Thus, for angles θ and 90◦ − θ, the range is the same. 17. r (t) = v(t) = −r0 ω sin ωti + r0 ω cos ωtj; v = |v(t)| =  2 2 ω = v/r0 ; a(t)  = r (t) = −r0 ω cos ωti − r0 ω sin ωtj



r02 ω 2 sin2 ωt + r02 ω 2 cos2 ωt = r0 ω

r02 ω 4 cos2 ωt + r02 ω 4 sin2 ωt = r0 ω 2 = r0 (v/r0 )2 = v 2 /r0 .  √ 18. (a) v(t) = −b sin ti + b cos tj + ck; |v(t)| = b2 sin2 t + b2 cos2 t + c2 = b2 + c2 √ t t√ ds √ 2 = b + c2 (b) s = 0 |v(t)|du = 0 b2 + c2 du = t b2 + c2 ; dt  d2 s (c) = 0; a(t) = −b cos ti−b sin tj; |a(t)| = b2 cos2 t + b2 sin2 t = |b|. Thus, d2 s/dt2 = 2 dt |a(t)|. a = |a(t)| =

19. Let the initial speed of the projectile be s and let the target be at (x0 , y0 ). Then vp (0) = s cos θi + s sin θj and vt (0) = 0. Using a(t) = −32j, we have  vp (t) = a dt = −32tj + c s cos θi + s sin θj = vp (0) = c vp (t) = s cos θi + (s sin θ − 32t)j rp (t) = st cos θi + (st sin θ − 16t2 )j + b.

y (x0,y0)

x0 tan θ θ x0

x

Since rp (0) = 0, b = 0 and rp (t) = st cos θi + (st sin θ − 16t2 )j. Also, vt (t) = −32tj + c and since vt (0) = 0, c = 0 and vt (t) = −32tj. Then rt (t) = −16t2 tj + b. Since rt (0) = x0 i + y0 j, bx0 i + y0 j and rt (t) = x0 i +(y0 − 16t2 )j. Now, the horizontal component of rp (t) will be x0 when t = x0 /s cos θ at which time the vertical component of rp (t) will be (sx0 /s cos θ) sin θ − 16(x0 /s cos θ)2 = x0 tan θ − 16(x0 /s cos θ)2 = y−) − 16(x0 /s cos θ)2 . Thus, rp (x0 /s cos θ) = rt (x0 /s cos θ) and the projectile will strike the target as it falls. 20. The initial angle is θ = 0, the initial height is 1024 ft, and the initial speed is s = 180(5280)/3600 = 264 ft/s. Then x(t) = 264t and y(t) = −16t2 + 1024. Solving y(t) = 0 we see that the pack hits the ground at t = 8 seconds. The horizontal distance tranvelled is x(8) = 2112 feet. From the figure in the text, tan α = 1024/2112 = 16/33 and α ≈ 0.45 radian or 25.87◦ . 21. By Problem 17, a = v 2 /v0 = 15302 /(4000 · 5280) ≈ 0.1108. We are given mg = 192, so m = 192/32 and we = 1192 − (192/32)(0.1108) ≈ 191.33 lb.

CHAPTER 12. VECTOR-VALUED FUNCTIONS

806

22. By problem 17, the centripetal acceleration is v 2 /r0 . Then the horizontal force is mv 2 /r0 . The vertical force is 32m. The resultant force is U = (mv 2 /r0 )i + 32mj. From the figure, we see that tan φ = (mv 2 /r0 )/32m = v 2 /32r0 . Using r0 = 60 and v = 44 we obtain tan φ = 442 /32(60) ≈ 1.0083 and φ ≈ 45.24◦ .

< 0, 32m>

φ



23. Solving x(t) = (v0 cos θ)t for t and substituting into y(t) − 12 gt2 + (v0 sin θ)t + s0 we obtain 1 y=− g 2



x v0 cos θ

2 + (v0 sin θ)

x g + s) = − 2 x2 + (tan θ)x + s0 , v0 cos θ 2v0 cos2 θ

which is the equation of a parabola. 24. Since the projectile is launched from ground level, s0 = 0. To find the maximum height we maximize y(t) = − 12 gt2 + (v0 sin θ)t. Solving y  (t) = −gt + v0 sin θ = 0, we see that t = (v0 /g) sin θ is a critical point. Since y  (t) = −g ≤ 0, H=y

v0 sin θ g

=

v02 sin2 θ v0 sin θ 1 v02 sin2 θ g = + v sin θ 0 2 g2 g 2g

is the maximum height. To find the range we solve y(t) = − 12 gt2 + (v0 sin θ)t = t(v0 sin θ − 1 2 gt) = 0. The positive solution to this equation is t = (2v0 sin θ)/g. The range is thus x(t) = (v0 cos θ)

v 2 sin 2θ 2v0 sin θ = 0 . g g

25. Letting r(t) = x(t)i + y(t)j + z(t)k, the equation dr/dt = v is equivalent to dx/dt = 6t2 x, dy/dt = −4ty 2 , dz/dt = 2t(z + 1). Separating the variables and integrating, we obtain x/x = 6t2 dt, dy/y 2 = −4tdt, dz/(z + 1) = 2tdt, and ln x = 2t3 + c1 , −1/y = 2t2 + c2 , ln(z + 1) + t2 + c3 . Thus, 3

r(t) = k1 e2t i +

2 1 j + (k3 et − 1)k. 2t2 + k2

26. We require the fact that dr/dt = v. Then d dp dr dL = (r × p = r + × p = τ + v × p = τ + v × mv = τ + m(v × v) = τ + 0 = τ. dt dt dt dt 27. (a) Since F is directed along r we have F = cr for some constant c. Then τ = r × F = r × (cr) = c(r × r) = 0. (b) If τ = 0 then dL/dt = 0 and L is constant.

12.4. CURVATURE AND ACCELERATION

807

28. (a) Since the cannon is pointing directly to the left, tha parmetric equations describing the path of the cannon ball are given by 1 x(t) = v0 t, y(t) = − gt2 + s0 2



2s0 . At that The cannon ball will touch the groun when y = 0, which occurs at t = g  

2s0 2s0 time, x is given by x = . Notice that this x value will be farther = −v0 g g to the left with increasing values of v0 . Therefore, the cannon ball travels farther with more gunpowder.  2s0 . This value (b) As shown in part (a), the cannon ball will touch the groun when t = g of t is independent of v0 . This occurs because v0 has no vertical component. (c) If the cannon ball is dropped, we have v0 = 0. Therefore, the parametric equations describing the cannon ball motion are given by 1 x(t) = 0, y(t) = − gt2 + s0 . 2  2s0 As before, y = 0 when t = . Therefore the cannon ball touches the ground at the g same time regardless of whether it is fired or dropped.

PROBLEMAS 3.4 12.4 Curvature and Acceleration 1. r (t) = −t sin ti + t cos tj + 2tk; sin t cos t 2 T=− √ i+ √ j+ √ k 5 5 5

|r (t)| =

 √ t2 sin2 t + t2 cos2 t + 4t2 = 5t;

√ 2. r (t) = et (− sin t + cos t)i + et (cos t + sin t)i + 2et k, √ |r (t)| = [et (sin2 t−2 sin t cos t+cos2 t)+e2t (cos2√t+2 sin t cos t+sin2 t)+2e2t ]1/2 = 4e2t = 2et ; 1 1 2 k T(t) = (− sin t + cos t)i + (cos t + sin t)j + 2 2 2  3. √ We assume a > 0. r (t) = −a sin ti + a cos tj + ck; |r (t)| = a2 sin2 t + a2 cos2 t + c2 = a2 + c 2 ; a sin t a cos t a cos t c dT a sin t T(t) − √ = −√ i+ √ j+ √ k; i− √ j, 2 + c2 2 + c2 2 + c2 2 + c2 dt a a a a a2 + c 2    2 2 2 2  dT    = a cos t + a sin t = √ a ; N = − cos ti − sin tj;  dt  2 a2 + a2 + c 2 a2 + c 2   c   i j k     a cos t c a sin t  = √c sin t i − √c cos t + √ a √ √ k; B = T × N =  − √ 2  2 2 2 2 2 a + c a + c a + c a2 + c 2 a2 + c 2 a2 + c 2     − cos t − sin t 0 √ 2 2 a/ a + c |dT/dt| a = √ = 2 κ= 2 2 r (t) a + c2 a +c

CHAPTER 12. VECTOR-VALUED FUNCTIONS

808 √

√ |r ; (1)| = 3; 1 T(t) = (1 + t2 + t4 )−1/2 (i + tj + t2 k), T(1) √ (i + j + k); 3 dT 1 t 2 4 −3/2 3 2 = − (1 + t + t ) (2t + 4t )i + [(1 + t + t)−1/2 − (1 + t2 + t)−3/2 (2t + 4t3 )]j dt 2 2 2 2 4 −1/2 t 2 4 −3/2 3 [2t(1 + t + t ) (1 + t + t ) (2t + 4t )]k; √   2  d 1 d 1 1 2 1 1 T(1) = − √ i + √ k,  T(1) = + = √ ; N(1) = − √ (i − k)k, dt dt 3 3 3 3 3 2     i j k  √ √ √  1 B(1) =  1/ √3 1/ 3 1/√3  = √ (i − 2j + k); 6  −1/ 2 0 1/ 2  √ √ √   d  2/ 3 2 κ =  T(1) = |r (1)| = √ = dt 3 3 √ 1 (3i − 3j + 2 2k). 5. From Example 1 in the text, a normal to the osculating plane is B(π/4) = 26 √ √ An equation The point on√the curve √ when t = π/4 is ( 2, 2, 3π/4). √ √ of the√plane is 3(x − √ 2) − 3(y − 2) + 2 2(z − 3π/4(= 0, 3x − 3y + 2 2z = 3π/2, or 3 2x − 3 2y + 4z = 3π. 4. r (t) = i + tj + t2 k;

|r (t)| =

1 + t 2 + t4 ,

6. From Problem 4, a normal to the osculating plane is B(1) = √16 (i − 2j + k). The point on the curve when t = 1 is (1, 1/2, 1/3). An equaiton of the plane is (x−1)−2(y −1/2)+(z −1/3) = 0 or x − 2y + z = 1/3. √ 7. v(t) = j + 2tk, |v(t)| = 1 + 4t2 ; a(t) = 2k; v · a = 4t, v × a = 2i, |v × a| = 2; 4t 2 , aN = √ aT = √ 2 1 + 4t 1 + 4t2 8. v(t) = −3 sin ti + 2 cos tj + k,  √  |v(t)| = 9 sin2 t + 4 cos2 t = 1 = 5 sin2 t + 4 sin2 t + 4 cos2 t + 1 = 5 sin2 +1; a(t) = −3 cos ti − 2 sin tj; v · a = 9 sin t cos cos t, t − 4 sin t cos t = 5 sin t√ √ 2 v × a = 2 sin ti − 3 cos tj + 6k, |v × a| = 4 sin +(cos2 t + 36 = 5 cos2 t + 8;  √ 5 sin t cos t cos2 t + 8 aT  , aN = sin2 t + 1 sin2 t + 1 √ 9. v(t) = 2ti + 2tj + 4tk, |v(t)| = 2 6t, t > 0; a(t) = 2i + 2j + 4k; v · a = 24t, v × a = 0; √ 24t aT = √ = 2 6, aN = 0, t > 0 2 6t √ 10. v(t) = 2ti − 3t2 j = 4t3 k, |v(t)| = t 4 + 9t2 + 16t4 , t >); a(t) = 2i −√ 6tj + 12t2 k; 3 5 4 3 2 2 v · a = 4t + 18t + 48t ; v × a =√−12t i − 16t j − 6t k, |v × a| = 2t 36t4 + 64t2 + 9; 4 + 18t2 + 48t4 2t 36t4 + 64t2 + 9 , aN = √ t>0 aT = √ 4 + 9t2 + 16t4 4 + 9t2 + 16t4 √ 11. v(t) = 2i + 2tj, |v(t)| = 2 1 + t2 ; a(t) = 2j; v × a = 4k, |v × a| = 4; 2t 2 , aN = √ aT = √ 2 1+t 1 + t2

12.4. CURVATURE AND ACCELERATION

809

√ 1 + t2 1 t 2t 1 − t2 12. v(t) = i+ j, |v(t)| = ; a(t) = − i+ j; 2 2 2 2 2 1+t 1+t 1+t (1 + t ) (1 + t2 )2 t − t3 1 1 2t + ; v×a= k, |v × a| = ; v·a=− (1 + t2 )3 (1 + t2 )3 (1 + t2 )2 (1 + t2 )2 2 3 2 2 t/(1 + t ) t a/(1 + t ) 1 aT = − √ , aN = √ =− = 2 2 2 2 2 (1 + t )3/2 (1 + t2 )3/2 1 + t )/(1 + t 1 + t /(1 + t ) 13. v(t) = −5 sin ti + 5 cos tj, |v(t)| = 5; a(t) = −5 cos ti − 5 sin tj; v × a = 25k, |v × a| = 25; aT = 0, aN = 5

v · a = 0,

  14. v(t) = sinh ti + cosh tj, |v(t)| = sinh t2 + cosh2 t a(t) = cosh ti + sinh tj v · a = 2 sinh t cosh t; v × a = (sinh2 t − cosh2 t)k = −k, |v × a| = 1; 2 sinh t cosh t 1 aT =  , aN =  2 2 2 sinh + cosh sinh + cosh2 √ 15. v(t) = et (i + j + k), √|v(t)| = 3e−t ; a(t) = e−t (i + j + k); v · a = −3e−2t ; v × a = 0, |v × a| = 0; aT = − 3e−t , aN = 0 √ 16. v(t) = i + 2j + 4k, |v(t)| = 21; a(t) = 0; v · a = 0, v × a = 0, |v × a| = 0; aT = 0, aN = 0  17. v(t) = −a sin ti + b cos tj + ck, |v(t)| = a2 sin2 t + b2 cos2 +c2 ; a(t) = −a cos ti − b sin tj; v × a = bc sin ti− ac cos tj + abk, |v × a| = b2 c2 sin2 t + a2 c2 cos2 t + a2 b2 b2 c2 sin2 t + a2 c2 cos2 t + a2 b2 |v × a| = κ= |v|3 (a2 sin2 t + b2 cos2 t + c2 )3/2  18. (a) v(t) = −a sin ti + b cos tj, |v(t)| = a2 sin2 t + b2 cos2 t; a(t) = −a cos ti − b sin tj; ab v × a = abk; |v × a| = ab; κ = 2 2 (a sin t + b2 cos2 t)3/2 (b) When a = b, |v(t)| = a, |v × a| = a2 , and κ = a2 /a3 = 1/a. 19. The equation of a line is v(t) = b + tc, when b and c are constant vectors. v(t) = c, |v(t)| = |c|; a(t) = 0; v × a = 0; κ = |v × a|/|v|3 = 0 20. v(t) = a(1 − cos t)i + a sin tj; v(π) = 2ai, |v(π)| = 2a; a(t) = a sin ti + a cos tj,  i j k   |v × a| 2a2 1  = = a(π) = −aj; |v × a| =  2a 0 0  = −2a2 k; |v × a| = 2a2 ; κ = 3 3 |v| 8a 4a  0 −a 0   21. v(t) = f  (t)i + g  (t)j, |v(t)| = [f  (t)]2 + [g  (t)]2 ; a(t) = f  (t)i + g  (t)j; v × a = [f  (t)g  (t) − g  (t)f  (t)]k, |v × a| = |f  (t)g  (t) − g  (t)f  (t)|; |v × a| |f  (t)g  (t) − g  (t)f  (t)| κ= = |v|3 ([f ”(t)]2 + [g  (t)]2 )3/2 22. For y = F (x), r = xi + F (x)j. We identify f (x) = x and g(x) = F (x) in Problem 21. Then f  (x) = 1, f  (x) = 0, g  (x) = F  (x), g  (x) = F  (x), and κ = |F  (x)|/(1 + [F  (x)]2 )3/2 .

CHAPTER 12. VECTOR-VALUED FUNCTIONS

810 23. F (x) = x2 , F  (x) = 2,

F (0) = 0,

F (1) = 1;

F  (0) = 2,

F  (1) = 2;

F  (x) = 2x,

F  (0) = 0, F  (1) = 2; 2 1 κ(0) = = 2; ρ(0) = ; 2 3/2 2 (1 + 0 )

2 2 = √ ≈ 0.18; 2 3/2 (1√+ 2 ) 5 5 √ 5 5 ≈ 5.59; Since 2 > 2/5 5, the curve is ”sharper” at (0, 0). ρ(1) = 2 κ(1) =

24. F (x) = x3 ,

F (−1) = −1,

F  (1/2) = 3/4;

F (1/2) = 1/8;

F  (x) = 6x, F  (−1) = −6,

F  (x) = 3x2 ,

F  (−1) = 3,

F  (1/2) = 3;

κ(−1) =

3 √ ≈ 0.19; 5 10 √ 5 10 ≈ 5.27; ρ(−1) = 3 3 3 125 ≈ 1.54; ρ( 12 ) = ≈ 0.65 κ( 12 ) = = 125/64 192 [1 + (3/4)2 ]3/2 Since 1.54 > 0.19, the curve is ”sharper” at (1/2, 1/8).

|−6| (1+32 )3/2

=

6 √

10 10

=

|F  (x)| . |1 + (F  (x))2 |3/2 2 . Now, F  (x)2x, F  (x) = 2, and (F  (x))2 = 4x2 so that κ = (1 + 4x2 )3/2 As x → ±∞, the denominator grows without bound. Therefore, κ(x) → 0 as x → ±∞.

25. Letting F (x) = x2 , we can use Problem 22 to get κ(x) =

y

x 26. (a)

√ 2t(t2 + 2) 3t(2t2 + 1) t4 + 4t2 + 1 √ ; − (t4 + t2 + 1)5/2 (t4 + t2 + 1)3/2 t4 + 4t2 + 1 critical numbers occur at t = −.271469, t = 0, and t = .271469.

(b) κ (t) =

(c) Maximum of 1.017182 occurs at t = −.271469 and t = .271469. 27. Since (c, F (c)) is an inflection point and F  exists on an interval containg c, we must have F  (c) = 0. Therefore, using the formula from Problem 22, we see that the curvature is zero. 28. We use the fact that T · N = 0 and T · T = N · N = 1. Then |a(t)|2 = a · a = (an N + at T) · (an N + at T) = a2N N · N + 2an at N · T + a2T T · T = a2N + a2T .

CHAPTER 12 IN REVIEW

811

PROBLEMAS DE REPASO DE LA UNIDAD 3

Chapter 12 in Review A. True/False 1. True; |v(t)| =

√

2

2. True; the curvature of a circle of radius a is κ = a1 . 3. True 4. False; consider r(t) = t2 i. In this case, v(t) = 2ti and a(t) = 2i. Since v · a = 4t, the velocity and acceleration vectors are not orthogonal for t = 0. 5. True

5.2 6. False; see Problem 20c in Section 14.2 7. True 8. True 9. False; consider r1 (t) = r2 (t) = i. 10. True,

dr dr dr d d |r(t)|2 = (r · r) = r · + · r = 2r · . dt dt dt dt dt

B. Fill in the Blanks 1. y = 4 2. 0 3. r (t) = 1, 2t, t2  so r (1) = 1, 2, 1 4. r (t) = 0, 2, 2t so r (1) = 0, 2, 2    i j j    √ 5. r (1) × r (1) =  1 2 1  = 2, −2, 2 so r (1) × r (1) = 12.  0 2 2  √ √ √ 12 2 r (1) × r (1) √ . = Since r (1)| = 6, we have κ(1) = = |r (1)|3 6 6 6   1 1, 2, 1 r (1) 2 1 = √ 6. T(1) =  = √ ,√ ,√ |r (1)| 6 6 6 6    2 1 1, 2t, t  r (t) 2t t2 =√ 7. T(t) =  = √ ,√ ,√ |r (t)| 1 + 4t2 + t4 1 + 4t2 + t4 1 + 4t2 + t4  1 + 4t2 + t4  2 4 −2(t + 2) −2(t − 1) 2t(2t2 + 1) , 4 , 4 So T (t) = . 4 2 3/2 2 3/2 (t + 4t (t + 4t + 1) (t + 4t2 + 1)3/2  + 1)  −6 −1 1 6 1 1 1  √ √ √ . , 0, and |T , 0, (1)| = = This gives T (1) = 6 + 6 = 3/2 63/2 6 6 3  6  −1 √ √1 , 0,  T (1) 1 −1 6 6 Therefore N(1) =  = =  √ , 0, √ . 1 √ |T (1)| ( 3) 2 2

CHAPTER 12. VECTOR-VALUED FUNCTIONS

812   i  1 8. B(1) = T(1) × N(1) =  √6 −1  √

j

√2 6

2

0

  k    1 −1 1 1 √ = √ ,√ ,√ 6  3 3 3 √1  2



9. A normal to the normal plane is T(1) = 1 3 ,

√1 , √2 , √1 6 6 6



so we can use n = 1, 2, 1 as a vector

normal to the plane. Since r(1) = 1, 1, the point (1, 1, 13 ) lies on the normal plane at t = 1. Thus an equation of the normal plane is (x − 1) + 2(y − 1) + (z − 13 ) = 0 or x + 2y + z = 1) 3 or 3x + 6y + 3z = 10   −1 √1 10. A normal to the osculating plane is B(1) = √13 , √ , . So we can use n = 1, −1, 1 as a 3 3 normal vector. Using the point (1, 1, 13 ), an equation of the osculating plane is (z − 1) − (y − 1) + (z − 13 ) = 0 or x − y + z = 13 or 3x − 3y + 3z = 1.

C. Exercises

π

√ π√ cos2 t + sin2 +1dt = 0 2dt = 2π √ √ √ √ t√ 2. r (t) = 5i + j + 7k; s(t) = 0 25 + 1 + 49du = 5 3t; s(3) = 15 3. Solving 5 3t = 80 3, √ we see that the distance traveled will be 80 3 when t = 16 or at the point (80, 17, 112).

1. r (t) = cos ti + sin tj + k;

3. r(3) = −27i + 8j + k;

s=

0

r (t) = −6ti = √

is x = −27 − 18t, y = 8 + t, z = 1 + t. 4.

2 + k; t+1

r (2) = −18i + j + k. The tangent line

5. z

z

y x

y

x

6.

d d d [r1 (t) × r2 (t)] = r1 (t) × r2 (t) + r1 (t) × r2 (t) dt dt dt = (t2 i + 2tj + t3 k) × (−i + 2tj + 2tk) + (2ti + 2j + 2t2 k) × [−ti + t2 j + (t2 + 1)k] = (4t2 − 2t4 )i − 3t3 j + (2t3 + 2t)k + (2t2 + 2 − 3t4 )i − (5t3 + 2t)j + (2t3 + 2t)k = (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k d d [r1 (t) × r2 (t)] = [(2t3 + 2t − t5 )i − (2t4 + t2 )j + (t4 + 2t2 )k] dt dt = (2 + 6t2 − 5t4 )i − (8t3 + 2t)j + (4t3 + 4t)k

CHAPTER 12 IN REVIEW 7.

813

d d d [r1 (t) · r2 (t)] = r1 (t) · r2 (t) + r1 (t) · r2 (t) dt dt dt = (cos ti − sin tj + 4t3 k) · (2ti + sin tj + 2e2t k) (− sin ti − cos tj + 12t2 k) · (t2 i + sin tj + e2t k) = (2t cos t − sin t cos t + 8t3 e2t − t2 sin t − sin t cos t + 12t2 e2t = 2t cos t − t2 sin t − 2 sin t cos t + 8t3 e2t + 12t2 e2t d d [r1 (t) · r2 (t)] = [t2 cos t − sin2 t + 4t3 e2t ] = −t2 sin t + 2t cos t − 2 sin t cos t + 8t3 e2t + 12t2 e2t dt dt

8.

d d [r1 (t) · (r2 (t) × r3 (t))] = r1 (t) · [r2 (t) × r3 (t)] + r (t) · [r2 (t) × r3 (t)] dt dt = r1 (t) · [(r2 (t) × r3 (t)) + (r2 (t) × r3 (t))] + r1 (t) · (r2 (t) × r3 (t)) = r1 (t) · (r2 (t) × r3 (t)) + r1 (t) · r2 (t) × r3 (t)) = r1 (t) · (r2 (t) × r3 (t))

9. We are given F = ma = 2j; v(0) = i + j + k. and r(0) = i + j. Then



2 2 v(t) = a(t)dt = jdt = tj + c m m i = j + k = v(0) = c

2 t+1 j+k v(t) = i + m

1 2 t + t j + tk + b r(t) = ti + m i + j = r(0) = b

1 2 t + t + 1 j + tk r(t) = (t + 1)i + m The parametric equations are x = t, y =

1 2 t + t + 1, z = t. m

10. y

x v

a

v(t) = i − 3t2 j, v(1) √ a(t) = −6tj, a(1) = −6j √ = i − 3j; |v(1)| = |i − 3j| = 1 + 9 = 10

CHAPTER 12. VECTOR-VALUED FUNCTIONS

814

11. v(t) = 6i + j + 2tk; a(t) = 2k. To find when the particle passes through the plane, we solve −6t + t + t2 = −4 or t2 − 5t + 4 = 0. This gives t = 1 and t = 4. v(1) = 6i + j + 2k, a(1) = 2k; v(4) = 6i + j + 8k, a(4) = 2k 12. We are given r(0) = i + 2j + 3k.

r(t) =

v(t)dt =

(−10ti + (3t2 − 4t)j + k)dt = −5t2 i + (t3 − 2t2 )j + tk + c i + 2j + 3k = r(0) = c

r(t) = (1 − 5t2 )i + (t3 − 2t2 + 2)j + (t + 3)k r(t) = −19i + 2j + 5k √ √ √  √ 13. v(t) = a(t)dt = ( 2 sin ti + 2 cos tj)dt = − 2 cos √ ti + 2 sin √tj + c; −i + j + √ k = v(π/4) = −i + j + c, c = k; v(t) = − 2 cos ti + 2 sin tj + k; √ ti− 2 cos tj+tk+b; i+2j+(π/4)k = r(π/4) = −i−j+(π/4)k+b, b = 2i+3j; r(t) = − 2 sin √ √ r(t) = (2 − 2 2 sin t)i + (3 − 2 cos t)j + tk; r(3π/4) = i + 4j + (3π/4)k √ 3 3 14. v(t) = ti + t2 j − tk; |v| = t t2 + 2, t > 0; a(t) = i + 2tj − k; v · a = t + 2t √ + t = 2t√+ 2t ; 3 2 2 √ 2t 2t + 2t t 2 2 + 2t v × a = t2 bi + t2 k, |v × at2 2; aT = √ = √ , aN = √ = √ ; 2 2 2 2 t t +2 t +2 t t +2 t +2 √ √ 2 t2 2 = κ= 3 2 3/2 2 t (t + 2) t(t + 2)3/2 

 15. r (t) = sinh = sinh 1i + cosh 1j + k;  ti + cosh tj + k, r (1)√ √ √ 2 2  |r (t)| = sinh t + cosh t + 1 = 2 cosh2 t = 2 cosh t; |r (1)| = 2 cosh 1; 1 1 1 1 T = √ tanh ti + √ j + √ sech tk, T(1) = √ (tanh 1i + j + sech 1k); 2 2 2 2 dT 1 1 d 1 1 2 = √ sech ti − √ sech t tanh tk; T(1) = √ sech2 1i − √ sech 1 tanh 1k, dt 2 2 2 dt 2  d  sech 1 1 2 2  T(1) == √ sech 1 + tanh +1 = √ sech 1; N(1) = sech 1i − tanh 1k;  dt  2 2 1 1 1 B(1) = T(1) × N(1) = − √ tanh 1i + √ (tanh2 1 + sech2 1)j − √ sech 1k 2 2 2 1 = √ (− tanh 1i + j − sech 1k) 2 √   d 1 (sech 1)/ 2 = sech2 1 κ =  T(1) /|r (1)| = √ dt 2 2 cosh 1

16. The parametric equations describing the path of the ball are √ x(t) = 66 cos(30◦ )t = 33 3ty(t) = −16t2 + 66 sin(30◦ )t + 148 = −16t2 + 33t + 148 The ball touches the ground when y(t) = 0 or −16t2 + 33t + 148 = 0. This occurs when t ≈ 4.243. The ball therefore strikes the ground √ at x(4.243) = 242.52 ft. The velocity of the √ ball at time t is v(t) = 33 3, −32t + 33. The impact velocity is given by v(4.243) = 33 3, −32(4.243) + 33 ≈ 57.158, −102.776. The impact speed is then |v(4.243)| ≈ 117.6 ft/s.

CÁLCULO VECTORIAL MATEMÁTICAS 3 MANUAL DE SOLUCIONES UNIDAD 4 Chapter 13 FUNCIONES DE VARIAS VARIABLES

Partial Derivatives PROBLEMAS 4.1

13.1

Functions of Several Variables

1. {(x, y)|(x, y) = (0, 0)} 2. {(x, y)|x = x ± 3y} 3. {(t, Y )|y = x2 } 4. {(x, y)|y ≥ −4} 5. {(s, t)|s, t any real numbers}  6. {(u, b)|(u, v(= (0, 0)} {(u, v)|u2 + v 2 = 1} 7. {(r, s)| |s| ≥ 1} 8. {(θ, φ) | tan θ tan φ = 1} kπ, k an integer}



{(θ, φ)

θ = π/2 + kπ, k an integer}



{(θ, φ) | φ = π/2 +

9. (u, v, w)|u2 + v 2 + q 2 ≥ 16 10. {(x, y, z)|x2 + y 2 + < 25 and z = 5} √ √ 11. (c); The domain of f (x, y) = x+ y − x is {(x, y)|x ≥ 0, y−x ≥ 0} = {(x, y)|x ≥ 0, y ≥ x} √ 12. (e); The domain of f (x, y) = xy is {(x, y)|xy ≥ 0} = {(x, y)|x ≥ 0, y ≥ 0 or x ≤ 0, y ≤ 0}     13. (b); The domain of f (x, y) = ln(x − y 2 ) is (x, y)|x − y 2 > 0 = (x, y)|x > y 2   x2 + y 2 − 1  is (x, y)|x2 + y 2 − 1 ≥ 0, y = x = 14. (h); The domain of f (x, y) = y−x   (x, y)|x2 + y 2 ≥ 1, y = x    15. (d); The domain of f (x, y) = xy − 1 is (x, y)| xy − 1 ≥ 0 = (x, y)| xy ≥ 1 815

CHAPTER 13. PARTIAL DERIVATIVES

816 16. (g); The domain of f (x, y) =

x4 + y 4 is {(x, y)|xy = 0} = {(x, y)|x = 0, y = 0} xy

17. (f ); The domain of f (x, y) = sin−1 (xy) is {(x, y)||xy| ≤ 1}    18. (a); The domain of f (x, y) = y − x2 is {(x, y)|y − x2 ≥ 0} = ∗x, y)y ≥ x2 19. {(x, y)|x ≥ 0 and y ≥ 0}

y

x

20. {(x, y)| x2 ≤ 1 and y 2 ≥ 4}



{(x, y)| x2 ≥ 1 and y 2 ≤ 4}  {(x, y)| |x| ≤ 1 and |y| ≥ 2} {(x, y)| |x| ≥ 1 and |y| ≤ 2}

y 2

1 x

21. {(x, y)|y − x ≥ 0} y

22. {(x, y)|xy ≥ −1} y

x

23. {z | z ≥ 10}

25. {w ||; −1 ≤ w ≤ 1} 4 27. f (2, 3) = 2 (2t − 1)dt = (t2 − t)|42 = 12 − 2 = 10 1 f (−1, 1) = −1 (2t − 1)dt = (t2 − t)|1−1 = 0 − 2 = −2

x

24. all real numbers

26. {x | w < 7}

13.1. FUNCTIONS OF SEVERAL VARIABLES f (5, −5) = ln

28. f (3, 0) = ln 9/9 = ln 1 = 0; 29. f (−1, 1, −1) = (−2)2 = 4;

817

1 25 = ln = − ln 2 25 + 25 2

f (2, 3, −2) = 22 = 4

√ √ √ 30. f ( 3, 3, 6) = 1/3 + 1/2 + 1/6 = 1;

f (1/4, 1/5, 1/3) = 16 + 25 + 9 = 50

31. A plane through the origin perpendicular to the xz-plane 32. A parabolic cylinder perpendicular to the yz-plane 33. The upper half of a cone lying above the xy-plane with axis along the positive z-axis 34. The upper half of a hyperboloid of two sheets with axis lying along the positive z-axis 35. The upper half of an ellipsoid

36. A hemisphere lying below the yy-plane

37. y = − 12 x + C

38. x = y 2 − c y

y

x

x

39. x2 − y 2 = 1 + c2

40. 4x2 + 9y 2 = 36 − c2 , −6 ≤ c ≤ 6 y

y

x x

CHAPTER 13. PARTIAL DERIVATIVES

818 41. y = x2 + ln c, c > 0

42. y = x + tan c, −π/2 < c < π/2 y

y

x

x

43. x2 /9 + z 2 /4 = c; elliptical cylinder 44. Setting f (x, y, z) equal to a constant√c, we have (x − 1)2 + (y − 2)2 + (z − 2)2 = c which is the equation of a sphere of radius c centered at (1, 2, 3). Therefore, the level curves are concentric spheres centered at (1, 2, 3).

45. x2 + 3y 2 + 6z 2 = c; ellipsoid

46. 4y − 2z + 1 = c; plane

47.

c=0

c0 z

z

y x

y

x

y

48. Setting x = −4, y = 2, and z = −3 in x2 /16 + y 2 /4 + z 2 /9 = c we obtain c = 3. The equation 2 2 2 the x-intercepts are of the √ surface is x /16 + y /4 + z /9 =√3. Setting y = z = 0 we find √ ±4 3. Similarly, the y-intercepts are ±2 3 and the z-intercepts are ±3 3.

13.1. FUNCTIONS OF SEVERAL VARIABLES

819

49. P

v

50. From V = s2 h we obtain h = V /s2 . 51. C(r, h) = πr2 (1.8) + πr2 (1) + 2πh(2.3) = 2.8πr2 + 4.6πrh 250 − xy . Thus, x+y

52. Let the height of the box be h. Then 2xy + 2xh + 2yh = 500 and h = 250xy + x2 y 2 . x+y

 2 53. V + πr2 g + 13 πr2 23 h = 11 9 πr h V = xyh =



54. From the figure, we see that t = x tan θ = x



z

y2 − z2 xz

= y2 − z2

 x θ t y θ

My2-z2

z

55. X = 2(156)(50) = 15, 600 sq cm √ √ 56. h(20, −6.67) + (10 20 − 20 + 10.5)(33 + 6.67) = (20 5 − 9.5)(39.67) ≈ 1397 kcal/m2 h 57. (a) The distance the water falls in time t is s(t) = 12 gt2 + vt where vis the velocity of the water at the top level (t = 0). The velocity of the water at time t is v(t) = gt + v. If t1 is the time it takes a cross-section of water to fall from the top level to the bottom level, then V = gt1 + v and t1 = (V − v)/g. The distance traveled in time t1 is   2  V −v V −v 1 1 h = gt21 + vt1 = g +v 2 2 g g Simplifying the equation we obtain 2gh = V 2 − v 2 . Now the rates at the top and bottom levels are Z = vπr2 and Q = V πr2 (recall that the flow rate is constant). Solving for 2 2 2 2 2 2 v and V and substituting into 2gh √ = V − v we obtain 2gh = (Q/πr ) − (Q/πR ) . πr2 R2 2gh . Solving for Q we find Q = √ R4 − r 4 (b) When r = 0.2 cm, R = 1 cm, and h = 10, Q ≈ 7.61 cm3 /s.

CHAPTER 13. PARTIAL DERIVATIVES

820

PROBLEMAS 13.2 Limits4.2and Continuity 1.

2.

lim

(x,y)→(5,−1)

(x2 + y 2 ) = 25 + 1 = 26

4−1 x2 − y = =3 2−1 (x,y)→(2,1) x − y lim

5x2 + y 2 5x2 = lim = 5. (x,y)→(0,0) x2 + y 2 (x,y)→(0,0) x2 5x2 + y 2 y2 = lim = 1. The limit does not exist. On x = 0, lim 2 2 (x,y)→(0,0) x + y (x,y)→(0,0) y 2

3. On y = 0,

lim

4.

4x2 + y 2 1 4+4 = = 4 4 16 + 16 4 (x,y)→(1,2) 16x + y

5.

4 − x2 − y 2 4−1−1 =1 = x2 + y 2 1+1 (x,y)→(1,1)

lim

lim

2x2 − y −y = lim = ∞. (x,y→(0,0) 2y 2 (x,y→(0,0) x2 + 2y 2 2x2 − y 2x2 = lim = 2.. The limit does not exist. On y = 0, lim (x,y→(0,0) x2 + 2y 2 (x,y→(0,0) x2

6. On x = 0,

lim

x2 y x3 x = 0. = lim = lim 2 4 +y (x,y→(0,0) x + x2 (x,y→(0,0) x2 + 1 x2 y x4 1 lim = lim = . The limit does not exist. On y = x2 , 4 2 4 4 2 (x,y→(0,0) x + y (x,y→(0,0) x + x

7. On y = x,

lim

(x,y→(0,0) x4

6xy 2 6x3 6x = lim = lim = 0. (x,y→(0,0) x2 + y 4 (x,y→(0,0) x2 + x4 (x,y→(0,0) 1 + x2 2 4 x y 6y lim = lim = 3. The limit does not exist. On x = y 2 , 4 2 4 (x,y→(0,0) x + y (x,y→(0,0) y + y 4

8. On y = x,

9. 10. 11.

lim

lim

x3 y 2 (x + y)3 = 1(4)(27) = 108

lim

6 6 xy =− = x2 − y 2 4−9 5

(x,y)→(1,2)

(x,y)→(2,3)

exy 1 = =1 1 (x,y)→(0,0) x + y + 1 lim

sin xy sin mx2 = lim 2 2 (x,y)→(0,0) x + y (x,y)→(0,0) (1 + m2 )x2 m sin mx2 m = lim = . 1 + m2 (x,y)→(0,0) 1 + m2 mx2 The limit does not exist.

12. On y = mx,

lim

13.2. LIMITS AND CONTINUITY 13. 14.

15.

lim

(x,y)→(2,2)

821

1 xy 4 = = x3 + y 2 8+4 3

lim

(x,y)→(π,π/4)

√ cos(3x + y) = cos(3π + π/4) = cos 13π/4 = − 2/2

1 x2 − 3y + 1 =− 3 (x,y)→(0,0) x + 5y − 3 lim

16. On y = mx,

x2 y 2 x2 m2 x2 m2 = lim = . 4 4 4 4 + 5y 1 + 5m4 (x,y)→(0,0) x + 5m x

lim

(x,y)→(0,0) x4

The limit does not exist. 17.

18.

19.

lim

(x,y)→(4,3)

x + 2y 4+6 = 4(9) = 360 x−y 4−3

x2 y 0 =0 = 1+0 (x,y)→(1,0) x+ y 3 lim

xy − x − y + 1 (x − 1)(y − 1) = lim 2 + y − 2x − 2y + 2 (x,y)→(1,1) (x − 1)2 + m2 (x − 1)2 On y − x = m(x − 1), (x − 1)(y − 1) (x − 1)m(x − 1) m lim = lim = . 2 2 2 2 2 2 1 + m2 (x,y)→(1,1) (x − 1) + m (x − 1) (x,y)→(1,1) (x − 1) + m (x − 1) The limit does not exist. lim

(x,y)→(1,1) x2

20. On x = 0, 21.

xy 2

lim

(x,y)→(0,3)

xy − 3y −3y = lim . The limit does not exist. x2 + y 2 − 6y + 9 (x,y)→(0,3) (y − 3)2

x3 y + xy 3 − 3x2 − 3y 2 xy(x2 + y 2 ) − 3(x2 + y 2 = lim x2 + y 2 x2 + y 2 (x,y)→(0,0) (x,y)→(0,0) lim

=

22. 23.

24.

lim

(x,y)→(0,0)

(xy − 3) = −3

y 3 + 2x3 4 8 − 16 = = −2 − 40 21 (x,y)→(−2,2) x + 5xy 2 lim

lim

(x,y)→(1,1)

ln(2x2 − y 2 ) = ln(2 − 1) = 0

sin−1 (x/y) sin−1 (1/2) π/6 1 = = = −1 (x − y cos−1 (−1) π 6 (x,y)→(1,2) cos In Problems 25-30 let x = r cos θ and y = r sin θ. Then x2 + y 2 = r2 and (x, y) → (0, 0) if lim

and only if r → 0. We also use the facts that | cos θ| ≤ 1 and | sin θ| ≤ 1 for all θ. 25.

(x2 − y 2 )2 (r2 cos2 θ − r2 sin2 θ)2 r4 (cos2 θ − sin2 θ)2 = lim = lim r→0 r→0 r2 r2 (x,y)→(0,0) x2 + y 2 lim

= lim r2 cos2 2θ = 0 r→0

CHAPTER 13. PARTIAL DERIVATIVES

822 26.

sin(3x2 + 3y 2 ) sin 3r2 = lim Use L’Hˆ opital’s Rule r→0 x2 + y 2 r2 (x,y)→(0,0) 6r cos 3r2 = lim 3 cos 3r2 = 3 = lim r→0 r→0 2r lim

27.

6xy 6r2 cos θ sin θ  √ = lim = lim 3|r| sin 2θ = 0 r→0 r→0 (x,y)→(0,0) r2 x2 + y 2

28.

x2 − y 2 r2 cos2 θ − r2 sin2 θ  √ = lim = lim |r| cos 2θ = 0 r→0 r→0 (x,y)→(0,0) r2 x2 + y 2

29.

x3 r3 cos3 θ = lim = lim r cos3 θ = 0 r→0 r→0 r2 (x,y)→(0,0) x2 + y 2

30.

x3 + y 3 r3 cos3 θ + r3 sin3 θ = lim = lim r(cos3 θ + sin3 θ) = 0 2 2 r→0 r→0 r2 (x,y)→(0,0) x + y

lim

lim

lim

lim

31. {(x, y) | x ≥ 0 and y ≥ −x} 32. {(x, y) | x = 0 and y = 0} 33. {(x, y) | y = 0 and x/y = π/2 + kπ, k and integer} 34. {(x, y) | x and y are real} 35. (a) For x2 + y 2 < 1, f (x, y) = 0 is continuous (b) For x ≥ 0, f (x, y) is not continuous since it is discontinuous at (2, 0). (c) For y > x, f (x, y) is not continuous since it is discontinuous at (2, 3). 36. (a) For y ≥ 3, f (x, y) is not continuous since it is not defined at (0, 3). (b) For |x| + |y| < 1, f (x.y) is discontinuous since it is not defined at (0, 0). (c) For (x − 2)2 + y 2 < 1, f (x, y) is discontinuous since it is not defined at (2, 0). 37. Since lim

(x,y)→(0,0)

f (x, y) =

6x2 y 3 6r5 cos2 θ sin3 θ = lim = lim 6r cos2 θ sin3 θ = 0 = f (0, 0) r→0 r→0 r4 (x,y)→(0,0) (x2 + y 2 )2 lim

the function is continuous at (0, 0). 38. Since f (x, 0) = 0 for all x and f (0, y) = 0 for all y, f (x, 0) and f (0, y) are continuous at x = 0 and y = 0, respectively. On y = x, lim

(x,y)→0,0)

f (x, y) =

so f (x, y) is not continuous at (0, 0).

x2 1 = , 4 (x,y)→(0,0) 2x2 + 2x2 lim

13.3. PARTIAL DERIVATIVES

823

39. Choose  > 0. Using x = r cos θ and y = r sin θ we have 3xy 2 3t cos θr2 sin2 θ 3 r cos θ sin2 θ. = 2x2 + 2y 2 2r2 2  x2 + y 2 < δ, we have Let δ = 2 3 . Now, whenever r =   3 3 2 3 3 3xy 2 2 | = |r cos θ sin θ| ≤ |r| < δ = | 2 = . 2x + 2y 2 2 2 2 2 3 Thus

3xy 2 = 0. + y2

lim

(x,y)→(0,0) x2

40. Choose  > 0. Using x = r cos θ and y = r sin θ we have x2 y 2 r2 cos2 θr2 sin2 θ = = r2 cos2 θ sin2 θ. x2 + y 2 r2    √ √  x2 y 2  2 2 = r2 cos2 θ sin2 θ ≤ r2 ≤ . Thus, Now, whenever r = x + y <  (for δ = ),  2 x + y2  x2 y 2 lim = 0. 2 (x,y)→(0,0) x + y 2 41. Where y = x, we have f (x, y) =

(x − y)(x2 + xy + y 2 ) x3 − y 3 = = x2 + xy + y 2 . x−y x−y

When y = x, we have x2 + xy + y 2 = x2 + x2 + x2 = 3x2 = f (x, y). Therefore, f (x, y) = x2 + xy + y 2 throughout the entire plane. Since x2 + xy + y 2 is a polynomial, f must be continuous throughout the plane and thus has no discontinuities.  42. Choose  > 0. Then for δ = , whenever 0 < (x − a)2 + (y − b)2 < δ, we have  |f (x, y) − b| = |y − b| ≤ (x − a)2 + (y − b)2 < δ = . Thus,

lim

(x,y)→(a,b)

y = b.

PROBLEMAS 4.3

13.3 1.

Partial Derivatives

7(x = x) + 8y 2 − 7x − 8y 2 7 x ∂z = lim = lim =7 →0 x ∂x →0

x 7x + 8(y + y)2 − 7x − 8y 2 16y y + 8( y)2 ∂z = lim = lim y→0 ∂y y→0

y

y = lim (16y + 8 y) = 16y y→0

CHAPTER 13. PARTIAL DERIVATIVES

824 2.

3.

(x + x)y − xy y x ∂z = lim = lim = y; x→0 x→0 ∂x

x

x ∂z x(y + y) − xy x y = lim = lim =x y→0 y→0 ∂y

y

y ∂z 3(x + x)2 y + 4x + x)y 2 − 3x2 y − 4xy 2 = lim ∂x x→0

x 3x2 y + 6x( x)y + 3( x)2 y + 4xy 2 + 4( x)y 2 − 3x2 y − 4xy 2 = lim x→0

x 6x( x)y + 3( x)2 y + 4( x)y 2 = lim = lim (6xy + 3( x)y + 4y 2 ) = 6xy + 4y 2 x→0 x→0

x ∂z 3x2 (y + y) + 4x(y + y)2 − 3x2 y − 4xy 2 = lim ∂y y→0

y 3x2 y + 3x2 y + 4xy 2 + 8xy y + 4x( y)2 − 3x2 y − 4xy 2 = lim y→0

y 3x2 y + 8xy y + 4x( y)2 = lim = lim (3x2 + 8xy + 4x y) = 3x2 + 8xy y→0 y→0

y

x+

x − ∂z x2 + x x + xy + ( x)y − x2 − x x − xy x+ +y x+y 4. = lim = lim x→0 ∂x x→0

x (x + x + y)(x + y) x y ( x)y = = lim x→0 (x + x + y)(x + y) x (x + y)2 x x − ∂z x2 + xy − x2 − xy − x y x + y + y x + y = lim = lim y→0 (x + y + y)(x + y) y ∂y y→0

y −x y x =− = lim y→0 (x + y + y)(x + y) y (x + y)2 5. zx = 2x − y 2 ; zy = −2xy + 20y 4 6. zx = −3x2 + 12xy 3 ; zy = 18x2 y 2 + 10y 7. zx = 20x3 y 3 − 2xy 6 + 30x4 ; zy = 15x4 y 2 − 6x2 y 5 − 4 8. zx = 3x2 y 2 sec2 (x3 y 2 ); zy = 2x3 sec2 (x3 y 2 ) √ 2 24y x 9. zx = √ ; z = − y (3y 2 + 1)2 x(3y 2 + 1) 10. zx = 12x2 − 10x + 8; zy = 0 11. zx = −(x3 − y 2 )−2 (3x2 ) = −3x2 (x3 − y 2 )−2 ; zy = −(x3 − y 2 )−2 (−2y) = 2y(x3 − y 2 )−2 12. zx = 6(−x4 + 7y 2 + 3y)5 (−4x) = −24x( − x4 + 7y 2 − 3y)5 ; zy = 6(−x4 + 7y 2 + 3y)5 (14y + 3) 13. zx = 2(cos 5x)(− sin 5x)(5) = −10 sin 5x cos 5x; zy = 2(sin 5y)(cos 5y)(5) = 10 sin 5y cos 5y

13.3. PARTIAL DERIVATIVES 14. zx = (2x tan−1 y 2 )ex 3

2

tan−1 y 2

825 ; zy =

3

2x2 y x2 tan−1 y2 e 1 + y4

3

15. fx = x(3x2 yex y + ex y ; fy = x4 ex y       θ 1 θ θ θ θ θ 2 2 16. fθ = φ cos ; fφ = φ cos − 2 + 2φ sin = −θ cos + 2φ sin φ φ φ φ φ φ φ (x + 2y)3 − (3x − y) 7y = ; (x + 2y)2 (x + 2y)2   (x2 − y 2 )2 y − xy 2(x2 − y 2 )2x 18. fx = = (x2 −y 2 )4  (x2 − y 2 )x − xy 2(x2 − y 2 )(−2y) fy = (x2 − y 2 )4 17. fx =

fy =

(x + 2y)(−1) − (3x − y)(2) −7x = (x + 2y)2 (x + 2y)2

−3x2 y − y 3 ; (x2 − y 2 )3 3xy 2 + x3 = 2 (x − y 2 )3

8u 15v 2 ; gv = 2 3 − 5v 4u + 5v 3 √ √ s r 1 1 20. hr = √ + 2 ; hx = − 2 − √ r s 2s r 2s r    y  y2  y √ √ 1 y/z y e + 1 ey/z ; wz = −yey/z − 2 = 2 ey/z 21. wx = √ ; wy = 2 x − y =2 x− z z z z x   1 xy 22. wx = xy + (ln xz)y = y + y ln xz; wy = x ln xz; wz = x z 19. gu =

4u2

23. Fu = 2uw2 − v 3 − vwt2 sin(ut2 ); Fv = −3uv 2 + w cos(ut2 ); Fx = 3(2x2 t)3 (4xt) = 16xt(2x2 t)3 = 128x7 t4 ; Ft = −2uvwt sin(ut2 ) + 64x8 t3 4 5

24. Gp = 2pq 3 e2r s 4 5 Gq = 3p2 q 2 e2r s 4 5 4 5 Gr = p2 q 3 (8r3 s5 )e2r s = 8p2 q 3 r3 s5 e2r s 4 5 4 5 Gs = p2 q 3 (10r4 s4 )e2r s = 10p1 q 3 r4 s4 e2r s 25. zy = 16x3 y 3 , zy (1, −1) = −16 26. zx = 12x2 y 4 , zx (1, −1) = 12 (x + y)18x − 18xy 18x2 = , fy (−1, 4) = 2. An equation of the tangent line is given (x + y)2 (x + y)2 by x = −1 and z + 24 = 2(y − 4). Parametric equations of the line are x = −1, y = 4 + t, z = −24 + 2t.

27. fy =

(x + y)18y − 18xy 18y 2 = , fx (−1, 4) = 32. An equation of the tangent line is given (x + y)2 (x + y)2 z + 24 , y = 4. by y = 4 and z + 24 = 32(x + 1). Symmetric equations of the line are x + 1 = 32

28. fx =

CHAPTER 13. PARTIAL DERIVATIVES

826 −x , zx (2, 2) = −2 29. zx =  9 − x− y 2 √ √ √ 3 30. zy =  , zy ( 2, 3) = − 2 2 2 9−x −y −y

31.

∂z ∂2z = yexy ; = y 2 exy ∂x ∂x2

32.

