Vector Calculus. Chapter 2

Transcription

1 Chaper Vecor Calculus. Elemenar. Vecor Produc. Differeniaion of Vecors 4. Inegraion of Vecors 5. Del Operaor or Nabla (Smbol 6. Polar Coordinaes

2 Chaper Coninued 7. Line Inegral 8. Volume Inegral 9. Surface Inegral 0. Green s Theorem. Divergence Theorem (Gauss Theorem. Sokes Theorem

3 . Elemenar Vecor Analsis Definiion. (Scalar and vecor Scalar is a quani ha has magniude bu no direcion. For insance mass, volume, disance Vecor is a direced quani, one wih boh magniude and direcion. For insance acceleraion, veloci, force

4 We represen a vecor as an arrow from he origin O o a poin A. O OA A or O a A The lengh of he arrow is he magniude of he vecor wrien as or. OA a 4

5 .. Basic Vecor Ssem Uni vecors,, Perpendicular o each oher In he posiive direcions of he aes have magniude (lengh 5

6 .. Magniude of vecors Le P (,, z. Vecor wih magniude (lengh OP p i j z k [,, z] OP p OP p z is defined b 6

7 .. Calculaion of Vecors. Vecor Equaion Two vecors are equal if and onl if he corresponding componens are equals Le a ai a Then a b a j a k b, andb bi b a b, a b j bk. 7

8 8. Addiion and Subracion of Vecors. Muliplicaion of Vecors b Scalars k b a j b a i b a b a ( ( ( k b j b i b b ( ( ( hen isascalar, If

9 Eample. Given p 5 i j - k and q 4 i - j k. Find a p q b p - q c Magniude of vecor p d q - 0 p 9

10 . Vecor Producs If a a i a j a k and b bi b j b k, Scalar Produc (Do produc or a b ab ab ab a. b a b cos, is heangle beween aandb 0

11 Vecor Produc (Cross produc i j k a b a a a or a b a bsin n b b b ab -ab i- ab - ab j ab -ab k

12 Whaever direcion of a b is, i has o be perpendicular o boh a and b. Le c a b Swing a o b The direcion of c is poined ino he page. hps:// Do produc and cross produc Visualizaion

13 Applicaion of Muliplicaion of Vecors a b a b a Given vecors and, projecion ono is defined b The vecor projecion of a on b is he uni vecor of b b he scalar projecion of a on b: Vecor projecion ab. proj a b b b ab. comp a b b ab. lengh ( l b Scalar projecion a comp b a b The scalar projecion of a on b is he magniude of he vecor projecion of a on b.

14 4

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18 b The area of riangle A ab. b a 8

19 9 c The area of parallelogram d The volume of erahedrone e The volume of parallelepiped a b a b A a b c 6 c c c b b b a a a 6 V a. b c a b c c c c b b b a a a V a. b c

20 0

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22 Eample. and. beween angle andhe, deermine., and Given b a b a a b k j i b k j i a -

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24 . Vecor Differenial Calculus Le A be a vecor depending on parameer u, A( u a ( u i a ( u j a ( u k The derivaive of A(u is obained b differeniaing each componen separael, z da da du da du da du i j z k du 4

25 A ( u. The nh derivaive of vecor is given b dn A dna dna dna i j k du dun n dun dun z d n A The magniude of is n du d n n du A dna n n n d a d a n n z du du du 5

26 Eample. If A u hence i-u j 5k da du d A du 6

27 Eample.4 The posiion of a moving paricle a ime is given b 4,, z 5. Obain The veloci and acceleraion of he paricle. The magniude of boh veloci and acceleraion a. 7

28 Soluion The parameer is, and he posiion vecor is r ( (4 i ( j ( 5 The veloci is given b dr 4i ( j ( d The acceleraion is dr j d 0 k (6 0 k.. k. 8

