The Formulas of Vector Calculus John Cullinan
|
|
- Martina Moody
- 5 years ago
- Views:
Transcription
1 The Fomuls of Vecto lculus John ullinn Anlytic Geomety A vecto v is n n-tuple of el numbes: v = (v 1,..., v n ). Given two vectos v, w n, ddition nd multipliction with scl t e defined by Hee is bief list of definitions v + w = (v 1,..., v n ) + (w 1,..., w n ) = (v 1 + w 1,..., v n + w n ) t v = t (v 1,..., v n ) = (tv 1,..., tv n ). (1) The dot poduct: v w = (v 1,..., v n ) (w 1,..., w n ) = n v i w i. i=1 (2) The length of vecto: v = v v = v v2 n. Vectos of length 1 e clled unit vectos. (3) The ngle θ between two vectos: v w = v w cos θ. Fo two vectos v, w with v, w 0, the enclosed ngle θ is hence given by n θ = cos 1 i=1 v iw i ( n i=1 v2 i )1/2 ( n i=1 w2 i )1/2 (4) Two vectos v, w e clled () pependicul o othogonl if v w = 0. We denote this by v w. (b) pllel if v w = v w. We denote this by v w. Note tht two vectos v, w e pllel if nd only if thee is scl t such tht v = t w. The zeo vecto 0 = (0,..., 0) is by definition pllel nd pependicul to evey vecto. Lines nd Plnes Let u, v be two vectos with v 0. Then u + t v = (u 1 + tv 1,..., u n + tv n ); t yields line in n. This line psses though the point u in the diection of v. Let u, v, w n be vectos whee v nd w e not pllel. Then u + s v + t w = (u 1 + sv 1 + tw 1,..., u n + sv n + tw n ) defines plne though the point u, spnned by v nd w. The notion of pependicul vecto yields nothe fom of epesenting plne in 3. Let 0 = (x 0, y 0, z 0 ) be fixed point in the plne nd n vecto pependicul to the plne. An bity point = (x, y, z) on the plne stisfies n ( 0 ) = 0. Moeove, the eqution n ( 0 ) = 0 defines plne in 3 fo ny fixed n, v 3, povided n 0. In 3, the epesenttion of line l(t) = (x, y, z) = u + t v cn be solved fo t in ech coodinte if none of the vlues v 1, v 2, v 3 e zeo: t = x u 1 v 1 t = y u 2 v 2 t = z u 3 v 3 This yields nothe epesenttion of line in 3, which is clled symmetic eqution: x u 1 = y u 2 = z u 3. v 1 v 2 v 3 The pmetic eqution epesenting plne cn be tnsfomed similly giving the stndd eqution of plne: x + by + cz = d, whee the noml vecto n is (, b, c). 1
2 oss Poduct At point p of ny plne in 3 thee is exctly one line pependicul to the plne. If v nd w spn the plne, vecto spnning this pependicul line is given by the coss poduct v w. Fo two vectos v, w 3 the coss poduct v w is defined by v w = (v 1, v 2, v 3 ) (w 1, w 2, w 3 ) = (v 2 w 3 v 3 w 2, v 3 w 1 v 1 w 3, v 1 w 2 v 2 w 1 ) Altentively, the coss poduct is given by the 3 3 deteminnt v w = v i j k 1 v 2 v. 3 w 1 w 2 w 3 v w is pependicul to both v nd w v, w nd v w e oiented ccoding to the ight hnd ule if θ is the ngle between v nd w, then v w = v w sin θ. The length of v w is equl to the e of the pllelogm spnned by v nd w. The coss poduct is (1) line: t (v w) = (t v) w = v (t w) nd u (v + w) = u v + u w. (2) nti-commuttive: v w = w v Anothe emkble popety of the coss poduct is the following: u (v w) = (u v) w. Geometiclly, u (v w) is the volume of the pllelepiped given by u, v, nd w. Sphees nd ylindes The vecto nottion llows fo n elegnt desciption of othe geometic objects such s sphees nd cylindes. Fo scl 0 nd vecto c 3, the eqution x c 2 = 2 yields sphee of dius centeed t c. Fo scl 0 nd vectos c, n 3 with n = 1, the eqution n (x c) 2 = 2 yields cylinde of dius centeed ound the line given by c + t n. Pmeteized uves in 3 Velocity, Speed, nd Acceletion Given thee diffeentible functions x, y, z : [, b], the vecto (t) = (x(t), y(t), z(t)) cn be undestood s descibing pticle moving though spce; the pticle s position depends on the (time) pmete t. This peception gives ise to the following definitions. We define the velocity of t time t to be (t) = (x (t), y (t), z (t)) speed of t time t to be (t) = [x (t)] 2 + [y (t)] 2 + [z (t)] 2 cceletion of t time t to be (t) = (x (t), y (t), z (t)). Pticles moving t constnt speed Suppose tht (t) = c fo ll t fo cuve (t). Diffeentiting (t) = c, we obseve tht (t) is othogonl to (t) fo ll t: (t) = 1 (t) (t) = 0 fo ll t. Smooth uves A cuve : [, b] 3 is clled smooth if its speed vnishes t most t the endpoints: (t) 0 fo ll t (, b). Aclength The clength l(, b) of cuve : [, b] 3 is given by l(, b) = b (t) dt A cuve : [, b] is clled pmeteized with espect to clength if (t) = 1 fo ll t [, b]. uvtue Let : [, b] 3 be pmeteized with espect to clength. Then κ(t) = (t) is clled the cuvtue of t t. Diffeentiting the eqution (t) = 1 shows tht (t) nd (t) e othogonl fo ll t. If κ(t) 0, the vecto n(t) = (t) (t) is well-defined unit vecto. 2
