dx was area under f ( x ) if ( ) 0


 Brett Hawkins
 5 years ago
 Views:
Transcription
1 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 xyplne). **Figue  get this imge in you hed, it s get visul! f xy,, long cuve, is the lue cutin, o fence ceted fom the xyplne up to The line integl of ( ) whee it hits the sufce, f (, ) xy, pictued in ed on top of the fence. The line integl is equl to the e of tht lue cutin. If f ( x) 0, then f ( xy, ) dsepesents the positive e of one side of the cutin. Scl Line Integls e used to epesent totl mss nd totl chge nd to find electic potentils. We ll use them in this section to solve wok done y foce field on pticle moving long pth nd to find mss nd cente of mss of thin wies in the shpe of cuves (cuve ). Figue  How cn we find the e of one of these lue veticl ectngles? s suc length Δ s  chnge in suc length So we just multiply f ( xy, ) Δ s (height) (length) This is 1 piece. We need to sum up ll the pieces which is Riemnn n Sum, (, ) i= 1 f x y Δs. As usul, when we tke the nume of i i i suintevls to ppoch then the Riemnn Sum tuns into the exct integl: Line Integl of f long cuve is This is the ig pictue ut we need to f ( xy, ) ds know how to ctully evlute these. * f (, ) xy is sufce. * is cuve (so mye they should e clled cuve integls). We ll need to define cuve. *ds is in suc length, so ou s s un fom wht to wht on ou limits of integtion? This is going to e ou new ttchment (like ddθ in pol, dzddθ in cylindicl, ρ sinφ dρ dθ dφ in spheicl) nd gin it s ecuse of how these things e mesued. We ll find ou new ttchment ds. Lowe cse s. This will llow us to evlute the line integl.
2 Mentioned in this textook ck in Section 10.3 on the fist pge of the lesson (ut elly intoduced ck in lculus II, section 6.4 in the Stewt lculus ook): The Length of cuve with pmetic equtions x = f ( t), y= g( t), t, is the limit of lengths of inscied polygons nd, fo the cse whee f nd g e continuous, we ived t the fomul dx dy L= + dt dt dt This is wht we ttch on fo ds. This is how suc lengths e mesued. So, when is defined pmeticlly (x nd y s functions of t): dx dy ds = + dt dt dt,, f ( xy) ds = f( xt () yt ()) + dt MEMORIZE THIS! dx dy dt dt ***Notice thee s no hee, t The new integl is in tems of t, dt. You will need to descie/wite cuve pmeticlly (in tems of t). When setting up line integls this is usully the most difficult pt!!!!!!!! You ve got to come up with pmetic epesenttion of cuve; cuve. You will e using t s you pmete. When you see the lowe cse s in ds, think of suc nd don t foget to ttch plce of the ds. dx dy + dt dt dt in A moe commonly used compct fom ises if we use vecto nottion to descie cuve : t = x t, y t, t ( ) ( ) ( ) f xt yt f t ( ( ), ( )) ( ( )) dx dy + x t + y t t dt dt *** (, ) = ( ( )) c () () () f xy ds f t t dt 1443 ds Notice if you function = 1 If ll heights = 1, it s (c length) 1 = c length. 1 t dt which is the c length fomul ck in Section 10.3, p. 708: L= t dt
3 Let s tlk out pmeteizing cuves. If cuve is given s y is function of x, so y g( x) = (like y = x ) The ook will choose x fo the pmete, x x, y g( x) consistent with how we ve pmeteized cuves: x t, y g( t) = = ut I will lwys use t s the pmete to sty = =. Exmple: If y = x, then I ll pmetize the cuve s x = t, y = t o t () = t, t Then we e ck to ( ) ( ( ) ( )) dx dy f xy, ds = f xt, yt + dt dt dt The sme ide cn e used if cuve is given x s function of y, like ou fist exmple. Pmeteize the following cuves : ) is the c of the cuve y x = e fom (1, 0) to ( e,1) ) consists of the top hlf of the cicle ( 1,0)to (,3) x + y = 1 fom (1, 0) to ( 1, 0) nd the line segment fom c) is the line segment fom ( 1,5,0) to (1,6,4) d) : the cicle x + y = 4 oiented counteclockwise
4 Exmple 1: Evlute xy ds, whee is the line segment fom ( ) ( ) So this is the line integl of sufce ( ) 0,1 to 3,4. ds efes to sucs f xy, = xy, long cuve, with espect to c length. Visulize Figue. If f ( x) 0, ou nswe will e the e of the cutin unde the sufce f ( xy, ) = xy, long cuve, the line segment fom ( 0,1 ) to ( 3,4 ). f t t dt ( ()) o ( (), ()) dx dy f xt yt + dt dt dt Anothe ppliction: IF the function epesents line density t point (x,y) of thin wie in the shpe of cuve, THEN the nswe epesents the totl mss of the wie. Exmple : Evlute 3 x y ds if is x y =, 0 x. If f ( x) 0, ou nswe will e the e of the cutin unde the sufce f ( x, y) 3 x =, long cuve, y x y =, 0 x. IF insted the function epesents line density t point (x,y) of thin wie in the shpe of cuve, THEN the nswe epesents the totl mss of the wie.