∂z ∂2z ∂3z 4 −4 = −2x4 y −3 ; = 6x y ; = −24x4 y −5 ∂y ∂y 2 ∂y 3

33. fx = 10xy 2 − 2y 3 ; fxy = 20xy − 6y 2 34. f (p, q) = ln(p + q) − 2 ln q, fq =

2 1 1 − , fqp = − p+q q (p + q)2

35. wt = 3u2 v 3 t2 , wtu 6uv 3 t2 ; wtuv 18uv 2 t2 36. wv = −

u2 sin(u2 v) u4 cos(u2 x) 3u4 cos(u2 v) ; wvv − ; wvvt = 3 3 t t t4 2

2

2

2

2

37. Fr = 2rer cos θ; Frθ − 2rer sin θ; Frθr − 2r(2rer ) sin θ − 2er sin θ = −2er (2r2 + 1) sin θ (s − t) − (s + t)(−1) 2s 4s = ; Htt = ; 2 2 (s − t) (s − t) (s − t)3 4 2 (s − t) − 4x(3)(s − t) −8s − 4t = = 6 (x − t) (s − t)4

38. Ht = Htts 39. 40.

∂z ∂2z ∂z ∂2z = −5x4 y 2 +8xy; = −60x3 y 2 +8y; = 6x5 −20x3 y 3 +4y 2 ; = −60x3 y 2 +8y ∂y ∂x∂y ∂x ∂y∂x ∂z 2x (1 + 4x2 y 2 )2 − 2x(8xy 2 ) 2y ∂z 2 − 8x2 y 2 ∂z = = = ; = ; 2 2 2 2 2 2 2 2 ∂y 1 + 4x y ∂x∂y (1 + 4x y ) (1 + 4x y ) ∂x 1 + 4x2 y 2 2 2 2 2 2 ∂z (1 + 4x y )2 − 2y(8x y) 2 − 8x y = = ∂y∂x (1 + 4x2 y 2 )2 (1 + 4x2 y 2 )2

41. wu = 3u2 v 4 − 8uv 2 t3 , wuv = 12u2 v 3 − 16uvt3 , wuvt = −48uvt2 ; wt = −12u2 v 2 t2 + v 2 , wtv = −24u2 vt2 + 2v, wtvu = −48uvt2 ; wv = 4u3 v 3 − 8u2 vt3 + 2vt, wvu = 12u2 v 3 − 16uvt3 , wvut = −48uvt2 42. Fη = 6η 2 (η 3 +ξ 2 +τ ) = 6η 5 +6η 2 ξ 2 +6η 2 τ, Fηξ = 12η 2 ξ, Fηξη = 24ηξ; Fξ = 4ξ(η 3 +ξ 3 +τ ) = 4η 3 ξ + 4xi3 + 4xiτ, Fξη = 12η 2 ξ, Fξηη 24ηξ; Fηη = 30η 4 + 12ηξ 2 , Fηηξ = 24ηξ 43. 2x + 2zzx = 0, zx = −x/z; 2y + 2zzy = 0, zy = −y/z 2x ; 2z − y 2 2yz 2zzy = y 2 zy + 2yz =⇒ (2z − y 2 )zy = 2yz =⇒ zy = 2z − y 2

44. 2zzx = 2x + y 2 zx =⇒ (2z − y 2 )zx = 2x =⇒ zx =

13.3. PARTIAL DERIVATIVES

827

vz − 2uv 3 ; 2z − uv 2 2 uz − 3u v 2zzv + 3u2 v 2 − uvzv − uz = 0 =⇒ (2z − uv)zv = uz − 3u2 v 2 =⇒ zv = 2z − uv

45. 2zzu + 2uv 3 − uvzu − vz = 0 =⇒ (2z − uv)zu = vz − 2uv 3 =⇒ zu =

46. sez zs + ez − test + 12s2 t = zs =⇒ (sez − 1)zs = tes − ez − 12s2 t =⇒ zs = sez zt − sest + 4s3 = zt =⇒ (sez − 1)zt = sest − 4s3 =⇒ zt =

sest − 4s3 sez − 1

test − ez − 12s2 t ; sez − 1

47. ax = y sin θ, Ay = x sin θ, Aθ = xy cos θ 48. Vh = (π/3)(r2 + rR + R2 ), Vr = (π/3)h(2r + R), VR = (π/3)h(r + 2R) 49.

50.

51.

52.

53.

∂u ∂2u = 2π(cosh 2πy + sinh 2πy) cos 2πx; = −4π 2 (cosh 2πy + sinh 2πy) sin 2πx; ∂x ∂x2 ∂2u ∂u = (2π sinh 2πy + 2π cosh 2πy) sin 2πx; = (4π 2 cosh 2πy + 4π 2 sinh 2πy) sin 2πx; ∂y ∂y 2 ∂2u ∂2u + 2 = −4π 2 (cosh 2πy + sinh 2πy) sin 2πx + 4π 2 (cosh 2πy + sinh 2πy) sin 2πx = 0 ∂x2 ∂y  nπ  ∂ 2 u  nπ  nπ ∂u n2 π 2 −(nπx/L) = − e−(nπx/L sin = e sin ; y; ∂x L L ∂x2 L2 L    2 2 2 nπ nπ nπ −(nπx/L) n π ∂u ∂ u = e cos = − 2 e−(nπx/L) sin y; y; ∂y L L ∂y 2 L L  nπ  n2 π 2  nπ  n2 π 2 −(nπx/L) ∂2u ∂2u − =0 + 2 = e sin e−(nπx/L) sin 2 2 2 ∂x ∂y L L L L 2x 2y ∂z ∂2z (x2 + y 2 )2 − 2x(2x) 2y 2 − 2x2 ∂z = 2 ; = 2 , = = , ∂x x + y 2 ∂x2 (x2 + y 2 )2 (x2 + y 2 ) ∂y x + y2 2 2 2 2 2 2 2 2 2 2 ∂ z (x + y )2 − 2y(2y) 2x − 2y ∂ z ∂ z 2y − 2x + 2x − 2y 2 = = ; + = =0 ∂y 2 (x2 + y 2 )2 (x2 + y 2 )2 ∂x2 ∂y 2 (x2 + y 2 )2 ∂z 2 2 2 2 2 2 = 2yex −y sin 2xy + 2xex −y cos 2xy = 2ex −y (x cos 2xy − y sin 2xy), ∂x ∂2z 2 2 2 2 = 2ex −y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy) + 4xex −y (x cos 2xy − y sin 2xy); ∂x2 ∂z − 2 − 2 − 2 = −2xex y sin 2xy − 2yex y cos 2xy = −2ex y (x sin 2xy + y cos 2xy), ∂y ∂2z − 2 − 2 = −2ex y (2x2 cos 2xy − 2xy sin 2xy + cos 2xy) + 4yex y (x sin 2xy + y cos 2xy); ∂y 2 − 2 ∂2z ∂2z + 2 =2ex y (−2xy sin 2xy + cos 2xy − 2y 2 cos 2xy + 2x2 cos 2xy − 2xy sin 2xy 2 ∂x ∂y − 2x2 cos 2xy + 2xy sin 2xy − cos 2xy + 2xy sin 2xy − 2y 2 cos 2xy) = 0 x ∂u =− 2 ; ∂x (x + y 2 + z 2 )3/2 ∂2u 2x2 − y 2 − z 2 = ; ∂x2 (x2 + y 2 + z 2 )5/2

y z ∂u ∂u =− 2 =− 2 ; ; ∂y (x + y 2 + z 2 )3/2 ∂z (x + y 2 + z 2 )3/2 ∂2u −x2 + 2y 2 − z 2 ∂2u −x2 − y 2 + 2z 2 = ; = ; ∂y 2 (x2 + y 2 + z 2 )5/2 ∂z 2 (x2 + y 2 + z 2 )5/2

CHAPTER 13. PARTIAL DERIVATIVES

828

∂2u ∂2u ∂2u 2x2 − y 2 − z 2 − x2 + 2y 2 − z 2 − x2 − y 2 + 2z 2 + + = =0 ∂x2 ∂y 2 ∂z 2 (x2 + y 2 + z 2 )5/2 54.

√ √ ∂2u ∂u √ 2 2 2 2 2 m2 +n2 x = m + n2 e m +n x cos my sin nz; = (m + n )e cos my sin nz; ∂x ∂x2 2 √ √ ∂u ∂ u 2 2 2 2 = −me m +n x sin my sin nz; = −m2 e m +n x cos my sin nz; ∂y ∂y 2 √ √ ∂u ∂2u 2 2 2 2 = ne m +n x cos my cos nz; = −n2 e m +n x cos my sin nz; 2 ∂z ∂z √ √ ∂2u ∂2u ∂2u 2 2 2 2 + 2 + 2 = (m2 + n2 )e m +n x cos my sin nz − m2 e m +n x cos my sin nz 2 ∂x ∂y ∂z

− n2 e 55.

56.

57.

√

m2 +n2 x

cos my sin nz = 0

∂2u ∂2u ∂u ∂u = cos at cos x, = −a sin at sin x, = − cos at sin x; = −a1 cos at sin x; ∂x ∂x2 ∂t ∂t2 ∂2u ∂2u a2 2 = a2 (− cos at sin x) = 2 ∂x ∂t ∂2u ∂u = − sin(x + at) + cos(x − at), = − cos(x + at) − sin(x − at); ∂x ∂x2 ∂2u ∂2u ∂u = −a sin(x + at) − a cos(x − at), = −a2 cos(x + at) − a2 sin(x − at); a2 2 = 2 ∂t ∂t ∂x ∂2u 2 2 −a cos(x + at) − a sin(x − at) = 2 ∂t 2x ∂C ∂2C 4x2 −1/2 −x2 /kt 2 / 2 = − t−/12 e−x kt , = t e − t−1/2 e−x /kt ; 2 2 2 ∂x kt ∂x k t kt 2 t−3/2 −x2 /kt k ∂ 2 C x2 −1/2 −x2 /kt t−1/2 −x2 /kt ∂C ∂C −1/2 x −x2 /kt =t e e e − ; = 2t e − = 2 2 ∂t kt 2 4 ∂x kt 2t ∂t

58. (a) Pv = −k(T /V 2 ) (b) P V = kt, P VT = k, VT = k/P (c) P V = kT, V = kTp , Tp = V /k  ∂u −gx/z, 0 ≤ x ≤ at = 59. (a) −gt, x > at ∂t For x > at, the motion is that of a freely falling body. ∂u (b) For x > at, = 0. For x > at, the string is horizontal. ∂x 60.

∂S = 0.0790975w0.425 h−0.275 ; Sh (60, 36) + 0.0790975(60)0.425 (36)−0.275 ≈ 0.1682 ∂h The approximate increase in skin-area as h increases from 36 to 37 inches is 0.1682 sq ft. ∂2z fx (x + Δx, y) − fx (x, y) = lim 2 Δx→0 ∂x Δx fy (x, y + Δy) − fx (x, y) ∂2z = lim (b) Δy→0 ∂y 2 Δy

61. (a)

13.3. PARTIAL DERIVATIVES

829

∂2z fy (x + Δx, y) − fy (x, y) = lim ∂x∂y Δx→0 Δx

(c)

62. Integrating zx = 2xy 3 + 2y + 1/x with respect to x, we obtain z = x2 y 3 + 2xy + ln x + φ(y). Then 3x2 y 2 + 2x + 1 = zy = 3x2 y 2 + 2x + φ (y). Since φ (y) = 1, φ(y) = y + C, and z = x2 y 3 + 2xy + ln x + y + C. 63. Consider the mixed partials: ∂ ∂2z = ∂y∂x ∂y



∂z ∂x

 = 2y

and

∂ ∂2z = ∂x∂y ∂x



∂z ∂y

 = 2x.

∂2z ∂2z ∂z ∂z , , , and are all continuous on an open set, we should have ∂x ∂y ∂y∂x ∂x∂y ∂2z ∂2z = on that set. But the mixed partials are equal only on the line y = x, which ∂y∂x ∂x∂y contains no open set in the plane. Therefore, such a function cannot exist.

Since

64. (a) There are 10 different third-order partial derivatives: Fxxx , Fxxy , Fxxz , Fxyy , Fxyz , Fxzz , Fyyy , Fyyz , Fyzz , Fzzz (b) Since the mixed partials are equal, the order in which differentiation occurs is irrelevant. The nth order partial derivatives are given by ∂nz ∂nz ∂nz ∂nz ∂nz , , , . . . , , . ∂xn ∂xn−1 ∂y ∂xn−2 ∂y 2 ∂x∂y n−2 ∂y n Hence, there are n + 1 different nth order partial derivatives. 65. (a) There slopes of the surface in the x and y directions are zero everywhere. This implies that the surface must have constant height everywhere. Therefore f must have the form f (x, y) = c. (b) Since the mixed partials are both zero, we have     ∂z ∂ ∂z = 0 and df rac∂∂y =0 ∂x ∂y ∂x ∂z ∂z is a function of y alone and is a function of x alone. Therefore, z ∂y ∂x has no term that depends on both x and y. Hence z is of the form z = g(x) + h(y) + c where g and h are twice continuously differentiable functions of a single variable.

which implies

66. The level curves suggest that the surface height is decreasing as we move slightly to the right ∂z < 0 and of the point, and increasing as we move slightly up from the point. This implies ∂x ∂z > 0. ∂y  ∂z  f (0 + Δx, 0) − f (0, 0) 0/2(Δx)2 67. = lim = 0; = lim  Δx→0 ∂x (0,0) Δx→0 Δx Δx  ∂z  f (0, 0 + Δy) − f (0, 0) 0/2(Δy)2 = lim =0 = lim  Δy→0 ∂y (0,0) Δy→0 Δy Δy

CHAPTER 13. PARTIAL DERIVATIVES

830

  ∂z y 5 − 4x2 y 3 − x4 ∂z  −x5 + 4x3 y 2 + xy 4 ∂z  ∂z 68. (a) = = ; = y; ; = −x ∂x (x2 + y 2 )2 ∂x (0,y) ∂y x2 + y 2 ) 2 ∂y (x,0) (b)

∂2z ∂2z ∂2z ∂2z = 1; = −1 =⇒ = ∂y∂x ∂x∂y ∂y∂x ∂x∂y

PROBLEMAS 4.4 13.4 Linearization and Differentials 1.

2.

3.

4.

5.

∂f ∂f = 4y 2 − 6x2 y so (1, 1) = −2 ∂x ∂x ∂f ∂f = 8zy − 2x3 so (1, 1) = 6 ∂y ∂y f (1, 1) = 2 The linearization is L(x, y) = 2 − 2(x − 1) + 6(y − 1) = −2x + 6y − 2 3x2 y ∂f ∂f =  (2, 2) = 3 so 3 ∂x ∂x 2 x y ∂f x3 ∂f =  (2, 2) = 1 so 3 ∂y ∂y 2 x y f (2, 2) = 4 The linearization is L(x, y) = 4 + 3(x − 2) + (y − 2) = 3x + y − 4  ∂f 353 x2 ∂f = x2 + y 2 +  (8, 15) = so 2 2 ∂x ∂x 17 x +y ∂f xy 120 ∂f = (8, 15) = so ∂y ∂y 17 x2 + y 2 120 f (8, 15) = 136 The linearization is L(x, y) = 136+ 353 17 (x−8)+ 17 (y −15) = ∂f π 3π  −3 ∂f = 3 cos x cos y so = , ∂x ∂x 4 4 2 ∂f π 3π  −3 ∂f = = 3 sin x sin y so , ∂y ∂y 4 4 2

π 3π  −3 f 4, 4 = The linearization is L(x, y) = 2 3 2 (π − 1) 2x ∂f ∂f = 2 (−1, 1) = −1 so ∂x x + y3 ∂x ∂f 3y 2 3 ∂f = 2 (−1, 1) = so ∂y x + y3 ∂y 2

−3 2

−

3 2

x−

π 4



−

3 2

y−

3π 4

353 120 17 x+ 17 y −136



=

−3 2 x

− 32 y +

3 f (−1, 1) = ln(2) The linearization is L(x, y) = ln(2) − (x + 1) + 32 (y − 1) = −x + y − 52 + ln(2) 2 ∂f ∂f π  2π = 3e−2y cos 3x so 0, 3 = 3e− 3 6. ∂x ∂x ∂f ∂f π  = −2e−2y sin 3x so 0, 3 = 0 ∂y ∂y  2π 2π f 0, π3 = 0 The linearization is L(x, y) = 3e− 3 (x − 0) = 3xe− 3 √ √ 7. Note that we are trying to approximate f (102, 80) where f (x, y) = x + 4 y. Since (102, 80) is reasonably close to (100, 81), we can use the linearization of f at (100, 81) to approximate

13.4. LINEARIZATION AND DIFFERENTIALS

831

the value at (102, 80). To do this, we compute ∂f 1 1 ∂f 1 1 1 ∂f ∂f = √ , (100, 81) = , = (100, 81) = = , and , 3/2 ∂x ∂x 20 ∂y ∂y 4(27) 108 2 x 4y f (100, 81) = 13 1 1 The linearization is L(x, y) = 13 + 20 (x − 100) + 108 (y − 81). For the approximation, we have 1 1 1 1 − 108 = 7069 L9102, 80) = 13 + 20 (102 − 100) + 108 (80 − 81) = 13 + 10 540 ≈ 13.0907 √ x 8. We are trying to approximate f (36, 63) where f (x, y) = √ . Use the linearization of f at the y point (36, 64). To do this, compute √ ∂f 1 1 ∂f 3 3 ∂f ∂f − x = √ √ , (36, 64) = , = 2y (35, 64) = − , and f (36, 64) = . 3/2 ; ∂x 96 ∂y ∂y 512 4 2 x y ∂x 1 3 3 (y − 64). For the approximation, we have The linearization if L(x, y) = + (x − 36) − 4 96 512 1 3 387 3 (63 − 64) = ≈ .7559. L(36, 63) = + (36 − 36) − 4 96 512 512 9. First, linearize f at (2, 2). To do this, compute ∂f ∂f ∂f ∂f = 2(x2 + y 2 )2x, (2, 2) = 64, = 2(x2 + y 2 )2y, (2, 2) = 64, and f (2, 2, ) = 64. ∂x ∂x ∂y ∂y The linearizationis L(x, y) = 64 + 64)x − 2) + 64)y − 2). For the approximation, we have L(1.95, 2.01) = 64 + 64(−0.05) + 64(0.01) ≈ 61.44. 

10. First, linearize f at 12 , 3 . To do this, compute

 ∂f 1  ∂f 1  ∂f ∂f π = −πy sin(πxy), = −πx sin(πxy), , 3 = 3π, , 3 = , and f 12 , 3 = 2 2 ∂x ∂x ∂y ∂y 2 0.   1 π The linearization is L(x, y) = 3π x − + (y − 3). For the approximation, we have 2 2 π L(0.52, 2.96) = 3π(0.02) + (−0.04) ≈ 0.1257. 2 11. dz = 2x sin 4ydx + 4x2 cos 4ydy   2 2 2 2 2 2 2 2 2 2 12. dz = x(2xex −y ) + ex −y dx − 2yxex −y dy = (2x2 + 1)ex −y dx − 2xyex −y dy 13. dz = 

2x 2x2 − 4y 3

dx − 

6y 2 2x2 − 4y 3

dy

14. dz = 45x2 y(5x3 y + 4y 5 )2 dx + (15x3 + 60y 4 )(5x3 y + 4y 5 )2 dy 15. df =

(s + 3t)2 − (2s − t) (s + t)(−1) − (2s − t)3 7t 7s ds + dt = − dt (s + 3t)2 (s + 3t)2 (s + 3t)2 (s + 3t)2

16. dg = 2r cos 3θdr − 3r2 sin 3θdθ 17. dw = 2xy 4 z −5 dx + 4x2 y 3 z −5 dy − 5x2 y 4 z −6 dz 2

2

2

18. dw = −2xe−z sin(x2 + y 4 )dx − 4y 3 e−z sin(x2 + y 4 )dy − 2ze−z cos(x2 + y 4 )dz 19. dF = 3r2 dr − 2s−3 ds − 2t−1/2 dt

CHAPTER 13. PARTIAL DERIVATIVES

832

20. dG = sin φ cos θdρ − ρ sin φ sin θdθ + ρ cos φ cos θdφ 21. w = ln u + ln v − ln s − ln t; dw = 22. dw = √

du dv ds dt + − − u v s t

v st2 s2 t u du − √ dv + √ ds + √ dt u 2 + s 2 t2 − v 2 u 2 + s 2 t2 − v 2 u 2 + s 2 t2 − v 2 u 2 + s 2 t2 − v 2

23. Δz = z(2.2, 3.9) − z(2, 4) = (6.6 + 15.6 + 8) − (6 + 16 + 8) = 0.1; dz = 3dx + 4dy When x = 2, y = 4, dx = 0.2, and dy = −0.1, dz = 3(0.2) + 4(−0.1) = 0.2 24. Δz = z(0.2, −0.1) − z(0, 0) = 2(0.2)2 (−0.1) + 5(−0.1) − 0 = −0.508; dz = 4xydx + (2x+ 5)dy When x = y = 0, dx = 0.2, and dy = −0.1, dz = 5(−0.1) = −0.5. 25. Δz = z(3.1, 0.8) − z(3, 1) = (3.1 + 0.8)2 − (3 + 1)2 = 15.21 − 16 = −0.79; dz = 2(x + y)dx + 2(x + y)dy. When x = 3, y = 1, dx = 0.1, and dy = −0.2, 2(3 + 1)(0.1) + 2(3 + 1)(0.2) = 0.8 − 1.6 = −0.8

dz =

26. Δz = z(0.9, 1.1) − z(1, 1) = [(0.9)2 + (0.9)2 (1.1)2 + 2] − [1 + 1 + 2] = 3.7901 − 4 = −0.2099; dz = (2x + 2xy 2 )dx + 2xy dy. When x = y = 1, dx = −0.1, and dy = 0.1, dz = 4(−0.1) + 2(0.1) = −0.2. 27. Δz = 5(x + Δx)2 + 3(y + Δy) − (x + Δx)(y + Δy) − (5x2 + 3y − xy) = 10xΔx + 5(Δx)2 + 3Δy − xΔy − yΔx − ΔxΔy = (10x − y)Δx + (3 − z)Δy + (5Δx)Δx − (Δx)Δy 1 = 5Δx, 2 = −Δx 28. Δz = 10(y + Δy)2 + 3(x + Δx) − (x + Δx) − (10y 2 + 3x − x2 ) = 20yΔy + 10(Δy)2 + 3Δx − 2xΔx − (Δx)2 = (3 − 2x)Δx + 20yΔy − (Δx)Δx + (10Δy)Δy 1 = −Δx, 2 = 10Δy 29. Δx = (x + Δx)2 (y + Δy)2 − x2 y 2 = [x2 + 2xΔx + (Δx)2 ][y 2 + 2yΔy + (Δy)2 ] − x2 y 2 = 2x2 yΔy + x2 (Δy)2 + 2xy 2 Δx + 4xy(Δx)Δy + 2x(Δx)(Δy)2 + y 2 (Δx)2 + 2y(Δx)2 Δy + (Δx)2 (Δy)2 = 2xy 2 Δx + 2x2 yΔy + [4xyΔy + 2x(Δy)2 + y 2 x]Δx + [x2 Δy + 2y(Δx)2 + (Δx)2 Δy]Δy 1 = 4xyΔy + 2x(Δy)2 + y 2 x, 2 = x2 Δy + 2y(Δx)2 + (Δx)2 Δy (Several other choices of 1 and 2 are possible.) 30. Δz = (x + Δx)3 − (y + Δy)3 − (x3 − y 3 ) = 3x2 Δx + 3x(Δx)2 + (Δx)3 − 3y 2 Δy − 3y(Δy)2 − (Δy)3 = 3x2 Δx − 3y 2 Δy + [3xΔx + (Δx)2 ] − [3yΔy + (Δy)2 ]Δy 1 = 3xΔx + (Δx)2 , 2 = −3yΔy − (Δy)2 31. R =

R 1 R2 R3 ; ΔR1 = ±0.009R1 , ΔR! = ±0.009R2 , ΔR3 = ±0.0009R3 R 2 R 3 + R 1 R 3 + R1 R2

13.4. LINEARIZATION AND DIFFERENTIALS

833

        R22 R32 R12 R32    |ΔR| ≈ |dR| ≤  (±0.009R1 ) +  (±0.009R2 ) 2 2 (R2 R3 + R1 R3 + R1 R2 ) (R2 R3 + R1 R3 + R1 R2 )   2 2   R R 1 2 +  (±0.009R3 ) (R2 R3 + R1 R3 + R1 R2 )2   R2 R3 + R1 R3 + R 1 R 2 = 0.009R = 0.009R R2 R3 + R1 R3 + R 1 R 2 The maximum percentage error is approximately 0.9%. 32. We are given ΔT = ±0.006T and ΔV = ±0.008V. Then    kT k kT kT (0.006) + (0.008) = P (0.014). |ΔP | ≈ |dP | =  (±0.006T ) − 2 (±0.008V ) ≤ V V V V Thus, the approximate maximum percentage error in P is 1.4%. (2r2 + R2 − R(2R) −R(4r) 2r2 − R2 4rR dR + mg dr = mg dR − mg dr 2 2 2 2 2 2 + 2 2 (2r + R ) 2r + R ) (2r R ) (2r+ R2 )2 When R = 4, r = 0.8, dR = 0.1, and dr = 0.1,

33. dT = mg



2(0.8)2 − 42 4(0.8)4 ΔT ≈ dT = mg (0.1) − (0.1) [2(0.8)2 + 42 ]2 [2(0.8)2 + 42 ]2   −1.472 − 1.28 = mg ≈ −0.009 mg. 298.598



The tension decreases. 34. V = πr2 h, dV = 2πrhdr + πr2 dh. When r = 5, h = 10, dr = 0.3, and dh = 0.5, ΔV ≈ dV = 2π(5)(1−)(0.3) + π(52 )(0.5) = 42.5π cm3 . Since V (5, 10) = 250π cm3 , V (5.3, 10.5) = V (5, 10) + ΔV ≈ V (5, 10) + dV = 250π + 42.5π = 292.5π cm3 . 35. V = lwh, dV = whdl + lhdq + lwdh. With dl = ±0.02l, dw = ±0.05w, and dh = ±0.08h, |ΔV | ≈ |dV | = |wh(±0.02l) + lh(±0.05w) + lw(±0.08h)| ≤ lwh(0.02 + 0.5 + 0.8) = 0.15V. The approximate percentage increase in volume is 15%. 36. S = 2lw + 2lh + 2wh, dS = (2w + 2h)dl + (2l + 2h)dw + (2l + 2w)dh With l = 3, w = 1, h = 2, sl = 0.06, dw = 0.05, and dh = 0.16, ΔS ≈ dS = (2 + 4)(0.06) + (6 + 4)(0.05) + (6 + 2)(0.16) = 2.14 ft2 . Since S(3, 1, 2) = 22 ft2 , the new surface area is approximately S(3, 1, 2) + dS = 24.14 ft2 .

CHAPTER 13. PARTIAL DERIVATIVES

834

37. dS = 0.1091(0.425)w−0.575 h0.725 dw + 0.1091(0.725)w0.425 h−0.275 dh With dw = ±0.03w and dh = ±0.05h, |ΔS| ≈ |dS| = 0.1091|0.425w−0.575 h0.725 (±0.03w) + 0.725w0.425 h−0.275 (±0.05h)| ≤ 0.1091[0.425w0.425 h0.725 (0.03)] + 0.1091[0.725w0.425 h0.725 (0.05)] = 0.1091w0.425 h0.725 (0.013 + 0.036) = 0.049S. The approximate maximum percentage error is 4.9%. 

2 1/2 1 38. Z = R + 100L − ; 1000c    2 −1/2   1 2 1 1 dZ = r + 100L − 2RdR + 2 100L − (100)dL 2 1000c 1000c     1 1 dC 2 100L − 1000c 1000c2   X 2 2 −1/2 = (R + X ) dC RdR + 1000XdL + 1000c2 With R = 4000, L + 0.4, C = 10−5 , dR = 25, d := 0.05, and dC = 1.1 × 10−5 − 10−5 = 10−6 , we have X = 300 and   300 2 2 −1/2 −6 dZ = (400 + 300 ) 10 400(25+ 100(300)(0.05) + 1000(10)−10 1 (10000 + 15000 + 3000) = 56 ohms. = 500 

2

The new impedance is approximately Z(400, 0.4, 10−5 ) + dZ = 500 + 56 = 556 ohms. 39. (a) If a function w = f (x, y, z) is differentiable at a point (x0 , y0 , z0 ), then the function L(x, y, z) = f (x0 , y0 , z0 )+fx (x0 , y0 , z0 )(x−x0 )+fy (x0 , y0 , z0 )(y−y0 )+fz (x0 , y0 , z0 )(z−z0 ) is a linearization of f at (x0 , y0 , z0 ).  (b) Let f (x, y, z) = x2 + y 2 + z 2 . Then we wish to approximate f (9.1, 11.75, 19.98). To do this, linearize f at (9, 12, 20). Compute x 9 ∂f ∂f = (9, 12, 20) = , 2 2 2 ∂x ∂x 25 x +y +z ∂f y 12 ∂f = (9, 12, 20) = , 2 2 2 ∂y ∂y 25 x +y +z ∂f z 4 ∂f = (9, 12, 20) = and f (9, 12, 20) = 25 , ∂z 5 x2 + y 2 + z 2 ∂z

13.4. LINEARIZATION AND DIFFERENTIALS

835

12 4 9 (x − 9) + (y − 12) + (z − 20). For the 25 25 5 12 4 9 approximation, we have L(9.1, 11.75, 19.98) = 25 + (0.1) + (−0.25) + (−0.02) = 25 25 5 24.9 The linearization is L(x, y, z) = 25 +

4.4.3, if f were differentiable at (0, 0), then f would have to be 40. According to Theorem 13.4.3, continuous at (0, 0). However, as shown in Problem 38 in Exercises 13.2, 4.2, f is not continuous at (0, 0). Therefore, f cannot be differentiable at (0, 0).

41. (a) The graph of z = f (x, y) is an inverted cone with vertex at the origin. Since the graph comes to a sharp ”point” at the origin, there is no possible increment formula for Δz that will work in every direction there.

(b) We show that the partial derivative fx does not exist at (0, 0). If h > 0,

√ √ h 2 + 0 2 − 02 + 0 2 f (0 + h) − f (0, 0) = h h √ 2 |h| h = =1 = h h

f (0 + h) − f (0, 0) = −1. h f (0 + h) − f (0, 0) does not exist. But this means fx does not exist at Therefore, lim h→0 h (0, 0) and thus f is not differentiable at (0, 0). But if h < 0, then

CHAPTER 13. PARTIAL DERIVATIVES

836 42.

Δz z

z x

y

x

y

Δy

Δx ΔV

y

x

z z

x

y

dV

43.

ΔV − dV

(a) From the figure we see that α = π − (θ + φ). Then xh = L cos θ + l cos α = L cos θ + l cos(π − θ − φ) = L cos θ − l cos(θ + φ) and yh = L sin θ − l sin α = L sin θ − l sin(π − θ − φ) = L sin θ − l sin(θ + φ).

L

θ

θ

φ

l

α

(x0,y0)

13.5. CHAIN RULE

837

(b) Using l sin(θ + φ) = ye − yh and l cos(θ + φ) = xe − xh , we have dxh = (−L sin θ + l sin(θ + φ))dθ + l sin(θ + φ)dφ = −yh dθ + (ye − yh )dφ dyh = (L cos θ − l cos(θ + φ))dθ − l cos(θ + φ)dφ = xh − (xe − xh )dφ. (c) One position has the lower arm reaching straight up, with the elbow on the x-axis, so that θ = 0 and φ = 270◦ . The other position has the lower arm reaching straight across, with the elbow on the y-axis, so that θ = φ = 90◦ . In both cases, (xh , yh ) = (L, L). In the first case, (xe , ye ) = (L, 0), and in the second case (xe , ye ) = (0, L). In general, the approximate maximum error in xh is |dxh | = | − hh dθ + (ye yh )dφ| ≤ L|dθ| + |ye − L||dφ| = (L + |ye − L|)

π . 180

Thus, in the first case the approximate maximum error is 2πL/180, while in the second case it is only πL/180. 44. (a) The horizontal and vertical components of velocity are v cos θ and v sin θ, respectively. The projectile strikes the cliff wall at time t = D/v cos θ. At this time its height is  2 D 1 1 D2 1 = D tan θ − g 2 sec2 θ. H = tv sin θ − gt2 = D tan θ − g 2 2 v cos θ 2 v   ∂H D2 d2 ∂H 2 2 2 dv + dθ = g 3 sec θdv + D sec θ − g 2 sec θ tan θ dθ (b) dH = ∂v ∂θ v v (c) When D = 100, g = 32, v = 100, and θ = 45◦ , we have H = 68 ft. (d) Taking |dv|≤ 1 and  we find  |dθ| ≤ π/180 1002 1002 π sec2 |dH| ≤ 32 sec2 4 (1) + 100 sec2 π4 − 32 3 100 1002 3.01 ft.

 π 4

tan π4

π 180



=

34π 16 + ≈ 25 45

(e) We have dH =

∂H ∂H ∂H + + ∂v ∂θ ∂D

=g

    D2 D2 D 2 2 2 2 sec θdv + D sec θ − g sec θ tan θ + tan θ − g sec θ dD. v3 v2 v2

With |dD| ≤ 2, we obtain dH ≤

34 34π 16 34π 18 + + = + ≈ 3.73 ft. 25 45 25 25 45

PROBLEMAS 4.5

13.5 1.

2.

Chain Rule

∂z dx ∂z dy 2x 2y dz (2t) + 2 (−2t−3 ) = + = 2 dt ∂x dt ∂y dt x + y2 x + y2 4xt − 4yt−3 = x2 + y 2 ∂z dx ∂z dy dz = + dt ∂x dt ∂y dt = (3x2 y − y 4 )(5e5t ) + (x3 − 4xy 3 ) (5 sec(t) tan(t))

CHAPTER 13. PARTIAL DERIVATIVES

838

∂z dx ∂z dy dz = + = −3 sin(3x + 4y)(2) − 4 sin(3x + 4y)(−1) dt ∂x dt ∂y dt 5π At t = π, x = and y = −5π 4 2      15π 15π dz  − 5π + 4 sin − 5π = −6 + 4 = −2 so = −6 sin dt t=π 2 2   dz −8 ∂z dx ∂z dy xy 4. = + = ye + xexy (3) dt ∂x dt ∂y dt (2t + 1)2 At t = 0, x = 4 and y = 5 dz  so = −40e20 + 12e20 = −28e20 dt t=0     1 1 2r 4r 2u dp r 2 r √ = (2u)− + = 5. − √ − 3 − du 2s + t (2s + t)2 u (2s + t)2 2 u 2s + t u3 (2s + t)2 2 u(2s + t)2

3.

y2 2xy 3xy 2 y 2 sin s 2xy cos s 3xy 2 sec2 s dr = 3 (− sin s) + 3 (cos s) − 4 (sec2 s) = − + − ds z z z z3 z3 z4 2 2 ∂z ∂x ∂z ∂y + = y 2 exy (3u2 ) + 2xyexy (1) 7. zu = ∂x ∂u ∂y ∂u 6.

2

2

= 3u2 y 2 exy + 2xyexy 2 2 ∂z ∂x ∂z ∂y + = y 2 exy (0) + 2xyexy (−2v) zv = ∂x ∂v ∂y ∂v = −4vxyexy

2

∂z ∂x ∂z ∂y + = 2x cos 4y(2uv 3 ) − 4x2 sin 4y(3u2 ) ∂x ∂u ∂y ∂u = 4uv 3 x cos 4y − 12u2 x2 sin 4y ∂z ∂x ∂z ∂y + = 2x cos 4y(3u2 v 2 ) − 4x2 sin 4y(3v 2 ) zv = ∂x ∂v ∂y ∂v = 6u2 v 2 x cos 4y − 12v 2 x2 sin 4y

8. zu =

9. zu = 4(4u3 ) − 10y[2(2u − v)(2)] = 16u3 − 40(2u − v)y zv = 4(−24v 2 ) − 10y[2(2u − v)(−1)] = −96v 2 + 20(2u − v)y  2 1 −2x 2y 2xv 2 v 2y + 10. zu = + 2 − 2 = 2 2 2 (x + y) v (x + y) u v(x + y) u (x + y)2    u 2v −2x 2y 4xv 2yu − 2 + zv = − =− 2 2 (x + y) v (x + y)2 u v (x + y)2 u(x + y)2 3 2 3 (u + v 2 )1/2 (2u)(−e−t sin θ) + (u2 + v 2 )1/2 (2v)(−e−t cos θ) 2 2 = −3u(u2 + v 2 )1/2 e−t sin θ − 3v(u2 + v 2 )1/2 e−t cos θ 3 3 wθ = (u2 + v 2 )1/2 (2u)e−t cos θ + (u2 + v 2 )1/2 (2v)(−e−t sin θ) 2 2 = 3u(u2 + v 2 )1/2 e−t cos θ − 3v(u2 + v 2 )1/2 e−t sin θ

11. wt =

13.5. CHAIN RULE

839

√ √ u/2 uv rs2 u v/2 uv rv 12. wr = (2r) + (2rs2 ) = √ +√ 1 + uv 1 + uv uv(1 + uv) uv(1 + uv) √ √ u/2 uv r2 su v/2 uv −sv (−2s) + (2r2 s) = √ +√ ws = 1 + uv 1 + uv uv(1 + uv) uv(1 + uv) 2

2

13. Ru = s2 t4 (ev ) + 2rst4 (−2uve−u ) + 4rs2 t3 (2uv 2 eu 2

2

2 2

v

2

2

) = s2 t4 ev − 4uvrst4 e−u + 8uv 2 rs2 t3 eu

2 2

2

2

Rv = s2 t4 (2uvev ) + 2rst4 (e−u ) + 4rs2 t3 (2u2 veu v ) = 2s2 t4 uvev + 2rst4 e−u + 8rs2 t3 u2 veu       t2 1/t 1 1 t 1 1 1 t2 √ √ 14. Qx = + 2+ + + = 2 2 2 2 2 P q t r 1 + (x/t) qt r(t + x2 ) 1−x p 1−x     −x/t2 2x 1 1 x 1 2t sin−1 x 2x − 3− = Qt = (2t sin−1 x) + − 3 + 2 P q t r 1 + (x/t) p qt r(t2 + x2 ) 2y xu u cosh rs y cosh rs 2x +  = +  15. wt =  2 2 2 2 2 2 u 2 x + y rs + tu 2 x + y x + y (rs + tu) u x2 + y 2 s st sinh rs 2y xs yst sinh rs 2x +  = −  wr =  2 2 2 2 2 2 rs + tu u 2 x +y 2 x +y x + y (rs + tu) u x2 + y 2 2y t −t cosh rs yt cosh rs 2x xt +  −  wu =  = 2 2 2 2 2 2 2 2 u 2 x + y rs + tu 2 x + y x + y (rs + tu) u x2 + y 2 16. sφ = 2pe3θ + 2q[− sin(φ + θ)] − 2rθ2 + 4(2) = 2pe3θ − 2q sin(φ + θ) − 2rθ2 + 8 sθ = 2p(3e3θ ) + 2q[− sin(φ + θ)] − 2r(2φθ) + 4(8) = 6pφe3θ − 2q sin(φ + θ) − 4rφθ + 32 17. (a) 3x2 − 2x2 (2yy  ) − 4xy 2 + y  = 0 =⇒ (1 − 4x2 y)y  = 4xy 2 − 3x2 =⇒ y  = (b) fx = 3x2 − 4xy 2 , fy = −4x2 y + 1; y  = −

4xy 2 − 3x2 1 − 4x2 y

4xy 2 − 3x2 3x2 − 4xy 2 = −4x2 y + 1 1 − 4x2 y

18. (a) 1 + 4yy  = ey y  =⇒ 1 = (ey − 4y)y  =⇒ y  =

ey

1 − 4y

1 1 = y 4y − ey e − 4y y cos xy 19. (a) y  = (cos xy)(xy  + y) =⇒ (1 − x cos xy)y  = y cos xy =⇒ y  = 1 − x cos xy (b) f (x, y) = y − sin xy; fx = −y cos xy, fy = 1 − x cos xy; y cos xy −y cos xy = y = − 1 − x cos xy 1 − x cos xy (b) f (x, y) = x + 2y 2 − ey ; fx = 1, fy = 4y − ey ; y  = −

2 (x + y)−1/3 (1 + y  ) = xy  + y =⇒ 2(x + y)−1/3 + 2(x + y)−1/3 y  = 3xy  + 3y 3   3y − 2(x + y)−1/3 =⇒ 2(x + y)−1/3 − 3x y  = 3y − 2(x + y)−1/3 =⇒ y  = 2(x + y)−1/3 − 3x 2 2 (b) f (x, y) = (x + y)2/3 − xy; fx = (x + y)−1/3 − y, fy = (x + y)−1/3 − x; 3 3 2 (x + y)−1/3 − y 3y − 2(x + y)−1/3 y  = − 23 = −1/3 − x 2(x + y)−1/3 − 3x 3 (x + y)

20. (a)

2 2

v

2 2

v

CHAPTER 13. PARTIAL DERIVATIVES

840 21. Fx = 2x, Fy = 2y, Fz = −2z;

2x x ∂z 2y y ∂z =− = ; =− = ∂x −2z z ∂y −2z z

(2/3)x−1/3 2 −1/3 2 2 ∂z z 1/3 x =− , Fy = y −1/3 , Fz = z −1/3 ; = − ; 3 3 3 ∂x (2/3)z −1/3 x1/3 (2/3)y −1/3 ∂z z 1/3 =− = − 1/3 −1/3 ∂y (2/3)z y

22. Fx =

23. F (x, y, z) = xy 2 z 3 + x2 − y 2 − 5z 2 , Fx = y 2 z 3 + 2x, Fy = 2xyz 3 − 2y, Fz = 3xy 2 z 2 − 10z y 2 z 3 + 2x y 2 z 3 + 2x 2xyz 3 − 2y 2xyz 3 − 2y ∂z ∂z =− = = − = ; ∂x 3xy 2 z 2 − 10z 10z − 3xy 2 z 2 ∂y 3xy 2 z 2 − 10z 10z − 3xy 2 z 2 −1/x z 1 1 1 ∂z =− = ; 24. F (x, y, z) = z − ln(xyz); Fx = − , Fy = − , Fz = 1 − ; x y z ∂x 1 − 1/z xz − x ∂z −1/y z =− = ∂y 1 − 1/z yz − y 25. Let y = x + at and z = x − at. Then u(x, t) = F (y) + G(z) and ∂u dF ∂y dG ∂z dF dG = + = + ; ∂x dy ∂x dz ∂x dy dz ∂u dF ∂y dG ∂z dF dG = + =a −a ; ∂t dy ∂t dz ∂t dy dz Thus, a2

∂2u d2 G ∂z d2 F d2 F ∂y d2 G + = = + ; ∂x2 dy 2 ∂x dz 2 ∂x dy 2 dz 2 ∂2u d2 G ∂z d2 F d2 G d2 F ∂y −a 2 = a2 2 + a2 2 . =a 2 2 ∂xt dy ∂t dz ∂t dy dz

2 2 ∂2u ∂2u 2d F 2d G = a + a = . ∂x2 dy 2 dz 2 ∂t2

26. Solving η = x + at and ξ = x − at for x and t, we obtain x = (η + ξ)/2 and t = (η − ξ)/2a. Then ∂u ∂x ∂u ∂t 1 ∂u 1 ∂u ∂u = + = − ∂ξ ∂x ∂ξ ∂t ∂ξ 2 ∂x 2a ∂t and

Setting

∂2u 1 ∂ 2 u ∂x 1 ∂ 2 u ∂t 1 ∂2u 1 ∂2u = − = − ∂η∂ξ 2 ∂x2 ∂η 2a ∂t2 ∂η 4 ∂x2 4a2 ∂t2 ∂2u 1 ∂2u ∂2u 1 ∂2u ∂2u = 0, we have − 2 2 = 0 or a2 2 = 2 . 2 ∂η∂ξ 4 ∂x 4a ∂t ∂x ∂t

27. With x = r cos θ and y = r sin θ ∂u ∂x ∂u ∂y ∂u ∂u ∂u = + = cos θ + sin θ ∂r ∂x ∂r ∂y ∂r ∂x ∂y ∂2u ∂ 2 u ∂y ∂2u ∂ 2 u ∂x ∂2u 2 cos θ + sin θ = = cos θ + sin2 θ ∂r2 ∂x2 ∂r ∂y 2 ∂r ∂x2 ∂y 2 ∂u ∂u ∂x ∂u ∂y ∂u ∂u ∂ 2 u ∂y = + = (r sin θ) + (−r sin θ) + 2 (r cos θ) ∂θ ∂x ∂θ ∂y ∂θ ∂y ∂y ∂y ∂θ ∂2u ∂2u ∂u ∂u cos θ + r2 2 sin2 θ − r sin θ − r2 2 cos2 θ. = −r ∂x ∂x ∂y ∂y

13.5. CHAIN RULE

841

∂2u ∂2u + 2 = 0, we have ∂x2 ∂y   ∂ 2 u 1 ∂u 1 ∂2u ∂u ∂2u ∂2u 1 ∂u 2 2 + 2 2 = cos θ + sin θ + cos θ + 2 sin θ + ∂r2 r ∂r r ∂θ ∂x2 ∂y r ∂x ∂y   1 ∂2u ∂2u ∂u ∂u + 2 −r + r2 2 sin2 θ − r sin θ + r2 2 cos2 θ r ∂x ∂x ∂y ∂y   2 1 ∂2u u ∂ ∂u 1 2 2 2 2 cos θ − cos θ = (cos θ + sin θ) + (cos θ + sin θ) + ∂x2 ∂y 2 ∂x r r   ∂u 1 1 + sin θ − sin θ ∂y r r ∂2u ∂2u = + 2 =0 ∂x2 ∂y

Using

28.

∂z dz ∂u ∂z dz ∂u = , = ∂x du ∂x ∂y du ∂y

29. Letting u = y/x in Problem 28, we have ∂z dz ∂u dz ∂u dz  y  ∂z dz +y =x +y =x x − 2 +y ∂x ∂y du ∂x du ∂y du x du

    1 y dz −y + = = 0. x du x x

30. We first compute ∂u ∂r ∂u x ∂u  = = ∂x ∂r ∂x ∂r x2 + y 2 y2 y2 ∂2u ∂2u ∂u x ∂u ∂ 2 u x2 = + = + 2 2 2 2 2 2 3/2 2 2 3/2 ∂x ∂r (x + y ) ∂r (x + y ) ∂r x + y 2 x2 + y 2 ∂r ∂2u ∂ 2 u y2 x2 ∂u = . ∂y 2 ∂r (x2 + y 2 )3/2 ∂r2 x2 + y 2 Then

∂2u ∂2u ∂u x2 + y 2 ∂ 2 u x2 + y 2 ∂ 2 u 1 ∂u . + 2 = + 2 2 = + 2 2 2 2 3/2 ∂x ∂y ∂r (x + y ) ∂r x + y ∂r2 r ∂r

31. We first compute ∂ ∂u = B erf ∂x ∂x



x √ 4kt



∂ =B ∂x



2 π



√ x/ 4kt

 0

e

−v 2

dv

1 −x2 /4kt 2 = B√ √ e π 4kt

2 1  x  −x2 /4kt 2 ∂2u x √ √ = B = −B √ √ e−x /4kt e − 3 ∂x2 2kt π 4kt 2k π kt   √       x/ 4kt x ∂ ∂u x ∂ 2 1 −3/2 2 −v 2 = B erf √ =B − t e dv = B √ − √ e−x/4kt ∂t ∂t ∂t π 0 2 π 4kt 2 k 2 x = −B √ √ e−x /4kt 3 2 π kt

Then k

∂2u ∂u . = ∂x2 ∂t

CHAPTER 13. PARTIAL DERIVATIVES

842

∂I dE ∂I dR 1 E dI = + = (2) − 2 (−1), dt ∂E dt ∂R dt R R 2 60 3/5 8 1 dI = + + = amp/min. = and when E = 60 and R = 50, dt 50 502 25 25 125

32. We are given dE/dt = 2 and dR/dt = −1. Then

33. Since the height of the triangle is x sin θ, the area is given by A = 12 xy sin θ. Then dA ∂A dx ∂A dy ∂A dθ 1 dx 1 dy 1 dθ = + + = y sin θ + x sin θ + xy cos θ . dt ∂x dt ∂y dt ∂θ dt 2 dt 2 dt 2 dt When x = 10, y = 8, θ = π/6, dz/dt = 0.3, dy/dt = 0.5, and dθ/dt = 0.1, √      1 1 3 1 1 (0.1) (0.3) + (10) (0.5) + (10)(8) 2 2 2 2 2 √ √ = 0.6 + 1.25 + 2 3 = 1.85 + 2 2 ≈ 5.31 cm2 /s.

dA 1 = (8) dt 2

34.

35.

(V − 0.0427)(0.08)dT /dt 0.08T (dV /dt) 3.6 dV dP = − + 3 dt (V − 0.0427)2 (V − 0.0427)2 V dt   3.6 dV dT 0.08T 0.08 + − = V − 0.0427 dt V3 (V − 0.0427)2 dt   dw dh dS = 0.1091 0.425w−0.575 h0.725 + 0.725w0.425 h−0.275 dt dt dt When w = 25, h = 29, dw/dt = 4.2, and dh/dt = 2, dS = 0.1091[0.425(25)−0.575 (29(0.725 (4.2) + 0.725(25)0.425 (29)−0.275 (2)] ≈ 0.5976in2 /yr. dt

36.

∂w dx ∂w dy ∂w dz xdx/dt + ydy/dt + zdz/dt dw −4x sin t + 4y cos t + 5z  = + + = = 2 2 2 dt ∂x dt ∂y dt ∂z dt x +y +z 16 cos2 t + 16 sin2 t + 25t2 −16 sin t cos t + 16 sin t cos t + 25t 25t √ = =√ 2 16 + 25t 16 + 25t2  dw  125π 125π/2 =√ = ≈ 4.9743 dt t=5π/2 64 + 625π 2 16 + 625π 2 /4

37. Since dT /dT = 1 and 0 = FT =

∂P = 0, ∂T

∂F ∂V ∂F ∂T ∂V ∂F/∂T 1 ∂F ∂P + + =⇒ =− =− . ∂P ∂T ∂V ∂T ∂T ∂T ∂T ∂F/∂V ∂T /∂V

38. (a) From the law of sines,

r 500 500 sin φ = so r = . sin φ sin(π − θ − φ) sin(θ + φ)

(b) r = 500 sin 75◦ / sin 137◦ ≈ 708 yds

13.5. CHAIN RULE

843

(c) Using the chain rule, we obtain ∂r dθ ∂r dφ dr = + dt ∂θ dt ∂φ dt sin(θ + φ) cos φ − sin φ cos(θ + φ) dφ 500 sin φ cos(θ + φ) dθ + 500 =− dt dt sin2 (θ + φ) sin2 (θ + φ) 500 sin φ cos(θ + φ) dθ =− dt sin2 (θ + φ) [sin θ cos φ + cos θ sin φ] cos φ − sin φ[cos θ cos φ − sin θ sin φ] dφ + 500 dt sin2 (θ + φ) 500 sin φ cos(θ + φ) dθ 500 sin θ dφ =− + . 2 dt sin (θ + φ) sin2 (θ + φ) dt When dθ/dt = 5◦ = 5π/180 and dφ/dt = −10π/180, we have     dr 500 sin 75◦ cos 137◦ 5π 10π 500 sin 62◦ =− − + ≈ −99.4yd/min. dt 180 180 sin2 137◦ sin2 137◦ The distance from C to A is decreasing. 39. (a) Using f = π, l = 6, V = 100, and c = 330, 000 we obtain f ≈ 380.04 cycles per second.    c A −1/2 1 A 1 A c ∂f = =− f and − 2 =− (b) We first note that ∂V 4π lV lv 4π lV V 2V 1 ∂f = − f. ∂l 2l   ∂f dV ∂f dl 1 dV 1 dl f 1 dV 1 dl df = + =− f − f =− + Then . dt ∂V dt ∂l dt 2V dt 2l dt 2 V dt l dt Using dV /dt = −10, dl/dt = 1, V = 100, and l = 6 we find     df f 1 1 f 1 1 =− (−10) + (1) = − − < 0. dt 2 100 6 2 6 10 The frequency is decreasing. 40. (a)

w w x x

w y

w z

y

z

y x

z x x

x

CHAPTER 13. PARTIAL DERIVATIVES

844 ∂w ∂w y ∂w dz dw = + + dx ∂x ∂y x ∂z dx (b) Using the formula from Part (a), we have

dw = (y 2 + 1) + (2xy − 2z) dx

  1 + (−2y)(ex ) x

41. z

z u

z v

u u t1 u t2 t1

t2

z w

w

v v t1

u u t4 t3 t3

t4

t1

v t2 t2

v v t4 t3 t3

w t1 w t2 t4

t1

w w t4 t3

t2

t3

∂z ∂z ∂u ∂z ∂v ∂z ∂w = + + ∂t2 ∂u ∂t2 ∂v ∂t2 ∂w ∂t2 ∂z ∂z ∂u ∂z ∂v ∂z ∂w = + + ∂t4 ∂u ∂t4 ∂v ∂t4 ∂w ∂t4 42. Since w = F (x, y, z, u) = 0, ∂w/∂x = 0. Also dx/dx = 1, ∂y/∂x = 0, and ∂z/∂x = 0. Then ∂w dx ∂u ∂z ∂u = Fx (x, y, z, u) + Fu (x, y, z, u) + FZ (x, y, z, u) + Fu (x, y, z, u) ∂x dx ∂x ∂x ∂x implies ∂u/∂x = −Fx (x, y, z, u)/Fu (x, y, z, u). Similarly, ∂u/∂y = −FY (x, y, z, u)/Fu (x, y, z, u) and ∂u/∂z = −FZ (x, y, z, u)/Fu (x, y, z, u). 43. Letting F (x, y, z, u) = −xyz + x2 yu + 2xy 3 u − u4 − 8 we find FZ = −yz + 2xyu + 2y 3 u, Fy = −xz + x2 u + 6xy 2 u, Fz = −xy, and Fu = x2 y + 2xy 3 − 4u3 . Then −yz + 2xyu + 2y 3 u ∂u =− 2 , ∂x x y + 2xy 3 − 4u3

−xz + x2 y + 6xy 2 u ∂u =− 2 , ∂y x y + 2xy 3 − 4u3

xy ∂u = 2 . ∂z x y + 2xy 3 − 4u3

44. (a) Let u = λx and v = λy. Then f (u, v) = λn f (x, y), and differentiating both sides with respect to λ, we have ∂f ∂u ∂f ∂v + = nλn−1 f (x, y) or xfu (u, v) + yfu (u, v) = nλn−1 f (x, y). ∂u ∂λ ∂v ∂λ Letting λ = 1, we have u = x and y = v, so xfx (x, y) + yfy (x, y) = nf (x, y). (b) f (λx, λy) = 4(λx)2 (λy 3 ) − 3(λx)(λy)4 + (λx)5 = λ5 f (x, y)

t4

13.6. DIRECTIONAL DERIVATIVE

845

(c) xfx + yfy = x(8xy 3 − 3y 4 + 5x4 ) + y(12x2 y 2 − 12xy 3 ) = 8x2 y 3 − 3xy 4 + 5x5 + 12x2 y 3 − 12xy 4 = 20x2 − 15xy 4 + 5x5 = 5(4x2 y 3 − 3xy 4 + x5 ) = 5f (x, y)   y y

 λy (d) By observing that f =f = λ0 f , we see that z = f xy is homogeneous λx x x of degree zero.

