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

2. 9.1 Vector Functions
Introduction A parametric curve in space or space curve is a set of ordered coordinates (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), g(t)> = f(t)i + g(t)j or r(t) = <f(t), g(t), h(t)> = f(t)i + g(t)j + h(t)k are vector functions. See Fig 9.1

3. Fig 9.1

4. Example 1: Circular Helix
Graph the curve by r(t) = 2cos ti + 2sin tj + tk, t  0
Solution x2 + y2 = (2cos t)2 + (2sin t)2 = 22 See Fig 9.2. The curve winds upward in spiral or circular helix.

5. Fig 9.2

6. Example 2 Graph the curve by r(t) = 2cos ti + 2sin tj + 3k
Solution x2 + y2 = (2cos t)2 + (2sin t)2 = 22, z = 3 See Fig 9.3.

7. Fig 9.3

8. Example 3
Find the vector functions that describes the curve C of the intersection of y = 2x and z = 9 – x2 – y2.
Solution Let x = t, then y = 2t, z = 9 – t2 – 4t2 = 9 – 5t2 Thus, r(t) = ti + 2tj +(9 – 5t2)k. See Fig 9.4.

9. Fig 9.4

10. Limit of a Vector Function
DEFINITION 9.1
If exist, then
Limit of a Vector Function

11. If , then (i) , c a scalar (ii) (iii)
THEOREM 9.1
If , then (i) , c a scalar (ii) (iii)
Properties of Limits

12. A vector function r is said to be continuous at t = a if
(i) r(a) is defined, (ii) limta r(t) exists, and
(iii) limta r(t) = r(a).
DEFINITION 9.2
Continuity
The derivative of a vector function r is (2)
for all t where the limits exists.
DEFINITION 9.3
Derivative of Vector Function

13. If , where f, g, and h are Differentiable, then Proof
THEOREM 9.2
If , where f, g, and h are
Differentiable, then
Differentiation of Components
Proof

14. 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 in (a, b), and the corresponding curve is called a smooth curve.

15. Geometric Interpretation of r’(t)
See Fig 9.5.

16. Fig 9.5

17. 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 – 2y2, −1  x  1 and r’(t) = −2sin 2ti + cos tj, r’(0) = j, r’(/6) =

18. Fig 9.6

19. Example 5
Find the tangent line to x = t2, y = t2 – t, z = −7t at t = 3
Solution x’ = 2t, y’ = 2t – 1, z’ = −7 when t = 3, and r(3) = 9i + 6j – 21k that is P(9, 6, –21), then we have x = 9 + 6t, y = 6 + 5t, z = –21 – 7t

20. If r is a differentiable vector function and s = u(t) is a
THEOREM 9.3
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
Chain Rule

21. Example 7
If r(s) = cos2si + sin2sj + e–3sk, s = t4, then

22. If r1 and r2 are differentiable vector functions and u(t)
THEOREM 9.4
If r1 and r2 are differentiable vector functions and u(t)
is a differentiable scalar function.
(i)
(ii)
(iii)
(iv)
Chain Rule

23. Integrals of Vector Functions

24. Example 8
If r(t) = 6t2i + 4e–2t j + 8cos 4t k, then where c = c1i + c2j + c3k.

25. Length of a Space Curve
If r(t) = f(t)i + g(t)j + h(t)k is a smooth function, then the length of this smooth curve over (a,b) is (3)

26. Arc Length as a Parameter
A curve in the plane or in space can be parameterized in terms of the arc length, r(s).
||r’(s)||=1, i.e. r’(s) is a unit tangent vector. (hint: the length of the curve from r(0) to r(s) is s).

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

28. 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

29. Example 1
Position vector: r(t) = t2i + tj + (5t/2)k. Graph the curve defined by r(t) and v(2), a(2).
Solution so that See Fig 9.7.