3 Noml nd Binoml Vectos If κ(t) 0, we cn define the Noml vecto of the cuve t time t to be n(t) Binoml vecto of the cuve t time t to be b(t) = (t) n(t) The binoml vecto is obviously unit vecto, so we cn pply the sme esoning s befoe to see tht b(t) nd b (t) e othogonl. On the othe hnd, diffeentiting b(t) = (t) n(t) we get: Hence b (t) is pllel to n(t). b (t) = (t) n(t) + (t) n (t) = (t) n (t). Tosion The eqution b (t) = τ(t) n(t) defines the tosion τ of the cuve t time t. Fenet Fme Wheneve κ(t) 0, the vectos (t), n(t), b(t) e mutully othogonl unit vectos. They spn the Fenet Fme. uvtue of uves NOT pmeteized with espect to Aclength Let : [, b] 3 be my smooth cuve. Its cuvtue κ(t) t time t is given by κ(t) = (t) (t) (t) 3. el-vlued Functions Domin nd nge A function f : n ssigns unique el vlue f(x 1,..., x n ) to ech point (x 1,..., x n ) of set D in n. The set D is clled the domin of f. The set = {f(x 1,..., x n ) (x 1,..., x n ) D} is clled the nge of f. Limits A el numbe L is sid to be the limit of f t (, b,... ) if fo ll sequences ( m, b m,... ) with lim m m =, lim m b m = b,..., the following holds: lim f( m, b m,... ) = L. m We denote this by lim f(x 1, x 2,... ) = L. (x 1,x 2,... ) (,b,... ) Equivlently, if fo evey el numbe ɛ > 0 thee is nothe el numbe δ > 0 such tht (x 1, x 2,... ) (, b,... ) < δ f(x 1, x 2,... ) L < ɛ. ontinuity A function f : n with domin D is sid to be continuous t (, b,... ) D if lim f(x 1, x 2,... ) = f(, b,... ) (x 1,x 2,... ) (,b,... ) Ptil Deivtives Let f : n be function with domin D nd (x 1,..., x n ) D. The ptil deivtive of f t (x 1,..., x n ) with espect to x i is given by the limit f f(x 1,..., x i + h,..., x n ) f(x 1,..., x i,..., x n ) = lim. x i h 0 h Diffeentibility A function f : n with domin D is clled diffeentible t (x 1,..., x n ) D if ll ptil deivtives exist nd e continuous t (x 1,..., x n ). Diectionl Deivitve Let f : n be function with domin D nd D. Suppose tht u is unit vecto in n. The diectionl deivtive of f t in the diection of u is given by d f( + tu) dt. t=0 Gdient Vecto Let f : n be function with domin D tht is diffeentible t D. The gdient vecto f of f t is given by f ( f = x 1,..., f ) x n 3
4 Diectionl Deivtives nd the Gdient Vecto Let f : n be function with domin D tht is diffeentible t D. Using the chin ule, the diectionl deivtive is given by d f( + tu) dt = f u t=0 Highe Deivtives nd liut s Theoem Let f : n be function with domin D nd suppose the ptil deivtives of f e themselves diffeentible. Then diffeentiting f x i with espect to x j is the sme s diffeentiting f x j with espect to x i : 2 f = 2 f. x i x j x j x i Globl nd Locl Extem A function f : n with domin D hs globl mximum t D if f(x) f() fo ll x D globl minimum t D if f(x) f() fo ll x D locl mximum t D if thee is disc centeed t such tht f(x) f() fo ll x locl minimum t D if thee is disc centeed t such tht f(x) f() fo ll x iticl Points Let f : D n be diffeentible. We cll point D citicl point if f = 0. If f hs n extemum t, then is citicl, but the convese is not necessily tue. Functions of Two Vibles Second Deivtive Test Let f : D 2 nd its deivtives be diffeentible nd let (x 0, y 0 ) D be citicl point of f. Let ( 2 f 2 ( f 2 ) 2) D(x 0, y 0 ) = x 2 y 2 f x y (x0,y 0) if D(x 0, y 0 ) > 0 nd 2 f x 2 < 0, then f hs mximum t (x 0, y 0 ). (x0,y 0) if D(x 0, y 0 ) > 0 nd 2 f x 2 > 0, then f hs minimum t (x 0, y 0 ). (x0,y 0) if D(x 0, y 0 ) < 0, then f hs sddle t (x 0, y 0 ). if D(x 0, y 0 ) = 0, then the second deivtive test gives no infomtion bout the ntue of the citicl point. The Double Integl Let = [, b] [c, d] nd let f : be continuous. The double integl of f ove is defined to be f(x, y) da = lim (x i x i 1 )(y j y j 1 )f(x i, yj ) P 0 whee P = P [,b] P [c,d] is ptition of, P = P [,b] P [c,d] is its nom, nd (x i, y j ) [x i 1, x i ] [y j 1, y j ]. i,j Fubini s Theoem Let f : 2 be continuous nd = [, b] [c, d]. The double integl is given by b d d b f(x, y) da = f(x, y) dydx = f(x, y) dxdy Level Sufces Let f(x, y, z) : 3 be function of thee vibles. The level set {(x, y, z) f(x, y, z) = c} fo constnt c genelly yields sufce in 3. c Tngent Plnes of Level Sufces onside the level set {(x, y, z) f(x, y, z) = c} of function f : 3. (1) The gdient vecto f(x 0, y 0, z 0 ) t point (x 0, y 0, z 0 ) is pependicul to the plne tngent to the level sufce t (x 0, y 0, z 0 ). 4 c
