Chpter 9 Vector Differentil Clculus, Grd, Div, Curl Kreyszig by YHLee;100510; 9-1 9.1 Vectors in 2 Spce nd 3 Spce Two kinds of quntities used in physics, engineering nd so on. A sclr : A quntity representing mgnitude. A vector : A quntity representing mgnitude nd direction. Terminl point A vector is represented by n rrow. The til is clled initil point. The hed (or the tip) is clled terminl point Initil point The distnce between the initil nd the terminl points is clled the distnce, mgnitude, or norm. A velocity is vector, v, nd its norm is v. A vector of length 1 is clled unit vector. Definition Equlity of Vectors Two vectors nd b re equl, = b, if they hve the sme length nd the sme direction. A trnsltion does not chnge vector. Components of Vector A vector is given with initil point P: ( x, y, z ) nd terminl point :(,, ) 1 1 1 The vector hs three components long x, y, nd z coordintes. =,, [ ] where 1 = x2 x1, 2 = y2 y1, 3 = z2 z1 The length is given by 2 2 2 = + + Q x y z. 2 2 2
Exmple 1 Components nd Length of Vector : 4, 0, 2 A vector with initil point P ( ) nd terminl point :( 6, 1, 2) Q. Kreyszig by YHLee;100510; 9-2 Its components re = 6 4 2, = 1 0 1, = 2 2 0 Hence = [ 2, 1, 0] = + + Its length is ( ) 2 2 2 2 1 0 5 Position vector A point A:(x, y, z) is given in Crtesin coordinte system. The position vector of the point A is vector drwn from the origin to the point A. r = x, y, z [ ] Theorem 1 Vectors s Ordered Rel Triple Numbers A vector in Crtesin coordinte system cn be represented by three rel numbers (,, ) which corresponds to three components. Hence = b mens tht 1 = b1, 2 = b2 nd 3 = b3., Vector Addition, Sclr Multipliction Definition Addition of two vectors = [ 1, 2, 3] + b = [ + b, + b, + b ] 1 1 2 2 3 3 nd b = [ b, b, b ] Prllelogrm rule Hed To Til rule
Kreyszig by YHLee;100510; 9-3 Bsic Properties of Vector Addition + b= b+ ( + b) + c = + ( b + c) + 0= 0+ = + = ( ) 0 : commuttive : ssocitive A vector hs the sme length s but with the opposite direction. Definition In sclr multipliction, vector is multiplied by sclr c. c = c, c, c [ ] c hs the sme direction with with incresed length for c >0, opposite direction for c <0 Bsic Properties of Sclr Multipliction c ( + b) = c+ cb ( c+ k) = c+ k ck ( ) = ( ck ) = ck 1 =, 0 = 0 1 =, ( ) b+ = b ( ) Unit Vectors ˆˆ i, j, k ˆ A vector cn be represented s =,, = i ˆ + ˆ j+ k [ ] ˆ using three unit vectors ˆi, ˆj, k ˆ long x, y, z xes ˆ 1,0,0 ˆ 0,1,0 k ˆ = 0,0,1 i = [ ], j = [ ], [ ]
Kreyszig by YHLee;100510; 9-4 Exmple 2 Let Vector Addition. Multipliction by Sclrs = [ 4, 0,1] 1 nd b = 2, 5, 3 Then = [ 4, 0, 1], 7 = [ 28, 0, 7] 4 + b = 6, 5, 3 2 2( b) = 2 2,5, = 2 2b 3 Exmple 3 iˆˆ, j, k ˆ Nottion for Vectors The two vectors in Exmple 2 re = 4ˆ i + kˆ ˆ ˆ 1 b = 2i 5j+ k 3 ˆ 9.2 Inner Product (Dot Product) Definition Inner Product (Dot Product) of Vectors is defined s b= bcosγ= b + b + b 1 1 2 2 3 3 0 γ π, the ngle between nd b Orthogonlity is orthogonl to b if b= 0. γ should be π /2 when 0 nd b 0 Theorem 1 Orthogonlity b= 0 if nd only if nd b re perpendiculr to ech other Length nd Angle = Then b b cosγ= = b b b