PROBLEMAS 4.6

13.6

Directional Derivative

1. ∇f = (2x − 3x2 y 2 )i + (4y 3 − 2x3 y)j 2

2

2. ∇f = 4xye−2x y i + (1 + 2x2 e−2x y )j 3. ∇F =

2xy 3xy 2 y2 i + j − k z3 z3 z4

4. ∇F = y cos yzi + (x cos yz − xyz sin yz)j − xy 2 sin yzk 5. ∇f = 2xi − 8yj; ∇f (2, 4) = 4i − 32j x3 − 4y 3 27 5 3x2 i+  j; ∇f (3, 2) = √ i − √ j 6. ∇f =  3 4 3 4 38 2 38 2 x y−y 2 x y−y 7. ∇F = 2xz 2 sin 4yi + 4x2 z 2 cos 4yj + 2x2 z sin 4yk √ √ 4π 4π 4π ∇F (−2, π/3, 1) = −4 sin i + 16 cos j + 8 sin k = 2 3i − 8j − 4 3k 3 3 3 3 1 2y 2z 4 2x i+ 2 j+ 2 k; ∇F (−4, 3, 5) = − i + j + k x2 + y 2 + z 2 x + y2 + z2 x + y2 + z2 25 25 25 √ √ f (x + h 3/2, y + h/2) − f (x, y) (x + h 3/2)2 + (y + h/2)2 )2x − y 2 = lim 9. Du f (x, y) = lim h→0 h→0 h h √ 2 2 √ √ h 3x + 3h /4 = hy + h /4 = lim ( 3x + 3h/4 + y + h/4) = 3x + y = lim h→0 h→0 h √ √ f (x + h 2/2, y + h 2/2) − f (x, y) 10. Du f (x, y) = lim h→0 h √ √ 3x + 34 2/2 − (y + h 2/2)2 − 3x + y 2 = lim h→0 h √ √ √ √ √ √ 3h 2/2 − h 2yh2 /2 = lim (3 2/2 − 2y − h/2) = 3 2/2 − 2y = lim h→0 h→0 h √ √ 3 1 15 3 2 6 3 5 i j; ∇f = 15x y i + 30x y j; ∇f (−1, 1) = 15i − 30j; Du f (−1, 1) = − 15 = 11. u = 2 2 2 √ 15 ( 3 − 2) 2 8. ∇F =

CHAPTER 13. PARTIAL DERIVATIVES

846

√ 2 2 i j; ∇f = (4 + y 2 )i(2xy − 5)j; ∇f (3, −1) = 5i − 11j; 12. u = 2 2 √ √ √ 5 2 11 2 − = −3 2 Du f (3, −1) = 2 2 √ √ 3 10 −y 1 10 x 1 i− j; ∇f = 2 i+ 2 j; ∇f (2, −2) = i + j; 13. u = 2 2 10 10 y x +y 4 4 √ x +√ √ 10 3 10 10 Du f (2, −2) = − =− 40 40 20 √

8 y2 x2 6 i + j; ∇f = i + j; ∇f (2, −1) = i + 4j; 10 10 (x + y)2 (x + y)2 3 16 19 = Du f (2, −1) = + 5 5 5 √ 15. u = (2i + j)/ 5; ∇f = 2y(xy + 1)i + 2x(xy + 1)j; ∇f (3, 2) = 28i + 42j; 42 98 2(28) Du f (3, 2) = √ + √ = √ 5 5 5 √ √ 16. u = −i; ∇F = 2x tan yi + x2 sec2 yj; ∇f (1/2, π/3) = 3i + j; Du f (1/2, π/3) = − 3 14. u =

1 1 17. u = √ j + √ k; ∇f = 2xy 2 (2z + 1)2 i2x2 y(2z + 1)2 j + 4x2 y 2 (2z + 1)k; ∇f (1, −1, 1) = 2 2 √ 12 6 18 18i − 18j + 12k; Du f (1, −1, 1) = − √ + √ = − √ = −3 2 2 2 2 1 2 1 2x 2y 2y 2 − 2x2 18. u = √ i − √ j + √ k; ∇f = 2 i − 2 j + k; ∇f (2, 4, −1) = 4i − 8j − 24k; z z z3 6 6 6 √ 16 24 4 Du f (2, 4, −1) = √ − √ − √ = −6 6 6 6 6 x2 + 4z y2 i+  j+  k; x2 y + 2y 2 z 2 x2 y + 2y 2 z x2 y + 2y 2 z ∇f (−2, 2, 1) = −i + j + k; Du f (−2, 2, 1) = −1

19. u = −k; ∇f = 

xy

√ 2 1 2 20. u = −(4i−4j+2k)/ 36 = − i+ j− k; ∇f = 2i−2yj+2zk; ∇f (4, −4, 2) = 2i+8j+4k; 3 3 3 8 4 16 4 − = Du f (4, −4, 2) = − + 3 3 3 3 √ 21. u = (−4i − j/ 17; ∇f = 2(x − y)i − 2(x − y)j; ∇f (4, 2) = 4i − 4j; 4 12 16 Du f (4, 2) = − √ + √ = − √ 17 17 17 √ 22. u = (−2i + 5j/ 29; ∇f = (3x2 − 5y)i − (5x − 2y)j; ∇f (1, 1) = −2i − 3j; 15 11 4 Du f (1, 1) = √ − √ = − √ 29 29 29 √ √ 2 2x 2x j 23. ∇f = 2e sin yi + e cos yj; ∇f (0, π/4) = 2i + 2  √ √ √  √ 2 1/2 The maximum Du is ( 2) + ( 2/2)2 = 5/2 in the direction 2i + ( 2/2)j.

13.6. DIRECTIONAL DERIVATIVE

847

24. ∇f = (xyex−y + yex−y i + (−xyex−y + xex−y j; ∇f (5, 5) = 30i − 20j √ 1/2  The maximum Du is 302 + (−20)2 = 10 13 in the direction 30i − 20j. 25. ∇f = (2x + 4z)i + 2z 2 j + (4x + 4yz)k; ∇f (1, 2, −1) = −2i + 2j − 4k √ 1/2  The maximum Du is (−2)2 + (2)2 + (−4)2 = 2 6 in the direction −2i + 2j − 4k. 26. ∇f = yzi + xzj + xyk; ∇f (3, 1, −5) = −5i − 15j + 3k 1/2 √  The maximum Du is (−5)2 + (−15)2 + (3)2 = 259 in the direction −5i − 15j + 3k. 2 2 2 2 2 2 27. ∇f = 2x sec(x + y )i+ 2y sec2 (x + y )j;  π/6 sec (π/3)(i + j) = 8 π/6(i + j) ∇f ( π/6, π/6) = 2  The minimum Du is −8 π/6(12 + 12 )1/2 = −8 π/3 in the direction −(i + j).

28. ∇f = 3x2 i − 3y 2 j; ∇f (2, −2) = 12i − 12j = 12(i − j) √ 1/2  The minimum Du is −12 12 + (−1)2 = −12 2 in the direction −(i − j) = −i + j. √ y √ √ 2 ze x 3 y √ i + xze j + √ k; ∇f (16, 0, 9) = i + 12j + k. The minimum Du is 29. ∇f = 8 3 2 x 2 z √   2 3 2 2 2 1/2 = − 83281/24 in the direction − i − 12j − k. − (3/8) + 12 + (2/3) 8 3 1 1 1 i + j − k; ∇f (1/2, 1/6, 1/3) = 2i + 6j − 3k x y z 1/2  The minimum Du is − 22 + 62 (−3)2 = −7 in the direction −2i − 6j + 3k.

30. ∇f =

31. Using implicit differentiation on 2x2 + y 2 = 9 we find y  =√−2x/y. At (2, 1) the slope of the tangent line is −2(2)/1 = −4. √Thus, u√= ±(i − 4j)/ √ 17. Now, ∇f = i + 2yj and ∇f (3, 4) = i + 8j. Thus, Du = ±(1/ 17 − 32 17) = ±31/ 17. x + 2y 3x + 3y − 1 2x + y − 1 √ √ 32. ∇f = (2x + y − 1)i + (x + 2y)j; Du f (x, y) = + √ = Solving 2 2 2 √ (3x + 3y − 1)/ 2 = 0 we see that Du is 0 for all points on the line 3x + 3y = 1.  33. (a) Vectors perpendicular to 4i + 3j are ±(3i − 4j). Take u = ±

 3 4 i− j . 5 5

√ 3 4 (b) u = (4i + 3j)/ 16 + 9 = i + j 5 5 4 3 (c) u = − i − j 5 5 34. D−u f (a, b) = ∇f (a, b) · (−u) = −∇f (a, b) · u = −Du f (a, b) = −6 35. (a) ∇f = (3x2 − 6xy 2 )i + (−6x2 y + 3y 2 )j 9x2 − 18xy 2 − 6x2 y + 3y 2 3(3x2 − 6xy 2 ) −6x2 y + 3y 2 √ √ √ + = Du f (x, y) = 10 10 10

CHAPTER 13. PARTIAL DERIVATIVES

848

3 3 (b) F (x, y) = √ (3x2 − 3xy 2 − 2x2 y + y 2 ); ∇F = √ [(6x − 6y 2 − 4xy)i + (−12xy − 10 10 2x2 + 2y)j]       3 3 3 1 2 √ √ Du F (x, y) = √ (6x − 6y − 4xy) + √ (−12xy − 2x2 + 2y) 10 10 10 10 3 1 9 = (3x − 3y 2 − 2xy) + (−6xy − x2 + y) = (27x − 27y 2 − 36xy − 3x2 + 3y) 5 5 5 12 5 α − β = 7 and Dv f (a, b) = 36. Let ∇f (a, b) = αi + βj. Then Du f (a, b) = ∇f (a, b) · u = 13 13 12 5 α − β = 3. Solving for α and β, we obtain α = 13 and β = −13/6. Thus, ∇f (a, b) · v = 13 13 ∇f (a, b) = 13i − (13/6)j. 37. 38. ∇f = 2x, −5y, |∇f | =



10x2 + 25y 2 = 10, 4x2 + 25y 2 = 100,

y2 x2 + =1 25 4

y

x

39. ∇T = 4xi + 2yj; ∇T (4, 2) = 16i + 4j. The minimum change in temperature (that is, the maximum decrease in temperature) is in the direction −∇T (4, 3) = −16i − 4j. 40. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x (t)i + y  (t)j. Since we want the direction of motion to be −∇T (x, y), we have x (t)i + y  (t)j = −∇T (x, y) = 4xi + 2yj. Separating variables in dx/dt = 4x, we obtain dx/x = 4dt, ln x = 4t + c1 , and x = C1 e4t . Separating variables in dy/dt = 2y, we obtain dy/y = 2dt, ln y = 2t + c2 , and y = C2 e2t . Since x(0) = 4 and y(0) = 2, we have x = 4e4t and y = 2e2t . The equation of the path is 4e4t i + 2e2t j for t ≥ 0, or eliminating the parameter, x = y 2 , y ≥ 0. 41. Let x(t)i + y(t)j be the vector equation of the path. At (x, y) on this curve, the direction of a tangent vector is x (t)i + y  (t)j. Since we want the direction of motion to be ∇T (x, y), we have x (t)i + y  (t)j = ∇T (x, y) = −4xi − 2yj. Separating variables in dx/dt = −4x, we obtain dx/x = −4dt, ln x = −4t + c1 , and x = C1 e−4t . Separating variables in dy/dt = −2y, we obtain dy/y = −2dt, ln y = −2t + c2 , and y = C2 e−2t . Since x(0) = 3 and y(0) = 4, we have x = 3e−4t and y = 4e−2t . The equation of the path is 3e−4t i + 4e−2t j, or eliminating the parameter, 16x = 3y 2 , y ≥ 0.

13.6. DIRECTIONAL DERIVATIVE

849

42. Substituing x = 0, y = 0, z = 1, and T = 500 into t = and T (x, y, z) =

500 . x2 + y 2 + z 2

k we see that k = 500 x2 + y 2 + z 2

1 2 2 1

1, −2, −2 = i − j − k 3 3 3 3 1000x 1000y 1000z ∇T = − 2 i− 2 j− 2 k 2 2 2 2 2 2 (x + y + z ) (x + y + z ) (x + y 2 + z 2 )2 750 750 500 i− j− k ∇T (2, 3, 3) = − 121 121 121       1 500 2 750 2 750 2500 Du T (2, 3, 3) = − − − − − = 3 121 3 121 3 121 363

(a) u =

(b) The direction of maximum increase is ∇T (2, 3, 3) = −

750 750 252 500 i− j− k= (−2i − 3j − 3k). 121 121 121 121

(c) The maximum rate of change of T is |∇T (2, 3, 3)| =

√ 250 √ 250 22 . 4+9+9= 121 121

Gmy Gm Gmx i+ 2 j= 2 (xi + yj) 2 3/2 2 3/2 +y ) (x + y ) (x + y 2 )3/2 The maximum and minimum values of Du U (x, y) are obtained when u is in the directions ∇U and −∇U , respectively. Thus, at a point (x,y), not (0,0), the directions of maximum and minimum increase in U are xi+yj and −xi−yj, respectively. A vector at (x, y) in the direction ±(xi+yj) lies on a line through the origin.

43. ∇U =

(x2

44. Since ∇f = fx (x, y)i + fy (x, y)j, we have ∂f /∂x = 3x2 + y 3 + yexy . Integrating, we obtain f (x, y) = x3 + xy 3 + exy + g(y). Then fy = 3xy 2 + xexy + g  (y) = −2y 2 + 3xy 2 + xexy . Thus, g  (y) = −2y 2 , g(y) = − 23 y 3 + c, and f (x, y) = x3 + xy 3 + exy − 23 + C. 45. ∇(cf ) =

∂ ∂ (cf )i + j = cfx i + cfy j = c(fx i + fy j) = c∇f ∂x ∂y

46. ∇(f + g) = (fx + gx )i + (fy + gy )j = (fx i + fy j) + (gx i + gy j) = ∇f + ∇g 47. ∇(f g) = (f gx + fx g)i + (f gy + fy g)j = f (gx i + gy j) + g(fx i + fy j) = f ∇g + g∇f     48. ∇(f /g) = (gfx − f gx )/g 2 i + (gfy − f gy )/g 2 j = g(fx i + fy j)/g 2 − f (gx i + gy j)/g 2 = g∇f /g 2 − f ∇g/g 2 = (g∇f − f ∇g)/g 2  x ∂r y ∂r x y = = x2 + y 2 so = and = 2 2 2 2 ∂x r ∂y r x +y x +y !x y" 1 r , This gives ∇r = = x, y = r r r r

49. r(x, y) =

CHAPTER 13. PARTIAL DERIVATIVES

850 50.

df ∂r ∂ (f (r)) df ∂r ∂ (f (r)) ∂ (f (r)) ∂ (f (r)) = and = so that ∇f (r) = ,  ∂x dr ∂x ∂y dr ∂y ∂x ∂y df ∂r df ∂r df ∂r ∂r = , =

,  = f  (r)∇r = f  (r)r/r dr ∂x dr ∂y dr ∂x ∂y

51. Let u = u1 i + u2 j and v = v1 i + v2 j. Dv f = (fx i+ fy j) · v = v1 fx + v2 fy  ∂ ∂ (v1 fx + v2 fy )i + (v1 fx + v2 fy )j · u = [(v1 fxx + v2 fyz )i + (v1 fxy + v2 fyy )j] · u Du Dv f = ∂x ∂y = u1 v1 fxx + u1 v2 fyx + u2 v1 fxy + u2 v2 fyy D − uf = (fx i + fy j) · u = u1 fx + u2 fy   ∂ ∂ (u1 fx + u2 fy )i + (u1 fx + u2 fy )j · v = [(u1 fxx + u2 fyx )i + (u1 fxy + u2 fyy )j] · v Dv Du f = ∂x ∂y = u1 v1 fxx + u2 v1 fyx + u1 v2 fxy + u2 v2 fyy Since the second partial derivatives are continuous, fxy = fyx and Du Dv f = Dv Du f. [Note that this result is a generalization fxy = fyx since Di Dj f = fyx and Dj Di f = fxy ]    i j k   ∂ ∂ ∂   52. ∇ × F =    ∂x ∂y ∂z   f f2 f3  1       ∂f3 ∂f2 ∂f3 ∂f1 ∂f2 ∂f1 = − − − i− j+ k ∂y ∂z ∂x ∂z ∂x ∂y

PROBLEMAS 4.7

13.7

Tangent Planes and Normal Lines

1. Since f (6, 1) = 4, the level curve is x − 2y = 4. ∇f = i − 2j; ∇f (6, 1) = i − 2j

y

x

2. Since f (1, 3) = 5, the level curve is y+2x = 5x or y = 3x, x = 0. y 1 ∇f = − 2 i + j; ∇f (1, 3) = −3i + j x x

y

x

13.7. TANGENT PLANES AND NORMAL LINES

851

3. Since f (2, 5) = 1, the level curve is y = x2 + 1. ∇f = −2xi + j; ∇f (2, 5) = −4i + j

y

x

4. Since f (−1, 3) = 10, the level curve is x2 + y 2 = 10. ∇f = 2xi + 2yj; ∇f (−1, 3) = −2i + 6j

y

x

5. Since f (−2, −3) = 2, the level curve is x2 /4 + y 2 /0 = 2 2y x x2 /8 + y 2 /18 = 1. ∇f = i + j; ∇f (−2, −3) = −i − 2 9

y

or 2 j 3

x

6. Since f (2, 2) = 2, the level curve is y 2 = 2x, x = 0. 2y y2 ∇f = − 2 i + j; ∇f (2, 2) = −i + 2j x x

y

x

852

CHAPTER 13. PARTIAL DERIVATIVES

7. Since f (1, 1) = −1, the level curve is (x − 1)2 − y 2 = −1 or y 2 − (x − 1)2 = 1. ∇f = 2(x − 1)i − 2yj; ∇f (1, 1) = −2j

y

x

8. Since f (π/6, 3/2) = 1, the level curve is y − 1 = sin x or −(y − 1) cos x 1 y = 1 + sin x, sin x = 0. ∇f = i+ j; 2 sin x sin x √ ∇f (π/6, 3/2) = − 3i + 2j

y

x

9. Since f (3, 1, 1) = 2, the level curve is y + z = 2 ∇f = j + k; ∇f (3, 1, 1) = j + k

z

2

2

y

x

10. Since f (1, 1, 3) = −1, the level curve is x2 + y 2 − z = −1 or z = 1 + x2 + y 2 . ∇f = 2xi + 2yj − k; ∇f (1, 1, 3) = 2i + 2j − k

z

y

x 11. Since F (3, 4, 0) = 5, the level curve is x2 + y 2 + z 2 = 25. x y z ∇F =  i+  j+  k; 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 3 4 ∇F (3, 4, 0) = i + j 4 5

z 5 5 y x

13.7. TANGENT PLANES AND NORMAL LINES

853

12. Since F (0, −1, 1) = 0, the level curve is x2 − y 2 + z = 0 or z = y 2 − x2 . ∇F = 2xi − 2yj + k; ∇F (0, −1, 1) = 2i + k

z

y x

 13. F (x, y, z) = x2 + y 2 − z; ∇F = 2xi + 2yj − k. We want ∇F = c 4i + j + 12 k or 2x = 4c, 2y = c, −1 = c/2. From the third equation c = −2. Thus, x = −4 and y = −1. Since z = x2 + y 2 = 16 + 1 = 17, the point on the surface is (−4, −1, −17). 14. F (x, y, z) = x3 + y 3 + z; ∇F = 3x2 i + 2yj + k. We want ∇F = c(27i + 8j + k) or 3x2 = 27c, 2y = 8c, 1 = c. From c = 1 we obtain x = ±3 and y = 4. Since z = 15 − x3 − y 2 = 15 − (±3)3 − 16 = −1 ∓ 27, the points on the surface are (3, 4, −28) and (−3, 4, 26). 15. F (x, y, z0 = x2 + y 2 + z 2 ; ∇F = 2xi + 2yj + 2zk. ∇F (−2, 2, 1) = −4i + 4j + 2k. The equation of the tangent plane is −4(x + 2) + 4(y − 2) + 2(z − 1) = 0 or −2x + 2y + z = 9. 16. F (x, y, z) = 5x2 − y 2 + 4z 2 ; ∇F = 10xi − 2yj + 8zk; ∇F (2, 4, 1) = 20i − 8j + 8k.The equation of the tangent plane is 20(x − 2) − 8(y − 4) + 8(z − 1) = 0 or 5x − 2y + 2z = 4. 17. F (x, y, z) = x2 − y 2 − 3z 2 ; ∇F = 2xi − 2yj − 6zk; ∇F (6, 2, 3) = 12i − 4j − 18k. The equation of the tangent plane is 12(x − 6) − 4(y − 2) − 18(z − 3) = 0 or 6x − 2y − 9z = 5. 18. F (x, y, z) = xy + yz + zx; ∇F = (y + z)i + (x + z)j + (y + x)k; ∇F (1, −3, −5) = −8i − 4j − 2k. The equation of the tangent plane is −8(x − 1) − 4(y + 3) − 2(z + 5) = 0 or 4x + 2y + z = −7. 19. F (x, y, z) = x2 + y 2 + z; ∇F = 2xi + 2yj + k; ∇F (3, −4, 0) = 6i − 8j + k. The equation of the tangent plane is 6(x − 3) − 8(y + 4) + z = 0 or 6x − 8y + z = 50. 20. F (x, y, z) = xz; ∇F = zi + xk; ∇F (2, 0, 3) = 3i + 2k. The equation of the tangent plane is 3(x − 2) + 2(z − 3) = 0 or 3x + 2z = 12. √ ∇F (π/2, π/4, −1 2) = 21. F (x, y,√z) = cos(2x+y)−z; ∇F = −2 sin(2x+y)i−sin(2x+y)j−k; √   √  √ 2 2 π π 1 j−k. The equation of the tangent plane is 2 x − = 2i+ + y− − z+√ 2 2 2 4 2    √ π  π √ 1 5π 0, 2 x − + y− − 2 z+√ + 1. = 0, or 2x + y − 2z = 2 4 4 2 22. F (x, y, z) = x2 y 3 + 6z; ∇F = 2xy 3 i + 3x2 y 2 j + 6k; ∇F (2, 1, 1) = 4i + 12j + 6k. The equation of the tangent plane is 4(x − 2) + 12(y − 1) + 6(z − 1) = 0 or 2x + 6y + 3z = 13. √ √ √ 2x 2y 23. F (x, y, z) = ln(x2 + y 2 ) − z; ∇F = 2 i+ 2 j − k; ∇F (1/ 2, 1/ 2, 0) = 2i + 2 2 x +y x  +y    √ √ √ 1 1 √ √ + 2 y− − (z − 0) = 2j − k. The equation of the tangent plane is 2 x − 2 2     √ √ √ 1 1 +2 y− √ − 2z = 0, or 2x + 2y − 2z = 2 2. 0, 2 x − √ 2 2

854

CHAPTER 13. PARTIAL DERIVATIVES

−2y 24. F (x, sin 4x − z; ∇F = 32e−2y cos 4xi − 16e−2y sin 4xj − k; ∇F (π/24, 0, 4) = √ y, z) = 8e 16 3i − 8j − k. The equation of the tangent plane is √ √ √ 2 3π 16 3(x − π/24) − 8(y − 0) − (z − 4) = 0 or 16 3x − 8y − z = − 4. 3

25. The gradient of F (x, y, z) = x2 + y 2 + z 2 is ∇F = 2xi + 2yj + 2zk, so the normal vector to the surface at (x0 , y0 , z0 ) is 2x0 i + 2y0 j + 2z0 k. A normal vector to the plane 2x + 4y + 6z = 1 is 2i + 4j + 6k. Since we want the tangent plane to be parallel to the given plane, we find c 2c, z0 = 3c. Now, (x0 , y0 , z0 ) is on so that 2x0 = 2c, 2y0 = 4c, 2z0 = 6c or x0 = c, y0 = √ 2 2 2 2 + (2c) + (3c) = 14c = 7 and c = ±1/ 2. Thus, the points on the surface the surface, so c √ √ √ √ √ √ are ( 2/2, 2, 3 2/2) and − 2/2, − 2, −3 2/2). 26. The gradient of F (x, y, z) = x2 − 2y 2 − 3z 2 is ∇F (x, y, z) = 2xi − 4yj − 6zk, so a normal vector to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i − 4y0 j − 6z0 k. A normal vector to the plane 8x+4y+6z = 5 is 8i+4j+6k. Since we want the tangent plane to be parallel to the given plane, we find c so that 2x0 = 8c, −4y0 = 4c, −6z0 = 6c or x0 = 4c, y0 = −c, √z0 = −c. 2 2 2 2(−c)2 − 3(−c) Now (x0 , y0 , z0 ) is on the surface, √ −√ √ = √ 11c √ = 33 and c = ± 3. Thus, √so (4c) the points on the surface are (4 3, − 3, − 3) and (−4 3, 3, 3). 27. The gradient of F (x, y, z) = x2 +4x+y 2 +z 2 −2z is ∇F = (2x+4)i+2yj+(2z−2)k, so a normal to the surface at (x0 , y0 , z0 ) is (2x0 + 4)i + 2y0 j + (2z0 − 2)k. A horizontal plane has normal ck for c = 0. Thus, we want 2x0 + 4 = 0, 2y0 = 0, 2z0 − 2 = c or x0 = −2, y0 = 0, z0 = c + 1. Since (x0 , y0 , z0 ) is on the surface, (−2)2 + 4(−2) + (c + 1)2 − 2(c + 1) = c2 − 5 = 11 and c = ±4. The points on the surface are (−2, 0, 5) and (−2, 0, −3). 28. The gradient of F (x, y, z) = x2 + 3y 2 + 4z 2 − 2xy is ∇F = (2x − 2y)i + (6y − 2x)j + 8zk, so a normal to the surface at (x0 , y0 , z0 ) is 2(x0 − y0 )i + 2(3y0 − x0 )j + 8z0 k. (a) A normal to the xz plane is cj for c = 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = c, 8z0 = 0 or x0 = y0 , 3y0 −x0 = c/2, z0 = 0. Solving the first two equations, we obtain 2 , z0 ) is on the surface, (c/4)2 +3(c/4)2 +4(0)√ −2(c/4)(c/4) = x0 = y0 = c/4. Since (x0 , y0√ √ 2 = 16 and c = ±16/ 2. Thus, the points on the surface are (4/ 2, 4/ 2, 0) and 2c /16 √ √ (−4 2, −4 2, 0). (b) A normal to the yz-plane is ci for c = 0. Thus, we want 2(x0 − y0 ) = c, 2(3y0 − x0 ) = 0, 8z0 = 0 or x0 − y0 = c/2, x0 = 3y0 , z0 = 0. Solving the first two equations, we obtain x0 = 3c/4 and y0 = c/4. Since (x0 , y0 , z0 )√is on the surface, (3c/4)2 + 3(c/4)2 + = 6c2 /16 4(0)2√− 2(3c/4)(c/4) √ √ = 16 and √ c = ±16 6. Thus, the points on the surface are (12/ 6, 4/ 6, 0) and (−12/ 6, −4/ 6, 0). (c) A normal to the xy-plane is ckfor c = 0. Thus, we want 2(x0 − y0 ) = 0, 2(3y0 − x0 ) = 0, 8z0 = c or x0 = y0 , 3y0 −x0 = 0, z0 = c/8. Solving the first two equations, we obtain x0 = y0 = 0. Since (x0 , y0 , z0 ) is on the surface, 02 +3(0)2 +4(c/8)2 −2(0)(0) = c2 /16 = 16 and c = ±16. Thus, the points on the surface are (0, 0, 2) and (0, 0, −2).

29. If (x0 , y0 , z0 ) is on x2 /a2 + y 2 /b2 + z 2 /c2 = 1, then x20 /a2 + y02 /b2 + z02 /c2 = 1 and x0 , y0 , z0 ) is on the plane xx0 /a2 + yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is

13.7. TANGENT PLANES AND NORMAL LINES

855

∇F (x0 , y0 , z0 ) = (2x − 0/a2 )i + (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i + (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel and the plane is tangent to the surface. 30. If (x0 , y0 , z0 ) is on x2 /a2 − y 2 /b2 + z 2 /c2 = 1, then x20 /b2 − y02 /b2 + z02 /c2 = 1 and (x0 , y0 , z0 ) is on the plane xx0 /a2 − yy0 /b2 + zz0 /c2 = 1. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = (2x0 /a2 )i − (2y0 /b2 )j + (2z0 /c2 )k. A normal to the plane is (x0 /a2 )i − (y0 /b2 )j + (z0 /c2 )k. Since the normal to the surface is a multiple of the normal to the plane, the normal vectors are parallel, and the plane is tangent to the surface. 31. F (x, y, z) = x2 + 2y 2 + z 2 ; ∇F = 2xi + 4yj + 2zk; ∇F (1, −1, 1) = 2i − 4j + 2k. Parametric equations of the line are x = 1 + 2t, y = −1 − 4t, z = 1 + 2t. 32. F (x, y, z) = 2x2 − 4y 2 − z; ∇F = 4xi − 8yj − k; ∇F (3, −2, 2) = 12i + 16j − k. Parametric equations of the line are x = 3 + 12t; y = −2 + 16t, z = 2 − t. 33. F (x, y, z) = 4x2 + 9y 2 − z; ∇F = 8xi + 18yj − k; ∇F (1/2, 1/3, 3) = 4i + 6j − k. Symmetric y − 1/3 z−3 x − 1/2 equations of the line are = = . 4 6 −1 34. F (x, y, z) = x2 + y 2 − z 2 ; ∇F = 2xi + 2yj − 2zk; ∇F (3, 4, 5) = 6i + 8j − 10k. Symmetric x−3 y−4 z−5 equations of the line are = = . 6 8 −10 35. Let F (x, y, z) = x2 + y 2 − z 2 . Then ∇F = 2xi + 2yj − 2zk and a normal to the surface at (x0 , y0 , z0 ) is x0 i + y0 j − z0 k. An equation of the tangent plane at (x0 , y0 , z0 ) is x0 (x − x0 ) + y0 (y − y0 ) − z0 (z − z0 ) = 0 or x0 x + y0 y − z0 z = x20 + y02 − z02 . Since (x0 , y0 , z0 ) is on the surface, z02 = x20 + y02 and x20 + y02 − z02 = 0. Thus, the equation of the tangent plane is x0 x + y0 y − z0 z = 0, which passes through the origin. 1 1 1 √ √ x+ y + z. Then ∇F = √ i+ √ j+ √ k and a normal to the surface 2 x 2 y 2 z 1 1 1 at (x0 , y0 , z0 ) is √ i + √ j + √ k. An equation of the tangent plane at (x0 , y0 , z0 ) is 2 x0 2 y0 2 z0 1 1 1 1 1 1 √ √ √ √ (x−x0 )+ √ (y−y0 )+ √ (z−z0 ) = 0 or √ x+ √ y+ √ z = x0 + y0 + z0 = 2 x0 2 y0 2 z0 x0 y0 z0 √ √ √ √ √ √ √ √ √ √ √ √ sum of the intercepts is x0 a + y0 a + z0 a = ( x0 + y0 + z0 ) a = √a. The a · a = a.

36. Let F (x, y, z) =

√

37. A normal to the surface at (x0 , y0 , z0 ) is ∇F (x0 , y0 , z0 ) = 2x0 i + 2y0 j + 2z0 k. Parametric equations of the normal line are x = x0 + 2x0 t, y = y0 + 2y0 t, z = z0 + 2z0 t. Letting t = −1/2, we see that the normal line passes through the origin. 38. The normal lines to F (x, y, z) = 0 and G(x, y, z) = 0 are Fx i+Fy j+Fz k and Gx i+Gy j+Gz k, respectively. These vectors are orthogonal if and only if their dot product is 0. Thus, the surfaces are orthogonal at P if and only if Fx Gx + Fy Gy + Fz Gz = 0. 39. We have F (x, y, z) = x2 + y 2 + z 2 and G(x, y, z) = x2 + y 2 − z 2 . ∇F = 2x, 2y, 2z =  0 except at the origin

856

CHAPTER 13. PARTIAL DERIVATIVES ∇G = 2x, 2y, −2z =  0 except at the origin Therefore, the gradient vectors are nonzero at each of the intersection points. Now Fx Gx + Fy Gy + Fx Gz = (2x)(2x) + (2y)(2y) + (2z)(−2z) = 4x2 + 4y 2 − 4z 2 = 4(x2 + y 2 + z 2 ) = 4(0) = 0 The second to last equality follows from the fact that the intersection points lie on both surfaces and hence satisfy the second equation x2 + y 2 − z 2 = 0.

40. Let F (x, y, z) = x2 − y 2 + z 2 − 4 and G(x, y, z) = 1/xy 2 − z. Then Fx Gx + Fy Gy + Fz Gz = (2x)(−1/x2 y 2 ) + (−2y)(−2/xy 3 ) + (2z)(−1) = −2/xy 2 + 4/xy 2 − 2z = 2(1/xy 2 − z). For (x, y, z) on both surfaces, F (x, y, z) = G(x, y, z) = 0. Thus, Fx Gx + Fy Gy + Fz Gz = 2(0) and the surfaces are orthogonal at points of intersection.

PROBLEMAS 4.8 of Multivariable Functions 13.8 Extrema 1. fx = 2x; fxx = 2; fxy = 0; fy = 2y; fyy = 2; D = 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 4 > 0 and fxx (0, 0) = 2 > 0, f (0, 0) = 5 is a relative minimum. 2. fx = 8x; fxx = 8; fxy = 0; fy = 16y; fyy = 16; D = 128. Solving fx = 0 and fy = 0, we obtain the critical point (0, 0). Since D(0, 0) = 128 > 0 and fxx (0, 0) = 8 > 0, f (0, 0) = 0 is a relative minimum. 3. fx = −2x + 8; fxx = −2; f xy = 0; fy = −2y + 6; fyy = −2; D = 4. Solving fx = 0 and fy = 0 we obtain the critical point (4, 3). Since D(4, 3) = 4 > 0 and fxx (4, 3) = −2 < 0, f (4, 3, ) = 25 is a relative maximum. 4. fx = 6x − 6; f xx = 6; f xy = 0; fy = 4y + 8; fyy = 4; D = 24. Solving fx = 0 and fy = 0, we obtain the critical point (1, −2). Since D(1, −2) = 24 > 0 and fxx (1, −2) = 6 > 0, f (1, −2) = −11 is a relative minimum. 5. fx = 10x + 20; fxx = 10; fxy = 0; fy = 10y − 10; fyy = 10; D = 100. Solving fx = 0 and fy = 0, we obtain the critical point (−2, 1). Since D(−2, 1) = 100 > 0 and fxx (−2, 1) = 10 > 0, f (−2, 1) = 15 is a relative minimum. 6. fx = −8x − 8; fxx = −8; fxy = 0; fy = −4y + 12; fyy = −4; D = 32. Solving fx = 0 and fy = 0, we obtain the critical point (−1, 3). Since D(−1, 3) = 32 > 0 and fxx (−1, 3) = −8 < 0, f (−1, 3) = 27 is a relative maximum. 7. fx = 12x2 − 12; fxx = 24x; fxy = 0; fy = 3y 2 − 3; fyy = 6y; D = 144xy. Solving fx = 0 and fy = 0, we obtain the critical points (−1, −1), (−1, 1), (1, −1), and (1, 1). Since D(−1, 1) = −144 < 0 and D(1, −1) = −144 < 0, these points do not give relative extrema. Since D(−1, −1) = 144 > 0 and fxx (−1, −1) = −24 < 0, f (−1, −1) = 10 is a

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS

857

relative maximum. Since D(1, 1) = 144 > 0 and fxx (1, 1) = 24 > 0, relative minimum.

f (1, 1) = −10 is a

8. fx = −3x2 + 27; fxx = −6x; fxy = 0; fy = 6y 2 − 24; fyy = 12y; D = −72xy. Solving fx = 0, fy = 0, we obtain the critical points (−3, −2), (−3, 2), (3, −2), and (3, 2). Since D(−3, −2) = −432 < 0 and D(3, 2) = −432 < 0, these points do no give relative extrema. Since D(−3, 2) = 432 > 0 and fxx (−3, 2) = 18 > 0, f (−3, 2) = 432 > 0 and fxx (3, −2) = −18 < 0, f (3, −2) = 89 is a relative maximum. 9. fx = 4x − 2y − 10; fxx = 4; fxy = −2; fy = 8y − 2x − 2; fyy = 8; D = 32 − (−2)2 = 28. Setting fx = 0 and fy = 0, we obtain 4x − 2y = 10 and 8y − 2x = 2 or 2x − y = 5 and 4y − x = 1. Solving, we obtain the critical point (3, 1). Since D(3, 1) = 28 > 0 and fxx (3, 1) = 4 > 0, f (3, 1) = −14 is a relative minimum. 10. fx = 10x + 5y − 10; fxx = 10; fxy = 5; fy = 10y + 5x − 5; fyy = 10; D = 100 − (5)2 = 75. Setting fx = 0 and fy = 0, we obtain 10x + 5y = 10 and 10y + 5x = 5 or 2x + y = 2 and 2y + x = 1. Solving, we obtain the critical point (1, 0). Since D(1, 0) = 75 > 0 and fxx (1, 0) = 10 > 0, f (1, 0) = 13 is a relative minimum. 11. fx = 2t − 8; fxx = 0; fxy = 2; fy = 2x − 5; fyy = 0; D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

D = 0 − 22 = −4. Since

12. fx = 2y + 6; fxx = 0; fxy = 2; fy = 2x + 10; fyy = 0; D(x, y) = −4 < 0 for all (x, y), there are no relative extrema.

D = 0 − 22 = −4. Since

13. fx = −6x2 + 6y; fxx = −12x; fxy = 6; fy = −6y 2 + 6x; fyy = −12y; D = 144xy − 36. Setting fx = 0 and fy = 0, we obtain −6x2 + 6y = 0 and −6y 2 + 6x = 0 or y = x2 and x = y 2 . Substituting x = y 2 into y = x2 , we obtain y = y 4 or y(y 3 − 1) = 0. Thus, y = 0 and y = 1. The critical points are (0, 0) and (1, 1). Since D(0, 0) = −36 < 0, (0, 0) does not give a relative extremum. Since D(1, 1) = 108 > 0 and fxx (1, 1) = −12 < 0, f (1, 1) = 12 is a relative maximum. 14. fx = 3x2 − 6y; fxx = 6x; fx y = −6; fy = 3y 2 − 6x; fyy = 6y; D = 36xy − 36. Setting fx = 0 and fy = 0, we obtain 3x2 − 6y = 0 and 3y 2 − 6x = 0 or x2 = 2y and y 2 = 2x. Substituting y = x2 /2 into y 2 = 2x we obtain x4 = 8x or x(x3 − 8) = 0. Thus, x = 0 and x = 2. The critical points (0, 0) and (2, 2). Since D(0, 0) = −36 < 0, f (0, 0) is not an extremum. Since D(2, 2) = 108 > 0 and fxx (2, 2) = 12 > 0, f (2, 2) = 19 is a relative minimum. 15. fx = y + 2/x2 ; fxx = −4/x3 ; fxy = 1; fy = x + 4/y 2 ; fyy = −8/y 3 ; D = 32/x3 Y 3 − 1. Setting fx = 0 and fy = 0 we obtain y + 2/x2 = 0 and x + 4/y 2 = 0. Substituting y = −2/x2 into x + 4/y 2 = 0 we obtain x + x4 = x(1 + x3 ) = 0. Since x = 0 is not in the domain of f, the only critical point is (−1, −2). Since D(−1, −2) = 3 > 0 and fxx (−1, −2) = 4 > 0, f (−1, −2) = 14 is a relative minimum. 16. fx = −6xy − 3y 2 + 36y; fxx = −6y; fxy = −6x − 6y + 36 = 6(6 − x − y); fy = −3x2 − 6xy + 36x; fyy = −6x; D = 36xy − 36(6 − x − y)2 . Setting fx = 0 and fy = 0 we obtain −6xy − 3y 2 + 36y = 0 and −3x2 − 6xy + 36x = 0 or −3y(2x + y − 12) = 0 and −3x(x + 2y − 12) = 0. Letting y = 0, the first equation is satisfied and the second

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858

equation becomes −3x(x − 12) = 0. Thus, (0, 0) and (12, 0) are critical points. Similarly, letting x = 0 we obtain the critical point (0, 12). Finally solving 2x + y = 12 and x + 2y = 12 we obtain the critical point (4, 4). Since D(0, 0) = −362 < 0, D(0, 12) = −362 < 0, and D(12, 0) = −362 < 0, none of these points give relative extrema. Since D(4, 4) = 432 > 0 and fxx (4, 4) = −24 < 0, f (4, 4) = 192 is a relative maximum. 17. fx = (xex + ex ) sin y; fxx = (xex + 2ex ) sin y; fxy = (xex + ex ) cos y; fy = xex cos y; Setting fx (x, y) = 0 fyy = −xex sin y; D = −xe2x (x + 2) sin2 y − e2x (x + 1)2 cos2 y. and fy (x, y) = 0 we obtain (xex + ex ) sin y = 0 and xex cos y = 0. Since ex > 0 for all x, we have (x + 1) sin y = 0 and x cos y = 0. When x = −1, we must have cos y = 0 or y = π/2 + kπ, k an integer. When x = 0, we must have sin y = 0 or y = kπ, k an integer. Thus, the critical points are (0, kπ) and (−1, π/2 + kπ), k an integer. Since D(0, kπ) = 0−cos2 kπ < 0, (0, kπ) does not give a relative extrema. Now, D(−1, π/2+kπ) = e−2 sin2 (π/2+kπ)−0 > 0 and fxx (−1, π/2+kπ) = e−1 sin(π/2+kπ). Since fxx (−1, π/2+kπ) is positive for k even and negative for k odd, f (−1, π/2 + kπ) = −e−1 are relative minima for k even, and f (−1, π/2 + kπ) = e−1 are relative maxima for k odd. 2

2

2

2

18. fx = (2x + 4)ey −3y+x +4x ; fxx = [(2x + 4)2 + 2]ey −3y+x +4x ; 2 2 2 2 fxy = (2x + 4)(2y − 3)ey −3y+x +4x ; fy = (2y − 3)ey −3y+x +4x ; 2 2 2 2 fyy = [(2y − 3)2 + 2]ey −3y+x +4x ; D = [(2x + 4)2 + 2][(2y − 3)2 + 2] · e2(y −3y+x +4x) − [(2x + 2 2 4)(2y − 3)]2 e2(y −3y+x +4x) . Setting fx = 0 and fy = 0 and using the fact that an exponential function is always positive, we obtain 2x + 4 = 0 and 2y − 3 = 0. Thus, a critical point is (−2, 3/2). Since D(−2, 3/2) = 4e2(9/4−9/2+4−8) > 0 and fxx (−2, 3/2) = 2e9/4−9/2+4−8 > 0, f (−2, 3/2) = e9/4−9/2+4−8 = e−25/4 is a relative minimum. 19. fx = cos x; fxx = − sin x; fxy = 0; fy = cos y; fyy = − sin y; D = sin x sin y. Solving fx = 0 and fy = 0, we obtain the critical points (π/2 + mπ, π/2 + nπ) for m and n integers. For m even and n odd or m odd and n even, D < 0 and no relative extrema result. For m and n both even, D > 0 and fxx < 0 and f (π/2 + mπ, π/2 + nπ) = 2 are relative maxima. For m and n both odd, D > 0 and f xx > 0 and f (π/2 + mπ, π/2 + nπ) = −2 are relative minima. 20. fx = y cos xy; fxx = −y 2 sin xy; fxy = −xy sin xy + cos xy; fy = x cos xy; fyy = −x2 sin xy; D = x2 y 2 sin2 xy − (−xy sin xy + cos xy)2 = 2xy sin xy cos xy − cos2 xy. Setting fx = 0 and fy = 0 we see that (0, 0) is a critical point. Also, solving cos xy = 0 we obtain xy = π/2 + kπ or y = π(1 + 2k)/2x for k an integer. Since D(0, 0) = −1 < 0, (0, 0) does not give a relative extrema. For any of the critical points (x, π(1 + 2k)/2x), D = 0 and no conclusion can be drawn from the second partials test. Since −1 ≤ sin xy ≤ 1 for all (x, y)f (x, π(1 + 2k)/2x) = −1 for k odd are relative minima and f (x, π(1 + 2k)/2x) = 1 for k even are relative maxima. 21. Let the numbers be x, y, and 21 − x − y. We want to maximize P (x, y) = xy(21 − x − y) = 21xy − x2 y − xy 2 . Now Px = 21y − 2xy − y 2 ; Pxx = −2y; Pxy = 21 − 2x − 2y; Py = 21x − x2 − 2xy; Pyy = −2x; D = 4xy − (21 − 2x − 2y)2 . Setting Px = 0 and Py = 0, we obtain y(21 − 2x − y) = 0 and x(21 − x − 2y) = 0. Letting x = 0 and y = 0, we obtain the critical points (0, 0), (0, 21), and (21, 0). Each of these results in P = 0 which is clearly not a maximum. Solving 21 − 2x − y = 0 and 21 − x − 2y = 0, we obtain the critical point (7, 7).

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS

859

Since D(7, 7) = 147 > 0 and Pxx (7, 7) = −14 < 0, P (7, 7) = 343 is a maximum. The three numbers are 7,7, and 7. 22. Let the sides of the base of the box by x and y. Then, since the volume of the box is 1, its height is 1/xy and S = 2xy + 2x(1/xy) + 2y(1/xy) = 2xy + 2/y + 2/x, x > 0, y > 0. Now Sx = 2y − 2/x2 ; Sxx = 4/x3 ; Sxy = 2; Sy = 2x − 2/y 2 ; Syy = 4/y 3 ; D = 16/x3 y 3 − 4. Setting Sx = 0 and Sy = 0 we obtain y = 1/x2 and x = 1/y 2 . The critical point is (1, 1). Since D(1, 1) = 12 > 0 and Sxx (1, 1) = 4 > 0, S(1, 1) = 6 is a minimum. The box is 1 foot on each side. 23. Let (x, y, 1 − x − 2y) be a point on the plane x + 2y + z = 1. We want to minimize f (x, y) = x2 + y 2 + (1 − x − 2y)2 . Now fx = 2x − 2(1 − x − 2y); fxx = 4; fxy = 4; fy = 2y − 4(1 − x − 2y); fyy = 10; D = 40 − 42 = 24. Setting fx = 0 and fy = 0 we obtain 2x−2(1−x−2y) = 0 and 2y−4(1−x−2y) = 0 or 2x+2y = 1 and 2x+5y = 2. Thus, (1/6, 1/3) is a critical point. Since D = 24 > 0 and fxx = 4 > 0 for all (x, y), f (1/6, 1/3) = 1/6 is a minimum. Thus, the point on the plane closest to the origin is (1/6, 1/3, 1/6). 24. Let (x, y, 1 − x − y) be a point on the plane x + y + z = 1. We want to minimize the square of the distance between the point and the plane. This is given by f (x, y) = (x − 2)2 + (y − 3)2 + (−x − y)2 = 2x2 + 2y 2 − 4x − 6y + 2xy + 13. fx = 4x − 4 + 2y; fxx = 4; fxy = 2; fy = 4y − 6 + 2x; fyy = 4; D = 16 − 22 = 12. Setting fx = 0 and fy = 0 we obtain 4x − 4 + 2y = 0 and 4y − 6 + 2x = 0 or 2x + y = 2 and x + 2y = 3. Thus, (1/3, 4/3) is a critical point. Since D = 12 > 0 and fxx = 4 > 0 for all (x, y), f (1/3, 4/3) =√25/3 is a minimum. Thus, the least distance between the point and the plane is 25/3 = 5/ 3. 25. Let (x, y, 8/xy) be a point on the surface. We want to minimize the square of the distance to the origin or f (x, y) = x2 + y 2 + 64/x2 y 2 . Now fx = 2x − 128/x3 y 2 ; fxx = 2 + 384/x4 y 2 ; fxy = 256/x3 y 3 ; fy = 2y − 128/x2 y 3 ; fyy = 2 + 384/x2 y 4 ; D = (2 + 384/x4 y 2 )(2 + 384/x2 y 4 ) − (256)2 /x6 y 6 . Setting fx = 0 and fy = 0 we obtain 2x − 128/x3 y 2 = 0 and 2y − 128/x2 y 3 = 0 or x4 y 2 = 64 and x2 y 4 = 64; x = 0, y = 0. This gives x4 y 2 = x2 y 4 , x2 y 2 (x2 − y 2 ) = 0 or x2 = y 2 . Thus, x6 = 64 and x = ±2. Similarly, y = ±2 and the critical points are (−2, −2), (−2, 2), (2, −2), and (2, 2). Since D(±2, ±2) = 48 > 0 and fxx (±2, ±2) = 8 > 0, f (±2, ±2) = 12 are minima. The points closest to√the origin √ are (−2, −2, 2), (−2, 2, −2), (2, −2, −2), and (2, 2, 2). The minimum distance is 12 = 2 3. 26. We will minimize the square of the distance between the lines. This is given by f (s, t) = [(3 + 2s) − t]2 + [(6 + 2s) − (4 − 2t)]2 + [(8 − 2s) − (1 + t)]2 = (2s − t + 3)2 + (2s + 2t + 2)2 + (−2s − t − 7)2 = 12s2 + 6t2 + 8st − 8s + 12t + 62. fs = 24s+8t−8; fss 24; fst = 8; ft = 12t+8s−12; ftt = 12; D = 24(12)−64 = 224. Solving 24s+8t−8 = 0 and 12t+8s−12 = 0 we obtain the critical point (0, 1). Since D(0, 1) = 224 > 0 points on√the and fss (0, 1) = 24 > 0, we see that f (0, 1) is a minimum. The corresponding  √ lines are (3, 6, 8) on L2 and (1, 2, 2) on L1 . The minimum distance is f (0, 1) = 56 = 2 14.

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860

27. We will maximize the square of the volume of the box in the first octant, V (x, y) = x2 Y 2 z 2 = x2 y 2 (c2 − c2 x2 /a2 − c2 y 2 /b2 ). Vx = 2c2 xy 2 −4c2 x3 y 2 /a2 −2c2 xy 4 /b2 ; Vxx = 2c2 y 2 −12c2 x2 y 2 /a2 −2c2 y 4 /b2 ; Vxy = 4c2 xy− 8c2 x3 y/a2 − 8c2 xy 3 /b2 ; Vy = 2c2 x2 y − 2c2 x4 /a2 − 4c2 x2 y 3 /b2 ; Vyy = 2c2 x2 − 2c2 x4 /a2 − 2 . Setting Vx = 0 and Vy = 0 we obtain xy 2 − 2x3 y 2 /a2 − 12c2 x2 y 2 /b2 ; D = Vxx Vyy − Vxy 4 2 2 4 2 2 3 2 2 2 2 2 2 2 xy /b = 0, x y−x y/a −2x y /b = 0, or, assuming x √> 0 and √ y > 0, 2b x +a y = a b . 2 2 2 2 Solving, we obtain x = a /3 and y = b /3. Thus, (a/ 3, b/ 3) is a critical point. Since √ √ 14 14 4 20 2 2 4 a b c >0 D(a/ 3, b/ 3) = (− b2 c2 )(− a2 c2 ) − (− abc2 )2 = 9 9 9 9 and vxx = −

√ √ 14 2 2 b c < 0, V (a/ 3, b/ 3) = a2 b2 c2 /27. The maximum volume is 9  √ √ √ 8 V (a/ 3, b/ 3) = 8 3abc/9.

28. Let a + b + c = k. Then c = k − a − b and we want to maximize V (a, b) = 4πab(k − a − b)/3 = 4π(kab − a2 b − ab2 )/3. 4π 8π 4π 4π (kb − 2ab − b2 ); Vaa = − ; Vab = (k − 2a − 2b); Vb = (ka − a2 − 2ab); 3 3 3 3 64x2 16x2 8π ab − (k − 2a − 2b)2 . Setting Va = 0 and Vb = 0 we obtain Vbb = − a; D = 3 9 9 kb − 2ab − b2 = 0 or ka − a2 − 2ab = 0, a = 0, b = 0, or 2a + b = k and a + 2b = k. Solving, we get a = b = k/3. Since D(k/3, k/3) = 16π 2 k 2 /27 > 0 and Vaa (k/3, k/3) = −8πk/9 < 0, the volume is maximized when a = b = k/3. Since c = k − a − b = k/3, a = b = c and the ellipsoid is a sphere. Va =

29. The perimeter is given by P = 2x + 2y + 2x sec θ and the area is a = 2xy + x2 tan θ. Solving P for 2y and substituting in A, we obtain A = P x − 2x2 (1 + sec θ) + x2 tan θ. Now Ax (x, θ) = P −4x(1+sec θ)2x tan θ; Axx (xθ) = −4(1+sec θ)+2 tan θ; Axθ (x, θ) = −4x sec θ tan θ + 2x sec2 θ; Aθ (xθ) = x2 sec θ(sec θ − 2 tan θ); Aθθ (x, θ) = 2x2 sec θ(tan θ − 2 sec2 θ + 1). We assume that x > 0 and 0 ≤ θ ≤ π/2.

x tanθ

x secθ

θ x

x

y

Setting Ax = 0 and Aθ = 0, we obtain P −4x(1+sec θ)+2x tan θ = 0 and x2 sec θ(sec θ−2 tan θ) = 0. We note from the second equation and the fact that sec θ = 0 for all θ that sec θ − 2√ tan θ = 0. Solving for θ, we obtain θ = 30◦ and solving Ax = 0 for x, we obtain x) = P/(4 + 2 3). Since √ √ √ √ is a D(x0 , 30◦ ) = (−2 3 + 2)(4x20 ( √3 − 5)/3 3) − 02 > 9 and Axx = 2 − 2 3 < 0, A(x0 , 30◦ ) √ sec θ, we obtain P = 2y+P ? 3. maximum. Letting x = P/(4+2 3) and θ = 30√◦ in P = 2x+2y+2x √ √ Thus, the area is maximized for x = P/(4 + 2 3), y = P ( 3 − 1)/2 3, and θ = 30◦

13.8. EXTREMA OF MULTIVARIABLE FUNCTIONS (x sin θ)(24 − 1 2x + x cos θ) = 24x sin θ − 2x sin θ + x2 sin 2θ. 2 Now Ax = 24 sin θ − 4x sin θ + x sin 2θ; Axx = −4 sin θ + sin 2θ; Axθ = 24 cos θ − 4x cos θ + 2x cos 2θ; Aθ = 24x cos θ − 2x2 cos θ + x2 cos 2θ; Aθθ = −24x sin θ + 2x2 sin θ − 2x2 sin 2θ.