5 (2) The tngent plne is given by the eqution f(x 0, y 0, z 0 ) (x x 0, y y 0, z z 0 ) = 0. Using Lgnge Multiplies to Find Extem with onstints Let f : 3 be diffeentible. To find the mximum nd minimum vlue of f subject to the constint g(x, y, z) = c, the gdients f nd g must be pllel. An lgoithm to find the mximum nd minimum vlues is hence given by: (1) Find ll points (x, y, z) such tht f = λ g, fo some λ, nd g(x, y, z) = c. (2) Evlute f t these points. The lgest (smllest) vlue is the mximum (minimum) of f subject to the constint g(x, y, z) = c. The Tiple Integl Let g : 3 be continuous nd = [, b] [c, d] [e, f]. The tiple integl of g ove is defined to be g(x, y, z) dv = lim (x i x i 1 )(y j y j 1 )(z k z k 1 )g(x i, yj, zk) P 0 i,j,k hin ule Let (t) be smooth cuve in n nd let f : n be function of sevel vibles. Then df dt = f (t). Moe genelly, suppose ech of the vibles x i is function of the vibles t 1,..., t m, then f t j = n i=1 f x i x i t j hnge of Vibles Fomul in Two Dimensions Let (x(u, v), y(u, v)) be pmete tnsfomtion with Jcobin mtix ( ) x/ u x/ v J = y/ u y/ v which mps S 2 into 2. Then f(x, y) dxdy = S f(u, v) det J dudv Pol oodintes The pmete tnsfomtion (x(, θ), y(, θ)) = ( cos θ, sin θ) expesses (x, y) in pol coodintes. Then f(x, y) dxdy = f(, θ) ddθ Sufce Ae Let f : 2 be diffeentible. The sufce e of f ove D is given by [ ] 2 [ ] 2 f f dxdy x y Pth Integls Let f : 2 be diffeentible nd let (t) pmeteize smooth cuve in 2. The pth integl of f long is given by f((t)) (t) dt S 5
6 Vecto Fields A vecto field F ssigns to ech point in domin n vecto in n. Line Integls Let F be vecto field on domin n nd let be smooth cuve in pmeteized by (t). The line integl of F long is given by F((t)) (t) dt onsevtive Vecto Fields nd the Potentil Function If F = f fo some function f : 2, then F is clled consevtive vecto field with potentil function f. Line Integls of onsevtive Vecto Fields Let F = f be consevtive on domin n. Fo ny continuous cuve in fom u to b which is pmeteized by (t), we hve F (t) dt = f(v) f(u). In pticul (1) If is closed then F dt = 0. (2) if is nothe continuous cuve in fom u to v pmeteized by s(t), then F s dt = F dt. The integl is sid to be pth-independent. Line Integls of Vecto Fields in 2 Let F = (P, Q) be vecto field on domin 2 nd let be smooth cuve in given by (x(t), y(t)). Then b F(P (x(t), y(t)), Q(x(t), y(t))) (x (t), y (t)) dt = P (x, y) dx + Q(x, y) dy onsevtive Vecto Fields in 2 If F = (P, Q) = f, then by liut s theoem, P y = Q x. Geen s Theoem Let be simply connected egion with positively-oiented boundy. Then Q P dx + Q dy = x P y dxdy Stokes Theoem If S is sufce in 3 with boundy S pmeteized by nd F is vecto field in 3, then F d = ( F) ds S Divegence (Guss ) Theoem If V is compct volume in 3 with boundy V = S nd F is vecto field in 3, then ( F) dv = F ds V S V 6
13.5. Torsion of a curve Tangential and Normal Components of Acceleration
13.5 osion of cuve ngentil nd oml Components of Acceletion Recll: Length of cuve '( t) Ac length function s( t) b t u du '( t) Ac length pmetiztion ( s) with '( s) 1 '( t) Unit tngent vecto '( t) Cuvtue:
More informationdx was area under f ( x ) if ( ) 0
13. Line Integls Line integls e simil to single integl, f ( x) dx ws e unde f ( x ) if ( ) 0 Insted of integting ove n intevl [, ] (, ) f xy ds f x., we integte ove cuve, (in the xy-plne). **Figue - get
More informationMATHEMATICS IV 2 MARKS. 5 2 = e 3, 4
MATHEMATICS IV MARKS. If + + 6 + c epesents cicle with dius 6, find the vlue of c. R 9 f c ; g, f 6 9 c 6 c c. Find the eccenticit of the hpeol Eqution of the hpeol Hee, nd + e + e 5 e 5 e. Find the distnce
More informationFriedmannien equations
..6 Fiedmnnien equtions FLRW metic is : ds c The metic intevl is: dt ( t) d ( ) hee f ( ) is function which detemines globl geometic l popety of D spce. f d sin d One cn put it in the Einstein equtions
More informationElectric Potential. and Equipotentials
Electic Potentil nd Euipotentils U Electicl Potentil Review: W wok done y foce in going fom to long pth. l d E dl F W dl F θ Δ l d E W U U U Δ Δ l d E W U U U U potentil enegy electic potentil Potentil
More informationπ,π is the angle FROM a! TO b
Mth 151: 1.2 The Dot Poduct We hve scled vectos (o, multiplied vectos y el nume clled scl) nd dded vectos (in ectngul component fom). Cn we multiply vectos togethe? The nswe is YES! In fct, thee e two
More informationMath 4318 : Real Analysis II Mid-Term Exam 1 14 February 2013
Mth 4318 : Rel Anlysis II Mid-Tem Exm 1 14 Febuy 2013 Nme: Definitions: Tue/Flse: Poofs: 1. 2. 3. 4. 5. 6. Totl: Definitions nd Sttements of Theoems 1. (2 points) Fo function f(x) defined on (, b) nd fo
More informationElectric Field F E. q Q R Q. ˆ 4 r r - - Electric field intensity depends on the medium! origin
1 1 Electic Field + + q F Q R oigin E 0 0 F E ˆ E 4 4 R q Q R Q - - Electic field intensity depends on the medium! Electic Flux Density We intoduce new vecto field D independent of medium. D E So, electic
More informationLecture 11: Potential Gradient and Capacitor Review:
Lectue 11: Potentil Gdient nd Cpcito Review: Two wys to find t ny point in spce: Sum o Integte ove chges: q 1 1 q 2 2 3 P i 1 q i i dq q 3 P 1 dq xmple of integting ove distiution: line of chge ing of
More informationr a + r b a + ( r b + r c)
AP Phsics C Unit 2 2.1 Nme Vectos Vectos e used to epesent quntities tht e chcteized b mgnitude ( numeicl vlue with ppopite units) nd diection. The usul emple is the displcement vecto. A quntit with onl