Kreyszig by YHLee;100510; 9-5 Bsic Properties of Inner Product ( q 1 + qb 2 ) c= q 1 c+ qb 2 c b= b 0 = 0 if nd only if = 0 ( + b) c = c + b c b b + b + b 2 2 2 2 + b + b = 2 + b ( ) : linerity : symmetry : positive definiteness : positive definiteness : distributive : Cuchy Schwrz inequlity : tringle inequlity : prllelogrm equlity Exmple 1 Inner Product. Angle between Vectors = 1, 2, 0 b = 3, 2, 1 re given Two vectors [ ] nd [ ] b= 13 + 2(2) + 01 1 2 2 2 = 1 + 2 + 0 5 2 ( ) 2 2 b = b b 3 + 2 + 1 14 b 1 1 γ= rccos = cos = 1.69061 = 96.865 b 5 14 Exmple 2 Work Done by Force A constnt force p is exerted on body. But the body is displced long vector d. Then the work done by the force in the displcement of the body is W = ( pcosα) d p d Inner product is used nicely here. Exmple 3 Component of Force in Given Direction Wht force in the rope will hold the cr on 25 o rmp. The weight of the cr is 5000 lb. Since the weight points downwrd, it cn be represented by vector s = 0, 5000, 0 [ ] cn be given by sum of two vectors = c + p Force exerted to the rope by the cr Force exerted to the rmp by the cr p = cos 65 2113 lb o From the figure, ( ) o A vector in the direction of the rope b = 1, tn25, 0 b The force on the rope p = 2113 lb b o
Kreyszig by YHLee;100510; 9-6 Projection (or component) of in the direction of b b p= cosγ= b p is the length of the orthogonl projection of onto b. Orthonorml Bsis The orthogonl unit vectors in Crtesin coordintes system form n orthonorml bsis for 3 spce. iˆˆ, j, k ˆ ( ) An rbitrry vector is given by liner combintion of the orthonorml bsis. The coefficients of vector cn be determined by the orthonormlity. v l i ˆ l ˆ = + j+ l k ˆ l v i, l v ˆ = = j, l = v kˆ ˆ = li ˆ i ˆ + l ˆ j i ˆ + lk i ˆ ˆ Exmple 5 Orthogonl Stright Lines in the Plne Find the stright line L 1 pssing through the point P: (1, 3) in the xy plne nd perpendiculr to the stright line L 2 : x 2y+2=0 The eqution of the stright line L2: b1x+ b2y = k b r = k, in vector form. b = 1, 2, 0, r = x, y, 0 From the next Exmple 6, b is norml to L 2. = 1, 2, 0 = 2, 1, 0 A norml vector to b [ ] is [ ] A line norml to is L : 2x+ y= c 1 L 1 psses through P: (1, 3) 2+3=c L : 2x+ y = 5 1 [ ] [ ]
Exmple 6 Norml Vector to Plne Find unit vector norml to the plne 4x+ 2y+ 4z= 7 Kreyszig by YHLee;100510; 9-7 Express the plne in vector form : r = c The unit vector Rewriting r = c : nˆ = c nˆ r = p Projection of r onto ˆn A position vector r is from the origin to point on the plne. The sme projection p for ny r ˆn should be surfce norml. : p= c/, constnt. Since = [ 4, 2, 4] is given, the surfce norml is obtined s 1 nˆ = 6 9.3 Vector Product (Cross Product, Outer Product) Definition The vector product of v = b nd b is defined s : nother vector Its mgnitude v = b = b sinγ : γ, ngle between nd b Its direction is perpendiculr to both nd b conforming to right hnded triple( or screw). Note tht b represents the re of the prllelogrm formed by nd b.
