1. Vector Calculus CHAPTER 9.1 ~ 9.4

2. Ch9.1~9.4_2 Contents  9.1 Vector Functions 9.1 Vector Functions  9.2 Motion in a Curve 9.2 Motion in a Curve  9.3 Curvature and Components of Acceleration 9.3 Curvature and Components of Acceleration  9.4 Partial Derivatives 9.4 Partial Derivatives

3. Ch9.1~9.4_3 9.1 Vector Functions  Introduction A parametric curve in space or space curve is a set of ordered (x, y, z), where x = f(t), y = g(t), z = h(t)(1)  Vector-Valued Functions Vectors whose components are functions of t, r(t) = = f(t)i + g(t)j orr(t) = = f(t)i + g(t)j + h(t)k are vector functions. See Fig 9.1

4. Ch9.1~9.4_4 Fig 9.1

5. Ch9.1~9.4_5 Example 1: Circular Helix Graph the curve by r(t) = 2cos ti + 2sin tj + tk, t  0 Solution x 2 + y 2 = (2cos t) 2 + (2sin t) 2 = 2 2 See Fig 9.2. The curve winds upward in spiral or circular helix.

6. Ch9.1~9.4_6 Fig 9.2

7. Ch9.1~9.4_7 Example 2 Graph the curve by r(t) = 2cos ti + 2sin tj + 3k Solution x 2 + y 2 = (2cos t) 2 + (2sin t) 2 = 2 2, z = 3 See Fig 9.3.

8. Ch9.1~9.4_8 Fig 9.3

9. Ch9.1~9.4_9 Example 3 Find the vector functions that describes the curve C of the intersection of y = 2x and z = 9 – x 2 – y 2. Solution Let x = t, then y = 2t, z = 9 – t 2 – 4t 2 = 9 – 5t 2 Thus, r(t) = ti + 2tj +(9 – 5t 2 )k. See Fig 9.4.

10. Ch9.1~9.4_10 Fig 9.4

11. Ch9.1~9.4_11 If exist, then DEFINITION 9.1 Limit of a Vector Function

12. Ch9.1~9.4_12 If, then (i), c a scalar (ii) (iii) THEOREM 9.1 Properties of Limits

13. Ch9.1~9.4_13 A vector function r is said to be continuous at t = a if (i) r(a) is defined, (ii) lim t  a r(t) exists, and (iii) lim t  a r(t) = r(a). DEFINITION 9.2 Continuity The derivative of a vector function r is (2) for all t which the limits exists. DEFINITION 9.3 Derivative of Vector Function

14. Ch9.1~9.4_14 Proof If, where f, g, and h are Differentiable, then THEOREM 9.2 Differentiation of a Components

15. Ch9.1~9.4_15 Smooth Curve  When the component functions of r have continuous first derivatives and r’(t)  0 for t in the interval (a, b), then r is said to be a smooth function, and the corresponding curve is called a smooth curve.

16. Ch9.1~9.4_16 Geometric Interpretation of r’(t) See Fig 9.5.

17. Ch9.1~9.4_17 Fig 9.5

18. Ch9.1~9.4_18 Example 4 Graph the curve by r(t) = cos 2t i + sin t j, 0  t  2 . Graph r’(0) and r’(  /6). Solution x = cos 2t, y = sin t, then x = 1 – 2y 2, −1  x  1 And r’(t) = −2sin 2ti + cos tj, r’(0) = j, r’(  /6) =

19. Ch9.1~9.4_19 Fig 9.6

20. Ch9.1~9.4_20 Example 5 Find the tangent line to x = t 2, y = t 2 – t, z = −7t at t = 3 Solution x’ = 2t, y’ = 2t – 1, z’ = −7 When t = 3, andr(3) = 9i + 6j – 21k that is P(9, 6, –21), then we have x = 9 + 6t, y = 6 + 5t, z = –21 – 7t

21. Ch9.1~9.4_21 Example 6  If r(t) = (t 3 – 2t 2 )i + 4tj + e -t k, then r’(t) = (3t 2 – 4t)i + 4j − e -t k, and r”(t) = (6t – 4)i + e -t k.