More informationWork, Potential Energy, Conservation of Energy. the electric forces are conservative: ur r
Wok, Potentil Enegy, Consevtion of Enegy the electic foces e consevtive: u Fd = Wok, Potentil Enegy, Consevtion of Enegy b b W = u b b Fdl = F()[ d + $ $ dl ] = F() d u Fdl = the electic foces e consevtive
More informationRELATIVE KINEMATICS. q 2 R 12. u 1 O 2 S 2 S 1. r 1 O 1. Figure 1
RELAIVE KINEMAICS he equtions of motion fo point P will be nlyzed in two diffeent efeence systems. One efeence system is inetil, fixed to the gound, the second system is moving in the physicl spce nd the
More informationOptimization. x = 22 corresponds to local maximum by second derivative test
Optimiztion Lectue 17 discussed the exteme vlues of functions. This lectue will pply the lesson fom Lectue 17 to wod poblems. In this section, it is impotnt to emembe we e in Clculus I nd e deling one-vible
More informationClass Summary. be functions and f( D) , we define the composition of f with g, denoted g f by
Clss Summy.5 Eponentil Functions.6 Invese Functions nd Logithms A function f is ule tht ssigns to ech element D ectly one element, clled f( ), in. Fo emple : function not function Given functions f, g:
More informationPhysics 505 Fall 2005 Midterm Solutions. This midterm is a two hour open book, open notes exam. Do all three problems.
Physics 55 Fll 5 Midtem Solutions This midtem is two hou open ook, open notes exm. Do ll thee polems. [35 pts] 1. A ectngul ox hs sides of lengths, nd c z x c [1] ) Fo the Diichlet polem in the inteio
More informationPhysics 604 Problem Set 1 Due Sept 16, 2010
Physics 64 Polem et 1 Due ept 16 1 1) ) Inside good conducto the electic field is eo (electons in the conducto ecuse they e fee to move move in wy to cncel ny electic field impessed on the conducto inside
More informationU>, and is negative. Electric Potential Energy
Electic Potentil Enegy Think of gvittionl potentil enegy. When the lock is moved veticlly up ginst gvity, the gvittionl foce does negtive wok (you do positive wok), nd the potentil enegy (U) inceses. When
More informationChapter 28 Sources of Magnetic Field
Chpte 8 Souces of Mgnetic Field - Mgnetic Field of Moving Chge - Mgnetic Field of Cuent Element - Mgnetic Field of Stight Cuent-Cying Conducto - Foce Between Pllel Conductos - Mgnetic Field of Cicul Cuent
More informationCollection of Formulas
Collection of Fomuls Electomgnetic Fields EITF8 Deptment of Electicl nd Infomtion Technology Lund Univesity, Sweden August 8 / ELECTOSTATICS field point '' ' Oigin ' Souce point Coulomb s Lw The foce F
More information1 Using Integration to Find Arc Lengths and Surface Areas
Novembe 9, 8 MAT86 Week Justin Ko Using Integtion to Find Ac Lengths nd Sufce Aes. Ac Length Fomul: If f () is continuous on [, b], then the c length of the cuve = f() on the intevl [, b] is given b s
More informationRadial geodesics in Schwarzschild spacetime
Rdil geodesics in Schwzschild spcetime Spheiclly symmetic solutions to the Einstein eqution tke the fom ds dt d dθ sin θdϕ whee is constnt. We lso hve the connection components, which now tke the fom using
More informationGeneral Physics II. number of field lines/area. for whole surface: for continuous surface is a whole surface
Genel Physics II Chpte 3: Guss w We now wnt to quickly discuss one of the moe useful tools fo clculting the electic field, nmely Guss lw. In ode to undestnd Guss s lw, it seems we need to know the concept
More informationu(r, θ) = 1 + 3a r n=1
Mth 45 / AMCS 55. etuck Assignment 8 ue Tuesdy, Apil, 6 Topics fo this week Convegence of Fouie seies; Lplce s eqution nd hmonic functions: bsic popeties, computions on ectngles nd cubes Fouie!, Poisson
More informationElectricity & Magnetism Lecture 6: Electric Potential
Electicity & Mgnetism Lectue 6: Electic Potentil Tody s Concept: Electic Potenl (Defined in tems of Pth Integl of Electic Field) Electicity & Mgnesm Lectue 6, Slide Stuff you sked bout:! Explin moe why
More informationPhysics 11b Lecture #11
Physics 11b Lectue #11 Mgnetic Fields Souces of the Mgnetic Field S&J Chpte 9, 3 Wht We Did Lst Time Mgnetic fields e simil to electic fields Only diffeence: no single mgnetic pole Loentz foce Moving chge
More informationTopics for Review for Final Exam in Calculus 16A
Topics fo Review fo Finl Em in Clculus 16A Instucto: Zvezdelin Stnkov Contents 1. Definitions 1. Theoems nd Poblem Solving Techniques 1 3. Eecises to Review 5 4. Chet Sheet 5 1. Definitions Undestnd the
More informationFI 2201 Electromagnetism
FI 1 Electomgnetism Alexnde A. Isknd, Ph.D. Physics of Mgnetism nd Photonics Resech Goup Electosttics ELECTRIC PTENTIALS 1 Recll tht we e inteested to clculte the electic field of some chge distiution.