Kreyszig by YHLee;100510; 9-8 In components b= b b, b b, b b [ ] 2 3 3 2 3 1 1 3 1 2 2 1 v = b cn be clculted s follows iˆ ˆj kˆ ˆ ˆ ˆ b = i j + k b b b b b b b b b 2 3 1 3 1 2 2 3 1 3 1 2 Exmple 1 Vector Product b = = [ 1, 1, 0] The vector product, [ 3, 0, 0] Exmple 2 Vector Products of the Stndrd Bsis Vectors Theorem 1 Generl Properties of Vector Products ( k) b= k( b) = kb : for every sclr k ( b+ c) = ( b) + ( c) : distributive ( + b) c = ( c) + ( b c) : distributive b = ( b) : nticommuttive b c b c : not ssocitive ( ) ( )
Exmple 3 Moment of Force Kreyszig by YHLee;100510; 9-9 A force p is exerted on point A. Point Q nd point A is connected by vector r. The moment m bout point Q is defined s m= pd, where d is the perpendiculr distnce from Q to L. m= p r sinγ In vector form m= r p : Moment vector Exmple 5 Velocity of Rotting Body A vector w cn describe rottion of rigid body. Its direction the rottion xis (right hnd rule) Its mgnitude = ngulr speed ω (rdin/sec) The liner speed t point P v =ωd w r sinγ w r In vector form v = w r
Sclr Triple Product Kreyszig by YHLee;100510; 9-10 The sclr triple product is defined s bc = b c ( ) ( ) It cn be clculted s b c = b b b ( ) c c c Theorem 2 Properties nd Applictions of Sclr Triple Products () The dot nd cross cn be interchnged ( b c) = ( b) c (b) Its bsolute vlue is the volume of the prllelepiped formed by, b nd c. (c) Any three vectors re linerly independent if nd only if their sclr tipple product is nonzero. Proof: () It cn be proved by direct clcultions b c = b c β β b c (b) ( ) cos ( cos ) re of the bse height of the prllelepiped (c) If three vectors re in the sme plne or on the sme stright line, b c. either the dot or cross product becomes zero in ( ) b c ( ) 0 Three vectors NOT in the sme plne or on the sme stright line. They re linerly independent. Exmple 6 Tetrhedron A tetrhedron is formed by three edge vectors, b = 0, 4, 1 c = 5, 6, 0 = [ 2, 0, 3] Find its volume., [ ], [ ] First, find the volume of the prllelepiped using sclr triple product. The volume of tetrhedron is 1/6 of tht of prllelepiped. Therefore, the nswer is 12.
9.4 Vector nd Sclr Functions nd Fields. Kreyszig by YHLee;100510; 9-11 A vector function gives vector vlue for point p in spce In Crtesin Coord. v = v( p) = v1( p), v2( p), v3( p) v( x, y, z) = v1( x, y, z), v2( x, y, z), v3( x, y, z) A sclr function gives sclr vlues : f = f( p) A vector function defines vector field. A sclr function defines sclr field. In Engineering Mening of field = Mening of function. The field implies sptil distribution of quntity. Exmple 1 Sclr function The distnce from fixed point p o to ny point p is sclr function, f ( p ). f ( p ) defines sclr field in spce It mens tht the sclr vlues re distributed in spce. ( ) = ( ) = ( ) + ( ) + ( ) 2 2 2 f p f x, y, x x xo y yo z zo In different coordinte system p nd f ( p) p hve different forms, but ( ) o is sclr function. f p hs the sme vlue. Direction cosines of the line from Not sclr function. p o to p depend on the choice of coordinte system.