30. We want to maximize A(x, θ)

861

=

2

x θ 24-2x

x sinθ

θ x cosθ

We assume 0 < x < 12 and =< θ < π/2. Setting Ax = 0 and Aθ = 0 we obtain 24 sin θ − 4x sin θ + 2x sin θ cos θ = 0 and 24x cos θ − 2x2 cos θ + x2 (2 cos2 θ − 1) = 0 or 12 − 2x + x cos θ = 0 and 2x cos2 θ − 2x cos θ + 2 cos θ − x = 0. Solving the first equation for cos θ and substituting into the − x = 0. Simplifying, we second equation, we obtain 2x(2 − 12/x)2 − 2x(12 − 12/x) + 24(2 √ − 12/x)√ find x = 8 and√cos θ = 1/2 or θ = 60◦ .√Since D(8, 60◦ ) = (−3 3/2)(−96 3) − (−12)2 = 288 > 0 and Axx = −3 3/2 < 0, A(8, 60◦ = 48 3) square inches is the maximum area. 2 2 31. fx = − x−1/3 , fy = − y −1/3 . Since fx = 0 and fy = 0 have no solutions, f (x, y) has no 3 3 critical points and Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 16 ≥ 16 − (x1/3 )2 − (y 1/3 )2 = f (x, y), and f (0, 0) = 16 is an absolute maximum. 32. fx = −4x3 y 2 ; fxx = −12x2 y 2 ; fxy = −8x3 y; fy = −2x4 y; fyy = −2x4 ; D = 24x6 y 2 − 64x6 y 2 = −40x6 y 2 . Setting fx = 0 and fy = 0 we see that (0, y) and (x, 0) are critical points for any x and y. Since, for any critical point, D = 0, Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 1 ≥ 1 − (x2 y)2 = f (x, y), and f (0, 0) = 1 is an absolute maximum. 33. fx = 10x; fxx = 10; fxy = 0; fy = 4y 3 ; fyy = 12y 2 ; D = 120y 2 . Solving fx = 0 and fy = 0 we obtain the critical point (0, 0). Since D(0, 0) = 0, Theorem 13.8.2 does not apply. However, for any (x, y), f (0, 0) = −8 ≤ 5x2 +y 4 −8 = f (x, y) and f (0, 0) = −8 is an absolute minimum. x y2 xy y ; fxx = 2 ; fxy = − 2 ; fy =  ; 34. fx =  (x + y 2 )3/2 (x + y 2 )3/2 x2 + y 2 x2 + y 2 x2 −xy x2 y 2 ; D = −( 2 )2 = 0. Since D = 0 for all (x, y), fyy = 2 2 3 2 2 3/2 (x + y ) (x + y ) (x + y 2 )3/2  Theorem 13.8.2 does not apply. However, for all (x, y), f (0, 0) = 0 ≤ x2 + y 2 = f (x, y), so f (0, 0) = 0 is an absolute minimum.

In Problems 35-38 we parameterize the boundary of R by letting x = cos t and y = sin t; 0 ≤ t ≤ 2π. Then, for (x(t), y(t)) on the boundary, we maximize or minimize F (t) = f (cos t, sin t) on [0, 2π]. √ 35. fx = 1; fy = 3. There √ of R. On the boundary we √ are no critical points on the interior consider F (t) = cos t + 3 sin t. Solving F  (t) = − sin t + 3 cos t = 0, we obtain critical points at t = π/3 2, and F (4π/3) = −2, we √ and t = 4π/3. Comparing F (0) = 1, F (π/3) = √ see that f (1/2, 3/2) = 2 is an absolute maximum and f (−1/2, − 3/2) = −2 is an absolute minimum.

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CHAPTER 13. PARTIAL DERIVATIVES

36. fx = y; fy = x. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corre1 sponding function value f (0, 0) = 0. On the boundary we consider F (t) = cos t sin t = sin 2t. 2 Solving F  (t) = cos 2t = 0, we obtain critical points at π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0, ) = 0, F (0) = 0, F√ (π/4) √= 1/2, F (3π/4) =√−1/2, F (5π/4) = 1/2, √ 2/2, 2/2) = f (− 2/2, − 2/2) = 1/2 are absolute and F (7π/4) = −1/2, we see that f ( √ √ √ √ maxima and f (− 2/2, 2/2) = f ( 2/2, − 2/2) = −1/2 are absolute minima. 37. fx = 2x + y; fy = x + 2y. Solving fx = 0 and fy = 0 we obtain the critical point (0, 0) with corresponding function value f (0, 0) = 0. On the boundary we consider F (t) = 1 cos2 t + cos t sin t + sin2 t = 1 + sin 2t. Solving F  (t) = cos 2t = 0 we obtain critical points at 2 π/4, 3π/4, 5π/4, and 7π/4. Comparing f (0, 0) = 0, F (0) = 1, F (π/4) = 3/2; √ √ F (3π/4) =√ 1/2, F (5π/4) = 3/2, and F (7π/4) = 1/2, we see that f ( 2/2, w/2) = √ f (− 2/2, − 2/2) = 3/2 are absolute maxima and f (0, 0) = 0 is an absolute minimum. 38. fx = −2x; fy = −6y + 4. Solving fx = 0 and fy = 0, we obtain the critical point (0, 2/3), which is inside R, with corresponding function value f (0, 2/3) = 5. On the boundary we consider F (t) = − cos2 t − 3 sin2 t + 4 sin t + 1. Solving F  (t) = 2 cos t sin t − 6 sin t cos t + 4 cos t = 4 cos t − 4 sin t cos t = 0, we obtain critical points at π/2 and 3π/2. Comparing f (0, 2/3) = 5, F (0) = 0, F (π/2) = 2, and F (3π/2) = −6, we see that f (0, −1) = −6 is an absolute minimum and f (0, 2/3) = 5 is an absolute maximum. 39. fx = 4; fy = −6. There are no critical points over the region R, so absolute extrema must occur on the boundary. We parameterize the boundary by x = 2 cos t and y = sin t for 0 ≤ t ≤ 2π. Considering F (t) = 8 cos t − 6 sin t we obtain F  (t) = −8 sin t − 6 cos t. Solving F  (t) = 0 we find tan t = −3/4. Using 1 + tan2 t = sec2 t we see that sec2 t = 25/16 and cos t = −4/5, t is in the second quadrant and sin t = 3/5. The corresponding points on the boundary of R are (8/5, −3/5) and (−8/5, 3/5). Comparing f (0) = F (2π) = f (2, 0) = 8, f (8/5, −3/5) = 10, and f (−8/5, 3/5) = −10 we see that the absolute minimum is f (−8/5, 3/5) = −10 and the absolute maximum is f (8/5, −3/5) = 10. 40. fx = y − 2; fy = x − 1. Solving fx = 0 and fy = 0 we obtain the critical point (1, 2) in the region. On x = 0, F (y) = f (0, y) = −y + 6, which has no critical points for 0 ≤ y ≤ 8. The endpoints of the interval are (0, 0) and (0, 8). On y = 0, G(x) = f (x, 0) = −2x + 6, which has no critical points for 0 ≤ x ≤ 4. The endpoints of the interval are (0, 0) and (4, 0). On y = −2x + 8, H(x) = f (x, −2x + 8) = x(−2x + 8) = x(−2x + 8) − 2x − (−2x + 8) + 6 = −2x2 + 8x − 2. Solving H  (x) = −4x + 8 = 0 we obtain x = 2. The corresponding point on the triangle is (2, 4). Comparing f (0, 0) = 6, f (0, 8) = −2; f (4, 0) = −2; f (2, 4) = 6, and f (1, 2) = 4 we see that absolute maxima are f (0, 0) = f (2, 4) = 6 and absolute minima are f (0, 8) = f (4, 0) = −2. 41. (a) fx = y cos xy; fy = x cos xy. Setting fx = 0 and fy = 0 we obtain y cos xy = 0 and x cos xy = 0. If y = 0 from the first equation, then necessarily x = 0 from the second equation. Thus, (0, 0) is a critical point. For x = 0 and y = 0 we have cos xy = 0 or xy = π/2. Thus, all points (x, π/2x) for 0 ≤ x ≤ π are also critical points. (b) Since 0 ≤ sin xy ≤ 1 for 0 ≤ x ≤ π and 0 ≤ y ≤ 1, f (x, y) = sin xy has absolute minima at any points for which sin xy = 0 and absolute maxima at any points for which

13.9. METHOD OF LEAST SQUARES

863

sin xy = 1. Thus, f (x, y) has absolute minima when xy = 0 or xy = π, that is, at the points (0, y), (x, 0), and (π, 1) which are in the region. Absolute maxima occur when xy = π/2 or along the curve y = π/2x in the region (c) z

1

y

x

42. We want to maximize P (x, y) = R(x, y) − C(x, y) = 108x − 8x2 + 192y − 6y 2 − 4xy − 20. Now Px = 108 − 16x − 4y; Pxx = −16; Pxy = −4; Py = 192 − 4x; Pyy = −12; D = 192 − 16 = 176. Setting Px = 0 and Py = 0 we obtain 108 − 16x − 4y = 0 and 192 − 12y − 4x = 0 or 4x + y = 27 and x + 3y = 48. Solving, we see that (3, 15) is a critical point. Since D(3, 15) = 175 > 0 and Pxx (3, 15) = −16 < 0, P (3, 15) = 1582 is the maximum profit 43. Since the volume of the box is 60, the height is 60/xy. Then C(x, y) = 10xy + 20xy + 2[2x60/xy + 2y60/xy] = 30xy + 240/y + 240/x. Cx = 30y − 240/x2 ; Cxx = 480/x3 , Cxy = 30; Cy = 30x − 240/y 2 ; cyy = 480/y 3 ; D = 4802 /x3 y 3 −900. Setting Cx = 0 and Cy = 0 we obtain 30y −240/x2 = 0 and 30x−240/y 2 = 0 or y = 8/x2 and x = 8/y 2 . Substituting the first equation into the second, we have x = x4 /8 or x(x3 − 8) = 0. Thus, (2, 2) is a critical point. Since D(2, 2) = 2700 > 0 and Cxx (2, 2) = 60 > 0, C(2, 2) is a minimum. Thus, the cost is minimized when the base of the box is 2 feet square and the height is 15 feet.

13.9 1.

4 #

Method of Least Squares xi = 14,

i=1

4 #

yi = 8,

i=1

4 #

xi yi = 30,

i=1

4 #

x2i = 54, m =

4(30) − 14(8) = 0.4, 4(54) − (14)2

x2i = 14, m =

4(34) − 6(14) = 2.6, 4(14) − (6)2

i=1

54(8) − 30(14) b= = 0.6, y = 0.4x + 0.6 4(54) − (14)2 2.

4 # i=1

xi = 6,

4 # i=1

yi = 14,

4 # i=1

xi yi = 34,

4 # i=1

14(14) − 34(6) b= = −0.4, y = 2.6x − 0.4 4(14) − (6)2

13.10. LAGRANGE MULTIPLIERS

865

10. The least-squares line is given by y = 2.0533x − 3837.115. Plugging 2020 in for x, we predict that the population will be 310.551 million.

PROBLEMAS 4.9

13.10

Lagrange Multipliers

1. f has constrained extrema where the level lines intersect the circle.

2. f has constrained extrema where the level curve intersect the line.

y

y

2

1 x

1

x

3. fx = 1; fy = 3; gx = 2x; gy = 2y. We need to solve 1 = 2λx, 3 = 2λy, x2 + y 2 − 1 = 0. Dividing the second equation by the first, we obtain √ 3 = y/x or y =√3x. Substituting √ into 2 2 + 9x = 1 or x = ±1/ 10. For x = 1/ 10, y = 3/ the third equation, we have x √ √ √ √ 10 and √ A constrained maximum is f (1/ 10, 3/ 10) + 10 and a for x = −1/ 10, y = −3/ 10. √ √ √ constrained minimum is f (−1/ 10, −3/ 10) = − 10. 4. fx = y; fy = x; gx = 1/2; gy = 1. We need to solve y = λ/2, x = λ, x/2 + y − 1 = 0. From the first two equations y = x/2. Substituting into the second equation, we have x = 1. Thus, f (1, 1/2) = 1/2 is a constrained extremum. Since (0, 1) satisfies the constraint and f (0, 1) = 0 < 1/2, f (1, 1/2) = 1/2 is a constrained maximum. 5. fx = y; fy = x; gx = 2x; gy = 2y. We need to solve y = 2λx, x = 2λy, x2 + y 2 − 2 = 0. Substituting the second equation into the first, we obtain y = 4λ2 y or y(4λ2 − 1) = 0. If y = 0, then from the second equation x = 0. Since g(0, 0) = −2 = 0, (0, 0) does not satisfy the constraint. Thus, λ = ±1/2 and y = ±x. Substituting into the third equation, we have 2x2 = 2 or x = ±1. Solutions of the system are x = 1, y = 1, λ = 1/2, x = −1, y = −1; λ = 1/2, x = 1, y = −1, λ = −1/2, and x = −1, y = 1, λ = −1/2. Thus, f (1, 1) = f (−1, −1) = 1 are constrained maxima and f (1, −1) = f (−1, 1) = −1 are constrained minima. 6. fx = 2x; fy = 2y; gx = 2; gy = 1. We need to solve 2x = 2λ, 2y = λ, 2x + y − 5 = 0. Substituting the second equation into the first, we find x = 2y. Substituting into the third equation, we have 4y + y − 5 = 0 or y = 1. A constrained extremum is f (2, 1) = 5. Since (0, 5) satisfies the constraint and f (0, 5) = 25 > 5, f (2, 1) = 5 is a constrained minimum. 7. fx = 6x; fy = 6y; gx = 1; gy = −1. We need to solve 6x = λ, 6y = −λ; x − y − 1 = 0. From the first two equations, we obtain x + y = 0. Solving this with the third equation, we

866

CHAPTER 13. PARTIAL DERIVATIVES obtain x = 1/2, y = −1/2. Thus, f (1/2, −1/2) = 13/2 is a constrained extremum. Since (1, 0) satisfies the constraint and f (1, 0) = 8 > 13/2, f (1/2 − 1/2) = 13/2 is a constrained minimum.

8. fx = 8x; fy = 4y; gx = 8x; gy = 2y. We need to solve 8x = 8λx, 4y = 2λy, 4x2 + y 2 − 4 = 0 or x(λ − 1) = 0, y(λ − 2) = 0, 4x2 + y 2 = 4. If x = 0, then from the third equation y = ±2. If y = 0, then from the third equation x = ±1. The cases λ = 1 and λ = 2 lead also to y = 0 and x = 0, respectively. Thus, f (0, 2) = f (0, −2) = 18 are constrained maxima and f (1, 0) = f (−1, 0) = 14 are constrained minima. 9. fx = 2x; fy = 2y; gx = 4x3 ; gy = 4y 3 . We need to solve 2x = 4λx3 , 2y = 4λy 3 , x4 + y 4 − 1 = 0 or x(2λx2 − 1) = 0, y(2λy 2 − 1) = 0, x4 + y 4 = 1. If x = 0, the from the 2 = 1 = 2λy 2 we have x2 = y 2 . third equation y = ±1. If y = 0, then x = ±1. From 2λx √ √ 4 4 Substituting into the third equation, we obtain x = ±1/ 2 and y = ±1/ 2. Solutions of √ 4 2, ±1/[4]2). Thus, f (0, ±1) = f (±1, 0) = 1 are the system are (0, ±1), (±1, 0), and (±1/ √ √ √ constrained minima and f (±1/ 4 2, ±1/ 4 2) = 2 are constrained maxima. 10. fx = 16x−8y; fy = 4y −8x; gx = 2x; gy = 2y. We need to solve 16x−8y = 2λx, 4y −8x = 2λy, x2 + y 2 − 10 = 0 or 8 − 47/x = λ, x2 + y 2 = 10. From the first two equations, we obtain 6 − 4y/x = −4x/y, 6(y/x) − 4(y/x)2 = −4, and 2(y/x)2 − 3(y/x) − 2 = 0. Factoring, we have (2y/x + 1)(y/x − 2) = 0. Then y = −x/2 and y = 2xs. √ Substituting y = −x/2 into the third equation, we have x2 + x2 /4 = 10 and x =√±2 2. Substituting 2 2 of the y = 2x into the we have √ equation, √ √ √ x √+ 4x = 10√ and x√ = ± 2. Solutions √ √ √ third (− 2, system √ (2 2, − 2), (−2 2, 2), ( 2, 2 2), and √ √−2 2). Thus, √ f (2√ 2, − 2) = √ are f (−2 2, 2) = 100 are constrained maxima and f ( 2, 2 2) = f (− 2, −2 2) = 0 are constrained minima. √ √ √ 11. fx = 3x2 y; fy = x3 ; gx = 1/2 x; gy = 1/2 y. We need to solve 3x2 y = λ/2 x, x3 = √ √ √ √ √ x + y − 1 = 0 or 6x5/2 y = λ, 2x3 y 1/2 = λ, x + y = 1. From the first two λ/2 y, √ √ equations, we obtain 3x5/2 y = x3 y 1/2 and 3 y = x. Substituting into the third equation, √ √ √ we have 3 y + y = 4 y = 1. Then, y = 1/16 and x = 9/16. Since (1/4, 1/4) satisfies the constraint and f (1/4, 1/4) = 1/256, f (9/16, 1/16) + 729/65, 536 is a constrained maximum. We also consider x = 0, which requires y = 1; and y = 0, which requires x = √ 1. Since x ≥ 0 √ and y ≥ 0, f (0, 1) = f (1, 0) = 0 ≤ x3 y = f (x, y) for all (x, y) which satisfy x + y = 1. Thus, f (0, 1) = 0 and f (1, 0) = 0 are constrained minima. 12. fx = y 2 ; fy = 2xy; gx = 2x; gy = 2y. We need to solve y 2 = 2λx, 2xy = 2λy, x2 + 2 2 y 2 − 27 = 0 or y 2 = 2λx, y(x √ − λ) = 0, x + y = 27. When y = 0 in the third equation, 2 we obtain x = 27 or x = ±3 3, and λ = 0. When x = λ in the second equation, we obtain 2 2 2 y 2 = 2x2 from the first √ x + 2x = 3x = 27 from the√third equation.√This gives √ equation and x = ±3 and y = ±3 2. Since f (±3√ 3, 0) = 0, we√see that f (−3, 3 2) = f (−3, −3 2) = −54 are constrained minima and f (3, 3 2) = f (3, −3 2) = 54 are constrained maxima. 13. Fx = 1; Fy = 2; Fz = 1; gx = 2x; gy = 2y; gz = 2z. We need to solve 1 = 2λx, 2 = 2λy; 1 = 2λz, x2 + y 2 + z 2 − 30 = 0. From the first and second equations, we obtain y = 2x. From the first and third equations, we obtain z √ = x. Substituting into √ the fourth √ 2 2 2 2 + 4x + x = 6x = 30. Thus, x = ± 5, y = ±2 5, z = ± 5.√Then, equation, we have x √ √ √ √ √ √ √ F ( 5, 2 5, 5) = 6 5 is a constrained maximum and F (− 5, −2 5, − 5) = −6 5 is a constrained minimum.

13.10. LAGRANGE MULTIPLIERS

867

14. Fx = 2x; Fy = 2y; Fx = 2z; gx = 1; fy = 2; gz = 3. We need to solve 2x = λ, 2y = 2λ, 2z = 3λ, x + 2y + 3z − 4 = 0. From the first and second equations, y = 2x. From the first and third equations, z = 3x. Substituting into the fourth equation, we have x+4x+9x = 14x = 4. Thus, x = 2/7, y = 4/7, andz = 6/7. Then, F (2/7, 4/7, 6/7) = 56/49 is a constrained extremum. Since (4, 0, 0) satisfies the constraint and F (4, 0, 0) = 16 > 56/49, F (2/7, 4/7, 6/7) = 56/49 is a constrained minimum. 15. Fx = yz; Fy = xz; Fz = xy; gx = 2x; gy = y/2; gz = 2z/9. We need to solve yz = 2λx, xz = λy/2, xy = 2λz/9 or xyz/2 = λx2 , xyz/2 = λy 2 /4, xyz/2 = λz 2 /9 along with x2 + y 2 /4 + z 2 /9 − 1 = 0 or x2 + Y 2 /4 + z 2 /9 = 1 for x > 0, y > 0, z > 0. From the first three equations and the fact that λ = 0, x2 = y 2 /4 = z 2 /9. Substituting into the 2 2 2 2 2 2 2 third equation, we √ obtain √ x +√x + x = 3x = 1, so x = 1/3, y =√4/3, and z = 3. √ Since 23/6, 1, 1) Thus, F ( 3/3, 2 3/3,√ 3) = 2 3/3 is√a constrained √ extremum. √ √ ( √ √ satisfies the constraint and F ( 23/6, 1, 1) = 23/6 < 2 3/3, F ( 3/3, 2 3/3, 3) = 2 3/3 is a constrained maximum. 16. Fx = yz; Fy = xz; Fz = xy; gx = 3x2 ; gy = 3y 2 ; gz = 3z 2 . We need to solve = 3λz 3 along yz + 3λx2 , xz = 3λy 2 , xy = 3λz 2 or xyz = 3λx3 , syz = 3λy 3 , xyz √ √ with 3 3 3 3 3 3 3 x + y + z√ − 24 = 0 or x + y + z = 24. Taking λ = 0 we see that ( 24, 0, 0), (0, 3 24, 0), and (0, 0, 3 24) satisfy the system. If λ = 0, the the first three equations imply x3 = y 3 = z 3 . Substituting into the fourth√ equation,√we obtain x3 +x3 +x3 = 3x√3 = 24 or x = 2. Then √ √ (2, 2, 2) 3 3 3 satisfies the system. Since 24 = 2 3, F (2 3, 0, 0) = F (0, 2 3 3, 0) = F (0, 0, 2 3 3) = 5 is a constrained minimum and F (2, 2, 2) = 13 is a constrained maximum. 17. Fx = 3x2 ; Fy = 3y 2 ; Fz = 3z 2 ; gx = 1; gy = 1; gz = 1. We need to solve 3x2 = λ, 3y 2 = λ, 3z 2 = λ, x + y + z − 1 = 0 for x > 0, y > 0, z > 0, and hence λ > 0. From the first three equations x2 = y 2 = z 2 , and since x, y, and z are positive, x = y = z. Then, from the fourth equation, x = y = z = 1/3 and F (1/3, 1/3, 1/3) = 1/9 is a constrained extremum. Since (1/2, 1/4, 1/4) satisfies the constraint and F (1/2, 1/4, 1/4) = 5/32 > 1/9, F (1/3, 1/3, 1/3) = 1/9 is a constrained minimum. 18. Fx = 8xy 2 z 2 ; Fy = 8x2 yz 2 ; Fz = 8x2 Y 2 z; gx = 2x; gy = 2y; gz = 2z. We need to solve 8xy 2 z 2 = 2λx, 8x2 yz 2 = 2λy, 8x2 y 2 z = 2λz or 4x2 y 2 z 2 = λx2 , 4x2 y 2 z 2 = λy 2 , 4x2 y 2 z 2 = λz 2 along with x2 + y 2 + z 2 − 9 = 0 or x2 + y 2 + z 2 = 9 for x > 0, y > 0, z > 0, and hence lambda > 0. From the first three equations, we see x2 = y 2 = z 2 . Substituting into 2 2 2 2 the third equation, we obtain √ √ x √+ x + x = 3x = 9. Thus, since x, y, and z are positive, √ a constrained extremum. Since (1, 2, 2) satisfies x = y = z = 3 and F ( 3, 3, 3) = 108 is √ √ √ the constraint and F (1, 2, 2) = 64 < 108, F ( 3, 3, 3) = 108 is a constrained maximum. 19. Fx = 2x; Fy = 2y; Fz = 2z; gx = 2; gy = 1; gz = 1; hx = −1; hy = 2; hz = −3. We need to solve 2x = 2λ−μ, 2y = λ+2μ, 2z = λ−3μ subject to 2x+y+z = 1, −x+2y−3z = 4. Solving the first three equations for x, y, and z, respectively, and substituting into the constraint equations, we obtain 2λ − μ + λ/2− 3μ/2 = 1, −λ + μ/2 + λ + 2μ − 3λ/2 + 9μ/2 = 4 or 6λ − 3μ = 2, −3λ + 14μ = 8. From this, we obtain λ = 52/75 and μ = 54/75. Then x = 1/3, y = 16/15, and z = −11/15. Thus, F (1/3, 16/15, −11/15) = 134/75 is a constrained minimum. 20. Fx = 2x; Fy = 2y; Fz = 2z; gx = 4; gy = 0; gz = 1; hx = 2x; hy = 2y; hz = −2z. We need to solve 2x = 4λ + 2xμ, 2y = 2yμ, 2z = λ − 2zμ subject to 4x + z = 7, z 2 = x2 + y 2 .

868

CHAPTER 13. PARTIAL DERIVATIVES Consider the second equation. If y = 0, then the constraint equations become 4x + z = 7 and z 2 = x2 . The solutions of these equations are x = z = 7/5 and x = −z = 7/3. In either case, the first and third equations can be solved for λ and μ. Thus, (7/5, 0, 7/5) and (7/3, 0, −7/3) are candidates for constrained extrema. Now, if y = 0, then from the second equations μ = 0. In this case, 2x = 4λ and 2z = λ or x = 4z. Then the first constraint equation becomes 16z + z = 17z = 7, so z = 7/17. Then, x = 28/17 and y 2 = z 2 − x2 < 0. Hence, the system has no solution when y = 0. Thus, F (7/5, 0, 7/5) = 98/25 is a constrained minimum and F (7/3, 0, −7/3) = 98/9 is a constrained maximum.

21. We want  to maximize A(x, yx y/2 subject to P (x, y) = Mx2+y2 x + y + x2+ y 2 − 4 = 0. Ax = y/2; Ay = x/2; y Px = 1 + x/ x2 + y 2 ; Py = 1 + y/ x2 + y 2 .We need to solve  y/2 = λ + λx/ x2 + y 2 , x/2 = λ + λy/ x2 + y 2 , x x + y + x2 + y 2 − 4 = 0 for x > 0, y > 0, and hence  λ > 0. Subtracting the second equation from the first, we have (y − x)/2 = λ(x − y)/ x2 + y 2 or   2 2 2 2 (y − x) = (y − x)(−2λ/ x + y ). Since −2λ/ x + y is negative, 1, and hence √ it cannot equal √ 2 = (2 + y − x = 0 or y √ = x. Substituting in the third equation gives 2x + 2x 2)x √ √ √ = 4. Thus, x = y = 4/(2 + 2) and this maximum area is A(4/(2 + 2), 4/(2 + 2)) = 4/(3 + 2 2). 22. Let the base of the box have dimensions x and y and let the height be z. We want to maximize V (x, y, z) = xyz subject to S(x, y, z) = xy + 2yz + 2xz − 75 = 0. Now Vx = yz; Vy = xz; Vz = xy; Sx = y + 2z; Sy = x + 2z; Sz = 2y + 2x. We need to solve yz = λ(y + 2z), xz = λ(x + 2z), xy = λ(2y + 2x), xy + 2yz + 2xz − 75 = 0 or xyz = λ(xy + 2xz), xyz = λ(xy + 2yz), xyz = λ(2yz + 2xz), xy + 2yz + 2xz = 75, for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have xy + 2xz = xy + 2yz, which gives xz = yz or x = y; and xy + 2yz = 2yz + 2xz, which gives xy = 2xz or y = 2z. Substituting x = y = 2z into the fourth equation, we obtain 4z 2 + 4z 2 + 4z 2 = 12z 2 = 75. Thus, z = 5/2cm and x = y = 5cm. When the box is closed, S(x, y, z) = 2(xy + yz + xz) − 75, Sx = 2(y + z), Sy = 2(x + z), Sz = 2(x + y), and we need to solve xyz = 2λ(xy + xz), xyz = 2λ(xy + yz), xyz = 2λ(yz + xz), 2(xy + yz + xz) = 75 for x > 0, y > 0, z > 0, and thus λ > 0. From the first three equations, we have xy+xz = xy+yz, which gives xz = yz or x = y; and xy+yz = yz +xz, which gives xy = xz or 2 2 2 2 y = z. Substituting x √ = y = z into the fourth equation, we obtain √ 2(x + x + x ) = 6x = 75. Thus, x = y = z = 5/ 2. The box is a cube with each side 5/ 2cm.  23. We want to maximize V (x, y) = 9πx + 3πy subject to S(x, y) = 9π + 6πx + 3π 9 + y 2 − 81π.  2 Now V 6π; Sy = 3πy/ 9 + y . We need to solve 9π = 6πλ, 3π = x = 9π; Vy = 3π; Sx =  3πλy/ 9 + y 2 , 9π + 6πx + 3 9 + y 2 − 81π = 0 for x > 0 and y > √ 0. From the first 5. From the third equation, λ = 3/2. Usingλ = 3/2 in the second equation gives y = 6/ √ equation, we √ have 6x + 3 9 +√36/5 = 72 or x = 12 − 9/2 5. The volume is maximum when x = 12 − 9/2 5m and y = 6/ 5m. 1 2 1 24. Ux = x−2/3 y 2/3 ; Uy = x1/3 y −1/3 ; gx = 1; gy = 6. We need to solve x−2/3 y 2/3 = 3 3 3 1 2 1/3 2/3 1/3 x y−1/3 = 6λ, x + 6y − 18 = 0 or y = 3λx y , x = 3λx2/3 y 1/3 , x + λ, 3 3 6y = 18. From the first two equations, y = x/3. Substituting into the third equation, we

13.10. LAGRANGE MULTIPLIERS

869

have x + 2x = 3x = 18. Thus, x = 6 and y = 2. Since (12, 1) satisfies the constraint and U (12, 1) = 121/3 , U (6, 2) = 61/3 22/3 = 241/3 is a constrained maximum. 25. We want to maximize z(x, y) = P − x − y subject to z 2 /xy 3 = k or (P − x − y)2 − kxy 3 = 0. Now zx = −1; zy = −1; gx = −2(P − x − y) − ky 3 ; gy = −2(P − x − y) − 3kxy 2 . We need to solve −1 = −2λ(P − x − y) − λky 3 , −1 = −2λ(P − x − y) − 3λkxy 2 , (P − x − y)2 = kxy 3 for x > 0, y > 0, and z > 0. From the first two equations, √ we have y = 3x. Substituting into the third equation, we obtain (P − 4x)2 = 27kx4 or 27kx2 = P − 4x. (Since z > 0, z = P − x − y = P − 4x > 0.) Using the quadratic formula and the fact that x > 0, we find  √ √ 16 + 4P 27k −2 + 4 + P 27k √ . = x= 27k 2 27k  √ √ Then the maximum value of z is P − 4x = P + 4(2 − 4 + P 27k/ 27k. −4 +



26. (a) See part (b). (b) Maximizing 1/(x21 · · · x2n ) is equivalent to minimizing the denominator F (x1 , . . . xn ) = x21 + · · · + x2n . The constraint is still x1 + · · · xn = 1, which we can write as g(x1 , . . . xn ) = x1 + · · · + xn − 1 = 0. Since ∂F/∂xi = 2xi and ∂g/∂xi = 1, we can get the equations 2x1 = λ,, 2x2 = λ, , . . . , 2xn = λ, x1 + · · · xn = 1. The solution is x1 = . . . = 1/2 ( and λ = 1/2n). 4 + y 4 =√1 to the origin. 27. f (x, y) is the square of the distance from a point on the graph of x√ 4 The points (0, ±1) and (±1, 0) are closest to the origin, while (±1/ 2, ±1/ 4 2) are farthest from the origin.

28. F (x, y, z) is the square of the distance from a point on the plane x + 2y + 3z = 4 to the origin. The point (2/7, 4/7, 6/7) is closest to the origin. 29. F is the square of the distance of points on the intersection of the planes 2x + y + z = 1 and −x + 2y − 3z = 4 from the origin. The point (1/3, 16/15, −11/15) is closest to the origin. 30. F is the square of the distance of points on the intersection of the plane 4x + z = 7 and the circular cone z 2 = x2 = y 2 . The point (7/5, 0, 7/5) is closest to the origin and the point (7/3, 0, −7/3) is farthest from the origin. 31. We want to minimize f (x, ) = x2 + y 2 subject to xy 2 = 1. Now fx = 2x; fy = 2y; gx = y 2 ; gy = 2xy. We need to solve 2x = λy 2 , 2y = 2λxy, xy 2 − 1 = 0 or 2xy = λy 3 , 2xy = 2λx2 y, xy 2 = 1 for x > 0, y > 0, and hence λ > 0. From the first two equations, we have y 3 = 2x2 y or y 2 = 2x2 . Substituting into the third equation gives 2x3 = 1 or x = 2−1/3 . Again, from the third equation we have y = 1/(2−1/3 )1/2 = 21/6 . Thus, the point closest to the origin is (2−1/3 , 21/6 ). Since the surface is F (x, y, z) = xy 2 − 1 = 0, ∇F = y 2 i + 2xyj is normal to the surface at (x, y, z). Thus, a normal to the surface at (2−1/3 , 21/6 , 0) is ∇F (2−1/3 , 21/6 , 0) is ∇F (2−1/3 , 21/6, 0) = 21/3 i + 2(2−1/3 )(21/6 )j = 21/3 i + 25/6 j = 22/3 (2−1/3 i + 21/6 j). Since ∇F (2−1/3 , 21/6 , 0) is a multiple of the vector from the origin to P (2−1/3 , 21/6 , 0), this vector is perpendicular to the surface.

CHAPTER 13. PARTIAL DERIVATIVES

870

1 −2/3 1/3 1/3 1 1 x y z ; Fy = x1/3 y −2/3 z 1/3 ; Fz = x1/3 y 1/3 z −2/3 ; gx = 1; gy = 3 3 3 1 −2/3 1/3 1/3 1 1/3 −2/3 1/3 1 1/3 1/3 −2/3 x y x y z y z = λ, z = λ, = 1; gz = 1. We need to solve x 3 3 3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 1/3 = 3λy, x y z = λ, x + y + z − k = 0 or x y Z1/3 = 3λx, x y z λz, x + y + z = k. From the first three equations, x = y = z. Substituting into the fourth equation, x + x + x = 3x = k. Thus, x = y = z = k/3 and F (k/3, k/3, k/3) = k/3 is a constrained maximum.

32. Fx =

33. For any x + y + z = k, by Problem 32,

√ 3

xyz ≤

x+y+z k = . 3 3

34. Distance from the xz-plane is measured by |y|. Alternatively, we will find the extreme values of F (x, y, z) = y 2 subject to g(x, y, z) = x2 + z 2 − 1 = 0 and h(x, y, z) = x + y + 2z − 4 = 0. Now Fx = 0; Fy = 2y; Fz = 0; gx = 2x; gy = 0; gz = 2z; hx = 1; hy = 1; hz = 2. We need to solve 0 = 2λx + μ, 2y = μ, 0 = 2λz + 2μ, x2 + z 2 = 1, x + y + 2z = 4. By inspection we see that if λ = 0, then μ = 0 and y = 0. Similarly, if μ = 0, then λ = 0 and y = 0. Substituting y = 0 into the fourth and fifth equations, we obtain the system x2 + z 2 = 1, x + 2z = 4, which is inconsistent. Thus, μ = 0 and λ = 0. Now, solving the first three equations for x, y, and z and substituting into the fourth and fifth equations, we obtain the system

μ 2μ μ2 μ2 μ + − = 4 or 5μ2 = 4λ2 , (λ − 5)μ = 8λ. + = 1, − 4λ2 λ2 2λ 2 λ 8λ 2 ) = 4λ2 Solving the second equation for μ and substituting into the first, we obtain 5( λ −5 √ or 80 = (λ − 5)2 . Thus, √ λ = 5 ± 4 5. From μ = 8λ/(λ − 5) we find that corresponding 2 values of μ are 8 ± 2 5. Since √ function F (x, y, z) = y is √ 2y = μ we see that the objective when μ = 8 + 2 5. Corresponding values of minimized when μ = 8 − 2 5 and maximized √ √ 8±2 5 μ 1 √ = ∓ √ ≈ |mp0.45, y − μ/2 = 4 ± 5 ≈ 6.24 and =− x, y, and z are x = − 2λ 10 ± 8 5 5 √ √ √ √ 1.76, z = −μ/λ = 2x = ∓2/ 5 ≈ ∓0.89. The closest point is (−1/ 5, 4 − 5, −2 5) or about (−4.5, 6.24, −0.89).

Chapter 13 in Review A. True/False 1. False; see Example 3 in Section 13.2 in the text. 2. False; (0,4.1) is in the domain of g but not in the domain of f .

36. From Problem 35,m = kπ and ds = dt.  π  π  1 Mx = yρds = kxyds = k (1 + cos t) sin tdtk − cos t + sin2 t  = 2k 2 C C 0 0  π π 2 2 My = intC xρds = intC kx ds = k (1 + cos t) dt = k (1 + 2 cos t + cos2 t)dt 0 0   π 1 1 3 = k t + 2 sin t + t + sin 2t  = kπ 2 4 2 0 3 2 3kπ/2 2k = ; y = Mx /m = = . The center of mass is (3/2, 2/π). x = My /m = kπ 2 kπ π

PROBLEMAS 15.2 Line 4.10 Integrals of Vector Fields

1.

2.

y

x

y

x

15.2. LINE INTEGRALS OF VECTOR FIELDS

3.

965

4.

y

y

x

5.

x

6.

y

y

x

7. Since each vector points in a northeasterly direction, the vector field must have positive i and j components. Therefore, the answer is (b). 8. Since each vector points in a northwesterly direction, the vector field must have negative i and positive j components. Therefore, the answer is (a).

x

966

CHAPTER 15. VECTOR INTEGRAL CALCULUS

9. Since each vector points in a southwesterly direction, the vector field must have negative i and j components. Therefore, the answer is (d). 10. Since each vector points in a southeasterly direction, the vector field must have positive i and negative j components. Therefore, the answer is (c). 11. Note that the k component of each vector is always positive. Therefore, the answer is (d). 12. Note that the i component of each vector is always positive. Therefore, the answer is (c). 13. Note that each vector points directly away from the origin. Therefore, the answer is (a). 14. Note that the i and j components of each vector are zero. Therefore, the answer is (b). F = e3t i − (e−4t )et j = e3t i − e−3t j; dr = (−2e−2t i + et j)dt; F · dr = (−2et − e−2t )dt; ln 2  ln 2  15.  3 19 1 31 F · dr = (−2et − e−2t )dt = (−2et + e−2t ) = − − (− ) = − 2 8 2 8 0 c 0 15. 16. F = 2(t)(t2 )i + t2 j = 2t3 i + t2 j; dr = (i + 2tj)dt; F · dr = 4t3 dt; 2 2 F · dr = 0 4t2 dt = t4 0 = 16 C

15.2. LINE INTEGRALS OF VECTOR FIELDS 32. ∇f (x, y) = (1 + 2 cos 5xy − 10xy sin 5xy)i + (−1 − 10x2 sin 5xy)j 16.

33. ∇f (x, y, z) = tan−1 zyi + 17.

xy xz j+ 2 2 k +1 y z +1

y2 z2

34. ∇f (x, y, z) = (1 − 2xyz 4 )i − x2 z 4 j − 4x2 yz 3 k 18.   2 2 35. ∇f (x, y, z) = e−y i + 1 + 2xye−y j + k 19.

20. 36. ∇f (x, y, z) =

2x 8y 3 18z 5 i + j + k x2 + 2y 4 + 3z 6 x2 + 2y 4 + 3z 6 x2 + 2y 4 + 3z 6

 37. ∇ x2 + 12 y 2 = 2xi + yj = F(x, y). Therefore, the answer is (b). 21.

38. 22. ∇

1

2x

2

 + y 2 − 4 = xi + 2yj = F(x, y). Therefore, the answer is (c).



23. 39. ∇ 2x + 12 y 2 + 1 = 2i + yj = F(x, y). Therefore, the answer is (d).

40. 24. ∇

1

2x

2

 + 13 y 3 − 5 = xi + y 2 j = F(x, y). Therefore, the answer is (a).

41. 25. φ(x, y) = sin x + y + cos y

42. 26. φ(x, y) = xe−y

27. φ(x, y) = x + y 2 − 4z 3 43.

44. 28. φ(x, y) = xy 2 z 3

969

(b) Use y = 4x for 0 ≤ x ≤ 2.  2  (2,8) (y 3 + 3x2 y)dx + (x3 + 3y 2 x + 1)dy = [(64x3 + 12x3 ) + (x3 + 48x3 + 1)(4)]dx (0,0)



0

2

= 0

2 (272x3 + 4)dx = (68x4 + 4x)0 = 1096

10. 11. Py = 12x3 y 2 = Qx throughout the plane and the vector field is a conservative field. φx = 29. 4x3 y 3 +3, φ = x4 y 3 +3x+g(y), φy = 3x4 y 2 +g  (y) = 3x4 y 2 +1, g(y) = y, φ = x4 y 3 +3x+y 12. Py = 6xy 2 = Qx throughout the plane and the vector field is a conservative field. φx = 30. 2xy 3 , φ = x2 y 3 + g(y), φy = 3x2 y 2 + g  (y) = 3x2 y 2 + 3y 2 , g(y) = y 3 , φ = x2 y 3 + y 3 13. Py = −2xy 3 sin xy 2 + 2y cos xy 2 , Qx = −2xy 3 cos xy 2 − 2y sin xy 2 throughout the plane and 31. the vector is not a conservative field.

CHAPTER 15. VECTOR INTEGRAL CALCULUS

974

14. Py = −4xy(x2 + y 2 + 1)−3 = Qx throughout the plane and the vector field is a conservative 32. 1 field. φx = x(x2 + y 2 + 1)−2 , φ = − (x2 + y 2 + 1)−1 + g(y), φy = y(x2 + y 2 + 1)−2 + g  (y) = 2 1 y(x2 + y 2 + 1)−2 , g(y) = 0, φ = − (x2 + y 2 + 1)−1 2 15. 33. Py = 1 = Qx throughout the plane and the vector field is a conservative field. φx = x3 + 1 1 1 1 y, φ = x4 + xy + g(y), φy = x + g  (y) = x + y 3 , g(y) = − y 4 , φ = x4 + xy + y 4 4 4 4 4

34. 16. Py = 4e2y , Qx = e2y throughout the plane and the vector field is not a conservative field. 35. Py = 0 = Qx , Px = 0 = Rx , Qz = −1 = Ry throughout 3-space and the vector field is a 17. conservative field. ∂g = 3y 2 − x, φx = 2x, φ = x2 + g(y, z)φy = ∂y g(y, z) = y 3 − yz + h(z), φ = x2 + y 3 − yz + h(z), φz = −y + h (z) = −y, h(z) = 0, φ = x2 + y 3 − yz 18. Py = 2x = Qx , Pz = 0 = Rx , Qz = −e−y = Ry throughout 3-space and the vector field is a 36. conservative field. ∂g = x2 − ze−y , φx = 2xy, φ = x2 y + g(y, z), φy = x2 + ∂y g = ze−y + h(z), φ = x2 y + ze−y + h(z), φz = e−y + h (z) = ey − 1, h(z) = −z, φ = x2 y + ze−y − z 19. Since Py = −e−y = Qx , F is conservative and C F · dr is independent of the path. Thus, instead of the given curve we may use the simpler curve C1 : y = x, 0 ≤ x ≤ 1. Then  (2x + e−y )dx + (4y − xe−y )dy W = 

C1 1

= 0

(2x + e

−x



1

)dx + 0

(4x − e−x )dx

1 1 = (x2 − e−x )0 + (2x2 + xe−x )0

Integration by parts

= [(1 − e−1 ) − (−1)] + [(2 + e−1 + e−1 ) − (1)] = 3 + e−1 .

CHAPTER 15. VECTOR INTEGRAL CALCULUS

970

45. 37.

46. 38.

y

y

x

47. 38.

x

48. 38.

y

x

y

x

15.3. INDEPENDENCE OF THE PATH

49. 41.

971

42. 50.

y

y

x

x

51. Let φ(x, y, z) = −c(x2 + y 2 + z 2 )−1/2 . Then 43. cx cy cz i+ 2 j+ 2 k (x2 + y 2 + z 2 )3/2 (x + y 2 + z 2 )3/2 (x + y 2 + z 2 )3/2 c(xi + yj + zk) = 2 (x + y 2 + z 2 )3/2 cr = 3 =F |r|

∇φ(x, y, z) =

52. Yes; if f and g differ by a constant, they will have the same gradient field. 44.

15.3

Independence of the Path 1 3 x + g(y), φy = 3 (2,2)  1 3 1 3 1 3 (2,2) 2 1 3 16  2 2 3  g (y) = y , g(y) = y , φ = x + y , (0,0) x dx + y dy = (x + y ) = 3 3 3 3 3

1. (a) Py = 0 = Qx and the integral is independent of path. φx = x2 , φ =

(0,0)

2 2 2 (2,2) 2 2 3  16 2 2 (b) Use y = x for 0 ≤ x ≤ 2. (0,0) x dx + y dy = 0 (x + x )dx = x  = 3 0 3

2. (a) Py = 2x = Qx and the integral is independent of path. φx = 2xy, φ = x2 y +g(y), φy = (2,4) (2,4) x2 + g  (y) = x2 , g(y) = 0, φ = x2 y, (1,1) 2xydx + x2 dy = x2 y (1,1) = 16 − 1 = 15 (b) Use y = 3x − 2 for 1 ≤ x ≤ 2. 2 (2,4) 2 2 2xydx + x2 dy = 1 [2x(3x − 2) + x2 (3)]dx = 1 (9x2 − 4x)dx = (3x3 − 2x2 )1 = 15 (1,1)

CHAPTER 15. VECTOR INTEGRAL CALCULUS

1002

−P fx − Qfy + R −fx i − fy j + k  ; F·n=  ; 1 + fx2 + fy2 1 + fx2 + fy2  dS = 1 + fx2 + fy2 dA −P fx − Qfy + R   F · ndS = 1 + fx2 + fy2 dA = (−P fx − Qfy + R)dA s R R 1 + fx2 + fy2

PROBLEMAS 4.11

15.7

Curl and Divergence

1. curlF = (x − y)i + (x − y)j; divF = 2z 2. curlF = −2x2 i + (10y − 18x2 )j + (4xz − 10z)k; divF = 0 3. curlF = 0; divF = 4y + 8z 4. curlF = (xe2y + ye−yz + 2xye2y )i − ye2y j + 3(x − y)2 k; divF = 3(x − y)2 − ze−yz 5. curlF = (4y 3 − 6xz 2 )i + (2x3 − 3x2 )k; divF = 6xy 3 6. curlF = −x3 zi + (3x2 yz − z)j + ( x2 y 2 − y − 15y 2 )k; divF = (x3 y − x) − (x3 y − x) = 0 2 7. curlF = (3e−z − 8yz)i − xe−z j; divF = e−z + 4z 2 − 3ye−z 8. curlF = (2xyz 3 + 3y)i + (y ln x − y 2 z 3 )j + (2 − z ln x)k; divF =

yz − 3z + 3xy 2 z 2 x

9. curlF = (xy 2 ey + 2xyey + x3 yez + x3 yzez )i − y 2 ey j + (−3x2 yzez − xex )k; divF = xyex + yex − x3 zez 10. curlF = (5xye5xy + e5xy + 3xz 3 sin xz 3 − cos xz 3 )i + (x2 y cos yz − 5y 2 e5xy )j + (−z 4 sin xz 3 − x2 z cos yz)k; divF = 2x sin yz 11. div r = 1 + 1 + 1 = 3    i j k   12. curl r =  ∂/∂x ∂/∂y ∂/∂z  = 0i − 0j + 0k = 0  x y z     i      j k    ∂ ∂ ∂ ∂ ∂ ∂   a a a 13. a×∇ =  1 2 3  = a2 ∂z − a3 ∂y i+ a3 ∂x − a1 ∂z j+ a1 ∂y − a2 ∂x k  ∂/∂x ∂/∂y ∂/∂z      i j k  ∂ ∂ ∂ ∂ ∂ ∂   − a3 a3 − a1 a1 − a2 (a × ∇) × r =  a2   ∂z ∂y ∂x ∂z ∂y ∂x    x y z = (−a1 − a1 )i − (a2 + a2 )j + (−a3 − a3 )k = −2a   ∂ ∂ ∂ + a2 + a3 14. ∇ × (a × r) = (∇ · r)a − (∇ · a)r = (1 + 1 + 1)a − a1 r ∂x ∂y ∂z = 3i − (a1 i + a2 j + a3 k) = 2a

15.7. CURL AND DIVERGENCE   ∂/∂x  15. ∇ · (a × r) =  a1  x   i  16. ∇ × r =  ∂/∂x  x

∂/∂y a2 y

j ∂/∂y y

1003 ∂/∂z a3 z

k ∂/∂z z

    = ∂ (a2 z − a3 y) − ∂ (a1 z − a3 x) + ∂ (a1 y − a2 x) = 0  ∂x ∂y ∂z 

    = 0; a × (∇ × r) = a × 0 = 0  

  i  17. r × a =  x  a1

 j k  y z  = (a3 y − a2 z)i − (a3 x − a1 z)j + (a2 x − a1 y)k; r · r = x2 + y 2 + z 2 a2 a3     i j k   ∂/∂y ∂/∂z  ∇ × [(r · r)a] =  ∂/∂x  (r · r)a1 (r · r)a2 (r · r)a3  = (2ya3 − 2za2 )i − (2xa3 − 2za1 )j + (2xa3 − 2ya1 )k = 2(r × a)

18. r · a = a1 x + a2 y + a3 z; r · r = x2 + y 2 + z 2 ; ∇ · [(r × r)a] = 2xa1 + 2ya2 + 2za3 = 2(r · a) 19. Let F = P (x, y, z)i + Q(x, y, z)j + R(x, y, z)k and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k. ∇ · (F + G) = ∇ · [(P + S)i + (Q + T )j + (R + U )k] = Px + Sx + Qy + Ty + Rx + Uz = (Px + Qy + Rz ) + (Sx + Ty + Uz ) = ∇ · F + ∇ · G 20. Let F = P (x, y, z)i  and G = S(x, y, z)i + T (x, y, z)j + U (x, y, z)k.  + Q(x, y, z)j + R(x, y, z)k   j j k   ∂/∂z  ∇ × (F + G) =  ∂/∂x ∂/∂y  P +S Q+T R+U  = (Ry + Uy − Qz − Tz )i − (Rx + Ux − Pz − Sz )j + (Qx + Tx − Py − Sy )k = (Ry − Qz )i − (Rx − Rz )j + (Qx − Py )k + (Uy − Tz )i − (Ux − Sz )j + (Tx − Sy )k =∇×F+∇×G 21. ∇ · (f F) = ∇ · (f P i + f Qj + f Rk) = f Px + P fx + f Qy + Qfy + f Rz + Rfz = f (Px + Qy + Rz ) + (P fx + Qfy + Rfz ) = f (∇ · F) + F · (∇f )   j  22. ∇ × (f F) =  ∂/∂x  fP

j ∂/∂y fQ

k ∂/∂z fR

     

= (f Ry + Rfy − f Qz − Qfz )i − (f Rx + Rfx − f Pz − P fz )j (f Qx + Qfx − f Py − P fy )k = (f Ry − f Qz )i − (f Rx − f Pz )j + (f Qx − f Py )k + (Rfy − Qfz )i − (Rfx − P fz )j + (Qfx − P fy )k

  i  f [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k] +  fx  P

j fy Q

k fz R

    = f (∇ × F) + (∇f ) × F  

CHAPTER 15. VECTOR INTEGRAL CALCULUS

1004

23. Assuming continuous second partial derivatives,   j  curl(gradf ) = ∇ × (fx i + fy j + fz k) =  ∂/∂x  fx

j ∂/∂y fy

k ∂/∂z fz

     

= (fzy − fyz )i − (fzx − fxz )j + (fyx − fxy )k = 0. 24. Assuming continuous second partial derivatives, div(curlF) = ∇ · [(Ry − Qz )i − (Rx − Pz )j + (Qx − Py )k] = (Ryx − Qzx − (Rxy − Pzy ) + (Qxz − Pyz ) = 0. 25. Let F = P (x, y, z)i + Q(x,  y, z)j + R(x, y, z)k and G = S(x, y, z)i + Y (x, y, z)j + U (x, y, z)k.  i j k    F × G =  P Q R  = (QU − RT )i − (P U − RS)j + (P T − QS)k  S T U  div (F × G) = (QUx + Qx U − RTx − Rx T ) − (P Uy + Py U − RSy − Ry S) + (P Tz + Pz T − QSz − Qz S) = S(Ry − Qz ) + T (Pz − Rx ) + U (Qx − Py ) − P (U − y − Tz ) − Q(Sz − Ux ) − R(Rx − Sy ) = G · (curlF) − F · (curlG) 26. Using Problems 20 and 23, curl(curlF + gradf ) = ∇ × (curlF + gradf ) = ∇ × (curlF) + ∇ × (gradf ) = curl(curlF) + curl(gradf ) = curl(curlF) + 0 = curl(curlF). 27. curl F = −8yzi − 2zj − xk; curl (curl F) = 2i − (8y − 1)j + 8zk 28. For F = P i + Qj + Rk, curl (curl F) = (Qxy − Pyy − Pzz + Rxz )i + (Ryz − Qzz − Qxx + Pyx )j + (Pzx − Rxx − Ryy + Qzy )k and −∇2 F + grad(divF) = −(Pxx + Pyy + Pzz )i − (Qxx + Qyy + Qzz )j − (Rxx + Ryy + Rzz )k + grad(Px + Qy + Rz ) = −Pxx i − Qyy j − Rzz k + (−Pyy − Pzz )i + (−Qxx − Qzz )j + (−Rxx − Ryy )k + (Pxx + Qyx + Rzx )i + (Pxy + Qyy + Rzy )i + (Pxz + Qyz + Rzz )k = (−P − P + Q + R)i + (−Qxx − Qzz + Pxy + Rzy )j + (−Rxx − Ryy + Pxz + Qyz )k. Thus, curl(curlF) = −∇2 F + grad(divF).