5 Just like odiny single integls, if f ( xy, ) is positive function f (, ) xy dsepesents the e of one side of the cutin (whose se is nd whose height ove the point ( x, y ) is (, ) If the sufce, f (, ) f xy. xy, dips elow the xyplne, it gives us the signed e of the cutin (e ove e elow) f xy, ds f xy, ove line cuve (lying in the xyplne). ( )  The line integl of function ( ) Let s tke function f nd cuve nd shift them up dimension. f ( xyz,, ) ds  The line integl of function f (,, ) It just dds z onto the fomul on the left:,,,, xyz ove cuve which my lie nywhee in spce. f ( xyz) ds = f( xt () yt () zt ()) + + dt o ( ()) dx dy dz dt dt dt sme s it ws efoe f t t dt Exmple 3: Evlute the line integl (with espect to x, y, nd z), z dx + x dy + y dz, whee is the given cuve x t y t z t 3 =, =, = fom ( ) ( ) 0, 0, 0 to 1,1,1. Notice the line integl is NOT with espect to c length (ds), ut insted with espect to x, y, nd z. Thee is no ds, ut dx, dy, nd dz insted. Becuse it is not mesued with sucs, we do not ttch dx dy dz + + o t. dt dt dt Insted, you ll need to clculte dx, dy, nd dz nd expess eveything in tems of pmete t.
6 If is given in septe pts, you need to set up two septe integls. 1 (, ) = (, ) + (, ) f xy ds f xy ds f xy ds 1 Thee my e moe thn two. These line integls we e looking t ight now depend on the pth they tke, so they must e delt with septely. They e clled Dependent of Pth. 1 3 You d get diffeent e if you took the line integl using cuve 3 insted of 1 +. Think of the e unde the cutin Exmple 4: Evlute the line integl, ( + ) x y z ds whee is the pt of the pol c fom ( 0,0 ) to ( 1,1 ) in the xyplne nd the line segment fom ( ) ( ) 1,1, 0 to 1,1,1. y= x in the 1 st qudnt Notice f ( xyz,, ) = x+ y z, not f ( xy, ), so it s not the e of the cutin. Notice cuve hs two pts nd lso notice we e with espect to c length ds. Wht could ou nswe men? If x + y z is the mss long cuve, then ou nswe epesents the totl mss. If x + y z is the chge density long, then ou nswe epesents the totl chge.