Exmple 3 Vector Field (Grvittion field) Kreyszig by YHLee;100510; 9-12 Newton's lw of grvittion c F = 2 r 2 2 2 o o o : r = ( x x ) + ( y y ) + ( z z ) The direction of F is from p to p o. F( x, y, z) defines vector field in spce. In vector form Define the position vector, r x x ˆ i + y y ˆ j+ z z k ( ) ( ) ( ) ˆ o o o : Its direction is from p o to p. Then c F = r 3 r Vector Clculus Convergence An infinite sequence of vectors (1), (2), (3),... lim ( n) = 0. n lim = :, limit vector ( n ) n converges to if Similrly, vector function v( t) lim v( t) l = 0. t t o limv( t) t t o = l hs the limit l t t o if Continuity v( t) is continuous t limv t = v t t to ( ) ( ) o t = t if o v( t) is continuous t t = to if nd only if its three components re continuous t t o v t = v t ˆi + v t ˆj+ v t kˆ ( ) ( ) ( ) ( )
Definition Derivtives of Vector Function v( t) is differentible t t if the limit exists v( t+δt) v( t) v' ( t) = lim Δ t 0 Δt v t Clled derivtive of ( ) Kreyszig by YHLee;100510; 9-13 v( t) is differentible t t if nd only if its three components re differentible t t. v' t = v1' t, v2' t, v3' t ( ) ( ) ( ) ( ) Differentition rules ( cv )' = cv ' ( u+ v)' = u + v ( u v)' = u v + u v ( u v)' = u v + u v uv w' = u ' v w + uv ' w + uv w' ( ) ( ) ( ) ( ) : c, constnt Exmple 4 Let v( t) Derivtive of Vector Function of Constnt Length v t be vector function with constnt length, ( ) ( ) 2 = ( ) ( ) = 2 v t v t v t c The totl derivtive ( v v )' = 2 v v ' = 0 v ' = 0 or v v ' = c. Prtil Derivtives of Vector Function Let the components with two or more vribles be differentible v = v t, t,... t ˆi + v t, t,... t ˆj+ v t, t,... t kˆ ( ) ( ) ( ) 1 1 2 n 2 1 2 n 3 1 2 n The prtil derivtive of v with respect to t m v v1 ˆ v2 ˆ v3 = i + j+ kˆ t t t t m m m m The second prtil derivtive 2 2 2 2 v v1 ˆ v2 ˆ v3 = i + j+ kˆ t t t t t t t t l m l m l m l m Exmple 5 Prtil derivtives
9.5 Curves. Arc length. Curvture. Torsion Kreyszig by YHLee;100510; 9-14 A mjor ppliction of vector clculus concerns curves nd surfces. A curve C cn be represented by vector function with prmeter t. r( t) = x( t), y( t), z( t) = x( t) i ˆ + y( t) ˆ j+ z( t) k ˆ : prmetric representtion The direction of the curve is determined by incresing vlues of t Another representtion of curve x, y = f x, z= g x ( ) ( ) projection of C onto xy plne projection of C onto xz plne Another representtion of C cn be given by n intersection of two surfces F x, y, z = 0, G x, y, z = 0 ( ) ( ) Exmple 1 Circle 2 2 The circle x + y = 4, z = 0 is given In prmetric representtion r t = x(), t y(), t z() t 2cos t, 2sin t, 0 ( ) [ ] [ ] : 0 t 2π Check ( ) ( ) 2 2 2 2 x y t t + 2cos + 2sin = 4 For t=0 : r ( 0) = [ 2, 0, 0] For t =π /2 : r ( π /2) = [ 0, 2, 0] As t increses, r( t) moves counterclockwise.