15.7. CURL AND DIVERGENCE

1005

29. fz = 6x + 4y − 9z; fxx = 6; fy = 10y + 4z; fyy = 10; fz = −9x − 16z; fzz = −16; ∇2 f = fxx + fyy + fzz = 6 + 10 − 16 = 0 30. Using Problem 21, ∇ · (f ∇f ) = f (∇ · ∇f ) + ∇f · ∇f = f (∇2 f ) + |∇f |2 . 31. fx = 6x + 4y − 9z; fxx = 6; fy = 10y + 4x; fyy = 10; fz = −9x − 16z; fzz = −16; ∇2 f + fx x + fy y + zz z = 6 + 10 − 16 = 0 ∂f (a − x)A = 3/2 2 ∂x [(x − a) + (y − b)2 + (z − c)2 ]  2 2 2 A 2(x − a) − (y − b) − (z − c)2 ∂ f = 5/2 ∂x2 [(x − a)2 + (y − b)2 + (z − c)2 ] ∂f (b − y)A = 3/2 2 ∂y [(x − a) + (y − b)2 + (z − c)2 ]  2 2 2 A 2(y − b) − (x − a) − (z − c)2 ∂ f = 5/2 ∂y 2 [(x − a)2 + (y − b)2 + (z − c)2 ] (c − z)A ∂f = 3/2 2 ∂z [(x − a) + (y − b)2 + (z − c)2 ]  A 2(z − c)2 − (x − a)2 − (y − b)2 ∂2f = 5/2 ∂z 2 [(x − a)2 + (y − b)2 + (z − c)2 ] 2 2 2 ∂ f ∂ f ∂ f Now + 2 + 2 = 0. Hence f is harmonic. ∂x2 ∂y ∂z   1 4xy 4xy 33. fx = − 2 =− 2 2 2 2 2 (x + y − 1) (x + y − 1)2 + 4y 2 4y 1+ 2 2 2 (x + y − 1) 12x4 y − 4y 5 + 8x2 y 3 − 8x2 y − 8y 3 − 4y [(x2 + y 2 − 1)2 + 4y 2 ]4y − 4xy[4x(x2 + y 2 − 1)] fxx = − = [(x2+ y 2 − 1) + 4y 2 ]2 [(x2 + y 2 − 1)2 + 4y 2 ]2  2(x2 + y 2 − 1) − 4y 2 1 2(x2 + y 2 − 1)2 fy = = 2 (x2 + y 2 − 1)2 (x + y 2 − 1)2 + 4y 2 4y 2 1+ 2 2 2 (x + y − 1) [(x2 + y 2 − 1)2 + 4y 2 ](−4y) − 2(x2 + y 2 − 1)2 [4y(x2 + y 2 − 1)2 + 8y] fyy = [(x2 + y 2 − 1)2 + 4y 2 ]2 4 5 2 3 −12x y + 4y − 8x y + 8x2 y + 8y 3 + 4y = [(x2 + y 2 − 1)2 + 4y 2 ]2 ∇2 f = fxx + fyy = 0 32.

34.

∂2f ∂f = 4x3 − 12xy 2 , = 12x2 − 12y 2 , ∂x ∂x2 ∂f ∂2f = −12x2 y + 4y 3 , = −12x2 + 12y 2 , ∂y ∂y 2 ∂2f ∂2f Now + 2 = 0. Hence f is harmonic. 2 ∂x ∂y

35. Using Problems 25 and 23, ˙ ∇ · F = div (∇f × ∇g) = ∇g (curl ∇f ) − ∇f · (curl g) = ∇g · 0 − ∇f · 0 = 0.

CHAPTER 15. VECTOR INTEGRAL CALCULUS

1006

36. Recall that a · (a × b) = 0. Then, using Problmems 25, 23, and 22, ∇ · F = (∇f × f ∇g) = f ∇g · (curl ∇f ) − ∇f · (curl f ∇g) = f ∇g · 0 − ∇f · (∇ × f ∇g) = −∇f · [f (∇ × ∇g) + (∇f × ∇g)] = −∇f · [f curl ∇g + (∇f × ∇g)] = −∇f · [f 0 + (∇f × ∇g)] = −∇f · (∇f × ∇g) = 0. 2 37. The surface is g(x, y) = x2 + y 2 + z 4z − 4 = 0. 2 ∇g = 2xi + 2yj + 8zk, |∇g| = 2 x + y 2 + 16z 2 ; 1 z=M1-x2/4-y2/4 xi + yj + 4zk 2xi + 2yj + 8zk = ; n=  2 x2 + y 2 + 16z 2 x2 + y 2 + 16z 2 2 y 12z(x2 + y 2 ) 2 2 R ∇ × F = (3x − 3y )k, (∇ × F) · n =  x2 + y 2 + 16z 2 2 r=2 Writing the equation of the surface as z =  2 2 1 − x /4 − y /4, we have x x , zy = −  = zx 4 1 − x2 /4 − y 2 /4 16 − 3x2 − 3y 2 y , and dS =  −  dA. 2 2 4 1 − x /4 − y /4 2 4 − x2 − y 2 Then, using polar coordinates,      12z(x2 − y 2 ) 16 − 3x2 − 3y 2   Flux = (∇ × F) · ∇dS = dA x2 + y 2 + 16z 2 2 4 − x2 − y 2 S R     6 1 − x2 /4 − y 2 /4(x2 − y 2 ) 16 − 3x2 − 3y 2 dA   = x2 + y 2 + 16 − 4x2 − 4y 2 4 − x2 − y 2 R  π/4  2   π/4  2  = 1 − r2 /4(r2 cos2 θ − r2 sin2 θ)rdrdθ 4 − r2 = 3r2 cos 2θdrdθ 0



0

π/4

= 0

2  π/4  3 4  π/4 r cos 2θ dθ = 12 cos 2θdθ = 6 sin 2θ|0 = 6.  4 0

0

0

0

    i j k  1 1  1  ∂/∂x ∂/∂y ∂/∂z 38. curl v = curl (ω × r) =   2 2 2 ω 2 z − ω 3 y ω3 x − ω 1 z ω1 y − ω 2 x  1 = [(ω1 + ω1 )i − (ω2 − ω2 )j + (ω3 + ω3 )k] = ω1 i + ω2 j + ω3 k = ω 2    i j k   39. curl F = −Gm1 m2  ∂/∂x ∂/∂y ∂/∂z   x/|r|3 y/|r|3 z/|r|3  = −Gm1 m2 [(−3yz/|r|5 + 3yz/|r|5 )i − (−3xz/|r|5 + 3xz/|r|5 )j + (−3xy/|r|5 + 3xy/|r|5 )k] =0 div F = −Gm1 m2



 −2x2 + y 2 + z 2 x2 − 2y 2 + z 2 x2 + y 2 − 2z 2 + + =0 |r|5/2 |r|5/2 |r|5/2

15.8. STOKES’ THEOREM

1007

40. (a) Expressing the vertical component of V in polar coordinates, we have 2r2 sin θ cos θ sin 2θ 2xy = = 2 2 2 4 (x + y ) r r2 Similarly,

x2 − y 2 r2 (cos2 θ − sin2 θ) cos 2θ = = . (x2 + y 2 )2 r4 r2

Since lim (sin 2θ)/r2 = lim (cos 2θ)/r2 , V ≈ Ai for r large or (x, y) far from the origin. r→∞ r→∞   x2 − y 2 2Axy (b) Identify P (x, y) = A 1 − 2 , and R(x, y) = 0, we , Q(x, y) = − 2 (x − y 2 )2 (x − y 2 )2 have Py =

2Ay(3x2 − y 2 ) 2Ay(3x2 − y 2 ) , Q = , x (x2 + y 2 )3 (x2 + y 2 )3

and Pz = Qz = Rx = Ry = 0.

Thus, curlV = (Ry − Qz )i + (Pz − Rx )j + (Qx − Py )k = 0 and V is irrotational. 2Ax(x2 − 3y 2 ) 2Ax(3y 2 − x2 ) , Qy = , and Rz = 0, ∇·F = Px +Qy +Rz = 0 2 2 3 (x + y ) (x2 + y 2 )3 and V is incompressible.

(c) Since Px =

41. We first note that curl (∂H/∂t) = ∂(curl H)/∂t and curl (∂E/∂t) = ∂(curl E)/∂t. Then, from Problem 30, −∇2 E = −∇2 E + 0 = −∇2 E + grad 0 = −∇2 E + grad (div E) = curl (curl E)     1 ∂ 1 ∂E 1 ∂H 1 ∂ 1 ∂2E curl H = − = curl − =− =− 2 c ∂t c ∂t c ∂t c ∂t c ∂t and ∇2 E = 2

1 2 2 c2 ∂ E/∂t .

Similarly,

2

−∇ H = −∇ H + grad (div H) = curl (curl H) = curl   1 ∂ 1 ∂H 1 ∂2H = − =− 2 c ∂t c ∂t c ∂t2 and ∇2 H =



1 ∂E c ∂t

 =

1 ∂ curl E c ∂t

1 2 ∂ H/∂t2 . c2

42. We note that div F = 2xyz − 2xyz + 1 = 1 = 0. If F = curl G, then div (curl G) = div F = 1. But, by problem 24, for any vector field G, div (curl G) = 0. Thus, F cannot be the curl of G.

15.8

Stokes’ Theorem

μ2 μ 2μ μ2 μ + − = 4 or 5μ2 = 4λ2 , (λ − 5)μ = 8λ. + 2 = 1, − 2 4λ λ 2λ 2 λ 8λ 2 ) = 4λ2 Solving the second equation for μ and substituting into the first, we obtain 5( λ−5 √ or 80 = (λ − 5)2 . Thus, √ λ = 5 ± 4 5. From μ = 8λ/(λ − 5) we find that corresponding 2 values of μ are 8 ± 2 5. Since √ function F (x, y, z) = y is √ 2y = μ we see that the objective when μ = 8 + 2 5. Corresponding values of minimized when μ = 8 − 2 5 and maximized √ √ 8±2 5 μ 1 √ = ∓ √ ≈ |mp0.45, y − μ/2 = 4 ± 5 ≈ 6.24 and =− x, y, and z are x = − 2λ 10 ± 8 5 5 √ √ √ √ 1.76, z = −μ/λ = 2x = ∓2/ 5 ≈ ∓0.89. The closest point is (−1/ 5, 4 − 5, −2 5) or about (−4.5, 6.24, −0.89).

PROBLEMAS DE REPASO DE LA UNIDAD 4

Chapter 13 in Review A. True/False 4.2 in the text. 1. False; see Example 3 in Section 13.2 2. False; (0,4.1) is in the domain of g but not in the domain of f . 3. True 4. True 5. False; consider z = y 2 .

CHAPTER 13 IN REVIEW

871

6. False; consider f (x, y) = xy at (0, 0). 7. False; ∇f is perpendicular to the level curve f (x, y) = c. 8. True 9. True 10. False; at a saddle point fx = fy = 0, but there is no extremum.

B. Fill in the Blanks 1.

1 3x2 + xy 2 − 3xy − 2y 3 3+1−3−2 =− = 2 2 5x − y 5−1 4 (x,y)→(1,1) lim

2. where x − y + 1 = 0 3. 3x2 + y 2 = 3(2)2 + (−4)2 = 28 4.

∂ ∂T ∂p ∂T ∂q T (p, q) = + = T p g ξ + T q hξ ∂ξ ∂p ∂ξ ∂q ∂ξ

5.

∂F dr ∂F ds d F (r, s) = + = Fr g  (w) + Fs h (w) dw ∂r dw ∂s dw

6. dg = gs Δs + gt Δt =

2 4s Δs − 3 Δt t2 t

7. fyyzx 8.

∂3f ∂y 2 ∂x

 ∂f ∂  y 9. Using the Fundamental Theorem of Calculus, we have (x, y) = F (t)dt = F (y) ∂y ∂y x    ∂  y ∂  x ∂  x ∂f (x, y) = − y F (t)dt = − F (t)dt = F (t)dt = −F (x) x y ∂x ∂x ∂x ∂x 10. ∇F (x0 , y0 , z0 ) = i + j + k ∂2 ∂ fx (x, y)g(y)h(z) = [fx (x, y)g  (y)h(z) + fxy (x, y)g(y)h(z)] ∂z∂y ∂z = fx (x, y)g  (y)h (z) + fxy (x, y)g(y)h (z)

11. Fx,y,z =

12. The distinct fourth-order partial derivatives are fxxxx , fxxxy , fxxyy , fxyyy , and fyyyy .

CHAPTER 13. PARTIAL DERIVATIVES

872

C. Exercises 1. zy = −x3 ye−x

2. zu = −

3. fr =

3

y

+ e−x

3

y

v sin uv = −v tan uv cos uv

3 2 3 3 r (r + θ2 )−1/2 ; frθ = − r2 θ(r3 + θ2 )−3/2 2 2

4.

∂f ∂2f = 2(2x + xy 2 )(2 + y 2 ) = 2(2 + y 2 )2 x; = 2(2 + y 2 )2 ∂x ∂x2

5.

∂z ∂2z = 3x2 y 2 sinh x2 y 3 ; = 9x4 y 4 cosh x2 y 3 + 6x2 y sinh x2 y 3 ∂y ∂y 2

6.

∂z ∂2z ∂3z 2 2 2 2 2 = −4y(ex + e−y ); = −8xyex ; 2 = −16x2 yex − 8yex ∂y ∂x∂y ∂ x∂y

7. Fs = 3s2 t5 v −4 ; Fst = 15s2 t4 v −4 ; Fstv = −60s2 t4 v −5

8.

xy x y ∂2w x ∂3w 2x 2 ∂w x 1 ∂4w = 2 + + ; =− 2 − 2 + ; 2 = 3; = 3 2 ∂z z y x ∂y∂z z y x ∂ y∂z y ∂x∂ y∂z y

9. ∇f = −

10. ∇F =

1 1 1 y 1 y x 1 i+ j=− 2 i+ 2 j; ∇f (1, −1) = i + j 2 2 2 2 2 2 2 x 1 + y /x x 1 + y /x x +y x +y 2 2

2x 9y 2 4(x2 − 3y 3 ) i− 4 j− k; ∇F (1, 2, 1) − 2i − 36j + 92k 4 z z z5

6 1 2 11. ∇f = (2xy − y 2 )i + (x2 − 2xy)j; u = √ i + √ j = √ (i + 3j); 40 40 10 1 1 2 2 2 Du f = √ (2xy − y + 3x − 6xy) = √ (3x − 4xy − y 2 ) 10 10 1 2 2y 2z 2 2x i+ 2 j+ 2 k; u = − i + j + k; 2 2 2 2 2 2 +y +z x +y +z x +y +z 3 3 3 −4x + 2y + 4z Du F = 3(x2 + y 2 + z 2 )

12. ∇F =

x2

CHAPTER 13 IN REVIEW 13. {(x, y)|(x + y)2 ≤ {(x, y)| |x + y| ≤ 1}

873 1}

=

14. {(x, y)|y > x, y = x + 1} y

y

x x

15. Δz = 2(x + Δx)(y + Δy) − (y + Δy)2 − (2xy − y 2 ) = 2xΔy + 2yΔx + 2ΔxΔy − 2yΔy − (Δy)2 16. Δz = (x + Δx)2 − 4(y + Δy)2 + 7(x + Δx) − 9(y + Δy) + 10 − (x2 − 4y 2 + 7x − 9y + 10) = 2Δx + (Δx)2 − 8yΔy − 4(Δy)2 + 7Δx − 9Δy 4x + 3y − (x − 2y)4 11y (4x + 3y)(−2) − (x − 2y)3 −11x = ; zy = = ; (4x + 3y)2 (4x + 3y)2 (4x + 3y)2 (4x + 3y)2 11y 11x dz = dx − dy (4x + 3y)2 (4x + 3y)2

17. zx =

18. Ax = 2y + 2z; Ay = 2x + 2z; Az = 2y + 2x; dA = 2(y + z)dx + 2(x + z)dy + 2(x + y)dz  √ √ √ 19. zy = 4y/ x2 + 4y 2 , zy (− 5, 1) = 4/3, z(− 5, 1) = 3. The line is given by x = − 5 and √ y−1 z−3 4 = . z − 3 = (y − 1). Symmetric equations of the line are x = − 5, 3 4 3 −−→ 1 1 20. The direction vector is P Q = 2i + 2j. ∇z = (y + 2x)i + xj. u = √ i + √ j; Du = ∇z · u = 2 2 √ √ √ √ (y + 2x + x)/ 2 = (y + 3x)/ 2; Du (2, 3) = 9/ 2. The slope of the tangent line is 9/ 2. 21. fx = 2xy 4 , fy = 4x2 y 3 . (a) u = i, Du (1, 1) = fx (1, 1) = 2 √ √ √ (b) u = (i − j/ 2, Du (1, 1) = (2 − 4)/ 2 = −2/ 2 (c) u = j, Du (1, 1) = fy (1, 1) = 4 22. (a)

∂w dx ∂w dxy ∂w dz dw = + + dt ∂x dt ∂y dt ∂z dt y z x 6 cos 2t +  (−8 sin 2t) +  15t2 = 2 2 2 2 2 2 2 x +y +z x +y +z x + y2 + z2 (6x cos 2t − 8y sin 2t + 15zt2 )  = x2 + y 2 + z 2

CHAPTER 13. PARTIAL DERIVATIVES

874 (b)

∂w dx ∂w dxy ∂w dz dw = + + dt ∂x dt ∂y dt ∂z dt   8r 6 x 2t y 2r z = cos +  sin 15t2 r3 + 2 2 2 2 2 2 2 2 2 2 r r t t x +y +z x +y +z x +y +z   6x 2t 8yr 2r cos + 2 sin + 15zt2 r3 r r t t  = x2 + y 2 + z 2

√ 1 π 23. F (x, y, z) = sin xy − z; ∇F = y cos xyi + x cos xyj − k; ∇F (1/2, 2π/3, 3/2) = i + j − k. 3 4 √ 1 1 2π 3 π ) = 0 or 4πx+3y −12z = The equation of the tangent plane is (x− )+ (y − )−(z − 3 2 4 3 2 √ 4π − 6 3. 24. We want to find a normal to the surface that is parallel to k. ∇F = (y −2)i+(x−2y)j+2zk. We need y − 2 = 0 and x − 2y = 0. The tangent√plane is parallel√to z = 2 when y = 2 and x = 4. In this case z 2 = 5. The points are (4, 2, 5) and (4, 2, − 5). 25. ∇F = 2xi + 2yj; The equation of the tangent plane is 6(x − 3) + 8(y − 4) = 0 or 3x + 4y = 25. 26. We want to minimize 1 3 Du f = u · ∇f = √ (i + j) · [(3x2 + 3y − 6x)i + (3x + 3y 2 )j] = √ (x2 + y − 2x + x + y 2 ) 2 2 or equivalently, we want to minimize F (x, y) = x2 − x + y 2 + y. Now Fx = 2x − 1; Fxx = 2; Fxy = 0; Fy = 2y + 1; Fyy = 2; D = 4. Solving Fx = 0 and Fy = 0 we obtain x = 1/2 and y = −1/2. Since D = 4 > 0 and Fxx = 2 > 0, F, and hence Du f, has a minimum at (1/2, −1/2). 27. We want to maximize v(x, y, z) = xyz subject to x + 2y + z = 6. Now Vx = yz; Vy = xz; VZ = xy; gx = 1; gy = 2; gz = 1. We need to solve yz = λ, xz = 2λ, xy = λ or xyz = λx, xyz = 2λy, xyz = λz along with x + 2y + z − 6 = 0 or x + 2y + z = 6. From the first three equations, we have x = 2y = z. Substituting into the fourth equation gives x + x + x = 3x = 6 or x = 2. Then y = 1 and z = 2 and V (2, 1, 2) = 4 is the maximum volume. c2 Dθ2 G c2 (b) dM = (θ2 dD + 2Dθdθ) G   c2 θ 2 2Dθ θ2 dθ 2Dθ dD dM = dD + dθ = + 2 , so dD + dθ = (c) We have M G M M Dθ2 Dθ2 D θ          dθ   dM   dD dθ   dD    ≤ 0.10 + 2(0.02) = 0.14 = 14%.  = ≤ + 2 + 2  θ  M  D   θ D

28. (a) M =

CHAPTER 13 IN REVIEW

875

√ 29. We√are given v = 14 5ry −1/2 , dr = −1, dy = 1, r = 20, and y = 25. Now, dv = √ 14 5y −1/2 dr − 7 5ry −3/2 dy and the approximate change in volume is √ √ √ Δv ≈ 14 5(25)−1/2 (−1) − 7 5(20)(25)−3/2 (1) = −98 5/25 ≈ −8.77cm/s. 30. Δf = 2xi + 2yj, Δf (3, 4) = 6i = 8j √ √ (a) Δf (1, −2)2i −√4j; u =√(2i − 4j) √20 = (i √ − 2j)/ 5; Du f (3, 4) = 6 5 − 16 5 = −10 5 = −2 5 √ (b) v = (6i + 8j)/ 100 = (3i + 4j)/5; Dv f (3, 4) = 18/5 + 32/5 = 10 R2 − r 2 . Then, after a straightforward but lengthy comR2 − 2rR cos(θ − φ) + r2 putaiton, we find

31. Let g(r, θ) =

gr = grr = gθθ =

(2r2 R + 2R3 ) cos(θ − φ) − 4rR2 (R2 − 2rR cos(θ − φ) + r2 )2 ,

8R4 cos2 (θ − φ) + (−12rR3 − 4r3 R) cos(θ − φ) − 4R4 + 12r2 R2 , [R2 − 2rR cos(θ − φ) + r2 ]3

(4r4 R2 − 4r2 R4 ) cos2 (θ − φ) + (2r5 R − 2rR5 ) cos(θ − φ) − 8r4 R2 + 8r2 R4 (R2 − 2rR cos θ + r2 )3

and r2 grr + rgr + gθθ . Then   π  π r2 π r 1 2 r Urr + rUr + Uθθ = g(r, θ)f (φ)dφ + g(r, θ)f (φ)dφ + g(r, θ)f (φ)dφ 2π π 2π π 2π π  π 1 = (r2 grr + rgr + gθθ )dφ = 0. 2π π αz βy αAxα y β βAxα y β = ; fy = Aβxα y β−1 = = ; x x y y xαzx − αz xα(αz/x) − αz α2 z − αz α(α − 1) fxx = = = = ; x2 x2 x2 x2 2 yβzy − βz yβ(βz − βz) β z − βz β(β − 1)z fyy = = = = ; y2 y2 y2 y2 αβz α(βz)/y = fxy = fyz = x xy

32. fx = Aαxα−1 y β =

33. Since D = 4(6) − 52 = −1 < 0, f (a, b) is not a relative extremum. 34. Since D = 2(7) − 02 = 14 > 0 and fxx = 2 > 9, f (a, b) is a relative minimum. 35. Since D = (−5)(−9) − 62 = 9 > 0 and fxx = −5 < 0, f (a, b) is a relative maximum. 36. Since D = (−2)(−8) − 42 = 0, no determination is possible.

CHAPTER 13. PARTIAL DERIVATIVES

876 37. Since x = L cos θ and y = L sin θ, 1 1 1 A = xy = L2 sin θ cos θ = l2 sin 2θ. 2 2 4

L

y

θ x

38. Substituting x = h cot φ into tan θ = h=

tan θ . 1 − tan θ cos φ

h and solving, we obtain 1+x

h φ

θ 1

x

39. A = xy − (y − 2z)(x − 2z) − z 2 = 2(x + y)z − 5z 2 40. We are given V (x, y, z) = xyz, x = 30, y = 40, z = 25, and dx = dy = −1 and dz = −1/2. Then dV = yzdx + dzdy + dydz, so the approximate volume of plastic is |dV | = 40(25)(1) + 30(25)(1) + 30(40)(1/2) = 2350cm3 .

    41. V = (2x)(2y)z = 4xy 4 − x2 − y 2 = 16xy − 4xy x2 + y 2 42. C(x, y, z) = 1.5(2xy + 2xz + 2yz + xz + 5yz) =

3 (2xy + 3xz + 7yz) 2

CÁLCULO VECTORIAL MATEMÁTICAS 3

Chapter 14

MANUAL DE SOLUCIONES

Multiple Integrals

UNIDAD 5 INTEGRALES MÚLTIPLES PROBLEMAS 5.1 14.1 The Double Integral 1. With f (x, y) = x + 3y + 1 and ΔAk = 1,   R

(x + 3y + 1)dA ≈ f (1/2, 1/2) + f (3/2, 1/2) + f (5/2, 1/2) + f (1/2, 3/2) + f (3/2, 3/2) + f (5/2, 3/2) + f (1/2, 5/2) + f (3/2, 5/2) = 3 + 4 + 5 + 6 + 7 + 8 + 9 + 10 = 52.

2. With f (x, y) = 2x + 4y and ΔAk = 1/4,   R

(2x + 4y)dA ≈

1 [2(3/2) + 4(1/2) + 2(2) + 4(1/2) + 2(5/2) + 4(1/2) + 2(1/2) + 4(1) 4 + 2(3/2) + 4(1) + 2(1) + 4(1) + 2(1/2) + 4(3/2) + 2(1) + 4(3/2) + 2(3/2)

+ 4(3/2) + 2(1) + 4(2) + 2(1/2) + 4(2) + 2(1/2) + 4(5/2)] 1 = (3 + 2 + 4 + 2 + 5 + 2 + 4 + 4 + 3 + 4 + 2 + 4 + 1 + 6 + 2 + 6 + 3 4 93 + 6 + 2 + 8 + 1 + 8 + 1 + 10) = . 4 3. (a) With f (x, y) = x + y, and ΔAk = 1,   (x + y) dA ≈ (−3/2 + 1/2) + (−1/2 + 1/2) + (1/2 + 1/2) + (3/2 + 1/2) R

+ (−3/2 + 3/2) + (−1/2 + 3/2) + (1/2 + 3/2) + (3/2 + 3/2) 1 16 = (−2 + 0 + 2 + 4 + 0 + 2 + 4 + 6) = = 8. 2 2 (b) With f (x, y) + y + 4 and ΔAk = 1,

877

CHAPTER 14. MULTIPLE INTEGRALS

878   R

(x + y) dA ≈ (−2 + 1) + (−1 + 1) + (0 + 1) + (1 + 1) + (−2 + 2) + (−1 + 2) + (0 + 2) + (1 + 2) = 8.

4. With   f (x, y) = xy and ΔAk = 1/4, 1 xydA ≈ [0(1/2) + (1/2)(1/2) + (−1/2)(1) + (0)(1) + (1/2)(1) 4 R + (1)(1) + (−1/2)(3/2) + (0)(3/2) + (1/2)(3/2) + (1)(3/2) + (−1/2)(2) + (0)(2) + (1/2)(2) + (1)(2) + (−1)(5/2) + (−1/2)(5/2) + (0)(5/2) + (1/2)(5/2) + (1)(5/2) + (3/2)(5/2) + (−1)(3) + (−1/2)(3) + (0)(3) + (1/2)(3) + (1)(3) + (3/2)(3) + (−1)(7/2) + (−1/2)(7/2)

y

4

1

x

+ (0)(7/2) + (1/2)(7/2) + (1)(7/2) + (3/2)(7/2)] = 73/16 5. 6. 7.

8.

 R

 R

 R

 R

10dA = 10 10dA = 10 10dA = 10

10dA = 10

 R

dA = 10(6) = 60

R

dA = 10(12) = 120

  



 1 2 dA = 10 π(2) = 10π R 4 

1 dA = 10 (5) R 2

  5 125 = 2 2

9. No, since x + 5y is negative at (3, −1) which is in R. 10. Yes, since x2 + y 2 is nonnegative on R. 11. 12. 13. 14. 15.

 R



10dA = 10

R

dA = 10(8) = 80



R

−5xdA = −5

R

(2x + 4y)dA = 2

R

(2x + 4y)dA =

R

(3x + 7y + 1)dA = 3

    

R

 R

xdA = −5(3) = −15

R

 R

 R

y dA −

f (x, y)dA =

xdA − 



R

R 2

R

 R

 R

 R1

 R1

 R

 

f (x, y)dA +

ydA = 2(3) + 4(7) = 34

ydA = 3 − 7 = −4

xdA + 7

(2 + y) dA =

f (x, y)dA = f (x, y)dA = −5. R2

18. Since 

xdA + 4

 

2

16. 17.



ydA +

y dA − 4

 R2

f (x, y)dA +

R

dA = 3(3) + 7(7) + 8 = 66

 

2

R



  R

dA − 4

  R

ydA −

R

y 2 dA

= −4(8) − 4(7) = −

f (x, y)dA = 4 + 14 = 18  R2

f (x, y)dA, 25 = 30 +

 R2

f (x, y)dA and

14.2. ITERATED INTEGRALS

PROBLEMAS 5.2 Integrals 14.2 Iterated 1.



dy = y + c1 (x)

2. By holding y fixed, (1 − 2y)dy = x − 2yx + c2 (y) 3. By  3  2  holding y fixed, x x √ √ 2 y + c2 (y) (6x y − 3x y)dx = 6 y−3 3 2 3 √ = 2x3 y − x2 y + c2 (y) 2 4. By  2  holding x fixed, y y 3/2 √ + c1 (x) − 3x (6x2 y − 3x y)dy = 6x2 2 (3/2) = 3x2 y 2 − 2xy 3/2 + c1 (x) 5. By holding x fixed,  1 ln |y + 1| dy = + c1 (x) x(y + 1) x 6. By  2  holding x fixed, x (1 + 10x − 5y 4 )dx = x + 10 − 5xy 4 + c2 (y) 2 = x + 5x2 − 5xy 4 + c2 (y) 7. By    holding y fixed, sin 4x (12y cos 4x − 3 sin y)dx = 12y − 3x sin y + c2 (y) 4 = 3y sin 4x − 3x sin y + c2 (y) 8. By  holding x fixed, tan 3xy sec2 3xydy = + c1 (x) 3x 9. By  holding y fixed,  y √ dy = y 2x + 3y + c2 (y) 2x + 3y 10. By  holding x fixed,  1 (2x + 5y)7 + c1 (x) (2x + 5y)6 dy = 5 7 (2x + 5y)7 + c1 (x) = 35  3 3 11. (6xy − 5ey ) dx = (3x2 y − 5xey ) −1 = (27y − 15ey ) − (3y + 5ey ) = 24y − 20ey −1

879

CHAPTER 14. MULTIPLE INTEGRALS

880 

2

12.

tan xy dy =

1



3x

3x 2 x3 exy = x2 exy 1 = x2 (e3x − ex )

y3

y 3 (8x3 y − 4xy 2 ) dx = (2x4 y − 2x2 y 2 ) √y = (2y 13 − 2y 8 ) − (2y 3 − 2y 3 ) = 2y 13 − 2y 8

13. 1

 14.

√



y

2x xy x x x 2 2 ln(x dy = + y ) = [ln(x2 + 4x2 ) − ln x2 ] = ln 5 2 2 x +y 2 2 2 0

2x

15. 

0 x

e2y/x dy =

16. x3



2 1 1 ln | sec xy| = ln | sec 2x − sec x| x x 1

sec y

17. tan y

3 2 x 2y/x x x x e 3 = (e2y/x − e2x /x ) = (e2 − e2x ) 2 2 2 x

sec y (2x + cos y) dx = (x2 + x cos y) tan y = sec2 y + sec y cos y − tan2 y − tan y cos y = sec2 y + 1 − tan2 y − sin y = 2 − sin y

 18.

1 √

y

y ln x dx

Integration by parts = y(x ln x −



π/2

19. x

1 x)|√

√

= y(0 − 1) − y( y ln

y

cos x sin3 ydy = cos x cos y



√

y−

√

√

y) = −y − y y



1 ln y − 1 2



  π/2  − sin2 y 2 − sin2 x 2 2 − − = 0 − cos x 3 3 x 3 3

cos2 (1 − cos2 x) 2 cos2 x cos2 x sin3 x 2 cos2 x + = + 3 3 3 3 1 cos2 x cos4 x 2 cos2 x − + = cos2 x − cos4 x = 3 3 3 3 1   1  1 sin xy cos xy x sin xy cos xy xy 2 + + y cos xydx = y = 20. 2y 2 1/2 2 2 1/2 1/2     x x sin 2 cos 2 sin x cos x − sin x2 cos x2 sin x cos x x x x + + + = − = 2 2 2 4 2 4 =



2



21. 1



x2 −x

(8x − 10y + 2) dy dx =

2

1  2

= 

1

= 1

2

x 2 (8xy − 5y + 2y) 2

dx

−x

[(8x3 − 5x4 + 2x2 ) − (−8x2 − 5x2 − 2x)] dx 2 (8x3 − 5x4 + 15x2 + 2x) dx = (2x4 − x5 + 5x3 + x2 ) 1

= 44 − 7 = 37

14.2. ITERATED INTEGRALS 



1

y

22. −1

0

√



y 1 (x + y)3 dy −1 0 −1 3 0  1   1  1 1 1 7 4 1 3 3 3 y =0 [(y + y) − (0 + y) ] dy = 7y dy = = 3 −1 3 −1 3 4 −1

(x + y)2 dx dy =

 √2−y2

2

23. −

0

881

√

2−y 2



1



y

(x + y)2 dx

√

 (2x − y) dx dy =

2

0 √

 0

√

= 0



π/4



cos x

24. 0



0

π



3y

25. 0



y

2



√

−



2



2 − y 2 ) − (2 − y 2 + y

  2 − y 2 ) dy

cos x  π/4 (1 + 4y tan2 x) dy dx = (y + 2y 2 tan2 x) = (cos x + 2 cos2 x tan2 x) dx 0 0 0  π/4   π/4 1 = (cos x + 2 sin2 x) dx = sin x + x − sin 2x 2 0 0 √ √ 2 2+π−2 2 π 1 + − = = 2 4 2 4 

2y sin πx

π/4

2

π



2



√

dy dx = 1

0

x

2y sin πx

2

dy



2

dx = 1

√ x  y sin πx dx = 2

2

2

0

1

x sin πx2 dx

2 1 1 1 1 2 cos πx = − (cos 4π − cos π) = − (1 − (−1)) = − =− 2π 2π 2π π 1



x

27. 1

(2 − y 2 − y

2−y



√2  2 2 2 4√ (−2 2 − y 2 ) dy = (2 − y 2 )3−2 = (0) − 23/2 = − 2 3 3 3 3 0



x

0

ln 3

2

1

3y  1 π 1 sin(2x + y) cos(2x + y) dx dy dy (sin 7y − sin 3y) dy 2 0 0 2 y  π  1 1 1 = − cos 7y + cos 3y 2 7 3  0   1 1 1 1 1 4 − (−1) + (−1) − − + =− = 2 7 3 7 3 21

26. 1



dy =

√2−y2 (x2 − xy) √ 2

= 



6ex+2y dy dx =

0



ln 3 1

x  ln 3 ln 3 x+2y e (3e3x − 3ex ) dx = (e3x − 3ex ) 1 dx = 1 0

= (27 − 9) − (e3 − 3e) = 18 − e3 + 3e 28.

 1  2y 0

0

e−y

2

dx dy =

1 0

1 2 2y 2 2 1 xe−y dy = 0 2ye−y dy = −e−y = −e−1 − (−1) = 1 − e−1 0

0

CHAPTER 14. MULTIPLE INTEGRALS

882 

3



2x+1

29. 0



x+1

1



y



0 9



2

x



0 1/2



y

32. 0



0

e





y 1

4



√

2ye



−x

1

6

 √25−y2 /2

35. 0



0

2

 √20−y2

36. y2

0



π

37. π/2





1 0



9 1

y 1  1 2 1 6 1 1 1 2 5/2 5 − (y − x ) = − (−y ) dy = − y = 5 5 30 30 0 0 0

x  9 9 π 1 π π −1 y tan dx = ln |x| = = ln 9 x x 0 4x 4 4 1 1 

1 √ dx dy = 1 − x2

x

34. 1

√

1/2

sin

−1

0

y  1/2 x dy = sin−1 y dy 0

Integration by parts

0

√ √   1/2  π + 6 3 − 12 3 1 π −1= = y sin−1 y + 1 − y 2 = + 2 6 2 12 0 y  e  e y dx dy = y ln x dy = y ln y dy Integration by parts x 1 1 1    e  1 2 1 2 1 2 1 2 1 1 y ln y − y = e − e − − = = (e2 + 1) 2 4 2 4 4 4 1

33. 1

3

dx dy =

1 dy dx = x+ y 2

31. 1

2 3/2

x(y − x )

30. 0

2x+1  3 √ 2 y − x dx = 2 ( x + 1 − 1) dx 0 0 x+1       3 2 2 16 10 =2 (x + 1)3/2 = 2 −3 − = 3 3 3 3 0 

1 √ dy dx = y−x

√x  4 dy dx = y e dx = (xe−x − e−x ) dx 1 1 1 4 = (−xe−x − e−x + e−x ) 1 = −4e−4 + e−1 

4

2 −x

1  (25 − y 2 ) − x2

xy

 √25−y2 /2 x sin−1  dx dy = dy 2 25 − y 0 0  6  6 π −1 1 dy = dy = π = sin 2 0 0 6 

√

x

6



  2  1 1 4 3 2 3/2 2 y dx dy = dy = (y 20 − y − y ) dy = − (20 − y ) − y 3 4 0 0 0 y2 √     √ 1 1 40 5 − 76 = − (64) − 4 − − (40 5) − 0 = 3 3 3 

2

20−y 2

0 e sin y dx dy = e sin y cos y π/2 0

Integration by parts



π

x

= (− cos y +



2



π

dy =

cos y cos y π )|π/2 e

π/2

(sin y − ecos y sin y) d

= (1 + e−1 ) − (0 + 1) = e−1

14.2. ITERATED INTEGRALS 

1



y 1/3

883 

2

38.

1

6x ln(y + 1) dx dy = 0

0

0

y1/3  2x ln(y + 1) dy = 2

1

3

0

0

y ln(y + 1) dy

Integration by parts 

 1 1 2 = y ln(y + 1) − y + y − ln(y + 1) 2 0 1 1 = (ln 2 + 1 − ln 2) − (0 − 0 + 0 − ln 1) = 2 2



2π



(cos x − sin y) dy dx =

39. π



x 0

2

2π π

x  (y cos x + cos y) dx = 0

2π π

(x cos x + cos x − 1) dx

Integration by parts 2π

= (cos x + x sin x + sin x − x)|π = (1 − 2π) − (−1 − π) = 2 − π



3



1/x

40. 1

0

1 dy dx = x+1



3 1

1/x   3  3 1 1 y 1 dx = − dx = dx x + 1 0 x x+1 1 x(x + 1) 1 3

= [ln x − ln(x + 1)]|1 = (ln 3 − ln 4) − (0 − ln 2) = ln 3/2



5π/12

√



2 sin 2θ

41. π/12



π/3

1





1+cos θ

42. 0

√2 sin 2θ 5π/12   5π/12  1 2 1 1 r r dr dθ = dθ = sin 2θ − dθ = − (cos 2θ + θ) 2 2 2 π/12 π/12 π/12  √ 1  √  √ 3 5π 3 5π 3 π 1 − + − + = − =− 2 2 12 2 12 2 6 

3 cos θ

r dr dθ =

5π/12

π/3 0

1+cos θ 1 2 r dθ 2 3 cos θ

π/3 1 1 (1 + 2 cos θ − cos2 θ) dθ = (θ + 2 sin θ − 4θ − 2 sin 2θ) 2 0 2 0 √ √ 1 π = (−π + 3 − 3) = − 2 2 

=

π/3

CHAPTER 14. MULTIPLE INTEGRALS

884 43.

44.

y=2x+1

y

y

x=My x=-My

x

x

45.

46.

y

y=x2+1

y

x=M16-y2

x

x

y=-x2

√x    4 1 2 1 2 x2 x y dx = x x− 47. x ydydx = dx 4 0 x/2 0 2 0 2 x/2    4  4 1 2 1 4 1 4 1 x − x dx = x − x5 = 2 8 8 40 0 0 32 128 = = 32 − 5 5  2  2  2  2y 2y 1 1 x3 y dy = y(8y 3 − y 6 )dy x2 ydxdy = 0 y2 0 3 0 3 y2    2  2 8 4 1 7 8 5 1 8 = y − y dy = y − y 3 3 15 24 0 0 256 32 32 − = = 3  5   √ 15 

4



√

x



2

4

x

Therefore 0

x/2

4

x2 ydydx =

2

0

2y

y2

x2 ydxdy

y

y=Mx y=1/2x

x

y

x=y2 x=2y

x

14.2. ITERATED INTEGRALS 

1

√



48.

1



0 1

√1−x2 2xy √ dx 2

2xdydx =

√

− 1−x2

0



1−x2

885

− 1−x

 2x

= 

0 1

=

4x





0



1

4 = 3   √1−y2 2xdxdy =

−1

−1

0

x2





1

Therefore,



√

1 − x2 + 2

1−

√

x2 dx

  1 − x2 dx

y=M1-x2

1 4 2 3/2 = − (1 − x ) 3 0



1−y 2

dy =

1 −1

0

1−x

1

y=-M1-x2

x

y

(1 − y 2 )dy

 1 1 = y − y 3 3 −1     1 1 4 = 1− − −1 + = 3 3 3   √ 2 2

√ − 1−x2

0

49.

1

y

x=0

x=M1-y2 x

1−y

2xdydx =

2xdxdy −1

0

3  2 2 x dy dx = x y dx = 3x2 dx = x3 −1 = 8 − (−1) = 9 −1 0 −1 −1 0 2   3 2  3  3  3 8 1 1 3 3 2 x + x dx dy = dy = 3 dy = 3y|0 = 9 dy = 3 3 0 −1 0 3 0 0 −1 



2



2



3

2



2



4

2

2

(2x + 4y) dx dy =

50. −2

2

−2  2

= 

4

−2





2

4

(2x + 4y) dy dx = 2

−2

2  4

= 2 3



π

51. 1

0

4  (x + 4xy) dy =

2

2

2

−2

[(16 + 16y) − (4 + 8y)] dy

2 (12 + 8y) dy = (12y + 4y 2 ) −2 = (24 + 16) − (−24 + 16) = 48

2 (2xy + 2y ) 2

−2



4

dx = 2

[(4x + 8) − (−4 + 8)] dx

4 8x dx = 4x2 2 = 64 − 16 = 48

 π    3  2 3 2 2 3π 2 x y − 4 cos y dx = x − 4 − (4) dx (3x y − 4 sin y) dy dx = 2 2 1 1 0   2  3  3 2 π 3 3π 2 x − 8 dx = x − 8x = 2 2 1 1   2   2 π 27π − 24 − − 8 = 13π 2 − 16 = 2 2 2



3



CHAPTER 14. MULTIPLE INTEGRALS

886 

π



0

3 1



(3x2 y − 4 sin y) dy dx =



3 (x3 y − 4x sin y) 1 dy =

π 0

π

= 0



π

[(27y − 12 sin y) − (y − 4 sin y)]dy

0

π (26y − 8 sin y)dy = (13y 2 + 8 cos y) 0

= (13π 2 − 8) − (8) = 13π 2 − 16  2   1 x2 4 dy = 8y ln |x + 1| − 2 8y ln 3 − 2 52. dxdy = dy y + 1 0 y +1 0 0 0 0 1 = (4y 2 ln 3 − 4 tan−1 y) 0 = 4 ln 3 − π  1  2     2 1  2 8y 4y 2 4 2x π − 2 − 2x tan−1 y = − dydx = dx x+1 y +1 x+1 x+1 2 0 0 0 0 0  π  2 = 4 ln |x + 1| − x2 = 4 ln 3 − π 4 0 β β 53. We use the fact that α kF (t)dt = k α F (t)dt. Then 





   

1



2



8y 2x − 2 x+1 y +1

d

c



∞



0

0



b

a

∞

54.



xye

1



d

f (x)g(y)dxdy = −(2x2 +3y 2 )

c



b

g(y) 

a

b

f (x)dx dy =

d

f (x)dx

a

c

  2 2 xe−2x ye−3y dxdy 0 0    ∞  ∞ −2x2 −3y 2 = xe dx · ye dy 0 0      ∞

g(y)dy



∞

dxdy =

a

=

lim

a→∞

0

xe

−2x2

b

dx ·

lim

b→∞

0

ye

−3y 2

 dy

   2 a 2 e−2x e−3y = lim − · lim − a→∞ b→∞ 4 6 0 

 

 2 2 1 1 −e−2a −e−3b = lim + · lim + a→∞ b→∞ 4 4 6 6     1 1 1 = · = 4 6 24 

PROBLEMAS 15.3 14.3 Evaluation of Double Integrals   1. R

x3 y 2 dA =



1 0



x 0

x3 y 2 dydx =

1 1 7 1 x = = 21 0 21



1 0

x  1 1 6 1 3 3 x y dx = x dx 3 3 0 0

y

x

1

x

14.3. EVALUATION OF DOUBLE INTEGRALS  



2.

2



R

2

(x + 1)dydx = 

0

x 2

= 

0

2

= 0

 =

0

4−x (xy + y) dx

y

x

[(4x − x2 + 4 − x) − (x2 + x)]dx

4-x

(2x − 2x2 + 4)dx x

 2 2 20 x − x3 + 4x = 3 3 0 2

 



3.



4−x

(x + 1)dA =

887



1

x2

(2x + 4y + 1)dA =

(2x + 4y + 1)dydx 

1

= 0  1

= 

0 1

= 0

y

x3

0

R

x

x 2 (2xy + 2y + y) dx 3

x2

2

x3

x

x

[(2x3 + 2x4 + x2 ) − (2x4 + 2x6 + x3 )]dx 3

2



6

(x + x − 2x )dx =

 1 1 4 1 3 2 7 x + x − x 4 3 7 0

25 1 1 2 + − = 4 3 7 84 x    1 x  1 4. xey dA = xey dydx = xey dx =

R



0

0

1

= 0

 =

 

(xex − x)dx x

2



Integration by parts



8

2

2xydydx = 

x3

0 2

= 0

x

x

2xydA = R

y

0

 1   1 2 1 1 xe − e − x = e − e − − (−1) = 2 2 2 0



5.