More informationB.A. (PROGRAMME) 1 YEAR MATHEMATICS
Gdute Couse B.A. (PROGRAMME) YEAR MATHEMATICS ALGEBRA & CALCULUS PART B : CALCULUS SM 4 CONTENTS Lesson Lesson Lesson Lesson Lesson Lesson Lesson : Tngents nd Nomls : Tngents nd Nomls (Pol Co-odintes)
More informationThe Divergence Theorem
13.8 The ivegence Theoem Back in 13.5 we ewote Geen s Theoem in vecto fom as C F n ds= div F x, y da ( ) whee C is the positively-oiented bounday cuve of the plane egion (in the xy-plane). Notice this
More informationELECTROSTATICS. 4πε0. E dr. The electric field is along the direction where the potential decreases at the maximum rate. 5. Electric Potential Energy:
LCTROSTATICS. Quntiztion of Chge: Any chged body, big o smll, hs totl chge which is n integl multile of e, i.e. = ± ne, whee n is n intege hving vlues,, etc, e is the chge of electon which is eul to.6
More informationLecture 10. Solution of Nonlinear Equations - II
Fied point Poblems Lectue Solution o Nonline Equtions - II Given unction g : R R, vlue such tht gis clled ied point o the unction g, since is unchnged when g is pplied to it. Whees with nonline eqution
More informationPartial Derivatives. Limits. For a single variable function f (x), the limit lim
Limits Prtil Derivtives For single vrible function f (x), the limit lim x f (x) exists only if the right-hnd side limit equls to the left-hnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationSTD: XI MATHEMATICS Total Marks: 90. I Choose the correct answer: ( 20 x 1 = 20 ) a) x = 1 b) x =2 c) x = 3 d) x = 0
STD: XI MATHEMATICS Totl Mks: 90 Time: ½ Hs I Choose the coect nswe: ( 0 = 0 ). The solution of is ) = b) = c) = d) = 0. Given tht the vlue of thid ode deteminnt is then the vlue of the deteminnt fomed
More informationAnswers to test yourself questions
Answes to test youself questions opic Descibing fields Gm Gm Gm Gm he net field t is: g ( d / ) ( 4d / ) d d Gm Gm Gm Gm Gm Gm b he net potentil t is: V d / 4d / d 4d d d V e 4 7 9 49 J kg 7 7 Gm d b E
More information1. The sphere P travels in a straight line with speed
1. The sphee P tels in stight line with speed = 10 m/s. Fo the instnt depicted, detemine the coesponding lues of,,,,, s mesued eltie to the fixed Oxy coodinte system. (/134) + 38.66 1.34 51.34 10sin 3.639
More informationEinstein Classes, Unit No. 102, 103, Vardhman Ring Road Plaza, Vikas Puri Extn., Outer Ring Road New Delhi , Ph. : ,
CA Vectos & Thei Repesenttion : CB MV VECTORS Vecto quntities e specified y definite mgnitude nd definite diections A vecto is genelly epesented y diected line segment, sy AB A is clled the initil point
More informationOn the Eötvös effect
On the Eötvös effect Mugu B. Răuţ The im of this ppe is to popose new theoy bout the Eötvös effect. We develop mthemticl model which loud us bette undestnding of this effect. Fom the eqution of motion
More informationAlgebra Based Physics. Gravitational Force. PSI Honors universal gravitation presentation Update Fall 2016.notebookNovember 10, 2016
Newton's Lw of Univesl Gvittion Gvittionl Foce lick on the topic to go to tht section Gvittionl Field lgeb sed Physics Newton's Lw of Univesl Gvittion Sufce Gvity Gvittionl Field in Spce Keple's Thid Lw
More informationSchool of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007
School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 4 Due on Sep. 1, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt
More informationEECE 260 Electrical Circuits Prof. Mark Fowler
EECE 60 Electicl Cicuits Pof. Mk Fowle Complex Numbe Review /6 Complex Numbes Complex numbes ise s oots of polynomils. Definition of imginy # nd some esulting popeties: ( ( )( ) )( ) Recll tht the solution
More informationThis immediately suggests an inverse-square law for a "piece" of current along the line.
Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iot-svt Lw T nvese-sque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line
More informationHigher Checklist (Unit 3) Higher Checklist (Unit 3) Vectors
Vectors Skill Achieved? Know tht sclr is quntity tht hs only size (no direction) Identify rel-life exmples of sclrs such s, temperture, mss, distnce, time, speed, energy nd electric chrge Know tht vector
More informationSolutions to Midterm Physics 201
Solutions to Midtem Physics. We cn conside this sitution s supeposition of unifomly chged sphee of chge density ρ nd dius R, nd second unifomly chged sphee of chge density ρ nd dius R t the position of
More information440-2 Geometry/Topology: Differentiable Manifolds Northwestern University Solutions of Practice Problems for Final Exam
440-2 Geometry/Topology: Differentible Mnifolds Northwestern University Solutions of Prctice Problems for Finl Exm 1) Using the cnonicl covering of RP n by {U α } 0 α n, where U α = {[x 0 : : x n ] RP
More informationJim Lambers MAT 280 Spring Semester Lecture 17 Notes. These notes correspond to Section 13.2 in Stewart and Section 7.2 in Marsden and Tromba.