Kreyszig by YHLee;100510; 9-15 Exmple 2 Ellipse An ellipse in the form of vector function r t = x(), t y(), t z() t = cos t, bsin t, 0 ( ) [ ] [ ] x= cos t, y= bsint 2 2 2 2 cos + sin x y 1 2 2 + b = : Conventionl form of n ellipse ( t) ( t) Exmple 3 stright line A stright line in the direction of unit vector b. It psses through point A. is position vector. A point on the line is given by position vector r t = + tb + tb, + tb, + tb ( ) [ ] 1 1 2 2 3 3 Curves Plne curve Twisted curve Simple curve Arc of curve : curve in plne. : the others. : curve without multiple points t which the curve intersects or touches itself. : portion between two points in the curve, simply clled 'curve'. Exmple 3 circulr helix A circulr helix in prmetric representtion r t = cos t, sin t, ct ( ) [ ] c>0 : right hnded screw c<0 : left hnded screw c=0 : circle
Kreyszig by YHLee;100510; 9-16 Tngent to Curve A tngent is stright line touching curve. A vector pssing through both P nd Q is 1 r ( t +Δ t ) r ( t ) Δt Tke the limit 1 r' ( t) = lim r t ( t t) r( t) Δ 0 t +Δ Δ Tngent vector of C t P The unit tngent vector r ' uˆ = : Direction of incresing t r ' The function for the tngent of C t P qw = r + wr ( ) ' : r nd r ', constnt vectors. w, prmeter Exmple 5 Tngent to n Ellipse Find the tngent to the ellipse 1 2 2 1 4 x + y = t P :( 2, 1/ 2) An ellipse, x y + = =2, b=1 b r t = 2cos t, sin t, 0 2 2 1 2 2 The prmetric presenttion ( ) [ ] At :( 2, 1/ 2) P t =π /4 r' t = 2sin t, cos t, 0 r ' π /4 = 2, 1/ 2, 0 The derivtive ( ) [ ] At P ( ) The tngent is qw ( ) = 2, 1/ 2 + w 2, 1/ 2 2( 1 w), ( 1/ 2)( 1+ w)
Kreyszig by YHLee;100510; 9-17 Length of Curve The length of the curve N l = lim r ( + nδt) r ( + ( n 1) Δt) Δ t 0 n = 1 N r ( ) ( ( 1) ) lim + n Δ t r n t + Δ Δt Δ t 0 Δt = Hence b n 1 r () t dt b b l = r () t dt = r r dt Arc Length s of Curve The upper limit is now vrible t st () r rdt : Arc length of C. dr r = dt (11) Liner element ds Differentite (11) 2 2 2 2 ds dr dr dx dy dz = + +, : rt () = xti () ˆ+ ytj () ˆ+ ztk () ˆ dt dt dt dt dt dt ( ) = ( ) + ( ) + ( ) 2 2 2 2 ds dx dy dz dr dr : dr = dx iˆ+ dy ˆ j + dz kˆ ds is clled the liner element of C. Arc Length s Prmeter The prmeter t cn be replced by the rc length s in the eqution of curve. For exmple, the unit tngent vector is given s uˆ s = r ' s ( ) ( ) where r' s+δs r' s r' ( s) = lim Δ s 0 Δs ( ) ( ) tngent vector mgnitude of the tngent vector Note tht ds r' ( s) = = 1 ds
Exmple 6 Circulr Helix. Circle. Arc Length s Prmeter. A helix rt () = [ cos t, sin t, ct] r () t = [ sin, t cos t, c] 2 2 r r = + c The rc length is t s= + c dt = t + c 0 2 2 2 2 Kreyszig by YHLee;100510; 9-18 Replce t by s in rt () s cos s ˆ sin s r i ˆj c s kˆ 2 2 = 2 2 + + 2 2 2 2 + c + b + c + c r * ( s), new function For circle, let c=0 then t = s/ s r r ˆ ˆ 1() s cos s i sin s = + j Curves in Mechnics. Velocity. Accelertion A curve C my represent pth of moving body. rt () : t is time in this cse v t = ( ) r()' t : velocity vector The mgnitude dr ds ds v = r ds dt dt Speed of the body long the curve, v is prllel to the tngent of C. ( v is the velocity of the moving body long C ) t = v' t = rt ( )" ( ) ( ) : ccelertion vector Tngentil nd Norml Accelertion In generl = + tn norm : tngentil nd norml ccelertion vectors Proof: dr () t dr ds vt () = : s, rc length dt ds dt = us ˆ( ), unit tngent vector on C Differentite the both sides 2 2 ˆ () d ˆ() ds du ds t us us ˆ() d s d us ˆ( ) = + 2 dt dt : us ˆ( ), tngentil vector., norml vector ds dt dt ds Norml ccelertion. Tngentil ccelertion.