0

(64x − x7 )dx =

0



8 xy dx 3 2

1

x

y

x

 2 1 32x2 − x8 = 96 8 0

x3

x

CHAPTER 14. MULTIPLE INTEGRALS

888   6. R

x √ dA = y





1

xy

x2 +1

−1



3−x2

1

(x

=2



−1

−1/2

 dydx =

3 − x2 − x



1 −1

3−x2 √ 2x y dx 2

y

x +1

x2+1

3

x2 + 1)dx

1 1 1 = 2[− (3 − x2 )3/2 − (x2 + 1)3/2 ] 3 3 −1 2 3/2 3/2 3/2 3/2 = − [(2 + 2 ) − (2 + 2 )] = 0 3   7. R



y dA = 1 + xy

1



0

1 0

y dxdy = 1 + xy



1 0

3-x2

x

1 ln(1 + xy) dy

y

0

1

1 ln(1 + y)dy = [(1 + y) ln(1 + y) − (1 + y)]|0

= 0

1

= (2 ln 2 − 2) − (−1) = 2 ln 2 − 1

1

 y  2 πx πx y dxdy = dy − cos y x y 0 1 1 R 0  2 y y = − cos πy + dy Integration by parts π π 1   2 y 1 y 2 = − 2 sin πy − 3 cos πy + π π 2π 1     1 1 2 1 3π 2 − 4 = − 3+ + − = 3 π π π 2π 2π 3 x     √3  x   √3  9. x2 + 1dA = x2 + 1dydx = y x2 + 1 dx R 0 −x 0  

8.

sin

πx dA = y



2



y2

√

3

(x 

0

= = π/4



√



3

2x 0

10.

y y2

x

y y=x

−x

=



x

sin



 

2

x2 + 1 + x





x2 + 1)dx x

√3 2 2 3/2 x2 + 1dx = (x + 1) 3

y=-x

0

2 3/2 14 (4 − 13/2 ) = 3 3 1



π/4

1 1 2 x 2

1 xdA = xdxdy = dy = 2 0 tan y 0 R tan y π/4  π4 1 1 2 = (2 − sec y)dy = (2y − tan y) 2 0 2 0  π 1 1 π −1 = − = 2 2 4 2



π4 0

y

(1 − tan2 y)dy π/4

x=tany

1

x

14.3. EVALUATION OF DOUBLE INTEGRALS  



11. R

(x + y)dA =

4



0



2

(x + y)dxdy +

0

2

889



0

4

(x + y)dxdy

2

 2  4  2 1 2 1 2 x + xy dy + x + xy dy = 2 2 0 0 0 2  2  4 4 2 (2 + 2y)dy + [(8 + 4y) − (2 + 2y)]dy = (2y + y 2 ) 0 + (6y + y 2 ) 0 = 



4

0

0

= 24 + 16 = 40  



12. R

(x + y)dA =

4



0

4 0

 (x + y)dxdy −

3



1

3 1

(x + y)dxdy

 4  3  3 1 2 1 2 x + xy dy − x + xy dy = 2 2 0 1 0 1     3   4 1 9 + 3y − + y dy (8 + 4y)dy − = 2 2 0 1 2 4 2 3 = (8y + 2y ) − (4y + y ) = 64 − (21 − 5) = 48 



4

0

 13. A =

3 0

2x−x



2

dydx =

−x

3 0

 =

1

(2x − x2 + x)dx

y

 3 3 2 1 3 9 x − x = 2 3 2 0

2x-x2 x

-x

14. Using symmetry,  A=2

 15. A =

1



4 1 4





2−y 2 y2

0

y

dxdy = 2



ex ln x

dydx =

4 1

1 0



 1 2 3 8 2 2 (2−y −y )dy = 2 2y − y = . 3 3 0

y2 2-y2 x

4

(ex − ln x)dx = (ex − x ln x + x)|1

y ex

= (e − 4 ln 4 + 4) − (e + 1) = e4 − e − 4 ln 4 + 3

lnx

10 4

x

CHAPTER 14. MULTIPLE INTEGRALS

890  16. A = 



4

√ (2− x)2

0 4

= 0



4−x

dydx =

√ (4 x − 2x)dx =

4 0



[4 − x − (2 −

√

y

x)2 ]dx

 4 8 3/2 16 x − x2 = 3 3 0

4-x

(2-Mx)2 x

 17. A =

1



−2x+3 x3

−2

 =



dydx =

1 −2

(−2x + 3 − x3 )dx

 1 1 4 7 63 −x + 3x − x = − (−14) = 4 4 4 −2

y

-2x+3

2

x3

x

18. Expressing y = −x2 + 3x and y = −2x + 4 as functions of 3 1√ 1 y, we have x = − 9 − 4y and x = 2 − y. 2 2 2    2  2−y/2  2  3 1 y − A= dxdy = 9 − 4y dy 2 − − √ 2 2 2 0 3/2− 9−4y/2 0    2  1 1 27 1 1 13 = y − y 2 − (9 − 4y)3/2 = − − − = 2 4 12 12 12 6 0

19. The correct integral is ( c).  2  √4−y2  V =2 (4 − y)dxdy = 2 −2



0

 = 2 2y 4 − y 2 + 8 sin−1 20. The correct integral is (b).

y

2-y/2

3/2-1/2 M9-4y

√ 4−y2  2  (4 − y)x dy = 2 (4 − y) 4 − y 2 dy −2 −2 0  2 y 1 + (4 − y 2 )3/2 = 2(4π − (−4π)] = 16π 2 3 −2 2

x

14.3. EVALUATION OF DOUBLE INTEGRALS  V =8

r 0

 √r2 −y2 0

(r2 − y 2 )1/2 dxdy = 8

 r y 2 2 y =8 (r − y )dy = 8 r − 3 0 0     2r3 r3 16 3 r = 8 r3 − =8 = 3 3 3 

r





r 0

891

√r2 −y2 (r2 − y 2 )1/2 x dy 0

3

21. Setting z = 0 we have y = 6 − 2x.  3  6−2x  V = (6 − 2x − y)dydx =

 6−2x 1 2 dx 6y − 2xy − y 2 0 0 0 0   3  3 1 (18 − 12x + 2x2 )dx = 6(6 − 2x) − 2x(6 − 2x) − (6 − 2x)2 dx = 2 0 0   3 2 = 18x − 6x2 + x3 = 18 3 3



y

6-2x

0

22. Setting z = 0 we have y ± 2. 32 (4 − y 2 )dydx = V = 0 0  3 16 dx = 16 0 3

x

3 0



 2 1 4y − y 3 dx 3

y

=

0

x

1 1 23. Solving for z, we have x = 2 − x + y. Setting z = 0, we 2 2 see that this surface (plane) intersects the xy-plane in the line y = x − 4. Since z(0, 0) = 2 > 0, the surface lies above the xy-plane over the quarter-circular region.   2  √4−x2  1 1 V = 2 − x + y dydx 2 2 0 0  √4−x2  2 1 1 = dx 2y − xy + y 2 2 4 0 0   2  1  1 = 2 4 − x2 − x 4 − x2 + 1 − x2 dx 2 4 0    2 1 1 3 −1 x 2 3/2 2 + (4 − x ) + x − x = x 4 − x + 4 sin 2 6 12 0   2 4 = 2π + 2 − − = 2π 3 3

y

M4-x2 2

2

x

CHAPTER 14. MULTIPLE INTEGRALS

892

y

24. Setting z = 0 we have y = 3. Using symmetry, 3  √3  √3  3 1 2 (3 − y)dydx = 2 (3y − y ) dx V =2 2 2 0 x2 0 x √   √3  3 9 9 1 4 1 5 2 3 x−x + x ( − 3x + x )dx = 2 =2 2 2 2 10 0 0 √   √ √ √ 9 9 24 3 . =2 3−3 3+ 3 = 2 10 5

3 x2

x

25. Note that z = 1 + x2 + y 2 is always positive. Then  3−3x  1  3−3x  1 1 3 2 2 2 V = (1 + x + y )dydx = dx y+x y+ y 3 0 0 0 0  1 [(3 − 3x) + x2 (3 − 3x) + 9(1 − x)3 ]dx = 

0

= 0

1

y

3-3x

(12 − 30x + 30x2 − 12x3 )dx

1 = (12x − 15x2 + 10x3 − 3x4 ) 0 = 4. 26. In the first octant, z = x + y is nonnegative. Then  √9−x2  3  √9−x2  3 1 V = (x + y)dydx = dx xy + y 2 2 0 0 0 0   3  9 1 2 2 = x 9 − x + − x dx 2 2 0   3 1 1 9 = − (9 − x2 )3/2 + x − x3 3 2 6 0   27 9 − = − (−9) = 18. 2 2

x

y

3

M9-x2

3

x

y 27. In the first octant z = 6/y is positive. Then  6 6x 5  6 dy 65 6 6 dy = 30 = 30 ln y|1 = 30 ln 6. 6 V = 1 0 dxdy = 1 1 y y y 0

1

5

x

14.3. EVALUATION OF DOUBLE INTEGRALS

893

28. Setting z = 0, we have x2 /4 + y 2 /16 = 1. Using symmetry, 

2



y

√

2 4−x2

1 (4 − x2 − y 2 )dydx 4 0 0 2√4−x2  2 1 =4 (4y − x2 y − y 3 ) dx 12 0 0  2   2 [8 4 − x2 − 2x2 4 − x2 − (4 − x2 )3/2 ]dx =4 3 0

V =4

4 2M4-x2

Trig substitution    x 1 x = 4 4x 4 − x2 + 16 sin−1 − x(2x2 − 4) 4 − x2 − 4 sin−1 2 4 2  2  1 x + x(2x2 − 20) 4 − x2 −4 sin 2 0 12  16π 4π 4π − − =4 − (0) = 16π. 2 2 2 29. Note that z = 4−y 2 is positive for |y| ≤ 1. Using symmetry, √2x−x2  2  √2x−x2  2 1 V =2 (4 − y 2 )dydx = 2 (4y − y 3 ) dx 3 0 0 0 0      1 2 2 2 2 = 2 0 4 2x − x − (2x − x ) 2x − x dx 3  2   1 (4 1 − (x − 1)2 − [1 − (x − 1)2 ] 1 − (x − 1)2 )dx =2 3 0

2

x

y 1

M2x-x2

u = x − 1, du = dx   1   1   11 1 1  =2 [4 1 − u2 − (1 − u2 ) 1 − u2 ]du = 2 1 − u2 + u2 1 − u2 du 3 3 3 −1 −1 Trig substitution   1  11  1 11 1 −1 2 2 2 u 1−u + sin u + x(2x − 1) 1 − u + sin u =2 6 6 24 24 −1     11 π 1 π 1 π 11 π 15 + − . =2 − − = 6 2 24 2 6 2 24 2 4

x

CHAPTER 14. MULTIPLE INTEGRALS

894

30. From z = 1 − x2 and z = 1 − y 2 we have 1 − x2 = 1 − y 2 or y = x (in the first octant). Thus, the surfaces intersect in the plane y = x. Using symmetry,  1  1  1 1   1 3 2 1 − y dydx = 2 V =2 y − y dx 3 0 x 0 x   1 2 1 − x + x3 dx =2 3 3 0   1 2 1 2 1 4 1 x− x + x = . =2 3 2 12 2

z

1

1

y

y=x

1

0

x

31. From z = 4−x−2y and z = x+y, we have 4−x−2y = x+y 3 or x = 2 − y.  2 4/3

V =

0

2−3y/2

0

z 4

[4 − x − 2y) − (x + y)]dxdy

2−3y/2  4x − x − 3xy dy = 0 0  2    4/3 3 3 3 = 4(2 − y) − 2 − y − 3 2 − y y dy 2 2 2 0   4/3  9 = 4 − 6y + y 2 dy 4 0   4/3 3 16 = = 4y − 3y 2 + y 3 4 9 0 

4/3



2

2

y

x=2-3y/2

x

32. Using symmetry, √9−x2  3  √9−x2  3 1 3 2 2 2 V =4 (9 − x − y )dydx = 4 [(9 − x )y − y ] 3 0 0 0 0  8 3 2 3/2 0 (9 − x ) dx Trig substitution = 3 3  81π x 8 x 243 8 243 π sin−1 ] = ( )= . = [− (2x2 − 45) 9 − x2 + 3 8 8 3 3 8 2 2

z 9

y

0

x

33. From z = x2 and z = −x + 2 we have x2 = −x + 2 or x = 1 (in the first octant). Then 1  5 1  5 1 2 1 3 2 V = (−x + 2 − x )dxdy = (− x + 2x − x ) dy 2 3 0 0 0 0  5 35 7 dy = . = 6 0 6

y=M

9-x2

z

y x

14.3. EVALUATION OF DOUBLE INTEGRALS

895

34. From 2z = 4 − x2 − y 2 and z = 2 − y we have z 4 − x2 − y 2 = 4 − 2y or x2 + (y − 1)2 = 1. We find the volume in√the first octant and use symmetry.    2  1−(y−1)2  1 2 1 2 V =2 2 − x − y − (2 − y) dxdy 2 2 0 0 √  1−(y−1)2  2 1 1 dy =2 − x3 − xy 2 + xy y 6 2 0 0 x  2   3/2 1 2 1 − y 1 − (y − 1)2 + y 1 − (y − 1)2 dy =2 − 1 − (y − 1)2 6 2 0   2  1 1 2 3/2 2 2 =2 − [1 − (y − 1) ] + (2y − y ) 1 − (y − 1) dy 6 2 0   2 1 1 =2 − [1 − (y − 1)2 ]3/2 + [1 − (y − 1)2 ]3/2 dy 6 2 0  2 2 [1 − (y − 1)2 ]3/2 dy Trig substitution = 3 0  2     2 3 2 3 π 3  π y−1 π −1 2 2 [2(y − 1) − 5] 1 − (y − 1) + sin (y − 1) = − = − − = 3 8 8 3 82 8 2 4 0 √ 35. Solving x = y 2 for y, we obtain y = x. Thus, y



2 0

y

4



2

f (x, y)dxdy =

0

2 √

0

x

f (x, y)dydx. x=y2

36. Solving x = Thus,   √

 f (x, y)dxdy =

0

x

√ 25 − y 2 for y, we obtain y = ± 25 − x2 .

25−y 2

5

−5



5 0

√



y x=M25-y2

25−x2

√ − 25−x2

f (x, y)dydx. x

37. Solving y = ex for x, we obtain x = ln y. Thus, 

3 0





ex 1

f (x, y)dydx =

e3 1



y

3 ln y

f (x, y)dxdy.

y=ex

3

x

CHAPTER 14. MULTIPLE INTEGRALS

896

38. Solving x = 3 − y and x = y/2 for y, we obtain y = 3 − x and y = 2x. Thus,  1  2x  3  3−x  2  3−y f (x, y)dxdy = f (x, y)dydx+ f (x, y)dydx. 0

y/2

0

0

1

y

x=y/2

0

x=3-y

x

√ 39. Solving y = 3 x and y = 2 − x for x, we obtain x = y 3 and x = 2 − y. Thus, 

1



0

√ 3



x

f (x, y)dydx+

0



2 1



2−x

f (x, y)dydx =

0

1



3 y=M x

2−y

f (x, y)dxdy.

y3

0

y

y=2-x

x

√ √ 40. Solving x = y and x = 2 − y for y, we obtain y = x2 and y = 2 − x2 . Thus, 

1 0



√



y

f (x, y)dxdy+

0

2



1

√



2−y

f (x, y)dxdy =

0

1



0

y

2−x2

f (x, y)dydx.

x2

x=My 1 x=M2-y



1



1

41. 0

x

x

2



y 1 3 x 1 + y 4 dy 1 + y 4 dydx = y 0 0 0 3 0  1   1  1 1 1 (1 + y 4 )3/2 = y 3 1 + y 4 dy = 1 3 0 3 6 0 1 √ (2 2 − 1) = 18 

1



y

  x 1 + y 4 dxdy = 2

x

1

y=x

x



1



2

42. 0

2y

e

−y/x

 dxdy = 

2 0

0

x/2 0

2

=



e

−y/x

 dydx =

(−xe−1/2 + x)dx =

2 0



2 0

−xe

x/2

−y/x

0

y x=2y

(1 − e−1/2 )xdx

2 1 −1/2 2 )x = 2(1 − e−1/2 ) = (1 − e 2 0

x

14.3. EVALUATION OF DOUBLE INTEGRALS 

2



43. 0



√ x cos x3/2 dxdy = cos x3/2 dydx = y cos x3/2 dx y2 0 0 0 0 4  4 √ 2 2 = x cos x3/2 dx = sin x3/2 = sin 8 3 3 0 0 

4

√



1

44.

1−x2

√

− 1−x2

−1

4



√



x

  x 1 − x2 − y 2 dydx =

1 =− 3 1



1

0

x

y x=y2

x

 √1−y2

−



1 −1

1 dydx = 1 + y4

45.

4

 2 2 √ 2 x 1 − x − y dxdy −1 − 1−y √ 1−y2  1 1 = [− (1 − x2 − y 2 )3/2 ] √ dy 3 2 1

−1



897

 

y y=M1-x2

x

1−y

y=-M1-x2

(0 − 0)dy = 0

1 0 1

= 0

y x dy 4 0 0 1+y 0 1 y 1 π −1 2 dy = tan y = 4 1+y 2 8 0



y

1 dxdy = 1 + y4



1

y y=x

x



4



46. 0



2 √

y

x 2 x3 + 1dxdy = y x3 + 1 dx 0 0 0 0 2  2  2 = x2 x3 + 1dx = (x3 + 1)3/2 9 

2



  x3 + 1dydx =

x2

0

2



y

x=My

0

52 2 = (93/2 − 13/2 ) = 9 9

47. fave

1 = A =

1 A



d c



b a

d c



2

1 xydxdy = A



2

1 (b − a )y dy = 2 A

d c



b x2 y dy 2 a

 d (b − a2 )y 2 4 c 2

1 (b2 − a2 )(d2 − c2 ) = A 4 But A = (b − a)(d − c), so fave =

(b + a)(d + c) (b2 − a2 )(d2 − c2 ) = 4(b − a)(d − c) 4

x

CHAPTER 14. MULTIPLE INTEGRALS

898

48. fave

1 = A

√



3

√ − 3

 √9−3y2 −

√

9−3y 2

9 − x2 − 3y 2 dxdy

√9−3y2 x 1 − 3y 2 x dy = √ 9x − − √ A − 3 3 2 

√

3

3

−

9−3y

    1 (9 − 3y 2 )3/2 2 2 2 = − 3y 9 − 3y 9 9 − 3y − A − √3 3     (9 − 3y 2 )3/2 + 3y 2 9 − 3y 2 dy − −9 9 − 3y 2 + 3    √3  1 9 − 3y 2 9 − 3y 2 2 2 2 − 3y + 9 − − 3y dy = 9 − 3y 9 − A − √3 3 3  √3  1 9 − 3y 2 (12 − 4y 2 )dy = A − √3 √ √ √ √  √  3  4 3y  3y 3y 3y 9 81 1 12 √ − √ (6y 2 − 9) 9 − 3y 2 + sin−1 9 − 3y 2 + sin−1 = √ A 2 2 3 3 8 9 3 3 3 − 3 √ 27π 3 = 2A 49. Let S be the solid with base R and height described by the function f (x, y). The volume of S is equal to the volume of the solid with base R and constant height fave .  d    d b 1 50. (a) cos 2π(x + y)dA = cos 2π(x + y)dxdy = [sin 2π(b + y) − sin 2π(a + b)]dy 2π c R c a  d 1 [(sin 2πb cos 2πy + cos 2πb sin 2πy) − (sin 2πa cos 2πy + cos 2πa sin 2πy)] dy = 2π c  d 1 (S1 cos 2πy + C1 sin 2πy)dy = 2π c 1 1 = [S1 (sin 2πd − sin 2πc) − C1 (cos 2πd − cos 2πc)] = (S1 S2 − C1 C2 ) 2 2 4π 4π  d    d b 1 sin 2π(x + y)dA = sin 2π(x + y)dxdy − [cos 2π(b + y) − cos 2π(a + y)]dy 2π R c a c  d 1 [(cos 2πb cos 2πy − sin 2πb sin 2πy) − (cos 2πa cos 2πy − sin 2πa sin 2πy)]dy =− 2π c  d 1 (C1 cos 2πy − S1 sin 2πy)dy =− 2π c 1 1 = − 2 [C1 (sin 2πd − sin 2πc) + S1 (cos 2πd − cos 2πc)] = − 2 (C1 S2 + S1 C2 ) 4π 4π (b) If b − a = n is an integer, then b = a + n and 

√

3

sin 2πb = sin 2π(a + n) = sin 2πa cos 2πn + cos 2πa sin 2πn = sin 2πa cos 2πb = cos 2π(a + n) = cos 2πa cos 2πn − sin 2πa sin 2πn = cos 2πa.

14.4. CENTER OF MASS AND MOMENTS

899

  In this case, S1 = 0 and C1 = 0, so cos 2π(x+y)dA = 0 and sinR sin 2π(x+y)dA = R 0. Similarly, is d − c is an integer, the double integrals are zero. (c) If both integrals are 0, then 0 = (S1 S2 − C1 C2 )2 + (C1 S2 + S1 C2 )2 = S12 S22 + C12 C22 + C12 S22 + S12 C22 = (S12 + C12 )(S22 + C22 ). Thus, either S12 + C12 = 0, in which case S1 = C1 = 0, or S22 + C22 = 0, in which case S2 = C2 = 0. Suppose S1 = C1 = 0, and b − a = k or b = a + k. We want to show that k is an integer. Consider S1 = sin 2πb − sin 2πa = sin 2π(a + k) − sin 2πa = sin 2πa cos 2πk + cos 2πa sin 2πk − sin 2πa C1 = cos 2πb − cos 2πa = cos 2π(a + k) − cos 2πa = cos 2πa cos 2πk − sin 2πa sin 2πk − cos 2πa S1 − C1 = (sin 2πa − cos 2πa) cos 2πk + (sin 2πa + cos 2πa) sin 2πk − (sin 2πa − cos 2πa) = (sin 2πa − cos 2πa)(cos 2πk − 1) + (sin 2πa + cos 2πa) sin 2πk. Since a is arbitrary we must have cos 2πk − 1 = 0 and sin 2πk = 0, which implies k is an integer. Similarly, if S2 = C2 = 0, d − c must be an integer.   cos 2π(x + y)dA = sin 2π(x + y)dA = 0 for k = 51. By Problem 50 (a) we have Rk Rk 1, 2, · · · , n. Then       cos 2π(x + y)dA = cos 2π(x + y)dA + · · · + cos 2π(x + y)dA = 0 + · · · + 0 = 0 R

R1

and  

Rn

  R

sin 2π(x + y)dA =

  R1

sin 2π(x + y)dA + · · · +

Rn

sin 2π(x + y)dA = 0 + · · · + 0 = 0.

Therefore by Problem 45 (c), at least one of the two sides of R must have integer length.

PROBLEMAS 5.4 of Mass and Moments 14.4 Center  1. m = =

3 0





4

xydxdy =

0

3 4y 2 0 = 36  3 4 2

My =



0

0

3

= 0

3 0

x ydxdy =

4  3 1 2 x y dy = 8ydy 2 0 0



3 0

4 1 3 x y dy 3 0

3 32 2 64 ydy = y = 96 3 3 0

y

3 x=4

x

CHAPTER 14. MULTIPLE INTEGRALS

900  Mx = 

3 0

0

4 0

3

=





2

xy dxdy =

3 0

4 1 2 2 x y 2 0

3 8 8y 2 dy = y 3 = 72 3 0

x = My /m = 96/36 = 8/3; y = Mx /m = 72/36 = 2. The center of mass is (8/3, 2).  2  4−2x  2  2 2 2 4−2x y 2. m = x dydx = x y0 dx = x2 (4 − 2x)dx 0

0

0

0



 2 4 3 1 4 8 32 x − x = −8= (4 − 22 − 2x3 )dx = = 4 3 2 3 3 0 4−2x 0  2  2  4−2x  3 x3 dydx = x3 y dx = x3 (4 − 2x)dx My = y=4-2x 0 0 0 0 0   2  2 2 = (4x3 − 2x4 )dx = x4 − x5 5 0 0 16 64 = = 16 − 2 5 5 4−2x  2  2  4−2x  2 1 2 2 1 x y x2 ydydx = dx = x2 (4 − 2x)2 dx Mx = 2 0 0 0 2 0 0   2   2 4 3 1 2 1 5 2 3 4 2 3 4 4 x −x + x = (16x − 16x + 4x )dx = 2 (4x − 4x x )dx = 2 2 0 3 5 0 0   32 32 32 − 16 + = =2 15 15 15 32/15 = 4/5. x = My /m = 16/5 = 6/5; y = Mx /m = 8/3 The center of mass is (6/5, 4/5). 

2

3. Since both the region and ρ are symmetric with respect to the line x = 3, x = 3. 6−y  3  6−y  3 m= 2ydxdy = 2xy 0 y 0 y  3  3 = 2y(6 − y − y)dy = (12y − 4y 2 )dy 0



 3 4 = 6y 2 − y 3 = 18 3 0   3  6−y Mx = 2y 2 dxdy = 0

y

0

6−y  2xy 2 dxdy = y

3

4

3

0

2y 2 (6 − y − y)dy =

3 = (4y − y ) 0 = 27 y = Mx /m = 27/18 = 3/2. The center of mass is (3, 3/2). 3

y

x=y

0

3

x

x=6-y

x



3 0

(12y 2 − 4y 3 )dy

14.4. CENTER OF MASS AND MOMENTS

901 y 3

x=y

4. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. Using symmetry,

x

 y 1 3 2 x + xy dy m= (x + y )dxdy = 3 0 0 0 0 3   3  3 1 3 1 4 y + y 3 dy = = y 3 dy = y 4 = 27 3 3 3 0 0 0  y    3 y  3  3 1 3 1 4 4 3 4 Mx = x y + xy 3 dy = y + y 4 dy = (x2 y + y 3 )dxdy = y dy 3 3 3 0 0 0 0 0 0 3 4 5 324 y = = 15 0 5 324/5 = 12/5. The center of mass is (0, 12/5). y = Mx /m = 27 

3



y

2

2



3



y y=x2 1

 x2 x 1 2 5. m = (x + y)dydx = xy + y dx 2 0 0 0 0    1  1 1 1 1 7 x4 + x5 = = x3 + x4 dx = 2 4 10 20 0 0  x2  1  x2  1 1 2 2 2 My = (x + xy)dydx = x y + xy dx 2 0 0 0 0    1  1 1 1 17 1 x5 + x6 = = x4 + x5 dx = 2 5 12 60 0 0  x2  1    1  x2  1 1 2 1 3 1 5 1 6 2 xy + y = x + x dx Mx = (xy + y )dydx = 2 3 2 3 0 0 0 0 0   1 1 6 1 11 x + x7 = = 12 21 84 0 17/60 11/84 = 17/21; y + Mx /m = = 55/147. x = My = m = 7/20 7/20 The center of mass is (17/21, 55/147). 

1

2



1

1

x

y 2

y=Mx

4

x

CHAPTER 14. MULTIPLE INTEGRALS

902

√x 1 2 6. m = (y + 5)dydx = ( y + 5y) dx 0 0 0 2 0    4  4 √ 1 2 10 3/2 1 92 x + 5 x dx = x + x = = 3 2 4 3 0 0  √x  4  √x  4 1 2 xy + 5xy dx My = (xy + 5x)dydx = 2 0 0 0 0   4 1 3 224 x + 2x5/2 = = 6 3 0   √x  4  √x  4  4 1 3 5 2 1 3/2 5 2 y + y dx = x + x dx Mx = (y + 5y)dydx = 3 2 3 2 0 0 0 0 0   4 2 5/2 5 2 364 = x + x = 15 4 15 0 224/3 364/15 = 56/23; y = Mx /m = = 91/115. x = My /m = 92/3 92/3 The center of mass is (56/23, 91/115). 

4



√

x



4

y 1

y=1-x2

1

7. The density is ρ = ky. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. Using symmetry, 1−x2  1   1  1−x2 1 2 y kydydx = 2k dx = k 01 (1 − x2 )2 dx m=2 0 0 2 0 0 1  1 2 1 5 2 4 3 =k (1 − 2x + x )dx = k(x − x + x ) 3 5 0 0   2 1 8 k =k 1− + = 3 5 15 1−x2  1  1  1−x2  1 1 3 2 2 y Mx = 2 ky dydx = 2k dx = k (1 − x2 )3 dx 3 3 0 0 0 0 0   1  1 2 2 3 1 7 2 4 6 3 5 = k (1 − 3x + 3x − x )dx = k x − x + x − x 3 0 3 5 7 0   3 1 2 32 k = k 1−1+ − = 3 5 7 105 32k/105 = 4/7. The center of mass is (0,4/7). y = Mx /m = 8k/15

x

y y=sin x 1

π

x

14.4. CENTER OF MASS AND MOMENTS 8. The density is ρ = kx.   π  sin x kxdydx = m= 0

0

π 0

sin x  kxy dx = 0

903

π 0

kx sin xdx

Integration by parts π

= k(sin x − x cos x)|0 = kπ  π  sin x  kx2 dydx = My =

sin x  π kx2 y dx = kx2 sin xdx Integration by parts 0 0 0 0 0 π = k(−x2 cos x + 2 cos x + 2x sin x) 0 = k[(π 2 − 2) − 2] = k(π 2 − 4) sin x  π  sin x  π 1 2 kxy Mx = kxydydx = dx 2 0 0 0 0  π  π 1 1 = kx sin2 xdx = kx(1 − cos 2x)dx 2 4 0  π   π 0 1 = k xdx − x cos 2xdx Integration by parts 4 0 0 π π 1 1 1 1 1 1 = k[ x2 − (cos 2x + 2x sin 2x) ] = k( π 2 ) = kπ 2 4 2 0 4 4 2 8 0 k(π 2 − 4) kπ 2 /8 = π − 4/π; y = Mx /m = = π/8. x = My /m = kπ kπ The center of mass is (π − 4/π, π/8).  9. m =

1



ex

y dydx =

0

0

3



1 0

1 1 4x 1 4 e = (e − 1) = 16 16 0  1  ex  3 xy dydx = My = 

0

0

1

π

ex  1 1 4 1 4x y dx = e dx 4 0 0 4

1 0

ex 1 4 xy dx 4 0

y=ex

y

1

1 4x xe dx Integration by parts = 0 4  1    1 1 4x 1 4x 1 3 4 1 1 4 xe − e = = 4 16 e + 16 = 64 (3e + 1) 4 4 16 0 x 1  1  ex 4  1 1 5 e  1 1 5x 1 5x 1 5 Mx = 0 0 y dydx = 0 y dx = 0 e dx = e (e − 1) = 5 0 5 25 0 25 4 4 5 3e + 1 16(e5 − 1) (3e + 1)/64 (e − 1)/25 = ; = x = My /m = y = M /m = x (e4 − 1)/16 4(e4 − 1) (e4 − 1)/16 25(e4 − 1) 4 5 3e + 1 16(e − 1) , ≈ (0.77, 1.76). The center of mass is ( 4 4(e − 1) 25(e4 − 1)

1

x

CHAPTER 14. MULTIPLE INTEGRALS

904

10. Since both the region and ρ are symmetric with respect to the y-axis, x = 0. √Using symmetry, 9−x2  3  √9−x2  3 2 2 m=2 x dydx = 2 x y dx 

0

0

3

=2 0

x2



0

9 − x2 dx

y

3

y=M9-x2

0

Trig substitution 3



 3  x 81π 81 81 π −1 x 2 2 (2x − 9) 9 − x + sin · = . =2 = 8 8 3 0 √ 4 2 8 √ 2 9−x  3  9−x2  3  3 1 2 2 Mx = 2 x y x2 ydydx = 2 dydx = x2 (9 − x2 )dx 2 0 0 0 0 0 3 1 162 = (3x2 − x5 = 5 0 5 162/5 = 16/5π. The center of mass is (0, 16/5π). y = Mx /m = 81π/8 y−y2  1 11. Ix = 2xy dxdy = x y dy = (y − y 2 )2 y 2 dy 0 0 0 0 0   1  1 1 1 1 1 y 5 − y 6 + y 7 = = (y 4 − 2y 5 + y 6 )dy = 5 3 7 105 0 0 

1



y−y 2



2

1

2 2

x

y 1 x=y-y2

x

 12. Ix =

1



√ x2

0

x

x2 y 2 dydx =



1 0

1 1 2 9/2 1 9 1 = ( x − x ) = 3 9 9 27 0

√

x  1 2 3 1 1 7/2 x y dx = (x − x8 )dx 3 3 2 0 x

y y=x2 y=Mx

1

1

13. Using symmetry,  π/2  cos x  2 ky dydx = 2k Ix = 2

cos x  π/2 1 3 2 y dx = k cos3 xdx 3 3 0 0 0 0 0 π/2  π/2 1 2 2 4 = k cos x(1 − sin2 x)dx = k(sin x − sin3 x) = k. 3 0 3 3 9 0 2



√

4−x2

3



y

π/2

√4−x2  1 4 1 2 y 14. Ix = y dydx = dx = (4 − x2 )2 dx 4 4 0 0 0 0 0  2   1 2 1 1 8 2 4 3 5 = (16 − 8x + x )dx = 16x − x + x 4 0 4 3 5 0     1 64 32 2 1 64 + = 32 − =8 1− + = 4 3 5 3 5 15 

x

1

y=cos x

π/2

2

x

y

2

y=M4-x2

2

x

14.4. CENTER OF MASS AND MOMENTS  15. Iy =

4 0

1 = 3





√

y



2

x ydxdy =

0

4 0

905

√y   1 3 1 4 3/2 1 4 5/2 x y dy = y ydy = y dy 3 3 0 3 0 0

y 4

 4 2 7/2 2 7/2 256 y (4 ) = = 7 21 21 0

x=My

2

 16. Iy =

1

17. Iy = 

x

x4 dydx =

 0

1 4 x y

1 2 11/2 1 7 3 x − x ) = 11 7 77 0



1 0

0





3 y

1

= =

x2

0

=(

√

3

√

x

(4x + 3x y)dxdy =

dx =

x2

1 0

x

y

(x9/2 − x6 )dx

y=x2

y=Mx

1

1



2



1 0

3 (x + x y) dy 4

y

3

y

x

x=y

1

4

(81 + 27y − 2y )dy

3

 1 27 2 2 5 941 81y + y − y = 2 5 10 0

18. The density is ρ = ky. Using symmetry, 1−x2  1  1−x2  1  1 1 2 2 2 kx y kx ydydx = 2 dx = k x2 (1 − x2 )2 dx Iy = 2 2 0 0 0 0 0 1  1 1 2 1 8k . =k (x2 − 2x4 + x6 )dx = k( x3 − x5 + x7 ) = 3 5 7 105 0 0 19. Using symmetry, √a2 −y2  a  √a2 −y2  a  a 1 2 x m=2 xdxdy = 2 dy = (a2 − y 2 )dy 0 0 0 2 0 0 a 1 3 2 2 = (a y − y ) = a3 . 3 3 0 √a2 −y2   a  √a2 y 2  a 1 1 a 2 3 4 x Iy = 2 x dxdy = 2 dy = (a − y 2 )2 dy 2 0 0 0 0 4 0 a  1 4 2 2 3 1 5 4 5 1 a 4 2 2 4 a (a − 2a y + y )dy = (a y − a y + y ) = = 2 0 2 3 5 15 0    Iy 4a5 /15 2 = = a Rg = m 2a3 /3 5

x

y 1

y=1-x2 x

1

y a

y=Ma2-x2

a

x

CHAPTER 14. MULTIPLE INTEGRALS

906

 a  a−x  a a−x a kdydx = 0 ky 0 dx = k 0 (a − x)dx = y 0 0 a 1 2 1 2 a k(ax − x ) = ka 2 2 y=a-x 0 a−x  a  a  a−x  a 1 1 ky 3 Ix = ky 2 dydx = dx = k (a − x)3 dx 3 0 0 0 0 3 0 a x a  a 1 3 2 2 1 4 1 4 1 3 2 3 3 3 ka (a − 3a x − x )dx = k(a x − a x + ax − x ) = = k 3 0  3 2 4 12 0   Ix ka4 /12 1 = = a Rg = m ka2 /2 6

20. m =

21. (a) Using symmetry,   b√a2 −x2 /a  4b3 a 2 a Ix = 4 0 y 2 dydx = 3 (a − x2 )3/2 dx x = a sin θ, dx = a cos θdθ 3a 0 0  π/2  π/2 1 4 4 (1 + cos 2θ)2 dθ = ab3 cos4 θdθ = ab3 3 3 4 0 0 π/2  1 3 3 1 1 1 1 1 3 π/2 (1 + cos 2θ + + cos 4θ)dθ = ab ( θ + sin 2θ + sin 4θ) = ab 3 2 2 3 2 2 8 0 0 =

ab3 π . 4

(b) Using symmetry,   a  b√a2 −x2 /a 4b a 2  2 x2 dydx = x a − x2 dx x = a sin θ, dx = a cos θdθ Iy = 4 a 0 0 0  π/2  π/2 1 (1 − cos2 2θ)dθ = 4a3 b sin2 θ cos2 θdθ = 4a3 b 4 0 0 π/2  π/2 1 1 1 a3 bπ 3 3 1 . (1 − − cos 4θ)dθ = a b( θ − sin 4θ) = =a b 2 2 2 8 4 0 0 1 3 1 ab π/πab = b. 2 2  1 3 1 (d) Rg = Iy /m = a bπ/πab = a 2 2 (c) Using m = πab, Rg =



Ix /m =

22. The equation of the ellipse is 9x2 /a2 + 4y 2 /b2 = 1 and the equation of the parabola is y = ±(9bx2 /8a2 − b/2). Letting Ie and Ip represent the moments of inertia of the ellipse and parabola, respectively, about the x-axis, we have  0  0  b√a2 −9x2 2a a b3 a 2 y dydx = (a2 − 9x2 )3/2 dx x = sin θ, dx = cos θdθ Ie = 2 3 12a 3 3 −a/3 0 −a/3  0 ab3 π b3 a 4 b3 a 3π = = cos4 θdθ = 3 12a 3 −π/3 36 16 192 and

14.4. CENTER OF MASS AND MOMENTS

907

 3  9b 2 2 2a/3 b − y dydx = x dx Ip = 2 3 0 2 8a2 0 0  3     2 b3 2a/3 b3 2a/3 243 4 729 6 9 27 = x − x 1 − 2 x2 dx = 1 − 2 x2 + dx 3 8 0 4a 12 0 4a 16a4 64a6  2a/3  8ab3 243 5 729 7 b3 32a b3 9 3 x − x = x − 2x + = . = 12 4a 80a4 64a6 12 105 315 0 ab3 π 8ab3 + . Then Ix = Ie + Ip = 192 315 

2a/3



b/2−9bx2 /8a2

2

1 1 4 ka . 23. From Problem 20, m = ka2 and Ix = 2  a−x12  a  a−x  a kx2 dydx = kx2 y dx = k Iy = 0

0



0

0

 4 1 3 1 4 1 4 ax − x = ka =k 3 4 12 0 1 4 1 1 ka + ka4 = ka4 I0 = Ix + Iy = 12 12 6

y a 0

a

x2 (a − x)dx

y=a-x

a

x

3 1 3 1 , and from Problem 16, Iy = . Thus, I0 = Ix + Iy = + = 24. From Problem 12, Ix = 27 77 27 77 158 . 2079 25. The density is ρ = k/(x2 + y 2 ). Using symmetry,  √2  6−y2  √2 6−y2 k I0 = 2 (x2 + y 2 ) 2 dxdy = 2 kx dy 2 2 x + y 2 y +2 0 0  √2 2 2 (6 − y 2 − y 2 − 2)dy = 2k 4y − y 3 = 2k 3 0 0 √   8√ 16 2 k. = 2k 2 = 3 3 

 26. I0 =

3



0



4 y

3



√

2



2

k(x + y )dxdy = k



3



0



y 2 x=6-y2

y +2

 4 1 3 2 x + xy dy 3 y

64 1 + 4y 2 − y 3 − y 3 dy 3 3 0   3 64 4 3 1 4 y + y − y = 73k =k 3 3 3

6

x=y2+2

x

y 3

x=y

=k

0

1 1 27. From Problem 20, m = ka2 , and from Problem 21, I0 = ka4 . 2 6    ka4 /6 1 = a. Then Rg = I0 /m = ka2 /2 3

4

x

CHAPTER 14. MULTIPLE INTEGRALS

908

28. Since the plate is homogeneous, the density is ρ = m/lw. Using symmetry,  w/2  l/2   l/2  w/2 m 2 4m 1 (x + y 2 )dydx = I0 = 4 dx x2 y + y 3 lw lw 3 0 0 0 0  l/2       lw3 4m wl3 4m l/2 w 2 w3 4m w 3 w3 l2 + w 2 x + x + x = + . = dx = =m lw 0 2 24 lw 6 24 lw 48 48 12 0

PROBLEMAS 5.5 14.5 Double Integrals in Polar Coordinates 1. Using symmetry,  π/2  3+3 sin θ  A=2 rdrdθ = 2 

−π/2

π/2

= −π/2

0

9(1 + sin θ)2 dθ = 9



π/2 −π/2

π/x −π/2

3+3 sin θ 1 2 r dθ 2

6

0

(1 + 2 sin θ + sin2 θ)dθ

3

polar axis

3

polar axis

 π/2 1 1 = 9 θ − 2 cos θ + θ − sin 2θ 2 4 −π/2     3π 3 π 27π − − =9 = 22 2 2 2 

2. Using symmetry,  π  2+cos θ  A=2 rdrdθ = 2

2+cos θ  π 1 2 r dθ = (2 + cos θ)2 dθ 0 0 0 2 0 0  π   π 1 1 2 (4 + 4 cos θ + cos θ)dθ = 4θ + 4 sin θ + θ + cos 2θ = 2 4 0 0     1 π 1 9π . = 4π + + − = 2 4 4 2 π

3. Solving r = 2 sin θ and r = 1, we obtain sin θ = 1/2 or θ = π/6. Using symmetry,  π/6  2 sin θ  π/2  1 A=2 rdrdθ + 2 rdrdθ 0

0

π/6

2

0

2 sin θ 1  π/2  π/6  π/2 π/6 1 2 1 2 2 r r dθ = dθ + 2 4 sin θdθ + dθ =2 2 0 π/6 2 0 π/6 0 0 √  π π  π √3 π 4π − 3 3 π/6 − + = = (2θ − sin 2θ)|0 + = − 2 6 3 2 3 6 8 sin 4θ  π/4  8 sin 4θ  π/4  π/4 1 2 1 r 4. A = rdrdθ = dθ = 64 sin2 4θdθ 2 2 0 0 0 0 0  π/4  1 1 θ− sin 8θ = 32 = 4π 2 16 0 

2

1 polar axis

polar axis

14.5. DOUBLE INTEGRALS IN POLAR COORDINATES

5. Using symmetry,  π/6  5 cos 3θ  V =2 4rdrdθ = 4 0



= 100

 6. V =

2π 0

1 =− 3

 7. V =

1 =− 3

 8. V = 

2 0

0

2 0 2π

  9 − r2 rdrdθ = (5

3/2

0



3 1 2π

0

 π

2π

=





0

 π/6 1 1 25π θ+ sin 6θ = 2 12 3 0



2π 0

0

π/6

0

5 cos 3θ  π/6 2 r dθ = 4 25 cos2 3θdθ 0 0

2 1 − (9 − r2 )3/2 dθ 3 0

2π 0

polar axis

3 1 2 3/2 − (16 − r ) dθ 3 1

√ √ 1 2π(15 15 − 7 7) 3/2 3/2 3/2 3/2 (7 − 15 )dθ = (15 − 7 )2π = 3 3

√

 r2 rdrdθ =

2π 0

polar axis

5 1 3 r dθ 3 0

250π 125 dθ = 3 3

polar axis

1+cos θ 1 3 r sin θ 9. V = (r sin θ)rdrdθ = dθ 3 0 0 0 0  π/2   1 π/2 1 1 = (1 + cos θ)3 sin θdθ = − (1 + cos θ)4 3 0 3 4 0 1 5 4 = − (1 − 2 ) = 12 4 

polar axis

√ 1 2π(27 − 5 5) 3/2 − 27)dθ = (27 − 5 )2π = 3 3

  2 16 − r rdrdθ =

5 0

2π

909

π/2



1+cos θ



π/2

1

2

1

polar axis

polar axis

CHAPTER 14. MULTIPLE INTEGRALS

910 10. Using symmetry,  π/2  cos θ  V =2 (2 + r2 )rdrdθ =

 cos θ 1 4 dθ r + r 4 0 0 0 0    π/2   π/2  1 1 1 + cos 2θ 2 2 4 2 ) dθ =2 cos θ + cos θ dθ = 2 cos θ + ( 4 4 2 0 0   π/2  1 1 1 2 cos2 θ + + cos 2θ + cos2 2θ dθ = 8 4 8 0  π/2  1 1 1 1 1 19π sin 4θ . = θ + sin 2θ + θ + sin 2θ + θ + = 2 8 8 16 64 32 0  π/2  3  π/2 1 2 3 1  π/2 r dθ = k 0 8dθ = 2kπ 11. m = 0 krdrdθ = k 0 1 2 1 2 3  π/2  3  π/2  3  π/2 1 r3 cos θ dθ kxrdrdθ = k r2 cos θdrdθ = k My = 3 0 1 0 1 0 1 π/2  π/2 3 polar 1 26 26 axis k sin θ k = k 26 cos θdθ = = 3 0 3 3 0 13 26k/3 = . x = My /m = 2kπ 3π Since the region and density function are symmetric about the ray θ = π/4, y = x = 13/3π and the center of mass is (13/3π, 13/3π). π/2



2

12. The interior of the upper-half circle is traced from θ = 0 to π/2. The density is kr. Since both the region and the density are symmetric about the polar axis, y = 0.  m=



π/2 0

cos θ



2

kr drdθ = k

0

π/2 0

cos θ  1 3 k π/2 r dθ = cos3 θdθ 3 3 0



0

=

k 4

0



Thus, x =

π/2 0

cos5 θdθ =

k 4



1

0

π/2  2 1 2k 2 + cos θ sin θ = 3 3 9 0  π/2  cos θ  (r cos θ)(r)(rdrdθ) = k My = k k = 3

1

π/2 0



cos θ 0

r3 cos θdrdθ = k

 π/2 1 2k 2 = sin θ − sin3 θ + sin5 θ 3 5 15 0

2k/15 = 3/5 and the center of mass is (3/5, 0). 2k/9



π/2 0

polar axis

cos θ 1 4 r cos θ dθ 4 0

14.5. DOUBLE INTEGRALS IN POLAR COORDINATES

911

3

polar axis

13. In polar coordinates the line x = 3 becomes r cos θ √ = 3 or r = 3 sec θ. The angle of inclination of the line y = 3x is π/3. 3 sec θ  π/3   π/3 1 m= r4 dθ 03 sec θ r2 rdrdθ = 4 0 0 0   81 π/3 81 π/3 sec4 θdθ = (1 + tan2 θ) sec2 θdθ = 4 0 4 0 π/3 √ 81 √ 1 81 √ 81 3 (tan θ + tan θ) ( 3 + 3) = = 3 = 4 3 4 2 0  π/3  3 sec θ  π/3  3 sec θ My = xr2 rdrdθ = r4 cos θdrdθ 0



0

π/3

= 0

0

0

3 sec θ  1 5 243 π/3 r cos θ dθ = sec5 θ cos θdθ 5 5 0 0

 243 √ 486 √ 243 π/3 (2 3) = sec4 θdθ = 3 = 5 0 5 5  π/3  3 sec θ  π/3  3 sec θ  yr2 rdrdθ = r4 sin θdθ = Mx =

3 sec θ 1 5 r sin θ 5 0 0 0 0 0 0  π/3   π/3 243 243 243 0π/3 tan θ sec4 θdθ = = sec5 θ sin θdθ = tan θ(1 + tan2 θ) sec2 θdθ 5 0 5 5 0  π/3   729 243 1 1 243 3 9 243 π/3 3 2 2 4 tan θ + tan θ ( + )= (tan θ + tan θ) sec θdθ = = = 5 0 5 2 4 5 2 4 4 0 √ 486 3/5 729/4 √ = 12/5; y = Mx /m = √ = x = My /m = 81 3/2 81 3/2 √ √ 3 3/2. The center of mass is (12/5, 3 3/2). π/3

polar axis

CHAPTER 14. MULTIPLE INTEGRALS

912

14. Since both the region and the density are symmetric about the x-axis, y = 0. Using symmetry, 4 cos 2θ  π/4  4 cos 2θ  π/4 1 2 m=2 r krdrdθ = 2k dθ 2 0 0 0 0 π/4  π/4 1 1 2 = 16k cos 2θdθ = 16k( θ + sin 4θ) = 2kπ 2 8 0 0  π/4  4 cos 2θ  π/4  4 cos 2θ  My = 2 kxrdrdθ = 2k r2 cos θdrdθ = 2k 0

=

128 k 3

128 k = 3

0

 

π/4 0 π/4 0

0

cos3 2θ cos θdθ =

128 k 3

0



π/4 0

π/4 0

4 cos 2θ 1 3 r cos θ dθ 3 0

(1 − 2 sin2 θ)3 cos θdθ

(1 − 6 sin2 θ + 12 sin4 θ − 8 sin6 θ) cos θdθ

π/4 12 8 128 k(sin θ − 2 sin3 θ + sin5 θ − sin7 θ) 3 5 7 0 √ √ √ √  2 2 3 2 2 1024 √ 128 k − + − = = 2k 3 2 2 10 14 105 √ √ 512 2 1024 2/105 = . x = My /m = 2kπ √ 105π The center of mass is (512 2/105π, 0) or approximately (2.20, 0). =

2

4 polar axis

15. The density is ρ = k/r.  π/2  2+2 cos θ  π/2  2+2 cos θ k m= rdrdθ = k drdθ r 0 2 0 2  π/2 π/2 =k 2 cos θdθ = 2k(sin θ)|0 = 2k 2+2 cos θ  π/2  2+2 cos θ  0π/2  2+2 cos θ  1 k x rdrdθ = k r cos θdrdθ = k 0π/2 r2 cos θdθ My = r 2 2 0 2 0 2  π/2  π/2 1 2 (8 cos θ + 4 cos θ) cos θdθ = 2k (2 cos2 θ + cos θ − sin2 θ cos θ)dθ = k 2 0 0 π/2 1 3π + 4 π 2 1 k = 2k( + ) = = 2k(θ + sin 2θ + sin θ − sin3 θ) 2 3 2 3 3 0

14.5. DOUBLE INTEGRALS IN POLAR COORDINATES  Mx =

π/2 0



2+2 cos θ 2

k y rdrdθ = k r



 π/2



2+2 cos θ 2

r sin θdrdθ = k 

913 π/2 0

2+2 cos θ 1 2 r sin θdθ 2 2

 π/2 1 1 4 2 2 3 = k (8 cos θ + 4 cos θ) sin θdθ = k −4 cos θ − cos θ 2 0 2 3 0    4 1 8 = k − −4 − = k 2 3 3 3π + 4 4 (3π + 4)k/3 8k/3 = ; y = Mx /m = = x = My /m = 2k 6 2k 3 The center of mass is ((3π + 4)/6, 4/3). 2+2 cos θ  π  2+2 cos θ  π  π 1 2 r 16. m = krdrdθ = k dθ = 2k (1 + cos θ)2 dθ 2 0 0 0 2 0 0  π = 2k (1 + 2 cos θ + cos2 θ)dθ 4 polar 0   π axis 1 1 = 2k θ + 2 sin θ + θ + sin 2θ = 3πk 2 4 0 2+2 cos θ  π  2+2 cos θ  π  2+2 cos θ  π 1 3 r My = kxrdrdθ = k r2 cos θdrdθ = k cos θdθ 0 0 0 0 0 3 0  π  π 8 8 = k (1 + cos θ)3 cos θdθ = k (cos θ + 3 cos2 θ + 3 cos3 θ + cos4 θ)dθ 3 0 3 0     π  3 3 3 1 1 8 θ + sin 2θ + (3 sin θ − sin3 θ) + θ + sin 2θ + sin 4θ = k sin θ + 3 2 4 8 4 32 0   15 8 π = 5πk = k 3 8 2+2 cos θ  π  2+2 cos θ  π  2+2 cos θ  π 1 3 Mx = r kyrdrdθ = k r2 sin θdrdθ = k sin θdθ 0 0 0 0 0 3 0  π  π 8 8 = k (1 + cos θ)3 sin θdθ = k (1 + 3 cos θ + 3 cos2 θ + cos3 θ) sin θdθ 3 0 3 0     π  8 15 1 8 1 32 = k − cos θ − 32 cos2 θ − cos3 θ − cos4 θ = k − − k = 3 4 3 4 4 3 0 5πk 32k/3 = 5/3; y = Mx /m = = 32/9π. The center of mass is (5/3, 32/9π). x = My /m = 3πk 3πk a  2π  a  2π  a  2π 1 4 2 2 2 3 r sin θ dθ 17. Ix = y krdrdθ = k r sin θdrdθ = k 4 0 0 0 0 0 0  2π  4  2π 4 4 1 1 ka kπa ka a polar θ − sin 2θ = sin2 θdθ = = axis 4 0 4 2 4 4 0 

π/2

CHAPTER 14. MULTIPLE INTEGRALS

914  18. Ix =

2π 0



 2π  a 1 r3 rdrdθ = sin2 θdrdθ 4 1 + r4 0 0 0 1+r a  2π  1 1 1 1 ln(1 + r4 ) sin2 θdθ = ln(1 + a4 ) θ − sin 2θ 4 4 2 4



2π

= 0

a

y2

0

π = ln(1 + a4 ) 4

a polar axis

0

19. Solving a = 2a cos θ, cos θ = 1/2 or θ = π/3. The density is k/r3 . Using symmetry,  π/3  2a cos θ  π/3  2a cos θ k cos2 θdrdθ Iy = 2 x2 3 rdrdθ = 2k r 0 a 0 a  π/3 = 2k (2a cos3 θ − a cos2 θ)dθ

a

2a

polar axis

0

 π/3  1 1 2 = 2ak 2 sin θ − sin3 θ − θ − sin 2θ 3 2 4 0  √  √ √ √ 5ak 3 akπ 3 π 3 − − = − = 2ak 3− 4 6 8 4 3 20. Solving 1 = 2 sin 2θ, we obtain sin 2θ = 1/2 or θ = π/12 and θ = 5π/12.  Iy =

5π/12 π/12

 =



2 sin 2θ 1

2

x sec θrdrdθ = 2 sin 2θ

5π/12 1 4

π/12



2

r 4

1

 dθ = 4

5π/12 π/12

5π12



2 sin 2θ 1

r3 drdθ

1

polar axis

sin4 2θdθ

π/12

 5π/12 3 1 1 θ − sin 4θ + sin 8θ =2 4 4 32 π/12  √ √   √ √  √ π 8π + 7 3 3 3 3 3 5π + − − − + = =2 16 8 64 16 8 64 16 

21. From Problem 17, Ix = kπa4 /4. By symmetry, Iy = Ix . Thus I0 = kπa4 /2. a polar axis

14.5. DOUBLE INTEGRALS IN POLAR COORDINATES  22. The density is ρ = kr. I0 =

π



θ



2

π

915 θ 1 5 r dθ 5 0

r (kr)rdrdθ = k 0   π 1 1 kπ 6 1 θ6 = θ5 dθ = k = k 5 0 5 6 30 0  3  1/r  3  1/r k 23. The density is ρ = k/r. I0 = r2 rdθdr = k f 2 dθdr r 1 0 1 0  3   3   1 1 2 r = 4k r2 =k dr = k r 2 1 1 2a cos θ  π  2a cos θ  π  π 1 r4 24. I0 = r2 krdrdθ = k dθ = 4ka4 cos4 θdθ 0 0 0 4 0  π0    3 3π 1 1 3kπa4 4 θ + sin 2θ + sin 4θ = 4ka4 = 4ka = 8 4 32 8 2 0  3  √9−x2   π 25. x2 + y 2 dydx = 03 |r|rdrdθ −3

0



0 π

0

0



π

= √



√ 2 2/2  1−y

26. y

0

0



y2 x2 + y 2

3  1 3 r dθ = 9 3 0

π/4 



dxdy = 

0



π/4

1 0 1

= 0

0

 28.

√

x

√

− x

√

 0

π−x2

2

2

sin(x + y )dydx =

3

polar axis

2a

polar axis r=3

0

dθ = 9π

r2 sin2 θ rdrdθ |r|

3 polar axis

r=1

r2 sin2 θdrdθ 1

π/4



π 0



√ 0

x

(sin r2 )rdrdθ

√ x π 1 = − cos r2 dθ 2 0 0  π 1 =− (−1 − 1)dθ = π 2 0 

polar axis

π

1  1 3 2 1 π/4 2 r sin θ dθ = = sin θdθ 3 3 0 0 0 π/4 1 1 1 π−2 = ( θ − sin 2θ) = 3 2 4 24 0 √ 1  1  1−y2  π/2  1  π/2 2 2 2 1 r2 e dθ 27. ex +y dxdy = er rdrdθ = 2 0 0 0 0 0 0  π/2 π(e − 1) 1 (e − 1)dθ = = 2 0 4 

1

polar axis

r=1

1

polar axis

r=Mπ

Mπ

polar axis

CHAPTER 14. MULTIPLE INTEGRALS

916



1

29.

√



√

0

4−x2

1−x2

x2 dydx = x2 + y 2

 

2 1

√



4−x2

0 π/2  2



π/2



1 2

= 0

1

r=2 1

r2 cos2 θ rdrdθ r2

= 0

x2 dydx x2 + y 2

2 polar axis

r cos2 θdrdθ

2  1 2 3 π/2 2 r cos θdθ = = cos2 θdθ 2 2 0 0 1  π/2  1 3 1 3π θ + sin 2θ = = 2 2 4 8 0 



1

 √2y−y2

30. 0

0

π/2

r=csc θ 2

2

(1 − x − y )dxdy 

π/4 

2 sin θ

= 0

0

r=2 sin θ

(1 − r2 )rdrdθ +



π/2  π/4

csc θ 0

polar axis

1

(1 − r2 )rdrdθ

 2 sin θ  csc θ  π/2  1 2 1 4 1 2 1 4 dθ + dθ r − r r − r = 2 4 2 4 0 π/4 0 0   π/2   π/4 1 1 2 4 2 4 csc θ − csc θ dθ (2 sin θ − 4 sin θ)dθ + = 2 4 0 π/4      π/2 3 1 1 1 1 1 3 θ − sin 2θ + sin 4θ + − cot θ − (− cot θ − cot θ) = θ − sin 2θ − 2 2 8 2 4 3 π/4      1 π 1 1 16 − 3π = − + + 0− − + = 8 2 4 12 24 



5

√



31.

π/4 



25−x2

π



(4x + 3y)dydx = −5

0



0 π



5 0

r=5

(4r cos θ + 3r sin θ)rdrdθ

0

= 0

5

5

5

(4r2 cos θ + 3r2 sin θ)drdθ

polar axis

 5 4 3 r cos θ + r3 sin θ dθ 3 0 0   π  π 500 500 cos θ + 125 sin θ dθ = sin θ − 125 cos θ = 250 = 3 3 0 0 

π



=

1

r=1

1

polar axis

14.5. DOUBLE INTEGRALS IN POLAR COORDINATES 

1

 √1−y2

32. 0

0



2

33. I = 

∞



0

 π/2  1 1 1  rdrdθ dxdy = 2 2 1+ x +y 0 0 1+r 1  π/2  1  π/2 1 = )drdθ = (1 − [r − ln(1 + r)] dθ 1 + r 0 0 0 0  π/2 π (1 − ln 2)dθ = (1 − ln 2) = 2 0 ∞

0

e

π/2

lim

=

t→∞

0

−(x2 +y 2 )



dxdy = 2

 R



1 1 − e−t + 2 2

  34.