7 Insted of integting ove 1 Do you think f ( x, y) ds= +? 3 1 +, wht if you just mke eeline fom ( ) ( ) 0,0,0 to 1,1,1 nd cll it 3? 3 1 Even though the cuves hve the sme endpoints, the vlues won t necessily e equl. The vlue of line integl depends not just on the endpoints of the cuve, ut lso on the pth it tkes. c ( + ) x y z ds whee is the pt of the pol 0,0 to 1,1 in the xyplne nd the line segment fom ( ) ( ) 1,1, 0 to 1,1,1. y = x in the 1 st qudnt fom ( ) ( ) Insted let s sy is the line segment fom ( 0,0,0 ) to ( 1,1,1 ). : t = ttt,,, 0 t 1 3 3() (hee is the pmeteiztion of the line segment) 3 = 1,1,1 = ( t) () t 1 ( ) ( ) } ds, f x y ds = t+ t t dt = Not the sme. 6 0 When this hppens we sy the integl is dependent of pth (fom stt to finish it depends on the pth it tkes to get thee). When scl function f tuns into vecto function F we will sy F is not consevtive vecto field. Lte, we ll integte functions tht e independent of pth. We ll get the sme nswe no mtte wht pth we tke fom stt to finish. We ll look fo quick nd esy wy (lte on). When scl function f tuns into vecto function F we will sy F is consevtive vecto field. So when the integl is dependent of pth you hve to do it in pieces. We ll e lening how to simplify this pocess lso (insted of hving to do it in unch of pieces).
8 So, in summy so f: Line integl of f (, ) xy ove line cuve lying in the xyplne.  Ae of the side of cutin.  If f ( xy, ) epesents line density t point (x,y) of thin wie in the shpe of cuve, then the nswe epesents the totl mss of the wie. Line Integls f (, ) xy ds= pmeticlly defined dx dy f ( xt (), yt ()) + dt dt dt vecto nottion o f () t t dt ( ) () 13, ( t) x( t), y( t) = Splitting into pts if piecewise 1 Line integl of f (,, ) xyz ove cuve which my lie nywhee in spce.,,,, f ( xyz) ds = f( xt () yt () zt ()) + + dt o ( ()) dx dy dz dt dt dt f t t dt Oienttion, Figue 8. Theoeticl Ide. Genelly, given pmetiztion x = xt (), y= yt (), t detemines n oienttion of cuve, with the positive diection coesponding to incesing vlues of the pmete t. If the line integl epesents the e of the side of the cutin, is it going to mtte which wy we tvese cuve? No. When we integte with espect to c length, ds, the vlue of the integl does not chnge when we evese the oienttion of the cuve (it doesn t mtte if you go fom to o to ). The e will e the sme vlue. (, ) = (, ) f xy ds f xy ds Ac length is just length; it s lwys positive. So this is tue of ds, ut not tue of dx o dy. When you chnge fom ds to dt you ve picked THE diection of the pth y you choice of t s.
9 ***Think of this s new section completely. Hee s n ppliction of line integls (don t eplce this with plin line integls). Appliction 1 ente of Mss If ρ ( x, y) epesents line density t point (, ) mss of the wie, m x y of thin wie shped like cuve, then the totl (, ) c ente of mss ( x, y) 1 x= x ρ ( x, y) ds m 1 y = y ρ ( x, y) ds m m= ρ x y ds Know these. Use you gphing clculto to clculte them quickly. Don t foget to ttch t dt fo ds. Let s ty one: Exmple 5: A wie tkes the shpe of the semicicle x + y = 1, y 0, nd it is thicke ne its se thn ne the top. Find the cente of mss of the wie if the line density t ny point is popotionl to its distnce fom the line y = 1. Agin, this is just n ppliction of line integls. Don t eplce this with plin line integls.