Exmple 7 Centripetl ccelertion. Centrifugl Force. The revolving pth of smll body rt () = Rcos( ω t) ˆi + Rsin( ωt) ˆj : ω, ngulr speed Kreyszig by YHLee;100510; 9-19 The velocity vector : v = r () t Rωsin( ω t) ˆi + Rωcos( ωt) ˆj The speed of the body : v = Rω Liner speed Angulr speed = : ω Distnce to the center 2 2 2 The ccelertion vector : = v Rω cos( ωt) ˆi Rω sin( ωt) ˆ j ωr Direction towrd the center 2 2 =ω r ωr : centripetl ccelertion If the body is to be in the circulr orbit during rottion, someone should supply the centripetl force to the body. Otherwise, it will devite from the orbit due to the centrifugl force. Exmple 8 Superposition of Rottions A projectile moves long meridin of the rotting erth. Find its ccelertion. The ngulr speed of the erth rottion is ω. The unit vector b lso rottes, bt ˆ = cos( ω ti ) ˆ + sin( ω t) ˆ j ( ) The projectile on the meridin with n ngulr speed of γ. Its position vector, rt () = Rcos( γ tb ) ˆ+ Rsin( γtk ) ˆ : R, rdius of the erth The derivtives v = r () t = Rcos( γt) bˆ γrsin( γ t) bˆ+γrcos( γt) kˆ ˆ ˆ 2 ˆ 2 = v = Rcos( γt) b 2γRsin( γt) b γ Rcos( γt) b γ Rsin( γt) kˆ Inserting ˆb nd ˆb 2 ˆ ˆ 2 = ωrcos( γt) b 2γRsin( γt) b γ r Centripetl ccelertion by the erth. Centripetl ccelertion by the rottion of P on the meridin. Coriolis ccelertion due to interction of two rottions. Coriolis ccelertion is in the direction opposite to the erth s rottion The projectile devites from the meridin towrd the erth s rottion direction in the Northern Hemisphere. (??) If the projectile is to be in the meridin during rottion, someone should supply the Coriolis ccelertion to the projectile. Otherwise, it will devite from the meridin in the direction opposite to the Coriolis ccelertion. Curvture nd Torsion (skip)
9.6 Clculus Review Kreyszig by YHLee;100510; 9-20 Chin Rule Let ll the functions be continuous nd hve continuous first prtil derivtives in their domins. Let every point (u, v) in B hs the corresponding point [ x(u,v), y(u,v), z(u,v) ] in D. The function defined in B, w = f x(,), u v y(,), u v z(,) u v [ ] hs first prtil derivtives w w x w y w z = + + u x u y u z u, w w x w y w z = + + v x v y v z v. Simple exmples : If w = f( x, y, z) nd x x( t), y y( t), z z( t) then w w x w y w z = + + t x t y t z t. If w = f( x) nd x x( t) then w w x = t x t. =, = = =, Men Vlue Theorem Let f( xyz,, ) be continuous nd hve continuous first prtil derivtive in domin D. Let two points, Po :( x0, y0, z 0) nd P:( x0 + h, y0 + k, z0 + l), be in D nd the line segment between the points be in D. Then, f f f f( x0 + h, y0 + k, z0 + l) f( x0, y0, z0) = h + k + l x y z The prtil derivtives re evluted t point on the line segment A simple exmple: For function of two vribles f f f( x0 + h, y0 + k) f( x0, y0) = h + k x y