Jim Lmbers MAT 28 Spring Semester 29- Lecture 7 Notes These notes correspond to Section 3.2 in Stewrt nd Section 7.2 in Mrsden nd Tromb. Line Integrls Recll from single-vrible clclus tht if constnt force
More informationMath Advanced Calculus II
Mth 452 - Advnced Clculus II Line Integrls nd Green s Theorem The min gol of this chpter is to prove Stoke s theorem, which is the multivrible version of the fundmentl theorem of clculus. We will be focused
More informationHomework: Study 6.2 #1, 3, 5, 7, 11, 15, 55, 57
Gols: 1. Undestnd volume s the sum of the es of n infinite nume of sufces. 2. Be le to identify: the ounded egion the efeence ectngle the sufce tht esults fom evolution of the ectngle ound n xis o foms
More information6. Numbers. The line of numbers: Important subsets of IR:
6. Nubes We do not give n xiotic definition of the el nubes hee. Intuitive ening: Ech point on the (infinite) line of nubes coesponds to el nube, i.e., n eleent of IR. The line of nubes: Ipotnt subsets
More informationElectronic Supplementary Material
Electonic Supplementy Mteil On the coevolution of socil esponsiveness nd behvioul consistency Mx Wolf, G Snde vn Doon & Fnz J Weissing Poc R Soc B 78, 440-448; 0 Bsic set-up of the model Conside the model
More informationPreviously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system
436-459 Advnced contol nd utomtion Extensions to bckstepping contolle designs Tcking Obseves (nonline dmping) Peviously Lst lectue we looked t designing nonline contolles using the bckstepping technique
More information( ) ( ) ( ) ( ) ( ) # B x ( ˆ i ) ( ) # B y ( ˆ j ) ( ) # B y ("ˆ ( ) ( ) ( (( ) # ("ˆ ( ) ( ) ( ) # B ˆ z ( k )
Emple 1: A positie chge with elocit is moing though unifom mgnetic field s shown in the figues below. Use the ight-hnd ule to detemine the diection of the mgnetic foce on the chge. Emple 1 ˆ i = ˆ ˆ i
More informationJim Lambers MAT 280 Spring Semester Lecture 26 and 27 Notes
Jim Lmbers MAT 280 pring emester 2009-10 Lecture 26 nd 27 Notes These notes correspond to ection 8.6 in Mrsden nd Tromb. ifferentil Forms To dte, we hve lerned the following theorems concerning the evlution
More informationnot to be republishe NCERT VECTOR ALGEBRA Chapter Introduction 10.2 Some Basic Concepts
44 MATHEMATICS Chpte 10 In most sciences one genetion tes down wht nothe hs built nd wht one hs estblished nothe undoes In Mthemtics lone ech genetion builds new stoy to the old stuctue HERMAN HANKEL 101
More information3.1 Magnetic Fields. Oersted and Ampere
3.1 Mgnetic Fields Oested nd Ampee The definition of mgnetic induction, B Fields of smll loop (dipole) Mgnetic fields in mtte: ) feomgnetism ) mgnetiztion, (M ) c) mgnetic susceptiility, m d) mgnetic field,
More informationContinuous Charge Distributions
Continuous Chge Distibutions Review Wht if we hve distibution of chge? ˆ Q chge of distibution. Q dq element of chge. d contibution to due to dq. Cn wite dq = ρ dv; ρ is the chge density. = 1 4πε 0 qi
More informationPX3008 Problem Sheet 1
PX38 Poblem Sheet 1 1) A sphee of dius (m) contins chge of unifom density ρ (Cm -3 ). Using Guss' theoem, obtin expessions fo the mgnitude of the electic field (t distnce fom the cente of the sphee) in
More informationof Technology: MIT OpenCourseWare). (accessed MM DD, YYYY). License: Creative Commons Attribution- Noncommercial-Share Alike.
MIT OpenouseWe http://ocw.mit.edu 6.1/ESD.1J Electomgnetics nd pplictions, Fll 25 Plese use the following cittion fomt: Mkus Zhn, Eich Ippen, nd Dvid Stelin, 6.1/ESD.1J Electomgnetics nd pplictions, Fll
More informationThe Regulated and Riemann Integrals
Chpter 1 The Regulted nd Riemnn Integrls 1.1 Introduction We will consider severl different pproches to defining the definite integrl f(x) dx of function f(x). These definitions will ll ssign the sme vlue
More informationSection 17.2 Line Integrals
Section 7. Line Integrls Integrting Vector Fields nd Functions long urve In this section we consider the problem of integrting functions, both sclr nd vector (vector fields) long curve in the plne. We
More informationMATH 13 FINAL STUDY GUIDE, WINTER 2012
MATH 13 FINAL TUY GUI, WINTR 2012 This is ment to be quick reference guide for the topics you might wnt to know for the finl. It probbly isn t comprehensive, but should cover most of wht we studied in
More informationMath 2263 Solutions for Spring 2003 Final Exam
Math 6 Solutions fo Sping Final Exam ) A staightfowad appoach to finding the tangent plane to a suface at a point ( x, y, z ) would be to expess the cuve as an explicit function z = f ( x, y ), calculate
More informationDEPARTMENT OF CIVIL AND ENVIRONMENTAL ENGINEERING FLUID MECHANICS III Solutions to Problem Sheet 3
DEPATMENT OF CIVIL AND ENVIONMENTAL ENGINEEING FLID MECHANICS III Solutions to Poblem Sheet 3 1. An tmospheic vote is moelle s combintion of viscous coe otting s soli boy with ngul velocity Ω n n iottionl
More informationProperties and Formulas