10 Appliction Wok Line Integls of Vecto Fields We ll use Vecto Line Integls to find the wok done y foce field, F, when moving pticle long pth. Don t confuse this with plin line integls of function f. Bck in section: 6.5 W = f ( x) dx Wok done y vile foce f ( ) 9.3 W = F D to long the xxis. x in moving pticle fom Wok done y constnt vecto foce F (such s W = F D cosθ electicl o gvittionl) in moving n oject fom point P to point Q, whee D = uuu PQ, the displcement vecto. Hee F cts on the oject nd we must wok ginst the foce field to move the oject. F θ P D Q Now, we e going to compute the wok done y foce field, F, in moving pticle long smooth cuve,. The fomul you e going to use is: W = F ( () t ) t dt Explntion of the fomul: F xyz o = PQR,, foce field in 3dim.(like gvittionl o electic foce fields) t = x t y t z t (defines cuve in spce) (,, ) F () (), (), () T = ( x, y, z) o T t is the unit tngent vecto, T() t = t ( ) () t diffeent points on cuve.. Think of the tngent vecto t = () t Divide the cuve into unch of sucs of length s ( ds) Δ. Δ s If Δ s is vey smll, then s the pticle moves long the cuve in the ppoximte diection of the unit tngent vecto T. t = T Δ s (W = F D so now F TΔs )
11 Wok done y the foce field F in moving the pticle long cuve is ppoximtely F Δ st o F TΔs. F T This dot poduct computes wok t ech point on cuve. Totl wok done in moving the pticle long is the sum of Δ s ll F ( xyz,, ) T( xyz,, ) Δs As lwys, s the nume of eqully spced sucs, n, the wok done y foce field F is the limit of Riemnn sums which ecomes the integl. Hee s visuliztion in dimensions: uve is in lck, vecto field F is in lue nd t cetin points on cuve in geen, nd the unit tngent vecto T is in ed Pticle is moving west. Pticle is moving est. Hee most of the dot poducts F T e Hee most of the dot poducts F T e negtive ecuse the ngles etween positive ecuse the ngles etween the vectos e otuse so the line the vectos e otuse so the line integl will e negtive. integl will e positive. The vecto field is moe often in the opposite The vecto field is moe often in the sme diection s the pth the oject tkes. diection s the pth the oject tkes. W= F x y z T x y z ds = F T ds (,, ) (,, ) Scleduces ck to egul lines Wok is the line integl with espect to c length of the tngentil component of the foce. Wow!
12 W= F x y z T x y z ds = F T ds But this fomul (,, ) (,, ) doesn t mtch the fomul I gve you W = F ( () t ) t dt so let s do little moe wok. If cuve is given y the vecto eqution ( t) x( t), y( t), z( t) ( t) T() t = (sect 10.4) () t =, then unit Tngent vecto F xyz T xyz ds s We cn ewite (,, ) (,, ) ( t) W = F( () t ) t dt = F( () t ) t dt 1443 t 1443 F 1443 ds T evited W = F d Wok done y foce field, F, in moving pticle long smooth cuve,. This occus in othe es of physics lso like clculting the flux coss plne cuve defined s the integl of the noml component of vecto field, the thn the tngentil component Definition: Let F e continuous vecto field defined on smooth cuve, given y vecto function ( t), t. Then the line integl of F long is: u F d = F t t dt = F T ds 1443 ( ()) d = WORK Appliction F t ( ( )) mens F ( xt ( ), yt ( ), zt ( )) This is the wok done y foce field, F, in moving pticle long smooth cuve,. Notice, this is vey simil to ou plin line integl, f ( () t ) t dt, the wok ppliction one is just missing the solute vlue nd hs the dot poduct. Don t mix them up!!
13 Exmple 6: F, F x, y = y, x Find the wok done y the foce field ( ) unit cicle centeed t the oigin. W = F ( () t ) t dt Need: ( t), F( ( t) ), ( t), nd F( ( t) ) ( t), in moving n oject fom ( 1, 0 ) to ( 0,1 ) on the
14 Why did we get negtive wok? When we get negtive wok, the vecto field must siclly e in the opposite diection s the pth the oject took. Moe of the dot poducts F T e negtive ecuse the ngle etween them is otuse. (Figue 1, no vectos stting on the (lue) cuve,, point oughly in the sme diection s the cuve. In fct most vectos point in the opposite diection). The foce field impedes the movement of the pticle long the cuve. Hee is ou Exmple 6 done y hnd (o just show them on Mthemtic): Find the wok done y the foce field F, F( x, y) = y, x unit cicle centeed t the oigin. ( 0,1 ) ( ),, in moving n oject fom ( ) ( ) 1, 0 to 0,1 on the ( 1, 0 ) Domin Rnge Point Vecto F = y, x ( 1, 0 ) 0, 1, ( 0,1 ) 1, 0, 0.7, Negtive π ecuse ou vectos in the vecto field e pointing in the opposite diection s ou pth fom ( 1, 0 ) to ( 0,1) on. 1 Hee, it mttes which wy we go (the diection of the pth) with vecto field F in this wok ppliction. F d = F d + Tvese uve in opposite diections one nswe is positive nd the othe negtive. But when we used c length, ds, ( length is length, it s lwys positive) it didn t mtte which f xyds, = f xyds, we we tvesed cuve : ( ) ( ). +