9.7 Grdient of Sclr Field. Directionl Derivtive Kreyszig by YHLee;100510; 9-21 Definition 1 Grdient The grdient of sclr function f( xyzis,, ) defined s f grd ˆ f ˆ f f = i + j+ kˆ : vector function x y z "del" opertor is defined s ˆ i ˆ = + j+ kˆ x y z : differentil opertor grd f = f Directionl Derivtive Definition 2 Directionl Derivtive Directionl derivtive of f t P in the direction of ˆb is defined by df f ( Q) f() P Df b = lim, ds s 0 s where Q is vrible point displced from P by ( sb ˆ) The line L in Crtesin Coordintes rs () = xsi () ˆ+ ysj () ˆ+ zsk () ˆ r ( s) = x ( s) i + y ( s ) j + z ( s ) k : this is unit vector ˆb The function f on C in prmetric representtion f xs (), ys (), zs () ( ) The directionl derivtive df f x f y f z f ˆi f ˆj f kˆ x iˆ y ˆj z kˆ = + + + + + + ds x s y s z s x y z s s s df (grd f) bˆ ds Exmple 1 Directionl derivtive 2 2 2 Find the directionl derivtive of f( xyz,, ) = 2x + 3y + z t the point P: (2,1,3) in the direction of = iˆ 2kˆ grdf = 4xiˆ+ 6yj ˆ+ 2zkˆ (grd f) ˆ ˆ ˆ (2,1,3) = 8i + 6 j+ 6k 1 2 The unit direction vector : ˆ = i ˆ k ˆ 5 5 1 The directionl derivtive : ˆ 2 ˆ Df ( ˆ ˆ ˆ i k i j k) = 8 + 6 + 6 = 1.789 5 5
Grdient Is Vector. Mximum Increse Kreyszig by YHLee;100510; 9-22 Theorem 1 Vector Chrcter of Grdient. For sclr function f() p = f(,,) x y z, its grdient is vector function nd its direction corresponds to the mximum increse of f on P. Its mgnitude nd direction re independent of the coordinte systems. proof: Df= bˆ grdf bˆ grdf cosγ grdf cosγ b The mximum directionl derivtive for bˆ grd f, or γ=0 grd f direction of mx. D g f. Its mx. vlue is grdf. The directionl derivtive f( Q) f( P) lim Δ s 0 Δs nd ˆb is fixed in spce. : independent of coordintes Therefore, grdf should be independent of coordinte system Grdient s Surfce Norml Vector A surfce S : f( x, y, z) = c A curve C : rt () = xti () ˆ+ ytj () ˆ+ ztk () ˆ Its derivtive : r () t = x () t ˆi + y () t ˆj+ z () t kˆ Tngent vector of the curve. If C is on S, the surfce eqution becomes f xt (), yt (), zt () = c. [ ] Differentite the surfce function w.r.t t f f f x + y + z = 0 : x' = dx/ dt, x y z (grd f) r = 0 : The tngent vector of C is lso tngent to S Tngent to C nd S. Surfce norml vector Theorem 2 Grdient s Surfce Norml Vector When surfce S is given by f( x, y, z) = c, grd f t point P on S represents the norml vector of S t P. Exmple 2 Grdient s surfce norml vector 2 2 2 A cone is given by z = 4( x + y ). Find unit surfce norml vector t P : (1,0,2) The surfce function is ( ) 4 x 2 + y 2 z 2 = 0 At P grd f = 8 x iˆ+ 8 yˆj 2 z kˆ grd f = 8iˆ 4kˆ 8 4 The unit surfce norml vector is nˆ = i ˆ k ˆ 80 80