29 A, he veloci of he paricle is dr ( 4i (( j (( d 4i 5j k. and he magniude of he veloci is 0( k dr ( 4 5 d 0. 9

30 A, he acceleraion of he paricle is d r( j (6( 0 k d j 6k. and he magniude of he acceleraion is d r( 6 d 65. 0

31 .. Differeniaion of Two Vecors A ( B( If boh and are vecors, hen a b c d u u d ( ca c da du du d ( A B da db du du du d db da ( A. B A.. B du du d da ( A B A db du B du du du

32 .. Parial Derivaives of a Vecor If vecor depends on more han one parameer, i.e A,,, (,,, (,,, (,,, ( k u u u a j u u u a i u u u a u u Au n z n n n

33 Parial derivaive of wih respec o is given b e..c. A u, k u u a j u u a i u u a u u A k u a j u a i u a u A z z

34 Eample.5 6, 6, 6 4, 6, 4 hen ( ( If vi u v F u v F k ui v F uk j u F vk j uvi v F k u u j i v u F k v u j v u uv i F - - 4

35 Eercise. - u v F u v F v F u F v F u F k v u j v u u vi F,,, hen ( ( If 5

36 .4 Vecor Inegral Calculus The concep of vecor inegral is he same as he inegral of real-valued funcions ecep ha he resul of vecor inegral is a vecor. If hen Au ( a b a ( u i a Au ( du b a b a ( u a j a ( u k a ( u dui z ( u duj b a a ( u duk. z 6

37 Eample ] [ 5] [ ] [ 4 5 ( 4 (. calculae, 4 5 ( 4 ( If 4 k j i k j i dk dj di Fd Fd k j i F Answer 7

38 Eercise ( (. calculae, 4 ( ( If k j i dk dj di Fd Fd k j i F Answer 8

39 .5 Del Operaor Or Nabla (Smbol Operaor is called vecor differenial operaor, defined as i j k. z 9

40 .5. Grad (Gradien of Scalar Funcions If,,z is a scalar funcion of hree variables and is differeniable, he gradien of is defined as grad i * isascalar funcion * isa vecor funcion j k. z 40

41 Eample.7 z z z z z z z z z z z,hence Given (,,. P a grad,deermine If Soluion 4

42 ( (( (( (( ( (( ( (( ( ( ( ((( wehave (,,, P A. ( ( ( z Therefore, k j i k j i k z z j z z i z z k j i 4

43 Eercise. If z z, deermine grad a poin P (,,. 4

44 Soluion 0. 6 (,,, P A Grad z hen, Given k j i z z 44

45 .5.. Grad Properies If A and B are wo scalars, hen ( A B AB ( AB A( B B( A 45

46 .5. Direcional Derivaive Direciona lderivaive of d ds wherea a. grad dr dr, in hedirecion ofais which isauni veco r in hedirecion ofdr. 46

47 Eample.8 Compue hedireciona lderivaive of z a hepoin (,, -in hedirecion ofhevecor A i j-4k. z 47

48 Soluion Direcional derivaive of in he direcion of d a. grad ds wheregrad i j k z Given z z, hence (z i (4 z j ( anda z k. a A. A 48

49 9. 4 ( hen, 4 given Also, (( (( ( (4(( ( ((( (,,-, A A k j i A k j i k j i 49

50 A Therefore, a i j- 4 k A Then, dφ a. ds i j- 4 k.(6 i 9j-k (6 (9-4 (

51 .5. Uni Normal Vecor Equaion (,, z consan is a surface equaion. Since (,, z consan, he derivaive of is zero; i.e. d dr cos dr.grad grad cos 0 0 5

52 This shows ha when (,, z consan, grad dr. grad dr z Vecor grad is called normal vecor o he surface (,, z consan 5

53 Uni normal vecor is denoed b Eample.9 Calculae he uni normal vecor a (-,, for z z 0. n. 5

54 Soluion Given z z 0. Thus 6. 4 and ( ( ( (-,,, A. ( ( ( - - k j i k j i k j z i z ( 6 6 is vecor normal uni The k j i k j i n 54

55 .5.4 Divergence of a Vecor...(. as defined is of divergence he, If z a a a A diva k a j a i a k z j i A diva A k a j a i a A z z z 55