15 I m going to sk this next question two diffeent wys so you cn see thei connection. It s just mtte of nottion! 1. Evlute + ( + ) = fom ( ) ( ) xy dx x y dy long the cuve y x 1,1 to, 4. F xy, = xy, x+ y long the cuve y x. Find the wok done y foce field ( ) W = F ( () t ) t dt = fom ( 1,1 ) to (, 4). Let s ty the fist one: 1. Evlute xy dx + ( x+ y) dy long the cuve y nything s we go so they cn see the connection to the second wy. = x fom ( 1,1 ) to (, 4). Do not simplify Notice these line integls of f long e with espect to x nd y insted of c length ds. Let s stt the second one until you see it is the sme s the fist one: F xy, = xy, x+ y long the cuve. Find the wok done y foce field ( ) y = x fom ( 1,1 ) to (, 4). W = F t t dt ( ( )) ( )
16 This is the connection etween line integls of scl fields nd line integls of vecto fields. The polem I sked two diffeent wys: xy dx + ( x+ y) dy could e witten s F d if F = xy, x+ y Textook Exmple 6 is: ydx+ zdy+ xdz which could e witten s F d if F = y, z, x scl vecto field Just think of tht d s the deivtive of ( t) x( t), y( t), z( t) = nd do the dot poduct. W = Pdx+ Qdy + Rdz = F d lin integl of line integl of scl fields vecto field F = = if F P( xyz,, ), Q( xyz,, ), R( xyz,, ), ( t) xt ( ), yt ( ), zt ( ) If they don t see it, hee s the long explntion: = = Suppose vecto field F on 3dim. is F P( xyz,, ), Q( xyz,, ), R( xyz,, ), ( t) xt ( ), yt ( ), zt ( ) F d = F t t dt ( ()) F t = P x t y t z t Q x t y t z t R x t y t z t F t ( ( )) ( (), (), ()), ( (), (), ()), ( (), (), ()) ( ( )) = P, Q, R fo ese. dx dy dz t =,, dt dt dt Let s gee to cll it dx dy dz F( () t ) t = P + Q + R dt dt dt So F t t dt ( ()) dx dy dz = P + Q + R dt dt dt dt = P dx + Q dy + R dz Which is the line integl long with espect to x, y, nd z. W= F d = Pdx+ Qdy+ Rdz line integl of line integl of vecto field F scl fields so given this nottion, you know F = P, Q, R.
Electric 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 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 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 informationThis immediately suggests an inversesquare law for a "piece" of current along the line.
Electomgnetic Theoy (EMT) Pof Rui, UNC Asheville, doctophys on YouTube Chpte T Notes The iotsvt Lw T nvesesque Lw fo Mgnetism Compe the mgnitude of the electic field t distnce wy fom n infinite line
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 information10 Statistical Distributions Solutions
Communictions Engineeing MSc  Peliminy Reding 1 Sttisticl Distiutions Solutions 1) Pove tht the vince of unifom distiution with minimum vlue nd mximum vlue ( is ) 1. The vince is the men of the sques
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 informationThe Formulas of Vector Calculus John Cullinan
The Fomuls of Vecto lculus John ullinn Anlytic Geomety A vecto v is n ntuple 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
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 informationWeek 8. Topic 2 Properties of Logarithms
Week 8 Topic 2 Popeties of Logithms 1 Week 8 Topic 2 Popeties of Logithms Intoduction Since the esult of ithm is n eponent, we hve mny popeties of ithms tht e elted to the popeties of eponents. They e
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 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 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 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 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 informationThe Area of a Triangle
The e of Tingle tkhlid June 1, 015 1 Intodution In this tile we will e disussing the vious methods used fo detemining the e of tingle. Let [X] denote the e of X. Using se nd Height To stt off, the simplest
More informationWe partition C into n small arcs by forming a partition of [a, b] by picking s i as follows: a = s 0 < s 1 < < s n = b.