917

(x + y)dA =

π/2



0

 =

0

dθ =

 

∞ 0

e

π/2 0

−r 2

 rdrdθ =

π/2 0

 (r cos θ + r sin θ)rdrdθ =

2 1 3 r (cos θ + sin θ) 3



t 1 −r2 lim − e dθ t→∞ 2 0

√

π 1 π dθ = ; I = 2 4 2

2 2 sin θ

π/2 0



π/2

2 sin θ

8 dθ = 3



π/2 0 π/2

0



2 2 sin θ

r2 (cos θ + sin θ)drdθ

(cos θ + sin θ − sin3 θ cos θ − sin4 θ)dθ

 π/2 1 3 8 1 3 ] sin θ − cos θ − sin4 θ + sin3 θ cos θ − θ + sin 2θ 3 4 4 8 16 0    8 1 3π 28 − 3π = 1− − − (−1) = 3 4 16 6 =

35. The volume of the cylindrical portion of the tank is Vc = π(4.2)2 19.3 ≈ 1069.56m3 . We take the equation of the ellipsoid to be x2 x2 5.15  + = 1 or z = ± (4.2)2 − x2 − y 2 . 2 2 (4.2) (5.15) 4.2 The volume of the ellipsoid is 

     5.15 10.3 2π 4.2 (4.2)2 − x2 − y 2 dxdy = [(4.2)2 − r2 ]1/2 rdrdθ 4.2 4.2 0 0  R  4.2   2π 1 1 2 10.3 10.3 2π [(4.2)2 − r2 ]3/2 − dθ = = (4.2)3 dθ 4.2 0 2 3 4.2 3 0 0

Ve = 2

=

2π 10.3 (4.2)3 ≈ 380.53. 3 4.2

The volume of the tank is approximately 1069.56 + 380.53 = 1450.09m3 . 36. (a) With b > 2 we have

CHAPTER 14. MULTIPLE INTEGRALS

918   C

IdA =

lim

R→∞



2π



0

R 0

r dr (r + c)b

u = r + c, du = dr

 2  R+c  R+c r −b u−c r1−b 1−b −b −c = πa du = πa (u − cu )du = πa ub 2−b 1 − b c c c  2−b    c c2 − b (R + c)2−b c(R + c)1−b − − = πa − πa b−2 b−1 b−2 b−1   1 πa c = − πa − . (b − 1)(b − 2)cb−2 (b − 2)(R + c)b−2 (b − 1)(R + c)b−1

  (b)

1 2

C



R+c

I(r)dA =

πa (b − 1)(b − 2)cb−2

(c) Identifying a = 68.585, b = 2.351, and c = 0.248 in part b we find that the total number of infections in the plane is approximately 741.25. 

  37. (a) P =

C

D(r)dA =

2π



0

R

D0 e−r/d rdrdθ = 2πD0

0



R 0

re−r/d dr

R = 2πD0 (−dre−r/d − d2 er/d ) = 2πdD0 [d − (R + d)e−R/d ] 0

(b) Using 

  C

rD(r)dA =

2π 0



R 0

rD0 e

−r/d

 rdrdθ = 2πD0

R 0

r2 e−r/d dr

R = 2πD0 (−2d3 e−r/d − 2d2 re−r/d − dr2 e−r/d ) 0

 2 2 2 −R/d = 2πdD0 2d − (R + 2dR + 2d )e

we have  rD(r)dA 2d2 − (R2 + 2dR + 2d2 )e−R/d  C = D(r)dA d − (R + d)e−R/d C (c) Letting R −→ ∞ in the result of parts (a) and (b) we find that the total population is 2πd2 D0 and the average commute for the total population is 2d2 /d = 2d. 38. In the first case, let the circle centered at (D/2, 0) be described by the equation r = D cos θ for −π/2 ≤ θ ≤ π/2 and assume that the snow is plowed to the origin. Then  

 D3 π/2 rdA = r drdθ = (1 − sin2 θ) cos θdθ 3 R −π/2 0 0  π/2  3 4D3 2D 1 . = = sin θ − sin3 θ 3 3 9 0 

π/2



D cos θ

2

14.6. SURFACE AREA

919

In the second case, let the circle centered at the origin be described by the equation r = D/2 for 0 ≤ θ ≤ 2π, and assume the snow is plowed to the origin. Then  

 R

rdA =

2π 0



D/2 0

r2 drdθ =

D/2 2π 3 πD3 r . = 3 12 0

3

4D /9 16 = ≈ 1.698, which means that plowing snow to one πD3 /12 3π point on the perimeter is approximately 69.8% more costly than plowing to the center. The ratio of these integrals is

PROBLEMAS 5.6 Area 14.6 Surface 1. Letting z = 0, we have 2x + 3y = 12. Using f (x, y) = z = 3 1 3 29 1 . 3− x− y we have fx = − , fy = − , 1+fx2 +fy2 = 2 4 2 4 16 Then √  6  6  4−2x/3  29 2 A= 29/16dydx = (4 − x)dx 4 3 0 0 0 6 √ √ √ 1 29 29 = (4x − x2 ) = (24 − 12) = 3 29. 4 3 4

y 4 y=4-2x/3

6 x

0

2. We see from the graph in Problem 1 that the plane is entirely above the region bounded by r = sin 2θ in the 1 3 first octant. Using f (x, y) = z = 3 − x − y we have 2 4 1 3 29 2 2 . Then fx = − , f y = − , 1 + f x + fy = 2 4 16 sin 2θ √  π/2  π/2  sin 2θ  29 1 2 r A= 29/16rdrdθ = dθ 4 2 0 0 0 0 π/2 √  π/2 √ √ 1 29 29 1 29π 2 = ( θ − sin 4θ) . sin 2θdθ = = 8 8 2 8 32 0 0 √ 3. Using f (x, y) = z = 16 − x2 we see that for 0 ≤ x ≤ 2 and 0 ≤ y ≤ 5, z.0. Thus, the surface is entirely above the x region. Now fx = − √ , fy = 0, 1 + fx2 + fy2 = 16 − x2 x2 16 1+ = and 2 16 − x 16 − x2 2  5  5  5 2 4 π 10π −1 x √ dy = . dxdy = 4 sin dy = 4 A= 2 4 0 3 16 − x 0 0 0 0 6

r=sin 2 θ

1

polar axis

y 5

x=2

2

x

CHAPTER 14. MULTIPLE INTEGRALS

920

4. The region in the xy-plane beneath the surface is bounded by the graph of x2 + y 2 = 2. Using f (x, y) = z = x2 + y 2 we have fx = 2x, fy = 2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then,  A=



2π 0

1 = 12

√

0 2π

 0



2

 1 + 4r2 rdrdθ =

2π 0

r=M2

M2 polar

√2 1 2 3/2 (1 + 4r ) dθ 12

axis

0

13π . (27 − 1)dθ = 3

5. Letting z = 0 we have x2 + y 2 = 4. Using f (x, y) = z = 4 − (x2 + y 2 ) we have fx = −2x, fy = −2y, 1 + fx2 + fy2 = 1 + 4(x2 + y 2 ). Then  A= =

2π 0

1 12



2

0 2π

 0

  2 1 + 4r rdrdθ =

(173/2 − 1)dθ =

2π 0

r=2

2 1 2 3/2 (1 + 4r ) dθ 3 0

2 polar axis

π (173/2 − 1). 6

6. The surfaces x2 + y 2 + z 2 = 2 and z 2 = x2 + y 2 intersect on the cylinder 2x2 + 2y 2 = 2 or x2 + y 2 = 1. There are portions of the sphere within the cone  both above and below the xy-plane. Using f (x, y) = 2 − x2 − y 2 we have x y , fy = −  , 1 + fx2 + fy2 = fx = −  2 − x2 − y 2 2 − x2 − y 2 2 . Then 2 − x2 − y 2

 √ 1 2π  1 √  2π  2 2 √ rdrdθ = 2 2 − 2 − r dθ A=2 2 2−r 0 0 0 0  2π √ √ √ √ =2 2 ( 2 − 1)dθ = 4π 2( 2 − 1).

r=1

1 polar axis

0



7. Using f (x, y) = z = 25 − x2 − y 2 we have x y , fy = −  , fx = −  25 − x2 − y 2 25 − x2 − y 2 25 1 + fx2 + fy2 = . Then 25 − x2 − y 2  5  √25−y2 /2 5  dxdy A= 25 − x2 − y 2 0 0 √25−y2 /2  5  5 25π x π =5 dy = . sin−1  dy = 5 2 6 25 − y 0 0 6 0

y 5

x=M25-y2/2

3

x

14.6. SURFACE AREA

921

8. In the first octant, the graph of z = x2 −y 2 intersects the xyplane in the line y = x. The surface is in the first octant for x > y. Using f (x, y) = z = x2 − y 2 we have fx = 2x, fy = −2y, 1 + fx2 + fy2 = 1 + 4x2 + 4y 2 . Then  A=

π/4 0

1 = 12



2



0



π/4 0

 1 + 4r2 rdrdθ =

π/4 0

r=2

2 1 (1 + 4r2 )3/2 dθ 12

2 polar axis

0

π (173/2 − 1). (173/2 − 1)dθ = 48

9. There are portions of the sphere within the cylinder both above and below the xy-plane. Using f (x, y) = z =  x a2 − x2 − y 2 we have fx = −  , fy = 2 1 − x2 − y 2 a2 y , 1 + fx2 + fy2 = 2 . Then, using − a − x2 − y 2 a2 − x2 − y 2 symmetry,

r=a sin θ

a

polar axis

a sin θ

 π/2  a 2 2 √ dθ rdrdθ = 4a − a −r a2 − r 2 0 0 0 0  π/2  π/2  (a − a 1 − sin2 θ)dθ = 4a2 (1 − cos θ)dθ = 4a

 A=2 2

0

π/2



a sin θ

0

π/2 π = 4a2 (θ − sin θ) 0 = 4a2 ( − 1) = 2a2 (π − 2). 2 10. There are portions of the cone within the cylinder both above and 1 2 below the xy-plane. Using f (x, y) = x + y 2 , we have fx = 2 x y 5  , fy =  , 1 + fx2 + fy2 = . Then, using 4 2 x2 + y 2 2 x2 + y 2 symmetry, 2 cos θ 

 π/2  2 cos θ √  π/2 1 2 5 A=2 2 rdrdθ = 2 5 r dθ 4 2 0 0 0 0  π/2  √  π/2 √ √ 1 1 θ + sin 2θ =4 5 cos2 θdθ = 4 5 = 5π. 2 4 0 0

r=2 cos θ

2

polar axis

CHAPTER 14. MULTIPLE INTEGRALS

922

11. There are portions of the surface in each octant with areas equal  to the area of the portion in the first octant. Using f (x, y) = z = a2 − y 2 y a2 . Then we have fx = 0, fy =  , 1 + fx2 + fy2 = 2 a − y2 a2 − y 2  A=8

a

y x=Ma2-y2

a

 √a2 −y2

x

a  dxdy a2 − y 2 √a2 −y2  a x  dy = 8a dy = 8a2 . a 2 − y 2 0 0

0

0



a

= 8a 0

12. √ From Example 1, the area of the portion of the hemisphere √ with x2 + y 2 = b2 is 2πa(a − 2 2 a − b ). Thus, the area of the sphere is A = 2 lim 2πa(a − a2 − b2 ) = 2(2πa2 ) = 4πa. b→a

13. The projection of the surface onto √ the xy-plane is shown in the graph. Using f (x, z) = y = a2 − x2 − z 2 we have x z , fz = − √ , 1 + fx2 + fx = − √ 2 2 2 2 a −x −z a − x2 − z 2 a2 fz2 = 2 . Then a − x2 − z 2 √ a2 −C12  2π  √a2 −c21  2π  a A= rdrdθ = a − a2 − r2 √ dθ √ 2 2 √ 2 2 2 a − r2 a −c2

0



=a

2π 0

0

z

r=Ma2-c12 r=Ma2-c22

x

a −c2

(c2 − c1 )dθ = 2πa(c2 − c1 ).

14. The surface area of the cylinder x2 + z 2 = a2 from y = c1 to y = c2 is the area of a cylinder of radius a and height c2 − c1 . This is 2πa(c2 − c1 ). 15. The equations of the spheres are x2 + y 2 + z 2 = a2 and x2 + y 2 + (z + a)2 = 1. Subtracting these equations, we obtain(z − a)2 − z 2 = 1 − a2 or −2az + a2 = 1 − a2 . Thus, the spheres intersect on the plane z = a − 1/2a. The region of integration is x2 + y 2 + (a − 1/2a)2 = a2 or r2 = 1 − 1/4a2 . The area is  2π  √1−1/4a2 √1−1/4a2 2 2 −1/r 2 2 1/2 A=a (a − r ) rdrdθ = 2πa[−(a − r ) ] 0 0 0 ⎞ ⎛     1/2  2 1/2 1 1 ⎠ = π. = 2πa ⎝a − a− = 2πa a − a2 − 1 − 2 4a 2a

16.

(a) Both states span 7 degrees of longitude and 4 degrees of latitude, but Colorado is larger because it lies to the south of Wyoming. Lines of longitude converge as they go north, so the east-west dimensions of Wyoming are shorter than those of Colorado.

14.7. THE TRIPLE INTEGRAL

923

 (b) We use the function f (x, y) = R2 − x2 − y 2 to describe the northern hemisphere, where R ≈ 3960 miles is the radius of the Earth. We need to compute the surface area over a polar rectangle P of the form θ1 ≤ θ ≤ θ2 , R cos φ2 ≤ r ≤ R cos φ1 . We have −x −y fx =  and fy =  2 2 2 2 2 2 R −x −y  R −x −y  x2 + y 2 so that 1 + fx2 + fy2 = 1+ 2 = R − x2 − y 2 R √ . 2 R − r2 Thus    A=

P

 1 + fx2 + fy2 dA =

θ2



θ1

R cos φ1 R cos φ2

√

φ2 φ1

R R θ2

θ1

R rdrdθ R2 − r 2

R cos φ2  = (θ2 − θ1 )R R2 − r2 = (θ2 − θ1 )R2 (sin φ2 − sin φ1 ). R cos φ1

The ratio of Wyoming to Colorado is then sin 45◦ − sin 41◦ ≈ 0.941. Thus Wyoming is about 6% sin 41◦ − sin 37◦ smaller than Colorado. (c) 97,914/104,247 ≈ 0.939, which is close to the theoretical value of 0.941. (Our formula for the area says that the area of Colorado is approximately 103,924 square miles, while the area of Wyoming is approximately 97,801 square miles.

PROBLEMAS 5.7 14.7 The Triple Integral 1.

1 42 1 2 dydz x ( + xy + xz) 2 −2 2 −2 2 −1 2 4 4 2  42 4 = 2 −2 (2y + 2z)dydz = 2 (y + 2yz) dz = 2 8zdz = 4z 2 2 = 48 42 1

(x + y + z)dxdydz = −1

−2



3



x





xy

2.

3



x

24xydzdydx = 1

1

2

1

1

xy   x 24xyz dydx = 13 (24x2 y 2 − 48xy)dydx 2

1

x  3 2 3 2 = (8x y − 24xy ) dx = (8x5 − 24x3 − 8x2 + 24x)dx 1 1 1   3 4 6 1552 8 3 14 4 2 x − 6x − x + 12x = 522 − = = 3 3 3 3 1 

3

CHAPTER 14. MULTIPLE INTEGRALS

924 

6





6−x

6−x−z

3. 0



0

1



0



1−x

√

0

4. 0



0

π/2



y2



y

5. 0

0

0

y

6−x 1 2 dydzdx = (6 − x − z)dzdx = (6z − xz − z ) dx 2 0 0 0 0   6  6 1 1 (18 − 6x + x2 )dx = 6(6 − x) − x(6 − x) − (6 − x)2 dx = 2 2 0 0   6 1 = 18x − 3x2 + x3 = 36 6 0 

6





6−x

6

√y  1  1−x 4x z dzdydx = x z dydx = x2 y 2 dydx 0 0 0 0 0 1−x  1   1 2 3 1 1 2 1 1 2 = x y dx = x (1 − x)3 dx = (x − 3x3 + 3x4 − x5 )dx 3 0 3 0 0 3 0  1  1 1 3 3 4 3 5 1 6 1 x − x + x − x = = 3 3 4 5 6 180 0 

2 3

1



1−x

2 4

y 2  π/2  y2  π/2 x x x cos dzdxdy = y cos dxdy = y 2 sin dy y y y 0 0 0 0  π/2 y 2 sin ydy Integration by parts = 0

π/2 = (−y 2 cos y + 2 cos y + 2y sin y) 0 = π − 2 √



2



6.

√

0



2 y

ex 0

√



2

xdzdxdy =

2

0



2 √

2

y

xex dxdy =

√

 0

2

2 1 x2 e (e4 − ey )dy 2 √ y

√2 √ √ √ √ 1 1 1 4 y = (ye − e ) = [(e4 2 − e 2 ) − (−1)] = (1 + e4 2 − e 2 ) 2 2 2 0 

1



1



2−x2 −y 2

7. 0

8.

0

0

2−x2 y2  1 1 2 2 xye dzdxdy = xye dxdy = (xye2−x −y − xy)dxdy 0 0 0 0 0  1   1  1 1 2−y2 1 2−x2 −y2 1 2 1 1−y2 1 − x y dy = − y + ye − ye − ye dy = 2 2 2 2 2 0 0 0   1   1 1−y2 1 2 1 2−y2 1 1 1 1 1 2 1 1 e − − e) − ( e − e = − y − e = = e2 − e 4 4 4 4 4 4 4 4 4 2 0 z



1



1

z

x 2  4  1/2 −1 y sin dxdz = 0 0 sin−1 xdxdz 0 0 0 0 0 2 2 x 0 x −y   √ 1/2 √ √  4  π π 3 1 4 −1 + − 1 dz = + 2 3 − 4 = 0 x sin x + 1 − x2 dz = 0 26 2 3 0  4  1/2  x2



1

dydxdz =

 4  1/2

Integration by parts

14.7. THE TRIPLE INTEGRAL   



9. D

zdV =

5 0



0

3 1

5

=





925 

y+2

zdxdydz =

y

3  2yz dz = 1

5



3

z

2xdydz 0

5

1

5 4zdz = 2z 2 0 = 50

5 0

3 y

x=y

3 x=y+2 x

10. Using    symmetry, (x2 + y 2 )dV = 2 D

2



x2

0

4

(x2 + y 2 )dzdydx

4−y (x + y )z dydx 2

2



4



0 2



x 4

=2 x2

0

4−y 0

 =2

z



4

2

2

0

2

2

2

4

2 x

3

(4x − x y + 4y − y )dydx

y

y=x2

4 1 2 2 4 3 1 4 =2 (4x y − x y + y − y ) dx 2 3 4 0 x2 2 4 5 8 3 64 5 7 1 9 23, 552 . = 2( x + x − x − x + x ) = 3 3 5 42 36 315 0 

2

2

z

11. The other five integrals are  4  2−x/2  4  4  z  (z−x)/2 f (x, y, z)dzdydx, 0 0 0 f (x, y, z)dydxdz, 0 0 x+2y  4  z/2  z−2y  4  4  (z−x)/2 f (x, y, z)dydzdx, 0 0 f (x, y, z)dxdydz, 0 02 x4 0 z−2y f (x, y, z)dxdzdy. 0 2y 0

4 z=2y z=x

2

y

x+2y=4

4 x

12. The √ other five integrals are  3  36−4y2 /3  3 f (x, y, z)dzdxdy, 03 02  √36−9x2 /21 f (x, y, z)dydxdz, 1 0 0  3  3  √36−4y2 /3 f (x, y, z)dxdydz, 1 0 0  3  3  √36−4y2 /3 f (x, y, z)dxdzdy, 02 13 0√36−9x2 /2 f (x, y, z)dydzdx. 0 1 0 13.

28 4  8  4  y1/3 (a) V = 0 x3 0 dzdydx (b) V = 0 0 0 dxdz 428 (c) V = 0 0 x2 dydxdz

z

3

y=M36-9x2 /2

3 2

x=M

36-4y2 /3

x

y

CHAPTER 14. MULTIPLE INTEGRALS

926

√ 14. Solving z = x and x + z = 2, we obtain x = 1, z = 1.  3  1  2−z  1  2−z  3 (a) V = 0 0 z2 dxdzdy (b) V = 0 z2 dydxdz 0 √ 31 x  3  2  2−x (c)V = 0 0 0 dzdxdy + 0 1 0 dzdxdy 15.

16.

z

z

5 3

x=2-2z/3

4

2

y

3

x

y

x=M9-y2

3 x

The region in the first octant is shown. 17.

18.

z

z

6

4 y=-M1-x2

2 y=M1-x2

2

y

x

2

2

y

y=M4-x2

x

19.

20.

z

z

1

2

3

3 1 x

2

y

x

y

14.7. THE TRIPLE INTEGRAL

927

√ 21. Solving x = y 2 and 4 − x = y 2 , we obtain x = 2, y = ± 2. Using symmetry,  V =2

0 3



=2 0

22. V =

2





4−y 2

dxdydz = 2

y2

 √2  2 3 4y − y dz = 2 3 0

√



0

2

0





√



3

4−x2



2

0

3

(4 − 2y 2 )dydz

√ 8 2 dz = 16 2. 3

2



y

x=4-y2

5

 dzdydx =

2 0

√



4−x2

0

x+y z dydx

z

2

0

 √4−x2 1 2 (x + y)dydx = dx = xy + y 2 0 0 0 0    2  2  1 1 1 3 2 2 3/2 2 = x 4 − x + (4 − x ) dx = − (4 − x ) + 2x − x 2 3 6 0 0     4 8 16 = 4− − − = 3 3 3 

√

4

x=y2

√

x

x+y 0

0

0

0

√



3

z

5



4−x2

2



2

2

y=M

y

4-x2

x

z

23. Adding the two equations, we obtain 2y = 8. Thus, the paraboloids intersect in the plane y = 4. Their intersection is a circle of √ radius 2. Using symmetry,  2  4−x2  8−x2 −z2 dydzdx V =4 

0

2



0

√

x2 +z 2

4−x2

=4 

0

0

2



2

(8 − 2x − 2x )dzdx = 4

2

2 0



2 z=M4-x2

8 y

2 x

 √4−x2 2 3 dx 2(4 − x )z − z 3 0 2

4 (4 − x2 )3/2 dx Trig substitution =4 0 3  2  16 x −1 x 2 2 = = 16π. − (2x − 20) 4 − x + 6 sin 3 8 2 0 24. Solving x = 2, y = x, and z = x2 + y 2 , we obtain the point (2,2,8).  2 x  2  x  x2 +y2 dzdydx = (x2 + y 2 )dydx V = 0



0

2

= 0

0

x  1 3 2 (x y + y ) dx = 3 0

25. We are given ρ(x, y, z) = kz.

0

2 0

z

(2,2,8)

0

2 4 3 1 4 16 x dx = x = . 3 3 0 3

2 x

2 y=x

y

CHAPTER 14. MULTIPLE INTEGRALS

928

y1/3  8 4 m= kzdxdzdy = k xz dzdy = k y 1/3 zdzdy 0 0 0 0 0 0 0 0 4   8  8  8 3 4/3 1 1/3 2 y z dy = 8k y y 1/3 dy = 8k =k = 96k 4 0 2 0 0 0 y1/3  8  4  y1/3  8 4  8 4 2 2 kz dxdzdy = k xz dzdy = k y 1/3 z 2 dzdy Mxy = 



8

4

0

y 1/3

0

0

0

0

0

0

0

0

0

0

0

0

0

8



4

4   8  8 3 4/3 1 1/3 3 64 64 1/3 =k y z dy = k k y y dy = = 256k 3 3 4 0 3 0 0 0 1/3 y  8  4  y1/3  8 4  8 4 = kyzdxdzdy = k xyz dzdy = i y 4/3 zdzdy 

Mxz





8

0

0

4   8  8 3 7/3 1 4/3 2 3072 =k y z dy = 8k y y 4/3 dy = 8k = 7 k 2 7 0 0 0 0 y1/3  8 4  8  4  y1/3  8 4 1 2 1 x z Myz = kxzdxdzdy = k dzdy = k y 2/3 zdzdy 2 0 0 0 0 2 0 0 0 0 4   8  8  8 3 5/3 1 2/3 2 384 1 2/3 y z dy = 4k y k y dy = 4k = = k 2 0 2 5 5 0 0 0 384k/5 3072k/7 256k = 4/5; y = Mxz /m = = 32/7; z = Mxy /m = = 8/3 x = Myz /m = 96k 96k 96k The center of mass is (4/5, 32/7, 8/3). 

8

26. We use the form of the integral in Problem 14(b) of this section. Without loss of generality, we take 1.   1ρ=2−z  1  2−z  1 3 m= dydxdz = 3dxdz = 3 (2 − z − z 2 )dz z2

0

0

0

z2

1 7 1 2 1 3 = 3(2z − z − z ) = 2 3 2 0  1  2−z  3  zdydxdz = Mxy =

Mxz

Myz

0

3  1  2−z yz dxdz = 3zdxdz 0 z2 0 0 z2 0 z2 0   1  1  1 2−z 1 3 1 4 5 2 3 2 xz dz = 3 (2z − z − z )dz = 3 z − z − z = =3 3 4 4 0 0 0 z2 3    1  2−z  3  1  2−z 1 2 9 1 2−z y dxdz = = ydydxdz = dxdz 2 0 2 0 z2 0 z2 0 0 z2 1  1 2 1 3 9 1 9 21 2 = (2 − z − z )dz = (2z − z − z ) = 2 0 2 2 3 4 0  1  2−z  3  1  2−z 3  1  2−z = xdydxdz = xy dxdz = 3xdxdz z2

0



1

=3 0

0

1

0



2−z

z2

0

0

z2

2−z 1  1 2 3 1 3 1 3 1 5 16 2 4 2 x dz = (4 − 4z + z − z )dz = (4z − 2z + z − z ) = 2 z 2 2 0 2 3 5 5 0

14.7. THE TRIPLE INTEGRAL

929

16/5 21/4 5/4 = 32/35, y = Mxz /m = = 3/2, z = Mxy /m = = 5/14. 7/2 7/2 7/2 The centroid is (32/35, 3/2, 5/14).

x = Myz /m =

27. The density is ρ(x, y, z) = ky. Since both the region and the density function are symmetric with respect to the xyand yz-planes, x = z = 0. Using symmetry, √  3  2  √4−x2  3  2 4−k2 m=4 kydzdxdy = 4k yz dxdy 0

0

0



3



2

= 4k 

0

0

y



4 − x2 dxdy = 4k 

3

= 4k 0

0



πydy = 4πk

3 0

0

z z=M4-x2

2 3

2

y

x

0

x  2 −1 x 2 y 4 − x + 2 sin dy 2 2 0

 3 1 2 y = 18πk 2 0

√4−x2  3 2  ky 2 dzdxdy = 4k y 2 z dxdy = 4k y 2 4 − x2 dxdy Mxz = 4 0 0 0 0 0 0 0 0   3  3  3    2 1 3 x x −1 2 2 2 y = 36πk. y 4 − x + 2 sin dy = 4k πy dy = 4πk = 4k 2 2 0 3 0 0 0 36πk = 2. The center of mass is (0,2,0). y = Mxz /m = 18πk 

3



2



√

28. The density is ρ(x, y, z) = kz.  1  x  y+2  m= kzdzdydx = k 0

= = =

1 k 2 1 k 2 1 k 6

x2 0  1 x 0

1



0

x x2

3



2

z

y+2 1 2 z dydx 2

(y + 2)2 dydx x 1 (y + 2)3 dx 3 x2

2

0

x2



1



0 1 0



4−x2

[(x + 2)3 − (x2 + 2)3 ]dx =

y=x

1

1 k 6



1 0

y=x2 x

[(x + 2)3 − (x6 + 6x4 + 12x2 + 8)]dx

  1 1 6 407 1 1 k = k (x + 2)4 − x7 − x5 − 4x3 − 8x = 6 4 7 5 840 0 y+2  1 x  1  x  y+2  1 x 1 3 1 Mxy = z kz 2 dzdydx = k dydx = k (y + 2)3 dydx 3 3 2 2 2 0 x 0 0 x 0 x 0 x  1  1 1 1 1 (y + 2)4 dx = k [(x + 2)4 − (x2 + 2)4 ]dx = k 3 0 4 12 0 x2  1 1 k [(x + 2)4 − (x8 + 8x6 + 24x4 + 32x2 + 16)]dx = 12 0   1 1 1 9 8 7 24 32 3 1493 1 5 k (x + 2) − x − x − − x − 16x = k = 12 5 9 7 5 3 1890 0

y

CHAPTER 14. MULTIPLE INTEGRALS

930

y+2  1 x 1 2 1 yz = kyzdzdydx = k dydx = k y(y + 2)2 dydx 2 0 x2 0 x2 0 0 x2 2 0  x  1 x  1 1 4 4 3 1 1 y + y + 2y 2 dx (y 3 + 4y 2 + 4y)dydx = k = k 2 0 x2 2 0 4 3 x2   1 4 4 1 1 = k − x8 − x6 − 74x4 + x3 + 2x2 dx 2 0 4 3 3   1 1 1 4 7 1 2 68 k = k − x9 − x7 − x5 + x4 + x3 = 2 36 21 20 3 3 315 0 y+2  1 x  1  x  y+2  1 x 1 2 1 xz = kxzdzdydx = k dydx = k x(y + 2)2 dydx 2 2 0 x2 0 0 x2 2 0 x 0 x  1  1 1 1 1 x(y + 2)3 dx = k = k [x(x + 2)3 − x(x2 + 2)3 ]dx 2 0 3 6 0 x2  1 1 [x4 + 6x3 + 12x2 + 8x − x(x2 + 2)3 ]dx = k 6 0   1 3 1 21 1 1 k = k x5 + x4 + 4x3 + 4x2 − (x2 + 2) = 6 5 2 8 80 

Mxz

Myz

1



x





y+2

1



x

0

21k/80 68k/315 = 441/814, y = Mxz /m = = 544/1221, x = Myz /m = 407k/840 407k/840 1493k/1890 = 5972/3663. The center of mass is (441/814,544/1221,5972/3663). z = Mxy /m = 407k/840  29. m =

1



−1

√

1−x2

√ − 1−x2



8−y 2+2y

z

(x + y + 4)dzdydx 8

y=-M1-x2 2

1 y=M1-x2

y

x z

30. Both the region and the density function are symmetric with respect to the xy- and√ yz-planes. Thus,  2  1+z2  √1+z2 −y2 z 2 dxdydz. m=4 −1

0

0

2 2

2 x

31. We are given ρ(x, y, z) = kz.

y=M1+x2 y

14.7. THE TRIPLE INTEGRAL

931

y1/3 1 3 3 Iy = kz(x + z )dxdzdy = k ( x z + xz ) dzdy 0 0 0 0 3 0 0   4  8  8 4 1 2 1 1/3 4 1 yz + y 1/3 z 3 dzdy = k yz + y z dy =k 3 6 4 0 0 0 0    8  8 8 4 2560 y + 64y 1/3 dy = k y 2 + 48y 4/3 = k =k 3 3 3 0 0 √  4 5 2560k/3 From Problem 25, m = 96k. Thus, Rg = Iy /m = = . 96k 3 

8



4



y 1/3

2



2

32. We are given ρ(x, y, z) = k.  1  2−z  3  Ix = k(y 2 + z 2 )dydxdz = k z2

0

0

1

8





2−z z2

0

4



 3  1  2−z 1 3 y + yz 2 dxdz = k (9 + 3z 2 )dxdz 3 2 0 z 0

2−z  1 2 (9x + 3xz ) dz = k (18 − 9z − 3z 2 − 3z 3 − 3z 4 )dz =k 0 0 z2   1 9 2 3 4 3 5 223 3 k = k 18z − z − z − z − z = 2 4 5 20 0  1  2−z  3  1  2−z  1 m= kdydxdz = k 3dxdz = 3k (2 − z − z 2 )dz 

0

1



z2

0

z2

0

 1 7 1 2 1 3 = 3k 2z − z − z = k 2 3 2 0    Ix 223k/20 223 = = Rg = m 7k/2 70  33. Iz = k 

0



0

=k =k

1



1



0



0

1

0

1−x

1−x

1−x

0



1−x−y 0

0

z

(x2 + y 2 )dzdydx

1

(x2 + y 2 )(1 − x − y)dydx 1 2

3

2

2

2

1

3

(x − X − x y + y − xy − y )dydx

y

y=1-x

x

 1−x 1 1 1 =k dx (x2 − x3 )y − x2 y 2 + (1 − x)y 3 − y 4 2 3 4 0 0   1  1 1 6 1 4 1 2 1 4 1 1 5 1 k 3 4 5 =k [ x − x + x + (1 − x) ]dx = k x − x + x − (1 − x) = 2 2 12 6 4 10 60 30 0 0 

1



34. We are given ρ(x, y, z) = kx.  1  2  4−z  Iy = kx(x2 + z 2 )dydxdz = k 0

0



=k

1 0

z



2 0

z

1 0

0

4−z 2 (x3 + xz 2 )y dxdz z

2

x

(x3 + xz 2 )(4 − 2z)dxdz

y=z 1

y=4-z 4

y

CHAPTER 14. MULTIPLE INTEGRALS

932

2 1 4 1 2 2 ( x + x z )(4 − 2z) dz =k 4 2 0 0  1  1 (4 + 2z 2 )(4 − 2z)dz = 4k (4 − 2z + 2z 2 − z 3 )dz =k 

1

0

0



 1 2 1 41 k = 4k 4z − z 2 + z 3 − z 4 = 3 4 3 0  35. We are given ρ(x, y, z) = k x2 + y 2 + z 2 . Both the region and the integrand are symmetric with respect to the yzand xz-planes.  5  √25−x2  5  2 2 2 2 2 Iz = 4 √ 2 2 k(x + y ) x + y + z dzdydx 0

0

z 5

x +y

5 y

5 x

36. We are given ρ(x, y, z) = kz. Both the region and the integrand are symmetric with respect to the xz- and xy-planes.  1  √1−x2  √1−z2 kx(x2 + z 2 )dydzdx Iy = 4 −1

0

0

y=M25-x2

z 1

z=M1-x2

y=M1-x2

1

1 x

PROBLEMAS 5.8Integrals in Other Coordinate Systems 14.8 Triple √ √ √ √ 1. x = 10 cos 3π/4 = −5 2; y = 10 sin 3π/4 = 5 2; (−5 2, 5 2, 5) √ √ 2. x = 2 cos 5π/6 = − 3; y = 2 sin 5π/6 = 1; (− 3, 1, −3) √ √ √ √ 3. x = 3 cos π/3 = 3/2; y = 3 sin π/3 = 3/2; ( 3/2, 3/2, −4) √ √ √ √ 4. x = 4 cos 7π/4 = 2 2; y = 4 sin 7π/4 = −2 2; (2 2, −2 2, 0) 5. x = 5 cos π/2 = 0; y = 5 sin π/2 = 5; (0, 5, 1) √ √ 6. x = 10 cos 5π/3 = 5; y = 10 sin 5π/3 = −5 3; (5, −5 3, 2)

√ 7. With x = 1 and y = −1 we have r2 = 2 and tan θ = −1. The point is ( 2, −π/4, −9) √ √ 8. With x = 2 3 and y = 2 we have r2 = 16 and tan θ = 1/ 3. The point is (4, π/6, 17). √ √ √ √ 9. With x = − 2 and y = 6 we have r2 = 8 and tan θ = − 3. The point is (2 2, 2π/3, 2). √ 10. With x = 1 and y = 2 we have r2 = 5 and tan θ = 2. The point is ( 5, tan−1 2, 7).

y

14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS

933

11. With x = 0 and y = −4 we have r2 = 16 and tan θ undefined. The point is (4, −π/2, 0). √ √ √ 12. With x = 7 and y = − 7 we have r2 = 14 and tan θ = −1. The point is ( 14, −π/4, 3). 13. r2 + z 2 = 25 14. r cos θ + r sin θ − z = 1 15. r2 − z 2 = 1 16. r2 cos2 θ + z 2 = 16 17. z = x2 + y 2 18. z = 2y 19. r cos θ = 5, z = 5 √ √ √ 20. tan θ = 1/ 3, y/z = 1/ 3, z = 3y, x > 0 21. The equations are r2 = 4, r+ z 2 = 16, and z = 0.  2π  2   2π  2  √16−r2 rdzdrdθ = r 16 − r2 drθ V = 0

0



2π

= 0

0

0

2  1 2 3/2 − (16 − r ) dθ = 3 0

z

z=M16-r2

4

0

2π 0

√ √ 2π (64 − 24 3) (64 − 24 3)dθ = 3 2 2

y

r=2

x

22. The equation is z = 10 − r2 .  2π  3  10−r2  V = rdzdrdθ = 0

0



2π



= 0

1

 3  9 2 1 4 r − r dθ = 2 4 0

2π 0

z



3 0

2π 0

10

r(9 − r2 )drdθ

z=10-r2

81 81π dθ = . 4 2 3 3

y

r=3 x

23. The equations are z = r2 , r = 5, and z = 0.  2π  5  r2  2π  5  V = rdzdrdθ = r3 drdθ = 

0

0

2π

= 0

0

0

0

2π 0

z

5 1 4 r dθ 4 0

z=r2

625π 625 dθ = 4 2 5

r=5

5 x

y

CHAPTER 14. MULTIPLE INTEGRALS

934

z 24. Substituting the first equation into the second, we see y=r2 that the surfaces intersect in the plane y = 4. Using polar 2 coordinates in the xz-plane, the equations of the surfaces become y = r2 and y = 12 r2 + 2.   2π  2  2  2π  2  r2 /2+2 2 r 2 + 2 − r drdθ x rdydrdθ = r V = y=r2/2+2 2 0 0 r2 0 0   2  2π  2π   2π  2  1 1 2dθ = 4π 2r − r3 drdθ = r2 − r4 dθ = = 2 8 0 0 0 0 0 √ z 25. The equation is √ z = a2 − r2 . By symmetry, x = y = 0.  2π  a  a2 −r2  2π  a  a rdzdrdθ = r a2 − r2 drdθ m= 0 0 0 0 0 a  2π  2π 1 3 1 2 a dθ = πa3 − (a2 − r2 )3/2 dθ = = a 3 3 3 0 0 0 x √a2 −r2  2π  a  √a2 −r2  2π  a 1 2 rz Mxy = zrdzdrdθ = drdθ 0 0 0 0 2 0 0   1 2π a r(a2 − r2 )drdθ = 2 0 0   a   1 2π 1 4 1 1 2π 1 2 2 1 4 a r − r dθ = a dθ = πa4 = 2 0 2 4 2 4 4 0 0 πa4 /4 = 3a/8. The centroid is (0, 0, 3a/8). z = Mxy /m = 2πa3 /3

2π 

m= =

0

k 2

4 0



2π

5

kxrdxdrdθ = k

0





4

25rdrdθ = 0

25k = 2  Myz =

0



0 2π

25k 2

2π  0



2π 0

4 0

y

z=Ma2-r2

a

4

r=4

5 1 2 rz drdθ 2

4 1 2 r dθ 2 0

0

4 5 x

8dθ = 200kπ 0 2π  4 0



5 0

2

kx rdxdrdθ = k



2π 0



4 0

5  2π  4 1 3 1 rx drdθ = k 125rdrdθ 3 3 0 0 0

4  2π 125 2 1 2000 1 r dθ = k kπ 1000dθ = = k 3 2 3 0 3 0 0 2000kπ/3 = 10/3, The center of mass of the given solid is (10/3, 0, 0). x = Myz /m = 200kπ 

2π

y

z

26. We use polar coordinates in the yz-plane. The density is ρ(x, y, z) = kz. By symmetry, y = z = 0. 

4

y

14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS

935

√ 27. The equation is z√= 9 − r2 and the density is ρ = k/r2 . When x = 2,√r = √ 5.  2π  5  9−r2 Iz = r2 (k/r2 )rdzdrdθ(k/r2 ) 0

0

2π 

 =k

0

√

2π

0  √



2π





0 2π

 =k

0

=k =k

z

2

√9−r2 rz drdθ

5

3

2

5

0

   r 9 − r2 − 2r drdθ

3 x

 √5 1 − (9 − r2 )3/2 − r2 dθ 3 0

28. The equation  2π  1 is  z1 = r and the density is ρ = kr. (y 2 + z 2 )(kr)rdzdrdθ Ix = 

=k

2π

0

0

1 0

2π

z

r





1 r



1



1

(r4 sin2 θ + r2 z 2 )dzdrdθ

 1 1 2 3 =k (r sin θ)z + r z drdθ 3 0 0 r   2π  1  1 1 =k r4 sin2 θ + r2 − r5 sin2 θ − r4 drdθ 3 3  1  2π0  0 1 5 2 1 3 1 6 2 1 6 r sin θ + r − r sin θ − r dθ =k 5 9 6 18 0 0   2π  1 1 sin2 θ + =k dθ 30 18 0  2π  1 1 1 13 θ− sin 2θ + θ = πk =k 60 120 18 90 

y

8 4 dθ = πk 3 3

0

0

z=M9-r2

3

4

2

0

1

z=r 1

y

x

√ 29. (a) x = (2/3) sin(π/2) sin(π/2) cos(π/6) = 3/3; y = (2/3) sin(π/2) sin(π/6) = 1/3; √ √ (b) With x = 3/3 and y = 1/3 we have r2 = 4/9 and tan θ = 3/3. The point is (2/3, π/6, 0). √ √ 2/4; y = 5√sin(5ı/4)√sin(2π/3) = −5 6/4; 30. (a) x = 5 sin(5π/4) cos(2π/3) √ =5 √ z = 5 cos(5π/4) = −5 2/2; (5 2/4, −5 6/4, −5 2/2) √ √ √ (b) With x = 5 2/4√and y = −5√ 6/4 we have r2 = 25/2 and tan θ = − 3. The point is (5/ 2, 2π/3, −5 2/2). √ 31. (a) x = 8 sin(π/4) cos(3π/4) = −4; y = 8 sin(π/4) sin(3π/4) = 4; z = 8 cos(π/4) = 4 2; √ (−4, 4, 4 2) √ √ (b) With x = −4 and y = 4 we have r2 = 32 and tan θ = −1. The point is (4 2, 3π/4, 4 2).

936

CHAPTER 14. MULTIPLE INTEGRALS

√ 32. (a) x = (1/3) sin(5π/3) cos(π/6) = −1/4;√y = (1/3) sin(5π/3) sin(π/6) = − 3/12; z = (1/3) cos(5π/3) = 1/6; (−1/4, − 3/12, 1/6) √ √ 2 (b) With x = −1/4 and √ y = − 3/12 we have r = 1/12 and tan θ = 3/3. The point is (1/2 3, π/6, 1/6). √ √ 33. (a) x = √ cos 0 = 2 2; y = 4 sin(3π/4) sin 0 = 0; z = 4 cos(3π/4) = −2 2; √ 4 sin(3π/4) (2 2, 0, −2 2) √ √ √ (b) With x = 2 2 and y = 0 we have r2 = 8 and tan θ = 0. The point is (2 2, 0, −2 2). √ 34. (a) x = 1 sin(11π/6) cos π = 1/2; y = 1 sin(11π/6) sin π = 0; z = 1 cos(11π/6) = 3/2; √ (1/2, 0, ( 3/2) √ (b) With x = 1/2 and y = 0 we have r2 = 1/4 and tan θ = 0. The point is (1/2, 0 3/2). 2 35. With √ x = −5, y = −5, and z = 0, we have ρ = 50, tan θ = 1, and cos φ = 0. The point is (5 2, π/2, 5π/4). √ √ √ z = 1, we have ρ2 = 5, tan θ = − 3, and cos φ = 1/ 5. 36. With z = 1, y√= − 3, and √ The point is ( 5, cos−1 1/ 5, −π/3). √ √ √ 2 37. With x = 3/2, √ y = 1/2, and z = 1, we have ρ = 2, tan θ = 1/ 3, and cos φ = 1/ 2. The point is ( 2, π/4, π/6). √ 38. With x = − 3/2, y = 0, and z = −1/2, we have ρ2 = 1, tan θ = 0, and cos φ = −1/2. The point is (1, 2π/3, 0). √ √ 39. With x = 3, y = −3, and z = 3 2, we have ρ2 = 36, tan θ = −1, and cos φ = − 2/2. The point is (6, π/4, −π/4) √ √ 40. With x = 1, y √ = 1, and z = − 6, we have ρ2 = 8, tan θ = 1, and cos φ = − 3/2 The point is (2 2, 5π/6, π/4).

41. ρ = 8 42. ρ2 = 4ρ cos φ; ρ = 4 cos φ √ 43. 4z 2 = 3x2 + 3y 2 + 3z 2 ; 4ρ2 cos2 φ = 3ρ2 ; cos φ = ± 3/2; φ = π/6, 5π/6 44. −x2 − y − z 2 = 1 − 2z 2 ; −ρ2 = 1 − 2ρ2 cos2 φ; ρ2 (2 cos2 φ − 1) = 1 45. x2 + y 2 + z 2 = 100 46. cos φ = 1/2; ρ2 cos2 φ = ρ2 /4; 4z 2 = x2 + y 2 + z 2 ; z 2 + y 2 = 3z 2 47. ρ cos φ = 2; z = 2 48. ρ(1 − cos2 φ) = cos φ; ρ2 − ρ2 cos2 φ = ρ cos φ; x2 + y 2 + z 2 − z 2 = z; z = x2 + y 2

14.8. TRIPLE INTEGRALS IN OTHER COORDINATE SYSTEMS

937

49. The

equations are φ = π/4 and ρ = 3. 3  2π  π/4  3  2π  π/4 1 ρ3 sin φ dφdθ V = ρ2 sin φdrhodφdθ = 3 0 0 0 0 0 0 π/4  2π  2π  π/4 9 sin φdφdθ = −9 cos φ dθ = 0 0 0 0   2π  √ √ 2 − 1 dθ = 9π(2 − 2) = −9 2 0

50. The equations are ρ = 2, θ = π/4, and θ = π/3. 2  π/3  π/2  2  π/3  π/2 1 3 2 ρ sin φ dφdθ ρ sin φdρdφdθ = 3 π/4 0 0 π/4 0 0  π/3  π/2 8 sin φdφdθ = 3 π/4 0 π/2  8 π/3 = − cos φ dθ 3 π/4 0  2π 8 π/3 (0 + 1)dθ = = 3 π/4 9

z

0



π/2



π/6

0

8 = 3 8 = 3

0



π/2

z

2 ρ=2

2

0



π/2 0

π/6 0

y

2

x

z ρ=2secφ

2

2 sec φ 1 2 ρ sin φ dφdθ 3 0



y

x

0

=

2

2

51. Using Problem 43, the equations are φ = π/6, θ = π/2, and ρ cos φ = 2.  π/2  π/6  2 sec φ V = ρ2 sin φdρdφdθ 0

ρ=3

3

sec3 φ sin φdφdθ =

8 3



π/2 0



π/6 0

sec2 φ tan φdφdθ

1

1

π/6  1 2 4 π/2 1 2 tan φ dθ = π dθ = 2 3 0 3 9

y

x

0

52. The equations are ρ = 1 and φ = π/4. We find the volume above the xy-plane and double. 1  2π  π/2  2π  π/2  1 1 3 2 ρ sin φ dφdθ ρ sin φdρdφdθ = 2 V =2 3 0 π/4 0 0 π/4 0 π/2  2π  π/2  2π  2π √ 2 2 2 2 = dθ sin φdφdθ = − cos φ = 3 0 3 3 2 π/4 0 0 π/4 √ 2π 2 = 3

z 1

ρ=1

1 x

1

y

CHAPTER 14. MULTIPLE INTEGRALS

938

53. By symmetry, x = y = 0. The equations are φ = π/4 and z ρ = 2 cos φ. 2 2 cos φ  2π  π/4  2 cos φ  2π  π/4 1 3 2 m= ρ sin φ ρ sin φdρdφdθ = dφdθ 3 0 0 0 0 0 0 π/4    8 2π 1 8 2π π/4 3 4 sin φ cos φdφdθ = − cos φ dθ = 3 0 3 0 4 0 1 0   2π  x 1 2 − 1 dθ = π =− 3 0 4  2π  π/4  2 cos φ  2π  π/4  2 cos φ zρ2 sin φdρdφdθ ρ3 sin φ cos φdρdφdθ Mxy = 0

0

0

0

0

ρ=2cosφ

1

y

0

2 cos φ  2π  π/4 π/4 1 4 cos5 φ sin φdφdθ = ρ sin φ cos φ dφdθ = 4 4 0 0 0 0 0 π/4   2π   2π 1 7 1 2 − 1 dθ = π − cos6 φ =− =4 6 3 0 8 6 0 0 7π/6 = 7/6. The centroid is (0, 0, 7/6). z = Mxy /m = π 

2π 

z

54. We are given density= kz. By symmetry, x = y = 0. The equation is ρ = 1.  m=

2π 0

= Mxy

1 k 4

π/2 0



=k =



2π

π/2 0



2π

1 0



0



1 1 4 ρ sin φ cos φ dφdθ 4

2π



0

π/2 0

0

ρ=1

ρ3 sin φ cos φdρdφdθ 1

π/2 0

0

1

x

sin φ cos φdφdθ

π/2  2π  2π 1 1 kπ 1 k sin2 φ dθ = k dθ = 4 0 2 8 4 0  2π  π/2  1 0  2π  = kz 2 ρ2 sin φdρdφdθ = k 0



0



0

kzρ2 sin φdρdφdθ = k



1

0

0

π/2 0



1 0

ρ3 cos2 φ sin φdρdφdθ

1  2π  π/2 1 5 1 2 ρ cos φ sin φ dφdθ = k =k cos2 φ sin φdφdθ 5 5 0 0 0 0 0 π/2  2π  2π 1 1 2 1 kπ − cos3 φ =− k (0 − 1)dθ = = k 5 0 3 15 15 0 0 2kπ/15 = 8/15. The center of mass is (0, 0, 8/15). z = Mxy /m = kπ/4 

2π



π/2

1

y

14.9. CHANGE OF VARIABLES IN MULTIPLE INTEGRALS

939

55. We are given density= k/ρ.  2π  cos−1 4/5  5 k 2 ρ sin φdρdφdθ m= 0 0 4 sec φ ρ 5  2π  cos−1 4/5 1 2 ρ sin φ =k dφdθ 2 0 0 

2π



z 5

ρ=4sec φ

4 sec φ

3

cos−1 4/5

1 = k (25 sin φ − 16 tan φ sec φ)dφdθ 2 0 0 cos−1 4/5  2π 1 k (−25 cos φ − 16 sec φ) dθ 2 0 0  2π 1 [−25(4/5) − 16(5/4) − (−25 − 16)]dθ = k 2 0  2π 1 dθ = kπ = k 2 0  2π  π  a 56. Iz = (x2 + y 2 )(kρ)ρ2 sin φdρdφdθ 0



2π

0



π

0



a

2

2

2

2

x

z

3

sin π cos θ + ρ sin φ sin θ)ρ sin φdρdφdθ 0 0 0 a  2π  π  2π  π  a 1 6 3 3 5 ρ sin φ dφdθ ρ sin φdρdφdθ = k =k 0 0 0 0 0 6 0 a     1 3 2π π 1 6 2π π 3 2 sin φdφdθ = ka (1 − cos φ) sin φdφdθ x = ka 6 6 0 0 0 0  π  2π   2π 4 4π 6 1 1 1 dθ = ka − cos φ + cos3 φ dθ = ka3 = ka3 6 3 6 3 9 0 0 0

=k

14.9

y

3

a 2

ρ=5

ρ=a

a

y

Change of Variables in Multiple Integrals

1. T : (0, 0) −→ (0, 0); (0, 2) −→ (−2, 8); (4, 0) −→ (16, 20); (4, 2) −→ (14, 28) 2. Writing x2 = v − u and y = v + u and solving for u and v, we obtain u = (y − √ x2 )/2 and 2 −1 v = (x +y)/2. Then the images under T are (1, 1) −→ (0, 1); (1, 3) −→ (1, 2); ( 2, 2) −→ (0, 2).

Vector Integral Calculus PROBLEMAS 15.1 Line5.9 Integrals 



1. C



π/4

2(5 cos t)(5 sin t)(−5 sin t) dt = −250

2xy dx = 0



√  π/4 1 125 2 3 sin t =− 3 6 0

π/4 0

sin2 t cos t dt

= −250   π/4   π/4  π/4 1 2 3 2xy dy = 2(5 cos t)(5 sin t)(5 cos t) dt = 250 cos t sin t dt = 250 − cos t 3 C 0 0 0  √  √ 125 2 250 1− = (4 − 2) = 3 4 6  π/4  π/4   2xy ds = 2(5 cos t)(5 sin t) 25 sin2 t + 25 cos2 t dt = 250 sin t cos t dt C

0

= 250 

3

2.

2



0

 π/4 1 125 2 sin t = 2 2 0 

1

(x + 2xy + 2x) dx = C

3

4



1

[8t + 2(2t)(t ) + 2(2t)]2 dt = 2 0



1

=2

(8t3 + 4t5 + 4t) dt

0

(8t3 + 4t5 + 4t) dt

0

  1 2 28 = 2 2t4 + t6 + 2t2 = 3 3 0   1  1 3 2 3 4 (x + 2xy + 2x) dy = [8t + 2(2t)(t ) + 2(2t)]2t dt = 2 (8t4 + 4t6 + 4t2 ) dt C

0



=2

 1 8 5 4 7 4 3 736 t + t + t = 5 7 3 105 0 957

0

CHAPTER 15. VECTOR INTEGRAL CALCULUS

958 

(x3 + 2xy 2 + 2x) ds = C



1

[8t3 + 2(2t)(t4 ) + 2(2t)]





1

4 + 4t2 dt = 8

0

0



t(1 + t2 )5/2 dt

 1 1 8 7/2 2 7/2 (1 + t ) = = 7 (2 − 1) 7 0   0  0 0 3. (3x2 + 6y 2 ) dx = [3x2 + 6(2x + 1)2 ] dx = (27x2 + 24x + 6) dx = (9x3 + 12x2 + 6x) −1 −1

C

−1

= −(−9 + 12 − 6) = 3  0  (3x2 + 6y 2 ) dy = [3x2 + 6(2x + 1)2 ]2 dx = 6 C −1  0  √ √ 2 2 (3x + 6y ) ds = [3x2 + 6(2x + 1)2 ] 1 + 4 dx = 3 5 −1

C

 8 8 8 x2 x2 56 dx = dx = dx = 3 2 /8 y 27x 27 27 1 8  8 C 2  18 x x2 8 4 2/3 4 −1/3 −1/3 x dy = dx = x dx = x = 3 2 y 27x /8 27 1 9 3 1 8 C 2  18  8  x x2  8 −1/3 2/3 3/2 −2/3 2/3 (1 + x ) ds = 1+x dx = x 1+x dx = 3 2 27 1 27x /8 1 C y 1 8 3/2 (5 − 23/2 ) = 27   2π  2π  2 2 2 2 t 2 5. (x + y )ds = (25 cos −25 sin t) 25 sin t + 25 cos tdt = 125 (cos2 t − sin2 t)dt 



4.