Vector Fields tht re Grdients of Sclr Fields (Potentils) Kreyszig by YHLee;100510; 9-23 A vector function cn be obtined by the grdient of sclr function. v P = grdf P ( ) ( ) ( ) f P is clled potentil. In this cse, the field by v( P) is conservtive field ( no loss of energy) 9.8 Divergence of Vector Field A differentible vector function v( x, y, z) = v ( x, y, z) iˆ+ v ( x, y, z) ˆj+ v ( x, y, z) kˆ The divergence of v is defined s v1 v2 v3 div v = + + x y z Using del opertor ˆ ˆ ˆ div v i j k ( v ˆ ˆ ˆ + + 1i + v2j+ v3k) v x y z : sclr function Theorem 1 Invrince of the divergence div v is sclr function independent of coordinte systems * * * In xyz spce with v1, v2, v 3 In xyz spce with v, v, v * * * v1 v2 v3 v1 v2 v3 div v = + + div v = + + * * * x y z x y z It will be proved in 10.7 * * * 1 3 3 If f is twice differentible f grd ˆ f ˆ f f = i + j+ kˆ x y z Tke the divergence 2 2 2 f f f 2 div(grd f) = + + f 2 2 2 x y z 2 div(grd f) = f : x y z 2 2 2 2 + + 2 2 2, Lplce opertor
Exmple 2 Flow of Compressible fluid. Physicl Mening of the Divergence Kreyszig by YHLee;100510; 9-24 A smll box B with volume Δ V =ΔxΔyΔ z. No source or sink in the volume. Fluid flows through the box B. The fluid velocity vector v = v ˆ i + v ˆ j+ v k ˆ Flux density is defined s u =ρ v = u i ˆ + u ˆ j+ u k ˆ Density of the fluid The flux density is the trnsfer of certin quntity cross unit re per unit time. Mss of fluid in this cse Loss of mss by outwrd flow through surfces of B (1) Loss of mss due to flow in y direction during time intervlδt ΔV ( u2) y+δ yδδ x zδ+ t { ( u2) y} ΔΔ x zδ t {( u2) y+δy ( u2) y} ΔΔ x zδ Δ t u2 Δt Δy = Δ u2 (2) Loss due to flow in x direction duringδt ΔV ( u1) x+δxδδ y zδ+ t { ( u1) x} ΔΔ y zδ Δ t u1 Δt Δx (3) Loss due to flow in z direction during Δt ΔV ( u3) z+δzδδ x yδ+ t { ( u3) z} ΔΔ x yδ Δ t u3 Δt Δz The totl loss of mss in B duringδt Δu1 Δu2 Δu3 ρδ ( V ) + + ΔVΔ t = Δt Δx Δy Δz t Reduced mss inside B. Loss by flow through six side surfces of B. ρ div u = t : continuity eqution, conservtion of mss ρ For stedy flow, = 0 div u = 0 t For constnt ρ (incompressible) div v = 0, condition of incompressibility
9.9 Curl of Vector Field Kreyszig by YHLee;100510; 9-25 The curl of vector function vx (, y, z) = vi ˆ ˆ 1 + vj 2 + vk 3ˆ ˆi j kˆ curl v = v = x y z v v v is defined s ˆ v ˆ ˆ 3 v2 i + v1 v3 j+ v2 v1 k y z z x x y Exmple 1 Curl of Vector Function A vector function is given, v = yziˆ+ 3zxj ˆ+ zkˆ, find the curl of v iˆ j kˆ v = 3xi ˆ+ yj ˆ+ 2zkˆ x y z yz 3zx z Exmple 2 Rottion of rigid body The rottion is described by the ngulr speed vector w. Its direction : right hnd rule Its mgnitude : ngulr speed, ω The liner speed of point on the body : v = w r Let the xis of rottion be z xis, w =ωkˆ ˆi j kˆ v = w r 0 0 ω ω yi ˆ+ωx ˆj x y z The curl of v iˆ j kˆ curl v = 2ωkˆ x y z ωy ωx 0 curlv = 2w Theorem 2 Grd, Div, Curl The curl of the grdient of sclr function is lwys zero ( f) = 0 : irrottionl The divergence of the curl of vector function is lwys zero v = ( ) 0 Theorem 3 curl v Invrince of the curl is vector independent of coordinte systems