Mth 255  Vector lculus II Notes 4.2 Pth nd Line Integrls We begin with discussion of pth integrls (the book clls them sclr line integrls). We will do this for function of two vribles, but these ides cn
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 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 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 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 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 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 informationUniversity of. d Class. 3 st Lecture. 2 nd
University of Technology Electromechnicl Deprtment Energy Brnch Advnced Mthemtics Line Integrl nd d lss st Lecture nd Advnce Mthemtic Line Integrl lss Electromechnicl Engineer y Dr.Eng.Muhmmd.A.R.Yss Dr.Eng
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 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 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 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 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 informationImproper Integrals. The First Fundamental Theorem of Calculus, as we ve discussed in class, goes as follows:
Improper Integrls The First Fundmentl Theorem of Clculus, s we ve discussed in clss, goes s follows: If f is continuous on the intervl [, ] nd F is function for which F t = ft, then ftdt = F F. An integrl
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 informationHomework 3 MAE 118C Problems 2, 5, 7, 10, 14, 15, 18, 23, 30, 31 from Chapter 5, Lamarsh & Baratta. The flux for a point source is:
. Homewok 3 MAE 8C Poblems, 5, 7, 0, 4, 5, 8, 3, 30, 3 fom Chpte 5, msh & Btt Point souces emit nuetons/sec t points,,, n 3 fin the flux cuent hlf wy between one sie of the tingle (blck ot). The flux fo
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 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 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 informationOn the diagram below the displacement is represented by the directed line segment OA.
Vectors Sclrs nd Vectors A vector is quntity tht hs mgnitude nd direction. One exmple of vector is velocity. The velocity of n oject is determined y the mgnitude(speed) nd direction of trvel. Other exmples
More information2.4 Linear Inequalities and Interval Notation
.4 Liner Inequlities nd Intervl Nottion We wnt to solve equtions tht hve n inequlity symol insted of n equl sign. There re four inequlity symols tht we will look t: Less thn , Less thn or
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 unth distnce is 1.4 nd the thmoon 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 informationSection 35 SHM and Circular Motion
Section 35 SHM nd Cicul Motion Phsics 204A Clss Notes Wht do objects do? nd Wh do the do it? Objects sometimes oscillte in simple hmonic motion. In the lst section we looed t mss ibting t the end of sping.
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 informationPreviously. Extensions to backstepping controller designs. Tracking using backstepping Suppose we consider the general system
436459 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 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 onevible
More information9.4 The response of equilibrium to temperature (continued)
9.4 The esponse of equilibium to tempetue (continued) In the lst lectue, we studied how the chemicl equilibium esponds to the vition of pessue nd tempetue. At the end, we deived the vn t off eqution: d
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 CuentCying Conducto  Foce Between Pllel Conductos  Mgnetic Field of Cicul Cuent
More informationChapter 2: Electric Field
P 6 Genel Phsics II Lectue Outline. The Definition of lectic ield. lectic ield Lines 3. The lectic ield Due to Point Chges 4. The lectic ield Due to Continuous Chge Distibutions 5. The oce on Chges in
More informationr = (0.250 m) + (0.250 m) r = m = = ( N m / C )
ELECTIC POTENTIAL IDENTIFY: Apply Eq() to clculte the wok The electic potentil enegy of pi of point chges is given y Eq(9) SET UP: Let the initil position of q e point nd the finl position e point, s shown
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 information13.5. 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 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 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 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 Coodintes)
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 information5: The Definite Integral
5: The Definite Integrl 5.: Estimting with Finite Sums Consider moving oject its velocity (meters per second) t ny time (seconds) is given y v t = t+. Cn we use this informtion to determine the distnce
More informationMath 8 Winter 2015 Applications of Integration
Mth 8 Winter 205 Applictions of Integrtion Here re few importnt pplictions of integrtion. The pplictions you my see on n exm in this course include only the Net Chnge Theorem (which is relly just the Fundmentl
More informationChapter 21: Electric Charge and Electric Field
Chpte 1: Electic Chge nd Electic Field Electic Chge Ancient Gees ~ 600 BC Sttic electicit: electic chge vi fiction (see lso fig 1.1) (Attempted) pith bll demonsttion: inds of popeties objects with sme
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 informationFluids & Bernoulli s Equation. Group Problems 9
Goup Poblems 9 Fluids & Benoulli s Eqution Nme This is moe tutoillike 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 informationSo, if we are finding the amount of work done over a nonconservative vector field F r, we do that long ur r b ur =
3.4 Geen s Theoem Geoge Geen: selftaught English scientist, 79384 So, if we ae finding the amount of wok done ove a nonconsevative vecto field F, we do that long u b u 3. method Wok = F d F( () t )
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 informationSURFACE TENSION. eedge 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 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 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 informationset is not closed under matrix [ multiplication, ] and does not form a group.