Popeties nd Fomuls Cpte 1 Ode of Opetions 1. Pefom ny opetion(s) inside gouping symols. 2. Simplify powes. 3. Multiply nd divide in ode fom left to igt. 4. Add nd sutt in ode fom left to igt. Identity
More information6. Gravitation. 6.1 Newton's law of Gravitation
Gvittion / 1 6.1 Newton's lw of Gvittion 6. Gvittion Newton's lw of gvittion sttes tht evey body in this univese ttcts evey othe body with foce, which is diectly popotionl to the poduct of thei msses nd
More informationLine Integrals. Partitioning the Curve. Estimating the Mass
Line Integrls Suppose we hve curve in the xy plne nd ssocite density δ(p ) = δ(x, y) t ech point on the curve. urves, of course, do not hve density or mss, but it my sometimes be convenient or useful to
More information10 m, so the distance from the Sun to the Moon during a solar eclipse is. The mass of the Sun, Earth, and Moon are = =
Chpte 1 nivesl Gvittion 11 *P1. () The un-th distnce is 1.4 nd the th-moon 8 distnce is.84, so the distnce fom the un to the Moon duing sol eclipse is 11 8 11 1.4.84 = 1.4 The mss of the un, th, nd Moon
More informationSchool of Electrical and Computer Engineering, Cornell University. ECE 303: Electromagnetic Fields and Waves. Fall 2007
School of Electicl nd Compute Engineeing, Conell Univesity ECE 303: Electomgnetic Fields nd Wves Fll 007 Homewok 3 Due on Sep. 14, 007 by 5:00 PM Reding Assignments: i) Review the lectue notes. ii) Relevnt
More informationAQA Maths M2. Topic Questions from Papers. Circular Motion. Answers
AQA Mths M Topic Questions fom Ppes Cicul Motion Answes PhysicsAndMthsTuto.com PhysicsAndMthsTuto.com Totl 6 () T cos30 = 9.8 Resolving veticlly with two tems Coect eqution 9.8 T = cos30 T =.6 N AG 3 Coect
More informationFluids & Bernoulli s Equation. Group Problems 9
Goup Poblems 9 Fluids & Benoulli s Eqution Nme This is moe tutoil-like thn poblem nd leds you though conceptul development of Benoulli s eqution using the ides of Newton s 2 nd lw nd enegy. You e going
More informationTHEORY OF EQUATIONS OBJECTIVE PROBLEMS. If the eqution x 6x 0 0 ) - ) 4) -. If the sum of two oots of the eqution k is -48 ) 6 ) 48 4) 4. If the poduct of two oots of 4 ) -4 ) 4) - 4. If one oot of is
More informationChapter 7. Kleene s Theorem. 7.1 Kleene s Theorem. The following theorem is the most important and fundamental result in the theory of FA s:
Chpte 7 Kleene s Theoem 7.1 Kleene s Theoem The following theoem is the most impotnt nd fundmentl esult in the theoy of FA s: Theoem 6 Any lnguge tht cn e defined y eithe egul expession, o finite utomt,
More information(a) Counter-Clockwise (b) Clockwise ()N (c) No rotation (d) Not enough information
m m m00 kg dult, m0 kg bby. he seesw stts fom est. Which diection will it ottes? ( Counte-Clockwise (b Clockwise ( (c o ottion ti (d ot enough infomtion Effect of Constnt et oque.3 A constnt non-zeo toque
More informationCHAPTER 7 Applications of Integration
CHAPTER 7 Applitions of Integtion Setion 7. Ae of Region Between Two Cuves.......... Setion 7. Volume: The Disk Method................. Setion 7. Volume: The Shell Method................ Setion 7. A Length
More informationdefined on a domain can be expanded into the Taylor series around a point a except a singular point. Also, f( z)
08 Tylo eie nd Mcluin eie A holomophic function f( z) defined on domin cn be expnded into the Tylo eie ound point except ingul point. Alo, f( z) cn be expnded into the Mcluin eie in the open dik with diu
More informationElectrostatics (Electric Charges and Field) #2 2010
Electic Field: The concept of electic field explains the action at a distance foce between two chaged paticles. Evey chage poduces a field aound it so that any othe chaged paticle expeiences a foce when
More information( dg. ) 2 dt. + dt. dt j + dh. + dt. r(t) dt. Comparing this equation with the one listed above for the length of see that
Arc Length of Curves in Three Dimensionl Spce If the vector function r(t) f(t) i + g(t) j + h(t) k trces out the curve C s t vries, we cn mesure distnces long C using formul nerly identicl to one tht we
More informationSURFACE TENSION. e-edge Education Classes 1 of 7 website: , ,
SURFACE TENSION Definition Sufce tension is popety of liquid by which the fee sufce of liquid behves like stetched elstic membne, hving contctive tendency. The sufce tension is mesued by the foce cting
More informationELECTRO - MAGNETIC INDUCTION
NTRODUCTON LCTRO - MAGNTC NDUCTON Whenee mgnetic flu linked with cicuit chnges, n e.m.f. is induced in the cicuit. f the cicuit is closed, cuent is lso induced in it. The e.m.f. nd cuent poduced lsts s
More informationDYNAMICS. Kinetics of Particles: Newton s Second Law VECTOR MECHANICS FOR ENGINEERS: Ninth Edition CHAPTER. Ferdinand P. Beer E. Russell Johnston, Jr.
Ninth E CHPTER VECTOR MECHNICS OR ENGINEERS: DYNMICS edinnd P. ee E. Russell Johnston, J. Lectue Notes: J. Wlt Ole Texs Tech Univesity Kinetics of Pticles: Newton s Second Lw The McGw-Hill Copnies, Inc.
More informationSPA7010U/SPA7010P: THE GALAXY. Solutions for Coursework 1. Questions distributed on: 25 January 2018.