22. Ch9.1~9.4_22 If r is a differentiable vector function and s = u(t) is a differentiable scalar function, then the derivatives of r(s) with respect to t is THEOREM 9.3 Chain Rule

23. Ch9.1~9.4_23 Example 7 If r(s) = cos2si + sin2sj + e –3s k, s = t 4, then

24. Ch9.1~9.4_24 If r 1 and r 2 be differentiable vector functions and u(t) A differentiable scalar function. (i) (ii) (iii) (iv) THEOREM 9.4 Chain Rule

25. Ch9.1~9.4_25 Integrals of Vector Functions

26. Ch9.1~9.4_26 Example 8 If r(t) = 6t 2 i + 4e –2t j + 8cos 4t k, then where c = c 1 i + c 2 j + c 3 k.

27. Ch9.1~9.4_27 Length of a Space Curve  If r(t) = f(t)i + g(t)j + h(t)k, then the length of this smooth curve is (3)

28. Ch9.1~9.4_28 Example 9 Consider the curve in Example 1. Since, from (3) the length from r(0) to r(t) is Using then (4) Thus

29. Ch9.1~9.4_29 9.2 Motion on a Curve  Velocity and Acceleration Consider the position vector r(t) = f(t)i + g(t)j + h(t)k, then

30. Ch9.1~9.4_30 Example 1 Position vector: r(t) = t 2 i + tj + (5t/2)k. Graph the curve defined by r(t) and v(2), a(2). Solution so that See Fig 9.7.

31. Ch9.1~9.4_31 Fig 9.7

32. Ch9.1~9.4_32 Note: ‖ v(t) ‖ 2 = c 2 or v ‧ v = c 2 a(t) ‧ v(t) = 0

33. Ch9.1~9.4_33 Example 2 Consider the position vector in Example 2 of Sec 9.1. Graph the velocity and acceleration at t =  /4. Solution Recall r(t) = 2cos ti + 2sin tj + 3k. thenv(t) = −2sin ti + 2cos tj a(t) = −2cos ti −2sin t j and

34. Ch9.1~9.4_34 Fig 9.8

35. Ch9.1~9.4_35 Centripetal acceleration  See Fig 9.9. For circular motion, a(t) is called the centripetal acceleration.  Fig 9.9

36. Ch9.1~9.4_36 Curvilinear Motion in the Plane  See Fig 9.10. Acceleration of gravity : −gj An initial velocity: v 0 = v 0 cos  i + v 0 sin  j from an initial height s 0 = s 0 j, then where v(0) = v 0, then c 1 = v 0. Therefore v(t) = (v 0 cos  )i + (– gt + v 0 sin  )j

37. Ch9.1~9.4_37 Integrating again and using r(0) = s 0, Hence we have (1) See Fig 9.11

38. Ch9.1~9.4_38 Fig 9.10

39. Ch9.1~9.4_39 Fig 9.11

40. Ch9.1~9.4_40 Example 3 A shell is fired from ground level with v 0 = 768 ft/s at an angle of elevation 30 degree. Find (a) the vector function and the parametric equations of the trajectory, (b) the maximum attitude attained, (c) the range of the shell (d) the speed of impact. Solution (a) Initially we have s 0 = 0, and (2)

41. Ch9.1~9.4_41 Example 3 (2) Since a(t) = −32j and using (2) gives (3) Integrating again, Hence the trajectory is (4) (b) From (4), we see that dy/dt = 0 when −32t + 384 = 0 or t = 12. Thus the maximum height H is H = y(12) = – 16(12)2 + 384(12) = 2304 ft

42. Ch9.1~9.4_42 Example 3 (3) (c) From (4) we see that y(t) = 0 when −16t(t – 24) = 0, or t = 0, 24. Then the range R is (d) from (3), we obtain the impact speed of the shell

43. Ch9.1~9.4_43 9.3 Curvature and Components of Acceleration  Unit Tangent We know r’(t) is a tangent vector to the curve C, then (1) is a unit tangent. Since the curve is smooth, we also have ds/dt = ||r’(t)|| > 0. Hence (2) See Fig 9.19.

44. Ch9.1~9.4_44 Fig 9.16

45. Ch9.1~9.4_45  Rewrite (3) as that is, (4) From (2) we have T = dr/ds, then the curvature of C at a point is (3) DEFINITION 9.4 Curvature

46. Ch9.1~9.4_46 Example 1 Find the curvature of a circle of radius a. Solution We already know the equation of a circle is r(t) = a cos ti + a sin tj, then We get Thus, (5)

47. Ch9.1~9.4_47 Fig 9.17

48. Ch9.1~9.4_48 Tangential and Normal Components  Since T is a unit tangent, then v(t) = ||v(t)||T = vT, then (6) Since T  T = 1 so that T  dT/dt = 0 (Theorem 9.4),we have T and dT/dt are orthogonal. If ||dT/dt||  0, then (7) is a unit normal vector to C at a point P with the direction given by dT/dt. See Fig 9.18.

49. Ch9.1~9.4_49 Fig 9.18

50. Ch9.1~9.4_50  The vector N is also called the principal normal. However  =║dT / dt║/ v, from (7) we have dT/dt =  vN. Thus (6) becomes (8) By writing (8) as a(t) = a N N + a T T (9) Thus the scalar functions a N and a T are called the tangential and normal components.