0

C



2π

= 125 



6.

0 π

(2x + 3y) d = C

2π 125 sin 2t = 0 cos 2tdt = 2 0

(6 sin 2t + 6 cos 2t)(−4 sin 2t) dt 0 π

−24 sin2 2t − 24 sin 2t cos 2t dt   0 π 1 = −24 (1 − cos 2t) − 24 sin 2t cos 2t dt 2 0 π = −12t + 6 sin2 2t − 12 sin2 2t 0 =

= −12π 

 7. C

z dx =

π/2 0

t(− sin t) dt

Integration by parts

π/2

 C

= (t cos t − sin t)|0 = −1  π/2 z dy = t cos t dt Integration by parts 0

π/2

= (t sin t + cos t)|0

=

π −1 2

0

15.1. LINE INTEGRALS

959

π/2 π2 1 2 z dz = t dt = t = 2 0 8 0 π/2  C √  z dx = t sin2 t + cos2 t + 1 dt = 2





0

C

8.

4xyz C 4xyz C 4xyz C

4xyz C

π/2 0

t dt =

√ π2 2 8

1 1 3 8 1 8 8 9 8 2 2 t (t )(2t)t dt = t = dx = 4 t dt = 3 3 27 27 0 01    01  1 1 3 16 2 2 t (t2 )(2t)2t dt = dy = 4 t7 dt = t8 = 3 3 3 3 1 0   01 0 1  1 3 16 16 16 t (t2 )(2t)2 dt = dz = 4 t6 dt = = 21 3 21 21 0 0   1   1  01   1 3 8 8 1 9 2 7 200 6 2 t (2t) t4 + 4t2 + 4 dt = t + t = ds = 4 t (t + 2) dt = 3 3 3 9 7 189 0 0 0 





π/2

1







9. Using x as the parameter, dy = dx and 

 C

= (2x + y) dx + xy dy =

2 −1

 =

(2x + x + 3 + x + 3x) dx =

 2 1 3 x + 3x2 + 3x = 21. 3 −1

10. Using x as the parameter, dy   2= 2x dx and  2 (2x + y) dx + xy dy = (2x + x + 1) dx + C

−1

 =



2

2 −1

2

2

(x2 + 6x + 3) dx

−1

x(x + 1)2x dx =

 2 2 5 141 x + x3 + x2 + x = . 5 5 −1



2

(2x4 + 3x2 + 2x + 1) dx

−1

11. From (−1, 2) to (2, 2) we use x as a parameter with y = 2 and dy = 0. From (2, 2) to (2, 5) we use y as a parameter with x = 2 and dx = 0. 

 C

(2x + y) d + xy dy =



2

5

(2x + 2) dx + −1

2

2 2 2y dy = (x2 + 2x) −1 + y 2 −1 = 9 + 21 = 30

12. From (−1, 2) to (−1, 0) we use y as a parameter with x = −1 and dx = 0. From (−1, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (2, 5) we use y as a parameter with x = 2 and dx = 0. 

 C

(2x + y) d + xy dy =



0 2

= 2 + 3 + 25 = 30



2

(−1)y dy +

5

2x dx + −1

0

0 2 2 1 2y dy = − y 2 + x2 −1 + y 2 0 2 2

CHAPTER 15. VECTOR INTEGRAL CALCULUS

960

13. Using x as a the parameter, dy = 2xdx. 

 C

y dx + x dy =

1 0

x2 dx +





1

x(2x) dx =

0

1 0

1 3x2 dx = x3 0 = 1

14. Using x as a the parameter, dydx. 

 C

y dx + x dy =



1

x dx +

0



1

x dx =

0

1 0

1 2x dx = x2 0 = 1

15. From (0, 0) to (0, 1) we use y as a parameter with x = dx = 0. From (0, 1) to (1, 1) we use x as a parameter with y = 1 and dy = 0. 

 C

y dx + x dy = 0 +

1

1 dx = 1 0

16. From (0, 0) to (1, 0) we use x as a parameter with y = dy = 0. From (1, 0) to (1, 1) we use y as a parameter with x = 1 and dx = 0. 

 C

 17.

y dx + x dy = 0 + 

9

1

1 dy = 1 0



1 (6x + 2y + 2) dx + 4xy dy = (6t + 2t ) t−1/2 dt + 2 C 4 9 = (2t3/2 + 2t3/2 ) = 460 2

2

9 4



√

4 tt dt +

9

(3t1/2 + 5t3/2 ) dt

4

4

18.



(−y 2 ) dx + xy dy = C

 19. C

2

(−t6 )2 dt + 0

2x3 y d + (3x + y) dy =





2

4x dx + 2y dy = C

1

2(y 6 )y2y dy +



1

−1  1

2 0

4t6 dt =

(3y 2 + y) dy =



1 −1

2 4 7 512 t = 7 0 7 (4y 8 + 3y 2 + y) dy

4 9 1 26 y + y 3 + y 2 = 9 2 9 −1

4(y 3 + 1)3y 2 dy +

−1 6

(2t)(t3 )3t2 dt = 0

−1

 =

20.



2

3

= 2y + 4y +



2 y 2 −1





2

2

2y dy = −1

(12y 5 + 12y 2 + 2y) dy

−1

= 165

21. From (−2, 0) to (2, 0) we use x as a parameter with y = dy = 0. From (2, 0) to (−2, 0) we parameterize the semicircle as x = 2 cos θ and y = 2 sin θ for 0 ≤ θ ≤ π.

15.1. LINE INTEGRALS



2

C

961



2

(x + y ) dx − 2xy dy =

2 −2

x

2





π

4(−2 sin θ dθ) −

dx + 0 π

π 0

8 cos θ sin θ(2 cos θ dθ)

2  1 3 x −8 (sin θ + 2 cos2 θ sin θ) dθ 3 −2 0   π 2 64 16 80 16 3 − 8 − cos θ − cos θ = − =− = 3 3 3 3 3 0 =

22. We start at (0, 0) and use x as a parameter. 

2

C



2

1



4

1



2

0

(x + x ) dx − 2 xx (2x dx) + (x2 + x) dx 0 0 1    0 √ 1 −1/2 x x x dx −2 2 1 1  1  0  1 3 3 = (x2 − 3x4 ) dx + x2 dx = (−3x4 ) dx = − x5 = − 5 5 0 1 0 0

(x + y ) dx − 2xy dy =

2

23. From (1, 1) to (−1, 1) and (−1, −1) to (1. − 1) we use x as a parameter with y = 1 and y = −1, respectively, and dy = 0. From (−1, 1) to (−1, −1) and (1, −1) to (1, 1) we use y as a parameter with x = −1 and z = 1, respectively, and dx = 0.  C

x2 y 3 dx − xy 2 dy =



−1 1

x2 (1) dx +



−1 1

−(−1)y 2 dy +



1 −1

x2 (−1)3 dx +



−1 1 1 1 1 3 1 3 1 3 1 3 8 = x + y − x − y =− 3 1 3 −1 3 −1 3 −1 3

1 −1

−(1)y 2 dy

24. From (2, 4) to (0, 4) we use x as a parameter with y = 4 and dy = 0. From (0, 4) to (0, 0) we use y as a parameter with x = dx = 0. From (0, 0) to (2, 4) we use y = 2x and dy = 2dx.  C

x2 y 3 dx − xy 2 dy =

 C

0 2





0

2

0 dy + 4



π

y dx − x dy =

3 sin t(−2 sin t) dt − 0  π = −6 dt = −6π

 Thus,

x2 (64) dx −

0

x2 (8x3 ) dx −



2 0

x(4x2 )2 dx

0 2 2 352 4 6 512 256 64 3 x + x − 2x4 0 = − + − 32 = − = 3 2 3 0 3 3 3

 25.



0

−C

y dx − x dy = 6π.



π 0

2 cos t(3 cos t) dt = −6

π 0

(sin2 t + cos2 ) dt

CHAPTER 15. VECTOR INTEGRAL CALCULUS

962  26. C

x2 y 3 + x3 y 2 dy =



1

−1  1

= −1  1

=

x2 (x12 ) dx + x

14



1

dx +



1

x3 (x8 )(4x3 ) dx

−1

4x14 dx

−1

1 5 15 x 15 −1

5x14 dx =

−1

5 2 5 + = = 15 15 3 27. We parameterize the line segment from (0, 0, 0) to (2, 3, 4) by x = 2t, y = 3y, z = 4t for 0 ≤ t ≤ 1. We parameterize the line segment from (2, 3, 4) to (6, 8, 5) by x = 2 + 2t, y = 3 + 5t, z = 4 + t, 0 ≤ t ≤ 1. 

 C

y dx + z dy + x dz =



1

0  1 0





1

3t(2 dt) + 0



(4 + t)(5 dt) +



(55t + 34) dt =

 1 55 2 123 t + 34t = 2 2 0

    2 5 5 2 t (3t2 dt) + t dt (3t) 4 2 0 0 0    2  2 3 4 3 5 5 3 15 4 15 2 3 = t + t + t = 56 3t + t + t dt = 4 2 4 4 2 0 0





C

(3 + 5t)(4 dt) 0

(2 + 4t) dt

0

28.

1

2t(4 dt) + 0

1 0

1

=



1

4t(3 dt) +

y dx + z dy + x dz =

2

t3 (3 dt) +



2



29. From (0, 0, 0) to (6, 0, 0) we use x as a parameter with y = dy = 0 and z = dz = 0. From (6, 0, 0) to (6, 0, 5) we use z as a parameter with x = 6 and dx = 0 and y = dy = 0. From (6, 0, 5) to (6, 8, 5) we use y as a parameter with x = 6 and dz = 0 and z = 5 and dz = 0.  6  5  8  y dx + z dy + z dz = 0 dx + 6 dz + 5 dy = 70 0

C

0

0

30. We parameterize the line segment from (0, 0, 0) to (6, 8, 0) by x = 6t, y = 8t, z = 0 for 0 ≤ t ≤ 1. From (6, 8, 0) to (6, 8, 5) we use z as a parameter with x = 6, dx = 0, and y = 8, dy = 0. 

 C

 31. C

y dx + z dy + z dz =

10x dx − 2xy 2 dy + 6xz dz =



1 0



1

5

8t(6 dt) + 0

10(t) dt −

0



1 0

2(t)(t2 )2 (2t) dt +

1 1 1 = 5t2 0 − 4t6 0 + 18t6 0 = 5 − 4 + 18 = 19

1 6 dz = 24t2 0 + 30 = 54 

1 0

6(t)(t3 )(3t2 ) dt

15.1. LINE INTEGRALS

963

32. Parametrize the line segments as follows: C1 : r1 (t) = ti + tj, 0 ≤ t ≤ 1 C2 : r2 (t) = i + j + tk, 0 ≤ t ≤ 1 C3 : r3 (t) = (1 − t)i + (1 − t)j + (1 − t)k, 0 ≤ t ≤ 1 We  then have   C1

3x dx − y 2 dy + z 2 dz =

1

0

1

3t dt −

0

t2 dt

7 3 1 − = 2 3 6  1   1 3(1)(0) dt − (1)2 (0) dt + 3x dx − y 2 dy + z 2 dz = =

 C2

0

0

1 0

t2 dt

1 = 3  1  1  1  2 2 2 3x dx − y dy + z dz = 3(1 − t)(−1) dt − (1 − t) (−1) dt + (1 − t)2 (−1) dt C3 0 0 0       3 1 1 3 = − − − + − =− 2 3 3 2 



2

33. C1

y dx + xydy =

1

2



(4t + 2) 2dt + 0



1

1

(2t + 1)(4t + 2)4dt = 0

0

(64t2 + 64t + 16)dt

 1 64 3 208 64 t + 32t2 + 16t = + 32 + 16 = 3 3 3 0 √3  √3  √3  √3  208 8 8 y 2 dx + xydy = 4y 4 (2t)dt + 2t4 (4t)dt = 16t5 dt = t6 = 72 − = 3 3 3 C2 1 1 1 1 e3  e3  e3   e3 1 2 8 8 (ln t)2 dt = (ln t)3 y 2 dx + xydy = 4(ln t)2 dt + 2(ln t)2 dt = t t t 3 C3 e e e e 208 8 = (27 − 1) = 3 3 √   2 √ 2 2 √ 2  √ 1 3 16 5 t = 34. C1 xyds = 0 t(2t) 1 + 4dt = 2 5 0 t dt = 2 5 3 3 0  2  2    1 xyds = t(t2 ) 1 + 4t2 dt = t3 1 + 4t2 dt u = 1 + 4t2 , du = 8tdt; t2 = (u − 1) 4 C2 0 0  17  17 1 1 1 (u − 1)u1/2 du = = (u3/2 − u1/2 )du 4 8 32 1 1  17  2 5/2 2 3/2 1 u − u = 32 5 3 1 √ 391 17 + 1 = √   3  3 120  √  3 √ 1 √ 16 5 xyds = (2t − 4)(4t − 8) 4 + 16dt = 16 5 (t − 2)2 dt 16 5 (t − 2)3 = 3 3 C3 2 2 2 C1 and C3 are different parameterization of the same curve, while C1 and C2 are different 

=

CHAPTER 15. VECTOR INTEGRAL CALCULUS

964 curves.

35. We are given ρ = kx. Then  m=

 C

ρ dx =

= k (t +



π

0 π sin t)|0

kx ds = k

π 0

(1 + cos t)



sin2 t + cos2 t dt = k



π 0

(1 + cos t) dt

= kπ.

36. From Problem 35,m = kπ and ds = dt.   π  π 1 Mx = yρds = kxyds = k (1 + cos t) sin tdtk − cos t + sin2 t = 2k 2 C C 0 0  π π 2 2 My = intC xρds = intC kx ds = k (1 + cos t) dt = k (1 + 2 cos t + cos2 t)dt 0 0   π 1 1 3 = k t + 2 sin t + t + sin 2t = kπ 2 4 2 0 3 2 3kπ/2 2k = ; y = Mx /m = = . The center of mass is (3/2, 2/π). x = My /m = kπ 2 kπ π

15.2

Line Integrals of Vector Fields

and

  

   D

dV =

EabcdV  = abc

   E

dV  = abc



4 π 3

 =

4 πabc. 3

30. Let u = xy and v = xy 1.4 . Then xy 1.4 = c =⇒ v = c; xy = b =⇒ u = b; xy 1.4 = d =⇒ v = d; xy = a =⇒ u = a. ∂(u, v) y x = 0.4xy 1.4 = 0.4v =⇒ ∂(x, y) = 5 = ∂(x, y) y 1.4 1.4xy 0.4 ∂(u, v) 2v  d    d b   5 5 dv 5 5 dA = dudv = (b − a) = (b − a)(ln d − ln c) dA = 2v 2v 2 v 2 R S c a c

PROBLEMAS DE REPASO DE LA UNIDAD 5

Chapter 14 in Review A. True/False 1. True; use ex

2

−y

2

5.2 = ex e−y and Problem 53 in Section 14.2

2. True 3. True 4. False; consider f (x, y) = x. 5. False; both the density function and the lamina must be symmetric about an axis. 6. True; the equation of the plane is θ =

π 4,

θ=

5π 4

CHAPTER 14. MULTIPLE INTEGRALS

948

B. Fill in the Blanks 



5

1. y 2 +1

5y 8y − x 3



5 dx = (8xy 3 − 5y ln x) y2 +1 = 40y 3 − 5y ln 5 − [8(y 2 + 1)y 3 − 5y ln(y 2 + 1)] = −8y 5 + 32y 3 + 5y ln

2. 16 

a

 b√1−x2 /a2

6. 7.

3. square  √ 2 2 c

√

−b 1−x2 /a2 −c  4  √x f (x, y)dydx 0 x/2 −a

y2 + 1 y 4. II

5. f (x, 4) − f (x, 2)

1−x /a −y 2 /b2

√

1−x2 /a2 −y 2 /b2

ρ(x, y, z)dzdydx y y=Mx y=x/2

1

4 x √ 8. x = 6 sin(5π/3) cos(5π/6) = 9/2; y = 6 sin(5π/3) sin(5π/6) = −3 3/2; z = 6 cos(5π/3) = 3 √ The point is (9/2, −3 3/2, 3). √ √ √ √ 9. r = 2 sin(π/4) = 2; θ = 2π/3; z = 2 cos(π/4) = 2; ( 2, 2π/3, 2)  4  √4−y y 10. f (x, y)dxdy 4 √ y=4-x 1

2

− 4−y

0

1

11. z = r2 ; ρ = cot φ csc φ 12. circle

C. Exercises 1. Holding x fixed,



12x2 e−4xy = −5xy + y + c1 (x) −4x = −3xe−4xy − 5xy + y + c1 (x)

(12x2 e−4xy − 5x + 1)dy =

2. Holding y fixed,  ln |3xy + 4| 1 dx = + c2 (y) 4 + 3xy 3y  y y y 2 sin xydx = −y cos xy|y3 = y(cos y 4 − cos y 2 ) 3. y3



ex

4. 1/x

ex x x dy = − = x2 − x/ex y2 y 1/x

x

CHAPTER 14 IN REVIEW 

2



2x

ye

5. 0

0

y−x

949 

dydx =

2

(ye 0  2

= 0

y−x

−e

y−x

2x ) dx

Integration by parts

0

(2xex − ex + e−x )dx

Integration by parts

2 = (2xex − 2ex − ex − e−x ) 0 = e2 − e−2 + 4 

4



4  4   4   4 1 y x −1 x 2 = − ln(16 + x − ) dx = tan 2 2 2 16 + x 16 + x 4 2 0 16 + x x 0 0     π 1 1 π 1 π 1 1 − ln 32 − 0 − ln 16 = + (ln 16 − ln 32) = + ln = 4 2 2 4 2 4 2 2

4

1 dydx = 16 + x2

6. 0



x

1



√

x

7. 0

x

sin y dydx = y



4

1



 

y2

0 1

= 0

y

sin y dxdy = y

(sin y − y sin y)dy



1 0

y sin y x dy y y2

Integration by parts

y

y=x

1

y=Mx

1 x

1

= (− cos y − s ∈ y + y cos y)|0 = (− cos 1 − sin 1 + cos 1) − (−1) = 1 − sin 1 

e2



8. e



ln xdydx =

0 5



1/x



π/2



cos θ

9. 0

0

0

e2 e

1/x e2  e2 1 1 1 2 2 ln xdx = (ln x) = (2x2 − 12 ) = y ln x dx = x 2 2 3 e e 0

cos θ  5  π/2 cos3 θdθdz 3r drdθdz = r dθdz = 0 0 0 0 0  π/2  5  5  π/2 1 (1 − sin2 θ) cos θdθdz = dz sin θ − sin3 θ = 3 0 0 0 0  5 2/3dz = 10/3 = 2



5



π/2

3

0



π/2



sin x



ln x

10. π/4

0

0

ln x  π/2  sin x y y e dydxdz = e dxdz = (x − 1)dxdz π/4 0 π/4 0 0  sin z  π/2    π/2  1 2 1 = x − x sin2 z − sin z dz = 2 2 π/4 π/4 0  √   π/2  1 1 π 1 2 π z − sin 2z + cos z − + = = − 4 8 8 16 8 2 π/4 √ π+2−8 2 = 16 

π/2



sin x

CHAPTER 14. MULTIPLE INTEGRALS

950

coordinates 11. Using polar coorindates, 

 



2π



8

5dA = R

5rdrdθ = 5 0

0

0

  12. Using symmetry,



 R

π



8  1 2 r dθ = 5 2 0

r=8

2π

32dθ = 320π.

8 polar axis

0



1+cos θ

π

1+cos θ 1 2 r dθ 2 0

r=1+cos θ

dA = 2 rdrdθ = 2 0 0 0 R  π = (1 + 2 cos θ + cos2 θ)dθ 0   π 1 1 = θ + 2 sin θ + θ + sin 2θ = 3π/2. 2 4 0

  13.

2π

(2x + y)dA = 

1 0

0



y 2 +1 2y

1

=



(2x + y)dxdy =

1 0

2 polar axis

y2 +1 (x + xy) dy

x=y2+1

2y

[(y 2 + 1)2 + (y 2 + 1)y − (4y 2 + 2y 2 )]dy

= (y 4 + y 3 − 4y 2 + y + 1)dy =



x

 1 1 5 1 4 4 3 1 2 37 y + y − y + y + y = 5 4 3 2 60 0

14. Substracting z = 6 − x − y from z = x + y, we obtain x + y = 3.     3  3−x  6−x−y  3  3−x 6−x−y xdV = xdzdydx = xz dydx R 0 0 x+y 0 0 x+y  3  3−x = (6x − 2x2 − 2xy)dydx 0

3



0 3

= =  = 0

 =

z

6

z=6-x-y

0



0

x=2y

y

2

3

3−x (6xy − 2x2 y − xy 2 ) dx

3 y

0

[6x(3 − x) − 2x2 (3 − x) − x(3 − x)2 ]dx (9x − 6x2 + x3 )dx

 3 9 2 1 27 x − 2x3 + x4 = 2 4 4 0

2 x

CHAPTER 14 IN REVIEW

951

√ 15. The circle x2 + y 2 = 1 intersects y = x at x√= 1/ 2. The y = x at x = 3/ 2. circle x2 + y 2 = 9 intersects    1/√2  √9−x2 1 1 dA = dydx √ 2 2 2 2 0 R x +y 1−x2 x + y  3/√2  √9−x2 1 + dydx √ 2 x + y2 x 1/ 2

y y=M9-x2

y=x

x

y

16. The circles are r = 1 and r = 3; the line is θ = π/4. 3    π/2  3  π/2 1 1 dA = rdrdθ = ln r dθ 2 2 2 x + y r R π/4 1 π/4 1  π/2 π ln 3dθ = ln 3 = 4 π/4

r=3

r=1

x

17. y

y=x2

4

2

x

y=-x2

z

18. The region is symmetric with respect to the xz- and yzplanes and is shown in the first octant.

z=x2+y2 1

1

y

1

x



1



√ 3

19. 0

y

y



2

cos x dxdy = 

1

1 0

x x3

0

=



2

cos x dydx =



1 0



x y cos x2 x3

(x cos x2 − x3 cos x2 )dx

y y=x

dx

1

y=x3 1

x

CHAPTER 14. MULTIPLE INTEGRALS

952 1  1 1 sin x2 = x2 (x cos x2 )dx 2 0 0

=

Integration by parts   1 1 2 1 1 2 2 x sin x + cos x = sin 1 − 2 2 2   0 1 1 1 1 sin 1 + cos 1 − = sin 1 − 2 2 2 2 1 − cos 1 = 2

z

20. The six   2  4−2x 0 4

  

0 8 4 4

 

0

0 8−z 0 8−y

4



8 4

 0

0

2





 

1

4

21. 1/2

y+z=8

2x+z=8

F (x, y, z)dzdxdy;

2x+y=4

y x

F (x, y, z)dxdzdy; 

8−2x−z 0

F (x, y, z)dydxdz; 8−2x−z

F (x, y, z)dydzdx.



x−x2

2



1

(4z + 1)dydxdz = 0

are:

4−y/2−z/2

0

√

integral

F (x, y, z)dxdydz;

8−2x 



the

8−2x−y

0

4

of

F (x, y, z)dzdydx;

4 4−y/2−z/2

0 4−z/2

0





0 2−y/2



forms

8−2x−y

8

0

0



2



1

(4z + 1)

= 0



1/2





1/2



(4z − 1)



x−2 dxdz

 2 1 1 − x− dxdz 4 2

Trig substitution

x − 1/2  x − 1/2 1 = x − x2 + sin−1 (4z + 1) 2 8 1/2 0  2 2  π π 5π = − 0 dz = (2z 2 + z) = (4z + 1) 16 16 8 0 0 2

 1

1/2

dz

22. The region is the portion of the sphere of radius 1 centered at the origin in the first octant and the octant below that. Using spherical coordinates, we have

CHAPTER 14 IN REVIEW 

1

√



 √1−x2 −y2

1−x2

0

0

−

√

1−x2 −y 2

953 (x2 + y 2 + z 2 )4 dzdydx =



π/2



0

π



0

1 0

ρ8 ρ2 sin φdρdφdθ

 1  π/2  π 1 1 1 ρ 1 sin φ dφdθ = sin φdφdθ = 11 11 0 0 0 0 0 π  π/2  π/2 1 1 π = − cos φ dθ = 2dθ = 11 0 11 0 11 

π/2



π



0

23. fx = y; fy = z, 1 + fx2 + x2y = 11 + x2 + y 2 . Using cylindrical coordinates,  A=

2π



0

1



 1+

0

r2 rdrdθ

1  1 1 2π 3/2 2π √ 2 3/2 (1 + r ) dθ = (2 2 − 1). (2 − 1)dθ = 3 3 0 3 0

2π

= 0

  3    √3   √3  2 6 2 2 2 3 2 24. V = 6 − y dydx = 6y − y dx = (18 − 6) − 6x − x dx 2 3 9 9 0 x2 0 0 x    √3  √3  √ √ 2 6 2 7 6√ 48 √ 2 3 3= 3 12x − 6x + x dx = 12x − 2x + x = 12 3 − 6 3 + = 9 63 7 7 0 0 

√

3



3





1



25. (a) V =

0

2x



x

 1−

x2 dydx

= 0

1 1 1 = − (1 − x2 )3/2 = 3 3 0  (b) V =

1 0



y y/2



1

 1 − x2 dxdy +

2

2x   2 y 1 − x dx = x



1

1 y/2

1 0

x



1 − x2 dx

 1 − x2 dxdy

y

26. We are given ρ = k(x2 + y 2 ).   1 x k(x2 + y 2 )dydx = k m= x3

0



=k

1 0

 =k





1 0

1 1 x4 + x− x5 − x9 dx 3 3



 x2 1 3 x y + y dx 3 x3

1

2

 1 1 5 1 1 1 k x + x7 − x6 − x10 = 5 21 6 30 21 0

y=x2 y=x3 1

x

CHAPTER 14. MULTIPLE INTEGRALS

954  My =

1



0



x x3

3



2

k(x + xy )dydx = k

1



0

 x2   1 1 3 1 1 x y + xy dx = k x5 + x7 − x6 − x10 dx 3 3 3 0 x3 3

 1 1 6 1 8 1 7 1 11 65k x + x − x − x =k = 1848 6 24 7 33 0  x2  1 x  1 1 2 2 1 4 2 3 Mx = x y + y dx k(x y + y )dydx = k 2 4 0 x3 0 x3   1 1 6 1 8 1 8 1 12 x + x − x − x =k dx 2 4 2 4 0   1 1 7 1 9 1 13 20k x − x − x =k = 14 36 52 819 0

65k/1848 20k/819 = 65/88; y = Mx /m = = 20/39 x = My /m = k/21 k/21 The center of mass is (65/88, 20/39).  x2 1 2 3 27. Iy = k(x + x y )dydx = k x y + x y dx 3 x3 0 0 x3    1  1 1 1 1 1 1 1 41 x7 + x9 − x8 − x12 = k =k x6 + x8 − x7 − x11 dx = k 3 3 7 27 8 36 1512 0 0 

1



x2

4



2 2

1



4

28. (a) Using symmetry,  a  √a2 −x2  √a2 −x2 −y2  V =8 dzdydx = 8 0

0

0

a 0

√

 0

a2 −x2

 a2 − x2 − y 2 dydx

Trig substitution  √a2 −x2  a   a 2 2 − x a y y π a2 − x2 −1 sin √ dx =8 a2 − x2 − y 2 + dx = 8 2 2 2 a2 − x2 0 0 0 2   a 1 3 4 2 = 2π a x − x = πa3 3 3 0 (b) Using symmetry,  2π  a  V =2

√

 2π  a  rdzdrdθ = 2 r a2 − r2 drdθ 0 0 0 0 0 a   2π 2 2π 3 4 1 − (a2 − r2 )3/2 dθ = a dθ = πa3 =2 3 3 3 0 0 0 a  2π  π  a  2π  π 1 3 2 ρ sin φ dφdθ (c) V = ρ sin φdρdφdθ = 3 0 0 0 0 0 0 π  2π  π  2π  1 1 1 2π 3 4 = a3 sin φdφdθ = −a3 cos φ dθ = 2a dθ = πa3 3 0 3 0 3 3 0 0 0 a2 −r 2

29. We use spherical coordinates.

CHAPTER 14 IN REVIEW  V =



2π



π/4 tan−1

0



2π



1/2

π/4

= tan−1 1/2

0

=

1 3





2π



0 2π

=9 0

3 sec φ 0

ρ2 sin φdρdφdθ

3 sec φ 1 3 ρ sin φ dφdθ 3 0

π/4 tan−1

955

1/2

27 sec3 φ sin φdφdθ = 9

π/4 1 2 tan φ 2 −1 tan

1/2

dθ =

9 2



2π



0



2π 0

1−

1 9





π/4 tan−1

1/2

tan φ sec2 φdφdθ

dθ = 8π

2 1 2 ρ sin φ dφdθ ρ sin φdρdφdθ = 30. V = 3 0 0 1 0 0 1 π/6    2π  π/6  2π  π/6  8 1 7 7 2π sin φ − sin φ dφdθ = sin φdφdθ = − cos φ dθ = 3 3 3 0 3 0 0 0 0 0

 √ √   √ 7 7π 3 3 7 2π − (−1) dθ = 1− 2π = (2 − 3) − = 3 0 2 3 2 3 

2π



π/6



2



2

2π



π/6

y

y

2

31. x = 0 =⇒ u = 0, v = −y =⇒ u = 0, −1 ≤ v ≤ 0 x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4 x = 1 =⇒ u = 2y, v = 1 − y 2 = 1 − u2 /4 y = 0 =⇒ u = 0, v = x2 =⇒ u = 0, 0 ≤ v ≤ 1 y = 1 =⇒ u = 2x, v = x2 − 1 = u2 /4 − 1 ∂(u, v) 2y 2x = −4(x2 + y 2 ) = ∂(x, y) 2x −2y

1

1 R

S 1

1

x

x

-1

1 ∂(x, y) =− ∂(u, v) 4(x2 + y 2 )      √ (x2 + y 2 ) 3 x2 + y 2 dA = (x2 + y 2 ) 3 v − =⇒

R

 2  1−u2 /4  1 dA = 1 v 1/3 dvdu 4(x2 + y 2 ) 4 0 u2 /4−1 S 1−u2 /4   2

 1 2 3 4/3 3 (1 − u2 /4)4/3 − (u2 /4 − 1)4/3 du = v du = 4 0 4 16 0 u2 /4−1  2

 3 = (1 − u2 /4)4/3 − (1 − u2 /4)4/3 du = 0 16 0 y

32. y = x =⇒ u + uv = v + uv2 =⇒ v = u 2 x = 2 =⇒ u + uv = 2 =⇒ v = (2 − u)/u y = 0 =⇒ v = 0 or u = −1 ∂(x, y) 1 − w u =1+u+v = we take v = 0 v 1+u ∂(u, v)

v

2

R

2

x

2

u

CHAPTER 14. MULTIPLE INTEGRALS

956 Using x = u + uv and y = v + uv we find

(x − y)2 = (u + uv − v − uv)2 = (u − v)2 = u2 − 2uv + v 2 x + y = u + uv + v + uv = u + v + 2uv 2

(x + y) + 2(x + y) + 1 = u2 + 2uv + v 2 + 2(u + v) + 1 = (u + v)2 + 2(u + v) + 1 = (u + v + 1)2 . Then   R

 

 1  2/(1+v) 1   (u + v + 1)dA = dA = dudv (x − y)2 + 2(x + y) + 1 0 v S u+v+1    1  1 2 1 2 1 − v dv = 2 ln(1 + v) − v = 2 ln 2 − . = 1 + v 2 2 0 0 1

CÁLCULO VECTORIAL MATEMÁTICAS 3 Chapter 10 MANUAL DE SOLUCIONES

Conics and Polar Coordinates APÉNDICE SECCIONES CÓNICAS 10.1

Conic Sections y

1. vertex: (0, 0) focus: (1, 0) directrix: x = −1 axis: y = 0

x

y

2. vertex: (0, 0) focus: (7/8, 0) directrix: x = − 78 axis: y = 0 x

y

3. vertex: (0, 0) focus: (0, −4) directrix: y = 4 axis: x = 0

x

668

10.1. CONIC SECTIONS 4. vertex:(0, 0) 1 focus: 0, 40 1 directrix: y = − 40 axis: x = 0

669 y

x

y

5. vertex: (0, 1) focus: (4, 1) directrix: x = −4 axis: y = 1

x

6. vertex: (−2, −3) focus: (−4, −3) directrix: x = 0 axis: y = −3

y

x

7. vertex: (−5, −1) focus: (−5, 2) directrix: y = 0 axis: x = −5

y

x

670 8. vertex:(2, 0)  focus: 2, − 14 directrix: y = 14 axis: x = 2

9. (y + 6)2 = 4(x + 5) vertex: (−5, −6) focus: (−4, −6) directrix: x = −6 axis: y = −6

10. (x + 3)2 = −(y + 2) vertex: (−3, −2) focus: (−3, −9/4) directrix: y = −7/4 axis: x = −3

2  11. x + 52 = 14 (y − 1) vertex: (−5/2, −1) focus: (−5/2, −15/16) directrix: y = −17/16 axis: x = − − 5/2

CHAPTER 10. CONICS AND POLAR COORDINATES y

x

y

x

y

x

y

x

10.1. CONIC SECTIONS 12. (x − 1)2 = 4(y − 4) vertex: (1, 4) focus: (1, 5) directrix: y = 3 axis: x = 1

671 y

x

13. (y − 4)2 = −2(x + 3) vertex: (3, 4) focus: (5/2, 4) directrix: x = 7/2 axis: y = 4

y

x

  14. (y − 2)2 = 4 x + 14 vertex: (−1/4, 2) focus: (3/4, 2) directrix: x = −5/4 axis: y = 2

y

x

15. x2 = 28 16. y 2 = −16x 17. y 2 = 10x 18. x2 = −40y 19. The parabola is of the form (y − k)2 = 4p(x − h) with (h, k) = (−2, −7) and p = 3. Thus the equation is (y + 7)2 = 12(x + 2). 20. The parabola is of the form (x−h)2 = 4p(y −5) with (h, k) = (2, 0) and p = 3, so the equation of the parabola is (x − 2)2 = 12y.

672

CHAPTER 10. CONICS AND POLAR COORDINATES

21. The parabola is of the form x2 = 4py with (−2)2 = 4p(8). Thus p = x2 = 12 y. 22. The parabola is of the form y 2 = 4px with 1 x. y 2 = 16

 1 2 4

= 4p(1) so p =

1 64 .

1 8

and the equation is

Thus the equation is

23. To find the x-intercept set y = 0. Solving 42 = 4(x + 1) gives x = 3. The x-intercept is (3, 0). To find the y-intercept set x = 0. Solving (y + 4)2 = 4 gives y = −4 ± 2. The y-intercepts are (0, −2) and (0, −6). √ 2 24. To find √set y = 0. Solving (x − 1) = 2 gives x = 1 ± 2. The x-intercepts are √ the x-intercept (1 + 2, 0) and (1 − 2, 0). To find the y-intercept set x = 0. Solving 1 = −2(y − 1) gives y = 12 . The y-intercept is (0, 1/2). y

25. center: (0,√ 0) foci: (0, ± 15) vertices: (0, ±4) endpoints of the √ minor axis: (±1, 0) 15 eccentricity: 4

26. center: (0, 0) foci: (±4, 0) vertices: (±5, 0) endpoints of the minor axis: (0, ±3) 4 eccentricity: 5

27.

y2 x2 + = 1 center: (0, 0) 16 9√ foci: (± 7, 0) vertices: (±4, 0) endpoints of the √ minor axis: (0, ±3) 7 eccentricity: 4

x

y

x

y

x

10.1. CONIC SECTIONS

28.

673

y2 x2 + = 1 center: (0, 0) 2 4 √ foci: (0, ± 2) vertices: (0, ±2) √ endpoints of the √ minor axis: (± 2, 0) 2 eccentricity: 2

29. center: (1,√ 3) foci: (1 ± 13, 3) vertices: (−6, 3), (8, 3) endpoints of the √ minor axis: (1, −3), (1, 9) 13 eccentricity: 7

30. center: (−1, 2)√ foci: (−1, 2 ± 11) vertices: (−1, −4), (−1, 8) endpoints of the √ minor axis: (−6, 2), (4, 2) 11 eccentricity: 6

y

x

y

x

y

x

CHAPTER 10. CONICS AND POLAR COORDINATES

674

31. center: (−5, −2)√ foci: (−5, −2 ± 15) vertices: (−5, −6), (−5, 2) endpoints of the √ minor axis: (−6, −2), (−4, −2) 15 eccentricity: 4

32. center: (3, −4)√ foci: (3, −4 ± 17) vertices: (3, −13), (3, 5) endpoints of the √ minor axis: (−5, −4), (11, −4) 17 eccentricity: 9



2 y + 12 =1 33. x + 4 center: (0, −1/2)√ foci: (0, −1/2 ± 3) vertices: (0, −5/2), (0, 3/2) endpoints of the √ minor axis: (−1, −1/2), (1, −1/2) 3 eccentricity: 2 2

y

x

y

x

y

x

10.1. CONIC SECTIONS 34.

35.

36.

675

(y − 4)2 (x + 2)2 + =1 2 72 center: (−2, 4)√ foci: (−2, 4 ± 70)√ vertices: (−2, 4 ± 6 2) √ endpoints of the √ minor axis: (−2 ± 2, 4) 35 eccentricity: 6

(y + 1)2 (x − 7)2 + =1 9 25 center: (2, −1) foci: (2, −5), (2, 3) vertices: (2, −6), (2, 4) endpoints of the minor axis: (−1, −1), (5, −1) 4 eccentricity: 5

(y − 1)2 (x + 1)2 + =1 5 9 center: (−1, 1) foci: (−1, −1), (−1, 3) vertices: (−1, −2), (−1, 4) √ endpoints of the minor axis: (−1 ± 5, 1) 2 eccentricity: 3

y

x

y

x

y

x

CHAPTER 10. CONICS AND POLAR COORDINATES

676

37.

(y + 3)2 x2 + =1 9 3 center: (0, −3) √ foci: (± 6, −3) vertices: (±3, −3) √ 3) endpoints of the minor axis: (0, −3 ± √ 6 eccentricity: 3

(y − 1/2)2 =1 3 center: (1, 1/2)√ foci: (1, 1/2 ± 2)√ vertices: (1, 1/2 ± 3) endpoints of the √ minor axis: (0, 1/2), (2, 1/2) 2 eccentricity: 3

38. (x − 1)2 +

y

x

y

x

39. The center is (0, 0) with the x-axis as the major axis. a = 5 and c = 3, so b = 4. Thus the y2 x2 + = 1. equation is 25 16 √ 40. The center is (0, 0) with the x-axis as the major axis. a = 9 and c = 2, so b = 77. Thus the y2 x2 + = 1. equation is 81 77 41. The center is (1, −3) with the x-axis as the major axis. a = 4 and b = 2. Thus the equation (x − 1)2 (y + 3)2 is + = 1. 16 4 42. The center is (1, −2) with the y-axis as the major axis. a = 4 and b = 3. Thus the equation (x − 1)2 (y + 2)2 is + = 1. 9 16 √ √ 43. The center is (0, 0) with the x-axis as the major axis. c = 2 and b = 3, so a = 11. Thus y2 x2 + = 1. the equation is 11 9

10.1. CONIC SECTIONS 44. The center is (0, 0) with the y-axis as the major axis. c = y2 x2 + = 1. the equation is 59 64

677 √

5 and a = 8, so b =

√

59. Thus

√ 45. The center is (0, 0) with the y-axis as the major axis. c = 3 thus 9 = a2 − b2 and a = 9 + b2 . y2 x2 = 1. The ellipse passes through the point Thus the equation is of the form 2 + b 9 + b2 √ √ √ (−1)2 (2 2)2 + = 1. Solving this for b, we obtain b = 3. Thus a2 = 12 and (−1, 2 2), thus 2 2 b 9+b x2 y2 the equation is + = 1. 3 12 46. The center is (0, 0) with the x-axis as the major axis and a = 5. The equation is of the form √ 19 x2 y 2 5 + 2 = 1. The ellipse passes through the point ( 5, 4) so + 2 = 1. Solving for b2 , we 25 b 25 b y2 x2 2 + = 1. obtain b = 20. Thus the equation of the ellipse is 25 20 47. The y-axis as the major axis with c = 3 and a = 4. Thus b = (y − 3)2 (x − 1)2 + =1 ellipse is 7 16

√

7 and the equation of the

48. The center is (15/2, 4) with the x-axis as the major axis. a = 1/2 and c = 7/2, thus b = (y − 4)2 (x − 15/2)2 = 1. + Thus the equation is (11/2)2 18

√

18.

y

49. center: (0, √ 0) foci: (± 41, 0) vertices: (±4, 0) asymptotes: y = ± 54 x √ 41 eccentricity: 4

x

y

50. center: (0, √ 0) foci: (± 8, 0) vertices: (±2, 0) asymptotes: y√= ±x eccentricity: 2

x

CHAPTER 10. CONICS AND POLAR COORDINATES

678

51.

52.

x2 y2 − = 1 center: (0, 0) 20 4 √ foci: (0, ±2 6)√ vertices: (0, ±2 5)√ asymptotes: y= ± 5x 6 eccentricity: 5

y

x

x2 y2 − = 1 center: (0, 0) 9 16 foci: (0, ±5) vertices: (0, ±3) asymptotes: y = ± 34 x 5 eccentricity: 3

53. center: (5,√ −1) foci: (5 ± 53, −1) vertices: (3, −1), (7, −1) asymptotes: y = −1 ± 72 (x − 5) √ 53 eccentricity: 2

y

x

y

x

10.1. CONIC SECTIONS 54. center: (−2,√ −4) foci: (−2 ± 35,√−4) vertices: (−2 ± 10, −4) 5x + 10 asymptotes: y = −4 ± √ 10  7 eccentricity: 2

55. center: (0, 4)√ foci: (0, 4 ± 37) vertices: (0, −2), (0, 10) asymptotes: y√= 4 ± 6x 37 eccentricity: 6

679 y

x

y

x

56. center: (−3, 1/4)√ foci: (−3, 1/4 ± 13) vertices: (−3, 9/4), (−3, −7/4) 1 asymptotes: y = ± 23 (x + 3) √ 2 13 eccentricity: 2

y

x

680

57.

58.

59.

CHAPTER 10. CONICS AND POLAR COORDINATES (y − 1)2 (x − 3)2 − = 1 center: (3, 1) 5 √ 25 foci: (3 ± 30,√1) vertices: (3 ± 5, 1) √ asymptotes: y√= 1 ± 5(x − 3) eccentricity: 6

y

x

(y − 1/2)2 (x + 1)2 − = 1 center: (−1, 1/2) 10 √ 50 foci: (−1 ± 60,√1/2) vertices: (−1 ± 10, 1/2) √ asymptotes: y√= 1/2 ± 5(x + 1) eccentricity: 6

(y − 1)2 (x − 2)2 − = 1 center: (2, 1) 6 √ 5 foci: (2 ± 11,√1) vertices: (2 ± 6, 1)  5 (x − 2) asymptotes: y = 1 ± 6  11 eccentricity: 6

y

x

y

x

10.1. CONIC SECTIONS 60.

(y − 3)2 (x − 8)2 − = 1 center: (8, 3) 25 √ 16 foci: (8 ± 41, 3) vertices: (13, 3), (3, 3) asymptotes: y√= 3 ± 4/5(x − 8) 41 eccentricity: 5

(x − 1)2 = 1 center: (1, 3) √1/4 foci: (1, 3 ± 5/2) vertices: (1, 2), (1, 4) asymptotes: y√= 3 ± 2(x − 1) 5 eccentricity: 2

61. (y − 3)2 −

681 y

x

y

x

62.

(x + 1)2 (y + 5)2 − = 1 center: (−1, −5) 18 4√ foci: (−1, −5 ± 22)√ vertices: (−1, −5 ± 18) √ asymptotes: y = −5 ± 218 (x + 1) √ 11 eccentricity: 3

y

63. The center is (0, 0) with the y-axis as the transverse axis. c = 4 and a = 2, thus b = x2 y2 − =1 The equation is 4 12 64. The center is (0, 0) with the y-axis as the transverse axis. c = 3 and a = 3/2, thus b =

x

√

12.

√ 3 3 . 2

CHAPTER 10. CONICS AND POLAR COORDINATES

682 The equation is

y2 x2 − =1 9/4 27/4

65. The center is (1, −3) with the y-axis as the transverse axis. c = 3 and a = 2, thus b = (x − 1)2 (y + 3)2 − =1 The equation is 4 5

√

5.

66. The center is (2, 2) with the y-axis as the transverse axis. c = 5 and a = 32, thus b = 4. The (y − 2)2 (x − 2)2 equation is − = 1. 9 16 67. The center is (−1, 3) with the y-axis as the transverse axis. a = 1 and the equation is of the √ (y − 3)2 (x + 1)2 form − = 1. The hyperbola passes through the point (−5, 3 + 5) thus 2 1 b √ (−5 + 1)2 (x + 1)2 2 2 2 = 1. (3 + 5 − 3) − = 1. Thus b = 4 and the equation is (y − 3) − b2 4 68. The center is (3, −5) with the y-axis as the transverse axis. a = 3 and the equation is of (y − 3)2 (x + 5)2 the form − = 1. The hyperbola passes through the point (1, −1) thus 9 b2 2 2 (6) (y − 3)2 (x + 5)2 (−4) 324 − 2 = 1. Thus b2 = and the equation is − = 1. 9 b 7 9 324/7 69. The center is (2, 4) with the y-axis as the transverse axis and a = 1. After solving the x+6 x asymptote given in the problem for y, we obtain y = = + 3. The equation of the 2 2 (x − 2)2 2 hyperbola is of the form (y − 4) − = 1. The asymptote equations for this hyperbola b2   x−2 −x + 2 x 2 and y − 4 = (these are also equivalent to y = + 4 − are y − 4 = and b b b  b  2 x y = − + 4+ ). Letting b equal 2 or -2 will yield one asymptote with the equation b b (x − 2)2 x = 1. y = + 3. In either case, the equation of the hyperbola is (y − 4)2 − 2 4 √ √ c 70. The y-axis is the conjuate axis. The center is (−5, 7) with b = 3 and = 10. Thus c = 10a. a (y + 7)2 2 2 2 2 2 2 = 1. Since c = b +a then 10a = 9+a and a = 1. Thus the equation is (x+5)2 − 9 71. We place the coordinate axes so that the origin is at the vertex of the parabola. The point (2, 2) lies on the parabola. Thus the equation is x2 = 2y with p = 1/2. The focus of this parabola occurs at the point (0, 1/2). Thus the light source is 6 inches from the vertex. 72. We place the coordinate axes so that the origin is at the vertex of the parabola. The point (10, 4) lies on the parabola. Thus the equation is x2 = 25y with p = 25/4. The focus is located at (0, 25/4). The eyepiece should be located 6.25 ft from the vertex. 73. We place the coordinate axes so that the origin is at the vertex of the parabola. The parabola is of the form x2 = 4py and contains the point (20, 1). Thus the equation of the parabola is x2 = 400y. The towers are located at x = 175 and x = −175. Hence the height of the towers

10.1. CONIC SECTIONS

683

is found by solving (175)2 = 400y. Solving this equation yields y = 76.5625. Therefore the towers are 76.5625 ft above the road. 74. We place the coordinate axes so that the origin is at the vertex of the parabola. The parabola is of the form x2 = 4py and contains the point (125, 75). Thus the equation of the parabola is x2 = 625 12 . We need to find the y-value of the point on the parabola when we are 50 ft from the tower or when x = 75f t. Hence this y-value is found by solving the equation 752 = 625 3 y which yield the solution y = 27 ft. The height of the cable above the roadway at a point 50 ft from one of the towers is 27 ft. 75. We place the coordinate axes so that the origin is at end of the pipe with the parabola in Quadrants 3 and 4. The equation is of the form x2 = 4py and the point (4, −2) lies on the parabola. Therefore the equation is x2 = −8y. The water hits the ground at y = −20. The point on the parabola with y-value -20 is found by solving x2 = −8(−20). This point is x = 12.65. Thus the water hits the ground 12.65 m from the point on the ground directly beneath the end of the pipe. 76. We place the coordinate axes with the x-axis along the ground and the y-axis to be through the dart √ thrower. Thus the dart was released at the point (0, 5) and hits the ground√at the point (10 10, 0). The parabola is of the form x2 = 4p(y−5) and contains the point (10 10, 0). Therefore the equation of the parabola is x2 = −200y. To find the height of the dart 10 ft from the thrower, we need to find the y-value of the point on the parabola corresponding to the x-value of 10. Hence, we need to solve the equation 102 = −200y which yields y = −1/2f t. So 10 feet from the thrower the dart will be 0.5 ft below the thrower or it will be 4.5 ft from the ground. 77. Taking the center of the ellipse to be at the origin, we have a = 3.6 × 107 and b = 3.52 × 107 . Since c2 = a2 − b2 , c2 = 12.96 × 1014 − 12.3904 × 1014 = 0.5696 × 1014 and c ≈ 0.75 × 107 . The perihelion or least distance is a − c ≈ 2.85 × 107 miles or 28.5 million miles. And the aphelion or greatest distance is a + c ≈ 4.35 × 107 miles or 43.5 million miles.

y 3.52x107 b

a

78. Using a = 3.6×107 and c = 0.75×107 , we compute the eccentricity e =

c

a-c

0.75 × 107 ≈ 0.20833. 3.6 × 107

79. From a = 1.67 × 109 and 4.25 × 108 we obtain c2 = a2 − b2 = 2.7889 × 1018 − 18.0625 × 106 = 2.78.89 × 1016 − 18.0625 × 1016 = 260.8275 × 1016 = 2.608275 × 1018 . Then c ≈ 1.615 × 109 and the eccentricity is

1.615 × 109 c ≈ ≈ 0.967. a 1.67 × 109

3.6x107 x

CHAPTER 10. CONICS AND POLAR COORDINATES

684

80. We place the coordinate axes so that the origin is at the center of the ellipse. The length of the major axis is 200+2(4000)+1000 = 9200 so that a = 4600. Therefore c = 4600 − (200 + 4000) √ = 400. Then b2 = a2 − c2 = 46002 − 4002 = (1000 21)2 , y2 x2 √ + =1 and the equation is 2 4600 (1000 12)2

P satellite

200 mi

1000 mi center of earth

center of elliptical orbit

81. We place the coordinate axes so that the origin is at the point across the base. Thus  midway   2 x 2 a = 5 and b = 15 so the equation of the doorway is y = 15 1 − . The height of the 25    32 doorway at a point on the base 4 ft from the center is y = 152 1 − = 12 ft. 25 82. The base of the room is an ellipse with a = 20 and b = 16. We place the coordinate axes so that the origin is at the point in the center of the room and let the major axes be in the x-direction. c2 = a2 − b2 , so c2 = 202 − 162 , which gives c = 12. Thus the foci occur at the points (−12, 0) and (12, 0). Therefore the listening and whispering posts occur along the center line on the longer part of the base 12 ft in each direction from the center.

83.

84.

center vertices foci

ellipse (0, 1) (−2, 1),√(2, 1), (0, −2), √ (0, 4) (0, 1 − 5), (0, 1 + 5)

center vertices foci

ellipse (1, 4) (−2, 4), √(4, 4), (1, 3),√(1, 5) (1 − 2 2, 4), (1 + 2 2, 4)

shifted ellipse (4, 1) (2.1), (6.1), √ (4. − 2),√(4.4) (4, 1 − 5), (4, 1 + 5) shifted ellipse (−4, 7) (−7, 7), √ (−1, 7), (−4, 6),√(−4.8) (−4 − 2 2, 7), (−4 + 2 2, 7)

x2 y2 − =1 144 25 (b) Conjugate hyperbolas have the same asymptotes and do not intersect.

85. (a)

(x − 3/2)2 (y + 5/2)2 − = 1, The 5 5 equations of the asymptotes are y = 7/2−x and x = 1+x. These lines are perpendicular, thus the hyperbola is a rectangular hyperbola.

86. (a) The equation of the hyperbola can be written as

(b) The hyperbola in Problem 50 is a rectangular hyperbola.

10.2. PARAMETRIC EQUATIONS

685

√ 87. Since a = 4 and b = 20, we have c2 = a2 + b2 = 16 + 20 = 36 and hence c = 6. Thus the foci occur at F1 = (−6, 0) and F2 = (6, 0). The line joining (−6, −5) and F2 is given by 5 5 x − . The ray of light travels southwest along this line. y= 12 2 √ √ 88. (a) The distance from (0, b) to (a, 0) is a2 + b2 . Thus R = a2 + b2 = r. √ √ 2 2 2 2 (b) From A = a + √ r and B = R − b = a + b + r − b = a + b + (A − a) − b = 2 2 A − (a + b) + a + b we see that      A − B = a + b − a2 + b2 = (a + b)2 − a2 + b2 = a2 + 2ab + b2 − a2 + b2 > 0. Thus, A > B.

10.2

Parametric Equations

1. t x y

-3 -5 6

-2 -3 2

t x y

0 1 0

π √6 3 2

2.

-1 -1 0 π √4 2 2

1/4

1/2

0 1 0

1 3 2 π 3 1 2

3/4

3.

2 5 6

3 7 12

π 2

0 1

5π 6 √ - 23

1/4

7π √4 2 2

1/2

4.

5. y

y

y

x x

x

6.

7. y

8. y

y

x

x

x