Prolem 2.3: Which of the following collections of 2 2 mtrices with rel entries form groups under [ mtrix ] multipliction? i) Those of the form for which c d 2 Answer: The set of such mtrices is not closed
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 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 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 ighthnd ule to detemine the diection of the mgnetic foce on the chge. Emple 1 ˆ i = ˆ ˆ i
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 information(a) CounterClockwise (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? ( CounteClockwise (b Clockwise ( (c o ottion ti (d ot enough infomtion Effect of Constnt et oque.3 A constnt nonzeo toque
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 2013 Outline 1 Riemnn Sums 2 Riemnn Integrls 3 Properties
More informationChapter 0. What is the Lebesgue integral about?
Chpter 0. Wht is the Lebesgue integrl bout? The pln is to hve tutoril sheet ech week, most often on Fridy, (to be done during the clss) where you will try to get used to the ides introduced in the previous
More informationCh 26  Capacitance! What s Next! Review! Lab this week!
Ch 26  Cpcitnce! Wht s Next! Cpcitnce" One week unit tht hs oth theoeticl n pcticl pplictions! Cuent & Resistnce" Moving chges, finlly!! Diect Cuent Cicuits! Pcticl pplictions of ll the stuff tht we ve
More information10 Vector Integral Calculus
Vector Integrl lculus Vector integrl clculus extends integrls s known from clculus to integrls over curves ("line integrls"), surfces ("surfce integrls") nd solids ("volume integrls"). These integrls hve
More informationRiemann Sums and Riemann Integrals
Riemnn Sums nd Riemnn Integrls Jmes K. Peterson Deprtment of Biologicl Sciences nd Deprtment of Mthemticl Sciences Clemson University August 26, 203 Outline Riemnn Sums Riemnn Integrls Properties Abstrct
More information7.1 Integral as Net Change and 7.2 Areas in the Plane Calculus
7.1 Integrl s Net Chnge nd 7. Ares in the Plne Clculus 7.1 INTEGRAL AS NET CHANGE Notecrds from 7.1: Displcement vs Totl Distnce, Integrl s Net Chnge We hve lredy seen how the position of n oject cn e
More informationDefinite Integrals. The area under a curve can be approximated by adding up the areas of rectangles = 1 1 +
Definite Integrls 5 The re under curve cn e pproximted y dding up the res of rectngles. Exmple. Approximte the re under y = from x = to x = using equl suintervls nd + x evluting the function t the lefthnd
More informationProperties of Integrals, Indefinite Integrals. Goals: Definition of the Definite Integral Integral Calculations using Antiderivatives
Block #6: Properties of Integrls, Indefinite Integrls Gols: Definition of the Definite Integrl Integrl Clcultions using Antiderivtives Properties of Integrls The Indefinite Integrl 1 Riemnn Sums  1 Riemnn
More informationPhysics 111. Uniform circular motion. Ch 6. v = constant. v constant. Wednesday, 89 pm in NSC 128/119 Sunday, 6:308 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, 89 pm in NSC 128/119 Sundy, 6:308 pm in CCLIR 468 Announcements
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 informationEnergy Dissipation Gravitational Potential Energy Power
Lectue 4 Chpte 8 Physics I 0.8.03 negy Dissiption Gvittionl Potentil negy Powe Couse wesite: http://fculty.uml.edu/andiy_dnylov/teching/physicsi Lectue Cptue: http://echo360.uml.edu/dnylov03/physicsfll.html
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 informationInterpreting Integrals and the Fundamental Theorem
Interpreting Integrls nd the Fundmentl Theorem Tody, we go further in interpreting the mening of the definite integrl. Using Units to Aid Interprettion We lredy know tht if f(t) is the rte of chnge of