51. Ch9.1~9.4_51 The Binormal  A third vector defined by B = T  N is called the binormal. These three vectors T, N, B form a right-hand set of mutually orthogonal vectors called the moving trihedral. The plane of T and T is called the osculating plane, the plane of N and B is called the rectifying plane. See Fig 9.19.

52. Ch9.1~9.4_52 Fig 9.19

53. Ch9.1~9.4_53 Example 2 The position vector r(t) = 2cos ti + 2sin tj + 3tk, find the vectors T, N and B, and the curvature. Solution Since from (1), Next we have

54. Ch9.1~9.4_54 Example 2 (2) Hence (3) gives N = – cos ti – sin tj Now, Finally using and

55. Ch9.1~9.4_55 Example 2 (3) From (4) we have

56. Ch9.1~9.4_56 Formula for a T, a N and Curvature  Observe then (10) On the other hand

57. Ch9.1~9.4_57 Since ||B|| = 1, it follows that (11) then (12)

58. Ch9.1~9.4_58 Example 3 The position vector r(t) = ti + ½t 2 j + (1/3)t 3 k is said to be a “twisted” cube”. Find the tangential and normal components of the acceleration at t. Find the curvature. Solution Since v  a = t + 2t 3 and From (10),

59. Ch9.1~9.4_59 Example 3 (2) Now and From (11) From (12)

60. Ch9.1~9.4_60 Radius of Curvatures   = 1/  is called the radius of curvature.  See Fig 9.20.

61. Ch9.1~9.4_61 9.4 Partial Derivatives  Functions of Two Variables See Fig 9.21. The graph of a function z = f(x, y) is a surface in 3-space.

62. Ch9.1~9.4_62 Fig 9.21

63. Ch9.1~9.4_63 Level Curves  The curves defined by f(x, y) = c are called the level curves of f. See Fig 9.22.

64. Ch9.1~9.4_64 Example 1 The level curves of f(x, y) = y 2 – x 2 are defined by y 2 – x 2 = c. See Fig 9.23. For c = 0, we obtain the lines y = x, y = −x.

65. Ch9.1~9.4_65 Level Surfaces  The level surfaces of w = F(x, y, z) are defined by F(x, y, z) = c.

66. Ch9.1~9.4_66 Example 2 Describe the level curves of F(x, y, z) = (x 2 + y 2 )/z. Solution For c  0, (x 2 + y 2 )/z = c, or x 2 + y 2 = cz. See Fig 9.24. ，

67. Ch9.1~9.4_67 Fig 9.24

68. Ch9.1~9.4_68 Partial Derivatives  For y = f(x), For z = f(x, y), (1) (2)

69. Ch9.1~9.4_69 Example 3 If z = 4x 3 y 2 – 4x 2 + y 6 + 1, find, Solution

70. Ch9.1~9.4_70 Alternative Symbols  If z = f(x, y), we have

71. Ch9.1~9.4_71 Higher-Order and Mixed Derivatives  If z = f(x, y), we have:  Second-order partial derivatives:  Third-order partial derivatives:  Mixed second-order partial derivatives:

72. Ch9.1~9.4_72 Alternative Symbols  If f has continuous second partial derivatives, then f xy = f yx (3)

73. Ch9.1~9.4_73 Example 4 If then

74. Ch9.1~9.4_74 If z = f(u, v) is differentiable, and u = g(x, y) and v = h(x, y) have continuous first partial derivatives, then (5) THEOREM 9.5 Chain Rule

75. Ch9.1~9.4_75 Example 5 If z = u 2 – v 3, u = e 2x – 3y, v = sin(x 2 – y 2 ), find and Solution Since then (6) (7)

76. Ch9.1~9.4_76 Special Case  If z = f(u, v) is differentiable, and u = g(t) and v = h(t) are differentiable, then (8) If z = f(u 1, u 2,…, u n ) and each variable u 1, u 2,…, u n are functions of x 1, x 2,…, x k, we have (9)

77. Ch9.1~9.4_77 Similarly, if u 1, u 2,…, u n are functions of a single variable t, then (10) These results can be memorized in terms of a tree diagram. See next page.

78. Ch9.1~9.4_78

79. Ch9.1~9.4_79 Example 6 If r = x 2 + y 5 z 3 and x = uve 2s, y = u 2 – v 2 s, z = sin(uvs 2 ), find  r/  s. Solution According to the tree diagram,

80. Ch9.1~9.4_80 Example 7 If z = u 2 v 3 w 4 and u = t 2, v = 5t – 8, w = t 3 + t ， find dz/dt. Solution Another approach: differentiate z = t 4 (5t – 8) 3 (t 3 + t) 4

81. Ch9.1~9.4_81 Zill 工程數學 ( 上 ) 茆政吉譯