SPA7U/SPA7P: THE GALAXY Solutions fo Cousewok Questions distibuted on: 25 Jnuy 28. Solution. Assessed question] We e told tht this is fint glxy, so essentilly we hve to ty to clssify it bsed on its spectl
More informationSection 3.2 Maximum Principle and Uniqueness
Section 3. Mximum Principle nd Uniqueness Let u (x; y) e smooth solution in. Then the mximum vlue exists nd is nite. (x ; y ) ; i.e., M mx fu (x; y) j (x; y) in g Furthermore, this vlue cn e otined y point
More informationMAGNETIC EFFECT OF CURRENT & MAGNETISM
TODUCTO MAGETC EFFECT OF CUET & MAGETM The molecul theo of mgnetism ws given b Webe nd modified lte b Ewing. Oested, in 18 obseved tht mgnetic field is ssocited with n electic cuent. ince, cuent is due
More information(read nabla or del) is defined by, k. (9.7.1*)
9.7 Gadient of a scala field. Diectional deivative Some of the vecto fields in applications can be obtained fom scala fields. This is vey advantageous because scala fields can be handled moe easily. The
More informationPhysics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 8-9 pm in NSC 128/119 Sunday, 6:30-8 pm in CCLIR 468
ics Announcements dy, embe 28, 2004 Ch 6: Cicul Motion - centipetl cceletion Fiction Tension - the mssless sting Help this week: Wednesdy, 8-9 pm in NSC 128/119 Sundy, 6:30-8 pm in CCLIR 468 Announcements
More informationCHAPTER 18: ELECTRIC CHARGE AND ELECTRIC FIELD
ollege Physics Student s Mnul hpte 8 HAPTR 8: LTRI HARG AD LTRI ILD 8. STATI LTRIITY AD HARG: OSRVATIO O HARG. ommon sttic electicity involves chges nging fom nnocoulombs to micocoulombs. () How mny electons
More informationPhysics 1502: Lecture 2 Today s Agenda
1 Lectue 1 Phsics 1502: Lectue 2 Tod s Agend Announcements: Lectues posted on: www.phs.uconn.edu/~cote/ HW ssignments, solutions etc. Homewok #1: On Mstephsics this Fid Homewoks posted on Msteingphsics
More informationReview of Mathematical Concepts
ENEE 322: Signls nd Systems view of Mthemticl Concepts This hndout contins ief eview of mthemticl concepts which e vitlly impotnt to ENEE 322: Signls nd Systems. Since this mteil is coveed in vious couses
More informationAdvanced Higher Maths: Formulae
: Fomule Gee (G): Fomule you bsolutely must memoise i ode to pss Advced Highe mths. Remembe you get o fomul sheet t ll i the em! Ambe (A): You do t hve to memoise these fomule, s it is possible to deive
More informationPhysics Tutorial V1 2D Vectors
Physics Tutoial V1 2D Vectos 1 Resolving Vectos & Addition of Vectos A vecto quantity has both magnitude and diection. Thee ae two ways commonly used to mathematically descibe a vecto. y (a) The pola fom:,
More informationTwo dimensional polar coordinate system in airy stress functions
I J C T A, 9(9), 6, pp. 433-44 Intentionl Science Pess Two dimensionl pol coodinte system in iy stess functions S. Senthil nd P. Sek ABSTRACT Stisfy the given equtions, boundy conditions nd bihmonic eqution.in
More informationPolynomials and Division Theory
Higher Checklist (Unit ) Higher Checklist (Unit ) Polynomils nd Division Theory Skill Achieved? Know tht polynomil (expression) is of the form: n x + n x n + n x n + + n x + x + 0 where the i R re the
More informationThings to Memorize: A Partial List. January 27, 2017
Things to Memorize: A Prtil List Jnury 27, 2017 Chpter 2 Vectors - Bsic Fcts A vector hs mgnitude (lso clled size/length/norm) nd direction. It does not hve fixed position, so the sme vector cn e moved
More information( ) D x ( s) if r s (3) ( ) (6) ( r) = d dr D x
SIO 22B, Rudnick dpted fom Dvis III. Single vile sttistics The next few lectues e intended s eview of fundmentl sttistics. The gol is to hve us ll speking the sme lnguge s we move to moe dvnced topics.
More information( ) ( ) Physics 111. Lecture 13 (Walker: Ch ) Connected Objects Circular Motion Centripetal Acceleration Centripetal Force Sept.
Physics Lectue 3 (Wlke: Ch. 6.4-5) Connected Objects Cicul Motion Centipetl Acceletion Centipetl Foce Sept. 30, 009 Exmple: Connected Blocks Block of mss m slides on fictionless tbletop. It is connected
More informationMark Scheme (Results) January 2008
Mk Scheme (Results) Jnuy 00 GCE GCE Mthemtics (6679/0) Edecel Limited. Registeed in Englnd nd Wles No. 4496750 Registeed Office: One90 High Holbon, London WCV 7BH Jnuy 00 6679 Mechnics M Mk Scheme Question
More informationAs is natural, our Aerospace Structures will be described in a Euclidean three-dimensional space R 3.
Appendix A Vecto Algeba As is natual, ou Aeospace Stuctues will be descibed in a Euclidean thee-dimensional space R 3. A.1 Vectos A vecto is used to epesent quantities that have both magnitude and diection.
More informationINDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Analysis Autumn 2012
Lecture 6: Line Integrls INDIAN INSTITUTE OF TECHNOLOGY BOMBAY MA205 Complex Anlysis Autumn 2012 August 8, 2012 Lecture 6: Line Integrls Lecture 6: Line Integrls Lecture 6: Line Integrls Integrls of complex
More information(A) 6.32 (B) 9.49 (C) (D) (E) 18.97
Univesity of Bhin Physics 10 Finl Exm Key Fll 004 Deptment of Physics 13/1/005 8:30 10:30 e =1.610 19 C, m e =9.1110 31 Kg, m p =1.6710 7 Kg k=910 9 Nm /C, ε 0 =8.8410 1 C /Nm, µ 0 =4π10 7 T.m/A Pt : 10
More information7.5-Determinants in Two Variables
7.-eteminnts in Two Vibles efinition of eteminnt The deteminnt of sque mti is el numbe ssocited with the mti. Eve sque mti hs deteminnt. The deteminnt of mti is the single ent of the mti. The deteminnt
More informationMATH Summary of Chapter 13
MATH 21-259 ummry of hpter 13 1. Vector Fields re vector functions of two or three vribles. Typiclly, vector field is denoted by F(x, y, z) = P (x, y, z)i+q(x, y, z)j+r(x, y, z)k where P, Q, R re clled
More information