30. Fig 9.7

31. Particle moves with constant speed
‖v(t)‖2 = c2 or v‧v = c a(t)‧v(t) = 0

32. 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. then v(t) = −2sin ti + 2cos tj a(t) = −2cos ti −2sin t j and

33. Fig 9.8

34. Centripetal acceleration
See Fig 9.9. For circular motion, a(t) is called the centripetal acceleration.
Fig 9.9

35. Curvilinear Motion in the Plane
See Fig Acceleration of gravity : −gj An initial velocity: v0 = v0 cos i + v0 sin j from an initial height s0 = s0 j, then where v(0) = v0, then c1 = v0. Therefore v(t) = (v0cos )i + (– gt + v0sin )j

36. Integrating again and using r(0) = s0, Hence we have (1) See Fig 9.11

37. Fig 9.10

38. Fig 9.11

39. Example 3
A shell is fired from ground level with v0 = 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 s0 = 0, and (2)

40. 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 = 0 or t = 12. Thus the maximum height H is H = y(12) = – 16(12) (12) = 2304 ft

41. 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

42. 9.3 Curvature and Components of Acceleration
Unit Tangent We know r’(t) is a tangent vector to the smooth 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.

43. Fig 9.16

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

45. 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)

46. Fig 9.17

47. 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.

48. Fig 9.18

49. The vector N is also called the principal normal
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) = aNN + aTT (9) Thus the scalar functions aN and aT are called the normal and tangential components.

50. 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 N is called the osculating plane, the plane of N and B is called the rectifying plane. See Fig 9.19.

51. Fig 9.19

52. 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

53. Example 2 (2) Hence (7) gives N = – cos ti – sin tj
Now, Finally using and

54. Example 2 (3)
From (4) we have

55. Formula for aT, aN and Curvature
Observe then (10) On the other hand

56. Since ||B|| = 1, it follows that (11) then (12)

57. Example 3
The position vector r(t) = ti + ½t2j + (1/3)t3k 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 + 2t3 and From (10),

58. Example 3 (2)
Now and From (11) From (12)

59. Radius of Curvatures  = 1/ is called the radius of curvature.
See Fig 9.20.

60. 9.4 Partial Derivatives
Functions of Two Variables See Fig The graph of a function z = f(x, y) is a surface in 3-space.

61. Fig 9.21

62. Level Curves
The curves defined by f(x, y) = c are called the level curves of f. See Fig 9.22.

63. Example 1
The level curves of f(x, y) = y2 – x2 are defined by y2 – x2 = c. See Fig For c = 0, we obtain the lines y = x, y = −x.

64. Level Surfaces
The level surfaces of w = F(x, y, z) are defined by F(x, y, z) = c.

65. Example 2 Describe the level curves of F(x, y, z) = (x2 + y2)/z.
Solution For c  0, (x2 + y2)/z = c, or x2 + y2 = cz. See Fig ，

66. Fig 9.24

67. Partial Derivatives
For y = f(x), For z = f(x, y), (1) (2)

68. Example 3
If z = 4x3y2 – 4x2 + y6 + 1, find ,
Solution

69. Alternative Symbols
If z = f(x, y), we have

70. 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:

71. Alternative Symbols
If f has continuous second partial derivatives, then fxy = fyx　　　　　　　　　　 (3)

72. Example 4
If then

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

74. Example 5 If z = u2 – v3, u = e2x – 3y, v = sin(x2 – y2), find and
Solution Since then (6) (7)

75. Special Case
If z = f(u, v) is differentiable, and u = g(t) and v = h(t) are differentiable, then (8) If z = f(u1, u2,…, un) and each variable u1, u2,…, un are functions of x1, x2,…, xk, we have (9)

76. Similarly, if u1, u2,…, un are functions of a single variable t, then
Similarly, if u1, u2,…, un are functions of a single variable t, then (10) These results can be memorized in terms of a tree diagram. See next page.

78. Example 6
If r = x2 + y5z3 and x = uve2s, y = u2 – v2s, z = sin(uvs2), find r/s.
Solution According to the tree diagram,

79. Example 7
If z = u2v3w4 and u = t2, v = 5t – 8, w = t3 + t， find dz/dt.
Solution Another approach: differentiate z = t4(5t – 8)3(t3 + t)4

80. Thank You !
Zill工程數學(上)茆政吉譯