56 Eample.0. (( (( (( (,,, poin A.. (,,. poin a deermine, If diva z z z a a a A diva diva k z zj i A z Answer 56

57 Eercise.4. 4 (,,, poin A. (,,. poin a deermine, If - diva z a a a A diva diva k z z j i A z Answer 57

58 Remarks Aisa vecor funcion, bu diva isascalar funcion. IfdivA 0, vecor Aiscalled solenoid vecor. 58

59 .5.5 Curl of a Vecor. ( b isdefined of curl he, If z z z a a a z k j i A A curl k a j a i a k z j i A A curl A k a j a i a A 59

60 Eample. If A ( 4 - z i ( deermine curla a (,, -. j - zk, 60

61 Soluion. 4 ( ( ( ( ( ( ( ( k j z z zi k z j z z z i z z z z z k j i A A curl

62 Eercise ( (( ( ( ( (( ( ( ( (,,-, A k j i k j i A curl (,,. poin a deermine, ( ( If A curl k z j z i z A - -

63 Answer. 6 5 (,,, A. ( ( ( k j i A curl k z j z z i z z A curl Remark funcion. isalsoa vecor and funcion isa vecor A curl A

64 .6 Polar Coordinaes Polar coordinae is used in calculus o calculae an area and volume of small elemens in eas wa. Les look a siuaions where des Cares Coordinae can be rewrien in he form of Polar coordinae. 64

65 .6. Polar Coordinae for Plane (r, θ d ds rcos rsin ds rdrd 65

66 .6. Polar Coordinae for Clinder (,, z z ds dv cos sin z z z ds d dz dv d d dz 66

67 .6. Polar Coordinae for Sphere (r,, z rsin cos rsin sin z rcos ds r sin d d dv r sin drd d r 67

68 Eample. (Volume Inegral Calculae FdVwhereF i z j k andv and V isaspace bounded b z 0, 9. z 4 - z 4 68

69 Soluion Since i is abou a clinder, i is easier if we use clindrical polar coordinaes, where cos, where 0 sin, z z, dv d,0,0z4. d dz 69

70 .7 Line Inegral Ordinar inegral f ( d, we inegrae along he -ais. Bu for line inegral, he inegraion is along a curve. f (s ds f (,, z ds A r O rdr B 70

71 .7. Vecor Field, FInegral Le a vecor field and r di The scalar produc F. dr ( F i F d Fd Fd Fdz. F F i F j F k z dj dzk. F. dr is wrien as j F k.( di dj dzk z z 7

72 Ifa vecor field isalong hecurve, C hen he lineinegral offalonghe curve C fromapoin A oanoher F. dr F d F d c F c poin Bisgiven b c c F dz. z 7

73 7 Eample.. if, 4, curve he along (4,, ob (0,0,0 froma. Calculae zk zj i F z dr F c -

74 74 Soluion. 4 4 And. 4 4 ( ( (4( ( (4 Given dk dj di dzk dj di dr k j i k j i zk zj i F - - -

75 75. 6 (8 6 8 ( 4 ( (4 (4 (4 ( 4 (4 4 4 (. Then d d d d d d d dk dj di k j i F dr ,, 4, 4 (4,,, and,a B 0. 0, 0, 0, 4 (0,0,0, A A

76 ( d F dr B A

77 77 Eercise ,, curve on he (,, (0,0,0oB froma. calculae, If - F dr z dr F zk zj i F B A c Answer

78 78 * Double Inegral * 4uni., ( inegrals. order boh in, ( Find. and 0, line asraigh b region bounded in 4, Given ( - R R da f da f R f Answer.6 Eample

79 .8 Volume Inegral.8. Scalar Field, F Inegral If V is a closed region and F is a scalar field in region V, volume inegral F of V is FdV Fdddz V V 79

80 Eample.4 Scalar funcion F defeaed in one cubic ha has been buil b planes 0,, 0,, z 0 and z. Evaluae volume inegral F of he cubic. z O 80