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 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 informationUnit 6. Magnetic forces
Unit 6 Mgnetic foces 6.1 ntoduction. Mgnetic field 6. Mgnetic foces on moving electic chges 6. oce on conducto with cuent. 6.4 Action of unifom mgnetic field on flt cuentcying loop. Mgnetic moment. Electic
More informationReview of Gaussian Quadrature method
Review of Gussin Qudrture method Nsser M. Asi Spring 006 compiled on Sundy Decemer 1, 017 t 09:1 PM 1 The prolem To find numericl vlue for the integrl of rel vlued function of rel vrile over specific rnge
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 information10. AREAS BETWEEN CURVES
. AREAS BETWEEN CURVES.. Ares etween curves So res ove the xxis re positive nd res elow re negtive, right? Wrong! We lied! Well, when you first lern out integrtion it s convenient fiction tht s true in
More informationChapter 9 Definite Integrals
Chpter 9 Definite Integrls In the previous chpter we found how to tke n ntiderivtive nd investigted the indefinite integrl. In this chpter the connection etween ntiderivtives nd definite integrls is estlished
More informationWeek 10: Line Integrals
Week 10: Line Integrls Introduction In this finl week we return to prmetrised curves nd consider integrtion long such curves. We lredy sw this in Week 2 when we integrted long curve to find its length.
More informationp(t) dt + i 1 re it ireit dt =
Note: This mteril is contined in Kreyszig, Chpter 13. Complex integrtion We will define integrls of complex functions long curves in C. (This is bit similr to [relvlued] line integrls P dx + Q dy in R2.)
More informationThe Wave Equation I. MA 436 Kurt Bryan
1 Introduction The Wve Eqution I MA 436 Kurt Bryn Consider string stretching long the x xis, of indeterminte (or even infinite!) length. We wnt to derive n eqution which models the motion of the string
More informationThis chapter is about energy associated with electrical interactions. Every
23 ELECTRIC PTENTIAL whee d l is n infinitesiml displcement long the pticle s pth nd f is the ngle etween F nd d l t ech point long the pth. econd, if the foce F is consevtive, s we defined the tem in
More informationI1 = I2 I1 = I2 + I3 I1 + I2 = I3 + I4 I 3
2 The Prllel Circuit Electric Circuits: Figure 2 elow show ttery nd multiple resistors rrnged in prllel. Ech resistor receives portion of the current from the ttery sed on its resistnce. The split is
More information= ΔW a b. U 1 r m 1 + K 2
Chpite 3 Potentiel électiue [18 u 3 mi] DEVOIR : 31, 316, 354, 361, 35 Le potentiel électiue est le tvil p unité de chge (en J/C, ou volt) Ce concept est donc utile dns les polèmes de consevtion d énegie
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 righthnd side limit equls to the lefthnd side limit, i.e., lim f (x) = lim f (x). x x + For two vribles
More informationChapters 4 & 5 Integrals & Applications
Contents Chpters 4 & 5 Integrls & Applictions Motivtion to Chpters 4 & 5 2 Chpter 4 3 Ares nd Distnces 3. VIDEO  Ares Under Functions............................................ 3.2 VIDEO  Applictions
More informationLecture 2e Orthogonal Complement (pages )
Lecture 2e Orthogonl Complement (pges ) We hve now seen tht n orthonorml sis is nice wy to descrie suspce, ut knowing tht we wnt n orthonorml sis doesn t mke one fll into our lp. In theory, the process
More informationSection 6: Area, Volume, and Average Value
Chpter The Integrl Applied Clculus Section 6: Are, Volume, nd Averge Vlue Are We hve lredy used integrls to find the re etween the grph of function nd the horizontl xis. Integrls cn lso e used to find
More informationMath 4318 : Real Analysis II MidTerm Exam 1 14 February 2013
Mth 4318 : Rel Anlysis II MidTem 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 information