81 Soluion FdV V z 0 0. z 0 0 z dz [ z] z 0 z 0 ddz 0 0 dddz 0 [ ] dz ddz 6 8

82 .9 Surface Inegral.9. Vecor Field, Inegral If vecor field inegral F F F defeaed on surface S, surface of S is defined as where S S n S F. ds S F. nds 8

83 Eample.5 Vecor field F i j k defeaed on surface S: z 9 and bounded b 0, 0, z 0 in he firs ocan. Evaluae F. ds. S 8

84 Soluion Given S: z 9is bounded b 0, 0, z 0in he s ocan. This refer o sphere wih cener a (0,0,0 and radius, r, in he s ocan. z O 84

85 So, grad S is S S i S j S k z i j zk, and S ( ( ( z z

86 Therefore, S n S S F. ds F. nds S i j zk 6 ( i j zk. ( i j k ( S i j zk ds ( z ds. S 86

87 Using polar coordinae of sphere, rsin cos sin cos rsin sin sin sin z rcos cos ds r sin d d 9sin d d where 0,. 87

88 F ds S. [(sin cos (sin sin [sin sin cos 0 0 (sin sin cos ][9sin ] dd sin sin sin cos ] dd

89 89 Eercise.7 ocan. 0in he 0and 0, 4, b bounded ofheregion isasurface and where on, Evaluae s z z S k z j i F S F ds S :8 6 Answer

90 .0 Green s Theorem If c is a closed curve in couner-clockwise on plane-, and given wo funcions P(, and Q(,, - dd ( Pd S Q P c Qd where S is he area of c. 90

91 Eample.6 Prove Green's Theorem for [( d ( d ] c which has been evaluaed b boundar ha defined as 0, 0and 4 in he firs quarer. Soluion C C O C 9

92 c Given [( d ( d ] where P and Q.We defined curve c as c, c and c. i For c : 0, d 0and 0 c c d ( Pd Qd ( d ( d 9

93 ii For c : 4,in he firs quarer from (,0 o (0,. This curve acuall a par of a circle. Therefore, i's more easier if we inegrae b using polar coordinae of plane, cos, sin, 0 d - sin d, d cos d 9

94 sin sin 8cos cos 8sin cos 8sin ( cos 8sin 4cos 8sin ( ] cos ( (sin ((cos sin ( (sin [((cos ( ( ( d d d d d d Qd Pd c c

95 iii For c : 0, d 0, 0 Pd Qd d d c ( ( c ( 0 d 0-4. ( Pd Qd 8 ( c

96 b Now, we evaluae Q P S - dd where Q and P. Again, because his is a par of he circle, we shall inegrae b using polar coordinae of plane, rcos, rsin wh ere 0r, 0 and dd ds rdrd. 96

97 S Q - P dd S ( - dd 0 r 0 ( -r sin sin 0 r - r 6 sin 0 - d 6 cos 0-6. rdrd 0 d 97

98 Therefore, ( Pd Qd Q P dd C S LHS RHS Green's Theorem has been proved. 98

99 . Divergence Theorem (Gauss Theorem If S is a closed surface including region V in vecor field V F divfdv S F. ds. f f f divf z 99

100 Eample.7 Prove Gauss' Theorem for vecor field, F i j z k in he region bounded b planes z 0, z 4, 0, 0and 4 in he firs ocan. 00

101 Soluion 4 z S S 4 S O S S 5 0

102 For his problem, he region of inegraion is bounded b 5 planes : S : z 0 S : z 4 S : 0 S4 : 0 S5 : 4 To prove Gauss' Theorem, we evaluae boh and F. ds, he answer should be he same. S V divfdv 0

103 We evaluae divfdv. Given F i j z k. V So, divf ( ( ( z z z. Also, divfdv ( z dv. V V The region is a par of he clinder. So, we inegrae b using polar coordinae of clinder, cos ; sin ; z z dv dd dz where 0,0,0z4. 0

104 Therefore, V ( z dv ( z 0 4 [ z z ] divfdv 0. V [0 ] (0 dd 0 (40 d d z dzd d dd 04

105 Now, we evaluae F. ds F. nds. S S i S : z 0, n- k, ds rdrd F i j 0k F. n ( i j.( - k 0 F. nds 0. S 05

106 ii S : z 4, n k, ds rdrd F i j (4 k i j 6 k. 6 F. n ( i j 6 k.( k 6. Therefore for S, 0r, 0 F nds S 0 r 0 6. rdrd 06

107 iii S : 0, n- j, ds ddz F i j z k F n i j zk j. (.( - -. Therefore for S, 0, 0z4 4 F. nds (- dzd S 0 z

108 iv S : 0, n- i, ds ddz 4 F. nds 0. S 4 F 0i j zk j zk F. n ( j zk.( i

109 v 5: 4, S ds d dz S i j and S S i j 5 n S 5 4 ( i j. B using polar coordinae of clinder : cos, sin, z z where for S5 :, 0, 0z 4, ds d dz 09

110 F. n ( i j z k. i j ( cos ( sin cos sin ; kerana. (cos sin. S 5 F. nds 4 0 z ((cos sin ( d dz 0

111 Finall, F. ds F. ds F. ds F. ds F. ds F. ds S S S S S S F. ds 0. S LHS RHS Gauss' Theorem has been proved.

112 . Sokes Theorem If F is a vecor field on an open surface S and boundar of surface S is a closed curve c, herefore S curlf ds F dr c i j k curlf F z f f f z

113 Eample.8 Surface S is he combinaion of iapar of he clinder 9beween z 0 and z 4for 0. iiahalf of he circlewih radius a z 4,and iii plane 0 If F zi j zk,prove Sokes' Theorem for his case.

114 Soluion z S 4 S We can divide surface Sas S : 9for0 z4and 0 S : z 4, half of he circle wih radius S : 0 C C O S 4

115 We can also mark he pieces of curve C as C : Perimeer of a half circle wih radius. C : Sraigh line from (-,0,0 o (,0,0. Le sa, we choose o evaluae S curlf ds firs. Given F zi j zk 5

116 6 So, ( ( ( ( ( ( ( k j z k z j z z z i z z z z z k j i F curl

117 B inegraing each par of he surface, ( i Forsurface S : 9, S i j and S ( ( 6 7

118 Then, S i j n ( i S 6 j and curlf n ( -z j ( -z. k i j 8

119 B using polar coordinae of clinder ( because S : 9 is a par of he clinder, cos, sin, z z ds d dz where, 0 and 0z4. 9

120 Therefore, curl F n ( -z sin -z sin ( - z; because Also, dsd dz 0

121 curl F ds curlf nds S S 4 z sin ( -z ddz ( -z -cos 0 ( -z( -(- dz -4 dz

122 (ii For surface surface is S : z 4 nk., normal vecor uni o he B using polar coordinae of plane, rsin, z 4 dan ds rdrd where 0r and 0.

123 S S 8 r 0 0 r r 0 0 curl F n ( z j k - k rsin curl F ds curl FndS ( rsin ( rdrd sin d dr

124 (iii For surface S : 0, normal vecor uni o he surface is ds d dz n -j The inegraion limis :. - and 0z4 So, curlf n (( -z j k ( -j z- 4

125 Then, curlf. ds curlf. nds S S 4 ( z- dzd - z 0 4. curlf. ds curlf. ds curlf. ds curlf. ds S S S S

126 Now, we evaluae F. dr for each pieces of he curve C. C i C is a half of he circle. Therefore, inegraion for C will be more easier if we use polar coordinae for plane wih radius r, ha is cos, sin dan z 0 where 0. 6

127 F zi j zk and (cos (sin j 9sin cos j dr di d j dzk - sin d i cos d j. 7

128 From here, F. dr 7sin cos d. F. dr 7sin cos C 0-9cos 8. 0 d 8

129 ii Curve C is a sraigh line defined as, 0 and z 0, where -. Therefore, F zi j zk 0 F. dr 0 C 9

130 F. dr F. dr F. dr C C C We alread show ha curlf. ds F. dr S C Sokes' Theorem has been proved. 0